May be there is an operator missing in the first function, h(x). I will solve this in two ways, 1) as if the h(x) = 5x and 2) as if h(x) = 5 + x
1) If h(x) = 5x and k(x) = 1/x
Then (k o h) (x) = k ( h(x) ) = k(5x) = 1/(5x)
2) If h(x) = 5 + x and k (x) = 1/x
Then (k o h)(x) =k ( h(x) ) = k (5+x) = 1 / [5 + x]
Answer:
Continuous random variable
Step-by-step explanation:
Answer:
5a-3b+12c
Step-by-step explanation:
(2a - 3b + 5c) - (-3a - 7c)
Distribute the minus sign
2a -3b +5c +3a +7c
Combine like terms
5a-3b+12c
<span>x=y+2
2y=x-1
substitute </span>x=y+2 into 2y=x-1
2y=x-1
2y=y+2 -1
2y - y = 1
y = 1
<span>x=y+2
</span><span>x=1+2
x=3
answer
x = 3 and y = 1</span>
Answer: the answer would be 64
Step-by-step explanation:
