
Both the numerator and denominator are continuous at

, which means the quotient rule for limits applies:

Perhaps you meant to write that

instead? In that case, you would have
Answer: 3.675 seconds
Step-by-step explanation:
Hi, when the object hits the ground, h=0:
h=−16t^2+48.6t+37.5
0=−16t^2+48.6t+37.5
We have to apply the quadratic formula:
For: ax2+ bx + c
x =[ -b ± √b²-4ac] /2a
Replacing with the values given:
a=-16 ; b=48.6; c=37.5
x =[ -(48.6) ± √(-48.6)²-4(-16)37.5] /2(-16)
x = [ -48.6 ± √ 4,761.96] /-32
x = [ -48.6 ± 69] /-32
Positive:
x = [ -48.6 + 69] /-32 = -0.6375
Negative:
x = [ -48.6 - 69] /-32 = 3.675 seconds (seconds can't be negative)
Feel free to ask for more if needed or if you did not understand something.
If r(x)= 11x, and c(x)=6x +20, then just put that into the equation.
So instead of p(x) = r(x) - c(x)
I would be p(x) = 11x - 6x + 20
Now solve.
p(x) = 5x + 20 is your final answer, so A.
Answer:
A.
B. b > 11
C.
D.
Step-by-step explanation:
<u>Given the following algebraic expression;</u>
A.
We would simplify the equation by multiplying all through by 4;
Divide both sides by -3;
B. 5b - 28 > 27
Rearranging the equation, we have;
5b > 27 + 28
5b > 55
Divide both sides by 5
b > 11
C.
Divide both sides by 13
D.
Collecting like terms, we have;
Divide both sides by -1
Step-by-step explanation:
(64y100)1/2
1/2×64y 1/2×100
32y50