1-6-2-1-3-3- 32 -2 MARKS-

Concept: 1 Mark Application: 1 Mark (1/6)^ 2 (1/3)^ 3/3^2 = {36 27}×1/9 = 9 ×1/9 = 1 ...

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Byju's Answer

Standard VIII

Mathematics

Division of a Polynomial by a Polynomial

1/6-2-1/3-3÷ ...

Question

${(\frac{1}{6})^{- 2} - (\frac{1}{3})^{- 3}} \div 3^{2}$ [2 MARKS]

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Solution

Concept: 1 Mark
Application: 1 Mark

$\frac{(\frac{1}{6})^{- 2} - (\frac{1}{3})^{- 3}}{3^{2}}$

$= {36 - 27} \times \frac{1}{9}$

$= 9 \times \frac{1}{9}$

$= 1$

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Similar questions

Q. Simplify

{[{(\frac{2}{3})}^{2}]}^{3} \times {(\frac{1}{3})}^{- 2} \times 3^{- 1} \times \frac{1}{6}

Q. Simplify:

{[{(\frac{2}{3})}^{2}]}^{3} \times {(\frac{2}{3})}^{- 4} \times 3^{- 1} \times \frac{1}{6}

Evaluate :

$(i) (3^{- 1} \times 9^{- 1}) \div 3^{- 2} (i i) (3^{- 1} \times 4^{- 1}) \div 6^{- 1} (i i i) (2^{- 1} + 3^{- 1})^{3} (i v) (3^{- 1} \div 4^{- 1})^{2} (v) (2^{2} + 3^{2}) \times {(\frac{1}{2})}^{2} (v i) (5^{2} - 3^{2}) \times {(\frac{2}{3})}^{- 3} (v i i) [{(\frac{1}{4})}^{- 3} - {(\frac{1}{3})}^{- 3}] \div {(\frac{1}{6})}^{- 3} (v i i i) {[{(- \frac{3}{4})}^{- 2}]}^{2} (i x) {{(\frac{3}{5})}^{- 2}}^{- 2} (x) (5^{- 1} \times 3^{- 1}) \div 6^{- 1}$

Simplify:

$(i) {{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}} \div {(\frac{1}{4})}^{- 3} (i i) (3^{2} - 2^{2}) \times {(\frac{2}{3})}^{- 3} (i i i) {(\frac{1}{2}) \times (- 4)^{- 1}}^{- 1} (i v) {[{{(\frac{- 1}{4})}^{2}}^{- 2}]}^{- 1} (v) {{(\frac{2}{3})}^{2}}^{3} \times {(\frac{1}{3})}^{- 4} \times 3^{- 1} \times 6^{- 1}$

(a). Choose the largest and smallest integer among the following:

$32, - 50, - 75, 120, - 365, 175$

(b). $S i m p l i f y : (- 3) + (- 2) + (- 5) + 7$

(c). $- 6 \div 3 = ?$

(d). $3 \times 6 \times 2 \times (- 40) = ?$ [4 MARKS]

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{(16)−2−(13)−3}÷32 [2 MARKS]

Concept: 1 Mark Application: 1 Mark (16)−2−(13)−332 ={36−27}×19 =9×19 =1

${(\frac{1}{6})^{- 2} - (\frac{1}{3})^{- 3}} \div 3^{2}$ [2 MARKS]

Concept: 1 Mark
Application: 1 Mark

$\frac{(\frac{1}{6})^{- 2} - (\frac{1}{3})^{- 3}}{3^{2}}$

$= {36 - 27} \times \frac{1}{9}$

$= 9 \times \frac{1}{9}$

$= 1$