Answer:
The answer is 67 degrees.
Step-by-step explanation:
The angle x is supplementary with 23 degrees because that angle is an alternate interior. Recall that is within a right angles as well so
90 - 23 = 67
Answer:
x=5
y=-2
Step-by-step explanation:
I just took this!
Hope I helped
1. √48 → √16 * √3 = 4√3
3. √12 → √4 * √3 = 2√3
7. √125 → √25 * √5 = 5√5
8. √20 → √4 * √5 = 2√5
9. √18 → √9 * √2 = 3√2
10. √32 → √16 * √2 = 4√2
11. √50 → √25 * √2 = 5√2
12. √27 → √9 * √3 = 3√3
![\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{}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ h=-16t^2+\stackrel{\stackrel{v_o}{\downarrow }}{65}t](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%7B%7D%7B%5Ctextit%7Binitial%20velocity%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h_o%3D%5Cstackrel%7B%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~%5Cdotfill%5C%5C%5C%5C%20h%3D-16t%5E2%2B%5Cstackrel%7B%5Cstackrel%7Bv_o%7D%7B%5Cdownarrow%20%7D%7D%7B65%7Dt)
now, take a look at the picture below, so for 2) and 3) is the vertex of this quadratic equation, 2) is the y-coordinate and 3) the x-coordinate.
Answer:
(-a,b)
Step-by-step explanation:
You use midpoint formula which is (x1+x2/2),(y1+y2/2)
