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3 years ago

the radius of a circle is 3 inches. what is the area of the sector bounded by a 90° arc? answer in simplest form

Mathematics

2 answers:

Kisachek [45]3 years ago

5 0

1/4 x Pi x 3 ^2

Quarter x Pi x 3 squared

amm18123 years ago

4 0

If the radius of the circle is 3 inches, then the area would be found by using the formula $\pi r^2$ . You plug it in, and the formula would get us that the area is $9 \pi$ square inches. Since $\frac{360}{90}$ would equal 4, you would do $\frac{9}{4}$ , which would equal $2 \frac{2}{5}$

SORRY IF MY TEX FAILS

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Given that lim x → 2 f ( x ) = 1 lim x → 2 g ( x ) = − 4 lim x → 2 h ( x ) = 0 limx→2f(x)=1 limx→2g(x)=-4 limx→2h(x)=0, find the

VashaNatasha [74]

Answer:

According what I can read, I have the following statements:

$\lim_{x \to 2} f(x) = 1$

$\lim_{x \to 2} g(x) = -4$

$\lim_{x \to 2} h(x) = 0$

a) Applying properties of limits

$\lim_{x \to 2} f(x) + 5g(x) = \lim_{x \to 2} f(x) + 5 \lim_{x \to 2} g(x) = 1 + 5*-4 = -19$

b) Applying properties of limits

$\lim_{x \to 2} g(x)^{3} = {(\lim_{x \to 2} g(x))}^{3} = (-4)^{3} = -64$

c) Applying properties of limits

$\lim_{x \to 2} \sqrt{f(x)} = \sqrt{\lim_{x \to 2} f(x)} = \sqrt{1} = 1$

d) Applying properties of limits

$\lim_{x \to 2} 4*g(x)*f(x) = 4*\lim_{x \to 2} g(x)*\lim_{x \to 2} f(x) = 4*-4*1 =-16$

e) Applying properties of limits

$\lim_{x \to 2} g(x)*h(x) = \lim_{x \to 2} g(x)*\lim_{x \to 2} h(x) = -4*0 =0$

f) Applying properties of limits

$\lim_{x \to 2} g(x)*h(x)*f(x) = \lim_{x \to 2} g(x)*\lim_{x \to 2} h(x)*\lim_{x \to 2} f(x = -4*0*1 =0$

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Naya [18.7K]

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Artyom0805 [142]

Answer:

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Step-by-step explanation:

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The graph of a proportional relationship passes through (6, 42)(6, 42) and (1, y)(1, y). Find y.

n200080 [17]

Answer:

y=7

Step-by-step explanation:

6/42 =1/y

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natita [175]

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