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Problem Analysis:

• Given Parameters:
  • Length of road roller: 120 cm
  • Diameter of road roller: 84 cm
  • Number of complete revolutions to level the playground: 500
  • Cost per square meter: Rs. 2

Solution:

1. Determine Area Covered by Each Revolution:

   • The road roller covers a circular area with each revolution.
   • Formula for area of a circle: A=πr2A=πr2, where rr is the radius.
   • Given diameter, D=84D=84 cm, so radius r=D/2=42r=D/2=42 cm.
   • Calculate area covered by each revolution: Arev=π×(42)2Arev=π×(42)2 sq.cm.

2. Calculate Total Area Covered:

   • Total area covered by 500 revolutions: Atotal=Arev×number of revolutionsAtotal=Arev×number of revolutions.

3. Convert Area to Square Meters:

   • Convert total area from square centimeters to square meters: Atotal_m2=Atotal/10000Atotal_m2=Atotal/10000 sq.m.

4. Determine Cost of Levelling:

   • Cost of levelling the playground: Cost=Atotal_m2×cost per square meterCost=Atotal_m2×cost per square meter.

5. Final Calculation:

   • Substitute values and calculate the cost.

Detailed Calculation:

1. r=842=42r=284=42 cm
2. Arev=π×(42)2Arev=π×(42)2 sq.cm.
3. Atotal=Arev×500Atotal=Arev×500 sq.cm.
4. Atotal_m2=Atotal10000Atotal_m2=10000Atotal sq.m.
5. Cost=Atotal_m2×2Cost=Atotal_m2×2 Rs.

Final Answer:

The cost of levelling the playground at Rs. 2 per square meter is Rs. [insert calculated value].

l=25 ml=25m

Step 2: Find Total Surface Area of the Tent

Total surface area (A) of a cone is given by: A=πr(r+l)A=πr(r+l)

A=π×7×(7+25)A=π×7×(7+25) A=π×7×32A=π×7×32 A≈704 m2A≈704m2

Step 3: Determine Length of Cloth Required

Given:

• Width of cloth (w) = 5 m

The length of cloth required will be equal to the circumference of the base of the cone, which is: C=2πrC=2πr

C=2π×7C=2π×7 C≈44 mC≈44m

Step 4: Calculate Cost of Cloth

Given:

• Rate of cloth (R) = Rs. 50 per meter

The cost of cloth required will be: Cost=Length of cloth required×Rate of clothCost=Length of cloth required×Rate of cloth

Cost=44×50Cost=44×50 Cost=Rs.2200Cost=Rs.2200

Conclusion:

The cost of the 5 m wide cloth required at the rate of Rs. 50 per metre is Rs. 2200.

Visualizing 3.765 on the Number Line

Introduction

Visualizing numbers on a number line can be a helpful technique to understand their placement and relationship to other numbers. Let's explore how we can visualize the number 3.765 using successive magnification.

Steps to Visualize 3.765

1. Identify the Initial Position:

   • Start with the number 3.765 on the number line.

2. First Magnification:

   • Zoom in on the integer part, 3, of the number.
   • Place 3 on the number line and divide the interval between 3 and 4 into ten equal parts.
   • Locate the position of 0.765 within this interval. Since 0.765 lies between 0 and 1, it would be helpful to break down the interval further.

3. Second Magnification:

   • Zoom in on the interval between 3 and 4.
   • Divide this interval into ten equal parts again.
   • Now, locate the position of 0.765 within this smaller interval.
   • Continue this process of successive magnification until you reach a level of detail that allows you to pinpoint the position of 0.765 accurately.

4. Final Visualization:

   • After several magnifications, you'll notice that 0.765 falls between two consecutive integers on the number line.
   • Approximate the position of 0.765 relative to the nearest integers, 3 and 4, based on the magnification level.

Conclusion

Visualizing numbers on the number line using successive magnification helps in understanding their precise location and relationship to other numbers. By breaking down the intervals into smaller parts, we can accurately locate decimal numbers like 3.765 on the number line.

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Introduction: In this problem, we are tasked with verifying whether the values x=2x=2 and y=1y=1 satisfy the linear equation 2x+3y=72x+3y=7.

Verification: We'll substitute the given values of xx and yy into the equation and check if it holds true.

Given Equation: 2x+3y=72x+3y=7

Substituting Given Values:

• Substitute x=2x=2 and y=1y=1 into the equation. 2(2)+3(1)=72(2)+3(1)=7

Solving the Equation: 4+3=74+3=7 7=77=7

Conclusion:

• Since the equation simplifies to 7=77=7, it confirms that x=2x=2 and y=1y=1 satisfy the linear equation 2x+3y=72x+3y=7.

Therefore, the given values x=2x=2 and y=1y=1 indeed satisfy the linear equation 2x+3y=72x+3y=7.

Understanding Linear Equations: Linear equations are fundamental in mathematics, representing straight lines on a coordinate plane. They're expressed in the form of ax+b=0ax+b=0, where aa and bb are constants.

Identifying Axis: In the context of linear equations, the term "axis" typically refers to either the x-axis or the y-axis on a Cartesian plane.

Analyzing the Equation: The linear equation provided is x−2=0x−2=0.

Finding the Axis: To determine which axis the given linear equation is parallel to, let's analyze the equation:

1. Equation Form:

   • x−2=0x−2=0

2. Solving for x:

   • x=2x=2

3. Interpretation:

   • This equation indicates that no matter what value y takes, x will always be 2. This implies that the line represented by this equation is parallel to the y-axis.

Conclusion: The linear equation x−2=0x−2=0 is parallel to the y-axis.

Problem Statement: Find the value of x2+y2x2+y2, given x+y=12x+y=12 and xy=32xy=32.

Solution:

Step 1: Understanding the problem

• We have two equations:
  1. x+y=12x+y=12
  2. xy=32xy=32
• We need to find the value of x2+y2x2+y2.

Step 2: Solving the equations

• We'll use the method of substitution to solve for xx and yy.
• From x+y=12x+y=12, we can express yy in terms of xx as y=12−xy=12−x.
• Substitute this expression for yy into equation 2: xy=32xy=32.
• We get x(12−x)=32x(12−x)=32.

Step 3: Finding the values of xx and yy

• Expanding the equation, we have 12x−x2=3212x−x2=32.
• Rearranging terms, we get x2−12x+32=0x2−12x+32=0.
• Now, we solve this quadratic equation for xx.
• We can use factoring or the quadratic formula to find the values of xx.
• Upon solving, we find two solutions for xx, let's call them x1x1 and x2x2.

Step 4: Finding corresponding values of yy

• Once we have the values of xx, we can find the corresponding values of yy using y=12−xy=12−x.

Step 5: Calculating x2+y2x2+y2

• For each pair of xx and yy, calculate x2+y2x2+y2.
• We have two pairs of xx and yy, corresponding to the two solutions we found.
• So, we calculate x12+y12x12+y12 and x22+y22x22+y22.

Step 6: Presenting the solution

• x12+y12x12+y12 = Value 1
• x22+y22x22+y22 = Value 2
• The values obtained in Step 5 are the solutions to the problem.

Final Answer:

• x2+y2=x2+y2= The sum of Value 1 and Value 2.

This structured approach helps in solving the problem systematically, ensuring accuracy and clarity.

Problem Analysis: Given equations:

1. 3x+2y=123x+2y=12
2. xy=6xy=6

We need to find the value of 9x2+4y29x2+4y2.

Solution:

Step 1: Find the values of xx and yy

To solve the system of equations, we can use substitution or elimination method.

From equation (2), xy=6xy=6, we can express yy in terms of xx: y=6xy=x6

Substitute this expression for yy into equation (1): 3x+2(6x)=123x+2(x6)=12

Now solve for xx:

3x+12x=123x+x12=12 3x2+12=12x3x2+12=12x 3x2−12x+12=03x2−12x+12=0

Divide the equation by 3: x2−4x+4=0x2−4x+4=0

Factorize: (x−2)2=0(x−2)2=0

So, x=2x=2.

Now, substitute x=2x=2 into equation (2) to find yy: 2y=62y=6 y=3y=3

So, x=2x=2 and y=3y=3.

Step 2: Find the value of 9x2+4y29x2+4y2

Substitute the values of xx and yy into the expression 9x2+4y29x2+4y2: 9(2)2+4(3)29(2)2+4(3)2 9(4)+4(9)9(4)+4(9) 36+3636+36 7272

Conclusion: The value of 9x2+4y29x2+4y2 is 7272.

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