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marusya05 [52]
3 years ago
15

Find two consecutive odd numbers such that the sum of three-sevenths of the first number and one-third of the second number is e

qual to thirty-eight.
29 and 31
39 and 41
49 and 51
Mathematics
2 answers:
Nastasia [14]3 years ago
7 0
2 consecutive odd numbers....x and x + 2

3/7x + 1/3(x + 2) = 38
3/7x + 1/3x + 2/3 = 38
3/7x + 1/3x = 38 - 2/3....multiply by common denominator of 21
9x + 7x = 798 -14
16x = 784
x = 784 / 16
x = 49

x + 2 = 49 + 2 = 51

ur numbers are 49 and 51

ch4aika [34]3 years ago
4 0
I think the answer is 29 and 31

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h (-√(2x + h) - √2x)

= h>0 <u>4</u><u>x</u><u> </u><u>+</u><u> </u><u>2</u><u>h</u><u> </u><u>-</u><u> </u><u>4</u><u>x</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

h(-√(2x + h) -√2x)

= h>0 <u>2</u><u>h</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

h(-√(2x+h) - √2x)

= h>0 <u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>2</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

-√(2x+h) - √2x

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