Question

11) Find the value of i) (4⁰ + 4¹) × 2²​

Answer 1

⇒ (4^0 + 4^1) × 2^2

⇒ ( 1 + 4) × 4     {a^0 = 1}

⇒   5 × 4

⇒   20

Answer 2

$\\\\→(4⁰ + 4¹) × 2²$

$→(1+4)×4$

$→5×4$

$→20$

$\\ \\ \\ \sf (4⁰ + 4¹) × 2² \bf= 20$

• $\tiny\begin{gathered}\begin{gathered}\\\\\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \bigstar \: \underline{\bf{}}\\ {\boxed{\begin{array}{c | c} \frac{ \: ~~~~~~~~~~\: \: \: \: \:\sf Laws \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }{ } &\frac{ \: ~~~~~~~~~~ \: \: \: \: \:\sf Example \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }{ }\\ \sf \bigstar{a}^{m} \times {a}^{n} = {a}^{m + n} & \sf {a}^{2} \times {a}^{3} = {a}^{2 + 3} = {a}^{6} \\ \\ \sf \bigstar{a}^{m} \div {a}^{n} = {a}^{m - n}& \sf {a}^{3} \div {a}^{2} = {a}^{3 - 2} = {a}^{1} \\ \\ \sf{\bigstar \: \: \: \: \: \: ( {a}^{m} ) ^{n} = {a}^{mn} } & \sf( {a}^{2} ) ^{3} = {a}^{2 \times 3} = {a}^{6} \\ \\ {\bigstar\sf a {}^{m} \times {n}^{m} = (ab) ^{m} } &\sf a {}^{2} \times {b}^{2} = (ab) ^{2}\\ \\ \sf\bigstar \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {a}^{0} = 1& \sf {2}^{0} = 1 \: \: \: \: \\ \\ \sf \bigstar \: \: \: \: {\dfrac{ {a}^{m} }{ {b}^{m} }= \left( \dfrac{a}{b} \right) ^{m} }& \sf{\dfrac{ {a}^{2} }{ {b}^{2} }= \left( \dfrac{a}{b} \right) ^{2} }\\\\\bigstar~~~~~~~ \sf x^{\frac{m}{n} }=\sqrt[n]{x^m}\sf = (\sqrt[n]{x})^m & \sf x^{\frac{2}{3} }=\sqrt[3]{x^2} = (\sqrt[n]{x})^m\\ \\\\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$