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3 years ago

5

You work for the city recreation department, which is hosting a 5 kilometer run/walk next week. Your coworker measured the cours

e and found that it is only 4.75 kilometers long. He asks you to move the finish line forward 0.25 kilometers. Your tape measure is marked in feet and inches. What is the distance he has asked you to move the finish line, to the nearest foot?

2 answers:

dezoksy [38]3 years ago

5 0

Answer:

Step-by-step explanation:

Troyanec [42]3 years ago

5 0

Answer:820

Step-by-step explanation: 1 kilometer=3280.84 ft Divide that by 4.

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Activity 4: Performance Task

Nookie1986 [14]

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>
<em>The sum of multiplies of 9 between 25 and 100 is 567</em>
<em>The sum of the first 20 terms is 630</em>
<em>The sum of the first 15 terms is 480</em>
<em>The sum of the first 32 terms is 3136</em>
<em>The sum of the first 27 terms is -486</em>
<em>The sum of the first 51 terms is 2193</em>

<em />

<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

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

$\mathbf{S_{11} = 726}$

<u />

<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:

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

Substitute known values

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

$\mathbf{S_{9} = 567}$

<u>(f) Sum of first 20 terms</u>

The given parameters are:

$\mathbf{a = 3}$

$\mathbf{d = 3}$

$\mathbf{n = 20}$

The sum of n terms of an AP is:

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

Substitute known values

$\mathbf{S_{20} = \frac{20}2(2*3 + (20 - 1)3)}$

$\mathbf{S_{20} = 630}$

<u>(f) Sum of first 15 terms</u>

The given parameters are:

$\mathbf{a = 4}$

$\mathbf{d = 4}$

$\mathbf{n = 15}$

The sum of n terms of an AP is:

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

Substitute known values

$\mathbf{S_{15} = \frac{15}2(2*4 + (15 - 1)4)}$

$\mathbf{S_{15} = 480}$

<u>(g) Sum of first 32 terms</u>

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

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

$\mathbf{S_{32} = 3136}$

<u>(g) Sum of first 27 terms</u>

The given parameters are:

$\mathbf{a = 8}$

$\mathbf{d = -2}$

$\mathbf{n = 27}$

The sum of n terms of an AP is:

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

Substitute known values

$\mathbf{S_{27} = \frac{27}2(2*8 + (27 - 1)*-2)}$

$\mathbf{S_{27} = -486}$

<u>(h) Sum of first 51 terms</u>

The given parameters are:

$\mathbf{a = -7}$

$\mathbf{d = 2}$

$\mathbf{n = 51}$

The sum of n terms of an AP is:

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

Substitute known values

$\mathbf{S_{51} = \frac{51}2(2*-7 + (51 - 1)*2)}$

$\mathbf{S_{51} = 2193}$

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Apple juice is sold in 2.5l cartons and 350ml boxes. a 2.5l carton sells for $4.15 and six 350ml boxes sell for $4.50. calculate

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Answer:

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