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2 years ago

Answer each question please! Brainliest!

Mathematics

2 answers:

7nadin3 [17]2 years ago

5 0

Answer:

The answer is: HAHA GOTCHA BIG BROTHER!!!!

(hehe)

Step-by-step explanation:

Nimfa-mama [501]2 years ago

3 0

Answer:

Answer: 64/3π -16/3)in²

Step-by-step explanation:

Area of segment equals area of sector minus area of isosceles triangle.

=0/360 x πr² -1/2r²sin (0)

Given; the length of chord, d=8√3in.

and the angle of the sector, 0=120

We can use the formula for calculating the length of a chord to find the radius of the circle.

D = 2r sin (0/2)

8√3= 2r (√3/2)

r=8in

Area of the segment= 120/360 x π x 8² - ½ x 8² sin (120)

= (64/3 π- 16 √3)in²

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chubhunter [2.5K]

Answer:

They are related becasue if you find the area of a parallelogram then divide it by two then you have the area of a triangle

Step-by-step explanation:

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3 years ago

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Fudgin [204]

Answer:

Step-by-step explanation:

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3 years ago

Evaluate: 9 + x/5 for x = 30 15 16 17 18

mart [117]

<u>Answer:</u>

$\bold{9+\frac{x}{5}=15}$

<u>Solution:</u>

Given: x=30

To solve: $9+\frac{x}{5}$

On substituting the value of x,

$9+\frac{30}{5}$

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2 years ago

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natali 33 [55]

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2 years ago

(a) Use the power series expansions for ex, sin x, cos x, and geometric series to find the first three nonzero terms in the powe

Fofino [41]

Answer:

a) $\mathbf{4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} ...}$

b) See Below for proper explanation

Step-by-step explanation:

a) The objective here is to Use the power series expansions for ex, sin x, cos x, and geometric series to find the first three nonzero terms in the power series expansion of the given function.

The function is $e^x + 3 \ cos \ x$

The expansion is of $e^x$ is $e^x = 1 + \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!} + ...$

The expansion of cos x is $cos \ x = 1 - \dfrac{x^2}{2!}+ \dfrac{x^4}{4!}- \dfrac{x^6}{6!}+ ...$

Therefore; $e^x + 3 \ cos \ x = 1 + \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!} + ... 3[1 - \dfrac{x^2}{2!}+ \dfrac{x^4}{4!}- \dfrac{x^6}{6!}+ ...]$

$e^x + 3 \ cos \ x = 4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} + \dfrac{x^3}{3!}+ ...$

Thus, the first three terms of the above series are:

$\mathbf{4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} ...}$

The series for $e^x + 3 \ cos \ x$ is $\sum \limits^{\infty}_{x=0} \dfrac{x^x}{n!} + 3 \sum \limits^{\infty}_{x=0} ( -1 )^x \dfrac{x^{2x}}{(2n)!}$

let consider the series; $\sum \limits^{\infty}_{x=0} \dfrac{x^x}{n!}$

$|\frac{a_x+1}{a_x}| = | \frac{x^{n+1}}{(n+1)!} * \frac{n!}{x^x}| = |\frac{x}{(n+1)}| \to 0 \ as \ n \to \infty$

Thus it converges for all value of x

Let also consider the series $\sum \limits^{\infty}_{x=0}(-1)^x\dfrac{x^{2n}}{(2n)!}$

It also converges for all values of x

7 0

3 years ago