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3 years ago

10

What is the scale factor use to create this dilation?

1 answer:

andrezito [222]3 years ago

3 0

Given Info: "The dashed triangle is the image of the solid triangle"

So the "before" is the large solid triangle and the "after" is the smaller dashed triangle.

The horizontal side of the solid triangle is 15 units. The corresponding side for the dashed triangle is 3 units. Dividing the two values gives: 3/15 = 1/5 = 0.2

Make this value negative. The reason why is due to the fact that we have a sort of reflection going on as we're scaling down the figure.

So the final answer is -0.2

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2 years ago

Find all values of x for which the series converges. (Enter your answer using interval notation.) [infinity] 9 x − 7 9 n n = 0 F

makkiz [27]

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The series converges to $\frac{1}{1-9x}$ for $\frac{-1}{9} < x < \frac{1}{9}$

Step-by-step explanation:

Given the series is $\sum_{n=0}^{\infty} 9^n x^n$

We have to find the values of x for which the series converges.

We know,

$\sum_{n=0}^{\infty} ar^{n-1}$ converges to (a) / (1-r) if r < 1

Otherwise the series will diverge.

Here, $\sum_{n=0}^{\infty} 9^n x^n = \sum_{n=0}^{\infty} (9x)^{n}$ is a geometric series with |r| = | 9x |

And it converges for |9x| < 1

Hence, the given series gets converge for $\frac{-1}{9} < x < \frac{1}{9}$

And geometric series converges to $\frac{a}{1-r}$

Here, a = 1 and r = 9x

Therefore, $\frac{a}{1-r} = \frac{1}{1-9x}$

Hence, the given series converges to $\frac{1}{1-9x}$ for $\frac{-1}{9} < x < \frac{1}{9}$

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