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IRINA_888 [86]
2 years ago
7

Express 1/243 as a power

Mathematics
1 answer:
nikdorinn [45]2 years ago
5 0

Answer:

logbase3(1/243)

Step-by-step explanation:

This is the equation you would use to solve.

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alukav5142 [94]
2x^2 - 4 = 68
add 4 to both sides
2x^2 = 72
divide both sides by 2
x^2 = 36
find the square root of each sides
x = 6, -6

ANSWER: the positive value of x is 6
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3 years ago
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AlexFokin [52]

Answer:

The answer is d.

Draw the ray from 12 downward

4 0
2 years ago
(m+2)(m+3)=(m+2)(m-2)
Anastaziya [24]
(m+2)(m+3)= (m+2)(m-2)
⇒ m^2+ 3m+ 2m+ 6= m^2 -4
⇒ 5m+ 6= -4 (m^2 on both sides cancels out)
⇒ 5m= -4-6
⇒ 5m= -10
⇒ m= -10/5
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The final answer is m=-2~
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levacccp [35]

Answer:

5 units

Step-by-step explanation:

You can use the distance formula or pythagorean theorem to solve this.

Original point A: (0,0)

New point B: (3, 4)

Distance formula: \sqrt{(x_{2}-x_{1})^{2}   +(y_{2}-y_{1})^{2} }

Plug the points in:

\sqrt{(3-0)^{2} +(4-0)^{2} } = 5

5 units

5 0
3 years ago
HELPPPP!!!
Hatshy [7]

Answer:

(x - 1)²/4² - (y - 2)²/2² = 1 ⇒ The bold labels are the choices

Step-by-step explanation:

* Lets explain how to solve this problem

- The equation of the hyperbola is x² - 4y² - 2x + 16y - 31 = 0

- The standard form of the equation of hyperbola is

  (x - h)²/a² - (y - k)²/b² = 1 where a > b

- So lets collect x in a bracket and make it a completing square and

  also collect y in a bracket and make it a completing square

∵ x² - 4y² - 2x + 16y - 31 = 0

∴ (x² - 2x) + (-4y² + 16y) - 31 = 0

- Take from the second bracket -4 as a common factor

∴ (x² - 2x) + -4(y² - 4y) - 31 = 0

∴ (x² - 2x) - 4(y² - 4y) - 31 = 0

- Lets make (x² - 2x) completing square

∵ √x² = x

∴ The 1st term in the bracket is x

∵ 2x ÷ 2 = x

∴ The product of the 1st term and the 2nd term is x

∵ The 1st term is x

∴ the second term = x ÷ x = 1

∴ The bracket is (x - 1)²

∵  (x - 1)² = (x² - 2x + 1)

∴ To complete the square add 1 to the bracket and subtract 1 out

   the bracket to keep the equation as it

∴ (x² - 2x + 1) - 1

- We will do the same withe bracket of y

- Lets make 4(y² - 4y) completing square

∵ √y² = y

∴ The 1st term in the bracket is x

∵ 4y ÷ 2 = 2y

∴ The product of the 1st term and the 2nd term is 2y

∵ The 1st term is y

∴ the second term = 2y ÷ y = 2

∴ The bracket is 4(y - 2)²

∵ 4(y - 2)² = 4(y² - 4y + 4)

∴ To complete the square add 4 to the bracket and subtract 4 out

   the bracket to keep the equation as it

∴ 4[y² - 4y + 4) - 4]

- Lets put the equation after making the completing square

∴ (x - 1)² - 1 - 4[(y - 2)² - 4] - 31 = 0 ⇒ simplify

∴ (x - 1)² - 1 - 4(y - 2)² + 16 - 31 = 0 ⇒ add the numerical terms

∴ (x - 1)² - 4(y - 2)² - 16 = 0 ⇒ add 14 to both sides

∴ (x - 1)² - 4(y - 2)² = 16 ⇒ divide both sides by 16

∴ (x - 1)²/16 - (y - 2)²/4 = 1

∵ 16 = (4)² and 4 = (2)²

∴ The standard form of the equation of the hyperbola is

   (x - 1)²/4² - (y - 2)²/2² = 1

4 0
3 years ago
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