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cupoosta [38]
2 years ago
13

What are the coordinates of R' for the dilation D(0.5. p)(PQRS)?

Mathematics
1 answer:
11111nata11111 [884]2 years ago
6 0

Answer:

R' = (-1,-0.5)

Step-by-step explanation:

See attachment for complete question.

From the attachment:

P = (-3,-3)

R = (1,2)

Dilation:

D_{(0.5,P)}

Required

Determine R'

First, subtract the coordinates of P from R. This means that R is measured from P.

R = (1-(-3),2-(-3))

R = (1+3,2+3)

R = (4,5)

Next, apply dilation factor 0.5

R' = R * 0.5

R' = (4,5) * 0.5

R' = (4* 0.5,5* 0.5)

R' = (2,2.5)

Lastly, measure R' from the origin by adding the coordinates of P to R'

R' = (2+(-3),2.5+(-3))

R' = (2-3,2.5-3)

R' = (-1,-0.5)

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pantera1 [17]

Answer:

Discount Percentage is 25%

Step-by-step explanation:

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4 0
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Long division<br> 3x+1/6x^6+5x^5+2x^4-9x^3+7x^2-10x+2
inna [77]
(6 x^{6}+5x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2) / (3x+1)

We divide first number from first parenthesis with first number from second parenthesis. Then the resulting number we multiply by all numbers in second parenthesis and substract from first parenthesis.

6 x^{6}/3x = 2 x^{5} \\  \\ 6 x^{6}+5x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2 - 6 x^{6} -2 x^{5} = 3x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2\\

We repeat previous steps until we run out of numbers:
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We are left with a number that has no x inside. This is remainder.
The final solution is sum of all these solutions and remainder:
(2 x^{5}+x^{4}+\frac{1}{3} x^{3} - \frac{28}{9} x^{2} +\frac{91}{27} x)+(-\frac{361}{81} )
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