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Vladimir79 [104]

2 years ago

13

In a square, doubling every side increases the area by 27 units. What is the length of side of the newly formed square? a) 3 b)

6 c) 9 18 d

2 answers:

MatroZZZ [7]2 years ago

6 0

<h3>Answer: B) 6</h3>

===========================================================

Explanation:

x = original side length

2x = double the side length

The old area is x^2. The new square area is (2x)^2 = 4x^2

new area = (old area) + 27

4x^2 = x^2 + 27

4x^2-x^2 = 27

3x^2 = 27

x^2 = 27/3

x^2 = 9

x = sqrt(9)

x = 3

The old original square has a side length of 3 units.

The new larger square has a side length of 2x = 2*3 = 6 units which is the final answer (choice B)

old area = 3^2 = 9

new area = 6^2 = 36

The jump from 9 to 36 is +27 to help confirm the answer.

slava [35]2 years ago

4 0

The answer is c I take the test

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\lim _{x\to 0}\left(\frac{2x\ln \left(1+3x\right)+\sin \left(x\right)\tan \left(3x\right)-2x^3}{1-\cos \left(3x\right)}\right)

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$\displaystyle \lim_{x\to 0}\left(\frac{2x\ln \left(1+3x\right)+\sin \left(x\right)\tan \left(3x\right)-2x^3}{1-\cos \left(3x\right)}\right)$

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$\displaystyle \lim_{x\to 0}\left(2\ln(1+3x)+\dfrac{6x}{1+3x}+\cos(x)\tan(3x)+3\sin(x)\sec^2(x)-6x^2}{3\sin(3x)}\right)$

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$\displaystyle \lim_{x\to0}\frac{2\ln(1+3x)}{3\sin(3x)} + \lim_{x\to0}\frac{6x}{3(1+3x)\sin(3x)} \\\\ + \lim_{x\to0}\frac{\cos(x)\tan(3x)}{3\sin(3x)} + \lim_{x\to0}\frac{3\sin(x)\sec^2(x)}{3\sin(3x)} \\\\ - \lim_{x\to0}\frac{6x^2}{3\sin(3x)}$

Now recall two well-known limits:

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Compute each remaining limit:

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$\displaystyle \lim_{x\to0}\frac{6x}{3(1+3x)\sin(3x)} = \frac23 \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}\frac{1}{1+3x} = \frac23$

$\displaystyle \lim_{x\to0}\frac{\cos(x)\tan(3x)}{3\sin(3x)} = \frac13 \times \lim_{x\to0}\frac{\cos(x)}{\cos(3x)} = \frac13$

$\displaystyle \lim_{x\to0}\frac{3\sin(x)\sec^2(x)}{3\sin(3x)} = \frac13 \times \lim_{x\to0}\frac{\sin(x)}x \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}\sec^2(x) = \frac13$

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Hope this answers your question.

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<h2>Hello!</h2>

The answer is:

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Have a nice day!

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I have calculus problems that I need help with.

aleksklad [387]

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