Answer:
A) x ≤ -2 and 0 ≤ x ≤ 3
Step-by-step explanation:
g(x) is decreasing when g'(x) is negative.
Use second fundamental theorem of calculus to find g'(x).
g(x) = ∫₋₁ˣ (t³ − t² − 6t) / √(t² + 7) dt
g'(x) = (x³ − x² − 6x) / √(x² + 7) (1)
To find when g'(x) is negative, first find where it is 0.
0 = (x³ − x² − 6x) / √(x² + 7)
0 = x³ − x² − 6x
0 = x (x² − x − 6)
0 = x (x − 3) (x + 2)
x = -2, 0, or 3
Check the intervals before and after each zero.
x < -2, g'(x) < 0
-2 < x < 0, g'(x) > 0
0 < x < 3, g'(x) < 0
3 < x, g'(x) > 0
g(x) is decreasing on the intervals x ≤ -2 and 0 ≤ x ≤ 3.
The first one is 52 because if you toke 80 and subtract it by28 it will give you x witch is 52 s=52
each friend will get 1/3 a liter of soda. Just divide 2 by 6 and that's your answer.
Hopefully this helps! Let me know if you have any more questions.
Answer:
6
Step-by-step explanation:
3-2x = -1.5x
Add 2x to both sides:
3 = 0.5x
Divide both sides by 0.5
X = 3/0.5
X = 6
The answer is 6
Answer:
I am sorry I don't know how to do it