Melody

Find the equation of a sine function that has a period of pi, amplitude 5, a vertical shift of zero and passes through (pi/6, 5/2).

\(y=asin(n(x-k))\)

a = amplitude = 5

period = 2pi/n,  so n=2

k is the horizontal phase shift which we have not been given so

y=5sin[2(x-k)]

and it passes through  (pi/6,  5/2)

sub to find k

\(\frac{5}{2} =5sin[2(\frac{pi}{6}-k)]\\~\\ \frac{1}{2} =sin[2(\frac{pi}{6}-k)]\\~\\ \frac{\pi}{6}=2(\frac{pi}{6}-k)\\~\\ \frac{\pi}{6}=\frac{\pi}{3}-2k\\~\\ k=\frac{\pi}{12}\\~\\ k\approx\;\;0.26\qquad (\mbox{ I added this for the benefit of the diagram)}\\~\\ so\\~\\ y=5sin[2(x-\frac{\pi}{12})] \)

The phase shift  pi/12 in the positive direction.

I forgot to include the graph

1- sin^2theta/ csc^2thea-1

This is the question that you asked. - it is probably not be the question that you intended.

Perhaps you should introduce some brackets.  

\(1- \frac{sin^2\theta}{csc^2\theta}-1= \frac{-sin^2\theta}{csc^2\theta}=-sin^4\theta \)

I really like your explanation Chris :)

show that 1+i is a solution for the equation x^2-2x+2=0

\((1+i)^2 =1^2+2i+i^2=1+2i-1=2i\\ -2(1+i)=-2-2i\\ so\\ (1+i)^2-2(1+i)+2\\ =2i-2-2i+2=0\\ \therefore\\ 1+i \quad \mbox{is a solution to }\;\;x^2-2x+2=0\)

If the graph of y = |x| is translated so that the point (1, 1) is moved to (1, 4), what is the equation of the new graph?

If you think about those 2 points, the new one is exactly 3 units directly above the old one so it is just the old y value + 3 more.

y=|x|+3

What square is the product of four consecutive odd integers ?

\((2n-3)(2n-1)(2n+1)(2n+3)=k^2\\ (4n^2-9)(4n^2-1)=k^2\\ 16n^4-40n^2+9=k^2\\ 16n^4-40n^2=k^2-9\\ 16(n^4-\frac{40}{16}n^2)=k^2-9\\ 16(n^4-\frac{10}{4}n^2+\frac{25}{16})=k^2-9+25\\ 16(n^2-\frac{5}{4})^2=k^2+16\\ 16(n^2-\frac{5}{4})^2=k^2+16\\ \)

We have a pythagorean triad here.

I mean      k,      4     and    4(n^2-5/4)      is a pythagorean triad

 If k=3 then

\(16(n^2-\frac{5}{4})^2=25\\ 4(n^2-\frac{5}{4})=5\\ (n^2-\frac{5}{4})=\frac{5}{4}\\ (n^2-\frac{5}{4})=\frac{5}{4}\\ n^2=0\\ n=0\)

So one solution is      \(-3*-1*1*3=9\)

YES I know, I did that the SUPER long way!

According to WolframAlpha, this is the ONLY integer solution. 

So 9 is the product of 4 consecutive odd integers.   

2x^2-5x-6=0

The number under the square root is called the discriminant and the symbol for it is a trangle.

It is +73 whench means there are 2 real roots :)

\(2x^2-5x-6=0\\ \triangle=25-4*2*-6\\ \triangle=25+48\\ \triangle=73\\ \)

y=sin(x+100°)+1

y=sin(x)

Move the curve left 100 degrees and you have   y=sin(x+100)      [That is a phase shift of -100 degrees.]

Move the curve up 1 unit and you get y=sin(x+100)+1

Here are the graphs 

how to find y value in projectile motion

It would be better if you give a full question and then it can be discussed. 

If you want just the y value can be discussed.

ROF LOL

Good luck to you, a pessimist won't loan you any in the first place because they will figure that they won't get it back.