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lyudmila [28]
3 years ago
10

How can I solve number 13 and 14?

Mathematics
2 answers:
Ede4ka [16]3 years ago
5 0
The top one can go by halves, and the bottom one can go by 1's. so for the top 1/2, 1, 1 1/2, 2, and 2 1/2 while the bottom would be 0 1 2 3 4... and so on
SSSSS [86.1K]3 years ago
3 0
13. \frac{1}{2}, 1, 1\frac{1}{2}, 2, 2\frac{1}{2}
14. \frac{1}{5},  \frac{2}{5},  \frac{3}{5},  \frac{4}{5}, 1, 1 \frac{1}{5}, 1 \frac{2}{5}, 1 \frac{3}{5}
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Prove the sum of three consecutive is divisible by three, that the sum of 5 consecutive integers is divisible by 5, but the sum
iren2701 [21]

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Proved

Step-by-step explanation:

Solving (a):

Let the numbers be: x, x + 1, x + 2.

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Sum = x + x + 1 + x + 2

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Sum = x + x + x + 1 + 2

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Result = \frac{Sum}{3}

Result = \frac{3x+3}{3}

Result = \frac{3(x+1)}{3}

Result = x + 1

<em>Hence, this is true because there is no fractional part after the division</em>

<em></em>

Solving (b):

Let the numbers be: x, x + 1, x + 2,x+3,x+4

Their sum is:

Sum = x + x + 1 + x + 2+x + 3 + x + 4

Collect Like Terms

Sum = x + x + x + x + x+1  + 2 + 3 + 4

Sum = 5x+10

Divide sum by 5.

Result = \frac{5x + 10}{5}

Result = \frac{5(x + 2)}{5}

Result = x + 2

<em>Hence, this is true because there is no fractional part after the division</em>

<em></em>

Solving (c):

Let the numbers be: x, x + 1, x + 2,x+3

Their sum is:

Sum = x + x + 1 + x + 2+x + 3

Collect Like Terms

Sum = x + x + x + x + 1  + 2 + 3

Sum = 4x+6

Divide sum by 4.

Result = \frac{4x + 6}{4}

Split

Result = \frac{4x}{4} + \frac{6}{4}

Result = x + 1.5

<em>The 1.5 means that the sum can not be divisible by 4</em>

5 0
3 years ago
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