Division Questions to Print

Introduction – Division questions to print:

In mathematics, division (÷) is the arithmetic operation which is the opposite of multiplication. Particularly, if ‘c’ times ‘b’ equals ‘a’, written as: c * b = a. Here ‘b’ is not 0, then ‘a’ divided by ‘b’ equals ‘c’, written as: = c.

From the above expression, ‘a’ is known as the dividend, ‘b’ the divisor and ‘c’ the quotient. Dividend = (Divisor * Quotient) + Remainder.  Let us see division questions to print. Please express your views of this topic Sum of Arithmetic Series Formula by commenting on blog.

Division Questions to Print:

Steps to solve division:

To make an arithmetic division operation with an accurate sign the below two steps will be very useful.

Step 1: The division of two or more numbers same signs will be positive sign.

a) Positive ÷ positive = positive

b) Negative ÷ negative = positive

Step 2: The division of two or more numbers unrelated signs will be negative sign.

a) Positive ÷ negative = negative

b) Negative ÷ positive = negative.

Example Problems – Division Questions to Print:

Problem 1:

Divide 1348 by 2

Solution:

674`-> ` Quotient

--------

2  | 1348

12  (-)

--------

14

14 (-)

--------

8

8 (-)

--------

0`->` Remainder

---------

Steps to solve the above problem:

Given: `448 / 2`

Step 1: Let us take first two digits which is 13. Then we know that 6 times 2 = 12.

Step 2: Now we have to get the next term. That is 4.

Step 3: So presently we have 14.

Step 4: We know that 7 times 2 = 14. Subtract 14 from 14 = 0. Then get the next term 8.

Step 6: So presently we have 8. 4 times 2 = 8.

Step 7: So the quotient is 674 and remainder is 0.

Problem 2:

Divide 3144 by 4

Solution:

786`->` Quotient

--------

4  | 3144

28  (-)

------

34

32 (-)

--------

24

24 (-)

-----------

0`->` Remainder

-----------

Here quotient is 786

Remainder is 0

Problem 3:

Divide 3580 by 5

Solution:

716`->` Quotient

--------

5  | 3580

35  (-)

------

80

80 (-)

--------

0`->` Remainder

---------

Here quotient is 716

Remainder is 0

Problem 4:

Divide 5438 by 8

Solution:

679.75`->` Quotient

--------

8  | 5438

48  (-)

----------

63

56 (-)

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

78

72 (-)

--------

60

56

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

40

40

-----------

0`-> ` Remainder

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

Here quotient is 679.75

Remainder is 0

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Practice problems – Division questions to print:

1) Divide 778 by 2

Ans: 389 (quotient)

2) Divide 638 by 3

Ans: 212.6

3) Divide 3025 by 5

Ans: 605

4) Divide 5432 by 6

Ans: 905.3

5) Divide 7821 by 9

Ans: 869

Altitude Problem

Introduction about altitude problem:

The altitude of shapes are the height of the 3d shapes. The altitude is the height measurements of the objects. The students will learn about the altitude of the 3Dimensional shapes. Altitude is the height which measures from the top of the object to the bottom of the object. The altitude problems can be solved with the help of some examples. Now we are going to see about the altitude problems. Having problem with Perimeter of a Trapezoid keep reading my upcoming posts, i will try to help you.

About Altitude Problems:

Now we are going to see about the altitude problems with the help of the tutor and also some of the concepts of the altitudes. The altitudes are present for every objects so that the height of the objects can be determined with the help of some of the formulas for the objects in the geometry.

Altitude of the square pyramid (height) = `(3V)/(a^2)` units

Altitude of the cylinder (h) = `V/pir^2` units

Altitude of the of the rectangular prism (h) = `(Volume)/(w * l)` units

These are some of the altitude formulas of some of the objects. All the objects in the geometry are having the height measurements. Now we see some of the altitude problems with some of the examples. Please express your views of this topic Volume of a Triangular Prism Formula by commenting on blog.

Problems in Altitude:

Example 1:

Find the altitude of the square pyramid which has the volume measurement is about 88 cm3 and the side length is about 6 cm?

Solution:

Now we are going to determine the altitude of the square pyramid as follows,

Volume of the given square pyramid (V) = 88 cm3

The side length measurement (a) = 6 cm

The formula to determine the altitude of the square pyramid as follows,

Altitude of the square pyramid (h) = `(3V)/(a^2)` units

= `(3 * 88)/(6^2)`

= `264/36`

= 7.33 cm

Altitude of the square pyramid (h) = 7.33 cm

Example 2:

Determine the altitude of the volume of the cylinder is 850 cm3 and radius measuerment is about 6 cm.

Solution:

Now we are going to see about the altitude of the cylinder with the help of the formulas as follows,

The volume of the cylinder (A) = 850 cm3

The radius of the cylinder (r) = 6 cm

The formula used for the cylinder is given as follows,

Altitude of the cylinder (h) = `V/(pi r^2)` units

= `(850)/(pi * 36) `

= `850/113.04`

Altitude of the cylinder (h) = 7.5 cm.

Website to Solve Math Problems

Introduction about website to solve math problems

There are many different website to solve math problems. Tutor vista is also one of the wesite. Tutor vista is the wonderful website to solve math problems. Many tutors with high qualification were found in the website to solve math problems. The tutors help the students with each and every steps any time. Many students were enjoying this website to solve math problems. Tutor vista website to solve math problems is the top of all the website that is specifically designed and activating all the time with live tutors to help on the math problems. I like to share this Interest Formulas with you all through my article.

Website to Solve Math Problems

Below are some of the model questions by the tutor vista website to solve math problems. Tutor vista helps with all the topics and different chapters in mathematics. For example we are going to see about the arithmetic problems.

Example 1: Add of 427 + 283

Solution

4 2 7                First add the ones digit 7 + 3 = 10
2 8 3 +             place 0 in ones digit and carry 1 to next digit and add with tens digit.
----------              So 1+2+8 =11 where 1 is the carry. Place 1 and carry 1 to hundred’s
7 1 0                digit. Add 1+4+2 = 7
----------
Example 2:  Subtract 83-64

Solution

8 3                 First subtract the ones digit 3 – 4.This cannot be subtracted.
6 4 -              so borrow one from tens digit hence one’s digit become 13
----------            13- 4 =9. Next subtract the tens digit 7 - 6 = 1
1 9
-----------
Example 3: Multiply of 1 8 3* 4

1 8 3             First multiply 4 with the ones digit 3. So 4x3 = 12
4 x          Place 2 and carry 1 as remainder to next digit
----------            Next multiply 4 with 8 we get 4x8 = 32 and add the carry 1
7 3 2            So 33 place 3 and carry 3 as remainder. Multiply 1x4 = 4 and
----------            add the carry 3. So 4+3 =7
Example 4: Solve arithmetic operation of divide 77 by 7

11
7)7 7        consider the first digit 7. We can divide 7 by7
7            So consider next digit 7. So 1times 7 is7
---------     So quotient 11
7
7
----------
0
----------

Understanding Cotangent Function is always challenging for me but thanks to all math help websites to help me out. 

Website to Solve Math Problems

The tutor vista also provides some model questions for your practice with the answer at the end. With the help of the examples shown above try the problems given below.

1. a) Add 38 + 236

b) Add 425 + 156

2.  a) Subtract 98 – 63

b) Subtract 372 – 32

3. a) Multiply 41 x 19

b) Multiply 248 x 3

4. a) Divide 546/ 2

b) Divide 186/6

Answers

1.  a) 274    b) 581

2.  a) 35     b) 340

3.  a) 779    b) 744

4. a) 273       b) 31

10th Grade Math Tutorial

Introduction of 10th grade math tutorial:-

In grade (10) means tenth grade is a year of education in all over the world. The tenth grade is the ten school year post kindergarten. Students are usually 15 - 16 years old. Tutorial is the new way for the students. Tutorial show the math concept and give you practice problems. Students work the problems in own place. The students first learning about the math definitions then use formula to workout the math problems. It is more helpful for test preparation and increases our own practice skills.

Topics in 10th Grade Math Tutorial:-

In the following basic topics involves the 10th grade math tutorial

Number theory
Mensuration
Theory of sets
Algebra
Applied Mathematics

Number Theory

In earlier classes, you might have come across various patterns of numbers like

1, 2, 3, 4, 5, ...

1, 3, 5, 7, 9, ...

1, 8, 27, 64, 125,...

These patterns are generally known as sequences. An arrangement of numbers of which

One number is designated as the first, another as second, another as third and so on is known as a sequence.

Mensuration

We do measurement in our daily life in many situations. For example, we measure the length of cloth for stitching, the area of a wall for painting, the perimeter of a plot for fencing,the volume of a water tank for filling.

Theory of sets

We have learnt about sets in the previous classes. All branches of Mathematics can be brought into the frame work of set theory.

Algebra

Algebra has been studied for many centuries. Babylonian, ancient Chinese and Egyptian mathematicians proposed and solved problems in words.

Applied Mathematics

Applied Mathematics represents a great stream gushing out irresistibly in many directions overcoming all resistances and all along retaining its originality. Is this topic Calculating Sample Standard Deviation hard for you? Watch out for my coming posts.

Example Problems for 10th Grade Math Tutorial:-

Problem 1:-

The nth term of a sequence is 7n – 3. Show that it is an A.P and find the first term and the common difference.

Solution:-

`t_n` = 7n-3

`t_1` = 7-3
= 4

`t_2` = 14-3
= 11

`t_3` = 21-3
= 18

`t_4` = 28-3
= 25

The sequence is 4, 11, 18, 25, ...

First term is 4, Common difference = 11 – 4 = 18 – 11 = 25 – 18 = 7

The given sequence is an A.P with C.D = 7

Problem 2:-

Find from the table cos 63° 29′

Solution:

Since cosine value decreases as the degree measure increases from 0° to 90°

63° 29′ = 63° 29′ + 5′
cos 63° 29′ = 0.4478

Difference of 5′ = 0.00013

cos 63° 29′ = 0.4478-0.0013
= 0.4465

One Meter Equals How many Yards

Introduction for One meter equals how many yards
Meters:

The metre (or meter), symbol m, is the base unit of length in the International System of Units (SI). it is defined as the distance travelled by light in a complete vacuum in 1/299,792,458 of a second.

Yards:

A yard (abbreviation: yd) is a unit of length in several different systems, including English units, Imperial units, and United States customary units. (Source: Wiki)

Formula for one meter equals how many yards

1 meter = 1.0936133 yards

One meter is equals to one point zero nine three six one three three yards

Please express your views of this topic Find Circumference of a Circle by commenting on blog.

Examples for One Meter Equals How many Yards :

Example 1  :

Find 51 meter equals how many yards?

Solution:

Step 1:

Formula for finding meters to yards

1 meter = 1.0936133 yards

Step 2:

So to find 51 meter equals how many yards

Step 3:

Multiply 51 with 1.0936133= 55.7742783

Step 4:

Therefore, 51 meters = 55.7742783 yards

Example 2 :

Find 171 meter equals how many yards?

Solution:

Step 1:

Formula for Finding meters to yards

1 meter = 1.0936133 yards

Step 2:

So to find 171 meter equals how many yards

Step 3:

Multiply 171 with 1.0936133= 187.0078743

Step 4:

Therefore, 171 meters = 187.0078743 yards

Example 3:

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Find 201 meter in yards?

Solution:

Step 1:

Formula for Finding meters to yards

1 meter = 1.0936133 yards

Step 2:

So to find 201 meter equals how many yards

Step 3:

Multiply 201 with 1.0936133= 219.8162733.

Step 4:

Therefore, 201 meter= 219.8162733 yards

Example 4:

Find 501 meter in yards?

Solution:

Step 1:

Formula for Finding meters to yards

1 meter = 1.0936133 yards

Step 2:

So to find 501 meter equals how many yards

Step 3:

Multiply 501 with 1.0936133= 547.9002633

Step 4:

Therefore, 501 meter= 547.9002633 yards

Example 5:

Find 1251 meter equals how many yards?

Solution:

Step 1:

Formula for Finding meters to yards

1 meter = 1.0936133 yards

Step 2:

So in find 1251 meter equals how many yards

Step 3:

Multiply 1251 with 1.0936133= 1368.1102383

Step 4:

Therefore, 1251 meters = 1368.1102383 yards

Example 6:

Convert 11 meters to meter?

Solution:

Step 1:

Formula for Finding meters to yards

1 meter = 1.0936133 yards

Step 2:

So to find 11 meters to yard

Step 3:

Multiply 11 with 1.0936133= 12.0297463 yards

Step 4:

Therefore, 11 meter= 12.0297463 yards

Practice Problems Meter in Yards:

Practice Problem 1:

Find 1001 meter in yards?

Answer:

1094.7069133

Practice Problem 2:

Find 1501 meter in yards?

Answer:

1641.5135633 yards

Rotate Vector by Angle

Introduction to rotate vector by angle:

When a vector is rotated by an angle the angle between the original vector and the newly formed vector can be calculated using the formula for calculating the dot product of the original vector and the newly formed vector. In the following article we will see in detail about the topic rotate vector by angle. Understanding Adding Vector Components is always challenging for me but thanks to all math help websites to help me out.

More about Rotate Vector by Angle:

In math the vectors are the geometric quantities with both the magnitude as well as the direction. When a vector is rotated by an angle a new vector is formed and the two vectors will be having the same vertex point. So the angle by which the vector is rotated is calculated by using the dot product of the vectors formula. And the formula for the dot product of the vectors is given by,

`A*B = |A| |B| Cos theta`

The formula can also be written as,

`Cos theta = (A*B)/(|A| |B|)`

So for calculating the angle θ between the two vectors will be

`theta = arccos [(A*B)/(|A| |B|)]`

Example Problem on Rotate Vector by Angle:

1. Calculate the angle between the vectors A = 1i + 3j + 3k and B = 2i - 2j + 4k.

Solution:

The angle between the vectors theta = `arccos [(A*B)/(|A| |B|)]`

`A*B = (2i + 3j + 3k)*(2i - 2j + 4k)`

`= 1*2 + 3*-21 + 4*3`

`= 2-6+12`

`= 8`

`|A| = sqrt(1^2 + 3^2 + 3^2) = sqrt(1+9+9) = sqrt (19)`

`|B| = sqrt(2^2 + (-2)^2 + 4^2) = sqrt(4+4+16) = sqrt (24)`

`theta = arccos [(A*B)/(|A| |B|)]`

`= arccos [(8)/(sqrt22*sqrt24)]`

`= arccos (8/22.98)`

`= arccos (0.348)`

`= "69.6 degrees"`


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Practice problems on rotate vector by angle:

1. Calculate the angle between the 3D vectors A = i - 3j + k and B = 2i - j + 4k having a common vertex point.

Answer: 58.2 degrees

2. Calculate the angle between the 3D vectors A = i - 2j + 3k and B = i + 6j - 3k having a common vertex point.

Answer: 142 degrees

Grow Cube Solution

Introduction to grow cube solution:

In geometry, a cube is a three-dimensional solid object bounded by six square faces, facets or sides, with three meeting at each vertex. The cube can also be called a regular hexahedron and is one of the five Platonic solids. It is a special kind of square prism, of rectangular parallelepiped. The cube is dual to the octahedron. (Source: Wikipedia).

Formulas Used in Grow Cube Solution:

Surface area of the cube is 6a^2.
Face diagonal of the cube is √2 a.
Space diagonal of the cube is √3 a.
Radius of the inscribed space is `a/2` .
The volume of the cube for the edge length a is a^3. Here a is the edge length of the cube.

Examples for Grow Cube Solution:

Example 1 to grow cube solution:

Find the surface area for the cube whose edge length is 3cm.

Solution:

The edge length of the cube is 3.

The formula for the surface area is 6a^2.

Surface area of cube =6a^2

Substitute the value of a in the formula a=3.

Surface area of cube =6(3^2)

Surface area of cube =6(9)

Surface area of cube =54.

The surface area for the cube is 54 cm^2.

Example 2 to grow cube solution:

The edge length of the cube is 4 cm. Evaluate the volume for the cube.

Solution:

The edge length of the cube is 4 cm.

The formula for volume of the cube is a^3.

Volume of the cube =a^3

Substitute the value of a in the formula, a=4.

Volume of the cube =4^3

Volume of the cube = 64

The volume of the cube is 64 cm^3, whose edge length is 4cm.

Example 3 to grow cube solution:

The edge length of the cube is 7 cm. Find the radius of the inscribed space.

Solution:

The edge length of the cube is 7 cm.

The formula for the radius of the inscribed space is 7 cm.

Radius of inscribed space= `a/2`

Substitute the value of a in the above formula, a=7.

Radius of inscribed space= `7/2`

Radius of inscribed space= 3.5

The radius of inscribed space for the edge length 7 cm is 3.5 cm. Between, if you have problem on these topics Help Solve Math Problems, please browse expert math related websites for more help on free math online help.

Practice Problem for Grow Cube Solution:

The cube edge length is 10 cm. Find the surface area for the cube.
Answer: 600 cm^2

Find the volume of the cube whose edge length is 9 cm.
Answer: 729 cm^3