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3 years ago

What is the volume of a sphere with a surface area of 16π ft²?

2 answers:

4 0

Sphere Surface Area = 4 • π • r²
For it to equal 16 PI, then radius must equal 2
4*PI*2*2 = 16 PI


Sphere Volume = 4/3 • π • r³

Sphere Volume = 4/3 • π • 2^3

Sphere Volume = 4/3 *PI * 8

Sphere Volume = 32 / 3 PI


Sphere Volume = 10.666 PI cubic feet AND I think that is answer B
which SHOULD read 10 (2/3) PI ft^3

____ [38]3 years ago

3 0

Answer:

the answer is 10 2/3. hope this helps

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This can be written as:

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3 years ago

What product have the same sign as ( -2 3/7 ) (-6/11) ?

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Answer:

1. -7/8

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2. They have different values but are the same sign.

3. They have the same value but are different signs.

4. They have the same value and the same sign.

Step-by-step explanation:

Multiplying fractions: multiply numerators and multiply denominators

1. 3/4 X -6/7 = -18/28 = -9/14

2. -2/7 X -3/7 = 6/49

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3 years ago

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7

salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in x = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in x = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in x = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

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3 years ago

An ion has a charge of +7. How far is it from being neutral?

rusak2 [61]

Answer:

The answer is B. -7

Step-by-step explanation:

the -7 cancels out the +7 which makes it neutral

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3 years ago

What is the volume of a sphere with a surface area of ​ 16π ​ft²?

What is the volume of a sphere with a surface area of 16π ft²?