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tiny-mole [99]
2 years ago
3

The landlord of an apartment building needs to purchase enough digits to label all of the apartments from 100 through 125 on the

first floor and 200 through 225 on the second floor. The digits can only be purchased in a package that contains one of each digit 0 through 9. How many packages must the landlord purchase?
Mathematics
1 answer:
ASHA 777 [7]2 years ago
6 0
The numbers 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 224, 225  contains 26 numbers 0, 52 numbers 1, 44 numbers 2, 6 numbers 3, 6 numbers 4, 6 numbers 5, 4 numbers 6, 4 numbers 7, 4 numbers 8, 4 numbers 9. The largest number is 52, hence the landlord should buy 52 packages.
Answer: 52 packages.

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Answer:

2.5 meters per second

Step-by-step explanation:

Given,

The speed of Charlotte = 9 km/h

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The age of two groups of yoga students are shown in the following dot plot
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The base of an aquarium with given volume V is made of slate and the sides are made of glass. If the slate costs seven times as
lubasha [3.4K]

Answer:

l = \sqrt[3]{\frac{2V}{7}}     b = \sqrt[3]{\frac{2V}{7}}       h = \sqrt[3]{\frac{49V}{4}}

Step-by-step explanation:

Represent the volume of the box with V and the dimensions with l, b and h.

The volume (V) is:

V = l * b * h

Make h the subject of the formula

h = \frac{V}{lb}

The surface area (S) of the aquarium is:

S = lb + 2(lh + bh)

Where lb represents the area of the base (i.e. slate):

The cost (C) of the surface area is:

C = 7 * lb + 1 * 2(lh + bh)

C = 7lb + 2(lh + bh)

C = 7lb + 2h(l + b)

Substitute \frac{V}{lb} for h in the above equation

C = 7lb + 2*\frac{V}{lb}(l + b)

C = 7lb + \frac{2V}{lb}(l + b)

C = 7lb + \frac{2V}{b} + \frac{2V}{l}

Differentiate with respect to l and with respect to b

C_l=7b - \frac{2V}{l^2} =0

C_b=7l - \frac{2V}{b^2} =0

To solve for b and l, we equate both equations and set l to b (to minimize the cost)

7b - \frac{2V}{l^2}=7l - \frac{2V}{b^2}

7l - \frac{2V}{l^2}=7b - \frac{2V}{b^2}

By comparison:

l =b

C_l=7b - \frac{2V}{l^2} =0 becomes

7l - \frac{2V}{l^2}=0

7l = \frac{2V}{l^2}

Cross Multiply

7l^3 = 2V

Solve for l

l^3 = \frac{2V}{7}

l = \sqrt[3]{\frac{2V}{7}}

Recall that: l =b

b = \sqrt[3]{\frac{2V}{7}}

Also recall that:

h = \frac{V}{lb}

h = \frac{V}{\sqrt[3]{\frac{2V}{7}}*\sqrt[3]{\frac{2V}{7}}}

h = \frac{V}{\sqrt[3]{\frac{4V^2}{49}}}

Apply law of indices

h = \sqrt[3]{\frac{49V^3}{4V^2}}

h = \sqrt[3]{\frac{49V}{4}}

The dimension that minimizes the cost of material of the aquarium is:

l = \sqrt[3]{\frac{2V}{7}}     b = \sqrt[3]{\frac{2V}{7}}       h = \sqrt[3]{\frac{49V}{4}}

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