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drek231 [11]
2 years ago
7

a rock is thrown from the top of a tall building. the distance, in feet, between the rock and the ground x seconds after it is t

hrown is given by f(x) =-16x^2-4x+382 How long after the rock is thrown is it 340 feet from the ground?
Mathematics
1 answer:
frez [133]2 years ago
3 0

Answer:

1.5 seconds

Step-by-step explanation:

Since f(x) represents the distance from the ground to the rock (in feet), we just need to find the values of x for which f(x) = 340. Substituting f(x) = 340 into f(x) = -16x² - 4x + 382 gives us:

340 = -16x² - 4x + 382

0 = -16x² - 4x + 42 (Subtract 340 from both sides)

16x² + 4x - 42 = 0  (Multiply entire equation by -1, "flip" equation)

8x² + 2x - 21 = 0    (Divide entire equation by 2)

(4x + 7)(2x - 3) = 0  (Factor LHS)

4x + 7 = 0 or 2x - 3 = 0 (Zero Product Property)

4x = -7 or 2x = 3 (Subtract -7 and add 3 from/to both sides, respectively)

x = -\frac{7}{4} = -1.75 or  x =\frac{3}{2} = 1.5

x = -1.75 is an extraneous solution because in the context of the question, you can't have negative seconds so therefore, the final answer is 1.5 seconds.

Hope this helps!

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\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

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  • Derivative Notation

Basic Power Rule:

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  2. 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

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