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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

Graph of the Equation 2x – 3y = 12

To draw the graph of the equation 2x−3y=122x−3y=12, let's 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.

Rewrite Equation in Slope-Intercept Form

2x−3y=122x−3y=12
−3y=−2x+12−3y=−2x+12
y=23x−4y=32x−4

Plotting the y-intercept and Slope

1. Y-intercept: When x=0x=0,
   y=23(0)−4y=32(0)−4
   y=−4y=−4
   So, the y-intercept is (0, -4).

2. Slope: The coefficient of xx is 2332, which represents the slope.
   For every increase of 1 in xx, yy increases by 2332.
   For every decrease of 1 in xx, yy decreases by 2332.

Plotting Points and Drawing the Graph

Now, let's plot some points to draw the graph:

• x = 3: y=23(3)−4=2−4=−2y=32(3)−4=2−4=−2
  Point: (3, -2)

• x = 6: y=23(6)−4=4−4=0y=32(6)−4=4−4=0
  Point: (6, 0)

• x = -3: y=23(−3)−4=−2−4=−6y=32(−3)−4=−2−4=−6
  Point: (-3, -6)

Plotting the Graph

With these points, we can draw a straight line passing through them.

Points where the Graph Intersects the Axes

X-axis

To find where the graph intersects the x-axis, we set y=0y=0 and solve for xx:

0=23x−40=32x−4
23x=432x=4
x=4×32x=24×3
x=6x=6

So, the graph intersects the x-axis at x=6x=6, which corresponds to the point (6, 0).

Y-axis

To find where the graph intersects the y-axis, we set x=0x=0 and solve for yy:

y=23(0)−4y=32(0)−4
y=−4y=−4

So, the graph intersects the y-axis at y=−4y=−4, which corresponds to the point (0, -4).

Summary

• X-axis intersection: (6, 0)
• Y-axis intersection: (0, -4)

This information helps us visualize and understand the behavior of the equation 2x−3y=122x−3y=12 on the coordinate plane.

Graph of 9x – 5y + 160 = 0

To graph the equation 9x – 5y + 160 = 0, we'll first rewrite it in slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept.

Step 1: Rewrite the equation in slope-intercept form

9x – 5y + 160 = 0

Subtract 9x from both sides:

-5y = -9x - 160

Divide both sides by -5 to isolate y:

y = (9/5)x + 32

Now we have the equation in slope-intercept form.

Step 2: Identify the slope and y-intercept

The slope (m) is 9/5 and the y-intercept (b) is 32.

Step 3: Plot the y-intercept and use the slope to find additional points

Now, let's plot the y-intercept at (0, 32). From there, we'll use the slope to find another point. The slope of 9/5 means that for every 5 units we move to the right along the x-axis, we move 9 units upwards along the y-axis.

So, starting from (0, 32), if we move 5 units to the right, we move 9 units up to get the next point.

Step 4: Plot the points and draw the line

Plot the y-intercept at (0, 32) and the next point at (5, 41). Then, draw a line through these points to represent the graph of the equation.

Finding the value of y when x = 5

To find the value of y when x = 5, we'll substitute x = 5 into the equation and solve for y.

9x – 5y + 160 = 0

9(5) – 5y + 160 = 0

45 – 5y + 160 = 0

Combine like terms:

-5y + 205 = 0

Subtract 205 from both sides:

-5y = -205

Divide both sides by -5 to solve for y:

y = 41

So, when x = 5, y = 41.

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.

Monomial and Binomial Examples with Degrees

Monomial Example (Degree: 82)

• Definition: A monomial is a mathematical expression consisting of a single term.
• Example: 5x825x82
  • Explanation:
    • The coefficient is 55.
    • The variable is xx.
    • The exponent is 8282.

Binomial Example (Degree: 99)

• Definition: A binomial is a polynomial with two terms.
• Example: 3x99+2x983x99+2x98
  • Explanation:
    • The first term: 3x993x99
      • Coefficient: 33
      • Variable: xx
      • Exponent: 9999
    • The second term: 2x982x98
      • Coefficient: 22
      • Variable: xx
      • Exponent: 9898

Additional Notes:

• Monomials have only one term, whereas binomials have two terms.
• The degree of a monomial is the sum of the exponents of its variables.
• The degree of a binomial is the highest degree of its terms.

Given:

• x2+y2+z2=83x2+y2+z2=83
• x+y+z=15x+y+z=15

To Find:

• x3+y3+z3−3xyzx3+y3+z3−3xyz

Approach:

1. Use the identity (x+y+z)3=x3+y3+z3+3(x+y)(y+z)(z+x)(x+y+z)3=x3+y3+z3+3(x+y)(y+z)(z+x).
2. Given x+y+z=15x+y+z=15, find x3+y3+z3x3+y3+z3.
3. Also, find (x+y)(y+z)(z+x)(x+y)(y+z)(z+x).
4. Substitute the values in the expression x3+y3+z3−3xyzx3+y3+z3−3xyz.

Step-by-Step Solution:

1. Find x3+y3+z3x3+y3+z3:

   • Using the identity (x+y+z)3=x3+y3+z3+3(x+y)(y+z)(z+x)(x+y+z)3=x3+y3+z3+3(x+y)(y+z)(z+x), where x+y+z=15x+y+z=15.
   • (15)3=x3+y3+z3+3(x+y)(y+z)(z+x)(15)3=x3+y3+z3+3(x+y)(y+z)(z+x)
   • 3375=x3+y3+z3+3(xy+yz+zx+3xyz)3375=x3+y3+z3+3(xy+yz+zx+3xyz) (Expanding (x+y+z)3(x+y+z)3)
   • x3+y3+z3=3375−3(xy+yz+zx)x3+y3+z3=3375−3(xy+yz+zx) (Subtracting 3xyz3xyz from both sides)

2. Find (x+y)(y+z)(z+x)(x+y)(y+z)(z+x):

   • Given x+y+z=15x+y+z=15, let's find xy+yz+zxxy+yz+zx.
   • Squaring x+y+z=15x+y+z=15:
     • (x+y+z)2=(15)2(x+y+z)2=(15)2
     • x2+y2+z2+2(xy+yz+zx)=225x2+y2+z2+2(xy+yz+zx)=225 (Expanding (x+y+z)2(x+y+z)2)
     • 83+2(xy+yz+zx)=22583+2(xy+yz+zx)=225 (Given x2+y2+z2=83x2+y2+z2=83)
     • xy+yz+zx=225−832=71xy+yz+zx=2225−83=71
   • Using (x+y)(y+z)(z+x)=(xy+yz+zx)+xyz(x+y)(y+z)(z+x)=(xy+yz+zx)+xyz:
     • (x+y)(y+z)(z+x)=71+xyz(x+y)(y+z)(z+x)=71+xyz

3. Substitute values into the expression:

   • x3+y3+z3−3xyz=3375−3(71)−3xyzx3+y3+z3−3xyz=3375−3(71)−3xyz
   • x3+y3+z3−3xyz=3375−213−3xyzx3+y3+z3−3xyz=3375−213−3xyz
   • x3+y3+z3−3xyz=3162−3xyzx3+y3+z3−3xyz=3162−3xyz

Final Answer:

• x3+y3+z3−3xyz=3162−3xyzx3+y3+z3−3xyz=3162−3xyz