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3 years ago

I need this asap someone help me plz

Mathematics

1 answer:

alexandr1967 [171]3 years ago

5 0

Answer:

660 nails

Step-by-step explanation:

Take all of john's nails and multiply by 4.4 because 176/40 is 4.4. Multiply 4.4 by the john's other nails. Add the products together and thats your answer.

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Find the difference -1/10 - 8/10 -9/10 9/10 -7/10 7/10

TEA [102]

The answer to this would be A: -9/10. Hope this helps!

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4 years ago

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The water in a reservoir decreased by 47% during a drought. Before the drought, the reservoir held 44,580 acre ft (AF) of water.

RSB [31]

44,580 x 47 %

44,580 x .47 = 20,952.6

So, the water decreased 20,952.6 ft

If you want the water level after the drought just subtract.

44,580 - 20,952.6 = 23,627.4 ft

Hope this helps. :)

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3 years ago

In every bar graph , line graph , or scatter plot , the vertical axis should include the following

alexgriva [62]

It should include zero

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3 years ago

Donna is running a track it takes her 10 minutes to run 6 laps if she keeps running at the same speed how long will it take her

kaheart [24]

To solve this problem we can make a proportion.
6 5
-- = ---
10 x

to solve this proportion we can cross multiply and divide.
5 times 10 is 50. divided by 6 is 8.3
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7 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