The answer is the last one 7/8 x -3/4 =6
Answer:
Should not be rejected.
Step-by-step explanation:
Answer:
see below the first three problems
Step-by-step explanation:
f(g(-2))
First, find g(-2) using function g(x). Then use that value as input for function f(x).
g(x) = -2x + 1
g(-2) = -2(-2) + 1
g(-2) = 5
f(x) = 5x
f(5) = 5(5)
f(5) = 25
f(g(-2)) = 25
g(h(3))
First, find h(3) using function h(x). Then use that value as input for function g(x).
h(x) = x^2 + 6x + 8
h(3) = 3^2 + 6(3) + 8 = 9 + 18 + 8
h(3) = 35
g(x) = -2x + 1
g(35) = -2(35) + 1 = -70 + 1
g(35) = -69
g(h(3)) = -69
f(g(3a))
First, find g(3a) using function g(x). Then use that value as input for function f(x).
g(x) = -2x + 1
g(3a) = -2(3a) + 1
g(3a) = -6a + 1
f(x) = 5x
f(-6a + 1) = 5(-6a + 1)
f(-6a + 1) = -30a + 5
f(g(3a)) = -30a + 5
The <em>correct answer</em> is:
We can use the associative property to write 70 as 7*10. We can do this same process to write 6000 as 6*1000. This gives us:
(7*10)*(6*1000)
Using the commutative property, we can rearrange the factors of this problem
(7*6)*(10*1000)
7*6 is in our multiplication tables, and we know that powers of 10 allow us to just add the correct number of zeros. This means we would have 7*6=42, and have 4 zeros, for 420000.
Given :Enzo wins 5 tickets from every game, and Beatriz wins 11 tickets from every game.We need to find the minimum number of games that Enzo could have played to win the same number of tickets.
The minimum number of games that Enzo could have played to win the same number of tickets Will be the least common multiple of 11 and 5.
The factors of 11 and 5 are
11=11x1
5= 5x1
Least common multiply = 11x5=5.
The minimum number of games that Enzo could have played to win the same number of tickets is 55.
