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Kazeer [188]
3 years ago
11

What whole number dimensions would allow the students to maximize the volume while keeping the surface area at most 160 square f

eet
Mathematics
1 answer:
ycow [4]3 years ago
6 0

Answer:

The whole number dimension that would allow the student to maximize the volume while keeping the surface area at most 160 square is 6 ft

Step-by-step explanation:

Here we are required find the size of the sides of a dunk tank (cube with open top) such that the surface area is ≤ 160 ft²

For maximum volume, the side length, s of the cube must all be equal ;

Therefore area of one side = s²

Number of sides in a cube with top open = 5 sides

Area of surface = 5 × s² = 180

Therefore s² = 180/5 = 36

s² = 36

s = √36 = 6 ft

Therefore, the whole number dimension that would allow the student to maximize the volume while keeping the surface area at most 160 square = 6 ft.

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Write the equation for the hyperbola with foci (–12, 6), (6, 6) and vertices (–10, 6), (4, 6).
fomenos

Answer:

\frac{(x--3)^2}{49} -\frac{(y-6)^2}{32}=1

Step-by-step explanation:

The standard equation of a horizontal hyperbola with center (h,k) is

\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1

The given hyperbola has vertices at (–10, 6) and (4, 6).

The length of its major axis is 2a=|4--10|.

\implies 2a=|14|

\implies 2a=14

\implies a=7

The center is the midpoint of the vertices (–10, 6) and (4, 6).

The center is (\frac{-10+4}{2},\frac{6+6}{2}=(-3,6)

We need to use the relation a^2+b^2=c^2 to find b^2.

The c-value is the distance from the center (-3,6) to one of the foci (6,6)

c=|6--3|=9

\implies 7^2+b^2=9^2

\implies b^2=9^2-7^2

\implies b^2=81-49

\implies b^2=32

We substitute these values into the standard equation of the hyperbola to obtain:

\frac{(x--3)^2}{7^2} - \frac{(y-6)^2}{32}=1

\frac{(x+3)^2}{49} -\frac{(y-6)^2}{32}=1

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~

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