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4 years ago

Convert: 15°C to °F -9 -31 59 164

Mathematics

1 answer:

Anton [14]4 years ago

6 0

15 is 59
-9 is 15.8 so 16
-31 is 28.8 so 29
59 is 138.2
164 is 327.2

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2 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.

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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:

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

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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:

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

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2 years ago

Write 4.55 as a mixed number and as a improper fraction

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Step-by-step explanation:

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4 years ago

A cylinder has a volume of 175 cubic units and a height of 7 units. The diameter of the cylinder is

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The answer is 10 units

The diameter of the cylinder (d) is the twice of its radius (r): d = 2r

The volume (V) of the cylinder with radius r and height h is:
V = π r² h

We have:
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175π = π * r² * 7
175 = r² * 7
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r = √25
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4 years ago

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