Thread: Need allitle help on these questions.

1. Need allitle help on these questions.

I'm having trouble on these questions trying to figure them out but think my work is wrong.

1.Which of the following is equavlant to (2z^3)/(8z^(n-2))
a. 1/4z^1-n
b. z^5-n
c. 1/4z^(5-n)

2.Find the range of negative squareroot (36-x^2)
so far I gotten (x+6)=-6 and (x-3)=6 so would it be y:y -6<y<6?

3. Find all solutions for sin^2=3cos^2
so would it be 0, pie/2, 1/3 ?

2. Originally Posted by goldenroll
I'm having trouble on these questions trying to figure them out but think my work is wrong.

1.Which of the following is equavlant to (2z^3)/(8z^(n-2))
a. 1/4z^1-n
b. z^5-n
c. 1/4z^(5-n)
1. choice is c, i wont explain it because i think you wrote in the answer? i get z^(5-n)/4
2.Find the range of negative squareroot (36-x^2)
so far I gotten (x+6)=-6 and (x-3)=6 so would it be y:y -6<y<6?

3. Find all solutions for sin^2=3cos^2
so would it be 0, pie/2, 1/3 ?
Number 3 is pi/3

sin^2x = 3(1 - sin^2x)
4sin^2x = 3
sinx = root of 3 / 2
x = 60
x = pi / 3

and i think your number 2 is right because
you switch x and y and then solver for 36 - y^2 under the square root to be greater than or equal to 0

3. Originally Posted by goldenroll
I'm having trouble on these questions trying to figure them out but think my work is wrong.

1.Which of the following is equavlant to (2z^3)/(8z^(n-2))
a. 1/4z^1-n
b. z^5-n
c. 1/4z^(5-n)

2.Find the range of negative squareroot (36-x^2)
so far I gotten (x+6)=-6 and (x-3)=6 so would it be y:y -6<y<6?

3. Find all solutions for sin^2=3cos^2
so would it be 0, pie/2, 1/3 ?
For Q1:

$\frac{2z^3}{8z^{n-2}}=\frac{1z^{3-(n-2)}}{4}=\frac{z^{5-n}}{4}$

Thus the answer is neither of the choices. If the question is written as (2z)^3, then the answer is B. If it's 2z^3, then my answer doesn't fit into any.

For Q3:

$sin^2x=3cos^2x$

Divide by $cos^2x$ on the both sides, you get: $tan^2x = 3$

Square root both sides, you get: $tan x$ = sqrt(3) or -sqrt(3)

Therefore the critical angle is pi/3. If you are looking angles between 0 and 2pi, the possible answers are pi/3, 2*pi/3, 4*pi/3, 5*pi/3.

Hope this helps.

4. $
\frac{2z^3}{8z^{n-2}}=\frac{1z^{3-(n-2)}}{4}=\frac{z^{5-n}}{4}
$
That's c isn't it?

It depends how you read the brackets .

5. Yes I think
c should be read as $\frac14\:z^{5-n}$