Answer:
P = (13h+k+6m) cm
Step-by-step explanation:
Given that,
The side lengths of a triangle are :
(4h + 2k) cm, (9h + 4m) cm, and (2m - k) cm
We need to find the perimeter of the triangle.
We know that,
Perimeter = sum of all sides
So,
P = (4h + 2k)+(9h + 4m)+(2m - k)
= (4h+9h)+(2k-k)+(4m+2m)
= 13h+k+6m
Hence, the perimeter of the triangle is (13h+k+6m) cm.
Step-by-step explanation:
(2ab-7a-4) - (7ab-4)
= 2ab - 7a - 4 - 7ab +4
(-) × (-) = (+)
(-) × (-) = (+)-(-4)= +4
then 4 will be cancelled
= 2ab - 7ab - 7a
= -5ab -7a
Answer: I think so
Step by step:
![\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.
![\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+65}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left( -\cfrac{65}{2(-16)}~~,~~0-\cfrac{65^2}{4(-16)} \right) \implies \left( \cfrac{65}{32}~,~0- \cfrac{4225}{-64}\right)](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvertex%20of%20a%20vertical%20parabola%2C%20using%20coefficients%7D%20%5C%5C%5C%5C%20h%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%2B65%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%20-%5Ccfrac%7B65%7D%7B2%28-16%29%7D~~%2C~~0-%5Ccfrac%7B65%5E2%7D%7B4%28-16%29%7D%20%5Cright%29%20%5Cimplies%20%5Cleft%28%20%5Ccfrac%7B65%7D%7B32%7D~%2C~0-%20%5Ccfrac%7B4225%7D%7B-64%7D%5Cright%29)
![\bf \left( \cfrac{65}{32}~,~0+ \cfrac{4225}{64}\right)\implies \left( \stackrel{seconds}{2\frac{1}{32}}~~,~~ \stackrel{feet~hight}{66\frac{1}{64}}\right)](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%28%20%5Ccfrac%7B65%7D%7B32%7D~%2C~0%2B%20%5Ccfrac%7B4225%7D%7B64%7D%5Cright%29%5Cimplies%20%5Cleft%28%20%5Cstackrel%7Bseconds%7D%7B2%5Cfrac%7B1%7D%7B32%7D%7D~~%2C~~%20%5Cstackrel%7Bfeet~hight%7D%7B66%5Cfrac%7B1%7D%7B64%7D%7D%5Cright%29)