3 years ago

CAN SOMEONE PLEASEEE HELP SIMPLIFY THESE TWO EXPONENT PROBLEMS I HAVE NO CLUE HOW

Mathematics

1 answer:

Alecsey [184]3 years ago

6 0

Answer:

20. $-\frac{y^{12} }{9x^{4} }$

Step-by-step explanation:

21. $3 x^{2} y^{2}$

You might be interested in

What is the slope of the line passing through (16, -2) and (-30, 6)?

Novay_Z [31]

$m=\frac{\Delta y}{\Delta x}=\frac{-2-6}{16-(-30)}=\frac{-8}{46}=\boxed{-\frac{4}{23}}$ .

Hope this helps.

5 0

3 years ago

Read 2 more answers

If twice a number, n, is 5 less than the number squared, which of the following equations could be used to properly solve for n?

hjlf

Answer:

2n - 5 = n^2 (or n squared)

3 0

3 years ago

Which expression is equivalent to 9 x squared minus 2 y + 3 x squared minus 3 y?

lyudmila [28]

Answer:

The correct answer is 11 x squared + 2 y + x squared minus 3 y

Step-by-step explanation:

Just took the test on Edg. Hope it helps :)

8 0

3 years ago

Read 2 more answers

I REALLY NEED HELP I WILL PUT CROWN IF YOU ANSWER CORRECTLY

faust18 [17]

Answer:

I think it would be $\frac{36}{4} -6$

Step-by-step explanation:

stomp me if its wrong <3

$\frac{36}{4} -6 = 9-6 = 3$

<em> hopefully this helps ~</em>

5 0

1 year ago

Read 2 more answers

How to solve this :') please help

blsea [12.9K]

Answer:

3/2

Step-by-step explanation:

sin(3π/4 - β) = sin(3π/4)cosβ - cos(3π/4)sinπ =

$\sin(\frac{3\pi}{4}-\beta)=\sin(\frac{3\pi}{4})\cos\beta-\cos\frac{3\pi}{4}\sin\beta\\=\frac{1}{\sqrt{2}}\cos\beta-(-\frac{1}{\sqrt{2}})\sin\beta\\\\=\frac{1}{\sqrt{2}}(\cos\beta +\sin\beta)\\\\\sin^2(\frac{3\pi}{4}-\beta)=\frac{1}{2}\cos\beta +\sin\beta)^2=\frac{1}{2}(\cos^2\beta +\sin^2\beta+2\sin\beta\cos\beta\\=\frac{1}{2}(1+\sin2\beta)=\frac{1}{2}(1-\frac{1}{5}) = \frac{2}{5}\\$

Use $\cot^2x=\csc^2x-1=\frac{1}{\sin^2x}-1$

$\cot^2(\frac{3\pi}{4}-\beta)=\frac{1}{\sin^2(\frac{3\pi}{4}-\beta)}-1 = \frac{1}{\frac{2}{5}}-1=\frac{5}{2}-1=\frac{3}{2}$

7 0

2 years ago