Answer:
Perimeter of triangle J= 18.0 units (3 s.f.)
Perimeter of triangle K= 19.0 units (3 s.f.)
Step-by-step explanation:
The length of each side of the triangle can be found using the distance formula below:

Please see the attached pictures for full solution.
Answer:
3 + 3p
Step-by-step explanation:
since there's no equal sign or no value that it shows all of that added together is equivalent to it's not really equal to anything. However if you're asking what that equation would look like if you simplified it down , then it would be 3 + 3p because you combine the like terms
Check the picture below.
based on the equation, if we set y = 0, we'd end up with 0 = 0.5(x-3)(x-k).
and that will give us two x-intercepts, at x = 3 and x = k.
since the triangle is made by the x-intercepts and y-intercepts, then the parabola most likely has another x-intercept on the negative side of the x-axis, as you see in the picture, so chances are "k" is a negative value.
now, notice the picture, those intercepts make a triangle with a base = 3 + k, and height = y, where "y" is on the negative side.
let's find the y-intercept by setting x = 0 now,
![\bf y=0.5(x-3)(x+k)\implies y=\cfrac{1}{2}(x-3)(x+k)\implies \stackrel{\textit{setting x = 0}}{y=\cfrac{1}{2}(0-3)(0+k)} \\\\\\ y=\cfrac{1}{2}(-3)(k)\implies \boxed{y=-\cfrac{3k}{2}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of a triangle}}{A=\cfrac{1}{2}bh}~~ \begin{cases} b=3+k\\ h=y\\ \quad -\frac{3k}{2}\\ A=1.5\\ \qquad \frac{3}{2} \end{cases}\implies \cfrac{3}{2}=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)](https://tex.z-dn.net/?f=%5Cbf%20y%3D0.5%28x-3%29%28x%2Bk%29%5Cimplies%20y%3D%5Ccfrac%7B1%7D%7B2%7D%28x-3%29%28x%2Bk%29%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bsetting%20x%20%3D%200%7D%7D%7By%3D%5Ccfrac%7B1%7D%7B2%7D%280-3%29%280%2Bk%29%7D%20%5C%5C%5C%5C%5C%5C%20y%3D%5Ccfrac%7B1%7D%7B2%7D%28-3%29%28k%29%5Cimplies%20%5Cboxed%7By%3D-%5Ccfrac%7B3k%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20a%20triangle%7D%7D%7BA%3D%5Ccfrac%7B1%7D%7B2%7Dbh%7D~~%20%5Cbegin%7Bcases%7D%20b%3D3%2Bk%5C%5C%20h%3Dy%5C%5C%20%5Cquad%20-%5Cfrac%7B3k%7D%7B2%7D%5C%5C%20A%3D1.5%5C%5C%20%5Cqquad%20%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B3%7D%7B2%7D%3D%5Ccfrac%7B1%7D%7B2%7D%283%2Bk%29%5Cleft%28-%5Ccfrac%7B3k%7D%7B2%7D%20%5Cright%29)

now, we can plug those values on A = (1/2)bh,
![\bf \stackrel{\textit{using k = -2}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-2)\left(-\cfrac{3(-2)}{2} \right)\implies A=\cfrac{1}{2}(1)(3) \\\\\\ A=\cfrac{3}{2}\implies A=1.5 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{using k = -1}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-1)\left(-\cfrac{3(-1)}{2} \right) \\\\\\ A=\cfrac{1}{2}(2)\left( \cfrac{3}{2} \right)\implies A=\cfrac{3}{2}\implies A=1.5](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Busing%20k%20%3D%20-2%7D%7D%7BA%3D%5Ccfrac%7B1%7D%7B2%7D%283%2Bk%29%5Cleft%28-%5Ccfrac%7B3k%7D%7B2%7D%20%5Cright%29%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%283-2%29%5Cleft%28-%5Ccfrac%7B3%28-2%29%7D%7B2%7D%20%5Cright%29%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%281%29%283%29%20%5C%5C%5C%5C%5C%5C%20A%3D%5Ccfrac%7B3%7D%7B2%7D%5Cimplies%20A%3D1.5%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Busing%20k%20%3D%20-1%7D%7D%7BA%3D%5Ccfrac%7B1%7D%7B2%7D%283%2Bk%29%5Cleft%28-%5Ccfrac%7B3k%7D%7B2%7D%20%5Cright%29%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%283-1%29%5Cleft%28-%5Ccfrac%7B3%28-1%29%7D%7B2%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B2%7D%282%29%5Cleft%28%20%5Ccfrac%7B3%7D%7B2%7D%20%5Cright%29%5Cimplies%20A%3D%5Ccfrac%7B3%7D%7B2%7D%5Cimplies%20A%3D1.5)
<span>Answer:
8.517 in word form is eight and five hundred seventeen thousandths or eight</span>
Answer:

Step-by-step explanation:
Move all terms to the left side and set equal to zero. Then set each factor equal to zero.