$\bf \begin{array}{llll} \textit{logarithm of factors} \\\\ log_a(xy)\implies log_a(x)+log_a(y) \end{array}~\hfill \begin{array}{llll} \textit{Logarithm of rationals} \\\\ log_a\left( \frac{x}{y}\right)\implies log_a(x)-log_a(y) \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \log_{12}\left( \cfrac{~~\frac{1}{2}~~}{8w} \right)\implies \begin{array}{llll} \log_{12}\left( \frac{1}{2} \right)&-&\log_{12}(8w)\\\\ \log_{12}(1)-\log_{12}(2)&-&[\log_{12}(8)+\log_{12}(w)] \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \log_{12}(1)-\log_{12}(2)-\log_{12}(8)-\log_{12}(w)~\hfill$

7 0

3 years ago

A bag contains four red marbles, two green ones, one lavender one, three yellows, and one orange marble. HINT (See Example 7.] H

Murljashka [212]

Answer: 35

Step-by-step explanation:

Given : A bag contains four red marbles, two green ones, one lavender one, three yellows, and one orange marble.

Total = 4+2+1+3+1=11

To find sets of four marbles include none of the red ones, we need to exclude red marbles when we count the total number of marbles.

Then, the total marbles(exclude red) =11-4=7

Now, the combination of 7 marbles taking 4 at a time is given by :-

$^7C_4=\dfrac{7!}{4!(7-4)!}=\dfrac{7\times6\times5\times4!}{4!3!}=35$

Hence, the number of sets of four marbles include none of the red ones = 35

4 0

2 years ago