Answer:
k = -7
Step-by-step explanation:
Solve for k:
-7 k - 44 = k + 12
Subtract k from both sides:
(-7 k - k) - 44 = (k - k) + 12
-7 k - k = -8 k:
-8 k - 44 = (k - k) + 12
k - k = 0:
-8 k - 44 = 12
Add 44 to both sides:
(44 - 44) - 8 k = 44 + 12
44 - 44 = 0:
-8 k = 12 + 44
12 + 44 = 56:
-8 k = 56
Divide both sides of -8 k = 56 by -8:
(-8 k)/(-8) = 56/(-8)
(-8)/(-8) = 1:
k = 56/(-8)
The gcd of 56 and -8 is 8, so 56/(-8) = (8×7)/(8 (-1)) = 8/8×7/(-1) = 7/(-1):
k = 7/(-1)
Multiply numerator and denominator of 7/(-1) by -1:
Answer: k = -7
Answer:
<em>Choice: B.</em>
Step-by-step explanation:
<u>Operations With Functions</u>
Given the functions:
![f(x)=\sqrt[3]{12x+1}+4](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%5B3%5D%7B12x%2B1%7D%2B4)
The function (g-f)(x) can be obtained by replacing both functions and subtracting them as follows:
![(g-f)(x)= \log(x-3)+6 - (\sqrt[3]{12x+1}+4)](https://tex.z-dn.net/?f=%28g-f%29%28x%29%3D%20%5Clog%28x-3%29%2B6%20-%20%28%5Csqrt%5B3%5D%7B12x%2B1%7D%2B4%29)
Operating:
![(g-f)(x)= \log(x-3)+6 - \sqrt[3]{12x+1}-4](https://tex.z-dn.net/?f=%28g-f%29%28x%29%3D%20%5Clog%28x-3%29%2B6%20-%20%5Csqrt%5B3%5D%7B12x%2B1%7D-4)
Joining like terms:
![\boxed{(g-f)(x)= \log(x-3) - \sqrt[3]{12x+1}+2}](https://tex.z-dn.net/?f=%5Cboxed%7B%28g-f%29%28x%29%3D%20%5Clog%28x-3%29%20-%20%5Csqrt%5B3%5D%7B12x%2B1%7D%2B2%7D)
Choice: B.
Assuming the order required is as n-> inf.
As n->inf, o(log(n+1)) -> o(log(n)) since the 1 is insignificant compared with n.
We can similarly drop the "1" as n-> inf, the expression becomes log(n^2+1) ->
log(n^2)=2log(n) which is still o(log(n)).
So yes, both are o(log(n)).
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