The problem can be translated into an equation that is something like 4/5 + 3/x = 1/2
we cannot have x equal to zero because the number can be infinite.
So the LCD here is 10x, so multiply both sides by that to get:
8x + 30 = 5x
Subtract 5x and 30 from both sides:
3x = -30
divide:
x = -10
The solution isn't zero so there is a solution.
Answer:
Step-by-step explanation:
Step 1: Given equation: 
To get u, by subtraction equality property subtract both sides of the equation by 
.
Step 2: By division equality property, divide both sides of the equation by t.
Therefore, 
.
So, in Emily’s physics class, she got .
T = 5, so after 5 years
p(t) = t^3 - 14t^2 + 20t + 120
Take derivative to find minimum:
p’(t) = 3t^2 - 28t + 10
Factor to solve for t:
p’(t) = (3t - 2)(t - 5)
0 = (3t - 2)(t - 5)
0 = 3t - 2
2 = 3t
2/3 = t
Plug 2/3 into original equation, this is a maximum. We want the minimum:
0 = t - 5
5 = t
Plug back into original:
5^3 - 14(5)^2 + 20(5) + 120
125 - 14(25) + 100 + 120
125 - 350 + 220
- 225 + 220
p(5) = -5
Answer:
1/14x+6
Step-by-step explanation:
I have already learn that
