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
A. ((5^3) / (2^2)(x^5)) * ((2^4) / (5^2)x) <span>(5 / (2^2)(x^5)) * ((2^4) / x) The 5^3 cancels out the 5^2 </span><span>(5 / (x^5)) * ((2^2) / x) The 2^4 cancels out the 2^2 </span> b. (7(a^3)(b^2) / 2(c^5)) / (49(a^2)b / (2^3)(c^20) <span>((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 </span><span>(a(b^2) / 2(c^5)) / (7b / (2^3)(c^20) The a^3 cancels out the a^2 </span><span>(a(b) / 2(c^5)) / (7 / (2^3)(c^20) The b^2 cancels out the b </span><span>(ab / 2) / (7 / (2^3)(c^15) The c^20 cancels out the c^5 </span><span>ab / (7 / (2^3)(c^15) The 2^3 cancels out the 2 </span> 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) <span>The y^6 cancels out the y^5 </span>(4(x^2) / 21(x^3)) * (7(x^4)y / y) <span>The 12 cancels out the 3 </span>(4(x^2) / 21) * (7xy / y) <span>The x^4 cancels out the x^3 </span>(4(x^2) / 3) * (xy / y) <span>The 21 cancels out the 7</span>
