To solve for this, we need to find for the value of x
when the 1st derivative of the equation is equal to zero (or at the
extrema point).
So what we have to do first is to derive the given
equation:
f (x) = x^2 + 4 x – 31
Taking the first derivative f’ (x):
f’ (x) = 2 x + 4
Setting f’ (x) = 0 and find for x:
2 x + 4 = 0
x = - 2
Therefore the value of a is:
a = f (-2)
a = (-2)^2 + 4 (-2) – 31
a = 4 – 8 – 31
a = - 35
Answer: 5x+7
Step-by-step explanation:
2x+3x+7
5x+7
Let's write some equations.
Mingwei's distance from Town A after

hours from 8:00 is

.
Ali's distance from Town B after

hours is

, since he doesn't start walking for 40 minutes.
When Mingwei's distance is twice Ali's, they've met up (since their distance from Town A is twice their distance from Town B).
So, this gives

, so

, so

, so the time is 10:40.
After

hours, Mingwei has traveled

kilometers while Ali has traveled sixty, so the distance between the towns is

kilometers.
Answer:
G(12) = 15
Step-by-step explanation:
When doing g(x) equations, all you do is substitute the value inside the parentheses to x in the equation.
So you get 3/4(12) + 6.
Then simplify to get g(12) = 15