Answer:
<h3>The value which is needed to be added to transform the quadratic expression as a perfect square
Trinomial is 16.</h3>
Answer:
0
Step-by-step explanation:
Answer:
y = -2
Step-by-step explanation:
Use the formula for slope: ![m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7By_%7B2%7D-y_%7B1%7D%7D%7Bx_%7B2%7D-x_%7B1%7D%7D)
"m" means slope, and we know it is -1.
Decide which points will be point 1 and point 2
Point 1 (-2, y) x₁ = -2 y₁ = y
Point 2 (0, -4) x₂ = 0 y₂ - -4
Substitute the "x" and "y" values into the formula
![m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7By_%7B2%7D-y_%7B1%7D%7D%7Bx_%7B2%7D-x_%7B1%7D%7D)
Simplify
Remember m = -1. Get rid of the fraction
Multiply both sides by 2
The "2"s cancel out on the right
Start isolating "y"
Add 4 to both sides
![2= -y](https://tex.z-dn.net/?f=2%3D%20-y)
Divide both sides by -1 to isolate "y"
-2 = y Answer
y = -2 Variable on left side
Volume of cube, V = edge^3
Let edge of cube#1 = (x-4) m, therefore volume of cube#1, v1 = (x-4)^3 m
Let edge of cube#2 = x m, therefore volume of cube#2, v2 = x^3 m
Diff. in volume (in m) = 1216 = v2-v1 = [ x^3 - (x-4)^3 ]
= x^3 - [(x-4)(x-4)(x-4)]
= x^3 - [<span>x^2 - 8x +16(x - 4)]
= </span> x^3 - [ x^3 - 12x^2 + 48x - 64 ]
= 12x^2 - 48x + 64
= 4 (3x^2 - 12x + 16)
Therefore 4 (3^2 - 12x + 16) = 1216
3x^2 - 12x + 16 = 1216/4 = 304
3x^2 - 12x - 288 = 0
3 (x^2 - 4x - 96) = 0
(x^2 - 4x - 96) = 0
(x - 12) (x + 8) =0
(x-12) = 0
Therefore x = 12 m
Edge of cube#2 = x m = 12m
Edge of cube#1 = (x-4) m = 8m