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Solution:

Step 1: Understand the Problem

To solve this problem, we need to find the ratio of the volumes of two spheres when the radius of one sphere is doubled.

Step 2: Use the Volume Formula for a Sphere

The volume VV of a sphere is given by the formula:

V=43πr3V=34πr3

Where:

• VV is the volume of the sphere
• rr is the radius of the sphere
• ππ is a constant approximately equal to 3.14159

Step 3: Determine the Ratios

Let's denote:

• V1V1 as the volume of the sphere with the original radius
• V2V2 as the volume of the sphere with the doubled radius

Given that the radius of the second sphere is twice the radius of the first sphere, we have:

r2=2r1r2=2r1

Step 4: Calculate the Ratios

Substituting the values into the volume formula, we get:

For the first sphere: V1=43πr13V1=34πr13

For the second sphere: V2=43π(2r1)3V2=34π(2r1)3

Now, we can find the ratio of their volumes:

Ratio of volumes=V2V1=43π(2r1)343πr13Ratio of volumes=V1V2=34πr1334π(2r1)3

=8r13πr13π=r13π8r13π

=81=8=18=8

Step 5: Conclusion

The ratio of the volumes of the two spheres is 8:18:1.

So, when the radius of a sphere is doubled, the ratio of their volumes becomes 8:18:1.

Problem Solving: Finding Height and Total Surface Area of a Cylinder

Given Information:

• Radius of the cylinder: r=7r=7 cm
• Volume of the cylinder: V=2002V=2002 cm³

Step 1: Finding the Height of the Cylinder

The formula for the volume of a cylinder is given by: V=πr2hV=πr2h

Where:

• VV is the volume of the cylinder
• rr is the radius of the cylinder
• hh is the height of the cylinder

Substituting the given values: 2002=π×(7)2×h2002=π×(7)2×h

2002=49π×h2002=49π×h

h=200249πh=49π2002

Now, calculate the value of hh:

h≈200249×3.14h≈49×3.142002

h≈2002153.86h≈153.862002

h≈12.99 cmh≈12.99cm

So, the height of the cylinder is approximately 12.99 cm12.99cm.

Step 2: Finding the Total Surface Area of the Cylinder

The formula for the total surface area of a cylinder is given by: A=2πrh+2πr2A=2πrh+2πr2

Where:

• AA is the total surface area of the cylinder
• rr is the radius of the cylinder
• hh is the height of the cylinder

Substituting the given values: A=2π×7×12.99+2π×(7)2A=2π×7×12.99+2π×(7)2

A=2π×7×12.99+2π×49A=2π×7×12.99+2π×49

A=2π×90.93+98πA=2π×90.93+98π

A=181.86π+98πA=181.86π+98π

A=279.86πA=279.86π

Now, calculate the value of AA:

A≈279.86×3.14A≈279.86×3.14

A≈878.66 cm2A≈878.66cm2

So, the total surface area of the cylinder is approximately 878.66 cm2878.66cm2.

Conclusion:

• The height of the cylinder is approximately 12.99 cm12.99cm.
• The total surface area of the cylinder is approximately 878.66 cm2878.66cm2.

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.

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.