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3 years ago

9

A square is inscribed in a circle. If the area of the square is 9 in^2, what is the ratio of the circumference of the circle to

the area of the square?

1 answer:

Alja [10]3 years ago

3 0

Answer:

√2π / 4

Step-by-step explanation:

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4 years ago

How much further up the wall does the ladder in figure 2 reach than

ASHA 777 [7]

The ladder in figure 2 reaches 3.2 feet further up than the ladder in figure 1.

Step-by-step explanation:

Step 1:

In both the given figures, the ladder, the wall and the floor form a right-angled triangle. The floor is the adjacent side, the wall is the opposite side and the ladder is the hypotenuse.

According to the Pythagorean theorem,

$a^{2}+b^{2}=c^{2}$ , where c is the length of the hypotenuse while a and b are the lengths of the other two sides.

Step 2:

For the ladder in figure 1, assume the distance from the floor to the ladder's top is x feet. So a = x, b = 8 and c = 10 (hypotenuse).

$a^{2}+b^{2}=c^{2}$ , $x^{2}+8^{2}=10^{2}$ , $x^{2}=100-64=36, x=\sqrt{36}=6 \text { feet }$ .

So the distance between the floor and the ladder's top is 6 feet.

Step 3:

For the ladder in figure 2, assume the distance between the floor and the ladder's top is y feet. So a = y, b = 4 and c = 10 (hypotenuse).

$a^{2}+b^{2}=c^{2}$ , $y^{2}+4^{2}=10^{2}$ , $y^{2}=100-16=48, x=\sqrt{84}=9.1651 \text { feet }$ .

So the distance between the floor and the ladder's top is 9.1651 feet.

Step 4:

The difference in heights = The wall height in figure 2 - the wall height in figure 1.

The difference in heights = 9.1651 feet - 6 feet = 3.1651 feet.

Rounding this off to the nearest tenth of a foot, we get 3.2 feet.

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3 years ago

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3 years ago

Read 2 more answers

4/5 divided by 1/3 plus 1/5 minus 3/5

r-ruslan [8.4K]

Answer:

$\frac{\frac{4}{5}}{\frac{1}{3}+\frac{1}{5}-\frac{3}{5}}=-12$

Step-by-step explanation:

Considering the expression

$\frac{\frac{4}{5}}{\frac{1}{3}+\frac{1}{5}-\frac{3}{5}}$

Solution Steps:

$\frac{\frac{4}{5}}{\frac{1}{3}+\frac{1}{5}-\frac{3}{5}}$

as

$\mathrm{Combine\:the\:fractions\:}\frac{1}{5}-\frac{3}{5}:\quad -\frac{2}{5}$

so

$=\frac{\frac{4}{5}}{\frac{1}{3}-\frac{2}{5}}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{\frac{b}{c}}{a}=\frac{b}{c\:\cdot \:a}$

$=\frac{4}{5\left(\frac{1}{3}-\frac{2}{5}\right)}$

join $\frac{1}{3}-\frac{2}{5}:\quad -\frac{1}{15}$

so

$=\frac{4}{5\left(-\frac{1}{15}\right)}$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$=\frac{4}{-5\cdot \frac{1}{15}}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{-b}=-\frac{a}{b}$

$=-\frac{4}{5\cdot \frac{1}{15}}$

$\mathrm{Multiply\:}5\cdot \frac{1}{15}\::\quad \frac{1}{3}$

so

$=-\frac{4}{\frac{1}{3}}$

$\mathrm{Simplify}\:\frac{4}{\frac{1}{3}}:\quad \frac{12}{1}$

so

$=-\frac{12}{1}$

$\mathrm{Apply\:rule}\:\frac{a}{1}=a$

$=-12$

Therefore

$\frac{\frac{4}{5}}{\frac{1}{3}+\frac{1}{5}-\frac{3}{5}}=-12$

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3 years ago

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Answer:

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Step-by-step explanation:

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