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3 years ago

5

Greg works at a video store. He earns $7 an hour, plus $5 for each membership he sells?

2 answers:

Katarina [22]3 years ago

8 0

A=7h+5m

...............

lutik1710 [3]3 years ago

3 0

Here, <span>a=amount earned
h=hours worked in one week
m= number of memberships sold in one week

Equation would be: a = 7h + 5m
[ Put the value of unity with their corresponding variables ]

Hope this helps! </span>

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Which of the following is the y-intercept of the line with the equation 2y=6x+4

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Factorise<br> 24p2 + pq - 23q²

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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

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

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

Read more about arithmetic progressions at:

brainly.com/question/13989292

4 0

2 years ago

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Can someone please help me with this?

Charra [1.4K]

Im pretty sure its 60

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3 years ago

(6x^3+9x-8)+(5x-9x^2+7

irinina [24]

Add like terms
I'm assuming that you forgot to put the parenthasees at the end

so exg
4x+8x=12x
2x^2+3x^2=5x^2
3x^2+3x^3=3x^2+3x^3
group like tems
6x^3+9x-8+5x-9x^2+7
6x^3-9x^2+14x-1

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3 years ago

Read 2 more answers