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netineya [11]
2 years ago
13

During the halftime performance for a football game, the color guard is planning a new formation in which seven members stand ar

ound a central point and stretch their flag to the person immediately to their left. 10. what is the measure of each exterior angle of the formation?
SAT
1 answer:
shtirl [24]2 years ago
5 0

There are different kinds of angles. The measure of each exterior angle of the formation is 51.4.

<h3>What is a exterior angle in math?</h3>

The term exterior angle is known to be the angle that is found between a side of a polygon and that of an extended adjacent side.

The shape of the halftime is regular heptagon.

Note that a  regular heptagon has 7 congruent sides and 7 congruent interior angles.

So the exterior angles are said to be congruent, based on the fact that angles supplementary to congruent angles are  regarded as congruent.

So Let take n as the measure of each exterior angle.

7n = 360

n= 360/n

To Solve for n =51.4

Therefore, The measure of each exterior angle of the formation is 51.4.

Learn more about exterior angle from

brainly.com/question/2546141

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Use series to approximate the definite integral i to within the indicated accuracy. i = 1/2 x3 arctan(x) dx 0 (four decimal plac
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The expression \int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx  is an illustration of definite integrals

The approximated value of the definite integral is 0.0059

<h3>How to evaluate the definite integral?</h3>

The definite integral is given as:

\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx

For arctan(x), we have the following series equation:

\arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 1}}{2n + 1}}

Multiply both sides of the equation by x^3.

So, we have:

x^3 * \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 1}}{2n + 1}}  * x^3

Apply the law of indices

x^3 * \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 1 + 3}}{2n + 1}}

x^3 * \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}}

Evaluate the product

x^3 \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}}

Introduce the integral sign to the equation

\int\limits^{1/2}_{0}  x^3 \arctan(x)\ dx =\int\limits^{1/2}_{0} \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}}

Integrate the right hand side

\int\limits^{1/2}_{0}  x^3 \arctan(x)\ dx =[ \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}} ]\limits^{1/2}_{0}

Expand the equation by substituting 1/2 and 0 for x

\int\limits^{1/2}_{0}  x^3 \arctan(x)\ dx =[ \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{(1/2)^{2n + 4}}{2n + 1}} ] - [ \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{0^{2n + 4}}{2n + 1}} ]

Evaluate the power

\int\limits^{1/2}_{0}  x^3 \arctan(x)\ dx =[ \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{(1/2)^{2n + 4}}{2n + 1}} ] - 0

\int\limits^{1/2}_{0}  x^3 \arctan(x)\ dx = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{(1/2)^{2n + 4}}{2n + 1}}

The nth term of the series is then represented as:

T_n = \frac{(-1)^n}{2^{2n + 5} * (2n + 4)(2n + 1)}

Solve the series by setting n = 0, 1, 2, 3 ..........

T_0 = \frac{(-1)^0}{2^{2(0) + 5} * (2(0) + 4)(2(0) + 1)} = \frac{1}{2^5 * 4 * 1} = 0.00625

T_1 = \frac{(-1)^1}{2^{2(1) + 5} * (2(1) + 4)(2(1) + 1)} = \frac{-1}{2^7 * 6 * 3} = -0.0003720238

T_2 = \frac{(-1)^2}{2^{2(2) + 5} * (2(2) + 4)(2(2) + 1)} = \frac{1}{2^9 * 8 * 5} = 0.00004340277

T_3 = \frac{(-1)^3}{2^{2(3) + 5} * (2(3) + 4)(2(3) + 1)} = \frac{-1}{2^{11} * 10 * 7} = -0.00000634131

..............

At n = 2, we can see that the value of the series has 4 zeros before the first non-zero digit

This means that we add the terms before n = 2

This means that the value of \int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx to 4 decimal points is

\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx = 0.00625 - 0.0003720238

Evaluate the difference

\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx = 0.0058779762

Approximate to four decimal places

\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx = 0.0059

Hence, the approximated value of the definite integral is 0.0059

Read more about definite integrals at:

brainly.com/question/15127807

5 0
2 years ago
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