Given that: Distance travelled = x km and total fare = ₹ y Total fare = Fare for first km + Fare for remaining distance Therefore, the equation: y = 8 + 5 × (x - 1) ⇒ y = 5x +3 For the graph: Putting x = 1, we have, y = 5 × 1 + 3 = 8 Putting x = 2, we have, y = 5 × 2 + 3 = 13 Putting x = 3, we have,Read more
Given that: Distance travelled = x km and total fare = ₹ y
Total fare = Fare for first km + Fare for remaining distance
Therefore, the equation: y = 8 + 5 × (x – 1) ⇒ y = 5x +3
For the graph:
Putting x = 1, we have, y = 5 × 1 + 3 = 8
Putting x = 2, we have, y = 5 × 2 + 3 = 13
Putting x = 3, we have, y = 5 × 3 + 3 = 18
Hence, A(1, 8), B(2, 13) and C[3, 18) are solutions of the equation.
The taxi fare in a city is as follows: For the first kilometre, the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking the distance covered as x km and total fare as Rs y, write a linear equation for this information, and draw its graph.
Given that: Distance travelled = x km and total fare = ₹ y Total fare = Fare for first km + Fare for remaining distance Therefore, the equation: y = 8 + 5 × (x - 1) ⇒ y = 5x +3 For the graph: Putting x = 1, we have, y = 5 × 1 + 3 = 8 Putting x = 2, we have, y = 5 × 2 + 3 = 13 Putting x = 3, we have,Read more
Given that: Distance travelled = x km and total fare = ₹ y
See lessTotal fare = Fare for first km + Fare for remaining distance
Therefore, the equation: y = 8 + 5 × (x – 1) ⇒ y = 5x +3
For the graph:
Putting x = 1, we have, y = 5 × 1 + 3 = 8
Putting x = 2, we have, y = 5 × 2 + 3 = 13
Putting x = 3, we have, y = 5 × 3 + 3 = 18
Hence, A(1, 8), B(2, 13) and C[3, 18) are solutions of the equation.
The taxi fare in a city is as follows: For the first kilometre, the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking the distance covered as x km and total fare as Rs y, write a linear equation for this information, and draw its graph.
Use suitable identities to find the following products: (x + 4) (x + 10)
(i) (x + 4)(x + 10) = x² + (4 + 10)x + 4 × 10 [∵ (x + a)(x + b) = x² + (a + b)x + ab ] = x² + 14x + 40
(i) (x + 4)(x + 10)
See less= x² + (4 + 10)x + 4 × 10 [∵ (x + a)(x + b) = x² + (a + b)x + ab ]
= x² + 14x + 40
Use suitable identities to find the following products: (x + 8) (x – 10)
(ii) (x + 8)(x + -10) = x² + (8 - 10)x + 8 × (-10) [∵ (x + a)(x + b) = x² + (a + b)x + ab ] = x² - 2x - 80
(ii) (x + 8)(x + -10)
See less= x² + (8 – 10)x + 8 × (-10) [∵ (x + a)(x + b) = x² + (a + b)x + ab ]
= x² – 2x – 80
Use suitable identities to find the following products: (3x + 4) (3x – 5)
(iii) (3x + 4)(3x - 5) = (3x)² + (4 - 5)3x + 4 × (-5) [∵ (x + a)(x + b) = x² + (a + b)x + ab ] = 9x² - 3x - 20
(iii) (3x + 4)(3x – 5)
See less= (3x)² + (4 – 5)3x + 4 × (-5) [∵ (x + a)(x + b) = x² + (a + b)x + ab ]
= 9x² – 3x – 20