Geometry Solving Ratios

Introduction to geometry solving ratios

Usually we come across with the ratio of two numbers, if two numbers are x and y then their ratio are given by x : y. Same thing can be extended to geometry of same kind like, between the lengths, areas, perimeters, surface areas, volume etc. These ratios in geometry will help us to know the enlargement or the shrinking of the sizes or shapes of a geometrical figure.

Now let us see few problems of this kind in the following examples.

Example Problems on Geometry Solving Ratios:
Ex 1: The ratio of two supplementary angles are in the ratio 3 : 7. Find the measure of each angle.

Soln: Let the angles be given by 3x : 7x.

Since their sum is supplementary, then 3x + 7x = 180.

Therefore, 10x = 180

`=>` x = 18.

Therefore the angles are 3(18) and 7(18). Hence, they are 540 and 1260.

Ex 2: If the lengths of the sides of a triangle are in the ratio 3 : 4 : 5. Find the area of the triangle if its perimeter is 144cm.

Soln: Given: Perimeter = 144 cm.

The ratios are 3x + 4x + 5x = 144

`=>` 12x = 144

x = 12.

Therefore, the sides of the triangle are given by 3(12), 4(12), 5(12). That is, 36cm, 48cm, 60cm.

To find the area, let us use the Heron’s formula.

Area = `sqrt[s(s-a)(s-b)(s-c)]` . The values of a, b, c can be taken as 36, 48, 60 respectively.

Here S = `[a+b+c]/2`

=`[36 + 48 + 60]/2`

= `144/2` = 72.

Therefore, the area is given by:

A = `sqrt(72(72-36)(72-48)(72-60))`

= 864 cm2.


Between, if you have problem on these topics Nets for Solid Figures, please browse expert math related websites for more help on Types of Sequences.

More Example Problem on Geometry Solving Ratios:

Ex 3: The total surface area of solid cylinder is 616 cm2. if the ratio between its curved surface area and its total surface area is 1 : 2, find volume of the cylinder. [Take `pi = 22/7` ]

Soln: Given: Total surface area = 2`pi` r(r + h)

= 616 cm2 -----------------(1)

Also, the ratios of curved surface area : total surface area = 1 : 2.

`=>` 2`pi` rh : 2`pi` r(r+h) = 1 : 2

`h/(r+h) = 1/2`

2h = h + r

`=>` h = r.

Therefore from (1), we get, 2`pi` r(r+r) = 616

4 `pi` r2 = 616

r2 = 49

r = h = 7.

Therefore, the volume of the cylinder = `pir^2h` = `22/7 xx 7^2 xx 7` = 1078 cm3.                                                

Perimeter of a Right Angle Triangle

Introduction about right angle triangle:

A right triangle or right-angled triangle is a triangle in which one angle is a right angle (that is, a 90 degree angle). The relation between the sides and angles of a right triangle is the basis for trigonometry. The length of boundary is called perimeter triangle. We can calculate the perimeter of right angle triangle by adding the all three sides. In this article we shall discus how to calculate the perimeter of right triangle.

Formula to Find the Perimeter of the Right Triangle:

Perimeter of triangle (P) = a + b + c units

a, b and c are the names of three sides.

Example Problems to Find the Perimeter of the Right Angle Triangle:

1.      Find the perimeter of right angle triangle whose sides are 3 cm, 4 cm and 5 cm.

Solution:

Step 1:  Given:

Side a = 3 cm

Side b = 4 cm

Side c = 5 cm

Step 2:       Perimeter of right angle triangle (P) = a + b + c units

= 3+ 4 + 5

Step 3:        Perimeter of right angle triangle (P) = 12 cm

2.      Find the perimeter of right angle triangle whose sides are 5 cm, 12 cm and 13 cm.

Solution:

Step 1:  Given:

Side a = 5 cm

Side b = 12 cm

Side c = 13 cm

Step 2:     Perimeter of right angle triangle (P) = a + b + c units

= 5 + 12 + 13

Step 3:    Perimeter of right angle triangle (P) = 30 cm

3.      Find the perimeter of right angle triangle whose sides are 9 cm, 12 cm and 15 cm.

Solution:

Step 1:  Given:

Side a = 9 cm

Side b = 12 cm

Side c = 15 cm

Step 2:    Perimeter of right angle triangle (P) = a + b + c units

= 9 + 12 + 15

Step 3:  Perimeter of right angle triangle (P) = 36 cm

I am planning to write more post on What is the Area of a Triangle, Exterior Angle Theorem. Keep checking my blog.

Practice problems to find the perimeter of the right angle triangle:

4.      Find the perimeter of right angle triangle whose sides are 6 cm, 8 cm and 10 cm.

Answer: perimeter of right angle triangle (P) = 24 cm

5.      Find the perimeter of right angle triangle whose sides are 4 cm, 2 cm and 4.47 cm.

Answer: perimeter of right angle triangle (P) = 10.47 cm

6.      Find the perimeter of right angle triangle whose sides are 5 cm, 10 cm and 11.18 cm.

Answer: perimeter of right angle triangle (P) = 26.18 cm

Odd Number Series

Introduction to odd number series:

Odd Number definition: A number ‘x’ which is not divided exactly by 2 then ‘x’ is odd number. For example 3 is odd number which is not divided exactly by 2. If it is divided by 2 then ‘x’ is even number. In another way we can find the odd number easily. Whether the last digit of number ends with 1, 3, 5, 7 and 9 then we can say that as odd number.

For example:  87 are odd or even?

The number end with 7 so according to our definition 87 is odd.

We write the odd number series from 1 to 100 as

1, 3, 5, 7 ... up to 99.
1 + 3 + 5 + 7 + … + 99
To find the nth term of number series formula is

tn = a + (n – 1)d

To find the sum of an number series formula is

Sn = n/2 [ 2a + (n – 1)d ]

Or

Sn = n/2 [ a + l ]

Here ‘n’ is number of term

‘a’ is first term of series

‘d’ is difference between two consecutive terms

‘l’ is Last term of the series

Example: 1 + 3 + 5 + … find 13th term of this series?

Solution:

a = 1;

d = 2nd term – first term = 3 – 1 = 2

n = 13

tn = a + (n – 1)d

= 1 + (13 – 1)2

= 1 + (12)2

= 1 + 24

= 25

t13 = 25

Example Problem on Odd Number Series

Find Sum of first three odd numbers?
Solution:

Let us take first three odd number is 1, 3, and 5.

Sum of three odd number = 1 + 3 + 5

= 9

List out odd numbers greater than 2 and smaller than 20.
Solution:

Which is not divisible by 2 then the number is called odd number.

Odd numbers = 3, 5, 7, 9, 11, 13, 15, 17 and 19.

8 odd numbers available in between the greater than 2 and smaller than 20.

Prove that sum of an even number and odd number as odd number?
Proof:

Let us take ‘2x’ is Even number and ‘2y + 1’ Odd number.

Sum of these even and odd numbers:

2x + (2y +1) = 2(x + y) + 1

Let N = x + y and we can write the sum as

2x + (2y +1) = 2N +1

Let,

x =1 then 2x = 2; y = 2 then 2y + 1 = 5;

2x + 2y + 1 = 2(1 + 2) +1

= 6 + 1

= 7 this is odd number.

Therefore “sum of an even and odd number is odd number”

Sum of three consecutive odd numbers are 21 then find their odd numbers?
Solution:

Let us take 2x + 1 is one odd number

Then consecutive three odd numbers = 2x + 1, 2x + 3, 2x + 5

Sum of these odd numbers = 21

2x + 1+ 2x + 3+ 2x + 5 = 21

6x + 9 = 21

6x = 12

x = 12 / 6

x = 2

Then    2x + 1 = 2 (2) + 1 = 5

2x + 3 = 2 (2) + 3 = 7

2x + 5 = 2 (2) + 5 = 9

The odd numbers are 5, 7, and 9.

3, 7, 11, … find the sum upto 15th term?
Solution:

a = 3; d = 7 – 3 = 4; n = 15

Formula is,

Sn = n/2 [ 2a + (n – 1)d ]

= 15/2 [2*3 + (15 – 1)*4]

= 15/2 [6 + 56]

= 15/2 [62]

S15 = 465

Between, if you have problem on these topics Even Number, please browse expert math related websites for more help on Convert Fraction to Percent.

Exercise on Odd Number Series

Practice problem on odd number series

The sum of two odd consecutive numbers is 16. Find their odd numbers?
Sum of two odd numbers is _______?
The sum of four odd numbers is 40. The three odd numbers is 9, 11 and 13 find the 4th odd number?
1 + 7 + 13 +…. Find the 20th term?
1 + 5 + 9 +….  Find sum up to 18th term?
Answer Key of odd number series

7, 9
Odd
7
115
630

Algebra Prime Numbers

Introduction to algebra prime number:

Prime numbers definition:

In arithmetic, a prime number is normal numerals that have accurately two separate normal numeral divisors 1 and itself. The prime number can be divided only during 1 and the same number. There are infinitely a lot of prime facts. The examples of first twenty-five prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97.

Example of Algebra Prime Numbers:

The examples of algebra prime numbers is given below that,

Example:

Locate the one to hundred prime numbers. 

Solution:

First one to ten prime numbers is given below that,

Here, 2, 3, 5, and 7.

Next leaven to twenty prime numbers is given below that,

Here, 11, 13, 17, and 19.

Next twenty-one to thirty prime numbers is given below that,

Here, 23 and 29.

Next thirty-one to forty prime numbers is given below that,

Here, 31 and 37.

Next forty-one to fifty prime numbers is given below that,

Here, 41, 43 and 47.

Next fifty-one to sixty prime numbers is given below that,

Here, 53 and 59.

Next sixty-one to seventy prime numbers is given below that,

Here, 61 and 67.

Next seventy-one to eighty prime numbers is given below that,

Here, 71, 73 and 79.

Next eighty-one to ninety prime numbers is given below that,

Here, 83 and 89.

Last ninety-one to hundred prime numbers is given below that,

Here, only having a one prime numbers. That is 97.

Between, if you have problem on these topics how to solve inequalities, please browse expert math related websites for more help on definite integral.

Another Example of Algebra Prime Numbers:

The another examples of algebra prime numbers is given below that,

Example:

Locate the prime numbers of following data. The data’s are 7, 9, 12, 13, 17, 21, 27, 29, 31, 32, 35, 37, 41, 45, 47 and 50.

Solution:

7 is a prime number.

9 is not a prime number.

12 is not a prime number.

13 is a prime number.

17 is a prime number.

21 is not a prime number.

27 is not a prime number.

29 is a prime number.

31 is a prime number.

32 is not a prime number.

35 is not a prime number.

37 is a prime number.

41 is a prime number.

45 is not a prime number.

47 is a prime number.

50 is not a prime number.

The Substitution Method Solver

Introduction to the substitution method solver:-

In algebra, a linear system of two equations can be solved by two methods. (1) Elimination method (2) substitution method.
If one equation is of the form ax + by +c = 0 and the other equation is of the form px + qy + d = 0 then we can use the substitution
methods by substituting x for y or y for x and solve the equations.
Step 1 : Take the equation ax + by = -c
ax   =  -by - c
x =  -by  - c
a     a
Step 2 : Substitute the value of x in the second equation px + qy +d=0 and find y.
Step 3 : Substitute the value of y in any one of the two given equation and solve for x
Alternatively you can substitute  the value of y in the step 1 equation also to find x
This is called  substitution method solver.

Solve Example by Substitution Methods:-

Here is a problem of linear equation which we are going to solve by substitution methods.
# Solve 2x + 3y = 8 and 5x + 2y = 9
Solution:-
Step 1 : Let us write the first equation 2x + 3y = 8
2x         = 8 - 3y
x          = ( 8 - 3y)/2
Step 2 : Let us substitute the x value in the equation  5x + 2y = 9
5( 8 - 3y)/2 + 2y = 9
( 40 - 15y)/2 + 2y = 9
20 - 7.5 y + 2y   = 9
20 - 5.5y           = 9
Step 3 : Let us subtract 20 from both sides    - 20                    - 20
______________________
- 5.5 y          = - 11
Step 4 : Let us divide both sides by  5.5 we get          -y          = -2
Step 5 : Cancelling negative sign on both sides  we get y = 2
Step 6 : Substituting the value of y = 2  for x = (8 - 3y) /2 we get  (8 - 3*2) / 2
= ( 8 - 6)/2
=  2/2
= 1
Answer  :  x = 1 and y = 2
Step 7 : Let us check our solution by substituting x = 1 and y = 2 in any one of the equation given in the problem
Let us check on 2x + 3y = 8
2(1) + 3(2) gives 2 + 6 = 8 which is the RHS
Hence our solution is correct.

Solving Problem on a Linear Equation by Substitution Method:-

Let us do another linear equation  problem by substitution method:
# Solve 3x + 2y = 7/4 and 2x + 3y  = 2
Solution:
Step 1 : Let us use the first equation to find the value of y
3x + 2y = 7/4
2y = 7/4 - 3x
y = 7/8 - 3x/2
Step 2 : Let us substitute the y value in the equation 2x + 3y = 2
2x + 3( 7  - 3x)  = 2
8      2
2x +  21  - 9x    = 2
8       2
2x - 4.5x       = 2  - 21
8
- 2.5x   = 16 - 21
8
- 2.5x  =  -5/8
- 2.5x  = - 0.625
Step 3 : Let us divide both sides by 2.5 we get        -x   =  - 0.25
x   = 0.25
x   = ¼
Step 4 : Let us substitute x = 1/4 in equation       y = 7  - 3x
8    2
y = 7 - 3*¼
8   2
y = 7 - 3
8   8
y = 4/8
y = ½
Answer : The solution for this problem is x = ¼  and y = ½
Let us check our solution to see if it is correct
Step 1 : substituting x = ¼ and y = ½ in one of the equations say 2x + 3y we get 2*¼ + 3 *½ = 1  + 3 =  2 RHS
2      2
Hence our solution is correct.

Ordered Number Pairs

Introduction to Ordered Number Pairs:

The numbers can be written in certain order inside the parenthesis like (x, y). Ordered pair is one of the branches of algebra in math. A number pair (x, y) is located in any point or an object in a plane is called as an ordered number pair. Ordered pairs consists of two terms such as x and y. Here, the position of the terms cannot be changed unless it is equal. Ordered pairs are otherwise called as co-ordinates.

Finding Ordered Number Pairs

Depending upon the given number of equations, the procedure should be changed to learn the difference of solving the ordered pairs.

1. If only one equation is given, we have to convert the equation in below format like,

y = a1x +c,

Substitute the values for x and find the value of y.

Here, x is said to be substituted value and y is an obtained value.

Therefore, many values can be obtained for different x values.

2. If it is more than two equations is given,

Solve the equations and find out the values of x and y, then (x, y) is an ordered pair.

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Worked Examples to Ordered Number Pairs

Problem 1:

Find the ordered number pairs of the equation y - 3x = 2

Solution:

Step 1:

Let us write the given equation as

y  - 3x = 2.

Step 2:

Change the above equation into y = 2 + 3x,

Substitute x = 0

y = 2 + 3(0),

y = 2

Therefore, the ordered pair (x, y) is (0, 1).

Step 3:


Substitute x = 1 in the equation, we get,

y = 2 + 3(1),

y = 2 + 3 = 5

Therefore, the ordered pair (x, y) is (1, 5).

Step 4:

Substitute x = 2 in the equation, we get,

y = 2 + 3(2)

y = 2 + 6 = 8

Therefore, the ordered pair (x, y) is (2, 8).

Step 5:

Substitute x = 3 in the equation we get,

y = 2 + 3(3)

y = 2 + 9 = 11

Therefore, the ordered pair (x, y) is (3, 11).

Step 6:

Substitute x = 4 in the equation, we get,

y = 2 + 3(4),

y = 2 + 12

Therefore the ordered pair (x, y) is (4, 14).

Problem 2:

Find the ordered pair of the equations

2x + 3y = 6

3x + 5y = 4

Solution:

Step 1:

Let's start with 2x + 3y = 6 for the variable x.

Step 2:

Move the 3y to the right hand side by subtracting 3y on both sides, like this,

2x = 6 – 3y

Step 3:

To isolate x value, we have to divide both sides of the equation by 2 nearer to the x variable on the left hand side. The last step is to divide both sides of the equation by 2, we get,

x =6/2-(3/2)y

x = 3 – (3/2)y

Step 4:

Next, let's solve 3x + 5y = 4 for the variable y.

Step 5:

Move the 3x to the right hand side by subtracting 3x on both sides, like this,

5y = 4 – 3x

Step 6:

To isolate the y value, we have to divide both sides of the equation by 5 nearer to the y variable on left hand side of the equation. The last step is to divide both sides of the equation by 6, we get,

y = 4/5 - (3/5)x

Step 7:

Now, plug the x result, x = 3 – (3/2)y in y, we get,

This gives y = 4/5 - (3/5)(3 – (3/2)y)

4/5 – 9/5 – (9/10)y

-9y/10 = 9/5 – 4/5

-9y/10 = 5/5

-9y/10 = 1

-9y = 10

y = - 10/9

Step 8:

Plug the value of y in the x equation we get,

x = 3 - (3/2)(- 10/9)

x = 3 + 30/18

x = 3 + 15/9

x = 3 + 5/3

x = (9 + 5)/3

x = 14/3

Therefore, the solution for ordered number pair is (14/3, -10/9).

Describe Two Laws of Exponents

Introduction to describe two laws of exponents:
Exponents are defined as one of the basis for mathematics. Exponents are also called as the power of the given functions. Simply it is defined as the number of times, the number is multiplied. For example 92 are called the exponent function. In this example, the number 2 is called the exponent and the number 9 is called as the base number.

Explanation for Describe Two Laws of Exponent

The explanations for describe two laws of exponents are given below,

There are many laws of exponents. They are as follows,

Describe two important laws of exponents:

If the multiplication of two exponents is given means, then we can add the power terms and thus produces the answers.
Law: Pa  `xx`   Pb  =  Pa+b

If the division of two exponents is given means, then we can bring the denominator function to the numerator and we can subtract the power terms and thus produces the answers.
Law: `p^a/p^b` = pa-b

Other laws of exponents:

If the numbers are given in the multiplication format and if it has the power terms means, then we can separate the power terms for each numbers and thus produces the answers.
Law: ( p `xx` q )a = pa `xx` qa

If the numbers are given in the division format and if it has the power terms means, then we can separate the power terms for each numbers and thus produces the answers.
Law: `(p/q)^a`    = `p^a/q^a`

Example Problem for Describe Two Laws of Exponents

Problem 1: Solve the given exponents function, 33 `xx` 34

Solution:

Step 1: Write the given functions,

33 `xx` 34

Step 2: Add the power of the given exponents, we get,

33 `xx` 34 = 33+4 

Step 3: Now find the value for the obtained result, we get,

37  = 2187

Thus, this is the required solution by using the laws of exponents.

Problem 2: Solve the given exponents function, `(2^4/2^2)` 

Solution:

Step 1: Write the given functions,

`(2^4/2^2)`

Step 2: In this we have to bring the denominator term to the numerator, we get,

24 - 2

Step 3: Subtract the terms which we obtained.

24 - 2 = 22

Step 4: Now find the value for the obtained result, we get,

22 = 4

Thus, this is the required solution by using the laws of exponents.

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Practice Problem for Describe Two Laws of Exponents

Problem 1: Solve the given exponents function, 42 `xx`  42

Answer: 256

Problem 2: Solve the given exponents function, `3^3/3^2`

Answer: 3

Equivalent Fractions Number Line

Introduction to equivalent fractions number line:

Fractions:

A fraction is a number that can represent part of a whole. The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on. A much later development was the common or "vulgar" fractions which are still used today (½, ?, ¾, etc.) and which consist of a numerator and a denominator.

Equivalent fractions:

Two or more fractions has the same value is known as equivalent fractions.

In this article we are going to see about equivalent fractions number line and  some problems based on equivalent fractions number line.

Equivalent Fractions Number Line :

The equivalent fractions number line used to show the equivalent fractions in chart. The following figure shows the equivalent fractions number line.

Fig(i) Equivalent fractions number line

The above figure shows the equivalent fraction number line for 1/8 .

By using this number line, we get equivalent fractions of 1/8 is 2/16. Let us see some problems on finding equivalent fractions .

Between, if you have problem on these topics how to multiply 3 fractions, please browse expert math related websites for more help on how to divide and multiply fractions.

Problems Equivalent Fractions Number Line :

Problem 1:

Find 2 equivalent fractions to 3/5

Solution:

Given , fraction 3/5

We need to find the equivalent fraction to `3/5`

Multiply 3/5 by 2 on both numerator and denominator,

`3/5` = `( 3 * 2 ) / (5 * 2)`

= `6/ 10`

Equivalent fraction to `3/5 ` = `6 / 10` .

Multiply `3/5` by 5 on both numerator and denominator,

`3/5` = `( 3 * 5) / ( 5 * 5 )`

= `15/ 25`

Equivalent fraction to `3/5` = `6 /10` = `15 /25`

Answer: 2 equivalent fractions of `3 / 5` is `6/ 10` , `15 /25`

Problem 2:

Find the 3 equivalent fractions of `2/15` .

Solution:

Given, The fraction `2/15` .

We need to find the equivalent fraction to `2/15`

Multiply `2/15` by 2 on both numerator and denominator,

`2/15` = `( 2 * 2 ) / (15 * 2)`

= `4/ 30`

Equivalent fraction to `2/15 ` = `4 / 30` .

Multiply  2/15 by 5 on both numerator and denominator,

`2/15 ` = `( 2*5) / ( 15*5 )`

= ` 10/ 75`

Equivalent fraction to `2/15` = `4 / 30` = `10/75`

Multiply ` 2/15` by 3 on both numerator and denominator,

`2/15` = `( 2* 3 ) / (15 * 3)`

= `6/ 45`

Answer: 3 equivalent fraction of `2/ 15` is `4/ 30` , `10/ 75` , `6 /45`

Problem 3:

Find the 3 equivalent fractions of `8/12` .

Solution:

Given, The fraction `8/12` .

We need to find the equivalent fraction to `8/12`

Multiply `8/12` by 2 on both numerator and denominator,

`8/12` = `( 8 * 2 ) / (12 * 2)`

= `16/ 24`

Equivalent fraction to `8/12 ` = `16 / 24` .

Multiply  8/12 by 5 on both numerator and denominator,

`8/12 ` = `( 8*5) / ( 12*5 )`

= ` 40/ 60`

Equivalen fraction to `8/12` = `16 / 24` = `40/60`

For another equivalent fraction, Multiply ` 8/12` by 3 on both numerator and denominator,

`8/12` = `( 8* 3 ) / (12 * 3)`

= `24/ 36`

Answer: 3 equivalent fraction of `8/ 12` is `16/ 24` , `40/ 60` , `24 /36`

Interest Expense Equation

Introduction :

In this article we will discuss about the interest expensive equation. The “  Interest expense equation ” gives the period of the interest rate. Interest rate is not always charged yearly. It will be charged every 6 months, Monthly, some times Daily!. The “ Interest rate” contains two types,  simple interest rate and Second compound interest rate. Here compound interest only denoted by the interest expense equation because compound interest gives the expensive interest amount. In summary, to calculate the  simple interest rate, just find out the interest rate for one period and then multifly the number of periods. To calculate the compound  interest rate or interest expense equation, first find out the interest rate for the first period, add its total  and then calculate the interest rate for the next period, and so on..,

Formula

Formula for interest expense Equation:

Formula for simple interest:

simple interest SI = pnr

where 

p  =  principal amount
n  =  period
r  =  interest rate


Formula for compound interest or interest expense equation:

compound interest  CI   =   P (1  +`r/100` )n

Where

p  =  principal amount
n  =  period
r  =  interest rate

I am planning to write more post on solving linear equations with 3 variables, what are some prime numbers. Keep checking my blog.

Examples

Example For interest expense equation:

Ricky  paid $2000 at 10%  in finance then after 4 years what amount he will get? .   Finance using simple interest and compound interest method. But ricky expect interest expense method.

Solution:

Here,

Principal amount  p = $2000

Period        n  =  4 years

Interest  rate        r =  10%

Simple interest SI = pnr

SI =2000*4*10/100 

=  800

Simple interest    =  800

Compound interest or interest expense CI   =   P (1  +` r/100`   )n

CI = 2000( 1+`10/100` )4

=  2000(`11/100` )4

=  2000( `11/10*11/10*11/10*11/10` )

=  2928

CI   =  $2928     

Total amount in simple interest  = 2000+800

Simple interest amount is   =$2800

But the compound interest amout is $2928 .

Interest expensively gives the compound interest.

So after 4 years Ricky got $2928.