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NikAS [45]
3 years ago
13

a rock is thrown in the air from the edge of a seaside cliff. it's height in feet is represented by f(x) = - 16(ײ-8x-9), where

is the numbers of seconds since the rock was thrown. the height of the rock is 0 feet when it hits water​
Mathematics
1 answer:
Ainat [17]3 years ago
4 0

Answer:

6 seconds

Step-by-step explanation:

Just took the test :)

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Answer:

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Solve 73 make sure to also define the limits in the parts a and b
Aleks04 [339]

73.

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a)

\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3

b)

Since we can't divide by zero, we need to find when:

x^4-2x^2+144=0

But before, we can factor the numerator and the denominator:

\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}

Now, we can conclude that the vertical asymptotes are located at:

\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}

so, for x = -3:

\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty

For x = 4:

\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty

4 0
1 year ago
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balandron [24]
First covert them to fractions with the same denominator so you can compare them (9/12 and 4/12), then subtract the numerators, and not the denominators (9/12-4/12), then whatever you get is the answer (5/12). If you can, simplify. 
7 0
3 years ago
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Rama09 [41]

Answer:

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Step-by-step explanation:

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Answer:

what is this

Step-by-step explanation:

7 0
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