<h3>Answer: The lines are perpendicular</h3>
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Work Shown:
Use the slope formula to find the slope of the line through the two given points.
m = (y2 - y1)/(x2 - x1)
m = (-4 - (-5))/(6 - (-7))
m = (-4 + 5)/(6 + 7)
m = 1/13
The slope of the line through the two given points is 1/13.
This is not equal to -13, which was the slope of the first line, so the two lines are not parallel.
However, the two lines are perpendicular because multiplying the two slope values leads to -1
(slope1)*(slope2) = (-13)*(1/13) = -1
note: It might help to think of -13 as -13/1 to help multiply the fractions.
<u>Answer</u>
√7 × √5
√3 × √2
<u>Explanation</u>
√9*√16 = 3 × 4
= 12
= 12/1 ⇒ It is a rational answer.
√7*√5 = √(7×5)
= √35
= 5.916079783.... ⇒ Irrational
√3*√2 = √(3×2)
= √6
= 2.449489743... ⇒ Irrational
4√2*√2 = 4 × √(2×2)
= 4 × √4
= 4 × 2
= 8
= 8/1 ⇒ It is rational
The expressions that would give irrational answers are: √7*√5 and√3*√2
The number 2.85 can be writen using the fraction 285/100 which is equal to 57/20 when reduced to lowest terms.
It is also equal to 2 17/20 when writen as a mixed number.
You can use the following approximate value(s) for this number:57/20 =~ 2 6/7 (if you admit a error of 0.250627%)2.85 =~ 2 5/6 (if you admit a error of -0.584795%)2.85 =~ 3 (if you admit a error of 5.263158%)
![\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.
