Solve Accuracy

Introduction to solve accuracy:

In mathematics, accuracy is defined as the most important topic. Accuracy is mainly used to find the approximate values. When we divide the two numbers, if the answer comes in decimal points means, then the obtained answer is rounded off to a nearest integer value. The concept of making the approximate numbers is known as the accuracy.  In this article, we are going to solve many examples about the accuracy. I like to share this Subtraction Word Problems with you all through my article.

Explanation to Solve Accuracy

The explanation to solve accuracy is given below the following section,

Example:

The value of the golden ratio is given by 1.61803399

The value of accuracy given for the golden ratio is 1.618

The concept of rounding off the value to its nearest integer is known as the accuracy. Understanding Basic Fractions is always challenging for me but thanks to all math help websites to help me out.

Example Problem to Solve Accuracy

Problem 1: What is the accuracy by solving the division problem, 6589 `-:` 3226

Solution:

Step 1: Given:

6589 `-:` 3226

Step 2: Divide:

6589 `-:` 3226

= 2.0424674

Step 3: Accuracy:

2.0424674

= 2.04 ( The obtained number is rounded off to nearest integer)

Result: Accuracy = 2.04

Thus, this is the obtained answer for solving the accuracy.

Problem 2: What is the accuracy by solving the division problem, 8964 `-:` 3645

Solution:

Step 1: Given:

8964 `-:` 3645

Step 2: Divide:

8964 `-:` 3645

= 2.459259

Step 3: Accuracy:

2.459259

= 2.45 ( The obtained number is rounded off to nearest integer)

Result: Accuracy = 2.45

Thus, this is the obtained answer for solving the accuracy.

Problem 3: What is the accuracy by solving the division problem, 7893 `-:` 4569

Solution:

Step 1: Given:

7893 `-:` 4569

Step 2: Divide:

7893 `-:` 4569

= 1.72751149

Step 3: Accuracy:

1.72751149

= 1.72( The obtained number is rounded off to nearest integer)

Result: Accuracy = 1.72

Thus, this is the obtained answer for solving the accuracy.

Practice Problem to Solve Accuracy

Problem 1: What is the accuracy by solving the division problem, 9785 `-:` 2456

Answer: 3.98

Problem 2: What is the accuracy by solving the division problem, 6987 `-:` 1265

Answer: 5.52

Distance between Houses

Introduction on Distance between Houses:

Distance can be measured by using the units called meter, kilometer, mile, nautical mile etc... Distance is the measure which is used to find the length between the two objects which may be movable object or immovable object. Having problem with Adding Fractions with Common Denominators keep reading my upcoming posts, i will try to help you.

Distance between houses: Through online we can see how to calculate the distance between two houses.

Examples on Distance between Houses:

Example 1:

Distance between two houses is 5 km. convert the distance into a meter.

Solution:

We know that 1 km = 1000 m.

Given:

Distance between two houses is 5 km.

We have to convert 5 km into meters.

So,

5 km = 5 * 1000 m

5 km = 5000 m.

Finally, distance between two houses is 5000 meter.

Example 2:

Distance between house A and house B is 1 kilometer, house B and house C is 3 kilometer. Find the distance between the house A and house C.

Solution:

We know that Distance between house A and house B is 1 kilometer.

Distance between house B and house C is 3 kilometer.

To find the distance between house A and house C, Add all the given distance.

Distance between house A and house C = Distance between house A and house B + Distance between house B and house C.

= 1 + 3

= 4 kilometer.

Distance between house A and house C = 4 kilometer.

Example 3:

Distance between House A and House B is 3 kilometer, distance between house A and house C is 7 kilometer. Calculate the distance between the house B and house C.

Solution:

We know that Distance between house A and house B is 3 kilometer.

Distance between house A and house C is 7 kilometer.

To find the distance between house B and house C, subtract the distance of house AC with house AB distance.

Distance between house B and house C = Distance between house A and house C - Distance between house A and house B.

= 7 - 3

= 4 kilometer.

Distance between house B and house C = 4 kilometer. Please express your views of this topic 9th cbse sample papers by commenting on blog.

Practice Problems on Distance between Houses:

Problem1:

Distance between two houses is 10 km. convert the distance into a meter.

Answer:

10,000 meter.

Problem 2:

Distance between house A and house B is 5 kilometer, house B and house C is 3 kilometer. Find the distance between the house A and house C.

Answer:

8 kilometer.

Root 3 Value

Introduction of root 3 value:

Root is the form of method in the radical, which has been used in mathematics. The radical form has been given as root √. In which the root radical symbol root "√" or otherwise `root(2)(x)` . The numerical significance inside the radical symbol or root symbol called as the radicand. There are many methods available in the rooting analysis they are, square root, cube root, fourth root and its up-to nth root. Here we are going to see about  root 3 value using the formula method and the example problems and practice problems related to the root.

Properties of Roots:

The product of two Variables in the roots which are in different variables in the root, it can be written with in a single root into as a product of those two rooted variables.
`"sqrt(x) xx` =`sqrt((x)(y))`     (By the First property)

The variable at the outside the root symbol, when it gets into root, it turns into a square of that value.
`(x)sqrt(y)` = `sqrt((x^2)(y))` ` (By the Second Property)`

In division, root of the whole division which can be written as a individual roots, as root in numerator with root separately and in denominator with root separately.
`sqrt((x)/(y))` = `sqrt(x)/sqrt(y)`        (By the Third property)

The number without the square value within the root as the result. It is because, When a perfect square value comes out of the root symbol.
`sqrt4` = 2     (By the Fourth Property)

Root 3 Value - Example Problems:

Root 3 value - Problem:

Calculate the root value 3

Solution:

The square root value for 3 is nearly equal value to the square values between 1 and 2, because 12 = 1  and 22 = 4

Step 1: Divide 3 by 1.

`3 / 1 ` = 3

Step 2: Take average for 3 and 1.

` (3 +1) / 2 ` = 2

Step 3: Divide 3  by 2

` 3 / 2 ` =  1.5

Step 4: Take average for the 1.5 and  2

`(1.5 + 2)/2 ` =  1.75

Step 5: Now check the result by taking square for that value.

1.75 2 =  3.0625

If the value is not equal to the requested value then repeat the steps  4, 3 and 5.

Step 6: Divide 3 by 1.75

`3/ 1.75` =  1.71428571

Step 7: Take average for 1.75 and  1.71428571

(1.71428571+ 1.75)/2 = 1.73214285

Step 8: Now check the result by taking square for that value.

1.73214285 2 =  3.00031885

This value is very closer to the given root value `sqrt (3) `

If the value is not equal to the requested value then repeat the steps  4, 3 and 5.Please express your views of this topic Operations on Rational Numbers by commenting on blog.

Root 3 Value - Practice Problems:

Problem 1:

Calculate the root value for 33

Answer: `sqrt 33` =  5.74456265

Problem 2:

Calculate the value for 333

Answer: `sqrt 333` =  18.2482876

Solving Initial Value Problem

Introduction for solving the initial value problem:

Initial value problem is the part of differential equation in mathematics. Initial value problem is an ordinary differential equation joint with particular value, called the initial condition, of the strange function at a known point in the area of the result. In this article we shall discuss  how to solve these initial value problems in the differential equation Having problem with Integral Calculator with Steps keep reading my upcoming posts, i will try to help you.

Example Problems on Solving the Initial Value Problem:

Example 1:

Solving the differential equation `(dy/dx)=4y` using initial condition `y(1)=2`

Solution:

Given differential equation is `(dy/dx)=4y` , initial condition is `y(1)=2`

Step 1:

4y = 0 when y = 0 Thus the constant y(x) = 0 is a particular solution.

Step 2:

Separate the variables on both sides,

`(dy)/y=4 dx`

Step 3:

Take the integral on both sides,

`ln y=4x+B`

step 4:

Take ln on both sides,

`y=e^B e^4x`

`y=Ce^4x`

where `C = eB` if y > 0, and `C = -eB` if y < 0.

Step 5:

General solution: `y(x)=Ce^(4x)` where C is any constant (C can be 0 from Step 1).

Step 6:

Substitute 1 for x and 2 for y

`2=Ce^4,`

`C=2e^(-4)`

The final solution is `y(x)=2e^(-4) e^(4x) =2e^(-(4-4x))`

Example 2:

Solve the initial value problem

Solution:

`y^'' -3y^' +2y=0` , `y(0)=1` ,  `y^' (0)=-1`

First we find the general solution to the differential equation. We can find the general solution to the homogeneous equation by guessing a solution in the form `y(t)=e^(rt)`

`y^'' -3y^' +2y=0`

`(r^2 -3r +2) e^(rt)`

`(r-1)(r-2) e^(rt)`

if we choose r = 1,2 then we find the general solution `y_c (t)=C_1 e^(-t)+C_2 e^(-2t)` for the homogeneous equation.

Now we find the solution corresponding to the initial conditions. We have the derivatives

`y_c (t)=C_1 e^(-t)+C_2 e^(-2t)`

`y(0)=C_1+C_2`

`y^'(t)=C_1 e^(-t)+C_2 e^(-2t)`

`y^'(0)=C_1-C_2`

We see that `C_1+C_2=1 and C_1-C_2=-1`

Solving the above equations we get `C_1=0` and `C_2=1`

Substitute these values we get the solution

`y^' (t)=e^(-t)`


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Practice Problems on Solving Initial Value Problems:

Problem 1:

Solve the initial value problem

`y^'' +3y^' +2=0` , `y(0)=1` , `y^' (0)=-1`

Answer: `y^(t)=e^(-t)`

Problem 2:

Solve the initial value problem

`y^'' +7y^' +12=0` , `y(0)=1` , `y^' (0)=-1`

Answer: `y^(t)=e^(-3t)`

Pi and Radius Math Examples

Introduction to pi and radius math examples:

Pi is the important symbol in math. Radius is the measure that is mainly used in circles.  Pi and radius are the terms that are mostly used in sphere, cone, and cylinder.  Because these figures are the combination of circle shapes.  In this article we have to discuss about “Pi and radius math examples”.

Radius: radius is the distance between the center  to the point on the circle shapes.

pi: it is constant term, the value is 3.14 or `22/7 `

Brief Description of Pi and Radius Math Examples
Example 1:

Find the area and circumference of the circle whose radius is equal to 7 cm.

Solution:

Given radius is equal to 7cm.

Area:

Area of the circle = p r^2 square units

Pi is represented by the symbol p = `(22)/(7)` or 3.14 and r= 7 cm

Therefore area =  `(22)/(7)` x 7 x 7

That is A= 154 square cm

Perimeter:

Perimeter of the circle = 2 p r square units

Therefore area =2x  `(22)/(7)` x 7

That is A= 44 cm

Example 2:

Find the area and volume of the sphere whose radius is equal to 7 cm.

Solution:

Given radius is equal to 7cm.

Area:

Area of the Sphere = 4 p r^2 square units

Pi is represented by the symbol p = `(22)/(7)` or 3.14 and r= 7 cm

Therefore area = 4 x  `(22)/(7)` x 7 x 7

That is A= 616 square cm

Volume:

Volume of the Sphere = `(4)/(3)` p r 3Cubic units

Therefore Volume =`(4)/(3)` x  `(22)/(7)` x 73

That is V= 1437.3 Cubic cm

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Example 3:

Find the area and volume of the cone whose radius is equal to 7 cm and slant height is equal to 10 cm and height is equal to 6 cm.

Solution:

Given radius is equal to 7 cm and slant height is equal to 10 cm and height is 6 cm.

Area:

Area of the Cone = p r (l + r) square units

Pi is represented by the symbol p =  `(22)/(7)` or 3.14 and r= 7 cm and l= 10 cm

Therefore area = `(22)/(7)` x 7 x (10+7)

That is A= 374 square cm

Volume:

Volume of the Cone = `(1)/(3)` p r^2h Cubic units

Therefore Volume =`(1)/(3)`x `(22)/(7)` x 72x6

That is V= 308 Cubic cm

These are the pi and radius problems in math.

Patterns in Math Fourth Grade

Introduction to patterns in math fourth grade:

A patterns, from the French patron, is a type of the theme of recurring events or objects, sometimes referred to as elements of a set. It can be a template or model which can be used with the generate things or parts of a thing. Let we see about the patterns in fourth grade problems like missing terms and sequences. Understanding formula for prime numbers is always challenging for me but thanks to all math help websites to help me out. (Source - Wikipedia )

Problems Based on Patterns in Math Fourth Grade

Example 1 in patterns in math fourth grade:

Find the missed out term in the given numbers 53,51, 49, ____, 45.
The above stated number is in descending order.

The known difference between the number   53-51 = 51 – 49 = 2.

Hence the given numbers are in descending order so we can subtract the numbers 2 from 49 means the answer is 47 .

Example 2 in math patterns in  fourth grade:

Find the missed term in the given numbers 63,61, 59, ____, 55.

The above stated number is in descending order.

The known difference between the number 63-61=61-59 = 2.

Hence the given numbers are in descending order so we can subtract the numbers 2 from 59 means the answer is 47


More Problems in Patterns for Math Fourth Grade

Example 1: 6 36  216  1296

Find the missing terms in the given patterns 6,36,216,1296, ____, _____
Given that:

6,36,216,1296, ____, _____

Solution:

Step 1:   The starting value of the pattern is 6

Step 2:   The next value of the pattern = 6* 6 = 36

Step 3:   The third value can be written as  36 * 6 = 216

Step 4:   The fourth value is 216 * 6 = 1296

Now the left out value can be written as

1296 * 6  = 7776

7776* 6 = 46656

So the correct pattern is 6,36,216,1296, 7776, 46656

The required solution is 6,36,216,1296, 7776, 46656.

Example 2: Find the missing terms in the given patterns  13,169,2197,28561, ____, _____
Given that:

13,169,2197,28561, ____, _____

Solution:

Step 1:   The starting value of the pattern is 13

Step 2:   The next value of the pattern = 13* 13 = 169

Step 3:   The third value can be written as  169 * 13 = 2197

Step 4:   The fourth value is 2197 * 13 = 28561

Now the left out value can be written as

28561 * 13 = 371293

371293* 13 = 4826809

So the correct pattern is  13,169,2197,28561,371293 ,4826809

The required solution is 13,169,2197,28561,371293 ,4826809

Substitute A with B

Introduction to substitute a with b:

Substitution is one of the methods used in math to find the value of unknown variable. When this process takes place some of the arithmetic operations like addition, subtraction, multiplication, division can be done. Finally the result will be a finite. Substituting a with b means the equation to find the value of a should have a variable b. Is this topic Linear Equation Solver hard for you? Watch out for my coming posts.

Example Problem – Substitute a with B.

Example 1 – Substitute a with b.

Find the value of a from the expressions 2a+10b = 18 and 5b = 20.

Solution:

The given expressions are 2a+10b = 18 and 5b = 20

To find the value of a it needs to calculate the value for b

5b = 20

Divide by 5 on both sides

b = 4.

Now plug in the value of b in the equation 2a+10b=18

2a+10(4) = 18

2a+40 = 18

Subtract 40 on both sides

2a+40 -40= 18-40

2a = -22

Divide by 2 on both sides

a = -11

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Two more Example Problem – Substitute a with B.

Example 2 – Substitute a with b:

Find the value of a from the expression 10a+4 = 14a+8b and 3b+4 = 10

Solution:

The given expressions are 10a+4 = 14a+8b and 3b+4 = 10

To find the value of a it needs to find the value of b from the equation 3b+4 = 10

Subtract 4 on both sides

3b+4 -4= 10-4

3b = 6

Divide by 3 on both sides

b = 2

Now plug in b = 2 in the equation 10a+4 = 14a+8b

10a+4 = 14a+8(2)

10a+4 = 14a +16

Subtract 14a on both sides

10a+4-14a = 14a +16-14a

-4a+4=16

Subtract 4 on both sides

-4a =12

Divide by -4 on both sides

-a = 3

a = -3.

Example 3 – Substitute a with b:

Find the value of a from the expressions 5a+10 = 12b and b = 17

Solution:

The given expressions are 5a+10 = 12b and b = 17

Now plug in b= 17 in the equation 5a+10 = 12b to find the value of a

5a+10 = 12(17)

5a+10 = 204

Subtract 10 on both sides

5a = 194

Divide by 5 on both sides

a = 38.8