Part A : y=$0.75m+$4.50
Part B : $8.25
![\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{64}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{0\qquad \textit{from the ground}}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Binitial%20velocity%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20~~~~~~%5Ctextit%7Bin%20feet%7D%20%5C%5C%5C%5C%20h%28t%29%20%3D%20-16t%5E2%2Bv_ot%2Bh_o%20%5Cend%7Barray%7D%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Cstackrel%7B64%7D%7B%5Ctextit%7Binitial%20velocity%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h_o%3D%5Cstackrel%7B0%5Cqquad%20%5Ctextit%7Bfrom%20the%20ground%7D%7D%7B%5Ctextit%7Binitial%20height%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h%3D%5Cstackrel%7B%7D%7B%5Ctextit%7Bheight%20of%20the%20object%20at%20%22t%22%20seconds%7D%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)
Check the picture below, it hits the ground at 0 feet, where it came from, the ground, and when it came back down.
Answer:
-2x +8
Step-by-step explanation:
The terms involving x are "like" terms, so their coefficients are added when the terms are combined.
= 32x -34x +8
= (32 -34)x +8 . . . . . x is factored out according to the distributive property
= -2x +8
<h3>Answer: Choice D</h3>
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Explanation:
The long way to do this is to multiply all the fractions out by hand, or use a calculator to make shorter work of this.
The shortest way is to simply count how many negative signs each expression has.
The rule is: if there are an even number of negative signs, then the product will be positive. Otherwise, the product is negative.
For choice A, we have 3 negative signs. The result (whatever number it is) is negative. Choice B is a similar story. Choice C is also negative because we have 1 negative sign. Choices A through C have an odd number of negative signs.
Only choice D has an even number of negative signs. The two negatives multiply to cancel to a positive. The negative is like undoing the positive. So two negatives just undo each other. This is why the multiplied version of choice D will be some positive number.
Or you can think of it as opposites. If you are looking up (positive direction) and say "do the opposite" then you must look down (negative direction). Then if you say "do the opposite", then you must look back up in the positive direction.
