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Airida [17]
3 years ago
14

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

Mathematics
1 answer:
Trava [24]3 years ago
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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Point G is the point (3,-1). Which point is 5
Georgia [21]

Answer:

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5 0
3 years ago
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Somebody please help so I can pass, please
ASHA 777 [7]
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
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Next, we are going to use the vertex to rewrite our quadratic equation:
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y=2(x-8)^2+(-72)
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8 0
3 years ago
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8 0
3 years ago
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62

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Happy learning!

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3 years ago
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