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2 years ago

Please tell me what is 136% of 115

Mathematics

2 answers:

Ronch [10]2 years ago

8 0

156.4 hope it helps......

valentina_108 [34]2 years ago

8 0

The answer is 156.4

Happy studying!

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What’s the minimum cuts needed to cut a circle into 8 equal parts

OlgaM077 [116]

Answer:

4 cuts

Step-by-step explanation:

you first do a vertical cut

then a horizontal cut

then two diagonal cuts

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2 years ago

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the train traveled at an average speed of 48 miles per hour for the first 2 hours and at 60 miles an hour for the next hours. Wh

Aleksandr [31]

Answer: 56 mph

Step-by-step explanation:

48 miles x 2 hours = 96 miles

60 miles x 4 hours = 240 miles

240 + 96 = 360 total miles

360 / 6 (hours) = 56 mph

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2 years ago

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3 years ago

A car dealership buys a car for $12,500 and sells it for $15,750. What is the percent markup( percent increase)

Natalka [10]

Answer:

About 1.26%

Step-by-step explanation:

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Hope this helped!

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2 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

2 years ago