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Ira Lisetskai [31]
3 years ago
15

Find the midpoint of the segment with the following endpoints (-1,3) and (6,-2)

Mathematics
2 answers:
Sliva [168]3 years ago
7 0

Answer:

(5/2, 1/2)

Step-by-step explanation:

the midpoint formula is

(x1 + x2/2) , (y1 + y2/2) <-- both the x1 and y1 coordinates also are divided by 2

so to solve for the midpoint, we take our endpoints and plug it into the formula

we will list the coordinates as the following to plug it into the formula:

x coordinate 1 = -1

x coordinate 2 = 6

y coordinate 1 = 3

y coordinate 2 = -2

substituting these values into the formula we get this:

x =(-1 + 6/2) , y = (3 + -2/2)

x = -1 + 6 = 5/2

we cant simplify 5/2 any further, so 5/2 is our x coordinate for the midpoint

y = 3 + (-2) = 1/2

1/2 is already simplified, so 1/2 is our y coordinate for the midpoint

in an ordered pair it would be (5/2, 1/2)

DENIUS [597]3 years ago
3 0

Answer:

(5/2 and 1/2)

Step-by-step explanation:

x1,y1=(-1,3)  x2,y2=(6,-2)

6-1/2,-2+3/2

= 5/2 and 1/2

Hope this helps

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If the factory used ⅔, and 1 batch is ⅓, the factory made 1 batch for each 3rd they used. ⅔, 2 batches

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Hello, we want to write 2 + 4 + 6 + ... + 20

for n = 1, 2n = 2*1 =2

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3 years ago
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Step-by-step explanation:

8 0
3 years ago
HELP PLEASE!! I DONT UNDERSTAND!!!!!!!!!! THANKS SO MUCH
8090 [49]

Hello, please consider the following.

We will multiply the numerator and denominator by

4+\sqrt{6x}

to get rid of the root in the denominator.

First of all, we cannot divide by 0, right? So, we need to make sure that the denominator is different from 0.

4-\sqrt{6x} =0\sqrt{6x}=4\\\\\text{Take the square}\\\\6x=4^2=16\\\\x=\dfrac{16}{6}=\dfrac{8}{3}

We need to take any x real number different from 8/3 then and simplify the expression.

Let's do it!

\begin{aligned}\dfrac{4}{4-\sqrt{6x}}&=\dfrac{4(4+\sqrt{6x})}{(4+\sqrt{6x})(4-\sqrt{6x})}\\\\&=\dfrac{4(4+\sqrt{6x})}{(4^2-\sqrt{6x}^2)}\\\\&=\dfrac{4(4+\sqrt{6x})}{(16-6x)}\\\\&=\dfrac{2(4+\sqrt{6x})}{(8-3x)}\\\\&\large \boxed{=\dfrac{8+2\sqrt{6x}}{8-3x}}\end{aligned}

Thank you

8 0
3 years ago
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