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3 years ago

PLEASE HELP FAST WHOEVER ANSWERS FIRST AND CORRECTLY GETS BRAINLIEST

1 answer:

Nutka1998 [239]3 years ago

7 0

After round-off to the nearest whole number, it would be:
250.311 → 250
4.65 → 5

So, now we need to divide 250 /5
It would be equal to 50

So, your final answer is 50

Hope this helps!

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PLEASE NEEP HELP ASAP:)

Sloan [31]

35%
---- Than cross multiply with ---- = 24500
100% 100 --------
100
Which ends up to be 245.

4 0

3 years ago

Read 2 more answers

<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

for your career project, you plan to design ans sell t-shirts. During the first 5 weeks of selling the shirts, the functions f(s

stepladder [879]

Using translation concepts, we have that:

The new profit equation is f(x) = (8s/3) - 20, which is a vertical stretch by a factor of 2 of the graph of f(s).

Selling 300 shirts, the new profit is of $780.

<h3>What is a translation?</h3>

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction either in it’s definition or in it’s domain. Examples are shift left/right or bottom/up, vertical or horizontal stretching or compression, and reflections over the x-axis or the y-axis.

For this problem, the profit function is given by:

f(s) = 4/3s - 20

To double the function, we multiply the slope by 2, hence the new profit equation is:

f(s) = 8/3s - 20.

This new function is a vertical stretch by a factor of 2 of the graph of f(s).

The new profit selling 300 shirts is given by:

f(300) = (8/3) x 300 - 20 = 8 x 100 - 20 = $780.

More can be learned about translation concepts at brainly.com/question/28373831

#SPJ1

4 0

1 year ago

Romeo will roll a 6-sided number cube, numbered 1-6, twice. What is the probability of rolling an odd number on the 1st roll and

elena-s [515]

1st rol= 3/6
2nd roll= 1/6

8 0

2 years ago

Answer al 4 question and I’ll give your Brainly

miv72 [106K]

Answer:

C-C-B-D

Step-by-step explanation:

Hope this helps and look out for people trying to take your points!

6 0

3 years ago