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3 years ago

A circle with radius 6 has a sector with a central angle of 48 what is the area of the sector

Mathematics

2 answers:

inn [45]3 years ago

5 0

Answer:

≈ 15.08 units²

Step-by-step explanation:

The area (A) of the sector is calculated as

A = area of circle × fraction of circle

= πr² × $\frac{48}{360}$ ( r is the radius )

= π × 6² × $\frac{48}{360}$

= 36π × $\frac{48}{360}$ = π × $\frac{48}{10}$ = $\frac{48\pi }{10}$ ≈ 15.08

Reil [10]3 years ago

3 0

Answer:

24/5 pi

Step-by-step explanation:

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Step-by-step explanation:

point slope formula: y-y1=m(x-x1)

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3 years ago

when the temperature is 68 F, the speed of sound in air is approximately 343 meters per second. Approximately how many kilometer

Digiron [165]

Hello!

Since we are given that the speed of sound in the air travels about 343 meters per second, we can multiply 343 meters per second by 5, and then we convert the meters to kilometers.

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2 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

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

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2 years ago