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Ulleksa [173]
3 years ago
13

I don’t know how to do this one. could someone help

Mathematics
1 answer:
DedPeter [7]3 years ago
6 0

Answer:

If it is being dilated from (0,0) the answer is below.  

Step-by-step explanation:

It depends on where it is scaled from.  If you choose a point directly to that points left it will be pushed tot he right, and vice versa if it is on its right.    Same for above and below

I am going to assume we are dilating from (0,0), but again, if there is another part of the question saying where it's from use this method for that as well.

So, from (0,0) the initial point is (-3, 11)  Scaling by a factor of 3 means it gets pushes AWAY so it is 3 times as far away.  (If it were a fraction or decimal it would be getting pulled closer)

Using (0,0) as the dilation point is the easiest because we just have to ask how far are the x and y coordinates away from (0,0) and any number is THAT number away from 0.  

We break this into the two coordinates too.  so how far is -3 from 0?  -3.  so we want a point 3 times as far away, so what is that?  -9.  Same for 11, 11 is 11 away from 0 so we want it 3 times as far, which is 33.

Now that we have the two coordinates  that is the answer.  

(-9, 33)

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<span>Solve.

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</span>x^2 + 4x + 4 = 18 \\  \\ x^2+4x+4-18= 0 \\  \\ x^2+4x-14=0 \\  \\ x_1_y_2=  \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} \qquad a= 1\qquad b= 4\qquad c= -14 \\  \\  \\  x_1_y_2=  \dfrac{-4\pm \sqrt{4^2-4(1)(-14)} }{2(1)} \\  \\  \\  x_1_y_2=  \dfrac{-4\pm \sqrt{16-(-56)} }{2}  \\  \\  \\  x_1_y_2=  \dfrac{-4\pm \sqrt{72} }{2}  \\  \\  \\  x_1=  \dfrac{-4+ \sqrt{72} }{2} \qquad\qquad x_2=  \dfrac{-4+ \sqrt{72} }{2}\\  \\  \\  x_1=  \dfrac{-4+ \sqrt{2^3*3^2} }{2} \qquad\qquad x_2=  \dfrac{-4- \sqrt{2^3*3^2} }{2}
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