(i) Taking Celsius on x-axis and Fahrenheit on y-axis, the linear equation is given by: y = (9/5)x + 32 For plotting the graph: Putting x = 0, we have, y = (9/5) × 0 + 32 = 32 putting x = 5, we have, y = (9/5) × 5 + 32 = 41 Putting x = 10, we have, y = (9/5) × 10 + 32 = 50 Hence, A(0, 100), B(5, 41)Read more
(i) Taking Celsius on x-axis and Fahrenheit on y-axis, the linear equation is given by: y = (9/5)x + 32
For plotting the graph:
Putting x = 0, we have, y = (9/5) × 0 + 32 = 32
putting x = 5, we have, y = (9/5) × 5 + 32 = 41
Putting x = 10, we have, y = (9/5) × 10 + 32 = 50
Hence, A(0, 100), B(5, 41) and c(10, 50) are the solutions of the equation.
(ii) If the temperature is 30° C, then
F = (9/5) × 30 + 32 = 54 + 32 = 86
Hence, if the temperature is 30°C, the temperature in Fahrenheit is 86°F.
(iii) If the temperature is 95°F, then
95 = (9/5)C + 32
⇒ 95 – 32 = (9/5)C
⇒ 63 × 5/9 = C
⇒ C = 35°
If the temperature is 95°F, the temperature in Celsius is 35°C.
(iv) If temperature is 0°C, then
F = (9/5) × 0 + 32 = 0 + 32 = 32
If the temperature is 0°F, then
0 = (9/5)C + 32
⇒ -32 = (9/5)C
⇒ -32 × 5/9 = C
⇒ – 160/9 = C
⇒ C = -17.8°
If the temperature is 0°C, the temperature in Fahrenheit is 32°F and if the temperature is 0°F, the temperature in Celsius is -17.8°C.
(v) Let x° be the temperature which is numerically the same in both Fahrenheit and Celsius, then
x = (9/5)x + 32
⇒ x – 32 = (9/5)x
⇒ (x – 32) × 5 = 9x
⇒ 5x – 160 = 9x
⇒ 4x = -160
⇒ x – 40°
Hence, -40° is the temperature which is numerically the same in both Fahrenheit and Celsius.
(i) Equation y = 3 can be represented in one variable on number line. (ii) For two variables representation of y = 3, we will use Cartesian plane. Now the equation: 0.x + y = 3 ⇒ y = 3 - 0.x Putting x = 1, we have, y = 3 - 0.1 = 3 Putting x = 2, we have, y = 3 - 0.2 = 3 Hence, A(1, 3) and B(2, 3) arRead more
(i) Equation y = 3 can be represented in one variable on number line.
(ii) For two variables representation of y = 3, we will use Cartesian plane. Now the equation:
0.x + y = 3
⇒ y = 3 – 0.x
Putting x = 1, we have, y = 3 – 0.1 = 3
Putting x = 2, we have, y = 3 – 0.2 = 3
Hence, A(1, 3) and B(2, 3) are the two solutions of the given equation.
(i) (0,2) Given equation: x - 2y = 4 In x- 2y = 4, putting x = 0 and y = 2, we have, 0 - 2 × 2 = -4 ≠ 4 Therefore, (0, 2) is not a solution of the equation.
(i) (0,2)
Given equation: x – 2y = 4
In x- 2y = 4, putting x = 0 and y = 2, we have, 0 – 2 × 2 = -4 ≠ 4
Therefore, (0, 2) is not a solution of the equation.
(v) (1,1) Given equation: x - 2y = 4 In x - 2y = 4, putting x = 1 and y = 1, we have, 1 - 2 × 1 = -1 ≠ 4 Hence, (1, 1) is not a solution of the equation.
(v) (1,1)
Given equation: x – 2y = 4
In x – 2y = 4, putting x = 1 and y = 1, we have, 1 – 2 × 1 = -1 ≠ 4
Hence, (1, 1) is not a solution of the equation.
In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius: F:(9/5) C+32
(i) Taking Celsius on x-axis and Fahrenheit on y-axis, the linear equation is given by: y = (9/5)x + 32 For plotting the graph: Putting x = 0, we have, y = (9/5) × 0 + 32 = 32 putting x = 5, we have, y = (9/5) × 5 + 32 = 41 Putting x = 10, we have, y = (9/5) × 10 + 32 = 50 Hence, A(0, 100), B(5, 41)Read more
(i) Taking Celsius on x-axis and Fahrenheit on y-axis, the linear equation is given by: y = (9/5)x + 32
For plotting the graph:
Putting x = 0, we have, y = (9/5) × 0 + 32 = 32
putting x = 5, we have, y = (9/5) × 5 + 32 = 41
Putting x = 10, we have, y = (9/5) × 10 + 32 = 50
Hence, A(0, 100), B(5, 41) and c(10, 50) are the solutions of the equation.
(ii) If the temperature is 30° C, then
F = (9/5) × 30 + 32 = 54 + 32 = 86
Hence, if the temperature is 30°C, the temperature in Fahrenheit is 86°F.
(iii) If the temperature is 95°F, then
95 = (9/5)C + 32
⇒ 95 – 32 = (9/5)C
⇒ 63 × 5/9 = C
⇒ C = 35°
If the temperature is 95°F, the temperature in Celsius is 35°C.
(iv) If temperature is 0°C, then
F = (9/5) × 0 + 32 = 0 + 32 = 32
If the temperature is 0°F, then
0 = (9/5)C + 32
⇒ -32 = (9/5)C
⇒ -32 × 5/9 = C
⇒ – 160/9 = C
⇒ C = -17.8°
If the temperature is 0°C, the temperature in Fahrenheit is 32°F and if the temperature is 0°F, the temperature in Celsius is -17.8°C.
(v) Let x° be the temperature which is numerically the same in both Fahrenheit and Celsius, then
See lessx = (9/5)x + 32
⇒ x – 32 = (9/5)x
⇒ (x – 32) × 5 = 9x
⇒ 5x – 160 = 9x
⇒ 4x = -160
⇒ x – 40°
Hence, -40° is the temperature which is numerically the same in both Fahrenheit and Celsius.
Give the geometric representations of y = 3 as an equation.
(i) Equation y = 3 can be represented in one variable on number line. (ii) For two variables representation of y = 3, we will use Cartesian plane. Now the equation: 0.x + y = 3 ⇒ y = 3 - 0.x Putting x = 1, we have, y = 3 - 0.1 = 3 Putting x = 2, we have, y = 3 - 0.2 = 3 Hence, A(1, 3) and B(2, 3) arRead more
(i) Equation y = 3 can be represented in one variable on number line.
See less(ii) For two variables representation of y = 3, we will use Cartesian plane. Now the equation:
0.x + y = 3
⇒ y = 3 – 0.x
Putting x = 1, we have, y = 3 – 0.1 = 3
Putting x = 2, we have, y = 3 – 0.2 = 3
Hence, A(1, 3) and B(2, 3) are the two solutions of the given equation.
Check which of the following are solution of the equation x – 2y = 4 and which are not: (0, 2)
(i) (0,2) Given equation: x - 2y = 4 In x- 2y = 4, putting x = 0 and y = 2, we have, 0 - 2 × 2 = -4 ≠ 4 Therefore, (0, 2) is not a solution of the equation.
(i) (0,2)
See lessGiven equation: x – 2y = 4
In x- 2y = 4, putting x = 0 and y = 2, we have, 0 – 2 × 2 = -4 ≠ 4
Therefore, (0, 2) is not a solution of the equation.
Check which of the following are solutions of the equation x – 2y = 4 and which are not: (4, 0)
(iii) (4,0) Given equation: x - 2y = 4 In x-2y = 4, putting x = 4 and y = 0, we have, 4 - 2 × 0 = 4 Hence, (4,0) is a solution of the equation.
(iii) (4,0)
See lessGiven equation: x – 2y = 4
In x-2y = 4, putting x = 4 and y = 0, we have, 4 – 2 × 0 = 4
Hence, (4,0) is a solution of the equation.
Check which of the following are solutions of the equation x – 2y = 4 and which are not: ( √2, 4√2)
(v) (1,1) Given equation: x - 2y = 4 In x - 2y = 4, putting x = 1 and y = 1, we have, 1 - 2 × 1 = -1 ≠ 4 Hence, (1, 1) is not a solution of the equation.
(v) (1,1)
See lessGiven equation: x – 2y = 4
In x – 2y = 4, putting x = 1 and y = 1, we have, 1 – 2 × 1 = -1 ≠ 4
Hence, (1, 1) is not a solution of the equation.