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3 years ago

Two similar polygons have areas of 16 square inches and 64 square inches. The ratio of a pair of corresponding sides is 1/4.

Mathematics

1 answer:

Luden [163]3 years ago

5 0

Your answer is A. True.

Hope this helps.

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John has six bills of paper money in the following denominations $1, $5, $10, $20, $50, and $100. If he selects 3 bills at a tim

densk [106]

the probability will be 0.55

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3 years ago

Rewrite with only sin x and cos x.

Annette [7]

Option A

$\cos 3 x=\cos x-4 \cos x \sin ^{2} x$

Solution:

Given that we have to rewrite with only sin x and cos x

Given is cos 3x

$cos 3x = cos(x + 2x)$

We know that,

$\cos (a+b)=\cos a \cos b-\sin a \sin b$

Therefore,

$\cos (x+2 x)=\cos x \cos 2 x-\sin x \sin 2 x$ ---- eqn 1

We know that,

$\sin 2 x=2 \sin x \cos x$

$\cos 2 x=\cos ^{2} x-\sin ^{2} x$

Substituting these values in eqn 1

$\cos (x+2 x)=\cos x\left(\cos ^{2} x-\sin ^{2} x\right)-\sin x(2 \sin x \cos x)$ -------- eqn 2

We know that,

$\cos ^{2} x-\sin ^{2} x=1-2 \sin ^{2} x$

Applying this in above eqn 2, we get

$\cos (x+2 x)=\cos x\left(1-2 \sin ^{2} x\right)-\sin x(2 \sin x \cos x)$

$\begin{aligned}&\cos (x+2 x)=\cos x-2 \sin ^{2} x \cos x-2 \sin ^{2} x \cos x\\\\&\cos (x+2 x)=\cos x-4 \sin ^{2} x \cos x\end{aligned}$

$\cos (x+2 x)=\cos x-4 \cos x \sin ^{2} x$

Therefore,

$\cos 3 x=\cos x-4 \cos x \sin ^{2} x$

Option A is correct

7 0

3 years ago

Solve the simultaneous equations: 3a-2b=14 12a+9b=39

kirill [66]

Use elimination
Multiply first equation by 4
12a - 8b = 56
12a + 9b = 39
Subtract both
-17b = 17
b = -1
Plug in -1 for b
3a - 2(-1) = 14
3a = 12, a = 4
Final answer: a = 4, b = -1

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3 years ago

What is the volume of a cube that has a side length of 5 centimeters? Recall the formula .

mixas84 [53]

The answer would be 125

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3 years ago

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What type of triangle has no congruent sides and one angle that has a measure greater than 90° but less than 180°?

ololo11 [35]

Answer:

D as it is the obtuse

Step-by-step explanation:

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3 years ago

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