y = 16 - x
Add 'x' to each side:
x + y = 16
Subtract 'y' from each side:
x = 16 - y
Wiki User
∙ 14y agoY=(x+8)^2-5
.25
No. A proportional relationship between "y" and "x" must be of the form:y = kx where "k" can be any constant. Thus, y = 16x would work perfectly. However, the additional "+4" makes it impossible to convert it to this form.
Ok, in order for anyone to give you the common factor of this, we need the format. is this -16x^2? If so, then the common factor would be x. x is the only thing that -16x^2, 1000x, and x have in common. If this were -16x^2 + 1000x + x =y then it would simplify to x(-16x+1001)=y. Ok, in order for anyone to give you the common factor of this, we need the format. is this -16x^2? If so, then the common factor would be x. x is the only thing that -16x^2, 1000x, and x have in common. If this were -16x^2 + 1000x + x =y then it would simplify to x(-16x+1001)=y.
16 + x = y
If: 16x-15-6x = 13 Then: 10x = 28 And: x = 2.8.
x=2.52
-29
7.5625
If your function is 4x^2+16x+16=0 then x=-2, if it is 4x^2-16x+16=0, then x=2
Let one of the number be x, and the other number be y (from the problem we see that x and y are positive numbers).Let x < y.So we have:x + y = 935y/x = 16y = 16xx + y = 935 (substitute 16x for y)x + 16x = 93517x = 935 (divide by 17 to both sides)x = 55y = 16x = 16(55) = 880Thus, the numbers are 55 and 880.
y = 4x² + 16x ; Take the derivative: y' = 8x + 16. Set equal to zero, and x = -2. Substitute into the original: y = 4(-2)² + 16*(-2) = -16, so the vertex is (-2,16). Or, factor to y = 4x(x + 4). The zeros of this are x = 0 and x = -4. Since a parabola is symmetric, the vertex is halfway between these, or x = -2, then substitute as above and get (-2,16).