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:
x=0.y=3
Step-by-step explanation:
5x-y=-3
y=5x+3
sub y=5x+3 into 6x+y=3,
6x+5x+3=3
11x=0
x=0
6(0)+y=3
y=3
Answer:
$27.83
Step-by step explanation:
No, it is not a valid inference because her classmates do not make up a random sample of the students in the school.
Answer:
Step-by-step explanation:
Given that
AB n GF = O
E ∈ ∠GOB
H ∈ ∠AOF
The we can say, using similar angles or triangles, that
BF = AG
EG = AH
BG = FA