<h3>
Answer: False</h3>
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Explanation:
I'm assuming you meant to type out
(y-2)^2 = y^2-6y+4
This equation is not true for all real numbers because the left hand side expands out like so
(y-2)^2
(y-2)(y-2)
x(y-2) .... let x = y-2
xy-2x
y(x)-2(x)
y(y-2)-2(y-2) ... replace x with y-2
y^2-2y-2y+4
y^2-4y+4
So if the claim was (y-2)^2 = y^2-4y+4, then the claim would be true. However, the right hand side we're given doesn't match up with y^2-4y+4
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Another approach is to pick some y value such as y = 2 to find that
(y-2)^2 = y^2-6y+4
(2-2)^2 = 2^2 - 6(2) + 4 .... plug in y = 2
0^2 = 2^2 - 6(2) + 4
0 = 4 - 6(2) + 4
0 = 4 - 12 + 4
0 = -4
We get a false statement. This is one counterexample showing the given equation is not true for all values of y.
Answer:
The answer is 54 cubes.
Step-by-step explanation:
<u>Answer</u>
12 units
<u>Explanation</u>
This can be solved using the properties of a circle.
DG and EG are secants f the circle.
DG×HG =EG×FG
(x+3+5)×5 = (x+6)×6
(x+8)5 = 6x + 36
5x +40 = 6x + 36
x = 4
DG = (x+3+5)
= 4 + 3 + 5
= 12 units
