Word Problems in Decimals Addition

In this lesson word problem are involving with decimals addition. That is nothing but part of a whole numbers and decimals numbers. In this word problems carefully notice the numbers and digits places. Let we see examples of decimals addition in word problems and then you will do the practice of following word problems.      

Example of Decimal Word Problems:

Let us see decimals addition in word problems

Decimals addition in word problem 1:

Q :At hotel, the men drunk 4.30 (litre) milk and women drunk 5.30(litre). Find total amount of milk?

Solution:

Given: men drunk 4.30(litre) milk, women drunk 5.30(litre) .Find total milk quantity?

Step 1: we need total (litre of) milk so here we should add the given values

4.30 + 5.30

Step 2: (4.30 + 5.30) = 9.60

Step 3: The answer is 9.60 litre.

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Decimals addition in word problem 2:

Q:Today I bought 6.25ml oil. And yesterday I brought 1.5ml oil. Find total amount of oil?

Solution:

Given: today 6.25ml oil, yesterday 1.5ml

Step 1: we need total (ml of) oil so here we should add the given values

6.25ml + 1.5ml

Step 2:  6.25ml + 1.5ml = 7.75ml

Step 3: total amount of oil is 7.75ml (Answer)

Decimals addition in word problem3:

Q: My dog is 4.25kg weight. My rabbit is 3.25kg weight. Find total number of weight?

Solution:

Given: dog 4.25kg weight, rabbit  3.25kg weight

Step 1: we need to find total number of weight.

(4.25kg+3.25kg)

Step 2: (4.25kg+3.25kg) = 7.50kg

Set3: the total number of weight =7.50kg

These are the examples of decimals addition of word problems.

Practices word problems of decimals addition.

1)      I bought 25 pen and my sister bought3.45 pen. Find total number of ?  

Answer: 5.95

2)      Jack has 10075 and William has50.34. How much money do they have together?      

Answer: $151.09.

3)      After buying some apples for 6255 jack has 62.55 left. How much money did jack have to begin with?      

Answer $125.10

Maximum and Minimum value

In mathematics, maxima and minima, known collectively as extrema (singular: extremum), are the largest value (maximum) or smallest value (minimum), that a function takes in a point either within a given neighbourhood (local extremum) or on the function domain in its entirety (global or absolute extremum). More generally, the maxima and minima of a set (as defined in set theory) are the greatest and least values in the set. Source – Wikipedia.

Steps to Find Maximum and Minimum value:

Find the maximum and minimum value of a function in calculus, the following rule has to be followed,

Step1: Find the f '(x), which is the delineation of the function f(x).

Step 2: Solve f '(x) =0, Find the roots a, b, c, etc. Then these are the applicants for maximum and minimum values.

Step 3: We examine the function at everyone of these values. Let us take x =c.

Step 4: Establish the symbol of f ' (x) for values of 'x' faintly less than 'c' and for 'x' faintly greater than 'c'.

Conclusion:

If f '(x) symbol modifies from positive to negative as 'x' raises. Then x =c is the point of maximum value.

If f '(x) symbol modifies from negative to negative as 'x' raises. Then x =c is the point of minimum value.

If f ' (x) sign not modifies as 'x' increases. Then x =c is the point of neither maximum nor minimum value.

Example problems for minimum and maximum value:

1) Find the maximum value of y = -3x^2 + x – 5

Solution:-

Y= -3x^2 + x – 5

= -3(x^2 – 1/3) x – 5

= -3(x – 1/6)^2 + 1/12– 5

11

= -3(x – 1/6)^2 – 4  12

11

Maximum value is – 4 12.

2) Find two nonnegative values which addition is 7 and so that the product of one value and the square of the other value is a maximum.

Solution:

Consider variables x and y represent two nonnegative numbers. The sum of the two numbers are given to be

7 = x + y ,

So that

y = 7 - x .

We desire to maximize the product

P = x y^2.

But, before we differentiate the right side, we will write it as a function of x only. Substitute for y getting

P = xy^2

= x (7-x)^2.

Now differentiate this equation using the product law and chain rule, getting

P' = x(2) (7-x)(-1) + (1) ( 7-x)^2

= (7-x) [-2x + (7-x)]

= (7-x) [7-3x]

= (7-x) (3)[ 3-x ]

= 0

for

x=7 or x=3 .

Note that because both x and y are nonnegative numbers and their sum is 9, it follows that 0 <=x<= 7. See the adjoining sign chart for P'.

If   x = 3 and y= 6 ,

then

P= 100 is the largest product.

Finding the Percentage of one number and another

In mathematics, a word problem for a set S with respect to a system of finite encoding of its elements is the algorithmic problem of deciding whether two given representatives represent the same element of the set. Percentage is a way of expressing a number as a fraction of 100. It is often denoted using the percent sign, "%". In this article we shall discuss finding the percentage of one number and another number example problem. (Source: Wikipedia)

Finding the percentage of one number and another example problem

To finding the percentage of one number and another following formula used

Percentage formula = Changes in value   x 100%
                                    Original value

Example 1:

Find how much percentage increase from the number 80 to 95.5?

Solution:

Increased value = new value – old value

Increase value = 95.5 – 80 = 15.5

Percentage increase = `15.5/80` x 100%

Increased percentage = 19.3%

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Example 2:

Find how much percentage decrease from the number 105 to 95.5?

Solution:

Decrease value = new value – old value

Decrease value = 95.5 – 105 = 9.5

Percentage decrease = `9.5/105` x100%

Decreased percentage = 9%

Example 3:

Jasmine last month earned salary `$` 1250. He earned this month salary `$` 1600. Find how many percentage increases in her salary amount.

Solution:

Jasmine increased salary = new salary – original salary

`$` 1600 –`$` 1250 = 350

Salary percentage increase = `350/1250` x100% = 28%

Jasmine increased salary percentage is 28%

Example 4:

Kishore last month earned salary `$` 1800. He earned this month salary `$` 1500. Find how much percentage decreased in his salary amount.

Solution:

Kishore decreased salary = new salary – original salary

`$` 1500 –`$` 1800 = 300

Salary percentage increase = `300/1800` x100%= 16.6%

Kishore decreases salary percentage is 16.6%

Finding the percentage of one number and another practice problem

Problem 1:

Joes last month earned salary `$` 1550. He earned this month salary` $` 1800. Find Percentage Increase in her salary amount.

Answer:

Joes increased salary percentage is 13.8%

Problem 2:

Ram last month earned salary `$` 1300. He earned this month salary `$` 600. Find how much percentage decreased in his salary amount.

Answer:

Ram decreases salary percentage is 53.8%

Truth Tables Solver

In mathematics, a truth table is one important topic in logic. It can be used to understand the basic concepts of Boolean algebra using the truth table. It can be the process of algebra of propositions. Truth tables are used for all input values; it will produce a result as true for the expression. Let us see some example problems for truth tables.

Truth tables solver - Unary operations:

Unary operations:

Two unary operations are

• Logic identity
• Logic negation

Logic identity:

It is used for an operation for one logical value. When the value of the proposition is true then it will produce the operands values as true otherwise it is false.

Truth Table for Logic identity:

In the truth table, T means operand is True and f means operand is false.

Using the truth tables solver, we solve the truth table for logic identity. First we enter the function A in the solver

   

Next we enter the compute button in the solver, then we get the truth table for function.



Logic negation:

It is also used for an operation for one logical value. When the value of the proposition is true then it will produce the operands values as false otherwise the value of the proposition is false then it will produce the operands as false.

Truth table for logic negation:

Logical negation can be written as not A or it can be denoted as ¬ A.

A ¬A
T  F
F  T

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Truth tables solver – Binary operations:

Binary operations:

Some of the binary operations are,

Logic conjunction:

It is used for an operation for two logical values. If A and B are the two values of the propositions, it will produce the result when both of the operands are true that produces a value of a propositions are true otherwise it is false.

Truth table for logic conjunction:

Truth table for logic conjunction can be written as A & B, or A . B

Using the truth tables solver, we solve the truth table for logic identity. First we enter the function A & B in the solver.

    

Next we enter the compute button in the solver, then we get the truth table for the given function.



Logic disjunction:

It is used for an operation for two logical values. If A and B are the two values of the propositions, it will produce the result when anyone of the operands are true that produces a value of a proposition is true otherwise it is false.

Truth table for logic disjunction:

Truth table for logic disjunction can be written as A || B or A + B

A B A & B
T T   T
F T   T
T F   T
F F   F

Logarithms

Logarithms are taught at an intermediate student of algebra who is familiar with the elementary concepts.Logarithm or  Log helps us to solve many complex arithmetical problems in simpler way. Natural logarithm or ln means logarithm to the base e. e is a irrational constant equal to 2.718281828. The natural log is denoted as ln(x) or log e (x). We know that e0=1. Here in this topic we are going to see about log in algebra 2. Natural logarithm is defined for all positive real numbers.The common logarithm are to the base 10 and is written as log(x) simply without mentioning the base. If the logarithm is to any other base then it should be clearly specified.

logarithms algebra 2 - Example problems:



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logarithms algebra 2 - Problem 1:

Solve the given logarithmic expression: ln 3x + ln 5 = 3

Solution:

Given logarithmic expression: ln 3x + ln 5 = 3

ln 3x + 1.6094 = 3                   ( The value of ln 5 = 1.6094)

Subtract by 1.6094 on both side in the above expression.

ln 3x + 1.6094 - 1.6094 = 3 - 1.6094

ln 3x =  1.3906

loge 3x = 1.3906                        (ln x = loge x)

( we know, y = logbP       and P = by)

Here, y = 1.3906         P = 3x                b = e

So,    3x = e^1.3906   = 4.0172

3x = 4.0172

Divided by 3 on both side in the above equation.

` (3x)/3` = `4.0172 / 3`

x = 1.3390

Answer: The value of x = 1.3390



logarithms algebra 2 - Problem 2:

Solve the given logarithmic expression: ln 5x + ln 3x = 0

Solution:

Given logarithmic expression: ln 5x + ln 3x = 0

ln 5x + ln 3x = 0                                         ln A + ln B = ln (A + B)

ln (5x * 3x) = 0                                         (The value of ln 5 = 1.6094)

ln 15x^2 =  0

loge 15x^2 =  0                                (ln x = loge x)

( we know, y = logbP       and P = by)

Here, y = 0       P =15x^2                b = e

So,    15x^2  = e0   = 1

15x^2  = 1

Divided by 15 on both side in  the above equation.

` (15x^2)/15` = `1 / 15`

x^2 = 0.066

x = `+- sqrt 0.066`

Answer: The value of x = ± 0.2569

logarithms algebra 2 - Problem 3:
 
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Solve the given equation and find the x value. 3^-x^2 = `1/27.`

Solution:

Given equation 3^-x^2 = `1/27` .                                  exponential form

log3`(1/27)` = -x^2                                                  logarithmic form

-x^2 = log3`(1/27)`

= log3 1 -  log3 27

= 0 - 3

= -3

x = 31/2.

Answer: The value of x = 31/2.

logarithms algebra 2 - Problem 4:

Solve the given equation and find the x value:  4 (a^-n) = 160

Solution:

Given equation,   4 (a^-n) = 160

a^-n = 40                                                    exponential form

log1.140 = 2x+3                                                logarithmic form

2x + 3 = `(ln 40)/(ln 1.1) `

= 38.7039 (approximately)

Subtract by 3 on both side in above equation

2x + 3 - 3 = 38.7039 - 3

2x = 35.7039

Both side divided by 2. So, we get

`(2x)/2` =` 35.7039/2`

x = 17.85

Answer: The value of x = 17.85 (approximately)

Examples of trig functions

Trig functions generally define the function of angles. To relate the angles of triangle to its length of sides is the common use of trig functions. Trig functions are generally applied in modeling periodic phenomena and study of triangles. Sine, cosine, tangent, cosecant, secant, and cotangent are the most memorable trig functions. Trig functions refer to tabulating the trig functions and trig identities in order. In  this article we shall discuss about examples involved in trig functions.

I like to share this Graphs of Trig Functions with you all through my article.

Examples of Trig Functions:

Let us see some of the examples problems for trig function.

Examples 1: Prove that, sin^4 x − 2sin^2 x cos^2 x + cos^4 x =  cos^2 (2x).

Proof: L.H.S. = sin^4 x − 2sin^2 x cos^2 x + cos^4 x,

=> (sin^2 x)^2 − 2sin^2 x cos^2 x + (cos^2 x)2,

We know that, (a + b)^2 = a^2 − 2ab + b^2, similarly,

=> (sin^2 x − cos^2 x)^2,

=> (sin^2 x −( 1 − sin^2 x))^2,

=> (sin^2 x − 1 + sin^2 x)^2,

=> (2sin^2 x − 1)^2,

=> (−1 (1 − 2sin^2 x))^2,

=> (1 − 2sin^2 x)^2,

=> (cos(2x))^2,

=> cos^2(2x).

=> R.H.S.

Hence Proved that, sin^4 x − 2sin^2 x cos^2 x + cos^4 x = cos^2(2x).


Examples 2: Prove that, tan^2 x + 1 = sec^2 x.

Proof: We know that, sin^2 x + cos^2 x = 1

Dividing the above equation by cos^2 x, we get,

=> `(sin^2x + cos^2x) / (cos^2x) = 1 / (cos^2x)` ,

=> `(sin^2x) / (cos^2x) + (cos^2x) / (cos^2x) = 1 / (cos^2x)` ,

By using the trigonometric functions; `sinx/cosx` = tan x, and `1/cosx` = secx, we get,

=> tan^2 x + 1 = sec^2 x

Hence proved that, tan^2 x + 1 = sec^2 x.

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Examples 3: Prove that, `(tan A + cot B)/(cotA + tanB) = tan A /tanB` .

Proof: L.H.S. = `(tan A + cot B)/(cotA + tanB)` ,

We know that, cot x = `1/tanx` , we get,

=> `((tan A + 1/tanB))/((1/tanA + tanB))` ,

By taking L.C.M. We get,

=> `(((tanAxxtanB + 1)/tanB))/(((1 + tanAxxTanB)/tanA))` ,

By taking reciprocal, we get,

=>  `((tanAxxtanB + 1)/tanB)xx(tanA/(1 + tanAxxTanB))` ,

By simplification, we get,

=> `tan A /tanB` ,

=> R.H.S.

Hence proved that, `(tan A + cot B)/(cotA + tanB) = tan A /tanB` .

Practice Problems of Trigonometric Functions:

Problem 1: Prove that, `(1 + tan x)/(1+cotx) = (sin x + tanx)/(1 + cosx)`.

Problem 2: Prove that, `cosx /(1+sinx)` = sec x − tan x.

Problem 3: Prove that, `(1+cosx)/(1-cosx)` = (csc x + cot x)2.

Algebra Fractional Exponents

In algebra we can utilize numbers with letters of the alphabet such like a,b,c, etc for any numerical values are select. thus, algebra is an expansion of arithmetic. Here the fractional exponents are a part of Algebra, that make an expression as am/n. In the fractional exponent the denominator that is identical to the index of the radical such as x1/n=`root(n)(x)`.

Algebra fractional exponents:

Algebra is really larger than basic algebra and when special rules of method are used also when process are formulate for more than numbers. In the exponential is the function e^x, where e is the digit such to the consequence e^x the equivalent its contain derived. The exponential function is utilized to constitution happening when a constant modify in the self-determining variable give the similar proportional vary in the dependent variable.

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exponential Function:

e^a*eb = e^a +b
e^a/eb = e^a-b
eln(x) = x
ln(e^x) =x
e^0=1

examples for algebra fractional exponents:

example 1:

How to Solve the fractional exponents x^5/3 * x^4/3.

Solution:

Step 1:   The given  equations x^5/3 * x^4/3.

Step 2:    Rule 1: x^y * xz = x^y+z

By using this rules for this function

Step 3:       x^5/3 * x^4/3 = x^5/3+4/3

Step 4:        x^5/3 * x^4/3 = x^5/3+4/3

Step 5:         = x^(5+4)/3

Step 6:        = x^9/3

So the solution is    x^5/3 * x^4/3 = x^9/3.

example 2:

How to Solve the fractional exponents  (51/2)^8

Solution:

Step 1:  (51/2)^8

Step 2:     Rule 2:      (x^y)z = x^yz

Step 3:  By using this rule in this function

Step 4:   (51/2)^8 = 51/2*8

Step 5:    (51/2)^8 = 5^4

So the solution is  (51/2)^8 =  625

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example 3:

How to Solve the fractional exponents  22/3 * 24/6

Solution:

Step 1: the given function is  22/3 * 24/6

Step 2: Rule 3:    x^y * x^z = x^y+z

Step 3:   By using this rule in this function is

Step 4:         22/3 * 24/6 = 22/3+4/6

Step 5:        22/3 * 24/6 = 2(4+4)/6

Step 6:        =  2^(4+4)/6

= 28/6 = 24/3

So the answer is 24/3

example 4:

How to Solve the fractional exponents  491/2

Solution:

Step 1:  the given equation  491/2

Step 2:   Rule 4: x1/n=`root(n)(x)`

Step 3: by using rule in this function

Step 5:   491/2  =`root(2)(49)`

Step 6: So the solution is  491/2  =`root(2)(49)`

example 5:

How to Solve the fractional exponents  92*0.5

Solution:

Step 1:  92*0.5

Step 2:  92*0..5 = 92 * 90.5

Step 3:  = 92 * 91/2

Step 4:  92*0.5 = 81 * 9

So the solution is 729