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guajiro [1.7K]
3 years ago
12

The midpoint of a line with the endpoints (-2, 0) and (-6, 8) is:

Mathematics
2 answers:
Lera25 [3.4K]3 years ago
8 0
Midpoint is (x1 + x2)/2 , (y1 + y2)/2
(-2,0)...x1 = -2 and y1 = 0
(-6,8)...x2 = -6 and y2 = 8
now we sub and solve
m = (-2 +(-6) / 2 , (8 + 0)/2
m = (-2 - 6)/2 , 8/2
m = -8/2 , 8/2
m = (-4,4) <==
Norma-Jean [14]3 years ago
5 0

Answer:

(-4,4)

Step-by-step explanation:

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Match each equation with its solution set.
taurus [48]

Answer:

1- The solution of I2x + 5I = 9 is {-7 , 2}

2- The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- The solution of I5 - xI = 6 is {-1 , 11}

4- The solution of I6x - 8I + 7 = 5 is ∅

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Step-by-step explanation:

* At first lets explain the meaning of IxI = a

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 magnitude of x is always positive

Ex: I-2I is 2

* Now lets find the solution of each equation

1- ∵ I2x + 5I = 9

∴ 2x + 5 = 9 ⇒ subtract 5 from both sides

∴ 2x = 4 ⇒ divide both sides by 2

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OR

∴ 2x + 5 = -9 ⇒ subtract 5 from both sides

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* The solution of I2x + 5I = 9 is {-7 , 2}

2- ∵ I2x + 7I + 2 = 11 ⇒ Subtract 2 from both sides

∴ I2x + 7I = 9

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∴ x = 1

OR

∴ 2x + 7 = -9 ⇒ subtract 7 from both sides

∴ 2x = -16 ⇒ divide both sides by 2

∴ x = -8

* The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- ∵ I5 - xI = 6

∴ 5 -x = 6 ⇒ subtract 5 from both sides

∴ -x = 1 ⇒ divide both sides by -1

∴ x = -1

OR

∴ 5 -x = -6 ⇒ subtract 5 from both sides

∴ -x = -11 ⇒ divide both sides by -1

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* The solution of I5 - xI = 6 is {-1 , 11}

4- ∵ I6x - 8I + 7 = 5 ⇒ Subtract 7 from both sides

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* The solution of I6x - 8I + 7 = 5 is ∅

5- ∵ Ix + 3I = 12

∴ x + 3 = 12 ⇒ subtract 3 from both sides

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OR

∴ x + 3 = -12 ⇒ subtract 3 from both sides

∴ x = -15

* The solution of Ix + 3I = 12 is {-15 , 9}

6- ∵ Ix - 3I = -12

- I  I never give negative answer

* The solution of Ix - 3I = -12 is ∅

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3 years ago
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lara31 [8.8K]

Answer:

6. \frac{29}{15} or 1\frac{14}{15}

7. \frac{133}{96} or 1\frac{37}{96}

3 0
3 years ago
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