Based on the graph given, it can be deduced from the computation that the value of y will be -8.
Point C is given as (-10 , -2) and D = (10 , 6)
The Slope of line CD = ( 6 - (-2) /( 10 - (-10) = 8/20 = 2/5
Since EF is parallel to line CD
, the slope of EF = 2/5
Equation y = (2/5)x + c
5y = 2x + 5c
It passes through F ( 6 , -4)
= 5(-4) = 2*6 + 5c
= -20 = 12 + 5c
5c = -32
5y = 2x - 32
Point E ( - 4 , y)
5y = 2(-4) - 32
5y = -8 - 32
5y = -40
y = -8
Therefore, the value of y is -8.
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Answer:
I think it is either A or D
So using a(2)=0 we can first solve for k by substituting t for 2
0 = (2-k)(2-3)(2-6)(2+3)
0 = (2-k)(-1)(-4)(5)
0 = (2-k)20
0 = 40 - 20k
-40 = -20k
k = 2
The next step would be to find all the 0s of a.
0 = (t-2)(t-3)(t-6)(t+3)
T = 2,3,6,-3
Then we find the product
2x3x6x-3 = -108
Since the problem asks for the absolute value, the answer is positive 108
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