Hi there! So joe had 1/2 of a bag of peanuts and he wants to share them equally, it would be 1/6! Answer B
The answer is not correct
Use this systems of equations to solve:
x = first antifreeze
y = second antifreeze
![\left \{ {{.2x + .12y = .18(15)} \atop {x + y = 15}} \right.](https://tex.z-dn.net/?f=%20%5Cleft%20%5C%7B%20%7B%7B.2x%20%2B%20.12y%20%3D%20.18%2815%29%7D%20%5Catop%20%7Bx%20%2B%20y%20%3D%2015%7D%7D%20%5Cright.%20)
Isolate y.
x + y = 15
Subtract x from both sides.
y = -x + 15
Substitute y into the other equation.
.2x + .12(-x + 15) = .18(15)
Simplify.
.2x - .12x + 1.8 = 2.7
Subtract 1.8 from both sides.
.08x = .9
Divide both sides by .08
x = 11.25
Substitute x in the equation that we isolated y in.
y = -11.25 + 15
y = 3.75
11.25 L of the first antifreeze and 3.75 L of the second.
Answer:
Option 4th is correct
![2 \cdot (x-4)(x+4)](https://tex.z-dn.net/?f=2%20%5Ccdot%20%28x-4%29%28x%2B4%29)
Step-by-step explanation:
Using difference of square:
![a^2-b^2=(a-b)(a+b)](https://tex.z-dn.net/?f=a%5E2-b%5E2%3D%28a-b%29%28a%2Bb%29)
GCF is the largest number that divide the given polynomial or expression.
Given the expression:
![2x^2-32](https://tex.z-dn.net/?f=2x%5E2-32)
We have to completely factor the expression.
GCF of
and 32 is, 2
then;
![2 \cdot (x^2-16)](https://tex.z-dn.net/?f=2%20%5Ccdot%20%28x%5E2-16%29)
Using difference of square
![2 \cdot (x-4)(x+4)](https://tex.z-dn.net/?f=2%20%5Ccdot%20%28x-4%29%28x%2B4%29)
therefore, the completely factor of
is, ![2 \cdot (x-4)(x+4)](https://tex.z-dn.net/?f=2%20%5Ccdot%20%28x-4%29%28x%2B4%29)