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3 years ago

Please help asap! Will give brainliest! Please read the question then answer correctly! No guessing.

Mathematics

2 answers:

Doss [256]3 years ago

7 0

Answer:

the answer is D

Step-by-step explanation:

x^2-12x+35=x^2-7x-5x+35=x*(x-7)-5(x-7)=(x-7)(x-5)

frutty [35]3 years ago

4 0

Answer:

(x - 5) (x - 7)

Step-by-step explanation:

To factor this trinomial, you must split the middle term (-12x) into two terms that can be added to get -12x, and multiplied to get 35:

$x^2$ - 12x + 35

$x^2$ -7x - 5x + 35

Group:

( $x^2$ -7x) (-5x + 35)

Take out the GCF (Greatest Common Factor):

x(x - 7) -5(x - 7)

(x - 5) (x - 7)

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Which statement describes the first step to solve the equation by completing the square?

neonofarm [45]

The first step I would take is
.. Multiply both sides of the equation by 1/4.

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You could start by adding 16 to both sides of the equation, to make the left side the square (2x +4)^2.

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2 years ago

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Which co-ordinates do you plot to draw the graph of x + 2y = 3 ?

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Just subisute values that will work
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3 years ago

If you placed $1,000.00 in a savings account with an interest rate of 4.5% month versus an account with 3.75% month, how much in

Setler [38]

The answer would be C. Hope this helps!

7 0

2 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

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Simplify foundation of alegebra

Eva8 [605]

1.) -8 + 10(6 + 4r)

First multiply the 10 into (6 + 4r)

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Distribute the - into (2m - 7m² - 6)

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3 years ago