Answer:
b+6
Problem:
If the average of b and c is 8, and d=3b-4, what is the average of c and d in terms of b?
Step-by-step explanation:
We are given (b+c)/2=8 and d=3b-4.
We are asked to find (c+d)/2 in terms of variable, b.
We need to first solve (b+c)/2=8 for c.
Multiply both sides by 2: b+c=16.
Subtract b on both sides: c=16-b
Now let's plug in c=16-b and d=3b-4 into (c+d)/2:
([16-b]+[3b-4])/2
Combine like terms:
(12+2b)/2
Divide top and bottom by 2:
(6+1b)/1
Multiplicative identity property applied:
(6+b)/1
Anything divided by 1 is that anything:
(6+b)
6+b
b+6
The relationship is proportional because it is in standard form
Answer:
5.12 seconds to the nearest hundredth.
Step-by-step explanation:
When the ball hits the ground h = 0 so
-16t^2 + 81t + 5 = 0
t = [ -81 +/- sqrt ((81^2 - 4 * -16 * 5)]/ (2*-16)
= (-81 +/- sqrt 6881) / -32
= 5.12 , -0.6
So the answer is 5.123 seconds.
