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Stolb23 [73]
3 years ago
14

The vertices of APQR are P(-3, 8) Q(-6,-4), and R(1, 1). If you reflect APQR across the x-axis, what will be the coordinates of

the vertices of the image AP'Q'R'?
Mathematics
1 answer:
Ymorist [56]3 years ago
4 0

Answer:(-3,-8),(-6,4),(1,-1)

Step-by-step explanation:

Given

Vertices of the triangle are

P(-3,8), Q(-6,-4),R(1,1)

Coordinates (a,b) when reflected about x-axis is given as (a,-b) i.e. Y coordinate changes its sign

\therefore P\rightarrow P'(-3,-8)\\\Rightarrow Q\rightarrow Q'(-6,4)\\\Rightarrow R\rightarrow R'(1,-1)

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3 years ago
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Answer:

1) 0, 180

2) 90

3) 3pi/2

4) pi/2, -3pi/2

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7) pi

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Step-by-step explanation:

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5 0
3 years ago
Find the radius of convergence, r, of the series. ? n2xn 7 · 14 · 21 · ? · (7n) n = 1
defon
I'm guessing the series is supposed to be

\displaystyle\sum_{n=1}^\infty\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}

By the ratio test, the series converges if the following limit is less than 1.

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7\cdot14\cdot21\cdot\cdots\cdot(7n)\cdot(7(n+1))}}{\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}}\right|

The first n terms in the numerator's denominator cancel with the denominator's denominator:

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7(n+1)}}{n^2x^n}\right|

|x^n| also cancels out and the remaining factor of |x| can be pulled out of the limit (as it doesn't depend on n).

\displaystyle|x|\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2}{7(n+1)}}{n^2}\right|=|x|\lim_{n\to\infty}\frac{|n+1|}{7n^2}=0

which means the series converges everywhere (independently of x), and so the radius of convergence is infinite.
3 0
3 years ago
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