If you write it out:

N = a*(10^5 ) + b*(10^4 ) + c*(10^3 ) + d*(10^2 ) + e*(10^1 ) + f*(10^0 )

Then you can count the number of times 5 appears as a,b,c,d,e,f.

How many numbers have a=5? 10^5 (because b-f can be anything)

How many numbers have b=5? 10^5 (because a and c-f can be anything)

...

And when you add them up you have 6*10^5.

Two steps. First show that such a g exists. Well, we can define one. That is, we must define what g(b) means. First, because f is surjective, there exists an a in A such that f(a) = b. Because f is injective, f(a) = f(a') if and only if a = a'. Therefore, we can define g(b) to be a such that f(a) = b (because there is always exactly one such a for which this is true).. Using this definition you canb then easily show that both identities are true.

Now uniqueness. Consider g and g'. We need to show that for all b in B, g(b) = g'(b). Well, consider f(g(b)) = b and that f(g'(b)) = b because of our assumption that both compositions give us identities. Then f(g(b)) = f(g'(b)). Since f is injective, this implies that g(b) = g'(b). Thus, the two functions g and g' are the same.

x/(x-1) = 1 + 1/(x-1)
So it's essentially 1/x shifted to the right 1 unit, and up 1 unit.
as x goes to 0^(+), 1/x goes to ∞ , and as x goes to 0^(-), 1/x goes to -∞.
Hence as x goes to 1^(+), x/(x-1)=1+1/(x-1) goes to ∞, and as x goes to 1^(-1), x/(x-1)=1+1/(x-1) goes to -∞.
(another way is to notice that the limit is "1/0", which in limits means + or - ∞ )

1/ln(x), ln is monotonically increasing and continuous, and ln(1)=0, hence as x goes to 1^(+), ln(x) goes to 0^(+), hence 1/ln(x) goes to ∞, and as x goes to 1^(-), ln(x) goes to 0^(-), so 1/ln(x) goes to -∞

So you have essentially "∞-∞" (since the signs match for each direction)

Here is a graph of the relevant functions: https://www.desmos.com/calculator/or1ri0qotw

Sketch: Since the function is quadratic (has a t^2), it will be a parabola. Since the factor before the t^2 is negative, the parabola will open up downwards like an upside down U. At t=0, you have y=40. You should make your "rough sketch" based on this info, or use wolfram alpha http://www.wolframalpha.com/input/?i=y%3D40%2B80t-5t^2

I assume you've seen this written out as y=y0+v0 t + (a/2)t^2. Compare your equation with this one. What is the initial height? initial velocity? acceleration? From this info, write down the equation for velocity v=v0+at. When the object is at its greatest height the velocity will be zero. Use that to solve for t in your velocity equation, the plug it back into the equation for height y.

c) plug in t=4 and solve for y. d) plug in y=120 and solve for t.

Since school is about learning, I will first direct you here to see if you might be able to do it by yourself with a bit of background information.

There aren't any secrets,just persistence,hard work and practice,practice and more practice.

I would suggest going over material @ http://www.khanacademy.org/

Practice some old exams so you can see what you can expect.

For this to be continuous, the graph must be connected the entire way. Currently, your graph looks something like this ( https://www.desmos.com/calculator/bpgmhby84b ) where the second section change with respect to whatever the value of a is.

We need this graph to be connected! For this to happen we need the whole 2ax graph to touch the other graph where it ends. This happens at the y-value of 8. So, we need 2ax to be equal to 8, or solve the equation 2ax = 8.

Currently, we have 2 variables. a and x so we can't solve it. But hold up, look at the graph. Do we know what x-value this connection has to occur at? It has to happen at x = 3 because that's the x-coordinate of where the graphs meet. So we plug that in

2ax = 8

2a(3) = 8

6a = 8

a = 8/6 = 1.33333

If you plug that in for a on the demos graph you will see that your whole graph is now connected

You really should have learned how to derivate before tackling these problems, else they will be near pointless.

The derivate of 5.00 + 10.00t + 2.00·t² with respect to t is 10.00 + 4.00t. If you derive that one more time you will get 4.00.

In general, the derivate of t^n with respect to t is n·t^n-1.

The derivate of f(t) + g(t) is the same as the sum of the derivate of f(t) and the derivate of g(t).

So, the derivate for t⁰ = 0, so the derivate of our first term 5.00 = 5.00·t⁰ is 5.00·0 = 0.

The derivate of t = t^1 is 1t^1-1 = 1t⁰ = 1, so the derivate of the second term 10.00·t is 10.00. The derivate of t² is 2t¹ = 2t, so the derivate of the third term 2.00·t^2 is 2.00·4·t = 4.00·t

Derivation could be seen as finding out "how quickly something changes". So, if you have a function f(t) that tells you the position at time t (which is what you have right now: θ = f(t) = 5.00 + 10.0·t + 2.00·t^2 ), then derivating that function will give you back a function f'(t) = 10.00 + 4.00·t that tells you how quickly the position changes eg. the speed at time t. If you then derivate the speed function f'(t) you will get back a function f''(t) = 4.00 that tells you how quickly the speed changes eg. acceleration at time t. In this case, since f''(t) is constant, the objects rotation acceleration is constant.

My way of explaining this isn't very good, please check out http://www.khanacademy.org/ for more information. Direct link to the first video on derivation: http://www.khanacademy.org/video/calculus--derivatives-1--new-hd-version?playlist=Calculus

There are definitely missing givens as has been discussed below, You can check this geogebra sketch you can play around with to see there are many different triangles that match those givens.

Sorry, should have been more clear.

N is any number below 1 million, or any number less than 6 digits (we can quickly note that 1 million is not a solution and safely ignore it because it doesn't have a 5).

We can represent N by the sum of its digits:

N = a(10^5 ) + b(10^4 ) + c(10^3 ) + d(10^2 ) + e(10^1 ) + f(10^0 )

Where a, b, c, d, e, f are any number from 0-9. Together, they give any and all numbers less than a million (note leading zeros allow for numbers with less than 6 digits).

You want to count how many times does a=5, plus how many times b=5, etc etc. If you add them up, you get how many times 5 appears (note this double counts numbers like 55, but since you're only asking for how many 5's, double counting is actually correct).

But how do you know how many numbers have a=5? Since b-f can be anything from 0-9, the number of possibilities is just (by the counting rule):

b * #c * #d * #e * #f = 10 * 10 * 10 * 10 * 10 = 10^5 .

Hope that makes sense.

For example, You have 100m of fencing to fence the largest area of grass in a rectangular shape. What length (L) and width (w) should the field be?

1) Write the function:

Well we are looking for MAXIMUM area so the equation will be to doi with the area

Area = L x w

But that has 2 variables and we only want 1... so we need to write one of them using the other. We can use the amount of fence we have:

100 = L + w + L + w

2L = 100 - 2w

L = 50 - w

So we can substitute in that value for L giving us our equation:

Area = (50 - w) x w

Area = -w^2 + 50w

This is what that looks like: https://www.desmos.com/calculator/dpgp2jsjdg

2) Differentiate

Area = -w^2 + 50w

we will write as:

f(w) = -w^2 + 50w (It means the same)

f'(w) = -2w + 50

3) set to 0:

0 = -2w + 50

4) Solve for 'w'

0 = -2w + 50

2w = 50

w = 25

5) Evaluate the result: Well we only have one answer, the width of the field being 25 meters, meaning the length is also 25 meters, and we end up with a square field with area of 625 m^(2).

I assume most of this should have been completed by now, so I will throw in some answers.
a) using technology...
b) 5800 years
c) t = 8267 ln(2) = 5730.247741689068 years
d)(5800 - 5730.247741689068)/5730.247741689068 = 1.21% error
2. N0 e^(-t/T) = N0 2^(-t/t_half)
e^(-t/T) = 2^(-t/t_half)
ln( e^(-t/T) ) = ln( 2^(-t/t_half) )
(-t/T) ln(e) = (-t/t_half) ln(2)
(-t/T) (t_half/-t) = ln(2)
t_half = T ln(2)

There are two solutions

y = -9 and y = 1

https://www.desmos.com/calculator/jblwcnrrfx

In the coordinate (3, -4) it is the y component we are interested in -4

So your lines parallel to the x-axis are passing through points y = -4 + 5 = 1, and -4 - 5 = -9

(3, 1) and (3, -9)

> Let me know if there's anything I can do to help you out.

Heh.

Double heh on the fraud that is the use of TI's overpriced hand-held computers in education. Tell your teacher to waste his/her own money on one of those and that you will just use your computer. For instance, see if this helps you solve it.

[; f'(x) = \frac {-1} {(x + 2)^2} ;]

Call the value of [; x ;] we want [; x_2 ;], then [; y_2 = \frac {1} {(x_2 + 2)} ;]. The two-point equation for a line is [; y - y_1 = m ( x - x_1 ) ;]. We know that the slope [; m ;] is equal to the derivative of [; f(x) ;] at [; x_2 ;], so

[; y_2 - y_1 = \frac {1} {(x_2 + 2)} - y_1 = f'(x_2) ( x_2 - x_1 ) = \frac {-1} {(x_2 + 2)^2} ( x_2 - x_1 ) ;]

[; \Rightarrow \frac {1} {(x_2 + 2)} - y_1 = \frac {-1} {(x_2 + 2)^2} ( x_2 - x_1 ) ;]

Both [; x_1 ;] and [; y_1 ;] are given , so just solve for [; x_2 ;].

Edit: Got rid of [; * ;]. Looked too much like convolution.

Plots

Is the domain you're looking at [0,2*pi] or [0,pi/2] [I think you left out a '/' ;-) ]. If it's the latter, then your bounds are correct.

FYI, Wolfram Alpha is a great resource for checking your work :-)

This might help you: http://www.khanacademy.org/math/probability/v/basic-probability

In short, since dice rolls are independent events (the previous roll has no impact on the current or next roll) you mulitply the individual probabilities.

Chance of getting a six on roll #1: 1/6

Chance of getting a six on roll #2: 1/6

Chance of getting a six on rolls 1 & 2: 1/36. Or ~ 3%

Good Luck.

And as for the steps, well, there aren't really any steps. You just have to learn to recognize things that are factorable. If you really can't figure it out, you can always use the old standby – the quadratic formula.

Damn, that shouldn't be written like that. Made me think it was an interval, rather than a finite set of naturals.

The proof is technical and relies on a good grasp of induction. I don't feel up to giving you an explanation, so I'll supply you with an alternate proof.

Let n > 0 be an arbitrary integer. Suppose that A is a subset of I_n and we have a bijection f from I_n to A. Since I_n is finite, we can list its elements: 1, 2, ..., n. By the definition of a bijection, the elements f(1), f(2), ..., f(n) are all distinct elements of A. Further, there are no other elements of A (again from definition of bijection, f(I_n) = A). Thus the set A has exactly n elements. There is only one subset of I_n with exactly n elements: the set I_n itself. We then have A = I_n.

I know a lot of people already helped but maybe my approach can add to that so here I go...

When you revolve around the x-axis you use π ∫ y^2 dx. Similarly, when you revolve around the y-axis you use π ∫ x^2 dy, by just swapping the x's and y's. So by writing y=e^-x as x=-ln(y) you can calculate the volume created by revolving around y-axis with π ∫ ln^2 (y) dy.

As for the boudaries, the area that is revolved around thr y-axis looks like this:

https://www.desmos.com/calculator/ceddjxrmwq

Where I divided it up into two parts, the bottom part creates a cylinder with height 1/2 and a radius of ln(2), of which the volume can be easily calculated. The upper part can then be calculated with the integral and the boundaries from y=1/2 to y=1.

I hope this helped :)

Forget that x is a variable; when you're integrating over y, it isn't.

So the first step is [; \int_0^1 a^2 y \cos{(ay)} dy ;] [wa]

(The a's are really x's, but it might help you think of them as constants.)

The result (after evaluating the limits) will only have x's in it, and you can integrate over x in the usual way.

That looks a lot more tractable!

So, we have 1/(x+4)(x+2)^2 = A/(x+2)^2 + B/(x+2) + C/(x+4).

Rewrite that as 1 = A(x+4) + B(x+2)(x+4) + C(x+2)^2 .

Substitute x = -4, and we have 1 = C(-2)^2 . Therefore C = 1/4.

Substitute x = -2, and we have 1 = A(2). Therefore A = 1/2.

Solving for B is more difficult, because we cannot set "x" as equal to some variable and eliminate the other terms easily. Try looking at the Wolfram link and scroll down to partial fraction expansion and click "show steps." You can expand our equation and set like powers as equal on both sides, like this:

Ax + 4A +Bx^2 +6Bx + 8B + Cx^2 + 4Cx + 4C = 1.

Group things up as:

(B + C)x^2 + (A + 6B + 4C)x + (4A + 8B + 4C)1 = 0x^2 + 0x + 1.

Set each power of "x" as equal to the other and we get a system of equations:

B + C = 0

A + 6B + 4C = 0

4A + 8B + 4C = 1.

Knowing that B + C = 0, or that B = -C, and we have that C = 1/4, we have B = -1/4.

You could use this system to solve for all three variables, but that's not as fun.

Can you give the entire original problem? As you gave it, your problem doesn't make much sense and you're dividing by zero a few times, unless I'm misreading it.

Check your negatives and make sure you are using brackets for the tan^2 (x) term.

Confirm your answer with WolframAlpha

This can easily be brute-forced, for example see this, which returns (among other things):

{8, 9, 5, 6} {11, 2, 3, 12} {1, 4, 7, 10, 13}

which can readily be verified as being correct.

Think first of a function like this:

f(x) = (x – 1)(x – 2)(x – 3)(x – 4)(x – 5)(x – 6)

That has zeros at 1, 2, 3, 4, 5, 6. That is, the graph contains the points (1, 0), (2, 0), (3, 0), (4, 0), (5, 0), (6, 0).

Now consider shearing it by adding x to the function.

g(x) = x + (x – 1)(x – 2)(x – 3)(x – 4)(x – 5)(x – 6)

This stretches the graph in a way that sends (1, 0) to (1, 1), sends (2, 0) to (2, 2), etc. That's because for x = 1 through 6, this new g(x) function is equal to

g(x) = x + 0

since exactly one of the six factors of the second term in the function becomes 0.

I suggest then that this new g(x) function is the 6th degree function that has a leading coefficient of 1 and contains those 6 points (1, 1), (2, 2), ... , (6, 6).

You can then figure out for yourself

g(7) = 7 + (7 – 1)(7 – 2)(7 – 3)(7 – 4)(7 – 5)(7 – 6)

g(7) = 7 + 6!

I've gotta go now. I'm leavin' on a jet plane ...

Here's the graph:

https://www.desmos.com/calculator/tcslgujubj

Assuming I'm understand the question correctly and that the sequence can be represented as 6+7x; x>=0. A simple summation gives the solution: http://www.wolframalpha.com/input/?i=sum+6%2B7x+from+0+to+87

EDIT: I guess my answer is wrong :( I've never seen an equation like that in my limited math teachings, the closest I could compare it to is breaking up the summation and converting it into a closed form. Which might(?) be what the P0+Pn, etc equation is after simplifying.

Edit 1: Just saw the part about x > 0. I think I fixed it.

Edit 2: Here's a graph in case you want a visual aid.

​

There's a few ways to do this. Here are some:

​

1. Use a double integral
   1. We're given the y-bounds as constants, so that must be the outer integral.
   2. X is bounded by the parabolic line, so our x-bound will be the solution for x.
      1. If y = x^2 then x = sqrt(y)
   3. Set up the double integral
      1. ∫∫ dx dy from [x1, x2] = [ 0, sqrt(y) ] and [y1, y2] = [1, 4]
2. Rotate the graph
   1. Swap the variables and solve for y
      1. x = y^2 ---> y = sqrt(x)
      2. If you graph this you'll see it's half of a horizontal parabola
   2. Since we swapped the variables, our bounds are now 1 <= x <= 4
   3. Set up the integral
      1. ∫sqrt(x) dx from [1, 4]
3. Use brute force to chip away at the unwanted parts.
   1. You can start with a square that covers the area you're interested in.
      1. ∫4 dx from [0, 2]
   2. Cut out a small rectangle to get rid of everything under the y = 1 line.
      1. ∫1 dx from [0, 1]
   3. Cut out the remaining area under the curve.
      1. ∫x^2 dx from [1, 2]
   4. Put everything together and multiply by 2 in order to include both sides of the parabola
      1. ∫4 dx from [0, 2] - ∫1 dx from [0, 1] - ∫x^2 dx from [1, 2]

There is a 'difference of differences' solution to this that is simple to see and understand, however it is a little long. So let me just say that 5 numbers can be linked by an order 4 equation.

That is to say:

y = ax^4 + bx^3 + cx^2 + dx + e

where y = a term in your sequence, x = term number, and a,b,c,d,e are constants to be determined.

term (x) : value(y)

0 : 764

1 : 778

2 : 753

3 : 767

4 : 742

We started numbering the terms from 0 to speed up the solution, but it can be done where term numbers are 1 to 5

So these are the equations we have to find the constants a,b,c,d,e

764 = a(0)^4 + b(0)^3 + c(0)^2 + d(0) + e

778 = a(1)^4 + b(1)^3 + c(1)^2 + d(1) + e

753 = a(2)^4 + b(2)^3 + c(2)^2 + d(2) + e

767 = a(3)^4 + b(3)^3 + c(3)^2 + d(3) + e

742 = a(4)^4 + b(4)^3 + c(4)^2 + d(4) + e

The above equations simplify to:

764 = a(0) + b(0) + c(0) + d(0) + e

778 = a(1) + b(1) + c(1) + d(1) + e

753 = a(16) + b(8) + c(4) + d(2) + e

767 = a(81) + b(27) + c(9) + d(3) + e

742 = a(256) + b(64) + c(16) + d(4) + e

5 equations to find 5 variables a,b,c,d,e

So we use simultaneous equations (or matrix) to solve.

because we started numbering our terms from 0 we get a quick solution for 'e' from the first equation

764 = a(0)^4 + b(0)^3 + c(0)^2 + d(0) + e

764 = 0 + 0 + 0 + 0 + e

e = 764

after that it is the long slog of algebra (or internet: http://math.cowpi.com/systemsolver/5x5.html) which gives:

a = -6.5

b = 52

c = -130

d = 98.5

e = 764

Your equation is:

y = -6.5x^4 + 52x^3 - 130x^2 + 98.5x + 764

https://www.desmos.com/calculator/mrkifgjlsp

does this help?

updated to include both upper and lower rectangles.

EDIT: Here's a nicer version you can manipulate.

For perimeter 40, the dimensions of the rectangle follow this rule:

2x + 2y = 40

That means

x + y = 20

So you could think of dimension y = 20 – x.

An expression for the area is product of the dimensions, so

area = x * y

area = x * (20 – x)

area = 20x – x^2

The graph of area as a function of x will be a parabola.

https://www.desmos.com/calculator/05fgd58q6p

Notice that area will be 0 when x(20 – x) = 0, which happens either when x = 0 or x = 20. So the graph will include (0, 0) and (20, 0). The graph of this area function is a parabola. The highest point of the parabola is halfway between those two zeros, because a parabola has bilateral symmetry.

So, the maximum area happens when x = 10.

x = 10, y = 20 – x = 10, Area = 100 cm^2.

The only thing I can think of is that the lines are perpendicular. If you solve for y in each of the equations, you'll see that the slopes are negative reciprocals.

To display the solution graphically, just use wolframalpha.

You can try the factor theorem: basically what you do is look at the constant term, 720. List all the integer factors of 720 into a list: 1, -1, 2, -2, 3, -3, 4, -4,... and so on. What you then need to do is to plug as many of those values as you can into the original polynomial, F(x) to see if one of the factors gives you 0. (It's literally like this: does F(1) = 0, does F(-1) = 0, does F(2) = 0, does F(-2) = 0, and so on until you actually find a root which does work). Once you find that root, A, you know that (x - A) is a factor of F(x) and so you can then use polynomial long division (or synthetic division, which is faster) in order to find the other roots. Doing this division will reduce the degree of your polynomial by one, which should make finding the remaining roots easier. If you haven't covered factor theorem yet, you can plot F(x) onto your graphing calculator or just go on Desmos and plot it there: https://www.desmos.com/calculator

Note that sin(x) and -sin(x) are symmetrical to each other about the x axis, and that the combined graphs are symmetrical about the y axis

See Graph: https://www.desmos.com/calculator/3eq89zs9ns

Therefore, it is enough to find the integral of sin(x) from 0 to pi/2 and multiply by 4 (assuming you want the full area).

This is a typical sort of question in New South Wales' year 11 advanced maths, which many would agree is the normal math level.

Anyhow, you and your peers need to dedicate an hour in experimenting with quadratics on a graphing calculator to understand them better (or do it old school and graph them yourselves, although you'll need more time).

Start by writing down a quadratic, graphing it, and solving for x by using the quadratic formula. Compare the solutions in the quadratic formula to the parabola on the graph. Now change the quadratic and redo the process of comparing the graph to the solutions.

Examples that should help you realize what the importance of the discriminant is would be

y=(x-1)^(2)+1

y=(x-1)^(2)

y=(x-1)^(2)-1

But expand each factor so that it's in the form ax^(2)+bx+c so that you can plug it into the formula.

Also I've found in my experience that hardly any students ever realize - even at the end of the HSC - that the quadratic formula can also be written as

x = -b/2a ± √(b^(2)-4ac)/2a

Which is essentially

x = number ± another number

Which is saying that since a parabola is symmetrical about its axis, its roots are hence an equal distance to the right and left of its axis. The axis is found at x=-b/2a and the roots will be found √(b^(2)-4ac)/2a to the right of it, and √(b^(2)-4ac)/2a to the left of it.

Finally, once you've all played around with this stuff and understand it, for the last question, just apply the quadratic formula as you normally would, but then you'll have the discriminant to be another quadratic in terms of k. You would've already learned what the discriminant does for various values, so then you basically are being asked the question - when does this quadratic in k have the values <0, =0 and >0, which is again just solving another quadratic like you usually do, with graphing and using the quadratic formula.

A graphing calculator that may help you: https://www.desmos.com/calculator

Put both lines in the form of y = mx + c , so

First one is y = x + 6

Second one is y = -x + 10

I use desmos graphing calculator.

Here is a screenshot showing the lines graphed and where they intersect --> screenshot

First, let's look at the domain. We have a quotient, so the denominator cannot be 0. It's also a square root, so it can't be negative.

We know when the denominator is 0, the graph tends to infinity.

So, let's first factorise the quadratic to get (x+2)(x-1). We know then that this will be 0 when x = -2 or 1 - so these are our vertical asymptotes. Now, the function will be undefined in the read numbers when the sqrt function is < 0.

At something like x = -3, we have sqrt(4) = 2 and f(x) = -4. So we are defined for x < -2. At x = 0, we have sqrt(-2) = undefined in R, and so we are undefined between -2 and 1. At x = 2, we have sqrt(4) = 2 and f(x) = 7/2 and is defined for x > 1.

To summarize so far, we have:

• A 1/x shaped graph (reciprocal)
• f(x) is negative for x < -2^(+)
• f(x) -> -∞ as x -> -2 (try this out for yourself)
• f(x) is undefined for -2 < x < 1
• f(x) is positive for x > 1
• f(x) -> ∞ as x -> 1^(-)

For the range, we consider what happens as we approach the endpoints of the interval(s) of the domain of the function. So, consider what happens as x -> -∞, x -> -2^(+), x -> 1^(-) and x -> ∞.

For graphing, I would substitute some values in and create a plot using what you already know.

As a reference - this is what the graph should look like.

I wasn't thinking correctly. I was thinking of pi/2.

But here: http://www.wolframalpha.com/input/?i=lim_x-%3E0cosx

On the x axis, it's at be zero no? cos IS the x axis, isn't it? Or did I become retarded after trig?

10^5 is the count of numbers with a=5.

Another 10^5 is the count of numbers with b=5 (because a, c-f, can be any number from 0-9, and the same counting rule can be used to figure this out).

Add all of them together and you get 10^5 + 10^5 + 10^5 + 10^5 + 10^5 + 10^5 , which is the number of times 5 appears as a + number of times 5 appears as b + number of times 5 appears as c + ...

Which comes out to 6*10^5 .

Getting back to you on the not-brute-force solution, here's how I solved it.

There is 1 occurence of 5 in the range 1-10, which is simply in the number 5. Let us therefore say f(0) = 1.

In the range 1-100, there is an occurence for 10*the number of occurrences there are in the range 1-10 (05, 15, 25, 35, ..., 95) plus 10 more for each of the numbers beginning with 5 (50, 51, 52, ..., 59). So 20 in total. f(1) = 20.

In the range 1-1000, it's all the occurrences from the range 1-100 but multiplied by 10 (because the numbers 05, 15, 25, 35, ..., 95 as well as 50, 51, 52, ..., 59) plus all the 500s, and there are 100 of those. So 300 in total. f(2) = 300.

As you can see, each 10^x has 10 times the occurrences of 10^(x-1) plus all the numbers where 5 is the leading digit, where you simply add on 10^x. In the end, you get

f(0) = 1 f(x) = 10 * f(x-1) + 10^x

(I tend to think about stuff in a recursive manner)

Now you evaluate f(5) (because f(x) solves for numbers up to 10^(x+1)) and you get your answer.

Nope. Notice how it doesn't touch the x-axis

Technically, there is no real answer. There are two complex x-intercepts, but I doubt that's what's being asked for.

Your answer to the first one is right. You are close on the second one, but are missing a few points :) Think about what the graph of sin(x) looks like or look at what kami_inu posted to see that is it 0 at more than just two x-values. You will probably want to use set-builder notation to express that domain here (that is, {x | some statement about x}) instead of interval notation.

This is a process called 'completing the square'. Make y=0:

> 4x^2 - 4x + 21 = 0

Divide all terms by 4, in order to get the 'a' in 'ax^2' equal to 1.

> x^2 - x + 21/4 = 0

Move the constant factor to the other side.

> x^2 - x = -21/4

Now, notice that (x - 0.5)^2 = (x^2 - x + 0.25). This means we can transform the x^2 - x in our equation into (x - 0.5)^2, provided we add the +0.25 onto the other side as well.

> (x - 0.5)^2 = -21/4 + 0.25

Now square root both sides, making sure not to forget the ±

> (x - 0.5) = ± √(-21/4 + 0.25)

Finally add 0.5 to both sides, leaving 'x' by itself on the left.

> x = 0.5 ± √(-21/4 + 0.25)

Notice how the expression under the square root is always negative. Negative square roots are undefined, meaning there are no solutions for x when y=0 in this equation. Plotting the graph confirms this.

I decided to see what Wolfram Alpha would say. This is the solution. I'm not sure how to get it, but hopefully that will give you some ideas.

I was fancying a guess that the person who assigned the problem intended the component functions to be differentiable everywhere, rather than the curve itself. Something like this

Also you should try Khan Academy if you need help with learning concepts. I'm in Calculus I, and it's been a life saver for me, and his coverage of algebra is even better than his coverage of calculus.

http://www.khanacademy.org/math/algebra/polynomials/v/factoring-trinomials-with-a-leading-1-coefficient

He can probably explain factoring a lot better than I can, I'm not great with teaching.

Though I prefer the other proofs provided here to the one in your book, I understand it might be the overall style of the proof that you want to learn. I'll give you an example using concrete numbers -- hopefully that'll help you abstract the concept to an arbitrary n.

I'll assume induction isn't giving you the problem, so for induction we'll assume that if there's a bijection from I_n to A and A is a subset of I_n that A = I_n. For concreteness, let's say n = 2. Now, for n+1, let's define our bijection f such that f(n+1) = f(3) = a; that is, 3 maps to some element in A that we'll call a. Let's remove a from A and call the resulting set A', so A' has 2 elements. Now if A' is a subset of I_2 -- that is, if the element a that we removed happened to be 3 -- then we already know from our induction assumption that A' = I_n. Since A' is simply A without the element a, we can add the element a back that to both sides of the equation and get A = I_{n+1]. That is, we add back the 3 we removed from A and also add it to the set of I_2, which is I_3. We're done with the first case!

But what if upon removing a from A we find that A' is not a subset of I_n? That is, the element a is not 3 -- it's one of the others. Let's redefine our bijection f to make it so that it does map to a and we can use the same reasoning as before. We know there's an element -- let's call it p -- in I_3 that maps to 3 in A since it's a bijection. We'll define a new bijection, g, that is equal to f(x) unless x = 3 or x = p. Let's simply flip them so that f(3) = a' = 3, and f(p) = a. Recall that our prior argument was to remove the element that n+1 (3) mapped to, and the proof succeeded if a = n+1 = 3. Well, g has that exact property, so just remove 3 from A to get A''. By our inductive hypothesis, A'' = I_n, and adding back the three gets A = I_{n+1} -- that is, adding back 3 to both sets gets you A = I_3. We're done!

Hope that doesn't make it more confusing :)

There is no such thing as "trigonometric differentiation" in it's own kind of label like you have given it.

"Trigonometric differentiation" is just taking a derivative of a function that varies sinusoidally. A derivative is:

Lim dx->0 f(x+h) - f(x) / h. "Trigonometric differentiation" as you define it is just a shortcut to reach this result.

So what you are looking for is anything that behaves in a sinusoidal manner. To be specific, anything that oscillates behaves this way; pendulums, springs, sound and so on.

The position function for a spring is x=(the distance that the spring was offset at the start)cos((angular velocity)(time)) dx/dt=(the distance that the spring was offset at the start)(-1)sin((angular velocity)(time))(angular velocity) etc

As per your specific question, "if they are simply math laws to further advance our knowledge?" Their is no math without potential application of math. Everything you will learn has real life applications. Just google the rules for specific examples.

This seems obvious, though the proof is unnecessarily complicated.

Let f: I_n --> A be a bijection. Then I_n and A have the same cardinality; in fact, |I_n|=|A|=n. Since A is a subset of I_n, we must have A=I_n.

> I'm thinking that the the lim x->2- and lim x->2+ must equal different values

Correct.

>but how do you have f(2) to not exist

Simply don't have f be defined when x = 2.

>AND have LHS and RHS equal approach a different value?

Define f to have different values on the left and right side of x = 2.

For example:

f(x) = 1, when x &lt; 2, f(x) = 3, when x &gt; 2.

Notice that f is not defined when x = 2.

You should graph this: https://www.desmos.com/calculator/6s6rxlo2af

• First problem:
  • If you go to -6^(-) (-6 from the left side), you will use the function piece on the top. Plug in a number less than, but super close to -6 to that function piece such as -6.0000001.... You will get 8/0, which is ∞ . Therefore, the limit does NOT exist on this section.
  • If you go to -6+ (-6 from the right side), you will use the function piece on the bottom. Plug in a number larger than, but super close to -6 such as -5.99999... to that function piece. You will get a very small number, which is super close to 0. Therefore, the limit is 0 on this piece.
  • A limit law states that if you the limit to a number from the left side is different than that of the right side, then the limit for that number is non-existent. As stated above, the limit to -6^(-) (DNE) is not the same as -6^(+) (0). Therefore, the limit to -6 does not exist.

https://www.desmos.com/calculator/k9bs9tt5wl

a) How long the rocket is in the air

What value for t makes h = 0 (again after t = 0) or, how lonbg before it hits the ground again?

0 = -5t^2 + 60t

5t^2 = 60t

5t = 60

t = ^(60)/5

t = 12

> I don't get how you can cut the area up into 8 parts and know what equations describe those parts though.

You've made a good first step getting the upper half of what we're dealing with. Now imagine what other symmetries there are. Can we flip the area over the y-axis too and get the same shape? What about over the line y = x? Each one we can say yes to leads to another multiplication by 2 and reduction of work in the integral.

Here's a visual of what's going on.

I'm guessing talking about Trig/Pre-calc, and not something more advanced, like Abstract Algebra?

If so, Khan Academy has a few videos on polar coordinates and other Pre-calc topics.

Khan has 8 videos on solids of revolution. I know it isn't an answer to your problem, but I guarantee you would be able to master this topic by even just skimming through most of the videos.

There are tons of videos here, specifically under the sub-section of Algebra: Logarithms. Any future subjects you may need to tackle are probably covered on this site as well.

This site is another good one. He only focuses on math but provides great examples and is very easy to follow along with.

Desmos Graph

First, you will have to recognize that because you're revolving functions of x around a horizontal axis, it will be easier to use the washer/disk method, where the thickness of the disk is dx. The bounds for your integral are correct, as these functions intersect at x = 0 and x = 1.

If your axis of rotation was y = 0, then the y value of your functions would represent the radius, or the distance from the axis. If your axis is now y = 5, how would you represent the distance between your function value for some x in [0, 1] and your axis y = 5?

Looking at the graph, we can say that for both functions, the expression 5 - y represents the distance from the axis 5 and the value y obtained from either y = 5x or y = 5sqrt(x). Substituting these values into 5 - y we obtain expressions for each radius:

5 - 5x 5 - 5sqrt(5)

Because you are revolving two curves, you will first have to determine which revolution is the "base" and which revolution is "drilling" into the solid formed by the first revolution. This will change based on the axis of revolution.

Looking at the graph, we see that the line y = 5x (green region) will be the "base", as it occupies the larger area in our specified region. The curve y = 5sqrt(x) will be our "drill bit", as when we revolve it with will subtract its volume from our "base". With this, we can designate which radius is the smaller one that will subtract from the larger solid.

R(x) = 5 - 5x r(x) = 5 - 5sqrt(5)

Finally, we can put these two radius equations in our integral for washer method.

\pi * \int_{a}^{b} R(x)^2 - r(x)^2 dx \pi * \int_{0}^{1} (5 - 5x)^2 - (5 - 5sqrt(5))^2 dx

I calculated this integral in Desmos so you can see your result there.

This is the function graphed

https://www.desmos.com/calculator/e7cle5dim2

You are right that it looks like there are multiple values for 'p' coresponding with each value for 'c' but let's check that.

P , C

1, 1

2, 3

3, 7

4, 13

5, 21

6, 5

7, 17

8, 5

9, 21

10, 13

11, 7

and yes there are repeats...

Just an interesting observation (not the solution).

Consider y = 2x^2 - 7x + 14 notice that when x = 2 that -7x + 14 drops off leaving y = 2x^2 or y = 8

Now consider y = x^3 - nx^2 + 2nx

The last terms can be written as: -nx^2 + n2x which again cancels out when x = 2 leaving y = x^3 or y = 8

So the point (2, 8) is always a point of intersection for these two functions regardless of the value of 'n'. (So there's your 1 point).

https://www.desmos.com/calculator/kz7atffgkk

(1, 6), (2, 11), (3, 18), (4, 27)

We can link 4 points with an order 3 function of the form:

y = ax^3 + bx^2 + cx + d

for some 'a', 'b', 'c', 'd' to be determined

So substituting in the known points (x, y) and we get

(6) = a(1)^3 + b(1)^2 + c(1) + d

(11) = a(2)^3 + b(2)^2 + c(2) + d

(18) = a(3)^3 + b(3)^2 + c(3) + d

(27) = a(4)^3 + b(4)^2 + c(4) + d

Which all simplifies down to be

6 = a + b + c + d

11 = 8a + 4b + 2c + d

18 = 27a + 9b + 3c + d

27 = 64a + 16b + 4c + d

4 variables and 4 simultaneous equations.

We can solve any a number of ways... like using an online solver: http://math.bd.psu.edu/~jpp4/finitemath/4x4solver.html

a = 0

b = 1

c = 2

d = 3

so the algebraic expression is

y = x^2 + 2x + 3

https://www.desmos.com/calculator/emquc3kuun

Here you go.

I've also checked this with desmos. Here is the graph confirming that this is the correct answer.

You can set k equal to 4/3, and you'll get the proper value for "a" ("x" in the desmos graph).

Here's what the first person means:

https://www.desmos.com/calculator/mgasfevxeb

As you can see, they are not equivalent equations and no matter what you do you can't use trig identities to show that one side is equivalent to the other.

Second thing, I noticed a few "rookie" mistakes that led you to find a proof. Line 11 is algebraically incorrect. You cannot cancel out cos(θ) that way. That's why you were able to find a "proof".

FWIW, here's something I made a while back that might help you visualize what the integration does, you can adjust the bounds and the number of intervals, and the function itself to see the lower and upper Riemann sums visualized. https://www.desmos.com/calculator/iqfytpafkw

We are looking for A,B,C,D in f(x)=A sin(Bx + C) + D such that it fits the graph.

First, we can easily find the amplitude of the sinusoidal wave, since, the maximal point is at 3, and the minimal point is at -1, so the amplitude is 4.
A is half the amplitude, so A=2
The horizontal line of symmetry is at y=(-1+3)/2=1, so D=1
1 full cycle is 2 units, so we know that B*2=2 π, so B=π
Finally, we have f(x)=2 sin(π x + C) + 1 = 2 sin(π (x + C/π)) + 1
We know that C/π is the horizontal displacement to the left, notice that if we displace the graph 1 unit to the right we'll have the graph of 2 sin(πx)+1, so C/π=1, so C=π
so we get
f(x) = 2 sin(π x + π) + 1
Which does indeed fit the graph

Yep you need the table of values for latitude and that range 24 to 2 degrees. Lets 'ASSUME' that it is also linear.

Let: M = depth in meters

(L) Latitude : water temp at M = 200m : water temp at M = 1000m

0 : (200, 24) : (1000, 2)

90 : (200, 2) : (1000, 2)

So from them we can calculate the two equations that are the upper and lower bounds from these points

(200, 24) : (1000, 2) gives us:

T = [-^(11)/400*M + ^(55)/2*] + 2

and

(200, 2) : (1000, 2) gives us:

T = [0M + 0] + 2

The difference in these is the different Latitudes (variable 'L'), 0 in the first equation and 90 in the second.

So we can factor that in as a coefficient to the [...] part to give us your equation:

T = (90 - L) x (-[^(11)/36000]M + [^(55)/180]) + 2

Which looks like this:

https://www.desmos.com/calculator/qc751ovzcs

Play with slider for 'L' for different latitudes.

Note: this solution ASSUMES a linear relationship between Latitude and water temperature at 200m

Note: this equation ONLY gives valid answers when: {200 ≤ M ≤ 1000} and {0 ≤ L ≤ 90}.

1. Factor and Cancel
2. Factor and Cancel
3. Add fractions using common denominator
4. Same as 3

If you want to check your answers, try www.wolframalpha.com the first one: http://www.wolframalpha.com/input/?i=%28x^2%2B2x-24%29%2F%28x^2-11x%2B28%29%3D

To understand, fix N at some value, say N = 1.96^2 . This is what the "when N = ..." clause means in your column headers. So now you have P = sqrt(p(1-p)) = sqrt(p - p^2 ). Check out Wolfram Alpha to see a graph of what this looks like.

Notice how the graph rises then falls and it's symmetric around the line p = .5.

There are formulas for differentiation that will be helpful here. Particularly, the power rule and chain rule. Power rule:

[; \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{n} \right )= n\,x^{n-1} ;]

Chain rule:

[; \left ( f \cdot g \right )' \, \left ( t \right )=f'(g(t))*g'(t) ;]

That f dot g thing means a function inside a function, i.e. f(g(t)). It basically means that when taking the derivative of a function with another function inside of it, you first take the derivative of the outer function (and leave the inner function inside and undifferentiated) and then multiply it by the derivative of the inner function.

If you've never learned any of this before, this may be confusing.

First, you should rewrite the equation as

[; \left ( x+2 \right )^{-1/2} ;]

This makes it easier to visualize how you use the power rule in this problem. By applying the power rule, you bring down the -1/2 and then subtract 1 from the exponent, giving you

[; \frac{-1 \left ( x+2 \right )^{-3/2} }{2} ;]

Now, you have to apply the chain rule to the "x+2" inside. Since the derivative of x is just 1, and the derivative of 2 is zero, you are actually just multiplying the equation by 1 so the answer I got in the previous step doesn't change. My answer can also be rewritten as what WolframAlpha says.

[; \frac{-1}{2\left ( x+2 \right )^{3/2}} ;]

If none of this made any sense to you, check out http://www.khanacademy.org and look up the calculus videos about the power rule and the chain rule. He is much better at explaining things than I am.

3 variables now...

For a graphical representation of this equation: https://www.desmos.com/calculator/zuodooron8

Have a play with the slider for 'b'.

With 3 variables you need 3 equations to find a solution.

Here is an explicit parametrisation of contour C:
(R*cos(t),R*sin(t)) for t in [0,pi)
-R((t-pi)/epsilon)-epsilon+t-pi,0) for t in [pi,pi+epsilon)
(R*cos(t-epsilon-2pi),-Rsin(t-epsilon-2pi)) for t in [pi+epsilon,2pi+epsilon)
(t-2pi,0) for t in [2pi+epsilon,2pi+R)

https://www.desmos.com/calculator/p6renlrczr

1.)

It just barely doesn't fit. You can see this graphically but here's how to do it analytically/algebraically:

You are given the semi-minor axis a = 9 ft, and the major axis (dividing this by two gives you the semi-major of b = 15 ft). For an ellipse the equation is:

(x/a)^(2) + (y/b)^(2) = 1

Thus this ellipse is represented by:

(x/15)^(2) + (y/9)^(2) = 1

The top of the truck is at y = 8.3 ft, which is below the 9 ft height of the tunnel so we can't just immediately claim it won't pass through; we have to get fancy. We can use the fact that y = 8.3 here to solve for what x values this line passes through the ellipse. Since the truck is 12 ft in width, these x values must be less than -6 or greater than 6 or the truck will not fit. Let's solve:

(x/15)^(2) + (8.3/9)^(2) = 1

(x/15)^(2) = 1 - (8.3/9)^(2)

x = ±15√[1 - (8.3/9)^(2)]

x ~ ± 5.8

The acquired x values do not fit our conditions for the truck passing safely, so it doesn't fit.

Just refreshed the page and saw you deleted that question but I'll leave this explanation.

2.)

The formula for the focal length of an ellipse given the semi-axes a and b is c = √[a^(2) - b^(2)]. It's fairly straight forward, if you remember to halve the length of the room to get the semi-major axis:

c = √[17^(2) - 8^(2)]

c = √[9(25)]

c = 3(5)

c = 15 ft

They should stand 15 ft from the center of the room.

Here are two blue semicircles that fit the description.

https://www.desmos.com/calculator/wryztcl4ou

Green curve shows 2nd derivative is positive between x = 1 and x = 3, and negative between x = 4 and x = 8. The second derivative is not defined at 1, 4, and 8, but your requirement refers to intervals not including the endpoints, because curvy parentheses were used, not square brackets.

First let f''(x) be 4-x. As you can see, it fits the given data - therefore, any function with that second derivative would fit.
f(x) is, therefore, 2x^2-x^3/6+Cx+C_2 , where C and C_2 are arbitrary constants (if you are not proficient in integration, you can just check that this function's second derivative is 4-x).
f(x)=2x^(2)-x^(3)/6
You can graph it using Desmos. Here is the plot.

For the waterway problem, see this diagram.

https://www.desmos.com/calculator/khfzi4ehzx

You know water is 1 foot below center of circle (4.5 ft deep when circle has radius 5.5 ft).

If the boat bottom is 1 foot below that, you can draw a right triangle with leg 2, hypotenuse 5.5, and then find the other leg of the right triangle.

Double that to get the maximum width of the boat.

This assumes the boats have wide flat bottoms.

Wow I'm an idiot I switched the Desmos to Degrees instead of Radians and it graphed it perfectly https://www.desmos.com/calculator/qsbfuzeygc. I appreciate the reply with switching it to Radians, in my high school they don't teach us anything about Radians we only used degrees, not sure if its a Canadian thing.

for the second one, the maximum area is 0.05 units^2. Not sure how to find this by hand, and it took me like an hour to figure out how to find this using Desmos. Here are the graphs I used: https://www.desmos.com/calculator/tr36schxpz . I know this post might not be very helpful/useful, but you might enjoy that graph...

You have all your numbers right. In terms of explaining:
B) It's less than 10 ft above the ground from when the ball is thrown to .154 seconds and after 2.03 seconds of being thrown.
C) It is more than 10 ft above the ground from .154 seconds of being thrown to 2.03 seconds of being thrown.
D) Solve for t in h(t)=20 and the explanation is the same as for part B
E) Same as part C except use the times that you find in part D

Try graphing the function here so that you can visualize what is happening.

ln ((1-x)/(3-x)) has two branches.

https://www.desmos.com/calculator/vqpvaris2i

ln (1-x) – ln (3-x) is the branch with positive values.

ln (x-1) – ln (x-3) is the branch with negative values.

When you take square root, you're going to have real answers only when the argument is positive, so that's why the first equation makes sense to use.

Here is an example.

Notice that Desmos lets you define piecewise functions by using brackets {} and putting the interval then colon : followed by the equation.

https://www.desmos.com/calculator/twejhoclpg

Here the original curve (red) runs from x = 0 to x = 20.

Then I created Y2 equation (green) using +1/20 as leading coefficient from x = 20 to x = 40.

Finally, the Y3 equation (blue) uses –1/40 as the leading coefficient and runs from x = 40 to about x = 72.1.

Touch the gray circle icons at left of equation 4 and equation 5 to toggle on mini tangent lines that show the smooth connections have the correct slopes.

Of course that should be: y = 1, not y = 3...

But I highly recommend using an online graphing tool like this one here called Desmos to check that your solution is correct.

inverse secant is not 1/cos^-1 x but rather cos^-1 (1/x)

Desmos: https://www.desmos.com/calculator/cznbs7l2jc

On the Ti-89 you can get it from going to the Math Menu and then the trig menu, and I presume it would be similar on the TI-84 (going through the math menu rather than looking for a button on your calculator).

For second question: Suppose a rectangle has a given area of 100 square units. To find the minimum possible perimeter, consider some examples.

x = 2, y = 50; perimeter = 2(2+50) = 104
x = 4, y = 25; perimeter = 2(4+25) = 58
x = 5, y = 20; perimeter = 2(5+20) = 50
x = 10, y = 10; perimeter = 2(10+10) = 40
x = 20, y = 5; perimeter = 2(20+5) = 50
x = 25, y = 4; perimeter = 2(25+4) = 58
x = 50, y = 2; perimeter = 2(50+2) = 104

https://www.desmos.com/calculator/a80hxjtibm

Above a graph of perimeter as a function of x, showing the above points on it. It appears that the minimum perimeter happens when x and y are the same, i.e. when the rectangle is a 10 by 10 square.

The perimeter function uses fact that xy = 100, so y = 100/x, and perimeter = 2x + 2y, so

P = 2x + 2y

P = 2x + 2(100/x)

Interesting question.

Maybe try solving for y:

y = 3 - 3x - x^2

If you graphed that, you'd get a downward-U shape called a parabola. Here's the graph I used and will be referencing throughout.

I'm also going to advise graphing the line f(x)=-x.

Why? Well, if the number pair from the parabola y ALSO fall on this line f, that means the will sum to zero. Number pairs on the parabola y that are BELOW this line f will sum to a negative number, and those that are ABOVE this line f will sum to a positive number. Think about what that line communicates about a number and its opposite, how the relative position to that line squares with what I just said before continuing. Play around on that graph, drag your cursor around the parabola; convince yourself this is true.

The further you get from that line f(x), the greater that sum becomes. What point on that parabola is the furthest away from that line? I have a good guess from looking at the graph (and you can probably use this to ballpark your guesses and arrive at the solution max(x+y)=4 just from here), but how to figure it out analytically?

Well, what if we "slide" that line y=-x up the y-axis? What would that mean? Do that by sliding b in y=-x+b. Notice how the line is like "sweeping" up that parabola, intersecting it at two places. If you get the furthest away from the original b=0, while still touching the parabola, it means your line will now just barely graze the parabola. This is called being tangent to it.

Our question is now: what line, with slope -1, is tangent to the parabola y?

This is a calculus question involving derivatives. Differentiate y:

dy/dx = -3 - 2x

Then, that is equal to -1 when:

-1 = -3 - 2x

2 = -2x

x = -1

Thus at (-1, 5), x+y is a maximum.

Play with 'a' in this visual if you'd like. It should help you see what's going on graphically.

The function will be continuous if there are no holes. You'll notice that the upper expression of the piecewise function is defined up to but not including 3, while the lower is defined from and including 3. Horizontally speaking, no holes are present. The only thing we need to see now is when this is true vertically speaking. To do this, set x = 3 for both expressions and equate them, solving for a:

(3)^(2) - 1 = 2(3)(a)

8 = 6a

a = 8/6

Just add the expressions according to how the domain is partitioned.

Let f(x) = x^2, for all x. Then f is continuous.

Let g(x) = x if x<0, and x^2+1 if x >= 0. Then g is not continuous at x = 0, but continuous everywhere else.

To add f and g, add the expressions for when x < 0, and for when x >= 0:

f(x) + g(x) = x^2 + x if x <0, and 2x^2 + 1 if x >= 0.

This is what the functions look like.

ln(3) is a constant, basically they want the integral from x=0 to x=ln(3) of cosh(2x)+sech^(2)(x)
This gives [(1/2)sinh(2x)+tanh(x)]0^ln(3)
Which gives
(1/2)sinh(2ln(3))+tanh(ln(3))-(1/2)sinh(2(0))-tanh(0)=(1/2)sinh(ln(9))+tanh(ln(3))
Using the definitions of sinh and tanh:
(1/2)(1/2)(e^ln(9) -e^-ln(9) ) + (e^ln(3) -e^-ln(3) )/(e^ln(3) +e^-ln(3) )
(1/4)(9-1/9) + (3-1/3)/(3+1/3)
(1/4)(80/9) + (8/3)/(10/3)
20/9 + 4/5
100/45 + 36/45
136/45

For the diagram, try approximating the area with a triangle: desmos

As for tan(), technically its domain is all real numbers EXCEPT for x = (2n + 1)*(pi/2), where n is an integer. The answer key is probably referring to the RESTRICTED DOMAIN of tan(x) so that its inverse, arctan(x) is actually a function that passes the vertical line test (if you graph arctan(x), its range is -pi/2 < x < pi/2, which is the domain of tan(x), and its domain is all real numbers, which is the range of tan(x)).

Graph: https://www.desmos.com/calculator/cs8ojjybve

If you go to the first function and delete the bracketed portion (the restriction), you can see tan(x) in its full periodic form (which, if we were to flip it across the line y = x, would fail the vertical line test, since there would just lots of "S" shaped graphs crossing everywhere along the y-axis).

I don't think you should have the 1 part in this. It should just be the sum of the squares. And there is a well known formula for the sum of squares

http://www.wolframalpha.com/input/?i=sum+x^2+from+x+%3D+1+to+n

Which you could prove by induction.

>0.65 (guilty) <-single juror making right decision: 0.8 need to convict 9..12
>0.35 (innocent) <-single one making right decision: 0.9 need to acquit 4..12

so it's:

>0.65*\sum_{i=9}^{12} 0.8^i 0.2^{12-i} (12 over i) + 0.35*\sum_{i=4}^{12} 0.9^i 0.1^{12-i} (12 over i) = wolfarmalpha = 0.86647

sounds kinda low? :S

the logic being: > 1. first you decide whether s/he's guilty or not since odds differ
> 2. you check how many people you need to make the right decision (first one needing 9..12) and sum
> 3. for every such group (say the 9) do the following:
> 3a. pick some 9 people from the 12: (12 over 9)=12!/(9!*(12-9)!)
> 3b. make them all pick right: 0.8^9
> 3c. make the others pick wrongly: (1-0.8)^(12-9)
> 4. sum it all :D

edit: formatting

Okay, thanks. Let's try this.

y = 4^(x^2 + 3)

ln(y) = ln(4^(x^2 + 3))

ln(y) = (x^2 + 3)ln4

Differentiate…

1/y * y' = (2x)ln4

Multiply by y (well, by what it equals).

y' = (2x)ln4 * 4^(x^2 + 3)

That doesn't seem like the answer Wolfram|Alpha proposes, so I'm still slightly confused (assuming I did that correctly).

http://www.wolframalpha.com/input/?i=derivative+of+4%5E%28x%5E2+%2B+3%29

Start by dividing the threshold total weight by the total number of men to get the threshold mean weight of each man. Then calculate the probability of the mean weight per person in a sample of size 36 being greater than that threshold mean weight (6000/36) given the population mean and the population standard deviation. However, keep in mind that you are drawing the mean weight of the men from a sampling distribution drawn from the underlying distribution and not the underlying distribution representing the weights of individual men.

I tried to write this in such a way as to give a hint as to the course of action to be taken without giving away much. If you still need help, this link should help or I can help out whenever I wake up.

What you will want to do is what they've said below, plug your value of 2 in.

Once you've got that, you have another term you want to get rid of; f(2/3).

Since this recursive function holds for almost all values of x (as you said, x = -1 is undefined), let's plug it back in. You get 1 - f(2/3) + 3f(2/(2/3 +1)) = 1 - f(2/3) + 3f(6/5) = 2

Let's see if f(6/5) reduces. Most likely it won't, so you're going to be stuck at the original statement of f(2) = 3f(2/3) - 5.

Usually, you can take recursive functions as above, and find a condition where you end up getting the terms to loop back. (There was a question like this earlier this week in the same style that had a closed answer).

Also, you can solve for the answer of f(1), as 4f(1) = 2 so f(1) = 1/2.

A recursive definition will have a base case, which prevents an infinite loop. For example, here is the recursive definition of the factorial function:

f(0) = 1 f(x) = f(x-1) * x

So to evaluate f(5), you'd go:

f(5) = f(4) * 5 = f(3) * 4 * 5 = f(2) * 3 * 4 * 5 = f(1) * 2 * 3 * 4 * 5 = f(0) * 1 * 2 * 3 * 4 * 5 = 1 * 1 * 2 * 3 * 4 * 5 = 120

So f(5) = 120.

Note that if there is no base case (e.g. if the factorial function was simply "f(x) = f(x-1) * x") there's no answer, because you'd just keep going deeper and deeper and never come up with an answer that doesn't involve f(x). This is a circular definition as opposed to a recursive one.

I used Rhinoceros for the graphic. There's a proof somewhere regarding the close-packing of spheres that states that the tetrahedron configuration minimizes the space taken up, I think.

Right-hand rule is a pretty touchy subject between teachers/students. Everyone swears their way is best. As someone with a poor spatial awareness in general, I find the following to work best:

It's known that the unit coordinate vectors follow: i x j = k (i is unit vector in x direction, j is unit in y direction, etc).

To find A x B = C:

1. Point all four fingers along A, like this picture: http://en.citizendium.org/images/thumb/e/e4/Right-hand-rule.jpg/350px-Right-hand-rule.jpg

2. Curl those four fingers towards B. In that picture, if B was pointing into the background, you kinda have to rotate your wrist 180 degrees to do this.

3. C points along the direction of your thumb.

I like this one because it's super simple, and easy to apply in general to cross products. But, that's just my dogma. Try and find one that works for you, and stick with it. Once you do, try it with i x j, and see what you get.

That's going to be a tricky one. Looks like you should start off with partial fraction decomposition and then just integrate by parts. Cymath, Symbolab, and Integral Calculator will be your best friends.

Yes, it would make more sense that way. Because the stability of the fixed points changes depending on the parameter. This process is known as bifurcation and is important when studying phase planes.

Usually when someone says numericaly plot the phase plane they mean calculate different trajectories with different initial conditions and possibly different parameter values. Usually you would pick parameter values in all of the different stability states of the parameters. The number and type of fixed points can and does change sometimes (see the bifurcation article).

Also, you may want to plot the direction field, which usually means plotting arrows at a few points in the direction of the derivatives.

This is what that looks like in python.

Most likely, the main part of your project is implementing the Runge-Kutta algorithm in C++.

Fair enough, I prefer f*i(x)=C*Πj∈{1..n}* (x-x*j) + *Πj ≠ i (x-x*j)/(xi-xj) for all real(well complex) C, as it provides you with a family of polynomials all sharing the property that fi(xi) = 1 and fi(xj*) = 0 when j ≠ i, and with your preference as a special case when C = 0

Also here's a little desmos graph showing that function, you can change the list of numbers(it should not contain duplicates), you can change i(it needs to be an integer from 1 to the length of the list), and you can change the constant C

I actually find these functions to be quite nice looking when you superimpose all functions for all i, here's how it looks (for the C=0) case

C2: y^2 + y(2x -5) + (-4x -5 +x^2 ) =0

y = -(1/2)(2x-5) ± (1/2)[(2x-5)^2 -4(-4x -5+x^2 )]^1/2

y = -(1/2)(2x-5) ± (1/2)[45-4x]^1/2

Check

45-4x >0 when x< 45/4

I = { x= 4 to x= 45/4}

A = ∫I y*+* - y*-*

In quadratic graphs like these, p is the distance shifted left or right (if p is negative, shifted right), and q is the scaling (so a greater q would give a more vertically stretched graph, a negative q would flip the graph, a q less than 1 would shrink it etc).

If the graph was just y = x^2, it would be a parabola facing upwards, centered on 0. This graph has been shifted two units to the right (so a p of -2), and inverted (so q must be negative). The exact value of q can then be found by substituting in known values and solving, e.g.

y = q(x+p)^2

-4 = q(0 - 2)^2

-4 = 4q

q = -1

(you know from the graph that when y = -4, x = 0, and, from the reasoning above, p = -2)

https://www.desmos.com/calculator/l7vxtxwjoq Play with the equation in this graphing calculator (change the p and q values and watch how it changes) to get a feel for how this works.