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professor190 [17]
3 years ago
6

Which steps would prove the circles similar?

Mathematics
1 answer:
Eddi Din [679]3 years ago
8 0

Answer:

You can translate smaller circle onto larger circle so centers line up and then dilate to prove similarity.

Step-by-step explanation:

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3 years ago
What is the answer to this solution. -7r−4≥ 4r+2
tia_tia [17]

Answer:

The solution is:

-7r-4\ge \:\:4r+2\quad \::\quad \:\begin{bmatrix}\mathrm{Solution:}\:&\:r\le \:\:-\frac{6}{11}\:\\ \:\:\mathrm{Decimal:}&\:r\le \:\:-0.54545\dots \:\\ \:\:\mathrm{Interval\:Notation:}&\:(-\infty \:\:,\:-\frac{6}{11}]\end{bmatrix}

Please check the attached line graph below.

Step-by-step explanation:

Given the expression

-7r-4\ge \:4r+2

Add 4 to both sides

-7r-4+4\ge \:4r+2+4

Simplify

-7r\ge \:4r+6

Subtract 4r from both sides

-7r-4r\ge \:4r+6-4r

Simplify

-11r\ge \:6

Multiply both sides by -1 (reverses the inequality)

\left(-11r\right)\left(-1\right)\le \:6\left(-1\right)

Simplify

11r\le \:-6

Divide both sides by 11

\frac{11r}{11}\le \frac{-6}{11}

Simplify

r\le \:-\frac{6}{11}

Therefore, the solution is:

-7r-4\ge \:\:4r+2\quad \::\quad \:\begin{bmatrix}\mathrm{Solution:}\:&\:r\le \:\:-\frac{6}{11}\:\\ \:\:\mathrm{Decimal:}&\:r\le \:\:-0.54545\dots \:\\ \:\:\mathrm{Interval\:Notation:}&\:(-\infty \:\:,\:-\frac{6}{11}]\end{bmatrix}

Please check the attached line graph below.

7 0
3 years ago
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