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3 years ago

If jon had 2,881 and tia had 4,163 how much more money tia has then jon

Mathematics

2 answers:

Sophie [7]3 years ago

4 0

Answer:

1282

Step-by-step explanation:

Sidana [21]3 years ago

3 0

Answer:

Tia has $1,282 more than Jon

Step-by-step explanation:

4,163-2881 = 1282

Tia has $1,282 more than Jon

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The price of a shoes as reduced from 60 to 45. by what percentage was the price of the shoes reduced?

sweet-ann [11.9K]

45 is 75% of 60 but if you’re looking for the difference subtract 75 from 100, so 25%

8 0

2 years ago

Calcula y grafica la siguiente fracción afín y cuadratica y=3×2

salantis [7]

Answer:

y = 6

Step-by-step explanation:

You just multiply the 3 and the 2 which equals 6

5 0

2 years ago

-(x+3) = 2x-6 what is x

Makovka662 [10]

Answer:

x=1

Step-by-step explanation:

So, first you make sure to multiply the parenthesis by the negative so then it would be,

-x-3=2x-6

Then you add the -3 from the right to the other side

so, -x=2x-3

Then you want all the X variables to one side so you subtract the 2x to both sides.

so, -3x=-3

divide both by -3

x=1

7 0

2 years ago

In a seminar of 1200 people, 65 percent were women, How many men were in the seminar?

Nookie1986 [14]

Answer:

420

Step-by-step explanation:

If there were 1200 people, and 65 percent were women, that means the rest of the crowd are men, which is 35 percent. By multiplying 35 percent by the overall population, 1200, (1200*35/100), you'll get 420 men.

4 0

2 years ago

Read 2 more answers

Activity 4: Performance Task

Nookie1986 [14]

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

The sum of multiplies of 6 between 8 and 70 is 390
The sum of multiplies of 5 between 12 and 92 is 840
The sum of multiplies of 3 between 1 and 50 is 408
The sum of multiplies of 11 between 10 and 122 is 726
The sum of multiplies of 9 between 25 and 100 is 567
The sum of the first 20 terms is 630
The sum of the first 15 terms is 480
The sum of the first 32 terms is 3136
The sum of the first 27 terms is -486
The sum of the first 51 terms is 2193

(a) Sum of multiples of 6, between 8 and 70

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

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

$\mathbf{S_{10} = 390}$

(b) Multiples of 5 between 12 and 92

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

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

$\mathbf{S_{16} = 840}$

(c) Multiples of 3 between 1 and 50

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

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

$\mathbf{S_{16} = 408}$

(d) Multiples of 11 between 10 and 122

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

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

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

(e) Multiples of 9 between 25 and 100

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

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

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

(f) Sum of first 20 terms

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

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

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

(f) Sum of first 15 terms

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

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

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

(g) Sum of first 32 terms

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

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

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

(g) Sum of first 27 terms

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

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

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

(h) Sum of first 51 terms

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

Read 2 more answers