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butalik [34]
3 years ago
15

The function shown models the horizontal and vertical distances, in feet, traveled by a person on a roller coaster. What horizon

tal distance is traveled to first reach a height of 200 feet? Use the drop-down menus to complete the statements. The person first reaches a height of 200 feet after traveling a horizontal distance between feet. The horizontal distance traveled to reach a height of 200 feet is approximately feet.
Mathematics
2 answers:
Pani-rosa [81]3 years ago
6 0

Answer:

20 and 25

23

Step-by-step explanation:

lora16 [44]3 years ago
6 0

Answer: The answer is 20 and 25 for the first part then the second part is 23.

Step-by-step explanation:

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If f(x) = 4x - 12, what is f(2)?<br> O A. -4<br> O'B. 4<br> O c. -20<br> D. 8<br> Helpppp plzzz
photoshop1234 [79]

Answer:

-4

Step-by-step explanation:

f(x) = 4x - 12

Let x=2

f(2) = 4*2 - 12

     = 8-12

    = -4

8 0
2 years ago
Quadrilateral army is rotated 90° about the origin
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8 0
2 years ago
The distance between two cities on a map is 3.5 centimeters. The map uses a scale in which 1 centimeter represents 20 kilometers
zhenek [66]
1 cm = 20 km

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8 0
3 years ago
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Simplify: 7^6 ÷ 7^2<br><br> A.) 7^3<br> B.) 7^4<br> C.) 7^8<br> D.) 7^12
juin [17]
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6 0
3 years ago
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Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n
aliya0001 [1]

Answer:

f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...

We first compute the n-th derivative of f(x)=\ln(1+2x), note that

f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\

Now, if we compute the n-th derivative at 0 we get

f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!

and so the Maclaurin series for f(x)=ln(1+2x) is given by

f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2  \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}

3 0
3 years ago
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