Hello I think the answer would be three.
We have to solve two inequations:
Equation 1:
30(x-1)≥0
(x-1)≥0/30
(x-1)≥0
x≥1 (solution 1)
Equation 2:
5x²≥0
x≥0
The solution is: solution1 <span>∩ solution2
Teherefore; x≥1</span>
Answer:
Step-by-step explanation:
The lady gets paid $8 per hour that the shop is opened, so multiply the pay rate ($8) by the amount of hours the shop is opened
$8 * 8 = $64
Now add the $64 onto the expense of running the shop
$64 + $240 = $304
Now take that amount away from Friday's sales
$530 - $304 = $ 226
Profit($226) = Income($530) - Expenses($304)
$226 = $226
=)
f(x)=x3−5
Replace f(x)
with y
.
y=x3−5
Interchange the variables.
x=y3−5
Solve for y
.
Since y
is on the right side of the equation, switch the sides so it is on the left side of the equation.
y3−5=x
Add 5
to both sides of the equation.
y3=5+x
Take the cube root of both sides of the equation to eliminate the exponent on the left side.
y=3√5+x
Solve for y
and replace with f−1(x)
.
Replace the y
with f−1(x)
to show the final answer.
f−1(x)=3√5+x
Set up the composite result function.
f(g(x))
Evaluate f(g(x))
by substituting in the value of g into f
.
(3√5+x)3−5
Simplify each term.
Remove parentheses around 3√5+x
.
f(3√5+x)=3√5+x3−5
Rewrite 3√5+x3
as 5+x
.
f(3√5+x)=5+x−5
Simplify by subtracting numbers.
.
Subtract 5
from 5
.
f(3√5+x)=x+0
Add x
and 0
.
f(3√5+x)=x
Since f(g(x))=x
, f−1(x)=3√5+x is the inverse of f(x)=x3−5
.
f−1(x)=3√5+x
