Problem 1: 2x > 4x - 6 problem 2: -3r < 10 - r problem 3: 5c - 4 > 8c + 2

1 answer:

8 0

<span>Problem 1: 2x > 4x - 6
2x < 6
x < 3

problem 2: -3r < 10 - r
-2r < 10
r > -5

problem 3: 5c - 4 > 8c + 2
3c < -6
c < -2</span>

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Somebody please help so I can pass, please

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First, we are going to find the vertex of our quadratic. Remember that to find the vertex $(h,k)$ of a quadratic equation of the form $y=a x^{2} +bx+c$ , we use the vertex formula $h= \frac{-b}{2a}$ , and then, we evaluate our equation at $h$ to find $k$ .

We now from our quadratic that $a=2$ and $b=-32$ , so lets use our formula:
$h= \frac{-b}{2a}$
$h= \frac{-(-32)}{2(2)}$
$h= \frac{32}{4}$
$h=8$
Now we can evaluate our quadratic at 8 to find $k$ :
$k=2(8)^2-32(8)+56$
$k=2(64)-256+56$
$k=128-200$
$k=-72$
So the vertex of our function is (8,-72)

Next, we are going to use the vertex to rewrite our quadratic equation:
$y=a(x-h)^2+k$
$y=2(x-8)^2+(-72)$
$y=2(x-8)^2-72$
The x-coordinate of the minimum will be the x-coordinate of the vertex; in other words: 8.

We can conclude that:
The rewritten equation is $y=2(x-8)^2-72$
The x-coordinate of the minimum is 8