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3 years ago

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There are four brands of smart phones that are most popular in United States. The number of people in United States that have ea

ch brand is given. Brand A: 4.48x10^4;Brand B: 3.84x10^7; Brand C:6.4x10^6; Brand D: 1.50x10^7

1 answer:

GREYUIT [131]3 years ago

5 0

So what is the question?

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48/64 Which choices are equivalent to the fraction below? Check all that apply. A. 8/6 B. 3/4 C. 6/8 D. 12/16 E. 9/12 F. 24/32

irinina [24]

Answer:

B, C, D, F

Step-by-step explanation:

to simplify the fraction 48/64, you need to divide both the numerator and denominator by their common factors. B is correct since you can divide both by 16, C is correct since both are divisible by 8, D is correct since both are divisible by 4, and F is correct since both are divisible by 2. A, however, would be correct only if the fraction were 64/48 and E would only be correct if 9 was a factor of 48 and 12 was a factor of 64, however neither haev such factors.

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2 years ago

Read 2 more answers

Can anyone help with the last problem about The Difference Quotient of a Function

Marysya12 [62]

Answer:

tbh i really dont know

Step-by-step explanation:

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2 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:

<u>Calculus</u>

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 <em>x</em> = 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 <em>x</em> = 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 <em>x</em> = 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

6 0

2 years ago

Larry’s family made a donation to the library. His mom gave half as much as larry. His dad gave 5 times as much as Larry’s siste

storchak [24]

Answer:

The family donated $52.50

Step-by-step explanation:

Larry's mom donated $7.50 ($15 ÷ $2=$7.50)

Larry's dad donated $25.00 ($5 x $5=$25.00)

Larry's sister donated $5.00 ($15 - $10=$5.00)

Larry donated $15.00

6 0

2 years ago

a child should no longer play in the kid zone area when they reach 4'4" tall joe is 54" tall is he allowed to play in the kid zo

Andreyy89

Answer: No.

Step-by-step explanation: Joe is 4'6" tall. Convert 54" to feet by dividing by 12. 54/12=4.5. Half of a foot is 6 inches.

6 0

2 years ago