Learn Mathematics with Free Lessons & Tips

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.

Finding Rational Numbers Between 1 and 2

Rational numbers are those that can be expressed as a fraction of two integers. Here's how we can find five rational numbers between 1 and 2.

Method 1: Using Averaging

1. Step 1: Average 1 and 2 to find the first rational number:

   • (1 + 2) / 2 = 3 / 2 = 1.5

2. Step 2: Repeat the process to find more rational numbers:

   • (1 + 1.5) / 2 = 2.5 / 2 = 1.25
   • (1.25 + 1.5) / 2 = 2.75 / 2 = 1.375
   • (1.25 + 1.375) / 2 = 2.625 / 2 = 1.3125
   • (1.3125 + 1.375) / 2 = 2.6875 / 2 = 1.34375

Method 2: Using Reciprocals

1. Step 1: Take the reciprocal of 2:

   • 1 / 2 = 0.5

2. Step 2: Repeat the process to find more rational numbers:

   • 1 / (2 + 1) = 1 / 3 ≈ 0.333
   • 1 / (3 + 1) = 1 / 4 = 0.25
   • 1 / (4 + 1) = 1 / 5 = 0.2
   • 1 / (5 + 1) = 1 / 6 ≈ 0.167

Summary:

• Rational numbers between 1 and 2: 1.5, 1.25, 1.375, 1.3125, 1.34375 (using averaging method)
• Rational numbers between 1 and 2: 0.5, 0.333, 0.25, 0.2, 0.167 (using reciprocal method)

These methods provide us with a variety of rational numbers between 1 and 2, demonstrating the flexibility and diversity of such numbers.

Solutions for 2x + 3y = 8

Introduction: In this problem, we're tasked with finding solutions to the equation 2x + 3y = 8. There are multiple solutions that satisfy this equation. Let's explore four of them:

Solution 1: Using Integer Values

• Choose a set of integer values for x and solve for y.
• Let's say x = 2.
• Substitute x = 2 into the equation: 2(2) + 3y = 8.
• Solve for y: 4 + 3y = 8.
• 3y = 8 - 4.
• 3y = 4.
• y = 4/3.
• So, one solution is (2, 4/3).

Solution 2: Using Fractional Values

• Choose fractional values for x and solve for y.
• Let's say x = 1/2.
• Substitute x = 1/2 into the equation: 2(1/2) + 3y = 8.
• Solve for y: 1 + 3y = 8.
• 3y = 8 - 1.
• 3y = 7.
• y = 7/3.
• Another solution is (1/2, 7/3).

Solution 3: Using a Variable for y

• Express y in terms of x and a constant.
• Rearrange the equation to isolate y: 3y = 8 - 2x.
• Divide both sides by 3: y = (8 - 2x)/3.
• So, a solution can be represented as (x, (8 - 2x)/3).

Solution 4: Using Graphical Method

• Graph the equation on a coordinate plane.
• Plot the points where the line intersects the x-axis and the y-axis.
• Determine the coordinates of these points as solutions.
• By plotting, we find that two points of intersection are (4, 0) and (0, 8/3).
• Thus, solutions are (4, 0) and (0, 8/3).

Conclusion: The equation 2x + 3y = 8 has multiple solutions, including both integer and fractional values of x and y. Additionally, solutions can also be represented using variables. Graphically, the solutions are the points where the line intersects the axes.

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.

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).

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.

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.

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