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-BARSIC- [3]
2 years ago
15

Could someone help me out?

Mathematics
1 answer:
Studentka2010 [4]2 years ago
3 0

The equation of the line is y = -11x + 232

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

The given parameters are:

Slope (m)= -11

Point (x1, y1) = (31, -109)

The linear equation is then calculated as:

y = m(x - x1) + y1

This gives

y = -11(x - 31) - 109

Evaluate the product

y = -11x + 341 - 109

Evaluate the like terms

y = -11x + 232

Hence, the equation of the line is y = -11x + 232

Read more about linear equations at:

brainly.com/question/14323743

#SPJ1

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What are the exact solutions of x2 + 5x + 1 = 0? (5 points)
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Answer:

x=\frac{-5\pm\sqrt{21}  }{2}

Step-by-step explanation:

We use the quadratic formula here which says for a quadratic equation ax^2+bx+c=0.

x= \frac{-b\pm\sqrt{b^2-4ac} }{2a}

Now in our case

a=1\\b=5\\c=1

so we have:

x=\frac{-5\pm\sqrt{5^2-4(1)(1)}  }{2(1)} =\frac{-5\pm\sqrt{21}  }{2}

\boxed{x=\frac{-5\pm\sqrt{21}  }{2} }

Which are our solutions.

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3 years ago
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Lunna [17]
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olga2289 [7]
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If I plug points P and Q into the slope formula, we get:

(1-5)/(-2+8) = -4/6 = -2/3

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tatyana61 [14]

Answer:

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Step-by-step explanation:

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