Answer:
(2n+6)^2
Step-by-step explanation:
To solve this, you can use the given formula: x^2 + 2xy + y^2
In this case, 4n^2 is x^2, 24n is 2xy, and 36 is y^2. The next step is:
(2n)^2 + 2(2n)(6) + (6)^2
Since this equation fits into this formula ( x^2 + 2xy + y^2), we can do:
(2n+6)(2n+6) =
(2n+6)^2
Hence, the answer is (2n+6)^2
Answer: C
Step-by-step explanation:
x-3y=5, y=x-7 *solve it
the answer is (8,1)
so..
C. (8,1) is a solution to both equations so it is a solution to the system
<span>If there has to be 2 men and 2 women, we know
that we must take a group of 2 men out of the group of 15 men and a group of 2
women out of the group of 20 women. Therefore, we have:
(15 choose 2) x (20 choose 2)
(15 choose 2) = 105
(20 choose 2) = 190
190*105 = 19950
Therefore, there are 19950 ways to have a group of 4 with 2 men and 2women.</span>
<span>If there has to be 1 man and 3 women, we know
that we must take a group of 1 man out of the group of 15 men and a group of 3
women out of the group of 20 women. Therefore, we have:
(15 choose 1) x (20 choose 3)
(15 choose 1) = 15
(20 choose 3) = 1140
15*1140 = 17100
Therefore, there are 17100 ways to have a group of 4 with 3 women and 1 man.</span>
<span>We now find the total outcomes of having a group
with 4 women.
We know this is the same as saying (20 choose 4) = 4845</span>
Therefore, there are 4845 ways to have a group of
4 with 4 women.
We now add the outcomes of 2 women, 3 women, and
4 women and get the total ways that a committee can have at least 2 women.
19950 + 17100 + 4845 = 41895 ways that there will
be at least 2 women in the committee
An equilateral triangle cross section can be obtained by cutting the cube by a plane defined by the midpoints of the three edges emanating from any one vertex.
