Define Hypothesis Math Term

Define hypothesis math term:

The term hypothesis refers the condition statement just after the word if. In the given conditional statement.

For example Suppose the condition is “in the given quadrilateral figure -  four sides are equal in measure then the shape is said to be square” the hypothesis of the given statement is “all four sides are equal in measure. In the given example,” all four sides are equal in measure. “Always, never or sometimes true.

Options:

1)Always

2)Never

3)Sometimes.

Answer is (3)

Solution:

Answer 3) sometimes it is correct because, it’s true if only the given shape is square, In case of other shapes like rectangle, rhombus then  condition is not satisfied.


Example 1)

Find the math term hypothesis of the given equation.

If x = 1 then x2 +3x +5 = 9

Option:

1)True

2)false 

Answer : 1)True.

Solution:

Substitute x value in the given equation,

X+3x+5 =9

1+3(1) +5 = 9

1+3+5 =9

9=9

Therefore the given condition is true.

Example 2:

Find the math term hypothesis in t he given statement is “Every  integer greater than 1 is product of prime number “

Solution:

Let us consider the if  x is true statement .

Therefore ,

X`=>` y  is a true statement  and y is a true  conditional statement.

Now,we need to find the series of the statement x1,x2 …..etc.,and then verifies the given conditional statement” Every  integer greater than 1 is product of prime number”.

X=x1

X1=x2,

X2=x3,

….

….

Xn-1 =xn

Xn =Y are ture.

Because assume that the given statement X is true then all repeated statement of the simple rule  Y  is true..

Example 3:

The conditional statement is “if n is an integer then n2 is even  and then n is odd.”

Solution:

N is odd  number .

so  n-1 is even number .

Therefore , we have,

n-1 =2k

where ,k is the constant .

n-1 = 2k ,k `in` Z

n=2k+1

Similarly we have by hypothesis,

  n2 =x;  x `in` Z

And then,

n2 = 4k2 +4k+1

 =2(2k2 +2k)+1

In the above equation, 2(2k2+2k) is even number.

So 2(2k2+2k) +1 is the odd number.Therefore our hypothesis is violated.

So  the integer number “n “ must be even number.

So,

X`=>` S =” n2  is even ”

Y  `=>` ” n is even ”

Not –Y `=>` ” n is odd ”


Practice problems related to the term hypothesis in math :


1) Prove that if m and n are even integers, then n +m is even.

2) Prove that if n is an odd integer, then n 2 is an odd integer.

Find Domain off

Introduction of Domain function

The domain of definition or simply the domain of a function is the set of "input" or argument values for which the function is defined. That is, the function provides an "output" or "value" for each member of the domain. The domain of cosine is the set of all real numbers, while the domain of the square root consists only of numbers greater than or equal to 0 when the Function is represented in an xy


Examples


Problem 1: Find the domain of the real valued linear function f given below.

f(x) = x + 1

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

x=1                    f(x) =1+1

By solving it we get

f(x) =2

The ordered pair is (1, 2)

x=2                            f(x) =2+1

By solving it we get

f(x) =3

The ordered pair is = (2, 3)

x=3         f(x) =3+1

By solving it we get

f(x) =4

The ordered pair is = (3, 4)

x=4                          f(x) =4+1

By solving it we get

f(x) =5

The ordered pair is = (4, 5)

x=5                            f(x) =5+1

By solving it we get

f(x) =6

The ordered pair is = (5, 6)

The domain function is 1, 2,3,4,5

Problem 2: Find the domain of the real valued linear function f given below.

f(x) = 2x + 1

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

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

By solving it we get

f(x) =3

The ordered pair is (1, 3)

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

By solving it we get

f(x) =5

The ordered pair is = (2, 5)

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

By solving it we get

f(x) =7

The ordered pair is = (3, 7)

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

By solving it we get

f(x) =9

The ordered pair is = (4, 9)

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

By solving it we get

f(x) =11

The ordered pair is = (5, 11)

The domain function is 1, 2,3,4,5


Practice problem


Problem : Find the domain of the real valued linear function f given below.

f(x) = 2x + 5

Vista Math Help

Introduction to vista math help:

Tutorvista is one of the online tutoring services providers; Tutorvista provides helps in various subjects like English, math, and science.  Tutor vista helps at any time around the clock regardless of place where they are located.  Student can learn and do homework from their home at any time. Here tutor to explain step by step so that the students can easily understand much easier. In this article we shall discuss tutorvista help in math problems.


Tutorvista math help example problem


Example 1:

Sum the polynomial  5x4 – 6x2 + 9x + 5 and 5x + 6x3 – 3x2 – 1.

Solution:

Use the associative property and distributive properties add real numbers

(5x4 – 6x2 + 9x + 5) + (6x3 – 3x2 + 5x – 1) = 5x4 + 6x3 – 6x2 – 3x2 + 9x + 5x + 5 – 1

= 5x4 + 6x3 – (6+3)x2 + (9+5)x + 4

= 5x4 + 6x3 – 9x2 + 14x + 4.

An another method for adding two polynomials

5x4 +0x3– 6x2 + 9x + 5

0x4 + 6x3 – 3x2 + 5x – 1

____________________
8x4 + 6x3– 9x2 + 14x + 4
____________________

Example 2:

Consider e1 and e2 are two events with a random trial such that P(E2)=0.25, P(E1 or E2)= 0.95 and P(E1 and E2) = 0.35, find the value of P(E1).

Solution:

Let P (E1) = x then,

P (E1 or E2) = P (E) + P (E2) – P (E1 and E2)

0.95 = x + 0.25 – 0.35

Adding the trial value

x = (0.95-0.25 + 0.35) = 0.65

Hence the value of P (E1) = 1.05

Example 3:

Find the increases percentage value from 35 to 50.5?

Solution:

Changes in value = new value – original value

50.5 – 35 = 15.5

Increased percentage = `15.5 / 35` x100%

Answer:

The increased percentage value is = 44.2%


Tutorvista math help practice problem


Problem:

Adding the polynomial 8x4 – 5x2 + 9x + 4 and 4x + 6x3 – 3x2 – 1.

Answer: 8x4 + 6x3– 8x2 + 13x + 3

Problem:

Find how many percentage increases from 30 to 55.5?

Answer: 85%

Problem:

Consider e1 and e2 are two events with a random trial such that P(E2)=0.35, P(E1 or E2) = 0.75 and P(E1 and E2) = 0.45, find the value of P(E1).

Answer:

P (E1) = 0.85

Dividing Word Problems

Introduction to dividing word problems:

Division is one of the arithmetic operations and also it is the inverse of multiplication. We know that the solution of dividing two rational numbers is another rational number when the divisor is not Zero. Division of any number by zero is not defined. Because of zero is multiplied by any particular number will always result in a product of zero. If we can enter these types of numbers into the calculator then error message is displayed.
Examples of dividing word problems:


Dividing word problem 1:

Jack invites 12 friends to his Wedding party. He distributes 72 balloons equally among them. How many balloons did each friend get?

Solution:

We can find the number of balloons per each by using the division method.

Therefore Total number of balloons= 72

Total person= 12

Therefore, `72/12` `= 6`

Answer: 6

Dividing word problem 2:

A board game has 45 total coins. There are equal number of coins of 5 different colors. How many coins are there of each color?

Solution:

We can find the number of coins per each color by using the division method.

 Total number of coins= 45

 Total number of colors= 5

So, `45/5` `= 9`

Here 45 is 9 times of 5

Answer: 9

Dividing word problem 3:

Uncle has planned 15 total prizes for the party. Each game would have 3 winners. How many games will be played at the party?

Solution:

`=` `15/ 3`

Here 15 has 5 times of 3

= 5

Answer: 5

Dividing word problem 4:

Dad bought 56 sweets for the party and distributed them equally among 7 gift pouches. How many sweets did she place in each pouch?

Solution:

= `56/ 7`

= 8

Here 56 is 8 times of 7.

Answer: 8


Practice problems of dividing word problems:


Mary has 48 dollar in her purse. Each toy costs 6 dollar. How many toys can she buy? Answer: 8
John divided 72 by 3 on his toy calculator. What quotient did the calculator display? Answer: 24
 Ton has 75 blocks. He makes 15 equal stacks. How many blocks are there in each stack?   Answer: 5

Alpha Math

Introduction to Alpha Math:

     Alpha is the first letter of the Greek alphabet and it is written as α . The word "mathematics" comes from the Greek, which means learning, study, science, and additionally came to have the narrower and more technical meaning "Mathematical Study", 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.


Example problems of Alpha math:


Here we will discuss about the alpha math,

Alpha math – Example problem: 1

  Prove that the condition that one root of ax2 + bx + c = 0 may be the square of the other is m3 + l2 n + ln2 = 3lmn

Proof:

Let the roots of the given equation be  α and α2

then,                 α +  α2 = `(m)/(-alpha)`

 α+  α2  =  α3 = `(n)/(alpha)`

Cubing both sides of above equation (α+ α2)3 =- (m3 / l3)

 α3 +  α6 + 3α3 (α+ α2 )= - (m3 / l3 )

           n/l + (n/l)2 +3n/l (-m/l)= -m3 / l3

that is m3 + l2 n + ln2 = 3lmn

Alpha math – Example problem: 2

  We form the quadratic equation whose roots are α and γ .

Proof:

Then x = α ,

           y = γ are the roots

Therefore x - α  = 0 and

              y - γ = 0

        (x -α) (y -γ) = 0


Some more example problems of Alpha math:


Alpha math – Example problem: 3

F the roots of x4 - 6x3 + 13x2 - 12x + 4 = 0 are α,α,γ,γ  then the values of  α,  γ= ?

Solution:

        2α + 2γ = 6

              α + γ  = 3

            α 2γ 2 = 4

                    αγ = 2

    α= 2  and γ = 1

Alpha math – Example problem: 4

Prove that sin2α + sin2 ( α + 60) + sin2 (  α - 60)  =`(3)/(2)`

L. H. S = sin2α + [ sin  ( α + 60) ]2 + [ sin (α - 60)]2

sin2α + [ sinαcos 60 + cos2αsin 60) ]2+ sinα cos60 - cosα sin60]2

sin2α+ 2( sin2αcos2 60 + cos2α sin2 60 )

sin2α+ 2 [`(1)/(4)`sin2α+ `(3)/(4)` cos2α ]

sin2α+ [ (sin2 α) / 2 ] + [ (3cos2α ) / 2 ]

`(3)/(4)` [ sin2α + cos2α ] = = R. H. S

Here we proved the L.H.S is equal to R.H.S.

Math Second Grade Fraction

Introduction to math second grade fraction:

Here we are going to learn identify the fraction, which shape show the fraction, Parts of a group, Word problems, Compare fractions, Order fractions, generally fraction contains two parts and these parts are divided by the line, top part is called numerator and bottom is called denominator. Hear the top part is the total number of shaded region and bottom part is total number of division.


Identifying the fractions - math second grade fraction:


Example problem:

Find the fractions of the shaded region?

Solution:


Here the circle is divided into 4 parts, so it will be come under the part of the denominator and 2 regions are shaded it will be come under the part of the numerator



So we can write in the form of fractions (shaded region) =`2/ 4`

Which shape illustrates the fraction- math second grade fraction:?

Which shape shows the fraction `3/8` ?



Solution:

Circle is divided into 3/8 .


Find the fractions from the parts of the group- math second grade fraction::


suppose the group (generally group means collection of objects) has many diagrams like triangle, square and rectangle means we can write the one shape in the form of fraction means we have to count the total diagrams in the particular group it will be come under the part of the denominator and count the particular shapes it will be write in the numerator part.

Example:

What fraction of the square in the below diagram?



Solution:

Word problems- math second grade fraction:

John’s family lives on an avenue with 3 houses. 2 of the houses are prepared of brick. How we can make the fractions (build the house prepared of brick)?

Solution;

here total number of house is 3so it will be come under the part of the denominator ,2 of the houses are made with brick so it will be come in the part of numerator .so we can write those fraction in the below form,

`2/ 3`

Math Geometry Activities

Introduction for math geometry activities:

The geometry activities are one of the most important branches of Mathematics. Math geometry activities gives the idea of various geometrical shapes and figures in our daily life such as articles in the houses, wells, buildings, bridges etc.

The word ‘Geometry’ means learns of properties of figures and shapes and the relationship between them. The geometry shapes are point, line, square, rectangle, triangle, and circle. From this we can learn that the math geometry activities have begun from ancient times. The math geometry activities example problems and practice problems are given below.


Example problems - math geometry activities:


Example problem 1:

In the ?ABC the angle B is bisected and the bisector meets AC in D. If ?ABC = 80° and ?BDC = 95°, find ?A and ?C.

Solution: Let us See the Figure



From ?BDC, 40° + 95° + ?C = 180°

After solving this, We get

? ?C = 180° - 135° = 45°

From ?ABC, ?A + ?B +?C = 180°

?A + 80° + 45° = 180°

After solving this, We get

?A = 180° - 125° = 55°.

Example problem 2:

If the angles of a triangle are in the ratio 1 : 3 : 6, find them.

Solution:

Let the angles be 1x, 3x, 6x.

Then 1x + 3x + 6x = 180°

After simplify this, We get

10x = 180° or x = 18°.

The angles are 1 × 18°, 3 × 18°, 6 × 18°,

After simplify this, We get

The angles are 18°, 54°, 108°.


Practice problems - math geometry activities:

Practice problem 1:

Find out the geometry equation of straight line passing through the given point (2, 3) and cutting off

Determine the equal intercepts along the positive directions of both the axes.

Answer: x + y = 5.

Practice problem 2:

Find the points on y-axis,then perpendicular distance from the straight line 4x - 3y - 12 = 0 is 3.

Answer: points are (0, 1) and (0, - 9)

Practice problem 3:

Find out the equation of straight line passing through the given intersection of the straight lines 2x + y = 8 and 3x - y = 2 and through the point (2, - 3).

Answer:  x = 2