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Sidana [21]
3 years ago
7

Hi, can someone please help, and possibly explain the answer please.

Mathematics
1 answer:
mr_godi [17]3 years ago
8 0

Answer:

x=\frac{-(-2)\±\sqrt{(-2)^2-4(3)(0)} }{2(3)}

Step-by-step explanation:

Quadratic formula: x=\frac{-b\±\sqrt{b^2-4ac} }{2a} when the equation is 0=ax^2+bx+c

The given equation is 1=-2x+3x^2+1. Let's first arrange this so its format looks like y=ax^2+bx+c:

1=-2x+3x^2+1

1=3x^2-2x+1

Subtract 1 from both sides of the equation

1-1=3x^2-2x+1-1\\0=3x^2-2x+0

Now, we can easily identify 3 as a, -2 as b and 0 as c. Plug these into the quadratic formula:

x=\frac{-b\±\sqrt{b^2-4ac} }{2a}\\x=\frac{-(-2)\±\sqrt{(-2)^2-4(3)(0)} }{2(3)}

I hope this helps!  

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Step-by-step explanation:

Given the original coordinate as C (-6, 9) and new coordinates after dilation as C' (-2, 3), we can see that the value of the new coordinates has reduces by a factor. Let the factor be k

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4 years ago
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