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zavuch27 [327]
3 years ago
15

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Mathematics
1 answer:
777dan777 [17]3 years ago
5 0

Answer:   (C) 13,120

<u>Step-by-step explanation:</u>

Given the sequence {4, 12, 36, 108, ... , 8748} we know that the first term (a) is 4 and the ratio (r) is \dfrac{12}{4}=3

Input the values above into the Sum formula:

S_n=\dfrac{a(1-r^n)}{1-r}\\\\S_8=\dfrac{4(1-3^8)}{1-3}\\\\.\quad=\dfrac{4(1-6561)}{-2}\\\\.\quad=\dfrac{4(-6560)}{-2}\\\\.\quad=-2(-6560)\\\\.\quad=13,120

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*Mastery Check. DO NOT ask for help. Find the area of the sector. Round your answer to the nearest tenth.
Juliette [100K]

The area of the sector is 472.6 ft²

<h3>Area of a Sector</h3>

For finding the area of the sector, you need apply the formula:

A=\pi r^2* \frac{\alpha }{360}, where:

r= radius

α= central angle

The question gives:

r= radius= 19 ft

α= central angle = 150°

Thus, the area of the sector will be:

A=\pi r^2* \frac{\alpha }{360}=\pi *19^2*\frac{150}{360} =472.6

Read more about the area of a sector here:

brainly.com/question/22972014

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2 years ago
You and your family attend your brother’s championship baseball game. Between innings you decide to go to the snack stand. You g
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Find the area of trapezium where area=5cm base=6cm height=8cm
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2 years ago
Which is the solution set of the inequality 15y-9&lt;36
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15y - 9 <36.
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How to know if a function is periodic without graphing it ?
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A function f(t) is periodic if there is some constant k such that f(t+k)=f(k) for all t in the domain of f(t). Then k is the "period" of f(t).

Example:

If f(x)=\sin x, then we have \sin(x+2\pi)=\sin x\cos2\pi+\cos x\sin2\pi=\sin x, and so \sin x is periodic with period 2\pi.

It gets a bit more complicated for a function like yours. We're looking for k such that

\pi\sin\left(\dfrac\pi2(t+k)\right)+1.8\cos\left(\dfrac{7\pi}5(t+k)\right)=\pi\sin\dfrac{\pi t}2+1.8\cos\dfrac{7\pi t}5

Expanding on the left, you have

\pi\sin\dfrac{\pi t}2\cos\dfrac{k\pi}2+\pi\cos\dfrac{\pi t}2\sin\dfrac{k\pi}2

and

1.8\cos\dfrac{7\pi t}5\cos\dfrac{7k\pi}5-1.8\sin\dfrac{7\pi t}5\sin\dfrac{7k\pi}5

It follows that the following must be satisfied:

\begin{cases}\cos\dfrac{k\pi}2=1\\\\\sin\dfrac{k\pi}2=0\\\\\cos\dfrac{7k\pi}5=1\\\\\sin\dfrac{7k\pi}5=0\end{cases}

The first two equations are satisfied whenever k\in\{0,\pm4,\pm8,\ldots\}, or more generally, when k=4n and n\in\mathbb Z (i.e. any multiple of 4).

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It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when k is any common multiple of 4 and 10/7. The least positive one would be 20, which means the period for your function is 20.

Let's verify:

\sin\left(\dfrac\pi2(t+20)\right)=\sin\dfrac{\pi t}2\underbrace{\cos10\pi}_1+\cos\dfrac{\pi t}2\underbrace{\sin10\pi}_0=\sin\dfrac{\pi t}2

\cos\left(\dfrac{7\pi}5(t+20)\right)=\cos\dfrac{7\pi t}5\underbrace{\cos28\pi}_1-\sin\dfrac{7\pi t}5\underbrace{\sin28\pi}_0=\cos\dfrac{7\pi t}5

More generally, it can be shown that

f(t)=\displaystyle\sum_{i=1}^n(a_i\sin(b_it)+c_i\cos(d_it))

is periodic with period \mbox{lcm}(b_1,\ldots,b_n,d_1,\ldots,d_n).
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