please answer these questions of math please do not spam
Answers
Solutions:-
i)
Given that
a⁶×a⁸
=>a^(6+8)
Since a^m × a^n = a^(m+n)
=> a¹⁴
a⁶ × a⁸ = a¹⁴
ii) Given that
x^5×x^-3
=> x^(5+(-3))
Since a^m × a^n = a^(m+n)
=> x^(5-3)
=>x²
x^5×x^-3 = x²
iv) Given that a²b³ × a⁵b²
It can be rearranged as
=> (a²×a⁵)×(b³×b²)
=> a^(2+5) × b^(3+2)
Since a^m × a^n = a^(m+n)
=> a⁷b⁵
a²b³×a⁵b² = a⁷b⁵
v) Given that
5x⁷×3x⁴
It can be rearranged as
=> 5×3×(x⁷×x⁴)
=> 15×(x^(7+4))
Since a^m × a^n = a^(m+n)
=> 15x¹¹
5x⁷×3x⁴ = 15x¹¹
vii) Given that
x^7y^-5 × x^-5y³
It can be rearranged as
=> x^7×x^-5 × y^-5×y^3
=> x^(7+(-5))× y^(-5+3)
Since a^m × a^n = a^(m+n)
=> x^(7-5)×y^-2
=> x^2 × y^-2
=> x^2/y^2
Since x^-n = 1/x^n
x^7y^-5 × x^-5y³ = x² y^-2 or x²/y²
viii) Given that.
x^-2 y^5 × x^0 y^-7
=> x^-2 y^5 × y^-7
Since a⁰ = 1
=> x^-2 × (y^5×y^-7)
=> x^-2 × y^(5-7)
Since a^m × a^n = a^(m+n)
=> x^-2 ×y^-2
=> (1/x²)×(1/y²)
Since x^-n = 1/x^n
=> 1/(x²y²)
x^-2 y^5 × x^0 y^-7 = x^-2 ×y^-2 or
1/(x²y²)
Used formulae:-
- a^m × a^n = a^(m+n)
- x^-n = 1/x^n
- a⁰ = 1