Question

please answer these questions of math please do not spam ​

Answer 1

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