Question 3, @satellite73 this one has 4 parts so it might be a little longer, if you need to go or want to finish tomorrow just let me know i wont mind (: thanks for all your help today btw
lets take a look
first part is real easy replace \(s\) by \(96\) and \(v\) by \(80\) and get \[H(t)=-16t^2+80t+96\]
second part is also easy the vertex is where the first coordinate is \(-\frac{b}{2a}\) which in our case is \(-\frac{80}{2\times (-16)}=\frac{5}{2}\)
or if you prefer decimals \(2.5\)
then the maximum height will be \[H(2.5)=-16\times (2.5)^2+80\times (2.5)+96\] which you compute using a calculator
I am sorry had to go to the bathroom give me a second to catch up (:
take your time
Okay, read it all and wow yeah part A was easy lol how did i not see that
part B i hope you get, if didn't leave out any steps
yes i understood that too dont worry (:
plus i used a calculator, so i know the answer is right
C want you to do \[-16t^2+80t+96=31+32.2t\]
it says use a table, i would use this http://www.wolframalpha.com/input/?i=-16t^2%2B80t%2B96%3D31%2B32.2t
ignore the negative solution because time does not go backwards round and say it is at about \(t=4\)
or you could just plug in 4 and say it is close
okay yeah thanks, anyway if you were to round the answer it would be a 4 right? (:
yes
and finally D is also easy it goes up until time is \(2.5\) because that was the first coordinate of the vertex since \(4>2.5\) it is on the way DOWN
Okay, thank you soooo much (: <3
yw you owe me a beer good luck!
when im old enough to get one then you will get your beer :P
oops i will settle for a root beer
lol xD well anyway thanks a bunch this really means a lot considering i have eoc exam on monday and now i can take it since you helped me with this, all i need to do now is go back and write down the answers, so yeah thanks xoxo
yw good luck don't study too late
thank you and haha will do (:
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