3.2.1 Squares, Square Roots, Cube and Cube Roots, PT3 Practice - Mathematics

3.2.1 Squares, Square Roots, Cube and Cube Roots, PT3 Practice

Question 1:

Calculate the values of the following:

\begin{array}{l} (a) \sqrt{\frac{50}{98}} \\ (b) \sqrt{1 \frac{17}{64}} \\ (c) \sqrt{81} - \sqrt{0.01} \\ (d) \sqrt{3.24} \end{array}

$\begin{array}{l} (a) \sqrt{\frac{50}{98}} \\ (b) \sqrt{1 \frac{17}{64}} \\ (c) \sqrt{81} - \sqrt{0.01} \\ (d) \sqrt{3.24} \end{array}$

Solution:

(a) \sqrt{\frac{50}{98}} = \sqrt{\frac{5025}{9849}} = \sqrt{\frac{25}{49}} = \frac{5}{7}

$(a) \sqrt{\frac{50}{98}} = \sqrt{\frac{5025}{9849}} = \sqrt{\frac{25}{49}} = \frac{5}{7}$

(b) \sqrt{1 \frac{17}{64}} = \sqrt{\frac{81}{64}} = \frac{9}{8} = 1 \frac{1}{8}

$(b) \sqrt{1 \frac{17}{64}} = \sqrt{\frac{81}{64}} = \frac{9}{8} = 1 \frac{1}{8}$

\begin{array}{l} (c) \sqrt{81} - \sqrt{0.01} = 9 - \sqrt{\frac{1}{100}} \\ = 9 - \frac{1}{10} \\ = 9 - 0.1 \\ = 8.9 \end{array}

$\begin{array}{l} (c) \sqrt{81} - \sqrt{0.01} = 9 - \sqrt{\frac{1}{100}} \\ = 9 - \frac{1}{10} \\ = 9 - 0.1 \\ = 8.9 \end{array}$

\begin{array}{l} (d) \sqrt{3.24} = \sqrt{3 \frac{24}{100}} = \sqrt{3 \frac{6}{25}} \\ = \sqrt{\frac{81}{25}} \\ = \frac{9}{5} = 1 \frac{4}{5} \end{array}

$\begin{array}{l} (d) \sqrt{3.24} = \sqrt{3 \frac{24}{100}} = \sqrt{3 \frac{6}{25}} \\ = \sqrt{\frac{81}{25}} \\ = \frac{9}{5} = 1 \frac{4}{5} \end{array}$

Question 2:

Calculate the values of the following:

\begin{array}{l} (a) \sqrt[3]{\frac{16}{250}} \\ (b) \sqrt[3]{- \frac{4}{256}} \\ (c) \sqrt[3]{0.008} \\ (d) \sqrt[3]{0.729} \end{array}

$\begin{array}{l} (a) \sqrt[3]{\frac{16}{250}} \\ (b) \sqrt[3]{- \frac{4}{256}} \\ (c) \sqrt[3]{0.008} \\ (d) \sqrt[3]{0.729} \end{array}$

Solution:

(a) \sqrt[3]{\frac{16}{250}} = \sqrt[3]{\frac{8}{125}} = \frac{2}{5}

$(a) \sqrt[3]{\frac{16}{250}} = \sqrt[3]{\frac{8}{125}} = \frac{2}{5}$

(b) \sqrt[3]{- \frac{4}{256}} = \sqrt[3]{- \frac{1}{64}} = - \frac{1}{4}

$(b) \sqrt[3]{- \frac{4}{256}} = \sqrt[3]{- \frac{1}{64}} = - \frac{1}{4}$

\begin{array}{l} (c) \sqrt[3]{0.008} = \sqrt[3]{\frac{8}{1000}} \\ = \frac{2}{10} \\ = 0.2 \end{array}

$\begin{array}{l} (c) \sqrt[3]{0.008} = \sqrt[3]{\frac{8}{1000}} \\ = \frac{2}{10} \\ = 0.2 \end{array}$

\begin{array}{l} (d) \sqrt[3]{- 0.729} = \sqrt[3]{- \frac{729}{1000}} \\ = - \frac{9}{10} \\ = - 0.9 \end{array}

$\begin{array}{l} (d) \sqrt[3]{- 0.729} = \sqrt[3]{- \frac{729}{1000}} \\ = - \frac{9}{10} \\ = - 0.9 \end{array}$

Question 3:

Find the value of

\sqrt[3]{3 \frac{3}{8}} + \sqrt{2 \frac{14}{25}} .

$\sqrt[3]{3 \frac{3}{8}} + \sqrt{2 \frac{14}{25}} .$

Solution:

\begin{array}{l} \sqrt[3]{3 \frac{3}{8}} + \sqrt{2 \frac{14}{25}} = \sqrt[3]{\frac{27}{8}} + \sqrt{\frac{64}{25}} \\ = \frac{3}{2} + \frac{8}{5} \\ = \frac{31}{10} = 3 \frac{1}{10} \end{array}

$\begin{array}{l} \sqrt[3]{3 \frac{3}{8}} + \sqrt{2 \frac{14}{25}} = \sqrt[3]{\frac{27}{8}} + \sqrt{\frac{64}{25}} \\ = \frac{3}{2} + \frac{8}{5} \\ = \frac{31}{10} = 3 \frac{1}{10} \end{array}$

Question 4:

Find the values of the following:

(a) 1 – (–0.3)³.

(b)

{(2.1 \div \sqrt[3]{27})}^{2}

${(2.1 \div \sqrt[3]{27})}^{2}$

Solution:

(a)

1 – (–0.3)³= 1 – [(–0.3) × (–0.3) × (–0.3)]

= 1 – (–0.027)

= 1 + 0.027

= 1.027

(b)

\begin{array}{l} {(2.1 \div \sqrt[3]{27})}^{2} = {(2.1 \div 3)}^{2} \\ = {(0.7)}^{2} \\ = 0.49 \end{array}

$\begin{array}{l} {(2.1 \div \sqrt[3]{27})}^{2} = {(2.1 \div 3)}^{2} \\ = {(0.7)}^{2} \\ = 0.49 \end{array}$

2 thoughts on “3.2.1 Squares, Square Roots, Cube and Cube Roots, PT3 Practice”

qin

May 24, 2020 at 3:18 pm | Reply

Question 5(b) isn’t 12÷6×(0.3×0.3×0.3) then the answer become 0.054?
- myhometuition
  
  June 23, 2020 at 11:49 am | Reply
  
  Thanks for pointing out our mistake.
  Correction had been made accordingly.

2 thoughts on “3.2.1 Squares, Square Roots, Cube and Cube Roots, PT3 Practice”

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