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3 years ago

PLEASE ANSWER PICTURE IS SHOWN

Mathematics

2 answers:

zysi [14]3 years ago

6 0

It's D. the function has exactly one x-intercept.

Hope I helped! ( Smiles )

Kruka [31]3 years ago

3 0

Your answer is c.the function has more than one x-intercept.

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Kathy buys a bicycle after a 30% markdown. The original price was $490.00. What did<br> she pay?

Umnica [9.8K]

Answer:

$343

Step-by-step explanation:

To get 30% of 490 we divide 30% by 1

30% ÷ 1 = 147

Now, we subtract 490 by 147

490 - 147 = 343

4 0

3 years ago

Original price: $119.00; Markdown: 70%

Leokris [45]

The answer would be $35.70. 70% off would be $83.30 off of $119.

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3 years ago

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Q25 A train travelling at a speed of 30km/h crosses a bridge 25m long in 12 sec. How long will it take to overtake a another tra

Makovka662 [10]

Answer:

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Step-by-step explanation:

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3 years ago

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

1/3 x4 need help asap

Leviafan [203]

Answer:

4/3 or 1 1/3

Step-by-step explanation:

7 0

3 years ago

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