Student Directed Learning

Introduction for student directed learning:

In this article, we will discuss about the student directed learning. In online, we will get direct help with clear explanation for mathematics subjects. The students who have doubts in the topics, so they access the online for help in math. Its also useful to workout the homework problems and doubt clarification. The online topic is sequences. It has two types

1. Arithmetic sequence

2. Geometric sequence.

The student directed learning formulas and example problems are given below.

Student Directed Learning Sequences for Formulas:

Definitions:

Arithmetic sequence means that, the sequence of a numbers such that the difference between two consecutive members of the sequence is a constant. Geometric sequence means that, the sequence of a numbers such that the ratio between two consecutive members of the sequence is a constant.

Formula for arithmetic sequence:

nth term of the sequence : `a_n` = `a_1` + (n - 1)d

Series of the sequence: `S_n` = `(n(a_1 + a_n))/2 `

Formula for geometric sequence:

nth term of the sequence:` a_n` = a1 * rn-1

Series of the sequence: `S_n` = `(a_1(1-r^n))/(1 - r)`

Student Directed Learning Sequences for Example Problems:

Example problem 1:

Find the 13th term of the given series 24, 27, 30, 33, 36,...

Solution:

First term of the series, a1 = 24

Difference of two consecutive terms, d = 27 - 24

n = 13

The formula to find the nth term of an arithmetic series, `a_n = a_1 + (n-1)d`

So, the 13th term of the series 24, 27, 30, 33, 36,... = 24 + (13 - 1) 2

= 24 + 12 * 2

= 24 + 24

After simplify this, we get

= 48

So, the 13th term of the sequence 24, 27, 30, 33, 36,... is 48.

Example problem 2:

Find out the 5th term of a geometric sequence if a1 = 90 and the common ratio (C.R) r = 2

Solution:

Use the formula `a_n = a_1 * r^(n-1)` that gives the nth term to find `a_5` as follows

`a_5 = a_1 * r^(5-1)`

= 90 * (2)4

= 90 * 16

After simplify this, we get

= 1440

The 5th term of a geometric sequence is 1440

The above examples are helpful to study of Student directed learning sequences.

Simple Solutions Math Answers

Introduction for math:

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. There is debate over whether mathematical objects such as numbers and points exist naturally or are human creations. Albert Einstein, on the other hand, stated that "as far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality.                                                                                                                                               source Wikipedia

Problems for Simple Solutions Math Answers

Problem 1:

Find the simple solutions.

3x-3=3

Solution:

the given function is 3x-3=3

add 3 on both sides of this equation.

3x+3-3=3+3

We get

3x=6

Divide by 3 on both sides of this equation

`(3/3)x` =`6/3`

We get

x=2

answers of this simple solution is x=2

Problem 2:

Find the simple solutions

5x-3=7

Solution:

the given function is 5x-3=7

add 3 on both sides of this equation.

5x+3-3=7+3

We get

5x=10

Divide by 5 on both sides of this equation

`(5/5)x` =`10/5`

We get

x=`10/5`

answers of this simple solution is x=2

Problem 3:

Find the simple solutions

2x-3=3

Solution:

the given function is 2x-3=3

add 4 on both sides of this equation.

2x+3-3=3+3

We get

2x=6

Divide by 2 on both sides of this equation

`(2/2)x` =`6/2`

We get

x=3

answers of this simple solution is x=3

Problem 4:

Find the simple solutions

2x+7=-21

Solution:

the given function is 2x+7=-21

subtract 7 on both sides of this equation.

2x+7-7=-21-7

We get

2x=28

Divide by 2 on both sides of this equation

`(2/2) x` =`28/2`

We get

x=14

answers of this simple solution is x=14

Some more Problems are Simple Solutions Math Answers:

Problem 5:

Find the area and perimeter of the rectangle,

Length of the rectangle=10 mm

Height of the rectangle = 12 mm

Solution:

Formula for area of the rectangle = length*height

Area of the rectangle = 12*10

=120 mm^2

Formula for perimeter of the rectangle =2*length + 2* height

=2*12+2*10

=24+20

=42 mm

answers

Area of the rectangle = 120mm^2

Perimeter of the rectangle = 42 mm

Problem 6:

Find the area and circumference of the triangle,

Radius of the circle =6 mm and `pi` = 3.14

Solution:

Formula for area of the circle = `pi` r2

Area of the circle = 3.14*62

=3.14*36

=113.04 mm^2

Formula for circumference of the circle =2 `pi` r

=2*3.14*6

=37.68 mm

answers

Area of the circle = 113.04 mm^2

Circumference of the circle = 37.68 mm

Very Small Numbers Learning

Introduction to very small numbers learning:

Let us study about very small numbers learning. The mathematics world is seemed to be fully composed of various kinds of numbers.Understanding List of Prime Numbers to 100 is always challenging for me but thanks to all math help websites to help me out.
The two basic categories that numbers will get split into are named as small numbers and small numbers.
The small numbers are the numbers that are seemed to be found to out by comparing each number with others. Some of the examples for very small numbers learning are given.
Very Small Numbers Learning:

Very small numbers learning – example 1:

Do solving the given number series (5 * 4), (4 - 2), (3 + 2.5), (2 * 0.2), (0.4 * 2.5), (46 - 10), `(0.8/0.2)` and find out the small number that are calculated in the series?

Solution:

The number series provided with the various operations for calculations is as follows: (5 * 4), (4 - 2), (3 + 2.5), (2 * 0.2), (0.4 * 2.5), (46 - 10), `(0.8/0.2)` .
The operations in the given number series are calculated as below:
(5 * 4), (4 - 2), (3 + 2.5), (2 * 0.2), (0.4 * 2.5), (46 - 10), `(0.8/0.2)`
(20), (2), (5.5), (0.4), (1), (36), (4)
20, 2, 5.5, 0.4, 1, 36, 4
To find the very small value in the given number series arrange the calculated values of the number series in the order smallest to the greatest values as follows:
0.4 < 1 < 2 < 4 < 5.5 < 20 < 36
Therefore in the given number series after calculation the very small number is found be as ‘0.4’.

Please express your views of this topic Number Place Value by commenting on blog.

Prepare for small numbers – example 2:

Do solving the given number series `(10/4)` , `(4/4)` , (3 * 6), (2 * 0.16), (-0.4 - 2.5), (5 + 10), (0.8 + (-0.8)) and find out the small number that are calculated in the series?

Solution:

The number series provided with the various operations for calculations is as follows: `(10/4)` , `(4/4)` , (3 * 6), (2 * 0.16), (-0.4 - 2.5), (5 + 10), (0.8 + (-0.8)).
The operations in the given number series are calculated as below:
`(10/4)` , `(4/4)` , (3 * 6), (2 * 0.16), (-0.4 - 2.5), (5 + 10), (0.8 + (-0.8))
(2.5), (1), (18), (0.32), (-2.9), (15), (0)
2.5, 1, 18, 0.32, -2.9, 15, 0
To find the very small value in the given number series arrange the calculated values of the number series in the order smallest to the greatest values as follows:
-2.9 < 0 < 0.32 < 1 < 2.5 < 18 < 15
Therefore in the given number series after calculation the very small number is found be as ‘-2.9’.


Prepare for small numbers – exercises:

Which one is very small even square number between 400 and 600? (Answer: 441)
Which is the very small cube number that lies between 1 and 600? (Answer: 8)
Mention the very small odd prime number from 600 and 800. (Answer: 601)

Solving Mental Math Problems

Solving Mental Math Problems

Mental math practice comprises arithmetical calculations using only the human brain, with no help from calculators, computers, or pen and paper. People use mental math when computing tools are not available, when it is faster than other means of calculation or in a competition context. Mental math often involves the use of specific techniques devised for specific types of problems. Source: Wikipedia.

Solving Mental Math Problems - Example Problems

Solving example problems

Example 1: Linda has $74 and Lisa has $99. How much did they has altogether?

Solution:

We have to add 74 and 99 to know the sum.

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

Step 2: Add the two tens' place digits: 9 + 7 = 16.

Step 3: The sum of the ones' place digits is a two-digit number so decrease the ones' place sum by 10: 13 - 10 = 3 and increase the tens' place sum by 1: 16 + 1 = 17.

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

Answer is $173.

Example 2: A pair of shoes cost is $57. A pant cost is $81. Find the difference between the pair of shoes and pant.

Solution:

We have to subtract 57 from 81 to know the difference.

Step 1: Subtract the two tens' place digits: 8 - 5 = 3

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

Decrease the answer for the tens' place by 1: 3 - 1 = 2 and increase the top ones' place value by 10: 1 + 10 = 11.

Step 3: Subtract the two ones' place values: 11 - 7 = 4

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

Answer is $24.

Example 3: A bucket cost is $73, oven cost is 5 times more than bucket cost. What is the oven cost?

Solution:

We have to multiply 73 by 5 to know the computer cost.

Step 1: Multiply the ones' digit by 5, 5 x 3 = 15

Step 2: Multiply the tens' digit by 5, 5 x 7 = 35

Step 3: Product of ones' place is two digits (15), so increase the product of tens' digit by 1 (15), 35 + 1 = 36

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

Answer is $365.

Example 4: Lisa had 72 candies. She gave 6 candies to each her friend. How many friends she had?

Solution:

We have to divide 72 by 6 to know the answer.

Step 1: Divide the tens' digit by 6, 7/6 = 1 with remainder 1

Step 2: Multiply remainder 1 by 10, 1 x 10 = 10 and add it to ones' place, 10 + 2 = 12

Step 3: Divide 12 by 6, 12/6 = 2

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

Answer is 12.

Solving Mental Math Problems - Practice Problems

Solving practice problems

Problem 1: Shan saved $73 and Sham saved $78. How much did they save altogether?

Problem 2: A shirt cost is $73, Pant cost is $99. Find the cost of difference of two items.

Problem 3: A wall clock costs 3 times as much as a watch. If the watch costs $51, what is the cost of wall clock?

Problem 4: Pond had 75 candies. He put the candies equally into 5 bags. How many candies were there in each bag?

Answer: 1) $151 2) $26 3) $153 4) 15

Trigonometric Identities Sum Help

Introduction to trigonometric identities sum help:

Trigonometry is arrived from the Greek word, trigonon = triangle and metron = measure. The father of trigonometry is Hipparchus. He designed the first trigonometric table. Trigonometry has wide range of applications in many fields like science, technology, astronomy etc. Identity is defined as an equation that is true for all probable values of its variables. Online help is one of the comfortable method of getting help from anywhere around the globe. Through online study, students can get about trigonometric identities sum. In this topic, we are going to see about, trigonometric identities sum help. I like to share this Pythagorean Trigonometric Identities with you all through my article.

Trigonometric Identities Sum Help - Trigonometric Identities:

The list of trigonometric identities sum are shown below,

Sum or difference of two angles:

sin (a ± b ) = sin a cos b ± cos a sin b

cos(a ± b) = cos a cos b ± sin a sin b

tan(a ± b) = `(tan a +- tan b)/ (1 +- tan a tan b)`

Sum and product formulas:

sin a + sin b = `2sin((a+b)/2)cos((a-b)/2)`

sin a - sin b = `2cos((a+b)/2) sin((a-b)/2)`

cos a + cos b = `2cos((a+b)/2) cos((a-b)/2)`

cos a – cos b = `-2sin((a+b)/2) sin((a-b)/2)`

Is this topic the Pythagorean theorem hard for you? Watch out for my coming posts.

Trigonometric Identities Sum Help: - Examples

Example 1:

Evaluate Sin 153

Solution:

Sin 153 = Sin (90+63)

= sin 90 cos 63 + cos 90 sin 63

= 1(0.454) + 0(0.891)

= 0.454 + 0

= 0.454

The answer is 0.454

Example 2:

Evaluate Cos 128

Solution:

Cos 128 = Cos (90 + 38)

= cos 90 cos 38 – sin 90 sin 38

= 0(0.788) – 1(0.616)

= 0 – 0.616

= -0.616

The answer is -0.616

Example 3:

Evaluate tan 38

Solution:

Tan 38 = Tan (45 - 7)

= `(tan 45 - tan 7)/(1+tan 45*tan 7)`

= `(1-0.123)/(1+(1*0.123))`

= `0.877/(1+0.123)`

=` 0.877/ 1.123`

= 0.781

The answer is 0.781

Example 4

Evaluate Cos 124

Solution:

Cos 124 = cos (90 + 34)

= cos 90 cos 34 - sin 90 sin 34

= 0(0.829) - 1(0.559)

= 0 – 0-0.559

= -0.559

The answer is -0.559

Example 5:

Evaluate, sin 50 - sin 40

Solution:

sin a - sin b = `2cos((a+b)/2)sin((a-b)/2)`

sin 50 - sin 40 = `2 cos((50+40)/2)sin((50-40)/2)`

= `2 cos (90/2) sin(10)/2`

= 2 cos 45 sin 5

= 2 (0.707)(0.087)

= 2 * 0.062

= 0.124

The answer is 0.124

Graphs and Histograms

Graphs are pictorial representation of data.Histogram is graphical representation of data in statistics.

Introduction to graphs and histograms:

A histogram graph is representation of a frequency distribution as a graph . The graph  consists of rectangles constructed with class intervals as bases and heights proportional to corresponding frequencies such that there is no gap between any two successive rectangles.

A histogram deals with continuous type of data.

Different types of histograms are:

Histogram of continuous grouped frequency distribution with equal class intervals.

Histogram of a continuous grouped frequency distribution with unequal class intervals.

Histogram when mid-points are given

Histogram for grouped frequency with inclusive classes(discontinuous class-intervals)

Steps for Construction of Histogram:

Choose a suitable scale on the x-axis and represent the class-limits on it

Choose a suitable scale on the y-axis and represent the corresponding frequencies on it

Draw rectangles with class intervals as bases and the respective frequencies as heights

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Example Histogram with Equal Class Interval:

Ex : 1 Constuct a histogram to represent the following data:

Marks
Number of

students

0-10    4
10-20    7
20-30    12
30-40    20
40-50    9
50-60    2
Sol:

Choose a scale

Step 1:Along x-axis 1 cm = 10 marks

Step 2:Along y-axis 1 cm = 5 students

Step3:Starting from 0, mark 10,20,30,40,50,60,70 on x-axis and 4,7,12,20,9,2 on the y-axis

Step4:Then we draw the rectangles with class intervals as bases and corresponding frequencies as heights.

Steps to Draw Histogram of Unequal Class Interval:
Choose a suitable scale on the x-axis and represent the class-limits on it
Determine a class interval which has the minimum class size. Let the minimum class size be h
Find the adjusted frequency of each class by using the formula :
Adjusted frequency of the class = `h/("Class - size of the class)` x frequency of the class

Choose suitable scale on the y-axis and represent the corresponding adjusted frequencies on it
Draw the rectangles , the width of rectangles will be according to class limit

Direction Vector of a Line

Introduction to direction vector of a line:

The direction vector is a vector point of direction and it indicates the direction of line. The direction vector of a line is based on real values of line segment equation. In math, the vector points are used in Euclidean space. Now we are going to see about direction vector of a line.

Explanation for Direction Vector of a Line
Direction vector:

The line is represented as `vecAB` and the arrow mark symbol is indicating the direction of line. The direction vectors are determined from line equation.

In Euclidean space, the direction vector of line D is determined by using line equation form real numbers that is the line equation form is ax + by + c = 0. Here a, b, and c are real numbers and the direction vector of line D is (-b, a). We can also consider the multiples of (-b, a) as direction vectors. Understanding empirical probability is always challenging for me but thanks to all math help websites to help me out.

More about Direction of Vector of Line

Example problems for direction vector of a line:

Problem 1: Find out the direction vector of a line segment D from line form.

4x + 3y + 1 = 0.

Solution:

The given line equation form is 4x + 3y + 1 = 0.

The line segment D has two end points AB.

The direction vector `vecAB` is determined from line form.

`vecAB` = (-3, 4).

Therefore, the direction vector of line is (-3, 4), (9, 16)…

Problem 2: Find out the direction vector of a line segment D from line form.

x - 2y + 4 = 0.

Solution:

The given line equation form is x - 2y + 4 = 0.

The line segment D has two end points AB.

The direction vector `vecAB` is determined from line form.

`vecAB` = (2, 1).

Therefore, the direction vector of line is (2, 1), (4,1)…

Exercise problems for direction vector of line:

1. Find out the direction vector of line D from 3x + 4y + 2 = 0.

Solution: The direction vector `vecAB` is (-4, 3).

2. Find out the direction vector of line D from 6x - 3y - 5 = 0.

Solution: The direction vector `vecAB` is (3, 6).