Answer:
[tex]\qquad\sf\implies \sf \:The\:swimmer\:will\:cover\:216 \: m \:in\:3\:minutes \\ \\ [/tex]
Step-by-step explanation:
Given that, speed of swimmer is 1.2 m/s
It means, a swimmer can cover 1.2 m in 1 second.
Now, we have to find how much distance it will cover in 3 minutes, i.e. 3 × 60 = 180 seconds
We know, more time, more distance covered.
So, it means, time taken and distance covered are in direct variation.
Let assume that distance covered by swimmer in 180 seconds be x m.
So, we have
[tex]\begin{array}{|c|c|c|} \\ \rm Distance \: covered \: (in \: m)\: &\rm 1.2 \: &\rm x\: \: \\ \\ \rm Time \: taken \: (in \:seconds) \: &\rm 1&\rm 180 \\ \end{array} \\ \\[/tex]
So, using law of direct variation, we have
[tex]\qquad\sf \: \dfrac{1.2}{1} = \dfrac{x}{180} \\ \\ [/tex]
[tex]\sf \: 1.2 = \dfrac{x}{180} \\ \\ [/tex]
[tex]\sf \: x = 180 \times 1.2 \\ \\ [/tex]
[tex]\qquad\sf\implies \sf \:x = 216 \: m \: \\ \\ [/tex]
Hence,
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]
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Answers & Comments
Answer:
[tex]\qquad\sf\implies \sf \:The\:swimmer\:will\:cover\:216 \: m \:in\:3\:minutes \\ \\ [/tex]
Step-by-step explanation:
Given that, speed of swimmer is 1.2 m/s
It means, a swimmer can cover 1.2 m in 1 second.
Now, we have to find how much distance it will cover in 3 minutes, i.e. 3 × 60 = 180 seconds
We know, more time, more distance covered.
So, it means, time taken and distance covered are in direct variation.
Let assume that distance covered by swimmer in 180 seconds be x m.
So, we have
[tex]\begin{array}{|c|c|c|} \\ \rm Distance \: covered \: (in \: m)\: &\rm 1.2 \: &\rm x\: \: \\ \\ \rm Time \: taken \: (in \:seconds) \: &\rm 1&\rm 180 \\ \end{array} \\ \\[/tex]
So, using law of direct variation, we have
[tex]\qquad\sf \: \dfrac{1.2}{1} = \dfrac{x}{180} \\ \\ [/tex]
[tex]\sf \: 1.2 = \dfrac{x}{180} \\ \\ [/tex]
[tex]\sf \: x = 180 \times 1.2 \\ \\ [/tex]
[tex]\qquad\sf\implies \sf \:x = 216 \: m \: \\ \\ [/tex]
Hence,
[tex]\qquad\sf\implies \sf \:The\:swimmer\:will\:cover\:216 \: m \:in\:3\:minutes \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]