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(1/3)^-1

We have:

(1/3)^-1 = (-6/1)^-1

= (3/1)¹ … [∵ (a/b)^-n = (b/a) ⁿ]

= 3

 538

A given number is said to be in standard form if it can be expressed as k × 10ⁿ, where k is a real number such that 1 ≤ k < 10 and n is a positive integer.

Then,

538 = 5.38 × 10²

10^2y = 25 

= 10^y = x 

= x² = 5² 

= x = 5 

= 1/x = 10^-y 

= 1/5

(23/25)°

= (23/25)° = 1

Because, by definition, we have a° = 1 for every integer.

{(3/2)^-1 ÷ (-2/5)^-1}

We know that,

= (3/2)^-1= (2/3)¹ … [∵ (a/b)^-n = (b/a) ⁿ]

= (-2/5)^-1 = (-5/2)¹ … [∵ (a/b)^-n = (b/a) ⁿ]

Now divide,

= {(2/3) ÷ (-5/2)}

= {(2/3) × (-2/5}

= {(2×-2) / (3×5)

= {-4/15}

{(4/3)^-1 – (1/4)^-1}^-1

We know that,

= (4/3)^-1= (3/4)¹ … [∵ (a/b)^-n = (b/a) ⁿ]

= (1/4)^-1 = (4/1)¹ … [∵ (a/b)^-n = (b/a) ⁿ]

Now subtract,

= {(3/4) – (4/1)}^-1

= {(3-16)/4}^-1 … [LCM of 4 and 1 is 4]

= {-13/4}^-1

= {-4/13}

(5^-1– 7^-1)^-1

We know that,

= (5)^-1= (1/5)¹ … [∵ (a/b)^-n = (b/a) ⁿ]

= (7)^-1 = (1/7)¹ … [∵ (a/b)^-n = (b/a) ⁿ]

Now subtract,

= {(1/5) – (1/7)}^-1

= {(7-5)/35}^-1 … [LCM of 5 and 7 is 35]

= {2/35}^-1

= {35/2}

Correct option is A) (- 5)¹⁶

(-5) x (-5) x (-5) x.... x(-5) (16 times) = (-5)¹⁶

\(\sqrt[3]{125\times27}\)

= (5³ × 3³) ^1/3 

= 5 × 3 

= 15

(256)^0.16 × (256)^0.09 

= (256) ^{0.16 + 0.09} 

= (256) ^0.25 

= ^{\(4^4\times\frac{1}{4}\)}

= 4

(i) Given (3/7)²

(3/7)² = (3/7) x (3/7)

= (9/49)

(ii) Given (7/9)³

(7/9)³ = (7/9) x (7/9) x (7/9)

= (343/729)

(iii) Given (-2/3)⁴

(-2/3)⁴ = (-2/3) x (-2/3) x (-2/3) x (-2/3)

= ((16/81)

Correct option is A) 12¹²

12¹⁶ x 12^-4 = 12^16-4 = 12¹²

(c) \(17\frac1{16}\)

Given exp. = 1 + \((8^2)^{-\frac12}+(2^5)^{\frac45}-(2^5)^{-\frac45}\)

= 1 + 8^–1 + 2⁴ – 2^–4

= \(1+ \frac18+16-\frac1{16}=\frac{16+2+256-1}{16}\)

^{= \(\frac{273}{16} = 17\frac1{16}\)}

{2-3(2-3)³} ³ 

= {2 – 3 (-1)³} ³ 

= {2 + 3}³ 

= 5³ 

= 125

Correct option is C) x²

\(\frac{x^{17}}{x^{15}}=\frac{x^{15}\times x^2}{x^{15}}\) = x²

Correct option is B) 5³

125 = 5 x 5 x 5 = 5³

\((\frac{2}{5})^{-3}=(\frac{5}{2})^3=\frac{5^3}{2^3}=\frac{125}{8}\)

(-2)^-5 = \(\frac{1}{(-2)^5}=\frac{1}{-32}=\frac{1\times(-1)}{-32\times(-1)}=\frac{-1}{32}\)

(-3)^-4 = \(\frac{1}{(-3)^4}=\frac{1}{(-1)^4\times(3)^4}=\frac{1}{(3)^4}=\frac{1}{81}\)

(d) \(1.\overline4\)

Given exp. = \(\frac{\big((2.5)^2\big)^{\frac12}\times\big((0.12)^2\big)^{\frac12}+1}{\big((0.3)^3\big)^{\frac13}\times\big((3^4)\big)^{\frac14}}\)

= \(\frac{2.5\times0.12+1}{0.3\times3}\) 

= \(\frac{0.3+1}{0.9} = \frac{1.3}{0.9}=\frac{13}{9}\) = \(1.\overline4\)

Correct option is D) mx

x + x + x +..... + x (m times) = x(1 + 1 + 1 + ....+ 1)(m times)

 = x.m = mx

(i) Given 5²ⁿ x 5³ = 5¹¹

= 5²ⁿ⁺³ = 5¹¹

On equating the coefficients, we get

2n + 3 = 11

⇒2n = 11- 3

⇒2n = 8

⇒ n = (8/2)

⇒ n = 4

(ii)  Given 9 x 3ⁿ = 3⁷

= (3)² x 3ⁿ = 3⁷

= (3)²⁺ⁿ = 3⁷

On equating the coefficients, we get

2 + n = 7

⇒ n = 7 – 2  

n = 5

(iii) Given 8 x 2ⁿ⁺² = 32

= (2)³ x 2ⁿ⁺² = (2)⁵      [since 2³ = 8 and 2⁵ = 32]
= (2)³⁺ⁿ⁺² = (2)⁵ 

On equating the coefficients, we get

3 + n + 2 = 5

⇒ n + 5 = 5

⇒ n = 5 -5

⇒ n = 0

(c) (1/16)

We know that,

= (2)^-1= (1/2)¹ … [∵ (a/b)^-n = (b/a) ⁿ]

= (4)^-1 = (1/4)¹ … [∵ (a/b)^-n = (b/a) ⁿ]

Now subtract,

= {(1/2) – (1/4)} ²

= {(2-1)/4}² … [LCM of 2 and 4 is 4]

= {1/4}²

= {1²/4²}

= {1/16}

(c) 15

We know that,

= (5)^-1= (1 /5)¹ … [∵ (a/b)^-n = (b/a) ⁿ]

= (3)^-1 = (1/3)¹ … [∵ (a/b)^-n = (b/a) ⁿ]

Now multiply,

= {(1/5) × (1/3)}^-1

= {(1×1)/ (5×3)}^-1

= {1/15}^-1

= {15/1}

= 15

(D) 24

We know that,

= (6)^-1= (1/6)¹ … [∵ (a/b)^-n = (b/a) ⁿ]

= (8)^-1 = (1/8)¹ … [∵ (a/b)^-n = (b/a) ⁿ]

• Now subtract,
• = {(1/6) – (1/8)}^-1

= {(4-3)/24}^-1 … [LCM of 6 and 8 is 24]

= {1/24}^-1

= {24/1}

= 24

4^x – 4^{x – 1} = 24

Let 4x = y

y - \(\frac{y}4\)= 24

4y – y = 96

y = 32

4^x = 32

2^2x = 2⁵ 

(2x)^x = \((2\times\frac{5}{2})^{\frac{5}{2}}\)

(5)^5/2 = \(25\sqrt5\)

Distance from earth to sun = 149600000000 m 

In standard form we have, 

149600000000 = 1496 × 100000000 

= 1.496 × 1000 × 100000000 

= 1.496 × 10³ × 10⁸ = 1.496 × 10¹¹ m.

Given, 

Mass of the earth = 5.97 × 10²⁴ kg 

Mass of the moon = 7.35 × 10²² kg 

Now, 

Mass of the earth = 5.97 × 10²⁴ = 5.97 × 10⁽²⁺²²⁾ = 5.97 × 10² × 10²² = 597 × 10²² 

So, 

We can also Wright the mass of the earth as 597 × 10²² kg 

Sum of the masses of the earth and the moon; 

= (597 × 10²²) + (7.35 × 10²²) = (597+7.35) × 10²² = 604.35 × 10²² kg 

= 6.0435 × 100 × 10²² = 6.0435 × 10² × 10²² = 6.0435 × 10⁽²⁺²²⁾ = 6.0435 × 10²⁴ kg

(i) 0.0006 = \(\frac{6}{10^4}=6\,\times10^{-4}\)

(ii) 0.00000083 = \(\frac{8^3}{10^3}\) = \(\frac{8.3\times10}{10^3}\) = 8.3 x 10^(1-8) = 8.3 x 10^-7

(iii) 0.0000000534 = \(\frac{534}{10^{10}}\) = \(\frac{5.34\times10^2}{10^{10}}\) = 5.34 x 10^(2-10) = 5.34 x 10^-8

(iv) 0.0027 = \(\frac{27}{10^4}\) = \(\frac{27\times10}{10^4}\) = 2.7 x 10^(1-4) = 2.7 x 10^-3

(v) 0.00000165 = \(\frac{165}{10^3}=\frac{1.65\times10^2}{10^3}\) = 1.65 x 10^(2-8) = 1.65 x 10^-6

(vi) 0.00000000689 = \(\frac{689}{10^{11}}=\frac{6.89\times10^2}{10^{11}}\) = 6.89 x 10^(2-11) = 6.89 x 10^-9

Thickness of one book = 20 mm 

Thickness of 5 books = 20 x 5 = 100 mm 

Thickness of one paper = 0.016 mm 

Thickness of 5 papers = 0.016 x 5 = 0.08 mm 

Total thickness of a stack = 100 + 0.08 = 100.08 mm 

= 100.08 x 10²/10² = 1.0008 x 10² mm

1 micron = \(\frac{1}{1000000}m=\) 1 × 10^-6 m.

i) 543.67

= (5 × 100) + (4 × 10) + (3 × 10⁰) + (6 / 10) + (7 / 10²)

= (5 × 10²) + (4 × 10) + (3 × 10⁰) + (6 × 10^-1) + (7 × 10^-2)   [∵ aⁿ = a^-n]

ii) 7054.243

= (7 × 1000) + (0 × 100) + (5 × 10) + (4 × 10⁰) + (2 / 10) + (4 / 100) + (3 / 1000)

= (7 × 10³) + (0 × 10²) + (5 × 10¹) + (4 × 10⁰) + (2 × 10^-1) + (4 × 10^-2) + (3 × 10^-3)

iii) 6540.305

= (6 × 1000) + (5 × 100) + (4 × 10) + (0 × 10⁰) + (3 / 10) + (0 / 100) + (5 / 1000)

= (6 × 10³) + (5 × 10²) + (4 × 10¹) + (0 × 10⁰) + (3 × 10^-1) + (0 × 10^-2) + (5 × 10^-3)

iv) 6523.450

= (6 × 1000) + (5 × 100) + (2 × 10) + (3 × 10⁰) + (4 / 10) + (5 / 100) + (0 / 1000)

= (6 × 10³) + (5 × 10²) + (2 × 10¹) + (3 × 10⁰) + (4 × 10^-1) + (5 × 10^-2) + (0 × 10^-3)

(i) 0.0000456

= 456 / 10000000 

= 456 × 10^-7

(ii) 0.000000529

= 529 / 1000000000 

= 529 × 10⁹

(iii) 0.0000000085

= 85 / 10000000000 

= 85 × 10¹⁰

(iv) 6020000000

= 602 × 10000000 

= 602 × 10⁷

(v) 35400000000

= 354 × 100000000 

= 354 × 10⁸

vi) 0.000437 × 10⁴

= 437 / 1000000 × 10⁴

= 437 × 10^-6 × 10⁴

= 437 × 10^(-6)+4

= 437 × 10^-2

(i) 4.37 × 10⁵

= 4.37 × 100000

= 437000

(ii) 5.8 × 10⁷

= 5.8 × 10000000

(iii) 32.5 × 10^-4

= 32.5 / 10⁴ = 32.5 / 10000

= 0.00325

(iv) 3.71529 × 10⁷

= 3.71529 × 10000000

= 37152900

(v) 3789 × 10^-5

= 3789 / 10⁵ = 3789 / 100000

= 0.03789

(vi) 24.36 × 10^-3

= 24.36 / 10³ = 24.36 / 1000

= 0.02436

(i) 6 × 6 × 6 × 6

= 6⁴

(ii) t × t = t²

(iii) b × b × b × b

= b⁴

(iv) 5 × 5 × 7 × 7 × 7

= 5² × 7³

(v) 2 × 2 × a × a

= 2² × a²

(vi) a × a × a × c × c × c × c × d

= a³ × c⁴ × d

(i) 2⁶

= 2 × 2 × 2 × 2 × 2 × 2 = 64

(ii) 9³

= 9 × 9 × 9 = 729

(iii) 11²

= 11 × 11 = 121

(iv) 5⁴

= 5 × 5 × 5 × 5 = 625.

(-8) ×(-8) × (-8) × (-8) × (-8)

The product of rational number multiplied several times by itself can be expressed in the power notations as,

(-8)⁵

(-1/6) ×(-1/6) × (-1/6)

The product of rational number multiplied several times by itself can be expressed in the power notations as,

(-1/6)³

(-4/3) ×(-4/3) × (-4/3) × (-4/3) × (-4/3)

The product of rational number multiplied several times by itself can be expressed in the power notations as,

(-4/3)⁵

(i) Given 7 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰

 = 7 x 10000 + 6 x 1000 + 0 x 100 + 4 x 10 + 5 x 1

= 70000 + 6000 + 0 + 40 + 5

= 76045

(ii) Given 5 × 10⁵ + 4 × 10⁴ + 2 × 10³ + 3 × 10⁰

= 5 x 100000 + 4 x 10000 + 2 x 1000 + 3 x 1

= 500000 + 40000 + 2000 + 3

= 542003

(iii) Given 9 × 10⁵ + 5 × 10² + 3 × 10¹

= 9 x 100000 + 5 x 100 + 3 x 10

= 900000 + 500 + 30

= 900530

(iv) Given 3 × 10⁴ + 4 × 10² + 5 × 10⁰

= 3 x 10000 + 4 x 100 + 5 x 1

= 30000 + 400 + 5

= 30405

(5/7) ×(5/7) × (5/7) × (5/7)

The product of rational number multiplied several times by itself can be expressed in the power notations as,

(5/7)⁴

(i) 279404

= 2 × 100000 + 7 × 10000 + 9 × 1000 + 4 × 100 + 0 × 10 + 4 × 1

= 2× 10⁵ + 7× 10⁴ + 9 × 10³ + 4 × 10² + 0 × 10¹ + 4 × 10⁰

(ii) 3006194

= 3 × 1000000 + 0 × 100000 + 0 × 10000 + 6 × 1000 + 1 × 100 + 9 × 10 + 4 × 1

= 3 × 10⁶ + 0 × 10⁵ + 0 × 10⁴ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 4 × 10⁰

(iii) 2806196

= 2 × 1000000 + 8 × 100000 + 0 × 10000 + 6 × 1000 + 1 × 100 + 9 × 10¹ + 6 × 10⁰

= 2 × 10⁶ + 8 × 10⁵ + 0 × 10⁴ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 6 × 10⁰

(iv) 120719

= 1 × 100000 + 2 × 10000 + 0 × 1000 + 7 × 100 + 1 × 10 + 9 × 1

= 1 × 10⁵ + 2 × 10⁴ + 0 × 10³ + 7 × 10² + 1 × 10¹ + 9 × 10⁰

(v) 20068

= 2 × 10000 + 0 × 1000 + 0 × 100 + 6 × 10 + 8 × 1

= 2 × 10⁴ + 0 × 10³ + 0 × 10² + 6 × 10¹ + 8 × 10⁰

Correct option is A) 1.2345 × 10⁴

Correct option is C) 1.353 × 10⁹ c.c. kms