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2 years ago

Lip balm is sold in hemispherical plastic containers. Given that the volume of lip balm in each container is 30 cm2, find \

Mathematics

1 answer:

Marina CMI [18]2 years ago

5 0

The Surface area of the container that is in contact with the lip balm ≈ 30.4 cm³.

<h3>What is the Volume of a Hemisphere?</h3>

A hemisphere is half of a sphere. Thus, the volume of a hemisphere is given as: V = = (2/3)πr³, where r is the radius of the hemisphere.

<h3>What is the Surface Area of a Hemisphere?</h3>

Surface area of a hemisphere is given as, SA = 2πr²

Given the following:

Volume of hemisphere = 30 cm³

i. Radius (r) of the container = radius of the hemisphere

2πr² = 30

r² = 30/2π

r² = 4.8

r ≈ 2.2 cm

ii. Surface area of the container that is in contact with the lip balm = surface area of hemisphere = 2πr²

Surface area = 2(3.14)(2.2²)

Surface area of the container that is in contact with the lip balm ≈ 30.4 cm³.

In summary:

i. The radius of the container is 2.2 cm

ii. The surface area of the container that is in contact with the lip balm ≈ 30.4 cm³.

Learn more about surface area of hemisphere on:

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