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True.

We know that, 1 hour = 60 minutes

1 minutes = 60 seconds

Therefore, 1 hour = 60 × 60 = 3600 seconds

False.

(-3)⁴ = -3 × -3 × -3 × -3 = 81

So, 81 ≠ – 12

As we know that,

⇒ a⁰ = 1

∴ 5⁰ = 1

False

1⁰ = 1

0¹ = 0

So, 1 × 0 = 0

(c) (x ÷ y)⁴

(By law of exponent: (a)^m÷(b)^m = (a÷b)^m)

The cells of a bacteria double itself every hour = 1 + 1 = 2 = 2¹

Total number of cell in 8 hours = 2^{1 x 8} = 2⁸

Given that, cells in human body = 1 trillon

1 trillon = 1000000000000

Standard form = 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10

Using the law of exponent, a^m x aⁿ = a^m+n

= 10¹²

∴ The standard form is 10¹²

We know that,

a ^{– b} = 1/a^b

∴ 5⁵× 5 ^{– 5} = 5⁵× 1/5⁵

⇒ 5⁵× 5 ^{– 5} = 1

For standard form, 0.000005678 

= 0.5678 x 10^-5

= 5.678 x 10^-5 x 10^-1

= 5.678 x 10^-6 

Hence, 5.678 x 10^-6 is the standard form of 0.000005678.

From the laws of exponents , we know that

a^m × aⁿ = a^(m+n)

And , a^m ÷ aⁿ = a^(m-n)

So ,we have , (2⁵ ÷ 2⁸) = 2^(5-8) = 2^-3

Now , 2^-3 × 2^–7= 2^(-3)+(-7) = 2^-10

∴ (2⁵ ÷ 2⁸) × 2^–7= 2^-10 = 1/2¹⁰ = 1/1024

∴ (2⁵ ÷ 2⁸) × 2^–7 = 1/1024

Given that, to find the product of 3.2 × 10⁶ and 4.1 × 10^–1

Product of 3.2 × 10⁶ and 4.1 × 10^–1 = (3.2 × 10⁶)( 4.1 × 10^–1)

= (3.2 × 4.1) × 10⁶ × 10^-1

Using the law of exponent, a^m x aⁿ = a^m+n

⇒ 13.12 × 10⁵

= 1.312 × 10⁵ × 10¹

= 1.312 × 10⁶

∴ The product of 3.2 × 10⁶ and 4.1 × 10^–1 in the standard form is 1.312 × 10⁶.

(i) 172 

= 1 x 100 +7 x 10 + 2 x 1

= 1 x 10² +7 x 10¹ + 2 x 10⁰

(ii) 5643 

= 5 x 1000 +6 x 100 +4 x 10 + 3 x 1

= 5 x 10³ + 6 x 10² + 4 x 10¹ + 3 x 10⁰

The distance from the sun to Pluto = 59,1,30,00,000 m

∴ standard form of 59,1,30,00,000 = 5913 × 10⁶

= 5.913 × 10³ × 10⁶

Using the law of exponent, a^m x aⁿ = a^m+n

= 5.913 × 10⁹

∴ The distance from the sun to Pluto is 5.913 × 10⁹

The distance travelled by birds = 15,000 km

∵ 1 km = 1000 m

∴ 15,000 km = 15000 × 1000 m

= 15000000 m

= 15 × 10⁶ m

= 1.5 × 10⁷ m

∴ The distance in metres is 1.5 × 10⁷ m

39,00,00,000 can be written as

39,00,00,000 = 39 × 10 × 10 × 10 × 10 × 10 × 10 × 10

= 39 × 10⁷

= 3.9 × 10¹ × 10⁷

Using the law of exponent, a^m x aⁿ = a^m+n

⇒ 39,00,00,000 = 3.9 × 10⁸

∴ The standard form of 39,00,00,000 is 3.9 × 10⁸

3⁸ = 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3

= 9 x 9 x 9 x 9

= 81 x 81 = 6561

and 8³ = 8 x 8 x 8 = 512

We know : 6561 > 512 or 3⁸ > 8³

∴3⁸ is greater than 8³.

a³b² = a³x b² = (a x a x a) x (b x b)

a²b³ = a² x a³ = (a x a ) x (b x b x b)

b²a³ = b² x a³ = (b x b) x (a x a x a)

b³a² = b³ x a² = (b x b x b) x (a x a)

In descending order, the numbers are arranged from largest to smallest. We have, 2^{2 + 3} = 2⁵=2 x 2 x 2 x 2 x 2 = 32.

(2²)³ => 2⁶= 2 x 2 x 2 x 2 x 2 x 2=64.

2 x 2² = 2¹⁺² => 2³ = 8.

3⁵/3²  = 3^5-2 => 3³ = 27.

3² x 3⁰ = 3²⁺⁰ => 3² = 9.

2³ x 5² =2 x 2 x 2 x 5 x 5 => 8 x 25 = 200.

Thus, the required descending order will be
(2² x 5²) > (2²)³  > 2²⁺³  > 3⁵/3²  > (3² x 3⁰ )  > (2 x 2²).

(i) 4³ = 4 x 4 x 4 = 64 and 3⁴ = 3 x 3 x 3 x 3= 81
Clearly 81 > 64
∴ 3⁴ is greater.

(ii) 5³ = 5 x 5 x 5 = 125
and 3⁵ = 3 x 3 x 3 x 3 x 3 = 243
Clearly 243 > 125
∴ 3⁵ is a greater.

(i) 3² x 3⁴ x 3⁸ 

= 3^{2 + 4 + 8} 

= 3¹⁴

(ii) (5²)³÷ 5³

= (5)⁶ ÷ 5³ 

= 5⁶/5³ 

= 5^{6 – 3} 

= 5³

1. Area of Kalahari, South Africa = 932,400 = 932400.00 [∴ standard form a x 10^k]

= 9324 x 10² 

= 9.324 x 10³ x 10² 

= 9.324 x 10⁶

2. Area of Thar, India = 199,430 

= 199430.00 

= 19943 x 10¹ 

= 1.9943 x 10⁴ x 10¹ 

= 1.9943 x 10⁵

3. Area of Gibson, Australia = 155,400

= 155400.00

= 1554 x 10² 

= 1.554 x 10³ x 10² 

= 1.554 x 10⁵

4. Area of Great Victoria, Australia = 647,500

= 647500.00

= 6475 x 10² 

= 6.475 x 10³ x 10⁵ 

= 6475 x 10⁵

5. Area of Sahara, North-Africa = 8,598,800

= 8598800.00

= 85988 x 10² 

= 8.5988 x 10⁴ x 10² 

= 85988 x 10⁶

Two numbers written in scientific notation can be compared. The number with the larger power of 10 is greater than the number with the smaller power of 10. If the powers of ten are the same, then the number with larger factor is the larger number.

Hence, required ascending order of the size will be Gibson, Australia < Thar, India < Great Victoria, Australia < Kalahari, South-Africa < Sahara, North-Africa.

Weight = 0.00000001 gram

∴ standard form of 0.00000001 gram = 0.1 × 10^-7 g

= 1 × 10^-1 × 10^-7 g

Using the law of exponent, a^m x aⁿ = a^m+n

= 1 × 10^-8 g

∴ The number in the standard form is 1 × 10^-8 g

Weight of new born bear = 4 kg

Weight increases by the power of 2 in 5 yr.

Weight of bear in 5 yr = (4)² = 16 kg

Standard form of 0.02543 m 

= 0.2543 x 10^-1 m 

= 2.543 x 10^-2 m 

Hence, standard form of 0.025434 s 2.543 x 10^-2 m.

The world population at the end of 20^th century = 6.1 × 10⁹ people

Usual form of 6.1 × 10⁹ = 6.1 × 1, 000, 000, 000

= 6, 100, 000, 000

6, 100, 000, 000 can be read as six billion one hundred million.

Annual sales of sugar in sugar factory = 3 billion 720 million kilograms = 3720000 kg

∴ standard form of 3720000 kg = 372 × 10 × 10 × 10 × 10 kg

= 372 × 10⁴ kg

= 3.72 × 10⁴ × 10² kg

Using the law of exponent, a^m x aⁿ = a^m+n

= 3.72 × 10⁶ kg

∴ The number in the standard form is 3.72 × 10⁶ kg

Given litres of water that flow through the river in a day = 230 billion = 230, 000, 000, 000 = 230 × 10⁹ litres

We know that 1 week = 7 days and 1 year = 365 days.

Litres of water that flow through that river in a week = 230 × 10⁹ × 7 = 1610 × 10⁹

In standard notation, 1610 × 10⁹ = 1.61 × 10³ × 10⁹

We know that by properties of exponents, a^m × aⁿ = a^{m + n}.

⇒ 1.61 × 10³ × 10⁹ = 1.61 × 10^{3 + 9} = 1.61 × 10¹

∴ Litres of water that flow through that river in a week = 1.61 × 10¹² litres

Litres of water that flow through that river in a year = 230 × 10⁹ × 365 = 83950 × 10⁹

In standard notation, 83950 × 10⁹ = 8.395 × 10⁴ × 10⁹

We know that by properties of exponents, a^m × aⁿ = a^{m + n}.

⇒ 8.395 × 10⁴ × 10⁹ = 8.395 × 10^{4 + 9} = 8.395 × 10¹

∴ Litres of water that flow through that river in a year = 8.395 × 10¹³ litres

Let W₁ be the work done by (x2³)machine.

Let W₂ be the work done by (x3³) machine.

Work done by both the machines is W_t = W₁ x W

W_t = 2³ x 3³

W_t = 2 x 2 x 2 x 3 x 3 x 3 = 8 x 27 = 216

If a single repeater has to be used, then the machine is of the form (x A^m) where A and m are both natural numbers.

216 = 6 x 6 x 6 = 6³

A = 6, m = 3.

Therefore, Shikha should use a (x6³) single repeater machine.

Given, 300 grams of a substance decrease to 300 × 2^-3 after 3 half-lives.

⇒ Evaluating 300 × 2^-3

We know by laws of exponents, a^-n = 1/aⁿ

⇒ 300 × 2^-3 

= 300 x 1/2² 

= 300 x 1/8 

= 75/2 

= 37.5 grams

∴ 3.75 grams of the substance are left.

(a) Single machine work = 100

Hook-up machine of prime base number that do the same work down by x 100

= 2² x 5²

= 4×25

= 100

(b) x 99 = 3² x 111 hook-up machine.

(c) x 37 machine cannot do the same work.

(d) x 1111 = 101 x 11 hook-up machine.

Let W₁ be the work done by first machine.

Let W₂ be the work done by second machine.

Work done by both the machines is W_t = W₁xW₂

W_t = 25² = 625

625 = 5x5x5x5 = 5²x5²

W₁xW₂ = 5²x5²

W₁ = W₂ = 5²

Therefore, Neha can use a hook-up of two (x5²) machines.

Two repeater machines that do the same work as (x 81) are (x 3⁴) and (x 9²).

Since, factor of 81 are 3 and 9.

In machine (a), (x 100 ²) = 10000 stretch. Since, it is two times the base machine.

In machine (b), (x 7 ⁵) = 16807 stretch.

Since, it is fair times the base machine.

In machine (c), (x 5⁷) = 78125 stretch.

Since, it is 7 times the base machine.

The machine produce x 1¹⁰⁰ =1

So, if we insert 2 cm long piece of chalk in that machine, the piece of chalk remains same.

Correct option is (B) 5184

\(3^4\times4^3=81\times64\) = 5184

Correct option is  B) 5184

By substituting the values of a and b in given equation, we get,

⇒ (-1)² × (2)²

⇒ 1 × 4 = 4

∴ The value of given expression is 4

The window overlooked a large garden and a playground at the back.

Correct option is (B) \((\frac{7}{5})^4\)

\(\frac{2401}{625}=\frac{7^4}{5^4}\) \(=(\frac{7}{5})^4\)

Correct option is  B) (\(\frac{7}{5}\))⁴ 

D. Her plump figure

According to Miss Beam, the real aim of the school was not so much to teach thought as it was to teach thoughtfulness and kindness to others.

He advised us to have faith in God.

According to Miss Beam, they do simple things at the school like teaching the children spelling, addition, subtraction, multiplication and writing. The rest is done by reading to them and through interesting talks, during which they have to sit still and keep their hands quiet.

B. scholastic methods

D. the parents were good enough to trust her.

7³ = 7 × 7 × 7 = 343

3⁷= = 3 × 3 × 3 × 3 × 3 × 3 × 3 = 2187

2187 > 343

Therefore, 3⁷ > 7³

(1)⁴ = 1 × 1 × 1 × 1 = 1

(1)⁵ = 1 × 1 × 1 × 1 × 1 = 1 

(1)⁷ = 1 × 1 × 1 × 1 × 1 × 1 × 1 = 1

(- 1)² = (-1) × (-1) = 1

(- 1)³ = (-1) × (1) × (-1) = – 1

(- 1)⁴ =(-1) × (-1) × (-1) × (-1) = 1

(- 1)⁵ = (-1) × (-1) × (-1) × (-1) × (- 1) = – 1

From the above illustrations, 

(i) It raised. to any power is 1. 

(ii) (-1) raised to even power is (- 1) and (-1) raised to an odd power is (-1). 

Thus (- 1)^m = 1 if ’m’ is even (- 1)^m 1 if ‘m’ is odd

[(-11)⁵]² = (-11)^5x2

[(-11)⁵]² = (-11)¹⁰

∴ [(-11)⁵]² = (-11)¹⁰

\(\frac {(-9)^{11}}{(-9)^7}\) = (-9)^11-7 = (-9)⁴

 \(\frac {2^9}{2^3}\) = 2^9-3 [∵ \( \frac {a^m}{a^n} = a^{m-n}\)]

= 2⁶

Correct option is  (b) salary