Learn Mathematics with Free Lessons & Tips

1. • =15 cm

2. Longest Pole Length:

   • The longest pole that can fit inside the room without protruding is equal to the diagonal length of the room.
   • Therefore, the longest pole length = 15 cm.

Conclusion: Hence, the longest pole that can be put in a room with dimensions l=10l=10 cm, b=10b=10 cm, and h=5h=5 cm is 15 cm.

Locating √3 on the Number Line

Introduction Locating √3 on the number line is an essential concept in mathematics, particularly in understanding irrational numbers and their placement in relation to rational numbers.

Understanding √3 √3 represents the square root of 3, which is an irrational number. An irrational number cannot be expressed as a fraction of two integers and has an infinite non-repeating decimal expansion.

Steps to Locate √3 on the Number Line

1. Identify Nearby Perfect Squares:

   • √3 lies between the perfect squares of 1 and 4.
   • √1 = 1 and √4 = 2.

2. Estimation:

   • Since 3 is between 1 and 4, the square root of 3 will be between 1 and 2.
   • By estimation, √3 is approximately 1.732.

3. Plotting √3 on the Number Line:

   • Start at 0 on the number line.
   • Move to the right until you reach approximately 1.732 units.

4. Final Position:

   • Mark the point on the number line corresponding to √3.

Conclusion Locating √3 on the number line involves understanding its position between perfect squares and accurately plotting its approximate value. This skill is fundamental for comprehending the continuum of real numbers and their relationships.

Are the square roots of all positive integers irrational?

Introduction: The question probes into the nature of square roots of positive integers, whether they are exclusively irrational or if there are exceptions.

Explanation: The statement that the square roots of all positive integers are irrational is false. While there are indeed many examples of square roots that are irrational, there are also instances where the square root of a positive integer results in a rational number.

Example:

• Square root of 4:
  • Integer: 4
  • Square root: √4 = 2
  • Nature: Rational

Explanation of the Example:

• The square root of 4 is 2, which is a rational number.
• This contradicts the notion that all square roots of positive integers are irrational.

Conclusion: In conclusion, not all square roots of positive integers are irrational. The square root of 4, for instance, is a rational number, demonstrating that exceptions exist to the notion that all such roots are irrational.

Decimal Expansions of Fractions

1. Decimal Expansion of 10/3:

• Calculation:

  • Divide 10 by 3.
  • The result will be 3.3333...

• Decimal Expansion:

  • 103=3.3‾310=3.3

2. Decimal Expansion of 7/8:

• Calculation:

  • Divide 7 by 8.
  • The result will be 0.875.

• Decimal Expansion:

  • 78=0.87587=0.875

3. Decimal Expansion of 1/7:

• Calculation:

  • Divide 1 by 7.
  • The result will be 0.142857142857...

• Decimal Expansion:

  • 17=0.142857‾71=0.142857

Conclusion:

• The decimal expansions for the given fractions are:
  • 103=3.3‾310=3.3
  • 78=0.87587=0.875
  • 17=0.142857‾71=0.142857

Finding Solutions of Line AB Equation

Given Information:

• Line AB is represented by the equation.
• A graph depicting Line AB is provided.

Procedure:

1. Identify Points on Line AB: Locate four points on the graph that lie on Line AB.
2. Determine Coordinates: Extract the coordinates of these points.
3. Substitute Coordinates: Substitute the coordinates into the equation of Line AB.
4. Verify Solutions: Confirm that the substituted coordinates satisfy the equation of Line AB.

1. Identify Points on Line AB:

• Locate four distinct points where the line intersects the axes or stands out on the graph.

2. Determine Coordinates:

• Note down the coordinates (x, y) of each identified point.

3. Substitute Coordinates:

• Use the coordinates obtained to substitute into the equation of Line AB.
• The equation of a line is typically in the form y = mx + b, where m is the slope and b is the y-intercept.

4. Verify Solutions:

• Confirm that the substituted coordinates satisfy the equation of Line AB.
• The substituted values should make the equation true when solved.

Example:

• Suppose the equation representing Line AB is y = 2x + 3.
• Points on the graph are (0, 3), (1, 5), (2, 7), and (-1, 1).
• Substituting these coordinates into the equation:
  • For (0, 3): 3 = 2(0) + 3 (True)
  • For (1, 5): 5 = 2(1) + 3 (True)
  • For (2, 7): 7 = 2(2) + 3 (True)
  • For (-1, 1): 1 = 2(-1) + 3 (True)
• All points satisfy the equation, confirming they lie on Line AB.

Conclusion:

• By following these steps, you can find solutions of the equation representing Line AB from the provided graph.

Writing a Linear Equation for Taxi Fare

Given Information:

• Initial fare: Rs 10 for the first kilometre
• Subsequent fare: Rs 6 per km
• Distance: xx km
• Total fare: Rs yy

Formulating the Linear Equation

Let's denote:

• xx: Distance travelled in kilometres
• yy: Total fare in rupees

Equation for Total Fare:

The total fare can be calculated as the sum of the initial fare and the fare for the subsequent distance.

So, the equation can be expressed as:

y=10+6(x−1)y=10+6(x−1)

Where:

• x−1x−1: Represents the distance after the first kilometre

Calculating Total Fare for 15 km

Now, let's substitute x=15x=15 into the equation to find the total fare for a 15 km journey.

y=10+6(15−1)y=10+6(15−1) y=10+6(14)y=10+6(14) y=10+84y=10+84 y=94y=94

Answer:

The total fare for a 15 km journey would be Rs. 94.

Problem Analysis:

Given the equation 2x−y=p2x−y=p and a solution point (1,−2)(1,−2), we need to find the value of pp.

Solution:

Step 1: Substitute the Given Solution into the Equation

Substitute the coordinates of the given solution point (1,−2)(1,−2) into the equation:

2(1)−(−2)=p2(1)−(−2)=p

Step 2: Solve for pp

2+2=p2+2=p 4=p4=p

Step 3: Final Result

p=4p=4

Conclusion:

The value of pp for the equation 2x−y=p2x−y=p when the point (1,−2)(1,−2) is a solution is 44.

Graph of the Equation x - y = 4

Graphing the Equation:

To draw the graph of the equation x−y=4x−y=4, we'll first rewrite it in slope-intercept form, which is y=mx+by=mx+b, where mm is the slope and bb is the y-intercept.

Given equation: x−y=4x−y=4

Rewriting in slope-intercept form:

y=x−4y=x−4

Now, let's plot the graph using this equation.

Plotting the Graph:

1. Find y-intercept:
   Set x=0x=0 in the equation y=x−4y=x−4
   y=0−4y=0−4
   y=−4y=−4
   So, the y-intercept is at the point (0,−4)(0,−4).

2. Find x-intercept:
   To find the x-intercept, set y=0y=0 in the equation y=x−4y=x−4.
   0=x−40=x−4
   x=4x=4
   So, the x-intercept is at the point (4,0)(4,0).

Drawing the Graph:

Now, plot the points (0,−4)(0,−4) and (4,0)(4,0) on the Cartesian plane and draw a straight line passing through these points. This line represents the graph of the equation x−y=4x−y=4.

Intersecting with the x-axis:

To find where the graph line meets the x-axis, we need to find the point where y=0y=0.

Substitute y=0y=0 into the equation x−y=4x−y=4:

x−0=4x−0=4

x=4x=4

So, when the graph line meets the x-axis, the coordinates of the point are (4,0)(4,0).

Graphing the Equation x + 2y = 6

To graph the equation x+2y=6x+2y=6, we'll first rewrite it in slope-intercept form (y=mx+by=mx+b):

x+2y=6x+2y=6 2y=−x+62y=−x+6 y=−12x+3y=−21x+3

Plotting the Graph

To plot the graph, we'll identify two points and draw a line through them:

1. Intercept Method:

   • y-intercept (when x = 0): y=−12(0)+3=3y=−21(0)+3=3 Therefore, the y-intercept is (0, 3).
   • x-intercept (when y = 0): 0=−12x+30=−21x+3 −12x=3−21x=3 x=−6x=−6 Therefore, the x-intercept is (-6, 0).

2. Slope Method: From the slope-intercept form y=−12x+3y=−21x+3, the slope is -1/2, meaning the line decreases by 1 unit in the y-direction for every 2 units in the x-direction.

Plotting the Points and Drawing the Line

Using the intercepts and the slope, we plot the points (0, 3) and (-6, 0), then draw a line through them.

Finding the Value of x when y = -3

Given y=−3y=−3, we substitute this value into the equation y=−12x+3y=−21x+3 and solve for x:

−3=−12x+3−3=−21x+3 −12x=−3−3−21x=−3−3 −12x=−6−21x=−6 x=−6×(−2)x=−6×(−2) x=12x=12

Conclusion

• The graph of the equation x+2y=6x+2y=6 is a straight line passing through points (0, 3) and (-6, 0).
• The value of xx when y=−3y=−3 is x=12x=12.

Perimeter Calculation for Rectangle with Given Area

Given Information:

• Area of the rectangle: 25x2−35x+1225x2−35x+12

Step 1: Determine the Dimensions

To calculate the perimeter of a rectangle, we need to know its length and width. We can find these dimensions using the area provided.

Step 2: Factorize the Area

Factorize the given quadratic expression 25x2−35x+1225x2−35x+12 to find its factors, which represent the possible lengths and widths of the rectangle.

Step 3: Use Factorization to Find Dimensions

Once the quadratic expression is factorized, identify the pairs of factors that, when multiplied, give the area of the rectangle. These pairs represent possible lengths and widths.

Step 4: Calculate Perimeter

With the length and width of the rectangle known, calculate the perimeter using the formula:

Perimeter=2×(Length+Width)Perimeter=2×(Length+Width)

Step 5: Finalize

Plug in the values of length and width into the perimeter formula to obtain the final result.

Let's proceed with these steps to find the perimeter.