A polygon with sides: 154-x, 50-2x, x+120, and x^2 + 7x.
Solve for x and give each angle, if necessary.
Actually it is not a polygon since it has a curve in it.
Doing a sketch, you should easily find the angle and intersect of the lines.
$\displaystyle *50-2x = x +120 $
$\displaystyle -70 = 3x $
$\displaystyle -\frac{70}{3} = x $
This is an example, it is NOT one of the vertexes.
For the parabola, I would use the derivative to find the slope at the intersect.
If needed ask for further explanations.
Hello magentarita,
If I were a betting person, I would bet that you meant to say angles instead of sides, because:
$\displaystyle 154-x + 50-2x + x+120 + x^2+7x = 360$
since the sum of the angles of a quadrilateral = 360 degrees.
Now, this all boils down to:
$\displaystyle x^2+5x-36=0$
$\displaystyle (x+9)(x-4)=0$
$\displaystyle x=-9 \ \ or \ \ x=4$
Using $\displaystyle x=-9$
$\displaystyle 154-(-9)=163$
$\displaystyle 50-2(-9)=68$
$\displaystyle (-9)+120=111$
$\displaystyle (-9)^2+7(-9)=18$
Now, you try using $\displaystyle x=4$ and see what happens.
I interpreted this to mean that these were the lengths of the sides, not equations of the lines. But I suspect that masters is right and they are really supposed to be measures of the angles so that they add to 360.
$\displaystyle *50-2x = x +120 $
$\displaystyle -70 = 3x $
$\displaystyle -\frac{70}{3} = x $
This is an example, it is NOT one of the vertexes.
For the parabola, I would use the derivative to find the slope at the intersect.
If needed ask for further explanations.