Two Step Equations with Integers

Introduction to two step equations with integers:

Solving two step equation with integers is nothing but solving the equations that has integers in two steps. just by simplifying the equations that involve integers and variables. here in resolving these two step equations with integers just we resolve the variables in the equations..The students will be able to solve one-step and two-step linear equations involving integers, fractions, and decimals. Let us see what is the two step equations with integers and the sample problems for two step equations.

Two Step Equations with Integers:

Solving Two-Step Equations - getting variable alone

1. Remove constant by adding opposite to both sides.

2. If variable is being divided by a number, multiply both sides by the number. If variable is being multiplied by a number, divide both sides by the number (remember dividing by a fraction is the same as multiplying by the reciprocal).Is this topic solving linear equations and inequalities hard for you? Watch out for my coming posts.

Example 1:

Simplify     `X/3` + 3 = 12

`X/3` + 3 – 3 = 12 – 3

`X/3` + 3 – 3 = 12 – 3

`X/3` = 9

`X/3` = 9

`X/3` · 3 = 9 ·3

X =  `9*3`

X = 27

Example 2:

Simplify     4X – 5 = 6

4X – 5 + 5 = 6 + 5

4X – 5 + 5 = 11

4X = 11

`(4X)/4` = `11/4`

`X` = `11/4`

X = 2.75.

Two Step Equations with Integers:

Example 3:

simplify -3- 9x = -21  keep the variables on one side and move the integers to other side.

-9x = -18  resolve by dividing

x  = 2

Example 4:

simplify 20 + 4x = 25

4x = 25 - 20

4x = 5

x = 20.

Example 5:

simplify

5x + 7 =14  keep the variables on one side and move the integers to other side.

5x = 7

x = `7/5 `

=1.4

Example 6:

simplify

`2/3` x - `1/(2x)` + 3 =15   keep the variables on one side and move the integers to other side.

`1/(6x)` + 3 =15

`1/(6x)` =12

x = 72

Example 7:

11=`2/(7m)` + 9

`11(7)` =`2/7` (7/1)m+9(7) Multiply each term by the Least Common Denominator (LCD)(7).

77-63 = 2m+63-63 Subtract 63 from both sides.

14/2 = `(2m)/2` Divide both coefficients by two

7=m

Properties of Limits

Introduction to properties of limits:

Let us see about the properties of limits. The majority fundamental concept of modern Calculus limit. Limit used to define the value that a function or order sequence approaches as the input or index approaches a few value. Properties of limits are important to calculus and also used to describe the continuity, the integrals and the derivates.

Definition:

The limits are the numerical values. The limit values are important for draw the specific diagram in the graph. Without the limits, the line cannot end at any session. The limits are may be positive or negative in the graphical representation. The limits are having some properties which are used in the mathematics.

Limit of numerical sequence:

Assume a mathematical sequence, a normal term of which advance to some number x at increasing an ordinal number n. In this case, that the mathematical series has a limit. This document has a more harsh definition: A number x is known as a limit of a numerical sequence.

This definition means, that x is a limit of a arithmetical order, if its general term approach unrestrictedly to x at increasing n. Geometrically it means, that for any > 0 it’s likely to find such a number N , that starting from n > N all terms of the sequence are located within an interval ( x – , x + ). A sequence, having a limit, is known as convergent; other hand - x divergent sequence.

First we will assume that `lim_(x->a)` f(x) and `lim_(x->a)g(x)` exist that c is any constant.

`lim_(x->a)[c f(x)] = clim_(x->a) f(x)`

In additional terms we can “factor” a multiplicative constant out of a limit.I have recently faced lot of problem while learning Calculus Solver, But thank to online resources of math which helped me to learn myself easily on net.

Properties of Limits:

The properties of limits are,

Assume,

`lim_(x->a)` f(x)= L and `lim_(x->a)` g(x) =M.

Where L and M indicates real numbers. Also assume c > 0 is a real number. Then

`lim_(x->a)` f(x) +g (x) =L+M.

`lim_(x->a)f(x) *g(x) = LM.`

`lim_(x->a)cf(x) = cL.`

Moreover M' 0 then,

`lim_(x->a)f(x) /g(x) = L/M.`

`lim_(x->a)1/g(x) = 1/L` .

These are the properties of limits used by the mathematics.

Decimal to Binary Formula

Introduction to decimal to binary formula:

In this section we have decimal to binary formula. Decimal to binary conversion is one of the most important conversions in math subject. Decimal to binary conversion is the very simple process compare than other conversion in math. Below we have some conversion based on decimal and binary numbers. Let us see about decimal to binary formula.

Example Problems for Decimal to Binary Formula:
Example problem 1: Can you convert the following decimal number into binary number with neat and clear explanation: 211

Solution:

Given 211

To convert the given number 211 into binary number we have to follow some conditions:

First we divide the given number by 2 until we get one or zero.

The neat step is given below:

Therefore, the binary value of 211 is 11010011.
Answer: The binary value of 211 is 11010011.

Example problem 2:  Can you convert the following binary number into decimal number with neat and clear explanation: 1101100

Solution:

Given 1101100

To convert the given binary number we have to multiply power of 2.

1101100 = 1 xx `2^6` + 1 xx `2^5` + 0 xx `2^4` + 1 xx `2^3 ` + 1 xx `2^2 ` + 0 xx `2^1` + 0 xx `2^0` = 64 + 32 +0 + 8 + 4+ 0 + 0 = 108

Therefore, the decimal number of 1101100 is 108.

Answer: The decimal number of 1101100 is 108.

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Practice Problems for Decimal to Binary Formula:

Practice problem 1: Can you convert the following decimal number into binary number: 56

Practice problem 2: Can you convert the following binary number into decimal number: 110011

Practice problem 3: Can you convert the following decimal number into binary number: 32

Practice problem 4: Can you convert the following binary number into decimal number: 100001

Practice problem 5: Can you convert the following decimal number into binary number: 78

Practice problem 6: Can you convert the following binary number into decimal number: 11111100

Solutions for decimal to binary formula:

Solution 1: 111000

Solution 2: 27

Solution 3: 100000

Solution 4: 33

Solution 5: 1001110

Solution 6: 252

Integral Calculus Calculator

Introduction to Integral Calculus Calculator:

The topics involved in calculus are functions, application of derivatives, limits, derivatives,  logarithm functions, trig functions and exponential functions. The two major sub divisions of calculus are differential calculus and integral calculus. In this article we shall discuss about Integral calculus calculator.  The following are the examples involved in Integral calculus calculator.

Integral Calculus Calculator Problems:


Example 1:  `int` 71ey dy

Solution: `int` 71ey dy = 71`int` ey dy

Integrating the equation, we get

`int` 71 ey dy  = 71 ey + C

Example 2:  Find `int_(-1)^1` log `((29-x)/(29+x))` dx

Solution: Let f(x) = log `((29-x)/(29+x))`

f(-x) = log`((29+x)/(29-x)) `

= log (29 + x) - log (29 - x)

= - [log(29 + x) - log(29 - x)]

= - [log`((29+x)/(29-x))` ] = -f(x)
Hence  f(-x) = -f(x), we can say that f(x) is an odd function

So, `int_(-1)^1` log`((29-x)/(29+x))`dx = 0


Example 3: Integrate the given equation 3x2+5x + 10 dx

Solution: ∫3x2+5x +10dx = ∫3x2dx +∫5x+∫10 dx

Integrating the above equation,we get 

=> 3x3+  5x2 + 10x + C
3      2     

Example 4:  Evaluate `int_0^1x e^xdx` 

Solution: Using the method of integration by parts,

As we know `int` udv = uv - `int` v du

Here u=x, du=dx, dv= exdx, v=ex

`int_0^1` x ex  dx = (x ex)10 - `int_0^1` ex  dx

= e - (ex)10

= e - (e- 1)

= 1

Example 5:  Find the integral of the given equation  23x3+4x2+2x dx

Solution: ∫3x3+4x2+2x dx 

=>∫ 3x3dx + ∫4x2 dx +∫2x dx 

Integrating the above equation

=> 3x4 +4x3 + 2x2  + c
4     3       2

Integral Calculus Calculator Practice Problems:
Problem 1: Find the integral of the given equation 4x2+2x+20 dx

Solution: 4x3 + 2x2 + 20x + c
3       2

Problem 2: Find the integral of the given equation x3+4x+50 dx

Solution: x4 + 4x2 + 50x + c
4       2

Problem 3: Find the integral of the given equation x2+2x+8 dx

Solution: x3 + 2x2 + 8x + c
3      2

Solving Online Absolute Minimum

Learn solving absolute minimum online:

The function f given by f(x) = x, x (0,1) has neither a maximum nor  a minimum value. If we replace (0,1) by the closed interval [0,1], then the function has a maximum value 1=f(1) and the minimum value 0 = f(0). This reveals the emphasis to be put on the interval on which the given function is defined. We could also note that, in the interval [0,1], f has neither a point of local maxima nor a point of local minima and so f has neither a local maximum value nor a local minimum value even though maximum and minimum values of f exists.

The maximum value 1 of f at x = 1 is called the absolute maximum value (greatest value ) of f on [0,1] and the minimum value 0 of f at x = 0 is called the absolute minimum value (least value) of f on [0,1].

Learn Theorems Used in Solving Absolute Minimum Online:

Let f be a continuous function on an interval I = [a,b]. Then, f has the absolute maximum value and f attains it at least once in I. Also, f has the absolute value and attains it at least once in I.

Let f be a differentiable function on I and let x0 be any interior point of I. then

1. If f attains its absolute maximum value at x0, then f ‘ (x0) = 0

2. If f attains its, then  absolute minimum value at x0, then f ‘ (x0) = 0

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Learn Rules in Solving Absolute Minimum Online:

We use the following rules to find the absolute maximum and minimum values of a function in a given interval.

Find all the points where f ’ takes the value zeros.
Take the endpoints of the interval
At all these points calculate the values of f.
Take the maximum and minimum values of f out of the values calculated in step 3. These will be the absolute maximum or minimum values.

Triangle with 3 Acute Angles

Introduction to triangle with 3 acute angles:

The triangle has three sides where the angles may differ and the name has been given to the triangle. The acute angle is nothing but the angle will be less than 90 degrees in its measure. The acute angle triangle contains all the three angles will be less than 90 degrees. Now we see about the triangle with 3 acute angles.

Triangle with 3 Acute Angles:

The triangle with three acute angles can be seen in an equilateral triangle. The triangles with three acute angles will be less than 90 degrees. The sum of these three acute angles in the triangle will be 180 degrees. In acute angled triangle, the angles will be less than 90 degree only. An example for acute angle will be shown below.

Problems on Triangle with 3 Acute Angles:

Example 1:

Find the third acute angle of the triangle where the other two acute angle measures are 57 and 77?

Solution:

The third acute angle of a triangle can be determined as follows,

The sum of the three acute angles will be 180 degrees in a triangle.

Let us assume the unknown triangle will be x

57 + 77 + x = 180

134 + x = 180

x = 180 - 134

x = 46.

Thus, the third acute angle measures about 46 degrees.

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Example 2:

Find the acute angle of a triangle where two angles of it are 55 degrees. Determine the third angle of an acute angled triangle?

Solution:

The two angles in the triangle are equal where its measurement is about 55 degrees.

Let us take the triangle measure and it can be calculated as explained below.

The angle which should be determine can be assumed as x or y.

55 + 55 + x = 180

110 + x = 180

x = 180 - 110

x = 70.

Since, this is an acute angle where its measure is less than 90 degrees.

Placing Fractions on a Number Line

Introduction of placing fraction on Number Line:

The number placed in the correct positions that have both positive and negative in a line is called as number line, where zero is placed at the center of the number line. Fractions can also be placed on the number line in correct place. Let see below the number line has been drawn show the number in correct place. Here we see how can point the correct placing of fractions on the number line.



Placing Fractions on the Number Line:

The fractions on the number line can mark the point in correct place easily.

Positive Fractions on the Number line:

Let take one positive fraction as `(13)/(4)` and see how to point that number in correct place on the number line.



Procedure for placing the fractions on the number line:

Step 1: Convert the fractions as mixed number.

The fraction  `(13)/(4)` = 3 ¼

Step 2: Keep the whole number 3 as it is.

Step 3: Dive the part in between 3 to 4 as 4 and mark the point as per the numerator of the fraction

Another Method:

Step 1: Keep the whole number as it

Step 2: Convert the fraction  `(1)/(4)` to decimal form.

The fraction of ¼ is equal to 0.25 decimal forms.

Step 3: Mark the point 0.2 after the whole number 3 as 3.25    

Placing Fractions on the Number Line:

Negative Fractions on the Number Line

Let take another example of negative fractions `(-7)/(2)`

Now we are not bothering about the negative sign, just placing the fraction in correct point on the number line of negative integer side. By placing fractions on the number line, as like as follow positive fraction on the number line as mentioned above.



Procedure to point the mixed numbers on the number line:

Step 1: Convert the fraction to mixed number.

`(-7)/(2)` = -3 ½

Step 2: Keep the whole number 3 as it is.

Step 3: Dive the part in between 3 to 4 as 2 and mark the point as per the number consists in numerator of the fraction

Another Method:

Step 1: Keep the whole number as it

Step 2: Convert the fraction `(1)/(2)` to decimal form.

The fraction of ½ is equal to 0.5 decimal forms.

Step 3: Placing the point 0.5 after the whole number 3 as 3.5