Show that- -89-1-5- -89- -711-89-1-5-711

We have, #10; #10; ( 89× 15)+( #10; 89× 710) #10; #10; ⇒( 845)+( 5690 #10;) #10; #10; ⇒( 1690)+( 5690 #10;) #10; #10; ...

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Byju's Answer

Standard XII

Mathematics

Factorial

Show that: ...

Question

Show that: $(- \frac{8}{9} \times \frac{1}{- 5}) + (- \frac{8}{9} \times - \frac{- 7}{11}) = - \frac{8}{9} \times (\frac{1}{- 5} + \frac{- 7}{11})$

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Solution

We have,

$(- \frac{8}{9} \times - \frac{1}{5}) + (- \frac{8}{9} \times - \frac{7}{10})$

$\Rightarrow (\frac{8}{45}) + (\frac{56}{90})$

$\Rightarrow (\frac{16}{90}) + (\frac{56}{90})$

$\Rightarrow \frac{72}{90}$

$\Rightarrow \frac{4}{5}$

Hence, this is the answer.

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Similar questions

Fill in the blanks:

$(i) - 4 \times \frac{7}{9} = \frac{7}{9} \times$ ___

$(i i) \frac{5}{11} \times \frac{- 3}{8} = \frac{- 3}{8} \times$ ___

$(i i i) \frac{1}{2} \times (\frac{3}{4} + \frac{- 5}{12}) = \frac{1}{2} \times$ ____ $+ (\times) \frac{- 5}{12}$

$(i v) \frac{- 4}{5} \times \frac{5}{7} \times$ ___ $= (\frac{- 4}{5} \times \frac{5}{7}) \times \frac{- 8}{9}$

Q. Evaluate:

(i) \frac{3}{7} + \frac{- 4}{9} - \frac{- 11}{7} - \frac{7}{9}

(i i) \frac{2}{3} + \frac{- 4}{5} - \frac{1}{3} - \frac{2}{5}

(i i i) \frac{4}{7} - \frac{- 8}{9} - \frac{- 13}{7} + \frac{17}{9}

source $(i) \frac{8}{9} + \frac{- 11}{6} (i i) 3 + \frac{5}{- 7} (i i i) \frac{1}{- 12} a n d \frac{2}{- 15} (i v) \frac{- 8}{19} + \frac{- 4}{57} (v) \frac{7}{9} + \frac{3}{- 4} (v i) \frac{5}{26} + \frac{11}{- 39} (v i i) \frac{- 16}{9} + \frac{- 5}{12} (v i i i) \frac{- 13}{8} + \frac{5}{36} (i x) 0 + \frac{- 3}{5} (x) 1 + \frac{- 4}{5} (x i) \frac{- 5}{16} + \frac{7}{24}$

Q. In a shop, different sizes of shoes sold in a week are 7, 9, 8, 6, 7, 10, 6, 7, 8, 9, 11, 5, 6, 7, 8, 5, 9, 10, 9, 11, 12, 8, 3, 3, 5, 4, 6, 7, 4, 6, 7, 8, 8, 6, 9. What is the difference of the number between shoes sold of size 6 and size 7?

Q. Twelve entries were submitted in a flower show competition. They were ranked by two judges as under:

Entries	:	1	2	3	4	5	6	7	8	9	10	11	12
Judge A	:	7	8	2	1	9	3	12	11	4	10	6	5
Judge B	:	6	4	1	3	11	2	12	10	5	9	7	8

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Show that: (−89×1−5)+(−89×−−711)=−89×(1−5+−711)

We have, (−89×−15)+(−89×−710) ⇒(845)+(5690) ⇒(1690)+(5690) ⇒7290 ⇒45 Hence, this is the answer.

Show that: $(- \frac{8}{9} \times \frac{1}{- 5}) + (- \frac{8}{9} \times - \frac{- 7}{11}) = - \frac{8}{9} \times (\frac{1}{- 5} + \frac{- 7}{11})$

We have,

$(- \frac{8}{9} \times - \frac{1}{5}) + (- \frac{8}{9} \times - \frac{7}{10})$

$\Rightarrow (\frac{8}{45}) + (\frac{56}{90})$

$\Rightarrow (\frac{16}{90}) + (\frac{56}{90})$

$\Rightarrow \frac{72}{90}$

$\Rightarrow \frac{4}{5}$

Hence, this is the answer.