So
2/6 of 318=106
so 106-1/2 or 106/2=53
So he has 53 boxes left
Calculating
the value of f(x) for the given interval.
For x = -
4, f(x) = f(- 4) = (- 4)^2 + 2 (- 4) + 3 = 11
For x =
6, f(x) = f(6) = (6)^2 + 2 (6) + 3 = 51
Now using
formula for the calculation of average rate of change of f(x) over the given
interval of [- 4, 6];
(f(b) –
f(a)) / b – a = (f(6) – f(- 4)) / 6 – (- 4) = (51 -11) / 10 = 4
<span>So option “E” is
correct.</span>
1/2 and 3/4 those are greater then 1/4 and less then 1
Step-by-step explanation:
![\log_{12}\left(\dfrac{\frac{1}{2}}{8w}\right)=\log_{12}\left(\dfrac{1}{16w}\right)\qquad\text{use}\ a^{-1}=\dfrac{1}{a}\\\\=\log_{12}(16w)^{-1}\qquad\text{use}\ \log_ab^n=n\log_ab\\\\=-1\log_{12}(16w)\qquad\text{use}\ \log_a(bc)=\log_ab+\log_ac\\\\=-\left(\log_{12}16+\log_{12}w\right)=-\log_{12}2^4-\log_{12}w\qquad\text{use}\ \log_ab^n=n\log_ab\\\\=-4\log_{12}2-\log_{12}w](https://tex.z-dn.net/?f=%5Clog_%7B12%7D%5Cleft%28%5Cdfrac%7B%5Cfrac%7B1%7D%7B2%7D%7D%7B8w%7D%5Cright%29%3D%5Clog_%7B12%7D%5Cleft%28%5Cdfrac%7B1%7D%7B16w%7D%5Cright%29%5Cqquad%5Ctext%7Buse%7D%5C%20a%5E%7B-1%7D%3D%5Cdfrac%7B1%7D%7Ba%7D%5C%5C%5C%5C%3D%5Clog_%7B12%7D%2816w%29%5E%7B-1%7D%5Cqquad%5Ctext%7Buse%7D%5C%20%5Clog_ab%5En%3Dn%5Clog_ab%5C%5C%5C%5C%3D-1%5Clog_%7B12%7D%2816w%29%5Cqquad%5Ctext%7Buse%7D%5C%20%5Clog_a%28bc%29%3D%5Clog_ab%2B%5Clog_ac%5C%5C%5C%5C%3D-%5Cleft%28%5Clog_%7B12%7D16%2B%5Clog_%7B12%7Dw%5Cright%29%3D-%5Clog_%7B12%7D2%5E4-%5Clog_%7B12%7Dw%5Cqquad%5Ctext%7Buse%7D%5C%20%5Clog_ab%5En%3Dn%5Clog_ab%5C%5C%5C%5C%3D-4%5Clog_%7B12%7D2-%5Clog_%7B12%7Dw)