Basic Math Facts

 Basic math facts article deals with the basic facts behind the various arithmetic operations. The method or steps need to be followed for solving the basic math problem.
                           The Basic math facts are:

• Addition Basic facts.
• Subtraction Basic facts
• Multuiplication Basic facts.
• Division Basic facts.

Basic math facts for addition and subtraction:

 Addition Basic facts:
           The basic math facts for the addition of two numbers.

1. Symbols for the both number should be same.
2. negative number  (+) negative number = negative number
3. Positive number  (+) postive number = positive number

Model problems:

1. 15+15

             Solution:
                          Here the sign for both are same. So add the two numbers
                                        = 15+15
                                        = 30 (positive number)
2. -15 -15
               Solution:
                       Here the both numbers are negative numbers, so add the two numbers.
                                       = -15-15
                                       = -30 (negative number)
Subtraction Basic facts:
           The basic math facts for the subtraction of two numbers.

1. Symbols for the numbers should be different.
2. postive number  (-) negative number = postive  number
3. Negative number  (-) postive number = negative number

Model problems:

1. 15-10

 Solution:
                          Here the sign for both are different. So subtract the two numbers
                                        = 15-10
                                        = 5 (positive number)
2. -15 +10
 Solution:
                       Here the both numbers are having different sign, so subtract the two numbers.
                                       = -15+10
                                      = -5 (negative number)

Basic math terms for multiplication and divsions:

Mulitplication Basic facts:
           The basic math facts for the multiplication of two numbers.

• Symbols for the both number may  be same or different
• negative number  (*) negative number = positive number
• Positive number  (*) postive number = positive number
• postive number (*) negative number = negative number

Model problems:
1.15*15
             Solution:
                          Here the sign for both are same. Therefore, answer is positive numbers
                                        = 15*15
                                        = 225 (positive number)
2.-15 *15
         Solution:
                       Here both numbers are  not of same sign

                                       = -15*15
                                       = -225 (negative number)
3.-15 * -15
Solution:
                      Here the sign for both are same. Therefore, answer is positive numbers
                                        = -15* -15
                                        = 225 (positive number)

Division Basic facts:
           The basic math facts for Division of two numbers.

1. Symbols for the both number may  be same or different
2. (negative number') /( negative number) = positive number
3. Positive number  / postive number = positive number
4. postive number / negative number = negative number

Model problems:
1.15/15
 Solution:
                          Here the sign for both are same. Therefore, answer is positive numbers
                                        = 15/15
                                        = 1 (positive number)
 2.-15 /15
 Solution:
                       Here both numbers are not of same sign
                                       = -15/15
                                       = - 1 (negative number)

3.-15 / -15
Solution:
                      Here the sign for both are same. Therefore, answer is positive numbers
                                        = -15/ -15
                             =  1 (positive number)

Solve Multiplying Trinomials

In Algebra trinomial is a polynomial having three terms .otherwise it consisting of three monomial. For an example.

8x + 5y + 3z having x, y z variable

4t + 6s2 + 3y3 Here t, s, y is a variable

5ts + 9t + 6s with t, s variables

Here we are going to study about how to solve multiply a trinomial and its example problems.
Solve example problems for multiplying trinomial

Example: 1

Multiplying trinomial with binomial

Solve (x2+2x+3) (3x+3)

Solution:

First we have to take 3x multiply with trinomials

3x3+6x2+9x

Next we take a 3

3x2+6x+9

Now combine the like terms

3x3+6x2+3x2+9x+6x+9

3x3+10x2+15x+9

This is the multiplying answer.

Example: 2

Multiplying trinomial with another trinomial

Solve (x2 +3x+3) (2x2 +6x+4)

Solution:

Here there are two trinomial is given

First we have to multiply the x2  term to the other trinomial we get

2x (2+2) +6x (2+1) + 4x2

= 2x4+6x3+4x2

Next we take the 3x term we get

3*2x(1+2)+ 3*6x2+12x

= 6x3+18x2+12x

Finally we take a constant term 3

= 6x2+18x+12

Now we combine all the terms we get

= 2x4+6x3+4x2 + 6x3+18x2+12x + 6x2+18x+12

Combine the like term x3

6x3+6x3 = 12 x3

Combine x2 term

= 4x2+18x2+6x2

= 28x2

Combine x term

12x+18x=30x

Therefore the final answer is

2x4+12x3+28x2+30x+12

Multiplying trinomial with same trinomial

Example: 3

Solve (x2+3x+2) (x2+3x+2)

Solution:

Now we have to expand the polynomial

Take x2 and multiply with all the terms we get

X4+3x3+2x2

Similarly take 3x and multiply with all the term we get

3x3+9x2+6x

Finally take constant 2 multiply with all the term we get

2x2+6x+4

Now we combine all the term

X4 +3x3 + 2x2 + 3x3 + 9x2 + 6x + 2x2 + 6x + 4

Combine the like terms we get

3x3+3x3= 6x3

Combine the x2 terms

2x2+2x2+9x2 = 13x2

Similarly combine the x term

6x+6x=12x

Finally constant 4

Combine all the terms

x4+6x3+13x2+12x+4

This is the multiplying answer for the given two trinomials.

Math Game Absolute Value

In math, the absolute value is also known as the modulus |x| of a real number x is x's arithmetic value without consider to its sign. So, for example, 5 is the absolute value of both 5 and −5.

A simplification of the absolute value for real numbers occurs in an extensive selection of math settings.

Properties of the math game absolute value:

The absolute value has the following four fundamental properties:

`|x| = sqrt(x^2)`                                         (1) Basic

`|x| \ge 0 `                                                  (2)     Non-negativity

`|x| = 0 \iff x = 0`                                  (3)     Positive-definiteness

`|xy| = |x||y|\,`                                        (4)     Multiplicativeness

`|x+y| \le |x| + |y|`                                  (5)     Subadditivity

Other important properties of the absolute value include:

` |-x| = |x|\, `                                             (6)     Symmetry

`|x - y| = 0 \iff x = y `                            (7)     Identity of indiscernible (equivalent to positive-definiteness)

`|x - y| \le |x - z| +|z - y|`                        (8)     Triangle inequality (equivalent to sub additivity)

`|x/y| = |x| / |y| \mbox{ (if } y \ne 0) \,`                          (9)      Preservation of division (equivalent to multiplicativeness)

`|x-y| \ge ||x| - |y||`                                  (10)     (equivalent to sub additivity)

If y > 0, two other useful properties concerning inequalities are:

`|x| \le y \iff -y \le a \le y`

` |x| \ge y \iff x \le -y \mbox{ or } y \le x`

Math game absolute value – Games:

Math game absolute value – Game 1:

Arrange the order of ascending -|-15|, |12|,|7|,|-99|,|-5|,|-8|, |-65|, |6|

Solution:

First we remove the modulus symbol

-15, 12, 7, 99, 5, 8, 65, 6

Then to arrange the given order

-15, 5, 6, 7, 8, 12, 65, 99
Math game absolute value – Game 2:

Arrange the order of descending -|-16|, |13|,|8|,|-9|,|-6|,|-7|, |-66|, |63|, -|21|, |-68|

Solution:

First we remove the modulus symbol

-16, 13, 8, 9, 6, 7, 66, 63, -21, 68

Then to arrange the given order

68, 66, 63, 13, 9, 8, 7, 6, -16, -21

Math game absolute value – Game 3:

Absolute Value

Arrange the descending order

Solution:

First we remove the modulus symbol

-81, -28,67,-59,98,71,38

Then to arrange the given order

98, 71, 67, 38, -28, -59, -81

Evaluate Each Expression

Evaluate each expression is very simple concept in math. When you substitute a particular value for each variable and then act the operations called evaluating expression. The mathematical expression is algebraic, it involves a finite sequence of variables and numbers and then algebraic operations. Evaluation is simplifying the expression.

Algebraic operations are:

Addition

subtraction

multiplication

division

raising to a power

Extracting a root.

I like to share this Evaluate Indefinite Integral with you all through my article.


Evaluating algebraic expressions are using PEMDAS method.

Rules for PEMDAS method:

1. First it performs operations inside the parenthesis.

2. Second exponents

3. Third multiplication and division from left to right

4. Finally addition and subtraction from left to right

Evaluate each expression for x=-5 and y=8

2x4-x2+y2

Solution:

Given expression is 2x4-x2+y2

Substitute x and y values in given expression

So,

=2(-5)4-(-5)2+ (8)2

=2(625)-(25) +64

=1250-25+64

=1314-25

=1289

Therefore answer is 1289

Evaluate the expression x3-x2y+9 for x=3, y=9

Solution:

Given expression x3-x2y+9

Plugging the x and y values in given expression

=33-329+9

=27-9*9+9

=27-81+9

=36-81

=-45

Therefore answer is -45

example problems for evaluating the expressions:

Evaluate each expression using PEMDAS method

3 * (9 + 12) - 62 `-:` 2 + 8

Solution:

Remember the PEMDAS rules

First parenthesis: (9+12) = (21)

3*21-62`-:` 2+8

Then exponents: 62 = 6 *6 = 36

3*21-36`-:` 2+8

Then multiplication and division: 3 * 21 = 63 and 36`-:` 2 = 18

63-18+8

Then addition: 63+8=71

71-18

Finally subtract the term

53

Therefore the answer is 53

Evaluate each expression for a = –2, b = 3, c = –4, and d = 4.

bc3 – ad

Solution:

= (3) (–4)3 – (–2) (4)

using PEMDAS exponent rule (-4)3=-4*-4*-4 = -64

= (3) (–64) – (–8)

then multiply the numbers 3*(-64) = -192

= –192 + 8

Add the numbers

= –184

(b + d)2

Solution:

= ((3) + (4))2

Add the inside numbers 3+4=7

= (7)2

using exponent rule 72=7*7=49

= 49

so, the answer is 49

a2b

Solution:

= (–2)2(3)

= (4) (3)

= 12

a – cd

Solution:

= (–2) – (–4)(4)

= (-2)-(-16)

= (-2) +16

= 16-2

= 14

b2 + d2

Solution:

= (3)2 + (4)2

= 9+16

= 25

Evaluate each expression For x = –3.

3x2 – 12x + 4

Solution:

= 3(–2)2 – 12(–2) + 4

= 3(4) + 24 + 4

= 12+24+4

=40

x4 + 3x3 – x2 + 6

Solution:

= (–3)4 + 3(–3)3 – (–3)2 + 6

= 81 + 3(–27) – (9) + 6

= 81 – 81 – 9 + 6

= –3

Solving Proportions

We will be solving proportions for x, by using our property that says the cross products of a proportion are equal to each other. If you remember the cross product we find them by multiplying a numerator one ratio time the denominator of the other. So just by looking over one proportion problems we will get the idea how exactly it is solved.

Say, x/15=21/45 here we will do the cross multiplication as 45 times x = 15 times 21 that is 45x= 315, now divide 45 both sides and we get x= 7. And this makes sense, 7 over 15 and 21/45 are equal fractions which is basically what proportions are.

One more question, 17/20=x/25. Here 17 times 25 equals to 20 times x. 425=20x, again dividing both the sides by 20 and we get the solution as x= 21.25. These types are nothing but theproportion solver.

We can learn this method only by practicing more by solving proportion examples. Thus the proportion examples are as follows, solving the proportion let us start with x/15=6/10. So here we have two ratios which are equal to each other. We need to find the value of x which will make the whole fraction proportionate to another.

We are going to find what this x number is going to represent so then this ratio in fact is equal to another ratio.  Using the cross product method will be used to solve this problem. we multiply two numbers which are diagonal from each other.10 times x=15 times 6.we get 10x=90, kindly remember there is no sign between the number and the variable it means they both have to multiply that here in this question will be 10 times x as 10 x. now by dividing 10 no both the sides,

we get the value of x as 9. Make sure if our solution is correct, simplify the problem by putting the value of x into the fraction, we will find the both fractions are proportionate to each other.

Let us take one more example, 3/15= y/50. Now again we have to solve this for the variable known as y. to do this we are going to find cross products. So 15 times y equals to 3 times 50, will be written as 15y=150, now dividing both the sides by 15 we get the value for y. and the value of y is 10.thus our final answer is y=10. And we can check that by plugging the value of y and simplifying shows that the fractions are in proportion.

Finctions Substituting

Substitution is the procedure of replace a changeable in a term with its real worth. If you are given an equation like 6x + 7 = 6+7, told that x = 1, and ask to find the value of the function what do you do? The first step is to replace with 1 for each 'x' in the problem. We get the expression as, 6(1) + 7 = 13

Functions substituting Example problems:

Problem 1: Find the functions substituting given below.

f(x) = 2x + 1

Substitute x=1, 2,3,4,5

Solution for  Functions substituting:

Step 1

x=1 f(1) =(2*1)+1

The answer is

f(1) =3

Step 2

x=2 f(2) =(2*2)+1

The answer is

F (2) =5

Step 3

x=3 f (3) =(2*3)+1

The answer is

F (3) =7

Step 4

x=4 f(4) =(2*4)+1

The answer is

F (4) =9

Step 5

x=5 f(4) =(2*5)+1

The answer is

F (4) =11

The substituting functions is 1, 2,3,4,5

F (1) = 3

F (2) =5

F (3) =7

F (4) = 9

F (5) = 11

Problem 2: Find the functions substituting given below.

f(x) = 3x +2

Substitute x=1, 2,3,4,5

Solution for  Functions substituting:

Step 1

x=1 f(1) =(3*1)+2

The answer is

f(1) =5

Step 2

x=2 f(2) =(3*2)+2

The answer is

f(2) =8

Step 3

x=3 f(3) =(3*3)+2

The answer is

f(3) =11

Step 4

x=4 f(4) =(3*4)+1

The answer is

f(4) =13

Step 5

x=5 f(5) =(3*5)+1

The answer is

f (5) =16

The substituting functions is 1, 2,3,4,5

F (1) = 5

F (2) =8

F (3) =11

F (4) =13

F (5) =16

Functions substituting Example problems:

Problem 3: Find the functions substituting given below.

F(x) = 2x +2

Substitute x=1, 2,3,4,5

Solution for  Functions substituting:

Step 1

x=1 f (1) =2*1+2

The answer is

f(1) =4

Step 2

x=2 f(2) =(2*2)+2

The answer is

f(2) =6

Step 3

x=3 f(3) =(3*2)+2

The answer is

f(3) =8

Step 4

x=4 f(4) =(2*4)+2

The answer is

f(4) = 10

Step 5

x = 5 f(5) = (2*5)+2

The answer is

f(5) = 12

The substituting functions  is 1, 2,3,4,5

F (1) = 4

F (2) = 6

F (3) = 8

F (4) = 10

F (5) =12

Problem 4.Find the functions substituting given below.

F(x) = 4x +4

Substitute x = 1, 2,3,4,5

Solution for Functions substituting

Step 1:

x=1 f (1) = (4*1)+4

The answer is

f(1) = 8

Step 2

x=2 f(2) = (4*2)+4

The answer is

f(2) =12

Step 3

x=3 f(3) =(4*2)+4

The answer is

f(3) =12

Step 4

x=4 f(4) =(4*4)+4

The answer is

f(4) =20

Step 5

x=5 f(5) =(4*5)+4

The answer is

f(5) =24

The substituting functions is 1, 2,3,4,5

F (1) =8

F (2)=6

F (3) =12

F (4) =20

F (5) =24

Math Word Problems Rate

Definition

Solving the Time, Speed, and Distance Triangle

The following formulas have been used if the speed is measured in knots, the distance in nautical miles, and the time in hours and/or tenths of hours (0.1 hour = 6 minutes).

Distance = Speed x Time

Speed = Distance ÷ Time

Time = Distance ÷ Speed

Math word problems rate - Examples

Math word problems rate - Example 1

A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?

Explanation:

Speed=`(600/(5*60))` = 2 m/sec.

Converting m/sec to km/hr (see important formulas section)

=`2*18/5`

=7.2 km/hr.

Math word problems rate - Example 2

The ratio between the speeds of two trains is 7: 8. If the second train is runs at 400 kms in 4 hours, then the speed of the first train is:

Explanation:

Let the speed of two trains be 7x and 8x km/hr.

Than, 8x=`400/4`

8x=100

X=12.5

Speed of first train = (7 x 12.5) km/hr = 87.5 km/hr.

Math word problems rate - Example 3

An aero plane covers with a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 1 hours it must travel at a speed of:

Explanation:

Distance =` (240 x 5)` = 1200 km.

Required speed=`(1200*3/5)` km/hr

=720km/hr

Math word problems rate - Example 4

A man completes a journey in 10 hours. He travels the first half of the journey of the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.

Explanation:

`(1/2)` `(x/21)` +`(1/2)` `(x/24)` =10

`(x/21)` +`(x/24)` =20

15x = 168 x 20

X(`168*` `20/15` `)` =224km

Math word problems rate - Example 5

A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot at 4 km/hr and partly of bicycle at 9 km/hr. The distance travelled on foot is:

Explanation:

Let the distance travelled on foot be x km.

Then, distance travelled on bicycle = (61 -x) km.

So, `x/4` + `(61-x)/9` =9

9x + `4(61 -x)` = 9 x 36

5x = 80

x = 16 km.

Math word problems rate - Practice Problem

Example

A man on tour of the travels at first 160 km in 64 km/hr and the next 160 km in 80 km/hr the average speed for the first 320 km of the tour is:

Answer=71.11 km/hr