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Educational Level Secondary school, High school and College
Program Purpose Provide step by step solutions of your problems using online calculators (online solvers)
Problem Source Your textbook, etc

1. Graph
1. General Graph for any f(x) and f(y)
2. Graph - Using Points and Slope
2.1 Line like x=-3y+5; 2x+y=1; x+2y<=5; x+y>=15
2.2 Line Using 2 Points like Point1=(3,-5) and Point2=(5,3)
2.3 Slope & point like Slope=7 and Point=(4,6)
2.4 Slope & Y-Intercept like Slope=2 and Y-Intercept=-4
2.5 Only a Point like plot a point (3,-5),(5,3)
2.6 Points on Number Line(X-axis) like -3,5,12,-7,0

3. Circle
4. Ellipse
5. Parabola
6. Hyperbola
7. Polar Graph

8. Statistics Graph
8.1 Histogram
8.2 Frequency Polygon
8.3 Frequency Curve
8.4 Less than type cumulative frequency curve
8.5 More than type cumulative frequency curve

2. Area
1. Circle
2. Semi-Circle
3. RegularHexagon
4. Square
5. Rectangle
6. Parallelogram
7. Rhombus
8. Trapezium
9. Scalene Triangle
10. Rightangle Triangle
11. Equilateral Triangle
12. Isoceles Triangle
13. Sector Segment

3. Volume
1. Cuboid
2. Cube
3. Cylinder
4. Cone
5. Sphere
6. Hemi-Sphere

4. Pythagoras Theorem
1. General Graph
1. `y = x`
2. `y <= sin(x)`
3. `y >= cos(x)`
4. `y = sqrt(x)`
5. `y <= sqrt(x^3-3x)`
6. `y >= |x|`
7. `y = x^3-x`
8. `y <= (e^x-e^(-x))/2`

1. `x = sin(y)`
2. `x <= cos(y)`
3. `x >= sqrt(y)`
4. `x = sqrt(y^3-3y)`
5. `x <= |y|`
6. `x >= sin(y)^3+cos(y)^3`
7. `x = y+sin(y)`
8. `x <= (sin(9y)+sin(10y))*sin(0.1y)`

2. Graph - Using Points and Slope
1. Lines
like x=-3y+5; 2x+y=1; x+2y<=5; x+y>=15

2. Line Using 2 Points
like Point1=(3,-5) and Point2=(5,3)

3. Slope & point
like Slope=7 and Point=(4,6)

4. Slope & Y-Intercept
like Slope=2 and Y-Intercept=-4

5. Only a Point
like plot a point (3,-5),(5,3)

6. Points on Number Line(X-axis)
like -3,5,12,-7,0

 
3. Circle
1. Circle-1
`X^2 + Y^2 = 9`

2. Circle-2
`(X+1)^2 + Y^2 = 12`

3. Circle-3
`X^2 + (Y-2)^2 = 15`

4. Circle-4
`(X+1)^2 + (Y-2)^2 = 9`

4. Ellipse
1. Ellipse-1
`X^2/4 + Y^2/9 = 9`

2. Ellipse-2
`(X+1)^2/4 + Y^2/9 = 12`

3. Ellipse-3
`X^2/4 + (Y-2)^2/9 = 15`

4. Ellipse-4
`(X+1)^2/4 + (Y-2)^2/9 = 9`

 
5. Parabola
1. Parabola-1
`Y = 3X^2`

2. Parabola-2
`Y = 3X^2 + 1`

3. Parabola-3
`Y = 3(X+1)^2`

4. Parabola-4
`Y = 3(X+1)^2 + 1`

5. Parabola-5
`X = 3Y^2`

6. Parabola-6
`X = 3Y^2 + 1`

7. Parabola-7
`X = 3(Y+1)^2`

8. Parabola-8
`X = 3(Y+1)^2 + 1`
6. Hyperbola
1. Hyperbola-1
`X^2/4 - Y^2/9 = 9`

2. Hyperbola-2
`2. (X+1)^2/4 - Y^2/9 = 12`

3. Hyperbola-3
`X^2/4 - (Y-2)^2/9 = 15`

4. Hyperbola-4
`(X+1)^2/4 - (Y-2)^2/9 = 9`

5. Hyperbola-5
`Y^2/4 - X^2/9 = 9`

6. Hyperbola-6
`(Y+1)^2/4 - X^2/9 = 12`

7. Hyperbola-7
`Y^2/4 - (X-2)^2/9 = 15`

8. Hyperbola-8
`(Y+1)^2/4 - (X-2)^2/9 = 9`

 
7. Polar Graph
1. `R = 4*cos(2*t)`
2. `R = 4*sin(2*t)`
3. `R = 2-4*sin(2*t)`
4. `R = 2-4*cos(2*t)`
5. `R = 2+4*cos(2*t)`
6. `R = 2+4*sin(2*t)`

8. Statistics Graph
1. Histogram
2. Frequency Polygon
3. Frequency Curve
4. Less than type cumulative frequency curve
5. More than type cumulative frequency curve
1. The frequency distribution of the marks obtained by 100 tudents in a test of Mathematics carrying 50 marks is given below.
Draw Histogram, Frequency Polygon, Frequency Curve, Less than type cumulative frequency curve and More than type cumulative frequency curve of the data.

Marks obtained0 - 910 - 1920 - 2930 - 3940 - 49
No of students815204512
1. Histogram 2. Frequency Polygon
3. Frequency Curve 4. Less than type cumulative frequency curve
5. More than type cumulative frequency curve
 
1. Circle
Area `(A) = pi r^2`
Circumference `(C) = 2 pi r = pi d`
Diameter `(d) = 2 r`


I know that for a circle Radius = 10 . From this find out Area of the circle.

`"Here radius " (r)=10 "(Given)"`

``

`"Diameter " (d)= 2 * r`

` = 2 * 10`

` = 20`


`"Perimeter" = 2 pi r`

` = 2 * 22/7 * 10`

` = 440/7`


`Area = pi r^2`

` = 22/7 * (10)^2`

` = 2200/7`
2. Semi-Circle
Area `(A) = 1/2 pi r^2`
Circumference `(C) = pi r = (pi d)/2`
Perimeter `(P) = pi r + 2 r`
Diameter `(d) = 2 r`


I know that for a Semi-Circle Radius = 10 . From this find out Area of the Semi-Circle.

`"Here radius "(r)=10" (Given)"`

``

`"Diameter "(d)= 2 r`

` = 2 * 10`

` = 20`

``

`"Circumference" = pi r`

` = 22/7 * 10`

` = 220/7`


`"Perimeter" = pi r + 2 r`

` = 22/7 * 10 + 2 * 10`

` = 220/7 + 20`

` = 360/7`


`"Area" = (pi r^2)/2`

` = (22/7 * (10)^2)/2`

` = 1100/7`
 
3. RegularHexagon
Perimeter `(P) = 6 a`
Area `(A) = sqrt(3)/4 xx 6 xx a^2`


I know that for a Regular Hexagon Side = 10 . From this find out Area of the Regular Hexagon.

`"Here Side "(a)=10" (Given)"`


`"Perimeter" = 6 a`

` = 6 * 10`

` = 60`


`"Area" = sqrt(3)/4 * 6 * a^2`

` = sqrt(3)/4 * 6 * 10^2`

` = 259.8076`
4. Square
Diagonal `(d) = sqrt(2) a`
Perimeter `(P) = 4a`
Area `(A) = a^2 = d^2/2`


I know that for a square Side(a) = 10 . From this find out Area of the square.

`"Here a" = 10" (Given)"`

`"Diagonal" = sqrt(2) * "a"`

` = sqrt(2) * 10`

` = 14.1421`


`"Perimeter" = 4 * "a"`

` = 4 * 10`

` = 40`


`"Area" = "a"^2`

` = 10^2`

` = 100`
 
5. Rectangle
Diagonal `(d) = sqrt(l^2 + b^2)`
Perimeter `(P) = 2(l+b)`
Area `(A) = l b`


I know that for a rectangle Length = 5 and Breadth = 12 . From this find out Area of the rectangle.

`"Here one-Side " (l) = 5" and other-Side " (b) = 12" (Given)"`

`"Diagonal" = sqrt(l^2 + b^2)`

` = sqrt(5^2 + 12^2)`

` = sqrt(25 + 144)`

` = sqrt(169)`

` = 13`


`"Perimeter" = 2 * "(sum of Sides)"`

` = 2 * (5 + 12)`

` = 34`


`"Area = Product of Sides"`

` = 5 * 12`

` = 60`
6. Parallelogram
Area `(A) = ah`
Perimeter `(P) = 2a + 2b`


I know that for a parallelogram a = 9 , b = 22 and h = 14 . From this find out Area of the parallelogram.

`"Here " a=9, b=22, h=14" (Given)"`


`"Perimeter" = 2 * (a + b)`

` = 2 * (9 + 22) `

` = 62`


`"Area" = a * h`

` = 9 * 14`

` = 126`
 
7. Rhombus
Radius `(r_1) = (d_1)/2`
Radius `(r_2) = (d_2)/2`
Side `(a) = sqrt(r_1^2 + r_2^2)`
Perimeter `(P) = 4 a`
Area `(SA) = (d_1 d_2)/2`


I know that for a rhombus d1 = 10 and d2 = 24 . From this find out Area of the rhombus.

`"Here, we have " d_1 = 10" and " d_2 = 24" (Given)"`


`a^2 = (d_1/2)^2 + (d_2/2)^2`

`a^2 = (10/2)^2 + (24/2)^2`

`a^2 = (5)^2 + (12)^2`

`a^2 = 169`

`a = 13`

`"Perimeter" = 4 * a`

` = 4 * 13`

` = 52`


`"Area" = 1/2 " (Product of diagonals)"`

` = 1/2 * d_1 * d_2`

` = 1/2 * 10 * 24`

` = 120`
8. Trapezium
Area `(A) = h/2 (a + b)`
Perimeter `(P) = a + b + c + d`


I know that for a trapezium a = 22 , b = 18 , c = 16 , d = 16 and h = 4 . From this find out Area of the trapezium.

`"Here " a=22, b=18, c=16, d=16, h=4" (Given)"`


`"Perimeter" = a + b + c + d`

` = 22 + 18 + 16 + 16 `

` = 72`


`"Area" = (a + b) * h/2`

` = (22 + 18) * 4/2`

` = 80`
 
9. Scalene Triangle
Perimeter `(P) = a+b+c`
`S = P/2 = (a+b+c)/2`
Area `(A) = sqrt(S (S - a) (S - b) (S - c))`


I know that for a scalene Triangles a = 3 , b = 4 and c = 5 . From this find out Area of the scalene Triangles.

`"Here " a=3, b=4, c=5" (Given)"`


`"We know that,"`

`"Perimeter" = a + b + c`

` = 3 + 4 + 5`

` = 12`

` `

`"Semi-Perimeter" = s = (a + b + c)/2`

` = 12/2`

` = 6`


`"Here " a=3, b=4, c=5" and semi-Perimeter" = 6`

`"We know that,"`

`"Area" = sqrt(s (s - a) (s - b) (s - c))`

` = sqrt(6 (6 - 3) (6 - 4) (6 - 5))`

` = 6`
10. Rightangle Triangle
Diagonal `(d) = sqrt(a^2 + b^2)`
Perimeter `(P) = a+b+c`
Area `(A) = 1/2(a b)`


I know that for a rightangle Triangles AB = 5 and BC = 12 . From this find out Area of the rightangle Triangles.

`"Here one Side" = 5" and other Side" = 12" (Given)"`

`"We know that,"`

`"In triangle ABC, by Pythagoras' theorem"`

`AC^2 = AB^2 + BC^2`

`AC^2 = 5^2 + 12^2`

`AC^2 = 25 + 144`

`AC^2 = 169`

`AC = 13`


`"Perimeter" = AB + BC + AC`

` = 5 + 12 + 13`

` = 30`

` `


`"Here base" = 5" and height = "12 `

`"We know that,"`

`"Area" = 1/2 * AB * BC`

` = 1/2 * 5 * 12`

` = 30`
 
11. Equilateral Triangle
Perimeter `(P) = 3 a`
Area `(A) = sqrt(3)/4 a^2`


I know that for a equilateral Triangles Side = 6 . From this find out Area of the equilateral Triangles.

`"Here " a = 6" (Given)"`


`"We know that,"`

`"Perimeter" = 3 * a`

` = 3 * 6`

` = 18`


`"We know that,"`

`"Area" = sqrt(3)/4 * a^2`

` = 1.732/4 * 6 * 6`

` = 15.5885`
12. Isoceles Triangle
Height `(h) = sqrt(a^2 - b^2/4)`
Perimeter `(P) = 2 a + b`
Area `(A) = (b h)/2`


I know that for a isoceles Triangles a = 5 and b = 6 . From this find out Area of the isoceles Triangles.

`"Here base " (b) = 6" and equal side " (a) = 5" (Given)"`


`"We know that,"`

`"Perimeter" = (2 * "equal Side") + "third Side"`

` = (2 * a) + b`

` = (2 * 5) + 6`

` = 16`


`"We know that,"`

`"Area" = 1/2 * "base" * "height"`

` = 1/2 * b * (sqrt(a^2 - b^2))/4`

` = 1/2 * 6 * sqrt(5^2 - 6^2)/4`

` = 1/2 * 6 * 4 `

` = 12`
 
13. Sector Segment
Length of the arc `= l = (pi r theta)/180`
Area of a minor sector `= (pi r^2 theta)/360`


I know that for a sector & segment Radius = 10 and angle of measure = 180 . From this find out length of arc of the sector & segment.

`"Here "r = 10" and " theta = 180" (Given)"`

`"Length of the arc " = l = (pi r theta)/180`

`=(22/7 * 10 * 180)/180`

`=31.4286`


`"Area of a minor sector "= (pi r^2 theta)/360`

`=(22/7 * 10^2 * 180)/360`

`=157.1429`
1. Cuboid
Diagonal `(d) = sqrt(l^2+ b^2+ h^2)`
Surface Area `(SA) = 2 (lb + bh + hl)`
Volume `(V) = lbh`



I know that for a cuboid Length = 3 , Breadth = 4 , and Height = 5 . From this find out Volume of the cuboid.

`"Here, we have " l = 3, b=4, h=5" (Given)"`

`"Diagonal"^2 = l^2 + b^2 + h^2`

`"Diagonal"^2 = 3^2 + 4^2 + 5^2`

`"Diagonal"^2 = 50`

`Diagonal = 7.0711`


`"Here, we have " l = 3, b=4, h=5" (Given)"`

`"Volume" = l * b * h`

` = 3 * 4 * 5`

` = 60`


`"Here, we have " l = 3, b=4, h=5" (Given)"`

`"Total Surface Area" = 2 (lb + bh + lh)`

` = 2 * (3 * 4 + 4 * 5 + 3 * 5)`

` = 2 * (47)`

` = 94`


`"Here, we have " l = 3, b=4, h=5" (Given)"`

`"Curved Surface Area" = 2h (l + b)`

` = 2 * 5 (3 + 4)`

` = 70`
2. Cube
diameter `(d) = sqrt(2) l`
Diagonal `= sqrt(3) l`
Surface Area `(SA) = 6 l^2`
Volume `(V) = l^3`


I know that for a cube Length = 3 . From this find out Volume of the cube.

`"Here, we have " l = 3`

`"Diagonal" = sqrt(3) l`

` = sqrt(3) * 3`

` = 5.1962`


`"Here, we have " l = 3`

`"Volume" = l^3`

` = 3^3`

` = 27`


`"Here, we have " l = 3`

`"Total Surface Area" = 6 l^2`

` = 6 * 3^2`

` = 6 * 9`

` = 54`


`"Here, we have " l = 3`

`"Curved Surface Area" = 4 l^2`

` = 4 * 3^2`

` = 4 * 9`

` = 36`
 
3. Cylinder
Curved Surface Area `(CSA) = 2 pi r h`
Total Surface Area `(TSA) = 2 pi r (r + h)`
Volume `(V) = pi r^2 h`


I know that for a cylinder Radius = 3 and Height = 10 . From this find out Curved Surface Area of the cylinder.

`"Here, we have Radius "(r) = 3" and Height "(h) = 10" (Given)"`


`"Volume" = pi r^2 h`

` = pi * 3^2 * 10`

` = 282.7433`


`"Total Surface Area" = 2 pi r (r+h)`

` = 2 * pi * 3 * (3 + 10)`

` = 245.0442`


`"Curved Surface Area" = 2 pi r h`

` = 2 * pi * 3 * 10`

` = 188.4956`
4. Cone
Height `(h) = sqrt(l^2 - r^2)`
Curved Surface Area `(CSA) = pi r l`
Total Surface Area `(TSA) = pi r (l + r)`
Volume `(V) = (pi r^2 h)/3`

I know that for a cone Radius = 3 and Length = 5 . From this find out Volume of the cone.

`"Here, we have Radius "(r) = 3" and Slant Height " (l) = 5" (Given)"`

`l^2 = r^2 + h^2`

`h^2 = l^2 - r^2`

`h^2 = 5^2 - 3^2`

`h^2 = 16`

`h = 4`


`"Volume" = (pi r^2 h)/3`

` = (pi * 3^2 * 4)/3`

` = 37.6991`


`"Total Surface Area" = pi r (l + r)`

` = pi * 3 (5 + 3)`

` = 75.3982`


`"Curved Surface Area" = pi r l`

` = pi * 3 * 5`

` = 47.1239`
 
5. Sphere
Surface Area `(SA) = 4 pi r^2`
Volume `(V) = 4/3 pi r^3`
diameter `(d) = 2r`


I know that for a sphere Radius = 3 . From this find out Volume of the sphere.

`"Here, we have Radius"(r) = 3" (Given)"`


`"Volume" = 4/3 pi r^3`

` = 4/3 * pi * 3^3`

` = 84.823`


`"Total Surface Area" = 4 pi r^2`

` = 4 * pi * 3^2`

` = 113.0973`


`"Curved Surface Area" = 4 pi r^2`

` = 4 * pi * 3^2`

` = 113.0973`
6. Hemi-Sphere
Curved Surface Area `(CSA) = 2 pi r^2`
Total Surface Area `(TSA) = 3 pi r^2`
Volume `(V) = 2/3 pi r^3`
diameter `(d) = 2r`


I know that for a Hemi-Sphere Radius = 3 . From this find out Volume of the Hemi-Sphere.

`"Here, we have Radius " (r) = 3" (Given)"`


`"Volume" = 2/3 pi r^3`

` = 2/3 * pi * 3^3`

` = 56.5487`


`"Total Surface Area" = 3 pi r^2`

` = 3 * pi * 3^2`

` = 84.823`


`"Curved Surface Area" = 2 pi r^2`

` = 2 * pi * 3^2`

` = 56.5487`
 
Pythagoras Theorem : In a right angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the remaining sides.


i.e. `AC^2 = AB^2 + BC^2`

Using Pythagoras Theorem, Find out AC when AB = 5 and BC = 12
Here AB=5 and BC=12 (Given)
We know that,
In triangle ABC, by Pythagoras' theorem
`AC^2 = AB^2 + BC^2`
`AC^2 = 5^2 + 12^2`
`AC^2 = 25 + 144`
`AC^2 = 169`
`AC = 13`

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