y=mx+b is the equation of a line;
m=slope , b= y-intercept
m=4 ; so we have : y=4x+b
We are give a set of points which it passes through, we can simply plug them in:
-2 = 4(3)+b (3 is the x and -2 is the y)
We get -2 = 12 +b .... -14=b
our final equation is : y=4x-14
Answer:
The Answer Is 74. Your Welcome
The zero product property tells us that if the product of two or more factors is zero, then each one of these factors CAN be zero.
For more context let's look at the first equation in the problem that we can apply this to:
![(x-3)(x+4)=0](https://tex.z-dn.net/?f=%28x-3%29%28x%2B4%29%3D0)
Through zero property we know that the factor
![(x-3)](https://tex.z-dn.net/?f=%28x-3%29)
can be equal to zero as well as
![(x+4)](https://tex.z-dn.net/?f=%28x%2B4%29)
. This is because, even if only one of them is zero, the product will immediately be zero.
The zero product property is best applied to
factorable quadratic equations in this case.
Another factorable equation would be
![2x^{2}+6x=0](https://tex.z-dn.net/?f=2x%5E%7B2%7D%2B6x%3D0)
since we can factor out
![2x](https://tex.z-dn.net/?f=2x)
and end up with
![2x(x+3)=0](https://tex.z-dn.net/?f=2x%28x%2B3%29%3D0)
. Now we'll end up with two factors,
![2x](https://tex.z-dn.net/?f=2x)
and
![(x+3)](https://tex.z-dn.net/?f=%28x%2B3%29)
, which we can apply the zero product property to.
The rest of the options are not factorable thus the zero product property won't apply to them.