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Goshia [24]
3 years ago
15

Write each expression as a single power of 10​

Mathematics
1 answer:
swat323 years ago
3 0

Answer:

<u>It</u><u> </u><u>is</u><u> </u>

<u>{10}^{5}</u>

Step-by-step explanation:

\frac{ {10}^{4}. {10}^{5}  . {10}^{6} }{ {10}^{3} . {10}^{7} }  =  \frac{ {10}^{(4 + 5 + 6)} }{ {10}^{(3 + 7)} }  \\  \\  =  \frac{ {10}^{15} }{ {10}^{10} }  \\  \\  =   {10}^{(15 - 10)}  \\  =  {10}^{5}

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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
Susan owns a small business there was a loss of $11 on Monday and a profit of $18 on Tuesday on Wednesday there was a loss of $7
Dmitry [639]

Answer:

B. $8 profit

Step-by-step explanation:

Loss is negative and profit is positive

Monday :      -11

Tuesday:     +18

Wednesday: -7

Thursday:    +8

                 ------------

                   +8

We have a profit of 8

6 0
2 years ago
Read 2 more answers
Simplify this expression.
nexus9112 [7]

Answer:

C

Step-by-step explanation:

If you would recall one of the laws of exponents, when dividing powers of the same base you subtract their exponents. So for this expression:

6⁷ ÷ 6⁵ = 6^(7-5) = 6² =36

4 0
3 years ago
Read 2 more answers
Question
Oksi-84 [34.3K]

Answer: 100 x 2^ t/9

Step-by-step explanation:

6 0
3 years ago
What is the volume of the rectangular prism shown?​
Yuliya22 [10]
The isn’t any pictures attached
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3 years ago
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