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Kaylis [27]
3 years ago
7

Circle P has center P(2,0) and a radius 20. Circle Q has center Q(0,4) and radius 2. Describe the rule for translating center Q

onto center P. Determine the scale factor for dilating circle Q so that it has the same radius as circle P. Are circles P & Q similar? Explain your answer
Mathematics
1 answer:
fomenos3 years ago
8 0
The circle the has a center P(2,0) is bigger than compared to the circle with the center Q(0,4) because the former has a radius 20, while the latter has a radius of only 2. You can view the graph here: 
https://www.desmos.com/calculator/od1pwqylou

(a) Describe the rule for translating center Q onto center P. By moving the center of circle Q overlapping the center of circle P, then we need to move it two units left and 4 units up.

(b) Determine the scale factor for dilating circle Q so that it has the same radius as circle P. In getting the scale factor, we need to compare the radius
= 2/10
= 1/10
So, the scale factor is 1/10

(c) Are circles P and Q similar? Explain your answer.
Yes, by applying a dilation value, circle Q can be transformed to circle P, vice versa.
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STatiana [176]

Answer:

x = 2

Step-by-step explanation:

Step 1: Write equation

3(x + 2) = 12

Step 2: Solve for <em>x</em>

  1. Distribute 3: 3x + 6 = 12
  2. Subtract 6 on both sides: 3x = 6
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Step 3: Check

<em>Plug in x to verify it's a solution.</em>

3(2 + 2) = 12

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luda_lava [24]

Answer:

45

Step-by-step explanation:

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katrin2010 [14]

Answer:

\boxed{\bold{Decimal \ Form: \ x \ = \ -0.125}}

\boxed{\bold{Fraction \ Form:x=-\frac{1}{8}}}

Step By Step Explanation:

Apply Fraction Cross Multiply: If \bold{\frac{a}{b}=\frac{c}{d}:then\ a\cdot \:d=b\cdot \:c}

\bold{\left(2x+1\right)\cdot \:5=2\left(x+2\right)}

\bold{\left(2x+1\right)\cdot \:5: \ 10x+5}

\bold{2\left(x+2\right): \ 2x+4}

\bold{10x+5=2x+4}

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\bold{10x+5-5=2x+4-5}

Simplify

\bold{10x=2x-1}

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\bold{10x-2x=2x-1-2x}

Simplify

\bold{8x=-1}

Divide Both Sides By 8

\bold{\frac{8x}{8}=\frac{-1}{8}}

Simplify

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Turn Fraction Into Decimal

\bold{-1 \ \div \ 8 \ = \ -0.125}

\bold{-0.125}

Regards,

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Charra [1.4K]

Answer:

the biconditional is a true statement

Step-by-step explanation:

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(If this is wrong I'm sorry I haven't done this kind of math in a while!)

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