Assuming the given is: 5u^7 - 21u^7
Since both algebraic terms are identical (both are using u^7), we can subtract the coefficients directly, as in: 5 - 21 = -16
Therefore, 5u^7 - 21u^7 = -16u^7

7 0

3 years ago

Solve the given system, using the substitution method.

alexandr1967 [171]

Y = 3x - 7 . . . . . (1)
6x - 2y = 12 . . . (2)

Putting (1) into (2) gives
6x - 2(3x - 7) = 12
6x - 6x + 14 = 12
14 = 12

Therefore, there is no solution.

5 0

2 years ago

Read 2 more answers

Suppose that three numbers are selected one by one, at random, and without replacement, from the set of integers {1, 2, 3, . . .

fiasKO [112]

A - the third number falls between the first two

B - the first number is smaller than the second

$P(A|B)=\dfrac{P(A\cap B)}{P(B)}\\\\ |\Omega|=n(n-1)(n-2)\\\\ |B|=\dfrac{n!}{2}\\\\ |A\cap B|=\dfrac{\dfrac{n!}{2}}{3}=\dfrac{n!}{6}\\\\ P(A\cap B)=\dfrac{\dfrac{n!}{6}}{n(n-1)(n-2)}=\dfrac{n!}{6n(n-1)(n-2)}=\dfrac{(n-3)!}{6}\\ P(B)=\dfrac{\dfrac{n!}{2}}{n(n-1)(n-2)}=\dfrac{n!}{2n(n-1)(n-2)}=\dfrac{(n-3)!}{2}\\\\ P(A|B)=\dfrac{\dfrac{(n-3)!}{6}}{\dfrac{(n-3)!}{2}}\\\\ P(A|B)=\dfrac{(n-3)!}{6}\cdot\dfrac{2}{(n-3)!}=\dfrac{1}{6}=\dfrac{1}{3}$

3 0

3 years ago