Question

How do you solve the following math questions? - Please explain your answers

Answer 1

16a. 5^3 could cancel 5^2 on denominator, same thing with 2^4 and 2^2. as a result, you should get 5*2^2/x^6 because when there is hidden 1 on x on the right side of the denominator

Answer 2

A. ((5^3) / (2^2)(x^5)) * ((2^4) / (5^2)x)
(5 / (2^2)(x^5)) * ((2^4) / x)  The 5^3 cancels out the 5^2
(5 / (x^5)) * ((2^2) / x)  The 2^4 cancels out the 2^2

b. (7(a^3)(b^2) / 2(c^5)) / (49(a^2)b / (2^3)(c^20)
((a^3)(b^2) / 2(c^5)) / (7(a^2)b / (2^3)(c^20) The 49 (also know as 7^2) cancels out the 7
(a(b^2) / 2(c^5)) / (7b / (2^3)(c^20) The a^3 cancels out the a^2
(a(b) / 2(c^5)) / (7 / (2^3)(c^20) The b^2 cancels out the b
(ab / 2) / (7 / (2^3)(c^15) The c^20 cancels out the c^5
ab / (7 / (2^3)(c^15) The 2^3 cancels out the 2

c. 10(x^3)(y^5) / (1 / -10(x^4)(y^3) 
Nothing in this equation will cancel out to simplify.

d. (12(x^2)(y^5) / 21(x^3)) * (7(x^4)y / 3(y^6))
(12(x^2) / 21(x^3)) * (7(x^4)y / 3y) The y^6 cancels out the y^5
(4(x^2) / 21(x^3)) * (7(x^4)y / y) The 12 cancels out the 3
(4(x^2) / 21) * (7xy / y) The x^4 cancels out the x^3
(4(x^2) / 3) * (xy / y) The 21 cancels out the 7