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3 years ago

Please help!

Mathematics

2 answers:

Papessa [141]3 years ago

6 0

Answer:

x = 1 and x = 2

x = 4 and x = -4

Step-by-step explanation:

Vertical asymptotes appear where the function does not have a value. This is most commonly when the denominator of a rational function is 0. Find the asymptotes by factoring the denominator and setting it equal to 0. Then solve for x.

First equation

x² - 3x + 2 factors into (x-1)(x-2)

When x-1 = 0, x = 1. When x-2=0, x = 2. The V.A. are at x = 1 and x = 2.

Second equation

x² - 16 factors into (x+4)(x-4)

When x+4= 0, x = -4. When x-4 = 0, then x = 4. The V.A. are at x = -4 and x = 4.

Amanda [17]3 years ago

3 0

B. x = -4

C. x = 4

Edgenuit.y2020 dudets

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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

- If IxI = a ⇒ then x = a or x = -a

- IxI never give a negative answer, because IxI means the

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

∴ x = 2

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

∴ 2x = -14 ⇒ divide both sides by 2

∴ x = -7

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

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

∴ I2x + 7I = 9

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

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∴ x = 1

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

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∴ x = -8

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

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

∴ -x = 1 ⇒ divide both sides by -1

∴ x = -1

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

∴ -x = -11 ⇒ divide both sides by -1

∴ x = 11

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

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

∴ x = 9

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

∴ x = -15

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

6- ∵ Ix - 3I = -12

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