I think it’s linear. no numbers repeat. i could be wrong but LOL
Check the picture below, so it reaches the maximum height at the vertex, let's check where that is
![h(t)=64t-16t^2+0 \\\\[-0.35em] ~\dotfill\\\\ \textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+64}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 64}{2(-16)}~~~~ ,~~~~ 0-\cfrac{ (64)^2}{4(-16)}\right)\implies \stackrel{maximum~height}{(2~~,~~\stackrel{\downarrow }{64})}](https://tex.z-dn.net/?f=h%28t%29%3D64t-16t%5E2%2B0%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ctextit%7Bvertex%20of%20a%20vertical%20parabola%2C%20using%20coefficients%7D%20%5C%5C%5C%5C%20h%28t%29%3D%5Cstackrel%7B%5Cstackrel%7Ba%7D%7B%5Cdownarrow%20%7D%7D%7B-16%7Dt%5E2%5Cstackrel%7B%5Cstackrel%7Bb%7D%7B%5Cdownarrow%20%7D%7D%7B%2B64%7Dt%5Cstackrel%7B%5Cstackrel%7Bc%7D%7B%5Cdownarrow%20%7D%7D%7B%2B0%7D%20%5Cqquad%20%5Cqquad%20%5Cleft%28-%5Ccfrac%7B%20b%7D%7B2%20a%7D~~~~%20%2C~~~~%20c-%5Ccfrac%7B%20b%5E2%7D%7B4%20a%7D%5Cright%29%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28-%5Ccfrac%7B%2064%7D%7B2%28-16%29%7D~~~~%20%2C~~~~%200-%5Ccfrac%7B%20%2864%29%5E2%7D%7B4%28-16%29%7D%5Cright%29%5Cimplies%20%5Cstackrel%7Bmaximum~height%7D%7B%282~~%2C~~%5Cstackrel%7B%5Cdownarrow%20%7D%7B64%7D%29%7D)
Answer:
31x+41y+6
Step-by-step explanation:
Distribute
27y+15x+14y+16x+6
Combine like terms
41y+31x+6
<span>40 units^2
The figure has the vertexes (2,7), (3,3), (3,-3), (-5,-5), (-6,-1), and (2,1). If you draw a line from (2,1) to (3,-3) you'll divide the figure into the rectangle and parallelogram. The total area will be the sum of the area of the parallelogram and the rectangle. The area of the parallelogram will be base times height, where the height is the distance between the line (2,7) to (2,1) and line (3,3) to (3,-3) which is 1 and the base is the length of the line (2,7) to (2,1) which is 6. So the parallelogram has an area of 6.
The length and width of the rectangle can be calculated from the Pythagorean theorem, so:
L = sqrt((2- -6)^2 + (1- -1)^2) = sqrt(8^2 + 2^2) = sqrt(64+4) = sqrt(68) = 2*sqrt(17)
W = sqrt((-6 - -5)^2 + (-1 - -5)^2) = sqrt(-1^2 + 4^2) = sqrt(1 + 16) = sqrt(17)
So the area of the rectangle is 2*sqrt(17)*sqrt(17) = 2*17 = 34.
And the total area is 34 + 6 = 40.</span>
