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andrey2020 [161]
3 years ago
10

Frank organized his jazz and rock CD Collection. Some CDs are double albums and are therefore wider . The regular CDs are 0.5 cm

wide. The double CDs are 1 cm wide. He found that his 125 CDs equal 75 cm. How many CDs are regular length and how many are double albums?
Mathematics
1 answer:
tester [92]3 years ago
4 0

Answer: 100 of the CDs are single album and 25 are double album.

Explanation: What you have to do first is find out how many single CDs which are .5 cm in length you can put in 125 with out going over 75 cm. Then the rest of the 25 are of course double album CDs and which equal out to be 125 CDs total and 75 cm total. (the double album CDs technically count asone by the way) I got my answer by first seeing how many double CDs can go in 125 without going over 75 cm. The max you can get is 75. So then you keep removing double album CDs until you get the magic number 125 CDs and 75 cm.

Hope I could help

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Solve for x, if 4x - 24 = 1 000.<br> Solve for x, if 7x = 343.<br> Solve for x, if x4 = 4 096.
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1) given ,

4x-24=1000

=> 4x = 1000+24

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Determine the dimensions of a rectangular solid (with a square base) with maximum volume if its surface area is 181.5 square cen
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Answer:

The base and height of the solid is 5.5cm

<em></em>

Step-by-step explanation:

Given

Surface\ Area = 181.5cm^2

Required

Determine the dimensions that maximizes the volume

Let the base dimension be x and the height be h

The volume is calculated as:

Volume =x^2 * h

Volume =x^2h

181.5 =x^2h

The surface area (S) is calculated as this:

S = 2(x^2 + xh + xh)

S = 2(x^2 + 2xh)

S = 2x^2 + 4xh

Substitute 181.5 for S

181.5 = 2x^2 + 4xh

Make h the subject:

4xh = 181.5 - 2x^2

h = \frac{181.5 - 2x^2}{4x}

Substitute h = \frac{181.5 - 2x^2}{4x} in Volume =x^2h

V = x^2(\frac{181.5 - 2x^2}{4x})

V = x(\frac{181.5 - 2x^2}{4})

V = \frac{1}{4}(x)(181.5 - 2x^2)

V = \frac{181.5x}{4} - \frac{2x^3}{4}

V = \frac{181.5x}{4} - \frac{x^3}{2}

To get the maximum, we differentiate V with respect to t and set the differentiation to 0

dV = \frac{181.5}{4} - \frac{3x^2}{2}

Set to 0

0 = \frac{181.5}{4} - \frac{3x^2}{2}

\frac{3x^2}{2} = \frac{181.5}{4}

Multiply through by 4

4 * \frac{3x^2}{2} = \frac{181.5}{4} * 4

2*3x^2 = 181.5

6x^2 = 181.5

x^2 = \frac{181.5}{6}

x^2 = 30.25

x = \sqrt{30.25

x = 5.5

Recall that:

h = \frac{181.5 - 2x^2}{4x}

h = \frac{181.5 - 2 * 5.5^2}{4 * 5.5}

h = \frac{181.5 - 60.5}{22}

h = \frac{121}{22}

h = 5.5

So, we have:

h = 5.5

x = 5.5

<em>Hence, the base and height of the solid is 5.5cm</em>

7 0
3 years ago
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