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
2/3 + n = 2n + 3 <== ur equation
2/3 - 3 = 2n - n
2/3 - 9/3 = n
-7/3 = n <== ur solution
or
2/3 + n = 2n + 3...multiply everything by common denominator of 3
2 + 3n = 6n + 9
2 - 9 = 6n - 3n
-7 = 3n
-7/3 = n
Answer:
H = 11m
Step-by-step explanation:
As we know, volume = length * breadth/width * height
Given,
length = 6m
Breadth/width = 3 1/2 or 3.5
Height = ?
So, l * b * h = v
=> 6m * 3.5m * h = 231m
=> 21m * h = 231m
=> h = 231m /21m = 11m
Therefore, height = 11m / 11metres
Hope this helps
Answer:
-1.28
Step-by-step explanation:
You would put these two equations into a graphing ccalculator:
y = -4x - 1
and
y= 5^x + 4
Wherver they intersect is your answer.
These two happen to intersect at (-1.28,4.13)
Since the equation has x-values, you would give the first number (x-coordinate) as your answer.
