Question

Simplify these thing below please. I am stuck again... Thank you

Answer 1

$9\sqrt{56 x^7 y^{12}}\qquad \begin{cases} 56=7\cdot 2\cdot 2\cdot 2\\ \qquad 7\cdot 2^2 \cdot 2\\ \qquad 2^2\cdot 14\\ x^7=x^{(3)(2)+1}\\ \qquad (x^3)^2\cdot x^1\\ y^{12}=y^{(6)(2)}\\ \qquad (y^6)^2 \end{cases}\hspace{5em} \begin{array}{llll} 9\sqrt{2^2(14)(x^3)^2 x (y^6)^2} \\\\\\ 9(2)(x^3)(y^6)\sqrt{14x} \\\\\\ {\Large \begin{array}{llll} 18x^3y^6\sqrt{14x} \end{array}} \end{array}$

Answer 2

Answer:

$\textsf{1.} \quad 18\;x^3\;y^{6}\sqrt{14x}$

$\textsf{2.} \quad -8\sqrt{2}$

 

Step-by-step explanation:

Question 1

Given expression:

$9\sqrt{56x^7y^{12}}$

$\textsf{Apply radical rule} \quad \sqrt{ab}=\sqrt{a}\sqrt{b}:$

$\implies 9\sqrt{56}\sqrt{x^7}\sqrt{y^{12}}$

Rewrite 56 as 4·14:

$\implies 9\sqrt{4 \cdot 14}\sqrt{x^7}\sqrt{y^{12}}$

$\textsf{Apply radical rule} \quad \sqrt{ab}=\sqrt{a}\sqrt{b}:$

$\implies 9\sqrt{4}\sqrt{14}\sqrt{x^7}\sqrt{y^{12}}$

Rewrite 4 as 2²:

$\implies 9\sqrt{2^2}\sqrt{14}\sqrt{x^7}\sqrt{y^{12}}$

Simplify:

$\implies 9\cdot 2\sqrt{14}\sqrt{x^7}\sqrt{y^{12}}$

$\implies 18\sqrt{14}\sqrt{x^7}\sqrt{y^{12}}$

$\textsf{Apply exponent rule} \quad \sqrt{a}=a^{\frac{1}{2}}:$

$\implies 18\sqrt{14}\;(x^7)^{\frac{1}{2}}\;(y^{12})^{\frac{1}{2}}$

$\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:$

$\implies 18\sqrt{14}\;x^{\frac{7}{2}}\;y^{\frac{12}{2}}$

$\implies 18\sqrt{14}\;x^{\frac{7}{2}}\;y^6$

Rewrite ⁷/₂ as 3 + ¹/₂

$\implies 18\sqrt{14}\;x^{(3+\frac{1}{2})}\;y^{6}$

$\textsf{Apply exponent rule} \quad a^{b+c}= a^b \cdot a^c:$

$\implies 18\sqrt{14}\;x^3 \; x^{\frac{1}{2}}\;y^{6}$

$\textsf{Apply exponent rule} \quad a^{\frac{1}{2}}=\sqrt{a}:$

$\implies 18\sqrt{14}\;x^3 \; \sqrt{x}\;y^{6}$

Rearrange:

$\implies 18\;x^3\;y^{6}\sqrt{14x}$

Question 2

Given expression:

$7\sqrt{32}-6\sqrt{72}$

Rewrite 32 as 16·2 and 72 as 36·2:

$\implies 7\sqrt{16 \cdot 2}-6\sqrt{36 \cdot 2}$

$\textsf{Apply radical rule} \quad \sqrt{ab}=\sqrt{a}\sqrt{b}:$

$\implies 7\sqrt{16}\sqrt{2}-6\sqrt{36}\sqrt{2}$

Rewrite 16 as 4² and 36 as 6²:

$\implies 7\sqrt{4^2}\sqrt{2}-6\sqrt{6^2}\sqrt{2}$

$\textsf{Apply radical rule} \quad \sqrt{a^2}=a, \quad a \geq 0:$

$\implies 7 \cdot 4\sqrt{2}-6\cdot 6\sqrt{2}$

Simplify:

$\implies 28\sqrt{2}-36\sqrt{2}$

$\implies -8\sqrt{2}$