![\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.
0 remaining 9 i’m just typing more bc it says it’s can’t be under 20 words
-2x+4y=4 Equation 1
x-2y=6 Equation 2
Solving by substitution method.
Isolate x from equation 2.
x=2y+6
Substitute value of x in equation 1
-2(2y+6)+4y=4
-4y-12+4y=4
-12=4
-16=0
Which is false.
Answer: No solution
Since there are two black queens out of 52 cards, there is a 2/52 chance of drawing a black queen first. This is equivalent to a 1/26 chance.
Now that we have removed a black queen, there are 51 cards left in the deck. 26 of them are red because we only took away a black card. This means that there is a 26/51 of drawing a red card next.
In order to find the probability of both of these happening, we multiply the two together. 1/26 * 26/51 = 26/1326. This reduces to 1/51. So, there is a 1/51 chance of drawing a black queen, then a red card.
