If x = -7, then x3 = -343 so that 2x3 = -686 y = 2 then 15 y = 30 So 2x3 + 15y = -686 + 30 = -656
y = -9
x+y
It can be simplified in the form of: 2x3+9x2-7x-34 = 0
x = -7, y = 2 ⇒ 2x3 + 15y = 2 x (-7)3 + 15 x 2 = 2 x (-343) + 15 x 2 = -656
y=xsquared-4x+2
Quadratic - the degree is two.
If x = -7, then x3 = -343 so that 2x3 = -686 y = 2 then 15 y = 30 So 2x3 + 15y = -686 + 30 = -656
y = -9
x+y
It can be simplified in the form of: 2x3+9x2-7x-34 = 0
x = -7, y = 2 ⇒ 2x3 + 15y = 2 x (-7)3 + 15 x 2 = 2 x (-343) + 15 x 2 = -656
6
No real solutions. But if you want to find the imaginaries: x=.5(25-i(337)^.5) y=.5(25-i(337)^.5) x^.5= the square root of x
No... whenever you have an exponent on your varaible your graph is not liner. this is a parabola.
y = x2 + 2mx + n complete the square y + m2 = x2 + 2mx + m2 + n = (x + m)2 + n (x + m)2 = y + m2 - n x + m = √(y + m2 - n) x = -m + √(y + m2 - n)
p(x) = x4 + 2x3 + x2 + 8x - 12When you go to graph this function, you make 'y' = p(x) .At the y-intercept, 'x' is zero.There, y = -12 .