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Arlecino [84]
3 years ago
11

Answer the questions, please I will give you 10 points

Mathematics
1 answer:
uranmaximum [27]3 years ago
8 0
A) 3 times 7x makes it a normal number of 21.
B) using PEMDAS you must multiply 5 times 5 to put 5 to its own power, then multiply by 2 to get 50, then add 4 getting 54.
C) 9 times 5y gets the normal number of 45, then subtract that from 80, you get 35.
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2/5x3/4 show your work if you get it right i will give u braniset. if u do for frre points i willl delete it
maw [93]
2x3
—— = 6/20 = 3/10
5x4
3 0
3 years ago
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The sales tax rate is 10%. If Julie buys a picnic table priced at $793.60,how much tax will she pay?
Elis [28]

Answer:

she will pay 79.36 in taxes

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
How do i write 12.009 in a word form?
tia_tia [17]
Your answer will be twelve and nine thousandths because in place value decimals are like tenths, hundredths, thousandths. In place value for regular numbers are ones, tens, hundreds and so on. So, when you have 12.009, you are going to separate the decimals with the numbers. Then, write down the number in words which would be twelve. Now, go to the decimal and write that down, which is 9 thousandths. Finally, combine the both of them to get: twelve and nine thousandths.

Hope this helps :) 
and good luck

5 0
3 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
What is the best next step in the construction of the perpendicular bisector of
Musya8 [376]

C, not sure but C. It makes sense to me & I got it right when I did it

4 0
3 years ago
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