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Take x-2 and insert it into 2x^2 + 3x-2 where the x is located
2x^2 + 3x-2
2(x-2)^2 + 3(x-2)-2
Now work out 2(x-2)^2 + 3(x-2)-2 also follow PEMDAS
2(x-2)^2 + 3(x-2)-2
Since (x-2)^2 is an Exponent, lets work with that first and expand (x-2)^2.
(x-2)^2
(x -2)(x-2)
x^2 -4x + 4
Now Multiply that by 2 because we have that in 2(x-2)^2
(x-2)^2 = x^2 -4x + 4
2(x-2)^2 = 2(x^2 -4x + 4)
2(x^2 -4x + 4) = 2x^2 - 8x + 8
2x^2 - 8x + 8
Now that 2(x-2)^2 is done lets move on to 3(x-2).
Use the distributive property and distribute the 3
3(x-2) = 3x - 6
All that is left is the -2
Now lets put it all together
2(x-2)^2 + 3(x-2)-2
2x^2 - 8x + 8 + 3x - 6 - 2
Now combine all our like terms
2x^2 - 8x + 8 + 3x - 6 - 2
Combine: 2x^2 = 2x^2
Combine: -8x + 3x = -5x
Combine: 8 - 6 - 2 = 0
So all we have left is
2x^2 - 5x
Answer:
RQ = 18
Step-by-step explanation:
KR is an angle bisector and divides the opposite side into segments that are proportional to the other two sides, that is
= 
, substitute values
= 
( cross- multiply )
45(38 - x) = 50x ← distribute left side
1710 - 45x = 50x ( add 45x to both sides )
1710 = 95x ( divide both sides by 95 )
18 = x
That is RQ = 18
72.5 as decimal or 72 1/2 as fraction
We know that
[scale factor]=1/2
[dilated area of circle]=[original area of circle]*[scale factor]²
so
[dilated area of circle]=[original area of circle]*[1/2]²
[dilated area of circle]=[original area of circle]*[1/4]
therefore
the answer is
<span>The area will be 1/4 the original</span>
