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disa [49]
3 years ago
11

Please answer this question now

Mathematics
2 answers:
anzhelika [568]3 years ago
5 0

Answer:

25.13

Step-by-step explanation:

son4ous [18]3 years ago
4 0

Answer:

C ≈ 25.13 feet

Step-by-step explanation:

A = πr²

16π = πr²

divide by π

16 = r²

r = 4

plug r into circumference equation:

C = 2πr

C = 2π(4)

C ≈ 25.13

hope this helps :)

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When Klorina swims with the current, she swims 10 km in 2 h. Against the current, she can swim only 8 km in the same time. How f
nasty-shy [4]

Answer:

Klorina's rate in still water is 4.5 km/h

Current's rate is 0.5 km/h

Step-by-step explanation:

Let

x km/h = Klorina's rate in still water

y km/h = current's rate

<u>With the current (current helps):</u>

Distance = 10 km

Time = 2 hours

Rate = x + y km/h

10=2(x+y)

<u>Against the current:</u>

Distance = 8 km

Time = 2 hours

Rate = x - y km/h

8=2(x-y)

Divide both equations by 2:

x+y=5\\ \\x-y=4

Add these equations:

x+y+x-y=5+4\\ \\2x=9\\ \\x=4.5\ km/h

Subtract these two equations:

(x+y)-(x-y)=5-4\\ \\x+y-x+y=1\\ \\2y=1\\ \\y=0.5\ km/h

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2 years ago
How could you use the property to compute 2x8x2 mentally
Sveta_85 [38]
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Mike saves $42 a week for 7 weeks. During his 8-day vacation, he plans to spend $8.25 each day for snacks. Which expression show
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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}

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3 years ago
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