This is the concept of application of quadratic expressions. Given that the height of the ball is modeled by the equation;
h=-7.3t^2+8.25t+2.1+5
The time taken for the ball to hit the ground will be given as falls;
-7.3t^2+8.25t+7.1=0
to solve for t we use the quadratic formula;
t=[-b+/-sqrt(b^2-4ac)]/(2a)
a=-7.3, b=8.25, c=2.1
t=[-8.25+/-sqrt[8.25^2+4*7.3*7.1]/(-2*7.3)
t= -0.572
or
t=1.702
since there is not negative time we take the time taken for the ball to hit the ground will be: t=1.702 sec
Answer:
The enrollment after 5 years is 10,724
Step-by-step explanation:
Generally, we can have the depreciation formula written as follows;
A = P(1 - r)^t
A is the number of enrollment in after a certain number of years t
P is the initial population which is 13,500
r is the rate of depreciation which is 4.5% = 4.5/100 = 0.045
t = 5 years
Substituting these values, we have it that;
A = 13,500(1-0.045)^5
A = 10,723.84
Answer:
175
Step-by-step explanation:
140
140/4=35
140+35=175
