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juin [17]
3 years ago
11

8 powered to the next third - 9•2 divided by 3

Mathematics
1 answer:
Svet_ta [14]3 years ago
4 0

Answer:

2³ = 2 · 2 · 2 = 8

   The phrase "two to the third power" or "two cubed" has a base of 2, an exponent of 3, and means that 2 is used as a factor 3 times. When math is typed on one line rather than written by hand, 2³ is sometimes written as 2^3 where the caret seperates the base from the exponent so the expression translates as "2 to the 3rd power."

Step-by-step explanation:

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Arlecino [84]

11 times the sun of a number and 15

11(y+15)

the answer is d

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2 years ago
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What is the area of the following circle?
zhannawk [14.2K]

Answer:

38.47 m2

Step-by-step explanation:

Area = 3.5 x 3.5 x 3.14

Area = 12.25 x 3.14

Area = 38.465 --> 38.47m2

Hope that helps!

3 0
3 years ago
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Lincoln invested $49,000 in an account paying an interest rate of 6\tfrac{1}{8}681​% compounded daily. Eli invested $49,000 in a
timofeeve [1]

Answer: 15856

Step-by-step explanation: 166786.9963 - 150930.6256

= 15856.3707

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6 0
2 years ago
Could i get some helpp
Korvikt [17]
The answer is 25......!
9 0
2 years ago
How to know if a function is periodic without graphing it ?
zhenek [66]
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).

The second two are satisfied whenever k\in\left\{0,\pm\dfrac{10}7,\pm\dfrac{20}7,\ldots\right\}, and more generally when k=\dfrac{10n}7 with n\in\mathbb Z (any multiple of 10/7).

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).
4 0
3 years ago
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