let's recall the remainder theorem.
we know that (x-1) is a factor, that means x -1 = 0 or x = 1.
since we know that (x-1) is a factor, then dividing the polynomial by it will give us a remainder of 0, which correlates with saying that f(1) = 0, in this case, so we can simply plug in "1" as the argument, knowing it gives 0.
![f(x)=3x^3+kx-11\\\\[-0.35em]~\dotfill\\\\\stackrel{0}{f(1)}=3(1)^3+k(1)-11\implies \stackrel{f(1)}{0}=3+k-11\implies 0=-8+k\implies 8=k](https://tex.z-dn.net/?f=f%28x%29%3D3x%5E3%2Bkx-11%5C%5C%5C%5C%5B-0.35em%5D~%5Cdotfill%5C%5C%5C%5C%5Cstackrel%7B0%7D%7Bf%281%29%7D%3D3%281%29%5E3%2Bk%281%29-11%5Cimplies%20%5Cstackrel%7Bf%281%29%7D%7B0%7D%3D3%2Bk-11%5Cimplies%200%3D-8%2Bk%5Cimplies%208%3Dk)
Answer:
The inequality would be true
Step-by-step explanation:
4(c+2) < 4c + 10
Distribute the 4 on the left
4c + 8 < 4c + 10
Subtract 4c from both sides
8 < 10
8 < 10 is true
Answer:
A(3) = -6 + (3 - 1) (5)
-6 + (2)(5)
-6 + (10)
4
A(4) = -6 + (4 - 1) (5)
-6 + (3)(5)
-6 + 15
9
A(10) = -6 + (10 - 1) (5)
-6 + (9)(5)
-6 + 45
39
There is the answer hope it helpd if you can mark me brainliest
