Check the picture below.
now, we have a triangle with all three sides, thus we can use Heron's Area Formula on the triangle.
![\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=10\\ b=26.695\\ c=22\\ s=29.3475 \end{cases} \\\\\\ A=\sqrt{29.3475(29.3475-10)(29.3475-26.695)(29.3475-22)} \\\\\\ A=\sqrt{29.3475(19.3475)(2.6525)(7.3475)}\implies A\approx \sqrt{11066.007} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill A\approx 105.195~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Ctextit%7BHeron%27s%20area%20formula%7D%20%5C%5C%5C%5C%20A%3D%5Csqrt%7Bs%28s-a%29%28s-b%29%28s-c%29%7D%5Cqquad%20%5Cbegin%7Bcases%7D%20s%3D%5Cfrac%7Ba%2Bb%2Bc%7D%7B2%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D10%5C%5C%20b%3D26.695%5C%5C%20c%3D22%5C%5C%20s%3D29.3475%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D%5Csqrt%7B29.3475%2829.3475-10%29%2829.3475-26.695%29%2829.3475-22%29%7D%20%5C%5C%5C%5C%5C%5C%20A%3D%5Csqrt%7B29.3475%2819.3475%29%282.6525%29%287.3475%29%7D%5Cimplies%20A%5Capprox%20%5Csqrt%7B11066.007%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20A%5Capprox%20105.195~%5Chfill)
Answer:
2.5 km/h
Step-by-step explanation:
Here we have a distance problem where we are given the distance the time time:
Rate = ?
Time = 2hrs
Distance = 5km
The simplest way to do this is to rearrange the formula to suit our need and plug in the numbers:

Now, plug in the numbers and solve:

Answer:
I think the answer is -32.
Step-by-step explanation:
I'm not sure.
-6.2 plus -4.8 is -11. add the 3.8 to get -7.2
-4.8 plus 3.8 is -1
-1 plus -6.2 is -7.2
-7.2=-7.2
Use substitution to find out
x=2
y=3
so
3=3(2)-4
3=6-4
3≠2
no, this point does NOT lie on the graph of y=3x-4