Answer:
1 ≥ t ≤ 3
Step-by-step explanation:
Given
h(t) = -16t² + 64t + 4
Required
Determine the interval which the bar is at a height greater than or equal to 52ft
This implies that
h(t) ≥ 52
Substitute -16t² + 64t + 4 for h(t)
-16t² + 64t + 4 ≥ 52
Collect like terms
-16t² + 64t + 4 - 52 ≥ 0
-16t² + 64t - 48 ≥ 0
Divide through by 16
-t² + 4t - 3 ≥ 0
Multiply through by -1
t² - 4t + 3 ≤ 0
t² - 3t - t + 3 ≤ 0
t(t - 3) -1(t - 3) ≤ 0
(t - 1)(t - 3) ≤ 0
t - 1 ≤ 0 or t - 3 ≤ 0
t ≤ 1 or t ≤ 3
Rewrite as:
1 ≥ t or t ≤ 3
Combine inequality
1 ≥ t ≤ 3
Answer:
Step-by-step explanation:
6/x^2 + 2 x = 7
2 x^3 - 7 x^2 = -6 (for x!=0)
(2 (x^3 - 3 x^2 + 3))/x^2 = 1
Answer:
yes
Step-by-step explanation:
The least common multiple is : 18
It is the lowest amount that can be reached.
Answer:
a number line that goes from 0 to 16. the whiskers range from 1 to 12, and the box ranges from 3 to 9. a line divides the box at 7.
