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2 years ago

Find the area of the shaded regions

Mathematics

1 answer:

Salsk061 [2.6K]2 years ago

4 0

your answer would be 254 I just did it and I got it right ✨

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Determine whether or not the following graph represents two quantities that are proportional to each other. Explain your reasoni

m_a_m_a [10]

Answer:

Not proportional.

Step-by-step explanation:

The values do not begin from a straight line at the origin. X begins at 2, not the origin.

4 0

3 years ago

Verify the following i.[ -3/4]3 = -27/64 ii. [-2/3]6 = 64/ 729

Juliette [100K]

Answer:

The relationships are correct and valid

Step-by-step explanation:

The correct question is as follows;

Verify the following

i.[ -3/4]^3 = -27/64

ii. [-2/3]^6 = 64/ 729

We have the solution as follows;

i) we have;

(-3/4)^3

That means ;

-3^3 = -3^3 = -27

4^3 = 64

so;

(-3/4)^3 = -27/64

ii) (-2/3)^6

(-2)^6 = 64

3^6 = 729

Thus;

(-2/3)^6 = 64/729

4 0

3 years ago

Simplify. (-1/2√2)(-6√18) 18 -18 -36

olga_2 [115]

Hey!

Your answer is solved below/above idk wherever just see in the picture.

8 0

3 years ago

Read 2 more answers

What is sin 0 when sec 0 = 5? 0 |(1,0) sin e = 2[?] [] Rationalize the denominator if necessary.

Komok [63]

Answer:

$Sin ~0 =\frac{2\sqrt{5} }{5}$

Step-by-step explanation:

$sec0=\sqrt{5}$

$cos~0=\frac{1}{sec0} =\frac{1}{\sqrt{5} }$

$sin ~0=\sqrt{1-cos^2~0}$

$=\sqrt{1-1/5}=\sqrt{5-1/5}$

$Sin~0=\sqrt{4/5} =2/\sqrt{5}*\sqrt{5}/\sqrt{5}$

$Sin~0=2\sqrt{5}/2=2\sqrt{5}/5$

--------------------------------

hope it helps...

have a great day!!!

7 0

3 years ago

Solve for x -2^2(-5x+10)=100

masha68 [24]

Answer:

$x=7$

Step-by-step explanation:

Hi there!

$-2^2(-5x+10)=100$

Solve -2^2:

$-4(-5x+10)=100$

Divide both sides by -4:

$\displaystyle \frac{ -4(-5x+10)}{-4}=\frac{100}{-4} \\\\-5x+10=-25$

Subtract 10 from both sides:

$-5x+10-10=-25-10\\-5x=-35$

Divide both sides by -5:

$\displaystyle \frac{-5x}{-5} = \frac{-35}{-5} \\\\x=7$

Therefore, x = 7.

I hope this helps!

3 0

3 years ago