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8090 [49]
3 years ago
15

Solve w^3 - 9w = 0 by factoring

Mathematics
1 answer:
madam [21]3 years ago
5 0

Answer:

w = 0 , -3 , 3

Step-by-step explanation:

w³ - 9w = 0

w(w² - 9) = 0

w(w² - 3²) = 0

w(w+3)(w-3) = 0            {a² - b² = (a + b)(a - b) }

w = 0   ;  w + 3 = 0                ; w - 3 = 0

                     w = -3              ;        w = 3

w = 0 , -3 , 3

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Find the vertices and foci of the hyperbola with equation quantity x plus one squared divided by sixteen minus the quantity of y
katrin2010 [14]

Answer:

The vertices are (3 , -5) , (-5 , -5)

The foci are (4 , -5) , (-6 , -5)

Step-by-step explanation:

* Lets study the equation of the hyperbola

- The standard form of the equation of a hyperbola with  

  center (h , k) and transverse axis parallel to the x-axis is

  (x - h)²/a² - (y - k)²/b² = 1

- The length of the transverse axis is 2 a

- The coordinates of the vertices are  (h  ±  a  ,  k)

- The coordinates of the foci are (h ± c , k), where c² = a² + b²

- The distance between the foci is  2c

* Now lets solve the problem

- The equation of the hyperbola is (x + 1)²/16 - (y + 5)²/9 = 1

* From the equation

# a² = 16 ⇒ a = ± 4

# b² = 9 ⇒ b = ± 3

# h = -1

# k = -5

∵ The vertices are (h + a , k) , (h - a , k)

∴ The vertices are (-1 + 4 , -5) , (-1 - 4 , -5)

* The vertices are (3 , -5) , (-5 , -5)

∵ c² = a² + b²

∴ c² = 16 + 9 = 25

∴ c = ± 5

∵ The foci are (h ± c , k)

∴ The foci are (-1 + 5 , -5) , (-1 - 5 , -5)

* The foci are (4 , -5) , (-6 , -5)

4 0
3 years ago
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D is the answer
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Calculando a expressão (-2)² + (+5)², obtemos qual resultado?
Contact [7]
This answer would be D) + 6.


Hope this helps if so please mark me as brainlest thank you !!
7 0
2 years ago
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