text
stringlengths 22
1.01M
|
|---|
# Math Expressions Grade 5 Unit 1 Lesson 4 Answer Key Strategies for Comparing Fractions
## Math Expressions Common Core Grade 5 Unit 1 Lesson 4 Answer Key Strategies for Comparing Fractions
Math Expressions Grade 5 Unit 1 Lesson 4 Homework
Compare
Strategies for Comparing Fractions Lesson 4 Math Expressions Question 1.
$$\frac{5}{8}$$ $$\frac{5}{9}$$
Explanation:
By comparing these two numbers (5/8) and (5/9). (5/8) is greater than (5/9).
Math Expressions Grade 5 Unit 1 Lesson 4 Answer Key Strategies for Comparing Fractions Question 2.
$$\frac{1}{5}$$ $$\frac{1}{4}$$
Explanation:
By comparing these two numbers (1/5) and (1/4). (1/5) is less than (1/4).
Answer Key Grade 5 Math Expressions Strategies for Comparing Fractions Question 3.
$$\frac{2}{5}$$ $$\frac{3}{5}$$
Explanation:
By comparing these two numbers (2/5) and (3/5). (2/5) is less than (3/5).
Question 4.
$$\frac{6}{8}$$ $$\frac{2}{3}$$
Explanation:
By comparing these two numbers (6/8) and (2/3). (6/8) is greater than (2/3).
Question 5.
$$\frac{10}{11}$$ $$\frac{11}{12}$$
Explanation:
By comparing these two numbers (10/11) and (11/12). (10/11) is less than (11/12).
Question 6.
$$\frac{3}{8}$$ $$\frac{5}{12}$$
Explanation:
By comparing these two numbers (3/8) and (5/12). (3/8) is less than (5/12).
Question 7.
$$\frac{5}{12}$$ $$\frac{4}{7}$$
Explanation:
By comparing these two numbers (5/12) and (4/7). (5/12) is less than (4/7).
Question 8.
$$\frac{1}{3}$$ $$\frac{4}{9}$$
Explanation:
By comparing these two numbers (1/3) and (4/9). (1/3) is less than (4/9).
Question 9.
$$\frac{1}{4}$$ $$\frac{2}{9}$$
Explanation:
By comparing these two numbers (1/4) and (2/9). (1/4) is greater than (2/9).
Question 10.
$$\frac{1}{12}$$ $$\frac{1}{15}$$
Explanation:
By comparing these two numbers (1/12) and (1/15). (1/12) is greater than (1/15).
Question 11.
$$\frac{7}{10}$$ $$\frac{11}{15}$$
Explanation:
By comparing these two numbers (7/10) and (11/15). (7/10) is less than (11/15).
Question 12.
$$\frac{12}{25}$$ $$\frac{51}{100}$$
Explanation:
By comparing these two numbers (12/25) and (51/100). (12/25) is less than (51/100).
Question 13.
During his first season on the school football team, Wade made 5 of the 9 field goals he tried. During his second season, he made 11 of the 15 field goals he tried. In which season did he make the greater fraction of the field goals he tried?
Wade tried 5/9 field goals in first season on the school football team.
Wade tried 11/15 field goals in second season on the school football team.
(5/9) < (11/15)
In second season Wade made the greater fraction (11/15) of the field goals.
Explanation:
During his first season on the school football team, Wade made 5 of the 9 field goals he tried. During his second season, he made 11 of the 15 field goals he tried. Comparing first season and second season he made the grater fraction of the field goals in second season.
Question 14.
Manuela bought $$\frac{11}{12}$$ yard of polka dot fabric and $$\frac{7}{9}$$ yard of flowered fabric. Which fabric did she buy more of?
(11/12) yard of polka dot fabric.
(7/9) yard of flowered fabric.
Manuela bought more polka dot fabric.
Explanation:
Manuela bought (11/12) yard of polka dot fabric and (7/9) yard of flowered fabric. (11/12) is greater than (7/9). So she bought more polka dot fabric compared to flowered fabric.
Question 15.
Of the 7 pens in Ms. Young’s desk, 3 are blue. Of the 9 pens in Mr. Fox’s desk, 5 are blue. Which teacher has a greater fraction of pens that are blue?
Ms. Young’s desk have (3/7) blue pens.
Mr. Fox’s desk have (5/9) blue pens.
(3/7) < (5/9)
Mr. Fox’s have greater fraction of pens that are blue.
Explanation:
Of the 7 pens in Ms. Young’s desk, 3 are blue. Of the 9 pens in Mr. Fox’s desk, 5 are blue. Compare both fraction (3/7) and (5/9). Mr. Fox’s have greater fraction of pens that are blue compared to Ms. Young’s desk.
Question 16.
Mr. Sommers spent 10 minutes of his 50-minute math period reviewing homework. Mr. Young spent 12 minutes
of his 60-minute math period reviewing homework. Which teacher spent a greater fraction of his math period reviewing homework?
Mr. Sommers spent (10/50) minutes.
Mr. Young spent (12/60) minutes.
(10/50) = (1/5)
(12/60) = (1/5)
Both teachers spent equally on math period reviewing homework.
Explanation:
Mr. Sommers spent 10 minutes of his 50-minute math period reviewing homework. Mr. Young spent 12 minutes
of his 60-minute math period reviewing homework. Both fractions are equal. So Both teachers spent equally on math period reviewing homework.
Math Expressions Grade 5 Unit 1 Lesson 4 Remembering
Complete.
Question 1.
$$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ = ____
(1/4) + (1/4) + (1/4) = (3/4)
Explanation:
Perform addition operation on these three fraction numbers (1/4), (1/4), (1/4). By adding these three faction numbers the sum is (3/4).
Question 2.
$$\frac{8}{9}$$ – $$\frac{4}{9}$$ = _____
(8/9) – (4/9) = (4/9)
Explanation:
Perform subtraction operation on these two fraction numbers (8/9), (4/9). Subtract (4/9) from (8/9) then the difference is (4/9).
Question 3.
$$\frac{4}{5}$$ + $$\frac{1}{5}$$ = ____
(4/5) + (1/5) = (5/5)
(4/5) + (1/5) = 1
Explanation:
Perform addition operation on these two fraction numbers (4/5), (1/5). By adding these two faction numbers the sum is 1.
Question 4.
$$\frac{3}{8}$$ + $$\frac{3}{8}$$ = _____
(3/8) + (3/8) = (6/8)
Explanation:
Perform addition operation on these two fraction numbers (3/8), (3/8). By adding these two faction numbers the sum is (6/8).
Write the multiplier or divisor for each pair of equivalent fractions.
Question 5.
$$\frac{5}{6}$$ = $$\frac{10}{12}$$
Multiplier = ________
2 x 5 = 10
2 x 6 = 12
(5/6) = (10/12)
(5/6) = (5/6)
Multiplier for the pair of above equivalent fraction is 2.
Explanation:
Multiplying the numerator and denominator of a fraction by the same number will produce an equivalent fraction. Consider the fraction (5/6). Multiply numerator and denominator with 2, then we get (5/6) =(10/12).
Question 6.
$$\frac{12}{15}$$ = $$\frac{4}{5}$$
Divisor = ________
3 x 4 = 12
3 x 5 = 15
(12/15) = (4/5)
(4/5) = (4/5)
Divisor for the pair of above equivalent fraction is 3.
Explanation:
Multiplying the numerator and denominator of a fraction by the same number will produce an equivalent fraction. Consider the fraction (4/5). Multiply numerator and denominator with 3, then we get (4/5) =(12/15).
Question 7.
$$\frac{3}{4}$$ = $$\frac{18}{24}$$
Multiplier = ________
6 x 3 = 18
6 x 4 = 24
(3/4) = (18/24)
(3/4) = (3/4)
Multiplier for the pair of above equivalent fraction is 6.
Explanation:
Multiplying the numerator and denominator of a fraction by the same number will produce an equivalent fraction. Consider the fraction (3/4). Multiply numerator and denominator with 6, then we get (3/4) =(18/24).
Question 8.
$$\frac{25}{50}$$ = $$\frac{5}{10}$$
Divisor = ________
5 x 5 = 25
5 x 10 = 50
(25/50) = (5/10)
(5/10) = (5/10)
Divisor for the pair of above equivalent fraction is 5.
Explanation:
Multiplying the numerator and denominator of a fraction by the same number will produce an equivalent fraction. Consider the fraction (5/10). Multiply numerator and denominator with 5, then we get (5/10) =(25/50).
Question 9.
$$\frac{1}{4}$$ = $$\frac{7}{28}$$
Multiplier = ________
7 x 1 = 7
7 x 4 = 28
(1/4) = (7/28)
(1/4) = (1/4)
Multiplier for the pair of above equivalent fraction is 7.
Explanation:
Multiplying the numerator and denominator of a fraction by the same number will produce an equivalent fraction. Consider the fraction (1/4). Multiply numerator and denominator with 7, then we get (1/4) =(7/28).
Question 10.
$$\frac{11}{22}$$ = $$\frac{1}{2}$$
Divisor = ________
11 x 1 = 11
11 x 2 = 22
(11/22) = (1/2)
(1/2) = (1/2)
Divisor for the pair of above equivalent fraction is 11.
Explanation:
Multiplying the numerator and denominator of a fraction by the same number will produce an equivalent fraction. Consider the fraction (1/2). Multiply numerator and denominator with 11, then we get (1/2) =(11/22).
Complete the chain of equivalent fractions.
Question 11.
$$\frac{2}{5}$$ = ___ = ____ = ___ = ____ = ____ = ____
Multiplier = ____
(2/5) = (4/10) = (6/15) = (8/20) = (10/25) = (12/30) = (14/35)
The multiplier for the above chain of equivalent fraction is 2, 3, 4, 5, 6, 7.
Explanation:
Multiplying the numerator and denominator of a fraction by the same number will produce an equivalent fraction. Consider the fraction (2/5). Multiply numerator and denominator with 2, 3, 4, 5, 6, 7 then we get (2/5) = (4/10) = (6/15) = (8/20) = (10/25) = (12/30) = (14/35). The multiplier for the above chain of equivalent fraction is 2, 3, 4, 5, 6, 7.
Question 12.
$$\frac{5}{9}$$ = ___ = ____ = ___ = ____ = ____ = ____
Divisor = ____
(5/9) = (10/18) = (15/27) = (20/36) = (25/45) = (30/54) = (35/63)
The divisor for the above chain of equivalent fraction are 2, 3, 4, 5, 6, 7.
Explanation:
Multiplying the numerator and denominator of a fraction by the same number will produce an equivalent fraction. Consider the fraction (5/9). Multiply numerator and denominator with 2, 3, 4, 5, 6, 7 then we get (5/9) = (10/18) = (15/27) = (20/36) = (25/45) = (30/54) = (35/63) . The multiplier for the above chain of equivalent fraction is 2, 3, 4, 5, 6, 7.
Solve.
Question 13.
Stretch Your Thinking Harry ate $$\frac{4}{8}$$ of a large pizza. Aidan ate $$\frac{1}{2}$$ of a small pizza. Harry said that since $$\frac{4}{8}$$ is equivalent to $$\frac{1}{2}$$, he and Aidan ate the same amount of pizza. Is he correct? Explain.
|
# Texas Go Math Kindergarten Lesson 15.4 Answer Key Quarter
Refer to our Texas Go Math Kindergarten Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Kindergarten Lesson 15.4 Answer Key Quarter.
## Texas Go Math Kindergarten Lesson 15.4 Answer Key Quarter
Essential Question
How do you identify, describe, and name quarters?
Explore
Explanation:
A quarter is worth 25 cents.
There are 5 quarters in the group
and there are 3 pennies.
Directions
Place a quarter to match each one shown at the top of the page. Tell what is alike about the quarters. Tell what is different about the quarters. Circle the quarters in the set at the bottom of the page. Write the number of quarters.
Share and Show
Question 1.
Explanation:
A quarter is worth 25 cents.
There are 3 quarters in the group
Question 2.
Explanation:
A quarter is worth 25 cents.
There are 10 quarters in the group
Question 3.
Explanation:
A quarter is worth 25 cents.
There are 8 quarters in the group
Directions
1-3. Count the quarters. Write the number that shows how many quarters.
Question 4.
Explanation:
A quarter is worth 25 cents.
There are 7 quarters in the bag
Question 5.
Explanation:
A quarter is worth 25 cents.
There are 4 quarters in the bag
Directions
4-5. Circle the quarters. Count and write the number of quarters.
Home Activity
• Show your child a quarter. Hove him or her tell you the name of the coin and describe both sides.
Problem Solving
Question 6.
Explanation:
Julia needs four quarters to buy a book
There are 4 quarters in the bag
so, Julia has sufficient money to buy a book
Question 7.
Explanation:
A quarter is worth 25 cents.
There are 2 quarters in the group
Directions
6. Julia needs four quarters to buy a book. Count the quarters to see if she has enough. Circle the quarters if she has enough. Write the number of quarters. 7. Choose the correct answer How many quarters are in this group?
### Texas Go Math Kindergarten Lesson 15.4 Homework and Practice
Question 1.
Explanation:
A quarter is worth 25 cents.
There are 9 quarters
Question 2.
Explanation:
A quarter is worth 25 cents.
In the bag there are 5 quarters
Directions
1. Count the quarters. Write the number that shows how many quarters. 2. Circle the quarters. Count and write the number of quarters.
Texas Test Prep
Lesson Check
Question 3.
Explanation:
A quarter is worth 25 cents.
There are 4 quarters
Question 4.
Explanation:
A quarter is worth 25 cents.
There are 3 quarters
Question 5.
|
Lecture 6 notes
# Lecture 6 notes - 4.3 CONDITIONAL PROBABILITY Ref Devore...
• Notes
• 3
This preview shows pages 1–2. Sign up to view the full content.
4.3 CONDITIONAL PROBABILITY Ref: Devore 7th Ed. Sections 2.4, 2.5. Suppose we toss a fair coin 5 times and it comes up heads all 5 times. Is the next toss more likely to be a tail? Ans: No. P [6 heads in a row | 5 heads in a row ] = 1 2 Definition . The conditional prob of event A given that event B has occurred is P ( A | B ) = P ( A B ) /P ( B ) Multiplication rule (Devore 7e, p. 69) P ( A B ) = P ( A | B ) P ( B ) = P ( B | A ) P ( A ) . Often P ( A | B ) or P ( B | A ) is known, and this rule allows us to calculate P ( A B ) . Example: Draw 2 cards from a pack. What is the probability both are hearts? A = 1st card is heart , B = 2nd card is heart . P (2 hearts) = P ( A B ) = P ( A ) P ( B | A ) = 13 52 × 12 51 = ˆ 13 2 ! / ˆ 52 2 ! Can extend multiplication rule. E.g. P ( A B C ) = P ( A ) P ( B | A ) P ( C | A B ) . e.g. Prob(5 hearts) = 13 52 × 12 51 × 11 50 × 10 49 × 9 48 = ˆ 13 5 ! / ˆ 52 5 ! Independence of Events We say events A and B are independent if P ( A B ) = P ( A ) P ( B ) . Rationale: If A,B independent, this implies P ( A | B ) = P ( A ) and P ( B | A ) = P ( B ) .
This preview has intentionally blurred sections. Sign up to view the full version.
This is the end of the preview. Sign up to access the rest of the document.
• Spring '05
• STAFF
{[ snackBarMessage ]}
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern
|
# Consider the sequence $x_1 = 0, x_n = \frac{1}{4} x_{n-1}^2 + 1$. Prove the limit exists by showing monotone and bounded, then find the limit
Consider the sequence $$x_1 = 0$$, $$x_n = \frac{1}{4} x_{n-1}^2 + 1$$. Prove the limit exists by showing monotone and bounded, then find the limit. I am having difficulty proving that is monotone (increasing) and bounded.
Proving this sequence is monotone by induction:
Base case: $$x_0 = 0, x_1 = 1$$
In general: $$x_n = \frac{1}{4} x_{n-1}^2 + 1$$ $$x_{n+1} = \frac{1}{4}x_n ^2 + 1 = \frac{1}{16}x_{n-1}^2 + \frac{5}{4}$$ How do I go on to prove that the sequence is increasing? Would it be easier to prove that it has an upper bound of 2 and use that?
To find the limit: $$L = lim_{n \to \infty} \frac{1}{4}x_n ^2 +1 = \frac{1}{4} (lim_{n \to \infty} x_n)^2 + 1 = \frac{1}{4}L^2 +1$$ Then solving $$L = \frac{1}{4}L^2 +1$$, I get L = 2
Note that $$x_n-x_{n-1}=(x_{n-1}/2-1)^2\ge0$$.
you need to show $$x_{n+1}-x_{n}\geq 0$$so $$x_n = \frac{1}{4} x_{n-1}^2 + 1\\ x_{n+1} = \frac{1}{4} x_{n}^2 + 1\\ x_{n+1} -x_n= \frac{1}{4} x_{n}^2-x_n + 1\\ x_{n+1} -x_n= \frac{1}{4}( x_{n}^2-4x_n + 4)=\frac{1}{4}( x_{n}-2)^2\geq 0$$
|
# Converting a Fraction to a Percent, and a Percent to a Fraction
Page content
## Converting a Percent to a Fraction
This is much easier than you may think: just put the number over 100 and simplify. For example, 5% = 5/100 = 1/25, and 150% = 150/100 = 1 ½.
Why does this work? If you think about it, the definition of a percent is really just “out of a hundred.” If you have 50% of something, what you’re really saying is that you have fifty out of a hundred, or 50/100.
## Converting a Fraction to a Percent – Method 1
Converting a fraction to a percent is more difficult, and there are two methods you can use to do it. In the first method, you would change the fraction so that it has 100 in its denominator. This method only works easily if the denominator is a factor of 100.
For example, let’s say that you were trying to convert 21/25 to a percent. You know that 25 X 4 = 100. In order to multiply the denominator by 4, you also have to multiply the numerator by 4. So 21/25 X 4/4 = 84/100. Because the definition of percent is “out of a hundred,” 84/100 is the same thing as 84%. So 21/25 = 84%.
You can do this with less complex fractions as well. For example, let’s say you wanted to convert 2/5 to a percent. You know that 5 X 20 = 100, so you can multiply the numerator and denominator by 20, which gives you 2/5 X 20/20 = 40/100 = 40%.
## Converting a Fraction to a Percent – Method 2
The first method seems simple enough, but remember that it only works if the denominator is a factor of a hundred. What do you do if the denominator is not a factor of 100? You use this second, two-step method:
Convert the fraction to a decimal.
Convert the decimal to a percent.
For example, to convert the fraction ¼ to a decimal, you would first convert it to a fraction (1 / 4 = .25). Then you would convert the decimal to a percent (.25 = 25%).
With these methods under your belt, you’ll be acing your math tests in no time!
## This post is part of the series: Fraction, Decimal, and Percent Conversions
Confused about how to do fraction, decimal, and percent conversions? It can get complicated, but if you learn each conversion separately, you’ll soon be converting like a pro!
|
# NCERT Solutions For Class 9th Maths Chapter 3 : Coordinate Geometry
CBSE NCERT Solutions For Class 9th Maths Chapter 3 : Coordinate Geometry. NCERT Solutins For Class 9 Mathematics. Exercise 3.1, Exercise 3.2, Exercise 3.3
### NCERT Solutions for Class IX Maths: Chapter 3 – Coordinate Geometry
Page No: 53
Exercise 3.1
1. How will you describe the position of a table lamp on your study table to another person?
To describe the position of a table lamp on the study table, we have two take two lines, a perpendicular and horizontal. Considering the table as a plane and taking perpendicular line as Y axis and horizontal as X axis. Take one corner of table as origin where both X and Y axes intersect each other. Now, the length of table is Y axis and breadth is X axis. From The origin, join the line to the lamp and mark a point. Calculate the distance of this point from both X and Y axes and then write it in terms of coordinates.
Let the distance of point from X axis is x and from Y axis is y then the the position of the table lamp in terms of coordinates is (x,y).
2. (Street Plan) : A city has two main roads which cross each other at the centre of the city. These two roads are along the North-South direction and East-West direction.
All the other streets of the city run parallel to these roads and are 200 m apart. There are 5 streets in each direction. Using 1cm = 200 m, draw a model of the city on your notebook. Represent the roads/streets by single lines. There are many cross- streets in your model. A particular cross-street is made by two streets, one running in the North – South direction and another in the East – West direction. Each cross street is referred to in the following manner : If the 2nd street running in the North – South direction and 5 th in the East – West direction meet at some crossing, then we will call this cross-street (2, 5). Using this convention, find:
(i) how many cross – streets can be referred to as (4, 3).
(ii) how many cross – streets can be referred to as (3, 4)
ALSO READ: NCERT Solutions for Class 6th Science Chapter 9 : Living Organisms and Their Surroundings
(i) Only one street can be referred to as (4, 3) as we see from the figure.
(ii) Only one street can be referred to as (3, 4) as we see from the figure.
Page No: 60
Exercise 3.2
1. Write the answer of each of the following questions:
(i) What is the name of horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane?
(ii) What is the name of each part of the plane formed by these two lines?
(iii) Write the name of the point where these two lines intersect.
ALSO READ: NCERT Solutions For Class 10th Maths Chapter 6 : Triangles
(i) The name of horizontal lines and vertical lines drawn to determine the position of any point in the Cartesian plane is x-axis and y-axis respectively.
(ii) The name of each part of the plane formed by these two lines x-axis and y-axis is quadrants.
(iii) The point where these two lines intersect is called origin.
2. See Fig.3.14, and write the following:
(i) The coordinates of B.
(ii) The coordinates of C.
(iii) The point identified by the coordinates (-3, -5).
(iv) The point identified by the coordinates (2, -4).
(v) The abscissa of the point D.
(vi) The ordinate of the point H.
(vii)The coordinates of the point L.
(viii) The coordinates of the point M.
(i) The coordinates of B is (-5, 2).
(ii) The coordinates of C is (5, -5).
(iii) The point identified by the coordinates (-3, -5) is E.
(iv) The point identified by the coordinates (2, -4) is G.
(v) Abscissa means x coordinate of point D. So, abscissa of the point D is 6.
(vi) Ordinate means y coordinate of point H. So, ordinate of point H is -3.
ALSO READ: NCERT Solutions for Class 6th History Chapter 4 : In The Earliest Cities
(vii) The coordinates of the point L is (0, 5).
(viii) The coordinates of the point M is (- 3, 0).
Page No: 65
Exercise 3.3
1. In which quadrant or on which axis do each of the points (-2, 4), (3, -1), (-1, 0), (1, 2) and (-3, -5) lie? Verify your answer by locating them on the Cartesian plane.
2. Plot the points (x, y) given in the following table on the plane, choosing suitable units of distance on the axes.
x -2 -1 0 1 3 y 8 7 -1.25 3 -1
|
<meta http-equiv="refresh" content="1; url=/nojavascript/">
You are viewing an older version of this Concept. Go to the latest version.
# Circle Graphs to Display Data
## Use percents to calculate the number of degrees needed for a circle graph.
0%
Progress
Practice Circle Graphs to Display Data
Progress
0%
Circle Graphs to Display Data
One day while Taylor was in the candy store, she saw a chart that her Dad had made sitting on the counter.
“What’s this mean?” she asked looking at the chart.
“That is a chart that shows our best sellers. Every other candy sells less than 10%, so I don’t usually include it. These are the top sellers. I keep track of our inventory each month and determine which candies were the top sellers. Then I create a graph of the data,” he explained.
“Where is the graph?”
“I could do that,” Taylor said smiling.
Taylor was so excited. She could finally put all of her math to work. She knew that a circle graph would be the best way to show the percentages. Here is the chart.
Lollipops - 55%
Licorice - 10%
Chocolates - 20%
Gummy Bears - 15%
Taylor started to work on the circle graph and she thought that she knew what she was doing, but then she got stuck. She couldn’t remember how to change each percentage into a number of degrees.
This is where you come in. It is your turn to help Taylor. Pay attention to this Concept and you will know how to create the circle graph in the end.
### Guidance
Creating a circle graph may seem tricky, but if you think about circle graphs it can become easier to figure out. First, notice that in the graphs at the end of the last Concept, that each percentage was converted to a specific number of degrees.
When you know the number of degrees that a percentage is equal to, you can use a protractor and a circle to draw it in exactly.
To figure this out, we have to figure out each percentage in terms of degrees. How do we do this?
First, we do this by creating a proportion. A percent is out of 100, so we can make a ratio out of any percent.
25% becomes 25100\begin{align*}\frac{25}{100}\end{align*}
15% becomes 15100\begin{align*}\frac{15}{100}\end{align*}
A circle is out of 360\begin{align*}360^ \circ\end{align*}. Since we are trying to figure out the number of degrees, we use a variable over 360 for the second ratio.
Here is a proportion for converting 25% to degrees.
25100=x360\begin{align*}\frac{25}{100} = \frac{x}{360}\end{align*}
Now we cross multiply and solve for the variable x\begin{align*}x\end{align*}. That will be the number of degrees.
100x100xx25%=25(360)=9000=90=90
Now if you were going to draw this on a circle graph, you could take a circle and your protractor and measure in a 90\begin{align*}90^\circ\end{align*} angle. That would equal 25% of the graph.
Now let's apply this.
The table shows the number of students in the seventh grade who are studying each foreign language. Make a circle graph that shows the data.
Foreign Language Number of Students Studying Language
Spanish 88
French 48
Italian 16
German 8
Step 1: Find the total number of seventh grade students studying a foreign language. Then find the percent of students studying each language.
88+48+16+8=160\begin{align*}88 + 48 + 16 + 8 = 160\end{align*}
Foreign Language Number of Students Studying Language Percent of Students Studying Language
Spanish 88 88160=1120=55%\begin{align*}\frac{88}{160} = \frac{11}{20} = 55 \%\end{align*}
French 48 48160=310=30%\begin{align*}\frac{48}{160} = \frac{3}{10} = 30 \%\end{align*}
Italian 16 16160=110=10%\begin{align*}\frac{16}{160} = \frac{1}{10} = 10 \%\end{align*}
German 8 8160=120=5%\begin{align*}\frac{8}{160} = \frac{1}{20} = 5 \%\end{align*}
Step 2: Find the measure of the central angle by multiplying 360\begin{align*}360^\circ\end{align*} by the percent.
Foreign Language Number of Students Studying Language Percent of Students Studying Language Degrees in Central Angle
Spanish 88 55% 55% of 360=0.55×360=198\begin{align*}360^\circ = 0.55 \times 360^\circ = 198^\circ\end{align*}
French 48 30% 30% of 360=0.30×360=108\begin{align*}360^\circ = 0.30 \times 360^\circ = 108^\circ\end{align*}
Italian 16 10% 10% of 360=0.10×360=36\begin{align*}360^\circ = 0.10 \times 360^\circ = 36^\circ\end{align*}
German 8 5% 5% of 360=0.05×360=18\begin{align*}360^\circ = 0.05 \times 360^\circ = 18^\circ\end{align*}
Step 3: Draw a circle with a compass. Draw one radius. Use that radius as a side of one central angle. Measure and draw the other central angles using a protractor.
Step 4: Label each sector with a title and percent and give a title to the entire circle graph.
Here is the final graph.
Convert each percent into degrees.
#### Example A
20%
Solution: 72\begin{align*}72^\circ\end{align*}
#### Example B
40%
Solution:144\begin{align*}144^\circ\end{align*}
#### Example C
75%
Solution:270\begin{align*}270^\circ\end{align*}
Here is the original problem once again. Use what you have learned to help Taylor make the circle graph.
One day while Taylor was in the candy store, she saw a chart that her Dad had made sitting on the counter.
“What’s this mean?” she asked looking at the chart.
“That is a chart that shows our best sellers. Every other candy sells less than 10%, so I don’t usually include it. These are the top sellers. I keep track of our inventory each month and determine which candies were the top sellers. Then I create a graph of the data,” he explained.
“Where is the graph?”
“I could do that,” Taylor said smiling.
Taylor was so excited. She could finally put all of her math to work. She knew that a circle graph would be the best way to show the percentages. Here is the chart.
Lollipops - 55%
Licorice - 10%
Chocolates - 20%
Gummy Bears - 15%
Taylor started to work on the circle graph and she thought that she knew what she was doing, but then she got stuck. She couldn’t remember how to change each percentage into a number of degrees.
First, we need to convert each percentage to a number of degrees. We can do this by multiplying each decimal by 360.
Lollipops .55×360=198\begin{align*}.55 \times 360 = 198^\circ\end{align*}
Licorice .10×360=36\begin{align*}.10 \times 360 = 36^\circ\end{align*}
Chocolates .20×360=72\begin{align*}.20 \times 360 = 72^\circ\end{align*}
Gummy Bears .15×360=54\begin{align*}.15 \times 360 = 54^\circ\end{align*}
Next, Taylor can use a protractor and a circle to create the circle graph. Here is her final work.
### Vocabulary
Here are the vocabulary words in this Concept.
Circle Graph
a visual display of data in a circle. A circle graph is created from percentages with the entire circle representing the whole. The sectors of the circle graph are divided according to degrees which are created out of 360\begin{align*}360^\circ\end{align*}.
Sector
the section of a circle graph. Each section is known as a sector. Each sector can be measured in degrees and given a percentage.
### Guided Practice
Here is the original problem once again.
Convert 30% into degrees.
First, we write a proportion.
30100=x360\begin{align*}\frac{30}{100} = \frac{x}{360}\end{align*}
Next, we cross multiply and solve for the variable.
10x100xx30%=30(360)=10800=108=108
### Video Review
Here is a video for review.
### Practice
Directions: Use what you have learned to tackle each problem.
1. The table shows the how much money the students in the seventh grade have raised so far for a class trip. Make a circle graph that shows the data.
Fundraiser Amount
Car wash $150 Book sale$175
Bake sale $100 Plant sale$75
2. Make a list of 5 popular ice cream flavors. Then survey your classmates asking them which of the 5 flavors is their favorite ice cream flavor. Use the data to make a circle graph.
3. Use a newspaper to locate a circle graph of some data. Then write five questions about the data.
Directions: Look at each percentage and then use a proportion to find the equivalent number of degrees. You may round your answer when necessary.
4. 12%
5. 25%
6. 28%
7. 42%
8. 19%
9. 80%
10. 90%
11. 34%
12. 15%
13. 5%
14. 10%
15. 78%
### Vocabulary Language: English
Sector
Sector
A sector of a circle is a portion of a circle contained between two radii of the circle. Sectors can be measured in degrees.
|
# CLASS-1FINDING SMALLER NUMBER
FINDING SMALLER NUMBER -
Finding Smaller number is a very primary and fundamental concept of learning math. Without making the concept of smaller number we cannot clear the concept of subtraction, and division methods.
Tick the smaller number in the following pairs –
From the above given problems we have given answer below
1) Here two number given, 15 & 37. Here 15 is previous number than 37, so 15 is smaller than 37. So, the correct answer is 15. We can denote like = 15 < 37 (smaller than)
2) Here two number given, 21 & 31. Here 21 is previous number than 31, so 21 is smaller than 31. So, the correct answer is 21. We can denote like = 21 < 31. (smaller than)
3) Here two number given, 85 & 54. Here 54 is previous number than 85, so 54 is smaller than 85. So, the correct answer is 54. We can denote like = 85 > 54 (smaller than)
4) Here two number given, 24 & 39. Here 24 is previous number than 39, so 24 is smaller than 39. So, the correct answer is 24. We can denote like = 24 < 39 (smaller than)
5) Here two number given, 27 & 41. Here 27 is previous number than 41, so 27 is smaller than 41. So, the correct answer is 27. We can denote like = 27 < 41 (smaller than)
6) Here two number given, 72 & 18. Here 18 is previous number than 72, so 18 is smaller than 72. So, the correct answer is 18. We can denote like = 72 > 18 (smaller than)
7) Here two number given, 30 & 12. Here 12 is previous number than 30, so 12 is smaller than 30. So, the correct answer is 12. We can denote like = 30 > 12 (smaller than)
8) Here two number given, 28 & 46. Here 28 is previous number than 46, so 28 is smaller than 46. So, the correct answer is 28. We can denote like = 28 < 46 (smaller than)
|
# Multiplying & Dividing Rational Numbers
An error occurred trying to load this video.
Try refreshing the page, or contact customer support.
Coming up next: Notation for Rational Numbers, Fractions & Decimals
### You're on a roll. Keep up the good work!
Replay
Your next lesson will play in 10 seconds
• 0:01 Rational Numbers
• 1:04 Multiplying
• 2:03 Dividing
• 3:58 Examples
• 5:33 Lesson Summary
Add to Add to
Want to watch this again later?
Log in or sign up to add this lesson to a Custom Course.
Timeline
Autoplay
Autoplay
Create an account to start this course today
Try it free for 5 days!
#### Recommended Lessons and Courses for You
Lesson Transcript
Instructor: Yuanxin (Amy) Yang Alcocer
Amy has a master's degree in secondary education and has taught math at a public charter high school.
After watching this video lesson, you will know how to multiply and divide rational numbers. You will know when you need to flip your number and when you don't.
## Rational Numbers
Sam works for this research company that gathers data for other companies for them to use. Part of Sam's job is to look at the numbers and then make certain calculations to give to the other companies so that they will better understand the information. The kinds of numbers that Sam sees for his job are called rational numbers. These are numbers that can be written as the fraction of two integers.
So, all simple fractions are rational numbers as are all integers because we can simply rewrite an integer as that number over 1. For example, 3/8 is a rational number as is the number 25 because it can be rewritten as 25/1. The proper form for any rational number is in fraction form. Unlike regular fractions where you want your bottom number to be larger than the top number, it is perfectly okay for the top number in a rational number to be larger than the bottom. So we can have rational numbers such as 8/7 and even 50/3.
## Multiplying
One of the calculations that Sam needs to make with his rational numbers is that of multiplying them together. Looking down at his papers, Sam sees 2/3*17/3 as one of his problems. How will he solve this problem?
Sam remembers from a long time ago when he was in school that to multiply two rational numbers together, all he has to do is multiply the tops together and the bottoms together and he will get his answer. Sometimes, he'll need to simplify his answer, too.
So multiplying the tops of his problem, he gets:
2*17 = 34
Multiplying the bottoms. He gets:
3*3 = 9
Putting it together, he gets 34/9 for his answer.
Can this be simplified? No. So 34/9 is his answer. Sam can leave his answer in this form even though the top number is larger than the bottom number because he is working with rational numbers.
## Dividing
The other type of calculation that Sam needs to make is that of dividing rational numbers. The next problem that Sam sees on his paper is this:
How does Sam solve this problem now? Sam remembers that to divide rational numbers, he can actually turn this problem into a multiplication problem by flipping the second rational number. So 7/8 becomes 8/7 and the division symbol turns into a multiplication symbol.
Before Sam goes ahead with his multiplication, he sees that he can actually simplify his problem a little bit. This is just like any other multiplication problem now, so if Sam can simplify his problem before going ahead with the multiplication, it will be easier for him to solve it.
Sam sees that he has an 8 on the top and a 6 on the bottom that he can simplify. Because this is a multiplication problem, Sam is able to take any number on the top and simplify it with any number on the bottom. It doesn't have to be in the same fraction. He can simplify these numbers by dividing both by 2. Remember that to simplify, you need to be able to divide both numbers by the same number. The 8 divided by 2 turns into a 4 and the 6 divided by a 2 turns into a 3.
So now the problem becomes
5/3*4/7
And Sam can go ahead with the multiplication of the tops and bottoms together. He gets:
5*4 = 20
3*7 = 21
Combining these two numbers, Sam gets an answer of:
20/21
This answer is already as simplified as it can get, so 20/21 is the answer that Sam will be giving to his customer.
## Examples
Sam has two more problems to go. See if you can help him with these next two problems.
11/12*4/3
To unlock this lesson you must be a Study.com Member.
Create your account
### Register for a free trial
Are you a student or a teacher?
What is your educational goal?
Back
Back
### Earning College Credit
Did you know… We have over 95 college courses that prepare you to earn credit by exam that is accepted by over 2,000 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.
To learn more, visit our Earning Credit Page
### Transferring credit to the school of your choice
Not sure what college you want to attend yet? Study.com has thousands of articles about every imaginable degree, area of study and career path that can help you find the school that's right for you.
Create an account to start this course today
Try it free for 5 days!
Support
|
## Tuesday, July 21, 2015
### Potential Flaw in The Second Solution for the problem $ab(a+b-10)+21a-3a^2+16b-2b^2=60$.
Solve in positive integers the equation $ab(a+b-10)+21a-3a^2+16b-2b^2=60$.
In this blog post, I will discuss why the last step in my attempt to look for the positive integers $a$ and $b$ doesn't sound so right.
Let's see from the point where we have gotten up to
$(a-2)(b-3)(a+b-5)=30$
$(a-2)(b-3)(a-2+b-3)=30$
Note that not only we have to consider the cases where:
1. $a-2\lt 0$ and $b-3\lt 0$,
2. $a-2\gt 0$ and $b-3\gt 0$,
We have to also consider the cases where
3. $a-2\lt 0$, and $b-3\gt 0$,
4. $a-2\gt 0$ and $b-3\lt 0$.
Then and only then we can safely list out all the possible solutions that we have found.
In my previous post (my last step), I only took care of the first two cases, and stupidly ignored the last two cases:
For the third case, we see that:
$a-2\lt 0$ and we note that $a\ne 2,\,b\ne 3$ or the LHS of the expression would become zero
and see where that leads us.
When $a-2\lt 0$,
$(\text{-ve})(b-3)(a-2+b-3)=30$
We can therefore set $b-3\gt 0$ to get:
$(-\text{ve})(+\text{ve})(-\text{ve}+\text{ve})=30$
In order to get the product of positive $30$, the last factor must yield a negative, i.e.
$a-2+b-3\lt 0$
$a\lt 5-b$
$0\lt a\lt 5-b$
This yields $b\lt 5$.
Combining both inequalities $b-3\gt 0$ and $b\lt 5$, this leads to $b=4$ and substituting this back into the equation
$(a-2)(b-3)(a+b-5)=30$ gives $a=7$.
Note that $b-3\lt 0$ has already been addressed in the first case.
For fourth case, we have:
We have to also take into consideration the possibility where $a-2\gt 0$:
When $a-2\gt 0$,
$(+\text{ve})(b-3)(a-2+b-3)=30$
We then let $b-3\lt 0$, so we have:
$(+\text{ve})(-\text{ve})(+\text{ve}+(-\text{ve}))=30$
In order to get the product of positive $30$, the last factor must yield a negative, i.e.
$a-2+b-3\lt 0$
$a\lt 5-b$
$0\lt a\lt 5-b$
This yields $b\lt 5$.
Combining both inequalities $b-3\lt 0$ and $b\lt 5$, this leads to $b=1,\,2$ of which there are no solutions for these two values of $b$.
For $a-2\gt 0$ and $b-3\gt 0$, that is just the last case, the one I covered in the last blog post. I will hence stop it from here, you can follow this link to check out the full solution.
All in all, we have to think and cover for ALL cases if we want to fully solving any given mathematical problem for best outcome.
|
# 2.4: Significant Figures in Calculations
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
⚙️ Learning Objectives
• Use significant figures correctly in arithmetical operations.
Calculators do just what is asked of them – no more and no less. However, they can sometimes get a little out of hand. If you multiply 2.49 by 6.3, you get an answer of 15.687, a value that ignores the number of significant figures in either number. Division with a calculator is even worse. When you divide 12.2 by 1.7, the answer you obtain is 7.176470588. Neither piece of data is accurate to nine decimal places, but the calculator does not know that. The human being operating the instrument has to make the decision about how the answer should be reported.
#### Rounding
Before dealing with the specifics of the rules for determining the significant figures in a calculated result, we need to be able to round numbers correctly. To round a number, first decide how many significant figures the number should have. Once you know that, round to that many digits, starting from the left. If the number immediately to the right of the last significant digit is less than 5, it is dropped and the value of the last significant digit remains the same. If the number immediately to the right of the last significant digit is greater than or equal to 5, the last significant digit is increased by 1.
Consider the measurement 207.518 m. As it stands, the measurement contains six significant figures. How would this number be rounded to a fewer number of significant figures? Follow the process as outlined in Table $$\PageIndex{1}$$.
Number of Significant Figures Rounded Value Reasoning
Table $$\PageIndex{1}$$: Rounding Examples
6
207.518 m ⇒ 207.518 m All digits are significant.
5
207.518 m ⇒ 207.52 m Round up, since the 6th digit is an "8" (≥5).
4
207.518 m ⇒ 207.5 m The last two digits are simply dropped, since the 5th digit is a "1" (<5).
3
207.518 m ⇒ 208 m Round up, since the 4th digit is a "5" (≥5).
2
207.518 m ⇒ 210 m ⇒ 2.1×102 m Round up, since the 3rd digit is a "7" (≥5). However, trailing zeros in a number that has no decimal point are ambiguous, so express in scientific notation.
1
207.518 m ⇒ 200 m ⇒ 2×102 m The last five digits are dropped, since the 2nd digit is a "0" (<5). The ones place and tens place are held with zeros. However, trailing zeros in a number that has no decimal point are ambiguous, so express in scientific notation.
Notice that the further a number is rounded, the less reliable that measurement becomes. An approximate value may be sufficient for some purposes, but scientific work requires a much higher level of detail.
It is important to be aware of significant figures when mathematically manipulating numbers. For example, dividing 125 by 307 on a calculator gives 0.4071661238… to an infinite number of digits. But do the digits in this answer have any practical meaning, especially when you are starting with numbers that have only three significant figures each? When performing mathematical operations, there are two rules for limiting the number of significant figures in an answer – one rule is for addition and subtraction, and one rule is for multiplication and division.
⚡️ Significant Figures and Rounding
In operations involving significant figures, the answer is reported in such a way that it reflects the reliability of the least precise operation. An answer is no more precise than the least precise number used to get the answer.
#### Multiplication and Division
Suppose you wanted to find the area of a rectangle and you measured the dimensions as 13.96 cm by 10.77 cm. Multiplying the two lengths together on a calculator yields an area of 150.349 cm2. Should the answer be reported as 150.349 cm2 or should it be rounded to 150.35 cm2 or 150.3 cm2 or even more?
Recall that all measurements have uncertainty. Assuming the last digit is the uncertain digit and may be estimated to the nearest ±0.01 cm, a second student may measure the dimensions of the rectangle as 13.95 cm by 10.76 cm. A third student may measure it as 13.97 cm by 10.78 cm. A fourth student may come up with 13.96 cm by 10.78 cm. The measurements are tabulated in the table below, along with the calculated areas.
If the calculated areas are examined more closely, it is easy to see that the hundreds place (the 1), tens place (the 5), and ones place (the 0) are all certain. They have no variation. We begin to see variation or uncertainty in the tenths place. Since significant figures are defined as all of the certain digits in a measurement plus one uncertain digit, the calculated areas should be rounded and reported to the nearest ±0.1 cm2
Student Length Width Calculated Area Reported Area
1 13.96 cm 10.77 cm 150.349 cm2 150.3 cm2
2 13.95 cm 10.76 cm 150.102 cm2 150.1 cm2
3 13.97 cm 10.78 cm 150.597 cm2 150.6 cm2
4 13.96 cm 10.78 cm 150.489 cm2 150.5 cm2
This is because the length and width each have four significant figures. Multiplying them together results in the area having four significant figures, i.e. three certain digits plus one uncertain digit. When the multiplied values have a different number of significant figures, the answer should be limited to the factor that has the lowest count of significant figures. The same rule applies to division.
⚡️ Rounding Rule for Multiplication and Division
The answer should be rounded so it contains the same number of significant figures as the measurement having the fewest number of significant figures.
✅ Example $$\PageIndex{1}$$
Write the answer for each expression using the appropriate number of significant figures.
1. $$\dfrac{346.1\;\mathrm{mi}}{5.3\;\mathrm h}=$$
2. $$14.58\;\mathrm{ft}\;\times\;5.73\;\mathrm{ft}$$
3. $$36\;\mathrm{in}\;\times26\;\mathrm{in}\;\times16\;\mathrm{in}$$
Solution
A
The calculated answer is 65.301887 mi/h. The reported answer should have two significant figures, since 5.3 h has the fewest significant figures (two) between the two measurements. 65 mi/h
B
The calculated answer is 83.5434 ft2. The reported answer should have three significant figures, since 5.73 ft has the fewest significant figures (three) between the two measurements. 83.5 ft2
C
The calculated answer is 14,976 in3. The reported answer should have two significant figures, since all three measurements have two significant figures. Since an answer of 15,000 in3 would represent an ambiguous number of significant figures, the answer should be reported in scientific notation with two significant figures. 1.5×104 in3
Now that we know how to report answers that involved multiplication and division, you may wonder if answers resulting from addition and subtraction are handled in the same manner. What if we wanted to know the combined distance of 385.17 m, 6.2 m, and 45.86 m? Notice that the first and last lengths were measured to the nearest ±0.01 m, while the second length was measured to the nearest ±0.1 m.
Original Measurements Change 1st Measurement Change 2nd Measurement Change 3rd Measurement
The table above shows what happens to the sum when the uncertain digit in each measurement is changed, one by one. Notice when the second measurement is changed from 6.2 m to 6.3 m that the sum also changes in the tenths place, going from 437.23 m to 437.33 m. In other words, the second measurement has the greatest impact on the calculated result since it is the least precise of the three measurements, ±0.1 m vs. ±0.01 m.
Therefore, the sum of 385.17 m, 6.2 m, and 45.86 m should be reported as 437.2 m.
This leads us to the rounding rule for addition and subtraction. While different, it is worded similarly to the rule for multiplication and division. Can you spot the difference? It is likely the rule with which you are most familiar.
⚡️ Rounding Rule for Addition and Subtraction
The answer should be rounded so it contains the same number of decimal places as the measurement having the fewest number of decimal places.
✅ Example $$\PageIndex{2}$$
1.
2. 1,027 mL + 611 mL + 363.06 mL
Solution
A
A calculator provides an answer of 267.31 kg, but because 8.4 kg is known to the nearest ±0.1 kg, the final answer should be expressed to the nearest ±0.1 kg. 267.3 kg
B
A calculator provides an answer of 2,001.06 mL, but because 1027 mL and 611 mL are both known to the nearest ±1 mL, the final answer should be expressed to the nearest ±1 mL. 2001 mL
✏️ Exercise $$\PageIndex{1}$$
Write the answer for each expression using the correct number of significant figures. As a reminder, calculators do not understand significant figures. You are the one who must apply the rules for calculated answers to a result obtained from a calculator.
1. $$\dfrac{165.110\;\mathrm g}{8.35\;\mathrm{mL}}=$$
2. $$8.6\;\mathrm g\;+\;32.06\;\mathrm g\;+\;88.7\;\mathrm g$$
3. $$255.0\;\mathrm{km}\;-\;99\;\mathrm{km}$$
4. $$44\;\mathrm{cm}\;\times\;43\;\mathrm{cm}$$
$$19.8\;\mathrm g/\mathrm{mL}$$
$$129.4\;\mathrm g$$
$$156\;\mathrm{km}$$
$$1.9\times10^3\;\mathrm{cm}^2$$
#### Calculations Involving Mixed Operations
In practice, chemists generally work with a calculator and carry all digits forward through subsequent calculations. When working on paper, however, we often want to minimize the number of digits we have to write out. Because successive rounding can compound inaccuracies, intermediate rounding needs to be handled correctly. When working on paper, always round an intermediate result so as to retain at least one more digit than can be justified and carry this number into the next step in the calculation. The final answer is then rounded to the correct number of significant figures at the very end.
✅ Example $$\PageIndex{3}$$
A cylinder filled with water was combined with 5.2 cm3 of water that had been drawn into a syringe. If the cylinder had an inner height of 16.8 cm and an inner radius of 1.86 cm, what is the combined volume of water? Report the answer with the correct number of significant figures. (Note: For a cylinder, $$V\mathit=\pi r^{\mathit2}h$$. You may use the $$\pi$$ function on your calculator or estimate the value of $$\pi$$ as 3.1416.)
Solution
Volume Explanation Result
Syringe
The volume of water in the syringe is 5.2 cm3, known to the nearest ±0.1 cm3 5.2 cm3
Cylinder
The volume of water in the cylinder = $$\pi r^{\mathit2}h$$ = $$3.141\underline6(1.8\underline6\;\mathrm{cm})^2(16.\underline8\;\mathrm{cm})$$ = $$18\underline2.594\;\mathrm{cm}^3$$.
The rules for multiplication and division apply. Since 16.8 cm and 1.86 cm both have three significant figures, the calculated volume has three significant figures, or 183 cm3.
Compounding of rounding errors in the final answer may be avoided by carrying extra digits along in the intermediate results. The uncertain digit is underlined (the ones place).
182.594 cm3
Combined
The sum of 5.2 cm3 and 182.594 cm3 is 187.794 cm3.
The rules for addition and subtraction apply. Since 5.2 cm3 is uncertain in the tenths place (±0.1 cm3) and 182.594 cm3 is uncertain in the ones place (±1 cm3), the sum is uncertain in the ones place (±1 cm3), or 187.794 cm3.
188 cm3
The volume should be reported as 188 cm3. Check out the video below for additional examples.
#### Summary
• Rounding
• If the number to be dropped is greater than or equal to 5, increase the number to its left by 1.
• If the number to be dropped is less than 5, there is no change.
• The rule in multiplication and division is that the final answer should have the same number of significant figures as the measurement having the fewest significant figures.
• The rule in addition and subtraction is that the final answer is should have the same number of decimal places as the measurement having the fewest decimal places.
This page is shared under a CK-12 license and was authored, remixed, and/or curated by Lance S. Lund (Anoka-Ramsey Community College), Melissa Alviar-Agnew, and Henry Agnew. Original source: https://www.ck12.org/c/chemistry/.
|
# Solution of trigonometric equations
The General solution of trigonometric equations of the type sin y = sin a, cos y = cos a and tan y = tan a
General Solution: The entire set of the values belonging to the unknown angle satisfies the equation. Also, it has all the specific solutions along with the principal solutions.
## General Solutions of a Trigonometric Equation
From the below-given diagram, we can observe that sin(π -θ) = sin θ & cos ( -θ) = cos θ.
We use the above description for finding the solutions of a few trigonometric equations.
## Now, Solve sin(x) = y for x.
Case 1: When the condition is -1≤y≤ 1, i.e, y’s value is between
(-1 & 1), so, there’s the solution.
sin(x) = y is the set of all the possible solutions which is:
x = sin-1(y) + 2 & x = −sin-1(y) + (2k+1) π, Where “k” can be taken as any integer; i.e, solutions for x is having sin-1(y) in addition to all the even multiples of the value π, together with the minus sin-1(y) & all the odd multiples regarding π.
Case 2: When -1 > y or y > 1, i.e, y’s value is very much large or very much small for the possible solution.
There’re no solutions.
## Solving Cos (x) = y for x.
Case 1: -1≤y≤ 1
cos(x) = y is the set of all the solutions which is equal to
x = ± cos-1(y) + 2,
Where “k” can be taken as any integer;
Case 2: When, -1 > y or y > 1
There exist no solutions.
## Solving tan(x) = y for x.
Tan(x) = y equation is the set of all solution to:
x = tan-1(y) + ,
Where the value of “k” can be taken as “any integer”
Example
Find the general solutions of 2 sinx = 1.
The equation is equivalent to Sinx =
Now, Sin ⁻¹ =
Hence the general solution is
x = ±π/4 + 2, where the value of k can be any integer.
## Trigonometric Equations with their General solutions
Trigonometrical Equation General Solution Sin θ = 0 θ = nπ Cos θ = 0 θ = nπ + π/2 Tan θ = 0 θ = nπ Sin θ = 1 θ = 2nπ + π/2 Cos θ = 1 θ = 2nπ Sin θ = a θ = nπ ± (-1)ⁿa Cos θ = a θ = 2nπ ± a Tan θ = a θ = nπ ± a Sin²θ = Sin²a θ = nπ ± a Cos²θ = Cos²a θ = nπ ± a Tan²θ= Tan²a θ = nπ ± a Sin θ = Sin aCos θ = Cos a θ = nπ ± a Sin θ = Sin aTan θ = Tan a θ = nπ ± a Tan θ = Tan aCos θ = Cos a θ = nπ ± a
|
one-step-equations-pre-algebra-moomoomath
# Interactive video lesson plan for: One step equations-pre algebra-moomoomath
#### Activity overview:
How to solve one step equations that include addition and subtraction
For a summary of the video and a practice link go to http://moomoomath.com
Transcript
One step equations with addition and subtraction. Let’s look at our problem. X minus five equals seven. What value minus five equals seven? I will add five to both sides, so X equals twelve. Now let’s take the same problem and slow it down. Let’s look at it slowly now, because this is subtraction in order to undo that I will add five to both sides, When working with equations you have to do the same operation to both sides. If I add five to the left I must add five to the right. Now, let’s simplify. Negative five plus five adds to zero, I’m left with on the left side. The value X on the right side I do the straight operation, seven plus five and that is equal to twelve. You will often see this with different sign numbers. I’m going to change it to X minus two is equal to negative ten. For this problem to undo and get the X by itself I will add two to the left and two to the right. The negative two and positive two still cancel and I’m left with X but on the right side I’m left with sign numbers. So negative ten plus two and when you have different sign you subtract and you have eight. Take the sign of the larger number which is eight. So X is equal to negative eight. In all equations you can go back, plug in your answer and check it. Twelve minus five does equal to seven, so that one checks, the second one negative eight minus two does add to negative ten. Thanks for visiting MooMooMath
* Check out my site
MooMooMath
http://www.moomoomath.com/
* Subscribe
* Pinterest
https://www.pinterest.com/moomoomath/
-~-~~-~~~-~~-~-
-~-~~-~~~-~~-~-
Tagged under: addition,algebra,math ,moomoomath,equations,subtraction, step,algebra ,How ,solve,math,educational,Equation,Pre-algebra (Field Of Study)
Clip makes it super easy to turn any public video into a formative assessment activity in your classroom.
Add multiple choice quizzes, questions and browse hundreds of approved, video lesson ideas for Clip
Make YouTube one of your teaching aids - Works perfectly with lesson micro-teaching plans
Play this activity
1. Students enter a simple code
2. You play the video
3. The students comment
4. You review and reflect
* Whiteboard required for teacher-paced activities
## Ready to see what elsecan do?
With four apps, each designed around existing classroom activities, Spiral gives you the power to do formative assessment with anything you teach.
Quickfire
Carry out a quickfire formative assessment to see what the whole class is thinking
Discuss
Create interactive presentations to spark creativity in class
Team Up
Student teams can create and share collaborative presentations from linked devices
Clip
Turn any public video into a live chat with questions and quizzes
### Spiral Reviews by Teachers and Digital Learning Coaches
@kklaster
Tried out the canvas response option on @SpiralEducation & it's so awesome! Add text or drawings AND annotate an image! #R10tech
Using @SpiralEducation in class for math review. Student approved! Thumbs up! Thanks.
@ordmiss
Absolutely amazing collaboration from year 10 today. 100% engagement and constant smiles from all #lovetsla #spiral
@strykerstennis
Students show better Interpersonal Writing skills than Speaking via @SpiralEducation Great #data #langchat folks!
|
# What is the slope of any line perpendicular to the line passing through (-3,1) and (-6,12)?
Apr 27, 2018
$\frac{3}{11}$
#### Explanation:
First, let's find the slope through the given two points: $\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$, where there are two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$:
$\frac{12 - 1}{- 6 - \left(- 3\right)}$
$\frac{11}{- 3}$
$- \frac{11}{3}$
Perpendicular slopes are opposite reciprocals of one another:
Opposites: put a negative sign in front of a positive number or take away a negative sign from a negative number (ex. $- 3 , 3$, $\frac{7}{11} , - \frac{7}{11}$
Reciprocals: flip the numerator and denominator (ex. $\frac{12}{5} , \frac{5}{12}$, $4 , \frac{1}{4}$)
The opposite of $- \frac{11}{3}$ is $\frac{11}{3}$
The reciprocal of $\frac{11}{3}$ is $\frac{3}{11}$
|
# Glossary for Lesson 2: Expressions
### Factors
The factors of a term or expression are numbers, variables, or combinations of numbers and variables that, when multiplied together, produce that term or expression. For example, 2 and 3 are factors of 6, and -1, 5, x, and y are factors of -5xy
### Commutative Property
For addition: The commutative property of addition states that changing the order in which numbers are added does not affect the final result. For example, 2+3=3+2. This is true no matter how many terms are being added together. For multiplication: The commutative property of multiplication states that changing the order in which numbers are multiplied does not affect the final result. For example, 2\cdot3=3\cdot2. This is true no matter how many terms are being multiplied together.
### Combining Like Terms
Like terms are terms which have the same variables with the same powers. 7xy^2 and -\frac {3}{2}xy^2 are like terms, while 7xy and -\frac {3}{2}xy^2 are not. Since the only thing that differs between like terms is the coefficient, they can be added together, and this is called combining like terms
### Polynomial
An expression made up of variables and/or constants that are combined with addition, subtraction, and multiplication and have non-negative integer exponents. Examples: -3x+4, 6y^4+7y^2-8y+13, 5xy^2z-11
### Degree of a Polynomial
To find the degree of a polynomial, we first need to find the degree of each of its terms. The degree of a term is the sum of the powers of all of the variables in that term. So, for instance, 7x^2 has a degree of 2, and -4xy^2z has a degree of 4. The degree of a polynomial is equal to the degree of its highest-degree term. Examples: 6y^4+7y^2-8y+13 has a degree of 4 or, in other words, is a fourth degree polynomial.
### Standard Form of a Polynomial
A polynomial is said to be in standard form when it is written with its highest-degree term first, with the rest of the terms written in order of descending degree. For example, 6y^4+7y^2-8y+13 is written in standard form, whereas 7y^2-8y+13+6y^4 is not.
### Exponent
Repeated multiplication can be written using exponential notation. For example, 3\cdot3\cdot3\cdot3\cdot3=3^5. Here, the number 3 is known as the base, and 5 is known as the exponent or the power. The base thus refers to the quantity being multiplied by itself, and the exponent tells us how many times this multiplication happens.
### Order of Operations
The order of operations establishes the order in which different mathematical procedures should be performed. This is often expressed with the acronym PEMDAS (or perhaps BEDMAS, BIDMAS, or BODMAS). PEMDAS, which stands for Parentheses Exponents Multiplication Division Addition Subtraction, reminds us that when simplifying expressions, we should start by simplifying all expressions inside parentheses, then apply exponents, then multiply or divide, and lastly add or subtract.
Subtraction and addition have the same level of priority, so when terms are being added or subtracted one after the other, we simply work from left to right. For example, if we had 1-4+7 , we would simplify it as follows: 1-4+7=-3+7=4.
Multiplication and division have the same sort of relationship: they have the same level of priority as one another, so when terms are being multiplied or divided one after the other, we simply work from left to right. For example, if we had 3\div 6\times 9 , we would simplify it as follows: 3\div 6\times 9=2\times 9=18. The order of operations gives us a set of general guidelines that we must apply carefully.
|
Question Video: Finding an Unknown Matrix Using Operations on Matrices | Nagwa Question Video: Finding an Unknown Matrix Using Operations on Matrices | Nagwa
# Question Video: Finding an Unknown Matrix Using Operations on Matrices Mathematics • First Year of Secondary School
## Join Nagwa Classes
Attend live Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher!
Consider the matrices π΄ = [0, β4 and 2, 1], π΅ = [β3, β7, 6 and 6, β4, 3], and πΆ = [7, β7 and β7, 2 and β7, 7]. Determine the matrix π that satisfies βπ^(π) = π΄Β² + (π΅πΆ)^(π).
06:28
### Video Transcript
Consider the matrices π΄ is the two-by-two matrix zero, negative four, two, one; π΅ is the three-by-two matrix negative three, negative seven, six, six, negative four, three; and πΆ is the two-by-three matrix seven, seven, negative seven, two, negative seven, seven. Determine the matrix π that satisfies negative π transpose is equal to π΄ squared plus π΅πΆ all transposed.
In this question, weβre given a matrix equation involving three known matrices π΄, π΅, and πΆ. We need to use this to determine the matrix π. To do this, letβs start by evaluating the right-hand side of this equation, and we can do this term by term. First, we need to evaluate π΄ squared. And we recall squaring a matrix means we multiply it by itself. π΄ squared is π΄ times π΄. Therefore, π΄ squared is given by the following expression.
And we recall to multiply two matrices together, we need to find the sum of the products of the corresponding entries in the rows of the first matrix and the columns of the second matrix. For example, to find the entry in row one, column one of matrix π΄ squared, we need to find the product of zero and zero and add this to the product of negative four and two. And we can evaluate this expression. Itβs equal to negative eight. This is the entry in row one, column one of matrix π΄ squared.
We can follow the same process to find the entry in row one, column two of matrix π΄ squared. Itβs equal to zero multiplied by negative four plus negative four multiplied by one, which we can calculate is equal to negative four. And we can follow the same procedure to find the final two elements of π΄ squared. We see that matrix π΄ squared is equal to the two-by-two matrix negative eight, negative four, two, negative seven. And now that we found an expression for matrix π΄ squared, letβs find an expression for the second term of the right-hand side of our equation. Thatβs the transpose of π΅πΆ. And to find the transpose of matrix π΅πΆ, letβs start by multiplying matrix π΅ by matrix πΆ. And although itβs not necessary, letβs check that we can multiply matrix π΅ by matrix πΆ.
Remember, to multiply two matrices together, we need the number of columns of the first matrix to be equal to the number of rows of the second matrix. We can see that these are both equal to three. Then the order of the products of these two matrices will be the number of rows of the first matrix by the number of columns of the second matrix. The product of these two matrices is a two-by-two matrix. And we find the products of these two matrices in the same way we did for the matrix π΄ squared.
We need to find the sum of the products of the corresponding elements of the rows of matrix π΅ with the columns of matrix πΆ. For example, the entry in row one, column one of matrix π΅πΆ will be negative three times seven plus negative seven times negative seven plus six multiplied by negative seven, which if we evaluate we see is equal to negative 14. So the entry in row one, column one of π΅πΆ is negative 14.
We can follow the same process to find the entry in row one, column two. Itβs negative three times seven plus negative seven times two plus six multiplied by seven, which if we calculate is equal to seven. We can do the same to find the remaining two entries of this matrix. π΅ times πΆ is the two-by-two matrix negative 14, seven, 49, 55. Now weβre ready to find an expression for the second term of the right-hand side of our equation. We need to take the transpose of matrix π΅πΆ. And to do this, letβs recall how we find the transpose of a matrix. This means we write the rows of the matrix as the columns of the new matrix. So the first column of the transpose of this matrix is found by using the first row of this matrix.
The first column of the transpose is negative 14, seven. And we can do the same to find the second column of this matrix. We use the second row of π΅πΆ. The second column is 49, 55. Therefore, weβve shown the transpose of π΅πΆ is the two-by-two matrix negative 14, 49, seven, 55. And now that we found π΄ squared and the transpose of π΅πΆ, we can evaluate the right-hand side of the equation. We just need to add π΄ squared to the transpose of matrix π΅πΆ.
Remember, to add two matrices of the same order together, we just need to add the corresponding entries together. For example, the entry in row one, column one of the sum of these two matrices will be negative eight plus negative 14, which we can calculate is negative 22. Similarly, the element in row one, column two of our matrix will be negative four plus 49, which we can calculate is 45. And we can continue this process to find the final two elements of the matrix. We have two plus seven is nine and negative seven plus 55 is 48. So our matrix is the two-by-two matrix negative 22, 45, nine, 48.
Now, we can substitute the matrix we found for the entire right-hand side of the equation weβre given. This gives us that negative π transpose is the two-by-two matrix negative 22, 45, nine, 48. And remember, we want to solve this equation for the matrix π, and thereβs a few different ways of doing this. For example, we could notice multiplying by the scalar and taking the transpose of a matrix does not change its order. Therefore, we can conclude that matrix π is also a two-by-two matrix. So we could write π as a two-by-two matrix and find all of the elements of matrix π. And this would work. However, we can solve this matrix equation directly.
First, weβll multiply both sides of our equation through by negative one. When we do this, on the left-hand side of our equation, we get negative one times negative one which is one multiplied by the transpose of π. And on the right-hand side of our equation, we get negative one multiplied by our matrix. And remember, to multiply a matrix by a scalar, we just multiply all of the entries of the matrix by our scalar. And since our scalar is negative one, this just means we switch the sign of all of the entries of the matrix. This gives us the two-by-two matrix 22, negative 45, negative nine, negative 48.
Now, we can solve the matrix π by taking the transpose of both sides of the equation. When we do this, on the left-hand side of the equation, we get the transpose of the transpose of π. And on the right-hand side of our equation, we get the transpose of the given two-by-two matrix. And we can then simplify this. First, for any matrix π, the transpose of the transpose of π is just equal to π. So the left-hand side of our equation is equal to π. π is just the transpose of the given two-by-two matrix. And therefore, we can find matrix π by taking the transpose of this matrix. π is the two-by-two matrix 22, negative nine, negative 45, negative 48.
## 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
|
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# The radius of the bore of a capillary tube is $r$ and the angle of contact is $\theta$. When the tube of sufficient length is dipped in the liquid, the radius of curvature of the meniscus of the liquid rising in the tube is(A) $r\sin \theta$(B) $rcos\theta$(C) $\dfrac{r}{{\sin \theta }}$ (D) $\dfrac{r}{{cos\theta }}$
Last updated date: 18th Sep 2024
Total views: 81k
Views today: 1.81k
Verified
81k+ views
Hint We proceed to solve this question by drawing a triangle in the meniscus region. Using properties of a triangle as well as trigonometry we find the radius of curvature of the meniscus. We can say that the meniscus formed is a concave meniscus because the liquid is rising in the tube. In the case of the convex meniscus, the liquid in the tube goes below the surface of the liquid.
Complete Step by step solution
The meniscus formed is a concave meniscus because from the question the liquid is rising in the capillary tube.
Pictorial representation of the meniscus
Here,
Radius of the capillary tube is represented by $r$
The same diagram can be made as follows
The radius of curvature of the meniscus is represented by $R$
The angle of contact is represented by $\theta$
From the angle of contact angle $x$ is equal to $\theta$
From the diagram, the bigger triangle is a combination of two right-angled triangles
Taking the triangle on the left side under consideration and using $cos\theta = \dfrac{{base}}{{hyp}}$ equal to $cos\theta = \dfrac{r}{R}$
$\Rightarrow R = \dfrac{r}{{cos\theta }}$
Hence radius of curvature of the meniscus is equal to $\dfrac{r}{{cos\theta }}$
Option (D) $\dfrac{r}{{cos\theta }}$is the correct answer.
Additional information A concave meniscus is formed when the adhesive force (between liquid and container) is greater than the cohesive force (intermolecular forces). Due to this, a concave meniscus is formed. Also, when the liquid is placed in a capillary tube it rises due to the adhesive force being greater than the cohesive force.
In the case of a liquid with a greater cohesive force than adhesive force, a convex meniscus is formed.
Note One might make the mistake of taking the meniscus as a convex meniscus. A convex meniscus is formed when the liquid in the capillary tube goes below the surface level of the remaining liquid. This question mentions that the liquid in the capillary tube is rising hence it is a concave meniscus.
|
# Undetermined Coefficients
I tried to change it to $y" + y = (14\sin x-28 \sin^3 x)$. The complementary solution is $C_1\cos x+C_2\sin x$ and the particular solution to $y" + y = 14 \sin x$ is $-7\sin x$. How do you find the particular solution to $y" + y = -28\sin^3 x$ ? What would be your guess?
Ans: $$c_1 \sin(x\sqrt2)+c_2\cos(s\sqrt2)-7\sin x-\sin(3x)$$
• You have the right idea, you need to convert $\sin(x)\cos(2x)$ into terms of sines and cosines. – Chee Han Feb 12 '17 at 6:58
Using double-angle formula, you can rewrite $\sin(x)\cos(2x)$ as $$\sin(x)\cos(2x) = \frac{1}{2}[\sin(3x) - \sin(x)].$$ Indeed, \begin{align*} \sin(3x) = \sin(2x + x) & = \sin(2x)\cos(x) + \cos(2x)\sin(x) \\ \sin(x) = \sin(2x-x) & = \sin(2x)\cos(x) - \cos(2x)\sin(x). \end{align*} Thus, this gives $$y'' + y = 7[\sin(3x) - \sin(x)].$$ Since there is no first derivative involved, normally one would guess a particular solution of the form $$y_p(x) = A\sin(3x) + B\sin(x).$$ Since $\sin(x)$ is a solution to the homogeneous equation $y''+y=0$, one should now guess a particular solution of the form $$y_p(x) = A\sin(3x) + Bx\sin(x) + Cx\cos(x).$$
• cos 2x = cos2 x – sin2 x or cos 2x = 2 cos2 x – 1 or cos 2x = 1 – 2 sin2 x are the only trig identities – Shay Feb 12 '17 at 7:16
• You mean you are only allowed to use those identities? Otherwise, en.wikipedia.org/wiki/List_of_trigonometric_identities – Chee Han Feb 12 '17 at 7:28
• where did the 14 go? – Shay Feb 12 '17 at 7:34
• Opps forgot about the 14, just added that. – Chee Han Feb 12 '17 at 7:35
• The answers shown but if I do it your way I get -9Asin3x-Bsinx+Asin3x+Bsinx=14sin3x-14sinx and I get -7/2 sin3x – Shay Feb 12 '17 at 7:45
|
# What does slope mean?
embizze | Certified Educator
Slope refers to the steepness of a line.
A line written in slope-intercept form is written as y=mx+b where m is the slope of the line. A positive slope indicates that the line rises as you move from left to right, while a negative slope indicates that the line falls as you scan from left to right. A horizontal line has zero slope, while a vertical line has no slope (the slope is undefined.)
Graphically the slope is the ratio of the rise over the run -- the vertical change divided by the horizontal change.
We can compute the slope of the line if we are given any two points on the line -- the slope is the difference in the y-values divided by the difference in the x-values. For example, the slope of the line containing (1,1) and (3,5) has slope (5-1)/(3-1)=2. (Note that we can go the other direction -- (1-5)/(1-3)=2)
Slope is the constant or average rate of change. For example, if a car is moving at 60mph, the distance function is a line with slope 60.
In calculus we find the slope of the line tangent to a curve at a point using teh derivative of the function.
studyfaq1 | Student
The slope of a line is the ratio of the amount that y increases as x increases some amount. Slope tells you how steep a line is, or how much y increases as x increases. The slope is constant (the same) anywhere on the line.
One way to think of the slope of a line is by imagining a roof or a ski slope. Both roofs and ski slopes can be very steep or quite flat. In fact, both ski slopes and roofs, like lines, can be perfectly flat (horizontal). You would never find a ski slope or a roof that was perfectly vertical, but a line might be.
We can usually visually tell which ski slope is steeper than another. Clearly, the three ski slopes get gradually steeper.
You could write the relationship like this:
Slope = (Change in height)/(Change in width)
Or:
Slope = rise/run
If y represents the vertical direction on a graph, and xrepresents the horizontal direction, then, this formula becomes:
Slope = (Change in y)/(Change in x)
Or:
In this equation, m represents the slope. The small triangles are read 'delta' and they are Greek letters that mean 'change.'
For the first ski slope example, the skier travels four units vertically and ten units horizontally. So, the first slope is m = 4/10.
The second ski slope involves a seven unit change vertically and ten units horizontally. So, the slope is m = 7/10. The second slope is steeper than the first because 7/10 is greater 4/10.
The third ski slope involves a seven unit change vertically and six units horizontally. So, the slope is m = 7/6. The third slope is the steepest of all.
## Slope on the Cartesian Plane
The Cartesian plane is a two-dimensional mathematical graph. When graphing on it, a line may not start at zero as in the ski slope examples. In fact, a line goes on forever at both ends. The slope of a line, however, is exactly the same everywhere on the line. So, you can choose any starting and ending point on the line to help you find its slope. It is also possible that you might be given a line segment, which is a section of a line that has a beginning and an end. Or, you might be given two points and you are expected to draw (or imagine) the line segment between them. In all these situations, finding the slope works the same way.
Just like with the ski slope, the goal is to find the change in height and the change in width. For the line segment in the image, you can simply count the squares on the grid.
Graph of a line segment
The difference in height between the two points is three units (three squares). The difference in width between the two points is two units (two squares). So, the slope of the line segment (the slope between the two points) is m = 3/2.
In mathematics class, you may memorize a formula to help you get the slope. The formula looks like this:
kandukurimaths | Student
straight line have one good property that rate of change is constant.means
ax+by+c=0 is general straight line equation. let x is independent variable and dependent variable and two arbitrary points (x1,y1),(x2,y2) then change in y coordinates by change in x coordinates is constant
slope=(y2-y1)/(x2-x1)
example y=2x
if we draw graph y=2x
lets take two points on line (1,2) and (2,4)
slope=(4-2)/(2-1)=2
from this example we need to observe two points
point one:
if you write straight line in the form of y=mx than m is slope
second point:
if see in graph we can form right angle triangle with given two points
and intersection of y=y1 and x=x2 that is (2,2)
using trigonometry function tan(theta)=(y2-y1)/(x2-x1)
any straight line equation can write in the form of y=mx+c
m is slope of line
if we know two points on line than we can calculate slope using (y2-y1)/(x2-x1)
if we know line making angle with x-axis is theta than we can calculate slope=tan(theta)
In example intercept zero but it wont effect the slope of line
thank you
|
# How to Solve Logarithmic Inequalities
When you solve logarithmic inequalities, you use the same rules for equations as you are used to.
Rule
### Logarithmicinequalities
aking the exponent on both sides of an inequality does not change the inequality. Thus, nothing happens to the inequality sign! $\begin{array}{llllllll}\hfill \mathrm{log}x&
Example 1
Solve the inequality $\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(x-4\right)\le 8$ $\begin{array}{llll}\hfill \mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(x-4\right)& \le 8\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {e}^{\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(x-4\right)}& \le {e}^{8}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x-4& \le {e}^{8}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& \le {e}^{8}+4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& \le \text{}2984.96\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Example 2
Solve the inequality $\mathrm{log}\phantom{\rule{-0.17em}{0ex}}\left(-2x+3\right)\ge 4$
$\begin{array}{llllll}\hfill \mathrm{log}\phantom{\rule{-0.17em}{0ex}}\left(-2x+3\right)& \ge 4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {e}^{\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(-2x+3\right)}& \ge {e}^{4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -2x+3& \ge {e}^{4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -2x& \ge {e}^{4}-3\phantom{\rule{2em}{0ex}}& \hfill & |÷\phantom{\rule{-0.17em}{0ex}}\left(-2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& \le \frac{{e}^{4}-3}{-2}=\frac{3-{e}^{4}}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ You need to flip the inequality sign when you divide by a negative number. You must have an equality sign in the last row as the answer is just changed to look nicer.
Example 3
Solve the inequality $\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)\le e$ $\begin{array}{llll}\hfill \mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)& \le e\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {e}^{\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)}& \le {e}^{e}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x-3& \le {e}^{e}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& \le {e}^{e}+3\approx \text{}18.15\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
|
Weighted Average
Weighted average is a mean calculated by giving values in a data set more influence according to some attribute of the data. It is an important section in CAT quantitative aptitude. Weighted average is an average in which each quantity to be averaged is assigned a weight, and these weightings determine the relative importance of each quantity on the average. Weightings are the equivalent of having that many like items with the same value involved in the average.
Weighted Average Formula
Observe that the average can be calculated directly only if the weights of all the factors are the same. If the values a, b, c, and d are the individual averages of each of the groups and n1, n2, n3 and n4 are the numbers of observations or number of values, then the Weighted Average is calculated by the below Weighted Average Formula:
The term “weight” stands for the relative importance attached to the different values.
1) In Class I there are 12 students of average age 20 years and in Class II there are 16 students of average age 23 years. What is the average age of both classes?
Solution:
=((12 ×20)+ (16 ×23))/((12+16)) = (240+368)/28 = 608/28 = 20.7 years
Note:
➜ Average always lies between the maximum and minimum value. It will be equal to the maximum or minimum value if all the values are equal
➜ Average is the result of the net surplus + net deficit
➜ If the value of each quantity is increased or decreased by the same quantity “q”, then the average will also increase or decrease by the quantity “q” respectively.
➜ If the value of each quantity is multiplied or divided by a quantity “q”, then the resultant average will also be multiplied or divided by “q” respectively.
➜ When the weights of all the quantities are the same, then the average can be calculated directly. If the weights are different, we need to use the concept of weighted average.
Short-cut Techniques in Weighted Average
Let us now see a useful shortcut technique to find the weighted average method with the help of an example.
2) The average of a batsman in 16 innings is 36. In the next innings, he is scoring 70 runs. What will be his new average?
a) 44 b) 38 c) 40 d) 48
Solution:
Conventionally solving
New average = (old sum+ new score)/(total number of innings) = ((16 ×36)+70)/((16+1)) = 38
Shortcut technique:
Step 1: Take the difference between the new score and the old average = 70-36= 34
Step 2: This 34 extra run is spread over 17 innings = 34/17=2
Step 3: Hence, the average increases by 2 = >36+2 = 38 is the new average
Let us try another question using the same technique.
3) The average marks of 19 children in a particular school is 50. When a new student with marks 75 joins the class, what will be the new average of the class?
Solution:
1) Take the difference between the old average and the new marks = 75-50=25
2) This score of 25 is distributed over 20 students => 25/20 = 1.25
3) Hence, the average increases by 1.25=> 50+1.25 = 51.25 Let us try another question where the average dips.
4) The average age of Mr. Mark’s 3 children is 8 years. A new baby is born. Find the average age of all his children?
Solution:
The new age will be 0 years. The difference between the old average and the new age = 0-8= -8
This age of 8 years is spread over 4 children => (-8/4= -2) Hence, the average reduces to 8-2= 6 years
Now let us look at a technique that will help us compute the new value when the average is given.
Take this question for example.
5) The average age of 29 students is 18. If the age of the teacher is also included the average age of the class becomes 18.2. Find the age of the teacher?
a) 28 b) 32 c) 22 d) 24
Solution:
Conventionally solving
Let the average age of the teacher =x
(29 × 18 + x × 1)/30
Solving for x, we get x = 24
Using the shortcut, based on the same method we used previously
Step 1: Calculate the change in average = 18.2-18 = 0.2
This change in 0.2 is reflected over a sample size of 30
New age is an increased by 30 * 0.2 = 6 years above the average 18+6= 24.
Think Quest- The average age of 26 students in an MBA school is 30. One student among these quits the school in between. Can you find the age of that student if the new average is 29.8?
Illustrations on Weighted Average
6) Average goals scored by 15 selected players in EPL is 16. The maximum goals scored by a player is 20 and the minimum is 12. Goals scored by players is between 12 and 20. What can be the maximum number of players who scored at least 18 goals?
a) 10 b) 5 c) 9 d) 6 e) none of these
Solution: Option (c)
To maximize the number of players who scored 18 and above number of goals, we will assume that only one person has scored 20. To counter him, we will have one person who will score 12 goals.
15-2 =13 players left
Now to maximize the 18 and above goals, for every two players who are scoring 18, we will have one player scoring 12. This is done, to arrive at the average of 16. We will have 8 players with a score of 18 and 4 players with a score of 12. The last player will have a score of 16 Thus, the maximum number of people with 18 and more goals = 9.
7) The average weight of a group of 8 girls is 50 kg. If 2 girls R and S replace P and Q, the new average weight becomes 48 kg. The weight of P= Weight of Q and the weight of R= Weight of S.Another girl T, is included in the group and the new average weight becomes 48 kg. Weight of T= Weight of R. Find the weight of P?
a) 48 kgs b) 52 kgs c) 46 kgs d) 56 kgs
Solution: Option (d)
8 x 50 +R+S-P-Q= 48×8 R+S-P-Q=-16
P+Q-R-S= 16 R=S and P=Q
P-R=8 One more person is included and the weight = 48 kg
Let the weight be a = (48 ×8+a)9/9 = 48
a=48 kg= weight of R weight of P= 48+8= 56 kg
Stay tuned with BYJU’S to get the latest notification on CAT 2022. Know the CAT syllabus, pattern and cut off only at BYJU’S.
|
# Subtracting Fractions With Different Denominators Made Easy!
//Subtracting Fractions With Different Denominators Made Easy!
Subtracting Fractions With Different Denominators Made Easy! 2017-08-15T03:23:31+00:00
# Subtracting Fractions
## Subtracting Fractions With Different Denominators Made Easy!
When subtracting fractions with different denominators, we follow the same process we used for adding unlike fractions. But since everybody doesn’t start with addition, we provide the same level of detail for subtraction.
First of all, when subtracting fractions with different denominators, the first step in the Rule says that we must change these fractions so that they have the “same denominator“.
Here are the steps for subtracting fractions with different denominators. We will break-down each step just like before to make sure you’ve got it. Then we will subtract some tougher numbers. And finally, we will help you pull everything together. Okay!
### So, here are the steps.
1. Build each fraction so that both denominators are equal. Remember, when subtracting fractions, the denominators must be equal. So we must complete this step first. What this really means is that you must find what is called a Common Denominator. Most of the time you will be required to work the problem using what’s called the Least Common Denominator (LCD). In either case you will build each fraction into an equivalent fraction.
2. Re-write each equivalent fraction using this new denominator
3. Now you can subtract the numerators, and keep the denominator of the equivalent fractions.
4. Re-write your answer as a simplified or reduced fraction, if needed.
But keep in mind, if you are doing homework, be sure to answer the problems in the form asked for in the assignment.
### –
Notice that the overall size of our point of reference
(The Whole) is EXACTLY the same.
Step #1 in our rule tells us that the denominators must be equal. And the easiest way to find a common denominator is to just multiply the denominators.
So let’s do that now…
2 x 3 = 6
The Common Denominator for 1/2 and 1/3 is 6
Step #2 – Re-write each equivalent fraction using this new denominator.
Since…
1/2 is equivalent to 3/6
And…
1/3 is equivalent to 2/6
We re-write our equation to read…
Subtract: 3/6 – 2/6
Now we are ready to do Step #3 – Subtract the numerators, and keep the denominator of the equivalent fractions (which is 6).
So, we end up with…
3/6 – 2/6 = (3 – 2 )/6 = 1/6
=
Finally, Step #4 – Re-write your answer as a simplified or reduced fraction, if needed.
In our example, the answer (1/6) is already in its simplest form. So, no further action is required!
That’s It!
A quick and easy way to subtract fractions with different denominators.
```Do you want to practice subtracting fractions?
```
|
# Sign Test for Rational Function Graphs
## Procedure for determining whether a function is positive or negative at a value.
Estimated10 minsto complete
%
Progress
Practice Sign Test for Rational Function Graphs
MEMORY METER
This indicates how strong in your memory this concept is
Progress
Estimated10 minsto complete
%
Sign Test for Rational Function Graphs
The asymptotes of a rational function provide a very rigid structure in which the function must live. Once the asymptotes are known you must use the sign testing procedure to see if the function becomes increasingly positive or increasingly negative near the asymptotes. A driving question then becomes how close does near need to be in order for the sign test to work?
### Sign Test for Rational Functions
Consider mentally substituting the number 2.99999 into the following rational expression.
f(x)=(x1)(x+3)(x5)(x+10)(x+2)(x4)(x3)\begin{align*}f(x)=\frac{(x-1)(x+3)(x-5)(x+10)}{(x+2)(x-4)(x-3)}\end{align*}
Without doing any of the arithmetic, simply note the sign of each term:
f(x)=(+)(+)()(+)(+)()()\begin{align*}f(x)=\frac{(+) \cdot (+) \cdot (-) \cdot (+)}{(+) \cdot (-) \cdot (-)}\end{align*}
The only term where the value is close to zero is (x3)\begin{align*}(x-3)\end{align*} but careful subtraction still indicates a negative sign. The product of all of these signs is negative. This is strong evidence that this function approaches negative infinity as x\begin{align*}x\end{align*} approaches 3 from the left.
Next consider mentally substituting 3.00001 and going through the same process.
f(x)=(+)(+)()(+)(+)()(+)\begin{align*}f(x)=\frac{(+) \cdot (+) \cdot (-) \cdot (+)}{(+) \cdot (-) \cdot (+)}\end{align*}
The product of all of these signs is positive which means that from the right this function approaches positive infinity instead. This technique is called the sign test. The sign test is a procedure for determining only whether a function is above or below the axis at a particular value.
The sign test helps you sketch and graph a function. Look at the following function:
f(x)=1(x+2)2(x1)\begin{align*}f(x)=\frac{1}{(x+2)^2 \cdot (x-1)}\end{align*}
Your first step for sketching this is to identify the vertical asymptotes. The vertical asymptotes occur at x=2\begin{align*}x=-2\end{align*} and x=1\begin{align*}x=1\end{align*}. Then, you use the asymptotes to perform a sign test. The points to use the sign testing procedure with are -2.001, -1.9999, 0.9999, 1.00001. The number of decimals does matter so long as the number is sufficiently close to the asymptote. Note that any real number squared is positive.
f(2.001)f(1.9999)f(0.9999)f(1.0001)=(+)(+)()==(+)(+)()==(+)(+)()==(+)(+)(+)=+\begin{align*}f(-2.001) &= \frac{(+)}{(+) \cdot (-)}=-\\ f(-1.9999) &= \frac{(+)}{(+) \cdot (-)}=-\\ f(0.9999) &= \frac{(+)}{(+) \cdot (-)}=-\\ f(1.0001) &= \frac{(+)}{(+) \cdot (+)}=+\end{align*}
Later when you sketch everything you will use your knowledge of zeroes and intercepts. For now, focus on just the portions of the graph near the asymptotes. Note that the graph below is NOT complete.
### Examples
#### Example 1
Earlier, you were asked how close the numbers need to be to perform the sign test. In order to truly answer the question about how close the numbers need to be, calculus should be used. For the purposes of PreCalculus, the testing number should be closer to the vertical asymptote than any other number in the problem. If the vertical asymptote occurs at 3 and 3.01 is in the problem elsewhere, do not choose 3.1 as a sign test number.
#### Example 2
Identify the vertical asymptotes and use the sign testing procedure to roughly sketch the nature of the function near the vertical asymptotes.
f(x)=(x+1)(x4)2(x1)(x+3)3100(x1)2(x+2)\begin{align*}f(x)=\frac{(x+1)(x-4)^2(x-1)(x+3)^3}{100(x-1)^2(x+2)}\end{align*}
Note that x=2\begin{align*}x=-2\end{align*} is clearly an asymptote. It may be initially unclear whether x=1\begin{align*}x=1\end{align*} is an asymptote or a hole. Just like holes have priority over zeroes, asymptotes have priority over holes. The four values to use the sign testing procedure are -2.001, -1.9999, 0.9999, 1.00001.
f(2.001)f(1.9999)f(0.9999)f(1.0001)=()(+)()(+)(+)()==()(+)()(+)(+)(+)=+=(+)(+)()(+)(+)(+)==(+)(+)(+)(+)(+)(+)=+\begin{align*}f(-2.001) &= \frac{(-) \cdot (+) \cdot (-) \cdot (+)}{(+) \cdot (-)}=-\\ f(-1.9999) &= \frac{(-) \cdot (+) \cdot (-) \cdot (+)}{(+) \cdot (+)}=+\\ f(0.9999) &= \frac{(+) \cdot (+) \cdot (-) \cdot (+)}{(+) \cdot (+)}=-\\ f(1.0001) &= \frac{(+) \cdot (+) \cdot (+) \cdot (+)}{(+) \cdot (+)}=+\end{align*}
A sketch of the behavior of this function near the asymptotes is:
#### Example 3
Create a function with two vertical asymptotes at 3 and -2 such that the function approaches positive infinity from both directions at both vertical asymptotes.
Earlier, there was a function that approached negative infinity from both sides of the asymptote. This occurred because the term was squared in the denominator. An even powered term will always produce a positive term.
f(x)=1(x3)2(x+2)2\begin{align*}f(x)=\frac{1}{(x-3)^2(x+2)^2}\end{align*}
#### Example 4
Create a function with three vertical asymptotes such that the function approaches negative infinity for large and small values of x\begin{align*}x\end{align*} and has an oblique asymptote.
There are an infinite number of possible solutions. The key is to create a function that may work and then use the sign testing procedure to check. Here is one possibility.
f(x)=x710(x1)2(x2)2(x4)2\begin{align*}f(x)=\frac{-x^7}{10(x-1)^2(x-2)^2(x-4)^2}\end{align*}
#### Example 5
Identify the vertical asymptotes and use the sign testing procedure to roughly sketch the nature of the function near the vertical asymptotes.
f(x)=(x2)3(x1)2(x+1)(x+3)x3(x+12)(x1)(x2)2\begin{align*}f(x)=\frac{(x-2)^3(x-1)^2(x+1)(x+3)}{x^3 \left(x+\frac{1}{2}\right)(x-1)(x-2)^2}\end{align*}
The vertical asymptotes occur at \begin{align*}x=0, -\frac{1}{2}\end{align*}. Therefore the \begin{align*}x\end{align*} values to sign test are -.001, 0.001, 3.999, 4.0001.
\begin{align*}f(-0.001) &= +\\ f(0.001) &= -\\ f(3.999) &= -\\ f(4.0001) &= +\end{align*}
### Review
Consider the function below for questions 1-4.
\begin{align*}f(x)=\frac{(x-2)^4(x+1)(x+3)}{x^3(x+3)(x-4)}\end{align*}
1. Identify the vertical asymptotes.
2. Will this function have an oblique asymptote? A horizontal asymptote? If so, where?
3. What values will you need to use the sign test with in order to help you make a sketch of the graph?
4. Use the sign test and sketch the graph near the vertical asymptotes.
Consider the function below for questions 5-8.
\begin{align*}g(x)=\frac{3(x-2)^2(x-1)^2(x+1)(x+3)}{15x^2(x+5)(x+1)(x-3)^2}\end{align*}
5. Identify the vertical asymptotes.
6. Will this function have an oblique asymptote? A horizontal asymptote? If so, where?
7. What values will you need to use the sign test with in order to help you make a sketch of the graph?
8. Use the sign test and sketch the graph near the vertical asymptote(s).
Consider the function below for questions 9-12.
\begin{align*}h(x)=\frac{9x^4-102x^3+349x^2-340x+100}{x^3-9x^2+24x-16}\end{align*}
9. Identify the vertical asymptotes.
10. Will this function have an oblique asymptote? A horizontal asymptote? If so, where?
11. What values will you need to use the sign test with in order to help you make a sketch of the graph?
12. Use the sign test and sketch the graph near the vertical asymptotes.
Consider the function below for questions 13-16.
\begin{align*}k(x)=\frac{2x^3-5x^2-11x-4}{3x^3+11x^2+5x-3}\end{align*}
13. Identify the vertical asymptotes.
14. Will this function have an oblique asymptote? A horizontal asymptote? If so, where?
15. What values will you need to use the sign test with in order to help you make a sketch of the graph?
16. Use the sign test and sketch the graph near the vertical asymptotes.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
End behavior End behavior is a description of the trend of a function as input values become very large or very small, represented as the 'ends' of a graphed function.
Horizontal Asymptote A horizontal asymptote is a horizontal line that indicates where a function flattens out as the independent variable gets very large or very small. A function may touch or pass through a horizontal asymptote.
Oblique Asymptote An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division.
Oblique Asymptotes An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division.
sign test The sign test is a procedure for determining only whether a function is above or below the $x$-axis at a particular $x$ value. The specific height is not calculated.
Vertical Asymptote A vertical asymptote is a vertical line marking a specific value toward which the graph of a function may approach, but will never reach.
|
# UVa 10885
## Summary
Find a set of 13 points on a 2-dimensional Cartesian plane, such that:
• the points have integer coordinates
• the distance between any two points is an integer
• no three points lie on the same straight line.
## Explanation
We can weaken the first two conditions. Suppose, that we have found a set of 13 points with rational coordinates and rational distances between every two of them. Let L be the LCM of denominators of all coordinates and distances. Then, by multiplying coordinates of every point by L, we'll get a set of points with integer coordiantes and integer distance between every two of them.
Now, let's impose an additional constraint on the set of points. We'll be looking for a set of points, all of which lie on the same circle. It's easy to see, that every three points from this set will not lie on the same line, because a straight line intersects a circle in at most two points.
As a simplification, we may further assume that the points lie on a circle of unit radius, whose center is at the origin. Then the coordinates of the points may be written in the following form:
${\displaystyle x_{i}=\cos \alpha _{i}}$, ${\displaystyle y_{i}=\sin \alpha _{i}}$, ${\displaystyle {\mbox{for}}i=1,\ldots ,13}$
The distance between i-th and j-th points is: ${\displaystyle {\sqrt {(\cos \alpha _{i}-\cos \alpha _{j})^{2}+(\sin \alpha _{i}-\sin \alpha _{j})^{2}}}={\sqrt {2-2\cos(\alpha _{i}-\alpha _{j})}}=2{\sqrt {\frac {1-\cos(\alpha _{i}-\alpha _{j})}{2}}}=2\sin \left({\frac {\alpha _{i}-\alpha _{j}}{2}}\right)}$
If we add the constraint, that every ${\displaystyle \sin {\frac {\alpha _{i}}{2}}}$ and ${\displaystyle \cos {\frac {\alpha _{i}}{2}}}$ are rational, then the points will have rational coordiantes, and distance between any two of them will be rational, too.
Let ${\displaystyle \sin \left({\frac {\alpha _{i}}{2}}\right)={\frac {p_{i}}{q_{i}}}}$, for some integers ${\displaystyle p_{i}}$ and ${\displaystyle q_{i}}$.
Then ${\displaystyle \cos \left({\frac {\alpha _{i}}{2}}\right)={\frac {\sqrt {q_{i}^{2}-p_{i}^{2}}}{q_{i}}}}$. This value will be rational if ${\displaystyle q_{i}^{2}-p_{i}^{2}=r_{i}^{2}}$, for some integer ${\displaystyle r_{i}}$.
So, what we are now really looking for if is just a set of pythagorean triples ${\displaystyle (r_{i},p_{i},q_{i})}$, that is integers, which satisfy the equation: ${\displaystyle r_{i}^{2}+p_{i}^{2}=q_{i}^{2}}$.
Having found a few of them (by e.g. brute force enumeration), you can express the coordinates of points:
${\displaystyle x_{i}={\frac {r_{i}^{2}-p_{i}^{2}}{q_{i}^{2}}}}$, ${\displaystyle y_{i}={\frac {2p_{i}r_{i}}{q_{i}^{2}}}}$.
## Output
0 180
0 3876
432 0
432 4056
1615 0
1615 4056
2047 180
2047 3876
2511 3528
2511 528
2880 1020
2880 3036
3136 2028
|
# Difference between revisions of "2009 AMC 12B Problems/Problem 21"
## Problem
Ten women sit in $10$ seats in a line. All of the $10$ get up and then reseat themselves using all $10$ seats, each sitting in the seat she was in before or a seat next to the one she occupied before. In how many ways can the women be reseated?
$\textbf{(A)}\ 89\qquad \textbf{(B)}\ 90\qquad \textbf{(C)}\ 120\qquad \textbf{(D)}\ 2^{10}\qquad \textbf{(E)}\ 2^2 3^8$
## Solution 1
Notice that either a woman stays in her own seat after the rearrangement, or two adjacent women swap places. Thus, our answer is counting the number of ways to arrange 1x1 and 2x1 blocks to form a 1x10 rectangle. This can be done via casework depending on the number of 2x1 blocks. The cases of 0, 1, 2, 3, 4, 5 2x1 blocks correspond to 10, 8, 6, 4, 2, 0 1x1 blocks, and so the sum of the cases is $$\binom{10}{0} + \binom{9}{1} + \binom{8}{2} + \binom{7}{3} + \binom{6}{4} + \binom{5}{5} = 1 + 9 + 28 + 35 + 15 + 1 = \boxed{89}.$$
## Solution 2
Let $S_n$ be the number of possible seating arrangements with $n$ women. Consider $n \ge 3,$ and focus on the rightmost woman. If she returns back to her seat, then there are $S_{n-1}$ ways to seat the remaining $n-1$ women. If she sits in the second to last seat, then the woman who previously sat there must now sit at the rightmost seat. This gives us $S_{n-2}$ ways to seat the other $n-2$ women, so we obtain the recursion $$S_n = S_{n-1}+S_{n-2}.$$
Starting with $S_1=1$ and $S_2=2,$ we can calculate $S_{10}=\boxed{89}.$
## Clarification of Solution 2
The seating possibilities of woman #$10$ become the two cases which we work out. $S_n$ was defined to be the number of different seating arrangements with $n$ women.
In the first "case" if woman #$10$ sits in the seat #$10$, this leads to a similar scenario, but with $9$ women instead. That means that for this case, there are a total of $S_{9}$ possible arrangements. We don't know how many exactly, but we are able to define it in terms of $S$.
During the second "case", woman #$10$ sits in seat #$9$. This time, woman #9 must go to seat #$10$, as she is the only other person who can go there. This leaves us with $8$ women, and we again represent this in terms of $S \Rightarrow S_8$.
Therefore, we can write $S_{10}$ in terms of $S_8$ and $S_9$, like so:
$$S_{10} = S_8 + S_9.$$
We can then generalize this to say
$$S_n = S_{n-1}+S_{n-2}.$$
Calculating $S_1 = 1$ and $S_2 =2,$ then following the recursive rule from above, we get $S_{10} = 89 \Rightarrow \boxed{\text{A}}$.
|
# Lesson video
In progress...
Hi am Mrs Dennett.
In this lesson we're going to be dividing a quantity in a ratio.
Tom and Mo sharing 300 pounds in the ratio, five to one how much do they each get? Let's draw a bar model to help us to answer this question using the information given.
Tom has five parts and Mo has one so Tom gets five boxes and Mo gets one box.
In total we're told that they shared 300 pounds.
So the sum of all the parts of the ratio, the sum of all the boxes must be 300 pounds.
We want to find out how much money they each gets.
We can see six equal boxes in total so we share the 300 pounds equally between the six boxes this gives us 50 pounds in each box.
Now we can work out how much Tom gets he has five boxes with 50 pound in each box so 50 times five is 250.
Mo has one box so Mo has 50 pounds and we state our answers Tom gets 250 pounds Mo gets 50 pounds.
At this point you could do a little check to make sure that 250 and 50 pounds makes the total that they were sharing 300 pounds and it does.
In this question a drink of juice is being made from cordial and water in the ratio two to seven.
Emma makes 360 millilitres of juice we have to work out how much water and cordial she uses.
Let's draw a bar model using the information that we've been given.
So we have cordial water in the ratio two to seven.
This means cordial has two boxes and water has seven boxes all of the boxes must be equal in size.
We know that Emma is making 360 millilitres of juice in total this means that the sum of all the boxes must be equal to 360 millilitres.
Let's work out how much cordial and how much water she has to use.
We can see that in total there are nine boxes so we have to share 360 millilitres equally between the nine boxes which gives us forty millilitres in each box.
We can now work out how much cordial and how much water Emma needs to use in her juice.
There are two boxes for cordial and each of them have got 40 millilitres in them.
So we do 40 millilitres times two which gives us 80 millilitres of cordial.
Now let's work out how much water she needs to use we can see that there are seven boxes for water and 40 millilitres in each box so this will give us 40 times seven which is 280 millilitres of water.
At this point we can do a quick check to make sure that 80 millilitres and 280 millilitres adds up to our 360 millilitres of juice in total and it does.
So Emma needs to use 80 millilitres of cordial and 280 millilitres of water to make her juice.
Here are some questions for you to try.
Pause the video to complete the task and restart when you were finished.
Did you notice that in the last question there are three parts to the ratio so you would need to draw three bars.
One bar with two boxes, one bar with five boxes and a final bar with one box.
Here is a question for to try.
Pause the video to complete the task and restart when you were finished.
You should have shared 300 millilitres equally between the six boxes to get 50 millilitres in each box.
From there, you can work out how much strawberry juice and how much Apple juice is needed.
Here is a question for you to try.
Pause the video complete the task and restart when you were finished.
When you draw your bar model you will see there are 20 boxes or 20 parts in total.
So we share the 100 squares equally between these boxes, giving us five squares in each box.
There are 45 red squares and 55 blue squares.
Then we're asked to find the difference which is 55 take away 45 which is 10 more blue squares.
Here are some questions for you to try.
Pause the videos to complete the questions and restart when you were finished.
Notice that in question four we're only asked for the number of dancers in the show which is forty nine.
For question five we draw up our model for the width and the length.
The perimeter of 32 includes two lengths and two widths.
We're only interested in one length and one width so we have 32 to get 16 this is the total of one length and one width.
This is divided by eight because there are eight boxes in total.
So 16 divided by eight would give us two centimetres in each box.
We can then find the width which is six centimetres and the length which is 10 centimetres.
Use these to find the area the area of a rectangle is given by a length times width 10 times six is 60 centimetres squared.
That was all for this lesson remember to take the exit quiz before you finish.
Thanks for watching.
|
# The combined resistance of two resistors R_(1 amd R_2 )in a parallel circuit is given by the formula R_(r= 1/(1/R_1 + 1/R_2 )) Simplify the formula. Help what is this talking about and how do I work it?
find R as a function fo r`1 and r2.
1/R= 1/r1 + `1/R2
R= ???
Get a common denominator on the right (r1r2), combine the fractions, then invert the left and right sides.
r= 1/(1/R_1 + 1/R_2 ) =
R_1 R_2/(R_1 + R_2)
## This formula is describing the combined resistance of two resistors, R1 and R2, when they are connected in parallel in an electrical circuit. Resistance is a measure of how easily electric current can flow through a material.
To simplify the formula and find R as a function of R1 and R2, you need to follow these steps:
2. To get rid of the fractions, find a common denominator for the terms on the right side. The common denominator is R1 * R2.
3. Multiply each fraction by the appropriate denominator to eliminate the fractions:
(1/R) * (R1 * R2/R1 * R2) = (1/R1) * (R1 * R2/R1 * R2) + (1/R2) * (R1 * R2/R1 * R2)
Simplifying:
R1 * R2 / (R1 * R2 * R) = (R2) / (R1 * R2) + (R1) / (R1 * R2)
4. Combine the fractions on the right side and simplify:
R1 * R2 / (R1 * R2 * R) = (R2 + R1) / (R1 * R2)
5. Multiply both sides of the equation by the common denominator (R1 * R2 * R) to eliminate the denominator on the left side:
R1 * R2 = (R2 + R1) * R
6. Now, divide both sides of the equation by (R2 + R1) to solve for R:
R = R1 * R2 / (R1 + R2)
Therefore, the simplified formula for the combined resistance R of two resistors R1 and R2 in a parallel circuit is:
R = R1 * R2 / (R1 + R2)
|
# Modulus Functions Equation
1. Sep 24, 2010
### Azure Skye
I have to solve the equation: |12x -9| = |3 -4x| +2x -1
I was wondering if it's supposed to be like:
12x -9 = 3 -4x +2x -1 and -12x +9 = 3 -4x +2x -1 ?
Or am I supposed to consider the positive and negative of 3 -4x also?
I didn't attend lessons for this and I know little about modulus functions. :/
2. Sep 24, 2010
### Mentallic
Actually what you do is you find what values of x make one of the absolute values more than zero, so lets do that now:
$$12x-9\geq 0$$
$$x\geq 3/4$$
Therefore for all $$x\geq 3/4$$ that inequality is positive.
For the other inequality, at x=3/4 we have 3-4x=0 and for any $$x\geq 3/4$$ we have $$3-4x\leq 0$$ So this means for all $$x\geq 3/4$$ we have the first inequality is more than zero, and the second is less than zero (so we take the negative of it). And things are just reversed when we consider when $$x<3/4$$.
3. Sep 24, 2010
### Azure Skye
So does the inequality formed from |12x -9| applies to |3 -4x|?
Then do I end up with something like 12x -9 = 0 +2x -1?
4. Sep 25, 2010
### Mentallic
Yes it does apply to the other inequality.
If you have |a|=|a+1| then you need to look at one of the absolute values. For the first, when $a\geq 0$ then you leave that as a, when $a<0$ you change that to -a (since the negative of a negative number is positive). But for the other inequality, when $a+1\geq 0$ or in other words, $a\geq -1$ then that absolute value is positive and thus left as a+1, so when $a<-1$ it becomes -(a+1)=-a-1.
So this means when a<-1 we have both absolute values are negative so we solve -(a)=-(a+1) and when a>0 we have both are positive, so we solve a=a+1. What about for $-1< a< 0$? Well we have that the first is negative, and the second is positive so we solve -a=a+1
The same idea applies to your problem, so no, you don't solve 12x-9=0+2x-1. Use the idea of cases when x>3/4 and x<3/4 (this problem is easier since you only need two cases). Also, you can add in the value of x=3/4 later or just include it into each inequality, it doesn't really matter.
5. Sep 25, 2010
### Azure Skye
Does this means that because |12x -9| is x > 3/4 and |3 -4x| is x< 3/4, when 12x -9 is positive, 3 -4x is negative and vice versa?
So when I consider x > 3/4, I get
12x -9 = 4x -3 +2x +1
x = 7/6
Is that correct or wrong?
|
## What your child will learn and do in Grade 1 Mathematics
In grade one, students work with whole numbers and place value. This includes grouping numbers into tens and ones as they learn to add and subtract up through 20. Activities in these areas include:
• Quickly and accurately adding numbers together that total up to 10 or fewer, and subtracting from numbers up through 10
• Adding a two-digit number and a one-digit number (45 + 9)
• Adding a two-digit number and a multiple of ten (54 + 20)
• Subtracting a multiple of 10 from a multiple of 10 (80 – 30)
• Comparing two-digit numbers based on place value using the symbols > (more than), equal to (=), and < (less than)
• Understanding the meaning of the equal sign (=) and determining if statements involving addition and subtraction are true or false (for example, which of the following statements are true? 3+3=6, 4+1=5+2)
• Measuring the lengths of objects using a smaller object (for example, measuring a pencil with paper clips)
• Putting objects in order from longest to shortest or shortest to longest
• Collecting and organizing data and answering questions about more, less, or how many
• Telling time and writing time to the hour and half-hour
• Understanding defining attributes of shapes (number of sides is defining; color is non-defining)
• Dividing circles and rectangles into halves and quarters
Helping your child learn outside of school:
• Play math games with your child to build fluency. For example, using a deck of cards, deal two cards and ask your child to add the two numbers before you do. Whoever says the total first, keeps the cards.
• Encourage your child to read and write numbers in different ways. For example, what are some ways that you can make the number 15? 15 can be 10+5, 7+8, 20-5, or 5+5+5.
• Have your child create story problems to represent addition, subtraction, and comparisons. For example, if you open a carton of eggs and take out seven, ask, “How many are left in the carton?”
• Have your child tell the time on the clock when you sit down to dinner or breakfast.
• Encourage your child to stick with it whenever a problem seems difficult.
• Can you do some easier problems and go back to this one after?
• What part of the problem is giving you trouble?
• Let's read the problem together and make sure we understand what it is asking.
• Can we draw a picture of the problem?
• Can we make up an easier problem that is similar to this? Then we can work our way up to this one.
• Let’s take a 10 minute break and come back to this one.
Websites
|
# diane draws abtuse, isoscules triangle with one of the angle mesuring 35. what i sthe messer of teh obtuse triangle
All angle = (110°, 35°, 35°)
Step-by-step explanation:
Given:
Triangle is a Obtuse isosceles triangle
One angle = 35°
Find:
All angle
Computation:
In the Obtuse isosceles triangle, one angle is obtuse and the other two angles are acute so, two equal angles are 35°
So,
Sum of angle property
x + 35° + 35° = 180°
x = 110°
Obtuse angle = 110°
All angle = (110°, 35°, 35°)
## Related Questions
In the figure, side AB is given by the expression (5x + 5)/(x + 3), and side BC is (3x + 9)/(2x - 4).The simplified expression for the area of rectangle ABCD is _______, and the restriction on x is ____.
The simplified expression for the area of rectangle ABCD is , and the restriction on x is x≠2 .
Step-by-step explanation:
Side AB = Width of rectangle = (5x + 5)/(x + 3)
Side BC = Length of rectangle = (3x + 9)/(2x - 4)
Area of Rectangle = Length * Width
Putting values:
Solving,
The restriction on x is that x ≠ 2, because if x =2 then denominator will be zero.
The simplified expression for the area of rectangle ABCD is , and the restriction on x is x≠2 .
99% confidence interval for the population standard deviation
99% = 2.58
how to find ?
Divide your confidence level by 2: . 95/2 = 0.475. Step 2: Look up the value you calculated in Step 1 in the z-table and find the corresponding z-value. The z-value that has an area of .
Reuben attached a wire between two poles on a hill as shown which is the closest to x the distance between the two poles
75 ft
Step-by-step explanation:
Here we are given a right angled triangle with a known angle of 20°, length of the hypotenuse to be 80 and we are to find the length of the base x.
For that, we can use the formula for cosine for which we need an angle and the lengths of base and hypotenuse.
So putting in the given values to get:
Therefore, the value of x is the closest to 75 ft.
Calculate two iterations of Newton's Method to approximate a zero of the function using the given initial guess. (Round your answers to three decimal places.) f(x) = x9 − 9, x1 = 1.6
Iteration 1:
Iteration 2:
Step-by-step explanation:
Formula for Newton's method is,
Given the initial guess as , therefore value of n = 1.
Also, .
Differentiating with respect to x,
Applying difference rule of derivative,
Applying power rule and constant rule of derivative,
Substituting the value,
Calculating the value of and
Calculating
Calculating ,
Substituting the value,
Therefore value after second iteration is
Now use as the next value to calculate second iteration. Here n = 2
Therefore,
Calculating the value of and
Calculating
Calculating ,
Substituting the value,
Therefore value after second iteration is
To calculate two iterations of Newton's Method, use the formula xn+1 = xn - f(xn)/f'(xn). Given an initial guess of x1 = 1.6 and the function f(x) = x9 - 9, calculate f(xn) and f'(xn) at x1 and then use the formula to find x2 and x3.
### Explanation:
To calculate two iterations of Newton's Method, we need to use the formula:
xn+1 = xn - f(xn)/f'(xn)
Given an initial guess of x1 = 1.6 and the function f(x) = x9 - 9, we can proceed as follows:
1. Calculate f(xn) at x1: f(1.6) = (1.6)9 - 9 = 38.5432
2. Calculate f'(xn) at x1: f'(1.6) = 9(1.6)8 = 368.64
3. Calculate x2: x2 = 1.6 - f(1.6)/f'(1.6) = 1.6 - 38.5432/368.64 = 1.494
4. Repeat the process to find x3 using the updated x2 as the initial guess.
brainly.com/question/31910767
#SPJ3
Which of the following is equivalent to 2 2/5
|
# function
## What is Composition of Functions ?
Here you will learn what is composition of functions with properties and examples. Let’s begin – What is Composition of Functions ? Let f : A $$\rightarrow$$ B & g : f : B $$\rightarrow$$ C be two functions. Then the function gof : f : A $$\rightarrow$$ C defined by (gof)(x) = g(f(x)) $$\forall$$ …
## Many One Function – Definition and Examples
Here you will learn what is many one function with definition and examples. Let’s begin – Many One Function Definition : A function f : A $$\rightarrow$$ B is said to be a many-one function if two or more elements of set A have the same image in B. Thus, f : A $$\rightarrow$$ B …
## One One Function (Injection) – Definition and Examples
Here you will learn what is one one function with definition and examples. Let’s begin – One One Function Definition Definition : A function f : A $$\rightarrow$$ B is said to be a one-one function or an injection if different elements of A have different images in B. Thus, f : A $$\rightarrow$$ B …
## Types of Functions in Maths – Domain and Range
Here you will learn types of functions in maths i.e polynomial function, logarithmic function etc and their domain and range. Let’s begin – Types of Functions in Maths (a) Polynomial function If a function is defined by f(x) = $$a_0x^n$$ + $$a_1x^{n-1}$$ + $$a_2x^{n-2}$$ + ….. + $$a_{n-1}x$$ + $$a_n$$ where n is a non …
## What is Inverse of a Function – Properties and Example
Here, you will learn what is inverse of a function, its properties and how to find the inverse of a function. Let’s begin – Inverse of a Function Let f : A $$\rightarrow$$ B be a one-one & onto function, then there exists a unique function g : B $$\rightarrow$$ A such that f(x) = …
## What is a Periodic Function – Definition and Example
Here, you will learn what is a periodic function with definition and example. Let’s begin – Periodic Function A function f(x) is called periodic if there exist a positive number T (T > 0), where T is the smallest such value called the period of the function such that f(x + T) = f(x), for …
## One One and Onto Function (Bijection) – Definition and Examples
Here, you will learn one one and onto function (bijection) with definition and examples. Let’s begin – What is Bijection Function (One-One Onto Function) ? Definition : A function f : A $$\rightarrow$$ B is a bijection if it is one-one as well as onto. In other words, a function f : A $$\rightarrow$$ B …
## Domain and Range of Greatest Integer Function
Here, you will learn domain and range of greatest integer function and properties of greatest integer function with example. Let’s begin – Greatest Integer Function or Floor Function For any real number x, we use the symbol [x] or $$\lfloor x \rfloor$$ to denote the greatest integer less than or equal to x. For example, …
## What is the Domain and Range of Modulus Function
Here, you will learn modulus function and what is the domain and range of modulus function. Let’s begin – The function f(x) defined by y = |x| = $$\begin{cases} x & \text{if}\ x \ge 0 \\ -x & \text{if}\ x < 0 \end{cases}$$ is called the modulus function. It is also called absolute value function. …
## How to Determine Odd Even Function
Here, you will learn what is odd even functions and how to determine if given function is odd or even with example. Let’s begin – Odd Even Function : Let a function f(x) such that both x and -x are in its domain then If f(-x) = f(x) then f is said to be an …
|
# What is the standard form of the equation of the parabola with a directrix at x=-9 and a focus at (8,4)?
Nov 1, 2017
The equation of the parabola is ${\left(y - 4\right)}^{2} = 17 \left(2 x + 1\right)$
#### Explanation:
Any point $\left(x , y\right)$ on the parabola is equidistant from the directrix and the focus.
Therefore,
$x - \left(- 9\right) = \sqrt{{\left(x - \left(8\right)\right)}^{2} + {\left(y - \left(4\right)\right)}^{2}}$
$x + 9 = \sqrt{{\left(x - 8\right)}^{2} + {\left(y - 4\right)}^{2}}$
Squaring and developing the ${\left(x - 8\right)}^{2}$ term and the LHS
${\left(x + 9\right)}^{2} = {\left(x - 8\right)}^{2} + {\left(y - 4\right)}^{2}$
${x}^{2} + 18 x + 81 = {x}^{2} - 16 x + 64 + {\left(y - 4\right)}^{2}$
${\left(y - 4\right)}^{4} = 34 x + 17 = 17 \left(2 x + 1\right)$
The equation of the parabola is ${\left(y - 4\right)}^{2} = 17 \left(2 x + 1\right)$
graph{((y-4)^2-34x-17)((x-8)^2+(y-4)^2-0.05)(y-1000(x+9))=0 [-17.68, 4.83, -9.325, 1.925]}
|
# What is the derivative of sin(3x)?
Jun 14, 2016
$3 \cos \left(3 x\right)$
#### Explanation:
The chain rule is a tool for differentiating composite functions, that is, a function inside a function.
Here, we have $\sin \left(3 x\right)$. This can be thought of as the function $3 x$ being put inside of the function $\sin \left(x\right)$.
When finding the derivative of such a function, the chain rule tells us that the derivative will be equal to the derivative of the outside function with the original inside function still inside of it, all multiplied by the derivative of the inside function.
So, for $\sin \left(3 x\right)$, the derivative the $\sin \left(x\right)$, the outside function, is $\cos \left(x\right)$.
So, the first part of the chain rule, the differentiated outside function with the inside function unchanged, gives us $\cos \left(3 x\right)$. Then, this is multiplied by the derivative of the inside function. The derivative of $3 x$ is $3$, so the overall derivative is $\cos \left(3 x\right) \times 3 = 3 \cos \left(3 x\right)$.
We can generalize this to all derivatives of sine functions:
$\frac{d}{\mathrm{dx}} \sin \left(f \left(x\right)\right) = \cos \left(f \left(x\right)\right) \cdot {f}^{'} \left(x\right)$
|
# Summing an arithmetic sequence
Say we were tasked to find the sum of the series from 51 through 375 of a series whose values are all separated by a common difference of six. That's a difference of 324 so it would take 324 hops if the hops were each one in size, but the hops are six times bigger than that so it takes six times fewer hops. 324/6 = (300+24)/6 = 50 + 4 = 54.
So note that there are 55 numbers in the series! Only one of the initial and last numbers corresponds to any of those hops (take your pick, it all depends on your perspective which one). In any case, there is one more stop than hop, considering that you are already at a stop before the first hop and you'll arrive at a stop after the last hop.
Option A:
We could subtract the first number from every number in the series, and end up with 55 of those subtracted terms.
So we could compute our sum as 55*51 + 6,12,18...318, 324
The first part is as easy as (here's an easy way to multiply any number by 51, write 51 as (100/2 + 1)): e.g. (55*(50+1)) = 55*50 + 55*1 = 55*100/2 + 55 = (55/2) * 100 + 55 = 2750 + 55 = 2805.
The second part could be thought of as 6 times the first 54 integers (1 through 54). So in that series there are 27 pairs each totaling 55. So our answer becomes 55*51 + 6*27*55 or if we factor out 55,
Answer = 55*(51+6*27) = 55* (51+162) = 55*213 = 5*11*213 = (2130+213)*5 = 5*2343 = 11715.
Option B:
Say... you know what, there are 55 numbers in the series, so I'm just going to keep the first number (51) outside the series and only sum up the other 54 numbers from 57 through 375 using the series formula. That's an even count (54) so comprises 27 pairs that each total 57+375 = 432.
Therefore the answer is
(this shows an easy way to multiply by 25 is to multiply by 100 and divide by four --- it doesn't matter in which order, whichever is easier for you)
51 + 27*432. = 51+ 25*432 + 2*432 = 51 + 432*100/4 + 864 = 915 + 108*100 = 10800 + 915 = 11715
\$48p/h
Isaak B.
Good (H.S. or College Math, Physics, Chem, EE Engineering) Cheap Tutor
3750+ hours
if (isMyPost) { }
|
# Difference between revisions of "2008 AMC 10B Problems/Problem 7"
## Problem
An equilateral triangle of side length $10$ is completely filled in by non-overlapping equilateral triangles of side length $1$. How many small triangles are required?
$\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 25\qquad\mathrm{(C)}\ 100\qquad\mathrm{(D)}\ 250\qquad\mathrm{(E)}\ 1000$
## Solution 1
The area of the large triangle is $\frac{10^2\sqrt3}{4}$, while the area of each small triangle is $\frac{1^2\sqrt3}{4}$. Dividing these two quantities results in $100$, therefore $\boxed{100} \mathrm{(C)}$ small triangles can fill the large one without overlap.
## Solution 2-4
$[asy] unitsize(0.5cm); defaultpen(0.8); for (int i=0; i<10; ++i) { draw( (i*dir(60)) -- ( (10,0) + (i*dir(120)) ) ); } for (int i=0; i<10; ++i) { draw( (i*dir(0)) -- ( 10*dir(60) + (i*dir(-60)) ) ); } for (int i=0; i<10; ++i) { draw( ((10-i)*dir(60)) -- ((10-i)*dir(0)) ); } [/asy]$
The number of triangles is $1+3+\dots+19 = \boxed{100}$.
Also, another way to do it is to notice that as you go row by row (from the bottom), the number of triangles decrease by 2 from 19, so we have: $19+17+15...+3+1 = \frac{19+1}{2}\cdot 10 = \boxed{100}$
A fourth solution is to notice that the small triangles are similar to the large triangle as they are both equilateral. Therefore, the ratio of their areas is the square of the ratio of their side lengths. Hence the ratio of their areas is $(1/10)^2=1/100$, so the answer is $\boxed{100}$.
## Solution 5
The side length of the large equilateral triangle is $10$. The number of equilateral triangles with side length $1$ that can fill it is $10^2$. This is equal to the sum of the first $10$ odd integers. The sum of the first $n$ odd integers is $n^2$. In general, an equilateral triangle with side length $n$ can be filled by $n^2$ triangles. The answer is $100\ \mathrm{(C)}$.
~mobius247
|
### Basics of Ohm’s Law
In 1827 the German physicist George Ohm published a pamphlet entitled “The Galvanic Circuit Investigated Mathematically”. It contained one of the first efforts to measure currents and voltages and to describe and relate them mathematically. One result was a statement of the fundamental relationship we now call Ohm’s Law.
Consider a uniform cylinder of conducting material, to which a voltage has been connected. The voltage will cause charge to flow, i.e. a current:
Ohm found that in many conducting materials, such as metal, the current is always proportional to the voltage. Since voltage and current are directly proportional, there exists a proportionality constant R, called resistance, such that:
This is Ohm’s Law. The unit of resistance (volts per ampere) is referred to as the ohm, and is denoted by the capital Greek letter omega, Ω.
We refer to a construction in which Ohm’s Law is obeyed as a resistor.
The ideal resistor relationship is a straight line through the origin:
Even though resistance is defined as R v/i , it should be noted that R is a purely geometric property, and depends only on the conductor shape and the material used in the construction. For example, it can be shown for a uniform resistor that the resistance is given by:
where l is the length of the resistor, and A is the cross-sectional area. The resistivity, r , is a constant of the conducting material used to make the resistor.
The circuit symbol for the resistor is shown below, together with the direction of current and polarity of voltage that make Ohm’s Law algebraically correct:
Example
Consider the circuit shown below.
The voltage across the 1 kΩ resistor is, by definition of an ideal voltage source,
v(t ) = 10 V . Thus, by Ohm’s Law, we get:
i1 v/R = 10/1000 = 0.01 A = 10 mA
i2 –v/R = -10/1000 = -0.01 A = -10 mA
Note that i2 = –i1 , as expected
### You May Also Like :
#### Capacitor Filter Operation
Capacitor filter. Fig. shows a typical capacitor filter circuit. It consists of a capacitor C placed across the rectifier output in parallel with load RL. The ...
|
Home Education How to Solve Various Types of Linear Equation?
# How to Solve Various Types of Linear Equation?
Algebra is an important part of mathematics. It is easy to learn by constant and regular practice. It is defined as a broad area of mathematics. It includes geometry, number theory, and analysis. It is the study of mathematical formulas. You need the constant guidance of teachers to solve algebraic equations. You need a reasoning power to understand various formulas of algebra. A linear equation is a part of algebra that is not difficult once you learn the basics. It is called a subset of equations. To solve a linear equation you require a pen and paper.
An equation having more than one variable can be solved by various methods of linear equation. In this equation X stands for a variable and the terms a, b stand for constants.
The linear equation is also known as a mathematical statement, having the symbol “=”.The linear equation has a degree of 1. It is called a linear combination of one or more than one term. It is written with two or one variable where a constant is present. After solving a linear equation if the values are plotted on the graph it gives a straight line.
There are many methods to solve a linear equation. The most common method to solve a linear equation is to take variables on one side, the left-hand side, and the numeric part on the right-hand side. An example of such kind of equation is x – 1 = 5 – 2x.
One step linear equation is represented by such type of equation- X+ 1 = 4. In such equations, the student should isolate the x term by subtracting one from both sides. After subtracting one from both the side we get the following equation X+ 1-1=4-1, and lastly we get x is equal to three.
In the case of solving two steps linear equation, we take the constant term on one side and x term on another side. The example for two-step linear equation is 3x + 4 = 10. In such equation we subtract 4 from both the side, 3x+ 4 – 4 = 10 – 4 and we get 3x + 0 = 6. In the next step we take the numerical part on the right-hand side, 3x = 6 – 0. We get finally 3 x is equal to zero. Now the numerical part is divided by 6 and we get finally x is equal to 2.
Cuemath is an online website that teaches students various mathematical formulas. It is a class that gives one-to-one riders to the students. It helps in developing the mind of the students so that they cultivate an attitude of problem-solving practice.
A linear equation with two negative numbers is difficult to solve. Before solving such equations we should keep in mind the following tips.
1. Negative X negative is equal to positive
2 positive X positive is equal to positive.
3. Positive X negative is equal to negative
Linear equations having multiple terms and variables require constant and regular guidance from a teacher. In this equation, we need to combine the like terms in the same way as we combine the constants. This is done by adding or subtracting the coefficient and keeping the same variable.
Another system of the linear equation having two equations with two variables is solved by reducing it to a single equation with a single variable. Solving an equation is not difficult if you know the basics. In the equations having only fractions, we use the least common denominator to solve the fractions. In this type, we multiply both sides of an equation by the least common denominator. We should keep in mind that if the variable in the denominator is a fraction then we should identify the variable values. This is solved the division by zero.
The next step involves a simplification of both the sides, clearing of the parentheses, and a combination of the like terms and then we divide both the sides of the equation by a coefficient. An important thing to be kept in mind while solving the variable equation is to verify your answer.
0 comment
|
# How do you solve 3(5^(x-1))=21?
$x = \setminus \frac{\setminus \ln 35}{\setminus \ln 5}$
#### Explanation:
Given that
$3 \left({5}^{x - 1}\right) = 21$
${5}^{x - 1} = \frac{21}{3}$
${5}^{x - 1} = 7$
Taking logs on both the sides as follows
$\setminus \ln {5}^{x - 1} = \setminus \ln 7$
$\left(x - 1\right) \setminus \ln 5 = \setminus \ln 7$
$x - 1 = \setminus \frac{\setminus \ln 7}{\setminus \ln 5}$
$x = \setminus \frac{\setminus \ln 7}{\setminus \ln 5} + 1$
$x = \setminus \frac{\setminus \ln 7 + \setminus \ln 5}{\setminus \ln 5}$
$x = \setminus \frac{\setminus \ln 35}{\setminus \ln 5}$
Jul 27, 2018
$x = 2.21$ (2.d.p)
#### Explanation:
$3 \left({5}^{x - 1}\right) = 21$
${5}^{x - 1} = 7$
${\log}_{10} {5}^{x - 1} = {\log}_{10} 7$
$\left(x - 1\right) {\log}_{10} 5 = {\log}_{10} 7$
$x - 1 = {\log}_{10} \frac{7}{\log} _ 10 5$
$x = 1 + {\log}_{10} \frac{7}{\log} _ 10 5$
$x = 2.21$ (2.d.p)
|
# Evaluate $\log_{5}{7^{\displaystyle -3\log_{7}{5}}}$
The logarithm of seven raised to the power of negative three times logarithm of five to base seven, to the base five is a logarithmic expression in arithmetic form, given in the math question. Now, let’s learn how to find the value of arithmetic logarithm expression by simplification.
### Minimize the Complexity of the function
The mathematical expression is formed by the combination of both logarithmic and exponential systems. It creates confusion in us while evaluating the expression and this type of expressions should be evaluated by reducing the complexity of the expression. Now, let us think about it logically.
The base number of the exponential expression inside the logarithmic expression is $7$ and the same base number is also base of logarithmic expression at exponent position. It creates an opportunity to eliminate the logarithmic expression from the exponent position.
$\implies$ $\log_{5}{7^{\displaystyle -3\log_{7}{5}}}$ $\,=\,$ $\log_{5}{7^{\displaystyle -3 \times \log_{7}{5}}}$
The factor $-3$ can be shifted to the exponent position of the number $5$ as per the power rule of logarithms.
$=\,\,$ $\log_{5}{7^{\displaystyle \log_{7}{5^{\displaystyle -3}}}}$
### Eliminate Log expression from whole expression
According to the fundamental rule of logarithms, the number $7$ raised to the power of logarithm of quantity to base $7$ is equal to the quantity. In this case, the quantity is the number $5$ is raised to the power of $-3$.
$=\,\,$ $\log_{5}{5^{\displaystyle -3}}$
### Find the Log expression by simplification
Now, use the power rule of logarithms one more time to shift the exponent $-3$ as a factor in the expression.
$=\,\,$ $(-3) \times \log_{5}{5}$
The logarithm of $5$ to the base $5$ is equal to one as per the logarithm of base rule.
$=\,\,$ $(-3) \times 1$
Finally, multiply the factors $-3$ and $1$ to find the product and also to evaluate the given logarithmic expression.
$=\,\,$ $(-3)$
$=\,\,$ $-3$
Latest Math Topics
Jun 26, 2023
###### Math Questions
The math problems with solutions to learn how to solve a problem.
Learn solutions
###### Math Worksheets
The math worksheets with answers for your practice with examples.
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.
|
We think you are located in United States. Is this correct?
# End of chapter exercises
## End of chapter exercises
Textbook Exercise 8.10
Calculate $$SV$$
$\begin{array}{rll} V\hat{T}U & = W\hat{V}T & \quad \text{(alt. } \angle \text{s}, WV \parallel TU) \\ \therefore TU & = VU = 35 & \quad \text{(isosceles } \triangle \text{)} \\ \frac{SW}{WT} & = \frac{SV}{VU} & \quad \text{(proportion Theorem)} \\ \therefore SV & = \frac{SW.VU}{WT} & \\ & = \frac{(10)(35)}{20} & \\ & = \text{17,5}\text{ units} & \\ \end{array}$
$$\frac{CB}{YB}=\frac{3}{2}$$. Find $$\frac{DS}{SZ}$$.
$\begin{array}{rll} DABC & \text{ is a parallelogram } & (DA \parallel CB \text{ and } DC \parallel AB)\\ DS &= SB & (\text{diagonals bisect})\\ & & \\ \dfrac{SZ}{ZB} &= \dfrac{CY}{YB} = \frac{3}{2} & (CS \parallel YZ)\\ & & \\ \dfrac{SZ}{SB} &= \dfrac{CY}{CB} = \frac{3}{5} & (CS \parallel YZ)\\ & & \\ \therefore SZ &= \frac{3}{5} SB & \\ & & \\ \therefore \dfrac{DS}{SZ} &= \dfrac{DS}{\frac{3}{5} SB} & (DS = SB) \\ & & \\ &= \frac{5}{3} & \end{array}$
Using the following figure and lengths, find $$IJ$$ and $$KJ$$ (correct to one decimal place).
$$HI= \text{20}\text{ m},KL= \text{14}\text{ m}, JL=\text{18}\text{ m}$$ and $$HJ=\text{32}\text{ m}$$.
$\begin{array}{rll} \dfrac{IJ}{LJ} & = \dfrac{HI}{KL} & \quad \text{(proportion Theorem)}\\ & & \\ IJ & = \dfrac{HI}{KL}(LJ) & \\ & & \\ & = \dfrac{20}{14}(18) & \\ & & \\ & = \dfrac{180}{7} & \\ & & \\ & = \text{25,7}\text{ m} & \end{array}$
$\begin{array}{rll} KJ & = \dfrac{LJ}{IJ}(HJ) & \\ & & \\ & = \dfrac{18}{\text{25,7}}(32) & \\ & & \\ & = \text{22,4}\text{ m} & \\ \end{array}$
Find $$FH$$ in the following figure.
$\begin{array}{rll} G\hat{F}H & = \hat{D} & \text{(corresp. } \angle \text{s}, GF \parallel ED) \\ G\hat{F}E & = F\hat{E}D & \text{(alt. } \angle \text{s}, GF \parallel ED) \\ \therefore F\hat{E}D & = \hat{D} & \\ \therefore EF &= FD = \text{45}\text{ cm} & \\ \frac{HF}{FD } &= \frac{21}{42} & \\ &= \frac{1}{2} & \\ \therefore HF & = \frac{1}{2}(45) & \\ & = \text{23,5}\text{ cm} & \end{array}$
In $$\triangle GHI$$, $$GH\parallel LJ$$, $$GJ\parallel LK$$ and $$\frac{JK}{KI}=\frac{5}{3}$$. Determine $$\frac{HJ}{KI}$$.
$\begin{array}{rll} L\hat{I}J & = G\hat{I}H &\\ J\hat{L}I & = H\hat{G}I & \text{(corresp. }\angle \text{s}, HG \parallel JL) \\ \therefore \triangle LIJ & \enspace ||| \enspace \triangle GIH & \text{(Equiangular }\triangle \text{s)} \end{array}$ $\begin{array}{rll} \frac{HJ}{JI}& =\frac{GL}{LI} & \left(\triangle LIJ \enspace ||| \enspace \triangle GIH\right)\\ \text{and }\frac{GL}{LI}& =\frac{JK}{KI} & \left(\triangle LIK \enspace ||| \enspace \triangle GIJ\right)\\ & =\frac{5}{3} \\ \therefore \frac{HJ}{JI}& =\frac{5}{3} \end{array}$ \begin{align*} \frac{HJ}{KI} & = \frac{HJ}{JI} \times \frac{JI}{KI} \end{align*} \begin{align*} JI & = JK+KI \\ & = \frac{5}{3}KI+KI \\ & = \frac{8}{3}KI \\ \frac{JI}{KI} & = \frac{8}{3} \\ & \\ \frac{HJ}{KI} & = \frac{HJ}{JI} \times \frac{JI}{KI} \\ & = \frac{5}{3}\times \frac{8}{3} \\ & = \frac{40}{9} \end{align*}
$$BF=\text{25}\text{ m}$$, $$AB=\text{13}\text{ m}$$, $$AD=\text{9}\text{ m}$$, $$DF=\text{18}\text{ m}$$.
Calculate the lengths of $$BC$$, $$CF$$, $$CD$$, $$CE$$ and $$EF$$, and find the ratio $$\frac{DE}{AC}$$.
$\begin{array}{rll} \frac{BC}{BF} & = \frac{AD}{AF} = \frac{9}{27} = \frac{1}{3} & (CD \parallel BA) \\ \therefore BC & = \frac{1}{3} \times 25 & \\ &= \text{8,3}\text{ m} & \\ & & \\ CF &= BF - BC & \\ &= 25 - \text{8,3} & \\ &= \text{16,7}\text{ m} & \\ & & \\ \frac{CD}{AB} &= \frac{DF}{AF} & (CD \parallel BA) \\ CD &= \frac{DF}{AF} \times AB & \\ &= \frac{18}{27} \times 13 & \\ &= \text{8,7}\text{ m} & \\ & & \\ \frac{CE}{CF} &= \frac{AD}{AF} & (DE \parallel AC) \\ CE &= \frac{AD}{AF} \times CF & \\ &= \frac{9}{27} \times \text{16,7} & \\ &= \text{5,6}\text{ m} & \\ & & \\ EF &= BF - (BC + CE) & \\ &= 25 - (\text{8,3} - \text{5,6}) & \\ &= \text{11,1}\text{ m} & \end{array}$
In $$\triangle XYZ$$, $$X\hat{Y}Z =\text{90}°$$ and $$YT \perp XZ$$. If $$XY = \text{14}\text{ cm}$$ and $$XT = \text{4}\text{ cm}$$, determine $$XZ$$ and $$YZ$$ (correct to two decimal places).
Use the theorem of Pythagoras to determine $$YT$$:
$\begin{array}{rll} \text{In } \triangle XTY, \quad YT^{2} &= XY^{2} - XT^{2} & (\text{Pythagoras}) \\ &= 14^{2} - 4^{2} & \\ &= 196 - 16 & \\ \therefore YT &= \sqrt{180} & \\ &= \sqrt{36 \times 5} & \\ &= 6 \sqrt{5} ~\text{cm} & \end{array}$
Use proportionality to determine $$XZ$$ and $$YZ$$:
$\begin{array}{rll} X\hat{Y}Z &= \text{90}° & \text{(given)} \\ YT &\perp XZ & \text{(given)} \\ \therefore \enspace \triangle XYT \enspace & ||| \enspace \triangle YZT \enspace ||| \enspace \triangle XZY & (\text{right-angled } \triangle \text{s}) \\ & & \\ \therefore \dfrac{YT}{TZ} &= \dfrac{XT}{YT} & (\triangle YZT \enspace ||| \enspace \triangle XYT ) \\ & & \\ \therefore YT^{2} &= TZ \cdot XT & \\ & & \\ \left( 6 \sqrt{5} \right)^{2} &= TZ \cdot 4 & \\ & & \\ \therefore TZ &= 45 & \\ & & \\ \text{And } XZ &= XT + TZ & \\ &= 4 + 45 & \\ &= \text{49}\text{ cm} & \\ & & \\ \text{In } \triangle XYZ, \quad YZ^{2} &= XZ^{2} - XY^{2} & (\text{Pythagoras}) \\ &= 49^{2} - 14^{2} & \\ \therefore YZ &= \sqrt{2205} & \\ &= \text{46,96}\text{ cm} & \end{array}$
Given the following figure with the following lengths, find $$AE$$, $$EC$$ and $$BE$$.
$$BC=15$$ cm, $$AB=4$$ cm, $$CD=18$$ cm, and $$ED=9$$ cm.
$\begin{array}{rll} B\hat{A}E & = C\hat{D}E & \text{(alt. } \angle \text{s}, AB \parallel CD) \\ A\hat{B}E & = D\hat{C}E & \text{(alt. } \angle \text{s}, AB \parallel CD) \\ A\hat{E}B & = D\hat{E}C & \text{(vert. opp. } \angle \text{s)}\\ \therefore \triangle AEB & \enspace ||| \enspace \triangle DEC & \text{(AAA)}\\ \therefore \frac{AE}{DE} & = \frac{AB}{DC} = \frac{4}{18} = \frac{2}{9} & (\triangle AEB \enspace ||| \enspace \triangle DEC) \\ & & \\ AE & = \dfrac{2}{9} DE & \\ & & \\ & = \dfrac{2}{9} (9) & \\ & & \\ & = \text{2}\text{ cm} & \\ \end{array}$
$\begin{array}{rll} \frac{EC}{BC} & = \dfrac{ED}{AD} = \dfrac{9}{11} & (AB \parallel CD) \\ & & \\ EC & = \dfrac{ED}{AD}(BC) & \\ & & \\ & = \dfrac{9}{11}(15) & \\ & & \\ & = \text{12,3}\text{ cm} & \\ \end{array}$
$\begin{array}{rll} \frac{BE}{BC} & = \dfrac{AE}{AD} = \dfrac{2}{11} & (AB \parallel CD) \\ & & \\ BE & = \dfrac{AE}{AD}(BC) & \\ & & \\ & = \dfrac{2}{11}(15) & \\ & & \\ & = \text{2,7}\text{ cm} & \\ \end{array}$
$$NKLM$$ is a parallelogram with $$T$$ on $$KL$$.
$$NT$$ produced meets $$ML$$ produced at $$V$$. $$NT$$ intercepts $$MK$$ at $$X$$.
Prove that $$\dfrac{XT}{NX} = \dfrac{XK}{MX}$$.
In $$\triangle TXK$$ and $$\triangle NXM$$:
$\begin{array}{rll} X\hat{T}K&= X\hat{N}M & (\text{alt. } \angle \text{s, } NK \parallel MV) \\ N\hat{X}M&= T\hat{X}K & (\text{vert. opp. } \angle \text{s}) \\ \therefore \triangle TXK & \enspace ||| \enspace \triangle NXM & (\text{AAA}) \\ \therefore \dfrac{TX}{NX} &= \dfrac{XK}{XM} & \end{array}$
Prove $$\triangle VXM \enspace ||| \enspace \triangle NXK$$.
In $$\triangle VXM \text{ and } \triangle NXK$$:
$\begin{array}{rll} \hat{V}&= X\hat{N}K & (\text{alt. } \angle \text{s, } NK \parallel MV) \\ M\hat{X}V&= K\hat{X}N & (\text{vert. opp. } \angle \text{s}) \\ \therefore \triangle VXM& \enspace ||| \enspace \triangle NXK & (\text{AAA}) \end{array}$
If $$XT = \text{3}\text{ cm}$$ and $$TV = \text{4}\text{ cm}$$, calculate $$NX$$.
$\begin{array}{rll} \dfrac{VX}{NX} &= \dfrac{XM}{XK} & (\triangle VXM \enspace ||| \enspace \triangle NXK \text{, proved in (b)}) \\ \text{But } \dfrac{XM}{XK} &= \dfrac{NX}{TX} & (\text{proved in (a)}) \\ \therefore \dfrac{VX}{NX} &= \dfrac{NX}{TX} & \\ NX^{2} &= VX.TX & \\ NX^{2}&= 3 \times 4 & \\ NX &= \sqrt{12} & \\ &= 2 \sqrt{3} \text{ cm} & \end{array}$
$$MN$$ is a diameter of circle $$O$$. $$MN$$ is produced to $$R$$ so that $$MN =2NR$$.
$$RS$$ is a tangent to the circle and $$ER \perp MR$$. $$MS$$ produced meets $$RE$$ at $$E$$.
Prove that:
$$SNRE$$ is a cyclic quadrilateral
$\begin{array}{rll} M\hat{S}N&= \text{90}° & (\angle \text{ in semi-circle}) \\ N\hat{R}E&=\text{90}° & (\text{given}) \\ \therefore SNRE&\text{ is a cyclic quad. } & (\text{ext. }\angle = \text{ opp. int. } \angle) \end{array}$
$$RS = RE$$
$\begin{array}{rll} N\hat{S}R&=\hat{M} = x & (\text{tangent/chord}) \\ \therefore E\hat{S}R&= \text{90}° - x & \\ \hat{E}&= \text{90}° - x & (MRE = \text{90}°, \enspace \hat{M} = x) \\ \therefore E\hat{S}R&= \hat{E} & \\ \therefore RS&= RE & (\text{isos. } \triangle) \end{array}$
$$\triangle MSN \enspace ||| \enspace \triangle MRE$$
In $$\triangle MSN$$ and $$\triangle MRE$$:
$\begin{array}{rll} \hat{M} &=\hat{M} & \\ M\hat{S}N &= \text{90}° & (\angle \text{ in semi-circle}) \\ M\hat{R}E &= \text{90}° & (\text{given}) \\ \therefore M\hat{S}N &= M\hat{R}E & \\ \therefore \triangle MSN& \enspace ||| \enspace \triangle MRE & (\text{AAA}) \end{array}$
$$\triangle RSN \enspace ||| \enspace \triangle RMS$$
In $$\triangle RSN$$ and $$\triangle RMS$$:
$\begin{array}{rll} \hat{R}&= \hat{R} & (\text{common } \angle) \\ R\hat{S}N&= \hat{M} & (\text{tangent/chord}) \\ \therefore \triangle RSN & \enspace ||| \enspace \triangle RMS & (\text{AAA}) \end{array}$
$$RE^{2} = RN.RM$$
$\begin{array}{rll} \dfrac{RS}{RN} &= \dfrac{RM}{RS} & \\ RS^{2} &= RN.RM & \\ \text{But } RS &= RE & \\ RE^{2} &= RN.RM & \end{array}$
$$AC$$ is a diameter of circle $$ADC$$. $$DB \perp AC$$.
$$AC = d, AD = c, DC = a \text{ and } DB = h$$.
Prove that $$h = \dfrac{ac}{d}$$.
$\begin{array}{rll} \triangle ADB& \enspace ||| \enspace \triangle DCB \enspace ||| \enspace \triangle ACD & (A\hat{D}C = \text{90}°, DB \perp AC) \\ \therefore \dfrac{DB}{AD}&= \dfrac{CD}{AC} & \\ \therefore \dfrac{h}{c}&= \dfrac{a}{d} & \\ \therefore h&= \dfrac{ac}{d} & \end{array}$
Hence, deduce that $$\enspace \dfrac{1}{h^{2}}=\dfrac{1}{a^{2}}+\dfrac{1}{c^{2}}$$.
$\begin{array}{rll} h^{2}&= \dfrac{a^{2}c^{2}}{d^{2}} & \\ \text{But } d^{2}&= a^{2} + c^{2} & (\text{In } \triangle ADC, \hat{D} = \text{90}°, \text{ Pythagoras}) \\ \therefore h^{2}&= \dfrac{a^{2}c^{2}}{a^{2}+c^{2}} & \\ \therefore \dfrac{1}{h^{2}}&= \dfrac{a^{2}+c^{2}}{a^{2}c^{2}} & \\ \dfrac{1}{h^{2}}&= \dfrac{a^{2}}{a^{2}c^{2}} + \dfrac{c^{2}}{a^{2}c^{2}} & \\ \dfrac{1}{h^{2}}&= \dfrac{1}{c^{2}} + \dfrac{1}{a^{2}} & \end{array}$
$$RS$$ is a diameter of the circle with centre $$O$$. Chord $$ST$$ is produced to $$W$$. Chord $$SP$$ produced meets the tangent $$RW$$ at $$V$$. $$\hat{\text{R}}_{1} = \text{50}°$$.
[NCS, Paper 3, November 2011]
Calculate the size of $$\text{W}\hat{\text{R}}\text{S}$$.
$$W\hat{R}S = \text{90}°\qquad (\text{tangent } \perp \text{ radius})$$
Find $$\hat{\text{W}}$$.
$\begin{array}{rll} R\hat{S}T & = \text{50}° & \qquad(\text{tangent chord th.})\\ \hat{W} & = \text{40}° & \qquad(\angle \text{ sum } \triangle) \end{array}$
OR
$\begin{array}{rll} \hat{T}_{1} & = \text{90}° & \qquad(\angle \text{s in semi-circle})\\ \hat{W} + \hat{R}_{1} & = \hat{T}_{1} & \qquad(\text{ ext. } \angle \triangle)\\ \hat{W} & = \text{40}° & \end{array}$
Determine the size of $$\hat{\text{P}}_1$$.
$\begin{array}{rll} \hat{R}_{2} & = \text{40}° & \qquad(\text{tangent } \perp \text{ radius})\\ \hat{P}_{1} & = \text{40}° & \qquad(\angle \text{s in same seg.}) \end{array}$
Prove that $$\hat{\text{V}}_1 = \text{P}\hat{\text{T}}\text{S}$$.
$\begin{array}{rll} \hat{P}_{1} & = \hat{W} & \qquad(= \text{40}°)\\ \text{WVPT is}& \text{ a} \text{ cyclic quadrilateral} &(\text{ext. } \angle = \text{ int. opp.})\\ \hat{V}_{1} & = P\hat{T}S & \qquad (\text{ext. } \angle \text{ cyclic quad.}) \end{array}$
OR
$\begin{array}{rll} \hat{T}_{1} & =\text{90}° & \qquad(\angle \text{s in semi-circle})\\ P\hat{T}S & = \text{90}° + \hat{T}_{2} & \\ \hat{T}_{2} & = \hat{S}_{1} & \qquad(\angle \text{s in same seg.})\\ P\hat{T}S & = \text{90}° + \hat{S}_{1} & \qquad(\text{ext. } \angle \triangle) \\ \hat{V}_{1} & = P\hat{T}S & \\ \end{array}$
OR
$\begin{array}{rll} \hat{P}_{2} & = \text{140}° & \qquad(\angle \text{s on str. line})\\ \hat{W} + \hat{P}_{2} & = \text{180}° & \\ \text{WVPT} &\text{is a cyclic quadrilateral} & \qquad(\text{opp. } \angle \text{s suppl})\\ \hat{V}_{1} & = P\hat{T}S & \qquad(\text{ext } \angle \text{ cyclic quad}) \end{array}$
OR
$\begin{array}{rll} \hat{V}_{1} & = \hat{R}_{1} + \hat{R}_{2} + \hat{S}_{1} & \qquad(\text{ext. } \angle \triangle)\\ \hat{V}_{1} & = \text{90}° + \hat{S}_{1} & \\ P\hat{T}S & = \text{90}° + \hat{T}_{2} & \\ \text{but } \hat{T}_{2} & = \hat{S}_{1} & \qquad(\angle \text{s in same seg.})\\ \hat{V}_{1} & = P\hat{T}S & \end{array}$
OR
In $$\triangle PTS$$ and $$\triangle WVS$$
$\begin{array}{rll} \hat{P}_{1} & = \hat{W} & \qquad(\text{40}°) \\ \hat{S}_{2} & \text{ is common} &\\ \hat{V}_{1} & = P\hat{T}S & \qquad (\angle \text{ sum } \triangle) \end{array}$
$$ABCD$$ is a cyclic quadrilateral and $$BC = CD.$$
$$ECF$$ is a tangent to the circle at $$C$$. $$ABE$$ and $$ADF$$ are straight lines.
Prove:
$$AC \text{ bisects } E\hat{A}F$$
$\begin{array}{rll} BC &= CD & (\text{given}) \\ \therefore B\hat{A}C &= D\hat{A}C & (\angle \text{s on equal chords}) \end{array}$
$$BD \parallel EF$$
$\begin{array}{rll} D\hat{C}F&= \hat{A_{2}} & (\text{tangent/chord}) \\ B\hat{D}C&= \hat{A_{1}} & (\angle \text{s on same chord}) \\ \hat{A_{1}} &= \hat{A_{2}} & (\text{proved in (a)}) \\ \therefore B\hat{D}C &= D\hat{C}F & \\ \therefore BD & \parallel EF & (\text{alt. } \angle \text{s are equal}) \end{array}$
$$\triangle ADC \enspace ||| \enspace \triangle CBE$$
In $$\triangle ADC$$ and $$\triangle CBE$$:
$\begin{array}{rll} \hat{A_{2}} &= D\hat{B}C & (\angle \text{s on chord } CD) \\ &= B\hat{C}E & (\text{alt. } \angle \text{s, } BD \parallel EF) \\ A\hat{D}C &= E\hat{B}C & (\text{ext. } \angle \text{ of a cyclic quad}) \\ \therefore \triangle ADC & \enspace ||| \enspace \triangle CBE & (\text{AAA}) \end{array}$
$$DC^{2} = AD.BE$$
$\begin{array}{rll} \dfrac{DC}{AD} &= \dfrac{BE}{BC} & (\triangle ADC \enspace ||| \enspace \triangle CBE) \\ \text{But } DC &= BC & (\text{given}) \\ \therefore \dfrac{DC}{AD} &= \dfrac{BE}{DC} & \\ \therefore DC^{2} &= AD.BE & \end{array}$
$$CD$$ is a tangent to circle $$ABDEF$$ at $$D$$. Chord $$AB$$ is produced to $$C$$. Chord $$BE$$ cuts chord $$AD$$ in $$H$$ and chord $$FD$$ in $$G$$. $$AC \parallel FD$$ and $$EF = AB$$. Let $$D_{4} = x$$ and $$D_{1} = y$$.
[NCS, Paper 3, November 2011]
Determine THREE other angles that are each equal to $$x$$.
$\begin{array}{[email protected]{\;}[email protected]{\quad}l} \hat{A} & = \hat{D}_{4} = x & (\text{tangent chord th.})\\ \hat{E}_{2} & = x & (\text{tangent chord th.})\\ \hat{D}_{2} & = \hat{A} = x & (\text{alt. } \angle \text{s, } CA \parallel DF) \end{array}$
Prove that $$\triangle \text{BHD} \enspace ||| \enspace \triangle \text{FED}$$.
In $$\triangle BHD$$ and $$\triangle FED$$
1. $$\hat{B}_{2} = \hat{F} \quad(\angle \text{s in same seg.})$$
2. $$\hat{D}_{3} = \hat{D}_{1} \quad(\text{chord subtends } = \angle \text{s})$$
$$\triangle BHD \enspace ||| \enspace \triangle FED \quad (\angle \angle \angle)$$
Hence, or otherwise, prove that $$AB \cdot BD = FD \cdot BH$$.
$\begin{array}{[email protected]{\;}[email protected]{\quad}l} \frac{FE}{BH} & = \frac{FD}{BD} & (||| \enspace \triangle\text{s})\\ \text{But } FE & = AB & (\text{given})\\ \frac{AB}{BH} & = \frac{FD}{BD} &\\ AB \cdot BD & = FD \cdot BH & \end{array}$
|
# How do you find the exact value of 3 over the square root of 8?
Sep 22, 2015
$\frac{3}{\sqrt{8}} = \frac{3}{2 \sqrt{2}}$
#### Explanation:
The expression can be simplified as
$\frac{3}{\sqrt{8}} = \frac{3}{\sqrt{4} \cdot \sqrt{2}} = \frac{3}{2 \sqrt{2}}$
However, it is not possible to make further progress as 2 and 3 are prime and the square root of 2 is an irrational number (it cannot be expressed as the ratio of two integers).
The irrationality of the square root of 2 is a famous result attributed to the Pythagorean Hipassus of Metapontum, who was allegedly executed (for blasphemy) for demonstrating a proof of this (see here).
As the reciprocal of a fraction is found by exchanging its numerator with its denominator, the reciprocal of some irrational number is also irrational.
The decimal expansion of an irrational number has a non-terminating non-recurring fractional part so there is no progress to be made by attempting that.
Expressions that include unresolved roots are known as surds (eg, see here).
That is, expression of this quantity as an exact number requires the use of a surd.
This might be achieved by multiplying through by $\frac{\sqrt{2}}{\sqrt{2}}$, which, it might be noted, is equal to $1$, that is, it is the identity under the operation of multiplication so that it will leave the value of the multiplicand unchanged.
$\frac{3}{2 \sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{3 \sqrt{2}}{2 \cdot 2} = \frac{3 \sqrt{2}}{4}$ (as noted by cal_F).
|
# Finding the maximum area of a quadrilateral when three points are given
I am working on problems in the chapter "Applications of Derivatives". I encountered the following problem:
Question:
Four points A,B,C, and D lie in that order on the parabola $$y=ax^2+bx+c$$. The coordinates of A,B, and D are $$(-2,3)$$,$$(-1,1)$$, and $$(2,7)$$ respectively. The coordinates of C for which the area of the quadrilateral ABCL is maximum is:
(A) $$(1/2,7/4)$$
(B) $$(1/2,-7/4)$$
(C) $$(-1/2,7/4)$$
(D) $$(-1/2,-7/4)$$
My Approach:
Introduction:
I found the equation of the parabola using the given coordinates. Then computed the area of the quadrilateral by considering an arbitrary value for point C. Then I found the coordinates of C by maximizing the value of A by using the concept of maxima. What I am looking for is an alternate proof or a property which relates the coordinates of third vertex of a quadrilateral when three points are fixed under certain constraints, to maximize the area.
We have been given the coordinates A$$(-2,3)$$,B$$(-1,1)$$ and D$$(2,7)$$.
Since these points lie on the given parabola $$y=ax^2+bx+c$$, on substituting the coordinates of A,B and D we get the following linear equations in variables $$a,b,$$ and $$c$$:
$$4a-2b+c=3$$
$$a-b+c=1$$
$$4a+2b+c=7$$
On solving these three equations, I got $$a=1,b=1,$$ and $$c=1$$. So the equation of the parabola is $$y=x^2+x+1$$.
In the following graph, the points A,B and D are fixed, whereas point C is a variable point on the parabola. We need to find the coordinates of point C such that the area of the quadrilateral is maximum.
In coordinate geometry, I came across the following formula to compute the area of a $$n$$ sided polygon when the coordinates are given:
Area A $$= \frac 1 2 \left( \left| {\begin{array}{cc}x_1 & x_2 \\y_1 & y_2 \\ \end{array} } \right|+\left| {\begin{array}{cc}x_2 & x_3 \\y_2 & y_3 \\ \end{array} } \right|+\left| {\begin{array}{cc}x_3 & x_4 \\y_3 & y_4 \\ \end{array} } \right|+ \dots +\left| {\begin{array}{cc}x_{n-1} & x_n \\y_{n-1} & y_n \\ \end{array} } \right|+\left| {\begin{array}{cc}x_n & x_1 \\y_n & y_1 \\ \end{array} } \right| \right)$$
where $$(x_1,y_1),(x_2,y_2),(x_3,y_3),\dots ,(x_{n-1},y_{n-1}),(x_n,y_n)$$ are coordinates of the $$n$$ vertices of the $$n$$ sided polygon taken in order. Here |.| denotes determinant of the matrix.
Now in the given question we are supposed to find the maximum area of a quadrilateral (4 sided polygon). So using the above formula, we get the following:
Area A $$= \frac 1 2 \left( \left| {\begin{array}{cc}x_1 & x_2 \\y_1 & y_2 \\ \end{array} } \right|+\left| {\begin{array}{cc}x_2 & x_3 \\y_2 & y_3 \\ \end{array} } \right|+\left| {\begin{array}{cc}x_3 & x_4 \\y_3 & y_4 \\ \end{array} } \right|+\left| {\begin{array}{cc}x_4 & x_1 \\y_4 & y_1 \\ \end{array} } \right| \right)$$
where where $$(x_1,y_1),(x_2,y_2),(x_3,y_3),$$ and $$(x_4,y_4)$$ are the coordinates of points A,B,C, and D respectively.
Now let the $$x$$-coordinate of point C be $$h$$. So its $$y$$-coordinate (from the equation of parabola) is $$h^2+h+1$$. So coordinates of point C be represented as $$(h^2+h+1)$$.
After substituting the values of the coordinates in the equation for area, I obtained the following equation:
Area A$$=\frac 1 2 -3h^2+3h+18$$
Area A attains the maximum value at $$h=1/2$$.
Hence the coordinates of point C is $$(1/2,7/4)$$ - option (A). My answer is correct.
Doubt:
Is there any other formal way of solving this problem? Is there any property for choosing the fourth vertex of the quadrilateral when three vertices are given, in order to get the maximum area under given constraints?
Credits to @Siddhant. After reading his answer, another method came to my mind and I wish to share it here.
The coordinates of points A,B and D have been given in the question. So these points are fixed. Point C is the variable point on the given parabola. We need to choose the appropriate value of C so that the area of the quadrilateral ABCD is maximum.
Let us consider the following diagram.
Area of the quadrilateral ABCD is nothing but the sum of areas of the triangles ABD and BCD. Since points A,B and D are fixed, the area of triangle ABD is constant for the given question. Since C is a variable point and is not moving parallel to the line BD, the area of triangle BCD is not constant.
Our objective is to maximize the area of the quadrilateral ABCD. Since area of ABD is fixed. We need to maximise the are of triangle BCD.
We know that, area of a triangle is given by $$\frac 1 2 bh$$ where $$b$$ and $$h$$ are the base and hypotenuse of the triangle respectively. In triangle BCD, let us consider the base as the line segment BD and the height represented by the green coloured line in the above diagram. Since B and D are fixed, the length of base BD is fixed. Now, in order to maximise the area of triangle BCD, it is sufficient to find the coordinates for the point C when it is at the maximum distance from the line segment BD(base of BCD).
Now let us find the equation of the line BD using the coordinates of B and D.
$$\frac{y-1}{x+1}=\frac 6 3 = 2$$ $$y-1=2(x+1)$$ $$2x-y+3=0$$
Next, let us compute the distance (or height $$h$$) of the variable point C$$(p,p^2+p+1)$$ from the base BD as shown below:
$$h=\frac{2p-(p^2+p+1)+3}{\sqrt{2^2+1^2}}$$ $$h=\frac{-p^2+p+2}{\sqrt 3}$$
We have obtained $$h$$ as a function of $$p$$. Now we need to maximize $$h$$. It is easy to see that the quadratic in $$p$$ attains the highest value at $$p=1/2$$.
Hence the coordinate for point C such that the area of quadrilateral ABCD is maximum is $$(1/2,7/4)$$.
• Nice geometric interpretation. Commented Oct 7, 2019 at 11:52
• Alternatively, you can require that the tangent to the parabola in $p$ is parallel to BD.
– chi
Commented Oct 7, 2019 at 18:07
Its not a completely different approach but just a way to reduce the matrix computations.
$$1^{st}$$ step - Get the parabola $$y(x) = x^2 + x + 1$$ which you got it with the points A,B,D.
Now, if you join BD, the total area you interested in maximizing would be $$Area(ABD)+Area(BCD)$$ triangles. As area cannot be negative, you need to maximize just Area(BCD) as points A,B,D are known and so is the value of Area(ABD).
Therefore computing the $$Area(BCD)$$ you can directly use the area of triangle formula =$$(1/2)det(A)$$ where $$A =$$ $$\left[ \begin{matrix} h & k & 1\\ -1 & 1 & 1 \\ 2 & 7 & 1\\ \end{matrix}\right]$$
This is a simple determinant which we get as $$Area(BCD) = (1/2)(3k-6h-9)$$. We need to maximize this following the constraint $$k=h^2 + h + 1$$. Just substitute for $$k$$ and differentiate with respect to $$h$$, we get $$h=1/2$$, and so $$k=7/4$$.
There is also a faster way of finding the value of coordinates of point $$C$$. After finding the equation of the curve. Just utilize the options, substitute each value of $$C$$. The one which satisfies is the correct answer. Luckily here only one option matches - option $$A$$. This may help you in exams where time counts!
• I agree your method is the simplest. Here it works great. But, I was seeing for method that work in all conditions. That's why I asked for a formal method and not just a method. Commented Oct 7, 2019 at 11:28
Substitute points $$( A,B,D )$$ into $$ax^2 + bx + c,$$ and solve using Cramer's Rule of three simultaneous equations in order to evaluate $$( a,b,c)=(1,1,1)$$.
The point $$(x_C,y_C)$$ can be found by drawing a tangent by differentiating $$y = ax^2+bx+c$$ w.r.t $$x$$ at future point C for maximum area.
This is so because maximum height of triangle $$(Area = base DB \cdot h_{max})$$ is between parallel lines.
Slope of DB =
$$2 ax_C +b= \dfrac{6}{3}=2 \rightarrow x_C=\dfrac12;$$
$$y_C= (\dfrac12)^2 + 1\cdot \, \frac12 + 1= \frac74.$$
|
# What is the solution of the system y=x-10, y=2x+5?
Nov 12, 2016
$x = - 15 \mathmr{and} y = - 25$
#### Explanation:
This is a perfect scenario for solving the two equations.
(which represent straight lines and the solution gives the point of intersection.)
$\textcolor{b l u e}{y = x - 10} \text{ and } \textcolor{red}{y = 2 x + 5}$
The two $y -$ values are equal!
$\textcolor{w h i t e}{\times \times \times \times \times \times x} \textcolor{b l u e}{y} = \textcolor{red}{y}$
Therefore:$\textcolor{w h i t e}{\times x} \textcolor{b l u e}{x - 10} = \textcolor{red}{2 x + 5}$
$\textcolor{w h i t e}{\times \times . \times x} - 10 - 5 = 2 x - x$
$\textcolor{w h i t e}{\times \times . \times x} - 15 = x \text{ } \leftarrow$ we have the x-value
$y = \left(- 15\right) - 10 = - 25 \text{ } \leftarrow$ from the first equation
Check in the second equation:
$y = 2 \left(- 15\right) + 5 = - 25$
$x = - 15 \mathmr{and} y = - 25$
|
Engineering Statistics ECIV 2305
1 / 10
# Engineering Statistics ECIV 2305 - PowerPoint PPT Presentation
## Engineering Statistics ECIV 2305
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Engineering Statistics ECIV 2305 Chapter 3 DISCRETE PROBABILITY DISTRIBUTIONS • 3.1 The Binomial Distribution • 3.2 The Geometric Distribution • 3.3 The Hypergeometric Distribution • 3.4 The Poisson Distribution 3.3 The Hypergeometric Distribution
2. Engineering Statistics ECIV 2305 Section 3.3 The Hypergeometric Distribution 3.3 The Hypergeometric Distribution
3. Normal kind item Special kind item • Consider the following example: N = total number of items =15 , r = Number of special kind items = 4 • If one item is chosen at random, the probability that it is of the special kind is p = r/N = 4/15 • If “n” items (OR “n” trials) are chosen at random with replacement, then the number of defective items chosen is a binomial RV , i.e. X~B(n, r/N) Examples of special kind items: - some black balls within white balls - some defective items within satisfactory items 3.3 The Hypergeometric Distribution
4. N = total number of items =15 , r = Number of special kind items = 4 • What if “n” items were selected without replacement? • The binomial distribution will not work. Instead, the number of defective items chosen will have a distribution called the hypergeometric distribution. 3.3 The Hypergeometric Distribution
5. The pmf of the hypergeometric distribution is: For example: If n = 7, r = 4 , and N = 15 ►► 0 ≤ x ≤ 4 If n = 7, r = 4 , and N = 10 ►► 1 ≤ x ≤ 4 With an expectation of : and a variance of : 3.3 The Hypergeometric Distribution
6. Example • Recall the example of the 500 chips (in sections 1.5 &1.7); nine of which are defective. Three chips are drawn at random without replacement. What is the probability that exactly one defective chip is drawn? N = 500 , r = 9 (this is the number of special kind items) n = 3 (three trials were performed) 3.3 The Hypergeometric Distribution
7. When the total number of items “N” (population size) is much larger than the sample size n, then sampling without replacement is very similar to sampling with replacement. This is because the reduction of the total population “N” by the items selected does not appreciably change the selection probabilities of additional items. ►►So, it becomes more convenient to use the binomial distribution; X~B(n, r/N) You can also use the hypergeometric, but it will be just more work. 3.3 The Hypergeometric Distribution
8. Example Milk Container Contents (page 182) • A box contains 16 milk containers. Six of them are underweight. • Five containers are chosen at random for inspection. • What is the distribution of the number of underweight milk containers in the sample chosen by the inspector? 3.3 The Hypergeometric Distribution
9. … continute Milk Container Contents example… 3.3 The Hypergeometric Distribution
10. Example Fish Tagging and Recapture (page 182) • A small lake contains 50 fish. Ten of them are tagged. • A fisherman catches eight fish. • What is the distribution of the number of tagged fish caught by the fisherman? 3.3 The Hypergeometric Distribution
|
# Factors of 39 | Pair Factors With Great Explanation
Contents
## Factors Of 39
Factors of 39 are the numbers multiplied together to form the result of 39. The fact that 2 and 7 are factors of 14 can be seen by multiplying 2 and 7. For example, 2 × 7= 14 as 2 and 7 are factors of 14. Similarly, we can find the factors of 39.
With this article, you will learn how to find factors of the number 39 and prime factors of 39 with detailed steps.
## What are the Factors of 39?
As a composite number, 39 has more than two factors, leaving it with factors other than 1 and 39.
Factors of 39
1, 3, 13, and 39
Smimilarly Factors of 24
1, 2, 3, 4, 6, 8, 12, and 24.
Additionally, we know that 39 is divisible by the factors listed above. The factors of a number can, however, be found in two ways: the division method and the factor pair method.
## What are the Factors of -39?
-39 has both additive inverse and negative factors, that is, negative numbers. A positive number is produced by dividing -39 by any negative number.
• (-39) ÷ (-1) = 39
• (-39) ÷ (-3) =13
• (-39) ÷ (-13) = 3
• (-39) ÷ (-39) = 1
Hence, factors of -39 include -1, -3, -13, and -39.
## How to Find the Factors of 39?
The following steps explain how to calculate the factors of 39.
Write the number 39 in your notebook first.
Think of 3 and 13 as two numbers that, when multiplied, give 39, for example, 3 x 13 = 39.
The prime number 3 only has two factors, which are 1 and the number itself. Therefore, it cannot be factorized further.
3 = 1 × 3
Furthermore, 13 is a prime number and cannot be further factored in.
13 = 1 × 13
As a result, 39 is factorized into 1 × 3 × 13.
Based on the above computation, there are 3 unique numbers: 1, 3, and 39.
## Factor pairs of 39
Every number has positive as well as negative factors. You can find the pair factors of 39 by multiplying the two numbers in a pair. They are as follows:
1 × 39 = 39; (1, 39)
3 × 13 = 39; (3, 13)
13 × 3 = 39; (13, 3)
39 × 1 = 39; (39, 1)
These are the positive pair factors of 39.
Thus, the factors of 39 are 1, 3, 13, and 39.
Next, let’s calculate 39’s negative pair factor:
(-1) × (-39) = 39
(-3) × (-13) = 39
(-13) × (-3) = 39
(-39) × (-1) = 39
As a result, the negative pair factors of 39 are (-1, -39), (-3, -13), (-13, 3) and (-39, -1).
## How to calculate Prime Factors of 39?
Here are the prime factors of 39 according to the prime factorization method.
We begin by dividing 39 by the smallest prime number, 2.
39 ÷ 2 = 19.5
Since fractions cannot be factors, hence 2 is not a prime factor for 39.
Continue with the next prime number, i.e. 3, 5, 7, etc.
39 ÷ 3 = 13,
As 3, and 13 are prime numbers, we cannot divide them.
As a result, the prime factorization of 39 is 3 × 13.
## Example:
### What are the common factors of 24 and 39?
Factors of 24 are,
1, 2, 3, 4, 6, 8, 12, and 24.
Similarly, factors of 39 are,
1, 3, 13, and 39.
Hence, the common factors are 1 and 3.
## FAQs
### 39 and 25 have what factors?
1, 3, 13, and 39 are the factors of 39.
Factors of 75= 1, 3, 5, 15, 25, 75.
### What is the greatest factor in 39?
1, 3, 13, and 39 are the factors of 39. Therefore, 39 is the greatest factor 39.
### The prime factor of 39 is?
39 has two prime factors, namely 3 and 39.
### How many factors are there in 36?
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, and 36.
### Among factors 38 and 48, what is common?
Factors of 38: 1, 2, 19, and 38.
1, 2, 3, 4, 6, 8, 12, 16, 24, and 48 are the factors of 48.
1, and 2 are the common factors of 38 and 48.
## Conclusion
I hope that with the above illustration you are now in a better position to understand the factors of 39.
Keep learning with ECTI.
|
Students can Download Maths Chapter 3 Algebra Ex 3.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.
## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.3
8th Maths Algebra Exercise 3.3 Question 1.
Expand
(i) (3m + 5)2
(ii) (5p – 1)2
(iii) (2n – 1)(2n + 3)
(iv) 4p2 – 25q2
Solution:
(i) (3m + 5)2
Comparing (3m + 5)2 with (a + b)2 we have a = 3m and b = 5
(a + b)2 = a2 + 2 ab + b2
(3m + 5)2 = (3m)2 + 2 (3m) (5) + 52
= 32m2 + 30m + 25 = 9m2 + 30m +25
(ii) (5p – 1)2
Comparing (5p – 1)2 with (a – b)2 we have a = 5p and b = 1
(a – b)2 = a2 – 2ab + b2
(5p – 1)2 = (5p)2 – 2 (5p) (1) + 12
= 52p2 – 10p + 1 = 25p2 – 10p + 1
(iii) (2n – 1)(2n + 3)
Comparing (2n – 1) (2n + 3) with (x + a) (x + b) we have a = -1; b = 3
(x + a) (x + b) = x2 + (a + b)x + ab
(2n +(- 1)) (2n + 3) = (2n)2 + (-1 + 3)2n + (-1) (3)
= 22n2 + 2 (2n) – 3 = 4n2 + 4n – 3
(iv) 4p2 – 25q2 = (2p)2 – (5q)2
Comparing (2p)2 – (5q)2 with a2 – b2 we have a = 2p and b = 5q
(a2 – b2) = (a + b)(a – b) = (2p + 5q) (2p – 5q)
Maths Chapter 3 Exercise 3.3 Class 8 Question 2.
Expand
(i) (3 + m)3
(ii) (2a + 5)3
(iii) (3p + 4q)3
(iv) (52)3
(v) (104)3
Solution:
(i) (3 + m)3
Comparing (3 + m)3 with (a + b)3 we have a = 3; b = m
(a + b)3 = a2 + 3a2b + 3 ab2 + b3
(3 + m)3 = 33 + 3(3)2 (m) + 3 (3) m2 + m3
= 27 + 27m + 9m2 + m3 = m3 + 9 m2 + 27m + 27
(ii) (2a + 5)3
Comparing (2a + 5)3 with (a + b)3 we have a = 2a, b = 5
(a + b)3 = a3 + 3a2b + 3ab2 + b3 = (2a)3 + 3(2a)2 5 + 3 (2a) 52 + 53
= 23a3 + 3(22a2) 5 + 6a (25) + 125
= 8a3+ 60a2 + 150a + 125
(iii) (3p + 4q)3
Comparing (3p + 4q)3 with (a + b)3 we have a = 3p and b = 4q
(a + b) 3 = a3 + 3a2b + 3ab2 + b3
(3p + 4q)3 = (3p)3 + 3(3p)2 (4q) + 3(3p)(4q)2 + (4q)3
= 33p3 +3 (9p2) (4q) + 9p (16q2) + 43q3
= 27p3 + 108p2q + 144pq2 + 64q3
(iv) (52)3 = (50 + 2)3
Comparing (50 + 2)3 with (a + b)3 we have a = 50 and b = 2
(a + b)3 = a3 + 3 a2b + 3 ab2 + b3
(50 + 2)3 = 503 + 3 (50)22 + 3 (50)(2)2 + 23
523 = 125000 + 6(2,500) + 150(4) + 8
= 1,25,000 + 15,000 + 600 + 8
523 = 1,40,608
(v) (104)3 = (100 + 4)3
Comparing (100 + 4)3 with (a + b)3 we have a = 100 and b = 4
(a + b)3 = a3 + 3 a2b + 3 ab2 + b3
(100 + 4)3 = (100)3 + 3 (100)2 (4) + 3 (100) (4)2 + (4)3
= 10,00,000 + 3(10000) 4 + 300 (16) + 64
= 10,00,000 + 1,20,000 + 4,800 + 64 = 11,24,864
Samacheer Kalvi 8 Maths Question 3.
Expand
(i) (5 – x)3
(ii) (2x – 4y)3
(iii) (ab – c)3
(iv) (48)3
(v) (97xy)3
Solution:
(i) (5 – x)3
Comparing (5 – x)3 with (a – b)3 we have a = 5 and b = x
(a – b)3 = a3 – 3a2b + 3ab2 – b3
(5 – x)3 = 53 – 3 (5)2 (x) + 3(5)(x2) – x3
= 125 – 3(25)(x) + 15x2 – x3 = 125
(ii) (2x – 4y)3
Comparing (2x – 4y)3 with (a – b)3 we have a = 2x and b = 4y
(a – b)3 = a3 – 3a2b + 3ab3 – b3
(2x – 4y)3 = (2x)3 – 3(2x)2 (4y) + 3(2x) (4y)2 – (4y)3
= 23x3 – 3(22x2) (4y) + 3(2x) (42y2) – (43y3)
= 8x3 – 48x2y + 96xy2 – 64y3
(iii) (ab – c)3
Comparing (ab – c)3 with (a – b)3 we have a = ab and b = c
(a – b)3 = a3 – 3a2b + 3ab2 – b3
(ab – c)3 = (ab)3 – 3 (ab)2 c + 3 ab (c)2 – c3
= a3b3 – 3(a2b2) c + 3abc2 – c3
= a3b3 – 3a2b2 c + 3abc2 – c3
(iv) (48)3 = (50 – 2)3
Comparing (50 – 2)3 with (a – b)3 we have a = 50 and b = 2
(a – b)3 = a3 – 3a2b + 3ab2 – b3
(50 – 2)3 = (50)3 – 3(50)2(2) + 3 (50)(2)2 – 23
= 1,25,000 – 15000 + 600 – 8 = 1,10,000 + 592
= 1,10,592
(v) (97xy)3
= 973 x3 y3 = (100 – 3)3 x3y3
Comparing (100 – 3)3 with (a – b)3 we have a = 100, b = 3
(a – b)3 = a3 – 3a2b + 3ab2 – b3
(100 – 3)3 = (100)3 – 3(100)2 (3) + 3 (100)(3)2 – 33
973 = 10,00,000 – 90000 + 2700 – 27
973 = 910000 + 2673
973 = 912673
97x3y3 = 912673x3y3
Samacheer Kalvi 8th Maths Book Question 4.
Simplify (i) (5y + 1)(5y + 2)(5y + 3)
(ii) (p – 2)(p + 1)(p – 4)
Solution:
(i) (5y + 1) (5y + 2) (5y + 3)
Comparing (5y + 1) (5y + 2) (5y + 3) with (x + a) (x + b) (x + c) we have x = 5y ; a = 1; b = 2 and c = 3.
(x + a) (x + b) (x + c) = x3 + (a + b + c) x2 (ab + bc + ca) x + abc
= (5y)3 + (1 + 2 + 3) (5y)2 + [(1) (2) + (2) (3) + (3) (1)] 5y + (1)(2) (3)
= 53y3 + 6(52y2) + (2 + 6 + 3)5y + 6
= 1253 + 150y2 + 55y + 6
(ii) (p – 2)(p + 1)(p – 4) = (p + (-2))0 +1)(p + (-4))
Comparing (p – 2) (p + 1) (p – 4) with (x + a) (x + b) (x + c) we have x = p ; a = -2; b = 1 ; c = -4.
(x + a) (x + b) (x + c) = x3 + (a + b + c) x2 + (ab + be + ca) x + abc
= p3 + (-2 + 1 + (-4))p2 + ((-2) (1) + (1) (-4) (-4) (-2)p + (-2) (1) (-4)
= p3 + (-5 )p2 + (-2 + (-4) + 8)p + 8
= p3 – 5p2 + 2p + 8
Samacheer Kalvi Guru 8th Maths Question 5.
Find the volume of the cube whose side is (x + 1) cm.
Solution:
Given side of the cube = (x + 1) cm
Volume of the cube = (side)3 cubic units = (x + 1)3 cm3
We have (a + b)3 = (a3 + 3a2b + 3ab2 + b3) cm3
(x + 1)3 = (x3 + 3x2 (1) + 3x (1)2 + 13) cm3
Volume = (x3 + 3x2 + 3x + 1) cm3
Samacheerkalvi.Guru 8th Maths Question 6.
Find the volume of the cuboid whose dimensions are (x + 2),(x – 1) and (x – 3).
Solution:
Given the dimensions of the cuboid as (x + 2), (x – 1) and (x – 3)
∴ Volume of the cuboid = (l × b × h) units3
= (x + 2) (x – 1) (x – 3) units3
We have (x + a) (x + b) (x + c) = x3 + (a + b + c) x2 + (ab + bc+ ca)x + abc
∴ (x + 2)(x – 1) (x – 3) = x3 + (2 – 1 – 3)x2 + (2 (-1) + (-1) (-3) + (-3) (2)) x + (2)(-1) (-3)
x3 – 2x2 + (-2 + 3 – 6)x + 6
Volume = x3 – 2x2 – 5x + 6 units3
|
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 3: Decimals Exercise – 3.1
Question: 1
Write each of the following as decimals:
(i) 8/100
Solution:
(i) 8/100
Mark the decimal point two places from right to left
Mark the decimal point one place from right to left
9/10 = 0.9
4/100
Mark the decimal point two places from right to left
4/100
= 0.04
Convert 2/10 and 6/1000 into decimals
2/10
Mark the decimal point one place from right to left
2/10
= 0.2
6/1000
Mark the decimal point three places from right to left
6/1000
= 0.006
23, 0.2, 0.006 are unlike decimals. So we convert them into like decimals.
Question: 2
Convert each of the following into fractions in the lowest form:
(i) 0.04
(ii) 2.34
(iii) 0.342
(iv) 17.38
(i) 0.04
= 0.04/1
= 4/100
= 1/25
(ii) 2.34
= 2.34/1
= 234/100
= 117/50
(iii) 0.342
= 0.342/1
= 342/1000
= 171/500
(iv) 17.38
=17.38/1
= 1738/100
= 869/50
Question: 3
Express the following fractions as decimals:
(i) 23/10
(i) 23/10
= 23/10
= 2.3
Question: 4
(i) 41.8, 39.24, 5.01 and 62.6
(ii) 18.03, 146.3, 0.829 and 5.324
(i) 148.65
(ii) 170.483
Question: 5
Find the value of:
(i) 9.756 - 6.28
(ii) 48.1- 0.37
(iii) 108.032 - 86.8
(iv) 100 - 26.32
(i) 3.476
(ii) 47.73
(iii) 21.232
(iv) 73.68
Question: 6
Take out 3.547 from 7.2
3.653
Question: 7
What is to be added to 36.85 to get 59.41?
Solution:
x + 36.85 = 59.41
x = 59.41-36.85
x = 22.56
Therefore 22.56 is added to 36.85 to get 59.41
Question: 8
What is to be subtracted from 17.1 to get 2.051?
Solution:
17.1- x = 2.051
17.1 = x + 2.051
x = 17.1 - 2.051
x = 15.049
Therefore 15.049 is subtracted from 17.1 to get 15.049
Question: 9
By how much should 34.79 be increased to get 70.15?
Solution:
34.79 + x = 70.15
x = 70.15 - 34.79
x = 35.36
Therefore 35.36 is increased to 70.15
Question: 10
By how much should 59.71 be decreased to get 34.58?
Solution:
59.71 – x = 34.58
59.71 - 34.58 = x
x = 25.13
Therefore 25.13 is decreased to get 34.58
|
Find the equation of the regression line for the given data. Then construct a scatter plot of the data and draw the regression line.
# Find the equation of the regression line for the given data. Then construct a scatter plot of the data and draw the regression line.
393k points
Find the equation of the regression line for the given data. Then construct a scatter plot of the data and draw the regression line. (The pair of variables have a significant correlation.) Then use the regression equation to predict the value of y for each of the given x-values, if meaningful. The number of hours 6 students spent for a test and their scores on that test are shown below.
Hours spent studying, x 1 2 3 4 4 5
Test score, y 42 47 52 50 62 69
(a) x = 3 hours
(b) x = 3.5 hours
(c) x = 12 hours
(d) x = 4.5 hours
linear regression
scatter plot
100 points
how did you get 161/3?
403.4k points
The find the line of regression use the following equations,
$$m = \frac{\bar{x}\bar{y}-\overline{xy}}{(\bar{x})^2-\overline{x^2}}$$
$$b = \bar{y}-m\bar{x}$$
$$\hat{y} = mx+b$$
now it is only a matter of finding $\bar{x}, \bar{y}, \overline{xy},\text{ and }\overline{x^2}$. So,
$$\bar{x}=\frac{1+2+3+4+4+5}{6}=\frac{19}{6}$$
$$\bar{y}=\frac{42+47+52+50+62+69}{6}=\frac{161}{3}$$
$$\overline{xy}=\frac{1(42)+2(47)+3(52)+4(50)+4(62)+5(69)}{6}=\frac{1085}{6}$$
$$\overline{x^2}=\frac{1^2+2^2+3^2+4^2+4^2+5^2}{6}=\frac{71}{6}$$
plugging these values into $m$ and $b$ gives,
$$m=\frac{(19/6)(161/3)-(1085/6)}{(19/6)^2-71/6}=\frac{392}{65}\approx 6.031$$
$$b=\frac{161}{3}-\frac{392}{65} \frac{19}{6}=\frac{2247}{65} \approx 34.569$$
and hence the line of regression is,
$$\hat{y}=6.031x+34.569$$
by plugging in the x values you will obtain the test score for any value of x. Remember that if the test score is above 100 that would be viewed as not meaningful.
Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$
Use LaTeX to type formulas and markdown to format text. See example.
• Answer the question above my logging into the following networks
### Post as a guest
• Your email will not be shared or posted anywhere on our site
•
Stats
Views: 3.1k
|
# Coordinate Geometry
## Geometry: Coordinate Geometry
Approach every question the same way to minimize mistakes. Start by reading the question carefully and identifying the bottom line. Then assess your options and choose the most efficient method to attack the problem. When you have an answer, loop back to verify that it matches your bottom line.
In the xy-plane, line l passes through the points (0, 0) and (2, 5). Line m is perpendicular to line l. What is the slope of line m?
Bottom line: slope m = ?
Assess your Options: You could draw out a graph and solve this visually, but that is a waste of time if you know the formula to find the slope of a line.
Attack the Problem: You are given the most information about line l, so start with that line. You should have the formula for slope memorized:
$slope=\frac{rise}{run}=\frac{\Delta y}{\Delta x}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
It is easiest just to think about slope as the change in y-values over the change in x-values. If you look up at the original points that you have been given, from zero the y-values go up to 5 and the x-values go up to 2. You now have 5 over 2.
The slope of line l is $\frac{5}{2}$.
At this point, some students will think they are finished and select answer (D). However, your bottom line was the slope of line m! The problem tells you that line m is perpendicular to line l. In order to find a perpendicular line, you must take the opposite reciprocal of the first line; in essence you must flip the sign (negative or positive) and the numbers (a fraction or whole number).
The slope of line m is $-\frac{2}{5}$.
(A) $-\frac{5}{2}$
(B) $-\frac{2}{5}$
(C) $\frac{2}{5}$
(D) $\frac{5}{2}$
|
# Unraveling 9.625 As A Fraction: A Step-By-Step Guide
Charlotte Miller
Updated on:
Unraveling 9.625 As A Fraction: A Step-By-Step Guide. Numbers play a fundamental role in our daily lives, allowing us to quantify and express various quantities accurately. Decimals are one such numeric representation that helps us convey precise values. In this blog post, we will delve into the world of decimals and explore the process of converting 9.625 into a fraction. By following the step-by-step guide outlined below, you’ll gain a clear understanding of how to simplify and express 9.625 as a fraction. Let’s begin!
Contents
## Understanding Decimals
Before we dive into converting 9.625 into a fraction, let’s first grasp the concept of decimals. Decimals are numeric values that represent fractions of a whole or divisions of a number. They consist of two primary components: the whole number part and the fractional part separated by a decimal point. In the case of 9.625, the whole number part is 9, and the fractional part is 625.
## Converting 9.625 To A Fraction: Step-By-Step Process
Converting 9.625 into a fraction involves a few simple steps. Let’s break down the process:
Step 1: Identify the decimal value.
In this case, the decimal value we are working with is 9.625.
Step 2: Determine the denominator.
The denominator represents the decimal place value. Since there are three decimal places in 9.625, the denominator will be 1000. (Remember, the denominator is always a power of 10 corresponding to the number of decimal places.)
Step 3: Write the decimal as a fraction.
To convert the decimal to a fraction, we place the decimal value over the determined denominator. In this case, 9.625 becomes 9625/1000.
## Simplifying The Fraction
The fraction 9625/1000 can be further simplified by dividing both the numerator and the denominator by their greatest common divisor (GCD). In this case, the GCD of 9625 and 1000 is 125. Dividing both by 125 gives us:
9625 ÷ 125 = 77
1000 ÷ 125 = 8
Therefore, the simplified fraction of 9.625 is 77/8.
## Decimal To Fraction Conversion In Different Forms
Now that we have successfully converted 9.625 into a fraction, let’s explore alternative representations of the same value.
Mixed Number Form: The fraction 77/8 can be expressed as a mixed number. Dividing the numerator (77) by the denominator (8) gives us 9 as the quotient, with a remainder of 5. Thus, the mixed number form is 9 5/8.
Percentage Form: To convert the fraction to a percentage, we multiply it by 100. So, 77/8 as a percentage is approximately 962.5%.
## Conclusion
Understanding the conversion of decimals into fractions is a valuable skill in mathematics. In this blog post, we have discussed how to convert 9.625 into a fraction and simplify it. The decimal 9.625 is equivalent to the fraction 77/8. By following the step-by-step guide provided, you can easily convert decimals into fractions, allowing for more precise calculations. Remember to practice this skill to reinforce your understanding. Stay tuned for more insightful content on mathematics and related topics.
We hope this blog post has provided a comprehensive understanding of converting 9.625 into a fraction. Keep exploring and enjoy your mathematical journey!
## FAQ
### What Is .625 As A Fraction Simplified Calculator?
0.625 as a fraction is 625/1000 which can be reduced to 5/8.
### What Is The Decimal For 9 5 8?
9 5/8 as a decimal is 9.63.
### What Is 7 4 5 In Fraction Form?
The mixed number 7 4/5 can be changed into the improper fraction 39/5.
### Is 0.625 The Same As 5 8?
As a decimal, is 0.625.
I Have Covered All The Following Queries And Topics In The Above Article
What Is 9.625 As A Fraction
9.625 As A Mixed Fraction
9.625 As A Fraction In Simplest Form
9.625 As A Fraction
How do you turn 9.625 into a fraction
How do you turn 9.625 into a fraction
|
# High School: Algebra
### Reasoning with Equations and Inequalities HSA-REI.C.9
9. Find the inverse of a matrix if it exists and use it to solve systems of linear equations (using technology for matrices of dimension 3 × 3 or greater).
This is where actual calculations take place. We've finally gotten to the fun stuff. Your students may think otherwise, but we know the truth.
Students should already know how to form a matrix equation of AX = B from a system of linear equations and be familiar with the concept of inverse matrices. They will use the inverse of a matrix in order to solve for the variables of the matrix that is formed from given equations.
Students should know that if the determinant of a square matrix is zero (ad – bc = 0), there is no inverse to the matrix. A matrix that takes the form
has an inverse of
Assuming we've already translated our equations -2x + y = 5 and -3x + y = -2 into matrix equation form, we can find the inverse matrix of the A matrix (the one with all the coefficients).
First, we take our matrix and find its determinant. That would be -2 × 1 – 1 × -3 = -2 + 3 = 1. The inverse matrix exists. Now, let's give our creation life! After switching around the numbers, we have to multiply by the inverse of the determinant, which is 11 or just 1. So our inverse matrix is fine as is.
So your students have found the inverse matrix. Now what? Well, the whole point of this process is to find the values of x and y. (They should write that down. That's important.) To do that, we can multiply the inverse matrix we found by the B matrix (the one with all the solutions). That should give us the values for x and y.
That gives us 1 × 5 + (-1 × -2) = 5 + 2 = 7 = x, and 3 × 5 + (-2 × -2) = 15 + 4 = 19 = y. If we plug those values back into the original linear equations, they should all hold up.
While this method might seem a bit cumbersome, students should appreciate that it's especially useful for systems of equations with many, many variables. On the other hand, the inverses for matrices larger than 2 × 2 should be calculated using technology. The whole point, though, is that they can help us solve for the values of the variables.
#### Drills
1. Solve the system of equations 32x + 14y = 4 and 32x – 10y – 3 = 0 using matrices.
x = 0.11, y = 0.042
The first step is to set up our matrix equation in the proper format. It should look something like this: . We can calculate the determinant to be (-10 × 32) – (14 × 32) = -768. Using it, we can find the inverse matrix and multiply it by the B matrix: . If we multiply them properly, we should end up with x = 0.11 and y = 0.042.
2. Given the system of equations 16x + 19y = 40 and 45x + 4y = 57, what are x and y equal to? Solve using matrices.
x = 1.167, y = 1.123
We can set up our matrices as . The determinant is 4 × 16 – 45 × 19 = -791, meaning that the inverse matrix multiplied by the solutions matrix is . If we carry out this operation, we end up with x = 1.167 and y = 1.123.
3. Given the equations 67x + 14y = 13 and 87x + 9y = 17, solve for x and y.
x = 0.197, y = -0.013
We can set up our matrices to be . The determinant of the coefficient matrix is 9 × 67 – 87 × 14 = -615. That means our inverse matrix multiplied by the solutions matrix is . Multiplying the two should give us x = 0.197 and y = -0.013.
4. You're determined to find the determinant of the following matrix. Being so determined, you're never going to give up, ever, until you find it on your own! That's just your personality, or at least we'll pretend it is. Estimate in minutes, how long it'll take you to find the determinant of the following matrix.
Forever
No, we aren't insulting you. The determinant can only be defined for square matrices. This matrix isn't square, so it will take you forever to do so. Sorry if that took you forever to figure out.
5. You're really out to prove yourself (especially if you got that last question wrong). However, this time around, you and a friend are out to see who will find the inverse of your specific matrix first. Which of you will find the inverse matrix first?
Neither will find one
Since we need the determinant to find the inverse, we cannot find the inverse of a non-square matrix (at least for the purposes of our discussion). There are exceptions to this rule, but they don't apply here for now. We're really sorry if that took your forever to figure out as well.
6. What is the inverse matrix of the coefficients of the two equations 15x + 15y = 3 and 14x + 14y = 7?
There is no inverse matrix
We shouldn't even have to draw out a matrix. By finding that our determinant 15 × 14 – 15 × 14 is equal to zero, we know there cannot be an inverse matrix. Yes, the matrix is square and no, it doesn't have an inverse. No, we aren't lying and yes, we are evil.
7. Using computer technology to help you solve the system of equations using matrices, find the values of x, y, and z respectively.
14x + 19y + 13z = 415
67x + 87y + 15z = 617
88x + 87y + 90z = 561
x = -85.7, y = 69.1, z = 23.2
This looks more brutal than gluing your hand to a charging rhino, but it's really not that bad. The computer will do most of the work for us; all we really need to do is set up the matrices properly, like this:
8. The determinant of a matrix is calculated by using which equation?
All other answers are not the equation for a determinant. If you haven't memorized it (learning is more important than memorizing, anyway), you can draw a generic 2 × 2 square matrix and label its numbers The determinant is the product of the top left and bottom right minus the product of the top right and bottom left, or adbc.
9. The inverse of a square 2 × 2 matrix is found by using which of the following?
By definition, this is how our inverse matrix is set-up. We're sorry if you got it wrong, but the only way to get that one right without answering (C) is to rewrite the definition. If you really want to do so, you can.
10. If the determinant of a matrix is found to be less than 0, then we can conclude that:
The inverse matrix exists
|
# Practice Test on Area and Perimeter of Square | Questions on Area and Perimeter of Square
Get the solutions for the Practice test on Area and Perimeter of Square here. Students can improve their math skills by solving the practice test on the area and perimeter of the square in 2D Mensuration. Scroll down this page to find the various methods to solve the problems on the Area and Perimeter of the square. Go through the questions and try to solve the problems using the area and perimeter of the square formulas.
### Formula for Area and Perimeter of Square
• Area of square = s × s = s²
• The perimeter of the square = 4s
• Diagonal of the square = √2 × a
### Word Problems on Area and Perimeter of Square
1. Find the area of the square whose side is 6cm.
Solution:
Given,
side = 6cm
Area of square = s × s = s²
A = 6cm × 6cm
A = 36 sq.cm
Thus the area of the square is 36 sq.centimeters.
2. The side of the square is 48m. Find the area and perimeter.
Solution:
Given that,
side of the square = 48m
To find the area of the square
We know that,
Area of square = s × s = s²
A = 48m × 48m
A = 2304 sq.m
Thus the Area of the square is 2304 sq.m.
Now find the perimeter of the square
P = 4a
P = 4(48)
P = 192 cm
Therefore the perimeter of the square is 192 cm.
3. One side of the square is 7m. Find
i. Area
ii. Perimeter
iii. Diagonal
Solution:
Given,
Side of the square = 7m
i. Area:
We know that,
Area of square = s × s = s²
A = 7m × 7m
A = 49 sq. meter
Thus the area of the square is 49 sq. m.
ii. Perimeter:
We know that,
The perimeter of the square = 4a
P = 4 × 7m
P = 28m
Thus the perimeter of the square is 28 meters.
iii. Diagonal:
We know that,
Diagonal of the square = √2 × a
D = √2 × 7m
D = 9.89 meters
Thus the diagonal of the square is 9.89 meters.
4. The side of the square is 60cm. Find the perimeter of the square.
Solution:
Given,
The side of the square is 60cm.
P = 4a
P = 4 × 60cm
P = 240cm.
Therefore the perimeter of the square is 240 centimeters.
5. The Perimeter of the square is 96cm. Find the side of the square.
Solution:
Given that,
The perimeter of the square is 96cm.
P = 4a
96cm = 4a
a = 96/4
a = 24cm
Thus the side of the square is 24cm.
6. The cost of cementing a square yard at ₹ 2 per Square Metre is ₹800. Find the cost of fencing it at a rate of ₹5 per meter.
Solution:
Given,
The cost of cementing a square yard at ₹ 2 per Square Metre is ₹800.
Let the side be x meters.
Rate of cementing = ₹ 2 per m²
Total cost = 800
Area for cementing = 800/2 = 400
x² = 400
x = 20
Perimeter of the yard = 4a
P = 4 × 20m = 80m
Rate of fencing = ₹5 per meter
Total cost of fencing the square yard is 80 × 5
= ₹400
Therefore the cost of fencing it at a rate of ₹5 per meter is ₹400.
7. What is the diagonal and perimeter of the square if the side is 4cm.
Solution:
Given,
The side of the square is 4cm.
Perimeter of the square = 4a
P = 4(4cm)
P = 16cm
We know that,
Diagonal of the square = √2 × a
D = √2 × a
D = √2 × 4
D = 5.65 cm.
Thus the diagonal of the square is 5.65cm.
8. The area of the square field is 144ft². Find the side, perimeter, and diagonal of the square field.
Solution:
Given that,
The area of the square field is 144ft²
Area of square = s × s = s²
144 = s²
s² = 144
s = √144 = 12ft
Thus the side of the square field is 12 ft.
The perimeter of the square = 4a
P = 4 × 12ft = 48ft
Thus the perimeter of the square field is 48 feet.
Diagonal of the square = √2 × a
D = √2 × a
D = √2 × 12
D = 16.9 feet
Therefore the diagonal of the square is 16.9 feet.
9. The diagonal of the square is 4√2cm. Find the area and perimeter of the square?
Solution: Given,
The diagonal of the square is 4√2cm.
Area of square = s × s = s²
A = 4 × 4 = 16 sq.cm
The perimeter of the square = 4a
P = 4 × 4 = 16cm
Thus the area and perimeter of the square is 16 sq. cm and 16 cm.
10. Find the perimeter of the square whose sides are
i. 2 cm
ii. 7cm
iii. 16cm
Solution:
Given,
i. a = 2 cm
We know that,
The perimeter of the square = 4a
P = 4(2cm)
P = 8cm
Thus the perimeter of the square is 8cm.
ii. a = 7cm
We know that,
The perimeter of the square = 4a
P = 4(7cm)
P = 28 cm
Thus the perimeter of the square is 28cm.
iii. a = 16cm
We know that,
The perimeter of the square = 4a
P = 4(16cm)
P = 64cm
Thus the perimeter of the square is 64 cm.
|
# How do you solve w/7+3=-1?
Jan 21, 2017
See the entire solution process below;
#### Explanation:
First, subtract $\textcolor{red}{3}$ from each side of the equation to isolate the $w$ term and keep the equation balanced:
$\frac{w}{7} + 3 - \textcolor{red}{3} = - 1 - \textcolor{red}{3}$
$\frac{w}{7} + 0 = - 4$
$\frac{w}{7} = - 4$
Now, multiply each side of the equation by $\textcolor{red}{7}$ to solve for $w$ while keeping the equation balanced:
$\frac{w}{7} \times \textcolor{red}{7} = - 4 \times \textcolor{red}{7}$
$\frac{w}{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}}} \times \cancel{\textcolor{red}{7}} = - 28$
$w = - 28$
|
# Decimal and Fractional Expansion | How to do Decimal Expansion? | How to Write Fractional Expansion?
In this article, you will learn about the Decimal and Fractional Expansion of a Decimal Number. Before Proceeding further know the definitions of Decimal, Fraction, and the Place Value Chart. A Decimal is any number in the base 10 number system and is used to separate units place from tenths place in decimal. The decimal point present in between separates the Whole Number Part and Decimal Part.
Also, Read:
### Decimal Expansion of a Number
Decimal Expansion of a Number is its representation in the base-10 system. In this System, each decimal place consists of digits 0-9 arranged such that each digit is multiplied by a power of 10 decreasing from left to right and a decimal place with 10^0 is the one’s place.
Decimal Expansion of Number 1423.25 is defined as
1423.25 = 1*103+4*102+2*101+3*100+2*10-1+5*10-2
= 1000+400+20+3+0.2+0.05
Decimal Expansion of a Number may terminate and in such case, the number is called a regular number or finite decimal. At times, the Decimal Expansion of a Number may become periodic and in such case, it is called a repeating decimal. However, the expansion may continue infinitely without repeating and it is called an irrational number.
### Fractional Expansion of Decimals
In the Expanded Form of Decimal Fractions, you will learn how to read and write the Decimal Numbers. Decimal Numbers can be written in expanded form using the Place Value Chart.
Let us understand the same by considering an example
384.264
384.264 = 3 × 100 + 8 × 10 + 4 × 1 + 2 × $$\frac { 1 }{10 }$$ + 6 × $$\frac { 1 }{100 }$$ + 4 × $$\frac { 1 }{1000 }$$
= 300+80+4+$$\frac { 2 }{10 }$$+$$\frac { 6 }{100 }$$+$$\frac { 4 }{1000 }$$
### Solved Examples on Decimal and Fractional Expansion
1. Write the decimal and fractional expansion of 334.252?
Solution:
In Decimal Expansion
334.252 = 3*100+3*10+4*1+2*$$\frac { 1 }{10 }$$+5*$$\frac { 1 }{100 }$$+2*$$\frac { 1 }{1000 }$$
= 300+30+4+$$\frac { 2 }{10 }$$+$$\frac { 5 }{100 }$$ + $$\frac { 2 }{1000 }$$
= 300+30+4+0.2+0.05+0.002
In Fractional Expansion
= 3*100+3*10+4*1+2*$$\frac { 1 }{10 }$$+5*$$\frac { 1 }{100 }$$+2*$$\frac { 1 }{1000 }$$
= 300+30+4+$$\frac { 2 }{10 }$$+$$\frac { 5 }{100 }$$ + $$\frac { 2 }{1000 }$$
2. Write the decimal and fractional expansion of 543.32?
Solution:
In Decimal Expansion
543.32 = 5*100+4*10+3*1+3*$$\frac { 1 }{10 }$$+2*$$\frac { 1 }{100 }$$
= 500+40+3+$$\frac { 3 }{10 }$$+$$\frac { 2 }{100 }$$
= 500+40+3+0.3+0.02
In Fractional Expansion
= 5*100+4*10+3*1+3*$$\frac { 1 }{10 }$$+2*$$\frac { 1 }{100 }$$
= 500+40+3+$$\frac { 3 }{10 }$$+$$\frac { 2 }{100 }$$
3. Write the Decimal and Fractional Expansion of 647.345?
Solution:
In Decimal Expansion
647.345 = 6*100+4*10+7*1+3*$$\frac { 1 }{10 }$$+4*$$\frac { 1 }{100 }$$+5*$$\frac { 1 }{1000 }$$
= 600+40+7+$$\frac { 3 }{10 }$$+$$\frac { 4 }{100 }$$+$$\frac { 5 }{1000 }$$
= 600+40+7+0.3+0.04+0.005
In Fractional Expansion
647.345 = 6*100+4*10+7*1+3*$$\frac { 1 }{10 }$$+4*$$\frac { 1 }{100 }$$+5*$$\frac { 1 }{1000 }$$
= 600+40+7+$$\frac { 3 }{10 }$$+$$\frac { 4 }{100 }$$+$$\frac { 5 }{1000 }$$
### FAQ’s on Decimal and Fractional Expansion
1. What is Decimal in Expanded Form?
Expanded form notation for the decimal numbers is the mathematical expression that shows the sum of the values of each digit in the number.
2. What is the Decimal Expansion of Number 164.38?
Decimal Number 164.38 can be written in expanded form by writing it as the sum of the place value of all the digits i.e. 1*100+6*10+4*1+3*$$\frac { 1 }{10 }$$+8*$$\frac { 1 }{100 }$$ = 100+60+4+0.3+0.08
3. What is the Fractional Expansion of Number 94.38?
Fractional Expansion of Number 94.38 is 9*10+4*10+3*$$\frac { 1 }{10 }$$+8*$$\frac { 1 }{100 }$$ which inturn results in 90+40+$$\frac { 3 }{10 }$$+$$\frac { 8 }{100 }$$
|
Q:
# Consider the relationship between the number of bids an item on eBay received and the item's selling price. The following is a sample of 5 items sold through an auction.Price in dollars 23 29 32 38 41Number of Birds 10 14 15 16 17A) Step 1 of 3: Draw a scatter plot of the given data.B)Step 2 of 3: Estimate the correlation in words: positive, negative, no correlation.C) Step 3 of 3: Calculate the correlation coefficient, r. Round your answer to three decimal places
Accepted Solution
A:
Answer:Step 1: Figure attachedStep 2: Positive correlationStep 3: r=0.9534Step-by-step explanation:Step 1For this step we can use excelWe click on Insert> Insert Scatterpplot(X,Y) or Bubblechart. We select the data and then we select the Layout 9. Then we can edit the title and the names for the x and y axis.The result obtained is on the Figure attached.Step 2As we can see on the plot obtained the slope obtained is positive since is increasing, so with this result we can conclude that the correlation would be positive and possible strong. We need to compute it in order to be sure of this.Step 3In order to calculate the correlation coefficient we can begin doing the following table:n x y xy x*x y*y1 23 10 230 529 1002 29 14 406 841 1963 32 15 480 1024 2254 38 16 608 1444 2565 41 17 697 1681 289$$n =5 \sum x \sum y \sum xy \sum x^2 \sum y^2$$n=5 $$\sum x = 163, \sum y = 72, \sum xy 2421, \sum x^2 =5519, \sum y^2 =1066$$ And in order to calculate the correlation coefficient we can use this formula:$$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$$$$r=\frac{5(2421)-(163)(72)}{\sqrt{[5(5519) -(163)^2][5(1066) -(72)^2]}}=0.9534$$So then the correlation coefficient would be r =0.9534
|
Consolidation of Steps 5-7 Year 3 Multiplication and Division Learning Video Clip – Classroom Secrets | Classroom Secrets
# Consolidation of Steps 5-7 Year 3 Multiplication and Division Learning Video Clip
## Step 5-7: Consolidation of Steps 5-7 Year 3 Multiplication and Division Learning Video Clip
Vodi and Efna are helping to get the village ready for winter by counting out the stores of food. Pupils will help Vodi and Efna divide the fish between the stores, work out how many baskets of vegetables each house can have, complete a calculation with missing numbers, discover a mistake Efna has made in her division and finally help Vodi to work out how many bags of nuts and seeds will be left over after being shared between the houses.
This Learning Video Clip has been designed as a consolidation tool for steps 5 and 7. It contains content relevant to all these steps and can be used in parts to recap a particular step or in on its own at the end of a teaching sequence for the included steps.
More resources for Spring Block 1.
Not a member? Sign up here.
Discussion points for teachers
1. Vodi solves the problem using place value counters. Do you agree?
Discuss what Vodi has shown. Discuss the place value chart and how many counters are in each row and their value.
Vodi is correct. He has correctly shown the division calculation using counters.
2. Use Vodi’s method to work out the answer to the new sum.
Discuss the method Vodi has used and use the same method to work out the answer using the new numbers.
88 ÷ 4 = 22
3. Help Efna to work out the answer.
Discuss the sum that Efna needs to solve. Discuss the methods that could be used.
11 baskets for each house.
4. How many bags of nuts can each house in Skara Brae use this winter?
Discuss and model the methods of working out the answer. Discuss facts we already know about the sum.
32 ÷ 8 = 4
5. Help Vodi by completing the missing numbers.
Discuss the method used and how to break down the calculation.
45 ÷ 3 can be written into parts; 30 ÷ 3 and 15 ÷ 3. The answer is 15.
6. What mistake has she made?
Discuss the number line and how Efna has tried to work out her answer. Discuss the error on the number line. Discuss how you know the answer is wrong.
Efna has counted incorrectly on the number line. The answer should be: 12 r1
7. Help Vodi to share them between the 8 houses. What is the remainder?
Discuss the different methods to use and discuss how they know there will be a remainder. Discuss if there were 49 before or after Vodi ate one and how this will affect the answer.
49 ÷ 8 = 6 r1
National Curriculum Objectives
This resource is available to play with a Premium Subscription.
|
# Solving equations and inequations
Remember when solving equations and to use the rule:
Change side, change operation
Look at the National 4 solving equations section before continuing.
### Example
Solve the equation $$- 2 - 3y = 11$$
$- 2 - 3y = 11$
$- 3y = 11 + 2$
Move the -2 over to the right hand side changes it to +2
-3 is multiplying on the left, so when moved to the right it divides
$- 3y = 13$
$y = \frac{{13}}{{ - 3}}\,or\,y = - 4\frac{1}{3}$
Now try the question below.
Question
Solve the equation $$35 = 5 - 6m$$ to find m.
$35 = 5 - 6m$
Move the -6m over to the left hand side, changing it to +6m.
$35 + 6m = 5$
Move the +35 to become -35
$6m = 5 - 35$
$6m = -30$
The 6 is multiplying on the left so now divide
$m = \frac{{ - 30}}{6} = \frac{{ - 5}}{1} = - 5$
|
# Calculate the Partial Sum of each Arithmetic Sequence or Series Given Two Terms
Problem 1. Find the sum of first 51 terms of an arithmetic sequence whose second and third terms are 14 and 18 respectively.
Solution:
Let a be the first term and d be the common difference of the arithmetic sequence. Then,
d=a3a2=18-14=4
Given a2=14.
[an=a+(n-1)d]
a+d=14
a+4=14
a=14-4=10
Using the formula, Snn[2a+(n-1)d], we get
S51=½⋅51⋅[2⋅10+(51-1)⋅4]
=½⋅51⋅(20+200)
=½⋅51⋅220
=5610.
Hence, the required sum is 5610.
P2. The 13th terms of an arithmetic sequence is 4 times its 3rd term. If its 5th term is 16, Find the sum of its first 10 terms.
Solution:
Let a be the first term and d be the common difference of the arithmetic sequence. Then, a13=4⋅a3 (Given).
[an=a+(n-1)d]
a+12d=4(a+2d)
a+12d=4a+8d
3a=4d … (1)
Also, a5=16 (Given)
a+4d=16 … (2)
Solving (1) and (2), we get
a+3a=16
4a=16
a=4
Putting a=4 in (1), we get
4d=3⋅4=12
d=3
Using the formula, Snn[2a+(n-1)d], we get
S10=½⋅10⋅[2⋅4+(10-1)⋅3]
=5⋅(8+27)
=5⋅35
=175.
Hence, the required sum is 175.
P3. The 16th term of an arithmetic sequence is 5 times its 3rd term. If its 10th term is 41, find the sum of its first 15 terms.
Solution:
Let a be the first term and d be the common difference of the arithmetic sequence. Then,
a6=5⋅a3 (Given)
[an=a+(n-1)d]
a+15d=5(a+2d)
a+15d=5a+10d
4a=5d … (1)
Also,
a10=41
a+9d=41
a=41-9d … (2)
Solving (1) and (2), we get
(1) … 4(41-9d)=5d
164-36d=5d
164=5d+36d
164=41d
d=4
(2) … a=41-9⋅4
=41-36
a=5
Using the formula, Snn[2a+(n-1)d], we get
S15=½⋅15⋅[2⋅5+(15-1)⋅4]
=15⋅½⋅(10+56)
=15⋅33
=495.
Hence, the required sum is 495.
Let’s read an important post Derivation of the partial sum formula of every Arithmetic Series.
P4. The 13th term of an arithmetic sequence is 4 times its 3rd term. If its 5th term is 16 then the sum of its first ten terms is
(a) 150 (b) 175 (c) 160 (d) 135
Solution: Let a be the first term and d be the common difference of the arithmetic sequence. Then,
Given a13=4a3.
[an=a+(n-1)d]
a+12d=4(a+2d)
a+12d=4a+8d
3a=4d … (1).
Given a5=16
a+4d=16 … (2)
Solving (1) and (2), we ge
a+3a=16
4a=16
a=4.
Putting a=4 in (1), we get
4d=3⋅4=12
d=3.
Using the formula, Snn[2a+(n-1)d]. we get
S10=½⋅10⋅[2⋅4+(10-1)⋅3]
=5⋅(8+27)
=5⋅35
=175.
Thus, the sum of its first 10 terms is 175.
|
# How do you write the inequality and solve given "twice a number increased by 3 is less than the number decreased by 4"?
Jun 2, 2017
See a solution process below:
#### Explanation:
To write the inequality we will start by defining "a number" as $n$.
Then, "twice a number" can be written as $2 n$.
"increased by 3" becomes: $2 n + 3$
"is less than" turns it into" $2 n + 3 <$
"the number decreased by 4" is: $n - 4$
Putting this together gives the inequality:
$2 n + 3 < n - 4$
To solve, first subtract $\textcolor{red}{3}$ and $\textcolor{b l u e}{n}$ from each side of the inequality to solve for $n$ while keeping the inequality balanced:
$- \textcolor{b l u e}{n} + 2 n + 3 - \textcolor{red}{3} < - \textcolor{b l u e}{n} + n - 4 - \textcolor{red}{3}$
$- \textcolor{b l u e}{1 n} + 2 n + 0 < 0 - 7$
$\left(- \textcolor{b l u e}{1} + 2\right) n < - 7$
$1 n < - 7$
$n < - 7$
Jun 2, 2017
Any number less than $- 7$ will make this inequality true.
$x < - 7$
#### Explanation:
Let the number be $x$
Write the left side as maths first:
$\textcolor{red}{\text{Twice a number ") color(blue)(" increased by 3 ") color(magenta)("is less than}}$
$\textcolor{red}{2 x} \textcolor{w h i t e}{w w w w w w w w w w w w w} \textcolor{b l u e}{+ 3} \textcolor{w h i t e}{w w w w w w w} \textcolor{m a \ge n t a}{<}$
Now do the same for the right side..
color(red)(2x)" " color(blue)(+ 3 ) " "color(magenta)( <)" "color(green)("the number decreased by 4"
color(red)(2x)" " color(blue)(+ 3 ) " "color(magenta)( <)" "color(green)(x-4)
We have the inequality, now solve it:
$2 x + 3 < x - 4$
$2 x - x < - 4 - 3$
$x < - 7$
|
# How to Read a Fraction
Fraction notation and operations may be the most abstract math monsters our students meet until they get to algebra. Before we can explain those frustrating fractions, we teachers need to go back to the basics for ourselves. First, let’s get rid of two common misconceptions:
• A fraction is not two numbers.
Every fraction is a single number. A fraction can be added to other numbers (or subtracted, multiplied, etc.), and it has to obey the Distributive Law and all the other standard rules for numbers. It takes two digits (plus a bar) to write a fraction, just as it takes two digits to write the number 18 — but, like 18, the fraction is a single number that names a certain amount of whatever we are counting or measuring.
• A fraction is not something to do.
A fraction is a number, not a recipe for action. The fraction 3/4 does not mean, “Cut your pizza into 4 pieces, and then keep 3 of them.” The fraction 3/4 simply names a certain amount of stuff, more than a half but not as much as a whole thing. When our students are learning fractions, we do cut up models to help them understand, but the fractions themselves are simply numbers.
## Learn to Read a Fraction
A fraction is a number that relates to whatever is defined as one whole thing. Before a fraction can have any meaning, something has to be defined as “one whole unit.” Of course, this is true of all numbers, not just fractions. Before three can have any meaning at all, one has to be defined. Are we talking about 3 miles or 3 bags of rice or 3 spaces on the number line? Natural numbers are defined in terms of whatever one is, so fractions are no different from other numbers in this respect.
At the beginning, students need to make extensive use of fraction models. Let’s not limit ourselves to the round food model, but it is a great place to start. For the rest of this post, you may imagine “one whole unit” to be a single pizza, cut up into fractional pieces.
The first confusing thing about fractions is vocabulary: those awful words numerator and denominator. To keep them straight, our students must learn how to read a fraction. Teach your children to read fractions from top to bottom, like the pages in a book:
• The top number of a fraction is its first name, which tells how many pieces you have.
• The line in a fraction means can be thought of as “divided by”, so 3/4 = 3 ÷ 4 and 5 ÷ 2 = 5/2.
• And the bottom number is the fraction’s last name or family name. It tells what size the pieces are.
Now, let’s go through that a little slower…
## Numerator: The Fraction’s First Name
The top number in a fraction is its first name — that is, the first thing you hear about the fraction. A fraction’s first name counts the number of pieces: 2/5 has two pieces, and 3/5 has three of them.
This makes it easy to count up the total number of pieces in a bunch of fractions:
$\frac{2}{5} + \frac{1}{5} + \frac{4}{5} = \frac{7}{5}$
And if you add another fifth, that would make 8/5. Add a few more to get 11/5, then maybe take some away:
$\frac{11}{5} - \frac{3}{5} - \frac{6}{5} = \frac{2}{5}\:,$ counting down this time.
You can keep counting up and down forever, as long as you are only talking about fifths.
Mnemonic: It may help your students to think “number-ator.” The numerator counts the number of pieces we have.
## Denominator: The Fraction’s Family Name
[Yes, this is out of order. We want our students to meet and get to know the fractions, so for beginners we focus on the names. Learning about the fraction bar comes later.]
The bottom number of a fraction is its last name, or family name. All fractions with the same last name are in the same family. The family name tells what size the pieces are — which is the same as saying how many of the pieces make one whole thing. (Remember that fractions are always defined in relationship to some whole thing.) With fourths, it takes four of them to make a whole. With sevenths, it takes seven of them to make a whole thing.
This is why, as the last name gets bigger, the fraction itself gets smaller. The more pieces your whole thing has been cut into, the smaller each piece will be.
$\frac{2}{4} > \frac{2}{7}$
…because both have two pieces, but 7ths are smaller pieces than 4ths.
A man is like a fraction whose numerator is what he is and whose denominator is what he thinks of himself. The larger the denominator, the smaller the fraction.
Mnemonic: It may help your students to think “down-nominator.” The denominator is the number down in the bottom of a fraction.
## Put Them Together, and What Do You Get?
So what number is the fraction? It is the number that names the amount of stuff in all those pieces.
• The fraction 3/4 names the amount of pizza in 3 pieces, when each piece is 1/4 the size of one whole pizza.
• The fraction 13/5 names the amount of pizza in 13 pieces, when each piece is 1/5 the size of one whole pizza.
• The fraction 8/8 names the amount of pizza in 8 pieces, when each piece is 1/8 the size of one whole pizza.
But it is important to remember that the fraction itself is NOT the same as those pieces of pizza. If I had 3/4 of a pizza, it would be the same amount of stuff even if it had not been cut up — for instance, if my daughter came through and cut out as much of the pizza as she wanted, without bothering to cut the rest apart. And the whole pizza is 8/8 of a pizza, even before the first cut is made.
As we work our way toward a solid understanding of fraction operations, we will often come back to our fraction models to ask these questions:
1. What size are the pieces?
2. How many of them do we have?
## The Line in a Fraction Can Be Read “Divided By”
Every fraction is also a division problem — or rather, it is the answer to a division problem. For instance, the fraction 7/3 names the amount of pizza each person would get, if we divided 7 pizzas evenly among 3 teenage boys:
$7 \div 3 = \frac{7}{3}$
Think of the division symbol “÷” as a miniature picture of a fraction. The dots stand for the numerator and denominator.
This is why the remainder in a division problem turns into a fraction. Consider those 3 teenage boys fighting over the 7 pizzas. It is easy to give them each 2 whole pizzas, but what shall we do with the remaining one? We have to cut it up, and each boy will get 1 ÷ 3 of it.
$7 \div 3 = \frac{7}{3} = 2 \frac{1}{3}$
Whatever our remainder is, that becomes the numerator of our remainder-fraction. The denominator is whatever we were dividing by.
Knowing that the line in a fraction means “divided by” will be important in solving complex fractions like this MathCounts puzzler:
$\frac{1}{1 + \frac{1}{1 + \frac{3}{4}}} = \; ?$
…or when our students meet algebra monster fractions like:
$\frac{16\left({x}^{2} - 81 \right)}{x + 9} = \; ?$
So we will keep reminding our students of this fact throughout middle school or junior high:
A fraction is can be written as a division problem.
## 27 thoughts on “How to Read a Fraction”
1. Article on fractions revealed overcomed certain misconcepts. The students have many misconception about fractions. After reading your article , I am able to deliver the concept very easily to my students. Thanks for giving such a good article.
1. Chris Opio says:
Hi my name isChris and I love your book in maths
2. The form of a/b does not automatically mean it is a fraction. So we must not teach children to say just because it is 2/3 so it is a fraction.
It is also debatable to say that 2/3 is just an one number. just like 23 can be expanded as 2 numbers as well. The form 2/3 is simply an invented notion to express part out of whole and that is it. This form is an most effective way of expressing part out of whole concept.
2/3 can be expressed as picking 2 out of 3 and this is another form of “fraction”, it is equivalent to say “cut equally 3 parts and 2 parts are chosen” in pure number sense but in real life experience the meaning might be different. so the form of a/b without any explanation at all, we are just not sure what it really means, it can be a fraction, can be a pure number, can be 2 chose out of 3 and can be 2 out of whole piece being cut, and can be a 2 compared to 3 as a ratio.
The article is confusing if it does not explain the form of a/b from a broad sense but instead try to narrowly define a/b as simply a fraction.
3. The next article in this series will discuss the various models we use to understand fractions, such as pieces of a whole thing (the model used in this article) or parts of a group (such as picking 2 out of 3).
For a discussion about expanding fractions into more than one number (like 23=20+3), keep your eyes open for the Alexandria Jones series on Egyptian fractions, coming in January 2008.
4. Nice explanation Denise! You must get a lot of students thinking about slices to repeatedly reinforce things ” …would be the same amount of stuff even if it had not been cut up…” What is the common mistake that your students make here? Do they just always think one out of two pieces is 1/2 even if the piece is only a quarter pizza?
Frank: Not sure why the big focus on what a/b means. Doesn’t it always mean a fraction? I’ve never heard it referred to as “parts of the whole” except as an explanation, which to me is just a long way of saying “fraction” or “division” anyway. a/b evaluates to a number all the time if a and b are numbers, doesn’t it?
I think it is instructive to teach a/b as a way of writing a single number, especially if you pair it eventually with numbers that cannot be written in decimal. eg, 0.333…. can be written exactly as 1/3. Not sure at which grade that occurs.
5. One problem my young students have is understanding how a fraction can be bigger than a whole unit. How can there be such a thing as 8/5, when you only have 5 pieces to begin with? (I wrote a blog post awhile back about my youngest daughter’s traumatic encounter with just such a concept.)
When I look at the confusing questions in that frustrating fractions quiz, I think most of them have to do with not understanding that a fraction is a single name for a certain amount of stuff, and thus it has to be treated as one (admittedly rather weird) number. Students who think of a fraction as two numbers will see no problem with a jumbled, mis-remembered rule like, “To add two fractions, I add the top numbers and add the bottom numbers.”
Also, by defining fractions as a certain amount of stuff, I hope to broaden my readers’ vision to see beyond the round food model. Many in my audience are homeschooling parents with no training in math other than what they studied in school years ago. If the teachers in Liping Ma’s study were so limited in their understanding of fractions, how much more limited are those of us who studied under such teachers.
6. I think it’s also useful to think of a fraction as either “a division problem” or “the answer to a division problem”. Thinking of a fraction as a certain representation of a given number or “amount of stuff” is useful too.
7. cicik says:
how to read 3/4 three fourths or three quarters, what about 6/7 – six sevenths or six over seven ? tx
8. Cicik, any of those readings will work, although I think “three fourths” and “six sevenths” communicate the meaning best. They are parallel to measurements like “three inches” or “six grams.” The “this over that” reading becomes more common in algebra, where you will find yourself working with very complex fractions.
9. monica says:
what??? Im 22 and dont even understand fractions…im so confused and trying to surff the internet to see if i can figure it out but nothing…
10. monica says:
okay, i dont know but heres the problem 7/12 ( , =) 2/3…i wrote the pizza circle out and 7/12 is greater…right? also i put it in the computer but it showing me wrong….?
11. Vic says:
Your explanation on fractions is really helpful for me to teach a struggling math student! 🙂 He understands it much faster when I used your method. Thank you!
12. morlon says:
What is the relationship between fractions of the same family. Using three family
proper fraction, improper fraction and mix fraction
13. Morlon, I am not sure quite what your questions means. Fractions of the same family (as I am using the term) means fractions with the same size pieces. In proper mathematical language, these are fractions with the same denominator — that is, the same bottom number.
For the other words you mentioned:
proper fraction = a fraction in which the numerator (top number) is smaller than the denominator
improper fraction = a fraction in which the numerator is not smaller than the denominator
mix fraction = I’m not sure, but it’s probably what I would call a “mixed number”, which means you write a whole number and a fraction together as an implied sum, like $3 \frac{1}{2}$.
1. Kim Beezhold says:
Thank you for using the proper names. Did you put a glossary in your new book?
Those technical words are very helpful for fractions.🙂
14. omar says:
i thank you for teaching mii more about fractions
15. Jacki says:
In the beginning of this post you say, “A fraction is not something to do”, but at the end of the post you say, “A fraction is a division problem”. Isn’t that contradictory?
16. You’re right, Jackie, that sounds like a contradiction. Perhaps I should have said it this way: “A fraction can be written as a division problem.”
17. Kelsey Hanigan says:
What a great explanation of fractions. Sometimes it helps to hear things in a different way to help you become a better teacher, thank you for that. I loved the quote from Tolstoy! I did find the section on “Put them together and what you get” to be a little confusing…maybe there is another way to explain that…what about a visual!
18. Favour says:
Please help me I know that 3/4 of 8 balls is 6 by 3*8 will give you 24/4 will give ,6 but how will I explain it practically?
1. One of the best ways to think through a fraction puzzle is what Ed Zaccaro calls “Think 1.” Can you think of what 1/4 of your set would be? Then use that to figure out how much would be three of those fourths.
19. Lauren Schweitzer says:
Learning how to read a fraction and knowing how to do the basic mathematical operation on it is a lesson that we should never miss and a must to learn.. It is a foundation that should be instilled on the minds of a young students. But sometimes doing it manually may result in a wrong answer. That is why I used a tool to confirm that the computation is correct. For checking if the calculation of fractions is correct, I used an app which shows the correct answer and display the step by step solution. It is actually a fraction calculator from http://www.fractioncalc.com. It is very useful in solving fraction and a very good tool to be used both by teacher and student for reference if their answer is correct and if their solution is right.
This site uses Akismet to reduce spam. Learn how your comment data is processed.
|
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 4.6: Summary
[ "article:topic", "authorname:thangarajahp", "license:ccbyncsa", "showtoc:yes" ]
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
#### Algebra
The algebra of a given object has rules and procedures specific to that object - the operations and rules of sets aren't exactly the same as the ones of numbers or statements. Taking mathematics in primary and secondary education, we are familiar with the algebra of numbers. Now we can apply those same thought processes to other areas of mathematics to prove statements or identities, and determine solutions for abstract problems.
Example $$\PageIndex{1}$$:
An "object" might be said to be "all numbers", or "the Universal set containing all numbers." In algebra, we know that we can do operations to numbers: addition, subtraction, multiplication, and division, among others.
These operations are closed to numbers: when we do addition with numbers, we will always receive a number as the result.
This is true of other objects also: statements, sets, and geometric shapes are all objects with their own distinct operations and properties.
In mathematics, we view algebra as the study of relationships between objects, as well as the study of groups of objects. We are already familiar with some relationships between specific quantities.
Example $$\PageIndex{2}$$:
In physics:
• $$F = ma$$, where $$F$$ is force, in newtons, $$m$$ is mass, in kg, and $$a$$ is acceleration
• $$v = \displaystyle \frac{d}{t}$$
We also know that these formulas can be manipulated algebraically - that is, that the formula defines the relationship between the elements of the formula, and can be used to determine any of the elements as required.
#### Some Objects, Operations, and Properties
Objects Numbers Statements Sets Geometric shapes Operations $$+,-,\cdot, \div$$ $$\neg,\vee,\wedge, \rightarrow$$ $$\subset, \subseteq, ^c, \cup, \cap$$ Reflection, rotation, translation Properties Closed Distributive Closed Distributive Closed Distributive Closed Distributive
Closed operations are those that generate an answer in the same group as the elements operated upon.
Distributive operations are those that, in a mathematical sentence, yield the same result when FOILed into a bracket as when not.
Example $$\PageIndex{3}$$:
$$\neg \left( A \vee B\right)$$ = $$\neg A \wedge \neg B$$
$$2 \left( 3 + 6 \right) = 2 \left( 3 \right) + 2 \left( 6 \right)$$
#### The Algebra of Geometry
In geometry, our operations are specific to the type of geometry we are doing. This means that the operations we use in 2-dimensional Euclidean geometry might not necessarily be the same as those we apply in 3-dimensional Euclidean or non-Euclidean geometry. For us, we will consider the 2-dimensional transformations as operations. Translations, reflections, and rotations follow the same rules as other operations: they are closed (when you move a shape, the result is a shape) as well as distributive (the order in which you move, rotate, or reflect a shape has no effect on the result).
|
2002 AMC 12A Problems/Problem 10
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
The following problem is from both the 2002 AMC 12A #10 and 2002 AMC 10A #17, so both problems redirect to this page.
Problem
Sarah places four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then pours half the coffee from the first cup to the second and, after stirring thoroughly, pours half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?
$\mathrm{(A) \ } \frac{1}{4}\qquad \mathrm{(B) \ } \frac13\qquad \mathrm{(C) \ } \frac38\qquad \mathrm{(D) \ } \frac25\qquad \mathrm{(E) \ } \frac12$
Solution
Solution 1
We will simulate the process in steps.
In the beginning, we have:
• $4$ ounces of coffee in cup $1$
• $4$ ounces of cream in cup $2$
In the first step we pour $4/2=2$ ounces of coffee from cup $1$ to cup $2$, getting:
• $2$ ounces of coffee in cup $1$
• $2$ ounces of coffee and $4$ ounces of cream in cup $2$
In the second step we pour $2/2=1$ ounce of coffee and $4/2=2$ ounces of cream from cup $2$ to cup $1$, getting:
• $2+1=3$ ounces of coffee and $0+2=2$ ounces of cream in cup $1$
• the rest in cup $2$
Hence at the end we have $3+2=5$ ounces of liquid in cup $1$, and out of these $2$ ounces is cream. Thus the answer is $\boxed{\text{(D) } \frac 25}$.
Solution 2
Let's consider this in steps. We have 4 ounces of coffee in the first cup. We hace 4 ounces of cream in the second cup. We take half of the coffee in the first cup(2 ounces), and add it to the second cup, yielding 6 ounces in total in the second cup(in a 1:2 ratio between coffee and cream, respecitvely). We then take half of the second cup and pour it into the first cup. $6/2=3$, so there is now 5 ounces in the first cup, 2 coffee and 3 the mixture. Remember that the mixture is in a 1:2 ratio between coffee and cream. So, coffee has one ounce and cream has 2 ounces. In total, there is 2 ounces in the 5 ounce first cup. Putting 2 over 5, we get the answer. Therefore, the answer is $\boxed{\text{(D) } \frac 25}$.
~MathKatana
Video Solution
2002 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 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 All AMC 12 Problems and Solutions
2002 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 16 Followed byProblem 18 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 All AMC 10 Problems and Solutions
|
# How do you solve 10 divided by 3?
Answer: 10 divided by 3 as a fraction is 10/3. Let us write 10 divided by 3 as a fraction. Explanation: 10 divided by 3 can be written as 10/3.
## What can 10 be divided by?
The factors of 10 are 1, 2, 5, and 10. You can also look at this the other way around: if you can multiply two whole numbers to create a third number, those two numbers are factors of the third. 2 x 5 = 10, so 2 and 5 are factors of 10. 1 x 10 = 10, so 1 and 10 are also factors of 10.
## What is 100 divided by 3 exactly?
The division 1÷3 is now 1÷10 which is equal to 0.1. so you see writing (in base ten) 100÷3=33.333333 does not mean that we cannot divide 100 into three equal parts.
## How do you write 19 divided by 3?
See also Is a dessert spoon the same size as a tablespoon?
Using a calculator, if you typed in 19 divided by 3, you’d get 6.3333. You could also express 19/3 as a mixed fraction: 6 1/3.
## How do you work out 98 divided by 7?
Multiply the newest quotient digit (4) by the divisor 7 . Subtract 28 from 28 . The result of division of 98÷7 98 ÷ 7 is 14 .
## What kind of fraction is 4 4?
The fraction 4/4 is an improper fraction. The definition of a proper fraction requires that the numerator be less than the denominator.
## What is unlike fraction?
Unlike fractions are fractions that have different denominators. Examples. The first fraction below has a denominator of two and the second fraction below has a denominator of three. Since the denominators are different, they are unlike fractions.
## Is 10 3 an improper fraction explain?
Yes, 103 is an improper fraction. An improper fraction is one in which the numerator of the fraction is larger than the…
## What is 3.3 as a percent?
Make use of the Free Decimal to Percent Calculator to change decimal value 3.3 to its equivalent percent value 0.033% in no time along with the detailed steps.
## How do you write 2 and a half in a fraction?
You would say “two and one half.” The other format is an improper fraction where the numerator is greater than the denominator (5/2). Mathematicians would say that is five halves. You will find both types of fractions in your problems. Both of those examples represent the same value (2 1/2 = 5/2).
## How do you write 2 halves as a fraction?
A half is 1/2 of the whole. 1/2 is 1 out of 2 equal parts. 👉 2 halves make one whole. As Nick is about to cut the pie, two more friends join them.
## What is 3/10 as a decimal and percent?
To convert fraction to decimal, divide 3/10, which would give you . 3 which is already a decimal. To convert decimal to percent, multiply by 100 which would give you 30, hence 30%.
## What is conjugate in math?
A math conjugate is formed by changing the sign between two terms in a binomial. For instance, the conjugate of x + y is x – y. We can also say that x + y is a conjugate of x – y. In other words, the two binomials are conjugates of each other.
## How many ways can you divide 100?
We can divide 100 into as many equal groups as the no. of factors. Therefore, we can share 100 marbles into equal groups in 9 ways.
## Which number is divided by 2?
All even numbers are divisible by 2. Therefore, a number is divisible by 2 if it has a 0, 2, 4, 6, or 8 in the ones place. For example, 54 and 2,870 are divisible by 2, but 2,221 is not divisible by 2.
## Is 12 divisible by 3 yes or no?
Since the answer to our division is a whole number, we know that 12 is divisible by 3. Hopefully now you know exactly how to work out whether one number is divisible by another.
|
# Chapter5Notes-VectorsinMechanics
```VECTORS
1. Vectors in Mechanics
We have seen already that speed and velocity are very similar concepts but that velocity is a
vector whilst speed is a scalar. Velocity, the vector quantity, contains more information as it also
indicates the direction of motion as well as the actual speed. We say that it has both magnitude
(size) and a direction. Speed contains only information about the magnitude. Our concept of
velocity is, however, still quite limited because we have only looked at how to represent the
direction by considering a change of sign. This has limited our work on velocity to acting along a
line in 1 dimension.
Consider, now, the following situation involving velocity in 2 dimensions:
A bird is flying due south at a speed of 4ms-1, but it is caught in a wind blowing east at 3ms-1.
What is the bird’s actual velocity?
Sketching a picture to show what happens to the birds flight over a 1 second interval. The bird
heads 3m east and 4m south.
N
Using Pythagoras’ Theorem the distance travelled in 1s is
d 32 4 2
d 5m
The speed of the bird over the ground is therefore 5ms-1
3m
θ
4m
The angle θ is given by
tan
4
3
tan 1 43
53.1o
Therefore the bearing is 90° + 53.1° = 143.1°.
So the birds velocity is 5ms-1 on a Bearing of 143.1°.
Now consider the following situations involving displacement and forces:
If you walk 3m east and then 4m south, how far and in what direction will you be from
A sledge is being pulled with forces of 3N to the east and 4N to the south. What is the
combined effect of these forces?
It is easy to deduce that the answer to both of these questions will contain a magnitude of 5 at an
angle of 143.1° from North.
2. Vector Notation
We can represent a vector in 3 different ways.
Consider the following situation to describe the velocity of our bird from our first example. It
had a seed of 5ms-1 on a Bearing of 143.1°.
We can represent its velocity as an arrow of length 5 units drawn at the correct angle
N
143.1
We could represent its velocity by splitting it into component parts
v = 3i – 4j
(where i represents the quantity in the horizontal direction and j represents the quantity in the
vertical direction)
We could represent its velocity as a column vector
3
v =
4
We may refer to the vector by a single letter ‘v’ or by using the labels at the end points
Eg. The vector joining A to B is denoted AB , whereas the distance from A to B is denoted AB (so
the arrow above the letters indicates a vector quantity rather than a scalar. It also indicates the
direction, from A to B).
Vectors that describe the same ‘movement’ or ‘resultant’ are equivalent. Equivalent vectors will
have the same magnitude AND the same direction. Vectors like this have no specific location
and are sometimes called ‘free vectors’. Vectors that are used to describe a position based on a
coordinate grid are called ‘position vectors’. So in the diagram below, AB PQ .
B
A
Q
P
Note that we can consider vectors in 3d by extending these ideas. It is obviously difficult to draw
vectors in 3d as they would need to be represented by a line pointing in 3d space, but you can
easily add an extra column to the column vector format. Using components we simply introduce
an extra letter k to represent the 3rd direction that we can move in) Eg v = 3i – 4j + 5k
3. Vector Arithmetic
Consider the vectors
AB = -3i + 4j
CD = 5i + 2j
a) Multiplying a vector by a number (scalar)
This changes the magnitude of the vector, but not its direction.
In in the diagram below, The vector AB has been multiplied by two:
PQ = 2 AB
Q
B
A
P
Using unit vector notation
we say that PQ = 2(-3i + 4j)
= -6i + 8j
Using column vector notation
3
we say that PQ = 2 =
4
6
8
In general, if AB k PQ , then AB and PQ are parallel, with AB being k times longer than
PQ .
Multiplying a vector by a negative number changes the magnitude of the vector, and reverses
its direction. So in the diagram below PQ = -2 AB
Q
B
A
P
In general the return journey from B to A is the exact opposite of the journey from A to B, so we
can say that AB BA .
To add two vectors together we place the beginning of one vector
at the end of the other. The resultant is the vector formed by
adding the two vectors together. In the diagram opposite,
OB OA AB
A
O
In the example you can see how we are
adding AB and CD together using a
diagram.
B
D
C
We can also add them together using unit
vector notation:
B
AB + CD = (-3i + 4j) + (5i + 2j)
= 2i + 6j
A
Or using column vector notation:
3 5 2
AB + CD =
4 2 6
b) Subtracting two vectors
To subtract two vectors we consider adding the ‘negative’ of the second vector to the first.
In the example you can see how to find
AB - CD using a diagram. We are in fact
finding AB + - CD
D
We can also add them together using unit
vector notation:
AB - CD = (-3i + 4j) - (5i + 2j)
= -8i + 2j
Or using column vector notation:
3 5 8
AB - CD =
4 2 2
Exercise 5B page 88 q 1 - 11
C
B
A
4. Vector Geometry
Activity 1
Vector Geometry powerpoint
We can use the rules for adding vectors to form a kind of ‘algebra of vectors’, known as vector
geometry.
B
C
Example 1 : The diagram shows a parallelogram OACB with
b
OA a and OB b . D is the midpoint of AC.
Express the following vectors in terms of a and b.
O
a) BC
b) AC
c) CA
d) OC
e) AB
D
A
a
f) OD .
a) Since OA and BC are equal and parallel, BC OA a
b) Since AC and OB are equal and parallel, AC OB b
c) Since CA is the return journey of AC, CA AC b
d) To get from O to C, we can go from O to A to C, so OC OA AC a b
e) Similarly, we have AB AC CB b a .
f) The journey from A to D is half the distance from A to C, so we have
OA 12 AC
a 12 b
S
Example 2 : The trapezium PQRS is such that PQ 2SR . X is
the midpoint of PQ, and Y is the midpoint of QR. If a and b
are defined as in the diagram opposite, express XY in terms of
a and b.
b
R
Y
a
P
X
Q
Now
XY XQ QY
12 PQ 12 QR
b 12 QP PS SR
b 12 (2b a b)
12 (a b)
Example 3 : In the triangle PQR, PQ a and PR b . The point X is on the line QR such that
QX : XR 2 : 3 . Find PX in terms of a and b.
It helps to draw a diagram.
PX PQ QX
Q
PQ 52 QR
2
PQ 52 QP PR
X
a
a (a b)
2
5
3
15 (3a 2b)
P
b
R
Exercise 5A page 82 q 1 - 11
5. Resolving Vectors
Activity 2
Resolving Vectors powerpoint
Use this activity to introduce the idea of splitting a vector up into its component parts – a
horizontal (i) part and a vertical (j) part
Example 1 : A vector R is of magnitude 16km and on a bearing of 030°. Taking i as a unit vector
due east and j as a unit vector due north, write R in the form ai + bj.
a 16sin 30
8
b 16 cos 30
8 3
R 8i 8 3 j
R
30
16
bj
ai
Example 2 : A vector R is of magnitude 20km and on a bearing of 210°. Taking i as a unit vector
due east and j as a unit vector due north, write R in the form ai + bj.
a -20sin600
= -10 3
b = -20cos600
= -10
R = -10 3 i – 10j
Example 3 : Two Forces P and Q with magnitude 4 and 5 respectively act as shown on a partcile
Q
5
P
30
o
4
60o
Find the magnitude and direction of the resultant force P + Q
It is not obvious from the diagram
what the precise size and direction
of the resultant force will be.
Q
5
P+Q
60o
P
o
4
30
Instead of solving this problem with a diagram, we resolve both forces into
component parts and use the following facts:
The vertical component of P + Q will be the same as the vertical component of P
plus the vertical component of Q
The horizontal component of P + Q will be the same as the horizontal component of
P plus the horizontal component of Q
Considering the vertical components
↑ 4sin30 + 5sin60 = 6.33
Considering the horizontal components → 4cos30 – 5cos60 = 0.96
So P + Q = 0.96i + 6.33j
The magnitude of the resultant is
The direction is
0.962 6.332 = 6.4
33
tan θ = 06..96
θ = 81.4o relative to the positive i direction
Exercise 5D page 95 q 1 - 12
6. Finding the magnitude and direction of a vector in 2d and in 3d
The magnitude of a vector a is its length or size. We denote the magnitude of vector a as a . We
can find the magnitude of a vector by using pythagoras. This works in both 2d and 3d. A unit
vector is a vector of length 1 unit, that points in the same direction as the original vector. To
change a vector into a unit vector simply divide it by its magnitude.
1
Example 1 : Find the magnitude of the vector 2 and find a unit vector that points in the
3
same direction.
Magnitude = 12 (2) 2 3 2
= 14
1
A unit vector in the same direction as 2
3
would be given by
1
14
2
14
3
14
Example 2 : Find the magnitude of the vector a = 3i + 4j. Hence find a unit vector in the
direction of a.
3i 4 j
4j
3i
Using Pythagoras’ Theorem,
a 2 32 42
25
a5
The magnitude is 5 units. A unit vector in the direction of a must have length 1, and therefore be
one-fifth of the length of a. The unit vector is therefore 15 (3i 4 j) .
7. Using Vectors in Mechanics
Example 1 : Anne attempts to swim directly across a river. If her swimming speed is 1 ms-1, and
the current is flowing at 2 ms-1, find her actual speed, and the angle which her path
makes with the bank of the river.
-1
2 ms
1 ms-1
1 ms-1
v
2 ms-1
The first diagram shows the two components of Anne’s speed. The second diagram shows how
the vectors are added, by placing the beginning of one vector at the end of the other. The
resultant vector is v.
To find the actual speed, we use Pythagoras’ Theorem.
v 2 12 22
v 5
2.24 ms-1 (2 d.p.)
To find the angle made with the bank,
1
2
26.6 (1 d.p.)
tan
Example 2 : A pilot wishes to fly his plane north-west, but a southerly wind of speed 60 km/h is
blowing. If the plane is capable of 400 km/h in still air, on what bearing does the
plane actually fly, and what will its actual speed be?
400 km/h
45
400 km/h
45
v
60 km/h
Using the cosine rule,
v 2 4002 602 2 400 60 cos135
v 444.46 km/h (2 d.p.)
Using the sine rule,
60 km/h
sin sin135
400
v
400sin135
sin
v
39.5, 140.5
The plane flies at 444.46 km/h, on a bearing of 360 – 39.5 = 320.5°.
Example 3 : The pilot now compensates for the southerly wind so that the plane does actually fly
due north-west. On what bearing must the pilot head the plane, and what is it speed?
We require the resultant vector to be in a north-west direction. So we have
400 km/h
v
45 60 km/h
By the sine rule,
sin sin 45
60
400
60sin 45
sin
400
6.1 (1 d.p.)
180 45
128.9
The plane must aim on a bearing of 180° + 128.9° = 308.9°.
To find the actual speed of the plane, we use the cosine rule.
v 2 4002 602 2 400 60 cos
v 440.17 km/h (2 d.p.)
The actual speed of the plane is 440.17 km/h.
Exercise 5E page 98 q 1 - 9
We can extend our work on kinematics into two or even three dimensions.
Example 4 : A particle is travelling with uniform velocity. The position vector of the particle is
3i 5 j m at time t 2 , and 7i 8 j m at time t 7 . Find the velocity of the particle.
change in displacement
time
(7i 8 j) (3i 5 j)
5
0.8i 2.6 j ms -1
v
The equations of uniform acceleration readily become vector equations, although the only one
needed in unit M1 is
v u at
Example 5 : A particle has velocity 3i 2 j ms-1 when t 0 , and velocity 7i 4 j ms-1 when
t 2 . Find the constant acceleration of the particle.
v u at
7i 4 j 3i 2 j 2a
a 2i j ms -2
Here are some more examples which use common vector methods.
Example 6 : A particle is moving with a constant velocity 12i 5 j ms-1. It passes through the
point with position vector 4i 5 j m at t 0 . Find the speed of the particle and its
distance from O when t 3 seconds.
speed 122 52
13 ms-1
The position vector of the particle at time t 3 is given by
r 4i 5 j 3(12i 5 j)
40i 20 j
distance 40 2 20 2
44.72 m (2 d.p.)
Example 7 : At noon a lighthouse keeper observes two ships A and B which have position vectors
4i 3j km and 4i 9 j km respectively, relative to the lighthouse O. (The unit
vectors i and j are directed due east and due north). The ships are moving with
constant velocities 4i 17 j km/h and 12i 5 j km/h respectively.
a) Write down in the form p qt the position vector of A and the position vector of
B at time t hours after noon.
b) Show that A and B will collide and find the time when this collision will occur
and the position vector of the point of collision.
c) In order to prevent the collision, at 12.15 pm ship A increases its speed and
changes its velocity to 16i 13j km/h. Ship B maintains its original velocity. Find
the distance between A and B at 12.30 pm now.
a) The position vectors are
rA (4i 3j) (4i 17 j)t
rB (4i 9 j) (12i 5 j)t
b) The ships will collide if they have the same position at the same time. We begin by assuming
this is the case at time t. We can then equate the i and j components of the position vectors.
For the i components,
4 4t 4 12t
16t 8
t 0.5
For the j components,
3 17t 9 5t
12t 6
t 0.5
As both ships have the same position vector at the same time, they do indeed collide. The time
of collision is 12.30 pm, and the position vector can be found by substituting t 0.5 into one
of the position vectors.
rA (4i 3j) 0.5(4i 17 j)
2i 11.5 j
Activity 1 : This can be demonstrated on Autograph. Set the axes from –20 to 20 in each
direction. For ship A, type in the equation x 4 4a, y 3 17a (don’t use t as the
variable for time). Under ‘draw option’ select 4 12 point. Type in the equation for
ship B in the same way. Then use the constant controller to manually increment a in
steps of 601 , starting at 0. The position of each ship will be plotted every minute.
c) The position vector of B is unchanged, but the new position vector of A at 12.30 pm is
rA (4i 3j) 0.25(4i 17 j) 0.25(16i 13 j)
i 10.5 j
The displacement from ship B to ship A is therefore 3i j , and the distance is given by
distance 32 12
3.16 km (2 d.p.)
Example 5 : A cruiser C is sailing due east at a constant speed of 20 km/h and a destroyer D is
sailing due north at a constant speed of 10 km/h. At noon C and D are at points with
position vectors 5i km and 20 j km respectively, relative to a fixed origin O. The
unit vectors i and j are due east and due north respectively.
a) Find the position vector of C relative to D at time t hours after noon.
b) Hence prove that C and D are closest together at 12.36 pm, and find the distance
between them at this time.
a) The position vectors at time t are
rC 5i 20ti
(20t 5)i
rD 20 j 10tj
(10t 20) j
The position vector of C relative to D is given by
rC rD (20t 5)i (10t 20) j
b) The distance d between the two vessels is simply the magnitude of this vector.
d 2 (20t 5) 2 (10t 20) 2
400t 2 200t 25 100t 2 400t 400
500t 2 600t 425
Differentiating to minimise d 2 ,
d d 2
dt
2
d d 2
dt 2
1000t 600
1000
The distance is a minimum when
1000t 600 0
t 0.6 hours
36 minutes
So the ships are closest at 12.36 pm. We know this is a minimum since the second derivative
is positive. The distance is given by
d 500t 2 600t 425
500 0.62 600 0.6 425
15.65 km (2 d.p.)
Activity 2 : Again, this can be demonstrated on Autograph.
Hard Extension Problem
Example 8 : At noon the position vectors and the velocity vectors of two ships A and B are
rA 2i j km
v A 3i j km/h
rB i 4 j km
v B 11i 3j km/h
a) Find the distance of closest approach of the ships, and the time at which this
occurs.
b) Find the times at which the ships are 5km apart.
a) The position vectors at time t are
rA 2i j (3i j)t
(3t 2)i (t 1) j
rB i 4 j (11i 3j)t
(11t 1)i (3t 4) j
The position vector of A relative to B is given by
rA rB (3t 2)i (t 1) j (11t 1)i (3t 4) j
(3 8t )i (5 2t ) j
The distance d between the two ships is given by
d 2 (3 8t ) 2 (5 2t ) 2
9 48t 64t 2 25 20t 4t 2
68t 2 68t 34
Differentiating to minimise d 2 ,
d d 2
dt
2
d d 2
dt 2
136t 68
136
The distance is a minimum when
136t 68 0
t 0.5 hours
So the ships are closest at 12.30 pm. We know this is a minimum since the second derivative
is positive. The distance is given by
d 68t 2 68t 34
68 0.52 68 0.5 34
4.12 km (2 d.p.)
b) We require
68t 2 68t 34 52
68t 2 68t 9 0
68 (68) 2 4 68 9
2 68
t 0.157, t 0.843
t
The times are 12.09 pm and 12.51 pm, to the nearest minute.
Activity 3 : Again, this can be demonstrated on Autograph.
```
|
education
# Vectors problem|How to solve vector in physics| how To prove; (I) A. [ B × C ] = B. [ C × A ] = C. [ A × B ], B. [ A × C ] = -A. [ B × C ] (III) and A. [ B × A ] = 0 is correct
Solving vector problem
Let’s solve maths(vectors)
Use A = 2i + 3j – 5k
B = 3i + j +2k
C = I – j + 3k
To prove;
(I) A. [ B × C ] = B. [ C × A ] = C. [ A × B ](II) B. [ A × C ] = -A. [ B × C ](III) A. [ B × A ] = 0 is correct
SOLUTIONS
Note: The coefficient of I, j and k are the ones multiplying themselves the I, j and k still remain the same. I.e alike factors multiply themselves
(I) A. [ B × C ] = B. [ C × A ] = C. [ A × B ]first we’ll solve for A. [ B × C ] then relate it to B. [ C × A ] after we’ll check if it’s true for C. [ A × B ]Let’s get down to business
Solve for A. [ B × C ]2i+3j-5k[ (3i +j+2k) × (i-j+3k) ]2i+3j-5k[ 3i×i +j×-j + 2k×3k ]2i+3j-5k[ 3i -j+6k ]2i × 3i + 3j×-j – 5k×6k
=6i -3j -30k
For B. (C × A)
3i+j+2k[ (i – j+3k) × (2i+3j-5k) ]3i+j+2k[ 2i-3j-15k ]3i×2i + j×-3j + 2k×-15k
=6i -3j -30k
For C. [A × B]i – j +3k [ (2i +3j -5k) × (3i + j +2k) ]i – j +3k [ 2i×3i +3j×j – 5k×2k ]i – j +3k [ 6i +3j – 10k ]i×6i – j×3j + 3k×-10k
= 6i -3j -30k
So therefore, A. [ B × C ] = B. [ C × A ] = C. [ A × B]
(II) B. [ A × C ] = -A. [ B × C ]Solve first for B. [ A × C ]3i + j + 2k [ (2i + 3j – 5k) × (i – j + 3k) ]3i + j + 2k [ 2i – 3j – 15k ]3i×2i + j×-3j + 2k×-15k
= 6i -3j -30k
Then solve for right hand side
-A. (B × C)
-[2i+3j-5k] [ (3i +j+2k) × (i-j+3k) ]-2i-3j+5k[ 3i×i +j×-j + 2k×3k ]-2i-3j+5k[ 3i -j+6k ]-2i × 3i – 3j×-j + 5k×6k
=-6i +3j + 30k
So therefore B. [ A × C ] is not equal to -A. [ B × C ]
(III) A. [ B × A ] = 0 is correct
As, A×B = ABsin theta
ABsin theta is a vector which is perpendicular to the plane having A vector and V vector which implies that it is also perpendicular to A vector.
And A = AB cos theta
NOTE: An angle is perpendicular only if it is 90°
Therefore,
A. [ B× A ] = AB cos theta. × AB son theta
Theta = 90°
= AB cos 90° × AB son 90°
= AB[0] × AB [1]Anything multiply by zero is what
Answer: sure it will be zero, so therefore it’s true
You can post your questions or solvings on any course via this website: emperorelectricalworks.com.ng
YOU CAN DO IT
### Oluwamuyide Peter
On the 4th of November I officially became a member of the exclusive 1st student with distinction after five years of no such record, in the history of The Polytechnic Ibadan, Faculty of Engineering to graduate with distinction as a DPP students since its establishment in 2011. My unrelenting power to solve problems, have made me to create a platform where student can get valid information anywhere, anyplace at anytime Evolving education world wide 🌎
|
# A lottery claims that 10% of tickets win a prize. How many tickets should you purchase to be more than 50% sure of winning a prize?
Oct 2, 2017
You have to buy at least $7$ tickets.
#### Explanation:
Let $p = 0.1$, which is the probability for winning a prize.
In this lottery tickets, the probability for NOT winning a prize is
$1 - p = 1 - 0.1 = 0.9$.
If you want more than 50% sure of winning at least one prize,
the number of tickets you have to buy $n$ satisfies the following inequation.
$1 - {\left(1 - p\right)}^{n} \ge 0.5$
$1 - {0.9}^{n} \ge 0.5$
${0.9}^{n} \le 0.5$.
Put n=1,2,… to the inequation you will find:
${0.9}^{6} = 0.531441 \ge 0.5$
${0.9}^{7} = 0.4782969 \le 0.5$
Thus, the minimum integer that meets ${0.9}^{n} \le 0.5$ is $n = 7$.
[What is the point?]
The concept I applied to this question is called complementary event.
https://en.wikipedia.org/wiki/Complementary_event
This idea is useful to evaluate the probability for having at least one event.
Consider a simpler case and you will notice the convenience.
[Example]$\textcolor{g r e e n}{\text{What is the odds to have at least one 6 when we roll three dices?}}$
If you solve this from the front, you need to consider three cases.
(a) three $6$s
(b) two $6$s and another number
(c) a $6$ and two other numbers
It will be complicated. Insted, you can solve it from the back door.
(1) If a dice is rolled, the probability of not having a $6$ is $1 - \frac{1}{6} = \frac{5}{6}$.
(2) When three dices are rolled, they are independent. So the probability of having no $6$s is ${\left(\frac{5}{6}\right)}^{3} = \frac{125}{216}$.
(3) Having at least one 6 is $\textcolor{red}{\text{complement}}$ to (2) and the probability is $1 - \frac{125}{216} = \frac{91}{216}$.
|
Menu
A
B
Home Page
# Week 11 Converting units
National curriculum content
• Convert between different units of metric measure (for example, kilometre and metre; centimetre and metre; centimetre and millimetre; gram and kilogram; litre and millilitre)
• Understand and use approximate equivalences between metric units and common imperial units such as inches, pounds and pints
• Solve problems involving converting between units of time
• Use all four operations to solve problems involving measurements such as length, mass and money using decimal notation, including scaling
Lesson objectives
1. Recap – Kilometers
2. Kilograms and kilometers
3. Millimeters and milliliters
4. Metric units
5. Imperial units
6. Converting units of time
7. Timetables
What we want children to know
• How to use a range of equipment when measuring.
• Understand which unit of measure to use.
• How to convert between different units of metric measure.
• How to convert between different units of imperial measure.
• How to convert between different units of time.
• Understand which operation to use when solving problems involving measurements.
What skills we want children to develop
Use knowledge to solve reasoning and problem solving questions such as:
Metric Units:
A 10 piece coin is 2mm thick.
Evie makes a pile of 10 pence coins worth £1.50.
What is the height of the pile of coins in centimeters?
Kilograms and Kilometers:
Amir buys 2,500 grams of potatoes and 2,000 grams of carrots.
Potatoes cost 78p.
Carrots cost £1.46.
He pays with a £5 note.
How much change does he get?
Converting Units of Time:
Teddy’s birthday is in March.
Amir’s birthday is in April.
Amir is 96 hours older than Teddy.
What dates could Teddy and Amir’s birthdays be?
Mathematical Talk
What does ‘kilo’ mean when used at the start of a word?
How would you convert a fraction of a kilometre to metres?
Would it be appropriate to measure your height in millimetres?
Which unit of measure would be best to measure; the height of a door frame, the length of a room, the width of a door?
Why do you think we still use imperial measures?
How precise should approximation be?
Is 0.75 hours the same as 75 minutes? Why or why not?
Top
|
# 1.2 Use the language of algebra (Page 5/18)
Page 5 / 18
Evaluate $3{x}^{2}+4x+1$ when $x=3.$
40
Evaluate $6{x}^{2}-4x-7$ when $x=2.$
9
## Indentify and combine like terms
Algebraic expressions are made up of terms. A term is a constant, or the product of a constant and one or more variables.
## Term
A term is a constant, or the product of a constant and one or more variables.
Examples of terms are $7,y,5{x}^{2},9a,\text{and}\phantom{\rule{0.2em}{0ex}}{b}^{5}.$
The constant that multiplies the variable is called the coefficient .
## Coefficient
The coefficient of a term is the constant that multiplies the variable in a term.
Think of the coefficient as the number in front of the variable. The coefficient of the term 3 x is 3. When we write x , the coefficient is 1, since $x=1·x.$
Identify the coefficient of each term: 14 y $15{x}^{2}$ a .
## Solution
The coefficient of 14 y is 14.
The coefficient of $15{x}^{2}$ is 15.
The coefficient of a is 1 since $a=1\phantom{\rule{0.2em}{0ex}}a.$
Identify the coefficient of each term: $17x$ $41{b}^{2}$ z .
14 41 1
Identify the coefficient of each term: 9 p $13{a}^{3}$ ${y}^{3}.$
9 13 1
Some terms share common traits. Look at the following 6 terms. Which ones seem to have traits in common?
$\begin{array}{cccccccccccccccc}5x\hfill & & & 7\hfill & & & {n}^{2}\hfill & & & 4\hfill & & & 3x\hfill & & & 9{n}^{2}\hfill \end{array}$
The 7 and the 4 are both constant terms.
The 5x and the 3 x are both terms with x .
The ${n}^{2}$ and the $9{n}^{2}$ are both terms with ${n}^{2}.$
When two terms are constants or have the same variable and exponent, we say they are like terms .
• 7 and 4 are like terms.
• 5 x and 3 x are like terms.
• ${x}^{2}$ and $9{x}^{2}$ are like terms.
## Like terms
Terms that are either constants or have the same variables raised to the same powers are called like terms .
Identify the like terms: ${y}^{3},$ $7{x}^{2},$ 14, 23, $4{y}^{3},$ 9 x , $5{x}^{2}.$
## Solution
${y}^{3}$ and $4{y}^{3}$ are like terms because both have ${y}^{3};$ the variable and the exponent match.
$7{x}^{2}$ and $5{x}^{2}$ are like terms because both have ${x}^{2};$ the variable and the exponent match.
14 and 23 are like terms because both are constants.
There is no other term like 9 x .
Identify the like terms: $9,$ $2{x}^{3},$ ${y}^{2},$ $8{x}^{3},$ $15,$ $9y,$ $11{y}^{2}.$
9 and 15, ${y}^{2}$ and $11{y}^{2},$ $2{x}^{3}$ and $8{x}^{3}$
Identify the like terms: $4{x}^{3},$ $8{x}^{2},$ 19, $3{x}^{2},$ 24, $6{x}^{3}.$
19 and 24, $8{x}^{2}$ and $3{x}^{2},$ $4{x}^{3}$ and $6{x}^{3}$
Adding or subtracting terms forms an expression. In the expression $2{x}^{2}+3x+8,$ from [link] , the three terms are $2{x}^{2},3x,$ and 8.
Identify the terms in each expression.
1. $9{x}^{2}+7x+12$
2. $8x+3y$
## Solution
The terms of $9{x}^{2}+7x+12$ are $9{x}^{2},$ 7 x , and 12.
The terms of $8x+3y$ are 8 x and 3 y .
Identify the terms in the expression $4{x}^{2}+5x+17.$
$4{x}^{2},5x,17$
Identify the terms in the expression $5x+2y.$
5 x , 2 y
If there are like terms in an expression, you can simplify the expression by combining the like terms. What do you think $4x+7x+x$ would simplify to? If you thought 12 x , you would be right!
$\begin{array}{c}\hfill 4x+7x+x\hfill \\ \hfill x+x+x+x\phantom{\rule{1em}{0ex}}+x+x+x+x+x+x+x\phantom{\rule{1em}{0ex}}+x\hfill \\ \hfill 12x\hfill \end{array}$
Add the coefficients and keep the same variable. It doesn’t matter what x is—if you have 4 of something and add 7 more of the same thing and then add 1 more, the result is 12 of them. For example, 4 oranges plus 7 oranges plus 1 orange is 12 oranges. We will discuss the mathematical properties behind this later.
Simplify: $4x+7x+x.$
Add the coefficients. 12 x
## How to combine like terms
Simplify: $2{x}^{2}+3x+7+{x}^{2}+4x+5.$
## Solution
Simplify: $3{x}^{2}+7x+9+7{x}^{2}+9x+8.$
$10{x}^{2}+16x+17$
Simplify: $4{y}^{2}+5y+2+8{y}^{2}+4y+5.$
$12{y}^{2}+9y+7$
## Combine like terms.
1. Identify like terms.
2. Rearrange the expression so like terms are together.
3. Add or subtract the coefficients and keep the same variable for each group of like terms.
#### Questions & Answers
How do you find divisible numbers without a calculator?
TAKE OFF THE LAST DIGIT AND MULTIPLY IT 9. SUBTRACT IT THE DIGITS YOU HAVE LEFT. IF THE ANSWER DIVIDES BY 13(OR IS ZERO), THEN YOUR ORIGINAL NUMBER WILL ALSO DIVIDE BY 13!IS DIVISIBLE BY 13
BAINAMA
When she graduates college, Linda will owe $43,000 in student loans. The interest rate on the federal loans is 4.5% and the rate on the private bank loans is 2%. The total interest she owes for one year was$1,585. What is the amount of each loan?
Sean took the bus from Seattle to Boise, a distance of 506 miles. If the trip took 7 2/3 hours, what was the speed of the bus?
66miles/hour
snigdha
How did you work it out?
Esther
s=mi/hr 2/3~0.67 s=506mi/7.67hr = ~66 mi/hr
Orlando
hello, I have algebra phobia. Subtracting negative numbers always seem to get me confused.
what do you need help in?
Felix
subtracting a negative....is adding!!
Heather
look at the numbers if they have different signs, it's like subtracting....but you keep the sign of the largest number...
Felix
for example.... -19 + 7.... different signs...subtract.... 12 keep the sign of the "largest" number 19 is bigger than 7.... 19 has the negative sign... Therefore, -12 is your answer...
Felix
—12
Thanks Felix.l also get confused with signs.
Esther
Thank you for this
Shatey
ty
Graham
think about it like you lost $19 (-19), then found$7(+7). Totally you lost just $12 (-12) Annushka I used to struggle a lot with negative numbers and math in general what I typically do is look at it in terms of money I have -$5 in my account I then take out 5 more dollars how much do I have in my account well-\$10 ... I also for a long time would draw it out on a number line to visualize it
Meg
practicing with smaller numbers to understand then working with larger numbers helps too and the song/rhyme same sign add and keep opposite signs subtract keep the sign of the bigger # then you'll be exact
Meg
Bruce drives his car for his job. The equation R=0.575m+42 models the relation between the amount in dollars, R, that he is reimbursed and the number of miles, m, he drives in one day. Find the amount Bruce is reimbursed on a day when he drives 220 miles
168.50=R
Heather
john is 5years older than wanjiru.the sum of their years is27years.what is the age of each
46
mustee
j 17 w 11
Joseph
john is 16. wanjiru is 11.
Felix
27-5=22 22÷2=11 11+5=16
Joyce
I don't see where the answers are.
Ed
Cindy and Richard leave their dorm in Charleston at the same time. Cindy rides her bicycle north at a speed of 18 miles per hour. Richard rides his bicycle south at a speed of 14 miles per hour. How long will it take them to be 96 miles apart?
3
Christopher
18t+14t=96 32t=96 32/96 3
Christopher
show that a^n-b^2n is divisible by a-b
What does 3 times your weight right now
Use algebra to combine 39×5 and the half sum of travel of 59+30
Cherokee
What is the segment of 13? Explain
Cherokee
my weight is 49. So 3 times is 147
Cherokee
kg to lbs you goin to convert 2.2 or one if the same unit your going to time your body weight by 3. example if my body weight is 210lb. what would be my weight if I was 3 times as much in kg. that's you do 210 x3 = 630lb. then 630 x 2.2= .... hope this helps
tyler
How to convert grams to pounds?
paul
What is the lcm of 340
Yes
Cherokee
How many numbers each equal to y must be taken to make 15xy
15x
Martin
15x
Asamoah
15x
Hugo
1y
Tom
1y x 15y
Tom
find the equation whose roots are 1 and 2
(x - 2)(x -1)=0 so equation is x^2-x+2=0
Ranu
I believe it's x^2-3x+2
NerdNamedGerg
because the X's multiply by the -2 and the -1 and than combine like terms
NerdNamedGerg
find the equation whose roots are -1 and 4
Ans = ×^2-3×+2
Gee
find the equation whose roots are -2 and -1
(×+1)(×-4) = x^2-3×-4
Gee
yeah
Asamoah
Quadratic equations involving factorization
there's a chatting option in the app wow
Nana
That's cool cool
Nana
Nice to meet you all
Nana
you too.
Joan
😃
Nana
Hey you all there are several Free Apps that can really help you to better solve type Equations.
Debra
Debra, which apps specifically. ..?
Nana
am having a course in elementary algebra ,any recommendations ?
samuel
Samuel Addai, me too at ucc elementary algebra as part of my core subjects in science
Nana
me too as part of my core subjects in R M E
Ken
at ABETIFI COLLEGE OF EDUCATION
Ken
ok great. Good to know.
Joan
5x + 1/3= 2x + 1/2
sanam
Plz solve this
sanam
5x - 3x = 1/2 - 1/3 2x = 1/6 x = 1/12
Ranu
Thks ranu
sanam
Erica
the previous equation should be 3x = 1/6 x=1/18
Sriram
for the new one 10x + 2x = 38 - 14
Sriram
12x = 24 x=2
Sriram
10x + 14 = -2x +38 10x + 2x = 38 - 14 12x = 24 divide both sides by the coefficient of x, which is 12 therefore × = 2
vida
a trader gains 20 rupees loses 42 rupees and then gains ten rupees Express algebraically the result of his transactions
a trader gains 20 rupees loses 42 rupees and then gains 10 rupees Express algebraically the result of his three transactions
vinaya
a trader gains 20 rupees loses 42 rupees and then gains 10 rupees Express algebraically the result of his three transactions
vinaya
a trader gains 20 rupees loses 42 rupees and then gains 10 rupees Express algebraically the result of his three transactions
vinaya
|
# Jeannie's scores on her first four tests were 80, 65, 87, and 75. What will she have to score on on her next test to obtain an average of at least 80 for the term?
Jun 3, 2018
$97$
#### Explanation:
We can rearrange the terms in the formula for average from
$\text{average"="sum of set"/"number of terms in the set}$
to
$\text{sum of set"="average"*"number of terms in the set}$
Since the average we aiming for is $80$ and that will be the fifth term, then the sum of set will be
$80 \cdot 5$ which is $400$, we can solve for the fifth term but finding the differnce between $400$ and the sum of the stated four terms.
$400 - \left(80 + 65 + 87 + 75\right)$ to which the answer is $97$
|
Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.3 | Set 1
• Difficulty Level : Expert
• Last Updated : 03 Mar, 2021
Find the average expenditure (in rupees) per household.
Solution:
Let the assumed mean (A) = 275
It’s seen that A = 275 and h = 50
So,
Mean = A + h x (Σfi ui/N)
= 275 + 50 (-35/200)
= 275 – 8.75
= 266.25
Which method did you use for finding the mean, and why?
Solution:
From the given data,
To find the class interval we know that,
Class marks (xi) = (upper class limit + lower class limit)/2
Now, let’s compute xi and fixi by the following
Here,
Mean = Σ fiui/N
= 162/ 20
= 8.1
Thus, the mean number of plants in a house is 8.1
We have used the direct method as the values of class mark xi and fi is very small.
Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
Let us assume mean (A) = 150
It’s seen that,
A = 150 and h = 20
So,
Mean = A + h x (Σfi ui/N)
= 150 + 20 x (-12/50)
= 150 – 24/5
= 150 = 4.8
= 145.20
Question 4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
Solution:
Using the relation (xi) = (upper class limit + lower class limit)/ 2
And, class size of this data = 3
Let the assumed mean (A) = 75.5
So, let’s calculate di, ui, fiui as following:
From table, it’s seen that
N = 30 and h = 3
So, the mean = A + h x (Σfi ui/N)
= 75.5 + 3 x (4/30
= 75.5 + 2/5
= 75.9
Therefore, the mean heart beats per minute for those women are 75.9 beats per minute.
Question 5. Find the mean of each of the following frequency distributions:
Solution:
Let’s consider the assumed mean (A) = 15
From the table it’s seen that,
A = 15 and h = 6
Mean = A + h x (Σfi ui/N)
= 15 + 6 x (3/40)
= 15 + 0.45
= 15.45
Question 6. Find the mean of the following frequency distribution:
Solution:
Let’s consider the assumed mean (A) = 100
From the table it’s seen that,
A = 100 and h = 20
Mean = A + h x (Σfi ui/N)
= 100 + 20 x (61/100)
= 100 + 12.2
= 112.2
Question 7. Find the mean of the following frequency distribution:
Solution:
From the table it’s seen that,
A = 20 and h = 8
Mean = A + h x (Σfi ui/N)
= 20 + 8 x (7/40)
= 20 + 1.4
= 21.4
Question 8. Find the mean of the following frequency distribution:
Solution:
Let’s consider the assumed mean (A) = 15
From the table it’s seen that,
A = 15 and h = 6
Mean = A + h x (Σfi ui/N)
= 15 + 6 x (5/40)
= 15 + 0.75
= 15.75
Question 9. Find the mean of the following frequency distribution:
Solution:
Let’s consider the assumed mean (A) = 25
From the table it’s seen that,
A = 25 and h = 10
Mean = A + h x (Σfi ui/N)
= 25 + 10 x (8/60)
= 25 + 4/3
= 79/3 = 26.333
Question 10. Find the mean of the following frequency distribution:
Solution:
Let’s consider the assumed mean (A) = 20
From the table it’s seen that,
A = 20 and h = 8
Mean = A + h x (Σfi ui/N)
= 20 + 8 x (5/40)
= 20 + 1
= 21
Question 11. Find the mean of the following frequency distribution:
Solution:
Let’s consider the assumed mean (A) = 20
From the table it’s seen that,
A = 20 and h = 8
Mean = A + h x (Σfi ui/N)
= 20 + 6 x (-9/20)
= 20 – 72/20
= 20 – 3.6
= 16.4
Question 12. Find the mean of the following frequency distribution:
Solution:
Let’s consider the assumed mean (A) = 60
From the table it’s seen that,
A = 60 and h = 20
Mean = A + h x (Σfi ui/N)
= 60 + 20 x (14/50)
= 60 + 28/5
= 60 + 5.6
= 65.6
Question 13. Find the mean of the following frequency distribution:
Solution:
Let’s consider the assumed mean (A) = 50
From the table it’s seen that,
A = 50 and h = 10
Mean = A + h x (Σfi ui/N)
= 50 + 10 x (-2/40)
= 50 – 0.5
= 49.5
My Personal Notes arrow_drop_up
|
# If the solve the problem
Question:
If $y=\sin (\sin x)$, prove that $: \frac{d^{2} y}{d x^{2}}+\tan x \cdot \frac{d y}{d x}+y \cos ^{2} x=0$
Solution:
Given,
$y=\sin (\sin x) \ldots \ldots .$ equation 1
To prove: $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\tan \mathrm{x} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y} \cos ^{2} \mathrm{x}=0$
We notice a second-order derivative in the expression to be proved so first take the step to find the second order derivative.
Let's find $\frac{d^{2} y}{d x^{2}}$
As $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)$
So, lets first find $\mathrm{dy} / \mathrm{dx}$
$\frac{d y}{d x}=\frac{d}{d x} \sin (\sin x)$
Using chain rule, we will differentiate the above expression
Let $t=\sin x \Longrightarrow \frac{d t}{d x}=\cos x$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} \frac{\mathrm{dt}}{\mathrm{dx}}$
$\frac{d y}{d x}=\cos t \cos x=\cos (\sin x) \cos x \ldots \ldots .$ equation 2
Again differentiating with respect to $x$ applying product rule:
$\frac{d^{2} y}{d x^{2}}=\cos x \frac{d}{d x} \cos (\sin x)+\cos (\sin x) \frac{d}{d x} \cos x$
Using chain rule again in the next step-
$\frac{d^{2} y}{d x^{2}}=-\cos x \cos x \sin (\sin x)-\sin x \cos (\sin x)$
$\frac{d^{2} y}{d x^{2}}=-y \cos ^{2} x-\tan x \cos x \cos (\sin x)$
[using equation $1: y=\sin (\sin x)$ ]
And using equation 2, we have:
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-y \cos ^{2} \mathrm{x}-\tan \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}$
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\mathrm{y} \cos ^{2} \mathrm{x}+\tan \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}=0 \ldots \ldots .$ proved
|
Sales Toll Free No: 1-855-666-7446
# Principal Properties of Right Triangles
Top
Sub Topics A triangle is a geometric figure which is made of three straight lines and the sum of its angles is equal to 180 degree. A triangle is right triangle or right angled triangle, if any one of its three angles is equal to 90 degree. The side opposite to the 90 degree angle or right angle is known as the hypotenuse. The properties of right triangle are properties that define a right angled triangle. The properties of right triangles are as follows:Area of Right Triangle: The right angled triangle area can be measured by using the formula, Area = $\frac{1}{2}$$\times x \times y, where ‘x’ and ‘y’ can be considered as the two sides of the triangle. This formula is used only for right angle triangle.Height or Altitude of Right Triangle: If an altitude from the vertex of the right angle is drawn to the opposite side or the hypotenuse, then two triangles are formed and both the triangles formed are similar to one another and the main triangle.The pythagoras theorem states that, if ‘h’ be the hypotenuse and ‘x’ and ‘y’ be the two sides of the triangle, then according to the pythagoras theorem, it is stated that h^2 = x^2 + y^2. So, according to the formula, hypotenuse’s square is equal to the sum of the square of other two sides of the triangle.Radius of Incircle and Circumcircle of Right Angled Triangle: According to this property, the radius of the incircle of the triangle can be found using the formula r = \frac{x + y - h}{2}, where ‘x’ and ‘y’ are the two sides of a triangle and 'h' is the hypotenuse. The radius of the circumcircle of a right angle triangle can be calculated using the formula r = \frac{h}{2}, where ‘h’ is the hypotenuse of the right angled triangle. ## Area of Right Triangle Back to Top We know that polygons are closed figures made up of just line segments. Now, all closed figures enclose some region within it. This enclosed region inclusive of the boundary of the polygon is called the area of the polygon. Area of any polygon is measured in terms of the squares of 1 unit each, that can accommodate in the figure. Now, polygons being rectilinear figures, it becomes rather easier to divide them into such unit squares. As polygons are made up of just line segments, depending on the number of sides each polygon has, they are put under different categories. The polygons with 3 sides are called triangles, those with 4 sides are quadrilaterals,those with 5 sides are pentagons and so on. A broad classification of each type is whether a polygon is regular with all sides equal or otherwise irregular. But of all the types of polygons, the triangles and the quadrilaterals are further classified into different groups. Let us learn about triangles first. The triangles are classified on the basis of measure of their angles, as well as, on the basis of the measure of their sides. We shall classify them first on the basis of their angles. On this basis, the triangles may be classified as follows: 1. Acute triangle 2. Right triangle 3. Obtuse triangle Coming back to area, all polygons enclose area. So, triangles also have some area. Let us try to understand area of a right triangle. We define area of right triangle as the region enclosed by a right triangle inclusive of its boundary. Area right triangle = \frac{1}{2}$$ \times \text{base} \times \text{height}$, where base and height are the two sides of the triangle containing the right angle.
Step 1: First, we have to check what kind of polygon we have, like regular polygon or irregular polygon.
Step 2: If we have regular polygon, then we use following formula for evaluation of inradius and circumradius:
Inradius of regular polygon = $\frac{1}{2}$$\times a \times \cot$$\frac{\pi}{n}$,
Here, a is side length and n is number of side of polygon.
Circumradius of regular formula = $\frac{1}{2}$$\times a \times \csc$$\frac{\pi}{n}$
Here, ‘a’ is the side length and n is the number of sides of polygon.
Step 3: If we have irregular polygon, then we should calculate inradius and circumradius of polygon manually by their graphic structure.
## Altitude of Right Triangle
Geometry altitude can be used to calculate the area of a triangle by using the formula $\frac{1}{2}$ bh, where b is the length of the base and h is the length of the altitude or height of the triangle. The altitudes of a triangle are related to the sides of the triangle through many theorems like equilateral triangle theorem, inradius theorems, area theorem and so on. There is one very important formula called Heron’s formula which relates to the sides and altitude of a triangle.
|
# Countable set
If you can count the things in a set, it is called a countable set. A set with one thing in it is countable, and so is a set with one hundred things in it. A set with all the natural numbers (counting numbers) in it is countable too. It's infinite but if someone counted forever they wouldn't miss any of the numbers. Sometimes when people say 'countable set' they mean countable and infinite. Georg Cantor coined the term.
## Examples of Countable Sets
a diagram illustrating the countability of the rationals
Countable sets include all sets with a finite number of members, no matter how many.
Countable sets also include some infinite sets, such as the natural numbers. Since it is impossible to actually count infinite sets, we consider them countable if we can find a way to list them all without missing any. The natural numbers have been nicknamed "the counting numbers" since they are what we usually use to count things with.
You can count the natural numbers with $({1, 2, 3...})\,\!$
You can count the whole numbers with $({0, 1, 2, 3...})\,\!$
You can count the integers with $({0, 1, -1, 2, -2, 3, -3...})\,\!$
You can count the square integer numbers with $({0, 1, 4, 9, 16, 25, 36, ...})\,\!$
### A more tricky example
The rational numbers are also countable, but it becomes tricky. We can't just list the fractions as (0/1, 1/1, -1/1, 2/1, -2/1...) since we will never get to 1/2 this way. However, we can list all of the rational numbers in the form of a table. The horizontal rows are the numerators while the vertical columns are the denominators. We can then zigzag through the list, starting at 1/1, then 2/1, then 1/2, then 1/3, then 2/2, then 3/1, then 4/1, then 3/2, then 2/3 etc. If we keep this up, we will get to all of the numbers on the table in due time! Each time we hit a rational number that is new for us, we add it to the list of counted numbers. We don't want to count numbers twice, so when we hit 3/6 we can skip it, because we already counted the rational number 1/2. In this way, we produce an infinite list with all the rational numbers. Therefore, the rational numbers are countable.
|
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 7.1: Line Graphs and Scatter Plots
Difficulty Level: At Grade Created by: CK-12
Learning Objectives
• Represent data that has a linear pattern on a graph.
• Represent data using a broken-line graph and represent two sets of data using a double line graph.
• Understand the difference between continuous data and discrete data as it applies to a line graph.
• Represent data that has no definite pattern as a scatter plot.
• Draw a line of best fit on a scatter plot.
• Use technology to create both line graphs and scatter plots.
Introduction
Each year the school has a fund raising event to collect money to support the school sport teams. This year the committee has decided that each class will make friendship bracelets and sell them for 2.00\begin{align*}\2.00\end{align*} each. To buy the necessary supplies to make the bracelets, each class is given40.00\begin{align*}\40.00\end{align*} as a start up fee. Create a table of values and draw a graph to represent the sale of 10 bracelets. If the class sells ten bracelets, how much profit will be made?
We will revisit this problem later in the lesson.
When data is collected from surveys or experiments, it can be displayed in different ways; tables of values, graphs, and box-and-whisker plots. The most common graphs that are used in statistics are line graphs, scatter plots, bar graphs, histograms, frequency polygons. Graphs are the most common way of displaying data because they are visual and allow you to get a quick impression of the data and determine if there are any trends in the data. You have probably noticed that graphs of different types are found regularly in newspapers, on websites, and in many textbooks.
If we think of independent and dependent variables in terms of the variables in an input/output machine – we can see that the input variable is independent of anything around it but the output variable is completely dependent on what we put into the machine. The input variable is the x\begin{align*}{\color{blue}x}\end{align*} variable and the output variable is the y\begin{align*}{\color{red}y}\end{align*} (or the f(x)\begin{align*}f(x)\end{align*}) variable.
If we apply this theory to graphing a straight line on a rectangular coordinate system, we must first determine which variable is the dependent variable and which one is the independent variable. Once this has been established, the ordered pairs can be plotted.
Example 1: If you had a job where you earned 9.00\begin{align*}\9.00\end{align*} an hour for every hour you worked up to a maximum of 30 hours, represent your earnings on a graph by plotting the money earned against the time worked. Solution: The dependent variable is the money earned and the independent variable is the number of hours worked. Therefore, money is on the y\begin{align*}y-\end{align*}axis and time is on the x\begin{align*}x-\end{align*}axis. The first step is to create a table values that represent the problem. The number pairs in the table of values will be the ordered pairs to be plotted on the graph. Time Worked (Hours) Money Earned 00\begin{align*}\0\end{align*}
1 9.00\begin{align*}\9.00\end{align*} 218.00\begin{align*}\18.00\end{align*}
3 27.00\begin{align*}\27.00\end{align*} 436.00\begin{align*}\36.00\end{align*}
5 45.00\begin{align*}\45.00\end{align*} 654.00\begin{align*}\54.00\end{align*}
Now that the points have been plotted, the decision has to be made as to whether or not to join them. Between every two points plotted on the graph are an infinite number of values. If these values are meaningful to the problem, then the plotted points can be joined. This data is called continuous data. If the values between the two plotted points are not meaningful to the problem, then the points should not be joined. This data is called discrete data. In the above problem, it is possible to earn 4.50\begin{align*}\4.50\end{align*} for working one-half hour and this value is meaningful for our problem. Therefore the data is continuous and the points should be joined. Now you know how to graph a straight line from a table of values. It is just as important to be able to graph a straight line from a linear function that models a problem. The equation of a straight line can be written in the form y=mx+b\begin{align*}y = mx + b\end{align*}, where m\begin{align*}m\end{align*} is the slope of the line and b\begin{align*}b\end{align*} is the y\begin{align*}y-\end{align*}intercept. Example 2: Draw a graph to model the linear function y=2x+5\begin{align*}y = 2x + 5\end{align*} Solution: The slope of the line is change in xchange in y.\begin{align*}\frac{change \ in \ x}{change \ in \ y}.\end{align*} The slope of this line is 21\begin{align*}\frac{2}{1}\end{align*}. The y\begin{align*}y-\end{align*}intercept is (0, 5). To graph this line, begin by plotting the y\begin{align*}y-\end{align*}intercept. From the y\begin{align*}y-\end{align*}intercept, move to the right one and up two. Plot this point. You can continue to move right one and up two in order to create more points on the line. Join the points with a smooth line by using a straight edge (ruler). If you found this difficult to do, you could make a table of values for the function by substituting values for x\begin{align*}x\end{align*} into the equation to determine values for y\begin{align*}y\end{align*}. Then you would plot the ordered pairs on the graph. Whichever way you plotted the points, the result would be a straight line graph. Let’s apply this method to an everyday problem. Example 3: Your school is having a teenage dance on Friday night. The dance will begin at 8:00 p.m. and will end at midnight. A DJ is hired to play the music. The cost of hiring the DJ is100\begin{align*}\100\end{align*} plus an additional 20.00\begin{align*}\20.00\end{align*} an hour. Using either a table of values or an equation, draw a graph that would represent the cost of hiring the DJ for the dance. How much would the school pay the DJ for playing music for the dance? Solution: An equation that would model this problem is y=20x+100\begin{align*}y = 20x + 100\end{align*}. To make the equation match the problem y\begin{align*}y\end{align*} can be replaced with c\begin{align*}c\end{align*} (cost) and x\begin{align*}x\end{align*} can be replaced with h\begin{align*}h\end{align*} (number of hours). Now the equation y=20x+100\begin{align*}y = 20x + 100\end{align*} becomes c=20h+100\begin{align*}c = 20h + 100\end{align*}. The DJ will play 4 hours of music and will be paid180.00\begin{align*}\180.00\end{align*}
Example 4: The total cost to lease a car is mostly dependent on the number of months you have the lease. The table of values below shows the cost and number of months for ten months of a lease. Plot the data points on a properly labeled xy\begin{align*}x-y\end{align*} axis. Draw the line all the way to the y\begin{align*}y-\end{align*}axis so that you can find the y\begin{align*}y-\end{align*}intercept. What could the y\begin{align*}y-\end{align*}intercept represent in this problem?
x(months)y()22100427006330083900104500\begin{align*}& \text{x(months)} && 2 && 4 && 6 && 8 && 10\\ & y(\) && 2100 && 2700 && 3300 && 3900 && 4500\end{align*} We will now return to the fund raising event that was presented in the introduction. You should be able to solve this problem now. Solution: Number of Bracelets Cost 040\begin{align*}\40\end{align*}
1 42\begin{align*}\42\end{align*} 244\begin{align*}\44\end{align*}
3 46\begin{align*}\46\end{align*} 448\begin{align*}\48\end{align*}
5 50\begin{align*}\50\end{align*} 652\begin{align*}\52\end{align*}
7 \begin{align*}\54\end{align*}
8 \begin{align*}\56\end{align*}
9 \begin{align*}\58\end{align*}
10 \begin{align*}\60\end{align*}
In this case the data is discrete. The graph tells that only whole numbers are meaningful for this problem and that selling ten bracelets would mean a profit of \begin{align*}\20.00\end{align*}. The sales indicate a total of \begin{align*}\60.00\end{align*} but this includes the start up money of \begin{align*}\40.00\end{align*}. Therefore \begin{align*}\60.00 - \40.00 = \20.00\end{align*} is the profit.
In all of the above examples, the type of line graph that was used was one that described a definite linear pattern. There is another type of line graph that is used when it is necessary to show change over time. This type of line graph is called a broken line graph. A line is used to join the values but the line has no defined slope.
Example 5: Joey has an independent project to do for his Physical Active Lifestyle class. He has decided to do a poster that shows the times recorded for running the 100 meter dash event over the last fifteen years. He has collected the following information from the local library.
Year Time (seconds) Year Time (seconds)
1995 11.3 2002 11.0
1996 11.2 2003 10.9
1997 11.2 2004 10.9
1998 11.2 2005 10.9
1999 11.2 2006 10.8
2000 11.2 2007 10.7
2001 11.2 2008 10.7
2009 10.5
Display the information that Joey has collected on a graph that he might use on his poster.
Solution:
From this graph, you can answer many of the following questions:
Questions
1. What was the fastest time for the 100m dash in the year 2000?
2. Between what two years was there the greatest decrease in the fastest time to complete the 100m dash?
3. As the years pass, why do think runners are completing the race in a faster time?
1. 11.2 seconds
2. Between 2001 and 2002; Between 2008 and 2009
3. The runners are living a healthier and more active life style.
A broken line graph can be extended to include two broken lines. This type of a line graph is very useful when you have two sets of data that relate to the same topic but are from two different sources. For example the deaths in a small town over the past ten years can be graphed on a broken line graph. To extend this data, natural deaths could be plotted along with the deaths that were the result of traffic accidents. With both lines on the same graph, comparing them would be made easier.
Example 6: Jane has operated an ice-cream parlor for many years. She has decided to retire and is anxious to sell her business. In order to show interested buyers the ice cream sales for the past two years, she has decided to show these sales on a double line graph. She will use the graph to show buyers what month had the highest sales, when the greatest change in sales occurs and to show them when an unexpected increase in sales occurs. Following is the information that Jane has recorded for the monthly sales during the years 2008 and 2009. Can you help Jane by using the double line graph to answer the questions?
Solution:
The month of August had the highest sales for both years. Between the months and August and September there is a great decrease in the ice cream sales. However, the month of December shows an unexpected increase in sales. This could be due to the holiday season.
Scatter Plots
Often, when real-world data is plotted, the result is a linear pattern. The general direction of the data can be seen, but the data points do not all fall on a line. This type of graph is a scatter plot. A scatter plot is often used to investigate the relationship (if one exists) between two sets of data. The data is plotted on a graph such that one quantity is plotted on the \begin{align*}x-\end{align*}axis and one quantity is plotted on the \begin{align*}y-\end{align*}axis. If the relationship does exist between the two sets of data, it will be visible when the data is plotted.
Example 1: The following graph represents the relationship between the price per pound of lobster and the number of lobsters sold. Although the points cannot be joined to form a straight line, the graph does suggest a linear pattern. What is the relationship between the cost per pound and the number of lobsters sold?
Solution:
From the graph, it is obvious that a relationship does exist between the cost per pound and the number of lobsters sold. When the cost per pound was low, the number of lobsters sold was high.
Example 2: The following scatter plot represents the sale of lottery tickets and the temperature.
Is there a relationship between the number of lottery tickets sold and the temperature?
Solution:
From the graph, it is clearly seen that there is no relationship between the number of lottery tickets sold and the temperature of the surrounding environment.
Example 3: The table below represents the height of ten children in inches and their shoe size.
\begin{align*}& \text{Height(in)} && 51 && 53 && 61 && 59 && 63 && 47 && 53 && 66 && 55 && 49\\ & \text{Shoe Size} && 2 && 4 && 6 && 5 && 7 && 1 && 3 && 9 && 4 && 2 \end{align*}
The information from the table can be displayed on a scatter plot.
Solution:
Yes, there is a relationship between the shoe size and the height of the child. Children who are short wear small-sized shoes and those who are taller wear larger shoes.
In this case, there is a direct relationship (correlation) between the shoe size and the height of the children. Correlation refers to the relationship or connection between two sets of data. The correlation between two sets of data can be weak, strong, negative, or positive, or in some cases there can be no correlation. The characteristics of the correlation between two sets of data can be readily seen from the scatter plot.
The scatter plot of the shoe sizes and the heights of the children show a strong, positive correlation. The scatter plot of the lottery tickets and the temperature showed no correlation.
If there is a correlation between the two sets of data on a scatter plot, then a straight line can be drawn so that the plotted points are either on the line or very close to it. This line is called the line of best fit. A line of best fit is drawn on a scatter plot so that it joins as many points as possible and shows the general direction of the data. When constructing the line of best fit, it is also important to keep, approximately, an equal number of points above and below the line. To determine where the line of best fit should be drawn, a piece of spaghetti can easily be rolled across the graph with the plotted points still being visible.
Returning to the scatter plot that shows the relationship between shoe sizes and the height of children, a line of best fit can be drawn to define this relationship.
In a later lesson, we will determine the equation of this line manually and by using technology.
Lesson Summary
In this lesson you learned how to represent data by graphing three types of line graphs-a straight line of the form \begin{align*}y = mx + b\end{align*}, a broken-line graph and a double line graph. You also learned about scatter plots and the meaning of correlation as it applies to a scatter plot. In addition, you saw the result of drawing a line of best fit on a scatter plot.
Points to Consider
• Is a double line graph the only representation used to compare two sets of data?
• Does the line of best fit have an equation that would model the data?
• Is there another representation that could be used instead of a broken line graph?
Review Questions
1. On the following graph circle the independent and dependent variables. Write a sentence to describe how the independent (input) variable is related to the dependent (output) variable in each graph.
2. Ten people were interviewed for a job at the local grocery store. Mr. Neal and Mrs. Green awarded each of the ten people, points as shown in the following table: \begin{align*}& \text{Mr.Neal} && 30 && 22 && 25 && 17 && 17 && 39 && 33 && 38 && 27 && 33\\ & \text{Mrs.Green} && 25 && 20 && 21 && 15 && 16 && 35 && 30 && 32 && 23 && 22 \end{align*} Draw a scatter plot to represent the above data. (You may use technology to do this).
3. The following data represents the fuel consumption of cars with the same size engine, when driven at various speeds. \begin{align*}& \text{Speed(km/h} && 48 && 99 && 64 && 128 && 112 && 88 && 120 && 106\\ & \text{Fuel Consumption(km/L)} && 7 && 14 && 9 && 18 && 16 && 13 && 17 && 15\end{align*}
1. Plot the data values.
2. Draw in the line of best fit.
3. Estimate the fuel consumption of a car travelling at a speed of 72 km/h.
4. Estimate the speed of a car that has a fuel consumption of 12 km/L.
4. Answer the questions by using the following graph that represents the temperature in \begin{align*}^\circ F\end{align*} for the first 20 days in July.
1. What was the coldest day?
2. What was the temperature on the hottest day? (Approximately)
3. What days appeared to have no change in temperature?
5. Answer the questions by using the following graph that represents the temperature in \begin{align*}^\circ F\end{align*} for the first 20 days in July in New York and in Seattle.
1. Which City has the warmest temperatures in July?
2. Which of the two cities seems to have temperatures that appear to be rising as the month progresses?
3. Approximately, what is the difference in the daily temperatures between the two cities?
6. The following graphs represent continuous and discrete data. Are the graphs labeled correctly with respect to these types of data? Justify your answer.
7. A car rental agency is advertising March Break specials. The company will rent a car for \begin{align*}\10\end{align*} a day plus a down payment of \begin{align*}\65\end{align*}. Create a table of values for this problem and plot the points on a graph. Using the graph, what would be the cost of renting the car for one week?
8. What type of graph would you use to display each of the following types of data?
1. The number of hours you spend doing Math homework each week for the first semester.
2. The marks you received in all your home assignments in English this year and the marks you received in all your home assignments in English last year
3. The cost of riding in a taxi cab that charges a base rate if \begin{align*}\5.00\end{align*} plus \begin{align*}\0.25\end{align*} for every mile you go.
4. The time in minutes that it takes you to walk to work each day for 10 days.
The dependent variable (distance) is increasing as the independent variable (time) is increasing.
1. Using the TRACE function will give the coordinates of the points
1. The fuel consumption of a car travelling at a speed of 72 km/h is approximately 10 L.
2. The speed of a car that has a fuel consumption of 12 km/L is approximately 85 km/h
1. The coldest day was July \begin{align*}7^{th}\end{align*}.
2. The hottest day was July \begin{align*}19^{th}\end{align*}.
3. There does not appear to be a change in temperature on July \begin{align*}1^{st}\end{align*} and \begin{align*}2^{nd}\end{align*}, July \begin{align*}10^{th}\end{align*} and \begin{align*}11^{th}\end{align*}, July \begin{align*}17^{th}\end{align*} and \begin{align*}18^{th}\end{align*}.
1. Seattle
2. Both cities appear to have rising temperature as the month progresses, but Seattle seems to have more hot days and on the \begin{align*}20^{th}\end{align*}, the temperature is still rising. The temperature in New York seemed to rise on the \begin{align*}19^{th}\end{align*} but on the \begin{align*}20^{th}\end{align*} the temperature appears to drop off.
3. There appears to be a difference of approximately 10 degrees between the temperatures of the cities
2. The first graph is labeled correctly as being continuous data. The amount of fuel remaining in your gas tank is plotted for each hour you drive. However, the amount of fuel in your gas tank decreases every minute/second you drive. All values on the graph are meaningful and therefore can be joined. This is continuous data. The second graph is also labeled correctly as being discrete data. The cost of CDs is plotted for each CD you purchase. The cost to you changes only when another CD is purchased. The values between the plotted points are not meaningful and therefore are not joined. This is discrete data.
3. \begin{align*}& \text{Number of Days} && 1 && 2 && 3 && 4 && 5\\ & \text{Cost(\)} && \75 && \85 && \95 && \105 && \115\end{align*} The cost of renting the car for one week (7 days) would be \begin{align*}\135.00\end{align*}. This is indicated on the graph by the horizontal line that is drawn from the \begin{align*}7^{th}\end{align*} day to the cost axis.
1. A scatter plot
2. A double line graph
3. A line graph
4. A broken-line graph
Answer Key for Review Questions (even numbers)
2.
Using the TRACE function will give the coordinates of the points
4.a. The coldest day was July \begin{align*}7^{th}\end{align*}.
b. The hottest day was July \begin{align*}19^{th}\end{align*}.
c. There does not appear to be a change in temperature on July \begin{align*}1^{st}\end{align*} and \begin{align*}2^{nd}\end{align*}, July \begin{align*}10^{th}\end{align*} and \begin{align*}11^{th}\end{align*}, July \begin{align*}17^{th}\end{align*} and \begin{align*}18^{th}\end{align*}.
6. The first graph is labeled correctly as being continuous data.
The amount of fuel remaining in your gas tank is plotted for each hour you drive. However, the amount of fuel in your gas tank decreases every minute/second you drive. All values on the graph are meaningful and therefore can be joined. This is continuous data.
The second graph is also labeled correctly as being discrete data. The cost of CDs is plotted for each CD you purchase. The cost to you changes only when another CD is purchased. The values between the plotted points are not meaningful and therefore are not joined. This is discrete data.
8. a. A scatter plot
b. A double line graph
c. A line graph
d. A broken-line graph
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Show Hide Details
Description
Tags:
Subjects:
|
# What is the solution to {:(–x+2y=4),(5x-3y=1):}?
Jan 13, 2017
The solution is $\left(x , y\right) = \left(2 , 3\right)$.
#### Explanation:
Each of these equations represents a line in 2D space. As with any pair of lines, they may cross, they may be parallel, or they may be the same line. Solving a pair of equations simultaneously means finding the $\left(x , y\right)$ point where the lines cross (if it exists).
We start by assuming there is a point $\left(x , y\right)$ that works for both equations
$\text{-} x + 2 y = 4$
$5 x - 3 y = 1$
If this is true, then we can rearrange each equation and combine the two equations together to help us narrow in on the coordinates of the $\left(x , y\right)$ point.
For example, if $\text{-} x + 2 y = 4$, then we have
$\textcolor{b l u e}{x = 2 y - 4}$
by solving for $x$. But, if this is the same $x$ that works for the other equation, we can substitute this expression for $x$ into the other equation like this:
$\text{ "5color(blue)x" } - 3 y = 1$
$5 \left(\textcolor{b l u e}{2 y - 4}\right) - 3 y = 1$
and we end up with an equation with just $y$. Thus, we can solve for $y$:
$10 y - 20 - 3 y = 1$
$\textcolor{w h i t e}{10 y - 20 -} 7 y = 21$
$\textcolor{w h i t e}{10 y - 20 - 7} \textcolor{red}{y = 3}$
So, this is the $y$-coordinate of the point that works for both lines. With this, we can now find the matching $x$-coordinate, by plugging in this value for $y$ into either of our starting equations:
$\text{-"x+2color(red)y" } = 4$
$\text{-} x + 2 \textcolor{red}{\left(3\right)} = 4$
$\text{-"x+6" } = 4$
$\text{-"x" "="-} 2$
$\implies x = 2$
That's it—we have found that the lines do cross, and the coordinates of the crossing point are $\left(x , y\right) = \left(2 , 3\right)$:
graph{(-x+2y-4)(5x-3y-1)=0 [-10, 10, -2, 8]}
|
Upcoming SlideShare
×
# Equivalent, simplifyng and comparing fractions
856 views
Published on
Published in: Technology
2 Likes
Statistics
Notes
• Full Name
Comment goes here.
Are you sure you want to Yes No
• Be the first to comment
Views
Total views
856
On SlideShare
0
From Embeds
0
Number of Embeds
105
Actions
Shares
0
28
0
Likes
2
Embeds 0
No embeds
No notes for slide
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• ### Equivalent, simplifyng and comparing fractions
1. 1. Equivalent, Simplifyingand Comparing Fractions By : YuHan
2. 2. Equivalent Fractions 1Equivalent Fractions are 2 fractions that havethe same value but a different Numerator andDenominator.
3. 3. Simplifying Fractions 1T Simplify Multiplication Fraction Problem : oYou can simplify a multiplication problem whenthe first fractions numerator is a multiple of thesecond fractions denominator. You can alsosimplify the second fractions numerator with thefirst fractions denominator.
4. 4. Simplifying Fractions 2Simplifying a Division Fraction Problem :You can simplify a division fraction problem bydoing the exact same steps as simplifying aMultiplication Fraction problem. But you have toflip the second fractions numerator anddenominator around before you simplify.
5. 5. Comparing Fractions 1Fractions with the same denominator is called aLike Fraction and fractions with differentdenominator is called a Unlike Fraction. T see owhich like fractions is greater, the fraction withthe greater numerator is bigger.
6. 6. Comparing Fractions 2T calculate which fractions with the same onumerator is bigger, you will have to see whichdenominator is smaller. The smaller thedenominator, the less parts the fraction has,making the numerator bigger. In a simpler form,The smaller the denominator, the larger thefraction is.
7. 7. Comparing Fractions 3How do you see which fraction is bigger whenthey both have a different denominator and anumerator? One of the ways is to first find thelowest common denominator of both thefractions and multiply the numerator by howmuch you multiplied to get the denominatorbelow. After you do those steps on both the -
8. 8. Comparing Fractions 4fractions, do the steps on the ComparingFractions 1 Slide to know which fraction isgreater.
|
# 7.05 Cross sections of prisms and pyramids
Lesson
A cross-section is the 2D shape we get when cutting straight through a 3D object. For example, the orange below (let's pretend it's a perfect sphere) has been cut along the vertical plane (straight up and down). The cross-section of the orange, where we can see inside, is a circle.
When we slice a piece of fruit, we see its cross-section.
#### Exploration
Let's use the applet below to explore cross-sections that are parallel to the base of a prism and pyramid with the same polygon base.
1. Drag the blue point in the 3D view up and down. What do you notice about the cross-section of the pyramid compared to the cross-section of the prism?
2. Move the slider at the bottom to change the number of sides. Do your observations change when you increase the number of sides for the base?
Now let's explore some cross-sections that are perpendicular to the base of a prism and pyramid with the same polygon base.
1. Drag the blue point in the 3D view to move the plane from left to right. What do you notice about the cross-section of the pyramid?
2. What do you notice about the cross-section of the prism?
3. Move the slider at the bottom to change the number of sides. Do your observations change when you increase the number of sides for the base?
4. How do the vertical cross-sections compare to the horizontal cross-sections of the same type of solid?
Recall that prisms have rectangular sides, and the shape on the top and the base is the same. The name of the base shape gives the prism its name.
Prisms
Triangular Square Rectangular Pentagonal Hexagonal Octagonal
Any cross-section taken parallel to the base in a prism is always the same shape and size as the base. In other words, we say that a prism has a uniform cross-section.
Recall that pyramids have triangular sides, and the shape of the base gives the prism its name.
Pyramids
Triangular Square Rectangular Pentagonal Hexagonal Octagonal
In a pyramid, any cross-section taken parallel to the base is always the same shape but is smaller in size than the base.
From the applets, we can see that three-dimensional shapes have more than one cross-section and they may or not be the same shape. It all depends on which way we cut them!
#### Practice questions
##### Question 1
We want to classify the following solid:
1. Does the shape have a uniform cross-section ?
Yes
A
No
B
2. The solid is a:
Triangular Pyramid
A
Rectangular Prism
B
Rectangular Pyramid
C
Triangular Prism
D
##### Question 2
Consider the solid in the adjacent figure.
1. If the solid is cut straight down below the dotted line, what cross-section results?
A pentagon
A
A triangle
B
A hexagon
C
A square
D
2. Does the solid above have a uniform cross-section?
Yes
A
No
B
3. What is the name of the solid?
A triangular prism
A
A rectangular prism
B
A hexagonal prism
C
A pentagonal prism.
D
##### Question 3
Which two of the objects below could have the following cross-section?
1. Triangular Pyramid
A
Cylinder
B
Pentagonal Pyramid
C
Pentagonal Prism
D
### Outcomes
#### MGSE7.G.3
Describe the two-dimensional figures (cross sections) that result from slicing three- dimensional figures, as in plane sections of right rectangular prisms, right rectangular pyramids, cones, cylinders, and spheres.
|
# Know Your Multiplication: Multiply by 3
In this worksheet, students learn to multiply by 3.
Key stage: KS 2
Curriculum topic: Number: Multiplication and Division
Curriculum subtopic: Use Multiplication/Division Facts (3, 4 and 8)
Difficulty level:
### QUESTION 1 of 10
This worksheet is about multiplying by 3.
Example
There are 9 pots.
Each pot has 3 flowers.
We want to work out how many flowers there are in total.
What is 9 times 3?
In total, there are 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 27 flowers.
What is 3 times 9?
3 × 9 = 27
9 × 3 = 27
Want to understand this further and learn how this links to other topics in maths?
Why not watch this video?
Here are 5 cones.
Each cone has 3 scoops of ice cream.
We want to work out how many scoops there are in total.
What is 3 times 5?
Here are 5 cars.
Each car has 3 persons.
We want to work out how many persons there are in total.
What is 5 times 3?
Here are 4 pots.
Each pot has 3 flowers.
We want to work out how many flowers there are in total.
What is 4 times 3?
Here are 7 cones.
Each cone has 3 scoops of ice cream.
We want to work out how many scoops there are in total.
What is 7 times 3?
Here are 6 cars.
Each car has 3 persons.
We want to work out how many persons there are in total.
What is 6 times 3?
Here are 9 pots.
Each pot has 3 flowers.
We want to work out how many flowers there are in total.
What is 9 times 3?
Here are 15 cars.
Each car has 3 persons.
We want to work out how many persons there are in total.
What is 15 times 3?
Here are 6 cones.
Each cone has 3 scoops of ice cream.
We want to work out how many scoops there are in total.
What is 6 times 3?
Here are 12 pots.
Each pot has 3 flowers.
We want to work out how many flowers there are in total.
What is 12 times 3?
Here are 10 cars.
Each car has 3 persons.
We want to work out how many persons there are in total.
What is 10 times 3?
• Question 1
Here are 5 cones.
Each cone has 3 scoops of ice cream.
We want to work out how many scoops there are in total.
What is 3 times 5?
15
EDDIE SAYS
3 x 5 = 15
5 + 5 + 5 = 15
• Question 2
Here are 5 cars.
Each car has 3 persons.
We want to work out how many persons there are in total.
What is 5 times 3?
15
EDDIE SAYS
5 x 3 = 15
3 + 3 + 3 + 3 + 3 = 15
• Question 3
Here are 4 pots.
Each pot has 3 flowers.
We want to work out how many flowers there are in total.
What is 4 times 3?
12
EDDIE SAYS
4 x 3 = 12
3 + 3 + 3 +3 = 12
• Question 4
Here are 7 cones.
Each cone has 3 scoops of ice cream.
We want to work out how many scoops there are in total.
What is 7 times 3?
21
EDDIE SAYS
7 x 3 = 21
3 + 3 + 3 + 3 + 3 + 3 + 3 = 21
• Question 5
Here are 6 cars.
Each car has 3 persons.
We want to work out how many persons there are in total.
What is 6 times 3?
18
EDDIE SAYS
6 x 3 = 18
3 + 3 + 3 + 3 + 3 + 3 = 18
• Question 6
Here are 9 pots.
Each pot has 3 flowers.
We want to work out how many flowers there are in total.
What is 9 times 3?
27
EDDIE SAYS
9 x 3 = 27
3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 27
• Question 7
Here are 15 cars.
Each car has 3 persons.
We want to work out how many persons there are in total.
What is 15 times 3?
45
EDDIE SAYS
15 x 3 = 45
3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3+ 3 = 45
• Question 8
Here are 6 cones.
Each cone has 3 scoops of ice cream.
We want to work out how many scoops there are in total.
What is 6 times 3?
18
EDDIE SAYS
6 x 3 = 18
3 + 3 + 3 + 3 + 3 + 3 = 18
• Question 9
Here are 12 pots.
Each pot has 3 flowers.
We want to work out how many flowers there are in total.
What is 12 times 3?
36
EDDIE SAYS
12 x 3 = 36
3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 36
• Question 10
Here are 10 cars.
Each car has 3 persons.
We want to work out how many persons there are in total.
What is 10 times 3?
30
EDDIE SAYS
10 x 3 = 30
3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 30
---- OR ----
Sign up for a £1 trial so you can track and measure your child's progress on this activity.
### What is EdPlace?
We're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them.
Get started
|
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# Evaluate $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$. What is the value of $\sin \left( {60^\circ + 30^\circ } \right)$. What can you conclude?
Last updated date: 14th Aug 2024
Total views: 390.3k
Views today: 5.90k
Verified
390.3k+ views
Hint: Here we will put the values of the sine and cosine functions in the first expression and find the required value. We will then find the values of the second expression by adding the angles and substituting the value of sine function of the obtained angle. Then we will compare the values of both the expressions. Then by using this we will get the required relationship between the given expressions.
Complete step by step solution:
First, we will find the value of the given expression i.e. $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$.
We know that the value of $\sin 30^\circ = \dfrac{1}{2},\sin 60^\circ = \dfrac{{\sqrt 3 }}{2},\cos 30^\circ = \dfrac{{\sqrt 3 }}{2},\cos 60^\circ = \dfrac{1}{2}$.
Now substituting all these values in the expression, we get
$\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ = \dfrac{{\sqrt 3 }}{2} \times \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{2} \times \dfrac{1}{2}$
Multiplying the terms, we get
$\Rightarrow \sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ = \dfrac{3}{4} + \dfrac{1}{4} = 1$…………………….$\left( 1 \right)$
Now we will find the value of $\sin \left( {60^\circ + 30^\circ } \right)$. Therefore, by simply adding the angle inside the bracket we can write the equation as
$\sin \left( {60^\circ + 30^\circ } \right) = \sin \left( {90^\circ } \right)$
We know that the value of $\sin \left( {90^\circ } \right) = 1$. Therefore, we get
$\Rightarrow \sin \left( {60^\circ + 30^\circ } \right) = 1$……………………..$\left( 2 \right)$
Now we can see from the equation $\left( 1 \right)$ and equation $\left( 2 \right)$ that the values as these equations are equations. Therefore, we get
$\Rightarrow \sin \left( {60^\circ + 30^\circ } \right) = \sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$
Hence from this we get this identity as
$\Rightarrow \sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$
Note:
In order to solve this question, we need to know the different values of the trigonometric function for different angles. Also in which quadrant which function is positive or negative. In the first quadrant, all the functions i.e. sin, cos, tan, cot, sec, cosec are positive. In the second quadrant, only the sin and cosec function are positive and all the other functions are negative. In the third quadrant, only tan and cot function is positive and in the fourth quadrant, only cos and sec function is positive.
|
# Linear Algebra/Definition and Examples of Vector Spaces/Solutions
## Solutions
{{{1}}}
This exercise is recommended for all readers.
Problem 3
Show that each of these is a vector space.
1. The set of linear polynomials ${\displaystyle {\mathcal {P}}_{1}=\{a_{0}+a_{1}x\,{\big |}\,a_{0},a_{1}\in \mathbb {R} \}}$ under the usual polynomial addition and scalar multiplication operations.
2. The set of ${\displaystyle 2\!\times \!2}$ matrices with real entries under the usual matrix operations.
3. The set of three-component row vectors with their usual operations.
4. The set
${\displaystyle L=\{{\begin{pmatrix}x\\y\\z\\w\end{pmatrix}}\in \mathbb {R} ^{4}\,{\big |}\,x+y-z+w=0\}}$
under the operations inherited from ${\displaystyle \mathbb {R} ^{4}}$.
Most of the conditions are easy to check; use Example 1.3 as a guide. Here are some comments.
1. This is just like Example 1.3; the zero element is ${\displaystyle 0+0x}$.
2. The zero element of this space is the ${\displaystyle 2\!\times \!2}$ matrix of zeroes.
3. The zero element is the vector of zeroes.
4. Closure of addition involves noting that the sum
${\displaystyle {\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\\w_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\\w_{2}\end{pmatrix}}={\begin{pmatrix}x_{1}+x_{2}\\y_{1}+y_{2}\\z_{1}+z_{2}\\w_{1}+w_{2}\end{pmatrix}}}$
is in ${\displaystyle L}$ because ${\displaystyle (x_{1}+x_{2})+(y_{1}+y_{2})-(z_{1}+z_{2})+(w_{1}+w_{2})=(x_{1}+y_{1}-z_{1}+w_{1})+(x_{2}+y_{2}-z_{2}+w_{2})=0+0}$. Closure of scalar multiplication is similar. Note that the zero element, the vector of zeroes, is in ${\displaystyle L}$.
This exercise is recommended for all readers.
Problem 4
Show that each of these is not a vector space. (Hint. Start by listing two members of each set.)
1. Under the operations inherited from ${\displaystyle \mathbb {R} ^{3}}$, this set
${\displaystyle \{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\in \mathbb {R} ^{3}\,{\big |}\,x+y+z=1\}}$
2. Under the operations inherited from ${\displaystyle \mathbb {R} ^{3}}$, this set
${\displaystyle \{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\in \mathbb {R} ^{3}\,{\big |}\,x^{2}+y^{2}+z^{2}=1\}}$
3. Under the usual matrix operations,
${\displaystyle \{{\begin{pmatrix}a&1\\b&c\end{pmatrix}}\,{\big |}\,a,b,c\in \mathbb {R} \}}$
4. Under the usual polynomial operations,
${\displaystyle \{a_{0}+a_{1}x+a_{2}x^{2}\,{\big |}\,a_{0},a_{1},a_{2}\in \mathbb {R} ^{+}\}}$
where ${\displaystyle \mathbb {R} ^{+}}$ is the set of reals greater than zero
5. Under the inherited operations,
${\displaystyle \{{\begin{pmatrix}x\\y\end{pmatrix}}\in \mathbb {R} ^{2}\,{\big |}\,x+3y=4{\text{ and }}2x-y=3{\text{ and }}6x+4y=10\}}$
In each item the set is called ${\displaystyle Q}$. For some items, there are other correct ways to show that ${\displaystyle Q}$ is not a vector space.
1. It is not closed under addition; it fails to meet condition 1.
${\displaystyle {\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\0\end{pmatrix}}\in Q\qquad {\begin{pmatrix}1\\1\\0\end{pmatrix}}\not \in Q}$
2. It is not closed under addition.
${\displaystyle {\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\0\end{pmatrix}}\in Q\qquad {\begin{pmatrix}1\\1\\0\end{pmatrix}}\not \in Q}$
3. It is not closed under addition.
${\displaystyle {\begin{pmatrix}0&1\\0&0\end{pmatrix}},\,{\begin{pmatrix}1&1\\0&0\end{pmatrix}}\in Q\qquad {\begin{pmatrix}1&2\\0&0\end{pmatrix}}\not \in Q}$
4. It is not closed under scalar multiplication.
${\displaystyle 1+1x+1x^{2}\in Q\qquad -1\cdot (1+1x+1x^{2})\not \in Q}$
5. It is empty, violating condition 4.
Problem 5
Define addition and scalar multiplication operations to make the complex numbers a vector space over ${\displaystyle \mathbb {R} }$.
The usual operations ${\displaystyle (v_{0}+v_{1}i)+(w_{0}+w_{1}i)=(v_{0}+w_{0})+(v_{1}+w_{1})i}$ and ${\displaystyle r(v_{0}+v_{1}i)=(rv_{0})+(rv_{1})i}$ suffice. The check is easy.
This exercise is recommended for all readers.
Problem 6
Is the set of rational numbers a vector space over ${\displaystyle \mathbb {R} }$ under the usual addition and scalar multiplication operations?
No, it is not closed under scalar multiplication since, e.g., ${\displaystyle \pi \cdot 1}$ is not a rational number.
Problem 7
Show that the set of linear combinations of the variables ${\displaystyle x,y,z}$ is a vector space under the natural addition and scalar multiplication operations.
The natural operations are ${\displaystyle (v_{1}x+v_{2}y+v_{3}z)+(w_{1}x+w_{2}y+w_{3}z)=(v_{1}+w_{1})x+(v_{2}+w_{2})y+(v_{3}+w_{3})z}$ and ${\displaystyle r\cdot (v_{1}x+v_{2}y+v_{3}z)=(rv_{1})x+(rv_{2})y+(rv_{3})z}$. The check that this is a vector space is easy; use Example 1.3 as a guide.
Problem 8
Prove that this is not a vector space: the set of two-tall column vectors with real entries subject to these operations.
${\displaystyle {\begin{pmatrix}x_{1}\\y_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}\\y_{2}\end{pmatrix}}={\begin{pmatrix}x_{1}-x_{2}\\y_{1}-y_{2}\end{pmatrix}}\qquad r\cdot {\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}rx\\ry\end{pmatrix}}}$
The "${\displaystyle +}$" operation is not commutative (that is, condition 2 is not met); producing two members of the set witnessing this assertion is easy.
Problem 9
Prove or disprove that ${\displaystyle \mathbb {R} ^{3}}$ is a vector space under these operations.
1. ${\displaystyle {\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\end{pmatrix}}={\begin{pmatrix}0\\0\\0\end{pmatrix}}\quad {\text{and}}\quad r{\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}rx\\ry\\rz\end{pmatrix}}}$
2. ${\displaystyle {\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\end{pmatrix}}={\begin{pmatrix}0\\0\\0\end{pmatrix}}\quad {\text{and}}\quad r{\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}0\\0\\0\end{pmatrix}}}$
1. It is not a vector space.
${\displaystyle (1+1)\cdot {\begin{pmatrix}1\\0\\0\end{pmatrix}}\neq {\begin{pmatrix}1\\0\\0\end{pmatrix}}+{\begin{pmatrix}1\\0\\0\end{pmatrix}}}$
2. It is not a vector space.
${\displaystyle 1\cdot {\begin{pmatrix}1\\0\\0\end{pmatrix}}\neq {\begin{pmatrix}1\\0\\0\end{pmatrix}}}$
This exercise is recommended for all readers.
Problem 10
For each, decide if it is a vector space; the intended operations are the natural ones.
1. The diagonal ${\displaystyle 2\!\times \!2}$ matrices
${\displaystyle \{{\begin{pmatrix}a&0\\0&b\end{pmatrix}}\,{\big |}\,a,b\in \mathbb {R} \}}$
2. This set of ${\displaystyle 2\!\times \!2}$ matrices
${\displaystyle \{{\begin{pmatrix}x&x+y\\x+y&y\end{pmatrix}}\,{\big |}\,x,y\in \mathbb {R} \}}$
3. This set
${\displaystyle \{{\begin{pmatrix}x\\y\\z\\w\end{pmatrix}}\in \mathbb {R} ^{4}\,{\big |}\,x+y+z+w=1\}}$
4. The set of functions ${\displaystyle \{f:\mathbb {R} \to \mathbb {R} \,{\big |}\,df/dx+2f=0\}}$
5. The set of functions ${\displaystyle \{f:\mathbb {R} \to \mathbb {R} \,{\big |}\,df/dx+2f=1\}}$
For each "yes" answer, you must give a check of all the conditions given in the definition of a vector space. For each "no" answer, give a specific example of the failure of one of the conditions.
1. Yes.
2. Yes.
3. No, it is not closed under addition. The vector of all ${\displaystyle 1/4}$'s, when added to itself, makes a nonmember.
4. Yes.
5. No, ${\displaystyle f(x)=e^{-2x}+(1/2)}$ is in the set but ${\displaystyle 2\cdot f}$ is not (that is, condition 6 fails).
This exercise is recommended for all readers.
Problem 11
Prove or disprove that this is a vector space: the real-valued functions ${\displaystyle f}$ of one real variable such that ${\displaystyle f(7)=0}$.
It is a vector space. Most conditions of the definition of vector space are routine; we here check only closure. For addition, ${\displaystyle (f_{1}+f_{2})\,(7)=f_{1}(7)+f_{2}(7)=0+0=0}$. For scalar multiplication, ${\displaystyle (r\cdot f)\,(7)=rf(7)=r0=0}$.
This exercise is recommended for all readers.
Problem 12
Show that the set ${\displaystyle \mathbb {R} ^{+}}$ of positive reals is a vector space when "${\displaystyle x+y}$" is interpreted to mean the product of ${\displaystyle x}$ and ${\displaystyle y}$ (so that ${\displaystyle 2+3}$ is ${\displaystyle 6}$), and "${\displaystyle r\cdot x}$" is interpreted as the ${\displaystyle r}$-th power of ${\displaystyle x}$.
We check Definition 1.1.
First, closure under "${\displaystyle +}$" holds because the product of two positive reals is a positive real. The second condition is satisfied because real multiplication commutes. Similarly, as real multiplication associates, the third checks. For the fourth condition, observe that multiplying a number by ${\displaystyle 1\in \mathbb {R} ^{+}}$ won't change the number. Fifth, any positive real has a reciprocal that is a positive real.
The sixth, closure under "${\displaystyle \cdot }$", holds because any power of a positive real is a positive real. The seventh condition is just the rule that ${\displaystyle v^{r+s}}$ equals the product of ${\displaystyle v^{r}}$ and ${\displaystyle v^{s}}$. The eight condition says that ${\displaystyle (vw)^{r}=v^{r}w^{r}}$. The ninth condition asserts that ${\displaystyle (v^{r})^{s}=v^{rs}}$. The final condition says that ${\displaystyle v^{1}=v}$.
Problem 13
Is ${\displaystyle \{(x,y)\,{\big |}\,x,y\in \mathbb {R} \}}$ a vector space under these operations?
1. ${\displaystyle (x_{1},y_{1})+(x_{2},y_{2})=(x_{1}+x_{2},y_{1}+y_{2})}$ and ${\displaystyle r\cdot (x,y)=(rx,y)}$
2. ${\displaystyle (x_{1},y_{1})+(x_{2},y_{2})=(x_{1}+x_{2},y_{1}+y_{2})}$ and ${\displaystyle r\cdot (x,y)=(rx,0)}$
1. No: ${\displaystyle 1\cdot (0,1)+1\cdot (0,1)\neq (1+1)\cdot (0,1)}$.
2. No; the same calculation as the prior answer shows a contition in the definition of a vector space that is violated. Another example of a violation of the conditions for a vector space is that ${\displaystyle 1\cdot (0,1)\neq (0,1)}$.
Problem 14
Prove or disprove that this is a vector space: the set of polynomials of degree greater than or equal to two, along with the zero polynomial.
It is not a vector space since it is not closed under addition, as ${\displaystyle (x^{2})+(1+x-x^{2})}$ is not in the set.
Problem 15
At this point "the same" is only an intuition, but nonetheless for each vector space identify the ${\displaystyle k}$ for which the space is "the same" as ${\displaystyle \mathbb {R} ^{k}}$.
1. The ${\displaystyle 2\!\times \!3}$ matrices under the usual operations
2. The ${\displaystyle n\!\times \!m}$ matrices (under their usual operations)
3. This set of ${\displaystyle 2\!\times \!2}$ matrices
${\displaystyle \{{\begin{pmatrix}a&0\\b&c\end{pmatrix}}\,{\big |}\,a,b,c\in \mathbb {R} \}}$
4. This set of ${\displaystyle 2\!\times \!2}$ matrices
${\displaystyle \{{\begin{pmatrix}a&0\\b&c\end{pmatrix}}\,{\big |}\,a+b+c=0\}}$
1. ${\displaystyle 6}$
2. ${\displaystyle nm}$
3. ${\displaystyle 3}$
4. To see that the answer is ${\displaystyle 2}$, rewrite it as
${\displaystyle \{{\begin{pmatrix}a&0\\b&-a-b\end{pmatrix}}\,{\big |}\,a,b\in \mathbb {R} \}}$
so that there are two parameters.
This exercise is recommended for all readers.
Problem 16
Using ${\displaystyle {\vec {+}}}$ to represent vector addition and ${\displaystyle \,{\vec {\cdot }}\,}$ for scalar multiplication, restate the definition of vector space.
A vector space (over ${\displaystyle \mathbb {R} }$) consists of a set ${\displaystyle V}$ along with two operations "${\displaystyle {\mathbin {\vec {+}}}}$" and "${\displaystyle {\mathbin {\vec {\cdot }}}}$" subject to these conditions. Where ${\displaystyle {\vec {v}},{\vec {w}}\in V}$,
1. their vector sum ${\displaystyle {\vec {v}}{\mathbin {\vec {+}}}{\vec {w}}}$ is an element of ${\displaystyle V}$. If ${\displaystyle {\vec {u}},{\vec {v}},{\vec {w}}\in V}$ then
2. ${\displaystyle {\vec {v}}{\mathbin {\vec {+}}}{\vec {w}}={\vec {w}}{\mathbin {\vec {+}}}{\vec {v}}}$ and
3. ${\displaystyle ({\vec {v}}{\mathbin {\vec {+}}}{\vec {w}}){\mathbin {\vec {+}}}{\vec {u}}={\vec {v}}{\mathbin {\vec {+}}}({\vec {w}}{\mathbin {\vec {+}}}{\vec {u}})}$.
4. There is a zero vector ${\displaystyle {\vec {0}}\in V}$ such that ${\displaystyle {\vec {v}}{\mathbin {\vec {+}}}{\vec {0}}={\vec {v}}\,}$ for all ${\displaystyle {\vec {v}}\in V}$.
5. Each ${\displaystyle {\vec {v}}\in V}$ has an additive inverse ${\displaystyle {\vec {w}}\in V}$ such that ${\displaystyle {\vec {w}}{\mathbin {\vec {+}}}{\vec {v}}={\vec {0}}}$. If ${\displaystyle r,s}$ are scalars, that is, members of ${\displaystyle \mathbb {R} }$), and ${\displaystyle {\vec {v}},{\vec {w}}\in V}$ then
6. each scalar multiple ${\displaystyle r\cdot {\vec {v}}}$ is in ${\displaystyle V}$. If ${\displaystyle r,s\in \mathbb {R} }$ and ${\displaystyle {\vec {v}},{\vec {w}}\in V}$ then
7. ${\displaystyle (r+s)\cdot {\vec {v}}=r\cdot {\vec {v}}{\mathbin {\vec {+}}}s\cdot {\vec {v}}}$, and
8. ${\displaystyle r{\mathbin {\vec {\cdot }}}({\vec {v}}+{\vec {w}})=r{\mathbin {\vec {\cdot }}}{\vec {v}}+r{\mathbin {\vec {\cdot }}}{\vec {w}}}$, and
9. ${\displaystyle (rs){\mathbin {\vec {\cdot }}}{\vec {v}}=r{\mathbin {\vec {\cdot }}}(s{\mathbin {\vec {\cdot }}}{\vec {v}})}$, and
10. ${\displaystyle 1{\mathbin {\vec {\cdot }}}{\vec {v}}={\vec {v}}}$.
This exercise is recommended for all readers.
Problem 17
Prove these.
1. Any vector is the additive inverse of the additive inverse of itself.
2. Vector addition left-cancels: if ${\displaystyle {\vec {v}},{\vec {s}},{\vec {t}}\in V}$ then ${\displaystyle {\vec {v}}+{\vec {s}}={\vec {v}}+{\vec {t}}\,}$ implies that ${\displaystyle {\vec {s}}={\vec {t}}}$.
1. Let ${\displaystyle V}$ be a vector space, assume that ${\displaystyle {\vec {v}}\in V}$, and assume that ${\displaystyle {\vec {w}}\in V}$ is the additive inverse of ${\displaystyle {\vec {v}}}$ so that ${\displaystyle {\vec {w}}+{\vec {v}}={\vec {0}}}$. Because addition is commutative, ${\displaystyle {\vec {0}}={\vec {w}}+{\vec {v}}={\vec {v}}+{\vec {w}}}$, so therefore ${\displaystyle {\vec {v}}}$ is also the additive inverse of ${\displaystyle {\vec {w}}}$.
2. Let ${\displaystyle V}$ be a vector space and suppose ${\displaystyle {\vec {v}},{\vec {s}},{\vec {t}}\in V}$. The additive inverse of ${\displaystyle {\vec {v}}}$ is ${\displaystyle -{\vec {v}}}$ so ${\displaystyle {\vec {v}}+{\vec {s}}={\vec {v}}+{\vec {t}}}$ gives that ${\displaystyle -{\vec {v}}+{\vec {v}}+{\vec {s}}=-{\vec {v}}+{\vec {v}}+{\vec {t}}}$, which says that ${\displaystyle {\vec {0}}+{\vec {s}}={\vec {0}}+{\vec {t}}}$ and so ${\displaystyle {\vec {s}}={\vec {t}}}$.
Problem 18
The definition of vector spaces does not explicitly say that ${\displaystyle {\vec {0}}+{\vec {v}}={\vec {v}}}$ (it instead says that ${\displaystyle {\vec {v}}+{\vec {0}}={\vec {v}}}$). Show that it must nonetheless hold in any vector space.
Addition is commutative, so in any vector space, for any vector ${\displaystyle {\vec {v}}}$ we have that ${\displaystyle {\vec {v}}={\vec {v}}+{\vec {0}}={\vec {0}}+{\vec {v}}}$.
This exercise is recommended for all readers.
Problem 19
Prove or disprove that this is a vector space: the set of all matrices, under the usual operations.
It is not a vector space since addition of two matrices of unequal sizes is not defined, and thus the set fails to satisfy the closure condition.
Problem 20
In a vector space every element has an additive inverse. Can some elements have two or more?
Each element of a vector space has one and only one additive inverse.
For, let ${\displaystyle V}$ be a vector space and suppose that ${\displaystyle {\vec {v}}\in V}$. If ${\displaystyle {\vec {w}}_{1},{\vec {w}}_{2}\in V}$ are both additive inverses of ${\displaystyle {\vec {v}}}$ then consider ${\displaystyle {\vec {w}}_{1}+{\vec {v}}+{\vec {w}}_{2}}$. On the one hand, we have that it equals ${\displaystyle {\vec {w}}_{1}+({\vec {v}}+{\vec {w}}_{2})={\vec {w}}_{1}+{\vec {0}}={\vec {w}}_{1}}$. On the other hand we have that it equals ${\displaystyle ({\vec {w}}_{1}+{\vec {v}})+{\vec {w}}_{2}={\vec {0}}+{\vec {w}}_{2}={\vec {w}}_{2}}$. Therefore, ${\displaystyle {\vec {w}}_{1}={\vec {w}}_{2}}$.
Problem 21
1. Prove that every point, line, or plane thru the origin in ${\displaystyle \mathbb {R} ^{3}}$ is a vector space under the inherited operations.
2. What if it doesn't contain the origin?
1. Every such set has the form ${\displaystyle \{r\cdot {\vec {v}}+s\cdot {\vec {w}}\,{\big |}\,r,s\in \mathbb {R} \}}$ where either or both of ${\displaystyle {\vec {v}},{\vec {w}}}$ may be ${\displaystyle {\vec {0}}}$. With the inherited operations, closure of addition ${\displaystyle (r_{1}{\vec {v}}+s_{1}{\vec {w}})+(r_{2}{\vec {v}}+s_{2}{\vec {w}})=(r_{1}+r_{2}){\vec {v}}+(s_{1}+s_{2}){\vec {w}}}$ and scalar multiplication ${\displaystyle c(r{\vec {v}}+s{\vec {w}})=(cr){\vec {v}}+(cs){\vec {w}}}$ are easy. The other conditions are also routine.
2. No such set can be a vector space under the inherited operations because it does not have a zero element.
This exercise is recommended for all readers.
Problem 22
Using the idea of a vector space we can easily reprove that the solution set of a homogeneous linear system has either one element or infinitely many elements. Assume that ${\displaystyle {\vec {v}}\in V}$ is not ${\displaystyle {\vec {0}}}$.
1. Prove that ${\displaystyle r\cdot {\vec {v}}={\vec {0}}}$ if and only if ${\displaystyle r=0}$.
2. Prove that ${\displaystyle r_{1}\cdot {\vec {v}}=r_{2}\cdot {\vec {v}}}$ if and only if ${\displaystyle r_{1}=r_{2}}$.
3. Prove that any nontrivial vector space is infinite.
4. Use the fact that a nonempty solution set of a homogeneous linear system is a vector space to draw the conclusion.
Assume that ${\displaystyle {\vec {v}}\in V}$ is not ${\displaystyle {\vec {0}}}$.
1. One direction of the if and only if is clear: if ${\displaystyle r=0}$ then ${\displaystyle r\cdot {\vec {v}}={\vec {0}}}$. For the other way, let ${\displaystyle r}$ be a nonzero scalar. If ${\displaystyle r{\vec {v}}={\vec {0}}}$ then ${\displaystyle (1/r)\cdot r{\vec {v}}=(1/r)\cdot {\vec {0}}}$ shows that ${\displaystyle {\vec {v}}={\vec {0}}}$, contrary to the assumption.
2. Where ${\displaystyle r_{1},r_{2}}$ are scalars, ${\displaystyle r_{1}{\vec {v}}=r_{2}{\vec {v}}\,}$ holds if and only if ${\displaystyle (r_{1}-r_{2}){\vec {v}}={\vec {0}}}$. By the prior item, then ${\displaystyle r_{1}-r_{2}=0}$.
3. A nontrivial space has a vector ${\displaystyle {\vec {v}}\neq {\vec {0}}}$. Consider the set ${\displaystyle \{k\cdot {\vec {v}}\,{\big |}\,k\in \mathbb {R} \}}$. By the prior item this set is infinite.
4. The solution set is either trivial, or nontrivial. In the second case, it is infinite.
Problem 23
Is this a vector space under the natural operations: the real-valued functions of one real variable that are differentiable?
Yes. A theorem of first semester calculus says that a sum of differentiable functions is differentiable and that ${\displaystyle (f+g)^{\prime }=f^{\prime }+g^{\prime }}$, and that a multiple of a differentiable function is differentiable and that ${\displaystyle (r\cdot f)^{\prime }=r\,f^{\prime }}$.
Problem 24
A vector space over the complex numbers ${\displaystyle \mathbb {C} }$ has the same definition as a vector space over the reals except that scalars are drawn from ${\displaystyle \mathbb {C} }$ instead of from ${\displaystyle \mathbb {R} }$. Show that each of these is a vector space over the complex numbers. (Recall how complex numbers add and multiply: ${\displaystyle (a_{0}+a_{1}i)+(b_{0}+b_{1}i)=(a_{0}+b_{0})+(a_{1}+b_{1})i}$ and ${\displaystyle (a_{0}+a_{1}i)(b_{0}+b_{1}i)=(a_{0}b_{0}-a_{1}b_{1})+(a_{0}b_{1}+a_{1}b_{0})i}$.)
1. The set of degree two polynomials with complex coefficients
2. This set
${\displaystyle \{{\begin{pmatrix}0&a\\b&0\end{pmatrix}}\,{\big |}\,a,b\in \mathbb {C} {\text{ and }}a+b=0+0i\}}$
The check is routine. Note that "1" is ${\displaystyle 1+0i}$ and the zero elements are these.
1. ${\displaystyle (0+0i)+(0+0i)x+(0+0i)x^{2}}$
2. ${\displaystyle {\begin{pmatrix}0+0i&0+0i\\0+0i&0+0i\end{pmatrix}}}$
Problem 25
Name a property shared by all of the ${\displaystyle \mathbb {R} ^{n}}$'s but not listed as a requirement for a vector space.
Notably absent from the definition of a vector space is a distance measure.
This exercise is recommended for all readers.
Problem 26
1. Prove that a sum of four vectors ${\displaystyle {\vec {v}}_{1},\ldots ,{\vec {v}}_{4}\in V}$ can be associated in any way without changing the result.
${\displaystyle {\begin{array}{rl}(({\vec {v}}_{1}+{\vec {v}}_{2})+{\vec {v}}_{3})+{\vec {v}}_{4}&=({\vec {v}}_{1}+({\vec {v}}_{2}+{\vec {v}}_{3}))+{\vec {v}}_{4}\\&=({\vec {v}}_{1}+{\vec {v}}_{2})+({\vec {v}}_{3}+{\vec {v}}_{4})\\&={\vec {v}}_{1}+(({\vec {v}}_{2}+{\vec {v}}_{3})+{\vec {v}}_{4})\\&={\vec {v}}_{1}+({\vec {v}}_{2}+({\vec {v}}_{3}+{\vec {v}}_{4}))\end{array}}}$
This allows us to simply write "${\displaystyle {\vec {v}}_{1}+{\vec {v}}_{2}+{\vec {v}}_{3}+{\vec {v}}_{4}}$" without ambiguity.
2. Prove that any two ways of associating a sum of any number of vectors give the same sum. (Hint. Use induction on the number of vectors.)
1. A small rearrangement does the trick.
${\displaystyle {\begin{array}{rl}({\vec {v}}_{1}+({\vec {v}}_{2}+{\vec {v}}_{3}))+{\vec {v}}_{4}&=(({\vec {v}}_{1}+{\vec {v}}_{2})+{\vec {v}}_{3})+{\vec {v}}_{4}\\&=({\vec {v}}_{1}+{\vec {v}}_{2})+({\vec {v}}_{3}+{\vec {v}}_{4})\\&={\vec {v}}_{1}+({\vec {v}}_{2}+({\vec {v}}_{3}+{\vec {v}}_{4}))\\&={\vec {v}}_{1}+(({\vec {v}}_{2}+{\vec {v}}_{3})+{\vec {v}}_{4})\end{array}}}$
Each equality above follows from the associativity of three vectors that is given as a condition in the definition of a vector space. For instance, the second "${\displaystyle =}$" applies the rule ${\displaystyle ({\vec {w}}_{1}+{\vec {w}}_{2})+{\vec {w}}_{3}={\vec {w}}_{1}+({\vec {w}}_{2}+{\vec {w}}_{3})}$ by taking ${\displaystyle {\vec {w}}_{1}}$ to be ${\displaystyle {\vec {v}}_{1}+{\vec {v}}_{2}}$, taking ${\displaystyle {\vec {w}}_{2}}$ to be ${\displaystyle {\vec {v}}_{3}}$, and taking ${\displaystyle {\vec {w}}_{3}}$ to be ${\displaystyle {\vec {v}}_{4}}$.
2. The base case for induction is the three vector case. This case ${\displaystyle {\vec {v}}_{1}+({\vec {v}}_{2}+{\vec {v}}_{3})=({\vec {v}}_{1}+{\vec {v}}_{2})+{\vec {v}}_{3}}$ is required of any triple of vectors by the definition of a vector space. For the inductive step, assume that any two sums of three vectors, any two sums of four vectors, ..., any two sums of ${\displaystyle k}$ vectors are equal no matter how the sums are parenthesized. We will show that any sum of ${\displaystyle k+1}$ vectors equals this one ${\displaystyle ((\cdots (({\vec {v}}_{1}+{\vec {v}}_{2})+{\vec {v}}_{3})+\cdots )+{\vec {v}}_{k})+{\vec {v}}_{k+1}}$. Any parenthesized sum has an outermost "${\displaystyle +}$". Assume that it lies between ${\displaystyle {\vec {v}}_{m}}$ and ${\displaystyle {\vec {v}}_{m+1}}$ so the sum looks like this.
${\displaystyle (\cdots \,({\vec {v}}_{1}+(\cdots +{\vec {v}}_{m}))\,\cdots )+(\cdots \,({\vec {v}}_{m+1}+(\cdots +{\vec {v}}_{k+1}))\,\cdots )}$
The second half involves fewer than ${\displaystyle k+1}$ additions, so by the inductive hypothesis we can re-parenthesize it so that it reads left to right from the inside out, and in particular, so that its outermost "${\displaystyle +}$" occurs right before ${\displaystyle {\vec {v}}_{k+1}}$.
${\displaystyle =(\cdots \,({\vec {v}}_{1}+(\,\cdots \,+{\vec {v}}_{m}))\,\cdots )+(\cdots ({\vec {v}}_{m+1}+(\cdots +{\vec {v}}_{k})\cdots )+{\vec {v}}_{k+1})}$
Apply the associativity of the sum of three things
${\displaystyle =((\,\cdots \,({\vec {v}}_{1}+(\cdots +{\vec {v}}_{m})\,\cdots \,)+(\,\cdots \,({\vec {v}}_{m+1}+(\cdots \,{\vec {v}}_{k}))\cdots )+{\vec {v}}_{k+1}}$
and finish by applying the inductive hypothesis inside these outermost parenthesis.
3. Problem 27
For any vector space, a subset that is itself a vector space under the inherited operations (e.g., a plane through the origin inside of ${\displaystyle \mathbb {R} ^{3}}$) is a subspace.
1. Show that ${\displaystyle \{a_{0}+a_{1}x+a_{2}x^{2}\,{\big |}\,a_{0}+a_{1}+a_{2}=0\}}$ is a subspace of the vector space of degree two polynomials.
2. Show that this is a subspace of the ${\displaystyle 2\!\times \!2}$ matrices.
${\displaystyle \{{\begin{pmatrix}a&b\\c&0\end{pmatrix}}\,{\big |}\,a+b=0\}}$
3. Show that a nonempty subset ${\displaystyle S}$ of a real vector space is a subspace if and only if it is closed under linear combinations of pairs of vectors: whenever ${\displaystyle c_{1},c_{2}\in \mathbb {R} }$ and ${\displaystyle {\vec {s}}_{1},{\vec {s}}_{2}\in S}$ then the combination ${\displaystyle c_{1}{\vec {v}}_{1}+c_{2}{\vec {v}}_{2}}$ is in ${\displaystyle S}$.
1. We outline the check of the conditions from Definition 1.1. Additive closure holds because if ${\displaystyle a_{0}+a_{1}+a_{2}=0}$ and ${\displaystyle b_{0}+b_{1}+b_{2}=0}$ then
${\displaystyle (a_{0}+a_{1}x+a_{2}x^{2})+(b_{0}+b_{1}x+b_{2}x^{2})=(a_{0}+b_{0})+(a_{1}+b_{1})x+(a_{2}+b_{2})x^{2}}$
is in the set since ${\displaystyle (a_{0}+b_{0})+(a_{1}+b_{1})+(a_{2}+b_{2})=(a_{0}+a_{1}+a_{2})+(b_{0}+b_{1}+b_{2})}$ is zero. The second through fifth conditions are easy. Closure under scalar multiplication holds because if ${\displaystyle a_{0}+a_{1}+a_{2}=0}$ then
${\displaystyle r\cdot (a_{0}+a_{1}x+a_{2}x^{2})=(ra_{0})+(ra_{1})x+(ra_{2})x^{2}}$
is in the set as ${\displaystyle ra_{0}+ra_{1}+ra_{2}=r(a_{0}+a_{1}+a_{2})}$ is zero. The remaining conditions here are also easy.
3. Call the vector space ${\displaystyle V}$. We have two implications: left to right, if ${\displaystyle S}$ is a subspace then it is closed under linear combinations of pairs of vectors and, right to left, if a nonempty subset is closed under linear combinations of pairs of vectors then it is a subspace. The left to right implication is easy; we here sketch the other one by assuming ${\displaystyle S}$ is nonempty and closed, and checking the conditions of Definition 1.1. First, to show closure under addition, if ${\displaystyle {\vec {s}}_{1},{\vec {s}}_{2}\in S}$ then ${\displaystyle {\vec {s}}_{1}+{\vec {s}}_{2}\in S}$ as ${\displaystyle {\vec {s}}_{1}+{\vec {s}}_{2}=1\cdot {\vec {s}}_{1}+1\cdot {\vec {s}}_{2}}$. Second, for any ${\displaystyle {\vec {s}}_{1},{\vec {s}}_{2}\in S}$, because addition is inherited from ${\displaystyle V}$, the sum ${\displaystyle {\vec {s}}_{1}+{\vec {s}}_{2}}$ in ${\displaystyle S}$ equals the sum ${\displaystyle {\vec {s}}_{1}+{\vec {s}}_{2}}$ in ${\displaystyle V}$ and that equals the sum ${\displaystyle {\vec {s}}_{2}+{\vec {s}}_{1}}$ in ${\displaystyle V}$ and that in turn equals the sum ${\displaystyle {\vec {s}}_{2}+{\vec {s}}_{1}}$ in ${\displaystyle S}$. The argument for the third condition is similar to that for the second. For the fourth, suppose that ${\displaystyle {\vec {s}}}$ is in the nonempty set ${\displaystyle S}$ and note that ${\displaystyle 0\cdot {\vec {s}}={\vec {0}}\in S}$; showing that the ${\displaystyle {\vec {0}}}$ of ${\displaystyle V}$ acts under the inherited operations as the additive identity of ${\displaystyle S}$ is easy. The fifth condition is satisfied because for any ${\displaystyle {\vec {s}}\in S}$ closure under linear combinations shows that the vector ${\displaystyle 0\cdot {\vec {0}}+(-1)\cdot {\vec {s}}}$ is in ${\displaystyle S}$; showing that it is the additive inverse of ${\displaystyle {\vec {s}}}$ under the inherited operations is routine. The proofs for the remaining conditions are similar.
|
Become a math whiz with AI Tutoring, Practice Questions & more.
HotmathMath Homework. Do It Faster, Learn It Better.
# Pythagorean Theorem
The Pythagorean Theorem is one of the fundamental principles of Euclidean geometry and defines the relationship between the three sides of a right triangle. It was first proven by an ancient Greek mathematician and religious leader named Pythagoras of Samos who believed that everything in the universe was comprised of and could be described by numbers. Modern science sure makes use of this idea!
The Pythagorean Theorem (sometimes called Pythagoras' Theorem) states that in a right triangle with legs of lengths a and b and a hypotenuse of length c, the area of the square whose side is the hypotenuse is equal to the sum of the areas on the squares on the other two sides. If that doesn't make much sense to you, consider the following diagram:
All the Pythagorean Theorem is saying is that the area of square c is equal to the combined areas of squares a and b. Or, the green square has the same area as the blue and red squares added together. That's not so bad, right?
## Expressing the Pythagorean theorem mathematically
Saying the Pythagorean Theorem in words can be difficult and you won't always have access to a diagram like the one above. Fortunately, it can be expressed as a simple equation:
${a}^{2}+{b}^{2}={c}^{2}$ , where c represents the length of the hypotenuse and a and b represent the lengths of the other two sides of the triangle.
The validity of this equation (sometimes called the Pythagorean Equation) has been proven countless times through various mechanisms, some of which date back thousands of years. Any three whole numbers satisfying this equation are called Pythagorean Triples. As long as you memorize this equation, you'll be prepared to apply the Pythagorean Theorem whenever it comes up.
Remember, however, that the Pythagorean Theorem only applies to right triangles. If you're working with a triangle that doesn't have an angle measuring exactly 90 degrees, you cannot apply the Pythagorean Theorem.
## Applying the Pythagorean theorem
As you know from your study of algebraic equations, you can manipulate both sides of an equation to isolate a specific variable and determine more information than you were originally provided with. The same thing applies to the Pythagorean Equation.
If you know the length of both legs of a right triangle but not the hypotenuse, you can use the following equation to solve for c:
$\sqrt{{a}^{2}}+\sqrt{{b}^{2}}=\sqrt{{c}^{2}}$
Likewise, if you know the length of the hypotenuse and one of the legs but not the other, you can use the following equations to get the length of the remaining leg:
$a=\sqrt{{c}^{2}-{b}^{2}}$ OR $b=\sqrt{{c}^{2}-{a}^{2}}$
Since all of these formulas were derived from the Pythagorean Equation, they only apply to right triangles. Eventually, you'll study the law of cosines and discover how to solve for any side of any triangle with the lengths of two sides and the angle between them. When you do, you might notice that the equation can be reduced to the Pythagorean Equation when you're working with a right triangle. Isn't it great when mathematics works out so perfectly?
## The converse of the Pythagorean theorem
While the most obvious application of the Pythagorean Theorem is to determine the length of the sides of a right triangle, the converse is also true. If a triangle has sides a, b, and c, and if ${a}^{2}+{b}^{2}={c}^{2}$ , then it must be a right triangle. Similarly, any triangle with three sides that do not satisfy the ${a}^{2}+{b}^{2}={c}^{2}$ equation cannot be a right triangle. This is a great way to identify right triangles when they aren't directly labeled for you.
## Practical applications of the Pythagorean theorem
Many students struggle with math concepts that feel abstract, often asking, "How am I ever going to use this?" The Pythagorean Theorem is frequently among the biggest offenders since the entire concept revolves around working with very specific triangles you wouldn't expect to find outside of your mathematics textbook.
The Pythagorean Theorem is important for two reasons. First, professionals in STEM fields often use it to describe distances or theoretical objects that haven't been created yet. For example, surveyors will use it to ensure their maps accurately depict the slope and height of geographic features they're describing while engineers utilize it to accurately predict how much material they'll need to design prototypes. There are also practical applications in activities such as flying a kite, though obviously, you can do that without knowing what a hypotenuse is.
Second, the Pythagorean Theorem represents a breakthrough in humanity's creative critical thinking skills. Pythagoras didn't have a modern high school math book to work with and still came up with a theory that can be applied to an infinite number of triangles. Later mathematicians were able to build on his work and discover hundreds of independent proofs and applications, firmly identifying mathematics as an outlet for creative thinking.
If you're considering a career in a STEM field, those abstract triangles might not remain abstract for too much longer. If you aren't, you can still look at the Pythagorean Theorem as an example of what can be achieved if you put your mind to it.
## Pythagorean theorem practice problems
a. If the two non-hypotenuse legs of a right triangle measure 3 and 4 meters, respectively, what is the perimeter of the triangle? (Remember that you find the perimeter by adding all three sides of the triangle together)
${3}^{2}+{4}^{2}={c}^{2}$
$9+16={c}^{2}$
$25={c}^{2}$
$c=\sqrt{25}$
$c=5\mathrm{meters}$
Now we can find the perimeter by adding all three sides together:
$\mathrm{Perimeter}=3+4+5=12\mathrm{meters}$
b. If a right triangle's hypotenuse measures 50 inches and one of its legs measures 14 inches, what is the length of the third leg?
${14}^{2}+{b}^{2}={50}^{2}$
$196+{b}^{2}=2500$
${b}^{2}=2304$
$b=\sqrt{2304}$
$b=48\mathrm{inches}$
c. What is the area of a right triangle with a hypotenuse measuring 70 centimeters and one leg measuring 42 centimeters? (The formula for the area of a triangle is ½ x base x height)
${42}^{2}+{b}^{2}={70}^{2}$
$1764+{b}^{2}=4900$
${b}^{2}=3136$
$b=\sqrt{3136}$
$b=56\mathrm{centimeters}$
Now we can find the area using the formula for the area of a triangle ( $A=\frac{1}{2}×\mathrm{base}×\mathrm{height}$ ):
$\mathrm{Area}=\frac{1}{2}×42×56=1176{\mathrm{cm}}^{2}$
d. The three sides of a triangle measure 10 feet, 20 feet, and 40 feet. Could this be a right triangle? Why or why not?
No, the numbers don't satisfy the equation ${a}^{2}+{b}^{2}={c}^{2}$
e. The three sides of a triangle measure 30 inches, 40 inches, and 50 inches. Could this be a right triangle? Why or why not?
${30}^{2}+{40}^{2}={50}^{2}$
$900+1600=2500$
$2500=2500$
Since the equation holds true, this can be a right triangle.
## Get help with the Pythagorean theorem today
The Pythagorean theorem is a challenging topic that many math students struggle with, so your student is far from alone if they need more help than their classroom teacher can offer. Fortunately, a private tutor can leverage your student's preferred learning style to deepen their understanding of the Pythagorean theorem whether that means using music to help them remember the formula or a crash course on how to perform the calculations involved. Please contact the friendly Educational Directors at Varsity Tutors today to learn more about the benefits of 1-on-1 instruction!
;
|
# How do you find the roots of x^3-12x+16=0?
May 29, 2017
$x = 2 \text{ with multiplicity 2}$
$x = - 4 \text{ with multiplicity 1}$
#### Explanation:
$\text{note that } {2}^{3} - 12 \left(2\right) + 16 = 0$
$\Rightarrow x = 2 \text{ is a root and " (x-2)" is a factor}$
$\Rightarrow {x}^{2} \left(x - 2\right) + 2 x \left(x - 2\right) - 8 \left(x - 2\right)$
$= \left(x - 2\right) \left({x}^{2} + 2 x - 8\right)$
$= \left(x - 2\right) \left(x + 4\right) \left(x - 2\right)$
$\Rightarrow {\left(x - 2\right)}^{2} \left(x + 4\right) = 0$
$\Rightarrow x = 2 \text{ multiplicity of 2}$
$\text{and " x=-4" multiplicity of 1}$
May 29, 2017
The roots are $4$ and $2$(double root)
#### Explanation:
Let $f \left(x\right) = {x}^{3} - 12 x + 16$
Then,
We see that
$f \left(2\right) = {2}^{3} - 12 \cdot 2 + 16 = 8 - 24 + 16 = 0$
Therefore,
$\left(x - 2\right)$ is a factor of $f \left(x\right)$
To find the other factors, we perform a long division
$\textcolor{w h i t e}{a a a a}$$x - 2$$\textcolor{w h i t e}{a a a a}$$|$${x}^{3} + 0 {x}^{2} - 12 x + 16$$\textcolor{w h i t e}{a a a a}$$|$${x}^{2} + 2 x - 8$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a}$${x}^{3} - 2 {x}^{2}$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a}$$0 + 2 {x}^{2} - 12 x$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a}$$+ 2 {x}^{2} - 4 x$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a}$$+ 0 - 8 x + 16$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a}$$- 8 x + 16$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a}$$- 0 + 0$
The quotient is
${x}^{2} + 2 x - 8 = \left(x - 2\right) \left(x + 4\right)$
Therefore,
${x}^{3} - 12 x + 16 = {\left(x - 2\right)}^{2} \left(x + 4\right)$ graph{x^3-12x+16 [-16.15, 15.88, -3.97, 12.05]}
|
# ADDING AND SUBTRACTING REAL NUMBERS
All real numbers can be represented on a number line as shown below.
A number line can be used to model addition and subtraction of real numbers.
To model addition of a positive number, move right. To model addition of a negative number, move left.
Subtraction :
To model subtraction of a positive number, move left. To model subtraction of a negative number, move right.
## Adding and Subtracting Numbers on a Number Line
Add or subtract using a number line.
Example 1 :
-3 + 6
Solution :
On a number line, start at 0 and move left to -3.
To add 6, move right 6 units.
-3 + 6 = 3
Example 2 :
-2 - (-9)
Solution :
On a number line, start at 0 and move left to -2.
To subtract -9, move right 9 units.
-2 - (-9) = 7
## Absolute Value of a Number
The absolute value of a number is its distance from zero on a number line. The absolute value of 5 is 5.
|5| = 5
|-5| = 5
Adding Numbers with the Same Sign :
Add the absolute values and use the sign of the numbers.
4 + 711 -4 + (-9)-13
Adding Numbers with Different Signs :
Subtract the absolute values and use the sign of the number with the greater absolute value.
-7 + 125 5 + (-14)-9
Two numbers are opposites, if their sum is 0. A number and its opposite are additive inverses and are the same distance from the zero. They have the same absolute value.
For example, 5 and -5 are opposites, because
5 + (-5) = 0
Additive inverse of 5 is -5 and additive inverse of -5 is 5.
And also, 5 and -5 have the same absolute value.
|5| = 5
|-5| = 5
Words :
The sum of a real number and its opposite is 0.
Numbers :
7 + (-7) = (-7) + 7 = 0
Algebra :
For any real number y,
y + (-y) + (-y) + y = 0
## Subtracting Real Numbers
Words :
Numbers :
2 - 6 = 2 + (-6) = -4
Algebra :
For two real numbers x and y,
x - y = x + (-y)
Kindly mail your feedback to v4formath@gmail.com
## Recent Articles
1. ### Angular Speed and Linear Speed
Dec 07, 22 05:15 AM
Angular Speed and Linear Speed - Concepts - Formulas - Examples
2. ### Linear Speed Formula
Dec 07, 22 05:13 AM
Linear Speed Formula and Examples
|
Percent decrease from 719 to 55
This page will answer the question "What is the percent decrease from 719 to 55?" and also show you how to calculate the percent decrease from 719 to 55.
Before we continue, note that the percent decrease from 719 to 55 is the same as the percentage decrease from 719 to 55. Furthermore, we will refer to 719 as the initial value (n) and 55 as the final value (f).
So what exactly are we calculating? The initial value is 719 and then a percent is used to decrease the initial value to the final value of 55. We want to calculate what that percent is!
Here are step-by-step instructions showing you how to calculate the percent decrease from 719 to 55.
First, we calculate the amount of decrease from 719 to 55 by subtracting the final value from the initial value, like this:
719 - 55
= 664
To calculate the percent of any number, you multiply the value (n) by the percent (p) and then divide the product by 100 to get the answer, like this:
(n × p) / 100 = Answer
In our case, we know that the initial value (n) is 719 and that the answer (amount of decrease) is 664 to get the final value of 55. Therefore, we fill in what we know in the equation above to get the following equation:
(719 × p) / 100 = 664
Next, we solve the equation above for percent (p) by first multiplying each side by 100 and then dividing both sides by 719 to get percent (p):
(719 × p) / 100 = 664
((719 × p) / 100) × 100 = 664 × 100
719p = 66400
719p / 719 = 66400 / 719
p = 92.3504867872045
Percent ≈ 92.3505
That's all there is to it! The percentage decrease from 719 to 55 is 92.3505%. In other words, if you take 92.3505% of 719 and subtract it from 719, then the difference will be 55.
The step-by-step instructions above were made so we could clearly explain exactly what a percent decrease from 719 to 55 means. For future reference, you can use the following percent decrease formula to calculate percent decreases:
((n - f)/n) × 100 = p
f = Final Value
n = Initial Value
p = Percent decrease
Once again, here is the math and the answer to calculate the percent decrease from 719 to 55 using the percent decrease formula above:
((n - f)/n) × 100
= ((719 - 55)/719) × 100
= (664/719) × 100
= 0.923504867872045 × 100
≈ 92.3505
Percent Decrease Calculator
Go here if you need to calculate another percent decrease.
Percent decrease from 719 to 56
Here is the next percent decrease tutorial on our list that may be of interest.
|
Basic Algebra/Introduction to Basic Algebra Ideas/Working With Negative Numbers
Positive
Negative
Lesson
Negative Numbers
A positive number is a number greater than zero.
A negative number is a number less than zero. You make a negative number by doing the negative operation on a positive number. You use the " – " sign for the negative operation. This sign is the same you use for subtracting.
Adding a negative number is the same as subtracting a positive number.
• ${\displaystyle 7+(-4)=7-4}$
• ${\displaystyle x+(-y)=x-y}$
Subtracting a negative number is the same as adding a positive number.
• ${\displaystyle 7-(-4)=7+4}$
• ${\displaystyle x-(-y)=x+y}$
Multiplying and Dividing
Multiplying a negative number by a positive number, or a positive number by a negative number makes the result negative.
• ${\displaystyle (-2)\times 3=-6}$
• ${\displaystyle 2\times (-3)=-6}$
Multiplying a negative number by a negative number makes the result positive.
• ${\displaystyle (-2)\times (-3)=6}$
You do the same for dividing.
• ${\displaystyle (-6)\div 3=-2}$
• ${\displaystyle 6\div (-3)=-2}$
• ${\displaystyle (-6)\div (-3)=2}$
Exponentiating
Exponentiating a negative number to an even (a number you can divide by two) power makes the result positive.
• ${\displaystyle (-3)^{2}=9}$
• ${\displaystyle (-x)^{2}=(-x)\times (-x)=x^{2}}$
Exponentiating a negative number to an odd (a number you can not divide by two) power makes the result negative.
• ${\displaystyle (-2)^{3}=-8}$
• ${\displaystyle (-x)^{3}=(-x)\times (-x)\times (-x)=x^{2}\times (-x)=-x^{3}}$
Order of Operations
The negative operation has the same precedence as multiplying and dividing.
• ${\displaystyle 3+8\div 4=3+2=5}$
• ${\displaystyle -3^{2}=-(3\times 3)=-9}$
• ${\displaystyle (-3)^{2}=(-3)\times (-3)=9}$
Example Problems
• ${\displaystyle 4+(-4)=0}$
• ${\displaystyle 4+(-7)=-3}$
• ${\displaystyle 0+(-2)=-2}$
• ${\displaystyle -5+7-2\times (-4)=10}$
Practice Problems
1
${\displaystyle 6+(-3)=}$
2
${\displaystyle 3+(-9)=}$
3
${\displaystyle -4\times 4=}$
4
${\displaystyle 4\times (-9)=}$
5
${\displaystyle -2\times (-4)=}$
6
${\displaystyle {\frac {-25}{5^{2}}}=}$
7
${\displaystyle -4\div 2=}$
« Basic AlgebraWorking With Negative Numbers » Variables and Expressions Solving Equations Using Properties of Mathematics
|
Home Mathematics Modeling mathematical ideas: developing strategic competence in elementary and middle school
# LESSON STUDY VIGNETTE: THE CATHEDRAL PROBLEM
For this lesson study, we decided to focus on one specific problem that was provided to the teachers as an opening problem for the day, the cathedral problem that is adapted from Burns, S. (2003):
Text Box 9.3 A Math Happening 9b: The Cathedral Problem
While building a medieval cathedral, it cost 37 guilders to hire 4 artists and 3 stonemasons, or 33 guilders for 3 artists and 4 stonemasons. What would be the expense of just 1 of each worker?
The following sections include the data analysis of teacher thinking that went into solving this problem. Each group was asked to create a poster representing their solution and each teacher was also asked to reflect on how they participated in the problem-solving process and how they would take this problem back to their classroom to present to their students. Photographs of all of the posters created by the teacher groups of the cathedral problem were taken and arranged in order on the day of completion.
Topics included reasoning up and down, direct and inverse thinking, unitizing, and, ratios and proportional thinking. The data from the posters were analyzed for content, connections between concepts, and any possible differences related to the time already spent in the seminar. Data were also analyzed from the teacher reflections for common themes as well as individual perspectives. First, we will present the analysis of the poster artifacts based on what the teachers in the respective groups shared followed by our analysis of the corresponding teacher reflections.
Poster Proofs to Document Visible Thinking
The first group, Figure 9.4 (left), argued that if one artist and one stonemason together made \$11, then the total for three of each would be \$33. However, they reasoned that because we know that \$33 is enough to pay those six workers plus another stonemason, then one stonemason and one artist must together make less than \$11. This group then used the guess-and-check method. They first assumed that the total for one artist and one stonemason was \$8. They tried the combination of \$1 for the cost of one artist and \$7 for the cost of one stonemason (\$8 total); however, they discovered that the total for 3 artists and 4 stonemasons was less than the needed \$33.
Figure 9.4 Poster artifacts showing guess and check, tabular/pictorial, linear addition. Source: Authors.
They tried other combinations but saw that the total cost was decreased; so, they abandoned the idea of an \$8 total. They then assumed that the total for one artist and one stonemason was \$9. However, their starting guess for the cost of one artist was \$2, not \$1. They found that the combination of three artists at \$3 each and four stonemasons at \$6 did total the needed \$33. However, when they used these amounts in the second scenario, they found that it did not work: four artists at \$3 each and three stonemasons at \$6 did not total the needed \$37. They then assumed that the total for one artist and one stonemason was \$10. Using the same logic, starting at \$1 per artist and \$9 per stonemason, then \$2 per artist and \$8 per stonemason, etc., they arrived at a solution of \$3 per artist and \$7 per stonemason, which they demonstrated would satisfy both requirements.
In all three guess-and-check calculations, they assumed that the artists would earn less than the stonemasons would; so, they arrived at the correct figures for the solutions but had the assignments to the two types of workers backward. They showed that four artists at \$3 each and three stonemasons at \$7 each would total \$33. However, the original question stated that the cost of \$33 applied to three artists and four stonemasons. So, although their logic was correct, they made a minor error in the interpretation.
The second group, Figure 9.4 (middle), presented tabular and pictorial representations of the two scenarios. There are also several indications that they chose the values of seven and \$3 for the costs of the two types of workers, but there is no clear explanation of how they arrived at that conclusion. At the top left of the poster, four rows of seven marks each are made to represent the artists; each group of seven is circled, showing that the cost of each of four artists is \$7.
To the right, there are three rows of three marks each, representing that each of three stonemasons earns \$3 each. There is no indication that any values other than the correct solution were considered. There is also no indication of exactly how the correct values were calculated. However, at the bottom of the poster, the expressions written seem to indicate that an algebraic solution using two simultaneous equations was employed.
The third group, Figure 9.4 (right), presented a linear addition solution using symbols to balance two equations. The top of the poster shows the two scenarios and the bottom of the poster shows the addition of these two. The top left of their poster shows 3 tens (squares) and 7 units (circles) to represent \$37. On the right of the equal sign, there are four A’s, for artists, and three S’s, for stonemasons. Directly below that is a similar configuration to represent a \$33 cost for three artists and four stonemasons. The lower left portion of the poster shows 6 tens (squares) and 10 units (circles) to represent \$70. This is the addition of the tens (squares) and units (circles) from the two equations on the top of the poster.
On the right side, there are seven A’s and seven S’s, which are the sum of the A’s and S’s from the two equations. This group reasoned that they now had a total of seven artists and seven stonemasons and a total of \$70. They circled one artist, one stonemason, and the group of ten units (circles) to show that one of each worker would cost \$10. This was one of the few groups who answered the question as written. They made no attempt to determine the individual costs for one artist or one stonemason. Members of this group were not unanimous about whether or not they should do so;
however, several members of this group were confident that the question merely asked for the cost for one artist and one stonemason together and that individual costs were not required.
The fourth group (not shown) listed two scenarios and then depicted the artists making \$7 each and the stonemasons making \$3 each. This seemed to be working backward. They reason that \$33 and \$37 added together equals \$70; simultaneously, they reason that three artists and four stonemasons added to four artists and three stonemasons results in seven of each type of worker. If seven artists and seven stonemasons cost \$70, the group reasons that one artist and one stonemason cost \$10.
An interesting approach to finding the individual cost for each type of workers follows. First, the group realizes that both scenarios have three artists and three stonemasons. One scenario has an extra artist, and the other scenario has an extra stonemason. Based on their conclusion that one artist and one stonemason cost \$10, they derive that three artists and three stonemasons cost \$30. Using this baseline, they argue that the scenario, which has the extra artist, is \$37, which is \$7 more than their baseline. Therefore, the artist must cost \$7. And, the scenario which has an extra stonemason costs \$33, which is \$3 more than their baseline. Therefore, the stonemason must cost \$3.
While each poster seemed to represent a different thinking behind the solution strategy, teachers were able to make important connections between their respective poster proofs. All five of the possible representation strategies were used by the groups: tables, pictures, graphs, numbers and symbols, and verbal descriptions. A review of the artifacts from the other three groups (B, C, D) revealed a similar variety of approaches. Some groups used a strictly algebraic strategy with the simultaneous equations 4a + 3s = 37 and 3a + 4s = 33.
However, there were a variety of other interesting approaches that were employed by the teachers in solving these equations beyond the traditional textbook approaches including substitution, linear addition, or matrix approaches. One group started off with finding ways to arrive at partitioning the number 37, including 25 + 12, 26 + 11, and 27 + 10, even though none of these contain one value which is divisible by 7 and another value which is divisible by 3. Their last attempt, though, 28 + 9, did factor correctly. They then used the same strategy to arrive at partitioning 33, using 19 + 14, 20 + 13, and finally arriving at the correct 21 + 12.
Another group, first wanted to do a comparison to determine who made more. They reasoned that the cost with an extra artist was greater than the cost with an extra stonemason, concluding that artists cost more. Using algebra, they found that the cost for one artist was \$4 greater than the cost of a stonemason. Their poster used this idea to show that an artist earns \$7 and a stonemason earns \$3. Another interesting observation involving parity was made by one of the groups because the total cost in either scenario was odd and the number of total workers in each scenario was odd, then the individual pay for each type of worker must be odd. If there are four workers of the same type, then their total pay will be an even integer. But, the three remaining workers must have an odd wage or else the total cost would be an even integer. Using the same logic in the second scenario, their poster showed that both types of workers must have an odd value for their daily pay.
As these poster illustrations clearly indicate, the teachers exhibited a wide variation in their thinking. Specifically, the comparison strategy generated a lot of interesting conversation that helped the instructors to bring a nice closure to this problem using proportional reasoning. Since this group demonstrated that one artist must cost \$4 more than a stonemason, starting from the first setup of four artists and three stonemasons that costs \$37, it became clear to the teachers that adding an artist and taking away a stonemason to this will yield five artists and two stonemasons that costs \$41.
Continuing this process, we have six artists and one stonemason that costs \$45 and doing this one more time yields seven artists that costs \$49 which immediately helps to solve for the cost of an artist which was \$7 and from that obtain the cost of a stonemason which was \$3. It was interesting to see such profound thinking from teachers for a problem that most students would normally do using a textbook approach of solving system of linear equations.
Related topics
|
Guided Lessons
Premium
# Introduction to Division with Remainders
In this lesson, your students will use visual and hands-on strategies to divide whole numbers with remainders.
Need extra help for EL students? Try theDivision SkitsPre-lesson.
EL Adjustments
GradeSubjectView aligned standards
No standards associated with this content.
Which set of standards are you looking for?
Need extra help for EL students? Try theDivision SkitsPre-lesson.
Students will be able to divide whole numbers with remainders using base ten blocks and standard algorithm.
The adjustment to the whole group lesson is a modification to differentiate for children who are English learners.
EL adjustments
(5 minutes)
• Using a projector, demonstrate how to divide with a remainder by building the solution to a problem with base ten cubes (e.g., 14 ÷ 4).
• Point out the RemainderIn your solution, and define the term (e.g., "After dividing a number into equal groups, the remainder is the whole number that is left over, or remaining.").
(10 minutes)
• Build another problem with base ten blocks (e.g., 35 ÷ 2).
• Draw the same problem in a maths notebook, using a projector to show students the process.
• Teach the standard algorithm for the same example and discuss how the algorithm is connected to the visual representation.
• Repeat with another example like 134 ÷ 3(build, draw, algorithm).
(15 minutes)
• Hand out base ten blocks so that every student has a set.
• Write a problem on the board and have students build it with a partner. Then have them draw their solution in their own maths notebook (e.g., 17 ÷ 5).
• Invite a volunteer to show their drawn solution to class.
• As a class, solve the same problem with the standard algorithm.
• Repeat with another example, like 53 ÷ 3(build, draw, algorithm). But this time, have students use the algorithm with their partner. Then, review as a class.
(20 minutes)
• Give each student a strip of three problems so that students seated near one another have different strips. Ensure that before copying & cutting the worksheet into strips, you have labeled each row (A, B, C).
• Instruct students to BuildThe first problem listed on their strip with blocks, DrawThe second, and solve the third with the Standard algorithm.
• Circulate the room as students work and provide support as needed.
• Organise a jigsaw review:
• Group students according to which strip of problems they solved and have them review their process and solutions as a group (e.g., students with the strip labeled "A" would check their work with other "A's"). Note: students may need to rebuild their solution to problem one as a group in order to check their work.
• Check in with each group as they review. (Optional: take photos of their base ten block solutions to display or add to maths notebooks.)
Support:
• Give students the same strip of problems as their seat partner so they can work together to solve them.
• Provide a pre-drawn solution and have students build it using blocks.
• Group students into smaller groups during the jigsaw review so that only 3–4 students are in each group. (Note: in this scenario, you will have more than one group reviewing each strip of problems.)
Enrichment:
• Have students write a word problem using one of the problems on their strips.
• Have students solve word problems with remainders (see optional materials for a sample worksheet).
(5 minutes)
• On a half sheet of paper, assign students from each jigsaw group a division problem (e.g., "If you were in group A, solve 18 ÷ 5"). This will ensure that students are solving a different problem than their seat partner, so that the assessment is completed independently.
• Have them write their problem at the top of their scratch paper, then instruct them to draw the problem on one half of the paper and solve it with the standard algorithm on the other. Collect and check for understanding.
(5 minutes)
• Display a division problem, using the standard algorithm, which is solved incorrectly (e.g., 62 ÷ 5 = 11 r7).
• Have students correct the problem by drawing it out and ask them to explain what was wrong with the other solution (e.g., "The remainder shouldn't be greater than the divisor.").
• Ask students to consider the role of remainders and discuss. (i.e., What happens to remainders after we solve a problem? Are remainders important? When might we use division and need to consider the remainder?)
### Add to collection
Create new collection
0
0Items
|
## Permutations and Combinations
### 3.1 A Grid of Numbers
Lesson materials located below the video overview.
Here’s a famous puzzle:
Starting at the top-left cell marked S and taking horizontal steps one place only to the right or vertical steps downwards only, how many different paths are to the location marked E?
Play with this puzzle for a while before reading on. As you play, perhaps contemplate the following questions:
1. Given the location of the point E, is the grid shown in the diagram unnecessarily large?
2. Marking in different paths from S to E is awfully complicated. One could first count paths to different cells first, ones easier to handle, and look for patterns. For example, how many distinct paths are there from S to any cell on the top row? Write the answers in those cells. How many distinct paths take you to any cell in the leftmost column? To cells in the second row? Second column? Third row?
3. If you are willing to trust patterns, can you make a good guess as to the answer to the original puzzle?
There are two ways to approach this puzzle.
APPROACH NUMBER 1: FORMULAS
Every path from S to E can be described by a sequence of letters R and D. For example, the path given in the diagram can be described by the sequence:
RDRRDDRRRDRR
This sequence contains eight Rs and four Ds. Moreover, any sequence of eight Rs and four Ds corresponds to a path from S to E.
Practice: Mark in on the diagram the paths given by DDRRRRRDRRRD and RRRRRRRRDDDD.
Thus the number of paths from S to E matches the number of ways to arrange twelve letters, eight Rs and four Ds. (That is, to label twelve slots with eight Rs and four Ds). The answer to the original puzzle is:
$$\dfrac{12!}{8!4!} = 495$$ paths.
Exercise 30: How many paths are there from S to the bottom-right cell of the grid?
Exercise 31: Suppose the cell E is $$a$$ steps to the right of S and $$b$$ steps down from S. Show that the number of paths from S to E is given by: $$\dfrac{(a+b)!}{a!b!}$$.
Number the rows 0, 1, 2, … with the top row being the zero-th row (for “zero steps to the right”) and number the columns 0, 1, 2, … with the leftmost column being the zero-th column (for “zero steps down”). The cell E in the original diagram has position: row 4, column 8. The number of paths to it is $$\dfrac{12!}{4!8!}$$. This is the number of ways to arrange 4 Rs and 8 Ds (RRRRDDDDDDDD).
In general, numbering rows and columns this way, the cell in row $$a$$ and column $$b$$ requires $$a$$ Rs and $$b$$ Ds to get to it and so the number of paths to it is:
$$\dfrac{(a+b)!}{a!b!}$$.
Notice that this formula is correct for the entries on the top row which require $$b=0$$ down steps, and for entries in the left column with $$a=0$$:
$$\dfrac{(a+0)!}{a!0!} = \dfrac{a!}{a!0!} = 1$$
$$\dfrac{(0+b)!}{0!b!} = \dfrac{b!}{0!b!} = 1$$
(Thank goodness we set $$0! = 1$$.) Notice for the top left entry right at position S we get:
$$\dfrac{(0+0)!}{0!0!} = \dfrac{0!}{0!0!} = 1$$.
Apparently the math suggests there is one way to start at S and end at S – just stay put!
APPROACH 2: PATTERNS
If you fill in the answers for the number of paths to each cell, the following grid of numbers appears:
(Exercise 32 suggests that the position labeled S should also be assigned the number 1.)
Exercise 33: Explain why the table is symmetrical about the southeast diagonal line.
We have the formula $$\dfrac{(a+b)!}{a!b!}$$ for the entry in the $$a$$-th row and $$b$$-th column (starting the counts at zero).
Have you noticed that each entry in an interior cell is the sum of two numbers – the number just above the cell and the number just to the left of the cell? This makes sense in terms of counting paths. Consider the circled cell. To reach this cell one can either first reach the cell just above – there are 15 ways to do this – and then step down, or reach the cell just to its left – there are 20 ways to accomplish this – and then step right. This gives a total of 15 + 20 = 35 paths to the circled cell.
Practice: Use this observation to fill the remainder of the table.
## Books
Take your understanding to the next level with easy to understand books by James Tanton.
BROWSE BOOKS
## Guides & Solutions
Dive deeper into key topics through detailed, easy to follow guides and solution sets.
BROWSE GUIDES
## Donations
Consider supporting G'Day Math! with a donation, of any amount.
Your support is so much appreciated and enables the continued creation of great course content. Thanks!
|
# Area of an equilateral triangle
The area of an equilateral triangle is $\frac{s^2\sqrt{3}}{4}$, where $s$ is the sidelength of the triangle.
## Proof
Method 1: Dropping the altitude of our triangle splits it into two triangles. By HL congruence, these are congruent, so the "short side" is $\frac{s}{2}$.
Using the Pythagorean theorem, we get $s^2=h^2+\frac{s^2}{4}$, where $h$ is the height of the triangle. Solving, $h=\frac{s\sqrt{3}}{2}$. (note we could use 30-60-90 right triangles.)
We use the formula for the area of a triangle, ${bh \over 2}$ (note $s$ is the length of a base), so the area is $$\boxed{\frac{s^2\sqrt{3}}{4}}$$
Method 2: (warning: uses trig.) The area of a triangle is $\frac{ab\sin{C}}{2}$. Plugging in $a=b=s$ and $C=\frac{\pi}{3}$ (the angle at each vertex, in radians), we get the area to be $\frac{s^2\sin{c}}{2}=\frac{s^2\frac{\sqrt{3}}{2}}{2}=\boxed{\frac{s^2\sqrt{3}}{4}}$
|
# HCF And LCM
## Highest common factor and Lowest common multiple ( HCF & LCM )
HCF & LCM :-
Let us see here some points related to Factors, Multiples, LCM and HCF :
1. Factors and Multiples : All the numbers that divide a number completely, i.e.,
without leaving any remainder, are called factors of that number. For example,
24 is completely divisible by 1, 2, 3, 4, 6, 8, 12, 24. Each of these numbers
is called a factor of 24 and 24 is called a multiple of each of these numbers.
2. LCM ( Lowest common multiple)
The least number which is exactly
divisible by each of the given numbers is called the least common multiple of
those numbers. For example, consider the numbers 3, 31 and 62 (2 x 31). The LCM
of these numbers would be 2 x 3 x 31 = 186.
To find the LCM of the given numbers, we express each number as a product of
prime numbers. The product of highest power of the prime numbers that appear in
prime factorization of any of the numbers gives us the LCM.
For example, consider the numbers 2, 3, 4 (2 x 2), 5, 6 (2 x 3). The LCM of
these numbers is 2 x 2 x 3 x 5 = 60. The highest power of 2 comes from prime
factorization of 4, the highest power of 3 comes from prime factorization of 3
and prime factorization of 6 and the highest power of 5 comes from prime
factorization of 5.
3. HCF (Highest common factor)
The greatest number that divides two
or more numbers is the highest common factor (HCF) for those numbers. For
example, consider the numbers 30 (2 x 3 x 5), 36 (2 x 2 x 3 x 3), 42 (2 x 3 x
7), 45 (3 x 3 x 5). 3 is the largest number that divides each of these numbers,
and hence, is the HCF for these numbers.
HCF is also known as Greatest Common Divisor (GCD).
HCF और LCM: –
आइए हम यहां फैक्टर्स, मल्टीपल्स, एलसीएम और एचसीएफ से संबंधित कुछ बिंदुओं को देखते हैं:
1. कारक और गुणक: वे सभी संख्याएँ जो किसी संख्या को पूर्ण रूप से विभाजित करती हैं, अर्थात, बिना किसी शेष को छोड़े, उस संख्या के कारक कहलाते हैं। उदाहरण के लिए, 24 1, 2, 3, 4, 6, 8, 12, 24 से पूरी तरह से विभाज्य है। इनमें से प्रत्येक संख्या को 24 का कारक कहा जाता है और 24 को इनमें से प्रत्येक संख्या का गुणक कहा जाता है।
2. एलसीएम (सबसे कम सामान्य एकाधिक): कम से कम संख्या जो दिए गए प्रत्येक संख्या से बिल्कुल विभाज्य है, उन संख्याओं में से सबसे कम संख्या वाली सामान्य बहु को कहा जाता है। उदाहरण के लिए, संख्या 3, 31 और 62 (2 x 31) पर विचार करें। इन संख्याओं का LCM 2 x 3 x 31 = 186 होगा।
दिए गए नंबरों के LCM को खोजने के लिए, हम प्रत्येक संख्या को अभाज्य संख्याओं के उत्पाद के रूप में व्यक्त करते हैं। किसी भी संख्या के अभाज्य गुणनखंड में दिखाई देने वाली अभाज्य संख्याओं की उच्चतम शक्ति का उत्पाद हमें LCM देता है।
उदाहरण के लिए, संख्या 2, 3, 4 (2 x 2), 5, 6 (2 x 3) पर विचार करें। इन संख्याओं का LCM 2 x 2 x 3 x 5 = 60 है। 2 की उच्चतम शक्ति 4 के अभाज्य गुणनखंड से आती है, 3 की उच्चतम शक्ति 3 के अभाज्य गुणनखंड से आती है और 6 की अभाज्य गुणनखंडन से। और 5 की उच्चतम शक्ति 5 के मुख्य कारक से आता है।
3. एचसीएफ (उच्चतम सामान्य कारक): दो या अधिक संख्याओं को विभाजित करने वाली सबसे बड़ी संख्या उन संख्याओं के लिए उच्चतम सामान्य कारक (एचसीएफ) है। उदाहरण के लिए, संख्या 30 (2 x 3 x 5), 36 (2 x 2 x 3 x 3), 42 (2 x 3 x 7), 45 (3 x 3 x 5) पर विचार करें। 3 सबसे बड़ी संख्या है जो इनमें से प्रत्येक संख्या को विभाजित करती है, और इसलिए, इन संख्याओं के लिए HCF है।
HCF को ग्रेटेस्ट कॉमन डिविज़र (GCD) के रूप में भी जाना जाता है।
|
<meta http-equiv="refresh" content="1; url=/nojavascript/">
You are reading an older version of this FlexBook® textbook: CK-12 Texas Instruments Geometry Teacher's Edition Go to the latest version.
# 6.2: Hanging with the Incenter
Difficulty Level: At Grade Created by: CK-12
This activity is intended to supplement Geometry, Chapter 5, Lesson 3.
ID: 11360
Time Required: 45 minutes
## Activity Overview
In this activity, students will explore the angle bisector theorem and discover that if a point is on the angle bisector of a segment, then the point is equidistant from the sides. This is an introductory activity, but students will need to know how to grab and move points, measure lengths, and construct the perpendicular bisector with Cabri Jr.
Topic: Triangles & Their Centers
• Angle Bisector Theorem
• Incenter
Teacher Preparation and Notes
Associated Materials
## Problem 1 – Exploring the Angle Bisector Theorem
Students will be exploring the distance from a point on the angle bisector to the segment. They will discover that if a point is on the angle bisector of an angle, then it is equidistant from the two segments of the angle.
Students will measure the lengths of segments using the Distance & Length tool (press GRAPH and select Measure > D.& Length). To drag a point, students will move the cursor over the point, press ALPHA, move the point to the desired location, and then press ALPHA again to release the point.
## Problem 2 – Exploring the Incenter of a Triangle
Students will need to open a new Cabri Jr. file by pressing $Y=$, selecting New, and answer no if asked to save.
Students are to create an acute triangle and find the angle bisector of all three angles of the triangle. Students should realize that they are concurrent and answer Questions 4–8 on the accompanying worksheet.
Students will need to find the distance from the incenter to the 3 sides of the triangle. Students may need to be reminded how to find the distance from a point to a line or segment. It is the perpendicular distance from the incenter to the side. Teachers should encourage the students to hide their angle bisectors. Be sure that all students understand that the perpendicular line created from the incenter to a side of the triangle is not the perpendicular bisector.
Students should discover that all of these distances are equal.
## Problem 3 – Extension
Problem 3 is an extension of this activity. Students will use the handheld and the angle bisectors to find the coordinates of fence posts on a plot of land. Note: This problem will be very hard for the average student.
Students must first set up coordinates for the 2 fences at (-1, 2), (1, 0), (1, 3), and (5, 3). Next, students should create segments for the two fences. From here, there are several approaches to solving this problem.
Using the incenter:
One approach is to find the incenter of the triangle formed by the intersection of the two fences (over the pond) and a segment joining the two fences. Students should construct a segment from (1, 0) to (5, 3) connecting the two fences. This is to create two angles that they can use to find angle bisectors. The point of intersection of the two angle bisectors is the incenter of the triangle. This point is therefore equidistant from the two fences.
However, we need two points to determine the line where the fence will be. Students will need to create a similar triangle that connects the two fences. To guarantee similarity, choose a segment that is parallel to the segment joining (1, 0) and (5, 3) using the Parallel tool (Press ZOOM, then scroll down to Parallel). Find the incenter of the second triangle and you can create the line through those two incenters. One point on the line should be at (5.24, 0) or (5.2, 0) and the other can be any other point on the line connecting the two incenters. The $y-$intercept is another convenient point to find the coordinates of (0, 2.2).
Using only an angle bisector:
Another approach is to find the point of intersection of the fences (if they were continued over the pond). Students can construct lines for the Fence 1 and Fence 2 and plot the intersection point of the lines. Then, students can use the Angle Bisector tool to find the angle bisector of this angle. This ray will be equidistant from Fence 1 and Fence 2. Students can plot a point on this ray and display and record the coordinates of this point. Then, move the point and record the new coordinates.
## Solutions
Position $1^{st}$ position $2^{nd}$ position $3^{rd}$ position $4^{th}$ position
$DF$ 2.25473 2.72284 1.85757 1.12521
$DE$ 2.25473 2.72284 1,85757 1.12521
2. They are equal.
3. Equidistant
4. They are concurrent.
5. Not possible
6. Not possible
7. Obtuse, Acute, and Right
8. They are all equal.
9. Sample answer: (5.2, 0) and (0, 2.2)
Feb 23, 2012
Nov 03, 2014
|
# Degree (Angles)|Definition & Meaning
## Definition
An angle with a measure of one degree is equal to one-third of a whole circle. It is most frequently used in mathematics and science, where it is represented by the sign °. Although radians and degrees are both valid ways to express angles, degrees are more frequently employed.
In a plane with only two dimensions, an angle that measures one degree is referred to as a degree angle. Itt is one of the most frequently employed metrics to express the overall magnitude of angles in two-dimensional shapes. In geometry, trigonometry, and many other subfields of mathematics, a degree angle is a standard unit of measurement.
It is common practice to express the angle in terms of an arc length, which refers to the total distance traveled around the circumference of a circle that the angle subtends. For instance, traveling a quarter of the length of a circle corresponds to a 90 degrees movement around the circle.
## Diagrammatic Representation of the Degree Angles
The following figure represents the one complete rotation of the circle, which is equal to 360 degrees. Each quarter is 90 degrees.
Figure 1 – Complete representation of degree angle
The different angles of the degree angle system are represented in the following figure.
Figure 2 – Different angles of degree system
The following plot shows the degrees that are present in each quarter. Each of the four quarters has a degree angle of 90. To get from point A to point B, you will need to turn through an angle of 180 degrees to accomplish what you set out to do.
Figure 3 – Degree representation for each quarter
## What Is an Angle in Degrees?
One of the most fundamental and crucial ideas in geometry is the concept of a degree angle. It measures both the size and the form of an angle. The measurement of an angle formed by two lines meeting at a common point is sometimes referred to as a “degree angle.” It can also be used to indicate the size of an angle created when an object is rotated by a certain degree.
The amount of rotation around a point equals one degree. The angle created by this rotation, which is expressed in degrees, is known as a degree angle. One degree of an angle is equivalent to 1/360th of a circle since a whole circle has 360 degrees.
The most frequent angle is a right angle, which is 90 degrees. Because it is used to determine the size and shape of angles and other objects, the degree angle plays a significant role in geometry.
In trigonometry and other mathematical applications, degrees are also used to measure angles. Everyday life also makes use of the degree angle. To ensure that a window or door frame fits properly, the angle of the frame is measured.
## Applications of Degree Units in Angle Measurements
There are many applications for the degree angle in mathematics and science, as it is a standard unit of measurement for angles.
The angles of various geometric figures, such as triangles and circles, are measured in degrees. Trigonometry is the branch of mathematics that uses measurements in terms of angles to determine other quantities, such as lengths, distances, and even slopes and heights.
Stars, planets, and other heavenly bodies can all have their positions and orientations determined using degree angles. Forces, velocities, and accelerations in physics are all measured in terms of degrees of angle.
Degree angles have practical applications beyond the scientific realm as well. Cake pan angles, for instance, can be measured in degrees for more precise results. In the same way, builders utilize degree angles to set the pitch of a building’s walls, roof, and other components.
Chair and table angles, as well as other furniture angles, can be measured in degrees. Clothing measurements, such as the angle of a neckline or sleeve, can also be expressed in degrees.
## Distinction Between Radians and Degrees
Radians and degrees are both angle measurement units that are used to calculate the size of angles and arcs in circles. The conversion factor is the primary distinction between both. Radians are the standard unit of angle measurement in mathematics and physics, and they are based on the radian, the natural unit of a circle.
A radian is basically the angle subtended by an arc on a circle where the length of the arc is equal to the circle’s radius.
In other words, if you draw a circle with a unit radius and an arc with a length of one on it, the angle subtended by the arc is one radian. This definition assumes that a whole circle has an angle of 2 radians. Conversely, degrees are the conventional unit of angle measurement in common use and are based on the ancient Babylonians’ sexagesimal system.
A degree is one-third of a whole circle. This signifies that the angle of a whole circle is 360°/180 is the conversion factor between radians and degrees. Because the measurement is based on the natural unit of a circle, radians are commonly used in mathematics and physics, where precision is critical. Because degrees are easy to use and understand, they are commonly utilized in everyday applications such as navigation.
Furthermore, radians and degrees both measure angles in distinct ways. Radians measure angle size based on the length of an arc on a circle, whereas degrees measure angle size based on a fraction of a full circle. This means that the same angle can be measured in radians and degrees in various ways.
## Numerical Example of Angle Conversion from Degrees to Radians
What is the equivalent of 180 degrees in radians?
### Solution
To convert it into radians, we know that:
angle in radians = angle in degrees x ($\pi$ radians / 180 degrees)
By putting values, we get:
180 degrees in radians = 180 x ($\pi$ / 180)
180 degrees in radians = $\pi$ radians
All images/mathematical drawings were created with GeoGebra.
|
# Every Line is a Graph of a Linear Equation
Videos to help Grade 8 students learn how to prove that any point on the graph of y = mx + b is on a line l and that any point on a line l is a point on the graph of y = mx + b.
Related Topics:
Lesson Plans and Worksheets for Grade 8
Lesson Plans and Worksheets for all Grades
More Lessons for Grade 8
Common Core For Grade 8
## New York State Common Core Math Module 4, Grade 8, Lesson 20
### Lesson 20 Student Outcomes
• Students know that any non-vertical line is the graph of a linear equation in the form of y = mx + b, where b is a constant.
• Students write the equation that represents the graph of a line.
### Lesson 20 Summary
• Write the equation of a line by determining the y-intercept, (0, b) and the slope, m, and replacing the numbers b and m into the equation y = mx + b
Lesson 20 Opening Exercise
Find the equations of the lines for graph 1 and graph 2.
Examples
Given the graph of a line, we want to be able to write the equation that represents it.
Which form of a linear equation do you think will be most valuable for this task, the standard form ax + by = c, or slope-intercept form y = mx + b.
Write the equation that represents the graph of the line shown below:
First, identify the y-intercept.
Now we must use what we know about slope to determine the slope of the line.
What fraction represents the slope of this line?
What must the equation of the line be?
Exercises 1 - 6
Write the equation that represents the line shown.
Use the properties of equality to change the equation from slope intercept form, y = mx + b, to standard form, ax + by = c, where a, b, and c are integers and a is not negative.
|
# What is a partial fraction expansion process?
Partial fraction expansion or a partial fraction decomposition is a process in which we can separate one complicated fraction into a sum of few smaller ones. This is a process that has many applications – most importantly in integration.
Let’s say you have a rational expression . You know that . Now you want to know how you can write this fraction in a form of sum of two different fractions with denominators and .
First you write your fraction in the following way.
Now you multiply whole equation with the common denominator:
Next step is to get rid of the brackets.
The easiest way to solve this is by using method of contrary coefficients- we’ll simply add these two equations.
Since we got one unknown we can easily find the other one.
When we put these values we got into the original equation
We get:
And that is it. We can easily check our solution by doing the inverse procedure.
What are the things you should pay attention to? First of all, the first example we made is a rational expression where we could factorize our denominator into product of linear factors. The procedure is quite different when you can’t factorize them like that.
One example of an expression like that is
First, we can extract from the denominator.
This is how you should do it. For every linear member you put a fraction with numerator that has only one letter. When the powers of the expression rise, so does your numerator. For quadratic expression you put , for cube expression you put and so on. Generally you always put in for the numerator polynomial with degree that is for smaller than the expression you are trying to separate from the given fraction.
From this step everything is the same. Now we have to multiply this whole equation with the denominator of the given fraction.
Next step is to get rid of the brackets and, again, equalize coefficients that multiply the unknowns with same powers.
, ,
, ,
One more thing can appear in these kind of tasks that is solved differently than these examples before. This is the case when you can write the given denominator as the product of linear polynomials, but two or more of them are equal.
For example .
Since we can express denominator as the product of finitely many linear polynomials, all expressions in the separate fractions’ numerators will be linear. Since we have third degree polynomial we have to go through all the powers to get to the third degree. This means that for the polynomial we’ll have three different fractions – those whose denominators are , and .
Shares
|
# 8th Class Mathematics Simple and Compound Intersest Compound Interest
## Compound Interest
Category : 8th Class
### Compound Interest
If the interest is calculated at the end of certain fixed period and principal for the next is the amount after adding the interest of the previous period to the principal.
Formula for Calculating Compound Interest
$A=P{{\left( 1+\frac{r}{100} \right)}^{n}}$
$C.I.=P\left[ {{\left( 1+\frac{r}{100} \right)}^{n}}-1 \right]$
Thus, $\mathbf{C}\mathbf{.I}\mathbf{.=A-P}$
Where,
P = Principal Amount
r = Interest Rate
n = Number of times the amount is compounded
A = Amount after time t
C.I. = Compound interest.
• The number which has very peculiar affinity for each other are called the amicable number. For example 220 and 284. All the factors of 220 less than it adds up to given 284.
• Zerah by the age of eight can calculate the mathematical calculation up to 8 to $\text{1}{{\text{6}}^{\text{th}}}$ power in about 30 seconds.
• Time between slipping on a peel and colliding with the pavement is 1 banano second.
• 365.25 days is equal to the 1 unicycle.
• 1 million bicycle is equal to the 2 mega cycles.
• The interest is either calculated quarterly, half yearly or annually.
• The simple interest is calculated with the help of the relation $\frac{P\times R\times T}{100}$.
• The compound interest is calculated with the help of the relation $C.I.=P\left[ {{\left( 1+\frac{r}{100} \right)}^{n}}-1 \right]$
• The amount is calculated with the help of the relation $A=P{{\left( 1+\frac{r}{100} \right)}^{n}}$ where n denotes the number of times the principal has been compounded.
• If the rate of interest decreases at a certain rates the amount is calculated with; the help of $A=P{{\left( 1-\frac{r}{100} \right)}^{n}}$.
Arwin lent Rs. 10,000 on first January 2000 to his friend Robert at 5% simple interest which amounts to Rs. 18000 after certain period of time. Find the date on which he will get the amount.
(a) $\text{3}{{\text{1}}^{\text{st}}}$ December 2016
(b) $\text{3}{{\text{1}}^{\text{st}}}$ December 2015
(c) $\text{3}{{\text{1}}^{\text{st}}}$ December 2014
(d) $\text{3}{{\text{1}}^{\text{st}}}$ December 2013
(e) None of these
Codi borrows Rs. 10000 for 120 days at 10% simple interest per year from his friend Lorentz for admission of his son in the school. Find the interest he has to pay to his friend after 120 days.
(a) Rs. 382.20
(b) Rs. 255.65
(c) Rs. 328.76
(d) Rs. 192.72
(e) None of thee
Johnson has Rs. 4000 he wants to invest it in two types of bond. The first bond pays him 7% and second pays an interest 9% per annum. He uses the first bond for 12 years and other for 6 years such that interest on first type is double to that of other. Find the amount in each type of bond.
(a) Rs. 1275, & Rs. 725
(b) Rs. 2250, & Rs. 1750
(c) Rs. 925, & Rs. 1075
(d) Rs. 1325, & Rs. 675
(e) None of these
If m, n, p are the three sums of money such that n is the simple interest on m and p is the simple interest on n for the same time period and at the same rate of interest. The relation among m, n, p is given by:
(a) ${{\text{n}}^{\text{2}}}=\text{mp}$
(b) ${{\text{m}}^{\text{2}}}=\text{np}$
(c) ${{p}^{\text{2}}}\text{=}\,\text{nm}$
(d) $\text{p}=\text{nm}$
(e) None of these
If the simple interest on a certain sum of money is one - ninth of the principal and rate of interest is equal to the time for which interest is found. Find the rate of interest on that sum of money.
(a) 10 %
(b) $\frac{10}{3}%$
(c) 5.5 %
(d) 7.25 %
(e) None of these
When the Interest is Compounded Annually but Rates are Different for Different Years
Let principal = Rs. P, time = 2 years and let the rate of interest be m% per annum (p.a.) during the first year and n% p. a. during the ${{\text{2}}^{\text{nd}}}$years
Then the formula of amount after 2 years. $=P\left( 1+\frac{{{R}_{1}}}{100} \right)\left( 1+\frac{{{R}_{2}}}{100} \right)$
This formula may similarly be extended for any number of years
When the Interest is Compounded Annually but Time is a Fraction
Let time is 2 years and 4 month, then
Amount $=\left[ P{{\left( 1+\frac{R}{100} \right)}^{2}}\left( 1+\frac{\frac{1}{3}\times R}{100} \right) \right]$
Compounded Half Yearly
Let principal = p, Rate = r%. Time = n years.
If compounded half yearly then, rate = $\frac{r}{2}$% per half year, time = 2n
Then amount $=P\times {{\left( 1+\frac{r}{2\times 100} \right)}^{2n}}$
Compound Interest = Amount - Principal
Compounded Quarterly
Let principal = p. Rate = r% per annum, Time = n years.
If compounded quarterly then, rate = $\frac{r}{4}$% per quarter, time = 4n
Then amount $A=P\times {{\left( 1+\frac{r}{2\times 100} \right)}^{4n}}$
Compound Interest = Amount - Principal
#### Other Topics
You need to login to perform this action.
You will be redirected in 3 sec
|
Te Kete Ipurangi
Communities
Schools
### Te Kete Ipurangi user options:
Level Five > Number and Algebra
# When Small Gets Bigger
Achievement Objectives:
Achievement Objective: NA5-3: Understand operations on fractions, decimals, percentages, and integers.
AO elaboration and other teaching resources
Specific Learning Outcomes:
Solve multiplication and division problems that involve fractions.
Description of mathematics:
Number Framework Stage 8
Required Resource Materials:
Paper strips
Scissors
When Small gets bigger (Material Master 8-16)
Activity:
Students’ experience with whole numbers leads them to expect that multiplying results in an answer that is “larger” and when dividing that the answer is “smaller” than the dividend. These are not always true for decimal fractions. For example 23.9 ÷ 0.0891 is larger than 23.9 not smaller.
#### Using Materials
Problem: “Maurice works at the supermarket. He has to cut up 4 kilograms of cheese into 0.5-kilogram packs for sale. How many packs will he make?” “What is 4 ÷ 0.5?”
Write 4 ÷ 0.5 on the board. Get the students to make a strip like this representing the 4 kilograms of cheese:
Then get them to cut the strip into “1-kilogram” pieces. Discuss how to create 0.5- kilogram packs. (Answer: Cut each piece in half.)
Do the cutting. How many pieces are there? (Answer: 8.)
Discuss why 4 ÷ 0.5 = 8. “Is it a surprise that the answer is more than 4?”
Examples. Use paper and cutting for word stories: 3 ÷ 0.5, 2 ÷ 0.25, 3.5 ÷ 0.5, 1 ÷ 0.25, 2.5 ÷ 0.5 ...
#### Using Imaging
Problem: “Morrie has 5 kilograms of chocolate. He puts the chocolate into separate packets of 0.25 kilograms. How many packets will Morrie make?” “What is 5 ÷ 0.25?”
Write 5 ÷ 0.25 on the board. Get the students to imagine what Morrie would do to 1 kilogram of chocolate. (Answer: He would make 4 packets.)
“How many bags can he make?” (Answer: 5 x 4 = 20.)
Write 5 ÷ 0.25 = 20 on the board.
Examples: Imagine how to solve these problems. Word stories and problems for:
3.5 ÷ 0.5, 1.25 ÷ 0.25, 8 ÷ 0.5, 2.5 ÷ 0.25 ...
#### Using Number Properties
Examples: Worksheet (Material Master 8–16).
Understanding Number Properties: How will you decide whether the answers to problems like 345.67 ÷ 1.008 or 2345.09 ÷ 0.012 get bigger or smaller?
#### Understanding Number Properties:
z ÷ y is always smaller than z when y is ...
z ÷ y is always bigger than z when y is ...
|
# How do you solve 2\frac { 1} { 3} + w = 4\frac { 2} { 9}?
Jun 26, 2017
See explanation
#### Explanation:
$2 \frac{1}{3} + w = 4 \frac{2}{9}$
$w = 4 \frac{2}{9} - 2 \frac{1}{3}$
$w = \frac{38}{9} - \frac{7}{3}$
$w = \frac{38}{9} - \frac{21}{9}$
$w = \frac{17}{9}$
$w = 1 \frac{8}{9}$
Aug 3, 2017
$1 \frac{8}{9}$
#### Explanation:
$2 \frac{1}{3} + w = 4 \frac{2}{9} \text{ } \leftarrow$ isolate the $w$ by subtracting $2 \frac{1}{3}$
$w = 4 \frac{2}{9} - 2 \frac{1}{3}$
You can work with the whole numbers and the fractions separately.
$w = 2 \frac{2 - 3}{9} \text{ } \leftarrow$ convert a whole number to $\frac{9}{9}$
$w = 1 \frac{9 + 2 - 3}{9}$
$w = 1 \frac{8}{9}$
|
# 7.5: Examples of Integral Energy Conservation
Example 7.1
Consider a flow in a long straight pipe. Initially the flow is in a rest. At time, $$t_0$$ the
Fig. 7.6 Flow in a long pipe when exposed to a jump in the pressure difference.
a constant pressure difference is applied on the pipe. Assume that flow is incompressible, and the resistance or energy loss is $$f$$. Furthermore assume that this loss is a function of the velocity square. Develop equation to describe the exit velocity as a function of time. State your assumptions.
Solution 7.1
The mass balance on the liquid in the pipe results in
$\label{unsteadyPipe:massIni} 0 = \overbrace{\int_V \dfrac{\partial \rho}{\partial t} dV}^{=0} + \overbrace{\int_A \rho\,U_{bn} dA}^{=0} + \int_A \rho\,U_{rn} dA \Longrightarrow \cancel{\rho}\cancel{A}\,U_{in} = \cancel{\rho}\cancel{A}\,U_{exit}$
There is no change in the liquid mass inside pipe and therefore the time derivative is zero (the same mass resides in the pipe at all time). The boundaries do not move and the second term is zero. Thus, the flow in and out are equal because the density is identical. Furthermore, the velocity is identical because the cross area is same. It can be noticed that for the energy balance on the pipe, the time derivative can enter the integral because the control volume has fixed boundaries. Hence,
$\label{unsteadyPipe:energyIni1} \begin{array}{rcl} \dot{Q} - \overbrace{\dot{W}_{shear}}^{=0} + & \overbrace{\dot{W}_{shaft}}^{=0} = \displaystyle \int_V \dfrac{d}{dt} \left( E_u + \dfrac{U^2}{2} + g\,z\right)\,\rho\, dV + \ & \displaystyle \int_S \left( h + \dfrac{U^2} {2} + g\,z \right) U_{rn}\, \rho \,dA + \displaystyle \int_S P U_{bn} dA \end{array}$ The boundaries shear work vanishes because the same arguments present before (the work, where velocity is zero, is zero. In the locations where the velocity does not vanished, such as in and out, the work is zero because shear stress are perpendicular to the velocity). There is no shaft work and this term vanishes as well. The first term on the right hand side (with a constant density) is
$\label{unsteadyPipe:fTerm} \rho \int_{V_{pipe}} \dfrac{d}{dt} \left( E_u + \dfrac{U^2}{2}+ \overbrace{g\,z}^{\scriptsize constant} \right)\,dV = \rho \,U\,\dfrac{d\,U}{dt} \overbrace{V_{pipe}}^{L\,\pi\,r^2} + \rho \, \int_{V_{pipe}} \dfrac{d}{dt} \left(E_u\right)\,dV$ where $$L$$ is the pipe length, $$r$$ is the pipe radius, $$U$$ averaged velocity. In this analysis, it is assumed that the pipe is perpendicular to the gravity line and thus the gravity is constant. The gravity in the first term and all other terms, related to the pipe, vanish again because the value of z is constant. Also, as can be noticed from equation (112), the velocity is identical (in and out). Hence the second term becomes
$\label{unsteadyPipe:sTerm} \int_A \left( h + \left(\cancelto{constant}{\dfrac{U^2}{2} + g\,z}\right) \right) \rho\,U_{rn} dA = \int_A \overbrace{\left( E_u + \dfrac{P}{\rho} \right)}^{h} \,\rho\,U_{rn} dA$ Equation (115) can be further simplified (since the area and averaged velocity are constant, additionally notice that $$U = U_{rn}$$) as
$\label{unsteadyPipe:sTerms} \int_A \left( E_u + \dfrac{P}{\rho} \right) \,\rho\,U_{rn} dA = {\Delta P\,U\, A} + \int_A \rho\,E_u\,U_{rn}\, dA$ The third term vanishes because the boundaries velocities are zero and therefore
$\label{unsteadyPipe:tTerm} \int_A P\,\,U_{bn} dA = 0$ Combining all the terms results in
$\label{unsteadyPipe:combinedEnergyIni} \dot{Q} = \rho \,U\,\dfrac{d\,U}{dt} \overbrace{V_{pipe}}^{L\,\pi\,r^2} + \rho \,\dfrac{d}{dt} \int_{V_{pipe}} E_u\,dV +\Delta P\,U \,dA + \int_A \rho\,E_u\,U\,dA$ equation (118) can be rearranged as
$\label{unsteadyPipe:combinedEnergy2} \overbrace{\dot{Q} - \rho \,\int_{V_{pipe}} \dfrac{d \left(E_u \right)}{dt} \,dV - \int_A \rho\,E_u\,U\,dA}^{-K\dfrac ParseError: EOF expected (click for details) Callstack: at (Bookshelves/Chemical_Engineering/Map:_Fluid_Mechanics_(Bar-Meir)/07:_Energy_Conservation/7.5:_Examples_of_Integral_Energy_Conservation), /content/body/div[2]/p[10]/span, line 1, column 2 = \rho \,{L\,\pi\,r^2}\, U\,\dfrac{d\, U }{dt} + \left( P_{in} - P_{out} \right)\,U$ The terms on the LHS (left hand side) can be combined. It common to assume (to view) that these terms are representing the energy loss and are a strong function of velocity square . Thus, equation (119) can be written as
$\label{unsteadyPipe:combinedEnergy1} - K\,\dfrac{U^2}{2} = \rho \,{L\,\pi\,r^2}\,U\, \dfrac{d\,U}{dt} + \left( P_{in} - P_{out} \right) \,U$ Dividing equation (120) by $$K\,U/2$$ transforms equation (??) to
$\label{unsteadyPipe:combinedEnergy} {U} + \dfrac{2\,\rho \,{L\,\pi\,r^2}}{K} \dfrac{d\,U}{dt} = \dfrac{2 \left( P_{in} - P_{out} \right) }{K}$ Equation (121) is a first order differential equation. The solution this equation is described in the appendix and which is
$\label{unsteadyPipe:sol1} U=\text{ e}^{-\left( \dfrac{t\,K}{2\,\pi \,{r}^{2}\,\rho\,L}\right)}\, \left( \dfrac{2\,\left( P_{in} - P_{out} \right) \mbox{\huge e}^{\left(\dfrac{t\,K}{2\,\pi \,{r}^{2}\,\rho\,L}\right)}}{K} + c \right) \,\text{ e}^{\left(\dfrac{2\,\pi \,{r}^{2}\,\rho\,t\,L}{K} \right)}$ Applying the initial condition, $$U(t=0) = 0$$ results in
$\label{unsteadyPipe:solWini} U=\dfrac{2\,\left( P_{in} - P_{out} \right)}{K} \left( 1- \mbox{\huge e}^{- \left(\dfrac{t\,K}{2\,\pi \,{r}^{2}\,\rho\,L}\right)}\right)$ The solution is an exponentially approaching the steady state solution. In steady state the flow equation (121) reduced to a simple linear equation. The solution of the linear equation and the steady state solution of the differential equation are the same.
$\label{unsteadyPipe:ss} U=\dfrac{2\,\left( P_{in} - P_{out} \right)}{K}$ Another note, in reality the resistance, K, is not constant but rather a strong function of velocity etc.). This function will be discussed in a greater extent later on. Additionally, it should be noted that if momentum balance was used a similar solution (but not the same) was obtained (why? hint the difference of the losses accounted for).
The following example combined the above discussion in the text with the above example (7.1).
Example 7.2
A large cylindrical tank with a diameter, $$D$$, contains liquid to height, $$h$$. A long pipe is connected to a tank from which the liquid is emptied. To analysis this situation,
Fig. 7.7 Liquid exiting a large tank trough a long tube.
consider that the tank has a constant pressure above liquid (actually a better assumption of air with a constant mass.). The pipe is exposed to the surroundings and thus the pressure is $$P_{atmos}$$ at the pipe exit. Derive approximated equations that related the height in the large tank and the exit velocity at the pipe to pressure difference. Assume that the liquid is incompressible. Assume that the resistance or the friction in the pipe is a strong function to the velocity square in the tank. State all the assumptions that were made during the derivations.
Solution 7.2
Fig. 7.8 Tank control volume for Example.
This problem can split into two control volumes; one of the liquid in the tank and one of the liquid in pipe. Analysis of control volume in the tank was provided previously and thus needed to be sewed to Example 7.1. Note, the energy loss is considered (as opposed to the discussion in the text). The control volume in tank is depicted in Figure 7.7.
## Tank Control Volume
The effect of the energy change in air side was neglected. The effect is negligible in most cases because air mass is small with exception the "spring'' effect (expansion/compression effects). The mass conservation reads
$\label{longPipeTank:massIni} \overbrace{\int_V \dfrac{\partial \rho}{\partial t} dV}^{=0} + \int_A \rho\,U_{bn} dA + \int_A \rho\,U_{rn} dA = 0$
The first term vanishes and the second and third terms remain and thus equation (125) reduces to
$\label{longPipeTank:massRTi} \cancel{\rho}\,U_{1}\,A_{pipe} = \cancel{\rho} \,U_3\,\overbrace{\pi\,R^2}^{A_{tank}} = \cancel{\rho} \,\dfrac{dh}{dt}\,\overbrace{\pi\,R^2}^{A_{tank}}$ It can be noticed that $$U_3 = dh/dt$$ and $$D=2\,R$$ and $$d=2\,r$$ when the lower case refers to the pipe and the upper case referred to the tank. Equation (126) simply can be written when the area ratio is used (to be changed later if needed) as
$\label{longPipeTank:massRT} U_{1}\,A_{pipe} = \dfrac{dh}{dt}\,{A_{tank}} \Longrightarrow U_1 = \left( \dfrac{R}{r} \right)^2 \dfrac{dh}{dt}$ The boundaries shear work and the shaft work are assumed to be vanished in the tank. Therefore, the energy conservation in the tank reduces to
$\label{unsteadyPipe:energyIni} \dot{Q} - \overbrace{\dot{W}_{shear}}^{=0} + \overbrace{\dot{W}_{shaft}}^{=0} = \dfrac{d}{dt} \displaystyle \int_{V_t} \left( E_u + \dfrac ParseError: EOF expected (click for details) Callstack: at (Bookshelves/Chemical_Engineering/Map:_Fluid_Mechanics_(Bar-Meir)/07:_Energy_Conservation/7.5:_Examples_of_Integral_Energy_Conservation), /content/body/div[4]/div[1]/p[5]/span[1], line 1, column 4 ^{U_1}\, \rho \,dA + \\ \displaystyle \int_{A_3} P \overbrace{U_{bn}}^{U_3} dA = \overbrace{\dfrac{d}{dt} \displaystyle \int_{V_t} E_u \rho\, dV + \int_{A_1} E_u\,\rho\,U_{rn}\,dA- \dot{Q}}^ {K\,\dfrac ParseError: EOF expected (click for details) Callstack: at (Bookshelves/Chemical_Engineering/Map:_Fluid_Mechanics_(Bar-Meir)/07:_Energy_Conservation/7.5:_Examples_of_Integral_Energy_Conservation), /content/body/div[4]/div[1]/p[5]/span[2], line 1, column 4 $ Similar arguments to those that were used in the previous discussion are applicable to this case. Using equation (38), the first term changes to
$\label{longPipeTank:unstadyT} \begin{array}{rl} \dfrac{d}{dt} \displaystyle \int_V \rho\,\left( \dfrac{U^2}{2} + g\,z\right) \, dV & \cong \rho\, \dfrac{d}{dt} \left( \left[ \dfrac ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Chemical_Engineering/Map:_Fluid_Mechanics_(Bar-Meir)/07:_Energy_Conservation/7.5:_Examples_of_Integral_Energy_Conservation), /content/body/div[4]/div[1]/p[6]/span, line 1, column 1 \,U_{rn}\, \cancel{\rho} \,dA = \rho\,P_1\,U_1\,A_1$ It is assumed that the exit velocity can be averaged (neglecting the velocity distribution effects). The second term can be recognized as similar to those by equation (45). Hence, the second term is
$\label{longPipeTank:secondOtherP} \int_A \left( \dfrac{U^2}{2\dfrac{}{}} + \overbrace{g\,z}^{z=0}\right)\, U_{rn}\, \rho \,dA \cong \dfrac{1}{2} \left( \dfrac{dh}{dt\dfrac{}{}} \dfrac{A_3}{A_1}\right)^2\, U_1 \, \rho \,A_1 = \dfrac{1}{2} \left( \dfrac{dh}{dt\dfrac{}{}} \dfrac{R}{r}\right)^2\, U_1 \, \rho \,A_1$ The last term on the left hand side is
$\label{longPipeTank:lTerm} \int_A P U_{bn} dA = P_{3}\,A\, \dfrac{dh}{dt}$ The combination of all the terms for the tank results in
$\label{longPipeTank:TenergyF} \dfrac{d}{dt} \left( \left[ \dfrac ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Chemical_Engineering/Map:_Fluid_Mechanics_(Bar-Meir)/07:_Energy_Conservation/7.5:_Examples_of_Integral_Energy_Conservation), /content/body/div[4]/div[1]/p[9]/span, line 1, column 1 \right)^2 \left( \dfrac{A_3}{A_1\dfrac{}{}}\right)^2\, U_1 \, \,A_1 + \,\dfrac{K_t}{2\,\rho} \left( \dfrac{dh}{dt\dfrac{}{}}\right)^2 = \dfrac{\left( P_3 - P_1 \right) } {\rho}$
## Pipe Control Volume
The analysis of the liquid in the pipe is similar to Example 7.1. The conservation of the liquid in the pipe is the same as in Example 7.1 and thus equation (112) is used
$\label{longPipeTank:massPini} U_{1} = U_{2}$
$\label{longPipeTank:combinedEnergy2} {U_p} + \dfrac{4\,\rho \,{L\,\pi\,r^2}}{K_p} \dfrac{d\,U_p}{dt} = \dfrac{2 \left( P_1- P_2 \right) }{K_p}$ where $$K_p$$ is the resistance in the pipe and $$U_p$$ is the (averaged) velocity in the pipe. Using equation (127) eliminates the $$U_p$$ as
$\label{longPipeTank:combinedEnergy1} \dfrac{dh}{dt} + \dfrac{4\,\rho \,{L\,\pi\,r^2}}{K} \dfrac{d^2\,h}{dt^2} = \left(\dfrac{R}{r} \right)^2 \dfrac{2 \left( P_1 - P_2 \right) }{K_{p}}$ Equation (137) can be rearranged as
$\label{longPipeTank:combinedEnergy} \dfrac{K_p}{2\,\rho} \left(\dfrac{r}{ R\dfrac{}{}} \right)^2 \left( \dfrac{dh}{dt\dfrac{}{}} + \dfrac{4\,\rho \,{L\,\pi\,r^2}}{K} \dfrac{d^2\,h}{dt^2} \right) = \dfrac{ \left( P_1 - P_2 \right) }{\rho}$
## Solution
The equations (138) and (134) provide the frame in which the liquid velocity in tank and pipe have to be solved. In fact, it can be noticed that the liquid velocity in the tank is related to the height and the liquid velocity in the pipe. Thus, there is only one equation with one unknown. The relationship between the height was obtained by substituting equation (127) in equation (138). The equations (138) and (134) have two unknowns ($$dh/dt$$ and $$P_1$$) which are sufficient to solve the problem. It can be noticed that two initial conditions are required to solve the problem. } The governing equation obtained by from adding equation (138) and (134) as
$\label{longPipeTank:govEq} \dfrac{d}{dt} \left( \left[ \dfrac ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Chemical_Engineering/Map:_Fluid_Mechanics_(Bar-Meir)/07:_Energy_Conservation/7.5:_Examples_of_Integral_Energy_Conservation), /content/body/div[4]/div[3]/p[2]/span, line 1, column 1 + \dfrac{g\,h}{2} \right] \right. \left. \overbrace{h\, A}^{V} \right) - \dfrac{1}{2}\left( \dfrac{dh}{dt\dfrac{}{}} \right)^2 \left( \dfrac{A_3}{A_1\dfrac{}{}}\right)^2 \, U_1 \, \,A_1 + \dfrac{K_t}{2\,\rho\dfrac{}{}} \left( \dfrac{dh}{dt\dfrac{}{}}\right)^2 \\ + \dfrac{K_p}{2\,\rho} \left(\dfrac{r}{R\dfrac{}{}} \right)^2 \left( \dfrac{dh}{dt\dfrac{}{}} + \dfrac{4\,\rho \,{L\,\pi\,r^2}}{K} \dfrac{d^2\,h}{dt^2} \right) = \dfrac{ \left( P_3 - P_2 \right) }{\rho}$
The initial conditions are that zero initial velocity in the tank and pipe. Additionally, the height of liquid is at prescript point as
$\label{longPipeTank:iniCon} \begin{array}{cc} h(0) = & h_0 \ \dfrac{dh}{dt} (0) = & 0 \end{array}$
The solution of equation can be obtained using several different numerical techniques. The dimensional analysis method can be used to obtain solution various situations which will be presented later on.
|
Lesson 5, Topic 2
In Progress
# Law 2: Division Law
Lesson Progress
0% Complete
$$\scriptsize a^m \div a^n = a^{m \: – \: n}$$
Similar to the Multiplication Law, the Division Law applies only when we have numbers with the same base numbers.
e.g.
27 ÷ 22
= 27-2
= 25
By expansion:
= $$\scriptsize 2^7 \: \div \:2^2 \\= \frac{2^7}{2^2} \\ = \frac{2 \: \times \: 2 \: \times \: 2\: \times \: 2\: \times \: 2\: \times \: 2\: \times \: 2}{2 \: \times \: 2}\\ = \frac{2 \: \times \: 2 \: \times \: 2\: \times \: 2\: \times \: 2\: \times \: \not{2}\: \times \: \not{2}}{\not{2}\: \times \: \not{2}}\\ \scriptsize = 2^5$$
### Example 1
Simplify the following by expansion
a. 76 ÷ 72
b. m9 ÷ m4
c. x7 ÷ x4
d. h10 ÷ h4
e. 1312 ÷139
f. 18h5 ÷ 9h4
g. 3x2 ÷ 21x7
Solution
a. $$\scriptsize 7^6 \: \div \:7^2 \\= \frac{7^6}{7^2} \\ = \frac{7 \: \times \: 7 \: \times \: 7\: \times \: 7\: \times \: \not{7}\: \times \: \not{7}}{\not{7} \: \times \: \not{7}}\\ \scriptsize = 7^4$$
b. $$\scriptsize m^9 \: \div \:m^4 \\= \frac{m^9}{m^4} \\ =\scriptsize \frac{m \: \times \: m \: \times \: m\: \times \: m\: \times \: m\: \times \: m\: \times \: m\: \times \: \not{m}\: \times \: \not{m}}{\not{m} \: \times \: \not{m}\: \times \: \not{m}\: \times \: \not{m}}\\ \scriptsize = m^5$$
c. $$\scriptsize x^7 \: \div \:x^4 \\= \frac{x^7}{x^4} \\ = \frac{x \: \times \: x \: \times \: x\; \times \: \not{x}\: \times \:\not{ x}\: \times \: \not{x}\: \times \: \not{x}}{\not{x} \: \times \: \not{x}\: \times \: \not{x}\: \times \: \not{x}}\\ \scriptsize = x^3$$
d. $$\scriptsize h^{10} \: \div \:h^4 \\= \frac{h^{10}}{h^4} \\ = \frac{h \: \times \: h \: \times \: h \: \times \: h \: \times \: h \: \times \:h\: \times \: \not{h} \: \times \:\not{ h}\: \times \: \not{h}\: \times \: \not{h}}{\not{h} \: \times \: \not{h}\: \times \: \not{h}\: \times \: \not{h}}\\ \scriptsize = h^6$$
e. $$\scriptsize 13^{12} \: \div \:13^9 \\= \frac{13^{12}}{13^9} \\ =\scriptsize \frac{13 \: \times \: 13 \: \times \: 13 \: \times \: \not{13} \: \times \: \not{13} \: \times \:\not{13} \: \times \: \not{13} \: \times \: \not{13}\: \times \: \not{13}\: \times \: \not{13} \: \times \: \not{13} \: \times \: \not{13}}{\not{13} \: \times \: \not{13}\: \times \: \not{13}\: \times \: \not{13} \: \times \: \not{13} \: \times \: \not{13}\: \times \: \not{13}\: \times \: \not{13} \: \times \: \not{13}}\\ \scriptsize = 13^3$$
f. $$\scriptsize 18 h^{5} \: \div \: 9 h^4 \\= \frac{18h^{5}}{9h^4} \\ = \frac{18 \: \times \: h \: \times \: \not{h} \: \times \: \not{h} \: \times \: \not{h} \: \times \: \not{h} }{9 \: \times \: \not{h} \: \times \: \not{h}\: \times \: \not{h}\: \times \: \not{h}}\\ \scriptsize = 2h$$
g. $$\scriptsize 3x^2 \: \div \: 21x^7 \\ = \frac{3x^2}{21x^7} \\ = \frac{3 \: \times \: \not{x} \: \times \: \not{x}}{21 \: \times \: x \: \times \: x \: \times \: x \: \times \: x \: \times \: x\: \times \: \not{x} \: \times \: \not{x}} \\ = \frac{1}{7x^5}$$
### Example 2
Simplify the following by the law of indices
a. 76 ÷ 72Â
b. m9 ÷ m
c. x7 ÷ x4Â
d. h10 ÷ h4 Â
e. 1312 ÷ 139
f. 18h5 ÷ 9h4 Â
g. 3x2 ÷ 21x7
Solution
a. 76 ÷ 72
= 76-2
= 74
same base
b. m9 ÷ m4
= m9-4
= m5
same base
c. x7 ÷ x4
= x7-4
= x3
same base
d. h10 ÷ h4
= h10-4
= h6
same base
e. 1312 ÷ 13
= 1312-9
= 133
same base
f. 18h5 ÷ 9h4
$$= \frac{18h^8}{9h^4}\\ = \scriptsize 2h^{5\:-\:4}\\ = \scriptsize 2h^1 \\= \scriptsize 2h$$
same base
g. 3x2 ÷ 21x7
= $$\frac{3}{21} \scriptsize x^2 \: \div \: x^7 \\= \frac{3}{21} \scriptsize x^{2 \: -\: 7}\\ = \frac{1}{7} \scriptsize x^{-5}$$
same base
error:
|
# Question: How Do You Use The Mode Formula?
## How do you find the mode in maths?
The mode is the number that appears the most.To find the mode, order the numbers lowest to highest and see which number appears the most often.Eg 3, 3, 6, 13, 100 = 3.The mode is 3..
## What is the formula of mean median and mode?
Median = Size of (n+12)th item. In case of even number of values. Median = average of n2th and n+22th item. The Mode: The mode is that value in a series of observation which occurs with greatest frequency.
## What is mode math example?
Mode: The most frequent number—that is, the number that occurs the highest number of times. Example: The mode of {4 , 2, 4, 3, 2, 2} is 2 because it occurs three times, which is more than any other number.
## How do you find the mean and mode?
Mean: Add up all the numbers of the set. Divide by how many numbers there are. Mode: The number that occurs the most.
## What is the mode if there is no mode?
In that case, you have to add those two numbers together and then divide by two to find the median. The mode of a data set refers to the number that occurs most often. If there is not a number that occurs more than any other, we say there is no mode for the data.
## What if there is 2 modes?
Mode – The mode is the number that appears the most. … If there are two numbers that appear most often (and the same number of times) then the data has two modes. This is called bimodal. If there are more than 2 then the data would be called multimodal.
## What does mean mode?
The “mode” is the value that occurs most often. If no number in the list is repeated, then there is no mode for the list.
## How do I calculate mean?
The mean is the average of the numbers. It is easy to calculate: add up all the numbers, then divide by how many numbers there are. In other words it is the sum divided by the count.
## How do you find the most common value?
Select a blank cell, here is C1, type this formula =MODE(A1:A13), and then press Enter key to get the most common number in the list. Tip: A1:A13 is the list you want to find most common number form, you can change it to meet your needs.
## What is the formula of mode?
In this article, we will try and understand the mode function, examples and explanations of each example along with the formula and the calculations. Where, L = Lower limit Mode of modal class. fm = Frequency of modal class….Mode Formula Calculator.Mode Formula =L + (fm – f1) x h / (fm – f1) + (fm – f2)=0 + (0 – 0) x 0 / (0 – 0) + (0 – 0)= 0
## How do I find most common entry in Excel?
Most Frequently Occurring WordThe MATCH function returns the position of a value in a given range. … To find the position of the most frequently occurring word (don’t be overwhelmed), we add the MODE function and replace A7 with A1:A7.Finish by pressing CTRL + SHIFT + ENTER.More items…
## How do you find the mean median and mode?
The mean means average. To find it, add together all of your values and divide by the number of addends. The median is the middle number of your data set when in order from least to greatest. The mode is the number that occurred the most often.
## How do you explain mode?
Mode means the number that occurs most frequently. To find the mode of the your test scores, for example, your teacher would list all your math test scores in order from smallest to largest or from smallest to largest, and then find the number that appears on the list most frequently. Mean, Median, Mode, etc.
## How do you find the mode of a column in Excel?
To find the most occurring value in Excel, use the MODE function and select the range you want to find the mode of. In our example below, we use =MODE(B2:B12) and since 2 students have scored 55 we get the answer as 55.
## How do you find the mode of text in Excel?
Select a blank cell you will place the most frequent value into, type the formula =INDEX(A2:A20,MODE(MATCH(A2:A20,A2:A20,0))) (A2:A20 is the list where you will find out the most frequent (mode for) text value from) into it, and then press the Ctrl + Shift + Enter keys.
## What is the math mode?
more … The number which appears most often in a set of numbers. Example: in {6, 3, 9, 6, 6, 5, 9, 3} the Mode is 6 (it occurs most often).
## What are the types of mode in statistics?
One mode: unimodal: 1, 2, 3, 3, 4, 5. Two: bimodal: 1, 1, 2, 3, 4, 4, 5. Three: trimodal: 1, 1, 2, 3, 3, 4, 5, 5. More than one (two, three or more) = multimodal.
## What is the difference between mode sngl and mode mult?
MULT function as the MODE. SNGL function returns the lowest mode, whereas the MODE. MULT function returns an array of all the modes.
## What is typical number in math?
The typical number is the one that has the most but it’s also surrounded by the most. If they’re all over, pick the one in the middle. If they are all even, pick the one in the middle.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.