I would say gaudy , it’s pretty descriptive
For this case what we must do is find a quadratic function that is already factored.
This is because in the factored quadratic equations, it is easier to observe the zeros of the function.
In this case, the zeros of the function represent the time at which the company did not make any profit.
We have the following equation:
p (t) = 40 (t - 3) (t + 2) (t - 5) (t + 3)
We observed that there was no gain in:
t = 3
t = 5
The other roots are discarded because they are negative
Answer:
a.p (t) = 40 (t - 3) (t + 2) (t - 5) (t + 3)
H = 2f / (m+1)
[multiply by (m+1)]
h(m+1) = 2f
[divide by 2]
f = (h (m + 1)) / 2
3b / (b+2) = 12 / (b+2)
[multiply by (b+2)]
3b = 12
[divide by 3]
b = 4
3 / (6x + 1) / 2 = 8 / (x + 4) / 3
[multiply both denominators to mike one denominator]
3 / 8(6x+1) = 8 / 3(x+4)
[expand brackets]
3 / (48x + 8) = 8 / (3x + 12)
[multiply by (48x + 8)]
3 = 8(48x + 8) / (3x+12)
[multiply by (3x+12)]
3(3x +12) = 4(48x + 8)
[simplify]
9x + 36 = 192x + 32
173x = 4
x = 4 / 173
