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3 years ago

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The length of the side of square A is 50% of the length of the square side expressing the area of the shaded area of square

A as a percentage of the area of square B.

1 answer:

KonstantinChe [14]3 years ago

8 0

Answer:

Let's call the side of square b x. Therefore, the side of square a is 0.5x, making its area 0.25x² and the area of square b is x². This means that the percentage is 25%.

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Help please, i am struggling on how to solve!

Elina [12.6K]

Answer:

$x=3$

Step-by-step explanation:

The hypotenuses are equal in length of both triangles.

$13=4x+1$

$13-1=4x$

$12=4x$

$12/4=x$

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3 years ago

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Solve the equation for the indicated variable<br> x - 3y for y

Colt1911 [192]

Answer:y=x/3

Step-by-step explanation:

x-3y=0

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HELP!!!brainlist<br> Drag a reason to each box to complete the flowchart proof.

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The complete proof statement and reason for the required proof is as follows:

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4 years ago

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10x+2y=10 in slope intercept form

faust18 [17]

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3 years ago

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Evaluate the following integral using trigonometric substitution

serg [7]

Answer:

The result of the integral is:

$\arcsin{(\frac{x}{3})} + C$

Step-by-step explanation:

We are given the following integral:

$\int \frac{dx}{\sqrt{9-x^2}}$

Trigonometric substitution:

We have the term in the following format: $a^2 - x^2$ , in which a = 3.

In this case, the substitution is given by:

$x = a\sin{\theta}$

So

$dx = a\cos{\theta}d\theta$

In this question:

$a = 3$

$x = 3\sin{\theta}$

$dx = 3\cos{\theta}d\theta$

So

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9-(3\sin{\theta})^2}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9 - 9\sin^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{\theta})}}$

We have the following trigonometric identity:

$\sin^{2}{\theta} + \cos^{2}{\theta} = 1$

So

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

Replacing into the integral:

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{2}{\theta})}} = \int{\frac{3\cos{\theta}d\theta}{\sqrt{9\cos^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{3\cos{\theta}} = \int d\theta = \theta + C$

Coming back to x:

We have that:

$x = 3\sin{\theta}$

So

$\sin{\theta} = \frac{x}{3}$

Applying the arcsine(inverse sine) function to both sides, we get that:

$\theta = \arcsin{(\frac{x}{3})}$

The result of the integral is:

$\arcsin{(\frac{x}{3})} + C$

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3 years ago