## Calculus: Early Transcendentals 8th Edition

a) $6\sqrt 2$ b) $48a^{5}b^{7}$ c) $\frac{x}{9y^{7}}$
a) First, you descompose the numbers inside the square root: $\sqrt 200$ - $\sqrt 32$ = $\sqrt (100* 2)$ -$\sqrt(16*2)$ Then, you take outside the square root of 100 (that is, 10) and 16 (that is, 4) : $\sqrt (100* 2)$ -$\sqrt(16*2)$ = 10 $\sqrt2$ - 4 $\sqrt 2$ After that, you apply the Common Factor having the $\sqrt2$ in common: 10 $\sqrt2$ - 4 $\sqrt 2$ = $\sqrt2 *(10-4)$ and realize the subtraction $\sqrt2 *(10-4)$ = 6 $\sqrt 2$ so $\sqrt 200$ - $\sqrt 32$ = 6 $\sqrt 2$ b) knowing that $(abc)^{n}$ =$(a^{n}b^{n}c^{n})$ we apply this to the square presented in the problem: $(3a^{3}b^{3})*(4ab^{2})^{2}$=$(3a^{3}b{3})*(4^{2}a^2b^{2*2}$)=$(3a^{3}b^{3})*(16a^2b^{2*2}$) then, taking away the parenthesis: $3a^{3}b^{3}*16a^2b^{4}$ and knowing that $a^n* a^m= a^{n+m}$ we resolve that $3*16*a^{3+2}*b^{3+4}$ = $48a^{5}b^{7}$ c) Knowing that $(\frac{x}{y})^n$=$(\frac{x^n}{y^n})$ and $(a^{x})^{n} = a^{x*n}$ Then: $(\frac{3x^{3/2}y^3}{x^2 y^{-1/2}})^{2}$ =$(\frac{3^{-2}x^{3*(-2)/2}y^{3*(-2)}}{x^{2*(-2)} y^{-1*(-2)/2}})$ =$(\frac{3^{-2}x^{-3}y^{-6}}{x^{-4)} y})$ then, having taken into account that $a^{-n} = \frac{1}{a^{n}}$ solve that: $(\frac{3^{-2}x^{-3}y^{-6}}{x^{-4)} y})$ = $(\frac{x^4}{3^2*x^3*y^6*y})$ and considering that $\frac{a^n}{a^m} = {a^{n-m}}$ we conclude that: $(\frac{x^4}{3^2*x^3*y^6*y}) = (\frac{x}{9*y^7})$