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Sure looks like a Feynman path integral (or "functional integral") for a quantum field theory. see e.g. http://www.scholarpedia.org/article/Path_integral#A_generalization:_The_field_integral

Btw. part of one of the millennium problems consists in showing that these are mathematically well defined (physicists usually don't bother about this).

For an central angle in standard position, this will happen at 135° and 225°.

The radius of the circle is r = sqrt(98) = 7*sqrt(2) so the coordinates of the intersection of the circle with the tangent lines are given by

p1 = r*<cos(135°), sin(135°)> = <-7,7>

p2 = r*<cos(225°), sin(225°)> = <7,-7>

We know that for a line of the form y=mx+b that m=1 since the gradient is 1.

Now, we find

p1: 7=-7+b => b=14

p2: -7=7+b => b=-14

Therefore,

y1 = x+14

y2 = x-14

https://www.desmos.com/calculator/45fhzetv96

I'm not entirely sure what you're trying to graph exactly, but based on your description I you're basically trying to make this common representation of the EM spectrum? In that case, something like [cos(10*ln(-x))] https://www.desmos.com/calculator/onifhydmoo) should do the trick. If you want a precise correspondence between wavelength in meters and the number line you need to throw in a few constants and such.
To answer your question about translating graphs, to mirror a graph about the y axis, you need to replace x with -x wherever you find it, and to move a graph n units to the right, replace it with x+n.
Finally, that "anomaly" you came across at e in your graph of cos(x^(16/x) ) is because the graph of x^(16/x) has a local maximum at that point (this is no coincidence, it has to do with calculus)

You should know that rectangles have 90 degrees in each corner, equilateral triangles have 60 degrees in each corner, isosceles triangles have two base angles the same (so PRQ = 67 in the third shape, and the angle sum of any triangle is 180.

Then you also need to know that angles in a right angle will add to 90 (so 90 - 39 = 51 in the first shape) and that angles on a straight line add to 180 (so 180 - 60 = 120 in the second shape).

Put in as much information as you can and you should be able to get the answers.

Answers are 69, 23 and 33

Not sure how to do this exactly, but check out this desmos link for how I'd do it. First we subtract 12.12 (the average y value) from each y value to get a graph which looks like Acos(Bx-C). Then divide by 1.68 (the highest y value) to get a graph which looks like cos(Bx-C). I was going to play around to find C and B but immediately B=0.5, C=0 looks like a very good fit.
So A=1.68, B=0.5, C=0, D=12.12 is an alright guess.

Check this out:
https://www.desmos.com/calculator/rnrxrqkgdt
The function x = a√sin(by) makes a series of "petals" along the y-axis, and switching y->r and x->theta can be thought of as wrapping the y axis around the origin, you can see how the lowest petal in the x,y form gets warped into the first petal in the r,theta form. There's a few youtube videos which show this graphically if you wanna have a look.

Put more effort in your post.

Anyway, this is a equation in two coordinates which you can plot using Desmos as

3x - 2y = x*y

As you can see, as x increases it asymptotes to y=3, same thing when x decreases. It's discontinuous at x = -2.

I'm pretty sure that's the textbook I used on my calculus course. I'd highly recommend checking out Z-Library for textbooks: I don't think I've every needed one that they didn't have for free as a PDF.

Of which we did Isolating a variable means rearranging an algebraic equation so that a different variable is on its own. The goal is to choose a sequence of operations that will leave the variable of interest on one side and put all other terms on the other side of the equal sign. For example, Isolate x in the following equation: x+4=12. To isolate x, we must get rid of the 4 term from the left side of the equation and move it to the right side of the equation. To do this, subtract 4 from both sides of the equation: x+4−4=12−4. Then we can see that 4−4=0, so the left side of the equation is simply x. The resulting expression is x = 8 Ref- brilliant

Ok, next attempt. We compute the vectors from the origin to the left and right vertex of the horizontal line labelled w as P=(x,y) and Q=(x+w,y).

(x+w)^2+y^2=r0^2

x^2+y^2=r^2

2*x*w+w^2=r0^2-r^2

x=(r0^2-r^2-w^2)/(2*w)

y=sqrt(r^2-x^2)

The unit vectors from the origin in direction to the vertices are now:

v=Q/sqrt(Q.Q)

w=P/sqrt(P.P)

From then on it is like in my original expression:

cross(v,z)=a and cross(z,w)=b

Let v0, v1 and w0, w1 be x and y coordinates of v and w respectively.

z = inverse([[-v1,v0],[w1,-w0]]).(a,b) =

1/(v0*w1-v1*w0) * [a*w0+b*v0, a*w1+b*v1]

This time tested with Geogebra: https://www.geogebra.org/calculator/tkzjezdx

The simplest way to solve this is to understand how to use Pascal's triangle according to the binomial theorem.

Have a look at Pascal's triangle:

https://www.geogebra.org/m/hcmgsfpu

Those rows of numbers that you see there are rows of coefficients. If you are unsure how to read Pascal's triangle, I'd suggest looking it up. It's useful.

But back to our problem: Pascal's triangle is for (x + y)^n . But we have (2x + 1)^n. Therefore, we can't just read the coefficient off immediately (even if 84 is there in the triangle; this would be wrong).

We must understand that every time x is multiplied by itself, so is 2. So x^2 will have 2^2 = 4 next to it: 4 x^2

This means, in order to read off Pascal's triangle, we must first divide 84 by 4 to "translate it" into (x + y)^n . 84 / 4 = 21.

Now we read it off. The row of coefficients that has 21 as a coefficient of x^2 is 7 rows from the top. This corresponds to n = 7, which is our answer.

Maths teacher here. The absolute best way I've found is SLOP - Shed Loads Of Practice.

I second corbettmaths, its an amazing resource.

One of my friends at a different school shared this with me which gives you loads of other resources - the exam stuff is mainly edexcel so check with your teachers of your exam board before you look at the papers.

https://www.amazon.com/dp/B08MBSJL8K

Folks having Difficulties or any sort tiff with algebras and equations and struggling with the X in your life , here is the solution for all your problems in simplified and sorted way. You get access to all the excercises in simplistic manner thus truly making Algebra Easy.

To steal the title of a book:

I Think You’ll Find It’s a Bit More Complicated Than That

Different countries record their statistics for what counts as a COVID death differently.

Also, there's a huge difference in the death rate between a modern, unstressed, health system and a non-functioning/non-available health system. Figures I've heard, but not verified, is that without a functioning and non-stressed health system the mortality rate could be closer to 10%.

Also, you don't just have direct COVID deaths. You also have deaths caused due to the impact that COVID has had on the health system, economy and society in general.

I can't quite make sense of their PR calculation, but this is how I get it to work:

S is vertically below P. Because of the symmetry, it should be fairly clear that S is on AC, and in fact that S is 1/4 of the way along AC. Also SP is half of h, ie 2sqrt(7).

R is on BC, 2cm from BC (1/4 of the way along), so that angle RSP is a right-angle and SR is at right angles to BC.

Because AB = 8, ratios should make it obvious that SR = 6. (Also BR = 2).

So, PR^2 = SP^2 + SR^2 which leads to PR = 8.

Now the triangle PRQ is right-angled with right-angle at R, and from the information given, and BR=2, it should be fairly obvious that RQ = 4.

So PQ^2 = PR^2 + RQ^2 = 8^2 + 4^2, leading to the required answer.

https://www.amazon.com/Trigonometry-Essentials-Practice-Workbook-Answers-ebook/dp/B011ZLN00E/ref=sr_1_2?ie=UTF8&amp;qid=1548941349&amp;sr=8-2&amp;keywords=trigonometry+practice+workbook

Go through this. You'll never forget it because there are so many problems. It'll take you two to three weekends if you really take your time.

Get the paperback.