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kirza4 [7]
3 years ago
9

1.) Solve 2x^2-12x+20=0 using the quadratic solution set? What is the solution set?

Mathematics
1 answer:
marshall27 [118]3 years ago
4 0
Please, share just ONE problem at a time.  Thanks.

<span>Solve 2x^2-12x+20=0:
Simplify this by dividing each term by 2:  x^2 - 6x + 10 = 0

Identify a, b and c:  a=1, b=-6 and c=10.  Then b^2=36.

Write out the solutions using the quadratic formula:

        6 plus or minus sqrt(36-40)
x = ---------------------------------------
                            2

sqrt(36-40) = sqrt(-4) = plus or minus i2

Then:
 
         6 plus or minus i2
x = --------------------------- (answer)
                  2</span>
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Can someone please help me with this? I'll give brainliest :)
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The slope of y = 3x - 4 on the interval [2, 5] is 3 and the slope of y = 2x^2-4x - 2 on the interval [2, 5] is 10

<h3>How to determine the slope?</h3>

The interval is given as:

x = 2 to x = 5

The slope is calculated as:

m = \frac{y_2 -y_1}{x_2-x_1}

<u>16. y = 3x - 4</u>

Substitute 2 and 5 for x

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So, we have:

m = \frac{11  - 2}{5 - 2}

m = \frac{9}{3}

Divide

m = 3

Hence, the slope of y = 3x - 4 on the interval [2, 5] is 3

<u>17. y = 2x^2-4x - 2</u>

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y = 2 * 5^2 - 4 * 5 - 2 = 28

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m = \frac{28  + 2}{5 - 2}

m = \frac{30}{3}

Divide

m = 10

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Read more about slopes at:

brainly.com/question/3605446

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