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3 years ago

Which factor do 64x^2 + 48x + 9 and 64x^2 - 9 have in common?

1 answer:

Ber [7]3 years ago

8 0

Start with the second expression. That factors easily into (8x-3)(8x+3).
Since 64x^2 + 48x + 9 has no negative signs, I'd bet that the common factor is (8x+3). Divide 64x^2 + 48x + 9 by that to verify.

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butalik [34]

Step-by-step explanation:

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3 years ago

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Lapatulllka [165]

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3 years ago

Find two number nearest to 8888888 which are exactly divisible by 2915

Marysya12 [62]

Step-by-step explanation:

Given problem is to find nearest number,

= 8888888/2915

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remainder= 1053

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3 years ago

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AnnZ [28]

For this case the vertex of the parabola is the point at which the parabola presents a minimum.

The parabola presents a minimum in this case, because the parabola opens upwards.

The vertex is then a minimum.

The vertex of the function is located in the following ordered pair:

(x, y) = (1, -1)

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8 0

4 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:

Given equation:

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

This is an equation for a horizontal hyperbola.

To complete the square for a hyperbola

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