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Pavlova-9 [17]
3 years ago
14

The rule (x, y)→(x−6, y−7) represents a translation 6 units _______ and 7 units________?

Mathematics
2 answers:
VikaD [51]3 years ago
8 0
6 units right and 7 units down is the correct answer
mestny [16]3 years ago
5 0
6 units right and 7 units down

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Someone please help me will give BRAILIEST!!!!!!
Kipish [7]
The answer would be 15
6 0
3 years ago
Read 2 more answers
Which of the following box and whisker plots correctly displays the data set 120 145 133 105 117 150 ?
blsea [12.9K]

Answer:

3rd answer down

Step-by-step explanation:

ok so you take the median of all the #'s which turns out to be 126.5 and that is where your bar should point to in the box, then you should start with 105, and your last bit of the line should end at 150, the start of your box should start at the median of 105 and 117, and the end of your box should stop at the median of 145 and 150.

7 0
3 years ago
What is the measure of EA? Need help ASAP!
Ghella [55]
The measurement is 51
8 0
4 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
What is the slope of the line that represents this relationship?
galben [10]

Answer:

undefine

Step-by-step explanation:

the slope does not show rise over run it just sgow one of them therefore it is undefine

7 0
4 years ago
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