text stringlengths 22 1.01M |
|---|
# How do you solve using elimination of 3x-2y=-1 and 3x-4y=9?
Jun 8, 2018
Multiply the second equation by $- 1$ , and add the two equations together to get:
$x = - \frac{11}{3}$ and $y = - 5$
#### Explanation:
Multiply the second equation by $- 1$ and add the two equations together:
$+ 3 x - 2 y = - 1$
$- 3 x + 4 y = - 9$
$0 + 2 y = - 10$
This eliminates the $x$ variable, so we can find $y$ :
$y = - 5$
Now substitute $y$ in either equation to obtain $x$ :
$3 x - 2 \cdot \left(- 5\right) = - 1$
$3 x = - 1 - 10 = - 11$
$x = - \frac{11}{3}$ |
# RD Sharma Solutions - Chapter 26 - Data Handling-IV (Probability), Class 8, Maths Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8
## Class 8: RD Sharma Solutions - Chapter 26 - Data Handling-IV (Probability), Class 8, Maths Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8
The document RD Sharma Solutions - Chapter 26 - Data Handling-IV (Probability), Class 8, Maths Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
All you need of Class 8 at this link: Class 8
PAGE NO 26. 14:
Question 1:
The probability that it will rain tomorrow is 0. 85. What is the probability that it will not rain tomorrow?
Ans.
Let A be the event of raining tomorrow. The probability that it will rain tomorrow, P(A), is 0. 85.
Since the event of raining tomorrow and not raining tomorrow are complementary to each other, the probability of not raining tomorrow is:
P(A) = 1 − P(A) = 1 − 0. 85 = 0. 15
Let A be the event of raining tomorrow. The probability that it will rain tomorrow, PA, is 0. 85.
Since the event of raining tomorrow and not raining tomorrow are complementary to each other, the probability of not raining tomorrow is:
= 1 - PA = 1 - 0. 85 = 0. 15
Question 2:
A die is thrown. Find the probability of getting:
(i) a prime number
(ii) 2 or 4
(iii) a multiple of 2 or 3
Ans.
When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5 and 6. Thus, the sample space will be as follows: S = {1,2,3,4,5,6}
(i) Let A be the event of getting a prime number. There are 3 prime numbers (2,3 and 5) in the sample space. Thus, the number of favourable outcomes is 3.
Hence, the probability of getting a prime number is as follows:P(A) = = 3/6 = 1/2
(ii) Let A be the event of getting a two or four. Two or four occurs once in a single roll. Therefore, the total number of favourable outcomes is 2.
Hence, the probability of getting 2 or 4 is as follows:P(A) = 2/6 = 1/3
(iii) Let A be the event of getting multiples of 2 or 3. Here, the multiples of 2 are 2, 4, 6 and the multiples of 3 are 3 and 6. Therefore, the favourable outcomes are 2, 3, 4 and 6.
Hence, the probability of getting a multiple of 2 or 3 is as follows:P(A) = = 4/6 = 2/3
PAGE NO 26. 15:
Question 3:
In a simultaneous throw of a pair of dice, find the probability of getting:
(i) 8 as the sum
(ii) a doublet
(iii) a doublet of prime numbers
(iv) a doublet of odd numbers
(v) a sum greater than 9
(vi) an even number on first
(vii) an even number on one and a multiple of 3 on the other
(viii) neither 9 nor 11 as the sum of the numbers on the faces
(ix) a sum less than 6
(x) a sum less than 7
(xi) a sum more than 7
(xii) at least once
(xiii) a number other than 5 on any dice.
Ans.
When a pair of dice is thrown simultaneously, the sample space will be as follows:S = {(1,1), (1,2), (1,3), (1,4),⋯(6,5), (6,6)} Hence, the total number of outcomes is 36.
(i) Let A be the event of getting pairs whose sum is 8. Now, the pairs whose sum is 8 are (2,6), (3,5), (4,4), (5,3) and (6,2).
Therefore, the total number of favourable outcomes is 5.
∴ P(A) == 5/36 (ii) Let A be the event of getting doublets in the sample space.
The doublets in the sample space are (1,1), (2,2), (3,3), (4,4), (5,5) and (6,6). Hence, the number of favourable outcomes is 6.
∴ P(A) = = 6/36 = 1/6 (iii) Let A be the event of getting doublets of prime numbers in the sample space.
The doublets of prime numbers in the sample space are (2,2), (3,3) and (5,5). Hence, the number of favourable outcomes is 3.
∴ P(A) = = 3/36 = 1/12
(iv) Let A be the event of getting doublets of odd numbers in the sample space. The doublets of odd numbers in the sample space are (1,1), (3,3) and (5,5).
Hence, the number of favourable outcomes is 3.
∴ P(A) = = 3/36 = 1/12
(v)Let A be the event of getting pairs whose sum is greater than 9. The pairs whose sum is greater than 9 are (4,6),(5,5), (5,6),(6,4),(6,5) and (6,6).
Hence, the number of favorable outcomes is 6.
∴ P(A) = = 6/36 = 1/6
(vi)Let A be the event of getting pairs who has even numbers on first in the sample space. The pairs who has even numbers on first are: (2,1), (2,2),…(2,6), (4,1),⋯,(4,6), (6,1),⋯(6,6).
Hence, the number of favourable outcomes is 18.
∴ P(A) = = 18/36 = 1/3
(vii)Let A be the event of getting pairs with an even number on one die and a multiple of 3 on the other.
The pairs with an even number on one die and a multiple of 3 on the other are (2,3), (2,6), (4,3), (4,6), (6,3) and (6,6). Hence, the number of favourable outcomes is 6.
∴ P(A) = = 6/36 = 1/6
(viii) Let A be the event of getting pairs whose sum is 9 or 11. The pairs whose sum is 9 are (3,6), (4,5), (5,4) and (6,3). And, the pairs whose sum is 11 are (5,6) and (6,5). Hence, the number of favourable outcomes is 6.
∴ P(A) = = 6/36 = 1/6
∴ P(sum of the pairs with neither 9 nor 11) = 1 − P(sum of the pairs having 9 or 11) = 1 − 16 = 56
(ix) Let A be the event of getting pairs whose sum is less than 6. The pairs whose sum is less than 6 are (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2) and (4,1). Hence, the number of favourable outcomes is 10.
∴ P(A) = = 10/36 = 5/18
(x) Let A be the event of getting pairs whose sum is less than 7.
The pairs whose sum is less than 7 are (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (4,1), (4,2) and (5,1).
Hence, the number of favourable outcomes is 15.
∴ P(A) = = 15/36 = 5/12
(xi)Let A be the event of getting pairs whose sum is more than 7.
The pairs whose sum is more than 7 are (2,6), (3,5), (3,6), (4,4), (4,5), (4,6), (5,3), (5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5) and (6,6).
Hence, the number of favourable outcomes is 15.
∴ P(A) = = 15/36 = 5/12
(xii) Incomplete question
(xiii)Let A be the event of getting pairs that has the number 5.
The pairs that has the number 5 are (1,5), (2,5), (3,5), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4) and (6,6).
Hence, the number of favourable outcomes is 11.
∴ P(A) = = 11/36
= 1−P(A) = 1−1136 = 25/36
Question 4:
Three coins are tossed together. Find the probability of getting:
(iii) at least one head and one tail
(iv) no tails
Ans.
When 3 coins are tossed together, the outcomes are as follows:S = {(h,h,h), (h,h,t), (h,t,h), (h,t,t), (t,h,h), (t,h,t), (t,t,h), (t,t,t)}Therefore, the total number of outcomes is 8.
(i) Let A be the event of getting triplets having exactly 2 heads. Triplets having exactly 2 heads: (h,h,t), (h,t,h), (t,h,h)Therefore, the total number of favourable outcomes is 3. P(A) = = 3/8
(ii) Let A be the event of getting triplets having at least 2 heads.
Triplets having at least 2 heads: (h,h,t), (h,t,h), (t,h,h), (h,h,h)
Therefore, the total number of favourable outcomes is 4.
P(A) = = 4/8 = 1/2
(iii) Let A be the event of getting triplets having at least one head and one tail.
Triplets having at least one head and one tail: (h,h,t), (h,t,h), (t,h,h), (h,h,t), (t,t,h), (t,h,t)
Therefore, the total number of favourable outcomes is 6.
P(A) = = 6/8 = 3/4
(iv) Let A be the event of getting triplets having no tail. Triplets having no tail: (h,h,h)
Therefore, the total number of favourable outcomes is 1.
P(A) = = 1/8
Question 5:
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
(i) a black king
(ii) either a black card or a king
(iii) black and a king
(iv) a jack, queen or a king
(v) neither a heart nor a king
(vii) neither an ace nor a king
(viii) neither a red card nor a queen.
(ix) other than an ace
(x) a ten
(xii) a black card
(xiii) the seven of clubs
(xiv) jack
(xvi) a queen
(xvii) a heart
(xviii) a red card
Ans.
(i) There are two black kings, spade and clover. Hence, the probability that the drawn card is a black king is: 2/52 = 1/26
(ii) There are 26 black cards and 4 kings, but two kings are already black. Hence, we only need to count the red kings. Thus, the probability is: (26+2)/52 = 7/13
(iii) This question is exactly the same as part (i) . Hence, the probability is: 2/52 = 1/26
(iv) There are 4 jacks, 4 queens and 4 kings in a deck. Hence, the probability of drawing either of them is: (4+4+4)/52 = 3/13
(v) This means that we have to leave the hearts and the kings out. There are 13 hearts and 3 kings (other than that of hearts). Hence, the probability of drawing neither a heart nor a king is: (52-13-3)/52 = 9/13
(vi) There are 13 spades and 3 aces (other than that of spades). Hence the probability is: (13+3)/52 = 4/13
(vii) This means that we have to leave the aces and the kings out. There are 4 aces and 4 kings. Hence, the probability of drawing neither an ace nor a king is: (52−-4−-4)/52 = 11/13.
(viii) This means that we have to leave the red cards and the queens out. There are 26 red cards and 2 queens (only black queens are counted since the reds are already counted among the red cards). Hence, the probability of drawing neither a red card nor a queen is: (52-26-2)/52 = 6/13
(ix) It means that we have to leave out the aces. Since there are 4 aces, then the probability is (52−4)/52 = 12/13
(x) Since there are four 10s, the probability is: 4/52 = 1/13
(xi) Since there are 13 spades, the probability is: 13/52 = 1/4
(xii) Since there are 26 black cards, the probability is: 26/52 = 1/2
(xiii) There is only one card named seven of clubs. Hence, the probability is 1/52.
(xiv) Since there are 4 jacks, the probability is: 4/52 = 1/13
(xv) There is only 1 card named ace of spade. Hence, the probability is 1/52.
(xvi) Since there are 4 queens, the probability is: 4/52 = 1/13
(xvii) Since there are 13 hearts, the probability is: 13/52 = 1/4
(xviii) Since there are 26 red cards, the probability is 26/52 = 1/2
Question 6:
An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.
Ans.
Number of red balls = 10
Number of white balls = 8
Total number of balls in the urn = 10 + 8 = 18
Therefore, the total number of cases is 18 and the number of favourable cases is 8.
∴ P(The ball drawn is white) = = 8/18 = 4/9
Question 7:
A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:
(i) white?
(ii) red?
(iii) black?
(iv) not red?
Ans.
Number of red balls = 3
Number of black balls = 5
Number of white balls = 4
Total number of balls = 3 + 5 + 4 = 12
Therefore, the total number of cases is 12.
(i) Since there are 4 white balls, the number of favourable outcomes is 4.
P(a white ball) = = 4/12 = 1/3
(ii) Since there are 3 red balls, the number of favourable outcomes is 3.
P(a red ball) = = 3/12 = 1/4
(iii) Since there are 5 black balls, the number of favourable outcomes is 5.
P(a black ball) = = 5/12
(iv) P(not a red ball) = 1−P(a red ball) = 1−1/4 = 3/4
Question 8:
What is the probability that a number selected from the numbers 1, 2, 3, . . . , 15 is a multiple of 4?
Ans.
There are 15 numbers from 1, 2,⋯,15. Hence, the total number of cases is 15. Again, the multiples of 4 are 4, 8 and 12. Therefore, the total number of favourable cases is 3.
∴ P(the number is a multiple of 4) = = 3/15 = 1/5
Question 9:
A bag contains 6 red, 8 black and 4 white balls. A ball is drawn at random. What is the probability that ball drawn is not black?
Ans.
Number of red balls = 6
Number of black balls = 8
Number of white balls = 4
Total number of balls = 6 + 8 + 4 = 18
∴ Total number of cases = 18
Again, number of balls that are not black = 18 − 8 = 10
Thus, the number of favourable cases is 10.
∴ P(the drawn ball is not black) = = 10/18 = 5/9
Question 10:
A bag contains 5 white and 7 red balls. One ball is drawn at random. What is the probability that ball drawn is white?
Ans.
Number of white balls = 5
Number of red balls = 7
Total number of balls = 5 + 7 = 12
∴ The total number of cases = 12
Again, there are 5 white balls. Therefore, the number of favourable outcomes is 5.
∴ P(the drawn ball is white) = = 5/12
Question 11:
A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:
(i) white
(ii) red
(iii) not black
(iv) red or white
Ans.
Number of red balls = 4
Number of black balls = 5
Number of white balls = 6
Total number of balls in the bag = 4 + 5 + 6 = 15
Therefore, the total number of cases is 15.
(i) Let A denote the event of getting a white ball. Number of favourable outcomes, i. e. number of white balls =
P(A) = = 6/15 = 2/5
(ii) Let B denote the event of getting a red ball. Number of favourable outcomes, i. e. number of red balls = 4
P(B) = = 4/15
(iii) Let C denote the event of getting a black ball. Number of favourable outcomes, i. e. number of black balls = 5
P(C) = = 5/15 = 1/3
Therefore, the probabilty of not getting a black ball is as follows:
= 1 − P(C) = 1 − 1/3 = 2/3
(iv) Let D denote the event of getting a red or a white ball. P(D) = = = 10/15 = 2/5
Question 12:
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:
(i) red
(ii) black
Ans.
Number of red balls = 3
Number of black balls = 5
Total number of balls = 3 + 5 = 8
(i) Let A be the event of drawing a red ball.
∴ P(A) = = 3/8
(ii) Let B be the event of drawing a black ball.
∴ P(B) = = 5/8
Question 13:
A bag contains 5 red marbles, 8 white marbles, 4 green marbles. What is the probability that if one marble is taken out of the bag at random, it will be
(i) red
(ii) white
(iii) not green
Ans.
Number of red marbles = 5
Number of white marbles = 8
Number of green marbles = 4
Total number of marbles in the bag = 5 + 8 + 4 = 17
∴ Total number outcomes = 17
(i) Let A be the event of drawing a red ball.
∴ P(A) = = 5/17
(ii) Let B be the event of drawing a white ball.
∴ P(B) = = 8/17
(iii) Let C be the event of drawing a green ball.
∴ P(C) = = 4/17
Now, the event of not drawing a green ball is:
= 1 − P(C) = 1 − 4/17 = 13/17
PAGE NO 26. 16:
Question 14:
If you put 21 consonants and 5 vowels in a bag. What would carry greater probability? Getting a consonant or a vowel? Find each probability.
Ans.
Number of consonants = 21
Number of vowels = 5
Total number of possible outcomes = 21 + 5 = 26
Let C be the event of getting a consonant and V be the event of getting a vowel.
∴ P(C) = = 21/26 And,
P(V) = = 5/26
Thus, the consonants have a greater probability.
Question 15:
If we have 15 boys and 5 girls in a class which carries a higher probability? Getting a copy belonging to a boy or a girl. Can you give it a value?
Ans.
Number of boys in the class = 15
Number of girls in the class = 5
Total number of students in the class = 15 + 5 = 20
∴ Number of possible outcomes = 20
Since the number of boys is more than the number of girls, boys will have a higher probability. Hence, there is the higher probability of getting a copy belonging to a boy. Let A be the event of getting a boy's copy and B be the event of getting a girl's copy.
∴ P(A) = 15/20 = 3/4
And, P(B) = 5/20 = 1/4
Question 16:
If you have a collection of 6 pairs of white socks and 3 pairs of black socks. What is the probability that a pair you pick without looking is
(i) white?
(ii) black?
Ans.
Number of pairs of white socks = 6
Number of pairs of black socks = 3
Total number of pairs of socks = 6 + 3 = 9
∴ Number of possible outcomes = 9
(i) Let A be the event of getting a pair of white socks.
∴ P(A) = 6/9 = 2/3
(ii) Let B be the event of getting a pair of black socks.
∴ P(B) = 3/9 = 1/3
Question 17:
If you have a spinning wheel with 3-green sectors, 1-blue sector and 1-red sector. What is the probability of getting a green sector? Is it the maximum?
Ans.
Number of green sectors in the wheel = 3
Number of blue sectors in the wheel = 1
Number of red sectors in the wheel = 1
Total number of sectors in the wheel = 3 + 1 + 1 = 5
∴ Number of possible outcomes = 5
Let A, B and C be the events of getting a green, blue and red sector, respectively.
∴ P(A) = 3/5, P(B) = 1/5 and P(C) = 1/5
Hence, the probability of getting a green sector is the maximum.
Question 18:
When two dice are rolled:
(i) List the outcomes for the event that the total is odd.
(ii) Find probability of getting an odd total.
(iii) List the outcomes for the event that total is less than 5.
(iv) Find the probability of getting a total less than 5?
Ans.
Possible outcomes when two dice are rolled:S = {(1,1), (1,2), (1,3), (1,4),⋯,(6,5), (6,6)}Therefore, the number of possible outcomes in the sample space is 36.
(i) The outcomes for the event that the total is odd:E = {(1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (3,6), (4,1), (4,3), (4,5), (5,2), (5,4), (5,6), (6,1), (6,3), (6,5)}
(ii) The number of favourable outcomes is 18.
∴ P(E) = 18/36 = 1/2
(iii) The outcomes for the event that total is less than 5:B = {(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)}
(iv) The number of favourable outcomes is 6.
∴ P(B) = 6/36 = 1/6
The document RD Sharma Solutions - Chapter 26 - Data Handling-IV (Probability), Class 8, Maths Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
All you need of Class 8 at this link: Class 8
Use Code STAYHOME200 and get INR 200 additional OFF
## RD Sharma Solutions for Class 8 Mathematics
88 docs
### Top Courses for Class 8
Track your progress, build streaks, highlight & save important lessons and more!
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
; |
Question
# A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The static coefficient of friction between the block and the slab is 0.60 while the kinetic coefficient is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If g = 9.8m/s2, then what will be the resulting acceleration of the slab?
Open in App
Solution
## Step1: Given dataGiven that a $40kg$ slab rests on a frictionless floor and a $10kg$ block rests on top of the slab.Also, the static coefficient of friction between the block and the slab is ${\mathrm{μ}}_{s}=0.60$Step2: Formula used$F=ma\left[whereF=force,m=mass,a=acceleration\right]$Step3: Calculating the accelerationLet us first analyze if these blocks move together.For that, we will be finding the acceleration of the total mass, which means the $massofslab+massofblock$, i.e., $\left(40+10\right)kg$.Now using the formula $F=ma$or we can write- $a=\frac{F}{m}$Now, given the value of $F=100N$ in the question and the total mass is $40+10=50kg$.So, $a=\frac{100}{50}=2m/{s}^{2}$Now the maximum frictional force is f${f}_{m}={\mathrm{μ}}_{s}×N$Also, $N={m}_{1}g$ so the maximum frictional force will be : ${f}_{s}={\mathrm{μ}}_{s}×{m}_{1}g$So, putting the values of friction coefficient ${\mathrm{μ}}_{s}=0.60$, mass of block is ${m}_{1}=10kg$ and $g=9.8m/{s}^{2}$we get- $fs=\left(0.60\right)\left(10\right)\left(9.8\right)=58.8N$Thus, we can see frictional force is smaller than the applied force and proves that the block and the slab do not move relative to each other.Now, for the block we have $F=\mathrm{μ}{m}_{1}g$Putting the values of the known terms, we get- $F=0.4×10×9.8=39.2N$Now, the resulting acceleration of the slab will be, $a=\frac{F}{m}=\frac{39.2}{40}=0.98m/{s}^{2}$Hence, the resulting acceleration of the slab is $0.98m/{s}^{2}.$
Suggest Corrections
11 |
Monday, July 15, 2024
Home > ICSE Solutions for Class 10 Maths > Linear Inequations Class 10 ICSE Maths Revision Notes Chapter 4 PDF
Linear Inequations Class 10 ICSE Maths Revision Notes Chapter 4 PDF
Hi students, Welcome to Amans Maths Blogs (AMB). In this article, you will get Linear Inequalities Class 10 Maths Revision Notes Chapter 4 PDF.
ICSE Class 10 Maths Chapter 4 Linear Inequations Important Concepts
What is an Inequality?
In mathematics, a statement that contains > (greater than), ≥ (greater than or equal to), < (less than), ≤ (less than or equal to), ≠ (not equal to) sign are called an inequality. The inequality defines the comparison between two numbers or other mathematical expressions.
For example:
(i) a < b means that a is less than b.
(ii) ab means that a is greater than b.
(iii) a ≤ b or a ⩽ b means that a is less than or equal to b.
(iv) a ≥ b or a ⩾ b means that a is greater than or equal to b.
(v) a ≠ b means that a is not equal to b.
What is an Inequations?
A mathematical statement that indicate the value of variables or an algebraic expression which is not equal to other value are called an inequation. These inequations are written using the inequalities signs, > (greater than), ≥ (greater than or equal to), < (less than), ≤ (less than or equal to), ≠ (not equal to).
For example:
(i) x < 5
(ii) y > 10
(iii) (3x + 6) ≤ 5
(iv) (2y – 9) ≥ 15
(v) (3k – 2) ≠ 5
What is an Linear Inequations?
A mathematical statements of any of the forms ax + b > 0, ax + b ≥ 0, ax + b < 0, ax + b ≤ 0 are called as linear inequations in variable x, where a, b are real numbers and a ≠ 0.
For example:
(i) x < 15
(ii) y > 12
(iii) (3x – 8) ≤ 15
(iv) (2y + 9) ≥ 15
(v) (3k + 2) ≠ 5
What are Replacement Set and Solution Set?
For an inequation, the set of elements or values from which the values of the variable are taken is called the replacement set or also known as domain of variable.
Now, the solution of an inequation depends upon the replacement set. So, an inequation may have one, many or no solution, depending upon the replacement set.
The solutions of inequation make a set of values of variable, and it is known as solution set of the inequation.
Every solution set is a subset of replacement set.
For example: we need to solve x + 2 > 7.
Depending the replacement set, the solutions set will be different.
If replacement set A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then solution of given inequation, means solution is S = {6, 7, 8, 9, 10}.
If replacement set B = {0, 1, 2, 3, 4, 5, 6, 7}, then solution of given inequation, means solution is S = {6, 7}.
If replacement set C = {0, 1, 2, 3, 4}, then solution of given inequation, means solution is S = φ (No Solution).
What are Inequality Rules or Postulates?
There are following inequality rules:
(i) An inequation is NOT changed if the same number is added to both sides of inequation.
For example:
x – 5 > 12
⇒ x – 5 + 5 > 12 + 5 (Adding 5 to both sides)
⇒ x > 17
(ii) An inequation is NOT changed if the same number is subtracted from both sides of inequation.
For example:
y + 12 < 15
⇒ y + 12 – 12 < 15 – 12 (Subtracting 12 from both sides)
⇒ y < 3
(iii) An inequation is NOT changed if the same positive number is multiplied to both sides of inequation.
For example:
x/5 < 2
⇒ x/5 * 5 < 2 * 5 (Multiplying 5 to both sides)
⇒ x < 10
(iv) An inequation is changed if the same negative number is multiplied to both sides of inequation. And, the inequality sign is reversed.
For example:
-x/3 < 4
⇒ -x/3 * (-3) > 4 * (-3) (Multiplying -3 to both sides)
⇒ x > -12
(v) An inequation is NOT changed if the same positive number divides both sides of inequation.
For example:
3x < 15
⇒ 3x / 3 < 15 / 3 (Both sides are divided by 3)
⇒ x < 5
(vi) An inequation is changed if the same negative number divides both sides of inequation. And, the inequality sign is reversed.
For example:
-3x < 21
⇒ -3x / -3 > 21 / -3 (Both sides are divided by -3)
⇒ x > -7
(vii) Any term of an inequation may be taken to other side with its sign changed without affecting the sign of inequality.
For example:
x + 10 < 51
⇒ x < 51 – 10 (10 is moved to RHS and it becomes negative without affecting inequality sign)
⇒ x < 41
S Chand ICSE Class 10 Maths Chapterwise Notes & Solutions
Chapter 1 : GST Notes & Solutions
| Notes | Exercise 1
Chapter 2 : Banking Notes & Solutions
| Notes | Exercise 2 | Revision Exercise
Chapter 3 : Shares & Dividends Notes & Solutions
| Notes | Exercise 3A | Exercise 3B | Revision Exercise
Chapter 4 : Linear Inequations in One Variable Notes & Solutions
| Notes | Exercise 4 | Revision Exercise
Chapter 5 : Quadratic Equations Notes & Solutions
| Notes | Exercise 5A | Exercise 5B | Exercise 5C| Exercise 5D| Exercise 5E| Revision Exercise
Chapter 6 : Ratios & Proportions Notes & Solutions
| Notes | Exercise 6A | Exercise 6B | Exercise 6C| Revision Exercise
Join Telegram to get PDF
AMAN RAJ
I am AMAN KUMAR VISHWAKARMA (in short you can say AMAN RAJ). I am Mathematics faculty for academic and competitive exams. For more details about me, kindly visit me on LinkedIn (Copy this URL and Search on Google): https://www.linkedin.com/in/ambipi/
error: Content is protected !! |
# The Power of (a + b + c)²: Unlocking the Potential of Algebraic Expansion
Algebra, with its intricate equations and complex formulas, can often seem daunting to students. However, understanding the fundamental concepts of algebra is crucial for success in mathematics and various other fields. One such concept is the expansion of (a + b + c)², which holds immense power in simplifying expressions and solving equations. In this article, we will explore the significance of (a + b + c)², its applications in real-world scenarios, and provide step-by-step examples to help you grasp this concept with ease.
## The Basics: Understanding (a + b + c)²
Before delving into the applications of (a + b + c)², let’s first understand what it represents. (a + b + c)² is an algebraic expression that denotes the square of the sum of three variables: a, b, and c. Mathematically, it can be expanded as:
(a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc
This expansion is derived using the distributive property of multiplication over addition. By multiplying each term within the parentheses by itself and then summing them up, we obtain the expanded form of (a + b + c)².
## Applications of (a + b + c)²
The expansion of (a + b + c)² finds its applications in various fields, including mathematics, physics, and computer science. Let’s explore some of these applications:
### 1. Simplifying Expressions
The expansion of (a + b + c)² allows us to simplify complex algebraic expressions. By substituting the variables with specific values, we can easily evaluate the expression. For example, consider the expression (2x + 3y + 4z)². By expanding it using the formula, we get:
(2x + 3y + 4z)² = (2x)² + (3y)² + (4z)² + 2(2x)(3y) + 2(2x)(4z) + 2(3y)(4z)
This expansion simplifies the expression and enables us to perform further calculations or solve equations more efficiently.
### 2. Solving Equations
The expansion of (a + b + c)² is particularly useful in solving quadratic equations. By rearranging the equation and applying the expansion, we can simplify it and find the roots. For instance, consider the equation x² + 5x + 6 = 0. By rewriting it as (x + 2)(x + 3) = 0 and expanding (x + 2)², we can easily determine the values of x that satisfy the equation.
### 3. Probability Calculations
In probability theory, (a + b + c)² is employed to calculate the probabilities of different outcomes. For example, consider a scenario where three dice are rolled simultaneously. To determine the probability of obtaining a sum of 10, we can expand (a + b + c)², where a, b, and c represent the outcomes of each dice roll. By counting the favorable outcomes and dividing them by the total possible outcomes, we can calculate the probability accurately.
### 4. Geometric Interpretation
The expansion of (a + b + c)² also has a geometric interpretation. It represents the sum of the squares of the three sides of a triangle, along with the sum of the products of the pairs of sides multiplied by two. This interpretation finds applications in geometry, where it helps in calculating areas, perimeters, and other properties of triangles.
## Step-by-Step Examples
Let’s now work through a few step-by-step examples to solidify our understanding of (a + b + c)²:
### Example 1:
Expand (2x + 3y + 4z)².
Solution:
(2x + 3y + 4z)² = (2x)² + (3y)² + (4z)² + 2(2x)(3y) + 2(2x)(4z) + 2(3y)(4z)
= 4x² + 9y² + 16z² + 12xy + 16xz + 24yz
### Example 2:
Solve the equation x² + 6x + 9 = 0.
Solution:
Rewrite the equation as (x + 3)² = 0.
By applying the square root property, we get:
x + 3 = 0
x = -3
Therefore, the solution to the equation is x = -3.
## Q&A
### Q1: Can (a + b + c)² be expanded for more than three variables?
A1: Yes, the expansion of (a + b + c)² can be generalized for any number of variables. The formula for expanding (a₁ + a₂ + a₃ + … + aₙ)² is:
(a₁ + a₂ + a₃ + … + aₙ)² = a₁² + a₂² + a₃² + … + aₙ² + 2(a₁a₂ + a₁a₃ + … + aₙ₋₁aₙ)
A2: The expansion of (a + b + c)² is a specific case of the binomial theorem, which provides a formula for expanding the power of a binomial. The binomial theorem states that:
(a + b)ⁿ = C(n, 0)aⁿb⁰ + C(n, 1)aⁿ⁻¹b¹ + C(n, 2)aⁿ⁻²b² + … + C(n, n-1)abⁿ⁻¹ + C(n, n)a⁰bⁿ
where C(n, r) represents the binomial coefficient, given by n! / (r!(n-r)!).
### Q3: Are there any real-world applications of (a + b + c)²?
A3: Yes, the expansion of (a + b + c)² finds applications in various real-world scenarios. For
Reyansh Sharma
Rеyansh Sharma is a tеch bloggеr and softwarе еnginееr spеcializing in front-еnd dеvеlopmеnt and usеr intеrfacе dеsign. With еxpеrtisе in crafting immеrsivе usеr еxpеriеncеs, Rеyansh has contributеd to building intuitivе and visually appеaling intеrfacеs.
Latest articles
Related articles |
·
# Using differentiation to determine future stock prices in layman’s terms….
Using differentiation to determine future stock prices in layman’s terms….
Step 1 - Given Equation:
The equation dS/dt = μ * St tells us something important: it shows that during a tiny moment, the change in the price of an asset (like a stock) is equal to a certain number μ (we call it the "drift") multiplied by the asset's current price St.
Step 2 - Dividing Both Sides:
If we divide both sides of the equation by St (the asset's current price), and then multiply both sides by a tiny amount of time dt, we get dSt/St = μ * dt or (1/ St) dt This equation is important because it links how the percentage change in the asset's price relates to the drift μ and the tiny time step dt.
Step 3 - Solving the Differential Equation:
Now we're going to solve this equation. Think of it like solving a puzzle to find out how the asset's price changes over time. By integrating the right side of the equation (which means adding up all the small changes), we get:
ln(St) + c1 = μ * t + c2
NB: If you differentiate ln (St) with respect to St you get (1/St)
Step 4 - Simplifying and Antilog:
If we subtract c1 from both sides of the equation, we get ln(St) = μ * t + (c2- c1).
Now, let's use a special operation called "antilog" on both sides. It's like the opposite of taking the natural logarithm. This operation turns "ln(St)" back into "St."
Step 5 - Getting the Final Equation:
So, by using the antilog on both sides, we get St = e^(μt + c2 - c1).
Now, we see that c2 - c1 can be replaced with just "c" (a single number).
Step 6 - Wrapping It Up:
When we have c = c2 - c1, and we set e^(c) equal to the initial price (S₀), the equation becomes St = S₀ * e^(μt). This means that the future price of the asset is the initial price times "e" raised to the power of the drift μ times t.
In simpler terms, the equation helps us predict how the asset's price changes over time using a drift rate μ. It's like having a math formula to see how an asset's price will grow over time based on a constant rate of change.
#StockPricePrediction #Differentiation #MathematicsInFinance#FinancialModeling #FutureStockPrices #MathConcepts #DriftRate#ContinuousTimeModeling #FinancialAnalysis #PredictiveModeling
Écrire commentaire
Commentaires: 0
FINANCE TUTORING, organisme de formation enregistré sous le numéro 24280185328
Florian CAMPUZAN
Tél: 0680319332 |
# Signed Number Representation in Computers
On paper, we can represent a negative number by prefixing it with a minus sign. But, in computers, we will have to encode the sign in the number itself. This article discusses three prominent methods used for representing signed numbers in computers. All three methods work by treating the leftmost digit as a kind of sign bit.
## Sign-Magnitude representation
At first glance, the problem of representing signed numbers in a computer might seem quite straightforward. The most obvious solution would be:
• Reserve one digit, possibly the leftmost digit, for representing the sign of the number.
• Use the remaining digits to represent its magnitude.
The convention is that a sign-bit of 0 indicates a positive number and a sign-bit of 1 indicate a negative number. This method is known as the sign-magnitude representation. It has been used in some computers in the past but it never gained much popularity.
Let us look at the example of a 4-digit number. After dedicating 1 bit for sign, the remaining 3 bits can represent numbers up to 7. This gives us a range of [-7..+7].
I will be using 4-bit numbers in most of the examples in this article. Real word sizes are larger, but using smaller words keeps the examples simpler. The principles presented here will work the same for more realistic word sizes like 32-bit or 64-bit.
Table 1: Sign-Magnitude representation of 4-bit numbers
Decimal Sign-Magnitude
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
0 1000
-1 1001
-2 1010
-3 1011
-4 1100
-5 1101
-6 1110
-7 1111
Sign-Magnitude representation is the easiest for humans to understand. But it cannot be implemented efficiently in hardware. For performing addition or subtraction we need to first check the sign and magnitude of the operands. The result of these checks determines which operation we need to perform, the order of the operands, and the sign of the result.
Let us say we want to add 0001 and 1010. Here the first number represents +1 and the second number represents -2. So -(2 – 1) = -1 is what we are looking for. The opcode of the instruction is Add. But we need to check the sign of the operands to see whether addition or subtraction is required. If subtraction is required, we have more work to do. Next, we have to check the magnitude of the numbers. Only then we know which number should be the minuend and which should be the subtrahend.
The previous paragraph might sound like a repetition of basic signed number arithmetic rules. While it might feel natural to us, computers can’t do it efficiently. 1’s complement and 2’s complement provide a better solution, as we will see.
Another problem with sign-magnitude representation is that it has two representations for zero. In our 4-bit example, both 0000 and 1000 represent zero. This would add a fair amount of complexity to the computer.
## 1’s complement representation
Positive numbers are represented in the same way in both 1’s complement and sign-magnitude. The leftmost digit of all positive numbers has a sign-bit of zero. The remaining digits represent the magnitude of the number. For negative numbers, we find the 1’s complement of the number and use that instead.
Complement forms exist for numbers of all radices or bases. There are two types of complements:
1. The Radix Complement.
Also known as r’s complement where r is the radix. So for binary numbers, r’s complement is the same as 2’s complement. For decimal numbers, r’s complement is the same as 10’s complement.
2. The Diminished Radix Complement.
Also known as (r-1)’s complement. This is the same as 1’s complement for binary numbers and 9’s complement for decimal numbers.
### How to calculate 1’s complement
The formula for finding (r-1)’s complement is rn-1-N. Where r is the radix, n is the number of digits in the number and N is the actual number.
One way to understand this formula would be like this. rn is the smallest n +1 digit integer. Hence rn-1 is the largest n digit integer. So to find (r-1)’s complement you just find the largest n digit integer and subtract N from it.
An example should make this clear. Let us say you want to find the (r-1)’s or 9’s complement of decimal number 55. Here n is 2 digits and the largest 2 digit integer is 99. So 9’s complement of 55 is 99 – 55 = 44. Similarly, 9’s complement of 555, would be 999 – 555 = 444.
Now let us look at a 1’s complement example. We want to find the 1’s complement of 610 = 1102. There are 3 digits in 110 and the largest 3 digit binary integer is 111. 1112 – 1102 = 12.
There is an easier way to find 1’s complement. Simply invert each bit in the number using not gates. That is, every 1 in the number becomes 0 and every 0 becomes 1. Thus 1’s complement of 110 is 001.
### Decimal to 1’s complement mapping for 4-bit numbers
You may have begun to notice that finding the complement of a number is like counting up from the bottom of a list of numbers.
Table 2: 1’s complement representation of 4-bit numbers
Decimal 1’s Complement
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
-7 1000
-6 1001
-5 1010
-4 1011
-3 1100
-2 1101
-1 1110
0 1111
Like the sign-magnitude table, the rows of this table are in the ascending order of the actual numbers used by the computer, not what those numbers represent. Getting familiar with this ordering will help in understanding how 1’s complement or 2’s complement work. The positive numbers plus one of the two zeros take up the top half of the table. The bottom half of the table is reserved for the other zero and the negative numbers.
Now, let us address the most important question. What do we gain by representing numbers this way? The benefit is that we can do additions or even subtractions on signed numbers without bothering to check the sign or magnitude of the numbers. Plus we need only adder units to perform both addition and subtraction
### 1’s Complement Addition
In 1’s complement, addition is done using a variant of modular arithmetic. In modular arithmetic, the values wrap around once we reach some specified maximum. A clock with only an hour handle and hour 12 labeled as 0 would be an example of modulo 12 arithmetic. Since there are 16 elements in our example list we would be doing modulo 16 arithmetic.
There is one unusual rule which we need to follow in 1’s complement addition. If there is a carry out from the most significant digit, it is not brought down to the left of the result as we usually do. Instead, it is added to the result. This might sound a bit confusing, so let us look at some examples.
### Examples
Let us say we want to add 5 and -2. In 1’s complement, these numbers are 0101 and 1101 respectively.
If we had done normal addition the result would have been 10010. If we had done true modular addition the result would have been 0010. But 1’s complement addition gives us a result of 0011. We can verify that 00112 is 310. Why the carry out is handled this way is explained in the How does it work? section. For the moment let us focus on the fact that we just added a negative number and a positive number, without bothering about which number has the higher magnitude and got the right result. Will this work if the negative number had the higher magnitude? What about -510 + 210 = 10102 + 00102 ? Let us see.
From the 1’s complement table given above, we can verify that the result 1100 is the 1’s complement representation of -3.
### Find the original number given its 1’s complement
We don’t need to look up a table every time to find what a 1’s complement number represents. There is a simple formula for finding the original number given its 1’s complement. We just need to find 1’s complement of the 1’s complement. This makes sense because we found the 1’s complement by subtracting the original number from 2n-1.
### 1’s Complement Subtraction
Let us look at how subtraction works. For example, what if we wanted 5 – 2. For subtraction, there is an extra step. First, we need to do to find the 1’s complement of the subtrahend. After that, it is the same addition we saw in the first example. The extra step does not complicate the hardware since it is a simple step required by all subtraction instructions.
Despite the advantages provided by 1’s complement, it is not used in any of the modern computers. Its main drawback is that it has two representations of zero. Modern computers use 2’s complement method since it does not have this limitation.
## 2’s Complement Representation
2’s complement is the method used for representing signed numbers in pretty much all modern computers. It is quite similar to 1’s complement. Once we know 1’s complement, there isn’t much new to learn.
### How to calculate
Remember that the formula for finding diminished radix or (r-1)’s complement was rn – 1 – N. Thus the formula for 1’s complement was 2n – 1 – N.
For radix complement or r’s complement, the formula is rn – N. Where r is the radix, n is the number of digits in the number and N is the number itself. Thus the formula for 2’s complement is 2n – N.
To find the 2’s complement of a 3-bit number like 110, we find the smallest 4-bit number which is 1000, and subtract 110 from it.
10002 − 1102 = 0102
Another method for finding 2’s complement is to find 1’s complement first and then add 1. Let us find the 2’s complement of 1102 using this method. 1’s complement of 1102 can be found using the bit inversion method mentioned earlier.
x = 110
1’s complement of x = 001 (Just invert each bit)
And now, to find the 2’s complement:-
0012 + 12 = 0102
There is one special case to consider when finding 2’s complement. If we try to find the 2’s complement of 0, we get a number that won’t fit our word size. In this case, the extra digit on the left must be discarded.
Table 3: 2’s complement representation of 4-bit numbers
Decimal 2’s Complement
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
-8 1000
-7 1001
-6 1010
-5 1011
-4 1100
-3 1101
-2 1110
-1 1111
In 2’s complement, there is only one representation of zero. A sign-bit of 0 means zero or a positive number. The numbers with a sign-bit of 1 represent only negative numbers. This means that there is an extra negative number in 2’s complement representation. In our 4-bit number example, the 2’s complement numbers range from [-8 .. +7].
### 2’s Complement Addition
Rules for addition in 2’s complement are similar to that for 1’s complement. The only difference is that, carry out, if any, from the most significant digit, is simply ignored. There is no need to increment the result by 1. This means that unlike in 1’s complement, in 2’s complement we do true modular arithmetic.
Please note that throwing away the carry to wrap around does not work in all cases of modular addition. For example, it will not work in the clock example mentioned earlier. The trick works in computers because of the ranges we use. If we are doing n digit addition, the largest number we support would be the largest n digit integer. In our 4-bit example, this would be 1111. If we add 1 to it and throw away the carry we get 0000.
### Example
Let us try to add 5 and -2 again, this time using 2’s complement. The 2’s complement numbers are 0101 and 1110 respectively.
The carry out from the most significant digit is ignored.
### 2’s Complement Subtraction
Subtraction is also similar in 2’s complement. The steps are, find the 2’s complement of the subtrahend and add the numbers using 2’s complement rules. The extra negative number which 2’s complement representation has, does pose difficulties. If we try to find the 2’s complement of the negative number with the largest magnitude, we will get the same number back. In our example, the 2’s complement of 1000 is 1000. We have to watch out for this special case.
## How does it work?
Let us take a look at how the complement methods work. I will be focusing on 2’s complement, the principles are roughly the same for 1’s complement. I will mention the differences between the two as we go along.
Let us look at another 4-bit, 2’s complement example. Let us say, we want to subtract +2 from +5. For reference, please take a look at table 3 showing the 2’s complement representation of 4-bit numbers. It is important to understand the order of numbers in that table. It may be a good idea to keep the table accessible in a different browser tab. The number which represents +5 is in 6th position in the table and the number is 0101.
Think of it this way, what we need to do is to go up two positions in the list and reach the number representing +3. But we don’t really like going up the list since that would mean having to subtract. We would like to reach our destination by going down the list, that is, by adding numbers. This is how we can achieve this. Go down the list instead of up. When we cross the last element of the list there will be a carry from the most significant digit. Throw away this carry so that we can jump back to the top of the list.
For example, if we are at the bottom of the list where the number is 1111 and we add one to it, we get 10000 which is a 5-bit number. But if we throw away the carry from the most significant digit the result would be 0000, which is the top of our list.
Our list consists of 16 numbers. If we start at the 6th position in the list and add 16 using modulo 16 addition, we will get back to where we started. What if we add 14 instead? We will reach the 4th position where the number is +3, which is where we want to be.
First, we have to figure out which number to add. We find the 2’s complement of the subtrahend, that is the number representing -2. This number is at the second last position of the list. Note that the distance between -2 and the top of the list is 2 when doing modulo 16 addition. Further note that 2 is the number that we want to subtract from 5. Now, all we need to do is a 2’s complement addition of the number representing -2 and the number representing 5 i.e. 11002 + 01012. There will be a carry out from the most significant digit. We cycle around the list by ignoring this carry and reach our destination of +3.
Check out the image below showing 2’s complement numbers depicted in a circle. We are allowed to move only in a clockwise direction in this circle.
### Carry out in 1’s complement addition
Remember that in 1’s complement addition we did not quite follow the modulo addition rules. When there was a carry out from the most significant digit, we did not throw it away. We incremented the result by one instead. The reason why this is required is that in 1’s complement there are two representations of zero. The first element in the list and the last element in the list are both zeros.
Let say, we add the 1’s complement representations of -1 and +2 and throw away the carry from the most significant digit. We may think we will get +1 as the result, but we won’t because there are two zeros between -1 and +1. So each time we make the jump from the zero at the bottom of the list to the zero at the top of the list we have to compensate for the jump by adding a 1 to the result.
So how do we detect that a 1’s complement addition has lead to a jump from one zero to the other? Remember that ignoring the carry from the most significant digit is the trick we use to jump from the bottom of the list to the top of the list. So, all we need to do is to check if there was a carry out from the most significant digit.
### Detecting overflow in 2’s complement addition
One very important point which we haven’t looked at so far is the question of overflow. There needs to be a way for the program to detect an overflow. Detecting overflow is a little tricky in 2’s complement.
The results of 2’s complement additions are often larger than our word length. But exceeding word length does not imply an overflow. We could still get a valid result after chopping off the extra bit from the left. On the other hand, we could get an overflow without actually exceeding our word length. For example, adding two positive numbers might produce a negative number as result.
To detect overflow, we need to look at both the carry in and carry out of the most significant digit. As it turns out, there is a simple rule for detecting overflow in 2’s complement addition. If the carry out and carry in for the most significant digit are different, then there is an overflow.
Let us look at some examples from our 4-bit 2’s complement representation to check if the rule holds up. It also gives us an opportunity to take a closer look at how 2’s complement additions work.
#### When both operands are positive
If both the operands are positive, the result must also be positive. If we add two positive numbers and get a negative number as the result then there is an overflow.
Example a: 210 + 310 = 00102 + 00112
We added 2 positive numbers and the result is also positive. There is no carry out or carry into the most significant digit, so the rule holds.
Example b: 310 + 510 = 00112 + 01012
We added two positive numbers and veered into territory reserved for negative numbers. +8 is too big to be represented as a signed number in a 4-bit word. Notice that there is a carry into the most significant digit but no carry out. This implies an overflow.
#### When both operands are negative
Remember that the negative numbers are in the bottom half of our list of numbers. When we add negative numbers there will always be a carry from the most significant digit. Each time we will be throwing away this carry so that we can cycle through the list. To avoid an overflow we must cross the positive territory at the top of the list and get back into negative territory.
You could think of the magnitude of a negative number as a distance from its 2’s complement to zero. This does not make sense for normal arithmetic but it does make sense for modular addition since addition wraps around the list. Take a look at the following examples, the binary numbers are in 2’s complement form.
If we take a negative number which is x distance away from 00002 and another negative number which is y distance away from 00002 and add the two using modular addition, we will get a number which is x + y distance away from 00002.
Example a: -110 + -210 = 11112 + 11102
Here we have two operands with small magnitude. This means that their 2’s complements are large numbers. The result easily gets past the top half of the list and reaches the bottom half. There is a carry of 1 coming in and out of the most significant digit, this implies that there is no overflow.
Example b: -710 + -510 = 10012 + 10112
We added two negative numbers and got a positive number as the result. Also, note that there is a carry out of the most significant digit but no carry coming into it.
#### When the operands have opposite sign
When the operands are of opposite sign, there won’t be an overflow since the result is guaranteed to have a lower magnitude than both the operands. If the operands are small enough to fit our word size, then the result will fit as well. |
Study Material
# Varignon’s theorem
## What is Varignon’s theorem?
Varignon’s theorem, in Mechanics, states that the sum of the moments produced by a system of concurrent forces with respect to a certain point is equal to the moment of the resultant force with respect to the same point. varignon’s theorem examples
For this reason this theorem is also known as the principle of moments .varignon’s theorem examples
Although the first to enunciate it was the Dutch Simon Stevin (1548-1620), the creator of the hydrostatic paradox, the French mathematician Pierre Varignon (1654-1722) was the one who later gave it its final form.
An example of how Varignon’s theorem works in Mechanics is the following: suppose that a simple system of two coplanar and concurrent forces 1 and 2 acts on a point (denoted in bold because of their vector character). These forces result in a net or resultant force, called R .
Each force exerts a torque or moment about a point O, which is calculated by the vector product between the position vector OP and the force F , where OP is directed from O to the point of concurrency P:
O1 = OP × 1
O2 = OP × 2
Since R = 1 + 2 , then:
O = OP × 1 + OP × 2 = O1 + O2
But since OP is a common factor, then, applying distributive property to the cross product:
O = OP × ( 1 + 2 ) = OP × R
Therefore, the sum of the moments or torques of each force with respect to point O is equivalent to the moment of the resultant force with respect to the same point. varignon’s theorem examples
## Statement and proof
Let a system of N concurrent forces, formed by 1 , 2 , 3 … N , whose lines of action intersect at point P (see figure 1), the moment of this force system O , with respect to at a point O is given by:
O = OP × 1 + OP × 2 + OP × 3 +… OP × N = OP × ( 1 + 2 + 3 +… N )
### Demonstration
To prove the theorem, use is made of the distributive property of the vector product between vectors.
Let the forces 1 , 2 , 3 … N be applied to points A 1 , A 2 , A 3 … A N and concurrent at point P. The moment resulting from this system, with respect to a point O, called O is the sum of the moments of each force with respect to that point:
O = ∑ OAi × i
Where the sum goes from i = 1 to i = N, since there are N forces. Since we are dealing with concurrent forces and since the vector product between parallel vectors is zero, it happens that:
PAi × i = 0
With the null vector denoted as 0 .
The moment of one of the forces with respect to O, for example that of the force i applied on A i , is written like this:
Oi = OAi × i
The position vector OAi can be expressed as the sum of two position vectors:
OAi = OP + PAi
In this way, the moment about O of the force i is:
Oi = ( OP + PAi ) × i = ( OP × i ) + ( PAi × i )
But the last term is null, as explained above, because PAi lies on the line of action of i , therefore:
Oi = OP × i
Knowing that the moment of the system with respect to point O is the sum of all the individual moments of each force with respect to said point, then:
O = ∑ Oi = ∑ OP × i
Since OP is constant it comes out of the sum:
O = OP × (∑ i )
But ∑ i is simply the net force or resultant force R , therefore it is immediately concluded that:
O = OP × R
## Example
Varignon’s theorem facilitates the calculation of the moment of the force F with respect to the point O in the structure shown in the figure, if the force is decomposed into its rectangular components and the moment of each of them is calculated:
## Applications of Varignon’s theorem
When the resultant force of a system is known, Varignon’s theorem can be applied to replace the sum of each of the moments produced by the forces that compose it by the moment of the resultant. varignon’s theorem examples
If the system consists of forces on the same plane and the point with respect to which the moment is to be calculated belongs to that plane, the resulting moment is perpendicular.
For example, if all the forces are in the xy plane, the moment is directed in the z axis and it only remains to find its magnitude and its sense, such is the case of the example described above.
In this case, Varignon’s theorem allows us to calculate the resulting moment of the system through the summation. It is very useful in the case of a three-dimensional force system, for which the direction of the resulting moment is not known a priori. varignon’s theorem examples
To solve these exercises, it is convenient to decompose forces and position vectors into their rectangular components, and from the sum of the moments, determine the components of the net moment.
## Exercise resolved
Using Varignon’s theorem, calculate the moment of the force F around the point O shown in the figure if the magnitude of F is 725 N.
x = 725 N ∙ cos 37 º = 579.0 N
y = – 725 NN ∙ sin 37 º = −436.3 N
Similarly, the position vector r directed from O to A has the components:
x = 2.5 m
y = 5.0 m
The moment of each component of the force about O is found by multiplying the force and the perpendicular distance.
Both forces tend to rotate the structure in the same direction, which in this case is clockwise, to which a positive sign is arbitrarily assigned: varignon’s theorem examples
Ox = F x ∙ r y ∙ sin 90º = 579.0 N ∙ 5.0 m = 2895 N ∙ m
Oy = F y ∙ r x ∙ sin (−90º) = −436.3 N ∙ 2.5 m ∙ (−1) = 1090.8 N ∙ m
The resulting moment about O is:
O = Ox + Oy = 3985.8 N ∙ m perpendicular to the plane and in a clockwise direction. varignon’s theorem examples |
# Placement Papers - Sutherland
## Why Sutherland Placement Papers?
Learn and practice the placement papers of Sutherland and find out how much you score before you appear for your next interview and written test.
## Where can I get Sutherland Placement Papers with Answers?
IndiaBIX provides you lots of fully solved Sutherland Placement Papers with answers. You can easily solve all kind of placement test papers by practicing the exercises given below.
## How to solve Sutherland Placement Papers?
You can easily solve all kind of questions by practicing the following exercises.
### Sutherland - Recent Sample Question Papers of Sutherland
Posted By : Priya Rating : +37, -6
1. Salaries of A, B and C are in the ratio 4 : 5 : 9. If their salaries are increased by 25%, 10% and 50% respectively, what will be the new ratio of their salaries?
6:5:13
10:11:27
5:6:15
11:20:27
None of these
Let their original salaries be 4k, 5k and 9k.
Now, new salaries are 1.25 4k : 1.15k : 1.59k = 5k : 5.5k : 13.5k = 10:11:27
2. The ratio of the present ages of A and B 9 : 5. Five years earlier the ratio of their was 2 : 1. What is the
average of their present age?
20 years
25 years
35 years
40 years
None of these
Solution:
let present age of A=9x
So present age of B=5x
According to question
9x-5/5x-5=2/1
solving this we get,
x=5 so average is 45+25/2=35 years.
3. A Shopkeeper wants to earn 12% profit on an item after giving 20% discount to the customer. By what percentage should he increase his marked price to arrive at the label price?
24%
32%
40%
16%
Let the current price be Rs. 100.
For getting 12% profit he should sell it at Rs. 112.
Let the label price be x.
i.e. he need to increase 40% on label price
4. A can do a work in 60 days and B can do the same work in 40 days. They work together for 12 days and then A goes away. In how many days will B finish the remaining work?
16 days
20 days
25 days
28 days
24 days
Work done by A and B in 12 days is
=12(1/60+1/40)=12 * 5/120=0.5
Remaining work = 1- [0.5]=0.5 work
B does 1/40 work in one day
so B does 0.5 work in 40 * 0.5 days = 20 days
5. A train crosses platforms of length 160 meters and 220 metres in 16 seconds and 20 seconds respectively. What is the length of the train?
60 metres
175 metres
80 metres
100 metres
120 metres
Let the length of the train 'x' and speed of train be 'V' ms-1.
{160+x}/16={220+x}/20=V
{160+x}/4={220+x}/5
800+5x=880+4x;
5x-4x=880-800
x=80 metres
7. 8423 + 3120 + 6543) / (1536 + 377 + 189) =?
5.8
14.6
8.6
18.3
17.2
(8423+3120+6543) is nearly equal to 18000
& (1536+377+189) is nearly equal to 2100.
So required answer is (18000) /2100 = 8.6.
8. (13% of 7439) * (3.23 % of 537) =?
16243
14002
18674
19874
16774
first we calculate 13% of 7439
10% of 7439 = 744 & its 3% is nearly 224 therefore 13% of 7439 = 744 + 224 = 968.
Similarly 3.23 % of 537 = 16.2 + 1 = 17.2. So we get answer as 17.2 x 968 = 16660 (approx).
Hence option E is the answer.
9. (96)2 / 79507 1/3 = ?
215
259
298
324
432
Solution:
9216/43=215(app.)
10. (15.28 x 3.56) /3.15=?
12
9
68
17
42
3.56/3.15=1 app
So answer will be little above than 15 so answer is 17.
Like this page? +37 -6
### Companies List:
#### Current Affairs 2021
Interview Questions and Answers |
Instasolv
IIT-JEE NEET CBSE NCERT Q&A
4.5/5
# NCERT Exemplar Class 8 Maths Chapter 1 Solutions: Rational Numbers
NCERT Maths Exemplar Solutions for Class 8 Chapter 1 – Rational Numbers is an ideal study material to learn and understand all the essential concepts of this chapter. Our subject matter experts have ensured that all of your confusions have been tackled while preparing the NCERT Exemplar Class 8 Maths Problems Solutions. We have used a very simple language in order to make these solutions more understanding and comprehensive.
NCERT Class 8 Maths Exemplar Chapter 1 has a total of 150 exemplar problems. It includes exercises from simple to complex levels in terms of deriving answers and applying fundamentals learnt from the chapter. You will come across many theoretical explanations as well as solved examples in the chapter. Topics like rational numbers, natural numbers, properties of commutativity, properties of associativity, reciprocals etc. have been covered in the NCERT Exemplar problems and solutions for Class 8 Maths Chapter 1.
The team at Instasolv is devoted towards enhancing your learning and providing quality guidance through the NCERT Maths Exemplar Solutions for Class 8 Chapter 1 – Rational Numbers. The solutions adhere to the latest CBSE curriculum and guarantee exceptional aid in terms of grasping concepts, revising the chapter and having a thorough preparation for your exams.
## Important Topics for NCERT Maths Exemplar Solutions Chapter 1 – Rational Numbers
Natural Numbers, Whole Numbers and Integers
You would have studied about different kinds of numbers in your junior classes. In this chapter, you will have a revision of all the previously learned concepts with regards to numbers and the applications that they go through. You will learn about natural numbers, real numbers, integers and rational numbers.
Rational Numbers
In this chapter, you will come across the term “rational numbers”. These are numbers that can be written in the form of p/q, and here both p and q are integers. Also, the denominator in this format i.e. q is a non zero quantity. You will also learn about various properties of rational numbers. Some of the discussed properties are –
1. Rational numbers are closed under addition based on the theory that by adding up two values which are rational we will get the answer that will also be rational.
2. Rational numbers are also closed under subtraction keeping the fact that on performing subtraction on any two rational numbers the outcome will also be a rational number.
3. Rational numbers are closed under multiplication again working on the fact that the product of two rational numbers will also be a rational number.
4. Rational numbers are not closed under division. Excluding zero, all the rational numbers are not closed under division.
Multiplicative Identity
In this chapter, you will learn about multiplicative identity. An identity which when multiplied to value, does not hamper its original value is known as a multiplicative identity. In the case of rational numbers, their multiplicative identity is -1.
### Exercise Discussion of NCERT Maths Exemplar Solutions Class 8 Chapter 1 – Rational Numbers
• The first exercise of NCERT Exemplar Class 8 for Maths Chapter 1 comprises multiple-choice questions. There are conceptual questions based on the explanations from the chapter as well as equations where you need to perform calculations.
• There are observation-based exemplar problems in the chapter like question number 100, where a series of numbers are provided to you and you are required to select the ones which are rational in nature.
• A similar problem is given in question 101 where again a series has been provided with the denominators in the multiples of four and you are required to find the one where the lowest value of four can exist.
• The chapter also has complex equations as in exemplar problem 106 where you need to perform various applications according to the rules that have been discussed in the chapter.
• There is also a fair share of word problems towards the end of the chapter like question number 122.
## Why Use NCERT Maths Exemplar Solutions Class 8 Chapter 1 – Rational Numbers by Instasolv?
• Our NCERT Class 8 Maths Exemplar problems and solutions are designed by subject experts who are aware of your learning requirements to score well in CBSE exams.
• The solutions are presented in a student-friendly manner to ensure proper understanding.
• All the essential topics have been dealt with carefully and every question has been tackled.
• We have ensured that you have all the essential study material in one place and have a stress free learning experience.
Instasolv’s NCERT Exemplar Solutions for Class 8 Chapter 1 – Rational numbers are extremely easy to access and are absolutely free of cost.
More Chapters from Class 8 |
# Simplifying Square Roots that Contain Variables
If you are looking to simplify square roots that contain numerals as the radicand, then visit our page on how to simplify square roots.
In this lesson, we are going to take it one step further, and simplify square roots that contain variables.
## Square Roots with Just One Variable
When you take the square root of a term that is "squared", your answer is the " base" of that term. Here's what I mean.
So, what happens if the variable is raised to a power other than 2?
Did you see how we rewrote the radicand but still maintained its value? Let's look at another example where a variable is raised to the sixth power.
Now, let's look at a simplifying a square root that contains the product of a number and a variable.
## Simplifying the Square Root of the Product of a Number and a Variable.
Let's look at another example where we simplify the square root of multiple factors. Take note of the variable z in this example. It must remain under the square root since it is z to the first power.
## More Simplifying Square Roots with Multiple Variables
One last example. If you have a variable that is raised to an odd power, you must rewrite it as the product of two squares - one with an even exponent and the other to the first power. Check out the variable x in this example.
## You may also like these topics!
• ### Learn How to Simplify a Square Root in 2 Easy Steps
Learning how to simplify a square root can be broken down into 2 easy steps. First think of the factors and determine if one of those factors is a perfect square.
• ### Everything You Need to Know About Radicals In Math
Your complete guide to studying radicals in math.
• ### Simple Steps for Adding Square Roots
Follow these simple steps to learn how to add square roots
• ### Using the Radical Sign or Symbol
The radical sign is used when taking the square root or the nth root of a number. |
# How do you simplify 2(sqrt3)-4(sqrt2)+6(sqrt3)+8(sqrt2)?
##### 2 Answers
May 29, 2017
See a solution below:
#### Explanation:
First, group like terms. In this problem the like terms are the radicals:
$2 \left(\sqrt{3}\right) + 6 \left(\sqrt{3}\right) - 4 \left(\sqrt{2}\right) + 8 \left(\sqrt{2}\right)$
Next, factor the common terms:
$\left(2 + 6\right) \left(\sqrt{3}\right) + \left(- 4 + 8\right) \left(\sqrt{2}\right)$
$8 \left(\sqrt{3}\right) + 4 \left(\sqrt{2}\right)$
Now, factor out the common term outside the radical:
$\left(4 \times 2\right) \left(\sqrt{3}\right) + 4 \left(\sqrt{2}\right)$
$4 \left(2 \left(\sqrt{3}\right) + \left(\sqrt{2}\right)\right)$
$4 \left(2 \sqrt{3} + \sqrt{2}\right)$
May 29, 2017
color(green)(4(sqrt2+2sqrt3)
#### Explanation:
$2 \left(\sqrt{3}\right) - 4 \left(\sqrt{2}\right) + 6 \left(\sqrt{3}\right) + 8 \left(\sqrt{2}\right)$
$\therefore = 2 \sqrt{3} + 6 \sqrt{3} - 4 \sqrt{2} + 8 \sqrt{2}$
$\therefore = 8 \sqrt{3} + 4 \sqrt{2}$
:.=color(green)(4(sqrt2+2sqrt3)
check:
:.color(green)(2(sqrt3)-4(sqrt2)+6(sqrt3)+8(sqrt2)=19.51326071
:.color(green)(4(sqrt2+2sqrt3)=19.51326071 |
# What is 201/101 as a decimal?
## Solution and how to convert 201 / 101 into a decimal
201 / 101 = 1.99
Fraction conversions explained:
• 201 divided by 101
• Numerator: 201
• Denominator: 101
• Decimal: 1.99
• Percentage: 1.99%
The basis of converting 201/101 to a decimal begins understanding why the fraction should be handled as a decimal. Fractions and decimals represent parts of a whole, sometimes representing numbers less than 1. Depending on the situation, decimals can be more clear. We don't say 1 and 1/2 dollar. We use the decimal version of \$1.50. Same goes for fractions. We will say 'the student got 2 of 3 questions correct'. So let’s dive into how and why you can convert 201/101 into a decimal.
201 / 101 as a percentage 201 / 101 as a fraction 201 / 101 as a decimal
1.99% - Convert percentages 201 / 101 201 / 101 = 1.99
## 201/101 is 201 divided by 101
The first step of teaching our students how to convert to and from decimals and fractions is understanding what the fraction is telling is. 201 is being divided into 101. Think of this as our directions and now we just need to be able to assemble the project! The numerator is the top number in a fraction. The denominator is the bottom number. This is our equation! To solve the equation, we must divide the numerator (201) by the denominator (101). Here's how our equation is set up:
### Numerator: 201
• Numerators represent the number of parts being taken from a denominator. Overall, 201 is a big number which means you'll have a significant number of parts to your equation. The bad news is that it's an odd number which makes it harder to covert in your head. Large two-digit conversions are tough. Especially without a calculator. Now let's explore X, the denominator.
### Denominator: 101
• Denominators represent the total number of parts, located below the vinculum or fraction bar. 101 is a large number which means you should probably use a calculator. But the bad news is that odd numbers are tougher to simplify. Unfortunately and odd denominator is difficult to simplify unless it's divisible by 3, 5 or 7. Ultimately, don't be afraid of double-digit denominators. So without a calculator, let's convert 201/101 from a fraction to a decimal.
## How to convert 201/101 to 1.99
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 101 \enclose{longdiv}{ 201 }$$
We will be using the left-to-right method of calculation. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.
### Step 2: Solve for how many whole groups you can divide 101 into 201
$$\require{enclose} 00.1 \\ 101 \enclose{longdiv}{ 201.0 }$$
How many whole groups of 101 can you pull from 2010? 101 Multiply this number by 101, the denominator to get the first part of your answer!
### Step 3: Subtract the remainder
$$\require{enclose} 00.1 \\ 101 \enclose{longdiv}{ 201.0 } \\ \underline{ 101 \phantom{00} } \\ 1909 \phantom{0}$$
If there is no remainder, you’re done! If you still have numbers left over, continue to the next step.
### Step 4: Repeat step 3 until you have no remainder
Remember, sometimes you won't get a remainder of zero and that's okay. Round to the nearest digit and complete the conversion. There you have it! Converting 201/101 fraction into a decimal is long division just as you learned in school.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals depend on the life situation you need to represent numbers. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. Same goes for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But each represent values in everyday life! Here are examples of when we should use each.
### When you should convert 201/101 into a decimal
Investments - Comparing currency, especially on the stock market are great examples of using decimals over fractions.
### When to convert 1.99 to 201/101 as a fraction
Time - spoken time is used in many forms. But we don't say It's '2.5 o'clock'. We'd say it's 'half passed two'.
### Practice Decimal Conversion with your Classroom
• If 201/101 = 1.99 what would it be as a percentage?
• What is 1 + 201/101 in decimal form?
• What is 1 - 201/101 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 1.99 + 1/2?
### Convert more fractions to decimals
From 201 Numerator From 101 Denominator What is 201/91 as a decimal? What is 191/101 as a decimal? What is 201/92 as a decimal? What is 192/101 as a decimal? What is 201/93 as a decimal? What is 193/101 as a decimal? What is 201/94 as a decimal? What is 194/101 as a decimal? What is 201/95 as a decimal? What is 195/101 as a decimal? What is 201/96 as a decimal? What is 196/101 as a decimal? What is 201/97 as a decimal? What is 197/101 as a decimal? What is 201/98 as a decimal? What is 198/101 as a decimal? What is 201/99 as a decimal? What is 199/101 as a decimal? What is 201/100 as a decimal? What is 200/101 as a decimal? What is 201/101 as a decimal? What is 201/101 as a decimal? What is 201/102 as a decimal? What is 202/101 as a decimal? What is 201/103 as a decimal? What is 203/101 as a decimal? What is 201/104 as a decimal? What is 204/101 as a decimal? What is 201/105 as a decimal? What is 205/101 as a decimal? What is 201/106 as a decimal? What is 206/101 as a decimal? What is 201/107 as a decimal? What is 207/101 as a decimal? What is 201/108 as a decimal? What is 208/101 as a decimal? What is 201/109 as a decimal? What is 209/101 as a decimal? What is 201/110 as a decimal? What is 210/101 as a decimal? What is 201/111 as a decimal? What is 211/101 as a decimal?
### Convert similar fractions to percentages
From 201 Numerator From 101 Denominator 202/101 as a percentage 201/102 as a percentage 203/101 as a percentage 201/103 as a percentage 204/101 as a percentage 201/104 as a percentage 205/101 as a percentage 201/105 as a percentage 206/101 as a percentage 201/106 as a percentage 207/101 as a percentage 201/107 as a percentage 208/101 as a percentage 201/108 as a percentage 209/101 as a percentage 201/109 as a percentage 210/101 as a percentage 201/110 as a percentage 211/101 as a percentage 201/111 as a percentage |
# Using Diagrams to Solve GMAT Rate Problems: Part 1
Diagrams are great! Like all types of scratch-work, diagrams can forestall cognitive fatigue because working a problem out on paper is much less demanding than doing all the work in your head. Diagrams can also help you to visualize relationships, and can make problems more concrete.
Generally though, we use diagrams to generate equations, which we still have to solve. So good scratch-work doesn’t substitute for other mathematical ability.
Today we’ll use speed problems as examples, since they’re the most common sort of GMAT rate problem, but we’ll talk about work-rate and other problems in a couple of days.
Let’s jump into an example!
## This is a pretty stock example:
• It involves exactly two travelers.
• Those travelers are moving in opposite directions. In this case they’re moving apart, but in other problems they’ll be moving toward one another.
• Those travelers are moving simultaneously, at least for the portion of their travels that concern us.
Train X and train Y pass one another traveling in opposite directions. Forty minutes later they are 100 miles apart. If train X’s constant speed is 30 miles per hour greater than train Y’s, how far does train X travel during that time?
A. 28
B. 40
C. 60
D. 72
E. 80
The diagram we’ll use for this problem is a RTD Table. It consists simply of three columns (for rate, time, and distance) and some number of rows. For this problem we’ll use three rows, one for each of the trains and one for their combined rate:
If you like, you can include multiplication and equality signs, to remind yourself how the values in each row are related to one another: RT=D. I’ve included those signs below in red, but in subsequent illustrations of the table I’m going to leave them out. I will include one more thing though. Circle or otherwise indicate the cell for which we’re solving. In this case, that’s X’s distance.
Begin by writing in whatever information is given to you. In this case that would include the combined distance traveled by the two trains, and the time traveled.
A few notes: (1) We didn’t include a distance for either train X or train Y, because we’re not given those. (2) For every story about simultaneous travel, the time value will be the same all the way down the column. (3) In this case that time value is 2/3 of an hour. Yes, it was given to us as 40 minutes, but since the speed is given to us in miles per hour, we need to change something to keep the units consistent. It’s often easier to change from a few large units (like hours) to many small units (like minutes) but in this case I chose to stick with hours. (4) If you’re consistent in your units, and if the units in the table are the same as in the answers, units don’t really matter any more. In math class, you’d probably represent the units, but I’m not going to bother.
Also capture the rate for each of the trains, using as few variables as possible. In this case, that might mean calling train X’s rate simply r and calling train Y’s rate r-30, since we’re given that train X’s constant speed is 30 miles per hour greater than train Y’s. (You could also have called train Y’s rate r and train X’s rate r+30.)
Next add the rates of the two trains to get their combined rate. Every time that you have a story about two travelers moving in opposite directions (whether toward each other or away from each other) and moving simultaneously (starting and stopping at the same time, or both moving for the entire portion of the journey that you’re concerned with) add the rates.
We now have a complete equation in the bottom row. Since RT=D, we can see that (2r-30)(2/3)=100. Solve for r:
(2r-30)(2/3)=100
(4r/3)-20=100
4r/3=120
4r=360
r=90
Notice that the diagram merely generated this equation for us; it didn’t solve the equation. Of course, our ultimate goal is not to solve for r but to solve for the distance traveled by X. Substitute 90 for r and complete the row for X.
## Couldn’t we do that without the table?
Yep. We could have simply noted that for any simultaneous travel problem (combined rate)(time)=(combined distance). We’re given the time (2/3) and combined distance (100 miles). If we can manage to represent the combined rates (2r-30) we should be able to solve for r. If we can also characterize what we’re solving for (2/3 of r) then we can answer the question.
Some people can do all of that without the table.
On the other hand, some people who can do all of that without the table in practice find that they need help organizing and translating information once they’re well into the test. So it’s worthwhile learning to use the table.
Another reason to learn to use the table is that some rate problems are a lot more subtle than this one. We’ll look at such problems in a couple of days.
## Author
• Michael Schwartz is really good at standardized tests. He’s earned multiple perfect scores on the GRE, GMAT, and LSAT. He’d rather have perfect pitch or be able to run low 1:40s for the 800 meters, but you take what you get. He has decades of teaching and curriculum-development experience. One of these days he might finish his dissertation and collect that Ph.D. in philosophy. Might. |
# EXAMPLE 5 Standardized Test Practice SOLUTION Because f (3) = 0, x – 3 is a factor of f (x). Use synthetic division. 3 1 – 2 – 23 60 3 3 – 60 1 1 – 20.
## Presentation on theme: "EXAMPLE 5 Standardized Test Practice SOLUTION Because f (3) = 0, x – 3 is a factor of f (x). Use synthetic division. 3 1 – 2 – 23 60 3 3 – 60 1 1 – 20."— Presentation transcript:
EXAMPLE 5 Standardized Test Practice SOLUTION Because f (3) = 0, x – 3 is a factor of f (x). Use synthetic division. 3 1 – 2 – 23 60 3 3 – 60 1 1 – 20 0
EXAMPLE 5 Use the result to write f (x) as a product of two factors. Then factor completely. f (x) = x 3 – 2x 2 – 23x + 60 The zeros are 3, – 5, and 4. Standardized Test Practice The correct answer is A. ANSWER = (x – 3)(x + 5)(x – 4) = (x – 3)(x 2 + x – 20)
EXAMPLE 6 Use a polynomial model BUSINESS The profit P (in millions of dollars ) for a shoe manufacturer can be modeled by P = – 21x 3 + 46x where x is the number of shoes produced (in millions). The company now produces 1 million shoes and makes a profit of \$25,000,000, but would like to cut back production. What lesser number of shoes could the company produce and still make the same profit?
EXAMPLE 6 Use a polynomial model SOLUTION 25 = – 21x 3 + 46x Substitute 25 for P in P = – 21x 3 + 46x. 0 = 21x 3 – 46x + 25 Write in standard form. You know that x = 1 is one solution of the equation. This implies that x – 1 is a factor of 21x 3 – 46x + 25. Use synthetic division to find the other factors. 1 21 0 – 46 25 21 21 –25 21 21 – 25 0
EXAMPLE 6 Use a polynomial model So, (x – 1)(21x 2 + 21x – 25) = 0. Use the quadratic formula to find that x 0.7 is the other positive solution. The company could still make the same profit producing about 700,000 shoes. ANSWER
GUIDED PRACTICE for Examples 5 and 6 Find the other zeros of f given that f (– 2) = 0. 7. f (x) = x 3 + 2x 2 – 9x – 18 SOLUTION Because f (– 2 ) = 0, x + 2 is a factor of f (x). Use synthetic division. – 2 1 2 – 9 – 18 – 2 0 18 1 0 – 9 0
GUIDED PRACTICE for Examples 5 and 6 Use the result to write f (x) as a product of two factors. Then factor completely. f (x) = x 3 + 2x 2 – 9x – 18 The zeros are 3, – 3, and – 2. = (x + 2)(x + 3)(x – 3) = (x + 2)(x 2 – 9 2 )
GUIDED PRACTICE for Examples 5 and 6 8. f (x) = x 3 + 8x 2 + 5x – 14 SOLUTION Because f (– 2 ) = 0, x + 2 is a factor of f (x). Use synthetic division. – 2 1 8 5 – 14 – 2 –12 14 1 6 – 7 0
GUIDED PRACTICE for Examples 5 and 6 Use the result to write f (x) as a product of two factors. Then factor completely. f (x) = x 3 + 8x 2 + 5x – 14 The zeros are 1, – 7, and – 2. = (x + 2)(x + 7)(x – 1) = (x + 2)(x 2 + 6x – 7 )
GUIDED PRACTICE for Examples 5 and 6 9. What if? In Example 6, how does the answer change if the profit for the shoe manufacturer is modeled by P = – 15x 3 + 40x ? SOLUTION 25 = – 15x 3 + 40x Substitute 25 for P in P = – 15x 3 + 40x. 0 = 15x 3 – 40x + 25 Write in standard form. You know that x = 1 is one solution of the equation. This implies that x – 1 is a factor of 15x 3 – 40x + 25. Use synthetic division to find the other factors. 1 15 0 – 40 25 15 15 –25 15 15 – 25 0
GUIDED PRACTICE for Examples 5 and 6 So, (x – 1)(15x 2 + 15x – 25) = 0. Use the quadratic formula to find that x 0.9 is the other positive solution. The company could still make the same profit producing about 900,000 shoes. ANSWER
Download ppt "EXAMPLE 5 Standardized Test Practice SOLUTION Because f (3) = 0, x – 3 is a factor of f (x). Use synthetic division. 3 1 – 2 – 23 60 3 3 – 60 1 1 – 20."
Similar presentations |
# Quick Answer: How Do You Find An Angle Of A Triangle?
## How do you find an angle?
The most common way to measure angles is in degrees, with a full circle measuring 360 degrees.
You can calculate the measure of an angle in a polygon if you know the shape of the polygon and the measure of its other angles or, in the case of a right triangle, if you know the measures of two of its sides..
## How do you find the outside angle of a triangle?
The measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles. (Non-adjacent interior angles may also be referred to as remote interior angles.) FACTS: An exterior ∠ is equal to the addition of the two Δ angles not right next to it.
## How do you find the size of an angle?
Subtract the given supplementary angle (its value in degrees) from 180 to calculate the size of the angle in question. Supplementary angles, or straight angles, are those whose sum adds up to 180 degrees.
## What is the external angle of a triangle?
An exterior (or external) angle is the angle between one side of a triangle and the extension of an adjacent side.
## How do you find the angle of a triangle given two sides?
ExampleStep 1 The two sides we know are Adjacent (6,750) and Hypotenuse (8,100).Step 2 SOHCAHTOA tells us we must use Cosine.Step 3 Calculate Adjacent / Hypotenuse = 6,750/8,100 = 0.8333.Step 4 Find the angle from your calculator using cos-1 of 0.8333:
## Which tool do you use to find the angles of a shape?
A protractor is the most common device used to measure angles.
## How do you find an angle without a protractor?
How to Calculate Angles Without a ProtractorMark Two Points on the Line Opposite the Angle.Measure the Line.Use the Sine Formula.Calculate the Angle.
## How do you find a missing angle?
We can use two different methods to find our missing angle:Subtract the two known angles from 180° :Plug the two angles into the formula and use algebra: a + b + c = 180°
## How do you find the measure of an angle in a triangle?
If we add all three angles in any triangle we get 180 degrees. So, the measure of angle A + angle B + angle C = 180 degrees. This is true for any triangle in the world of geometry. We can use this idea to find the measure of angle(s) where the degree measure is missing or not given.
## How do you find an angle of a triangle with 3 sides?
“SSS” is when we know three sides of the triangle, and want to find the missing angles….To solve an SSS triangle:use The Law of Cosines first to calculate one of the angles.then use The Law of Cosines again to find another angle.and finally use angles of a triangle add to 180° to find the last angle.
## What is the measure of an obtuse angle?
An obtuse angle has a measurement greater than 90 degrees but less than 180 degrees. However, A reflex angle measures more than 180 degrees but less than 360 degrees. So, an angle of 190 degrees would be reflex. Comment on Anthony Fritz’s post “A 190 degree angle is not obtuse.
## What affects the size of an angle?
Can you answer it? The direction of one side affects the size of an angle.
## How do you find the side and side of an angle?
“SAS” is when we know two sides and the angle between them. use The Law of Cosines to calculate the unknown side, then use The Law of Sines to find the smaller of the other two angles, and then use the three angles add to 180° to find the last angle. |
# 1994 AIME Problems/Problem 6
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
The graphs of the equations
$y=k, \qquad y=\sqrt{3}x+2k, \qquad y=-\sqrt{3}x+2k,$
are drawn in the coordinate plane for $k=-10,-9,-8,\ldots,9,10.\,$ These 63 lines cut part of the plane into equilateral triangles of side length $\tfrac{2}{\sqrt{3}}.\,$ How many such triangles are formed?
## Solution
### Solution 1
We note that the lines partition the hexagon of the six extremal lines into disjoint unit regular triangles, and forms a series of unit regular triangles along the edge of the hexagon.
$[asy] size(200); picture pica, picb, picc; int i; for(i=-10;i<=10;++i){ if((i%10) == 0){draw(pica,(-20/sqrt(3)-abs((0,i))/sqrt(3),i)--(20/sqrt(3)+abs((0,i))/sqrt(3),i),black+0.7);} else{draw(pica,(-20/sqrt(3)-abs((0,i))/sqrt(3),i)--(20/sqrt(3)+abs((0,i))/sqrt(3),i));} } picb = rotate(120,origin)*pica; picc = rotate(240,origin)*pica; add(pica);add(picb);add(picc); [/asy]$
Solving the above equations for $k=\pm 10$, we see that the hexagon in question is regular, with side length $\frac{20}{\sqrt{3}}$. Then, the number of triangles within the hexagon is simply the ratio of the area of the hexagon to the area of a regular triangle. Since the ratio of the area of two similar figures is the square of the ratio of their side lengths, we see that the ratio of the area of one of the six equilateral triangles composing the regular hexagon to the area of a unit regular triangle is just $\left(\frac{20/\sqrt{3}}{2/\sqrt{3}}\right)^2 = 100$. Thus, the total number of unit triangles is $6 \times 100 = 600$.
There are $6 \cdot 10$ equilateral triangles formed by lines on the edges of the hexagon. Thus, our answer is $600+60 = \boxed{660}$.
### Solution 2
There are three types of lines: horizontal, upward-slanting diagonal, and downward-slanting diagonal. There are $21$ of each type of line, and a unit equilateral triangle is determined by exactly one of each type of line. Given an upward-slanting diagonal and a downward-slanting diagonal, they determine exactly two unit equilateral triangles as shown below. $[asy] size(60); pair u=rotate(60)*(1,0),d=rotate(-60)*(1,0),h=(1,0); draw((0,0)--4*u^^-2*h+4*u--(-2*h+4*u+4*d)); draw(u--2*u+d,dotted); draw(3*u--3*u-h,dotted); [/asy]$ Therefore, if all horizontal lines are drawn, there will be a total of $2\cdot 21^2=882$ unit equilateral triangles. Of course, we only draw $21$ horizontal lines, so we are overcounting the triangles that are caused by the undrawn horizontal lines. In the below diagram, we draw the diagonal lines and the highest and lowest horizontal lines. $[asy] size(200); pair u=rotate(60)*(2/sqrt(3),0),d=rotate(-60)*(2/sqrt(3),0),h=(2/sqrt(3),0); for (int i=0;i<21;++i) {path v=(-20/sqrt(3),0)-2*u+i*u--(0,20)--(20/sqrt(3),0)+2*d-i*d;draw(shift(0,-2*i)*v);} for (int i=0;i<21;++i) {path v=rotate(180)*((-20/sqrt(3),0)-2*u+i*u--(0,20)--(20/sqrt(3),0)+2*d-i*d);draw(shift(0,2*i)*v);} draw((-15,-10)--(15,-10)); draw((-15,10)--(15,10)); [/asy]$
We see that the lines $y=-21,-20,\dots, -11$ and $y=11,12,\dots,21$ would complete several of the $882$ unit equilateral triangles. In fact, we can see that the lines $y=-21,-20,\dots,-11$ complete $1,2,(1+3),(2+4),(3+5),(4+6),\dots,(9+11)$ triangles, or $111$ triangles. The positive horizontal lines complete the same number of triangles, hence the answer is $882-2\cdot 111=\boxed{660}$.
### Solution 3 Elementary Counting
Picturing the diagram in your head should give you an illustration similar to the one above. The distance from parallel sides of the center hexagon is 20 units, and by extending horizontal lines to the sides of the hexagon. This shows that for every side of the hexagon there are 10 spaces. Therefore the side length of the biggest triangle (imagine one of the overlapping triangles in the Star of David) is 30. The total number of triangles in the hexagon can be found by finding the number of triangles in the extended triangle and subtracting the 3 corner triangles. This gives us $30^2 - 10^2 - 10^2- 10^2 = 600$. That is the number of triangles in the hexagon. The remaining triangles form in groups of 10 on the exterior of each side. $600 + 6 * 10 = \boxed{660}$.
-jackshi2006 |
Question Video: Determining Conditional Probabilities Involving Dice | Nagwa Question Video: Determining Conditional Probabilities Involving Dice | Nagwa
# Question Video: Determining Conditional Probabilities Involving Dice Mathematics • Third Year of Secondary School
## Join Nagwa Classes
Two dice are rolled to give a pair of numbers. Given that both numbers are greater than 1, what is the probability that they are both equal to 2?
03:35
### Video Transcript
Two dice are rolled to give a pair of numbers. Given that both numbers are greater than one, what is the probability that they are both equal to two?
Now we could apply a little bit of logic to calculate the probability here. However, the phrase “given that” tells us that we’re actually working with a conditional probability. And in this case, we actually have a formula that we can use. The formula tells us that the probability of an event 𝐴 occurring given that an event 𝐵 has already occurred is equal to the probability of 𝐴 intersection 𝐵 over the probability of 𝐵. Remember, this vertical lines simply means given that. And this symbol that looks a little bit like a lowercase n is the intersection. When we talk about the intersection, we’re talking about 𝐴 and 𝐵 occurring.
So we’re going to define our two events to start with. We’re going to say that 𝐴 is the event that both dice are equal to two. Then 𝐵 is the event that both numbers are greater than one. And that’s great because the probability of 𝐴 given 𝐵 is now the probability that both dice are equal to two given that both numbers are greater than one. We now need to calculate the probability of 𝐴 and 𝐵 or 𝐴 intersection 𝐵 and the probability of 𝐵.
And so we’re going to recall that for two independent events, let’s now all those 𝐶 and 𝐷, the probability of 𝐶 intersection 𝐷 occurring is equal to the probability of 𝐶 times the probability of 𝐷. Now we chose 𝐶 and 𝐷 very specifically here. The outcomes we’re now interested in is the score we get on each individual dice. Note that the score on one dice doesn’t affect the score on the other. And so rolling each dice is an independent event.
Let’s begin with the probability of 𝐵. We know that this is both numbers being greater than one. Well, that means each dice could have the numbers two, three, four, five, or six on them. The probability that the score on one dice is greater than one then is five-sixths. So the the probability that the score is greater than one on both dice is five-sixths times five-sixths. When we multiply fractions, we simply multiply their two denominators and their two numerators. So we get 25 over 36.
But what about the probability of 𝐴 intersection 𝐵? This is the probability that both dice are equal to two and that both numbers are greater than one. Well, two is always greater than one. And so actually the probability of 𝐴 intersection 𝐵 here is the same as the probability of 𝐴. The probability that the number on one dice is equal to two is one-sixth. So the probability that both dice are equal to two is one-sixth times one-sixth. And that’s equal to one over 36. And we’re now ready to use the conditional probability formula.
The probability of 𝐴 given that 𝐵 has already occurred then is one over 36 all divided 25 over 36. We know that to divide by a fraction, we simply multiply by the reciprocal of that fraction. So we get one over 36 times 36 over 25. Then we spot that we can cross cancel by dividing through by a common factor of 36. We then have one over one times one over 25. And that’s equal to one over 25. Given that both numbers are greater than one then, the probability that they’re both equal to two is one out of 25.
## Join Nagwa Classes
Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!
• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions |
# NCERT Solutions for Class 8 Maths Chapter 11 :Direct and Inverse Proportion CBSE Exercise 11.1
In this page we have NCERT book Solutions for Class 8th Maths Chapter 11 :Direct and Inverse Proportion for Exercise 11.1. Hope you like them and do not forget to like , social share and comment at the end of the page.
Question 1
Following are the car parking charges near a railway station upto
4 hours Rs 60
8 hours Rs 100
12 hours Rs 140
24 hours Rs 180
Check if the parking charges are in direct proportion to the parking time.
We know that two quantities are in direct proportion if whenever the values of one quantity increase, then the value of another quantity increase in such a way that ratio of the quantities remains same
Here The charges are not increasing in direct proportion to the parking time because
4/60 ≠ 8/100 ≠ 12/140 ≠ 24/180
Question 2
A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.
Parts of red pigments 1 4 7 12 20 Part of base 8
The ratio of Parts of red pigments and part of base = 1/8
Parts of red pigments 1 4 7 12 20 Part of base 8 a b c d
Now Parts of red pigments and part of base are in direct proportion
Now according to law of direct proportion
4/a =1/8
a=32
7/b= 1/8
b=56
12/c=1/8
c=96
20/d=1/8
d= 160
So results is
Parts of red pigments 1 4 7 12 20 Part of base 8 32 56 96 160
Question 3
In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?
Let a be the red pigment part required with 1800ml of base,the as per law of direct proportion
1/75= a/1800
a=24
Question 4
A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Let x bottles will be filled in 5 hours,
Now hours and amount of bottles are in direct proportion
Then as per law of direct proportion
840/6= a/5
a=700
Question 5
A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?
Let x be the actual length of bacteria
Now length of bacteria and enlargement are in direct proportion, so
5/50000= x/1
Or x=1/10000 cm
Now let us assume y be the length when it is enlarged 20,000
Again as per law of direct proportion
y/20000= 5/50000
y= 2cm
Question 6
In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?
The model of the ship and actual ship are in direct proportion
Let x be the length of the model of the ship, then
9/12= x/28
X= 21 cm
Question 7
Suppose 2 kg of sugar contains 9 × 106 crystals. How many sugar crystals are there in (i) 5 kg of sugar? (ii) 1.2 kg of sugar?
Amount of sugar in kg is direct proportion to amount of crystals
1. Let x be the crystals in 5k,then 2/(9×106)= 5/x
x=45 × 54 crystals
2. Let y be the crystals in 1.2k,then 2/(9×106)= 1.2/x
x=10.8 × 53 crystals
Question 8
Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?
Answer: Length of scale and distance covered are in direct proportion
Let x be the scale in map
1/18-=x/72
x=4 cm
Question 9
A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time
(i) the length of the shadow cast by another pole 10 m 50 cm high
(ii) the height of a pole which casts a shadow 5m long.
Answer: Length of the pole and length of shadow are in direct proportion
Also let us convert everything in meter
1. Let x be the length of the shadow cast by another pole 10 m 50 cm high 6.5/3.2= 10.5/x
Or x =6 m
2. Let x be the length of the shadow cast by another pole 10 m 50 cm high 6.5/3.2= 5/x
Or x =8.75 m
Question 10
A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
As speed is same, then distance travelled and time taken will be in direct proportion
So let us assume x be the distance travelled in 5 hours or 300 minutes
14/25= x/300
x=168 km |
Significant digits
# Significant digits
## Significant digits
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Significant digits
2. 1. Non-zero digits are ALWAYS significant. Example- 123456789 2. Zeros that are between other significant digits are ALWAYS significant. Example- 3004 3. Final zeros that occur before or after the decimal are ALWAYS significant. Example- 78.00 4. Zeros used for spacing the decimal are NOT significant. Example- 0.003 RULES FOR SIGNIFICANT DIGITS
3. EXAMPLES
4. 2804 0.00305 2.84 30.24 0.0029 6.89 4.6 X 104 0.001 75.00 6.0 50000 123.056 360 100.04 HOW MANY SIGNIFICANT DIGITS?
5. When you have numbers to add/subtract you will sometimes need your answer to have the correct number of significant digits. Here is the rule: Look at all of the numbers to add/subtract and find the number with the LEAST number of sig. digs. AFTER the decimal. This is how many sig. digs. should be after the decimal in your answer. Example - 24.686 2.343 3.21 30.239 ANSWER = 30.24 ADDING & SUBTRACTING
6. When you have numbers to multiply/divide you will sometimes need your answer to have the correct number of significant digits. Here is the rule: Look at all of the numbers to multiply/divide and find the number with the LEAST number of sig. digs. This is how many sig. digs. that you should have in your answer. Example - 3.22 x 2.1 = 6.762 ANSWER = 6.8 MULTIPLYING & DIVIDING
7. 1.2 + 1.26 = • 2.46 • Final Answer: 2.5 • 2.476-1.72 = • 0.756 • Final Answer: 0.76 • 10.72694321 – 10.1 = • 0.62694321 • Final Answer: 0.6 EXAMPLE PROBLEMS FOR +/-
8. 1.53 x 2.5 • 3.825 • Final Answer: 3.8 • 10.0 / 5.0 = • 2 • Final Answer: 2.0 • 10.2 x 1000 • 102,000 • Final Answer: 10.2 x 10³ EXAMPLE PROBLEMS FOR X/÷ |
# What is the derivative? sqrt( (ax) / (a^2 + x^2)
Oct 4, 2017
$\frac{d}{\mathrm{dx}} \sqrt{\frac{a x}{{a}^{2} + {x}^{2}}} = \frac{1}{2 \sqrt{a x}} . \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ \left(\frac{3}{2}\right)$
#### Explanation:
We seek:
$\frac{d}{\mathrm{dx}} \sqrt{\frac{a x}{{a}^{2} + {x}^{2}}}$
We need to apply the chain rule and the quotient rule. For clarity we will perform explicit substitutions for the product rule and the chain rule, with practice this can be done in situ:
Let us use the following variables:
Let { (u,=ax, ), (v,=a^2+x^2, ), (w,=u/v, =(ax) / (a^2 + x^2) ), (y,=sqrt(w),=sqrt((ax) / (a^2 + x^2))) :}
Then let us start some differentiation, we start by differentiating the two parts of the quotient $u$ and $v$ wrt $x$:
$\frac{\mathrm{du}}{\mathrm{dx}} = a$ ..... [A]
$\frac{\mathrm{dv}}{\mathrm{dx}} = 2 x$ ..... [B]
And also, we can differentiate $y$ wrt $w$:
$\frac{\mathrm{dy}}{\mathrm{dw}} = \frac{1}{2} {w}^{- \frac{1}{2}} = \frac{1}{2 \sqrt{w}}$ ..... [C}
So we can apply the quotient rule to find the derivative of $w$ wrt $x$:
$\frac{\mathrm{dw}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$ ........ "The Quotient Rule"
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{\left({a}^{2} + {x}^{2}\right) \left(a\right) - \left(a x\right) \left(2 x\right)}{{a}^{2} + {x}^{2}} ^ 2 \setminus \setminus \setminus \setminus$ Using [A] and [B]
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{{a}^{3} + a {x}^{2} - 2 a {x}^{2}}{{a}^{2} + {x}^{2}} ^ 2$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ 2$ ..... [D]
We can use the chain rule to get:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dw}} . \frac{\mathrm{dw}}{\mathrm{dx}}$
$\setminus \setminus \setminus \setminus \setminus = \frac{1}{2 \sqrt{w}} . \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ 2 \setminus \setminus \setminus \setminus$ from [C] & [D]
Now we can replace $w$ with our earlier substitution:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{\left(\frac{u}{v}\right)}} . \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ 2$
$\setminus \setminus \setminus \setminus \setminus = \frac{1}{2 \sqrt{\left(\frac{a x}{{a}^{2} + {x}^{2}}\right)}} . \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ 2$
$\setminus \setminus \setminus \setminus \setminus = \frac{1}{2} . \sqrt{\frac{{a}^{2} + {x}^{2}}{a x}} . \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ 2$
$\setminus \setminus \setminus \setminus \setminus = \frac{1}{2 \sqrt{a x}} . {\left({a}^{2} + {x}^{2}\right)}^{\frac{1}{2}} . \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ 2$
$\setminus \setminus \setminus \setminus \setminus = \frac{1}{2 \sqrt{a x}} . \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ \left(\frac{3}{2}\right)$
With practice we can just write these derivatives by applying the rules as we go and using the chain rule implicitly
$y = {\left(\frac{a x}{{a}^{2} + {x}^{2}}\right)}^{\frac{1}{2}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {\left(\frac{a x}{{a}^{2} + {x}^{2}}\right)}^{- \frac{1}{2}} \frac{d}{\mathrm{dx}} \left(\frac{a x}{{a}^{2} + {x}^{2}}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2} {\left(\frac{a x}{{a}^{2} + {x}^{2}}\right)}^{- \frac{1}{2}} \frac{\left({a}^{2} + {x}^{2}\right) \left(a\right) - \left(2 x\right) \left(a x\right)}{{a}^{2} + {x}^{2}} ^ 2$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2} {\left(\frac{a x}{{a}^{2} + {x}^{2}}\right)}^{- \frac{1}{2}} \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ 2$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2} {\left(a x\right)}^{- \frac{1}{2}} \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ \left(\frac{3}{2}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2 {\left(a x\right)}^{\frac{1}{2}}} \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ \left(\frac{3}{2}\right)$
Giving us the same result with much lkess work. |
# Ages of Three Children puzzle facts for kids
Kids Encyclopedia Facts
The Ages of Three Children puzzle is a logic puzzle which on first inspection seems to have insufficient information to solve, but which rewards those who persist and examine the puzzle critically.
## The puzzle
A census taker approaches a woman leaning on her gate and asks about her children. She says, "I have three children and the product of their ages is seventy–two. The sum of their ages is the number on this gate." The census taker does some calculation and claims not to have enough information. The woman enters her house, but before slamming the door tells the census taker, "I have to see to my eldest child who is in bed with measles." The census taker departs, satisfied.
The problem can be presented in different ways, giving the same basic information: the product, that the sum is known, and that there is an oldest child (e.g. their ages adding up to today's date, or the eldest being good at chess).
Another version of the puzzle gives the age product as thirty–six, which leads to a different set of ages for the children.
## Solutions
### for 72
The prime factors of 72 are 2, 2, 2, 3, 3; in other words, 2 × 2 × 2 × 3 × 3 = 72
This gives the following triplets of possible solutions;
Age one Age two Age three Total (Sum)
1 1 72 74
1 2 36 39
1 3 24 28
1 4 18 23
1 6 12 19
1 8 9 18
2 2 18 22
2 3 12 17
2 4 9 15
2 6 6 14
3 3 8 14
3 4 6 13
Because the census taker knew the total (from the number on the gate) but said that he had insufficient information to give a definitive answer; thus, there must be more than one solution with the same total.
Only two sets of possible ages add up to the same totals:
A. 2 + 6 + 6 = 14
B. 3 + 3 + 8 = 14
In case 'A', there is no 'eldest child' - two children are aged six (although one could be a few minutes or around 9 to 12 months older and they still both be 6). Therefore, when told that one child is the eldest, the census-taker concludes that the correct solution is 'B'.
### for 36
The prime factors of 36 are 2, 2, 3, 3 This gives the following triplets of possible solutions;
Age one Age two Age three Total (Sum)
1 1 36 38
1 2 18 21
1 3 12 16
1 4 9 14
1 6 6 13
2 2 9 13
2 3 6 11
3 3 4 10
Using the same argument as before it becomes clear that the number on the gate is 13, and the ages 9, 2 and 2.
Ages of Three Children puzzle Facts for Kids. Kiddle Encyclopedia. |
#### Thank you for registering.
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
CART 0
• 0
MY CART (5)
Use Coupon: CART20 and get 20% off on all online Study Material
ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.
There are no items in this cart.
Continue Shopping
# Chapter 15: Properties of Triangles Exercise – 15.2
### Question: 1
Two angles of a triangle are of measures 150° and 30°. Find the measure of the third angle.
### Solution:
Let the third angle be x
Sum of all the angles of a triangle = 180°
105° + 30° + x = 180°
135° + x = 180°
x = 180° – 135°
x = 45°
Therefore the third angle is 45°
### Question: 2
One of the angles of a triangle is 130°, and the other two angles are equal. What is the measure of each of these equal angles?
### Solution:
Let the second and third angle be x
Sum of all the angles of a triangle = 180°
130° + x + x = 180°
130° + 2x = 180°
2x = 180°– 130°
2x = 50°
x = 50/2
x = 25°
Therefore the two other angles are 25° each
### Question: 3
The three angles of a triangle are equal to one another. What is the measure of each of the angles?
### Solution:
Let the each angle be x
Sum of all the angles of a triangle =180°
x + x + x = 180°
3x = 180°
x = 180/3
x = 60°
Therefore angle is 60° each
### Question: 4
If the angles of a triangle are in the ratio 1: 2: 3, determine three angles.
### Solution:
If angles of the triangle are in the ratio 1: 2: 3 then take first angle as ‘x’, second angle as ‘2x’ and third angle as ‘3x’
Sum of all the angles of a triangle=180°
x + 2x + 3x = 180°
6x = 180°
x = 180/6
x = 30°
2x = 30° × 2 = 60°
3x = 30° × 3 = 90°
Therefore the first angle is 30°, second angle is 60° and third angle is 90°
### Question: 5
The angles of a triangle are (x − 40) °, (x − 20) ° and (1/2 − 10) °. Find the value of x.
### Question: 6
The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10°. Find the three angles.
### Solution:
Let the first angle be x
Second angle be x + 10°
Third angle be x + 10° + 10°
Sum of all the angles of a triangle = 180°
x + x + 10° + x + 10° +10° = 180°
3x + 30 = 180
3x = 180 - 30
3x = 150
x = 150/3
x = 50°
First angle is 50°
Second angle x + 10° = 50 + 10 = 60°
Third angle x + 10° +10° = 50 + 10 + 10 = 70°
### Question: 7
Two angles of a triangle are equal and the third angle is greater than each of those angles by 30°. Determine all the angles of the triangle
### Solution:
Let the first and second angle be x
The third angle is greater than the first and second by 30° = x + 30°
The first and the second angles are equal
Sum of all the angles of a triangle = 180°
x + x + x + 30° = 180°
3x + 30 = 180
3x = 180 - 30
3x = 150
x = 150/3
x = 50°
Third angle = x + 30° = 50° + 30° = 80°
The first and the second angle is 50° and the third angle is 80°
### Question: 8
If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.
### Solution:
One angle of a triangle is equal to the sum of the other two
x = y + z
Let the measure of angles be x, y, z
x + y + z = 180°
x + x = 180°
2x = 180°
x = 180/2
x = 90°
If one angle is 90° then the given triangle is a right angled triangle
### Question: 9
If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.
### Solution:
Each angle of a triangle is less than the sum of the other two
Measure of angles be x, y and z
x > y + z
y < x + z
z < x + y
Therefore triangle is an acute triangle
### Question: 10
In each of the following, the measures of three angles are given. State in which cases the angles can possibly be those of a triangle:
(i) 63°, 37°, 80°
(ii) 45°, 61°, 73°
(iii) 59°, 72°, 61°
(iv) 45°, 45°, 90°
(v) 30°, 20°, 125°
### Solution:
(i) 63°, 37°, 80° = 180°
Angles form a triangle
(ii) 45°, 61°, 73° is not equal to 180°
Therefore not a triangle
(iii) 59°, 72°, 61° is not equal to 180°
Therefore not a triangle
(iv) 45°, 45°, 90° = 180°
Angles form a triangle
(v) 30°, 20°, 125° is not equal to 180°
Therefore not a triangle
### Question: 11
The angles of a triangle are in the ratio 3: 4: 5. Find the smallest angle
### Solution:
Given that
Angles of a triangle are in the ratio: 3: 4: 5
Measure of the angles be 3x, 4x, 5x
Sum of the angles of a triangle =180°
3x + 4x + 5x = 180°
12x = 180°
x = 180/12
x = 15°
Smallest angle = 3x
=3 × 15°
= 45°
### Question: 12
Two acute angles of a right triangle are equal. Find the two angles.
### Solution:
Given acute angles of a right angled triangle are equal
Right triangle: whose one of the angle is a right angle
Measured angle be x, x, 90°
x + x + 180°= 180°
2x = 90°
x = 90/2
x = 45°
The two angles are 45° and 45°
### Question: 13
One angle of a triangle is greater than the sum of the other two. What can you say about the measure of this angle? What type of a triangle is this?
### Solution:
Angle of a triangle is greater than the sum of the other two
Measure of the angles be x, y, z
x > y + z or
y > x + z or
z > x + y
x or y or z > 90° which is obtuse
Therefore triangle is an obtuse angle
### Question: 14
AC, AD and AE are joined. Find
∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA
∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA
### Solution:
We know that sum of the angles of a triangle is 180°
Therefore in ∆ABC, we have
∠CAB + ∠ABC + ∠BCA = 180° — (i)
In ∆ACD, we have
∠DAC + ∠ACD + ∠CDA = 180° — (ii)
In ∆AEF, we have
∠FAE + ∠AEF + ∠EFA = 180° — (iv)
Adding (i), (ii), (iii), (iv) we get
∠CAB + ∠ABC + ∠BCA + ∠DAC + ∠ACD + ∠CDA + ∠EAD + ∠ADE + ∠DEA + ∠FAE + ∠AEF +∠EFA = 720°
Therefore ∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA = 720°
### Question: 15
Find x, y, z (whichever is required) from the figures given below:
### Solution:
(i) In ∆ABC and ∆ADE we have:
x = 40°
∠AED = ∠ACB (corresponding angles)
y = 30°
We know that the sum of all the three angles of a triangle is equal to 180°
x + y + z = 180° (Angles of ∆ADE)
Which means: 40° + 30° + z = 180°
z = 180° - 70°
z = 110°
Therefore, we can conclude that the three angles of the given triangle are 40°, 30° and 110°.
(ii) We can see that in ∆ADC, ∠ADC is equal to 90°.
We also know that the sum of all the angles of a triangle is equal to 180°.
Which means: 45° + 90° + y = 180° (Sum of the angles of ∆ADC)
135° + y = 180°
y = 180° – 135°.
y = 45°.
We can also say that in ∆ABC, ∠ABC + ∠ACB + ∠BAC is equal to 180°.
(Sum of the angles of ∆ABC)
40° + y + (x + 45°) = 180°
40° + 45° + x + 45° = 180° (y = 45°)
x = 180° –130°
x = 50°
Therefore, we can say that the required angles are 45° and 50°.
(iii) We know that the sum of all the angles of a triangle is equal to 180°.
Therefore, for △ABD:
∠ABD +∠ADB + ∠BAD = 180° (Sum of the angles of ∆ABD)
50° + x + 50° = 180°
100° + x = 180°
x = 180° – 100°
x = 80°
For ∆ABC:
∠ABC + ∠ACB + ∠BAC = 180° (Sum of the angles of ∆ABC)
50° + z + (50° + 30°) = 180°
50° + z + 50° + 30° = 180°
z = 180° – 130°
z = 50°
Using the same argument for ∆ADC:
∠ADC + ∠ACD + ∠DAC = 180° (Sum of the angles of ∆ADC)
y +z + 30° =180°
y + 50° + 30° = 180° (z = 50°)
y = 180° – 80°
y = 100°
Therefore, we can conclude that the required angles are 80°, 50° and 100°.
(iv) In ∆ABC and ∆ADE we have:
y = 50°
Also, ∠AED = ∠ACB (Corresponding angles)
z = 40°
We know that the sum of all the three angles of a triangle is equal to 180°.
Which means: x + 50° + 40° = 180° (Angles of ∆ADE)
x = 180° – 90°
x = 90°
Therefore, we can conclude that the required angles are 50°, 40° and 90°.
### Question: 16
If one angle of a triangle is 60° and the other two angles are in the ratio 1: 2, find the angles
### Solution:
We know that one of the angles of the given triangle is 60°. (Given)
We also know that the other two angles of the triangle are in the ratio 1: 2.
Let one of the other two angles be x.
Therefore, the second one will be 2x.
We know that the sum of all the three angles of a triangle is equal to 180°.
60° + x + 2x = 180°
3x = 180° – 60°
3x = 120°
x = 120/3
x = 40°
2x = 2 × 40
2x = 80°
Hence, we can conclude that the required angles are 40° and 80°.
### Question: 17
It one angle of a triangle is 100° and the other two angles are in the ratio 2: 3. find the angles.
### Solution:
We know that one of the angles of the given triangle is 100°.
We also know that the other two angles are in the ratio 2: 3.
Let one of the other two angles be 2x.
Therefore, the second angle will be 3x.
We know that the sum of all three angles of a triangle is 180°.
100° + 2x + 3x = 180°
5x = 180° – 100°
5x = 80°
x = 80/5
2x = 2 ×16
2x = 32°
3x = 3×16
3x = 48°
Thus, the required angles are 32° and 48°.
### Question: 18
In △ABC, if 3∠A = 4∠B = 6∠C, calculate the angles.
### Solution:
We know that for the given triangle, 3∠A = 6∠C
∠A = 2∠C — (i)
We also know that for the same triangle, 4∠B = 6∠C
∠B = (6/4)∠C — (ii)
We know that the sum of all three angles of a triangle is 180°.
Therefore, we can say that:
∠A + ∠B + ∠C = 180° (Angles of ∆ABC) — (iii)
On putting the values of ∠A and ∠B in equation (iii), we get:
2∠C + (6/4)∠C +∠C = 180°
(18/4) ∠C = 180°
∠C = 40°
From equation (i), we have:
∠A = 2∠C = 2 × 40
∠A = 80°
From equation (ii), we have:
∠B = (6/4)∠C = (6/4) × 40°
∠B = 60°
∠A = 80°, ∠B = 60°, ∠C = 40°
Therefore, the three angles of the given triangle are 80°, 60°, and 40°.
### Question: 19
Is it possible to have a triangle, in which
(i) Two of the angles are right?
(ii) Two of the angles are obtuse?
(iii) Two of the angles are acute?
(iv) Each angle is less than 60°?
(v) Each angle is greater than 60°?
(vi) Each angle is equal to 60°
### Solution:
(i) No, because if there are two right angles in a triangle, then the third angle of the triangle must be zero, which is not possible.
(ii) No, because as we know that the sum of all three angles of a triangle is always 180°. If there are two obtuse angles, then their sum will be more than 180°, which is not possible in case of a triangle.
(iii) Yes, in right triangles and acute triangles, it is possible to have two acute angles.
(iv) No, because if each angle is less than 60°, then the sum of all three angles will be less than 180°, which is not possible in case of a triangle.
Proof:
Let the three angles of the triangle be ∠A, ∠B and ∠C.
As per the given information,
∠A < 60° … (i)
∠B< 60° … (ii)
∠C < 60° … (iii)
On adding (i), (ii) and (iii), we get:
∠A + ∠B + ∠C < 60°+ 60°+ 60°
∠A + ∠B + ∠C < 180°
We can see that the sum of all three angles is less than 180°, which is not possible for a triangle.
Hence, we can say that it is not possible for each angle of a triangle to be less than 60°.
(v) No, because if each angle is greater than 60°, then the sum of all three angles will be greater than 180°, which is not possible.
Proof:
Let the three angles of the triangle be ∠A, ∠B and ∠C. As per the given information,
∠A > 60° … (i)
∠B > 60° … (ii)
∠C > 60° … (iii)
On adding (i), (ii) and (iii), we get:
∠A + ∠B + ∠C > 60°+ 60°+ 60°
∠A + ∠B + ∠C > 180°
We can see that the sum of all three angles of the given triangle are greater than 180°, which is not possible for a triangle.
Hence, we can say that it is not possible for each angle of a triangle to be greater than 60°.
(vi) Yes, if each angle of the triangle is equal to 60°, then the sum of all three angles will be 180° , which is possible in case of a triangle.
Proof:
Let the three angles of the triangle be ∠A, ∠B and ∠C. As per the given information,
∠A = 60° … (i)
∠B = 60° …(ii)
∠C = 60° … (iii)
On adding (i), (ii) and (iii), we get:
∠A + ∠B + ∠C = 60°+ 60°+ 60°
∠A + ∠B + ∠C =180°
We can see that the sum of all three angles of the given triangle is equal to 180°, which is possible in case of a triangle. Hence, we can say that it is possible for each angle of a triangle to be equal to 60°.
### Question: 20
In ∆ABC, ∠A = 100°, AD bisects ∠A and AD perpendicular BC. Find ∠B
### Solution:
Consider ∆ABD
We know that the sum of all three angles of a triangle is 180°.
Thus,
∠ABD + ∠BAD + ∠ADB = 180° (Sum of angles of ∆ABD)
Or,
∠ABD + 50° + 90° = 180°
∠ABD =180° – 140°
∠ABD = 40°
### Question: 21
In ∆ABC, ∠A = 50°, ∠B = 100° and bisector of ∠C meets AB in D. Find the angles of the triangles ADC and BDC
### Solution:
We know that the sum of all three angles of a triangle is equal to 180°.
Therefore, for the given △ABC, we can say that:
∠A + ∠B + ∠C = 180° (Sum of angles of ∆ABC)
50° + 70° + ∠C = 180°
∠C= 180° –120°
∠C = 60°
∠ACD = ∠BCD =∠C2 (CD bisects ∠C and meets AB in D. )
∠ACD = ∠BCD = 60/2= 30°
Using the same logic for the given ∆ACD, we can say that:
∠DAC + ∠ACD + ∠ADC = 180°
50° + 30° + ∠ADC = 180°
If we use the same logic for the given ∆BCD, we can say that
∠DBC + ∠BCD + ∠BDC = 180°
70° + 30° + ∠BDC = 180°
∠BDC = 180° – 100°
∠BDC = 80°
Thus,
For ∆ADC: ∠A = 50°, ∠D = 100° ∠C = 30°
∆BDC: ∠B = 70°, ∠D = 80° ∠C = 30°
### Question: 22
In ∆ABC, ∠A = 60°, ∠B = 80°, and the bisectors of ∠B and ∠C, meet at O. Find
(i) ∠C
(ii) ∠BOC
### Solution:
We know that the sum of all three angles of a triangle is 180°.
Hence, for △ABC, we can say that:
∠A + ∠B + ∠C = 180° (Sum of angles of ∆ABC)
60° + 80° + ∠C= 180°.
∠C = 180° – 140°
∠C = 140°.
For △OBC,
∠OBC = ∠B2 = 80/2 (OB bisects ∠B)
∠OBC = 40°
∠OCB =∠C2 = 40/2 (OC bisects ∠C)
∠OCB = 20°
If we apply the above logic to this triangle, we can say that:
∠OCB + ∠OBC + ∠BOC = 180° (Sum of angles of ∆OBC)
20° + 40° + ∠BOC = 180°
∠BOC = 180° – 60°
∠BOC = 120°
### Question: 23
The bisectors of the acute angles of a right triangle meet at O. Find the angle at O between the two bisectors.
### Solution:
We know that the sum of all three angles of a triangle is 180°.
Hence, for ∆ABC, we can say that:
∠A + ∠B + ∠C = 180°
∠A + 90° + ∠C = 180°
∠A + ∠C = 180° – 90°
∠A + ∠C = 90°
For ∆OAC:
∠OAC = ∠A2 (OA bisects LA)
∠OCA = ∠C2 (OC bisects LC)
On applying the above logic to △OAC, we get:
∠AOC + ∠OAC + ∠OCA = 180° (Sum of angles of ∆AOC)
∠AOC + ∠A2 + ∠C2 = 180°
∠AOC + ∠A + ∠C2 = 180°
∠AOC + 90/2 = 180°
∠AOC = 180° – 45°
∠AOC = 135°
### Question: 24
In ∆ABC, ∠A = 50° and BC is produced to a point D. The bisectors of ∠ABC and ∠ACD meet at E. Find ∠E.
### Solution:
In the given triangle,
∠ACD = ∠A + ∠B. (Exterior angle is equal to the sum of two opposite interior angles.)
We know that the sum of all three angles of a triangle is 180°.
Therefore, for the given triangle, we can say that:
### Question: 25
In ∆ABC, ∠B = 60°, ∠C = 40°, AL perpendicular BC and AD bisects ∠A such that L and D lie on side BC. Find ∠LAD
### Solution:
We know that the sum of all angles of a triangle is 180°
Therefore, for ∆ABC, we can say that:
### Question: 26
Line segments AB and CD intersect at O such that AC perpendicular DB. It ∠CAB = 35° and ∠CDB = 55°. Find ∠BOD.
### Solution:
We know that AC parallel to BD and AB cuts AC and BD at A and B, respectively.
∠CAB = ∠DBA (Alternate interior angles)
∠DBA = 35°
We also know that the sum of all three angles of a triangle is 180°.
Hence, for △OBD, we can say that:
∠DBO + ∠ODB + ∠BOD = 180°
35° + 55° + ∠BOD = 180° (∠DBO = ∠DBA and ∠ODB = ∠CDB)
∠BOD = 180° - 90°
∠BOD = 90°
### Question: 27
In Figure, ∆ABC is right angled at A, Q and R are points on line BC and P is a point such that QP perpendicular to AC and RP perpendicular to AB. Find ∠P
### Solution:
In the given triangle, AC parallel to QP and BR cuts AC and QP at C and Q, respectively.
∠QCA = ∠CQP (Alternate interior angles)
Because RP parallel to AB and BR cuts AB and RP at B and R, respectively,
∠ABC = ∠PRQ (alternate interior angles).
We know that the sum of all three angles of a triangle is 180°.
Hence, for ∆ABC, we can say that:
∠ABC + ∠ACB + ∠BAC = 180°
∠ABC + ∠ACB + 90° = 180° (Right angled at A)
∠ABC + ∠ACB = 90°
Using the same logic for △PQR, we can say that:
∠PQR + ∠PRQ + ∠QPR = 180°
∠ABC + ∠ACB + ∠QPR = 180° (∠ABC = ∠PRQ and ∠QCA = ∠CQP)
Or,
90°+ ∠QPR =180° (∠ABC+ ∠ACB = 90°)
∠QPR = 90° |
## Adding Mixed Numbers With Common Denominators
### Learning Outcomes
• Model addition of mixed numbers with a common denominator
• Add mixed numbers with a common denominator
## Model addition of two mixed numbers with a common denominator
So far, we’ve added and subtracted proper and improper fractions, but not mixed numbers. Let’s begin by thinking about addition of mixed numbers using money.
If Ron has $1$ dollar and $1$ quarter, he has $1\Large\frac{1}{4}$ dollars.
If Don has $2$ dollars and $1$ quarter, he has $2\Large\frac{1}{4}$ dollars.
What if Ron and Don put their money together? They would have $3$ dollars and $2$ quarters. They add the dollars and add the quarters. This makes $3\Large\frac{2}{4}$ dollars. Because two quarters is half a dollar, they would have $3$ and a half dollars, or $3\Large\frac{1}{2}$ dollars.
$1\Large\frac{1}{4}$+$2\Large\frac{1}{4}$
$\text{________}$
$3\Large\frac{2}{4}=\normalsize3\Large\frac{1}{2}$
When you added the dollars and then added the quarters, you were adding the whole numbers and then adding the fractions.
$1\Large\frac{1}{4}+\normalsize2\Large\frac{1}{4}$
We can use fraction circles to model this same example:
$1\Large\frac{1}{4}+\normalsize2\Large\frac{1}{4}$ Start with $1\Large\frac{1}{4}$ one whole and one $\Large\frac{1}{4}$ pieces $1\Large\frac{1}{4}$ Add $2\Large\frac{1}{4}$ more. two wholes and one $\Large\frac{1}{4}$ pieces $+ 2\Large\frac{1}{4}$ The sum is: three wholes and two $\Large\frac{1}{4}$‘s $3\Large\frac{2}{4} = \normalsize 3\Large\frac{1}{2}$
Doing the Manipulative Mathematics activity “Model Mixed Number Addition/Subtraction” will help you develop a better understanding of adding and subtracting mixed numbers.
### Example
Model $2\Large\frac{1}{3}+\normalsize1\Large\frac{2}{3}$ and give the sum.
Solution:
We will use fraction circles – whole circles for the whole numbers and $\Large\frac{1}{3}$ pieces for the fractions.
two wholes and one $\Large\frac{1}{3}$ $2\Large\frac{1}{3}$ plus one whole and two $\Large\frac{1}{3}$ s $+ 1\Large\frac{2}{3}$ sum is three wholes and three $\Large\frac{1}{3}$ s $3\Large\frac{3}{3} = 4$
This is the same as $4$ wholes. So, $2\Large\frac{1}{3}+\normalsize1\Large\frac{2}{3}=4$.
### Example
Model $1\Large\frac{3}{5}+\normalsize2\Large\frac{3}{5}$ and give the sum as a mixed number.
### Try It
Model, and give the sum as a mixed number. Draw a picture to illustrate your model.
$2\Large\frac{5}{6}+\normalsize1\Large\frac{5}{6}$
Model, and give the sum as a mixed number. Draw a picture to illustrate your model.
$1\Large\frac{5}{8}+\normalsize1\Large\frac{7}{8}$
## Add mixed numbers with like denominators
Modeling with fraction circles helps illustrate the process for adding mixed numbers: We add the whole numbers and add the fractions, and then we simplify the result, if possible.
### Add mixed numbers with a common denominator: Method 1
Step 1. Add the whole numbers.
Step 2. Add the fractions.
Step 3. Simplify, if possible.
### Example
Add: $3\Large\frac{4}{9}+\normalsize2\Large\frac{2}{9}$
Solution:
$3\Large\frac{4}{9}+\normalsize2\Large\frac{2}{9}$ Add the whole numbers. $\color{red}{3} \Large\frac{4}{9}$ $\Large\frac{+\color{red}{2}\frac{2}{9}}{\color{red}{5}}$ Add the fractions. $3\color{red}{ \Large\frac{4}{9}}$ $\Large\frac{+2\color{red}{ \Large\frac{2}{9}}}{5\color{red}{ \Large\frac{6}{9}}}$ Simplify the fraction. $\color{red}{5\Large\frac{6}{9}}=5\Large\frac{2}{3}$
### Try It
In the example above, the sum of the fractions was a proper fraction. Now we will work through an example where the sum is an improper fraction.
### Example
Find the sum: $9\Large\frac{5}{9}+\normalsize5\Large\frac{7}{9}$
### Try It
In the following video we show another example of how to add two mixed numbers.
## Another way to add mixed numbers with like denominators
An alternate method for adding mixed numbers is to convert the mixed numbers to improper fractions and then add the improper fractions. This method is usually written horizontally.
### Add mixed numbers with a common denominator: Method 2
Step 1. Convert the mixed numbers to improper fractions.
Step 2. Add the improper fractions.
Step 3. Rewrite as a mixed number.
Step 4. Simplify, if possible.
### Example
Add by converting the mixed numbers to improper fractions: $3\Large\frac{7}{8}+4\Large\frac{3}{8}$.
### Try It
The table below compares the two methods of addition, using the expression $3\Large\frac{2}{5}+\normalsize6\Large\frac{4}{5}$ as an example. Which way do you prefer?
Mixed Numbers Improper Fractions
$\begin{array}{}\\ \\ \hfill 3\frac{2}{5}\hfill \\ \hfill \frac{+6\frac{4}{5}}{9\frac{6}{5}}\hfill \\ \hfill 9+\frac{6}{5}\hfill \\ \hfill 9+1\frac{1}{5}\hfill \\ \hfill 10\frac{1}{5}\hfill \end{array}$ $\begin{array}{}\\ \hfill 3\frac{2}{5}+6\frac{4}{5}\hfill \\ \hfill \frac{17}{5}+\frac{34}{5}\hfill \\ \hfill \frac{51}{5}\hfill \\ \hfill 10\frac{1}{5}\hfill \end{array}$
## Contribute!
Did you have an idea for improving this content? We’d love your input. |
# The area under a curve
We will begin our discussions by considering the simple geometric problem of determining the area underneath the graph of a function. In fact, we will later see how to use the same ideas to compute the volume of many different solids. These issues were of great importance to ancient mathematicians and led to the discovery of many of the ideas we'll talk about.
The area of a triangle
Of course, we all know that the area of a triangle is equal to half the product of the base and height of the triangle. However, to illustrate our methods, we will consider this simple example from another perspective.
To be more specific, we will compute the area underneath the graph as x varies from 0 to 1.
Our method for computing this area will be first to approximate this area by the area of some rectangles which lie underneath the graph. Then we will increase the number of these rectangles so that our approximation becomes better and better. If you drag the red dot in the following figure, you can vary how many rectangles will be used in the approximation. As more rectangles are added, the area of the triangle is filled up by the rectangles. This means that the area of all the rectangles will be a good approximation to the area of the triangle.
Let's get started by building the rectangles. We will use n to denote the number of rectangles we are going to use. Since the rectangles will span the interval from x = 0 to x = 1, we will break this interval into n pieces of width as in the figure.
The points are
Let us now consider one of the rectangles as in the figure. We know that the width of the rectangle is So that the rectangle fits underneath the graph, its upper left corner should lie on the graph which is
This means that the height of the rectangle is and hence its area is since
To put everything together now, the area of all the rectangles is
since the first rectangle we consider has left endpoint while the last rectangle has left endpoint
Notice that each term has a common factor so that we may write
If you remember, the sum we need to evaluate is simply that of an arithmetic sequence: in particular, the sum
In our case, k = n - 1 so that
Now consider what happens as the number of rectangles n becomes very large. Then the term becomes very small and we see that which is as we expect since we know that the area of the triangle is equal to
To summarize, we have computed the area of a triangle in a roundabout way. We approximated it by the area of some rectangles and observed that as the number of rectangles gets very large, the appproximation becomes more precise. We can, and will, use this method to find some areas that we do not already know.
The triangle revisited
Now we would like to go back and revisit our triangle. Instead of computing the are under the graph as x varies from 0 to 1, we would like to allow the right endpoint of the interval to vary. It might not be immediately clear to you why we want to do this, but we will see a very interesting relationship if we do.
In the following figure, you can vary the right endpoing by dragging the ball around.
We will use the notation A(x) to denote the function which measures the area under the curve between 0 and x. We could, of course, determine A(x) as we did above. However, let's use the fact that this is measuring the area of a right isosceles triangle where the length of the legs is x. This means that
Now we have something very interesting: we are considering the area under the graph and we have found that this area is Notice that the derivative
In other words, the derivative of the area function is the original function which defines the graph. Is this merely a coincidence? Sorry, but you'll have to wait and see.
The area under a parabola
In the previous example, we did not really need a new technique since we know the area of a triangle. However, in this example, we will find the area of a region which cannot be computed through conventional means. Namely, we will compute the area under the parabola as x varies from 0 to 1.
As in the example with the triangle, we will approximate this area with the area of many rectangles. Then we will let the number of rectangles increase so that we have a better approximation.
As before, we divide the interval from 0 to 1 into n pieces of width The endpoints of these sub-intervals are, as before,
In this case, the height of the rectangles are given by and so the area of a rectangle is We then find that the sum of the areas of all the rectangles is
In order to evaluate this, we need to be able to sum a sequence consisting of squares. We have discussed this sum earlier: recall that we found
In the case we are now looking at, the sum ends at
This says that
Notice that as n becomes very large, the two terms on the right become very small and Since the area of the rectangles provide a better and better approximation as n becomes large, we can say that the area under the parabola between and is equal to
Here we were able to compute an area which we could not have found through standard geometry. This will be a very useful technique for us throughout the course. If you are uncomfortable with any of the ways in which we have used the sigma notation above, you should write out the sums represented by the notation and verify the ways in which the notation has been manipulated.
More generally: Let us now explore an area function similar to the one we considered for the area of a triangle. In particular, will represent the area under the parabola between 0 and x. In the following diagram, you may explore this function by dragging the ball around.
As before, we will divide the interval from 0 to x into n pieces each of width The corresponding points will be denoted by where The height of the rectangles will again be so that the area of a rectangle will be
Now if we sum the area of the rectangles, we obtain
As is now familiar, as n becomes very large, we find that the rectangles provide a better and better approximation for the area and hence
Once again, we see a remarkable relationship: measures the area under the graph of the function and we see that
The area under an exponential
Finally, we will consider the area under the graph of over the interval from 0 to x. The following diagram will illustrate this function.
A large part of this computation proceeds just as in the previous example. We devide the interval into n pieces of width The points defined by these sub-intervals are Now the height of the rectangles will be given by the value of he function at so that the height is This means that the area is If we now sum all the areas of the rectangles, we have
You may recognize this last sum: it is a geometric series (meaning that the ratio of one term to its predecessor is constant) and it may be easily evaluated. For instance, suppose we want to add
The trick is to multiply the sum by the ratio r:
If we subtract the second expression from the first, we find that
which means that the sum
If we apply this to our example, we have
Now let's consider what happens when we have a lot of rectangles. This means that n is very large or rather that is very close to 0.
To study this situation, we will consider the linear approximation of the functoin at the point Since we have it follows that both and This means that the linear approximation is and that this approximation is good when is close to 0. In other words, when t is close to 0.
Now when n becomes very large, we have that is very close to 0. This means that In other words,
and so
Once again, we have found that the area function has the original function as its derivative. This is still more evidence for the Fundamental Theorem of Calculus which we will soon meet.
Summary
Let's summarize what we have done. To find the area under the graph of a function between and , we divide the interval into n pieces. These pieces each have width and we call the left endpoint of these subintervals
Now we form a rectangle whose height is and whose area is If we sum all the areas of these rectangles, we have
Finally, as we consider more and more rectangles, the quantity gives a better and better approximation so that we may write the area as
We will see many expressions like this throughout the term. |
# PUZZLEVILLA
or
Remember me
### Register
OR
100 Coins on a Table Puzzle
Difficulty Level
You have a 100 coins laying flat on a table, each with a head side and a tail side. 10 of them are heads up, 90 are tails up. You can't feel, see or in any other way find out which side is up. Split the coins into two piles such that there are the same number of heads in each pile.
We have 100 coins in which 10 of them are heads up and 90 are tails up. Now make two piles A & B by selecting random coins.
Pile A by picking any 90 coins and
Pile B with remaining 10 coins
Now flip all the coins in pile B. It will result in same number of heads in both piles. :)
Lets discuss the solution in detail
Pile A : 90 coins
Let the number of heads in Pile A = n
The number of tails in Pile A = 90 - n
Pile B : 10 coins
Number of heads in Pile B = (Total number of heads - Number of heads in Pile A) = 10 - n
Number of tails in Pile B = (Total no: if coins in Pile B - Number of heads in Pile B) = 10 - (10-n) = n
So we can conclude, Number of heads in Pile A = Number of tails in Pile B = n
Now flip over all the coins in pile B. We now have the same number of heads in both piles.
Guest Said:Posted On 2017-05-08
k
Guest Said:Posted On 2016-05-03
just split them up
Guest Said:Posted On 2016-01-22
Since a coin has both head and tail hence randomly split all the 100 coins into two halves so that each pile will have equal no. of heads and tails
Guest Said:Posted On 2015-12-31
If n = 0, Pile A has no heads up. This means Pile B has all 10 coins with heads up. Flip all coins in Pile B would give Pile B with no heads up. That is, both Pile A and Pile B have the same number (= 0) of heads up !
Guest Said:Posted On 2015-11-30
if n = 0 your solution is wrong. Sorry man.
Submit |
How do you graph h(x) = -2(x-4)(x+2) ?
Jan 16, 2017
Identify the intercepts, vertex and axis from the formula...
Explanation:
Given:
$h \left(x\right) = - 2 \left(x - 4\right) \left(x + 2\right)$
We can see several properties of the curve from this formula:
• The multiplier of the leading (${x}^{2}$) term is $- 2$, which will result in an inverted parabola with vertex pointing upwards.
• The two $x$ intercepts are at $x = 4$ and $x = - 2$, that is $\left(4 , 0\right)$ and $\left(- 2 , 0\right)$.
• Since a parabola is symmetric about its axis, its axis will be midway between these two $x$ intercepts, at $x = 1$.
• The vertex lies at the intersection of the axis with the parabola, so we can find it by substituting $x = 1$ into the formula:
$y = - 2 \left(\textcolor{b l u e}{1} - 4\right) \left(\textcolor{b l u e}{1} + 2\right) = - 2 \left(- 3\right) \left(3\right) = 18$
So the vertex is at $\left(1 , 18\right)$
• We can find the $y$ intercept by substituting $x = 0$ to find:
$y = - 2 \left(\textcolor{b l u e}{0} - 4\right) \left(\textcolor{b l u e}{0} + 2\right) = - 2 \left(- 4\right) \left(2\right) = 16$
That is $\left(0 , 16\right)$
Here's a graph of the parabola with the features we found indicated:
graph{(y+2(x-4)(x+2))(12(x-4)^2+y^2-0.04)(12(x+2)^2+y^2-0.04)(12(x-1)^2+(y-18)^2-0.04)(x-1)(12x^2+(y-16)^2-0.04) = 0 [-4, 6, -5, 21]} |
# Aptitude Algebraic Expressions Test Paper 1
1) If x2+= 34,x+is equal to
1. 3
2. 4
3. 5
4. None of these
Explanation:
Adding 2 to the L.H.S and R.H.S of the equation:
x2+2+=34+2[∵(a+b)2=a2+b2+2ab]
(x+)2= 36
(x+)= +6,-6
2) The value of x in the equation: 16x += 8 is
Explanation:
16x+=8
=8
16x2+1=8x
=>16x2+1-8x=0
=>(4x-1)2=0
=>x =
3) The equation whose roots are 4 and 5, is
1. x2+9x-20=0
2. x2+9x+20=0
3. x2-9x+20=0
4. x2-9x-20=0
Explanation:
The required equation is:
x2-(sum of roots)x+product of roots=0
=>x2-(4+5)x+4 ×5=0
=>x2-9x+20=0
4) The perimeter of a rectangle is 48 meters, and its area is 135 m2. The sides of the rectangle are
1. 15 m, 9m.
2. 19m, 5m.
3. 45m, 3m.
4. 27m, 5m.
Explanation:
Let the sides of the rectangle be x and y meters respectively. Then
2x + 2y = 48
2 (x+y) = 48
x+y = 24
y = 24 -x
We have, xy = 135
So, x (24 - x) = 135
24x - x2 = 135
x2 -24x + 135 = 0
x2 - 9x - 15x + 135 = 0
(x-9) (x -15) = 0
x = 9 m or 15 m
So, when x = 9m, y = 24 - 9 = 15m
And when x = 15, y = 24 -15 = 9 m
So, the sides of rectangle are 15m and 9m.
5) The roots of the equation (x + 3) (x - 3) = 160 are
1. + 13
2. 13, 13
3. + 12
4. 12, 12
Explanation:
(x + 3) (x - 3) = 160
x2 - 9 = 160
x2 = 169
x = +13, - 13
Aptitude Algebraic Expressions Test Paper 2
Aptitude Algebraic Expressions Test Paper 3
Aptitude Algebraic Expressions Test Paper 4 |
# 8.4: Hanging Scales 6
Difficulty Level: At Grade Created by: CK-12
Look at the pictures of the scales below. Can you write equations to represent what you see on each scale? Can you figure out the value of each letter? In this concept, we will practice working with equations that represent what we see on scales. We will then practice solving these systems of equations.
### Guidance
In order to solve the problem above, use the problem solving steps.
• Start by describing what information is given.
• Then, identify what your job is. In these problems, your job will be to figure out the value of each of the three variables.
• Next, make a plan for how you will solve. In these problems, write equations to represent the scales first. Then, solve the system of equations.
• Then, solve the problem.
• Finally, check your solution. Make sure that your solution causes each scale to have the correct weight.
#### Example A
Write equations. Figure out the weights of the blocks.
Solution:
We can use problem solving steps to help.
\begin{align*}& \mathbf{Describe:} && \text{There are three scales with blocks.}\\ &&& \text{A: Two} \ x \ \text{and two} \ z \ \text{blocks. They weigh 20 pounds.}\\ &&& \text{B: One} \ x, \ \text{one} \ y, \ \text{and one} \ z \ \text{block. They weigh 15 pounds.}\\ &&& \text{C: One} \ x \ \text{and two} \ y \ \text{blocks. They weigh 14 pounds.}\\ & \mathbf{My \ Job:} && \text{Use the scales as clues. Figure out the weights of the blocks.}\\ & \mathbf{Plan:} && \text{Write equations, one for each scale.}\\ &&& A: z+x+z+x=20; \ B: x+y+z=15; \ C:x+y+y=14\\ &&& \text{Solve the equations.}\\ & \mathbf{Solve:} && A: z+x+z+x=20. \ \text{There are two of each block, so} \ z+x=10\\ &&& \text{B}: (z + x) +y = 15. \ \text{Replace} \ (z + x) \ \text{with} \ 10.\\ &&& \quad \quad 10 + y = 15, \ \text{and}\\ &&& \quad \quad y = 15 - 10, \ \text{or} \ 5 \ \text{pounds.}\\ &&& C: x+y+y=14. \ \text{Replace each} \ y \ \text{with} \ 5.\\ &&& \quad \quad x + 10 = 14, \ \text{and}\\ &&& \quad \quad x = 14 - 10, \text{or} \ 4 \ \text{pounds}\\ &&& A: z + x = 10. \ \text{Replace} \ x \ \text{with} \ 4. \ \text{Then}\ z + 4 = 10.\\ &&& \quad \quad z = 10 - 4, \ \text{or} \ 6 \ \text{pounds.}\\ & \mathbf{Check:} && \text{Replace each block with its weight. Check that the total equal the number of}\\ &&& \text{pounds shown on the scales.} \\ &&& A: 6+4+6+4=20; \ B: 4+5+6=15; \ C: 4+5+5=14.\end{align*}
#### Example B
Write equations. Figure out the weights of the blocks.
Solution:
We can use problem solving steps to help.
\begin{align*}& \mathbf{Describe:} && \text{There are three scales with blocks.}\\ &&& \text{D: One} \ x, \ \text{one} \ y, \ \text{and one} \ z \ \text{block. They weigh 19 pounds.}\\ &&& \text{E: Two} \ y \ \text{and two} \ z \ \text{blocks. They weigh 24 pounds.}\\ &&& \text{F: One} \ z \ \text{and two} \ x \ \text{blocks. They weigh 24 pounds.}\\ & \mathbf{My \ Job:} && \text{Use the scales as clues. Figure out the weights of the blocks.}\\ & \mathbf{Plan:} && \text{Write equations, one for each scale.}\\ &&& D: x+y+z=19; \ E: y+z+y+z=24; \ F:x+x+z=24\\ &&& \text{Solve the equations.}\\ & \mathbf{Solve:} && E: y+z+y+z=24. \ \text{There are two of each block, so} \ y+z=12\\ &&& \text{D}: x+(y+z) = 19. \ \text{Replace} \ (y+z) \ \text{with} \ 12.\\ &&& \quad \quad x+12=19, \ \text{and}\\ &&& \quad \quad x = 19 - 12, \ \text{or} \ 7 \ \text{pounds.}\\ &&& F: x+x+z=24. \ \text{Replace each} \ x \ \text{with} \ 7.\\ &&& \quad \quad 14 + z = 24, \ \text{and}\\ &&& \quad \quad z = 24 - 14, \text{or} \ 10 \ \text{pounds}\\ &&& E: y+z=12. \ \text{Replace} \ z \ \text{with} \ 10. \ \text{Then}\ y+10=12.\\ &&& \quad \quad y = 12 - 10, \ \text{or} \ 2 \ \text{pounds.}\\ & \mathbf{Check:} && \text{Replace each block with its weight. Check that the total equal the number of}\\ &&& \text{pounds shown on the scales.} \\ &&& D:7+2+10=19; \ E: 2+10+2+10=24; \ F: 7+7+10=24.\end{align*}
#### Example C
Write equations. Figure out the weights of the blocks.
Solution:
We can use problem solving steps to help.
\begin{align*}& \mathbf{Describe:} && \text{There are three scales with blocks.}\\ &&& \text{G: One} \ x, \ \text{one} \ y, \ \text{and one} \ z \ \text{block. They weigh 20 pounds.}\\ &&& \text{H: Two} \ x \ \text{and two} \ y \ \text{blocks. They weigh 26 pounds.}\\ &&& \text{I: One} \ y \ \text{and two} \ z \ \text{blocks. They weigh 22 pounds.}\\ & \mathbf{My \ Job:} && \text{Use the scales as clues. Figure out the weights of the blocks.}\\ & \mathbf{Plan:} && \text{Write equations, one for each scale.}\\ &&& G:x+y+z=20; \ H: y+x+y+x=26; \ I:y+z+z=22\\ &&& \text{Solve the equations.}\\ & \mathbf{Solve:} && H: y+x+y+x=26. \ \text{There are two of each block, so} \ y+x = 13\\ &&& \text{G}: (x+y) + z = 20. \ \text{Replace} \ (x + y) \ \text{with} \ 13.\\ &&& \quad \quad 13 + z = 20, \ \text{and}\\ &&& \quad \quad z = 20 - 13, \ \text{or} \ 7 \ \text{pounds.}\\ &&& I: y + z + z = 22. \ \text{Replace each} \ z \ \text{with} \ 7.\\ &&& \quad \quad y + 14 = 22, \ \text{and}\\ &&& \quad \quad y = 22 - 14, \text{or} \ 8 \ \text{pounds}\\ &&& H: y+x=13. \ \text{Replace} \ y \ \text{with} \ 8. \ \text{Then}\ 8 + x = 13.\\ &&& \quad \quad x = 13 - 8, \ \text{or} \ 5 \ \text{pounds.}\\ & \mathbf{Check:} && \text{Replace each block with its weight. Check that the total equal the number of}\\ &&& \text{pounds shown on the scales.} \\ &&& G: 5+7+8=20; \ H: 8+5+8+5=26; \ I: 8+7+7=22.\end{align*}
#### Concept Problem Revisited
We can use problem solving steps to help.
\begin{align*}& \mathbf{Describe:} && \text{There are three scales with blocks.}\\ &&& \text{A: Two} \ x \ \text{and two} \ y \ \text{blocks. They weigh 26 pounds.}\\ &&& \text{B: One} \ x, \ \text{one} \ y, \ \text{and one} \ z \ \text{block. They weigh 22 pounds.}\\ &&& \text{C: One} \ x \ \text{and two} \ z \ \text{blocks. They weigh 24 pounds.}\\ & \mathbf{My \ Job:} && \text{Use the scales as clues. Figure out the weights of the blocks.}\\ & \mathbf{Plan:} && \text{Write equations, one for each scale.}\\ &&& A: x + y + x + y = 26; \ B: x + y + z = 22; \ C:x + z + z = 24\\ &&& \text{Solve the equations.}\\ & \mathbf{Solve:} && A: x + y + x + y = 26. \ \text{There are two of each block, so} \ x + y = 13\\ &&& \text{B}: (x + y) + z = 22. \ \text{Replace} \ (x + y) \ \text{with} \ 13.\\ &&& \quad \quad 13 + z = 22, \ \text{and}\\ &&& \quad \quad z = 22 - 13, \ \text{or} \ 9 \ \text{pounds.}\\ &&& C: x + z + z = 24. \ \text{Replace each} \ z \ \text{with} \ 9.\\ &&& \quad \quad x + 18 = 24, \ \text{and}\\ &&& \quad \quad x = 24 - 18, \text{or} \ 6 \ \text{pounds}\\ &&& A: x + y = 13. \ \text{Replace} \ x \ \text{with} \ 6. \ \text{Then}\ 6 + y = 13.\\ &&& \quad \quad y = 13 - 6, \ \text{or} \ 7 \ \text{pounds.}\\ & \mathbf{Check:} && \text{Replace each block with its weight. Check that the total equal the number of}\\ &&& \text{pounds shown on the scales.} \\ &&& A: 6 + 7 + 6 + 7 = 26; \ B: 6 + 7 + 9 = 23; \ C: 6 + 9+ 9 = 24.\end{align*}
### Vocabulary
In math, an unknown is a letter that stands for a number that we do not yet know the value of. In this concept, the blocks that we did not know the weights of were unknowns. An equation is a math sentence that tells us two quantities that are equal. In this concept, we wrote equations with unknowns to represent what we saw on the scales. A system of equations is a set of equations that represents a given problem. Since we wrote multiple equations for each problem in this concept, we wrote a system of equations for each problem.
### Guided Practice
Write equations. Figure out the weights of the blocks.
1.
2.
3.
1. J: \begin{align*}x + y + z = 22\end{align*}; K: \begin{align*}x + x + y = 25\end{align*}; \begin{align*}L: x + z + x+ z= 26\end{align*}
\begin{align*}x= 8, \ y = 9, \ z = 5\end{align*}
2. M: \begin{align*}y + x + y + x = 32\end{align*}; N: \begin{align*}z + z + x = 28\end{align*}; P: \begin{align*}x+ y+ z = 25\end{align*}
\begin{align*}x= 10, \ y = 6, \ z = 9\end{align*}
3. Q: \begin{align*}x + y + z = 28\end{align*}; R: \begin{align*}y + z + y+ z= 34\end{align*}; S: \begin{align*}x + x+ y= 32\end{align*}
\begin{align*}x= 11, \ y = 10, \ z = 7\end{align*}
### Practice
Write equations. Figure out the weights of the blocks.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Show Hide Details
Description
Difficulty Level:
Tags:
Date Created:
Jan 18, 2013
Mar 23, 2016
# We need you!
At the moment, we do not have exercises for Hanging Scales 6.
Files can only be attached to the latest version of Modality |
Rearranging Formulas Worksheets, Questions and Revision
GCSE 4 - 5GCSE 6 - 7KS3AQAEdexcelOCRWJECAQA 2022Edexcel 2022OCR 2022WJEC 2022
Rearranging Formulas
A formula is a way of giving a mathematical relationship between different things, expressed using algebra. For example, the relationship between speed, distance and time can be captured by the simple formula,
$s=\dfrac{{d}}{{t}}$.
You will need to be able to answer 5 key question types when rearranging formulas, these are shown below.
Make sure you are happy with the following topics before continuing.
Level 4-5 GCSE KS3
Question type 1: Simple linear formula
The formula for the circumference of a circle is $C=2\pi r$. Here the circumference is the subject of the formula.
Making a different letter the subject of a formula means manipulating the formula to get that letter by itself on one side of the equation.
Example: For the equation $C=2\pi \textcolor{blue}{r}$, make $\textcolor{blue}{r}$ the subject.
Step 1: Dividing both sides of the equation by $2\pi$
\begin{aligned}(\textcolor{maroon}{\div 2\pi})\,\,\,\,\,\,\,\,\, C &= 2\pi \textcolor{blue}{r} \,\,\,\,\,\,\,\,\, (\textcolor{maroon}{\div 2\pi}) \\ \dfrac{C}{2\pi} &= \dfrac{\cancel{2\pi} \textcolor{blue}{r}}{\cancel{2\pi}} \end{aligned}
Step 2: Rewrite the formula in its final form, typically having the subject on the left hand side.
$\textcolor{blue}{r} = \dfrac{C}{2\pi}$
Level 4-5 GCSE KS3
Question type 2: Formulas involving fractions
Some questions will include making a letter that appears in a fraction the subject of the formula.
$d=\dfrac{(u+v)t}{2}$
This formula describes the relation that distance, $d$, initial velocity, $u$, final velocity $v$ and time $t$ given constant acceleration. We can manipulate this formula to make $u$ the subject.
Example: For the equation $d=\dfrac{(\textcolor{blue}{u}+v)t}{2}$, make $\textcolor{blue}{u}$ the subject.
Step 1: Multiplying both sides by $2$,
\begin{aligned}(\textcolor{maroon}{\times 2})\,\,\,\,\,\,\,\,\, d & =\dfrac{(\textcolor{blue}{u}+v)t}{2} \,\,\,\,\,\,\,\,\, (\textcolor{maroon}{\times 2}) \\ 2d & =(\textcolor{blue}{u}+v)t \end{aligned}
Step 2: Dividing both sides of the equation by $t$,
\begin{aligned}(\textcolor{maroon}{\div t})\,\,\,\,\,\,\,\,\, 2d & =(\textcolor{blue}{u}+v)t \,\,\,\,\,\,\,\,\, (\textcolor{maroon}{\div t}) \\ \dfrac{2d}{t} & =\textcolor{blue}{u}+v\end{aligned}
Step 3: Finally subtracting $v$ and rewriting so $\textcolor{blue}u$ is on the left hand side,
\begin{aligned}(\textcolor{maroon}{-v})\,\,\,\,\,\,\,\,\, \dfrac{2d}{t} & =\textcolor{blue}{u}+v \,\,\,\,\,\,\,\,\, (\textcolor{maroon}{-v}) \\ \textcolor{blue}{u} & = \dfrac{2d}{t} -v\end{aligned}
Level 4-5 GCSE KS3
Level 4-5 GCSE KS3
Question type 3: Formulas involving squares
Sometimes the required subject can appear as a square. For example, if we are asked to rearrange the formula for the area of a circle to make $r$ the subject. Firstly, recall that the formula for the area of a circle is,
$A=\pi \textcolor{blue}{r^2}$
Example: For the equation $A=\pi \textcolor{blue}{r^2}$, make $\textcolor{blue}{r}$ the subject.
Step 1: Dividing both sides of the equation by $\pi$, we get,
$\dfrac{A}{\pi}=\textcolor{blue}{r^2}$
Now, $r$ is on its own on one side but it’s not technically the subject, since it is squared.
Step 2: If we now square root both sides of the equation, and since $r$ is always positive, we get,
$\textcolor{blue}{r}=\sqrt{\dfrac{A}{\pi}}$
Level 4-5 GCSE KS3
Question type 4: Formulas involving square roots
Conversely, some questions may have the subject appear in a square root.
Example: Rearrange the following formula $d=\sqrt{\dfrac{3\textcolor{blue}{h}}{2}}$ to make $\textcolor{blue}{h}$ the subject.
Step 1: Squaring both sides of the equation, we get,
$d^2=\dfrac{3\textcolor{blue}{h}}{2}$
Step 2: Next, multiply both sides by $2$ to get,
$2d^2=3\textcolor{blue}{h}$
Step 3: Finally, dividing both sides by $3$ gives us,
$\textcolor{blue}{h}=\dfrac{2d^2}{3}$
Level 4-5 GCSE KS3
Level 6-7 GCSE
Question type 5: Formulas when the letter appears twice
In some instances the required subject will appear more than once in the given formula. In these examples we factorise the terms involving the subject.
Example: Rearrange the following formula $H=2\textcolor{blue}{R}-g\textcolor{blue}{R}$ to make $\textcolor{blue}{R}$ the subject.
Step 1: We take out a factor of $R$ from both terms i.e. we factorise, and get
$H=\textcolor{blue}{R}(2-g)$
Step 2: Now we divide by $(2-g)$, to get
$\textcolor{blue}{R}=\dfrac{H}{2-g}$
Level 6-7 GCSE
Example Questions
In order to make $\textcolor{blue}{m}$ the subject, we will multiply both sides by $t$ and get:
$Ft=\textcolor{blue}{m}v$
Then, if we divide both sides of the equation by $v$, and rewrite it so that $m$ is on the left hand side, we get,
$\textcolor{blue}{m}=\dfrac{Ft}{v}$
Firstly, we will multiply both sides of the equation by 2 to get rid of the fraction:
$2A=(\textcolor{blue}{a}+b)h$
Then, if we divide both sides by $h$, we get
$\dfrac{2A}{h}=\textcolor{blue}{a}+b$
Finally, subtracting $b$ from both sides, we get
$\textcolor{blue}{a}=\dfrac{2A}{h}-b$
Step 1: Multiplying both sides by $r^2$, we get
$F\textcolor{blue}{r}^2=kq$
Step 2: Next, divide both sides by $F$ to get
$\textcolor{blue}{r}^2=\dfrac{kq}{F}$
Step 3: Finally, square rooting both sides gives us,
$\textcolor{blue}{r}=\sqrt{\dfrac{kq}{F}}$
Multiplying both sides of the equation by $x+c$, we get
$\textcolor{blue}{x}=\dfrac{a(\textcolor{blue}{x}+c)}{b}$
Next, multiply both sides by $b$ and expanding the right-hand side, to get
$b\textcolor{blue}{x}=a\textcolor{blue}{x}+ac$
Subtracting $ax$ from both sides and factorising,
$b\textcolor{blue}{x}-a\textcolor{blue}{x}=\textcolor{blue}{x}(b-a)=ac$
Finally, dividing both sides by $(b-a)$ gives us,
$\textcolor{blue}{x}=\dfrac{ac}{b-a}$
Multiplying both sides of the equation by $b-4$, we get
$a(\textcolor{blue}{b}-4)=(3-2\textcolor{blue}{b})$
Next, expanding the left-hand side, to get
$a\textcolor{blue}{b}-4a=3-2\textcolor{blue}{b}$
Collecting all the terms with a factor of $b$ on one side of the equation,
$a\textcolor{blue}{b}+2\textcolor{blue}{b}=3+4a$
Factorising $b$ out of the left-hand side and dividing by the terms left we find,
$\textcolor{blue}{b}=\dfrac{3+4a}{a+2}$
Related Topics
Level 1-3GCSEKS3
Level 4-5GCSEKS3
Expanding Brackets
Level 1-3Level 4-5GCSEKS3
Factorising
Level 4-5Level 6-7GCSE
Surds
Level 6-7Level 8-9GCSE
Worksheet and Example Questions
(NEW) Rearranging Formulas Exam Solutions - MME
Level 4-5 GCSENewOfficial MME
Level 4-5 GCSE
Level 6-7 GCSE
You May Also Like...
GCSE Maths Revision Cards
Revise for your GCSE maths exam using the most comprehensive maths revision cards available. These GCSE Maths revision cards are relevant for all major exam boards including AQA, OCR, Edexcel and WJEC.
From: £8.99
GCSE Maths Revision Guide
The MME GCSE maths revision guide covers the entire GCSE maths course with easy to understand examples, explanations and plenty of exam style questions. We also provide a separate answer book to make checking your answers easier!
From: £14.99
GCSE Maths Predicted Papers 2022 (Advance Information)
GCSE Maths 2022 Predicted Papers are perfect for preparing for your 2022 Maths exams. These papers have been designed based on the new topic lists (Advance Information) released by exam boards in February 2022! They are only available on MME!
From: £5.99
Level 9 GCSE Maths Papers 2022 (Advance Information)
Level 9 GCSE Maths Papers 2022 are designed for students who want to achieve the top grades in their GCSE Maths exam. Using the information released in February 2022, the questions have been specifically tailored to include the type of level 9 questions that will appear in this year's exams.
£9.99 |
# why use standard deviation instead of variance
Deviation just means how far from the normal The Standard Deviation is a measure of how spread
out numbers are. Its symbol is The formula is easy: it is the square root of the Variance. So now you ask, What is the Variance? The heights (at the shoulders) are: 600mm, 470mm, 170mm, 430mm
and 300mm. Find out the Mean, the Variance, and the Standard Deviation. so the mean (average) height is 394 mm. Let's plot this on the chart:
To calculate the Variance, take each difference, square it, and then average the result: So the Variance is 21,704 And the Standard Deviation is just the square root of Variance, so: And the good thing about the Standard Deviation is that it is useful. Now we can show which heights are within one Standard Deviation (147mm) of the Mean: So, using the Standard Deviation we have a standard way of knowing what is normal, and what is extra large or extra small. Rottweilers are tall dogs. And Dachshunds are a bit short. but don't tell them! But. there is a small change with Sample Our example has been for a Population (the 5 dogs are the only dogs we are interested in). But if the data is a Sample (a selection taken from a bigger Population), then the calculation changes!
All other calculations stay the same, including how we calculated the mean. Think of it as a correction when your data is only a sample. Here are the two formulas, explained at Looks complicated, but the important change is to divide by N-1 (instead of N ) when calculating a Sample Variance. When we measure the variability of a set of data, there are two closely linked statistics related to this: the б and, which both indicate how spread-out the data values are and involve similar steps in their calculation. However, the major difference between these two statistical analyses is that the standard deviation is the square root of the variance. In order to understand the differences between these two observations of statistical spread, one must first understand what each represents: Variance represents all data points in a set and is calculated by averaging the squared deviation of each mean while the standard deviation is a measure of spread around the mean when the central tendency is calculated via the mean. As a result, the variance can be expressed as the average squared deviation of the values from the means or [squaring deviation of the means] divided by the number of observations and standard deviation can be expressed as the square root of the variance.
To fully understand the difference between these statistics we need to understand the calculation of the variance. The steps to calculating the sample variance are as follows: Calculate the sample mean of the data. Find the difference between the mean and each of the data values. Square these differences. Add the squared differences together. Divide this sum by one less than the total number of data values. The mean provides the center point or of the data. The differences from the mean help to determine the deviations from that mean. Data values that are far from the mean will produce a greater deviation than those that are close to the mean. The differences are squared because if the differences are added without being squared, this sum will be zero. The addition of these squared deviations provides a measurement of total deviation. The division by one less than the sample size provides a sort of mean deviation.
This negates the effect of having many data points each contribute to the measurement of spread. As stated before, the standard deviation is simply calculated by finding the square root of this result, which provides the absolute standard of deviation regardless of a total number of data values. When we consider the variance, we realize that there is one major drawback to using it. When we follow the steps of the calculation of the variance, this shows that the variance is measured in terms of square units because we added together squared differences in our calculation. For example, if our sample data is measured in terms of meters, then the units forБ a variance would be given in square meters. In order to standardize our measure of spread, we need to take the square root of the variance. This will eliminate the problem of squared units, and gives us a measure of the spread that will have the same units as our original sample. There are many formulas in mathematical statistics that have nicer looking forms when we state them in terms of variance instead of standard deviation.
• Views: 282
why does hair grow faster on one side
why do you use the chi square statistic
why do we use box and whisker plots
why do we use analysis of variance
why do we take log of data
why do we need to collect data
why is normal distribution important in statistical analysis |
## 275- A problem from Homework 2
September 21, 2008
Quite a few of you had difficulties with problem 18 of the Additional and Advanced Exercises for Chapter 10, so I am posting a solution here.
The problem asks to derive the trigonometric identity $\sin(A-B)=\sin A\cos B-\cos A\sin B$ by forming the cross product of two appropriate vectors.
In problems that involve trigonometry or geometry, it is convenient to begin with vectors that have some clear geometric meaning related to the problem at hand, so it seems natural to consider the vectors $\vec u=(\cos A,\sin A,0)$ and $\vec v=(\cos B,\sin B,0)$.
These are two vectors in the plane, but we look at them as vectors in 3-D, as we should, since we want to look at their cross product, and this is only defined for vectors in 3-D.
So: $\vec u$ is a vector of size 1 ($\|\vec u\|=\sqrt{\cos^2 A+\sin^2 A+0^2}=1$) that forms an angle of $A$ radians with the $x$-axis (measured counterclockwise). Similarly, $\vec v$ is a vector of size 1 that forms an angle of $B$ radians with the $x$-axis (measured counterclockwise).
Now: We need to analyze the angle between $\vec u$ and $\vec v$, which seems to be the technical point of this exercise, so let’s do this very carefully. This angle is the angle measured starting at $\vec u$ and moving counterclockwise until we find $\vec v$, is usually $B-A$, but it may be $2\pi-A+B$ if, for example, $\vec u$ and $\vec v$ are vectors in the first quadrant and $B. (Although we can “ignore'' this case since the sine function is periodic with period $2\pi$.)
Similarly: The direction of $\vec u\times\vec v$ is obtained by the Right-Hand Rule, meaning $\vec u\times \vec v$ is a vector perpendicular to the plane spanned by $\vec u$ and $\vec v$ (the $xy$-plane), but it may be a positive (or zero) multiple of ${\bf k}$, or a negative multiple of ${\bf k}$, depending on whether the angle between $\vec u$ and $\vec v$ is smaller than (or equal to) $\pi$, or larger than $\pi$.
The magnitude of $\vec u\times \vec v$ is $\|\vec u\|\|\vec v\|\sin\theta$, where $\theta$ is either the angle $\alpha$ between $\vec u$ and $\vec v$, or $2\pi-\alpha$, whichever is between $\null0$ and $\pi.$
Putting these two bits of information (about direction and magnitude) together, we find that $\vec u\times\vec v=(0,0,\sin(B-A))$ if $0\le B-A\le\pi$. If $B-A>\pi$, then $\vec u\times\vec v=(0,0,-\sin(2\pi+A-B)$ but $\sin(2\pi+\alpha)=\sin(\alpha)=-\sin(-\alpha)$ for any $\alpha$, so also in this case $\vec u\times\vec v=(0,0,\sin(B-A))$.
Finally, $\vec u\times\vec v$, component-wise, is found by computing the formal determinant $\left|\begin{array}{ccc}{\bf i}&{\bf j}&{\bf k}\\ \cos A&\sin A&0\\ \cos B&\sin B&0\end{array}\right|=(0,0,\cos A\sin B-\sin A\cos B)$. Comparing this expression with the one above, we find the desired identity.
(Actually, we find it with the roles of $A$ and $B$ reversed, but this is of course irrelevant. And of course this deduction only works for angles between $\null0$ and $2\pi$, but the identity is true in all other cases as well, thanks to the periodicity properties of sine and cosine.)
In this second talk I proved the equivalence of Bagaria’s, and Goldstern-Shelah’s formulation of the bounded forcing axiom for a poset ${\mathbb P}$ that preserves $\omega_1$.
We presented several characterizations of club subsets of ${\mathcal P}_{\omega_1}(X)$ for $X$ uncountable. We then defined when a forcing notion is proper and provided some basic examples of proper forcings, namely ccc, $\sigma$-closed forcings, and their products. |
Home > CALC > Chapter 1 > Lesson 1.2.1 > Problem1-21
1-21.
Given the following functions, compute the given values.
1. $f ( x ) = \left\{ \begin{array} { c c c } { 4 - 3 x } & { \text { for } } & { x \leq 1 } \\ { x ^ { 2 } } & { \text { for } x > 1 } \end{array} \right.$
Find $f (0)$, $f (1)$, and $f (3)$.
1. $f ( x ) = \left\{ \begin{array} { c c } { \sqrt { x } } & { \text { for } x < 3 } \\ { 3 - x } & { \text { for } x \geq 3 } \end{array} \right.$
Find $f (1)$, $f (3)$, and $f (9.4)$.
1. $f ( x ) = \left\{ \begin{array} { l l l } { - x } & { \text { for } } & { x \leq 0 } \\ { \frac { 5 } { x } } & { \text { for } } & { 0 < x \leq 1 } \\ { 6 - 2 x } & { \text { for } } & { x > 1 } \end{array} \right.$
Find $f (−3)$, $f (0)$, $f (0.5)$ and $f (4)$.
1. Sketch a graph of $f(x)$ in part (c) above.
$x = 1$ can be called, informally, the Boundary Point of the piecewise function $f(x)$. This is because $x ≤ 1$ is the location where the left and the right pieces switch.
$f(0)$ and $f(1)$ are on the left side of the Boundary Point, $f(3)$ is on the right side of the Boundary Point.
$f(0) = 4 - 3(0) = 4$ LEFT PIECE
$f(1) = 4 - 3(1) = 1$ LEFT PIECE
$f(3) = (3)^² = 9$ RIGHT PIECE
The Boundary Point is $x ≥ 3$. So $f(1)$ lies on the left side of the piecewise function while $f(3)$ and $f(9.4)$ lie on the right.
This function has two boundary points and three pieces: a left piece, a middle piece and a right piece. On which piece do the points $f(-3)$, $f(0)$, $f(0.5)$ and $f(4)$ lie? |
The gcf of 12 and 42 is the biggest positive integer that divides the numbers 12 and 42 without a remainder. Order out, it is the greatest typical factor that 12 and also 42. Here you can find the gcf the 12 and 42, along with a full of three techniques for computer it. In addition, we have actually a calculator girlfriend should check out. Not only have the right to it recognize the gcf of 12 and 42, but also that of 3 or more integers including twelve and forty-two because that example. Keep analysis to discover everything about the gcf (12,42) and also the terms concerned it.
You are watching: What is the gcf of 12 and 42
What is the GCF the 12 and 42
If you simply want to recognize what is the greatest typical factor the 12 and also 42, it is 6. Usually, this is written asgcf(12,42) = 6
The gcf that 12 and also 42 deserve to be obtained like this:
The factors of 12 room 12, 6, 4, 3, 2, 1. The factors of 42 room 42, 21, 14, 7, 6, 3, 2, 1. The common components of 12 and 42 room 6, 3, 2, 1, intersecting the two sets above. In the intersection components of 12 ∩ components of 42 the greatest facet is 6. Therefore, the greatest usual factor of 12 and 42 is 6.
Taking the over into account you likewise know just how to uncover all the usual factors of 12 and 42, not just the greatest. In the next section we display you just how to calculate the gcf the twelve and also forty-two by way of two more methods.
How to discover the GCF that 12 and also 42
The greatest typical factor of 12 and also 42 have the right to be computed by utilizing the least common multiple aka lcm that 12 and also 42. This is the easiest approach:
gcf (12,42) = frac12 imes 42lcm(12,42) = frac50484 = 6
Alternatively, the gcf the 12 and 42 deserve to be uncovered using the prime factorization of 12 and also 42:
The element factorization that 12 is: 2 x 2 x 3 The element factorization of 42 is: 2 x 3 x 7 The prime factors and multiplicities 12 and 42 have in common are: 2 x 3 2 x 3 is the gcf of 12 and also 42 gcf(12,42) = 6
In any kind of case, the easiest means to compute the gcf of 2 numbers prefer 12 and 42 is by utilizing our calculator below. Keep in mind that that can also compute the gcf of more than two numbers, be separated by a comma. Because that example, go into 12,42. The calculation is carried out automatically.
comparable searched terms on our site also include:
Use the GCF that 12 and 42
What is the greatest common factor that 12 and 42 provided for? Answer: the is useful for to reduce fractions like 12 / 42. Just divide the nominator and the denominator through the gcf (12,42) to reduce the fraction to shortest terms.
frac1242 = fracfrac126frac426 = frac27.
Properties the GCF of 12 and also 42
The most crucial properties of the gcf(12,42) are:
Commutative property: gcf(12,42) = gcf(42,12) Associative property: gcf(12,42,n) = gcf(gcf(42,12),n) n eq 0 hinspacein hinspacemathbbZ
The associativity is specifically useful to gain the gcf of three or more numbers; our calculator renders use that it.
To amount up, the gcf the 12 and also 42 is 6. In usual notation: gcf (12,42) = 6.
If you have been in search of gcf 12 and 42 or gcf 12 42 climate you have come to the exactly page, too. The same is the true if you typed gcf because that 12 and also 42 in your favorite search engine.
Note that you can uncover the greatest common factor of numerous integer pairs including twelve / forty-two by utilizing the search form in the sidebar that this page.
Questions and also comments regarded the gcf the 12 and 42 space really appreciated. Use the type below or send us a mail to get in touch.
See more: Thing 1 And Thing 2 Characters, Thing One And Thing Two
Please struggle the share buttons if our article about the greatest typical factor that 12 and 42 has been useful to you, and also make sure to bookmark ours site. |
Limits of sequences
Properties of convergent sequences
The limit value is exclusively determined by the behavior of the terms in its close neighborhood
Bounded sequences
Properties of convergent sequences
The limit value is exclusively determined by the behavior of the terms in its close neighborhood
a) We use the definition of convergence that states, a number L is the limit of a sequence if for every e > 0 there exist a natural number n0(e) such that
| an - L | < e for all n > n0(e),
to examine the behavior of the sequence {an} as n tends to infinity.
Considering the geometric interpretation of the above inequality we can write
- e < an - L < + e for all n > n0(e),
or L - e < an < L + e for all
shown on the number line (see the example sequence (5) in the previous section)
Thus, if a number L is the limit of a sequence {an} then in the interval from L - e to L + e, of the length 2e, symmetrical with respect to L, lie all terms of the sequence beginning with the one which is far enough, indexed by n0, so lie infinitely many terms, and it holds to every however small interval 2e. This (infinite) part of the length 2e of a convergent sequence is usually called tail.
Outside of this interval there is finite but any number of initial terms of the sequence, but they don't have any influence on existence of the limit or its value. The value of the limit stays unchanged if these initial terms are, substituted, dropped or are replaced by another. Thus, the beginning part of the convergent sequence outside of the interval 2e is a finite or discrete set of numbers.
Therefore, the limit value is exclusively determined by the behavior of the terms in its close neighborhood.
Bounded sequences
A sequence is bounded above if there is a number M such that an < M for all n.
It is bounded below if there is a number m such that an > m for all n
If a sequence is bounded above and bellow it is called bounded sequence.
Thus, a sequence is bounded if there exist a number M > 0 such that | an | < M for all n.
A sequence which is not bounded is called unbounded.
b) From the fact that outside the interval of the length 2e lie only finite number of terms of a sequence it follows that every convergent sequence is bounded, that is
there exists a number M > 0 such that | an | < M for all n.
Therefore, all terms of a convergent sequence can be closed, in the interval containing finite number of the initial terms outside the interval 2e, and in the interval 2e itself.
Any convergent sequence is bounded (both above and below).
c) If {an} is a convergent sequence, then every subsequence of that sequence converges to the same limit.
Let from a convergent sequence extracted is infinitely many terms, an1, an2, . . . , anp, . . . , using any principle, for example extracted is every other term, then they make new infinite sequence called subsequence. Then, the extracted subsequence converges to the same limit as the original sequence.
That is, if | an - L | < e for all n > n0(e), then | anp - L | < e for all np > n0(e)
since the natural numbers np are contained within n, and np ® oo when n ® oo .
d) The limit L can belong to a sequence but should not. By its definition the limit does not belong to a sequence since the definition says, the limit is a value that is approached increasingly closely by a sequence as n tends to infinity.
e) Terms of a sequence which are far enough can be considered as the approximate value of the limit L.
This way defined is every irrational number as the limit of the two infinite sequences, {an} and {bn} of rational numbers that satisfies following conditions.
1) The sequence {an} is increasing (nondecreasing) and the {bn} is decreasing (nonincreasing), i.e.,
a1 < a2 < . . . < an < an + 1 < . . . and b1 > b2 > . . . > bn > bn + 1 > . . .
2) None of the terms of {an} is greater then any of the terms of {bn},
an < bn where n = 1, 2, 3, . . .
that is, every number an lies to the left of every number bn on the number line.
3) The difference bn - an of a two terms of the given sequences of the same index n can become arbitrary small as n tends to infinity,
lim (bn - an) = 0 as n ® oo
as is shown on the number line below.
Therefore, according to the given conditions the following must hold.
There is only one real number L determined by these two sequences such that none of the terms of {an} is greater then L (an < L) and that none of the terms of {bn} is less then L (bn > L).
Thus, both sequences, {an} and {bn} approach the same limit L from below and above, i.e.,
For example, the circumference of the circle is the limit to which the circumferences of inscribed and circumscribed polygons increase and decrease, respectively, as the number of sides increases to infinity.
Calculus contents B |
Edit Article
How to Use Art to Teach Math
Art and math do not just produce static results but rather provide part of the dynamic interface of culture. For one article to attempt how to use art to teach math is just mind boggling! However, this guide contains some central principles and ideas for using art to teach math. With it you should be able to branch off from painting into other art and media such as interior design and decor, pottery, weaving, dance, cinema, video, digital art and fractals, photography, sculpture, wardrobe and costume design, jewelry and accessories, fashion, makeup, lighting, printmaking, commercial art and graphic design, comics and poetry and writing, drawing and engraving, music, theatre, graffiti, architecture, landscaping and beyond.[1] To get you started, this guide looks at a fresh interplay between (acrylic +/or digital) painting and the theory of Neutral Operations. While these projects won't be suitable for all ages and stages of math education, they will hopefully provide a range of ideas you can build off and explore, depending on where you or your students are in their knowledge of math and art.
Part 1 Use Painting to Teach Math
1. 1
Realize that there are many motions involved when creating a painting. These can be called "operations" or "operators" in the mathematical sense, as they are like verbs in language. Exploring these concepts with painting and math-related projects can be really helpful for students in creating mental connections between art and math.
2. 2
Start with this basic computation project. Paint a square and paint one diagonal between opposite corners. Since the sides of the square are equal, they may be assigned a unit length of 1 and by the Pythagorean Theorem of a^2 + b^2 = c^2, the diagonal c = the square root of 2. However, if the equal sides are a different length, say pi = π, then the diagonal c equals π * sqrt(2). We can use the above yellow box method to approximate the value of the square root of 2, to be about 1.414:
• You can then use the article Find the Area of a Square Using the Length of its Diagonal to find that the area = (d^2)/2 = [(π*sqrt(2))^2]/2 = π^2. The area of this square is π^2 square units and the sides are π units long. But you already knew that the area was π^2 because side = π and side^2 = area. π^2 = the area. The square root of this area is also an area, and its sides are sqrt(π) * sqrt(π) because its area = π.. Sqrt(π) = approx. 1.772,453,850,905,52 (with decimal commas inserted to make it easier to read that it's accurate to 10^-14) (where "^" means exponentiation, as is used in Excel).
• As an extension of the project, center and paint the internal π square.
• Question: Is the inner square's diagonal equal to sqrt(π)*sqrt(2)?
• Question: Can you draw the Circle and Diameter that the original side π is the ratio between, since C = π*D, or C/D = π?
• Question: If radius r is set equal to 1, then the sides of the outer square also = 4πr, or 2 circumferences of 2πr, correct?
3. The Garthwaite Curve, A Spiral of Spheroids or Wavicles (Wave-Particles)
3
Do this Neutral Operations painting project. Start with the equation π*d = π+d = C.
• Algebraically, solve for d:
• π*d = π+d;
• (π*d)-d = (π+d)-d;
• d*(π-1) = π by factoring the left and simplifying the right sides of the equation;
• d*(π-1)/(π-1) = π/(π-1);
• d = π/(π - 1) and d has been solved for in terms of π and 1;
• d = 3.14159265358979 / 2.14159265358979; d = 1.46694220692426, solved.
• π * d = C; 3.14159265358979 * 1.46694220692426 = 4.60853486051405 AND
• π + d = C; 3.14159265358979 + 1.46694220692426 = 4.60853486051405 √,
• i.e. π * d = π + d = C and the "Neutral Circle" has been discovered!!
• Let d=1.47 and C=4.61, given a ruler in centimeters.
• Then draw and paint the Neutral Circle. Be aware also that because n*(π*d) = n*(π+d) = n*C, that any size or number of circles may be obtained!
• Can you find another 'Neutral Circle' by solving for π*d = π-d = C? Can you find a third 'Neutral Circle by solving for π*d = π^d = C? Yes, it's solvable! Also try π*d = d^π = C as is solvable.
• See article Create a Spirallic Spin Particle Path or Necklace Form or Spherical Border
4. 4
Try this imagining project: Imagine a painting in as much detail as you can, in groups of gradated pixels. Then count to 100, but do it in random groups: count from 17 up to 31, 82 down to 65, 0 up to 16, 54 down to 32, 83 up to 100 and 64 down to 55, resting for 15 seconds between groups. Activate then rest your mind, eg. in conversing and meditation periods, sleep and dreams; this is where the "seeds" of both art and mathematical ideas come from.
5. 5
Do this refocusing project: Look at some flowers in a clear jar or vase half full of water, but concentrate only upon the colors where the water, jar and stems are. Can you see the shadows of the stems on the inner walls of the jar? Do you know what the incidence and refraction angles of the light are as caused by first the glass jar and then the water? Move your senses, especially the eyes; here is how one one perceives the data's pattern(s).
• Now create teams to look at phone book pages to find each 4 number suffix starting with a value greater than 4 on the page and highlight them with a yellow marker. Are they less than half the numbers on the whole page?
• Now use a black marker to blacken all the phone number suffixes that start with a 7 or 9 and study the entire resulting pattern. Then check out the reverse side of the page where the markers have seeped through and see what information has been covered.
6. 6
Try this control project: Take a brush and dip it into water and put at first the least amount of pigment possible on it and make a circle on some paper, then add a little more pigment each time and make each circle just a tiny bit thicker and randomly offset from each other -- how many different size lines did you make? Stroke, usually brushstroke, but it could be with another applicator; here is how the data is "massaged" by the algorithm(s).
• Construct an infinite programming loop and follow the value of the increment, if it increments by -r each time, 10 times. How long does it take for raindrops to become smooth again when they fall into a puddle if -6.28 is negative 2 pi r, the circumference of a circle of random radius r? (r's initial value = -1)
7. 7
Do this energy flows project: Breathe and be aware of your own subtle body rhythm(s); this is the essential space that flows between operators and values, data objects, pattern pieces, etc. and allows energy flows to occur. After several other projects, do this one again. Then, put on some salsa or jazz or music with a complex beat +/or melodic structure and create your own random calligraphic movements in response to it by writing freeform cursively in "jazz or zen calligraphy". Such random bursts of energy do occur in physics and mathematics but they are relatively rare.
• You can translate this to the digital medium and create overlapping "burst functions f(x)" of variable x to create this sort of effect in Microsoft Excel, plotted in a smoothline scatter chart to obtain wonderful linear precision and exact symbolic copies, i.e. the fundaments of language. Bézier curves are also used in font design and typography. See perhaps if you can invent a painting medium that will react to a laser light pen? There must be some way to do that initially via a darkroom approach one would guess. Woodcarving is also done with lasers now, from which block prints can be fashioned. See article Acquire Bézier Curves Using Excel
8. 8
Tackle this popularity project: Travel to the art subject, especially for landscapes/cityscapes, etc.; this has to do with the transportability of the algorithm -- its utility. Think of which computer app, which is really just 1's and 0's, is the most popular program ever. And what is the most popular painting or picture or film ever? Is there any commonality between the two? Did Van Gogh paint Paris so often because he thought the paintings would sell well?
9. 9
Do this inputs and memory project: Rest and eat, etc., and engage in other activities which may input to your work; this represents part of the growth of the art or math work from seeds to full fruiting orchard. The idea is to be able to incorporate into the art regular and peculiar behavior patterns, just "being human" and the demographics of reality. Work on some math homework exercises for a little while and chew a very flavorful gum. Does the gum flavor help to remember the lesson better? In other words, step out of your demographic by changing your internal state.
10. 10
Do this time project: Create a painting using your fingers and fingernails for brushes but just the tips, very gently, slowly and carefully. Now find the solution to this problem as quickly as you can because time is of the essence, not accuracy: A man is twice as old as his son. In 21 years, how much older will he be than his son? Learn that time is what you make of it. Sometimes a reasonably good ballpark estimate is close enough for survival's sake.
11. 11
Do this correction project: Correct the previous finger-painting artwork work by cloth lifting, smudging, overpainting, etc. with a fine brush; this the detail correcting work in any discipline that throws out the unfeasible and unworkable portions. How many wrong answers did you get to the above man and his son math problem before you got to the correct answer of 15 years? Learn that trial and error is a tried and true method of problem solving although it may not win many prizes for being lightning quick.
12. 3 nested For Next Loops Mathematical Art by Behrooz Rezvani
12
Do this looping project: Layer the work; this is like the looping functions and nested loops of programming algorithms. If you don't know about layering paintings or nested loops, do the research until you do. Try applying some clear gel to your work and seeing if you like the effect. Try adding a time value to a clock function in your nested loop and seeing if you can get your computer to tell time decently.
13. 13
Do this expression project: Teach math by using art by using verbs to describe the various actions that are taken from the beginning business side of shopping for materials, and then final marketing and selling side as well. Most artists shop for discounts but high quality work demands high quality materials that do not come cheap. Learn that managing money economically is necessary to surviving decently for most people. Draw simple icons or pictographs expressing the business transactions, with the verb in quotes underneath and a formula example from math as well. Modify the following article for art materials shopping: Use an App to Budget While Grocery Shopping
14. 14
Do this marketing project: Market to a unique and specialized aesthetic; this takes great skill in the "performance" (creation) of the piece; many artists enjoy classical music or complex jazz or improv harmonics to create by because it is conducive to achieving great statements with just the sufficient and necessary amount of detail, never producing anything overwrought. Just as E=mc^2 is simple and profound, so is the painting of Christian Rohlfs. Of course, Van Gogh had no idea that one of his paintings would sell for \$53 million, but he may very well have valued it that highly, given what he sacrificed in life for his work -- something of society and his sanity, to be sure. Many people are now able to produce likenesses of his work, but his unique take on reality at the time was something he suffered greatly to achieve.
• Do this project: Divide your canvas into the ratio 61.8% by 38.2%, twice, both vertically and horizontally and dramatize your subject(s) accordingly. This is one expression of the Golden Ratio, Phi (⦶), used at least since the Greeks in sculpture and architecture, and also by Michelangelo, Da Vinci and many other Masters. Computing the ratio with exactitude requires considerable effort on an artist's part, as it is (1±SQRT(5))/2 and is transcendental.
• Likewise, many mathematicians suffer for their work, as their theories underlying great elegance can be extremely complex to apply fully. That is the value -- the universality, just as in physics, it is the universality of the principle or law that is so greatly admired. Some play the norm: There is also much to be said for marketing to a mundane popular sexual and violent aesthetic that placates the masses. No one is saying you will sell your artwork for a million bucks but you never know unless you try.
Pine cones demonstrate phyllotaxis and the Golden Ratio, Phi ⦶ = -.618034 or 1.618034
15. 15
Do this ratio (proportions) project: Read and do the wikiHow article, Do Common Ratio Analysis of the Financials and invent a drawing or painting which demonstrates with curves each ratio (lines are a type of curve). Be aware that financial ratio analysis is critical to the survival of many banks and commercial enterprises. Then, after a week, go back and look at the curves and see if you can match each artwork to its proper ratio. A knowledge of spreadsheets and charting in Excel will be gained in completing this project.
Contour Graph of Financial Ratios
16. 16
Do this operations project: Think about the "verbs" or "operators" of math then: some common ones are +, -, *, /, :, exp(onentiation), log, etc. You can find more by googling "LaTex symbols". In art, we add in various ways, subtract/correct/erase in certain ways, create multiples and divide up spaces (creating proportions or ratios thereby). See the article Make Math Symbols on Your Mac (OS X) and the even better article How to Use LaTeX for Text Formatting
17. 17
Try this artistic math project: "Multiply" green over an area by covering many pixels with it or creating multiple plies of pigment. "Add" more yellow and blue to the composition perhaps to "balance the equation". "Subtracting" brightness (whites, yellows, light greys, etc.) will "reduce" the foreground. "Dividing" or breaking up a recognizable symbol or pattern bestirs a sense of trauma, alarm or tragedy perhaps -- it depends on how the contrast is handled -- Van Gogh often used outlining subjects to "positive" (often psychologically emotional) effect..
18. Van Gogh's 'Irises', Upper Rt Excel Transparency
18
Use composition to teach equations: A painted or mathematical surface pattern is an 'ar+Ranged' set of data elements, from the pixels of pointillism and digital art to the zones of Cubism and the lush brushwork of Impressionism. The "Domain and Range of a Function" are established concepts in pre-Calculus (research it if need be). A "Surface" may be comprised of a manifold of many like curves or contours, like the warping of gnarled redwood curves forms the surface of a beautiful table where roots or branches have grown over the years. By copying the works and curves of the Masters, an appreciation for their skill is acquired. (Notice the image is transparent on an Excel grid, for copying precision.) See article Paint Photos or Copy Masters Using XL Transparency
• One of the uses of the Calculus is to determine the arc length or curve and another use is to determine the area under a curve or between two curves, which are useful capabilities for artists. In the Theory of Neutral Operations, if addition is held neutral to multiplication between the two numbers 5 and 5/4, then 5 + 5/4 = 25/4 and so does 5 * 5/4 = 25/4. The two verb-states of Addition and Multiplication, or actions, or operations (or even functions), are held neutral to each other in the equation between the two constant values. This is a special state, and a special set of numbers, the Neutral Set, is thereby created. Members of the Neutral Set for a+b=a*b would include: {(2,2/1),(3,3/2),(4,4/3),...,(100,100/99)}.
• Artwork has been created using various Neutral Relations and the Neutral Set. That is using math to teach art. But once one sees that Nature can choose between ADDING a LENGTH (distance) or creating an AREA (surface) by MULTIPLYING, one can see the particular pattern of the Neutral Set is commonplace in Nature. Another neutral operation is to solve ab + cd = ab * cd = e, where ab and ccd are two rectangular areas, but with a little more thought and work might just as easily be triangles or combinations of forms.
Organic Valve, w/ Four Neutral Relations: a+b = a*b, a-b = a*b, a+b = a/b, a-b = a/b
See article Create Floral and Other Images with Trig and Neutral Operations and notice how the left top part will close over and meet the right opening, to form a spheroid, or allow for organic articulation. Neutral Operations and the spirals of radius r = angle ⊖ theta have a definite role in biology.
1. 1
Do this neutral project: Paint a length of 4 units and another length extending at right angles from the base of the first length equal to 4/3 of a unit. 4 + 4/3 = 16/3 and 4 * 4/3 = 16/3 too. Imagine that the rectangle formed by constructing the other two sides is filled in of its area, so that 4 * 4/3 =16/3 is the area is true. Now the length of 16/3 equals the area of 16/3 if and only if the mass used to fill the entire length equals the mass used to fill the entire area, which is possible in Nature, and can be represented both in art and math. Do so now by carefully measuring out equal masses.
2. 2
Observe and copy Nature, as Da Vinci recommended an artist do. Through this exploration, you can gain the hard-won lessons that reality has forged from the infinite possibilities Nature had before her, given all the random collisions of particles and sub-particles. Van Gogh did something new with the new pigments industry provided: he "informed" nature with purified chemicals, and thus even helped to inspire Kandinsky and Abstract Art, which in turn played a role in inspiring Jazz. In rendering a still life, one may bring in for comparison the differing chemistry of the subject matter and the representative media by using chemical equations, and for art events, chemical processes which definitely have a mathematical aspect.
3. "Warped Spheroidal Chain Asymptotes", by Chris Garthwaite - an XL trig chart.
3
Paint contours and asymptotes as accurately as possible, then see what you can achieve digitally using trigonometry. Refer to How to Create Spheroidal Asymptotes and Skewed Sphere Ring.
4. 4
Think of Chaos and Order when painting energy pattern flows and learn about Chaos Theory in math. Just as the gnarly knots of redwood seem to capture jetties of liquid water or fog the coastal trees often live in, the fractal spirals and standing waves of modern Chaos Theory in math seem to capture Nature's patterns of growth and decay. Spirals are everywhere: in quark movement in and out of dimensional planes, in the DNA double helix, in our body's organ development, in lightwaves and all electromagnetic radiation, and in the formation of the great galaxies and black holes.
• One of the very simplest formulas, besides Identity (1=1), is r=⊖, radius r = angle theta, i.e. as the radius increments so does the angle grow, so the result is a spiral. That is, if the length of radius r = .001, .002, .003 .... then the angle grows from .001º to .002º to .003º and on around the circle, again and again. But the statement itself, the equation, could not be more simple: r=⊖ ... 3 symbols, from which one gets the idea of uniform growth and decay, depending upon the starting value.
• Thus, by showing your students the simple tightly-woven spiral, you can use art to teach one of the profound formulas in all of mathematics! And it could come by basketweaving just as easily as from painting. In fact, given that the number pi is transcendental and infinite, it is not known whether there is such a thing as a perfect circle or whether it is actually a kind of spiral ... perfection is rather rare (it exists probably only as a Platonic Ideal and/or in Heaven in the Mind of God, and that is still a probability as far as is known). Many fractal algorithms are based upon self-iterative spirals.
5. 5
Learn and teach Form = Formula. (The etymology and definitions of "form" and "formula" are referred to in the References and Citations Section, below.) And the commutative principle applies: Formula = Form. Another way to state this may simply be: Art = Math and Math = Art. A deep derivation of the two word's etymologies would be revealing. Basically, start off with articulation=method, which is really pretty close to ⊖=r. Valuable lessons can be learned by researching the etymological derivations of words.
• In terms of painting, suppose the subject is a Vase V of Flowers F on a Table T in front of Wallpaper W, with a light source L. L is "ambient" to V+F+T+W and there is also shadow, -l = -v-f (as the painting does not show any shadow of the table or wall). So if Ambience A= L-l = V+F+T+W-v-f, then all the subjects are really composed of light or shadow for the purposes of the painting. However, because the flowers are partially translucent, -f on V and T is not the absence of the hue but prismatically it's altered slightly. Light passing through a red rose has the red subtracted from it, so it tends to be more blue-green. Analyzing in this way, the full value of A can be arrived at for each section of the design, by carefully projecting rays from the light source, L.
• What is often true, however, is that there are secondary sources of light and reflected light within a given environment, so keen observation of reality gives the practiced eye an advantage. That said, a given artist may desire to color the ambiance emotionally with "the blues" of despondency, or a very bright and cheery yellow! Van Gogh was known for his very pleasing still lifes of sunflowers and irises, set against colorful walls, as created very strong moods in observers, even though the detailed curves and contours of the plants themselves were perhaps only roughly rendered. To add such emotionalism mathematically is to change the chroma key of the painting and also to favor a rather blurry lens as far as the ray-tracing goes. However, even given the lack of detail, Van Gogh's brushwork resembles little flames often, and his paintings seem passionately full of emotional fire! This he accomplished by painting with upstrokes rather than downstrokes it's thought.
• The point is, some emotional or psychological effects that seem to lack definiteness and mathematical precision may be accomplished by using the new "fuzzy math" perhaps, where 2+2=5, approximately, or exactly -- so long as one is counting the answer by 5's. In fuzzy math, each of the two's may count as a "half-integer" of about 2.5 but where the only integer known is 5. 2.5 is a guesstimate; it is a theoretical value only in this system, rather than one with which calculations may be made. Therefore, the problem fails to be answerable according to the postulates of the system. In terms of pigments, black is made of all pigments, so the theory that the absence of light should be painted by adding a little black to the pigment the shadow falls upon is a rather curious one. Pigments are additive but light is subtractive.
Part 2 Considering the Mathematical Basis for Artistic Concepts
1. 1
Consider the types of brushstrokes one may apply. A brushstroke meeting the surface is an arbiter of pattern, just as in math an algorithm (or perhaps an individual series or function) is an arbiter of pattern. The book, "The Tao of Painting", illustrates Chinese brushwork for the beginner. Many Asian paintings contain a lot of very dynamic spatial form, i.e. chi energy! Either a brushstroke or algorithm may be empty -- take on the value of zero to work with existing canvas colors or even a negative -- picking up color off the canvas!
2. 2
Consider hue and/or value. Color Theory contains a lot of mathematical information. Colors have wavelength frequencies, which are measured numerically in hertz. Labelling colors with the hertz can be instructive for students. One can also google "RGB Color Codes Chart" and get the HTML, HEX(adecimal) and RGB codes for a given color in a color picker online now. RGB color space or RGB color system, constructs all the colors from the combination of the Red, Green and Blue colors.[2]
• RGB ≡ Red, Green, Blue
• The red, green and blue use 8 bits each, which have integer values from 0 to 255. This makes 256*256*256 = 256^3 = 16,777,216 possible colors.
• Each pixel in the LCD monitor displays colors this way, by combination of red, green and blue LEDs (light emitting diodes).
• When the red pixel is set to 0, the LED is turned off. When the red pixel is set to 255, the LED is turned fully on.
• Any value between them sets the LED to partial light emission.
• Considering that numbers equal colors, and they may also take on rational or even real values in spectrometry, ask yourself some questions. Can colors equal the square root of negative one however? Can they be imaginary? Yes, because raw electricity's spark can be a color and i, which is the symbol for the square root of negative one, is an imaginary number used in electronics. Or so it could be submitted. Students can be taught to calculate in hexadecimal or use an online converter. The systems color pickers are based on can be taught and are quite interesting. There is considerable consumer value in color picking in interior design and decor.
3. 3
Look at parallel ideals between art and math. Consider that in art, the ideal is (perhaps) the constructed aesthetic. In math, the ideal is the applied logic of theory (perhaps). One way of thinking about what the two disciplines may have in common is that their mutual output is PATTERN and FIT. FIT is a question of PROXIMITY and SPACE(s), which are concerns in Topology, regarding surfaces and much more in math. Pattern Recognition is an entire branch of study and science under Computer Science and also has to do with the analysis of intelligence. Art, fundamentally, is either Pattern Creation or Symbol Creation, or both. Math uses symbols once to describe the process and again to describe the data or form, generally speaking.
4. Trigonometric Iris, by C. Garthwaite
4
Use gradation precision in art color theory to teach about fractions. When mixing 2 or more colors, an artist might use "a half inch" out of the tube of pthalo blue pigment with "a quarter inch" of naples yellow pigment to achieve as blended with the palette knife a certain green, and yet the final green on the canvas may have a center and then gradations towards the edges, or be brighter at the edges and be graded towards the center, or have a side brighter and be gradated towards another side or curve. These are essentially fractions. By having the student construct gradation charts, especially in watercolors, they are learning a very important skill and technique of control.
5. 5
Consider symbols. Take for example the common portrait, used by student and genius alike. What similar event is there in math? In math, there are formulas which are famous for their makers: Pythagoras, Newton, Einstein, et al. Can one find the Pythagorean Theorem in a painting? Perhaps, but it is probably more obviously present via the calipers used by the Greeks in their great sculptures to calculate perfect proportions for their gods. And, without the triangle of that theorem, there would be no such thing as perspective! The Chinese were painting very small men in front of large mountains a long, long time ago. There are layers of horses in the cave paintings of the Cro-Magnon and also people depicted by the Egyptians. It isn't perspective itself, but it is the interim idea working up to it.
• That is, tied up in the very nature of layers and numbers is also the idea of distance, and so its symbol. Can one use the layers of the Great Pyramid at Giza to teach Math and Symbology? Yes. It's right there on the ubiquitous dollar bill! It's also in the fundamental idea of (decorated) clothing and status -- right back to animal skins and papyrus rolls, which were used for both art and math, i.e. trade and education and gifting.
6. 6
Take a minute to consider some of the processes art and math share in common as one matures:
• 01) Learn basic skills for the pattern to create, e.g. a digital portrait using Bézier curves;
• 03) Learn about the various theories and theorems (+ axioms, postulates, corollaries, etc);
• 04) Do exercises, i.e. practice a lot to get much better;
• 05) Imitate the masters and read everything - including the online state of the art;
• 06) Think about areas that aren't explored yet, where you might make a contribution;
• 07) Share progress with other professionals, per your specific area of expertise;
• 08) Do more research, bringing in areas by analogy and metaphor - widen horizons;
• 09) Invent own system and perhaps also symbology/vocabulary/style and create works!
• 10) Organize the business and marketing and then conduct business professionally;
• 11) Write a book or blog about what's been learned;
• 12) Consider teaching and other ways of connecting with questioners;
• 13) Get involved with children to see them bring fresh ideas to the work!
Part 3 Using the Graphic and Math Capabilities of Microsoft Excel
1. 1
Explore some options for connecting math and art through the computer. There are many examples of this kind of project, but the following one is a great place to start.
2. 2
Open a new workbook in Microsoft Excel.
3. 3
Set Preferences. Be mindful that these settings will affect your future XL work.
• General - Set Show this number of recent documents to 15; set Sheets in new workbook to 3; this editor works with Body Font, in a font size of 12; set your preferred file path/location;
• View - Check Show formula bar by default; check Indicators only, and comments on hover for Comments; show All for objects; Show row and column headings, Show outline symbols, Show zero values, Show horizontal scroll bar, Show vertical scroll bar, Show sheet tabs;
• Edit - Check all; Display 0 number of decimal places; set Interpret as 21st century for two-digit years before 30; Uncheck Automatically convert date system;
• AutoCorrect - Check all
• Chart - In Chart Screen Tips, check Show chart names on hover, and check Show data marker values on hover; leave the rest unchecked;
• Calculation - Automatically checked; Limit iteration to 100 Maximum iterations with a maximum change of 0.0001, unless goal seeking (which is not anticipated for this project), then .000 000 000 000 01 (w/o spaces); check Save external link values;
• Error checking - Check all and this editor uses dark green or red to flag errors;
• Save - Check all; set to 5 minutes;
• Compatibility - check Check documents for compatibility
• Ribbon - All checked, except Hide group titles is unchecked.
• View - Show Gridlines -- unchecked
4. 4
[Optional - Go to Tools-Macro-Record New Macro]
5. 5
Click on the Media Browser tool icon along the top of the Toolbar.
6. 6
Select Shapes.
7. 7
Select the S-shaped modifiable curve bly clicking on it with the mouse.
8. 8
Make a snaky and spiral shape by clicking where a new curve is wanted on the spreadsheet.
9. 9
Click on the curve and select menu item Format Shape.
• Do Line - thickness 40 and gradient style Linear and make the right color yellow with the left color dark blue.
10. 10
Make sure that your shape resembles the one above.
11. Anderson tartan, as woven by an XL macro, then placed in aligned box border fill and heart fill
11
Optionally, halt your macro and inspect the macro to see the data points and actions recorded!
• Learn how to use basic trigonometry to create graphic objects in Excel via articles such as the following: Create a Sin and Cos Circle in Excel and the guides in the Microsoft Excel Imagery category.
• The macro code that would result would be something like the following:
• Sub Macro1()
• ' Macro1 Macro
• ' squiggle
• ' (many nodes were cut out of this code to avoid extraneous length in this article)
• ' Keyboard Shortcut: Option+Cmd+s
• With ActiveSheet.Shapes.BuildFreeform(msoEditingAuto, 265, 77)
• 102.6666141732,_ 181.6666929134, 128.3333070866, 167, 152
• 175.6666929134, _ 170.6666929134, 198.1666929134, 177, 219
• 239.8333070866, 184 _ , 267.6666929134, 205, 277
• .AddNodes msoSegmentCurve, msoEditingAuto, 226, 286.3333070866, 281, 285,
• 303 _ , 275
• .ConvertToShape.Select
• End With
• With Selection.ShapeRange.Line
• .Visible = msoTrue
• .Weight = 30
• End With
• Selection.ShapeRange.Line.Visible = msoTrue
• Selection.ShapeRange.Line.Visible = msoTrue
• Selection.ShapeRange.Line.Visible = msoTrue
• Selection.ShapeRange.Line.Visible = msoTrue
• End Sub
Community Q&A
200 characters left
Article Info
Categories: Mathematics | Graphics
Thanks to all authors for creating a page that has been read 133,187 times. |
# Shell Method Vs Disk Method
When it comes to calculus, there are specific methods used to calculate areas and volumes of different shapes. The two most commonly used methods for finding volumes of three-dimensional shapes are the shell method and the disk method. Both methods have their advantages and disadvantages depending on the shape being analyzed. In this article, we will discuss the differences between the shell method and the disk method and their applications.
What is the Shell Method?
The shell method is a technique used to calculate the volume of a solid of revolution by integrating cylindrical shells. Solid of revolution refers to a three-dimensional shape obtained by rotating a two dimension shape around a central axis.
The shell method is most commonly used to calculate the volume of a cylinder or a frustrum, a cone with the top cut off. The formula for the shell method is V=2π ∫ x( f(x)-g(x))dx, where V represents the volume of the solid, f(x) and g(x) are the two equations of the curves that enclose the solid, and x is the variable of integration. The integral is taken between the bounds of the solid.
For a better understanding of the shell method, let’s take the example of finding the volume of a cylinder with radius r and height h, using the shell method.
First, we will slice the cylinder into small circular strips parallel to its base. When we place these strips at the top of each other, they form a cylindrical shell.
Next, we will calculate the volume of each cylindrical shell. The volume of one cylindrical shell can be calculated by multiplying its height, which is r, and its circumference, which is 2π. Therefore, the volume of one cylindrical shell will be 2πr.
Now, to calculate the volume of the entire cylinder, we will add up the volumes of all these cylindrical shells. The set of infinite shells will be 2πrh if the height of the cylinder is h.
What is the Disk Method?
The disk method is another method used to calculate the volume of a solid of revolution. This method involves slicing the solid into thin disk-shaped segments or slices, then adding up the volume of each slice. This method is most commonly used in calculating the volume of a sphere or a cone.
To understand the disk method, let’s take the example of finding the volume of a sphere with radius r.
First, we will slice the sphere into thin disks, parallel to its base. We begin by cutting the sphere in half, creating a cross-section. Next, we will slice the half-sphere into circular disks perpendicular to the base.
The volume of one disk can be calculated by multiplying the area of its base, which is πr², by its thickness, which is smaller than the radius. Hence, the volume of one disk is πr²x, where x is the thickness of the disk.
Now to get the volume of the whole sphere, the set of infinite disks, we will integrate the volume of one disk using the definite limits of the function, which will be written as V=π∫(R²-x²)dx.
The Difference between Shell Method and Disk Method
Even though both methods can be used to calculate the volume of a solid of revolution, there are significant differences between them.
In the shell method, the shape is divided into cylindrical shells that are integrated along the axis of rotation. As a result, this method works best when the axis of revolution is on the side of the shape. The shell method is also ideal when dealing with shapes that have difficult cross-sections, such as cut-out sections or asymmetrical shapes.
On the other hand, the disk method divides the shape into a series of slices, which are then measured and added together. This method works best when the axis of revolution is located in the center of the shape. The disk method is ideal for dealing with shapes that have a clear circular cross-section.
The keywords that can be used to optimize this article are: Shell Method, Disk Method, Calculus, Solids of Revolution, Volume of a Solid, Cylinder, Cone, Sphere.
Conclusion
The shell method and the disk method are both essential techniques used in calculus to calculate the volume of a solid of revolution. The shell method is more appropriate when the axis of rotation is on the side of the shape, while the disk method is ideal when the axis of rotation is in the center of the solid. When dealing with complex shapes with varying cross-sections, the shell method proves more useful. The disk method, on the other hand, excels at handling shapes with clear circular cross-sections. Understanding the differences between these two methods can be helpful in calculating volumes of three-dimensional shapes. |
The Evolution of the
R E A L N U M B E R S
7
# RATIONAL NUMBERS
THE RATIONAL NUMBERS are the numbers of ordinary arithmetic. They are the whole numbers, the fractions, the mixed numbers, and decimals. Those are the numbers with whose names we count and measure.
Why are they called rational? We have seen that every fraction has the same ratio to 1 as the numerator has to the denominator:
ab : 1 = a : b.
A number that has the same ratio to 1 as two natural numbers -- whose relationship to 1 we can always name -- we say is rational.
Now we can write any number of arithmetic as a fraction and thus show that ratio to 1. A whole number, such as 6, we can write as ; we can write any mixed number as a fraction; and we can write any decimal as a fraction.
The rational numbers are simply the numbers of arithmetic.
(In algebra, those numbers of arithmetic are extended to their negative images. See Topic 2 of Precalculus.)
Problem 1. Which of these numbers are rational?
1 5 38 6¼ .005 9.2 1.6340812437
All of them!
Problem 2. Write each of the following as a fraction.
5 = 51 6¼ = 25 4 .35 = 35 100 9.2 = 9210 1.732 = 17321000
Problem 3. To what does the word "rational" refer?
The ratio of two natural numbers.
The number line
We need the numbers of arithmetic for measuring. Therefore we can think of them as naming a distance from 0 along the number line.
But will those rational numbers account for every distance from 0? Will every length be a rational number of units? To pursue that question, we have the following theorem:
Theorem. Any two rational numbers have the same ratio as two natural numbers.
That is true because:
Fractions with the same denominator have the same ratio
as their numerators.
And we can always express two fractions with the same denominator.
Example 1. 25 : 35 = 2 : 3
25 is two thirds of 35 .
We could prove that by multiplying both fractions by their common denominator. (Lesson 3.)
Example 2. 23 : 56
We can make the denominators the same.
23 = 46
Therefore,
23 : 56 = 46 : 56 = 4 : 5.
Example 3. 23 : 58
In this example, we can choose a common denominator, 3 × 8 = 24. We can then obtain the numerators by cross-multiplying:
We can always express the ratio of two fractions by cross-multiplying. Cross-multiplying gives the numerators of the common denominator.
Example 4. 45 : 79 = 36 : 35
Example 5. = 10 : 3
Example 6. Explicitly, what ratio has 12 to 1 34 ?
Explicitly means to verbally name that ratio.
Answer. 12 : 1 34 = 12 : 74 = 4 : 14 = 2 : 7
Explicitly, then, 12 is two sevenths of 1 34 .
Example 7. .3 is to 1.24 in the same ratio as which two natural numbers?
Answer. We can "clear of decimals" by multiplying both numbers by the same power of 10; in this case, 100:
.3 : 1.24 = 30 : 124 = 15 : 62, upon dividing by 2.
We have now established the theorem:
Any two rational numbers have the same ratio
as two natural numbers
.
Example 8. A photograph measures 2½ inches by 3½ inches. You want to enlarge it so that the shorter side will be 10 inches. How long will the larger side be?
Solution. Proportionally,
2½ inches : 3½ inches = 52 : 72 = 5 : 7.
So the question is:
5 : 7 = 10 inches : ? inches
Now, 5 has been multiplied by 2. Therefore, 7 will also be multiplied by 2. (Lesson 3.) The longer side will be 14 inches.
Problem 4. Show that these rational numbers have the same ratio as two natural numbers.
a) 59 : 79 = 5 : 7. The denominators are the same.
b) 15 3 : 16 3 = 15 : 16 c) 12 : 34 = 24 : 34 = 2 : 3
d) 25 : 37 = 14 : 15. Cross-multiply. e) 12 : 13 = 3 : 2
f) 38 : 7 10 = 30 : 56 = 15 : 28 g) 49 : 23 = 12 : 18 = 2 : 3
h) 2 : 12 = 4 : 1 i) 56 : 7 = 5 : 42 j) 23 : 1 = 2 : 3 k) 1 : 12 = 2 : 1 l) 85 : 1 = 8 : 5 m) 1 : 85 = 5 : 8
n) 1 : 3 12 = 1 : 72 = 2 : 7
o) 6 78 : 5 = 55 8 : 5 = 55 : 40 = 11 : 8
p) 2 34 : 3 12 = 11 4 : 72 = 22 : 28 = 11 : 14
Problem 5. Explicitly, what ratio has
a) 12 to 2? 12 : 2 = 1 : 4. 12 is one fourth of 2.
b) 43 to 29 ? 43 : 29 = 36 : 6 = 6 : 1. 43 is six times 29 .
c) 1 14 to 12 ? 1 14 : 12 = 54 : 12 = 54 : 24 = 5 : 2. 1 14 is two and a half times 12 .
Problem 6. Show that these rational numbers have the same ratio as two natural numbers.
a) .2 : .3 = 2 : 3 b) .2 : .03 = 20 : 3 c) 2 : .03 = 200 : 3
d) .025 : 1 = 25 : 1000 = 1 : 40 e) .025 : .01 = 25 : 10 = 5 : 2
f) 6.1 : 6.01 = 610 : 601
Problem 7. A loaf of bread weighs 1 13 pounds, and you
want to cut off half a pound; where will you cut the loaf?
(Hint: What ratio has half a pound to 1 13 pounds?)
12 : 1 13 = 12 : 43 = 3 : 8. Cut three eighths of the loaf.
Problem 8.
a) Corresponding to every rational number, is there a distance from 0
a) on the number line? Yes.
b) Corresponding to every distance from 0, is there a rational number?
a) Hmmm. Is there?
Next Topic: Measurement: Geometry and arithmetic
Please make a donation to keep TheMathPage online.
Even \$1 will help. |
# Use the Product Rule to Simplify the Expression by tutorcircleteam
VIEWS: 5 PAGES: 4
• pg 1
``` Use the Product Rule to Simplify the Expression
Use the Product Rule to Simplify the Expression
We use the product rule to simplify the expression where expression is a
combination of different kind of variables, numbers and operations like addition,
subtraction, multiplication and division.Now, we discuss how product rule
simplify different kind of expressions:
Combination of algebraic and exponential: If expression is a combination of
algebraic and exponential function, then with the help of product rule we can
easily solve differentiation of that expression-
d (x.ex) = x. d (ex) + ex.d (x)
dx dx dx
= x.ex + ex.1
= ex(x+1)
Know More About Antiderivative of a x
Combination of algebraic and trigonometric: If expression is a combination of
algebraic and trigonometric function, then-
d (x2.sin x) = x2. d (sin x) + sinx.d (x2)
dx dx dx
= x2.(cos x) + sin x .(2x)
= x(x.cos x + 2.sin x)
Combination of algebraic and logarithm: If expression is a combination of
algebraic and logarithm function, then-
d (x.ln x) = x. d (ln x) + ln x.d (x)
dx dx dx
= x.(1) + ln x.(1)
x
= 1+ ln x
Combination of trigonometric: If expression is a combination of two
trigonometric functions, then-
d (sin x. cosx) = sin x. d (cos x) + cos x.d (sin x)
dx dx dx
= sin x.(-sin2x) + cos x.(cos x)
= cos2x – sin2x
= cos2x
Combination of trigonometric and exponential: If expression is a combination of
trigonometric and exponential function, then-
d (sin x.ex) = sin x. d (ex) + ex. d (sin x)
dx dx dx
= sin x. (ex) + ex.(cos x)
= ex (sinx + cos x)
These are different expressions which we simplify by product rule.
Thank You
TutorCircle.com
```
To top |
# Linear Equations in Two Variables Class 9 Notes
## CBSE Class 9 Maths Linear Equations In Two Variables Notes:-
Get the complete notes on linear equations in two variables class 9 here. In this article, you are going to study the basics of linear equations involving one variable, two variables, and so on. Also, learn how to graph the linear equations, and how to find the solutions for linear equations in detail.
### Linear equation in one variable
When an equation has only one variable of degree one, then that equation is known as linear equation in one variable.
• Standard form: ax+b=0, where a and b ϵ R & a ≠0
• Examples of linear equation in one variable are :
– 3x-9 = 0
– 2t = 5
To know more about Linear equations in one variable, visit here.
### Linear equation in 2 variables
When an equation has two variables both of degree one, then that equation is known as linear equation in two variables.
Standard form: ax+by+c=0, where a, b, c ϵ R & a, b ≠0
Examples of linear equations in two variables are:
– 7x+y=8
– 6p-4q+12=0
To know more about Linear equations in 2 variables, visit here.
## Examples of Linear Equations
### The solution of linear equation in 2 variables
A linear equation in two variables has a pair of numbers that can satisfy the equation. This pair of numbers is called as the solution of the linear equation in two variables.
• The solution can be found by assuming the value of one of the variable and then proceed to find the other solution.
• There are infinitely many solutions for a single linear equation in two variables.
## Graph of a Linear Equation
### Graphical representation of a linear equation in 2 variables
• Any linear equation in the standard form ax+by+c=0 has a pair of solutions in the form (x,y), that can be represented in the coordinate plane.
• When an equation is represented graphically, it is a straight line that may or may not cut the coordinate axes.
To know more about Graphical representation of a linear equation, visit here.
### Solutions of Linear equation in 2 variables on a graph
• A linear equation ax+by+c=0Â is represented graphically as a straight line.
• Every point on the line is a solution for the linear equation.
• Every solution of the linear equation is a point on the line.
### Lines passing through the origin
• Certain linear equations exist such that their solution is (0, 0). Such equations when represented graphically pass through the origin.
• The coordinate axes namely x-axis and y-axis can be represented as y=0 and x=0, respectively.
### Lines parallel to coordinate axes
• Linear equations of the form y=a, when represented graphically are lines parallel to the x-axis and a is the y-coordinate of the points in that line.
• Linear equations of the form x=a, when represented graphically are lines parallel to the y-axis and a is the x-coordinate of the points in that line.
#### 1 Comment
1. thanks for the help |
# Math Insight
### Critical points, monotone increase and decrease
A function is called increasing if it increases as the input $x$ moves from left to right, and is called decreasing if it decreases as $x$ moves from left to right. Of course, a function can be increasing in some places and decreasing in others: that's the complication.
We can notice that a function is increasing if the slope of its tangent is positive, and decreasing if the slope of its tangent is negative. Continuing with the idea that the slope of the tangent is the derivative: a function is increasing where its derivative is positive, and is decreasing where its derivative is negative.
This is a great principle, because we don't have to graph the function or otherwise list lots of values to figure out where it's increasing and decreasing. If anything, it should be a big help in graphing to know in advance where the graph goes up and where it goes down.
And the points where the tangent line is horizontal, that is, where the derivative is zero, are critical points. The points where the graph has a peak or a trough will certainly lie among the critical points, although there are other possibilities for critical points, as well.
Further, for the kind of functions we'll deal with here, there is a fairly systematic way to get all this information: to find the intervals of increase and decrease of a function $f$:
• Compute the derivative $f'$ of $f$, and solve the equation $f'(x)=0$ for $x$ to find all the critical points, which we list in order as $x_1 < x_2 < \ldots < x_n$.
• (If there are points of discontinuity or non-differentiability, these points should be added to the list! But points of discontinuity or non-differentiability are not called critical points.)
• We need some auxiliary points: To the left of the leftmost critical point $x_1$ pick any convenient point $t_o$, between each pair of consecutive critical points $x_i,x_{i+1}$ choose any convenient point $t_i$, and to the right of the rightmost critical point $x_n$ choose a convenient point $t_n$.
• Evaluate the derivative $f'$ at all the auxiliary points $t_i$.
• Conclusion: if $f'(t_{i+1})>0$, then $f$ is increasing on $(x_i,x_{i+1})$, while if $f'(t_{i+1}) <0$, then $f$ is decreasing on that interval.
• Conclusion: on the ‘outside’ interval $(-\infty,x_o)$, the function $f$ is increasing if $f'(t_o)>0$ and is decreasing if $f'(t_o) <0$. Similarly, on $(x_n,\infty)$, the function $f$ is increasing if $f'(t_n)>0$ and is decreasing if $f'(t_n) <0$.
It is certainly true that there are many possible shortcuts to this procedure, especially for polynomials of low degree or other rather special functions. However, if you are able to quickly compute values of (derivatives of!) functions on your calculator, you may as well use this procedure as any other.
Exactly which auxiliary points we choose does not matter, as long as they fall in the correct intervals, since we just need a single sample on each interval to find out whether $f'$ is positive or negative there. Usually we pick integers or some other kind of number to make computation of the derivative there as easy as possible.
It's important to realize that even if a question does not directly ask for critical points, and maybe does not ask about intervals either, still it is implicit that we have to find the critical points and see whether the functions is increasing or decreasing on the intervals between critical points.
#### Examples
Find the critical points and intervals on which $f(x)=x^2+2x+9$ is increasing and decreasing: Compute $f'(x)=2x+2$. Solve $2x+2=0$ to find only one critical point $-1$. To the left of $-1$ let's use the auxiliary point $t_o=-2$ and to the right use $t_1=0$. Then $f'(-2)=-2 <0$, so $f$ is decreasing on the interval $(-\infty,-1)$. And $f'(0)=2>0$, so $f$ is increasing on the interval $(-1,\infty)$.
Find the critical points and intervals on which $f(x)=x^3-12x+3$ is increasing, decreasing. Compute $f'(x)=3x^2-12$. Solve $3x^2-12=0$: this simplifies to $x^2-4=0$, so the critical points are $\pm 2$. To the left of $-2$ choose auxiliary point $t_o=-3$, between $-2$ and $+2$ choose auxiliary point $t_1=0$, and to the right of $+2$ choose $t_2=3$. Plugging in the auxiliary points to the derivative, we find that $f'(-3)=27-12>0$, so $f$ is increasing on $(-\infty,-2)$. Since $f'(0)=-12 <0$, $f$ is decreasing on $(-2,+2)$, and since $f'(3)=27-12>0$, $f$ is increasing on $(2,\infty)$.
Notice too that we don't really need to know the exact value of the derivative at the auxiliary points: all we care about is whether the derivative is positive or negative. The point is that sometimes some tedious computation can be avoided by stopping as soon as it becomes clear whether the derivative is positive or negative.
#### Exercises
1. Find the critical points and intervals on which $f(x)=x^2+2x+9$ is increasing, decreasing.
2. Find the critical points and intervals on which $f(x)=3x^2-6x+7$ is increasing, decreasing.
3. Find the critical points and intervals on which $f(x)=x^3-12x+3$ is increasing, decreasing. |
Lesson 2, Topic 5
In Progress
Electromotive Force (E.M.F)
Lesson Progress
0% Complete
E.m.f in Series:
When cells are connected in series, the positive terminal of one is connected to the negative terminals of the other. With this arrangement, a larger emf is obtained in the circuit with improved current.
Etotal = E1 + E2 + E3
rtotal = r1 + r2 + r3
Where r = the internal resistance of the cell.
E.m.f in Parallel:
For a parallel connection, all positive terminals are joined together and all negative terminals are joined together.
The e.m.f of one of the cell is equal to the total e.m.f E1 = E2 = E3
Example 1:
A cell of e.m.f 1.5V is connected to a resistor of resistance 3Ω. Calculate the current flowing.
Solution:
Emf = 1.5V, R= 3Ω, I =?
V = IR
I = $$\frac{V}{R}$$
I = $$\frac{1.5 V}{3 \Omega}$$
I = 0.5A
Example 2:
A battery of emf 24v and internal resistance 4Ω is connected to a resistor of 32Ω. What is the terminal p.d of the battery?
Solution:
Whether you are asked to or not, always sketch a circuit diagram before solving any question on current electricity.
First, calculate the current, I
I = $$\frac {E}{R + r}$$
= $$\frac {24}{32 + 4 }$$
= $$\frac {24}{36 }$$
= $$\frac {2}{3} \scriptsize A$$
or 0.67A
∴ Terminal p.d, V = IR
= 0.67 x 32
= 21.3v
Evaluation Question:
1. A cell of e.m.f 1.5V and internal resistance 2.5Ω is connected in series with an ammeter of resistance 0.5Ω and a load of resistance 7.0Ω. Calculate the current in the circuit.
2. A cell of e.m.f 1.5V is connected in series with a resistor of resistance 3.0Ω. A voltmeter connected across the cell registers 0.9V, calculate the internal resistance of the cell.
error: |
# 1.1 homework interval and set notation
## 1.1 homework interval and set notation
Inequality: interval notation: inequality: interval notation: Sep 169:20 PM Domain / Range: Domain: Set of all possible values of x Range: Set of all possible values of y.Hence the solution in interval notation is (-∞, 5/12) U (7/12, ∞).If an element x is a member of the set S, we write.1 Lesson WWhat You Will Learnhat You Will Learn Represent intervals using interval notation.This set is all numbers between –2 and positive infinity.Find the domain of the function.You can use braces { } to represent a set by listing its members or elements.In some other cases, it could be literally impossible to write.(Enter your answer using interval notation.1 Sets of Real Numbers and The Cartesian Coordinate Plane 5 Example 1.Express the following interval in set notation.Homework Equations i) x^3 + x^2 > 2x ii) l 2 1.1 homework interval and set notation - x l =< 4.Describe the domain and range of the graph using an 1.1 homework interval and set notation inequality, set notation, and interval notation.This same set can be written in set 1.1 homework interval and set notation notation: {x ∈ R ∣ x < 5 or 10 < x < 20 or x > 100} Finally, note that if the characterization of the set is rather complex, the set notation becomes preferable to the interval one, which would require a great number of intervals in the union.Then describe its end behavior.Each object in the set is called an element of the set.Fxj 1 1} (1,∞) arenthesis: open circle {B 𝑥 | 𝑥≤2} (−∞,2] racket: closed circle.Learn vocabulary, terms, and more with flashcards, games, and other study tools Number Line Set Notation Interval Notation {P 𝑥 | 𝑥>1} (1,∞) arenthesis: open circle {B 𝑥 | 𝑥≤2} (−∞,2] racket: closed circle.Answers to practice exercises can be found on pages 15-17 Student's Solutions Manual for Algebra for College Students (7th Edition) Edit edition.Learn vocabulary, terms, and more with flashcards, games, and other study tools Intermediate Algebra for College Students (7th Edition) answers to Chapter 1 - Section 1.(Enter your answer in interval notation.) h(x) = 1/(sqrt(x^2-4x)^1/4) Answer: (-infinity,0)U(4,infinity) h(x) = 1/(sqrt(x^2-6x)^1/4) Answer: (-infinity,0)U(6,infinity) 13.
#### Changing equilibrium homework, set 1.1 homework and notation interval
Homework 1 Real Analysis Joshua Ruiter March 23, 2018 Note on notation: When I use the symbol ˆ, it does not imply that the subset is proper.Cumulative Review 4 Set Notation.1 Basic Set Theory and Interval Notation.Write solution set using in-terval notation.Express the following sets of numbers using interval notation.This notation can be read as x interval notation.X í4 62/87,21 The set includes all real numbers less than or equal to ±4.Set-builder & Interval Notation.The empty set is the set that contains no elements.1 - Algebraic Expressions, Real Numbers, and Interval Notation - Exercise Set - Page 12 1 including work step by step written by community members like you.Video for all operations of sets Extra practice/interactive links: 1., ISBN-10: 0-13417-894-7, ISBN-13: 978-0-13417-894-3, Publisher: Pearson.Another commonly used, and arguably the most concise, method for describing inequalities and solutions to inequalities is called interval notation.The x-axis is the perpendicular bisector of the line segment through A (2, 5.Elements in a set 1.1 homework interval and set notation do not “repeat”.Find the domain of the function.That is, for each k2N, there is a covering.3) – Represent inequalities using interval notation.1 Sets and Set Operations Express sets using interval notation; Required Reading 0.With this convention, sets are built with parentheses or brackets, each having a distinct meaning.1: Set builder and Interval Notation.Types of sets and set notation practice 2.Methods of Describing Sets: Sets may be described in many ways: by roster, by set-builder notation, by interval notation, by graphing on a number line, and/or by Venn diagrams..Using Interval Notation In mathematics, a collection of objects is called a set.A set is a collection of unique elements.Write each set of numbers in set -builder and interval notation, if 1.1 homework interval and set notation possible.1: Write the set in interval notation and graph the interval Name: Homework # 1 Solve the given inequality.For sets A and B, A is called a subset of B, denoted , if every element of A is also an element of B Exercise 1: (4 pts) Graph the sets and express each set in the interval notation: (i) {x1-15x sign.In set -builder notation this set can be described as { x | x í4, x }.Graph of fx x() 3:=− +2 Domain: Inequality: _____ Set Notation: _____ Interval Notation: 1-1.Union and Intersection of sets 3.A set is a collection of objects.For each of these, state the least upper bound and greatest lower bounds, if these exist.Stitz-Zeager Prerequisites - pages 3-12.Practice Exercises Stitz-Zeager Prerequisites - pages 13-14.Video for interval notation/set builder notation/roster notation 2.In writing AˆX, I mean only that a2A =)a2X, leaving open the possibility that and the total length of these intervals is 1 (1 ˘)k.Textbook Authors: Blitzer, Robert F.If an element x is not a member of the set S, we write. |
# GRAPHING QUADRATIC EQUATION AND FIND THE NATURE OF ROOTS
Graphing Quadratic Equation and Find the Nature of Roots :
In this section, you will learn, how to examine the nature of roots of a quadratic equation using its graph.
To obtain the roots of the quadratic equation
ax2 + bx + c = 0
graphically, we first draw the graph of
y = ax2 +bx +c
The solutions of the quadratic equation are the x coordinates of the points of intersection of the curve with x-axis.
## Graphing Quadratic Equation and Find the Nature of Roots - Questions
To find the questions i and ii, please visit the page "How to Find Nature of Solution of Quadratic Equation with Graph"
To find the questions iii and iv, please visit the page "Finding Nature of Quadratic Equation by Graphing"
Question 1 :
Graph the following quadratic equations and state their nature of solutions.
(v) x2 - 6x + 9 = 0
Solution :
Draw the graph for the function y = x2 - 6x + 9
Let us give some random values of x and find the values of y.
x-4-3-2-101234 x216941014916 -6x-24-18-12-60-6-12-18-24 +9999999999 y101494101
Points to be plotted :
(-4, 1) (-3, 0) (-2, 1) (-1, 4) (0, 9) (1, 4) (2, 1) (3, 0) (4, 1)
To find the x-coordinate of the vertex of the parabola, we may use the formula x = -b/2a
x = -(-6)/2(1) = 6/2 = 3
By applying x = 3, we get the value of y.
y = 32 - 6(3) + 9
y = 9 - 18 + 9
y = 0
Vertex (3, 0)
The graph of the given parabola intersect the x-axis at the one point. Hence it has real and equal roots.
(vi) (2x - 3)(x + 2) = 0
Solution :
(2x - 3)(x + 2) = 0
2x2 + 4x - 3x - 6 = 0
2x2 + x - 6 = 0
Let us give some random values of x and find the values of y.
y = 2x2 + x - 6
x-4-3-2-101234 2x23218820181832 x-4-3-2-101234 -6-6-6-6-6-6-6-6-6-6 y2290-5-6-441530
Points to be plotted :
(-4, 22) (-3, 9) (-2, 0) (-1, -5) (0, -6) (1, -4) (2, 4) (3, 15) (4, 30)
To find the x-coordinate of the vertex of the parabola, we may use the formula x = -b/2a
x = -1/2(2) = 1/4
By applying x = 1/4, we get the value of y.
y = 2(1/4)2 + (1/4) - 6
y = 2(1/16) + (1/4) - 6
y = -45/8
Vertex (1/4, -45/8)
The graph of the given parabola intersects the x-axis at two points. Hence it has two real and unequal roots.
After having gone through the stuff given above, we hope that the students would have understood, how to examine the nature of roots of a quadratic equation using its graph.
Apart from the stuff given in this sectionif you need any other stuff in math, please use our google custom search here.
You can also visit the following web pages on different stuff in math.
WORD PROBLEMS
Word problems on simple equations
Word problems on linear 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 |
# U-Substitution
### U-Substitution
You may start to notice that some integrals cannot be integrated by normal means. Therefore, we introduce a method called U-Substitution. This method involves substituting ugly functions as the letter "u", and therefore making our integrands easier to integrate. We will use this technique to integrate many different functions such as polynomial functions, irrational functions, trigonometric functions, exponential functions and logarithmic functions. We will also integrate functions with a combination of different types of functions.
Basic concepts: Chain rule, Antiderivatives,
#### Lessons
Pre-requisite:
* Differential Calculus –“Chain Rule”
* Integral Calculus –“Antiderivatives”
Note:
The main challenge in using the $u-Substitution$ is to think of an appropriate substitution.
- Question: how to choose $u$?
- Answer: choose $u$ to be some function in the integrand whose differential also occurs!
hint:
$u$ is usually the inside of a function, for example:
- the inside a power function: $( u )^{10}$
- the inside a radical function: $\sqrt{u}$
- the inside of an exponential function: $e^u$
- the inside of a logarithmic function: $\ln$? $(u)$
- the inside of a trigonometric function: $\sin$ $(u)$
• Introduction
Introduction to u-Substitution
$\cdot$ What is $u$-Substitution?
$\cdot$ Exercise: Find $\int (5x^4-6x) \cos (x^5-3x^2)dx$.
- How to pick "$u$"?
- How to verify the final answer?
• 1.
Integrate: Polynomial Functions
$\int-7x(6x^2+1)^{10}dx$
• 2.
a)
$\int\frac{x^6}{{^3}\sqrt{(8-5x^7)^2}}dx$
b)
$\int\sqrt{6-3x}$ $dx$
• 3.
Integrate: Exponential Functions
$\int e^{2x}dx$
• 4.
Integrate: Logarithmic Functions
a)
$\int \frac{(\ln x)^3}{x}dx$
b)
$\int \frac{dx}{x \ln x}$
• 5.
Integrate: Trigonometric Functions
a)
$\int \sin ^3 x \cos x\; dx$
b)
$\int \sec ^2 x(\tan x-1)^{100}\;dx$
• 6.
Not-So-Obvious U-Substitution
a)
$\int \sqrt{x^3-8}x^5dx$
b)
$\int {^3}\sqrt{1+x^2}x^5dx$
c)
$\int \frac{1+x}{1+x^2}dx$
d)
$\int \cot x$ $dx$
• 7.
Evaluate Definite Integrals in Two Methods
Evaluate: $\int_{-1}^{2} \sqrt{6-3x} dx$
a)
Introduction to definite integrals.
b)
Method 1: evaluate the definite integral in terms of “$x$”.
c)
Method 2: evaluate the definite integral in terms of “$u$".
d)
Method 1 VS. Method 2.
• 8.
Evaluate Definite Integrals
Evaluate: $\int_{0}^{\frac{\pi}{3}} \frac{\sin \theta}{\cos ^2 \theta}d \theta$
• 9.
Definite Integral: Does Not Exist (DNE)
Evaluate: $\int_{1}^{5} \frac{dx}{(x-3)^2}$ |
# TIME VALUE OF MONEY (TVM)
Size: px
Start display at page:
Download "TIME VALUE OF MONEY (TVM)"
Transcription
1 TIME VALUE OF MONEY (TVM) INTEREST Rate of Return When we know the Present Value (amount today), Future Value (amount to which the investment will grow), and Number of Periods, we can calculate the rate of return with this formula: Example: = The Present Value is \$10,000, the Future Value is \$299,599.22, and there are 30 periods. Confirm that the annual compound interest rate is 12%. FV = 299, PV = 10,000 m = 30 =,., i = i =.12 Page 1 of 15
2 Calculating the Rate (i) in an Ordinary Annuity Using the PVOA equation, we can calculate the interest rate (i) needed to discount a series of equal payments back to the present value. In order to solve for (i), we need to know the present value amount, the amount of the equal payments, and the length of time (n). Exercise #9. Sylvia has an investment account that shows a balance of \$2, on January 1, She wants to make five withdrawals of \$700 each on December 31 of years 2011 through Sylvia wants the account to have a balance of \$0 on December 31, In order to proceed with her plans, what annual interest rate does Sylvia need on her account, assuming that annual interest earnings are added to the principal on December 31 of each year? PVOA= \$2, \$700 \$700 \$700 \$700 \$700 1 year 1 year 1 year 1 year 1 year 1/1/11 12/31/11 12/31/12 12/31/13 12/31/14 12/31/ n = 5 years; i =??% per year Calculation of Exercise #9 using the PVOA Table The interest rate for the ordinary annuity described above can be computed with the following equation: PVOA = PMT times [ PVOA factor for n = 5 years; i =?? per year ] \$2, = \$700 times [ PVOA factor for n = 5 years; i =?? per year ] \$2, / \$700 = [ PVOA factor for n = 5 years; i =?? per year ] = [ PVOA factor for n = 5 years; i =?? per year ] = PVOA factor for n = 5 years; i = 12% per year Let's review this calculation. We insert into the equation the components that we know: the present value, payment amount, and the number of periods. In line four, we calculate our factor to be We now know both the PVOA factor (3.605) and the number of years (n = 5). We go to the PVOA Table and look across the n = 5 row until we come to the factor Tracking up the column, we see that the factor is in the column with the heading of 12%. Since the periods in question are annual periods, the answer of i = 12% means the investment has to earn 12% per year. *12% times the previous account balance Page 2 of 15
3 Exercise #10. Matt is moving to Texas and needs to borrow \$5,616 on January 1, His budget will allow him to make quarterly payments of \$800 on the first day of January, April, July, and October. Matt's loan includes 8 quarterly payments; the first payment is due on April 1, What is the rate (compounded quarterly) that Matt will be paying (and the lender will be receiving) under this arrangement? Before calculating the interest rate, we organize the information on a timeline: PVOA= \$5,616 \$800 \$800 \$800 \$800 \$800 \$ months 3 months 3 months 3 months 3 months 1/1/11 4/1/11 7/1/11 10/1/11 1/1/12 10/1/12 1/1/ n = 8 three-month periods; i =?? per quarter Calculation of Exercise #10 using the PVOA Table The number of periods/payments in the ordinary annuity described above can be computed with the following PVOA equation: PVOA = PMT times [ PVOA factor for n = 8 quarters; i =?? per quarter ] \$5,616 = \$800 times [ PVOA factor for n = 8 quarters; i =?? per quarter ] \$5,616 / \$800 = [ PVOA factor for n = 8 quarters; i =?? per quarter ] 7.02 = [ PVOA factor for n = 8 quarters; i =?? per quarter ] 7.02 = PVOA factor for n = 8 quarters; i = 3% per quarter Let's review this calculation. We insert into the equation the components that we know: the present value, the recurring payment amount, and the number of periods. In line four, we calculate our factor to be We now know both the PVOA factor (7.02) and the number of periods (n = 8). We go to the PVOA Table and look across the n = 8 row until we come to the factor Tracking up the column, we see that we are in the 3% column. Since the periods in question are quarterly periods, the answer of i = 3% means the loan has an annual rate of 12% per year (3% times 4 quarters per year). Page 3 of 15
4 Effective Annual Rate of Interest (Effective Yield) To convert a nominal rate to an equivalent effective rate: = + i = Nominal or stated interest rate m = Number of compounding periods per year Example: What effective rate will a stated annual rate of 6% yield when compounded semi-annually? = +. =. Continuous compounding Where e = = Example: Find the value at the end of 2 years of Fred Moreno s R100 deposit in an account paying 8% annual interest compounded continuously. =.. =. Page 4 of 15
5 Present value of a Perpetuity = Example: Ross Clark wishes to endow a chair in finance at his alma mater. The university indicated that it requires R per year to support the chair, and the endowment would earn 10% per year. Determine the amount Ross must give the university to fund the chair. =. Mixed Stream = Use ordinary present value or future value formulas for each stream and then sum them up. Page 5 of 15
6 NUMBER OF PERIODS You can determine the number of periods required for an initial investment to grow to a specified amount with this formula: = + PV = present value, the amount you invested FV = future value, the amount your investment will grow to i = interest per period Example: You put \$10,000 into a savings account at a 9.05% annual interest rate compounded annually. How long will it take to double your investment? =,.,.. =. n = 8 years Annuity Payments Calculate Payments When Present Value Is Known The Present Value is an amount that you have now, such as the price of property that you have just purchased or the value of equipment that you have leased. When you know the present value, interest rate, and number of periods of an ordinary annuity, you can solve for the payment with this formula: =, Page 6 of 15
7 PVA = Present Value of an ordinary annuity (payments are made at the end of each period) i = interest per period n = number of periods Example: You can get a \$150,000 home mortgage at 7% annual interest rate for 30 years. Payments are due at the end of each month and interest is compounded monthly. How much will your payments be? PVA = 150,000, the loan amount i = interest per month (.07 / 12) n = 360 periods (12 payments per year for 30 years) payment = Calculate Payments When Future Value Is Known The Future Value is an amount that you wish to have after a number of periods have passed. For example, you may need to accumulate \$20,000 in ten years to pay for college tuition. When you know the future value, interest rate, and number of periods of an ordinary annuity, you can solve for the payment with this formula: =, FVA = Future Value of an ordinary annuity (payments are made at the end of each period) i = interest per period n = number of periods Example: In 10 years, you will need \$50,000 to pay for college tuition. Your savings account pays 5% interest compounded monthly. How much should you save each month to reach your goal? FVA = 50,000, the future savings goal i = interest per month (.05 / 12) n = 120 periods (12 payments per year for 10 years) payment = Page 7 of 15
8 PRESENT VALUE Present Value Of A Single Amount The relationship between the present value and future value can be expressed as: = + Table use =, PV = Present Value FV = Future Value i = Interest Rate Per Period n = Number of Compounding Periods m = Compounding periods per year Example: You want to buy a house 5 years from now for \$150,000. Assuming a 6% interest rate compounded annually, how much should you invest today to yield \$150,000 in 5 years? FV = 150,000 i =.06 n = 5 m = 1 =, +. =,. Page 8 of 15
9 Present Value of a Mixed Stream Finding the present value of a mixed stream of cash flows is similar to finding the future value of a mixed stream. We determine the present value of each future amount and then add all the individual present values together to find the total present value. Present Value of Annuities The Present Value of an Ordinary Annuity could be solved by calculating the present value of each payment in the series using the present value formula and then summing the results. A more direct formula is: Page 9 of 15
10 =, PVA = Present Value of an Ordinary Annuity PMT = Amount of each payment i = Discount Rate Per Period n = Number of Periods Example 1: What amount must you invest today at 6% compounded annually so that you can withdraw \$5,000 at the end of each year for the next 5 years? PMT = 5,000 i =.06 n = 5 Use tables =,, PVA = 21, Application: Loan amortization =, Example: You want to determine the equal annual end-of-year payments necessary to amortise fully a R6000, 10% interest loan over 4 years. =. Page 10 of 15
11 Present Value of an Annuity Due The Present Value of an Annuity Due is identical to an ordinary annuity except that each payment occurs at the beginning of a period rather than at the end. Since each payment occurs one period earlier, we can calculate the present value of an ordinary annuity and then multiply the result by (1 + i). PMT = Payments i = Discount Rate Per Period =, + Example: What amount must you invest today a 6% interest rate compounded annually so that you can withdraw \$5,000 at the beginning of each year for the next 5 years? PMT = 5,000 i =.06 n = 5 =,. Page 11 of 15
12 FUTURE VALUE Future Value Of A Single Amount Future Value is the amount of money that an investment made today (the present value) will grow to by some future date. Since money has time value, we naturally expect the future value to be greater than the present value. The difference between the two depends on the number of compounding periods involved and the going interest rate. The relationship between the future value and present value can be expressed as: Formula: = + Using tables =, FV = Future Value PV = Present Value i = Interest Rate Per Period n = Number of Compounding Periods m = Compounding periods per year Example: You can afford to put \$10,000 in a savings account today that pays 6% interest compounded annually. How much will you have 5 years from now if you make no withdrawals? PV = 10,000 i =.06 n = 5 m = 1 FV = 13, Page 12 of 15
13 Example 2: Another financial institution offers to pay 6% compounded semiannually. How much will your \$10,000 grow to in five years at this rate? Interest is compounded twice per year so you must divide the annual interest rate by two to obtain a rate per period of 3%. Since there are two compounding periods per year, you must multiply the number of years by two to obtain the total number of periods. PV = 10,000 i =.06 / 2 =.03 n = 5 * 2 = 10 FV = 13, Future Value of Annuities The Future Value of an Ordinary Annuity could be solved by calculating the future value of each individual payment in the series using the future value formula and then summing the results. A more direct formula is: =, FVA = Future Value of an Ordinary Annuity PMT = Amount of each payment i = Interest Rate Per Period n = Number of Periods Example: What amount will accumulate if we deposit \$5,000 at the end of each year for the next 5 years? Assume an interest of 6% compounded annually. PV = 5,000 i =.06 n = 5 =,. Page 13 of 15
14 Application: Deposits needed to accumulate a future sum =, Example: You want to determine the equal annual end-of-year deposits required to accumulate R30000 at the end of 5 years, given an interest rate of 6%. =. Future Value of an Annuity Due The Future Value of an Annuity Due is identical to an ordinary annuity except that each payment occurs at the beginning of a period rather than at the end. Since each payment occurs one period earlier, we can calculate the present value of an ordinary annuity and then multiply the result by (1 + i)., =, + Example: What amount will accumulate if we deposit \$5,000 at the beginning of each year for the next 5 years? Assume an interest of 6% compounded annually. PV = 5,000 i =.06 n = 5, =,. Future Value of a Mixed Stream Determining the future value of a mixed stream of cash flows is straightforward. We determine the future value of each cash flow at the specified future date and then add all the individual future values to find the total future value. Page 14 of 15
15 Page 15 of 15
### Chapter The Time Value of Money
Chapter The Time Value of Money PPT 9-2 Chapter 9 - Outline Time Value of Money Future Value and Present Value Annuities Time-Value-of-Money Formulas Adjusting for Non-Annual Compounding Compound Interest
More information
### Chapter 4. Time Value of Money. Copyright 2009 Pearson Prentice Hall. All rights reserved.
Chapter 4 Time Value of Money Learning Goals 1. Discuss the role of time value in finance, the use of computational aids, and the basic patterns of cash flow. 2. Understand the concept of future value
More information
### Chapter 4. Time Value of Money. Learning Goals. Learning Goals (cont.)
Chapter 4 Time Value of Money Learning Goals 1. Discuss the role of time value in finance, the use of computational aids, and the basic patterns of cash flow. 2. Understand the concept of future value
More information
### Future Value. Basic TVM Concepts. Chapter 2 Time Value of Money. \$500 cash flow. On a time line for 3 years: \$100. FV 15%, 10 yr.
Chapter Time Value of Money Future Value Present Value Annuities Effective Annual Rate Uneven Cash Flows Growing Annuities Loan Amortization Summary and Conclusions Basic TVM Concepts Interest rate: abbreviated
More information
### DISCOUNTED CASH FLOW VALUATION and MULTIPLE CASH FLOWS
Chapter 5 DISCOUNTED CASH FLOW VALUATION and MULTIPLE CASH FLOWS The basic PV and FV techniques can be extended to handle any number of cash flows. PV with multiple cash flows: Suppose you need \$500 one
More information
### 1.3.2015 г. D. Dimov. Year Cash flow 1 \$3,000 2 \$5,000 3 \$4,000 4 \$3,000 5 \$2,000
D. Dimov Most financial decisions involve costs and benefits that are spread out over time Time value of money allows comparison of cash flows from different periods Question: You have to choose one of
More information
### CHAPTER 2. Time Value of Money 2-1
CHAPTER 2 Time Value of Money 2-1 Time Value of Money (TVM) Time Lines Future value & Present value Rates of return Annuities & Perpetuities Uneven cash Flow Streams Amortization 2-2 Time lines 0 1 2 3
More information
### Chapter 5 Time Value of Money 2: Analyzing Annuity Cash Flows
1. Future Value of Multiple Cash Flows 2. Future Value of an Annuity 3. Present Value of an Annuity 4. Perpetuities 5. Other Compounding Periods 6. Effective Annual Rates (EAR) 7. Amortized Loans Chapter
More information
### Time Value of Money. 15.511 Corporate Accounting Summer 2004. Professor S. P. Kothari Sloan School of Management Massachusetts Institute of Technology
Time Value of Money 15.511 Corporate Accounting Summer 2004 Professor S. P. Kothari Sloan School of Management Massachusetts Institute of Technology July 2, 2004 1 LIABILITIES: Current Liabilities Obligations
More information
### Chapter 6 Contents. Principles Used in Chapter 6 Principle 1: Money Has a Time Value.
Chapter 6 The Time Value of Money: Annuities and Other Topics Chapter 6 Contents Learning Objectives 1. Distinguish between an ordinary annuity and an annuity due, and calculate present and future values
More information
### Week 4. Chonga Zangpo, DFB
Week 4 Time Value of Money Chonga Zangpo, DFB What is time value of money? It is based on the belief that people have a positive time preference for consumption. It reflects the notion that people prefer
More information
### Important Financial Concepts
Part 2 Important Financial Concepts Chapter 4 Time Value of Money Chapter 5 Risk and Return Chapter 6 Interest Rates and Bond Valuation Chapter 7 Stock Valuation 130 LG1 LG2 LG3 LG4 LG5 LG6 Chapter 4 Time
More information
### 2. How would (a) a decrease in the interest rate or (b) an increase in the holding period of a deposit affect its future value? Why?
CHAPTER 3 CONCEPT REVIEW QUESTIONS 1. Will a deposit made into an account paying compound interest (assuming compounding occurs once per year) yield a higher future value after one period than an equal-sized
More information
### Chapter 6. Discounted Cash Flow Valuation. Key Concepts and Skills. Multiple Cash Flows Future Value Example 6.1. Answer 6.1
Chapter 6 Key Concepts and Skills Be able to compute: the future value of multiple cash flows the present value of multiple cash flows the future and present value of annuities Discounted Cash Flow Valuation
More information
### Integrated Case. 5-42 First National Bank Time Value of Money Analysis
Integrated Case 5-42 First National Bank Time Value of Money Analysis You have applied for a job with a local bank. As part of its evaluation process, you must take an examination on time value of money
More information
### The Time Value of Money
The Time Value of Money Time Value Terminology 0 1 2 3 4 PV FV Future value (FV) is the amount an investment is worth after one or more periods. Present value (PV) is the current value of one or more future
More information
### Problem Set: Annuities and Perpetuities (Solutions Below)
Problem Set: Annuities and Perpetuities (Solutions Below) 1. If you plan to save \$300 annually for 10 years and the discount rate is 15%, what is the future value? 2. If you want to buy a boat in 6 years
More information
### Discounted Cash Flow Valuation
6 Formulas Discounted Cash Flow Valuation McGraw-Hill/Irwin Copyright 2008 by The McGraw-Hill Companies, Inc. All rights reserved. Chapter Outline Future and Present Values of Multiple Cash Flows Valuing
More information
### Chapter 4: Time Value of Money
FIN 301 Homework Solution Ch4 Chapter 4: Time Value of Money 1. a. 10,000/(1.10) 10 = 3,855.43 b. 10,000/(1.10) 20 = 1,486.44 c. 10,000/(1.05) 10 = 6,139.13 d. 10,000/(1.05) 20 = 3,768.89 2. a. \$100 (1.10)
More information
### Chapter 4. The Time Value of Money
Chapter 4 The Time Value of Money 1 Learning Outcomes Chapter 4 Identify various types of cash flow patterns Compute the future value and the present value of different cash flow streams Compute the return
More information
### Chapter 6. Time Value of Money Concepts. Simple Interest 6-1. Interest amount = P i n. Assume you invest \$1,000 at 6% simple interest for 3 years.
6-1 Chapter 6 Time Value of Money Concepts 6-2 Time Value of Money Interest is the rent paid for the use of money over time. That s right! A dollar today is more valuable than a dollar to be received in
More information
### FIN 3000. Chapter 6. Annuities. Liuren Wu
FIN 3000 Chapter 6 Annuities Liuren Wu Overview 1. Annuities 2. Perpetuities 3. Complex Cash Flow Streams Learning objectives 1. Distinguish between an ordinary annuity and an annuity due, and calculate
More information
### Chapter 6. Learning Objectives Principles Used in This Chapter 1. Annuities 2. Perpetuities 3. Complex Cash Flow Streams
Chapter 6 Learning Objectives Principles Used in This Chapter 1. Annuities 2. Perpetuities 3. Complex Cash Flow Streams 1. Distinguish between an ordinary annuity and an annuity due, and calculate present
More information
### Solutions to Problems: Chapter 5
Solutions to Problems: Chapter 5 P5-1. Using a time line LG 1; Basic a, b, and c d. Financial managers rely more on present value than future value because they typically make decisions before the start
More information
### How to calculate present values
How to calculate present values Back to the future Chapter 3 Discounted Cash Flow Analysis (Time Value of Money) Discounted Cash Flow (DCF) analysis is the foundation of valuation in corporate finance
More information
### Chapter 4. Time Value of Money
Chapter 4 Time Value of Money Learning Goals 1. Discuss the role of time value in finance, the use of computational aids, and the basic patterns of cash flow. 2. Understand the concept of future value
More information
### 1. If you wish to accumulate \$140,000 in 13 years, how much must you deposit today in an account that pays an annual interest rate of 14%?
Chapter 2 - Sample Problems 1. If you wish to accumulate \$140,000 in 13 years, how much must you deposit today in an account that pays an annual interest rate of 14%? 2. What will \$247,000 grow to be in
More information
### THE VALUE OF MONEY PROBLEM #3: ANNUITY. Professor Peter Harris Mathematics by Dr. Sharon Petrushka. Introduction
THE VALUE OF MONEY PROBLEM #3: ANNUITY Professor Peter Harris Mathematics by Dr. Sharon Petrushka Introduction Earlier, we explained how to calculate the future value of a single sum placed on deposit
More information
### Topics. Chapter 5. Future Value. Future Value - Compounding. Time Value of Money. 0 r = 5% 1
Chapter 5 Time Value of Money Topics 1. Future Value of a Lump Sum 2. Present Value of a Lump Sum 3. Future Value of Cash Flow Streams 4. Present Value of Cash Flow Streams 5. Perpetuities 6. Uneven Series
More information
### The Time Value of Money C H A P T E R N I N E
The Time Value of Money C H A P T E R N I N E Figure 9-1 Relationship of present value and future value PPT 9-1 \$1,000 present value \$ 10% interest \$1,464.10 future value 0 1 2 3 4 Number of periods Figure
More information
### TIME VALUE OF MONEY PROBLEM #4: PRESENT VALUE OF AN ANNUITY
TIME VALUE OF MONEY PROBLEM #4: PRESENT VALUE OF AN ANNUITY Professor Peter Harris Mathematics by Dr. Sharon Petrushka Introduction In this assignment we will discuss how to calculate the Present Value
More information
### Appendix C- 1. Time Value of Money. Appendix C- 2. Financial Accounting, Fifth Edition
C- 1 Time Value of Money C- 2 Financial Accounting, Fifth Edition Study Objectives 1. Distinguish between simple and compound interest. 2. Solve for future value of a single amount. 3. Solve for future
More information
### CHAPTER 5 INTRODUCTION TO VALUATION: THE TIME VALUE OF MONEY
CHAPTER 5 INTRODUCTION TO VALUATION: THE TIME VALUE OF MONEY 1. The simple interest per year is: \$5,000.08 = \$400 So after 10 years you will have: \$400 10 = \$4,000 in interest. The total balance will be
More information
### FIN 5413: Chapter 03 - Mortgage Loan Foundations: The Time Value of Money Page 1
FIN 5413: Chapter 03 - Mortgage Loan Foundations: The Time Value of Money Page 1 Solutions to Problems - Chapter 3 Mortgage Loan Foundations: The Time Value of Money Problem 3-1 a) Future Value = FV(n,i,PV,PMT)
More information
### Finding the Payment \$20,000 = C[1 1 / 1.0066667 48 ] /.0066667 C = \$488.26
Quick Quiz: Part 2 You know the payment amount for a loan and you want to know how much was borrowed. Do you compute a present value or a future value? You want to receive \$5,000 per month in retirement.
More information
### Regular Annuities: Determining Present Value
8.6 Regular Annuities: Determining Present Value GOAL Find the present value when payments or deposits are made at regular intervals. LEARN ABOUT the Math Harry has money in an account that pays 9%/a compounded
More information
### 2 The Mathematics. of Finance. Copyright Cengage Learning. All rights reserved.
2 The Mathematics of Finance Copyright Cengage Learning. All rights reserved. 2.3 Annuities, Loans, and Bonds Copyright Cengage Learning. All rights reserved. Annuities, Loans, and Bonds A typical defined-contribution
More information
### 5. Time value of money
1 Simple interest 2 5. Time value of money With simple interest, the amount earned each period is always the same: i = rp o We will review some tools for discounting cash flows. where i = interest earned
More information
### Key Concepts and Skills. Multiple Cash Flows Future Value Example 6.1. Chapter Outline. Multiple Cash Flows Example 2 Continued
6 Calculators Discounted Cash Flow Valuation Key Concepts and Skills Be able to compute the future value of multiple cash flows Be able to compute the present value of multiple cash flows Be able to compute
More information
### Time Value of Money, Part 5 Present Value aueof An Annuity. Learning Outcomes. Present Value
Time Value of Money, Part 5 Present Value aueof An Annuity Intermediate Accounting II Dr. Chula King 1 Learning Outcomes The concept of present value Present value of an annuity Ordinary annuity versus
More information
### Discounted Cash Flow Valuation
Discounted Cash Flow Valuation Chapter 5 Key Concepts and Skills Be able to compute the future value of multiple cash flows Be able to compute the present value of multiple cash flows Be able to compute
More information
### first complete "prior knowlegde" -- to refresh knowledge of Simple and Compound Interest.
ORDINARY SIMPLE ANNUITIES first complete "prior knowlegde" -- to refresh knowledge of Simple and Compound Interest. LESSON OBJECTIVES: students will learn how to determine the Accumulated Value of Regular
More information
### Chapter 4 Time Value of Money ANSWERS TO END-OF-CHAPTER QUESTIONS
Chapter 4 Time Value of Money ANSWERS TO END-OF-CHAPTER QUESTIONS 4-1 a. PV (present value) is the value today of a future payment, or stream of payments, discounted at the appropriate rate of interest.
More information
### Chapter 2 Applying Time Value Concepts
Chapter 2 Applying Time Value Concepts Chapter Overview Albert Einstein, the renowned physicist whose theories of relativity formed the theoretical base for the utilization of atomic energy, called the
More information
### Ch. Ch. 5 Discounted Cash Flows & Valuation In Chapter 5,
Ch. 5 Discounted Cash Flows & Valuation In Chapter 5, we found the PV & FV of single cash flows--either payments or receipts. In this chapter, we will do the same for multiple cash flows. 2 Multiple Cash
More information
### TVM Applications Chapter
Chapter 6 Time of Money UPS, Walgreens, Costco, American Air, Dreamworks Intel (note 10 page 28) TVM Applications Accounting issue Chapter Notes receivable (long-term receivables) 7 Long-term assets 10
More information
### Main TVM functions of a BAII Plus Financial Calculator
Main TVM functions of a BAII Plus Financial Calculator The BAII Plus calculator can be used to perform calculations for problems involving compound interest and different types of annuities. (Note: there
More information
### Exercise 6 8. Exercise 6 12 PVA = \$5,000 x 4.35526* = \$21,776
CHAPTER 6: EXERCISES Exercise 6 2 1. FV = \$10,000 (2.65330 * ) = \$26,533 * Future value of \$1: n = 20, i = 5% (from Table 1) 2. FV = \$10,000 (1.80611 * ) = \$18,061 * Future value of \$1: n = 20, i = 3%
More information
### Chapter 7 SOLUTIONS TO END-OF-CHAPTER PROBLEMS
Chapter 7 SOLUTIONS TO END-OF-CHAPTER PROBLEMS 7-1 0 1 2 3 4 5 10% PV 10,000 FV 5? FV 5 \$10,000(1.10) 5 \$10,000(FVIF 10%, 5 ) \$10,000(1.6105) \$16,105. Alternatively, with a financial calculator enter the
More information
### Appendix. Time Value of Money. Financial Accounting, IFRS Edition Weygandt Kimmel Kieso. Appendix C- 1
C Time Value of Money C- 1 Financial Accounting, IFRS Edition Weygandt Kimmel Kieso C- 2 Study Objectives 1. Distinguish between simple and compound interest. 2. Solve for future value of a single amount.
More information
### Real estate investment & Appraisal Dr. Ahmed Y. Dashti. Sample Exam Questions
Real estate investment & Appraisal Dr. Ahmed Y. Dashti Sample Exam Questions Problem 3-1 a) Future Value = \$12,000 (FVIF, 9%, 7 years) = \$12,000 (1.82804) = \$21,936 (annual compounding) b) Future Value
More information
### International Financial Strategies Time Value of Money
International Financial Strategies 1 Future Value and Compounding Future value = cash value of the investment at some point in the future Investing for single period: FV. Future Value PV. Present Value
More information
### Time Value of Money Concepts
BASIC ANNUITIES There are many accounting transactions that require the payment of a specific amount each period. A payment for a auto loan or a mortgage payment are examples of this type of transaction.
More information
### PRESENT VALUE ANALYSIS. Time value of money equal dollar amounts have different values at different points in time.
PRESENT VALUE ANALYSIS Time value of money equal dollar amounts have different values at different points in time. Present value analysis tool to convert CFs at different points in time to comparable values
More information
### Principles of Managerial Finance INTEREST RATE FACTOR SUPPLEMENT
Principles of Managerial Finance INTEREST RATE FACTOR SUPPLEMENT 1 5 Time Value of Money FINANCIAL TABLES Financial tables include various future and present value interest factors that simplify time value
More information
### Time-Value-of-Money and Amortization Worksheets
2 Time-Value-of-Money and Amortization Worksheets The Time-Value-of-Money and Amortization worksheets are useful in applications where the cash flows are equal, evenly spaced, and either all inflows or
More information
### Chapter F: Finance. Section F.1-F.4
Chapter F: Finance Section F.1-F.4 F.1 Simple Interest Suppose a sum of money P, called the principal or present value, is invested for t years at an annual simple interest rate of r, where r is given
More information
### 3. Time value of money. We will review some tools for discounting cash flows.
1 3. Time value of money We will review some tools for discounting cash flows. Simple interest 2 With simple interest, the amount earned each period is always the same: i = rp o where i = interest earned
More information
### Discounted Cash Flow Valuation
BUAD 100x Foundations of Finance Discounted Cash Flow Valuation September 28, 2009 Review Introduction to corporate finance What is corporate finance? What is a corporation? What decision do managers make?
More information
### PowerPoint. to accompany. Chapter 5. Interest Rates
PowerPoint to accompany Chapter 5 Interest Rates 5.1 Interest Rate Quotes and Adjustments To understand interest rates, it s important to think of interest rates as a price the price of using money. When
More information
### TIME VALUE OF MONEY PROBLEM #7: MORTGAGE AMORTIZATION
TIME VALUE OF MONEY PROBLEM #7: MORTGAGE AMORTIZATION Professor Peter Harris Mathematics by Sharon Petrushka Introduction This problem will focus on calculating mortgage payments. Knowledge of Time Value
More information
### E INV 1 AM 11 Name: INTEREST. There are two types of Interest : and. The formula is. I is. P is. r is. t is
E INV 1 AM 11 Name: INTEREST There are two types of Interest : and. SIMPLE INTEREST The formula is I is P is r is t is NOTE: For 8% use r =, for 12% use r =, for 2.5% use r = NOTE: For 6 months use t =
More information
### The values in the TVM Solver are quantities involved in compound interest and annuities.
Texas Instruments Graphing Calculators have a built in app that may be used to compute quantities involved in compound interest, annuities, and amortization. For the examples below, we ll utilize the screens
More information
### CHAPTER 9 Time Value Analysis
Copyright 2008 by the Foundation of the American College of Healthcare Executives 6/11/07 Version 9-1 CHAPTER 9 Time Value Analysis Future and present values Lump sums Annuities Uneven cash flow streams
More information
### NPV calculation. Academic Resource Center
NPV calculation Academic Resource Center 1 NPV calculation PV calculation a. Constant Annuity b. Growth Annuity c. Constant Perpetuity d. Growth Perpetuity NPV calculation a. Cash flow happens at year
More information
### CHAPTER 6 DISCOUNTED CASH FLOW VALUATION
CHAPTER 6 DISCOUNTED CASH FLOW VALUATION Answers to Concepts Review and Critical Thinking Questions 1. The four pieces are the present value (PV), the periodic cash flow (C), the discount rate (r), and
More information
### FinQuiz Notes 2 0 1 4
Reading 5 The Time Value of Money Money has a time value because a unit of money received today is worth more than a unit of money to be received tomorrow. Interest rates can be interpreted in three ways.
More information
### In Section 5.3, we ll modify the worksheet shown above. This will allow us to use Excel to calculate the different amounts in the annuity formula,
Excel has several built in functions for working with compound interest and annuities. To use these functions, we ll start with a standard Excel worksheet. This worksheet contains the variables used throughout
More information
### CHAPTER 4. The Time Value of Money. Chapter Synopsis
CHAPTER 4 The Time Value of Money Chapter Synopsis Many financial problems require the valuation of cash flows occurring at different times. However, money received in the future is worth less than money
More information
### Mathematics. Rosella Castellano. Rome, University of Tor Vergata
and Loans Mathematics Rome, University of Tor Vergata and Loans Future Value for Simple Interest Present Value for Simple Interest You deposit E. 1,000, called the principal or present value, into a savings
More information
### Applying Time Value Concepts
Applying Time Value Concepts C H A P T E R 3 based on the value of two packs of cigarettes per day and a modest rate of return? Let s assume that Lou will save an amount equivalent to the cost of two packs
More information
### The Time Value of Money
The following is a review of the Quantitative Methods: Basic Concepts principles designed to address the learning outcome statements set forth by CFA Institute. This topic is also covered in: The Time
More information
### How To Use Excel To Compute Compound Interest
Excel has several built in functions for working with compound interest and annuities. To use these functions, we ll start with a standard Excel worksheet. This worksheet contains the variables used throughout
More information
### USING THE SHARP EL 738 FINANCIAL CALCULATOR
USING THE SHARP EL 738 FINANCIAL CALCULATOR Basic financial examples with financial calculator steps Prepared by Colin C Smith 2010 Some important things to consider 1. These notes cover basic financial
More information
### CHAPTER 6 Accounting and the Time Value of Money
CHAPTER 6 Accounting and the Time Value of Money 6-1 LECTURE OUTLINE This chapter can be covered in two to three class sessions. Most students have had previous exposure to single sum problems and ordinary
More information
### Chapter 4. The Time Value of Money
Chapter 4 The Time Value of Money 4-2 Topics Covered Future Values and Compound Interest Present Values Multiple Cash Flows Perpetuities and Annuities Inflation and Time Value Effective Annual Interest
More information
### ANNUITIES. Ordinary Simple Annuities
An annuity is a series of payments or withdrawals. ANNUITIES An Annuity can be either Simple or General Simple Annuities - Compounding periods and payment periods coincide. General Annuities - Compounding
More information
### MAT116 Project 2 Chapters 8 & 9
MAT116 Project 2 Chapters 8 & 9 1 8-1: The Project In Project 1 we made a loan workout decision based only on data from three banks that had merged into one. We did not consider issues like: What was the
More information
### CHAPTER 5 INTRODUCTION TO VALUATION: THE TIME VALUE OF MONEY
CHAPTER 5 INTRODUCTION TO VALUATION: THE TIME VALUE OF MONEY Answers to Concepts Review and Critical Thinking Questions 1. The four parts are the present value (PV), the future value (FV), the discount
More information
### FINA 351 Managerial Finance, Ch.4-5, Time-Value-of-Money (TVM), Notes
FINA 351 Managerial Finance, Ch.4-5, Time-Value-of-Money (TVM), Notes The concept of time-value-of-money is important to know, not only for this class, but for your own financial planning. It is a critical
More information
### TIME VALUE OF MONEY. In following we will introduce one of the most important and powerful concepts you will learn in your study of finance;
In following we will introduce one of the most important and powerful concepts you will learn in your study of finance; the time value of money. It is generally acknowledged that money has a time value.
More information
### Time Value of Money Problems
Time Value of Money Problems 1. What will a deposit of \$4,500 at 10% compounded semiannually be worth if left in the bank for six years? a. \$8,020.22 b. \$7,959.55 c. \$8,081.55 d. \$8,181.55 2. What will
More information
### hp calculators HP 20b Time value of money basics The time value of money The time value of money application Special settings
The time value of money The time value of money application Special settings Clearing the time value of money registers Begin / End mode Periods per year Cash flow diagrams and sign conventions Practice
More information
### Solutions to Problems
Solutions to Problems P4-1. LG 1: Using a time line Basic a. b. and c. d. Financial managers rely more on present value than future value because they typically make decisions before the start of a project,
More information
### Chapter 4: Time Value of Money
Chapter 4: Time Value of Money BASIC KEYS USED IN FINANCE PROBLEMS The following two key sequences should be done before starting any "new" problem: ~ is used to separate key strokes (3~N: enter 3 then
More information
### Chapter 5 Discounted Cash Flow Valuation
Chapter Discounted Cash Flow Valuation Compounding Periods Other Than Annual Let s examine monthly compounding problems. Future Value Suppose you invest \$9,000 today and get an interest rate of 9 percent
More information
### Chapter 3 Mathematics of Finance
Chapter 3 Mathematics of Finance Section 3 Future Value of an Annuity; Sinking Funds Learning Objectives for Section 3.3 Future Value of an Annuity; Sinking Funds The student will be able to compute the
More information
### Present Value Concepts
Present Value Concepts Present value concepts are widely used by accountants in the preparation of financial statements. In fact, under International Financial Reporting Standards (IFRS), these concepts
More information
### The Interest Rate: A loan, expressed as a percentage of the amount loaned per year.
Interest Rates Time Value of Money The Interest Rate Simple Interest Amortizing a Loan The Interest Rate: A loan, expressed as a percentage of the amount loaned per year. Interest rate is the "price" of
More information
### Dick Schwanke Finite Math 111 Harford Community College Fall 2013
Annuities and Amortization Finite Mathematics 111 Dick Schwanke Session #3 1 In the Previous Two Sessions Calculating Simple Interest Finding the Amount Owed Computing Discounted Loans Quick Review of
More information
### FinQuiz Notes 2 0 1 5
Reading 5 The Time Value of Money Money has a time value because a unit of money received today is worth more than a unit of money to be received tomorrow. Interest rates can be interpreted in three ways.
More information
### How To Calculate A Pension
Interests on Transactions Chapter 10 13 PV & FV of Annuities PV & FV of Annuities An annuity is a series of equal regular payment amounts made for a fixed number of periods 2 Problem An engineer deposits
More information
### Topics Covered. Ch. 4 - The Time Value of Money. The Time Value of Money Compounding and Discounting Single Sums
Ch. 4 - The Time Value of Money Topics Covered Future Values Present Values Multiple Cash Flows Perpetuities and Annuities Effective Annual Interest Rate For now, we will omit the section 4.5 on inflation
More information
### Finance 331 Corporate Financial Management Week 1 Week 3 Note: For formulas, a Texas Instruments BAII Plus calculator was used.
Chapter 1 Finance 331 What is finance? - Finance has to do with decisions about money and/or cash flows. These decisions have to do with money being raised or used. General parts of finance include: -
More information
### Present Value and Annuities. Chapter 3 Cont d
Present Value and Annuities Chapter 3 Cont d Present Value Helps us answer the question: What s the value in today s dollars of a sum of money to be received in the future? It lets us strip away the effects
More information
### rate nper pmt pv Interest Number of Payment Present Future Rate Periods Amount Value Value 12.00% 1 0 \$100.00 \$112.00
In Excel language, if the initial cash flow is an inflow (positive), then the future value must be an outflow (negative). Therefore you must add a negative sign before the FV (and PV) function. The inputs
More information
### Module 5: Interest concepts of future and present value
file:///f /Courses/2010-11/CGA/FA2/06course/m05intro.htm Module 5: Interest concepts of future and present value Overview In this module, you learn about the fundamental concepts of interest and present
More information
### Texas Instruments BAII Plus Tutorial for Use with Fundamentals 11/e and Concise 5/e
Texas Instruments BAII Plus Tutorial for Use with Fundamentals 11/e and Concise 5/e This tutorial was developed for use with Brigham and Houston s Fundamentals of Financial Management, 11/e and Concise,
More information
### The explanations below will make it easier for you to use the calculator. The ON/OFF key is used to turn the calculator on and off.
USER GUIDE Texas Instrument BA II Plus Calculator April 2007 GENERAL INFORMATION The Texas Instrument BA II Plus financial calculator was designed to support the many possible applications in the areas
More information |
# Reflection across the x-axis: y = -f(x)
Now Playing:Reflection across the x axis– Example 1
Intros
1. An Experiment to Study "Reflection Across the X-axis"
Sketch and compare: $y = {\left( {x - 4} \right)^3}$
VS.
$- y = {\left( {x - 4} \right)^3}$
2. Sketch both quadratic functions on the same set of coordinate axes.
3. Compared to the graph of $y = {\left( {x - 4} \right)^3}$:
• the graph of $- y = {\left( {x - 4} \right)^3}$ is a reflection in the ___________________.
Examples
1. Reflection Across the X-axis
Given the graph of $y = f\left( x \right)$ as shown, sketch:
1. $y = - f\left( x \right)$
2. In conclusion:
$\left( y \right) \to \left( { - y} \right)$: reflection in the ____________________ $\Rightarrow$ all $y$ coordinates ______________________________.
Relationship between two variables
Notes
The concept behind the reflections about the x-axis is basically the same as the reflections about the y-axis. The only difference is that, rather than the y-axis, the points are reflected from above the x-axis to below the x-axis, and vice versa.
## Reflection Across the X-Axis
Before we get into reflections across the y-axis, make sure you've refreshed your memory on how to do simple vertical and horizontal translations.
## How to Reflect Over X-Axis:
One of the most basic transformations you can make with simple functions is to reflect it across the x-axis or another horizontal axis. In a potential test question, this can be phrased in many different ways, so make sure you recognize the following terms as just another way of saying "perform a reflection across the x-axis":
1) Graph $y = -f(x)$
2) Graph $-f(x)$
3) Reflect over $x$ axis
In order to do this, the process is extremely simple: For any function, no matter how complicated it is, simply pick out easy-to-determine coordinates, divide the y-coordinate by (-1), and then re-plot those coordinates. That's it!
The best way to practice drawing reflections across the y-axis is to do an example problem:
Example:
Given the graph of $y = f(x)$ as shown, sketch $y = -f(x)$.
Remember, the only step we have to do before plotting the $-f(x)$ reflection is simply divide the y-coordinates of easy-to-determine points on our graph above by (-1). When we say "easy-to-determine points" what this refers to is just points for which you know the x and y values exactly. Don't pick points where you need to estimate values, as this makes the problem unnecessarily hard. Below are several images to help you visualize how to solve this problem.
Step 1: Know that we're reflecting across the x-axis
Since we were asked to plot the –$f(x)$ reflection, is it very important that you recognize this means we are being asked to plot the reflection over the x-axis. When drawing reflections across the $x$ and $y$ axis, it is very easy to get confused by some of the notations. So, make sure you take a moment before solving any reflection problem to confirm you know what you're being asked to do.
Step 2: Identify easy-to-determine points
Remember, pick some points (3 is usually enough) that are easy to pick out, meaning you know exactly what the x and y values are. In this case, let's pick (-2 ,-3), (-1 ,0), and (0,3).
Step 3: Divide these points by (-1)
While the $x$ values remain the same, all we need to do is divide the $y$ values by (-1)!
Step 4: Plot the new points
And that's it! Simple, right?
## What is the Axis of Symmetry:
In some cases, you will be asked to perform horizontal reflections across an axis of symmetry that isn't the x-axis. But before we go into how to solve this, it's important to know what we mean by "axis of symmetry". The axis of symmetry is simply the horizontal line that we are performing the reflection across. It can be the x-axis, or any horizontal line with the equation $y$ = constant, like $y$ = 2, $y$ = -16, etc.
## How to Find the Axis of Symmetry:
Finding the axis of symmetry, like plotting the reflections themselves, is also a simple process. In this case, all we have to do is pick the same point on both the function and its reflection, count the distance between them, divide that by 2, and count that distance away from one of the graphs. This is because, by it's definition, an axis of symmetry is exactly in the middle of the function and its reflection.
The best way to practice finding the axis of symmetry is to do an example problem.
Example:
Find the axis of symmetry for the two functions shown in the images below.
Again, all we need to do to solve this problem is to pick the same point on both functions, count the distance between them, divide by 2, and then add that distance to one of our functions. Let's pick the origin point for these functions, as it is the easiest point to deal with.
Now, by counting the distance between these two points, you should get the answer of 2 units. The last step is to divide this value by 2, giving us 1. Now we know that our axis of symmetry is exactly one unit below the top function's origin or above the bottom functions origin. Looking at the graph, this gives us $y$ = 5 as our axis of symmetry! Let's take a look at what this would look like if there were an actual line there:
And that's all there is to it! You may learn further on how to graph transformations of trigonometric functions and how to determine trigonometric functions from their graphs in other sections. |
top of page
Search
# How to factorize quadratic equations of the form ax^2 + bx + c, where a ≠ 1
In this note, you will learn:
· How to factorize quadratic equations of the form ax^2 + bx + c, (where a ≠ 1)
Previously, we discussed on how to factorize quadratic equations in the form of x^2 + bx + c, where c > 0 ( and b < 0). For those who missed it, be sure to check it out from the link above!
Today, we will be dealing with another version of quadratic equations which is the factorization of quadratic equations of the form ax^2 + bx + c, where a ≠ 1.
Like the previous article, we are going to apply the guess and check together with logic and reasoning to factorize the problem in hand. However, we will have to be careful since there are more possible pairs of combinations to consider! How so you may ask? Let’s try a practice question for you to visualize.
## How to factorize quadratic equations of the form ax^2 + bx + c, where a ≠ 1
For this example, let’s take 2x^2 + 15x + 18 as the problem we are trying to solve here.
First step is similar to the previous case, so explanation is not required, but if you are still unsure, do refer back to the previous example!
As shown in the diagram above, for the second step, there should be a sequence of thought process.
For example, if we were to take the factors of 18 as 1 and 18, we can see the it is impossible to obtain +15x since 18 > 15, and no sum will yield the desired bx term.
Then what about the factors 2 and 9?
Similarly, this option is also not viable as (2x) (2) + (x)(9) = 13x and (2x) (9) + (x)(2) = 20x. The answers obtained are either larger or smaller than our desired bx term.
Let’s see how the last option is able to help us yield our desired bx term. The factors 3 and 6 when multiplied with x and 2x is able to give us (2x) (3) + (x)(6) = 12x and (2x) (6) + (x)(3) = 15x.
From this, we are able to see that one of the combinations is able to yield our desired answer, hence this is why we chose it as our most viable option.
Finally, we can cross multiply the values in the upper left quadrant and we will check if we end up with the desired bx term that we were supposed to obtain. In this case, we did it! The answer after factorizing 2x^2 + 15x + 18 is equals to (2x + 3) (x + 6)!
And that’s all for today, students! Math Lobby hopes that after this article, you have a clear understanding on the methods of how to factorize quadratic equations using the multiplication frame!
If you have any pending questions, please do go on to our Facebook page, Instagram or contact us directly at Math Lobby! We have certified mathematics tutors to aid you in your journey to becoming a better student!
As always: Work hard, stay motivated and we wish all students a successful and enjoyable journey with Math Lobby!
*
*
* |
Cross Multiplication Method
Chapter 3 Class 10 Pair of Linear Equations in Two Variables
Serial order wise
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Ex 3.5 ,1 Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method (iv) x – 3y – 7 = 0 3x – 3y – 15 = 0 x – 3y – 7 = 0 3x – 3y – 15 = 0 x – 3y – 7 = 0 Comparing with a1x + b1y + c1 = 0 ∴ a1 = 1, b1 = –3, c1 = –7 3x – 3y – 15 = 0 Comparing with a2x + b2y + c2 = 0 ∴ a2 = 3, b2 = –3, c2 = –15 a1 = 1, b1 = –3, c1 = –7 & a2 = 3, b2 = –3, c2 = –15 𝒂𝟏/𝒂𝟐 𝑎1/𝑎2 = 1/3 𝒃𝟏/𝒃𝟐 𝑏1/𝑏2 = (−3)/(−3) 𝑏1/𝑏2 = 1 𝒄𝟏/𝒄𝟐 𝑐1/𝑐2 = (−7)/(−15) 𝑐1/𝑐2 = 7/15 Since 𝒂𝟏/𝒂𝟐 ≠ 𝒃𝟏/𝒃𝟐 We have a unique solution Solving x – 3y – 7 = 0 …(1) 3x – 3y – 15 = 0 …(2) Using cross-multiplication 𝑥/(−3 ×(−15) − (−3) ×(−7) ) = 𝑦/(3 ×(−7) − 1 × (−15) ) = 1/(1 ×(−3) − 3 ×(−3) ) 𝑥/((45) − 21 ) = 𝑦/(−21 + 15 ) = 1/(−3 + 9 ) 𝑥/(24 ) = 𝑦/(−6 ) = 1/6 Now, 𝒙/𝟐𝟒 = 𝟏/𝟔 x = 24/6 ∴ x = 4 𝒚/(−𝟔) = 𝟏/𝟔 y = (−6)/6 ∴ y = – 1 Therefore, x = 4, y = –1 is the solution of our equation |
# How do we analyse the deflection of simply supported beams using calculus.
In our previous article on deflection in simply supported beams, we looked at calculating deflection using standard tables. In this article, however, our focus will be on calculating deflection using calculus.
## Relationship Between Deflection, Slope and Radius of Curvature
As we mentioned in our previous articles, a beam bends when it’s subject to loading. In the image below, you can see a simple beam before being loaded, then the same beam after being loaded.
To derive an expression relating to the slope, deflection, and radius of curvature of a beam, let us consider a small portion (labelled AB) of a beam bent into an arc, shown the image below:
Let ds = Length of beam AB
C= = the center of the arc (into which the beam has been bent),
= Angle which the tangent at A makes with XX – axis, and +d = Angle which the tangent at B makes with XX-axis
We then find, from the geometry of the figure, that:
If the coordinates of point A are x and y, then:
Differentiating the above equation with respect to x, we obtain
From bending theory of a beam, the expression relating the radius of curvature, moment of inertia, elastic modulus and moment is given by:
Now, if we substitute 1R=d2ydx2 into the above equation, we will have
Units in the above equation include:
E = elastic modulus in N/m2
I = moment of inertia in m4
M = moment in Nm
y and x are taken in m.
This equation is the fundamental expression from which the slope and deflection in a beam can be calculated. Slope and deflection expressions can be obtained by carrying out double integration and applying boundary conditions of the beam to solve the expressions.
## Sign Conventions
To find out the slope and deflection of a centre line of a beam at any point proper, there is need to use sign conventions. The following sign conventions will be used:
1. x is positive when measured towards the right.
2. y is negative when measured downwards.
3. M (bending moment) is negative when hogging.
4. Slope is negative when the rotation is clockwise.
## Double integration Method
As we’ve seen, the fundamental equation for obtaining slope and deflection in beam is
Note that first integration of the above equation gives the value of slope (dydx). The second integration of the expression gives the value of the deflection (y).
## Simply supported beam of span l carrying a point load at mid span.
Let‘s consider a simply supported beam AB of span l carrying a point load W at mid-span C, shown in Fig 19. Since the load is symmetrically applied, the maximum deflection (ymax) will occur at mid span.
The support reactions at A and B equals = W2
Consider the left half AC of the span. The bending moment at any section XX in AC distance. x from A is given by,
First integration gives,
Where C1 is the constant of integration.
Applying boundary conditions: at
Thus we have,
Substituting the value of C1 gives
Slope (dydx) at the support A, i.e when x = 0
Therefore,
Integrating the slope equation once more, gives
and again using the boundary conditions:
Hence, we have
which is the deflection equation. Substitute x = l2 to obtain the mid-span deflection
Hence, deflection at C, equals
## Example of a 4m span beam
A simply supported beam with a 4 m span carries a uniformly distributed load of 20 kN/m on the whole span. Calculate the maximum slope and deflection of the beam. Take E = 2 105 N/mm2 and I = 5000 104 mm4.
Length of beam = 4 m
UDL = 20 kN/m = 20000 N/m
E = 2 105 N/mm2 = 2 1011 N/m2
I = 5000 104 mm4 = 0.00005 m2
Slope is given by wl324EI = 20000 4324 2 1011 0.00005 = 5.3310-3 rad
Maximum deflection (occurring at midspan) is given by ymax= 5wl4384EI = 520000443842 1011 0.00005 = 0.00667 m
Keep an eye out for future articles on more engineering calculations, tips and explanations.
### Interested in our courses?
Interested in civil or mechanical engineering? Find out more about all the civil engineering courses we have available by clicking here, and the mechanical engineering courses by clicking here.
Diploma in Civil Engineering
Diploma in Mechanical Technology
Diploma in Mechanical Engineering
Diploma in Renewable Energy
Diploma in Material Science
Diploma in Sustainable Construction
Diploma in Structural Engineering
Diploma in Thermodynamics
Diploma in Building and Construction Engineering
Diploma in Thermofluids
Higher International Certificate in Civil Engineering
Higher International Diploma in Civil Engineering
Higher International Diploma in Mechanical Engineering
Higher International Certificate in Mechanical Engineering
Alternatively, you can view all our online engineering courses here.
## Recent Posts
### Understanding the mechanisms involved in a voltage amplifier
Understanding the mechanisms involved in a voltage amplifier In our previous articles, we discussed analogue signals and their characteristics. We’re going to dive a little deeper into analogue signals and see how a voltage amplifier works. What is an operational amplifier? Operational amplifiers, or op-amps, are one of the basic building blocks of analogue electrical […]
### Characteristics of analogue electronics
Characteristics of analogue electronics In our previous articles, we’ve discussed subjects such as the different types of semiconductors and how they work. We’re now going to focus on analogue electronics—what it is and its characteristics. What is an analogue signal? An analogue signal is continuous; in other words, it can have an infinite number of […]
### The structure and mechanism of a Bipolar transistor
The structure and mechanism of a Bipolar transistor In our previous article, we discussed the operation of metal oxide transistors, and now we’re going to have a look at bipolar transistors. We’ll look at their mechanism and how they work. What is a bipolar transistor? A bipolar junction transistor is a semiconductor device which can […] |
Select Page
We can use the simplification of algebraic expressions to get simpler expressions to work with. You can simplify expressions by applying the distributive property of multiplication to remove grouping signs such as parentheses and combine like terms.
In this article, we will look at a summary of simplifying algebraic expressions. In addition, we will explore several simplification examples with answers to fully master this topic.
##### ALGEBRA
Relevant for
Exploring examples of simplifying algebraic expressions.
See examples
##### ALGEBRA
Relevant for
Exploring examples of simplifying algebraic expressions.
See examples
## Summary of simplification of algebraic expressions
The simplification of algebraic expressions allows us to obtain simpler expressions that can be manipulated with ease. To simplify algebraic expressions, we can apply the distributive property to remove parentheses and other grouping signs, and we can combine like terms.
### Distributive property
The distributive property tells us how to eliminate grouping signs by distributing the multiplication of a number to all the internal terms of the parentheses:
### Combine like terms
Like terms are algebraic terms that have the same variables raised to the same power. For example, } and are like terms since they have the same variable (x) raised to the same power (2). Similarly, the terms and are also like terms since they have the same variables raised to the same power.
## Simplification of algebraic expressions – Examples with answers
The following examples have their respective solutions, so you can study them carefully and master the process of simplifying algebraic expressions.
### EXAMPLE 1
Simplify the algebraic expression: .
We have the variable x and we have constant terms. Therefore, we combine with terms with variables and combine the constant terms:
### EXAMPLE 2
Simplify the algebraic expression: .
We have a parenthesis so we start by using the distributive property to distribute the 2 and remove the parentheses:
Now we combine like terms. We combine the variables and the constant terms:
### EXAMPLE 3
Simplify the algebraic expression: .
In this case, we have the variable x with a power of 1 and with a power of 2, so we combine like terms separately for the power of 1 and for the power of 2. We also combine the constant terms separately:
### EXAMPLE 4
Simplify the algebraic expression: .
We start by applying the distributive property to remove the parentheses:
Now combine like terms. We combine terms with the variable x with different powers separately:
### EXAMPLE 5
Simplify the algebraic expression: .
We start by removing the parentheses using the distributive property:
In this case, we have terms with both the x variable and the y variable. We have to combine terms that have the same variables raised to the same powers:
### EXAMPLE 6
Simplify the algebraic expression: .
We start by removing both parentheses using the distributive property:
We have to combine the terms that have the same variables raised to the same power:
### EXAMPLE 7
Simplify the algebraic expression: .
We start by removing the parentheses using the distributive property:
Here we have several terms with the variable x with different powers. We have to make sure to combine only the terms that have the same power:
## Simplifying algebraic expressions – Practice problems
In addition to the solved examples, the following problems help to master the simplification of algebraic expressions. If you have trouble solving these exercises, you can carefully study the solved examples above. |
UK
January 24, 2019
Global Assignment Help
( 42959 views )
## How to Solve a Cubic Equation Problem?
When you are assigned with a cubic equation problem in your assignment by your maths professor, then it may look unsolvable for you in the beginning. However, the easy ways for solving it has been already introduced hundreds of years ago. We know that finding the solution is a bit difficult, but with the right approach, even the most complex and trickiest cubic equation can be solved easily. Here are a few strategies to tackle with it.
### Way 1: Solve It with Quadratic Formula
Cubic equation are in the form of ax3+bx2+cx+d=0
If you see that the equation is not in standard form, then do the basic arithmetic calculations to get the cubic equation.
On the other hand, if the equation contains a constant, then you need to follow a different approach.
Divide the Equation with an X
Since your equation has an X variable in it, one X can be factored out in the form of:
X(ax2+bx+c)
Use the Quadratic Formula to Solve it
The obtained equation is in quadratic form. This means that now you can easily find the values of a, b, and c just by placing them in the quadratic formula, i.e.,
### Way 2: Finding Solutions with Integers
The way maintained above is quite convenient because you don’t have to implement any new mathematical formula or trick. But, it won’t help you with all the cubic equations.
Like, if your equation has a non-zero value for d, then you need to follow the step mentioned below.
Example: 2x3+5x2+15x+6=0
In this the value of d is 6, so you can’t implement the above method.
Here you have to find the factors of a and d
A=2=2*1
D=6=6*1, 3*2
Factors of a=1,2 and of d=1,2,3,6.
Divide the factors of a by d
The next step is to make a list of the values that you get after dividing each factor of a by each factor of d.
1,1/2,1/3,1/6 and 2,1,2/3,3
Next, add the negatives to it
-1,-1/2,-1/3,-1/6 and -2,-1,-2/3,-3
Once you have got the list of the values, you can easily find the answers to your cubic equation quickly by plugging each integer manually and finding which one equals zero. However, if you don’t want to calculate the answers by this method, then you can go with another technique called synthetic division.
### Way 3: Follow a Discriminant Approach
For this method of finding a cubical equation solution, you have to deal with the coefficients of the terms in the equation.
For example: if your equation is x3-3x2+3x-1, we would write a=1, b=-3, c=3, and d=1. Don't forget that when a variable doesn't have a coefficient, it's implicitly assumed that its coefficient is 1.
So, these are the three different ways to solve a cubic equation easily. Well, solving maths assignments require too much efforts and patience. And, if because of your extracurricular activities you don’t have enough time to complete them, then you can opt for our online assignment help service.
### How Can Our Online Writing Services Help You With Your Maths Assignment?
After a rigorous round of interviews, Global Assignment Help has hired a highly-skilled team of expert writers that is capable of offering the finest maths assignment help to students. By seeking assistance from us, your paper will be properly written by experienced professionals writers. They make sure that there are no traces of plagiarism in the document as they only deliver genuine work within the given time limit. If you think sample assignments can help you with your assignment, then we also offer free samples on variety of topics which you can easily go through on our website. Moreover, our customer care executives are available 24*7 to assist you regarding your issues at any odd hour of the day or night. |
No new puzzle this month. Let’s get straight into the answer from last month. Judging from the answers, this one wasn’t as explained as well as it could have been. So, we’ll go through what was in our brains when we wrote it.
There is a sequence that starts with: 2, 4, 6, 8 and is followed by a blank. There are many numbers that could follow. We weren’t aware of this when we wrote the puzzle but at oeis.org there are about 200 sequences, many obscure, some irrelevant, that start with 2, 4, 6, 8. I’m going to include the puzzle solution factoring in all of these potential, mostly obscure, sequences. From analysis of these obscure sequences, none continue to 3, 6, 13 or 26. This will be relevant later.
Of all the numbers that can follow they fall into three categories. The number could be larger, smaller or the same as the last. What we are trying to do is to reduce our options and we’re using a technique known as equivalence partitioning.
Initial thoughts guesses:
• 2, 4, 6, 8, 6
• 2, 4, 6, 8, 8
• 2, 4, 6, 8, 10
Our first four numbers show that the sequence isn’t a decreasing sequence. The smaller (6) and equal (8) numbers prove that the sequence doesn’t have to be increasing. The larger number, if you chose 10, tells you nothing new. But we can do better with our first guess:
• 2, 4, 6, 8, 3
If 3 is in the sequence, then we have a non-increasing sequence that allows odd numbers and isn’t an obscure sequence.
• 2, 4, 6, 8, 3, -1
If we follow that up with -1 and it is allowed, then we can have any number in the sequence and it will be allowed. If -1 is not allowed. We can have any non-negative number. Follow this up with a 0. If we allow 0, then we do allow any non-negative number. If 0 is not in the sequence, then our sequence has to be made of positive numbers.
• 2, 4, 6, 8, 3, -1 –> any number
• 2, 4, 6, 8, 3, -1x, 0 –> any non-negative number
• 2, 4, 6, 8, 3, -1x, 0x –> any positive number
Let’s go back to the 3 and if that is not in the sequence. We should follow up with a 13. If we allow 13 then we know that odd numbers are ok, but smaller numbers are not. We also know that it isn’t an obscure sequence but we still not sure if the sequence is a greater than or a greater than or equal to. So we follow that up with another 13.
• 2, 4, 6, 8, 3x, 13, 13 –> Any number that is greater to or equal to the prior
• 2, 4, 6, 8, 3x, 13, 13x –> Any number that is greater than the prior.
What if 13 isn’t allowed. We are the point where we know that odd numbers are not allowed but that’s all we know. So we should follow that up with a 6. We need to know if smaller numbers are supported.
• 2, 4, 6, 8, 3x, 13x, 6 –> Any even number
If the 6 is not successful we try an 8. If the 8 is in the sequence we have a greater than or equal to situation. When the 8 is not in the sequence we follow up with a 26. We need to know if the sequence is one of the obscure ones or just an ever increasing even number
• 2, 4, 6, 8, 3x, 13x, 6x, 8 –> any even number that is greater to or equal to the prior
• 2, 4, 6, 8, 3x, 13x, 6x, 8x, 26 –> any even number that is greater than the prior
• 2, 4, 6, 8, 3x, 13x, 6x, 8x, 26x –> a special pattern from oeis.org
How do we deal with the special patterns? That same way we have before, what’s the number that tells us the most about what we don’t know yet. What challenges our assumptions the most.
Of the potential next numbers it could be 10, 12, 14, 16, 18, 20, 22, 24, 30, 32, 38, 50 and 212. But when waited for number of sequences that have that number next the order changes to: 10, 12, 16, 20, 24, 22, 14, 18, 30, 32, 38, 50 and 212. There are 37 sequences that continue to 10. 27 that continue to 12 and the remainder have 5 or fewer options, most only have 1.
We continue the process as before, whittling away. No prizes are awarded this month.
For those that want to do some analysis on the oeis.org dataset, you can do the following:
Get all the things:
``for (( c=0; c<=854; c += 10 )) do wget -k "http://oeis.org/search?q=2,4,6,8&start=\$c&fmt=data" -O \$c.html;done``
Merge all the things:
``cat *.html > merged.html``
Open up sublime, and find all occurances of `<tt>`, expand selection to the line, cut the selection. New document, paste. Remove the `<tt>` and `</tt>`.
Now you have the dataset. F5 will sort the lines. Remove all those that don’t start with 2, 4, 6, 8. You can use a regex find here `^2, 4, 6, 8`.
Enjoy. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Kinetic Energy
## The energy of moving matter. KE = 1/2 mass X velocity 2
Estimated2 minsto complete
%
Progress
Practice Kinetic Energy
Progress
Estimated2 minsto complete
%
Kinetic Energy
What could these four photos possibly have in common? Can you guess what it is? All of them show things that have kinetic energy.
### Defining Kinetic Energy
Kinetic energy is the energy of moving matter. Anything that is moving has kinetic energy—from atoms in matter to stars in outer space. Things with kinetic energy can do work. For example, the spinning saw blade in the photo above is doing the work of cutting through a piece of metal.
### Calculating Kinetic Energy
The amount of kinetic energy in a moving object depends directly on its mass and velocity. An object with greater mass or greater velocity has more kinetic energy. You can calculate the kinetic energy of a moving object with this equation:
\begin{align*}\mathrm{Kinetic\; Energy\; (KE)=\frac{1}{2} mass \times velocity^2}\end{align*}
This equation shows that an increase in velocity increases kinetic energy more than an increase in mass. If mass doubles, kinetic energy doubles as well, but if velocity doubles, kinetic energy increases by a factor of four. That’s because velocity is squared in the equation.
Let’s consider an example. The Figure below shows Juan running on the beach with his dad. Juan has a mass of 40 kg and is running at a velocity of 1 m/s. How much kinetic energy does he have? Substitute these values for mass and velocity into the equation for kinetic energy:
\begin{align*}\text{KE}=\frac{1}{2} \times 40 \ \text{kg} \times (1\frac{\text{m}}{\text{s}})^2=20 \ \text{kg} \times \frac{\text{m}^2}{\text{s}^2}=20 \ \text{N} \cdot \text{m},\end{align*} or \begin{align*}20 \ \text{J}\end{align*}
Notice that the answer is given in joules (J), or N • m, which is the SI unit for energy. One joule is the amount of energy needed to apply a force of 1 Newton over a distance of 1 meter.
What about Juan’s dad? His mass 80 kg, and he’s running at the same velocity as Juan (1 m/s). Because his mass is twice as great as Juan’s, his kinetic energy is twice as great:
\begin{align*}\text{KE}=\frac{1}{2} \times 80 \ \text{kg} \times (1 \frac{\text{m}}{\text{s}})^2=40 \ \text{kg} \times \frac{\text{m}^2}{\text{s}^2}=40 \ \text{N} \cdot \text{m}\end{align*}, or \begin{align*}40 \ \text{J}\end{align*}
Q: What is Juan’s kinetic energy if he speeds up to 2 m/s from 1 m/s?
A: By doubling his velocity, Juan increases his kinetic energy by a factor of four:
\begin{align*}\text{KE}=\frac{1}{2} \times 40 \ \text{kg} \times (2 \frac{\text{m}}{\text{s}})^2=80 \ \text{kg} \times \frac{\text{m}^2}{\text{s}^2}=80 \ \text{N} \cdot \text{m}\end{align*}, or \begin{align*}80 \ \text{J}\end{align*}
### Summary
• Kinetic energy (KE) is the energy of moving matter. Anything that is moving has kinetic energy.
• The amount of kinetic energy in a moving object depends directly on its mass and velocity. It can be calculated with the equation: \begin{align*}\text{KE}=\frac{1}{2}\text{mass} \times \text{velocity}^2\end{align*}.
### Vocabulary
• kinetic energy: Energy of moving matter.
### Practice
At the following URL, review kinetic energy and how to calculate it. Then take the quiz at the bottom of the Web page. Be sure to check your answer. http://www.physicsclassroom.com/class/energy/u5l1c.cfm
### Review
1. What is kinetic energy?
2. The kinetic energy of a moving object depends on its mass and its
1. volume.
2. velocity.
3. distance.
4. acceleration.
3. The bowling ball in the Figure below is whizzing down the bowling lane at 4 m/s. If the mass of the bowling ball is 7 kg, what is its kinetic energy?
### Vocabulary Language: English
kinetic energy
kinetic energy
Energy of moving matter. |
Miracle Learning Center
# Equations
## 16 May Equations
Maths is one of the most useful subjects in the world and is mandatory to learn for all the students around the world. Maths is connected with a lot of other subjects and has connections with many professions in society. But sadly it is one of the most disliked subjects among the students.
The word equation comes from the word equate. Is simply refers to a statement that asserts the equality of two given expressions. Mathematically speaking, the presence of an equal sign (=) distinguishes an equation. It is to be interpreted as signifying that the expression to the right of the sign is equal to the expression on the left side.
Equations do not necessarily have to contain an unknown value as one would normally expect. Their main purpose is not just to solve problems to find the value of an unknown. Rather, they also provide critical alternate approaches to problem solving. For example, the equation tanx = sinx/cosx does not have any unknown value that needs to be found. It simply gives the assertion that it is mathematically correct to substitute tanx with sinx/cosx. In trigonometry, an equation of this nature is referred to as an identity. In order for an equation to maintain its validity, ALL arithmetic laws must be strictly observed. This is important especially when commutating terms of the equation from one side of the equal sign to another. Failure to do so will render the equation false.
Mathematical syllabigenerally consists a lot of equations that needs solving. These usually incorporate a lot of other mathematical principles such as trigonometry and calculus. Solving equations therefore must be mastered in order to tackle these principles. Students must learn to isolate the value of interest in the equation, commonly referred to as the ‘unknown’. This is done through mathematical operations and laws. Once the unknown is isolated on one side of the equal sign, consolidate and combine the remaining expression on the other side until it is its simplest form. Some equations can not be simplified to a single and simple value. This is however mathematically acceptable as long as the ‘unknown’ is isolated in its simplest form on one side.
If you also want to achieve great marks in Maths and want to become the topper of your class then come join the premium classes of secondary maths tuition, primary mathematics tuition, JC maths tuition today. We assure you that you will never find our classes to be boring. |
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# For $0<\theta <\dfrac{\pi }{2}$ , the solution(s) of $\sum\limits_{m=1}^{6}{\text{cosec}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)}\text{cosec}\left( \theta +\dfrac{m\pi }{4} \right)=4\sqrt{2}$ is (are):a)$\dfrac{\pi }{4}$b)$\dfrac{\pi }{6}$ c)$\dfrac{\pi }{12}$ d) $\dfrac{5\pi }{12}$
Last updated date: 19th Jun 2024
Total views: 403.5k
Views today: 6.03k
Verified
403.5k+ views
Hint: We have a trigonometric expression as: $\sum\limits_{m=1}^{6}{\text{cosec}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)}\text{cosec}\left( \theta +\dfrac{m\pi }{4} \right)=4\sqrt{2}$
We can write $\sqrt{2}=\text{cosec}\dfrac{\pi }{4}$ . As the expression contains $\text{cosec}\theta$ , try to convert the expression in terms of $\sin \theta$ . Then, we can write $\sin \dfrac{\pi }{4}$ as $\sin \left[ \left( \theta +\dfrac{m\pi }{4} \right)-\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right) \right]$
Later on, by using the identity: $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ , split the term $\sin \left[ \left( \theta +\dfrac{m\pi }{4} \right)-\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right) \right]$. Now, simplify the whole expression by cancelling the terms to get an equation in terms of $\cot \theta$ . Now, expand the summation given by putting values of m and cancel out to the terms to get a simplified equation. Now, using various trigonometric identities, find the value of $\theta$
We have:
$\sum\limits_{m=1}^{6}{\text{cosec}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)}\text{cosec}\left( \theta +\dfrac{m\pi }{4} \right)=4\sqrt{2}......(1)$
As we know that: $\text{cosec}\dfrac{\pi }{4}=\sqrt{2}$
So, we can write equation (1) as:
$\sum\limits_{m=1}^{6}{\text{cosec}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)}\text{cosec}\left( \theta +\dfrac{m\pi }{4} \right)=4\text{cosec}\dfrac{\pi }{4}......(2)$
As we know that: $\text{cosec}\theta =\dfrac{1}{\sin \theta }$ , so we can write equation (2) as:
\begin{align} & \sum\limits_{m=1}^{6}{\dfrac{1}{\text{sin}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)\text{sin}\left( \theta +\dfrac{m\pi }{4} \right)}}=\dfrac{4}{\text{sin}\dfrac{\pi }{4}} \\ & \sum\limits_{m=1}^{6}{\dfrac{\text{sin}\dfrac{\pi }{4}}{\text{sin}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)\text{sin}\left( \theta +\dfrac{m\pi }{4} \right)}}=4......(3) \\ \end{align}
Now, we can write: $\sin \dfrac{\pi }{4}=\sin \left[ \left( \theta +\dfrac{m\pi }{4} \right)-\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right) \right]$ in equation (3), we get:
$\sum\limits_{m=1}^{6}{\dfrac{\sin \left[ \left( \theta +\dfrac{m\pi }{4} \right)-\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right) \right]}{\text{sin}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)\text{sin}\left( \theta +\dfrac{m\pi }{4} \right)}}=4......(4)$
Now, by applying identity: $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$, we can write equation (4) as:
$\sum\limits_{m=1}^{6}{\dfrac{\left[ \sin \left( \theta +\dfrac{m\pi }{4} \right)\cos \left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)-\cos \left( \theta +\dfrac{m\pi }{4} \right)\sin \left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right) \right]}{\text{sin}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)\text{sin}\left( \theta +\dfrac{m\pi }{4} \right)}}=4......(5)$
Now, by expanding the equation (5), we get:
\begin{align} & \sum\limits_{m=1}^{6}{\dfrac{\sin \left( \theta +\dfrac{m\pi }{4} \right)\cos \left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)}{\text{sin}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)\text{sin}\left( \theta +\dfrac{m\pi }{4} \right)}}-\dfrac{\cos \left( \theta +\dfrac{m\pi }{4} \right)\sin \left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)}{\text{sin}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)\text{sin}\left( \theta +\dfrac{m\pi }{4} \right)}=4 \\ & \sum\limits_{m=1}^{6}{\dfrac{\cos \left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)}{\text{sin}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)}}-\dfrac{\cos \left( \theta +\dfrac{m\pi }{4} \right)}{\text{sin}\left( \theta +\dfrac{m\pi }{4} \right)}=4......(6) \\ \end{align}
Since $\dfrac{\cos \theta }{\sin \theta }=\cot \theta$ , we can write equation (6) as:
$\sum\limits_{m=1}^{6}{\cot \left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)}-\cot \left( \theta +\dfrac{m\pi }{4} \right)=4......(7)$
Now, expand the summation by putting the values of m, we get:
\begin{align} & \Rightarrow \left[ \cot \left( \theta +\dfrac{\left( 1-1 \right)\pi }{4} \right)-\cot \left( \theta +\dfrac{\pi }{4} \right) \right]+\left[ \cot \left( \theta +\dfrac{\left( 2-1 \right)\pi }{4} \right)-\cot \left( \theta +\dfrac{2\pi }{4} \right) \right] \\ & \text{ }+.....+\left[ \cot \left( \theta +\dfrac{\left( 6-1 \right)\pi }{4} \right)-\cot \left( \theta +\dfrac{6\pi }{4} \right) \right]=4 \\ & \Rightarrow \cot \theta -\cot \left( \theta +\dfrac{\pi }{4} \right)+\cot \left( \theta +\dfrac{\pi }{4} \right)-\cot \left( \theta +\dfrac{2\pi }{4} \right) \\ & \text{ }+.....+\cot \left( \theta +\dfrac{5\pi }{4} \right)-\cot \left( \theta +\dfrac{6\pi }{4} \right)=4 \\ & \Rightarrow \cot \theta -\cot \left( \theta +\dfrac{6\pi }{4} \right)=4 \\ & \Rightarrow \cot \theta -\cot \left( \theta +\dfrac{3\pi }{2} \right)=4......(8) \\ \end{align}
As we know that: $\cot \left( \dfrac{3\pi }{2}+\theta \right)=-\tan \theta$
So, we can write equation (8) as:
$\Rightarrow \cot \theta +\tan \theta =4......(9)$
Now, write $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ and $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ in equation (9), we get:
\begin{align} & \Rightarrow \cot \theta +\tan \theta =4 \\ & \Rightarrow \dfrac{\cos \theta }{\sin \theta }+\dfrac{\sin \theta }{\cos \theta }=4 \\ & \Rightarrow {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =4\sin \theta \cos \theta ......(10) \\ \end{align}
As we know that: ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ and $2\sin \theta \cos \theta =\sin 2\theta$, so we can write equation (10) as:
$\Rightarrow 1=2\sin 2\theta ......(11)$
Now, solving for $\theta$, we can write equation (11) as:
\begin{align} & \Rightarrow \sin 2\theta =\dfrac{1}{2} \\ & \Rightarrow \sin 2\theta =\sin \dfrac{\pi }{6}\text{ or }\sin \dfrac{5\pi }{6} \\ & \Rightarrow 2\theta =\dfrac{\pi }{6}\text{ or }\dfrac{5\pi }{6} \\ & \Rightarrow \theta =\dfrac{\pi }{12}\text{ or }\dfrac{5\pi }{12} \\ \end{align}
So, the correct answer is “Option C and D”.
Note: For a given trigonometric expression, it is always easier to convert the expression in terms of sine and cosine. Also, if a summation expression is given, always try to expand the summation by putting the values of the variable and cancel out the terms if possible. |
# Advanced Volume on the ISEE
#### Middle and Upper Level
LESSON GOAL: Solve multi-step problems that involve the volume of a figure.
ISEE Middle-Level Volume Question:
The surface area of a cube is 150 centimeters. What is its volume?
A) 5 cm3
B) 25 cm3
C) 100 cm3
D) 125 cm3
The solution:
Step 1: Work backwards from the surface area to determine the length of each side of your cube. Since the cube has 6 faces, we can divide the surface area by 6 to get the area of each face. Find its square root to get the side length:
150 ÷ 6 = 25; 25 = 5
Helpful tip: Since you don’t get a calculator, the ISEE usually provides numbers that are easy to work with, such as perfect squares. If you find yourself doing a complicated calculation, double check your work before continuing.
Step 2: Find the volume by taking the side length, 5, to the third power. The answer is D.
5 × 5 × 5 = 125
Helpful tip: Don’t stop before you’re finished! The test is tricky and often offers partial solutions, like A and B, as answer choices.
ISEE Upper-Level Volume Question:
The height of the cylinder shown is 5 times its diameter. The formula used to find the volume of a cylinder is V = πr2h or V = r2 where r is the radius of the cylinder and h is the height of the cylinder. If the diameter of the cylinder is 4 in., what is its volume, in inches3?
A) 80π
B) 100π
C) 160π
D) 320π
Helpful tip: If you see an unfamiliar figure on the test and you don’t know the formula, don’t panic! The ISEE often gives the formula for a less common figure within the problem. That said, it’s important to memorize common formulas, such as the volume of cubes and rectangular prisms.
Step 1: Use the information in the problem to figure out each variable in the formula:
r = half the diameter: 4 ÷ 2 = 2 in.
h = 5 times the diameter: 4 × 5 = 20 in.
Helpful tip: Don’t mix up diameter and radius! The test often gives you one of these measures when you need the other one to solve the problem.
Step 2: Plug each variable into the equation and solve: V = r2 = 22 x 20 x π = 80π. The correct answer is A. |
# week 5 DQ3
label Statistics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
In what way has the learning from this course helped you to effectively test hypothesis, draw conclusions, and provide recommendations to management for resolving the problem taken for study? Consider what you have actually achieved in the research study.
Jul 5th, 2015
In a research study, the most effective ways to solve a problem is to form the hypothesis and test it. Here the the steps to effectively form a hypothesis and test it for validity.
1. Identify the null hypothesis for both one-tailed and two-tailed tests. The null hypothesis is a parameter equals zero although there are exceptions. A null hypothesis is u1-u2 =0 or u1= u2. For a one tailed test , the null hypothesis is either a parameter is greater than or equal to zero.
2. The second step is to specify the α level that is known as the significance level. The values are 0.05 and 0.01.
3. The third step is to compute the probability value ( p-value). The p-value is the probability of obtaining a sample statistic as different or more different from parameter specified in the null hypothesis given that the null hypothesis is true.
4. The last step is to compare the probability value with the α level. If the probability value is lower than the α level, then you reject the null hypothesis. The lower the probability value ( p-value), the more confidence you will have that the null hypothesis is false. If your p-value is higher than the conventional α level of 0.05, then most scholars will consider your findings inconclusive. For example, if the p-value is 0.005, this p-value is lower than your α level of 0.05. Therefore, you would reject the null hypothesis, and is more confidence that the null hypothesis is false. If you fail to reject the null hypothesis , that does not mean that you will support for the null hypothesis. It just means that you do not have sufficiently strong data to reject it.
Simply, to make the right decision, you need to form the null hypothesis and alternate hypothesis. If your p-value is smaller than the α level, then you reject the null hypothesis and accept the alternate hypothesis. If the p-value is greater than 0.05 α level, then you do not reject the null hypothesis.
Jul 5th, 2015
...
Jul 5th, 2015
...
Jul 5th, 2015
Nov 17th, 2017
check_circle |
# The Hard Way
Here’s an interesting problem from a homework assignment in a class on stochastic processes: A woman standing at an intersection wants to cross the street. She will only cross if she feels that she can get all the way to the other side before a car arrives. Cars arrive randomly at a particular rate, but she can see them far enough off to determine if she has time to cross. On average, how long will the woman need to wait before she can cross the street?1Ross, Sheldon M., Introduction to Probability Models, Academic Press, 2010.
While the original problem provides more details (e.g. an exact description of the rate at which cars arrive, and a parameter that describes how cautious the woman is as a function of how long she requires the gap between cars to be), it is possible to learn something about mathematical thinking without going into those details.
### The Hard Way
The basic goal of the problem is to find an average, or expected, value for the amount of time that the woman must wait. One approach would be to attempt to calculate this average directly. The general idea when calculating an average is to multiply each possible outcome by how likely that outcome is, and add up the results.2This may look different from what you were taught in school, which may have been to add up all of the outcomes, then divide by the total number of outcomes. However, it turns out that this is the same thing. Suppose you want to find the average of the five numbers 4, 4, 5, 7, and 10. If you add them all up, you get 30. Dividing by 5 (the number of outcomes), you get 6. So the average is 6. On the other hand, the probability of getting a 4 is 2 out of 5 (0.4) (there are a total of five numbers, two of which are fours), and the probabilities of each of the other numbers is 1 out of 5 (0.2). We then calculate the average as $$(0.4)4+(0.2)(5+7+10) = 6$$. We get the same result in either case.
For instance, if we know that a bag of M&M’s contains 24 candies with probability 10%, 25 candies with probability 85%, and 26 candies with probability 5%, then the average number of candies per bag is given by
$$24(0.10)+25(0.85)+26(0.05) = 24.95.$$
This tells us that the average bag of M&M’s contains about 24.95 candies (which we would normally round up).
So if we want to know how long the woman has to wait before she can cross the street, we need to figure out what the possible outcomes are (i.e. what are the possible amounts of time that she could wait), and how likely each of those outcomes is. Once we figure those out, we just have to do the multiplication and add everything up.
#### What are the possible outcomes?
In general, the woman could wait for any period of time. If no cars are coming, then she might be able to cross immediately. If a car is coming, then she is going to have to wait for it to go by, then either cross or wait some more. Cars arrive at random, so she could wait for one second, one minute, one hour, an infinite amount of time, or anything in between.
On the other hand, the situation can be broken down a little bit. Each time a car goes by, the woman is either going to cross the road as soon as it passes, or wait for the next car. That means that we only need to determine how many cars pass before she can cross. She can cross immediately, or after the first car goes by, or after the second car goes by, and so on.
Basically, the possible outcomes are the number of cars that go by before she crosses.
#### How likely is each outcome?
The problem actually gives the rate at which cars go by, and this rate is related to the probability of each outcome. Again, it is helpful to break things down a bit.
When the woman arrives at the intersection, there is a certain probability that she will be able to cross the street immediately. This probability is related to the rate at which the cars go by, and is given in the problem. The exact details are not important, so I will spare you the notation.
If she can’t cross immediately, then she is going to have to wait for at least one car to go by, then decide whether or not to cross. After the first car goes by, there is a certain probability that she will be able to cross, and a certain probability that she will have to wait. It turns out that these probabilities are the same as when she first arrived, times an extra factor to take into account the fact that she has already waited for the first car to pass.
As we start figuring out the probabilities of waiting for more and more cars, we should discover that there is a pattern. Again, the details aren’t really important—the important bit is that the probability of needing to wait for a certain number of cars to pass goes down as the number of cars increases. That is, she is less likely to have to wait for 100 cars than just 2.
We will take advantage of this pattern in order to determine the probability of waiting for a certain number of cars to go by before the woman can cross.
#### Putting it together.
The general idea is that we are going to have to multiply some number of cars by the probability that we have to wait for that number of cars, then add up all of these products. As the woman could, in theory, wait for an infinite number of cars to pass, we have to add up an infinite number of things.
Moreover, while there is a helpful pattern for determining the probabilities, this pattern depends on a random variable (the length of time between cars going by). In order to come to terms with these probabilities, we are going to have to integrate.
There are mathematical techniques for getting a handle on infinite sums (they aren’t actually as scary as they sound), and integration isn’t too terribly difficult once you learn about it, but these techniques are relatively advanced. Most students won’t see them until college, if ever. Calculus will get us the correct solution, but there has to be a better way!
### The Clever Way
The hard way of solving the problem takes a naive approach, and attacks the problem with advanced mathematical techniques. There is another approach which requires that the problem solver be clever.
The basic logic follows from the question “If I toss a fair coin and it comes up heads, what is the probability that the next toss of the coin will be heads?” If you have taken any probability at all, you probably know that the result of the second toss doesn’t depend on the result of the first toss. They are independent events. That means that the second toss has an equal probability of being heads or tails, no matter what the first toss was.
Another question that we could ask is “If the first toss is tails, how many times will I have to toss the coin until it comes up heads?” It turns out that the number of tosses before I get a head does not depend upon the first toss at all. If I start tossing a fair coin right now, I expect to toss it an average of two times before I get a head. If I toss the coin and get a tail on the first toss, I expect to toss the coin an additional two times before getting a head. In fact, no matter how many times I have tossed the coin, and no matter what the previous tosses were, I can always expect to toss the coin two more times to get a head.
This actually relates to the woman crossing the street. When she first arrives at the intersection, she will expect to wait for a certain amount of time. This waiting time is the sum of two different possible waiting times: if she cross immediately, she doesn’t wait any time at all; otherwise, she waits for the first car to pass, then expects to wait for exactly the same amount of time that she would have expected to wait in the first place. Figure out the probability of either waiting or not waiting, and it isn’t too hard to find the expected waiting time.
The English here is a little tricky, so let me use a little bit of notation, which may clear things up a bit. Let’s call the amount of time that the woman expects to wait $$E[T]$$.3The notation $$E[X]$$ is a pretty common way of writing “the expected value of X,” which means more or less the same thing as the average of X. The notation $$Pr\{\hbox{an event}\}$$ is a way of writing “the probability that an event will occur.” So $$Pr\{\hbox{no wait}\}$$ is “the probability of there being no wait,” which, in the context of the woman waiting to cross the road, is the probability that she will not have to wait for any cars to pass.
$$E[T] = (0\hbox{ min})Pr\{\hbox{no wait}\}+([\hbox{wait until first car goes by}]+E[T])Pr\{\hbox{wait}\}$$.
The problem gives numbers for the amount of time that the woman has to wait for the first car to go by, and enough information to figure out the probability of the woman not having to wait. Given those numbers, the problem reduces to something that can be done after high school algebra.
### Which Way?
There are at least two ways of solving this problem, which I have labeled “the hard way,” and “the clever way.” Both methods ultimately render exactly the same result. However, when I typed up my results, the hard way required two pages of writing, and some dense mathematical notation. The clever way took only one paragraph and some simple algebra.
This dichotomy of methodologies is pretty common, both in mathematics, and in the wider world. There is often a straightforward-but-complicated way of solving a problem, and a clever-but-simple way of solving the same problem. One can jury-rig a quick fix which gets the job done, but could break; or one can engineer a lasting solution.
Finding simple, elegant, clever solutions is at the heart of mathematics. Every mathematician that is worth her salt prefers to use as little notation as possible, and to find solutions that depend on a clever bit of logic rather than a brute-force attack on the problem. Why do things the hard way when they can be done the clever way?
This entry was posted in Probability and tagged . Bookmark the permalink. |
## Step 1
We still have different denominators (bottom numbers), though, therefore we require to acquire a usual denominator. This will certainly make the bottom number match. Multiply the denominators with each other first. Now, multiply each numerator by the other term"s denominator.
You are watching: 1/3 plus 1/4
Now we multiply 1 by 4, and get 4, then we multiply 3 through 4 and also get 12.
Now because that the second term. You main point 1 by 3, and get 3, climate multiply 3 by 4 and also get 12.
This offers us a brand-new problem that looks favor so:
4
12
+3
12
## Step 2
Since our denominators match, we can include the numerators.
4 + 3 = 7
That gives us an answer of
7
12
## Step 3
Can this portion be reduced?
First, us attempt to divide it through 2...
Nope! So currently we shot the next greatest prime number, 3...
Nope! So currently we try the next biggest prime number, 5...
Nope! So now we shot the next biggest prime number, 7...
Nope! So currently we shot the next greatest prime number, 11...
See more: How Did Senator J. William Fulbright Build Support For The Doves In Congress?
No good. 11 is larger than 7. Therefore we"re done reducing.
There you have it! Here"s the final answer come 1/3 + 1/41
3
+1
4
=7
12
1/4 + 1/2
1/4 - 1/2
1/4 + 1/3
1/4 - 1/3
1/4 + 2/3
1/4 - 2/3
1/4 + 1/4
1/4 - 1/4
1/4 + 2/4
1/4 - 2/4
1/4 + 3/4
1/4 - 3/4
1/4 + 1/5
1/4 - 1/5
1/4 + 2/5
1/4 - 2/5
1/4 + 3/5
1/4 - 3/5
1/4 + 4/5
1/4 - 4/5
1/4 + 1/6
1/4 - 1/6
1/4 + 2/6
1/4 - 2/6
1/4 + 3/6
1/4 - 3/6
1/4 + 4/6
1/4 - 4/6
1/4 + 5/6
1/4 - 5/6
1/4 + 1/7
1/4 - 1/7
1/4 + 2/7
1/4 - 2/7
1/4 + 3/7
1/4 - 3/7
1/4 + 4/7
1/4 - 4/7
1/4 + 5/7
1/4 - 5/7
1/4 + 6/7
1/4 - 6/7
1/4 + 1/8
1/4 - 1/8
1/4 + 2/8
1/4 - 2/8
1/4 + 3/8
1/4 - 3/8
1/4 + 4/8
1/4 - 4/8
1/4 + 5/8
1/4 - 5/8
1/4 + 6/8
1/4 - 6/8
1/4 + 7/8
1/4 - 7/8
1/4 + 1/9
1/4 - 1/9
1/4 + 2/9
1/4 - 2/9
1/4 + 3/9
1/4 - 3/9
1/4 + 4/9
1/4 - 4/9
1/4 + 5/9
1/4 - 5/9
1/4 + 6/9
1/4 - 6/9
1/4 + 7/9
1/4 - 7/9
1/4 + 8/9
1/4 - 8/9
1/4 + 1/10
1/4 - 1/10
1/4 + 2/10
1/4 - 2/10
1/4 + 3/10
1/4 - 3/10
1/4 + 4/10
1/4 - 4/10
1/4 + 5/10
1/4 - 5/10
1/4 + 6/10
1/4 - 6/10
1/4 + 7/10
1/4 - 7/10
1/4 + 8/10
1/4 - 8/10
1/4 + 9/10
1/4 - 9/10
1/4 + 1/11
1/4 - 1/11
1/4 + 2/11
1/4 - 2/11
1/4 + 3/11
1/4 - 3/11
1/4 + 4/11
1/4 - 4/11
1/4 + 5/11
1/4 - 5/11
1/4 + 6/11
1/4 - 6/11
1/4 + 7/11
1/4 - 7/11
1/4 + 8/11
1/4 - 8/11
1/4 + 9/11
1/4 - 9/11
1/4 + 10/11
1/4 - 10/11
1/4 + 1/12
1/4 - 1/12
1/4 + 2/12
1/4 - 2/12
1/4 + 3/12
1/4 - 3/12
1/4 + 4/12
1/4 - 4/12
1/4 + 5/12
1/4 - 5/12
1/4 + 6/12
1/4 - 6/12
1/4 + 7/12
1/4 - 7/12
1/4 + 8/12
1/4 - 8/12
1/4 + 9/12
1/4 - 9/12
1/4 + 10/12
1/4 - 10/12
1/4 + 11/12
1/4 - 11/12 |
# 11.7: Understand Slope of a Line (Part 1)
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$
( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\id}{\mathrm{id}}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\kernel}{\mathrm{null}\,}$$
$$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$
$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$
$$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$
$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$
$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vectorC}[1]{\textbf{#1}}$$
$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$
$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$
$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$
##### Learning Objectives
• Use geoboards to model slope
• Find the slope of a line from its graph
• Find the slope of horizontal and vertical lines
• Use the slope formula to find the slope of a line between two points
• Graph a line given a point and the slope
• Solve slope applications
##### be prepared!
Before you get started, take this readiness quiz.
1. Simplify: $$\dfrac{1 − 4}{8 − 2}$$. If you missed this problem, review Example 4.6.12.
2. Divide: $$\dfrac{0}{4}, \dfrac{4}{0}$$. If you missed this problem, review Example 7.5.5.
3. Simplify: $$\dfrac{15}{-3}, \dfrac{−15}{3}, \dfrac{−15}{−3}$$. If you missed this problem, review Example 4.6.11.
As we’ve been graphing linear equations, we’ve seen that some lines slant up as they go from left to right and some lines slant down. Some lines are very steep and some lines are flatter. What determines whether a line slants up or down, and if its slant is steep or flat?
The steepness of the slant of a line is called the slope of the line. The concept of slope has many applications in the real world. The pitch of a roof and the grade of a highway or wheelchair ramp are just some examples in which you literally see slopes. And when you ride a bicycle, you feel the slope as you pump uphill or coast downhill.
## Use Geoboards to Model Slope
In this section, we will explore the concepts of slope.
Using rubber bands on a geoboard gives a concrete way to model lines on a coordinate grid. By stretching a rubber band between two pegs on a geoboard, we can discover how to find the slope of a line. And when you ride a bicycle, you feel the slope as you pump uphill or coast downhill.
We’ll start by stretching a rubber band between two pegs to make a line as shown in Figure $$\PageIndex{1}$$.
Figure $$\PageIndex{1}$$
Does it look like a line?
Now we stretch one part of the rubber band straight up from the left peg and around a third peg to make the sides of a right triangle as shown in Figure $$\PageIndex{2}$$. We carefully make a 90° angle around the third peg, so that one side is vertical and the other is horizontal.
Figure $$\PageIndex{2}$$
To find the slope of the line, we measure the distance along the vertical and horizontal legs of the triangle. The vertical distance is called the rise and the horizontal distance is called the run, as shown in Figure $$\PageIndex{3}$$.
Figure $$\PageIndex{3}$$
To help remember the terms, it may help to think of the images shown in Figure $$\PageIndex{4}$$.
Figure $$\PageIndex{4}$$
On our geoboard, the rise is 2 units because the rubber band goes up 2 spaces on the vertical leg. See Figure $$\PageIndex{5}$$.
What is the run? Be sure to count the spaces between the pegs rather than the pegs themselves! The rubber band goes across 3 spaces on the horizontal leg, so the run is 3 units.
Figure $$\PageIndex{5}$$
The slope of a line is the ratio of the rise to the run. So the slope of our line is $$\dfrac{2}{3}$$. In mathematics, the slope is always represented by the letter m.
##### Definition: Slope of a line
The slope of a line is m = $$\dfrac{rise}{run}$$.
The rise measures the vertical change and the run measures the horizontal change.
What is the slope of the line on the geoboard in Figure $$\PageIndex{5}$$?
$\begin{split} m &= \dfrac{rise}{run} \\ m &= \dfrac{2}{3} \\ The\; line\; &has\; slope\; \dfrac{2}{3} \ldotp \end{split}$
When we work with geoboards, it is a good idea to get in the habit of starting at a peg on the left and connecting to a peg to the right. Then we stretch the rubber band to form a right triangle.
If we start by going up the rise is positive, and if we stretch it down the rise is negative. We will count the run from left to right, just like you read this paragraph, so the run will be positive.
Since the slope formula has rise over run, it may be easier to always count out the rise first and then the run.
##### Example $$\PageIndex{1}$$:
What is the slope of the line on the geoboard shown?
Solution
Use the definition of slope.
$m = \dfrac{rise}{run}$
Start at the left peg and make a right triangle by stretching the rubber band up and to the right to reach the second peg. Count the rise and the run as shown.
The rise is 3 units. $m = \dfrac{3}{run}$
The run is 4 units. $m = \dfrac{3}{4}$
The slope is $$\dfrac{3}{4}$$.
##### Exercise $$\PageIndex{1}$$:
What is the slope of the line on the geoboard shown?
$$\frac{4}{3}$$
##### Exercise $$\PageIndex{2}$$:
What is the slope of the line on the geoboard shown?
$$\frac{1}{4}$$
##### Example $$\PageIndex{2}$$:
What is the slope of the line on the geoboard shown?
Solution
Use the definition of slope.
$m = \dfrac{rise}{run}$
Start at the left peg and make a right triangle by stretching the rubber band to the peg on the right. This time we need to stretch the rubber band down to make the vertical leg, so the rise is negative.
The rise is −1.
$m = \dfrac{−1}{run}$
The run is 3.
$\begin{split} m &= \dfrac{−1}{3} \\ m &= − \dfrac{1}{3} \end{split}$
The slope is $$− \dfrac{1}{3}$$.
##### Exercise $$\PageIndex{3}$$:
What is the slope of the line on the geoboard?
$$-\frac{2}{3}$$
##### Exercise $$\PageIndex{4}$$:
What is the slope of the line on the geoboard?
$$-\frac{4}{3}$$
Notice that in the first example, the slope is positive and in the second example the slope is negative. Do you notice any difference in the two lines shown in Figure $$\PageIndex{6}$$.
Figure $$\PageIndex{6}$$
As you read from left to right, the line in Figure A, is going up; it has positive slope. The line Figure B is going down; it has negative slope.
Figure $$\PageIndex{7}$$
##### Example $$\PageIndex{3}$$:
Use a geoboard to model a line with slope $$\frac{1}{2}$$.
Solution
To model a line with a specific slope on a geoboard, we need to know the rise and the run.
Use the slope formula. $$m = \dfrac{rise}{run}$$ Replace m with $$\dfrac{1}{2}$$. $$\dfrac{1}{2} = \dfrac{rise}{run}$$
So, the rise is 1 unit and the run is 2 units.
Start at a peg in the lower left of the geoboard. Stretch the rubber band up 1 unit, and then right 2 units.
The hypotenuse of the right triangle formed by the rubber band represents a line with a slope of $$\dfrac{1}{2}$$.
##### Exercise $$\PageIndex{5}$$:
Use a geoboard to model a line with the given slope: m = $$\dfrac{1}{3}$$.
##### Exercise $$\PageIndex{6}$$:
Use a geoboard to model a line with the given slope: m = $$\dfrac{3}{2}$$.
##### Example $$\PageIndex{4}$$:
Use a geoboard to model a line with slope $$\dfrac{−1}{4}$$.
Solution
Use the slope formula. $$m = \dfrac{rise}{run}$$ Replace m with $$− \dfrac{1}{4}$$. $$- \dfrac{1}{4} = \dfrac{rise}{run}$$
So, the rise is −1 and the run is 4.
Since the rise is negative, we choose a starting peg on the upper left that will give us room to count down. We stretch the rubber band down 1 unit, then to the right 4 units.
The hypotenuse of the right triangle formed by the rubber band represents a line whose slope is $$− \dfrac{1}{4}$$.
##### Exercise $$\PageIndex{7}$$:
Use a geoboard to model a line with the given slope: m = $$\dfrac{−3}{2}$$.
##### Exercise $$\PageIndex{8}$$:
Use a geoboard to model a line with the given slope: m = $$\dfrac{−1}{3}$$.
## Find the Slope of a Line from its Graph
Now we’ll look at some graphs on a coordinate grid to find their slopes. The method will be very similar to what we just modeled on our geoboards.
To find the slope, we must count out the rise and the run. But where do we start?
We locate any two points on the line. We try to choose points with coordinates that are integers to make our calculations easier. We then start with the point on the left and sketch a right triangle, so we can count the rise and run.
##### Example $$\PageIndex{5}$$:
Find the slope of the line shown:
Solution
Locate two points on the graph, choosing points whose coordinates are integers. We will use (0, −3) and (5, 1).
Starting with the point on the left, (0, −3), sketch a right triangle, going from the first point to the second point, (5, 1).
Count the rise on the vertical leg of the triangle. The rise is 4 units. Count the run on the horizontal leg. The run is 5 units. Use the slope formula. $$m = \dfrac{rise}{run}$$ Substitute the values of the rise and run. $$m = \dfrac{4}{5}$$
The slope of the line is $$\dfrac{4}{5}$$.
Notice that the slope is positive since the line slants upward from left to right.
##### Exercise $$\PageIndex{9}$$:
Find the slope of the line:
$$\frac{2}{5}$$
##### Exercise $$\PageIndex{10}$$:
Find the slope of the line:
$$\frac{3}{4}$$
##### HOW TO: FIND THE SLOPE FROM A GRAPH
Step 1. Locate two points on the line whose coordinates are integers.
Step 2. Starting with the point on the left, sketch a right triangle, going from the first point to the second point.
Step 3. Count the rise and the run on the legs of the triangle.
Step 4. Take the ratio of rise to run to find the slope.
$m = \dfrac{rise}{run}$
##### Example $$\PageIndex{6}$$:
Find the slope of the line shown:
Solution
Locate two points on the graph. Look for points with coordinates that are integers. We can choose any points, but we will use (0, 5) and (3, 3). Starting with the point on the left, sketch a right triangle, going from the first point to the second point.
Count the rise – it is negative. The rise is −2. Count the run. The run is 3. Use the slope formula. $$m = \dfrac{rise}{run}$$ Substitute the values of the rise and run. $$m = \dfrac{-2}{3}$$ Simplify. $$m = - \dfrac{2}{3}$$
The slope of the line is $$− \dfrac{2}{3}$$.
Notice that the slope is negative since the line slants downward from left to right.
What if we had chosen different points? Let’s find the slope of the line again, this time using different points. We will use the points (−3, 7) and (6, 1).
Starting at (−3, 7), sketch a right triangle to (6, 1).
Count the rise. The rise is −6. Count the run. The run is 9. Use the slope formula. $$m = \dfrac{rise}{run}$$ Substitute the values of the rise and run. $$m = \dfrac{-6}{9}$$ Simplify the fraction. $$m = - \dfrac{2}{3}$$
The slope of the line is $$− \dfrac{2}{3}$$.
It does not matter which points you use—the slope of the line is always the same. The slope of a line is constant!
##### Exercise $$\PageIndex{11}$$:
Find the slope of the line:
$$-\frac{4}{3}$$
##### Exercise $$\PageIndex{12}$$:
Find the slope of the line:
$$-\frac{3}{5}$$
The lines in the previous examples had y-intercepts with integer values, so it was convenient to use the y-intercept as one of the points we used to find the slope. In the next example, the y-intercept is a fraction. The calculations are easier if we use two points with integer coordinates.
##### Example $$\PageIndex{7}$$:
Find the slope of the line shown:
Solution
Locate two points on the graph whose coordinates are integers. (2, 3) and (7, 6) Which point is on the left? (2, 3)
Starting at (2, 3), sketch a right angle to (7, 6) as shown below.
Count the rise. The rise is 3. Count the run. The run is 5. Use the slope formula. $$m = \dfrac{rise}{run}$$ Substitute the values of the rise and run. $$m = \dfrac{3}{5}$$
The slope of the line is $$\dfrac{3}{5}$$.
##### Exercise $$\PageIndex{13}$$:
Find the slope of the line:
$$\frac{5}{4}$$
##### Exercise $$\PageIndex{14}$$:
Find the slope of the line:
$$\frac{3}{2}$$
## Find the Slope of Horizontal and Vertical Lines
Do you remember what was special about horizontal and vertical lines? Their equations had just one variable.
• horizontal line y = b; all the y -coordinates are the same.
• vertical line x = a; all the x -coordinates are the same.
So how do we find the slope of the horizontal line y = 4? One approach would be to graph the horizontal line, find two points on it, and count the rise and the run. Let’s see what happens in Figure $$\PageIndex{8}$$. We’ll use the two points (0, 4) and (3, 4) to count the rise and run.
Figure $$\PageIndex{8}$$
What is the rise? The rise is 0. What is the run? The run is 3. What is the slope? $$m = \dfrac{rise}{run} \tag{11.4.21}$$ $$m = \dfrac{0}{3} \tag{11.4.22}$$ $$m = 0 \tag{11.4.23}$$
The slope of the horizontal line y = 4 is 0.
All horizontal lines have slope 0. When the y-coordinates are the same, the rise is 0.
##### Definition: Slope of a Horizontal Line
The slope of a horizontal line, y = b, is 0.
Now we’ll consider a vertical line, such as the line x = 3, shown in Figure $$\PageIndex{9}$$. We’ll use the two points (3, 0) and (3, 2) to count the rise and run.
Figure $$\PageIndex{9}$$
What is the rise? The rise is 2. What is the run? The run is 0. What is the slope? $$m = \dfrac{rise}{run} \tag{11.4.24}$$ $$m = \dfrac{2}{0} \tag{11.4.25}$$
But we can’t divide by 0. Division by 0 is undefined. So we say that the slope of the vertical line x = 3 is undefined. The slope of all vertical lines is undefined, because the run is 0.
##### Definition: Slope of a Vertical Line
The slope of a vertical line, x = a, is undefined.
##### Example $$\PageIndex{8}$$:
Find the slope of each line: (a) x = 8 (b) y = −5
Solution
(a) x = 8
This is a vertical line, so its slope is undefined.
(b) y = −5
This is a horizontal line, so its slope is 0.
##### Exercise $$\PageIndex{15}$$:
Find the slope of the line: x = −4.
undefined
##### Exercise $$\PageIndex{16}$$:
Find the slope of the line: y = 7.
0
## Use the Slope Formula to find the Slope of a Line between Two Points
Sometimes we need to find the slope of a line between two points and we might not have a graph to count out the rise and the run. We could plot the points on grid paper, then count out the rise and the run, but there is a way to find the slope without graphing.
Before we get to it, we need to introduce some new algebraic notation. We have seen that an ordered pair (x, y) gives the coordinates of a point. But when we work with slopes, we use two points. How can the same symbol (x, y) be used to represent two different points?
Mathematicians use subscripts to distinguish between the points. A subscript is a small number written to the right of, and a little lower than, a variable.
• (x1, y1) read x sub 1, y sub 1
• (x2, y2) read x sub 2, y sub 2
We will use (x1, y1) to identify the first point and (x2, y2) to identify the second point. If we had more than two points, we could use (x3, y3), (x4, y4), and so on.
To see how the rise and run relate to the coordinates of the two points, let’s take another look at the slope of the line between the points (2, 3) and (7, 6) in Figure $$\PageIndex{10}$$.
Figure $$\PageIndex{10}$$
Since we have two points, we will use subscript notation.
$\begin{split} x_{1}, y_{1} \qquad &x_{2}, y_{2} \\ (2, 3) \qquad &(7, 6) \end{split}$
On the graph, we counted the rise of 3. The rise can also be found by subtracting the y-coordinates of the points.
$\begin{split} y_{2} &- y_{1} \\ 6 &- 3 \\ &\; 3 \end{split}$
We counted a run of 5. The run can also be found by subtracting the x-coordinates.
$\begin{split} x_{2} &- x_{1} \\ 7 &- 2 \\ &\; 5 \end{split}$
We know $$m = \dfrac{rise}{run} \tag{11.4.26}$$ So $$m = \dfrac{3}{5} \tag{11.4.27}$$ We rewrite the rise and run by putting in the coordinates. $$m = \dfrac{6 - 3}{7 - 2} \tag{11.4.28}$$ But 6 is the y-coordinate of the second point, y2 and 3 is the y-coordinate of the first point y1. So we can rewrite the rise using subscript notation. $$m = \dfrac{y_{2} - y_{1}}{7 - 2} \tag{11.4.29}$$ Also 7 is the x-coordinate of the second point, x2 and 2 is the x-coordinate of the first point x2. So we rewrite the run using subscript notation. $$m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}} \tag{11.4.30}$$
We’ve shown that m = $$\dfrac{y_{2} − y_{1}}{x_{2} − x_{1}}$$ is really another version of m = $$\dfrac{rise}{run}$$. We can use this formula to find the slope of a line when we have two points on the line.
##### Definition: Slope Formula
The slope of the line between two points (x1, y1) and (x2, y2) is
$m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}} \tag{11.4.31}$
Slope is y of the second point minus y of the first point
over
x of the second point minus x of the first point.
##### Example $$\PageIndex{9}$$:
Find the slope of the line between the points (1, 2) and (4, 5).
Solution
We’ll call (1, 2) point #1 and (4, 5) point #2. $$\begin{split} x_{1}, y_{1} \qquad &x_{2}, y_{2} \\ (1, 2) \qquad &(4, 5) \end{split}$$ Use the slope formula. $$m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}} \tag{11.4.32}$$ Substitute the values in the slope formula: y of the second point minus y of the first point $$m = \dfrac{5 - 2}{x_{2} - x_{1}} \tag{11.4.33}$$ x of the second point minus x of the first point $$m = \dfrac{5 - 2}{4 - 1} \tag{11.4.34}$$ Simplify the numerator and the denominator. $$m = \dfrac{3}{3} \tag{11.4.35}$$ m = 1
Let’s confirm this by counting out the slope on the graph.
The rise is 3 and the run is 3, so
$\begin{split} m &= \dfrac{rise}{run} \\ m &= \dfrac{3}{3} \\ m &= 1 \end{split}$
##### Exercise $$\PageIndex{17}$$:
Find the slope of the line through the given points: (8, 5) and (6, 3).
1
##### Exercise $$\PageIndex{18}$$:
Find the slope of the line through the given points: (1, 5) and (5, 9).
1
How do we know which point to call #1 and which to call #2? Let’s find the slope again, this time switching the names of the points to see what happens. Since we will now be counting the run from right to left, it will be negative.
We’ll call (4, 5) point #1 and (1, 2) point #2. $$\begin{split} x_{1}, y_{1} \qquad &x_{2}, y_{2} \\ (4, 5) \qquad &(1, 2) \end{split}$$ Use the slope formula. $$m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}} \tag{11.4.36}$$ Substitute the values in the slope formula: y of the second point minus y of the first point $$m = \dfrac{2 - 5}{x_{2} - x_{1}} \tag{11.4.37}$$ x of the second point minus x of the first point $$m = \dfrac{2 - 5}{1 - 4} \tag{11.4.38}$$ Simplify the numerator and the denominator. $$m = \dfrac{-3}{-3} \tag{11.4.39}$$ m = 1
The slope is the same no matter which order we use the points.
##### Example $$\PageIndex{10}$$:
Find the slope of the line through the points (−2, −3) and (−7, 4).
Solution
We’ll call (−2, −3) point #1 and (−7, 4) point #2. $$\begin{split} x_{1}, y_{1} \qquad &x_{2}, y_{2} \\ (-2, -3) \qquad &(-7, 4) \end{split}$$ Use the slope formula. $$m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}} \tag{11.4.40}$$ Substitute the values. y of the second point minus y of the first point $$m = \dfrac{4 - (-3)}{x_{2} - x_{1}} \tag{11.4.41}$$ x of the second point minus x of the first point $$m = \dfrac{4 - (-3)}{-7 - (-2)} \tag{11.4.42}$$ Simplify. $$m = \dfrac{7}{-5} \tag{11.4.43}$$ m = $$- \dfrac{7}{5}$$
Let’s confirm this on the graph shown.
$\begin{split} m &= \dfrac{rise}{run} \\ m &= \dfrac{-7}{5} \\ m &= - \dfrac{7}{5} \end{split}$
##### Exercise $$\PageIndex{19}$$:
Find the slope of the line through the pair of points: (−3, 4) and (2, −1).
-1
##### Exercise $$\PageIndex{20}$$:
Find the slope of the line through the pair of points: (−2, 6) and (−3, −4). |
Home | | Maths 5th Std | Exercise 1.2 (Area of the rectangle and square)
# Exercise 1.2 (Area of the rectangle and square)
Text Book Back Exercises Questions with Answers, Solution : 5th Maths : Term 3 Unit 1 : Geometry : Exercise 1.2 (Area of the rectangle and square)
Exercise 1.2
1. The length of the side of each square is given below. Find its area.
(i) 10 metres
(ii) 5 cm
(iii) 15 metres
(iv) 16 cm
(i) 10 metres
Side = 10m
Area of the square = a2
= 102
=100sq.m
(ii) 5 cm
Side = 5cm
Area of the square = a2
= 52
= 25 sq.cm
(iii) 15 metres
Side = 15cm
Area of the square = a2
= 152
= 225sq.m
(iv) 16 cm
Side = 16cm
Area of the square = a2
= 162
=256sq.cm
2. Find the area of the following rectangles.
(i) length = 6 cm and breadth = 3 cm
Area = 1 × b
= 6×3
= 18 sq.cm
(ii) length = 7 m and breadth = 4 m
Area = 1 × b
= 7 × 4
= 28 sq.cm
(iii) length = 8 cm and breadth = 5 cm
Area = 1 × b
= 8×5
= 40 sq.cm
(iv) length = 9 m and breadth = 6 m
Area = 1 × b
= 9 × 6
= 54 sq.cm
3. If the cost of 1 sq.m of a plot is ₹ 800, then find the total cost of the plot that is 15 m long and 10 m breadth.
Length of the plot = 15m
Breadth of the plot = 10m
Area of the plot = 1 × b
= 15×10 =150sq.m
Cost of 1 sq.m = ₹ 800
Cost of 150 sq.m = 800 × 150 = 1,20,000
Answer : Cost of 150sq.m of a plot = ₹ 1,20,000
4. The side of a square is 6 cm. The length of a rectangle is 10 cm and its breadth is 4 cm. Find the perimeter and area of both the square and rectangle.
Step :1
Side of a square = 6cm
Area of a square = a2
= 62 = 36 sq.cm
Perimeter of a square = 4a
= 4 × 6 = 24cm
Step:2
length of a rectangle = 10cm
Breadth of a rectangle = 4cm
Area of a rectangle = 1 × b
= 10 × 4 = 40 sq.cm
Perimeter of rectangle = (2 ×1) + (2 × b)
= (2 × 10) + (2 × 4)
= 20 + 8 = 28cm
Area of the square = 36 sq.cm
Perimeter of a square = 24cm
Area of a rectangle = 40sq.cm
Perimeter of a rectangle = 28cm
5. What will be the labour cost of laying the floor of an assembly hall which is 14 m long and 10 m breadth if the cost of laying is ₹ 60 per sq.m?
Length of the assembly hall = 14m
Breadth of an assembly hall = 10m
Area of a rectangle hall = 1 × b
= 14 × 10 = 140 sq.m
Cost of laying floor for 1sq.m = ₹ 60’
Cost of laying floor for 140 sq.m = 140 × 60 =8400
Answer: Cost of laying floor for 140 sq.m = ₹ 8400
Exercise 1.2
1. (i) 100 sq.m (ii) 2.5 sq.m (iii) 40 sq.cm (iv) 54 sq.m
2. (i) 18 sq.cm (ii) 28 sq.m (iii) 40 sq.cm (iv) 54 sq.m
3. 1,20,000
4. 24cm; 36 sq.cm; 28cm; 140 sq.cm
5. 8,400
Activity 2
Using the grid sheet find the area and perimeter of the rectangles and squares. Each square is 1 sq. cm.
a) Area of square = a2
a2 = 4
a = 2
Perimeter = 4a
= 4 × 2
= 8 units
b) Area of square = a2
a2 = 9
a = 3
Perimeter = 4a
= 4 × 3
= 12 units
c) Area of square = 1 × b
= 5 × 2
= 10sq.cm
Perimeter = 2 ( 1+b)
= 2 (5 +2)
= 14 units
d) Area of rectangle = 1 × b
= 5 × 3
= 15sq.cm
Perimeter =2( l +b )
= 2 (5 +3)
=16 units
Tags : Geometry | Term 3 Chapter 1 | 5th Maths , 5th Maths : Term 3 Unit 1 : Geometry
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
5th Maths : Term 3 Unit 1 : Geometry : Exercise 1.2 (Area of the rectangle and square) | Geometry | Term 3 Chapter 1 | 5th Maths |
## Graphing a Circle
Graphing circles requires two things: the coordinates of the center point, and the radius of a circle. A circle is the set of all points the same distance from a given point, the center of the circle. A radius, $r$, is the distance from that center point to the circle itself.
On a graph, all those points on the circle can be determined and plotted using coordinates.
## Circle Equations
Two expressions show how to plot a circle: the center-radius form and the standard form. Where $x$ and $y$ are the coordinates for all the circle's points, $h$ and $k$ represent the center point's $x$ and $y$ values, with $r$ as the radius of the circle
The center-radius form looks like this:
### Standard Equation of a Circle
The standard, or general, form requires a bit more work than the center-radius form to derive and graph. The standard form equation looks like this:
In the general form, $D$, $E$, and $F$ are given values, like integers, that are coefficients of the $x$ and $y$ values.
If you are unsure that a suspected formula is the equation needed to graph a circle, you can test it. It must have four attributes:
1. The $x$ and $y$ terms must be squared
2. All terms in the expression must be positive (which squaring the values in parentheses will accomplish)
3. The center point is given as , the $x$ and $y$ coordinates
4. The value for $r$, radius, must be given and must be a positive number (which makes common sense; you cannot have a negative radius measure)
The center-radius form gives away a lot of information to the trained eye. By grouping the $h$ value with the , the form tells you the $x$ coordinate of the circle's center. The same holds for the $k$ value; it must be the $y$ coordinate for the center of your circle.
Once you ferret out the circle's center point coordinates, you can then determine the circle's radius, $r$. In the equation, you may not see ${r}^{2}$, but a number, the square root of which is the actual radius. With luck, the squared $r$ value will be a whole number, but you can still find the square root of decimals using a calculator.
Try these seven equations to see if you can recognize the center-radius form. Which ones are center-radius, and which are just line or curve equations?
Only equations 1, 3, 5 and 6 are center-radius forms. The second equation graphs a straight line; the fourth equation is the familiar slope-intercept form; the last equation graphs a parabola.
## How To Graph a Circle Equation
A circle can be thought of as a graphed line that curves in both its $x$ and $y$ values. This may sound obvious, but consider this equation:
Here the $x$ value alone is squared, which means we will get a curve, but only a curve going up and down, not closing back on itself. We get a parabolic curve, so it heads off past the top of our grid, its two ends never to meet or be seen again.
Introduce a second $x$-value exponent, and we get more lively curves, but they are, again, not turning back on themselves.
The curves may snake up and down the $y$-axis as the line moves across the $x$-axis, but the graphed line is still not returning on itself like a snake biting its tail.
To get a curve to graph as a circle, you need to change both the $x$ exponent and the $y$ exponent. As soon as you take the square of both $x$ and $y$ values, you get a circle coming back unto itself!
Often the center-radius form does not include any reference to measurement units like mm, m, inches, feet, or yards. In that case, just use single grid boxes when counting your radius units.
### Center At The Origin
When the center point is the origin of the graph, the center-radius form is greatly simplified:
For example, a circle with a radius of 7 units and a center at looks like this as a formula and a graph:
## How To Graph A Circle Using Standard Form
If your circle equation is in standard or general form, you must first complete the square and then work it into center-radius form. Suppose you have this equation:
Rewrite the equation so that all your $x$-terms are in the first parentheses and $y$-terms are in the second:
You have isolated the constant to the right and added the values ${?}_{1}$ and ${?}_{2}$ to both sides. The values ${?}_{1}$ and ${?}_{2}$ are each the number you need in each group to complete the square.
Take the coefficient of $x$ and divide by 2. Square it. That is your new value for ${?}_{1}$:
Repeat this for the value to be found with the $y$-terms:
Replace the unknown values ${?}_{1}$ and ${?}_{2}$ in the equation with the newly calculated values:
Simplify:
You now have the center-radius form for the graph. You can plug the values in to find this circle with center point and a radius of $5.385$ units (the square root of 29):
### Cautions To Look Out For
In practical terms, remember that the center point, while needed, is not actually part of the circle. So, when actually graphing your circle, mark your center point very lightly. Place the easily counted values along the $x$ and $y$ axes, by simply counting the radius length along the horizontal and vertical lines.
If precision is not vital, you can sketch in the rest of the circle. If precision matters, use a ruler to make additional marks, or a drawing compass to swing the complete circle.
You also want to mind your negatives. Keep careful track of your negative values, remembering that, ultimately, the expressions must all be positive (because your $x$-values and $y$-values are squared).
### Next Lesson:
Instructor: Malcolm M.
Malcolm has a Master's Degree in education and holds four teaching certificates. He has been a public school teacher for 27 years, including 15 years as a mathematics teacher.
### 20+ Math Tutors in Ashburn, VA
Get better grades with tutoring from top-rated private tutors. Local and online.
15 chapters | 149 lessons
Tutors online
### Find a math tutor in Ashburn, VA
Learn faster with a math tutor. Find a tutor locally or online. |
# Solve a linear-quadratic system by graphing
## Presentation on theme: "Solve a linear-quadratic system by graphing"— Presentation transcript:
Solve a linear-quadratic system by graphing
EXAMPLE 1 Solve a linear-quadratic system by graphing Solve the system using a graphing calculator. y2 – 7x + 3 = 0 Equation 1 2x – y = 3 Equation 2 SOLUTION STEP 1 Solve each equation for y. y2 – 7x + 3 = 0 2x – y = 3 y2 = 7x – 3 – y = – 2x + 3 y = x – 3 Equation 1 y = 2x – 3 Equation 2
EXAMPLE 1 Solve a linear-quadratic system by graphing STEP 2 Graph the equations y = y = and y = 2x – 3 7x – 3, Use the calculator’s intersect feature to find the coordinates of the intersection points. The graphs of and y = 2x – 3 intersect at (0.75, 21.5). The graphs of and y = 2x – 3 intersect at (4, 5). y = – 7x – 3, y = 7x – 3,
EXAMPLE 1 Solve a linear-quadratic system by graphing ANSWER The solutions are (0.75, – 1.5) and (4, 5). Check the solutions by substituting the coordinates of the points into each of the original equations.
Solve a linear-quadratic system by substitution
EXAMPLE 2 Solve a linear-quadratic system by substitution Solve the system using substitution. x2 + y2 = 10 Equation 1 y = – 3x + 10 Equation 2 SOLUTION Substitute –3x + 10 for y in Equation 1 and solve for x. x2 + y2 = 10 Equation 1 x2 + (– 3x + 10)2 = 10 Substitute for y. x2 + 9x2 – 60x = 10 Expand the power. 10x2 – 60x + 90 = 0 Combine like terms. x2 – 6x + 9 = 0 Divide each side by 10. (x – 3)2 = 0 Perfect square trinomial x = 3 Zero product property
EXAMPLE 2 Solve a linear-quadratic system by substitution To find the y-coordinate of the solution, substitute x = 3 in Equation 2. y = – 3(3) + 10 = 1 ANSWER The solution is (3, 1). CHECK You can check the solution by graphing the equations in the system. You can see from the graph shown that the line and the circle intersect only at the point (3, 1).
Step each equation for y.
GUIDED PRACTICE for Examples 1 and 2 1. x2 + y2 = 13 y = x – 1 SOLUTION x2 + y2 = 13 STEP 1 Equation 1 y = x – 1 Equation 2 Step each equation for y. x2 + y2 = 13 y2 = 13 – x2 + y = 13 – x2 y = x – 1 Equation 1 Equation 2
GUIDED PRACTICE for Examples 1 and 2 y = x – 1. Graph the equation y = 13 – x2 , and STEP 2 Use the calculator’s intersect feature to find the coordinates of the intersection points. The graphs of y = 13 – x2 and y = x – 1. intersect at (3, 2). The graphs (–2,–3). The solutions are (3,2) and(–2, –3). Check the solutions by substituting the coordinates of the points into each of the original equations.
Solve each equation for y.
GUIDED PRACTICE for Examples 1 and 2 2. x2 + 8y2 – 4 = 0 y = 2x + 7 STEP 1 Solve each equation for y. 8y2 = – x2 + 4 y = 2x + 1 y = x2 + 4 8 Equation 1 Equation 2 STEP 2 Graph the equation and y = 2x + 7. y = x2 + 4 8 Use the calculator’s intersect feature to find the coordinates of the intersection points. The graphs does not intersect at any point so, no solutions.
Solve each equation for y.
GUIDED PRACTICE for Examples 1 and 2 3. y2 + 6x – 1 = 0 y = – 04x + 2.6 STEP 1 Solve each equation for y. y2 + 6x – 1 = 0 y = – 0.4x + 2.6 y = + – 6 x +1 Equation 1 Equation 2 STEP 2 Graph the equation y = – 6 x +1 , and y = – 0.4x + 2.6 Use the calculator’s intersect feature to find the coordinates of the intersection points. The graphs intersect at the points (–1.57, 3.23) and (–22.9, 11.8).
Substitute 0.5x – 3 for y in Equation 2 and solve for x.
GUIDED PRACTICE for Examples 1 and 2 4. y = 0.5x – 3 x2 + 4y2 – 4 = 0 SOLUTION Substitute 0.5x – 3 for y in Equation 2 and solve for x. x2 + 4y2 – 4 = 0 Equation 2 x2 + 4 (0.5x – 3)2 – 4 = 0 Substitute for y. x2 + y (0.25x2 – 3x + 9) – 4 = 0 Expand the power. 2x2 – 12x + 32 = 0 Combine like terms. x2 – 6x + 16 = 0 Divide each side by 2. This equation has no solution.
Substitute – x – 1 for y in Equation 1 and solve for x.
GUIDED PRACTICE for Examples 1 and 2 5. y2 – 2x – 10 = 0 y = x 1 SOLUTION Substitute – x – 1 for y in Equation 1 and solve for x. y2 – 2x2 – 10 = 0 Equation 1 (– x – 1)2 – 2x – 10 = 0 Substitute for y. x x – 2x – 10 = 0 Expand the power. x2 – 9 = 0 Combine like terms. x2 = 9 Add 9 to each side. x = ±3 simplify.
GUIDED PRACTICE for Examples 1 and 2 To find the y-coordinate of the solution, substitute x = 3 and x = 3 in equation 2. y = –3 –1 = – 4 y = 3 –1 = 2 The solutions are (3, –4), and (–3, 2) ANSWER
Substitute 4x – 8 for y in Equation 2 and solve for x.
GUIDED PRACTICE for Examples 1 and 2 6. y = 4x – 8 9x2 – y2 – 36 = 0 SOLUTION Substitute 4x – 8 for y in Equation 2 and solve for x. 9x2 – y2 – 36 = 0 Equation 2 9x2 – (4x – 8)2 – 36 = 0 Substitute for y. 9x2 –(16x2 – 64x + 64) – 36 = 0 Expand the power. – 7x2 + 64x – 100 = 0 Combine like terms. 7x2 – 64x = 0 Divide each side by –1. (7x – 50) (x –2) = 0 simplify.
To find the y-coordinate of the solution, substitute.
GUIDED PRACTICE for Examples 1 and 2 7x – 50 = 0 or x – 2 = 0 x = 7 50 or x = 2 To find the y-coordinate of the solution, substitute. x = 7 50 and x = 2 in Equation 1 y = – 8 = 7 50 144 y = 4(2) – 8 = 0 The solutions are (2, 0), and , ANSWER 7 50 144 |
# AP 10th Class Maths 6th Chapter Triangles Exercise 6.3 Solutions
Well-designed AP 10th Class Maths Textbook Solutions Chapter 6 Triangles Exercise 6.3 offers step-by-step explanations to help students understand problem-solving strategies.
## Triangles Class 10 Exercise 6.3 Solutions – 10th Class Maths 6.3 Exercise Solutions
Question 1.
State which pairs of triangles in given figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
i)
Solution:
In ∆ABC and ∆PQR
∠A = ∠P = 60°
∠B = ∠Q = 80°
∠C = ∠R = 40°
By Angle-Angle-Angle criterion of similarity
∆ABC ~ ∆PQR.
ii)
Solution:
In ∆ABC and ∆QRP
By side-side-side criterion of similarity ∆ABC ~ ∆QRP.
iii)
Solution:
In ∆LMP and ∆DEF
$$\frac{\mathrm{LM}}{\mathrm{DE}}$$ = $$\frac{2.7}{4}$$
$$\frac{\mathrm{MP}}{\mathrm{EF}}$$ = $$\frac{2}{5}$$
$$\frac{\mathrm{PL}}{\mathrm{DF}}$$ = $$\frac{3}{6}$$ = $$\frac{1}{2}$$
$$\frac{2.7}{4}$$ ≠ $$\frac{2}{5}$$ ≠ $$\frac{1}{2}$$
So, $$\frac{\mathrm{LM}}{\mathrm{DE}}$$ ≠ $$\frac{\mathrm{MP}}{\mathrm{EF}}$$ ≠ $$\frac{\mathrm{PL}}{\mathrm{DF}}$$
Therefore, ∆LMP ~ ∆DEF.
iv)
Answer:
In ∆NML and ∆PQR
$$\frac{\mathrm{NM}}{\mathrm{PQ}}$$ $$=\frac{2.5}{5}$$ = $$\frac{1}{2}$$
∠M = ∠Q = 70°
$$\frac{\mathrm{NM}}{\mathrm{PQ}}$$ = $$\frac{\mathrm{ML}}{\mathrm{QR}}$$ = $$\frac{1}{2}$$ and ∠M = ∠Q = 70°
Therefore, side-angle-side criterion of similarity, ∆NML ~ ∆PQR.
v)
Solution:
In ∆ABC and ∆DEF
∠A = ∠F = 80°
$$\frac{\mathrm{AB}}{\mathrm{DE}}$$ = $$\frac{2.5}{7}$$
$$\frac{\mathrm{BC}}{\mathrm{EF}}$$ = $$\frac{3}{6}$$ = $$\frac{1}{2}$$
$$\frac{\mathrm{AC}}{\mathrm{DF}}$$ = $$\frac{x}{5}$$
$$\frac{\mathrm{AB}}{\mathrm{DE}}$$ ≠ $$\frac{\mathrm{BC}}{\mathrm{EF}}$$ ≠ $$\frac{\mathrm{AC}}{\mathrm{DF}}$$ and ∠A = ∠F = 80°
Therefore, ∆ABC ~ ∆DEF
vi)
Solution:
In ∆DEF,
∠D + ∠E + ∠F = 180°
70° + 80° + ∠F = 180°
∠F = 180° – 150° = 30°
In ∆PQR,
∠P + ∠Q + ∠R = 180°
∠P + 80° + 30° = 180°
∠P = 180° – 110° = 70°
∠D = ∠P = 70°
∠E = ∠Q = 80°
∠F = ∠R = 30°
Therefore, Angle-Angle-Angle criterion of similarity ∆DEF ~ ∆PQR.
Question 2.
In figure, ∆ODC ~ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
Solution:
Given ∆ODC ~ ∆OBA
∠BOC = 125° and ∠CDO = 70°
∠BOC + ∠COD = 180°
∠COD = 180° – ∠BOC
∠COD = 180° – 125° = 55°
∠AOB = ∠COD = 55°
(Vertically opposite angles)
∠OAB + ∠OBA + ∠AOB = 180°
∠OAB + 70° + 55° = 180°
∠OAB= 180° – 125° = 55°
Therefore, ∠DOC = 55°,
∠DCO = ∠OAB = 55°
∠OAB = 55°.
Question 3.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that $$\frac{\mathrm{OA}}{\mathrm{OC}}$$ = $$\frac{\mathrm{OB}}{\mathrm{OD}}$$.
Solution:
Given in ABCD trapezium, AB || DC and diagonals AC, BD intersect at ‘O’.
AB || DC.
So, ∠OAB = ∠OCD
(Alternate interior angles)
∠AOB = ∠COD (Vertically opposite angles)
∠OBA = ∠ODC (Alternate interior angles)
In ∆AOB and ∆COD,
∠OAB = ∠OCD
∠AOB = ∠COD
∠OBA = ∠ODC
By Angle-Angle-Angle criterion of similarity
∆MOB ~ ∆COD.
If two triangles are similar, then their corresponding sides are in proportion.
Therefore, $$\frac{\mathrm{OA}}{\mathrm{OC}}$$ = $$\frac{\mathrm{OB}}{\mathrm{OD}}$$.
Question 4.
In given figure, $$\frac{\mathbf{Q R}}{\mathbf{Q S}}$$ = $$\frac{\text { QT }}{\text { PR }}$$ and ∠1 = ∠2. Show that ∆PQS ~ ∆TQR.
Answer:
Given $$\frac{\mathrm{QR}}{\mathrm{QT}}$$ = $$\frac{\mathrm{QS}}{\mathrm{PR}}$$
$$\frac{\mathrm{QR}}{\mathrm{QT}}$$ = $$\frac{\mathrm{QS}}{\mathrm{PR}}$$
⇒ $$\frac{\mathrm{QT}}{\mathrm{QR}}$$ = $$\frac{\mathrm{PR}}{\mathrm{QS}}$$ → (1)
But we know, in ∆PQR, sides opposite to equal angles are equal.
∴ PQ = QR → (2)
Put (2) in (1) ⇒ $$\frac{Q T}{Q R}$$ = $$\frac{\mathrm{PQ}}{\mathrm{QS}}$$
⇒ $$\frac{\mathrm{PQ}}{\mathrm{QT}}$$ = $$\frac{\mathrm{QS}}{\mathrm{QR}}$$
Thus in ∆PQR and ∆TQR,
$$\frac{\mathrm{PQ}}{\mathrm{QT}}$$ = $$\frac{\mathrm{QS}}{\mathrm{QR}}$$ and ∠PQS = ∠TQR
By Side-Angle-Side criterion similarity
∆PQS ~ ∆TQR.
Question 5.
S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.
Solution:
Given in ∆RPQ and ∆RTS
∠P = ∠RTS (Angle)
∠R = ∠PRQ (Angle)
By Angle-Angle-Angle similarity
∆RPQ ~ ∆RTS.
Question 6.
In given figure, if ∆ABE $$\cong$$ ∆ACD, show that ∆ADE ~ ∆ABC.
Solution:
Given ∆ABE $$\cong$$ ∆ACD
If two triangles are congruent, then their corresponding sides and angles are equal.
That is AB = AC (Side)
BE = CD (Side)
AE = AD (Side)
In ∆ADE and ∆ABC,
$$\frac{\mathrm{AB}}{\mathrm{AC}}$$ = $$\frac{\mathrm{AD}}{\mathrm{DE}}$$ and ∠BAC = ∠DAE.
By Side-Angle-Side similarity criterion
∆ADE ~ ∆ABC.
Question 7.
In the given figure, altItudes AD and CE of ∆ABC intersect each other at the point P. Show that:
i) ∆AEP ~ ∆CDP
ii) ∆ABD ~ ∆CBE
iii) ∆AEP ~ ∆ADB
iv) ∆PDC ~ ∆BEC
Solution:
Given AD and CE are altitudes of ∆ABC intersect at P.
i) In ∆AEP and ∆CDP,
∠AEP = ∠CDP = 90°
∠APE = ∠CPD (Vertically opposite angles)
By Angle-Angle criterion of similarity,
∆AEP ~ ∆CDP.
ii) In ∆ABD and ∆CBE
∠ADB = ∠CEB = 90°
∠ABD = ∠CBE (Common angle)
By Angle-Angle criterion of similarity,
∆ABD ~ ∆CBE.
iii) In ∆AEP ~ ∆ADB
∠AEP = ∠ADB = 90°
∠PAE = ∠DAB (Common angle)
By Angle-Angle criterion of similarity,
∆AEP ~ ∆ADB.
iv) In ∆PDC ~ ∆BEC
∠CDP = ∠BEC = 90°
∠PCD = ∠ECB (Common angle)
By Angle-Angle criterion of similarity,
∆APDC ~ ∆BEC.
Question 8.
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ~ ∆CFB.
Solution:
Given E is a point on the side AD produced of a parallelogram ABCD.
BE and CD intersect at F.
In ∆ABE and ∆CFB.
∠A = ∠C (Opposite angles are equal)
∠AEB = ∠CBF (Alternate interior angles)
By Angle-Angle criterion of similarity,
∆ABE ~ ∆CFB.
Question 9.
In given figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
i) ∆ABC ~ ∆AMP
ii) $$\frac{\mathrm{CA}}{\mathrm{PA}}$$ = $$\frac{\mathrm{BC}}{\mathrm{MP}}$$
Solution:
i) Given ∆ABC and ∆AMP are two right triangles.
∠B = ∠M = 90°
In ∆ABC and ∆AMP,
∠ABC = ∠AMP = 90°
∠CAB = ∠MAP (Common angle)
By Angle-Angle criterion of similarity
∆ABC ~ ∆AMP
ii) If two triangles are similar, then their corresponding sides are in proportion.
Therefore, $$\frac{\mathrm{CA}}{\mathrm{PA}}$$ = $$\frac{\mathrm{BC}}{\mathrm{MP}}$$.
Question 10.
CD and GH are respectively the bisec-tors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC ~ ∆FEG, show that:
i) $$\frac{\mathbf{C D}}{\mathbf{G H}}$$ = $$\frac{A C}{F G}$$
ii) ∆DCB ~ ∆HGE
iii) ∆DCA ~ ∆HGF
Solution:
Given CD and GH are bisectors of ∠ACB and ∠EGF and ∆ABC – ∆FEG.
If two triangles are similar, then their corresponding sides are in proportion and corresponding angles are equal.
i.e., ∠A = ∠F, ∠B = ∠E and ∠C = ∠G
i) If two triangles are similar, then their corresponding sides are in proportion.
$$\frac{\mathrm{AC}}{\mathrm{FG}}$$ = $$\frac{\mathrm{CD}}{\mathrm{GH}}$$
ii) In ∆DCB and ∆HGE
$$\frac{\angle C}{2}$$ = $$\frac{\angle G}{2}$$
∠BCD – ∠HGE (Angle)
∠B = ∠E
By Angle-Angle criterion of similarity
∆ADCB ~ ∆HGE.
iii) In ∆DCA and ∆HGF
∠A = ∠F (Angle)
∠ACD = ∠FGH (Angle)
By Angle-Angle criterion of similarity
∆DCA ~ ∆HGF.
Question 11.
In given figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF.
Solution:
Given in ∆ABC, AB = AC, AD ⊥ BC and EF ⊥ AC.
If AB = AC, then ∠B = ∠C.
Angles which are equal to the opposite sides are equal.
In ∆ABD, ∆ECF
∠ADB = ∠EFC = 90° (Angle)
∠ABD = ∠ECF (Angle)
By Angle-Angle criterion of similarity
∆ABD ~ ∆ECF.
Question 12.
Sides AB and BC and median AD of a triangle ABC are respectively propor-tional to sides PQ and QR and medi-an PM of ∆PQR (see figure). Show that ∆ABC ~ ∆PQR.
Solution:
Given $$\frac{\mathrm{AB}}{\mathrm{PQ}}$$ = $$\frac{\mathrm{BC}}{\mathrm{QR}}$$ = $$\frac{\mathrm{AD}}{\mathrm{PM}}$$
∆ABC and ∆PQR, where AD and PM are medians.
In ∆ABD and ∆PQM,
By side-side-side criterion of similarity
∆ABD ~ ∆PQM.
If two triangles are similar, then their corresponding sides are proportion and corresponding angles are equal.
∠ABD = ∠PQM
Now, in ∆ABC and ∆PQR
$$\frac{\mathrm{AB}}{\mathrm{PQ}}$$ = $$\frac{\mathrm{BD}}{\mathrm{QM}}$$ and ∠B = ∠Q
By Side-Angle-Side criterion of similarity
∆ABC ~ ∆PQR.
Question 13.
D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD.
Solution:
Given in ∆ABC, ∠ADC = ∠BAC.
In ∆ABC and ∆DAC
∠BAC = ∠CDA
∠ACB = ∠DCA (Common angle)
By Angle-Angle criterion of similarity
∆ABC ~ ∆DAC.
If two triangles are similar, then their corresponding sides are proportion.
$$\frac{\mathrm{BC}}{\mathrm{AC}}$$ = $$\frac{\mathrm{AC}}{\mathrm{DC}}$$
AC∙AC = CB∙CD
Therefore, AC2 ≠ CB∙CD.
Question 14.
Sides AB and AC and median AD of a triangle ABC are respectively propor-tional to sides PQ and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆PQR.
Solution:
In ∆ABC and ∆PQR,
$$\frac{\mathrm{AB}}{\mathrm{PQ}}$$ = $$\frac{\mathrm{AC}}{\mathrm{PR}}$$ = $$\frac{\mathrm{AD}}{\mathrm{PM}}$$
Extend AD to DE such that AD = DE and join BE and CE.
Similarly, extend PM to ML such that PM = ML and join QL and RL.
In quadrilateral ABEC diagonals bisect each other. So, ABEC, is a parallelogram. Similarly, quadrilateral PQLR also a parallelogram.
By side-side-side criterion of similarity
∆ABE ~ ∆PQL
So, ∠BAE = ∠QPL → (1)
Similarly, ∆CAE, ∆PLR
$$\frac{\mathrm{AC}}{\mathrm{PR}}$$ = $$\frac{\mathrm{AD}}{\mathrm{PM}}$$ = $$\frac{\mathrm{CE}}{\mathrm{LR}}$$
$$\frac{\mathrm{AC}}{\mathrm{PR}}$$ = $$\frac{2 \mathrm{AD}}{2 \mathrm{PM}}$$ = $$\frac{\mathrm{CE}}{\mathrm{LR}}$$
Therefore, $$\frac{\mathrm{AC}}{\mathrm{PR}}$$ = $$\frac{A E}{P L}$$ = $$\frac{\mathrm{CE}}{\mathrm{LR}}$$
By side-side-side criterion of similarity
∆CAE ~ ∆PLR
So, ∠CAE = ∠RPL → (2)
By adding (1) and (2)
∠BAE + ∠CAE = ∠QPL + ∠RPL
∠A = ∠P
In ∆ABC, ∆PQR,
$$\frac{A B}{P Q}$$ = $$\frac{A C}{P R}$$ and ∠A = ∠P
By Side-Angle-Side criterion of similarity
∆ABC ~ ∆PQR.
Question 15.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution:
Let AB is the length of the pole is 6 m.
and length of its shadow BC = 4 m.
DE is the height of the tower = x m.
Length of its shadow EF = 24 m.
In ∆ABC, ∆DEF
∠B = ∠E = 90°
∠ACB = ∠DFE (sun rays can make equal angles)
By Angle-Angle criterion of similarity
∆ABC ~ ∆DEF
then, corresponding sides are in pro-portion.
⇒ x = 6 × 7 = 42 m.
Therefore, height of the tower is 42 m.
Question 16.
If AD and PM are medians of triangles ABC and PQR, respectively where ∆ABC ~ ∆PQR, prove that $$\frac{A B}{P Q}$$ = $$\frac{A D}{P M}$$.
Solution:
Given ∆ABC ~ ∆PQR, AD and PM are medians of triangles ABC and PQR.
So, ∠B = ∠Q (Corresponding angles)
and $$\frac{A B}{P Q}$$ = $$\frac{B C}{Q R}$$ = $$\frac{A C}{P R}$$
By side-side-side criterion of similarity
∆ABD ~ ∆PQM
Corresponding sides are in proportion.
Therefore, $$\frac{A B}{P Q}$$ = $$\frac{A D}{P M}$$. |
# Solve by Factoring 5/(3x+2)=3/(2x)
Move to the left side of the equation by subtracting it from both sides.
Simplify .
To write as a fraction with a common denominator, multiply by .
To write as a fraction with a common denominator, multiply by .
Write each expression with a common denominator of , by multiplying each by an appropriate factor of .
Multiply and .
Multiply and .
Reorder the factors of .
Combine the numerators over the common denominator.
Simplify the numerator.
Multiply by .
Apply the distributive property.
Multiply by .
Multiply by .
Subtract from .
To write as a fraction with a common denominator, multiply by .
To write as a fraction with a common denominator, multiply by .
Write each expression with a common denominator of , by multiplying each by an appropriate factor of .
Multiply and .
Multiply and .
Reorder the factors of .
Combine the numerators over the common denominator.
Simplify the numerator.
Multiply by .
Apply the distributive property.
Multiply by .
Multiply by .
Subtract from .
Find the LCD of the terms in the equation.
Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.
Since contain both numbers and variables, there are four steps to find the LCM. Find LCM for the numeric, variable, and compound variable parts. Then, multiply them all together.
Steps to find the LCM for are:
1. Find the LCM for the numeric part .
2. Find the LCM for the variable part .
3. Find the LCM for the compound variable part .
4. Multiply each LCM together.
The LCM is the smallest positive number that all of the numbers divide into evenly.
1. List the prime factors of each number.
2. Multiply each factor the greatest number of times it occurs in either number.
Since has no factors besides and .
is a prime number
The number is not a prime number because it only has one positive factor, which is itself.
Not prime
The LCM of is the result of multiplying all prime factors the greatest number of times they occur in either number.
The factor for is itself.
occurs time.
The LCM of is the result of multiplying all prime factors the greatest number of times they occur in either term.
The factor for is itself.
occurs time.
The LCM of is the result of multiplying all factors the greatest number of times they occur in either term.
The Least Common Multiple of some numbers is the smallest number that the numbers are factors of.
Multiply each term by and simplify.
Multiply each term in by in order to remove all the denominators from the equation.
Simplify .
Rewrite using the commutative property of multiplication.
Cancel the common factor of .
Factor out of .
Cancel the common factor.
Rewrite the expression.
Cancel the common factor of .
Cancel the common factor.
Rewrite the expression.
Simplify .
Apply the distributive property.
Simplify the expression.
Multiply by .
Multiply by .
Multiply by .
Multiply by .
Add to both sides of the equation.
Solve by Factoring 5/(3x+2)=3/(2x)
Scroll to top |
# GRE Math : How to find an angle in a parallelogram
## Example Questions
### Example Question #1 : How To Find An Angle In A Parallelogram
In a given parallelogram, the measure of one of the interior angles is 25 degrees less than another. What is the approximate measure rounded to the nearest degree of the larger angle?
Possible Answers:
101 degrees
102 degrees
103 degrees
78 degrees
77 degrees
Correct answer:
103 degrees
Explanation:
There are two components to solving this geometry puzzle. First, one must be aware that the sum of the measures of the interior angles of a parallelogram is 360 degrees (sum of the interior angles of a figure = 180(n-2), where n is the number of sides of the figure). Second, one must know that the other two interior angles are doubles of those given here. Thus if we assign one interior angle as x and the other as x-25, we find that x + x + (x-25) + (x-25) = 360. Combining like terms leads to the equation 4x-50=360. Solving for x we find that x = 410/4, 102.5, or approximately 103 degrees. Since x is the measure of the larger angle, this is our answer.
### Example Question #2 : How To Find An Angle In A Parallelogram
Figure is a parallelogram.
Quantity A: The largest angle of .
Quantity B:
Which of the following is true?
Possible Answers:
Quantity A is larger.
Quantity B is larger.
The two quantities are equal.
The relationship cannot be determined.
Correct answer:
The two quantities are equal.
Explanation:
By using the properties of parallelograms along with those of supplementary angles, we can rewrite our figure as follows:
Recall, for example, that angle is equal to:
, hence
Now, you know that these angles can all be added up to . You should also know that
Therefore, you can write:
Simplifying, you get:
Now, this means that:
and . Thus, the two values are equal.
### Example Question #3 : How To Find An Angle In A Parallelogram
Figure is a parallelogram.
What is in the figure above?
Possible Answers:
Cannot be computed from the data given.
Correct answer:
Explanation:
Because of the character of parallelograms, we know that our figure can be redrawn as follows:
Because it is a four-sided figure, we know that the sum of the angles must be . Thus, we know:
Solving for , we get:
Tired of practice problems?
Try live online GRE prep today. |
# Limit of sinx/x as x approaches 0: Formula, Proof
The limit of sinx/x as x approaches 0 is 1, that is, the limit formula of sinx/x when x tends to 0 is given by
limx→0 $\dfrac{\sin x}{x} = 1$.
Let us now prove that the limit of sinx/x is equal to 1 when x goes to 0.
## Limit of sinx/x as x→0 by Sandwich or Squeeze Theorem
Prove that limx→0 $\dfrac{\sin x}{x} = 1$.
Solution:
We know that for all real values of x,
sin x ≤ x and x ≤ tan x.
This implies that
sin x ≤ x ≤ tan x
Dividing by sinx we get that
$1 \leq \dfrac{x}{\sin x} \leq \dfrac{\tan x}{\sin x}$
⇒ $1 \leq \dfrac{x}{\sin x} \leq \dfrac{1}{\cos x}$ as we know tanx=sinx/cosx.
Now, taking the limit x→0 on both sides we get that
limx→0 1 ≤ limx→0 $\dfrac{x}{\sin x}$ ≤ limx→0 $\dfrac{1}{\cos x}$
⇒ 1 ≤ limx→0 $\dfrac{x}{\sin x}$ ≤ $\dfrac{1}{\cos 0}$
⇒ 1 ≤ limx→0 $\dfrac{x}{\sin x}$ ≤ 1 as the value of cos0 is 1.
So by the Squeeze theorem of limits, we conclude that
limx→0 $\dfrac{x}{\sin x}$ = 1
By the division rule of limits,
limx→0 $\dfrac{\sin x}{x}$ = 1.
So the value of the limit of sinx/x is equal to 1 when x tends to 0, and it is obtained by the Squeeze theorem (or Sandwich theorem) of limits.
Main Article: Limit: definition, formulas and examples
## Lim x→0 sinx/x by l’Hopital’s Rule
To find the value of limx→0 sinx/x, first notice that $\dfrac{\sin 0}{0}=\dfrac{0}{0}$, so the given limit is an indeterminate form. Thus, by l’Hopital’s rule, we have that
limx→0 $\dfrac{\sin x}{x}$
= limx→0 $\dfrac{(\sin x)’}{x’}$ where $’$ denoted the first order derivative with respect to x.
= limx→0 $\dfrac{\cos x}{1}$
= limx→0 cosx
= cos 0
= 1.
Hence the limit of sinx/x by l’Hopital’s rule is equal to 1 when x approaches 0.
You can read: Proofs of all Limit Properties
Proofs of all Limit formulas
## Solved Examples
Now, as an application of the limit formula of sinx/x when x tends to zero, we will evaluate a few limits.
limx→0 $\dfrac{\sin 2x}{x}$
= limx→0 $\Big( \dfrac{\sin 2x}{2x} \times 2 \Big)$
=2 limx→0 $\dfrac{\sin 2x}{2x}$
[Let t=3x. Then t→0 when x→0]
=2 limx→0 $\dfrac{\sin t}{t}$
=2 × 1 as the limit of sinx/x is 1 by above when x→0.
=2
So the limit of sin2x/x is equal to 2 when x tends to 0.
limx→0 $\dfrac{\sin 3x}{x}$
= limx→0 $\Big( \dfrac{\sin 3x}{2x} \times 3 \Big)$
=3 limt→0 $\dfrac{\sin t}{t}$, where t=3x, so that t→0 when x→0
=3 × 1 = 3.
Therefore, the limit of sin3x/x is equal to 3 when x tends to 0.
limx→0 $\dfrac{\sin 3x}{\sin 2x}$
= limx→0 $\Big( \dfrac{3}{2} \times \dfrac{\sin 3x}{3x} \times \dfrac{2x}{\sin 2x} \Big)$
= $\dfrac{3}{2}$ limx→0 $\dfrac{\sin 3x}{3x}$ limx→0 $\dfrac{2x}{\sin 2x}$ by the properties of limits.
[Let t=3x and z=2x. Then both t, z→0 when x→0]
= $\dfrac{3}{2}$ × 1 × $\dfrac{1}{\lim\limits_{z \to 0}\frac{\sin z}{z}}$
= $\dfrac{3}{2}$ × 1 × $\dfrac{1}{1}$
= $\dfrac{3}{2}$.
So the limit of sin3x/2x is equal to 3/2 when x tends to 0.
## FAQs
Q1: What is the limit of sinx/x when x→0?
Answer: The limit of sinx/x when x→0 is equal to 1. In other words, limx→0 sinx/x = 1.
Q2: What is limx→0 sinx/x?
Answer: The value of limx→0 sinx/x is equal to 1.
Q3: What is the limit of sin2x/x when x→0?
Answer: The limit of sin2x/x when x→0 is equal to 2. That is, limx→0 sin2x/x = 1.
Q4: What is the limit of sin3x/x when x→0?
Answer: The limit of sin3x/x when x→0 is equal to 3. That is, limx→0 sin3x/x = 1.
Q5: What is the limit of sin2x/sin3x when x→0?
Answer: The limit of sin2x/sin3x is equal to 2/3 when x tends to 0.
Q3: What is the limit of sin3x/2x when x→0?
Answer: The limit of sin3x/sin2x is equal to 3/2 when x tends to 0.
Share via: |
# Number Formed by Any Power – Definition, Rules, Examples | How to Raise a Number by any Power?
A power tells how many times the base is used as a factor. The number formed by any power will be the multiple of the given number. The sum of the digits of the product will be equal to the factors of the given number. Before you start solving the problems students must remember that there are some rules to find a product. Know the Procedure on how to raise a number by any power, examples of numbers formed by any power, etc.
## How to Raise a Number by any Power?
Follow the simple guidelines listed below to obtain Number Formed by any Power. They are as follows
• Firstly, identify the base and power in the given number.
• Later, write down the base as many times as the power and place a multiplication symbol in between them.
• Multiply the numbers and find the result.
### Rules on Numbers Raised by any Powers
1. Any number raised to the power of zero, except zero, equals one.
a0Â = 1
2. Any number raised to the power of one equals the number itself.
a¹ = a
3. When bases are equal powers should be added.
### Examples of Number Raised by any Power
Example 1.
Write the number 3 formed by any power.
Solution:
We have to write the number formed any power for 3.
Let us take random multiples of 3 to check the result.
First multiple any number with 3 and then add the result and check whether it is multiple of 3 like 3, 6, 9.
3¹ = 3 → 3 + 0 = 3 → 3 is a multiple of 3.
3³ = 27 → 2 + 7 = 9 → 9 is a multiple of 3.
3 × 2 = 6 → 6 is a multiple of 3.
3 × 10 = 30 → 3 + 0 = 3 → 3 is a multiple of 3.
3 × 18 = 54 → 5 + 4 = 9 → 9 is a multiple of 3.
3 × 42 = 126 → 1 + 2 + 6 = 9 → 9 is a multiple of 3.
We observe that the number formed by any power is 3, 6, 9.
The sum of the digits of a number formed by 3 will be the multiples of 3.
Example 2.
Write the number 9 formed by any power.
Solution:
We have to write the number formed any power for 9.
Let us take random multiples of 9 to check the result.
First multiple any number with 9 and then add the result and check whether it is a multiple of 9.
9¹ = 9
9² = 9 × 9 = 81 = 8 + 1 = 9 → 9 is a multiple of 9.
9³ = 9 × 9 × 9 = 729 = 7 + 2 + 9 = 18 = 1 + 8 = 9 → 9 is a multiple of 9.
94 = 9 × 9 × 9 × 9 = 6561 = 6 + 5 + 6 + 1 = 18 = 1 + 8 = 9 → 9 is a multiple of 9.
9 × 12 = 108 = 1 + 0 + 8 = 9 → 9 is a multiple of 9.
9 × 428 = 3852 = 3 + 8 + 5 + 2 = 18 = 1 + 8 = 9 → 9 is a multiple of 9.
We observe that the number formed by any power is 9.
The sum of the digits of a number formed by any power for 9 is 9.
Example 3.
Write the number 5 formed by any power.
Solution:
We have to write the number formed any power for 5.
Let us take random multiples of 5 to check the result.
Multiply the numbers with any power and then check whether it is a multiple of 5.
We know that the multiplies of 5 will have 0 or 5 in its last digit.
5¹ = 5 → 5 is a multiple of 5.
5² = 5 × 5 = 25 → 5 is a multiple of 5.
5³ = 5 × 5 × 5 = 125 → 5 is a multiple of 5.
56 = 5 × 5 × 5 × 5 × 5 × 5 = 15625 → 5 is a multiple of 5.
All the values above are having 5 in their last digit which means the above numbers are multiples of 5.
Example 4.
Write the number 2 formed by any power.
Solution:
We have to write the number formed any power for 2.
Let us take random multiples of 2 to check the result.
Multiply the numbers with any power and then check whether it is a multiple of 2.
We know that the multiplies of 2 will have 0, 2, 4, 6, 8 at its last digit.
2¹ = 2 → 2 is a multiple of 2.
2² = 2 × 2 = 4 → 4 is a multiple of 2.
2³ = 2 × 2 × 2 = 8 → 8 is a multiple of 2.
2 × 126 = 252 = 252 is a multiple of 2.
All the values above are having 2, 4, 8 in their last digit which means the above numbers are multiples of 2.
Example 5. Write the number 6 formed by any power.
Solution:
We have to write the number formed any power for 6.
Multiply the numbers with any power and then check whether it is a multiple of 5.
Let us take random multiples of 6 to check the result.
6 × 50 = 300 = 3 + 0 + 0 = 3
6 × 23 = 138 = 1 + 3 + 8 = 12 = 1 + 2 = 3
6 × 68 = 408 = 4 + 8 = 12 = 1 + 2 = 3
6² = 6 × 6 = 36 = 3 + 6 = 9
6³ = 6 × 6 × 6 = 216 = 2 + 1 + 6 = 9
We observe that the number formed by any power is 3, 9.
The sum of the digits of a number formed by 6 will be the multiples of 3.
Scroll to Top
Scroll to Top |
# What are the perfect squares up to 200?
## What are the perfect squares up to 200?
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 2, 5, 8, 10, 13, 17, 18, 20, 25, 26, 29, 32, 50, 65, 85, 125, 130, 145, 170, 185, 200, 3, 6, 9, 11, 12, 14, 17, 18, 19, 21, 22, 24….Square Number.
square partitions
100 1116
150 6521
200 27482
### What are the perfect squares in math?
The first 12 perfect squares are: {1, 4, 9, 25, 36, 49, 64, 81, 100, 121, 144…} Perfect squares are used often in math. Try to memorize these familiar numbers so that you can recognize them as they are used in many math problems. The first five squares of the negative integers are shown below.
#### How many perfect squares are there below 200?
Answer: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 2, 5, 8, 10, 13, 17, 18, 20, 25, 26, 29, 32, 50, 65, 85, 125, 130, 145, 170, 185.
IS 200 a perfect square in math?
200 is not a perfect square, hence its square root is an irrational number.
How do you find the perfect squares between 200 and 300?
Square numbers from 200 to 300 are 225 , 256 , 289
1. Solution: Given that, We have to write the square numbers from 200 to 300.
2. Square numbers from 200 to 300. Thus square numbers from 200 to 300 are 225 , 256 , 289.
3. Learn more: Numbers between 200 and 300 that are exactly divisible by 6,8,9. brainly.in/question/12371583.
## What are the first 100 perfect squares?
They are 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900 and 961.
### How do you find perfect squares?
You can also tell if a number is a perfect square by finding its square roots. Finding the square root is the inverse (opposite) of squaring a number. If you find the square root of a number and it’s a whole integer, that tells you that the number is a perfect square. For instance, the square root of 25 is 5.
#### WHY IS 100 a perfect square?
100 is a perfect square. Because 10 * 10 = 100.
IS 200 a perfect square or cube?
Is 200 a Perfect Cube? The number 200 on prime factorization gives 2 × 2 × 2 × 5 × 5. Here, the prime factor 5 is not in the power of 3. Therefore the cube root of 200 is irrational, hence 200 is not a perfect cube.
How many perfect squares are there between 100 and 200?
So the perfect square numbers from 100 to 200 are 100, 121, 144, 169 and 196.
## What are the square numbers between 100 and 200?
So, the squares of 11,12,13 and 14 lie between 100 and 200.
### What are perfect squares?
In other words, the perfect squares are the squares of the whole numbers such as 1 or 1 2, 4 or 2 2, 9 or 3 2, 16 or 4 2, 25 or 5 2 and so on. Also, get the perfect square calculator here. Perfect Squares from 1 to 100 Below shows the list of perfect squares from 1 to 100 along with their factors (product of integers).
#### How to memorize the perfect squares by heart?
Learning the perfect squares by heart is undoubtedly a challenge. The trick, however, is to memorize a few at a time. Begin with the first 25 square numbers, then move on to 50, and raise the bar to 100 with the help of these squaring number charts available in three different number ranges 1-25, 1-50, and 1-100.
How many numbers can you square?
For example, 2 x 2 is the same as 2 2, which is read ‘two squared.’ Now, you can square any number. For example, 7.4 2, read ‘seven and four-tenths squared,’ is the same as 7.4 x 7.4. However, it’s only when you square whole numbers, which are numbers without decimals or fractions, that you can get a perfect square.
What is the perfect square of 25?
The neat part is that the perfect square number forms a perfect square image. Take a look at the diagram appearing here, which shows that 5 2 equals the perfect square of 25. Visually, it makes sense too, when you see that it’s essentially a 5×5 grid. This can really help if you’re more comfortable with diagrams than with numbers. |
# 1998 AHSME Problems/Problem 16
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
The figure shown is the union of a circle and two semicircles of diameters $a$ and $b$, all of whose centers are collinear. The ratio of the area, of the shaded region to that of the unshaded region is
$\mathrm{(A) \ } \sqrt{\frac ab} \qquad \mathrm{(B) \ }\frac ab \qquad \mathrm{(C) \ } \frac{a^2}{b^2} \qquad \mathrm{(D) \ }\frac{a+b}{2b} \qquad \mathrm{(E) \ } \frac{a^2 + 2ab}{b^2 + 2ab}$
## Solution
To simplify calculations, double the radius of the large circle from $\frac{a + b}{2}$ to $a + b$. Each region is similar to the old region, so this should not change the ratio of any areas.
In other words, relabel $a$ and $b$ to $2a$ and $2b$.
The area of the whole circle is $A_{big} = \pi\cdot (a + b)^2$
The area of the white area is about $\frac{A_{big}}{2}$, which is the bottom half of the circle. However, you need to subtract the little shaded semicircle on the bottom, and add the area of the big unshaded semicircle on top. Thus, it is actually $A_{white} = \frac{1}{2}\pi\cdot (a+b)^2 - \frac{1}{2}\pi a^2 + \frac{1}{2}\pi b^2$
Factoring gives $A_{white} = \frac{\pi}{2}\cdot (a^2 + 2ab + b^2 - a^2 + b^2)$
Simplfying the inside gives $A_{white} = \frac{\pi}{2}\cdot (2ab + 2b^2)$
$A_{white} = \pi(ab + b^2)$
With similar calculations, or noting the symmetry of the situation, $A_{grey} = \pi(ab + a^2)$
The desired ratio is thus $\frac{ab + a^2}{ab + b^2} = \frac{a(a+b)}{b(a+b)}$, which is option $\boxed{B}$.
1998 AHSME (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions |
# STA 1625 Rasmussen College Es
Gaming and probability are intertwined. In fact, there is an entire discipline in mathematics called “Game Theory.” In this assessment, you will apply discrete and continuous probability distributions to games. There will be two parts: the first will concentrate on discrete probability distributions, and the second part will focus on continuous probability distributions.
##### Instructions
###### Part One – Data Table
The theoretical probability of rolling a fair six-sided die is 1/6 for any specific single outcome, such as rolling a one. You want to test the theoretical probability by running an experiment.
In this experiment, you need to roll a six-sided die 25 times. Record the outcome of each die roll. Create a discrete probability distribution using your outcomes as the probability. For example, if you rolled 4 fives out of your 25 total rolls, your probability would be 4/25.
• x – 1 – 2 – 3 – 4 – 5 – 6
• P(x)
###### Part Two – Discrete Probability Distribution
After filling in the table above with your experimental probability, answer the following questions. Show all work for full credit. Calculations should be performed in Excel while answers including an explanation of steps using proper terminology are provided in a separate document.
1. What is the expected outcome for rolling a six-sided die using the discrete probability distribution table above?
2. What is the probability of rolling an even number according to the discrete probability distribution table above?
• How does this compare to the theoretical probability of 0.5?
• Explain why you think there is a difference between the theoretical probability and the experimental probability you found.
3. Create a binomial probability distribution based on the discrete probability distribution table above where a success is rolling an even number. Answer the following questions:
• How do you know this is a Binomial Probability Distribution? Explain by showing how this example fits all four properties of a Binomial Probability Distribution.
• Define n,p,q.
• What is the probability that you will roll exactly 12 even numbers?
• What is the probability that you will roll at least 12 even numbers?
• Find the expected number of even numbers that you will roll.
###### Part Three – Continuous Probability Distribution
Dice are a common tool used in several board games. One board game which utilizes two dice is Monopoly. While the outcomes of rolling two dice in this game would be a discrete random variable, we are interested in looking at a continuous random variable associated with Monopoly and its respective probability.
The time it takes to finish a game of Monopoly is normally distributed with a mean of 120 minutes and a standard deviation of 30 minutes. Using this premise, answer the following questions. Show all work for full credit. Calculations should be performed in Excel while answers including an explanation of steps using proper terminology are provided in a separate document.
1. Explain why this is a continuous probability distribution instead of a discrete probability distribution.
2. What is the probability that a game lasts less than 45 minutes?
3. What is the probability that a game lasts more than 160 minutes?
4. What is the z-score of a game that lasts exactly 105 minutes?
### ORDER THIS OR A SIMILAR PAPER AND GET 20% DICOUNT. USE CODE: GET2O
Posted in Uncategorized
# STA 1625 Rasmussen College Es
Data can be collected from many different sources. In this assignment, you will perform a short survey of 10 people. From this data, you will create a frequency chart, histogram, and pie chart. You will also perform several statistical calculations.
##### Instructions
###### Part One – Survey Data
For this assignment, you will need to ask 10 people the following 5 questions and record their answers in an Excel spreadsheet:
• What is their first name?
• What is their age?
• Are they currently a student?
• How many hours of sleep do they average in a night?
• What is their favorite ice cream flavor?
Please Note: Some of the data collected will be used in the following module assignments. Make sure to ask all five questions and record all the data for each question.
###### Part Two – Graphs & Calculations
After you collect data from 10 people on the 5 questions above, complete the following in the same Excel spreadsheet used to collect the data:
• Create two frequency charts:
• The first frequency chart will be for their favorite ice cream flavor.
• The second frequency chart will be for their age and should contain between 5 and 10 class intervals.
• Create a histogram for age.
• Label the axes, title, and bars.
• Create a pie chart for favorite ice cream flavor.
• Label the key and percentage of each section.
• Calculate the following statistical values on the age variable:
• Measures of center – mean, median, mode, and midrange.
• Measures of variation – range and standard deviation.
###### Part Three – Presentation
After you have created your graphs and performed your calculations, create a presentation with the following information on each slide:
• Slide 1: Title slide with assignment title, name, course, date, instructor name.
• Slide 2: Provide a copy of the raw data in table format collected from 10 people.
• Slide 3: Copy and paste your two frequency charts.
• Slide 4: Copy and paste your histogram.
• What conclusions can you draw?
• Slide 5: Copy and paste your pie chart.
• What conclusions can you draw?
• Slide 6: Share the measures of center for age.
• Slide 7: Share the measures of variation for age.
• Slide 8: Conclusion – Summarize the data you collected.
• Were there any interesting outcomes you weren’t expecting?
• Was there any data that was helpful to draw further conclusions?
• What benefits did you gain from visualizing the data?
• What benefits did you gain from performing statistical calculations on the data?
### ORDER THIS OR A SIMILAR PAPER AND GET 20% DICOUNT. USE CODE: GET2O
Posted in Uncategorized
# STA 1625 Rasmussen College Es
You are recently hired at a sleep research company to perform statistical calculations. Your boss has assigned you two hypothesis tests to run.
Instructions
###### Part One – Hypothesis Tests
For the assignment, complete hypothesis tests of the two situations listed below by:
• Constructing an 8-step hypothesis test, showing all calculations,
• Including both the critical value method and the p-value method in your calculations.
• Using the correct terminology and interpretation of the resulting confidence interval for each test.
###### Hypothesis Test – Situation 1
It has been reported that the mean amount of sleep that adults receive is 7 hours per night. Your boss has asked you to look into this statistic by running your own study. You gather 49 adults and find the following information. The mean amount of sleep was 6.2 hours with a standard deviation of 1.8 hours. Based on your results, you claim that adults receive less than 7 hours of sleep. Test your claim with a 0.05 significance level.
###### Hypothesis Tests – Situation 2
Your boss also would like you to look at average sleep for a specific age range (between 35 and 44 years of age). It has been previously reported that 38.3% of adults in this age range receive the recommended amount of sleep per night (at least 7 hours). You decide to test this claim by gathering 32 adults between the ages of 35 and 44 years old. In the sample, you find that 15 of the participants receive at least 7 hours of sleep. Test the claim that more than 38.3% of adults get the recommended amount of sleep per night with a significance level of 0.01.
###### Part Two – Hypothesis Errors
After you have completed your hypothesis tests, in a separate paragraph,
• Explain the possible error(s) that could have occurred and what they imply for each test.
• Explain how you might prevent the error(s) from occurring in future hypothesis tests.
### ORDER THIS OR A SIMILAR PAPER AND GET 20% DICOUNT. USE CODE: GET2O
Posted in Uncategorized
# STA 1625 Rasmussen College Es
##### Overview
Statistics covers a wide range of topics from basic calculations of mean, median, and mode to finding the probability of a normal variable. In the last week of the course, review the concepts we have discussed and reflect on your experiences in this course.
##### Instructions
In your initial post, reflect on the following questions and provide explanations and examples:
• What concept(s) did you find to be the most enjoyable? Why?
• What concept(s) do you think will be most applicable in your future career path? Explain with examples.
• What concept(s) do you think will be most applicable in your everyday life? Explain with examples.
• Reflect on your peer’s post and identify any answers that you disagree with.
• Can you relate to any of the examples they provided?
• Or are there any other examples that your classmate may have missed in their reflection?
• Or reflect on the similarities and differences among your postings. |
# The Secant of a Circle: Exploring its Definition, Properties, and Applications
A circle is a fundamental geometric shape that has fascinated mathematicians and scientists for centuries. One of the key concepts associated with circles is the secant, which plays a crucial role in various mathematical and real-world applications. In this article, we will delve into the definition, properties, and applications of the secant of a circle, providing valuable insights and examples along the way.
## What is a Secant?
Before we dive into the specifics of the secant of a circle, let’s first understand what a secant is in general. In mathematics, a secant is a line that intersects a curve at two or more distinct points. In the context of a circle, a secant is a line that intersects the circle at two distinct points, creating a chord.
## The Secant of a Circle
When we talk about the secant of a circle, we are referring to a line that intersects the circle at two distinct points, extending beyond the circle on both sides. This line is commonly referred to as a secant line. The points where the secant line intersects the circle are known as the points of intersection.
The length of the secant line is an important property to consider. It is defined as the distance between the two points of intersection. This length can vary depending on the position of the secant line relative to the circle.
### Secant Line Length Formula
To calculate the length of a secant line, we can use the following formula:
Secant Line Length = 2 * Radius * sec(θ/2)
Here, θ represents the angle between the secant line and the radius of the circle at one of the points of intersection. The function sec(θ/2) denotes the secant of half the angle.
## Properties of the Secant of a Circle
The secant of a circle possesses several interesting properties that are worth exploring. Let’s take a closer look at some of these properties:
### 1. Intersecting Chords Theorem
One of the most important properties of the secant of a circle is the Intersecting Chords Theorem. According to this theorem, when two chords intersect inside a circle, the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.
This theorem can be expressed mathematically as:
AB * BC = CD * DE
where AB and BC represent the segments of one chord, and CD and DE represent the segments of the other chord.
### 2. Secant-Secant Power Theorem
Another important property of the secant of a circle is the Secant-Secant Power Theorem. This theorem states that when two secant lines intersect outside a circle, the product of the lengths of one secant line and its external segment is equal to the product of the lengths of the other secant line and its external segment.
Mathematically, this theorem can be expressed as:
AB * BC = DE * EF
where AB and BC represent the lengths of one secant line and its external segment, and DE and EF represent the lengths of the other secant line and its external segment.
### 3. Tangent-Secant Power Theorem
The Tangent-Secant Power Theorem is another important property associated with the secant of a circle. This theorem states that when a tangent and a secant line intersect at a point outside the circle, the square of the length of the tangent is equal to the product of the lengths of the secant line and its external segment.
Mathematically, this theorem can be expressed as:
AB^2 = BC * BD
where AB represents the length of the tangent line, BC represents the length of the secant line, and BD represents the length of the external segment of the secant line.
## Applications of the Secant of a Circle
The secant of a circle finds applications in various fields, including mathematics, physics, and engineering. Let’s explore some of these applications:
### 1. Trigonometry
In trigonometry, the secant function is defined as the reciprocal of the cosine function. It is denoted as sec(θ) and represents the ratio of the hypotenuse to the adjacent side in a right triangle.
The secant function is widely used in trigonometric calculations, such as finding the lengths of sides and angles in triangles, solving trigonometric equations, and analyzing periodic functions.
### 2. Optics
In optics, the secant of a circle is used to calculate the focal length of a lens. The focal length is the distance between the lens and the point where parallel rays of light converge or diverge after passing through the lens.
By measuring the distance between the lens and the image formed by the lens, along with the distance between the lens and the object, the secant of the angle of incidence can be calculated. This information is then used to determine the focal length of the lens.
### 3. Engineering
In engineering, the secant of a circle is used in structural analysis to calculate the deflection of beams and columns under load. The deflection is the degree to which a structural element bends or deforms under applied load.
By considering the secant of the angle of deflection, engineers can determine the amount of deformation that will occur in a structural element and design accordingly to ensure structural integrity.
## Summary
The secant of a circle is a fundamental concept in mathematics with various properties and applications. It is a line that intersects a circle at two distinct points, creating a chord. The length of the secant line can be calculated using the secant function. The secant of a circle has properties such as the Intersecting Chords Theorem, Secant-Secant Power Theorem, and Tangent-Secant Power Theorem. It finds applications in trigonometry, optics, and engineering, among other fields.
## Q&A
### 1. What is the difference between a secant and a chord?
A secant is a line that intersects a curve at two or more distinct points, while a chord is a line segment that connects two points on a curve, such as a circle. In the context of a circle, a secant is a line that intersects the circle at two distinct points, |
Lesson: Simplify Fractions
9 Views
4 Favorites
Lesson Objective
SWBAT simplify fractions.
Lesson Plan
AIM:
Element/Time
Teacher
Student expected response
Do Now
(8:30 – 8:40)/(10:00 –10:10)
· Walk around and check students do now responses or HW – remediate as necessary.
· PLAY MUSIC???
· Come in silently, begin work right away
· Work on 2-4 spiraled review problems
· Work on Mental Math
· Check HW when finished
HW Check
· Allow students to check both homework assignments by placing answers on the board
· Check and remediate student HW
Mental Math
(8:40 – 8:50) /(10:10 – 10:20)
Review 1 – 2 strategies for mental math
Read string of problems for mental math
· Share strategies and answers for mental math
· Mentally calculate long string of numbers focusing on fact fluency
Mini Lesson
(8:50 – 9:25)/(10:20 – 10:55)
Opening/Hook:
· What does it mean if two fractions are equivalent?
Intro:
· Today we are going review a little bit by going back to working with fractions.
· As we know every fraction has an infinite number of fractions that are different names for the same amount.
· However, every fraction only has one fraction that is the “simplest form” of the fraction. Today we are going to work on two different ways to simplify fractions. Just so you know some people call this reducing fractions (label)
· Clarify that a simplified or reduced fraction is the same amount – it is equivalent!!
Station Teaching (groups will be boys and girls??)
Robin: Teach reducing fractions using prime factorization
Notes:
· First find the prime factorization of the numerator and the denominator.
· Then cross out whatever they share
· Multiply whatever is left on the top and bottom – and you have your fraction that you have left is your simplified fraction.
· Model example:
= =
Ivory: Simplifying using GCF
· Find the GCF of numerator and denominator
· Divide both the numerator and the denominator by the GCF
· Model example:
÷ =
GCF
Guided:
· Students write down the method that they find the easiest to do.
· Discuss for students – sometimes when the numbers are familiar GCF method is easier but when the numbers are unfamiliar it is easy to use prime factorization.
· Students use that method to simplify 3 fractions.
Independent:
· 5- 10 simplifying fractions problems.
Closing:
· What did we learn about today?
· Why did we learn it?
· What will your work look like?
· What are some common mistakes?
Right is right: They are different ways to name the same amount.
Students take notes on:
Aim: Simplify fractions
Key Points:
Simplifying fractions means to make the fraction as simple as possible.
Ex: 4/8 is really ½ in its simplest form.
2 ways to simplify fractions
Prime Factorization Divide by GCF
4 Questions:
ü We learned how to simplify fractions.
ü It’s important because it will help us when we work on computing with fractions, but also because it accepted as a math rule to always right fractions in their simplest form.
ü Work will have the prime factorization crossed out or dividing the fraction by the GCF.
ü Forgetting that you must divide the top and bottom by the same number or not actually going all the way down to the simplest form.
Interim Review
(9:25 – 9:35)/(10:55 -11:05)
· Show models to students – addition means combining, putting together, joining. When you read the following problems, ask yourself if it is doing one of these things – if yes, then it is an addition problem.
Students will read through 5-6 word problems (some with very challenging numbers)
Students will ask themselves the questions – is this problem about combining, putting together or joining –
Students will determine if the problem is addition or something else.
Skills Time
(9:35- 9:50) /(11:05-11:25)
Froehlich???
OOO
Kaiyonna, Ronginea, Ashley, Maurice, Tilaya
Ivory???
Use sticky notes to demonstrate the law of multiplying exponents with like basese
Exponents
Marlin, Anthony, Akirah, Tilaya, Sammy, Matthew
Whole Group:
· Must Do: Menu of Problems for the Week.
· Should Do: Organize Your Math Notes, Study – Use sticky notes on your Journals to review important ideas/concepts,
· Could Do: Versatiles, Math Game Choice, Various Center Activities,
Lesson Resources
CW Simplest Form Classwork 4 Exit Ticket Simplify Fractions Assessment 2 |
Notes On Scalar Product - CBSE Class 12 Maths
Vector multiplication is of two types. One is dot or scalar product and the other is cross or vector product. The name itself indicates the result. If we use scalar product, the result is a scalar, that is, a real number. If we use vector product, the result is a vector. Dot product of vectors $\stackrel{→}{\text{a}}$ and $\stackrel{→}{\text{b}}$: $\stackrel{→}{\text{a}}$ . $\stackrel{→}{\text{b}}$ = |$\stackrel{→}{\text{a}}$| || cos ϴ, where 0o ≤ ϴ ≤ 180o If $\stackrel{→}{\text{a}}$ = $\stackrel{→}{\text{0}}$ or = $\stackrel{→}{\text{0}}$ $\stackrel{→}{\text{a}}$ . $\stackrel{→}{\text{b}}$ = 0 Properties of dot product: (1) $\stackrel{→}{\text{a}}$ . $\stackrel{→}{\text{b}}$ ∈ R (2) If $\stackrel{→}{\text{a}}$ . $\stackrel{→}{\text{b}}$ = 0 ⇔ $\stackrel{→}{\text{a}}$ ⊥ $\stackrel{→}{\text{b}}$ (3) When θ = 0° , $\stackrel{→}{\text{a}}$ . = |$\stackrel{→}{\text{a}}$| || (i) $\stackrel{→}{\text{a}}$ . $\stackrel{→}{\text{a}}$ = |$\stackrel{→}{\text{a}}$|2 (4) When θ = 180°, $\stackrel{→}{\text{a}}$ . = - |$\stackrel{→}{\text{a}}$| || (5) If θ is the angle between the vectors $\stackrel{→}{\text{a}}$ and $\stackrel{→}{\text{b}}$$\stackrel{}{\text{}}$$\stackrel{}{\text{}}$ Cos θ = $\frac{\text{}\stackrel{→}{\text{a}}\text{.}\stackrel{→}{\text{b}}}{\text{|}\stackrel{_}{\text{a}}\text{| |}\stackrel{_}{\text{b}}\text{|}}$ For unit vectors $\stackrel{^}{\text{i}}$ , $\stackrel{^}{\text{j}}$ and : $\stackrel{^}{\text{i}}$ . $\stackrel{^}{\text{i}}$ = |$\stackrel{^}{\text{i}}$|2 = 1 $\stackrel{^}{\text{j}}$ . $\stackrel{^}{\text{j}}$ = ||2 = 1 $\stackrel{^}{\text{k}}$ . $\stackrel{^}{\text{k}}$ = ||2 = 1 For mutually perpendicular unit vectors $\stackrel{^}{\text{i}}$ , $\stackrel{^}{\text{j}}$ and : $\stackrel{^}{\text{i}}$ . $\stackrel{^}{\text{j}}$ = 0 $\stackrel{^}{\text{j}}$ . $\stackrel{^}{\text{k}}$ = 0 $\stackrel{^}{\text{k}}$ . $\stackrel{^}{\text{i}}$ = 0
#### Summary
Vector multiplication is of two types. One is dot or scalar product and the other is cross or vector product. The name itself indicates the result. If we use scalar product, the result is a scalar, that is, a real number. If we use vector product, the result is a vector. Dot product of vectors $\stackrel{→}{\text{a}}$ and $\stackrel{→}{\text{b}}$: $\stackrel{→}{\text{a}}$ . $\stackrel{→}{\text{b}}$ = |$\stackrel{→}{\text{a}}$| || cos ϴ, where 0o ≤ ϴ ≤ 180o If $\stackrel{→}{\text{a}}$ = $\stackrel{→}{\text{0}}$ or = $\stackrel{→}{\text{0}}$ $\stackrel{→}{\text{a}}$ . $\stackrel{→}{\text{b}}$ = 0 Properties of dot product: (1) $\stackrel{→}{\text{a}}$ . $\stackrel{→}{\text{b}}$ ∈ R (2) If $\stackrel{→}{\text{a}}$ . $\stackrel{→}{\text{b}}$ = 0 ⇔ $\stackrel{→}{\text{a}}$ ⊥ $\stackrel{→}{\text{b}}$ (3) When θ = 0° , $\stackrel{→}{\text{a}}$ . = |$\stackrel{→}{\text{a}}$| || (i) $\stackrel{→}{\text{a}}$ . $\stackrel{→}{\text{a}}$ = |$\stackrel{→}{\text{a}}$|2 (4) When θ = 180°, $\stackrel{→}{\text{a}}$ . = - |$\stackrel{→}{\text{a}}$| || (5) If θ is the angle between the vectors $\stackrel{→}{\text{a}}$ and $\stackrel{→}{\text{b}}$$\stackrel{}{\text{}}$$\stackrel{}{\text{}}$ Cos θ = $\frac{\text{}\stackrel{→}{\text{a}}\text{.}\stackrel{→}{\text{b}}}{\text{|}\stackrel{_}{\text{a}}\text{| |}\stackrel{_}{\text{b}}\text{|}}$ For unit vectors $\stackrel{^}{\text{i}}$ , $\stackrel{^}{\text{j}}$ and : $\stackrel{^}{\text{i}}$ . $\stackrel{^}{\text{i}}$ = |$\stackrel{^}{\text{i}}$|2 = 1 $\stackrel{^}{\text{j}}$ . $\stackrel{^}{\text{j}}$ = ||2 = 1 $\stackrel{^}{\text{k}}$ . $\stackrel{^}{\text{k}}$ = ||2 = 1 For mutually perpendicular unit vectors $\stackrel{^}{\text{i}}$ , $\stackrel{^}{\text{j}}$ and : $\stackrel{^}{\text{i}}$ . $\stackrel{^}{\text{j}}$ = 0 $\stackrel{^}{\text{j}}$ . $\stackrel{^}{\text{k}}$ = 0 $\stackrel{^}{\text{k}}$ . $\stackrel{^}{\text{i}}$ = 0
Previous
Next
➤ |
# Multiplication (by 3+ digits)
We discuss the algorithm for multiplying multi-digit numbers in detail in the lesson Multiplication (2x2 digits). Once you start working with longer numbers, the process continues:
Example
$1734 \times 231 =$
First, set the problem up vertically. It's most efficient to put the number with the most digits on top:
$$\begin{array}{r} &1734\\\times\!\!\!\!\!\!&231\\ \hline \end{array}$$
We start by focusing on the ones digit of the multiplier (the number on the bottom). Some students cover or cross out the tens digit in this phase so that they don't get confused. If students are struggling to remember which number they are multiplying by, we recommend that they circle the digit that they are working with (rather than crossing anything out).
Then, we multiply the ones digit of the multiplier (in red below) times each digit in the multiplicand (upper number) and put the answer values in the appropriate columns in the answer line (don't forget to carry as necessary -- we're going to leave those marks out of the example for clarity):
$$\begin{array}{r} &1734\\\times\!\!\!\!\!\!&23\color{red}{1}\\ \hline &\quad \color{red}{1734} \end{array}$$
So far, this is just like multiplying by one digit. Then comes the trickier "stacking part."
Now, we need to multiply the tens digit of the multiplier (bottom number) times each digit of the multiplicand (top number). But that digit is in the tens column, so although it is a 4, it represents 40. That means all of our answers are 10 times greater than they look. So, we slide our answers over by one place value (we'll break this down below). We slide our answer digits over by putting a placeholder zero in the ones column of the second row of answers (no whole number times 40 will give you an answer less than 10).
$$\begin{array}{r} &1734\\\times\!\!\!\!\!\!&231\\ \hline & 1734\\&0 \end{array}$$
Then we multiply 3 times each of the numbers on the top row, putting the answers in the first available slots in the second answer row:
$$\begin{array}{r} &1734\\\times\!\!\!\!\!\!&3\color{purple}{3}\color{black}1\\ \hline & 1734\\&\color{purple}{5202}0 \end{array}$$
Now we have two rows of answers (one from multiplying the ones digit of the multiplier and one from multiplying the tens digit of the multiplier). Next we multiply the hundreds place of the multiplier by each number in the multiplicand. First, we have to start a new answer row, and because this is the hundreds column, we start the row with two placeholder zeros.
$$\begin{array}{r} &1734\\\times\!\!\!\!\!\!&\color{green}{2}\color{black}31\\ \hline & 1734\\&52020\\&\color{green}{3468}00 \end{array}$$
$$\begin{array}{r} &1734\\\times\!\!\!\!\!\!&\color{green}{2}\color{black}31\\ \hline & \color{red}{1734}\\&+\color{purple}{52020}\\&+\color{green}{3468}00\\\hline &\mathbf{400554} \end{array}$$
You can see the pattern. Every additional digit in the multiplier yields another row in the answer section and an additional place holder zero in that row. Know this rule, you can multiple numbers of any size, just make sure to line up the digits in the answer rows that addition doesn't get to complicated!
Hint: Students often find it helpful to put a an X or a line through their placeholder zeros so that it's easy to see how many they've written down. If students struggle to keep track of their placeholders, try having them put lines or Xs through those placeholder zeros to make them distinct from regular zeros that result multiplication that equals zero.
$$\begin{array}{r} &1734\\\times\!\!\!\!\!\!&\color{green}{2}\color{black}31\\ \hline & \color{red}{1734}\\&+\color{purple}{5202}\!\bcancel{0}\!\!\\&+\color{green}{3468}\!\bcancel{0}\!\!\!\bcancel{0}\!\!\!\\\hline &\mathbf{400554} \end{array}$$ |
# What is 10 percent of 100000000 + Solution with Free Steps
10 percent of 100000000 results in 10000000. The solution may be found by multiplying 100000000 by the factor of 0.10.
Suppose that the GDP of a country is 100000000 dollars. Due to Covid-19, the country had to import a lot of vaccines and consequently had to take a lot of loans. They also lost a lot of tourism revenue. According to the figures, they post a GDP growth of ten percent. Since you already know that the 10 percent of 100000000 is equal to 10000000, so you can easily tell that their revenue growth for this year was 10000000. Adding both values, you can compute that the total GDP this year was 110000000 dollars.
## What Is 10 percent of 100000000?
10% of 100000000 is equal to the number 10000000. This result will be obtained by multiplying the factor 0.10 by 100000000.
The solution to the question 10% of 100000000 can be found by multiplying 10/100 fractional values with the number 100000000. This answer may be further reduced to get the 10% of 800, which is 10000000.
## How To Calculate 10% of 100000000?
We may calculate what portion of 100000000 equals 10% by using the easy mathematical steps estimating 10 percent of 100000000 shown below:
### Step 1
Writing 10 percent of 100000000 in mathematical form:
10 percent of 100000000 = 10% x 100000000
### Step 2
Substitute the % symbol with the fraction 1/100:
10 percent of 100000000 = ( 10 x 1/100 ) x 100000000
### Step 3
Rearranging ( 10 x 1/100 ) x 100000000:
10 percent of 100000000 = ( 10 x 100000000 ) / 100
### Step 4
Multiplying 10 with 100000000:
10 percent of 100000000 = ( 1000000000 ) / 100
### Step 5
Dividing 9600 by 100:
10 percent of 100000000 = 10000000
So, the 10 percent of 100000000 is equal to 10000000.
10 percent of 100000000 can be depicted by using the following picture.
Figure 1: Pie Chart of 10 percent of 100000000
The orange slice of the pie chart represents the 10 percent of 100000000, equal to 10000000. The green slice of the pie chart represents 90 percent of 100000000, equal to 90000000. The total area of the pie chart represents 100 percent of 100000000, equal to 100000000.
The percentage is a special scale of zero to one hundred used to compare differently scaled values with one another.
All the Mathematical drawings/images are created using GeoGebra. |
# Proof of Limit Rule of an Exponential function
The literals $a$ and $b$ are two constants, and $x$ is a variable. A function in terms of $x$ is denoted by $f(x)$ in mathematics. An exponential function in terms of constant $b$ and function $f(x)$ is written as $b^{\displaystyle f(x)}$ mathematically. In this case, $a$ is a value of the variable $x$.
Now, let’s start deriving the limit rule of an exponential function in calculus mathematically.
### Define the Limit of a function
The limit of a function $f(x)$ as the input $x$ approaches $a$ is written as follows.
$\displaystyle \large \lim_{x \,\to\, a}{\normalsize f(x)}$
Take, the limit of the function as $x$ tends to $a$ is equal to $L$.
$\implies$ $L \,=\, \displaystyle \large \lim_{x \,\to\, a}{\normalsize f(x)}$
Now, find the limit of the function as $x$ closer to $a$ by direct substitution method.
$\implies$ $L \,=\, f(a)$
### Define the Limit of an Exponential function
Now, write the limit of an exponential function as $x$ approaches $a$ in mathematical form.
$\displaystyle \large \lim_{x \,\to\, a}{\normalsize b^{\displaystyle f(x)}}$
Use the direct substitution method and evaluate the limit of the exponential function as $x$ approaches $a$.
$\implies$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize b^{\displaystyle f(x)}} \normalsize \,=\, b^{\displaystyle f(a)}$
### Limit Rule of an Exponential function
According to the first step, $L = f(a)$. So, replace the value of $f(a)$ by $L$ in the above mathematical equation.
$\implies$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize b^{\displaystyle f(x)}} \normalsize \,=\, b^{\displaystyle L}$
Actually, it is taken that $L \,=\, \displaystyle \large \lim_{x \,\to\, a}{\normalsize f(x)}$ in the first step.
$\,\,\, \therefore \,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize b^{\displaystyle f{(x)}}}$ $\,=\,$ $b^{\, \displaystyle \large \lim_{x \,\to\, a} \, {\normalsize f{(x)}}}$
Therefore, it is proved that the limit of an exponential function is equal to the limit of the exponent with same base. Thus, the limit rule of an exponential function is proved mathematically in calculus.
Latest Math Topics
Jun 26, 2023
###### Math Questions
The math problems with solutions to learn how to solve a problem.
Learn solutions
Practice now
###### Math Videos
The math videos tutorials with visual graphics to learn every concept.
Watch now
###### Subscribe us
Get the latest math updates from the Math Doubts by subscribing us. |
### Theory:
Consider the following example:
Ram has gone out walking every day. In the first two weeks, he has walked the following distances on each of the days.
$$4$$, $$5$$, $$7$$, $$2$$, $$6$$, $$7$$, $$5$$, $$4$$, $$7$$, $$6$$, $$4$$, $$3$$, $$5$$ and $$4$$ miles respectively.
Construct a frequency distribution table and find the range of the data collected.
Step $$1$$: Let us arrange the numbers in ascending or descending order.
$$2$$, $$3$$, $$4$$, $$4$$, $$4$$, $$4$$, $$5$$, $$5$$, $$5$$, $$6$$,$$6$$, $$7$$, $$7$$, $$7$$
Step $$2$$: Let us construct a frequency distribution table of the ordered data.
Distance(in miles) Frequency $$2$$ $$1$$ $$3$$ $$1$$ $$4$$ $$4$$ $$5$$ $$3$$ $$6$$ $$2$$ $$7$$ $$3$$
Step $$3$$: Find the range of the data
From the tabular column, the highest value is $$7$$, and the lowest value is $$2$$.
$$\text{Range} = \text{Highest value} - \text{Lowest value}$$
$$= 7 - 2$$
$$= 5$$
The range of the given set of data is $$5$$. |
### 62nd Putnam 2001
Problem A4
Points X, Y, Z lie on the sides BC, CA, AB (respectively) of the triangle ABC. AX bisects BY at K, BY bisects CZ at L, CZ bisects AX at M. Find the area of the triangle KLM as a fraction of that of ABC.
Solution
Use vectors. Take some origin O and let A be the vector OA. Similarly for the other points. We use repeatedly the fact that if the point P lies on the line QR, then for some k, P = kQ + (1 - k)R.
For some k we have: X = kB + (1-k)C. Hence M = X/2 + A/2 = A/2 + kB/2 + (1-k)C/2. Now Z is a linear combination of M and C and also of A and B, so it must be (2M - (1 - k)C)/(1 + k) = (A + kB)/(1 + k). Hence L = Z/2 + C/2 = (A + kB + (1 + k)C)/(2 + 2k).
Y is a combination of B and L and of A and C. So it must be (A + (1 + k)C)/(2 + k). Hence K = B/2 + Y/2 = (A + (2 + k)B + (1 + k)C)/(4 + 2k). X is a combination of A and K and also of B and C, so it must be ( (2 + k)B + (1 + k)C)/(3 + 2k). But it is also kB + (1-k)C, so k2 + k - 1 = 0, or k = (√5 - 1)/2.
At this point, this approach degenerates into a slog (although a straightforward one). By repeated use of the relation k2 = 1 - k (and hence 1/(k+1) = k, 1/(2+k) = k2), we find that 2K = k2A + B + kC, 2L = kA + k2B + C, 2M = A + kB + k2C.
Now the area of ABC is (A x B + B x C + C x A)/2, and the area of KLM is (K x L + L x M + M x K)/2.
Expanding the latter gives the answer. |
# 2022 AIME I Problems/Problem 4
## Problem
Let $w = \dfrac{\sqrt{3} + i}{2}$ and $z = \dfrac{-1 + i\sqrt{3}}{2},$ where $i = \sqrt{-1}.$ Find the number of ordered pairs $(r,s)$ of positive integers not exceeding $100$ that satisfy the equation $i \cdot w^r = z^s.$
## Solution 1
We rewrite $w$ and $z$ in polar form: \begin{align*} w &= e^{i\cdot\frac{\pi}{6}}, \\ z &= e^{i\cdot\frac{2\pi}{3}}. \end{align*} The equation $i \cdot w^r = z^s$ becomes \begin{align*} e^{i\cdot\frac{\pi}{2}} \cdot \left(e^{i\cdot\frac{\pi}{6}}\right)^r &= \left(e^{i\cdot\frac{2\pi}{3}}\right)^s \\ e^{i\left(\frac{\pi}{2}+\frac{\pi}{6}r\right)} &= e^{i\left(\frac{2\pi}{3}s\right)} \\ \frac{\pi}{2}+\frac{\pi}{6}r &= \frac{2\pi}{3}s+2\pi k \\ 3+r &= 4s+12k \\ 3+r &= 4(s+3k). \end{align*} for some integer $k.$
Since $4\leq 3+r\leq 103$ and $4\mid 3+r,$ we conclude that \begin{align*} 3+r &\in \{4,8,12,\ldots,100\}, \\ s+3k &\in \{1,2,3,\ldots,25\}. \end{align*} Note that the values for $s+3k$ and the values for $r$ have one-to-one correspondence.
We apply casework to the values for $s+3k:$
1. $s+3k\equiv0\pmod{3}$
2. There are $8$ values for $s+3k,$ so there are $8$ values for $r.$ It follows that $s\equiv0\pmod{3},$ so there are $33$ values for $s.$
There are $8\cdot33=264$ ordered pairs $(r,s)$ in this case.
3. $s+3k\equiv1\pmod{3}$
4. There are $9$ values for $s+3k,$ so there are $9$ values for $r.$ It follows that $s\equiv1\pmod{3},$ so there are $34$ values for $s.$
There are $9\cdot34=306$ ordered pairs $(r,s)$ in this case.
5. $s+3k\equiv2\pmod{3}$
6. There are $8$ values for $s+3k,$ so there are $8$ values for $r.$ It follows that $s\equiv2\pmod{3},$ so there are $33$ values for $s.$
There are $8\cdot33=264$ ordered pairs $(r,s)$ in this case.
Together, the answer is $264+306+264=\boxed{834}.$
~MRENTHUSIASM
## Solution 2
First we recognize that $w = \operatorname{cis}(30^{\circ})$ and $z = \operatorname{cis}(120^{\circ})$ because the cosine and sine sums of those angles give the values of $w$ and $z$, respectively. By De Moivre's theorem, $\operatorname{cis}(\theta)^n = \operatorname{cis}(n\theta)$. When you multiply by $i$, we can think of that as rotating the complex number $90^{\circ}$ counterclockwise in the complex plane. Therefore, by the equation we know that $30r + 90$ and $120s$ land on the same angle.
This means that $$30r + 90 \equiv 120s \pmod{360},$$ which we can simplify to $$r+3 \equiv 4s \pmod{12}.$$ Notice that this means that $r$ cycles by $12$ for every value of $s$. This is because once $r$ hits $12$, we get an angle of $360^{\circ}$ and the angle laps onto itself again. By a similar reasoning, $s$ laps itself every $3$ times, which is much easier to count. By listing the possible values out, we get the pairs $(r,s)$: $$\begin{array}{cccccccc} (1,1) & (5,2) & (9,3) & (13,1) & (17,2) & (21,3) & \ldots & (97,1) \\ (1,4) & (5,5) & (9,6) & (13,4) & (17,5) & (21,6) & \ldots & (97,4) \\ (1,7) & (5,8) & (9,9) & (13,7) & (17,8) & (21,9) & \ldots & (97,7) \\ [-1ex] \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ (1,100) & (5,98) & (9,99) & (13,100) & (17,98) & (21,99) & \ldots & (97,100) \end{array}$$ We have $25$ columns in total: $34$ values for the first column, $33$ for the second, $33$ for the third, and then $34$ for the fourth, $33$ for the fifth, $33$ for the sixth, etc. Therefore, this cycle repeats every $3$ columns and our total sum is $(34+33+33) \cdot 8 + 34 = 100 \cdot 8 + 34 = \boxed{834}$.
~KingRavi
## Video Solution
~Steven Chen (www.professorchenedu.com)
~ThePuzzlr
~MRENTHUSIASM
## Video Solution
~AMC & AIME Training |
Calendar Capers
Choose any three by three square of dates on a calendar page...
Have You Got It?
Can you explain the strategy for winning this game with any target?
Pair Sums
Five numbers added together in pairs produce: 0, 2, 4, 4, 6, 8, 9, 11, 13, 15 What are the five numbers?
Number Daisy
Age 11 to 14 Challenge Level:
We received lots of excellent solutions to this problem. Well done to students from Raines School, Lancaster Lane School, Risley Primary School, Garden International School, Glan-Y-Mor, Selside School, Kings School, Beijing City International School, Cirencester Kingshill School, Highfields School, Winster C of E School and Balfour Junior School for all finding solutions which give numbers greater than 25.
Alison managed to find numbers 1 to 46 with this daisy: 1 in centre, 2,4,12,19,8 around the edge.
Ria's method for making a daisy with numbers 1 to 28 can be found here.
Simon and Will made a daisy with numbers 1 to 37 and explain their method below:
To begin, we used the same flower as NRICH. We put 1 in the middle because we decided that it would be the most used number. We put 2 in next to make 2 and 3. Then we put a 4 in and we could make numbers 5,6,7. We wrote the numbers we could make underneath the flower. We then added 8 to make 8,9,10. We put 8 next to the 2 and we then could make numbers up to 15. We then added 12 and put it next to the 4 to make 16. The 10 was the last number to be added. We added 8 + 10 + 12 + 4 + 2 + 1 to make the biggest number 37. We then tried to make all the numbers in between and it worked.
Here are some more solutions.
Emma's daisy makes all the numbers 1 to 38:
Tamsin's daisy makes all the numbers from 1 to 40:
Carolyn's daisy makes all the numbers from 1 to 43:
Stephen's daisy makes all the numbers from 1 to 45:
Students from Riccarton High School found another daisy that makes all the numbers from 1 to 45:
They checked this worked by entering the six numbers in the cells of a spreadsheet and reproduced the numbers in six columns with the total on the right. It was then necessary to go through deleting some numbers on each line to give the totals 1 to 45 as in Ali's solution below:
1 21 8 9 2 4 Total 1 1 2 2 1 2 3 4 4 1 4 5 2 4 6 1 2 4 7 8 8 1 8 9 9 10 9 2 11 1 9 2 12 1 8 4 13 1 9 4 14 9 2 4 15 1 9 2 4 16 8 9 17 1 8 9 18 8 9 2 19 1 8 9 2 20 21 21 1 21 22 21 2 23 1 21 2 24 21 4 25 1 21 4 26 21 2 4 27 1 21 2 4 28 21 8 29 21 9 30 1 21 9 31 21 9 2 32 1 21 9 2 33 1 21 8 4 34 1 21 9 4 35 21 9 2 4 36 1 21 9 2 4 37 21 8 9 38 1 21 8 9 39 21 8 9 2 40 1 21 8 9 2 41 21 8 9 4 42 1 21 8 9 4 43 21 8 9 2 4 44 1 21 8 9 2 4 45 |
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# If p is prime number what is the lcm of $p,{p^2},{p^3}$
Last updated date: 24th Jul 2024
Total views: 348.9k
Views today: 10.48k
Verified
348.9k+ views
Hint: In the question we are told that the p is a prime number. A prime number is the number which has factors itself and 1.
In this case p has the factors p and 1. Follow the procedure of factorization and find lcm stepwise.
In this case $p,{p^2},{p^3}$
P is a prime number. We can find the factors of each term as
$p = p \times 1 \\ {p^2} = p \times p \times 1 \\ {p^3} = p \times p \times p \times 1 \;$
In this case we can observe the factors of the given terms and see if any common factors exist.
Taking common factors in all $p,{p^2},{p^3}$ and continuing till the end.
The common in three and two of them will be multiplied only once and the remaining all factors will be multiplied to the result.
This will help us In getting the answers by applying the basic definition of lowest common multiple.
It is a multiple which can be divided by all the given numbers and will be the lowest of such categories to exist.
So, finally solving,
We get the final solution as ${p^3}$ .
${p^3}$ is the lcm of given numbers $p,{p^2},{p^3}$ .
So, the correct answer is “ ${p^3}$ ”.
Note: In the process of finding the lcm we must factorise the components. If in case hcf is given and we need to find the lcm product of two numbers = product of lcm and hcf.
Understand that hcf is a factor and lcm is multiple of given numbers. Almost every time lcm > hcf. |
>> Actually two kinds of multiplication. The first one is very predictable. It's called "scalar multiplication." Works like this, suppose we have a matrix A defined this way. For scalar multiplication we multiply times each item within the matrix. 2A means 2 times the entries here for the matrix and so we just double all of them. 2 times negative 2, negative 4. 2 times 1, 2. 2 times 3, 6. 2 times negative 1, negative 2. 2 times 0, 0. 2 times negative 4, negative 8. Highly predictable kind of thing. Now with this idea, we can actually write the expressions and simplify expressions in this matrix format. If this is matrix B then we can evaluate or simplify 2B minus 3A. Now we do it like this. 2B means 2 times matrix B, that's from up here, minus 3 times A. Well matrix A is the one from over here. And now we do the scalar multiplication business, double all of these entries and we get these. Now when we slide over here, we actually have two techniques that we can use. Minus 3, you see it's minus 3 times this. Now I have elected to take the technique where I'm gonna multiply 3 times all of these and then write minus, you see 3 times all of these. The other technique would be to say minus 3 times all of these and add the two matrices which ever way you'd like to do it is okay. But again, I'm doing this, I'm saying, "Okay, it's easier for me to multiply 3 times all these entries and then it's easy for me to subtract down here rather than add." So I'm thinking of it this way, 3 times negative 2, negative 6. 3 times 1, 3. 3 times 3, 9. 3 times negative 1, negative 3. I'm sure you get the picture. Now we have minus between these two and we subtract corresponding entries. So we would have 0 minus negative 6. That's 0 plus 6 or 6 for that entry. Then 4 minus 3 is 1. Then negative 8 minus 9, negative 8 with negative 9, that's negative 17. And then let's see, 6 minus negative 3, that's 6 plus 3, 9. Then 0 minus 0 is 0. And then negative 2 minus negative 12, that's negative 2 plus 12 or 10. Okay. |
fanyattehedzg
2021-12-26
Evaluate the integrals.
$\int \frac{{\left({\mathrm{tan}}^{-1}x\right)}^{5}}{\left(1+{x}^{2}\right)}dx$
Lakisha Archer
Step 1
We have to evaluate the integral:
$\int \frac{{\left({\mathrm{tan}}^{-1}x\right)}^{5}}{\left(1+{x}^{2}\right)}dx$
This integral will be solved by substitution method since derivative of one function is present in the integral.
Assuming,
$t={\mathrm{tan}}^{-1}x$
Differentiating with respect to x, we get
$\frac{dt}{dx}=\frac{d\left({\mathrm{tan}}^{-1}x\right)}{dx}$
$=\frac{1}{1+{x}^{2}}$
$dt=\frac{1}{1+{x}^{2}}dx$
Step 2
Substituting above values in the integral,
$\int \frac{{\left({\mathrm{tan}}^{-1}x\right)}^{5}}{\left(1+{x}^{2}\right)}dx=\int {t}^{5}dt$
$=\frac{{t}^{5+1}}{5+1}+C$ (since $\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C$)
$=\frac{{t}^{6}}{6}+C$
$=\frac{1}{6}{\left({\mathrm{tan}}^{-1}x\right)}^{6}+C$
Where, C is an arbitrary constant.
We have substituted value of t for last step.
Hence, value of integral is $\frac{1}{6}{\left({\mathrm{tan}}^{-1}x\right)}^{6}+C$.
David Clayton
$\int \frac{{\left({\mathrm{tan}}^{-1}x\right)}^{5}}{\left(1+{x}^{2}\right)}dx=\int \frac{{\mathrm{arctan}}^{5}\left(x\right)}{{x}^{2}+1}dx$
$\int \frac{{\mathrm{arctan}}^{5}\left(x\right)}{{x}^{2}+1}dx$
Substitution $u=\mathrm{arctan}\left(x\right)⇒\frac{du}{dx}=\frac{1}{{x}^{2}+1}⇒dx=\left({x}^{2}+1\right)du:$
$=\int {u}^{5}du$
Integral of a power function:
$\int {u}^{n}du=\frac{{u}^{n+1}}{n+1}$ at n=5:
$=\frac{{u}^{6}}{6}$
Reverse replacement $u=\mathrm{arctan}\left(x\right):$
$=\frac{{\mathrm{arctan}}^{6}\left(x\right)}{6}$
$\int \frac{{\mathrm{arctan}}^{5}\left(x\right)}{{x}^{2}+1}dx$
$=\frac{{\mathrm{arctan}}^{6}\left(x\right)}{6}+C$
karton
$\int \frac{\left({\mathrm{tan}}^{-1}x{\right)}^{5}}{\left(1+{x}^{2}\right)}dx=\int \frac{{\mathrm{arctan}}^{5}\left(x\right)}{{x}^{2}+1}dx$
Transform the expression
$\int {t}^{5}dt$
Use $\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1},n\ne -1$ to evaluate the integral
$\frac{{t}^{6}}{6}$
Substitute back
$\frac{\mathrm{arctan}\left(x{\right)}^{6}}{6}$
$\frac{\mathrm{arctan}\left(x{\right)}^{6}}{6}+C$ |
# Find the Cubes of the Following Number by Column Method 72 . - Mathematics
Sum
Find the cubes of the following number by column method 72 .
#### Solution
We have to find the cube of 72 using column method. We have: $a = 7 and b = 2$
Column Ia3 Column II $a = 7 \text{ and } b = 2$ Column III $3 \times a \times b^2$ Column IVb3 $7^3 = 343$ $3 \times a^2 \times b = 3 \times 7^2 \times 2 = 294$ $3 \times a^2 \times b = 3 \times 7^2 \times 2 = 294$ $2^3 = 8$ +30 +8 +0 8 373 302 84 373 2 4 8
Thus, cube of 72 is 373248.
Is there an error in this question or solution?
Chapter 4: Cubes and Cube Roots - Exercise 4.1 [Page 9]
#### APPEARS IN
RD Sharma Class 8 Maths
Chapter 4 Cubes and Cube Roots
Exercise 4.1 | Q 19.3 | Page 9
Share |
## Wednesday, April 9, 2014
### An old post on how to simplfy square roots, an important concept
In a first year algebra class, students will encounter problems involving simplifying square roots. But many times the teacher simply doesn't explain the process by which this is accomplished in a manner that students can understand. I will explain a way that will make simplifying square roots easier for students of any ability level.
The first method I use when teaching students how to calculate square root is to look to see if the number is a perfect square first. I suggest students memorize the perfect squares from 1 to 25 as follows: 1, 4, 9, 16, 25, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625. So if you are asked to calculate the square root of 441, you automatically know the answer is 21 or -21, since a negative number times another negative number yields a positive number.
For non-perfect squares or larger numbers that you are unsure of, if the number is a perfect square, I suggest using a factor tree. For example, suppose you want to simplify a square root of 48. It is not a perfect square since 6 times 6 equals 36 and 7 times 7 equals 49. No whole number times itself equals 48. So break it down into factors. I always suggest trying to find a perfect square as one of the factors and in this case, 16 times 3 equals 48 and 16 is a perfect square. Remember: Since you are dealing with square root, the factors are also square root. So the square root of 48 equals the square root of 16 times the square root of 3. Three is a prime number so you cannot break down 3 any farther using a factor tree. We know that the square root of 16 is 4, so the answer is 4 times the square root of 3.
Another more difficult example---say we need to find the square root of 2025. With such a large number, most people won't know if this is a perfect square, so use the factor tree. Know that any number ending in 5 is divisible by 5. So 5 times 405 is 2025. But 405 can be broken down into factors, using the same rule, therefore 5 times 81 is 405. Now we have the square root of 5 times the square root of 5 times the square root of 81. Notice then that the square root of 5 times the square root of 5 equals the square root of 25. Now this problem becomes simple because you notice we have two perfect squares here, 25 and 81. The square root of 25 is 5 and square root of 81 is 9. So the answer is 5 times 9, which is 45 and -45, since -45 times -45 equals 2025. This problem is actually a perfect square, but if you do not recognize it as such, you can use the factor tree method I just described.
When dealing with the square root of a negative number, imaginary numbers come into play. The square root of -1 equals an imaginary number denoted as "i". So in the above problems if we had the square root of -48, you have square root of -1 times square root of 3 times square root of 16. The answer is 4i times the square root of 3, and -4i times square root of 3. If we have the square root of -2025, the answer is simply 45i and -45i.
Hope my method helps you when trying to figure out the square root of both positive and negative numbers. |
# Slide Rule Using Common and Natural Log Scales
2,508
19
8
Posted
## Introduction: Slide Rule Using Common and Natural Log Scales
The slide rules that took men to the Moon were made without computers, so these instructions are based on computerless production. There are still books in existence that contain the log scales necessary without the use ot computers.
I never used anything except circular slide rules, so this Instructable will give directions on how to make those kind.
Numbers people do not always do crafts well, so for the sake of people who craft well, the numbers to make a Natural Log scale for a slide rule would be helpful and are included at Step 7.
I don't do pictures well, and I speak too much math and too much verbose. I will try to appreciate questions to help me clarify, if the questions are polite. An internet search ought to produce plenty of pictures of "circular slide rules".
The slide rule is based on logarithmic scales, and the accuracy of the slide rule is only as good as the precision of the markings.
1. Mark the uniform log scale.
2. Mark the C scale from the log scale.
3. Mark the Inverse, Square, Cube and / or identical D scales.
4. Mark the Natural Log scale, optional.
## Step 1: Quick Overview of the Math of Logarithms
According to programming conventions, * means multiplication and ^ means exponents, so
2 * 3 = 6
and
3 ^ 2 = 9
The slide rule is based entirely on the logarithmic scale. Logarithms are not really intimidating because they are easily calculated with calculators or software. For now.
log(2) = .30103
10 ^ .30103 = 2
log(3) = .4771
10 ^ .4771 = 3
log(50) = 1.699
10 ^ 1.699 = 50
A logarithm is defined as:
when
a ^ x = y
then
x*log(a) = log(y)
For example, because
2 ^ 3 = 8
then
3 * log(2) = log(8)
and
3 = log(8) / log(2).
Logarithms are usually only roughly equal, they usually require several places for accuracy.
10 ^ .48 = 3.01995
but
10 ^ .4771212 = 2.9999996
## Step 2: Calculations
Choose a precision of N from 1 to 9. 3 is convenient.
If N = 4 then
log(3) = .4771
when N = 2 then
log(3) = .48
The circumference will be determined by the precision N you choose and the length of the scale or markings you choose. If the precision N = 3 and the smallest unit is 1 mm, then the circumference will be
(10 ^ 3) * 1 mm = 1 meter
Because the circumference equates to PI * diameter, the diameter of the circle would be
(10 / PI)* 10 ^ (N - 1) = 3.183 * 10 ^ (N - 1)
in whichever units are being used.
When N = 3 and the units are millimeters, the diameter will be
(10 / PI) * 10 ^ (N-1) = 3.183 * 10 ^ 2
= 318.3 mm
which would be
318.3 / 25.4 inches = 12.53 inches
The diameter would then be 12.53 inches and the radius would be 6.265 inches.
A circle can be formed accurately by fastening the measuring stick in the center of the circle and then marking the radius (half of the diameter) in many directions. There are 360 degrees in a circle.
On the log scale, 360/10 = 36 degrees would be the size of each .1 increment on the log scale.
360 / 100 = 3.6 for each .01 and
360 / 100 = .36 for each .001.
On the blank circle, measure 10 ^ N units for the circumference, using measuring tape or a yardstick. If it is too large, then evenly shave off some of the circumference, but be sure to keep it a perfect circle. If it is too small, then a new circle must be cut.
Start marking the circumference and the face of the blank circle when the circumference is exactly 1,000 units. Paper is better for finding the perfect dimensions and for making patterns.
It is convenient to mark all the .001 increments of the Log scale as 1 millimeter each.
To make sure the slide rule is exactly the right diameter, it would be convenient to use a tape measure or a measuring stick to verify the circumference of the circle being used before beginning to make the markings.
Using a tape measure would also be a convenient way to mark off the 1 millimeter increments around the outside of the circle for the Log Scale.
## Step 3: Mark the Circle
There are 360 degrees in a circle.
On the log scale, 360/10 = 36 degrees would be the size of each .1 increment on the log scale.
360 / 100 = 3.6 for each .01 and
360 / 100 = .36 for each .001.
On the blank circle, measure 10 ^ N units for the circumference, using measuring tape or a yardstick. If it is too large, then evenly shave off some of the circumference, but be sure to keep it a perfect circle. If it is too small, then a new circle must be cut.
Start marking the circumference and the face of the blank circle when the circumference is exactly 1,000 units. Paper is better for finding the perfect dimensions and for making patterns.
It is convenient to mark all the .001 increments of the Log scale as 1 millimeter each.
To make sure the slide rule is exactly the right diameter, it would be convenient to use a tape measure or a measuring stick to verify the circumference of the circle being used before beginning to make the markings. Using a tape measure would also be a convenient way to mark off the 1 millimeter increments around the outside of the circle for the Log Scale.
## Step 4: Protractors
This is also a convenient way to make a protractor, which makes it a lot easier to make gears.
If we use 1 mm per each degree,
360 millimeters / 3.141592654 = 114.6 mm diameter, or 57.3 mm radius.
pi = 3.141592654
Millimeters are convenient units, but as long as they are all uniform, any unit would serve.
## Step 5: Two Dials or Two Cursor Hands?
I personally like the slide rules that have an inner turning dial somewhat better than the ones that have two clear cursors like clock hands. Using two cursor hands requires that they be turned on the face of the slide rule without moving at all relative to the other cursor hand.
If there is a rotating inner circle, then there would be a D scale on the outside (of the inner circle) to line up exactly with the C scale on the inside (of the outer ring).
Fractions work much better with two dials that can rotate relative to each other.
## Step 6: Scales: C (&D) B, K, CI
In my opinion, the log scale should be the scale on the outside of the largest outside ring of the slide rule, because it requires the highest precision and because it is the foundation on which all the other scales rest.
It makes it easier to read the scales when the smallest increments are the shortest length of marking and the 5s are one length and the 10s are another length.
The range of the C scale is 0 to 1 in increments of .1 and .01 (and .001 if large enough). The C scale is 10 raised to the log scale, so that 2 on the C scale would line up with .301 on the Log scale, and 7 on the C scale with .845.
The D scale is exactly the same as the C scale and is only used on slide rules with two dials the rotate independently.
The CI scale is the inverse of the C, so 5 on the C scale would line up with 2 on CI and 2 on the C scale would line up with 5 on CI. Also, 3 would line up with 333, and 4 would line up with 25.
The B scale is C ^ 2 and the K scale is C ^ 3. With a LogLog scale, these scales are less necessary.
Slide rules assume mental calculation of exponents.
## Step 7: Natural Log Scale
The Natural Log scale is especially useful to verify interest rates being charged. The Natural Log scale can also calculate exponents directly.
On the Natural Log Scale, e ^ .02, e ^ .2, e ^ 2, e ^ 20 all line up, one above the other, under 2 on the C scale. Decimals and magnitudes are calculated mentally. Attached are some tables of natural logs, but it would be a good idea to verify the numbers with an old book of mathematical tables.
ln(LL scale) = C scale
The Natural Log scale can be a spiral for the artistically adept. It is also shown as nesting circles with steps up.
The characteristics (exponents) are left up to mental calculations, but it is easy to calculate 10 ^ 3. To find the characteristic or magnitude of the exponent of e only requires following the line down in value to e ^ (10N) which is 1 on the C scale, 0 on log scale, to easily see the exponent of e that applies. For example, e ^ .1 = 1.1052 and e ^ .04 = 1.0408.
PRIME NUMBERS
From the logs of primes, other values can be calculated. There are about 200 prime numbers between 1 and 1,000.
A slide rule can be created with only a table of the common and natural logs of prime numbers. The other values can be calculated with each prime number and its common (base 10) or natural (base e) logarithm.
For example:
4.2 = (2 x 3 x 7) / 10
= 10 ^ [log(2) + log(3) + log(7) - log(10)]
= 10 ^ [.30103 + .47712 +.8451 - 1]
4.2 = 10 ^ .62345
However, people who most need a slide rule may be more prone to make mistakes in calculating from prime numbers.
## Step 8: Master Model
To increase precision, measure out at a large scale, then scale down to a smaller scale.
Make a very large scale outer "donut" template from which to form the portable, personal slide rules.
In the center, have a pin to put the blank slugs on to make into more portable slide rules.
## Recommendations
• ### 3D CAM and CNC Class
464 Enrolled
• ### Make it Move Contest
We have a be nice policy.
## Questions
I had teachers who used a slide rule but no one ever taught us to use one. We just pushed buttons. Do you know a web site that I could use to learn? Got to be very user friendly. My mom tried to teach me once. The results were not pretty--we ended up not talking for a few days. Really not good memories.
2 replies
I did have to use the instruction books that came with mine to learn some functions. But the basic rule of thumb to use when learning anything new is to start with what you know.
Start with an equation like 2 X 3 = 6. It is best to use different numbers when learning, rather than 2 X 2. Point one cursor to 1 and the other to 2, then move the cursor from 1 to 3.
If the cursors did not move relative to each other, the cursor that pointed to 2 should now be pointing to 6.
If your circular slide rule has a dial so that the C and D scales move relative to each other, then point one "1" cursor to 2, then on the same scale that has the "1" pointing to 2, the 3 should be pointing to 6.
This also shows that 6 / 3 = 2 and 6 / 2 = 3.
For the other scales (functions), I look at what lines up with 2. If 5 lines up with 2, then it is an inverse scale, and 4 will line up with 2.5. Be careful because this scale is reversed from the others.
If 2 lines up with 4 and 3 lines up with 9, it is a "B" scale, which means it shows squares, and square roots. If 2 lines up with .3, it is an "L" or log10 scale.
For the sine, cosine and tangent scales, I had to make a short list of sines and cosines for 30, 45, and 60 degrees.
Just play! That is the best way to learn.
Or you could use these sites:
Eric's Slide Rules
http://www.sliderule.ca/manual.htm
Greg's Slide Rules
Ron Manley's Slide Rules site
http://www.sliderules.info/a-to-z/manuals.htm
Or YouTube "how to use a slide rule"
Thanks. This is very helpful. I actually feel like I can learn this. Thanks!
I love the pictures, microsoft paint all the way! |
What is a 3 3 identity matrix?
Linear Algebra. Find the 3×3 Identity Matrix 3. 3. The identity matrix or unit matrix of size 3 is the 3x⋅3 3 x ⋅ 3 square matrix with ones on the main diagonal and zeros elsewhere.
What is a 3 by 2 matrix?
When we describe a matrix by its dimensions, we report its number of rows first, then the number of columns. Matrix A is therefore a ‘3 by 2’ matrix, which is written as ‘3×2. Matrix B has 2 rows and 3 columns. We call numbers or values within the matrix ‘elements.
What is unit matrix of order 3?
3× 3 Identity Matrix This is known as the identity matrix of order 3 or unit matrix of order 3 × 3. Identity Matrix is donated by In × n, where n × n shows the order of the matrix. A × I n × n = A, A = any square matrix of order n × n.
What is the rank of null matrix?
Since the null matrix is a zero matrix, we can use the fact that a zero matrix has no non-zero rows or columns, hence, no independent rows or columns. So, we have found out that the rank of a null matrix is 0.
What will be the rank of matrix?
How to Find Matrix Rank. The maximum number of linearly independent vectors in a matrix is equal to the number of non-zero rows in its row echelon matrix. Therefore, to find the rank of a matrix, we simply transform the matrix to its row echelon form and count the number of non-zero rows.
What is the rank of unit matrix of order 3 by 3?
The rank of a 3×3 unit matrix is 3.
How to find the 3×3 identity matrix 3?
The identity matrix or unit matrix of size 3 3 is the 3x⋅3 3 x ⋅ 3 square matrix with ones on the main diagonal and zeros elsewhere. In this case, the identity matrix is ⎡ ⎢⎣1 0 0 0 1 0 0 0 1⎤ ⎥⎦ [ 1 0 0 0 1 0 0 0 1].
Which is an example of an identity matrix?
Here, the 2 x 2 and 3 x 3 identity matrix is given below: Identity Matrix is donated by I n X n, where n X n shows the order of the matrix. A X I n X n = A, A = any square matrix of order n X n.
When do you get an identity matrix after multiplying two inverse matrices?
The above is 2 x 4 matrix as it has 2 rows and 4 columns. 3) We always get an identity after multiplying two inverse matrices. If we multiply two matrices which are inverses of each other, then we get an identity matrix.
Is the identity matrix A square or square matrix?
The “identity” matrix is a square matrix with 1 ‘s on the diagonal and zeroes everywhere else. Multiplying a matrix by the identity matrix I (that’s the capital letter “eye”) doesn’t change anything, just like multiplying a number by 1 doesn’t change anything.
What is a 3 3 identity matrix? Linear Algebra. Find the 3×3 Identity Matrix 3. 3. The identity matrix or unit matrix of size 3 is the 3x⋅3 3 x ⋅ 3 square matrix with ones on the main diagonal and zeros elsewhere. What is a 3 by 2 matrix? When we describe a matrix… |
Search 72,991 tutors
0 0
## how do i solve 1-3(X+1)+7(X-6)>0
IM SO CONFUSED
To solve this inequality, you first need to simplify the left side.
You can do this by first applying the distributive property; you will distribute the coefficients outside of the parentheses to the terms inside the parentheses through multiplication.
For the first set of parentheses you multiply -3 by X and by 1. Your result will be -3x and -3. Since you have distributed, you can eliminate the parentheses.
Let's use the distributive property on the second parentheses. Multiply 7 by X and -6. Your result is 7x - 42. Again, you can eliminate the parentheses.
The next step is to combine your like terms. Like terms are terms that have the same variable and the same degree (exponent).
The like terms on the left side are the constants: 1, -3, and -42. Let's combine the negative integers first. -42 and -3 make -45. Add 1 and your result is -44
You can rewrite the inequality to read -3x +7x - 44 >0
Now combine the remaining like terms: -3x and 7x. The commutative property allows us to switch the order of these terms without changing the result: 7x -3x. The result is 4x.
Almost there! We want to solve for the variable x, so we need to isolate it, or get it by itself on one side of the inequality. To do that, we will use inverse operations to "undo" what is happening to the x.
X is being multiplied by 4 and X is subtracting 44.
To "undo" subtracting 44, you should add 44 to both sides of the inequality. 4x -44 + 44 > 0 + 44. The constants eliminate each other on the left, and the right side becomes 44.
4x > 44
X is being multiplied by 4. To "undo" multiplying by 4, you should divide both sides of the inequality by 4.
4x /4 > 44/4 =
x > 11
This is your solution set. To check your work, substitute a value greater than 11 into your inequality. If the inequality is true, values than 11 work. You should also substitute a value less than 11 into your inequality. That should make the inequality false.
Check: Let x = 20
1 - 3(12+1) + 7(12-6) > 0 =
1 - 3(13) + 7(6) > 0 =
1- 39 + 42 > 0 =
-38 + 42 > 0 =
4 > 0 is true, so x can be greater than 11.
Check: Let x = 10
1 -3(10+1) + 7(10-6) >0 =
1- 3(11) + 7(4) > 0 =
1 - 33 + 28 > 0 =
-32 + 28 > 0 =
-4 > 0 is not true. Therefore, x cannot be less than 11.
I hope this helped! |
# How do you write the standard form of a line given (-6, -3) and (-8, 4)?
Jan 19, 2017
$\textcolor{red}{7} x + \textcolor{b l u e}{2} y = \textcolor{g r e e n}{- 48}$
#### Explanation:
First, we can use the point-slope formula to obtain the equation of the line. To do this we must first find the slope of the line which can be done because two points are provided.
The slope can be found by using the formula: $m = \frac{\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}}{\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}}$
Where $m$ is the slope and ($\textcolor{b l u e}{{x}_{1} , {y}_{1}}$) and ($\textcolor{red}{{x}_{2} , {y}_{2}}$) are the two points on the line.
Substituting the values from the problem gives:
$m = \frac{\textcolor{red}{4} - \textcolor{b l u e}{- 3}}{\textcolor{red}{- 8} - \textcolor{b l u e}{- 6}}$
$m = \frac{7}{-} 2 = - \frac{7}{2}$
The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$
Where $\textcolor{b l u e}{m}$ is the slope and $\textcolor{red}{\left(\left({x}_{1} , {y}_{1}\right)\right)}$ is a point the line passes through.
Substituting the calculate slope and the first point gives:
$\left(y - \textcolor{red}{- 3}\right) = \textcolor{b l u e}{- \frac{7}{2}} \left(x - \textcolor{red}{- 6}\right)$
$\left(y + \textcolor{red}{3}\right) = \textcolor{b l u e}{- \frac{7}{2}} \left(x + \textcolor{red}{6}\right)$
Now we can transform this into the standard form.
The standard form of a linear equation is:
$\textcolor{red}{A} x + \textcolor{b l u e}{B} y = \textcolor{g r e e n}{C}$
where, if at all possible, $\textcolor{red}{A}$, $\textcolor{b l u e}{B}$, and $\textcolor{g r e e n}{C}$are integers, and A is non-negative, and, A, B, and C have no common factors other than 1
$y + \textcolor{red}{3} = \left(\textcolor{b l u e}{- \frac{7}{2}} \times x\right) + \left(\textcolor{b l u e}{- \frac{7}{2}} \times \textcolor{red}{6}\right)$
$y + \textcolor{red}{3} = - \frac{7}{2} x - \frac{7 \times 6}{2}$
$\textcolor{red}{2} \left(y + 3\right) = \textcolor{red}{2} \left(- \frac{7}{2} x - \frac{42}{2}\right)$
$2 y + 6 = \left(\textcolor{red}{2} \times - \frac{7}{2} x\right) - \left(\textcolor{red}{2} \times \frac{42}{2}\right)$
$2 y + 6 = \left(\cancel{\textcolor{red}{2}} \times - \frac{7}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} x\right) - \left(\cancel{\textcolor{red}{2}} \times \frac{42}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}\right)$
$2 y + 6 = - 7 x - 42$
$2 y + 6 + \textcolor{red}{7 x} - \textcolor{b l u e}{6} = - 7 x - 42 + \textcolor{red}{7 x} - \textcolor{b l u e}{6}$
$\textcolor{red}{7 x} + 2 y + 6 - \textcolor{b l u e}{6} = - 7 x + \textcolor{red}{7 x} - 42 - \textcolor{b l u e}{6}$
$\textcolor{red}{7 x} + 2 y + 0 = 0 - 48$
$\textcolor{red}{7} x + \textcolor{b l u e}{2} y = \textcolor{g r e e n}{- 48}$ |
# How do you factor the square root of 175?
## How do you factor the square root of 175?
25 can be shown as 52. Thus, the simplified radical form of the square root of 175 is 5√7.
### Is 175 square number?
175 is not a perfect square.
#### Which of the following is a factor of 175?
The factors of 175 are 1, 5, 7, 25, 35 and 175.
What is 175 Simplified?
Rewrite 175 as 52⋅7 5 2 ⋅ 7 . Factor 25 25 out of 175 175 .
What is the cube of 175?
Cube root of 175 can be represented as 3√175. The value of cube root of one is 175. The nearest previous perfect cube is 125 and the nearest next perfect cube is 216 . Cube root of 175 can be represented as 3√175.
## Is 175 rational or irrational?
175 is a rational number because it can be expressed as the quotient of two integers: 175 ÷ 1.
### What is the factoring method?
Factoring is the process by which we go about determining what we multiplied to get the given quantity. A common method of factoring numbers is to completely factor the number into positive prime factors. A prime number is a number whose only positive factors are 1 and itself.
#### What are multiples of 175?
The first 5 multiples of 175 are 175, 350, 525, 700, 875. The sum of the first 5 multiples of 175 is 2625 and the average of the first 5 multiples of 175 is 525. Multiples of 175: 175, 350, 525, 700, 875, 1050, 1225, 1400, 1575, 1750 and so on.
What is the LCM of 175?
The first few multiples of 125 and 175 are (125, 250, 375, 500, . . . ) and (175, 350, 525, 700, 875, 1050, 1225, . . . ) respectively….LCM of 125 and 175.
1. LCM of 125 and 175
4. FAQs
What is the cube root of 17576 through estimation?
26
∴3√17576 = 26.
## How to simplify the square root of 175?
To simplify the square root of 175 means to get the simplest radical form of √175. List the factors of 175 like so: Identify the perfect squares* from the list of factors above: Divide 175 by the largest perfect square you found in the previous step: Calculate the square root of the largest perfect square:
### What are the factors of 175?
In other words, the numbers that are multiplied together in pairs resulting in the number 175 are the factors of 175. As the number 175 is a composite number, it has many factors other than 1 and 175. Therefore, the factors of 175 are 1, 5, 7, 25, 35 and 175.
#### Is 175 a perfect square?
Is 175 a perfect square? 175 is a perfect square if the square root of 175 equals a whole number. As we have calculated further down on this page, the square root of 175 is not a whole number. 175 is not a perfect square.
What is the justification for taking out the square root?
The justification for taking out the square root of any number is this theorem to help simplify √a*b = √a * √b. The square root of a number is equal to the number of the square roots of each factor. Is 175 A Prime Number? |
# Factors of 824: Prime Factorization, Methods, and Examples
The factors of 824 are numbers that, when divided by 824, give zero as the remainder. The given number’s factors can be positive and negative.
### Factors of 824
Here are the factors of number 824.
Factors of 824: 1, 2, 4, 8, 103, 206, 412, and 824.
### Negative Factors of 824
The negative factors of 824 are similar to its positive aspects, just with a negative sign.
Negative Factors of 824: –1, -2, -4, -8, -103, -206, -412, and -824
### Prime Factorization of 824
The prime factorization of 824 is the way of expressing its prime factors in the product form.
Prime Factorization: 2 x 2 x 2 x 103
In this article, we will learn about the factors of 824 and how to find them using various techniques such as upside-down division, prime factorization, and factor tree.
## What Are the Factors of 824?
The factors of 824 are 1, 2, 4, 8, 103, 206, 412, and 824. These numbers are the factors as they do not leave any remainder when divided by 824.
The factors of 824 are classified as prime numbers and composite numbers. The prime factors of the number 824 can be determined using the prime factorization technique.
## How To Find the Factors of 824?
You can find the factors of 824 by using the rules of divisibility. The divisibility rule states that any number, when divided by any other natural number, is said to be divisible by the number if the quotient is the whole number and the resulting remainder is zero.
To find the factors of 824, create a list containing the numbers that are exactly divisible by 824 with zero remainders. One important thing to note is that 1 and 824 are the 824’s factors as every natural number has 1 and the number itself as its factor.
1 is also called the universal factor of every number. The factors of 824 are determined as follows:
$\dfrac{824}{1} = 824$
$\dfrac{824}{2} = 412$
$\dfrac{824}{4} = 206$
$\dfrac{824}{8} = 103$
Therefore, 1, 2, 4, 8, 103, 206, 412, and 824 are the factors of 824.
### Total Number of Factors of 824
For 824, there are eight positive factors and eight negative ones. So in total, there are 16 factors of 824.
To find the total number of factors of the given number, follow the procedure mentioned below:
1. Find the factorization/prime factorization of the given number.
2. Demonstrate the prime factorization of the number in the form of exponent form.
3. Add 1 to each of the exponents of the prime factor.
4. Now, multiply the resulting exponents together. This obtained product is equivalent to the total number of factors of the given number.
By following this procedure, the total number of factors of X is given as:
Factorization of 824 is 1 x 2$^3$ x 103.
The exponent of 1, and 103 is 1 where as 2 is 3.
Adding 1 to each and multiplying them together results in 16.
Therefore, the total number of factors of 824 is 16. Eight are positive, and eight factors are negative.
### Important Notes
Here are some essential points that must be considered while finding the factors of any given number:
• The factor of any given number must be a whole number.
• The factors of the number cannot be in the form of decimals or fractions.
• Factors can be positive as well as negative.
• Negative factors are the additive inverse of the positive factors of a given number.
• The factor of a number cannot be greater than that number.
• Every even number has 2 as its prime factor, the smallest prime factor.
## Factors of 824 by Prime Factorization
The number 824 is a composite number. Prime factorization is a valuable technique for finding the number’s prime factors and expressing the number as the product of its prime factors.
Before finding the factors of 824 using prime factorization, let us find out what prime factors are. Prime factors are the factors of any given number that are only divisible by 1 and themselves.
To start the prime factorization of 824, start dividing by its most minor prime factor. First, determine that the given number is either even or odd. If it is an even number, then 2 will be the smallest prime factor.
Continue splitting the quotient obtained until 1 is received as the quotient. The prime factorization of 824 can be expressed as:
824 = 2 x 2 x 2 x 103
## Factors of 824 in Pairs
The factor pairs are the duplet of numbers that, when multiplied together, result in the factorized number. Factor pairs can be more than one depending on the total number of factors given.
For 824, the factor pairs can be found as:
1 x 824 = 824
2 x 412 = 824
4 x 206 = 824
8 x 103 = 824
The possible factor pairs of 824 are given as (1, 824), (2, 412), (4, 206), and (8, 103).
All these numbers in pairs, when multiplied, give 824 as the product.
The negative factor pairs of 824 are given as:
1 x -824 = 824
-2 x -412 = 824
-4 x -206 = 824
-8 x -103 = 824
It is important to note that in negative factor pairs, the minus sign has been multiplied by the minus sign, due to which the resulting product is the original positive number. Therefore, 1, -2, -4, -8, -103, -206, -412, and -824 are called negative factors of 824.
The list of all the factors of 824, including positive as well as negative numbers, is given below.
Factor list of 824: 1, -1, 2, -2, 4, -4, 8, -8, 103, -103, 206, -206, 412, -412, 824, and -824
## Factors of 824 Solved Examples
To better understand the concept of factors, let’s solve some examples.
### Example 1
How many factors of 824 are there?
### Solution
The total number of Factors of 824 is 16.
Factors of 824 are 1, 2, 4, 8, 103, 206, 412, and 824.
### Example 2
Find the factors of 824 using prime factorization.
### Solution
The prime factorization of 824 is given as:
824 $\div$ 2 = 412
412 $\div$ 2 = 206
206 $\div$ 2 = 103
103 $\div$ 103 = 1
So the prime factorization of 824 can be written as:
2 x 2 x 2 x 103 = 824 |
Intro to Subtraction Math Lesson Plan - Grades K-3
1%
It was processed successfully!
WHAT IS SUBTRACTION?
Subtraction is taking away one group from a bigger group. When you subtract you are finding how many are left. You can subtract by counting how many are left. You can also subtract by counting back.
To better understand subtraction…
WHAT IS SUBTRACTION?. Subtraction is taking away one group from a bigger group. When you subtract you are finding how many are left. You can subtract by counting how many are left. You can also subtract by counting back. To better understand subtraction…
## LET’S BREAK IT DOWN!
### Count to find how many cars in all.
Counting tells you how many you have in all. There are toy cars on a table. Say one number for each car. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. There are 10 cars on the table.
Count to find how many cars in all. Counting tells you how many you have in all. There are toy cars on a table. Say one number for each car. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. There are 10 cars on the table.
### Count how many baby chicks are left.
There are baby chicks in a nest. Count the chicks to see how many in all. 1, 2, 3, 4, 5. There are 5 chicks in all. Sasha takes 1 chick out of the nest. How many chicks are left? 1, 2, 3, 4. There are 4 chicks left.
Count how many baby chicks are left. There are baby chicks in a nest. Count the chicks to see how many in all. 1, 2, 3, 4, 5. There are 5 chicks in all. Sasha takes 1 chick out of the nest. How many chicks are left? 1, 2, 3, 4. There are 4 chicks left.
### Use “–” and “=” to show subtraction.
A “–” shows a group being taken away from another group. 5 – 1 = 4. The “–” shows that 1 chick was taken away from a group of 5. The “=” shows that 5 takeaway 1 is equal to 4. There are 4 chicks left.
Use “–” and “=” to show subtraction. A “–” shows a group being taken away from another group. 5 – 1 = 4. The “–” shows that 1 chick was taken away from a group of 5. The “=” shows that 5 takeaway 1 is equal to 4. There are 4 chicks left.
### Use fingers to subtract donuts.
You made 8 donuts. Your family at 3. How many donuts are left? Show 8 with your fingers. Put 3 fingers down. Count how many fingers you still have up:1, 2, 3, 4, 5. There are 5 donuts left. 8 – 3 = 5.
Use fingers to subtract donuts. You made 8 donuts. Your family at 3. How many donuts are left? Show 8 with your fingers. Put 3 fingers down. Count how many fingers you still have up:1, 2, 3, 4, 5. There are 5 donuts left. 8 – 3 = 5.
### Count back to subtract balloons.
There were 10 balloons. Then 3 balloons popped. How many balloons are left? Start at 10. Count back 3: 9, 8, 7. There are 7 balloons left. 10 – 3 = 7.
Count back to subtract balloons. There were 10 balloons. Then 3 balloons popped. How many balloons are left? Start at 10. Count back 3: 9, 8, 7. There are 7 balloons left. 10 – 3 = 7.
## INTRO TO SUBTRACTION VOCABULARY
To put together. If you add 3 and 2, you find how many there are in all.
Subtraction
To take away. If you take away 2 from 3, you find how many are left after 2 are taken away from 3.
Count on
To count in increasing counting order. 1, 2, 3, 4, 5, 6, …
Count back
To count in decreasing counting order. 9, 8, 7, 6, 5, …
Greater
A number is greater than another number if you count it second when counting.
Lesser
A number is less than another number if you count it first when counting.
## INTRO TO SUBTRACTION DISCUSSION QUESTIONS
### What does subtraction mean?
Subtraction means to take away. When I subtract 3 from 5, I take away 3 from 5.
### How can you use counters to show subtraction?
I can use counters to show how many there are to start. Then I can take away counters to show how many I subtract. Then I can count the amount that is left. Or I can count back from the start number as I take away the number of counters to subtract.
### Look at the number sentence 6 – 4 = 2. What does the – mean? What does the 2 mean?
The – means that 4 is taken away from 6. The 2 means that there are 2 left over. 6 take away 4 is 2.
### There are 4 balls in a bin. 1 is taken out. How many balls are left in the bin?
4 – 1 = 3. There are 3 balls left.
### How do you know what number to take away?
I always take away the smaller (lesser) number. The words in the problem tell me what number is there to start and what is taken away.
X
Success
We’ve sent you an email with instructions how to reset your password.
Ok
x
3 Days
Continue to Lessons
30 Days
Get 30 days free by inviting other teachers to try it too.
Share with Teachers
Get 30 Days Free
By inviting 4 other teachers to try it too.
4 required
*only school emails accepted.
Skip, I will use a 3 day free trial
Thank You!
Enjoy your free 30 days trial |
##### answer this one. show work. algebra 1
Algebra Tutor: None Selected Time limit: 1 Day
Apr 22nd, 2015
Mary,
First look at the place where the line crosses the x-axis and where it crosses the y-axis. The x-intercept is at the point (5/2,0) and the y-intercept is at the point (0,-1). From these two points we can calculate the slope of the line, which should be positive since the line has an upward incline. Mathematically, the slope is defined as the change is y divided by the change in x: we can calculate each using the two points (5/2,0) and (0,-1).
slope = [0 - (-1)]/[5/2 - 0] = 1/(5/2) = 2/5 = m and the standard equation of a line is y = m*x + b. We now know that
y = (2/5)*x + b. We need to determine b. Notice from that graph that when x = 0, y = -1. Plug these two values into the previous equation and you get
-1 = (2/5)*0 + b, which means that b = -1 since (2/5)*0 = 0.
So, the relationship between x and y as shown in the line is best described mathematically as y = (5/3)*x - 1.
Apr 21st, 2015
algebra part.PNG these are the options look over it again.
Apr 22nd, 2015
Whoops. 5/2 should have been 3/2. Let me redo here.
The x-intercept is at the point (3/2,0) and the y-intercept is at the point (0,-1). From these two points we can calculate the slope of the line, which should be positive since the line has an upward incline. Mathematically, the slope is defined as the change is y divided by the change in x: we can calculate each using the two points (3/2,0) and (0,-1).
slope = [0 - (-1)]/[3/2 - 0] = 1/(3/2) = 2/3 = m and the standard equation of a line is y = m*x + b. We now know that
y = (2/3)*x + b. We need to determine b. Notice from that graph that when x = 0, y = -1. Plug these two values into the previous equation and you get
-1 = (2/3)*0 + b, which means that b = -1 since (2/3)*0 = 0.
So, the relationship between x and y as shown in the line is best described mathematically as y = (2/3)*x - 1.
In terms of the options given, it would be: Option D. Sorry about that. Let me know if you find any other issues.
Apr 22nd, 2015
...
Apr 22nd, 2015
...
Apr 22nd, 2015
May 27th, 2017
check_circle |
show that
9/8<summation(1/(n^3))<5/4
2. Originally Posted by nzm88
show that
9/8<summation(1/(n^3))<5/4
Do you mean $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^3}$ is bounded between $\displaystyle \frac{9}{8}$ and $\displaystyle \frac{5}{4}$ ?
Consider $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^3}= 1 + \frac{1}{2^3} + \frac{1}{3^3} + ... > 1 + \frac{1}{2^3 } = 1 + \frac{1}{8} = \frac{9}{8}$ , the lower bound .
Now , we are going to find the upper bound :
$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^3}$
$\displaystyle = 1 + \sum_{n=2}^{\infty} \frac{1}{n^3}$
Consider $\displaystyle 0 < n^3 - n < n^3 , n \geq 2$
$\displaystyle \frac{1}{ n^3 - n} > \frac{1}{n^3}$
$\displaystyle \sum_{n=2}^{\infty}\frac{1}{ n^3 - n} > \sum_{n=2}^{\infty}\frac{1}{n^3}$
$\displaystyle \sum_{n=2}^{\infty} \left [ \frac{1}{n(n^2 - 1)} \right ] > \sum_{n=2}^{\infty}\frac{1}{n^3}$
$\displaystyle \sum_{n=2}^{\infty} \left [ \frac{1}{2} \left ( \frac{1}{ n -1} + \frac{1}{n + 1} \right ) - \frac{1}{n} \right ] > \sum_{n=2}^{\infty}\frac{1}{n^3}$
The left hand side is a telescoping series , the value is $\displaystyle \frac{1}{4}$
Therefore ,
$\displaystyle \frac{1}{4} > \sum_{n=2}^{\infty}\frac{1}{n^3}$
$\displaystyle 1 + \frac{1}{4 } = \frac{5}{4} > 1 + \sum_{n=2}^{\infty}\frac{1}{n^3} = \sum_{n=1}^{\infty}\frac{1}{n^3}$
3. Originally Posted by simplependulum
Do you mean $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^3}$ is bounded between $\displaystyle \frac{9}{8}$ and $\displaystyle \frac{5}{4}$ ?
Consider $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^3}= 1 + \frac{1}{2^3} + \frac{1}{3^3} + ... > 1 + \frac{1}{2^3 } = 1 + \frac{1}{8} = \frac{1}{9}$ , the lower bound .
Typo: $\displaystyle 1+ \frac{1}{8}= \frac{8}{9}$, of course.
Now , we are going to find the upper bound :
$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^3}$
$\displaystyle = 1 + \sum_{n=2}^{\infty} \frac{1}{n^3}$
Consider $\displaystyle 0 < n^3 - n < n^3 , n \geq 2$
$\displaystyle \frac{1}{ n^3 - n} > \frac{1}{n^3}$
$\displaystyle \sum_{n=2}^{\infty}\frac{1}{ n^3 - n} > \sum_{n=2}^{\infty}\frac{1}{n^3}$
$\displaystyle \sum_{n=2}^{\infty} \left [ \frac{1}{n(n^2 - 1)} \right ] > \sum_{n=2}^{\infty}\frac{1}{n^3}$
$\displaystyle \sum_{n=2}^{\infty} \left [ \frac{1}{2} \left ( \frac{1}{ n -1} + \frac{1}{n + 1} \right ) - \frac{1}{n} \right ] > \sum_{n=2}^{\infty}\frac{1}{n^3}$
The left hand side is a telescoping series , the value is $\displaystyle \frac{1}{4}$
Therefore ,
$\displaystyle \frac{1}{4} > \sum_{n=2}^{\infty}\frac{1}{n^3}$
$\displaystyle 1 + \frac{1}{4 } = \frac{5}{4} > 1 + \sum_{n=2}^{\infty}\frac{1}{n^3} = \sum_{n=1}^{\infty}\frac{1}{n^3}$
5. Originally Posted by HallsofIvy
Typo: $\displaystyle 1+ \frac{1}{8}= \frac{8}{9}$, of course.
Is this another typo?
its $\displaystyle \frac{9}{8}$.
6. Originally Posted by TWiX
Is this another typo?
its $\displaystyle \frac{9}{8}$.
fractional dyslexia ... many are afflicted.
7. Originally Posted by skeeter
fractional dyslexia ... many are afflicted.
Do not use hard words
Use simple words
My English is horrible
8. HaHa , I am just wondering what if my post is published . |
# Introduction to Fractions
Contributor: Delaine Thomas. Lesson ID: 12784
Can you eat an entire pizza yourself? You can't eat it or share it unless you cut it into slices. Dividing things into parts makes fractions, and you will need to know that for your own pizza parlor!
categories
## Fractions and Operations
subject
Math
learning style
Visual
personality style
Lion, Beaver
Intermediate (3-5)
Lesson Type
Quick Query
## Lesson Plan - Get It!
Audio:
Put on your detective hat, follow the clues, and try to figure out what you will be learning today! Don't worry if you don't catch on to the video right away; it's a bit strange!
Attention Getter to Fractions - TEASe
That’s right — you are going to learn about fractions!
A fraction is a part of a whole. For example, if I had a chocolate pie and I cut it into 8 slices, and I ate 1 piece of the pie, I could say that I ate 18 of the chocolate pie.
Watch Introduction to Fraction for Kids, from Smart Learning for All, so you can learn more about the parts of a fraction:
Use what you learned from the video and take a few seconds to study the circle below:
• First, how many parts are in the whole circle?
That’s right! There are 3 equal parts in the circle.
• Is this number the numerator or the denominator?
Yes, it is the denominator, because the denominator tells us how many parts are in the whole thing. So we write it under the fraction bar:
• How many of the parts are red?
If you said, "1," you are correct. There is only 1 red part. So 1 is the numerator because it tells us how many parts of the whole we are talking about.
Let’s try another one.
• How many parts are in this circle?
There are 6 parts in the circle.
• How many are red?
• Which number, the 6 or the 4, is the numerator?
The numerator tells us how many parts we are talking about, so those are the parts that are colored red. In this case, there are 4 parts out of the 6 that are red.
Our fraction looks like this:
Fractions are very useful in our everyday lives.
• Can you think of some ways fractions might be useful to you?
Share your ideas with you parent or teacher, then move on to the Got It? section to have some fun with worksheets!
## Elephango's Philosophy
We help prepare learners for a future that cannot yet be defined. They must be ready for change, willing to learn and able to think critically. Elephango is designed to create lifelong learners who are ready for that rapidly changing future. |
Share
Explore BrainMass
# Linear Algebra
### Algebra: Solving and graphing systems of equations
I have a few questions I need help with. What are the steps to solving a system of equation by elimination? What are the steps to solving a system of equation by substitution? What are the steps to solving a system of equation by graphing? Once you have graph a system of equation, how do you check your answer? Whe
### Equations
Provide Details using addition/elimination.....Strategy for the addition Method. 3x+5y=-11 x-2y=11 Solving each system by substitution or addition 2x-x=3 x=3y-5 solve by substitution, indicate, independent, dependent, or inconsistent 3y=x+5 3x-9y=-10 solve by addition, Indicate, independent, dependent, or inconsi
### Methods of Solving Systems of Equations
When solving a system of equations, what are the 3 possible outcomes? Describe how you can determine each of the possible outcomes using one of the algebraic methods. Do not look at the graph. Do not look at the individual equations, just look at the outcome, the "answer" you get after using substitution or elimination/a
### Systems of Equations
Systems of equations can be solved by graphing or by using substitution or elimination. What are the pros and cons of each method? Which method do you like best? Why? What circumstances would cause you to use a different method? Review examples 2, 3, and 4 in section 8.4 of the text. How does the author determine what the
### Real-Life Application of a Linear Equation
Linear equations provide information on how quickly data is rising or falling, known as the slope of the equation. Find a real-world application of a linear equation and discuss the meaning of the equation. Explain how the values of the slope affect the overall meaning of the equation.
### Prove that Binomial[2n, n] = Sum from k=0 to n of Binomial[n,k]^2
Prove that (2n choose n) = the sum from k=0 to n of (n choose k) squared.
### Algebra
Sue is offered a new job selling furniture where she makes a base pay of \$1500 a month plus commission. According to a current employee, she should be able to earn \$30,000 in sales monthly. To maintain her standard of living she must make \$3000 total monthly. What commission rate must she earn to maintain her standard of livi
### Functions and Linear Equations
Could you explain to me the difference between functions and linear equations? I need to know if all linear equations are functions and if there are any instances in which a linear equation is not a function. Could you create an equation of a non linear function and provide 2 inputs so they can be evaluated? also Could you e
### Inconsistent or dependent system of equations
Solve the system of linear equations. If the unique solution does not exist state weather the system is inconsistent or dependent. 1. -x - y = 4 x -y = -8 2. 3x - 2y = 23 x +5y = -15 3. 4x + 2y = 4 y = 2 - 2x 4. 4x - 3y = 9 2x + y = 12 5. 5x + 4y = 40 x + 2y = 11
### Probability, Linear Programming and Statistics
Show your work. 1. (3 points) Find the equation of the line shown: 2. (6 points) A bank loaned \$15,000, some at an annual rate of 16% and some at an annual rate of 10%. If the income from these loans was \$1800, how much was loaned at 10%? 3. (3 point) Write the augmented matrix of the system: 2x1 -3x2 +
### Systems of Equations with Three Variables
Given Π0 = 0.5Π0 + 0.3Π1 +0.2 Π2 Π1 = 0.4Π0 + 0.4Π1 + 0.3Π2 Π2 = 0.1Π0 + 0.3Π1 + 0.5Π2 Where 1 = Π0 + Π1 + Π2 The solution should yield: Π0 = 21/62 Π1 = 23/62 Π2 = 18/62 Please see the attached file for the fully formatted
### Graphing Equations and Inequalities
Plot (- 1, 4) on the coordinate axes; I cannot insert the letter X which belongs on the side of the graph just like the letter Y is above. Y The numbers are 1through 10; also a negative 1-10;
### Linear Equation Word Problems
Please see the attached file for the fully formatted problems. 1. Translate the sentence to an equation and solve it. Six less than eight times a number is equal to that number added to one. 2. The length of a rectangular mailing label is 3 centimeters less than twice the width. The perimeter is 54 centimeters. Find the d
### System of differential equations
Please see the attached file for the fully formatted problems. consider the following system: dX/dt = AX + B Where A= (1 4) (0 2) And B = (3,6) a. Determine the equilibrium point X* of the system. b. Is the equilibrium point stable or unstable? c. Draw precisely the phase portrait of the system.
### Exponential Functions : Population Growth
The U.S. Census Bureau International Database gives the population of the ten most- populated countries in the year 2000 and population predictions for 2050 for these countries: Chinahad 1.2 billion people in 2000 and is predicted to have a population of 1.3 billion in 2050. The U.S. had a population of 284 million in 2000 with
### Semidirect product or split extention
If H and K are subgroup of G, with K a normal subgroup of G, K intersect H=1 and KH=G, then G is called a semidirect product (or split extention) of K by H. If sigma=(12) in S_n, n>=2, show that s_n is a semidirect product of A_n by <sigma> Show that the dihedral group D_n=<a,b|a^n=b^2=1,b^-1ab=a^-1> is a semidirect produc
### Calculations for solving systems of equations
Please help with the following problems. Provide step by step calculations. 3. Solve for X, Y, and Z in the following systems of three equations: a. X + 2Y + Z = 6 X + Y = 4 3X + Y + Z = 8 b. 10X + Y + Z = 12 8X + 2Y +Z = 11 20X - 10Y - 2Z = 8 c. 22X + 5Y + 7Z = 12 10X + 3Y + 2Z = 5
### Linear Equations: Slopes and Intercepts
PLEASE SEE ATTACHED FILE FOR FULLY FORMATTED QUESTIONS 1. Find the slope of the line. ANSWER: 2. Find the slope of the line that goes through the pair of points: (2, 5), (6, 10). 3. Graph the line through (2, 3) with slope . 4. Draw L1 through (-2, 1) with slope , and draw L2 through (-2, 1) with sl
### Solving equations and system of equations
The financial institution has asked your firm to place some spot advertisements. The financial institution wants you to find the best allocation of ads between 2 of the local media outlets. The company you have chosen to handle the transaction has estimated the average number of potential new customers reached per spot anno
### Algebra - Linear equations.
Please explain and answer questions. Thanks 1. Equations of Lines in Slope-Intercept Form Find the slope and y-intercept for each line. 3x + 5y + 10 = 0 2. Write each equation in slope-intercept form y- 5 = - ¾ (x+1) 3. Interest rates. A credit manager rates each applicant for a car loan on a scale of
### Linear Equations
Please help with the following problems. There are three methods to solving Linear Systems with two Equations. They are the Graph method, the Elimination method, and the Substitution method. When would you use each method? What makes each method better than the other methods?
### Systems of Equations Application Word Problems
1-2 pages Details: The financial institution has asked your firm to place some spot advertisements. The financial institution wants you to find the best allocation of ads between 2 of the local media outlets. The company you have chosen to handle the transaction has estimated the average number of potential new customers
### Solving Systems of Linear Equations and Word Problems
1.) Solving Systems by Graphing and Substitution Investing her bonus. Donna invested her \$33,000 bonus and received a total of \$970 in interest after one year. If part of the money returned 4% and the remainder 2.25%, then how much did she invest at each rate? 2.) Ticket sa
### 1 page plus graph as attachment---Solve equations and systems of equations in two variables, formulate real-world problems in terms of linear equations and solve these systems by several commonly applied methods.
The financial institution was very pleased with your presentation and has asked you to place the advertising ads. You have found out the best way to ensure the best outcome for your customer is through the use of systems of equations. After reading up on systems of equations, you have found out how powerful they can be for solvi
### Linear Equations, Linear Programming, Word Problems and Measures of Central Tendency
Show your work. 1. (3 points) Find the equation of the line shown: 2. (6 points) A bank loaned \$15,000, some at an annual rate of 16% and some at an annual rate of 10%. If the income from these loans was \$1800, how much was loaned at 10%? 3. (3 point) Write the augmented matrix of the system: 2x1 -3x2 + x3 =
### Solving Systems of Linear Equations
Solve each step by graphing y = -2/3x 2x+3y = 5 Answer: Solve each system by substitution. Determine whether the equations are independent, dependent or inconsistent x= y + 3 3x - 2y =4 Answer: Solve each system by substitution. Determine whether the equations are independent, dependent or inconsistent 2x -
### Graphing Linear Equations
Please see the attached file for the fully formatted problems. 1. a) Determine if the ordered pair (2, ¼) is a solution of the equation (1/8)x + 3y = 1 b) Explain why or why not the pair is a solution. 2. a)Graph or describe fully the graph of the equation y = (2/3)x + 1. b) Show the table method with values for x o
### Basic Algebra : Mean, Median, Mode and Linear Equations
Fiona kept the following records of her phone bills for 12 months: \$42, \$37, \$51, \$41, \$58, \$44, \$42, \$35, \$53, \$58, \$46, \$57 Find the median of Fiona's monthly phone bills. (a) Find the mean of the following set of numbers. 12, 16, 19, 22, 28, 30 (b) Find the mode of the following set of numbers. 11, 1 |
# Multi Step Equations With Rational Numbers 7th Grade Worksheets
A Logical Amounts Worksheet will help your youngster be more familiar with the principles associated with this percentage of integers. In this particular worksheet, students are able to resolve 12 various problems associated with rational expression. They will learn how to flourish several numbers, class them in couples, and determine their products and services. They will also training simplifying realistic expression. When they have perfected these concepts, this worksheet might be a valuable tool for continuing their studies. Multi Step Equations With Rational Numbers 7th Grade Worksheets.
## Rational Amounts can be a percentage of integers
There are 2 forms of phone numbers: rational and irrational. Rational phone numbers are described as entire figures, whereas irrational numbers usually do not recurring, and get an endless amount of digits. Irrational phone numbers are no-absolutely no, no-terminating decimals, and sq beginnings which are not best squares. These types of numbers are not used often in everyday life, but they are often used in math applications.
To define a rational amount, you need to realize just what a logical variety is. An integer is a complete variety, plus a reasonable variety is a ratio of two integers. The proportion of two integers is definitely the quantity on the top divided up with the quantity on the bottom. For example, if two integers are two and five, this would be an integer. However, there are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They can be manufactured right into a small fraction
A reasonable variety includes a denominator and numerator which are not absolutely nothing. Because of this they can be indicated like a fraction. Along with their integer numerators and denominators, realistic phone numbers can furthermore have a unfavorable worth. The unfavorable worth should be located on the left of along with its absolute benefit is its extended distance from no. To streamline this example, we shall point out that .0333333 can be a portion that can be published as being a 1/3.
In addition to unfavorable integers, a realistic quantity may also be produced right into a small fraction. By way of example, /18,572 can be a reasonable variety, when -1/ is just not. Any fraction comprised of integers is rational, so long as the denominator fails to consist of a and will be published as an integer. Similarly, a decimal that ends in a point is yet another reasonable amount.
## They create perception
Even with their title, logical figures don’t make a lot feeling. In mathematics, they are single entities with a unique span on the number collection. Consequently if we add up some thing, we are able to buy the dimensions by its rate to its initial quantity. This contains true even though you will find unlimited reasonable amounts between two specific figures. In other words, numbers should make sense only if they are ordered. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
In real life, if we want to know the length of a string of pearls, we can use a rational number. To find the duration of a pearl, by way of example, we might add up its width. A single pearl is 15 kilograms, which is actually a rational number. Furthermore, a pound’s bodyweight means ten kilograms. As a result, we should be able to divide a lb by 10, without the need of concern yourself with the length of an individual pearl.
## They can be depicted as a decimal
If you’ve ever tried to convert a number to its decimal form, you’ve most likely seen a problem that involves a repeated fraction. A decimal quantity may be written as a multiple of two integers, so four times several is equal to 8. A comparable problem requires the repetitive portion 2/1, and both sides needs to be separated by 99 to have the right answer. But how would you make the conversion? Here are a few good examples.
A logical variety will also be printed in great shape, such as fractions along with a decimal. A great way to stand for a realistic variety in a decimal is usually to separate it into its fractional equivalent. You will find 3 ways to divide a logical number, and every one of these approaches brings its decimal comparable. One of these brilliant techniques is always to break down it into its fractional equivalent, and that’s what’s called a terminating decimal. |
Search IntMath
Close
# 3. Solving Non-linear Inequalities
Another method of solving inequalities is to express the given inequality with zero on the right side and then determine the sign of the resulting function from either side of the root of the function.
The steps are as follows:
1. Rewrite the inequality so that there is a zero on the right side.
2. Find all linear factors of the function.
3. To find the critical values, set each linear function to zero and solve for x.
4. Determine the sign of the function in the intervals formed by the critical values.
5. The solution will be those intervals in which the function has the correct signs satisfying the inequality.
### Example 1
Solve the inequality x2 − 3 > 2x
First, we rearrange the inequality with a zero on the right:
x2 − 2x − 3 > 0
which can be factored to give:
(x + 1)(x - 3) > 0
Setting both factors to zero, we get:
(x + 1) = 0 and (x - 3) = 0
x = -1 and x = 3
Therefore the critical values are
x = -1 and x = 3.
These critical values divide the number line into 3 intervals:
x < -1,
-1 < x < 3, and
x > 3.
Next, we need to determine the sign (plus or minus) of the function in each of the 3 intervals.
For the first interval, x < -1,
The value of (x + 1) will be negative (substitute a few values of x less than -1 to check),
The value of (x − 3) will also be negative
So in the interval x < -1, the value of the function x2 − 2x − 3 will be
negative × negative = positive
We continue doing this for the other 2 intervals and summarise the results in this table:
Interval (x + 1) (x - 3) sign of f(x) x < -1 − − + -1 < x < 3 + − − x > 3 + + +
We are solving for
(x + 1)(x - 3) > 0
The intervals that satisfy this inequality will be those where f(x) has a positive sign.
Hence, the solution is: x < -1 or x > 3.
Here is the graph of our solution:
### Example 2
Solve the inequality x3 − 4x2 + x + 6 < 0
Using methods learnt in earlier chapters (see Remainder and Factor Theorems), the expression can be factored to give:
(x + 1)(x − 2)(x − 3) < 0
Setting the factors to zero, we get:
(x + 1) = 0
(x − 2) = 0
(x − 3) = 0
So x = −1, x = 2, or x = 3.
Therefore the critical values are
x = −1, x = 2 and x = 3.
These critical values divide the number line into 4 intervals:
x < −1,
−1 < x < 2,
2 < x < 3 and
x > 3.
Next, we determine the sign of the function by the following method:
Interval (x + 1) (x − 2) (x − 3) sign of f(x) x < −1 − − − − −1 < x < 2 + − − + 2 < x < 3 + + − − x > 3 + + + +
Since we want f(x) < 0, the intervals that satisfy this inequality will be those with a negative sign.
Hence, the solution is: x < −1 or 2 < x < 3.
Here is the graph of our solution:
### Example 3
Solve the inequality ((x-2)^2(x+3))/(4-x) < 0
The critical values are
x = −3, x = 2 and x = 4.
These critical values divide the number line into 4 intervals.
Hence, the signs of the function are:
Interval ((x−2)^2(x+3))/(4−x) sign of f(x) x < −3 − −3 < x < 2 + 2 < x < 4 + x > 4 −
Since we want f(x) < 0, the intervals that satisfy this inequality will be those with a negative sign.
Hence, the solution is: x < −3 or x > 4.
Here's the graph of the solution.
### Exercises
1. Solve x^3-2x^2+x >= 0
x^3−2x^2+x >= 0
x(x^2−2x+1) >= 0
x(x−1)^2 >= 0
Our critical values are: x=0 or 1.
Testing as above, we find our solution set is: x>=0.
Here's the solution graph:
2. The mass m, in Mg, of fuel in a rocket after launch is
m = 2000 − t2 − 140t
where t is measured in minutes. During what time is the mass of fuel greater than 500\ "Mg"?
We need to solve: 2000 − t^2− 140t > 500.
1500 − t^2− 140t > 0
t^2+ 140t − 1500 < 0
(t + 150)(t − 10) < 0
So −150 < t < 10.
But t cannot be negative, so 0 ≤ t < 10.
The solution graph is:
3. The object distance p, and the image distance q, for a camera of focal length 3 cm is given by:
p=(3q)/(q-3)
For what values of q is p > 12 cm?
We set up the inequality as follows:
(3q)/(q−3)>12
The logical first step would be to multiply both sides by (q − 3). However, we can only do that if (q − 3) is positive. (Otherwise, if (q − 3) is negative, the > sign would need to change to <.)
Of course, q ≠ 3, since we cannot have 0 on the bottom of a fraction.
So we have (if q > 3):
3q > 12(q − 3)
Divide both sides by 3:
q > 4(q − 3)
Expand brackets:
q > 4q − 12
Add 12 to both sides and subtract q from both sides:
12 > 4qq
12 > 3q
4 > q
Reverse sides (and reversing the > sign as well):
q < 4
So including the condition q > 3, our final answer is:
3 < q < 4
The graph of the solution: |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.