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PtichkaEL [24]
3 years ago
14

Which transformation was performed on PQRS to form P'QʻR'S'? (1 point)

Mathematics
1 answer:
REY [17]3 years ago
3 0

Given:

A transformation was performed on PQRS to form P'QʻR'S'.

To find:

The transformation.

Solution:

The given graph shows two similar polygons. It means, dilation was performed.

From the given graph it is clear that,

PS=2 units

P'S'=1 units

Now,

\text{Scale factor}=\dfrac{P'S'}{PS}

\text{Scale factor}=\dfrac{1}{2}

A dilation factor of \dfrac{1}{2} was performed on PQRS to form P'QʻR'S'.

Therefore, the correct option is (3).

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Solve 2.02W = -3.636
ValentinkaMS [17]

Answer: W=-1.8

Step-by-step explanation:

Divide both sides by 2.02 and you are left with W=-1.8.

Hope this helps!

7 0
3 years ago
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If the diameter of a circle is 3 and pi=3.14, what is the circumference? Im confused on how to get the circumference.
Afina-wow [57]

Answer:

9.426 units

Step-by-step explanation:

Given data

Diameter= 3 units

Radius= 1.5 units

The circumference is given as

C=2πr

C=2*3.142*1.5

C=9.426 units

Hence the circumference is 9.426 units

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
If you knew that the vertical intercept for a straight line was 15, that the slope was -.5, and that the independent variable ha
Finger [1]
Y=-.5x+15 plug in 8 for x and you get 11
7 0
3 years ago
30 power of 2 muilpity by pie
Ulleksa [173]

Answer: 30^2 * pi = 2827.43.

7 0
3 years ago
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