So easyies way is find all multiplues of 6 that are less than 50
6
12
18
24
30
36
42
48
and not 42
6
12
18
24
30
36
48
more than 25
30
35
48
do not have a 3 in them
48
the answer is 48
There are two ways to do this.
The first way is to algebraically find (f+g)(x) first and plug in x = 5 later. Doing that method leads us to
(f+g)(x) = f(x) + g(x)
(f+g)(x) = 6x+3 + x-4
(f+g)(x) = 7x-1
(f+g)(5) = 7(5)-1
(f+g)(5) = 34
OR
you can compute f(5) and g(5) first, then add up those sub-results to get
f(x) = 6x+3
f(5) = 6(5)+3
f(5) = 33
g(x) = x-4
g(5) = 5-4
g(5) = 1
Adding up these results gives: (f+g)(5) = f(5) + g(5) = 33+1 = 34
Either way, the final answer is 34
Answer:
X= -7/6, 17/8
Step-by-step explanation:
Foil it, then use quadratic to find zeros/roots.
B. 10.1%
(2)(12)($88.18) ÷ [($1100)(18+1)]
(2,116.32) ÷ (20,900)
= 0.1012
= 10.1%
It is 22 my guy because i think it is 22
