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3 years ago

Triangle EFG is dilated by a scale factor of 5.

Mathematics

1 answer:

miv72 [106K]3 years ago

5 0

B because the scale factor multiples the coordinate by that number

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Is the function f(x)= <img src="https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Bx%7D" id="TexFormula1" title="\sqrt[5]{x}" alt="\sqrt[5

katen-ka-za [31]

Answer:stop being lazy

Step-by-step explanation:

7 0

3 years ago

The area of a rectangle room is 750 square feet. The width of the room is 5 feet less than the length of the room. Which equatio

olga_2 [115]

By definition, the area of a rectangle is given by:
A = (width) * (long)
Substituting values we have:
750 = (y-5) * (y)
Rewriting we have:
750 = y ^ 2 - 5y
y ^ 2 - 5y - 750 = 0
Answer:
An equation that can be used to solve for And the length of the room is:
y ^ 2 - 5y - 750 = 0

5 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'm confused on six...

Nadusha1986 [10]

A. 5:3 and 6 to 4

b. no because it is a very good price for what is in the fruit salad .

<3

3 0

3 years ago

Felix found the sum of two prime numbers and one composite number to be 45. The difference between the greatest and least number

erica [24]

I think it could be 13,17, and 15.
If I am wrong then I am sorry

3 0

3 years ago