Answer :
Answer:
distance covered in (x + 5) hours = (x² + 7x + 10) km
distance covered in 1 hour :
[tex] = \frac{ {x}^{2} + 7x + 10}{x + 5} [/tex]
x² + 7x + 10 = x² + 2x + 5x + 10
= x(x + 2) + 5(x + 2)
= (x + 2)(x + 5)
[tex] = \frac{(x + 2)(x + 5)}{x + 5} [/tex]
= (x + 2) km
Appropriate Question:
Varsha travelled with constant speed for (x + 5) hours, covering a total distance represented by the expression (x² + 7x + 10) km. How far does she travel in 1 hour?
Answer:
She travel (x + 2) km in 1 hour.
Step-by-step explanation:
Given that,
[tex]\sf\: Distance \: travelled \: in \: (x + 5) \: hours = x^2 + 7x + 10 \: km \\ [/tex]
So,
[tex]\sf\: Distance \: travelled \: in \:1 \: hour = \dfrac{x^2 + 7x + 10}{x + 5} \: km \\ [/tex]
On splitting the middle term in numerator, we get
[tex]\sf\: Distance \: travelled \: in \:1 \: hour = \dfrac{x^2 + 5x + 2x+ 10}{x + 5} \: km \\ [/tex]
[tex]\sf\: Distance \: travelled \: in \:1 \: hour = \dfrac{x(x + 5) + 2(x+ 5)}{x + 5} \: km \\ [/tex]
[tex]\sf\: Distance \: travelled \: in \:1 \: hour = \dfrac{(x + 5)(x+ 2)}{x + 5} \: km \\ [/tex]
[tex]\implies\sf\: Distance \: travelled \: in \:1 \: hour = (x+ 2) \: km \\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\: Distance \: travelled \: in \:1 \: hour = (x+ 2) \: km \: } \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Alternative Method:
Given that, Distance travelled in (x + 2) hours is (x² + 7x + 10) km.
Now, we have to find in how much she travel in 1 hour.
We know, less time, less distance travelled.
So, it means, time and distance travelled are in direct variation.
Let assume that distance travelled in 1 hour be y km
So, we have
[tex]\begin{array}{|c|c|c|} \\ \rm Distance \:travelled \:\: &\rm x^2 + 7x + 10\: &\rm y\: \: \\ \\ \rm Time \: taken \: in\: hours \: &\rm x + 5&\rm 1 \\\end{array}\\[/tex]
So, using law of direct variation, we have
[tex]\sf\: \dfrac{ {x}^{2} + 7x + 10 }{x + 5} = \dfrac{y}{1} \\ [/tex]
[tex]\sf\: y = \dfrac{ {x}^{2} + 7x + 10 }{x + 5} \\ [/tex]
[tex]\sf\: y = \dfrac{ {x}^{2} + 5x + 2x + 10 }{x + 5} \\ [/tex]
[tex]\sf\: y = \dfrac{ x(x + 5) + 2(x + 5) }{x + 5} \\ [/tex]
[tex]\sf\: y = \dfrac{ (x + 5)(x + 2) }{x + 5} \\ [/tex]
[tex]\implies\sf\:y = x + 2 \\ [/tex]
Hence, she travel (x + 2) km in 1 hour.
[tex]\rule{190pt}{2pt}[/tex]
[tex]\bf\: Additional\: Information:[/tex]
[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]