Solve Equation Numerically

In mathematics, to solve an equation is to find what values (numbers, functions, sets, etc.) fulfill a condition stated in the form of an equation (two expressions related by equality). These expressions contain one or more unknowns, which are free variables for which values are sought that cause the condition to be fulfilled. To be precise, what are sought are often not necessarily actual values, but, more in general, mathematical expressions (Source from Wikipedia). Here are going to solve equation numerically and its examples.

Solving equation numerically - Example problems:

Solve equation numerically - Example: 1

To solve the following equation,

2x+3+3x =8

Solution:

Add the like terms

Here the like terms is 2x and 3x

2x+3x=5x

Therefore the equation is

5x+3=8

Add both sides -3

5x+3-3 =8-3

5x=5

Divide both sides 5

There fore the value of x=1

Solve equation numerically -  Example: 2

Solve the following equation by substitution method

x-y =3

2x+4=6

Solution:

x-y =3…………….1

2x+4=6……………2

From the equation 1 can we write

x=y+3……………..3

Substitute x value in equation 1

Therefore

2x+4=6

2(y+3) =6

2y+6=6

Add both sides -6

2y+6-6=6-6

2y=0

y=0

Substitute y=0 in equation 3

x=y+3

x=0+3

x=3

Therefore the solution is x=3, y=0

Solve equation numerically - Example:3

To solve the following expression,

5x+3 = 2x+1

Move the 3 to the right hand side by subtracting 3 from both sides, like this:

From the left hand side:
3 - 3 = 0

The answer is 5x

From the right hand side:
1 - 3 = -2

The answer is -2+2x

Now, the equation reads:

5x = -2+2x

Move the 2x to the left hand side by subtracting 2x from both sides, like this:

From the left hand side:

5x - 2x = 3x

The answer is 3x

From the right hand side:

2x - 2x = 0

The answer is -2

Now, the equation reads:

3x = -2

To isolate the x, we have to divide both sides of the equation by the other variables around the x on the left side of the equation.

The solution to your equation is:

x =-2/3

Geometry Practice Trapezoids

Geometry is the mathematics that deals with  measurement, properties and relationships of lines, points, angles, surfaces, and solids. In geometry trapezoid is a shape that has a pair of opposite sides of parallel, the oter two sides would not be parallel. The shape of the trapezoid differs but it should satisfy the property of pair of opposite sides are parallel. Let us take several geometry practice trapezoid probblems.

Trapezoid

Geometry practice trapezoids:

Problem for Geometry practice trapezoids

Example 1 :

Find the area of the trapezoid of base b1 = 5cm, b2 = 6 cm and height of the trapezoid is 7 cm.

Trapezoid

Solution:

The b1= 5cm, b2 = 6 cm and height h = 7 cm.
The area of a trapezoid is one-half the height h times the sum of the base lengths b1 and b2 .

Area of the trapezoid = `1/2 ` (b1 +b2)* h square units.

= `1/2` (5 + 6) * 7

= `1/2` * 11 * 7

= `77/2`

= 38.5 cm2

Area of the trapezoid is equal to 38.5 cm2

Example 2:

Find the area of the trapezoid of base b1 = 4cm, b2 = 7 cm and height of the trapezoid is 7 cm.

Trapezoid

Solution:

The b1= 4cm, b2 = 7 cm and height h = 7 cm.
The area of a trapezoid is one-half the height h times the sum of the base lengths b1 and b2 .

Area of the trapezoid = `1/2 ` (b1 +b2)* h square units.

= `1/2` (4 + 7) * 7

= `1/2` * 11 * 7

= `77/2`

= 38.5 cm2

Area of the trapezoid is equal to 38.5 cm2

More problems on geometry practice trapezoids:

Example 3:

Find the perimeter of the trapezoid  if the four sides of the trapezoid are given as 5 cm, 7 cm, 4 cm, 6 cm.

Trapezoid
TRAPEZOID

solution:

Given the four sides of the trapezoid as  a = 5cm, b = 7cm, c = 4cm, d = 6cm.

The perimeter of the trapezoid (P)= a+b+c+d square units

= 5 + 7 + 4 +6

= 22 cm2

Example 4:

Find the perimeter of the trapezoid  if the four sides of the trapezoid are given as 6 cm, 7 cm, 4 cm, 5 cm.

Trapezoid
TRAPEZOID

solution:

Given the four sides of the trapezoid as  a = 6cm, b = 7cm, c = 4cm, d = 5cm.

The perimeter of the trapezoid (P)= a+b+c+d square units

= 6 + 7 + 4 + 5

= 22 cm2

Learn Online the Chain Rule

Rule for solving the derivative of a composition of two functions.

If y is a function of   u   and   u is a function of   x, then y is a function of x.

The chain rule notify us how to find the derivative of  y   with respect to x

Online:

Through online to help you learn better and faster. In fact they not only help you with accepting Math concepts better but also help you with your Math homework and assignments.

Chain rule formula:

f (x) = g (h(x))

f’(x) = g’ (h(x) h’(x)

(dy) / (dx) = (dy)/(du) xx (du)/(dx)

Learn online the Chain Rule - Examples:

Learn online the Chain Rule - Example 1:

Differentiate y = (3x + 1)2

D (3x + 1)2 = 2(3x +1) 2-1 D (3x + 1)

= 2 (3x + 1) (3)

= 6 (3x + 1).

Learn online the Chain Rule - Example 2:

Differentiate y = (1 -4x + 7x5) 30

The outer is the 30th power and the inner is (1 – 4x +7x5). Differentiate the 30th power first, leaving (1 – 4x +7x5) unchanged. Then differentiate (1 – 4x +7x5). Thus,

D ( 1– 4x + 7x5) = 30 ( 1-4x+7x5) 30-1 D( 1-4x+7x5)

= 30 (1-4x+7x5) 20-1 (-4+35x4)

= 30(35x4 – 4)(1-4x+7x5)29.

= (4x+x-5)1/2

Learn online the Chain Rule - Example 3:

Differentiate the following:

Y = sin(x2+3)

Let u = x2 +3 so y = sin u

(du)/(dx) =2x         (dy)/(du) = cos u

(dy)/(dx)  = cos ux 2x

(dy)/(dx) = 2xcos(x2+3)

Example 4:

Y = ln (3x3-4x+2)

U = 3x3 - 4x + 2         y = ln u

(du)/(dx) = 9x2 -4          (dy)/(du) = 1/u

(dy)/(dx)  = 1/u xx (9x2 – 4)

(dy)/(dx) = (9x^2 - 4)/(3x^3 - 4x + 2)

Note that (dy)/ (dx) = (u')/u

In general if

Y = ln f(x) then (dy)/(dx) = (f'(x))/f(x)

This is a useful result to remember

Example 5:

Differentiate the following:

Y = sin 2 4x

Now sin 2 4x = (sin 4x)2

Let u = sin 4x                  y = u2

(du)/(dx) = 4 cos 4x       (dy)/(du) =2u

(dy)/(dx) = 2u x 4cos 4x

(dy)/(dx)  = 8 sin 4x cos 4x

Learn online the Chain Rule - Practice Problems:

Practice problem 1:

Differentiate y = sqrt (3e^x + 4)

Answer:

(3e^x)/ (2(sqrt(3e^x + 4)))

Practice problem 2

Differentiate  y = (3x2+1)2

Answer:

36x3 + 12x

Inequality Notation

Inequalities are algebraic equations and it variable x cannot be equal. Inequalities have the different values. In general, inequality notation is used to describe intervals. The inequality notations are,

Greater than

Less than

= Greater than or equal to

= Less than or equal to

The inequality also has chained inequality notation abc which means ab and bc.

Inequality Notation Properties:

The inequality notation “a  b" is read as "a is less than b"
The inequality notation "a  b" is read as "a is greater than b"


These notations define as the sense of the inequality.

(a) If the signs of inequality point in the same direction the same logic is used

(b) If the signs of inequality point in the opposite direction then it is opposite logic/sense.

Inequality Notation for intervals:

Example:

Use inequality notation to illustrate each set.

All the Y in the interval (-4, 8)
X is nonnegative


Solution:

The inequality notation is,

-4  y  8,

0  x

Example problems on inequality:

Example 1:

Solve the inequality 4x+ 2  3

Solution:

4x+ 2  3

Subtract 2 from both sides

4x+2 -2  3-2

4x  1

Divide both sides by 4

4x/4  1/4

x1/4

This means that ANY number greater than ¼ will make the equation 4x+ 2  3 true!

Example 2:

Solve the inequality y + 12  24

Solution:
Simply subtract 12 from each sides:

y + 12 -12   24 -12

y  12


This means that ANY number less than 12 will make the equation y + 12  24 true!

Examples 3:

Solve the inequality       x-35 and 2x+14 16

Solution:

Step 1: Solve the first inequality

X-35

Add 3 on both sides,

x-3 + 3 5+3

x8

Step 2: solve the second inequality

2x+14 16

Subtract 14 from both sides,

2x+14-14  16 -14

2x12

Divide by 2 on both sides

x6

The final solution is: x8 and x6

This means that 6x8.

Product Profitability Model

Let us see about product profitability model,

This pattern provides the helps finance, product management, and operations teams review the overall profitability of specific products quarterly and annually. By conduct a product profitability model, your company can recognize the direct and indirect costs associated with increasing and contribution a definite product and evaluate a product's donation to the foot line.

Reveals differences in product profitability model:

Let us see about differences in product profitability model,

Our product profitability spreadsheet model include the means cost centers to work out product profitability.
It combines direct costing and ‘activity-based costing’ to pathway cost of goods, royalties, and operating expense for marketing, product development, and customer hold up.
Size of support belongings and hold up issue to recognize areas for enhancement that reduce support costs and improve customer satisfaction.
All costs do not having products to acquire a useful compute of product profitability.
Product profitability analysis often uncovers unexpected results, particularly in organization with many products.



Details About the product profitability model:

The model computes gross margin and gross margin % for each product by subtracting cost of sales, which consists of cost of goods and royalties.

Cost of goods is optionally individually as cost per unit, cost as a % of income, or fixed costs for every product.
Royalties paid are optionally particular as cost per unit, cost as a % of income or fixed costs for each product.

Resources of Product profitability model:

The most important limitation of product profitability analysis is reasonable with planned factors, such as income growth rate, and long-term potential of products that is not reflect in present profitability.
Where results of the analysis differ from specialist opinions, we recommend that  carefully explore the reasons for the divergence. The experts may not have an accurate sensitivity of fixed cost, or the profitability analysis may have missed factors that the experts recognize.

Natural Exponential Function

An exponentiation is relating two numbers, the base a and the exponent n. The n value is a positive integer and exponentiation corresponds to frequent multiplication or a product of n factors.The exponent is written as superscript to the right of the base to the nth power.  The power p is defined as when n is negative integer for non-zero. When the base a is positive real number, p is defined for exponential function.

Formulas for natural exponential function

ex       =  `(x)/(1!)`       +  `(x^(2))/(2!) + .................`

e0 = 1

d(ex) = ex

In trigonometry function the natural exponential function,

cos (x) =`(e^(ix) + e^(-ix))/(2)`

sin (x) = `(e^(ix) - e^(-ix))/(2i)`


In complex number the natural exponential function:

The symbol eiθ or exp (iθ) (called exponential of iθ) is defined by

eiθ = cos θ + i sin θ

This relation is known as Euler’s formula.

If z ≠ 0 then z = r (cos θ + i sin θ) = reiθ. This is called the exponential

form of the complex number z. By straight forward multiplication of

eiθ1 = (cos θ1 + i sin θ1) and eiθ2 = cos θ2 + i sin θ2

we have eiθ1.eiθ2 = ei(θ1+ θ2)

Example problems for natural exponential function

Problem for natural exponnential function 1:

Expand the term e2x    in natural exponential function.

Solution:

The given function is e2x we have to expand the term using the formula

Solution:

ex       =  `(x)/(1!)`       +  `(x^(2))/(2!) + ....................`

e2x    =   1    +     `(2x)/(1!)`       +  `(2x^(2))/(2!) + ..............................`

=  1    +     `(2x)/(1!)`       +  `4x^(2)/(2!) + .......................`

=  1   +   2x     +   2 x2                +   ..................

Problem for natural exponnential function 2:

Expand the term e4x    in natural exponential function.

Solution:

The given function is e4x we have to expand the term using the formula

ex       =  `(x)/(1!)`       +  `(x^(2))/(2!) + ...................`

e4x    =   1    +     `(4x)/(1!)`       +  `(4x^(2))/(2!) + .........................`

=  1    +     `(4x)/(1!)`       +  `16x^(2)/(2!) + ............................`

=  1   +   4x     +   8 x2                +   ..................