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igor_vitrenko [27]
3 years ago
11

The drawing shows a stack of alternating cups. The cups are 9.2 cm high. How many cups will fit in a door that is 211.77 cm high

?
Mathematics
1 answer:
KIM [24]3 years ago
7 0

Answer:

The total number of whole cups that we can fit in the dispenser is 25

Step-by-step explanation:

It is given that the height of each cup is 20 cm.

But when we stack them one on top of the other, they only add a height of 0.8 to the stack.  

The stack of cups has to be put in a dispenser of height 30 cm.

So we need o find out how many cups can fit in the dispenser.

Since the first cup is 20 cm high, the height cannot be reduced. So the space to fit in the remaining cups in the stack is only 30-20 cm as that’s the remaining space in the dispenser

So,

30 - 20 = 10 cm

To stack the other cups we have 10 cm of height remaining

As we know that addition of each adds 0.8 cm to the stack, the total number of cups that can be fit in the dispenser can be calculated by the following equation. Let the number of cups other than the first cup be denoted by ‘x’.

10 + 0.8x = 30

0.8x = 20

x = 25

The total number of cups that we can fit in dispenser is 25

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An arithmetic progression is simply a progression with a common difference among consecutive terms.

  • <em>The sum of multiplies of 6 between 8 and 70 is 390</em>
  • <em>The sum of multiplies of 5 between 12 and 92 is 840</em>
  • <em>The sum of multiplies of 3 between 1 and 50 is 408</em>
  • <em>The sum of multiplies of 11 between 10 and 122 is 726</em>
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  • <em>The sum of the first 27 terms is -486</em>
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<u>(a) Sum of multiples of 6, between 8 and 70</u>

There are 10 multiples of 6 between 8 and 70, and the first of them is 12.

This means that:

\mathbf{a = 12}

\mathbf{n = 10}

\mathbf{d = 6}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{10} = \frac{10}2(2*12 + (10 - 1)6)}

\mathbf{S_{10} = 390}

<u>(b) Multiples of 5 between 12 and 92</u>

There are 16 multiples of 5 between 12 and 92, and the first of them is 15.

This means that:

\mathbf{a = 15}

\mathbf{n = 16}

\mathbf{d = 5}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{16} = \frac{16}2(2*15 + (16 - 1)5)}

\mathbf{S_{16} = 840}

<u>(c) Multiples of 3 between 1 and 50</u>

There are 16 multiples of 3 between 1 and 50, and the first of them is 3.

This means that:

\mathbf{a = 3}

\mathbf{n = 16}

\mathbf{d = 3}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{16} = \frac{16}2(2*3 + (16 - 1)3)}

\mathbf{S_{16} = 408}

<u>(d) Multiples of 11 between 10 and 122</u>

There are 11 multiples of 11 between 10 and 122, and the first of them is 11.

This means that:

\mathbf{a = 11}

\mathbf{n = 11}

\mathbf{d = 11}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{16} = \frac{11}2(2*11 + (11 - 1)11)}

\mathbf{S_{11} = 726}

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<u>(e) Multiples of 9 between 25 and 100</u>

There are 9 multiples of 9 between 25 and 100, and the first of them is 27.

This means that:

\mathbf{a = 27}

\mathbf{n = 9}

\mathbf{d = 9}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

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\mathbf{S_{9} = \frac{9}2(2*27 + (9 - 1)9)}

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The given parameters are:

\mathbf{a = 3}

\mathbf{d = 3}

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The sum of n terms of an AP is:

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Substitute known values

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\mathbf{d = 4}

\mathbf{n = 15}

The sum of n terms of an AP is:

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Substitute known values

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\mathbf{S_{15} = 480}

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The given parameters are:

\mathbf{a = 5}

\mathbf{d = 6}

\mathbf{n = 32}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{32} = \frac{32}2(2*5 + (32 - 1)6)}

\mathbf{S_{32} = 3136}

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The given parameters are:

\mathbf{a = 8}

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\mathbf{n = 27}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

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The given parameters are:

\mathbf{a = -7}

\mathbf{d = 2}

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The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

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