Check the picture below.
so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?
now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2 = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.
![\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Binitial%20velocity%20in%20feet%7D%20%5C%5C%5C%5C%20h%28x%29%20%3D%20-16x%5E2%2Bv_ox%2Bh_o%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Ctextit%7Binitial%20velocity%7D%260%5C%5C%20%5Cqquad%20%5Ctextit%7Bof%20the%20object%7D%5C%5C%20h_o%3D%5Ctextit%7Binitial%20height%7D%26%5C%5C%20%5Cqquad%20%5Ctextit%7Bof%20the%20object%7D%5C%5C%20h%3D%5Ctextit%7Bobject%27s%20height%7D%260%5C%5C%20%5Cqquad%20%5Ctextit%7Bat%20%22t%22%20seconds%7D%5C%5C%20x%3D%5Ctextit%7Bseconds%7D%261%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%200%3D-16%281%29%5E2%2B0x%2Bh_o%5Cimplies%200%3D-16%2Bh_o%5Cimplies%2016%3Dh_o%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20h%28x%29%20%3D%20-16x%5E2%2B16~%5Chfill)
quick info:
in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².
Answer:
I will answer this for you and the links are so annyoing fr
Step-by-step explanation:
give me a couple of minutes to figure it out then I will post the answer!
Answer:
Step-by-step explanation:
Add all them up to get 2500
Then you divide 900 by 2500 to get 0.36
Finally you multiple 0.36 by 100 to get 36%
Answer:
No
Step-by-step explanation:
Reduce each fraction to find if they are equal to the same ratio.
2/4 reduces to 1/2
14/20 reduces to 7/10
They do not form a proportion.
Answer:
Corresponding angle theorem, vertical angle theorem, and the transitive property of congruence.
Step-by-step explanation:
Considering a set of parallel lines cut by a transversal. (Refer the attached image).
Now that lines J and K are parallel, then by "corresponding angle theorem"
By "vertical angle theorem"
Using, "transitive property of congruence"
And that is our required proof. In this whole proof we have used "corresponding angle theorem", "vertical angle theorem", and the "transitive property of congruence".
