Is when you divide long numbers
Any of the 11 man can be chosen, and combined with any of the 8 women.
Assume we select man1. The selected committee can be:
(m1,w1), (m1,w2), (m1,w3), (m1,w4), (m1,w5), (m1,w6), (m1,w7), (m1,w8),
so there are 8 committees selections with man1 in them.
we could repeat the same procedure for the remaining 10 men, and get 8 committees where each of them is a member.
so there are 11*8=88 ways of choosing 1 man and 1 woman.
Answer: 88
Answer:
Step-by-step explanation:
Easiest way would be to substitute points into the given equations and plot your answers.
E.g. if you substitute points into your first equation you get:
If x = 0, y = -4(0) +2 = 2
If x = 1, y = -4(1) +2 = -2
If x = -1, y = -4(-1) +2 = 6
Etc.
Second equation:
If x = 0, y= -(0) -1 = -1
If x = 1, y = -1 -1 = -2
If x = -1, y = -(-1) -1 = 0
Pretty much just plug in values for x into the equations to find y.
Once you have your points, draw a line :)
Where the two lines intercept would be the solution.
(3x^3 + 4x²) + (3x^3 – 4x^2 – 9x) =
We have to combine like terms, terms that have x raised to the same power:
First, eliminate the parenthesis:
3x^3 + 4x² + 3x^3 – 4x^2 – 9x =
We have 2 terms raised to the second power (4x^2 and -4 x^2) simply add and subtract.
3x^3+3x^3+4x^2-4x^2-9x
6x^3-9x
Answer:
5115
Step-by-step explanation:
a1(1-(r)^n)/1-r
5(1-(2.00)^10)/1-2.00
