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2 years ago

[ BRAINLIEST TO CORRECT ] please answer

Mathematics

2 answers:

Katyanochek1 [597]2 years ago

5 0

Answer:

OK good job

Step-by-step explanation:

welcome free brainly pls??? bless u

LUCKY_DIMON [66]2 years ago

3 0

Answer:\

Free answer?

Step-by-step explanation:

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Which point-slope equation represents a line that passes through (3, –2) with a slope of negative StartFraction 4 Over 5 EndFrac

Schach [20]

Answer:

y + 2 = (-4/5)(x - 3)

Step-by-step explanation:

Your slope is -4/5. Your line passes through (3, -2).

Applying the point slope formula, we get:

y + 2 = (-4/5)(x - 3)

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4 years ago

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3 years ago

write a multiplication problem using two decimals that has a product greater than 0.1 but less than 0.2

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3 years ago

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Put these numbers in order from greatest the least 28/13 -0.4 1 6/20

tiny-mole [99]

Answer:

Greatest to least

28/13, 1, 6/20, -0.4

Step-by-step explanation:

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3 years ago

Identify the standard form of the equation by completing the square.

OLEGan [10]

Answer:

$\dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Step-by-step explanation:

<u>Given equation</u>:

$4x^2-9y^2-8x+36y-68=0$

This is an equation for a horizontal hyperbola.

<u>To complete the square for a hyperbola</u>

Arrange the equation so all the terms with variables are on the left side and the constant is on the right side.

$\implies 4x^2-8x-9y^2+36y=68$

Factor out the coefficient of the x² term and the y² term.

$\implies 4(x^2-2x)-9(y^2-4y)=68$

Add the square of half the coefficient of x and y inside the parentheses of the left side, and add the distributed values to the right side:

$\implies 4\left(x^2-2x+\left(\dfrac{-2}{2}\right)^2\right)-9\left(y^2-4y+\left(\dfrac{-4}{2}\right)^2\right)=68+4\left(\dfrac{-2}{2}\right)^2-9\left(\dfrac{-4}{2}\right)^2$

$\implies 4\left(x^2-2x+1\right)-9\left(y^2-4y+4\right)=36$

Factor the two perfect trinomials on the left side:

$\implies 4(x-1)^2-9(y-2)^2=36$

Divide both sides by the number of the right side so the right side equals 1:

$\implies \dfrac{4(x-1)^2}{36}-\dfrac{9(y-2)^2}{36}=\dfrac{36}{36}$

Simplify:

$\implies \dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Therefore, this is the standard equation for a horizontal hyperbola with:

center = (1, 2)
vertices = (-2, 2) and (4, 2)
co-vertices = (1, 0) and (1, 4)
$\textsf{Asymptotes}: \quad y = -\dfrac{2}{3}x+\dfrac{8}{3} \textsf{ and }y=\dfrac{2}{3}x+\dfrac{4}{3}$
$\textsf{Foci}: \quad (1-\sqrt{13}, 2) \textsf{ and }(1+\sqrt{13}, 2)$

4 0

2 years ago