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2 years ago

5

A triangle ABC is reflected across the y-axis and then translated 5 units down to triangle A'B'C'. How are the side lengths of t

riangle ABC related to the side lengths of triangle A'B'C'?

1 answer:

Oksi-84 [34.3K]2 years ago

3 0

Answer:

d, c, a, b

d= 32, 56

c = 34, 64

a = 85, 92

b = 23, 55

Step-by-step explanation:

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qwelly [4]

Answer is (a)
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2 years ago

Jessica hit a golf ball 150.75 yards. Kayla hit a golf ball 130.25 yards. How much farther did Jessica hit a golf ball?

Lapatulllka [165]

Answer:

Jessica hit the golf ball 20.50 yards farther than Kayla.

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2 years ago

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Which of the following statements is true in a one-way ANOVA? a. The critical value of the test will be a value obtained from th

Musya8 [376]

Answer:

d) All of the above

Step-by-step explanation:

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The correct option here is (d).

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3 years ago

78 ÷ 8 is equivalent to 78.00 ÷ 8. We can write zeros in the tenths place and hundredths place without changing the value. Use t

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Step-by-step explanation:

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2 years ago

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Express the function as the sum of a power series by first using partial fractions. f(x) = 8 x2 − 4x − 12 f(x) = ∞ n = 0 find th

alexandr1967 [171]

I'm guessing the function is

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which, split into partial fractions, is equivalent to

$\dfrac1{x-6}-\dfrac1{x+2}$

Recall that for $|x|$ we have

$\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n$

With some rearranging, we find

$\dfrac1{x-6}=-\dfrac16\dfrac1{1-\frac x6}=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n$

valid for $\left|\dfrac x6\right|$ , or $|x|$ , and

$\dfrac1{x+2}=\dfrac12\dfrac1{1-\left(-\frac x2\right)}=\displaystyle\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n$

valid for $\left|-\dfrac x2\right|$ , or $|x|$ .

So we have

$f(x)=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n-\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n$

$f(x)=\displaystyle-\sum_{n=0}^\infty\left(\frac{x^n}{6^{n+1}}+\frac{(-x)^n}{2^{n+1}}\right)$

$f(x)=\displaystyle-\sum_{n=0}^\infty\frac{1+3(-3)^n}{6^{n+1}}x^n$

Taken together, the power series for $f(x)$ can only converge for $|x|$ , or $-2$ .

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2 years ago