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3 years ago

What is the midpoint of the segment shown below

Mathematics

1 answer:

Softa [21]3 years ago

6 0

Answer:

(-1, -1/2)

Step-by-step explanation:

To find the x coordinate of the midpoint, add the x coordinates and divide by 2

( -11+9)/2 = -2/2 = -1

To find the y coordinate of the midpoint, add the y coordinates and divide by 2

(0+-1)/2 = -1/2

(-1, -1/2)

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3 years ago

What are the zeros of the quadratic function f(x) = 6x2 + 12x - 7?<br>

mina [271]

Answer:

$\frac{-6+\sqrt{78} }{6}$ and $\frac{-6-\sqrt{78} }{6}$

Step-by-step explanation:

We are asked that what are the zeros of the quadratic function f(x) = 6x² +12x -7.

So, we have to find the roots of the equation, f(x) = 6x² +12x -7 =0 ...... (1)

Since the quadratic function can not be factorized, so we have to apply Sridhar Acharya's formula.

This formula gives if, ax² +bx +c =0, the the two roots of the equation are

$\frac{-b+\sqrt{b^{2}-4ac } }{2a} , \frac{-b-\sqrt{b^{2}-4ac } }{2a}$

Therefore, in our case 'a' being 6, 'b' being 12 and 'c' being -7, the two roots of the equation (1) will be

$\frac{-12+\sqrt{12^{2}-4*6*(-7) } }{2*6}, \frac{-12-\sqrt{12^{2}-4*6*(-7) } }{2*6}$

= $\frac{-12+\sqrt{312} }{12},\frac{-12-\sqrt{312} }{12}$

= $\frac{-6+\sqrt{78} }{6}$ and $\frac{-6-\sqrt{78} }{6}$

Hence, x= $\frac{-6+\sqrt{78} }{6}$

and x= $\frac{-6-\sqrt{78} }{6}$

(Answer)

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4 years ago

More slope pls helpppppp

noname [10]

<h3>Answer: C) 2</h3>

Explanation:

Recall that the general point-slope form is $y - y_1 = m(x - x_1)$ where m is the slope and $(x_1, y_1)$ is the point the line goes through. We see that the 2 and the m match up, so m = 2 is the slope.

Side note: We can rewrite that given equation into y - (-5) = 2(x - (-1)) to see that the line goes through the point (-1, -5)

5 0

3 years ago

Read 2 more answers