3 years ago

Arrange the following fraction In ascenending order 7/13,7/9,7/15,7/19

Mathematics

1 answer:

Tems11 [23]3 years ago

8 0

Answer:

$\frac{7}{19},\frac{7}{15},\frac{7}{13},\frac{7}{9}$

Step-by-step explanation:

Since the numerators are equal, arrange the denominators. The fraction with the biggest denominator is the smallest fraction.

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1/1×3 + 1/3×5 + ... + 1/47×49 HELP PLZ

Natalka [10]

Answer:

24/49

Step-by-step explanation:

Let's add the terms and see if there's a pattern

$\dfrac{1}{1\times 3}+\dfrac{1}{3\times 5}=\dfrac{5+1}{1\times 3\times 5}=\dfrac{2}{5}\quad\text{sum of 2 terms}\\\\\dfrac{2}{5}+\dfrac{1}{5\times 7}=\dfrac{14+1}{5\times7}=\dfrac{3}{7}\quad\text{sum of 3 terms}$

Suppose we say the sum of n terms is (n/(2n+1)), the next term in the series will be 1/((2n+1)(2n+3)) and adding that to the presumed sum gives ...

$\dfrac{n}{2n+1}+\dfrac{1}{(2n+1)(2n+3)}=\dfrac{n(2n+3)+1}{(2n+1)(2n+3)}=\dfrac{2n^2+3n+1}{(2n+1)(2n+3)}\\\\=\dfrac{(2n+1)(n+1)}{(2n+1)(2n+3)}=\dfrac{n+1}{2n+3}\text{ matches }\dfrac{(n+1)}{2(n+1)+1}$