Exponents Test - 9

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

Then,

540 = 5.4 × 10²

[{(-1/2)²}^-2]^-1= [{(-1²/2²)}^-2]^-1

= [{1/4}^-2]^-1

= [{4} ²]^-1

= [16^]-1

= [1/16]

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

Then,

Diameter of Earth = 13000000m

= (1.3 × 10⁷)m (in standard form)

We know that,

= (-1/2)^-6=(-2/1)⁶ … [∵ (a/b)^-n =(b/a) ⁿ]

= (-2)⁶

= 64

(2/3)^-5=(3/2)⁵

= (3⁵/2⁵)

= (243/32)

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

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

Then,

6500000 = 6.5 × 10⁶

We know that,

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

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

Now add,

= {(1/6) + (2/3)}^-1

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

= {5/6}^-1

= {6/5}

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}

We know that,

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

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

Now subtract,

= {(4/3) – (4/1)} ^-1

= {(4-12)/3}^-1 …[LCM of 3 and 1 is 3]

= {-8/3}^-1

= {-3/8}

(5/6)⁰ =1

By definition, we have a⁰=1 for every integer.

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

Then,

Distance between Earth and Moon = 390000000 m

= (3.9 × 10⁸)m (in standard form)

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

We know that,

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

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

= (1/4)^-2 =(4/1)² … [∵ (a/b)^-n =(b/a) ⁿ]

Now add,

= (2)²+ (3)²+(4)²

= 4 + 9 + 16

= 29

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

Then,

Number of stars in a galaxy = 500000000000

= (5 × 10¹¹)(in standard form)