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timurjin [86]
3 years ago
9

2. The larger circle of an annulus has a diameter of 10 cm, and the smaller circle has a diameter of 7.6 cm. What is the area of

the annulus?
Mathematics
1 answer:
Alik [6]3 years ago
5 0

Answer:

Area = 3.77 cm^2

Step-by-step explanation:

The diagram using the given statement is attached below

Diameter of larger circle is 10 so radius R1 = 10/2= 5

diameter of smaller circle is 7.6, radius (r1)= 3.8

We need to find the area of the shaded region

Area of the circle =\pi r^2

Area of annulus =\pi R1-\pi r1

=\pi (5)-\pi (3.8)

Area = 3.77 cm^2

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Solve 12g=12(2/3g-1)+11
DedPeter [7]
Solve for g:12 g = 12 ((2 g)/3 - 1) + 11
Put each term in (2 g)/3 - 1 over the common denominator 3: (2 g)/3 - 1 = (2 g)/3 - (3)/3:12 g = 12 (2 g)/3 - (3)/3 + 11
(2 g)/3 - (3)/3 = (2 g - 3)/3:12 g = 12 (2 g - 3)/3 + 11
12/3 = (3×4)/3 = 4:12 g = 4 (2 g - 3) + 11
4 (2 g - 3) = 8 g - 12:12 g = 8 g - 12 + 11
Grouping like terms, 8 g - 12 + 11 = 8 g + (11 - 12):12 g = 8 g + (11 - 12)
11 - 12 = -1:12 g = 8 g + -1
Subtract 8 g from both sides:12 g - 8 g = (8 g - 8 g) - 1
12 g - 8 g = 4 g:4 g = (8 g - 8 g) - 1
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Divide both sides of 4 g = -1 by 4:(4 g)/4 = (-1)/4
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4 years ago
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Use the definition of a Taylor series to find the first four nonzero terms of the series for f(x) centered at the given value of
Black_prince [1.1K]

Answer:

The first four nonzero terms of the Taylor series of \frac{7}{x + 1} around a=2 are:

f\left(x\right)\approx P\left(x\right) = \frac{7}{3}- \frac{7}{9}\left(x-2\right)+\frac{7}{27}\left(x-2\right)^{2}- \frac{7}{81}\left(x-2\right)^{3}+\frac{7}{243}\left(x-2\right)^{4}

Step-by-step explanation:

The Taylor series of the function <em>f </em>at <em>a </em>(or about <em>a</em> or centered at <em>a</em>) is given by

f\left(x\right)=\sum\limits_{k=0}^{\infty}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k

To find the first four nonzero terms of the Taylor series of \frac{7}{x + 1} around a=2 you must:

In our case,

f\left(x\right) \approx P\left(x\right) = \sum\limits_{k=0}^{n}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k=\sum\limits_{k=0}^{4}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k

So, what we need to do to get the desired polynomial is to calculate the derivatives, evaluate them at the given point, and plug the results into the given formula.

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Evaluate the function at the point: f\left(2\right)=\frac{7}{3}

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Evaluate the function at the point: \left(f\left(2\right)\right)^{\prime }=- \frac{7}{9}

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Evaluate the function at the point: \left(f\left(2\right)\right)^{\prime \prime }=\frac{14}{27}

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Evaluate the function at the point: \left(f\left(2\right)\right)^{\prime \prime \prime }=- \frac{14}{27}

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Evaluate the function at the point: \left(f\left(2\right)\right)^{\prime \prime \prime \prime }=\frac{56}{81}

Apply the Taylor series definition:

f\left(x\right)\approx\frac{\frac{7}{3}}{0!}\left(x-\left(2\right)\right)^{0}+\frac{- \frac{7}{9}}{1!}\left(x-\left(2\right)\right)^{1}+\frac{\frac{14}{27}}{2!}\left(x-\left(2\right)\right)^{2}+\frac{- \frac{14}{27}}{3!}\left(x-\left(2\right)\right)^{3}+\frac{\frac{56}{81}}{4!}\left(x-\left(2\right)\right)^{4}

The first four nonzero terms of the Taylor series of \frac{7}{x + 1} around a=2 are:

f\left(x\right)\approx P\left(x\right) = \frac{7}{3}- \frac{7}{9}\left(x-2\right)+\frac{7}{27}\left(x-2\right)^{2}- \frac{7}{81}\left(x-2\right)^{3}+\frac{7}{243}\left(x-2\right)^{4}

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