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kondor19780726 [428]
3 years ago
8

What is the image of 6, -7 after dilation by scale factor of 3 centered at the origin

Mathematics
1 answer:
earnstyle [38]3 years ago
4 0

Answer:

(18,−21)

Step-by-step explanation:

Dilation: multiply each coordinate

(x,y)\rightarrow (3x,3y)

(x,y)→(3x,3y)

Dilations multiply by scale factor

(\color{red}{6},\color{blue}{-7}) \rightarrow (3\cdot \color{red}{6},3\cdot \color{blue}{-7})=\mathbf{(18,-21)}

(6,−7)→(3⋅6,3⋅−7)=(18,−21)

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3 years ago
5x +6y = 20 (1)
slega [8]
(2) + (2)

=> 8x - 6y = -46 ———(3)

(1) + (3)

=> 5x + 6y + (8x - 6y) = 20 - 46

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=> 13x + 0y = -26

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=> x = -26/13

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2 years ago
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Find the values of k so that each remainder is three. <br> 10. (x^2+ 5x + 7) = (x + k)
Goryan [66]

Answer:

k=1\text{ or } k=4

Step-by-step explanation:

We can use the Polynomial Remainder Theorem. It states that if we divide a polynomial P(x) by a <em>binomial</em> in the form (x - a), then our remainder will be P(a).

We are dividing:

(x^2+5x+7)\div(x+k)

So, a polynomial by a binomial factor.

Our factor is (x + k) or (x - (-k)). Using the form (x - a), our a = -k.

We want our remainder to be 3. So, P(a)=P(-k)=3.

Therefore:

(-k)^2+5(-k)+7=3

Simplify:

k^2-5k+7=3

Solve for <em>k</em>. Subtract 3 from both sides:

k^2-5k+4=0

Factor:

(k-1)(k-4)=0

Zero Product Property:

k-1=0\text{ or } k-4=0

Solve:

k=1\text{ or } k=4

So, either of the two expressions:

(x^2+5x+7)\div(x+1)\text{ or } (x^2+5x+7)\div(x+4)

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Step-by-step explanation:

For it to be a function, each value of x has to map onto one value of y.

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