GDP was a linear function of time (year) during the period of 1985-99.
The slope is positive.
Equation is: GDP(y)=(y-1985)*0.032*577+577=577*[(y-1985)*0.032+1] check: if y=1985, then GDP=577 .
So, we need to solve for "y" this equation: GDP=4000=577*[(y-1985)*0.032+1]then, y=[4000/577-1]/0.032+1985 = 2170.4
we solveGDP=2200=577*[(y-1985)*0.032+1] [4 is replaced by 2.2] y=[2200/577-1]/0.032+1985=2072.9 [almost 2073]
Answer:
can u attach a question its not there
Answer:
No solutions.
Step-by-step explanation:
If you normalize the second one by dividing by 2, you get:
y = 5/2x+4.
Now you see that both have the same slope (5/2). They are parallel lines that never intersect.
It's about 54.3 (repeating)
vi is going in the positive direction (up). (That's my choice). a (acceleration) is going in the minus direction (down). The directions could be reversed.
Givens
vi = 160 ft/s
vf = 0 (the rocket stops at the maximum height.)
a = - 9.81 m/s
t = ????
Remark
YOu have 4 parameters between the givens and what you want to solve. Only 1 equation will relate those 4. Always always list your givens with these problems so you can pick the right equation.
Equation
a = (vf - vi)/t
Solve
- 32 = (0 - 160)/t Multiply both sides by t
-32 * t = - 160 Divide by -32
t = - 160/-32
t = 5
You will also need to solve for the height to answer part B
t = 5
vi = 160 m/s
a = - 32
d = ???
d = vi*t + 1/2 a t^2
d = 160*5 + 1/2 * - 32 * 5^2
d = 800 - 400
d = 400 feet
Part B
You are at the maximum height. vi is 0 this time because you are starting to descend.
vi = 0
a = 32 m/s^2
d = 400 feet
t = ??
formula
d = vi*t + 1/2 a t^2
400 = 0 + 1/2 * 32 * t^2
400 = 16 * t^2
400/16 = t^2
t^2 = 25
t = 5 sec
The free fall takes the same amount of time to come down as it did to go up. Sort of an amazing result.
