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# Mathematics | Closure of Relations and Equivalence Relations
Prerequisite : Introduction to Relations, Representation of Relations
Combining Relations :
As we know that relations are just sets of ordered pairs, so all set operations apply to them as well. Two relations can be combined in several ways such as –
• Union – consists of all ordered pairs from both relations. Duplicate ordered pairs removed from Union.
• Intersection – consists of ordered pairs which are in both relations.
• Difference – consists of all ordered pairs only in , but not in .
• Symmetric Difference – consists of all ordered pairs which are either in or but not both.
There is another way two relations can be combined that is analogous to the composition of functions.
Composition – Let be a relation from to and be a relation from to , then the composite of and , denoted by , is the relation consisting of ordered pairs where and for which there exists an element such that and .
• Example – What is the composite of the relations and where is a relation from to with and is a relation from to with ?
• Solution – By computing all ordered pairs where the first element belongs to and the second element belongs to , we get –
Composition of Relation on itself :
A relation can be composed with itself to obtain a degree of separation between the elements of the set on which is defined.
Let be a relation on the set .
The powers where are defined recursively by -
and .
Theorem – Let be a relation on set A, represented by a di-graph. There is a path of length , where is a positive integer, from to if and only if .
Important Note : A relation on set is transitive if and only if for
Closure of Relations :
Consider a relation on set . may or may not have a property , such as reflexivity, symmetry, or transitivity.
If there is a relation with property containing such that is the subset
of every relation with property containing , then is called the closure of
with respect to .
We can obtain closures of relations with respect to property in the following ways –
1. Reflexive Closure – is the diagonal relation on set . The reflexive closure of relation on set is .
2. Symmetric Closure – Let be a relation on set , and let be the inverse of . The symmetric closure of relation on set is .
3. Transitive Closure – Let be a relation on set . The connectivity relation is defined as – . The transitive closure of is .
Example – Let be a relation on set with . Find the reflexive, symmetric, and transitive closure of R.
Solution –
For the given set, . So the reflexive closure of is
For the symmetric closure we need the inverse of , which is
.
The symmetric closure of is-
For the transitive closure, we need to find .
we need to find until . We stop when this condition is achieved since finding higher powers of would be the same.
Since, we stop the process.
Transitive closure,
Equivalence Relations :
Let be a relation on set . If is reflexive, symmetric, and transitive then it is said to be a equivalence relation.
Consequently, two elements and related by an equivalence relation are said to be equivalent.
Example – Show that the relation
is an equivalence relation. is the congruence modulo function. It is true if and only if divides .
Solution – To show that the relation is an equivalence relation we must prove that the relation is reflexive, symmetric and transitive.
1. Reflexive – For any element , is divisible by .
. So, congruence modulo is reflexive.
2. Symmetric – For any two elements and , if or i.e. is divisible by , then is also divisible by .
. So Congruence Modulo is symmetric.
3. Transitive – For any three elements , , and if then-
. So, is transitive.
4. Since the relation is reflexive, symmetric, and transitive, we conclude that is an equivalence relation.
Equivalence Classes :
Let be an equivalence relation on set .
We know that if then and are said to be equivalent with respect to .
The set of all elements that are related to an element of is called the
equivalence class of . It is denoted by or simply if there is only one
relation to consider.
Formally,
Any element is said to be the representative of .
Important Note : All the equivalence classes of a Relation on set are either equal or disjoint and their union gives the set .
The equivalence classes are also called partitions since they are disjoint and their union gives the set on which the relation is defined
• Example : What are the equivalence classes of the relation Congruence Modulo ?
• Solution : Let and be two numbers such that . This means that the remainder obtained by dividing and with is the same.
Possible values for the remainder-
Therefore, there are equivalence classes –
GATE CS Corner Questions
Practicing the following questions will help you test your knowledge. All questions have been asked in GATE in previous years or in GATE Mock Tests. It is highly recommended that you practice them.
References –
Composition of Relations – Wikipedia
Discrete Mathematics and its Applications, by Kenneth H Rosen
This article is contributed by Chirag Manwani. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
# GATE CS Corner Company Wise Coding Practice
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# 7(h+3)=6(h-3)
How exactly do you work this out?
7(h + 3) = 6(h - 3)
Starting on the left hand side of the equation, distribute the 7 by multiplying it by each term inside the parenthesis (h + 3):
7(h + 3) = 7·h + 7·3 = 7h + 21
Looking at the right hand side of the equation, distribute the 6 by multiplying it by each term inside the parenthesis (h - 3):
6(h - 3) = 6·h - 6·3 = 6h - 18
Now we arrive at the following:
7(h + 3) = 6(h - 3)
7·h + 7·3 = 6·h - 6·3
7h + 21 = 6h - 18
The goal is to solve for the unknown variable, h. To do so, we want to isolate it to one side of the equation by moving all term that contain this variable to one side of the equation and moving all other terms to the other side of the equation, combining like terms as we go.
With this, we first subtract 6h from both sides of the equation and then subtract 21 from both sides of the equation:
7h + 21 - 6h = 6h - 18 - 6h
7h - 6h + 21 = 6h - 6h - 18
1h + 21 = 0 - 18
h + 21 = - 18
h + 21 - 21 = -18 - 21
h + 0 = -39
h = -39
Hi Annie -
Are you trying to solve for h here? If so, you first should simplify the equation:
7(h + 3) = 6(h - 3) distribute here
7h + 21 = 6h - 18
Now move all of the terms containing h to one side and everything else to the other side (remember, you can do whatever you want to the equation as long as whatever you do to one side you do to the other). So, to move all of the terms containing h to one side and everything else to the other, let's start by subtracting 6h from both sides:
(7h + 21) -6h = (6h - 18) - 6h
Rearranging a little and simplifying, we get:
7h - 6h + 21 = 6h - 6h - 18
h + 21 = -18
Now, let's subtract 21 from each side:
h + 21 - 21 = -18 - 21
h = -39
Which is the final answer. As a check, you can plug in this value for h in the original equation, and see if you get a true statement:
7(h + 3) = 6(h - 3)
7(-39 + 3) = 6(-39 - 3)
7(-36) = 6(-42)
-252 = -252
Since -252 does in fact equal -252, we know that we have the correct answer for h. If we had the wrong answer for h, we would have something like -13 = 156 or something, which is not true.
Hope this helps!
Mike T. |
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# ABCD is a rectangle if $\angle BPC = {124^o}$ Calculate: $\angle BAP{\text{ and }}\angle ADP$.
Last updated date: 13th Jun 2024
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Hint: In this particular question use the concept that the adjacent sides of the rectangle always makes a 90 degrees and the diagonals of the rectangle are always equal and also bisect each other so use these concepts to reach the solution of the question.
Given data:
$\angle BPC = {124^o}$................. (1)
Now as we see from the figure that APC is the diagonal of the rectangle which is a straight line so it makes a straight angle with the point p.
Now as we know that the straight angle = 180 degrees.
So, $\angle BPC + \angle BPA = {180^o}$
Now from equation (1) we have,
$\Rightarrow {124^o} + \angle BPA = {180^o}$
$\Rightarrow \angle BPA = {180^o} - {124^o} = {56^o}$............ (2)
Now as we know that in a rectangle the length of the diagonals are always equal and also bisect each other.
Therefore, PA = PB = PC = PB.
So triangles ABP, BCP, CDP, and DAP are an isosceles triangle.
Now as we know that in a isosceles triangle angles opposite to equal sides are always equal.
So in triangle ABP, as AP = BP, so $\angle PBA = \angle BAP$.............. (3)
Now as we know that in a triangle the sum of all angles are equal to 180 degrees.
So in triangle ABP we have,
$\Rightarrow \angle PBA + \angle BAP + \angle BPA = {180^o}$
Now from equation (2) and (3) we have,
$\Rightarrow \angle BAP + \angle BAP + {56^o} = {180^o}$
$\Rightarrow \angle BAP = \dfrac{{{{180}^o} - {{56}^o}}}{2}$
$\Rightarrow \angle BAP = \dfrac{{{{124}^o}}}{2} = {62^o}$
$\Rightarrow \angle PBA = \angle BAP = {62^o}$
Now as we know in a rectangle opposite sides are equal and parallel to each other.
Therefore, $\angle PBA = \angle PDC = {62^o}$............. (4) (alternate angles)
Now as we know that in a rectangle adjacent sides of the rectangle always makes a 90 degrees.
Therefore, $\angle ADC = {90^o}$
$\Rightarrow \angle ADP + \angle PDC = {90^o}$
Now from equation (4) we have,
$\Rightarrow \angle ADP + {62^o} = {90^o}$
$\Rightarrow \angle ADP = {90^o} - {62^o} = {28^o}$
So, the correct answer is “${28^o}$”.
Note: Whenever we face such types of questions the key concept we have to remember is that always recall all the properties of rectangle and isosceles triangle which is stated above and always recall that if a line join end points of a two parallel lines then the opposite angle create by this line is always equal and are called as alternate angles. |
S k i l l
i n
A L G E B R A
35
SIMULTANEOUS LINEAR EQUATIONS
Section 2:
Section 3:
THIS LESSON DEPENDS on Lesson 9: Linear equations.
Example 1. Solve simultaneously for x and y:
x + y = 10 x − y = 2
This means that we must find values of x and y that will solve both equations. We must find two numbers whose sum is 10 and whose difference is 2.
The two numbers, obviously, are 6 and 4:
6 + 4 = 10 6 − 4 = 2
Let us represent the solution as the ordered pair (6, 4).
Now, these two equations --
x + y = 10 x − y = 2
-- are linear equations (Lesson 33). Hence, the graph of each one is a straight line. Here are the two graphs:
The solution to the simultaneous equations is their point of intersection. Why? Because that coördinate pair solves both equations. (Lesson 33.) That point is the one and only point on both lines.
Example 2. Solve simultaneously for x and y.
2x + y = 4 x − y = −1
Solution. In this case, the solution is not obvious. Here is a general strategy for solving simultaneous equations:
When one pair of coefficients are negatives of one another,
add the equations vertically, and that unknown will cancel.
We will then have one equation in one unknown, which we can solve.
Upon adding those equations, the y's cancel:
2x + y = 4 x − y = −1 __________________________________ 3x = 3 x = 33 x = 1
To solve for y, the other unknown :
Substitute the value of x in one of the original equations.
2· 1 + y = 4 y = 4 − 2 y = 2
If we report the solution as an ordered pair, then the solution is (1, 2). Those are the coördinates of the point of intersection of the two lines.
This method of solving simultaneous equations is called the method of addition.
Example 3. Solve the same system of equations by the method of substitution.
1) 2x + y = 4 2) x − y = −1
Here is the method of substitution:
Solve one of the equations for one unknown in terms of the other.
Then, substitute that in the other equation.
That will yield one equation in one unknown, which we can solve.
Let us solve equation 1) for y:
1) y = 4 − 2x And now, substitute this for y in equation 2): 2) x − (4 − 2x) = −1 This equation has only the unknown x: x − 4 + 2x = −1 3x = −1 + 4 3x = 3 x = 1 To find y, substitute x = 1 in line 1): y = 4 − 2· 1 y = 2
Problem 1 . Solve this system of two equations in two unknowns.
Solve it by the method of addition. Solve it again by the method of substitution.
x − y = 2 2x + y = 10
Do the problem yourself first!
To solve for y:
Substitute x = 4 in either of the original equations; for example, in the bottom equation:
2· 4 + y = 10 y = 10 − 8 y = 2.
The method of substitution:
x − y = 2 2x + y = 10.
Solve the top equation for x:
x = 2 + 2 x = 4.
Problem 2.
a) What is the graph of two linear equations?
Two straight lines.
b) Where on the graph do we find the solution?
As their point of intersection.
Section 2: More examples. Cramer's Rule: The method of determinants
Next Lesson: Word problems that lead to simultaneous equations
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# 5.3: Probability Computations for General Normal Random Variables
### Learning Objectives
1. To learn what a standard normal random variable is.
2. To learn how to use Figure 12.2 "Cumulative Normal Probability" to compute probabilities related to a standard normal random variable.
### Definition
A standard normal random variable is a normally distributed random variable with mean μ = 0 and standard deviation σ = 1. It will always be denoted by the letter Z.
The density function for a standard normal random variable is shown in Figure 5.9 "Density Curve for a Standard Normal Random Variable".
Figure 5.9 Density Curve for a Standard Normal Random Variable
To compute probabilities for Z we will not work with its density function directly but instead read probabilities out of Figure 12.2 "Cumulative Normal Probability" in Chapter 12 "Appendix". The tables are tables of cumulative probabilities; their entries are probabilities of the form P(Z<z). The use of the tables will be explained by the following series of examples.
### Example 4
Find the probabilities indicated, where as always Z denotes a standard normal random variable.
1. P(Z < 1.48).
2. P(Z< −0.25).
Solution:
1. Figure 5.10 "Computing Probabilities Using the Cumulative Table" shows how this probability is read directly from the table without any computation required. The digits in the ones and tenths places of 1.48, namely 1.4, are used to select the appropriate row of the table; the hundredths part of 1.48, namely 0.08, is used to select the appropriate column of the table. The four decimal place number in the interior of the table that lies in the intersection of the row and column selected, 0.9306, is the probability sought: P(Z<1.48)=0.9306.
Figure 5.10 Computing Probabilities Using the Cumulative Table
1. The minus sign in −0.25 makes no difference in the procedure; the table is used in exactly the same way as in part (a): the probability sought is the number that is in the intersection of the row with heading −0.2 and the column with heading 0.05, the number 0.4013. Thus P(Z < −0.25) = 0.4013.
### Example 5
Find the probabilities indicated.
1. P(Z > 1.60).
2. P(Z > −1.02).
Solution:
1. Because the events Z > 1.60 and Z ≤ 1.60 are complements, the Probability Rule for Complements implies that
P(Z>1.60)=1P(Z1.60)
Since inclusion of the endpoint makes no difference for the continuous random variable Z, P(Z1.60)=P(Z<1.60), which we know how to find from the table. The number in the row with heading 1.6 and in the column with heading 0.00 is 0.9452. Thus P(Z<1.60)=0.9452 so
P(Z>1.60)=1P(Z1.60)=10.9452=0.0548
Figure 5.11 "Computing a Probability for a Right Half-Line" illustrates the ideas geometrically. Since the total area under the curve is 1 and the area of the region to the left of 1.60 is (from the table) 0.9452, the area of the region to the right of 1.60 must be 10.9452=0.0548.
Figure 5.11 Computing a Probability for a Right Half-Line
1. The minus sign in −1.02 makes no difference in the procedure; the table is used in exactly the same way as in part (a). The number in the intersection of the row with heading −1.0 and the column with heading 0.02 is 0.1539. This means that P(Z<1.02)=P(Z1.02)=0.1539, hence
P(Z>1.02)=1P(Z1.02)=10.1539=0.8461
### Example 6
Find the probabilities indicated.
1. P(0.5<Z<1.57).
2. P(2.55<Z<0.09).
Solution:
1. Figure 5.12 "Computing a Probability for an Interval of Finite Length" illustrates the ideas involved for intervals of this type. First look up the areas in the table that correspond to the numbers 0.5 (which we think of as 0.50 to use the table) and 1.57. We obtain 0.6915 and 0.9418, respectively. From the figure it is apparent that we must take the difference of these two numbers to obtain the probability desired. In symbols,
P(0.5<Z<1.57)=P(Z<1.57)P(Z<0.50)=0.94180.6915=0.2503
Figure 5.12 Computing a Probability for an Interval of Finite Length
1. The procedure for finding the probability that Z takes a value in a finite interval whose endpoints have opposite signs is exactly the same procedure used in part (a), and is illustrated in Figure 5.13 "Computing a Probability for an Interval of Finite Length". In symbols the computation is
P(2.55<Z<0.09)==P(Z<0.09)P(Z<2.55)0.53590.0054=0.5305
Figure 5.13 Computing a Probability for an Interval of Finite Length
The next example shows what to do if the value of Z that we want to look up in the table is not present there.
### Example 7
Find the probabilities indicated.
1. P(1.13<Z<4.16).
2. P(5.22<Z<2.15).
Solution:
1. We attempt to compute the probability exactly as in Note 5.20 "Example 6" by looking up the numbers 1.13 and 4.16 in the table. We obtain the value 0.8708 for the area of the region under the density curve to left of 1.13 without any problem, but when we go to look up the number 4.16 in the table, it is not there. We can see from the last row of numbers in the table that the area to the left of 4.16 must be so close to 1 that to four decimal places it rounds to 1.0000. Therefore
P(1.13<Z<4.16)=1.00000.8708=0.1292
2. Similarly, here we can read directly from the table that the area under the density curve and to the left of 2.15 is 0.9842, but −5.22 is too far to the left on the number line to be in the table. We can see from the first line of the table that the area to the left of −5.22 must be so close to 0 that to four decimal places it rounds to 0.0000. Therefore
P(5.22<Z<2.15)=0.98420.0000=0.9842
The final example of this section explains the origin of the proportions given in the Empirical Rule.
### Example 8
Find the probabilities indicated.
1. P(1<Z<1).
2. P(2<Z<2).
3. P(3<Z<3).
Solution:
1. Using the table as was done in Note 5.20 "Example 6"(b) we obtain
P(1<Z<1)=0.84130.1587=0.6826
Since Z has mean 0 and standard deviation 1, for Z to take a value between −1 and 1 means that Z takes a value that is within one standard deviation of the mean. Our computation shows that the probability that this happens is about 0.68, the proportion given by the Empirical Rule for histograms that are mound shaped and symmetrical, like the bell curve.
2. Using the table in the same way,
P(2<Z<2)=0.97720.0228=0.9544
This corresponds to the proportion 0.95 for data within two standard deviations of the mean.
3. Similarly,
P(3<Z<3)=0.99870.0013=0.9974
which corresponds to the proportion 0.997 for data within three standard deviations of the mean.
### Key Takeaways
• A standard normal random variable Z is a normally distributed random variable with mean μ = 0 and standard deviation σ = 1.
• Probabilities for a standard normal random variable are computed using Figure 12.2 "Cumulative Normal Probability".
### Exercises
1. Use Figure 12.2 "Cumulative Normal Probability" to find the probability indicated.
1. P(Z < −1.72)
2. P(Z < 2.05)
3. P(Z < 0)
4. P(Z > −2.11)
5. P(Z > 1.63)
6. P(Z > 2.36)
2. Use Figure 12.2 "Cumulative Normal Probability" to find the probability indicated.
1. P(Z < −1.17)
2. P(Z < −0.05)
3. P(Z < 0.66)
4. P(Z > −2.43)
5. P(Z > −1.00)
6. P(Z > 2.19)
3. Use Figure 12.2 "Cumulative Normal Probability" to find the probability indicated.
1. P(−2.15 < Z < −1.09)
2. P(−0.93 < Z < 0.55)
3. P(0.68 < Z < 2.11)
4. Use Figure 12.2 "Cumulative Normal Probability" to find the probability indicated.
1. P(−1.99 < Z < −1.03)
2. P(−0.87 < Z < 1.58)
3. P(0.33 < Z < 0.96)
5. Use Figure 12.2 "Cumulative Normal Probability" to find the probability indicated.
1. P(−4.22 < Z < −1.39)
2. P(−1.37 < Z < 5.11)
3. P(Z < −4.31)
4. P(Z < 5.02)
6. Use Figure 12.2 "Cumulative Normal Probability" to find the probability indicated.
1. P(Z > −5.31)
2. P(−4.08 < Z < 0.58)
3. P(Z < −6.16)
4. P(−0.51 < Z < 5.63)
7. Use Figure 12.2 "Cumulative Normal Probability" to find the first probability listed. Find the second probability without referring to the table, but using the symmetry of the standard normal density curve instead. Sketch the density curve with relevant regions shaded to illustrate the computation.
1. P(Z < −1.08), P(Z > 1.08)
2. P(Z < −0.36), P(Z > 0.36)
3. P(Z < 1.25), P(Z > −1.25)
4. P(Z < 2.03), P(Z > −2.03)
8. Use Figure 12.2 "Cumulative Normal Probability" to find the first probability listed. Find the second probability without referring to the table, but using the symmetry of the standard normal density curve instead. Sketch the density curve with relevant regions shaded to illustrate the computation.
1. P(Z < −2.11), P(Z > 2.11)
2. P(Z < −0.88), P(Z > 0.88)
3. P(Z < 2.44), P(Z > −2.44)
4. P(Z < 3.07), P(Z > −3.07)
9. The probability that a standard normal random variable Z takes a value in the union of intervals (−∞, −a] ∪ [a, ∞), which arises in applications, will be denoted P(Z ≤ −a or Za). Use Figure 12.2 "Cumulative Normal Probability" to find the following probabilities of this type. Sketch the density curve with relevant regions shaded to illustrate the computation. Because of the symmetry of the standard normal density curve you need to use Figure 12.2 "Cumulative Normal Probability" only one time for each part.
1. P(Z < −1.29 or Z > 1.29)
2. P(Z < −2.33 or Z > 2.33)
3. P(Z < −1.96 or Z > 1.96)
4. P(Z < −3.09 or Z > 3.09)
10. The probability that a standard normal random variable Z takes a value in the union of intervals (−∞, −a] ∪ [a, ∞), which arises in applications, will be denoted P(Z ≤ −a or Za). Use Figure 12.2 "Cumulative Normal Probability" to find the following probabilities of this type. Sketch the density curve with relevant regions shaded to illustrate the computation. Because of the symmetry of the standard normal density curve you need to use Figure 12.2 "Cumulative Normal Probability" only one time for each part.
1. P(Z < −2.58 or Z > 2.58)
2. P(Z < −2.81 or Z > 2.81)
3. P(Z < −1.65 or Z > 1.65)
4. P(Z < −2.43 or Z > 2.43)
1. 0.0427
2. 0.9798
3. 0.5
4. 0.9826
5. 0.0516
6. 0.0091
1.
1. 0.1221
2. 0.5326
3. 0.2309
2.
1. 0.0823
2. 0.9147
3. 0.0000
4. 1.0000
3.
1. 0.1401, 0.1401
2. 0.3594, 0.3594
3. 0.8944, 0.8944
4. 0.9788, 0.9788
4.
1. 0.1970
2. 0.01980
3. 0.0500
4. 0.0020 |
# McGraw Hill My Math Grade 3 Chapter 2 Lesson 2 Answer Key Patterns in the Addition Table
All the solutions provided in McGraw Hill Math Grade 3 Answer Key PDF Chapter 2 Lesson 2 Patterns in the Addition Table will give you a clear idea of the concepts.
## McGraw-Hill My Math Grade 3 Answer Key Chapter 2 Lesson 2 Patterns in the Addition Table
Study the addition table for number patterns. Look for sets of numbers that follow a certain order.
Math in My World
Example 1
Danny colored a pattern of squares from left to right on the downward diagonal in yellow. Describe the pattern.
Even Numbers
Finish Danny’s pattern of even numbers. Color the squares.
0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 There is a pattern of add 2.
When you add 2 to an even number, the sum is an even number.
0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 There is a pattern of add 2.
When you add 2 to an even number, the sum is an even number.
Odd Numbers
Start with the green square. Color the pattern of odd numbers on the downward diagonal in green. Write the numbers.
1, ___, __, ____, ____, ____, ___, ___, ___, ____
There is a pattern of add ___. When you add ____ to an odd number, the sum is an ___ number.
1,3,5,7,9,11,13,15,17,19.
There is a pattern of add 2. When you add 2 to an odd number, the sum is an odd number.
Example 2
1. What pattern of numbers do you see in the diagonal of yellow boxes?
2. Look at the circled sum. Follow left and above to the circled addends.
3. Draw a triangle around the sum in the addition table that has the same addends. Follow left and above to its addends. Complete the number sentences.
3 + 4 = 7
4 + 3 = 7
The two number sentences are an example of the Commutative property.
Guided Practice
Describe Danny’s new pattern in the addition table below.
Question 1.
When ____ is added to a number, the sum is that number.
0
Explanation:
When 0 is added to a number, the sum is that number.
Question 2.
This is an example of the ____ Property of Addition.
Identity
Explanation:
Identity property of addition states that if you add zero to a number, the sum is the same as the given number.
Talk Math
How do you find patterns in numbers?
Independent Practice
Question 3.
Shade a diagonal of numbers blue that show the sums equal to 8.
Explanation:
diagonal of numbers blue that show the sums equal to 8 are
0 + 8 = 8, 1 + 7 = 8, 2 +6 = 8, 3 + 5 = 8, 4 + 4 = 8, 5 + 3 = 8, 6 + 2 = 8, 7 + 1 = 8, 8 + 0 = 8
Question 4.
Shade a diagonal of numbers green that show the sums equal to 5.
Explanation:
The diagonal of numbers green that show the sums equal to 5 are
0 + 5 = 5, 1 + 4 = 5, 2 + 3 = 5, 3 + 2 =5, 4 + 1 = 5, 5 +0 = 5
Question 5.
Shade a row of numbers yellow that represent sums with one addend of 4.
Explanation:
row of numbers yellow that represent sums with one addend of 4 are
4 +0 = 4, 4 + 1= 5, 4 + 2 = 6, 4 + 3 = 7 , 4 + 4 = 8, 4 + 5 = 9, 4 +6 = 10
Question 6.
Shade a column of numbers pink that represent sums with one addend of 6.
Explanation:
a column of numbers pink that represent sums with one addend of 6 are
6 + 0= 6, 6 + 1 =7, 6 + 2= 8 , 6+ 3 = 9 ,6 + 4 = 10
Question 7.
Shade two squares purple that each represent the sum of 3 and 9. What property does this show?
Explanation:
The sum of 3 and 9 is 12
3 + 9 = 12 , 9 + 3 = 12
This is commutative property of addition
Question 8.
Circle two squares that each represent a sum of 0 and 10. What two properties does this show?
Explanation:
The sum of 10 and 0 = 10
10 + 0= 10 this is identity property
10 + 0 = 10 , 0 + 10 = 10 this is commutative property
Question 9.
Shade two addends red that have a sum of 8. Complete the number sentence. Write the greater addend first.
____ + ___ = 8
Explanation:
two addends red that have a sum of 8 is 7 and 1
____ + ____ = 8
Explanation:
The other two addends are 1 and 7.
1 + 7 = 8, 7 + 1 = 8 this shows commutative property of addition.
Problem Solving
Question 10.
Mathematical PRACTICE 5 Use Math Tools Mario stacked 8 boxes. She had no more boxes to stack. Find the total number of boxes stacked. Shade two squares that each represent the sum. Write two number sentences.
____________________
What two properties does this show?
____________________
____________________
Two properties are Commutative and identity property of addition.
Explanation:
Mario stacked 8 boxes.
She had no more boxes to stack.
total number of boxes stacked are 8
8 +0 = 8 , Identity
8 + 0 = 8 , 0 + 8 = 8 , Commutative property
Question 11.
Pedro ran 3 miles on Sunday and 2 miles on Monday. Find the total number of miles he ran. Shade two squares that each represent the sum. Write two number sentences.
____________________
What property does this show?
____________________
Commutative property
Explanation:
Pedro ran 3 miles on Sunday and 2 miles on Monday
the total number of miles he ran is 3 + 2 = 5 miles
3 + 2 = 5, 2 +3 = 5, Commutative
HOT Problems
Question 12.
Mathematical PRACTICE 4 Model Math Write a real-world problem for which you can use the addition table and the Commutative Property of Addition to solve. Then solve.
In real-world it is important to solve faster. Commutative property and addition table can help us solve an operation faster.
Explanation:
The commutative property tells us that the result of an addition or multiplication is always the same, regardless of the order of the elements with which it operates.
The order of the addends does not change the result, no matter the order in which you put the addends, you will always get the same result.
Question 13.
? Building on the Essential Question How can addition patterns help me add mentally?
Addition pattern help me to know what kind of number should result when finding the sum of two addends.
Explanation:
You can use place value to help add by increasing the digit in one place at a time.
### McGraw Hill My Math Grade 3 Chapter 2 Lesson 2 My Homework Answer Key
Practice
Question 1.
Shade a diagonal of odd numbers red.
Explanation:
Odd numbers are 1,3,5,7,9,11,13,15,17,19
Question 2.
Shade a diagonal of even numbers yellow.
Explanation:
Even numbers are 0,2,4,6,8,10,12,14,16,18,20
Question 3.
Circle two squares that each represent a sum of 3 and 4. This shows the Commutative Property of Addition.
3 + 4 = ___ and 4 + 3 = ____
Explanation:
3 + 4 = 7 and 4 + 3 = 7
Question 4.
Circle the two addends which make a sum of the shaded 12. Write the number sentence.
Explanation:
The sum of 8 + 4 = 12
The two addends are 8 and 4
Question 5.
Shade a diagonal of numbers green that show the sums equal to 9.
Explanation:
The sum of 1 + 3 + 5 = 9
Question 6.
Shade a row of numbers yellow that represents sums with one addend of 10.
Explanation:
10 + 0 = 10 as 0 is the one which shows the sum = 10 with one addend of 10
It shows identity property
Problem Solving
Question 7.
Mathematical PRACTICE 3 Justify Conclusions Jasmine had 11 friends over at her house. Every time the doorbell rang, 2 more friends arrived. The doorbell rang 3 times. How many friends did Jasmine have over altogether?
17
Explanation:
Jasmine had 11 friends over at her house
Every time the doorbell rang, 2 more friends arrived.
The doorbell rang 3 times.
so 2 + 2+ 2 = 6
sum of 11 and 6 = 17
Total number of friends did Jasmine have over altogether is 17
Question 8.
Steve colors the following sums on an addition table. If he continues the pattern, will the numbers continue to be even? Explain.
12, 14, 16, 18
Yes
Explanation:
When you add 2 to an even number, the sum is an even number.
As the above pattern is an even number pattern it continues to be even.
Test Practice
Question 9.
Danielle is saving for a bicycle. Her last 4 bank deposits are shown in the table. If the pattern continues, how much will her next bank deposit be?
A. $2 B.$16
C. $17 D.$19 |
### Learning Outcomes
• Use the zero product principle to solve quadratic equations that can be factored
Often the easiest method of solving a quadratic equation is by factoring. Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation. Note that we will not spend a lot of time explaining how to factor in this section. You may want to seek help if you do not feel confident about factoring.
Solving by factoring depends on the zero-product property which states that if $a\cdot b=0$, then $a=0$ or $b=0$, where a and b are real numbers or algebraic expressions.
The process of factoring a quadratic equation depends on the leading coefficient, whether it is 1 or another integer. We will look at both situations; but first, we want to confirm that the equation is written in standard form, $a{x}^{2}+bx+c=0$, where a, b, and c are real numbers and $a\ne 0$. The equation ${x}^{2}+x - 6=0$ is in standard form.
We can use the zero-product property to solve quadratic equations in which we first have to factor out the greatest common factor (GCF) and for equations that have special factoring formulas as well, such as the difference of squares, which we will see later in this section.
### The Zero-Product Property and Quadratic Equations
The zero-product property states
$\text{If }a\cdot b=0,\text{ then }a=0\text{ or }b=0$
where a and b are real numbers or algebraic expressions.
A quadratic equation is an equation containing a second-degree polynomial; for example,
$a{x}^{2}+bx+c=0$
where a, b, and c are real numbers, and if $a\ne 0$, it is in standard form.
In the quadratic equation ${x}^{2}+x - 6=0$, the leading coefficient, or the coefficient of ${x}^{2}$, is $1$. We have one method of factoring quadratic equations in this form.
### Reminder: Given a quadratic equation with the leading coefficient of 1, factor it
1. Find two numbers whose product equals c and whose sum equals b.
2. Use those numbers to write two factors of the form $\left(x+k\right)\text{ or }\left(x-k\right)$, where k is one of the numbers found in step $1$. Use the numbers exactly as they are. In other words, if the two numbers are $1$ and $-2$, the factors are $\left(x+1\right)\left(x - 2\right)$.
3. Solve using the zero-product property by setting each factor equal to zero and solving for the variable.
### Example
Factor and solve the equation: ${x}^{2}+x - 6=0$.
In the following video, we provide more examples of factoring to solve quadratic equations.
In our next example, we will solve a quadratic equation that is written as a difference of squares.
### Example
Solve the difference of squares equation using the zero-product property: ${x}^{2}-9=0$.
## Solving Quadratics with a Leading Coefficient $\ne1$
Recall that when the leading coefficient is not $1$, we factor a quadratic equation using a method called grouping, which requires four terms. With the equation in standard form, let us review the grouping procedures:
1. With the quadratic in standard form, $a{x}^{2}+bx+c=0$, multiply $a\cdot c$.
2. Find two numbers whose product equals $ac$ and whose sum equals $b$.
3. Rewrite the equation replacing the $bx$ term with two terms using the numbers found in step $2$ as coefficients of x.
4. Factor the first two terms and then factor the last two terms. The expressions in parentheses must be exactly the same to use grouping.
5. Factor out the expression in parentheses.
6. Set the expressions equal to zero and solve for the variable.
### Example
Use grouping to factor and solve the quadratic equation: $4{x}^{2}+15x+9=0$.
The following video contains another example of solving a quadratic equation using factoring by grouping.
Sometimes, we may be given an equation that does not look like a quadratic at first glance. In our next examples we will solve a cubic polynomial equation where the GCF of each term is x and can be factored. The result is a quadratic equation that we can solve.
### Example
Solve the equation by factoring: $-3{x}^{3}-5{x}^{2}-2x=0$.
In this last video example, we solve a quadratic equation with a leading coefficient of -1 using a shortcut method of factoring and the zero product principle.
### Summary
You can find the solutions, or roots, of quadratic equations by setting one side equal to zero, factoring the polynomial, and then applying the Zero Product Property. The Principle of Zero Products states that if $ab=0$, then either $a=0$ or $b=0$, or both a and b are $0$. Once the polynomial is factored, set each factor equal to zero and solve them separately. The answers will be the set of solutions for the original equation.
Not all solutions are appropriate for some applications. In many real-world situations, negative solutions are not appropriate and must be discarded. |
# Section 6.7 - Applications
```Motion Problem
The distance formula: d = rt or r = d/t or t = d/r
You may want to find the following table helpful
in solving motion problem.
Distance
Quantity 1/Case 1
Quantity 2/ Case 2
Rate/speed
Time
Example 1
The top speed of car A is 33 mph less than the top
speed of car B. At their top speeds, car B can travel 6
miles in the same time that car A can travel 5 miles.
Find the top speed of each car.
Let x be the top speed of car A
Then (x + 33) is the top speed of car B
Solution
Distance
(miles)
Car A
Car B
Rate/speed
(mph)
5
6
The top speed of car A is
The top speed of car B is
Time
(hours)
x
5
x
x + 33
6
x 33
Example 2
A fisherman travels 9 miles downstream with the current in
the same time that he travels 3 miles upstream against the
current. If the speed of the current is 6 mph, what is the
speed at which the fisherman travels in still water?
Let x be the speed at which the fisherman travels in still water.
Distance
(miles)
Downstream
Upstream
Rate/speed
(mph)
9
3
Time
(hours)
x+6
9
x6
x–6
3
x6
Share-Work Problems
Rate of work:
If a job can be completed in t units of time,
then the rate of work is 1/t
Similar to the distance formula, we have the
following
Work completed = rate
or
W = rt
x
time
Example 3
Joe can paint a house in 3 hours.
Sam can paint the same house in 5 hours.
How long does it take them to paint together?
Example 4
The CUT-IT-OUT lawn mowing company consists
of two people: Tina and Bill. If Bill cuts a lawn by
himself, he can do it in 4 hrs. If Tina cuts the same
lawn herself, it takes her an hour longer than Bill.
How long would it take if they worked together?
Ratios
a
The ratio of a to b is
or a to b or a:b
b
A Proportion is a statement that two ratios
are equal. In other words, a proportion has
the form
a c
b d
Examples
Solve the following proportions
11
z
3 123
w2 1
4w
6
6 c 1
c
5
Example
Billie earns \$412 for a 40-hour week. If she
missed 10 hours of work last week, how much
did she get paid?
Similar Triangles
x
b
a
c
y
z
Example
A tree casts a shadow 18 ft long at the same time
as a woman 5 feet tall casts a shadow 1.5 feet
long. Find the height of the tree.
``` |
# A right-angled triangle has three angles, alpha, beta and theta with theta=90. How do I prove that sinalpha=cosbeta?
Mar 31, 2017
See below
#### Explanation:
A right-angled triangle has three angles, $\alpha$, $\beta$ and $\theta$, with one of those being ${90}^{\text{o}}$. Let's say $\theta = 90$.
Now, we're trying to prove that $\sin \alpha = \cos \beta$. Since all the angles in a triangle add up to $180$, and one of these angles is already $90$, then we can say that $\alpha = 90 - \beta$ and $\beta = 90 - \alpha .$
Thus $\sin \alpha = \cos \left(90 - \alpha\right)$. Using the compound angle formula, $\cos \left(a - b\right) = \cos \cos b + \sin a \sin b$, we can say that $\cos \left(90 - \alpha\right) = \cos \alpha \cos 90 + \sin \alpha \sin 90$.
$\cos 90 = 0 , \sin 90 = 1 \therefore \cos \alpha \cos 90 - \sin \alpha \sin 90 = 0 \cos \alpha + 1 \sin \alpha = \sin \alpha$
$\cos \left(90 - \alpha\right) = \sin \alpha$
$\cos \left(90 - \alpha\right) = \cos \beta$
$\therefore \cos \beta = \sin \alpha$ ΟΕΔ |
Number System Conversion
In this tutorial, you will learn how to convert numbers from one base system to another. We will see all possible sets of conversions among the four most used number systems. For example, Binary to Decimal, Hexadecimal to Octal, Decimal to Hexadecimal, etc.
Contents:
Decimal to Binary, Octal, and Hexadecimal Conversion
The process for conversion from Decimal to any other base follows these basic steps.
• For a given number in decimal, we divide the integer part of the number by radix value, which we have to convert into and go on dividing till we get zero as quotient.
• We keep a track of all the remainders at all stages and then write them in the reverse order that is from bottom-up, which gives us the integer part of the number in the required base.
• Due to writing in reverse order, the first remainder becomes the least significant bit (LSB) and the last remainder becomes the most significant bit (MSB).
• To obtain the fractional part, we multiply the entire fractional part with the radix and write down the integer part obtained from it.
• We can repeat this multiplication step with the newly obtained fractional part from the previous multiplication, till we get the fractional part as zero and get the perfect fractional equivalent.
• We write down the digits in the order we performed the multiplications. Usually, getting a number with its fractional part as zero takes long and we stop after 3-4 digits of precision.
The following figure shows Decimal to Octal, Hexadecimal, and Binary Conversion with the same number as an example.
As can be seen from the figure,
• While converting to octal, hexadecimal, and binary, we divided the number by 8, 16, and 2 respectively, and wrote down the remainders in reverse order as our integer part.
• For the fractional part, we have multiplied with the respective bases and have written them down in the order of our performed multiplication.
Binary, Octal, and Hexadecimal to Decimal Conversion
To convert any number from any other base to decimal, we simply express the number as its product of digits and corresponding radix-powers of its current base and add the obtained values. For example,
• Octal to decimal: – We multiply the digits with powers of 8 according to their positions.
(1456)8 = 1 x 83 + 4 x 82 + 5 x 81 + 6 x 80
= 512 + 256 + 40 + 6
= (814)10
• Binary to decimal: – We multiply the digits with powers of 2 according to their positions.
(100110.101)2 = 1 x 25 + 0 x 24 + 0 x 23 + 1 x 22 + 1 x 21 + 0 x 20 + 1 x 2-1 + 0 x 2-2 + 1 x 2-3
= 32 + 0 + 0 + 4 + 2 + 0 + 0.5 + 0 + 0.125
= (38.625)10
• Hexadecimal to decimal: – We multiply the digits with powers of 16 according to their positions.
(6D4.B0)16 = 6 x 162 + 13 x 161 + 4 x 160 + 11 x 16-1 + 0 x 16-2
= 1536 + 48 + 4 + 0.6875 + 0
= (1748.6875)10
Binary to Octal Conversion
For converting a binary number to octal we make use of the fact that all octal digits can be perfectly expressed with 3 binary bits. The steps that we follow are: –
• For the integer part: We make groups of 3 bits beginning from the last bit, moving towards higher bits.
• For fractional part: We begin making groups of 3 bits just after the radix point and move towards the right.
• We then write down the corresponding octal value for each group of those three bits.
For example,
1. (110111001)2 can be written in octal as: – 1 1 0 | 1 1 1 | 0 0 1 = (671)8
2. (011110.010101)2 can be written in octal as: – 0 1 1 | 1 1 0 |.| 0 1 0 | 1 0 1 = (36.25)8
• In case, any group is left with lesser number of bits, more zeroes can be added in front of it (for the integer part) and at the end (for the fractional part), to make that group of 3 bits.
For example,
1. (11100111.11)2 can be written in octal as: – 1 1|1 0 0 | 1 1 1 |.| 1 1 when regrouped becomes 0 1 1 | 1 0 0 | 1 1 1 |.| 1 1 0 = (347.6)8
Octal to Binary Conversion
To convert from octal to binary system, we write down the corresponding values of each digit of octal in its binary 3-bits form. This is a reverse operation of what we did in the previous section. The following figure shows how to convert a number from octal to binary.
As is clear from the figure,
• Each octal digit is converted into its 3-bit binary equivalent and obtained values are written all together to give the binary equivalent of the octal number.
Here is a table to always refer to while converting from octal to binary or vice-versa.
Positional Values(22 21 20) 4 2 1 4 2 1 4 2 1 4 2 1 4 2 1 4 2 1 4 2 1 4 2 1
Binary 000 001 010 011 100 101 110 111
Octal 0 1 2 3 4 5 6 7
As shown in the table,
• The positional values given by 4 2 1, can be used to calculate binary equivalents of octal numbers by writing 1 for that positional value which contributes to the sum of that digit. For example, 7=4+2+1, so we can write it as 111, etc.
For converting a number from binary to hexadecimal, we make groups as we did for octal, but instead of 3-bits, we make 4-bit groups. Here are the steps to convert:
• We make groups of 4-bit, from the right-most bit (LSB) to the left-most bit (MSB) for the integer part, and write down the corresponding hexadecimal values for each group.
• For the fractional part, we begin grouping after the radix point and move towards the right.
• Also, we can add zeroes in front of a group (for the integer part) and zeroes to the end of a group (for the fractional part), to make it a 4-bit group.
For example: –
1. (1011110011.11101)2 can be written in hexadecimal as: –
0010|1111|0011|.|1110|1000 = (2F3.E8)16
2. (11010011.1010)2 can be written in hexadecimal as: –
1101|0011|.|1010 = (D3. A)16
For hexadecimal to binary conversion, we write down each hexadecimal digit’s 4-bit binary equivalent, and group the results together to give the final converted value. Here is a figure that shows hexadecimal to binary conversion.
As can be seen from the figure,
• Each digit was converted individually from hexadecimal to binary and written together to give the binary form of the hexadecimal number.
Here is a table to refer to while converting a number from hexadecimal to binary and vice-versa.
Positional Values(23 22 21 20) 8 4 2 1 8 4 2 1 8 4 2 1 8 4 2 1 8 4 2 1 8 4 2 1 8 4 2 1 8 4 2 1
Binary 0000 0001 0010 0011 0100 0101 0110 0111
Hexadecimal 0 1 2 3 4 5 6 7
Binary 1000 1001 1010 1011 1100 1101 1110 1111
Hexadecimal 8 9 A B C D E F
As shown in the table,
• The binary equivalent of hexadecimal digits can be remembered easily by adding the positional values of 8 4 2 1 by using 1 for only those positions, which add up to the sum of that digit. For example, 9=8+0+0+1, hence we write it as 1001.
Octal numbers cannot be directly converted into hexadecimal form. So we convert it to an intermediate decimal or binary form. Using decimal as an intermediate form, our calculations increase, so we use binary for quick conversion. The steps are shown below using an example:
1st step: Convert the number into binary by converting each octal digit individually into its 3-bit binary code: – (110 111 010 . 011 100)
2nd step: Group by 4-bits from the right for the integer part and from the left for the fractional part: – (0001) (1011) (1010) . (0111) (0000)
3rd step: Write the corresponding hexadecimal digits to each 4-bit group: – (1BA.70)16
Hexadecimal numbers also need an intermediate form to convert into octal numbers. Here are the steps to follow:
To convert (A2B6.83)16 into octal:
1st step: Convert the hexadecimal number into binary form by expanding each digit individually into its 4-bit binary code: – (1010 0010 1011 0110 . 1000 0011)
2nd step: From the right group the digits into 3-bits for the integer part and start after the radix point for the fractional part: – (001) (010) (001) (010) (110) (110) . (100) (000) (110)
3rd step: Convert each 3-bit group into its equivalent octal digit: – (121266.406)8
Key Points to Remember
Here are some key points listed, which are to be remembered in “Number System Conversion”.
• A decimal number can be converted to any base by dividing the integer part by that base repeatedly till the remainder is zero and writing the remainders in reverse order.
• From any base, a number can be converted into decimal form by expanding the number as its digits and radix-power product, followed by adding up the values.
• Binary can be converted into Octal and Hexadecimal by grouping into 3 and 4 bits respectively and writing the corresponding digits.
• Octal or hexadecimal is converted back into binary by expanding each digit’s 3-bit or 4-bit binary codes respectively.
• Octal and Hexadecimal numbers cannot be interconverted directly; they need to be converted first either into decimal or binary.
• For grouping any number, we move from LSB to MSB for the integer part, and for the fractional part, we start grouping after the radix point till the last fractional digit.
• On converting a number from a lower base to a higher one, the number of digits decreases for the integer part and increases for the fractional part.
• On converting from a higher base to a lower one, the number of digits increases for the integer part and decreases for the fractional part.
If you find any mistake above, kindly email to [email protected] |
# The Rational Numbers Honors Math – Grade 7. Problem of the Day A small submarine descended to a depth of mi and then ascended mi. These changes can be.
## Presentation on theme: "The Rational Numbers Honors Math – Grade 7. Problem of the Day A small submarine descended to a depth of mi and then ascended mi. These changes can be."— Presentation transcript:
The Rational Numbers Honors Math – Grade 7
Problem of the Day A small submarine descended to a depth of mi and then ascended mi. These changes can be written as signed (positive or negative) numbers. Let’s recall some words for positive and negative numbers. POSITIVENEGATIVELet’s write these signed numbers. Read: negative 2 and ¼. Read: positive three eighths or three eighths.
A Rational Number is a number that can be written in fractional form,, where a and b are integers and. Here are some examples of rational numbers. Rational Number Form All integers are rational numbers. Integers – the set of whole numbers and their opposites VOCAB ALERT Natural Numbers Whole Numbers
Rational Numbers Integers -98 -4 Whole Numbers 0 Natural or Counting Numbers 1, 2, 3, 4, … Set of numbers starting with 1 and counting up by ones. Set of natural numbers and 0 Set of whole numbers and their opposites. Set of numbers that can be expressed as the quotient of two integers.
Identify the set(s) of numbers to which each belongs. 7-100
Rational Numbers can be represented in different forms. To write a mixed number as an improper fraction… 1.Multiply the denominator by the whole number part (integer). 2.Add the product to the numerator. 3.The denominator stays the same! x 63 + If the mixed number is negative, the improper fraction will be negative!
Write in the form.
A Rational Number in the form,, can be written as a decimal. To write a fraction as a decimal, follow one of these three methods… InspectionEquivalent FractionsDivision If the fraction, has a denominator of 10, 100, 1000, etc…, say the fraction and write the decimal. If the fraction can be renamed as an equivalent fraction with a denominator of 10, 100 or 1000, rename the fraction and use inspection. If the previous methods do not work, divide the numerator by the denominator. Can be used all the time. Seven hundredths Fifty-six hundredths
Write as a decimal. If the quotient has a remainder of zero, the result is a terminating decimal.
Write as a decimal. If the quotient has a remainder of zero, the result is a terminating decimal.
Write as a decimal. If the quotient has a digit or a group of digits that repeat without end, the result is a repeating decimal. To show that one or more digits repeat in a decimal, use an ellipsis or an overbar.
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# Linear independence
### Linear independence
#### Lessons
We say that a set of vectors {$v_1, \cdots , v_p$} in $\Bbb{R}^n$ is linearly independent if:
$v_1 x_1+v_2 x_2+\cdots+v_p x_p=0$
gives only the trivial solution. In other words, the only solution is:
We say that a set of vectors {$v_1, \cdots , v_p$} in $\Bbb{R}^n$ is linearly dependent if:
$v_1 x_1+v_2 x_2+\cdots+v_p x_p=0$
gives a non-trivial solution. In other words, they are linearly dependent if it has a general solution (aka has free variable).
We can determine if the vectors is linearly independent by combining all the columns in matrix (denoted as A) and solving for
$Ax=0$
Fast way to tell if 2 or more vector are linearly dependent
1. The vectors are multiples of one another
2. There are more vectors than there are entries in each vector.
3. There is a zero vector
• Introduction
Linear Independence Overview:
a)
Linearly independent
• Definition of linear independence
• Trivial solution
b)
Linearly dependent
• Definition of linear dependence
• Non-trivial solutions
c)
Fast ways to determine linear dependence
• Vectors are multiples of one another
• # of Vectors > # of Entries in each vector
• Zero Matrix
• 1.
Determining Linear independence by solving
Determine if the following vectors are linearly independent:
• 2.
Determine if the matrix is linearly independent by solving $Ax=0$:
• 3.
Determining Linear dependence by inspection
Determine by inspection if the following vectors are linearly dependent:
a)
b)
c)
d)
• 4.
Linear dependence/independence with unknown constant
Find the value(s) of $k$ for which the vectors are linearly dependent. |
# The Simple Math Equation: 200 Divided By 5
## Introduction
When it comes to math, we all have our strengths and weaknesses. Some of us may excel in algebra or calculus, while others struggle with simple arithmetic. Regardless of where you stand, we all need to start with the basics. In this article, we’ll be discussing one of the simplest math equations: 200 divided by 5.
### What is Division?
Before we dive into the specifics of 200 divided by 5, let’s first discuss what division is. Division is the mathematical operation of separating a number into equal parts. For example, if you have 10 apples and you want to divide them into 5 equal groups, you would have 2 apples in each group.
### The Equation: 200 Divided by 5
Now that we have a basic understanding of division, let’s talk about the equation at hand: 200 divided by 5. This equation is asking us to divide 200 into 5 equal parts. To solve this equation, we can use a few different methods. One of the most common methods is long division. We can start by dividing 2 (the first digit of 200) by 5. Since 5 does not go into 2, we move onto the next digit, which is 20. 5 goes into 20 four times (5 x 4 = 20), so we write 4 above the 0 in 20. We then subtract 20 from 20, which leaves us with 0. We bring down the next digit, which is 0, and divide 50 by 5. 5 goes into 50 ten times (5 x 10 = 50), so we write 10 above the 0 in 50. We subtract 50 from 50, which leaves us with 0. Finally, we bring down the last digit, which is 0, and divide 0 by 5. Since 5 cannot go into 0, we have our final answer: 200 divided by 5 equals 40.
### Other Methods for Dividing
While long division may be the most common method for dividing, there are other methods that can be used as well. For example, we can use mental math to solve simple division problems like 200 divided by 5. To do this, we can use multiplication. We know that 5 times 40 equals 200, so we can simply reverse the equation to get our answer: 200 divided by 5 equals 40.
### Practical Applications of Division
Now that we know how to solve 200 divided by 5, let’s talk about some practical applications of division. Division is used in a variety of real-world scenarios, from splitting a pizza with friends to calculating a company’s profits. For example, let’s say you have a pizza with 8 slices and 4 people want to split it evenly. You would use division to figure out how many slices each person gets. In this case, you would divide 8 (the total number of slices) by 4 (the number of people), which would give you 2. Each person would get 2 slices of pizza.
### Conclusion
In conclusion, 200 divided by 5 is a simple math equation that can be solved using a variety of methods. While long division may be the most common method, mental math can also be used in some cases. Division is a fundamental math skill that has practical applications in many aspects of our lives. By mastering the basics, we can build a strong foundation for more complex math concepts in the future. |
## Piecewise-Defined Functions
### Learning Outcomes
• Write piecewise defined functions.
• Graph piecewise-defined functions.
Sometimes, we come across a function that requires more than one formula in order to obtain the given output. For example, in the toolkit functions, we introduced the absolute value function $f\left(x\right)=|x|$. With a domain of all real numbers and a range of values greater than or equal to 0, absolute value can be defined as the magnitude, or modulus, of a real number value regardless of sign. It is the distance from 0 on the number line. All of these definitions require the output to be greater than or equal to 0.
If we input 0, or a positive value, the output is the same as the input.
$f\left(x\right)=x\text{ if }x\ge 0$
If we input a negative value, the output is the opposite of the input.
$f\left(x\right)=-x\text{ if }x<0$
Because this requires two different processes or pieces, the absolute value function is an example of a piecewise function. A piecewise function is a function in which more than one formula is used to define the output over different pieces of the domain.
We use piecewise functions to describe situations in which a rule or relationship changes as the input value crosses certain “boundaries.” For example, we often encounter situations in business for which the cost per piece of a certain item is discounted once the number ordered exceeds a certain value. Tax brackets are another real-world example of piecewise functions. For example, consider a simple tax system in which incomes up to $10,000$ are taxed at $10%$, and any additional income is taxed at $20\%$. The tax on a total income, $S$ , would be $0.1S$ if ${S}\le10,000$ and $1000 + 0.2 (S - 10,000)$ , if $S> 10,000$ .
### A General Note: Piecewise Functions
A piecewise function is a function in which more than one formula is used to define the output. Each formula has its own domain, and the domain of the function is the union of all these smaller domains. We notate this idea like this:
$f\left(x\right)=\begin{cases}\text{formula 1 if x is in domain 1}\\ \text{formula 2 if x is in domain 2}\\ \text{formula 3 if x is in domain 3}\end{cases}$
In piecewise notation, the absolute value function is
|x|=\begin{cases}\begin{align}x&\text{ if }x\ge 0\\ -x&\text{ if }x<0\end{align}\end{cases}
### How To: Given a piecewise function, write the formula and identify the domain for each interval.
1. Identify the intervals for which different rules apply.
2. Determine formulas that describe how to calculate an output from an input in each interval.
3. Use braces and if-statements to write the function.
### Example: Writing a Piecewise Function
A museum charges $5 per person for a guided tour with a group of 1 to 9 people or a fixed$50 fee for a group of 10 or more people. Write a function relating the number of people, $n$, to the cost, $C$.
### Example: Working with a Piecewise Function
A cell phone company uses the function below to determine the cost, $C$, in dollars for $g$ gigabytes of data transfer.
C\left(g\right)=\begin{cases}\begin{align}{25} \hspace{2mm}&\text{ if }\hspace{2mm}{ 0 }<{ g }<{ 2 }\\ { 25+10 }\left(g - 2\right) \hspace{2mm}&\text{ if }\hspace{2mm}{ g}\ge{ 2 }\end{align}\end{cases}
Find the cost of using 1.5 gigabytes of data and the cost of using 4 gigabytes of data.
### How To: Given a piecewise function, sketch a graph.
1. Indicate on the $x$-axis the boundaries defined by the intervals on each piece of the domain.
2. For each piece of the domain, graph on that interval using the corresponding equation pertaining to that piece. Do not graph two functions over one interval because it would violate the criteria of a function.
### Example: Graphing a Piecewise Function
Sketch a graph of the function.
f\left(x\right)=\begin{cases}\begin{align}{ x }^{2} \hspace{2mm}&\text{ if }\hspace{2mm}{ x }\le{ 1 }\\ { 3 } \hspace{2mm}&\text{ if }\hspace{2mm} { 1 }<{ x }\le 2\\ { x } \hspace{2mm}&\text{ if }\hspace{2mm}{ x }>{ 2 }\end{align}\end{cases}
### Try It
Graph the following piecewise function.
f\left(x\right)=\begin{cases}\begin{align}{ x}^{3} \hspace{2mm}&\text{ if }\hspace{2mm}{ x }<{-1 }\\ { -2 } \hspace{2mm}&\text{ if } \hspace{2mm}{ -1 }<{ x }<{ 4 }\\ \sqrt{x} \hspace{2mm}&\text{ if }\hspace{2mm}{ x }>{ 4 }\end{align}\end{cases}
### Try It
You can use an online graphing tool to graph piecewise defined functions. Watch this tutorial video to learn how.
Graph the following piecewise function with an online graphing tool.
f\left(x\right)=\begin{cases}\begin{align}{ x}^{3} \hspace{2mm}&\text{ if }\hspace{2mm}{ x }<{-1 }\\ { -2 } \hspace{2mm}&\text{ if } \hspace{2mm}{ -1 }<{ x }<{ 4 }\\ \sqrt{x} \hspace{2mm}&\text{ if }\hspace{2mm}{ x }>{ 4 }\end{align}\end{cases}
### Q&A
Can more than one formula from a piecewise function be applied to a value in the domain?
No. Each value corresponds to one equation in a piecewise formula.
## Contribute!
Did you have an idea for improving this content? We’d love your input. |
# Video: Finding the Missing Coefficient of a Polynomial given One of Its Factors
Find the value of π given that 2π₯Β³ + ππ₯Β² β 21π₯ β 36 is divisible by (π₯ + 4).
02:11
### Video Transcript
Find the value of π given that two π₯ cubed plus ππ₯ squared minus 21π₯ minus 36 is divisible by π₯ plus four.
Here, weβve been given a polynomial with a missing coefficient. π is the coefficient of π₯ squared. And weβre told that this cubic polynomial is divisible by the linear expression π₯ plus four. In other words, π₯ plus four is a factor of the polynomial two π₯ cubed plus ππ₯ squared minus 21π₯ minus 36. We can therefore answer this question by applying the factor theorem. This tells us that if π₯ plus π is a factor of the polynomial π of π₯, then π of negative π will be equal to zero, and the converse is also true. If π of negative π is equal to zero, then we know that π₯ plus π will be a factor of π of π₯.
So, we can define π of π₯ to be our cubic polynomial. And as π₯ plus four is a factor of this polynomial, we know that π of negative four must be equal to zero. By substituting negative four for π₯ and setting the resulting expression equal to zero, we now have an equation which we can solve in order to find the value of π. Two multiplied by negative four cubed plus π multiplied by negative four squared minus 21 multiplied by negative four minus 36 is equal to zero.
By evaluating each of these powers of negative four and then simplifying the equation by grouping like terms, we arrive at the equation 16π minus 80 is equal to zero. We can then add 80 to each side of the equation and finally divide by 16 to give π equals 80 over 16. We can cancel by a factor of eight, first of all, giving 10 over two. And 10 over two is, of course, simply equal to five. So, by using the factor theorem, weβve found the value of π, the missing coefficient, in a polynomial π of π₯. π is equal to five. |
## Topic outline
• ### Lowest Common Multiple (LCM)
#### Lowest Common Multiple
The Lowest Common Multiple (LCM) of two or more given number is the lowest (or smallest or least) of their common multiples.
Example: LCM of 12 and 15.
Multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144, 156, 168, 180, 192, 204, 216, 228, 240, 252…
Multiples of 15 are 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195, 210, 225, 240
Common Multiples of 12 and 15 are 60, 120, 180, 240…
Lowest Common Multiples of 12 and 15 is 60. (60 is the smallest number that both the numbers are factors of this number.)
#### LCM by Prime Factorization Method
Example: Find LCM of 16, 24 and 56
Step1: Find out the prime factors of the given numbers.
Prime factor of 24 = 2 x 2 x 2 x 3 = 23 x 3
Prime factor of 56 = 2 x 2 x 2 x 7 = 23 x 7
Step2: Identify the maximum number of occurrence of each prime number. Here, 24 (2 has occurred 4 times), 31 (3 has occurred 1 time) and 71 (7 has occurred 1 time)
Step3: Now multiply the outcomes.
2 x 2 x 2 x 2 x 3 x 7 = 336
#### LCM by Division Method
Step 1: Find the prime factors of the two or more numbers by dividing the numbers by least prime number till we get 1. (Choose the least prime that at least divide one of the given numbers)
Step 2: Then multiply all the prime factors to get the LCM.
Example: Find LCM of 20, 25 and 30
LCM of 20, 25 and 30 = 2 x 2 x 3 x 5 x 5 = 300 |
# Probability Distribution | Binomial-Poisson-Normal Distribution | Business Statistics Notes | B.Com Notes Hons & Non Hons | CBCS Pattern
## BUSINESS STATISTICS NOTESB.COM 2ND AND 3RD SEM NEW SYLLABUS (CBCS PATTERN)Probability and Probability DistributionsPart B: Probability Distribution - Binomial - Poisson - Normal
### Meaning of Binomial Distributions
The binomial distribution also known as ‘Bernoulli Distribution’ is associated with the name of a Swiss mathematician James Bernoulli also known as Jacques or Jakob (1654-1705). Binomial distribution is a probability distribution expressing the probability of one set of dichotomous alternatives, i.e., success or failure.
This distribution has been used to describe a wide variety of processes in business and the social sciences as well as other areas. The type of process which gives rise to this distribution is usually referred to as Bernoulli trial or as a Bernoulli process. The mathematical model for a Bernoulli process is developed under a very specific set of assumption involving the concept of a series of experimental trials. These assumptions are:
1) An experiment is performed under the same conditions for a fixed number or trials, say, n.
2) In each trial, there are only two possible outcomes of the experiment.
3) The probability of a success denoted by p remains constant from trial to trial. The probability of a failure denoted by q is equal to (1 – p). If the probability of success is not the same in each trial, we will not have binomial distribution.
4) The trials are statistically independent, i.e., the outcomes of any trial or sequence of trials do not affect the outcomes of subsequent trials.
### Application of Binomial distribution
Binomial distribution is applicable when the trials are independent and each trial has just two outcomes success and failure. It is applied in coin tossing experiments, sampling inspection plan etc.
### Fitting a Binomial Distribution
When a binomial distribution is to be fitted to observe data, the following procedure is adopted:
1) Determine the values of p and q. If one of these values is known the other can be found out by the simple relationship p = (1 – q) and q = (1 – p). When p and p are equal the distribution is symmetrical, for p and q may be interchanged without altering the value of any term, and consequently terms equidistant from the two ends of the series are equal. If p and q are unequal, the distribution is skew. If p is less than ½ the distribution is positively skewed and when p is more than ½ the distribution is negatively skewed.
2) Expand the binomial (q + p)n. The power n is equal to one less than the number of terms in the expanded binomial. Thus, when two coins are tossed (n = 2) there will be three terms in the binomial. Similarly, when four coins are tossed (n = 4) there will be five terms, and so on.
3) Multiply each term of the expanded binomial by N (the total frequency), in order to obtain the expected frequency in each category.
### MEANING OF POISSON DISTRIBUTION
Poisson distribution is a discrete probability distribution and is very widely used in statistical work. It was developed by a French mathematician, Simeon Denis Poisson (1781-1840), in 1837. Poisson distribution may be expected in cases where the chance of any individual event being a success is small. The distribution is used to describe the behaviour of rare events such as the number of accidents on road, no. of printing mistake in a book, etc., and has been called “the law of improbable events”. In recent years the statisticians have had a renewed interest in the occurrence of comparatively rare events, such as serious floods, accidental release of radiation from a nuclear reactor, and the like.
The Poisson distribution is defined as:
### Role of the Poisson distribution
The Poisson distribution is used in practice in a wide variety of problems where there are infrequently occurring events with respect to time area, volume or similar units. Some practical situations in which Poisson distribution can be used are given below:
1. It is used in quality control statistics to count the number of defects of an item.
2. In biology to count the number of bacteria.
3. In physics to count the number of particles emitted from a radioactive substance.
4. In insurance problems to count the number of casualties
5. In waiting-time problems to count the number of incoming telephone calls or incoming customers.
6. Number of traffic arrivals such as trucks at terminals, aeroplanes in airports, ships at docks, and so forth.
7. In determining the number of deaths in a district in a given period, say, a year, by a rare disease.
8. The number of typographical errors per page in typed material, number of deaths as a result of road accidents, etc.
9. In problems dealing with the inspection of manufactured products with the probability that any one piece is defective is very small and the lots are very large.
10. To model the distribution of the number of persons joining a queue (a line) to receive a service or purchase of a product.
## ALSO READ: CHAPTER WISE BUSINESS STATISTICS NOTESUnit 1: Statistical Data and Descriptive Statisticsa. Nature and Classification of datab. Measures of Central Tendencyc. Measures of Variationd. Skewness, Moments and KurtosisUnit 2: Probability and Probability Distributionsa. Theory of Probabilityb. Probability distributions:Binomial-Poisson-NormalUnit 3: Simple Correlation and Regression Analysisa. Correlation Analysisb. Regression AnalysisUnit 4: Index NumbersUnit 5: Time Series AnalysisUNIT 6: Sampling Concepts, Sampling Distributions and EstimationPAST EXAM SOLVED QUESTION PAPERS Non-CBCS Pattner Question Paper:1. Business Statistics Question Papers (From 2014 onwards) B.Com 3rd Sem2. All Question Papers are Available HereCBCS Pattern Question Paper 1. Business Statistics (Hons) - 2020 2021 (Held in 2022)Business Statistics Solved Question Papers1. Business Statistics Solved Question Papers (Non-CBCS Pattern)2. Business Statistics Solved Questions Papers (CBCS Pattern): 2020 2021 2022CHAPTER WISE PRACTICAL QUESTION BANK:Chapter wise practical question bank is available in our mobile application. Download DTS App for Chapter Wise practical question bank.
### MEANING OF NORMAL DISTRIBUTION
The binomial and the Poisson distributions are the most useful theoretical distributions for discrete variables, i.e., they relate to the occurrence of distinct events. In order to have mathematical distribution suitable for dealing with quantities whose magnitude is continuously variable, a continuous distribution is needed. The normal distribution, also called the normal probability distribution, happens to be most useful theoretical distribution for continuous variables. Many statistical date concerning business and economic problems are displayed in the form of normal distribution.
The normal distribution was first described by Abraham De Moivre (1667-1754) as the limiting form of the binomial model in 1733. Normal distribution was rediscovered by Gauss in 1809 and by Laplace in 1812. Both Gauss and Laplace were led to the distribution by their work on the theory of errors of observations arising in physical measuring processes particularly in astronomy. Throughout the 18th and 19th centuries, various efforts were made to establish the normal model as the underlying law ruling all continuous random variables, - thus the name normal. These efforts failed because of the false premises. The normal model has, nevertheless, become the most important probability model in statistical analysis.
The normal distribution is an approximation to binomial distribution. Whether or not p is equal to q, the binomial distribution tends to the form of the continuous curve and when n becomes large at least for the material part of the range. As a matter of fact, the correspondence between the binomial and the curve is surprisingly close even for comparatively low values of n, provided that p and q are fairly near equality. The limiting frequency curve obtained as n becomes large is called the normal frequency curve or simply the normal curve.
The normal curve is represented in several forms. The following is the basic from relating to the curve with mean and standard deviation .
### Properties of the Normal Distribution
The following are the important properties of the normal curve and the normal distribution.
1) The normal curve is “bell-shaped” and symmetrical in its appearance. If the curves were folded along its vertical axis, the two halves would coincide. The number of cases below the mean in a normal distribution is equal to the number of cases above the mean, which makes the mean and median coincide. The height of the curve for a positive deviation of 3 units is the same as the height of the curve for negative deviation of 3 units.
2) The height of the normal curve is at its maximum at the mean. Hence the mean and mode of the normal distribution coincide. Thus for a normal distribution mean, median and mode are all equal.
3) There is one maximum point of the normal curve which occurs at the mean. The height of the curve declines as we go in either direction from the mean. The curve approaches nearer and nearer to the base but it never touches it, i.e., the curve is asymptotic to the base on either side. Hence its range is unlimited or infinite in both directions.
4) Since there is only one maximum point, the normal curve is Unimodal, i.e., it has only one mode.
5) The points of inflexion, i.e., the points where the change in curvature occurs are mean + sd.
6) As distinguished from Binomial and Poisson distribution where the variable is discrete, the variable distributed according to the normal curve is a continuous one.
7) The first and third quartiles are equidistant from the median.
8) The mean deviation is 4th or more precisely 0.7979 of the standard deviation.
9) The area under the normal curve distributed as follows: -
### Significance of the Normal Distribution
The normal distribution is mostly used for the following purposes:-
1. To approximate or “fit” a distribution of measurement under certain conditions.
2. To approximate the binomial distribution and other discrete or continuous probability distributions under suitable conditions.
3. To approximate the distribution of means and certain other quantities calculated from samples, especially large samples. |
# Thread: An expression with square roots
1. ## An expression with square roots
How would you calculate
$\displaystyle (\sqrt {2} - \sqrt {5}) \cdot \sqrt{ 7 + 2 \sqrt {10}}$ ?
2. Hello, p.numminen!
How would you calculate: .$\displaystyle (\sqrt {2} - \sqrt {5}) \cdot \sqrt{ 7 + 2 \sqrt {10}}$ ?
There is a very sneaky trick we can pull . . .
First, we notice that: .$\displaystyle 7 + 2\sqrt{10}$ .just happens to equal $\displaystyle \left(\sqrt{2} + \sqrt{5}\right)^2$. .
Check it out!
Then: .$\displaystyle \sqrt{7 + 2\sqrt{10}} \:=\:\sqrt{\left(\sqrt{2} + \sqrt{5}\right)^2} \:=\:\sqrt{2} + \sqrt{5}$
So the problem becomes: .$\displaystyle \left(\sqrt{2} - \sqrt{5}\right)\left(\sqrt{2} + \sqrt{5}\right) \;=\;\left(\sqrt{2}\right)^2 - \left(\sqrt{5}\right)^2 \;\;=\;2 - 5 \;\;=\;\;\boxed{-3}$
3. We can simplify $\displaystyle \sqrt{7+2\sqrt{10}}$ by simultaneous equations as well.
Let $\displaystyle \sqrt{7+2\sqrt{10}}=A+B\sqrt{10}$
Squaring, we get $\displaystyle 7+2\sqrt{10}=A^2+10B^2+2AB\sqrt{10}$
Now by matching up coefficients, we get two equations:
$\displaystyle A^2+10B^2=7$ ...[1]
$\displaystyle 2AB=2 \implies AB=1$ ...[2]
$\displaystyle B=\frac{1}{A}$ ...[2']
Now substituting into [1]:
$\displaystyle A^2+10\left(\frac{1}{A} \right)^2=7$
$\displaystyle A^4+10=7A^2$
$\displaystyle A^4-7A^2+10=0$
Now we make the substitution $\displaystyle u=A^2$, then it becomes:
$\displaystyle u^2-7u+10=0$
$\displaystyle (u-2)(u-5)=0$
$\displaystyle u=5$ or $\displaystyle u=2$
Then $\displaystyle A^2=5$ or $\displaystyle A^2=2$
So $\displaystyle A=\pm\sqrt{5}$ or $\displaystyle A=\pm \sqrt{2}$
So the solutions are $\displaystyle \{(A,B)\sqrt{5},\frac{1}{\sqrt{5}}),(-\sqrt{5},-\frac{1}{\sqrt{5}}),(\sqrt{2},\frac{1}{\sqrt{2}}), (-\sqrt{2},-\frac{1}{\sqrt{2}})\}$
Let's try the first one; then we have:
$\displaystyle \sqrt{7+2\sqrt{10}}=\sqrt{5}+\frac{1}{\sqrt{5}} \cdot \sqrt{10}=\sqrt{5}+\sqrt{2}$
You can check this by equating both sides as Soroban did. Note that $\displaystyle A=\sqrt{2}$, $\displaystyle B=\frac{1}{\sqrt{2}}$ also works, but the other two solutions Do Not. They are the negative solutions. That's why you have to check the final result to see if it fits!
Of course it then follows that $\displaystyle (\sqrt{2}-\sqrt{5})(\sqrt{2}+\sqrt{5})=2-5=-3$
4. Sometimes, one try to guess what's the expression which powered to 2 gives us the result.
Note that
$\displaystyle \sqrt{7+2\sqrt{10}}=\sqrt{\left(2+2\sqrt{10}+5\rig ht)}$
Now, you can see the expression $\displaystyle (a+b)^2$ which is hidden there.
Of course, always remember that $\displaystyle \sqrt{x^2}=|x|$ |
# GMAT Gurus Speak Out: Permutation and Combination Basics
Aiming for a 700+ on the GMAT? You never know when a challenging combination or permutation question will pop up three-quarters of the way through your exam to wreck havoc on your score. This advanced concept is not as commonly tested as algebra fundamentals or number properties, but it’s definitely worth knowing the basics in case you do see it.
The Fundamental Counting Principle states that if an event has x possible outcomes and a different independent event has y possible outcomes, then there are xy possible ways the two events could occur together. For example, how many three-digit integers have either 6 or 9 in the tens digit and 1 in the units digit?
To solve, we need to find the possible outcomes for each digit (hundreds, tens, and units) and multiply them. Each digit has 10 possible values (0 through 9). The hundreds digit can be any of these except 0 (since a three-digit number cannot begin with 0). The tens digit has only 2 options (6 or 9). The units digit has only 1 possibility (1). Therefore, the total number of possibilities is 9 x 2 x 1 = 18.
Permutations are sequences. In a sequence, order is important. How many different ways can four people sit on a bench? For the first spot on the bench, we have 4 to choose from. For the next spot we’ll have 3, for the third spot we’ll have 2, and the last remaining person will take the final spot. Therefore, there are 4 x 3 x 2 x 1 = 24 ways. Harder permutations problems will require you to use this formula:
n = the number of options
r = the number chosen from those options
For example, how many possible options are there for the gold, silver, and bronze medals out of 12 athletes? Here n = 12 and r = 3. Since the order in which the athletes finish matters, we know to use the Permutation formula:
n! / (n – r)! = 12! / (12 – 3)! = 12! / 9! = 12 x 11 x 10 = 1,320 options
Combinations are groups. Order doesn’t matter. The Combination formula is only slightly different from the Permutation formula:
Let’s say Dominic took 10 photos. He wants to put 7 of them on Facebook. How many groups of photos are possible?
n! / r! (n – r)! = 10! / 7! (10 – 7)! = 10! / 7! 3!
= 10 x 9 x 8 / 3 x 2 x 1
= 720 / 6
= 120 different groups
Remember to ask yourself whether order matters in the problem, and don’t forget the Fundamental Counting Principle! The GMAT may also combine one or more of these concepts in a longer Word Problem to make the question more challenging, but if you can remember these basics, you’ll be good to go!
Plan on taking the GMAT soon? We have GMAT prep courses starting all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
Vivian Kerr is a regular contributor to the Veritas Prep blog, providing tips and tricks to help students better prepare for the GMAT and the SAT. |
Graphing Systems of Linear Equations
Learning Objective(s)
· Solve a system of linear equations by graphing.
· Determine whether a system of linear equations is consistent or inconsistent.
· Determine whether a system of linear equations is dependent or independent.
· Determine whether an ordered pair is a solution of a system of equations.
· Solve application problems by graphing a system of equations.
Introduction
Recall that a linear equation graphs as a line, which indicates that all of the points on the line are solutions to that linear equation. There are an infinite number of solutions. If you have a system of linear equations, the solution for the system is the value that makes all of the equations true. For two variables and two equations, this is the point where the two graphs intersect. The coordinates of this point will be the solution for the two variables in the two equations.
Systems of Equations
The solution for a system of equations is the value or values that are true for all equations in the system. The graphs of equations within a system can tell you how many solutions exist for that system. Look at the images below. Each shows two lines that make up a system of equations.
One Solution No Solutions Infinite Solutions If the graphs of the equations intersect, then there is one solution that is true for both equations. If the graphs of the equations do not intersect (for example, if they are parallel), then there are no solutions that are true for both equations. If the graphs of the equations are the same, then there are an infinite number of solutions that are true for both equations.
When the lines intersect, the point of intersection is the only point that the two graphs have in common. So the coordinates of that point are the solution for the two variables used in the equations. When the lines are parallel, there are no solutions, and sometimes the two equations will graph as the same line, in which case we have an infinite number of solutions.
Some special terms are sometimes used to describe these kinds of systems.
The following terms refer to how many solutions the system has.
o When a system has one solution (the graphs of the equations intersect once), the system is a consistent system of linear equations and the equations are independent.
o When a system has no solution (the graphs of the equations don’t intersect at all), the system is an inconsistent system of linear equations and the equations are independent.
o If the lines are the same (the graphs intersect at all points), the system is a consistent system of linear equations and the equations are dependent. That is, any solution of one equation must also be a solution of the other, so the equations depend on each other.
The following terms refer to whether the system has any solutions at all.
o The system is a consistent system of linear equations when it has solutions.
o The system is an inconsistent system of linear equations when it has no solutions.
We can summarize this as follows: o A system with one or more solutions is consistent. o A system with no solutions is inconsistent. o If the lines are different, the equations are independent linear equations. o If the lines are the same, the equations are dependent linear equations.
Example Problem Using the graph of y = x and x + 2y = 6, shown below, determine how many solutions the system has. Then classify the system as consistent or inconsistent and the equations as dependent or independent. The lines intersect at one point. So the two lines have only one point in common, there is only one solution to the system. Because the lines are not the same the equations are independent. Because there is just one solution, this system is consistent. Answer The system is consistent and the equations are independent.
Advanced Example Problem Using the graph of y = 3.5x + 0.25 and 14x – 4y = -4.5, shown below, determine how many solutions the system has. Then classify the system as consistent or inconsistent and the equations as dependent or independent. The lines are parallel, meaning they do not intersect. There are no solutions to the system. The lines are not the same, the equations are independent. There are no solutions. Therefore, this system is inconsistent. Answer The system is inconsistent and the equations are independent.
Advanced Question Which of the following represents dependent equations and consistent systems? A) B) C) D) Show/Hide Answer
Verifying a Solution
From the graph above, you can see that there is one solution to the system y = x and x + 2y = 6. The solution appears to be (2, 2). However, you must verify an answer that you read from a graph to be sure that it’s not really (2.001, 2.001) or (1.9943, 1.9943).
One way of verifying that the point does exist on both lines is to substitute the x- and y-values of the ordered pair into the equation of each line. If the substitution results in a true statement, then you have the correct solution!
Example Problem Is (2, 2) a solution of the system y = x and x + 2y = 6? y = x 2 = 2 TRUE (2, 2) is a solution of y = x. x + 2y = 6 2 + 2(2) = 6 2 + 4 = 6 6 = 6 TRUE (2, 2) is a solution of x + 2y = 6. Since the solution of the system must be a solution to all the equations in the system, check the point in each equation. Substitute 2 for x and 2 for y in each equation. Answer (2, 2) is a solution to the system. Since (2, 2) is a solution of each of the equations in the system, (2, 2) is a solution of the system.
Example Problem Is (3, 9) a solution of the system y = 3x and 2x – y = 6? y = 3x 9 = 3(3) TRUE (3, 9) is a solution of y = 3x. 2x – y = 6 2(3) – 9 = 6 6 – 9 = 6 -3 = 6 FALSE (3, 9) is not a solution of 2x – y = 6. Since the solution of the system must be a solution to all the equations in the system, check the point in each equation. Substitute 3 for x and 9 for y in each equation. Answer (3, 9) is not a solution to the system. Since (3, 9) is not a solution of one of the equations in the system, it cannot be a solution of the system.
Example Problem Is (−2, 4) a solution of the system y = 2x and 3x + 2y = 1? y = 2x 4 = 2(−2) 4 = −4 FALSE (−2, 4) is not a solution of y = 2x. 3x + 2y = 1 3(−2) + 2(4) = 1 −6 + 8 = 1 2 = 1 FALSE (−2, 4) is not a solution of 3x + 2y = 1. Since the solution of the system must be a solution to all the equations in the system, check the point in each equation. Substitute −2 for x and 4 for y in each equation. Answer (−2, 4) is not a solution to the system. Since (−2, 4) is not a solution to either of the equations in the system, (−2, 4) is not a solution of the system.
Remember, that in order to be a solution to the system of equations, the value of the point must be a solution for both equations. Once you find one equation for which the point is false, you have determined that it is not a solution for the system.
Which of the following statements is true for the system 2x – y = −3 and y = 4x – 1? A) (2, 7) is a solution of one equation but not the other, so it is a solution of the system B) (2, 7) is a solution of one equation but not the other, so it is not a solution of the system C) (2, 7) is a solution of both equations, so it is a solution of the system D) (2, 7) is not a solution of either equation, so it is not a solution to the system Show/Hide Answer
Graphing as a Solution Method
You can solve a system graphically. However, it is important to remember that you must check the solution, as it might not be accurate.
Example Problem Find all solutions to the system y – x = 1 and y + x = 3. First, graph both equations on the same axes. The two lines intersect once. That means there is only one solution to the system. The point of intersection appears to be (1, 2). Read the point from the graph as accurately as possible. y – x = 1 2 – 1 = 1 1 = 1 TRUE (1, 2) is a solution of y – x = 1. y + x = 3 2 + 1 = 3 3 = 3 TRUE (1, 2) is a solution of y + x = 3. Check the values in both equations. Substitute 1 for x and 2 for y. (1, 2) is a solution. Answer (1, 2) is the solution to the system y – x = 1 and y + x = 3. Since (1, 2) is a solution for each of the equations in the system, it is the solution for the system.
Example Problem How many solutions does the system y = 2x + 1 and −4x + 2y = 2 have? First, graph both equations on the same axes. The two equations graph as the same line. So every point on that line is a solution for the system of equations. Answer The system y = 2x + 1 and −4x + 2y = 2 has an infinite number of solutions.
Which point is the solution to the system x – y = −1 and 2x – y = −4? The system is graphed correctly below. A) (−1, 2) B) (−4, −3) C) (−3, −2) D) (−1, 1) Show/Hide Answer
Graphing a Real-World Context
Graphing a system of equations for a real-world context can be valuable in visualizing the problem. Let’s look at a couple of examples.
Example Problem Andres was trying to decide which of two mobile phone plans he should buy. One plan, TalkALot, charged a flat fee of \$15 per month for unlimited minutes. Another plan, FriendFone, charged a monthly fee of \$5 in addition to charging 20¢ per minute for calls. To examine the difference in plans, he made a graph: If he plans to talk on the phone for about 70 minutes per month, which plan should he purchase? Look at the graph. TalkALot is represented as y = 15, while FriendFone is represented as y = 0.2x + 5. The number of minutes is listed on the x-axis. When x = 70, TalkALot costs \$15, while FriendFone costs about \$19. Answer Andres should buy theTalkALot plan. Since TalkALot costs less at 70 minutes, Andres should buy that plan.
Note that if the estimate had been incorrect, a new estimate could have been made. Regraphing to zoom in on the area where the lines cross would help make a better estimate.
Paco and Lisel spent \$30 going to the movies last night. Paco spent \$8 more than Lisel. If P = the amount that Paco spent, and L = the amount that Lisel spent, which system of equations can you use to figure out how much each of them spent? A) P + L = 30 P + 8 = L B) P + L = 30 P = L + 8 C) P + 30 = L P − 8 = L D) L + 30 = P L − 8 = P Show/Hide Answer
Summary
A system of linear equations is two or more linear equations that have the same variables. You can graph the equations as a system to find out whether the system has no solutions (represented by parallel lines), one solution (represented by intersecting lines), or an infinite number of solutions (represented by two superimposed lines). While graphing systems of equations is a useful technique, relying on graphs to identify a specific point of intersection is not always an accurate way to find a precise solution for a system of equations. |
Question
1. # Parallel Perpendicular Or Neither
Have you ever been in a geometry class and been confused when the teacher said something like this, “This line is parallel perpendicular or neither?” It can seem daunting to figure it out, but with a few simple steps you can easily find out if two lines are parallel, perpendicular, or neither. In this blog post we will explore how to determine if two lines are parallel, perpendicular, or neither by using the slope of the line. We will also look at some examples to help solidify your understanding of how to identify these different types of lines.
## What is the difference between parallel and perpendicular?
There are many ways to think about the difference between parallel and perpendicular lines, but perhaps the most straightforward way to think about it is in terms of angles. Parallel lines have angles that are equal to each other, while perpendicular lines have angles that are complementary to each other (meaning they add up to 90 degrees).
Another way to think about it is in terms of slope. Parallel lines have the same slope, while perpendicular lines have slopes that are opposite reciprocals of each other (meaning one line has a positive slope and the other has a negative slope, or one line has a negative slope and the other has a positive slope).
Ultimately, whether two lines are parallel or perpendicular can be determined by looking at their equations. If the equations are identical, then the lines are parallel. If the equations are not identical but are still similar ( meaning they have the same variables but different constants), then the lines are perpendicular.
## When to use each
When it comes to parallel and perpendicular lines, there are certain instances where one is preferred over the other. In general, parallel lines are used when two or more lines need to be the same distance apart from each other, while perpendicular lines are used when two lines need to intersect at a 90-degree angle. However, there are also some instances where neither parallel nor perpendicular lines are needed, such as when creating a freeform shape.
Here are some specific examples of when to use each:
– When creating a graph, use parallel lines to plot points that are the same distance apart from each other. This will make it easier to see any patterns that may emerge.
– When measuring something with a ruler or tape measure, use perpendicular lines to ensure accuracy. This is because it’s easier to line up the ruler or tape measure at a 90-degree angle than it is at any other angle.
– When drawing a picture or diagram, you may sometimes want to use neither parallel nor perpendicular lines in order to create a more freeform look. This can be especially effective if you’re trying to depict something organic, like a tree or clouds.
## Examples of when you would use each
There are many examples of when you would use each of these types of lines in geometry. For instance, if you were drawing a rectangle, you would use two parallel lines and two perpendicular lines. If you were drawing a square, you would use four perpendicular lines. If you were drawing a triangle, you would use three straight lines that are neither parallel nor perpendicular to each other.
## How to remember which is which
There are a few simple tricks you can use to keep parallel and perpendicular lines straight in your mind. First, remember that parallel lines never intersect, while perpendicular lines always intersect at right angles. You can also think of parallel lines as two trains running side by side on the same track – they’ll never crash into each other. On the other hand, perpendicular lines are like two cars driving on a road – they will meet at some point if they keep going long enough.
Another way to keep these concepts straight is to think about the prefixes “para-” and “per-“. “Para-” means “beside”, so Parallel lines run beside each other. “Per-” means “through”, so Perpendicular lines go through each other.
## Conclusion
We hope that this article has helped you understand the differences between parallel, perpendicular, and neither. With the knowledge of these three concepts in hand, you should be able to quickly determine which type of line is applicable in a variety of situations. Just remember that parallel lines never intersect while perpendicular lines always do and if no intersection occurs then it’s neither!
2. Parallel, perpendicular, or neither? It’s a question that’s often asked when it comes to various shapes, lines, and angles. But what does it all mean, and when does it apply?
At its most basic, parallel and perpendicular refer to the relationship between two lines. Parallel lines are lines that are always the same distance apart and never touch. Perpendicular lines are lines that intersect at a 90-degree angle.
But parallel and perpendicular aren’t just limited to lines. They can also be used to describe the relationship between two planes. A plane is a flat surface that extends in all directions, like a tabletop or a piece of paper. Parallel planes are planes that are always the same distance apart, and perpendicular planes are planes that intersect at a 90-degree angle.
In addition to lines and planes, parallel and perpendicular can also be used to describe the relationship between two angles. An angle is created when two lines intersect. A parallel angle is an angle that is the same size, while a perpendicular angle is an angle that intersects at a 90-degree angle.
So when it comes to parallel, perpendicular, or neither, the answer will depend on what two elements are being compared. If the elements are lines, then the answer is either parallel or perpendicular. If the elements are planes, then the answer is either parallel or perpendicular. And if the elements are angles, then the answer is either parallel or perpendicular.
No matter what type of elements are being compared, understanding parallel and perpendicular will help you identify the relationships between them. |
# How do you solve the system of equations algebraically 2x+5y=4, 3x+6y=5?
Jul 11, 2018
Solution: $x = \frac{1}{3} , y = \frac{2}{3}$
#### Explanation:
2 x+ 5 y = 4 ; (1) , 3 x+ 6 y = 5 ;(2) . Multiplying equation (1) by
$3$ and equation (2) by $2$ we get ;
6 x+ 15 y = 12 ; (3) , 6 x+ 12 y = 10 ;(4) . Subtracting equation
(4) from equation (3) we get,
(6 x+ 15 y)-(6 x+12 y)= 12-10; or
$\cancel{6 x} + 15 y - \cancel{6 x} - 12 y = 2 \mathmr{and} 3 y = 2 \therefore y = \frac{2}{3}$ Putting
$y = \frac{2}{3}$ in equation (2) we get, $3 x + 6 \cdot \frac{2}{3} = 5 \mathmr{and} 3 x + 4 = 5$ or
$3 x = 1 \mathmr{and} x = \frac{1}{3} \therefore x = \frac{1}{3} , y = \frac{2}{3}$
Solution: $x = \frac{1}{3} , y = \frac{2}{3}$ [Ans] |
# Right Angle – Definition, Examples, Facts
Welcome to Brighterly, the best place for children to learn and explore math concepts in a fun and engaging way! In this adventure, we’re going to dive deep into the world of right angles and unveil their secrets. So buckle up, and let’s start our exciting journey into the realm of right angles!
## What is a Right Angle?
In geometry, a right angle is an angle that measures precisely 90 degrees. This special angle is formed when two straight lines intersect, creating four equal angles. Right angles are commonly found in squares, rectangles, and other geometric shapes. They play a crucial role in many mathematical concepts and real-world applications, such as architecture, engineering, and even art!
## Right Angle Shape
A shape that has a right angle is called a right-angled shape. Some common right-angled shapes include squares, rectangles, and right-angled triangles. These shapes have at least one angle that measures exactly 90 degrees. You can identify a right angle by looking for a small square in the corner, which is often used to represent a 90-degree angle.
## Right Angle Triangle
A right angle triangle is a triangle with one angle measuring exactly 90 degrees. The side opposite the right angle is called the hypotenuse and is always the longest side of the triangle. The other two sides are known as the adjacent and opposite sides. Right angle triangles have unique properties and are used extensively in trigonometry, geometry, and various real-life applications.
## Right Angle Triangle Formula
The most famous formula related to right angle triangles is the Pythagorean theorem. This theorem states that the square of the length of the hypotenuse (c) is equal to the sum of the squares of the other two sides (a and b). In other words, `a² + b² = c²`. This formula helps us find the length of any side of a right angle triangle if we know the lengths of the other two sides.
## Right Angle Triangle Properties
Some key properties of right angle triangles include:
1. The sum of the angles in a triangle always adds up to 180 degrees, and since one angle is 90 degrees, the other two angles must be acute (less than 90 degrees).
2. The side opposite the right angle is the hypotenuse and is always the longest side.
3. The Pythagorean theorem is always true for right angle triangles.
## Right Angle Triangle Area
To find the area of a right angle triangle, you can use the following formula: `Area = ½ × base × height`. The base and height are the two sides that form the right angle, and they are perpendicular to each other.
## Right Angle Isosceles Triangle
A right angle isosceles triangle is a special type of right angle triangle where the two sides that form the right angle are equal in length. In this case, the two acute angles will also be equal, each measuring 45 degrees.
## How to Draw a Right Angle Using A Protractor?
To draw a right angle using a protractor, follow these steps:
1. Draw a straight line on a piece of paper.
2. Place the protractor’s center hole on the endpoint of the line, making sure the base line of the protractor aligns with the drawn line.
3. Locate the 90-degree mark on the protractor and draw a small dot.
4. Remove the protractor and draw a straight line connecting the endpoint of the first line to the dot. You now have a right angle!
## Practice Problems
1. In a right angle triangle, the lengths of two sides are 3 cm and 4 cm. Find the length of the hypotenuse.
2. If the hypotenuse of a right angle triangle measures 13 cm and one side measures 5 cm, what is the length of the other side?
3. Calculate the area of a right angle triangle with a base of 6 cm and a height of 8 cm.
4. Determine the angles of a right angle isosceles triangle.
## Conclusion
Throughout this informative and engaging article, we have discovered the fascinating properties of right angles, right angle triangles, and how they form the backbone of numerous mathematical concepts. As we journeyed through the world of right angles, we learned about their shapes, applications, and unique features that make them essential in both simple and complex mathematical problems.
By mastering these fundamental concepts, your child will be well-equipped to tackle more advanced topics in mathematics with confidence and enthusiasm. Here at Brighterly, we are dedicated to providing your child with an exciting and enjoyable learning experience that will spark their curiosity and inspire a lifelong love for math.
Remember, the adventure doesn’t end here! Keep exploring and learning with Brighterly as we continue to uncover the wonders of mathematics together. Happy learning!
## Frequently Asked Questions on Right Angle
### Why is a right angle called a right angle?
The term “right angle” comes from the Latin word “rectus,” which means “straight” or “upright.” A right angle is called a right angle because it is formed when two straight lines intersect, creating an angle that appears straight or upright.
### Can a triangle have more than one right angle?
No, a triangle cannot have more than one right angle. The sum of the angles in a triangle is always 180 degrees, and if a triangle had two right angles, the sum would be greater than 180 degrees, which is not possible.
### How can I find the area of a right angle triangle if I only know the length of the hypotenuse?
You cannot determine the area of a right angle triangle with only the length of the hypotenuse. You need to know the lengths of the two sides that form the right angle (base and height) to calculate the area.
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# Arithmetic Sequence Explicit Formula
## Arithmetic Sequence Explicit Formula
To get the mean of the data given to students, use the Arithmetic Sequence Explicit Formula. The most popular applications of this Arithmetic Sequence Explicit Formula are in statistics and economics. The arithmetic progression To determine the nth term in the arithmetic sequence, an explicit formula called the Arithmetic Sequence Explicit Formula is used. Each formula contains restrictions on how it can be applied to simplify things for pupils. Students are aware that nothing is faultless and that everything contains faults. The explicit formula for arithmetic sequence also has quite a lot of limitations, much like other formulas.
## What Is Arithmetic Sequence Explicit Formula?
The Arithmetic Sequence Explicit Formula is used by students to determine the Arithmetic Sequence Explicit of a series of numbers. This formula is used by students in statistics. Statistics is used by students to study and understand the data and organise it correctly so they can answer any questions regarding it.
### Derivation of Arithmetic Sequence Explicit Formula
Students will gain a better understanding of why it is crucial that they study the Arithmetic Sequence Explicit Formula by learning how it was derived. Students will have a better understanding of the Arithmetic Sequence Explicit Formula origins and possible applications by studying its derivation. Students will be able to adapt the method in this way to tackle additional problems quickly, such as different types of sequences using conventional formulas.
### Solved Example
Extramarks has provided a number of instances of the Arithmetic Sequence Explicit Formula that have been solved so that students can learn how to correctly use it and how to creatively use it to find the best and simplest way to solve a problem. Having some already solved examples or questions will help students understand how to answer. They will also be able to answer questions more accurately and faster during their exams. Students have the chance to utilise Extramarks’ resources to their fullest potential. Resources on Extramarks are easily accessible to students. Practice sheets, study notes, revision notes, NCERT solutions, previous year papers, answer keys, and study material for specific exams like NEET, JEE, CUET, etc. are some of the resources offered by Extramarks. Hence, students must use the Extramarks provided resources to study and understand the topics, formulas, and concepts of any chapter of any subject. |
# Video: Pack 3 β’ Paper 2 β’ Question 17
Pack 3 β’ Paper 2 β’ Question 17
04:54
### Video Transcript
The expression 12π₯ squared minus five π₯ minus two to the power of negative one multiplied by nine π₯ squared minus four can be simplified to the form ππ₯ plus π over ππ₯ plus π, where π, π, π, and π are integers. Find π, π, π, and π.
These questions are fairly common in higher exam papers. However, this one looks very different from the ones we usually come across. We can rewrite it slightly so that it looks similar to those we know by considering what the power of the first bracket means.
A negative power gives us the reciprocal of an expression or number. Remember the reciprocal of π₯ is just found by dividing one by π₯. That means π₯ to the power of negative π¦ is equal to one over π₯ to the power of π¦. For instance, two to the power of negative four is equal to one over two to the power of four or one over 16. Seven to the power of negative two is one over seven squared which is one over 49.
In this case, 12π₯ squared minus five π₯ minus two to the power of negative one is the same as one over 12π₯ squared minus five π₯ minus two to the power of one. Anything to the power of one is itself. So weβll disregard that power at this point. That means the expression in our question becomes one over 12π₯ squared minus five π₯ minus two multiplied by nine π₯ squared minus four.
Remember when weβre multiplying fractions, we can write the second fraction as nine π₯ squared minus four over one without changing its value. Finally, we can multiply these two expressions. Finally, multiplying the numerator of the first fraction by the numerator of the second gives us nine π₯ squared minus four. Similarly, multiplying the denominators gives us 12π₯ squared minus five π₯ minus two.
Now, this is where the expression starts to look recognizable. Whenever we have an algebraic fraction which needs simplifying, itβs sensible to find common factors. To do this, we first factorize both the numerator and the denominator. Letβs start with the numerator. We know this factorizes into two brackets since there are no common factors in each part of the expression. In fact, since itβs two square numbers with a minus sign between them, itβs called the difference of two squares.
To factorize an expression, itβs the difference of two squares. For example, π squared minus π squared. Itβs π plus π multiplied by π minus π. Nine π₯ squared minus four factorizes to three π₯ plus two multiplied by three π₯ minus two. Now, letβs factorize the denominator. Since two is a prime number, the only possible numbers that could be in the final part of each bracket are one and two. Weβll worry about the signs of these in a moment.
Now, letβs consider what the first term in each bracket could be. The factor pairs of 12π₯ squared are 12π₯ and π₯, six π₯ and two π₯, or four π₯ and three π₯. There is a little trick here. If we look at the expression nine π₯ squared minus four, that factorizes to three π₯ plus two multiplied by three π₯ minus two. When we simplify our fraction, weβre hoping that there are going to be some common factors. That means there should be one bracket in each of the expressions thatβs the same.
One of the brackets in the factorized form of 12π₯ squared minus five π₯ minus two must have both a three π₯ and a two in it then. It also means that the first bracket must have that other factor pair of 12π₯ squared. It must have four π₯ in it. Once weβve selected the numbers in each of our brackets, letβs expand it back out using the FOIL method to check our signs.
Four π₯ multiplied by three π₯ is 12π₯ squared. Four π₯ multiplied by two is eight π₯. One multiplied by three π₯ is three π₯. And one multiplied by two is two. In fact, we know it needs to be negative two, which tells us that one of the signs for either one or two is going to be negative. To make negative five π₯, we need negative eight π₯ plus three π₯. That means that the sign in the first bracket is a positive and the sign in the second bracket is a negative.
Now that we fully factorized both the numerator and the denominator, letβs put them back into our algebraic fraction. Notice that both the numerator and the denominator share the common factor of three π₯ minus two. This means we can divide through by three π₯ minus two. Our fraction simplifies to three π₯ plus two over four π₯ plus one.
Remember the question was asking us to find the value of π, π, π, and π. π is three, π is two, π is four, and π is one. |
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Pulley and Constraint Relations - JEE Important Topic
An Introduction to Pulley and Constraints
Last updated date: 23rd Mar 2023
Total views: 62.7k
Views today: 0.39k
We all have seen pulleys in our day-to-day life and also sometimes used them in various places such as well for pulling water. It is a very powerful method which includes a wheel and a string and uses small force to pull large objects. But we all know that there are some forces such as gravity or friction which restrict or oppose the motion of any body. In classical mechanics, these restrictions are called constraints. The type of motion which has constraints is called constraint motion.
In this article, we are going to understand pulley, constraints, relation between pulley and constraints, and problems related to pulley and constraints relation.
What is a Pulley?
A pulley is defined as a machine made up of metal or wood which uses a wheel and a rope for lifting heavy objects by applying small force. A pulley works when a force is applied on one side of the rope while the other side is connected to a heavy weight which needs to be pulled. On applying force, the rope moves downward through the wheel and the weight on the other side gets pulled up. This phenomenon is commonly used in pulling water from the wells.
Pulley System
What are Constraints?
The restrictions or opposition to the motion of any object caused by different factors which reduce its free movement are called constraints. The type of motion which has constraints is called constraint motion.
According to the constraint relation, the sum of products of all tensions in the strings and velocities of blocks connected to the strings is equal to 0. Also, it can be said that the total power by tension is 0. This relation can be represented as follows:
$\sum{T.\text{ }\overline{v}=0}$
If we differentiate the above equation with respect to time while keeping the velocity vector constant, we get another relation which is given as
$\sum{T.\text{ }\overline{a}=0}$
This is so because, when a velocity vector is kept constant, the sum of products of all tensions in the strings and the accelerations of blocks connected to the strings is equal to 0.
Relation Between Pulley and Constraints
According to the pulley and constraints relation, when it is given that the pulley is massless, then the tensions present on both the sides of the pulley are equal. This relation can be represented as follows:
$2{{T}_{P}}=T+T=2T$
Pulley with Tensions in Strings
Problems Related to Pulley and Constraint Relation
Problem-1: It is given that the pulley in the below diagram is massless and the mass of block A and B is 7kg and 9kg, respectively. Calculate the acceleration of the two blocks and tensions in string 1 and string 2.
Pulley with Masses A and B
Solution: The tensions in the strings are depicted below.
Pulley with Masses A and B with tension T1 and T2
Now, let $a$ be the acceleration of the two blocks and ${{T}_{1}}$ and ${{T}_{2}}$ be the tensions in the strings 1 and 2, respectively.
The formula for tension is
$T = mg + ma$
So, the tension in string 1 due to block A is
${{T}_{1}} = 7\times 10+7\times a$
${{T}_{1}} = 70 + 7a$
And, the tension in string 1 due to block B is
${{T}_{1}} = 9\times 10 + 9\times a$
${{T}_{1}} = 90 + 9a$
Also,
${{T}_{1}} - 70 = 7a$ and $90- {{T}_{1}} = 9a$
On subtracting both the equations, we get
$20 = 16a$
$a = 1.25mlse{{c}^{2}}$
Now, substitute the value of a in any of the equations.
${{T}_{1}} = 7 + 7\times 1.25$
${{T}_{1}} = 96.25N$
As the given pulley is massless,
${{T}_{2}} = 2{{T}_{1}}$
Thus,
${{T}_{2}} = 192.5N$
Summary
A pulley is a machine made up of metal or wood which uses a wheel and a rope for lifting heavy objects by applying small force. The restrictions on the motion of any object caused by different factors which reduce its free movement are called constraints. The pulley and constraints relation is given by the equation $2{{T}_{P}} = T + T = 2T$. In order to calculate the acceleration or tension in two blocks connected to a pulley, first use the formula of tension and then use the pulley constraint relation. On further solving, the acceleration and tension of the two blocks can be calculated.
Competitive Exams after 12th Science
FAQs on Pulley and Constraint Relations - JEE Important Topic
1. Define pulley.
A pulley is defined as a machine made up of metal or wood which uses a wheel and a rope for lifting heavy objects by applying small force. A pulley works when a force is applied on one side of the rope while the other side is connected to a heavy weight which needs to be pulled. On applying force, the rope moves downward through the wheel and the weight on the other side gets pulled up. This phenomenon is commonly used in pulling water from the wells.
2. Give any four everyday examples of the pulley.
Some everyday examples of pulley include the following:
• Lift uses a pulley to pull the passengers.
• Theatre curtains use pulley for pulling and dropping down the curtains.
• Gym equipment also contain pulley for exercises.
• Construction
3. What are constraints? Also, explain constraints relation.
The restrictions or opposition on the motion of any object caused by different factors which reduces its free movement are called constraints. The type of motion which has constraints is called constraint motion. |
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TO DO A CERTAIN WORK, B WOULD TAKE 3 TIMES AS LONG AS A AND C TOGETHER AND C TWICE AS LONG AS A AND B TOGETHER.THE THREE MEN TOGETHER COMPLETE THE WORK IN 10 DAYS. HOW LONG WOULD EACH TAKE SEPARATELY?
[Reveal] Spoiler:
A,B,C --24, 40 AND 30 DAYS RESPECTIVELY
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02 Aug 2009, 08:30
Let a,b and c be the units of work done by A, B and C respectively.
B WOULD TAKE 3 TIMES AS LONG AS A AND C TOGETHER => 3b = a+ c
C TWICE AS LONG AS A AND B TOGETHER => 2c = a +b
b = (a+c)/3
Substituting in the 2 equation we get
2c = a + (a+c)/3
6c = 3a + a + c
5c = 4a
a/c = 5/4
a:b:c = 5:3:4
Since all three can complete the work in 10 days
Total number of unit of wrk = (5+3+4)10 = 120
Number of days taken by A to complete the work = 120/5 = 24 days
Number of days taken by B to complete the work = 120/3 = 40 days
Number of days taken by A to complete the work = 120/4 = 30 days
Re: WORK DONE [#permalink] 02 Aug 2009, 08:30
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# WORK DONE
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Local Maxima And Minima
Maxima and Minima are one of the most common concepts in differential calculus. A branch of Mathematics called “Calculus of Variations” deals with the maxima and the minima of the functional. The calculus of variations is concerned with the variations in the functional, in which small change in the function leads to the change in the functional value.
The first variation is defined as the linear part of the change in the functional, and the second part of the variation is defined in the quadratic part. Functional is expressed as the definite integrals which involve the functions and their derivatives.
The functions that maximize or minimize the functional are can be found using the Euler – Lagrange of the calculus of variations. These two Latin maxima and minima words basically mean the maximum and minimum value of a function respectively, which is quite evident. The maxima and minima are collectively called “Extrema”. Here, we assume our function to be continuous for its entire domain. Before knowing how to find maxima and minima, we should first learn about derivatives. Assuming that you all know how to find derivatives, let us go ahead and learn about some curves. What are the curves?
Also, learn:
What are the Curves?
Figure 1: Curves
Figure 2: Value of a Function
A curve is defined as one-dimensional continuum. In figure 1, that curve is graph of a function $$f ~in~ x$$. $$f(x)$$ represents the value of function at $$x$$. The value of $$f$$ when $$x = a$$, will be $$f(a)$$ . Similarly, for $$B,~ C ~and ~D$$. You can refer fig. 2 to understand this. From the figure it is quite clear that the value of the given function has its maximum value at x=b, i.e. $$f(b)$$.
Interval of a function plays a very important role to find extreme values of a function. If the interval for which the function $$f$$ is defined in $$R$$, then we can’t talk about maxima and minima of $$f$$. We can understand it logically that though $$f(b)$$ appears to have the maximum value, we can’t be sure it has the largest value till we have seen the graph for its entire domain.
Local Maxima and Minima
We may not be able to tell whether $$f(b)$$ is the maximum value of $$f$$, but we can give some credit to point . We can do this by declaring $$B$$ as the local maximum for function $$f$$ . These are also called relative maxima and minima. These local maxima and minima are defined as:
• If $$f(a) \leq f(x)$$ for all $$x$$ in $$P’s$$ neighborhood (within the distance nearby $$P$$ , where $$x = a$$ ), $$f$$ is said to have a local minimum at $$x = a$$.
• If $$f(a) \geq f(x)$$ for all in $$P’s$$ neighborhood (within the distance nearby $$P$$, where $$x = a$$), $$f$$ is said to have a local maximum at $$x = a$$.
In the above example, $$B ~and~ D$$ are local maxima and $$A ~and~ C$$ are local minima. Local maxima and minima are together referred to as Local extreme.
Let us now take a point $$P$$, where $$x = a$$ and try to analyze the nature of the derivatives. There are total of four possibilities:
• If $$f'(a) = 0$$ , the tangent drawn is parallel to $$x -axis$$ , i.e. slope is zero. There are three possible cases:
• The value of $$f$$ , when compared to the value of $$f$$ at $$P$$ , increases if you move towards right or left of $$P$$ (Local minima: look like valleys)
• The value of $$f$$, when compared to the value of $$f$$ at $$P$$, decreases if you move towards right or left of $$P$$ (Local maxima: look like hills)
• The value of $$f$$ , when compared to the value of $$f$$ at $$P$$ , increases and decreases as you move towards left and right respectively of $$P$$(Neither: looks like a flat land)
• If, the tangent is drawn at a negative slope. The value of f'(a), at p, increases if you move towards left of and decreases if you move towards the right of . So, in this case, also, we can’t find any local extrema.
• If, the tangent is drawn at a positive slope. The value of f'(a), at P, increases if you move towards the right of and decreases if you move towards left of . So, in this case, we can’t find any local extrema.
• $$f’$$ doesn’t exist at point $$P$$, i.e. the function is not differentiable at $$P$$. This normally happens when the graph of $$f$$ has a sharp corner somewhere. All the three cases discussed in the previous point also hold true for this point.
To remember this, you can refer the Table 1.
Table 1: Various possibilities of derivatives of a function
Nature of f'(a) Nature of Slope Example Local Extremum f'(a) > 0 Positive Neither f'(a) < 0 Negative Neither f'(a) = 0 Zero Local Minimum Local Maximum Neither Not Defined Not Defined Local Minimum Local Maximum Neither
What is Critical Point?
In mathematics, a Critical point of a differential function of a real or complex variable is any value in its domain where its derivative is 0. We can hence infer from here that every local extremum is a critical point but every critical point need not be a local extremum. So, if we have a function which is continuous, it must have maxima and minima or local extrema. This means that every such function will have critical points. In case the given function is monotonic, the maximum and minimum values lie at the endpoints of the domain of the definition of that particular function.
Maxima and minima are hence very important concepts in the calculus of variations, which helps to find the extreme values of a function. You can use these two values and where they occur for a function using the first derivative method or the second derivative method.
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# Statistics - Hypothesis Testing a Proportion
A population proportion is the share of a population that belongs to a particular category.
Hypothesis tests are used to check a claim about the size of that population proportion.
## Hypothesis Testing a Proportion
The following steps are used for a hypothesis test:
1. Check the conditions
2. Define the claims
3. Decide the significance level
4. Calculate the test statistic
5. Conclusion
For example:
• Population: Nobel Prize winners
• Category: Born in the United States of America
And we want to check the claim:
"More than 20% of Nobel Prize winners were born in the US"
By taking a sample of 40 randomly selected Nobel Prize winners we could find that:
10 out of 40 Nobel Prize winners in the sample were born in the US
The sample proportion is then: $$\displaystyle \frac{10}{40} = 0.25$$, or 25%.
From this sample data we check the claim with the steps below.
## 1. Checking the Conditions
The conditions for calculating a confidence interval for a proportion are:
• The sample is randomly selected
• There is only two options:
• Being in the category
• Not being in the category
• The sample needs at least:
• 5 members in the category
• 5 members not in the category
In our example, we randomly selected 10 people that were born in the US.
The rest were not born in the US, so there are 30 in the other category.
The conditions are fulfilled in this case.
Note: It is possible to do a hypothesis test without having 5 of each category. But special adjustments need to be made.
## 2. Defining the Claims
We need to define a null hypothesis ($$H_{0}$$) and an alternative hypothesis ($$H_{1}$$) based on the claim we are checking.
The claim was:
"More than 20% of Nobel Prize winners were born in the US"
In this case, the parameter is the proportion of Nobel Prize winners born in the US ($$p$$).
The null and alternative hypothesis are then:
Null hypothesis: 20% of Nobel Prize winners were born in the US.
Alternative hypothesis: More than 20% of Nobel Prize winners were born in the US.
Which can be expressed with symbols as:
$$H_{0}$$: $$p = 0.20$$
$$H_{1}$$: $$p > 0.20$$
This is a 'right tailed' test, because the alternative hypothesis claims that the proportion is more than in the null hypothesis.
If the data supports the alternative hypothesis, we reject the null hypothesis and accept the alternative hypothesis.
## 3. Deciding the Significance Level
The significance level ($$\alpha$$) is the uncertainty we accept when rejecting the null hypothesis in a hypothesis test.
The significance level is a percentage probability of accidentally making the wrong conclusion.
Typical significance levels are:
• $$\alpha = 0.1$$ (10%)
• $$\alpha = 0.05$$ (5%)
• $$\alpha = 0.01$$ (1%)
A lower significance level means that the evidence in the data needs to be stronger to reject the null hypothesis.
There is no "correct" significance level - it only states the uncertainty of the conclusion.
Note: A 5% significance level means that when we reject a null hypothesis:
We expect to reject a true null hypothesis 5 out of 100 times.
## 4. Calculating the Test Statistic
The test statistic is used to decide the outcome of the hypothesis test.
The test statistic is a standardized value calculated from the sample.
The formula for the test statistic (TS) of a population proportion is:
$$\displaystyle \frac{\hat{p} - p}{\sqrt{p(1-p)}} \cdot \sqrt{n}$$
$$\hat{p}-p$$ is the difference between the sample proportion ($$\hat{p}$$) and the claimed population proportion ($$p$$).
$$n$$ is the sample size.
In our example:
The claimed ($$H_{0}$$) population proportion ($$p$$) was $$0.20$$
The sample proportion ($$\hat{p}$$) was 10 out of 40, or: $$\displaystyle \frac{10}{40} = 0.25$$
The sample size ($$n$$) was $$40$$
So the test statistic (TS) is then:
$$\displaystyle \frac{0.25-0.20}{\sqrt{0.2(1-0.2)}} \cdot \sqrt{40} = \frac{0.05}{\sqrt{0.2(0.8)}} \cdot \sqrt{40} = \frac{0.05}{\sqrt{0.16}} \cdot \sqrt{40} \approx \frac{0.05}{0.4} \cdot 6.325 = \underline{0.791}$$
You can also calculate the test statistic using programming language functions:
### Example
With Python use the scipy and math libraries to calculate the test statistic for a proportion.
import scipy.stats as stats
import math
# Specify the number of occurrences (x), the sample size (n), and the proportion claimed in the null-hypothesis (p)
x = 10
n = 40
p = 0.2
# Calculate the sample proportion
p_hat = x/n
# Calculate and print the test statistic
print((p_hat-p)/(math.sqrt((p*(1-p))/(n))))
Try it Yourself »
### Example
With R use the built-in prop.test() function to calculate the test statistic for a proportion.
# Specify the sample occurrences (x), the sample size (n), and the null-hypothesis claim (p)
x <- 10
n <- 40
p <- 0.20
# Calculate the sample proportion
p_hat = x/n
# Calculate and print the test statistic
(p_hat-p)/(sqrt((p*(1-p))/(n)))
Try it Yourself »
## 5. Concluding
There are two main approaches for making the conclusion of a hypothesis test:
• The critical value approach compares the test statistic with the critical value of the significance level.
• The P-value approach compares the P-value of the test statistic and with the significance level.
Note: The two approaches are only different in how they present the conclusion.
### The Critical Value Approach
For the critical value approach we need to find the critical value (CV) of the significance level ($$\alpha$$).
For a population proportion test, the critical value (CV) is a Z-value from a standard normal distribution.
This critical Z-value (CV) defines the rejection region for the test.
The rejection region is an area of probability in the tails of the standard normal distribution.
Because the claim is that the population proportion is more than 20%, the rejection region is in the right tail:
The size of the rejection region is decided by the significance level ($$\alpha$$).
Choosing a significance level ($$\alpha$$) of 0.05, or 5%, we can find the critical Z-value from a Z-table, or with a programming language function:
Note: The functions find the Z-value for an area from the left side.
To find the Z-value for a right tail we need to use the function on the area to the left of the tail (1-0.05 = 0.95).
### Example
With Python use the Scipy Stats library norm.ppf() function find the Z-value for an $$\alpha$$ = 0.05 in the right tail.
import scipy.stats as stats
print(stats.norm.ppf(1-0.05))
Try it Yourself »
### Example
With R use the built-in qnorm() function to find the Z-value for an $$\alpha$$ = 0.05 in the right tail.
qnorm(1-0.05)
Try it Yourself »
Using either method we can find that the critical Z-value is $$\approx \underline{1.6449}$$
For a right tailed test we need to check if the test statistic (TS) is bigger than the critical value (CV).
If the test statistic is bigger than the critical value, the test statistic is in the rejection region.
When the test statistic is in the rejection region, we reject the null hypothesis ($$H_{0}$$).
Here, the test statistic (TS) was $$\approx \underline{0.791}$$ and the critical value was $$\approx \underline{1.6449}$$
Here is an illustration of this test in a graph:
Since the test statistic was smaller than the critical value we do not reject the null hypothesis.
This means that the sample data does not support the alternative hypothesis.
And we can summarize the conclusion stating:
The sample data does not support the claim that "more than 20% of Nobel Prize winners were born in the US" at a 5% significance level.
### The P-Value Approach
For the P-value approach we need to find the P-value of the test statistic (TS).
If the P-value is smaller than the significance level ($$\alpha$$), we reject the null hypothesis ($$H_{0}$$).
The test statistic was found to be $$\approx \underline{0.791}$$
For a population proportion test, the test statistic is a Z-Value from a standard normal distribution.
Because this is a right tailed test, we need to find the P-value of a Z-value bigger than 0.791.
We can find the P-value using a Z-table, or with a programming language function:
Note: The functions find the P-value (area) to the left side of Z-value.
To find the P-value for a right tail we need to subtract the left area from the total area: 1 - the output of the function.
### Example
With Python use the Scipy Stats library norm.cdf() function find the P-value of a Z-value bigger than 0.791:
import scipy.stats as stats
print(1-stats.norm.cdf(0.791))
Try it Yourself »
### Example
With R use the built-in pnorm() function find the P-value of a Z-value bigger than 0.791:
1-pnorm(0.791)
Try it Yourself »
Using either method we can find that the P-value is $$\approx \underline{0.2145}$$
This tells us that the significance level ($$\alpha$$) would need to be bigger than 0.2145, or 21.45%, to reject the null hypothesis.
Here is an illustration of this test in a graph:
This P-value is bigger than any of the common significance levels (10%, 5%, 1%).
So the null hypothesis is kept at all of these significance levels.
And we can summarize the conclusion stating:
The sample data does not support the claim that "more than 20% of Nobel Prize winners were born in the US" at a 10%, 5%, or 1% significance level.
Note: It may still be true that the real population proportion is more than 20%.
But there was not strong enough evidence to support it with this sample.
## Calculating a P-Value for a Hypothesis Test with Programming
Many programming languages can calculate the P-value to decide outcome of a hypothesis test.
Using software and programming to calculate statistics is more common for bigger sets of data, as calculating manually becomes difficult.
The P-value calculated here will tell us the lowest possible significance level where the null-hypothesis can be rejected.
### Example
With Python use the scipy and math libraries to calculate the P-value for a right tailed hypothesis test for a proportion.
Here, the sample size is 40, the occurrences are 10, and the test is for a proportion bigger than 0.20.
import scipy.stats as stats
import math
# Specify the number of occurrences (x), the sample size (n), and the proportion claimed in the null-hypothesis (p)
x = 10
n = 40
p = 0.2
# Calculate the sample proportion
p_hat = x/n
# Calculate the test statistic
test_stat = (p_hat-p)/(math.sqrt((p*(1-p))/(n)))
# Output the p-value of the test statistic (right tailed test)
print(1-stats.norm.cdf(test_stat))
Try it Yourself »
### Example
With R use the built-in prop.test() function find the P-value for a right tailed hypothesis test for a proportion.
Here, the sample size is 40, the occurrences are 10, and the test is for a proportion bigger than 0.20.
# Specify the sample occurrences (x), the sample size (n), and the null-hypothesis claim (p)
x <- 10
n <- 40
p <- 0.20
# P-value from right-tail proportion test at 0.05 significance level
prop.test(x, n, p, alternative = c("greater"), conf.level = 0.95, correct = FALSE)\$p.value
Try it Yourself »
Note: The conf.level in the R code is the reverse of the significance level.
Here, the significance level is 0.05, or 5%, so the conf.level is 1-0.05 = 0.95, or 95%.
## Left-Tailed and Two-Tailed Tests
This was an example of a right tailed test, where the alternative hypothesis claimed that parameter is bigger than the null hypothesis claim.
You can check out an equivalent step-by-step guide for other types here:
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# LESSON 6-3 COMPOUND INEQUALITIES
## Presentation on theme: "LESSON 6-3 COMPOUND INEQUALITIES"— Presentation transcript:
LESSON 6-3 COMPOUND INEQUALITIES
Objective: To solve and graph inequalities containing “and” or “or”.
Real-World Connection
Do you like to swim? Does your family have a swimming pool? Determining how to keep the water in a pool chemically balanced is one way this skill is used in real-world situations.
Vocabulary compound inequality – two inequalities joined by the word and or the word or solution for and inequalities – (Intersection) any number that makes both inequalities true solution for or inequalities – (Union) any number that makes either inequality true
STEPS for AND problems Write the compound inequality as two inequalities joined by and. Solve each inequality. Simplify and write the solutions as one statement. Graph the solution.
Translate the verbal phrase into an inequality
Translate the verbal phrase into an inequality. Then graph the inequality. a. All real numbers that are greater than – 2 and less than 3. Inequality: – 2 < x < 3 Graph: b. All real numbers that are less than 0 or greater than or equal to 2. x < 0 or x ≥ 2 Inequality: Graph:
All real numbers that are less than –1 or greater than or equal to 4.
c. Inequality: x < –1 or x ≥ 4 d. All real numbers that are greater than or equal To –3 and less than 5. Inequality: x ≥ –3 and x < 5 = –3 ≤ x < 5
Write and graph a real-world inequality:
CAMERA CARS A crane sits on top of a camera car and faces toward the front. The crane’s maximum height and minimum height above the ground are shown. Write and graph a compound inequality that describes the possible heights of the crane.
Write and graph a real-world inequality:
SOLUTION Let h represent the height (in feet) of the crane. All possible heights are greater than or equal to 4 feet and less than or equal to 18 feet. So, the inequality is 4 ≤ h ≤ 18.
You try… Solve each inequality and graph the solution.
1) -6 < 3x < 15 2) -3 < 2x-1 < 7
You try… 3) 7 < -3a + 1 < 13 Solution: -4 < n < -2
You try… 4) The books were priced between \$3.50 and \$6.00, inclusive.
Solution: < b < 6
To solve OR problems, solve each inequality separately.
5) Solve –2x + 7 > 3 or 3x – 4 > 5.
You try… 6) Write an inequality that represents all real numbers that are at most –5 or at least 3. Graph your solution.
Remember, when written like this, it is an AND problem!
7) Graph 3 < 2m – 1 < 9 Remember, when written like this, it is an AND problem! 3 < 2m – 1 AND 2m – 1 < 9 Solve each inequality. Graph the intersection of 2 < m and m < 5. 5 -
Which inequalities describe the following graph?
-2 -1 -3 o y > -3 or y < -1 y > -3 and y < -1 y ≤ -3 or y ≥ -1 y ≥ -3 and y ≤ -1 Answer Now
Which is equivalent to -3 < y < 5?
y > -3 or y < 5 y > -3 and y < 5 y < -3 or y > 5 y < -3 and y > 5 Answer Now
Which is equivalent to x > -5 and x ≤ 1?
Summary Re-write the compound inequality into two problems and solve.
For and problems, combine the solutions into one statement. For or problems, solve each separately. Graph.
Homework: Study for quiz Wednesday, 1/15 Quiz covers chapter 6 sections 1-3
Explain the difference between “and” vs. “or” compound inequalities.
Exit Ticket Explain the difference between “and” vs. “or” compound inequalities. How do you solve a compound inequality? |
# Partial fractions
```Partial fractions
An algebraic fraction such as
partial fractions. Specifically
3x + 5
can often be broken down into simpler parts called
− 5x − 3
2x2
2
1
3x + 5
=
−
− 5x − 3
x − 3 2x + 1
2x2
In this unit we explain how this process is carried out.
In order to master the techniques explained here it is vital that you undertake plenty of practice
exercises so that they become second nature.
After reading this text, and/or viewing the video tutorial on this topic, you should be able to:
• explain the meaning of the terms ‘proper fraction’ and ‘improper fraction’
• express an algebraic fraction as the sum of its partial fractions
Contents
1
1. Introduction
2
2. Revision of adding and subtracting fractions
2
3. Expressing a fraction as the sum of its partial fractions
3
4. Fractions where the denominator has a repeated factor
5
5. Fractions in which the denominator has a quadratic term
6
6. Dealing with improper fractions
7
c mathcentre August 7, 2003
1. Introduction
An algebraic fraction is a fraction in which the numerator and denominator are both polynomial expressions. A polynomial expression is one where every term is a multiple of a power
of x, such as
5x4 + 6x3 + 7x + 4
The degree of a polynomial is the power of the highest term in x. So in this case the degree is
4.
The number in front of x in each term is called its coefficient. So, the coefficient of x4 is 5.
The coefficient of x3 is 6.
Now consider the following algebraic fractions:
x
x2 + 2
x3 + 3
x4 + x2 + 1
In both cases the numerator is a polynomial of lower degree than the denominator. We call
these proper fractions
With other fractions the polynomial may be of higher degree in the numerator or it may be of
the same degree, for example
x4 + x2 + x
x+4
3
x +x+2
x+3
and these are called improper fractions.
Key Point
If the degree of the numerator is less than the degree of the denominator the fraction is said to
be a proper fraction
If the degree of the numerator is greater than or equal to the degree of the denominator the
fraction is said to be an improper fraction
2. Revision of adding and subtracting fractions
We now revise the process for adding and subtracting fractions. Consider
1
2
−
x − 3 2x + 1
In order to add these two fractions together, we need to find the lowest common denominator.
In this particular case, it is (x − 3)(2x + 1).
c mathcentre August 7, 2003
2
We write each fraction with this denominator.
2
2(2x + 1)
1
x−3
=
and
=
x−3
(x − 3)(2x + 1)
2x + 1
(x − 3)(2x + 1)
So
2
1
2(2x + 1)
x−3
−
=
−
x − 3 2x + 1
(x − 3)(2x + 1) (x − 3)(2x + 1)
The denominators are now the same so we can simply subtract the numerators and divide the
result by the lowest common denominator to give
2
1
4x + 2 − x + 3
3x + 5
−
=
=
x − 3 2x + 1
(x − 3)(2x + 1)
(x − 3)(2x + 1)
Sometimes in mathematics we need to do this operation in reverse. In calculus, for instance,
or when dealing with the binomial theorem, we sometimes need to split a fraction up into its
component parts which are called partial fractions. We discuss how to do this in the following
section.
Exercises 1
Use the rules for the addition and subtraction of fractions to simplify
a)
2
3
+
x+1 x+3
b)
5
3
−
x−2 x+2
c)
4
2
−
2x + 1 x + 3
d)
1
2
−
3x − 1 6x + 9
3. Expressing a fraction as the sum of its partial fractions
In the previous section we saw that
2
1
3x + 5
−
=
x − 3 2x + 1
(x − 3)(2x + 1)
3x + 5
. How can we get this back to its component parts ?
(x − 3)(2x + 1)
By inspection of the denominator we see that the component parts must have denominators of
x − 3 and 2x + 1 so we can write
3x + 5
A
B
=
+
(x − 3)(2x + 1)
x − 3 2x + 1
where A and B are numbers. A and B cannot involve x or powers of x because otherwise the
terms on the right would be improper fractions.
The next thing to do is to multiply both sides by the common denominator (x − 3)(2x + 1).
This gives
(3x + 5)(x − 3)(2x + 1)
A(x − 3)(2x + 1) B(x − 3)(2x + 1)
=
+
(x − 3)(2x + 1)
x−3
2x + 1
Then cancelling the common factors from the numerators and denominators of each term gives
3x + 5 = A(2x + 1) + B(x − 3)
Now this is an identity. This means that it is true for any values of x, and because of this we
can substitute any values of x we choose into it. Observe that if we let x = − 12 the first term
on the right will become zero and hence A will disappear. If we let x = 3 the second term on
the right will become zero and hence B will disappear.
3
c mathcentre August 7, 2003
If x = −
1
2
1
3
− +5 = B − −3
2
2
7
7
= − B
2
2
from which
B = −1
Now we want to try to find A.
If x = 3
14 = 7A
so that A = 2.
Putting these results together we have
3x + 5
A
B
=
+
(x − 3)(2x + 1)
x − 3 2x + 1
2
1
=
−
x − 3 2x + 1
which is the sum that we started with, and we have now broken the fraction back into its
component parts called partial fractions.
Example
3x
as the sum of its partial fractions.
(x − 1)(x + 2)
Observe that the factors in the denominator are x − 1 and x + 2 so we write
Suppose we want to express
A
B
3x
=
+
(x − 1)(x + 2)
x−1 x+2
where A and B are numbers.
We multiply both sides by the common denominator (x − 1)(x + 2):
3x = A(x + 2) + B(x − 1)
This time the special values that we shall choose are x = −2 because then the first term on the
right will become zero and A will disappear, and x = 1 because then the second term on the
right will become zero and B will disappear.
If x = −2
−6 = −3B
−6
B =
−3
B = 2
c mathcentre August 7, 2003
4
If x = 1
3 = 3A
A = 1
Putting these results together we have
1
2
3x
=
+
(x − 1)(x + 2)
x−1 x+2
and we have expressed the given fraction in partial fractions.
Sometimes the denominator is more awkward as we shall see in the following section.
Exercises 2
Express the following as a sum of partial fractions
2x − 1
2x + 5
3
a)
b)
c)
(x + 2)(x − 3)
(x − 2)(x + 1)
(x − 1)(2x − 1)
d)
1
(x + 4)(x − 2)
4. Fractions where the denominator has a repeated factor
Consider the following example in which the denominator has a repeated factor (x − 1)2 .
Example
3x + 1
as the sum of its partial fractions.
(x − 1)2 (x + 2)
There are actually three possibilities for a denominator in the partial fractions: x − 1, x + 2 and
also the possibility of (x − 1)2 , so in this case we write
Suppose we want to express
C
A
B
3x + 1
+
=
+
(x − 1)2 (x + 2)
(x − 1) (x − 1)2 (x + 2)
where A, B and C are numbers.
As before we multiply both sides by the denominator (x − 1)2 (x + 2) to give
3x + 1 = A(x − 1)(x + 2) + B(x + 2) + C(x − 1)2
(1)
Again we look for special values to substitute into this identity. If we let x = 1 then the first
and last terms on the right will be zero and A and C will disappear. If we let x = −2 the first
and second terms will be zero and A and B will disappear.
If x = 1
4 = 3B
so that
B=
4
3
If x = −2
5
9
We now need to find A. There is no special value of x that will eliminate B and C to give us
A. We could use any value. We could use x = 0. This will give us an equation in A, B and C.
Since we already know B and C, this would give us A.
−5 = 9C
5
so that
C=−
c mathcentre August 7, 2003
But here we shall demonstrate a different technique - one called equating coefficients. We take
equation 1 and multiply-out the right-hand side, and then collect up like terms.
3x + 1 = A(x − 1)(x + 2) + B(x + 2) + C(x − 1)2
= A(x2 + x − 2) + B(x + 2) + C(x2 − 2x + 1)
= (A + C)x2 + (A + B − 2C)x + (−2A + 2B + C)
This is an identity which is true for all values of x. On the left-hand side there are no terms
involving x2 whereas on the right we have (A + C)x2 . The only way this can be true is if
A+C =0
This is called equating coefficients of x2 . We already know that C = − 59 so this means that
A = 59 . We also already know that B = 43 . Putting these results together we have
3x + 1
5
4
5
=
+
−
(x − 1)2 (x + 2)
9(x − 1) 3(x − 1)2 9(x + 2)
and the problem is solved.
Exercises 3
Express the following as a sum of partial fractions
a)
5x2 + 17x + 15
(x + 2)2 (x + 1)
x
2
(x − 3) (2x + 1)
b)
c)
x2 + 1
(x − 1)2 (x + 1)
5. Fractions in which the denominator has a quadratic term
Sometimes we come across fractions in which the denominator has a quadratic term which
cannot be factorised. We will now learn how to deal with cases like this.
Example
Suppose we want to express
(x2
5x
+ x + 1)(x − 2)
as the sum of its partial fractions.
Note that the two denominators of the partial fractions will be (x2 +x+1) and (x−2). When the
denominator contains a quadratic factor we have to consider the possibility that the numerator
can contain a term in x. This is because if it did, the numerator would still be of lower degree
than the denominator - this would still be a proper fraction. So we write
5x
Ax + B
C
=
+
(x2 + x + 1)(x − 2)
x2 + x + 1 x − 2
As before we multiply both sides by the denominator (x2 + x + 1)(x − 2) to give
5x = (Ax + B)(x − 2) + C(x2 + x + 1)
One special value we could use is x = 2 because this will make the first term on the right-hand
side zero and so A and B will disappear.
c mathcentre August 7, 2003
6
If x = 2
10
7
Unfortunately there is no value we can substitute which will enable us to get rid of C so instead
we use the technique of equating coefficients. We have
10 = 7C
and so
C=
5x = (Ax + B)(x − 2) + C(x2 + x + 1)
= Ax2 − 2Ax + Bx − 2B + Cx2 + Cx + C
= (A + C)x2 + (−2A + B + C)x + (−2B + C)
We still need to find A and B. There is no term involving x2 on the left and so we can state
that
A+C =0
10
10
Since C =
we have A = − .
7
7
The left-hand side has no constant term and so
−2B + C = 0
But since C =
so that
B=
C
2
10
5
then B = . Putting all these results together we have
7
7
10
x + 57
− 10
5x
7
7
=
+
(x2 + x + 1)(x − 2)
x2 + x + 1 x − 2
=
10
−10x + 5
+
2
7(x + x + 1) 7(x − 2)
=
5(−2x + 1)
10
+
2
7(x + x + 1) 7(x − 2)
Exercises 4
Express the following as a sum of partial fractions
a)
x2 − 3x − 7
(x2 + x + 2)(2x − 1)
b)
13
(2x + 3)(x2 + 1)
c)
x
(x2 − x + 1)(3x − 2)
6. Dealing with improper fractions
So far we have only dealt with proper fractions, for which the numerator is of lower degree than
the denominator. We now look at how to deal with improper fractions.
Consider the following example.
Example
Suppose we wish to express
7
4x3 + 10x + 4
in partial fractions.
x(2x + 1)
c mathcentre August 7, 2003
The numerator is of degree 3. The denominator is of degree 2. So this fraction is improper.
This means that if we are going to divide the numerator by the denominator we are going to
divide a term in x3 by one in x2 , which gives rise to a term in x. Consequently we express the
partial fractions in the form:
C
D
4x3 + 10x + 4
= Ax + B + +
x(2x + 1)
x
2x + 1
Multiplying both sides by the denominator x(2x + 1) gives
4x3 + 10x + 4 = Ax2 (2x + 1) + Bx(2x + 1) + C(2x + 1) + Dx
Note that by substituting the special value x = 0, all terms on the right except the third will be
zero. If we use the special value x = − 12 all terms on the right except the last one will be zero.
If x = 0
4=C
1
If x = −
2
1
4 10
+4 = − D
− −
8
2
2
1
1
− −5+4 = − D
2
2
1
1
= − D
−1
2
2
D = 3
Special values will not give A or B so we shall have to equate coefficients.
4x3 + 10x + 4 = Ax2 (2x + 1) + Bx(2x + 1) + C(2x + 1) + Dx
= 2Ax3 + Ax2 + 2Bx2 + Bx + 2Cx + C + Dx
= 2Ax3 + (A + 2B)x2 + (B + 2C + D)x + C
Now look at the term in x3 .
2A = 4
so that
A=2
Now look at the term in x2 . There is no such term on the left. So
A + 2B = 0
A = −2B
so that
so that
2
B = − = −1
2
Putting all these results together gives
4
3
4x3 + 10x + 4
= 2x − 1 + +
x(2x + 1)
x 2x + 1
and the problem is solved.
Exercise 5
Express the following as a sum of powers of x and partial fractions
a)
x3 + 1
x2 + 1
b)
2x4 + 3x2 + 1
x2 + 3x + 2
c mathcentre August 7, 2003
c)
7x2 − 1
x+3
8
Exercise 1
5x + 11
a)
(x + 1)(x + 3)
Exercise 2
1
1
+
a)
x+2 x−3
b)
2x + 16
(x − 2)(x + 2)
c)
10
(2x + 1)(x + 3)
b)
3
1
−
x−2 x+1
c)
3
6
1
1
−
d)
−
x − 1 2x − 1
6(x − 2) 6(x + 4)
d)
11
(3x − 1)(6x + 9)
Exercise 3
1
1
3
3
2
2
−
b)
+
+
−
a)
2
2
x + 2 (x + 2)
x+1
49(x − 3) 7(x − 3)
49(2x + 1)
1
1
1
c)
+
+
2
2(x − 1) (x − 1)
2(x + 1)
Exercise 4
3
2x + 1
−
a) 2
x + x + 2 2x − 1
Exercise 5
−x + 1
a) x + 2
x +1
9
b)
4
2x − 3
− 2
2x + 3 x + 1
b) 2x2 − 6x + 17 +
c)
6
45
−
x+1 x+2
−2x + 3
6
+
2
7(x − x + 1) 7(3x − 2)
c) 7x − 21 +
62
x+3
c mathcentre August 7, 2003
``` |
# What is 3/541 Simplified?
Are you looking to calculate how to simplify the fraction 3/541? In this really simple guide, we'll teach you exactly how to simplify 3/541 and convert it to the lowest form (this is sometimes calling reducing a fraction to the lowest terms).
To start with, the number above the line (3) in a fraction is called a numerator and the number below the line (541) is called the denominator.
So what we want to do here is to simplify the numerator and denominator in 3/541 to their lowest possible values, while keeping the actual fraction the same.
To do this, we use something called the greatest common factor. It's also known as the greatest common divisor and put simply, it's the highest number that divides exactly into two or more numbers.
In our case with 3/541, the greatest common factor is 1. Once we have this, we can divide both the numerator and the denominator by it, and voila, the fraction is simplified:
3/1 = 3
541/1 = 541
3 / 541
As you can see, 3/541 cannot be simplified any further, so the result is the same as we started with. Not very exciting, I know, but hopefully you have at least learned why it cannot be simplified any further!
So there you have it! You now know exactly how to simplify 3/541 to its lowest terms. Hopefully you understood the process and can use the same techniques to simplify other fractions on your own. The complete answer is below:
3/541
## Convert 3/541 to Decimal
Here's a little bonus calculation for you to easily work out the decimal format of the fraction we calculated. All you need to do is divide the numerator by the denominator and you can convert any fraction to decimal:
3 / 541 = 0.0055
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "What is 3/541 Simplified?". VisualFractions.com. Accessed on August 2, 2021. https://visualfractions.com/calculator/simplify-fractions/what-is-3-541-simplified/.
• "What is 3/541 Simplified?". VisualFractions.com, https://visualfractions.com/calculator/simplify-fractions/what-is-3-541-simplified/. Accessed 2 August, 2021.
### Preset List of Fraction Reduction Examples
Below are links to some preset calculations that are commonly searched for:
## Random Fraction Simplifier Problems
If you made it this far down the page then you must REALLY love simplifying fractions? Below are a bunch of randomly generated calculations for your fraction loving pleasure: |
# Introduction to Trig Ratios
The major functions of trigonometric ratios are sine, cosine, tangent, cosecant, secant and cotangent.
These functions are used to find the relationship between the angle and side of a right angled triangle.
Some specific angles are:
• 0° and 90°
• 45°
• 30° and 60°
## Trigonometric Ratios of 0° and 90°
In ΔABC, right – angled at B, and ∠BAC = θ
So, from ΔABC, we have
Sin θ = BC / AC
Cos θ = AB / AC
Tan θ = BC / AB
### Case I: ∠A is becoming small
If ∠A is made smaller and smaller in the ΔABC, till it becomes zero. As ∠A gets smaller and smaller, the length of the BC decreases. The point C gets closer to point B, and finally when A becomes very close to 0°, AC becomes almost the same as AB.
When ∠A is very close to 0°, BC gets very close to 0 and so the value of
Sin A = BC / AC is very close to 0.
Also, when A is very close to 0°, AC is same as AB and so the value of
Cos A = AB/AC is very close to 1.
From the above discussion, we have
Sin 0° = 0
Cosec 0° = 1 / Sin 0° = 1/0 = not defined
Cosec 0° = ∞
Cos 0° = 1
Sec 0° = 1 / Cos 0° = 1/1 = 1
Sec 0° = 1
Using Sin and Cos values, we can find Tan 0°
Tan 0° = Sin 0° / Cos 0° = 0
Tan 0° = 0
Also, Cot 0° = 1 / Tan 0° = 1/0 = not defined
Cot 0° = ∞
### Case II: ∠A is becoming large
Now, let’s see when ∠A is made larger and larger in ΔABC till it becomes 90°. As ∠A gets larger and larger, ∠C gets smaller and smaller. So, the length of the side AB goes on decreasing. The point A gets closer to point B. Finally when ∠A is very close to 90°, ∠C becomes very close to 0° and the side AC almost coincides with side BC.
When ∠C is very close to 0°, A is very close to 90°, side AC is nearly the same as side BC.
So, Sin A is very close to 1.
From above discussion we get,
Sin 90° = 1
Cosec 90° = 1
Cos 90°= 0
Sec 90° = ∞
Tan 90°= ∞
Cot 90° = 0
## Trigonometric Ratios of 45°
In ΔABC, right angled at B, if one angle is 45, then the other angle by angle sum property of triangle will also be 45.
∠A = ∠C = 45°
So, BC = AB (Isosceles triangle property)
Let, AB = BC = ‘a’
Then by Pythagoras theorem, AC2 = AB2 + BC2
AC2 = a2 + a2 = 2a2
AC = a√2.
Using formulas for trigonometric ratios:
Sin 45° = Side opposite to angle 45° = a/a√2 = 1/√2
Hypotenuse
Cos 45° = Side adjacent to angle 45° = a/a√2 = 1/√2
Hypotenuse
Tan 45° = Side opposite to angle 45° = a/a = 1
Side adjacent to angle 45°
Also, Cosec 45° = √2, Sec 45° = √2, Cot 45° = 1
## Trigonometric Ratios of 30° and 60°
Let ΔABC, be an equilateral triangle. So, ∠A = ∠B = ∠C = 60°
Draw perpendicular AD from A to the side BC.
Now, ΔABD ≅ ΔACD (by ASA)
Therefore, BD = DC
∠BAD = ∠CAD (by CPCT)
Consider, ΔABD
A = 30, B = 60, D = 90
Let AB = x
So, BD = x/2
And we will find the length of AD by Pythagoras theorem.
AB2 = AD2 + BD2
AB2 – BD2 = AD2
x2 – x2/4 = AD2
AD2 = 3x2/4
AD = x√3/2
Using formulas for trigonometric ratios:
Sin 30° = Side opposite to angle 30° =x/2 / x = 1/2
Hypotenuse
Cos 30° = Side adjacent to angle 30° = x√3/2 / x = √3/2
Hypotenuse
Tan 30° = Side opposite to angle 30° = x/2 / x√3/2 = 1/√3
Side adjacent to angle 30°
Also, Cosec 30° = 2, Sec 30° = 2/√3, Cot 30° = √3
Similarly,
Sin 60° = √3/2
Cosec 60° = 2/√3
Cos 60° = ½
Sec 60° 2
Tan 60° = √3
Cot 60° = 1/√3
Now try it yourself! Should you still need any help, click here to schedule live online session with e Tutor!
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### Reference Links:
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# 9.1: Exploring Geometric Sequences
Difficulty Level: At Grade Created by: CK-12
This activity is intended to supplement Calculus, Chapter 8, Lesson 1.
ID: 10992
Time required: 20 minutes
## Activity Overview
In this activity, students will explore geometric series. They will consider the effect of the value for the common ratio and first term using the TRANSFRM APP for the TI-84. Students will graphically analyze geometric series using graphs.
Topic: Sequences and Series
• Explore geometric sequences
Teacher Preparation and Notes
• This activity serves as a nice introduction to geometric series. Students will need the TRANSFRM APP on their TI-84. The accompanying worksheet can add depth to the questions and helpful key press explanation.
• An extension would be to solve infinite geometric series that converge using sigma notation and the limit of the partial sum formula.
• Students should make sure to turn off all Stat Plots before beginning the activity.
• To exit the Transformation App, go to the App menu, press enter on Transfrm and then select Uninstall.
• To download the Transformation Graphing Application, go to http://www.education.ti.com/calculators/downloads/US/Activities/Detail?id=10992 and select Transformation Graphing from the Applications header.
Associated Materials
## Example of Geometric Sequence
Students are shown the path of a ball that is bouncing. Show that the common ratio of the heights is approximately the same.
2.84.0=0.72.02.80.711.42.0=0.7\begin{align*} \frac{2.8}{4.0} =0.7 && \frac{2.0}{2.8} \approx 0.71 && \frac{1.4}{2.0} = 0.7\end{align*}
## Changing the Common Ratio
For the first part of this activity, students explore geometric sequences graphically by varying the value of r\begin{align*}r\end{align*}, the common ratio. Pressing the right or left arrows will change the value of r\begin{align*}r\end{align*} and updates the graph each time. Note that students will not see a graph for A\begin{align*}A\end{align*} values less than zero.
They should see that r\begin{align*}r-\end{align*}values between 0 and 1 could model the heights of the ball in the example.
Inquiry questions
• Why do you think the r\begin{align*}r-\end{align*}value is called the common ratio? [Answer: The ratios of consecutive terms are the same; They have that in common.]
• What would happen if you added all the terms of this sequence? For what common ratio conditions do you think the sum will diverge (get larger, and not converge to some number)?
Changing the Initial Value and the Common Ratio
Students will now change the equation in the Transfrm App to explore the changes in the graph of the general series, an=a1rn1\begin{align*}a_n = a_1 \cdot r^{n - 1}\end{align*}. Students are use the up and down arrow to move from one variable to another and use the left and right arrows to change the value of the variable. They should be sure to try negative and positive values for B\begin{align*}B\end{align*} for various values of A\begin{align*}A\end{align*}. (Y1=BA(X1)\begin{align*}(Y1 = B*A^{\land}(X - 1)\end{align*}
Inquiry question
• Which variable seems to have a more profound effect on the sequence? Explain.
## Extension – Deriving and Applying the Partial Sum Formula
On the worksheet, students are shown the derivation of the formula for the sum of a finite geometric series. You may need to explain in detail the substitution in the third line. For example, a2=ra1\begin{align*}a_2 = r \cdot a_1\end{align*}, so then a3=ra2=r(ra1)=r2a1\begin{align*}a_3 = r \cdot a_2 = r \cdot (r \cdot a_1) = r^2 \cdot a_1\end{align*}. In the fourth line, r\begin{align*}r\end{align*} is multiplied by both sides, changing rn1\begin{align*}r^{n - 1}\end{align*} to rn\begin{align*}r^n\end{align*}.
Students can apply the formula to find the sum of the series given on the worksheet.
Inquiry questions:
• If you cut a piece of paper in half, then cut one of the halves in half and repeated the process, what is the sum of all these fractional pieces? [Answer: one whole]
• However, if you gave your teacher one penny on the first day of class, 2 on the next, then 4, 8, 16, 32, ... would this number converge?
• How would the formula for a finite geometric sequence look if r\begin{align*}r\end{align*} is a fractional number and n\begin{align*}n\end{align*} is increasing?
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# How do you find the quotient of (9m^3-6m^2+3m+2)div(m-1) using long division?
Jul 21, 2018
$9 {m}^{2} + 3 m + \frac{2}{m - 1}$
#### Explanation:
$\left(9 {m}^{3} - 6 {m}^{2} + 3 m + 2\right) \div \left(m - 1\right)$
$\textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{w h i t e}{\ldots \ldots \ldots \ldots} 9 {m}^{2} + 3 m$
$m - 1 | \overline{9 {m}^{3} - 6 {m}^{2} + 3 m + 2}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots .} \underline{9 {m}^{3} - 9 {m}^{2}}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} 3 {m}^{2} - 3 m$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} \underline{3 {m}^{2} - 3 m}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} 0 + 0 + 2$
$\frac{9 {m}^{3} - 6 {m}^{2} + 3 m + 2}{m - 1} = 9 {m}^{2} + 3 m + \frac{2}{m - 1}$ |
GeoGebra Classroom
# Inscribed Angle Theorem
## Activity #1
Recall that in a circle, an INSCRIBED ANGLE is an angle whose vertex lies on the circle, and its sides are chords that interact the circle at two distinct points. In the activity below, the blue angle is an INSCRIBED ANGLE that intercepts the red arc. The red angle is a central angle (of the circle) that also intercepts the same red arc. In fact, this applet was designed so that both the inscribed angle and central angle always intercept the same red arc. 1) Drag the green slider all the way to the right in the applet below and watch what happens. 2) Now drag the green slider all the way back. Move any one or more of the blue and/or red points around and repeat step (1).
Answer each one of the questions below so they can be submitted in the final activity. Use either the PDf or a piece of paper. Answer in complete sentences: 1) How does the measure of any central angle of a circle compare with the measure of its intercepted arc? 2) According to what you've observed in the activity above, how does the measure of the inscribed angle compare with the measure of the central angle (that intercepts the same arc?) 3) Use your results from (1) and (2) to describe how one could find the measure of an inscribed angle given the measure of the arc it intercepts.
## Activity 2. Inscribed Angle meets the Diameter
In this activity, the central angle always remains a straight angle (180 degrees). Therefore the intercepted arc is a semicircle. Click on the pink checkbox to show the inscribed angle. Notice how the inscribed angle and central angle both intercept the same arc. Use the inscribed angle theorem (activity 1) you've just learned to make a conjecture, an educated guess, as to what the measure of the inscribed angle in this applet should be. Be sure to move points B, C, and the pink vertex of the inscribed angle around as well. (You can also change the radius of the circle if you wish.) |
# Binary numbers and logical operators
We have looked at simple numbers and operations before. In this article you will learn how numbers work inside the computer and a some of magic to go along with that 🙂
More detailed: While this is not directly useful in web applications or most desktop applications, it is very useful to know.
In this article you will learn how to use binary numbers in Python, how to convert them to decimals and how to do bitwise operations on them.
Related course:
Complete Python Bootcamp: Go from zero to hero in Python
## Binary numbers
At the lowest level, the computer has no notion whatsoever of numbers except ‘there is a signal’ or ‘these is not a signal’. You can think of this as a light switch: Either the switch is on or it is off.
This tiny amount of information, the smallest amount of information that you can store in a computer, is known as a bit. We represent a bit as either low (0) or high (1).
To represent higher numbers than 1, the idea was born to use a sequence of bits. A sequence of eight bits could store much larger numbers, this is called a byte. A sequence consisting of ones and zeroes is known as binary. Our traditional counting system with ten digits is known as decimal.
Lets see that in practice:
```# Prints out a few binary numbers. print int('00', 2) print int('01', 2) print int('10', 2) print int('11', 2)```
The second parameter 2, tells Python we have a number based on 2 elements (1 and 0). To convert a byte (8 bits) to decimal, simple write a combination of eight bits in the first parameter.
```# Prints out a few binary numbers. print int('00000010', 2) # outputs 2 print int('00000011', 2) # outputs 3 print int('00010001', 2) # outputs 17 print int('11111111', 2) # outputs 255```
How does the computer do this? Every digit (from right to left) is multiplied by the power of two.
The number ‘00010001‘ is (1 x 2^0) + (0 x 2^1) + (0 x 2^2) + (0 x 2^3) + (1 x 2^4) + (0 x 2^5) + (0 x 2^6) + (0 x 2^7) = 16 + 1 = 17. Remember, read from right to left.
The number ‘00110010’ would be (0 x 2^0) + (1 x 2^1) + (0 x 2^2) + (0 x 2^3) + (1 x 2^4) + (1 x 2^5) + (0 x 2^6) + (0 x 2^7) = 32+16+2 = 50.
Try the sequence ‘00101010’ yourself to see if you understand and verify with a Python program.
## Logical operations with binary numbers
Binary Left Shift and Binary Right Shift
Multiplication by a factor two and division by a factor of two is very easy in binary. We simply shift the bits left or right. We shift left below:
Bit 4 Bit 3 Bit 2 Bit 1
0 1 0 1
1 0 1 0
Before shifting (0,1,0,1) we have the number 5 . After shifting (1,0,1,0) we have the number 10. In python you can use the bitwise left operator (<<) to shift left and the bitwise right operator (>>) to shift right.
```inputA = int('0101',2) print "Before shifting " + str(inputA) + " " + bin(inputA) print "After shifting in binary: " + bin(inputA << 1) print "After shifting in decimal: " + str(inputA << 1)```
Output:
```Before shifting 5 0b101 After shifting in binary: 0b1010 After shifting in decimal: 10```
## The AND operator
Given two inputs, the computer can do several logic operations with those bits. Let’s take the AND operator. If input A and input B are positive, the output will be positive. We will demonstrate the AND operator graphically, the two left ones are input A and input B, the right circle is the output:
In code this is as simple as using the & symbol, which represents the Logical AND operator.
```# This code will execute a bitwise logical AND. Both inputA and inputB are bits. inputA = 1 inputB = 1 print inputA & inputB # Bitwise AND```
By changing the inputs you will have the same results as the image above. We can do the AND operator on a sequence:
```inputA = int('00100011',2) # define binary sequence inputA inputB = int('00101101',2) # define binary sequence inputB print bin(inputA & inputB) # logical AND on inputA and inputB and output in binary```
Output:
`0b100001 # equals 00100001`
This makes sense because if you do the operation by hand:
```00100011 00101101 -------- Logical bitwise AND 00100001```
## The OR operator
Now that you have learned the AND operator, let’s have a look at the OR operator. Given two inputs, the output will be zero only if A and B are both zero.
To execute it, we use the | operator. A sequence of bits can simply be executed like this:
```inputA = int('00100011',2) # define binary number inputB = int('00101101',2) # define binary number print bin(inputA) # prints inputA in binary print bin(inputB) # prints inputB in binary print bin(inputA | inputB) # Execute bitwise logical OR and print result in binary```
Output:
```0b100011 0b101101 0b101111```
## The XOR operator
This is an interesting operator: The Exclusive OR or shortly XOR.
To execute it, we use the ^ operator. A sequence of bits can simply be executed like this:
```inputA = int('00100011',2) # define binary number inputB = int('00101101',2) # define binary number print bin(inputA) # prints inputA in binary print bin(inputB) # prints inputB in binary print bin(inputA ^ inputB) # Execute bitwise logical OR and print result in binary```
Output:
```0b100011 0b101101 0b1110```
## 7 thoughts on “Binary numbers and logical operators”
1. - February 20, 2016
How to do a logical NOT operation using this way?
1. - February 22, 2016
You could use the ~ operator if you want the complements. An overview of the operators is here: https://wiki.python.org/moin/BitwiseOperators
2. - August 29, 2016
print “After shifting in binary: ” + bin(inputA << 1)
Did you know the << is getting displayed in HTML code.
2. - June 1, 2015
why is this happening???
>>> A = int(‘00100011’,2)
>>> B = int(‘00101101’,2)
>>> print bin(A & B)
0b100001
>>> print bin(A and B)
0b101101
1. - June 1, 2015
They are different operators, in the same way that * and + are different operators. The first one compares the bits, the second one compares both statements and should be used as if (a and b) then; In detail:
The first is a bitwise logical and. It compares the bits, if both are 1 it will be 1, otherwise 0. This means the computer simply looks if there is ‘high’ on both numbers at the same position.
``` 0b100011 0b101101 ------------- 0b100001 ```
The second is not a bitwise logical and operator and simply the normal and operator. If A and B then true.
I hope I explained well, if you have any questions feel free to ask. Hope you enjoy my site 🙂
1. - July 22, 2015
wish they made a quick and to the point tutorial for all langauges as you have done.
1. - October 16, 2015
Agreed. These tuts are amazing. They are concise, yet packed full of vital information. This is the only tutorial series I’ve found that is either A) able to go into bitwise operation or B) done it well at all. |
### APTITUDE – problems on SQUARE AND CUBE ROOTS
Formula:
The Product of two same numbers in easiest way as follow.
Example:let us calculate the product of 96*96
Solution: Here every number must be compare with the 100.
See here the given number 96 which is 4 difference with the 100.
so subtract 4 from the 96 we get 92 ,then the square of the
number 4 it is 16 place the 16 beside the 92 we get answer
as 9216.
9 6
– 4
————–
9 2
————–
4*4=16
9 2 1 6
therefore square of the two numbers 96*96=9216.
Example: Calculate product for 98*98
Solution: Here the number 98 is having 2 difference when compare
to 100 subtract 2 from the number then we get 96 square the
number 2 it is 4 now place beside the 96 as 9604
9 8
– 2
————-
9 6
————-
2*2=4
9 6 0 4.
so, we get the product of 98*98=9604.
Example: Calculate product for 88*88
Solution: Here the number 88 is having 12 difference when compare
to 100 subtract 12 from the 88 then we get 76 the square of the
number 12 is 144 (which is three digit number but should place
only two digit beside the 76) therefore in such case add one to
6 then it becomes 77 now place 44 beside the number 77 we will get
7744.
88
-12
————
76
———–
12*12=144
76
+ 144
——————–
7744
——————–
Example: Find the product of the numbers 46 *46?
Solution:consider the number 50=100/2. Now again go comparision with
the number which gets when division with 100.here consider the number
50 which is nearer to the number given. 46 when compared with the
number 50 we get the difference of 4. Now subtract the number 4 from
the 46, we get 42. As 50 got when 100 get divided by 2.
so, divided the number by 2 after subtraction.
42/2=21 now square the the number 4 i.e, 4*4=16
just place the number 16 beside the number 21
we get 2116.
4 6
4
—————-
4 2 as 50 = 100/2
42/2=21
now place 4*4=16 beside 21
2 1 1 6
Example: Find the product of the numbers 37*37
Solution:
consider the number 50=100/2
now again go comparision with the number which gets when
division with 100.
here consider the number 50 which is nearer to the number given.
37 when compared with the number 50 we get the difference of 13.
now subtract the number 13 from the 37, we get 24.
as 50 got when 100 get divided by 2.
so, divided the number by 2 after subtraction.
24/2=12
now square the the number 13 i.e, 13*13=169
just place the number 169 beside the number 21
now as 169 is three digit number then add 1 to 2 we get
1t as 13 then place 69 beside the 13
we get 1369.
3 7
1 3
—————–
2 4 as 50 = 100/2
24/2=12
square 13* 13=169
1 2
+ 1 6 9
———————–
1 3 6 9
————————-
Example: Find the product of 106*106
Solution: now compare it with 100 ,
The given number is more then 100
then add the extra number to the given number.
That is 106+6=112
then square the number 6 that is 6*6=36
just place beside the number 36 beside the 112,then
we get 11236.
1 0 6
+ 6
———————
1 1 2
——————–
now 6* 6=36 place this beside the number 112, we get
1 1 2 3 6
Square root: If x2=y ,we say that the square root of y
is x and we write ,√y=x.
Cube root: The cube root of a given number x is the number
whose cube is x. we denote the cube root of x by x1/3 .
Examples:
1.Evaluate 60841/2 by factorization method.
Solution: Express the given number as the product of prime
factors. Now, take the product of these prime factors choosing
one out of every pair of the same primes. This product gives the
square root of the given number.
Thus resolving 6084 in the prime factors ,we get 6084
2 6024
2 3042
3 1521
3 507
13 169
13
6084=21/2 *31/2 *131/2
60841/2=2*3*13=78.
2.what will come in place of question mark in each of the following
questions?
i)(32.4/?)1/2 = 2
ii)86.491/2 + (5+?1/2)2 =12.3
Solution: 1) (32.4/x)1/2=2
Squaring on both sides we get
32.4/x=4
=>4x=32.4
=>x=8.1
ii)86.491/2 + √(5+x2)=12.3
solutin:86.491/2 + (5+x1/2 )=12.3
9.3+ √(5+x1/2 )=12.3
=> √(5+x1/2 ) =12.3-9.3
=> √(5+x1/2 )=3
Squaring on both sides we get
(5+x1/2 )=9
x1/2 =9-5
x1/2 =4
x=2.
3.√ 0.00004761 equals:
Solution: √ (4761/108)
√4761/√ 108
. 69/10000
0.0069.
4.If √18225=135,then the value of
√182.25 + √1.8225 + √ 0.018225 + √0.00018225.
Solution: √(18225/100) +√(18225/10000) +
√(18225/1000000) +√(18225/100000000)
=√(18225)/10 + (18225)1/2/100 +
√(18225)/1000 + √(18225)/10000
=135/10 + 135/100 + 135/1000 + 135/10000
=13.5+1.35+0.135+0.0135=14.9985.
5.what should come in place of both the question
marks in the equation (?/ 1281/2= (162)1/2/?) ?
Solution: x/ 1281/2= (162)1/2/x
=>x1/2= (128*162)1/2
=> x1/2= (64*2*18*9)1/2
=>x2= (82*62*32)
=>x2=8*6*3
=>x2=144
=>x=12.
6.If 0.13 / p1/2=13 then p equals
Solution: 0.13/p2=13
=>p2=0.13/13
=1/100
p2=√(1/100)
=>p=1/10
therefore p=0.1
7.If 13691/2+(0.0615+x)1/2=37.25 then x is equals to:
Solution
37+(0.0615+x)1/2=37.25(since 37*37=1369)
=>(0.0615+x)1/2=0.25
Squaring on both sides
(0.0615+x)=0.0625
x=0.001
x=10-3.
8.If √(x-1)(y+2)=7 x& y being positive whole numbers then
values of x& y are?
Solution: √(x-1)(y+2)=7
Squaring on both sides we get
(x-1)(y+2)=72
x-1=7 and y+2=7
therefore x=8 , y=5.
9.If 3*51/2+1251/2=17.88.then what will be the
value of 801/2+6*51/2?
Solution: 3*51/2+1251/2=17.88
3*51/2+(25*5)1/2=17.88
3*51/2+5*51/2=17.88
8*51/2=17.88
51/2=2.235
therefore 801/2+6 51/2=(16*51/2)+6*1/25
=4 51/2+6 51/2
=10*2.235
=22.35
10.If 3a=4b=6c and a+b+c=27*√29 then Find c value is:
Solution: 4b=6c
=>b=3/2*c
3a=4b
=>a=4/3b
=>a=4/3(3/2c)=2c
therefore a+b+c=27*291/2
2c+3/2c+c=27*291/2
=>4c+3c+2c/2=27*291/2
=>9/2c=27*291/2
c=27*291/2*2/9
c=6*291/2
11.If 2*3=131/2 and 3*4=5 then value of 5*12 is
Solution:
Here a*b=(a2+b2)1/2
therefore 5*12=(52+122)1/2
=(25+144)1/2
=1691/2
=13
12.The smallest number added to 680621 to make
the sum a perfect square is
Solution: Find the square root number which
is nearest to this number
8 680621 824
64
162 406
324
1644 8221
6576
1645
therefore 824 is the number ,to get the nearest
square root number take (825*825)-680621
therefore 680625-680621=4
hence 4 is the number added to 680621 to make it
perfect square.
13.The greatest four digit perfect square number is
Solution: The greatest four digit number is 9999.
now find the square root of 9999.
9 9999 99
81
189 1819
1701
198
therefore 9999-198=9801 which is required number.
14.A man plants 15376 apples trees in his garden and arranges
them so, that there are as many rows as there are apples trees
in each row .The number of rows is.
Solution: Here find the square root of 15376.
1 15376 124
1
22 53
44
244 976
976
0
therefore the number of rows are 124.
15.A group of students decided to collect as many paise from
each member of the group as is the number of members. If the
total collection amounts to Rs 59.29.The number of members
in the group is:
Solution: Here convert Money into paise.
59.29*100=5929 paise.
To know the number of member ,calculate the square root of 5929.
7 5929 77
49
147 1029
1029
0
Therefore number of members are 77.
16.A general wishes to draw up his 36581 soldiers in the form
of a solid square ,after arranging them ,he found that some of
them are left over .How many are left?
square root of given number the remainder will be the left man
1 36581 191
1
29 265
261
381 481
381
100(since remaining)
Therefore the left men are 100.
17.By what least number 4320 be multiplied to obtain number
which is a perfect cube?
Solution: find l.c.m for 4320.
2 4320
2 2160
2 1080
2 540
2 270
3 135
3 45
3 15
5
4320=25 * 33 * 5
=23 * 33 * 22 *5
so make it a perfect cube ,it should be multiplied by 2*5*5=50 |
# The learning nerd!
On a mission to learn everything!
# Binary addition and subtraction — February 15, 2015
## Intro
This is a continuation of the learning binary series. Be sure to read learning binary before reading this post.
The basic values column
1000 | 100 | 10 | 1
Example
1 1 1 5 9 5 + 3 5 6 4 _ _ _ _ 5 1 5 9
We take the first digit and carry that. While the second digit remains below.
The digit you carried is added to the number.
5 + 4 = 9 > 9 Is less then 10 so we don’t carry over.
9 + 5 = 15 > 15 is higher then 10. We need to carry a 1 over into the 100’s column.
1 + 5 + 5 = 11 > 11 is higher then 10. We need to carry a 1 over into the 1000’s column.
1 + 1 + 3 = 5 > 5 is less then 10 so we don’t carry over.
Binary addition is the same concept. Except for your using binary numbers 1 and 0.
#### The five basic addition rules
0 + 0 = 0
1 + 0 = 1
10 + 0 = 10
1 + 1 = 10 > 1 + 1 = 2 > 10 = 2 in binary
10 + 1 = 11
Example
1 1 1 1 0 1 1 + 1 0 1 1 _ _ _ _ _ 1 0 1 1 0
1 + 1 = 10 > We carry the 1
1 + 1 + 1 = 10 + 1 = 11 > We carry the 1
1 + 0 = 1
1 + 1 = 10 > Since there is no more columns to carry. We put 10.
#### You can double check
1011 = 11
1011 = 11
11 + 11 = 22
Answer 10110 = 22
# Binary subtraction
Example
3 4 16 3 3 5 6 4 − 2 3 4 9 0 _ _ _ _ _ _ 1 0 0 7 4
4 – 0 = 4
Borrow 1 to make 16 – 9 = 7
Borrow to make 4 – 4 = 0
Borrow to make 3 – 3 = 0
3 – 2 = 1
10074
Onto binary subtraction
#### The four basic subtraction rules
0 – 0 = 0
1 – 0 = 1
1 – 1 = 0
10 – 1 = 1
0 1 10 10 1 0 1 0 1 − 1 1 1 0 _ _ _ _ _ _ 0 0 1 1 1
1 – 0 = 1
Borrow 1 to make 10 – 1 = 1
Borrow 1 to make 10 – 1 = 1
Borrow 1 to make 1 – 1 = 0
Borrow 0 to make 0
#### You can double check
10101 = 21
1110 = 14
21 – 14 = 7
Answer 111 = 7
I hope you enjoy my tutorial on how to subtract and add binary numbers. The next guide in this series will be how to multiply and divide binary numbers.
Categorised as: Binary
1. Domenica says:
If I saw this post back in October when I had to learn assembly code this would have been amazing helpful. The professor over complicated things. I am bookmarking this for future reference! |
# How do you solve the system 2x-3y=-24 and x+6y=18 using substitution?
Jul 15, 2017
See a solution process below:
#### Explanation:
Step 1) Solve the second equation for $x$:
$x + 6 y = 18$
$x + 6 y - \textcolor{red}{6 y} = 18 - \textcolor{red}{6 y}$
$x + 0 = 18 - 6 y$
$x = 18 - 6 y$
Step 2) Substitute $\left(18 - 6 y\right)$ for $x$ in the first equation and solve for $y$:
$2 x - 3 y = - 24$ becomes:
$2 \left(18 - 6 y\right) - 3 y = - 24$
$\left(2 \times 18\right) - \left(2 \times 6 y\right) - 3 y = - 24$
$36 - 12 y - 3 y = - 24$
$36 + \left(- 12 - 3\right) y = - 24$
$36 + \left(- 15 y\right) = - 24$
$36 - 15 y = - 24$
$- \textcolor{red}{36} + 36 - 15 y = - \textcolor{red}{36} - 24$
$0 - 15 y = - 60$
$- 15 y = - 60$
$\frac{- 15 y}{\textcolor{red}{- 15}} = \frac{- 60}{\textcolor{red}{- 15}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 15}}} y}{\cancel{\textcolor{red}{- 15}}} = 4$
$y = 4$
Step 3) Substitute $4$ for $y$ in the solution to the second equation at the end of Step 1 and calculate $x$:
$x = 18 - 6 y$ becomes:
$x = 18 - \left(6 \cdot 4\right)$
$x = 18 - 24$
$x = - 6$
The solution is: $x = - 6$ and $y = 4$ or $\left(- 6 , 4\right)$
Jul 15, 2017
color(blue)(y=4,x=-6
#### Explanation:
$\therefore 2 x - 3 y = - 24 - - - - \left(1\right)$
$x + 6 y = 18 - - - - \left(2\right)$
:.
in (2) $x = 18 - 6 y$
substitute color(blue)(x=18-6y in (1)
$\therefore 2 \left(\textcolor{b l u e}{18 - 6 y}\right) - 3 y = - 24$
$\therefore 36 - 12 y - 3 y = - 24$
$\therefore - 15 y = - 24 - 36$
multiply both sides by$- 1$
$\therefore 15 y = 24 + 36$
$\therefore 15 y = 60$
:.color(blue)(y=4
substitute color(blue)(y=4 in (2)
$\therefore x + 6 \left(\textcolor{b l u e}{4}\right) = 18$
$\therefore x + 24 = 18$
$\therefore x = 18 - 24$
:.color(blue)(x=-6
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)(check:
substitute color(blue)(y=4 andcolor(blue)(x=-6 in (2)
:.(color(blue)(-6))+6(color(blue)(4))=18))
$\therefore - 6 + 24 = 18$
:.color(blue)(18=18 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 4.4: Triangle Congruence Using ASA, AAS, and HL
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Use the ASA Congruence Postulate, AAS Congruence Theorem, and the HL Congruence Theorem.
• Complete two-column proofs using SSS, SAS, ASA, AAS, and HL.
## Review Queue
1. Write a two-column proof.
Given: \begin{align*}\overline{AD} \cong \overline{DC}, \overline{AB} \cong \overline{CB}\end{align*}
Prove: \begin{align*}\triangle DAB \cong \triangle DCB\end{align*}
2. Is \begin{align*}\triangle PON \cong \triangle MOL\end{align*}? Why or why not?
3. If \begin{align*}\triangle DEF \cong \triangle PQR\end{align*}, can it be assumed that:
a) \begin{align*}\angle F \cong \angle R\end{align*}? Why or why not?
b) \begin{align*}\overline{EF} \cong \overline{PR}\end{align*}? Why or why not?
Know What? Your parents changed their minds at the last second about their kitchen layout. Now, they have decided they to have the distance between the sink and the fridge be 3 ft, the angle at the sink \begin{align*}71^\circ\end{align*} and the angle at the fridge is \begin{align*}50^\circ\end{align*}. You used your protractor to measure the angle at the stove and sink at your neighbor’s house. Are the kitchen triangles congruent now?
## ASA Congruence
Like SAS, ASA refers to Angle-Side-Angle. The placement of the word Side is important because it indicates that the side that you are given is between the two angles.
Consider the question: If I have two angles that are \begin{align*}45^\circ\end{align*} and \begin{align*}60^\circ\end{align*} and the side between them is 5 in, can I construct only one triangle? We will investigate it here.
Investigation 4-4: Constructing a Triangle Given Two Angles and Included Side Tools Needed: protractor, pencil, ruler, and paper
1. Draw the side (5 in) horizontally, halfway down the page. The drawings in this investigation are to scale.
2. At the left endpoint of your line segment, use the protractor to measure the \begin{align*}45^\circ\end{align*} angle. Mark this measurement and draw a ray from the left endpoint through the \begin{align*}45^\circ\end{align*} mark.
3. At the right endpoint of your line segment, use the protractor to measure the \begin{align*}60^\circ\end{align*} angle. Mark this measurement and draw a ray from the left endpoint through the \begin{align*}60^\circ\end{align*} mark. Extend this ray so that it crosses through the ray from Step 2.
4. Erase the extra parts of the rays from Steps 2 and 3 to leave only the triangle.
Can you draw another triangle, with these measurements that looks different? The answer is NO. Only one triangle can be created from any given two angle measures and the INCLUDED side.
Angle-Side-Angle (ASA) Congruence Postulate: If two angles and the included side in one triangle are congruent to two angles and the included side in another triangle, then the two triangles are congruent.
The markings in the picture are enough to say \begin{align*}\triangle ABC \cong \triangle XYZ\end{align*}.
Now, in addition to SSS and SAS, you can use ASA to prove that two triangles are congruent.
Example 1: What information would you need to prove that these two triangles are congruent using the ASA Postulate?
a) \begin{align*}\overline{AB} \cong \overline{UT}\end{align*}
b) \begin{align*}\overline{AC} \cong \overline{UV}\end{align*}
c) \begin{align*}\overline{BC} \cong \overline{TV}\end{align*}
d) \begin{align*}\angle B \cong \angle T\end{align*}
Solution: For ASA, we need the side between the two given angles, which is \begin{align*}\overline{AC}\end{align*} and \begin{align*}\overline{UV}\end{align*}. The answer is b.
Example 2: Write a 2-column proof.
Given: \begin{align*}\angle C \cong \angle E, \overline{AC} \cong \overline{AE}\end{align*}
Prove: \begin{align*}\triangle ACF \cong \triangle AEB\end{align*}
Statement Reason
1. \begin{align*}\angle C \cong \angle E, \overline{AC} \cong \overline{AE}\end{align*} Given
2. \begin{align*}\angle A \cong \angle A\end{align*} Reflexive PoC
3. \begin{align*}\triangle ACF \cong \triangle AEB\end{align*} ASA
## AAS Congruence
A variation on ASA is AAS, which is Angle-Angle-Side. Recall that for ASA you need two angles and the side between them. But, if you know two pairs of angles are congruent, then the third pair will also be congruent by the \begin{align*}3^{rd}\end{align*} Angle Theorem. Therefore, you can prove a triangle is congruent whenever you have any two angles and a side.
Be careful to note the placement of the side for ASA and AAS. As shown in the pictures above, the side is between the two angles for ASA and it is not for AAS.
Angle-Angle-Side (AAS or SAA) Congruence Theorem: If two angles and a non-included side in one triangle are congruent to two corresponding angles and a non-included side in another triangle, then the triangles are congruent.
Proof of AAS Theorem:
Given: \begin{align*}\angle A \cong \angle Y, \angle B \cong \angle Z, \overline{AC} \cong \overline{XY}\end{align*}
Prove: \begin{align*}\triangle ABC \cong \triangle YZX\end{align*}
Statement Reason
1. \begin{align*}\angle A \cong \angle Y, \angle B \cong \angle Z, \overline{AC} \cong \overline{XY}\end{align*} Given
2. \begin{align*}\angle C \cong \angle X\end{align*} \begin{align*}3^{rd}\end{align*} Angle Theorem
3. \begin{align*}\triangle ABC \cong \triangle YZX\end{align*} ASA
By proving \begin{align*}\triangle ABC \cong \triangle YZX\end{align*} with ASA, we have also shown that the AAS Theorem is valid. You can now use this theorem to show that two triangles are congruent.
Example 3: What information do you need to prove that these two triangles are congruent using:
a) ASA?
b) AAS?
c) SAS?
Solution:
a) For ASA, we need the angles on the other side of \begin{align*}\overline{EF}\end{align*} and \begin{align*}\overline{QR}\end{align*}. Therefore, we would need \begin{align*}\angle F \cong \angle Q\end{align*}.
b) For AAS, we would need the angle on the other side of \begin{align*}\angle E\end{align*} and \begin{align*}\angle R\end{align*}. \begin{align*}\angle G \cong \angle P\end{align*}.
c) For SAS, we would need the side on the other side of \begin{align*}\angle E\end{align*} and \begin{align*}\angle R\end{align*}. So, we would need \begin{align*}\overline{EG} \cong \overline{RP}\end{align*}.
Example 4: Can you prove that the following triangles are congruent? Why or why not?
Solution: Even though \begin{align*}\overline{KL} \cong \overline{ST}\end{align*}, they are not corresponding. Look at the angles around \begin{align*}\overline{KL}, \angle K\end{align*} and \begin{align*}\angle L\end{align*}. \begin{align*}\angle K\end{align*} has one arc and \begin{align*}\angle L\end{align*} is unmarked. The angles around \begin{align*}\overline{ST}\end{align*} are \begin{align*}\angle S\end{align*} and \begin{align*}\angle T\end{align*}. \begin{align*}\angle S\end{align*} has two arcs and \begin{align*}\angle T\end{align*} is unmarked. In order to use AAS, \begin{align*}\angle S\end{align*} needs to be congruent to \begin{align*}\angle K\end{align*}. They are not congruent because the arcs marks are different. Therefore, we cannot conclude that these two triangles are congruent.
Example 5: Write a 2-column proof.
Given: \begin{align*}\overline{BD}\end{align*} is an angle bisector of \begin{align*}\angle CDA, \angle C \cong \angle A\end{align*}
Prove: \begin{align*}\triangle CBD \cong \angle ABD\end{align*}
Solution:
Statement Reason
1. \begin{align*}\overline{BD}\end{align*} is an angle bisector of \begin{align*}\angle CDA, \angle C \cong \angle A\end{align*} Given
2. \begin{align*}\angle CDB \cong \angle ADB\end{align*} Definition of an Angle Bisector
3. \begin{align*}\overline{DB} \cong \overline{DB}\end{align*} Reflexive PoC
3. \begin{align*}\triangle CBD \cong \triangle ABD\end{align*} AAS
## Hypotenuse-Leg Congruence Theorem
So far, the congruence postulates we have learned will work on any triangle. The last congruence theorem can only be used on right triangles. Recall that a right triangle has exactly one right angle. The two sides adjacent to the right angle are called legs and the side opposite the right angle is called the hypotenuse.
You may or may not know the Pythagorean Theorem (which will be covered in more depth later in this text). It says, for any right triangle, this equation is true:
\begin{align*}(leg)^2+(leg)^2=(hypotenuse)^2\end{align*}. What this means is that if you are given two sides of a right triangle, you can always find the third.
Therefore, if you know that two sides of a right triangle are congruent to two sides of another right triangle, you can conclude that third sides are also congruent.
HL Congruence Theorem: If the hypotenuse and leg in one right triangle are congruent to the hypotenuse and leg in another right triangle, then the two triangles are congruent.
The markings in the picture are enough to say \begin{align*}\triangle ABC \cong \triangle XYZ\end{align*}.
Notice that this theorem is only used with a hypotenuse and a leg. If you know that the two legs of a right triangle are congruent to two legs of another triangle, the two triangles would be congruent by SAS, because the right angle would be between them. We will not prove this theorem here because we have not proven the Pythagorean Theorem yet.
Example 6: What information would you need to prove that these two triangles are congruent using the: a) HL Theorem? b) SAS Theorem?
Solution:
a) For HL, you need the hypotenuses to be congruent. So, \begin{align*}\overline{AC} \cong \overline{MN}\end{align*}.
b) To use SAS, we would need the other legs to be congruent. So, \begin{align*}\overline{AB} \cong \overline{ML}\end{align*}.
## AAA and SSA Relationships
There are two other side-angle relationships that we have not discussed: AAA and SSA.
AAA implied that all the angles are congruent, however, that does not mean the triangles are congruent.
As you can see, \begin{align*}\triangle ABC\end{align*} and \begin{align*}\triangle PRQ\end{align*} are not congruent, even though all the angles are. These triangles are similar, a topic that will be discussed later in this text.
SSA relationships do not prove congruence either. In review problems 29 and 30 of the last section you illustrated an example of how SSA could produce two different triangles. \begin{align*}\triangle ABC\end{align*} and \begin{align*}\triangle DEF\end{align*} below are another example of SSA.
\begin{align*}\angle B\end{align*} and \begin{align*}\angle D\end{align*} are not the included angles between the congruent sides, so we cannot prove that these two triangles are congruent. Notice, that two different triangles can be drawn even though \begin{align*}\overline{AB} \cong \overline{DE}\end{align*}, \begin{align*}\overline{AC} \cong \overline{EF}\end{align*}, and \begin{align*}m \angle B = m \angle D\end{align*}.
You might have also noticed that SSA could also be written ASS. This is true, however, in this text we will write SSA.
## Triangle Congruence Recap
To recap, here is a table of all of the possible side-angle relationships and if you can use them to determine congruence or not.
Side-Angle Relationship Picture Determine Congruence?
SSS
Yes
\begin{align*}\triangle ABC \cong \triangle LKM\end{align*}
SAS
Yes
\begin{align*}\triangle ABC \cong \triangle XYZ\end{align*}
ASA
Yes
\begin{align*}\triangle ABC \cong \triangle XYZ\end{align*}
AAS (or SAA)
Yes
\begin{align*}\triangle ABC \cong \triangle YZX\end{align*}
HL
Yes, Right Triangles Only
\begin{align*}\triangle ABC \cong \triangle XYZ\end{align*}
SSA (or ASS) NO
AAA NO
Example 7: Write a 2-column proof.
Given: \begin{align*}\overline{AB} \ || \ \overline{ED}, \angle C \cong \angle F, \overline{AB} \cong \overline{ED}\end{align*}
Prove: \begin{align*}\overline{AF} \cong \overline{CD}\end{align*}
Solution:
Statement Reason
1. \begin{align*}\overline{AB} \ || \ \overline{ED}, \angle C \cong \angle F, \overline{AB} \cong \overline{ED}\end{align*} Given
2. \begin{align*}\angle ABE \cong \angle DEB\end{align*} Alternate Interior Angles Theorem
3. \begin{align*}\triangle ABF \cong \triangle DEC\end{align*} ASA
4. \begin{align*}\overline{AF} \cong \overline{CD}\end{align*} CPCTC
Example 8: Write a 2-column proof.
Given: \begin{align*}T\end{align*} is the midpoint of \begin{align*}\overline{WU}\end{align*} and \begin{align*}\overline{SV}\end{align*}
Prove: \begin{align*}\overline{WS} \ || \ \overline{VU}\end{align*}
Solution:
Statement Reason
1. \begin{align*}T\end{align*} is the midpoint of \begin{align*}\overline{WU}\end{align*} and \begin{align*}\overline{SV}\end{align*} Given
2. \begin{align*}\overline{WT} \cong \overline{TU}, \overline{ST} \cong \overline{TV}\end{align*} Definition of a midpoint
3. \begin{align*}\angle STW \cong \angle UTV\end{align*} Vertical Angle Theorem
4. \begin{align*}\triangle STW \cong \triangle VTU\end{align*} SAS
5. \begin{align*}\angle S \cong \angle V\end{align*} CPCTC
6. \begin{align*}\overline{WS} \ || \ \overline{VU}\end{align*} Converse of the Alternate Interior Angles Theorem
Prove Move: At the beginning of this chapter we introduced CPCTC. Now, it can be used in a proof once two triangles are proved congruent. It is used to prove the parts of congruent triangles are congruent in order to prove that sides are parallel (like in Example 8), midpoints, or angle bisectors. You will do proofs like these in the review questions.
Know What? Revisited Even though we do not know all of the angle measures in the two triangles, we can find the missing angles by using the Third Angle Theorem. In your parents’ kitchen, the missing angle is \begin{align*}39^\circ\end{align*}. The missing angle in your neighbor’s kitchen is \begin{align*}50^\circ\end{align*}. From this, we can conclude that the two kitchens are now congruent, either by ASA or AAS.
## Review Questions
For questions 1-10, determine if the triangles are congruent. If they are, write the congruence statement and which congruence postulate or theorem you used.
For questions 11-15, use the picture to the right and the given information below.
Given: \begin{align*}\overline{DB} \bot \overline{AC}, \overline{DB}\end{align*} is the angle bisector of \begin{align*}\angle CDA\end{align*}
1. From \begin{align*}\overline{DB} \bot \overline{AC}\end{align*}, which angles are congruent and why?
2. Because \begin{align*}\overline{DB}\end{align*} is the angle bisector of \begin{align*}\angle CDA\end{align*}, what two angles are congruent?
3. From looking at the picture, what additional piece of information are you given? Is this enough to prove the two triangles are congruent?
4. Write a 2-column proof to prove \begin{align*}\triangle CDB \cong \triangle ADB\end{align*}.
5. What would be your reason for \begin{align*}\angle C \cong \angle A\end{align*}?
For questions 16-20, use the picture to the right and the given information.
Given: \begin{align*}\overline{LP} \ || \ \overline{NO}, \overline{LP} \cong \overline{NO}\end{align*}
1. From \begin{align*}\overline{LP} \ || \ \overline{NO}\end{align*}, which angles are congruent and why?
2. From looking at the picture, what additional piece of information can you conclude?
3. Write a 2-column proof to prove \begin{align*}\triangle LMP \cong \triangle OMN\end{align*}.
4. What would be your reason for \begin{align*}\overline{LM} \cong \overline{MO}\end{align*}?
5. Fill in the blanks for the proof below. Use the given and the picture from above. Prove: \begin{align*}M\end{align*} is the midpoint of \begin{align*}\overline{PN}\end{align*}
Statement Reason
1. \begin{align*}\overline{LP} \ || \ \overline{NO}, \overline{LP} \cong \overline{NO}\end{align*} Given
2. Alternate Interior Angles
3. ASA
4. \begin{align*}\overline{LM} \cong \overline{MO}\end{align*}
5. \begin{align*}M\end{align*} is the midpoint of \begin{align*}\overline{PN}\end{align*}
Determine the additional piece of information needed to show the two triangles are congruent by the given postulate.
1. AAS
2. ASA
3. ASA
4. AAS
5. HL
6. SAS
Write a 2-column proof.
1. Given: \begin{align*}\overline{SV} \bot \overline{WU}\end{align*} \begin{align*}T\end{align*} is the midpoint of \begin{align*}\overline{SV}\end{align*} and \begin{align*}\overline{WU}\end{align*} Prove: \begin{align*}\overline{WS} \cong \overline{UV}\end{align*}
2. Given: \begin{align*}\angle K \cong \angle T\end{align*}, \begin{align*}\overline{EI}\end{align*} is the angle bisector of \begin{align*}\angle KET\end{align*} Prove: \begin{align*}\overline{EI}\end{align*} is the angle bisector of \begin{align*}\angle KIT\end{align*}
1.
Statement Reason
1. \begin{align*}\overline{AD} \cong \overline{DC}, \ \overline{AB} \cong \overline{CB}\end{align*} Given
2. \begin{align*}\overline{DB} \cong \overline{DB}\end{align*} Reflexive PoC
3. \begin{align*}\triangle DAB \cong \triangle DCB\end{align*} SSS
2. No, only the angles are congruent, you need at least one side to prove the triangles are congruent.
3. (a) Yes, CPCTC
(b) No, these sides do not line up in the congruence statement.
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# Real Numbers (Chapter 1, Exercise 1.2) – Class 10 Maths Solutions NCERT
The Chapter 1 of Class 10 Mathematics based on NCERT syllabus is about the Real Numbers. This page contains the Solution of Exercise 1.2, Chapter 1 of Class 10 Mathematics. The Exercise 1.2 of Class 10 Mathematics consists of 7 questions which are based on The Fundamental Theorem of Arithmetic.
The solution of all the 7 questions of Exercise 1.2 of Class 10, Chapter 1 is given below.
## Question 1: Express each number as a product of its prime factors:
### i. 140
Solution:
The prime factors of $140$ are
$140=2\times 2\times 5\times 7$
$\Rightarrow 140=2^2\times 5\times 7$
### ii. 156
Solution:
The prime factors of $156$ are
$156=2\times 2\times 3\times 13$
$\Rightarrow 156=2^2\times 3\times 13$
### iii. 3825
Solution:
The prime factors of $3825$ are
$3825=3\times 3\times 5\times 5\times 17$
$\Rightarrow 3825=3^2\times 5^2\times 17$
### iv. 5005
Solution:
The prime factors of $5005$ are
$5005=5\times 7\times 11\times 13$
### v. 7429
Solution:
The prime factors of $7429$ are
$7429=17\times 19\times 23$
## Question 2: Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
### i. 26 and 91
Solution:
The prime factors of $26$ are
$26=2\times 13$
The prime factors of $91$ are
$91=7\times 13$
Therefore,
$HCF=13$
$LCM=2\times 7\times 13=182$
Verification:
We know that
$LCM\times HCF=a\times b$
$\therefore LHS=LCM\times HCF$
$\Rightarrow LHS=183\times 13$
$\Rightarrow LHS=2366$
Now
$RHS=a\times b$
$\Rightarrow RHS=26\times 91$
$\Rightarrow RHS=2366$
Therefore, $LHS=RHS$
Hence verified.
### ii. 510 and 92
Solution:
The prime factors of $510$ are
$510=2\times 3\times 5\times 17$
The prime factors of $92$ are
$92=2\times 2\times 23$
Therefore,
$HCF=2$
$LCM=2\times 2\times 2\times 3\times 3\times 5\times 17=23460$
Verification:
We know that
$LCM\times HCF=a\times b$
$\therefore LHS=LCM\times HCF$
$\Rightarrow LHS=23460\times 2$
$\Rightarrow LHS=46920$
Now
$RHS=a\times b$
$\Rightarrow RHS=510\times 92$
$\Rightarrow RHS=46920$
Therefore, $LHS=RHS$
Hence verified.
### iii. 336 and 54
Solution:
The prime factors of $336$ are
$336=2\times 2\times 2\times 2\times 3\times 7$
The prime factors of $54$ are
$54=2\times 3\times 3\times 3$
Therefore,
$HCF=2\times 3=6$
$LCM=2\times 2\times 2\times 2\times 3\times 3\times 3\times 7=3024$
Verification:
We know that
$LCM\times HCF=a\times b$
$\therefore LHS=LCM\times HCF$
$\Rightarrow LHS=3024\times 6$
$\Rightarrow LHS=18144$
Now
$RHS=a\times b$
$\Rightarrow RHS=336\times 54$
$\Rightarrow RHS=18144$
Therefore, $LHS=RHS$
Hence verified.
## Question 3: Find the LCM and HCF of the following integers by applying the prime factorisation method.
### i. 12, 15 and 21
Solution:
The prime factorisation of $12$ is
$12=2\times 2\times 3$
The prime factorisation of $15$ is
$15=3\times 5$
And the prime factorisation of $21$ is
$21=3\times 7$
Therefore,
$HCF=3$
$LCM=2\times 2\times 3\times 5\times 7=420$
### ii. 17, 23 and 29
Solution:
All the three numbers 17, 23 and 29 are prime numbers.
Therefore,
$HCF=1$ and
$LCM=17\times 23\times 29=11339$
### iii. 8, 9 and 25
Solution:
The prime factorisation of $8$ is
$8=2\times 2\times 2$
The prime factorisation of $9$ is
$9=3\times 3$
And the prime factorisation of $25$ is
$25=5\times 5$
Therefore,
$HCF=1$ $LCM=8\times 9\times 25=1800$
## Question 4: Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
Given,
$a=306$
$b=657$
And $HCF=9$
We know that,
$LCM\times HCF =a\times b$
$\Rightarrow LCM=\frac{a\times b}{HCF}$
$\Rightarrow LCM=\frac{306\times 657}{9}$
$\Rightarrow LCM=22338$
Therefore, $LCM (306,\, 657)$ is $22338$.
## Question 5: Check whether $6^n$ can end with the digit $0$ for any natural number $n$.
Solution:
$6^n$ can be written as
$6^n=(2\times 3)^n$
$6^n=2^n\times 3^n$
We know that, any number that ends with $0$ is divisible by $5$. Therefore, the prime factors of such number must have $5$.
Here, the factors of $6^n$ does not have $5$. Therefore, the number $6^n$ cannot end with $0$.
## Question 6: Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
The first number
$7\times 11\times 13+13$
$=(7\times 11\times 13)+13$
$=13\times (7\times 11+1)$
$=13\times 78$
$=2\times 3\times 13\times 13$
The second number
$7\times 6\times 5\times 4\times 3\times 2\times 1+5$
$=5\times (7\times 6\times 4\times 3\times 2\times 1+1)$
$=5\times (1008+1)$
$=5\times 1009$
In both the above numbers, all the factors are prime numbers. Therefore, the given numbers are composite numbers.
## Question 7: There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Solution:
Time taken by Sonia$=18$ minutes
Time taken by Ravi$=12$ minutes
The time after which Sonia and Ravi will meet at the starting point is the LCM of $18$ and $12$.
The prime factorisation of $18$ is
$18=2\times 3\times 3$
The prime factorisation of $12$ is
$12=2\times 2\times 3$
$\therefore LCM=2\times 2\times 3\times 3=36$
Therefore, Sonia and Ravi will meet at the starting point after $36$ minutes.
I hope the solutions of Class 10, Exercise 1.2, Chapter 1 of Mathematics have helped you. If you have any issues/queries related to the NCERT Solutions of Class 10 Mathematics, Exercise 1.2, feel free to contact me at [email protected] or [email protected] or fill the form here. |
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# A particle executes SHM on a line $8cm$ long. Its K.E and P.E will be equal when its distance from the mean position is A. $4cm$B. $2cm$C. $\sqrt[2]{2}cm$D. $\sqrt 2$
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Hint: SHM stands for Simple Harmonic Motion, which is characterised as a motion in which the restoring force is proportional to the body's displacement from its mean location. This restoring force often moves in the direction of the mean location. The acceleration of a particle in simple harmonic motion is given by $a\left( t \right){\text{ }} = {\text{ }} - {\omega ^{2\;}}x\left( t \right)$ . Here, is the particle's angular velocity.
Complete step-by-step solution:
Simple harmonic motion is an oscillatory motion in which the particle's acceleration is directly proportional to its displacement from the mean position at any stage. It's a form of oscillatory motion that's a little different.
Simple Harmonic Motions (SHM) are oscillatory and periodic in nature, but not all oscillatory motions are SHM. The harmonic motion of all oscillatory motions, the most common of which is simple harmonic motion, is known as oscillatory motion (SHM).
The displacement, velocity, acceleration, and force in this form of oscillatory motion differ (with respect to time) in a way that can be represented by either sine (or) cosine functions, collectively known as sinusoids.
Now, coming to the question;
$K.E = P.E$
$\Rightarrow \dfrac{1}{2}m{\omega ^2}({A^2} - {x^2}) = \dfrac{1}{2}m{\omega ^2}{x^2}$
$\Rightarrow {A^2} - {x^2} = {x^2}$
$\Rightarrow 2{x^2} = {A^2}$
$\therefore x = \dfrac{A}{{\sqrt 2 }}$
$2A = 8$
$A = \dfrac{8}{2} = 4cm$
$x = \dfrac{A}{{\sqrt 2 }}$
$x = \dfrac{4}{{\sqrt 2 }} = \sqrt[2]{2}$
So, the correct option is: (C) $\sqrt[2]{2}cm$.
Note:Simple Harmonic Motion is a very useful and important method for understanding the properties of sound waves, light waves, and alternating currents. Any non-simple harmonic oscillatory motion can be expressed as a superposition of several harmonic oscillatory motions of different frequencies. |
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# 4.1: One Step at a Time
Difficulty Level: At Grade Created by: CK-12
This activity is intended to supplement Algebra I, Chapter 3, Lesson 1.
The following equations are examples of one step equations:
x+3=8x4=28x=40x3=2\begin{align*}x + 3 = 8 && x - 4 = -2 && 8x = 40 && \frac{x}{3} = 2\end{align*}
Is there a rule for solving one-step equations? To find out, solve several one-step equations with your calculator and look for a pattern. To start, clear out any functions from the Y=\begin{align*}Y=\end{align*} screen.
## Problem 1 – Addition equations
One way to solve an equation is by substitution, or trying different values for the variable until you find one that makes the equation true. Your calculator can help you solve the equation x+3=8\begin{align*}x + 3 = 8\end{align*} by substitution.
• Press Y=\begin{align*}Y=\end{align*} to access lists.
• Enter the expression from the left side of the equation into Y1\begin{align*}Y1\end{align*}.
• Enter the expression from the right side of the equation into Y2\begin{align*}Y2\end{align*}.
Use the Table feature to test different values for x\begin{align*}x\end{align*}.
• Press 2nd\begin{align*}2^{nd}\end{align*} [WINDOW] to access the Table Settings menu.
• Change the independent (Indpnt) variable setting from Auto to Ask, as shown.
• Press 2nd\begin{align*}2^{nd}\end{align*} [GRAPH] to access the table.
1st\begin{align*}1^{st}\end{align*} column: values for x\begin{align*}x\end{align*}
2nd\begin{align*}2^{nd}\end{align*} column: value of the left side of the equation, x+3\begin{align*}x + 3\end{align*}
3rd\begin{align*}3^{rd}\end{align*} column: value of the right side of the equation, 8\begin{align*}8\end{align*}
What value of x\begin{align*}x\end{align*} will make the two sides of the equation equal and the equation true? Enter guesses in the x\begin{align*}x\end{align*} column.
Use substitution to solve each equation. Enter the left side of each equation in Y1\begin{align*}Y1\end{align*} and the right side of each equation in Y2\begin{align*}Y2\end{align*}. Then use Table to look for the value of x\begin{align*}x\end{align*} that makes the equation true. Enter these in like the first part of this problem.
1. a. x+30=80x=\begin{align*}x + 30 = 80 \quad x = \underline{\;\;\;\;\;\;\;}\end{align*}
b. x+(3)=19x=\begin{align*}x + (-3) = 19 \quad x = \underline{\;\;\;\;\;\;\;}\end{align*}
c. 73.3=4.3+xx=\begin{align*}73.3 = 4.3 + x \quad x = \underline{\;\;\;\;\;\;\;}\end{align*}
d. x+4.5=2.5x=\begin{align*}x + 4.5 = 2.5 \quad x = \underline{\;\;\;\;\;\;\;}\end{align*}
Write two one-step addition equations of your own. Use substitution to solve them.
2. ==\begin{align*}\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \quad \underline{\;\;\;\;\;\;\;} = \underline{\;\;\;\;\;\;\;\;\;} \quad \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \quad \underline{\;\;\;\;\;\;\;} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
Look for a pattern in the equations you solved and their solutions.
3. a. The solution to x+3=8\begin{align*}x + 3 = 8\end{align*} is x=5\begin{align*}x = 5\end{align*}.
What operation can you perform with 8\begin{align*}8\end{align*} and 3\begin{align*}3\end{align*} to get 5\begin{align*}5\end{align*}?
b. Try this pattern on the other equations and solutions. Does it work? Give an example.
Listen as your teacher explains the Subtraction Property of Equality. This is what caused the pattern you found. You can use the Subtraction Property of Equality to solve addition equations.
m+5m+55m=1=15=4\begin{align*}m + 5 & = 1 \\ m + 5 - 5 & = 1 - 5 \\ m & = - 4\end{align*}
Fill in the boxes to solve each equation.
4. a. \begin{align*}2 + q & = 11 \\ 2 - \Box + q & = 11 - \Box \\ q & = \Box\end{align*} b. \begin{align*}t + 11 & = 10 \\ t + 11 - \Box & = 10 - \Box \\ t & = \Box\end{align*}
c. \begin{align*}n + 32 & = 5 \\ n + 32 - \Box & = 5 - \Box \\ n & = \Box\end{align*}
d. \begin{align*}p + 17 & = 0 \\ p + 17 - \Box & = 0 - \Box \\ p & = \Box\end{align*}
## Problem 2 – Multiplication equations
You can solve an addition equation by subtracting from both sides, because subtraction “undoes” addition. But what about other types of equations?
Use substitution to solve each equation. Enter the left side of each equation in \begin{align*}Y1\end{align*} and the right side of each quation in \begin{align*}Y2\end{align*}. Then use the table to look for the value of \begin{align*}x\end{align*} that makes the equation true.
5. a. \begin{align*}5x = 75 \quad x = \underline{\;\;\;\;\;\;\;}\end{align*}
b. \begin{align*}-7x = 28 \quad x = \underline{\;\;\;\;\;\;\;}\end{align*}
c. \begin{align*}4x = 52 \quad x = \underline{\;\;\;\;\;\;\;}\end{align*}
d. \begin{align*}-5x = 48 \quad x = \underline{\;\;\;\;\;\;\;}\end{align*}
Write two one-step multiplication equations of your own. Use substitution to solve them.
6. \begin{align*}& \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} && \underline{\;\;\;\;\;\;\;} = \underline{\;\;\;\;\;\;\;\;\;} \\ & \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} && \underline{\;\;\;\;\;\;\;} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
Look for a pattern in the equations you solved and their solutions.
7. a. The solution to \begin{align*}5x = 75\end{align*} is \begin{align*}x = 15\end{align*}.
What operation can you perform with \begin{align*}75\end{align*} and \begin{align*}5\end{align*} to get \begin{align*}15\end{align*}?
b. Try this pattern on the other equations and solutions. Does it work? Give an example.
Listen as your teacher explains the Division Property of Equality. This is what caused the pattern you found. You can use the Division Property of Equality to solve multiplication equations.
\begin{align*}7n & = 56 \\ \frac{7n}{7} & = \frac{56}{7} \\ n & = 8\end{align*}
Fill in the boxes to solve each.
8. a. \begin{align*}8q & = 64 \\ \frac{8q}{\Box} & = \frac{64}{\Box} \\ q & = \Box \end{align*}
b. \begin{align*}6t & = -120 \\ \frac{6t}{\Box} & = \frac{-120}{\Box} \\ t & = \Box \end{align*}
c. \begin{align*}2n & = 2 \\ \frac{2n}{\Box} & = \frac{2}{\Box} \\ n & = \Box \end{align*}
d. \begin{align*}-3p & = 48 \\ \frac{-3p}{\Box} & = \frac{48}{\Box} \\ p & = \Box \end{align*}
## Problem 3 – Inverse Operations
When one operation undoes another, they are called inverse operations. When two operations are inverse operations, either one undoes the other.
Inverse Operations
addition \begin{align*}\Leftrightarrow\end{align*} subtraction
multiplication \begin{align*}\Leftrightarrow\end{align*} division
9. a. What operation would you use to undo addition?
b. What operation would you use to undo subtraction?
c. What operation would you use to undo multiplication?
d. What operation would you use to undo division?
10. Write a rule to solve any one-step equation.
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How do you answer scientific notation?
How do you write 0.00125 in scientific notation?
To obtain the actual numeric value we started with we need to move the decimal point three places to the left. Thus we write this as , $0.00125 = 1.25 \times 10^ – 3$ in scientific notation. Therefore the solution for the expression is $0.00125 = 1.25 \times 10^ – 3$.
How do you write 0.005 in scientific notation?
1. Ex. 1: 7.3201×104.
2. Ex. 2: 3.58×10−2.
3. The given value in scientific notation standard form: 5×10−3.
How do you write 172 in scientific notation?
How do you write 172 in scientific notation? Answer: You write it as 1.72 × 102 with superscript or 1.72 × 10^2 using a caret.
How do you write 2300000000 in scientific notation?
Written in scientific notation, 2,300,000,000 is 2.3 x 10.
What are 3 examples of scientific notation?
• Example: 700. Why is 700 written as 7 × 102 in Scientific Notation ?
• Example: 4,900,000,000. 1,000,000,000 = 109 ,
• Example: 0.0055 is written 5.5 × 10-3
• Example: 3.2 is written 3.2 × 100
• Example: Suns, Moons and Planets.
• Examples:
• Example: 0.00012 is written 120 × 10.
What is the scientific notation of 796000?
We can represent this movement as positive 5 such that the correct scientific notation for 796,000 is 7.96×105 7.96 × 10 5 .
How do you write 2000000 in scientific notation?
Therefore: 2000000 = 2 × 106. This can also be expressed as 2 × 10^6 , using the caret symbol, or as 2e+6, which is called 2000000 in e-notation, further discussed in the section ahead.
What is the scientific notation of 872?
872 written in scientific notation is 8.72 × 102.
How do you write 0.0034 in scientific notation?
Therefore, the scientific notation of $0.034$ is $3.4 \times 10^ – 2$. Note: We will follow the following steps to write a decimal number in the form of scientific notation.
How do you write 0.00000055 in scientific notation?
If you want to change 0.00000055 to scientific notation, change the number to be between 1 and 10 (5.5). Then, because the decimal was moved 7 places to the right, we multiply by 10-7 to get 5.5·10-7.
What is the scientific notation of 5050?
5,050 written in scientific notation is 5.05 × 103.
What is the scientific notation of 62000000?
62,000,000 written in scientific notation is 6.2 × 107.
What is the fraction 3 260 in scientific notation to 3 significant figures?
So we have to convert this fraction into a scientific notation and such that it contains three significant figures. Right? So three divided by 260 will give us 0.0115.
How do you write 0.00024 in scientific notation?
0.00024 in standard form will be a 24 × 10 5b 2.4 × 103c 2.4 × 10 4d 24 × 10 3.
What is the exponent for 2300000?
The exponent form of 2300000 is 2.3 × 106.
What is 2300000 km written in scientific notation?
Scientific notation is based on powers of ten. The number 2,300,000 is written as 2.3 X 106 . Note.
What are the 4 rules of scientific notation?
The base should be always 10. The exponent must be a non-zero integer, that means it can be either positive or negative. The absolute value of the coefficient is greater than or equal to 1 but it should be less than 10. Coefficients can be positive or negative numbers including whole and decimal numbers.
How do you write 1001 in scientific notation?
How do you write 1001 in scientific notation? Answer: You write it as 1.001 × 103 with superscript or 1.001 × 10^3 using a caret.
How do you write 30000 in scientific notation?
It can be larger than one. But it needs to be less than 10. And the way we find that number is to take whatever number we’re converting and move its decimal point, so that way it is between one and 10. So 30000 could be written as 30000.0.
How do you write 12.3 in scientific notation?
Scientific notation means that you write a numeral as a number multiplied by 10 to a power. For example, we can write 123 as 1.23 × 10², 12.3 × 10¹, or 123 × 10⁰. Standard scientific notation puts one nonzero digit before the decimal point.
How do you write 90 in scientific notation?
How do you write 90 in scientific notation? Answer: You write it as 9 × 101 with superscript or 9 × 10^1 using a caret.
How do we write scientific notation explain with example?
Scientific notation is a way of writing very large or very small numbers. A number is written in scientific notation when a number between 1 and 10 is multiplied by a power of 10. For example, 650,000,000 can be written in scientific notation as 6.5 ✕ 10^8.
What is the scientific notation of 28000000?
28,000,000 written in scientific notation is 2.8 × 107. |
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# Evaluate the integral:
Question:
Evaluate the integral:
$\int \frac{x+1}{x^{2}+x+3} d x$
Solution:
$I=\int \frac{x+1}{x^{2}+x+3} d x$
As we can see that there is a term of $x$ in numerator and derivative of $x^{2}$ is also $2 x$. So there is a chance that we can make substitution for $x^{2}+x+3$ and I can be reduced to a fundamental integration.
As, $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}+\mathrm{x}+1\right)=2 \mathrm{x}+1$
$\therefore$ Let, $x=A(2 x+1)+B$
$\Rightarrow x=2 A x+A+B$
On comparing both sides -
We have,
$2 \mathrm{~A}=1 \Rightarrow \mathrm{A}=1 / 2$
$A+B=0 \Rightarrow B=-A=-1 / 2$
Hence,
$I=\int \frac{\frac{1}{2}(2 x+1)-\frac{1}{2}}{x^{2}+x+3} d x$
$\therefore I=\frac{1}{2} \int \frac{2 x+1}{x^{2}+x+3} d x-\frac{1}{2} \int \frac{1}{x^{2}+x+3} d x$
Let, $I_{1}=\frac{1}{2} \int \frac{2 x+1}{x^{2}+x+3} d x$ and $I_{2}=\frac{1}{2} \int \frac{1}{x^{2}+x+3} d x$
Now, $I=I_{1}-I_{2} \ldots .$ eqn 1
We will solve $I_{1}$ and $I_{2}$ individually.
As $I_{1}=\frac{1}{2} \int \frac{2 x+1}{x^{2}+x+3} d x$
Let $u=x^{2}+x+3 \Rightarrow d u=(2 x+1) d x$
$\therefore \mathrm{I}_{1}$ reduces to $\frac{1}{2} \int \frac{\mathrm{du}}{\mathrm{u}}$
Hence,
$\mathrm{I}_{1}=\frac{1}{2} \int \frac{\mathrm{du}}{\mathrm{u}}=\frac{1}{2} \log |\mathrm{u}|+\mathrm{C}\left\{\because \int \frac{\mathrm{dx}}{\mathrm{x}}=\log |\mathrm{x}|+\mathrm{C}\right\}$
On substituting the value of $u$, we have:
$I_{1}=\frac{1}{2} \log \left|x^{2}+x+3\right|+C \ldots . .$ eqn 2
As, $I_{2}=\frac{1}{2} \int \frac{1}{x^{2}+x+3} d x$ and we don't have any derivative of function present in denominator. $\therefore$ we will use some special integrals to solve the problem.
As denominator doesn't have any square root term. So one of the following two integrals will solve the problem.
i) $\int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C$ ii) $\int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C$
Now we have to reduce $I_{2}$ such that it matches with any of above two forms.
We will make to create a complete square so that no individual term of $x$ is seen in denominator.
$\therefore I_{2}=\frac{1}{2} \int \frac{1}{\mathrm{x}^{2}+\mathrm{x}+3} \mathrm{dx}$
$\Rightarrow I_{2}=\frac{1}{2} \int \frac{1}{\left\{x^{2}+2\left(\frac{1}{2}\right) x+\left(\frac{1}{2}\right)^{2}\right\}+3-\left(\frac{1}{2}\right)^{2}} d x$
Using: $a^{2}+2 a b+b^{2}=(a+b)^{2}$
We have:
$I_{2}=\frac{1}{2} \int \frac{1}{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{11}}{2}\right)^{2}} d x$
$\mathrm{I}_{2}$ matches with $\int \frac{1}{\mathrm{x}^{2}+\mathrm{a}^{2}} \mathrm{dx}=\frac{1}{\mathrm{a}} \tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{C}$
$\therefore I_{2}=\frac{1}{2}\left\{\frac{1}{\left(\frac{\sqrt{11}}{2}\right)} \tan ^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{11}}{2}}\right)+C\right\}$
$\Rightarrow I_{2}=\frac{1}{\sqrt{11}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{11}}\right)+C \ldots$ eqn 3
From eqn 1:
$I=I_{1}-I_{2}$
Using eqn 2 and eqn 3:
$I=\frac{1}{2} \log \left|x^{2}+x+3\right|+\frac{1}{\sqrt{11}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{11}}\right)+C$ |
The concept of Shares and Dividends are explained in the previous articles with example problems. In this article, we will provide the Worksheet on Basic Concept on Shares and Dividends with different types of questions. We request the students of 10 grade to go through the questions and find the solution for them. After attempting the solved examples of shares and dividends you can check the answer by clicking on the provided link. So, that you can know how much you have understood from the chapter shares and dividends.
Also Practice:
Shares and Dividends Maths Problems
Practice the problems given below to score better grades in the exams.
Example 1.
Find the market price of 6% of Rs 200 share when a person gets a dividend of Rs 65 by investing a Rs 1230.
Solution:
Given that
Amount of dividend = Rs 65
The rate of dividend = 6%
Therefore total face value = 65 × 200/6
= 2166.6
If the face value is Rs 200 then Market value = 1230 × 200/2166.6
= 113.5
Example 2.
Mano buys 20 shares of face value Rs 100 available at Rs 140.
(i) what is her investment?
(ii) If the dividend is 6%. What is the annual income?
(iii) If he wants to increase his annual income by Rs 130, how many extra shares should he buy?
Solution:
Given that
The face value = Rs 100
(i) Given that market value = Rs 140
And number of shares = 20
Therefore investment = number of shares × market value
= 20 × 140
= Rs 2800
(ii) We have income per share = 6% of face value
= (6/100) × 100
= Rs. 6
Therefore annual income = 6 × 20
= Rs 120
(iii) Therefore new annual income = 120 + 130 = Rs 250
Therefore number of shares = 250/6 = 41.6
Therefore, number of extra share to be increased = 41.6 – 20
= 21.6
Example 3.
How much money will be required to buy 400 shares of Rs 19.50 shares at a premium of Rs 1?
Solution:
Number of shares to be bought = 400
Rs 19.50 shares at a premium of Rs 1 means
The nominal value of the share is 19.50
Its marked value = 19.50 + 1 = 20.50
Money required by 1 share = Rs 20.50
Money required to buy 400 shares = 400 × 20.50 = 8200
Example 4.
How much money will be required to buy 600 Rs 40 shares at a discount of Rs 12.50
Solution:
Number of shares to be bought = 600
Rs 20 shares at a discount of Rs 12.50 means
The nominal value of the share is Rs 40 and
Market value = Rs 40 – Rs 12.50 = Rs 27.5
Money required to buy 600 = 600 × 27.50 = 16,500
Example 5.
A person buys 230 shares at a nominal value of Rs 10 each, which he sells at Rs 14.50 each. Find his profit and profit per cent.
Solution:
Nominal value of 230 shares = 10 × 230 = 2300
Market value of 230 shares = 14.50 × 230 = 3335
His profit = Rs 3335 – Rs 2300 = 1035
Profit = 1035/2300 × 100 = 45
Example 6.
A man invests Rs 700 in buying Rs 4 shares and when they are selling at a premium of Rs 2.15 he sells all the shares. Find his profit and profit percent
Solution:
Nominal value of 1 share = Rs 2
Market value of 1 share = Rs 2 + 10 = 12
Total money invested = Rs 600
Number of shares purchased = 600/12 = 50
Market value of 57.14 shares = 50 × 12 = 600
His profit = 600 – Rs 50 = 550
= 550/600 × 100 = 91.66
Example 7.
A man invests Rs 6600 in buying shares of a company of face value of rupees hundred each at a premium of 20% If he earns Rs 800 at the end of the year as the dividend.
(i) Find the number of shares in the company
(ii) The dividend percent per share
Solution:
Total investment = Rs 6600
Nominal value of 1 share = 100
Nominal value = face value
Marked value of 1 share = 20 % of 100 =
20/100 × 100 = 20
Number of shares per chased = 6600/20 = 330
Normal value of 330 shares = 330 × 100 = 33000
Let dividend percentage = Y %
Then Y % of Rs 33000 = Rs 800
= Y/100 × 33000 = 800
Y/100 = 800/33000
Y = 800/33000 × 100
Y = 2.42
Example 8.
A company declares a 6 percent dividend to the shareholders. If a receives Rs 1330 as his dividend, find the nominal value of his shares.
Solution:
Given that
Rate of dividend = 7%
Dividend = 1330
Let nominal value of his share = x
We know that
Dividend = Rate of dividend × nominal value
So, 1330 = 6/100 × x
x = 1330 × 100/7 = 19000
19000 × 100 = 1900000
Nominal value of his share = 1900000
Example 9.
A man invests a certain sum on buying 10% Rs 140 shares at a 5 % premium.
(i) Find his income from one share
(ii) The number of shares bought to have an income from the dividend Rs 5430
(iii) sum invested
Solution:
(i) Dividend on one share = 10% of Rs 140
= 10/100 × 140 = 14
So, the income from one share is = Rs 14
(ii) Number of shares bought by the man = annual income/dividend on one share
Annual income = 5430
= 5430/10 = 543
(iii) since the man bought shares of Rs 543 at 5 % Premium, the Market value of one share = Rs (1 + 5/100) × 543
Rs(105/100) × 543
= 570.15
His total investment = number of shares × Market value of one share
= 543 × 570.15
= 309591.45
Example 10.
Nikitha bought 300, Rs 50 shares at Rs 150 each. |
## Filters
Clear All
Given : and Area of shaded region = area of BCDB + are of ADCA Hence, the correct answer is B.
The area of the shaded region is to be found. Required area =ar(DOC)+ar(DOA) The region to the left of the y-axis is half of the circle with radius 4 units and centre origin. Area of the shaded region to the left of y axis is ar(1) = For the region to the right of y-axis and above x axis The parabola and the circle in the first quadrant intersect at point Remaining area is 2ar(2)...
Hence the required area Therefore the correct answer is B.
We have to find the area of the shaded region OCBAO Ar(OCBAO)=2ar(OCBO) For the fist quadrant In the first quadrant, the curves intersect at a point Area of the unshaded region in the first quadrant is The total area of the shaded region is- = Area of half circle - area of the shaded region in the first quadrant
We have to find the area of the shaded region ABC ar(ABC)=ar(ACLM)-ar(ALB)-ar(BMC) The lines intersect at points (1,2), (4,3) and (2,0) Area of the region bounded by the lines is 3.5 units
Equation of line joining A and B is Equation of line joining B and C is Equation of line joining A and C is ar(ABC)=ar(ABL)+ar(LBCM)-ar(ACM) ar(ABC)=8+5-6=7 Therefore the area of the triangle ABC is 7 units.
We have to find the area of the shaded region In the first quadrant y=|x|=x Area of the shaded region=2ar(OADO) The area bounded by the curves is 1/3 units.
We need to find the area of the shaded region ABCD ar(ABCD)=4ar(AOB) Coordinates of points A and B are (0,1) and (1,0) Equation of line through A and B is y=1-x The area bounded by the curve is 2 units.
We have to find the area of the shaded region BAOB O is(0,0) The line and the parabola intersect in the second quadrant at (-1,1) The line y=x+2 intersects the x axis at (-2,0) The area of the region enclosed by the parabola the line and the -axis is 5/6 units.
We have to find the area of the shaded region The given ellipse and the given line intersect at following points Since the shaded region lies above x axis we take y to be positive The required area is
We have to find the area of the shaded region COB The two curves intersect at points (2,3) and (4,12) Required area is
We have to find the area of the shaded region OBA The curves y=mx and y2=4ax intersect at the following points The required area is
The graph of y=sinx is as follows We need to find the area of the shaded region ar(OAB)+ar(BCD) =2ar(OAB) The bounded area is 4 units.
y=|x+3| the given modulus function can be written as x+3>0 x>-3 for x>-3 y=|x+3|=x+3 x+3<0 x<-3 For x<-3 y=|x+3|=-(x+3) Integral to be evaluated is
the area of the region lying in the first quadrant and bounded by and . The required area (ABCD) = The area of the shaded region is 7/3 units
the area between the curves and . The curves intersect at A(1,1) Draw a normal to AC to OC(x-axis) therefore, the required area (OBAO)= area of (OCAO) - area of (OCABO) Thus the area of shaded region is 1/6 units
The area bounded by the curev and -axis The area of the required region = area of ABCD Hence the area of the shaded region is 624.8 units
The area bounded by the curve and -axis The area of the required region = area of ABCD Hence the area of shaded region is 7/3 units
Simplifying the equation we get, or Thus it becomes :
Exams
Articles
Questions |
We have already discussed the problems on integration of functions in standard forms and the problems involving combinations of these functions. Integrals of certain functions can not be obtained directly if they are not in one of the standard forms, but may be reduced to standard forms by proper substitution. The method of evaluating an integral by reducing it into standard form by a proper substitution is called integration by substitution.The method of substitution is very much essential while integrating composite functions.
Let us consider two functions such that g(x) = f(h(x)), where h(x) is a continuously differentiable function.
Then to evaluate $\int$ f(h(x)) . h'(x). dx,
we substitute U = h ( x )
then h'(x) dx = dU
Substituting this in the above integral we get,
$\int$ f(h(x)) . h'(x). dx = $\int$ f(U) . dU
After evaluating this integral we substitute back the value of U.
## U Substitution Formula
The following formulae are derived by using U substitution.
1. Evaluate $\int$ ( a x + b )n dx
Proof: Let U = ax + b,
=> dU = a . dx
=> dx = $\frac{dU}{a}$
$\int$ ( ax + b )n dx = $\int$ Un . $\frac{dU}{a}$
= $\frac{U^{n+1}}{a(n+1)}$ + C
= $\int$ $\frac{(ax+b)^{n+1}}{a(n+1)}$ + C, where n $\ne$ -1
2. Evaluate $\int$ eax+b dx
Proof : Let U = ax + b
=> dU = a . dx
=> dx = $\frac{dU}{a}$
$\int$ eax+b dx = $\int$ eU . $\frac{dU}{a}$
= $\frac{1}{a}$ $\int$ eU . dU
= $\frac{1}{a}$ eU
= $\frac{1}{a}$ e(ax+b) + C
Following above procedures we have the following formulae by U substitution.
3. $\int$ $\frac{1}{ax+b}$ dx = $\frac{1}{a}$ ln | ax + b | + C
4. $\int$ abx+c dx = $\frac{1}{b}$ $\frac{a^{bx+c}}{log\;a}$ + C
5. $\int$ sin ( ax + b ) dx = - $\frac{1}{a}$ cos ( ax + b ) + C
6. $\int$ cos ( ax + b ) dx = $\frac{1}{a}$ sin ( ax + b ) + C
7. $\int$ sec2 (ax + b ) dx = $\frac{1}{a}$ tan ( ax + b ) + C
8. $\int$ cossec2 (ax + b ) dx = - $\frac{1}{a}$ cot ( ax + b ) + C
9. $\int$ sec (ax + b ) . tan ( ax + b ) dx = $\frac{1}{a}$ sec ( ax + b ) + C
10. $\int$ cosec (ax + b ) . cot ( ax + b ) dx = - $\frac{1}{a}$ cosec ( ax + b ) + C
## Integration by U Substitution
### We follow the following steps while integrating by U Substitution :
1. We first identify the function and its derivative in the given integral.
2. Assume the function as U and its derivative along with dx as dU.
3. When the integral is a function in terms of U, integrate by standard methods.
4. Substitute U as the function in Step 2 and express the answer in terms of x.
### 1. Integrals of the form $\int$ $\frac{f'(x)}{f(x)}$. dx
Substituting U = f ( x ), we get, f ' ( x ) dx = dU
Therefore the given integral will become
$\int$ $\frac{dU}{U}$ = ln{U} + C
= ln {f(x)| + C
### 2. Integrals of the form $\int$ f(x). f'(x). dx
Substituting U = f( x), we get f ' (x) dx = dU
Therefore, the given integral will become,
$\int$ U dU = $\frac{U^{2}}{2}$ + C
= $\frac{(f(x))^{2}}{2}$ + C [ substituting U = f ( x ) ]
### Solved Examples
Question 1: Evaluate $\int$ $\sqrt{x^{2}+x+3}$ (2x + 1 ) dx
Solution:
Let I = $\int$ $\sqrt{x^{2}+x+3}$ (2x + 1 ) dx
Substitute U = x2 + x + 3
Differentiating both sides, we get
dU = ( 2x + 1 ) dx
Putting in the given integral I we get,
I = $\int$ $\sqrt{x^{2}+x+3}$ (2x + 1 ) dx
= $\int$ $\sqrt{U}$ dU
= $\frac{U^{\frac{1}{2}+1}}{\frac{1}{2}+1}$
= $\frac{U^{\frac{3}{2}}}{\frac{3}{2}}$
= 2 $\frac{U^{\frac{3}{2}}}{3}$
= $\frac{2}{3}$ $( x^{2}+x+3)^{\frac{3}{2}}$ [ substituting U = f ( x ) ]
Therefore, $\int$ $\sqrt{x^{2}+x+3}$ (2x + 1 ) dx = $\frac{2}{3}$ $(x^{2}+x+3)^{\frac{3}{2}}$ + C
Question 2: Evaluate $\int$ $\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$ dx
Solution:
We have, I = $\int$ $\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$ dx
Substituting, U = ex + e-x ,
we get, dU = ( ex - e-x ) dx
Substituting in the given integral,
we get, I = $\int$ $\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$ dx
= $\int$ $\frac{dU}{U}$
= ln | U | + C
= ln | ex + e-x | + C
$\int$ $\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$ = ln | ex + e-x | + C
## U Substitution Examples
### Solved Examples
Question 1: Evaluate $\int$ $\frac{cos\;x-sin\;x}{cos\;x+sin\;x}$ dx
Solution:
We have I = $\int$ $\frac{cos\;x-sin\;x}{cos\;x+sin\;x}$ dx
Let U = cos x + sinx
Therefore, dU = ( - sin x + cos x ) dx
Substituting in the given integral we get,
I = $\frac{dU}{U}$
= ln | U | + C
= ln | cos x + sin x | + C
$\int$ $\frac{cos\;x-sin\;x}{cos\;x+sin\;x}$ dx = ln | cos x + sin x | + C
Question 2: Evaluate $\int$ cot x . dx
Solution:
We have I = $\int$ cot x . dx
= $\int$ $\frac{cos\;x}{sin\;x}$ dx
Let U = sin x, we get dU = cos x dx
Therefore, I = $\int$ $\frac{cos\;x}{sin\;x}$ dx
= $\int$ $\frac{dU}{U}$
= ln | U | + C
= ln | sin x | + C
$\int$ cot x . dx = ln | sin x | + C
Question 3: Evaluate $\int$ sec x . dx
Solution:
We have I = $\int$ sec x . dx
Multiplying the numerator and the denominator by ( sec x + tan x ), we get,
I = $\int$ sec x . $\frac{sec\;x+tan\;x}{sec\;x+tan\;x}$ dx
Let U = sec x + tan x
Therefore, dU = ( sec x. tan x + sec2 x ) dx
= sec x ( tanx + sec x ) dx
Substituting in the given integral, we get,
I = $\int$ sec x . $\frac{sec\;x+tan\;x}{sec\;x+tan\;x}$ dx
= $\int$ $\frac{dU}{U}$
= ln | U | + C
= ln | sec x + tan x | + C
$\int$ sec x . dx = ln | sec x + tan x | + C
Question 4: Evaluate $\int$ sin3 x dx
Solution:
Let I = $\int$ sin3 x dx
= $\int$ sin2 x . sinx . dx
= $\int$ ( 1 - cos2 x ) sin x . dx
Let U = cos x
we get, dU = - sin x . dx
=> sin x . dx = - dU
Substituting in the given integral, we get
I = $\int$ ( 1 - cos2 x ) sin x . dx
= $\int$ ( 1 - U2 ) ( - dU )
= $\int$ ( U2 -1 ) dU
= $\frac{U^{3}}{3}$ - U + C
= $\frac{sin^{3}x}{3}$ - sin x + C
$\int$ sin3 x dx = $\frac{sin^{3}x}{3}$ - sin x + C. |
Lesson Objectives
• Demonstrate an understanding of how to Solve a Linear Equation in One Variable
• Demonstrate an understanding of the Multiplication Property of Equality
• Demonstrate an understanding of how to find the LCD for a group of Fractions
• Demonstrate an understanding of multiplication with decimals by powers of 10
• Learn how to Solve Linear Equations in One Variable with Fractions
• Learn how to Solve Linear Equations in One Variable with Decimals
## How to Solve Linear Equations in One Variable with Fractions or Decimals
In the previous lesson, we showed how to solve any linear equation in one variable using a four-step process. In this lesson, we will take things one step further and learn how to solve a linear equation in one variable when fractions or decimals are involved.
### Solving Equations with Fractions
Let's suppose we encountered the following equation: $$\frac{1}{2}x + \frac{3}{4}=\frac{7}{4}$$ We could choose to use our four-step procedure with the fractions involved, however, there is an easier approach. We can clear our equation of fractions by multiplying both sides of the equation by the LCD of all fractions. In this case, we have denominators of 2, 4, and 4. Our LCD will be 4, so we can multiply both sides of our equation by 4 and clear the equation of fractions. Remember, we are allowed to multiply both sides of an equation by the same non-zero number and not change the solution: $$4\left(\frac{1}{2}x + \frac{3}{4}\right)=4 \cdot \frac{7}{4}$$ Let's use our distributive property on the left side: $$\require{cancel}4 \cdot \frac{1}{2}x + 4 \cdot \frac{3}{4}=4 \cdot \frac{7}{4}$$ Now we can do some simplifying and our equation will be cleared of fractions: $$2\cancel{4}\cdot \frac{1}{\cancel{2}}x + \cancel{4}\cdot \frac{3}{\cancel{4}}=\cancel{4}\cdot \frac{7}{\cancel{4}}$$ $$2x + 3=7$$ Now we have an equivalent equation, meaning it has the same solution, and it has been cleared of all fractions. Let's use our four-step process and obtain our solution: $$2x + 3=7$$ Isolate 2x on the left side: $$2x + 3 - 3=7 - 3$$ $$2x=4$$ Isolate the variable x: $$\frac{2}{2}x=\frac{4}{2}$$ $$\frac{\cancel{2}}{\cancel{2}}x=\frac{2\cancel{4}}{\cancel{2}}$$ $$x=2$$ Check: $$\frac{1}{2}x + \frac{3}{4}=\frac{7}{4}$$ $$\frac{1}{2}\cdot 2 + \frac{3}{4}=\frac{7}{4}$$ $$\frac{1}{\cancel{2}}\cdot \cancel{2}+ \frac{3}{4}=\frac{7}{4}$$ $$1 + \frac{3}{4}=\frac{7}{4}$$ $$\frac{4}{4}+ \frac{3}{4}=\frac{7}{4}$$ $$\frac{7}{4}=\frac{7}{4}$$ Let's look at a few examples.
Example 1: Solve each equation. $$\frac{3}{2}x + \frac{8}{3}=\frac{10}{3}+ \frac{5}{2}x$$ Let's begin by clearing the equation of fractions. Our denominators here are 2, 3, 3, and 2. Our LCD will be 6, so we will multiply both sides of the equation by the LCD of 6:
$$6 \left(\frac{3}{2}x + \frac{8}{3}\right)=6 \left( \frac{10}{3}+ \frac{5}{2}x \right)$$ Let's simplify each side. We will begin with the left side of the equation: $$6 \cdot \frac{3}{2}x + 6 \cdot \frac{8}{3}$$ $$3\cancel{6}\cdot \frac{3}{\cancel{2}}x + 2\cancel{6}\cdot \frac{8}{\cancel{3}}$$ Our left side becomes: $$9x + 16$$ Let's now simplify the right side: $$6 \cdot \frac{10}{3}+ 6 \cdot \frac{5}{2}x$$ $$2\cancel{6}\cdot \frac{10}{\cancel{3}}+ 3\cancel{6}\cdot \frac{5}{\cancel{2}}x$$ Our right side becomes: $$20 + 15x$$ Now we can put the two sides together. We have an equation which has been cleared of fractions: $$9x + 16=20 + 15x$$ Let's solve our equation using the four-step process: $$9x + 16 - 16=20 + 15x - 16$$ $$9x=4 + 15x$$ $$9x - 15x=4 + 15x - 15x$$ $$-6x=4$$ Now we can isolate our variable x. Let's divide both sides of the equation by (-6): $$\frac{-6}{-6}x=\frac{4}{-6}$$ $$\frac{\cancel{-6}}{\cancel{-6}}x=\frac{2\cancel{4}}{-3\cancel{6}}$$ We have our solution: $$x=-\frac{2}{3}$$ Check: $$\frac{3}{2}x + \frac{8}{3}=\frac{10}{3}+ \frac{5}{2}x$$ $$\frac{3}{2}\cdot -\frac{2}{3}+ \frac{8}{3}=\frac{10}{3}+ \frac{5}{2}\cdot -\frac{2}{3}$$ Left side: $$\frac{\cancel{3}}{\cancel{2}}\cdot -\frac{\cancel{2}}{\cancel{3}}+ \frac{8}{3}» -1 + \frac{8}{3}$$ Right side: $$\frac{10}{3}+ \frac{5}{\cancel{2}}\cdot -\frac{\cancel{2}}{3}» \frac{10}{3}- \frac{5}{3}$$ $$-1 + \frac{8}{3}=\frac{10}{3}- \frac{5}{3}$$ $$\frac{-3}{3}+ \frac{8}{3}=\frac{10}{3}- \frac{5}{3}$$ $$\frac{5}{3}=\frac{5}{3}$$ Example 2: Solve each equation. $$-\frac{10}{3}\left(-\frac{7}{2}x + 1 \right)=\frac{95}{3}$$ When we encounter an equation with fractions and parentheses are involved, we want to remove parentheses before we clear our fractions: $$-\frac{10}{3}\cdot -\frac{7}{2}x -\frac{10}{3}\cdot 1=\frac{95}{3}$$ $$\frac{35}{3}x -\frac{10}{3}=\frac{95}{3}$$ We can clear the fractions by multiplying both sides of the equation by the LCD, which is 3: $$3 \left(\frac{35}{3}x - \frac{10}{3}\right)=3 \cdot \frac{95}{3}$$ Let's use our distributive property to clean up the left side: $$3 \cdot \frac{35}{3}x - 3 \cdot \frac{10}{3}=3 \cdot \frac{95}{3}$$ Now we can simplify: $$\cancel{3}\cdot \frac{35}{\cancel{3}}x - \cancel{3}\cdot \frac{10}{\cancel{3}}=\cancel{3}\cdot \frac{95}{\cancel{3}}$$ $$35x - 10=95$$ Let's now solve this equation using our four-step process: $$35x - 10=95$$ Isolate the variable term: $$35x - 10 + 10=95 + 10$$ $$35x=105$$ Isolate the variable: $$\frac{35}{35}x=\frac{105}{35}$$ $$\frac{\cancel{35}}{\cancel{35}}x=\frac{3\cancel{105}}{\cancel{35}}$$ $$x=3$$ Check: $$-\frac{10}{3}\left(-\frac{7}{2}x + 1 \right)=\frac{95}{3}$$ $$-\frac{10}{3}\left(-\frac{7}{2}\cdot 3 + 1 \right)=\frac{95}{3}$$ $$-\frac{10}{3}\left(\frac{-21}{2}+ 1 \right)=\frac{95}{3}$$ $$-\frac{10}{3}\left(\frac{-21}{2}+ \frac{2}{2}\right)=\frac{95}{3}$$ $$-\frac{10}{3}\left( -\frac{19}{2}\right)=\frac{95}{3}$$ $$-\frac{5\cancel{10}}{3}\cdot -\frac{19}{\cancel{2}}=\frac{95}{3}$$ $$\frac{95}{3}=\frac{95}{3}$$
### Solving Equations with Decimals
When we encounter an equation with decimals involved, we can clear the decimals by multiplying both sides of the equation by the appropriate power of 10. Recall that when we multiply by 10 or a power of 10, we move the decimal point one place to the right for each zero in the power of 10. To clear the decimals from an equation we look for the largest number of decimal places. We then multiply by an appropriate power of 10 to clear that decimal and all other decimals will be cleared. Let's take a look at an example.
Example 3: Solve each equation.
-16.8 - 5.5x = 8 + 6.9x
The largest number of decimal places (places after the decimal point) is 1. This means we can multiply both sides of the equation by 10 and clear the decimals:
10(-16.8 - 5.5x) = 10(8 + 6.9x)
10 • (-16.8) + 10 • (-5.5x) = 10 • 8 + 10 • 6.9x
-168 + -55x = 80 + 69x
Our equation is cleared of decimals, now we can solve the equation using the four-step process:
Isolate the variable term:
-168 + 168 - 55x = 80 + 168 + 69x
-55x = 248 + 69x
-55x - 69x = 248 + 69x - 69x
-124x = 248
Isolate the variable: $$\frac{-124}{-124}x=\frac{248}{-124}$$ $$\frac{\cancel{-124}}{\cancel{-124}}x=\frac{2 \cancel{248}}{-\cancel{124}}$$ $$x=-2$$ Check:
-16.8 - 5.5x = 8 + 6.9x
-16.8 - 5.5(-2) = 8 + 6.9(-2)
-16.8 + 11 = 8 - 13.8
-5.8 = -5.8
#### Skills Check:
Example #1
Solve each equation. $$-\frac{3}{2}x - \frac{4}{5}x=\frac{529}{60}$$
Please choose the best answer.
A
$$x=-\frac{15}{8}$$
B
$$x=\frac{12}{13}$$
C
$$x=-\frac{7}{4}$$
D
$$x=-2$$
E
$$x=-\frac{23}{6}$$
Example #2
Solve each equation. $$-\frac{17}{6}\left(\frac{9}{4}x + 1\right)=\frac{5}{3}x + \frac{617}{32}$$
Please choose the best answer.
A
$$x=\frac{3}{2}$$
B
$$x=-\frac{11}{4}$$
C
$$x=-\frac{5}{4}$$
D
$$x=\frac{27}{13}$$
E
$$x=-6$$
Example #3
Solve each equation. $$-2.36 + 0.9x=x - 2.2$$
Please choose the best answer.
A
$$x=2$$
B
$$x=6.4$$
C
$$x=-1.6$$
D
$$x=6.7$$
E
$$x=4.23$$
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# Commutative Property
In mathematics, commutative property or commutative law explains that order of terms doesn’t matter while performing arithmetic operations.Â
Commutative property is applicable only for addition and multiplication processes. Thus, it means we can change the position or swap the numbers when adding or multiplying any two numbers. This is one of the major properties of integers.
For example: 1+2 = 2+1 and 2 x 3 = 3 x 2.Â
Commutative Property: A + B = B + A Â (Addition) A x B = B x AÂ Â (Multiplication)
## What is Commutative Property?
As we already discussed in the introduction, as per the commutative property or commutative law, when two numbers are added or multiplied together, then a change in their positions does not change the result.Â
### Examples
• 2+3 = 3+2 = 5
• 2 x 3 = 3 x 2 = 6
• 5 + 10 = 10 + 5 = 15
• 5 x 10 = 10 x 5 = 50
So, there can be two categories of operations that obeys commutative property:
• Commutative property of multiplication
### History
Although the official use of commutative property began at the end of the 18th century, it was known even in the ancient era.
The word, Commutative, originated from the French word ‘commute or commuter’ means to switch or move around, combined with the suffix ‘-ative’ means ‘tend to’. Therefore, the literal meaning of the word is tending to switch or move around. It states that if we swipe the positions of the integers, the result will remain the same.
According to the commutative property of addition, when we add two integers, the answer will remain unchanged even if the position of the numbers are changed.
Let A and B be the two integers, then;
A + B = B + A
### Examples of Commutative Property of Addition
• 1 + 2 = 2 + 1 = 3
• 3 + 8 = 8 + 3 = 11
• 12 + 5 = 5 + 12Â = 17
## Commutative Property of Multiplication
As per the commutative property of multiplication, when we multiply two integers, the answer we get after multiplication will remain the same, even if the position of the integers are interchanged.
Let A and B be the two integers, then;
A × B = B × A
### Examples of Commutative Property of Multiplication
• 1 × 2 = 2 × 1 = 2
• 3 × 8 = 8 × 3 = 24
• 12 × 5 = 5 × 12 = 60
## Important Facts Of Commutative Property
• Commutative property is only applicable for two arithmetic operations: Addition and Multiplication
• Changing the order of operands, does not change the resultÂ
• Commutative property of addition: A + B = B + A
• Commutative property of multiplication: A.B = B.A
## Other Properties
The other major properties of addition and multiplication are:
Now, observe the other properties as well here:
Associative Property of Addition and Multiplication
According to the associative law, regardless of how the numbers are grouped, you can add or multiply them together, the answer will be the same. In other words, the placement of parentheses does not matter when it comes to adding or multiplying.
Hence,
A + (B + C) = (A + B) + C A.(B.C) = (A.B).C
Examples:Â
• 1 + (2+3) = (1+2) + 3 → 6
• 3 x (4 x 2) = (3 x 4) x 2 → 24
Distributive Property of Multiplication
The distributive property of Multiplication states that multiplying a sum by a number is the same as multiplying each addend by the value and adding the products then.
According to the Distributive Property, if a, b, c are real numbers, then:
a x (b + c) = (a x b) + (a x c)
Example:
• 2 x (5 + 8) = (2 x 5) + (2 x 8)
• 2 x (13) = 10 + 16
• 26 = 26
There are certain other properties such as Identity property, closure property which are introduced for integers.
## Non-Commutative Property
Some operations are non-commutative. By non-commutative, we mean the switching of the order will give different results. The mathematical operations, subtraction and division are the two non-commutative operations. Unlike addition, in subtraction switching of orders of terms results in different answers.
Example: 4 – 3 = 1 but 3 – 4 = -1Â which are two different integers.
Also, the division does not follow the commutative law. That is,
6 ÷ 2 = 3
2 ÷ 6 = 1/3
Hence, 6 ÷ 2 ≠ 2 ÷ 6
## Solved Examples on Commutative Property
Example 1: Which of the following obeys commutative law?
1. 3 × 12
2. 4 + 20
3. 36 ÷ 6
4. 36 – 6
5. -3 × 4
Solution: Options 1, 2 and 5 follow the commutative law
Explanation:
1. 3 × 12 = 36 and
    12 x 3 = 36
=> 3 x 12 = 12 x 3 (commutative)
1. 4 + 20 = 24 and
     20 + 4 = 24
     => 4 + 20 = 20 + 4 (commutative)
1. 36 ÷ 6 = 6 and
     6 ÷ 36 = 0.167
=> 36 ÷ 6 ≠6 ÷ 36 (non commutative)
1. 36 − 6 = 30 andÂ
      6 – 36 = – 30
=> 36 – 6 ≠6 – 36 (non commutative)
1. −3 × 4 = -12 and
       4 x -3 = -12
=> −3 × 4 = 4 x -3 (commutative)
Q.2: Prove that a+ b = b+a if a = 10 and b = 9.
Sol: Given that, a = 10 and b = 9Â
LHS = a+b = 10 + 9 = 19  ……(1)
RHS = b + a = 9 + 10 = 19 ……(2)
By equation 1 and 2, as per commutative property of addition, we get;
LHS = RHS
Hence, proved.
Q.3: Prove that A.B = B.A, if A = 4 and B = 3.
Sol: Given, A = 4 and B = 3.
A.B = 4.3 = 12 ….. (1)
B.A = 3.4 = 12 …..(2)
By eq.(1) and (2), as per the commutative property of multiplication, we get;
LHS = RHS
A.B = B.A
Hence, proved.
## Practice Questions
Find which of the following is the commutative property of addition and multiplication.
• 3 + 4 = 4 + 3
• 10 x 7 = 7 x 10
• 8 x 9 = 9 x 8
• 6 + 4 = 4 + 6
To solve more problems on properties of math, download BYJU’S – The Learning App from Google Play Store and watch interactive videos.
## Frequently Asked Questions – FAQs
Q1
### What is commutative property? Give examples.
In Mathematics, a commutative property states that if the position of integers are moved around or interchanged while performing addition or multiplication operations, then the answer remains the same.
Examples are:
4+5 = 5+4 and 4 x 5 = 5 x 4
9 + 2 = 2 + 9 and 9 x 2 = 2 x 9
Q2
### What is commutative property of addition?
When two numbers are added together, then if we swap the positions of numbers, the sum of the two remains the same.
For example, 3+4 = 4+3 = 7
Q3
### What is the commutative property of multiplication?
When two numbers are multiplied together and if we interchange their positions, then the product of the two remains the same.
For example, 5 x 3 = 3 x 5 = 15
Q4
### What are the major four properties in Maths?
The four major properties in Maths are:
Identity property, commutative property, associative property and Distributive property
Q5
### What is the difference between commutative and associative property?
Commutative property holds regardless of order of numbers while addition or multiplication. Whereas associative property holds regardless of grouping of numbers. |
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# Area of Sectors and Segments
## Area of parts of a circle.
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Area of Sectors and Segments
What if you were given a circle with two radii in which the region between those two radii was shaded? How could you find the area of that shaded region of the circle? After completing this Concept, you'll be able to use the formula for the area of a circle's sector to solve problems like this one.
### Guidance
A sector of a circle is the area bounded by two radii and the arc between the endpoints of the radii. If $r$ is the radius and $\widehat{AB}$ is the arc bounding a sector, then the area of the sector is $A=\frac{m \widehat{AB}}{360^\circ} \cdot \pi r^2$ .
A segment of a circle is the area of a circle that is bounded by a chord and the arc with the same endpoints as the chord. The area of a segment is $A_{segment}=A_{sector}-A_{\triangle ABC}$
#### Example A
Find the area of the blue sector. Leave your answer in terms of $\pi$ .
In the picture, the central angle that corresponds with the sector is $60^\circ$ . $60^\circ$ is $\frac{1}{6}$ of $360^\circ$ , so this sector is $\frac{1}{6}$ of the total area. $area \ of \ blue \ sector=\frac{1}{6} \cdot \pi 8^2=\frac{32}{3} \pi$
#### Example B
The area of a sector is $8\pi$ and the radius of the circle is 12. What is the central angle?
Plug in what you know to the sector area formula and then solve for the central angle, which we will call $x$ .
$8 \pi &= \frac{x}{360^\circ} \cdot \pi 12^2\\8 \pi & =\frac{x}{360^\circ} \cdot 144 \pi\\8 &= \frac{2x}{5^\circ}\\x &= 8 \cdot \frac{5^\circ}{2}=20^\circ$
#### Example C
Find the area of the blue segment below.
The area of the segment is the area of the sector minus the area of the isosceles triangle made by the radii. If we split the isosceles triangle in half, each half is a 30-60-90 triangle, where the radius is the hypotenuse. The height of $\triangle ABC$ is 12 and the base is $2 \left(12 \sqrt{3}\right)=24 \sqrt{3}$ .
$A_{sector} &= \frac{120}{360} \pi \cdot 24^2 && A_\triangle =\frac{1}{2} \left(24 \sqrt{3} \right)(12)\\&= 192 \pi && \quad \ =144 \sqrt{3}$
The area of the segment is $A=192 \pi - 144 \sqrt{3} \approx 353.8$ units.
### Guided Practice
1. The area of a sector is $135 \pi$ and the arc measure is $216^\circ$ . What is the radius of the circle?
2. Find the area of the shaded region. The quadrilateral is a square.
3. Find the area of the blue sector of $\bigodot A$ .
1. Plug in what you know to the sector area formula and solve for $r$ .
$135 \pi &= \frac{216^\circ}{360^\circ} \cdot \pi r^2\\135 &= \frac{3}{5} \cdot r^2\\\frac{5}{3} \cdot 135 &= r^2\\225 &= r^2 \rightarrow r=\sqrt{225}=15$
2. The radius of the circle is 16, which is also half of the diagonal of the square. So, the diagonal is 32 and the sides would be $\frac{32}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}=16 \sqrt{2}$ because each half of a square is a 45-45-90 triangle.
$A_{circle} &= 16^2 \pi =256 \pi\\A_{square} &= \left(16 \sqrt{2} \right)^2 = 256 \cdot 2=512$
The area of the shaded region is $256 \pi -512 \approx 292.25$
3. The right angle tells us that this sector represents $\frac{1}{4}$ of the circle. The area of the whole circle is $A=\pi 8^2=64\pi$ . So, the area of the sector is $\frac{1}{4}64\pi = 16\pi$ .
### Practice
Find the area of the blue sector or segment in $\bigodot A$ . Leave your answers in terms of $\pi$ . Round any decimal answers to the nearest hundredth.
Find the radius of the circle. Leave your answer in terms of $\pi$ .
Find the central angle of each blue sector. Round any decimal answers to the nearest tenth.
1. Find the area of the sector in $\bigodot A$ . Leave your answer in terms of $\pi$ .
2. Find the area of the equilateral triangle.
3. Find the area of the segment. Round your answer to the nearest hundredth.
4. Find the area of the sector in $\bigodot A$ . Leave your answer in terms of $\pi$ .
5. Find the area of the right triangle.
6. Find the area of the segment. Round your answer to the nearest hundredth.
### Vocabulary Language: English Spanish
chord
chord
A line segment whose endpoints are on a circle.
circle
circle
The set of all points that are the same distance away from a specific point, called the center.
diameter
diameter
A chord that passes through the center of the circle. The length of a diameter is two times the length of a radius.
pi
pi
(or $\pi$) The ratio of the circumference of a circle to its diameter.
The distance from the center to the outer rim of a circle.
Arc
Arc
An arc is a section of the circumference of a circle.
arc length
arc length
In calculus, arc length is the length of a plane function curve over an interval.
A radian is a unit of angle that is equal to the angle created at the center of a circle whose arc is equal in length to the radius.
Scale Factor
Scale Factor
A scale factor is a ratio of the scale to the original or actual dimension written in simplest form.
Sector
Sector
A sector of a circle is a portion of a circle contained between two radii of the circle. Sectors can be measured in degrees.
Sector of a Circle
Sector of a Circle
A sector of a circle is the area bounded by two radii and the arc between the endpoints of the radii. |
Functions and Graphs
In the new pattern of all B-school entrance exams, different exam boards giving an importance in this area. An average of 4 to 5 questions in almost all the top level exams (CAT,XAT etc)can expect from this chapter. XAT contains a large number of questions related to the user defined functions than the pure mathematical functions. A clear theoretical awareness of the concept of functions is required to solve most of the questions.
Here we are going through the following areas in Module 1.
• Basic concepts of sets and relations
• Definition of functions
• Domain, co domain and range of a function
• Understanding functions
• Value of a function
• Operations in functions
• Classifying functions
• Graphical representation
• Maxima and minima in quadratic function
Basic concepts and Terminologies
Cross product or Cartesian product of two sets
Let us consider two non empty sets A and B. A= {a, b, c} and B= {1, 2}
Cross product of set A and B means set of all different possible mappings from set A to B. Set of all possible mappings can arrange as set of ordered pairs and this set of all relations can be expressed as A*B(read as A cross B)
A*B={(a,1), (a,2) , (b,1) , (b,2) , (c,1) , (c,2)}
Number of elements in A*B is the product of number of elements from set A and number of elements from set B.
A*B is also called the Cartesian product of the sets A and B.
Definition:
If A and B are two non-empty sets, then the Cartesian product of A and B is,
A*B = {(x , y) : x ε A and y ε B }
Either A or B is an empty set the A*B is also an empty set.
If (x, y) = (a, b) then x=a and y = b
Relation
A relation from set A to set B is the subset of Cartesian product A*B. Each of the selected ordered pairs from A * B describes a relation between the first element and the second element.
Domain, Co domain and Range of a relation
In a relation defined from set A to set B, the set of all first elements of the ordered pairs is called the domain of the relation and the set of all second elements (images) in the ordered pair is called the range of the relation. Set B is called the co-domain of the relation
i.e. Range ⊆ co-domain
Example
Let A= {1, 2, 3, 4} B= {1, 2, 3, 4, 5} and R is a relation defined from set A to set B
R= {(x, y): y = x+2, x ε A and y ε B}
Then R= {(1, 3), (2, 4), (3, 5)}
Domain of R = {1, 2, 3}
Range of R = {3, 4, 5}
Co-domain of R = {1, 2, 3, 4, 5}
Functions:
Functions are basically relations between two sets A and B.
When considering a relation from set A to set B, if each element in set A has one and only one image in set B, then the relation is called a function.
Let f is a function from set A to B then f(a) = b , where a ε A and b ε B and b is called image of a.
Illustrated examples:
A= {1, 2, 3, 4} and B= {a, b, c, d}
• f1 = {(1,a),(2,a),(3,a),(4,a)} is a function. i.e. each element in set A has one and only one image in set B.
Domain of f1 = {1, 2, 3, 4}
Range of f1 = {a}
• f2={(1,a),(2,b),(3,c),(4,d)} is a function.
Domain of f2 = {1, 2, 3, 4}
Range of f2 = {a, b, c, d}
• f3 = {(1, a), (2, b), (3, c} is not a function because the element 1 and 2 from set A does not have any image in set B.
• f4 = {(1, a),(1,b), (2, b),(2,b),(3,a), (4, d} is not a function because the element 4 in set A have more than one image in set B.
Value of a function:
Value of a function at certain input means the value of f(x) at certain value of x.
e.g. Let f(x) = 2x2 + 1
Then f (1) = 2(1)2 + 1 = 3
f (2) = 2(2)2 + 1 = 5
f (n+1) = 2(n+1)2 + 1
= 2(n2+2n+1) + 1
= 2n2+4n+2+1
= 2n2+4n+3
Operations on functions:
Functions are following the basic mathematical operations such as addition, subtraction, multiplication, division etc.
Let f(x) and g(x) are two different functions in x
(f + g)(x) = f(x) + g(x)
(f - g)(x) = f(x) - g(x)
(f * g)(x) = f(x) * g(x)
(f/g)(x) = f(x)/g(x) where g(x) ≠ 0
Illustrated examples
If f(x) = 2x and g(x) = x + 1 then
• (f + g)(x) = f (x) + g (x)
= 2x + x + 1
= 3x + 1
• (f-g)(x) = f(x) - g(x)
= 2x - (x+1)
= x-1
• (f * g)(x) = f(x) * g(x)
= 2x(x+1)
= 2x2+2x
• (f/g)(x) = f(x)/g(X) = 2x/x+1
Real Function:
Let f be a function from set A to set B. If A and B are the subset of the set of real numbers then f is a real function.
In this topic we are dealing only with real functions.
Classification of functions and corresponding graphical representation
Functions can be mainly classified into two
• Algebraic functions
• Transcendental functions
Algebraic functions
A function that can be expressed in algebraic forms is called an algebraic function.
E.g.: 4x+3, x2 - 2/x2 + 4 , 5x2 - 7x+1
Algebraic functions mainly classified into the following functions
• Polynomial function
A function f(x) can be expressed as polynomial function if and only if f can b expressed in the form
f(x) = a0 + a1x + a2x2 +.... + anxn , where n is a whole number and a1, a2,... an are real numbers
E.g. f(x) = 3x4+4x3-2x+1
g(x) = √3 x2- x + √2
Both the domain and range of a polynomial function is the set of all real numbers(R)
2. Identity function
If a function in the form f(x) = x, that is the input itself is the output, then it is called an Identity function
Both the domain and range of the identity function is the set of all real numbers.
3. Constant function
If all the input of a function have a common output i.e. f(x) = c, where c is a constant then the function is constant.
Domain = R Range = {c}
E.g. Equations for X and Y axis are the good examples for constant function
Y= 0 in the linear equation for representing X axis. i.e. for any value of x there should be one and only one value for y i.e. y = 0
E.g. f(x) = 3
i.e. y = f(x) = 3
X -4 -3 -2 -1 0 1 >2 y 3 3 3 3 3 3 3
4. Rational functions:
Functions of the form f(x)/g(x) is called a rational function, where f(x) and g(x) are polynomial functions and g(x) ≠ 0
E.g. f(x) = 1/x
Domain = R - {0}
Range = R - {0}
5. Irrational functions:
If an algebraic function containing any irrational terms then the function is irrational.
E.g. y = √x
Here x ≥ 0 Domain = [0, ∞)
y = √x ≥ 0 Range = [0, ∞)
6. Linear Functions:
A function in the form f(x) = ax + b, where a and b are any two real numbers and a≠ 0 is called a linear function.
Graph of any linear function is a straight line. Both the domain and range of a linear function is the set of all real numbers.
E.g. f(x) = 2x+1
Domain = R
Range = R
y= f(x) = 2x+1
X -2 -1 0 1 2 y -3 -1 1 3 5
A function of the form f(x) = ax2+bx+c, where a, b and c are constants (Real numbers) and a≠ 0 is called a quadratic function.
Graph of a quadratic function is a parabola.
E.g. f(x) = x2
i.e. y = x2
X -3 -2 -1 0 1 2 3 y 9 4 1 0 1 4 9
• The lowest or highest point of the graph is called the vertex of the parabola
• Line of symmetry of the graph of a quadratic function with respect to x is y- axis and this is simply called the axis of the parabola
Maximum and minimum value of a quadratic function
f(x) = ax2 + bx + c is the standard form of a quadratic function.
When a > 0, then the parabola open upward. Then the function will have a minimum value.
When a < 0, then the parabola open downward. Then the function will have a maximum value.
Vertex of the parabola is [-b/2a,f(-b/2a)] or [-b/2a, (4ac- b2)/4a]
The maximum or minimum value of the quadratic function = 4ac- b2/4a This is occurred when x = When a > 0, function has a minimum value. When a < 0, function has a maximum value.
E.g. Find the vertex of the parabola y = 2x2 + 3x + 4
a = (coefficient of x2 ) = 2
b = (coefficient of x) = 3
c = (constant term) = 4
x coordinates of vertex = -b/2a = -3/4
y coordinates of vertex = (4ac -b2)/4a
=(4*2*4)-32/ 4 * 2
=32-9/8
=23/8
Vertex of the parabola = ((-3/4) , (23/8))
E.g. Find the maximum and minimum value of the function f(x) = x2 +2x - 1
Solution:
Coefficient of x2 = 1. This is positive
the maximum value is
Hence we can find the minimum value.
Minimum value of the function = 4ac- b2/4a
=(4*1*(-1))- 22/4*1
= -4-4/4*1
=-8/4 = -2
8. Modulus function (Absolute value function)
A function f of the form f(x) = for each x ε R is called a modulus function.
If 'x' is a non negative real number, then f(x) = x
If 'x' is a negative number then f(x) = -x
i.e.|4| = 4
|-4| = -(-4) = 4
Domain of the modulus function is the set of all real numbers but the range is the set of all non negative real numbers.
i.e. Domain = R
Range = R+ or [0,∞)
Graph of f(x) =
x -2 -1 0 1 2 Y= |x| 2 1 0 1 2
9. Greatest integer function
A greatest integer function f(x) = [x] is defined as the value of the greatest integer, less than or equal to x
E.g. [3.2] =3
[5.9] = 5
[-7.2] = -8
Domain =R (Real numbers)
Range = Z (set of integers)
10. Smallest integer function
Smallest integer function f(x) = {x} is defined as the value of smallest integer greater than or equal to x.
E.g. {3.1} = 4
{-7.4}= -7
Domain = R
Range = Z
11. Odd Function
If f (-x) = - f(x) then f is an odd function
E.g. If f(x) = x + 2x3
Then
i.e. f(-2) = -f(2)
f is an odd function
12. Even Function
If f(-x) = f(x) then f is an even function
E.g. If f(x) = 2x2+1
f (1) = 2(1)2+1 = 3
and f(-1) = 2(-1)2+1 = 3
i.e. f(-1) = f(1)
f is an even function
Transcendental functions
A function which cannot be expressed algebraically is called a transcendental function.
Main transcendental functions are explained below.
• Trigonometric function
A function defined in the form of a trigonometric ratio is called a trigonometric function
E.g. f(x) = Sin x
g(x) = Cos x
etc
E.g. f(x)= Sin x
Sin x is a periodic function with period 2π.
Domain =R
Range = [-1,1]
• Logarithmic function
A function of the form f(x) = log x is called a logarithmic function
E.g f(x) = logex
Domain = R+
Range = R
Main Objectives under this module:
• More understanding of the graphs
• Transformations of a graph
• Vertical shift
• Horizontal shift
• Stretch and Shrink
• Changing directions
• Maximum and Minimum value of a function
• Application of derivatives in Maxima and Minima
More understanding on the graph of a function.
Transformations of a graph
• Vertical shift:
Through the sufficient changes in the function, its corresponding graph can move upward or downward.
Consider the graph of the function f(x) = x2.
• f(x) → f(x) + k, where 'k' is a positive real constant.
Then the graph moves 'k' units upward.
• f(x) → f(x) - k, where 'k' is a positive real constant.
Then the graph moves 'k' units downward.
• Horizontal shift:
• f(x) → f(x - k), where 'k' is a positive real constant.
Then the graph moves 'k' units towards right.
• f(x) → f(x + k), where 'k' is a positive real constant.
Then the graph moves 'k' units towards left.
• Shrink and stretch:
• f(x) → k f(x), where k > 1
Then the graph will shrink 'k' times along y axis
• f(x) → 1/k * f(x), where k > 1
Then the graph will stretch 'k' units along y axis.
• Changing direction:
f(x) → - f(x)
Then the graph will revolve 180° about x- axis.
Maxima and Minima
Here we are discussing the maximum and minimum value of a function. Its conventional application is lying in the area of differential calculus, even though we can think about some of the simple methods irrespective of the application of differentiation for finding the maxima and minima.
Examples:
• What is the minimum value of the function x2 + 6x + 12 ?
Solution:
x2 + 6x + 12 = x2 + 6x + 9 + 3
= (x + 3)2 + 3
Now the minimum value for (x +3)2 can be equal to zero, when x = -3.
Hence the minimum value of the function is 0 + 3 = 3.
Alternate method 1: (Formula)
Apply the formula 4ac- b2/4a for finding the minimum value of the given function.
a = 1
b = 6
c = 12
Minimum value = (4*1*12 - 62)/4*1 = 3
Alternate method 2: (differentiation)
Let y = x2 + 6x + 12
y' = dy/dx = 2x + 6
y'' = d2y/dx2 = 2
y'' > 0, hence the function has a minimum value.
Equate y' to zero to get the value of 'x' at which the given function has a minimum value.
2x + 6 = 0
x = -3
y = (-3)2 + 6(-3) + 12 = 3
Hence the minimum value of y = 3
• What are the maximum and minimum values of the function y = x2+ 2x+1/x2+ x+1
Solution:
Method 1:
y = x2+ 2x+1/x2+ x + 1
y (x2 x+1) = x2+ 2x+1
(y - 1) + (y - 2) x + (y - 1) = 0
As 'x' is a real number, b2- 4ac ≥ 0
(y - 2)2 - 4 (y -1) 2 ≥ 0
y (3y - 4) ≤ 0
y ≥ 0 and 3y - 4 ≤ 0
y ≥ 0 and y ≤ 4/3
Hence the minimum value of the function is '0' and the maximum value of the function is 4/3.
Method 2:
Application of derivatives:
Quotient rule of differentiation: d[U/V]/dx = (U'V-UV')/V2
Equate the derivative of the function to zero to get the values at which function has the minimum and maximum values.
d[(x2+ 2x + 1)/(x2+ x+1)]/dx = (2x+2)(x2 + x + 1) - (x2 + 2x + 1)(2x+1)/(x2+x+1)2 = 0
x2-1 = 0
x = ± 1
When x = -1 → y = 0
When x = 1 → y = 4/3
The minimum value of the function is 0.
The maximum value of the function is 4/3.
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limsup-liminf
# limsup-liminf - Limit sup and limit inf Introduction In...
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Limit sup and limit inf. Introduction In order to make us understand the information more on approaches of a given real sequence a n n 1 , we give two definitions, thier names are upper limit and lower limit. It is fundamental but important tools in analysis. Definition of limit sup and limit inf Definition Given a real sequence a n n 1 , we define b n sup a m : m n and c n inf a m : m n . Example 1 1 n n 1 0,2,0,2,... , so we have b n 2 and c n 0 for all n . Example 1 n n n 1 1,2, 3,4,... , so we have b n and c n for all n . Example n n 1 1, 2, 3,... , so we have b n n and c n for all n . Proposition Given a real sequence a n n 1 , and thus define b n and c n as the same as before. 1 b n , and c n n N . 2 If there is a positive integer p such that b p , then b n n N . If there is a positive integer q such that c q , then c n n N . 3 b n is decreasing and c n is increasing. By property 3, we can give definitions on the upper limit and the lower limit of a given sequence as follows. Definition Given a real sequence a n and let b n and c n as the same as before. (1) If every b n R , then inf b n : n N is called the upper limit of a n , denoted by lim n sup a n . That is, lim n sup a n inf n b n . If every b n , then we define lim n sup a n . (2) If every c n R , then sup c n : n N is called the lower limit of a n , denoted by lim n inf a n .
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That is, lim n inf a n sup n c n .
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# Patterns in Maths – Rules and Examples
Sep 17, 2022
## Key Concepts
Use a rule to extend a number pattern, identify features of the number pattern, and use the number pattern to solve a problem.
## Introduction
### Patterns
A pattern is a series of objects or numbers which repeat in a pre-designed manner.
### Number rules:
Example1: A function machine used a rule to change Lucy’s numbers into different numbers. The table shows Lucy’s numbers and the function machine’s changed numbers.
What is the function machine’s number of Lucy’s number 4?
What is the function machine’s number of Lucy’s number 5?
Use the rule to complete the table. Describe the feature of the pattern.
From the above table, the number 1 changes to 7, 2 changes to 14, and 3 changes to 21.
The rule: Multiply by 7.
So, 4 changes to 28, 5 changes to 35, and 6 changes to 42
The function machine’s numbers 7, 14, 21, 28, 35, 42, …
One of the features of this pattern is all of the function machine’s numbers are multiples of 7.
Example2: There are 3 vertices in 1 triangle and 6 vertices in 2 triangles. How many vertices are there in 3 triangles, 4 triangles?
Solution:
Rule: Multiply by 3
### What have we Learnt:
A pattern is a series of objects or numbers which repeat in a pre-designed manner.
### Concept map:
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […] |
# Real Interval is Bounded in Real Numbers
## Theorem
Let $\R$ be the real number line considered as an Euclidean space.
Let $a, b \in \R$.
Let $\mathcal I$ be one of the following real intervals:
$\displaystyle \openint a b$ $:=$ $\displaystyle \set {x \in \R: a < x < b}$ Open Real Interval $\displaystyle \hointr a b$ $:=$ $\displaystyle \set {x \in \R: a \le x < b}$ Half-Open (to the right) Real Interval $\displaystyle \hointl a b$ $:=$ $\displaystyle \set {x \in \R: a < x \le b}$ Half-Open (to the left) Real Interval $\displaystyle \closedint a b$ $:=$ $\displaystyle \set {x \in \R: a \le x \le b}$ Closed Real Interval
where $b \ge a$.
Then $\mathcal I$ is bounded in $\R$.
## Proof
Consider the open $\epsilon$-ball $B_\epsilon \paren a$ where $\epsilon = b + 1 - a$.
As $b \ge a$ we have that $b + 1 > a$ and so $\epsilon > 0$.
Let $x \in \mathcal I$.
Then, whatever type of real interval $\mathcal I$ actually is, $z \ge a$ and $x \le b$.
As $\epsilon > 0$ it follows that $x > a - \epsilon$.
Also:
$\displaystyle x$ $\le$ $\displaystyle b$ $\displaystyle$ $=$ $\displaystyle a + b - a$ $\displaystyle$ $<$ $\displaystyle a + \left({b + 1 - a}\right)$ $\displaystyle$ $=$ $\displaystyle a + \epsilon$
That is:
$a - \epsilon < x < a + \epsilon$
and so:
$x \in B_\epsilon \paren a$
The result follows by definition of bounded space.
$\blacksquare$ |
## Differentiation of Implicit Function
In this page differentiation of implicit function we are going to see some examples to understand where we have to apply this method.
Definition of implicit function:
When the relation between x and y is given by an equation in the form of f(x,y) = 0 and the equation is not easily solvable for y, then y is said to be implicit function.
Example 1:
Find dy/dx when x⁴ + 5 xy + y⁴ = 2
Solution:
Differentiating with respect to x on both sides
4x³ + 5 [x (dy/dx) + y (1) ] + 4 y³ = 0
4x³ + 5[x (dy/dx) + y ] + 4 y³ = 0
4x³ + 5x (dy/dx) + 5y + 4 y³ = 0
5x (dy/dx) = -4x³ - 5y - 4 y³
dy/dx = (-4x³ - 5y - 4 y³)/5x
dy/dx = -(4x³ + 5y + 4 y³)/5x
dy/dx = -(4x³/5x + 5y/5x + 4 y³/5x)
dy/dx = -(4x²/5 + y/x + 4y³/5x)
differentiation of implicit function
Example 2:
Find dy/dx when y tan x - y² cos x + 2x = 0
Solution:
Differentiating with respect to x
y (sec² x) + tan x (dy/dx) - [ y² (- sin x) + cos x (2y)(dy/dx) ] + 2 (1) = 0
y sec² x + tan x (dy/dx) - [ - y² sin x + 2y cos x (dy/dx) ] + 2 = 0
y sec² x + tan x (dy/dx) + y² sin x - 2y cos x (dy/dx) + 2 = 0
tan x (dy/dx) - 2y cos x (dy/dx) = -y sec² x - y² sin x - 2
tan x (dy/dx) - 2y cos x (dy/dx) = -[y sec² x + y² sin x + 2]
dy/dx [tan x - 2y cos x] = -[y sec² x + y² sin x + 2]
dy/dx = -[y sec² x + y² sin x + 2]/[tan x - 2y cos x]
Related Topics
Quote on Mathematics
“Mathematics, without this we can do nothing in our life. Each and everything around us is math.
Math is not only solving problems and finding solutions and it is also doing many things in our day to day life. They are:
It subtracts sadness and adds happiness in our life.
It divides sorrow and multiplies forgiveness and love.
Some people would not be able accept that the subject Math is easy to understand. That is because; they are unable to realize how the life is complicated. The problems in the subject Math are easier to solve than the problems in our real life. When we people are able to solve all the problems in the complicated life, why can we not solve the simple math problems?
Many people think that the subject math is always complicated and it exists to make things from simple to complicate. But the real existence of the subject math is to make things from complicate to simple.”
Differentiation of Implicit Function to First Principles
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Simplifying Expressions with Rational Exponents Worksheet |
# GRE Algebra | Rules of Exponent
• Last Updated : 24 Apr, 2019
In algebraic expression xm, x is the base and m is the exponent. For all positive number of x except x=1, if an equation contain xm = xn then it will be only possible when m = n.
Here are basic rules of Exponents:
1. If a number raised to the power zero then it should be equal to 1.
`x0 = 1 `
Example:
`20 = 1 `
2. A negative exponent is the same as the reciprocal of the positive exponent.
`x-m = 1/xm `
Example:
`2-4 = 1/24 = 1/16 `
3. If two powers have the same base then we can multiply the powers. When we multiply two powers we add their exponents.
`(xm) (xn) = xm+n `
Example:
`(23)(24) = 27 = 128 `
4. If two powers have the same base then we can divide the powers. When we divide powers we subtract their exponents.
`xm/xn = xm-n = 1/xn-m `
Example:
`34/32 = 32 = 9 `
5. If two powers have different base but same exponent then we multiply the base of powers and exponent will remain same.
`(xm)(ym) = (xy)m `
Example:
`3242 = 122 = 144 `
6. If base is a fraction then the exponent of the power multiply with numerator and denominator separately.
`(x/y)m = xm/ym `
Example:
`(2/3)2 = 22/32 = 4/9 `
7. If power has an exponent then both the exponents multiplied.
`(xm)n = xmn `
Example:
`(32)3 = 36 = 729 `
Avoid the common mistakes like below:
1. `xmyn ≠ (xy)m+n `
Here bases are not same so we cannot add the exponents.
2. `(xm)n ≠ xmxn `
Here exponents should be multiplied not added according to rule.
3. `(x + y)m ≠ xm + ym `
Have a look at (x + y)2 = x2 + 2xy + y2
My Personal Notes arrow_drop_up |
# Fraction Division via Rectangles
I appreciate Christopher Danielson’s post on common numerator fraction division because it’s important to examine how various algorithms work and how we can help our students become more flexible with their thinking. It’s not surprising that I teach fraction division using rectangles, and I really believe the kids seem to grasp it better because it’s visual.
But before we do fraction division, I ask kids about whole number division. What is 8 ÷ 2? What is 15 ÷ 5? Eventually we settle on something like: asking what is 8 divided by 2 is the same as asking how many groups of 2 are in 8.
Then we apply the same question to 3/4 ÷ 2/3 as “how many groups of 2/3 are in 3/4?
I guide them through this process:
Let’s draw out 3/4 and 2/3 on paper.
Half of them draw circles. Awful, drunk, ill-behaved circles.
Let’s use grid paper instead to draw our rectangles. I think you can show 3/4 much more accurately on grid paper than on a circle.
Please draw 2 rectangles of the same size. (By doing this, we are really dividing two fractions using the common denominator strategy. Christopher writes about it here.)
What size do you think? Does it matter?
Shade the first one to show 3/4 and the second one to show 2/3.
They mess up. They might draw a 1 x 4 rectangle, shade in 3 to show 3/4. But they don’t quite know how to shade in 2/3 of a 1 x 4.
So maybe we should think about the size of the rectangle more carefully. Look at the problem again. Three-fourths divided by two-thirds. Hmmm… What dimensions should our rectangles have so it’s easy to divide into fourths and thirds.
This prompt is enough for someone to say, Draw a 4 by 3 rectangle!
Bingo! I’m drawing these with you. Okay, so two rectangles of 4 by 3 — or 3 by 4 — doesn’t matter.
I’m shading in 3/4 on the first one and 2/3 on the second one.
So our question is: How many groups of 2/3 are in 3/4?
Because I colored mine in, can you help me ask the question again using colors instead?
Someone responds, How many pinks are in the greens?
Yeah. And how many little squares are pink? Okay, eight.
So, I’m going over to the green here and round up 8 pink squares.
I’m able to round up one group of 2/3 (pink) in the 3/4 (green).
Someone says, There’s one left over.
How much is this one little green square left over worth?
Right! 1/8 because we called 8 little boxes as one, so 1 little box must be 1/8.
Our answer then is 1 and 1/8.
A few students say, I get it.
How do we know that our answer of 1 1/8 is correct?
Okay, we’ll use a calculator.
I purposely use an online calculator where I’m entering the fractions as they appear. I don’t need to distract them right now with decimals or talk about parentheses. This is from CalculatorSoup.
Let’s do this again. Now with a mixed number just for fun. Let’s do 1 1/2 ÷ 2/5.
How many rectangles are we drawing?
What dimensions should they be?
Oh, but we have more than 1 whole here, so…?
We should have something like this then.
They say, How many groups of orange are in blue?
So let me round up the groups of orange that are in the blue.
I got three. And the leftover is? Right, three. Three out of…?
More students say, Three-fourths! Three and three-fourths. I get this!
This online calculator from Calcul allows for entries of mixed numbers.
Okay, your turn to do one all by yourself.
Please do 2/3 ÷ 1/2. (Same one Christopher used.)
I think these kids’ papers show understanding.
While these are not there yet.
I don’t know. But it seems that drawing pictures and doing more visual stuff start to disappear in middle school. Below is our textbook’s treatment of “dividing fractions and mixed numbers” — Chapter 5, Section 7 — the full 3 pages before the Exercises. Notice the two circles at the start of the section — that’s pretty much it. And circles are great if you have denominators of 2, 4, and 8.
I think if I can get my kids to first see the answer, then I can sell them the other algorithms — like multiply the reciprocal — and not come across as a fraud.
I also want to point out that I normally see this visual below for division of fractions. My way is different than this — I deliberately ask kids to draw 2 rectangles whose dimensions are the denominators.
Thank you to Rachel Emily Tabak for creating this accompanying worksheet, 18 – Frac division rectangles.
1. Posted July 19, 2014 at 6:03 pm | Permalink
Hi Fawn,
Awesome write up. I am particularly impressed by with fabulous job you did making the page do everything you wanted it to, ie, pictures, calculator snap shot, student work and text pages. Thanks for giving me something to think about.
• Fawn Nguyen
Posted July 19, 2014 at 6:13 pm | Permalink
Hi Amy. Don’t let Andrew Stadel see your comment. He thinks I’m technologically challenged. (I think he’s not very nice.) I can take pictures with my cell phone and crop. Yes, I can. Thanks, Amy.
2. Posted July 19, 2014 at 6:03 pm | Permalink
Fawn,
You’ve gone and done it again! You’ve illustrated an extremely conceptual way to show division of fractions. I’ve shown fraction division in a similar way, but only drew the dividend, not the divisor. The thought process was the same, but the answer didn’t jump out at me as clearly as SEEING the divisor right there. Duh! I also never thought of using graph paper and having the students reason why it’s efficient to use the two denominators as the dimensions of each rectangle that represents one whole. It is SOOO easy to see how many 2/5ths go into 1 1/2! Duh! I’ve been doing this a long time, and every day, following your blog and others, I learn a new approach. I love math! …and your blog!
• Fawn Nguyen
Posted July 19, 2014 at 6:30 pm | Permalink
And I sincerely appreciate your enthusiastic support, Elaine! I forgot to include a problem we did where the dividend is smaller than the divisor.
We did 2/5 ÷ 3/4. They knew what to draw.
So when I threw this slight curve ball at them, some of the kids switched the problem around as if division were commutative! Oy. But I think they got it now.
Many thanks, Elaine.
3. l hodge
Posted July 19, 2014 at 6:04 pm | Permalink
I like your rectangle method quite a bit. Talking about color is very clever as well. What do you think of eventually moving to a 12 x 1 rectangle instead of 3×4, and then moving to the number line as students become more proficient?
I like the ease of your method – much easier than a number line approach. But, I also like eventually moving to the number line to more firmly establish the value of the fractions we are dealing with.
• Fawn Nguyen
Posted July 19, 2014 at 7:12 pm | Permalink
The Common Core is very big on the number line, and I agree that we need to help kids understand the what/how/where/why of the number line. I just would not teach division of fractions using the number line though, as you say the rectangle is much easier and the answer pops out at them. It would be silly to draw pictures (or use the number line) every time they need to divide fractions, so my use of rectangles is just so they can see the answer for what it is instead of blindly trusting that multiplying by the reciprocal somehow gives the same answer.
Earlier this year I had kids order fractions on the number line. It’s tough, but I’m not giving up. I know some teachers are successful with using the number line to teach adding and subtracting integers. Thanks so much!
• l hodge
Posted July 19, 2014 at 7:15 pm | Permalink
The rectangle method has cleverly and convincingly turned 3/4 divided by 2/3 into 9 divided by 8. I like that simplicity and think this is an excellent method.
But, I also think this method allows students to avoid thinking of 3/4 & 2/3 as values. I doubt the students are thinking of the areas of the rectangles as 1 and the areas of the squares as 1/12 – i.e. thinking of 3/4 & 2/3 as values.
After awhile, if you use two 1×12 rectangles instead of two 3×4, you have almost moved to thinking of fractions as lengths. Put those two 1 x 12 rectangles right above a number line (between 0 & 1) and you pretty much have the number line approach. Now it is clear that the length of each rectangle is 1, the length of the green part is 3/4, the length of each square is 1/12, where these values are relative to 0 & 1, etc.
I don’t know when or if the number line should be brought in for dividing fractions. But, I see a lot of students that don’t really seem to think of fractions as values. As a result, these students have very poor rational number sense. They also often view the same procedures and concepts as completely separate when using rational numbers instead of whole numbers.
Anyway, hope this doesn’t come across in a negative way. I wish there were more folks like you putting out really good stuff for the rest of us to steal.
• Fawn Nguyen
Posted July 19, 2014 at 7:18 pm | Permalink
Not negative at all! If anything I really appreciate your making me think about this deeply. I also think you’re giving this rectangle thing more credit than it deserves. It is what it is. Not perfect. Not “the” way to teach anything. It’s just my way of seeing things.
When kids do not connect the dots, it’s our job as their teachers to help them do exactly that. I’m just having kids estimate a lot too before doing any calculation to build that number sense and operation sense. With fractions, I’m just happy right now that they can tell which fraction is bigger. So if they could tell, then I want them to think about taking a smaller number and dividing by a bigger number — or the money you have to divide up equally among the people is less than the number of people, then how much does each person get?
Baby steps. Thanks so much again!
4. Posted July 19, 2014 at 6:05 pm | Permalink
This blew my mind. It seems so simple, and yet, I’ve never seen this approach before.
And it is SO MUCH BETTER than the textbook version. I don’ know that the textbook makes it obvious that there is 1/3 remaining.
Also, I like that you throw those couple of mistakes up there. It makes sense that students would have a little trouble with the 1/3 part.
• Fawn Nguyen
Posted July 19, 2014 at 7:13 pm | Permalink
My textbook, after showing the two circles divided neatly into fourths, went right into the algorithm of multiplying by the reciprocal. Blehh.
I blew Nathan Kraft’s mind. Woohoo. Thanks, buddy.
5. Posted July 19, 2014 at 6:05 pm | Permalink
You and your rectangles. I love it. And this lesson is solid, too. As usual, I’m impressed.
• Fawn Nguyen
Posted July 19, 2014 at 7:13 pm | Permalink
Among the locals I’m crowned the Rectangle Queen. Now you know. Thanks much, Matt.
6. Posted July 19, 2014 at 6:06 pm | Permalink
Fawn – Thanks so much for sharing. I truly appreciate your ideas and how nicely you lay them out for the rest of us. I spent quite a bit of time with students visually representing division earlier in the year (though I find your way more intuitive). Students seemed to have it. I then “sold them an algorithm” – How do you branch off to the multiply reciprocal algorithm? – and all seemed great. Recently, I worked division back in and a strong student looked at me as if it was never discussed. Makes me feel good…yeah right, but I was wondering if you would go back to rectangles or do what at this point.
• Fawn Nguyen
Posted July 19, 2014 at 7:14 pm | Permalink
After the rectangles, I’d show them common denominator division since they’re already finding common denominators for adding and subtracting. I’d also do multiply the reciprocal. These are easy sell after they SEE the rectangles, I think. Thank you so much, Frank, for your kind words.
7. Debbie Boden
Posted July 19, 2014 at 6:07 pm | Permalink
Fawn, this is so great! I’m going to do it tonight with my adult basic math class! And thank you for the extra one in the comments too!
• Fawn Nguyen
Posted July 19, 2014 at 7:22 pm | Permalink
Thank you, Debbie, and you’re welcome.
8. Rebecca
Posted July 19, 2014 at 6:08 pm | Permalink
From your first example. “Someone says, There’s one left over.
How much is this one little green square left over worth?”
Q: Did any of your students say 1/12? I could see that being a common mistake since there is one green square out of the original 12 left over. If so, how did you address?
• Fawn Nguyen
Posted July 19, 2014 at 7:13 pm | Permalink
If they did, I didn’t hear them, but for sure in this first example, some are lost and probably thought 1/12. I posted examples of kids not seeing the remainder correctly. It’s emphasizing what we just gathered up and called “one.” We circled 8 boxes and we called that one whole, so 1 of those boxes must represent 1/8. It’s about asking kids to see what was defined as 1. Thanks, Rebecca.
9. Posted July 19, 2014 at 6:09 pm | Permalink
Hey Fawn! I opened my review of multiplying and dividing rational expressions with your fraction prompt! I love, love, love that the reluctant mathematicians are the first to answer, “how many one-halves are in three-fourths?” So cool.
Cheers!
10. Posted July 19, 2014 at 6:10 pm | Permalink
This is indeed beautiful stuff. I like the suggestion of eventually turning these into 1 by (common denominator) rectangles to make the connection to the number line (and help explain the method whose illustration you show at the very end).
I also wonder why textbooks seem so focused on showing just one method. Mostly I think it’s because 50 years ago, we needed kids to learn the most computationally efficient method we could find, so that they could compute quickly. Now if we want quick computation, we use a calculator, and so it’s time for curricula to adapt by showing methods like yours that help kids understand more instead of focusing so much on efficiently reaching answers.
Tatiana would be so happy to see more kids working with grids!
• Fawn Nguyen
Posted July 19, 2014 at 6:13 pm | Permalink
Hi Josh! I really need to try extending this into 1 x n rectangles. I’m looking into more ways to teach using the number line as the Common Core stresses it throughout. I think you’re right about the need for textbooks to do what they do because we didn’t have easy access to calculators back then. We also tend to teach the same way that we were taught. I only learn of different ways to think about things from specific courses that I’ve taken and in settings where I’m encouraged to come up with different strategies. (I’d gained so much from our Math Teachers’ Circle training and the subsequent workshops that we’ve held.)
For the last 9 years I’ve required my students to keep a quadrille-ruled composition notebook and much of our work is also done on graph paper. But if you see Tatiana, please let her know that I think of her often when we use grids.
Thank you, Josh.
11. Roger
Posted July 19, 2014 at 6:10 pm | Permalink
Using rectangle area models to teach fraction operations is one of the best models I have ever used with students.
I was introduced to this model in the early 1980s by a professor at Ball State University (in Indiana). She used a 11 pin by 11 pin geoboard and geo-islands (since her death in 1987 the pieces were sold as fraction islands).
Thanks for sharing, Roger
• Fawn Nguyen
Posted July 19, 2014 at 6:12 pm | Permalink
Algebra through Visual Patterns put out by the Math Learning Center (Portland, OR) is one of the best workshops I’ve attended. That really was the first time I started using rectangular models and really never looked back. Thank you, Roger.
12. jess
Posted July 19, 2014 at 6:11 pm | Permalink
i love your method of teaching division of fraction!! im going do it with my primary students
• Fawn Nguyen
Posted July 19, 2014 at 6:12 pm | Permalink
Yay!! Thanks for telling me that, Jess.
• jess
Posted July 19, 2014 at 7:16 pm | Permalink
btw, i faced question when the the first fraction is smaller then the 2nd one. how to solve it?
• jess
Posted July 19, 2014 at 7:16 pm | Permalink
it’s okay. i guess i already know the answer when i look through again ur explanation. just that it’s quite hard for me to explain well to my students .. Tq anyway!
13. mathmom
Posted August 8, 2014 at 11:39 am | Permalink
Do you have a good way of using the rectangles to show why “invert and multiply” is a convenient “shortcut” for doing fraction division, once they get what it means?
• Fawn
Posted August 11, 2014 at 6:53 am | Permalink
Hi mathmom. The rectangles serve as a visual model for fraction division, but to explain why “invert and multiply” work, I’d go back to using unit fractions and the language of division. For example, 6 divided by 1/3 can be asked as “how many unit lengths of 1/3 [on the number line] are in 6 units?” This question can be translated to the multiplication problem of 6 = ? x 1/3, or “how many copies of 1/3 to equal 6?” Students reason that it takes 3 copies of 1/3 (definition of a third) to make 1 whole, so 6 x 3 = 18, and this product 6 x 3 is the solution for the 6 divided by 1/3. With ample opportunities to work and reason about these problems, students will see why “invert and multiply” works.
Thank you for dropping in!
14. Beth
Posted April 24, 2015 at 3:12 pm | Permalink
My favorite part:
I guide them through this process: Let’s draw out 3/4 and 2/3 on paper.
Half of them draw circles. Awful, drunk, ill-behaved circles.
Such a great visual!
15. Namgay
Posted March 10, 2017 at 12:33 am | Permalink
Can any one solve this with grid or rectangless 2/3-3/4
16. Posted September 29, 2017 at 6:41 pm | Permalink
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• […] I’m showing my 6th graders how to divide fractions by drawing rectangles. […]
• By Day 8: Adding/Subtracting Fractions | Quadrant Dan on September 8, 2014 at 6:06 pm
[…] spend a few days review fraction operations. So I started digging in the Blogosphere. I came across Fawn Nguyen’s Post on dividing fractions and built off the same idea for addition and subtraction. The lesson went […]
• By Day 9: Pairs of Angles | Quadrant Dan on September 9, 2014 at 1:58 pm
[…] in Algebra we went over multiplying and dividing fractions. I followed along with Fawn, the day went OK. We used rectangles to model addition, subtraction, multiplication, AND division […]
• By Day 12 | axes of symmetry on September 18, 2014 at 3:55 pm
[…] 6th Grade – We drew some rectangles to represent fraction division. […]
• By The Ladder of Abstraction. | trigotometry on June 8, 2015 at 4:43 pm
[…] Nguyen presented a fabulous idea for visually representing fractions that I used as inspiration for teaching fraction operations to my students. To start the unit, I […]
• […] Fraction Division via Rectangles […]
• By Highlights from 2016-2017 | Count It All Joy on June 7, 2017 at 6:32 pm
[…] New Stuff in 6th Grade – This is the 4th time I’ve taught 6th grade, but I haven’t found ways to go about teaching the concepts as quickly as I have with 8th grade. However, I tried a few new things this year with 6th grade that I enjoyed. I used differentiated review stations that I learned about from Michelle at TMC 16. I used Julie’s ideas for teaching solving equations and Fawn’s ideas for dividing fractions. […]
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## RD Sharma class 9 maths Solutions Factorization of Polynomials
### RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 5.1
Question 1.
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer:
Solution:
(i) 3x2 – 4x + 15,
(ii) y2 + 2$$\sqrt { 3 }$$ are polynomial is one variable. Others are not polynomial or polynomials in one variable.
Question 2.
Write the coefficient of x2 in each of the following:
Solution:
Coefficient of x2,
in (i) is 7
in (ii) is 0 as there is no term of x2 i.e. 0 x2
Question 3.
Write the degrees of each of the following polynomials:
(i) 7x3 + 4x2 – 3x + 12
(ii) 12 – x + 2x3
(iii) 5y –Â $$\sqrt { 2 }$$
(iv) 7
(v) 0
Solution:
(i) Degree of the polynomial 7x3 + 4x2Â – 3x + 12 is 3
(ii) Degree of the polynomial 12 – x + 2x3 is 3
(iii) Degree of the polynomial 5y – $$\sqrt { 2 }$$is 1
(iv) Degree of the polynomial 7 is 0
(v) Degree of the polynomial 0 is 0 undefined.
Question 4.
(i) x + x2 + 4
(ii) 3x – 2
(iii) 2x + x2 [NCERT]
(iv) 3y
(v) t2 + 1
(v) 7t4 + 4t3 + 3t – 2
Solution:
(i)Â x + x2 + 4 It is a quadratic polynomial.
(ii) 3x – 2 : It is a linear polynomial.
(iii) 2x + x2: It is a quadratic polynomial.
(iv) 3y It is a linear polynomial.
(v) t2+ 1 It is a quadratic polynomial.
(vi) 7t4 + 4t3 + 3t – 2 It is a biquadratic polynomial.
Question 5.
Classify the following polynomials as polynomials in one-variable, two-variables etc.
(i) x2-xy +7y2
(ii) x2 – 2tx + 7t2 – x + t
(iii) t3 -3t2 + 4t-5
(iv) xy + yz + zx
Solution:
(i) x2 – xy + 7y2: It is a polynomial in two j variables x, y.
(ii) x2 – 2tx + 7t2 – x + t: It is a polynomial in two variables in x, t.
(iii) t3 – 3t2 + 4t – 5 : It is a polynomial in one variable in t.
(iv) xy +yz + zx : It is a polynomial in 3 variables in x, y and
Question 6.
Identify polynomials in the following:
Solution:
Question 7.
Identify constant, linear, quadratic and cubic polynomials from the following polynomials:
Solution:
(i) f(x) = 0 : It is a constant polynomial as it has no variable.
(ii) g(x) = 2x3 – 7x + 4 : It is a cubic polynomial.
(iii) h(x) = -3x + $$\frac { 1 }{ 2 }$$ : It is a linear polynomial.
(iv) p(x) = 2x2 – x + 4 : It is a quadratic polynomial.
(v) q(x) = 4x + 3 : It is linear polynomial.
(vi) r(x) = 3x3 + 4x2 + 5x – 7 : It is a cubic polynomial.
Question 8.
Give one example each of a binomial of degree 35 and of a monomial of degree 100.  [NCERT]
Solution:
Example of a binomial of degree 35 = 9x35 + 16
Example of a monomial of degree 100 = 2y100
### RD Sharma class 9 maths Solutions Factorization of Polynomials Chapter 6 Exercise 6.1
RD Sharma class 9 maths Solutions Factorization of Polynomials Chapter 6 Exercise 6.1 Q 1. |
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# System of equations
A system of equations is made up of 2 or more equations that we have to find a common solution. The solution of a system of equations is called the common solution of the equations. In other words, the solution of a system is the point where the recipes are cut:
The three methods to solve systems of equations:
• Substitution method.
• Elimination method.
• Equalization method.
## Methods for solving systems of equations
### Substitution method
In the substitution method, one unknown is solved in one of the equations and it is substituted in the other, obtaining an equation of one unknown. Their solution is substituted in the other equation.
### Elimination method
In the elimination method the two equations are prepared so that one of the unknowns has the same coefficient in both. By subtracting or adding them, an equation is obtained without that unknown.
### Equalization method
In the equalization method, the same unknown is solved in the two equations and the results are equaled to obtain an equation with one unknown.
## Exercises of system of equations
1. Find out the unknowns using the substitution method:
## Ejercicios de sistemas de ecuaciones
2. Find the values of x and y using the equalization method:
## Ejercicios de sistemas de ecuaciones
3. Apply the reduction method and solve the system:
## Ejercicios de sistemas de ecuaciones
### Factors of 36
The number 36 has 9 factors and is a composite number. To calculate the factors of 36 we divide,...
### Factors of 35
The number 35 has 4 factors and is a composite number. To calculate the factors of 35 we divide,...
### Factors of 34
The number 4 has 4 factors and is a composite number. To calculate the factors of 34 we divide,...
### Factors of 33
The number 33 has 4 factors and is a composiite number. To calculate the factors of 33 we divide,...
### Factors of 32
The number 32 has 6 factors and is a composite number. To calculate the factors of 32 we divide,...
### Factors of 31
The number 31 has 2 factors and is a prime number. To calculate the factors of 31 we divide,...
### Factors of 30
The number 30 has 8 factors and is a composite number. To calculate the factors of 30 we divide,...
### Factors of 29
The number 29 has 2 factors and is a prime number. To calculate the factors of 29 we divide,...
### Factors of 28
The number 28 has 5 factors and is a composite number. To calculate the factors of 28 we divide,...
### Factors of 27
The number 27 has 4 factors and is a composite number. To calculate the factors of 27 we divide,... |
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# A path of width 3.5m runs around a semi-circular grassy plot whose perimeter is 72m. Find the area of the path (Use $\pi =\dfrac{22}{7}$).
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Hint: Area of circle is $\pi {{r}^{2}}$ where ‘r’ is the radius of the circle and $\pi =\dfrac{22}{7}$. Perimeter is defined as the continuous line forming the boundary of a closed geometric figure.
Here, we have a path which runs around a semi-circular plot as shown in diagram:-
Width of path = 3.5m
Perimeter of plot (semi-circle BC) =72m
As we know that perimeter of any semi-circle can be given as $\pi r+2r$,
where r= Radius of inner semi-circle
$\pi =\dfrac{22}{7}$
Hence, Perimeter of plot as shown in diagram =72
\begin{align} & \pi r+2r=72 \\ & \dfrac{22r}{7}+\dfrac{2r}{1}=72 \\ & \dfrac{22r+14r}{7}=72 \\ \end{align}
or $\dfrac{36r}{7}=72$
So, we get
$r=\dfrac{72\times 7}{36}=14m$.
Now, we can determine the radius of semicircle AD or outer part of the path be OC+CD i.e. r+3.5.
Hence, R=r+3.5m
Where r=14m
So, R=14+3.5=17.5m
As we have to determine area of path which can be given by relation as,
Area of path (ABCD) =Area of outer semi-circle (AD)-Area of inner semi-circle (BC)
We have area of semicircle=$\dfrac{\pi {{r}^{2}}}{2}$
where r is the radius of a semi-circle.
Hence, area of path can be given as
Area of path = $\dfrac{\pi {{R}^{2}}}{2}-\dfrac{\pi {{r}^{2}}}{2}$
where R=17.5m and r=14m and $\pi =\dfrac{22}{7}$.
Area of path = $\dfrac{22}{7}\times \dfrac{1}{2}\times \left( {{R}^{2}}-{{r}^{2}} \right)=\dfrac{11}{7}\left( {{17.5}^{2}}-{{14}^{2}} \right)$
We have algebraic identity $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$
So, area of path can be rewritten as
Area of path = $\dfrac{11}{7}\left( 17.5-14 \right)\left( 17.5+14 \right)=\dfrac{11}{7}\times 3.5\times 31.5=11\times 0.5\times 31.5$
Area of path$=173.25{{m}^{2}}$
Note: One can go wrong while writing the perimeter of the semi-circle of the plot. He/she may apply the formula ‘$\pi r$’ in place of $\pi r+2r$ which will give the wrong answer. Hence, take care while writing the perimeter of the semi-circle. |
## Exploding Dots
### 3.5 (Optional) Multiplication by 10
Let’s answer one of the previous practice questions here.
Why must the answer to $$26417 \times 10$$ look like the original number with a zero tacked on to its end?
I remember being taught this rule in school: to multiply by ten tack on a zero. For example,
$$37 \times 10 = 370$$
$$98989 \times 10 = 989890$$
$$100000 \times 10 = 1000000$$
and so on.
This observation makes perfect sense in the dots-and-boxes thinking.
Here’s the number $$26417$$ again in a $$1 \leftarrow 10$$ machine.
Here’s $$26417 \times 10$$.
Now let’s perform the explosions, one at a time. (We’ll need an extra box to the left.)
We have that $$2$$ groups of ten explode to give $$2$$ dots one place to the left, and $$6$$ groups of ten explode to give $$6$$ dots one place to the left, and $$4$$ groups of ten explode to give $$4$$ dots one place to the left, and so on. The digits we work with stay the same. In fact, the net effect of what we see is all digits shifting one place to the left to leave zero dots in the ones place.
Indeed it looks like we just tacked on a zero to the right end of $$26417$$. (But this is really because of a whole lot of explosions.)
3. a) What must be the answer to $$476 \times 10$$? To $$476 \times 100$$?
b) What must be the answer to $$9190 \div 10$$? To $$3310000 \div 100$$?
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# Parent Function: Definition, Examples & Graphs
## What is a Parent Function?
Every function in the Cartesian plane stems from a particular parent function.
For example, every linear function can be generated from the parent function f(x) = x; Every other possible linear function of the form y = mx + b is a child function of this parent. Together, parent functions and child functions make up families of functions.
To put this another way, every function in a family is a transformation of a parent function. For example, the function f(x) = 2x is the linear parent function vertically stretched by a factor of 2; Instead of the function passing through (1, 1) the graph passes through (2, 1):
## 1. Absolute value parent function
The absolute value function is an even function with the parent p(x) = |x|.
Characteristics:
• Domain: (-∞, ∞).
• Range: [0, ∞]; If x ≥ 0, then f(x) = x and if x <0, then f(x) = -x.
• Inverse Function: f(x) = x, for x≥ 0.
• General form: f(x) = a|b(x – h) + k.
## 2. Constant Parent Function
The constant function is an even function that has the parent f(x) = c.
The graph depends on the value of c. For example, the following graph shows two constant functions where c = 3 (red) and c = 2.5 (blue):
Characteristics:
## 3. Cube Root Parent Function
The cube root function is an odd function that has the parent
f(x) = ∛x.
Characteristics:
## 4. Cubic Polynomial Parent Function
Cubic functions are odd functions. The parent is: f(x) = x3.
The cubic parent function is strictly increasing, which basically means it’s always headed upwards.
Characteristics:
## 5. Exponential Parent Function
The exponential function has no restrictions: inputs can be real numbers or imaginary numbers. The parent function is either f(x) = ex or f(x) = 10x.
Characteristics:
## 6. Linear function
The linear function is an odd function with the parent:
f(x) = x.
Characteristics:
• Domain: (-∞, ∞).
• Range: (-∞, ∞).
• Inverse Function: g(x) = x.
• General form: y = mx + b (m ≠ 0). Alternative: Ax + By + C = 0 or y – y0 = m(x – x0). For an overview of the different forms, see: Standard Form.
## 7. Logarithmic function
Parent: f(x) = logbx; Where b is the base.
For example, the parent f(x) = log2x is different from the parent f(x) = log10x.
The parent function for the natural logarithm function is ln(x).
Characteristics: |
Decimals, Expanded Form, Operations on Decimals
# Decimals, Expanded Form, Operations on Decimals
## Decimals
A fraction with denominator 10, 100, 1000, etc. are known as a decimal fraction.
Decimal fraction with denominator 10 is known as one-tenth (1/10), written as 0.1 in decimals and is read as zero point one.
The decimal fraction with denominator 100 is known as one-hundredth (1/100), written as 0.01 and is read as zero point zero one.
The decimal fraction with denominator 1000 is known as one-thousandth (1/1000), written as 0.001 and is read as zero point zero zero one.
A decimal number has a whole number part and a fractional part (decimal part) which are separated by a decimal point.
For example, in 46.835, 46 is the whole number part and 835 is the fractional part or decimal part.
### Expanded Form of Decimals
Consider the decimal 726.385
The expanded form of 726.385 is 726.385 = 7 × 100 + 2 × 10 + 6 × 1 + 3 × 1/10 + 8 × 1/100 + 5 × 1/1000 and can be read as seven hundred twenty-six point three eight five.
Example: Expand the following.
a. 495.672 b. 754.23
Solution:
a. 495.672 = 4 × 100 + 9 × 10 + 5 × 1 + 6 × 1/10 + 7 × 1/100 + 2 × 1/1000
b. 754.23 = 7 × 100 + 5 × 10 + 4 × 1 + 2 × 1/10 + 3 × 1/100
## Converting Decimals into Fractions
To convert decimals into fractions, write the number without the decimal point in
the numerator and write 10 or powers of 10 in the denominator according to the decimal places in the decimal and write this fraction in the simplest form.
Example: 0.6 = 6/10 = 3/5; 0.02 = 2/100 = 1/50; 3.125 = 3125/1000 = 25/8
## Converting Fractions into Decimals
1. When the denominator of the given fraction is 10 or a multiple of 10, count from extreme right to the left, mark the decimal point after as many digits of the numerator
as there are zeroes in the denominator.
For example, 46/10 = 4.6; 7054/100 = 70.54; 563/1000 = 0.563
2. When the denominator can be expressed as multiples of 10, multiply numerator and
denominator of a fraction by a suitable number such that the denominator becomes 10
or power of 10 and then follow the above steps.
For example, 1/2 = (1 × 5)/(2 × 5) = 5/10 = 0.5
## Like and Unlike Decimals
All the decimals having the same number of decimal places are called like decimals and all the decimals having different number of decimal places are called unlike decimals.
For example,
a. 4.57, 18.78, 127.09 are like decimals. b. 7.5, 4.85, 2.605 are unlike decimals.
The unlike decimals can be converted into like decimals by adding the required number of zeroes in the extreme right of the decimal.
Example: Convert 2.43, 4.6, 3.475 and 8.4237 into like decimals.
Solution: 2.43 can be written as 2.4300, 4.6 can be written as 4.6000 and 3.475 can be
written as 3.4750.
Thus, 2.4300, 4.6000, 3.4750 and 8.4237 are like decimals.
## Comparison of Decimals
While comparing two decimals, first we compare the whole number parts of both
the decimals. The decimal with greater whole number part is greater.
When whole number parts of both the decimal numbers are the same, then we compare the tenths part. If tenths parts of both the decimal numbers are the same, then we compare the hundredths part and so on.
Example: Compare and arrange the following numbers in ascending order:
76.63, 62.335, 63.36, 62.33
Solution: 62.330 < 62.335; 62.335 < 63.36; 63.36 < 76.63
Ascending order: 62.33, 62.335, 63.36, 76.63
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# Quadratic Equation cannot have more than Two Roots
We will discuss here that a quadratic equation cannot have more than two roots.
Proof:
Let us assumed that α, β and γ be three distinct roots of the quadratic equation of the general form ax$$^{2}$$ + bx + c = 0, where a, b, c are three real numbers and a ≠ 0. Then, each one of α, β and γ will satisfy the given equation ax$$^{2}$$ + bx + c = 0.
Therefore,
aα$$^{2}$$ + bα + c = 0 ............... (i)
aβ$$^{2}$$ + bβ + c = 0 ............... (ii)
aγ$$^{2}$$ + bγ + c = 0 ............... (iii)
Subtracting (ii) from (i), we get
a(α$$^{2}$$ - β$$^{2}$$) + b(α - β) = 0
⇒ (α - β)[a(α + β) + b] = 0
⇒ a(α + β) + b = 0, ............... (iv) [Since, α and β are distinct, Therefore, (α - β) ≠ 0]
Similarly, Subtracting (iii) from (ii), we get
a(β$$^{2}$$ - γ$$^{2}$$) + b(β - γ) = 0
⇒ (β - γ)[a(β + γ) + b] = 0
⇒ a(β + γ) + b = 0, ............... (v) [Since, β and γ are distinct, Therefore, (β - γ) ≠ 0]
Again subtracting (v) from (iv), we get
a(α - γ) = 0
⇒ either a = 0 or, (α - γ) = 0
But this is not possible, because by the hypothesis a ≠ 0 and α - γ ≠ 0 since α ≠ γ
α and γ are distinct.
Thus, a(α - γ) = 0 cannot be true.
Therefore, our assumption that a quadratic equation has three distinct real roots is wrong.
Hence, every quadratic equation cannot have more than 2 roots.
Note: If a condition in the form of a quadratic equation is satisfied by more than two values of the unknown then the condition represents an identity.
Consider the quadratic equation of the general from ax$$^{2}$$ + bx + c = 0 (a ≠ 0) ............... (i)
Solved examples to find that a quadratic equation cannot have more than two distinct roots
Solve the quadratic equation 3x$$^{2}$$ - 4x - 4 = 0 by using the general expressions for the roots of a quadratic equation.
Solution:
The given equation is 3x$$^{2}$$ - 4x - 4 = 0
Comparing the given equation with the general form of the quadratic equation ax^2 + bx + c = 0, we get
a = 3; b = -4 and c = -4
Substituting the values of a, b and c in α = $$\frac{- b - \sqrt{b^{2} - 4ac}}{2a}$$ and β = $$\frac{- b + \sqrt{b^{2} - 4ac}}{2a}$$ we get
α = $$\frac{- (-4) - \sqrt{(-4)^{2} - 4(3)(-4)}}{2(3)}$$ and β = $$\frac{- (-4) + \sqrt{(-4)^{2} - 4(3)(-4)}}{2(3)}$$
⇒ α = $$\frac{4 - \sqrt{16 + 48}}{6}$$ and β =$$\frac{4 + \sqrt{16 + 48}}{6}$$
⇒ α = $$\frac{4 - \sqrt{64}}{6}$$ and β =$$\frac{4 + \sqrt{64}}{6}$$
⇒ α = $$\frac{4 - 8}{6}$$ and β =$$\frac{4 + 8}{6}$$
⇒ α = $$\frac{-4}{6}$$ and β =$$\frac{12}{6}$$
⇒ α = -$$\frac{2}{3}$$ and β = 2
Therefore, the roots of the given quadratic equation are 2 and -$$\frac{2}{3}$$.
Hence, a quadratic equation cannot have more than two distinct roots.
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# Chapter 8: Further Topics in Algebra - PowerPoint PPT Presentation
Chapter 8: Further Topics in Algebra. 8.1 Sequences and Series 8.2 Arithmetic Sequences and Series 8.3 Geometric Sequences and Series 8.4 The Binomial Theorem 8.5 Mathematical Induction 8.6 Counting Theory 8.7 Probability. 8.7 Probability. Basic Concepts
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Presentation Transcript
8.1 Sequences and Series
8.2 Arithmetic Sequences and Series
8.3 Geometric Sequences and Series
8.4 The Binomial Theorem
8.5 Mathematical Induction
8.6 Counting Theory
8.7 Probability
Basic Concepts
• An experiment has one or more outcomes. The outcome of rolling a die is a number from 1 to 6.
• The sample space is the set of all possible outcomes for an experiment. The sample space for a dice roll is {1, 2, 3, 4, 5, 6}.
• Any subset of the sample space is called an event. The event of rolling an even number with one roll of a die is {2, 4, 6}.
Probability of an Event E
In a sample space with equally likely outcomes, the probability of an event E, written P(E), is the ratio of the number of outcomes in sample space S that belong to E, n(E), to the total number of outcomes in sample space S, n(S). That is,
Example A single die is rolled. Give the probability
of each event.
(a) E3 : the number showing is even
(b) E4 : the number showing is greater than 4
(c) E5 : the number showing is less than 7
(d) E6 : the number showing is 7
Solution The sample space S is {1, 2, 3, 4, 5, 6} so
n(S) = 6.
(a) E3 = {2, 4, 6} so
(b) E4= {5, 6} so
Solution
(c) E5 = {1, 2, 3, 4, 5, 6} so
(b) E6 = Ø so
• For an event E, P(E) is between 0 and 1 inclusive.
• An event that is certain to occur always has probability 1.
• The probability of an impossible event is always 0.
• The set of all outcomes in a sample space that do not belong to event E is called the complement of E, written E´. If S = {1, 2, 3, 4, 5, 6} and E = {2, 4, 6} then E´ = {1, 3, 5}.
• Probability concepts can be illustrated with Venn diagrams. The rectangle represents the sample space in an experiment. The area inside the circle represents event E; and the area inside the rectangle but outside the circle, represents event E´.
Example A card is drawn from a well-shuffled
deck, find the probability of event E, the card is
an ace, and event E´.
Solution There are 4 aces in the deck of 52
cards and 48 cards that are not aces. Therefore
The odds in favor of an event E are expressed as the
ratio of P(E) to P(E´) or as the fraction
Example A shirt is selected at random from a dark
closet containing 6 blue shirts and 4 shirts that are
not blue. Find the odds in favor of a blue shirt
being selected.
SolutionE is the event “blue shirt is selected”.
Solution The odds in favor of a blue shirt are
or 3 to 2.
Probability of the Union of Two Events
For any events E and F,
Example One card is drawn from a well-shuffled
deck of 52 cards. What is the probability of each
event?
(a) The card is an ace or a spade.
(b) The card is a 3 or a king.
Solution (a) P(ace or space) = P(ace) + P(spade)
(b) P(3 or K) = P(3) + P(K) – P(3 and K)
Properties of Probability
1.
2. P(a certain event) = 1;
3. P(an impossible event) = 0;
4.
5.
An experiment that consists of
• repeated independent trials,
• only two outcomes, success and failure, in
each trial,
is called a binomial experiment.
Let the probability of success in one trial be p.
Then the probability of failure is 1 – p.
The probability of r successes in n trials is given by
Example An experiment consists of rolling a die 10
times. Find the probability that exactly 4 tosses result
in a 3.
Solution Here , n = 10 and r = 4. The required probability is |
Study Guide
# Polynomial Division and Rational Expressions - Polynomial Division
## Polynomial Division
Polynomials have tons of similarities to integers, and throughout this section we'll be coming back a lot to the integer analogy to make our points. We'll look at polynomial division in a few different situations, each of which can be related to integer division. See? We did it already.
A lot of ideas in this section come from the section on fractions, so it's a good idea to make sure you're comfortable with that stuff before diving into this section. Slipping into your Snuggie might help you feel more comfortable as well. If anyone's willing, you may also want to think about requesting a plate of apple wedges and a Capri Sun.
We'll be using fraction notation to express polynomial division. For example:
(3x2 + 9x) divided by (5x – 2) is written as .
We'll treat these expressions like fractions. The number on top is still called the numerator, the number on the bottom is still called the denominator, and we still aren't allowed to divide by zero. Bummer.
To start with, we'll look at polynomial division where the divisor goes into the dividend with no remainder. As a matter of fact, we can guarantee that all the division problems in this particular unit will work out evenly. Absolutely no leftovers. Our apologies to your dog.
For polynomial division problems, there are three main things to do:
1. Cancel any common monomial factors that show up in both the numerator and denominator.
2. Factor all polynomials and cancel any binomial or trinomial factors that appear in both the numerator and denominator.
3. Use long division.
• ### Common Factors
The easiest case of polynomial division is when the divisor is a single term that happens to be a common factor of the dividend. You'll love when this happens. For the sake of the integer analogy, we'll do a couple of examples with integers first. You'll probably notice that we're not doing the integer problems in the most efficient way. That's not because we're being lazy or unsure of what we're doing. By approaching them in a certain manner, we'll be doing them in a way that will translate nicely to polynomials. Ah, there's a method to our madness...
### Sample Problem
Write as a sum of two integers.
The fraction is the sum of the two fractions and . Since and , we have:
### Sample Problem
Write as a sum of integers.
This is like the last problem, except we'll be splitting the fraction into a sum of four other fractions. Once you're done, try splitting an atom. Or your pants. Anything, really, as long as you're getting in some good splitting practice.
Performing division on each term, we find:
Now we'll do something similar with polynomials. Take five, integers. If the divisor is a common factor of the dividend, then the divisor divides (goes evenly into, with nothing left over) every term of the dividend, meaning we can break up the polynomial like we broke up the fractions above. Contrary to what you may have heard, breaking up isn't that hard to do.
### Sample Problem
Find .
We can split up the terms of the dividend and rewrite this as .
Simplifying each term gives us the final answer: 4x + 8. Just cancel out those x's in the denominators.
Another way to look at this sort of problem is to factor. We'll let the integers be the opening act again. They did such a great job when they opened for Coldplay last weekend that they deserve to be given another opportunity.
### Sample Problem
Write as a sum of two integers.
Pull out the common factor of 5 from the numerator and the denominator.
Then we can write the fraction as and cancel the 5 from the top and bottom to find .
Good job canceling the 5, but don't assume you'll never hear from it again. You're still on its email list.
After a bit of practice, you can skip the step of physically writing down where you pull the common factor out of each term...as long as you can do it mentally and still wind up with the correct answer, that is. If you can master this skill, who knows? You may be able to start doing all kinds of things mentally. Don't go all Chronicle on us, though.
When asked to divide a polynomial by a single term, you can use whichever method you like. You'll find the same answer whether you break up the polynomial first and then simplify the resulting terms, or pull out the common factor first. Don't do both at the same time, or you may disrupt the space-time continuum.
• ### Factoring
Factoring isn't only for when the divisor is a single term. We can use factoring to divide one polynomial by another polynomial, as long as the polynomial in the denominator is a factor of the polynomial in the numerator. We're sorry. We know you're keeping track of a lot of nomials.
For example, we can simplify by factoring 10 as 2 × 5 and canceling 2 from the top and bottom, and we can do similar things with polynomials.
When factoring, keep an eye out for common factors that are one term. No sense in making more work for yourself if you can avoid it. We know what a workaholic you are, but tone it down a bit.
### Sample Problem
Find the quotient: .
The polynomial on top factors as (6x + 5)(6x – 5), so we can rewrite the expression as:
Those all match, so now we can cancel the factor of (6x + 5) from the top and bottom to get our final answer:
(6x – 5)
### Sample Problem
What is 4x4 + 4x2 + 1 divided by 2x2 + 1?
We want to find:
The numerator factors as (2x2 + 1)(2x2 + 1), so we can rewrite the problem like this:
Then we cancel the factor of (2x2 + 1) from the top and bottom:
2x2 + 1
### Sample Problem
Find the quotient: .
The numerator and denominator look like something should cancel, right? Remember, we have that guarantee that the problems in this section will work out evenly, so we can take that to the bank. Actually, don't try taking that to the bank. They'll give you a weird look.
The first thing we're supposed to do is look for common factors, but there are no common factors shared by both the numerator and the denominator. They could probably learn a lot about sharing if they watched The Wiggles once in a while. Anyway, what we can do is pull out a common factor of 2 in the numerator to simplify that guy at least. Hopefully, doing this will shed some light on where we go next. We can now rewrite our original problem as:
The expressions 5x – 1 and 1 – 5x are negatives of each other. We're getting closer, we can feel it already:
(1 – 5x) = (-1)(5x – 1)
If we factor out -1 from the denominator, we have:
Now we can cancel 5x – 1 from the top and bottom:
When asked to divide polynomials, look for cases like this where expressions in the numerator and denominator are negatives of each other. Then factor out -1 and force them to get along.
Factoring works on multivariable polynomials too. Oh, happy day. We were starting to miss our old friends y and z.
• ### Long Division
After looking for single-term factors and finishing all our factoring, the next thing we do to divide polynomials is use good old-fashioned long division. You know, old-fashioned like the donut: delicious, time-tested, and kind of weird to think about in too much detail.
First, here's a reminder of how long division works with integers.
### Sample Problem
Find 611 ÷ 13.
Using long division notation, we set up the division problem as .
First off, 13 goes into 6 a grand total of zero times, no matter how forcefully you may try to squeeze it in there. We write 0 above 6, subtract 13 × 0 from 6, and bring down the next digit in the dividend:
Moving briskly along, 13 goes into 61 four times, so we write the digit 4 on top of the digit 1, subtract 13 × 4 = 52 from 61, and bring down the final 1:
Finally, 13 goes into 91 seven times (13 × 7 = 91):
We conclude that 611 ÷ 13 = 47. We can check this answer by multiplying 13 and 47 and making sure we find 611. Try it. It works! As advertised, there's no remainder. We hope this is building some trust between us. If only we could persuade you to fall backward and let us catch you.
Now we'll do some long division with polynomials.
### Sample Problem
Find using long division.
First we set up the long division, making sure the terms of each polynomial are written in decreasing order by exponent, like so:
Then we look at the first term of the divisor and see how many times it goes into the first term of the dividend:
Hmm...x2 goes into x5 a total of x3 times (that is, x2 × x3 = x5). We're dealing with exponents, so it's not the same as figuring out how many times the number 2 goes into the number 5. Here, we're subtracting exponents rather than dividing one number by another. That's why exponents look like they're floating away, because we "take them away."
We write x3 on top of x5:
We subtract (x3)(x2 + 4) = x5 + 4x3 from the dividend, the same as with normal long division, and then we "bring down" all of the remaining terms of the dividend:
Now we have a new polynomial. It even still has that new polynomial smell.
Next, we see how many times the first term of the divisor goes into the first term of this new polynomial:
Since is 3x, x2 goes into 3x3 a total of 3x times.
We subtract (3x)(x2 + 4) = 3x3 + 12x, and drop down the remaining terms. Whoa, careful. Don't drop them down so fast. These things are fragile.
Now we see that x2 goes into 7x2 a total of 7 times, and subtract 7(x2 + 4). We have 0 left over, so we're finished. You didn't think we'd finish all of that without a remainder, did you? Oh ye of little faith.
To make it official, the final answer is:
x3 + 3x + 7
Check the answer to the above example by multiplying (x2 + 4) and (x3 + 3x + 7). You should get x5 + 7x3 + 7x2 + 12x + 28.
In the example above, there were a few places where we wrote 0x2 or 0x. We did this so that like terms would line up nicely in the division problem. Obviously, 0x2 is the same as 0, but see how much easier it is to keep track of everything when you do it our way? Come to the Shmoop side...
Sometimes problems involve polynomials that "skip terms." Sort of like how some people—not you, obviously—skip classes, but without the negative repercussions. The polynomial 5x2 – 9, for example, "skips" the x term. A more mathematical and impressive way to say this is "the coefficient on the x term is 0." Aren't you impressed? We thought so. An even more impressive way to say this is "the coefficient of the first-degree term is 0." The most impressive way to say this is backwards and in Russian while doing a handstand. However, we don't want to put any undue pressure on you, so saying it the first way will be fine.
We can rewrite the polynomial like this without really changing it:
5x2 – 9 = 5x2 + 0x – 9
If the dividend skips terms when you're setting up polynomial long division, write them down anyway to help yourself keep things straight. If the dividend is 5x2 – 9, write 5x2 + 0x – 9 instead. We might need that column for subtraction somewhere in the middle of the long division. You'll be glad you stuck it in there. |
# How to Construct a Triangle(With Steps, Diagrams & Examples)
This post is also available in: हिन्दी (Hindi)
A triangle is a three-sided 2D shape formed by three lines known as its sides. A triangle has three vertices and three angles. The construction of triangles is one of the essential part of geometry.
Let’s understand how to construct a triangle with examples.
## How to Construct a Triangle?
A triangle has 6 basic elements – 3 sides and 3 angles. To construct any unique triangle, you should know at least 3 elements.
The following are the basic requirements to construct a unique triangle.
• All three sides are given (SSS – Side side side)
• Two sides and included angle are given (SAS – Side angle side)
• Two angles and the included side is given (ASA – Angle side angle)
• The measure of the hypotenuse and a side is given in the right triangle (RHS – Right angle hypotenuse side)
Let’s understand these constructions in detail.
### How to Construct a Triangle With SSS Property
When the length of three sides of the triangle is given, then the following are the steps to construct the required triangle.
Step 1: Draw a line segment $\text{AB}$, of length equal to the longest side of the triangle
Step 2: Now using a compass and ruler take the measure of the second side and draw an arc
Step 3: Again take the measure of the third side and cut the previous arc at a point $\text{C}$
Step 4: Now join the endpoints of the line segment to point $\text{C}$ and get the required $\triangle \text{ABC}$
### How to Construct a Triangle With SAS Property
When the length of the two sides and the angle included between them are given, then the following are the steps to construct the triangle.
Step 1: Draw a line segment $\text{AB}$, of length equal to the longest side of the triangle, using a ruler
Step 2: Put the centre of the protractor on one end of a line segment (say $\text{A}$) and measure the given angle. Join the points and construct a ray, such that the ray is nearer to the line segment $\text{AB}$
Step 3: Take the measure of another given side of the triangle using a compass and a ruler
Step 4: Place the compass at point $\text{A}$ and cut the ray at another point, $\text{C}$
Step 5: Join the other end of the line segment, i.e., $\text{B}$ to the point $\text{C}$
The $\triangle \text{ABC}$ is constructed.
### How to Construct a Triangle With ASA Property
When the measures of two angles and the side included between them are given of a triangle, then it is said to be $\text{ASA}$ congruency. The following are the steps to draw a triangle with $\text{ASA}$ property.
Step 1: Draw a line segment $\text{AB}$, of length equal to the given side of the triangle, using a ruler
Step 2: At one endpoint of the line segment (say $\text{A}$) measure one of the given angles and draw a ray $\text{AR}$
Step 3: At another endpoint of the line segment (i.e., $\text{B}$) measure the other angle using a protractor and draw the ray $\text{BQ}$, such that it cuts the previous ray at a point $\text{P}$
Step 4: Join the previous point $\text{P}$, with both the endpoints $\text{A}$ and $\text{B}$ of the line segment $\text{AB}$, to get the required triangle
### How to Construct a Triangle With RHS Property
When the hypotenuse side and any one of the other two sides of the right triangle are given, then it is $\text{RHS}$ property. The following are the steps to draw a triangle with $\text{RHS}$ property.
Step 1: Draw the line segment $\text{AB}$, equal to the measure of the other side
Step 2: At one endpoint, say $\text{AB}$, of line-segment measure the angle equal to $90^{\circ}$ degrees and draw a ray, $\text{AR}$
Step 3: Measure the length of the hypotenuse and draw an arc to cut the ray $\text{AR}$ at a point $\text{P}$
Step 4: Join the points $\text{P}$ and $\text{B}$ to get the required right triangle
## How to Construct a Triangle – Special Cases
After understanding the steps for the construction of general triangles, let’s now understand how to construct special triangles when
• sum of the two sides is given
• difference between the two sides is given
### How to Construct a Triangle With Base Angle and the Sum of the Other Two Sides
Let’s now understand the procedure of construction of a triangle where the base of a triangle, its base angle, and the sum of the other two sides are given.
For constructing a $\triangle \text{ABC}$ such that base $\text{BC}$, base angle $\angle \text{B}$ and the sum of the other two sides, i.e. $\text{AB} + \text{AC}$ are given, the following steps of construction are used.
Step 1: Draw the base $\text{BC}$ of $\triangle \text{ABC}$ as given and construct $\angle \text{XBC}$ of the given measure at $\text{B}$
Step 2: Keeping the compass at point $\text{B}$ cut an arc from the ray $\text{BX}$ such that its length equals $\text{AB} + \text{AC}$ at point $\text{P}$ and join it to $\text{C}$
Step 3: Now measure $\angle \text{BPC}$ and from $\text{C}$, draw an angle equal to $\angle \text{BPC}$
### How to Construct a Triangle With Base Angle and the Difference Between the Other Two Sides
Let’s now understand the procedure of construction of a triangle where the base of a triangle, its base angle, and the difference between the other two sides are given.
For constructing $\triangle \text{ABC}$ such that base $\text{BC}$, base angle $\angle \text{B}$ and difference of the other two sides, i.e. $\text{AB} – \text{AC}$ or $\text{AC} – \text{AB}$ is given, then for constructing triangles such as these two cases can arise
• $\text{AB} \gt \text{AC}$
• $\text{AC} \gt \text{AB}$
The following steps of construction are followed for the two cases:
#### Steps of Construction When $\text{AB} \gt \text{AC}$
Step 1: Draw the base $\text{BC}$ of $\triangle \text{ABC}$ as given and construct ∠XBC of the required measure at B as shown.
Step 2: From the ray, BX cut an arc equal to AB – AC at point P and join it to C as shown
Step 3: Draw the perpendicular bisector of PC and let it intersect BX at point A as shown:
Step 4: Join AC, ∆ABC is the required triangle.
#### Steps of Construction When $\text{AC} \gt \text{AB}$
Step 1: Draw the base BC of ∆ABC as given and construct ∠XBC of the required measure at B as shown.
Step 2: On the ray BX cut an arc equal to AB – AC at point P and join it to C. In this case, P will lie on the opposite side to the ray BX. Draw the perpendicular bisector of PC and let it intersect BX at point A as shown
Step 3: Join the points A and C, and hence ∆ABC is the required triangle.
## Practice Problems
1. Construct a $\triangle \text{PQR}$ in which $\text{PQ } = 5.4 \text{cm}$, $\angle \text{Q} = 60^{\circ}$ and $\text{PR} – \text{PQ} = 2.3 \text{cm}$.
2. Construct a $\triangle \text{XYZ}$ in which $\angle \text{Y} = 45^{\circ}$, $\angle \text{Z} = 75^{\circ}$ and $\text{XY} + \text{YZ} + \text{ZX} = 12 \text{cm}$.
3. Construct a right-angled triangle whose be is $3.8 \text{cm}$ and hypotenuse is $5.6 \text{cm}$.
4. Construct a $\triangle \text{ABC}$ in which $\angle \text{B} = 60^{\circ}$, $\angle \text{C} = 30^{\circ}$ and the length of the perpendicular from the vertex $\text{A}$ is $5.3 \text{cm}$.
## FAQs
### What is a triangle construction?
Drawing a triangle with specific dimensions considering all the related properties of triangles is called triangle construction. Construction of any type of triangle can be done with the help of a ruler, a compass, or a protractor.
### Which criterion is used for the construction of a triangle?
To construct a triangle, the following criteria are used.
a) SSS criterion: A triangle in which all three sides are known.
b) ASA criterion: A triangle in which two angles and one side are known.
c) SAS criterion: A triangle in which two sides and one angle are known.
d) RHS criterion: A triangle in which the hypotenuse and another side are known.
### How do you construct a triangle with 3 sides?
Draw a line with the length of the longest side. Draw two arcs from the two endpoints of the line drawn such that they intersect each other. Join the intersecting point with the vertices of the longest side.
## Conclusion
Drawing a triangle with specific dimensions considering all the related properties of triangles is called triangle construction. You can construct a unique triangle using a ruler, a compass, or a protractor. The basic criteria used to construct triangles are SSS, SAS, ASA, and RHS criteria. |
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Tyler has \$14.79 in his sock bank.
# Consumer Math (Calculating Compound Interest to the Penny)
This Math quiz is called 'Consumer Math (Calculating Compound Interest to the Penny)' and it has been written by teachers to help you if you are studying the subject at middle school. Playing educational quizzes is a fabulous way to learn if you are in the 6th, 7th or 8th grade - aged 11 to 14.
It costs only \$12.50 per month to play this quiz and over 3,500 others that help you with your school work. You can subscribe on the page at Join Us
As you may recall, there are two basic forms of interest, i.e., simple interest and compound interest. For this quiz you will only be dealing with compound interest.
In a separate quiz, you would calculate the compound interest to the whole dollar, meaning when you had to multiply the interest by its power, you were to round that amount to the nearest 100th. In this quiz, you will be required to calculate the interest by the full string of its power or to the penny. [NOTE: You will need to use a calculator to do the problems here.] This will enable you to more accurately determine the amount of interest and the full amount to be paid back or that will be left in savings at the end of the term.
So to just quickly refresh you on what “interest” is in the monetary world of saving and borrowing, it is the amount of extra money you earn or you have to pay back.
As we will be only dealing with compound interest calculations in this quiz, the formula that will be used is: A = P(1 + r)t.
A = The amount of money (including the accrued interest) after __ years/months or the compound amount.
P = The principal saved or owed.
r = The interest rate earned per year
t = The time period of the loan or amount saved (notice that the time is put into the “power” position)
Okay, let’s work out one problem together.
You have to borrow \$850.00 at an interest rate of 3.28% for 2 years. Using the compound interest formula that will read as follows:
A = 850(1 + .0328)2
850(1.0328)2
(1.0328 x 1.0328) = 1.0666758 (You need to use this entire percentage string to do the problem.)
850 x 1.0666758 = 906.67443 (Round this number to the nearest penny making it \$906.67.)
\$906.67 - \$850.00 = \$56.67
\$56.67 is the compound interest over two years and
A = \$906.67 is the full amount that will have to be paid back over the two years.
Had you rounded the percentage off to 1.07, the full amount to pay off would have come to \$909.50. It is an approximate amount but not the “exact” amount. When figuring out compound interest to the exact amount, you must use the percentage string.
1.
Zachery took out a personal loan of \$4,000.00 at an interest rate of 9%, compounded annually. He will pay the full amount back in 3 years. What will be the full amount of money Zachery will have to pay back and how much of that will be the compound interest?
Compound Interest Accrued: \$1,080.12; Full Amount to Pay-Off Loan: \$5,080.12
Compound Interest Accrued: \$1,008.12; Full Amount to Pay-Off Loan: \$5,008.12
Compound Interest Accrued: \$1,180.12; Full Amount to Pay-Off Loan: \$5,180.12
Compound Interest Accrued: \$1,260.12; Full Amount to Pay-Off Loan: \$5,260.12
The compound formula is A = P(1 + r)t. Substituting the letters for numbers we get:
4,000(1 + .09)3
(1 + .09)3 = (1.09 x 1.09 x 1.09) = 1.295029
4,000 x 1.295029 = \$5,180.116 (Rounded to the nearest penny so \$5,180.12 is the full amount needed to pay off the loan after 3 years)
\$5,180.12 - \$4,000.00 = \$1,180.12 (is the compound interest accrued over 3 years)
Solution: Zachery’s loan accrued \$1,180.12 in compound interest over the 3 years and the full amount that he will have to pay back is \$5,180.12.
2.
The Fuller Law Firm has \$2,600,000.00 in the bank which is earning 8.75% interest, compounded annually. If the firm does not touch this account, how much compound interest will it earn in 3 years and what will be the new amount of this bank account?
Compound Interest Earned: \$763,960.30; Amount in Savings: \$3,363,960.30
Compound Interest Earned: \$743,960.30; Amount in Savings: \$3,343,960.30
Compound Interest Earned: \$703,960.30; Amount in Savings: \$3,303,960.30
Compound Interest Earned: \$963,960.30; Amount in Savings: \$3,563,960.30
The compound formula is A = P(1 + r)t. Substituting the letters for numbers we get:
2,600,000(1 + .0875)3
(1 + .0875)3 = (1.0875 x 1.0875 x 1.0875) = 1.2861386
2,600,000 x 1.2861386 = \$3,343,960.30 (is the amount in this account after 3 years)
\$3,343,960.30 - \$2,600,000.00 = \$743,960.30 (is the compound interest earned over 3 years)
Solution: The Fuller Law Firm will earn \$743,960.30 in compound interest in 3 years and the full amount in its bank account will be \$3,343,960.30.
3.
Tyler has \$14.79 in his sock bank. His parents will pay him 50% interest, compounded, if he doesn’t touch the sock for 1 month. How much compound interest will accrue over 1 month and what will be the full amount that Tyler will get for not touching the money in his sock?
Compound Interest Earned: \$7.38; Amount in Savings: \$22.17
Compound Interest Earned: \$7.39; Amount in Savings: \$22.18
Compound Interest Earned: \$7.40; Amount in Savings: \$22.19
Compound Interest Earned: \$7.41; Amount in Savings: \$22.20
The compound formula is A = P(1 + r)t. Substituting the letters for numbers we get:
14.79(1 + .5)1
(1 + .5)1 = (1.5)
14.79 x 1.5 = \$22.185 (Rounded to the nearest penny makes it \$22.19 and this is the full amount Tyler will get after 1 month)
\$22.19 - \$14.79 = \$7.40 (is the compound interest accrued over 1 month)
Solution: Tyler will have earned \$7.40 in compound interest over the 1 month and the full amount that he will have in his sock will be \$22.19.
4.
Billy’s parents opened up a savings account for him when he was born. They put \$2,000.00 into the account where it has been earning 6% interest, compounded annually, for 12 years. How much compound interest has the account earned/accrued and what amount should be in the account now?
Compound Interest Earned: \$2,324.39; Amount in Savings: \$4,324.39
Compound Interest Earned: \$2,224.39; Amount in Savings: \$4,224.39
Compound Interest Earned: \$2,124.39; Amount in Savings: \$4,124.39
Compound Interest Earned: \$2,024.39; Amount in Savings: \$4,024.39
The compound formula is A = P(1 + r)t. Substituting the letters for numbers we get:
2,000(1 + .06)12
(1 + .06)12 = (1.06 x 1.06 x 1.06 x 1.06 x 1.06 x 1.06 x 1.06 x 1.06 x 1.06 x 1.06 x 1.06 x 1.06) = 2.012196
2,000 x 2.012196 = \$4,024.392 (Rounded to the nearest penny it is \$4,024.39 and it is the amount in the savings account after 12 years)
\$4,024.39 - \$2,000.00 = \$2,024.39 (is the compound interest earned over 12 years)
Solution: Billy’s savings account has earned \$2,024.39 in compound interest in 12 years and the full amount in the account should now be \$4,024.39.
5.
Travis opened up a Christmas savings account and put in \$1,500.00. The account will earn 16.4% interest, compounded annually. How much compound interest to the nearest rounded penny will he earn in 1 year and what will be the new amount of his Christmas savings account?
Compound Interest Earned: \$546.00; Amount in Savings: \$2,046.00
Compound Interest Earned: \$446.00; Amount in Savings: \$1,946.00
Compound Interest Earned: \$344.00; Amount in Savings: \$1,846.00
Compound Interest Earned: \$246.00; Amount in Savings: \$1,746.00
The compound formula is A = P(1 + r)t. Substituting the letters for numbers we get:
1,500(1 + .164)1
(1 + .164)1 = (1.164)
1,500 x 1.164 = \$1,746.00 (is the amount in savings after 1 year)
\$1,746.00 - \$1,500.00 = \$246.00 (is the compound interest earned over 1 year)
Solution: Travis will earn \$246.00 in compound interest in 1 year and the full amount in his Christmas savings account will be \$1,746.00.
6.
The public aquatic center took out a 4 year loan in the amount of \$50,000.00 to buy all equipment. The loan is earning 2.97% interest, compounded annually. How much compound interest will accrue over 4 years and what will be the full amount that the public aquatic center will have to pay back?
Compound Interest Accrued: \$5,909.90; Full Amount to Pay-Off Loan: \$55,909.90
Compound Interest Accrued: \$6,009.90; Full Amount to Pay-Off Loan: \$56,009.90
Compound Interest Accrued: \$6,109.90; Full Amount to Pay-Off Loan: \$56,109.90
Compound Interest Accrued: \$6,209.90; Full Amount to Pay-Off Loan: \$56,209.90
The compound formula is A = P(1 + r)t. Substituting the letters for numbers we get:
50,000(1 + .0297)4
(1 + .0297)4 = (1.0297 x 1.0297 x 1.0297 x 1.0297) = 1.1241979
50,000 x 1.1241979 = \$56,209.895 (Rounded to the nearest penny so \$56,209.90 is the full amount owed on the loan after 4 years)
\$56,209.90 - \$50,000.00 = \$6,209.90 (is the compound interest accrued over 4 years)
Solution: The public aquatic center’s loan accrued \$6,209.90 in compound interest over the 4 years and the full amount that it will have to pay back is \$56,209.90.
7.
The newly engaged couple took out a loan to pay for their wedding and honeymoon. They borrowed \$15,000.00 at 3% and will pay it back in 2 years. How much compound interest will accrue over 2 years and what will be the full amount that the couple will have to pay back?
Compound Interest Accrued: \$913.50; Full Amount to Pay-Off Loan: \$15,913.50
Compound Interest Accrued: \$943.50; Full Amount to Pay-Off Loan: \$15,943.50
Compound Interest Accrued: \$983.50; Full Amount to Pay-Off Loan: \$15,983.50
Compound Interest Accrued: \$993.50; Full Amount to Pay-Off Loan: \$15,993.50
The compound formula is A = P(1 + r)t. Substituting the letters for numbers we get:
15,000(1 + .03)2
(1 + .03)2 = (1.03 x 1.03) = 1.0609
15,000 x 1.0609 = \$15,913.50 (is the full amount owed on the loan after 2 years)
\$15,913.50 - \$15,000.00 = \$913.50 (is the compound interest accrued over 2 years)
Solution: The couple’s loan accrued \$913.50 in compound interest over the 2 years and the full amount that they will have to pay back is \$15,913.50.
8.
Jamison deposited \$645.00 into a savings account that is earning 3.9% interest, compounded annually. How much compound interest to the nearest rounded penny will he earn in 2 years and what will be the new amount of his savings?
Compound Interest Earned: \$50.29; Amount in Savings: \$695.29
Compound Interest Earned: \$51.29; Amount in Savings: \$696.29
Compound Interest Earned: \$52.29; Amount in Savings: \$697.29
Compound Interest Earned: \$53.29; Amount in Savings: \$698.29
The compound formula is A = P(1 + r)t. Substituting the letters for numbers we get:
645(1 + .039)2
(1 + .039)2 = (1.039 x 1.039) = 1.079521
645 x 1.079521 = \$696.29104 (Round this number to the nearest penny so you will have \$696.29 as the amount in savings after 2 years)
\$696.29 - \$645.00 = \$51.29 (is the compound interest earned over 2 years)
Solution: Jamison will earn \$51.29 in compound interest and his savings in 2 years will be \$696.29.
9.
Brett has \$11,500 in his savings account that is earning 6.5% interest, compounded annually. How much compound interest will he earn in 5 years and what will be the new amount of his savings account?
Compound Interest Accrued: \$4,256.00; Full Amount in Savings: \$15,756.00
Compound Interest Accrued: \$4,056.00; Full Amount in Savings: \$15,556.00
Compound Interest Accrued: \$3,856.00; Full Amount in Savings: \$15,356.00
Compound Interest Accrued: \$3,556.00; Full Amount in Savings: \$15,056.00
The compound formula is A = P(1 + r)t. Substituting the letters for numbers we get:
11,500(1 + .065)5
(1 + .065)5 = (1.065 x 1.065 x 1.065 x 1.065 x 1.065) = 1.3700866
11,500 x 1.3700866 = \$15,755.995 (Rounded to the nearest penny so \$15,756.00 is the full amount in the savings account after 5 years)
\$15,756.00 - \$11,500.00 = \$4,256.00 (is the compound interest accrued over 5 years)
Solution: Brett’s savings account accrued \$4,256.00 in compound interest over the 5 years and the full amount that he now has in savings is \$15,756.00.
10.
Mabel has a retirement account with \$49,000.00. It is earning 4.32% interest, compounded annually. If she doesn’t touch her account for 7 years, how much compound interest will it earn and what will be the new balance of her retirement account?
Compound Interest Earned: \$15,882.34; Amount in Savings: \$64,882.34
Compound Interest Earned: \$14,882.34; Amount in Savings: \$63,882.34
Compound Interest Earned: \$16,882.34; Amount in Savings: \$65,882.34
Compound Interest Earned: \$18,882.34; Amount in Savings: \$67,882.34
The compound formula is A = P(1 + r)t. Substituting the letters for numbers we get:
49,000(1 + .0432)7
(1 + .0432)7 = (1.0432 x 1.0432 x 1.0432 x 1.0432 x 1.0432 x 1.0432 x 1.0432) = 1.3445375
49,000 x 1.3445375 = \$65,882.337 (Rounded to the nearest penny it is \$65,882.34 and it is the amount in the retirement account after 7 years)
\$65,882.34 - \$49,000.00 = \$16,882.34 (is the compound interest earned over 7 years)
Solution: Mabel’s retirement account will earn \$16,882.34 in compound interest in 7 years and the full amount in the retirement account will be \$65,882.34. |
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# What is the formula to solve the general form of quadratic equation and what is its discriminant value?
Last updated date: 25th Jul 2024
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Answer
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Hint: In this problem, we can see the formula to solve the general form of the quadratic equation and what its discriminant value is. We should know that the general quadratic equation is of the form $a{{x}^{2}}+bx+c=0$, where a, b, c are constant terms and a is not equal to zero. here we can see the formula which is used to solve the quadratic equation and its discriminant values.
Complete step by step answer:
Here we can see the formula to solve the general form of the quadratic equation and what its discriminant value is.
We should know that the general quadratic equation is of the form,
$a{{x}^{2}}+bx+c=0$
We can solve this type of quadratic equation using the quadratic formula,
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
We can also write it as,
$x=\dfrac{-b\pm \sqrt{D}}{2a}$
Where, the discriminant is denoted as $\Delta$,
$\Delta ={{b}^{2}}-4ac$
We should know that,
If the discriminant value is equal 0, $\Delta =0$
Then we will have two real and equal roots (i.e. one real root).
If the discriminant value is greater than 0, $\Delta >0$
Then we will have two real and unequal roots.
If the discriminant value is less than 0, $\Delta <0$
Then we will have no real roots (imaginary roots).
Therefore, the formula to solve a general quadratic equation $a{{x}^{2}}+bx+c=0$ is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, where the discriminant is $\Delta ={{b}^{2}}-4ac$.
Note: We should always remember the condition of the discriminants which tells the type of roots, when the discriminant is equal to zero, the equation has one real root, when the discriminant is less than 0, it has no real roots and when the discriminant is greater than 0, then it has two real roots. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 2.6: Chapter 2 Review
Difficulty Level: At Grade Created by: CK-12
## Symbol Toolbox
if-thentherefore not\begin{align*}& \rightarrow && \text{if-then} && \sim \ \text{not}\\ & \therefore && \text{therefore}\end{align*}
## Keywords and Vocabulary
Inductive Reasoning
• Inductive Reasoning
• Conjecture
• Counterexample
Conditional Statements
• Conditional Statement (If-Then Statement)
• Hypothesis
• Conclusion
• Converse
• Inverse
• Contrapositive
• Biconditional Statement
Deductive Reasoning
• Logic
• Deductive Reasoning
• Law of Detachment
• Law of Contrapositive
• Law of Syllogism
Algebraic & Congruence Properties
• Reflexive Property of Equality
• Symmetric Property of Equality
• Transitive Property of Equality
• Substitution Property of Equality
• Subtraction Property of Equality
• Multiplication Property of Equality
• Division Property of Equality
• Distributive Property
• Reflexive Property of Congruence
• Symmetric Property of Congruence
• Transitive Property of Congruence
Proofs about Angle Pairs & Segments
• Right Angle Theorem
• Same Angle Supplements Theorem
• Same Angle Complements Theorem
## Review
Match the definition or description with the correct word.
1. 5=x\begin{align*}5 = x\end{align*} and y+4=x\begin{align*}y + 4 = x\end{align*}, then 5=y+4\begin{align*}5 = y +4\end{align*} — A. Law of Contrapositive
2. An educated guess — B. Inductive Reasoning
3. 6(2a+1)=12a+12\begin{align*}6(2a + 1) = 12a +12\end{align*} — C. Inverse
4. 2,4,8,16,32,\begin{align*}2, 4, 8, 16, 32, \ldots\end{align*} — D. Transitive Property of Equality
5. AB¯¯¯¯¯¯¯¯CD¯¯¯¯¯¯¯¯\begin{align*}\overline{AB} \cong \overline{CD}\end{align*} and CD¯¯¯¯¯¯¯¯AB¯¯¯¯¯¯¯¯\begin{align*}\overline{CD} \cong \overline{AB}\end{align*} — E. Counterexample
6. pq\begin{align*}\sim p \rightarrow \sim q\end{align*} — F. Conjecture
7. Conclusions drawn from facts. — G. Deductive Reasoning
8. If I study, I will get an “A\begin{align*}A\end{align*}” on the test. I did not get an A\begin{align*}A\end{align*}. Therefore, I didn’t study. — H. Distributive Property
9. A\begin{align*}\angle A\end{align*} and B\begin{align*}\angle B\end{align*} are right angles, therefore AB\begin{align*}\angle A \cong \angle B\end{align*}. — I. Symmetric Property of Congruence
10. 2 disproves the statement: “All prime numbers are odd.” — J. Right Angle Theorem
K. Definition of Right Angles
## Texas Instruments Resources
In the CK-12 Texas Instruments Geometry FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9687.
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# Maximum and minimum values of quadratics
Lesson
We've looked at quadratics and how we can graph them as parabolas. We also looked at groups of different features of these quadratics, including concavity, turning points, intercepts, gradient, as well as their symmetrical shape.
In this chapter, we are going to look more at turning points. A quadratic will have either a minimum or a maximum value at its turning point.
## Maximum and Minimum Values
Parabolas that are concave up have a minimum value. This means their $y$y value will never be less than a certain value.
Similarly, parabolas that are concave down have a maximum value. This means their $y$y value will never be more than a certain value.
Remember!
The minimum or maximum value always occurs at the axis of symmetry.
## Finding the Equation of the Axis of Symmetry
It's easy to see a maximum and minimum values when we look at graphs. However, we can also find these minimum or maximum values algebraically. To do this, we need to find the axis of symmetry
$x=\frac{-b}{2a}$x=b2a
Remember, the general form of a quadratic is $y=ax^2+bx+c$y=ax2+bx+c, so we just need to substitute in the relevant values. For example, let's find the equation of the axis of symmetry for $y=2x^2-4x+7$y=2x24x+7.
In this example, $a=2$a=2 and $b=-4$b=4, so let's substitute them into the formula:
$x$x $=$= $\frac{-\left(-4\right)}{2\times2}$−(−4)2×2 $=$= $\frac{4}{4}$44 $\therefore$∴ $x$x $=$= $1$1
## Finding the Turning Point
1. Determine whether the parabola is concave up or down. This will tell us whether we're finding a minimum or maximum value.
2. Find the $x$x value of the turning point using the axis of symmetry formula.
3. Substitute the value from step 2 into the original equation to find the $y$y value.
#### Examples
##### Question 1
Consider the equation $y=2x^2$y=2x2.
1. Is every value of $y$y positive?
No
A
Yes
B
No
A
Yes
B
2. The graph of $y=2x^2$y=2x2 is given below.
State the axis of symmetry of the parabola.
3. What is the minimum value of $y$y?
$y$y$=$=$\editable{}$
##### Question 2
Consider the curve $y=2\left(x-3\right)^2+3$y=2(x3)2+3.
1. Is the graph of the curve concave up or concave down?
Concave up
A
Concave down
B
Concave up
A
Concave down
B
2. What is the axis of symmetry of the parabola?
3. What is the minimum value of the graph?
$y$y$=$=$\editable{}$
4. At which value of $x$x does the minimum value occur?
$x$x $=$= $\editable{}$
5. If the graph was shifted $4$4 units downwards, what would be the new minimum value?
##### Question 3
Consider the equation $y=\left(x-12\right)\left(x-6\right)$y=(x12)(x6)
1. State the zeros of the function on the same line, separated by a comma.
2. Find the axis of symmetry.
3. The graph of the function is:
Concave up
A
Concave down
B
Concave up
A
Concave down
B
4. The graph has a:
Maximum value
A
Minimum value
B
Maximum value
A
Minimum value
B
5. Determine the minimum $y$y value of the function.
### Outcomes
#### NA6-4
Find optimal solutions, using numerical approaches
#### 91028
Investigate relationships between tables, equations and graphs |
# Math auto solver
In this blog post, we discuss how Math auto solver can help students learn Algebra. Our website can solving math problem.
## The Best Math auto solver
One tool that can be used is Math auto solver. The best step by step calculator is the one that you can use for calculations. There are many online calculators that you can access with just a few clicks. You can also download one for your phone so that you can have it in your hand whenever you need to do any kind of math. This will help you save time and make sure that you get things done as efficiently as possible. It is also a great way to be able to do math whenever and wherever you are. Some people even use it as a way to practice their skills so that they can make sure that they don’t fall behind when they are in school or at work. If you want to find the best step by step calculator, then look no further than this one!
More complicated types of differential equations use more than one variable to describe how one quantity changes with respect to another. Differential equations can be solved using several different methods depending on their specific characteristics. A common approach for solving linear differential equations is through the use of a computer program known as a solver. Solvers are used to find numerical solutions to problems where one quantity must be changed in order for another quantity to change in proportion to it. Solvers are also used to solve different types of differential equations. Linear differential equations are some of the most common types of differential equations because they lend themselves well to mathematical modeling and other applications that require simple, linear relationships between variables.
Partial fraction decomposition (PFD) is a method for solving simultaneous equations. It gives the solution of A * B = C in terms of A and B, and C = A * B. If we have two equations, A * B = C and A + B = C, then PFD gives us an equation of the form (A * B) - (A + B) = 0. The PFD algorithm solves the system by finding a solution to the following equation: A(B - C) = 0 This can be expressed as a simpler equation in terms of partial fractions as: B - C / A(B - C) = 0 This solution is called a "mixed" or "mixed-order" solution. Mixed-order solutions typically have less accuracy than higher-order solutions, but are much faster to compute. The PFD solver computes mixed-order solutions based on an interpolation scheme that interpolates between values of a function at points where it crosses zero. This scheme makes the second derivative zero on these points, and therefore the interpolant will be quadratic on these points. These points are computed iteratively so that they become increasingly accurate while computing time is reduced. Typically, linear systems like this are solved by double-differencing or Taylor's series expansion to approximate the second derivative term at
The disparities between minority groups and the majority is a major problem in the United States. Exact statistics on how many minorities are unemployed and how many people of lower income are living in poverty are hard to track, but it’s clear that there is still much to be done. One way that the inequality gap can be closed is by encouraging more minorities to go into STEM fields. This will not only help them to earn more money, but it will also give them more recognition in the workplace and make it easier for them to get raises and promotions. Another way that inequality can be closed is by improving access to education. If more minorities have access to quality education, they will be less likely to end up stuck in low-paying jobs or trapped in poverty.
Second, you can use a book that has math problems in it as a reference when you are studying. Third, you can practice by doing math problems on your own or with your friends. Fourth, you can also take online math classes if they are available to you. Fifth, you can take practice tests to see where your strengths and weaknesses lie. Sixth, you can talk with a teacher about any questions that you have about the subject matter. Finally, if all else fails, just keep trying!
In one word, it’s amazing. It doesn’t have ads like other annoying apps. And the steps show in the free version is enough to understand. Still, if you need more detailed answer then you should buy the premium membership. I really recommend this app. And also, it’s very fast and simple to use.
Maeve Gonzales
I don't personally use this app because I no longer attend school, however I help my nephew with his homework and I recommended that he tries the app for help. It's a very impressive application that solves math problems and shows the steps as well. So, it's not necessarily cheating if it helps you learn too. In the real world, people use search engines and calculators. I know complex math problems teach critical thinking skills but let's be honest, most people will never use them. It's time to stop teaching kids the "manual" way and teach them to use these apps because this, and some artificial intelligence is the future.
Helen Howard
Point slope form solver Math solver app Answers for math word problems Algebra 2 help online free Help with my math Math answers website |
# Semi-circle function
• Sep 1st 2008, 08:27 AM
sjenkins
Semi-circle function
The graph of y=(3x-x)^.5 is a semi-circle with diameter coincident with the positive x-axis, radius 1.5, center at (1.5,0). create a function passing through (3.5,5) and intersecting the x-axis at x=2 and x=5.
• Sep 1st 2008, 11:38 AM
Soroban
Hello, sjenkins!
Quote:
The graph of y=(3x-x)^.5 is a semi-circle with diameter coincident
with the positive x-axis, radius 1.5, center at (1.5,0).
. . All this is all true . . . but who needs it ??
Create a function passing through (3.5, 5) and intersecting the x-axis at x=2 and x=5.
Make a sketch . . .
Code:
| | (3½,5) | * | | | - - + - * - - - * - - | 2 5
We can see that a parabola will work.
From the two x-intercepts, we have: .$\displaystyle f(x) \:=\:a(x-2)(x-5)$
To pass through $\displaystyle \left(\frac{7}{2},\:5\right)$, we have: .$\displaystyle a\left(\frac{7}{2}-2\right)\left(\frac{7}{2} - 5\right) \:=\:5$
. . $\displaystyle a\left(\frac{3}{2}\right)\left(-\frac{3}{2}\right) \:=\:5 \quad\Rightarrow\quad -\frac{9}{4}\,a \:=\:5 \quad\Rightarrow\quad a \:=\:-\frac{20}{9}$
Therefore: .$\displaystyle f(x) \:=\;-\frac{20}{9}(x-2)(x-5) \quad\Rightarrow\quad\boxed{ f(x) \;=\;-\frac{20}{9}x^2 + \frac{140}{9}x - \frac{200}{9}}$ |
# Graphs and Equations of Circles Book: Math 3 (Green Book) Section: 5.2.
## Presentation on theme: "Graphs and Equations of Circles Book: Math 3 (Green Book) Section: 5.2."— Presentation transcript:
Graphs and Equations of Circles Book: Math 3 (Green Book) Section: 5.2
Standard Equation of a Circle What do you think we need to write an equation of a circle? Radius Center Point on a Circle
Lets See What We Know The point (6, 2) lies on a circle whose center is at the origin. Find the radius. What formula do we need to use in order to find the radius? DISTANCE FORMULA!!!!!!!
Lets See What We Know The point (-2, -2) lies on a circle whose center is at (4, -5). Find the radius.
Lets Make Some Observations Whats the Radius? Where is the center? Here is the equation, how is it related to the circle?
Lets Make Some Observations Whats the Radius? Where is the center? Here is the equation, how is it related to the circle?
So we have determined… Let the center of a circle be the point (h,k) and the radius be r From our observations we have determined that the standard equation of a circle is:
Write the Equation of the Circle
Graph the Circle
Write the Equation of the Circle Center: originRadius: 5 Center: (4, -2)Radius:
Homework GREEN BOOK Pg. 182 #1 – 22 WS
9. 10. 11. From the worksheet
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Understanding proportions is like having a secret formula to solve real-life mathematical problems. Whether you’re scaling a recipe, converting units of measurement, or determining discounts while shopping, proportions play a crucial role. In this blog post, we’ll demystify proportions and equip you with the skills to tackle them with confidence.
## Introduction
Proportions are fundamental to mathematics and have practical applications in various fields. They allow us to compare quantities, make predictions, and solve everyday problems. Let’s dive into the world of proportions and unlock their power.
## Definition of Proportions
Proportions are relationships between two or more ratios. In simple terms, they help us understand how different quantities relate to each other. A proportion is represented by an equation that states that two ratios are equal. For example, 2:4 = 5:10 is a proportion that shows the relationship between two sets of numbers.
## Solving Proportion Problems
To solve proportion problems, we use a simple technique called cross-multiplication. This involves multiplying the numerator of one ratio by the denominator of the other ratio and vice versa. We’ll guide you through step-by-step examples, reinforcing your understanding of this method. With practice, you’ll confidently solve proportion problems in no time.
Example of Solving a Proportion Problem
Imagine you’re following a recipe that calls for 4 cups of flour to make 8 servings, but you need to adjust the recipe to serve 10 people. How much flour will you need? This is a classic proportion problem.
1. Set up the proportion: The original recipe is 4 cups for 8 servings, so our ratios are 4/8 for the original and x/10 for the adjusted recipe, where x is the amount of flour we need for 10 servings.
2. Represent the proportion as an equation: We express this relationship as 4:8 = x:10.
3. Cross-multiply to find x: Multiply the numerator of one fraction by the denominator of the other fraction. Thus, 4 times 10 = 8 times x.
4. Solve for x: 40 = 8x simplifies to x = 5 after dividing both sides by 8.
5. Interpret the result: You will need 5 cups of flour to serve 10 people according to the recipe.
By following this step-by-step approach, you can apply the same method to solve various proportion problems, from scaling recipes to calculating distances or converting units.
## Real-Life Examples of Proportions
Proportions are not just theoretical concepts; they have practical applications in our everyday lives. Let’s explore some real-life examples where proportions come into play:
• Comparing ratios in cooking recipes: Adjusting ingredient quantities based on the desired serving size.
• Scaling a drawing or a map: Enlarging or reducing the size of a drawing or map while maintaining its proportions.
• Converting currencies or units of measurement: Determining equivalent values when converting between different systems.
• Determining discounts or sales tax in shopping: Calculating the final price after applying discounts or including sales tax.
Through these examples, you’ll see how proportions help us navigate various scenarios encountered in daily life.
## Conclusion
Congratulations! You’ve unlocked the power of proportion calculations. Remember, proportions are all around us, playing a crucial role in various aspects of our lives. By grasping the fundamentals, solving real-life problems becomes more manageable. Keep practicing, exploring new scenarios, and applying proportions to diverse situations—it’s a skill that will serve you well beyond the classroom.
Now, armed with your new knowledge, go forth and conquer proportion problems with confidence. Proportions are your secret weapon for solving mathematical puzzles and navigating the world with precision and accuracy. Happy calculating! |
Study S1 Mathematics Maths - Number Patterns - Geniebook
# Number Patterns
In this chapter, we will be discussing the below mentioned topics in detail:
• Common Difference (Direct)
• Common Difference (Indirect)
• General Term
• Find a formula for general term given a number pattern
• Find n given a particular iteration in the pattern
• Determine whether a given iteration is part of a number pattern
## Formula for the General Term, $$T_n$$
For any number sequence where each term differs from the successive term by a constant quantity,
$$T_n = a + d \;(n\; – 1)$$
Where, $$T_n$$ is the $$n^{th}$$ term in the sequence,
$$a$$ is the $$1^{st}$$ term in the sequence, and
$$d$$ is the common difference in quantity between successive terms.
Let’s understand this with the help of some examples:
Example 1:
Consider the following number sequence:
$$1^{st}\;term$$ $$2^{nd}\;term$$ $$3^{rd}\;term$$ $$4^{th}\;term$$ $$5^{th}\;term$$
For this sequence, it starts with $$1$$, then $$+3$$ to each term to get the next term.
\begin{align*} T_n &= 1 + 3 (n\; – 1)\\ &=1 + 3 n\; – 3\\ &=3n-2 \end{align*}
## Number Patterns
A number pattern is a sequence of figures linked by a specific rule.
Example 1:
Consider the following number pattern:
How would the next two figures look like?
Solution:
#### Number Of Lines
$$1$$ $$1$$ $$4$$
$$2$$
$$2$$
$$4 + 3 = 7$$
$$3$$
$$3$$
$$4 + 3 + 3 = 10$$
$$4$$
$$4$$
$$4 + 3 + 3 + 3 = 13$$
$$5$$
$$5$$
$$4 + 3 \;(4) = 16$$
$$6$$
$$6$$
$$4 + 3 \;(5) = 19$$
$$n$$
$$n$$
\begin{align*} 4 + 3(n – 1) &= 4 + 3n – 3\\ &= 3n + 1 \end{align*}
Hence, the number of lines in the $$n^{th}$$ figure, $$L_n = 3n + 1$$
Example 2:
The diagram shows some patterns made from floor tiles.
Find an expression, in terms of $$n$$, for the total number of tiles, $$T_n$$, in Figure $$n$$.
Solution:
#### Number Of Tiles
$$1$$
\begin{align*} 1 && && && && =1 && && = \frac{1\times2}{2} \end{align*}
$$2$$
\begin{align*} 1+2 && && && =3 && = \frac{2\times3}{2} \end{align*}
$$3$$
\begin{align*} 1+2+3 && && =6 && = \frac{3\times4}{2} \end{align*}
$$4$$
\begin{align*} 1+2+3+4 &&=10 && = \frac{4\times5}{2} \end{align*}
In Figure $$n$$
$$T_n = \frac{n \;(n+1)}2$$
Question 1:
The International Space Station (ISS) consists of oval shaped Space Pods and rectangular Solar Panels. The first 3 iterations of the ISS are as shown.
Find a formula, in terms of $$n$$, for
1. the total number of Space Pods, $$A$$, in iteration $$n$$, and
2. the total number of Space Panels, $$n$$, in iteration $$n$$.
Solution:
1. In Iteration 1, there is $$1$$ Space Pod.
In Iteration 2, there are $$2$$ Space Pods and so on.
Hence, in Iteration n, the number of Space Pods would be $$A = n$$
1. In Iteration 1, there are $$4$$ Space Panels; in Iteration 2, there are $$6$$ Space Panels; and in Iteration 3, there are $$8$$ Space Panels and so on.
Hence, in Iteration n, the number of Space Panels would be
\begin{align*} B &= 4 + 2 \;(n \;– 1)\\ &= 4 + 2n \;– 2\\ &= 2n + 2 \end{align*}
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# Adguary Calwile Laura Rogers Autrey~ 2nd Per. 3/14/11
## Presentation on theme: "Adguary Calwile Laura Rogers Autrey~ 2nd Per. 3/14/11"— Presentation transcript:
Adguary Calwile Laura Rogers Autrey~ 2nd Per. 3/14/11
Finding the area under a curve: Riemann, Trapezoidal, and Simpson’s Rule Adguary Calwile Laura Rogers Autrey~ 2nd Per. 3/14/11
Introduction to area under a curve
Before integration was developed, people found the area under curves by dividing the space beneath into rectangles, adding the area, and approximating the answer. As the number of rectangles, n, increases, so does the accuracy of the area approximation.
Introduction to area under a curve (cont.)
There are three methods we can use to find the area under a curve: Riemann sums, the trapezoidal rule, and Simpson’s rule. For each method we must know: f(x)- the function of the curve n- the number of partitions or rectangles (a, b)- the boundaries on the x-axis between which we are finding the area
Riemann Sums There are three types of Riemann Sums Right Riemann: Left Riemann: Midpoint Riemann:
Right Riemann- Overview
Right Riemann places the right point of the rectangles along the curve to find the area. The equation that is used for the RIGHT RIEMANN ALWAYS begins with: And ends with Within the brackets!
Right Riemann- Example
Remember: Right Only Given this problem below, what all do we need to know in order to find the area under the curve using Right Riemann? 4 partitions
Right Riemann- Example
For each method we must know: f(x)- the function of the curve n- the number of partitions or rectangles (a, b)- the boundaries on the x-axis between which we are finding the area
Right Riemann- Example
Right Riemann TRY ME! Volunteer:___________________ 4 Partitions
Did You Get It Right? n=4
Left Riemann- Overview
Left Riemann uses the left corners of rectangles and places them along the curve to find the area. The equation that is used for the LEFT RIEMANN ALWAYS begins with: And ends with Within the brackets!
Left Riemann- Example Remember: Left Only
Given this problem below, what all do we need to know in order to find the area under the curve using Left Riemann? 4 partitions
Left Riemann- Example For each method we must know:
f(x)- the function of the curve n- the number of partitions or rectangles (a, b)- the boundaries on the x-axis between which we are finding the area
Left Riemann- Example
Volunteer:___________
Left Riemann- TRY ME! Volunteer:___________ 3 Partitions
Did You Get My Answer? n=3
Midpoint Riemann- Overview
Midpoint Riemann uses the midpoint of the rectangles and places them along the curve to find the area. The equation that is used for MIDPOINT RIEMANN ALWAYS begins with: And ends with Within the brackets!
Midpoint Riemann- Example
Remember: Midpoint Only Given this problem below, what all do we need to know in order to find the area under the curve using Midpoint Riemann? 4 partitions
Midpoint Riemann- Example
For each method we must know: f(x)- the function of the curve n- the number of partitions or rectangles (a, b)- the boundaries on the x-axis between which we are finding the area
Midpoint Riemann- Example
Midpoint Riemann- TRY ME
Volunteer:_________ 6 partitions
Correct??? n=6
Trapezoidal Rule Overview
Trapezoidal Rule is a little more accurate that Riemann Sums because it uses trapezoids instead of rectangles. You have to know the same 3 things as Riemann but the equation that is used for TRAPEZOIDAL RULE ALWAYS begins with: and ends with Within the brackets with every“ f ” being multiplied by 2 EXCEPT for the first and last terms
Trapezoidal Rule- Example
Remember: Trapezoidal Rule Only Given this problem below, what all do we need to know in order to find the area under the curve using Trapezoidal Rule? 4 partitions
Trapezoidal Example For each method we must know:
f(x)- the function of the curve n- the number of partitions or rectangles (a, b)- the boundaries on the x-axis between which we are finding the area
Trapezoidal Rule- Example
Trapezoidal Rule- TRY Me
Volunteer:_____________ 4 Partitions
Trapezoidal Rule- TRY ME!!
n=4
Simpson’s Rule- Overview
Simpson’s rule is the most accurate method of finding the area under a curve. It is better than the trapezoidal rule because instead of using straight lines to model the curve, it uses parabolic arches to approximate each part of the curve. The equation that is used for Simpson’s Rule ALWAYS begins with: And ends with Within the brackets with every “f” being multiplied by alternating coefficients of 4 and 2 EXCEPT the first and last terms. In Simpson’s Rule, n MUST be even.
Simpson’s Rule- Example
Remember: Simpson’s Rule Only Given this problem below, what all do we need to know in order to find the area under the curve using Simpson’s Rule? 4 Partitions
Simpson’s Example For each method we must know:
f(x)- the function of the curve n- the number of partitions or rectangles (a, b)- the boundaries on the x-axis between which we are finding the area
Simpson’s Rule- Example
Simpson’s Rule TRY ME! Volunteer:____________ 4 partitions |
# Common Core: 1st Grade Math : Understanding Place Value
## Example Questions
### Example Question #101 : Understand That A Two Digit Number Represents Tens And Ones: Ccss.Math.Content.1.Nbt.B.2:
What digit is in the tens place?
Explanation:
The tens place is always the second number from the right.
### Example Question #102 : Understand That A Two Digit Number Represents Tens And Ones: Ccss.Math.Content.1.Nbt.B.2:
What digit is in the tens place?
Explanation:
The tens place is always the second number from the right.
### Example Question #103 : Understand That A Two Digit Number Represents Tens And Ones: Ccss.Math.Content.1.Nbt.B.2:
What digit is in the tens place?
Explanation:
The tens place is always the second number from the right.
### Example Question #104 : Understand That A Two Digit Number Represents Tens And Ones: Ccss.Math.Content.1.Nbt.B.2:
What digit is in the tens place?
Explanation:
The tens place is always the second number from the right.
### Example Question #105 : Understand That A Two Digit Number Represents Tens And Ones: Ccss.Math.Content.1.Nbt.B.2:
What digit is in the tens place?
Explanation:
The tens place is always the second number from the right.
### Example Question #106 : Understand That A Two Digit Number Represents Tens And Ones: Ccss.Math.Content.1.Nbt.B.2:
What digit is in the tens place?
Explanation:
The tens place is always the second number from the right.
### Example Question #107 : Understand That A Two Digit Number Represents Tens And Ones: Ccss.Math.Content.1.Nbt.B.2:
What digit is in the tens place?
Explanation:
The tens place is always the second number from the right.
### Example Question #108 : Understand That A Two Digit Number Represents Tens And Ones: Ccss.Math.Content.1.Nbt.B.2:
What digit is in the ones place?
Explanation:
The ones place is always the first number on the right.
### Example Question #1 : 11 To 19 Are Composed Of A Ten And One, Two, Three, Four, Five, Six, Seven, Eight, Or Nine Ones: Ccss.Math.Content.1.Nbt.B.2.B
What number is ten and one?
Explanation:
Ten and one means the same thing as . When we add .
### Example Question #2 : 11 To 19 Are Composed Of A Ten And One, Two, Three, Four, Five, Six, Seven, Eight, Or Nine Ones: Ccss.Math.Content.1.Nbt.B.2.B
What number is ten and three? |
# And binary values
To understand binary numbers, begin by recalling elementary school math. And binary values we first learned about numbers, we were taught that, in the decimal system, things are organized into columns: H T O 1 9 3 such that "H" is the hundreds column, "T" is the tens column, and "O" is the ones column. So the number "" is 1-hundreds plus 9-tens plus 3-ones. As you know, the decimal system uses the digits to represent numbers. The binary system works under the exact same principles as the decimal system, only it operates in base 2 rather than and binary values In other words, instead of columns being.
Therefore, it would shift you one column to the left. For example, "3" in binary cannot and binary values put into one and binary values. What would the binary number be in decimal notation? Click here to see the answer Try converting these numbers from and binary values to decimal: Since 11 is greater than 10, a one is put into the 10's column carriedand a 1 is recorded in the one's column of the sum. Thus, the answer is Binary addition works on the same principle, but the numerals are different.
Begin with one-bit binary addition:. In binary, any digit higher than 1 puts us a column to the left as would 10 in decimal notation. Record the 0 in the ones column, and carry the 1 to the twos column to get an answer of " The process is the same for multiple-bit binary numbers: Record the 0, carry the 1.
Add 1 from carry: Multiplication in the binary system works the same way as in the decimal system: Follow the same rules as in decimal division. For the sake of simplicity, throw away the remainder. Converting from decimal to binary notation is slightly more difficult conceptually, but can easily be done once you know how through the use of algorithms. Begin by thinking of a few examples. Almost as intuitive is the number 5: Then we just put this into columns.
This process continues until we have a remainder of 0. Let's take a look at how it works. To convert the decimal number and binary values to binary, we would find the largest power of 2 less than 75, which is Subtract 8 from 11 to get 3.
Thus, and binary values number is And binary values this algorithm a bit more formal gives us: Find the largest power of two in D. Let this equal P. Put a 1 in binary column P. Subtract P from D. And binary values zeros in all columns which don't have ones. This algorithm is a bit awkward. Particularly step 3, "filling in the zeros.
Now that we have an algorithm, we can use it to convert numbers from decimal to binary relatively and binary values. Our first step is to find P. Subtracting leaves us with Subtracting 1 from P gives us 4. Next, subtract 16 from 23, to get 7. Subtract 1 from P gives us 3. Subtract 1 from P to get 1. Subtract 1 from P to get 0. Subtract 1 from P to get P is now less than zero, so we stop. Another algorithm for converting decimal to binary However, this is not the only approach possible.
We can start at the right, rather than the left. This gives us the rightmost digit as a starting point. Now we need to do the remaining digits. One idea and binary values to "shift" them. It is also easy to see that multiplying and dividing by 2 shifts everything by one column: Similarly, multiplying by 2 shifts in the other direction: Take the number Dividing by 2 gives Since we divided the number by two, we "took out" one power of two.
Also note that a1 is essentially "remultiplied" by two just by putting it in front of a[0], so it is automatically fit into the correct column. Now we can subtract 1 from 81 to see what remainder we still must place Dividing 80 by 2 gives We can divide by two again to get This is even, so we put a 0 in the 8's column. And binary values we already knew how to convert from binary to decimal, we can easily verify our result.
These and binary values work well for non-negative integers, but how do we indicate negative numbers in the binary system? Before we investigate negative numbers, we note that the computer uses a fixed number of "bits" or binary digits.
An 8-bit and binary values is 8 digits long. For this section, we will work with 8 bits. The simplest way to indicate negation is signed magnitude. To indicatewe would simply put a "1" rather than a "0" as the first bit: In one's complement, positive numbers are represented as usual in regular binary. However, negative numbers are represented differently. To negate a number, replace all zeros with ones, and ones with zeros - flip the bits.
Thus, 12 would beand would be As in signed magnitude, the leftmost bit indicates the sign 1 is negative, 0 is positive. To compute the value of a negative number, flip the bits and translate as and binary values. Begin with and binary values number in one's complement. Add 1 if the number is negative. Twelve would be represented asand as To verify this, let's subtract 1 fromto get And binary values we flip the and binary values, we getor 12 in decimal.
In this notation, "m" indicates the total number of bits. Then convert back to decimal numbers.
The binary number system is an alternative to the decimal base number system that we use every day. Binary numbers are important because using them instead of the decimal system simplifies the design of computers and related technologies.
The simplest definition of the binary number system is a system of numbering that uses only two digits—0 and 1—to represent numbers, instead of using the digits 1 through 9 plus 0 to represent numbers. To translate between decimal numbers and binary numbers, you can use a chart like the one to the left.
Notice how 0 and 1 are the same in either system, but starting at 2, things change. For example, decimal 2 looks like 10 in the binary system. The 0 equals zero as you would expect, but the 1 actually represents 2. In every binary number, the first digit starting from the right side can equal 0 or 1. But if the second digit is 1, then it represents the number 2. If it is 0, then it is just and binary values.
The third digit can equal 4 or 0. The fourth digit can equal 8 or and binary values. If you write down the decimal values of each of the digits and binary values then add them up, you have the decimal value of the binary number.
In the case of binary 11, there is a 1 in the first position, which and binary values 1 and then another 1 in and binary values second position, so and binary values equals 2. As numbers get larger, new digits are added to the left. To determine the value of a digit, count the number of digits to the left of it, and multiply that number times 2.
For example, for the digital numberto determine the value of the 1, count the number and binary values digits to the left of the 1 and multiply that number times 2. The total value of binary is 4, since the numbers to the left of the 1 are both 0s. Now you know how to count digital numbers, but how do you add and subtract them? Binary math is similar to decimal math. Adding binary numbers looks like that in the box to the and binary values above. To add these binary numbers, do this: Start from the right side, just as in ordinary math.
Write a 1 down in the solution area. According to our rule, that equals 0, so write 0 and carry the 1 to the next column. Any time you have a column that adds up to decimal 3, you write down a 1 in the solution area and carry a 1. In the fifth column you have only the 1 that you carried over, so you write down 1 in the fifth column of the solution.
Computers rely on binary numbers and binary math because it greatly simplifies their tasks. Since there are only two possibilities 0 and 1 for each digit and binary values than 10, it is easier to store or manipulate the numbers.
Computers need a large number of transistors to accomplish all this, but and binary values is still easier and less expensive to do things with binary numbers rather than decimal numbers. The original computers were used primarily as calculators, but later they were used to manipulate other forms of information, such as words and pictures. In each case, engineers and programmers sat down and decided how they were going to represent a new type of information in binary form.
The chart shows the most popular way to translate the alphabet into binary numbers only the first six letters are shown. Although it is pretty complicated to do so, sounds and pictures can also be converted into binary numbers, too. The result is a huge array of binary numbers, and the volume of all this data is one reason why image files on a computer are so large, and why it is relatively slow to view video or download audio over an internet connection. Binary Numbers and Binary Math. |
# Pythagorean Theorem
The Pythagorean Theorem is a geometrical expression used often in math and physics. It used to 2 2 2 find the unknown side of a right triangle. The exponential form of this theorem a + b = c . That is the equation you use when you are looking for the unknown side of a right triangle, and it is what Ill demonstrate on the attached exhibit. The upside down capital L in the bottom of the left hand corner indicates that sides A & B are the legs of the triangle. Since we know side A = 5 inches and B = 3 inches we may fill that in to 2 2 2 or equation for step one. 1) 5 + 3 = c What the theorem will help us find is the c side of this triangle. 2. 25 + 9 = c All we do is distribute 5 to the second power and 3 to the second power as seen is step two. Next, we add these two numbers together to get 34, 25+9=34, in step three. 3. 25+9=34 Then, in step four we find the square root of 34. 4. 34 In step five we see that 5. 83 is the unknown side of the right triangle. 5. c= 5. 83 We found this answer by using the Pythagorean Theorem as taught in geometrical form.
This theorem may also be summed up by saying that the area of the square on the hypotenuse, or opposite side of the right angle, of a right triangle is equal to sum of the areas of the squared on the legs. The Pythagorean Theorem was a studied by many people and groups. One of those people being Euclid. Sometimes the Pythagorean Theorem is also referred to as the 47th Problem of Euclid. It is called this because it is included by Euclid in a book of numbered geometric problems. In the problem Euclid studied he would always use 3, 4, and 5 as the sides of the right triangle.
He did this because 5 x 5 = 3 x 3 + 4 x 4. The angle opposite the side of the legs was the right angle, it had a length of 5. The 3:4:5 in the right triangle was known as a Pythagorean triple or a three digits that could be put in a right triangle successfully. These three numbers were also whole numbers and were used in the Egyptian string trick, which I will talk about later. This Pythagorean triple, 3:4:5, are the smallest integer series to have been formed, and the only consecutive numbers in that group that is important. These numbers can be, and often were, studied from a philosophical stand point.
The symbolic meanings of the 3:4:5 triple told by modern writers such as Manly P. Hall say 3 stands for spirit, 4 stands for matter, and 5 stands for man. Using Halls study the symbolism of this arrangement is as follows: Matter (4) lays upon the plane of Earth and Spirit (3) reaches up to the Heaven and they are connected by Man (5) who takes in both qualities. A process similar to that of Euclid’s 47th Problem was the Egyptian string trick. Egyptians were said to have invented the word geometry (geo = earth, metry = measuring. The Egyptians used the 3:4:5 right triangle to create right triangles when measuring there fields after the Nile floods washed out there old boundary markers. The Egyptians used the same theory of Euclid, 5 x 5 = 3 x 3 + 4 x 4, to get there boundaries marked correctly. Although Euclid and the Ancient Egyptians studied the theorem, the true inventor of it ( or the person most people believed invented it first ) was Pythagoras of Samos and his group the Pythagoreans. Pythagoras was a man born in 580 B. C. on the island of Samos, in the Aegean Sea.
It is said Pythagoras was a man that spent his life traveling the world in search of wisdom. This search for wisdom led him to settle in Corona, a Greek colony in southern Italy, in about 530 B. C. Here Pythagoras gained famous status for his group known as the Brotherhood of Pythagoreans. This group devoted there lives to the study of mathematics. The group, as led by Pythagoras, could be described as almost cult-like because that it had symbols, rituals, and prayers. The group was also cult-like because of there odd ways of not writing down any of there discoveries.
It was also said that Pythagoras himself sacrificed a hecatomb, or an ancient Greek ritual of 100 oxen, when he discovered the Pythagorean Theorem. The group was also said to have vowed to secrecy. One day the Pythagoreans discovered irrational numbers. They referred to these numbers as algon or unutterable. They were so shocked by these numbers they killed a member of the group that mentioned them in public. The group believed in many things had to do with numbers. They said all things are numbers, and also numbers rule the universe, Pythagoreans believed that numbers were divine.
He also thought numbers one through ten, those of a decade, were especially sacred. Pythagoreans also thought that numbers had characteristics: 2 was female, 3 was male, odd numbers were good, and even numbers were evil. This belief by the Pythagoreans led to many discoveries including the Pythagorean Theorem. The Pythagoreans first discovered numbers could be associated with shapes. Numbers six, ten, and fifteen were all triangular numbers because they can be arranged in equilateral triangles. This study of numbers and shapes eventually led to the discovery of many different and important theory having to do with geometry.
Although, nobody is really positive who invented the theorem, Pythagoras or the Pythagoreans? This is unknown because of there vow so secrecy and there neglect of writing there discoveries down. So now nobody is even sure if Pythagoras had anything to do with the discovery. The puzzle of who invented the Pythagorean Theorem, Pythagoras or his followers, is so confusing because the group studied such a wide variety of different topics. Studies of the group includes many geometric proofs, astronomy, and music. The Pythagoreans believed that all these things had to do with numbers.
If you would ask my opinion on the theorem, I would have to say the original inventors were the Egyptians. They used the theorem, but the Pythagoreans were the first ones to write about and describe it more thoroughly. I think that the theorem was an important discovery for the future . I say this because the theorem was studied by so many great thinkers. The theorem is complex but simple because it is easy to use with right angles after you learn it, but it also has many philosophical meanings and parts to it. In all I think the Pythagorean Theorem is a confusing, but an important part to the past, present, and future of geometry.
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# Right Triangles Demystified
Last updated on J by Arikaran Kumar
Right Triangles
## What is a Right Triangle?
A right triangle is a special type of triangle in which one of the angles measures When we do have a right triangle, the measure of the remaining two angles that are not the right angle CANNOT be equal to , as the sum of all interior angles of a triangle must be equal to . A angle is usually notated by the symbol. Here are some examples:
Figure #1
Figure #2
Figure #3
In each of these triangles, the angle is located where the symbol is.
BONUS TIP:
In a right triangle, the longest side is known as the hypotenuse. The hypotenuse will ALWAYS be the side opposite of the angle. (Since the largest angle in a Right Triangle is always and the longest side is always opposite the largest interior angle). The two shorter sides are known as legs.
## What is a special right triangle?
Now that we know about right triangles, let’s talk about two special types of right triangles. Each triangle will have either of the following measures for its three angles.
- - OR - -
### 30° – 60° – 90°
In a - - right triangle, one angle measures , one angle measures , and one angle measures . The sides of the triangle opposite to each angle have a ratio of , respectively.
• The side opposite to is equal to
• The side opposite to is equal to
• The side opposite to is equal to
Figure #4
### 45° – 45° – 90°
In a - - right triangle, one angle measures and two angles measure . The sides of the triangle opposite to each angle have a ratio of .
• The sides opposite to each angle is equal to
• The side opposite to is equal to
Figure #5
Using these properties, we can easily solve for the remaining sides of a right triangle if given only one side measurement and knowing that we have a special right triangle.
## Fun facts about right triangles:
• Right triangle has many synonyms. Right-angled triangle, orthogonal triangle, rectangled triangle; the common thread, a triangle having one of the angles measuring .
• The foundations of trigonometry can be traced to the study of the relationship between the sides and the angles, other than the , of a right triangle.
## Examples
Find the missing side,, of each given triangle.
#1
Figure #6
For the problem in Figure #6, we are only given one side of the triangle. To solve for the missing side, , let’s first solve for the remaining angle. We can do that by setting up the following equation:
Where is the value of the missing angle.
Simplifying we get
Solving for we get
Since , we can see that we have a - - triangle. Using this information, we can solve for the missing side .
The side we are given is opposite of . The side opposite of is equal to . Therefore,
Solving for we get
Rationalizing we get
The side is equal to . So,
#2
Figure #7
In Figure #7, we will once again begin with solving for the missing angle. We’ll call it . So,
Solving for we get
Since , we can see that we have a - - triangle. Using this information, we can solve for the missing side .
The side we are given is opposite of . The side opposite of is equal to . Therefore,
Solving for we get
Rationalizing, we get
Simplifying, we get
The side is equal to . So,
Author: Mr. Vernon Sullivan, is a tutor at FPLA, a premier 1-on-1 tutoring center HQ in Miami FL. He teaches Algebra, Geometry, Pre-Cal, ACT, SAT, SSAT, HSPT, PERT, ASVAB and other test prep programs.
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Home MATHEMATICS TOPIC 2: AREA AND PERIMETER ~ MATHEMATICS FORM 4
# TOPIC 2: AREA AND PERIMETER ~ MATHEMATICS FORM 4
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#### AREA AND PERIMETER
Area of any Triangle
The Formula for the Area of any Triangle
Derive the formula for the area of any triangle
Area of triangle is given by½bh, whereby b is the base of the triangle and h is the height of the given triangle. Consider the illustrations below:
From the figure above, we see where the base and height are located.
Applying the Formula to find the Area of any Triangle
Apply the formula to find the area of any triangle
Example 1
The base of a triangle is 12cm long. If the corresponding height is 7cm, find the area of the triangle.
Solution
Consider the figure below:
The area of a triangle is given by½bh.
Area =½×12cm×7cm
Area = 42cm2
Therefore area of a triangle is 42cm2
Example 2
The lengths of two sides of a triangle are 6cm and 8cm. Find the area of a triangle if the included is
Solution:
Consider the triangle below, name it triangle ABC.
The area of a triangle above is given by½b×h
So, Area = ½× 8cm× 6cm × sin 45°
=24cm2× sin45°
= 16.97cm2
Therefore the area of ABC = 16.97cm2
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Previous articleTOPIC 1: COORDINATE GEOMETRY ~ MATHEMATICS FORM 4
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# Boat Trailer Wiring Y Harness
• Y Harness
• Date : October 31, 2020
## Boat Trailer Wiring Y Harness
Trailer Wiring
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Boat Trailer Wiring Y HarnessThe Way to Add Up the Intersection of a Venn Diagram I bet it was not in your mind to ask the question,which statement belongs in the intersection of the Venn diagram? It can be because you understand it's to do with triangles. However, what if it's not triangles that you're interested in? A Venn diagram is a diagram that shows the connections between an infinite number of sets, where one element represents each set. The diagram shows what happens when you take two places and add or remove elements from them. The Venn diagram is used to illustrate what happens when two sets are combined, when a single set is divided and when the same set is multiplied. Let's take a peek at the intersection of a Venn diagram. The junction of a Venn diagram is the set of points that are included between all elements of the sets. Each point is a set element itself. There are five potential intersections - two sets containing exactly two components, two sets containing three elements, three sets comprising four elements, five sets containing five components, and seven sets containing six elements. If you place the two places we've only looked in - two elements - and one pair containing two elements, then the intersection will be exactly 1 point. On the other hand, if you remove the one element and place the empty set rather, the intersection becomes just two things. So, the very first matter to consider is if one set contains the elements of another set. If one set contains the elements of another set, then the group contains exactly 1 element. To be able to determine whether a set includes the elements of another group, look at the intersection of the set and the set which contains the elements of the set you're working to determine. If a single set is divided and another group is multiplied, then the intersection of both sets that are contained between those two sets is obviously one point. The second aspect to consider is if two sets are the same or different. When two sets are exactly the same, they share the same intersection with each other. If two sets are exactly the same, their intersection are also the same. The next thing to consider is whether a single set is odd or even. When two places are , the intersection will be even, and when they're odd, the junction will be odd. Finally, when two places are mixed, then they will be mixed in this manner that their intersection is not unique. When you know the three things, you may easily understand what happens when you add up the intersection of the Venn diagram. You can also see exactly what happens if you eliminate the intersection points and split the set. |
# Lesson 4
Making the Moves
Let’s draw and describe translations, rotations, and reflections.
### 4.1: Reflection Quick Image
Here is an incomplete image. Your teacher will display the completed image twice, for a few seconds each time. Your job is to complete the image on your copy.
### 4.2: Make That Move
Your partner will describe the image of this triangle after a certain transformation. Sketch it here.
### 4.3: A to B to C
Here are some figures on an isometric grid. Explore the transformation tools in the tool bar. (Directions are below the applet if you need them.)
1. Name a transformation that takes Figure A to Figure B. Name a transformation that takes Figure B to Figure C.
2. What is one sequence of transformations that takes Figure A to Figure C? Explain how you know.
Translate
1. Select the Vector tool.
2. Click on the original point and then the new point. You should see a vector.
3. Select the Translate by Vector tool.
4. Click on the figure to translate, and then click on the vector.
Rotate
1. Select the Rotate around Point tool.
2. Click on the figure to rotate, and then click on the center point.
3. A dialog box will open. Type the angle by which to rotate and select the direction of rotation.
Reflect
1. Select the Reflect about Line tool.
2. Click on the figure to reflect, and then click on the line of reflection.
Experiment with some other ways to take Figure $$A$$ to Figure $$C$$. For example, can you do it with. . .
• No rotations?
• No reflections?
• No translations?
### Summary
A move, or combination of moves, is called a transformation. When we do one or more moves in a row, we often call that a sequence of transformations. To distinguish the original figure from its image, points in the image are sometimes labeled with the same letters as the original figure, but with the symbol $$’$$ attached, as in $$A’$$ (pronounced “A prime”).
• A translation can be described by two points. If a translation moves point $$A$$ to point $$A’$$, it moves the entire figure the same distance and direction as the distance and direction from $$A$$ to $$A’$$. The distance and direction of a translation can be shown by an arrow.
For example, here is a translation of quadrilateral $$ABCD$$ that moves $$A$$ to $$A’$$.
• A rotation can be described by an angle and a center. The direction of the angle can be clockwise or counterclockwise.
For example, hexagon $$ABCDEF$$ is rotated $$90^\circ$$ counterclockwise using center $$P$$.
• A reflection can be described by a line of reflection (the “mirror”). Each point is reflected directly across the line so that it is just as far from the mirror line, but is on the opposite side.
For example, pentagon $$ABCDE$$ is reflected across line $$m$$.
### Glossary Entries
• clockwise
Clockwise means to turn in the same direction as the hands of a clock. The top turns to the right. This diagram shows Figure A turned clockwise to make Figure B.
• counterclockwise
Counterclockwise means to turn opposite of the way the hands of a clock turn. The top turns to the left.
This diagram shows Figure A turned counterclockwise to make Figure B.
• image
An image is the result of translations, rotations, and reflections on an object. Every part of the original object moves in the same way to match up with a part of the image.
In this diagram, triangle $$ABC$$ has been translated up and to the right to make triangle $$DEF$$. Triangle $$DEF$$ is the image of the original triangle $$ABC$$.
• reflection
A reflection across a line moves every point on a figure to a point directly on the opposite side of the line. The new point is the same distance from the line as it was in the original figure.
This diagram shows a reflection of A over line $$\ell$$ that makes the mirror image B.
• rotation
A rotation moves every point on a figure around a center by a given angle in a specific direction.
This diagram shows Triangle A rotated around center $$O$$ by 55 degrees clockwise to get Triangle B.
• sequence of transformations
A sequence of transformations is a set of translations, rotations, reflections, and dilations on a figure. The transformations are performed in a given order.
This diagram shows a sequence of transformations to move Figure A to Figure C.
First, A is translated to the right to make B. Next, B is reflected across line $$\ell$$ to make C.
• transformation
A transformation is a translation, rotation, reflection, or dilation, or a combination of these.
• translation
A translation moves every point in a figure a given distance in a given direction.
This diagram shows a translation of Figure A to Figure B using the direction and distance given by the arrow.
• vertex
A vertex is a point where two or more edges meet. When we have more than one vertex, we call them vertices.
The vertices in this polygon are labeled $$A$$, $$B$$, $$C$$, $$D$$, and $$E$$. |
0X00F2 to decimal
0X00F2 is a hexadecimal (hex) number. We can tell it is a hex number because it starts with 0X. Hexadecimal numbers like 0X00F2 are not often used in daily life, but we see them used for certain things such as html colors, shortening binary numbers, computer error codes, and math exercises.
The hexadecimal number system has 16 symbols (base 16) instead of the decimal system which has 10 numbers (base 10). The hex symbols are 0 1 2 3 4 5 6 7 8 9 A B C D E F where A=10, B=11, C=12. D=13, E=14, and F=15.
Like we said above, the 0X indicates that it is a hex number and that is the only purpose of 0X. Thus, to convert a hex number such as 0X00F2 to decimal, we only need to look at the symbols after 0X which are 00F2.
To convert the hex number 0X00F2 to decimal (or any other hexadecimal number for that matter), you follow these steps:
Step 1) Multiply the last digit by 1, Multiply the second to last digit by 16, Multiply the third to the last digit by 16 × 16, Multiply the fourth to the last digit by 16 × 16 × 16, Multiply the fifth to the last digit by 16 × 16 × 16 × 16 and so on until all the digits are used.
Step 2) Add up all the products you got from Step 1 to get the answer to 0X00F2 in decimal.
Here is the math using using the steps above showing you how to convert 0X00F2 to decimal.
2 × 1 = 2
F × 16 = 240
0 × 16 × 16 = 0
0 × 16 × 16 × 16 = 0
2 + 240 + 0 + 0 = 242
That is all there is to it. Here is the answer to 0X00F2 in decimal:
242
Hex to Decimal Converter
Here you can convert another hexadecimal number to decimal. Remember hex numbers include numbers 0 through 9 and letters A through F.
0X
0X00F3 to decimal
Go here for the next hex number on our list that we have converted to decimal. |
• What you will learn in this lesson on Probability?
A bag contains apples and no other fruits. Take out one.
• What is the chance it is an apple?
The probability is 100% i.e., 1. A bag contains 3 apples and 3 oranges.
• What is the chance it is an apple?
50%, i.e. 1/2 as each type of fruit has equal chance or probability of being taken out. A dice having six faces with numbers marked from 1 to 6 is tossed.
• What is the chance that face with number 5 comes up?
1/6, since each numbered-face stands equal chance of coming up
• In the dice above, what is the chance an even number comes up?
Hmm, needs thought. 2, 4 and 6 are the even numbers on a dice. So, the probability an even number will turn up is 3/6 i.e. ½.
Now, face this scud: Take 2 balls out of a bag in which there are 3 red and 4 green balls.
• What is the probability that the two balls are red-colored?
To face this scud, you need to arm yourself with the details of Permutations and Combinations concepts. We will discuss them in light of solving Probability concepts.
## PROBABILITY:
Or, if you wish to capture a terse overview of each Probability Formula, then go through each of the following header-links. You can also click the header-links to take you to the page on the specific Probability formula:
## DEFINITION OF PROBABILITY:
If E denotes any Event that is required to happen, then probability the event E happens, denoted as P(E) is: (Number of outcomes favorable for the Event E)/ (Total number of Outcomes)
Example:
When a dice is tossed, the probability a prime number appears is 3/6 = ½, P (E) +P (Ê) = 1
Where P (E) denotes the probability that an Event E happens and P (Ê) denotes the probability that the Event E does not happen.
Then: P (E) +P (Ê) = 1
Example:
When a single fair coin is tossed we know: P (H) + P (T) = 1
Let P (E) indicate P (H) and therefore P (T) will indicate P (Ê), then P (H) + P (T) = 1
can be also written as P (E) +P (Ê) = 1; P (AT LEAST ONE) = 1– P(NONE)
Example:
When two unbiased coins are tossed at the same time, the probability that at least one heads will appear is =
1 – P (No Heads will appear) = 1 – ¼ = ¾
## INDEPENDENT EVENTS:
Two events A, B are said to be Independent Events if they occur without affecting the probability of occurrence of each other. If two events A, B are independent events then the probability the two events happen at the same time is equal to the multiplication of the respective probability values of the two events, i.e.
P (A and B) = P (A) × P (B)
### Example:
When two coins are tossed, the probability that Heads will appear on the first coin and tails on the second is:
P (H and T) = P (H) × P (T) = (½) × (½) = ¼
MUTUALLY EXCLUSIVE EVENTS: Two events A, B are said to be mutually exclusive or just exclusive, if each one of the two events occurs preventing (excluding) the other from taking place at the same time.
### Example:
On a single fair coin, each one of the two Heads and Tails happens preventing (excluding) the other from taking place at the same time.
## RANDOM EXPERIMENT:
Any activity on doing which the result cannot be certainly stated is called a Random Experiment.
### Examples:
1. Toss an unbiased coin and you can only guess what will happen, not certainly state which of the two Heads, Tails will happen.
Tossing a coin is therefore a random experiment.
2. Toss an unbiased dice, you can only guess which of the numbers from 1 to 6 will show up, but not surely bet on one number.
Tossing a dice is therefore a random experiment
## EXHAUSTIVE EVENTS:
What all numbers can you say may occur on tossing a dice? Numbers from 1 to 6. The outcomes, i.e. numbers from 1 to 6 form the list of all outcomes that are likely to happen, so they are collectively called Exhaustive Events (what all are available to occur). On a coin, Heads and Tails form exhaustive events.
## EQUALLY LIKELY EVENTS:
Toss an unbiased coin. You have no reason to prefer any one of the two outcomes (Heads, Tails) over the other. So, each one has the same chance of taking place. Two or more events which have the same chance or probability of occurrence are called equally likely events.
### Examples:
On tossing an unbiased coin, each of the two outcomes: Heads, Tails have the same chance i.e. probability of occurrence, so they are called equally likely events.
On tossing a fair dice, each of the outcomes, i.e. numbers from 1 to 6 have the same chance i.e. probability of occurrence, so they are called equally likely events.
## SAMPLE SPACE:
All the outcomes that are likely to happen when an experiment is done form the set called Sample Space.
### Example:
The sample space on tossing a coin is {H, T}
The sample space on tossing a dice is {1, 2, 3, 4, 5, 6} |
# Introduction
To calculate the relative error of a solution x to a system of linear equations Mx = b, we must determine a way of measuring the relative error of a matrix and a vector. To do this, we need an equivalent version of the absolute value of a real number. In this topic, we look at various definitions of norms of vectors and extensions of these definitions to matrices.
# Background
Useful background for this topic includes:
# The 2-Norm
Because the Euclidean norm takes the square root (1/2) of the sum of the squares (2), it is also known as 2-norm. In this class, we will always refer to the Euclidean norm by this name. It is also written as ||⋅||2.
# Norms of Matrices
Given a linear circuit, we may define a system of linear equations Gv = i where G is a matrix of conductances, v is an unknown vector of voltages, and i is a vector of currents.
Given a vector of currents, we may determine the size of the error by taking the 2-norm of the error vector, which you have seen in your course on linear algebra. You have probably also seen these three properties of a norm ||⋅||2, as well:
1. ||x||2 ≥ 0 and ||x||2 = 0 if and only if x = 0.
2. ||a x||2 = |a| ||x||2 for any scalar a.
3. ||x + y||2 ≤ ||x||2 + ||y||2
However, as we are dealing with matrices, one further condition we would like to satisfy is that the norm of the identity matrix is 1, that is, |||Id|||2 = 1.
Thus, our next goal is to try to define a technique of defining the error in the matrix G. For this, we need a norm on matrices. We will denote a matrix norm by three bars and therefore, to summarize our usage:
|a|
Absolute value of a scalar a.
||v||2
Norm of a vector v.
|||M|||2
Norm of a matrix M.
The first norm which comes to mind is probably to take the square root of the sum of the squares of all of the entries of the matrix. Unfortunately, this fails our second desirable property of a norm, namely, for an n × n identity matrix Id, it is quite easy to determine that |||Id||| is the square root of n.
# A Better Definition
Given a matrix M, let us define the norm of the matrix to be the maximum amount which it stretches any vector. For example, the vector (1, 0)T which has norm 1, when multiplied by the matrix M
has norm 5, while (0, 1)T multiplied by this matrix has norm 4. In fact, you can show that there exists a unit vector v such that ||Mv||2 ≈ 6.092681.
It can be shown that this definition of a matrix norm satisfies all of the four conditions which we have placed on a norm. We could define the matrix norm as
Unfortunately, this is difficult to calculate, however, it can be shown that this is equivalent to:
The 2-norm of a matrix M is the square root of the maximum eigenvalue of MMT. In Matlab:
>> sqrt( max( eig( M*M' ) ) )
We could solve for this using the technique given previously to find the maximum eigenvalue, however, we should note the following:
• Calculating MMT is an O(n3) operation.
• Calculating Mv is an O(n2) operation.
Therefore, if we calculate M(MTv) instead of calculating (MMT)v, we have an O(n2) operation instead of an O(n3) operation.
# Other Vector Norms
There are two other definitions of norms on vectors:
• ||v||1: the sum of the absolute values of the entries of v
• ||v||: the maximum of the absolute values of the entries of v
It can be shown that these have similar properties of the 2-norm, that is, if a sequence converges in the 2-norm, then it converges in the 1-norm and the ∞-norm, and vice versa.
These are also easier to calculate than the 2-norm: no squaring, no square root.
# The 1- and ∞-Matrix Norms
It can be shown that the 1-norm of a matrix M, if we use the same definition from above, may be given by: the maximum column sum of the absolute values of the entries of the matrix, or using Matlab:
>> max( sum( abs( M ) ) )
It can be shown that the ∞-norm of a matrix M, if we use the same definition from above, may be given by: the maximum row sum of the absolute values of the entries of the matrix, or using Matlab:
>> max( sum( abs( M' ) ) )
# The Use of the Norm
The most powerful tool which the above matrix norm gives us is the property:
where * may be any one of 1, 2, or ∞.
# The 2 Norm
As you saw in linear algebra, the 2 norm (or Euclidean norm) of a vector is the square root of the sum of the squares of the entries of the vector:
or in Matlab:
norm( v, 2 )
sqrt( sum( v.^2 ) )
The 2 norm |||M|||2 of a matrix M is the square root of the maximum eigenvalue of MMT. We could find this maximum eigenvalue using the technique described in the previous topic, and in Matlab, we could calculate:
norm( M, 2 )
sqrt( max( eig( M*M' ) ) ) % all the eigenvalues of M*M' are positive and real if M is real
# The 1 Norm
The 1 norm of a vector is defined as the sum of the absolute values of the entries of v:
or in Matlab:
>> norm( v, 1 );
>> sum( abs( v ) );
The corresponding 1 norm |||M|||1 of a matrix M is defined as the maximum absolute column sum. In Matlab, this is given by:
>> norm( M, 1 )
>> max( sum( abs( M ) ) )
# The ∞ Norm
The ∞ norm of a vector is defined as the maximum absolute entry of v:
or in Matlab:
>> norm( v, Inf )
>> max( abs( v ) )
The corresponding ∞ norm |||M||| of a matrix M is defined as the maximum absolute row sum. In Matlab, this is given by:
>> norm( M, Inf )
>> max( sum( abs( M' ) ) )
# The Condition Number of a Matrix
The condition number of a matrix M is defined as cond(M) = |||M||| |||M-1|||. For each of the three norms, 1, 2, and ∞ there is a corresponding condition number for the matrix.
# Error Analysis
Given a system of linear equations Mx = b, we never have the actual values of the matrix M and we can never store them exactly, even if we could. Therefore, there must always be a relative error for our matrix M, and if we define ΔM to be this error, the ratio |||ΔM|||/|||M||| is a measurement of the relative error the matrix M.
Thus, we may state the relative error of our solution x is proportional to the relative error of M according to the following formula:
This statement holds true regardless of the norm used (1, 2, or ∞), so long as the same norm is used in all cases, including the calculation of the condition number. Usually using the 1 norm is sufficient.
# Example 1
What are the 1 and ∞ norms of the matrix
?
The three absolute column sums are 10, 12, and 10, and thus the 1 norm is |||M|||1 = 12.
The three absolute row sums are 6, 13, and 13, and thus the ∞ norm is |||M||| = 13.
# Example 2
What is the largest 1 or ∞ norm of the L matrix found during the process of the PLU decomposition of an n × n matrix?
As all the entries of L are less than or equal to 1, it follows that the maximum 1 and ∞ norms are n while the minimum 1 and ∞ norms are 1.
# Example 3
Using the 1-norm, what is the maximum error in solving for x of the system Mx = b where M is the same matrix as in Question 1, |||ΔM|||1 = 0.003 and b = (2, -1, 1)T.
Using Matlab, we can solve Mx = b to get that x = (1,1,1)T. Also, we find that the condition number of the matrix M is 22.5, and therefore, we get that ||Δx||1 = 22.5⋅0.003⋅3/12 = 0.016875.
# Question 1
What are the 1, 2, and ∞ norms of the vector v = (1, -2, 3, 4, -5)T?
# Question 2
What are the 1 and ∞ norms of the matrix
# Question 3
Use Matlab to calculate the condition number of the matrix [1 2 3; 4 5 6; 7 8 10] for 1, 2, and ∞ using two techniques each: one using cond and the other using the formulae we have seen. The inverse of a matrix may be found by entering M^-1.
# Applications to Engineering
To be filled in later.
# Matlab
The norm and cond functions find the norm and condition number of a matrix:
>> M = rand( 5, 5 ); % create a random matrix
>> norm( M, 1 )
>> cond( M, 1 )
>> norm( M, 2 )
>> cond( M, 2 )
>> norm( M, Inf )
>> cond( M, Inf )
For very large matrices, this can be very expensive, so Matlab provides two estimators of these numbers, normest and condest:
>> normest( M, 1 )
>> condest( M, 1 )
>> normest( M, 2 )
>> condest( M, 2 )
>> normest( M, Inf )
>> condest( M, Inf )
# Maple
The norm and condition number of a matrix may be found using the Norm and ConditionNumber functions:
M := <<-6, 12, 2.4, 0>|<-1, 2, 10.4, 1>|<3.25, 1, -1.8, 14.8>|<10.25, 0, 2, 1.2>>;
LinearAlgebra[Norm]( M, 1 );
LinearAlgebra[ConditionNumber]( M, 1 );
LinearAlgebra[Norm]( M, 2 );
LinearAlgebra[ConditionNumber]( M, 2 );
LinearAlgebra[Norm]( M, infinity );
LinearAlgebra[ConditionNumber]( M, infinity );
For more help on either of these routines or on the LinearAlgebra package, enter:
?LinearAlgebra
?LinearAlgebra,Norm
?LinearAlgebra,ConditionNumber |
# 85 12.1 Flow Rate and Its Relation to Velocity
### Summary
• Calculate flow rate.
• Define units of volume.
• Describe incompressible fluids.
• Explain the consequences of the equation of continuity.
Flow rate ${Q}$ is defined to be the volume of fluid passing by some location through an area during a period of time, as seen in Figure 1. In symbols, this can be written as
${Q\:=}$ ${\frac{V}{t}},$
where ${V}$ is the volume and ${t}$ is the elapsed time.
The SI unit for flow rate is ${\text{m}^3\text{/s}},$ but a number of other units for ${Q}$ are in common use. For example, the heart of a resting adult pumps blood at a rate of 5.00 liters per minute (L/min). Note that a liter (L) is 1/1000 of a cubic meter or 1000 cubic centimeters ( ${10^{-3}\text{ m}^3}$ or ${10^3\text{ cm}^3}$ ). In this text we shall use whatever metric units are most convenient for a given situation.
### Example 1: Calculating Volume from Flow Rate: The Heart Pumps a Lot of Blood in a Lifetime
How many cubic meters of blood does the heart pump in a 75-year lifetime, assuming the average flow rate is 5.00 L/min?
Strategy
Time and flow rate ${Q}$ are given, and so the volume ${V}$ can be calculated from the definition of flow rate.
Solution
Solving ${Q=V/t}$ for volume gives
${V=Qt.}$
Substituting known values yields
$\begin{array}{lcl} {V} & {=} & {(\frac{5.00\text{ L}}{1\text{ min}})(75\text{ y})(\frac{1\text{ m}^3}{10^3\text{L}})(5.26\times10^5\frac{\text{min}}{\text{y}})} \\ {} & {=} & {2.0\times10^5\text{ m}^3.} \end{array}$
Discussion
This amount is about 200,000 tons of blood. For comparison, this value is equivalent to about 200 times the volume of water contained in a 6-lane 50-m lap pool.
Flow rate and velocity are related, but quite different, physical quantities. To make the distinction clear, think about the flow rate of a river. The greater the velocity of the water, the greater the flow rate of the river. But flow rate also depends on the size of the river. A rapid mountain stream carries far less water than the Amazon River in Brazil, for example. The precise relationship between flow rate ${Q}$ and velocity ${\bar{v}}$ is
${Q=A\bar{v}},$
where ${A}$ is the cross-sectional area and ${\bar{v}}$ is the average velocity. This equation seems logical enough. The relationship tells us that flow rate is directly proportional to both the magnitude of the average velocity (hereafter referred to as the speed) and the size of a river, pipe, or other conduit. The larger the conduit, the greater its cross-sectional area. Figure 1 illustrates how this relationship is obtained. The shaded cylinder has a volume
${V=Ad},$
which flows past the point $\text{P}$ in a time ${t}.$ Dividing both sides of this relationship by ${t}$ gives
${\frac{V}{t}}$ ${=}$ ${\frac{Ad}{t}}.$
We note that ${Q=V/t}$ and the average speed is ${v\bar{v}=d/t}.$ Thus the equation becomes ${Q=A\bar{v}}.$
Figure 2 shows an incompressible fluid flowing along a pipe of decreasing radius. Because the fluid is incompressible, the same amount of fluid must flow past any point in the tube in a given time to ensure continuity of flow. In this case, because the cross-sectional area of the pipe decreases, the velocity must necessarily increase. This logic can be extended to say that the flow rate must be the same at all points along the pipe. In particular, for points 1 and 2,
$\begin{array}{c} {Q_1=Q_2} \\ {A_1\bar{v}_1=A_2\bar{v}_2.} \end{array}$ $\rbrace$
This is called the equation of continuity and is valid for any incompressible fluid. The consequences of the equation of continuity can be observed when water flows from a hose into a narrow spray nozzle: it emerges with a large speed—that is the purpose of the nozzle. Conversely, when a river empties into one end of a reservoir, the water slows considerably, perhaps picking up speed again when it leaves the other end of the reservoir. In other words, speed increases when cross-sectional area decreases, and speed decreases when cross-sectional area increases.
Since liquids are essentially incompressible, the equation of continuity is valid for all liquids. However, gases are compressible, and so the equation must be applied with caution to gases if they are subjected to compression or expansion.
### Example 2: Calculating Fluid Speed: Speed Increases When a Tube Narrows
A nozzle with a radius of 0.250 cm is attached to a garden hose with a radius of 0.900 cm. The flow rate through hose and nozzle is 0.500 L/s. Calculate the speed of the water (a) in the hose and (b) in the nozzle.
Strategy
We can use the relationship between flow rate and speed to find both velocities. We will use the subscript 1 for the hose and 2 for the nozzle.
Solution for (a)
First, we solve ${Q=A\bar{v}}$ for ${\bar{v}_1}$ and note that the cross-sectional area is ${A=\pi{r}^2},$ yielding
${\bar{v}_1\:=}$ ${\frac{Q}{A_1}}$ ${=}$ ${\frac{Q}{\pi{r}_1^2}}.$
Substituting known values and making appropriate unit conversions yields
${\bar{v}_1\:=}$ ${\frac{(0.500\text{ L/s})(10^{-3}\text{ m}^3\textbf{/L})}{\pi(9.00\times10^{-3}\text{m})^2}}$ ${=\:1.96\text{ m/s}}.$
Solution for (b)
We could repeat this calculation to find the speed in the nozzle ${\bar{v}_2},$ but we will use the equation of continuity to give a somewhat different insight. Using the equation which states
${A_1\bar{v}_1=A_2\bar{v}_2},$
solving for ${\bar{v}_2}$ and substituting ${\pi{r}^2}$ for the cross-sectional area yields
${\bar{v}_2\:=}$ ${\frac{A_1}{A_2}}$ ${\bar{v}_1\:=}$ ${\frac{\pi{r}_1^2}{\pi{r}_2^2}}$ ${\bar{v}_1\:=}$ ${\frac{r_1^2}{r_2^2}}$ ${\bar{v}_1}.$
Substituting known values,
${\bar{v}_2\:=}$ ${\frac{(0.900\text{ cm})^2}{(0.250\text{ cm})^2}}$ ${1.96\text{ m/s}=25.5\text{ m/s}}.$
Discussion
A speed of 1.96 m/s is about right for water emerging from a nozzleless hose. The nozzle produces a considerably faster stream merely by constricting the flow to a narrower tube.
The solution to the last part of the example shows that speed is inversely proportional to the square of the radius of the tube, making for large effects when radius varies. We can blow out a candle at quite a distance, for example, by pursing our lips, whereas blowing on a candle with our mouth wide open is quite ineffective.
In many situations, including in the cardiovascular system, branching of the flow occurs. The blood is pumped from the heart into arteries that subdivide into smaller arteries (arterioles) which branch into very fine vessels called capillaries. In this situation, continuity of flow is maintained but it is the sum of the flow rates in each of the branches in any portion along the tube that is maintained. The equation of continuity in a more general form becomes
${n_1A_1\bar{v}_1=n_2A_2\bar{v}_2},$
where ${n_1}$ and ${n_2}$ are the number of branches in each of the sections along the tube.
### Example 3: Calculating Flow Speed and Vessel Diameter: Branching in the Cardiovascular System
The aorta is the principal blood vessel through which blood leaves the heart in order to circulate around the body. (a) Calculate the average speed of the blood in the aorta if the flow rate is 5.0 L/min. The aorta has a radius of 10 mm. (b) Blood also flows through smaller blood vessels known as capillaries. When the rate of blood flow in the aorta is 5.0 L/min, the speed of blood in the capillaries is about 0.33 mm/s. Given that the average diameter of a capillary is ${8.0\:\mu},$ calculate the number of capillaries in the blood circulatory system.
Strategy
We can use ${Q=A\bar{v}}$ to calculate the speed of flow in the aorta and then use the general form of the equation of continuity to calculate the number of capillaries as all of the other variables are known.
Solution for (a)
The flow rate is given by ${Q=A\bar{v}}$ or ${\bar{v}=\frac{Q}{\pi{r}^2}}$ for a cylindrical vessel.
Substituting the known values (converted to units of meters and seconds) gives
${\bar{v}\:=}$ ${\frac{(5.0\text{ L/min})(10^{-3}\text{ m}^3\textbf{/L})(1\text{ min/}60\text{ s})}{\pi(0.010\text{ m})^2}}$ ${=\:0.27\text{ m/s.}}$
Solution for (b)
Using ${n_1A_1\bar{v}_1=n_2A_2\bar{v}_1},$ assigning the subscript 1 to the aorta and 2 to the capillaries, and solving for ${n_2}$ (the number of capillaries) gives ${n_2=\frac{n_1A_1\bar{v}_1}{A_2\bar{v}_2}}.$ Converting all quantities to units of meters and seconds and substituting into the equation above gives
${n_2\:=}$ ${\frac{(1)(\pi)(10\times10^{-3}\text{ m})^2(0.27\text{ m/s})}{(\pi)(4.0\times10^{-6}\text{ m})^2(0.33\times10^{-3}\text{ m/s})}}$ ${=\:5.0\times10^9\text{ capillaries}}.$
Discussion
Note that the speed of flow in the capillaries is considerably reduced relative to the speed in the aorta due to the significant increase in the total cross-sectional area at the capillaries. This low speed is to allow sufficient time for effective exchange to occur although it is equally important for the flow not to become stationary in order to avoid the possibility of clotting. Does this large number of capillaries in the body seem reasonable? In active muscle, one finds about 200 capillaries per ${\text{mm}^3},$ or about ${200\times10^6}$ per 1 kg of muscle. For 20 kg of muscle, this amounts to about ${4\times10^9}$ capillaries.
# Section Summary
• Flow rate ${Q}$ is defined to be the volume ${V}$ flowing past a point in time ${t},$ or ${Q=\frac{V}{t}}$ where ${V}$ is volume and ${t}$ is time.
• The SI unit of volume is ${\text{m}^3}.$
• Another common unit is the liter (L), which is ${10^{-3}\text{ m}^3}.$
• Flow rate and velocity are related by ${Q=A\bar{v}}$ where ${A}$ is the cross-sectional area of the flow and ${\bar{v}}$ is its average velocity.
• For incompressible fluids, flow rate at various points is constant. That is,
$\begin{array}{c} {Q_1=Q_2} \\ {A_1\bar{v}_1=A_2\bar{v}_2} \\ {n_1A_1\bar{v}_1=n_2A_2\bar{v}_2.} \end{array}$ $\rbrace$
### Conceptual Questions
1: What is the difference between flow rate and fluid velocity? How are they related?
2: Many figures in the text show streamlines. Explain why fluid velocity is greatest where streamlines are closest together. (Hint: Consider the relationship between fluid velocity and the cross-sectional area through which it flows.)
3: Identify some substances that are incompressible and some that are not.
### Problems & Exercises
1: What is the average flow rate in ${\text{cm}^3\text{/s}}$ of gasoline to the engine of a car traveling at 100 km/h if it averages 10.0 km/L?
2: The heart of a resting adult pumps blood at a rate of 5.00 L/min. (a) Convert this to ${\text{cm}^3\text{/s}}.$ (b) What is this rate in ${\text{m}^3\text{/s}}?$
3: Blood is pumped from the heart at a rate of 5.0 L/min into the aorta (of radius 1.0 cm). Determine the speed of blood through the aorta.
4: Blood is flowing through an artery of radius 2 mm at a rate of 40 cm/s. Determine the flow rate and the volume that passes through the artery in a period of 30 s.
5: The Huka Falls on the Waikato River is one of New Zealand’s most visited natural tourist attractions (see Figure 3). On average the river has a flow rate of about 300,000 L/s. At the gorge, the river narrows to 20 m wide and averages 20 m deep. (a) What is the average speed of the river in the gorge? (b) What is the average speed of the water in the river downstream of the falls when it widens to 60 m and its depth increases to an average of 40 m?
6: A major artery with a cross-sectional area of ${1.00\text{ cm}^2}$ branches into 18 smaller arteries, each with an average cross-sectional area of ${0.400\text{ cm}^2}.$ By what factor is the average velocity of the blood reduced when it passes into these branches?
7: (a) As blood passes through the capillary bed in an organ, the capillaries join to form venules (small veins). If the blood speed increases by a factor of 4.00 and the total cross-sectional area of the venules is ${10.0\text{ cm}^2},$ what is the total cross-sectional area of the capillaries feeding these venules? (b) How many capillaries are involved if their average diameter is ${10.0\:\mu\text{m}}?$
8: The human circulation system has approximately ${1\times10^9}$ capillary vessels. Each vessel has a diameter of about ${8\:\mu\text{m}}.$ Assuming cardiac output is 5 L/min, determine the average velocity of blood flow through each capillary vessel.
9: (a) Estimate the time it would take to fill a private swimming pool with a capacity of 80,000 L using a garden hose delivering 60 L/min. (b) How long would it take to fill if you could divert a moderate size river, flowing at ${5000\text{ m}^3\text{/s}},$ into it?
10: The flow rate of blood through a ${2.00\times10^{-6}\text{ -m}}$ -radius capillary is ${3.80\times10^9\text{ cm}^3\text{/s}}.$ (a) What is the speed of the blood flow? (This small speed allows time for diffusion of materials to and from the blood.) (b) Assuming all the blood in the body passes through capillaries, how many of them must there be to carry a total flow of ${90.0\text{ cm}^3\text{/s}}?$ (The large number obtained is an overestimate, but it is still reasonable.)
11: (a) What is the fluid speed in a fire hose with a 9.00-cm diameter carrying 80.0 L of water per second? (b) What is the flow rate in cubic meters per second? (c) Would your answers be different if salt water replaced the fresh water in the fire hose?
12: The main uptake air duct of a forced air gas heater is 0.300 m in diameter. What is the average speed of air in the duct if it carries a volume equal to that of the house’s interior every 15 min? The inside volume of the house is equivalent to a rectangular solid 13.0 m wide by 20.0 m long by 2.75 m high.
13: Water is moving at a velocity of 2.00 m/s through a hose with an internal diameter of 1.60 cm. (a) What is the flow rate in liters per second? (b) The fluid velocity in this hose’s nozzle is 15.0 m/s. What is the nozzle’s inside diameter?
14: Prove that the speed of an incompressible fluid through a constriction, such as in a Venturi tube, increases by a factor equal to the square of the factor by which the diameter decreases. (The converse applies for flow out of a constriction into a larger-diameter region.)
15: Water emerges straight down from a faucet with a 1.80-cm diameter at a speed of 0.500 m/s. (Because of the construction of the faucet, there is no variation in speed across the stream.) (a) What is the flow rate in ${\text{ cm}^3\text{/s}}?$ (b) What is the diameter of the stream 0.200 m below the faucet? Neglect any effects due to surface tension.
16: Unreasonable Results
A mountain stream is 10.0 m wide and averages 2.00 m in depth. During the spring runoff, the flow in the stream reaches ${100,000\text{ m}^3\text{/s}}.$ (a) What is the average velocity of the stream under these conditions? (b) What is unreasonable about this velocity? (c) What is unreasonable or inconsistent about the premises?
## Glossary
flow rate
abbreviated Q, it is the volume V that flows past a particular point during a time t, or Q = V/t
liter
a unit of volume, equal to 10−3 m3
### Solutions
Problems & Exercises
1:
${2.78\text{ cm}^3\text{/s}}$
3:
27 cm/s
5:
(a) 0.75 m/s
(b) 0.13 m/s
7:
(a) ${40.0\text{ cm}^2}$
(b) ${5.09\times10^7}$
9:
(a) 22 h
(b) 0.016 s
11:
(a) 12.6 m/s
(b) ${0.0800\text{ m}^3\text{/s}}$
(c) No, independent of density.
13:
(a) 0.402 L/s
(b) 0.584 cm
15:
(a) ${127\text{ cm}^3\text{/s}}$
(b) 0.890 cm |
# Compound Interest Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Maths Solutions
Compound Interest Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Foundation Maths Solutions. In this article you will learn how to solve Multiple Choice Questions on Compound Interest very easily. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics.
## Compound Interest Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Maths Solutions
Board ICSE Publications Goyal brothers Prakshan Subject Maths Class 9th Chapter-2 Compound Interest Writer RS Aggrawal Book Name Foundation Topics Solution of MCQs Academic Session 2024-2025
### How to Solve MCQs on Compound Interest Easily
To solve mcqs on Compound Interest use formula or without formula and calculate unknown value Also focus on growth and deprecitaion formula in some cases. At last check the option available which is most suitable after calculation.
#### Multiple Choice Questions
Compound Interest Class 9 RS Aggarwal Goyal Brothers ICSE Foundation Maths Solutions.
Page- 41,42
##### Que-1: The compound interest on Rs 3750 for 2 years at 8% p.a., compounded annually is :
(a) Rs604 (b) Rs614 (c) Rs624 (d) Rs642
Solution- (c) Rs624
Reason: A = P * (1+R/100)^n
A = 3750 * (1+8/100)^2
CI = A- Principal = 624Rs.
##### Que-2: A man invests Rs46875 at 4% p.a. compound interest for 3 years. The interest for the first year will be :
(a) Rs1785 (b) Rs1587 (c) Rs1875 (d) Rs1758
Solution- (c) Rs1875
Reason: It is given that
Principal= 46875
Rate of interest = 4% p.a.
(i) Interest for the first year= Prt/100
Substituting the values
=(46875×4×1)/100
= 1875.
##### Que-3: A man deposits Rs10000 in a cooperative bank for 3 years at 9% p.a. If interest is compounded annually, then the amount he will get from the bank after 3 years is :
(a) Rs12950.29 (b) Rs12905.29 (c) Rs12059.29 (d) Rs12095.29
Solution- (a) Rs12950.29
Reason: Principal amount (P) is Rs.10,000
rate (r) is 9% or 0.09
years (t) is 3
A = 10000(1 + 0.09/1)^(1*3)
= 10000(1 + 0.09)^3
= 10000(1.09)^3
≈ 10000(1.295029)
≈ Rs.12,950.29.
##### Que-4: Rs16000 is deposited in a bank for three years. The rate of compound interest for first and second year are 8% and 12% respectively. At the end of third year the amount becomes Rs21384. The rate of interest for third year will be :
(a) 7% (b) 10% (c) 11% (d) 12%
Solution- (b) 10%
Reason: Initial deposit (P) = Rs16,000
Interest rate for the first year = 8%
Interest rate for the second year = 12%
Final amount after three years = Rs. 21,384
A1 = P(1+r1/100)
where r1 is the interest rate for the first year.
𝐴1 = 16000(1+8/100)
𝐴1 = 16000(1+0.08)
𝐴1 = 16000×1.08
A1 = 17280
A2 = A1(1+r2/100)
where r2 is the interest rate for the second year.
𝐴2 = 17280(1+12/100)
𝐴2 = 17280(1+0.12)
𝐴2 = 17280×1.12
𝐴2 = 19353.60
A3 = A2(1+r3/100)
We know that A3 = 21384.
So, 21384 = 19353.60(1+𝑟3/100)
To find 1+(𝑟3/100):
1+(𝑟3/100) = 21384/19353.60
1+(𝑟3/100) = 1.105
Solving for r3:
𝑟3/100 = 1.105 − 1
𝑟3/100 = 0.105
𝑟3 = 0.105 × 100
𝑟3 = 10%.
##### Que-5: A man borrows Rs5000 at 12% compound interest p.a., interest payable every six months. He pays back Rs1800 at the end of every six months. The third payment he has to make at the end of 18 months in order to clear the entire loan will be :
(a) Rs2024.60 (b) Rs2204.60 (c) Rs2240.60 (d) Rs2402.60
Solution- (a) Rs2024.60
Reason: For first year,
P = 5000
R =12%
T = 1/2 year
I = PTR/100
I = 5000×12×1/100×1/2
A = P + I
A = 5000+ 300
A = 5300
Man repaid = 5300 – 1800
= 3500
For next six month,
P = 3500
R = 12 %
T = 1/2
I = PTR/100
= 3500×12×1/100×1/2
= 210
Now we know that
A = P + I
= 3500 + 210
= 3710
Man repaid = 3710- 1800
= 1910
For last six months ,
P = 1910
R = 12
T = 1/2
I = PTR/100
= 114.6
A = P + I
= 1910+ 114.6
= 2024.6
##### Que-6: Thecompound interest for the second year on Rs8000 invested for 3 years at 10% p.a. is :
(a) Rs780 (b) Rs880 (c) Rs980 (d) Rs1080
Solution- (b) Rs880
Reason:For the first year
P = Rs8,000
N = 1year
R = 10 %
We have S.I. = PNR/100 = (8,000×1×10)/100 = Rs800
And Amount at the end of first year P+S.I. = Rs8,000 + Rs800 = Rs8,800
Now, for the second year
P = Rs8,800
N = 1year
R = 10 %
We have S.I. = PNR/100 = (8,800×1×10)/100 = Rs880
Thus, Compound interest for the second year = Rs880.
##### Que-7: A person took a loan of Rs6000 from a bank and agreed to pay back the amount along with interest in 2 years. If the rate of compound interest for the first year is 10% and second year is 12%, the amount he had to pay after 2 years will be :
(a) Rs 7329 (b) Rs 7932 (c) Rs 7292 (d) Rs 7392
Solution- (d) Rs7392
Reason: Principal amount (P) = Rs. 6000
Interest rate for the first year (r1) = 10%
𝐴1 = 6000(1+10/100)
𝐴1 = 6000(1+0.10)
𝐴1 = 6000×1.10 = 6600
Amount at the end of the first year (A1) = Rs. 6600
Interest rate for the second year (r2) = 12%
𝐴2 = 6600(1+12/100)
𝐴2 = 6600(1+0.12)
𝐴2 = 6600×1.12
𝐴2 = 7392.
##### Que-8: Nikita invests Rs 6000 for 2 years at a certain rate of interest compounded annually. At the end of the first year, it amounts to Rs6720. The rate of interest p.a. is :
(a) 8% (b) 10% (c) 12% (d) 14%
Solution- (c) 12%
Reason: P = Rs. 6000, Amount at the end of the first year = Rs. 6720
S.I. for first year = Rs. (6720 – 6000) = Rs. 720
Let r% be the rate of interest p.a.
720 = (6000×𝑟×1)/100
r = 720/60
r = 12%.
##### Que-9: The compound interest on Rs8640 for 3 years at 8% p.a. is :
(a) Rs2345 (b) Rs3245 (c) Rs3425 (d) Rs3452
Solution- (a) Rs2345
Reason:
##### Que-10: If the interest is compounded half-yearly, then, C.I. when the principal is Rs7400, the rate of interest is 5% p.a. and the duration is one year, is :
(a) Rs373.63 (b) Rs374.63 (c) Rs373.36 (d) Rs373
Solution- (b) Rs374.63
Reason:It is given that
Principal (P) = 7400
Rate of interest (r) = 5%
Period (n) = 1 year
We know that
A = P(1+r/(2×100))^2×n
Substituting the values
= 7400(1+5/200)^2
By further calculation
= 7400 × 205/200 × 205/200
= 7774.63
C.I. = A – P
C.I. = 7774.63 – 7400
C.I. = 374.63.
##### Que-11: The simple interest on a sum of money for 2 years at 4% per annum is Rs340. The compound interest on this sum for one year payable half-yearly at the same rate is :
(a) Rs170.70 (b) Rs107.70 (c) Rs171.70 (d) Rs270.70
Solution- (c) Rs171.70
Reason: Given : I = Rs340, T = 2 Years and R = 4%
∴P = (I×100)/(R×T)
= (Rs340×100)/(4×2) = Rs.4250
C.I. = P(1+r/(2×100))^n×2 − P
= Rs.4250(1+4/(2×100))^1×2 − Rs.4250
= Rs.4421.70 − Rs.4250
= Rs.171.70
##### Que-12: The compound interest on a certain sum of money at 5% p.a. for two years is Rs246. The simple interest on the same sum for three years at 6% p.a. will be :
(a) Rs432 (b) Rs430.50 (c) Rs432.75 (d) Rs431.75
Solution- (a) Rs432
Reason: CI = Rs246, R = 5%, T = 2 years
CI = A – P
246 = P[1+5/100]^2 – P
246 = P|(21/20)^2 – 1|
246 = P[41/400]
P = (246×400)/41
= Rs2400
Now, P = Rs2400, R = 6%, T = 3 years
SI = (2400×6×3)/100
= Rs432.
##### Que-13: Ramesh wants to get Rs6050 from a bank after 2 years. If the bank gives 10% p.a. compound interest, then the amount of money he has to keep now in the bank is :
(a) Rs5500 (b) Rs5000 (c) Rs5600 (d) Rs5800
Solution- (b) Rs5000
Reason: Amount = Rs. 6,050
Time = 2 years
Rate of interest = 10% p.a.
Let assume that Principal = x
So,
[(100 + Rate of interest)/100]Time = (Amount/Principal)
⇒ [(100 + 10)/100]2 = (6050/ x)
⇒ (110/100) × (110/100) = (6050/ x)
⇒ (121/100) = (6050/x)
By cross multiplying
⇒ x = 6050 × (100/121)
⇒ x = 5,000
##### Que-14: The difference between the compound interest and simple interest on a certain sum deposited for 2 years at 5% p.a. is Rs12. The sum will be :
(a) Rs4500 (b) Rs4600 (c) Rs4800 (d) Rs5000
Solution- (c) Rs4800
Reason:N = 2years
R = 5 %
We have S.I. = (PNR)/100 = (P×2×5)/100 = 10P/100 = 0.1P
And on interest being compounded for 2 years and R = 5 %,
Amount = P(1+R/100)^N
= P(1+5/100)²
= P×(1.05)²
= 1.1025P
So, C.I. = A−P = 1.1025P − P = 0.1025P
Given, C.I. − S.I = Rs12
= 0.1025P−0.1P = Rs12
= 0.0025P = Rs12
P = Rs4,800.
##### Que-15: At what rate of compound interest p.a. will Rs20000 amount to Rs26620 in 3 years ?
(a) 4% (b) 6% (c) 8% (d) 10%
Solution- (d) 10%
Reason: Amount = Principal (1 + r/100)^n
= 26620 = 20000 (1+r/100)³
( 1+ r%)³ = 26620/20000
= ( 1 +r%)³ = 1331 / 1000
= ( 1+ r %) = ³√ 1331 / 1000
= ( 1+ r%) = 11/10
= ( 1 + r/100 ) = 11/10
(100+r)/100 = 11 /10
= (100 + r) = (11×100)/10
= 100 + r = 110
= rate = 110 – 100
= 10%
##### Que-16: In what time will Rs5000 amount to Rs5832 at 8% rate of compound interest p.a. ?
(a) 2 years (b) 4 years (c) 6 years (d) 8 years
Solution- (a) 2 years
Reason: Principal = Rs. 5,000
Amount = Rs. 5,832
Rate = 8% p.a.
Amount = Principal × (1 + 𝑅/100)^n
⇒ 5832 = 5000 × (1 + 8/100)^n
⇒ 5832/5000 = (108/100)^n
⇒ (729/625) = (27/25)^n
⇒ (27/25)² = (27/25)^n
⇒ Time = 2 years
##### Que-17: A machine depreciates at the rate of 10% of its value at the beginning of the year. If the present value of the machine is Rs8000, its value after three years will be :
(a) Rs5382 (b) Rs5832 (c) Rs5238 (d) Rs5638
Solution- (b) Rs5832
Reason: machine value = ₹8000
machine depreciation every year 10%
first year -> 10% of 8000 = 800
value is = 8000 – 800 = ₹7200
second year again 10% decrease
value is = 10% of 7200 = 720
so 7200 – 720 = 6480
final third year also 10% decrease
value is = 10% of 6480 = 648
final value is = 6480 – 648 = ₹5832
##### Que-18: The present population of a town is 200000. The population will increase by 10% in the first year and 15% in the second year. The population of the town after two years will be :
(a) Rs253000 (b) Rs235000 (c) Rs203500 (d) Rs352000
Solution- (a) Rs253000
Reason: Population after n years = Present population x (1+𝑟/100)^𝑛
Present population = 2,00,000
After first year, population = = 2,00,000 x (1+10/100)^1
= 2,00,000 × 11/10
= 2,20,000
Population after two years = 2,20,000 ×(1+15/100)^1
= 253000
Thus, the population after two years is 2,53,000.
##### Que-19: A machine depreciates at the rate of 12% of its value at the beginning of a year. The machine was purchased for Rs10000 and is sold for Rs7744. The number of years, that the machine was used is :
(a) 2 (b) 4 (c) 6 (d) 8
Solution- (a) 2
Reason: V = 7744
P = 10000
r = 12%
n = ?
V = P(1-r/100)^n
7744 = 10000(1-12/100)^n
7744/10000 = (100-12/100)^n
7744/10000 = (88/100)^n
7744/10000 = 0.88^n
0.7744 = 0.88^n
0.2552^2 = 0.127^n {approx}
n = 2 years.
##### Que-20: The value of a machine depreciates every year at a constant rate. If the values of the machine in 2006 and 2008 are Rs 25000 and 19360 respectively, then the annual rate of depreciation is :
(a) 8% (b) 10% (c) 12% (d) 14%
Solution- (c) 12%
Reason: P = 25000 in 2006
V = 19360 in 2008
n = 2008 – 2006 = 2 years
V = P(1+r/100)^n
19360 = 25000(1+r/100)^2
19360/25000 = (1+r/100)^2
0.7744 = (1+r/100)^2
√0.7744 = 1+r/100
0.88 = 1+r/100
1 – 0.88 = r/100
0.12 = r/100
r = 1.2 x 100
r = 12%.
– : End of Compound Interest Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Maths Solutions : —
Thanks |
# Horizontal and vertical
The usage of the inter-related terms horizontal and vertical varies with context, There are important symmetries and asymmetries between the two terms which change as one goes from two to three dimensions, from a flat earth scenario to the spherical earth one.
## In two dimensions
In two dimensions 1. The vertical direction is designated. 2. The horizontal is normal to the vertical. Through any point P, there is exactly one vertical and exactly one horizontal. Alternatively, one can start by designating the horizontal direction.
In the context of a two-dimensional orthogonal Cartesian coordinate system on a Euclidean plane, to say that a line is horizontal or vertical, an initial designation has to be made. One can start off by designating the vertical direction, usually labelled the Y direction.[1] The horizontal direction, usually labelled the X direction,[2] is then automatically determined. Or, one can do it the other way around, i.e., nominate the x-axis, in which case the y-axis is then automatically determined. There is no special reason to choose the horizontal over the vertical as the initial designation: the two directions are on par in this respect.
The following hold in the two-dimensional case:
a) Through any point P in the plane, there is one and only one vertical line within the plane and one and only one horizontal line within the plane. This symmetry breaks down as one moves to the three-dimensional case.
b) A vertical line is any line parallel to the vertical direction. A horizontal line is any line normal to a vertical line.
c) Horizontal lines do not cross each other.
d) Vertical lines do not cross each other.
Not all of these elementary geometric facts are true in the 3-D context.
## In three dimensions
In the three-dimensional case, the situation is more complicated as now one has horizontal and vertical planes in addition to horizontal and vertical lines. Consider a point P and designate a direction through P as vertical. A plane which contains P and is normal to the designated direction is the horizontal plane at P. Any plane going through P, normal to the horizontal plane is a vertical plane at P. Through any point P, there is one and only one horizontal plane but a multiplicity of vertical planes. This is a new feature that emerges in three dimensions. The symmetry that exists in the two-dimensional case no longer holds.
## Observations
### The plumb line and spirit level
Spirit level bubble on a marble shelf tests for horizontality
In physics, in engineering, and in construction, the direction designated as vertical is usually that along which a plumb-bob hangs. Alternatively, a spirit level that exploits the buoyancy of an air bubble and its tendency to go vertically upwards may be used to test for horizontality. Modern rotary laser levels that can level themselves automatically and are robust sophisticated instruments work on the same fundamental principle.[3][4]
### The flat earth approximation
In the flat earth scenario,[5] where the earth is notionally a large (infinite) flat surface with gravitational field at a right angle to the surface, the earth surface is horizontal and any plane parallel to the earth surface is also horizontal. Vertical planes, e.g., walls, may be parallel to each other or they may intersect at a vertical line. Horizontal surfaces do not intersect. Furthermore, a plane cannot both be a horizontal plane at one place and a vertical plane somewhere else.
### The spherical Earth
Strictly, vertical walls are never parallel on the surface of a spherical planet
When the curvature of the earth is taken into account, the concepts of vertical and horizontal take on yet another meaning. On the surface of a smoothly spherical, homogenous, non-rotating planet, the plumb bob picks out as vertical the radial direction. Strictly speaking, it is now no longer possible for vertical walls to be parallel: all verticals intersect. This fact has real practical applications in construction and civil engineering, e.g., the tops of the towers of a suspension bridge are further apart than at the bottom. [6]
On a spherical planet, horizontal planes intersect. In the example shown, the blue line represents the tangent plane at the North pole, the red the tangent plane at an equatorial point. The two intersect at a right angle.
Also, horizontal planes can intersect when they are tangent planes to separated points on the surface of the earth. In particular, a plane tangent to a point on the equator intersects the plane tangent to the North Pole at a right angle. (See diagram). Furthermore, the equatorial plane is parallel to the tangent plane at the North Pole and as such has claim to be a horizontal plane. But it is. at the same time, a vertical plane for points on the equator. In this sense, a plane can, arguably, be both horizontal and vertical, horizontal at one place, and vertical at another.
### Further complications
For a spinning earth, the plumb line deviates from the radial direction as a function of latitude.[7] Only at the North and South Poles does the plumb line align with the local radius. The situation is actually even more complicated because earth is not a homomogenous smooth sphere. It is a non homogenous, non spherical, knobbly planet in motion, and the vertical not only need not lie along a radial, it may even be curved and be varying with time. On a smaller scale, a mountain to one side may deflect the plumb bob away from the true zenith.[8]
On a larger scale the gravitational field of the earth, which is at least approximately radial near the earth, is not radial when it is affected by the moon at higher altitudes.[9][10]
### Walls and floors
Brick wall showing horizontal mortar courses and vertical perpendiculars. By contrast, a level floor only contains horizontal lines.
On a (horizontal) floor, one can draw a horizontal line but not a vertical line in the sense of a plumb bob line. But on a (vertical) wall, one can draw both vertical and horizontal lines. In this sense, a vertical wall allows more options. This is reflected in the tools a bricklayer uses: a plumb line for verticality and a spirit level to check that the mortar courses are horizontal. On the other hand, in contrast to a wall a horizontal floor allows more options when one considers compass directions. One can draw on a floor lines going north, south, east, and west, in fact, along any compass direction. A wall allows fewer options. For instance, on a wall which runs along a longitude, an insect cannot crawl east.
### In the classroom
The y-axis on the wall is vertical but the one on the table is horizontal
In the 2-dimension case, as mentioned already, the usual designation of the vertical coincides with the y-axis in co-ordinate geometry. This convention can cause confusion in the classroom. For the teacher, writing perhaps on a white board, the y-axis really is vertical in the sense of the plumbline verticality but for the student the axis may well lie on a horizontal table.
### Independence of horizontal and vertical motions
Neglecting the curvature of the earth, horizontal and vertical motions of a projectile moving under gravity are independent of each other.[11] Vertical displacement of a projectile is not affected by the horizontal component of the launch velocity, and, conversely, the horizontal displacement is unaffected by the vertical component. The notion dates at least as far back as Galileo.[12]
When the curvature of the earth is taken into account, the independence of the two motion does not hold. For example, even a projectile fired in a horizontal direction (i.e., with a zero vertical component) may leave the surface of the spherical earth and indeed escape altogether.[13]
## References and notes
1. For an example of identification of the y-axis with the vertical, and the x-axis with the horizontal, see G.S.Rehill's Interactive Maths Series, "Building a Strong foundation in Mathematics", accessible on http://www.mathsteacher.com.au/year8/ch15_graphs/05_hor/ver.htm.
2. For a definition of "Horizontal axis" see Math Dictionary at http://www.icoachmath.com/math_dictionary/Horizontal_Axis.html
3. See Laser Levels
4. For how a spirit level works, see http://www.physicsforums.com/showthread.php?t=562730
5. See Theory and Problems of Theoretical Mechanics" by Murray R Spiegel, 1987, pg 62
6. Encyclopedia.com. In very long bridges, it may be necessary to take the earth's curvature into account when designing the towers. For example, in the New York's Verrazano Narrows Bridge, the towers, which are 700 ft (215 m) tall and stand 4260 ft (298 m) apart, are about 1.75 in (4.5 cm) farther apart at the top than they are at the bottom.
7. "Working in the Rotating Reference Frame of the Earth" (PDF).<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
8. Such a deflection was measured by Nevil Maskelyne. See Maskelyne, N. (1775). "An Account of Observations Made on the Mountain Schiehallion for Finding Its Attraction". Phil. Trans. Royal Soc. 65 (0): 500–542. doi:10.1098/rstl.1775.0050. Charles Hutton used the observed value to determine the density of the earth.
9. Cornish, Neil J. "The Lagrangian Points" (PDF). Montana State University – Department of Physics. Retrieved 29 July 2011.<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
10. For an example of curved field lines, see The gravitational field of a cube by James M. Chappell, Mark J. Chappell, Azhar Iqbal, Derek Abbott for an example of curved gravitational field. arXiv:1206.3857 [physics.class-ph] (or arXiv:1206.3857v1 [physics.class-ph] for this version)
11. Salters Hornerns Advanced Physics Project, As Student Book, Edexcel Pearson, London, 2008, p. 48.
12. See Galilleo's discussion of how bodies rise and fall under gravity on a moving ship in his Dialogue Concerning the Two Chief World Systems(trans. S. Drake). University of California Press, Berkeley, 1967, pp. 186–187.
13. See Harris Benson University Physics, New York 1991, page 268.
### General references
• Brennan, David A.; Esplen, Matthew F.; Gray, Jeremy J. (1998), Geometry, Cambridge: Cambridge University Press, ISBN 0-521-59787-0
• Murray R Spiegel, (1987), Theory and Problems of Theoretical Mechanics,Singapore, Mcgraw Hill's: Schaum's, ISBN 0-07-084357-0 |
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# 5.2: The Matrix of a Linear Transformation I
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Learning Objectives
1. Find the matrix of a linear transformation with respect to the standard basis.
2. Determine the action of a linear transformation on a vector in $$\mathbb{R}^n$$.
In the above examples, the action of the linear transformations was to multiply by a matrix. It turns out that this is always the case for linear transformations. If $$T$$ is any linear transformation which maps $$\mathbb{R}^{n}$$ to $$\mathbb{R}^{m},$$ there is always an $$m\times n$$ matrix $$A$$ with the property that $T\left(\vec{x}\right) = A\vec{x} \label{matrixoftransf}$ for all $$\vec{x} \in \mathbb{R}^{n}$$.
Theorem $$\PageIndex{1}$$: Matrix of a Linear Transformation
Let $$T:\mathbb{R}^{n}\mapsto \mathbb{R}^{m}$$ be a linear transformation. Then we can find a matrix $$A$$ such that $$T(\vec{x}) = A\vec{x}$$. In this case, we say that $$T$$ is determined or induced by the matrix $$A$$.
Here is why. Suppose $$T:\mathbb{R}^{n}\mapsto \mathbb{R}^{m}$$ is a linear transformation and you want to find the matrix defined by this linear transformation as described in [matrixoftransf]. Note that $\vec{x} =\bigg( \begin{array}{c} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \end{array} \bigg) = x_{1}\bigg( \begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \end{array} \bigg) + x_{2}\bigg( \begin{array}{c} 0 \\ 1 \\ \vdots \\ 0 \end{array} \bigg) +\cdots + x_{n}\bigg( \begin{array}{c} 0 \\ 0 \\ \vdots \\ 1 \end{array} \bigg) = \sum_{i=1}^{n}x_{i}\vec{e}_{i}$ where $$\vec{e}_{i}$$ is the $$i^{th}$$ column of $$I_n$$, that is the $$n \times 1$$ vector which has zeros in every slot but the $$i^{th}$$ and a 1 in this slot.
Then since $$T$$ is linear, \begin{aligned} T\left( \vec{x} \right)&=&\sum_{i=1}^{n}x_{i}T\left( \vec{e}_{i}\right) \\ &=&\bigg( \begin{array}{ccc} | & & | \\ T\left( \vec{e}_{1}\right) & \cdots & T\left( \vec{e}_{n}\right) \\ | & & | \end{array} \bigg) \bigg( \begin{array}{c} x_{1} \\ \vdots \\ x_{n} \end{array} \bigg) \\ &=& A\bigg( \begin{array}{c} x_{1} \\ \vdots \\ x_{n} \end{array} \bigg)\end{aligned} The desired matrix is obtained from constructing the $$i^{th}$$ column as $$T\left( \vec{e}_{i}\right) .$$ Recall that the set $$\left\{ \vec{e}_1, \vec{e}_2, \cdots, \vec{e}_n \right\}$$ is called the standard basis of $$\mathbb{R}^n$$. Therefore the matrix of $$T$$ is found by applying $$T$$ to the standard basis. We state this formally as the following theorem.
Theorem $$\PageIndex{2}$$: Matrix of a Linear Transformation
Let $$T: \mathbb{R}^{n} \mapsto \mathbb{R}^{m}$$ be a linear transformation. Then the matrix $$A$$ satisfying $$T\left(\vec{x}\right)=A\vec{x}$$ is given by $A= \bigg( \begin{array}{ccc} | & & | \\ T\left( \vec{e}_{1}\right) & \cdots & T\left( \vec{e}_{n}\right) \\ | & & | \end{array} \bigg)$ where $$\vec{e}_{i}$$ is the $$i^{th}$$ column of $$I_n$$, and then $$T\left( \vec{e}_{i} \right)$$ is the $$i^{th}$$ column of $$A$$.
The following Corollary is an essential result.
Corollary $$\PageIndex{1}$$: Matrix and Linear Transformation
A transformation $$T:\mathbb{R}^n\rightarrow \mathbb{R}^m$$ is a linear transformation if and only if it is a matrix transformation.
Consider the following example.
Example $$\PageIndex{1}$$: The Matrix of a Linear Transformation
Suppose $$T$$ is a linear transformation, $$T:\mathbb{R}^{3}\rightarrow \mathbb{ R}^{2}$$ where $T\bigg( \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \bigg) =\bigg( \begin{array}{r} 1 \\ 2 \end{array} \bigg) ,\ T\bigg( \begin{array}{r} 0 \\ 1 \\ 0 \end{array} \bigg) =\bigg( \begin{array}{r} 9 \\ -3 \end{array} \bigg) ,\ T\bigg( \begin{array}{r} 0 \\ 0 \\ 1 \end{array} \bigg) =\bigg( \begin{array}{r} 1 \\ 1 \end{array} \bigg)$ Find the matrix $$A$$ of $$T$$ such that $$T \left( \vec{x} \right)=A\vec{x}$$ for all $$\vec{x}$$.
Solution
By Theorem [thm:matrixoflineartransformation] we construct $$A$$ as follows: $A = \bigg( \begin{array}{ccc} | & & | \\ T\left( \vec{e}_{1}\right) & \cdots & T\left( \vec{e}_{n}\right) \\ | & & | \end{array} \bigg)$
In this case, $$A$$ will be a $$2 \times 3$$ matrix, so we need to find $$T \left(\vec{e}_1 \right), T \left(\vec{e}_2 \right),$$ and $$T \left(\vec{e}_3 \right)$$. Luckily, we have been given these values so we can fill in $$A$$ as needed, using these vectors as the columns of $$A$$. Hence, $A=\bigg( \begin{array}{rrr} 1 & 9 & 1 \\ 2 & -3 & 1 \end{array} \bigg)$
In this example, we were given the resulting vectors of $$T \left(\vec{e}_1 \right), T \left(\vec{e}_2 \right),$$ and $$T \left(\vec{e}_3 \right)$$. Constructing the matrix $$A$$ was simple, as we could simply use these vectors as the columns of $$A$$. The next example shows how to find $$A$$ when we are not given the $$T \left(\vec{e}_i \right)$$ so clearly.
Example $$\PageIndex{2}$$: The Matrix of Linear Transformation: Inconveniently
Defined
Suppose $$T$$ is a linear transformation, $$T:\mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$$ and $T\bigg( \begin{array}{r} 1 \\ 1 \end{array} \bigg) =\bigg( \begin{array}{r} 1 \\ 2 \end{array} \bigg) ,\ T\bigg( \begin{array}{r} 0 \\ -1 \end{array} \bigg) =\bigg( \begin{array}{r} 3 \\ 2 \end{array} \bigg)$ Find the matrix $$A$$ of $$T$$ such that $$T \left( \vec{x} \right)=A\vec{x}$$ for all $$\vec{x}$$.
Solution
By Theorem [thm:matrixoflineartransformation] to find this matrix, we need to determine the action of $$T$$ on $$\vec{e}_{1}$$ and $$\vec{e}_{2}$$. In Example [exa:matrixoflineartransformation], we were given these resulting vectors. However, in this example, we have been given $$T$$ of two different vectors. How can we find out the action of $$T$$ on $$\vec{e}_{1}$$ and $$\vec{e}_{2}$$? In particular for $$\vec{e}_{1}$$, suppose there exist $$x$$ and $$y$$ such that $\bigg( \begin{array}{r} 1 \\ 0 \end{array} \bigg) = x\bigg( \begin{array}{r} 1\\ 1 \end{array} \bigg) +y\bigg( \begin{array}{r} 0 \\ -1 \end{array} \bigg) \label{matrixvalues}$
Then, since $$T$$ is linear, $T\bigg( \begin{array}{r} 1 \\ 0 \end{array} \bigg) = x T\bigg( \begin{array}{r} 1 \\ 1 \end{array} \bigg) +y T\bigg( \begin{array}{r} 0 \\ -1 \end{array} \bigg)$
Substituting in values, this sum becomes $T\bigg( \begin{array}{r} 1 \\ 0 \end{array} \bigg) = x\bigg( \begin{array}{r} 1 \\ 2 \end{array} \bigg) +y\bigg( \begin{array}{r} 3 \\ 2 \end{array} \bigg) \label{matrixvalues2}$
Therefore, if we know the values of $$x$$ and $$y$$ which satisfy [matrixvalues], we can substitute these into equation [matrixvalues2]. By doing so, we find $$T\left(\vec{e}_1\right)$$ which is the first column of the matrix $$A$$.
We proceed to find $$x$$ and $$y$$. We do so by solving [matrixvalues], which can be done by solving the system $\begin{array}{c} x = 1 \\ x - y = 0 \end{array}$
We see that $$x=1$$ and $$y=1$$ is the solution to this system. Substituting these values into equation [matrixvalues2], we have $T\bigg( \begin{array}{r} 1 \\ 0 \end{array} \bigg) = 1 \bigg( \begin{array}{r} 1 \\ 2 \end{array} \bigg) + 1 \bigg( \begin{array}{r} 3 \\ 2 \end{array} \bigg) = \bigg( \begin{array}{r} 1 \\ 2 \end{array} \bigg) + \bigg( \begin{array}{r} 3 \\ 2 \end{array} \bigg) = \bigg( \begin{array}{r} 4 \\ 4 \end{array} \bigg)$
Therefore $$\bigg( \begin{array}{r} 4 \\ 4 \end{array} \bigg)$$ is the first column of $$A$$.
Computing the second column is done in the same way, and is left as an exercise.
The resulting matrix $$A$$ is given by $A = \bigg( \begin{array}{rr} 4 & -3 \\ 4 & -2 \end{array} \bigg)$
This example illustrates a very long procedure for finding the matrix of $$A$$. While this method is reliable and will always result in the correct matrix $$A$$, the following procedure provides an alternative method.
Procedure $$\PageIndex{1}$$: Finding the Matrix of Inconveniently Defined Linear Transformation
Suppose $$T:\mathbb{R}^{n}\rightarrow \mathbb{R}^{m}$$ is a linear transformation. Suppose there exist vectors $$\left\{ \vec{a}_{1},\cdots ,\vec{a}_{n}\right\}$$ in $$\mathbb {R}^{n}$$ such that $$\bigg( \begin{array}{ccc} \vec{a}_{1} & \cdots & \vec{a}_{n} \end{array} \bigg) ^{-1}$$ exists, and $T \left(\vec{a}_{i}\right)=\vec{b}_{i}$ Then the matrix of $$T$$ must be of the form $\bigg( \begin{array}{ccc} \vec{b}_{1} & \cdots & \vec{b}_{n} \end{array} \bigg) \bigg( \begin{array}{ccc} \vec{a}_{1} & \cdots & \vec{a}_{n} \end{array} \bigg) ^{-1}$
We will illustrate this procedure in the following example. You may also find it useful to work through Example [exa:2x2inconvenientmatrixoflintransf] using this procedure.
Example $$\PageIndex{3}$$: Matrix of a Linear Transformation
Given
Inconveniently
Suppose $$T:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$$ is a linear transformation and $T\bigg( \begin{array}{r} 1 \\ 3 \\ 1 \end{array} \bigg) =\bigg( \begin{array}{r} 0 \\ 1 \\ 1 \end{array} \bigg) ,T\bigg( \begin{array}{r} 0 \\ 1 \\ 1 \end{array} \bigg) =\bigg( \begin{array}{r} 2 \\ 1 \\ 3 \end{array} \bigg) ,T\bigg( \begin{array}{r} 1 \\ 1 \\ 0 \end{array} \bigg) =\bigg( \begin{array}{r} 0 \\ 0 \\ 1 \end{array} \bigg)$ Find the matrix of this linear transformation.
Solution
By Procedure [proc:findingmatrixoflineartransformation], $$A= \bigg( \begin{array}{rrr} 1 & 0 & 1 \\ 3 & 1 & 1 \\ 1 & 1 & 0 \end{array} \bigg) ^{-1}$$ and $$B=\bigg( \begin{array}{rrr} 0 & 2 & 0 \\ 1 & 1 & 0 \\ 1 & 3 & 1 \end{array} \bigg)$$
Then, Procedure [proc:findingmatrixoflineartransformation] claims that the matrix of $$T$$ is $C= BA^{-1} =\bigg( \begin{array}{rrr} 2 & -2 & 4 \\ 0 & 0 & 1 \\ 4 & -3 & 6 \end{array} \bigg)$
Indeed you can first verify that $$T(\vec{x})=C\vec{x}$$ for the 3 vectors above:
$\bigg( \begin{array}{ccc} 2 & -2 & 4 \\ 0 & 0 & 1 \\ 4 & -3 & 6 \end{array} \bigg) \bigg( \begin{array}{c} 1 \\ 3 \\ 1 \end{array} \bigg) =\bigg( \begin{array}{c} 0 \\ 1 \\ 1 \end{array} \bigg) ,\ \bigg( \begin{array}{ccc} 2 & -2 & 4 \\ 0 & 0 & 1 \\ 4 & -3 & 6 \end{array} \bigg) \bigg( \begin{array}{c} 0 \\ 1 \\ 1 \end{array} \bigg) =\bigg( \begin{array}{c} 2 \\ 1 \\ 3 \end{array} \bigg)$ $\bigg( \begin{array}{ccc} 2 & -2 & 4 \\ 0 & 0 & 1 \\ 4 & -3 & 6 \end{array} \bigg) \bigg( \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \bigg) =\bigg( \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \bigg)$
But more generally $$T(\vec{x})= C\vec{x}$$ for any $$\vec{x}$$. To see this, let $$\vec{y}=A^{-1}\vec{x}$$ and then using linearity of $$T$$: $T(\vec{x})= T(A\vec{y}) = T \left( \sum_i \vec{y}_i\vec{a}_i \right) = \sum \vec{y}_i T(\vec{a}_i) \sum \vec{y}_i \vec{b}_i = B\vec{y} = BA^{-1}\vec{x} = C\vec{x}$
Recall the dot product discussed earlier. Consider the map $$\vec{v}$$$$\mapsto$$ $$\mathrm{proj}_{\vec{u}}\left( \vec{v}\right)$$ which takes a vector a transforms it to its projection onto a given vector $$\vec{u}$$. It turns out that this map is linear, a result which follows from the properties of the dot product. This is shown as follows. \begin{aligned} \mathrm{proj}_{\vec{u}}\left( k \vec{v}+ p \vec{w}\right) &=&\left( \frac{(k \vec{v}+ p \vec{w})\cdot \vec{u}}{ \vec{u}\cdot \vec{u}}\right) \vec{u} \\ &=& k \left( \frac{ \vec{v}\cdot \vec{u}}{\vec{u}\cdot \vec{u}}\right) \vec{u}+p \left( { 0.05in}\frac{\vec{w}\cdot \vec{u}}{\vec{u}\cdot \vec{u}}\right) \vec{u} \\ &=& k \; \mathrm{proj}_{\vec{u}}\left( \vec{v}\right) +p \; \mathrm{proj} _{\vec{u}}\left( \vec{w}\right) \end{aligned}
Consider the following example.
Example $$\PageIndex{4}$$: Matrix of a Projection Map
Let $$\vec{u} = \bigg( \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \bigg)$$ and let $$T$$ be the projection map $$T: \mathbb{R}^3 \mapsto \mathbb{R}^3$$ defined by $T(\vec{v}) = \mathrm{proj}_{\vec{u}}\left( \vec{v}\right)$ for any $$\vec{v} \in \mathbb{R}^3$$.
1. Does this transformation come from multiplication by a matrix?
2. If so, what is the matrix?
Solution
1. First, we have just seen that $$T (\vec{v}) = \mathrm{proj}_{\vec{u}}\left( \vec{v}\right)$$ is linear. Therefore by Theorem [thm:matrixlintransf], we can find a matrix $$A$$ such that $$T(\vec{x}) = A\vec{x}$$.
2. The columns of the matrix for $$T$$ are defined above as $$T(\vec{e}_{i})$$. It follows that $$T(\vec{e}_{i}) = \mathrm{proj} _{\vec{u}}\left( \vec{e}_{i}\right)$$ gives the $$i^{th}$$ column of the desired matrix. Therefore, we need to find $\mathrm{proj}_{\vec{u}}\left( \vec{e}_{i}\right) = \left( \frac{\vec{e}_{i}\cdot \vec{u}}{\vec{u}\cdot \vec{u}}\right) \vec{u}$ For the given vector $$\vec{u}$$, this implies the columns of the desired matrix are $\frac{1}{14}\bigg( \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \bigg) , \frac{2}{14}\bigg( \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \bigg) , \frac{3}{14}\bigg( \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \bigg)$ which you can verify. Hence the matrix of $$T$$ is $\frac{1}{14}\bigg( \begin{array}{rrr} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{array} \bigg)$ |
# Angle between two lines
Finding the angle between two lines using a formula is the goal of this lesson. When two lines intersect in a plane, their intersection forms two pairs of opposite angles called vertical angles.
If the two lines are not perpendicular and have slopes m1 and m2, then you can use the following formula to find the angle between the two lines.
$$tan \ θ = \left\lvert \frac{m_2 - m_1}{1 + m_1m_2} \right\rvert$$
## How to derive the formula to find the measure of the angle between two lines
First, notice that when two lines intersect, one of the two pairs is acute and the other pair is obtuse. The angle between two lines is defined as the smallest of these angles or the acute angle denoted by θ.
We are going to use the inclinations of the two lines to find the angle between the two lines. You may need to review the lesson about inclination of line.
Second, we need to show that θ = θ2 - θ1
Using triangle ABC, we know that the sum of the angles in that tiangle is equal to 180 degrees.
θ + θ1 + x = 180 (equation 1)
Furthermore, x + θ2 = 180 (equation 2) since x and θ2 form a straight line.
Replace 180 with x + θ2 in equation 1.
We get θ + θ1 + x = x + θ2
Subtract x from both sides
θ + θ1 + x - x = x - x + θ2
θ + θ1 = θ2
Subtract θ1 from both sides
θ + θ1 - θ1 = θ2 - θ1
θ = θ2 - θ1
Now, you can use the formula for the tangent of the difference of two angles.
tan θ = tan ( θ2 - θ1)
$$tan \ θ = \frac{tan \ θ_2 - tan \ θ_1}{1 + tan \ θ_1 \ tan \ θ_2 }$$
Third, in the lesson about inclination of a line, we learned that tan θ = m
Thus, tan θ1 = m1 and tan θ2 = m2
After substituting m1 for tan θ1 and m2 for tan θ2 in the equation immediately above, we get:
$$tan \ θ = \left\lvert \frac{m_2 - m_1}{1 + m_1m_2} \right\rvert$$
Notice the use of the absolute value to ensure that tan θ is equal to a positive number.
## Exercises about finding the angle between two lines
1) Find the angle between the following two lines.
Line 1: 3x -2y = 4
Line 2: x + 4y = 1
Solution
Put 3x - 2y = 4 into slope-intercept form so you can clearly identify the slope.
3x - 2y = 4
2y = 3x - 4
y = 3x / 2 - 4/2
y = (3/2)x - 2
Put x + 4y = 1 into slope-intercept form so you can clearly identify the slope.
x + 4y = 1
4y = -x + 1
y = -x/4 + 1/4
y = (-1/4)x + 1/4
The slopes are 3/2 and -1/4 or 1.5 and -0.25. It does not matter which one is m1 or m2. You will get the same answer.
Let m1 = 1.5 and m2 = -0.25
$$tan \ θ = \frac{1.5 - -0.25}{1 + 1.5 \times -0.25}$$
$$tan \ θ = \frac{1.5 + 0.25}{1 + -0.375}$$
$$tan \ θ = \frac{1.75}{0.625} = 2.8$$
θ = tan-1(2.8) = 70.35 degrees
2) Find the angle between the following two lines.
Line 1: x - 4y + 4 = 0
Line 2: 3x - 4y - 8 = 0
Solution
Put x - 4y + 4 = 0 into slope-intercept form so you can clearly identify the slope.
x - 4y + 4 = 0
4y = x + 4
y = x / 4 + 4/4
y = (1/4)x + 1
Put 3x - 4y - 8 = 0 into slope-intercept form so you can clearly identify the slope.
3x - 4y - 8 = 0
4y = 3x - 8
y = 3x/4 - 8/4
y = (3/4)x - 2
The slopes are 3/4 and 1/4 or 0.75 and 0.25. Again, it does not matter which one is m1 or m2. You will get the same answer.
Let m1 = 0.75 and m2 = 0.25
$$tan \ θ = \frac{0.25 - 0.75}{1 + 0.25 \times 0.75}$$
$$tan \ θ = \frac{- 0.50}{1 + 0.1875}$$
$$tan \ θ = \frac{- 0.50}{1.1875} = -0.4211$$
tan θ = |-0.4211| = 0.4211
θ = tan-1(0.4211) =
22.84 degrees
100 Tough Algebra Word Problems.
If you can solve these problems with no help, you must be a genius!
Recommended |
# Hyperbolas
## Collection of points that share a constant difference between the distances between two foci.
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Hyperbolas
Hyperbolas are relations that have asymptotes. When graphing rational functions you often produce a hyperbola. In this concept, hyperbolas will not be oriented in the same way as with rational functions, but the basic shape of a hyperbola will still be there.
Hyperbolas can be oriented so that they open side to side or up and down. One of the most common mistakes that you can make is to forget which way a given hyperbola should open. What are some strategies to help?
### Graphing Hyperbolas
A hyperbola has two foci. For every point on the hyperbola, the difference of the distances to each foci is constant. This is what defines a hyperbola. The graphing form of a hyperbola that opens side to side is:
\begin{align*}\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\end{align*}
A hyperbola that opens up and down is:
\begin{align*}\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\end{align*}
Notice that for hyperbolas, \begin{align*}a\end{align*} goes with the positive term and \begin{align*}b\end{align*} goes with the negative term. It does not matter which constant is larger.
When graphing, the constants \begin{align*}a\end{align*} and \begin{align*}b\end{align*} enable you to draw a rectangle around the center. The transverse axis travels from vertex to vertex and has length \begin{align*}2a\end{align*}. The conjugate axis travels perpendicular to the transverse axis through the center and has length \begin{align*}2b\end{align*}. The foci lie beyond the vertices so the eccentricity, which is measured as \begin{align*}e=\frac{c}{a}\end{align*}, is larger than 1 for all hyperbolas. Hyperbolas also have two directrix lines that are \begin{align*}\frac{a^2}{c}\end{align*} away from the center (not shown on the image).
The focal radius is \begin{align*}a^2+b^2=c^2\end{align*}.
### Examples
#### Example 1
Earlier, you were asked how to determine the direction that a hyperbola opens. The best strategy to remember which direction the hyperbola opens is often the simplest. Consider the hyperbola \begin{align*}x^2-y^2=1\end{align*}. This hyperbola opens side to side because \begin{align*}x\end{align*} can clearly never be equal to zero. This is a basic case that shows that when the negative is with the \begin{align*}y\end{align*} value then the hyperbola opens up side to side.
#### Example 2
Put the following hyperbola into graphing form, list the components, and sketch it.
\begin{align*}9x^2-4y^2+36x-8y-4=0\end{align*}
\begin{align*}9(x^2+4x)-4(y^2+2y)&=4\\ 9(x^2+4x+4)-4(y^2+2y+1)&=4+36-4\\ 9(x+2)^2-4(y+1)^2&=36\\ \frac{(x+2)^2}{4}-\frac{(y+1)^2}{9}&=1\end{align*}
Shape: Hyperbola that opens horizontally.
Center: (-2, -1)
\begin{align*}a=2\end{align*}
\begin{align*}b=3\end{align*}
\begin{align*}c=\sqrt{13}\end{align*}
\begin{align*}e=\frac{c}{a}=\frac{\sqrt{13}}{2}\end{align*}
\begin{align*}d=\frac{a^2}{c}=\frac{4}{\sqrt{13}}\end{align*}
Foci: (-2+\begin{align*}\frac{\sqrt{13}}{2}\end{align*}, -1), ( -2-\begin{align*}\frac{\sqrt{13}}{2}\end{align*} , -1)
Vertices: (-4, -1), (0, -1)
Equations of asymptotes: \begin{align*}\pm \frac{3}{2}(x+2)= (y+1)\end{align*}
Note that it is easiest to write the equations of the asymptotes in point-slope form using the center and the slope.
Equations of directrices: \begin{align*}y=-2 \pm \frac{4}{\sqrt{13}}\end{align*}
#### Example 3
Find the equation of the hyperbola with foci at (-3, 5) and (9, 5) and asymptotes with slopes of \begin{align*}\pm \frac{4}{3}\end{align*}
The center is between the foci at (3, 5). The focal radius is \begin{align*}c=6\end{align*}. The slope of the asymptotes is always the rise over run inside the box. In this case since the hyperbola is horizontal and \begin{align*}a\end{align*} is in the \begin{align*}x\end{align*} direction the slope is \begin{align*}\frac{b}{a}\end{align*}. This makes a system of equations.
\begin{align*}\frac{b}{a}&=\pm \frac{4}{3}\\ a^2+b^2&=6^2\end{align*}
When you solve,you get \begin{align*}a=\sqrt{13}, \ b=\frac{4}{3} \sqrt{13}\end{align*}
\begin{align*}\frac{(x-3)^2}{13}-\frac{(y-5)^2}{\frac{16}{9} \cdot 13}=1\end{align*}
#### Example 4
Find the equation of the conic that has a focus point at (1, 2), a directrix at \begin{align*}x=5\end{align*}, and an eccentricity equal to \begin{align*}\frac{3}{2}\end{align*}. Use the property that the distance from a point on the hyperbola to the focus is equal to the eccentricity times the distance from that same point to the directrix:
\begin{align*}\overline{PF}=e \overline{PD}\end{align*}
This relationship bridges the gap between ellipses which have eccentricity less than one and hyperbolas which have eccentricity greater than one. When eccentricity is equal to one, the shape is a parabola.
\begin{align*}\sqrt{(x-1)^2+(y-2)^2}=\frac{3}{2}\sqrt{(x-5)^2}\end{align*}
Square both sides and rearrange terms so that it is becomes a hyperbola in graphing form.
\begin{align*}x^2-2x+1+(y-2)^2&=\frac{9}{4}(x^2-10x+25)\\ x^2-2x+1-\frac{9}{4}x^2+\frac{90}{4}x-\frac{225}{4}+(y-2)^2&=0\\ -\frac{5}{4}x^2+\frac{92}{4}x+(y-2)^2&=\frac{221}{4}\\ -5x^2+92x+4(y-2)^2&=221\\ -5\left(x^2-\frac{92}{5}x\right)+4(y-2)^2&=221\\\end{align*}
\begin{align*}-5\left(x^2-\frac{92}{5}x+\frac{92^2}{10^2}\right)+4(y-2)^2&=221-\frac{2116}{5}\\ -5\left(x-\frac{92}{10}\right)^2+4(y-2)^2&=-\frac{1011}{5}\\ \left(x-\frac{92}{10}\right)^2-(y-2)^2&=\frac{1011}{100}\\ \frac{\left(x-\frac{92}{10}\right)^2}{\left(\frac{1011}{100}\right)}-\frac{(y-2)^2}{\left(\frac{1011}{100}\right)}&=1 \end{align*}
#### Example 5
Given the following graph, estimate the equation of the conic.
Since exact points are not marked, you will need to estimate the slope of asymptotes to get an approximation for \begin{align*}a\end{align*} and \begin{align*}b\end{align*}. The slope seems to be about \begin{align*}\pm \frac{2}{3}\end{align*}. The center seems to be at (-1, -2). The transverse axis is 6 which means \begin{align*}a=3\end{align*}.
\begin{align*}\frac{(x+1)^2}{9}-\frac{(y+2)^2}{4}=1\end{align*}
### Review
Use the following equation for #1 - #5: \begin{align*}x^2+2x-4y^2-24y-51=0\end{align*}
1. Put the hyperbola into graphing form. Explain how you know it is a hyperbola.
2. Identify whether the hyperbola opens side to side or up and down.
3. Find the location of the vertices.
4. Find the equations of the asymptotes.
5. Sketch the hyperbola.
Use the following equation for #6 - #10: \begin{align*} -9x^2-36x+16y^2-32y-164=0 \end{align*}
6. Put the hyperbola into graphing form. Explain how you know it is a hyperbola.
7. Identify whether the hyperbola opens side to side or up and down.
8. Find the location of the vertices.
9. Find the equations of the asymptotes.
10. Sketch the hyperbola.
Use the following equation for #11 - #15: \begin{align*} x^2-6x-9y^2-54y-81=0\end{align*}
11. Put the hyperbola into graphing form. Explain how you know it is a hyperbola.
12. Identify whether the hyperbola opens side to side or up and down.
13. Find the location of the vertices.
14. Find the equations of the asymptotes.
15. Sketch the hyperbola.
To see the Review answers, open this PDF file and look for section 9.5.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Asymptotes An asymptote is a line on the graph of a function representing a value toward which the function may approach, but does not reach (with certain exceptions).
Conic Conic sections are those curves that can be created by the intersection of a double cone and a plane. They include circles, ellipses, parabolas, and hyperbolas.
conjugate axis For a hyperbola with equation of the form $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ or $\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$, the conjugate axis travels perpendicular to the transverse axis through the center and has length $2b$.
foci The foci of a hyperbola are the two points that define the hyperbola. For every point on the hyperbola, the difference of the distances to each focus is constant.
hyperbola A hyperbola is a conic section formed when the cutting plane intersects both sides of the cone, resulting in two infinite “U”-shaped curves.
hyperbolas A hyperbola is a conic section formed when the cutting plane intersects both sides of the cone, resulting in two infinite “U”-shaped curves.
transverse axis For a hyperbola with equation of the form $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ or $\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$, the transverse axis travels from vertex to vertex and has length $2a$. |
# The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 per m^2. - Mathematics
The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 per m2.
#### Solution
It is given that
Length (l) of room = 5 m
Breadth (b) of room = 4 m
Height (h) of room = 3 m
It can be observed that four walls and the ceiling of the room are to be white-washed. The floor of the room is not to be white-washed.
Area to be white-washed = Area of walls + Area of ceiling of room
= 2lh + 2bh + lb
= [2 × 5 × 3 + 2 × 4 × 3 + 5 × 4] m2
= (30 + 24 + 20) m2
= 74 m2
Cost of white-washing per m2 area = Rs 7.50
Cost of white-washing 74 m2 area = Rs (74 × 7.50)
= Rs 555
Concept: Surface Area of a Cuboid
Is there an error in this question or solution?
#### APPEARS IN
NCERT Class 9 Maths
Chapter 13 Surface Area and Volumes
Exercise 13.1 | Q 2 | Page 213 |
## Engage NY Eureka Math 2nd Grade Module 2 Lesson 2 Answer Key
### Eureka Math Grade 2 Module 2 Lesson 2 Problem Set Answer Key
Find the length of each object using one centimeter cube.
Mark the endpoint of each centimeter cube as you measure.
Question 1.
The picture of the eraser is about __4__ centimeters long.
The picture of the eraser is about 4 centimeters long,
Explanation:
As shown in the picture above the eraser is 4 centimeters long.
Question 2.
The picture of the calculator is about __6__ centimeters long.
The picture of the calculator is about 6 centimeters long,
Explanation:
As shown in the picture the calculator is about 6 centimeters long.
Question 3.
The length of the picture of the envelope is about ___10___ centimeters.
The length of the picture of the envelope is about 10 centimeters,
Explanation:
As shown above the length of the picture of the envelope is
Question 4.
Jayla measured her puppet’s legs to be 23 centimeters long.
The stomach is 7 centimeters long, and the neck and head
together are 10 centimeters long. What is the total length of the puppet?
The total length of the puppet is 40 centimeters,
Explanation:
Given Jayla measured her puppet’s legs to be 23 centimeters long.
The stomach is 7 centimeters long, and the neck and head
together are 10 centimeters long.
Therefore the total length of the puppet is
23 + 7 + 10 = 40 centimeters.
Question 5.
Elijah begins measuring his math book with his centimeter cube.
He marks off where each cube ends. After a few times, he decides
this process is taking too long and starts to guess where the cube
would end and then mark it.
Explain why Elijah’s answer will be incorrect.
________________________________________________________
Explanation:
Elijah’s answer is incorrect, why because not every
space is exactly one centimeter cube. So Elijah guess
work is incorrect.
### Eureka Math Grade 2 Module 2 Lesson 2 Exit Ticket Answer Key
Matt measured his index card using a centimeter cube.
He marked the endpoint of the cube as he measured.
He thinks the index card is 10 centimeters long.
a. Is Matt’s work correct? Explain why or why not.
______Yes Matt’s work is correct,_____
_____________________________________________
_____________________________________________
Yes, Matt’s work is correct,
Explanation:
Given Matt measured his index card using a centimeter cube.
He marked the endpoint of the cube as he measured.
He thinks the index card is 10 centimeters long, Yes Matt
is correct why because if we count he exactly measured 10
centimeters so Matt is correct.
b. If you were Matt’s teacher what would you tell him?
_____________________________________________
_____________________________________________
_____________________________________________
I would tell him his measuring method is correct,
Explanation:
If I were Matt’s teacher I would you tell him that his method
of measuring index card using a centimeter cube and
marking the endpoint of the cube as he measured is correct.
### Eureka Math Grade 2 Module 2 Lesson 2 Homework Answer Key
Use the centimeter square at the bottom of the next page to
measure the length of each object. Mark the endpoint of the
square as you measure.
Question 1.
The picture of the glue is about ___4_____ centimeters long.
The picture of the glue is about __4__ centimeters long,
Explanation:
As shown the picture of the glue is about 4 centimeters long.
Question 2.
The picture of the lollipop is about ___6____ centimeters long.
The picture of the lollipop is about 6 centimeters long,
Explanation:
As shown above the picture of the lollipop is about
5 centimeters long.
Question 3.
The picture of the scissors is about ___5_____ centimeters long.
The picture of the scissors is about ___5_____ centimeters long,
Explanation:
As shown the picture of the scissors is about 5 centimeters long.
Question 4.
Samantha used a centimeter cube and the mark and move
forward strategy to measure these ribbons. Use her work to
a. How long is the red ribbon? _____10_____ centimeters long.
The red ribbon is 10 centimeters long,
Explanation:
Samantha used a centimeter cube and the marker and moved
forward strategy to measure, So the red ribbon is
10 centimeters long.
b. How long is the blue ribbon? _____7______ centimeters long.
Blue ribbon is 7 centimeters long,
Explanation:
Samantha used a centimeter cube and the marker and moved
forward strategy to measure, So the blue ribbon is
7 centimeters long.
c. How long is the yellow ribbon? ____5_______ centimeters long.
The yellow ribbon is 5 centimeters long,
Explanation:
Samantha used a centimeter cube and the marker and moved
forward strategy to measure, So the yellow ribbon is
5 centimeters long.
d. Which ribbon is the longest? Red Blue Yellow
Among the three Red ribbon is the longest,
Explanation:
As per Samantha’s measurements
red ribbon is 10 centimeters long,
blue ribbon is 7 centimeters long and
yellow ribbon is 5 centimeters long, So
among the three Red ribbon is the longest.
e. Which ribbon is the shortest? Red Blue Yellow
Among the three Yellow ribbon is 5 centimeters and shortest,
Explanation:
As per Samantha’s measurements
red ribbon is 10 centimeters long,
blue ribbon is 7 centimeters long and
yellow ribbon is 5 centimeters long, So
among the three Yellow ribbon is the shortest.
f. The total length of the ribbons is __22___ centimeters. |
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# CLASS_11_MATHS_SOLUTIONS_NCERT
## Class XI Chapter 12 –
Class XI Chapter 12 – Introduction to Three Dimensional Geometry Maths ______________________________________________________________________________ 2 2 2 DA 1 2 23 14 925 9 43 Here, AB = CD = 6, BC = AD = 43 Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal. Therefore, ABCD is a parallelogram. Hence, the given points are the vertices of a parallelogram. Question 4: Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, - 1). Solution 4: Let P (x, y, z) be the point that is equidistant from points A(1, 2, 3) and B(3, 2, –1). Accordingly, PA = PB 2 2 PA PB x 1 y 2 z 3 x 3 y 2 z 1 2 2 2 2 2 2 ⇒ x 2 – 2x + 1 + y 2 – 4y + 4 + z 2 – 6z + 9 = x 2 – 6x + 9 + y 2 – 4y + 4 + z 2 + 2z + 1 ⇒ –2x –4y – 6z + 14 = –6x – 4y + 2z + 14 ⇒ – 2x – 6z + 6x – 2z = 0 ⇒ 4x – 8z = 0 ⇒ x – 2z = 0 Thus, the required equation is x – 2z = 0. Question 5: Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (-4, 0, 0) is equal to 10. Solution 5: Let the coordinates of P be (x, y, z). The coordinates of points A and B are (4, 0, 0) and (–4, 0, 0) respectively. It is given that PA + PB = 10. 2 2 2 2 2 2 x 4 y z x 4 y z 10 4 10 4 2 2 2 2 2 2 x y z x y z On squaring both sides, we obtain Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.
Class XI Chapter 12 – Introduction to Three Dimensional Geometry Maths ______________________________________________________________________________ 4 100 20 4 4 2 2 2 2 2 2 2 2 2 x y z x y z x y z x 8x 16 y z 100 20 x 8x16 y z x 8x16 y z 2 2 2 2 2 2 2 2 2 2 2 2 20 x 8x 16 y z 100 16x 2 2 2 5 x 8x 16 y z 25 4x On squaring both sides again, we obtain 25 (x 2 + 8x + 16 + y 2 + z 2 ) = 625 + 16x 2 + 200x ⇒ 25x 2 + 200x + 400 + 25y 2 + 25z 2 = 625 + 16x 2 + 200x ⇒ 9x 2 + 25y 2 + 25z 2 – 225 = 0 Thus, the required equation is 9x 2 + 25y 2 + 25z 2 – 225 = 0. Exercise 12.3 Question 1: Find the coordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) in the ratio (i) 2:3 internally, (ii) 2:3 externally. Solution 1: (i) The coordinates of point R that divides the line segment joining points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m: n are mx2 nx1 my2 ny1 mz2 nz1 , , mn mn mn Let R (x, y, z) be the point that divides the line segment joining points (–2, 3, 5) and (1, –4, 6) internally in the ratio 2:3 21 32 24 33 2635 x , y ,andz= 23 23 23 -4 1 27 i.e., x= , y , and z= 5 5 5 4 1 27 Thus, the coordinates of the required point are - , , 5 5 5 (ii) The coordinates of point R that divides the line segment joining points P (x1, y1, z1) and Q (x2, y2, z2) externally in the ratio m: n are mx2 nx1 my2 ny1 mz2 nz1 , , mn mn mn Let R (x, y, z) be the point that divides the line segment joining points(–2, 3, 5) and (1, –4, Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.
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Bell J.L. A primer of infinitesimal analysis (2ed., CUP, 2008)(ISBN 0521887186)(O)(138s)_MCat_
Stability Of Solitary Waves Of A Generalized Two-Component ...
IIT-JEE 2010 - Career Point
Teaching & Learning Plans - Project Maths
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t - Eötvös Loránd University
103 Trigonometry Problems
ICfJff' - Al Kossow's Bitsavers
The Philosophical magazine; a journal of theoretical ... - Index of
MARYLAND Orbital Maneuvering
THE MATHS TEACHER'S HANDBOOK - Arvind Gupta
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March 2011 - Career Point
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The Cubic Spline - STEM2
XT â MATHS Grade 11 â Equations
SOME ELEMENTARY INEQUALITIES IN GAS DYNAMICS EQUATION
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A Nonlinear Heat Equation with Temperature-Dependent ... - UFRJ
Optimal Bounds on the Kuramoto-Sivashinsky Equation Felix Otto ...
Rudin's Principles of Mathematical Analysis: Solutions to ... - MIT |
# Why is A(A+B) = A [Absorption Law]?
Could you proof it to me that A(A+B) = A?
AA + BA [AA = A]
A + AB
Then what?
• Use a truth table. – Yuval Filmus Jan 17 '17 at 12:00
I presume you are looking for a way to prove the identity using a calculus. So far you have used distributivity an idempotency.
Recall that A = A1 so you get A1+AB and you can use distributivity again, this time in the other direction. Then two obvious steps.
Here is a proof: $$A \stackrel{(1)}= A \cdot 1 \stackrel{(2)}= A \cdot (1+B) \stackrel{(3)}= A \cdot 1 + A \cdot B \stackrel{(4)}= A + A \cdot B \stackrel{(5)}= A \cdot A + A \cdot B \stackrel{(6)}= A \cdot (A+B).$$
Axioms used:
(1),(4) multiplicative identity
(2) absorption
(3),(6) distributivity
(5) idempotence
I always liked the $\min, \max$ definitions of $\cdot$ and $+$, since some courses in boolean algebra just give those laws and ask you to accept them. [Boolean Algebra: Basic Operations]
\begin{align*} x \land y &= x \cdot y = \min(x,y)\\ x \lor y &= x + y = \max(x,y) \end{align*} where $0 \leq x,y \leq 1$.
So $A(A+B)$ becomes: $$\min(A, \max(A,B))$$
• Case 1: If $A \leq B$ then
\begin{align*} \min(A, \color{red}{\max(A,B)}) &= \min(A, \color{red}{B})\\ &= \min(A, B) = A\\ \end{align*}
• Case 2: If $A \geq B$ follows the same way as the first one.
• A(A+B) as The AND operation distributes
• over the OR operation
• for all x, y, z ∈ B
• x • (y + z) = (x • y) + (x • z) so
• A(A+B) = AA+AB Idempotent x • x = x
• = A+AB
• = A(1+B) as 1+B=1 so
• = A(1) as A.1=A so
• = A
• I don't see any significant difference between this answer and Raiyan's existing answer. – John L. Feb 22 '19 at 8:14
• the question was "Could you proof it to me that A(A+B) = A?" not "Could you proof it to me that A=A(A+B) ?" is it right? – Mohamed Fathy Alhedhed Feb 23 '19 at 5:07 |
# Boolean Math (OR Logic) – CISSP Domain 3
Today we are going to take another look at some Boolean mathematics. In particular, we’re going to focus on the OR operational logic, as covered in Domain 3 of the CISSP common body of knowledge (CBK).
As a brief recap, Boolean math is based on a binary system. As such, there are only two values that we can have. The value will either be a 1, or it will be a 0. Where 1 is equal to a True condition and 0 is equal to a False condition.
The OR operation is symbolized by the symbol, and when you see this symbol on a Boolean math problem, then you will need to apply the OR operational logic.
Next, we want to define how the OR logic works. To apply the OR logic, we need to compare between two value and check to see if at least one of them is set to True, or 1. If one (or both) of the values is equal to 1, then the result is also equal to 1.
Below is the truth table for the OR logic. Again, one of the nice things about dealing with binary, is that it’s easy to map out the different combinations of 0 and 1.
Similar to the last Boolean article, there’s a good possibility that the CISSP exam may require you to solve a Boolean math problem. So, we are going to use the example problem below and apply the OR logic to solve the problem.
Example Boolean Problem:
X: 0 1 1 0 0 1 0 1
Y: 1 0 1 0 0 1 1 0
__________________________
X ∨ Y:
Whenever you’re faced with a Boolean problem, the first thing you want to do is verify what type of logic to apply. In the example above, we see that it is using the symbol, so we’re going to use the OR logic here.
Next, we compare the X and Y values in each column. So, starting with the first column we’ll reference the OR logic and ask ourselves, “Is at least one of the values equal to 1?”. If so, then the result will be 1, otherwise, the result will be 0.
Continue to work down each column in the problem and apply the OR operational logic to each pair of X/Y values, and the final answer should look like this:
You can create additional practice problems by making a random set of 0’s and 1’s for X and Y, and then solve them using the OR logic. Hopefully this has been helpful for identifying how the OR logic works or if you’re faced with this type of question on your CISSP exam.
This has been a quick look at the OR Boolean logic. There is an audio/video version of this material here, for anyone who prefers that format. If you’re interested in security fundamentals, we have a Professionally Evil Fundamentals (PEF) channel that covers a variety of technology topics. We also answer general basic questions in our Knowledge Center. Finally, if you’re looking for a penetration test, training for your organization, or just have general security questions please Contact Us.
The post Boolean Math (OR Logic) – CISSP Domain 3 appeared first on Professionally Evil Insights.
*** This is a Security Bloggers Network syndicated blog from Professionally Evil Insights authored by Bill McCauley. Read the original post at: https://secureideas.com/blog/2020/10/boolean-math-or-logic-cissp-domain-3.html |
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