Lines and Angles Math

Introduction to lines and angles math:

In math geometry the lines and angles are important tools. If any object in ideal, that is called as line and it is represented as straight curve. The angle is related with line that is the cross-section of two-line is create the angle and that intersection point is called as vertex. Here we see about types of line and angle in math.


Explanation for lines in math


Lines:

If we want to project the real world model in geometry means we can use the lines for straight object representation. This straight object is represented with small height and width and it is infinitely long. The Euclidean math geometry use the line commonly.

Types of lines:

Parallel lines:

The separate two coplanar lines are called as parallel lines that are this line does not intersect with each other. The distance between the two lines are equal and it is represented as | |. The parallel lines are illustrating the two postulates.

Corresponding angles postulate: The angles are same when the transversal is cut the two coplanar lines. So we can say that lines are parallel.

Parallel lines postulate: If the transversal cut the two parallel lines, the same measure angles are present.

Perpendicular lines:

When the angle of 90 degree on two lines that two lines are said to be as perpendicular lines. This perpendicular line is represented as ‘_’. The perpendicular line does not have any postulates but solve some theorems. They are perpendicular lines and slopes theorem, perpendicular to parallel theorem and two perpendiculars theorem.


Explanation for angles in math


Angles:

The above diagram is illustrating the angle by lines. The vertex of angle is A. The meeting line of two lines is called as arms of the angle and the arm points are used for naming the angle. When one angle arm is moved to the other arm is represent the size of an angle.

In math the degree is used for angle size measurement and in geometry the protractor is used. Adjacent angles, complementary angles ad supplementary angles are pairs of angles.

Doing Math Problems

Introduction of doing math problems:

Math is very important term in our day to day life. We have different concepts in math. And commonly formulas are used to solve the complex problems .We have various types of formulas in math. Algebra is the one of the important terms in mathematics. Here we are going to see some example math problems.

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Doing math problems:


Doing simple algebra problem:

Example 1:

g- 43 = 604.

Solution:

Step 1: g - 43 = 604.

Step 2: g - 43+ 43 = 604 + 43. (Add 43on both the sides).

Step 3: g = 647(so, the value of g is 647).

Doing simplification in algebra:

Example 2:

Given:

20x-5(12x-10x+2) +4x.

Solution:

Step 1: First we need to simplify the brackets. So we multiply (12x-10x+2) with 5

Step 2: So,20x-60x+50x-10+4x

Step 3: It can be written as 20x-60x+50x-10+4x

Step 4: So, the answer is 14x-10.

Doing parenthesis problem in math :

Example 3:

Given :

(14/2) * (45-9)

Solution :

Step 1: First we need to simplify the parenthesis ,so

Step 2: We get 14/2 = 7 and (45-9) =36.

Step 3: Here we need to multiply 7*36.

Step 4: So, the answer is 252.


More math problems


Doing inequality problem in math

Example 4:

Solve for 4(a+1) <2a br="">
Solution:

Step 1: It can be written as 4a+4<2a 2a="" both="" br="" on="" sides="" the="" ubtract="">
Step 2:4a+4-2a<2a-2a br="">
Step 3:2a+4<3 4="" both="" br="" on="" sides="" subtract="" the="">
Step 4:2a+4-4<3-4 .="" br="">
Step 5:2a<-1 .so="" a="-1/2.<br">


Doing subtracting algebraic expression in math:

Example 5:

Given:

Subtract 24ab – 10b – 18a from 30ab + 12b + 14a.

Solution:

30ab + 12b + 14a – (24ab – 10b – 18a)

Step1: 30ab + 12b + 14a – 24ab + 10b + 18a

Step2:30ab – 24ab + 12b + 10b + 14a + 18a

Step 3:6ab + 22b + 32a.

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Adding polynomials in math :

Example 6:

(6x+7y) + (2x+1y)

Solution:

Step 1: First we need to clear the parenthesis so, 6x+7y+2x+1y.

Step 2: Now we need to combine the like terms, so 6x+2x+7y+1y.

Step 3: Here we need to add the like terms, so 8x+8y.

Step 4: So the answer is 8(x+y).

Do My Math Problems For Me

Introduction:

Do my math problems for me

Mathematics is a very biggest subject which is very important for our life. Math is not only a subject to know and study but it teaches us what is life is and how does math approaches our life every time. Math problem are very interest to solve. Many different topics of math problem are there to find and solve. Math problem have different sequence of steps, methods, functions every thing. Many chapters are under the math. Now let us do some of the math problem in basics of mathematics like addition subtraction division multiplication etc.

Example math problem:


Find the sum of 5482 + 2451


Solution

5 4 8 2

2 4 5 1 +

------------

7 9 3 3

2. Find the difference of 7853 – 3521

Solution

7 8 5 3

3 5 2 1 –

------------------

4 3 3 2

3. Find the value of x in the equation 5x + 4 = 34

Solution

5x + 4 = 34

Subtract 4 on both sides

5x + 4 – 4 = 34 – 4

5x = 30

Divide 5 on both sides

5x/5 = 30 / 5

X = 6

4 .Use PEMDAS rule to solve the term

(5*5) + 9 – 8 / 2 *2 + 1

Solution

25 + 9 – 8 / 2 * 2 + 1 ------------(Parenthesis first)

25 + 9 – 8 / 4 + 1------------------(No exponents so net multiplication )

25 + 9 – 2 +1 ---------------------(Division)

34-3---------------------------------(Addition)

31

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Do my math problem:


Find the sum of 5487 + 2145

Find the difference of 987 – 123

Find the value of x in this equation 7x -2 = 61

John has 10 pens with him. He gave 3pens to his sister Rosy. Again he gave 2 pens to his friend Jane. Now how many

pens he will have at present.

Divide 845 by 5

Use PEMDAS rule to solve the term

6+2*3+5-(9/3)

Answers

1.7632

2.864

3.9

4. 5

5. 169

6.14

Algebraic Equations

Algebraic Equations Definition is given as equations that are a sum of number of terms where each term is the multiplication of some constant term and positive integral power of some variables. Algebraic equations can be divided into following categories as:
Linear equations: these include all equations with degree 1 which means the maximum power of variables can be 1 only. For example: x+5=12.

Quadratic equations: these are the equations of degree 2. Graphically these are represented by a U shaped parabola. Mathematically they are represented as: ax2 + bx + c=0 where a  0.
Exponential equations: these are equations where a term includes variable as the power of constant value for example: ex + e =0. Or 2x + x =2 etc.

Logarithmic equations: these equations involves a logarithmic term which is given as: loga x. Here ‘a’ cannot be 1 and a >0, x>0.  Let us see some methods of How to Solve Algebra Equations through some Algebraic Equations Examples now.

1) Linear equations: these equations are very easy to solve. Solution of the equation means to find out the value of variable that satisfies the given equation. Linear equations can be solved by isolating the variable of the equation. For ex:
2x+ 6 = 5.
to isolate x we can move 6 to RHS to get: 2x=5-6 = -1.
now a can be moved to RHS as: x= -1/2.

2) Quadratic equations: there are many methods to solve these equations like factoring, completing the square, quadratic formula and graphing. Here we will be quadratic formula to solve a quadratic equation.
The formula to solve a quadratic equation ax2 + bx + c=0 is given as:
x =
Example: solve 2x2 – 5x +2 =0.
By using above formula we have: a= 2, b =-5 and c =2.
x =

x =
x =
x=
So we get two solutions here as: x=(5+3)/4 and (5-3)/4 or 2, ½.

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3) Exponential equations: to solve these we first isolate exponential term and then take logarithm  at both the sides of equation.
Example: e2x - 3ex+2=0.
This can be written as: (ex-1)(ex-2)=0
Equate both terms of product by zero:
ex - 1=0
ex =1
=> taking logarithm on both sides we get: ln(ex) = ln(1)
=> x (ln(e)) = 0
=>x.1=0. Hence x=0.
ex - 2=0
Similarly solving this we get x=ln(2)=0.69314 approx.

4) Logarithmic Algebra Equation is solved by taking exponential.
Example: ln(x)=3
eln(x) = e3
x=e3.
x=20.085537 approx.

Grade 10th Math

Introduction to grade 10th math:

Mathematics is an essential tool used in throughout the entire world in many areas, like science, medicine, engineering, and social sciences. The 10th grade mathematics covers number system, algebra 1, algebra 11, pre calculus, calculus, statistics, geometry, trigonometry, and advanced trigonometry. In this article we shall discuss about 10th grade math problems and its solutions.

Grade 10th math example problem:


Example 1:

Solving the coefficient of the equation

`(1-(2)/((x-4))) = 8`

Solution:

This equation in the form of rational expression, multiplying the expression by the denominator value (x -4)
(x - 4) (1 - 2 / (x - 4)) = (x - 4)8

Simplifying the equation
(x - 4) - 2 = 8 (x - 4)

Multiplying the factors and arrange the like terms.
x - 6 = 8x - 32

Adding the value -8x to both sides and simplifying the equation

-7x - 6 = - 32

Adding the value + 6 to both sides
-5x = - 25

Solving for x

x = 25 /5.

x=5.

Example 2: Solve the inequality -4 < 2(x+4)-2<18 nbsp="" p="">
Solution:

Given inequalities

-4 < 2(x+4)-2<18 nbsp="" p="">
Multiplying the factor values and arranges the like term

-4<2x nbsp="" p="">-4<2x nbsp="" p="">Subtracting the value 6 on both sides of equation

-4-6<2x nbsp="" p="">-10<2x nbsp="" p="">
Dividing all terms by 2

-5
Conclusion:

The solution consists of all real number the interval (-5, 6)

Example:

A hemispherical bowl of radius 20 cm is contains with soap cream paste. This ice cream paste is made into cylindrical cakes each of radius 2 cm and height 2 cm, how many cakes do we get?

Solution:

Radius of the hemisphere = 20 cm;

Volume of the hemisphere =`3/ 2` `pi` r3 cubic units;

Therefore Volume of the soap paste =`3/2` x `pi ` x 20 x 20 x 20 cm3.

Radius of the cake = 2 cm;

Height of the soap cake = 2 cm;

Volume of the cake =pi x 2 x 2 x 2cm3.

Number of soap cakes = Volume of the soap cream paste / Volume of the cake

=   `(2/3 pi * 20*20*20)/(pi*2*2*2)`

Therefore we get 667 cakes out of the given paste.

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Grade 10th math practice problem:


Problem 1:

Solving the coefficient of the equation `(1-2/ (x-3)) ` =6

Answer:

x=13/5

Problem 2:

Solve the inequality -2< 2(x+6)-3<17 nbsp="" p="">Answer:

The solution consists of all real number the interval (-6, 4)

Learn Odds in Math

Introduction to learn odds in math:

A number which is not divisible by two is called as odd number in math. This is considered as whole number. The odd numbers are represented in set as { 1, 3, 5, 7, 9,……} and it does not contain multiples of two in math. We can learn the odds by identify the last digit that is if the last digit is 1,3 ,5 ,7 and so on means that number is odd.


Explanation for odd in math


Learn odd in math:

We can learn odd number’s arithmetic operations by using some results. They are,

In addition and subtraction:

If we add the even number and odd number means the resultant value is odd.For example add the numbers 8 + 1 = 9.

If we add the two odd numbers means the resultant value is even. For example add the 5 + 7 = 12.

In multiplication:

If we multiply the even and odd numbers means the resultant value is even. For example multiply the 6 x 7 = 42.

If we multiply the two odd numbers means the resultant value is odd. For example multiply the numbers 3 and 5 and we get the answer as odd value 15.

In other words:

The odd numbers are also called as gnomonic numbers in math. It is in the form as n = 2k + 1. Here integer is k. We can learn odd numbers by using this form. If we want to identify the odd numbers means divide that number by 2 and if we get the remainder, that number is odd.

The odd numbers are generated by using function as `(x(1 + x))/(1 + x)`  = x + 3x2 + 5x3 +......

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Learn odd numbers with example:


Problem 1: Find the odd numbers  from given set

{ 3, 6, 14, 7, 21, 34, 22, 47 }.

Answer:

The odd numbers are { 3, 7, 21, 47 }. Because these numbers are ended with 1,3, 7 and it does not divided by two.

Problem 2: Add and multiply the following odd numbers:

3 + 9 = 12.

53 + 7 = 60.

33 x 5 = 165.

Exercise problem for odd numbers:

Find the odd number from set { 190, 321, 453, 628 }.

Answer: { 321, 453 }.

Definition Reflexive Math

Introduction:

Reflexive is one of the properties in math. It is one of the postulates in equality. Reflexive property means anything equals to itself that is a = a where a is the set of numbers or number. Using reflexive property many geometric proofs were proved. Since it has equality we can find the answer for the given expressions.

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Example problems - Definition reflexive math:

Example 1 - Definition reflexive math:

Use the reflexive property to find the answer for x and the expressions are x+9 and 2x+8.

Solution:

The given expressions are x+9 and 2x+8

We the reflexive property anything equals to same thing

So, the expression can be written as x+9=2x+8

Add –x on both sides in the above equation

X+9+(-x) =2x+8+(-x)

X+9-x = 2x+8-x

9=x+8

Now subtract 8 on both sides

9-8=x+8-8

1=x

The value of x is 1.

Example 2 - Definition reflexive math:

Use the reflexive property to find the answer for y and the expressions are 2y+8 and y+8.

Solution:

The given expressions are 2y+8 and y+8.

We the reflexive property anything equals to same thing

So, the expression can be written as 2y+8 = y+8.

Add –y on both sides in the above equation

2y+8 +(-y)= y+8+(-y)

2y+8-y=y+8-y

y+8=8

Now subtract 8 on both sides

y+8-8=8-8

y=0

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More Example problems - Definition reflexive math:


Example 3 - Definition reflexive math:

Use the reflexive property to find the answer for x and the expressions are 3x+9 and 2x+8.

Solution:

The given expressions are 3x+9 and 2x+8

We the reflexive property anything equals to same thing

So, the expression can be written as 3x+9=2x+8

Add –xx on both sides in the above equation

3x+9+(-2x) =2x+8+(-2x)

3x+9-2x = 2x+8-2x

x+9 =8

Now subtract 9 on both sidesx

x+9-9=8-9

x=-1

The value of x is -1.

Example 4 - Definition reflexive math:

Use the reflexive property to find the answer for y and the expressions are 2y+8 and4y+8.

Solution:

The given expressions are 2y+8 and 4y+8.

We the reflexive property anything equals to same thing

So, the expression can be written as 2y+8 = 4y+8.

Add –2y on both sides in the above equation

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

8=2y+8

Now subtract 8 on both sides

8-8 =2y+8-8

2y=0

So, y=0.