Reposts include the address of the original question,. NOT the question itself.
First you need to figure out the gradient of the first line. Can you do that?
the gradient of the second line is the negative reciprocal of the first, that is m
Sub the pint in to find b
Add m and b together.
There is no one last step.
It depends on what your fist steps are.
But only one of those is really possible i guess.
NOTE dividing by one does nothing.
You can do at least some of this yourself!
Come back with your attempt
\(x^{log_y{x}}=2 \qquad \qquad \qquad y^{log_x{y}}=2^4\\ x^{\frac{log x}{log y}}=2 \qquad \qquad \qquad y^{\frac{log y}{log x}}=2^4\\ log\;{x^{\frac{log x}{log y}}}=log\;{2} \qquad \quad \log{y^{\frac{log y}{log x}}}=\log{2^4}\\ {\frac{log x}{log y}}*log\;x=log\;{2} \qquad \quad {\frac{log y}{log x}}*\log\;y=4\log\;2\\ \frac{({log x})^2}{log y}=log\;{2} \qquad \quad \qquad \frac{(log y)^2}{4log x}=\log\;2\\~\\ so\\ 4{{(log x})^3}=(log y)^3\\ \sqrt[3]{4}=\frac{logy}{logx}\\ log_xy = \sqrt[3]{4}\)
Oh I have to solve for x...
\(\frac{logx}{log y}=\frac{1}{\sqrt[3]{4}}\\ log_y x=4^{(-1/3)}\\ y^{log_y x}=y^{4^{(-1/3)}}\\ x=y^{4^{(-1/3)}}\\ \)
Thanks for your input grs75.
If you did it yourself then you have more skill and knowledge than you are giving yourself credit for.
Thanks Hectictar
Sorry guest, I did not realize that you were asking a question.
The angle sum of a triangle is 180 degrees.
(3 sides -2) *180 = 180
Let the number of sides of a polygon be n
The angle sum will be (n-2)*180
So what is the angle sum of a six-sided polygon (hexagon) ?
Can you take it from there?
the degree of f(x) is 4
the degree of g(x) is also 4
so the degree of the product will be 8
the little one is 4' by 9'
the side lengths of the big one are 5 times that so 20 by 45 feet
You can that it from there.
+118725 
