Thank you for posting your question here. I hope the answer below will help.
Vo=110 feet per second
<span>ho=2 feet </span>
<span>So, h(t) = -16t^2 +110t +2 </span>
<span>Take the derivative: h'(t) = 110 -32t </span>
<span>The maximum height will be at the inflection when the derivative crosses the x-axis aka when h'(t)=0. </span>
<span>So, set h'(t)=0 and solve for t: </span>
<span>0 = 110 -32t </span>
<span>-110 = -32t </span>
<span>t=3.4375 </span>
<span>t=3.44 seconds </span>
Answer:
interquartile range
Step-by-step explanation:
Answer:
I believe it is C but then again I'm not sure good luck though
<span>You only need an estimate? Great. Think of the necessary material as 6 yards, so ten costumes would take 60 yards. Then think of the fabric on hand as 240 yards, so taking 60 yards away from that would leave [an estimated] 180 yards of fabric.
For an equation, use:
F" = F' - 5.8 C
Where F' is the initial fabric stock, F" is the remaining fabric, and C is the count of costumes.
In exact numbers, the formula works out to:
F" = 238.2 - (5.8 * 10)
F" = 180.2
The estimate was very close.
Good luck.</span>
