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Finding the Base Area of a Right Circular Cylinder

Understanding the Problem To find the base area of a right circular cylinder, we need to utilize the given information about its circumference.

Given Information:

• Circumference of the base: 110 cm

Solution Steps:

1. Determine the Radius:
   • The circumference of a circle CC is given by the formula: C=2πrC=2πr.
   • Given C=110C=110 cm, we can rearrange the formula to solve for the radius rr: 110=2πr110=2πr Solving for rr: r=1102πr=2π110
2. Calculate the Base Area:
   • The formula for the area AA of a circle is: A=πr2A=πr2.
   • Plug in the value of rr obtained from step 1 into the formula: A=π(1102π)2A=π(2π110)2 Simplify: A=π(11024π2)A=π(4π21102) A=11024πA=4π1102
3. Final Calculation:
   • Calculate the value of AA: A=121004πA=4π12100 A≈3035.5πA≈π3035.5 A≈964.88A≈964.88 sq. cm (rounded to two decimal places)

Conclusion:

• The base area of the right circular cylinder is approximately 964.88964.88 square centimeters.

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].

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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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.

Determining the Value of k

Introduction: To find the value of k when (x – 1) is a factor of the polynomial 4x^3 + 3x^2 – 4x + k, we'll utilize the Factor Theorem.

Factor Theorem: If (x – c) is a factor of a polynomial, then substituting c into the polynomial should result in zero.

Procedure:

1. Substitute x=1x=1 into the polynomial to make (x – 1) a factor.
2. Equate the result to zero.
3. Solve for k.

Step-by-Step Solution:

1. Substitute x=1x=1:

   • 4(1)3+3(1)2–4(1)+k=04(1)3+3(1)2–4(1)+k=0
   • 4+3–4+k=04+3–4+k=0

2. Solve for k:

   • 3+k=03+k=0
   • k=−3k=−3

Conclusion: The value of k when (x – 1) is a factor of the given polynomial is k=−3k=−3.

Solution: Finding Values of a and b

Given Problem: If x3+ax2–bx+10x3+ax2–bx+10 is divisible by x2–3x+2x2–3x+2, we need to find the values of aa and bb.

Solution Steps:

Step 1: Determine the factors of the divisor

Given divisor: x2–3x+2x2–3x+2

We need to find two numbers that multiply to 22 and add up to −3−3.

The factors of 22 are 11 and 22.

So, the factors that add up to −3−3 are −2−2 and −1−1.

Hence, the divisor factors are (x–2)(x–2) and (x–1)(x–1).

So, the divisor can be written as (x–2)(x–1)(x–2)(x–1).

Step 2: Use Remainder Theorem

If f(x)=x3+ax2–bx+10f(x)=x3+ax2–bx+10 is divisible by (x–2)(x–1)(x–2)(x–1), then the remainder when f(x)f(x) is divided by x2–3x+2x2–3x+2 is zero.

According to Remainder Theorem, if f(x)f(x) is divided by x2–3x+2x2–3x+2, then the remainder is given by f(2)f(2) and f(1)f(1) respectively.

Step 3: Find the value of aa

Substitute x=2x=2 into f(x)f(x) and equate it to 00 to find the value of aa.

f(2)=23+a(2)2–b(2)+10f(2)=23+a(2)2–b(2)+10

0=8+4a–2b+100=8+4a–2b+10

18=4a–2b18=4a–2b

4a–2b=184a–2b=18

Step 4: Find the value of bb

Substitute x=1x=1 into f(x)f(x) and equate it to 00 to find the value of bb.

f(1)=13+a(1)2–b(1)+10f(1)=13+a(1)2–b(1)+10

0=1+a–b+100=1+a–b+10

11=a–b11=a–b

a–b=11a–b=11

Step 5: Solve the equations

Now we have two equations:

1. 4a–2b=184a–2b=18
2. a–b=11a–b=11

We can solve these equations simultaneously to find the values of aa and bb.

Step 6: Solve the equations

Equation 1: 4a–2b=184a–2b=18

Divide by 2: 2a–b=92a–b=9

Equation 2: a–b=11a–b=11

Step 7: Solve the system of equations

Adding equation 2 to equation 1: (2a–b)+(a–b)=9+11(2a–b)+(a–b)=9+11

3a=203a=20

a=203a=320

Substitute a=203a=320 into equation 2: 203–b=11320–b=11

b=203–11b=320–11

b=20–333b=320–33

b=−133b=3−13

Step 8: Final values of aa and bb

a=203a=320

b=−133b=3−13

So, the values of aa and bb are a=203a=320 and b=−133b=3−13 respectively.