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 = 100.08x10² mm Class 8 Maths Chapter 11 Exercise 12.2 Solution in Video for moreRead more
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 = 100.08×10² mm
Class 8 Maths Chapter 11 Exercise 12.2 Solution in Video
1 micron = 1/1000000 = 1/10⁶ = 1x10⁻⁶ m Class 8 Maths Chapter 11 Exercise 12.2 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
1 micron = 1/1000000 = 1/10⁶ = 1×10⁻⁶ m
Class 8 Maths Chapter 11 Exercise 12.2 Solution in Video
3.02×10⁻⁶ = 3.02/10⁶ = 0.00000302 Class 8 Maths Chapter 11 Exercise 12.2 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
3.02×10⁻⁶ = 3.02/10⁶ = 0.00000302
Class 8 Maths Chapter 11 Exercise 12.2 Solution in Video
0.0000000000085 = 0.0000000000085 x 10¹²/10¹² = 8.5x10⁻¹² Class 8 Maths Chapter 11 Exercise 12.2 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
0.0000000000085 = 0.0000000000085 x 10¹²/10¹² = 8.5×10⁻¹²
Class 8 Maths Chapter 11 Exercise 12.2 Solution in Video
25xt⁻⁴ / 5⁻³x10xt⁻⁸ (t≠0) = 5²xt⁻⁴/5⁻³x5x2xt⁻⁸ = 5²⁻(⁻³)⁻¹xt⁻⁴⁻⁸/2 [∵ aᵐ÷aⁿ=aᵐ⁻ⁿ] = 5²⁺³⁻¹xt⁻⁴⁺⁸/2 = 5⁴xt⁴/2 = 625/2 t⁴ Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
(5/8)⁻⁷ x (8/5)⁻⁴ = 5⁻⁷/8⁻⁷ x 8⁻⁴⁻(⁻⁷) [∵ (a/b)ᵐ =aᵐ/bᵐ] = 5⁻⁷⁻(⁻⁴) x 8⁻⁴⁻⁷ [∵ aᵐ÷aⁿ=aᵐ⁻ⁿ] = 5⁻⁷⁺⁴x8⁻⁴⁺⁷ = 5⁻³x8³ =8³/8³ [∵ a⁻ᵐ=1/aᵐ] = 512/125 Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
(5/8)⁻⁷ x (8/5)⁻⁴ = 5⁻⁷/8⁻⁷ x 8⁻⁴⁻(⁻⁷) [∵ (a/b)ᵐ =aᵐ/bᵐ]
= 5⁻⁷⁻(⁻⁴) x 8⁻⁴⁻⁷ [∵ aᵐ÷aⁿ=aᵐ⁻ⁿ]
= 5⁻⁷⁺⁴x8⁻⁴⁺⁷ = 5⁻³x8³ =8³/8³ [∵ a⁻ᵐ=1/aᵐ]
= 512/125
Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video
5ᵐ÷5⁻³=5⁵ ⇒ 5ᵐ⁻(⁻³)= 5⁵ [∵aᵐ÷aₙ=aᵐ⁻ᵐ] ⇒ 5ᵐ⁺³ =5⁵ Comparing exponents both sides, we get ⇒ m+3=5 ⇒ m=5-3 ⇒ m=2 Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
5ᵐ÷5⁻³=5⁵
⇒ 5ᵐ⁻(⁻³)= 5⁵ [∵aᵐ÷aₙ=aᵐ⁻ᵐ]
⇒ 5ᵐ⁺³ =5⁵
Comparing exponents both sides, we get
⇒ m+3=5
⇒ m=5-3
⇒ m=2
Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video
8⁻¹ x5³/2⁻⁴ = (2³)⁻¹ x 5³/2⁻⁴ = 2⁻³x5³/2⁻⁴ [∵(aᵐ)ⁿ=aᵐˣⁿ] = 2⁻³(⁻⁴)x5³= 2⁻³⁺⁴x5³ [∵ aᵐ÷aⁿ=aᵐ⁻ⁿ] = 2x125=250 Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
(3⁰+4⁻¹)x2² = (1+1/4)x2² [∵ a⁻ᵐ =1/aᵐ] = (4+1/4) x 2²=5/4x2²=5/2²x2² = 5x2²⁻² [∵ aᵐ÷aⁿ=aᵐ⁻ⁿ] = 5x2⁰ = 5x1 = 5 [∵ a⁰=1] Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
(-4)⁵ ÷(-4)⁸ = (-4)⁵⁻⁸ = (-4)⁻³ = 1/(-4)⁸ Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
(-4)⁵ ÷(-4)⁸ = (-4)⁵⁻⁸
= (-4)⁻³ = 1/(-4)⁸
Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video
In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack?
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 = 100.08x10² mm Class 8 Maths Chapter 11 Exercise 12.2 Solution in Video for moreRead more
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 = 100.08×10² mm
Class 8 Maths Chapter 11 Exercise 12.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
Express the number appearing in the following statements in standard form: 1 micron is equal to 1/1000000 m.
1 micron = 1/1000000 = 1/10⁶ = 1x10⁻⁶ m Class 8 Maths Chapter 11 Exercise 12.2 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
1 micron = 1/1000000 = 1/10⁶ = 1×10⁻⁶ m
Class 8 Maths Chapter 11 Exercise 12.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
Express the following numbers in usual form: 3.02×10⁻⁶
3.02×10⁻⁶ = 3.02/10⁶ = 0.00000302 Class 8 Maths Chapter 11 Exercise 12.2 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
3.02×10⁻⁶ = 3.02/10⁶ = 0.00000302
Class 8 Maths Chapter 11 Exercise 12.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
Express the following numbers in standard form: 0.0000000000085
0.0000000000085 = 0.0000000000085 x 10¹²/10¹² = 8.5x10⁻¹² Class 8 Maths Chapter 11 Exercise 12.2 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
0.0000000000085 = 0.0000000000085 x 10¹²/10¹² = 8.5×10⁻¹²
Class 8 Maths Chapter 11 Exercise 12.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
Simplify: 25xt⁻⁴ / 5⁻³x10xt⁻⁸ (t≠0)
25xt⁻⁴ / 5⁻³x10xt⁻⁸ (t≠0) = 5²xt⁻⁴/5⁻³x5x2xt⁻⁸ = 5²⁻(⁻³)⁻¹xt⁻⁴⁻⁸/2 [∵ aᵐ÷aⁿ=aᵐ⁻ⁿ] = 5²⁺³⁻¹xt⁻⁴⁺⁸/2 = 5⁴xt⁴/2 = 625/2 t⁴ Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
25xt⁻⁴ / 5⁻³x10xt⁻⁸ (t≠0) = 5²xt⁻⁴/5⁻³x5x2xt⁻⁸ = 5²⁻(⁻³)⁻¹xt⁻⁴⁻⁸/2 [∵ aᵐ÷aⁿ=aᵐ⁻ⁿ]
= 5²⁺³⁻¹xt⁻⁴⁺⁸/2 = 5⁴xt⁴/2 = 625/2 t⁴
Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
Evaluate: (5/8)⁻⁷ x (8/5)⁻⁴
(5/8)⁻⁷ x (8/5)⁻⁴ = 5⁻⁷/8⁻⁷ x 8⁻⁴⁻(⁻⁷) [∵ (a/b)ᵐ =aᵐ/bᵐ] = 5⁻⁷⁻(⁻⁴) x 8⁻⁴⁻⁷ [∵ aᵐ÷aⁿ=aᵐ⁻ⁿ] = 5⁻⁷⁺⁴x8⁻⁴⁺⁷ = 5⁻³x8³ =8³/8³ [∵ a⁻ᵐ=1/aᵐ] = 512/125 Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
(5/8)⁻⁷ x (8/5)⁻⁴ = 5⁻⁷/8⁻⁷ x 8⁻⁴⁻(⁻⁷) [∵ (a/b)ᵐ =aᵐ/bᵐ]
= 5⁻⁷⁻(⁻⁴) x 8⁻⁴⁻⁷ [∵ aᵐ÷aⁿ=aᵐ⁻ⁿ]
= 5⁻⁷⁺⁴x8⁻⁴⁺⁷ = 5⁻³x8³ =8³/8³ [∵ a⁻ᵐ=1/aᵐ]
= 512/125
Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
Find the value of m for which 5ᵐ÷5⁻⁻³=5⁵
5ᵐ÷5⁻³=5⁵ ⇒ 5ᵐ⁻(⁻³)= 5⁵ [∵aᵐ÷aₙ=aᵐ⁻ᵐ] ⇒ 5ᵐ⁺³ =5⁵ Comparing exponents both sides, we get ⇒ m+3=5 ⇒ m=5-3 ⇒ m=2 Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
5ᵐ÷5⁻³=5⁵
⇒ 5ᵐ⁻(⁻³)= 5⁵ [∵aᵐ÷aₙ=aᵐ⁻ᵐ]
⇒ 5ᵐ⁺³ =5⁵
Comparing exponents both sides, we get
⇒ m+3=5
⇒ m=5-3
⇒ m=2
Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
Evaluate: 8⁻¹ x5³/2⁻⁴
8⁻¹ x5³/2⁻⁴ = (2³)⁻¹ x 5³/2⁻⁴ = 2⁻³x5³/2⁻⁴ [∵(aᵐ)ⁿ=aᵐˣⁿ] = 2⁻³(⁻⁴)x5³= 2⁻³⁺⁴x5³ [∵ aᵐ÷aⁿ=aᵐ⁻ⁿ] = 2x125=250 Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
8⁻¹ x5³/2⁻⁴ = (2³)⁻¹ x 5³/2⁻⁴ = 2⁻³x5³/2⁻⁴ [∵(aᵐ)ⁿ=aᵐˣⁿ]
= 2⁻³(⁻⁴)x5³= 2⁻³⁺⁴x5³ [∵ aᵐ÷aⁿ=aᵐ⁻ⁿ]
= 2×125=250
Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
Find the value of: (3⁰+4⁻¹)x2²
(3⁰+4⁻¹)x2² = (1+1/4)x2² [∵ a⁻ᵐ =1/aᵐ] = (4+1/4) x 2²=5/4x2²=5/2²x2² = 5x2²⁻² [∵ aᵐ÷aⁿ=aᵐ⁻ⁿ] = 5x2⁰ = 5x1 = 5 [∵ a⁰=1] Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
(3⁰+4⁻¹)x2² = (1+1/4)x2² [∵ a⁻ᵐ =1/aᵐ]
= (4+1/4) x 2²=5/4×2²=5/2²x2²
= 5×2²⁻² [∵ aᵐ÷aⁿ=aᵐ⁻ⁿ]
= 5×2⁰ = 5×1 = 5 [∵ a⁰=1]
Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
Simplify and express the result in power notation with positive exponent:(-4)⁵ ÷(-4)⁸
(-4)⁵ ÷(-4)⁸ = (-4)⁵⁻⁸ = (-4)⁻³ = 1/(-4)⁸ Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/
(-4)⁵ ÷(-4)⁸ = (-4)⁵⁻⁸
= (-4)⁻³ = 1/(-4)⁸
Class 8 Maths Chapter 11 Exercise 12.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-12/