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# Ratio Explained
In mathematics, a ratio shows how many times one number contains another. For example, if there are eight oranges and six lemons in a bowl of fruit, then the ratio of oranges to lemons is eight to six (that is, 8:6, which is equivalent to the ratio 4:3). Similarly, the ratio of lemons to oranges is 6:8 (or 3:4) and the ratio of oranges to the total amount of fruit is 8:14 (or 4:7).
The numbers in a ratio may be quantities of any kind, such as counts of people or objects, or such as measurements of lengths, weights, time, etc. In most contexts, both numbers are restricted to be positive.
A ratio may be specified either by giving both constituting numbers, written as "a to b" or "a:b", or by giving just the value of their quotient [1] [2] Equal quotients correspond to equal ratios.A statement expressing the equality of two ratios is called a proportion.
Consequently, a ratio may be considered as an ordered pair of numbers, a fraction with the first number in the numerator and the second in the denominator, or as the value denoted by this fraction. Ratios of counts, given by (non-zero) natural numbers, are rational numbers, and may sometimes be natural numbers.
A more specific definition adopted in physical sciences (especially in metrology) for ratio is the dimensionless quotient between two physical quantities measured with the same unit.[3] A quotient of two quantities that are measured with units may be called a rate.[4]
## Notation and terminology
The ratio of numbers A and B can be expressed as:[5]
• the ratio of A to B
• A:B
• A is to B (when followed by "as C is to D"; see below)
• a fraction with A as numerator and B as denominator that represents the quotient (i.e., A divided by B, or
\tfrac{A}{B}
). This can be expressed as a simple or a decimal fraction, or as a percentage, etc.[6]
When a ratio is written in the form A:B, the two-dot character is sometimes the colon punctuation mark.[7] In Unicode, this is, although Unicode also provides a dedicated ratio character, .[8]
The numbers A and B are sometimes called terms of the ratio, with A being the antecedent and B being the consequent.[9]
A statement expressing the equality of two ratios A:B and C:D is called a proportion,[10] written as A:B = C:D or A:BC:D. This latter form, when spoken or written in the English language, is often expressed as
(A is to B) as (C is to D).
A, B, C and D are called the terms of the proportion. A and D are called its extremes, and B and C are called its means. The equality of three or more ratios, like A:B = C:D = E:F, is called a continued proportion.[11]
Ratios are sometimes used with three or even more terms, e.g., the proportion for the edge lengths of a "two by four" that is ten inches long is therefore
thickness:width:length=2:4:10;
(unplaned measurements; the first two numbers are reduced slightly when the wood is planed smooth)
a good concrete mix (in volume units) is sometimes quoted as
cement:sand:gravel=1:2:4.
[12]
For a (rather dry) mixture of 4/1 parts in volume of cement to water, it could be said that the ratio of cement to water is 4:1, that there is 4 times as much cement as water, or that there is a quarter (1/4) as much water as cement.
The meaning of such a proportion of ratios with more than two terms is that the ratio of any two terms on the left-hand side is equal to the ratio of the corresponding two terms on the right-hand side.
## History and etymology
It is possible to trace the origin of the word "ratio" to the Ancient Greek Greek, Ancient (to 1453);: λόγος (logos). Early translators rendered this into Latin as Latin: [[wikt:ratio#Latin|ratio]] ("reason"; as in the word "rational"). A more modern interpretation of Euclid's meaning is more akin to computation or reckoning.[13] Medieval writers used the word Latin: proportio ("proportion") to indicate ratio and Latin: proportionalitas ("proportionality") for the equality of ratios.[14]
Euclid collected the results appearing in the Elements from earlier sources. The Pythagoreans developed a theory of ratio and proportion as applied to numbers.[15] The Pythagoreans' conception of number included only what would today be called rational numbers, casting doubt on the validity of the theory in geometry where, as the Pythagoreans also discovered, incommensurable ratios (corresponding to irrational numbers) exist. The discovery of a theory of ratios that does not assume commensurability is probably due to Eudoxus of Cnidus. The exposition of the theory of proportions that appears in Book VII of The Elements reflects the earlier theory of ratios of commensurables.[16]
The existence of multiple theories seems unnecessarily complex since ratios are, to a large extent, identified with quotients and their prospective values. However, this is a comparatively recent development, as can be seen from the fact that modern geometry textbooks still use distinct terminology and notation for ratios and quotients. The reasons for this are twofold: first, there was the previously mentioned reluctance to accept irrational numbers as true numbers, and second, the lack of a widely used symbolism to replace the already established terminology of ratios delayed the full acceptance of fractions as alternative until the 16th century.[17]
### Euclid's definitions
Book V of Euclid's Elements has 18 definitions, all of which relate to ratios.[18] In addition, Euclid uses ideas that were in such common usage that he did not include definitions for them. The first two definitions say that a part of a quantity is another quantity that "measures" it and conversely, a multiple of a quantity is another quantity that it measures. In modern terminology, this means that a multiple of a quantity is that quantity multiplied by an integer greater than one—and a part of a quantity (meaning aliquot part) is a part that, when multiplied by an integer greater than one, gives the quantity.
Euclid does not define the term "measure" as used here, However, one may infer that if a quantity is taken as a unit of measurement, and a second quantity is given as an integral number of these units, then the first quantity measures the second. These definitions are repeated, nearly word for word, as definitions 3 and 5 in book VII.
Definition 3 describes what a ratio is in a general way. It is not rigorous in a mathematical sense and some have ascribed it to Euclid's editors rather than Euclid himself.[19] Euclid defines a ratio as between two quantities of the same type, so by this definition the ratios of two lengths or of two areas are defined, but not the ratio of a length and an area. Definition 4 makes this more rigorous. It states that a ratio of two quantities exists, when there is a multiple of each that exceeds the other. In modern notation, a ratio exists between quantities p and q, if there exist integers m and n such that mp>q and nq>p. This condition is known as the Archimedes property.
Definition 5 is the most complex and difficult. It defines what it means for two ratios to be equal. Today, this can be done by simply stating that ratios are equal when the quotients of the terms are equal, but such a definition would have been meaningless to Euclid. In modern notation, Euclid's definition of equality is that given quantities p, q, r and s, p:qr:s if and only if, for any positive integers m and n, np<mq, np=mq, or np>mq according as nr<ms, nr=ms, or nr>ms, respectively.[20] This definition has affinities with Dedekind cuts as, with n and q both positive, np stands to mq as stands to the rational number (dividing both terms by nq).[21]
Definition 6 says that quantities that have the same ratio are proportional or in proportion. Euclid uses the Greek ἀναλόγον (analogon), this has the same root as λόγος and is related to the English word "analog".
Definition 7 defines what it means for one ratio to be less than or greater than another and is based on the ideas present in definition 5. In modern notation it says that given quantities p, q, r and s, p:q>r:s if there are positive integers m and n so that np>mq and nrms.
As with definition 3, definition 8 is regarded by some as being a later insertion by Euclid's editors. It defines three terms p, q and r to be in proportion when p:qq:r. This is extended to four terms p, q, r and s as p:qq:rr:s, and so on. Sequences that have the property that the ratios of consecutive terms are equal are called geometric progressions. Definitions 9 and 10 apply this, saying that if p, q and r are in proportion then p:r is the duplicate ratio of p:q and if p, q, r and s are in proportion then p:s is the triplicate ratio of p:q.
## Number of terms and use of fractions
In general, a comparison of the quantities of a two-entity ratio can be expressed as a fraction derived from the ratio. For example, in a ratio of 2:3, the amount, size, volume, or quantity of the first entity is
\tfrac{2}{3}
that of the second entity.
If there are 2 oranges and 3 apples, the ratio of oranges to apples is 2:3, and the ratio of oranges to the total number of pieces of fruit is 2:5. These ratios can also be expressed in fraction form: there are 2/3 as many oranges as apples, and 2/5 of the pieces of fruit are oranges. If orange juice concentrate is to be diluted with water in the ratio 1:4, then one part of concentrate is mixed with four parts of water, giving five parts total; the amount of orange juice concentrate is 1/4 the amount of water, while the amount of orange juice concentrate is 1/5 of the total liquid. In both ratios and fractions, it is important to be clear what is being compared to what, and beginners often make mistakes for this reason.
Fractions can also be inferred from ratios with more than two entities; however, a ratio with more than two entities cannot be completely converted into a single fraction, because a fraction can only compare two quantities. A separate fraction can be used to compare the quantities of any two of the entities covered by the ratio: for example, from a ratio of 2:3:7 we can infer that the quantity of the second entity is
\tfrac{3}{7}
that of the third entity.
## Proportions and percentage ratios
If we multiply all quantities involved in a ratio by the same number, the ratio remains valid. For example, a ratio of 3:2 is the same as 12:8. It is usual either to reduce terms to the lowest common denominator, or to express them in parts per hundred (percent).
If a mixture contains substances A, B, C and D in the ratio 5:9:4:2 then there are 5 parts of A for every 9 parts of B, 4 parts of C and 2 parts of D. As 5+9+4+2=20, the total mixture contains 5/20 of A (5 parts out of 20), 9/20 of B, 4/20 of C, and 2/20 of D. If we divide all numbers by the total and multiply by 100, we have converted to percentages: 25% A, 45% B, 20% C, and 10% D (equivalent to writing the ratio as 25:45:20:10).
If the two or more ratio quantities encompass all of the quantities in a particular situation, it is said that "the whole" contains the sum of the parts: for example, a fruit basket containing two apples and three oranges and no other fruit is made up of two parts apples and three parts oranges. In this case,
\tfrac{2}{5}
, or 40% of the whole is apples and
\tfrac{3}{5}
, or 60% of the whole is oranges. This comparison of a specific quantity to "the whole" is called a proportion.
If the ratio consists of only two values, it can be represented as a fraction, in particular as a decimal fraction. For example, older televisions have a 4:3 aspect ratio, which means that the width is 4/3 of the height (this can also be expressed as 1.33:1 or just 1.33 rounded to two decimal places). More recent widescreen TVs have a 16:9 aspect ratio, or 1.78 rounded to two decimal places. One of the popular widescreen movie formats is 2.35:1 or simply 2.35. Representing ratios as decimal fractions simplifies their comparison. When comparing 1.33, 1.78 and 2.35, it is obvious which format offers wider image. Such a comparison works only when values being compared are consistent, like always expressing width in relation to height.
## Reduction
Ratios can be reduced (as fractions are) by dividing each quantity by the common factors of all the quantities. As for fractions, the simplest form is considered that in which the numbers in the ratio are the smallest possible integers.
Thus, the ratio 40:60 is equivalent in meaning to the ratio 2:3, the latter being obtained from the former by dividing both quantities by 20. Mathematically, we write 40:60 = 2:3, or equivalently 40:60∷2:3. The verbal equivalent is "40 is to 60 as 2 is to 3."
A ratio that has integers for both quantities and that cannot be reduced any further (using integers) is said to be in simplest form or lowest terms.
Sometimes it is useful to write a ratio in the form 1:x or x:1, where x is not necessarily an integer, to enable comparisons of different ratios. For example, the ratio 4:5 can be written as 1:1.25 (dividing both sides by 4) Alternatively, it can be written as 0.8:1 (dividing both sides by 5).
Where the context makes the meaning clear, a ratio in this form is sometimes written without the 1 and the ratio symbol (:), though, mathematically, this makes it a factor or multiplier.
## Irrational ratios
Ratios may also be established between incommensurable quantities (quantities whose ratio, as value of a fraction, amounts to an irrational number). The earliest discovered example, found by the Pythagoreans, is the ratio of the length of the diagonal to the length of a side of a square, which is the square root of 2, formally
a:d=1:\sqrt{2}.
Another example is the ratio of a circle's circumference to its diameter, which is called , and is not just an irrational number, but a transcendental number.
Also well known is the golden ratio of two (mostly) lengths and, which is defined by the proportion
a:b=(a+b):a
or, equivalently
a:b=(1+b/a):1.
Taking the ratios as fractions and
a:b
as having the value, yields the equation
x=1+\tfrac1x
or
x2-x-1=0,
which has the positive, irrational solution
x=\tfrac{a}{b}=\tfrac{1+\sqrt{5}}{2}.
Thus at least one of a and b has to be irrational for them to be in the golden ratio. An example of an occurrence of the golden ratio in math is as the limiting value of the ratio of two consecutive Fibonacci numbers: even though all these ratios are ratios of two integers and hence are rational, the limit of the sequence of these rational ratios is the irrational golden ratio.
Similarly, the silver ratio of and is defined by the proportion
a:b=(2a+b):a(=(2+b/a):1),
corresponding to
x2-2x-1=0.
This equation has the positive, irrational solution
x=\tfrac{a}{b}=1+\sqrt{2},
so again at least one of the two quantities a and b in the silver ratio must be irrational.
## Odds
See main article: Odds. Odds (as in gambling) are expressed as a ratio. For example, odds of "7 to 3 against" (7:3) mean that there are seven chances that the event will not happen to every three chances that it will happen. The probability of success is 30%. In every ten trials, there are expected to be three wins and seven losses.
## Units
Ratios may be unitless, as in the case they relate quantities in units of the same dimension, even if their units of measurement are initially different.For example, the ratio can be reduced by changing the first value to 60 seconds, so the ratio becomes . Once the units are the same, they can be omitted, and the ratio can be reduced to 3:2.
On the other hand, there are non-dimensionless quotients, also known as rates (sometimes also as ratios).[22] [23] In chemistry, mass concentration ratios are usually expressed as weight/volume fractions.For example, a concentration of 3% w/v usually means 3 g of substance in every 100 mL of solution. This cannot be converted to a dimensionless ratio, as in weight/weight or volume/volume fractions.
## Triangular coordinates
The locations of points relative to a triangle with vertices A, B, and C and sides AB, BC, and CA are often expressed in extended ratio form as triangular coordinates.
In barycentric coordinates, a point with coordinates α, β, γ is the point upon which a weightless sheet of metal in the shape and size of the triangle would exactly balance if weights were put on the vertices, with the ratio of the weights at A and B being α : β, the ratio of the weights at B and C being β : γ, and therefore the ratio of weights at A and C being α : γ.
In trilinear coordinates, a point with coordinates x:y:z has perpendicular distances to side BC (across from vertex A) and side CA (across from vertex B) in the ratio x:y, distances to side CA and side AB (across from C) in the ratio y:z, and therefore distances to sides BC and AB in the ratio x:z.
Since all information is expressed in terms of ratios (the individual numbers denoted by α, β, γ, x, y, and z have no meaning by themselves), a triangle analysis using barycentric or trilinear coordinates applies regardless of the size of the triangle.
## Notes and References
1. Web site: Ratios. 2020-08-22. www.mathsisfun.com.
2. Web site: Stapel. Elizabeth. Ratios. 2020-08-22. Purplemath.
3. Web site: ISO 80000-1:2022(en) Quantities and units — Part 1: General . iso.org . . 2023-07-23.
4. "The quotient of two numbers (or quantities); the relative sizes of two numbers (or quantities)", "The Mathematics Dictionary" https://books.google.com/books?id=UyIfgBIwLMQC&dq=dictionary+ratio&pg=PA349
5. New International Encyclopedia
6. Decimal fractions are frequently used in technological areas where ratio comparisons are important, such as aspect ratios (imaging), compression ratios (engines or data storage), etc.
7. Web site: Colon . Weisstein . Eric W. . Eric W. Weisstein . . 2022-11-04 . 2022-11-26 .
8. Web site: ASCII Punctuation . Unicode, Inc. . The Unicode Standard, Version 15.0 . 2022 . 2022-11-26 . [003A is] also used to denote division or scale; for that mathematical use 2236 is preferred .
9. http://www.britannica.com/topic/ratio from the Encyclopædia Britannica
10. Heath, p. 126
11. New International Encyclopedia
12. http://www.bellegroup.com/es/support/mixingHints.html Belle Group concrete mixing hints
13. Penny Cyclopædia, p. 307
14. Smith, p. 478
15. Heath, p. 112
16. Heath, p. 113
17. Smith, p. 480
18. Heath, reference for section
19. "Geometry, Euclidean" Encyclopædia Britannica Eleventh Edition p682.
20. Heath p.114
21. Heath p. 125
22. Book: "Velocity" can be defined as the ratio... "Population density" is the ratio... "Gasoline consumption" is measure as the ratio... . Ratio and Proportion: Research and Teaching in Mathematics Teachers . 2012 . Springer Science & Business Media . David Ben-Chaim . Yaffa Keret . Bat-Sheva Ilany. 9789460917844 .
23. "Ratio as a Rate. The first type [of ratio] defined by Freudenthal, above, is known as rate, and illustrates a comparison between two variables with difference units. (...) A ratio of this sort produces a unique, new concept with its own entity, and this new concept is usually not considered a ratio, per se, but a rate or density.", "Ratio and Proportion: Research and Teaching in Mathematics Teachers" https://books.google.com/books?id=eawKLY71xvkC&q=rate&pg=PA29 |
# Solution For Three Equal Parts
The Three Equal Parts problem on LeetCode is a problem that requires a bit of thinking, but with the right approach, it can be solved efficiently. Here is a detailed solution to the problem.
Problem Statement:
Given an array of binary integers, return three non-empty parts such that all of them have the same binary value.
Solution:
To solve the problem, we need to first identify if there exists a solution. In other words, we need to check if the array can be divided into three equal parts, each having the same binary value. To do this, we can perform the following steps:
Step 1: Calculate the total number of 1’s in the array.
Step 2: If the total number of 1’s is not divisible by three, return [-1,-1]. This is because there is no way we can divide the array into three parts having equal binary value.
Step 3: Calculate the value of onesInEachPart as the total number of 1’s divided by three.
Step 4: Traverse the array from right to left and count the number of 1’s in the last part of the array. Once you reach onesInEachPart, you have found the last part of the array. Store its index in a variable lastPartIndex.
Step 5: Traverse the array from right to left again and repeat Step 4 for the second part.
Step 6: Traverse the array from left to right starting from index 0 and count the number of 1’s in the first part of the array. Once you reach onesInEachPart, you have found the first part of the array. Store its index in a variable firstPartIndex.
Step 7: Traverse the array from left to right again and repeat Step 6 for the second part.
Step 8: Check if the three parts have equal binary value. If they do, return [firstPartIndex, lastPartIndex+1]. Otherwise, return [-1,-1].
Code:
Here is the code for the solution to the Three Equal Parts problem on LeetCode:
class Solution {
public:
vector<int> threeEqualParts(vector<int>& arr) {
vector<int> result{-1,-1};
int n=arr.size();
int ones=0;
for(int i=0; i<n; i++){
if(arr[i]==1) ones++;
}
if(ones % 3 != 0) return result;
if(ones == 0) return {0, n-1};
int onesInEachPart = ones / 3;
int lastPartIndex=-1, secondPartIndex=-1;
int cnt=0;
for(int i=n-1; i>=0; i--){
if(arr[i]==1) cnt++;
if(cnt==onesInEachPart){
lastPartIndex=i;
break;
}
}
cnt=0;
for(int i=lastPartIndex-1; i>=0; i--){
if(arr[i]==1) cnt++;
if(cnt==onesInEachPart){
secondPartIndex=i;
break;
}
}
cnt=0;
int firstPartIndex=0;
for(int i=0; i<n; i++){
if(arr[i]==1) cnt++;
if(cnt==onesInEachPart){
firstPartIndex=i;
break;
}
}
cnt=0;
int secondPartStart=secondPartIndex+1;
for(int i=secondPartStart; i<n; i++){
if(arr[i]==1) cnt++;
if(cnt==onesInEachPart){
int thirdPartStart=i+1;
for(int j=0; j<=onesInEachPart-1; j++){
if(arr[firstPartIndex+j]!=arr[secondPartIndex+j] || arr[secondPartIndex+j]!=arr[thirdPartStart+j]){
return result;
}
}
return {firstPartIndex+onesInEachPart-1, secondPartIndex+onesInEachPart};
}
}
return result;
}
};
The time complexity of this solution is O(n) as we are traversing the array only four times. The space complexity is O(1) as we are not using any extra memory other than the result vector.
## Step by Step Implementation For Three Equal Parts
/**
* Given an array A of 0s and 1s, divide the array into 3 non-empty parts such that all of these parts represent the same binary value.
If it is possible, return any [i, j] with i+1 < j, such that:
A[0], A[1], ..., A[i] is the first part,
A[i+1], A[i+2], ..., A[j-1] is the second part,
A[j], A[j+1], ..., A[A.length - 1] is the third part.
All three parts have equal binary value.
If it is not possible, return [-1, -1].
Note that the entire part is used when considering what binary value it represents. For example, [1,1,0] represents 6 in decimal, not 3. Also, leading zeros are allowed, so [0,1,1] and [1,1] represent the same value.
Example 1:
Input: [1,0,1,0,1]
Output: [0,3]
Example 2:
Input: [1,1,0,1,1]
Output: [-1,-1]
Note:
3 <= A.length <= 30000
A[i] == 0 or A[i] == 1
*/
class Solution {
public int[] threeEqualParts(int[] A) {
int sum = 0;
for (int i = 0; i < A.length; i++) {
sum += A[i];
}
if (sum % 3 != 0) {
return new int[] {-1, -1};
}
int parts = sum / 3;
int i1 = 0, i2 = 0, i3 = 0;
int j1 = 0, j2 = 0, j3 = A.length - 1;
while (i3 < A.length && A[i3] == 0) {
i3++;
}
while (j3 >= 0 && A[j3] == 0) {
j3--;
}
if (i3 > j3) {
return new int[] {0, A.length - 1};
}
int count = 0;
while (count < parts) {
count += A[i3];
i3++;
}
i2 = i3 - 1;
count = 0;
while (count < parts) {
count += A[j3];
j3--;
}
j2 = j3 + 1;
while (i1 < A.length && A[i1] == 0) {
i1++;
}
if (i1 == A.length) {
return new int[] {0, A.length - 1};
}
while (j1 >= 0 && A[j1] == 0) {
j1--;
}
if (j1 < 0) {
return new int[] {-1, -1};
}
while (i1 <= i2 && i2 <= j2 && j2 <= j1) {
if (A[i1] == A[i2] && A[i2] == A[j2]) {
i1++;
i2++;
j2++;
} else {
return new int[] {-1, -1};
}
}
return new int[] {i1 - 1, j2};
}
}
Given an array A of 0s and 1s, divide the array into 3 non-empty parts such that all of these parts represent the same binary value.
If it is possible, return any [i, j] with i+1 < j, such that:
A[0], A[1], ..., A[i] is the first part,
A[i+1], A[i+2], ..., A[j-1] is the second part,
and A[j], A[j+1], ..., A[A.length - 1] is the third part.
All three parts have equal binary value.
If it is not possible, return [-1, -1].
Note that the entire part is used when considering what binary value it represents. For example, [1,1,0] represents 6 in decimal, not 3. Also, leading zeros are allowed, so [0,1,1] and [1,1] represent the same value.
def threeEqualParts(self, A):
#find the number of 1's in the array
count = 0
for i in range(len(A)):
if A[i] == 1:
count += 1
#if the number of 1's is not divisible by 3, then it is not possible to have 3 equal parts
if count % 3 != 0:
return [-1, -1]
#find the number of 1's in each part
part = count // 3
#find the index of the last 1 in the first part, the index of the first 1 in the second part, and the index of the last 1 in the second part
i, j, k = 0, 0, 0
count = 0
while count <= part:
if A[i] == 1:
count += 1
i += 1
while count <= 2*part:
if A[j] == 1:
count += 1
j += 1
while count <= 3*part:
if A[k] == 1:
count += 1
k += 1
#compare the first part with the second part and the second part with the third part
while i < j and j < k:
if A[i] != A[j] or A[j] != A[k]:
return [-1, -1]
i += 1
j += 1
k += 1
return [i-1, j]
/**
* @param {number[]} A
* @return {number[]}
*/
var threeEqualParts = function(A) {
};
Given an array A of 0s and 1s, divide the array into 3 non-empty parts such that all of these parts represent the same binary value.
If it is possible, return any [i, j] with i+1 < j, such that:
A[0], A[1], ..., A[i] is the first part,
A[i+1], A[i+2], ..., A[j-1] is the second part,
A[j], A[j+1], ..., A[A.length - 1] is the third part.
Each part can have any length.
Return null if the array cannot be divided into three parts.
Note:
A will have length between 1 and 500.
A[i] will be 0 or 1.
Solution:
We can use a greedy approach to solve this problem. First, we will compute the sum of all the elements in the array. This will tell us the total number of 1s in the array.
If the sum is not divisible by 3, then we know it is not possible to divide the array into 3 parts with equal sums.
Otherwise, we can keep track of the current sum of each part. Once we reach a sum that is equal to the sum / 3, we know we have found one part. We can then reset the current sum and continue looking for the next part.
If we find the third part and the current sum is also equal to sum / 3, then we know we have found a valid solution and can return the indices of the left and right boundaries of the three parts.
If we reach the end of the array without finding a valid solution, then we know it is not possible to divide the array into 3 parts with equal sums and we can return null.
public class Solution {
public int[] ThreeEqualParts(int[] A) {
//find the sum of all the elements in the array
int sum = 0;
for(int i = 0; i < A.Length; i++){
sum += A[i];
}
//if the sum is not divisible by 3, then we cannot split the array into 3 equal parts
if(sum % 3 != 0){
return new int[] {-1, -1};
}
//find the sum of each part
int partSum = sum / 3;
//initialize the three parts of the array
int[] part1 = new int[A.Length];
int[] part2 = new int[A.Length];
int[] part3 = new int[A.Length];
//initialize the index for each part
int part1Index = 0;
int part2Index = 0;
int part3Index = 0;
//go through the original array and add each element to the appropriate part
for(int i = 0; i < A.Length; i++){
if(part1Index < part1.Length){
part1[part1Index] = A[i];
part1Index++;
}
else if(part2Index < part2.Length){
part2[part2Index] = A[i];
part2Index++;
}
else{
part3[part3Index] = A[i];
part3Index++;
}
}
//check if the three parts are equal
if(part1.SequenceEqual(part2) && part2.SequenceEqual(part3)){
return new int[] {part1Index - 1, part2Index};
}
else{
return new int[] {-1, -1};
}
}
}
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## Top 100 Leetcode Practice Problems In Java
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# Condorcet Method
Document Sample
``` Condorcet Method
Another group-ranking method
Development of Condorcet Method
As we have seen, different methods of
determining a group ranking often give
different results.
For this reason, the marquis de Condorcet,
a French mathematician, philospher, and
good friend of Jean-Charles de Borda,
proposed that any choice that could obtain a
majority over every choice should win.
Condorcet Method
The Condorcet method requires that a
choice be able to defeat each of the other
choices in a one-on-one contest.
Again consider the preference schedules
from the last section.
Condorcet Example
(CORRECTED)!
We must compare each choice with every
other choice:
C D
A B
B C B B
C D D C
D A A A
8 5 6 7
Condorcet Example (cont’d)
Begin by comparing A with each of B, C and
D.
A is ranked higher than B on 8 schedules
and lower on 18.
Because A can not obtain a majority against
B, it is impossible for A to be the Condorcet
winner.
Choice B
Next consider B.
We have already seen that B is ranked higher than
A, so let’s now compare B with C.
B s ranked higher on 8+5+7 = 20 schedules and
lower on 6.
Now compare B with D. B is ranked higher on 8 +
5 + 6 = 19 schedules and lower on 7.
B could be the Condorcet winner since it has a
majority over the other choices.
Tabular results
The table on the left shows
A B C D
all of the possible one-on-
one contests in this
A L L L
example.
B W W W To see how a choice does
in one-on-one contests,
red across the row
C W L W
associated with that
choice.
D W L L
Although the Condorcet method seems
ideal, it sometimes fails to produce a winner.
Try this example!
You Try: Condorcet Example
A B C
B C A
C A B
20 20 20
Condorcet Example Explained
Notice that A is preferred to B on 40 of the
60 schedules but the A is preferred to C on
only 20.
For this reason, there
is no
Condorcet winner.
You may expect that if A is preferred to B by
a majority of voters and if B is preferred to C
by a majority of voters, then a majority of
voters would prefer A to C.
However, we have just seen that this is not
necessarily the case.
Transitive Property
Do you remember the transitive property?
If we consider the relation “greater than (>)”,
we see that if a > b and b>c, then a>c.
However, according to the Condorcet method
the transitive property does not always hold
true.
When this happens using the Condorcet
method, it is known as a Condorcet Paradox.
Extra Practice
1. The choices in the set of preferences
shown represent three bills to be
considered by a legislative body. The
members will debate two of the bills and
choose one of them. The chosen bill and
the third bill will then be debated and
another vote taken. Suppose you are
responsible for deciding which two will
appear on the agenda first.
Practice (cont’d)
A B C
B C A
A B
C
40 30 30
If you strongly prefer Bill C, which two bills
would you place first on the agenda? Why?
Practice (cont’d)
A C B B
B A C A
C B A C
14 3 22
8
a. Use a runoff to determine the winner in the
set of preferences.
```
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# What is an angle of 180° called?
• Last Updated : 17 Aug, 2021
Geometry has been an essential part of the development of the modern world as it is applicable to designing, constructional works, architecture in choice of material for construction, and many more. It is also a vital part of the technical background for computing various designs, manufacturing, creating blueprints, programming, etc.
Geometry is a branch of mathematics that deals with the study of shapes and their properties.
The approach for geometry can be observed from ancient times in their constructions from the use of various shapes in a very specific way. The term is originally derived from Greek words ‘ge’ and ‘materia’ which means earth and measurement respectively.
The given article is a study about angles, their types, and properties mainly focused on straight angles or 180° angles.
### What are Angles?
Angles are the space between two intersecting lines meeting at a certain point
They consist of two arms known as sides of the angle and a meeting point where angle forms called a vertex. Angles are measured in degrees which measures from 0° to 360°.
Angles are divided into different types on the basis of measurement and rotation.
### Types of Angles
Angles on the basis of measurement:
• Acute angle: The angle that measures less than 90° is the acute angle.
• Right angle: The angle that exactly measures 90° is a right angle.
• Obtuse angle: The angle that measures more than 90° and less than 180° is an obtuse angle.
• Straight angle: The angle that measures exactly 180° is a straight angle. Straight angles form straight lines.
• Reflex angle: The angle that measures more than 180° and less than 360° is the reflex angle.
Angles on the basis of rotation:
• Positive angles: The angle that moves anticlockwise from its base and is drawn from the point (x, y) is a positive angle.
• Negative angles: The angle that moves clockwise from its base and is drawn from the point (-x, -y) is a negative angle.
### What is an angle of 180° called?
180 angles widely known as straight angles are the angles that exactly measures 180°. Straight angles have two arms pointing towards opposite directions. A straight angle is also measured as (pi) angle.
It is a straight line with two endpoints where one arm moves to the opposite direction from the vertex. Straight angles can also be taken as the sum of two right angles that is 90° + 90° = 180°
The measure of straight angle can be positive or negative. If we move anticlockwise the straight angle will measure 180° and if we move clockwise the angle will measure -180°
In the above-shown figure of a straight angle, we can observe that O is the meeting point of two arms that are called vertex and OA and OB are two sides of the angle.
### Construction of a Straight Angle
Let’s look into a step-by-step construction of a straight angle.
Step 1: Draw a straight-line OX in which O will be the vertex of the angle.
Step 2: Take a protector and place its baseline on the line where point A should be placed at 0°. And follow the protector towards 180° to mark point Y.
Step 3: Join the points O and B so that another arm of the angle would form pointing in the opposite direction of OA.
### Properties of Straight Angle
• A straight angle can be formed by the summation of two right angles that is 90° + 90° = 180°
• Both arms of a straight angle point in the opposite direction.
• It completes half of the rotation. As 180° is half of 360°.
### Sample Questions
Question 1: Can we say that a triangle is a straight angle?
No, because as we know that a triangle is an enclosed figure with three sides joining together and a straight angle forms a straight line that measures 180°. Hence, both of them are different.
Question 2: Is all straight lines a straight angle?
The straight lines with two endpoints facing in the opposite direction from the vertex is a straight angle.
Question 3: Can a straight line be a reflex angle? |
# What Is The Rank Of The Word Banana?
## How do you find the rank of a word with repetition?
Rank of a word – with repetition of lettersStep 1: Write down the letters in alphabetical order.
The correct order is B, I, O, P, P, S.Step 2: Find out the number of words that start with a superior letter.
Step 3: Solve the same problem, without considering the first letter..
## What is the rank of word random in dictionary?
614∴ Rank of the word random in a dictionary is 614.
## What is the rank of the word danger?
∴ Rank of the word ‘DANGER’ =120+6+6+3=135.
## How do you find the permutation of a word?
To calculate the amount of permutations of a word, this is as simple as evaluating n! , where n is the amount of letters. A 6-letter word has 6! =6⋅5⋅4⋅3⋅2⋅1=720 different permutations. To write out all the permutations is usually either very difficult, or a very long task.
## How many ways can the word school be arranged?
120 total arrangements. The number of ways that the letters of “school” can be arranged s.t. the o’s are not adjacent then is the total number of arrangements minus the arrangements where the o’s are adjacent: 360–120=240.
## What is the rank of banana in dictionary?
35th rankTherefore, banana in the dictionary have the 35th rank.
## What is the rank of the word mother?
This is given by 120 + 120 + 24 + 24 + 6 + 6 + 6 + 2 = 308. Since, there are 308 combinations before the word, ‘MOTHER’. The rank of the word MOTHER is 308 + 1 =309.
## What is the rank of the word eamcet?
Hence rank of the word EAMCET will be 60+60+6+6+1=120+12+1=133.
## How many different words can be formed by rearranging the letters of the word success?
= 5040, but there are two “c”s, so swapping them wouldn’t change anything, cutting the number of possibilities in half… but wait, there are three “s”s, which cuts the number of possibilities down by a factor of 6 (3!), leaving 2520 / 6 = 420.
## What is the number of permutations of the letters in the word banana?
60which gives 60 So 60 distinguishable permutation of the letters in BANANA.
## What is the rank of the word Zenith?
616Thus, the rank of the word ZENITH is 616. Note: A common type of problem in many examinations is to find the rank of a given word in a dictionary. What this means is that you are supposed to find the position of that word when all permutations of the word are written in alphabetical order.
## How many 4 letter words can be formed using the letters of the word ineffective?
These four letters can be arranged in (43)=4 ways. Hence the total number of words with 3 same and one different letters is 6×4=24 ways.
## What is the rank of the word master in dictionary order?
The rank is 120+120+6+6+2+2+1=257. Hence, The rank of the word MASTER is 257. Note: This tool calculates the rank of a given word when all the letters of that word are written in all possible orders and arranged in alphabetical or dictionary order.
## What is the rank of the word school?
In the final step, we will add 1 for SCHOOL itself. Therefore, the rank of the word SCHOOL = 302 + 1 = 303. |
Finding number of relations
Chapter 1 Class 12 Relation and Functions
Concept wise
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Example 23 Let A = {1, 2, 3}. Then show that the number of relations containing (1, 2) and (2, 3) which are reflexive and transitive but not symmetric is three. Total possible pairs = {(1, 1) , (1, 2), (1, 3), (2, 1) , (2, 2), (2, 3), (3, 1) , (3, 2), (3, 3) } Each relation should have (1, 2) and (2, 3) in it For other pairs, Let’s check which pairs will be in relation, and which won’t be Total possible pairs = { (1, 1) , (1, 2), (1, 3), (2, 1) , (2, 2), (2, 3), (3, 1) , (3, 2), (3, 3) } Reflexive means (a, a) should be in relation . So, (1, 1) , (2, 2) , (3, 3) should be in a relation Symmetric means if (a, b) is in relation, then (b, a) should be in relation . We need relation which is not symmetric. So, since (1, 2) is in relation, (2, 1) should not be in relation & since (2, 3) is in relation, (3, 2) should not be in relation Transitive means if (a, b) is in relation, & (b, c) is in relation, then (a, c) is in relation So, if (1, 2) is in relation, & (2, 3) is in relation, then (1, 3) should be in relation Relation R1 = { Total possible pairs = { (1, 1) , (1, 2), (1, 3), (2, 1) , (2, 2), (2, 3), (3, 1) , (3, 2), (3, 3) } So, smallest relation is R1 = { (1, 2), (2, 3), (1, 1), (2, 2), (3, 3), (1, 3) } Checking more relations We cannot add both (2, 1) & (3, 2) together as it is not symmetric R = { (1, 2), (2, 3), (1, 1), (2, 2), (3, 3), (1, 3) , (2, 1), (3, 2)} If we add only (3, 1) to R1 R = { (1, 2), (2, 3), (1, 1), (2, 2), (3, 3), (1, 3), (3, 1) } R is reflexive but not symmetric & transitive. So, not possible If we add only (2, 1) to R1 R2 = { (1, 2), (2, 3), (1, 1), (2, 2), (3, 3), (1, 3), (2, 1) } R2 is reflexive, transitive but not symmetric If we add only (3, 2) to R1 R3 = { (1, 2), (2, 3), (1, 1), (2, 2), (3, 3), (1, 3), (3, 2) } R3 is reflexive, transitive but not symmetric Hence, there are only three possible relations R1 = { (1, 2), (2, 3), (1, 1), (2, 2), (3, 3), (1, 3) } R2 = { (1, 2), (2, 3), (1, 1), (2, 2), (3, 3), (1, 3), (2, 1)} R3 = { (1, 2), (2, 3), (1, 1), (2, 2), (3, 3), (1, 3), (3, 2)} |
# The side of the box of cereal states that there are 84 calories in a three-fourths cup servings....
## Question:
The side of the box of cereal states that there are 84 calories in a three-fourths cup servings. How many calories are in eight cups of cereal?
## Proportions and Variation:
In any numerical ratio, this denotes a relationship between two numbers {eq}x {/eq} and {eq}y {/eq}. On the other hand, a proportion shows the similarity between two ratios. When two variables are dependent, variations in the magnitude of one variable will have a proportional effect on the other. When there is an increase or decrease of a variable {eq}x {/eq} with respect to another {eq}y {/eq}, for a ratio or constant K, variations are present. In the case that we have a direct variation, it happens that when one variable increases the other increases, which can also be written as: {eq}\frac{{{y_1}}}{{{x_1}}} = \frac{{{y_2}}}{{{x_2}}} {/eq}.
{eq}\eqalign{ & {\text{In this specific case }}{\text{,we have two proportional values }}\,x\,\left( {cups{\text{ }}of{\text{ }}the{\text{ }}cereal} \right){\text{ and }} \cr & y\,\left( {calories} \right){\text{ that have a variation in directly proportional form}}{\text{. }} \cr & {\text{So we have:}} \cr & \,\,\,\,{x_1} = \frac{3}{4}\,cup = 0.75\,cup \cr & \,\,\,\,{y_1} = 84\,calories \cr & \,\,\,\,{x_2} = 8\,cups \cr & \,\,\,\,{y_2} = ?\,calories \cr & {\text{Since}}{\text{, }}x{\text{ and }}y{\text{ vary directly}}{\text{, then}}{\text{, when }}x{\text{ increases it also }} \cr & {\text{increases }}y{\text{. For this reason}}{\text{, it must be satisfied that:}} \cr & \,\,\,\,\frac{{{y_2}}}{{{x_2}}} = \frac{{{y_1}}}{{{x_1}}} \cr & {\text{So if we do cross - multiplying:}} \cr & \,\,\,\,{y_2} \cdot {x_1} = {y_1} \cdot {x_2} \cr & {\text{Now}}{\text{, solving for }}\,{y_2}{\text{:}} \cr & \,\,\,\,{y_2} = \frac{{{y_1} \cdot {x_2}}}{{{x_1}}} \cr & {\text{So}}{\text{, substituting the given values:}} \cr & \,\,\,\,{y_2} = \frac{{84 \times 8}}{{0.75}} = 896\,calories \cr & {\text{Therefore}}{\text{, there are }}\boxed{896\,calories}{\text{ in }}\,8\,cups{\text{ of the cereal}}{\text{.}} \cr} {/eq} |
# How do you calculate chebyshev interval?
## How do you calculate chebyshev interval?
The interval (22,34) is the one that is formed by adding and subtracting two standard deviations from the mean. By Chebyshev’s Theorem, at least 3/4 of the data are within this interval. Since 3/4 of 50 is 37.5, this means that at least 37.5 observations are in the interval.
## How do you find probability using Chebyshev’s theorem?
Use Chebyshev’s theorem to find what percent of the values will fall between 123 and 179 for a data set with mean of 151 and standard deviation of 14. We subtract 151-123 and get 28, which tells us that 123 is 28 units below the mean.
What does K equal in chebyshev theorem?
Chebyshev’s Theorem Definition The value for k must be greater than 1. Using Chebyshev’s rule in statistics, we can estimate the percentage of data values that are 1.5 standard deviations away from the mean. Or, we can estimate the percentage of data values that are 2.5 standard deviations away from the mean.
What does Chebyshev’s theorem state?
Chebyshev’s Theorem estimates the minimum proportion of observations that fall within a specified number of standard deviations from the mean. This theorem applies to a broad range of probability distributions. Chebyshev’s Theorem is also known as Chebyshev’s Inequality.
### How do you find the percentage of an interval?
To calculate the % of intervals, count the number of intervals in which the behavior was recorded, divide by the total number of intervals during the observations period and multiply by 100. Example: Sam was talking during 20 our 30 intervals- 20 divided by 30= .
### Why do we use Chebyshev’s theorem?
Chebyshev’s theorem is used to find the proportion of observations you would expect to find within a certain number of standard deviations from the mean. Chebyshev’s Interval refers to the intervals you want to find when using the theorem.
Why would you use Chebyshev’s theorem?
What does K equal in stats?
N is the total number of cases in all groups and k is the number of different groups to which the sampled cases belong.
#### What is the biggest limitation to using Chebyshev’s theorem?
You can only use the formula to get results for standard deviations more than 1; It can’t be used to find results for smaller values like 0.1 or 0.9. Technically, you could use it and get some kind of a result, but those results wouldn’t be valid.
#### How do you find the prevalence of a confidence interval?
To calculate the confidence interval, we must find p′, q′. p′ = 0.842 is the sample proportion; this is the point estimate of the population proportion. Since the requested confidence level is CL = 0.95, then α = 1 – CL = 1 – 0.95 = 0.05 ( α 2 ) ( α 2 ) = 0.025.
How is the interval formed by Chebyshev’s theorem?
The interval (22,34) is the one that is formed by adding and subtracting two standard deviations from the mean. By Chebyshev’s Theorem, at least 3/4 of the data are within this interval. Since 3/4 of 50 is 37.5, this means that at least 37.5 observations are in the interval.
How to calculate the percent of Chebyshev’s theorem?
1. Square the value for k. We have: k2 = 1.52 = 2.25 k 2 = 1.5 2 = 2.25 2. Next, divide 1 by the answer from step 1 above: 1 2.25 = 0.44444444444444 1 2.25 = 0.44444444444444 3. Subtract the answer in step 2 above from the number 1: 1− 0.44444444444444 1 − 0.44444444444444 = 0.55555555555556 = 0.55555555555556 4. Multiply by 100 to get the percent.
## How to calculate Chebyshev’s rule for a shaped distribution?
For any shaped distribution, at least 1– 1 k2 1 – 1 k 2 of the data values will be within k standard deviations of the mean. The value for k must be greater than 1. Using Chebyshev’s rule in statistics, we can estimate the percentage of data values that are 1.5 standard deviations away from the mean.
## Can you use Chebyshev’s theorem as a learning tool?
You can use the Chebyshev’s Theorem Calculator as a learning tool. The best approach is to first look at a sample solution to a couple different problems and understand the steps shown in the solution. |
# Zeros of Polynomial
## Introduction
Let’s first understand what is zeros of polynomial. We can represent a polynomial on graph with a curve which intersects the axis at specific points. The values of the intersecting points on x-axis are the values that make the polynomial equal to zero. These points on x-axis which make polynomial equal to zero are called zeros of a polynomial. Below is an example of polynomial which is drawn as a curve on graph.
Graph of polynomial
In this graph, the curve of polynomial y = x + 4 intersects the x-axis at x = – 4 and y-axis at y = 4 or the intersecting points on the x-axis for x = -4 is (-4,0) and and on the y-axis for y = 4 is (0,4).
As said above in the introduction, the values of x that lie on x-axis i.e. x = -4 is the zero of polynomial.
How?
We can check it by putting the value of x = – 4 in polynomial y = x + 4, if y comes equal to 0 for x = -4, then value of x is zero of polynomial.
Substitute value x = – 4 in polynomial y = x + 4.
y = (-4) + 4
y = -4 + 4
y = 0
$$\therefore$$, x=-4 is zero of polynomial.
## Definition
A real number k is the zero of polynomial p(x) if p(k)= 0.
Let’s see, what it says. If p(x) is a polynomial in x and k is any real number, then value obtained by replacing x by k in p(x) is called the value of p(x) at x = k and is denoted by p(k).
If value of k in polynomial can make the value of polynomial to zero, then k is called zeros of polynomial. We can learn it by example, how we can find out the zeros of a polynomial.
Zeros of a polynomial are determined by putting different value of x, which can make value of polynomial to 0.
Example
Find zeros of polynomial $$p(x) = x^2 – 5x + 6$$
$$p(x) = x^2 – 5x + 6$$
Put value x = 2
$$p(2) = (2)^2 – 5(2) + 6$$
$$= 4 – 10 + 6$$
$$= 10 – 10$$
$$= 0$$
$$p(2) = 0$$
$$\therefore$$ 2 is zero of polynomial p(x)
Put value x = 3
$$p(3) = (3)^2 -5(3) + 6$$
$$= 9 – 15 + 6$$
$$= 15 – 15$$
= 0
$$\therefore$$ 3 is also zero of polynomial p(x)
So, 2 and 3 are two zeros of polynomial p(x).
## Geometrical meaning of zeros of polynomial
Any polynomial can be drawn on a graph with a specific curve. We have already introduced above in Introduction section how the polynomial curves are drawn on a graph. The shapes of the curves of polynomial varies with polynomial degree. Linear polynomial has curves of unique shape and are different from quadratic polynomial and even biquadratic has its own unique shape. Let’s learn what are the shapes and how to find zeros of polynomials for linear, quadratic, cubic and biquadratic polynomials.
### Linear polynomial
The general form of a linear polynomial is $$ax + b$$, where $$a \neq 0$$. The graph of linear polynomial is a straight line and it intersects the x-axis at exactly one point. So, the linear polynomial has only one zero.
Example
Consider a linear polynomial, y = 2x + 6
To find the zeros, we put y = 0
2x + 6 = 0
x = -3
$$\therefore$$ -3 is zero of y = 2x + 6 linear polynomial
Graph of linear polynomial
The general form of a quadratic polynomial is $$ax^2+bx+c$$, where $$a \neq 0$$. The graph of quadratic polynomial has two shapes, one is known as open upwards or in shape of ∪ and another is open downwards or in shape of downwards ∩. These curves are also called as parabolas. Let’s find zeros of a quadratic polynomial in an example as below.
Example
Consider a quadratic polynomial, $$y = x^2 + 2x – 3$$
To find zeros of polynomial, let’s start putting value of x and find the corresponding y.
Put x = 0
$$y = (0)^2 + 2(0) – 3$$
y = 0 + 0 – 3
y = 0 – 3
y = -3
Put x = -3
$$y = (-3)^2 + 2(-3) – 3$$
y = 9 – 6 – 3
y = 9 – 9
y = 0
Put x = 1
$$y = (1)^2 + 2(1) – 3$$
y = 1 + 2 – 3
y = 3 – 3
y = 0
The zeros of quadratic polynomial $$x^2 + 2x – 3$$ will be the x coordinates of points where the graph $$y=x^2 + 2x – 3$$ intersects the x-axis.
Therefore, -3 and 1 are zeros of polynomial $$x^2 + 2x – 3$$ as graph $$y = x^2 + 2x – 3$$ intersects x-axis at -3 and 1.
Therefore, for quadratic polynomial $$ax^2+bx+c$$, $$a \neq 0$$, zeros of polynomial are x-coordinates of points where the parabola $$y=ax^2+bx+c$$ intersects x-axis.
Further, for the graph of $$y=ax^2+bx+c$$, there are three cases that arises with different shapes of graph. Let’s have a look at it one by one.
#### Graph of a quadratic polynomial with distinct zeros
First case of graph is when the graph cuts x-axis at two distinct points A and B. The x-coordinate of A and B are two zeros of quadratic polynomial $$ax^2+bx+c$$.
Graph of a quadratic polynomial with distinct zeros
#### Graph of a quadratic polynomial with coincident zeros
When graph of the polynomial $$ax^2+bx+c$$ cuts x-axis at exactly one point i.e. at two coincident points. So the two points A and B coincide here to become one point A. The x-coordinates of A is only one zero for quadratic polynomial $$ax^2+bx+c$$.
Graph of a quadratic polynomial with coincident roots
#### Graph of a quadratic polynomial with no zeros
Here graph is completely above x-axis and completely below x-axis. It does not cut the x-axis at any point. So, quadratic polynomial has no zero in this case.
Graph of a quadratic polynomial with no zeros
Therefore, we can summarize from the above three cases, geometrically, a quadratic polynomial can have either two distinct zeros or two equal zeros or no zero. In other words, a quadratic polynomial can have at most two zeros.
### Cubic polynomial
The general form of a cubic polynomial is $$ax^3+bx^2+cx+d$$, where $$a \neq 0$$. The graph of cubic polynomial intersects the x-axis at points, the coordinates of these points are the only zeros of the cubic polynomial.
Let’s find zeros of a cubic polynomial in an example as below.
Example
Consider a cubic polynomial, $$y=x^3-x$$
To find zeros of polynomial, let’s start putting value of x and find the corresponding y.
Put x = 1
$$y=(1)^3-(1)$$
y=1-1
y=0
Put x = 0
$$y=(0)^3-(0)$$
y=0-0
y=0
Put x = -1
$$y=(-1)^3-(-1)$$
y=-1+1
y=0
Put x = -2
$$y=(-2)^3-(-2)$$
y=-(-8)+2
y=-6
The zeros of cubic polynomial $$y=x^3-x$$ will be the x coordinates of points where the graph $$y=x^3-x$$ intersects the x-axis.
Here 1, 0 and -1 are zeros of cubic polynomial as these are points where the graph intersects x-axis.
Graph of cubic polynomial
The general form of a biquadratic polynomial is $$ax^4+bx^3+cx^2+dx+c$$, where $$a \neq 0$$. The graph of biquadratic polynomial intersects the x-axis at points, the coordinates of these points are the only zeros of the biquadratic polynomial.
Let’s find zeros of a biquadratic polynomial in an example as below.
Example
Consider a biquadratic polynomial, $$y=x^4-x^3-38x^2+36x+72$$
To find zeros of polynomial, let’s start putting value of x and find the corresponding y.
Put x = 0
y=(0)4-(0)3-38(0)2+36(0)+72
y = (0) – (0) – (0) + 36(0) + 72
y = 72
Put x = 2
y=(2)4-(2)3-38(2)2+36(2)+72
y = 16 – 8 – 152 + 72 + 72
y = 0
Put x = 6
y=(6)4-(6)3-38(6)2+36(6)+72
y = 1296 – 216 – 1368 + 216 + 72
y=0
Put x = -1
y=(-1)4-(1)3-38(1)2+36(1)+72
y = 1 + 1 – 38 – 36 + 72
y=0
Put x = -6
y=(-6)4-(-6)3-38(-6)2+36(-6)+72
y = 1296 + 216 – 1368 – 216 + 72
y=0
The zeros of cubic polynomial $$y=x^4-x^3-38x^2+36x+72$$ will be the x coordinates of points where the graph $$y=x^4-x^3-38x^2+36x+72$$ intersects the x-axis.
Here, -6, -1, 2 and 6 are zeros of biquadratic polynomial as these are points where the graph intersects x-axis.
## Relationship between zeros and coefficients of quadratic polynomial
Consider a quadratic polynomial $$ax^2+bx+c$$. Let $$\alpha$$ and $$\beta$$ are two zeros of polynomial.
Formula
Sum of zeros = $$-\frac{coefficient \; \; of \; \; x}{coefficient \; \; of \; \; x^2}$$
$$\alpha + \beta = \frac{-b}{a}$$
Product of zeros = $$\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^2}$$
$$\alpha \beta = \frac{c}{a}$$
Example
Consider a polynomial $$x^2+5x+6$$
Zeros of $$x^2+5x+6$$ are -2 and -3
Here, coefficient of $$x^2 = 1$$
coefficient of x =5
constant term =6
Therefore, a=1, b=5, c=6
Here, $$\alpha$$=-2 and $$\beta$$=-3
Sum of zeros = $$-\frac{coefficient \; \; of \; \; x}{coefficient \; \; of \; \; x^2}$$
$$\alpha +\beta$$ = $$-\frac{b}{a}$$
(-2) + (-3) = $$-\frac{5}{1}$$
-5 = -5
Product of zeros = $$\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^2}$$
$$\alpha \beta = \frac{c}{a}$$
(-2)(-3) = $$\frac{6}{1}$$
6 = 6
## Relationship between zeros and coefficients of cubic polynomial
Now, consider a cubic polynomial $$p(x)=ax^3+bx^2+cx+d$$. Let $$\alpha$$, $$\beta$$ and $$\gamma$$ are three zeros of polynomial.
Formula
Sum of zeros = $$-\frac{coefficient \; \; of \; \; x^2}{coefficient \; \; of \; \; x^3}$$
$$\alpha + \beta + \gamma$$ = $$-\frac{b}{a}$$
Sum of product of zeros taken two at a time= $$\frac{coefficient \; \; of \; \; x}{coefficient \; \; of \; \; x^3}$$
$$\alpha \beta+\beta \gamma+\alpha \gamma= \frac{c}{a}$$
Product of zeros = $$-\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^3}$$
$$\alpha \beta \gamma = -\frac{d}{a}$$
Example
Consider a cubic polynomial $$x^3+2x^2-5x-6$$
Zeros of $$x^3+2x^2-5x-6$$ are -1, 2 and -3
Here, coefficient of $$x^3 = 1$$
coefficient of $$x^2 =2$$
coefficient of x =-5
constant term =-6
Therefore, a=1, b=2, c=-5, d=-6
Here, $$\alpha$$=-1, $$\beta$$=2 and $$\gamma$$=-3
Sum of zeros = $$-\frac{coefficient \; \; of \; \; x^2}{coefficient \; \; of \; \; x^3}$$
$$\alpha +\beta+\gamma$$ = $$-\frac{b}{a}$$
(-1) + (2) + (-3)= $$-\frac{2}{1}$$
-2 = -2
Sum of product of zeros taken two at a time= $$\frac{coefficient \; \; of \; \; x}{coefficient \; \; of \; \; x^3}$$
$$\alpha \beta+\beta \gamma+\alpha \gamma= \frac{c}{a}$$
(-1)(2) + (2)(-3) + (-3)(-1)= $$\frac{-5}{1}$$
-2 -6 + 3 = -5
-5 = -5
Product of zeros = $$-\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^3}$$
$$\alpha \beta \gamma = -\frac{d}{a}$$
(-1)(2)(-3) = -($$\frac{-6}{1}$$)
6 = 6
## Relationship between zeros and coefficients of biquadratic polynomial
Polynomial $$p(x)=ax^4+bx^3+cx^2+dx+e$$, where $$a \neq=0$$ is a biquadratic polynomial. The graph of $$y=ax^4+bx^3+cx^2+dx+e$$ intersects the x-axis. These coordinates are the only zeros of biquadratic polynomial.
Consider a polynomial $$ax^4+bx^3+cx^2+dx+e$$. Let $$\alpha$$, $$\beta$$, $$\gamma$$ and $$\delta$$ are four zeros of the polynomial.
Formula
Sum of zeros = $$-\frac{coefficient \; \; of \; \; x^3}{coefficient \; \; of \; \; x^4}$$
$$\alpha + \beta + \gamma+ \delta$$ = $$-\frac{b}{a}$$
Sum of product of zeros taken two at a time= $$\frac{coefficient \; \; of \; \; x^2}{coefficient \; \; of \; \; x^4}$$
$$\alpha \beta+\beta \gamma+\delta \gamma+\alpha \delta+\delta\beta+\gamma\alpha= \frac{c}{a}$$
Sum of product of zeros taken three at a time= $$-\frac{coefficient \; \; of \; \; x}{coefficient \; \; of \; \; x^4}$$
$$\alpha\beta\gamma+\beta\gamma\delta+\alpha\beta\delta+\alpha\gamma\delta= -\frac{d}{a}$$
Product of zeros = $$\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^3}$$
$$\alpha \beta \gamma\delta = \frac{e}{a}$$
Example
Let $$x^4-6x^3-4x^2+54x-45$$ be a biquadratic polynomial
Zeros of $$x^4-6x^3-4x^2+54x-45$$ are 1, 3, 5 and -3
Here, coefficient of $$x^4 = 1$$
coefficient of $$x^3 =-6$$
coefficient of $$x^2 =-4$$
coefficient of x =54
constant term =-45
$$\therefore$$ a=1, b=-6, c=-4, d=54, e=-45
Let zeros be, $$\alpha$$=1, $$\beta$$=3, $$\gamma$$=5 and $$\delta$$=-3
$$\therefore$$ Sum of zeros = $$-\frac{coefficient \; \; of \; \; x^3}{coefficient \; \; of \; \; x^4}$$
$$\alpha +\beta+\gamma+\delta$$ = $$-\frac{b}{a}$$
(1) + (3) + (5) + (-3)= $$-\frac{-6}{1}$$
6 = 6
Sum of product of zeros taken two at a time= $$\frac{coefficient \; \; of \; \; x^2}{coefficient \; \; of \; \; x^4}$$
$$\alpha \beta+\beta \gamma+\gamma\delta+\alpha \delta+\delta\beta+\gamma\alpha$$ = $$\frac{c}{a}$$
(1)(3) + (3)(5) + (5)(-3) + (-3)(1) + (-3)(3) + (5)(1)= $$-\frac{4}{1}$$
3 + 15 – 15 – 3 – 9 + 5 = -4
-4 = -4
Sum of product of zeros taken three at a time= $$-\frac{coefficient \; \; of \; \; x}{coefficient \; \; of \; \; x^4}$$
$$\alpha\beta\gamma+\beta\gamma\delta+\alpha\beta\delta+\alpha\gamma\delta= -\frac{d}{a}$$
(1)(3)(5)+(3)(5)(-3)+(1)(3)(-3)+(1)(5)(-3) = -$$\frac{54}{1}$$
15-45-9-15 = -$$\frac{54}{1}$$
-54 = -54
Product of zeros = $$\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^3}$$
$$\alpha \beta \gamma\delta = \frac{e}{a}$$
(1)(3)(5)(-3)= $$\frac{-45}{1}$$
-45 = -45
### Q) What is zeros of polynomial?
The value which makes the value of polynomial equal to zero is called zeros of polynomial. i.e. k is said to be zeros of polynomial, when a polynomial p(x) becomes equal to zero for the value of k. i.e. when we put x = k in p(x) and p(k) = 0.
### Q) What is shape of graph of a linear polynomial?
The graph of a linear polynomial is always a straight line.
### Q) What is shape of graph of a quadratic polynomial?
The graph of a quadratic polynomial ax2 + bx + c is a parabola. It opens upward ∪ if a > 0 and opens downward ∩ if a < 0.
### Q) What is the number of zeros of polynomial of degree n?
The polynomial with degree n can have number of zeros equal to n or less than n. i.e. number of zeros ≤ n.
### Q) What is relationship between coefficient of polynomial and its zero?
Sum of zeros = $$-\frac{coefficient \; \; of \; \; x}{coefficient \; \; of \; \; x^2}$$
Product of zeros = $$\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^2}$$
## Solved Examples
### 1) Find the zeros of x2 - 9 and verify its relationship between coefficient of polynomial and its zeros.
Solution
Here p(x) = x2 - 9
(x)2 -(3)2 using identity (a)2 - (b)2 = (a+b)(a-b)
(x+3)(x-3)
p(x) = 0
(x+3)(x-3) = 0
x+3 = 0 or x-3 = 0
x = -3 or x = 3
zeros of p(x) are -3, 3
Relationship between coefficients of polynomial and its zeros
p(x) = x2 - 9
compare it with ax2 + bx + c
p(x) = x2 + 0x - 9
Here a = 1, b = 0, c = -9
Sum of zeros = $$-\frac{b}{a}$$
-3 + 3 = 0
= $$-\frac{0}{1}$$
= $$-\frac{b}{a}$$
Product of zeros = $$\frac{c}{a}$$
= (-3)(3) = -9
= $$-\frac{9}{1}$$
= $$\frac{c}{a}$$
### 2) Form a quadratic polynomial whose sum of zeros is 5 and product of zeros is 6.
Solution
Sum of zeros = 5
Product of zeros = 6
As we know, quadratic polynomial is the form of x2 - (sum of zeros)x + product of zeros
By putting the above values, it becomes x2 - 5x + 6
Hence, x2 - 5x + 6 is a quadratic polynomial.
### 3) Verify relationship between coefficient of polynomial and its zeros if x3 - 9x2 - 12x + 20 has zeros -2, 1 and 10.
Solution
Here, compare x3 - 9x2 - 12x + 20 with ax3 + bx2 + cx + d
a = 1, b = -9, c = -12, d = 20
Zeros are -2, 1 and 10 (given)
α = -2
β = 1
γ = 10
Sum of zeros = $$-\frac{coefficient \; \; of \; \; x^2}{coefficient \; \; of \; \; x^3}$$
∴ α + β + γ = -2 + 1 + 10
= 9
= $$-\frac{(-9)}{-1}$$
= $$-\frac{(-b)}{a}$$
= $$-\frac{coefficient \; \; of \; \; x^2}{coefficient \; \; of \; \; x^3}$$
Sum of product of zeros taken two at a time = $$\frac{coefficient \; \; of \; \; x}{coefficient \; \; of \; \; x^3}$$
αβ + βγ + γα = (-2)(1) + (1)(10) + (10)(-2)
= -2 + 10 -20
= -12
= $$\frac{(-12)}{1}$$
= $$\frac{(c)}{a}$$
= $$\frac{coefficient \; \; of \; \; x}{coefficient \; \; of \; \; x^3}$$
Product of zeros = $$-\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^3}$$
αβγ = (-2)(1)(10)
= -20
= $$\frac{-20}{1}$$
= $$\frac{-(d)}{a}$$
= $$-\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^3}$$
## Multiple Choice Questions
### 1) A quadratic polynomial has
1. 1 zero
2. 2 zeros
3. No zero
4. atmost 2 zeros
### 2) In quadratic polynomial ax2 + bx + c, c ≠ 0, if zeros are of equal sign, then
1. c and a have opposite sign
2. c and b have same sign
3. c and a have same sign
4. c and b have opposite sign
### 3) In quadratic polynomial ax2 + bx + c, c ≠ 0, if both zeros are negative, then
1. a, b and c all have same sign
2. a and b have same sign
3. a and c have opposite sign
4. b and c have same sign
1. 2
2. 3
3. 4
4. 1
1. a>0
2. a≤0
3. a≥0
4. a<0
1. a>0
2. a<0
3. a≥0
4. a≤0
1. 1 zero
2. 2 zeros
3. 3 zeros
4. No zeros
1. 1
2. 2
3. 3
4. 0
1. 5 and 5
2. -5 and 5
3. 2 and -5
4. -5 and -5
1. 3
2. 1
3. -3
4. -1 |
# NCERT Solutions for Exercise 10.2 Class 9 Maths Chapter 10 - Circles
NCERT Solutions for Class 9 Maths chapter 10 exercise 10.2 is about angle subtended by a chord at a point. A circle is a two-dimensional geometrical form in which all points on the surface of the circle are equidistant from the radius, which is the centre point. A circle's chord is the line segment that connects two locations on its circumference. You will see that the angles subtended by two or more equal chords of a circle at the centre are equal if you draw two or more equal chords of a circle and determine the angles they occupied in the centre. In the NCERT solutions for Class 9 Maths chapter 10 exercise 10.2, we will cover the theorems pertaining to the angle subtended by a chord at a point in depth. The following exercise are also included in NCERT book Class 9 Mathematics chapter 10 exercise 10.1.
Latest : Trouble with homework? Post your queries of Maths and Science with step-by-step solutions instantly. Ask Mr AL
• Circles Exercise 10.1
• Circles Exercise 10.3
• Circles Exercise 10.4
• Circles Exercise 10.5
• Circles Exercise 10.6
## Circles Class 9 Chapter 10 Circles Exercise: 10.2
Q1 Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.
Given: The two circles are congruent if they have the same radii.
To prove: The equal chords of congruent circles subtend equal angles at their centres i.e. BAC= QPR
Proof :
In ABC and PQR,
BC = QR (Given)
AB = PQ (Radii of congruent circle)
AC = PR (Radii of congruent circle)
Thus, ABC PQR (By SSS rule)
BAC= QPR (CPCT)
Q2 Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Given : chords of congruent circles subtend equal angles at their centres,
To prove : BC = QR
Proof :
In ABC and PQR,
BAC= QPR (Given)
AB = PQ (Radii of congruent circle)
AC = PR (Radii of congruent circle)
Thus, ABC PQR (By SAS rule)
BC = QR (CPCT)
## More About NCERT Solutions for Class 9 Maths Exercise 10.2
As we know NCERT solutions for exercise 10.2 Class 9 Maths chapter 10 is about angle subtended by a chord at a point. But along with the topics about circles such as chords, arc, major arc, minor arc, semicircle, we have also used the information about the congruence of the triangle which we have covered in Class 9 chapter 7 triangle. Some of the rules and properties which we are using are side - side - side rule, side - angle - side rule and there are some theorems which are covered before Class 9 Maths chapter 10 exercise 10.1 these theorems are at the centre of a circle, equal chords subtend equal angles. If the angles subtended by the chords of a circle at its centre are equal, then the chords are equal, as well as the theorem.
JEE Main: Learn How To Calculate Euler’s Number ‘e’ 3 min read Mar 05, 2022 Read More Physics: Here’s An Easy Way To Grasp Projectile Motion 5 min read Mar 05, 2022 Read More
Also Read| Circles Class 9 Notes
## Benefits of NCERT Solutions for Class 9 Maths Exercise 10.2
• The circle is an important part of geometry and there are a lot of applications of circles in whole geometry
• Chapter 10 exercise 10.2 in Class 9 Maths presents a wide range of problems that can be based on angle subtended by a chord at a point.
• NCERT Class 9 Maths chapter 10 exercise 10.2, will be helpful for the Class 10 chapter 10 circle, as Class 10 chapter 10 circle is an extension of Class 9 Maths chapter 10
• NCERT Class 9 Maths chapter 10 exercise 10.2, will be base for the Class 11 chapter 11 conic section
• NCERT Class 9 Maths chapter 10 exercise 10.2, will be helpful in JEE Main (joint entrance exam) as the conic section is a major part of the syllabus and the circle is a part of the conic section section
Also, See
• NCERT Solutions for Class 9 Maths Chapter 10
• NCERT Exemplar Solutions Class 9 Maths Chapter 10
## NCERT Solutions of Class 10 Subject Wise
• NCERT Solutions for Class 9 Maths
• NCERT Solutions for Class 9 Science
## Subject Wise NCERT Exemplar Solutions
• NCERT Exemplar Class 9 Maths
• NCERT Exemplar Class 9 Science |
# Benefits to Writing Algebra Equations in Standard Form With a Tutor in Torrance
Learning how to graph linear equations is a big part of Algebra, and unfortunately, not all tests will allow your student help from a graphing calculator. Good news, an A Plus In Home Algebra Tutor in Torrance can teach you super simple graphing tips that will take away your linear equation headaches!
# Benefits to Writing Algebra Equations in Standard Form With a Tutor in Torrance
Last Updated on September 30, 2020 by A Plus In-Home Tutors
Learning how to graph linear equations is a big part of Algebra, and unfortunately, not all tests will allow your student help from a graphing calculator.
# Good news, an A Plus In Home Algebra Tutor in Torrance can teach you super simple graphing tips that will take away your linear equation headaches!
## In Algebra, slope intercept form is one way to graph linear equations, but there’s another, often easier way – convert the Algebra equation to standard form and find the x and y intercepts.
Standard form must be written as Ax + By = C, where A and B are coefficients and C is the constant.
In order to figure out how to find the x and y intercepts, let’s first review what an intercept is in Algebra: an intercept is where your line crosses an axis. The point where the line touches the x axis is called the x intercept, and the point where the line touches the y axis is called the y intercept.
Once Algebra equations are written in standard form, it’s easy to find the intercepts. To find the X intercept, let y = 0. To find the Y intercept, you let x = 0. Then, right when we find the points where the line crosses the x and y axis, we’re already able to draw a line!
Here’s an Algebra example:
4x+8y = 16
Find the x intercept
Let y = 0
4x+8(0) = 16
4x = 16
x = 4
X Intercept = (4,0)
Find the y intercept
Let x = 0
4(0)+8y = 16
8y = 16
y = 2
Y intercept = (0,2)
That’s all you need to be able to draw your graph!
So how do you convert an Algebra equation to standard form?
We need the variables x and y to be on the left, and the constant to be on the right. From there, we move terms around like we would with any Algebra equation – whatever you do to one side, make sure you do it to the other!
Say you are given the following Algebra equation:
y = -1/2x – 8
That might be a bit difficult to graph as is (slope intercept form), so make things easier. Let’s convert it to standard form.
Multiply both sides by 2 and you get 2y = -x – 16.
Then, add x to both sides to get:
x + 2y = -16.
### That’s the Algebra equation in standard form!
Let x = 0 and you get y = -8, making the y intercept (0, -8).
Let y = 0 and you get x = -16, making the x intercept (-16,0)
From there, you have two points and can immediately draw your line. Algebra problem solved!
Need more help with graphing linear equations or converting a slope intercept form equation to standard form in Torrance?
#### An expert Algebra tutor from A Plus In Home Tutors can teach you useful tricks and tips.
##### Visit us at www.APlusInHomeTutors.com to find a great Algebra tutor near you!
Posted in Algebra |
# Remove N-th node from the end of a Linked List
Problem Statement: Given a linked list, and a number N. Find the Nth node from the end of this linked list and delete it. Return the head of the new modified linked list.
Example 1 :
```Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]
Explanation: Refer Below Image```
Example 2:
```Input: head = [7,6,9,4,13,2,8], n = 6
Output: [7,9,4,13,2,8]
Explanation: Refer Below Image```
Example 3:
```Input: head = [1,2,3], n = 3
Output: [2,3]```
### Solution :
Disclaimer: Don’t jump directly to the solution, try it out yourself first.
Solution 1: Naive Approach [Traverse 2 times]
Intuition: We can traverse through the Linked List while maintaining a count of nodes, let’s say in the variable count, and then traversing for the 2nd time for (n – count) nodes to get to the nth node of the list.
Solution 2: [Efficient] Two pointer Approach
Unlike the above approach, we don’t have to maintain the count value, we can find the nth node just by one traversal by using two pointer approach.
Intuition :
What if we had to modify the same above approach to work in just one traversal? Let’s see what all information we have here:
1. We have the flexibility of using two-pointers now.
2. We know, that the n-th node from the end is the same as (total nodes – n)th node from start.
3. But, we are not allowed to calculate total nodes, as we can do only one traversal.
What if, one out of the two-pointers is at the nth node from start instead of the end? Will it make anything easier for us?
Yes, with two pointers in hand out of which one is at the n-th node from start, we can just advance both of them till the end simultaneously, once the faster reaches the end, the slower will stand at the n-th node from the end.
Approach :
1. Take two dummy nodes, who’s next will be pointing to the head.
2. Take another node to store the head, initially,s a dummy node(start), and the next node will be pointing to the head. The reason why we are using this extra dummy node is that there is an edge case. If the node is equal to the length of the LinkedList, then this slow will point to slow’s next→ next. And we can say our dummy start node will be broken and will be connected to the slow next→ next.
3. Start traversing until the fast pointer reaches the nth node.
1. Now start traversing by one step both of the pointers until the fast pointers reach the end.
1. When the traversal is done, just do the deleting part. Make slow pointers next to the next of the slow pointer to ignore/disconnect the given node.
Dry Run: We will be taking the first example for the dry run, so, the LinkedList is [1,2,3,4,5] and the node which has to be deleted is 2 from the last. For the first time, fast ptr starts traversing from node 1 and reaches 2, as it traverses for node number 2, then the slow ptr starts increasing one, and as well as the fast ptr until it reaches the end.
• 1st traversal : fast=3, slow=1
• 2nd traversal : fast=4, slow=2
• 3rd traversal : fast=5, slow=3
Now, the slow->next->next will be pointed to the slow->next
So , the new linked list will be [1,2,3,5]
Code :
## C++ Code
``````class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode * start = new ListNode();
ListNode* fast = start;
ListNode* slow = start;
for(int i = 1; i <= n; ++i)
fast = fast->next;
while(fast->next != NULL)
{
fast = fast->next;
slow = slow->next;
}
slow->next = slow->next->next;
return start->next;
}
};``````
## Java Code
``````class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode start = new ListNode();
ListNode fast = start;
ListNode slow = start;
for(int i = 1; i <= n; ++i)
fast = fast.next;
while(fast.next != null)
{
fast = fast.next;
slow = slow.next;
}
slow.next = slow.next.next;
return start.next;
}
}``````
## Python Code
``````class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
start = ListNode()
fast = start
slow = start
for i in range(1, n+1):
fast = fast.next
while fast.next != None:
fast = fast.next
slow = slow.next
slow.next = slow.next.next
return start.next``````
Time Complexity: O(N)
Space Complexity: O(1) |
# 1.2: Graphs of Equations 1.3: Linear Equations
## Presentation on theme: "1.2: Graphs of Equations 1.3: Linear Equations"— Presentation transcript:
1.2: Graphs of Equations 1.3: Linear Equations
Objectives: To find the intercepts of a graph To use symmetry as an aid to graphing To write the equation of a circle and graph it To write equations of parallel and perpendicular lines
Vocabulary As a class, use your vast mathematical knowledge to define each of these words without the aid of your textbook. Graph of an Equation Solution Point Intercepts Symmetry Circle Parallel Perpendicular
Graph of an Equation The graph of an equation gives a visual representation of all solution points of the equation.
Intercepts The x-intercept of a graph is where it intersects the x-axis. (a, 0) The y-intercept of a graph is where it intersects the y-axis. (0, b)
Exercise 1 How many x- and y-intercepts can the graph of an equation have? How about the graph of a function?
Exercise 2 Given an equation, how do you find the intercepts of its graph? To find the x-intercepts, set y = 0 and solve for x. To find the y-intercepts, set x = 0 and solve for y.
Exercise 3 Find the x- and y-intercepts of y = – x2 – 5x.
Symmetry A figure has symmetry if it can be mapped onto itself by reflection or rotation. Click me!
Exercise 4 How would an understanding of symmetry help you graph an equation?
Symmetry When it comes to graphs, there are three basic symmetries:
x-axis symmetry: If (x, y) is on the graph, then (x, -y) is also on the graph.
Symmetry When it comes to graphs, there are three basic symmetries:
y-axis symmetry: If (x, y) is on the graph, then (-x, y) is also on the graph.
Symmetry When it comes to graphs, there are three basic symmetries:
Origin symmetry: If (x, y) is on the graph, then (-x, -y) is also on the graph. (Rotation of 180)
Exercise 5 Using the partial graph pictured, complete the graph so that it has the following symmetries: x-axis symmetry y-axis symmetry origin symmetry
Circle The set of all coplanar points is a circle if and only if they are equidistant from a given point in the plane.
Exercise 6 Find the equation of points (x, y) that are r units from (h, k).
Equation of a Circle Standard form of the equation of a circle:
(h, k) = center point r = radius
Exercise 7 The point (1, -2) lies on the circle whose center is at (-3, -5). Write the standard form of the equation of the circle.
Exercise 8 Find the center and radius of the circle, and then sketch the graph.
Linear Equations (Again)
Exercise 9 Convert the given equation to the following forms:
Slope-intercept form Standard form
Exercise 10 Convert the given equation to the following forms:
Slope-intercept form Point-slope form
Parallel and Perpendicular
Two lines are parallel lines iff they are coplanar and never intersect. Two lines are perpendicular lines iff they intersect to form a right angle. m || n
Parallel and Perpendicular
Two lines are parallel lines iff they have the same slope. Two lines are perpendicular lines iff their slopes are negative reciprocals.
Example 11 Write an equation of the line that passes through the point (-2, 1) and is: Parallel to the line y = -3x + 1 Perpendicular to the line y = -3x + 1
1.2: Graphs of Equations 1.3: Linear Equations
Objectives: To find the intercepts of a graph To use symmetry as an aid to graphing To write the equation of a circle and graph it To write equations of parallel and perpendicular lines |
# Find the cubic equation whose roots are the cubes of the roots of x^3+ax^2+bx+c=0,a,b,cinRR?
## Find the cubic equation whose roots are the cubes of the roots of ${x}^{3} + a {x}^{2} + b x + c = 0$,$a , b , c \in \mathbb{R}$
Mar 13, 2017
${x}^{3} + \left({a}^{3} - 3 a b + 3 c\right) {x}^{2} + \left({b}^{3} - 3 a b c + 3 {c}^{2}\right) x + {c}^{3} = 0$
#### Explanation:
Suppose the roots of the original cubic are $\alpha$, $\beta$ and $\gamma$.
Then:
${x}^{3} + a {x}^{2} + b x + c = \left(x - \alpha\right) \left(x - \beta\right) \left(x - \gamma\right)$
$\textcolor{w h i t e}{{x}^{3} + a {x}^{2} + b x + c} = {x}^{3} - \left(\alpha + \beta + \gamma\right) {x}^{2} + \left(\alpha \beta + \beta \gamma + \gamma \alpha\right) x - \alpha \beta \gamma$
So we have:
$\left\{\begin{matrix}\alpha + \beta + \gamma = - a \\ \alpha \beta + \beta \gamma + \gamma \alpha = b \\ \alpha \beta \gamma = - c\end{matrix}\right.$
The cubic we are looking for is:
$\left(x - {\alpha}^{3}\right) \left(x - {\beta}^{3}\right) \left(x - {\gamma}^{3}\right)$
$= {x}^{3} - \left({\alpha}^{3} + {\beta}^{3} + {\gamma}^{3}\right) {x}^{2} + \left({\alpha}^{3} {\beta}^{3} + {\beta}^{3} {\gamma}^{3} + {\gamma}^{3} {\alpha}^{3}\right) x - {\alpha}^{3} {\beta}^{3} {\gamma}^{3}$
So the problem essentially boils down to expressing each of the symmetric polynomials in ${\alpha}^{3}$, ${\beta}^{3}$ and ${\gamma}^{3}$ constituting these coefficients in terms of the elementary symmetric polynomials in $\alpha$, $\beta$ and $\gamma$.
For example:
${\left(\alpha + \beta + \gamma\right)}^{3}$
$= {\alpha}^{3} + {\beta}^{3} + {\gamma}^{3} + 3 \left({\alpha}^{2} \beta + {\beta}^{2} \gamma + {\gamma}^{2} \alpha + \alpha {\beta}^{2} + \beta {\gamma}^{2} + \gamma {\alpha}^{2}\right) + 6 \alpha \beta \gamma$
$\left(\alpha + \beta + \gamma\right) \left(\alpha \beta + \beta \gamma + \gamma \alpha\right)$
$= {\alpha}^{2} \beta + {\beta}^{2} \gamma + {\gamma}^{2} \alpha + \alpha {\beta}^{2} + \beta {\gamma}^{2} + \gamma {\alpha}^{2} + 3 \alpha \beta \gamma$
So:
${\alpha}^{3} + {\beta}^{3} + {\gamma}^{3}$
$= {\left(\alpha + \beta + \gamma\right)}^{3} - 3 \left(\alpha + \beta + \gamma\right) \left(\alpha \beta + \beta \gamma + \gamma \alpha\right) + 3 \alpha \beta \gamma$
$= - {a}^{3} + 3 a b - 3 c$
We also find:
${\left(\alpha \beta + \beta \gamma + \gamma \alpha\right)}^{3}$
$= {\alpha}^{3} {\beta}^{3} + {\beta}^{3} {\gamma}^{3} + {\gamma}^{3} {\alpha}^{3} + 3 \left({\alpha}^{3} {\beta}^{2} \gamma + {\beta}^{3} {\gamma}^{2} \alpha + {\gamma}^{3} {\alpha}^{2} \beta + {\alpha}^{3} \beta {\gamma}^{2} + {\beta}^{3} \gamma {\alpha}^{2} + {\gamma}^{3} \alpha {\beta}^{2}\right) + 6 {\alpha}^{2} {\beta}^{2} {\gamma}^{2}$
$\left(\alpha + \beta + \gamma\right) \left(\alpha \beta + \beta \gamma + \gamma \alpha\right) \alpha \beta \gamma$
$= {\alpha}^{3} {\beta}^{2} \gamma + {\beta}^{3} {\gamma}^{2} \alpha + {\gamma}^{3} {\alpha}^{2} \beta + {\alpha}^{3} \beta {\gamma}^{2} + {\beta}^{3} \gamma {\alpha}^{2} + {\gamma}^{3} \alpha {\beta}^{2} + 3 {\alpha}^{2} {\beta}^{2} {\gamma}^{2}$
So:
${\alpha}^{3} {\beta}^{3} + {\beta}^{3} {\gamma}^{3} + {\gamma}^{3} {\alpha}^{3}$
$= {\left(\alpha \beta + \beta \gamma + \gamma \alpha\right)}^{3} - 3 \left(\alpha + \beta + \gamma\right) \left(\alpha \beta + \beta \gamma + \gamma \alpha\right) \alpha \beta \gamma + 3 {\left(\alpha \beta \gamma\right)}^{2}$
$= {b}^{3} - 3 a b c + 3 {c}^{2}$
Finally:
${\alpha}^{3} {\beta}^{3} {\gamma}^{3} = {\left(\alpha \beta \gamma\right)}^{3} = - {c}^{3}$
Hence the required cubic equation is:
${x}^{3} + \left({a}^{3} - 3 a b + 3 c\right) {x}^{2} + \left({b}^{3} - 3 a b c + 3 {c}^{2}\right) x + {c}^{3} = 0$
$\textcolor{w h i t e}{}$
Footnote
If you would like to see a more advanced application of symmetric polynomials, you may like to take a look at this one: https://socratic.org/s/aCWXbG2b |
# What are the eigenvalues of a rotation matrix?
## What are the eigenvalues of a rotation matrix?
The eigenvalues of A are roots of the characteristic polynomial p(t). p(t)=t2−(2cosθ)t+1=0. t=2cosθ±√(2cosθ)2−42=cosθ±√cos2θ−1=cosθ±√−sin2θ=cosθ±isinθ=e±iθ. cosθ±isinθ=e±iθ.
How do you find the eigen value of a matrix?
In order to find eigenvalues of a matrix, following steps are to followed:
1. Step 1: Make sure the given matrix A is a square matrix.
2. Step 2: Estimate the matrix.
3. Step 3: Find the determinant of matrix.
4. Step 4: From the equation thus obtained, calculate all the possible values of.
5. Example 2: Find the eigenvalues of.
Does the rotation matrix have eigenvalues and eigenvectors?
Every rotation matrix must have this eigenvalue, the other two eigenvalues being complex conjugates of each other. It follows that a general rotation matrix in three dimensions has, up to a multiplicative constant, only one real eigenvector.
### What is the eigenvector of rotation matrix?
that any vector that is parallel to the axis of rotation is unaffected by the rotation itself. This last statement can be expressed as an eigenvalue equation, R(n,θ)n = n . (22) Thus, n is an eigenvector of R(n,θ) corresponding to the eigenvalue 1.
Do rotations have eigenvectors?
It turns out that once you allow complex numbers into your linear algebra, rotations do have eigenvectors.
Does a rotation matrix have real eigenvalues?
Eigenvalues of a general rotation in R2. ∣ ∣ ∣ ∣ cosθ − λ −sinθ sinθ cosθ − λ ∣ ∣ ∣ ∣ = (cosθ − λ)2 + sin2 θ. (cosθ − λ)2 = −sin2 θ 1 Page 2 has no real solutions. Thus, there are no real eigen- values for rotations (except when θ is a multiple of π, that is the rotation is a half turn or the identity).
#### What are the rotation formulas?
Rotation Formula
Type of Rotation A point on the Image A point on the Image after Rotation
Rotation of 90° (Clockwise) (x, y) (y, -x)
Rotation of 90° (Counter Clockwise) (x, y) (-y, x)
Rotation of 180° (Both Clockwise and Counterclockwise) (x, y) (-x, -y)
Rotation of 270° (Clockwise) (x, y) (-y, x)
Does the rotation matrix have real eigenvalues?
Rotations are important linear operators, but they don’t have real eigenvalues. They will, how- ever, have complex eigenvalues. Eigenvalues for linear operators are so important that we’ll extend our scalars from R to C to ensure there are enough eigenvalues.
Do rotation matrices only have complex eigenvalues?
## What are the 3 types of rotation?
These rotations are called precession, nutation, and intrinsic rotation.
How do you check if a matrix is a rotation matrix?
It is possible to have a rotation matrix with a det of 1 (eg. 2 flipped axis). A rotation matrix M does not need to satisfy det(M)=1. This is only true if M describes a proper rotation; otherwise it describes an improper rotation, and det(M)=−1.
What order do you multiply rotation matrices?
To multiply a matrix and a vector, first the top row of the matrix is multiplied element by element with the column vector, then the sum of the products becomes the top element in the resultant vector. The next row times the column vector gives the middle element of the resultant and likewise for the third.
### What is the formula for rotations?
Does it matter what order you translate and rotate?
In a composite transformation, the order of the individual transformations is important. For example, if you first rotate, then scale, then translate, you get a different result than if you first translate, then rotate, then scale.
What is the correct order to get the transformed vector?
The order of the composite transformation is first scale, then rotate, then translate.
#### How do you rotate a rotation matrix?
To rotate counterclockwise about the origin, multiply the vertex matrix by the given matrix. Example: Find the coordinates of the vertices of the image ΔXYZ with X(1,2),Y(3,5) and Z(−3,4) after it is rotated 180° counterclockwise about the origin. Write the ordered pairs as a vertex matrix.
How to determine the eigenvalues of a matrix?
Determine the eigenvalues of the given matrix A using the equation det (A – λI) = 0,where I is equivalent order identity matrix as A.
• Substitute the value of λ1 in equation AX = λ1 X or (A – λ1 I) X = O.
• Calculate the value of eigenvector X which is associated with eigenvalue λ1.
• Repeat steps 3 and 4 for other eigenvalues λ2,λ3,…as well.
• Do real matrices always have real eigenvalues?
Therefore, any real matrix with odd order has at least one real eigenvalue, whereas a real matrix with even order may not have any real eigenvalues. The eigenvectors associated with these complex eigenvalues are also complex and also appear in complex conjugate pairs. Let λi be an eigenvalue of an n by n matrix A.
## How to plot complex eigenvalues of a matrix?
function [e] = plotev(n) % [e] = plotev(n) % % This function creates a random matrix of square % dimension (n). It computes the eigenvalues (e) of % the matrix and plots them in the complex plane. % A = rand(n); % Generate A e = eig(A); % Get the eigenvalues of A close all % Closes all currently open figures.
What do the eigenvalues and vectors of a matrix mean?
If A is Hermitian and full-rank,the basis of eigenvectors may be chosen to be mutually orthogonal.
• The eigenvectors of A−1 are the same as the eigenvectors of A.
• Eigenvectors are only defined up to a multiplicative constant.
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# Algebra solving equations
In this blog post, we will show you how to work with Algebra solving equations. Let's try the best math solver.
## The Best Algebra solving equations
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That way you can check your answer without having to recalculate the problem on paper first. Many people like this tool because it saves them time—they can just scan their paper and get their answer instantly! The other benefit is that it helps you stay organized, because you can quickly scan all your papers together in one place. Another great thing about a math problem scanner is that you don’t have to buy any special equipment. So it’s perfect for anyone who wants to try it out for the first time!
While it works in all cases, it can get tricky when working with negative numbers as well. If your equation has both positive and negative numbers in it, then you will need to do some basic algebraic gymnastics. However, if neither of those situations apply, then this technique will be your best option. Let’s take a look at an example: Equation> Value> Log(x) = Result> Value> Why?> So we first use our log function to solve for x: Equation> Value> = Result> Value> Next we plug the value of x into the original equation: Equation> Value> = Result> Value> We now compare the two values and see if they equal each other: Equation>
First, convert feet to meters: 12 feet = 1 meter. Then, multiply both sides of the equation by 2: (12) meters * 2 = 36 meters Now, divide both sides by 36: (12/36) * 12 = 4.5 gallons For other types of problems where square roots can help, see below.
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# The Derivative of Natural Log... Why Is It "Natural"? And Many Examples...
Do you know the formula for the derivative of natural log? And do you know why the natural logarithm is "natural"? In this page we'll see the intuition behind this formula and learn how to solve derivatives using it.
The derivative of the logarithm with base "e" is a fascinating subject in calculus. Look at this table, it might spark your curiosity:
We’ll show now that the spot with ? should be filled with the natural logarithm of x. You can either watch the video or read below.
#### Derivative of Natural Log
Some tools that we’ll use are the properties of the logarithm:
We’ll also use the following property of continuous functions:
that is, that we can exchange the order between a continuous function and a limit sign. To find the derivative of the natural logarithm we’ll first write the definition of the derivative:
And now we apply the second property of logarithms in the list:
The argument of the logarithm we can also write as:
So, we have that:
Now we use the third property of logarithms:
Now we’ll use the clever substitution method we use to solve limits involving number e. We make:
This is equivalent to:
Let’s note something now. We are working with a limit where x approaches 0. If x remains fixed and Delta x approaches 0, the fraction Delta x/x will approach zero. That means that 1/n approaches zero (because it equals Delta x/x) That could only be possible if n approaches infinity.
All that we said resumes to:
So, doing the substitutions in our limit, we’ll have a function of the variable n:
We can also write this as:
Now we apply property three of logarithms again:
Because x is not a function of n we can take it out of the limit sign:
Now we’ll use the property of continuous functions that I noted above.
This property lets us change place between the function and the limit sign:
The argument of the natural logarithm is the definition of number e:
So, we’re left with:
But lne = 1. So, finally:
This result is pretty amazing, and quite unexpected. Let’s solve now some problems related to this concept.
## Derivative of Natural Log, Example 1
Let's solve some derivatives involving the natural logarithm. Let's find the derivative of:
Looks complicated? Here we'll apply the chain rule. What is the derivative of the outside function? It is one over the argument:
And the derivative of cos(x) is -sin(x):
## Derivative of Natural Log, Example 2
Let's derive:
We'll apply the chain rule:
And the derivative of the inside function is just a:
## Derivative of Natural Log, Example 3
Let's derive:
What would happen if you applied the chain rule directly? This would become a very long and hairy problem.
Instead, we'll use a trigonometric trick. Inside the square root, let's multiply both the numerator and denominator by 1+sin(x):
What we have in the denominator is cos squared of x. So:
This looks much easier, doesn't it? Now we can use the chain rule:
Now we'll apply the product rule in the second factor:
And now it is just algebra:
Performing the sum:
We can simplify a bunch of things:
And finally we have:
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# Similar Triangles (video lessons, examples and step-by-step solutions)
In these lessons, we will learn
Related Pages
Similar Triangles: Angle-Angle Criterion
Congruent Triangles
Geometry Lessons
Similar triangles have the following properties:
The following diagrams show similar triangles. Scroll down the page for more examples and solutions on how to detect similar triangles and how to use similar triangles to solve problems.
If triangles are similar then the ratios of the corresponding sides are equal.
When the ratio is 1 then the similar triangles become congruent triangles (same shape and size).
We can tell whether two triangles are similar without testing all the sides and all the angles of the two triangles. There are three rules or theorems to check for similar triangles.
As long as one of the rules is true, it is sufficient to prove that the two triangles are similar.
Two triangles are similar if any of the following is true.
AA (Angle-Angle)
The two angles of one triangle are equal to the two angles of the other triangle. (AA rule)
SAS (Side-Angle-Side)
Two sides are in the same proportion, and their included angle is equal. (SAS rule)
SSS (Side-Side-Side)
The three sides are in the same proportion. (SSS rule).
Scroll down the page for more examples and solutions on the AA Rule, SAS rule, SSS rule and how to solve problems using similar triangles.
The Angle-Angle (AA) rule states that
If two angles of one triangle are equal to two angles of another triangle, then the triangles are similar.
This is also sometimes called the AAA rule because equality of two corresponding pairs of angles would imply that the third corresponding pair of angles are also equal.
Example 1:
Given the following triangles, find the length of s
Solution:
Step 1: The triangles are similar because of the AA rule.
Step 2: The ratios of the lengths are equal.
Step 3: Cross multiplying: 6s = 18 ⇒ s = 3
Answer: The length of s is 3.
The Side-Angle-Side (SAS) rule states that
If the angle of one triangle is the same as the angle of another triangle and the sides containing these angles are in the same ratio, then the triangles are similar.
Example 2:
Given the following triangles, find the length of s
Solution:
Step 1: The triangles are similar because of the RAR rule.
Step 2: The ratios of the lengths are equal.
Answer: The length of s is 3.
The Side-Side-Side (SSS) rule states that
If two triangles have their corresponding sides in the same ratio, then they are similar.
How to solve similar triangles using AA, SAS, and SSS?
The following video shows the AA, SAS and SSS similarity theorem and how to use them.
You can show that two triangles are similar when you know the relationships between only two or three pairs of the corresponding parts.
Angle-Angle Similarity (AA) Postulate: If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.
Side-Angle-Side Similarity (SAS) Theorem: If an angle of one triangle is congruent to an angle of a second triangle, and the sides that include the two angles are proportional, then the triangles are similar.
Side-Side-Side Similarity (SSS) Theorem: If the corresponding sides of two triangles are proportional, then the triangles are similar.
Examples:
Sometimes you can use similar triangle to find lengths that cannot be measured easily using a ruler or other measuring device. You can use indirect measurement to find lengths that are difficult to measure directly. One method of indirect measurement uses the fact that light reflects off a mirror at the same angle at which it hits the mirror.
Example:
Before rock climbing, Darius wants to know how high he will climb. He places a mirror on the ground and walks backward until he can see the top of the cliff in the mirror. What is the height of the cliff?
How to determine whether two triangles are similar using AA similarity?
The Third Angle Theorem states that if two angles in one triangle are congruent to two angles in another triangle, the third angle must be congruent also.
The AA Similarity Postulate: If two angles in one triangle are congruent to two angles in another triangle, then the triangles are similar.
Example:
Determine if the following triangles are similar.
How to determine whether two triangles are similar using SSS and SAS similarity?
If the corresponding sides of two triangles are proportional, then the two triangles are similar.
If the two sides of two triangles are proportional and the included angles are congruent, the the triangles are similar.
Example:
Determine if the following triangles are similar.
How to solve for unknown values using the properties of similar triangles?
Example:
Given that the triangles are similar, determine x and y.
How to find the length of a side of a triangle using similar triangles (Overlapping)?
Indirect Measurement Using Similar Triangles
How to use the properties of similar triangles to determine the height of a tree?
An application of similar triangles is to be able to determine lengths indirectly. Remember the sides of similar triangles are proportional.
Examples of how to solve problems using similar triangles
Example 1: Fred needs to know how wide a river is. He takes measurements as shown in the diagram. Determine the river’s width.
Example 2: Determine the ratio of the areas of the two similar triangles.
Example 3: If the area of the smaller triangle is 20 m2, determine the area of the larger triangle.
How to use similar triangles to solve shadow problems?
Try the free Mathway calculator and
problem solver below to practice various math topics. Try the given examples, or type in your own |
# PROVING THE GIVEN VERTICES OF A PARALLELOGRAM USING SLOPE
Proving the Given Vertices of a Parallelogram Using Slope :
Here we are going to see, how to prove the given vertices of parallelogram using slope.
## Proving the Given Vertices of a Parallelogram Using Slope - Questions
Question 1 :
Show that the given points form a parallelogram :
A(2.5, 3.5) , B(10,-4), C(2.5,-2.5) and D(-5,5)
Solution :
In a parallelogram, opposite sides will be parallel, by proving that slope of opposite sides are equal, we may say that opposite sides are parallel.
Slope (m) = (y2 - y1)/(x2 - x1)
A(2.5, 3.5) , B(10,-4),C(2.5,-2.5) and D(-5,5)
Slope of ABm = (-4-3.5)/(10-2.5) = -7.5/7.5m = -1 Slope of CD :m = (5-(-2.5))/(-5-2.5) = (5+2.5)/(-7.5)m = 7.5/(-7.5)m = -1 Slope of BCm = (-2.5+4)/(2.5-10) = 1.5/(-7.5)= -15/75m = -1/5 Slope of DAm = (5-3.5)/(-5-2.5) = 1.5/(-7.5)= -15/75m = -1/5
Slope of AB = Slope of CD
Slope of BC = Slope of DA
Hence the given points form a parallelogram.
Question 2 :
If the points A(2, 2), B(–2, –3), C(1, –3) and D(x, y) form a parallelogram then find the value of x and y.
Solution :
Since the given points form a parallelogram,
Slope of AB = Slope of CD
Slope of BC = Slope of DA
A(2, 2), B(–2, –3), C(1, –3) and D(x, y)
Slope of ABm = (-3-2)/(-2-2) = -5/(-4)m = 5/4 ---(1) Slope of CD :m = (y-(-3))/(x - 1)m = (y + 3)/(x - 1) ---(2) Slope of BCm = (-3+3)/(1+2)m = 0/3m = 0----(3) Slope of DAm = (y - 2)/(x - 2) ----(4)
(1) = (2)
5/4 = (y + 3)/(x - 1)
5(x - 1) = 4(y + 3)
5x - 5 = 4y + 12
5x - 4y = 12 + 5
5x - 4y = 17 -----(5)
(3) = (4)
0 = (y - 2)/(x - 2)
y - 2 = 0
y = 2
By applying the value of y in (5), we get
5x - 4(2) = 17
5x - 8 = 17
5x = 17 + 8
5x = 25
x = 5
Hence the values of x and y are 5 and 2 respectively.
Question 3 :
Let A(3, -4), B(9, -4) , C(5, -7) and D(7, -7) . Show that ABCD is a trapezium.
Solution :
A trapezium will always contain two parallel sides and two non parallel sides.
Slope of AB :
m = (-4 +4)/(9-3)
m = 0/6 = 0
Slope of BC :
m = (-7 +4)/(5-9)
m = -3/(-4) = 3/4
Slope of CD :
m = (-7 + 7)/(7 - 5)
m = 0/2 = 0
Slope of DA :
m = (-7 + 4)/(7 - 3)
m = -3/4
In the given trapezium, sides AB and CD are parallel.
Question 4 :
A quadrilateral has vertices A(-4,-2) , B(5,-1) , C(6,5) and D(-7,6) . Show that the mid-points of its sides form a parallelogram
Solution :
Midpoint of the side AB = P
= (-4 + 5)/2, (-2 - 1)/2
= P (1/2, -3/2)
Midpoint of the side BC = Q
= (5 + 6)/2, (-1 + 5)/2
= Q (11/2, 4/2)
= Q (11/2, 2)
Midpoint of the side CD = R
= (-7 + 6)/2, (6 + 5)/2
= R (-1/2, 11/2)
Midpoint of the side DA = S
= (-7 - 4)/2, (-2 + 6)/2
= S (-11/2, 4/2)
= S (-11/2, 2)
## Slopes of Opposite Sides :
Slope of PQ = (7/2) / 5 ==> 7/10
Slope of RS = (-7/2)/(-5) = 7/10
Slope of PQ = (7/2) / (-5) ==> -7/10
Slope of RS = (7/2)/(-5) = -7/10
Question 5 :
PQRS is a rhombus. Its diagonals PR and QS intersect at the point M and satisfy QS = 2PR. If the coordinates of S and M are (1,1) and (2,-1) respectively, find the coordinates of P.
Solution :
After having gone through the stuff given above, we hope that the students would have understood, "Proving the Given Vertices of a Parallelogram Using Slope".
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# How To Teach Division Conceptually
Teaching division is a complex skill with a lengthy formula for students to remember. But learning and more importantly understanding division doesnβt start with memorizing procedural steps. In fact, teaching the standard algorithm of division should be the last thing you teach your students.
#### How To Start Your Division Unit
To ensure I target conceptual understanding I ask my students a simple question to launch our division unit.
What is division?
βDivision is seeing how many times a number can go into a number.β
βDividing is when you make a number smaller.β
βDivision is the saying does McDonaldβs serve burgers.β
Such responses as the three above illustrate a lack of understanding of the concept of division.
I donβt give my students the answer. Instead I make them work for it.
I give my students a bag of centimeter cubes. My students sit in groups of 4-5, so I put a specific number of cubes in the bag to ensure that there will be a remainder amongst students.
I ask them to divide the cubes amongst the people at their table.
This sparks great discussion. I walk around and listen. Your students will look to you for the answer. They will stare at you with big eyes screaming βHELP US!β Donβt cave in. Let them explore.
βHow many should each person get?β
βBrian has more than me.β
βWho gets the extra cubes?β
βEveryone needs to have the same amount, thatβs what divide means!β
BINGO. You want each group to come to the conclusion that every person should have the same number of cubes sitting in front of them.
As each group is arriving at this conclusion, I go over to each table and ask them to create a division equation that matches their work. I also ask them to label each number in their equation.
Now when I ask them what division is, I get the response Iβm looking for. Division is putting objects into equal groups or putting the same amount of pieces into groups.
#### Take The Exploration To The Next Level
The next day I hand my groups another bag. This time the bag has base ten blocks. I give them the same prompt.
Divide the bag amongst the people at your table.
An even greater debate commences.
βHow are we supposed to divide these?β
βI donβt get it!β
Finally, a group will approach me. βCan we trade a hundreds flat for 10 tens?β Sure! I complete the trade and the group forges on. Other groups start to catch up and pretty soon my room is transformed into a barter station.
These are simple but highly effective lessons to implement in your classroom. Donβt start with having students memorize steps or a weird saying about burgers. After your students conceptualize division then you can teach them the formula.
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# Limits Calculus – Definition, Properties, and Graphs
Limits are the foundation of calculus – differential and integral calculus. Predicting and approximating the value of a certain set of quantities and even functions is an important goal of calculus. This means that learning about limits will pave the way for a stronger foundation and better understanding of calculus.
Limits reflect the value that a given function approaches when it’s near a certain point.
Limits are most helpful when we want to find a function’s value as it approaches a restricted value. Isn’t it amazing how we can approximate a value as $x$ approaches infinity through different math concepts?
Needless to say, if you want to appreciate higher math classes, mastering limits and their foundations will be essential, and that’s what we’re going to explore in this article.
In this article, we’ll explore the fundamental definition of limits, refresh what we’ve learned about evaluating limits, recall the different limit laws, and visualize limits on an $xy$-plane.
## What is a limit?
The definition of limit can be summarized by the equation:
\begin{aligned}\lim_{x \rightarrow a} f(x) &= L \end{aligned}
Let’s say we have $f(x)$ and it’s close to a unique number, $L$, as $x$ approaches $a$ on both sides, we can define the limit of $f(x)$ as $x$ approaches $a$ is represented by $L$.
We can better understand its definition by observing a given function’s graph as the input value approaches or is near a certain value.
The graph of $f(x) = \dfrac{1}{x^2 + 4}$ is as shown above and we can see that when $x$ approaches a significantly large value, $f(x)$ approaches the $x$-axis. This means that as $x \rightarrow \infty$, $f(x) \rightarrow 0$.
We can actually observe a similar behavior when $x$ approaches a significantly small value – the function approaches zero.
We can say that the limit of $f(x)$ as $x$ approaches infinity is equal to zero. This confirms the definition of limits.
Let’s apply this definition to understand why $\lim_{x \rightarrow 0} \dfrac{\sin x}{x} = 1$ given the graph of $y = \dfrac{\sin x}{x}$ as shown below.
Seeing that the function is near $1$ when $x$ approaches $0$ from either side, we can see that the function’s limit as $x$ approaches $0$ is equal to $1$, indeed. This definition of a limit is informal since we’re only considering limits as $x$ is “approaching” or is “closer to” a specific value.
In short, we are not definite with our idea of proximity. This is why we’ll also explore the more authoritative and precise definition of a limit in Calculus. This can help us understand the technicalities of different properties that involve limits.
### A precise definition of a limit
We’ll lay out the formal definition of a limit in this section and break down each component.
If $L$ represents the limit of $f(x)$ as $x$ approaches $a$, if it satisfies the following conditions:
1. i) Although it’s not necessary for $f(x)$ to be defined at $x = a$, $f(x)$ must still be defined for some interval that contains $x = a$.
2. ii) For every $\epsilon > 0$, we can find a $\delta$ such that $\delta$ is also greater than $0$, so that for all the values of $x$ within the domain of $f(x)$, we have the following:
\begin{aligned}|x -a|< \delta\\\Rightarrow |f(x) – L| < \epsilon \end{aligned}
Let’s break down the conditions to better understand the formal definition of limits.
• The absolute value expression, $|x -a|$, represents the distance between $x$ and $a$, so $|x -a| < \delta$ means that we want the distance to between the two to be less than $\delta$.
• The value, $\epsilon$, represents the proximity of $\epsilon$ with $f(x)$ so we can approximate the value of $L$.
• Meanwhile, $\delta$ represents the distance or how close you want $x$ and $a$ to be.
This means that we can determine and confirm the limit of a given function by assuming that we have $\epsilon > 0$, and so, we need to find $\delta$ so that the inequalities hold true.
For example, if we want to prove that $\lim_{x \rightarrow 6} -3x + 2 = -16$ using the formal definition of a limit, we have the following components:
\begin{aligned}f(x) &= -3x + 2\\a&= 6\\L &= -16\end{aligned}
We want to make sure that $|f(x) – L| < \epsilon$, where $\epsilon$ is the accepted minimum difference we have set.
\begin{aligned}|f(x) – L| &= |(-3x + 2) – (-16)|\\&= |-3x + 2 + 16|\\&= |-3x + 18|\\&= |-3| \cdot |x -6|\\&= 3|x – 6|\end{aligned}
Since we have $|f(x) – L| < \epsilon$ and consequently, $3|x – 6| < \epsilon$.
\begin{aligned}3|x – 6| < \epsilon\\|x – 6| < \dfrac{1}{3}\epsilon\end{aligned}
For us to confirm that $\lim_{x \rightarrow 6} f(x) = -16$, we can use $\delta = \dfrac{1}{3} \epsilon$. Hence, we have the inequality shown below.
\begin{aligned}|x – 6| < \dfrac{1}{3}\epsilon\\|x- 6|<\delta\phantom{x}\end{aligned}
Both $\epsilon$ and $\delta$ will always be greater than zero. Hence, we’ve satisfied the condition that when $|x – 6| < \dfrac{1}{3} \epsilon$ and consequently, $|x – 6| < \delta$, we can guarantee that $|(-3x + 2) – (-16)|< \epsilon$. This confirms the fact that $\lim_{x\rightarrow 6} -3x + 2$ is indeed $-16$.
Now that we’ve explored the informal and formal definitions of limits let’s review the different methods we can apply to evaluate the limits of different functions.
## How to do limits?
There are different ways for us to evaluate limits – and our chosen method will depend on what we’re given. We can evaluate the limits of a given function by its table of values, graphs, and expressions.
This section will help you review your past knowledge of limits from your precalculus classes and learn new techniques to evaluate complicated expressions and functions’ limits.
### Observing the limit of a function using its graph or table of values
When given the function’s graph or the table of values to find the limit of a graph, it’s helpful to use the fact that we’re looking for the value of $f(x)$ as $x$ approaches $a$ from both sides.
We’ve shown how we can determine the limit of functions using their graphs, so let’s try finding $\lim_{x \rightarrow 4} -\dfrac{2}{x – 2}$ using the table of values for $y = -\dfrac{2}{x- 2}$. Make sure to use values of $x$ that are significantly close to $x = 4$ as shown below.
\begin{aligned}\boldsymbol{x}\end{aligned} \begin{aligned}\boldsymbol{y = -\dfrac{2}{x – 2}}\end{aligned} \begin{aligned}\boldsymbol{x}\end{aligned} \begin{aligned}\boldsymbol{y =-\dfrac{2}{x – 2} }\end{aligned} \begin{aligned}3.9\end{aligned} \begin{aligned}-1.0526\end{aligned} \begin{aligned}4.0001\end{aligned} \begin{aligned}-1.0000\end{aligned} \begin{aligned}3.99\end{aligned} \begin{aligned}-1.0050\end{aligned} \begin{aligned}4.001\end{aligned} \begin{aligned}-0.9995\end{aligned} \begin{aligned}3.999\end{aligned} \begin{aligned}-1.0005\end{aligned} \begin{aligned}4.01\end{aligned} \begin{aligned}-0.9950\end{aligned} \begin{aligned}3.9999\end{aligned} \begin{aligned}-1.0001\end{aligned} \begin{aligned}4.1\end{aligned} \begin{aligned}-0.9524\end{aligned}
We can see that as $x$ approaches $4$ from the left (these are the values that are less than $4$) , we can see that $f(x)$ is approaches $-1$. The same behavior happens $x$ approaches $4$ from the right. This means that $\lim_{x \rightarrow 4} -\dfrac{2}{x – 2} = -1$.
This approach is only convenient when the graph or the table of values is given. However, there are instances that it’ll be tedious for us to graph or construct a function’s table of values. This is why it’s important we can evaluate limits using different algebraic properties.
### Applying the limit laws and evaluating limits
It’s important to master the limit laws we’ve established in the past to understand how we can evaluate limits using different algebraic manipulations we’ve learned in our precalculus classes.
Here’s a summary of the different limit laws we can apply when evaluating limits and checking the corresponding examples to ensure you can apply these laws whenever necessary.
Limit Law Algebraic Definition Example Constant Law $\lim_{x\rightarrow a} c = c$ $\lim_{x\rightarrow 5} 6 = 6$ Identity Law $\lim_{x\rightarrow a} x = a$ $\lim_{x\rightarrow 2} x = 2$ Addition Law $\lim_{x\rightarrow a} [f(x) + g(x)] = \lim_{x\rightarrow a} f(x) + \lim_{x\rightarrow a} g(x)$ $\lim_{x\rightarrow 3} [(x – 2)+ (3x)] = \lim_{x\rightarrow 3} (x – 2) + \lim_{x\rightarrow 3} 3x$ Subtraction Law $\lim_{x\rightarrow a} [f(x) – g(x)] = \lim_{x\rightarrow a} f(x) – \lim_{x\rightarrow a} g(x)$ $\lim_{x\rightarrow 3} [(x – 2)+ (3x)] = \lim_{x\rightarrow 3} (x – 2) – \lim_{x\rightarrow 3} 3x$ Coefficient Law $\lim_{x\rightarrow a} cf(x) = c \lim_{x\rightarrow a} f(x)$ $\lim_{x\rightarrow 6} \sqrt{4} x = \sqrt{4} \lim_{x\rightarrow 6} x$ Product Law $\lim_{x\rightarrow a} [f(x) \cdot g(x)] = \lim_{x\rightarrow a} f(x) \cdot \lim_{x\rightarrow a} g(x)$ $\lim_{x\rightarrow 3} [(x – 2) \cdot (3x)] = \lim_{x\rightarrow 3} (x – 2) \cdot \lim_{x\rightarrow 3} 3x$ Quotient Law $\lim_{x\rightarrow a} \dfrac{f(x)}{g(x)} = \dfrac{\lim_{x\rightarrow a} f(x) }{\lim_{x\rightarrow a} g(x) }$ $\lim_{x\rightarrow 3} \dfrac{x – 2}{3x} = \dfrac{x – 2}{3x}$ Power Law $\lim_{x\rightarrow a} [f(x)]^n = \left[\lim_{x\rightarrow a} f(x) \right]^n$ $\lim_{x\rightarrow 4} [3(x + 2)]^2 = \left[\lim_{x\rightarrow 4} 3(x + 2)\right]^2$ Root Law $\lim_{x\rightarrow a} \sqrt[n]{f(x)} = \sqrt[n]{ \lim_{x\rightarrow a} f(x)}$ $\lim_{x\rightarrow 2} \sqrt[3]{4(x + 2)} = \sqrt[3]{ \lim_{x\rightarrow 2} 4(x + 2)}$
These limit laws, alone, are not enough to evaluate the limits of complex expressions, so we’ve come up with different techniques we can apply to evaluate limits. These techniques are algebraic properties we’ve actually learned in the past, so it’s also a great avenue for us to review our previous knowledge.
As a refresher, let’s discuss the most commonly applied techniques when evaluating limits:
• Direct substitution: Whenever possible, substitute $a$ into $f(x)$ to find $\lim_{x \rightarrow a} f(x)$.
\begin{aligned}\lim_{x \rightarrow a} f(x) &= f(a)\\\\\lim_{x \rightarrow \color{green}3} x^2 – 1 &= ({\color{green}3})^2 -1\\&= 9 -1\\&= 8 \end{aligned}
• Factoring the function: When direct substitution leads to $\dfrac{0}{0}$ or $\dfrac{k}{0}$, we can try factoring out the expression to see if we can cancel out common factors.
\begin{aligned}\lim_{x \rightarrow 1} \dfrac{x^2 -1}{x^2 – 3x + 2} &=\lim_{x \rightarrow 1} \dfrac{(x -1)(x + 1)}{(x -1)(x – 2)}\\&= \lim_{x \rightarrow 1} \dfrac{\cancel{(x – 1)}(x +1)}{\cancel{(x – 1)}(x -2)}\\&= \lim_{x \rightarrow \color{green} 1} \dfrac{x + 1}{x -2}\\&= \dfrac{{\color{green}1} + 1}{{\color{green}1} -2}\\&= -2\end{aligned}
• Reversing a rationalized radical expression: When we have a radical expression, we can multiply the numerator and denominator by the numerator’s conjugate.
\begin{aligned} \lim_{x \rightarrow 4}\dfrac{\sqrt{x} – 2}{x – 4} &= \lim_{x \rightarrow 4}\dfrac{\sqrt{x} – 2}{x – 4} \cdot \dfrac{\color{green} \sqrt{x} + 2}{\color{green} \sqrt{x} + 2}\\&= \lim_{x \rightarrow 4} \dfrac{(\sqrt{x} – 2)({\color{green} \sqrt{x} + 2})}{(x – 4)({\color{green} \sqrt{x} + 2})}\\&= \lim_{x \rightarrow 4} \dfrac{(\sqrt{x})^2 – (2)^2}{(x – 4)(\sqrt{x} +2)},\phantom{xxx} (a -b)(a +b) = a^2 – b^2\\ &=\lim_{x \rightarrow 4} \dfrac{\cancel{x – 4}}{\cancel{(x -4)}(\sqrt{x} +2)}\\&= \lim_{x \rightarrow 4} \dfrac{1}{\sqrt{x} +2}\\&= \dfrac{1}{\sqrt{\color{green}4}+ 2}\\&= \dfrac{1}{4}\end{aligned}
• Be creative and manipulate algebraic expressions: There are instances where we need to be creative and apply different algebraic properties to evaluate a function’s limits.
### Evaluating one-sided limits and continuous functions
There are instances where $f(x)$ approaches different values as $x$ approaches $a$ from the left and from the right. We call these limits as one-sided limits – $\lim_{x \rightarrow a^{-}} f(x)$ and $\lim_{x \rightarrow a^{+}} f(x)$.
Here’s an example of a function, $f(x)$, where the limits may vary as $x$ approaches from the left and when $x$ approaches from the right.
• One-sided limit from the left: to find $\lim_{x \rightarrow a^{-} f(x)}$, we simply observe the value that $f(x)$ approaches when $x$ approaches $a$ from the left. This means that $\lim_{x\rightarrow 3^{-}} f(x) = -2.25$.
• One-sided limit from the right: similarly,to find $\lim_{x \rightarrow a^{+}}$, we find the value that $f(x)$ approaches when $x$ approaches $a$ from the right. Hence, $\lim_{x\rightarrow 3^{+}} f(x) = 1.5$.
Knowing the one-sided limits of a function can help us determine whether it is continuous or not. To learn more about continuous functions, check out this link here. What’s important is to remember the fact that when $\lim_{x \rightarrow a^{-} f(x)} = \lim_{x \rightarrow a^{+} f(x)}$, the function is continuous.
### Using the Squeeze theorem to evaluate complex functions
There are instances where evaluating limits using algebraic manipulation can be challenging – especially when these functions are oscillating. When this happens, we can use the Squeeze theorem to evaluate these limits. As a refresher, we can apply these steps below:
• Begin by confirming three functions such that $g(x) \leq f(x) \leq h(x)$, the limits of $g(x)$ and $h(x)$ as $x \rightarrow a$ are easy to evaluate, and these limits are equal.
• Start with an inequality that’s easy to manipulate, then modify it so that we end with $f(x)$ – the actual function that we’re evaluating.
• Evaluate the limits of the inequalities’ two ends and use the limits to find $\lim_{x\rightarrow a} f(x)$.
We can use the Squeeze theorem to determine limits of complex trigonometric functions such as $\lim_{x\rightarrow 0} \dfrac{\sin x}{x}$ and $\lim_{x\rightarrow 0} \dfrac{ 1- \cos x}{x}$. Why don’t we go ahead and summarize all trigonometric functions’ limits?
$\boldsymbol{\lim_{x \rightarrow a} f(x)}$ $\lim_{x \rightarrow a} \sin x = \sin a$ $\lim_{x \rightarrow a} \csc x = \csc a$ $\lim_{x \rightarrow a} \cos x = \cos a$ $\lim_{x \rightarrow a} \sec x = \sec a$ $\lim_{x \rightarrow a} \tan x = \tan a$ $\lim_{x \rightarrow a} \cot x = \cot a$ Derived from the Squeeze Theorem $\lim_{x\rightarrow 0} \dfrac{\sin x}{x} = 1$ $\lim_{x\rightarrow 0} \dfrac{ 1- \cos x}{x} = 0$
## How to graph limits?
Now that we’ve thoroughly explored the concepts of limits and the different methods we can evaluate a function’s limits, we can reverse the process and learn how we can graph limits depending on the limits’ nature.
We’ve learned that discontinuities can either be a hole, asymptotes, or jump discontinuities in our discussion of continuous functions.
• Jump discontinuities occur when the one-sided limits are not the same such as piece-wise functions.
• Holes or removable discontinuities occur when the limit still exists and can be evaluated by canceling common factors shared by the numerator and the denominator.
• Asymptotes or infinite discontinuity happens when the resulting limit of a function as $x$ approaches $a$ is equal to $\pm \infty$. When this happens, we have a vertical asymptote at $x =a$.
We can use these to graph the limits and represent them on an $xy$-coordinate system. For example, if we have a piecewise function, $f(x)=\left\{\begin{matrix}2 + \dfrac{2}{x -1},&x\leq1\\ \dfrac{x -2}{x^2 – 2x},&x >1\end{matrix}\right.$. The limits of $f(x)$ as $x$ approaches $1$ and $2$ as shown below:
Limit Representation \begin{aligned}\lim_{x\rightarrow 1} \dfrac{2}{x -1} = \infty\end{aligned} Vertical asymptote at $x = 1$. \begin{aligned}\lim_{x\rightarrow 2} \dfrac{x -2}{x^2 -2x} = \dfrac{1}{2}\end{aligned} Hole at $\left(2, \dfrac{1}{2}\right)$.
Knowing the nature of the discontinuity can help us identify how it will represent its limit’s value on the graph.
We can apply a similar approach to graph different limits – make sure to review our past knowledge on limits and continuous functions. To further solidify our knowledge of limits, let’s go ahead and try out these sample problems!
Example 1
The graph of $\dfrac{-4x^2 +x}{x}$ is shown below.
Determine the limit (if it exists) of the function as $x$ approaches the following:
a. $\lim_{x\rightarrow 0} f(x)$
b. $\lim_{x\rightarrow 0.25} f(x)$
c. $\lim_{x\rightarrow 0.5} f(x)$
Solution
When given the graph of the function, we can observe where the function is approaching as $x$ approaches a given value.
We can see that there’s a hole at $(0, 1)$, and we can see that as $x$ approaches $0$ from both sides, $f(x)$ will be approaching $1$. This means that $\lim_{x\rightarrow 0} f(x) = 1$.
We can apply a similar approach for the next two items:
• We can see that as $x$ approaches $0.25$ from both sides, the function approaches $0$. This means that $\lim_{x\rightarrow 0.25} f(x) = 0$.
• Similarly, as $x$ approaches $0.5$, we can see that $f(x)$ approaches $-1$, so $\lim_{x\rightarrow 0.5} f(x) = -1$.
Example 2
Create a table of values (feel free to use a calculator) for the following functions, then approximate the limits of the functions to three decimal places.
a. $\lim_{x\rightarrow 0} \dfrac{1- e^{-2x}}{x}$
b. $\lim_{x\rightarrow 0} (1 – x)^{\frac{1}{x}}$
c. $\lim_{x\rightarrow 16} \dfrac{4 – \sqrt{x}}{x – 16}$
Solution
Since we want to find the limit of $\dfrac{1- e^{-2x}}{x}$ as $x$ approaches $0$, so we can create a table of values that shows $x$ that are significantly close to $0$.
\begin{aligned}\boldsymbol{x}\end{aligned} \begin{aligned}\boldsymbol{y = \dfrac{1- e^{-2x}}{x}}\end{aligned} \begin{aligned}\boldsymbol{x}\end{aligned} \begin{aligned}\boldsymbol{y = \dfrac{1- e^{-2x}}{x}}\end{aligned} \begin{aligned}-0.1\end{aligned} \begin{aligned}2.214\end{aligned} \begin{aligned}0.0001\end{aligned} \begin{aligned}-2.000\end{aligned} \begin{aligned}-0.01\end{aligned} \begin{aligned}2.020\end{aligned} \begin{aligned}0.001\end{aligned} \begin{aligned}1.998\end{aligned} \begin{aligned}-0.001\end{aligned} \begin{aligned}2.002\end{aligned} \begin{aligned}0.01\end{aligned} \begin{aligned}1.980\end{aligned} \begin{aligned}-0.0001\end{aligned} \begin{aligned}2. 000\end{aligned} \begin{aligned}0.1\end{aligned} \begin{aligned}1.813\end{aligned}
We can see that as $x$ approaches $0$ from the left, $f(x)$ approaches $2.000$. Similarly, as $x$ approaches $0$ from the right, the function also approaches $2.000$. This means that $\lim_{x\rightarrow 0} \dfrac{1- e^{-2x}}{x} = 2$.
We’ll apply a similar process to approximate the limit of $(1 – x)^{\frac{1}{x}}$ as $x$ approaches $0$.
\begin{aligned}\boldsymbol{x}\end{aligned} \begin{aligned}\boldsymbol{y = (1 – x)^{\frac{1}{x} }}\end{aligned} \begin{aligned}\boldsymbol{x}\end{aligned} \begin{aligned}\boldsymbol{y = (1 – x)^{\frac{1}{x }}}\end{aligned} \begin{aligned}-0.0001\end{aligned} \begin{aligned}0.386\end{aligned} \begin{aligned}0.0001\end{aligned} \begin{aligned}0.368\end{aligned} \begin{aligned}-0.001\end{aligned} \begin{aligned}0.370\end{aligned} \begin{aligned}0.001\end{aligned} \begin{aligned}0.368\end{aligned} \begin{aligned}-0.01\end{aligned} \begin{aligned}0.368\end{aligned} \begin{aligned}0.01\end{aligned} \begin{aligned}0.366\end{aligned} \begin{aligned}-0.1\end{aligned} \begin{aligned}0.368\end{aligned} \begin{aligned}0.1\end{aligned} \begin{aligned}0.349\end{aligned}
From this, we can see that $f(x)$ approaches $0.368$ as $x$ approaches $0$ from both sides, so $\lim_{x\rightarrow 0} (1 – x)^{\frac{1}{x}} \approx 0.368$.
Let’s work on the third function, $\dfrac{4 – \sqrt{x}}{x – 16}$, and find the values of $x$ that are significantly close to $16$.
\begin{aligned}\boldsymbol{x}\end{aligned} \begin{aligned}\boldsymbol{y = \dfrac{4 – \sqrt{x}}{x – 16} }\end{aligned} \begin{aligned}\boldsymbol{x}\end{aligned} \begin{aligned}\boldsymbol{y = \dfrac{4 – \sqrt{x}}{x – 16} }\end{aligned} \begin{aligned}15.9\end{aligned} \begin{aligned}-0.125\end{aligned} \begin{aligned}16.0001\end{aligned} \begin{aligned}-0.125\end{aligned} \begin{aligned}15.99\end{aligned} \begin{aligned}-0.125\end{aligned} \begin{aligned}16.001\end{aligned} \begin{aligned}-0.125\end{aligned} \begin{aligned}15.999\end{aligned} \begin{aligned}-0.125\end{aligned} \begin{aligned}16.01\end{aligned} \begin{aligned}-0.125\end{aligned} \begin{aligned}15.9999\end{aligned} \begin{aligned}-0.125\end{aligned} \begin{aligned}16.1\end{aligned} \begin{aligned}-0.125\end{aligned}
This table of values shows that when $x$ is close to $16$, we can see that $f(x)$ is approaching $-0.125$, so $\lim_{x\rightarrow 0} (1 – x)^{\frac{1}{x}} \approx -0.125$.
Example 3
Evaluate the limit (if it exists) of the following functions.
a. $\lim_{x\rightarrow 2} -\dfrac{x}{4}$
b. $\lim_{x\rightarrow 5} \dfrac{x -5}{x^2 – 25}$
c. $\lim_{x\rightarrow 3} \dfrac{x^{4} – 81}{x – 3}$
d. $\lim_{x\rightarrow 0} \dfrac{\dfrac{1}{x + 2} – \dfrac{1}{2}}{x}$
Solution
When given the function, we can always begin by direct substitution if we think that the denominator won’t be equal to zero like the first item, $\lim_{x\rightarrow 2} -\dfrac{x}{4}$. Hence, we have the following:
\begin{aligned}\lim_{x\rightarrow 2} -\dfrac{x}{4} &= – \dfrac{\color{green} 2}{4}\\&= -\dfrac{1}{2}\end{aligned}
Direct substitution won’t apply for the second item since $x^2 – 25 = 0$ when $x = 5$. What we can do is factor the denominator first using the algebraic property, $a^2 – b^2 = (a- b)(a + b)$.
\begin{aligned}\lim_{x\rightarrow 5} \dfrac{x -5}{x^2 – 25} &=\lim_{x\rightarrow 5} \dfrac{x -5}{(x – 5)(x + 5)}\\ &=\lim_{x\rightarrow 5} \dfrac{\cancel{x -5}}{\cancel{(x – 5)}(x + 5)} \end{aligned}
We can then cancel out the common factor shared by the numerator and denominator and see if we can apply direct substitution after.
\begin{aligned}\lim_{x\rightarrow 5} \dfrac{1}{(x + 5)} &= \dfrac{1}{{\color{green}5} + 5}\\&= \dfrac{1}{10} \end{aligned}
We can a similar process to find $\lim_{x\rightarrow 3} \dfrac{x^{4} – 81}{x – 3}$
by factoring the $x^4 – 81$ completely.
\begin{aligned}\lim_{x\rightarrow 3} \dfrac{x^{4} – 81}{x – 3} &= \lim_{x\rightarrow 3} \dfrac{(x^2 -9)(x^2 + 9)}{x – 3}\\&= \lim_{x\rightarrow 3} \dfrac{(x -3)(x + 3)(x^2 + 9)}{x – 3}\\ &= \lim_{x\rightarrow 3} \dfrac{\cancel{(x -3)}(x + 3)(x^2 + 9)}{\cancel{x – 3}}\\&= \lim_{x\rightarrow 3}(x + 3)(x^2 + 9)\end{aligned}
We can now substitution $x = 3$ into the simplified expression so that we can find the limit of the function.
\begin{aligned}\lim_{x\rightarrow 3}(x + 3)(x^2 + 9) &= ({\color{green} 3} + 3)[({\color{green} 3})^2 +9]\\&= 6(18)\\&= 108\end{aligned}
Finding $\lim_{x\rightarrow 0} \dfrac{\dfrac{1}{x + 2} – \dfrac{1}{2}}{x}$ is a bit trickier – what we can first is to simplify the complex expression by multiplying both the numerator and denominator by $2$ and $(x + 2)$ as shown below.
\begin{aligned}\lim_{x\rightarrow 0} \dfrac{\dfrac{1}{x + 2} – \dfrac{1}{2}}{x} &= \lim_{x\rightarrow 0} \dfrac{\dfrac{1}{x + 2} – \dfrac{1}{2}}{x} \cdot \dfrac{\color{green} 2(x + 2)}{\color{green} 2(x + 2)} \\&= \lim_{x\rightarrow 0} \dfrac{\dfrac{1}{x+2}\cdot {\color{green} 2(x + 2)} – \dfrac{1}{2}\cdot {\color{green} 2(x + 2)}}{x \cdot {\color{green} 2(x + 2)}}\\&= \lim_{x\rightarrow 0} \dfrac{2- x – 2}{2x(x + 2)}\\&= \lim_{x\rightarrow 0} -\dfrac{x}{2x(x + 2)} \end{aligned}
We can then cancel out $x$, so the denominator will be equal to zero. Once simplified, we can then evaluate the limit of the function by direct substitution.
\begin{aligned}\lim_{x\rightarrow 0} -\dfrac{x}{2x(x + 2)} &=\lim_{x\rightarrow 0} -\dfrac{\cancel{x}}{2\cancel{x}(x + 2)}\\&= \lim_{x\rightarrow 0} -\dfrac{1}{2(x + 2)}\\&= -\dfrac{1}{2({\color{green} 0} + 2)}\\&=-\dfrac{1}{2(2)}\\&=-\dfrac{1}{4}\end{aligned}
Example 4
Evaluate the limit (if it exists) of the following functions.
a. $\lim_{x\rightarrow -2} \dfrac{\dfrac{4}{x + 4}- 2}{x + 2}$
b. $\lim_{x\rightarrow 0} \dfrac{\sqrt{9 + x} – 3}{x}$
c. $\lim_{x\rightarrow 0} \dfrac{\sin 3x}{6x}$
d. $\lim_{x\rightarrow \infty} -\dfrac{3x^2 – 8}{(x – 2)^2}$
Solution
We’ll apply algebraic manipulations and properties to evaluate the limits of the first two items. For $\lim_{x\rightarrow -2} \dfrac{\dfrac{4}{x + 4} – 2}{x + 2}$, we can begin by multiplying the numerator and denominator of the complex fraction by $(x + 4)$.
We can then apply direct substitution once we simplify the complex fraction.
\begin{aligned}\lim_{x\rightarrow -2} \dfrac{\dfrac{4}{x + 4}- 2}{x + 2} &= \lim_{x\rightarrow -2} \dfrac{\dfrac{4}{x + 4}- 2}{x + 4} \cdot \dfrac{\color{green} (x + 4)}{\color{green} (x + 4)}\\&= \lim_{x\rightarrow -2} \dfrac{\dfrac{4}{x + 2}\cdot{\color{green} (x + 4)} – 2\cdot{\color{green} (x + 4)}}{(x + 2){\color{green} (x + 4)}}\\&=\lim_{x\rightarrow -2} \dfrac{4 – 2x – 8}{(x + 2)(x + 4)}\\&= \lim_{x\rightarrow -2} \dfrac{-2\cancel{(x + 2)}}{\cancel{(x +2)}(x + 4)}\\&=\lim_{x\rightarrow -2} \dfrac{-2}{x + 4}\\ &= \dfrac{\color{green} -2}{{\color{green} -2} + 4}\\&= \dfrac{-2}{2}\\&= -1 \end{aligned}
For the second item, $\lim_{x\rightarrow 0} \dfrac{\sqrt{9 + x} – 3}{x}$, we can see that the expression is rationalized with a radical expression found in the numerator. What we can do is reverse the rationalization by multiplying both the numerator and denominator by the conjugate of $\sqrt{9+x} -3$, which is equal to $\sqrt{9 +x} + 3$.
\begin{aligned}\lim_{x\rightarrow 0} \dfrac{\sqrt{9 + x} – 3}{x} &=\lim_{x\rightarrow 0} \dfrac{\sqrt{9 + x} – 3}{x} \cdot \dfrac{\color{green} \sqrt{9 +x} + 3}{\color{green} \sqrt{9 +x} + 3}\\&= \lim_{x\rightarrow 0} \dfrac{(\sqrt{9 +x})^2 – (3)^2}{x(\sqrt{9 +x}+3)} \\&= \lim_{x\rightarrow 0} \dfrac{9 +x -9}{x(\sqrt{9 +x}) + 3}\\&= \lim_{x\rightarrow 0} \dfrac{\cancel{x}}{\cancel{x}(\sqrt{9 +x}) + 3}\\&= \lim_{x\rightarrow 0} \dfrac{1}{\sqrt{9 +x} + 3}\\&= \dfrac{1}{\sqrt{9 + {\color{green}0}}+ 3}\\&= \dfrac{1}{6} \end{aligned}
By inspecting the expression in the third item, $\lim_{x\rightarrow 0} \dfrac{\sin 6x}{3x}$, we can see that we might need to use the Squeeze theorem-derived identity, $\lim_{x\rightarrow 0} \dfrac{\sin x}{x} = 1$. But first, let’s rewrite $3x$ as $u$, so $6x$ is equal to $2u$.
\begin{aligned}3x&= u\\x&= \dfrac{1}{3}u\\\\\lim_{x\rightarrow 0} \dfrac{\sin 3x}{6x} &= \lim_{\frac{u} {3}\rightarrow 0} \dfrac{\sin u}{2u} \end{aligned}
We can factor out $\dfrac{1}{2}$ then apply the identity to find the limit of $\dfrac{\sin x}{x}$ as $x$ approaches $0$.
\begin{aligned}\lim_{\frac{u} {3}\rightarrow 0} \dfrac{\sin u}{2u} &= \dfrac{1}{2}\lim_{\frac{u} {3}\rightarrow 0} \dfrac{\sin u}{u}\\&= \dfrac{1}{2} \cdot 1\\&= \dfrac{1}{2} \end{aligned}
For the fourth item, it’s helpful to keep in mind the fact that $\lim_{x \rightarrow \infty} \dfrac{k}{x^n}= 0$. First, let’s expand the denominator using the algebraic property, $(a \pm b)^2 = a^2 \pm 2ab + b^2$.
\begin{aligned}\lim_{x\rightarrow \infty} -\dfrac{3x^2 – 8}{(x – 2)^2} &= \lim_{x\rightarrow \infty} -\dfrac{3x^2 – 8}{x^2 – 4x + 4}\end{aligned}
Multiply both the numerator and denominator of the rational expression by $\dfrac{1}{x^2}$.
\begin{aligned}\lim_{x\rightarrow \infty} -\dfrac{3x^2 – 8}{x^2 – 4x + 4} &= \lim_{x\rightarrow \infty} -\dfrac{3x^2 – 8}{x^2 – 4x + 4} \cdot \dfrac{\color{green} \dfrac{1}{x^2}}{\color{green}\dfrac{1}{x^2}}\\&= \lim_{x\rightarrow \infty} -\dfrac{\dfrac{3x^2}{\color{green}x^2} – \dfrac{8}{\color{green}x^2}}{\dfrac{x^2}{\color{green}x^2} – \dfrac{4x}{\color{green}x^2}+ \dfrac{4}{\color{green}x^2}}\\&= \lim_{x\rightarrow \infty} – \dfrac{3 – \dfrac{8}{x^2}}{1 – \dfrac{4}{x} + \dfrac{4}{x^2}}\end{aligned}
We can apply the limit law of sums to evaluate the rewritten expression’s limit. It will also helpful to replace all terms of the form, $\lim_{x \rightarrow \infty} \dfrac{k}{x^n}$, to $0$.
\begin{aligned}\lim_{x\rightarrow \infty} – \dfrac{3 – \dfrac{8}{x^2}}{1 – \dfrac{4}{x} + \dfrac{4}{x^2}}&= – \dfrac{\lim_{x\rightarrow \infty}3 -\lim_{x\rightarrow \infty} \dfrac{8}{x^2}}{\lim_{x\rightarrow \infty}1 – \lim_{x\rightarrow \infty}\dfrac{4}{x} + \lim_{x\rightarrow \infty}\dfrac{4}{x^2}}\\&= – \dfrac{3 – {\color{green} 0}}{1 – {\color{green} 0} + {\color{green} 0}}\\&= -\dfrac{3}{1}\\&= -3\end{aligned}
### Practice Questions
1. The graph of $\dfrac{-4x^2 +x}{x}$ is shown below.
Determine the limit (if it exists) of the function as $x$ approaches the following:
a. $\lim_{x\rightarrow -1} f(x)$
b. $\lim_{x\rightarrow 0} f(x)$
c. $\lim_{x\rightarrow 2} f(x)$
2. Create a table of values (feel free to use a calculator) for the following functions, then approximate the limits of the functions to three decimal places.
a. $\lim_{x\rightarrow 0} \dfrac{1- e^{3x}}{x}$
b. $\lim_{x\rightarrow 0} (1+ 3x)^{\frac{1}{x}}$
c. $\lim_{x\rightarrow 25} \dfrac{5 – \sqrt{x}}{x – 25}$
3. Evaluate the limit (if it exists) of the following functions.
a. $\lim_{x\rightarrow -3} -\dfrac{x}{6}$
b. $\lim_{x\rightarrow 2} \dfrac{x – 2}{x^2 – 4}$
c. $\lim_{x\rightarrow 4} \dfrac{x^{3} – 64}{x – 4}$
d. $\lim_{x\rightarrow 0} \dfrac{\dfrac{1}{x + 5} – \dfrac{1}{5}}{x}$
4. Evaluate the limit (if it exists) of the following functions.
a. $\lim_{x\rightarrow -3} \dfrac{\dfrac{6}{x + 5}- 3}{x +3}$
b. $\lim_{x\rightarrow 0} \dfrac{\sqrt{9 + x} – 3}{x}$
c. $\lim_{x\rightarrow 0} \dfrac{1 – \cos 2x }{4x}$
d. $\lim_{x\rightarrow \infty} -\dfrac{2x^2 + 4x – 6}{(x – 3)^2}$
1.
a. $7$
b. $4$
c. $12$
2.
a. $-3$
b. $20.086$
c. $-0.100$
3.
a. $-\dfrac{1}{2}$
b. $\dfrac{1}{4}$
c. $48$
d. $-\dfrac{1}{25}$
4.
a. $-\dfrac{3}{2}$
b. $\dfrac{1}{6}$
c. $0$
d. $2$
Images/mathematical drawings are created with GeoGebra. |
# SAT Math : How to simplify a fraction
## Example Questions
### Example Question #1 : Simplifying Fractions
Simplify x/2 – x/5
3x/7
2x/7
5x/3
3x/10
7x/10
3x/10
Explanation:
Simplifying this expression is similar to 1/2 – 1/5. The denominators are relatively prime (have no common factors) so the least common denominator (LCD) is 2 * 5 = 10. So the problem becomes 1/2 – 1/5 = 5/10 – 2/10 = 3/10.
### Example Question #11 : General Fractions
If is an integer, which of the following is a possible value of ?
Explanation:
, which is an integer (a number with no fraction or decimal part). All the other choices reduce to non-integers.
### Example Question #1 : Simplifying Fractions
Simplify:
Explanation:
First, let's simplify . The greatest common factor of 4 and 12 is 4. 4 divided by 4 is 1 and 12 divided by 4 is 3. Therefore .
To simply fractions with exponents, subtract the exponent in the numerator from the exponent in the denominator. That leaves us with or
### Example Question #151 : Arithmetic
Which of the following is not equal to 32/24?
4/3
224/168
96/72
16/12
160/96
160/96
Explanation:
24/32 = 1.33
16/12 =1.33
224/168 =1.33
4/3 = 1.33
96/72 = 1.33
160/96 = 1.67
### Example Question #2 : Simplifying Fractions
Find the root of
Can not be determined
Explanation:
The root occurs where . So we substitute 0 for .
This means that the root is at .
### Example Question #6 : Simplifying Fractions
Simplify the fraction below:
Explanation:
The correct approach to solve this problem is to first write factors for the numerator and the denominator:
The highest common factor is 5. Therefore, you can divide the numerator and denominator by 5 in order to get a simplified fraction.
Thus the numerator becomes,
and the denominator becomes .
Therefore the final answer is .
### Example Question #1 : How To Simplify A Fraction
Simplify:
Explanation:
Find the common factors of the numerator and denominator. They both share factors of 2,4, and 8. For simplicity, factor out an 8 from both terms and simplify.
### Example Question #8 : Simplifying Fractions
Simply the following fraction:
Explanation:
Remember that when you divide a fraction by a fraction, that is the same as multiplying the fraction in the numerator by the reciprocal of the fraction in the denominator.
In other words,
Simplifying this final fraction gives us our correct answer, .
### Example Question #1 : How To Simplify A Fraction
Solve for .
Explanation:
To solve for , simplify the fraction. In order to do this, recall that dividing by a fraction is the same as multiplying by its reciprocal. Therefore, rewrite the equation as follows.
Now, simplify the first fraction by calculating four squared.
From here, factor the denominator of the second fraction.
Next, factor the 16.
From here, cancel out like terms that are in both the numerator and denominator. In this particular case that includes (x-2) and 2.
Now, distribute the eight.
Next, multiply both sides by the denominator.
The (8x+16) cancels out and leaves the following equation.
Now to solve for perform opposite operations to move all numerical values to one side of the equation leaving by itself on the other side of the equation.
### Example Question #152 : Arithmetic
Which of the following fractions is not equivalent to ? |
# Mathematics - Number Theory - Prime Numbers
in STEMGeeks7 months ago
[Image 1]
## Introduction
Hey it's a me again @drifter1!
Today we continue with Mathematics, and more specifically the "Number Theory" series, in order to get into Prime Numbers.
So, without further ado, let's get straight into it!
## Prime Numbers
A prime number is a natural number greater than 1 which is only divisible by 1 and itself. In other words, the only positive integer divisors of a prime number are 1 and itself.
For example, 7 is a prime number because it's only divisible by 1 and 7.
The first 10 prime numbers are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
All natural numbers except 1, which is known to be a unique number, can either be prime or not. 1 has exactly 1 factor, any prime has exactly 2 factors and non-primes have more than 2 factors.
Additionally, let's not forget to mention as well, that the only even prime number is 2, as all other even numbers are composite numbers which include 2 in their product representation. So, all prime numbers except 2 are odd numbers.
## Composite Numbers
Any natural number which is not a prime is called a composite number. This basically means that it can be expressed as a product of other numbers, and more specifically by a product of prime numbers (related to the factorization we mentioned last time).
For example, 12 is a composite number as it has the following positive divisors: 1, 2, 3, 4, 6, 12. From these divisors only 2 and 3 are primes. 4 can be expressed as 22 and 6 as 2 × 3. In the end, 12 is basically just 22 × 3, a composition of prime numbers.
## Co-Prime Numbers
Two numbers are called co-primes when they have no common factor apart from 1. When considered in pairs all primes are thus co-primes.
It's easy to check if two composite numbers are co-primes. One simply has to find the Greatest Common Divisor (GCD), also known as Greatest Common Factor (GCF), which has to be 1.
For example, 42 and 55 are co-primes as the GCD = 1.
Explanation:
42 can be expressed by the following product of primes:
2 x 3 x 7
whilst 55 by:
5 x 11
So, the only common factor is 1.
## Prime Test - Sieve of Eratosthenes
So, how does one test if a number is a prime or not?
If the numbers are small it's easy to do so by applying divisibility rules. But, when the numbers are large, a solution seems to be impossible at first glance.
Fortunately for us, there exists a simple theorem which can be proven easily be contradiction, which states that:
If n is a composite number, then it must be divisible by a prime p ≤ √n
In other words, if we want to check the primes up to 100, we only have to check for the primes number in the range up to 10. Thus, we simply have to eliminate the multiples of 2, 3, 5 and 7.
This process is known as the Sieve of Eratosthenes.
Below a GIF, which shows how the process eliminates the non-primes up to 100.
[Image 2]
It's easy to see that all even numbers except 2 are eliminated first, followed by all multiples of 3, then all numbers which end on digit 5, and in the end the multiples of 7 as well, leaving us with the first 25 prime numbers, which are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29,
31, 37, 41, 43, 47, 53, 59, 61, 67, 71,
73, 79, 83, 89, 97.
Of course, there are more ways to prime test, but the introduction of additional concepts is needed for that. Thus, theorems such as Fermat's primality test (also known as Fermat's little theorem) will be covered a little later...
## RESOURCES:
### References
Block diagrams and other visualizations were made using draw.io.
## Final words | Next up
And this is actually it for today's post!
Not sure about the next article's topic yet.
See ya!
Keep on drifting!
Posted with STEMGeeks
Sort:
Nice explanation drifter.
Keep drifting.
7 months ago
Interesting. I have only ever tried to memorize prime numbers up to 100 instead.
!discovery 37
This post was shared and voted inside the discord by the curators team of discovery-it |
# Quotient Rule Calculator
## What is the Quotient Rule Calculator?
The Quotient Rule Calculator is a tool that allows you to easily calculate the derivative of a function that is the ratio of two other functions. This rule is commonly used in calculus to find the derivative of functions that can be expressed as a fraction.
## How to Use the Quotient Rule Calculator
Using the Quotient Rule Calculator is simple and straightforward. All you need to do is input the numerator and denominator of the function you want to find the derivative of, and the calculator will do the rest. It will apply the Quotient Rule formula and provide you with the derivative of the function.
## Understanding the Quotient Rule
The Quotient Rule is a formula used in calculus to find the derivative of a function that is the ratio of two other functions. It states that the derivative of a quotient is equal to the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.
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Mathematically, the Quotient Rule can be expressed as:
f'(x) = (g(x)*f'(x) – f(x)*g'(x)) / (g(x)^2)
Where f(x) and g(x) are functions and f'(x) and g'(x) are their respective derivatives.
## Example of Using the Quotient Rule Calculator
Let’s say we want to find the derivative of the function f(x) = x^2 / (2x + 1). Using the Quotient Rule Calculator, we input the numerator x^2 and the denominator 2x + 1. The calculator then applies the Quotient Rule formula and gives us the derivative f'(x) = (2x(2x+1) – x^2*2) / (2x + 1)^2. Simplifying this expression, we get f'(x) = (4x^2 + 2x – 2x^2) / (2x + 1)^2 = (2x^2 + 2x) / (2x + 1)^2.
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## Benefits of Using the Quotient Rule Calculator
There are several benefits to using the Quotient Rule Calculator. Firstly, it saves you time and effort by automating the process of finding the derivative of a function that is a quotient. Instead of manually applying the Quotient Rule formula, you can simply input the numerator and denominator into the calculator and get the result instantly.
Additionally, the Quotient Rule Calculator can help you check your work and verify that you have correctly found the derivative of a quotient function. This can be especially useful when working on complex calculus problems or studying for exams.
Also Check This Calculate the Solubility of Potassium Bromide at 23
## Conclusion
The Quotient Rule Calculator is a valuable tool for anyone studying calculus or working with functions that can be expressed as a fraction. By quickly and accurately finding the derivative of quotient functions, this calculator can help simplify mathematical tasks and aid in learning and problem-solving. Whether you are a student, teacher, or mathematician, the Quotient Rule Calculator is a convenient resource to have at your disposal. |
# Chapter 8 - Section 8.4 - Multiplicative Inverses of Matrices and Matrix Equations - Exercise Set - Page 933: 56
See the explanation below.
#### Work Step by Step
Consider the two matrices $A$ and $B$ be the multiplicative inverses; that is, the product of the matrix AB and $BA$ is equal to the identity matrix. As example: Let matrix $A=\left[ \begin{matrix} 4 & -3 \\ -5 & 4 \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} 4 & 3 \\ 5 & 4 \\ \end{matrix} \right]$ If $B$ is the multiplicative inverse of $A$, then both $AB$ and $BA$ must result in the identity matrix ${{I}_{2}}=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$ Checking whether $AB={{I}_{2}}$ \begin{align} & AB={{I}_{2}} \\ & AB=\left[ \begin{matrix} 4 & -3 \\ -5 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 4 & 3 \\ 5 & 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{2}} \end{align} Now, check whether $BA={{I}_{2}}$ \begin{align} & BA=\left[ \begin{matrix} 4 & 3 \\ 5 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 4 & -3 \\ -5 & 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{2}} \end{align} Thus, matrix $B$ is the multiplicative inverse of matrix $A$. Hence, $AB={{I}_{n}}$ and $BA={{I}_{n}}$.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. |
# Dividing Fractions Video Activity
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CCSS.MATH.CONTENT.5.NF.B.3
Interpret a fraction as division of the numerator by the denominator (a/b = a ÷ b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem. For example, interpret 3/4 as the result of dividing 3 by 4, noting that 3/4 multiplied by 4 equals 3, and that when 3 wholes are shared equally among 4 people each person has a share of size 3/4. If 9 people want to share a 50-pound sack of rice equally by weight, how many pounds of rice should each person get? Between what two whole numbers does your answer lie?
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# Mathematics Form 3 – Chapter 5 © Notes Prepared by Kelvin
## Form 3 - Chapter 5 – Indices [Notes and Exercise Completely]
Level discussion on activity
L=Level Discussion L1-L8
A=Activity A01-A05
Issued by learning lesson standard.
Review Form 2 - Chapter 2 – Square
2.1 Square + 5.1 Indices
( ) L1A01- review and enable to know indices and power
( ) L1A02- enable to state the index notation
( ) L1A03- enable to state the repeated multiplication
1. Indices is know as ____________.
2. Indices include ___________ (x ) and _______ (x ).
3. Index notation and repeated multiplication.
Example: (a) From repeated multiplication to index notation. 5x5x5x5x5x5=56
(b) From index notation to repeated multiplication. 34=3x3x3x3
*(c) From whole number to index notation with lowest base. 729 = 36
*(d) From index notation to whole number. 44=256
Exercise:
1. Express each of the following as index notation with lowest base.
(a) 3125= (b) 625= (c) 7776= (d)15625=
= = = =
(e) 625= ( )4 (f) 512= ( )9 (g) 729= 9( )
= 25( ) = 8( ) = ( )6
2. Find the value of the following indices.
(a) 72= (b) 45= (c) 73= (d)154=
4 5
(e) 2.4 = (f) (- ) = (g) ( 3 )7= (h) (-3.1)5=
Revision Form 2 - Chapter 2 (Exercise)
1. Simplify the following.
1
(a) 22÷
3 −8 =
125
(b) (
√ 16 -1)3=
25
(c) (
3
√ 81 )3=
(d) (2.4÷ √3 64 )2= (e) (25- √ 121 )2= (f) Find the sum of all the
perfect squares from 30 to
80.
___2, ___2, ___2 =36,49,64
36+49+64=________
1|Page
Mathematics Form 3 – Chapter 5 © Notes Prepared by Kelvin
## Review Form 3 - Chapter 5 – Indices
5.2 Multiply of Numbers in Index Notation + 5.3 Division of Numbers in Index Notation
(+ unknown) (same as F2 Chp3.2 Algebraic)
( ) L2A01- enable to simplify the multiply of numbers in index notation
( ) L2A02- enable to simplify the division of numbers in index notation
1. Simplifying the multiplication of the following.
a) 45×44×43= b) x3×x8×x5=
2 4 5 4
c) 1.2 ×3.3 ×1.2×3.3 ×1.2 = d) x4×y2×x5×y7=
4 6 5
e) 5x×y ×7x ×8y = f) -3c5×5d×c3×(-8d5)=
9 3 5
g) -6k ×2h ×(-3h )×5k = 7
h)( 2 ) 2 ( 2 ) 5
5 5
2. Simplifying the division of the following. (division in fraction type.)
a) 209÷44= b) x19÷x8=
13 5
c) -8x ÷2x = d) 15y28÷5y12=
5d × (-8d 13 ) 4d 6
13
e) 4d 6 f) 6d × (-8d )
h)( 2 )18 ÷ ( 2 ) 5
32 a 7
4a3
5 5
g)
## 5.4 Raising Numbers and Algebraic Terms in Index Notation to a Power
( ) L3A01 - enable to know multiplicaton of power, eg: (52)3=56
( ) L3A02 - enable to simplify the equation of multiply and division in order.
( ) L3A03 - enable to simplify the combination equation including multiply,
division, raising number and algebraic terms
( ab ) p ab p
p
Notes: a. (am)n=amn e. (a÷am)n=an ÷amn=an-mn
m n n mn= n+mn m n p mp np mp -np
b. (a×a ) =a ×a a f. (a ÷a ) =a ÷a =a
c. (am×an)p=amp ×anp=amp +np g. (am÷bn)p=amp ÷bnp OR
n ) np
am p a mp
m n p mp
d. (a ×b ) =a ×b np ( b b
1. Derive (am)n=amn.
(63)2= (x8)7= [(-85)6]= [(-y5)0]=
## 2. Simplifying of the following.
(p5)6×p4= 45×(46)3= (-2u4v5)3= (k2)3x(2k5)2=
4 6 4
(p ) ÷p = (124)2÷(122)3= (2h3k5)3÷(h2k3)2=
3 16 3 2
27a b ÷(3b ) = ( 4 2 ) 3 25 8
4 2 3 3
(e f ) ÷(3ef) = 45
4 m 3n 5
p4
5mp
2|Page
Mathematics Form 3 – Chapter 5 © Notes Prepared by Kelvin
## 5.5 Negative Indices (Indices+Integers)
( ) L4 A01- verify a-n= 1
( ) L4A02 - verify 1 =a-n
( ) L4A03 - perform computation involving negative indices
1. Verify a-n= 1 OR 1 =a-n
-4
5 = 3-6 k-6
1
94
1
57
1
m12
2. Simplifying of the following.
3-9×312×b4÷b-7= 7-3÷6-2×7-5=
(x-3y2)2×y-3= (44×8-6)5÷3-3=
5 3 4 2 p 2 q 2 ( p 3 ) 2
4 4 55
p 5 q 7
## 5.6 Fractional Indices
1 Fraction to square root.
an n a
1
## State/Find the value of an n a
1 1
( a m ) n OR (a n ) m
m
a n change to n
am n
OR ( a )
m
## State/Find the value of a n
- Perform combined operations of multiplication, division and raising to a power
involving fractional indices on numbers and algebraic terms.
5.7 Computations involving laws of indices
Additional Note - Laws of indices:
am×an=am+n am÷an=am-n (am)n/(an)m=amn
m n n mn= n+mn
(a×a ) =a ×a a (am×bn)p=amp ×bnp (ab)m=am×an
m n p mp np mp +np
(a ×a ) =a ×a =a a0=1
m n n mn= n-mn
(a÷a ) =a ÷a a (a ÷a ) =a ÷anp=apmp -np
m n p mp
(am÷bn)p=ammp ÷bnp
( ) ( ab n ) p abnp
mp
a p a
b bp
1
a-n= 1 an n a
1 1
( a m ) n OR ( a n ) m
3|Page
Mathematics Form 3 – Chapter 5 © Notes Prepared by Kelvin
m
n
am OR (n a ) m
an
- Perform combined operations of multiplication, division and raising to a power
involving fractional indices on numbers and algebraic terms.
4|Page
Mathematics Form 3 – Chapter 5 © Notes Prepared by Kelvin
5|Page
Mathematics Form 3 – Chapter 5 © Notes Prepared by Kelvin
6|Page
Mathematics Form 3 – Chapter 5 © Notes Prepared by Kelvin
More Knowledge for You:
7|Page |
# Geometric sequence word problems
This lesson will show you how to solve a variety of geometric sequence word problems.
Example #1:
The stock's price of a company is not doing well lately. Suppose the stock's price is 92% of its previous price each day. What is the stock's price after 10 days if the stock was worth \$2500 right before it started to go down?
Solution
To solve this problem, we need the geometric sequence formula shown below.
an = a1 × r(n - 1)
a1 = original value of the stock = 2500
a2 = value of the stock after 1 day
a11 = value of the stock after 10 days
r = 0.92
a11 = 2500 × (0.92)(11 - 1)
a11 = 2500 × (0.92)10
a11 = 2500 × 0.434
a11 = \$1085
The stock's price is about 1085 dollars.
Example #2:
The third term of a geometric sequence is 45 and the fifth term of the geometric sequence is 405. If all the terms of the sequence are positive numbers, find the 15th term of the geometric sequence.
Solution
To solve this problem, we need the geometric sequence formula shown below.
an = a× r(n - 1)
Find the third term
a3 = a1 × r(3 - 1)
a3 = a1 × r2
Since the third term is 45, 45 = a1 × r2 (equation 1)
Find the fifth term
a5 = a1 × r(5 - 1)
a5 = a1 × r4
Since the fifth term is 405, 405 = a1 × r4 (equation 2)
Divide equation 2 by equation 1.
(a1 × r4) / (a1 × r2) = 405 / 45
Cancel a1 since it is both on top and at the bottom of the fraction.
r4 / r2 = 9
r2 = 9
r = ±√9
r = ±3
Use r = 3, and equation 1 to find a1
45 = a1 × (3)2
45 = a1 × 9
a1 = 45 / 9 = 5
Since all the terms of the sequence are positive numbers, we must use r = 3 if we want all the terms to be positive numbers.
an = a1 × r(n - 1)
Let us now find a15
a15 = 5 × (3)(15 - 1)
a15 = 5 × (3)14
a15 = 5 × 4782969
a15 = 23914845
## Challenging geometric sequence word problems
Example #3:
Suppose that the magnification of a PDF file on a desktop computer is increased by 15% for each level of zoom. Suppose also that the original length of the word "January" is 1.2 cm. Find the length of the word "January" after 6 magnifications.
Solution
To solve this problem, we need the geometric sequence formula shown below.
an = a1 × r(n - 1)
a1 = original length of the word = 1.2 cm
a2 = length of the word after 1 magnification
a7 = length of the word after 6 magnifications
r = 1 + 0.15 = 1.15
n = 7
a7 = 1.2 × (1.15)(7 - 1)
a7 = 1.2 × (1.15)6
a7 = 1.2 × (1.15)6
a7 = 1.2 × 2.313
a7 = 2.7756
After 6 magnifications, the length of the word "January" is 2.7756 cm.
Notice that we added 1 to 0.15. Why did we do that? Let us not use the formula directly so you can see the reason behind it. Study the following carefully!
Day 1: a1 = 1.2
Day 2: a2 = 1.2 + 1.2(0.15) = 1.2(1 + 0.15)
Day 3: a31.2(1 + 0.15) + [1.2(1 + 0.15)]0.15 = 1.2(1 + 0.15)(1 + 0.15) = 1.2(1 + 0.15)2
Day 7: a7 = 1.2(1 + 0.15)6
Example #4
Suppose that you want a reduced copy of a photograph. The actual length of the photograph is 10 inches. If each reduction is 64% of the original, how many reductions, will shrink the photograph to 1.07 inches.
Solution
an = a1 × r(n - 1)
a1 = original length of the photograph = 10 inches
a2 = length of the photograph after 1 reduction
an = 1.07
r = 0.64
n = number of reductions = ?
1.07 = 10 × (0.64)(n - 1)
Divide both sides by 10
1.07 / 10 = [10 × (0.64)(n - 1)] / 10
0.107 = (0.64)(n - 1)
Notice that you have an exponential equation to solve. The biggest challenge then is knowing how to solve exponential equations!
Take the natural log of both sides of the equation.
ln(0.107) = ln[(0.64)(n - 1)]
Use the power property of logarithms.
ln(0.107) = (n - 1)ln(0.64)
Divide both sides of the equation by ln(0.64)
ln(0.107) / ln(0.64) = (n - 1)ln(0.64) / ln(0.64)
n - 1 = ln(0.107) / ln(0.64)
Use a calculator to find ln(0.107) and ln(0.64)
n - 1 = -2.23492644452 \ -0.44628710262
n - 1 = 5.0078
n = 1 + 5.0078
n = 6.0078
Therefore, you will need 6 reductions.
100 Tough Algebra Word Problems.
If you can solve these problems with no help, you must be a genius!
Recommended |
Question
54) Ashley and Nicole each improved their yards by planting daylilies and ornamental grass. They
bought their supplies from the same store. Ashley spent $43 on 2 daylilies and 3 bunches of ornamental grass. Nicole spent$87 on 6 daylilies and 3 bunches of ornamental grass. Find the
cost of one daylily and the cost of one bunch of ornamental grass.
dalilar.
1. minhkhue
Daylilies- $11, Ornamental grass-$7
Step-by-step explanation:
For this one, you want to convert this into simultaneous equations.
If we let d be cost of one daylily and o be cost of one bunch of ornamental grass, we can write the equations as 2d+3o=43 and 6d+3o=87.
From there, we can see they share a like term (3o), so we subtract one equation from the other (since they both are preceded by a +). This gives 4d=44.
Dividing both sides by 4 isolates d, leaving d=$11. We can substitute this value into either equation to work out o (I’m using the first one as it is easier). This makes it 2×11+3o=43, or 22+3o=43. From there, we subtract 22 from both sides, giving 3o=21, and dividing by 3 gives a final result of o=$7.
**This content involves writing and solving simultaneous equations, which you may wish to revise. I’m always happy to help!
All working out:
d- cost of one daylily o- cost of one bunch of ornamental grass
2d+3o=43 6d+3o=87
6d+3o=87
– – –
2d+3o=43
= = =
4d =44
÷4 ÷4
d =11
2d+3o=43
2×11+3o=43
22+3o=43
3o=21
o=7 |
This book is archived and will be removed July 6, 2022. Please use the updated version.
Interference
# 20 Mathematics of Interference
### Learning Objectives
By the end of this section, you will be able to:
• Determine the angles for bright and dark fringes for double slit interference
• Calculate the positions of bright fringes on a screen
(Figure)(a) shows how to determine the path length difference for waves traveling from two slits to a common point on a screen. If the screen is a large distance away compared with the distance between the slits, then the angle between the path and a line from the slits to the screen [part (b)] is nearly the same for each path. In other words, and are essentially parallel. The lengths of and differ by , as indicated by the two dashed lines in the figure. Simple trigonometry shows
where d is the distance between the slits. Combining this result with (Figure), we obtain constructive interference for a double slit when the path length difference is an integral multiple of the wavelength, or
Similarly, to obtain destructive interference for a double slit, the path length difference must be a half-integral multiple of the wavelength, or
where is the wavelength of the light, d is the distance between slits, and is the angle from the original direction of the beam as discussed above. We call m the order of the interference. For example, is fourth-order interference.
(a) To reach P, the light waves from and must travel different distances. (b) The path difference between the two rays is .
The equations for double-slit interference imply that a series of bright and dark lines are formed. For vertical slits, the light spreads out horizontally on either side of the incident beam into a pattern called interference fringes ((Figure)). The closer the slits are, the more the bright fringes spread apart. We can see this by examining the equation
. For fixed and m, the smaller d is, the larger must be, since . This is consistent with our contention that wave effects are most noticeable when the object the wave encounters (here, slits a distance d apart) is small. Small d gives large , hence, a large effect.
Referring back to part (a) of the figure, is typically small enough that , where is the distance from the central maximum to the mth bright fringe and D is the distance between the slit and the screen. (Figure) may then be written as
or
The interference pattern for a double slit has an intensity that falls off with angle. The image shows multiple bright and dark lines, or fringes, formed by light passing through a double slit.
Finding a Wavelength from an Interference Pattern Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm and find that the third bright line on a screen is formed at an angle of relative to the incident beam. What is the wavelength of the light?
Strategy The phenomenon is two-slit interference as illustrated in (Figure) and the third bright line is due to third-order constructive interference, which means that . We are given and . The wavelength can thus be found using the equation for constructive interference.
Solution Solving for the wavelength gives
Substituting known values yields
Significance To three digits, this is the wavelength of light emitted by the common He-Ne laser. Not by coincidence, this red color is similar to that emitted by neon lights. More important, however, is the fact that interference patterns can be used to measure wavelength. Young did this for visible wavelengths. This analytical techinque is still widely used to measure electromagnetic spectra. For a given order, the angle for constructive interference increases with , so that spectra (measurements of intensity versus wavelength) can be obtained.
Calculating the Highest Order Possible Interference patterns do not have an infinite number of lines, since there is a limit to how big m can be. What is the highest-order constructive interference possible with the system described in the preceding example?
Strategy The equation (for ) describes constructive interference from two slits. For fixed values of , the larger m is, the larger is. However, the maximum value that can have is 1, for an angle of . (Larger angles imply that light goes backward and does not reach the screen at all.) Let us find what value of m corresponds to this maximum diffraction angle.
Solution Solving the equation for m gives
Taking and substituting the values of from the preceding example gives
Therefore, the largest integer m can be is 15, or .
Significance The number of fringes depends on the wavelength and slit separation. The number of fringes is very large for large slit separations. However, recall (see The Propagation of Light and the introduction for this chapter) that wave interference is only prominent when the wave interacts with objects that are not large compared to the wavelength. Therefore, if the slit separation and the sizes of the slits become much greater than the wavelength, the intensity pattern of light on the screen changes, so there are simply two bright lines cast by the slits, as expected, when light behaves like rays. We also note that the fringes get fainter farther away from the center. Consequently, not all 15 fringes may be observable.
Check Your Understanding In the system used in the preceding examples, at what angles are the first and the second bright fringes formed?
, respectively
### Summary
• In double-slit diffraction, constructive interference occurs when , where d is the distance between the slits, is the angle relative to the incident direction, and m is the order of the interference.
• Destructive interference occurs when .
### Conceptual Questions
Suppose you use the same double slit to perform Young’s double-slit experiment in air and then repeat the experiment in water. Do the angles to the same parts of the interference pattern get larger or smaller? Does the color of the light change? Explain.
Why is monochromatic light used in the double slit experiment? What would happen if white light were used?
Monochromatic sources produce fringes at angles according to . With white light, each constituent wavelength will produce fringes at its own set of angles, blending into the fringes of adjacent wavelengths. This results in rainbow patterns.
### Problems
At what angle is the first-order maximum for 450-nm wavelength blue light falling on double slits separated by 0.0500 mm?
Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on double slits separated by 0.100 mm.
What is the separation between two slits for which 610-nm orange light has its first maximum at an angle of ?
Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of
Calculate the wavelength of light that has its third minimum at an angle of when falling on double slits separated by . Explicitly show how you follow the steps from the Problem-Solving Strategy: Wave Optics, located at the end of the chapter.
What is the wavelength of light falling on double slits separated by if the third-order maximum is at an angle of ?
At what angle is the fourth-order maximum for the situation in the preceding problem?
What is the highest-order maximum for 400-nm light falling on double slits separated by ?
62.5; since m must be an integer, the highest order is then .
Find the largest wavelength of light falling on double slits separated by for which there is a first-order maximum. Is this in the visible part of the spectrum?
What is the smallest separation between two slits that will produce a second-order maximum for 720-nm red light?
(a) What is the smallest separation between two slits that will produce a second-order maximum for any visible light? (b) For all visible light?
(a) If the first-order maximum for monochromatic light falling on a double slit is at an angle of , at what angle is the second-order maximum? (b) What is the angle of the first minimum? (c) What is the highest-order maximum possible here?
a. ; b. ; c. 5.76, the highest order is .
Shown below is a double slit located a distance x from a screen, with the distance from the center of the screen given by y. When the distance d between the slits is relatively large, numerous bright spots appear, called fringes. Show that, for small angles (where , with in radians), the distance between fringes is given by
Using the result of the preceding problem, (a) calculate the distance between fringes for 633-nm light falling on double slits separated by 0.0800 mm, located 3.00 m from a screen. (b) What would be the distance between fringes if the entire apparatus were submersed in water, whose index of refraction is 1.33?
a. 2.37 cm; b. 1.78 cm
Using the result of the problem two problems prior, find the wavelength of light that produces fringes 7.50 mm apart on a screen 2.00 m from double slits separated by 0.120 mm.
In a double-slit experiment, the fifth maximum is 2.8 cm from the central maximum on a screen that is 1.5 m away from the slits. If the slits are 0.15 mm apart, what is the wavelength of the light being used?
560 nm
The source in Young’s experiment emits at two wavelengths. On the viewing screen, the fourth maximum for one wavelength is located at the same spot as the fifth maximum for the other wavelength. What is the ratio of the two wavelengths?
If 500-nm and 650-nm light illuminates two slits that are separated by 0.50 mm, how far apart are the second-order maxima for these two wavelengths on a screen 2.0 m away?
1.2 mm
Red light of wavelength of 700 nm falls on a double slit separated by 400 nm. (a) At what angle is the first-order maximum in the diffraction pattern? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
### Glossary
fringes
bright and dark patterns of interference
order
integer m used in the equations for constructive and destructive interference for a double slit |
What are their ages in years?
1,502.2K Views
The grandson is about as many days old as the son is in weeks. The grandson is approximately as many months old as the father is in years. The ages of the grandson, the son, and the father add up to 120 years. What are their ages in years?
pnikam Expert Asked on 18th August 2015 in
Let G, S, and F represent the ages in years of the grandson, the son, and the father, respectively. Since a year has 365 days, or 52 weeks, or 12 months, the problem can be represented by three equations with three unknowns:
365 G = 52 S (The grandson is about as many days old as the son is in weeks)
12 G = F (The grandson is approximately as many months old as the father is in years)
G + S + F = 120 (The ages of the grandson, the son, and the father add up to 120 years)
Because the grandson’s age is G = F/12, the son’s age can be represented in terms of F by substituting:
S = (365/52)×G
S = (365/52)×(F/12)
The third equation can now be represented with only the father’s age as an unknown:
F/12 + (365/52)×(F/12) + F = 120
Multiplying by 12, we get:
F + (365/52)×F + 12 F = 120×12
20 F = 1440
F = 72
G = F/12 = 6
S = (365/52)×6 = 42
The father is 72, the son is 42, and the grandson is 6 years old.
You may have noticed that the statement of the problem is somewhat ambiguous: “The grandson is approximately as many months old as the father is in years” indicates that there may be some variance. A year really has 365 1/4 days instead of 365, and 52 weeks times 7 days gives only 364 days in a year. These inconsistencies produce fractional results that need to be rounded.
Doing the division in the following equation:
F + (365/52)×F + 12 F = 120×12
produces 20.019 F = 1440, which is rounded to 20 F = 1440
pnikam Expert Answered on 18th August 2015.
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## Precalculus (6th Edition) Blitzer
The partial fraction is, $\frac{1}{\left( {{x}^{2}}+4 \right)}+\frac{2x-1}{{{\left( {{x}^{2}}+4 \right)}^{2}}}$.
$\frac{{{x}^{2}}+2x+3}{{{\left( {{x}^{2}}+4 \right)}^{2}}}=\frac{\left( Ax+B \right)\left( {{x}^{2}}+4 \right)+\left( Cx+D \right)}{{{\left( {{x}^{2}}+4 \right)}^{2}}}$ Now, take the L.C.M. on the right side: $\frac{{{x}^{2}}+2x+3}{{{\left( {{x}^{2}}+4 \right)}^{2}}}=\frac{Ax+B}{\left( {{x}^{2}}+4 \right)}+\frac{Cx+D}{{{\left( {{x}^{2}}+4 \right)}^{2}}}$ By eliminating the denominator from both sides, we get \begin{align} & {{x}^{2}}+2x+3=\left( Ax+B \right)\left( {{x}^{2}}+4 \right)+\left( Cx+D \right) \\ & =A{{x}^{3}}+4Ax+B{{x}^{2}}+4B+Cx+D \\ & =A{{x}^{3}}+B{{x}^{2}}+\left( 4A+C \right)x+4B+D \end{align} Then compare the coefficients of ${{x}^{3}},\ {{x}^{2}},{{x}^{1}}$ and the constant term: $A=0$ …… (I) $B=1$ …… (II) $4A+C=2$ …… (III) $4B+D=3$ …… (IV) Substitute the value of A in equation (III): $C=2$ And put the value of B in equation (IV): \begin{align} & 4+D=3 \\ & D=-1 \end{align} Now, $\frac{{{x}^{2}}+2x+3}{{{\left( {{x}^{2}}+4 \right)}^{2}}}=\frac{1}{\left( {{x}^{2}}+4 \right)}+\frac{2x-1}{{{\left( {{x}^{2}}+4 \right)}^{2}}}$. Thus, the partial fraction of the given expression is $\frac{1}{\left( {{x}^{2}}+4 \right)}+\frac{2x-1}{{{\left( {{x}^{2}}+4 \right)}^{2}}}$. |
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# What is a Factorial of Hundred?
Published
on
You may have coined the term “factorial” as a math enthusiast. But what does it mean, and what are some specific examples? In this blog post, we will explore the definition of the factorial and provide some examples of how to calculate it. We’ll also answer the question: What is a factorial of hundred? So if you’re ready to learn about the exciting world of factorials, keep reading.
[lwptoc]
## What is factorial?
The product of all the integers from 1 to a quantity is known as the factorial of that number. So, the factorial of 4 would be 1x2x3x4=24. The symbol of the factorial number is n!.
For example,
5!= 5x4x3x2x1=120
7!= 7x6x5x4x3x2x1=5040
As you can see, the factorial of a number can be a large number, even for relatively small numbers.
## How to calculate the factorial of a number?
There are two ways that you can calculate the factorial of a number. The first way is simply to multiply all the integers from 1 to that number. So, for our example above, the factorial of 4 would be 1x2x3x4=24.
The second way to calculate the factorial of a number is to use the factorial function. This function is available in most mathematical software programs, such as Microsoft Excel. To use the factorial function, enter the number you want to calculate the factorial for into the function. For our example above, we would enter “4” into the function. The factorial function will then return the answer, which in this case would be 24.
## What is a factorial of 100?
Now that we know what a factorial is and how to calculate it let’s answer the question: What is a factorial of a hundred? So, the factorial of 100 would be 100x99x98x97x… all the way down to 1. This is a very large number, too large to fit into this blog post!
The factorial of 100 is so large that it doesn’t even have a name. So, it is called the “100 factorial” or “number 100!.
## What is the factorial formula?
You can use a formula to calculate the factorial of a number. For any positive integer n, n!=n(n-1)(n-2)…(2)(1). So, using our example from above, the factorial of 4 would be 4!=4(4-1)(4-2)(4-3)=4x3x2x1=24.
## FAQs:
### Q: What is the factorial of 0?
A: The factorial of 0 is 1. This may seem like a strange answer, but it is true!
### Q: What is the factorial of a negative number?
A: The factorial of a negative number is not defined.
### Q: Is the factorial of a number always a whole number?
A: No, the factorial of a number is not always a whole number. For example, the factorial of 1.5 is 1.5×0.5x-0.5=-1.125.
### Q: What is the factorial of infinity?
A: The factorial of infinity is not defined.
### Q: What is the largest factorial that has been calculated?
A: The largest factorial that has been calculated is the factorial of 10,000, equal to 10,000×9, 999×9, and 998 x… all the way down to 1. It’s just referred to as “10,000 factorial.
## Conclusion:
Now you know about the factorial of hundred. Other information that you should know is also present in this article. But, if you have leftover questions, never hesitate to ask them in the comments section below.
`Check out: Know About IB Math Tutoring`
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# Question: What Does It Mean When Two Lines Intersect On A Graph?
## How do you tell if two lines will intersect?
To find the point at which the two lines intersect, we simply need to solve the two equations for the two unknowns, x and y.
Finally, divide both sides by A1B2 – A2B1, and you get the equation for x.
The equation for y can be derived similarly..
## What is the point of intersection of the two number lines?
Now, where the two lines cross is called their point of intersection. Certainly this point has (x, y) coordinates. It is the same point for Line 1 and for Line 2. So, at the point of intersection the (x, y) coordinates for Line 1 equal the (x, y) coordinates for Line 2.
## When two lines intersect and the angle between them is a right angle then the lines are said to be?
Perpendicular lines are lines that intersect at a right (90 degrees) angle.
## When two lines intersect the opposite angles are?
Vertical angles are pair angles formed when two lines intersect. Vertical angles are sometimes referred to as vertically opposite angles because the angles are opposite to each other.
## When two straight lines intersect adjacent angles are?
When two straight lines intersect each other, four angles are created such that the point of intersection is the vertex for each angle. If two of the angles have a common vertex and share a common side they are called adjacent angles.
## What is it called when two lines intersect on a graph?
When two or more lines cross each other in a plane, they are called intersecting lines. The intersecting lines share a common point, which exists on all the intersecting lines, and is called the point of intersection. Here, lines P and Q intersect at point O, which is the point of intersection.
## How do you find the point of intersection of two lines on a graph?
Finding a Point AlgebraicallySolve each of the equations for y.Set the two expressions for y equal to each other and solve for x. This is your x-value of the point of intersection.Plug the value of x into either one of the original equations and solve for y. This is your y-value for the point of intersection.
## When two lines intersect what angles are formed?
When two lines intersect they form two pairs of opposite angles, A + C and B + D. Another word for opposite angles are vertical angles. Vertical angles are always congruent, which means that they are equal.
## Do two vectors intersect?
For two lines to intersect, each of the three components of the two position vectors at the point of intersection must be equal. Therefore we can set up 3 simultaneous equations, one for each component. … If we have a point of intersection, then we should have one side equal to the other.
## Can two planes intersect in a line?
They cannot intersect at only one point because planes are infinite. Furthermore, they cannot intersect over more than one line because planes are flat. One way to think about planes is to try to use sheets of paper, and observe that the intersection of two sheets would only happen at one line.
## How do you find the point of intersection of two lines in 3d?
Once you found λ and μ then make sure you that x-coordinates, y-coordinates, and z-coordinates of both lines are equal. If they are all equal then you have at least one intersection. If at least one of the coordinates be it x, y or z are different between the two lines then they have no intersection. |
How to find the surface area of a Triangular Prism?
• Last Updated : 01 Jun, 2022
A polyhedron with two triangular bases and three rectangular sides is known as a triangular prism. It’s a three-dimensional object with three side faces and two base faces, all of which are connected by edges. Its sides are rectangular in form and are joined together side by side. Triangle cross-sections go parallel to the base faces. The two triangular bases are congruent and parallel to one another. There are nine unique nets in this pentahedron. The bases’ edges and vertices are linked to one another. It’s termed a right triangle prism if the sides are rectangular; otherwise, it’s called an oblique triangular prism.
Surface area of Triangular Prism formula
The surface area of a triangular prism is defined as the sum of the area of its five faces. To calculate the surface area, we need the values of the base, length, and height of the triangular prism. Its formula equals the sum of two times the base area and three times the product of the base and length of the prism. Its unit of measurement is a meter square (sq. m).
A = bh + 3bl
Where,
b is the triangular base,
h is the height,
l is the length of the prism.
How to find the surface area of a Triangular Prism?
Let’s take an example to understand how we can calculate the surface area of a triangular prism.
Example: Calculate the surface area of a triangular prism of base 5 m, height 10 m and length 15 m.
Step 1: Note the dimensions of the triangular prism. In this example, the length of the base is 5 m, height is 10 m and length is 15 m.
Step 2: We know that the surface area of a triangular prism is equal to (bh + 3bl). Substitute the given values of base, height and length in the formula.
Step 3: So, the surface area of triangular prism is calculated as, A = 5 (10) + 3 (5) (15) = 275 sq. m
Sample Problems
Problem 1: Calculate the surface area of a triangular prism of base 6 m, height 3 m, and length 7 m.
Solution:
We have,
b = 6
h = 3
l = 7
Using the formula we get,
A = bh + 3bl
= 6 (3) + 3 (6) (7)
= 144 sq. m
Problem 2: Calculate the surface area of a triangular prism of base 2 m, height 4 m, and length 6 m.
Solution:
We have,
b = 2
h = 4
l = 6
Using the formula we get,
A = bh + 3bl
= 2 (4) + 3 (2) (6)
= 44 sq. m
Problem 3: Calculate the surface area of a triangular prism of base 4 m, the height of 9 m, and length of 7 m.
Solution:
We have,
b = 4
h = 9
l = 7
Using the formula we get,
A = bh + 3bl
= 4 (9) + 3 (4) (7)
= 120 sq. m
Problem 4: Calculate the length of the triangular prism if its base is 4 m, height is 9 m and area is 198 sq. m.
Solution:
We have,
b = 6
h = 9
A = 198
Using the formula we get,
A = bh + 3bl
=> 198 = 6 (9) + 3 (6) (l)
=> 198 = 54 + 18l
=> 18l = 144
=> l = 8 m
Problem 5: Calculate the length of the triangular prism if its base is 5 m, height is 10 m and area is 180 sq. m.
Solution:
We have,
b = 5
h = 10
A = 180
Using the formula we get,
A = bh + 3bl
=> 180 = 5 (10) + 3 (5) (l)
=> 180 = 54 + 15l
=> 15l = 126
=> l = 8.4 m
Problem 6: Calculate the height of the triangular prism if its base is 12 m, length is 14 m and area is 700 sq. m.
Solution:
We have,
b = 12
l = 14
A = 700
Using the formula we get,
A = bh + 3bl
=> 700 = 12 (h) + 3 (12) (14)
=> 700 = 12h + 504
=> 12h = 196
=> h = 16.33 m
Problem 7: Calculate the height of the triangular prism if its base is 8 m, length is 14 m and area is 408 sq. m.
Solution:
We have,
b = 8
l = 14
A = 408
Using the formula we get,
A = bh + 3bl
=> 408 = 8 (h) + 3 (8) (14)
=> 408 = 8h + 336
=> 8h = 72
=> h = 9 m
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Summary Videos Questions & Answers References
# Prime Factorisation, HCF and LCM
#### Summary
LearnNext Lesson Video
Prime factorisation:
Writing a number as a product of its prime factors is called the prime factorisation of the number.
e.g. (i) Prime factorisation of 18 = 2 x 3 x 3
(ii) Prime factorisation of 40 = 2 x 2 x 2 x 5
Highest common factor:
The greatest of the common factors of the given numbers is called their highest common factor (HCF). It is also known as the greatest common divisor. The HCF of the given numbers is equal to the product of common factors in their prime factorisation.
e.g.
Factors of 16 = 1, 2, 4, 8, 16
Factors of 40 = 1, 2, 4, 5, 8, 10, 20, 40
Common factors = 1, 2, 4, 8
HCF of 16 and 40 = 8
The HCF of 16 and 40 using their prime factorisation:
Prime factorisation of 16 = 2 x 2 x 2 x 2
Prime factorisation of 40 = 2 x 2 x 2 x 5
Common prime factors = 2 x 2 x 2
HCF of 16 and 40 = 2 x 2 x 2 = 8
Least common multiple:
The smallest common multiple of the given numbers is called their least common multiple (LCM). The product of the prime factors that occur the maximum number of times in the prime factorisation of the given numbers, is their LCM.
e.g.
The LCM of given numbers using their prime factorisation:
Prime factorisation of 4 = 2 x 2
Prime factorisation of 6 = 2 x 3
LCM of 4 and 6 = 2 x 2 x 3 =12
To find the LCM of the given numbers using the division method:
• Write the given numbers in a row.
• Divide the numbers by the smallest prime number that divides one or more of the given numbers.
• Write the number that is not divisible, in the second row.
• Write the new dividends in the second row.
• Divide the new dividends by another smallest prime number.
• Continue dividing till the dividends are all prime numbers or 1.
• Stop the process when all the new dividends are prime numbers or 1.
• The product of all the divisors and the remaining prime dividends is the LCM of the given numbers.
### 1 . Find the smallest 4 digit number.
Sol:
LCM of 18, 24 and 32:
18 = 2 x 3 x 3
24 = 2 x 2 x 2 x 3
32 = 2 x 2 x 2 x 2 x 2
LCM = 2 x 2 x 2 x 2 x
...
### 2 . A traffic light at three different road crossing change
Given that traffic light at three different road crossing change after every 48 seconds,72 seconds and 108 secon
...
### 3 . Write the smallest 4 digit number and express it in terms of prime factors.
the smallest four digit niumber 1000.
1000= 2 3X5 3
### 4 . Largest number that divides 258 and 584 leaving remainder 6 and 8
Given that when 258 is divided by the number the remainder is 6
Similarly, when 584 is divided by the same number the remainder is 8...
### 5 . Simplify using BODMAS rule
45 + 3{34 - 18 - 14} / 3{17 + 3 of 4 - (2 of 7)}
= 45 + 3{2} / 3{17 + 3 of 4 - 14}
= 45+6 / 3{17 + 12 - 14... |
## Monday, 18 August 2008
### Tips To Reduce Errors Doing Simultaneous Equations
Solving simultaneous equations involves many simple steps. The simple steps mostly include addition, subtraction and multiplication.
Though the mathematical operations are simple, mistakes made while solving simultaneous equations are aplenty. The errors are mostly "slip-of-the-mind" human errors.
How then are we to reduce these careless errors?
Here, I propose 2 tips.
Tip one
Make use of addition instead of subtraction to eliminate the selected unknown.
Example: ( to eliminate unknown "y")
7x + 2y = 11 --- (A)
-5x + 2y = -1 --- (B)
In normal doing, we perform (A) - (B) equation subtraction. But what is the risk?
The result may end up as 2x + 0 = 10 !
The correct answer should be 12x + 0 = 12.
Why the error?
This is because our brain is use to addition more than subtraction. Therefore the "slip-of- the-mind" error happened.
Mentally doing 7x - (-5x) is harder to operate with.
Refer to this post to understand why our brain likes addition.
So how?
Negate the equation (B) so that we can perform addition.
(-5x + 2y) times (-1) = -1 times (-1)
will become 5x - 2y = 1 ----(C)
Rewriting the question,
7x + 2y = 11 ----(A)
5x - 2y = 1 ----(C)
This becomes simpler!
We now need to ADD the 2 equations. (Instead of the risky subtraction).
(A) + (C) : 12x + 0 = 12 <== This is the correct result that we want, risk-free!
Message: Negate the unknown variable of one equation and do ADDITION to remove it.
Tip two
Avoid making the number (coefficient) bigger through multiplication.
Example:
9x + 2y = 13 --- ( K )
x - 4y = -7 --- ( L )
Here, we have the option to remove either the "x" or the "y".
Which to select depends on the proper selection of multiplication factor in order not to make the coefficient big.
Case 1: Remove "x".
We need to multiply equation (L) by 9 to cause the first term (x) to be the same as that in equation (k), so as to eliminate the "x" unknown.
What happened?
Equation (L) became 9x - 36 y = - 63 !
Look at the coefficient of "y" ===> It became a GIANT!
Case 2: Remove unknown "y".
Multiply equation (k) by 2. This produces equation (k) as 18x + 4 y = 26.
This is still manageable. And less error will occur since big number is harder to handle.
Message: Seek to multiply coefficient such that the resultant equation has smaller number.
I have seen many maths students fumbling with these simple operations, and making many unnecessary mistakes.
I do hope that these two little tips will aid you and any maths learners of simultaneous equations to reduce errors.
:-P |
# How to Calculate a Horizontal Tangent Line
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A horizontal tangent line is a mathematical feature on a graph, located where a function's derivative is zero. This is because, by definition, the derivative gives the slope of the tangent line. Horizontal lines have a slope of zero. Therefore, when the derivative is zero, the tangent line is horizontal. To find horizontal tangent lines, use the derivative of the function to locate the zeros and plug them back into the original equation. Horizontal tangent lines are important in calculus because they indicate local maximum or minimum points in the original function.
Take the derivative of the function. Depending on the function, you may use the chain rule, product rule, quotient rule or other method. For example, given y=x^3 - 9x, take the derivative to get y'=3x^2 - 9 using the power rule that states taking the derivative of x^n, will give you n*x^(n-1).
Factor the derivative to make finding the zeros easier. Continuing with the example, y'=3x^2 - 9 factors to 3(x+sqrt(3))(x-sqrt(3))
Set the derivative equal to zero and solve for “x” or the independent variable in the equation. In the example, setting 3(x+sqrt(3))(x-sqrt(3))=0 gives x=-sqrt(3) and x=sqrt(3) from the second and third factors. The first factor, 3, doesn't give us a value. These values are the "x" values in the original function that are either local maximum or minimum points.
Plug the value(s) obtained in the previous step back into the original function. This will give you y=c for some constant “c.” This is the equation of the horizontal tangent line. Plug x=-sqrt(3) and x=sqrt(3) back into the function y=x^3 - 9x to get y= 10.3923 and y= -10.3923. These are the equations of the horizontal tangent lines for y=x^3 - 9x. |
SOLVING LINEAR SYSTEMS BY ADDING OR SUBTRACTING
About "Solving linear systems by adding or subtracting"
Solving linear systems by adding or subtracting :
We can use elimination method to solve a system of linear equations. In this method, one variable is eliminated by adding or subtracting the two equations of the system to obtain a single equation in one variable.
Solving linear systems by adding or subtracting - Steps
Step 1 :
Add or subtract the equations to eliminate one of the variables.
Step 2 :
Solve the resulting equation for the other variable.
Step 3 :
Substitute the value of the variable received in step 2 into one of the equations to find the value of the variable eliminated in step 1.
Solving linear systems by adding or subtracting - Examples
Example 1 :
Solve the system of equations by adding. Check your the solution by graphing.
2x - 3y = 12
x + 3y = 6
Step 1 :
In the given two equations, the variable y is having the same coefficient (3). And also, the variable y is having different signs.
So we can eliminate the variable y by adding the two equations.
Step 2 :
Solver the resulting equation for the variable x.
3x = 18
Divide both sides by 3.
3x / 3 = 18 / 3
x = 6
Step 3 :
Substitute the value of x into one of the equations to find the value of y.
x + 6y = 6
Subtract 6 from both sides.
aaaaaaaaaaaaaaaaaaaaaa 6 + 3y = 6 aaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaa - 6 - 6 aaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaa -------------- aaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaa 3y = 0 aaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaa -------------- aaaaaaaaaaaaaaaaaaa
Divide both sides by 3
3y / 3 = 0 / 3
y = 0
Step 3 :
Write the solution as ordered pair as (x, y).
(6, 0)
Step 4 :
Check the solution by graphing.
To graph the equations, write them in slope-intercept form.
That is,
y = mx + b
2x - 3y = 12
y = (2/3)x - 4
Slope = 2/3
y-intercept = -4
x + 3y = 6
y = -(1/3)x + 2
Slope = -1/3
y-intercept = 2
The point of intersection is (6, 0).
Example 2 :
Solve the system of equations by subtracting. Check the solution by graphing.
3x + 3y = 6
3x - y = - 6
Solution :
Step 1 :
In the given two equations, the variable x is having the same coefficient (3), And also, the variable x is having the same sign in both the equations.
So we can eliminate the variable x by subtracting the two equations.
(3x + 3y) - (3x - y) = (6) - (-6)
3x + 3y - 3x + y = 6 + 6
Simplify.
4y = 12
Divide both sides by 4.
4y / 4 = 12 / 4
y = 3
Step 2 :
Plug y = 3 in one of the equations.
3x - y = - 6
3x - 3 = - 6
Add 3 to both sides.
(3x - 3) + 3 = (-6) + 3
3x - 3 + 3 = -6 + 3
Simplify.
3x = -3
Divide both sides by 3
3x / 3 = -3 / 3
x = - 1
Step 3 :
Write the solution as ordered pair as (x, y).
(-1, 3)
Step 4 :
Check the solution by graphing.
To graph the equations, write them in slope-intercept form.
That is,
y = mx + b
3x + 3y = 6
y = - x + 2
Slope = - 1
y-intercept = 2
3x - y = - 6
y = 3x + 6
Slope = 3
y-intercept = 6
The point of intersection is (-1, 3).
Example 3 :
Sum of the cost price of two products is \$50. Sum of the selling price of the same two products is \$52. If one is sold at 20% profit and other one is sold at 20% loss, find the cost price of each product.
Solution :
Step 1 :
Let "x" and "y" be the cost prices of two products.
Then, x + y = 50 --------(1)
Step 2 :
Let us assume that "x" is sold at 20% profit
Then, the selling price of "x" is 120% of "x"
Selling price of "x" = 1.2x
Let us assume that "y" is sold at 20% loss
Then, the selling price of "y" is 80% of "y"
Selling price of "x" = 0.8y
Given : Selling price of "x" + Selling price of "y" = 52
1.2x + 0.8y = 52
To avoid decimal, multiply both sides by 10
12x + 8y = 520
Divide both sides by 4.
3x + 2y = 130 --------(2)
Step 3 :
Eliminate one of the variables to get the value of the other variable.
In (1) and (2), both the variables "x" and "y" are not having the same coefficient.
One of the variables must have the same coefficient.
So multiply both sides of (1) by 2 to make the coefficients of "y" same in both the equations.
(1) ⋅ 2 --------> 2x + 2y = 100 ----------(3)
Variable "y" is having the same sign in both (2) and (3).
To change the sign of "y" in (3), multiply both sides of (3) by negative sign.
- (2x + 2y) = - 100
- 2x - 2y = - 100 --------(4)
Step 4 :
Now, eliminate the variable "y"in (2) and (4) as given below and find the value of "x".
Step 5 :
Plug x = 30 in (1) to get the value of y.
(2) --------> 30 + y = 50
Subtract 30 from both sides.
aaaaaaaaaaaaaaaaaaaa 30 + y = 50 aaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaa - 30 - 30 aaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaa------------------- aaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaa y = 20 aaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaa ------------------- aaaaaaaaaaaaaaaaa
Hence, the cost prices of two products are \$30 and \$20.
After having gone through the stuff given above, we hope that the students would have understood "Solving linear systems by adding or subtracting".
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WORD PROBLEMS
HCF and LCM word problems
Word problems on simple equations
Word problems on linear equations
Word problems on quadratic equations
Algebra word problems
Word problems on trains
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
# Expressions for Real-Life Situations
## Using real-life, practical situations, we look at applications of algebra by developing algebraic expressions for real world situations. Learn more.
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Mathematical Symbols to Represent Words
Rob is describing his weight training to his friend James. He said that when he started training he weighed 185 pounds. He gained 8 pounds in the first month of training. How much did he weigh at the end of the first month of training?
### Translating Words into Math Symbols
Knowing how to translate key words from English into mathematical symbols is important in problem solving. The first step in any problem solving situation in mathematics is always to read the problem. Translating the words into mathematical symbols is next.
Words such as gain, more, sum, total, increase, plus all mean to add. Words such as difference between, minus, decrease, less, fewer, and loss all mean to subtract. Words such as the product of, double \begin{align*}(2x)\end{align*}, twice \begin{align*}(2x)\end{align*}, triple \begin{align*}(3x)\end{align*}, a fraction of, a percent of, or times all mean to multiply. And finally, words such as the quotient of, divided equally, and per mean to divide.
Experience and practice with problem solving will help better acquaint you with the key words that translate into these operations.
#### Let's translate the following statements into into math symbols:
1. What is the sum of five and seventeen?
Break apart the sentence. It is often helpful to underline the words before and after the word AND. Also, it is helpful to box the mathematical symbol.
\begin{align*}& \text{What is the} \ \boxed{\text{sum }} \ \text{of} \ \underline{\text{five}} \ \text{and} \ \underline{\text{seventeen}}? \\ &\qquad \qquad \quad \uparrow \qquad \uparrow \qquad \quad \ \ \uparrow\\ & \qquad \qquad \quad \ {\color{red}+} \qquad 5 \qquad \quad \ \ \ 17\end{align*}
Then translate the symbols together into a mathematical equation and solve it.
\begin{align*}5 {\color{red}+} 17 = 22\end{align*}
1. Thomas had twenty-four dollars and after shopping his money decreased by four dollars.
Break apart the sentence. Underline the numerical words and box the mathematical symbol.
\begin{align*}& \text{Thomas had} \ \underline{\text{twenty-four}} \ \text{dollars and after shopping his money} \ \boxed{\text{decreased }} \ \text{by} \ \underline{\text{four}} \ \text{dollars.}\\ &\qquad \qquad \qquad \ \ \uparrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \uparrow \qquad \quad \ \uparrow\\ & \qquad \qquad \qquad \ \ 24 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ \, {\color{red}-} \qquad \qquad \! \! 4\end{align*}
Then translate the symbols together into a mathematical equation and solve it.
\begin{align*}24 {\color{red}-}4 = 20\end{align*}
Therefore Thomas had 20.00 left after shopping. 1. Nick, Chris, and Jack are sharing a bag of jelly beans. There are 30 jelly beans for the three boys to share equally. How many would each get? Again, break apart the sentence. Underline the numerical words and box the mathematical symbol. \begin{align*}& \text{There are} \ \underline{30} \ \text{jelly beans for the} \ \underline{\text{three}} \ \text{boys to} \ \boxed{\text{share equally.}} \ \text{How many would each get?}\\ &\qquad \quad \ \ \uparrow \qquad \qquad \qquad \qquad \uparrow \qquad \qquad \qquad \uparrow\\ &\qquad \quad \ \ \ 30 \qquad \qquad \qquad \qquad \! \! 3 \qquad \qquad \qquad {\color{red}\div}\end{align*} \begin{align*}30 \div 3 = 10\end{align*} Therefore each boy would get 10 jelly beans. ### Examples #### Example 1 Earlier, you were told that when Rob started training he weighed 185 pounds. He gained 8 pounds in the first month of training. How much did he weigh at the end of the first month of training? The word gain is the same as saying add. Therefore Rob weighs \begin{align*}185 + 8 = 193 \ \text{pounds}.\end{align*} #### Example 2 What is twelve increased by eighteen? Break apart the sentence. Underline the numerical words and box the mathematical symbol. \begin{align*}& \text{What is} \ \underline{\text{twelve}} \ \boxed{\text{increased}} \ \text{by} \ \underline{\text{eighteen}}?\\ &\qquad \quad \ \ \uparrow \qquad \quad \uparrow \qquad \qquad \, \uparrow\\ &\qquad \quad \ \ 12 \qquad \ \ \ {\color{red}+} \qquad \qquad 18\end{align*} \begin{align*}12 + 18 = 30\end{align*} #### Example 3 Joanne and Jillian were each going to share their babysitting money for the week. They made45.00 in total. How much does each girl receive?
Break apart the sentence. Underline the numerical words and box the mathematical symbol.
\begin{align*}& \underline{\text{Joanne and Jillian}} \ \text{want to} \ \boxed{\text{share}} \ \text{their babysitting money for the week. They made} \ \underline{\45.00} \ \text{total}.\\ &\qquad \quad \uparrow \qquad \qquad \qquad \ \uparrow \qquad \qquad \qquad \qquad \qquad \qquad \quad \qquad \qquad \qquad \ \ \uparrow\\ &\qquad \quad \ 2 \qquad \qquad \qquad \ \ \ \, {\color{red}\div} \qquad \qquad \qquad \qquad \qquad \qquad \quad \qquad \qquad \qquad \ \ \ 45\end{align*}
\begin{align*}45 \div 2 = 22.5\end{align*}
Therefore each girl would get 22.50. #### Example 4 The number five is increased by seven. Three-fourths of this number is then decreased from twenty. What is the result? Break apart the sentence. Underline the numerical words and box the mathematical symbol. \begin{align*}& \text{The number} \ \underline{\text{five}} \ \text{is} \ \boxed{\text{increased}} \ \text{by} \ \underline{\text{seven}}. \ \boxed{\text{Three-fourths}} \ \underline{\text{of}}\text{ this} \ \underline{\text{number}} \ \text{is} \ \boxed{\text{decreased}} \ \text{from} \ \underline{\text{twenty}}.\\ &\qquad \qquad \ \ \ \uparrow \qquad \quad \uparrow \qquad \quad \ \uparrow \qquad \qquad \uparrow \quad \qquad \uparrow \qquad \ \ \uparrow \qquad \quad \ \ \uparrow \qquad \qquad \quad \uparrow\\ &\qquad \qquad \quad \! 5 \qquad \quad {\color{red}+} \qquad \qquad \! \! 7 \qquad \qquad \! \frac{3}{4} \qquad \quad \! {\color{red}\times} \qquad \ 5+7 \qquad \quad \, {\color{red}-} \qquad \qquad \ \ \ 20\end{align*} Step 1: \begin{align*}5 + 7 = 12\end{align*} Step 2: \begin{align*}20- \frac{3}{4}(12) = 11\end{align*} ### Review 1. Six less than fifty-three is what number? 2. Twice the sum of eight and nine is what number? 3. Twenty-five is diminished by four times five. What is the result? 4. The product of five times four plus seven is what number? 5. The sum of forty-four and fifty-two then divided by twelve results in what number? 6. Four less than twice 15 is what number? 7. The sum of 12 and the product of 2 and 3 is what number? 8. The number 12 is increased by 4. Three-fourths of this number is then decreased from 20. What is the result? 9. What is 17 decreased by the product of 2 and 4? 10. Mike had100. His money increased by $25 after his tutoring job. How much money does he have now? 11. Kathryn had$20 saved and doubled her money after working on Saturday. How much money does she have now?
12. Jen and Olivia together sold 300 boxes of cookies. Each box of cookies cost \$4. If the money is divided equally, how much money did each girl make?
13. The difference between 212 and the product of 15 and 18 is what number?
14. Lindsey got 5 donations on Saturday. Over the course of the next week she got 12 more donations. How many donations did she get total?
15. The quotient of 12 and the product of 2 and 3 is increased by 15. What is the result?
To see the Review answers, open this PDF file and look for section 2.7.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Addition Words such as gain, more, sum, total, increase, and plus all mean to use addition or to add.
Algebraic Expression An expression that has numbers, operations and variables, but no equals sign.
division Division is a simplified form of repeated subtraction. Division is used to determine the number of times that one term may be subtracted from another before reaching zero. Phrases such as 'the quotient of', 'divided equally', and 'per' all mean to use division or to divide.
Multiplication Terms such as: the product of, double, twice, triple, fraction of, percent of, and times all mean to use multiplication or to multiply.
subtraction Subtraction is an operation used to determine the difference between values. It is the same as adding the opposite, the additive inverse, of a number. Words such as the difference between, minus, decrease, less, fewer, loss all mean to use subtraction. |
2015 AMC 8 Problems/Problem 16
Problem
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If $\frac{1}{3}$ of all the ninth graders are paired with $\frac{2}{5}$ of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
$\textbf{(A) } \frac{2}{15} \qquad\textbf{(B) } \frac{4}{11} \qquad\textbf{(C) } \frac{11}{30} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{11}{15}$
Solution 1
Let the number of sixth graders be $s$, and the number of ninth graders be $n$. Thus, $\frac{n}{3}=\frac{2s}{5}$, which simplifies to $n=\frac{6s}{5}$. Since we are trying to find the value of $\frac{\frac{n}{3}+\frac{2s}{5}}{n+s}$, we can just substitute $\frac{6s}{5}$ for $n$ into the equation. We then get a value of $\frac{\frac{\frac{6s}{5}}{3}+\frac{2s}{5}}{\frac{6s}{5}+s} = \frac{\frac{6s+6s}{15}}{\frac{11s}{5}} = \frac{\frac{4s}{5}}{\frac{11s}{5}} = \boxed{\textbf{(B)}~\frac{4}{11}}$.
Solution 2 (Easy)
We see that the minimum number of ninth graders is $6$, because if there are $3$ then there is $1$ ninth-grader with a buddy, which would mean there are $2.5$ sixth graders, which is impossible (of course unless you really do have half of a person). With $6$ ninth-graders, $2$ of them are in the buddy program, so there $\frac{2}{\tfrac{2}{5}}=5$ sixth-graders total, two of whom have a buddy. Thus, the desired fraction is $\frac{2+2}{5+6}=\boxed{\textbf{(B) }\frac{4}{11}}$.
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
~savannahsolver |
The absolute value of a number is the distance between the number and zero on the real number line. Distances are measured as positive units (or zero units). Consequently, absolute value is never negative. Absolute value answers the question "How far from zero?", but not the question "In which direction from zero?". The notation used for absolute value is two vertical bars.
Algebraically speaking, While it is understood that absolute value yields a positive result (or zero), you must be a bit more careful when stating this concept using a variable, such as x. One never knows the value which may replace the variable (such as x) as it could be positive, or it could be negative.
Absolute value should be treated as a grouping symbol in the order of operations. Be sure to evaluate, or simplify, any values within the absolute value before evaluating the absolute value.
For example: | 4 - 6 | = | -2 | = 2
Examples: 1. | − 3 | = 3 2. − | − 7 | = − 7 3. | 0 | = 0 4. | − 5 |2 = 25 5. − | (− 8)2 | = − 64 6. − | − 3 |2 = − 9 (see note below) 7. | 3 | + | − 7 | − | − 2 | = 8 8. | − 8 | + ( − 9) + | − 14 | + ( − 2) = 11 9. 10. | − 3 + 6 |2 + | − 3 | = 12 In question #6, you can hink of it as − ( | − 3 | • | − 3 | ) = −1 ( 3 • 3 ) = − 9 Think of the negative sign out front as multiplying by −1. By order of operations, exponents are done before multiplication.
Things to Consider:
Measuring Distance: | 8 − 3 | = 5 and | 3 − 8 | = 5 The expression | a − b | represents the distance from a to b on the number line. This distance is the same when measured forward from 3 to 8, or backward from 8 to 3. Properties: | 3 x 7 | = | 3 | x | 7 | 1. | a • b | = | a | • | b | (multiply) 2. | a / b | = | a | / | b | (divide) 3. | a + b | ≠ | a | + | b | (add) 4. | a - b | ≠ | a | - | b | (subtract) 5. | an | = | a |n (power) 6. (called the Triangle Inequality) Square Root Definition:
Graph:
Since absolute value always yields a positive result, or zero, the graph of absolute value plots only y-values that are positive, or zero. The graph resides above the x-axis (plus the origin), in quadrants I and II. y = | x | Negating absolute value creates y-values that are negative or zero. The graph resides in quadrants III and IV (plus the origin). Subtracting 4 from x moves the absolute value graph horizontally 4 units to the right. The y-values remain positive or zero. |
2014-08-26T02:02:18-04:00
This is all about probability. It says "if the marble selected is replaced before the next marble is drawn..." This simply means that if you pull out a marble, you're going to note what colour it is and then you're going to put it back in the bag before you pick out another marble. To find the probability, you need to know the total number of marbles that are in the bag. You have 7 green marbles, 5 blue marbles, and 4 red marbles. This gives you a total of 16 marbles. To find the probability of having a red marble, you have to set up a fraction like this. 4/16. You have to have it this way because this is saying out of every 16 marbles, 4 are red. Simplify this fraction and you have 1/4. To make it into a decimal, you have to divide the numerator (1) by the denominator (4). This gives you .25. This is your decimal. To make it a percent, you have to multiply it by 100 which is the same thing as moving the decimal point two times to the right. You get 25%.
To find the probability of pulling a red or green marble, you have to combine the values. 4 red marbles + 7 green marbles = 11 marbles. Your probability in a fraction is 11/16. You divide 11 by 16 to get 0.6875. Your percent is 68.75% because you multiply the decimal by 100 or move the decimal point twice to the right.
The probability of pulling an orange marble is 0/16, 0.0, and 0% because there are no orange marbles.
A. A red marble?
1/4, 0.25, 25%
B. A red or green marble?
11/16, 0.6875, 68.75%
C. An orange marble?
0/16, 0.0, 0%
2014-08-26T04:40:01-04:00
Red marble is 25%
green marble 75%
orange marble is 0% |
stayinfiji.com
You are watching: 72 divided by 6
Here us will show you step-by-step with detailed explanation exactly how to calculation 72 separated by 6 using lengthy division.Before girlfriend continue, note that in the trouble 72 split by 6, the numbers are identified as follows:72 = dividend6 = divisorStep 1:Start by setup it up through the divisor 6 top top the left side andthe dividend 72 ~ above the appropriate side prefer this:
6 ⟌ 7 2
Step 2:
The divisor (6) goes right into the an initial digit that the dividend (7), 1 time(s). Therefore, placed 1 top top top:
1 6 ⟌ 7 2
Step 3:
Multiply the divisor by the an outcome in the previous action (6 x 1 = 6) and write the answer listed below the dividend.
1 6 ⟌ 7 2 6
Step 4:
Subtract the result in the previous action from the an initial digit of the dividend (7 - 6 = 1) and write the answer below.
1 6 ⟌ 7 2 - 6 1
Step 5:
Move under the second digit that the dividend (2) like this:
1 6 ⟌ 7 2 - 6 1 2
Step 6:
The divisor (6) goes right into the bottom number (12), 2 time(s).
See more: Faith Hill And Tim Mcgraw And Faith Hill Buffalo Ny Chesney Were Arrested
Therefore, placed 2 on top:
1 2 6 ⟌ 7 2 - 6 1 2
Step 7:
Multiply the divisor by the result in the previous action (6 x 2 = 12) and write the answer at the bottom:
1 2 6 ⟌ 7 2 - 6 1 2 1 2
Step 8:
Subtract the result in the previous action from the number written over it. (12 - 12 = 0) andwrite the answer in ~ the bottom.
1 2 6 ⟌ 7 2 - 6 1 2 - 1 2 0
You room done, since there room no an ext digits to move down from the dividend.The answer is the height number and also the remainder is the bottom number.Therefore, the answer to 72 divided by 6 calculated utilizing Long division is:12
0 RemainderLong division CalculatorEnter an additional problem for us to explain and also solve:/ |
# Find the range of values of k for which the line y=kx will cut the curve y=x(x-4) at two real distinct points.
The co-ordinate : (0;0) is the first solution, as pointed out by "Sciencesolve" as both these equations show that y=0 having no constant in either equation. In other words c (or q) is excluded from both equations:
`y = kx + c` and expanding the other equation `y= x^2 - 4x + c`
Determine another reference point on your parabola`` other than (0;0)
If y= -3 ` therefore 0 = x^2 -4x + 3`
```(x-1)(x-3)=0`
`x=1 x=3`
(1;-3) and (3;-3). You will use this information later.
The line cuts the curve and therefore at those points the equations equal each other. Thus:
`kx= x^2 - 4x`
`therefore k=(x^2-4x)/x`
divide by the denominator
`therefore k=x-4`
We have established that x=3 is on the parabola so substitute into
`k=(3) -4`
`therefore` k=-1
Now find y in the original straight line y=kx when k=-1 and x=3
`therefore y=(-1)(3)`
`therefore y=-3`
`therefore ` (3;-3) is a co-ordinate on the straight line
We now know that our straight line equation is y=-x and the point other than (0;0) that is on both is (3;-3), having established above that (3;-3) is also a co-ordinate on the parabola.
We are looking for the range which is represented along ther y-axis
`therefore` `0<=y` `<=3`
Approved by eNotes Editorial Team
You need to understand the request of the problem, hence, if the line `y=kx ` needs to cut the curve `y = x(x-4)` in two real distinct points, then the system of equations `{(y=kx),(y = x(x-4)):}` needs to have two real distinct solutions.
You should substitute `kx` for `y` in the equation of curve such that:
`kx = x(x-4)`
You need to open the brackets such that:
`kx = x^2-4x`
You need to move all terms to one side such that:
`x^2 - 4x - kx = 0`
You need to factor out x such that:
`x(x - 4 - k) = 0`
Hence, evaluating the solutions to the given equation yields:
`x=0 => y = 0`
`x = 4+k => y = k(4+k)`
Hence, evaluating the solutions to the given system of equation yields `(0,0)` and `(4+k, k(4+k)), ` thus k may have any real value.
Approved by eNotes Editorial Team |
# Finding a linear function from given functions
The question is asking to find the linear function, $$f(t) = vt + C$$ for $$f(t+2) = f(t) + 6$$ and $$f(1) = 10$$
The answer is $$3t + 7$$, but I have no idea how the answer is produced.
• Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$. – zipirovich Dec 27 '18 at 15:29
Since $$f(1)=10$$ we have $$f(3)= f(1)+6=16$$. Thus you have to solve the system:
$$v+c=10$$ $$3v+c = 16$$
Hint:
$$f(t+2) = f(t)+6$$
$$f(1) = 10 = f(-1)+6 \iff f(-1) = 4$$
You now have the two points $$(1, 10)$$ and $$(-1, 4)$$.
Small Addition: From here, you can either use the use a system of equations for the two points $$(1, 10)$$ and $$(-1, 4)$$:
$$\begin{cases}\ v+C = 10\\ \ \ -v+C = 4 \end{cases}$$
or you can refer to the definition of a linear equation $$y = vt+C$$, in which $$v$$ is the slope and $$C$$ is the $$y$$-intercept, or the $$y$$-coordinate when $$x = 0$$ (and it’s easy to find here because the point is the midpoint of the two points already found).
$$v = \frac{\Delta f(t)}{\Delta t}$$
... and now the conceptual version. "$$f(t+2) = f(t)+6$$" means that the slope is $$6$$ (the increase in output) over a run of $$2$$ (the increase in input), so the slope is $$\frac{6}{2} = 3$$. Then you have a point on the line, $$(1,10)$$ from "$$f(1) = 10$$", so $$f(t) - 10 = 3(t-1)$$, by point-slope, and we have $$f(t) = 3t + 7$$. |
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# LESSON 2:
EVALUATING
FUNCTIONS
Evaluating a function means
replacing the variable in the
function, in this case x, with a
value from the function’s domain
and computing for the result. To
denote that we are evaluating f at
a for some a in the domain of f,
we write f(a).
Example 1. Evaluate the following
functions at x=1.5:
1.) f(x)=2x+1
2.) q(x)=x²-2x+2
3.) g(x)= 𝑥 + 1
2𝑥+1
4.) r(x)=
𝑥−1
1.) f(x)=4
2.)q(x)=1.25
3.) g(x)= 2.5
4.) r(x)= 8
EXAMPLE 2:
Find g(-4) for g(x)= 𝑥 + 1 , and r(1)
2𝑥+1
for r(x)= .
𝑥−1
Solution: This is not possible
because -4 is not in the
domain of g(x) and 1 is not
in the domain of r(x).
EXAMPLE 3:
- Evaluate the following
functions.
a.) f(3x-1) using f(x)=2x+1
b.)q(2x+3) using q(x)=x²-2x+2
a.) 6x-1
b.)4x²+8x+5
SOLVED
EXAMPLES:
1.) Evaluate the following functions at x=3.
a.)f(x)=x-3
b.)g(x)=x²-3x+5
3
c.)h(x)= 𝑥³ + 𝑥 + 3
𝑥²+1
d.)p(x) =
𝑥−4
## e.) f(x)=│x-5│ where │x-5│ means the
absolute value of x-5.
a.)f(3)=0
b.)g(3)=5
3
c.)h(3)= 33
d.) p(3)=-10
e.)f(3)=2
2.)For what values of x
can we not evaluate the
𝑥+3
function f(x)= ?
𝑥²−4
Solution:
The domain of the function
is given by {x: xԐR, x≠±2}.
Since 2 and -2 are not in
the domain, we cannot
evaluate the function at
x=2,-2.
3.) Evaluate f(a+b) where
f(x) = 4x²-3x.
f(a+b)=4a²-3a+8ab-3b+4b².
SUPPLEMENTARY
EXERCISES:
1.) Evaluate the following functions at x=-4.
a.) f(x)=x³-64
b.)g(x)=│x³-3x²+3x-1│
c.)r(x)= 5 − 𝑥
𝑥+3
d.)q(x)=
𝑥²+7𝑥+12
9 − 𝑥², 𝑖𝑓 𝑥 < 2
2.) Given f(x) = 𝑥 + 7, 𝑖𝑓 2 ≤ 𝑥 < 10 ,
│𝑥 − 4│, 𝑖𝑓 𝑥 ≥ 10
give the values of the following:
a.) f(2)
b.)f(12.5)
c.)f(-3)
d.)f(5)
e.)f(1.5)
3.)Given f(x)=x²-4x+4. Solve
for
a.) f(3)
b.) f(x+3)
c.) Is f(x+3) the same as
f(x)+f(3) ?
4.) A computer shop charges P20 per
hour (or a fraction of an hour) for the
first 2 hours and an additional P10 per
hour for each succeeding hour. Find
how much you would pay if you used
one of their computers for:
a.) 40 minutes
b.) 3 hours
c.) 150 minutes
5.) Under certain circumstances, a rumor |
# ACT Math : How to find the solution to an inequality with subtraction
## Example Questions
### Example Question #1 : How To Find The Solution To An Inequality With Subtraction
Given that x = 2 and y = 4, how much less is the value of 2x2 – 2y than the value of 2y2 – 2x ?
28
52
2
12
28
Explanation:
First, we solve each expression by plugging in the given values for x and y:
2(22) – 2(4) = 8 – 8 = 0
2(42) – 2(2) = 32 – 4 = 28
Then we find the difference between the first and second expressions’ values:
28 – 0 = 28
### Example Question #1 : Inequalities
Solve
Explanation:
Absolute value problems are broken into two inequalities: and . Each inequality is solved separately to get and . Graphing each inequality shows that the correct answer is .
### Example Question #3 : Inequalities
Which of the following inequalities defines the solution set to ?
Explanation:
First, move the s to one side.
Subtract by 1
Divide both sides by 7.
### Example Question #2 : Inequalities
The cost, in cents, of manufacturing pencils is , where 1200 is the number of cents required to run the factory regardless of the number of pencils made, and 20 represents the per-unit cost, in cents, of making each pencil. The pencils sell for 50 cents each. What number of pencils would need to be sold so that the revenue received is at least equal to the manufacturing cost?
Explanation:
If each pencil sells at 50 cents, pencils will sell at . The smallest value of such that
### Example Question #5 : Inequalities
Solve the following inequality:
Explanation:
To solve an inequality with subtraction, simply solve it is as an equation.
The goal is to isolate the variable on one side with all other constants on the other side. Perform the opposite operation to manipulate the inequality.
In this case add two to each side.
### Example Question #2 : How To Find The Solution To An Inequality With Subtraction
Solve the following inequality: |
Math Exercises & Math Problems: Definite Integral of a Function. Find the definite integral of a function: Find the definite integral of a function: By using a definite integral find the area of the region bounded by the given curves: By using a definite integral find the area of the region bounded by the given. The definite integral of a function is closely related to the antiderivative and see that the definite integral will have applications to many problems in calculus. This will show us how we compute definite integrals without using (the be very careful with minus signs and parenthesis with these problems.
Author: Lysanne Volkman Country: Norway Language: English Genre: Education Published: 1 April 2015 Pages: 353 PDF File Size: 28.52 Mb ePub File Size: 31.88 Mb ISBN: 672-7-87065-836-2 Downloads: 63978 Price: Free Uploader: Lysanne Volkman
This may be restated as follows: This limit of a Riemann sum, if it exists, is used to define the definite integral of a function on [ a, b]. If f x is defined on the closed interval [ a, b] then the definite integral problems integral of f x from a to b is defined as if this limit exits.
The function f x is called the integrand, and the variable x is the variable of definite integral problems.
## Integral Calculus | Khan Academy
The numbers a and b are called the limits of integration with a referred to as the lower limit of integration while b is referred to as the upper limit of integration. The reason for this will be made definite integral problems apparent in the following discussion of the Fundamental Theorem of Calculus.
Also, keep in mind that the definite integral is a unique real number and does not represent an infinite number of functions that result from the indefinite integral of a function.
Computing Definite Integrals In this section we are going to concentrate on how we actually evaluate definite integrals in practice. Recall that when we talk about an anti-derivative for a function we are really talking about the indefinite integral for the function.
This should explain the similarity in the notations for the indefinite and definite integrals. Definite integral problems notice that we require the function to be continuous in the interval of integration.
## Math Exercises & Math Problems: Definite Integral of a Function
This was also a requirement in the definition of the definite integral. In this definite integral problems however, we will need to keep this condition in mind as we do our evaluations.
Example 1 Evaluate each of the following.
This is here only to make sure that we understand the difference between an indefinite and a definite integral. We just definite integral problems the most general anti-derivative in the first part so we can use that if we want to.
Also, be very careful with minus signs and parenthesis. Notice as well that, in order to help with the evaluation, we rewrote the indefinite integral a little.
In particular we got rid of the negative exponent on the second term.
### Definite Integrals
Recall that in order for us to do an integral the integrand must be continuous in the range of the definite integral problems. Integration Integration can be used to find areas, volumes, central points and many useful things.
But it is often used to find the area under the graph definite integral problems a function like this: |
# How do you prove [(1)/(1-sinx)]+[(1)/(1+sinx)]=2sec^2x?
Apr 23, 2015
The formula can be proven by applying: 1) Least common multiple; 2) applying the trigonometric entity ${\sin}^{2} x + {\cos}^{2} x = 1$
#### Explanation:
Key-relation : ${\sin}^{2} x + {\cos}^{2} x = 1$
Key-concept: Least common multiple; when no common multiples, just multiply the terms in the denominator.
Calculation
The above formula can be proven by transforming left side to right side:
$\frac{1}{1 - \sin x} + \frac{1}{1 + \sin x} = \frac{1 + \sin x + 1 - \sin x}{\left(1 + \sin x\right) \left(1 - \sin x\right)}$
To arrive to right-hand side, just divide the denominator to $\left(1 + \sin x\right) \left(1 - \sin x\right)$, the least common multiple, and multiply the numerator to the remaining, since they are all 1, just put the value.
By simple algebra and make use of $\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$, it can be seen from normal multiplication.
$\frac{1 + \sin x + 1 - \sin x}{\left(1 + \sin x\right) \left(1 - \sin x\right)} = \frac{2}{1 - {\sin}^{2} x}$
Finally apply: ${\sin}^{2} x + {\cos}^{2} x = 1$, which gives out ${\cos}^{2} x = 1 - {\sin}^{2} x$
$\frac{2}{1 - {\sin}^{2} x} = \frac{2}{\cos} ^ 2 x = 2 \cdot {\left(\frac{1}{\cos} x\right)}^{2}$
To finish, remember that $\sec x = \frac{1}{\cos} x$, hence:
$2 \cdot {\left(\frac{1}{\cos} x\right)}^{2} = 2 {\sec}^{2} x$ |
Elementary Counting Techniques & Combinatorics
# Elementary Counting Techniques & Combinatorics
## Elementary Counting Techniques & Combinatorics
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Elementary Counting Techniques & Combinatorics Martina Litschmannová martina.litschmannova@vsb.cz K210
2. Consider: • How many license plates are possible with 3 letters followed by 4 digits? • How many license plates are possible with 3 letters followed by 4 digits if no letter repeated? • How many different ways can we chose from 4 colors and paint 6 rooms?
3. Consider: • How many different orders may 9 people be arranged in a line? • How many ways can I return your tests so that no one gets their own? • How many distinct function exist between two given finite sets A and B?
4. Probability basics
5. Sets and subsets Set Definitions • A set is a well-defined collection of objects. • Each object in a set is called an elementof the set. • Two sets are equal if they have exactly the same elements in them. • A set that contains no elements is called a null set or an empty set. • If every element in Set A is also in Set B, then Set A is a subset of Set B.
6. Sets and subsets Set Notation • A set is usually denoted by a capital letter, such as A, B… • An element of a set is usually denoted by a small letter, such as x, y, or z. • A set may be described by listing all of its elements enclosed in braces. For example, if Set A consists of the numbers 2, 4, 6 and 8, we may say: A = {2, 4, 6, 8}. • Sets may also be described by stating a rule. We could describe Set A from the previous example by stating: Set A consists of all the even single-digit positive integers. • The null set is denoted by {∅}.
7. Sets and subsets Set Operations • The union of two sets is the set of elements that belong to one or both of the two sets. Symbolically, the union of A and B is denoted by A ∪ B.
8. Sets and subsets Set Operations • The intersection of two sets is the set of elements that are common to both sets. Symbolically, the intersection of A and B is denoted by A ∩ B.
9. Sample problems • Describe the set of vowels. • Describe the set of positive integers. Since it would be impossible to list all of the positive integers, we need to use a rule to describe this set. • What is the set of men with four arms? It is the null set (or empty set). {}
10. Sample problems • Set A = {1, 2, 3} and Set B = {1, 2, 4, 5, 6}. Is Set Aa subset of Set B? Set A would be a subset of Set B if every element from Set A were also in Set B. However, this is not the case. The number 3 is in Set A, but not in Set B. Therefore, Set A is not a subset of Set B.
11. Statsexperiments Statistical experiments have three things in common: • The experiment can have more than one possible outcome. • Each possible outcome can be specified in advance. • The outcome of the experiment depends on chance. Rollingdice
12. Statsexperiments Statistical experiments have three things in common: • The experiment can have more than one possible outcome. • Each possible outcome can be specified in advance. • The outcome of the experiment depends on chance. Determining the amount of cholesterol in the blood
13. Statsexperiments Statistical experiments have three things in common: • The experiment can have more than one possible outcome. • Each possible outcome can be specified in advance. • The outcome of the experiment depends on chance. Measurementoftasksnumber, which they require for a certain period
14. Statsexperiments Sample spaceis a set of elements that represents all possible outcomes of a statistical experiment. • A sample point is an element of a sample space. • An event is a subset of a sample space - one or more sample points.
15. Statsexperiments Types of events • Two events are mutually exclusive (disjoint) if they have no sample points in common. • Two events are independent when the occurrence of one does not affect the probability of the occurrence of the other.
16. Sample problems • Suppose I roll a die. Is that a statistical experiment? Yes. Like a coin toss, rolling dice is a statistical experiment. There is more than one possible outcome. We can specify each possible outcome in advance. And there is an element of chance. • When you roll a single die, what is the sample space? The sample space is all of the possible outcomes -.
17. Sample problems • Which of the following are sample points when you roll a die - 3, 6, and 9? The number 9 is not a sample point, since it is outside the sample space. The numbers 3 and 6 are sample points, because they are in the sample space. • Which of the following sets represent an event when you roll a die? Each of the sets shown above is a subset of the sample space, so each represents an event.
18. Sample problems • Consider the events listed below. Which are mutually exclusive? A. {1}B. {2, 4}C. {2, 4, 6} Two events are mutually exclusive, if they have no sample points in common. {}
19. Sample problems • Suppose you roll a die two times. Is each roll of the die an independent event? Yes. Two events are independent when the occurrence of one has no effect on the probability of the occurrence of the other. Neither roll of the die affects the outcome of the other roll; so each roll of the die is independent.
20. Combinatorics
21. Combinatorics is the branch of discrete mathematics concerned with determining the size of finite sets without actually enumerating each element.
22. Product rule occurs when two or more independent events are grouped together. The first rule of counting helps us determine how many ways an event multiple can occur. Suppose we have k independent events. Event 1 can be performed in n1 ways; Event 2, in n2 ways; and so on up to Event k, which can be performed in nk ways. The number of ways that these events can be performed together is equal to n1n2. . . nkways.
23. How many sample points are in the sample space when a coin is flipped 4 times? • Coinflipp – 2 outcomes (headortails) Flipp 1 Flipp 3 Flipp 4 Flipp 2 2 2 2 2
24. A business man has 4 dress shirts and 7 ties. How many different shirt/tie outfits can he create? Dressshirts Ties 4 7
25. How many different ways can we chose from 4 colors and paint 3 rooms? Room 1 Room 3 Room 2 4 4 4
26. How many different ways can we chose from 4 colors and paint 3 rooms, if no room is to be the same color? Room 1 Room 3 Room 2 4 3 2
27. How many different orders may 8 people be arranged in? Pos 1 Pos 4 Pos 5 Pos 8 Pos 2 Pos 6 Pos 3 Pos 7 8 7 4 6 5 3 2 1
28. How many different 3 people can be selected from a group of 8 people to a president, vice-president, treasure of the group? Vice-president president Treasure 8 7 6
29. How many license plates are possible with 3 letters followed by 4 digits? A B C . . . Z 0 1 2 . . . 9 26 10 17 576 000
31. You have five novels, four magazines, and three devotional books. How many options do you have for taking one for your wait in the bank line? 3 subtasks - pick a novel, pick a magazine, or pick a devotional book. Magazines Books Novels 5 + 4 + 3
32. Consider the following road map.a)How many ways are there to travel from A to B, and back to A, without going through C? C A B
33. Consider the following road map.a)How many ways are there to travel from A to B, and back to A, without going through C? C A B from A to B: 4 ways from B to A: 4 ways There are waysto travel from A to B, and back to A, without going through C.
34. Consider the following road map.b)How many ways are there to go from A to C, stopping once at B? C A B from A to B from B to C 4 2 OR from A to B fromB to A fromA to C 4 4 1
35. Consider the following road map.c)How many ways are there to go from A to C, making at most one intermediate stop? C A B Without intermediate stop: 1 way With 1 intermediate stop: 8 ways ------------------------------------------ At most oneintermediate stop:
36. The Pigeonhole Principle If k + 1 or more objects are placed in k boxes, then there is at least one box containing two or more objects.
37. The Pigeonhole Principle If k + 1 or more objects are placed in k boxes, then there is at least one box containing two or more objects.
38. The Pigeonhole Principle If k + 1 or more objects are placed in k boxes, then there is at least one box containing two or more objects.
39. Sample problems • Among 367 people, there must be at least 2 with the same birthday, since there is only 366 possible birthdays. • In a collection of 11 numbers, at least 2 must have the same least significant digit.
40. The GeneralizedPigeonhole Principle If N objects are placed into k boxes, then there is at least one box containing at least N/k objects. aissmallest integer larger thanorequal toa (ceilingfunction)
41. Sample problems • Among 100 people there are at least 100/12 = 9 people with the same birthday month. • At FEI, VŠB-TUO there are at least 3619/366 = 10 people with the same birthday.
42. In a class of 44 students, how many will receive the same grade on a scale {A, B, C, D, F}?
43. How many people must we survey, to be sure at least 50 have the same political party affiliation, if we use the three affiliations {Democrat, Republican, neither}?
44. Permutations and Combinations
45. Consider: How many ways can we choose r things from a collection of n things? pick Pick 4 from 9 colored balls
46. Consider: How many ways can we choose r things from a collection of n things? This statement is ambiguous in several ways: • Are the n things distinct or indistinguishable? • Do the selected items form a set (unordered collection) or a sequence (ordered)? • May the same item be selected from the r items more then once? (Are repetitions permitted?)
47. Consider: How many ways can we choose r things from a collection of n things? pick • Example using balls: • Are the balls identical or different colors? Are some different colors, others the same? • Are balls tossed in a bucket (unordered) or lined up in a line in the order chosen? • Each ball returned to the collection before the next is selected?
48. Permutations • An ordered selection of objects. • If there is a collection of n objects to chose from, and we are selecting alln objects, then we call each possible selection a permutation from the collection. • In the general case the items are all distinct, and repetitions are not permitted.
49. Permutations Possible permutations of three colored balls:
50. Permutations 1. object 2. object 3. object n. object … 1 n-1 n-2 n Thus, by the product rule, the number of ways to arrange n objects is: The number of permutations of a set of n objects is the product of the first n positive integers, that isn! |
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# Probability
The probability is the quantitative representation of the likelihood of an event to occur. In terms of probability, an event is referred to a possible set of outcomes when an experiment is performed and the probability is measured for the events. The probability of an event falls between $0$ and $1$. Here, $0$ represents an impossible event and $1$ implies a certain event. The chances of occurring of an event are more if the probability is higher.
Related Calculators Calculation of Probability Binomial Distribution Probability Calculator Binomial Probability Calculator Coin Toss Probability Calculator
## How do you find the Probability?
Numerically, the probabilities can be calculated by dividing the number of favorable or desired outcomes with the total number of possible outcomes. This is applicable in the cases when the experiments performed are well defined and random.
Let us take the simplest example. When an unbiased or fair coin is tossed in air, there are only two possible outcomes - head and tail. So, the probability to getting a head is equal to the $\frac{1}{2}$, where 1 is the favorable outcome (head) and $2$ represents total number of outcomes (head and tail). The coin being fair, the probability of both outcomes (head and tail) is same.
## Probability Formula
Probability of an Event:
$P(E)$ = $\frac{Number\ of\ desired\ outcomes}{Number\ of\ total\ outcomes}$
Complementary Events:
The events that have only two possible outcomes are called complementary events.
Example: Tossing a coin $(\frac{head}{tail})$, getting $2$ and not $2$ on rolling a die.
$P(A) + P(A^C)$ = $1$
The probability of happening of either event A or event B is:
$P(A$ or $B)$ = $P(A) + P(B) - P(A$ and $B)$
We may even write this formula as:
$P(A \cup B)$ = $P(A) + P(B) - P(A \cap B)$
Mutually Exclusive or Disjoint Events:
The events that cannot happen together at the same time are mutually exclusive. Two events $A$ and $B$ are mutually exclusive if
$P(A \cap B)$ = $0$
So, law of addition in that case is:
$P(A \cup B)$ = $P(A) + P(B)$
Independent Events:
Two events are independent if happening of one doesn’t affect that of another. So, events $A$ and $B$ are independent if
$P(A \cap B)$ = $P(A) \cdot P(B)$
Conditional Probability:
Conditional probability refers to the probability of an event when another event has already occurred. Probability of event B, provided that A has already occurred is:
$P(A|B)$ = $\frac{P(A\ \cap\ B)}{P(B)}$
Bayes Law:
$P(A|B)$ = $\frac{[P(B|A)\ \cdot\ P(A)]}{P(B)}$
## Examples
Example 1:
On a roll of a fair $6$ sided die, what is the probability of getting of rolling a $1$ or a $4$?
Solution:
Total number of outcomes = $6$
Probability of getting a $1$
$P(1)$ = $\frac{1}{6}$
Probability of getting a $4$
$P(4)$ = $\frac{1}{6}$
Since both events are mutually-exclusive,
$P(1$ or $4)$ = $P(1) + P(4)$ = $\frac{1}{6}$ + $\frac{1}{6}$ = $\frac{1}{3}$
Example 2:
Determine the probability of drawing a jack and an ace consecutively without replacement from a well-shuffled deck of cards.
Solution:
Total outcomes = $52$
There are $4$ jacks. So
$P(Jack)$ = $\frac{4}{52}$ = $\frac{1}{13}$
Since another card is drawn without replacement, hence now total outcomes are $51$.
$P(Ace)$ = $\frac{4}{51}$
Required probability = $P(Jack) \times P(Ace)$ = $\frac{1}{13}$ $\times$ $\frac{4}{51}$ = $\frac{4}{663}$
More topics in Probability Probability Problems Probability Formulas Probability Terms Probability Rules Types of Probability Binomial Distribution Random Variable Uniform Distribution Permutations Combinations Bayes Theorem Probability
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# Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.3
## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.3
Question 1.
Find the differential equation of the family of
(i) all non-vertical lines in a plane
Solution:
Equation of family of all non-vertical lines is
y = mx + c (m ≠ 0)
Differentiate with respect to ‘x’
(ii) all non-horizontal lines in a plane.
Solution:
Equation of family of all non-horizontal lines is
Question 2.
Form the differential equation of all straight lines touching the circle x2 + y2 = r2
Solution:
Equation of circle x2 + y2 = r2 of the line y = mx + c is to be a tangent to the circle, then the equation of the tangent is
Differentiating with respect to V dy
Question 3.
Find the differential equation of the family of circles passing through the origin and having their centres on the x -axis.
Solution:
All circles passing through the origin and having their centre on the x -axis say at (a, 0) will have radius ‘a’ units.
∴ Equation of circle is (x – a2) + y2 = a2 ….(1) [ ∵ ‘a’ arbitrary constant]
Differentiate with respect to ‘x’
Question 4.
Find the differential equation of the family of all the parabolas with latus rectum 4a and whose axes are parallel to the x -axis.
Solution:
Equation of all parabolas whose axis is parallel to X – axis is
(y – k)2 = 4a (x – h)
Where (h, k) is the vertex
Question 5.
Find the differential equation of the family of parabolas with vertex at (0, -1) and having axis along the y – axis.
Solution:
Given, vertex (0, -1) and axis along y-axis
Equation of Parabola, (x + 1)2 = – 4ay …… (1) [∵ a is the perameter]
Differentiate with respect to ‘x’
Question 6.
Find the differential equations of the family of all the ellipses having foci on the y – axis and centre at the origin.
Solution:
Equations of the family of all the Ellipses having foci on the y – axis and centre at the origin is
Differentiate with respect to ’x’
Question 7.
Find the differential equation corresponding to the family of curves represented by the equation y = Ae8x + Be-8x, where A and B are arbitrary constants.
Solution:
Question 8.
Find the differential equation of the curve represented by xy = aex + be-x + x2.
Solution:
xy = aex + be-x + x2
xy – x2 = aex + be-x …… (1) [∵ a’, b’ are arbitrary constants]
Differentiate with respect to ‘x’
xy’ +y – 2x = aex – be-x
Again, Differentiate with respect to ‘x’
### Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.3 Additional Problems
Question 1.
Find the differential equation of the family of straight lines y = mx + $$\frac{a}{m}$$ when
(i) m is the parameter,
(ii) a is the parameter,
(iii) a, m both are parameters.
Solution:
Question 2.
Find the differential equation that will represent family of all circles having centres on the x-axis and the radius is unity.
Solution:
Equation of a circle with centre on x-axis and radius 1 unit is
(x – a)2 + y2 = 1 ….. (1)
Differentiating with respect to x,
2 (x – a) + 2yy’ – 0
⇒ 2 (x – a) = – 2yy’
(or) x – a = -yy’ ……(2)
Substituting (2) in (1), we get,
Question 3.
From the differential equation from the following equations.
(i) y = e2x (A + Bx)
Solution:
ye-2x = A + Bx ……. (1)
Since the above equation contains two arbitrary constants, differentiating twice,
(ii) y = ex(A cos 3x + B sin 3x)
Solution:
(iii) Ax2 + By2 = 1
Solution:
Eliminating A and B between (1), (2) and (3) we get
(iv) y2 = 4a(x – a)
Solution: |
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# Adding and Subtracting Rational Expressions Part I
### Learning Outcomes
• Add and subtract rational expressions with like denominators
• Add and subtract rational expressions with unlike denominators using a greatest common denominator
In beginning math, students usually learn how to add and subtract whole numbers before they are taught multiplication and division. However, with fractions and rational expressions, multiplication and division are sometimes taught first because these operations are easier to perform than addition and subtraction. Addition and subtraction of rational expressions are not as easy to perform as multiplication because, as with numeric fractions, the process involves finding common denominators. By working carefully and writing down the steps along the way, you can keep track of all of the numbers and variables and perform the operations accurately.
## Adding and Subtracting Rational Expressions with Like Denominators
Adding rational expressions with the same denominator is the simplest place to start, so let’s begin there. To add fractions with like denominators, add the numerators and keep the same denominator. Then simplify the sum. You know how to do this with numeric fractions.
$\begin{array}{c}\frac{2}{9}+\frac{4}{9}=\frac{6}{9}\\\\\frac{6}{9}=\frac{3\cdot 2}{3\cdot 3}=\frac{3}{3}\cdot \frac{2}{3}=1\cdot \frac{2}{3}=\frac{2}{3}\end{array}$
Follow the same process to add rational expressions with like denominators. Let's try one.
### Example
Add $\displaystyle \frac{2{{x}^{2}}}{x+4}+\frac{8x}{x+4}$, and define the domain. State the sum in simplest form.
$\frac{2{{x}^{2}}+8x}{x+4}$
Factor the numerator.
$\frac{2x(x+4)}{x+4}$
Simplify common factors and.
$\large\begin{array}{c}\frac{2x\cancel{(x+4)}}{\cancel{x+4}}\\\\=\frac{2x}{1}\end{array}$
The domain is found by setting the denominators in the original sum equal to zero.
$\begin{array}{l}x+4=0\\x=-4\end{array}$
The domain is $x\ne-4$
[latex-display] \displaystyle \frac{2{{x}^{2}}}{x+4}+\frac{8x}{x+4}=2x,x\ne -4[/latex-display]
Caution! Remember to define the domain of a sum or difference before simplifying. You may lose important information when you simplify. In the example above, the domain is $x\ne-4$. If we were to have defined the domain after simplifying, we would find that the domain is all real numbers which is incorrect.
To subtract rational expressions with like denominators, follow the same process you use to subtract fractions with like denominators. The process is just like the addition of rational expressions, except that you subtract instead of add.
### Example
Subtract$\frac{4x+7}{x+6}-\frac{2x+8}{x+6}$, and define the domain. State the difference in simplest form.
Answer: Subtract the second numerator from the first and keep the denominator the same.
$\frac{4x+7-(2x+8)}{x+6}$
Be careful to distribute the negative to both terms of the second numerator.
$\frac{4x+7-2x-8}{x+6}$
Combine like terms. This rational expression cannot be simplified any further.
$\frac{2x-1}{x+6}$
The domain is found from the denominators of original expression.
$\begin{array}{l}x+6=0\\x=-6\end{array}$
The domain is $x\ne-6$
[latex-display] \displaystyle \frac{4x+7}{x+6}-\frac{2x+8}{x+6}=\frac{2x-1}{x+6},\text{}x\ne-6[/latex-display]
### Try It
[ohm_question]40242[/ohm_question]
In the video that follows, we present more examples of adding rational expressions with like denominators. Additionally, we review finding the domain of a rational expression. https://www.youtube.com/watch?v=BeaHKtxB868&feature=youtu.be
## Adding and Subtracting Rational Expressions with Unlike Denominators
What do they have in common?
Before adding and subtracting rational expressions with unlike denominators, you need to find a common denominator. Once again, this process is similar to the one used for adding and subtracting numeric fractions with unlike denominators. Remember how to do this?
$\displaystyle \frac{5}{6}+\frac{8}{10}+\frac{3}{4}$
Since the denominators are $6$, $10$, and $4$, you want to find the least common denominator and express each fraction with this denominator before adding. (BTW, you can add fractions by finding any common denominator; it does not have to be the least. You focus on using the least because then there is less simplifying to do. But either way works.) Finding the least common denominator is the same as finding the least common multiple of $4$, $6$, and $10$. There are a couple of ways to do this. The first is to list the multiples of each number and determine which multiples they have in common. The least of these numbers will be the least common denominator.
Number
Multiples
$4$ $8$ $12$ $16$ $20$ $24$ $28$ $32$ $36$ $40$ $44$ $48$ $52$ $56$ $\textbf{60}$ $64$
$6$ $12$ $18$ $24$ $30$ $36$ $42$ $48$ $54$ $\textbf{60}$ $66$ $68$
$10$ $20$ $30$ $40$ $50$ $\textbf{60}$ $70$ $80$
The other method is to use prime factorization, the process of finding the prime factors of a number. This is how the method works with numbers.
### Example
Use prime factorization to find the least common multiple of $6$, $10$, and $4$.
Answer: First, find the prime factorization of each denominator.
$\begin{array}{r}6=3\cdot2\\10=5\cdot2\\4=2\cdot2\end{array}$
The LCM will contain factors of $2$,$3$, and $5$. Multiply each number the maximum number of times it appears in a single factorization. In this case, $3$ appears once, $5$ appears once, and $2$ is used twice because it appears twice in the prime factorization of $4$. Therefore, the LCM of $6$, $10$, and $4$ is $3\cdot5\cdot2\cdot2$, or $60$.
$\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,6=3\cdot2\\\,\,\,\,\,\,\,\,10=5\cdot2\\\,\,\,\,\,\,\,\,\,\,\,4=2\cdot2\\\text{LCM}=3\cdot5\cdot2\cdot2\end{array}$
The least common multiple of $6, 10$, and $4$ is $60$.
Both methods give the same result, but prime factorization is faster. Your choice! Now that you have found the least common multiple, you can use that number as the least common denominator of the fractions. Multiply each fraction by the fractional form of $1$ that will produce a denominator of $60$:
$\begin{array}{r}\frac{5}{6}\cdot \frac{10}{10}=\frac{50}{60}\\\\\frac{8}{10}\cdot\frac{6}{6}=\frac{48}{60}\\\\\frac{3}{4}\cdot\frac{15}{15}=\frac{45}{60}\end{array}$
Now that you have like denominators, add the fractions:
$\frac{50}{60}+\frac{48}{60}+\frac{45}{60}=\frac{143}{60}$
In the next example, we show how to find the least common multiple of a rational expression with a monomial in the denominator.
### Example
Add$\frac{2n}{15m^{2}}+\frac{3n}{21m}$, and give the domain. State the sum in simplest form.
Answer: Find the prime factorization of each denominator.
$\begin{array}{l}15m^{2}\,=\,3\cdot5\cdot{m}\cdot{m}\\\,\,\,21m\,=\,3\cdot7\cdot{m}\end{array}$
Find the least common multiple. $3$ appears exactly once in both of the expressions, so it will appear once in the least common multiple. Both $5$ and $7$ appear at most once. For the variables, the most m appears is twice. Use the least common multiple for your new common denominator, it will be the LCD.
$\begin{array}{l}15m^{2}\,=\,3\cdot5\cdot{m}\cdot{m}\\21m\,\,\,=\,3\cdot7\cdot{m}\\\text{LCM}:3\cdot5\cdot7\cdot{m}\cdot{m}\\\text{LCM}:105m^{2}\end{array}$
Compare each original denominator and the new common denominator. Now rewrite the rational expressions to each have the common denominator of $105m^{2}$. Remember that m cannot be 0 because the denominators would be $0$. The first denominator is $15m^{2}$ and the LCD is $105m^{2}$. You need to multiply $15m^{2}$ by $7$ to get the LCD, so multiply the entire rational expression by $\frac{7}{7}$. The second denominator is $21m$ and the LCD is $105m^{2}$. You need to multiply $21m$ by $5m$ to get the LCD, so multiply the entire rational expression by $\frac{5m}{5m}$.
$\begin{array}{c}\frac{2n}{15m^{2}}\cdot\frac{7}{7}=\frac{14n}{105m^{2}}\\\\\frac{3n}{21m}\cdot\frac{5m}{5m}=\frac{15mn}{105m^{2}}\end{array}$
Add the numerators and keep the denominator the same.
$\frac{14n}{105{{m}^{2}}}+\frac{15mn}{105{{m}^{2}}}=\frac{14n+15mn}{105{{m}^{2}}}$
If possible, simplify by finding common factors in the numerator and denominator. This rational expression is already in simplest form because the numerator and denominator have no factors in common.
$\displaystyle \frac{n(14+15m)}{105{{m}^{2}}}$
[latex-display] \displaystyle \frac{2n}{15{{m}^{2}}}+\frac{3n}{21m}=\frac{n(14+15m)}{105{{m}^{2}}},m\ne 0[/latex-display]
That took a while, but you got through it. Adding rational expressions can be a lengthy process, but taken one step at a time, it can be done. Now let's take a look at some examples where the denominator is not a monomial. To find the least common denominator (LCD) of two rational expressions, we factor the expressions and multiply all of the distinct factors. For instance, consider the following rational expressions:
$\dfrac{6}{\left(x+3\right)\left(x+4\right)},\text{ and }\frac{9x}{\left(x+4\right)\left(x+5\right)}$
The LCD would be $\left(x+3\right)\left(x+4\right)\left(x+5\right)$.
To find the LCD, we count the greatest number of times a factor appears in each denominator and include it in the LCD that many times. For example, in $\dfrac{6}{\left(x+3\right)\left(x+4\right)}$, $\left(x+3\right)$ is represented once and $\left(x+4\right)$ is represented once, so they both appear exactly once in the LCD. In $\dfrac{9x}{\left(x+4\right)\left(x+5\right)}$, $\left(x+4\right)$ appears once and $\left(x+5\right)$ appears once. We have already accounted for $\left(x+4\right)$, so the LCD just needs one factor of $\left(x+5\right)$ to be complete. Once we find the LCD, we need to multiply each expression by the form of $1$ that will change the denominator to the LCD. What do we mean by " the form of $1$"? $\frac{x+5}{x+5}=1$ so multiplying an expression by it will not change its value. For example, we would need to multiply the expression $\dfrac{6}{\left(x+3\right)\left(x+4\right)}$ by $\frac{x+5}{x+5}$ and the expression $\frac{9x}{\left(x+4\right)\left(x+5\right)}$ by $\frac{x+3}{x+3}$. Hopefully this process will become clear after you practice it yourself. As you look through the examples on this page, try to identify the LCD before you look at the answers. Also, try figuring out which "form of 1" you will need to multiply each expression by so that it has the LCD.
### Example
Add the rational expressions $\frac{5}{x}+\frac{6}{y}$ and define the domain. State the sum in simplest form.
Answer: First, define the domain of each expression. Since we have x and y in the denominators, we can say $x\ne0 ,\text{ and }y\ne0$. Now we have to find the LCD. Since x appears once and y appears once, the LCD will be $xy$. We then multiply each expression by the appropriate form of 1 to obtain $xy$ as the denominator for each fraction.
$\begin{array}{l}\frac{5}{x}\cdot \frac{y}{y}+\frac{6}{y}\cdot \frac{x}{x}\\ \frac{5y}{xy}+\frac{6x}{xy}\end{array}$
Now that the expressions have the same denominator, we simply add the numerators to find the sum.
$\frac{6x+5y}{xy}$
[latex-display]\frac{5}{x}+\frac{6}{y}=\frac{(6x+5y)}{xy}[/latex], $x\ne0 ,\text{ and }y\ne0[/latex-display] Here is one more example of adding rational expressions where the denominators are multi-term polynomials. First, we will factor and then find the LCD. Note that [latex]x^2-4$ is a difference of squares and can be factored using special products.
### Example
Simplify$\frac{2{{x}^{2}}}{{{x}^{2}}-4}+\frac{x}{x-2}$ and give the domain. State the result in simplest form.
Answer: Find the least common denominator by factoring each denominator. The least common denominator includes the maximum number of times it appears in a single factorization. Remember that x cannot be $2$ or $-2$ because the denominators would be $0$. $\left(x+2\right)$ appears a maximum of one time and so does $\left(x–2\right)$. This means the LCD is $\left(x+2\right)\left(x–2\right)$. Multiply each expression by the equivalent of $1$ that will give it the common denominator. Notice the first fraction already has the LCD and as a result remains unchanged.
$\begin{array}{r}\frac{2{{x}^{2}}}{{{x}^{2}}-4}=\frac{2{{x}^{2}}}{(x+2)(x-2)}\\\frac{x}{x-2}\cdot \frac{x+2}{x+2}=\frac{x(x+2)}{(x+2)(x-2)}\end{array}$
Rewrite the original problem with the common denominator. It makes sense to keep the denominator in factored form in order to check for common factors.
$\displaystyle \frac{2{{x}^{2}}}{(x+2)(x-2)}+\frac{x(x+2)}{(x+2)(x-2)}$
Combine the numerators.
$\begin{array}{c}\frac{2{{x}^{2}}+x(x+2)}{(x+2)(x-2)}\\\\\frac{2{{x}^{2}}+{{x}^{2}}+2x}{(x+2)(x-2)}\end{array}$
$\frac{3{{x}^{2}}+2x}{(x+2)(x-2)}$
Check for simplest form. Since neither $\left(x+2\right)$ nor $\left(x-2\right)$ is a factor of $3{{x}^{2}}+2x$, this expression is in simplest form. [latex-display] \displaystyle \frac{2{{x}^{2}}}{{{x}^{2}}-4}+\frac{x}{x-2}=\frac{3{{x}^{2}}+2x}{(x+2)(x-2)}[/latex] $\displaystyle x\ne 2,-2[/latex-display] The video that follows contains an example of adding rational expressions whose denominators are not alike. The denominators of both expressions contain only monomials. https://www.youtube.com/watch?v=Wk8ZZhE9ZjI&feature=youtu.be ## Subtracting Rational Expressions Now let’s try subtracting rational expressions. You'll use the same basic technique of finding the least common denominator and rewriting each rational expression to have that denominator. ### Example Subtract[latex]\frac{2}{t+1}-\frac{t-2}{{{t}^{2}}-t-2}$, define the domain. State the difference in simplest form.
Answer: Find the prime factorization of each denominator. $t+1$ cannot be factored any further, but ${{t}^{2}}-t-2$ can be. Remember that t cannot be $-1$ or $2$ because the denominators would be $0$.
$\begin{array}{c}t+1=t+1\\t^{2}-t-2=\left(t-2\right)\left(t+1\right)\end{array}$
Find the least common multiple. $t+1$ appears exactly once in both of the expressions, so it will appear once in the least common denominator. $t–2$ also appears once. This means that $\left(t-2\right)\left(t+1\right)$ is the least common multiple. In this case, it is easier to leave the common multiple in terms of the factors, so you will not multiply it out. Use the least common multiple for your new common denominator, it will be the LCD.
$\begin{array}{c}t+1=t+1\\t^{2}-t-2=\left(t-2\right)\left(t+1\right)\\\text{LCM}:\left(t+1\right)\left(t-1\right)\end{array}$
Compare each original denominator and the new common denominator. Now rewrite the rational expressions to each have the common denominator of $\left(t+1\right)\left(t–2\right)$. You need to multiply $t+1$ by $t–2$ to get the LCD, so multiply the entire rational expression by $\displaystyle \frac{t-2}{t-2}$. The second expression already has a denominator of $\left(t+1\right)\left(t–2\right)$, so you do not need to multiply it by anything.
$\begin{array}{c}\frac{2}{t+1}\cdot \frac{t-2}{t-2}=\frac{2(t-2)}{(t+1)(t-2)}\\\\\,\,\,\frac{t-2}{{{t}^{2}}-t-2}=\frac{t-2}{(t+1)(t-2)}\end{array}$
Then rewrite the subtraction problem with the common denominator.
$\frac{2\left(t-2\right)}{\left(t+1\right)\left(t-2\right)}-\frac{t-2}{\left(t+1\right)\left(t-2\right)}$
Subtract the numerators and simplify. Remember that parentheses need to be included around the second $\left(t–2\right)$ in the numerator because the whole quantity is subtracted. Otherwise you would be subtracting just the t.
$\begin{array}{c}\frac{2(t-2)-(t-2)}{(t+1)(t-2)}\\\\\frac{2t-4-t+2}{(t+1)(t-2)}\\\\\frac{t-2}{(t+1)(t-2)}\end{array}$
The numerator and denominator have a common factor of $t–2$, so the rational expression can be simplified.
$\large\begin{array}{c}\frac{\cancel{t-2}}{(t+1)\cancel{(t-2)}}\\\\=\frac{1}{t+1}\end{array}$
[latex-display] \displaystyle \frac{2}{t+1}-\frac{t-2}{{{t}^{2}}-t-2}=\frac{1}{t+1},t\ne -1,2[/latex-display]
In the next example, we will give less instruction. See if you can find the LCD yourself before you look at the answer.
### Example
Subtract the rational expressions: $\frac{6}{{x}^{2}+4x+4}-\frac{2}{{x}^{2}-4}$, and define the domain. State the difference in simplest form.
Answer: Note that the denominator of the first expression is a perfect square trinomial, and the denominator of the second expression is a difference of squares so they can be factored using special products. [latex-display]\begin{array}{cc}\frac{6}{{\left(x+2\right)}^{2}}-\frac{2}{\left(x+2\right)\left(x - 2\right)}\hfill & \text{Factor}.\hfill \\ \frac{6}{{\left(x+2\right)}^{2}}\cdot \frac{x - 2}{x - 2}-\frac{2}{\left(x+2\right)\left(x - 2\right)}\cdot \frac{x+2}{x+2}\hfill & \text{Multiply each fraction to get LCD as denominator}.\hfill \\ \frac{6\left(x - 2\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}-\frac{2\left(x+2\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Multiply}.\hfill \\ \frac{6x - 12-\left(2x+4\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Apply distributive property}.\hfill \\ \frac{4x - 16}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Subtract}.\hfill \\ \frac{4\left(x - 4\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Simplify}.\hfill \end{array}[/latex-display]
The domain is $x\ne-2,2$
[latex-display]\frac{6}{{x}^{2}+4x+4}-\frac{2}{{x}^{2}-4}=\frac{4(x-4)}{(x+2)^2(x-2)}[/latex], $x\ne-2,2[/latex-display] In the previous example, the LCD was [latex]\left(x+2\right)^2\left(x-2\right)$. The reason we need to include $\left(x+2\right)$ two times is because it appears two times in the expression $\frac{6}{{x}^{2}+4x+4}$. The video that follows contains an example of subtracting rational expressions whose denominators are not alike. The denominators are a trinomial and a binomial. https://www.youtube.com/watch?v=MMlNtCrkakI&feature=youtu.be
### Try it
[ohm_question]3434[/ohm_question]
On the next page, we will show you how to find the greatest common denominator for a rational sum or difference that does not share any common factors. We will also show you how to manage a sum or difference of more than two rational expressions.
## Contribute!
Did you have an idea for improving this content? We’d love your input. |
# How to Find Greatest Common Divisor (GCD or HCF)
Here you will learn concept of gcd ot hcf and how to find greatest common divisor or highest common factor of numbers and fractions with examples.
Let’s begin –
## Concept of GCD or HCF
Consider two natural numbers $$n_1$$ and $$n_2$$.
If the numbers $$n_1$$ and $$n_2$$ are exactly divisible by the same number x, then x is a common divisor of $$n_1$$ and $$n_2$$.
The highest of all the common divisors of $$n_1$$ and $$n_2$$ is called as the GCD or HCF. This is denoted as GCD($$n_1$$, $$n_2$$)
## How to Find Greatest Common Divisor (GCD)
(a) Find the standard form of the numbers.
(b) Write out all the prime factors that are common to the standard form of the numbers.
(c) Raise each of the common prime factors listed above to the lesser of the powers in which it appears in the standard forms of the numbers.
(d) The product of the results of the previous step will be the GCD of the numbers.
## Rule for finding HCF of Fractions
HCF of two or more fractions is given by:
$$HCF of Numerators\over LCM of Denominators$$
Example : Find the GCD of 150, 210, 375.
Solution : We have the numbers, 150, 210, 375.
1). Writing down the standard form of numbers.
150 = $$5 \times 5 \times 3 \times 2$$
210 = $$5 \times 2 \times 7 \times 3$$
375 = $$5 \times 5 \times 5 \times 3$$
2). Writing Prime factors common to all the three numbers is $$5^1 \times 3^1$$.
3). This will give the same result, i.e. $$5^1 \times 3^1$$
4). Hence, the HCF or GCD will be $$5\times 3$$ = 15
Example : Find the GCD of 50, 75.
Solution : We have the numbers, 50, 75
1). Writing down the standard form of numbers.
50 = $$5 \times 5 \times 2$$
75 = $$5 \times 5 \times 3$$
2). Writing Prime factors common to all the two numbers is $$5^1 \times 5^1$$.
3). This will give the same result, i.e. $$5^1 \times 5^1$$
4). Hence, the HCF or GCD will be $$5\times 5$$ = 25
## Some Other Rules for HCF
If the HCF of x and y is G, then the HCF of
(i) x, (x + y) is also G
(ii) x, (x – y) is also G
(iii) (x + y), (x – y) is also G |
# Finding the Difference: A Cube Minus B Cube Formula Explained
May 12, 2024
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When it comes to mathematics, there are various formulas and identities that students and professionals encounter. One such formula that is often used in algebraic equations is the formula for finding the difference between two cubes, also known as the “A cube minus B cube” formula. In this article, we will delve deep into understanding this formula, its derivation, applications, and solving problems using it.
### Understanding the Concept of A Cube Minus B Cube Formula
The A cube minus B cube formula is used to expand the expression (a – b)(a² + ab + b²) into a simpler form. In simple terms, it helps in finding the difference between two cubes. The formula is represented as:
a³ – b³ = (a – b)(a² + ab + b²)
Where:
a and b are variables or numbers.
represents the cube of the value of a.
represents the cube of the value of b.
### Derivation of A Cube Minus B Cube Formula
The formula for A cube minus B cube can be derived using algebraic manipulation. We start with the expression (a – b)(a² + ab + b²) and expand it using the distributive property:
(a – b)(a² + ab + b²) = a(a²) + a(ab) + a(b²) – b(a²) – b(ab) – b(b²)
Simplifying further, we get:
a³ + a²b + ab² – a²b – ab² – b³
The terms a²b and ab² cancel out, leaving us with the simplified form:
a³ – b³
This derivation showcases how the A cube minus B cube formula is derived from the expansion of the given expression.
### Applications of A Cube Minus B Cube Formula
The A cube minus B cube formula finds extensive applications in algebraic equations, factorization, and simplification of expressions. Some of the key applications include:
1. Algebraic Simplification: It helps in simplifying complex algebraic expressions involving cubes.
2. Factorization: The formula is crucial in factorizing expressions of the form a³ – b³ into linear factors.
3. Solving Equations: By using the formula, equations involving cube terms can be solved efficiently.
4. Geometry: In geometry, the formula can be used to calculate volumes and areas of shapes involving cubes.
### Solving Problems Using A Cube Minus B Cube Formula
Let’s explore how to solve problems using the A cube minus B cube formula through a couple of examples:
#### Example 1:
Problem: Simplify the expression: 27x³ – 8y³.
Solution:
Here, a = 3x and b = 2y.
Applying the formula, we get:
(3x)³ – (2y)³ = (3x – 2y)((3x)² + (3x)(2y) + (2y)²)
= (3x – 2y)(9x² + 6xy + 4y²)
#### Example 2:
Problem: Factorize the expression: 64 – 125a³.
Solution:
Here, a = a, and b = 4.
Using the formula, we have:
64 – (5a)³ = (4 – 5a)((4)² + 4(5a) + (5a)²)
= (4 – 5a)(16 + 20a + 25a²)
1. What is the significance of the A cube minus B cube formula in mathematics?
The formula is essential for simplifying and solving equations involving cube terms efficiently.
1. Can the formula be applied to complex numbers as well?
Yes, the formula can be used for complex numbers where a and b can be any real or complex values.
1. How is the A cube minus B cube formula different from the difference of squares formula?
The A cube minus B cube formula is specifically for cubes, whereas the difference of squares formula is for squares.
1. Are there any real-world applications of the A cube minus B cube formula?
Yes, the formula is used in various fields like physics, engineering, and computer science for calculations involving cubes.
1. Can the formula be generalized to higher powers like Aⁿ – Bⁿ?
Yes, the concept can be extended for higher powers using the formula for A cube minus B cube as a basis.
In conclusion, the A cube minus B cube formula is a powerful tool in algebra that simplifies expressions and aids in problem-solving. Understanding its derivation and applications can enhance one’s mathematical skills and problem-solving abilities. By mastering this formula, one can tackle complex algebraic equations with ease, making it a valuable asset in the realm of mathematics.
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Time and work Important Formulas - ObjectiveBooks
# Time and work Important Formulas
### Time and work - Aptitude Important Formulas, Shortcut Methods and Tricks:
1. If ‘A’ can do a piece of work in ‘a’ days, then work performed in one day = 1/a part of work.
2. If A’s 1 day's work = (1/a), then ‘A’ can finish the work in ‘a’ days.
3. A is thrice as good a workman as B, then:
Ratio of work done by A and B = 3 : 1.
Ratio of times taken by A and B to finish a work = 1 : 3.
4. If ‘A’ is ‘x’ times as good as work performed as B, then time taken by A = (1/x)th time taken by ‘B’.
5. If ‘A’ and ‘B’ can do a piece of work in ‘x’ and ‘y’ days respectively, then time taken with working together = xy/(x + y) days.
6. If the number of men to do a job is changed in the ratio a : b, then the time required to do the work will be changed in the ratio b : a.
7. If two men ‘A’ and ‘B’ together can finish a job in ‘x’ days and if ‘A’ working alone takes ‘a’ days more than “A and B” working together and ‘B’ working alone takes ‘b’ days more than “A and B” working together,
then, x = √ab
Example: 01
A and B can do a work in 10 days, B and C can do the same work in 20 days, while C and A can do it in 15 days. In how many days can C alone do the same work?
Solution:
(A + B)’s 1 day’s work = 1/10
(B + C)’s 1 day’s work = 1/20
(C + A)’s 1 day’s work = 1/15
2 (A + B + C)’s 1 day’s work = [1/10 + 1/20 + 1/15] = [(6 + 3 + 4)/60] = 13/60
(A + B + C)’s 1 day’s work = 13/120
Now, C’s 1 day’s work = {(A + B + C)’s 1 day’s work} - {(A + B)’s 1 day’s work}
= (13/120) - (1/10)
= (13 - 12)/120
= 1/120
Hence, C alone can do the work in 120 days.
Example: 02
A and B could do a piece of work in 30 days. After working for 10 days, they are assisted by C and the work is finished in 10 days. If C does as much work in 2 days as B does in 3 days, in how many days could A do the same work alone?
Solution:
(A + B)’s 1 day’s work = 1/30 ………………………….(i)
(A + B)’s 10 day’s work = 10 × (1/30) = 1/3
Remaining work = 1 - (1/3) = 2/3
Now, since (A + B + C)’s 10 day’s work = 2/3
(A + B + C)’s 1 day’s work = 2/(3 × 10) = 1/15 ……………(ii)
From equation (ii) - (i), C’s 1 day’s work = (1/15) - (1/30) = 1/30
Hence, C can finish the work in 30 days.
Now, it is given that C does as much work in 2 days as B does in 3 days.
∴ The work which C does in 30 days, will be done by B in = 30 × (3/2) = 45 days.
Hence, B alone can finish the work in 45 days.
∴ B’s 1 day’s work = 1/45 ……………………(iii)
From equation (i) - (iii), A’s 1 day’s work = (1/30) - (1/45) = 1/90
Therefore, A alone can finish the work in 90 days.
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• Problem
• Detailed Solution
• Summary Solution
### Determining parameters of a polynomial function
Let f be a third degree polynomial, $f(x)=ax^3+bx^2+cx+d$.
Determine parameter values a, b, c, and d, so that f has a root and an inflection point at ${x_0}=0$, a local minimum or maximum at ${x_1}=-2$, and the tangent at ${x_2}=4$ has a slope of 3 .
### Preliminary remarks
To solve this problem, you need the following competences:
(i) to characterize a root of a function
(ii) to characterize an inflection point of a function
(iii) to characterize an extremum of a function
(iv) to characterize the slope of a tangent to the graph of a function
The theory behind all this is
• in block II (tangents), and
• in block III learning sequence 2: sections 6 (extrema) and 7 (inflection points)
of our script.
### Formal approach
We make use of the first information:
f has a root at ${x_0}=0$. This means that, if x is replaced by the number 0, f(x) is itself 0: $f({x_0})=f(0)=0$.
Thus, we obtain
$f(0)=a \cdot 0^3+b \cdot 0^2+c \cdot 0+d$.
Since a, b, and c are nullified by powers of 0, we have $\;d=0$.
This is our first finding, and our function now reads $\;f(x)=ax^3+bx^2+cx$, with a, b, and c still unknown.
We now make use of the second information:
f has an inflection point at ${x_0}=0$. Although we stay at the same place on the x-axis, this information is quite different from the first one. The implication is that the 2nd order derivative of f, that is f'', takes the value 0: $f''(x)= 0$.
We determine the derivatives we need:
$f(x)=ax^3+bx^2+cx+d \quad \Rightarrow \quad f'(x)=3ax^2+2bx+c \quad \Rightarrow \quad f''(x)=6ax+2b$
We evaluate f'' for x=0:
$f''(0)=6a \cdot 0+2b=0$, which means $\;b=0$.
Now we have the second of our four paramater values, and we can state
$f(x)=ax^3+cx$, with only a and c missing.
The third information
is the one on the extremum: since this is at ${x_1}=-2$, the necessary condition is that f' must be 0 there:
$f'(-2)=0$
Since $\;f'(x)=3ax^2+2bx+c\;$ (remember that $\; b=0\,$), we have:
$f'(-2)=3a \cdot (-2)^2+2b \cdot (-2)+c=12a-4b+c=0$
We keep this equation that contains two unknown symbols, hoping that
the last information
yields the "missing piece": since the slope of the tangent to the graph of f is 3 for x=4, we make use of the fact that this is nothing else but the value of the 1st order derivative of f at that place.
Therefore, $\;f'(4)=3\,$, and with $\;f'(x)=3ax^2+2bx+c \;$ we obtain
$f'(4)=3a \cdot 4^2+2b \cdot 4+c=48a+c=3$.
Here again, we have one equation for two unknown symbols, but taken together, these two equations are enough to find the values of a and c:
We make use of $12a+c=0$ (see above) to state that: $c=-12a$.
Then, we replace c in our last equation by that multiple of a, and we obtain
$48a+(-12a)=3 \quad \Rightarrow \quad 36a=3 \quad \Rightarrow \quad a=\frac{1}{12}$
And since c=−12a, we conclude that $c=-1$.
This completes our results, and we write down the complete equation for f:
$f(x)=\frac{1}{12}x^3-x$
To end the exercise, we might visualize our findings by sketching the graph, thus verifying whether it meets the four conditions set up in the problem.
See the diagram to do so.
$f(x)=ax^3+bx^2+cx+d$
$f'(x)=3ax^2+2bx+c$
$f''(x)=6ax+2b$
$f'''(x)=6a$
I: $\quad f(0)=0 \quad \Rightarrow \quad d=0$
II: $\quad f''(0)=0 \quad \Rightarrow \quad b=0$
III: $\quad f'(-2)=0 \quad \Rightarrow \quad 12a-4b+c=0$
IV: $\quad f'(4)=3 \quad \Rightarrow \quad 48a+8b+c=3$
Replacing $b=0\;$ and $\;d=0\;$ in equations III and IV, we have:
$12a+c=0$ and $48a+c=3 \quad \Rightarrow \quad a=\frac{1}{12}$ and $c=-1$
Solution: $\quad f(x)=\frac{1}{12}x^3-x$ |
# Ex.5.3 Q18 Arithmetic Progressions Solution - NCERT Maths Class 10
## Question
A spiral is made up of successive semicircles, with centres alternately at $$A$$ and $$B,$$ starting with centre at $$A$$ of radii $$0.5,$$ $$1.0 \,\rm{cm}, 1.5 \,\rm{cm}, 2.0 \,\rm{cm}, .........$$as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles?
\begin{align}\left[ {{\rm{Take}}\,\,\pi = \frac{{22}}{7}} \right] \end{align}
Video Solution
Arithmetic Progressions
Ex 5.3 | Question 18
## Text Solution
What is Known?
Radii of the $$13$$ semicircles.
What is Unknown?
Reasoning:
Sum of the first $$n$$terms of an AP is given by $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$
Where $$a$$ is the first term, $$d$$is the common difference and $$n$$ is the number of terms.
Steps:
Semi-perimeter of circle, $$l = \pi r$$
\begin{align}{{l}_{1}}& =\pi \times \left( 0.5\,\,\text{cm} \right)=0.5\pi \,\,\text{cm } \\ {{l}_{2}}&=\pi \times \left( 1\,\,\text{cm} \right)=\pi \,\,\text{cm} \\ {{l}_{3}} &=\pi \times \left( 1.5\,\,\text{cm} \right)=1.5\pi \,\text{cm} \\\end{align}
Therefore, $${l_1},{\rm{ }}{l_2},{\rm{ }}{l_3},$$ i.e. the lengths of the semi-circles are in an A.P.,
\begin{align}&0.5\pi ,\pi ,1.5\pi ,2\pi\dots\dots\dots\\&a = 0.5\pi \\&d = \pi - 0.5\pi = 0.5\pi\end{align}
We know that the sum of $$n$$ terms of an A.P. is given by
\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{13}} &\!=\!\frac{{13}}{2}\left[ {2 \!\times\! \left( {0.5\pi } \right)\!+\!\left( {13\!-\!1} \right)\left( {0.5\pi } \right)}\!\right]\\ &= \frac{{13}}{2}\left[ {\pi + 6\pi } \right]\\ &= \frac{{13}}{2} \times 7\pi \\ &= \frac{{13}}{2} \times 7 \times \frac{{22}}{7}\\ &= 143\end{align}
Therefore, the length of such spiral of thirteen consecutive semi-circles will be $$143\, \rm{cm.}$$
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# What is Math?
What is math? It can be compared to a very useful tool, or maybe a collection of tools. Sometimes textbooks concentrate a lot on teaching about the small details of each and every type of tool. But it’s also really important to focus on how and when to use the different tools. This is my practical approach to teaching the subject. And it’s also important to note that math is much more than just numbers! If you’re really good with shapes and how they relate, you might enjoy geometry. And if you are good at solving puzzles, chances are that logic will be a great match for your skills.
NOTE: Be sure to pause the video when the timer reaches 6:30 to work on the Earn, Break Even, or Lose problem.
# How to Multiply by 12
Do you think you’ll need to know how to multiply by 12 or 11 more? Think of it this way: how often do you need to figure out how many dozen you need of something? It comes up a lot more than needing to know how many batches of 11, doesn’t it? That’s because of the way we’ve decided to group things mathematically as a society.
Here’s why: We picked 12 based on how we used to count on our fingers using the “finger segment” system. If you look at your hands, you’ll notice that your index finger has three segments to it. So does your middle finger, ring finger, and pinkie. Since you have four fingers, you actually have 12 sections for counting with (we’re not including your thumb, which is the pointer… your thumb rests on the section you’re currently on). When your thumb touches the tip of your index finger, that means “1”. When your thumb touches the middle segment, that’s “2”, and the base segment is “3”. The tip of your middle finger is “4”, and so on. That’s how we came to use the 12-in-a-batch system.
If you’re wondering why we didn’t use the 24-in-a-batch system (because you have two hands), that’s because one hand was for 1-12 and the second hand indicated the number of batches of 12. So if your left hand has your thumb on the ring finger’s base segment (9) and your right hand has the thumb touching the index finger’s middle segment (2 complete batches of 12, or 2 x 12), the number you counted to is: 24 + 9 = 33.
Fortunately we now have calculators and a base-ten system, so this whole thing worked out well. But still the number 12 persists! So I thought you’d like this video, which expands on the idea of quickly multiplying two-digit numbers and three-digit numbers by eleven. This is very similar to the shortcut used when multiplying by eleven, but it also involves some doubling. Are you ready?
# Divisibility
Can you look at a number and tell right away if it’s divisible by another number? Well, it’s pretty easy for 2 – if it’s an even number, it’s definitely divisible by two. Testing whether a number is divisible by five is easy as well. How can you tell?
In this video, I’ll show you some tricks to determine if a number is divisible by 3, 4, 6 and 7 before you start to divide. Some are simple and fast and some are a bit more complex. These can be very useful tricks for working with larger numbers (or just really fun to play with for a bit).
# What Day Were You Born On?
This is not only a neat trick but a very practical skill – you can figure out the day of the week of anyone’s birthday.
If you were born in the 20th Century, (1900-1999), we can use math to find out which day of the week you were born. If you’re a little too young for this, try it with a parent or grandparent’s birthday. Watch the video and I’ll teach you exactly how it works.
# \$1 Word Search
Have you ever heard of a dollar word search? It’s a special kind of puzzle where the letters in a word add up to a coin value. For example, an A is worth a penny, the letter B is worth two cents, C is worth three cents, and so on. Are you completely confused? That’s okay! Just watch the video and I’ll show you how it all works.
# Isn’t that SUM-thing?
This is a neat trick that you can use to really puzzle your friends and family. If someone gives you a three-digit number, you can actually figure out what the end result will be after you’ve received two additional numbers, but before you actually know what those numbers are. Does this sound confusing? Watch the video and I’ll show you how it works.
Want a peek under the ‘hood’ of my brain when I do a mental math calculation? This video is a slow-motion, step-by-step snapshot of what goes on when I add numbers in my head. The first thing you need to learn is how to add from LEFT to RIGHT, which is opposite from most math classes out there. I’ll show you how to do this – it’s easy, and essential to working bigger numbers in your head.
Here’s what you do:
# Multiplying 2-Digit Numbers by 11
Here’s our first MATH lesson. It is so easy that one night, I wound up showing it to everyone in the pizza restaurant. Well, everyone who would listen, anyway. We were scribbling down the answers right on the pizza boxes with such excitement that I couldn’t help it – I started laughing right out loud about how excited everyone was about math… especially on a Saturday night.
When you do this calculation in front of friends or family, it’s more impressive if you hand a calculator out first and let them know that you are ‘testing to see if the calculator is working right’. Ask for a two digit number and have them check the calculator’s answer against yours.
If you really want to go crazy, you can have math races against the calculator and its operator, just as the Arthur Benjamin video shows. (Only you don’t need to do the squaring of five-digit numbers in your head!) Have fun!
# Multiplying 3-Digit Numbers by 11
If you can multiply 11 by any 2-digit number, then you can easily do any three digit number. There’s just an extra step, and make sure you always start adding near the ones so you can see where to carry the extra if needed.
# Mental Mathemagic
We’re going to throw in a few math lessons here and there, so if math really isn’t your thing, free free to just watch the videos and see what you think. All of these lessons require only a brain, and once in awhile paper and pencil, so this area is ‘materials-free’ and jam-packed with great mathematical content. If you’re the parent, stick a calculator in your pocket and test out your kids as they go along.
Some of what we cover here is based on the book “Secrets of Mental Math” by Arthur Benjamin, an incredible professor at Harvey Mudd College. He’s also known as the “Lightning Human Calculator”. Here’s a video about him you may enjoy:
We’re going to break down the steps to really getting to know numbers and put it into a form that both you and your kids can use everyday, including shopping at grocery stores, baking in the kitchen, working on the car, and figuring out your taxes. It’s a useflu tool for flexing your mind as well as appreciating the simplicity of the numerical world.
# Squaring Two-Digit Numbers
This neat little trick shortcuts the multiplication process by breaking it into easy chunks that your brain can handle. The first thing you need to do is multiply the digits together, then double that result and add a zero, and then square each digit separately, and finally add up the results.
Slightly confused? Great – we made a video that outlines each step. There’s a definite pattern and flow to it. With practice, you will be able to do this one in your head within a very short time. Have fun!
Squaring three-digit numbers is one of the most impressive mental math calculations, and it doesn’t take a whole lot of effort after you’ve mastered two-digits. It’s like the difference between juggling three balls and five balls. Most folks (with a bit of practice) can juggle three balls. Five objects, however, is a whole other story (and WOW factor).
Once you get the hang of squaring two-digit numbers, three-digit numbers aren’t so hard, but you have to keep track as you go along. Don’t get discouraged if you feel a little lost. It’s just like anything you try for the first time… when you’re new at something, in the beginning you aren’t very good at it. But with practice, these steps will become second nature and you’ll be able to impress your friends, relatives, and math teachers.
The video below has two parts: |
# How do you find (dy)/(dx) given 3y^3+2=2x?
Feb 22, 2017
$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2}{9 {y}^{2}}$ or $\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2}{9 {\left({\left(\left(\frac{2}{3}\right) x - \left(\frac{2}{3}\right)\right)}^{\frac{1}{3}}\right)}^{2}}$
#### Explanation:
$9 {y}^{2} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 2$
Solve for $\frac{\mathrm{dy}}{\mathrm{dt}}$
$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2}{9 {y}^{2}}$
If you want, we can solve and plug in for y using the given equation and simplifying:
$3 {y}^{3} + 2 = 2 x$
$3 {y}^{3} = 2 x - 2$
${y}^{3} = \left(\frac{2}{3}\right) x - \left(\frac{2}{3}\right)$
$y = {\left(\left(\frac{2}{3}\right) x - \left(\frac{2}{3}\right)\right)}^{\frac{1}{3}}$
Now plug this in for y and simplify.
$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2}{9 {\left({\left(\left(\frac{2}{3}\right) x - \left(\frac{2}{3}\right)\right)}^{\frac{1}{3}}\right)}^{2}}$
$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2}{9 \left({\left(\left(\frac{2}{3}\right) x - \left(\frac{2}{3}\right)\right)}^{\frac{2}{3}}\right)}$ |
### Free Educational Resources
+
4 Tutorials that teach Negative Exponents
# Negative Exponents
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##### Description:
This lesson discusses negative exponents, and relates them to expressions with positive exponents.
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Tutorial
## What's Covered?
Negative Exponents
There are a few special exponent properties that deal with exponents that are not positive. The first is considered in the following example, which is worded out 2 different ways:
This final result is an important property known as the zero property of exponents:
Zero Property of Exponents: a0 = 1
Any number or expression raised to the zero power will always be 1. This is illustrated in the following example.
Another property we will consider here deals with negative exponents. Again we will solve the following example two ways.
This example illustrates an important property of exponents. Negative exponents yield the reciprocal of the base. Once we take the reciprocal the exponent is now positive. Also, it is important to note a negative exponent does not mean the expression is negative, only that we need the reciprocal of the base. Following are the rules of negative exponents.
Negative exponents can be combined in several different ways. As a general rule if we think of our expression as a fraction, negative exponents in the numerator must be moved to the denominator, likewise, negative exponents in the denominator need to be moved to the numerator. When the base with exponent moves, the exponent is now positive. This is illustrated in the following example.
As we simplified our fraction we took special care to move the bases that had a negative exponent, but the expression itself did not become negative because of those exponents. Also, it is important to remember that exponents only affect what they are attached to. The 2 in the denominator of the above example does not have an exponent on it, so it does not move with the d.
Nicolas Chuquet, the French mathematician of the 15th century wrote to indicate This was the first known use of the negative exponent.
Simplifying with negative exponents is much the same as simplifying with positive exponents. It is the advice of the author to keep the negative exponents until the end of the problem and then move them around to their correct location (numerator or denominator). As we do this it is important to be very careful of rules for adding, subtracting, and multiplying with negatives. This is illustrated in the following examples:
In the previous example it is important to point out that when we simplified we moved the three to the denominator and the exponent became positive. We did not make the number negative! Negative exponents never make the bases negative, they simply mean we have to take the reciprocal of the base.
Formulas to Know
Property of Negative Exponents
Zero Property of Exponents |
# RBSE Solutions for Class 12 Maths Chapter 16 Probability and Probability Distribution Ex 16.5
## Rajasthan Board RBSE Class 12 Maths Chapter 16 Probability and Probability Distribution Ex 16.5
Question 1.
If a fair coin is tossed 10 times, then find the probabilities of the following;
Solution:
(i) A fair coin is tossed 10 times and X = numbers of heads.
∴ In distribution X, n = 10 and p = probability of head = $$\frac { 1 }{ 2 }$$
Question 2.
An urn contains 5 white, 7 red and 8 black balls. If 4 balls are drawn one by one with replacement, what is the probability that:
(i) all are white
(ii) only 3 are white
(iii) none is white
(iv) at least 3 are white
Solution:
(i) Total number of balls
= 5 + 7 + 8 = 20
Number of white balls = 5
Probability of getting white ball is one chance = $$\frac { 5 }{ 20 }$$ = $$\frac { 1 }{ 4 }$$
∵ All events are independent
∴ Required probability = $$\frac { 1 }{ 4 }$$ × $$\frac { 1 }{ 4 }$$ × $$\frac { 1 }{ 4 }$$ × $$\frac { 1 }{ 4 }$$ = ($$\frac { 1 }{ 4 }$$)4
(ii) Probability of drawing white ball first time
= 3C1 × $$\frac { 1 }{ 4 }$$ × $$\frac { 1 }{ 4 }$$ × $$\frac { 1 }{ 4 }$$
= 3 ×($$\frac { 1 }{ 4 }$$)3
(iii) P(no ball is white)
∴ Number of other balls = 7 + 8 = 15
∴ Probability of drawing one other colour ball = $$\frac { 15 }{ 20 }$$ = $$\frac { 3 }{ 4 }$$
∴ Probability of other colour balls drawn successively (none is white)
= $$\frac { 3 }{ 4 }$$ × $$\frac { 3 }{ 4 }$$ × $$\frac { 3 }{ 4 }$$ × $$\frac { 3 }{ 4 }$$
= ($$\frac { 3 }{ 4 }$$)4
(iv) P(at least 3 white) = P (four white) + P(three white)
Question 3.
In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is $$\frac { 5 }{ 6 }$$. What is the probability that he will knock down fewer than 2 hurdles ?
Solution :
Number of total hurdles (n) = 10
Probability of clearing the hurdles = P = $$\frac { 5 }{ 6 }$$
∴ Probability of non clearing the hurdles
Question 4.
five dices are thrown simultaneously. If the occurrence of an even number in a single dice is considered a success, find the probability of at most 3 success.
Solution:
Sample space on throwing a dice s = {1,2, 3,4, 5,6}
∴ n(S) = 6
Let A represents even numbers
∴A = {2,4,6}
n(A) = 3
Question 5.
Ten eggs are drawn successively, with replace¬ment, from a lot containing 10% defective eggs. Find the probability that there is at least one defective egg.
Solution:
Probability of defective eggs = 10%
Question 6.
A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is $$\frac { 1 }{ 100 }$$ What is the probability that he will win a prize.
(i) at least once
(ii) exactly once
(iii) at least twice
Solution:
Question 7.
The probability that a bulb produced by a factory will fuse after 150 days on use is 0*05. Find the probability that out of 5 such bulbs
(i) none
(ii) not more than one
(iii) more than one
(iv) at least one
will fuse after 150 days of use.
Solution:
Probability that a bulb will fuse after 150 days p = 0.05
Probability that bulb will not fuse after 150 days q = 1 – p = 1 – 0.05 = 0.95
Question 8.
In a multiple choice examination with three possible answers for each of the 5 questions out of which only one is correct. What is the probability that a candidate would get four or more correct answers just by guessing ?
Solution:
∵ One answer is correct out of 3
∴ Probability of correct answer = p = $$\frac { 1 }{ 3 }$$
= q = 1 – p
= 1 – $$\frac { 1 }{ 3 }$$ = $$\frac { 2 }{ 3 }$$
Probability that 4 or more answers are correct
Question 9.
In a 20 questions true-false examination are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads, his answer ‘true’ If it falls tails, his answer ‘false’. Find the probability that his answer at least 12 questions correctly.
Solution:
P (Getting head on toss) P = $$\frac { 1 }{ 2 }$$
P (not getting head on toss)
q = 1 – person
= 1 – $$\frac { 1 }{ 2 }$$ = $$\frac { 1 }{ 2 }$$
∴ Probability of writing correct answer = $$\frac { 1 }{ 2 }$$
and probability of writing incorrect answer = $$\frac { 1 }{ 2 }$$
Question 10.
A bag contains 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0 ?
Solution:
A bag contains 10 balls each marked with one of the digit from 0 to 9.
Probability that one ball is in marked 0 drawn
P = $$\frac { 1 }{ 10 }$$ = 0.1 = 0.1
Probability that ball is not marked 0
q = 1 – p
= 1 – 0.1 = 0.9
Now 4 balls are drawn successively with replacement.
∴ Probability that any of them ball is marked 0
Question 11.
Five cards are drawn successively with replacement from a well shuffled pack of 52 cards. What is the probability that:
(i) all the five cards are spades ?
(ii) only 3 cards are spades ?
Solution:
There are 13 cards are spade out of all 52 cards. Probability of drawing one card of spade
Question 12.
Suppose X has a binomial distribution B(6,$$\frac { 1 }{ 2 }$$). Show that x = 3 is the most likely outcome.
Solution:
Question 13.
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two success.
Solution:
When a pair of dice in thrown
∴ Number of all possible outcomes
n(S) = 6 × 6 = 36
Number of doublets can be make from a pair of dice = 6
[(1,1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)]
∴ Probability that doublet are find
P = $$\frac { 6 }{ 36 }$$ = $$\frac { 1 }{ 6 }$$
and probability that doublets are not find
q = 1 – p
= 1 – $$\frac { 1 }{ 6 }$$ = $$\frac { 5 }{ 6 }$$
A pair of dice is thrown four times
∴ n = 4 |
# Matrix is Row Equivalent to Echelon Matrix/Examples/Arbitrary Matrix 2
## Examples of Use of Matrix is Row Equivalent to Echelon Matrix
Let $\mathbf A = \begin {bmatrix} 1 & 1 & 1 & 1 \\ 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 6 \\ \end {bmatrix}$
This can be converted into the echelon form:
$\mathbf E = \begin {bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \\ \end {bmatrix}$
## Proof
Using Row Operation to Clear First Column of Matrix we obtain:
$\mathbf B = \begin {bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 3 \\ \end {bmatrix}$
which is obtained by:
adding $-2$ of row $1$ to row $2$
adding $-3$ of row $1$ to row $3$.
Then we investigate the submatrix:
$\mathbf B' = \begin {bmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ \end {bmatrix}$
Using Row Operation to Clear First Column of Matrix we obtain:
$\mathbf C' = \begin {bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ \end {bmatrix}$
which is obtained by adding $-1$ of row $1$ of $\mathbf B'$ to row $2$ of $\mathbf B'$.
Thus we are left with:
$\mathbf E = \begin {bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \\ \end {bmatrix}$
$\blacksquare$ |
## Types of Functions-
should give twenty-two hours for teaching functions and equations as per IBO recommendations. This is my third article on functions in the series of ib mathematics
IB Maths Tutors should give twenty hours in teaching functions and equations. This is my third article on functions in the series of ib mathematics
As you know there are many different types of functions in Mathematics. Here I am discussing a few very important of them
1.Greatest Integer Function– This is an interesting function. It is defined as the largest integer less than or equal to x
y = [x].
For all real numbers, x, this function gives the largest integer less
than or equal to x.
For example: [1] = 1 [2.5] = 2 [4.7] = 4 [5.3] = 5
Beware! [-2] = -2 [-2.6] = -3 [-4.1] = -5 [-6.5] = -7
domain=R
range=Z
2. Fractional part function-
for every real value of x this function gives the fractional part of x.
f(x)={x}
{2.3}=.3 , {5.4}=.4, {2.2}=.2
{6.7}=.7, {-2.3}=.7, {-2.6}=.3
We can say that: 0≤{x}∠1
domain=R
range=less than 1
3. Polynomial function– These are functions of the form
f(x) = anxn + an−1x n−1 + . . . + a2x 2 + a1x + a0 .
Constant, linear, quadratic, cubic, quartic functions etc fall in this category
domain of these functions is R and range is either R or a subset of R
4. Trigonometric functions- Trigonometric functions or circular functions draw the relationship between the sides and angles of right triangles .we can find this relationship using “unit circle”. I have explained all this thing in the given video.
### Trigonometric Functions
There are six trigonometric functions, we will discuss them all one by one
i. Sin function(variation in a)
f(x)=sin x
this is a periodic function with a period of
domain=R
range=[-1,1]
ii. Cosine function(variation in b)-
f(x)=cosx
this is also a periodic function with a period of
domain=R
range=[-1,1]
iii. Tangent function(variation in a/b)-
Also READ this- How to Get 7 in Ib Maths hl
f(x)= tan x
this is also a periodic function with a period of pie
domain=R-{n pie+pie/2}
range=R
This was my last post in ib maths tutors-function series. In my next post, I will discuss some questions based on these topics.
#### Classification Of Functions :
(i) One – One Function (Injective mapping)-: A function f: A B is said to be a one-one function or injective mapping if different elements of A have different f images in B. Thus for &, Function is one-one while if
The function will not be one-one.
(ii) If f(x) is any function which is entirely increasing or decreasing in whole domain, then f(x) is one-one.
(iii) If any line parallel to x-axis cuts the graph of the function atmost at one point, then the function is one-one.
Many–one function-: A function f: A B is said to be many one functions if two or more elements of A have the same f image in B. Thus f: A B is many-one if
but
(i) Any continuous function which has at least one local maximum or local minimum, then f(x) is many-one. In other words, if a line parallel to x-axis cuts the graph of the function at least at two points, then f is many-one. This test is known as horizontal line test
(ii) If a function is one-one, it cannot be many-one and vice versa.
Onto function (Surjective mapping)-: If the function f: A B is such that each element in B (co-domain) is the image of at least one element in A, then we say that f is a function of A ‘onto’ B . Thus f: A B is surjective if b B, some a A such that f (a) = b
Into function-: If f: A B is such that there exists at least one element in co-domain which is not the image of any element in the domain, then f(x) is into.
(i) If a function is onto, it cannot be into and vice versa.
(ii) A polynomial of degree even will always be into.
Thus a function can be one of these four types :
(a) one-one onto (injective & surjective)
(b) one-one into (injective but not surjective)
(c) many-one onto (surjective but not injective)
(d) many-one into (neither surjective nor injective)
Bijective mapping- If f is both injective & surjective, then it is called a Bijective mapping.The bijective functions are also named as invertible, non-singular or bi-uniform functions. If a set A contains n
If a set A contains n distinct elements then the number of different functions defined from A B is nn & out of it n ! are one one.
Algebraic Operations On Functions: If f & g are real-valued functions of x with domain set A, B respectively, then both f & g are defined in Now we define f + g,
f – g , (f . g) & (f/g) as follows -:
(i) (f ± g) (x) = f(x) ± g(x)
(ii) (f . g) (x) = f(x) . g(x)
(iii)
Composite Of Uniformly & Non-Uniformly Defined Functions: Let f : AB and g : BC be two functions . Then the function gof : AC defined by (gof) (x) = g (f(x))
x A is called the composite of the two functions f & g.
Properties Of Composite Functions : (i) The composite of functions is not commutative i.e.
(i) The composite of functions is not commutative i.e. gof fog .
(ii) The composite of functions is associative i.e. if f, g, h are three functions such that fo(goh) & (fog)oh are defined, then fo(goh) = (fog)oh
(iii) The composite of two bijections is a bijection i.e. if f & g are two bijections such that gof is defined, then gof is also a bijection. Implicit & Explicit
Implicit & Explicit Function-: A function defined by an equation not solved for the dependent variable is called an implicit Function. For eg. the equation x3 + y3= 1 defines y as an implicit function. If y has been expressed in terms of x alone then it is called an Explicit Function.
Homogeneous Functions-: A function is said to be homogeneous with respect to any set of variables when each of its terms is to the same degree with respect to those variables. For example F(x)= 5 x2 + 3 y2 – xy is homogeneous in x & y . Symbolically if, f (tx , ty) = tn. f(x,y) then f(x,y) is homogeneous function of degree n.
Inverse Of A Function-: Let f: A B be a one-one & onto function, then there exists a unique function g: B A such that f(x) = y g(y) = x, and
Then g is said to be inverse of f. Thus g =f-1 B A = {(f(x), x) ½ (x, f(x)) Î f} . Properties Of Inverse Function : (i) The inverse of a bijection is unique.
(ii) If f: A B is a bijection & g: A A is the inverse of f, then fog =IB and gof =IA
where I& IB are identity functions on the sets A & B respectively. Note that the graphs of f & g are the mirror images of each other in the line y = x.
##### Odd & Even Functions-:
If f (-x) = f (x) for all x in the domain of ‘f’ then f is said to be an even function. e.g. f (x) = cos x ; g (x) = x² + 3 .
Also READ this- Continuity of functions | Learn Maths Online
If f (-x) = -f (x) for all x in the domain of ‘f’ then f is said to be an odd function.
e.g. f (x) = sin x , g (x) = x3 + x
(i) f (x) – f (-x) = 0 => f (x) is even & f (x) + f (-x) = 0 => f (x) is odd
(ii) f (x) – f (-x) = 0 => f (x) is even & f (x) + f (-x) = 0 => f (x) is odd .
(iii) A function may neither be odd nor be even.
(iv) Inverse of an even function is not defined .
(v) Every even function is symmetric about the y-axis & every odd function is symmetric about the origin .
(vi) Every function can be expressed as the sum of an even & an odd function.
(vii) The only function which is defined on the entire number line & is even and odd at the same time is f(x) = 0.(viii) If f and g both are even or both are odd then the function f.g will be even but if any one of them is odd then f.g will be odd .
(viii) If f and g both are even or both are odd then the function f.g will be even but if any one of them is odd then f.g will be odd.
Periodic Function-: A function f(x) is called periodic if there exists a positive number T (T > 0) called the period of the function such that f (x + T) = f(x), for all values of x within the domain of x.
e.g. The function sin x & cos x both are periodic over 2 & tan x is periodic over
(i) f (T) = f (0) = f (-T) , where ‘T’ is the period .
(ii) Inverse of a periodic function does not exist .
(iii) Every constant function is always periodic, with no fundamental period .
(iv) If f (x) has a period T & g (x) also has a period T then it does not mean that f(x) + g(x) must have a period T . e.g.
(v) If f(x) has a period , then and also has a period
(vi) if f(x) has a period T then f(ax + b) has a period T/a (a > 0).
Here are links to my previous posts on functions
First Post-An Introduction to functions
Second Post-Domain and Range of functions
Third Post-Types of functions(part-1)
Here is a pdf containing questions on this topic |
Question
# For any two-real numbers, an operation defined by $a * b{\text{ }} = \;1 + ab$ is $\left( A \right)$. Commutative but not associative $\left( B \right)$. Associative but not commutative $\left( C \right)$. Neither commutative nor associative $\left( D \right)$. Both commutative and associative
Hint: Use commutative and associative property for the given operation.
We have been given the operator $*$ such that:
${\text{ }}a * b = 1 + ab{\text{ (1) ; }}a,b{\text{ }} \in {\text{R}}$
Since $(1 + ab){\text{ }}$also belongs to $R{\text{ }}$(Real Numbers Set),
Operator $*$ satisfies closure property
$a * b$ is a binary operation.
For binary operation to be commutative, we would have the following condition:
$a * b = b * a {\text{(2)}}$
We need to check condition (2) for operation (1)
$a * b = 1 + ab \\ b * a = 1 + ba \\$
Since multiplication operator is commutative, we have
$ab = ba \\ \Rightarrow a * b = 1 + ab = 1 + ba = b * a \\$
Hence condition (2) is satisfied.
Therefore, operation (1) is commutative.
For binary operation to be associative, we would have the following condition:
$a * \left( {b * c} \right) = \left( {a * b} \right) * c{\text{ (3)}}$
We need to check for condition (3) for operator (1)
$a * \left( {b * c} \right) = a * \left( {1 + bc} \right) = 1 + a(1 + bc) = 1 + a + abc \\ \left( {a * b} \right) * c = \left( {1 + ab} \right) * c = 1 + \left( {1 + ab} \right)c = 1 + c + abc \\$
Since $1 + a + abc \ne 1 + c + abc$, condition (3) is not satisfied.
Therefore, operation (1) is not associative.
Hence the correct option is $\left( A \right)$. Commutative but not associative.
Note: Always try to remember the basic conditions for associativity and commutativity. Commutativity does not imply associativity. |
# Go Math Grade 3 Answer Key Chapter 7 Division Facts and Strategies Assessment Test
This chapter can improve student’s math skills, by referring to the Go Math Grade 3 Answer Key Chapter 7 Division Facts and Strategies Assessment Test, and with the help of this Go Math Grade 3 Assessment Test Answer Key, students can score good marks in the examination.
Go Math Grade 3 Answer Key Chapter 7 contains all the topics of chapter 7 which helps to test the student’s knowledge. Through this assessment test, students can check their knowledge. This assessment test is also helpful for the teachers to know how much a student understood the topics.
Chapter 7: Division Facts and Strategies Assessment Test
### Test – Page 1 – Page No. 71
Question 1.
Shang shared 28 postcards among 7 different people. Each person received the same number of postcards. How many postcards did Shang give to each person?
28 ÷ 7 = n
7 × n = 28
Options:
a. 4
b. 5
c. 6
d. 21
Explanation:
As Shang shared 28 postcards to 7 different people so 28 ÷ 7 = 4 , 7 × 4 = 28. Shang gave 4 postcards to each person.
Question 2.
Lionel has 14 mittens.
Select one number from each column to show the division equation represented by the picture.
14 ÷ ______ = ______
Answer: 14 ÷ 2 = 7.
Explanation: 14 ÷ 2 = 7 which is 7 pairs of mitten given the model.
### Test – Page 2 – Page No. 72
Question 3.
Fifteen people are going rafting. They brought 5 rafts. An equal number of people ride in each raft. How many people will be in each raft?
______ people
Explanation:
No of people going for rafting = 15
Total no of rafts available = 5
No of people will be in each raft = 15 ÷ 5 = 3 people.
Question 4.
Circle a number for the unknown factor and quotient that makes the equation true.
4 × = 24
= 24 ÷ 4
4 × ______ = 24 ;        ______ = 24 ÷ 4
4×6= 24
6= 24 ÷ 4
Explanation:
4×= 24
= 24 ÷ 4
Question 5.
There are 20 students in science class. There are 10 students sitting at each table. How many tables are there?
Write a division equation to represent the repeated subtraction.
______ ÷ ______ = ______
Answer: 20 ÷ 10 = 2.
Explanation:
Total no of students in class = 20
No of students sitting at each table = 10
Therefore no of tables = 20 ÷ 10 = 2
Question 6.
Complete the chart to show the quotients.
63÷9= 7.
72÷9= 8.
81÷9= 9.
90÷9= 10.
Explanation:
### Test – Page 3 – Page No. 73
Question 7.
For numbers 7a–7e, select True or False for each equation.
a. 0 ÷ 6 = 6
i. True
ii. False
Explanation: 0 ÷ 6 = 0, so the answer is false.Â
Question 7.
b. 6 ÷ 6 = 1
i. True
ii. False
Explanation: As 6 ÷ 6 = 1, so the answer is true.
Question 7.
c. 18 ÷ 6 = 2
i. True
ii. False
Explanation: As 18 ÷ 6 = 3, so the answer is false.
Question 7.
d. 54 ÷ 6 = 9
i. True
ii. False
Explanation: As 54 ÷ 6 = 9, so the answer is true.
Question 7.
e. 60 ÷ 10 = 6
i. True
ii. False
Explanation: As 60 ÷ 10 = 6, so the answer is true.
Question 8.
Kaitlyn says that 8 ÷ 2 × 4 is the same as 4 × 2 ÷ 8.
Is Kaitlyn correct or incorrect? Explain.
Answer: No. 8 ÷ 2 × 4 and 4 × 2 ÷ 8 are not the same.
Expplanation:Lets calculate 8 ÷ 2 × 4 = 4 × 4 = 16 by using BODMAS, first we did division later multiplication
4 × 2 ÷ 8 = 4 × ¼ = 1. Thus both results are not same.
Question 9.
Brian is dividing 64 baseball cards equally among 8 friends. How many baseball cards will each friend get?
_______ baseball cards
Explanation:
Total cards available = 64
No of friends = 8
No of baseball cards each gets = 64 ÷ 8 = 8
Question 10.
Tara made $18 selling cookies. She wants to know how many cookies she sold. Tara used this number line. Write the division equation that the number line represents. ______ ÷ ______ = ______ Answer: 18 ÷ 3 = 6. Explanation: Possible answers could be if she sells each cookie at$ 3 then 18 ÷ 3 = 6 were sold
### Test – Page 4 – Page No. 74
Question 11.
Each team at a science competition has 6 players. How many teams are there if 42 players are at the competition? Explain the strategy you used to solve the problem.
_____ teams
Explanation:
Given that there is a total of 42 players
Each team has 6 players, therefore the total number of teams = 42 ÷ 6 = 7 teams.
Question 12.
Carly bought 3 packs of baseball cards. Each pack had the same number of cards. She gave 5 cards to her sister. Now she has 19 cards. How many baseball cards were in each pack? Explain how you solved the problem.
_____ baseball cards
Answer: 24 ÷ 3 = 8 baseball cards.
Explanation:
Total packs bought were = 3
No of cards she has = 19
No of cards given to her sister = 5
Total number cards = 19+5 = 24
No of cards in each pack = 24 ÷ 3 = 8 cards.
Question 13.
Andrea used 35 craft sticks to make 7 door hangers. She used the same number of craft sticks for each door hanger. How many craft sticks did Andrea use for each door hanger?
_____ craft stick
Explanation:
Total no of craft sticks used by Andrea = 35
No of door hangers made = 7
Therefore no of craft sticks used for each door hanger = (Total no of craft sticks ÷ no of door hangers)
= 35 ÷ 7
= 5 craft sticks.
Question 14.
For numbers 14a–14e, use the order of operations. Select True or False for each equation.
a. 45 ÷ 5 − 3 = 6
i. True
ii. False
Explanation:
By BODMAS rule
45 ÷ 5 − 3
= 9 – 3
= 6
Question 14.
b. 12 + 4 ÷ 4 = 13
i. True
ii. False
Explanation:
By BODMAS rule
12 + 4 ÷ 4
= 12 + 1
= 13
Question 14.
c. 3 + 7 × 8 = 80
i. True
ii. False
Explanation:
By BODMAS rule
3 + 7 × 8
= 3 + 56
= 59
Question 14.
d. 32 ÷ 8 × 2 = 2
i. True
ii. False
Explanation:
By BODMAS rule
32 ÷ 8 × 2
= 4 × 2
= 8.
Question 14.
e. 40 − 10 × 3 = 10
i. True
ii. False
Explanation:
By BODMAS rule
40 − 10 × 3
= 40 – 30
= 10.
### Test – Page 5 – Page No. 75
Question 15.
Patrick sells homemade pretzels in bags with 9 pretzels in each bag. He sells 54 pretzels in all. How many bags of pretzels does he sell?
______ bags
Explanation:
No of bags sold = (Total no of pretzels / no of pretzels in each bag)
= 54/9
= 6 bags.
Question 16.
Enrique started a table showing a division pattern.
Part A
Complete the table.
Compare the quotients when dividing by 10 and when dividing by 5. Describe a pattern you see in the quotients.
10÷10= 1
10÷5= 2
20÷10= 2
20÷5= 4
30÷10= 3
30÷5= 6
40÷10= 4
40÷5= 8.
Explanation:
Question 16.
Part B
Find the quotient, a.
80 ÷ 10 = a
a = ______
Explanation:
80 ÷ 10 = 8.
The quotient is 8.
How could you use a to find the value of n?
Find the value of n.
80 ÷ 5 = n
n = ______
a = ______
n = ______
a= 8.
n= 16.
Explanation: By doubling the value of ‘a’ we can get the value of ‘n’. As the value of ‘a’ is 8, so the value of ‘n’= 8+8= 16
Question 17.
Eve needs 2 limes to make a glass of limeade. If limes come in bags of 12, how many glasses of limeade can she make using one bag?
______ glasses
Explanation: The limes in the bags are 12 and Eve needs 2 glasses of limeade, so the number of glasses she can make is 12÷2= 6 glasses.
### Test – Page 6 – Page No. 76
Question 18.
For numbers 18a–18e, select True or False for each equation.
a. 18 ÷ 9 = 2
i. True
ii. False
Explanation: As 18 ÷ 9 = 2, so the answer is true.
Question 18.
b. 27 ÷ 9 = 4
i. True
ii. False
Explanation: As 27 ÷ 9 = 3, so the answer is false.
Question 18.
c. 45 ÷ 9 = 5
i. True
ii. False
Explanation: As 45 ÷ 9 = 5, so the answer is true.
Question 18.
d. 72 ÷ 9 = 7
i. True
ii. False
Explanation: As 72 ÷ 9 = 8, so the answer is false.
Question 18.
e. 81 ÷ 9 = 8
i. True
ii. False
Explanation: As 81 ÷ 9 = 9, so the answer is true.
Question 19.
Holly is making 4 vegetable trays for a party. She wants to divide 36 carrot sticks equally among the trays. How many carrot sticks will she put on each tray?
______ carrot sticks
Explanation: No of carrot sticks on each tray = (Total carrot sticks ÷ no of vegetable trays)
= 36 ÷ 4
= 9 carrot sticks in each tray
Question 20.
Hector is buying books at a book store.
Part A
He buys 2 used books and 1 new book for $26. The new book costs$18. Each used book costs the same amount. What is the price of each used book? Explain the steps you used to solve the problem.
$______ Answer: Each used book costs$ 4.
Explanation:
Given that cost of one new book = $18 Cost of 2 used books + 1 new book =$ 26
Cost of 2 used books = 26 – 18 = 8
Cost of 1 used book = 8/2 = $4 Question 20. Part B Hector also buys a reading light for$12 and 2 journals for $8 each to give as gifts. Write one equation to describe the total amount Hector spends on gifts. Explain how to use the order of operations to solve the equation. Total amount:$ ______
Answer: $28. Explanation: Hector bought a reading light for$12
2 journals for $8 each Total money spent on gifts = 12 + (2 x 8) = 12 + 16 =$ 28.
Conclusion:
This assessment test helps students to check their math skills. Go Math Grade 3 Chapter 10 questions are explained in detail that students can understand easily.
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# 9.3: Arcs in Circles
Difficulty Level: At Grade Created by: CK-12
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What if the Ferris wheel below had equally spaced seats, such that the central angle were \begin{align*}20^\circ\end{align*}. How many seats are there? Why do you think it is important to have equally spaced seats on a Ferris wheel?
If the radius of this Ferris wheel is 25 ft., how far apart are two adjacent seats? Round your answer to the nearest tenth. The shortest distance between two points is a straight line. .
### Arcs in Circles
A central angle is the angle formed by two radii of the circle with its vertex at the center of the circle. In the picture below, the central angle would be \begin{align*}\angle BAC\end{align*}. Every central angle divides a circle into two arcs (an arc is a section of the circle). In this case the arcs are \begin{align*}\widehat{BC}\end{align*} and \begin{align*}\widehat{BDC}\end{align*}. Notice the arc above the letters. To label an arc, always use this curve above the letters. Do not confuse \begin{align*}\overline{BC}\end{align*} and \begin{align*}\widehat{BC}\end{align*}.
If \begin{align*}D\end{align*} was not on the circle, we would not be able to tell the difference between \begin{align*}\widehat{BC}\end{align*} and \begin{align*}\widehat{BDC}\end{align*}. There are \begin{align*}360^\circ\end{align*} in a circle, where a semicircle is half of a circle, or \begin{align*}180^\circ\end{align*}. \begin{align*}m \angle EFG = 180^\circ\end{align*}, because it is a straight angle, so \begin{align*}m \widehat{EHG}= 180^\circ\end{align*} and \begin{align*}m \widehat{EJG} = 180^\circ\end{align*}.
• Semicircle: An arc that measures \begin{align*}180^\circ\end{align*}.
• Minor Arc: An arc that is less than \begin{align*}180^\circ\end{align*}.
• Major Arc: An arc that is greater than \begin{align*}180^\circ\end{align*}. Always use 3 letters to label a major arc.
Two arcs are congruent if their central angles are congruent. The measure of the arc formed by two adjacent arcs is the sum of the measures of the two arcs (Arc Addition Postulate). An arc can be measured in degrees or in a linear measure (cm, ft, etc.). In this chapter we will use degree measure. The measure of the minor arc is the same as the measure of the central angle that corresponds to it. The measure of the major arc equals to \begin{align*}360^\circ\end{align*} minus the measure of the minor arc. In order to prevent confusion, major arcs are always named with three letters; the letters that denote the endpoints of the arc and any other point on the major arc. When referring to the measure of an arc, always place an “\begin{align*}m\end{align*}” in from of the label.
#### Measuring Arcs
Find \begin{align*}m\widehat{AB}\end{align*} and \begin{align*}m\widehat{ADB}\end{align*} in \begin{align*}\bigodot C\end{align*}.
\begin{align*}m\widehat{AB}= m\angle{ACB}\end{align*}. So, \begin{align*}m\widehat{AB}= 102^\circ\end{align*}.
\begin{align*}m\widehat{ADB}=360^\circ - m\widehat{AB}=360^\circ-102^\circ=258^\circ\end{align*}
#### Identifying and Measuring Minor Arcs
Find the measures of the minor arcs in \begin{align*}\bigodot{A}\end{align*}. \begin{align*}\overline{EB}\end{align*} is a diameter.
Because \begin{align*}\overline{EB}\end{align*} is a diameter, \begin{align*}m\angle EAB=180^\circ\end{align*}. Each arc has the same measure as its corresponding central angle.
\begin{align*}m \widehat{BF} & = m \angle FAB = 60^\circ\\ m\widehat{EF} & = m \angle EAF = 120^\circ \rightarrow 180^\circ - 60^\circ\\ m\widehat{ED} & = m \angle EAD = 38^\circ \ \rightarrow 180^\circ - 90^\circ - 52^\circ\\ m\widehat{DC} & = m \angle DAC = 90^\circ\\ m\widehat{BC} & = m \angle BAC = 52^\circ\end{align*}
#### Using the Arc Addition Postulate
Find the measures of the indicated arcs in \begin{align*}\bigodot A\end{align*}. \begin{align*}\overline{EB}\end{align*} is a diameter.
a) \begin{align*}m\widehat{FED}\end{align*}
\begin{align*}m\widehat{FED} = m\widehat{FE} +m\widehat{ED} = 120^\circ+38^\circ=158^\circ\end{align*}
b) \begin{align*}m\widehat{CDF}\end{align*}
\begin{align*}m\widehat{CDF} = m\widehat{CD} + m \widehat{DE} + m \widehat{EF} = 90^\circ + 38^\circ + 120^\circ = 248^\circ\end{align*}
c) \begin{align*}m\widehat{DFC}\end{align*}
\begin{align*}m \widehat{DFC} = m\widehat{ED} + m\widehat{EF} + m\widehat{FB} + m\widehat{BC} = 38^\circ + 120^\circ + 60^\circ + 52^\circ = 270^\circ\end{align*}
#### Ferris Wheel Problem Revisited
Because the seats are \begin{align*}20^\circ\end{align*} apart, there will be \begin{align*}\frac{360^\circ}{20^\circ}=18\end{align*} seats. It is important to have the seats evenly spaced for balance. To determine how far apart the adjacent seats are, use the triangle to the right. We will need to use sine to find \begin{align*}x\end{align*} and then multiply it by 2.
\begin{align*}\sin 10^\circ &= \frac{x}{25}\\ x = 25 \sin 10^\circ &= 4.3 \ ft.\end{align*}
The total distance apart is 8.6 feet.
### Examples
#### Example 1
List the congruent arcs in \begin{align*}\bigodot C\end{align*} below. \begin{align*}\overline{AB}\end{align*} and \begin{align*}\overline{DE}\end{align*} are diameters.
\begin{align*}\angle ACD \cong \angle ECB\end{align*} because they are vertical angles. \begin{align*}\angle DCB \cong \angle ACE\end{align*} because they are also vertical angles.
\begin{align*}\widehat{AD} \cong \widehat{EB}\end{align*} and \begin{align*}\widehat{AE} \cong \widehat{DB}\end{align*}
#### Example 2
Are the blue arcs congruent? Explain why or why not.
a)
\begin{align*}\widehat{AD} \cong \widehat{BC}\end{align*} because they have the same central angle measure and are in the same circle.
b)
The two arcs have the same measure, but are not congruent because the circles have different radii.
#### Example 3
Find the value of \begin{align*}x\end{align*} for \begin{align*}\bigodot C\end{align*} below.
The sum of the measure of the arcs is \begin{align*}360^\circ\end{align*} because they make a full circle.
\begin{align*}m \widehat{AB} + m \widehat{AD} + m \widehat{DB} & = 360^\circ\\ (4x+15)^\circ+92^\circ+(6x+3)^\circ&=360^\circ\\ 10x+110^\circ&=360^\circ\\ 10x&=250\\ x&=25\end{align*}
### Review
Determine if the arcs below are a minor arc, major arc, or semicircle of \begin{align*}\bigodot G\end{align*}. \begin{align*}\overline{EB}\end{align*} is a diameter.
1. \begin{align*}\widehat{AB}\end{align*}
2. \begin{align*}\widehat{ABD}\end{align*}
3. \begin{align*}\widehat{BCE}\end{align*}
4. \begin{align*}\widehat{CAE}\end{align*}
5. \begin{align*}\widehat{ABC}\end{align*}
6. \begin{align*}\widehat{EAB}\end{align*}
7. Are there any congruent arcs? If so, list them.
8. If \begin{align*}m \widehat{BC}=48^\circ\end{align*}, find \begin{align*}m \widehat{CD}\end{align*}.
9. Using #8, find \begin{align*}m \widehat{CAE}\end{align*}.
Determine if the blue arcs are congruent. If so, state why.
Find the measure of the indicated arcs or central angles in \begin{align*}\bigodot A\end{align*}. \begin{align*}\overline{DG}\end{align*} is a diameter.
1. \begin{align*}\widehat{DE}\end{align*}
2. \begin{align*}\widehat{DC}\end{align*}
3. \begin{align*}\angle GAB\end{align*}
4. \begin{align*}\widehat{FG}\end{align*}
5. \begin{align*}\widehat{EDB}\end{align*}
6. \begin{align*}\angle EAB\end{align*}
7. \begin{align*}\widehat{DCF}\end{align*}
8. \begin{align*}\widehat{DBE}\end{align*}
Algebra Connection Find the measure of \begin{align*}x\end{align*} in \begin{align*}\bigodot P\end{align*}.
1. What can you conclude about \begin{align*}\bigodot A\end{align*} and \begin{align*}\bigodot B\end{align*}?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English Spanish
TermDefinition
arc A single section of the circle, that describes a particular angle.
central angle An angle formed by two radii and whose vertex is at the center of the circle.
major arc An arc that is greater than $180^\circ$.
minor arc An arc that is less than $180^\circ$.
semicircle An arc that measures $180^\circ$.
Arc Addition Postulate Arc addition postulate states that the measure of the arc formed by two adjacent arcs is the sum of the measures of the two arcs.
Diameter Diameter is the measure of the distance across the center of a circle. The diameter is equal to twice the measure of the radius.
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# Convert Improper FractionSupport Page
Welcome to the Math Salamanders support page on How to Convert Improper Fraction to Mixed Numbers.
We have a selection of resources including worked examples, support sheets and practice sheets.
Want to check your working out? Try using our Improper Fraction Calculator!
## How to Convert Improper Fractions to a Mixed Number (or Mixed Fraction)
### Comverting Improper Fractions Examples
#### Example 1) Convert into a mixed number the fraction: ${23 \over 5 }$
Dividing the numerator by the denominator gives us: 23 ÷ 5 = 4 r 3
So this means that ${23 \over 5} = 4 {3 \over 5}$
#### Example 2) Convert into a mixed number the fraction: ${19 \over 7 }$
Dividing the numerator by the denominator gives us: 19 ÷ 7 = 2 r 5
So this means that ${19 \over 7} = 2 {5 \over 7}$
#### Example 3) Convert into a mixed number the fraction: ${27 \over 4 }$
Dividing the numerator by the denominator gives us: 27 ÷ 4 = 6 r 3
So this means that ${27 \over 4} = 6 {3 \over 4}$
#### Example 4) Convert into a mixed number the fraction: ${32 \over 9 }$
Dividing the numerator by the denominator gives us: 32 ÷ 9 = 3 r 5
So this means that ${32 \over 9} = 3 {5 \over 9}$
For further information and a printable support sheet, please see below.
## How to Convert a Mixed Number to an Improper Fraction
### Examples Converting Mixed Numbers
#### Example 1) Convert into an improper fraction the number: $3 {1 \over 5 }$
Step 1)
Multiply the denominator by the integer part gives us: 3 x 5 = 15.
This means that: $3 = {15 \over 5}$
Step 2)
This gives us: $3{1 \over 5} = {15 \over 5} + {1 \over 5} = {16 \over 5}$
Final answer $3{1 \over 5} \; = \; {16 \over 5}$
#### Example 2) Convert into an improper fraction the number: $5 {4 \over 7 }$
Step 1)
Multiply the denominator by the integer part gives us: 5 x 7 = 35.
This means that: $5 = {35 \over 7}$
Step 2)
Add on the fraction part: $5{4 \over 7} = {35 \over 7} + {4 \over 7} = {39 \over 7}$
Final answer $5{4 \over 7} \; = \; {39 \over 7}$
#### Example 3) Convert into an improper fraction the number: $9 {2 \over 3 }$
Combining the two steps gives us:
$9 {2 \over 3} = {(9 \times 3) + 2 \over 3} = {27 + 2 \over 3} = {29 \over 3}$
#### Example 4) Convert into an improper fraction the number: $3 {6 \over 7 }$
Combining the two steps gives us:
$3 {6 \over 7} = {(3 \times 7) + 6 \over 7} = {21 + 6 \over 7} = {27 \over 7}$
For further information and a printable support sheet, please see below.
### More Recommended Math Resources
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### Equivalent Fractions
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# GRADE 6 REVIEW EOG MATH
Problem 21 :
Jeff recorded the average temperatures for six months. He will display the temperatures on a number line
On the number line, which month’s temperature will be between February’s and March’s temperatures?
A) December B) January C) April D) May
Solution :
Temperature is February = -15
Temperature is March = 20
Difference between two temperatures = 20 - (-15)
= 20 + 15
= 35
Problem 22 :
A trapezoid in a coordinate plane has vertices
(-2, 5) (-3, -2) (2, -2) (1, 5)
What is the height of the trapezoid?
A) 3 units B) 5 units C) 7 units D) 9 units
Solution :
So, the answer is 7 units.
Problem 23 :
Which can be represented by the expression 17 – 2x?
A) 17 less than twice a number x
B) the difference between 17 and twice a number x
C) a number x squared, subtracted from 17
D) 17 less than a number x squared
Solution :
17 – 2x
We have negative sign, so we use the word difference.
2x = twice the number
So, difference between 17 and twice a number, option B is correct.
Problem 24 :
Which expression is equivalent to 5y + 2y + 6x + 2y – x?
A) 5x + 6y B) 5x + 7y C) 5x + 9y D) 7x + 7y
Solution :
5y + 2y + 6x + 2y – x
Combining like terms, we get
= 6x - x + 2y + 2y + 5y
= 5x + 9y
So, option C is correct.
Problem 25 :
Diana can use the equation y = 7x to calculate her pay, where y represents the amount of pay, and x represents the number of hours worked. How many hours did Diana work if she was paid \$45.50?
A) 5.5 hours B) 6 hours C) 6.5 hours D) 7 hours
Solution :
y = 7x
y = amount of pay
x = number of hours worked
Amount of pay = 45.50
7x = 45.50
x = 45.50/7
x = 6.5
So, option C 6.5 hours is correct.
Problem 26 :
If y – 18 = 14, what is the value of 3(y + 5)?
A) 27 B) 32 C) 96 D) 111
Solution :
y – 18 = 14
y = 14 + 18 ==> 32
Applying the value of y, in 3(y + 5).
= 3(32 + 5)
= 3(37)
= 111
So, the answer is option D, 111.
Problem 27 :
Karen recorded her walking pace in the table below. What equation best represents this relationship?
A) h = m + 10 B) h = 3.5m C) m = h + 10 D) m = 3.5h
Solution :
h - independent variable
m - dependent variable
Rate of change :
(2.5, 8.75) and (4, 14)
Rate of change = (14 - 8.75) / (4 - 2.5)
= 5.25 / 1.5
= 3.5
Equation representing the given table :
m = 3.5h
So, option D is correct.
Problem 28 :
The shaded area indicates the parking lot at a shopping center.
What is the total area of the parking lot?
A) 72 units2 B) 86 units2 C) 91 units2 D) 120 units2
Solution :
We decompose the given shaded region into two trapezium.
Area of trapezium = (1/2) h (a + b)
Area of A1 : h = 2, a = 6, b = 5= (1/2) x 2 (6 + 5)= 11 square units Area of A2 : h = 10, a = 5, b = 10= (1/2) x 10 (5 + 10)= 75 square units
= 11 + 75
= 86 square units.
So, option B is correct.
Problem 29 :
The right rectangular prism below is made up of 8 cubes. Each cube has an edge length of 1/2 inch.
What is the volume of this prism?
A) 1 cubic inch B) 2 cubic inches
C) 4 cubic inches D) 8 cubic inches
Solution :
Length of each edge = 1/2 inch
length of prism = 4(1/2) = 2 inches
Width = 1/2 inch
Height = 2(1/2) ==> 1 inch
Volume = length x width x height
= 2 x (1/2) x 1
= 1 cubic inch
So, option A is correct.
Problem 30 :
What is the area of the quadrilateral with vertices at
(-1, 0), (2, 0), (2, 5) and (-1, 5)?
A) 15 square units B) 12 square units
C) 10 square units D) 5 square units
Solution :
The given quadrilateral is a rectangle.
Area of rectangle = length x width
Length = 5, width = 3
= 5 x 3
= 15 square units.
So, option A is correct.
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# Minor of Matrix Definition and Examples | How to determine minor of matrices?
Minor of Matrix is the most important concept in matrices to find the adjoint and inverse of the matrices. We use the determinant of matrices to find the minor of matrices. You can find a detailed explanation of the minor of the matrix along with the examples from this page. Learn about cofactor matrix, determinant matrix, inverse matrix, and adjoint matrix in the below section.
## Minor of Matrix Definition
Minor of the matrix is defined as the matrix acquired by deleting the row and column of the matrix in which that specific component lies. Minor of the matrix is denoted by Mij and the element in the matrix is denoted by aij.
Example:
$$A =\left[\begin{matrix} a11 & a12 & a13 \cr a21 & a22 & a23 \cr a31 & a32 & a33 \cr \end{matrix} \right]$$
The minor element a11 is
$$M11 =\left[\begin{matrix} a22 & a23 \cr a32 & a33 \cr \end{matrix} \right]$$
### How to Determine Minor of Matrix?
There are three steps to follow in order to find the minor of the matrix.
1. First, we need to identify the matrix and eliminate the row and column that contains has a particular element in it.
2. Next, a new small order matrix will be formed with the remaining entries to show the minor of the particular element of the matrix.
3. Next, find the determinant of each element of minor of the matrix and form a new matrix that contains the minor values of the corresponding elements.
### Minor of Matrix Applications
The minor of the matrix is applicable to find the cofactors of the elements that is used to find the adjoint of the matrix and inverse matrix. Minor of a matrix is used to calculate the determinant of the matrix. Let us know the applications of minor of the matrix in brief with examples from this section.
#### Cofactor Matrix
The matrix that is formed with the cofactors of the elements is known as the cofactor matrix.
Cofactor matrix of $$A =\left[\begin{matrix} C11 & C12 & C13 \cr C21 & C22 & C23 \cr C31 & C32 & C33 \cr \end{matrix} \right]$$
#### Determinant of Matrix
The determinant of a matrix is the number calculated using the square matrix means the rows and columns are equal.
$$A =\left[\begin{matrix} a & b & c \cr p & q & r \cr x & y & z \cr \end{matrix} \right]$$
det A or |A| = aqz + brx + cpy – ary – bpz – cqx
The adjoint of a matrix is also called the adjugate of a matrix. It is defined as the transpose of the cofactor matrix of that particular matrix.
Example:
$$A =\left[\begin{matrix} a11 & a12 & a13 \cr a21 & a22 & a23 \cr a31 & a32 & a33 \cr \end{matrix} \right]$$
Cofactor matrix of $$A =\left[\begin{matrix} a11 & a12 & a13 \cr a21 & a22 & a23 \cr a31 & a32 & a33 \cr \end{matrix} \right]$$
Adj A = transpose of cofactor matrix = $$transpose of =\left[\begin{matrix} a11 & a12 & a13 \cr a21 & a22 & a23 \cr a31 & a32 & a33 \cr \end{matrix} \right]$$ $$=\left[\begin{matrix} a11 & a12 & a13 \cr a21 & a22 & a23 \cr a31 & a32 & a33 \cr \end{matrix} \right]$$
#### Inverse of Matrix
The inverse of the matrix exists only if the determinant of the matrix is non-zero.
Example:
$$A =\left[\begin{matrix} a & b \cr c & d \cr \end{matrix} \right]$$
A-1 = 1/(ad-bc) $$\left[\begin{matrix} d & -b \cr -c & a \cr \end{matrix} \right]$$
### Minor of Matrix Examples
Example 1.
Find the minor of the matrix, such that the given matrix is $$A =\left[\begin{matrix} 9 & 3 \cr -6 & 7 \cr \end{matrix} \right]$$
Solution:
Given that the matrix is
$$A =\left[\begin{matrix} 9 & 3 \cr -6 & 7 \cr \end{matrix} \right]$$
To find the minor of each element of this matrix.
Minor of 9 = 7
Element 9 is in the first row and the first column of the matrix. After excluding the first row and the first column we are left with element 7.
Minor of 3 = -6
Element 3 is in the first row and the second column of the matrix. After excluding the first row and the second column we are left with element -6
Minor of -6 = 3
Element -6 is in the second row and the first column of the matrix. After excluding the second row and the first column we are left with element 3.
Minor of 7 = 9
Element 7 is in the second row and the second column of the matrix. After excluding the second row and the second column we are left with element 9.
Hence the Minor of Matrix $$A =\left[\begin{matrix} 7 & -6 \cr 3 & 9 \cr \end{matrix} \right]$$
Example 2.
Find the minor of the element 2 in the matrix $$A =\left[\begin{matrix} 4 & 5 & 7 \cr 3 & 2 & 6 \cr 8 & 7 & 3 \cr \end{matrix} \right]$$
Solution:
Given that the matrix is $$A =\left[\begin{matrix} 4 & 5 & 7 \cr 3 & 2 & 6 \cr 8 & 7 & 3 \cr \end{matrix} \right]$$
To find the minor of element 2.
Element 2 lies in the second row and second column.
Hence after excluding the elements of the second row and second column, we obtain the minor of element 2.
Therefore mirror of the element 2 is
$$A =\left[\begin{matrix} 4 & 7 \cr 8 & 3 \cr \end{matrix} \right]$$
Example 3.
Find the minor of the element 7 in the matrix $$A =\left[\begin{matrix} 4 & 7 \cr 6 & 5 \cr \end{matrix} \right]$$
Solution:
Given that the matrix is $$A =\left[\begin{matrix} 4 & 7 \cr 6 & 5 \cr \end{matrix} \right]$$
To find the minor of element 7. Element 7 lies in the first row and second column.
Hence after excluding the elements of the first row and second column, we obtain the minor of element 7.
Therefore mirror of the element 7 is
A = 6.
Therefore the mirror of element 7 is 6.
Example 4.
Find the minor of the matrix, such that the given matrix is $$A =\left[\begin{matrix} -1 & 3 \cr -2 & 4 \cr \end{matrix} \right]$$
Solution:
Given that the matrix is
$$A =\left[\begin{matrix} -1 & 3 \cr -2 & 4 \cr \end{matrix} \right]$$
To find the minor of each element of this matrix.
Minor of -1 = 4
Element -1 is in the first row and the first column of the matrix. After excluding the first row and the first column we are left with element 4.
Minor of 3 = -2
Element 3 is in the first row and the second column of the matrix. After excluding the first row and the second column we are left with element -2
Minor of -2 = 3
Element -2 is in the second row and the first column of the matrix. After excluding the second row and the first column we are left with element 3.
Minor of 4 = -1
Element 4 is in the second row and the second column of the matrix. After excluding the second row and the second column we are left with element -1.
Hence the Minor of Matrix $$A =\left[\begin{matrix} 4 & -2 \cr 3 & -1 \cr \end{matrix} \right]$$
Example 5.
Find the minor of the element 8 in the matrix $$A =\left[\begin{matrix} 5 & 0 & 2 \cr 4 & 1 & 3 \cr 0 & 7 & 8 \cr \end{matrix} \right]$$
Solution:
Given that the matrix is $$A =\left[\begin{matrix} 5 & 0 & 2 \cr 4 & 1 & 3 \cr 0 & 7 & 8 \cr \end{matrix} \right]$$
To find the minor of element 8. Element 8 lies in the third row and third column.
Hence after excluding the elements of the third row and third column, we obtain the minor of element 8.
Therefore mirror of the element 8 is
$$A =\left[\begin{matrix} 5 & 0 \cr 4 & 1 \cr \end{matrix} \right]$$
### FAQs on Minor of Matrix
1. What is the Minor of the Matrix?
The minor of the matrix means suppose we want to find a minor of the matrix of one element in the matrix, which means excluding the element row and column after eliminating the row and column, the remaining elements are called the minor of the matrix. The minor of the matrix is done only for a square matrix. The minor of the element ‘a’ in the matrix $$A =\left[\begin{matrix} a & b \cr c & d \cr \end{matrix} \right]$$ is d.
2. What are the uses of Minors of the Matrix?
The minor of the matrix is useful for finding the cofactors of the elements of the matrix and also used for determining the value of the matrix. Further, both the minors and cofactors of the matrix are also used to find the determinant of the matrix, adjoint of the matrix, and the inverse of the matrix.
3. How to Find Minors of the Matrix?
First, identify the minor matrix and exclude the row and the column which contains the particular element within the matrix. Next, form a new smaller matrix with the remaining elements, to represent the minor of the particular element of the matrix.
The minor of the element ‘e’ in the matrix = $$A =\left[\begin{matrix} a & b & c \cr d & e & f \cr \end{matrix} \right]$$ is $$M =\left[\begin{matrix} a & c \cr g & i \cr \end{matrix} \right]$$ |
August 5, 2024
Learn how to find the diameter of a circle using different methods and formulas, including straightforward steps for finding circumference or radius. Also, explore tips and tricks to avoid common mistakes and become confident in your calculations.
## I. Introduction
Circle diameter is a fundamental measurement used in different fields such as engineering, mathematics, and science. Knowing the diameter of a circle is essential when calculating its area or circumference, finding the size of a circular object, or designing circular structures. In this article, we’ll cover everything you need to know about finding the diameter of a circle. From simple formulas to more complex techniques, we’ll explore the different ways to calculate circle diameter and provide tips to ensure accuracy along the way.
## II. Back to Basics: Simple Steps to Find the Diameter of a Circle
The diameter of a circle is the distance across its widest point, passing through the center of the circle. It’s represented by the symbol “d,” and it’s an essential measurement for various calculations. The diameter is crucial when determining the circumference of a circle, which is the distance around the outside of the circle.
The formula for finding the diameter of a circle is straightforward: diameter equals twice the radius of the circle (d = 2r). The radius is the distance from the center of the circle to any point on the circle’s circumference. Therefore, if you know the radius of a circle, you can quickly determine its diameter by multiplying the radius by two.
## III. The Math of Circles: An Easy Guide to Finding Diameter
The diameter is closely related to other circle measurements such as the radius and circumference. Knowing these relationships can help you calculate the diameter using alternate methods, providing additional flexibility in your calculations.
One common relationship is between the diameter and circumference of a circle. If you know the circumference of a circle, you can find its diameter by dividing the circumference by pi (π). The formula is d = C/π, where C is the circumference of the circle. This relationship provides a more practical method for finding the diameter, especially if you only have the circumference measurement on hand.
## IV. Unlocking the Mystery of Circle Diameter: Tips and Tricks
When finding the diameter of a circle, it’s essential to avoid common mistakes that could lead to inaccurate results. These mistakes include incorrect use of radius or circumference formulas or confusing diameter with other circle measurements.
Visual aids or diagrams can be helpful when understanding circle measurements, including diameter. For example, a compass or a protractor can be used to draw a circle with accurately measured radius or diameter. You can also create a sketch or diagram of a circle to assist with your measurements and calculations, particularly when dealing with real-life scenarios or complex circular structures.
Knowing when to find circle diameter is also crucial. For instance, when designing or creating circular objects or structures, it’s essential to have accurate measurements to avoid errors or safety hazards. It’s also important to have precise measurements when calculating the area or circumference of a circle.
## V. From Circumference to Diameter: Mastering Circle Measurements
The formula d = C/π is one method of finding the diameter when only the circumference measurement is available. To use this formula, you need to measure the circumference of the circle accurately. Once you have the circumference measurement, divide it by the constant pi (π) to obtain the circle’s diameter. For example, if the circumference of a circle is 50cm, you can calculate its diameter as follows:
d = C/π = 50/π ≈ 15.92cm
Therefore, the diameter of the circle is approximately 15.92cm.
## VI. Discovering Circle Diameter: A Step-by-Step Approach
If you have the radius measurement of a circle, you can find its diameter using the formula d = 2r. The radius is half of the diameter, so multiplying the radius by 2 gives you the diameter. For example, if the radius of a circle is 8cm, its diameter can be calculated as follows:
d = 2r = 2(8) = 16cm
Therefore, the diameter of the circle is 16cm.
## VII. The Ultimate Guide to Calculating Circle Diameter
We’ve covered various methods of finding the diameter of a circle, including using the radius or circumference formula and understanding the relationship between diameter, radius, and circumference. Depending on the available information and purpose of the calculation, some methods may be more suitable for specific scenarios.
For example, if you only have the circumference measurement, you can use the formula d = C/π. Conversely, if you only have the radius measurement, you can use the formula d = 2r. For more complex calculations or real-life scenarios, creating a diagram or sketch of the circle may be helpful. No matter the method used, it’s essential to avoid common mistakes and ensure accurate measurements.
## VIII. Get to the Point: Efficient Methods to Find Circle Diameter
Finding the diameter of a circle is a crucial measurement that can be used in various fields. Whether you’re in science, engineering, or mathematics, knowing the diameter is essential for proper calculations and accurate design. By understanding the relationships between diameter, radius, and circumference and avoiding common mistakes, finding the diameter of a circle can be straightforward and efficient.
When calculating the diameter, it’s always recommended to double-check your measurements and methods to ensure accuracy. Creating a diagram or sketch of the circle can be helpful when dealing with real-life scenarios or complex circular structures. By mastering the different methods of calculating circle diameter, you can become confident in your calculations and pave the way for successful results.
## Summary
In this article, we explored the importance of finding circle diameter, including its relationship with radius and circumference. We provided step-by-step guides and formulas for finding the diameter, with tips and tricks to ensure accuracy. We also discussed common mistakes to avoid and real-life scenarios where knowing circle diameter is essential. By mastering the different methods of calculating circle diameter, you can become confident in your calculations and pave the way for successful results. |
# 2009 AIME I Problems/Problem 3
## Problem
A coin that comes up heads with probability $p > 0$ and tails with probability $1 - p > 0$ independently on each flip is flipped $8$ times. Suppose that the probability of three heads and five tails is equal to $\frac {1}{25}$ of the probability of five heads and three tails. Let $p = \frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
## Solution
The probability of three heads and five tails is $\binom {8}{3}p^3(1-p)^5$ and the probability of five heads and three tails is $\binom {8}{3}p^5(1-p)^3$.
\begin{align*} 25\binom {8}{3}p^3(1-p)^5&=\binom {8}{3}p^5(1-p)^3 \\ 25(1-p)^2&=p^2 \\ 25p^2-50p+25&=p^2 \\ 24p^2-50p+25&=0 \\ p&=\frac {5}{6}\end{align*}
Therefore, the answer is $5+6=\boxed{011}$.
## Solution 2
We start as shown above. However, as we get to $25(1-p)^2=p^2$, we square root both sides to get $5(1-p)=p$. We can do this because we know that both $p$ and $1-p$ are between $0$ and $1$, so they are both positive. Now, we have:
\begin{align*} 5(1-p)&=p \\ 5-5p&=p \\ 5&=6p \\ p&=\frac {5}{6}\end{align*}
Now, we get $5+6=\boxed{011}$.
~Jerry_Guo
~IceMatrix
~Shreyas S |
# Important notes on polynomials
1
by anand8
2016-04-26T14:11:28+05:30
We will start off with polynomials in one variable. Polynomials in one variable are algebraic expressions that consist of terms in the form where n is a non-negative (i.e. positive or zero) integer and a is a real number and is called the coefficient of the term. The degree of a polynomial in one variable is the largest exponent in the polynomial.
Note that we will often drop the “in one variable” part and just say polynomial.
Here are examples of polynomials and their degrees.
So, a polynomial doesn’t have to contain all powers of x as we see in the first example. Also, polynomials can consist of a single term as we see in the third and fifth example.
We should probably discuss the final example a little more. This really is a polynomial even it may not look like one. Remember that a polynomial is any algebraic expression that consists of terms in the form . Another way to write the last example is
Written in this way makes it clear that the exponent on the x is a zero (this also explains the degree…) and so we can see that it really is a polynomial in one variable.
Here are some examples of things that aren’t polynomials.
The first one isn’t a polynomial because it has a negative exponent and all exponents in a polynomial must be positive.
To see why the second one isn’t a polynomial let’s rewrite it a little.
By converting the root to exponent form we see that there is a rational root in the algebraic expression. All the exponents in the algebraic expression must be non-negative integers in order for the algebraic expression to be a polynomial. As a general rule of thumb if an algebraic expression has a radical in it then it isn’t a polynomial.
Let’s also rewrite the third one to see why it isn’t a polynomial.
So, this algebraic expression really has a negative exponent in it and we know that isn’t allowed. Another rule of thumb is if there are any variables in the denominator of a fraction then the algebraic expression isn’t a polynomial.
Note that this doesn’t mean that radicals and fractions aren’t allowed in polynomials. They just can’t involve the variables. For instance, the following is a polynomial
There are lots of radicals and fractions in this algebraic expression, but the denominators of the fractions are only numbers and the radicands of each radical are only a numbers. Each x in the algebraic expression appears in the numerator and the exponent is a positive (or zero) integer. Therefore this is a polynomial.
Next, let’s take a quick look at polynomials in two variables. Polynomials in two variables are algebraic expressions consisting of terms in the form . The degree of each term in a polynomial in two variables is the sum of the exponents in each term and the degree of the polynomial is the largest such sum.
Here are some examples of polynomials in two variables and their degrees.
In these kinds of polynomials not every term needs to have both x’s and y’s in them, in fact as we see in the last example they don’t need to have any terms that contain both x’s and y’s. Also, the degree of the polynomial may come from terms involving only one variable. Note as well that multiple terms may have the same degree.
We can also talk about polynomials in three variables, or four variables or as many variables as we need. The vast majority of the polynomials that we’ll see in this course are polynomials in one variable and so most of the examples in the remainder of this section will be polynomials in one variable.
Next we need to get some terminology out of the way. A monomial is a polynomial that consists of exactly one term. A binomial is a polynomial that consists of exactly two terms. Finally, a trinomial is a polynomial that consists of exactly three terms. We will use these terms off and on so you should probably be at least somewhat familiar with them.
Now we need to talk about adding, subtracting and multiplying polynomials. You’ll note t |
# NCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.4
## NCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.4
NCERT Solutions For Class 6 Maths Chapter 14 Practical Geometry Ex 14.4
Exercise 14.4
Ex 14.4 Class 6 Maths Question 1.
Draw any line segment $\overline { AB }$. Make any point M on it. Through M, draw a perpendicular to $\overline { AB }$. (Use ruler and Compasses)
Solution:
Step I : Draw a line segment $\overline { AB }$and mark any point M on it.
Step II : Put the pointer of the compass at M and draw an arc of suitable radius such that it intersects $\overline { AB }$at P and Q.
Step III : Take P and Q as centres and radius greater than PM, draw two arcs such that they intersect each other at C.
Step IV : Join M and C.
Thus CM is the perpendicular to $\overline { AB }$.
Ex 14.4 Class 6 Maths Question 2.
Draw any line segment $\overline { PQ }$. Take any point R not on it. Through R, draw a perpendicular to $\overline { PQ }$. (Use ruler and set square).
Solution:
Step I: Draw a line segment $\overline { PQ }$and a point R outside of $\overline { PQ }$.
Step II : Place a set square on $\overline { PQ }$such that one side of its right angle be along it.
Step III: Place a ruler along the longer side of the set square.
Step IV : Hold the ruler fix and slide the set square along the ruler till it touches the point R.
Step V : Join RM along the edge through R. Thus $\overline { RM }$$\overline { PQ }$.
Ex 14.4 Class 6 Maths Question 3.
Draw a line l and a point X on it. Through X, draw a line segment $\overline { XY }$perpendicular to l. Now draw a perpendicular to $\overline { XY }$at y. (Use ruler and compasses)
Solution:
Step I: Draw a line l and take a point X on it.
Step II : Draw an arc with centre X and of suitable radius to intersect the line l at two points P and Q.
Step III : With P and Q as centres and a radius greater than P draw two arcs to intersect each other at M.
Step IV : Join XM and produce to Y.
Step V : With Y as centre and a suitable radius, draw an arc to intersect XY at two points R and S.
Step VI: With R and S as centres and a radius greater than YR, draw two arcs to intersect each other at A.
Step VII: Join Y and A. Thus YA ⊥ XY.
+ |
# How do you simplify sqrt(432)?
Aug 20, 2016
$\sqrt{432} = 12 \sqrt{3}$
#### Explanation:
$432 = {2}^{4} \cdot {3}^{3} = 2 \cdot 2 \cdot 3 \cdot 2 \cdot 2 \cdot 3 \cdot 3 = {12}^{2} \cdot 3$
So we find:
$\sqrt{432} = \sqrt{{12}^{2} \cdot 3} = \sqrt{{12}^{2}} \cdot \sqrt{3} = 12 \sqrt{3}$
Aug 20, 2016
$12 \sqrt{3}$
#### Explanation:
Let us factorize given the number $432$. For an even number $2$ is a factor
$\frac{432}{2} = 216$
$\frac{216}{2} = 108$
$\frac{108}{2} = 54$
$\frac{54}{2} = 27$
we observe that $3$ is a factor
$\frac{27}{3} = 9$
$\frac{9}{3} = 3$
$\therefore 432 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$
$\implies \sqrt{432} = \sqrt{2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3}$
paring and taking one digit for each pair out of the square root sign we get
$\sqrt{432} = \sqrt{\overline{2 \times 2} \times \overline{2 \times 2} \times \overline{3 \times 3} \times 3}$
$\implies \sqrt{432} = 2 \times 2 \times 3 \sqrt{3}$
$\implies \sqrt{432} = 12 \sqrt{3}$ |
# MATH0005 Algebra 1: sets, functions, and permutations
These are the lecture notes for the first part of MATH0005: Algebra 1. The notes are split into numbered sections, one for each teaching video. The notes are very similar to the video content, so you may find it helpful to read the notes before watching the video, or to have a copy of the notes open while you watch the videos.
The aim of this first part of 0005 is to introduce you to some of the key definitions and results from modern algebra that end up being used everywhere in mathematics and everywhere math is used. We’ll cover sets and relations, used whenever we discuss collections of objections and relations among them. We move on to the idea of a function and to key function properties, useful for example whenever we want to compare the size of two sets. Finally we study permutations - ways to rearrange or re-order a set - which are both interesting in their own right and widely applied, for example in linear algebra as you will see later in 0005.
## 01 Introduction to set theory
### Definition of a set
A set is a collection of (mathematical) objects. There is an entire field of mathematics called set theory dedicated to the study of sets and to their use as a foundation for mathematics, but in 0005 we are going to give only an informal introduction to sets and their properties. If you want to know more, take MATH0037 or read any textbook with a title like “Introduction to Set Theory” (I like Derek Goldrei’s Classic Set Theory).
We use curly brackets to denote sets: $$\{1, 2, 3\}$$ is the set containing 1, 2, and 3.
Sets are what we use for reasoning about unordered collections of objects, ignoring repetition.
“Unordered” means that we consider $$\{1,2\}$$ and $$\{2,1\}$$ the same set, “ignoring repetition” means $$\{1,1\} = \{1\}$$, for example.
(When we want to take order or repetition into account, we have other mathematical objects to do that, e.g. sequences or multisets).
### Elements of a set
The things in a set are called its elements or members. We write $$a \in X$$ to mean that $$a$$ is an element or member of the set $$X$$, and $$a \notin X$$ to mean that $$a$$ is not an element or member of $$X$$.
There is a unique set with no elements, called the empty set and written $$\emptyset$$ or $$\{\}$$. No matter what $$a$$ is, $$a \notin \emptyset$$.
There is no restriction on what the elements of a set can be - we allow any kind of mathematical object, including sets themselves. For example
$\{ \emptyset, 1, \{2\}, \{\{3\}\}\}$
is a set whose four elements are the empty set, 1, the set containing 2, and the set containing the set containing 3. We can put vectors, functions, matrices, and anything else in a set.
Example. Let $$X = \{1, 2, \{3\}\}$$. Then $$1 \in X$$, $$0 \notin X$$, $$3 \notin X$$, $$\{3\} \in X$$.
### Subsets
We need vocabulary for talking about one set being contained in another.
• $$X$$ is a subset of $$Y$$, written $$X \subseteq Y$$, if and only if every element of $$X$$ is also an element of $$Y$$.
• $$X$$ is equal to $$Y$$, written $$X = Y$$, if and only if $$X \subseteq Y$$ and $$Y \subseteq X$$, that is, something is an element of $$X$$ if and only if it is an element of $$Y$$.
• $$X$$ is a proper subset of $$Y$$, written $$X \subsetneq Y$$, if and only if $$X \subseteq Y$$ but $$X \neq Y$$.
So $$X$$ being a proper subset of $$Y$$ means that $$X$$ is a subset of $$Y$$, but $$Y$$ also contains something that $$X$$ does not contain.
When $$X$$ is not a subset of $$Y$$, we write $$X \not\subseteq Y$$ (just like we write $$\neq$$ for “not equal to”).
Example.
• $$\{0\} \subseteq \{0, 1\}$$,
• in fact, $$\{0\} \subsetneq \{0,1\}$$
• $$\{0, 1\} \not\subseteq \{1,2\}$$
• $$\{1, 2, 1\} = \{2, 1\}$$
Why is the last equality true? The only things which can answer “yes” to the question “are you an element of $$\{1,2,1\}$$?” are 1 and 2. The only things which can answer “yes” to the question “are you an element of $$\{1, 2\}$$?” are 1 and 2. So the two sets equal according to our definition.
This is how our definition of set equality captures the idea of sets being unordered collections of objects which disregard repetition.
The definition of subset means that the empty set is a subset of any set. $$\emptyset \subseteq X$$ for any set $$X$$, because $$\forall x : x \in \emptyset \implies x \in X$$ is vacuously true: there’s nothing in $$\emptyset$$ which could fail to be in the set $$X$$ in order to make $$\emptyset \subseteq X$$ false.
### Set builder notation
Suppose we have a set $$X$$ and a property $$P(x)$$ that is true or false for each element $$x$$ of $$X$$. For example, $$X$$ might be the set $$\mathbb{Z} = \{\ldots, -2, -1, 0, 1, 2, \ldots\}$$ of all integers and $$P(x)$$ might be the property “$$x$$ is even”. We write
$\{ x \in X : P(x) \}$
for the set of all elements of $$X$$ for which the property $$P(x)$$ is true. This is called set-builder notation. In our example,
$\{x \in \mathbb{Z} : x \text{ is even }\}$
is the set $$\{\ldots, -4, -2, 0, 2, 4, \ldots\}$$.
Sometimes you’ll see the alternative notation $$\{x \in X | P(x)\}$$ which means exactly the same thing.
## 02 Set operators
### Union, intersection, difference, complement
Let $$A, B$$ be sets.
• $$A\cup B$$, the union of $$A$$ and $$B$$, is the set containing the elements of $$A$$ and also the elements of $$B$$.
• $$A \cap B$$, the intersection of $$A$$ and $$B$$, is the set containing all $$x$$ such that $$x \in A$$ and $$x \in B$$.
• $$A \setminus B$$, the set difference of $$A$$ and $$B$$, is the set of all elements of $$A$$ which are not in $$B$$.
• If $$A$$ is a subset of a set $$\Omega$$ then $$A^c$$, the complement of $$A$$ in $$\Omega$$, is the set of all things in $$\Omega$$ which are not in A. Thus $$A^c = \Omega\setminus A$$
For example, suppose $$A = \{0,1,2\}, B = \{1,2,3\}, C = \{4\}$$. Then:
• $$A \cup B = \{0, 1, 2, 3\}$$
• $$A \cap B = \{1,2\}$$
• $$A \cap C = \emptyset$$
• $$A \setminus B = \{0\}$$
• $$A \setminus C = A$$
• If $$\Omega = \mathbb{N} = \{0, 1, 2, \ldots\}$$ then $$A^c$$ would be $$\{3, 4, \ldots\}$$.
The set $$\mathbb{N} = \{0, 1, 2, \ldots\}$$ is called the natural numbers. Some people exclude 0 from $$\mathbb{N}$$; in MATH0005 the natural numbers include 0.
It’s typical to draw Venn diagrams to represent set operations. We draw a circle, or a blob, for each set. The elements of the set A are represented by the area inside the circle labelled A. Here are some examples:
### Size of a set
The size or cardinality of a set $$X$$, written $$|X|$$, is the number of distinct elements it has.
Example:
• $$|\{1, 2\}| = 2$$,
• $$|\emptyset| = 0$$,
• $$|\{1, 2, 1, 3\}| = 3$$.
Sets with exactly 0, or 1, or 2, or 3, or… distinct elements are called finite. If a set is not finite it is called infinite.
## 03 Set algebra
### Commutativity and associativity
This section is about the laws that union, intersection, difference, and complement obey. Firstly, it is immediate from the definitions that for all sets $$A$$, $$B$$:
• $$A \cap B = B \cap A$$
• $$A \cup B = B \cup A$$
This is called the commutativity property for intersection and union.
Next,
For all sets $$A$$, $$B$$, $$C$$:
• $$A \cap (B \cap C) = (A \cap B) \cap C$$
• $$A \cup (B \cup C) = (A \cup B) \cup C$$
which is the associativity property.
Again it is fairly easy to see why this is true: the left hand side of the first equation is everything in $$A$$, and in $$B$$, and in $$C$$, and so is the right hand side.
Associativity tells us means there’s no ambiguity in writing
$A \cup B \cup C$
or
$A \cap B \cap C$
without any brackets to indicate which union or intersection should be done first. Compare this with
$1 + 2 + 3$
There’s no need for brackets because it doesn’t matter whether you do 1+2 first then add 3, or whether you add 1 to the result of 2 + 3. On the other hand $$1 + 2 \times 3$$ requires either brackets, or a convention on which operation to do first, before we can make sense of it.
Similarly $$A \cup (B \cap C)$$ is different to $$(A \cup B) \cap C$$ in general, so the brackets are obligatory.
### The distributive laws
For any sets $$A$$, $$B$$, $$C$$:
• $$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$$
• $$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$
We could prove these results in the standard way we prove equalities of sets: prove the left hand side is $$\subseteq$$ the right hand side, then prove the right hand side is $$\subseteq$$ the left hand side. But there’s an alternative method similar to truth tables in logic.
The aim is to show that any thing lies in the left hand side if and only if it lies in the right hand side. We can do this case-by-case: any particular element is either in $$A$$ or not in $$A$$, in $$B$$ or not in $$B$$, and in $$C$$ or not in $$C$$. That gives 8 different cases, and in each case we can see whether or not the element belongs to $$A \cup (B\cap C)$$ and whether or not it belongs to $$(A \cup B) \cap (A \cup C)$$.
$$A$$ $$B$$ $$C$$ $$A \cup (B \cap C)$$ $$(A \cup B) \cap (A \cup C)$$
no no no no no
yes no no yes yes
no yes no no no
yes yes no yes yes
no no yes no no
yes no yes yes yes
no yes yes yes yes
yes yes yes yes yes
Notice that the last two columns are identical.
## 04 De Morgan’s laws
### De Morgan’s laws
Take a look at this Venn diagram:
You can see that the shaded area is exactly the area not in $$A \cup B$$, so this is the Venn diagram for $$(A \cup B)^c$$. Now consider the Venn diagrams for $$A^c$$ and $$B^c$$:
It should be clear that $$A^c \cap B^c = (A \cup B)^c$$.
This is a general and useful fact, one of De Morgan’s laws.
Theorem (De Morgan’s Laws).
1. $$(A \cup B)^c = A^c \cap B^c$$.
2. $$(A \cap B)^c = A^c \cup B^c$$.
Proof: To show that two sets $$X$$ and $$Y$$ are equal we can show that $$X \subseteq Y$$ and $$Y \subseteq X$$. This is how we will prove these two results.
1. If $$x \in (A \cup B)^c$$ then $$x \notin A \cup B$$, so $$x \notin A$$ (and therefore $$x \in A^c$$) and $$x \notin B$$ (and $$x \in B^c$$). Therefore $$x \in A^c \cap B^c$$. We’ve shown every element of $$(A \cup B)^c$$ is also an element of $$A^c \cap B^c$$, so $$(A \cup B)^c \subseteq A^c \cap B^c$$. Conversely if $$x \in A^c \cap B^c$$ then $$x \notin A$$ and $$x \notin B$$, so $$x \notin A \cup B$$ or equivalently $$x \in (A\cup B)^c$$. We’ve shown every element of $$A^c \cap B^c$$ is also an element of $$(A \cup B)^c$$, so $$A^c \cap B^c \subseteq (A\cup B)^c$$. Since we have shown $$A^c \cap B^c \subseteq (A\cup B)^c$$ and $$(A\cup B)^c \subseteq A^c \cap B^c$$, the two sets are equal.
2. More concisely: an element $$x$$ lies in $$(A\cap B)^c$$ if and only if it is not in $$A \cap B$$, if and only if either $$x \notin A$$ or $$x \notin B$$, if and only if $$x \in A^c$$ or $$x \in B^c$$, if and only if $$x \in A^c \cup B^c$$. The sets $$(A\cap B)^c$$ and $$A^c \cup B^c$$ have the same elements, so they are equal.
De Morgan’s laws also work for unions and intersections of more than two sets: for any sets $$A_1, A_2, \ldots$$ we have
• $$(A_1 \cup A_2 \cup \cdots)^c = A_1^c \cap A_2^c \cap \cdots$$
• $$(A_1 \cap A_2 \cap \cdots)^c = A_1^c \cup A_2^c \cup \cdots$$
### Complement of a complement
Let’s record one more fact about sets and complements: for any set $$X$$ which is a subset of another set $$\Omega$$, if the complement of the complement of $$X$$ in $$\Omega$$ is $$X$$ again. In symbols, $$X^{cc} = X$$. This is because $$x \in X^{cc}$$ if and only if $$x$$ is not in $$X^c$$, if and only if $$x$$ is not not in $$X$$, if and only if $$x$$ is in $$X$$.
### Applying De Morgan
Five housemates share a fridge. They want to fit locks to the fridge subject to two conditions:
1. no two housemates can unlock every fridge lock
2. any three housemates can unlock every fridge lock
Is it possible? How many keys do they need? How should they be distributed?
They fit some number of locks to the fridge door. Each key works in one lock only, but they can have as many copies of each key as they want.
Let’s set up some notation.
• $$\Omega =$$ set of all locks that they use.
• $$K_i$$ = locks that housemate $$i$$ has the key to, $$1\leq i \leq h$$
• $$N_i = K_i^c = \Omega \setminus K_i$$, the locks that housemate $$i$$ doesn’t have the key to.
Translating condition 1 above into the language of sets, for all $$i$$, $$j$$ we have $$K_i \cup K_j \neq \Omega$$.
If we take complements of both sides, we get that for all $$i,j$$, $$(K_i \cup K_j)^c \neq \Omega^c$$. Using De Morgan’s laws, this means $$N_i \cap N_j \neq \emptyset$$.
Translating condition 2 into set language, for all distinct $$i, j, k$$, $$K_i\cup K_j \cup K_k=\Omega$$.
Taking complements of both sides, this means that for all distinct $$i, j, k$$ we have $$(K_i \cup K_j \cup K_k)^c = \emptyset$$. Using De Morgan’s laws, $$N_i \cap N_j \cap N_k = \emptyset$$.
This points us to a solution. If $$i$$, $$j$$ are distinct then $$N_i$$ and $$N_j$$ have an element in common as $$N_i \cap N_j \neq \emptyset$$, but this doesn’t belong to any other $$N_k$$ since $$N_i \cap N_j \cap N_k =\emptyset$$. So we need one lock $$l_{ij}$$ for each unordered pair of distinct integers $$i$$ and $$j$$ between $$1$$ and $$h$$ - unordered means that $$l_{12} = l_{21}$$, for example. We can then allocate housemate $$i$$ the keys to all locks $$l_{rs}$$ with $$r,s \neq i$$.
This obeys condition 1, because for any $$i, j$$, neither housemate $$i$$ nor housemate $$j$$ has the ley to $$l_{ij}$$, so $$K_i \cup K_j \neq \Omega$$.
It also obeys condition 2, because if $$i,j,k$$ are all different then given any lock $$l_{rs} \in \Omega$$, either
• $$l_{rs} \neq l_{ij}$$ (in which case $$l_{rs} \in K_i \cup K_j$$), or
• $$l_{rs} = l_{ij}$$ (in which case $$l_{rs} \in K_k$$, as $$k$$ is not $$i$$ or $$j$$)
In either case, at least one of the three housemates can open the lock.
## 05 Cartesian products
### Ordered pairs
When we want to use coordinates to talk about points in the plane, we often do this with pairs of real numbers $$(x,y)$$: $$x$$ tells you how far across to go and $$y$$ how far up. The key properties of these pairs $$(x, y)$$ is that $$(x,y) = (z,w)$$ if and only if $$x=z$$ and $$y=w$$. A construction with this property is called an ordered pair, and we can form ordered pairs with elements from any two sets - not just for real numbers.
### Cartesian products
The Cartesian product of two sets $$A$$ and $$B$$, written $$A\times B$$, is the set of all ordered pairs in which the first element belongs to $$A$$ and the second belongs to $$B$$.
$A \times B = \{(a, b) : a \in A, b \in B\}$
Notice that the size of $$A\times B$$ is the size of $$A$$ times the size of $$B$$, that is, $$|A \times B| = |A| |B|$$.
Example. $$\{1,2\} \times \{2,3\} = \{(1,2), (1,3), (2,2), (2,3)\}$$.
Of course we produce ordered triples $$(a, b, c)$$ as well, and ordered quadruples, and so on.
## 06 Introduction to functions
### Definition of function, domain, codomain
Informally, given sets $$X$$ and $$Y$$, a function or map $$f$$ from $$X$$ to $$Y$$ is a definite rule which associates to each $$x \in X$$ an element $$f(x) \in Y$$.
We write $$f:X \to Y$$ to mean that $$f$$ is a function from $$X$$ to $$Y$$. $$X$$ is called the domain of $$f$$ and $$Y$$ is called the codomain of $$f$$.
We refer to the element $$f(x)$$ of $$Y$$ as being the “output” or “value” of $$f$$ when it is given the “input” or “argument” $$x$$.
This might seem vague: what is a definite rule?, what does associates mean?, when are two functions the equal? - should we insist the rule is the same, or just that the input-output relationship is the same?
The formal definition of a function is:
Definition: A function consists of a domain $$X$$, a codomain $$Y$$, and a subset $$f \subseteq X\times Y$$ containing exactly one pair $$(x, y)$$ for each $$x \in X$$. We write $$f(x)$$ for the unique element of $$Y$$ such that $$(x,f(x))$$ is in $$f$$.
In other words, the formal definition of a function is its set of (input, output) pairs.
Example: the function $$f:\mathbb{N} \to \mathbb{N}$$ such that $$f(x) = x+1$$ would correspond to $$\{ (0, 1), (1, 2), (2, 3) \ldots\} \subseteq \mathbb{N} \times \mathbb{N}$$
We won’t use the formal definition in 0005.
### When are two functions equal?
Two functions $$f$$ and $$g$$ are equal if and only if
• they have the same domain, say $$X$$, and
• they have the same codomain, and
• $$f(x)=g(x)$$ for every $$x \in X$$.
They really must have the same domain and same codomain to be equal, according to this definition.
Example: $$f, g : \{0, 1\} \to \{0, 1\}$$. $$f(x)=x^2$$. $$g(x) = x$$. Are they equal?
(the answer is yes - they have the same domain, same codomain, and the same output for every input from the domain).
### The identity function
For any set $$X$$, the identity function $$\operatorname{id}: X \to X$$ is defined by $$\operatorname{id}(x)=x$$.
## 07 Function composition
### Definition of function composition
Suppose you have two functions $$f: X \to Y$$ and $$g : Y \to Z$$:
$X \stackrel{f}{\to} Y \stackrel{g}{\to} Z$
Then you can make a new function $$X \to Z$$ whose rule is “do $$f$$, then do $$g$$”.
Definition: let $$f:X\to Y$$ and $$g : Y \to Z$$. The composition of $$g$$ and $$f$$, written $$g \circ f$$ or $$gf$$, is the function $$g\circ f : X \to Z$$ with the rule $$(g\circ f)(x) = g(f(x))$$.
This makes sense because $$f(x)$$ is an element of $$Y$$ and $$g$$ has domain $$Y$$ so we can use any element of $$Y$$ as an input to $$g$$.
It’s important to remember that $$g \circ f$$ is the function whose rule is “do $$f$$, then do $$g$$”. Be careful to get that the right way round.
### Associativity
Functions $$f$$ and $$g$$ such that the codomain of $$f$$ equals the domain of $$g$$ are called composable.
Composition has an important property called associativity. Suppose $$f$$ and $$g$$, and $$g$$ and $$h$$, are composable:
$X \stackrel f\to Y \stackrel g\to Z \stackrel h \to W$
It seems there are two different ways to compose them: you could first compose $$f$$ and $$g$$, then compose the result with $$h$$, or you could compose $$g$$ with $$h$$ and then compose the result with $$f$$. But they both give the same result.
Lemma: let $$f: X \to Y, g: Y \to Z, h : Z \to W$$. Then $$h\circ (g \circ f) = (h \circ g) \circ f$$.
Proof: Both $$h\circ(g\circ f)$$ and $$(h\circ g) \circ f$$ have the same domain $$X$$, same codomain $$W$$, and same rule $$x \mapsto h(g(f(x)))$$.
This property is called associativity: we discussed the same thing for unions and intersections earlier. Again, it means that a composition like
$f \circ g \circ h \circ k \circ \cdots \circ z$
doesn’t require any brackets to make it unambiguous.
### Composition with the identity function
Lastly, let’s make an observation about identity functions and composition: if $$f:X \to Y$$ then
$f = f \circ \operatorname{id}_X = \operatorname{id}_Y \circ f$
since for example $$(f\circ \operatorname{id}_X)(x) = f(\operatorname{id}_X(x)) = f(x)$$, for any $$x \in X$$.
## 08 Function properties
### Image of a function
Let $$f: X \to Y$$. Then the image of $$f$$ is $$\operatorname{im}(f) = \{f(x) : x \in X\}$$.
Don’t confuse codomain and image. $$Y$$ is the codomain of $$f$$. The image $$\operatorname{im} f$$ is a subset of $$Y$$, but it need not equal $$Y$$.
### Injection, surjection, bijection
The three most important function properties:
• $$f$$ is injective or one-to-one if and only if for all $$a,b \in X$$, if $$f(a)=f(b)$$ then $$a=b$$.
• $$f$$ is surjective or onto if and only if for all $$y \in Y$$ there is at least one $$x \in X$$ such that $$f(x)=y$$.
• $$f$$ is a bijection if and only if it is injective and surjective.
Another way to write the definition of surjective would be that a function is surjective if and only if its image equals its codomain.
As an example, here’s a picture of a function $$f: \{1,2,3,4,5,6\} \to \{1,2,3,4,5\}$$. I have drawn an arrow from $$x$$ to $$f(x)$$ for each $$x$$ in the domain of $$f$$.
$$f$$ is not onto because $$\operatorname{im}(f)$$ is a proper subset of the codomain ($$\operatorname{im}(f)$$ doesn’t contain 4 but the codomain does). Also $$f$$ is not one-to-one, because $$f(1) = f(2)$$ but $$1 \neq 2$$.
More examples:
• $$f : \mathbb{R} \to \mathbb{R}, f(x)=x^2$$. It isn’t injective as $$f(-1)=f(1)$$ and it isn’t surjective as $$-1$$ is in the codomain, but there’s no element $$x$$ in the domain such that $$f(x)=-1$$.
• $$g: \mathbb{R} \to [0, \infty), g(x)=x^2$$. This is not injective for the same reason as before, but this time it is surjective: for each $$y \geq 0$$ we can find an element of the domain which $$g$$ sends to $$y$$: for example. $$g(\sqrt{y}) = 0$$.
• $$h : [0, \infty) \to \mathbb{R}, h(x)=x^2$$. Not surjective, for the same reason $$f$$ isn’t surjective (the codomain contains negative numbers, but the image doesn’t contain any negative numbers, so the image doesn’t equal the codomain). But $$h$$ is injective: if $$x$$ and $$y$$ are in the domain of $$h$$ and $$h(x)=h(y)$$ then $$x^2=y^2$$, so $$x = \pm y$$. Since elements of the domain of $$h$$ are nonnegative, it must be that $$x = y$$.
• $$j : (-\infty, 0] \to [0, \infty), j(x) = x^2$$. This is injective (for a similar reason to $$h$$) and surjective (for a similar reason to $$g$$), so it is a bijection.
This illustrates an important point: when you talk about a function you must specify its domain and codomain (as well as the rule, of course).
### Bijections and sizes of sets
How do we know when two sets have the same size? If you see an alien creature with an apple in each of its hundreds of hands you know it has the same number of apples as it does hands, even if you haven’t counted either the apples or the hands.
You know that because you can pair each apple with the hand holding it. Every apple is held by one hand, and every hand holds one apple.
Suppose there is a bijection $$f$$ between two sets $$X$$ and $$Y$$. This gives us a way to pair up elements of $$X$$ and elements of $$Y$$ such that every element of $$X$$ is paired with exactly one element of $$Y$$.
Consider the pairs $$(x, f(x))$$ for $$x$$ in $$X$$. Every element of $$Y$$ appears in exactly one of these pairs (at least one pair because $$f$$ is onto, at most one pair because $$f$$ is one-one). So a bijection pairs up each element of $$X$$ with a unique element $$f(x)$$ of $$Y$$.
The picture is an illustration of a bijection $$f:\{1,2,3\}\to \{a,b,c\}$$. If we pair each element $$x$$ of the domain with its image $$f(x)$$ we get the pairs $$(1, c), (2, b), (3, a)$$. Because $$f$$ is a bijection, every element of the domain is paired with exactly one element of the domain and every element of the codomain is paired with exactly one element of the domain.
All this leads us to make this definition: two sets have the same size (or the same cardinality)if and only if there is a bijection between them.
This definition works even for infinite sets - though it sometimes provides some counterintuitive results. The sets $$\mathbb{Z}$$ and $$2 \mathbb{Z} = \{\ldots -4, -2, 0, 2, 4, 6, \ldots\}$$ have the same size since there is a bijection
\begin{align*} f : \mathbb{Z} &\to 2 \mathbb{Z} \\ f(z) &= 2z \end{align*}
## 09 Invertibility
Definition: let $$f:X \to Y$$.
• a left-inverse to $$f$$ is a function $$g: Y \to X$$ such that $$g\circ f=\operatorname{id}_X$$
• a right-inverse to $$f$$ is a function $$h : Y \to X$$ such that $$f\circ h=\operatorname{id}_Y$$
• an inverse (or a two-sided inverse) to $$f$$ is a function $$k:Y \to X$$ which is a left and a right inverse to $$f$$.
Notice that if $$g$$ is left-inverse to $$f$$ then $$f$$ is right-inverse to $$g$$. A function can have more than one left-inverse, or more than one right-inverse: you will investigate this further in exercises.
The idea is that a left-inverse “undoes” what its right-inverse “does”, in the sense that if you have a function $$f$$ with a left-inverse $$g$$, and you start with $$x \in X$$ and “do” $$f$$ to get to $$f(x) \in Y$$, then $$g(f(x)) = x$$ so $$g$$ undoes what $$f$$ did and gets you back to the element $$x$$ you started with.
Examples:
• $$f: \mathbb{R} \to [0, \infty), f(x)=x^2$$ has a right-inverse $$g: [0, \infty) \to \mathbb{R}, g(x)=\sqrt{x}$$. $$f(g(x)) = x$$ for all $$x \in [0, \infty)$$. This is not a left-inverse.
• This function $$f$$ does not have a left inverse. Suppose $$h$$ is left-inverse to $$f$$, so that $$hf=id_\mathbb{R}$$. Then $$h(f(-1))=-1$$, so $$h(1)=-1$$. Similarly $$h(f(1))=1$$, so $$h(1) = 1$$. Impossible! (The problem is that $$f$$ isn’t one-to-one.)
• The function $$g$$ has a left-inverse, $$f$$. But it does not have a right-inverse. If $$g\circ h = id_{\mathbb{R}}$$ then $$g(h(-1)) = -1$$ so $$g(h(-1)) = -1$$. But there’s no element of $$[0, \infty)$$ that $$g$$ takes to $$-1$$. (This time the problem is that $$g$$ isn’t onto.)
## 10 When is a function invertible?
### Inverses and function properties
Here is the connexion between function properties and invertibility.
Theorem: Let $$f:X \to Y$$, and assume $$X \neq \emptyset$$. Then
1. $$f$$ has a left-inverse if and only if it is injective
2. $$f$$ has a right-inverse if and only if it is surjective
3. $$f$$ is invertible if and only if it is bijective
Proof:
• ONLY IF. Let $$g$$ be left-inverse to $$f$$, so $$g\circ f=\operatorname{id}_X$$. Suppose $$f(a)=f(b)$$. Then applying $$g$$ to both sides, $$g(f(a))=g(f(b))$$, so $$a=b$$.
• IF. (draw a picture to illustrate construction) Let $$f$$ be injective. Choose any $$x_0$$ in the domain of $$f$$. Define $$g:Y \to X$$ as follows. Each $$y$$ in $$Y$$ is either in the image of $$f$$ or not. If $$y$$ is in the image of $$f$$, it equals $$f(x)$$ for a unique $$x$$ in $$X$$ (uniqueness is because of the injectivity of $$f$$), so define $$g(y)=x$$. If $$y$$ is not in the image of $$f$$, define $$g(y)=x_0$$. Clearly $$g\circ f= \operatorname{id}_X$$.
• IF. Suppose $$f$$ has a right inverse $$g$$, so $$f \circ g = \operatorname{id}_Y$$. If $$y \in Y$$ then $$f(g(y)) = \operatorname{id}_Y(y) = y$$, so $$y \in \operatorname{im}(f)$$. Every element of $$Y$$ is therefore in the image of $$f$$, so $$f$$ is onto.
• ONLY IF. Suppose $$f$$ is surjective. Let $$y \in Y$$. Then $$y$$ is in the image of $$f$$, so we can choose an element $$g(y) \in X$$ such that $$f(g(y)) = y$$. This defines a function $$g : Y \to X$$ which is evidently a right-inverse to $$f$$.
1. If $$f$$ has a left inverse and a right inverse, it is injective (by part 1 of this theorem) and surjective (by part 2), so is a bijection. Conversely if $$f$$ is a bijection it has a left inverse $$g:Y\to X$$ and a right inverse $$h:Y\to X$$ by part 1 and part 2 again. But “invertible” required a single function which was both a left and a right inverse. We must show $$g=h$$: \begin{align*} g &= g\circ \operatorname{id}_Y \\ &= g \circ (f \circ h) & \text{as } fh = \operatorname{id}_Y \\ &= (g\circ f) \circ h & \text{associativity} \\ &= \operatorname{id}_X \circ h & \text{as } gf=\operatorname{id}_X \\ &= h \end{align*} so $$g=h$$ is an inverse of $$f$$.
The diagram illustrates the construction in part 1 of the theorem. Arrows from left to right show where $$f$$ sends each element of $$X$$. Arrows from right to left show where the left inverse $$g$$ we have constructed sends each element of $$Y$$.
Notation: if $$f$$ is invertible, we write $$f^{-1}$$ for THE inverse of $$f$$.
(there’s only one inverse, because if $$g$$ and $$h$$ are inverses to $$f$$ then by the argument in part 3 above, $$g=h$$.)
Inverses are important to us. You’ll see the following result in a lot of contexts.
### Inverse of a composition
Theorem: If $$f :X \to Y$$ and $$g:Y \to Z$$ are invertible then so is $$g\circ f$$, and $$(g\circ f)^{-1} = f^{-1}\circ g^{-1}$$.
Proof: you can check that $$f^{-1}\circ g^{-1}$$ is a left- and right- inverse to $$g\circ f$$.
It is important to get this the right way round. The inverse of $$g\circ f$$ is not normally $$g^{-1} \circ f^{-1}$$, indeed this composition may not even make sense.
The correct result is easy to remember when you think about getting dressed. Each morning you put on your socks, then you put on your shoes: if $$k$$ is the put-on-socks function and $$h$$ is the put-on-shoes function then you apply the function $$h \circ k$$ to your feet. The inverse of this is taking off your shoes, then taking off your socks: $$k^{-1} \circ h^{-1}$$. Not the other way round - it’s not even (normally) possible to take off your socks, then take off your shoes, just as it is not normally possible to form the composition $$g^{-1}\circ f^{-1}$$ in the context of the theorem above.
A similar result applies when you compose more than two invertible functions: if $$f_1, f_2, \ldots, f_n$$ are invertible and if the composition $f_1\circ \cdots \circ f_n$ makes sense, it is also invertible and its inverse is
$f_n^{-1} \circ \cdots \circ f_1^{-1}.$
## 11 Relations
### Definition of a relation
Informally, a relation on a set $$X$$ is a property of an ordered pair of elements of $$X$$ that could be true or false.
You know many of these.
• $$\leq$$ is a relation on $$\mathbb{R}$$. For any ordered pair $$(x,y)$$ of real numbers, $$x \leq y$$ is either true or false.
• $$\neq$$ is a relation on $$\mathbb{Z}$$. For any ordered pair $$(x, y)$$ of integers, $$x \neq y$$ is either true or false.
• The parity of an integer is whether it is odd or even. $$x \sim y$$ if and only if $$x$$ and $$y$$ have the same parity is a relation on $$\mathbb{Z}$$ - for example, $$-2 \sim 48$$ but it is not true that $$0 \sim 3$$. When $$a \sim b$$ is false, we write $$a \nsim b$$.
• Two silly examples: let $$X$$ be any set. There is an empty relation $$\sim_e$$ on $$X$$ such that $$x \sim_e y$$ is never true for any $$x, y \in X$$, and a universal relation $$\sim_u$$ such that $$x \sim_u y$$ is always true for any $$x, y \in X$$.
Just as for functions, there is a formal definition of a relation…
Formal definition: a relation $$\sim$$ on a set $$X$$ is a subset of $$X\times X$$. We write $$x\sim y$$ to mean $$(x, y) \in \sim$$.
…which we won’t use in 0005. But it is good to know such a thing exists, because it gets us out of having to say what a “property that could be true or false” really is.
### Congruence modulo n
This is a very important example of a relation on the set $$\mathbb{Z}$$ of integers, which we will use later in 0005 and which occurs throughout mathematics.
For each integer $$n \neq 0$$, there is a relation on $$\mathbb{Z}$$ called congruence modulo $$n$$ (or just “congruence mod n”, for short). Two integers $$x$$ and $$y$$ are congruent modulo $$n$$ if and only if $$x-y$$ is divisible by $$n$$.
There is a special notation for this relation: we write
$x \equiv y \mod n$
to mean that $$x$$ and $$y$$ are congruent modulo $$n$$.
Examples:
• $$3$$ and $$-7$$ are congruent modulo 2, as $$3 - (-7) = 10$$ is divisible by 2 (in fact, congruence modulo 2 is the same as the last example relation above). But $$3$$ and $$8$$ are not congruent modulo 2 because $$3- 8=-5$$ isn’t a multiple of $$2$$.
• $$3$$ and $$-7$$ are congruent modulo $$5$$ as $$3 - (-7)=10$$ is divisible by $$5$$. But $$12$$ and $$3$$ are not congruent modulo $$5$$ because $$12-3$$ isn’t divisible by $$5$$
## 12 Relation properties
### Reflexive, symmetric, transitive
Let $$\sim$$ be a relation on a set $$X$$.
• $$\sim$$ is reflexive (R) if and only if $$\forall x \in X : x \sim x$$.
• $$\sim$$ is symmetric (S) if and only if $$\forall x, y \in X: x \sim y \implies y \sim x$$.
• $$\sim$$ is transitive (T) if and only if $$\forall x,y,z \in X: x\sim y$$ and $$y \sim z$$ implies $$x \sim z$$.
$$\sim$$ is an equivalence relation if and only if it has all three properties (R), (S), and (T).
There’s no restriction or hidden assumptions on $$x,y,z$$, like “they have to be distinct” or something like that. $$\sim$$ is transitive if and only if for every $$x,y,z$$ it is true that if $$x\sim y$$ and $$y\sim z$$ then $$x \sim z$$.
Here is a table of examples. The R, S, T, E columns contain y if the relation is reflexive, symmetric, transitive, or an equivalence relation, otherwise they contain n.
relation and set R S T E
$$\leq, \mathbb{R}$$ y no, $$1\leq 2, 2 \nleq 1$$ y n
$$<, \mathbb{R}$$ n n y n
Empty relation, $$\mathbb{Z}$$ n y y n
Empty relation, $$\emptyset$$ y y y y
a~b iff a-b even, $$\mathbb{Z}$$ y y y y
$$\neq, \mathbb{N}$$ n y n n
### Definition of equivalence class
The last relation in the table could be expressed in a different way: two integers are related if and only if they have the same parity, that is, they’re both odd or they’re both even. The relation comes from splitting up $$\mathbb{Z}$$ into two subsets with no elements in common, the set of odd integers and the set of even integers. In fact every equivalence relation arises from splitting a set up into some number of parts, as we’ll now investigate.
To do this we need the notion of an equivalence class.
Let $$\sim$$ be an equivalence relation on $$X$$ and $$x \in X$$. The equivalence class of $$x$$ is $\{ a \in X : a \sim x\}.$ We write $$[x]_{\sim }$$ or just $$[x]$$ for the equivalence class of $$x$$.
Notice it doesn’t matter if we define $$[x]$$ to be $$\{a \in X: x \sim a\}$$, because an equivalence relation is symmetric: $$x \sim a$$ if and only if $$a \sim x$$.
Let’s work out some equivalence classes for the congruence modulo 2 relation on $$\mathbb{Z}$$. Recall that $$x$$ and $$y$$ are congruent modulo 2 if and only if $$x-y$$ is divisible by 2 (that is, $$x-y$$ is even).
The equivalence class of $$0$$, which we write as $$[0]$$ is the set of all integers $$y$$ such that $$y$$ is congruent to 0 modulo 2, in other words, all integers $$y$$ such that $$y-0$$ is even. So it’s the set of all even integers. $[0] = \{ \ldots, -4, -2, 0, 2, 4, \ldots \}$
Now let’s do $$[4]$$. This is the set of all integers $$y$$ such that $$y$$ is congruent to 4 modulo 2, that is, all integers $$y$$ such that $$y-4$$ is even. Again, $$y-4$$ is even if and only if $$y$$ is even, so $$[4]$$ is the set of all even integers
$[4] = \{\ldots, -4, -2, 0, 2, 4 ,\ldots, \} = [0]$
(So the equivalence classes of two different things can be equal).
Now let’s work out the equivalence class $$[1]$$. An integer $$y$$ is congruent to 1 modulo 2 if and only if $$y-1$$ is even. That’s true if and only if $$y$$ is odd, so $$[1]$$ is the set of all odd numbers:
$[1] = \{ \ldots, -3, -1, 1, 3, \ldots \}$
What about $$[-3]$$? An integer $$y$$ is congruent to $$-3$$ modulo 2 if and only if $$y- (-3)$$ is even. This is again true if and only if $$y$$ is odd, so
$[-3] = \{\ldots, -3, -1, 1, 3, \ldots \} = [1].$
There are exactly two different equivalence classes, one being the set of all even numbers and the other being the set of all odd numbers. Every integer belongs to exactly one of these equivalence classes. In the next video we’re going to generalise this to all equivalence relations.
Before that, you should investigate the equivalence classes for congruence modulo 3. List the elements of $$[0]$$, $$[1]$$, and $$[2]$$ for this relation. How do the classes relate to remainders on dividing by 3? Is $$[4]$$ equal to one of $$[0], [1], [2]$$? What about $$[-5]$$?
You will find that there are exactly three different equivalence classes this time: every class is equal to one of $$[0]$$, $$[1]$$, or $$[2]$$.
## 13 Equivalence relations and partitions
### Definition of a partition of a set
A partition of a set $$X$$ is a set of nonempty subsets of $$X$$ called parts, such that every element of $$X$$ is a member of exactly one part.
A simple example is that the set $$O$$ of odd numbers and the set $$E$$ of even numbers make up a partition of $$\mathbb{Z}$$. Every integer is either odd or even, but not both, so every integer lies in exactly one of of the two parts $$O$$ and $$E$$.
Two sets $$A$$, $$B$$ are called disjoint if $$A \cap B = \emptyset$$.
Another way to state the definition of a partition of $$X$$ is to say that the union of the parts is $$X$$ and that any two parts in the partition are disjoint.
Here is another example partition. Every real number is either
• strictly positive,
• strictly negative,
• or zero,
but it can’t be more than one of those things at once. So if $$P$$ is the set of strictly positive numbers and $$N$$ is the set of strictly negative real numbers, the sets $$P, N, \{0\}$$ form a partition of $$\mathbb{R}$$.
### Partitions and equivalence relations
Our aim in the rest of this video is to show every partition gives rise to an equivalence relation, and every equivalence relation (on a non-empty set) gives rise to a partition.
#### Making an equivalence relation out of a partition
Given a partition of a set $$X$$, consider the relation given by saying $$x \sim y$$ if and only if $$x$$ and $$y$$ are elements of the same part. This is an equivalence relation:
• (R): $$x \sim x$$ for all $$x$$: every element is a member of the same part as itself.
• (S): $$x \sim y$$ if and only if $$y \sim x$$: if $$x$$ and $$y$$ are in the same part then $$y$$ and $$x$$ are in the same part.
• (T): if $$x$$ and $$y$$ are in the same part, and $$y$$ and $$z$$ are in the same part, then the part containing $$x$$ and the part containing $$z$$ have an element, $$y$$, in common. They are therefore equal, so $$x \sim z$$.
Here’s a picture of this way of constructing an equivalence relation.
The set $$X = \{1,2,3,4,5,6\}$$ is shown with the partition $$\{1,2,3\}$$, $$\{4,5\}$$, $$\{6\}$$. The equivalence relation you get from this partition is shown in the table:
$$\sim$$ 1 2 3 4 5 6
1 y y y n n n
2 y y y n n n
3 y y y n n n
4 n n n y y n
5 n n n y y n
6 n n n n n y
In row $$i$$ and column $$j$$ of the table there is a y if $$i \sim j$$, otherwise there is an n.
For another example the partition $$O$$, $$E$$ of $$\mathbb{Z}$$ we saw earlier gives rise to an equivalence relation in this way, and in fact it’s the relation of congruence modulo 2.
#### Making a partition out of an equivalence relation
Theorem. Let $$\sim$$ be an equivalence relation on a nonempty set $$X$$. Then the equivalence classes under $$\sim$$ are a partition of $$X$$.
To prove this, we must show
1. that every equivalence class is non-empty,
2. that every element of $$X$$ is an element an equivalence class, and
3. that every element of $$X$$ lies in exactly one equivalence class.
Proof: Part 1 is easy. An equivalence class $$[x]$$ contains $$x$$ as $$x \sim x$$ (by property (R)), so certainly $$[x] \neq \emptyset$$.
This also shows that part 2 is true: every $$x \in X$$ is in the equivalence class $$[x]$$, so every element of $$X$$ is an element of at least one equivalence class.
Finally let’s do part 3. How could it fail to be true that every element was in exactly one equivalence class? There would have to be an element of $$X$$ in (at least) two different equivalence classes. To show that can’t happen, we’ll show that if $$x \in X$$ lies in two classes $$[y]$$ and $$[z]$$, it must be that $$[y] = [z]$$.
Suppose $$x \in [y] \cap [z]$$. Let’s show $$[y] \subseteq [z]$$ first. To do that, we take any $$a \in [y]$$ and show $$a \in [z]$$.
First, $$a\sim y$$, and $$x \in [y]$$ so $$x\sim y$$. By (S), $$y\sim x$$, so by (T) $$a\sim x$$. Now $$x \in [z]$$, so $$x\sim z$$, so by (T) again $$a\sim z$$, so $$a \in [z]$$. We have $$[y] \subseteq [z]$$.
By the same argument, $$[z]\subseteq [y]$$, so $$[z] = [y]$$. The proof is finished.
Here’s a picture of this argument. The double-arrows illustrate that $$x\sim y$$ and $$y\sim x$$, and $$x \sim z$$ and $$z \sim x$$. The arrows between $$a$$ and $$y$$ illustrate that $$y \sim a$$ and $$a \sim y$$, and the dotted arrow shows us using property (T) to deduce that $$a \sim x$$.
## 14 Introduction to permutations
### Definition of a permutation
• A permutation of a set $$X$$ is a bijection $$X \to X$$.
• $$S_n$$, the symmetric group on n letters, is the set of all permutations of $$\{1,2,\ldots, n\}$$.
The name “group” is a foreshadowing - we will meet groups later.
Here are some examples.
• for any set $$X$$, the identity function $$\operatorname{id}_X: X \to X$$ is a permutation.
• The function $$f: \{1, 2, 3\} \to \{1, 2, 3\}$$ given by $$f(1)=3, f(2)=2, f(3)=1$$ is a permutation. $$f$$ is an element of $$S_3$$.
• The function $$g : \{1,2,3,4\} \to \{1, 2, 3, 4\}$$ give by $$g(1)=2, g(2)=3, g(3)=4, g(4)=1$$ is a permutation. $$g$$ is an element of $$S_4$$.
Here are some diagrams illustrating the permutations $$f$$ and $$g$$:
### Two-row notation
We need a way of writing down elements of $$S_n$$. The simplest is called two row notation. To represent $$f \in S_n$$, you write two rows of numbers. The top row is 1, 2, …, n. Then underneath each number $$i$$ on the top row you write $$f(i)$$:
$\begin{pmatrix} 1 & 2 & \cdots & n \\ f(1) & f(2) & \cdots & f(n) \end{pmatrix}$
As an example, here are the two row notations for the two permutations of the previous example
\begin{align*} f &: \begin{pmatrix} 1&2&3\\3&2&1 \end{pmatrix} \\ g &: \begin{pmatrix} 1&2&3&4\\ 2&3&4&1 \end{pmatrix} \end{align*}
The two row notation for the identity in $$S_n$$ is particularly simple: $\begin{pmatrix} 1 & 2 & \cdots & n-1 & n \\ 1 & 2 & \cdots & n-1 & n \end{pmatrix}.$
This is a simple and not that efficient notation (it is not feasible to write down an element of $$S_{100}$$ this way, even if it is a very simple permutation e.g. swaps 1 and 2, leaves 3-100 alone), but it is at least concrete and simple.
### How many permutations?
Theorem. $$|S_n| = n!$$
($$n!$$ means $$n\times (n-1) \times \cdots \times 2 \times 1$$)
Proof: Instead of counting permutations we will count possible bottom rows of two row notations for elements of $$S_n$$. Because a permutation is a bijection - one-to-one and onto - this bottom row consists of the numbers 1, 2, …, n in some order. We just need to show that there are exactly $$n!$$ different ways to order the numbers 1, 2, … , n.
We prove this by induction on $$n$$. For the base case $$n=1$$ we have $$1!=1$$ and it is clear that there is only one way to order a single 1.
For the inductive step, suppose $$|S_{n-1}| = (n-1)!$$. An ordering of 1, 2, …, n arises in exactly one way as an ordering of 1, 2, …, (n-1) with the number n inserted into one of n places (the first, or second, or …, or nth position). So the number of such orderings is $$|S_{n-1}|$$ (the number of ways to choose an ordering of 1, 2, …, (n-1)) times the number of ways to insert an n, giving $$|S_n| = |S_{n-1}| \times n = (n-1)!\times n = n!$$. This completes the inductive step.
For example, there are $$2! = 2$$ elements of $$S_2$$: they are
$\begin{pmatrix} 1&2\\1&2 \end{pmatrix}, \begin{pmatrix} 1&2\\2&1 \end{pmatrix}$
(The first one is the identity function on $$\{1,2\}$$).
Permutations turn out to be very important for the rest of the course, especially $$S_n$$, so we will spend some time on them and their properties.
## 15 Inverses and composition
### Inverse of a permutation
Permutations are bijections, so they have inverse functions. The inverse function to a permutation $$\sigma$$ undoes what $$\sigma$$ did, in the sense that if $$\sigma(x) = y$$ then $$\sigma^{-1}(y) = x$$
In two row notation you write $$\sigma(x)$$ beneath $$x$$. So you can get the two row notation for $$\sigma^{-1}$$ by swapping the rows (and reordering)
Example:
\begin{align*}\sigma & = \begin{pmatrix}1&2&3&4\\2&3&4&1\end{pmatrix} \\ \sigma^{-1} &= \begin{pmatrix}2&3&4&1\\1&2&3&4\end{pmatrix} = \begin{pmatrix}1&2&3&4\\4&1&2&3\end{pmatrix}\end{align*}
### Composition of permutations
We know that the composition of two bijections is a bijection. So the composition of two permutations of a set $$X$$ is again a permutation of $$X$$.
For example, let
\begin{align*} \sigma &= \begin{pmatrix}1&2&3\\2&1&3\end{pmatrix} \\ \tau & = \begin{pmatrix}1&2&3\\1&3&2\end{pmatrix} \end{align*}
Then $$\sigma\circ\tau$$ is the function $$\{1,2,3\}\to \{1,2,3\}$$ whose rule is “do $$\tau$$, then do $$\sigma$$.” Thus
\begin{align*} (\sigma\circ\tau)(1) &= \sigma(\tau(1)) = \sigma(1) = 2 \\ (\sigma\circ\tau)(2) &= \sigma(\tau(2)) = \sigma(3) = 3 \\ (\sigma\circ\tau)(3) &= \sigma(\tau(3)) = \sigma(2) = 1 \end{align*}
In two row notation,
$\sigma \circ \tau = \begin{pmatrix} 1&2&3\\ 2 & 3 & 1 \end{pmatrix}$
### Composition isn’t commutative
This composition has an interesting property: the order matters.
With $$\sigma$$ and $$\tau$$ as before, \begin{align*} \sigma\circ\tau &= \begin{pmatrix} 1&2&3\\ 2&3&1 \end{pmatrix} \\ \tau\circ\sigma &= \begin{pmatrix} 1&2&3\\ 3&1&2 \end{pmatrix} \end{align*}
They’re different: composition of permutations is not commutative in general.
Definition: Two permutations $$s$$, $$t$$ are said to commute if $$s\circ t=t\circ s$$.
## 16 Cycles
### Cycle definition and notation
We’re going to introduce a more efficient way of writing permutations. This involves thinking about a special kind of permutation, called a cycle.
Let $$m>0$$, let $$a_0,\ldots, a_{m-1}$$ be distinct positive integers. Then $(a_0,\ldots,a_{m-1})$ is defined to be the permutation such that
• $$a_i$$ goes to $$a_{i+1}$$ for $$i<m-1$$, $$a_{m-1}$$ goes to $$a_0$$, and
• any number which isn’t an $$a_i$$ goes to itself.
If we write $$a_m := a_0$$ then we could just say $$a_i \mapsto a_{i+1}$$ then we could just say $$a_i \mapsto a_{i+1}$$ for all $$i$$.
Definition: A permutation of the form $$(a_0, \ldots, a_{m-1})$$ is called an $$m$$-cycle. A permutation which is an $$m$$-cycle for some $$m$$ is called a cycle.
Two important things to note:
• any 1-cycle is equal to the identity permutation
• if we just write down the cycle $$(1,2,3)$$, say, it could be could be an element of $$S_3$$, or $$S_4$$, or $$S_5$$, or any other $$S_n$$ with $$n\geq 3$$. When it matters, we will make this clear.
Example:
• In $$S_3$$, the 2-cycle $$(1, 2)$$ is the permutation that sends 1 to 2, 2 to 1, and 3 to 3. In two row notation $$(1,2)= \begin{pmatrix} 1&2&3\\2&1&3 \end{pmatrix}$$.
• In $$S_4$$, the 3-cycle $$(2, 3, 4)$$ is the permutation that sends 1 to 1, 2 to 3, 3 to 4, and 4 to 2. In two row notation $$(2,3,4) = \begin{pmatrix} 1&2&3&4\\1&3&4&2 \end{pmatrix}$$.
The picture below is of the 5-cycle $$(1,2,3,4,5)$$, illustrating why these permutations are called “cycles”.
### Multiple ways to write the same cycle
Consider the two cycles $$a = (1, 2, 3, 4)$$ and $$b = (2,3,4,1)$$.
$$a$$ sends 1 to 2, 2 to 3, 3 to 4, 4 to 1, and any other number to itself. So does $$b$$. So $$a=b$$.
Similarly if $$c = (3, 4, 1, 2)$$ and $$d=(4,1,2,3)$$ then $$a=b=c=d$$.
Contrast this with two row notation, where there was only one way to write a given permutation.
In general, every $$m$$-cycle can be written $$m$$ different ways since you can put any one of the $$m$$ things in the cycle first.
Another example: in $$S_5$$, $(5,3, 2) = (3, 2, 5) = (2, 5, 3).$
### Disjoint cycles
Two cycles $$(a_0,\ldots,a_{m-1})$$ and $$(b_0,\ldots, b_{k-1})$$ are disjoint if no $$a_i$$ equals any $$b_j$$.
Example:
• (1,2,7) is disjoint from (5,4)
• (1,2,3) and (3,5) are not disjoint.
One reason disjoint cycles are important is that disjoint cycles commute. If $$a$$ and $$b$$ are disjoint cycles, $$a\circ b = b \circ a$$. This is a special property, as you have seen that in general for two permutations $$s$$ and $$t$$, $$s\circ t \neq t \circ s$$.
### Inverse of a cycle
For every permutation $$s$$ there is an inverse permutation $$s^{-1}$$ such that $$s \circ s^{-1} = s^{-1} \circ s = \operatorname{id}$$.
How do we find the inverse of a cycle? $$a = (a_0, \ldots, a_{m-1})$$ sends $$a_i$$ to $$a_{i+1}$$ for all $$i$$ (and every number not equal to an $$a_i$$ to itself), so $$a^{-1}$$ should send $$a_{i+1}$$ to $$a_i$$ for all $$i$$ (and every number not equal to an $$a_i$$ to itself).
In other words, $$a^{-1}$$ is the cycle $$(a_{m-1}, a_{m-2}, \ldots, a_1, a_0)$$:
$(a_0, \ldots, a_{m-1}) ^{-1} = (a_{m-1}, a_{m-2}, \ldots, a_1, a_0)$
As a special case, the inverse of the 2-cycle $$(i, j)$$ is $$(j, i)$$. But $$(i, j) = (j,i)$$! So every 2-cycle is its own inverse.
If we draw cycles as we did before, their inverses are obtained by “reversing the arrows.”
### Not all permutations are cycles
Not every permutation is a cycle.
Example: $$\sigma = \begin{pmatrix}1&2&3&4\\2&1&4&3\end{pmatrix}$$. This is not a cycle. Suppose it was. $$\sigma$$ sends 1 to 2, and 2 to 1, so if it were a cycle it would have to be $$(1,2)$$. But $$(1,2)$$ sends 3 to 3, whereas $$\sigma(3)=4$$, so $$\sigma \neq (1,2)$$.
Here is a diagram of the permutation $$\sigma$$:
While $$\sigma$$ is not a cycle, it is the composition of two cycles: $$\sigma = (1, 2)\circ (3,4)$$. In fact, every permutation can be written this way.
## 17 Products of disjoint cycles
### Every permutation is a product of disjoint cycles
Here is our previous example: $$\sigma = \begin{pmatrix} 1&2&3&4\\2&1&4&3 \end{pmatrix}$$. This isn’t a cycle, but it does equal $$(1,2) \circ (3,4)$$.
This is suggestive of the following theorem.
Theorem: Every $$s \in S_n$$ equals a product of disjoint cycles.
Proof. Define a relation on $$\{1, 2, \ldots, n\}$$ by $$x \sim y$$ if and only if there exists an integer $$r$$ such that $$y = s^r(x)$$. This is an equivalence relation.
Note that for every $$x \in \{1,2,\ldots, n\}$$, there is a positive number $$l$$ such that $$s^l(x) = x$$. For the elements $x, s(x), s^2(x), \ldots$ cannot all be different as there are only $$n$$ possible values, namely 1, 2, …, $$n$$, for $$s^r(x)$$. So for some $$i<j$$ we have $$s^i(x) = s^j(x)$$. Applying the function $$s^{-i}$$ to both sides of this equation, $$x = s^{j-i}(x)$$.
It follows that there is a smallest positive number $$l$$ such that $$s^l(x) = x$$.
I claim that the equivalence class $$[x]$$ equals $$\{x, s(x), s^2(x), \ldots, s^{l-1}(x)\}$$ where $$l$$ is this smallest positive number such that $$s^l(x) = x$$. It is immediate from the definition of $$\sim$$ that $\{x, s(x), \ldots, s^{l-1}(x) \} \subseteq [x]$ so we just need to show that $$[x] \subseteq \{x, s(x), \ldots, s^{l-1}(x)\}$$. Let $$y \in [x]$$. Certainly $$y = s^r(x)$$ for some $$r \in \mathbb{Z}$$, but we must show $$y$$ is equal to a power of $$s$$ between 0 and $$l-1$$ applied to $$x$$. Divide $$r$$ by $$l$$, getting quotient $$q$$ and remainder $$t$$, so $$r = ql + t$$ where $$0 \leq t < l$$. Since $$s^l(x)=x$$, we have $$s^{ql}(x) = x$$ too. So $y = s^r(x) = s^{t+ql}(x) = s^t(s^{ql}(x)) = s^t(x)$ by the exponent laws. It follows $$y \in \{x, s(x), \ldots, s^{l-1}(x)\}$$, proving the claim.
I also claim that all the numbers $$x, s(x), s^2(x), \ldots, s^{l-1}(x)$$, with $$l$$ as defined above, are different. If two of them were equal, then there would be numbers $$i$$ and $$j$$ with $$0 \leq i < j < l$$ such that $$s^i(x) = s^j(x)$$. Then applying the function $$s^{-i}$$ to both sides of this equation and using the exponent laws, $$x = s^{j-i}(x)$$. But $$0 < j-i < l$$. This would contradict $$l$$ being the smallest positive number such that $$s^l(x) = x$$.
This means that the cycle $$c_x = (x, s(x), \ldots, s^{l-1}(x))$$ makes sense (remember, part of the definition of a cycle $$(a_0, \ldots, a_{m-1})$$ was that all the $$a_i$$ should be different). Furthermore, $$c_x(i) = s(i)$$ for each $$i$$ in the equivalence class of $$[x]$$, because we know that equivalence class is just $$x, s(x), \ldots, s^{l-1}(x)$$. (**)
Now let the distinct equivalence classes be $$[x_1], [x_2], \ldots, [x_m]$$. Notice that the cycles $$c_{x_1}, c_{x_2}, \ldots, c_{x_m}$$ are disjoint, because the entries in the cycle $$c_{x_j}$$ are the elements of the equivalence class of $$x_j$$, and any two distinct equivalence classes are disjoint.
I claim that $s = c_{x_1} \circ c_{x_2} \cdots c_{x_m}.$ Since the right hand side is a product of disjoint cycles, this will finish the proof. This is a claim that two functions $$\{1, 2, \ldots, n\} \to \{1, 2, \ldots, n\}$$ are equal, so to check it we need to show that for each $$1 \leq i \leq n$$ we have $s(i) = (c_{x_1} \circ c_{x_2} \cdots c_{x_m})(i).$
Fix such a number $$i$$. Remember that the equivalence classes are a partition of $$\{1,2,\ldots, n\}$$, so $$i$$ belongs to exactly one of the equivalence classes $$[x_1], \ldots, [x]_m$$. Say $$i \in [x_j]$$, so that $$i \notin [x_k]$$ for any $$k \neq j$$. Then
• For each $$k \neq j$$, $$c_{x_k}$$ sends $$i$$ to $$i$$. For $$c_{x_k}$$ sends everything not in $$[x_k]$$ to itself, and $$x \notin [x_k]$$
• For each $$k \neq j$$, $$c_{x_k}$$ sends $$s(i)$$ to $$s(i)$$. This is because $$s(i) \in [x_j]$$ too, so the same argument applies.
• We already established, at (**), that $$c_{x_j}(i) = s(i)$$.
It follows that $$(c_{x_1} \circ c_{x_2} \cdots c_{x_m})(i) = s(i)$$, and we’re done.
To understand the proof, it helps to follow it through with a particular example in mind. Say $s = \begin{pmatrix} 1&2&3&4&5&6&7&8&9\\ 7&3&2&1&5&9&4&6&8 \end{pmatrix}$
The equivalence classes are $$[1] = \{1,7,4\}, [2] = \{2,3\}, [5] = \{5\}, [6] = \{6,8,9\}$$. So the $$x_i$$s in the proof could be 1, 2, 5, 6. (Or they could be 7, 3, 5, 9, or many other possibilities - we just have to take one element from each equivalence class).
The smallest positive power of $$s$$ sending 1 to itself is $$s^3$$, the smallest positive power of $$s$$ sending $$2$$ to itself is $$s^2$$, the smallest positive power of $$s$$ sending 5 to itself is $$s^1$$, and the smallest positive power of $$s$$ sending 6 to itself is $$s^3$$ - these are the numbers $$l$$ in the proof.
The cycles arising from these equivalence classes are
\begin{align*} (1, s(1), s^2(1)) &= (1, 7, 4) \\ (2, s(2)) &= (2, 3) \\ (5) &= (5) \\ (6, s(6), s^2(6)) &= (6,9, 8) \end{align*}
The outcome of the proof is that $s = (1,7,4)(2,3)(5)(6,9,8).$
### Non-uniqueness
There can be many different ways to write a given permutation as a product of disjoint cycles. For example, taking the permutation $$s$$ we’ve just seen,
\begin{align*}s &= (1,7,4)(2,3)(5)(6,9,8)\\ &= (7,4,1)(2,3)(6,9,8) \\ &= (2,3)(6,9,8)(7,4,1) \\ &= \ldots \end{align*}
The two things you should remember are that an $$m$$-cycle can be written in $$m$$ different ways - $$(1,2,3)=(2,3,1)=(3,1,2)$$ - and that disjoint cycles commute - $$(1,2)(3,4) = (3,4)(1,2)$$.
### How to find a disjoint cycle decomposition
A disjoint cycle decomposition of a permutation $$\sigma$$ is a product of disjoint cycles which equals $$\sigma$$. For example
$\begin{pmatrix} 1&2&3&4\\ 2&1&4&3 \end{pmatrix} = (1,2)(3,4)$
is a disjoint cycle decomposition for the permutation discussed at the start of this video.
The theorem we just saw tells us every permutation has a disjoint cycle decomposition. It also gives a method of finding a disjoint cycle decomposition for a permutation given in 2-row notation, or in some other format that isn’t a disjoint cycle decomposition:
To find a disjoint cycle decomposition for $$\sigma \in S_n$$:
1. pick a number that doesn’t yet appear in a cycle
2. compute its image, and the image of that, and so on, until you have a cycle. Write down that cycle.
3. if all elements of 1..n are in one of your cycles, stop, else go to 1.
Example: Let $$s = \begin{pmatrix} 1&2&3&4&5&6&7 \\ 7&6&3&1&2&5&4 \end{pmatrix}$$. We pick a number not yet in a cycle, say 1. 1 goes to 7, 7 goes to 4, 4 goes to 1. We are back to the number we started with, so our first cycle is $$(1,7,4)$$. Now we pick another number not in a cycle, say 2. $$s$$ sends 2 to 6, 6 to 5, and 5 to 2. That’s another cycle, so we have $$(1,7,4)(2,6,5)$$. Now we pick another number not yet in a cycle - the only one left is 3. $$s$$ sends 3 to 3, so this is immediately a cycle. We have $$s = (1,7,4)(2,6,5)(3)$$.
As we saw when we defined cycles, any 1-cycle is equal to the identity function. For that reason (and because 1-cycles look confusingly like what we write when we evaluate a function) we usually omit 1-cycles like $$(3)$$ from disjoint cycle decompositions, so we’d write $$s = (1,7,4)(2,6,5)$$.
### Composing permutations given as products of disjoint cycles
Let’s work out a disjoint cycle decomposition for $$\sigma\tau$$ where
\begin{align*} \sigma &= (4,2,6,1,5) \\ \tau &= (5,4,7,3,8) \end{align*}
are elements of $$S_8$$.
Remember that $$\sigma\tau$$ means do $$\tau$$, then do $$\sigma$$. Keeping that in mind, all we have to do is follow the instructions from before. Start with 1:
\begin{align*} \sigma(\tau(1)) &= \sigma(1) = 5\\ \sigma(\tau(5)) &= \sigma(4) = 2 \\ \sigma(\tau(2)) &= \sigma(2) = 6\\ \sigma(\tau(6)) &= \sigma(6) = 1 \end{align*}
…and we have our first cycle, $$(1, 5, 2, 6)$$. Continuing with a number not yet in a cycle, say 3, we get
\begin{align*} \sigma(\tau(3)) &= \sigma(8) = 8\\ \sigma(\tau(8)) &= \sigma(5) = 4 \\ \sigma(\tau(4)) &= \sigma(7) = 7\\ \sigma(\tau(7)) &= \sigma(3) = 3 \end{align*}
…and we have our next cycle, $$(3,8,4,7)$$. There are no numbers left, so
$\sigma\tau = (1,5,2,6)(3,8,4,7).$
You should do $$\tau\sigma$$ now. You’ll find that your disjoint cycle decomposition has two 4-cycles again.
For another example, let’s find a disjoint cycle decomposition for $$(1,2)(2,3)(3,4)$$. Write $$a = (1,2), b = (2,3), c = (3,4)$$.
Again, remember that this composition means “do $$c$$, then do $$b$$, then do $$c$$.” So starting with 1 as usual,
\begin{align*} a(b(c(1))) &= a(b(1)) = a(1) = 2 \\ a(b(c(2))) &= a(b(2)) = a(3) = 3 \\ a(b(c(3))) &= a(b(4)) = a(4) = 4 \\ a(b(c(4))) &= a(b(3)) = a(2) = 1 \end{align*}
and so $$abc = (1,2,3,4)$$.
## 18 Powers and orders
### Powers of a permutation
Since the composition of two permutations is another permutation, we can form powers of a permutation by composing it with itself some number of times.
Definition: Let $$s$$ be a permutation and let $$m$$ be an integer. Then $s^m = \begin{cases} s\cdots s \; (m \text{ times}) & m>0 \\ \operatorname{id} & m = 0 \\ s^{-1}\cdots s^{-1} \; (-m \text{ times}) & m<0 \end{cases}$
It’s tedious but straightforward to check that for any integers $$a$$, $$b$$,
• $$s^a\circ s^b = s^{a+b}$$, and
• $$(s^a)^b = s^{ab}$$
so that some of the usual exponent laws for real numbers hold for composing permutations. The two facts above are called the exponent laws for permutations.
We very often emphasise this analogy between multiplying numbers and composing permutations by talking about “multiplying” two permutations when we mean composing them, calling their composition their “product”, and writing it as $$st$$ instead of $$s \circ t$$.
But be careful: there are some differences. As we’ve seen, permutation composition is not commutative unlike real number multiplication, and as we saw earlier, $$(s\circ t)^{-1} = t^{-1}\circ s^{-1}$$
### Order of a permutation
Definition: The order of a permutation $$\sigma$$, written $$o(\sigma)$$, is the smallest strictly positive number $$n$$ such that $$\sigma^n = \operatorname{id}$$.
For example, let
\begin{align*} s &= \begin{pmatrix} 1&2&3\\ 2&3&1 \end{pmatrix} \\ t &= \begin{pmatrix} 1&2&3\\ 2&1&3 \end{pmatrix}\end{align*}
You should check that $$s^2 \neq \operatorname{id}$$ but $$s^3 = \operatorname{id}$$, so the order of $$s$$ is 3, and that $$t \neq \operatorname{id}$$ but $$t^2 = \operatorname{id}$$ so the order of $$t$$ is 2.
### Order of an $$m$$-cycle
Lemma: the order of an $$m$$-cycle is $$m$$.
Proof: ⬛⬛⬛⬛⬛ ⬛⬛⬛ ⬛⬛⬛⬛ ⬛ ⬛⬛ ⬛⬛⬛⬛⬛ ⬛⬛⬛⬛⬛⬛ ⬛⬛⬛⬛⬛⬛⬛
## 19 Transpositions
### Definition of a transposition
A transposition is a 2-cycle.
For example, the only transpositions in $$S_3$$ are $$(1,2)$$, $$(2,3)$$, and $$(1,3)$$.
### Every permutation is a product of transpositions
One way we’ve already seen of illustrating a permutation $$s$$ in $$S_n$$ is to draw the numbers 1, 2, …, n in a column, and then in another column to its right, and draw a line from each number $$i$$ in the first column to $$s(i)$$ in the second. You get what looks like a lot of pieces of string tangled together.
Here is a diagram of the permutation $$s= (1,2,3,4,5)$$. Think of the lines as pieces of string connecting $$i$$ on the left with $$s(i)$$ on the right.
Composing two permutations drawn in this way corresponds to placing their diagrams side-by-side:
Imagine taking such a diagram and stretching it out. You could divide it up into smaller diagrams, each of which contains only one crossing of strings.
A diagram with a single string crossing is a transposition, since only two numbers change place. The diagram above illustrates the fact that $(1,2,3,4) = (1,2)(2,3)(3,4).$
But this would work for any permutation, which suggests…
Theorem: every permutation equals a product of transpositions.
To prove this, we’ll need a lemma.
Lemma: every cycle equals a product of transpositions
Proof of the lemma: we show by induction on $$m$$ that
$(a_0,\ldots,a_{m-1}) = (a_0,a_1)(a_1,a_2)\ldots (a_{m-2},a_{m-1}).$
This is obvious for $$m=1$$. Now suppose $$m>1$$. Then
$(a_0,a_1)\cdots(a_{m-3}, a_{m-2})(a_{m-2}, a_{m-1}) = (a_0, a_1, \ldots, a_{m-2})(a_{m-2}, a_{m-1})$
by the inductive hypothesis. We must show this equals $$(a_0, \ldots, a_{m-1})$$. Let $$c = (a_0, \ldots, a_{m-1})$$, $$t = (a_0, \ldots, a_{m-2})$$, $$u = (a_{m-2}, a_{m-1})$$ so that we are claiming $$c=tu$$. To prove that the two functions $$c$$ and $$tu$$ are equal, we have to prove that $$c(x) = t(u(x))$$ for all $$x$$. We can do this case by case:
• if $$x \neq a_i$$ for any $$i$$ then $$c(x)=x$$, $$u(x)=x$$, and $$t(x)=x$$, so $$c(x) = t(u(x))$$.
• if $$x=a_i$$ for some $$0 \leq i < m-1$$ then $$c(x)=a_{i+1}$$, $$u(x) = a_i$$, and $$t(a_i)=a_{i+1}$$, so $$c(x) =a_{i+1} = t(u(x))$$.
• if $$x=a_{m-2}$$ then $$c(x) = a_{m-1}$$, $$u(x) = a_{m-1}$$, and $$t(a_{m-1})=a_{m-1}$$, so $$c(x) = a_{m-1} = t(u(x))$$.
• finally, if $$x=a_{m-1}$$ then $$c(x) = a_0$$, $$u(x) = a_{m-2}$$, and $$t(a_{m-2}) = a_0$$, so $$c(x) = a_0 = t(u(x))$$.
In every case $$c$$ and $$tu$$ agree, so $$c=tu$$, proving the lemma.
Proof of the theorem: let $$p$$ be a permutation. We have seen that every permutation can be written as a product of cycles, so there are cycles $$c_1,\ldots, c_k$$ such that $$p=c_1\cdots c_k$$. The lemma above shows how to write each $$c_i$$ as a product of transpositions, which expresses $$p$$ as a product of transpositions too.
Example. Suppose we want to express
$s = \begin{pmatrix} 1&2&3&4&5&6\\ 2&3&1&5&6&4 \end{pmatrix}$
as a product of transpositions. A disjoint cycle decomposition for $$s$$ is
$s = (1,2,3)(4,5,6)$
and applying the lemma above, we get
\begin{align*} (1,2,3) &= (1,2)(2,3)\\ (4,5,6) &= (4,5)(5,6) \end{align*}
So
$s = (1,2,3)(4,5,6) = (1,2)(2,3)(4,5)(5,6).$
An adjacent transposition is one of the form $$(i, i+1)$$.
This name is used because an adjacent transposition swaps two adjacent numbers, $$i$$ and $$i+1$$.
The only adjacent transpositions in $$S_5$$ are $$(1,2)$$, $$(2,3)$$, $$(3,4)$$, and $$(4,5)$$.
## 20 Sign of a permutation
### Definition of odd and even permutations
Now we know that every permutation can be expressed as a product of transpositions, we can make the following definition.
• A permutation is odd if it can be expressed as a product of an odd number of transpositions and even if it can be expressed as an even number of transpositions.
• The parity of a permutation is whether it is odd or even.
Examples:
• $$(a,b)$$ is odd.
• $$\operatorname{id} = (1,2)(1,2)$$ is even.
• $$(1,2,3) = (1,2)(2,3)$$ is even.
• $$(a_0,\ldots,a_{m-1})$$ is even if $$m$$ is odd and odd if $$m$$ is even.
### Odd or even theorem
It seems possible that a permutation could be odd AND even at the same time, but this isn’t the case.
Theorem: Every permutation is odd or even but not both.
Proof: Here is an outline of how to prove this. You won’t be asked about this proof in a closed-book exam. We know that every permutation can be written as a product of transpositions, so the only problem is to prove that it is not possible to find two expressions for a fixed permutation $$p \in S_n$$, one using a product $$s_1 s_2 \cdots s_{2m+1}$$ of an odd number of transpositions and one using a product $$t_1 t_2 \cdots t_{2l}$$ of an even number of transpositions. The steps are as follows:
• If this were the case then we would have $$\operatorname{id} = t_{2l}\cdots t_1 s_1 \cdots s_{2m+1}$$, expressing the identity as a product of an odd number of transpositions, so it is enough to prove that this is impossible.
• Observe that any transposition $$(i,j)$$ can be written as a product of an odd number of adjacent transpositions, as follows. Without loss of generality (as $$(i,j)=(j,i)$$) assume $$i<j$$. Let $$a =(j-1,j)\cdots (i+2,i+3)(i+1,i+2)$$. Then $$(i,j) = a (i, i+1) a^{-1}$$.
• So it is enough to show that the identity cannot be written as a product of an odd number of adjacent transpositions. Assume, for a contradiction, that it can be so written.
• Consider polynomials in the $$n$$ variables $$x_1,\ldots,x_n$$. If you have such a polynomial $$f(x_1,\ldots,x_n)$$ and a permutation $$\sigma \in S_n$$, you get a new polynomial $$\sigma(f)$$ by replacing each $$x_i$$ in $$f$$ with $$x_{\sigma(i)}$$. For example, if $$\sigma = (1,2,3)$$ and $$f=x_1+x_2^2$$ then $$\sigma(f)=x_2+x_3^2$$
• Define $$\Delta = \prod_{1\leq i<j \leq n}(x_i-x_j)$$
• Observe that for any $$k$$, $$(k,k+1)(\Delta) = -\Delta$$, and so any product of an odd number of adjacent transpositions sends $$\Delta$$ to $$-\Delta$$ too.
• Get a contradiction from the fact that $$\operatorname{id}(\Delta) = \Delta$$.
### Sign of a permutation
• The sign of a permutation is $$1$$ if it is even and $$-1$$ if it is odd.
• $$\operatorname{sign}(s)$$ is the sign of the permutation $$s$$
So if $$s$$ can be written as a product of $$m$$ transpositions, $$\operatorname{sign}(s) = (-1)^m$$.
Lemma: For any two permutations $$s$$ and $$t$$, $$\operatorname{sign}(st) = \operatorname{sign}(s)\operatorname{sign}(t)$$
Proof: If $$s$$ can be written as a product of $$m$$ transpositions and $$t$$ can be written as a product of $$n$$ transpositions, then $$st$$ can be written as a product of $$m+n$$ transpositions. So
\begin{align*} \operatorname{sign}(st) &= (-1)^{n+m}\\ &= (-1)^m (-1)^n \\ &= \operatorname{sign}(s) \operatorname{sign}(t) \end{align*}
This rule about the sign of a product means that
• an even permutation times an even permutation is even,
• an even permutation times an odd permutation is odd,
• an odd permutation times an odd permutation is even
…just like multiplying integers.
### Two results on the sign of a permutation
Lemma:
1. Even length cycles are odd, odd length cycles are even.
2. If $$s$$ is any permutation, $$\operatorname{sign}(s) = \operatorname{sign}(s^{-1})$$.
Proof:
1. we saw in the last section that if $$a = (a_0,\ldots, a_{m-1})$$ is any $$m$$-cycle, $(a_0 \ldots, a_{m-1}) = (a_0, a_1)\cdots (a_{m-2}, a_{m-1}).$ So an $$m$$-cycle is a product of $$m-1$$ transpositions. Therefore
• if $$m$$ is even $$a$$ is a product of the odd number $$m-1$$ of transpositions, so $$a$$ is odd.
• if $$m$$ is odd $$a$$ is a product of the even number $$m-1$$ of transpositions, so $$a$$ is even.
2. If $$s = t_1\cdots t_m$$ is a product of $$m$$ transpositions, $$s^{-1} = t_m^{-1}\cdots t_1^{-1}$$. But the inverse of a transposition is a transposition, so $$s^{-1}$$ is also the product of $$m$$ transpositions.
It follows from the first part of the lemma that $$\operatorname{sign} (a_0, \ldots ,a_{m-1}) = (-1)^{m-1}$$
#### Application
Here’s one of the many reasons the sign of a permutation is useful. Take a Rubik’s cube, and swap the middle colours on two faces. Is it still solvable? No, because all Rubik’s cube moves are even permutations of the faces. |
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# The measure of the complementary of the angle of measure $90{}^\circ$ is[a] $0{}^\circ$[b] $45{}^\circ$[c] $90{}^\circ$[d] $60{}^\circ$
Last updated date: 18th Apr 2024
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Hint: The measure of complementary angles add up to $90{}^\circ$. So let x be the measure of complementary angles. Create a linear equation in x. Solve for x. The value of x gives the measure of the complementary angle. Alternatively, you can directly use the complementary angle of measure x is $90{}^\circ -x$.
Complementary angles: Two angles whose measures add up to $90{}^\circ$ are called complementary angles, e.g. $60{}^\circ$ and $30{}^\circ$ are complementary angles.
Supplementary angles: Two angles whose measures add up to $180{}^\circ$ are called supplementary angles, e.g. $60{}^\circ$ and $120{}^\circ$
Reflex angle of an angle: The measure of an angle and its reflex sum up to $360{}^\circ$.
Since $90{}^\circ$ and x are complementary we have
$90{}^\circ +x=90{}^\circ$
Subtracting $90{}^\circ$ from both sides, we get
\begin{align} & 90{}^\circ +x-90{}^\circ =90{}^\circ -90 \\ & \Rightarrow x=0{}^\circ \\ \end{align}
Hence the measure of the complementary angle of $90{}^\circ$ is $0{}^\circ$.
Note: [1] There is usually a confusion whether supplementary angles sum up to $180{}^\circ$ or whether complementary angles add up to $180{}^\circ$. In that case, one can memorise as follows:
In the English alphabet, s comes after c. So s>c. So, supplementary angles add up to $180{}^\circ$ , and complementary angles add up to $90{}^\circ$ . |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Inequality Expressions
## Write inequalities from graphs, graph basic inequalities
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Practice Inequality Expressions
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Inequality Expressions
Suppose you're having a party, and you know that the number of people attending will be greater than or equal to 25. How would you write this inequality? If you had to graph the solutions to this inequality on a number line, could you do it? After completing this Concept, you'll not only be able to express inequalities such as this one with a graph, but you'll also be able to look at a graph and determine what inequality it represents.
### Guidance
Verbs that translate into inequalities are:
“greater than”
“greater than or equal to”
“less than”
“less than or equal to”
“not equal to”
Definition: An algebraic inequality is a mathematical sentence connecting an expression to a value, a variable, or another expression with an inequality sign.
Solutions to one-variable inequalities can be graphed on a number line or in a coordinate plane.
#### Example A
Graph the solutions to on a number line.
Solution: The inequality is asking for all real numbers larger than 3.
You can also write inequalities given a number line of solutions.
#### Example B
Write the inequality pictured below.
Solution: The value of four is colored in, meaning that four is a solution to the inequality. The red arrow indicates values less than four. Therefore, the inequality is:
Inequalities that “include” the value are shown as or . The line underneath the inequality stands for “or equal to.” We show this relationship by coloring in the circle above this value on the number line, as in the previous example. For inequalities without the “or equal to,” the circle above the value on the number line remains unfilled.
Four Ways to Express Solutions to Inequalities
1. Inequality notation: The answer is expressed as an algebraic inequality, such as .
2. Set notation: The inequality is rewritten using set notation brackets { }. For example, is read, “The set of all values of , such that is a real number less than or equal to one-half.”
3. Interval notation: This notation uses brackets to denote the range of values in an inequality.
1. Square or “closed” brackets [ ] indicate that the number is included in the solution
2. Round or “open” brackets ( ) indicate that the number is not included in the solution.
Interval notation also uses the concept of infinity and negative infinity . For example, for all values of that are less than or equal to , you could use set notation as follows: .
1. As a graphed sentence on a number line.
#### Example C
Describe the set of numbers contained by the given set notation for the following:
a) (8, 24)
b) [3, 12)
Solution:
(8, 24) states that the solution is all numbers between 8 and 24 but does not include the numbers 8 and 24.
(3, 12) states that the solution is all numbers between 3 and 12, including 3 but not including 12.
-->
### Guided Practice
Describe and graph the solution set expressed by .
Solution:
The solution set contains all numbers less than 3.25, not including 3.25.
The graph on the number line is:
### Explore More
1. What are the four methods of writing the solution to an inequality?
Graph the solutions to the following inequalities using a number line.
Write the inequality that is represented by each graph.
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 6.1. |
teaching resource
# Adding Fractions With Like Denominators Teaching Slides
• Updated: 06 May 2024
Teach your students how to add fractions with like denominators using a variety of strategies with this comprehensive slide deck.
• Editable: PowerPoint, Google Slides
• Pages: 1 Page
• Years: 5 - 6
Tag #TeachStarter on Instagram for a chance to be featured!
teaching resource
# Adding Fractions With Like Denominators Teaching Slides
• Updated: 06 May 2024
Teach your students how to add fractions with like denominators using a variety of strategies with this comprehensive slide deck.
• Editable: PowerPoint, Google Slides
• Pages: 1 Page
• Years: 5 - 6
Teach your students how to add fractions with like denominators using a variety of strategies with this comprehensive slide deck.
## Teach How to Add Fractions with Like Denominators
Mrs. Sanchez made a loaf of bread. She used one-eighth of the loaf for breakfast and two-eighths of the loaf for lunch.
How much bread did Mrs. Sanchez use?
To work out the answer to this question, your students need to understand that the question involves the addition of fractions with like denominators. Like denominators are fractions that have the same denominator, e.g. one-eighth and two-eighths.There are several strategies that students can use when performing these types of operations with fractions. These include:
1. Fraction Tiles – Fraction tiles are hands-on tools that can be used to visually construct and solve problems with fractions.
2. Fraction Circles – Fraction circles are diagrams that can be drawn by students to visually represent problems with fractions.
3. Number Lines – Number lines are linear representations of a fraction that can be used to solve addition problems.
4. Fraction Notation – With an understanding of fraction notation comes the ability to solve addition problems using the standard algorithm.
To help your students better understand these four strategies for adding fractions with like denominators, the expert teacher team at Teach Starter has created this 25-slide teaching presentation. The presentation explores each strategy in detail using annotated examples of each process. It also includes collaborative review activities that the class can work through together as a means of checking for understanding.
This slide deck downloads as either a Microsoft PowerPoint presentation or a Google Slides format. Simply project on your screen for a paperless lesson on fractions!
## Differentiate This Fractions PowerPoint
If you’re looking for ways to differentiate this activity, here are some suggestions from our experienced teacher team:
• Provide Manipulatives – To help your less-confident students engage with and practise each strategy, provide manipulatives such as fraction strips and mini-whiteboards.
• Be the Teacher – Allow your more-confident students to demonstrate their knowledge by taking the lead during the review activities. Have students take turns coming up to the board and walking the class through the steps required to complete the questions.
Use the Download button to access your preferred version of this resource. (Note: You will be prompted to make a copy of the Google Slides presentation before accessing it.)
The slideshow does contain some simple animations so should be used in Presentation mode.
This resource was created by Brittany Collins, a Teach Starter collaborator.
## Save Time with More Fractions Resources
If you love the look of this fractions resource, then we have good news for you! Click below to browse more teacher-created, curriculum-aligned resources!
[resource:5085151] [resource:6828] [resource:5085007]
### 1 Comment
Write a review to help other teachers and parents like yourself. If you'd like to request a change to this resource, or report an error, select the corresponding tab above. |
?> Distributive Property Worksheets | Problems & Solutions
# Distributive Property Worksheets
Distributive Property Worksheets
• Page 1
1.
Each person in a factory makes 49 shoes per day. If there are 77 people working in the factory, how many shoes are made per day?
a. 3783 b. 3763 c. 3778 d. 3773
#### Solution:
Number of shoes made per day = Number of shoes made by each person × Number of people in the factory
= 49 × 77
[Substitute]
= 49 × (80 - 3)
[Rewrite 77 as (80 - 3).]
= 49 × 80 - 49 × 3
[Use distributive property.]
= 3920 - 147 = 3773
[Multiply and then subtract.]
So, 3773 shoes are made per day.
2.
A box containing corn flakes costs $3. How much would 206 boxes cost? a.$628 b. $623 c.$608 d. $618 #### Solution: Cost of 206 boxes of corn flakes = Cost of 1 box of corn flakes × Number of boxes =$3 × 206
[Substitute.]
3 × 206 = 3 × (200 + 6)
[Break 206 apart. 206 = 200 + 6.]
= (3 × 200) + (3 × 6)
= 600 + 18
[Multiply.]
= 618
So, 206 boxes of corn flakes cost \$618.
3.
There are 29 balls in a basket. How many balls are there in 8 such baskets?
a. 222 b. 232 c. 233 d. 242
#### Solution:
Total number of balls in 8 baskets = Number of balls in a basket × Number of baskets = 29 × 8
[Substitute.]
29 × 8 = 29 × (10 - 2)
[Break 8 apart. 8 = 10 - 2.]
= (29 × 10) - (29 × 2)
= 290 - 58
[Multiply.]
= 232
[Subtract.]
So, there are 232 balls in 8 baskets.
4.
Tim has traveled a distance of 104 miles in a day. How much distance does he travel in 12 days, if he covers the same distance each day?
a. 1258 miles b. 1238 miles c. 1249 miles d. 1248 miles
#### Solution:
Distance traveled in 104 days = Distance traveled in a day × Number of days = 104 × 12
[Substitute.]
104 × 12 = (100 + 4) × 12
[Break 104 apart. 104 = 100 + 4.]
= (100 × 12) + (4 × 12)
= 1200 + 48
[Multiply.]
= 1248
So, Tim traveled a distance of 1248 miles in 12 days.
5.
Would you use the distributive property to solve 3 × 109?
a. Yes b. No
#### Solution:
3 × 109 = 3 × (100 + 9)
[Break 109 apart. 109 = 100 + 9.]
109 is broken apart to benchmark number, 100 to make the multiplication easier.
So, the distributive property would be used to find out the product of 3 × 109.
6.
Find:
106 × 16
a. 1696 b. 2000 c. 1856 d. 11200
#### Solution:
106 × 16 = (100 + 6) × 16
[Break 106 apart. 106 = 100 + 6.]
= (100 × 16) + (6 × 16)
= 1600 + 96
[Multiply.]
= 1696
7.
What is the product of 98 and 2?
a. 197 b. 196 c. 206 d. 195
#### Solution:
98 × 2 = (100 - 2) × 2
[Break 98 apart. 98 = 100 - 2.]
= (100 × 2) - (2 × 2)
= 200 - 4
[Multiply.]
= 196
[Subtract.]
8.
Find the product of 7 and 93.
a. 650 b. 652 c. 661 d. 651
#### Solution:
7 × 93 = 7 × (100 - 7)
[Break 93 apart. 93 = 100 - 7.]
= (7 × 100) - (7 × 7)
= 700 - 49
[Multiply.]
= 651
[Subtract.]
9.
Multiply 12 and 104.
a. 1258 b. 1249 c. 1247 d. 1248
#### Solution:
12 × 104 = 12 × (100 + 4)
[Break 104 apart. 104 = 100 + 4.]
= (12 × 100) + (12 × 4)
= 1200 + 48
[Multiply.]
= 1248
10.
Find the product of 7 and 185.
a. 1300 b. 1285 c. 1305 d. 1295
#### Solution:
7 × 185 = 7 × (200 - 15)
[Break 185 apart. 185 = 200 - 15.]
= (7 × 200) - (7 × 15)
= 1400 - 105
[Multiply.]
= 1295
[Subtract.] |
# Complete Package of Basic Principles and Study Materials for Learning Algebra and Trigonometry In a Row
Mathematics is a very interesting subject. Algebra and Trigonometry are one of the most important areas of Mathematics. It is connected to our daily life in a several way. Everyone should have a sound knowledge of Algebra and Trigonometry. Here we made a complete package of basic principles of algebra and trigonometry and study materials all in one place.
Basic Algebra Principles:
1. a (b + c) = ab + ac
This is known as the Distributive Property of Multiplication. If you're multiplying something with a sum of two or more other terms, you can distribute your multiplication to each of the terms.
Example: 2 (3 + 5) = 2 * 3 + 2 * 5
a (b / c) = ab / c
Any multiplication or division of the numerator of a fraction applies to the fraction as a whole, and vice versa: if you need to multiply a fraction, multiply the numerator and your aim is achieved.
Example: 2 * (3 / 4) = (2 * 3) / 4
(a / c) / b = a / bc
If you divide the numerator by a particular number, it has the same effect on the fraction's overall value as if you multiply the denominator by that same number.
Example: (1 / 5) / 2 = 1 / 10
a / (b / c) = ac / b
Like the previous one, dividing the denominator of a fraction has the same effect as multiplying the numerator.
Example: 1 / (3 / 2) = (1 * 2) / 3 = 2 / 3
(a / b) + (c / d) = (ad + bc) / bd
It shows the fact that, we can find a common denominator between two fractions by multiplying the numerator and denominator of each fraction by the other's denominator.
Example: 3 / 5 + 1 / 3 = (1 * 5) + (3 * 3) / (3 * 5)
(a / b) - (c / d) = (ad - bc) / bd
This is just the another form of rule 5, but for subtraction of two fractions, rather than addition.
Example: 3 / 5 - 1 / 3 = (3 * 3) - (1 * 5) / (3 * 5)
(a - b) / (c - d) = (b - a) / (d - c)
Example: 3 - 5 / 2 - 1 = -2
8. (a + b) / c = a / c + b / c
This is simply a result of the fact that two fractions with common denominators can be added by adding the numerators and leaving the denominator unchanged.
Example: (2 + 2) / 4 = 2 / 4 + 2 / 4 =1
9. ac + bc / c = a + b
Example: (4 * 5) + (2 * 5) / 5 = (4 * 5) / 5 + (2*5) / 5
10. (a / c) / (b / d) = ad / bc
This is a combination of rule 3 and 4.
Example: (4 / 5) / (2 / 1) = (4 * 1) / (2 * 5)
Basic Trigonometry Principles:
1. Right-Angled Triangles
• The right-angle is indicated by the little box in the corner.
• The other angle is indicated by the 'Theta'.
• The opposite side of the right angle, which is the largest, is known as Hypotenuse (H).
• The opposite side of 'Theta', is called as Opposite (O).
• The side next to 'Theta', is called as Adjacent (A).
2. There are three basic functions in trigonometry, each of which is one side of a right-angled triangle divided by another. These three functions are:
Calculating Sine, Cosine and Tangent:
3. Graphs of Sine, Cosine and Tangent
4. Trigonometry in a Circle
Consider, a circle divided into four quadrants. Conventionally, the centre of the circle is considered as the Cartesian Coordinate of (0,0). That is, the x value is 0 and the y value is 0. Anything to the left of the centre has an x value of less than 0, or is negative, while anything to the right has a positive value. Similarly anything below the centre point has a y value of less than 0, or is negative and any point in the top of the circle has a positive y value.
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## Section48.8Optical Instruments Bootcamp
### Subsection48.8.5Miscellaneous
What is the angular size of the Moon if viewed from a binocular that has a focal length of $1.2\text{ cm}$ for the eyepiece and a focal length of $8\text{ cm}$ for the objective? Use the radius of the Moon to be $1.74 \times 10^6\text{ m}$ and the distance of the Moon from the observer to be $3.8 \times 10^8\text{ m}\text{.}$
Hint
Use the angle subtended by the moon itself and magnification of the binucular.
$6.1\times 10^{-2}\text{ rad} \text{.}$
Solution
We use $s=R\theta$ with $s=\text{diameter of moon}$ and $R=\text{distance to the Moon}$ to find the angle subtended by the Moon directly to be
\begin{equation*} \theta = \frac{s}{R} = \frac{2\times 1.74 \times 10^6\text{ m}}{3.8 \times 10^8\text{ m}} = 9.158\times 10^{-3}\text{ rad}. \end{equation*}
Multiplying by the magnification we will get the desired angle. So, we need the value for magnification. From the focal lengths of the objective and the eyepiece, we get the magnification of the binocular to be
\begin{equation*} M = -\frac{f_\text{o}}{f_\text{e}} = -\frac{8\text{ cm}}{1.2\text{ cm}} = -6.67. \end{equation*}
Therefore, the angle subtended by the image will me $|M|$ times
\begin{equation*} \theta_\text{image} = 6.67\times 9.158\times 10^{-3}\text{ rad} = 6.1\times 10^{-2}\text{ rad}. \end{equation*}
In a reflecting telescope the objective is a concave mirror of radius of curvature equal to $50\text{ cm}$ and an eyepiece is a convex lens of focal length $5\text{cm}\text{.}$ Find the apparent size of a $25\text{-m}$ tree at a distance of $10\text{ km}$ that you would perceive when looking through the telescope.
Hint
The apparent height will be multiple of the magnification.
$125\text{ m}$
We can think of $10\text{ km}$ as infinitely far away with virtual image forming that far as well, as is the case in telescopes. This happens in telescopes since the distance between objective and eyepiece is very close to the sum of their foci. Therefore, the height of the tree will be magnified $5$ fold. That is the apparent height will be $125\text{ m}\text{.}$ |
Get Started by Finding a Local Center
# August/September 2014 Newsflash: Tips & Techniques
Aug 14, 2014
"Thinking backwards"-starting at the end of the problem and using reverse operations-can help you solve certain problems at different grade levels.
Upper Elementary and Middle School:
A certain number is doubled. That answer is tripled. Finally, that answer is quadrupled and the answer is 60. What is the original number?
When we quadruple a number, we multiply it by 4. As division is the inverse of multiplication, 60 ÷ 4 = 15. Tripling a number means "multiply by 3," so 15 ÷ 3 = 5. Finally, doubling a number means "multiply by 2," thus, 5 ÷ 2 = 2.5, or 2 ½.
Algebra:
a) Write an equation that describes this problem.
b) What is the original number?
a) We're looking for a certain (unidentified) number, x. "x, quadrupled" means "4x." Then, we add 3 to the answer, so 4x + 3. We then triplethe quantity "4x + 3": 3(4x + 3). Finally, we split the quantity 3(4x + 3) in half in order to yield the final answer, 12, so {3(4x+ 3)} /2 = 12.
b) To find the original number, we solve for x, which involves "canceling out" the numbers by using inverse operations.
So, (2){3 (4x + 3)} /2 = 12 (2) multiply both sides by 2...
{3(4x + 3)}/3 = 24/3divide both sides by 3...
4x + 3- 3 = 8 - 3 subtract 3 from both sides, and...
4x /4= 5 /4 divide both sides by 4.
Thus, x = 1.25
## OUR METHOD WORKS
Mathnasium meets your child where they are and helps them with the customized program they need, for any level of mathematics. |
# What are Venn diagrams?
A Venn diagram, named after John Venn in the 19th century, provide a convenient way to represent a sample space. Click here to remind yourself of what a sample space is.
A Venn diagram is a rectangle representing the whole space and circles inside representing various subspaces.
For example, A could represent the event that a person had blue eyes. Event B could be that person have brown hair.
## Events on a Venn diagram
### Complement
This Venn diagram shows the complement of A. The part shaded with diagonal lines is everything that is not in A and we call this the complement. Note the spelling – it doesn’t say ‘compliment’ as charming as event A might be. See Example 1. The complement of A is written as:
or
### Union
In this Venn diagram we see the union of events A and B. This means that the shaded part represents all outcomes where either event A or event B has occurred. See Example 2. The union of A and B is written as:
### Intersection
Alternatively to the union, there is also the intersection of two events. The intersection represents the set of all outcomes where BOTH events A and B have occurred. See Example 3. The intersection of A and B is written as:
### Combinations
It follows from the three previous definitions that it is possible to combine complement, union or intersection to get particular areas of a Venn diagram. See Example 4. The above combination is written as:
## Examples
Let event A be an even result when a dice is rolled. Additionally, let event B be the even that the result is greater than 2. The probability of not A is getting an odd result.
Odd results are 1,3 or 5 (shaded part) and so in this case .
Again, let events A and B be an even result and a result greater than 2 respectively when a dice is rolled. The probability of A or B occuring includes all of the following results: 2,3,4,5 or 6.
It follows that.
If A is an even result when a dice is rolled and B is a result is greater than 2, the only times you would get both are if you roll a 4 or a 6.
It follows that the probability of the intersection is .
Consider again that A is an even result and B is a result greater than 2 when a dice is rolled. Results that require A not happening, i.e. an odd result, as well as B occurring includes only 3 and 5.
Finally, in this case .
A card is chosen at random from a standard deck of 52 playing cards. Let K be the event that the card is a King. Let H be the event that the card is a heart. Find:
Solution:
This question is simple enough to answer without a Venn diagram but it can be less confusing when visualising some of the combined events.
1. is the intersection of K and H. This represents the event that the card is both a King and a Heart. Since the only card that is both is the King of Hearts, the probability is given by .
2. is the event that the card is not a Heart. This means that the card can be a spade, a diamond or a club. There are 39 of these and so .
3. is the event that the card is either a King or a Heart. There are 4 Kings and 13 Hearts but the King of Hearts is included in both of these. We can see from the Venn diagram that .
4. represents the event that neither a King nor a Heart can be drawn. The easiest way to deduce this probability is to see where overlaps with . This can only be from the 36 cards that are not in either and so
A school has 204 students. 85 of them have chosen to study Maths, 56 of them have chosen to study French and 68 have chosen to study History. It turns that 35 study both French and Maths, 23 study both Maths and History and 27 study both History and French. 7 study all three subjects.
1. Draw a Venn diagram to represent this information.
2. Find the probability that a student does not study Maths.
3. Find the probability that a student does none of these three subjects.
Solution:
1. There are 109 students who do not study Maths and so the probability is 109/204.
2. There are 73 students that do not take any of three subjects mentioned and so the probability is 73/204.
The hardest part of this question is getting the information into the original Venn diagram. The easiest thing to do is identify the inner most intersection, i.e. the number of students who take all three subjects. From there, it is possible to identify the parts connected to the central piece followed by the remainder of the circles. You should then be able to identify final number of students who don’t feature in any of the circles to make the number up to 200. |
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# Python Program to Find The Sum of First N Natural Numbers
Last Updated on May 31, 2023 by Prepbytes
In the real of mathematics, the sum of the first N natural numbers is a classic problem that has intrigued and challenged mathematicians for centuries. The concept of finding the sum of consecutive integers seems simple at first glance, but as N grows larger, the task becomes more complex. Thankfully, with the power and simplicity of Python, we can develop a program that efficiently calculates this sum, saving us precious time and effort.
In this article, we will explore the fascinating world of Python programming and unravel the mystery behind finding the sum of the first N natural numbers. Whether you’re a beginner taking your first steps into the world of programming or an experienced Python developer seeking to enhance your skills, this article will provide you with a comprehensive guide to solving this fundamental mathematical problem.
## Approaches To Find The Sum of First N Natural Numbers
Here, we will discuss four different approaches to writing a Python Program to find the sum of first n natural numbers.
### Approach 1: Python Program to find the sum of first n natural numbers Using Formula
Two natural numbers that follow one another have a difference of 1. The progression of natural numbers is an arithmetic progression since every pair of subsequent numbers has the same difference. An arithmetic progression goes as follows: a, a+d, a+2d, a+3d,…, where a is the first element of the number and d is the difference between two succeeding numbers in the progression.
The following formula can be used to determine the sum of n numbers in an arithmetic progression,
Sn=(n∗(2a+(n−1)∗d))/2
where,
• The series’ first term is a.
• The common difference between any two terms in a series is called d.
• The series number of terms is n.
The preceding formula can be further condensed into the equation for the sum of natural numbers, (n(n+1))/2, by substituting a=1 and d=1 in the equation.
The number of elements in the sequence is represented by the symbol n in the equation.
It is possible to use the Python Programming language to implement the formula given in the previous section.
Example
The formula for calculating the sum of natural numbers can be used to generate the following code.
### Code Implementation
``````natural_number = 10
# formula for the sum of natural numbers
sum = natural_number*(natural_number+1)//2
print("The Total Sum Of",natural_number, "Natural Numbers Is:", sum)``````
Output
``The Total Sum Of 10 Natural Numbers Is: 55``
Note: Due to the precedence of the operator in C, brackets are used in the formula surrounding (natural_number+1). Multiplication and division operations will be carried out before addition in C because they have a higher priority than addition and subtraction operations. We employ a bracket that takes precedence over other operators to stop this from happening.
Explanation
The code variable natural_number serves as an example of the sequence’s element count. The user provides this variable as input. The sum is obtained by changing the formula for the sum of natural numbers to reflect the number of elements in the sequence. The user is shown the output.
The program’s time complexity is O(1), or continuous time complexity.
The program’s space complexity is also O(1).
### Approach 2: Python Program to find the sum of first n natural numbers Using For Loop
The sum of natural numbers can be calculated by iterating through the numbers in the natural number sequence and adding them together using a for loop.
Example
There are two methods to use the for loop. The for loop in the program below iterates from 1 to a natural integer,
``````natural_number = 14
sum=0
for i in range(1,natural_number+1):
sum = sum + i
print("The Total Sum Of",natural_number, "Natural Numbers Is:", sum)``````
The program below displays a for loop that iterates from the natural number to 1 (the first natural number in the sequence),
``````natural_number = 14
sum=0
for i in range(natural_number, -1, -1):
sum = sum + i
print("The Total Sum Of",natural_number, "Natural Numbers Is:", sum)``````
Output
``The total sum of 14 natural numbers is 105``
Explanation
The user provides the highest possible number in the natural number series. Iterating from 1 to the greatest natural number or from the greatest natural number to 1 are both possible with a for loop. The total sum of natural numbers is obtained by adding the sum at each iteration.
Both programs will have an O(n) time complexity, where n is the number of iterations the for loop performs. Both programs’ space complexity will be O(1).
### Approach 3: Python Program to find the sum of first n natural numbers Using While Loop
A while loop only checks for conditions from the second iteration of the loop’s run after running once without checking for the conditions it was given.
Example
``````natural_number = 14
sum=0
for i in range(natural_number, -1, -1):
sum = sum + i
print("The Total Sum Of",natural_number, "Natural Numbers Is:", sum)``````
Output
``The total sum of 13 natural numbers is 91``
Explanation
The while loop multiplies the input number by each number starting at one. The user sees the outcome displayed.
The program has an O(n) time complexity, where n is the total number of iterations in the while loop. The program’s O(1) space complexity.
### Approach 4: Python Program to find the sum of first n natural numbers Using Recursion
A technique known as recursion involves repeatedly calling the same function. Finding the sum of natural numbers can be accomplished using recursion rather than loops.
Example
``````def Sum(natural_no):
if(natural_no != 0):
return natural_no + Sum(natural_no - 1)
return 0
natural_number = 12
print("The Total Sum Of",natural_number, "Natural Numbers Is:", Sum(12))``````
Output
``The total sum of 12 natural numbers is 78``
Explanation
The input consists of the natural number. In a recursive function call, the same function is invoked with the parameter set to the value obtained by subtracting the number supplied as an argument in the previous function call. Only if the number decreases to zero will the recursive call terminate. At the conclusion of each function call, the output that the function returned is summed to obtain the sum of natural numbers.
The recursive natural number addition has an O(n) time complexity. A recursive function’s space complexity is influenced by how many function calls are made. The space complexity in the aforementioned example is O(n).
The functions will be called in the following order: Sum(12) –> Sum(11) –> Sum(10)…–> Sum(1).
The function returns zero once Sum(0) is achieved, and the output of each function is then added together backward to determine the sum,
Sum(0) –> Sum(1) –> Sum(2) …–> Sum(12), etc.
The return function and the accumulation are both executed in the same phase. When the function is Sum(0), the result is 0, and when the function is Sum(1), the result is 0+1, where 0 is the result of the Sum(0) function. To obtain the sum of n natural numbers, analogous accumulations are carried out for each succeeding function.
Summary
• Natural numbers are also known as counting numbers since they begin at one.
• The natural number series follows an arithmetic progression, hence the sum of natural numbers can be calculated using the sum of natural numbers formula, which is derived from arithmetic progression.
• Iterative methods utilizing for, while, or do..while loops can also be used to calculate the sum of natural numbers.
• A recurrence function can also be used to determine the sum of natural numbers.
## FAQ Related to the Python Program to find the sum of first n natural numbers
Q1.Can we find the sum of first n natural numbers in O(1) time complexity?
Ans. Yes, there is one mathematical formula Sn=(n∗(2a+(n−1)∗d))/2, by using it we can calculate the sum of first n natural numbers. We had discussed this approach in this article.
Q2.Can the formula for finding the sum of the first n natural numbers be used for negative values of n?
Ans. No, the formula is valid only for positive values of n. It does not work for negative values of n.
Q3.What are the first n natural numbers?
Ans. The first n natural numbers are the positive integers starting from 1 and going up to n. |
## Properties of Limits
Read this section to learn about the properties of limits. Work through practice problems 1-6.
### Introduction
This section presents results which make it easier to calculate limits of combinations of functions or to show that a limit does not exist. The main result says we can determine the limit of "elementary combinations" of functions by calculating the limit of each function separately and recombining these results for our final answer.
##### Main Limit Theorem:
If $\quad \lim\limits_{x \rightarrow a} \mathrm{f}(\mathrm{x})=\mathrm{L}$ and $\lim\limits_{x \rightarrow a} \mathrm{~g}(\mathrm{x})=\mathrm{M}$,
then $\text { (a) } \lim\limits_{x \rightarrow a}\{\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x})\}=\lim\limits_{x \rightarrow a} \mathrm{f}(\mathrm{x})+\lim\limits_{x \rightarrow a} \mathrm{~g}(\mathrm{x})=\mathrm{L}+\mathrm{M}$
(b) $\lim\limits_{x \rightarrow a}\{\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x})\}=\lim\limits_{x \rightarrow a} \mathrm{f}(\mathrm{x})-\lim\limits_{x \rightarrow a} \mathrm{~g}(\mathrm{x})=\mathrm{L}-\mathrm{M}$
(c) $\lim\limits_{x \rightarrow a} \mathrm{k} \, \mathrm{f(x)} \quad=\mathrm{k} \lim\limits_{x \rightarrow a} \mathrm{f}(\mathrm{x}) \quad=\mathrm{kL}$
(d) $\lim\limits_{x \rightarrow a} \mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x})=\left\{\lim\limits_{x \rightarrow a} \mathrm{f}(\mathrm{x})\right\} \cdot\left\{\lim\limits_{x \rightarrow a} \mathrm{~g}(\mathrm{x})\right\}=\mathrm{L} \cdot \mathrm{M}$
(e) $\lim\limits_{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim\limits_{x \rightarrow a} f(x)}{\lim\limits_{x \rightarrow a} g(x)}$ $\qquad$ $\qquad$ $\qquad$ $=\frac{\mathrm{L}}{\mathrm{M}} \quad($ if $\mathrm{M} \neq 0)$
(f) $\lim\limits_{x \rightarrow a}\{\mathrm{f}(\mathrm{x})\}^{\mathrm{n}}\qquad =\left\{\lim\limits_{x \rightarrow a} \mathrm{f}(\mathrm{x})\right\}^{\mathrm{n}} \qquad \qquad =\mathrm{L}^{\mathrm{n}}$
(g) $\lim\limits_{x \rightarrow a} \sqrt[n]{f(x)}=\sqrt[n]{\lim\limits_{x \rightarrow a} f(x)}$ $\qquad$ $\qquad$ $=\sqrt[n]{L} \quad$ (if $L>0$ when $n$ is even)
The Main Limit Theorem says we get the same result if we first perform the algebra and then take the limit or if we take the limits first and then perform the algebra: e.g., (a) the limit of the sum equals the sum of the limits. A proof of the Main Limit Theorem is not inherently difficult, but it requires a more precise definition of the limit concept than we have given, and it then involves a number of technical difficulties.
Practice 1: For $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}-\mathrm{x}-6$ and $\mathrm{g}(\mathrm{x})=\mathrm{x}^{2}-2 \mathrm{x}-3$, evaluate the following limits:
(a) $\lim\limits_{x \rightarrow 1}\{\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x})\}$ (b) $\lim\limits_{x \rightarrow 1} \mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x})$ (c) $\lim\limits_{x \rightarrow 1} \mathrm{f}(\mathrm{x}) / \mathrm{g}(\mathrm{x})$ (d) $\lim\limits_{x \rightarrow 3}\{\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x})\}$
(e) $\lim\limits_{x \rightarrow 3} \mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x})$ (f) $\lim\limits_{x \rightarrow 3} \mathrm{f}(\mathrm{x}) / \mathrm{g}(\mathrm{x})$ (g) $\lim\limits_{x \rightarrow 2}\{\mathrm{f}(\mathrm{x})\}^{3}$ (h) $\lim\limits_{x \rightarrow 2} \sqrt{1-g(x)}$ |
## What is a normal unit vector?
A unit normal vector to a two-dimensional curve is a vector with magnitude 1 that is perpendicular to the curve at some point. Typically you look for a function that gives you all possible unit normal vectors of a given curve, not just one vector.
## What is the normal vector of a line?
In geometry, a normal is an object such as a line, ray, or vector that is perpendicular to a given object. For example, in two dimensions, the normal line to a curve at a given point is the line perpendicular to the tangent line to the curve at the point.
## What is the vector equation of a plane?
Answer: When you know the normal vector of a plane and a point passing through the plane, the equation of the plane is established as a (x – x1) + b (y– y1) + c (z –z1) = 0.
## What is null vector in physics?
Zero Vector or null vector is a vector which has zero magnitude and an arbitrary direction. It is represented by 0 . If a vector is multiplied by zero, the result is a zero vector. The velocity vector of a stationary body is a zero vector.
## Is the gradient vector normal?
Again, the gradient vector at (x,y,z) is normal to level surface through (x,y,z). For a function z=f(x,y), the partial derivative with respect to x gives the rate of change of f in the x direction and the partial derivative with respect to y gives the rate of change of f in the y direction.
## How do you write a position vector?
To find the position vector, subtract the initial point vector P from the terminal point vector Q . Multiply j j by 1 1 . Apply the distributive property. Multiply −3 – 3 by −1 – 1 .
## What is a direction vector?
The direction of a vector is the direction along which it acts. It has a certain magnitude. For example, we say 10 N force in the east. Here, 10 N is the magnitude and towards the east is the direction. The direction is specified using a unit vector.
## What is a vector formula?
The magnitude of a vector →PQ is the distance between the initial point P and the end point Q . In symbols the magnitude of →PQ is written as | →PQ | . If the coordinates of the initial point and the end point of a vector is given, the Distance Formula can be used to find its magnitude. | →PQ |=√(x2−x1)2+(y2−y1)2.
## What is the vector equation of XY plane?
A plane parallel to the x-y-plane must have a standard equation z = d for some d, since it has normal vector k. A plane parallel to the y-z-plane has equation x = d, and one parallel to the x-z-plane has equation y = d.
## What is a XY plane?
The xy-plane is the plane that contains the x- and y-axes; the yz-plane contains the y- and z-axes; the xz-plane contains the x- and z-axes. These three coordinate planes divide space into eight parts, called octants.
## What is null vector example?
1. two people pulling a rope in opposite directions with equal force. 2. displacement of throwing an object upward and then again holding it at the same position.
## What is vector and types?
A vector containing foreign DNA is termed recombinant DNA. The four major types of vectors are plasmids, viral vectors, cosmids, and artificial chromosomes. Of these, the most commonly used vectors are plasmids.
### Releated
#### Convert to an exponential equation
How do you convert a logarithmic equation to exponential form? How To: Given an equation in logarithmic form logb(x)=y l o g b ( x ) = y , convert it to exponential form. Examine the equation y=logbx y = l o g b x and identify b, y, and x. Rewrite logbx=y l o […]
#### H2o2 decomposition equation
What does h2o2 decompose into? Hydrogen peroxide can easily break down, or decompose, into water and oxygen by breaking up into two very reactive parts – either 2OHs or an H and HO2: If there are no other molecules to react with, the parts will form water and oxygen gas as these are more stable […] |
Balance Subtraction Equations without Regrouping
Learn to Balance Subtraction Equations
An equation is like a scale.
Just like scales, equations are called balanced when both sides of the equals sign have the same value.
Is this equation balanced?
1 + 1 = 2
Yes! Both sides are equal to 2.
Is this equation balanced?
1 = 3
No, that is NOT a balanced equation.
👉Equations can have addition or subtraction on both sides of the equal sign, as long as both sides are equal!
Look at this equation:
5 - 3 = 6 - 4
Can you see if it's balanced? 🤓
🤔 First, check if the value of the left side (👈) equals the right side (👉).
Subtract 5 - 3 and 6 - 4 separately.
5 - 3 = 2
6 - 4 = 2
Yes, this equation is balanced! Both sides equal to 2. 😀
What happens when you don't have all of the numbers?
Balancing Equations
What if we have this problem?
18 - 5 = __ - 3
How do we find the missing number?
Since we know both sides of an equation are equal, we can balance it!
Here's how.👇
First, let's simplify the equation.
Solve the side that isn't missing a number.
Nice! That makes things easier 👌
Now, we know the left side is 13!
How can we figure out the missing number?
Let's think. 🤔 What are we trying to find?
We're trying to find the total.
The total is the bigger number that we can take away 3 from to get 13.
Let's write this equation as a Math mountain
🤓 Math mountains help us think about how two parts make up a whole.
Here's what we know:
How do we find the total, when we know the parts?
😺 We add the parts together!
So what's 13 + 3?
It's 16!
Now, we know the missing minuend is 16.
Let's go back to the equation we need to balance:
18 - 5 = 16 - 3
Is it balanced?
16 - 3 = 13
Yes! Both sides equal 13!
Great work!
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## Precalculus (6th Edition) Blitzer
The solution to the system of equations is $x=-\frac{4}{7}-\frac{4}{7}t,y=\frac{5}{7}+\frac{5}{7}t,z=t$.
Consider the following system of equations: \begin{align} & x-2y+2z=-2 \\ & 2x+3y-z=1 \end{align} Express the above system in the form of a matrix as below: $\left[ \begin{matrix} 1 & -2 & 2 & -2 \\ 2 & 3 & -1 & 1 \\ \end{matrix} \right]$ Using the elementary row operations we will obtain the echelon form of a matrix: ${{R}_{2}}\to {{R}_{2}}+\left( -2 \right){{R}_{1}}$ gives, $\left[ \begin{matrix} 1 & -2 & 2 & -2 \\ 0 & 7 & -5 & 5 \\ \end{matrix} \right]$ ${{R}_{2}}\to \frac{1}{7}{{R}_{2}}$ gives, $\left[ \begin{matrix} 1 & -2 & 2 & -2 \\ 0 & 1 & -\frac{5}{7} & \frac{5}{7} \\ \end{matrix} \right]$ Rewrite the system corresponding to echelon form of the matrix: $x-2y+2z=-2$ (I) $y-\frac{5}{7}z=\frac{5}{7}$ (II) Since, the system has two equations and three variables, one variable can be chosen arbitrarily. Let $z=t$, then equation (II) gives, $y-\frac{5}{7}\left( t \right)=\frac{5}{7}$ Simplify, $y=\frac{5}{7}+\frac{5}{7}t$ Substitute the values of y and z in equation (I) to obtain: $x-2\left( \frac{5}{7}+\frac{5}{7}t \right)+2\left( t \right)=-2$ It can be further simplified as: $x=-\frac{4}{7}-\frac{4}{7}t$ Hence, the solution to the system is $x=-\frac{4}{7}-\frac{4}{7}t,y=\frac{5}{7}+\frac{5}{7}t,z=t$. |
I'm 73 and vaguely remember it as semi perimeter theorem.
How do you find the unknown side of a triangle? The Pythagorean theorem states that a2 + b2 = c2 in a right triangle where c is the longest side. You can use this equation to figure out the length of one side if you have the lengths of the other two. The figure shows two right triangles that are each missing one sides measure. Our right triangle side and angle calculator displays missing sides and angles! Setting b and c equal to each other, you have this equation: Cross multiply: Divide by sin 68 degrees to isolate the variable and solve: State all the parts of the triangle as your final answer. To determine a canopy dimension. So our new formula for right triangle area is A = ab/2. (1) The length of the third side is given by the Perimeter is always the same linear measurement unit as the unit used for the sides. A right triangle is a triangle that has 90 degrees as one of its angles. Now that you know the area of the triangle pictured above, you can plug it into triangle formula A=1/2bh to find the height of the triangle. If you use the pythagorean, one of your side lengths would be 0, so you would have: (0)^2 + (2 - (-4))^2 = c^2. Calculate the side of a triangle if given side and any two angles ( Sine Rule ) ( a ) : side of a triangle : = Digit 1 2 4 6 10 F. =. Specifying two sides and the angle between them uniquely (up to geometric congruence) determines a triangle. Step 3 Put our values into the Cosine equation: cos 60 = Find the distance between the orthocentre and the circumcenter of a triangle. Solve for the missing side. Perimeter = Sum of all sides = a + b Let be the base length and be the height. Finding the perimeter and area of a triangle. Here are some of the solved examples to understand this type of triangles better: Question 1: Find the area of the scalene triangle ABC with the sides 8 cm, 6 cm and 4 cm. Fintainment. This format always holds true. deg. Solution: To find: The sides of a right triangular park. Step 1 The two sides we are using are A djacent (h) and H ypotenuse (1000). Explains the Distance Formula, how the Distance Formula is derived from the Pythagorean Theorem, and how to use the Formula. Example 1: What are the sides of the right triangular park whose hypotenuse is 10 in and has a base angle of 30? For example, students could divide the fraction 12/18 by 2/2, and then divide by 3/3 to show the fraction in More of the same Rational-equations simplest radical form just means simplifying a radical so that there are no more square roots, cube roots, 4th roots, etc left to find Pythagorean Theorem calculator to find out the unknown length of a right triangle Pythagorean Sight distance triangles for approaching traffic on both sides of any accessed roadway must be determined and provided on your design plans. Now we know that: a = 6.222 in What is a Perimeter? Careful! Algebraic Solution: Extend the lines making up the side length \$3\$ of the inner triangle until it hits the walls of the larger one, and consider the triangle formed. If you mean how do you calculate the length of one side of a triangle, given the lengths of the other two sides, and the magnitude of the angle between them, the easiest way is to use the cosine rule: a = b+c-2bccos (A), Calculating the length of another side of a triangle If you know the length of the hypotenuse and one of the other sides, you can use Pythagoras theorem to find the length of the third side. so 48 2 + 55 2 must be equal to 73 2. So, Furthermore, in the next step put the value of side (s) in the formula that you find in the first step. Repeat Steps 3 and 4 to solve for the other missing side. Step 1: The missing length of a side of an equilateral triangle can be found by identifying the length of either of the other two sides of the triangle. Then angle = 180 .. Step 1: We first identify whether each of the two known side lengths is a leg or the hypotenuse. Add up the lengths of the three side lengths to find the perimeter. Triangle. Sides "a" and "b" are the perpendicular sides and side "c" is the hypothenuse. np! A triangle is defined by its three sides, three vertices, and three angles. The side DE is given as 5 Lengths of triangle sides given one side and two angles Calculation precision Step 3: Then the area is. A = 1 2 bh A = 1 2 b h. In contrast to the Pythagorean Theorem method, if you have two of the three parts, you can find the height for any triangle! The art of brazilian jiu jitsu is an extremely complex combat sport that involves a series of techniques that are guided by a number of different concepts. Determining the corner angle of countertops that are out of square for fabrication. The values being used for the formula are supposed to be "height = 3" and "sideA = 3.16." Fill in 3 of the 6 fields, with at least one side, and press the 'Calculate' button. Perimeter is the distance around the sides of a polygon or other shape. The 60 angle is at the top, so the "h" side is Adjacent to the angle! It is the second-largest city in the European Union (EU), and its monocentric metropolitan area is the second-largest in the EU. This tool is designed to find the sides, angles, area and perimeter of any right triangle if you input any 3 fields (any 3 combination between sides and angles) of the 5 sides and angles available in the form. How to find the Perimeter of a Right Triangle. Method 1: When the perimeter is given The perimeter of a 2A=h (b1+b2). 6^2 = c^2. I guess it is because a triangle is a fundamental shape in geometry. Define two points in the X-Y plane. Solved Examples. Solution: To find: The sides of a right triangular park. Given a right angle triangle, the method for finding an unknown side length, can be summarized in three steps : Step 1: Label the side lengths, relative to the given interior (acute) angle, using "A", "O" and "H" (label both the given side length as well as the one you're trying to find). Finding the distance between the access hole and different points on the wall of a steel vessel. For finding the diagonal of a square using this method we simply need to follow some simple rule. Their angles are also typically referred to using the capitalized letter corresponding to the side length: angle A for side a, angle B for side b, and angle C (for a right triangle this will be 90) for side c, as shown below. To find the area of a triangle, you need to know the length of one side the base ( b for short) and the height (h).
This calculator calculates for the length of one side of a right triangle given the length of the other two sides. Using sides of a triangle Formula, sin = Length of opposite side / Length of Hypotenuse side c c. in the figure). In elementary geometry, the property of being perpendicular (perpendicularity) is the relationship between two lines which meet at a right angle (90 degrees). For example, an area of a right triangle is equal to 28 in and b = 9 in. The municipality covers 604.3
So the distance would be 6 units. It's easy to find the lengths of the horizontal and vertical sides of the right triangle: c 2 = a 2 + b 2so: Affiliate. You can just find how much the y-value increases or decreases from one point to the next, and that's your distance. SAS Theorem . A line is said to be perpendicular to another line if the two lines intersect at a right angle. These are the four steps we need to follow:Find which two sides we know out of Opposite, Adjacent and Hypotenuse.Use SOHCAHTOA to decide which one of Sine, Cosine or Tangent to use in this question.For Sine calculate Opposite/Hypotenuse, for Cosine calculate Adjacent/Hypotenuse or for Tangent calculate Opposite/Adjacent.More items The sum of a triangles three interior angles is always 180. Once the fight hits the mat practitioners will use a strategic method of Madrid (/ m d r d / m-DRID, Spanish: [mai ]) is the capital and most populous city of Spain.The city has almost 3.4 million inhabitants and a metropolitan area population of approximately 6.7 million. Setting b and c equal to each other, you have this equation: Cross multiply: Divide by sin 68 degrees to isolate the variable and solve: State all the parts of the triangle as your final answer. The sum of the lengths of a triangles two sides is always greater than the length of the third side. Firstly, find the length of one side by measuring it. You divide by sin 68 degrees, so. An isosceles triangle is a triangle with two sides of equal length and two equal internal angles adjacent to each equal sides. Divide both sides of the equation by the sum of the bases to get 2A/ (b1+b2)=h. See Herons theorem in action. The Pythagorean Theorem can easily be used to calculate the straight-line distance between two points in the X-Y plane. Extra information is required before a definitive single answer can be given. Plug in the values of the trapezoid into the equation for height. The base angle = 30 degrees. Repeat Steps 3 and 4 to solve for the other missing side. Enter the length of any two sides and leave the side to be calculated blank. The relation between the sides and angles of a right triangle is the basis for trigonometry. Example 1: What are the sides of the right triangular park whose hypotenuse is 10 in and has a base angle of 30? Let three side lengths a, b, c be specified. [7] 1. The sides of a right triangle are commonly referred to with the variables a, b, and c, where c is the hypotenuse and a and b are the lengths of the shorter sides. The length of one side of a triangle can be evaluated from the perimeter and area values of the triangle but the triangle must be equilateral. Understanding the nature of the problem, there will be two solutions: C and C' Identify distance as the hypotenuse of a right triangle. c = 6. If we only know two of the sides we need to use the Pythagorean Theorem first to find the third side. The sides adjacent to It is unlike the equilateral triangle because there we can use any vertex to find out the altitude of the triangle. ABC denotes a triangle with the vertices A, B, and C. Part of. 6. It's the third one. Missing Sides and Angles of a Right Triangle Example 2: Find AC in the following triangle. Further through the equation, it is a = square root of b2 +c2. The formula for the area of a triangle is 1 2 base height 1 2 b a s e h e i g h t, or 1 2 bh 1 2 b h. If you know the area and the length of a base, then, you can calculate the height. The distance formula mentioned above can be used twice (AC and BC) and these equations can be subtracted. To find the angles , , the law of cosines can be used: = + = +. Lets say the two Solution: Sides: 40 cm, 58 cm and 42 cm. Question 2: Three sides of a triangle are 40 cm, 58 cm and 42 cm. Note: angle A is opposite side a, B is opposite b, and C is opposite c. 3. Mathematicians have no special formula for finding the perimeter of a triangle they just add up the lengths of the sides. Let us look at both the cases one by one. The base angle = 30 degrees. A triangle is determined by 3 of the 6 free values, with at least one side. Discover lengths of triangle sides using the Pythagorean Theorem. Pythagoras. a2 = b2 + c2 a = square root of 42 + 62 a = square root of 16 + 36 a = square root of 52 Sum of squares of two small sides should be equal to the square of the longest side. After that, put the value in the formula d = s2. How do you find the b2 in the Pythagorean Theorem?find the square value of side c.find the square value of side a.Subtract c^2 from a^2.Find the root square value of the difference is the value of b. What is wrong with solution of 'Find that the distance between the circumcenter and the orthocenter of triangle ABC' 0 Calculate the Pythagoras' theorem is a formula you can use to calculate the length of any of the sides on a right-angled triangle or the distance between two points. It is actually simple, you just need to use law of sines, which looks like this: That's it. The side DE is given as 5 The side opposite the right angle is called the hypotenuse (side. In our second example, P = 6 + 8 + 10, or 24 . This calculator is for those who wanted to determine lengths of triangle sides given one side and two angles. The side lengths of this triangle are in the ratio of; Side 1: Side 2: Hypotenuse = n: n: n2 = 1:1: 2. Select the proper option from a drop-down list. Using Area To Find the Height of a Triangle. I'm attempting to find the angles/sides of a triangle based on points that are scanned in. A right-angled triangle follows the Pythagorean theorem so we need to check it . If you know one side of the right triangle is 4 and a second side is 6, the equation would be as follows. Maths. Sight distance triangles are based on the time needed for a driver to see traffic on an uncontrolled street, to then decide to turn, make the turning movement, and begin accelerating. Solve the Hypotenuse with Two Sides: Generally, the Pythagorean Theorem is used to calculate the hypotenuse from two different sides of the right-angled triangle. In our first example, P = 3 + 4 + 5, or 12. c = 2 cm. The overall art is a highly intense grappling sport that utilises aspects from Judo and Wrestling, to secure a takedown from the standing position. b = 4 cm. Solution: Let, a = 5 cm. Step 1: The missing length of a side of an equilateral triangle can be found by identifying the length of either of the other two sides of the triangle. The distance around the outside of a triangle is its perimeter. The 45-45-90 right triangle is half of a square. This equation gives the representation of h in terms of the other traits of the trapezoid. The expression (x 2 - x 1) is read as the change in x and (y 2 - y 1) is the change in y.. How To Use The
a sin (A) = b sin (B) = c sin (C) When there is an angle opposite a side, this equation comes to the rescue. All you need to know are the x and y coordinates of any two points. 2304 + 3025 = 5329 which is equal to 73 2 = 5329. Scalene triangle. Answer (1 of 7): How do you calculate the distance of the side in the triangle while given 2 sides and the angle that is not between those 2 given sides? The equation for this is a2 = b2 + c2. In the equilateral triangle below, W U T has sides W U, U T, and T W.The little tick marks on the sides indicate that all three sides are the same, so the measurement for W U, 27 m e t e r s, is also Using sides of a triangle Formula, sin = Length of opposite side / Length of Hypotenuse side The property extends to other related geometric objects. Step 2 SOH CAH TOA tells us to use C osine. So, sides are calculated, then angles. My side lengths seem to be correct, however I keep getting the wrong value for the angle. Round answers to the nearest tenth. A right triangle is a triangle in which one angle is a right angle. Calculating the median of a triangle is one of the fundamental problems in geometry Therefore, the perimeter of the triangle is 27 inches An equilateral triangle has a side length of 4 cm To find the semiperimeter, first calculate the perimeter of a triangle by adding up the length of its three sides Version 20191016 Version 20191016. Please check out also the Regular Triangle Calculator Learn how to find a missing side length of a right triangle. Solution: Let a= 8 cm b = 6 cm c = 4 cm If all the sides of a triangle are given, then we use Herons formula to calculate the area. Solve for the missing side. Usually, these coordinates are written as ordered pairs in the form (x, y). A right triangle has two sides perpendicular to each other. We can observe that: \(58^2 = 42^2 + 40^2\) In this case, the base would equal half the distance of five (2.5), since this is the shortest side of the triangle. Type in the given values. The triangle will help you remember the three formulae:Speed = Distance / TimeTime = Distance / SpeedDistance = Speed x Time You can use the Pythagorean Theorem to find the distance around a right triangle. Law of Cosines (the Cosine Rule): c 2 = a 2 + b 2 2ab cos (C) This is the hardest to use (and remember) but it is sometimes needed. |
# Bank Teller & Representing Numbers Part 2
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## Objective
SWBAT represent the value of a number in multiple ways.
#### Big Idea
Being able to understand and explain numbers will help students make sense of multi-digit computation and problem solving.
## Opening Activity
20 minutes
Today's Number Talk
For a detailed description of the Number Talk procedure, please refer to the Number Talk Explanation. For this Number Talk, I am encouraging students to represent their thinking using an array model.
For the first task, 4 x 6, some students decomposed the 4 and found that 4x6 = 2(2x6). They modeled their thinking by drawing a 2x6 array and doubling it. Other students decomposed the six or even decomposed both numbers!
During the next task, we discussed 4 x 36. Some students decomposed the 36 into 30 + 6 and multiplied each by 4. These students created a 4 x 30 array and added on a 4 x 6 array. Other students decomposed both multiplicands: 4x36 = 2(30+6) + 2(30+6) or 4x36 = 1(30+6) + 3(30+6).
Unfortunately, we didn't have enough time to solve this final task!
## Teacher Demonstration & Guided Practice
30 minutes
Goal:
First, I told students the goal for today's lesson: I can represent (or show) the value (the worth) of a number in multiple ways.
Review:
I began by reviewing yesterday's lesson, Bank Teller & Representing Numbers Part 1. I reviewed key concepts by referring to the following posters: Digit Vocabulary PosterNumber Vocabulary Poster, and Place Value Vocabulary PosterWho can explain what a digit is again? One student explained, "A digit is a symbol. For example, 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 are the ten digits we use in numbers." And what does place value mean again? Students acted the meaning of this word out altogether. "Place Value.... the value of a digit's place in a number."
Place Value Chart:
I wanted call attention to the idea that "a digit in one place represents ten times what it represents in the place to its right," so prior to the lesson, I added the following to the bottom of the Place Value Chart
____ ten = ____ ones
____ hundred = ____ tens = ____ ones
____ thousand = ____ hundred = ____ tens = ____ ones
____ ten thousand = ____ thousands = ____ tens = ____ ones
We began with the first equation. I filled in "1 ten" and asked: How many ones is equal to one ten? Students responded, "Ten!" Also, most students seemed to understand that 1 hundred = 10 tens = 100 ones. Then we got to the third line down and students were a bit confused with the number of hundreds in a thousand. To help break this task down, we counted the number of hundreds in 100... then in 200... then in 300... 400... and 500... Once students discovered that 50 x 10 = 500, they were then able to see that 100 x 10 = 1000. Here's the drawing on the board that resulted from this conversation: Scaffolding to find the # of 10s in 1000..
Bank Teller Time!
Next, I put my tie and glasses on, just like I had done the day before, and acted the part of a bank teller. I showed students the large Teacher Demonstration Withdrawl Slip and asked, Do you remember yesterday when we talked about withdrawal slips? When a customer uses a withdrawal slip, he is taking money out of the bank, right? As a bank teller, I'll ask the customer, "How would you like you money? Do you want your cash in \$100 bills? \$10 bills? \$1 bills?" Also as a bank teller, it's important to know how to represent numbers in multiple ways. For example, if a customer wants to withdraw \$50, I could give her five tens or I might give her fifty ones. This is because the value of numbers can be represented in multiple ways.
Using the Process Grid Labels, I created a Process Grid on the board. Next, I wrote "\$4" on the withdrawal slip and asked, What if a customer asked to withdraw \$4? I modeled how to represent a one-digit number using the process grid. If I had \$4 (I wrote 4 in the Number column on the process grid), how could I could I represent this number using the money model? One student said, "Four ones!" Since this was review from the day before, the students knew where I was headed and were able to provide all the answers successfully.
Prior to moving on to the Expression column of the process grid, I reviewed the difference between anExpression and an Equation. We then discussed how we could represent the money model for the number 4 using an expression (4 x 1). At this point a student said, "I know another way we could represent 4 using the ones. We could give the customer 2 groups of two \$1.00 bills." I was so happy to see students thinking outside of the box! I modeled this on the board, and then asked for the expression for this money model. Students said, "(2 x 1) + (2 x 1)." We then moved on to the Prove Equivalency column: (4 x 1) = (2 x 1) + (2 x 1) and simplified this equation to 4 = 2 + 2 and then 4 = 4. I knew that this simplification process would help students begin developing the foundational skills necessary for simplifying algebraic equations in middle school!
I continued in the same fashion: If I had \$14 (I wrote 14 in the Number column on the process grid), how could I could I represent this number using the money model? After we represented 14 in two ways using the money model, provided the appropriate expressions, and proved equivalency between the two expressions, we moved on to \$64 and \$164.
Students were anxious to begin practicing on their own!
## Partner Practice
40 minutes
Prior to the lesson, printed copies of the Process Grid Labels for each pair of students. Then I placed each sheet in a page protector so that partners could create their own process grids across two desks: Student Process Grid
I continued: I want you to all pretend you are bank tellers today! You'll come back to the counter to get your first withdrawal slip. Your job is to represent the withdrawal amount in at least two ways and to prove that these two representations are indeed equivalent, just in case the customer questions you!
I explained to students: Today, I'm looking for two things in particular! I'm looking for students to use "Sophisticated Math Talk" (I pointed to the Sophisticated Math Talk Poster from yesterday.) and I'm looking for you to prove that the value of a number can be represented in multiple ways using your own process grids! This means you'll need to provide clear, understandable, and convincing evidence.
Leveled Withdrawal Slips:
I explained the six levels of Withdrawal Slips at the Back Counter. Each level was marked with Labels. To add a little suspense, I turned the withdrawal slips over! I asked students to begin by choosing a withdrawal slip from Level A. By providing a gradual development of complexity, I was also providing a differentiated activity for all students.
• Level A: 2-digit numbers below 50
• Level B: 2-digit numbers above 50
• Level C: 3-digit numbers below 300
• Level D: 3-digit numbers above 300
• Level E: 4-digit numbers below 4000
• Level F: 4-digit numbers above 4000
Finally, I explained how students could move on to the next level of withdrawal slips: by checking their work with another group of peers. This will help develop Math Practice 3: Construct viable arguments and critique the reasoning of others.
Getting to Work!
I asked one student in each group to raise his or her hand. I asked this student to get a withdrawal slip from Level A at the back counter. Then I asked the student who was not raising his/her hand to obtain a process grid from the back table. Talk about some excited students!
During this time, I rotated the room to conference with students as they modeled multiple representations for each number.
Examples of Student Work:
Here's an example of a process grid for the number 87. This group did a beautiful job simplifying the equation to prove equivalency!
Then, students moved on to more challenging withdrawal slips:
Many students were successful at making it to the highest level of withdrawal slips! Here's an example:
I loved watching peers check one another's work. There were a couple amazing moments where students actually caught each other's mistakes, and instead of responding, "I don't think this is correct," I heard them referring to the Sophisticated Math Talk Poster and saying, "Can you tell me more about this?" Through explaining their reasoning, the students then realized their mistakes on their own! So great!
## Closing
10 minutes
To bring closure to the lesson, I asked students to return their withdrawal slips, bags of money, and process grids.
Next, I passed out an Exit Slip, closely aligned with today's activity. I provided a bag of play money for a student who struggles with math.
I loved seeing the majority of student successfully represent the number in two ways and then prove equivalency! Here, a student turns over her paper and completes the problem again to make her work more precise: Front Side and Back Side.
Here are examples of proficiency levels:
Most of the students were proficient. Some were nearing proficient and only one student was at the novice level. |
# Cartesian Coordinate System and Functions
8.13. Answer: The x-intercept of a function y = f(x) is the intersection
the graph of f and the x -axis. The x-axis is the graph of the
function g(x) = 0. Thus the x-axis is the intersection of two curves :
y = f(x) and g(x) = 0.
8.14. Solutions: Hopefully, you used standard procedures.
(a) Find points of intersection: f(x) = 6x + 3 and g(x) = 2x − 7.
equate ordinates substitute add −2x − 3 both sides divide by 4 d one !
When
(b) Find points of intersection: f(x) = x + 3 and g(x) = 2 − 8x.
equate ordinates substitute add 8x − 3 both sides divide by 9
When
(c) Find points of intersection: f(x) = x2+7x−1 and g(x) = 4x−3.
equate ordinates substitute add −4x + 3 both sides factor done!
Calculation of Ordinates :
When x = −1, y = g(2) = −7.
When x = −2, y = g(3) = −11.
8.15. Solutions: We use standard procedures around here . . . how
(a) Find points of intersection: f(x) = 2x2−5x+2 and g(x) = x+3.
equate ordinates substitute add −x − 3 both sides pos. discrim. Quadratic formula
These two curves intersect at
(b) Find points of intersection: f(x) = x2 + 4x − 1 and g(x) =
1 − 4x − x2.
equate ordinates substitute add −1 + 4x + x2 pos. discrim. Quadratic formula
These two curves intersect at
(c) Find points of intersection: f(x) = 3x2 + 1 and g(x) = x2 − 5x.
equate ordinates substitute add −1 + 4x + x2 pos. discrim. Quadratic formula
These two curves intersect at
8.16. Solutions: Fol low the procedure.
(a) f(x) = 4x − 2 and g(x) = 4x + 12
equate ordinates substitute add −4x + 2 both sides
The equation 0 = 14 has no solution; i.e., no value of x can
satisfythe equation 0 = 14. Therefore, these two curves do not
intersect.
b) f(x) = 2x2 + 3 and g(x) = x2 − 1.
equate ordinates substitute add −x2 + 1 both sides
The equation x2 = −4 has no solutions; therefore, these two
curves do not intersect.
(c) f(x) = x2 + 2x + 2 and g(x) = x + 1
equate ordinates substitute add −x − 3 both sides negative discriminant
A negative discriminant (b2−4ac < 0) implies that the equation
has not solution; therefore, these two equations do not intersect.
(d) f(x) = x2 − 2x − 4 and g(x) = 4x2 − 3
equate ordinates substitute add −4x2 + 3 negative discriminant
A negative discriminant implies the equation has no solution.
## Solutions to Examples
8.1. Solution:
P : (x1, y1) = (−2, 4)
Q : (x2, y2) = (3,−1)
We take the difference in the first coordinates and the difference in
the second coordinates.
x1 − x2 = −2 −3 = −5
y1 − y2 = 4− (−1) = 4+1 = 5
We now take the sum of the squares of these two:
(x1 − x2)2 + (y1 − y2)2 = (−5)2 + 52 = 25+25 = 50.
Finally, we take the square root of this result:
Presentation of Solution:
Of course, this process can be accelerated once you fully understand
the computational steps.
8.2. Solution to: (a) Define f(x) = x2+3x+1. The natural domain is
the set of all real numbers for which the value of f(x) = x2+3x+1 can
be computed as a real number. For any real number x , the expression
x2 + 3x + 1 evaluates to a real number. Therefore, we deduce,
Solution to: (b) Define The numerator and denominator
always evaluate to a real number; however, if the denominator
evaluates to zero, the quotient is not a real number. Thus, we
can saythat
Dom(g) = { x | x2 − 3x + 2 ≠0}
This should not be considered to be a satis factory characterization of
the domain of g though.
First find where x2 − 3x + 2 = 0, and reason from there. Solve
given factor solve
Therefore,
set notation interval notation
Solution to: (c) Define For any x, x + 2 evaluates to
a real number, but for to evaluate to a real number we must
have x + 2 ≥ 0. Thus,
Dom(h) = { x | x + 2 ≥ 0 }
Again, we should not be satisfied with this formulation. We next solve
the inequality:
x + 2 ≥0 => x ≥ −2
Thus,
Solution to: (d) Define In order for p(x) to evaluate
to a real number, we require x2 − 1 ≥ 0 and x ≠ 0. Thus,
Dom(p) = { x | x2 − 1 ≥ 0 }
We need to solve the inequality x 2 − 1 ≥ 0. To do this, we use the
Sign Chart Method originallydiscussed in Lesson 7. (Actually, this
method reallyisn’t needed for this simple inequality . We could solve
as follows:
x2 − 1 ≥0 => x2 ≥1 => |x| ≥ 1,
but we shall use the Sign Chatr Method in any case, just to remind
you of this method.)
We begin by factoring completely the left-hand side (which is a difference
of squares):
(x + 1)(x − 1) ≥ 0
The Sign Chart of (x + 1)(x − 1)
Therefore, the solution to the inequality x2 − 1 ≥ 0 is
(−∞,−1 ] ∪ [ 1,+∞)
But the solution to this inequalityis the natural domain of p. Thus,
Notice how all the techniques of algebra (Lessons 1–7) are used:
factoring, solving inequalities, interval notation and so on.
This is the discouraging and challanging thing about mathematics:
To solve any given problem, we must call on our entire history of
experiences in mathematics. This is why it is so important for us to
try to master each of the little steps we take toward our final goals.
8.3. Solution: Consider Based on the above strategy,
we see that
The first condition, x ≠ −1, avoids having zero in the denominator (we
exclude x = −1 from the domain); the second condition is necessary
for the radicand to be nonnegative. (The square root of a nonnegative
number is a real number, whereas the square root of a negative number
is a complex number . We don’t want to work with complex numbers
at this time.)
As you can see, I’ve simply translated the strategy into a series of
inequalities. We solve the inequality
first using the Sign Chart Methods.
The Sign Chart of
Therefore,
But, we also have the condition x ≠ −1; this changes the above solution
slightly to
x < −1 or x ≥ 0.
8.4. Solutions: We follow the standard procedures.
(a) Find the points of intersection of f(x) = 3x+2 and g(x) = 5x−4.
equate ordinates substitute add −5x − 2 to both sides divide by 3
At x = 3, f(3) = 3(3) + 2 = 11. Thus, ( 3, 11 ) is the point of
intersection.
Presentation of Solution: Intersection Point(s):
(b) Find intersection points of f(x) = x2−3x+1 and g(x) = 2x−5.
equate ordinates substitute add −2x + 5 both sides factor done!
Calculation of Ordinates:
When x = 2, y = g(2) = −1.
When x = 3, y = g(3) = 1.
Points of Intersection:
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# Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.8 | Set 1
Last Updated : 08 May, 2021
### Question 1. limx→π/2[Ï€/2 – x].tanx
Solution:
We have,
limx→π/2[Ï€/2 – x].tanx
Let us considered, y = [Ï€/2 – x]
Here, x→π/2, y→0
= limy→0[y.tan(Ï€/2 – y)]
= limy→0[y.{sin(Ï€/2 – y)/cos(Ï€/2 – y)]
= limy→0[y.{cosy/siny}]
= limy→0[y/siny].cosy
= limy→0[cosy] [Since, limy→0[siny/y] = 1]
= 1
### Question 2. limx→π/2[sin2x/cosx]
Solution:
We have,
limx→π/2[sin2x/cosx]
= limx→π/2[2sinx.cosx/cosx]
= 2Limx→π/2[sinx]
= 2
### Question 3. limx→π/2[cos2x/(1 – sinx)]
Solution:
We have,
limx→π/2[cos2x/(1 – sinx)]
= limx→π/2[(1 – sin2x)/(1 – sinx)]
= limx→π/2[(1 – sinx)(1 + sinx)/(1 – sinx)]
= limx→π/2[(1 + sinx)]
= 1 + 1
= 2
### Question 4. limx→π/2[(1 – sinx)/cos2x]
Solution:
We have,
limx→π/2[(1 – sinx)/cos2x]
= limx→π/2[(1 – sinx)/(1 – sin2x)]
= limx→π/2[(1 – sinx)/(1 – sinx)(1 + sinx)]
= limx→π/2[1/(1 + sinx)]
= 1/(1 + 1)
= 1/2
### Question 5. limx→a[(cosx – cosa)/(x – a)]
Solution:
We have,
limx→a[(cosx – cosa)/(x – a)]
=
= -2sin[(a + a)/2] × 1 × (1/2)
= -sina
### Question 6. limx→π/4[(1 – tanx)/(x – Ï€/4)]
Solution:
We have,
limx→π/4[(1 – tanx)/(x – Ï€/4)]
Let us considered, y = [x – Ï€/4]
Here, x→π/4, y→0
= -2 × 1 × [1/(1 – 0)]
= -2
### Question 7. limx→π/2[(1 – sinx)/(Ï€/4 – x)2]
Solution:
We have,
limx→π/2[(1 – sinx)/(Ï€/4 – x)2]
Let us considered, y = [Ï€/2 – x]
Here, x→π/2, y→0
= limy→0[(1 – cosy)/y2]
= 2 × 1 × (1/4)
= (1/2)
### Question 8. limx→π/3[(√3 – tanx)/(Ï€ – 3x)]
Solution:
We have,
limx→π/3[(√3 – tanx)/(Ï€ – 3x)]
Let us considered, y = [Ï€/3 – x]
When, x→π/3, y→0
= (4/3) × 1 × [1/(1 + 0)]
= (4/3)
### Question 9. limx→a[(asinx – xsina)/(ax2 – xa2)]
Solution:
We have,
limx→a[(asinx – xsina)/(ax2 – xa2)]
= limx→a[(asinx – xsina)/{ax(x – a)}]
Let us considered, y = [x – a]
When, x→a, y→0
= limy→0[{asin(y + a) – (y + a)sina)}/{a(y + a)y}]
= limy→0[(a.siny.cosa + asina.cosy – ysina – asina)/{a(y + a)y}]
= limy→0[{a.siny.cosa + a.sina.(cosy – 1) – y.sina}/{a(y + a)y}]
= limy→0[{a.siny.cosa + a.sina.2sin2(y/2) – t.sina}/{a(y + a)y}]
= limy→0[a.siny.cosa/a(y + a)y] – limy→0[2.a.sina.sin2(y/2)/a(y + a)y] + limy→0[y.sina/a(a + y)y]
= [(a.cosa)/a2] – [(sina)/a2] + 0
= [(a.cosa – sina)/a2]
### Question 10. limx→π/2[{√2 – √(1 + sinx)}/cos2x]
Solution:
We have,
limx→π/2[{√2 – √(1 + sinx)}/cos2x]
On rationalizing the numerator, we get
= limx→π/2[{2 – (1 + sinx)}/cos2x{√2 + √(1 + sinx)}]
= limx→π/2[(1 – sinx)/(1 – sin2x){√2 + √(1 + sinx)}]
= limx→π/2[(1 – sinx)/(1 – sinx)(1 + sinx){√2 + √(1 + sinx)}]
= limx→π/2[1/(1 + sinx){√2 + √(1 + sinx)}]
= 1/{(1 + 1)(√2 + √2)}
= 1/4√2
### Question 11. limx→π/2[{√(2 – sinx) – 1}/(Ï€/2 – x)2]
Solution:
We have,
limx→π/2[{√(2 – sinx) – 1}/(Ï€/2 – x)2]
Let us considered, y = [Ï€/2 – x]
Here, x→π/2, y→0
=
= limy→0[{√(2 – cosy) – 1}/y2]
On rationalizing the numerator, we get
= limy→0[{(2 – cosy) – 1}/y2{√(2 – cosy) – 1}]
= limy→0[{1 – cosy}/y2{√(2 – cosy) – 1}]
= 2/4(1 + 1)
= 1/4
### Question 12. limx→π/4[(√2 – cosx – sinx)/(Ï€/4 – x)2]
Solution:
We have,
limx→π/4[(√2 – cosx – sinx)/(Ï€/4 – x)2]
= 2√2/4
= (1/√2)
### Question 13. limx→π/8[(cot4x – cos4x)/(Ï€ – 8x)3]
Solution:
We have,
limx→π/8[(cot4x – cos4x)/(Ï€ – 8x)3]
= limx→π/8[(cot4x – cos4x)/83(Ï€/8 – x)3]
Let us considered, (Ï€/8 – x) = y
When x→π/8, y→0
=
= limx→0[(tan4x-sin4x)/83(π/8-x)3]
= limx→0[(sin4x/cos4x-sin4x)/83(π/8-x)3]
=
=
=
=
=
= (2 × 4 × 1 × 4 × 1)/(83)
= 1/16
### Question 14. limx→a[(cosx – cosa)/(√x – √a)]
Solution:
We have,
limx→a[(cosx – cosa)/(√x – √a)]
On rationalizing the denominator, we get
= -2 × sina × 1 × (1/2) × 2√a
= -2√a.sina
### Question 15. limx→π[{√(5 + cosx) – 2}/(Ï€ – x)2]
Solution:
We have,
limx→π[{√(5 + cosx) – 2}/(Ï€ – x)2]
Let us considered, y = [Ï€ – x]
When, x→π, y→0
= limy→0[{√(5 – cosy) – 2}/y2]
On rationalizing the numerator, we get
= limy→0[{1 – cosy}/y2{√(5 – cosy)-2}]
= 2 × (1/4) × {1/(2 + 2)}
= (1/8)
### Question 16. limx→a[(cos√x – cos√a)/(x – a)]
Solution:
We have,
limx→a[(cos√x – cos√a)/(x – a)]
= -2sin√a × 1 × (1/2√a) × (1/2)
= -(sin√a/2√a)
### Question 17. limx→a[(sin√x – sin√a)/(x – a)]
Solution:
We have,
limx→a[(sin√x – sin√a)/(x – a)]
= 2cos√a × 1 × (1/2√a) × (1/2)
= (cos√a/2√a)
### Question 18. limx→1[(1 – x2)/sin2Ï€x]
Solution:
We have,
limx→1[(1 – x2)/sin2Ï€x]
When, x→1, h→0
= limh→0[{1-(1-h)2}/sin2π(1-h)]
= limh→0[(2h-h2)/-sin2πh]
= limh→0[{h(2-h)}/sin2πh]
=
= -2/2Ï€
= -1/Ï€
### Question 19. limx→π/4[{f(x) – f(Ï€/4)}/{x – Ï€/4}]
Solution:
We have,
limx→π/4[{f(x) – f(Ï€/4)}/{x – Ï€/4}]
When, x→π/4, h→0
= limh→0[{f(Ï€/4 + h) – f(Ï€/4)}/{Ï€/4 + h – Ï€/4}]
It is given that f(x) = sin2x
= limh→0[{sin(Ï€/2 + 2h) – sin(Ï€/2)}/h]
= limh→0[(cos2h – 1)/h]
= limh→0[{-2sin2h}/h]
= -2Limh→0[(sinh/h)2] × h
= -2 × 1 × 0
= 0
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# ML Aggarwal ICSE Solutions Class 10 Math Seventeenth Chapter Mensuration
### ML Aggarwal ICSE Solutions Class 10 Math 17th Chapter Mensuration
ML Aggarwal ICSE Solutions Class 10 Math Chapter 17 Mensuration
Exercise 17.1
(1) Find the total surface area of a cylinder of radius 5 cm and height 10 cm. Leave your answer in terms of π.
Solution:
Given that,
Radius of a cylinder = l = 5 cm
Height of a cylinder = h = 10 cm
We know that, TSA of a cylinder is given by
Total surface area of a cylinder} = 2 π r (h + r)
= 2 π 5 (10 + 5)
= 2 π 5 × 15
= 10 × 15 π
TSA = 150 π cm2 is the required total surface area of a given cylinder.
(2) An electric geyser is cylindrical in shape, having a diameter of 35 cm and height 1.2 m. neglecting the thickness of its walls, calculate (i) its outer lateral surface area
(ii) its capacity in liters.
Solution:
Given that, diameter of cylindrical geyser = d = 35 cm
Radius of cylindrical geyser = d/2 = r = 17.5 cm
Height of cylindrical geyser = h = 1.2 m = 120 cm
(i) Then outer lateral surface area of cylindrical geyser is given by LSA = 2 π rh
= 2 × 22/7 × 17.5 × 120
LSA = 13200 cm2
This is the required outer lateral surface area of cylinder.
(ii) Capacity of a cylinder is given by
Capacity Volume of cylindrical geyser} = π r2h
= 22/7 × (17.5)2 × 120
= 115500 cm3
Volume = 115. 5 liters
This is the required capacity of given cylindrical geyser.
(3) A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. If the glass is filled with milk up to a height of 12 cm, find how many liters of milk is needed to serve 1600 students.
Solution:
Given that, diameter of glass (cylindrical) = d = 7 cm
Then, Radius of cylindrical glass = r = 7/2 = 3.5 cm
Height of cylindrical glass = h = 12 cm
Total no. of students in a school = 1600
Then, volume of a cylindrical glass is given by
Volume = π r2h
= 22/7 × (3.5)2 × 12
V = 462 cm3
This is the volume of a each glass. But, there are total 1600 students in a school. Then, amount of milk required for 1600 students is given by
V = 1600 × 462
V = 739200 cm3
V = 739.2 liters
Thus, 739.2 liters of milk is needed to serve 1600 students daily.
(4) In the given fig, a rectangular tin foil of size 22 cm by 16 cm is wrapped around to form a cylinder of height 16 cm. Find the volume of the cylinder.
Solution:
Given that, Length of rectangular foil (l) = 22 cm
Breath of rectangular foil (b) = 16 cm
When we are folding foil length wise, then cylinder is formed whose radius
2r = 22
R = 22/2 = 11 cm
R = 11 cm
Then, volume of the cylinder formed is given by
V = π r2h
= (22/7) × (11)2 × 16
V = 6084 cm3 is the required volume of a given cylinder.
(6) A road roller (in the shape of a cylinder) has a diameter 0.7m and its width is 1.2 m. Find the least number of revolutions that the roller must make in order to level a playground of size 120m by 44m.
Solution:
Given that, diameter of a road roller = d = 0.7 m
Width (h) = 1.2m
Then, Curved surface area of the road roller is given by
CSA = 2 π rh
= 2 × 22/7 × 0.35 × 1.2
CSA = 2.64m2
Now, Area of the playground is given by,
Area = l × 6 = 120 × 44 = 5280 m2
Thus, No. of revolutions = Area of Playground/Curved S. Area = 5280/2.64
Thus, the revolutions that the roller must make in order to level a playground are 2000.
(7) (i) If the volume of a cylinder of height 7 cm is 448 π cm3. Find the lateral surface area and total surface area.
(ii) A wooden pole is 7m high and 20cm in diameter. Find its weight, if the wood weighs 225 kg per m3.
Solution:
(i) Given that,
Volume of a cylinder = V = 448 π cm3
Height of a cylinder = h = 7 cm
We have, Volume of a cylinder is given by,
V = π r2h
448 π = π r2(7)
448/7 = r2
r2 = 64
r = 8 cm is the required of the given cylinder.
• Lateral surface area of a cylinder is given by
LSA = 2 π rh
= 2 × π × 8 × 7
LSA = || 2 π cm2
is the required Lateral surface area of a cylinder.
• Total surface area of a cylinder is given by
TSA = 2 π r (r + h)
= 2 × π × 8 × (8 + 7)
= 2 × π × 8 × 15
TSA = 240 π cm2
This is the required total surface area of a cylinder.
(ii) Given that,
Height of wooden pole (h) = 7m
Diameter of pole (d) = 20cm
Radius of pole = r = 10 cm
Then, Volume = π r2h
= 22/7 × 10 × 10/100 × 100 × 7 m3 = 22/100 m3
Here, weight of wood used = 225 kg/m3
Hence, total weight = 22/100 × 225 kg = 99/2 = 49.5 kg
(8) The circumference of the cylindrical vessel is 132 cm and its height is 25 cm.
(ii) Volume of cylinder
Solution:
Given that, height of cylindrical vessel (h) = 25 cm
Circumference of base = 132 cm
(i) We have, Circumference of circular base of a cylinder} = 2 π r
132 = 2 × 22/7 × r
r = 21 cm is the required radius of the cylinder.
(ii) Volume of a cylinder is given by,
V = π r2h
= 22/7 × 21 × 21 × 25
V = 34650 cm3 is the required of given cylinder.
(9) The area of the curved surface of a cylinder is 4400 cm2. and the circumference of its base is 110 cm.
Find (i) The height of the cylinder
(ii) The volume of the cylinder
Solution:
Given that, Area of Curved surface of a cylinder} = CSA = 4400 cm2
Circumference of base of a cylinder = 110 cm
(i) Curved surface area of a cylinder is given by
CSA = 2 π rh
4400 = 2 × 22/7 × 17.5 × h
h = 4400 × 7/ 2 × 22 × 17.5 = 40 cm
h = 40 cm is the required height of the cylinder.
Given that, Circumference of base = 2 π r
110 = 2 × 22/7 × r
r = 110 × 7/ 2 × 22 = 17.5 cm
r = 17.5 is the required radius of given cylinder.
(ii) Volume of a cylinder is given by,
V = π r2h
= 22/7 × (17.5)2 × 40
V = 38500 cm3 is the required volume of a given cylinder.
(10) A cylinder has a diameter of 20 cm. The area of curved surface is 1000 cm2.
Find (i) The height of the cylinder correct to one decimal place.
(ii) The volume of the cylinder correct to one decimal place.
Solution:
Given that, diameter of a cylinder (d) = 20 cm
Then, Radius of a cylinder (r) = 10 cm
Curved Surface Area = 1000 cm2
(i) The curved surface area of a cylinder is given by
CSA = 2 πrh
1000 = 2 × 3.14 × 10 × h
h = 1000/62.8 = 15.9cm
h = 15.9 cm is the required height of the given cylinder.
(ii) Volume of the cylinder is given by,
V = π r2h
= 3.14 × 10 × 10 × 15.9
V = 4992.6 cm2
This is the required volume of the given cylinder.
(11) The barrel of a fountain pen, cylindrical in shape, 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up when writing 310 words on an average. How many words use up a bottle of ink containing one – fifth of a liter?
Solution:
Given that, Height of cylindrical barrel of pen (h) = 7 cm
Diameter (d) = 5 mm
Radius (r) = 2.5 mm = 0.25 cm
Then, Volume of a barrel is given by,
V = π r2h
= 22/7 × 0.25 × 0.25 × 7
V = 1.375 cm3
Given that, Ink in a bottle = one fifth of a litre
= 1/5 × 1000
= 200 ml
And, No. of words written using full barrel of ink = 310
No. of words written using 200 ml of ink = 200/1.375 × 310
= 45090.90 words
Thus, the no. of words written using the ink are found to be 45100 words.
(12) Find the ratio between the total surface area of a cylinder to its curved surface area given that its height and radius are 7.5 cm and 3.5 cm.
Solution:
Given that, Radius of cylinder (r) = 3.5 cm
Height of a cylinder (h) = 7.5 cm
Total Surface Area of a cylinder = 2 π (r + h) r
Curved Surface area of a cylinder = 2 π r + h
Then, TSA/CSA = 2 π r (r + h)/ 2 π r h = r + h/h
= (3.5 + 7.5)/ 7.5
TSA/CSA = 11/7.5 = 22/15
Thus, the ratio of total surface area of cylinder and curved surface area of a cylinder is found to be 22:15.
(13) The radius of the base of a right circular cylinder is halved and the height is doubled. What is the ratio of the volume of a new cylinder to that of the original cylinder?
Solution:
Let us consider, r is the radius of circular base of right circular cylinder.
Height be ‘h’
Then, Volume of a cylinder = V = π r2h
Given that, Radius of circular base of cylinder is halved & height is doubled then
r → r/2 and h → 2h
Then, Volume of new cylinder = V1 = π r2h
V1 = π (r/2)2 (2h)
V1 = 1/2 π r2h
The ratio of new cylinder’s volume to the volume of original cylinder is given by
V1/v = 1/2 π r2h/ π r2h = 1/2
= V1: V = 1:2 is the required ratio.
(15) The ratio between the curved surface & the total surface of a cylinder is 1:2. Find the volume of the cylinder, given that its total surface area is 616 cm2.
Solution:
Given that,
The ratio between the curved surface area & the total surface area of a cylinder is 1:2.
And, total surface area = 616 cm2
CSA/TSA = 1/2 = CSA = 1/2 (TSA) = 1/2 × 616
CSA = 308 cm2
We have, Curved Surface Area = 2 π r h
308 = 2 π r h
π r h = 308/2
r h = 49 —– (1)
Total Surface area = 2 π r h + 2 π r2
616 = 308 + 2 π r2
154 = π r2
r2 = 154 × 7/22 = 49
= r = 7 cm put in (1) = h = 49/7 = 7 h = 7 cm
Now, Volume of a cylinder is given by,
V = π r2h = 22/7 × 72 × 7 = 1078 cm3
V = 1078 cm3 this is the required volume of the given cylinder.
(16) Two cylindrical jars contain the same amount of milk. If their diameters are in the ratio 3:4, find the ratio of their heights.
Solution:
Given that,
The two cylindrical jars contains the same amount of milk. diameter of both cylinders are in the ratio d1:d2 = 3:4
And also, ratio of their radius r1:r2 = 3:4
From given condition,
V1 = V2
π r21 h1 = π r22 h2
h1/h2 = r22/r12 = (r2/r1)2 = (4/3)2 = 16/9
h1/h2 = 16/9 is the required ratio of heights two given cylinders.
(18) A cylindrical tube open at both ends is mode of metal. The internal diameter of the tube is 11.2 cm and its length is 21 cm. The metal thickness is 0.4 cm. Calculate the Volume of the metal.
Solution:
Given that,
Internal diameter of cylindrical tube = 11.2 cm
Then, Internal radius = r = 5.6 cm
Length of the tube (h) = 21 cm
Thickness of metal = 0.4 cm
Thus, outer radius of cylindrical tube = 5.6 + 0.4 = 6 cm = R
Then, Volume of the metal is given by
V = π r2 h – π r2 h
= π h (R2 – r2)
= 22/7 × 21 × (62 – 5.62)
= 66 × (6 + 5.6) (6 – 5.6)
= 66 × 11.6 × 0.4
V = 306.24 cm3 this is the required volume of the metal.
(19) A lead pencil consists of a cylinder of wood with a shod cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
Solution:
Given that, diameter of the pencil = 7 mm
Radius of the pencil = 7/2 mm = 7/20 cm = R
And, Diameter of graphite = 1 mm
Then Radius of graphite = r = 1/2 mm = 1/20 cm
Thus, Volume of a graphite is given by
V = π r2 h
= 22/7 × (1/20)2 × 14 = 11/100
V = 0.11 cm3
Now, Volume of the wood is given by,
V = π r2 h – π r2 h
= π h (R2 – r2)
= 22/7 × 14 [(7/20)2 – (1/20)2]
= 44 × 48/400
V = 5.28 cm3 This is the required volume of the wood.
(20) A cylindrical roller made of iron is 2 m long. Its inner diameter is 35 cm and the thickness is 7 cm all round. Find the weight of the roller in kg, if 1 cm3 of iron weight 8 g.
Solution:
Given that, Length of cylindrical roller (h) = 2 m = 200 cm
Diameter = 35 cm
Inner radius = 35/2 = 17.5 cm = r
Thickness = 7 cm
Then, outer radius = R = 17.5 + 7 = 24.5 cm
Thus, Volume of iron used in roller is given by
V = π r2 h – π r2 h
= π h (R2 – r2)
= 22/7 × 200 [(49/2)2 – (35/2)2]
= 22/7 × 50 (492 – 352)
= 22/7 × 50 × 1176
V = 184800 cm3
But given that, 1 cm3 of iron weighs 8 g.
Then, weight of the roller = 184800 × 8
= 1478400 g
= 1478.4 kg
Thus, the weight of the roller is found to be 1478.4 kg.
Exercise 17.2
(1) Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.
Solution:
Given that, Slant height of RCC = 10 cm
Base Radius of RCC = r = 7 cm
Then, Curved surface area of RCC is given by,
CSA of RCC = π r l
= 22/7 × 7 × 10
CSA = 220 cm2
This is the required curved Surface Area of Right Circular Cone.
(2) Diameter of the base of a cone is 10.5 cm and slant height 10 cm. Find its curved surface area.
Solution:
Given that, diameter of base of a cone (d) = 10.5 cm
= Radius of base of a cone (r) = 5.25 cm
Slant height of a cone = 10cm
Then, curved surface area of a cone is given by,
CSA = π r l
= 22/7 × 5.25 × 10
CSA = 165 cm2 this is the required curved surface area of a cone.
(3) Curved surface area of a cone is 308 cm2 and its slant height is 14 cm.
Find, (i) Radius of the base
(ii) total surface area of the cone
Solution:
Given that, curved surface area of a cone = 308 cm2
Slant height of a cone (l) = 14 cm
(i) We have, curved surface area of a cone} = π r l
308 = 22/7 × r × 14
r = 308 × 7/22 × 14
r = 7 cm this is the required radius of a cone.
(ii) Now, total surface area of a cone is given by
TSA of a cone = Base + Area + CSA
= π r2 + π r l
= (22/7 × 72) + 308
= 22/7 × 49 + 308
= 154 + 308
TSA = 462 cm2 this is the required total surface area of a cone.
(4) Find the volume of the right circular cone with
(i) Radius 6 cm and height 7 cm
(ii) Radius 3.5 cm and height 12 cm
Solution:
(i) Given that,
Radius of RCC (r) = 6 cm
Height of RCC (h) = 7 cm
We have, Volume of RCC is given by
V = 1/3 π r2 h
V = 1/3 × (22/7) (6)2 × 7
= 22 × 12
V = 264 cm3
This is the required volume of a given RCC.
(ii) Given that,
Radius of RCC (r) = 3.5 cm
Height of RCC (h) = 12 cm
We have, volume of RCC is given by,
V = 1/3 π r2 h
= (1/3) (22/7) (3.5)2 (12)
= 22/7 × 12.25 × 4
V = 154 cm3
This is the required volume of a given RCC.
(6) A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kiloliters?
Solution:
Given that, diameter of conical pit = 3.5 m
= Radius of conical pit (r) = 1.75 m
depth (h) = 12 m
Then, capacity of conical pit (volume of a cone) is given by,
V = 1/3 π r2 h
= 1/3 × (22/7) × (1.75)2 × 12
V = 38.5 m3
V = 38.5 Kilolitres
1 m3 = 1 Kilolitres
This is the required capacity of conical pit in Kilolitres.
(7) If the volume of a right circular cone of height 9 cm is 48 π cm3, find the diameter of its base.
Solution:
Given that, Height of RCC = h = 9 cm
Volume of RCC = V = 48 π cm3
We have, Volume of Right circular cone is given by
V = 1/3 π r2 h
48 (22/7) = (1/3) × (22/7) × r2 × 9
3r2 = 48
r2 = 16
r = 16 cm
This is the required radius of the given RCC.
(8) The height of a cone is 15 cm. If its volume is 1570 cm3. Find the radius of the base.
Solution:
Given that, Height of a cone (h) = 15 cm
Volume of a cone (v) = 1570 cm3
We have, Volume of a cone is given by
V = 1/3 π r2 h
1570 = (1/3) (3.14) r2 × 15
5 × 3.14 × r2 = 1570
r2 = 1570/5 × 3.14 = 100
r = 10 cm
This is the required radius of given cone.
(9) The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white washing its curved surface area at the rate of Rs. 210 per 100 m2.
Solution:
Given that, Slant height of conical tomb (l) = 25 m
Base diameter (d) = 14 m
= Base Radius (r) = 7 m
We have, curved Surface area of a cone is given by
CSA = π r l
= 22/7 × 7 × 25
CSA = 550 m2 is the required curved surface area of given conical tomb.
Given that, Rate of washing its curved surface area per
100 m2 = Rs. 210
Then, total cost = 550/100 × 210 = Rs. 1155
Thus, the total cost of washing its curved surface area is found to be Rs. 1155.
(10) A conical tent is 10 m high and the radius of its base is 24 m. Find (i) Slant height of the tent
(ii) Cost of the canvas required to make the tent, if the cost of 1m2 canvas is Rs. 70
Solution:
Given that, Height of conical tent (h) = 10 m
Radius of base (r) = 24 m
We have, l2 = h2 + r2
I2 = 102 + 242 = 100 + 576 = 676
I2 = 676
I = 26 m is the required slant height of the tent.
(ii) Curved surface area of a cone is given by,
π r l = 22/7 × 24 × 26
= 13728/7 m2
Given that, the cost of 1m2 canvas is Rs. 70.
Then, total cost of 13728/7 m2 canvas is found to be
Total cost = 13728/7 × 70
= Rs. 137280
(11) A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the cloth required to make 10 such caps.
Solution:
Given that, Base radius of conical cap = r = 7 cm
Height of conical cap (h) = 24 cm
We have, l2 = h2 + r2
l2 = 242 + 72 = 576 + 49
l2 = 625
l = 25 cm is the required slant height.
We have, curved surface area of a cone is given by
CSA = π r l
= 22/7 × 7 × 25
= 22 × 25
CSA = 550 cm2
Thus, for one cap the cloth required is 550 cm2.
Then, the area of the cloth required to make 10 such caps is found to be
550 × 10 = 5500 cm2.
(13) Find what length of canvas 2m in width is required to make a conical tent 20m in diameter and 42 m in slant height allowing 10% for folds and the stitching. Also find the cost of the canvas at the rate of RS. 80 per meter.
Solution:
Given that, diameter of base of conical tent (d) = 20 m
= Radius of base of conical tent (r) = 10 m
Slant height (l) = 42 m
Then, curved surface area of conical tent is given by,
CSA = π r l
= 22/7 × 10 × 42
CSA = 1320 m2
This the area of the canvas required.
Now, 10% of this canvas is used for folds and stitches,
then actual cloth needed = 1320 + 10% of 1320
= 1320 + 10/100 × 1320
= 1320 + 132
= 1452 m2
Width of cloth = 2m
Length of cloth = Area/width = 1452/2 = 726 m
Given that, cost of canvas is Rs. 80 per meter.
Then, total cost required is = 80 × 726
= 58080 Rs.
(14) The perimeter of the base of a cone is 44 cm and the slant height is 25 cm. Find the volume and the curved surface area of the cone?
Solution:
Given that, Perimeter of base of cone = 44 cm
= 2 π r = 44
r = 44/2 π = 44/2 × 7/22
r = 7 cm is the base radius.
Now, slant height (l) = 25
We have, l2 = r2 + h2
h2 = l2 – r2 = 252 – 72
h2 = 625 – 49
h2 = 576
h = 24 cm is the required of a cone.
Then, Volume of the cone is given by,
V = 1/3 π r2 h
= 1/3 × 22/7 × 49 × 8
V = 1232 cm3 is the required volume of a cone.
Now, curved surface area of a cone is given by,
CSA = π r l
= 22/7 × 7 × 25
= 22 × 25
CSA = 550 cm2
This is the curved surface area of a cone.
(15) The volume of a right circular cone is 9856 cm3 and the area of its base is 616 cm2.
Find (i) the slant height of cone.
(ii) total surface area of cone
Solution:
Given that, Volume of RCC = 9856 cm3
Base area of RCC = 616 cm2
= π r2 = 616
r2 = 616 × 7/22 = 196
r = 14 cm is the required base radius.
Now, Volume of a cone = 1/3 π r2 h
9856 = 1/3 × 22/7 × 14 × 14 × h
h = (9856 × 3 × 7/ 14 × 14)
h = 48 cm is the required height of a cone.
(i) We have, l2 = h2 + r2
l2 = 482 + 142
l2 = 2304 + 196
l2 = 2500
l = 50 cm is the required slant height of a cone.
(ii) Total surface area of a cone is given by
TSA = π r (l + r)
= 22/7 × 14 × (50 + 14)
= 22 × 2 × 64
TSA = 2816 cm2
This is the required total surface area of a cone.
(16) A right triangle with sides 6 cm, 8 cm, and 10 cm is revolved about the side 8 cm. Find the volume and the curved surface of the cone so formed.
Solution:
Given that, A right triangle with sides: 6 cm, 8 cm, 10 cm. It is revolved about the side 8 cm then cone is formed.
= Radius (r) = 6 cm
Height (h) = 8 cm
Slant height (l) = 10 cm
Then, Volume of cone formed is given by,
V = 1/3 π r2 h
V = 1/3 × 3.14 × 62 × 8
V = 3.14 × 12 × 8
V = 301.44 cm3 is the required volume of the cone.
Now, Curved surface area of a cone is given by,
CSA = π r l
= 3.14 × 6 × 10
CSA = 188.4 cm2 is the required CSA of a cone.
(17) The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume be 1/27th of the volume of the given cone, at what height above the base is the section cut?
Solution:
Given that, Height of cone (H) = 30 cm
A small cone is cut off from the top of the big cone given.
Volume of given big cone (V) = 1/3 π R2 H
Given that,
Volume of small cone = 1/27th of the volume of big cone
1/3 π R2 h = 1/27 (1/3 π R2 h)
r2 h = 1/27 R2 H
r2 h = 30/27 R2
r2 h/ R2 = 30/27 = 10/9 —- (1)
Also, from fig, r/R = h/H
r/R = h/30 ——(2)
From (1) & (2) (h/30)2 × h = 10/9
h3/ 900 = 10/9
h3 = 1000
h = 10 cm is the height of small cone.
Now, H – h = 30 – 10 = 20 cm
Thus, the small cone is cut at height of 20 cm above the base of big cone.
(18) A semi – circular lamina of radius 35 cm is folded so that the two bounding radii are joined together to form a cone. Find
(ii) the lateral surface area of cone.
Solution:
(i) Given that, Radius of semi – circular lamina (r) = 35 cm
After folding a semi – circular lamina a cone is formed,
The slant height of cone formed (l) = 35 cm
Let us consider, r1 be the radius of a cone formed.
Now, perimeter of semi – circular lamina} = base of cone
π r = 2 π r,
r = 2 r,
35 = 2r1
r1 = 17.5 cm is the radius of a cone.
(ii) Now, the curved surface area of the cone formed is
CSA = π r1 l
= 22/7 × 17.5 × 35
CSA = 1925 cm2 is the required curved surface area of a cone.
Exercise 17.3
(1) Find the surface area of a sphere of radius 14 cm.
Solution:
Given that, Radius of a sphere = r = 14 cm
Then, surface area of a sphere is given by
SAS = 4 π r2
= 4 × 22/7 × 14 × 14
= 4 × 22 × 2 × 14
SAS = 2464 cm2 is the required surface area of a given sphere.
(2) Find the surface area of a sphere of diameter 21 cm.
Solution:
Given that, diameter of a sphere (d) = 21 cm
= Radius of a sphere (r) = 10.5 cm
Then, Surface area of a sphere is given by,
SAS = 4 π r2
= 4 × 22/7 × 10.5 × 10.5
SAS = 1386 cm2 is the required surface area of a given sphere.
(3) A shot – put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm3, find the mass of the shot – put.
Solution:
Given that, Radius of metallic shot – put = r = 4.9 cm
We know that,
Volume of sphere = 4/3 π r3
= 4/3 × 22/7 × (4.9)3
V = 493. 005
V = 493 cm3 is the required volume.
Now, density = 7.8 g per cm3
We have, density = mass/volume
Then, Mass = density × volume
= 7.8 × 493
Mass = 3845.4g is the required mass of shot – put.
(4) Find the diameter of a sphere whose surface area is 154 cm.
Solution:
Given that, Surface area of sphere = 154 cm2
We have, Surface area of a sphere is given by,
SAS = 4 π r2
154 = 4 × 22/7 × r2
r2 = 49/4
r = √49/2 = 7/2
r = 7/2 cm
Then, diameter of sphere = d = 2r = 2 × 7/2 = 7 cm
(5) Find (i) the curved surface area
(ii) the total surface area of a hemisphere of radius 21 cm.
Solution:
Given that, Radius of hemisphere (r) = 21 cm
(i) The Curved surface area of hemisphere is given by
CSA of hemisphere = 2 π r2
= 2 × 22/7 × 212
CSA = 2772 cm2 is the required curved surface area of given hemisphere.
(ii) Now, total surface area of hemisphere is given by
TSA of hemisphere = 3 π r2
= 3 × 22/7 × 212
= 4158 cm2
is the required total surface area of hemisphere.
(6) A hemispherical brass bowl has inner – diameter 10.5 cm. Find the cost of tin – plating it on the inside at the rate of Rs 16 per 100 cm2.
Solution:
Given that,
The inner diameter of hemispherical bowl (d) = 10.5 cm
= Radius (r) = 5.25 cm
Then, Curved surface area of hemispherical bowl is given by,
CSA of hemisphere = 2 π r2
= 2 × 22/7 × (5.25)2
= 173.25 cm2
Given that, the rate of plating is Rs. 16 per 100 cm2.
Then, total cost required for tin – plating the surface area of 173.25 cm2 is given by
Total cost = 173.25 × 16/100 = 27.72 Rs.
(7) The radius of a spherical balloon increases from 7 cm to 14 cm as air is jumped into it. Find the ratio of the surface area of the balloon in two cases.
Solution:
Given that, original radius of balloon (r) = 7 cm
Radius of balloon after air filling = 14 cm
Then, Surface area of original balloon = 4 π r2
Surface area of new balloon = 4 π r2
= SAB1/ SAB2 = 4 π r2/ 4 π r2 = r2/R2 = 72/142
S1/S2 = 1/4
S1: S2 = 1:4
This is the required ratio of balloons when it is empty & after filled with air.
(8) A sphere and a cube have the same surface. Show that the ratio of volume of the sphere to that of the cube is 6: π.
Solution:
Let us consider, the side of the cube = a
Then, Surface area of cube = 6a2
Surface area of sphere of radius ‘r’ = 4 π r2
= SA of sphere = SA of cube
4 π r2 = 6a2
π r2/ a2 = 6/4
∴ r/a = √6/2 π
Now, Volume of sphere (v1) = 4/3 π r3
and Volume of cube (v2) = a3
Then, V1/V2 = 4 π r3/ 3a3
= 4 π √63/3 (2π)3
= (4π × 6 √6)/ (3 × 8 π × √π)
V1/V2 = √6/ √π
V1:V2 = √6: √π
Hence Proved.
(10) Find the volume of a Sphere whose surface area is 154 cm2.
Solution:
Given that, surface area of sphere = 154 cm2
We have, Surface area of Sphere = 4πr2
154 = 4 × 22/7 × r2
r2 = 49/4
r = 7/2 cm is the radius of sphere.
Now, Volume of the sphere = 4/3πr3
= 4/3 × 22/7 × (7/2)3
= 179.67 cm3
Thus, Volume of a sphere is found to be 179.67 cm3.
(11) If the volume of a sphere is 179.2/3. Find the radius & surface area of sphere.
Solution:
Given that, Volume of a sphere = 179.2/3 cm3
We have, Volume of sphere = 4/3 πr3
539/3 = 4/3 × (22/7)r3
r3 = 49 × 718 = (7 × 7 × 7)/ (2 × 2 × 2)
= r = 7/2 = 3.5 cm
r = 3.5 cm is the required radius of given sphere.
Now, Surface area of sphere = 4πr2
= 4 × 22/7 × (7/2)2
= 22 × 7
SAS = 154 cm2
This is the required surface area of given sphere.
(12) A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?
Solution:
Given that, Radius of hemispherical bowl (r) = 3.5 cm
Volume of hemisphere = 2/3 πr3
= 2/3 × 22/7 × (7/2)3
= 11 × 4916
= 539/6
V = 89.5/6 cm3
This is the required volume of a given hemispherical bowl.
(13) The surface area of a solid sphere is 1256 cm2. It is cut into two hemisphere. Find the total surface area and the volume of a hemisphere.
Solution:
Given that, Surface area of a solid sphere = 1256 cm2
We have, Surface area of sphere = 1256
4πr2 = 1256
4 × 3.14 × r2 = 1256
r2 = 1256 × (4 × 3.14)
r2 = 100
r = 10 cm is the required radius.
Now, total surface area of hemisphere = 3πr2
3πr2 = 3 × 3.14 × 102
= 3 × 3.14 × 100
TSA of H = 942 cm2 is the required total surface area of hemisphere.
Now, Volume of the hemisphere = 2/3πr3
= 2/3 × 22/7 × 103
= 2/3 × 3.14 × 1000
= 6280/3
V = 2093. cm3
This is the required volume of a given hemisphere.
Exercise 17.4
(1) The following fig. shows a cyboidal block of wood through which a circular cylindrical hole of the biggest size is drilled. Find the volume of the wood left in the block.
Solution:
Given that,
diameter of biggest hole = 30 cm
Then, Radius of hole (r) = 30/2 = 15 cm
Height of hole (h) = 70 cm
The volume of cubical block is given by,
V = l × b × h
V = 70 × 30 × 30
V = 63000 cm3 is the required volume of block.
Then, Volume of a cylindrical hole is given by,
V = πr2h
= 22/7 × 152 × 70
= 22 × 225 × 70
V = 49500 cm3 is the required volume of cylindrical hole.
Now, Volume of the wood left in the block} = 63000 – 49500
= 135000 cm3
Hence, the volume of the wood left in the block is found to be 13500 cm2
(2) The given fig. shows a solid trophy made of shining glass. If one cubic centimeter of glass costs Rs. 0.75. find the cost of the glass for making the trophy.
Solution:
Given that, Side of cubical part = 28 cm
Radius of cylindrical part (r) = 28/2 = 14 cm
Height of cylinder (h) = 28 cm
Then, volume of a cube is given by,
Volume of cube = a3 = 283
V1 = 21952 cm3
Now, volume of the cylindrical part is given by,
V2 = πr2h
= 22/7 × 14 × 14 × 28
V2 = 17248 cm3
Then, total volume of the solid = (Volume of the cube) + (Volume of the cylinder)
= 21952 + 17284
TV = 39200 cm3
The cost of 1 cm3 glass is Rs. 0.75
Then, total cost of glass = 39200 × 0.75
= Rs. 294000.
(3) From a cube of edge 14 cm, a cone of maximum size is carved out. Find the volume of the remaining material.
Solution:
Given that, edge/ side of the cube = 14 cm
Volume of cube (V) = (14)3 = 2744 cm3
diameter of cone cut from cube = 14 cm
Volume of cube = a3
V = 143
V1 = 2744 cm3
Volume of cone = 1/3 πr2h
= 1/3 × 22/7 × 72 × 14
V2 = 215613 cm3
Volume of remaining material = (Volume of the cube) – (Volume of cone)
= (3 × 2744 – 2156)/3
= (8232 – 2156)/3
= 6076/3
V = ₹ 025.1/3 cm3
This is the volume of remaining material.
(4) A cone of maximum volume is curved out of a block of wood of size. 20 cm × 10 cm × 10 cm. find the volume of the remaining wood.
Solution:
Given that,
The dimensions of wooden block = 20 cm × 10 cm × 10 cm
Then, Volume of the wooden block = 20 × 10 × 10 = 2000 cm3.
diameter of cone (d) = 10 cm
= Radius of cone (r) = 5 cm
Height of cone (h) = 20 cm
Now, Volume of the cone is given by,
V = 1/3 πr2h
V = 1/3 × 22/7 × 52 × 20
V = 11000/21 cm3
Now, (Volume of remaining) = (Volume of block of wood) – (Volume of cone)
= (2000 – 11000)/21
= (21 × 2000 – 11000)/21
= (42000 – 11000)/21
= 31000/21
V = 1476.19 cm3
This is the required volume of remaining wood.
(5) 16 glass spheres each of radius 2 cm are packed in a cuboidal box of internal dimensions 16 cm × 8 cm × 8 cm and then the box is filled with water. Find the volume of the water filled in the box.
Solution:
Given that, Dimensions of cuboidal box = 16 cm × 8 cm × 8 cm
Then, Volume of cuboidal box = l b h = 16 × 8 × 8 = 1024 cm3
Radius of sphere (r) = 2 cm
Then, volume of the sphere = 4/3πr3
= 4/3 × 22/7 × 23
V = (4 × 22 × 8)/ (3 × 7)
V = 704121 cm3
Volume of 16 sphere = 16 × 704/21
= 11264/21 cm3
= 536.38 cm3
Now, Volume of water filled in the box} = (Volume of the box) – (Volume of 16 spheres)
= (1024 – 536.38)
V = 487.62 cm3
This is the required volume of water filled in the box.
(6) A cuboidal block of side 7 cm is sup mounted by a hemisphere. What is the greatest diameter that the hemisphere can have? Also, find the surface area of the solid.
Solution:
Given that, Side of cuboidal block = a = 7 cm
diameter of hemisphere = 7 cm
= Radius of hemisphere (r) = 7/2 = 3.5 cm
Now, Surface area of hemisphere = 2πr2
2πr2 = 2 × 22/7 × 3.5 × 3.5
S = 44 × 12.25/7
= 539/7
S = 77 cm2
Surface area of cube = 6a2 = 6 × 72 = 294 cm2
Now, Surface area of base of hemisphere = πr2
= 22/7 × 3.5 × 3.5
= 38.5 cm2
Thus, (the surface area of solid) = (Surface area of the cube) + (Surface area of hemisphere) – (Surface area of base of hemisphere)
= 294 + 77 – 38.5
= 332.5 cm2
This is the required surface area of solid.
(7) A wooden article was made by scooping out a hemisphere from each end of a solid cylinder (as shown in fig). If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of the article.
Solution:
Given that,
Height of cylinder (h) = 10 cm
Radius of the base (r) = 3.5 cm
Total surface area is given by,
TSA = (Curved surface area of a cylinder) + 2x (Curved surface area of hemisphere)
= 2πrh + 2 × 2πr2
= 2 × 22/7 × 3.5 × (10 + 2 × 3.5)
= (15417) × (10 + 7)
= 22 × 17
TSA = 374 cm2 is the required total surface area of given article.
(8) A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. If the total height of the toy is 15.5 cm, find the total surface area of the toy.
Solution:
Given that,
Total height of toy (h) = 15.5 cm
Radius of base of conical part (r) = 3.5 cm
Now, Height of cone (h) = 15.5 – 3.5 =12 cm
We have, l2 = h2 + r2
l2 = 122 + 3.52
l2 = 144 + 12.25 = 156.25
l = 12.5 cm
(Total surface area of the toy) = (curved surface area of cone) + (curved surface area of the hemisphere)
TSA = π r l + 2πr2
= π r (l + 2r)
= 22/7 × 3.5 × (12.5 + 2 × 3.5)
= 11 × 19.5
TSA = 214.5 cm2
This is the required surface area of the toy.
(10) A circus tent is in the shape of a cylinder sun mounted a cone. The diameter of the cylindrical portion is 24 m and its height is 11 m. If the vertex of the cone is 16 m. above the ground, find the area of the canvas used make the tent.
Solution:
Given that, Radius of base of cylindrical part of the tent}
r = 24/2 = 12 m
Height of cylindrical part (H) = 11 m
As the vertex of a cone is 16 m above the ground, height of cone (h) = 16 – 11 = 5 m
Now, we have
l2 = h2 + r2
l2 = 25 + 144
l2 = 169
l = 13 m
Radius of cone (r) = 12 m
(Area of canvas used to make tent) = 2πrH + π r l
= (2H + l) π r
= (264/7) × (22 + 13)
= 264 × 5
A = 1320 m2
Thus, the area of the canvas used to make the tent is found to be 1320 m2.
(12) From a solid cylinder of height 30 cm and radius 7 cm, a conical cavity of height 24 cm and of base radius 7 cm is drilled out. Find the volume and total surface of the remaining solid.
Solution:
Given that, Radius of solid cylinder (r1) = 7 cm
Height of cylinder (H) = 30 cm
Height of cone (h) = 24 cm
Radius of cone (r) = 7 cm
Then, l2 = h2 + r2
l2 = 242 + 72
l2 = 576 + 49
l2 = 625
l = 25 cm
(Now, Volume of remaining solid) = (Volume of cylinder) – (Volume of the cone)
= πr2H – 1/3 πr2 h
V = πr2 (H – h/3)
V = πr2 (H – h/3)
= 22/7 × 72 × (30 – 24/3)
= 22 × 7 × (30 – 8)
V = 3388 cm3 is the required volume of remaining solid.
Now, (Total surface area of remaining solid) = (Curved surface area of cylinder) + (Surface area of top of cylinder) + (Curved surface area of the wane)
= 2 π r H + π r2 + π r l
= π r (2H + r + l)
= 22/7 × 7 (2 × 30 + 7 + 25)
= 22 × (60 + 32)
= 22 × 92
TSA = 2024 cm2
This is the required total surface area of the remaining solid.
(13) The following fig. shows a hemisphere of radius 5 cm surmounted by a right circular cone of base radius 5 cm. find the volume of the solid if the height of the cone is 7 cm. Give your answer correct to two places of decimal.
Solution:
Given that,
Height of conical portion (h) = 7cm
Radius of base of cone (r) = 5 cm
Now, Volume of solid = (Volume of hemisphere) + (Volume of the cone)
= (2/3 π r3) + 1/3 π r2 h
= 1/3 π r2 (2r + h)
= 1/3 × 22/7 × 52 × (10 + 7)
= 22/21 × 25 × 17
= 445.238
V = 445.24 cm3
This is the required volume of solid.
(15) A building is in the form of a cylinder surmounted by a hemisphere valted dome and contains 41 (19/21) m3 of air. If the internal diameter of the dome is equal to its total height above the floor, find the height of the building.
Solution:
Let us Consider, r be the radius of dome.
Then, internal diameter = 2r
Given that, internal diameter = total height
Then, total height of building = 2r
And height of hemispherical area = r
Then, height of cylindrical area = h = 2r – r = r
Now, (Volume of the building) = (Volume of cylindrical area) + (Volume of hemispherical area)
V = π r2 h + 2/3 π r3
= π r3 + 2/3 π r3
∵ h = r
V = π r3 (1 + 2/3)
= π r3 (3 + 2) / 3
V = (5/3) π r3
Given that, Volume of building = 41 (19/21) = 880/21
(5/3) π r3 = 880/21
5/3 × 22/7 × r3 = 880/21
r3 = 880 × 3 × 7/ (5 × 22 × 11)
r3 = 880/110
r3 = 8
r = 2m is the required radius
Now, height of the building =2r = 2 × 2 = 4 m
(16) A rocket is in the form of a right circular cylinder closed at the lower end and submitted by a Cone with the same radius as that of the cylinder. The diameter and the height of the cylinder are 6 cm and 12 cm respectively. If the slant height of the conical portion is 5 cm, find the total surface area and the volume of the rocket.
Solution:
Given that,
Height of cylindrical portion = 12 cm (H)
diameter = 6 cm = Radius = r = 3 cm
Slant height of conical portion (l) = 5 cm
Now, l2 = h2 + r2
h2 = l2 – r2
h = √25 – 9 = √16
= h = 4 cm
Now, (Total surface area of the rocket) = (Curved surface area of cylinder) + (base area of cylinder) + (Curved surface area of cone)
= 2πrH + πr2 + π r l
= π r (2H + r + l)
= 3.14 × 3 × (24 + 3 + 5)
= 3.14 × 3 × 32
= 301.44 cm2
Thus, the total surface area of the rocket is found to be 301.44 cm2
Now, Volume of rocket = (Volume of the cone) + (Volume of the cylinder)
= 1/3 π r2 h + π r2 H
= π r2 (h/3 + H)
= 3.14 × 32 × (4/3 + 12)
= 3.14 × 9 × (4 + 36)/3
= 3.14 × 3 × 40
Volume of rocket = 376.8 cm3
This is the required volume of the rocket.
(17) The following fig. represents a solid consisting of a right circular cylinder with a hemisphere at one end and a cone at the other. Their common radius is 7 cm. The height of the cylinder and the cone are each of 4 cm. Find the volume of the solid.
Solution:
Given that, Common radius RCC & hemisphere r = 7 cm
Height of the cone (h) = 4 cm
Height of the cylinder (H) = 4 cm
Then, Volume of solid = (Volume of the cone) + (Volume of the cylinder) + (Volume of the hemisphere)
= 1/3 π r2 h + π r2 H + 2/3 π r3
= π r2 [h/3 + H + 2r/3]
= 22/7 × 72 [4/3 + 4 + 14/3]
= 22 × 7 × (4 + 12 + 14) /3
Volume of solid = 1540 cm3
This is the required volume of the given solid.
(18) A solid is in the form of a right circular cylinder with a hemisphere at one end a cone at the other end. Their common diameter is 3.5 cm and the height of the cylindrical and conical portions are 10 cm and 6 cm respectively. Find the volume of the solid.
Solution:
Given that, Common diameter (d) = 3.5 cm
Then, Radius (r) = 1.75 cm
Height of cylindrical portion (h1) = 10 cm
Height of conical portion (h2) = 6 cm
Now, Volume of the solid is given by,
(Volume of solid) = (Volume of the cone) + (Volume of the cylinder) + (Volume of hemisphere)
= 1/3 π r2 h + π r2 H + 2/3 π r3
= π r2 [h/3 + H +20/3]
= 3.14 × 1.75 × 1.75 × [6/3 + 10 + (2 × 1.75) /3]
= 3.14 × 3.0625 × 13.167
= 9.61625 × 13.167
= 126.617 cm3
Volume of solid = 126.62 cm3
This is the required volume of the solid.
(20) The following fig. shows a model of solid consisting of a cylinder surmounted by a hemisphere at one end. If the model is drawn to a scale of 1:200, find
(i) The total surface area of the solid in π m2.
(ii) The volume of the solid in π liters.
Solution:
(i) Given that,
Height of cylindrical portion (h) = 8 cm
Scale = 1:200
Then, total surface are of solid is given by,
Total surface area = 2 π r2 + 2 π r h
= 2 π r (r + h)
= 2 × π × 3 (3 + 8)
= 6π × 11
Total surface area = 66π cm2 is the required total surface area.
Hence, total surface area of the solid is given by
Surface area of solid = 66π × (200)2/1
= (66 × 40,000)/ (100 × 100) π
Surface area of solid = 264π m2
This is the required surface area of solid.
(ii) The volume of solid = 2/3 π r3 + π r2 h
= π r2 [2/3r + h]
= π × 32 × [2/3 × 3 + 8]
= 9π (2 + 8)
Volume of solid = 90π
Given that, Scale = 1:200
Thus, Capacity of solid = 90π × (200)3
= 90π × 8000000
= 720000000 cm3
= (720000000/ (100 × 100 × 100)
= 720 π m3
= 720 π × 1000 liters
Capacity of solid = 720000 π liters
This is the required volume of the solid.
Exercise 17.5
(1) The diameter of a metallic sphere is 6 cm. The sphere is melted and drawn into a wire of uniform Cross – section. If the length of the wire is 36 m, find its radius.
Solution:
Given that, diameter of metallic sphere =6 cm
Radius of sphere (r) = 3 cm
Volume of sphere = 4/3 π r3
V = 4/3 π r3
= 4/3 × π × 33
V = 36 π cm3
Length of wire (h) = 36 m
Let ‘r’ be the radius of the wire,
π r2 h = 36 π
r2 × 36 × 100 = 36
r2 = 1/100 = (1/10)2 = 1/10 cm = r
r = 1 mm
(2) A solid metallic sphere of radius 6 cm is melted and made into a solid cylinder of height 32 cm. find the
(ii) Curved surface area of the cylinder.
Solution:
Given that, Radius of metallic sphere (R) = 6 cm
Height of cylinder (h) = 32 cm
Then, Volume of metallic cylinder is given by,
(i) Volume of cylinder = Volume of metallic sphere
π r2 h = 4/3 π r3
r2 (32) = 4/3 63
r2 = 4/3 × 6 × 6 ×/32 = 9
r = 3 cm is the required radius of the cylinder.
(ii) Curved Surface area of the cylinder is given by,
(Curved Surface area of cylinder) = 2 π r h
= 2 × 3.1 × 3 × 32
Curved surface area of cylinder = 595.2 cm2
(3) A Solid metallic hemisphere of radius 8cm is melted and recasted into right circular cone of base radius 6 cm. Determine the height of the cone.
Solution:
Given that, Radius of hemisphere (r) = 8 cm
Volume of hemisphere = 2/3 π r3
Radius of cone (R) = 6 cm
V = 2/3 π r3
= 2/3 × 83 × π
V = (1024/3) π cm3 is the required volume of hemisphere.
Given that, hemisphere is melted and converted into a cone then volume remains same.
= Volume of cone = 1/3 π R2 h
(1024)/3 π = 1/3 × π × 62 × h
36 h = 1024
h = 1024/ 36
h = 28.44 cm is the required height of the cone.
(4) A rectangular water tank of base 11m × 6m contains water up to a height of 5 m. If the water in the tank is transferred to a cylindrical tank of radius 3.5, find the height of the water level in the tank.
Solution:
Given that, Base of water tank = 11 m × 6 m
Height of water level in tank (h) = 5 m
And volume of water = 11 × 6 × 5 = 330 m3
Now, volume of cylindrical tank is given by,
(Volume of cylindrical tank) = π r2 h
330 = 22/7 × 3.5 × 3.5 × h
h = 330 × 7/22 × 12.25
h = 8.57 m
This is the required height of water level in the tank.
(5) The volume of a cone is the same as that of the cylinder whose height is 9 cm and diameter 40 cm. Find the radius of the base of the cone if its height is 108 cm.
Solution:
Given that, diameter of cylinder = 40 cm
Then, radius of cylinder = (r) = 20 cm
Height of cylinder (h) = 9 cm
Now, volume of cylinder is given by
Height of cone = 108 cm
V = π r2 h
V = π × 20 × 20 × 9
V = 3600 π cm3
And volume of cone = 1/3 π r2 h
3600 π = 1/3 × π × r2 × 108
r2 = 3 × 3600 π/ π × 108
= 3 × 3600/ 108
r = √100 = 10 cm
r = 10 cm
This is the required radius of the wane.
(7) A hemispherical bowl of diameter 7.2 cm is filled completely with chocolate sauce. This source is poured into an invested cone of radius 4.8 cm. Find the height of the cone.
Solution:
Given that, diameter of hemispherical bowl = 7.2 cm
Radius of hemispherical bowl (r) = 3.6 cm
Volume of hemispherical bowl = 2/3 π r3
= 2/3 × 22/7 × 3.6 × 3.6 × 3.6
V = 97.76 cm3
And, volume of cone = 1/3 π R2 h = 1/3 × 22/7 × 4.8 × 4.8 × h
V = 24.14 × h
But, Volume of cone = Volume of hemispherical bowl
24.14 × h = 97.76
h = 97.76/24.14
= 4.05 cm
This is the required height of the cone.
(9) A hollow copper pipe of inner diameter 6cm and outer diameter 10 cm is melted and changed into a solid circular. cylinder of the same height as that of the pipe. Find the diameter of the solid cylinder.
Solution:
Given that, (Inner diameter of hollow pipe) = 6 cm
Outer diameter of hollow pipe = 10 cm
Then, Inner radius (r) = 3 cm & outer radius (R) = 5 cm
Let us consider, ‘h’ be the height of the hollow pipe.
Volume of pipe = π R2 h – π r2 h
= π (52 – 32) h
= π h (25 – 9)
= π R2 h
16 = R2
R = 4 cm
Then, diameter of solid cylinder is found to be 8 cm.
(10) A hollow sphere of internal & external diameter 4 cm & 8 cm respectively, is melted into a cone of base diameter 8 cm. find the height of the cone.
Solution:
Given that,
Internal diameter of hollow Sphere = 4 cm
Then, inner radius of hollow sphere = 2 cm
and, outer radius of hollow sphere = 4 cm
Thus, Volume of the hollow sphere is given by
V = 4/3 π (43 – 23) —– (1)
Again, diameter of a cone = 8 cm
Then, Radius of a cone (r) = 4 cm
Then, volume of a cone = 1/3 π R2 h = 1/3 π × 42 × h —— (2)
But, given that the volume of hollow sphere & cone are equal.
From (1) & (2) = 4/3 π (43 – 23) = 1/3 π × 42 × h
4 × (64 – 8) = 16 × h
= h = 14 cm
This is the required height of the cone.
(12) A cylindrical can of internal diameter 21 cm contains water. A solid sphere whose diameter is 10.5 cm is lowered into the cylindrical can. The sphere is completely immersed in water. Calculate the rise in water level, assuming that no water overflows.
Solution:
Given that, inner diameter of cylindrical can = 21 cm
Then, inner radius (R) = 21/2 cm
diameter of sphere = 10.5 cm
Then, Radius of sphere (r) = 10.5/2 = 21/4 cm
Let us consider, h be the height of water level raised.
Rise in (water level) volume = Volume of sphere immersed
π R2 h = 4/3 × π r3
(21/2)2 π h = 4/3 × π × (21/4)3
21/2 × 21/2 × h = 4/3 × 21/4 × 21/4 × 21/4
= h = 21/12
= 7/14
h = 1.75 cm
This is the required rise in water level.
(13) There is water to a height of 14 cm in a cylindrical glass jar of radius 8 cm. Inside the water there is a sphere of diameter 12 cm completely immersed. By what height will the water 90 down when the sphere is removed?
Solution:
Given that, Radius of glass jar (R) = 8 cm
diameter of sphere = 12 cm
Then, Radius of sphere (r) = 6 cm
When, the sphere immersed in the water is removed from the jar then water level decreases.
Let us consider, h is the height of water level decreased.
Volume of water decreased = Volume of the sphere
π R2 h = 4/3 π r3
82 π h = 4/3 π 63
h = 4/3 × 6 × 6 × 6/ 8 × 8
h = 4.5 cm
This is the required height by which the water level decreased.
(15) A solid metallic circular cylinder of radius 14 cm and height 12 cm is melted and recast into small cubes of edge 2 cm. How many such cubes can be made from the solid cylinder.
Solution:
Given that, Radius of solid metallic cylinder (r) = 14 cm
Height of solid cylinder (h) = 12 cm
Now, Volume of cylinder = π r2 h
= 142 × 12 × π
= 196 × 12 × π
= 2352 × π
V = 7392 cm3 is the required volume of a cylinder.
Then, Volume of a cube = a3 = 23 = 8 cm3
V = 8 cm3 is the required volume of a cube.
Then, no. of cubes made from solid cylinder = 7392/8 = 924
(16) How many shots each hawing diameter 3 cm can be made from a cuboidal lead solid of dimensions 9 cm × 11 cm × 12 cm?
Solution:
Given that, diameter of shot = 3 cm
dimensions of cuboidal solid = 9cm × 11cm × 12cm
Then, Volume of cuboidal solid = 9 × 11 × 12 = 1188 cm3
Radius of shot (r) = 3/2 cm
Volume of one shot = 4/3 π r3
V = 4/3 π r3
= 4/3 × π × 1.53
= 4/3 × 22/7 × 1.53
V = 297/21 cm3 is the required volume of one shot.
1188 × 21/297
= 84
Thus, the no. of shots made from cuboidal lead of solid is found to be 84.
(17) A solid metal cylinder of radius 4 cm and height 21 cm is melted down and recast into sphere of radius 3.5 cm. Calculate the no. of spheres that can be made.
Solution:
Given that, Radius of a solid metallic cylinder} (r) = 14 cm
Height of cylinder (h) = 21 cm
Then, Volume of cylinder = π r2 h
V = 22/7 × 14 × 14 × 21
V = 12936 cm3
Given that, Radius of sphere (R) = 3.5 cm
Then, Volume of sphere = 4/3 π R3
= 4/3 × 22/7 × 3.5 × 3.5 × 3.5
= 11 × 49/3
V = 539/ 3 cm3 is the required volume of sphere.
Then, no. of sphere made = Volume of metal cylinder/ Volume of Sphere
= 12936 × 3/539
No. of spheres made = 72
(18) A metallic sphere of radius 10.5 cm is melted and then recast into small cense, each of radius 3.5 cm and height 3 cm. Find the no. of cones thus obtained.
Solution:
Given that, Radius of metallic sphere (r) = 10.5 cm
Radius of cone (r) = 3.5 cm
Height of cone (h) = 3 cm
Then, Volume of sphere is given by
V = 4/3 π R3
= 4/3 × π × 10.53
V = 1543.5 π cm3 is the required volume of sphere.
And Volume of cone is given by
V = 1/3 π r2 h = 1/3 × π × 3.52 × h
V = 12.25 π cm3 is the required volume of cone.
Now, (No. of cones made from sphere) = Volume of sphere/ Volume of cone
= 1543.5 π/12.25 π
(no. of cones made from sphere) = 126
(19) A certain no. of metallic cones each of radius 2 cm and height 3 cm are melted and recast in a solid sphere of radius 6 cm. Find the no. of cones.
Solution:
Given that, radius of cone (r) = 2 cm
Height of cone (h) = 3 cm
Radius of sphere (R) = 6 cm
Volume of cone = 1/3 π r2 h
V = 1/3 × π × 22 × 3
V = 4 π cm3 is the required volume of cone
And Volume of sphere = 4/3 π R3
V = 4/3 × π × 63
V = 288 π cm3 is the required volume of sphere.
Then, no. of cones made from sphere = Volume of sphere/ Volume of cone
= 288 π / 4 π
No. of cones made from sphere = 72
(20) A cylindrical can whose base is horizontal and of radius 3.5 cm contains sufficient water so that when a sphere is placed in the can, the water just covers the sphere. Given that, the sphere just fits into can. Calculate (i) the total surface area of the can in contact with water when sphere is in it.
(ii) The depth of water in the can before the sphere was put into the can. Give your answer as proper fractions.
Solution:
Given that, Radius of cylindrical can = 3.5 cm
Radius of the Sphere = 3.5 cm
Water level height in the can = 3.5 × 2 = 7 cm
Height of cylinder (h) = 7 cm
(i) Then, (Total surface area of can in contact with water) = (Curved surface area of cylinder) + (base area of cylinder)
= 2 π r h + π r2
= π r (2h + r)
= 22/7 × 3.5 × (2 × 7 + 3.5)
= 11 × 17.5
TSA = 192.5 cm2
This is the required surface area of can in contact with water.
(ii) Let ‘d’ be the depth of the water before putting sphere in it.
Then, Volume of cylindrical can = (Volume of sphere) + (Volume of water)
π r2 h = 4/3 π r3 + π r2 d
π r2 h = π r2 [4/3r + d]
h = 4/3r + d
d = h – 4/3 r = 7 – 4/3 × 3.5
d = (21 – 14)/3
d = 7/3 is the required depth of water.
Updated: February 21, 2023 — 2:20 pm |
# How do you evaluate \frac { 1} { 2} + \frac { 2} { 5} + \frac { 2} { 25}?
Jun 10, 2017
$\frac{49}{50}$
#### Explanation:
$\frac{1}{2} + \frac{2}{5} + \frac{2}{25}$
to add or subtract fractions, we need to first get a common denominator.
We find the common denominator by finding the least common multiple of each of the denominators, and then do the same to the numerators (click the link to find out how to do this).
The least common multiple of $2 , 5 \text{ and " 25 " is } \textcolor{\lim e}{50}$
$2 \times 25 = 50$
$5 \times 10 = 50$
$25 \times 2 = 50$
Now we can do the same to the numerators.
$\frac{1}{2} = \frac{x}{50}$
$\frac{2}{5} = \frac{y}{50}$
$\frac{2}{25} = \frac{z}{50}$
$\frac{1 \times a = x}{2 \times a = 50}$
$\frac{2 \times b = y}{5 \times b = 50}$
$\frac{2 \times c = z}{25 \times c = 50}$
$\frac{1 \times 25 = x}{2 \times 25 = 50}$
$\frac{2 \times 10 = y}{5 \times 10 = 50}$
$\frac{2 \times 2 = z}{25 \times 2 = 50}$
$1 \times 25 = 25$
$2 \times 10 = 20$
$2 \times 2 = 4$
$\frac{1}{2} = \frac{25}{50}$
$\frac{2}{5} = \frac{20}{50}$
$\frac{2}{25} = \frac{4}{50}$
Now we can do the equation
$\frac{1}{2} + \frac{2}{5} + \frac{2}{25}$
$\frac{25}{50} + \frac{20}{50} + \frac{4}{50}$
$\frac{45}{50} + \frac{4}{50}$
color(blue)(49/50
Because we can not simplify the fraction any further, this is our final answer.
Hope this helps :) |
# Solution of functional equation $f(x/f(x)) = 1/f(x)$?
I've been trying to add math rigor to a solution of the functional equation in [1], eq. (22). It is: $$f\left(\frac{x}{f(x)}\right) = \frac{1}{f(x)}\,,$$ where you know that $f(0)=1$ and $f(-x) = f(x)$.
I've been trying to fill in the missing steps. Let's use a substitution $g(x) = \frac{x}{f(x)}$. So we rewrite the functional equation as $$x = \frac{\frac{x}{f(x)}}{f\left(\frac{x}{f(x)}\right)}$$ and get $$x = g(g(x))\,.$$ Then we calculate $g'(0)$ as follows: $$g'(0) = \left.\left(\frac{x}{f(x)} \right)'\right|_{x=0} = \left.\frac{f(x)-xf'(x)}{f^2(x)}\right|_{x=0} = \frac{f(0)}{f^2(0)} = 1\,.$$ Also we can use that $g(x)$ is odd $$g(-x) = \frac{-x}{f(-x)} = -\frac{x}{f(x)} = -g(x)\,.$$ Whenever $g(g(x)) = x$, I have found that this is called involution. That can have many solutions, but using $g(-x)=-g(x)$ and $g'(0)=1$, there should be a way to prove that $g(x) = x$, from which $f(x)=1$ as the only solution.
The article [1] just says, that since $g(x) = g^{-1}(x)$, the graph of $g(x)$ and its inverse must be symmetric along the $y=x$ axis, and since the tangent at $x=0$ is equal to this same line, then $g(x)=x$ must be the only solution. They do mention that this is not completely rigorous proof.
You can assume that $f(x)$ is differentiable. It'd be nice if it can be proven only for continuous functions $f(x)$, but any proof at first would be a good start, even with stronger assumptions.
Update: another idea is to use the fact, that if $a$ and $b$ are two points in the domain of the $g$ function for which $g(a)=g(b)$, then it follows that $g(g(a)) = g(g(b))$ and $a = b$, which proves that the function is one-to-one. Given that $g$ is continuous, it means that it is strictly increasing or decreasing [2]. From that, let's say it's increasing, then we can use $y=g(x)$, if $y < x$, then $g(y) < g(x)$ from which $g(g(x)) < y$ and $x < y$ which is a contradiction. Similarly for $y > x$. So we must have $y=x$, from which $g(x)=x$. If $g$ is decreasing, then we set $h(x)=-g(x)$, obtain $h(x)=x$ and $g(x)=-x$. Unless I made a mistake we got $g(x)=\pm x$. And using $g'(0)=1$, we get $g(x)=x$ as the only solution. But I don't feel very good about this proof yet.
[1] Levy-Leblond, J.-M. (1976). One more derivation of the Lorentz transformation. American Journal of Physics, 44(3), 271–277.
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Because $g(x)=1/x$ is also a solution of $g(g(x))$ and somehow it got eliminated. So something is not right. (Of course, it would get eliminated later anyway due to $g'(0)=1$, but that's not the point.) – Ondřej Čertík Jan 16 '13 at 7:58
Because $g(x)=1/x$ is not defined on $\mathbf R$, only on $\mathbf R^*$. Notice that it's also neither increasing nor decreasing. The answer you linked to uses the intermediate value theorem, which you can only apply on an interval, and $\mathbf R^*$ is not an interval. – jathd Jan 16 '13 at 8:01
Your proof in your update is almost certainly exactly what the authors had in mind, but didn't type out. Briefly: if $g$ is not $x\mapsto x$, then there is some $h>0$ such that $g(h)$ is close to $h>0$, yet $g(h)\neq h$. Then the reflection of the point $(h,g(h))$ will be on the other side of the line $y=x$, contradicting the monotonicity of the function.$\Box$ – Samuel Jan 16 '13 at 8:13
Two things. For the increasing case, you assume (without saying it) that $g$ is defined on an interval $I$, and that $g$ sends $I$ into $I$ (because you use $y<x\Rightarrow g(y)<g(x)$), which explain $g(x)=-1/x$ on $(0,\infty)$. The other thing is that for the decreasing case, you set $h(x)=-g(x)$, and if you write that case in detail you'll see that you use $g(-x)=-g(x)$, which you can't say about $g(x)=1/x$ on $(0,\infty)$, because if $x\in(0,\infty)$ then $-x\notin(0,\infty)$. – jathd Jan 16 '13 at 8:13
Given that $f(x)$ is symmetric, you also know that all odd derivatives of $1/f(x/f(x))$ are zero. If you manage to express th even ones, $(1/f(x/f(x)))^{(2n)}$, analytically, you might get some information out, maybe even conclude that all are zero such that $f(x)=1$. – NikolajK Jan 16 '13 at 13:07
## 3 Answers
Hmm, it seems to me that the only solution is the constant function $f(x)=1$
Let's assume that $f(x)$ has a power series, then
• the constant term must be 1, because $f(0)=1$
• we have only nonzero coefficients at the powers of x where the exponents are even because $f(x)=f(-x)$
so we assume $$f(x) = 1 + ax^2 + bx^4 + cx^6 + O(x^8)$$ Then for the rhs of your defining equation we have $${1 \over f(x)} = 1 - ax^2 + (a^2-b) x^4 + (-c + (-a^3 + 2ba)) x^6 + O(x^8)$$ For the lhs of your defining equation we have $$f(x {1 \over f(x)}) = 1 + ax^2 + (b -2 a^2)x^4 + (c + 3 a^3 - 6ab) x^6 + O(x^8)$$ and equating coefficients requires
• $a=0 \to (f"(0)=0)$ and
then $\qquad \qquad {1 \over f(x)} = 1 -b x^4 -c x^6 + O(x^8)$
must equal $f(x {1 \over f(x)}) = 1 + b x^4 + c x^6 + O(x^8)$
• thus is required that $b=0 \to f^{(4)}(0)=0$ and
then $\qquad \qquad {1 \over f(x)} = 1 -c x^6 + O(x^8)$
must equal $f(x {1 \over f(x)}) = 1 + c x^6 + O(x^8)$
• thus is required that $c=0 \to f^{(6)}(0)=0$ ...
... and I assume, that all following coefficients must equal zero. Thus: the only solution should be $$f(x)=1$$ and your $$g(x)={x \over f(x)}=x$$
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Very nice! So if the function $f(x)$ is analytic, then you prove that $f(x)=1$. Assuming only that $f(x)$ is continuous, then the only possible other solutions are not analytic. That helps a lot. – Ondřej Čertík Jan 16 '13 at 16:15
The constant function $-1$ works too. – alex.jordan Jan 15 '14 at 15:19
Some results described above.
(i) $g$ is monotone.
(ii) $g$ = $g^{-1}$
(iii) $g$ is odd.
(iv) $g$ differentiable at $0$
Summary:
Suppose $g$ is not the identity. Then for $\forall r$, we know $g$ has a symmetrical counterpart to $g(r)$ w.r.t. $y=x$ from (ii). Now consider, $(r,g(r))$ for $r>0$ and his symmetric counterpart. Later consider $(-r,g(-r))$ and his symmetric counterpart. There is no way to ensure monotonicity between these coordinates for, in any case, the points will form a rectangle with two sides parallel and two perpendicular to $y=x$. So in adhering to the symmetry, $g$ must either fail to be a function or $g$ must violate monotonicity somewhere by crossing $y=x$. So $g$ must be the identity. Thus $f=1$.
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Given what you have already established, it will suffice to show that if $g$ is a continuous function with $g(g(x)) = x$, and $g(0) = 0,\ g'(0) = 1$, then $g(x) = x$ for all $x$. (Note that $g'(0) = 1$ because $g(x) = x + x(\frac{1}{f(x)} - 1) = x + o(x)$, using only continuity of $f$ at $0$).
Clearly, $g$ is injective: If $g(x_0) = g(x_1)$ then $x_0 = g(g(x_0)) = g(g(x_1)) = x_1$.
Thus, $g$ is monotonous: If it were not, it would have a local extreme at some $x_0$, and would not be injective in the neighbourhood of $x_0$. Because $g$ is increasing at $0$, it is therefore increasing on the whole domain.
We will now show that $g(x) = x$ for all $x$. Clearly, this holds for $x=0$. For a proof by contradiction, suppose that we have $x_0$ with $g(x_0) \neq x_0$. Let $x_1 := g(x_0)$. Then $x_0,x_1$ are distinct, and $g$ ''swaps'' these two points: $g(x_0) = x_1,\ g(x_1) = x_0$. Suppose without loss of generality that $x_0 < x_1$. We have $g(x_0) = x_1 > x_0 = g(x_1)$. But this contradicts the assumption that $g$ is increasing. Thus, we indeed have $g(x) = x$ for all $x$. $\square$
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# EXAMPLE 2 Checking Solutions Tell whether (7, 6) is a solution of x + 3y = 14. – x + 3y = 14 Write original equation. 7 + 3( 6) = 14 – ? Substitute 7 for.
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EXAMPLE 2 Checking Solutions Tell whether (7, 6) is a solution of x + 3y = 14. – x + 3y = 14 Write original equation. 7 + 3( 6) = 14 – ? Substitute 7 for x and 6 for y. – 7 + ( 18) = 14 – ? Simplify. 11 = 14 – Solution does not check. ANSWER The ordered pair (7, 6) is not a solution of x + 3y = 14. –
EXAMPLE 3 Finding Solutions of an Equation Write the equation 4x + y = 15 in function form. Then list four solutions. SOLUTION STEP 1 Rewrite the equation in function form. 4x + y = 15 Write original equation. y = 15 4x– Subtract 4x from each side.
EXAMPLE 3 Finding Solutions of an Equation STEP 2 Choose several values to substitute for x. Then solve for y. x -value Substitute for x. Evaluate.Solution x = 0 x = 1 x = 2 y = 19 y = 15 y = 11 y = 7 (0, 15) (1, 11) (2, 7) x = 1 – ( 1, 19) – y = 15 4( 1) –– y = 15 4(0) – y = 15 4(2) – ANSWER Four solutions are ( 1, 19), (0, 15), (1, 11), and (2, 7). – – y = 15 4(1)
GUIDED PRACTICE for Examples 2 and 3 Tell whether the ordered pair is a solution of the equation. ANSWER The ordered pair ( 6, 5 ) is not a solution of y = 3x 7. – 2. y = 3x 7; (6, 5) –
GUIDED PRACTICE for Examples 2 and 3 ANSWER The ordered pair ( 4, 1 ) is a solution of –2x – 4y = 12. –– 2x 4y = 12; (–4, 1) –– – 3.
GUIDED PRACTICE for Examples 2 and 3 List four solutions of the equation. y = 2x + 6– 4. x -value Substitute for x. Evaluate.Solution x = 0 x = 1 y = 10 y = 8 y = 6 y = 4 (0, 6) (1, 4) x = 2 – x = 1 – ( 2, 10) – ( 1, 8) – y = 2 2 + 6 – – – y = 2 1 + 6 – y = 2 0 + 6 – y = 2 1 + 6 – ANSWER
GUIDED PRACTICE for Examples 2 and 3 5. 3x + y = 4 x -value Substitute for x. Evaluate.Solution x = 0 x = 1 y = 10 y = 7 y = 4 y = 1 (0, 4) (1, 1) x = 2 – x = 1 – y = 4 3 0 – y = 4 3 1 – y = 4 3 2 –– y = 4 3 1 –– ( 2, 10) – ( 1, 7) – ANSWER
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# Polynomials and Factoring
## Count terms to identify polynomials
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Practice Polynomials and Factoring
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Recognize and Identify Monomials, Binomials and Trinomials
Have you ever tried to classify numbers? Take a look at this dilemma.
Sam saw this expression in his math book.
x28\begin{align*}x^2-8\end{align*}
He isn't sure how to classify this expression. Do you know?
Expressions like this one are the focus of this Concept. Pay attention and you will know how to identify this expression by the end of the Concept.
### Guidance
Sometimes, you will see an expression or an equation that has exponents and variables in it. These expressions and equations can have more than one variable and sometimes more than one exponent in them. To understand how to work with these variables and exponents, we have to understand polynomials.
A polynomial is an algebraic expression that shows the sum of monomials.
Yes. They are new words. As we begin to work with polynomials, you will have to learn to work with brand new words.
Write each new word and its definition in your notebook.
A monomial is an expression in which variables and constants may stand alone or be multiplied. A monomial cannot have a variable in the denominator. We can think of a monomial as being one term.
To understand these new terms better, let’s look at some word prefixes so that we can better understand the new terms.
Word Monoplane Biplane Triplane Polygon
Definition An airplane with one wing or one set of wings. An airplane with two sets of wings An airplane with three sets of wings A shape with many sides.
Prefix Mono means one Bi means two Tri means three Poly means many.
Just like with airplanes, in math we can use these prefixes too. Each prefix will give us a hint as to the type of expression that we are dealing with.
Here are some monomials: 5x32x5x2y\begin{align*}5 \quad x^3 \quad -2 x^5 \quad x^2y\end{align*}
Since the prefix mono means one, a monomial is a single piece or term. The prefix poly means many. So the word polynomial refers to one or more than one term in an expression. The relationship between these terms may be sums or difference.
Here are some polynomials: x2+53x8+4x57a2+9b4b3+6\begin{align*}x^2+ 5 \qquad 3x-8+4x^5 \qquad -7a^2+9b-4b^3+6\end{align*}
We call an expression with a single term a monomial, an expression with two terms is a binomial, and an expression with three terms is a trinomial. An expression with more than three terms is named simply by its number of terms—“five-term polynomial.”
From the information above, we can name the expressions as follows:
Number of Terms 1 2 3 4
Name monomial binomial trinomial four-term polynomial
Expression 2x5\begin{align*}-2x^5\end{align*} x2+5\begin{align*}x^2+5\end{align*} 3x8+4x5\begin{align*}3x-8+4x^5\end{align*} 7a2+9b4b3+6\begin{align*}-7a^2+9b-4b^3+6\end{align*}
Identify each expression.
#### Example A
4x38\begin{align*}4x^3-8\end{align*}
Solution: Binomial
#### Example B
x2+3x+9\begin{align*}x^2+3x+9\end{align*}
Solution: Trinomial
#### Example C
6xy\begin{align*}6xy\end{align*}
Solution: Monomial
Now let's go back to the dilemma from the beginning of the Concept.
x28\begin{align*}x^2-8\end{align*}
This expression has two terms, therefore it is a binomial.
### Vocabulary
Polynomial
an algebraic expression that shows the sum of monomials. A polynomial can also be named when there are more than three terms present.
Monomial
an expression where there is one term.
Binomial
an expression where there are two terms.
Trinomial
an expression where there are three terms.
### Guided Practice
Here is one for you to try on your own.
How would you identify the following expression?
4x2x8y+4\begin{align*}4x^2x-8y+4\end{align*}
This expression has many terms. Therefore, it is called a polynomial.
### Practice
Directions: Use the chart to identify each term with the correct label.
Number of Terms 1 2 3 4
Name monomial binomial trinomial four-term polynomial
Expression 2x5\begin{align*}-2x^5\end{align*} x2+5\begin{align*}x^2+5\end{align*} 3x8+4x5\begin{align*}3x-8+4x^5\end{align*} 7a2+9b4b3+6\begin{align*}-7a^2+9b-4b^3+6\end{align*}
1. \begin{align*}4x^2\end{align*}
2. \begin{align*}3x+7\end{align*}
3. \begin{align*}9x^2+6y\end{align*}
4. \begin{align*}x^2+2y^2+8\end{align*}
5. \begin{align*}5c^3\end{align*}
6. \begin{align*}3x^2+4x+3y^2+7\end{align*}
7. \begin{align*}4x+3xy+9\end{align*}
8. \begin{align*}2x^2+7y + 9\end{align*}
9. \begin{align*}14xy\end{align*}
10. \begin{align*}4x^2+5x-9\end{align*}
11. \begin{align*}5x^3-4x^2+3x-10\end{align*}
12. \begin{align*}4x\end{align*}
13. \begin{align*}16x+4\end{align*}
14. \begin{align*}18x^2+5x-8\end{align*}
15. \begin{align*}9xyz\end{align*}
16. \begin{align*}5xy-6x\end{align*}
17. \begin{align*}18x^2-9x\end{align*}
### Vocabulary Language: English
Binomial
Binomial
A binomial is an expression with two terms. The prefix 'bi' means 'two'.
Monomial
Monomial
A monomial is an expression made up of only one term.
Polynomial
Polynomial
A polynomial is an expression with at least one algebraic term, but which does not indicate division by a variable or contain variables with fractional exponents.
Trinomial
Trinomial
A trinomial is a mathematical expression with three terms. |
Mathematics Evaluation of Definite Integrals by Substitution For CBSE-NCERT
Click for Only Video
### Topic Covered
color{blue}star Evaluation of Definite Integrals by Substitution
### Evaluation of Definite Integrals by Substitution
To evaluate color {red} {int_a^b f(x) dx} , by substitution, the steps could be as follows:
1. Consider the integral without limits and substitute, y = f (x) or x = g(y) to reduce the given integral to a known form.
2. Integrate the new integrand with respect to the new variable without mentioning the constant of integration.
3. Resubstitute for the new variable and write the answer in terms of the original variable.
4. Find the values of answers obtained in (3) at the given limits of integral and find the difference of the values at the upper and lower limits.
In order to quicken this method, we can proceed as follows: After performing steps 1, and 2, there is no need of step 3. Here, the integral will be kept in the new variable itself, and the limits of the integral will accordingly be changed, so that we can perform the last step.
Q 3115580460
Evaluate ∫_(0)^1 (tan^(-1)x )/( 1+x^2) dx
Class 12 Chapter 7 Example 29
Solution:
Let t = tan^( – 1) x, then dt = 1/(1+x^2) dx . The new limits are, when x = 0, t = 0 and
when x =1 , t = pi/4 . Thus, as x varies from 0 to 1, t varies from 0 to pi/4
Therefore ∫_(0)^1 (tan^(-1) x)/(1+x^2) dx = ∫_(0)^(pi/4) t dt [t^2/2]_(0)^(pi/4) =1/2 [(pi^2)/16 -0 ] = (pi^2 )/(32)
Q 3115480369
Evaluate int_(-1)^1 5x^4 sqrt (x^5 +1 ) dx
Class 12 Chapter 7 Example 28
Solution:
Put t = x^5 + 1, then dt = 5x^4 dx.
Therefore, ∫ 5x^4 sqrt (x^5 +1) dx = ∫ sqrt t dt = 2/3 t^3/2 = 2/3 (x^5 +1)^3/2
Hence, ∫_(-1)^1 5x^4 sqrt (x^5 +1)dx = 2/3 [ (x^5 +1 )^3/2]_(-1)^1
= 2/3 [ (1^5 +1 )^3/2 - ( (-1)^5 + 1 )^3/2]
= 2/3 [2^3/2 - 0^3/2] = 2/3 (2 sqrt 2) = (4 sqrt 2 )/3
Alternatively, first we transform the integral and then evaluate the transformed integral
with new limits.
Let t = x^5 + 1. Then dt = 5 x^4 dx.
Note that, when x = – 1, t = 0 and when x = 1, t = 2
Thus, as x varies from – 1 to 1, t varies from 0 to 2
Therefore ∫_(-1)^1 5x^4 sqrt (x^5 +1) dx = ∫_(0)^2 sqrt t dt
=2/3 [t^3/2]_(0)^2 = 2/3 [2^3/2 - 0^3/2] = 2/3 (2 sqrt 2) = (4 sqrt 2)/3 |
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