Learn Trigonometry Test Questions

Introduction to learn trigonometry test questions:
Here we are going to see the article as learn trigonometry test questions, generally trigonometry is used to study about the triangle especially right angle triangle ,the main purpose of trigonometry is  used to find the sides and angle of a right angle triangle with the help of a trigonometric functions  such as sin ,cos, tan ,secant ,cosecant and cot. Let us start to learn some of the trigonometry test questions.


Example 1-learn trigonometry test questions:

Suppose A and B are positive acute angles and the value of cos A = `12/15` , and sin B =` 8/10` , what is the value of Sin (A + B)?

Solution:

`cos A = 12/15` , from that we find the value of sin A

Already we know that the `cos A= (adjacent side)/("hypotenuse") `

So adjacent is 12 and hypotenuse is 15

Use the Pythagorean Theorem here to find the unknown side it will be shown in below,

`(Opposite side)^ 2= ("hypotenuse")^ 2- (adjacent)^2 `

So adjacent is 12 and hypotenuse is 15

`"= (15)^2-(12)^2=225-144=81=9^2`

Opposite side =9

So `sinA =(opposite side)/("hypotenuse") `

The value of `sin A is 9/15`

Similarly

`sin B = 8/10` , from that we find the value of sin B

Already we know that the `sin B = (opposite side)/("hypotenuse") `

So opposite side is 8 and hypotenuse is 10 now use the Pythagorean Theorem to find the adjacent values

`(Adjacent side)^2 ` `= (10)^2-(8)^2=100-64=36=6^2`

So the value of adjacent side is `6/10`

`cos B=6/10`

`"cos A = 12/15, sin B=8/10, sin A =9/15, cos B = 6/10`


sin (A+B) = sinAcosB + cosAsinB
sin (A+B)= `(9/15) xx (6/10) + (12/15) (8/10)`

sin (A+B)= `(54/150) + (96/150) =1`

Example 2-learn trigonometry test questions:

Suppose the adjacent side of a right angle triangle is 6cm and opposite side of a triangle is 8cm find the hypotenuse of a triangle?

Solution:

Here the adjacent side of a right angle triangle is given as 6cm and opposite side of a right angle triangle is given as 8cm

We know that formula for Pythagorean Theorem

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

Here AC is the hypotenuse

AB is the opposite side and BC is the adjacent side of a triangle

Plug those values in the above formula means we get the hypotenuse value

`AC^2=6^2+8^2`

`AC^2=36+64=100`

`AC^2= (10)^2`

So the value of hypotenuse is 10cm

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Some of the trigonometry test questions with the answer key:


                 `"sin^3A+cos^3A `
1) prove   _______________     = 1- sin A Cos A
                    sin A + cos A

2)Suppose the opposite side of a right angle triangle is 15cm and hypotenuse of a triangle is 17cm find the adjacent side of a triangle?

Answer:

8cm

Learn Independent Events

Learn Independent Events
An event is a one or more possible outcomes of a certain experiment. An event is called independent event if one event does not affect the other event. For example, choosing a 7 and 8 in the deck of card with replacement is two independent events. An event consisting of more than one simple event is called compound event. In this lesson we will learn about probability of independent events.


Learn Independent Events – Learn Example Problems


Example 1: A box contains 10 bulbs, in which 5 are red color and 5 are orange color bulbs. Two bulbs are drawn one by one with replacement of first bulb. What is the probability of drawing a red and orange bulb successively?
Solution:

Let S = Sample space, n(S) = 5 + 5 = 10

A be the event of drawing a red ball first, n(A) = 5

B be the event of drawing a orange bulb second, n(B) = 5

P(A) = `(n(A))/(n(S))` =` 5/10` = `1/2`

P(B) = `(n(B))/(n(S))` = `5/10` = `1/2`

P(A and B) = P(A) · P(B) = `1/2` · `1/2` = `1/4`

P(A and B) = `1/4` .

Example 2: A box contains 15 candies, in which 5 are lemon candies, 5 are pineapple candies, and 5 are orange candies. Three candies are drawn one by one with replacement of previous candy. What is the probability of drawing a lemon, pineapple, and orange candy successively?
Solution:

Let S = Sample space, n(S) = 5 + 5 + 5 = 15

A be the event of drawing a lemon candy first, n(A) = 5

B be the event of drawing a pineapple candy second, n(B) = 5

C be the event of drawing a orange candy third, n(C) = 5

A, B, and C are independent event, so one event does not affect the other events.

P(A) = `(n(A))/(n(S))` =` 5/15` = `1/3`

P(B) = `(n(B))/(n(S))` = `5/15` = `1/3`

P(C) = `(n(C))/(n(S))` = `5/15` = `1/3`

P(A and B and C) = P(A) · P(B) · P(C) = `1/3` · `1/3` · `1/3` = `1/27`

P(A and B and C) = `1/27` .

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Learn Independent Events – Practice Problems


Problem 1: A box contains 25 bulbs, in which 15 are red color and 10 are orange color bulbs. Two bulbs are drawn one by one with replacement of first bulb. What is the probability of drawing a red and orange bulb successively?

Problem 2: A box contains 25 candies, in which 10 are lemon candies, 10 are pineapple candies, and 5 are orange candies. Three candies are drawn one by one with replacement of previous candy. What is the probability of drawing a lemon, pineapple, and orange candy successively?

Answer: 1) ` 6/25` 2) `4/125`

Learn Intercept Group

Introduction for learn intercept group:

The intercept group has different types, one is x intercept and second one is y intercept and last one is z intercept. X intercept means point crosses the x axis and y intercept means point crosses the y axis of the line. When x = 0 and y= 0 that intercept is called as z intercept. Let us learn about the intercept group example problems and practice problems are given below.

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Learn intercept group example problems:


Example 1: Find the x intercept of the line, 2.5x + 5y = 20

Solution:

The slope intercept form is y = mx + b, where m is the slope of the line.

Given equation is in the form of ax + by = c. To find the x- intercept Plug y = 0 in the equation

Here x intercept, so y = 0

2.5x + 5(0) =20

2.5x = 20

Divide by 2.5 on both sides.

` (2.5x)/(2.5)` = ` (20)/(2.5)`

After simplify this, we get

x = 8

x intercept = 8

Example 2: Find the y intercept of the line, 12x + 6.5y = 45.5

Solution:

The slope intercept form is y = mx + b, where m is the slope of the line.

Given equation is in the form of ax + by = c. To find the y- intercept Plug x = 0 in the equation

Here y intercept, so x = 0

12(0) + 6.5y = 45.5

0 + 6.5y = 45.5

6.5y = 45.5

Divide by 6.5 on both sides.

`(6.5y)/(6.5)` = ` (45.5)/(6.5)`

After finding this, we get

y = 7

y intercept = 7

Example 3: Solve the z intercept of the given equation: 0.5x - 1.5y + 3.5z = 35

Solution:

The given equation is 0.5x - 1.5y + 3.5z = 35. Let us find z intercept.

Substitute x = 0 and y = 0

0.5(0) - 1.5(0) + 3.5z = 35

0 - 0 + 3.5z = 35

After simplify this, we get

3.5z = 35

Divided 3.5 on both sides, we get

`(3.5z)/(3.5)` = `(35)/(3.5)`

After simplify this, we get

z intercept = 10

These example problems are very helpful to learn of intercept group.


Learn intercept group practice problems:


Problem 1: Find the x intercept of the line, 2.5x + 1.5y = 40

Answer: x intercept = 16

Problem 2: Find the y intercept of the line, 5x + 14.5y = 58

Answer: y intercept = 4

Problem 3: Solve the z intercept of the given equation: 6x - 4y + 125z = 250

Answer: z intercept = 2

How to Learn My Numbers

Introduction for how to learn my numbers:
A numbers is a mathematical object used in counting and measuring. The notational symbols which represent a number is called a numeral, but in common usage the word number is used for both the abstract object and the symbol, as well as for word for the number. In addition to their used in counting and measuring, numerals are often used for labels (telephone numbers), for ordering (serial numbers), and for codes. In this article we shall dicuss  about how to learn my numbers. (Source.Wikipedia)


Classifications learn my numbers:


Types:

1. Natural numbers

2. Integers

3. Rational numbers

4. Real numbers

5. Complex numbers

1. Natural numbers:

The numbers 1, 2, 3… are called learn my natural numbers. They are also called counting learn numbers since they are used for counting objects.

2. Integers:

The numbers 0, 1, -1, 2, -2 … are called learn integers of which 1, 2, 3 … are called positive integers and -1, -2, -3… are called negative integers. The collection of all integers is denoted by the letter Z. Thus Z = {…, -3, -2, -1, 0, 1, 2, 3…}.

3. Rational numbers:

A number of the form t/r where t and r are integers and r ? 0 is called a rational number. The collection of all my rational numbers is denoted by Q. A rational number t/r is said to be in the proper form if r is a positive integer and t and r have no common factor other than 1.

Example, rational numbers `7/13` ,` 2/7`

Real numbers:

Combination for rational and irrational numbers are called as real numbers

Example: 42, -55/98

Complex numbers:

Learn complex number is of the form s + it where ‘s’ and ‘t’ are real numbers and i is called the imaginary unit, having the property that i2 = - 1. If z = s+ it then s  is called the real part of z, denoted by Re(z) and t is called the imaginary part of z and is denoted by Im(z). Examples forf complex numbers are 9 - i2,


Examples for learn my numbers:


Example 1:

Determine the rational number represented by 77.0.

Solution:

Let x =77.0. Then x = 0.777777…

? 100x = 77.777777…

? 100x - x = (77.777777…) - (0.777777…) = 77.0000…

? 99x = 77 or x = `77/99` =`7/9`

Example 2:

Write the real and imaginary parts of the following numbers:

(i) 6 - i 3 (ii)`93/2` i

Solution:

(i) Let z = 6 - i 3 ; Re(z) = 6, Im(z) = - 3

(ii) Let z =`93/2` i ; Re(z) = 0, Im(z) =`93/2`

Learn Roman Numerals Properly

Introduction to Roman numerals:

Roman numerals are a numerals system of ancient Rome based on letters of the alphabet, which are combined to signify the sum (or in some cases, the difference) of their numbers. The first ten Roman numerals are as follows:

I, II, III, IV, V, VI, VII, VIII, IX, and X.

In this be article we are going to learn Roman numerals properly. (Source: Wikipedia)


Learn Roman numerals properly:


All Roman numerals should be formed by combining the symbols that the Romans used.

That is I , V , X , L , and C are the letters used as Roman numerals.

Learning rules of Roman numerals properly:

In order to learn Roman numerals properly the best way is to start from the rules.

Rule 1:

Add that number if one or more letters are placed subsequent to another letter of greater number.

Examples are:

III = 1+1+1 = 3

VIII = 5+1+1+1 = 8

Rule 2:

Subtract that number if a letter is placed previous to another letter of greater number.

Examples are:

IV = 4 (5-1 = 4)

XC = 90 (100-10 =90)

XLI = 41 (50-10+1 = 41)

Rule 3:

We should only subtract exponents of 10 (I = 100, X = 101, C = 102)

Examples are:

IX = 9 (10-1 = 9)

XL = 40 (50-10 =40)

CD = 400 (500-100 = 400)

Rule 4:

We shouldn’t subtract more than 1 number from another number.

XXXVIII = 38 (10+10+10 + 5 + 1 + 1 + 1 = 18)

We shouldn’t write 38 as IIXXXX.


Learning example problems for Roman numerals properly:


Learning problem 1:

Write 7 in decimal number.

Solution:

The given decimal number can be written as 7 = 5+1+1

Note that the numbers given must be splitted only with the fundamental numerals I , V , X , L , C , D , M

Therefore 7 = 5+1+1

V  I  I

Where V indicates decimal number 5

I indicate decimal number 1

In roman numerals 7 can be written as VII.

Learning problem 2:

Write 19 in decimal number

Solution:

The given decimal number can be written as 19 = 10+(10-1)

Therefore 19 = 10+(10-1)

Where X indicates decimal number 10

I indicate decimal number 1

In roman numerals 19 can be written as XIX.

Learning problem 3:

Write 44 in decimal number

Solution:

The given decimal number can be written as 44 = (50-10)+(5-1)

Therefore 44 = 50+(5-1)

Where L indicates decimal number 50

X indicates decimal number 10

V indicates decimal number 5

I indicate decimal number 1

In roman numerals 44 can be written as XLIV.

How to Learn Multiplication Fast

Introduction to how to learn multiplication fast:
Multiplication is one of the basic operations in mathematics. Multiplication can be done using math tables. Multiplying a small number is a simple task, using math table, but when we consider for a larger number, this process is tedious. We have introduced a lot of easy ways for multiplication to learn as fast as possible. Here we are going to see how to learn multiplication fastly.


Methods to learn multiplication fast:


Many new trends in mathematics are introduced how to learn multiplication fast. These fast methods save our computational processing time during multiplication of complex problems.

Here we use some of the fast ways to learn how to do multiplication effectively,

Methods to learn multiplication fast for two digit number less than 20:

While multiplying any two digit number that is less than 20, follow the following steps,

Let us consider, 12 * 14

Choose the largest number in the front.
Here the largest number is 14. Hence the sum becomes 14 * 12.
Add the largest number (14) with the last digit of the lowest number (2).
Hence it becomes 14 + 2 = 16.
Multiply 10 with the resulting answer.
Hence we get 16 * 10 = 160.
Now multiply the last terms of two given numbers.
Hence we get, 4 * 2 =8.
Add 160 + 8 = 168.
Thus we got the right answer as 168.

Thus we learn how to multiply fastly for two digit number less than 20.

Another example:

Let us consider, 15 * 17

Choose the largest number in the front.
Here the largest number is 17. Hence the sum becomes 17 * 15.
Add the largest number (17) with the last digit of the lowest number (5).
Hence it becomes 17 + 5 = 22.
Multiply 10 with the resulting answer.
Hence we get 22 * 10 = 220.
Now multiply the last terms of two given numbers.
Hence we get, 7 * 5 =35.
Add 220 + 35 = 255.
Thus we got the right answer as 255.


Methods to learn multiplication fast for any number with multiples of 9:

While multiplying for any number with multiples of 9, follow the following steps,

Let us consider, 64 *99

Take 64, add two zeros with 64, because we have two 9. (If three 9, add 3 zeros).
Hence it becomes 6400.
Now subtract 6400 with the original number 64.
Hence we get, 6400 – 64 = 6336.
Thus we got the right answer as 6336 in a fast way.

Another example:

Let us consider, 34 *99

Take 64, add two zeros with 34, because we have two 9. (If three 9, add 3 zeros).
Hence it becomes 3400.
Now subtract 3400 with the original number 34.
Hence we get , 3400 – 34 = 3366.
Thus we got the right answer as 3366 in a fast way.

Thus we learn how to multiply fastly any number with multiples of 9.

Learn How to Count Percentage

Introduction to learn how to count percentage

In mathematics, a percentage is a way of expressing a number as a fraction of 100 (per cent meaning "per hundred" in French). It is often denoted using the percent sign, "%", or the abbreviation "pct". For example, 45% (read as "forty-five percent") is equal to 45 / 100, or 0.45. Counting percentage means the process of expressing the value of a number in 100. (Source: From Wikipedia).

Here we are going to learn how to count percentage.

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Example problems to learn counting percentage

Here we will learn some example problems to count percentage.

Example 1

Write the fraction `30/100` as a percentage

Solution

Here 30 is expressed as a fraction of 100. To write `30/100` as a percent we have to multiply `30/100` by 100.

We, get `30/100` * 100 = 30

So, 30/100 is equal to 30 percentage or 30%

Example 2

Count the percent of 30 in 80

Solution

Here we have to count the percent of 30 in 80.

Let x be the percentage,

So, `30/80` = `x/100`

x = `30/80` * 100

x = `300/8`

x = 37.5

So, 30 is 37.5 percent of 80

Example 3

What is 30 percent of 20.

Solution

Here we have to find the value of 30 percent in 20

Let the number be x

So, `x/20` = `30/100`

x = `30/100` * 20

x = `600/100`

x = 6

So, 30 percent of 20 is 6.


Few more examples to learn counting percentage

Here we will learn few more problems to count percentage.

Example 1: Write the decimal 0.25 as fraction

Solution

0.25 can be written as 25/100 in fraction

To convert this fraction into percentage we have to multiply the fraction by 100

Doing so, `25/100` * 100 = 25

So, 0.25 is 25 percent.

Example 2: Write the proportion 32:100 as a percent

Solution

The proportion 32:100 can be written as `32/100` in fractional form.

Now, `32/100` can be written as `32/100` * 100 = 32 percent

So, 32:100 is equal to 32 percent.

Example 3: 25 is 4 percent of what number.

Solution

Let the number be x

So, `25/x` = `4/100`

This equation can be written as, x = `25/4` * 100

x = 25 * 25

x = 625

So, 25 is 4 percent of 625.

Easy Solving Math Problems

Introduction to easy solving math problems:

Mathematics is the vast area, which involves both simple problems and complex problems. Solving simple problems is very easy, because simple problems involves simple basic concepts and formulas. Once we understand the basic concepts and formulas, the simple problems are easy to solve.
In this article of easy solving math problems, some easy math problems are solved and in addition, easy practice problems are given.

Easy solving math problems:


Example 1:

Multiply:  ( x + 1 ) ( x + 4 )

Solution:

( x + 1 ) ( x + 5 )  =  x ( x + 5 ) + 1 ( x + 5 )

=  x2 + 5x + 1x + 5

=  x2 + 6x + 5

Example 2:

Calculate the perimeter of rectangle with the length and width are 12 cm and 14 cm.

Solution:

Perimeter   =  2( l+ w )

=  2( 12 + 14 )

=  2 (26)

=  52 cm

Example 3:

Find the area of a triangle with base of 10 m and a height of 4 m.

Solution:

Area of  triangle  = ½ b h

= ½ (11) (4)

= 22 m2

Example 4:

Find the volume of cube with the side length of 13 m.

Solution:

Volume of cube  =  a3

= 133

= 2197 m3

Example 5:

Simplify: 6x + 2(x – 4)

Solution:

6x + 2(x – 4)  =  6x + 2x - 8

=  8x - 8

=  8( x - 1 )

Example 6:

Find the range for the given set of numbers:

{ 7 , 12 , 14 , 17 , 20 }

Solution:

Range   =  Maximum value - Minimum value

=  27 - 10

=  17

Example 7:

Find the Mean of the given data set: { 4, 10, 12, 18 }

Solution:

Mean   =  ( 4 + 10 + 12 + 18 ) / 4

=  44 / 4

= 11

Example 8:

Find the Circumference of a circle with radius 22 cm. Use 3.14 as pie value.

Solution:

Circumference of circle  = 2`pi`r

= 2 (3.14) 22

= 138.16 cm

Example 9:

Michael bought a doll for `$` 13 and he sold it for `$` 17. What is his gain?

Solution:

Cost price of doll   =  $ 13

Selling price of Clock  =  $ 17

Profit  or  Gain   =  Selling price - Cost Price

=  17 - 13

=  $ 4


Easy solving practice problems:


1) Multiply:  ( x + 2 ) ( x + 3 )

Answer: x2 + 5x + 6

2) Simplify: 2x + 3(x – 2)

Answer: 5x - 6

3) Find the volume of cube with the side length of 5 m.

Answer: 125 m3

4) Find the Mean of the data set: { 1, 3, 6, 10 }

Answer: 5

5) Find the Circumference of a circle with radius 11 cm. Use 3.14 as pie value.

Answer: 69.08 cm

1st Grade Math Fractions

Introduction of 1st grade math fractions:

The 1st grade math fractions are nothing but the fractions that is made to have the meaning of the numerator and the denominator. The numerator and the denominator play the important role in the fractions which gives the reducing of the fractions. Let us see the fraction that involves in the first grade. The 1st grade math fraction gives the simple fraction which gives the basic ideas of the fraction.

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1st grade math fractions:


This gives the each area mentioned in the red color which is the equal fractions or the equal amount. Therefore we have an idea ½ is made equal to the 2/4 and ½ is made equal to 4/8. For our reference when an apple is made to cut exactly down the middle, into two equally pieces, the first piece is identical to the one half of the apple. When the another apple is made into 4 equal pieces, then the two pieces of that apple shows the identical part of the apple that shows ½ , Hence we can say that ½ is made equal to 2/4. The fractions that are made to have the part whole number relationship.

The part whole number of a fraction of the number like 1/5 can be done through the whole number that requires the five equal parts and the one of these parts that is being considered. Let us have the briefing of the quotient or the ratio. The quotient is nothing but the term that has the 2/3 in which the 2 is the numerator and the 3 is the denominator. The method can be mentioned as 2/3. The ratio is made to have the briefing that gives the situation on it. Hence there are two boys or the every three girls in the team hence in this case two-thirds of the teams are boys.

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Examples for 1st grade math fractions:


Example 1: Sum up the two fractions (1/2 + ½)

Solution: ½ + ½ = 2/2.

Answer = 1.

Example 2: Sum up the two fractions (3/2 + ½)

Solution: 3/2 + ½ = 4/2.

Answer = 2.

Help Solve Math Problems

Help in solving math problems:

In this article we help how to solve the math problem in step by step. In math problems solving is the processes of simplification. We can say that, mathematics is mostly involved and experienced the process called solving. The math problem involves some operation known as monomial, polynomial degree equations solve method. Now we help you to solve some monomial, polynomial as a linear equation and the solving them and explained step by step for clear understanding.

Having problem with factoring polynomials help keep reading my upcoming posts, i will try to help you.

Help in solving monomial math problems:


Math help to solve the monomial problems

Solve: 12x + 9

Step 1: First write the equation, in this equation we have to find the variable of x

12x + 9

Step 2: Put 12x + 9 = 0

12x + 9 = 0

Step 3: subtract 9 on both side

12x + 9 – 9 = 0 -9

12x = -9

Step 4: divide 12 on both side

`(12x)/(12)` = `(-9)/(12)`

x = ` -3/4`

Let us we help you in another similar math problems.

Solve: 12x - 9

Step 1: First write the equation, in this equation we have to find the variable of x

12x - 9

Step 2: Put 12x - 9 = 0

12x - 9 = 0

Step 3: Add 9 on both side

12x - 9 + 9 = 0 + 9

12x = 9

Step 4: divide 12 on both side

`(12x)/(12)` = `(9)/(12)`

x = `3/4`

Therefore value of x = `3/4`

Is this topic Help Solving Math Problems hard for you? Watch out for my coming posts.

Help in solving polynomial math problems:


Math help for the polynomial math problem:

Example: solve by factoring x^2 - 20x + 100 = 0

Solution:

Step 1: First write the equation, in this equation we have to find the variable of x

x^2 - 20x + 100 = 0

Step 2: Split the middle value -20 into -10 and -10 when we add both, we get -20, If we multiply -10 with -10 we get 100, therefore the factor is

x^2 -10x -10x + 100 = 0

Step 4: now take the common values out.

x(x- 10) -10(x -10) = 0

(x -10) (x – 10) = 0

Step 5: now separate the values and it is equal to zero

x – 10 = 0 →(1) ;  x -10 = 0 → (2)

Step 6: Add 10 on both side of the equation (1) and (2)

For equation (1)

x -10 + 10 = 0 + 10

x = 10

For equation (2) the same

x = 10

The answer for the math problem is x =10


Practice problems:


1) Solve: 3x + 9 = 0

2) Factor x^2 + 3x + 2 = 0

Answer:

1) x = -3

2) x = -1, -2

Learn Acceleration

Introduction to learn acceleration:-

Learn acceleration is the rate of alteration of velocity with respect to time. In math word acceleration can be define the rate at which a point speeds up or speeds down. The unit of measurement used for acceleration as per SI units is m/s^2. In math, the word acceleration is used to represent an increase in speed and a decrease in speed is known as deceleration. Galileo was the first to discover that objects falling to earth have a constant acceleration which is known as acceleration due to gravity, denoted by ‘g’ and has a value of 9.79 m/s^2.

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Types of learn Acceleration:-


There are two types of learn acceleration:-

• Instantaneous acceleration

• Average acceleration.

The ratio of the change in velocity and time interval is known as Average acceleration and instantaneous acceleration is the change in velocity at one time.

Acceleration can also be determined using the formula F= m A.

Where F = force,

m is the mass of the object.

A is the acceleration.

This is according to the Newton’s second law of motion, which states that a body experiences a force ‘F’ while experiencing an Acceleration ‘A’, given by the equation F= mA where m is the mass of the body.

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Example problems for learn acceleration:-


Learn Acceleration is calculated using the formula A= Change in velocity/ Time

Learn acceleration problem1:-

A car accelerates from 15 m/s to 20 m/s in 5 seconds. What is the car’s acceleration?

Solution:-

Acceleration = Change in velocity/Time

A = 20 – 15/ 5

= 5/5

= 1

And on simplifying, we get

= 1 m / s^2

Learn acceleration problem2:-

A car accelerates from 30 m/s to 50 m/s in 10 seconds. What is the car’s acceleration?

Solution:-

Acceleration = Change in velocity/Time

A = 50 – 30/ 10

= 20/10

= 2

And on simplifying, we get

= 2 m / s^2.

Learn Numbers for Kids

Learn numbers for kids –Introduction:

A number is a mathematical object used in counting and measuring. A notational symbol which represents a number is called a numeral, but in common usage the word number is used for both the abstract object and the symbol, as well as for the word for the number. In addition to their use in counting and measuring, numerals are often used for labels. (Source Wiki)



Learn numbers for kids – Types:


Natural numbers

Integers

Rational numbers

Real numbers

Complex numbers

Computable numbers

Natural Numbers:

The natural Numbers are the numbers. This is the most common numbers are the natural numbers.

That is: one, two, three, four, …..n this is the natural numbers

Integers:

The integers are if less than 0 and same as the positive side.

That is: -n . . . . . . -4, -3, -2, -1, 0, 1, 2, 3, 4 . . . , +n

Rational Numbers:

A rational number is a digit that can be in the form y/z where y and z are digits and z is not equal to zero.

A rational number is:

X/Y

Where y is not zero

Real Numbers:

The real numbers are including all of the calculating numbers. Real numbers are usually write using decimal numbers, in which a decimal point is located to the right of the number with place value one.

That is: 123.5567

Complex numbers

Complex numbers of the form i{y}, where y is a non–zero real number are called imaginary numbers.

a+ bi, bi = imaginary number

Computable numbers

The computable numbers are the set of all real numbers.


Learn numbers for kids – Examples


Learn numbers for kids – Example 1:

Adding the two real number 325.87 + 687.45

325.87
+687.45
1013.32
Answer:

1013.32

Learn numbers for kids – Example 2:

Multiplying natural numbers and the complex numbers 3 (4 + 8i)

Solution:

3 x (4 + 8i)

= 12 + 24i

Answer:

12 + 24i

Learn numbers for kids – Practice Problems:

Multiply the two integer numbers: -8 x 64

Answer:

-512

Integer Length in Math

Introduction:

Definition integer:

Integers are the numbers with either positive or negative sign. Integer includes natural numbers and whole numbers also. Fractions are numbers that does not include in integer. Integer can be represented from negative numbers to positive numbers.

Example   …….-3, -2, -1, 0, 1, 2, 3, ………………

Integer Length in math:

Integer length is the number of digits of an integer. For example consider the number 54876. In this example the length of the given number is 5 since it consists of 5 digits. Integer length is nothing but the number of bits. Integer length is the number of place digits in a given number.

I like to share this Integer Division with you all through my article.

Example of Integer Length


Find the integer length of the number given below.

542316

Solution

Here the number of bits of numbers given is 6. So the integer length of the number 542316 is 6.

There is also the possibility of increase in integer length in math.

In the case of performing arithmetic operations like addition and multiplication the integer length will increase.

Example 1. Find the sum of 548 + 736 and find the integer length of the answer obtained.

Solution

5 4 8
7 3 6 +
----------
1 2 8 4
-------------

Here the length of the integer is 4.

Example 2. Find the product of 578 * 21

Solution:

5 7 8

2 1 *
------------------
5 7 8
1 1 5 6
--------------------
1 2 1 3 8
--------------------

Here the integer length is 5

Example 3. Find the difference 9875 – 2456

Solution:

9 8 7 5

2 4 5 6 -

----------------
7 4 1 9
------------------

Here the integer length is 4


PROBLEMS ON INTEGER LENGTH


Find the integer length of the following problem after performing the given arithmetic operation.

Add a) 5894 + 654
b) 124356+54120

2.  Subtract a)6548-3546

b)7812 - 120

3.  Multiply a) 456*23

b) 1596*324

4.  Divide   a) 625/5

b)1100/11

Answers:

1. a) 4     b)6

2. a) 4     b)4

3. a)5      b)6

4. a)2      b)3

Grade 9 Math

Introduction of grade nine fraction math:

The grade nine fraction maths is nothing but the fraction that involves the division, multiplication and subtraction. The grade nine fraction maths gives the most basic ideas of the fractions. Thus the fraction of the math gives the various topics and understandings of the math. The grade nine fraction math gives you various formation of the reduction of the fraction.

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Grade nine fraction math:


Grade nine fraction math are the one which makes the form that gives the changes that shows the form in the fraction method. When the method gives the comparing of the two terms. The comparisons of the terms are the one which gives the changes that makes the form of the terms in the comparison of the two fractions. Hence we can find the largest and the smallest terms when compared between the two terms.


Examples for grade nine fraction math:


Example 1: Compare the two fractions 3/4 and 1/6?

Solution: When comparing the two fractions we found that 3/4 is greater than the 1/6.

Example 2: Compare the two fractions 4/5 and 2/2?

Solution: When comparing the two fractions we found that 4/5 is lesser than the 2/2.

Example 3: Compute the fractions with regrouping in the fractions like “1/80 + 4/80 + 9/80 + 6/80” is made to regrouping as “1/80+9/80 +6/80+4/80” the second terms are considered as simpler one these terms leads to the sum of ten [(1+9)/80, (6+4)/80] this made easier to keep the track off.

Example4: Compute the fractions with regrouping like “3x/80 + 2y/80 + 4x/80 + y/80 = 7” hence the regrouping of the x and y terms which can be made simpler.

This can be regrouped as 7x/80 + 3y/80 = 7.

Example 5: The 2 39 reduced to a fraction is obtained as the terms, 2/39 this terms can be prepared through the terms 2/39 = 2/39.

1) The given fraction 2/39 is not made to reduce to the lowest terms. Hence the fractions cannot reduce to the lowest terms.

2) The above terms cannot be reduced with the common terms.

Example 6: The 45 reduced to a fraction is obtained as the terms, 45/100 this terms can be prepared through the terms 45/100 = 9/20.

1) The given fraction 45/100 is not made to reduce to the lowest terms. Hence the fractions are reduced to the lowest terms.

2) The above terms can be reduced with the common terms like 5.

3) The value 5 is the greatest common divisor or the greatest common factor of the both numbers 45 and 100.

4) Hence the 45/100 can be reduced through the value five which results to the 9/20.

What Math Property Is

Introduction to properties in math:

Here we are going to see about the properties in math. There are many number of properties present in math. Among the many properties we often use associative, commutative and distributive property.  We are going to see about all the properties. The properties are,

Property 1: Reflexive property.
Property 2: Symmetric property.
Property 3: Transitive property.
Property 4: Commutative property.
Property 5: Associative property.
Property 6: Distributive property.
Property 7: Addition property.
Property 8: Multiplication property.
Property 9: Additive identity.
Property 10: Multiplicative identity.

Math properties


Properties in math:

Here we are going to see about the properties in detail.

Associative Property:

In this property, when we change the order for calculation it does not changes the answer.
The grouping of the numbers is not taken into consideration.
This property can be applied for the addtion operation and the multiplication operation.
The associative property is given as,
(x + y) + z = x + (y + z)

(x * y) * z = x * (y * z)

Commutative Property:

In this property the order for the addition or multiplication is reversed or interchanged it does not affect the resultant answer.
This property can be applied in addition operation and multiplication operation.
The commutative property is given as,
x + y = y + x

x * y = y * x

Distributive Property:

In this property it combines both the addition and multiplication.
The parenthesis is used to separate the operations of the addition and multiplication.
The distributive property is given as,
x (y +z) = x y + x z

Addition Property:

In this property we add the same non zero numbers to both sides.
Let us consider a as the non zero number then the addition property is given as,
x +a = y +a.

Multiplication Property:

This property is also same as the addition property instead to add the numbers here we are going to multiply the numbers.
The multiplication property is given as,
x * a = y *a.

Reflexive Property:

When the number is equal on both sides then it is referred as reflexive property.
The reflexive property in math is given as,
x =x.

Transitive Property:

When the two number are equal to the number in the third position then it is referred as transitive property in math.

Additive Identity:

When a zero is added to the number then we get the same number as the answer thus it is referred as the additive identity.
The additive identity property in math is given as,
x +0 = x.

Multiplication Identity:

When one is multiplied with a number we get the same number as the answer the multiplicative identity in math is given as,
x (1) = x.


Example Problems


Example problem for property in math:

Problem 1: Simplify: 4 (x+ 5) – 3x

Solution:

4(x + 5) – 3x = 4x +4*5 -3x --------- (according to the property of distributivity)

= 4x + 20 – 3x -------- (by simplifying 4*5 = 20)

= 4x – 3x + 20 -------- (according to the property of commutativity)

= x (4- 3) + 20 ------- (according to the property of distributivity)

= x (1) + 20 -------- (by simplifying 4-3 = 1)

= x +20.

Hence the answer is x +20.

Problem 2: Simplify: 2p + 5q – 6p

Solution:

2p + 5q – 6p  = 2p – 6p + 5q -------- (according to the property of commutativity)

= (2p – 6p) + 5q ------ (according to the property of associativity)

= p (2 – 6) + 5q -------- (according to the property of distributivity)

= p (-4) + 5q ---------- (simplifying 2 -6 = -4)

= -4p + 5q.

Hence the answer is -4p +5q.

Third Grade Learn

Introduction on learning third grade Math:

Improving their calculating knowledge and improving their math’s sense using the third grade math. Third grade learn topics given below,

Place value
Comparing and ordering numbers
Counting by numbers other than one
Addition
Relationship between additon and Subtraction
Subtraction
Rounding and Estimation
Multiplication Facts
Multipication
Relation ship of Mulitplication and Division
Division Facts
Division
Fractions
Decimals
Consumer Math
Mesurement - Time
Measurement - Metric
Geometry
Roman Numerals
Sequences and Patterns

I like to share this Rounding Decimals with you all through my article.

Calculate the dollars and cents.


1. Add the total, 5dollars and 5dollars

Sum of the dollars = 5 + 5 = 10dollars.

2. Add the total, 6dollars, 2dollars and 4dollars.

Sum of the dollars = 6 + 2 + 4 = 12dollars.

3. Add the total, 8dollars and 9dollars

Sum of the dollars = 8 + 9 = 17dollars.

4. Add the total, 2cents and 5cents

Sum of the cents = 2 + 5 = 7 cents.

5. Add the total, 3 cents and 2 cents

Sum of the cents = 3 + 2 = 5 cents.


Learning Addition problems for third grade math


1. Add the two numbers,              1 + 1 = 2

2. Add the two numbers,              4 + 4 = 8

3. Add the two numbers,              6 + 8 = 16

4. Add the two numbers,              5 + 5 = 10

5. Add the two numbers,              7 + 8 = 15

6. Add the two numbers,              50 + 50 = 100

7. Add the two numbers,              100 + 100 = 200

8. Add the two numbers,              200 + 100 = 300

9. Add the two numbers,              300 + 100 = 400

10. Add the two numbers,              200 + 300 = 500

Learning Subtraction problems for third grade math:

1. Subtract the two numbers,        4 – 2 = 2

2. Subtract the two numbers,        6 – 5 = 1

3. Subtract the two numbers,          8 – 5 = 3

4. Subtract the two numbers,          9 – 3 = 5

5. Subtract the two numbers,          8 – 2 = 6

6. Subtract the two numbers,          15 – 5 = 10

7. Subtract the two numbers,          75 – 25 = 50

8. Subtract the two numbers,          100 – 25 = 75

9. Subtract the two numbers,          150 – 25 = 125

10. Subtract the two numbers,          200 – 25 = 175

Learning Multiplication problems for third grade  math.

1. Multiply the two numbers,         2 * 2 = 4

2. Multiply the two numbers,         5 * 5 = 25

3. Multiply the two numbers,         3 * 4 = 12

4. Multiply the two numbers,         6 * 2 = 12

5. Multiply the two numbers,         4 * 3 = 12

6. Multiply the two numbers,         12 * 12 = 144

7. Multiply the two numbers,         10 * 10 = 100

8. Multiply the two numbers,         20 * 10 = 200

9. Multiply the two numbers,         30 * 10 = 300

10. Multiply the two numbers,         40 * 10 = 400

Understanding What is a Acute Angle is always challenging for me but thanks to all math help websites to help me out.

Practice problems on learning third grade math


1. Problem for addition.

Solve the two numbers with addition,

A) 8 + 9

B) 5+6

2. Problem for subtraction.

Solve the two numbers with subtraction,

A) 6 – 2

B) 8 – 4.

3. Problem for multiplication.

Solve the two numbers with multiplication,

A) 3 * 2

B) 2 * 4.

Doing Math Online

Introduction to doing math online:

Mathematics is the study of quantity, structure, space, and change. Mathematicians seek out patterns, formulate new conjectures, and establish truth by rigorous deduction from appropriately chosen axioms and definitions. Mathematics is used throughout the world as an essential tool in many fields, including natural science, engineering, medicine, and the social sciences. (Source: From Wikipedia).Now, we are going to see some of the math problems online.

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Online math problems:


Example problem 1:

Simplify the expression: 20x + 6y - 14x - 10y - 2x - 2z

Solution:

20x + 6y - 14x - 10y - 2x - 2z

Add the like terms in the given expression

(20 - 14 - 2) x + (6 + (-10)) y - 2z

Simplify the expression

4 x - 4y - 2z

So, the answer is 4 x - 4y - 2z.

Example problem 2:

Evaluate the expression (-8 + y) × 5 + 40 ÷ 8 - y when y = 8.

Solution: Here, we substitute the value of 8 in the place of y,

(-8 + y) × 5 + 40 ÷ 8 - y becomes

(-8 + 8) × 5 + 40 ÷ 8 - 8 = 0 × 5 + 40 ÷ 8 – 8

= 0 + 5 - 8

= -3.

So, the answer is -3.


Few more online math problems:


Example problem 3:

Simplify the equation: 6x + 8 = 9x – 31

Solution:

6x + 8 = 9x – 31

Subtract 8 on both sides of the equation

6x + 8 - 8 = 9x – 31 – 8

6x = 9x – 39

Subtract 9x on both sides of the equation

6x – 9x = 9x – 39 – 9x

-3x = -39

Divide by -3 on both sides of the equation

`(-3x) / -3 = -39 / -3`

x = 13

So, the simplified answer is 13.

Example problem 4:

Find the distance between the two points (5, 10), (15, 17)?

Solution: Let d be the distance between A and B.

Then d (A, B) = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

= `sqrt((15 - 5)^2 + (17 - 10)^2)`

= `sqrt(10^2 + 7^2)`

= `sqrt(100+49)`

= `sqrt(149)`

= 12.2 units.

So, the distance between the given points is 12.2 units.

I have recently faced lot of problem while learning algebraic expression word problems, But thank to online resources of math which helped me to learn myself easily on net.

Online practice math problems:


1)      Simplify the expression:  1x - 11y + 9x - 6y. (Answer: 10 x - 17y)

2)      Solve for the variable x:  2(x + 1) = 32 + x (Answer: x = 30).

3)      Solve for the variable x:  2x + 12 = 12 (Answer: x = 0).

Learn Online Speed

Introduction :-

Distance word problems, repeatedly entitled as uniform rate problems, engage something travelling at set and stable speed or as well moving at an average speed.

The formulas that used in speed and distance problems are two basic formulas that relate distance, speed and time.

`Speed = (Distance) / ( Time)`

`Distance = Speed * Time.`

Now lets see the some online solved and practice problems that would be helpful in learning speed problems.

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Online learning Solved Problem:-

Problem 1:- If Rocky walks at 14 kilometer per hour instead of 10 kilometer per hour, he would have walked 40 kilometer extra. The actual distance traveled by him is:

Solution:-

Given

Rocky travels 40 kilometer extra if he walks 14 kilometer per hour instead of 10 kilometer per hour.

Let the actual distance travelled by rocky be X.

We know the formula that relates distance speed and time is

`Time = (distance) / (speed.)`

Rocky travels x kilometer at the rate of 10 kilometer per hour.

Rocky travels 40 more kilometers if he travels at the rate of 14 kilometer per hour.

`Time1 = X/ 10`

`Time 2 = X + 40 / 14.`

Time in both cases are same so time1 = time2.

So

`X/ 10 = X+ 40 / 14.`

By solving this we can find the actual distance travelled by Rocky.

`X/ 10 = X + 40 / 14.`

Multiply by 14 on both sides of the above equation.

`14 * X/ 10 = (X + 40) / 14 * 14`

Crossing out 14 and 14 on one side we get the simplified as

`14 * X/ 10 = (X + 40)`

Now multiply by 10 on both side of the above equation.

`10 * (14 * X) / 10 = 10 * (X + 40).`

Crossing out 10 and 10 on one side we get the simplified as

`14 *X = 10 * (X + 40).`

`14X = 10X + 10*40.`

`14 X = 10X + 400.`

Now subtract 10X on both sides of the above equation.

`14X - 10X = 10X + 400 -10X.`

`4X = 400`

Now divide it by 4 on both sides.

`(4X)/ 4 = 400/ 4.`

By simplifying it we get the answer as

`X = 100`

The actual distance walked by him was 100 miles.


Online learning practice Problem:-

Problem:- 1

Alex travels first 160 meters at 64 meters per hour and the next 160 meters at 80 meters per hour. Find the average speed for first 320 meters.

Answer:- 7.11 meters per hour.

Problem:- 2

Rosy travels first 100 meters at 25 meters per hour and the next 100 meters at 50 meters per hour. Find the average speed for first 100 meters.

Answer:- 6 meters per hour.

Math Problem Help

Introduction to math problem help:

Mathematics is the study of quantity, structure, space, and change. Mathematicians seek out patterns, formulate new conjectures, and establish truth by rigorous deduction from appropriately chosen axioms and definitions. Mathematics is used throughout the world as an essential tool in many fields, including natural science, engineering, medicine, and the social sciences. (Source: From Wikipedia). Now, we are going to see some of the problems help in math. From these problems, we can get clear idea about how to solve the math problems easily.

Having problem with How to Find the Equation of a Parabola keep reading my upcoming posts, i will try to help you.

Some of the math problems help:


Example 1: Simplify the equation: 4x + 2 = 5(x – 1)

Solution:

4x + 2 = 5(x – 1)

4x + 2 = 5x – 5

Subtract 2 on both sides of the equation

4x + 2 - 2 = 5x – 5 – 2

4x = 5x – 7

Subtract 5x on both sides of the equation

4x – 5x = 5x – 7 – 5x

-1x = -7

Divide by -1 on both sides of the equation

`(-1x) / -1 = (-7) / -1`

x = 7

So, the simplified answer is 7.

Example 2: Evaluate the expression (7 + p) × 3 + 18 ÷ 3 – 2p when p = 1.

Solution: Here, we substitute the value of 1 in the place of p,

(7 + p) × 3 + 18 ÷ 3 – 2p becomes

(7 + 1) × 3 + 18 ÷ 3 – 2(1) = 8 × 3 + 18 ÷ 3 – 2

= 24 + 6 - 2

= 28.

So, the answer is 28.


Few more math problems help:


Example 3: Solve the inequality: 10x – 7 > 23

Solution:

10x – 7 > 23

Add 7 on both side of the inequality

10x – 7 + 7 > 23 + 7

10x > 30

Divide by 10 on both side of the inequality

`(10x) / 10 > 30 / 10`

x > 3

So, the solution is (3, infinity).

Example 4:

Find the distance between the two points (1, 6), (3, 1)?

Solution: Let d be the distance between A and B.

Then d (A, B) = `sqrt [(x2 - x1)^2 + (y2 - y1)^2]`

= `sqrt [(3 -1)^2 + (1 - 6)^2]`

=` sqrt [ (2)^2 + (-5)^2]`

=` sqrt [4 + 25]`

= `sqrt [29]`

= 5.38

So, the distance between the given points is 5.4 units.


Practice math problems help:


1)      Simplify the expression: 11 x - 2y + 15x - 5y. (Answer: 26 x - 7y)

2)      Solve for the variable p:  4(p + 1) = 10 + p (Answer: p = 2).

3)      Solve for the variable x:  7x + 21 = 14 (Answer: x = -1).

Definition Mental Math

Definition of Mental Math
Definition: Mental math practice comprises arithmetical calculations using only the human brain, with no help from calculators, computers, pen and paper. People use mental math when those tools are not available, when it is faster than other means of calculation or in a competition context. Mental math calculation often involves the use of specific techniques devised for specific types of problems.  In this lesson we will discuss about definition of mental math with example problems.

Source wikipedia.


Definition of Mental Math – Example Problems


These example problems will shows you to how to do the mental math.

Example 1: Sofia has $94 and Jack has $49. How much dollar did they has altogether?

Solution:

We have to add 94 and 48 to know the sum.

Step 1: Add the two ones' place digits of 94 and 49: 4 + 9 = 13.

Step 2: Add the two tens' place digits of 94 and 49: 9 + 4 = 13.

Step 3: Sum of the ones' place digits is two-digit number, so decrease the ones' place sum by 10: 13 - 10 = 3 and increase the tens' place sum by 1: 13 + 1 = 14.

Step 4: Combine the tens' and ones' place sums.

Answer is $143.

Example 2: A cricket ball cost is $18. A bat cost is $65. Find the difference between the cost of ball and bat.

Solution:

We have to subtract 18 from 65 to know the difference.

Step 1: Subtract the two tens' place digits of 18 and 65: 6 - 1 = 5

Step 2: Bottom ones' digit is larger than the top ones' digit.

So, decrease the answer of the tens' place by 1: 5 - 1 = 4, and increase the top ones' place value by 10: 5 + 10 = 15.

Step 3: Subtract the two ones' place values: 15 - 8 = 7

Step 4: Combine the tens' and ones' place value.

Answer is $47.

Example 3: A ball cost is $8, what is the cost of 45 balls?

Solution:

We have to multiply 45 by 8 to know the answer.

Step 1: Multiply the ones' digit of 45 by 8: 45 x 8 = 40

Step 2: Multiply the tens' digit of 45 by 8: 45 x 8 = 32

Step 3: Product of ones' place is two digits (40), so increase the product of tens' digit by 4: 32 + 4 = 36

Step 4: Combine the ones' and twos' place value

Answer is $360.

Example 4: Jessica had 72 chocolates. She gave 4 chocolates to each her friend. How many friends she had?

Solution:

We have to divide 72 by 4 to know the answer.

Step 1: Divide the tens' digit of 72 by 4: 72 / 4 = 1 (Remainder 3)

Step 2: Multiply the remainder 3 by 10, 3 x 10 = 30, and add it to ones' place of 72, 30 + 2 = 32

Step 3: Divide 32 by 4, 32 / 4 = 8

Step 4: Combine the ones' and twos' place values

Answer is 18.

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Definition of Mental Math – Practice Problems


Problem 1: Sofia has $85 and Jack has $35. How much dollar did they has altogether?

Problem 2: A cricket ball cost is $15. A bat cost is $70. Find the difference between the cost of ball and bat.

Problem 3: A ball cost is $7, what is the cost of 53 balls?

Problem 4: Jessica had 51 chocolates. She gave 3 chocolates to each her friend. How many friends she had?

Answer: 1) $120 2) $55 3) $371 4) 17