Answer:
2147.60
Step-by-step explanation:
Pair up the terms into separate groups. Then factor each group individually (pull out the GCF). Once that is finished, you factor out the overall GCF to complete the full factorization.
8r^3 - 64r^2 + r - 8
(8r^3 - 64r^2) + (r - 8)
8r^2(r - 8) + (r - 8)
8r^2(r - 8) + 1(r - 8)
(8r^2 + 1)(r - 8)
So the final answer is (8r^2 + 1)(r - 8)
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Edit:
Problem 1b) Follow the same basic steps as in part A
28v^3 + 16v^2 - 21v - 12
(28v^3 + 16v^2) + (-21v - 12)
4v^2(7v + 4) + (-21v - 12)
4v^2(7v + 4) - 3(7v + 4)
(4v^2 - 3)(7v + 4)
The answer to part B is (4v^2 - 3)(7v + 4)
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Second Edit:
I apologize for the first edit. I misread what you were asking initially. Here is problem 2A. We follow the same basic steps as in 1a) and 1b). You'll need to rearrange terms first
27mz - 12nc + 9mc - 36nz
27mz + 9mc - 12nc - 36nz
(27mz + 9mc) + (-12nc - 36nz)
9m(3z + c) + (-12nc - 36nz)
9m(3z + c) -12n(c + 3z)
9m(3z + c) -12n(3z + c)
(9m - 12n)(3z + c)
3(3m - 4n)(3z + c)
Answer with Step-by-step explanation:
We are given that
We have to explain that why the function is discontinuous at x=2
We know that if function is continuous at x=a then LHL=RHL=f(a).
LHL=Left hand limit when x <2
Substitute x=2-h
where h is small positive value >0
Right hand limit =RHL when x> 2
Substitute
x=2+h
LHL=RHL=
f(2)=1
Hence, function is discontinuous at x=2
