Question

Алгебра!!!срочно нужно!!7КЛАСС Выполните действия:b⁶×b⁹×b=n⁷×n⁶=5⁷:5⁴=c⁹:c=Выполните действия:(K⁴)⁵= (3²)⁵=Выполните возведение в степень:(1 p¹ q3 )⁴= СРОЧНО НУЖНО!!​

Answer 1

Відповідь:

Покрокове пояснення:

1)$b^{16}$

2)$n^{13}$

3)$5^{3}=125$

4)$c^{8}$

5)$K^{20}$

6)$3^{10}$

7)$p^{4}q^{12}$

Answer 2

Algebraic Operations

Let's solve the given algebraic expressions step by step.

1. For the expression b⁶ × b⁹ × b = n⁷ × n⁶: - We can simplify this expression by adding the exponents when multiplying variables with the same base. - In this case, we have b⁶ × b⁹ × b = b^(6+9+1) = b^16. - Similarly, n⁷ × n⁶ = n^(7+6) = n^13. - Therefore, b⁶ × b⁹ × b = n⁷ × n⁶ simplifies to b^16 = n^13.

2. For the expression 5⁷ : 5⁴ = c⁹ : c: - When dividing variables with the same base, we subtract the exponents. - In this case, 5⁷ : 5⁴ = 5^(7-4) = 5³. - Similarly, c⁹ : c = c^(9-1) = c^8. - Therefore, 5⁷ : 5⁴ = c⁹ : c simplifies to 5³ = c^8.

3. For the expression (K⁴)⁵ + (3²)⁵: - To raise a power to another power, we multiply the exponents. - In this case, (K⁴)⁵ = K^(4×5) = K^20. - Similarly, (3²)⁵ = 3^(2×5) = 3^10. - Therefore, (K⁴)⁵ + (3²)⁵ simplifies to K^20 + 3^10.

4. For the expression (1+p¹+q³)⁴: - To raise a binomial to a power, we can use the binomial theorem or expand the expression. - Using the binomial theorem, the expanded form of (1+p¹+q³)⁴ is: (1⁴) + 4(1³)(p¹) + 6(1²)(p¹)² + 4(1)(p¹)³ + (p¹)⁴ + 4(1³)(q³) + 6(1²)(p¹)(q³) + 4(1)(p¹)²(q³) + 4(1²)(q³)² + 6(1)(p¹)(q³)² + 4(1)(q³)³ + (q³)⁴. - Simplifying this expression will give you the final result.

Please let me know if you need further assistance with algebraic operations!