A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?
Answer:

Step-by-step explanation:
Given:
A stuntman jumping off a 20-m-high building is modeled by the equation
-----------(1)
A high-speed camera is ready to making film between 15 m and 10 m above the ground
when the stuntman is 15m above the ground.
height
Put height value in equation 1





We know that the time is always positive, therefore 
when the stuntman is 10m above the ground.
height
Put height value in equation 1







Therefore ,time interval of camera film him is 
Answer:
x = 1
Step-by-step explanation:
Solve for x over the real numbers:
-1 + 2 + 1/x + 1/x = 3
-1 + 2 + 1/x + 1/x = 1 + 2/x:
1 + 2/x = 3
Bring 1 + 2/x together using the common denominator x:
(x + 2)/x = 3
Multiply both sides by x:
x + 2 = 3 x
Subtract 3 x + 2 from both sides:
-2 x = -2
Divide both sides by -2:
Answer: x = 1
Answer:sorry this probably is t the most helpful but the closest i could get was 399 lbs. it’s is st$497.7 for one and $$497.8.
Step-by-step explanation:
Answer:
u = 1
Step-by-step explanation:
Step-by-step explanation:
3(-8m-1) + 5(m+8) (remove the parentheses)
-24m - 3 + 5m + 40 (collect like terms and calculate)
-19m + 37