A cube because it’s already square n can be cut in square cross sections
The points of intersection are at (3, 6) and (-1, -2).
Since both of these equations have y isolated, we can set them equal to each other:
2x=x²-3
We want all of the variables on one side, so subtract 2x:
2x-2x = x²-3-2x
0=x²-3-2x
Write the quadratic in standard form:
0=x²-2x-3
This is easily factorable, as there are factors of -3 that will sum to -2. -3(1)=-3 and -3+1=-2:
0=(x-3)(x+1)
Using the zero product property we know that either x-3=0 or x+1=0; therefore x=3 or x=-1.
Substituting this into the first equation (it is simpler):
y=2(3) = 6
y=2(-1)=-2
Therefore the coordinates are (3, 6) and (-1, -2).
<span>quadrilateral, sum of interior angles = 360
</span><span>ratio 1 : 2 : 5 : 4
so 1+2+5+4 = 12
360 / 12 = 30
1 x 30 = 30
2 x 30 = 60
5 x 30 = 150
4 x 30 =120
answer
</span><span>
</span><span>The four angles of the quadrilateral are 30,60,150 and 120</span>
15x-20<130
15x<130+20
15x<150
X=150/15
x<10