The answer is true. Your welcome
Given a group of
n object. We want to make a selection of
k objects out of the n object. This can be done in
C(n, k) many ways, where
![C(n, k)= \frac{n!}{k!(n-k)!}](https://tex.z-dn.net/?f=C%28n%2C%20k%29%3D%20%5Cfrac%7Bn%21%7D%7Bk%21%28n-k%29%21%7D%20)
,
where k!=1*2*3*...(k-1)*k
Thus, we can do the selection of 3 cd's out of 5, in C(5,3) many ways,
where
![C(5, 3)= \frac{5!}{3!2!}= \frac{5*4*3*2*1}{3*2*1*2*1}= \frac{5*4}{2}=10](https://tex.z-dn.net/?f=C%285%2C%203%29%3D%20%5Cfrac%7B5%21%7D%7B3%212%21%7D%3D%20%5Cfrac%7B5%2A4%2A3%2A2%2A1%7D%7B3%2A2%2A1%2A2%2A1%7D%3D%20%5Cfrac%7B5%2A4%7D%7B2%7D%3D10%20%20%20)
Answer: 10
<span>The general form of quadratic equation with real
coefficients and leading coefficient 1, has x = -bi as a root
=> x = <u>-b + </u></span><u>√ b^2 – 4 ac</u><span>
2a
It is also written as:
=> ax^2 + bx + c = 0
Quadratic equation involves unknown numbers which is x, the numbers which a, b
and c are called coeffecients.
There are also quadratic factorization where you factor the polynomial give to
be able to get the value of the equation.</span>
No it is not, there is no common difference. <span />
Let x be equal to the number of drinks Yasmine consumed.
Jose had 2 times that drink so his number of consumed drink would be represented by 2x.
Sally had 3 fewer drink than Jose so her number of consumed drinks would be represented by 2x-3.
Altogether, the three of them consumed 72 drinks so your equation would be:
x+2x+(2x-3)=72
add like terms together:
5x-3=72
have the term with x be alone on one side of the equation, in this case by adding three to both sides:
5x=75
now divide both sides by five for the value of x and your answer is.....
x=15