Question

3/7 line between................ (a)4/9,5/9 (b)43/99,4/9(c)42/99,4/9(d)41/99,41/9​

Answer 1

$\frac{3}{7}$ lies in the intervals given in options C and D.

Step-by-step explanation:

Option A.

If $\frac{3}{7}$ lies between $\frac{4}{9}$ and $\frac{5}{9}$

the  $(\frac{3}{7}-\frac{4}{9})>0$

Therefore, $\frac{3}{7}-\frac{4}{9}=\frac{27-28}{63}$

= $-\frac{1}{63}$ which is negative that means $\frac{4}{9}>\frac{3}{7}$

Therefore, $\frac{3}{7}$ doesn't lie between $\frac{4}{9}$ and $\frac{5}{9}$

Option B.

Similarly, $\frac{3}{7}-\frac{43}{99}=\frac{297-301}{63}$

= $-\frac{4}{693}$

Remainder is negative again therefore, $\frac{43}{99}>\frac{3}{7}$

and doesn't lie between $\frac{43}{99}$ and $\frac{4}{9}$

Option C.

In this option $\frac{3}{7}-\frac{42}{99}=\frac{297-294}{63}$

= $\frac{3}{63}$>0

Now $\frac{4}{9}-\frac{3}{7}=\frac{28-27}{63}$

= $\frac{1}{63}>0$

Therefore, $\frac{3}{7}$ lies between $\frac{42}{99}$ and $\frac{4}{9}$

Option D.

$\frac{3}{7}-\frac{41}{99}=\frac{297-287}{63}$

= $\frac{10}{63}$

$\frac{41}{9}-\frac{3}{7}=\frac{287-27}{63}$

= $\frac{260}{63}>0$

Therefore, $\frac{3}{7}$ lies between $\frac{41}{99}$ and $\frac{41}{9}$

Options C and D are the correct options.

Learn more about the fractions from

https://brainly.in/question/14155218

Answer 2

Answer:

42/99 MARK ME AS Brainiest  

Step-by-step explanation:

3/7=0.428571⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

(a) 49=0.444….59=0.555…

(b) 4399=0.434343…49=0.4444….

(c) 49=0.4444…4299=0.424242…

(d) 4199=0.0414141…4299=0.424242…