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denpristay [2]
3 years ago
10

The distance between Cincinnati,Ohio and Charlotte,North Carolina,is about 336 miles. The distance between Cincinnati and Chicag

o,Illinois,is about 247 miles. If perry drove from Charlotte to Chicago by way of Cincinnati, find the distance he drove?
Mathematics
1 answer:
navik [9.2K]3 years ago
6 0

336+247=583miles

charlotte-chicago

charlotte-ohio=336

ohio- Chicago-247

so charlotte-chicago=583miles

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The coordinates of the vertexes of a triangle are A(3,3) B(6,6) C(9,3). What will be the new coordinates after a dilation
yuradex [85]

Answer:

A

Step-by-step explanation:

Dilation by 3 from origin causes each point to be multiplied by 3.

(3,3) * 3 = (9, 9)

(6, 6) * 3 = (18, 18)

(9, 3) * 3 = (27, 9)

which is A

8 0
3 years ago
Which person has more potential energy? Noah who jumped out of the plane at 100 m or Clara at 50 m in the air? Explain.
olga_2 [115]

Answer:

Noah

Step-by-step explanation:

Potential energy is the amount of energy that can be transfered to kinetic. This means that the higher person will have more potential energy.

4 0
3 years ago
I’m struggling to answer this question on my test, can someone explain it to me?
Flauer [41]
Correct answer is Option A: 18/48 = 21/56

You’ll have to divide the numerator by the denominator of each fraction to see whether they have the same quotient:

18/48 = 0.375
21/56 = 0.375


The other options are not proportional.

B) 20/28 = 15/18:

20/28 = 0.714
15/18 = 0.833

Thus, 20/28 ≠ 15/18

C). 3/5 = 13/10
3/5 = 0.6
13/10 = 1.3

Hence, 3/5 ≠ 13/10

D) 7/10 = 10/7
7/10 = 0.7
10/7 = 1.43

Thus, 7/10 ≠ 10/7

These solutions prove that the correct answer is A) 18/48 = 21/56

Please mark my answers as the Briainliest if you find my explanations helpful :)
4 0
3 years ago
Solve using Fourier series.
Olin [163]
With 2L=\pi, the Fourier series expansion of f(x) is

\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L
\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx

where the coefficients are obtained by computing

\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx
\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx

\displaystyle a_n=\frac1L\int_0^{2L}f(x)\cos\dfrac{n\pi x}L\,\mathrm dx
\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx

\displaystyle b_n=\frac1L\int_0^{2L}f(x)\sin\dfrac{n\pi x}L\,\mathrm dx
\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx

You should end up with

a_0=0
a_n=0
(both due to the fact that f(x) is odd)
b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)

Now the problem is that this expansion does not match the given one. As a matter of fact, since f(x) is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of f(x), which is to say we're actually considering the function

\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3

and enforcing a period of 2L=2\pi. Now, you should find that

\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)

The value of the sum can then be verified by choosing x=0, which gives

\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)
\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots

as required.
5 0
3 years ago
(i) 809, 708, 607, _____, _ class 4<br>__class 4​
schepotkina [342]

Answer:

809, 708, 607, 506, 405, 304, 203, 102, 1, -100

Step-by-step explanation:

809, 708, 607, _____, _______

First term = 809

Second term = 708

Third term = 607

Difference between first term and second term = 809 - 708

= 101

Difference between second term and third term = 708 - 607

= 101

Therefore, the common difference is 101

Fourth term = 607 - 101

= 506

Fifth term = 506 - 101

= 405

Sixth term = 405 - 101

= 304

Seventh term = 304 - 101

= 203

Eighth term = 203 - 101

= 102

Ninth term = 102 - 101

= 1

Tenth term = 1 - 101

= - 100

809, 708, 607, 506, 405, 304, 203, 102, 1, -100

4 0
2 years ago
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