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2 years ago

Amy runs 4 miles in 31 minutes how many minutes does she take per mile

Mathematics

1 answer:

Hitman42 [59]2 years ago

4 0

Answer:

about 8 minutes per mile

Step-by-step explanation:

a= t/d

31/4= 7.75

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Luna entered a raffle for a new pair of running shoes. She purchased 20 tickets and was told that she had a 20% chance of winnin

xenn [34]

Answer:

<h3>100 tickets</h3>

Step-by-step explanation:

Probability of winning = amount of tickets purchased/Total number of tickets

20% = 20/total number of tickets

20/100 = 20/total number of tickets

cross multiply

20 * total number of tickets = 20 * 100

20 * total number of tickets = 100

total number of tickets sold = 100 tickets

6 0

3 years ago

Read 2 more answers

Someone please help. i will mark as brainliest if right :)

DIA [1.3K]

Answer:

The other factor is 5x-2

Step-by-step explanation:

Using the factorisation method:

Two numbers that multiply to give -10 and add to give +3:

+5 and -2

$5x^{2} +5x-2x-2$

$5x(x+1) -2(x+1)$

$(x+1)(5x-2)$

So the other factor is 5x-2

I'm not sure about the second part of the question but I hope you understood how to deal with the first part.

5 0

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

3 years ago

If Jim was suppose to be at work at 8:45 but did not get there until 9:15. How late in minutes is he?

gavmur [86]

30 minutes 45-15=30 ok

3 0

3 years ago

A satellite moves at a speed of 27,950 kilometers per hour. About how much faster is one satellite than the other? Explain how t

Marianna [84]

27,950 k * X Kilometers
_____________________________ All you have to do now is cross multiply
1 hr 1hr

6 0

3 years ago