Share & grow the world's knowledge!
We want to connect the people who have knowledge to the people who need it, to bring together people with different perspectives so they can understand each other better, and to empower everyone to share their knowledge.
Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by °C = a °F + b. Find a and b, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit and water boils at 100 degrees Celsius and 212 degrees Fahrenheit. (Hint: When °C = 0, °F = 32 and when °C = 100, °F = 212. Use this information to find a and b and thus, the linear relationship between °C and °F.)
Substitute the temperature coordinates into the linear relationship C = aF + b. For the melting point, 0 = 32a + b, which means b = -32a. For the boiling point, 100 = 212a + b. Substituting b into the second equation yields 100 = 212a - 32a, which simplifies to 100 = 180a. This gives a = 100/180 = 5Read more
Substitute the temperature coordinates into the linear relationship C = aF + b. For the melting point, 0 = 32a + b, which means b = -32a. For the boiling point, 100 = 212a + b. Substituting b into the second equation yields 100 = 212a – 32a, which simplifies to 100 = 180a. This gives a = 100/180 = 5/9. Using a to find b gives b = -160/9. The values are a = 5/9 and b = -160/9.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2 Introduction to Linear Polynomials (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-2/
See lessWrite a polynomial of degree 3 in the variable x, in which the coefficient of the x2 term is –7.
A polynomial of degree 3 must have a maximum exponent of 3 on its variable. The general expression can be represented as ax3 + bx2 + cx + d where a cannot be zero. Following the specific instruction that the coefficient of the x2 term must be exactly -7, we can choose arbitrary values for the otherRead more
A polynomial of degree 3 must have a maximum exponent of 3 on its variable. The general expression can be represented as ax3 + bx2 + cx + d where a cannot be zero. Following the specific instruction that the coefficient of the x2 term must be exactly -7, we can choose arbitrary values for the other coefficients. An appropriate and simple polynomial satisfying these conditions is x3 – 7×2 + x + 1.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2 Introduction to Linear Polynomials (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-2/
See lessIf we multiply a number by 5/2 and add 2/3 to the product, we get –7/12. Find the number.
Let the required number be represented by the variable x. Translating the word problem into a mathematical equation results in 5/2 x + 2/3 = -7/12. To solve for x, subtract 2/3 from both sides of the expression, making 5/2 x = -7/12 - 8/12, which combines to -15/12. Multiplying both sides by 2/5 isoRead more
Let the required number be represented by the variable x. Translating the word problem into a mathematical equation results in 5/2 x + 2/3 = -7/12. To solve for x, subtract 2/3 from both sides of the expression, making 5/2 x = -7/12 – 8/12, which combines to -15/12. Multiplying both sides by 2/5 isolates the variable, resulting in x = (-15/12) multiplied by (2/5), which simplifies perfectly to -1/2.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2 Introduction to Linear Polynomials (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-2/
See lessIf you have rupees 800 and you save rupees 250 every month, find the amount you have after (i) 6 months (ii) 2 years. Express this as a linear pattern.
The savings situation forms a linear pattern represented by the equation y = 800 + 250m, where m represents the number of months passed. For part (i), substituting m = 6 months into the expression results in 800 + 250(6) = 2300 rupees. For part (ii), 2 years is equivalent to 24 months. SubstitutingRead more
The savings situation forms a linear pattern represented by the equation y = 800 + 250m, where m represents the number of months passed. For part (i), substituting m = 6 months into the expression results in 800 + 250(6) = 2300 rupees. For part (ii), 2 years is equivalent to 24 months. Substituting m = 24 into the formula results in 800 + 250(24) = 6800 rupees altogether.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2 Introduction to Linear Polynomials (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-2/
See lessThe digits of a two-digit number differ by 3. If the digits are interchanged and the resulting number is added to the original number, we get 143. Find both the numbers.
Let the digits be x and y. Since they differ by 3, we can assume y = x - 3. The original two-digit number is 10x + y, which simplifies to 11x - 3. Interchanging the digits creates the new number 10y + x, which simplifies to 11x - 30. Adding these two expressions gives (11x - 3) + (11x - 30) = 143. TRead more
Let the digits be x and y. Since they differ by 3, we can assume y = x – 3. The original two-digit number is 10x + y, which simplifies to 11x – 3. Interchanging the digits creates the new number 10y + x, which simplifies to 11x – 30. Adding these two expressions gives (11x – 3) + (11x – 30) = 143. This reduces to 22x – 33 = 143, meaning 22x = 176, so x = 8. The numbers are 85 and 58.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2 Introduction to Linear Polynomials (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-2/
See less