Question

multiply the sum of -3 upon 11and 5 upon 22 by the product of 4 upon 7 and -14 upon 9​

Answer 1

Answer:

Solution :-

Sum of :-

\frac{ - 3}{11} + \frac{5}{22}

11

−3

+

22

5

L.C.M of 11 , 22 is 22.

( \frac{ - 3}{11} \times \frac{2}{2} ) + ( \frac{5}{22} )(

11

−3

×

2

2

)+(

22

5

)

\frac{ - 6}{22} + \frac{5}{22}

22

−6

+

22

5

= \frac{ - 6 + 5}{22}=

22

−6+5

= \frac{ - 1}{22}=

22

−1

\sf\therefore Sum\: of \dfrac{-3}{11} , \dfrac{5}{22} = \dfrac{-1}{22}∴Sumof

11

−3

,

22

5

=

22

−1

Product of :-

\frac{4}{7} \times \frac{ - 14}{9}

7

4

×

9

−14

= \frac{4 \times - 14}{7 \times 9}=

7×9

4×−14

= \frac{ - 56}{63} = \frac{-8}{9}=

63

−56

=

9

−8

\sf\therefore Product\: of \dfrac{4}{7},\dfrac{-14}{9} = \dfrac{-8}{9}∴Productof

7

4

,

9

−14

=

9

−8

Product of :-

\frac{ - 1}{22} \times \frac{ - 8}{9}

22

−1

×

9

−8

= \frac{ - 1 \times - 8}{22 \times 9}=

22×9

−1×−8

= \frac{8}{198}=

198

8

= \sf\frac{4}{99}

99

4

Since ,

multiply \: the \: sum \: of \: \frac{ - 3}{11}and \frac{5}{22}by \: product \: of \: \frac{4}{7} and \frac{ - 14}{9}multiplythesumof

11

−3

and

22

5

byproductof

7

4

and

9

−14

= \sf\frac{4}{99}

99

4

Answer 2

Step-by-step explanation:

$-3 \div 11 + 5 \div 22(4 \div 7 \times - 14 \div 9)$

$- 6 + 5 \div 22(4 \times - 2 \div 9)$

$- 1 \div 22( - 8 \div 9)$

$- 1 \div 11 \times - 4 \div 9$

$\frac{4}{9}$

is answer