1 22-322 32-223 22- 334 -22-125 12-526 12-207 70-128 61-409 100-2010 51- 31

(1) 22 + 32 = 2 times; 2 + 3 times; 3= 4 + 9 = 13 (2) 32 minus; 22 = 3 times; 3 minus; 2 times; 2 = 9 minus; 4 = 5 (3) 22 times; 3 ...

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Byju's Answer

Standard VII

Mathematics

Properties of Multiplication of Integers

1 22+322 32-2...

Question

(1) 2² + 3²

(2) 3² − 2²

(3) 2² × 3³

(4) (−2)² × (1)²

(5) 1² + 5²

(6) 1² + 2⁰

(7) 7⁰ + 1²

(8) 6¹ − 4⁰

(9) 10⁰ − 2⁰

(10) 5¹× 3¹

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Solution

(1) 2² + 3² = 2 × 2 + 3 × 3= 4 + 9 = 13

(2) 3² − 2² = 3 × 3 − 2 × 2 = 9 − 4 = 5

(3) 2² × 3³ = 2 × 2 × 3 × 3 × 3 = 4 × 27 = 108

(4) (−2)² × (1)² = (−2) ×(−2) × 1 × 1 = 4 × 1 = 4

(5) 1² + 5² = 1 × 1 + 5 × 5 = 1 + 25 = 26

(6) 1² + 2⁰ = 1 × 1 + 1 = 1 + 1 = 2

(7) 7⁰ + 1² =1 + 1 × 1 = 1 + 1 = 2

(8) 6¹ − 4⁰= 6 − 1 = 5

(9) 10⁰ − 2⁰ =1 − 1 = 0

(10) 5¹× 3¹ = 5 × 3 = 15

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Similar questions

(1) 4² + 2⁴− 3²

(2) 3³ − 2²+ 2⁰

(3) 5³ − 2² + 4⁰

(4) 6¹ + 5⁰ − 4¹

(5) (−2)² + 1² − 2²

Q. Simplify:
(i)

(3^{2} + 2^{2}) \times {(\frac{1}{2})}^{3}

(ii)

(3^{2} - 2^{2}) \times {(\frac{2}{3})}^{- 3}

(iii)

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

(iv)

(2^{2} + 3^{2} - 4^{2}) \div {(\frac{3}{2})}^{2}

Q. Evaluate:

lim n \to \infty \frac{1}{n^{4}} [1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + . . . . . . (1^{2} + 2^{2} + 3^{2} + . . . . n^{2})]

$1 + \frac{1^{2} + 2^{2}}{2!} + \frac{1^{2} + 2^{2} + 3^{2}}{3!} + \frac{1^{2} + 2^{2} + 3^{2} + 4^{2}}{4!} + . . . \infty = \frac{b}{6} e$ . Find $b$

1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + . . . .

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(1) 22 + 32 (2) 32 − 22 (3) 22 × 33 (4) (−2)2 × (1)2 (5) 12 + 52 (6) 12 + 20 (7) 70 + 12 (8) 61 − 40 (9) 100 − 20 (10) 51 × 31

(1) 2² + 3²

(2) 3² − 2²

(3) 2² × 3³

(4) (−2)² × (1)²

(5) 1² + 5²

(6) 1² + 2⁰

(7) 7⁰ + 1²

(8) 6¹ − 4⁰

(9) 10⁰ − 2⁰

(10) 5¹× 3¹