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Gelneren [198K]
2 years ago
12

What is the solution set to the equation x^2+6x+3=0

Mathematics
1 answer:
marin [14]2 years ago
7 0

Answer:

That is a quadratic equation in which

a = 1

b = 6

c = 3

The quadratic formula is:

x = [-b +- sqr root (b^2 -4*a*c)] / 2*a

x = [-6 +- sqr root (36 -4 * 1 * 3)] / 2

x1 = [-6 + sqr root (24)] / 2

x2 = [-6 - sqr root (24)] / 2

Step-by-step explanation:

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polet [3.4K]
-7/5 + 3 hope this helps
5 0
3 years ago
Which transformation(s) can be used to map one triangle onto the other? Select two options.
myrzilka [38]

Answer:

what are the options?

Step-by-step explanation:

7 0
3 years ago
Find the value of a $3,000 certificate in 2 years, if the interest rate is 10% compounded annually.
ser-zykov [4K]

Answer:

600$

Step-by-step explanation:

3000×2×.10=600

the 10 percent needs to be changed to .10

5 0
2 years ago
Factor completely 4x^5-20x^3
nikdorinn [45]

Answer:

<em>4x^3(x^2 - 5)</em>

Step-by-step explanation:

First, try to factor a common factor.

GCF of 4 and -20 is 4.

GCF of x^5 and x^3 is x^3.

Factor out 4x^3.

4x^5 - 20x^3 =

= 4x^3(x^2 - 5)

3 0
3 years ago
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
Read 2 more answers
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