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NCERT Solutions Class 5 Maths Chapter 11 -Area and its Boundary
Area and its Boundary
PAGE: 146
Whose Slice is Bigger?
Parth and Gini bought aam pappad (dried mango slice) from a shop. Their pieces looked like these.
Both could not make out whose piece was bigger.
1. Suggest some ways to find out whose piece is bigger. Discuss.
Solution:
The length of piece A is 6 cm.
So 6 squares of side 1 cm can be arranged along its length.
The width of piece A is 5 cm.
So 5 squares can be arranged along its width.
Altogether 30 squares can be arranged on it.
So, the area of piece A = 6 cm × 5 cm = 30 square cm
2. Altogether how many squares can be arranged on it? ________
Solution:
30
3. So the area of piece A = ________ square cm
So the area of piece A = 30 square cm
4. In the same way find the area of piece B.
Solution:
Now, the area of piece B = 11 cm × 3 cm = 33 square cm
5. Who had the bigger piece? How much bigger?
Solution:
So, the area of piece B is bigger than the area of piece A.
Difference in area of piece A and piece B = 33 − 30 = 3 square cm
So, piece B is 3 square cm bigger than piece A.
Cover with stamps:
This stamp has an area of 4 square cm. Guess how many such stamps will cover this big rectangle.
Solution:
a) Measure the yellow rectangle. It is ________ cm long.
Solution:
14cm
b) How many stamps can be placed along its length? ________
Solution:
7 stamps
c) How wide is the rectangle? ________ cm
Solution:
8cm
d) How many stamps can be placed along its width? ________
Solution:
4 stamps
e) How many stamps are needed to cover the rectangle? ________
Solution:
28 stamps
f) How close was your earlier guess? Discuss.
Solution:
It was pretty close.
g) What is the area of the rectangle? ________ Square cm
Solution:
Area of rectangle = 7 × 4 = 28 square cm
h) What is the perimeter of the rectangle? ________ cm
Solution:
Length of the rectangle = 14 cm
Breadth of the rectangle = 8 cm
A rectangle has 2 lengths and 2 breadths.
So, perimeter of the rectangle = Sum of all its sides = Length of its boundary = 14 cm + 8 cm + 14 cm + 8 cm = 44 cm
The perimeter of rectangle is 44 cm.
Practice time:
a) Arbaz plans to tile his kitchen floor with green square tiles. Each side of the tile is 10 cm. His kitchen is 220 cm in length and 180 cm wide. How many tiles will he need?
Solution:
Given length of kitchen = 220 cm
Width of the kitchen = 180 cm
Each side of tile = 10 cm
Area of floor = length × width
= 220 × 180 = 39600 square cm
Area of a tile = side × side
= 10 × 10 = 100 square cm
Number of tiles = area of floor/ area of a tile
= 39600/100
= 396 tile.
b) The fencing of a square garden is 20 m in length. How long is one side of the garden?
Solution:
Given perimeter of garden = 20 m which has to be fenced.
Length of one side = perimeter/4
= 20/5
= 5m.
c) A thin wire 20 centimeters long is formed into a rectangle. If the width of this rectangle is 4 centimeters, what is its length?
Solution:
Given perimeter of a rectangle = 20 cm
Width of a rectangle = 4 cm
We know that Perimeter of rectangle = 2 (length + breadth)
20 = 2 length + 2 breadth
2 length = 20 – 2 breadth
2 length = 20 – 2 × 4
2 length = 20 – 8
Length = 12/2
= 6 cm
d) A square carom board has a perimeter of 320 cm. How much is its area?
Solution:
Given perimeter of carom board is 320 cm
We know that perimeter of square = 4 × side
Side = perimeter/4
Side = 320/4
= 80 cm
We know that area of square = side × side
= 80 × 80
= 6400 square cm
e) How many tiles like the triangle given here will fit in the white design? Area of design = ________ square cm
Solution:
6 triangular tiles will fit in to the given white design.
Now, area of 1 such triangular tile = 12 square cm
Area of 6 triangular tiles that make this design = 6 × 12 = 3 square cm
Area of design = 3 square cm
f) Sanya, Aarushi, Manav and Kabir made greeting cards. Complete the table for their cards:
Solution:
Perimeter of Sanya’s card = Sum of all its sides = 10 + 8 + 10 + 8 = 36 cm
Length of Manav’s card = 11 cm
Perimeter of his card = 44 cm
We have to find the width of Manav’s card.
Perimeter of card = Sum of all its sides = 11 + 11 + sum of 2 other sides = 22 + sum of 2 other sides
Now, sum of two other sides = 44 − 22 = 22 cm
The two other sides of the greeting cards are width.
So, width of Manav’s card = 22 ÷ 2 = 11 cm
Width of Aarushi’s card = 8 cm
Area of the card = 80 square cm
Now, we have to find length of the card.
Area of card = Length of card × 8 cm = 80 square cm
So, on dividing the area of card by its width, we can get its length.
Therefore, length of Aarushi’s card = 80 ÷ 8 = 10 cm
My belt is longest:
Take a thick paper sheet of length 14 cm and width 9 cm. You can also use an old postcard.
1. What is its area? What is its perimeter?
Solution:
Length of paper sheet = 14 cm
Breadth of paper sheet = 9 cm
Area of the sheet = 14 cm × 9 cm = 126 square cm
Perimeter of the sheet = 14 cm + 9 cm + 14 cm + 9 cm = 46 cm
2. Now cut strips of equal sizes out of it. Using tape join the strips, end to end, to make a belt.
How long is your belt? _____
Solution:
84 cm
3. What is its perimeter _____?
Solution:
90 cm
4. Whose belt is the longest in the class? _____
Solution:
Strips of the least width will make the longest belt in the class.
Discuss:
1. Why did some of your friends get longer belts than others?
Solution:
Because they made belts from thinner strips than others.
2. Is the area of your belt the same as the area of the postcard? Why or why not?
Solution:
Area of the belt of 3 cm wide strip = length × breadth
= 3 × 42 = 126 square cm
Yes, the area of my belt is same as the area of post card. Because every area of post card used for making belt.
3. What will you do to get a longer belt next time?
Solution:
By making thinner belts I can get the longer belt.
Share the land:
Nasreena is a farmer who wants to divide her land equally among her three children — Chumki, Jhumri and Imran. She wants to divide the land so that each piece of land has one tree. Her land looks like this.
1. Can you divide the land equally? Show how you will divide it. Remember each person has to get a tree. Colour each person’s piece of land differently
Solution:
Total number of boxes = 90
Hence one person share is = 90/3
= 30 boxes.
The division can be done as shown in the given figure:
2. If each square on this page is equal to 1 square metre of land, how much land will each of her children get? ________ Square m
Solution:
30 square meter.
3. Chumki, Jhumri and Imran need wire to make a fence.
Who will need the longest wire for fencing? _________
Solution:
Perimeter of Chumki’s land = 9 + 2 + 3 + 2 + 6 + 4 = 26 m
Perimeter of Jhumri’s land = 6 + 3 + 2 + 3 + 4 + 6 = 24 m
Perimeter of Imraan’s land = 8 + 5 + 3 + 2 + 5 + 3 = 26 m
So it is clear that Chumki and Imraan need longest wire of fencing.
4. How much wire in all will the three need? ___________
Solution:
Total length of wire = 26 m + 24 m + 26 m = 76 m.
Solution:
B. Draw a square of 9 square cm. Write A on it. Draw another square with double the side. Write B on it
Solution:
1. The perimeter of square A is __________ cm.
Solution:
12 cm
2. The side of square B is __________ cm.
Solution:
6 cm
3. The area of square B is __________ square cm.
Solution:
36 cm
4. The area of square B is __________ times the area of square A.
Solution:
4 times
5. The perimeter of square B is __________ cm.
Solution:
24 cm
6. The perimeter of square B is __________ times the perimeter of square A.
Solution:
2 times
Save the Birds:
There are two beautiful lakes near a village. People come for boating and picnics in both the lakes. The village Panchayat is worried that with the noise of the boats the birds will stop coming. The Panchayat wants motor boats in only one lake. The other lake will be saved for the birds to make their nests.
a) How many cm is the length of the boundary of Lake A in the drawing? _____________ (use thread to find out)
Solution:
When we measure the boundary of Lake A with the help of thread, it comes out to be around 30 cm.
b) What is the length of the boundary of Lake B in the drawing?
Solution:
When we measure the boundary of Lake B with the help of thread, it comes out to be around 25 cm.
c) How many kilometers long is the actual boundary of Lake A?
Solution:
Here, the scale is 1 cm = 1 km
So, length of the actual boundary of lake A = 30 km
d) How many kilometers long is the actual boundary of Lake B?
Solution:
Here, the scale is 1 cm = 1 km
So, length of the actual boundary of lake B = 25 km
e) A longer boundary around the lake will help more birds to lay their eggs. So which lake should be kept for birds? Which lake should be used for boats?
Solution:
The boundary around the lake A is more than the boundary around the lake B. So, Lake A should be kept for birds and Lake B should be used for boats.
f) Find the area of Lake B on the drawing in square cm. What is its actual area in square km?
Solution:
Lake B has 15 fully filled squares and 11 half-filled or more than half – filled squares. Area of 15 fully filled squares = 15 × 15 = 225 square cm
We consider the area of every single half – filled or more than half – filled square as 1 square cm
Thus, the area of 11 such squares = 11 × 11 = 121 square cm
So, total area of lake B = 225 + 121 = 346 square cm
We are given 1 cm on drawing = 1 km on ground
Therefore, 346 square cm on drawing = 346 km
King’s Story:
The King was very happy with carpenters Cheggu and Anar. They had made a very big and beautiful bed for him. So as gifts the king wanted to give some land to Cheggu, and some gold to Anar. King’s Story Cheggu was happy. He took 100 metres of wire and tried to make different rectangles. He made a 10 m × 40 m rectangle. Its area was 400 square metres. So he next made a 30 m × 20 m rectangle.
1. What is its area? Is it more than the first rectangle?
Solution:
Area of rectangle = 30 × 20 = 600 square m.
Yes, it is more than first rectangle.
2. What other rectangles can he make with 100 metres of wire? Discuss which of these rectangles will have the biggest area.
Solution:
Following rectangles are possible:
5 × 45 = 225 square m
15 × 35 = 525 square m
25 × 25 = 625 square m
The square will have biggest area.
3. Cheggu’s wife asked him to make a circle with the wire. She knew it had an area of 800 square metres. Why did Cheggu not choose a rectangle? Explain.
Solution:
Because none of the rectangle will have area 0f 800 square meter.
4. So Anar also tried many different ways to make a boundary for 800 square metres of land. He made rectangles A, B and C of different sizes. Find out the length of the boundary of each. How much gold wire will he get for these rectangles?
i. Gold wire for A = _________ metres
Solution:
As the area of rectangle A is shown as 40 m × 20 m.
So, the length of rectangle = 40 m Width of rectangle = 20 m
Gold wire for A = 40 + 20 + 40 + 20 = 120 metres
ii. Gold wire for B = _________ metres
Solution:
As the area of rectangle B is shown as 80 m ×10 m.
So, the length of rectangle = 80 m Width of rectangle = 10 m
Gold wire for B = 80 + 10 + 80 + 10 = 180 metres
iii. Gold wire for C = ____________ metres
Solution:
As the area of rectangle C is shown as 800 m × 1 m.
So, the length of rectangle = 800 m Width of rectangle = 1 m
Gold wire for C = 800 + 1 + 800 + 1 = 1602 metres
iv. So he will get ____________ metres of gold wire!!
Solution:
As the area of rectangle D is shown as 8000 m × 0.1 m.
So, the length of rectangle = 8000 m
Width of rectangle = 0.1 m
Gold wire for C = 8000 + 0.1 + 8000 + 0.1 = 16000 + 0.2 = 16000.2 metres.
So, he will get 16000.2 metres of gold wire. |
# 2.1.1: Five Number Summary and Box Plots Part 1
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Box plots (also called box-and-whisker plots or box-whisker plots) give a good graphical image of the concentration of the data. They also show how far the extreme values are from most of the data. A box plot is constructed from five values: the minimum value, the first quartile, the median, the third quartile, and the maximum value. We use these values to compare how close other data values are to them.
To construct a box plot, use a horizontal or vertical number line and a rectangular box. The smallest and largest data values label the endpoints of the axis. The first quartile marks one end of the box and the third quartile marks the other end of the box. Approximately the middle 50 percent of the data fall inside the box. The "whiskers" extend from the ends of the box to the smallest and largest data values. The median or second quartile can be between the first and third quartiles, or it can be one, or the other, or both. The box plot gives a good, quick picture of the data.
You may encounter box-and-whisker plots that have dots marking outlier values. In those cases, the whiskers are not extending to the minimum and maximum values.
Consider, again, this dataset.
1; 1; 2; 2; 4; 6; 6;.8; 7.2; 8; 8.3; 9; 10; 10; 11.5
The first quartile is two, the median is seven, and the third quartile is nine. The smallest value is one, and the largest value is 11.5. The following image shows the constructed box plot.
See the calculator instructions on the TI web site or in the appendix.
The two whiskers extend from the first quartile to the smallest value and from the third quartile to the largest value. The median is shown with a dashed line.
It is important to start a box plot with a scaled number line. Otherwise the box plot may not be useful.
##### Example 2.1.1.1
The following data are the heights of 40 students in a statistics class.
59; 60; 61; 62; 62; 63; 63; 64; 64; 64; 65; 65; 65; 65; 65; 65; 65; 65; 65; 66; 66; 67; 67; 68; 68; 69; 70; 70; 70; 70; 70; 71; 71; 72; 72; 73; 74; 74; 75; 77
Construct a box plot with the following properties; the calculator instructions for the minimum and maximum values as well as the quartiles follow the example.
• Minimum value = 59
• Maximum value = 77
• Q1: First quartile = 64.5
• Q2: Second quartile or median= 66
• Q3: Third quartile = 70
1. Each quarter has approximately 25% of the data.
2. The spreads of the four quarters are 64.5 – 59 = 5.5 (first quarter), 66 – 64.5 = 1.5 (second quarter), 70 – 66 = 4 (third quarter), and 77 – 70 = 7 (fourth quarter). So, the second quarter has the smallest spread and the fourth quarter has the largest spread.
3. $$\text{Range} = \text{maximum value} - \text{the minimum value} = 77 - 59 = 18$$
4. Interquartile Range: $$IQR = Q_{3} – Q_{1} = 70 - 64.5 = 5.5$$.
5. The interval 59–65 has more than 25% of the data so it has more data in it than the interval 66 through 70 which has 25% of the data.
6. The middle 50% (middle half) of the data has a range of 5.5 inches.
##### Calculator
To find the minimum, maximum, and quartiles:
Enter data into the list editor (Pres STAT 1:EDIT). If you need to clear the list, arrow up to the name L1, press CLEAR, and then arrow down.
Put the data values into the list L1.
Press STAT and arrow to CALC. Press 1:1-VarStats. Enter L1.
Press ENTER.
Use the down and up arrow keys to scroll.
Smallest value = 59.
Largest value = 77.
Q1: First quartile = 64.5.
Q2: Second quartile or median = 66.
Q3: Third quartile = 70.
To construct the box plot:
Press 4: Plotsoff. Press ENTER.
Arrow down and then use the right arrow key to go to the fifth picture, which is the box plot. Press ENTER.
Arrow down to Xlist: Press 2nd 1 for L1
Arrow down to Freq: Press ALPHA. Press 1.
Press Zoom. Press 9: ZoomStat.
Press TRACE, and use the arrow keys to examine the box plot.
##### Exercise 2.1.1.1
The following data are the number of pages in 40 books on a shelf. Construct a box plot using a graphing calculator, and state the interquartile range.
136; 140; 178; 190; 205; 215; 217; 218; 232; 234; 240; 255; 270; 275; 290; 301; 303; 315; 317; 318; 326; 333; 343; 349; 360; 369; 377; 388; 391; 392; 398; 400; 402; 405; 408; 422; 429; 450; 475; 512
$$IQR = 158$$
For some sets of data, some of the largest value, smallest value, first quartile, median, and third quartile may be the same. For instance, you might have a data set in which the median and the third quartile are the same. In this case, the diagram would not have a dotted line inside the box displaying the median. The right side of the box would display both the third quartile and the median. For example, if the smallest value and the first quartile were both one, the median and the third quartile were both five, and the largest value was seven, the box plot would look like:
In this case, at least 25% of the values are equal to one. Twenty-five percent of the values are between one and five, inclusive. At least 25% of the values are equal to five. The top 25% of the values fall between five and seven, inclusive.
##### Example 2.1.1.2
Test scores for a college statistics class held during the day are:
99; 56; 78; 55.5; 32; 90; 80; 81; 56; 59; 45; 77; 84.5; 84; 70; 72; 68; 32; 79; 90
Test scores for a college statistics class held during the evening are:
98; 78; 68; 83; 81; 89; 88; 76; 65; 45; 98; 90; 80; 84.5; 85; 79; 78; 98; 90; 79; 81; 25.5
1. Find the smallest and largest values, the median, and the first and third quartile for the day class.
2. Find the smallest and largest values, the median, and the first and third quartile for the night class.
3. For each data set, what percentage of the data is between the smallest value and the first quartile? the first quartile and the median? the median and the third quartile? the third quartile and the largest value? What percentage of the data is between the first quartile and the largest value?
4. Create a box plot for each set of data. Use one number line for both box plots.
5. Which box plot has the widest spread for the middle 50% of the data (the data between the first and third quartiles)? What does this mean for that set of data in comparison to the other set of data?
• Min = 32
• Q1 = 56
• M = 74.5
• Q3 = 82.5
• Max = 99
• Min = 25.5
• Q1 = 78
• M = 81
• Q3 = 89
• Max = 98
1. Day class: There are six data values ranging from 32 to 56: 30%. There are six data values ranging from 56 to 74.5: 30%. There are five data values ranging from 74.5 to 82.5: 25%. There are five data values ranging from 82.5 to 99: 25%. There are 16 data values between the first quartile, 56, and the largest value, 99: 75%. Night class:
2. The first data set has the wider spread for the middle 50% of the data. The IQRfor the first data set is greater than the IQR for the second set. This means that there is more variability in the middle 50% of the first data set.
##### Exercise 2.1.1.2
The following data set shows the heights in inches for the boys in a class of 40 students.
66; 66; 67; 67; 68; 68; 68; 68; 68; 69; 69; 69; 70; 71; 72; 72; 72; 73; 73; 74
The following data set shows the heights in inches for the girls in a class of 40 students.
61; 61; 62; 62; 63; 63; 63; 65; 65; 65; 66; 66; 66; 67; 68; 68; 68; 69; 69; 69
Construct a box plot using a graphing calculator for each data set, and state which box plot has the wider spread for the middle 50% of the data.
IQR for the boys = 4
IQR for the girls = 5
The box plot for the heights of the girls has the wider spread for the middle 50% of the data.
##### Example 2.1.1.3
Graph a box-and-whisker plot for the data values shown.
10; 10; 10; 15; 35; 75; 90; 95; 100; 175; 420; 490; 515; 515; 790
The five numbers used to create a box-and-whisker plot are:
• Min: 10
• Q1: 15
• Med: 95
• Q3: 490
• Max: 790
The following graph shows the box-and-whisker plot.
##### Exercise 2.1.1.3
Follow the steps you used to graph a box-and-whisker plot for the data values shown.
0; 5; 5; 15; 30; 30; 45; 50; 50; 60; 75; 110; 140; 240; 330
The data are in order from least to greatest. There are 15 values, so the eighth number in order is the median: 50. There are seven data values written to the left of the median and 7 values to the right. The five values that are used to create the boxplot are:
• Min: 0
• Q1: 15
• Med: 50
• Q3: 110
• Max: 330
### To find the IQR and create a box plot on the TI-83/84:
1. Go into the STAT menu, and then Choose 1:Edit
Figure 2.1.1.5 : STAT Menu on TI-83/84
2. Type your data values into L1. If L1 has data in it, arrow up to the name L1, click CLEAR and then press ENTER. The column will now be cleared and you can type the data in.
3. Go into the STAT menu, move over to CALC and choose 1-Var Stats. Press ENTER, then type L1 (2nd 1) and then ENTER. This will give you the summary statistics. If you press the down arrow, you will see the five-number summary.
4. To draw the box plot press 2nd STAT PLOT.
Figure 2.1.1.6 : STAT PLOT Menu on TI-83/84
5. Use Plot1. Press ENTER
Figure 2.1.1.7 : Plot1 Menu on TI-83/84 Setup for Box Plot
6. Put the cursor on On and press Enter to turn the plot on. Use the down arrow and the right arrow to highlight the boxplot in the middle of the second row of types then press ENTER. Set Data List to L1 (it might already say that) and leave Freq as 1.
7. Now tell the calculator the set up for the units on the x-axis so you can see the whole plot. The calculator will do it automatically if you press ZOOM, which is in the middle of the top row.
Figure 2.1.1.8 : ZOOM Menu on TI-83/84
Then use the down arrow to get to 9:ZoomStat and press ENTER. The box plot will be drawn.
Figure 2.1.1.9 : ZOOM Menu on TI-83/84 with ZoomStat
#### References
1. Data from West Magazine.
## Review
Box plots are a type of graph that can help visually organize data. To graph a box plot the following data points must be calculated: the minimum value, the first quartile, the median, the third quartile, and the maximum value. Once the box plot is graphed, you can display and compare distributions of data.
Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars.
##### Exercise 2.5.4
Construct a box plot below. Use a ruler to measure and scale accurately.
##### Exercise 2.5.5
Looking at your box plot, does it appear that the data are concentrated together, spread out evenly, or concentrated in some areas, but not in others? How can you tell?
More than 25% of salespersons sell four cars in a typical week. You can see this concentration in the box plot because the first quartile is equal to the median. The top 25% and the bottom 25% are spread out evenly; the whiskers have the same length.
## Bringing It Together
##### Exercise 2.5.6
Santa Clara County, CA, has approximately 27,873 Japanese-Americans. Their ages are as follows:
Age Group Percent of Community
0–17 18.9
18–24 8.0
25–34 22.8
35–44 15.0
45–54 13.1
55–64 11.9
65+ 10.3
1. Construct a histogram of the Japanese-American community in Santa Clara County, CA. The bars will not be the same width for this example. Why not? What impact does this have on the reliability of the graph?
2. What percentage of the community is under age 35?
3. Which box plot most resembles the information above?
1. For graph, check student's solution.
2. 49.7% of the community is under the age of 35.
3. Based on the information in the table, graph (a) most closely represents the data.
## Glossary
Box plot
a graph that gives a quick picture of the middle 50% of the data
First Quartile
the value that is the median of the of the lower half of the ordered data set
Frequency Polygon
looks like a line graph but uses intervals to display ranges of large amounts of data
Interval
also called a class interval; an interval represents a range of data and is used when displaying large data sets
Paired Data Set
two data sets that have a one to one relationship so that:
• both data sets are the same size, and
• each data point in one data set is matched with exactly one point from the other set.
Skewed
used to describe data that is not symmetrical; when the right side of a graph looks “chopped off” compared the left side, we say it is “skewed to the left.” When the left side of the graph looks “chopped off” compared to the right side, we say the data is “skewed to the right.” Alternatively: when the lower values of the data are more spread out, we say the data are skewed to the left. When the greater values are more spread out, the data are skewed to the right. |
You then have to add 6.5 to 36 to get the number of small dogs competing, which is 42.5. 1). This one comes from a first grader's homework. Fifth graders are also working on geometric understandings like area, perimeter, and three-dimensional shapes. Think back to that problem about the dogs at the dog show and use the same logic to solve this problem. Elementary Mathematics Questions and answers. All of the numbers in every row and column add up to 15! Includes basic questions about math asked by elementary level students. Not too bad, right? Ask probing questions that require students to explain, elaborate or clarify their thinking. Rely more on themselves to determine whether something is mathematically correct. Since we know that there are two zombies for every three humans and that 2 + 3 = 5, we can divide 85 by 5 to figure out that in total, there are 17 groups of humans and zombies. This test focuses on the essential content knowledge needed for teaching elementary mathematics. 30 Questions You'd Need to Ace to Pass 6th Grade Geography. Most straightforward math questions can be solved by thinking clearly and breaking things down into logical steps. If $$\Large \frac{a}{b}=\frac{2}{3}\ and\ \frac{b}{c}=\frac{4}{5}$$ , then the ratio $$\Large \frac{a+b}{b+c}$$ equal to: $$\Large \frac{a}{b}=\frac{2}{3}\ and\ \frac{b}{c}=\frac{4}{5}\ (given)\ or\ \frac{c}{b}=\frac{5}{4}$$. 3). Follow the question's "help" link, try your answer, then see our solution. Another big part of fifth grade math is proportional reasoning, or gaining a better understanding of fractions, decimals, and percentages. Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively. Yikes. Finding the answer to this final question will require using fractions. The first kind is the content knowledge needed to do the work of the student curriculum, such as solving a math problem similar to one students would solve. On the following pages you will find quizzes that cover several important elementary math topics. Discover thousands of math skills covering pre-K to 12th grade, from counting to calculus, with infinite questions that adapt to each student's level. If $$\Large x \circledast y = 3x+2y$$, then $$\Large 2 \circledast 3+3 \circledast 4$$ is equal to. Then value of $$\Large \frac{3xy}{2 \left(x^{2}-y^{2}\right) }$$ will be; 4). Answer : 256635. © 2020 Galvanized Media. Math symbols: Jokes: Forum: About us: Links: Contact us: Site map Program of Lessons Help Solution Answers Help Solution Answers Help Solution Answers. All you have to do is subtract $1.00 from$1.10 and then divide that answer, $0.10 by 2, to get your final answer,$0.05. YouTuber MindYourDecisions adapted this mind-boggling math question from a similar one found on an elementary school student's homework in China. Once you've divided 192 by 3 to get 64, you can put the decimal place back where it belongs and get your final answer of 0.64. Elementary Math; Elementary Math. Practice By clicking on the practice tab, students can use past competitions as a tool to practice and learn about different problems. Browse through all study tools. Every question has a worked solution. Try to figure out what all of the equations have in common. Download our free ACCUPLACER PDF with 30 questions that cover both the math and English sections of the ACCUPLACER. Enter your email address to get the best tips and advice. Since the patch of pads doubles in size every day, the lake would be half covered just one day before it was covered entirely. Mathematics Practice Test Page 3 Question 7 The perimeter of the shape is A: 47cm B: 72cm C: 69cm D: 94cm E: Not enough information to find perimeter Question 8 If the length of the shorter arc AB is 22cm and C is the centre of the circle then the circumference of the circle is: If you choose to use a calculator, be sure it is permitted, is working on test day, and has reliable batteries. Either way, very few individuals—even math teachers—have been able to find the correct answer to this problem. This problem shouldn't be too difficult to solve if you play a lot of sudoku. In order to be a good instructor to your fifth graders, it is helpf… Types of Questions Within the context of open-ended mathematical tasks, it is useful to group questions into four main categories (Badham, 1994). Solved mathematics questions with detailed explanations for easy understanding on various topics. If $$\Large a\ast b=2 \left(a+b\right)$$, then $$\Large 5\ast 2$$ is equal to: $$\Large a \ast b = 2 \left(a+b\right)$$, 8). Basic Math Questions and Answers Test your understanding with practice problems and step-by-step solutions. The formula used in each equation is 4x = Y. Students. Two numbers x and y (x > y) such that their sum is equal to three times their difference. You might need to ask your kids for help on this one. The 47 one-point questions and 5 two-point questions measure two kinds of content knowledge. Learn to reason mathematically. Learn vocabulary, terms, and more with flashcards, games, and other study tools. If $$\Large a\ast b=a+b+ab$$, then $$\Large 3\ast 4-2\ast 3$$ is equal to: 10). no. Elementary school students are tested often in the various methods of mathematics, such as multiplication, division and algebra. In order to solve this problem, you simply need to subtract the exponents (4-2) and solve for 32, which expands into 3 x 3 and equals 9. These online tests are designed to work on computers, laptops, iPads, and other tablets. In order to do that, in case you forgot, you have to flip the fraction and switch from division to multiplication, thus getting 3 x 3 = 9. When Best Life first wrote about this deceiving question, we had to ask a mathematician to explain the answer! Quizzes make learning fun! Answers to the questions are provided and located at the end of each page. If $$\Large p \times q = p+q+\frac{p}{q}$$, the va1ue of $$\Large 8 \times 2$$ is: $$\Large 8 \times 2 = 8 + 2 + \frac{8}{2}$$. Don't let the fact that 8.563 has fewer numberrs than 4.8292 trip you up. Albert: I don't know when Cheryl's birthday is, but I know that Bernard doesn't not know too. "Guys, I'm tired of living through history.". This problem comes straight from a standardized test given in New York in 2014. … If it takes 48 days for the patch to cover the entire lake, how long would it take for the patch to cover half of the lake? They were created by Khan Academy math experts and reviewed for curriculum alignment by experts at both Illustrative Mathematics and Khan Academy. Start studying WGU: C109 Elementary Math Methods Practice Questions. It's unclear why Cheryl couldn't just tell both Albert and Bernard the month and day she was born, but that's irrelevant to solving this problem. These questions may be used in your classroom to help students feel comfortable with the types of assessment formats they will experience. August 14 August 15 August 17. Hence, it would be best if you gave the utmost care to see that your child understands these topics adequately. 3rd Grade Math Tests. So, 41 = 4, 42 = 16, 43 = 64, and 44 = 256. Experts are answering questions even after midnight Therefore, if you are stuck on a math problem or homework at the last minute or if it is 10 PM, or 11 PM and you have not found a solution to your math problems, you will get help immediately. Get answer explanations for each question and begin your preparation for the ACCUPLACER college placement test. Since one measurement includes the cat's height and subtracts the turtle's and the other does the opposite, you can essentially just act like the two animals aren't there. If $$\Large x = 7-4\sqrt{3}$$, then the value of $$\Large x+\frac{1}{x}$$. Primary Maths (Grades 4 and 5) - Free Questions and Problems With Answers Grade 4 and 5 maths questions and problems to test the understanding of maths concepts and procedures are presented. An actual ACT Mathematics Test contains 60 questions to be answered in 60 minutes. This Sampler is not intended as a review – it is to familiarize students with format. If you flip the image upside down, you'll see that what you're dealing with is a simple number sequence. 1st Grade Math Tests. Both kinds of questions are necessary to develop mathematical proficiency. If the graph of y = f(x) is transformed into the graph of 2y - 6 = - 4 f(x - 3), point (a , b) on the graph of … This tricky math problem went viral a few years back after it appeared on an entrance exam in Hong Kong… for six-year-olds. This question comes directly from a second grader's math homework. Add spice to the day. When you draw a slanted line in the upper left quadrant of a "+," it becomes the number 4 and the equation thusly becomes 5 + 545 + 5 = 555. Answer : 554521. Elementarymath.edc.org is dedicated to giving K–6 teachers, families, and researchers ideas and materials for: This … You're forgiven if you don't remember exactly how exponents work. Trust us, we get it. CLAT Maths Questions are generally called Word problems which belongs Average, Percentage, Ratio and Proportion, Profit and Loss, Compound Interest, Simple Interest, Clock, HCF and LCM, Circle, Volume, Height and Distance … When you add together the values given for S, B, and G, the sum comes out to 40, and making the missing letter J (which has a value of 14) makes the other diagonal's sum the same. Now you have 9 – 9 + 1, and from there you can simply work from left to right and get your final answer: 1. Of course, it's not actually possible for half a dog to compete in a dog show, but for the sake of this math problem let's assume that it is. While logic might lead you to believe that your math skills have naturally gotten better as you've aged, the unfortunate reality is that, unless you've been solving algebra and geometry problems on a daily basis, the opposite is more likely the case. This mistake could make your mask useless. Other questions are meant to guide students to the heart of the mathematics. In order to figure out how many small dogs are competing, you have to subtract 36 from 49 and then divide that answer, 13, by 2, to get 6.5 dogs, or the number of big dogs competing. This problem, mathematically speaking, is very similar to one of the other ones on this list. If you're having trouble reading that, see here: "Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Let’s face it, math can get a little heavy sometimes so let’s lighten it up with a few quotes that have amused us over the years: Therefore, all you have to do is add the two measurements—170 cm and 130 cm—together and divided them by 2 to get the table's height, 150 cm. Thankfully, though, the New York Times explains step-by-step how to get to July 16, and you can read their detailed deduction here. Elementary Mathematics Questions and Answers for CLAT 2020 Exam, Download PDF For Free at Smartkeeda Asking better questions can open new doors for students, helping to promote mathematical thinking and encouraging classroom discourse. However, since the number alongside x is negative, we need x to be negative as well in order to get to 0. All you have to do is add a 0 to the end of 8.563 and then add like you normally would. Bestlifeonline.com is part of the Meredith Health Group. In order to solve this problem, you need to understand how the area of a parallelogram works. Therefore, x has to be -3. Adding two decimals together is easier than it looks. Unless you don't want your interviewee to work for you. Fifth grade is an important year for math. Here you'll find necessary information and be able to deepen your knowledge in all basic sections of elementary mathematics: arithmetic, algebra, geometry, trigonometry, functions and graphs, principles of analysis and so on. Using PEMDAS (an acronym laying out the order in which you solve it: "parenthesis, exponents, multiplication, division, addition, subtraction"), you would first solve the addition inside of the parentheses (1 + 2 = 3), and from there finish the equation as it's written from left to right. Start your class with a question, … Elementary Mathematics 5 Solution : (a) 650263 − 95742 = 554521 Here, 650263 is minuend, 95742 is subtrahend and the difference is 554521. Teachers. In order to solve this seemingly simple problem, you need to remove the decimal from 1.92 and act like it isn't there. If $$\Large a\ast b = 2a-3b+ab$$, then $$\Large 3\ast 5+5\ast 3$$ is equal to: 2). (b) 836041 − 579406 = 256635 Here, 836041 is minuend, 579406 is subtrahend and the difference is 256635. Try free elementary algebra questions for ACCUPLACER. This question was used in China to identify gifted 5th graders. Bernard: At first I don't know when Cheryl's birthday is, but I know now. The quizzes are interactive and easy to use. It was sneakily included in the legislation. Don't believe us? This question was used in China to identify gifted 5th … ACCUPLACER Math Study Guide – Free Algebra Questions. Live smarter, look better, and live your life to the absolute fullest. There is no need to download any app for these activities. The Students with Disabilities in Mathematics: Frequently Asked Questions-This is a PDF document. If you already know how the area of a parallelogram and the area of a triangle are related, then adding 79 and 10 and subsequently subtracting 72 and 8 to get 9 should make sense—but if you're still confused, then check out this YouTube video for a more in-depth explanation. Such questions help students: Work together to make sense of mathematics. You will receive incredibly detailed scoring results at the end of your SSAT Elementary Level Math practice test to help you identify your strengths and weaknesses. 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# Coprime Numbers
Created by: Team Maths - Examples.com, Last Updated: August 19, 2024
## Coprime Numbers
Coprime numbers, also known as relatively prime numbers, are pairs of numbers that share no common factors other than 1. This concept is a cornerstone of number theory and plays a vital role in various mathematical applications, from cryptography to algorithm design. Through engaging examples and clear explanations, our guide aims to illuminate the significance of coprime numbers, making this fundamental mathematical concept both accessible and intriguing.
## What are Coprime Numbers?
Coprime numbers are two or more integers that have no common divisors except for 1. In other words, their greatest common divisor (GCD) is 1. This definition underscores the importance of understanding divisibility and factors in mathematics. Coprime numbers are essential in areas such as fraction simplification and the Chinese Remainder Theorem, showcasing their broad applicability. Through the study of coprime numbers, students can gain deeper insights into the fabric of mathematics, developing a stronger foundation for advanced mathematical exploration and problem-solving.
## What is the Best Example of a Coprime Number?
A prime example of coprime numbers is the pair 8 and 15. To understand why these two numbers are coprime, we examine their factors. The factors of 8 are 1, 2, 4, and 8, while the factors of 15 are 1, 3, 5, and 15. The only common factor between 8 and 15 is 1, which means they share no other divisors. This characteristic defines them as coprime numbers.
The significance of using 8 and 15 as an example lies in their clear demonstration of the concept. Neither number is prime themselves, yet they do not share any prime factors, illustrating that coprime numbers do not have to be prime. This pair exemplifies the essence of coprimalityβ a mutual lack of divisibility by any number other than 1, highlighting the beauty of number theory in describing unique relationships between different integers. This example not only aids in grasping the concept of coprime numbers but also underscores the diversity and depth found within mathematical relationships.
## How to Find Co-prime Numbers?
To identify co-prime numbers, follow these simple steps:
1. List Factors: For each number in the pair, list out all its positive factors, including 1 and the number itself.
2. Identify Common Factors: Compare the lists of factors for each number to identify any common factors.
3. Check for GCD: If the only common factor between the two numbers is 1, then they are co-prime. This means their Greatest Common Divisor (GCD) is 1.
4. Use GCD Function: Alternatively, use a GCD function available in many calculators or programming languages. If the GCD of the two numbers is 1, they are co-prime.
Example: Consider the numbers 9 and 28. The factors of 9 are 1, 3, and 9, while the factors of 28 are 1, 2, 4, 7, 14, and 28. The only common factor is 1, so 9 and 28 are co-prime.
## Co-prime Numbers List
Coprime Numbers Factors of First Number Factors of Second Number
5 and 8 1, 5 1, 2, 4, 8
7 and 10 1, 7 1, 2, 5, 10
9 and 28 1, 3, 9 1, 2, 4, 7, 14, 28
14 and 25 1, 2, 7, 14 1, 5, 25
21 and 22 1, 3, 7, 21 1, 2, 11, 22
Here’s a list showcasing pairs of co-prime numbers, illustrating the diversity of these number pairs:
• 5 and 8 (Factors of 5: 1, 5; Factors of 8: 1, 2, 4, 8)
• 7 and 10 (Factors of 7: 1, 7; Factors of 10: 1, 2, 5, 10)
• 9 and 28 (Factors of 9: 1, 3, 9; Factors of 28: 1, 2, 4, 7, 14, 28)
• 14 and 25 (Factors of 14: 1, 2, 7, 14; Factors of 25: 1, 5, 25)
• 21 and 22 (Factors of 21: 1, 3, 7, 21; Factors of 22: 1, 2, 11, 22)
Each pair in this list shares no common factors other than 1, exemplifying their co-primality. This diversity underscores that co-prime numbers can be found across the number spectrum, offering infinite possibilities for exploration within mathematics.
## Properties of Co-Prime Numbers
Co-prime numbers, also known as relatively prime numbers, share distinct characteristics that make identifying them straightforward. Here are some key properties:
1. Universal Co-Primality with One: Every number is co-prime with 1, as the only shared divisor is 1 itself.
2. Prime Pair Co-Primality: Any pair of prime numbers are co-prime because their only common factor is 1. For instance, 2 and 3 have factors of 1 and themselves, making them co-prime.
3. Successive Integer Co-Primality: Consecutive numbers, like 2 and 3 or 14 and 15, are always co-prime, as their highest common factor (HCF) is 1.
4. Co-Prime Sums and Products: The sum and product of any two co-prime numbers are also co-prime to each other. For example, 2 and 3 are co-prime, making their sum (5) and product (6) co-prime.
5. Even Number Exclusion: Two even numbers cannot be co-prime since they both share 2 as a common factor.
6. Exclusion of Numbers Ending in 0 and 5: Numbers ending in 0 and 5 are not co-prime to each other due to their divisibility by 5, such as 10 and 15.
## Co-Prime Numbers from 1 to 100
Identifying co-prime pairs among the first 100 numbers involves applying the properties mentioned above. While listing every co-prime pair within this range is extensive, understanding the outlined properties allows for quick identification. For instance:
• (2, 3): Prime pair, hence co-prime.
• (4, 5): Successive integers, thus co-prime.
• (7, 12): No common factors other than 1, making them co-prime.
• (14, 15): Consecutive numbers, automatically co-prime.
• (21, 22): Following the consecutive number rule, they are co-prime.
These examples illustrate the application of co-prime properties, emphasizing patterns like the relationship between prime pairs, consecutive integers, and specific divisibility rules to identify co-prime numbers effectively.
## Co-Prime Numbers and Twin Prime Numbers
Co-prime numbers and twin prime numbers represent intriguing concepts in number theory, each with unique characteristics and applications. Below is a table comparing these two types of numbers:
Aspect Co-Prime Numbers Twin Prime Numbers
Definition Two or more integers that share no common factors other than 1. A pair of prime numbers that differ by 2.
Example 8 and 15 (Factors of 8: 1, 2, 4, 8; Factors of 15: 1, 3, 5, 15) (11, 13) and (17, 19)
Key Characteristic Do not need to be consecutive or prime. Must be consecutive primes.
Mathematical Relevance Important in algorithms, such as those used in cryptography. Highlight the distribution of prime numbers.
Use Simplifying fractions, Chinese Remainder Theorem. Studying patterns in prime numbers, conjectures in number theory.
## Facts about Coprime Numbers
Coprime numbers hold several fascinating properties that make them a subject of study in mathematics. Here are some key facts:
1. Greatest Common Divisor (GCD): The GCD of any two coprime numbers is always 1.
2. Not Necessarily Prime: Coprime pairs can consist of non-prime numbers, such as 8 and 15.
3. Key in Simplification: Coprime numbers are crucial for simplifying fractions to their lowest terms.
4. Cryptographic Significance: They play a critical role in key generation processes in encryption algorithms.
5. Ubiquitous in Equations: Coprime numbers are often used in solving Diophantine equations, which seek integer solutions.
Coprime numbers are a fundamental concept in mathematics, providing deep insight into the structure and properties of integers. Their study not only enriches our understanding of number theory but also finds practical applications in areas ranging from cryptography to algorithm design, showcasing the vast applicability of mathematical principles in solving real-world problems.
Text prompt |
## Section7.1Definition and Notation
### Subsection7.1.1Fundamentals
###### Definition7.1.1. Function.
A function from a set $A$ into a set $B$ is a relation from $A$ into $B$ such that each element of $A$ is related to exactly one element of the set $B\text{.}$ The set $A$ is called the domain of the function and the set $B$ is called the codomain.
The reader should note that a function $f$ is a relation from $A$ into $B$ with two important restrictions:
• Each element in the set $A\text{,}$ the domain of $f\text{,}$ must be related to some element of $B\text{,}$ the codomain.
• The phrase “is related to exactly one element of the set $B$” means that if $(a, b)\in f$ and $(a, c)\in f\text{,}$ then $b = c\text{.}$
Let $A = \{-2, -1,0, 1, 2\}$ and $B = \{0, 1, 2, 3, 4\}\text{,}$ and if $s = \{(-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4)\}\text{,}$ then $s$ is a function from $A$ into $B\text{.}$
Let $\mathbb{R}$ be the real numbers. Then $L = \{(x, 3x) \mid x \in \mathbb{R}\}$ is a function from $\mathbb{R}$ into $\mathbb{R}\text{,}$ or, more simply, $L$ is a function on $\mathbb{R}\text{.}$
It is customary to use a different system of notation for functions than the one we used for relations. If $f$ is a function from the set $A$ into the set $B\text{,}$ we will write $f:A \rightarrow B\text{.}$
The reader is probably more familiar with the notation for describing functions that is used in basic algebra or calculus courses. For example, $y =\frac{1}{x}$ or $f(x) =\frac{1}{x}$ both define the function $\left\{\left.\left(x,\frac{1}{x}\right)\right| x\in \mathbb{R}, x\neq 0\right\}\text{.}$ Here the domain was assumed to be those elements of $\mathbb{R}$ whose substitutions for $x$ make sense, the nonzero real numbers, and the codomain was assumed to be $\mathbb{R}\text{.}$ In most cases, we will make a point of listing the domain and codomain in addition to describing what the function does in order to define a function.
The terms mapping, map, and transformation are also used for functions.
###### Definition7.1.4. The Set of Functions Between Two Sets.
Given two sets, $A$ and $B\text{,}$ the set of all function from A into B is denoted $B^A\text{.}$
The notation used for sets of functions makes sense in light of Exercise 7.
One way to imagine a function and what it does is to think of it as a machine. The machine could be mechanical, electronic, hydraulic, or abstract. Imagine that the machine only accepts certain objects as raw materials or input. The possible raw materials make up the domain. Given some input, the machine produces a finished product that depends on the input. The possible finished products that we imagine could come out of this process make up the codomain.
We can define a function based on specifying the codomain element to which each domain element is related. For example, $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = x^2$ is an alternate description of $f= \left\{\left.\left(x, x ^2\right)\right|x \in \mathbb{R}\right\}\text{.}$
###### Definition7.1.6. Image of an element under a function.
Let $f:A \rightarrow B\text{,}$ read “Let $f$ be a function from the set $A$ into the set $B\text{.}$” If $a \in A\text{,}$ then $f(a)$ is used to denote that element of $B$ to which $a$ is related. $f(a)$ is called the image of $a\text{,}$ or, more precisely, the image of $a$ under $f\text{.}$ We write $f(a) = b$ to indicate that the image of $a$ is $b\text{.}$
In Example 7.1.5, the image of 2 under $f$ is 4; that is, $f(2) = 4\text{.}$ In Example 7.1.2, the image of $-1$ under $s$ is 1; that is, $s(-1) = 1\text{.}$
###### Definition7.1.7. Range of a Function.
The range of a function is the set of images of its domain. If $f:X \rightarrow Y\text{,}$ then the range of $f$ is denoted $f(X)\text{,}$ and
\begin{equation*} f(X) = \{f(a) \mid a \in X\} = \{b \in Y \mid \exists a \in X\textrm{ such that } f(a) = b\}\text{.} \end{equation*}
Note that the range of a function is a subset of its codomain. $f(X)$ is also read as “the image of the set $X$ under the function $f$” or simply “the image of $f\text{.}$”
In Example 7.1.2, $s(A)= \{0, 1, 4\}\text{.}$ Notice that 2 and 3 are not images of any element of $A\text{.}$ In addition, note that both 1 and 4 are related to more than one element of the domain: $s(1) = s(-1) = 1$ and $s(2) = s(-2) = 4\text{.}$ This does not violate the definition of a function. Go back and read the definition if this isn't clear to you.
In Example 7.1.3, the range of $L$ is equal to its codomain, $\mathbb{R}\text{.}$ If $b$ is any real number, we can demonstrate that it belongs to $L(\mathbb{R})$ by finding a real number $x$ for which $L(x) = b\text{.}$ By the definition of $L\text{,}$ $L(x) = 3x\text{,}$ which leads us to the equation $3x = b\text{.}$ This equation always has a solution, $\frac{b}{3}\text{;}$ thus $L(\mathbb{R}) = \mathbb{R}\text{.}$
The formula that we used to describe the image of a real number under $L\text{,}$ $L(x) = 3x\text{,}$ is preferred over the set notation for $L$ due to its brevity. Any time a function can be described with a rule or formula, we will use this form of description. In Example 7.1.2, the image of each element of $A$ is its square. To describe that fact, we write $s(a) = a^2$ ($a \in A$), or $S:A \rightarrow B$ defined by $S(a) = a^2\text{.}$
There are many ways that a function can be described. Many factors, such as the complexity of the function, dictate its representation.
Suppose a survey of 1,000 persons is done asking how many hours of television each watches per day. Consider the function $W: \{0,1,\ldots , 24\} \rightarrow \{0,1,2,\ldots ,1000\}$ defined by
\begin{equation*} W(t) = \textrm{the number of persons who gave a response of } t \textrm{ hours} \end{equation*}
This function will probably have no formula such as the ones for $s$ and $L$ above.
Consider the function $m:\mathbb{P} \rightarrow \mathbb{Q}$ defined by the set
\begin{equation*} m = \{(1, 1), (2, 1/2), (3, 9), (4, 1/4), (5, 25), . . . \} \end{equation*}
No simple single formula could describe $m\text{,}$ but if we assume that the pattern given continues, we can write
\begin{equation*} m(x) =\left\{ \begin{array}{cc} x^2 & \textrm{if } x \textrm{ is odd} \\ 1/x & \textrm{if } x \textrm{ is even} \\ \end{array} \right. \end{equation*}
### Subsection7.1.2Functions of Two Variables
If the domain of a function is the Cartesian product of two sets, then our notation and terminology changes slightly. For example, consider the function $C:\mathbb{N} \times \mathbb{N}\rightarrow \mathbb{N}$ defined by $C\left(\left(n_1,n_2\right)\right)=n_1^2+n_2^2- n_1n_2+10\text{.}$ For this function, we would drop one set of parentheses and write $C(4, 2) = 22\text{,}$ not $C((4, 2)) = 22\text{.}$ We call $C$ a function of two variables. From one point of view, this function is no different from any others that we have seen. The elements of the domain happen to be slightly more complicated. On the other hand, we can look at the individual components of the ordered pairs as being separate. If we interpret $C$ as giving us the cost of producing quantities of two products, we can imagine varying $n_1$ while $n_2$ is fixed, or vice versa.
### Subsection7.1.3SageMath Note
There are several ways to define a function in Sage. The simplest way to implement $f$ is as follows.
Sage is built upon the programming language Python, which is a strongly typed language and so you can't evaluate expressions such as f('Hello'). However a function such as $f\text{,}$ as defined above, will accept any type of number, so a bit more work is needed to restrict the inputs of $f$ to the integers.
A second way to define a function in Sage is based on Python syntax.
### Subsection7.1.4Non-Functions
We close this section with two examples of relations that are not functions.
Let $A = B = \{1, 2, 3\}$ and let $f= \{(1, 2), (2, 3)\}\text{.}$ Here $f$ is not a function from $A$ into $B$ since $f$ does not act on, or “use,” all elements of $A\text{.}$
Let $A = B = \{1,2, 3\}$ and let $g = \{(1, 2), (2, 3), (2, 1),(3, 2)\}\text{.}$ We note that $g$ acts on all of $A\text{.}$ However, $g$ is still not a function since $(2, 3) \in g$ and $(2, 1) \in g$ and the condition on each domain being related to exactly one element of the codomain is violated.
### Exercises7.1.5Exercises for Section 7.1
###### 1.
Let $A = \{1, 2, 3, 4\}$ and $B = \{a, b, c, d\}\text{.}$ Determine which of the following are functions. Explain.
1. $f \subseteq A \times B\text{,}$ where $f = \{(1, a), (2, b), (3, c), (4, d)\}\text{.}$
2. $g\subseteq A\times B\text{,}$ where $g = \{(1, a), (2, a), (3,b), (4,d)\}\text{.}$
3. $h \subseteq A \times B\text{,}$ where $h = \{(1, a), (2, b), (3, c)\}\text{.}$
4. $k \subseteq A\times B\text{,}$ where $k = \{(1, a), (2, b), (2, c), (3, a), (4, a)\}\text{.}$
5. $L\subseteq A\times A\text{,}$ where $L = \{(1, 1), (2, 1), (3, 1), (4, 1)\}\text{.}$
1. Yes
2. Yes
3. No
4. No
5. Yes
###### 2.
Let $A$ be a set and let $S$ be any subset of $A\text{.}$ Let $\chi_S: A\to \{0,1\}$ be defined by
\begin{equation*} \chi_S(x)= \left\{ \begin{array}{cc} 1 & \textrm{if } x\in S \\ 0 & \textrm{if } x\notin S \\ \end{array} \right. \end{equation*}
The function $\chi_S$ is called the characteristic function of $S\text{.}$
1. If $A = \{a, b, c\}$ and $S = \{a, b\}\text{,}$ list the elements of $\chi_S$ .
2. If $A = \{a, b, c, d, e\}$ and $S = \{a, c, e\}\text{,}$ list the elements of $\chi_S\text{.}$
3. If $A = \{a, b, c\}\text{,}$ what are $\chi_{\emptyset}$ and $\chi_A\text{?}$
###### 3.
Find the ranges of each of the relations that are functions in Exercise 1.
1. Range of $f=f(A)=\{a,b,c,d\}=B$
2. Range of $g=g(A)=\{a,b,d\}$
3. Range of $L=L(A)=\{1\}$
###### 4.
Find the ranges of the following functions on $\mathbb{Z}\text{:}$
1. $g = \{(x, 4x +1)|x \in \mathbb{Z}\}\text{.}$
2. $h(x) = \textrm{the least integer that is greater than or equal to } \sqrt{|x|}\text{.}$
3. $P(x) = x + 10\text{.}$
###### 5.
Let $f:\mathbb{P}\to \mathbb{P}\text{,}$ where $f(a)$ is the largest power of two that evenly divides $a\text{;}$ for example, $f(12)=4,f(9)=1,\text{and} f(8)=8\text{.}$ Describe the equivalence classes of the kernel of $f\text{.}$
###### 6.
Let $U$ be a set with subsets $A$ and $B\text{.}$
1. Show that $g:U\to \{0,1\}$ defined by $g(a)=\min \left(C_A(a),C_B(a)\right)$ is the characteristic function of $A\cap B\text{.}$
2. What characteristic function is $h:U\to \{0,1\}$ defined by $h(a)=\max \left(C_A(a),C_B(a)\right)\text{?}$
3. How are the characteristic functions of $A$ and $A^c$ related?
###### 7.
If $A$ and $B$ are finite sets, how many different functions are there from $A$ into $B\text{?}$
For each of the $\lvert A \rvert$ elements of $A\text{,}$ there are $\lvert B \rvert$ possible images, so there are $\lvert B \rvert\cdot \lvert B \rvert\cdot \ldots \cdot \lvert B \rvert=\left\lvert B \rvert^{\lvert A \rvert}\right.$ functions from $A$ into $B\text{.}$
Let $f$ be a function with domain $A$ and codomain $B\text{.}$ Consider the relation $K \subseteq A \times A$ defined on the domain of $f$ by $(x, y) \in K$ if and only if $f(x) = f(y)\text{.}$ The relation $K$ is called \textit{ the kernel of f}.
1. Prove that $K$ is an equivalence relation.
2. For the specific case of $A = \mathbb{Z}\text{,}$ where $\mathbb{Z}$ is the set of integers, let $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be defined by $f(x) = x^2\text{.}$ Describe the equivalence classes of the kernel for this specific function. |
# Instantaneous velocity: Definition, Formula, Examples, Calculation
The velocity is classified as average velocity and instantaneous velocity based on the time interval used to calculate it. In this article, we will go over instantaneous velocity in great detail.
Contents:
## What is the Instantaneous velocity?
Instantaneous velocity is the velocity of an object at a specific instant in time. It is an average velocity calculated for the smallest interval of time (dt→0). At a specific instant, it is a ratio of the smallest change in position (d𝑥) to its respective smallest time interval (dt).
During the motion, the object undergoes acceleration and retardation, or sometimes it moves with a constant velocity. Thus only average velocity is not able to describe the motion of the object at every instant of time.
In the above velocity-time graph, the Vaverage indicates the average velocity of the object during its total travel. While Va and Vb indicates the instantaneous velocities of the object at time ta and tb respectively.
## Instantaneous velocity formula:
• By using the method of limits, the instantaneous velocity (IV) of the object at a time ‘t’ is given by,
V =\lim_{\Delta t \to 0} \frac{S_{(t + \Delta t)}-S_{(t)}}{\Delta t}
V(t) =\frac{dS}{dt}
## How to find the Instantaneous velocity?
The following are the two methods to calculate the IV:-
1] Analytical method:
When the equation for the position of the object is provided as a function of time, then in such cases the instantaneous velocity can be calculated by any of the following two methods.
a) Limits: Here the velocity is calculated for the very smallest time interval (Δ t → 0).
V =\lim_{\Delta t \to 0} \frac{S_{(t + \Delta t)}-S_{(t)}}{\Delta t}
b) Differentiation: The instantaneous velocity is also calculated by taking the derivative of the displacement equation with respect to time ‘t’.
V(t) =\frac{dS}{dt}
2] Graphical method:
This method is used to calculate the instantaneous velocity from a position-time graph (Discussed below).
### Instantaneous velocity from a graph:
Following are the two cases to calculate the IV from the position-time graph:-
Case 1] When position-time graph is linear:
As shown in the below figure, if the profile of the position-time graph is linear then it indicates that the object is moving with a constant velocity.
In such cases, the instantaneous velocity at any instant is equal to the average velocity.
Vinstantaneous = Vaverage
Therefore for the above graph, the instantaneous velocity is equal to the slope of the position-time profile.
V_{\text{instantaneous}} = V_{\text{average}} = \text{Slope(1-2)} = \frac{x_{2}-x_{1}}{t_{2}-t_{1}}
Case 2] When position-time graph is non-linear:
The position-time profile in the below figure is non-linear hence the object has different values for instantaneous velocities.
In this case, the instantaneous velocity (IV) is equal to the slope of the tangent drawn to the position-time graph at that instant.
For such a profile, the IV is found by using the following steps:-
a] Locate the position of the object at the time t:
b] At the marked point, draw a tangent line to the curve:
c] Find the slope of the tangent:
\text{Slope(tangent 1-2)} = \frac{x_{2}-x_{1}}{t_{2}-t_{1}}
As the instantaneous velocity is equal to the slope of the tangent, Therefore,
V_{\text{instantaneous}} = \frac{x_{2}-x_{1}}{t_{2}-t_{1}}
## Examples:
1] If the displacement of the object is given by (3t² + 5t) m, then find the velocity of the object at t = 3.5 seconds.
Given:
S = (3t² + 5t) m
t = 3.5 seconds
Solution:-
Method 1: Differentiation
The instantaneous velocity of the object is given by,
V =\frac{dS}{dt}
V =\frac{d}{dt}(3t^{2}+5t)
V = 6t + 5
At t = 3.5 seconds, the instantaneous velocity is given by,
V = 6(3.5) + 5
V = 26 m/s
Method 2: Limit
The instantaneous velocity of the object is given by,
V =\lim_{\Delta t \to 0} \frac{S_{(t + \Delta t)}-S_{(t)}}{\Delta t}
= \lim_{\Delta t \to 0} \frac{[3(t + \Delta t)^{2} +5(t + \Delta t)] – [3t^{2}+5t]}{\Delta t}
= \lim_{\Delta t \to 0} \frac{3t^{2}+6t\Delta t+3\Delta t^{2}+5t+5\Deltat-3t^{2}-5t}{\Delta t}
= \lim_{\Delta t \to 0}\frac{3\Delta t^{2}+6t\Delta t+5\Delta t}{\Delta t}
= \lim_{\Delta t \to 0} 3\Delta t+6t+5
V = 6t + 5
The equation indicates the instantaneous velocity (IV) of the object as a function of time. The IV at t = 3.5 seconds is given by,
V = 6(3.5) + 5
V = 26 m/s
2] The below graph shows the change in position of the object with respect to time t, find the velocity of the object at time t = 4 seconds.
Given:
t = 4 seconds
Solution:-
Step 1] Locate the point on the curve that indicates the position of the object at t = 4 seconds:
Step 2] Draw a tangent to the curve at point a:
Step 3] Find the slope of the tangent:
The slope of the tangent represents the velocity of the object at t = 4 seconds.
V_{t=4} = \text{Slope(1-2)} = \frac{x_{2}-x_{1}}{t_{2}-t_{1}}
From above graph,
x2 = 0.8 m, t2 = 6 seconds, x1 = 0.2 seconds, t1 = 3 seconds
V_{t=4} = \text{Slope(1-2)} = \frac{0.8-0.2}{6-3}
Vt=4 = 0.2 m/s
This is the IV of the object at t = 4 seconds.
## FAQs:
1. What is importance of an Instantaneous velocity?
Instantaneous velocity (IV) is important to describe the motion of the moving object at a particular instant.
2. Why I get instantaneous velocity negative?
The sign of the instantaneous velocity depends on the direction of the motion of the object. Thus if the object is moving in a negative direction, then the I.V. of the object becomes negative.
3. What is an example of instantaneous velocity?
Assuming that the motorcycle is traveling straight, the speedometer on a motorcycle will display the instantaneous velocity of the motorcycle at each instant.
4. How does velocity differ from instantaneous velocity?
Velocity indicates an average value of velocity over a certain distance while instantaneous velocity describes a velocity at a specific instant.
5. Can instantaneous velocity be constant?
The instantaneous velocity changes during acceleration/deceleration and changes in direction, and it remains constant when an object moves with constant speed in the same direction.
6. Can instantaneous velocity become zero?
The instantaneous velocity can be zero. At the instant when motion starts or ends, the object has zero instantaneous velocity.
7. Why is instantaneous velocity not always higher than average velocity?
The average velocity indicates the average value for the velocity of the object. Thus the values for instantaneous velocity are sometimes greater and sometimes lower than the average velocity.
You may like to read this:
Pratik is a Graduated Mechanical engineer. He enjoys sharing the engineering knowledge learned by him with people. |
Giáo trình
# Precalculus
Mathematics and Statistics
## Graphs of the Other Trigonometric Functions
Tác giả: OpenStaxCollege
We know the tangent function can be used to find distances, such as the height of a building, mountain, or flagpole. But what if we want to measure repeated occurrences of distance? Imagine, for example, a police car parked next to a warehouse. The rotating light from the police car would travel across the wall of the warehouse in regular intervals. If the input is time, the output would be the distance the beam of light travels. The beam of light would repeat the distance at regular intervals. The tangent function can be used to approximate this distance. Asymptotes would be needed to illustrate the repeated cycles when the beam runs parallel to the wall because, seemingly, the beam of light could appear to extend forever. The graph of the tangent function would clearly illustrate the repeated intervals. In this section, we will explore the graphs of the tangent and other trigonometric functions.
# Analyzing the Graph of y = tan x
We will begin with the graph of the tangent function, plotting points as we did for the sine and cosine functions. Recall that
The period of the tangent function is$\text{\hspace{0.17em}}\pi \text{\hspace{0.17em}}$because the graph repeats itself on intervals of$\text{\hspace{0.17em}}k\pi \text{\hspace{0.17em}}$where$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$is a constant. If we graph the tangent function on$\text{\hspace{0.17em}}-\frac{\pi }{2}\text{\hspace{0.17em}}$to$\text{\hspace{0.17em}}\frac{\pi }{2},\text{\hspace{0.17em}}$we can see the behavior of the graph on one complete cycle. If we look at any larger interval, we will see that the characteristics of the graph repeat.
We can determine whether tangent is an odd or even function by using the definition of tangent.
Therefore, tangent is an odd function. We can further analyze the graphical behavior of the tangent function by looking at values for some of the special angles, as listed in [link].
$x$ $-\frac{\pi }{2}$ $-\frac{\pi }{3}$ $-\frac{\pi }{4}$ $-\frac{\pi }{6}$ 0 $\frac{\pi }{6}$ $\frac{\pi }{4}$ $\frac{\pi }{3}$ $\frac{\pi }{2}$ $\mathrm{tan}\left(x\right)$ undefined $-\sqrt{3}$ –1 $-\frac{\sqrt{3}}{3}$ 0 $\frac{\sqrt{3}}{3}$ 1 $\sqrt{3}$ undefined
These points will help us draw our graph, but we need to determine how the graph behaves where it is undefined. If we look more closely at values when$\text{\hspace{0.17em}}\frac{\pi }{3}we can use a table to look for a trend. Because$\text{\hspace{0.17em}}\frac{\pi }{3}\approx 1.05\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}\frac{\pi }{2}\approx 1.57,\text{\hspace{0.17em}}$we will evaluate$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$at radian measures$\text{\hspace{0.17em}}1.05as shown in [link].
$x$ 1.3 1.5 1.55 1.56 3.6 14.1 48.1 92.6
As$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$approaches$\text{\hspace{0.17em}}\frac{\pi }{2},\text{\hspace{0.17em}}$the outputs of the function get larger and larger. Because$\text{\hspace{0.17em}}y=\mathrm{tan}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$is an odd function, we see the corresponding table of negative values in [link].
$x$ −1.3 −1.5 −1.55 −1.56 $\mathrm{tan}\text{\hspace{0.17em}}x$ −3.6 −14.1 −48.1 −92.6
We can see that, as$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$approaches$\text{\hspace{0.17em}}-\frac{\pi }{2},\text{\hspace{0.17em}}$the outputs get smaller and smaller. Remember that there are some values of$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$for which$\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x=0.\text{\hspace{0.17em}}$For example,$\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{\pi }{2}\right)=0\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{3\pi }{2}\right)=0.\text{\hspace{0.17em}}$At these values, the tangent function is undefined, so the graph of$\text{\hspace{0.17em}}y=\mathrm{tan}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$has discontinuities atAt these values, the graph of the tangent has vertical asymptotes. [link] represents the graph of$\text{\hspace{0.17em}}y=\mathrm{tan}\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$The tangent is positive from 0 to$\text{\hspace{0.17em}}\frac{\pi }{2}\text{\hspace{0.17em}}$and from$\text{\hspace{0.17em}}\pi \text{\hspace{0.17em}}$to$\text{\hspace{0.17em}}\frac{3\pi }{2},\text{\hspace{0.17em}}$corresponding to quadrants I and III of the unit circle.
# Graphing Variations of y = tan x
As with the sine and cosine functions, the tangent function can be described by a general equation.
We can identify horizontal and vertical stretches and compressions using values of$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}B.\text{\hspace{0.17em}}$The horizontal stretch can typically be determined from the period of the graph. With tangent graphs, it is often necessary to determine a vertical stretch using a point on the graph.
Because there are no maximum or minimum values of a tangent function, the term amplitude cannot be interpreted as it is for the sine and cosine functions. Instead, we will use the phrase stretching/compressing factor when referring to the constant$\text{\hspace{0.17em}}A.$
Features of the Graph of y = Atan(Bx)
• The stretching factor is$\text{\hspace{0.17em}}|A|.$
• The period is$\text{\hspace{0.17em}}P=\frac{\pi }{|B|}.$
• The domain is all real numbers$\text{\hspace{0.17em}}x,$where$\text{\hspace{0.17em}}x\ne \frac{\pi }{2|B|}+\frac{\pi }{|B|}k\text{\hspace{0.17em}}$such that$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$is an integer.
• The range is$\text{\hspace{0.17em}}\left(\mathrm{-\infty },\infty \right).$
• The asymptotes occur at$\text{\hspace{0.17em}}x=\frac{\pi }{2|B|}+\frac{\pi }{|B|}k,\text{\hspace{0.17em}}$where$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$is an integer.
• $y=A\mathrm{tan}\left(Bx\right)\text{\hspace{0.17em}}$is an odd function.
## Graphing One Period of a Stretched or Compressed Tangent Function
We can use what we know about the properties of the tangent function to quickly sketch a graph of any stretched and/or compressed tangent function of the form$\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{tan}\left(Bx\right).\text{\hspace{0.17em}}$We focus on a single period of the function including the origin, because the periodic property enables us to extend the graph to the rest of the function’s domain if we wish. Our limited domain is then the interval$\text{\hspace{0.17em}}\left(-\frac{P}{2},\frac{P}{2}\right)\text{\hspace{0.17em}}$and the graph has vertical asymptotes at$\text{\hspace{0.17em}}±\frac{P}{2}\text{\hspace{0.17em}}$where$\text{\hspace{0.17em}}P=\frac{\pi }{B}.\text{\hspace{0.17em}}$On$\text{\hspace{0.17em}}\left(-\frac{\pi }{2},\frac{\pi }{2}\right),\text{\hspace{0.17em}}$the graph will come up from the left asymptote at$\text{\hspace{0.17em}}x=-\frac{\pi }{2},\text{\hspace{0.17em}}$cross through the origin, and continue to increase as it approaches the right asymptote at$\text{\hspace{0.17em}}x=\frac{\pi }{2}.\text{\hspace{0.17em}}$To make the function approach the asymptotes at the correct rate, we also need to set the vertical scale by actually evaluating the function for at least one point that the graph will pass through. For example, we can use
because$\text{\hspace{0.17em}}\mathrm{tan}\left(\frac{\pi }{4}\right)=1.$
Given the function$\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{tan}\left(Bx\right),\text{\hspace{0.17em}}$graph one period.
1. Identify the stretching factor,$\text{\hspace{0.17em}}|A|.$
2. Identify$\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$and determine the period,$\text{\hspace{0.17em}}P=\frac{\pi }{|B|}.$
3. Draw vertical asymptotes at$\text{\hspace{0.17em}}x=-\frac{P}{2}\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}x=\frac{P}{2}.$
4. For$\text{\hspace{0.17em}}A>0,\text{\hspace{0.17em}}$the graph approaches the left asymptote at negative output values and the right asymptote at positive output values (reverse for$\text{\hspace{0.17em}}A<0$).
5. Plot reference points at$\text{\hspace{0.17em}}\left(\frac{P}{4},A\right),\text{\hspace{0.17em}}$$\left(0,0\right),\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}\left(-\frac{P}{4},-A\right),\text{\hspace{0.17em}}$and draw the graph through these points.
Sketching a Compressed Tangent
Sketch a graph of one period of the function$\text{\hspace{0.17em}}y=0.5\mathrm{tan}\left(\frac{\pi }{2}x\right).$
First, we identify$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}B.$
Because$\text{\hspace{0.17em}}A=0.5\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}B=\frac{\pi }{2},\text{\hspace{0.17em}}$we can find the stretching/compressing factor and period. The period is$\text{\hspace{0.17em}}\frac{\pi }{\frac{\pi }{2}}=2,\text{\hspace{0.17em}}$so the asymptotes are at$\text{\hspace{0.17em}}x=±1.\text{\hspace{0.17em}}$At a quarter period from the origin, we have
$f(0.5)=0.5tan( 0.5π 2 ) =0.5tan( π 4 ) =0.5$
This means the curve must pass through the points$\text{\hspace{0.17em}}\left(0.5,0.5\right),$$\left(0,0\right),$and$\text{\hspace{0.17em}}\left(-0.5,-0.5\right).\text{\hspace{0.17em}}$The only inflection point is at the origin. [link] shows the graph of one period of the function.
Sketch a graph of$\text{\hspace{0.17em}}f\left(x\right)=3\mathrm{tan}\left(\frac{\pi }{6}x\right).$
## Graphing One Period of a Shifted Tangent Function
Now that we can graph a tangent function that is stretched or compressed, we will add a vertical and/or horizontal (or phase) shift. In this case, we add$\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}D\text{\hspace{0.17em}}$to the general form of the tangent function.
$f\left(x\right)=A\mathrm{tan}\left(Bx-C\right)+D$
The graph of a transformed tangent function is different from the basic tangent function$\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$in several ways:
Features of the Graph of y = Atan(BxC)+D
• The stretching factor is$\text{\hspace{0.17em}}|A|.$
• The period is$\text{\hspace{0.17em}}\frac{\pi }{|B|}.$
• The domain is$\text{\hspace{0.17em}}x\ne \frac{C}{B}+\frac{\pi }{|B|}k,$where$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$is an integer.
• The range is$\text{\hspace{0.17em}}\left(\mathrm{-\infty },-|A|\right]\cup \left[|A|,\infty \right).$
• The vertical asymptotes occur at$\text{\hspace{0.17em}}x=\frac{C}{B}+\frac{\pi }{2|B|}k,$where$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$is an odd integer.
• There is no amplitude.
• $y=A\text{\hspace{0.17em}}\mathrm{tan}\left(Bx\right)\text{\hspace{0.17em}}$is and odd function because it is the qoutient of odd and even functions(sin and cosine perspectively).
Given the function$\text{\hspace{0.17em}}y=A\mathrm{tan}\left(Bx-C\right)+D,\text{\hspace{0.17em}}$sketch the graph of one period.
1. Express the function given in the form$\text{\hspace{0.17em}}y=A\mathrm{tan}\left(Bx-C\right)+D.$
2. Identify the stretching/compressing factor,$\text{\hspace{0.17em}}|A|.$
3. Identify$\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$and determine the period,$\text{\hspace{0.17em}}P=\frac{\pi }{|B|}.$
4. Identify$\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$and determine the phase shift,$\text{\hspace{0.17em}}\frac{C}{B}.$
5. Draw the graph of$\text{\hspace{0.17em}}y=A\mathrm{tan}\left(Bx\right)\text{\hspace{0.17em}}$shifted to the right by$\text{\hspace{0.17em}}\frac{C}{B}\text{\hspace{0.17em}}$and up by$\text{\hspace{0.17em}}D.$
6. Sketch the vertical asymptotes, which occur atwhereis an odd integer.
7. Plot any three reference points and draw the graph through these points.
Graphing One Period of a Shifted Tangent Function
Graph one period of the function$\text{\hspace{0.17em}}y=-2\mathrm{tan}\left(\pi x+\pi \right)\text{\hspace{0.17em}}-1.$
• Step 1. The function is already written in the form$\text{\hspace{0.17em}}y=A\mathrm{tan}\left(Bx-C\right)+D.$
• Step 2.$\text{\hspace{0.17em}}A=-2,\text{\hspace{0.17em}}$so the stretching factor is$\text{\hspace{0.17em}}|A|=2.$
• Step 3.$\text{\hspace{0.17em}}B=\pi ,\text{\hspace{0.17em}}$so the period is$\text{\hspace{0.17em}}P=\frac{\pi }{|B|}=\frac{\pi }{\pi }=1.$
• Step 4.$\text{\hspace{0.17em}}C=-\pi ,\text{\hspace{0.17em}}$so the phase shift is$\text{\hspace{0.17em}}\frac{C}{B}=\frac{-\pi }{\pi }=-1.$
• Step 5-7. The asymptotes are at$\text{\hspace{0.17em}}x=-\frac{3}{2}\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}x=-\frac{1}{2}\text{\hspace{0.17em}}$and the three recommended reference points are$\text{\hspace{0.17em}}\left(-1.25,1\right),\text{\hspace{0.17em}}$$\left(-1,-1\right),\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}\left(-0.75,-3\right).\text{\hspace{0.17em}}$The graph is shown in [link].
• Analysis
Note that this is a decreasing function because$\text{\hspace{0.17em}}A<0.$
How would the graph in [link] look different if we made$\text{\hspace{0.17em}}A=2\text{\hspace{0.17em}}$instead of$\text{\hspace{0.17em}}-2?$
It would be reflected across the line$\text{\hspace{0.17em}}y=-1,\text{\hspace{0.17em}}$becoming an increasing function.
Given the graph of a tangent function, identify horizontal and vertical stretches.
1. Find the period$\text{\hspace{0.17em}}P\text{\hspace{0.17em}}$from the spacing between successive vertical asymptotes or x-intercepts.
2. Write$\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{tan}\left(\frac{\pi }{P}x\right).$
3. Determine a convenient point$\text{\hspace{0.17em}}\left(x,f\left(x\right)\right)\text{\hspace{0.17em}}$on the given graph and use it to determine$\text{\hspace{0.17em}}A.$
Identifying the Graph of a Stretched Tangent
Find a formula for the function graphed in [link].
The graph has the shape of a tangent function.
• Step 1. One cycle extends from –4 to 4, so the period is$\text{\hspace{0.17em}}P=8.\text{\hspace{0.17em}}$Since$\text{\hspace{0.17em}}P=\frac{\pi }{|B|},\text{\hspace{0.17em}}$we have$\text{\hspace{0.17em}}B=\frac{\pi }{P}=\frac{\pi }{8}.$
• Step 2. The equation must have the form$f\left(x\right)=A\mathrm{tan}\left(\frac{\pi }{8}x\right).$
• Step 3. To find the vertical stretch$\text{\hspace{0.17em}}A,$we can use the point$\text{\hspace{0.17em}}\left(2,2\right).$
• Because$\text{\hspace{0.17em}}\mathrm{tan}\left(\frac{\pi }{4}\right)=1,\text{\hspace{0.17em}}$$A=2.$
This function would have a formula$\text{\hspace{0.17em}}f\left(x\right)=2\mathrm{tan}\left(\frac{\pi }{8}x\right).$
Find a formula for the function in [link].
$g\left(x\right)=4\mathrm{tan}\left(2x\right)$
# Analyzing the Graphs of y = sec x and y = cscx
The secant was defined by the reciprocal identity$\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}x=\frac{1}{\mathrm{cos}\text{\hspace{0.17em}}x}.\text{\hspace{0.17em}}$Notice that the function is undefined when the cosine is 0, leading to vertical asymptotes at$\text{\hspace{0.17em}}\frac{\pi }{2},\text{\hspace{0.17em}}$$\frac{3\pi }{2},\text{\hspace{0.17em}}$etc. Because the cosine is never more than 1 in absolute value, the secant, being the reciprocal, will never be less than 1 in absolute value.
We can graph$\text{\hspace{0.17em}}y=\mathrm{sec}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$by observing the graph of the cosine function because these two functions are reciprocals of one another. See [link]. The graph of the cosine is shown as a dashed orange wave so we can see the relationship. Where the graph of the cosine function decreases, the graph of the secant function increases. Where the graph of the cosine function increases, the graph of the secant function decreases. When the cosine function is zero, the secant is undefined.
The secant graph has vertical asymptotes at each value of$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$where the cosine graph crosses the x-axis; we show these in the graph below with dashed vertical lines, but will not show all the asymptotes explicitly on all later graphs involving the secant and cosecant.
Note that, because cosine is an even function, secant is also an even function. That is,$\text{\hspace{0.17em}}\mathrm{sec}\left(-x\right)=\mathrm{sec}\text{\hspace{0.17em}}x.$
As we did for the tangent function, we will again refer to the constant$\text{\hspace{0.17em}}|A|\text{\hspace{0.17em}}$as the stretching factor, not the amplitude.
Features of the Graph of y = Asec(Bx)
• The stretching factor is$\text{\hspace{0.17em}}|A|.$
• The period is$\text{\hspace{0.17em}}\frac{2\pi }{|B|}.$
• The domain is$\text{\hspace{0.17em}}x\ne \frac{\pi }{2|B|}k,\text{\hspace{0.17em}}$where$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$is an odd integer.
• The range is$\text{\hspace{0.17em}}\left(-\infty ,-|A|\right]\cup \left[|A|,\infty \right).$
• The vertical asymptotes occur at$\text{\hspace{0.17em}}x=\frac{\pi }{2|B|}k,\text{\hspace{0.17em}}$where$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$is an odd integer.
• There is no amplitude.
• $y=A\mathrm{sec}\left(Bx\right)\text{\hspace{0.17em}}$is an even function because cosine is an even function.
Similar to the secant, the cosecant is defined by the reciprocal identity$\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x=\frac{1}{\mathrm{sin}\text{\hspace{0.17em}}x}.\text{\hspace{0.17em}}$Notice that the function is undefined when the sine is 0, leading to a vertical asymptote in the graph at$\text{\hspace{0.17em}}0,\text{\hspace{0.17em}}$$\pi ,\text{\hspace{0.17em}}$etc. Since the sine is never more than 1 in absolute value, the cosecant, being the reciprocal, will never be less than 1 in absolute value.
We can graph$\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$by observing the graph of the sine function because these two functions are reciprocals of one another. See [link]. The graph of sine is shown as a dashed orange wave so we can see the relationship. Where the graph of the sine function decreases, the graph of the cosecant function increases. Where the graph of the sine function increases, the graph of the cosecant function decreases.
The cosecant graph has vertical asymptotes at each value of$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$where the sine graph crosses the x-axis; we show these in the graph below with dashed vertical lines.
Note that, since sine is an odd function, the cosecant function is also an odd function. That is,$\text{\hspace{0.17em}}\mathrm{csc}\left(-x\right)=\mathrm{-csc}x.$
The graph of cosecant, which is shown in [link], is similar to the graph of secant.
Features of the Graph of y = Acsc(Bx)
• The stretching factor is$\text{\hspace{0.17em}}|A|.$
• The period is$\text{\hspace{0.17em}}\frac{2\pi }{|B|}.$
• The domain is$\text{\hspace{0.17em}}x\ne \frac{\pi }{|B|}k,\text{\hspace{0.17em}}$where$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$is an integer.
• The range is$\left(-\infty ,-|A|\right]\cup \left[|A|,\infty \right).$
• The asymptotes occur at$\text{\hspace{0.17em}}x=\frac{\pi }{|B|}k,\text{\hspace{0.17em}}$where$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$is an integer.
• $y=A\mathrm{csc}\left(Bx\right)\text{\hspace{0.17em}}$is an odd function because sine is an odd function.
# Graphing Variations of y = sec x and y= csc x
For shifted, compressed, and/or stretched versions of the secant and cosecant functions, we can follow similar methods to those we used for tangent and cotangent. That is, we locate the vertical asymptotes and also evaluate the functions for a few points (specifically the local extrema). If we want to graph only a single period, we can choose the interval for the period in more than one way. The procedure for secant is very similar, because the cofunction identity means that the secant graph is the same as the cosecant graph shifted half a period to the left. Vertical and phase shifts may be applied to the cosecant function in the same way as for the secant and other functions.The equations become the following.
$y=A\mathrm{sec}\left(Bx-C\right)+D$
$y=A\mathrm{csc}\left(Bx-C\right)+D$
Features of the Graph of y = Asec(BxC)+D
• The stretching factor is$\text{\hspace{0.17em}}|A|.$
• The period is$\text{\hspace{0.17em}}\frac{2\pi }{|B|}.$
• The domain is$\text{\hspace{0.17em}}x\ne \frac{C}{B}+\frac{\pi }{2|B|}k,$where$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$is an odd integer.
• The range is$\text{\hspace{0.17em}}\left(-\infty ,-|A|\right]\cup \left[|A|,\infty \right).$
• The vertical asymptotes occur at$\text{\hspace{0.17em}}x=\frac{C}{B}+\frac{\pi }{2|B|}k,$where$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$is an odd integer.
• There is no amplitude.
• $y=A\mathrm{sec}\left(Bx\right)\text{\hspace{0.17em}}$is an even function because cosine is an even function.
Features of the Graph of y = Acsc(BxC)+D
• The stretching factor is$\text{\hspace{0.17em}}|A|.$
• The period is$\text{\hspace{0.17em}}\frac{2\pi }{|B|}.$
• The domain is$\text{\hspace{0.17em}}x\ne \frac{C}{B}+\frac{\pi }{2|B|}k,$where$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$is an integer.
• The range is$\text{\hspace{0.17em}}\left(-\infty ,-|A|\right]\cup \left[|A|,\infty \right).$
• The vertical asymptotes occur at$\text{\hspace{0.17em}}x=\frac{C}{B}+\frac{\pi }{|B|}k,$where$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$is an integer.
• There is no amplitude.
• $y=A\mathrm{csc}\left(Bx\right)\text{\hspace{0.17em}}$is an odd function because sine is an odd function.
Given a function of the form$\text{\hspace{0.17em}}y=A\mathrm{sec}\left(Bx\right),\text{\hspace{0.17em}}$graph one period.
1. Express the function given in the form$\text{\hspace{0.17em}}y=A\mathrm{sec}\left(Bx\right).$
2. Identify the stretching/compressing factor,$\text{\hspace{0.17em}}|A|.$
3. Identify$\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$and determine the period,$\text{\hspace{0.17em}}P=\frac{2\pi }{|B|}.$
4. Sketch the graph of$\text{\hspace{0.17em}}y=A\mathrm{cos}\left(Bx\right).$
5. Use the reciprocal relationship between$\text{\hspace{0.17em}}y=\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}y=\mathrm{sec}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$to draw the graph of$\text{\hspace{0.17em}}y=A\mathrm{sec}\left(Bx\right).$
6. Sketch the asymptotes.
7. Plot any two reference points and draw the graph through these points.
Graphing a Variation of the Secant Function
Graph one period of$\text{\hspace{0.17em}}f\left(x\right)=2.5\mathrm{sec}\left(0.4x\right).$
• Step 1. The given function is already written in the general form,$\text{\hspace{0.17em}}y=A\mathrm{sec}\left(Bx\right).$
• Step 2.$\text{\hspace{0.17em}}A=2.5\text{\hspace{0.17em}}$so the stretching factor is$\text{\hspace{0.17em}}\text{2}\text{.5}\text{.}$
• Step 3.$\text{\hspace{0.17em}}B=0.4\text{\hspace{0.17em}}$so$\text{\hspace{0.17em}}P=\frac{2\pi }{0.4}=5\pi .\text{\hspace{0.17em}}$The period is$\text{\hspace{0.17em}}5\pi \text{\hspace{0.17em}}$units.
• Step 4. Sketch the graph of the function$\text{\hspace{0.17em}}g\left(x\right)=2.5\mathrm{cos}\left(0.4x\right).$
• Step 5. Use the reciprocal relationship of the cosine and secant functions to draw the cosecant function.
• Steps 6–7. Sketch two asymptotes at$\text{\hspace{0.17em}}x=1.25\pi \text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}x=3.75\pi .\text{\hspace{0.17em}}$We can use two reference points, the local minimum at$\text{\hspace{0.17em}}\left(0,2.5\right)\text{\hspace{0.17em}}$and the local maximum at$\text{\hspace{0.17em}}\left(2.5\pi ,-2.5\right).\text{\hspace{0.17em}}$[link] shows the graph.
• Graph one period of$\text{\hspace{0.17em}}f\left(x\right)=-2.5\mathrm{sec}\left(0.4x\right).$
This is a vertical reflection of the preceding graph because$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$is negative.
Do the vertical shift and stretch/compression affect the secant’s range?
Yes. The range of$\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{sec}\left(Bx-C\right)+D\text{\hspace{0.17em}}$is$\left(-\infty ,-|A|+D\right]\cup \left[|A|+D,\infty \right).$
Given a function of the form$\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{sec}\left(Bx-C\right)+D,\text{\hspace{0.17em}}$graph one period.
1. Express the function given in the form$\text{\hspace{0.17em}}y=A\text{\hspace{0.17em}}\mathrm{sec}\left(Bx-C\right)+D.$
2. Identify the stretching/compressing factor,$\text{\hspace{0.17em}}|A|.$
3. Identify$\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$and determine the period,$\text{\hspace{0.17em}}\frac{2\pi }{|B|}.$
4. Identify$\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$and determine the phase shift,$\text{\hspace{0.17em}}\frac{C}{B}.$
5. Draw the graph of$\text{\hspace{0.17em}}y=A\text{\hspace{0.17em}}\mathrm{sec}\left(Bx\right)\text{\hspace{0.17em}}.$but shift it to the right by$\text{\hspace{0.17em}}\frac{C}{B}\text{\hspace{0.17em}}$and up by$\text{\hspace{0.17em}}D.$
6. Sketch the vertical asymptotes, which occur at$\text{\hspace{0.17em}}x=\frac{C}{B}+\frac{\pi }{2|B|}k,$where$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$is an odd integer.
Graphing a Variation of the Secant Function
Graph one period of$\text{\hspace{0.17em}}y=4\mathrm{sec}\left(\frac{\pi }{3}x-\frac{\pi }{2}\right)+1.$
• Step 1. Express the function given in the form$\text{\hspace{0.17em}}y=4\mathrm{sec}\left(\frac{\pi }{3}x-\frac{\pi }{2}\right)+1.$
• Step 2. The stretching/compressing factor is$|A|=4.$
• Step 3. The period is
• Step 4. The phase shift is
• Step 5. Draw the graph of $\text{\hspace{0.17em}}y=A\mathrm{sec}\left(Bx\right),$but shift it to the right by$\text{\hspace{0.17em}}\frac{C}{B}=1.5\text{\hspace{0.17em}}$and up by$\text{\hspace{0.17em}}D=6.$
• Step 6. Sketch the vertical asymptotes, which occur at$\text{\hspace{0.17em}}x=0,x=3,$and$\text{\hspace{0.17em}}x=6.\text{\hspace{0.17em}}$There is a local minimum at$\text{\hspace{0.17em}}\left(1.5,5\right)\text{\hspace{0.17em}}$and a local maximum at$\text{\hspace{0.17em}}\left(4.5,-3\right).\text{\hspace{0.17em}}$[link] shows the graph.
• Graph one period of$\text{\hspace{0.17em}}f\left(x\right)=-6\mathrm{sec}\left(4x+2\right)-8.$
The domain of$\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$was given to be all$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$such that$\text{\hspace{0.17em}}x\ne k\pi \text{\hspace{0.17em}}$for any integer$\text{\hspace{0.17em}}k.\text{\hspace{0.17em}}$Would the domain of$\text{\hspace{0.17em}}y=A\mathrm{csc}\left(Bx-C\right)+D\text{\hspace{0.17em}}\text{be}\text{\hspace{0.17em}}x\ne \frac{C+k\pi }{B}?$
Yes. The excluded points of the domain follow the vertical asymptotes. Their locations show the horizontal shift and compression or expansion implied by the transformation to the original function’s input.
Given a function of the form$\text{\hspace{0.17em}}y=A\mathrm{csc}\left(Bx\right),\text{\hspace{0.17em}}$graph one period.
1. Express the function given in the form$\text{\hspace{0.17em}}y=A\mathrm{csc}\left(Bx\right).$
2. $\text{\hspace{0.17em}}|A|.$
3. Identify$\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$and determine the period,$\text{\hspace{0.17em}}P=\frac{2\pi }{|B|}.$
4. Draw the graph of$\text{\hspace{0.17em}}y=A\mathrm{sin}\left(Bx\right).$
5. Use the reciprocal relationship between$\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$to draw the graph of$\text{\hspace{0.17em}}y=A\mathrm{csc}\left(Bx\right).$
6. Sketch the asymptotes.
7. Plot any two reference points and draw the graph through these points.
Graphing a Variation of the Cosecant Function
Graph one period of$\text{\hspace{0.17em}}f\left(x\right)=-3\mathrm{csc}\left(4x\right).$
• Step 1. The given function is already written in the general form,$\text{\hspace{0.17em}}y=A\mathrm{csc}\left(Bx\right).$
• Step 2.$\text{\hspace{0.17em}}|A|=|-3|=3,$so the stretching factor is 3.
• Step 3.$\text{\hspace{0.17em}}B=4,$so$\text{\hspace{0.17em}}P=\frac{2\pi }{4}=\frac{\pi }{2}.\text{\hspace{0.17em}}$The period is$\text{\hspace{0.17em}}\frac{\pi }{2}\text{\hspace{0.17em}}$units.
• Step 4. Sketch the graph of the function$\text{\hspace{0.17em}}g\left(x\right)=-3\mathrm{sin}\left(4x\right).$
• Step 5. Use the reciprocal relationship of the sine and cosecant functions to draw the cosecant function.
• Steps 6–7. Sketch three asymptotes at$\text{\hspace{0.17em}}x=0,\text{\hspace{0.17em}}x=\frac{\pi }{4},\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}x=\frac{\pi }{2}.\text{\hspace{0.17em}}$We can use two reference points, the local maximum at$\text{\hspace{0.17em}}\left(\frac{\pi }{8},-3\right)\text{\hspace{0.17em}}$and the local minimum at$\text{\hspace{0.17em}}\left(\frac{3\pi }{8},3\right).$[link] shows the graph.
• Graph one period of$\text{\hspace{0.17em}}f\left(x\right)=0.5\mathrm{csc}\left(2x\right).$
Given a function of the form$\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{csc}\left(Bx-C\right)+D,\text{\hspace{0.17em}}$graph one period.
1. Express the function given in the form$\text{\hspace{0.17em}}y=A\mathrm{csc}\left(Bx-C\right)+D.$
2. Identify the stretching/compressing factor,$\text{\hspace{0.17em}}|A|.$
3. Identify$\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$and determine the period,$\text{\hspace{0.17em}}\frac{2\pi }{|B|}.$
4. Identify$\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$and determine the phase shift,$\text{\hspace{0.17em}}\frac{C}{B}.$
5. Draw the graph of$\text{\hspace{0.17em}}y=A\mathrm{csc}\left(Bx\right)\text{\hspace{0.17em}}$but shift it to the right by and up by$\text{\hspace{0.17em}}D.$
6. Sketch the vertical asymptotes, which occur at$\text{\hspace{0.17em}}x=\frac{C}{B}+\frac{\pi }{|B|}k,$where$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$is an integer.
Graphing a Vertically Stretched, Horizontally Compressed, and Vertically Shifted Cosecant
Sketch a graph of$\text{\hspace{0.17em}}y=2\mathrm{csc}\left(\frac{\pi }{2}x\right)+1.\text{\hspace{0.17em}}$What are the domain and range of this function?
• Step 1. Express the function given in the form$\text{\hspace{0.17em}}y=2\mathrm{csc}\left(\frac{\pi }{2}x\right)+1.$
• Step 2. Identify the stretching/compressing factor,$\text{\hspace{0.17em}}|A|=2.$
• Step 3. The period is$\text{\hspace{0.17em}}\frac{2\pi }{|B|}=\frac{2\pi }{\frac{\pi }{2}}=\frac{2\pi }{1}\cdot \frac{2}{\pi }=4.$
• Step 4. The phase shift is$\text{\hspace{0.17em}}\frac{0}{\frac{\pi }{2}}=0.$
• Step 5. Draw the graph of$\text{\hspace{0.17em}}y=A\mathrm{csc}\left(Bx\right)\text{\hspace{0.17em}}$but shift it up$\text{\hspace{0.17em}}D=1.$
• Step 6. Sketch the vertical asymptotes, which occur at$\text{\hspace{0.17em}}x=0,x=2,x=4.$
• The graph for this function is shown in [link].
Analysis
The vertical asymptotes shown on the graph mark off one period of the function, and the local extrema in this interval are shown by dots. Notice how the graph of the transformed cosecant relates to the graph of$\text{\hspace{0.17em}}f\left(x\right)=2\mathrm{sin}\left(\frac{\pi }{2}x\right)+1,$shown as the orange dashed wave.
Given the graph of$\text{\hspace{0.17em}}f\left(x\right)=2\mathrm{cos}\left(\frac{\pi }{2}x\right)+1\text{\hspace{0.17em}}$shown in [link], sketch the graph of$\text{\hspace{0.17em}}g\left(x\right)=2\mathrm{sec}\left(\frac{\pi }{2}x\right)+1\text{\hspace{0.17em}}$on the same axes.
# Analyzing the Graph of y = cot x
The last trigonometric function we need to explore is cotangent. The cotangent is defined by the reciprocal identity$\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}x=\frac{1}{\mathrm{tan}\text{\hspace{0.17em}}x}.\text{\hspace{0.17em}}$Notice that the function is undefined when the tangent function is 0, leading to a vertical asymptote in the graph at$\text{\hspace{0.17em}}0,\pi ,\text{\hspace{0.17em}}$etc. Since the output of the tangent function is all real numbers, the output of the cotangent function is also all real numbers.
We can graph$\text{\hspace{0.17em}}y=\mathrm{cot}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$by observing the graph of the tangent function because these two functions are reciprocals of one another. See [link]. Where the graph of the tangent function decreases, the graph of the cotangent function increases. Where the graph of the tangent function increases, the graph of the cotangent function decreases.
The cotangent graph has vertical asymptotes at each value of$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$where$\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x=0;\text{\hspace{0.17em}}$we show these in the graph below with dashed lines. Since the cotangent is the reciprocal of the tangent,$\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$has vertical asymptotes at all values of$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$where$\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x=0,\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}x=0\text{\hspace{0.17em}}$at all values of$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$where$\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$has its vertical asymptotes.
Features of the Graph of y = Acot(Bx)
• The stretching factor is$\text{\hspace{0.17em}}|A|.$
• The period is$\text{\hspace{0.17em}}P=\frac{\pi }{|B|}.$
• The domain is$\text{\hspace{0.17em}}x\ne \frac{\pi }{|B|}k,\text{\hspace{0.17em}}$where$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$is an integer.
• The range is$\text{\hspace{0.17em}}\left(-\infty ,\infty \right).$
• The asymptotes occur at$\text{\hspace{0.17em}}x=\frac{\pi }{|B|}k,\text{\hspace{0.17em}}$where$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$is an integer.
• $y=A\mathrm{cot}\left(Bx\right)\text{\hspace{0.17em}}$is an odd function.
# Graphing Variations of y = cot x
We can transform the graph of the cotangent in much the same way as we did for the tangent. The equation becomes the following.
$y=Acot( Bx−C )+D$
Properties of the Graph of y = Acot(Bx−C)+D
• The stretching factor is$\text{\hspace{0.17em}}|A|.$
• The period is$\text{\hspace{0.17em}}\frac{\pi }{|B|}.$
• The domain is$\text{\hspace{0.17em}}x\ne \frac{C}{B}+\frac{\pi }{|B|}k,$where$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$is an integer.
• The range is$\text{\hspace{0.17em}}\left(\mathrm{-\infty },-|A|\right]\cup \left[|A|,\infty \right).$
• The vertical asymptotes occur at$\text{\hspace{0.17em}}x=\frac{C}{B}+\frac{\pi }{|B|}k,$where$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$is an integer.
• There is no amplitude.
• $y=A\mathrm{cot}\left(Bx\right)\text{\hspace{0.17em}}$is an odd function because it is the quotient of even and odd functions (cosine and sine, respectively)
Given a modified cotangent function of the form$\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{cot}\left(Bx\right),$graph one period.
1. Express the function in the form$\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{cot}\left(Bx\right).$
2. Identify the stretching factor,$\text{\hspace{0.17em}}|A|.$
3. Identify the period,$\text{\hspace{0.17em}}P=\frac{\pi }{|B|}.$
4. Draw the graph of$\text{\hspace{0.17em}}y=A\mathrm{tan}\left(Bx\right).$
5. Plot any two reference points.
6. Use the reciprocal relationship between tangent and cotangent to draw the graph of$\text{\hspace{0.17em}}y=A\mathrm{cot}\left(Bx\right).$
7. Sketch the asymptotes.
Graphing Variations of the Cotangent Function
Determine the stretching factor, period, and phase shift of$\text{\hspace{0.17em}}y=3\mathrm{cot}\left(4x\right),\text{\hspace{0.17em}}$and then sketch a graph.
• Step 1. Expressing the function in the form$\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{cot}\left(Bx\right)\text{\hspace{0.17em}}$gives$\text{\hspace{0.17em}}f\left(x\right)=3\mathrm{cot}\left(4x\right).$
• Step 2. The stretching factor is$\text{\hspace{0.17em}}|A|=3.$
• Step 3. The period is$\text{\hspace{0.17em}}P=\frac{\pi }{4}.$
• Step 4. Sketch the graph of$\text{\hspace{0.17em}}y=3\mathrm{tan}\left(4x\right).$
• Step 5. Plot two reference points. Two such points are$\text{\hspace{0.17em}}\left(\frac{\pi }{16},3\right)\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}\left(\frac{3\pi }{16},-3\right).$
• Step 6. Use the reciprocal relationship to draw$\text{\hspace{0.17em}}y=3\mathrm{cot}\left(4x\right).$
• Step 7. Sketch the asymptotes,$\text{\hspace{0.17em}}x=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\frac{\pi }{4}.$
• The orange graph in [link] shows$\text{\hspace{0.17em}}y=3\mathrm{tan}\left(4x\right)\text{\hspace{0.17em}}$and the blue graph shows$\text{\hspace{0.17em}}y=3\mathrm{cot}\left(4x\right).$
Given a modified cotangent function of the form$\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{cot}\left(Bx-C\right)+D,\text{\hspace{0.17em}}$graph one period.
1. Express the function in the form$\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{cot}\left(Bx-C\right)+D.$
2. Identify the stretching factor,$\text{\hspace{0.17em}}|A|.$
3. Identify the period,$\text{\hspace{0.17em}}P=\frac{\pi }{|B|}.$
4. Identify the phase shift,$\text{\hspace{0.17em}}\frac{C}{B}.$
5. Draw the graph of$\text{\hspace{0.17em}}y=A\mathrm{tan}\left(Bx\right)\text{\hspace{0.17em}}$ shifted to the right by$\text{\hspace{0.17em}}\frac{C}{B}\text{\hspace{0.17em}}$and up by$\text{\hspace{0.17em}}D.$
6. Sketch the asymptotes$\text{\hspace{0.17em}}x=\frac{C}{B}+\frac{\pi }{|B|}k,$where$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$is an integer.
7. Plot any three reference points and draw the graph through these points.
Graphing a Modified Cotangent
Sketch a graph of one period of the function$\text{\hspace{0.17em}}f\left(x\right)=4\mathrm{cot}\left(\frac{\pi }{8}x-\frac{\pi }{2}\right)-2.$
• Step 1. The function is already written in the general form$\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{cot}\left(Bx-C\right)+D.$
• Step 2.$\text{\hspace{0.17em}}A=4,$so the stretching factor is 4.
• Step 3.$\text{\hspace{0.17em}}B=\frac{\pi }{8},$so the period is$\text{\hspace{0.17em}}P=\frac{\pi }{|B|}=\frac{\pi }{\frac{\pi }{8}}=8.$
• Step 4.$\text{\hspace{0.17em}}C=\frac{\pi }{2},$so the phase shift is$\text{\hspace{0.17em}}\frac{C}{B}=\frac{\frac{\pi }{2}}{\frac{\pi }{8}}=4.$
• Step 5. We draw$\text{\hspace{0.17em}}f\left(x\right)=4\mathrm{tan}\left(\frac{\pi }{8}x-\frac{\pi }{2}\right)-2.$
• Step 6-7. Three points we can use to guide the graph are$\text{\hspace{0.17em}}\left(6,2\right),\left(8,-2\right),\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}\left(10,-6\right).\text{\hspace{0.17em}}$We use the reciprocal relationship of tangent and cotangent to draw$\text{\hspace{0.17em}}f\left(x\right)=4\mathrm{cot}\left(\frac{\pi }{8}x-\frac{\pi }{2}\right)-2.$
• Step 8. The vertical asymptotes are$\text{\hspace{0.17em}}x=4\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}x=12.$
• The graph is shown in [link].
# Using the Graphs of Trigonometric Functions to Solve Real-World Problems
Many real-world scenarios represent periodic functions and may be modeled by trigonometric functions. As an example, let’s return to the scenario from the section opener. Have you ever observed the beam formed by the rotating light on a police car and wondered about the movement of the light beam itself across the wall? The periodic behavior of the distance the light shines as a function of time is obvious, but how do we determine the distance? We can use the tangent function.
Using Trigonometric Functions to Solve Real-World Scenarios
Suppose the function$\text{\hspace{0.17em}}y=5\mathrm{tan}\left(\frac{\pi }{4}t\right)\text{\hspace{0.17em}}$marks the distance in the movement of a light beam from the top of a police car across a wall where$\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$is the time in seconds and$\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$is the distance in feet from a point on the wall directly across from the police car.
1. Find and interpret the stretching factor and period.
2. Graph on the interval$\text{\hspace{0.17em}}\left[0,5\right].$
3. Evaluate$\text{\hspace{0.17em}}f\left(1\right)\text{\hspace{0.17em}}$and discuss the function’s value at that input.
1. We know from the general form of$\text{\hspace{0.17em}}y=A\mathrm{tan}\left(Bt\right)\text{\hspace{0.17em}}$that$\text{\hspace{0.17em}}|A|\text{\hspace{0.17em}}$is the stretching factor and$\text{\hspace{0.17em}}\frac{\pi }{B}\text{\hspace{0.17em}}$is the period.
We see that the stretching factor is 5. This means that the beam of light will have moved 5 ft after half the period.
The period is$\text{\hspace{0.17em}}\frac{\pi }{\frac{\pi }{4}}=\frac{\pi }{1}\cdot \frac{4}{\pi }=4.\text{\hspace{0.17em}}$This means that every 4 seconds, the beam of light sweeps the wall. The distance from the spot across from the police car grows larger as the police car approaches.
2. To graph the function, we draw an asymptote at$\text{\hspace{0.17em}}t=2\text{\hspace{0.17em}}$and use the stretching factor and period. See [link]
3. period:$\text{\hspace{0.17em}}f\left(1\right)=5\mathrm{tan}\left(\frac{\pi }{4}\left(1\right)\right)=5\left(1\right)=5;\text{\hspace{0.17em}}$after 1 second, the beam of has moved 5 ft from the spot across from the police car.
Access these online resources for additional instruction and practice with graphs of other trigonometric functions.
# Key Equations
Shifted, compressed, and/or stretched tangent function $y=A\text{\hspace{0.17em}}\mathrm{tan}\left(Bx-C\right)+D$ Shifted, compressed, and/or stretched secant function $y=A\text{\hspace{0.17em}}\mathrm{sec}\left(Bx-C\right)+D$ Shifted, compressed, and/or stretched cosecant function $y=A\text{\hspace{0.17em}}\mathrm{csc}\left(Bx-C\right)+D$ Shifted, compressed, and/or stretched cotangent function $y=A\text{\hspace{0.17em}}\mathrm{cot}\left(Bx-C\right)+D$
# Key Concepts
• The tangent function has period$\text{\hspace{0.17em}}\pi .$
• $f\left(x\right)=A\mathrm{tan}\left(Bx-C\right)+D\text{\hspace{0.17em}}$is a tangent with vertical and/or horizontal stretch/compression and shift. See [link], [link], and [link].
• The secant and cosecant are both periodic functions with a period of$\text{\hspace{0.17em}}2\pi .\text{\hspace{0.17em}}$$f\left(x\right)=A\mathrm{sec}\left(Bx-C\right)+D\text{\hspace{0.17em}}$gives a shifted, compressed, and/or stretched secant function graph. See [link] and [link].
• $f\left(x\right)=A\mathrm{csc}\left(Bx-C\right)+D\text{\hspace{0.17em}}$gives a shifted, compressed, and/or stretched cosecant function graph. See [link] and [link].
• The cotangent function has period$\text{\hspace{0.17em}}\pi \text{\hspace{0.17em}}$and vertical asymptotes at$\text{\hspace{0.17em}}0,±\pi ,±2\pi ,....$
• The range of cotangent is$\text{\hspace{0.17em}}\left(-\infty ,\infty \right),\text{\hspace{0.17em}}$and the function is decreasing at each point in its range.
• The cotangent is zero at$\text{\hspace{0.17em}}±\frac{\pi }{2},±\frac{3\pi }{2},....$
• $f\left(x\right)=A\mathrm{cot}\left(Bx-C\right)+D\text{\hspace{0.17em}}$is a cotangent with vertical and/or horizontal stretch/compression and shift. See [link] and [link].
• Real-world scenarios can be solved using graphs of trigonometric functions. See [link].
# Section Exercises
## Verbal
Explain how the graph of the sine function can be used to graph$\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x.$
Since$\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$is the reciprocal function of$\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$you can plot the reciprocal of the coordinates on the graph of$\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$to obtain the y-coordinates of$\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$The x-intercepts of the graph$\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$are the vertical asymptotes for the graph of$\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x.$
How can the graph of$\text{\hspace{0.17em}}y=\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$be used to construct the graph of$\text{\hspace{0.17em}}y=\mathrm{sec}\text{\hspace{0.17em}}x?$
Explain why the period of$\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$is equal to$\text{\hspace{0.17em}}\pi .$
Answers will vary. Using the unit circle, one can show that$\text{\hspace{0.17em}}\mathrm{tan}\left(x+\pi \right)=\mathrm{tan}\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$
Why are there no intercepts on the graph of$\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x?$
How does the period of$\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$compare with the period of$\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}x?$
The period is the same:$\text{\hspace{0.17em}}2\pi .$
## Algebraic
For the following exercises, match each trigonometric function with one of the graphs in [link].
$f\left(x\right)=\mathrm{tan}\text{\hspace{0.17em}}x$
$f\left(x\right)=\mathrm{sec}\text{\hspace{0.17em}}x$
IV
$f\left(x\right)=\mathrm{csc}\text{\hspace{0.17em}}x$
$f\left(x\right)=\mathrm{cot}\text{\hspace{0.17em}}x$
III
For the following exercises, find the period and horizontal shift of each of the functions.
$f\left(x\right)=2\mathrm{tan}\left(4x-32\right)$
$h\left(x\right)=2\mathrm{sec}\left(\frac{\pi }{4}\left(x+1\right)\right)$
period: 8; horizontal shift: 1 unit to left
$m\left(x\right)=6\mathrm{csc}\left(\frac{\pi }{3}x+\pi \right)$
If$\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x=-1.5,\text{\hspace{0.17em}}$find$\text{\hspace{0.17em}}\mathrm{tan}\left(-x\right).$
1.5
If$\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}x=2,\text{\hspace{0.17em}}$find$\text{\hspace{0.17em}}\mathrm{sec}\left(-x\right).$
If$\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x=-5,\text{\hspace{0.17em}}$find$\text{\hspace{0.17em}}\mathrm{csc}\left(-x\right).$
5
If$\text{\hspace{0.17em}}x\mathrm{sin}\text{\hspace{0.17em}}x=2,\text{\hspace{0.17em}}$find$\text{\hspace{0.17em}}\left(-x\right)\mathrm{sin}\left(-x\right).$
For the following exercises, rewrite each expression such that the argument$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$is positive.
$\mathrm{cot}\left(-x\right)\mathrm{cos}\left(-x\right)+\mathrm{sin}\left(-x\right)$
$-\mathrm{cot}x\mathrm{cos}x-\mathrm{sin}x$
$\mathrm{cos}\left(-x\right)+\mathrm{tan}\left(-x\right)\mathrm{sin}\left(-x\right)$
## Graphical
For the following exercises, sketch two periods of the graph for each of the following functions. Identify the stretching factor, period, and asymptotes.
$f\left(x\right)=2\mathrm{tan}\left(4x-32\right)$
stretching factor: 2; period:asymptotes:
$\text{\hspace{0.17em}}h\left(x\right)=2\mathrm{sec}\left(\frac{\pi }{4}\left(x+1\right)\right)\text{\hspace{0.17em}}$
$m\left(x\right)=6\mathrm{csc}\left(\frac{\pi }{3}x+\pi \right)$
stretching factor: 6; period: 6; asymptotes:
$j\left(x\right)=\mathrm{tan}\left(\frac{\pi }{2}x\right)$
$p\left(x\right)=\mathrm{tan}\left(x-\frac{\pi }{2}\right)$
stretching factor: 1; period:asymptotes:
$f\left(x\right)=4\mathrm{tan}\left(x\right)$
$f\left(x\right)=\mathrm{tan}\left(x+\frac{\pi }{4}\right)$
Stretching factor: 1; period:asymptotes:
$f\left(x\right)=\pi \mathrm{tan}\left(\pi x-\pi \right)-\pi$
$f\left(x\right)=2\mathrm{csc}\left(x\right)$
stretching factor: 2; period:asymptotes:
$f\left(x\right)=-\frac{1}{4}\mathrm{csc}\left(x\right)$
$f\left(x\right)=4\mathrm{sec}\left(3x\right)$
stretching factor: 4; period:asymptotes:
$f\left(x\right)=-3\mathrm{cot}\left(2x\right)$
$f\left(x\right)=7\mathrm{sec}\left(5x\right)$
stretching factor: 7; period:asymptotes:
$f\left(x\right)=\frac{9}{10}\mathrm{csc}\left(\pi x\right)$
$f\left(x\right)=2\mathrm{csc}\left(x+\frac{\pi }{4}\right)-1$
stretching factor: 2; period:asymptotes:
$f\left(x\right)=-\mathrm{sec}\left(x-\frac{\pi }{3}\right)-2$
$f\left(x\right)=\frac{7}{5}\mathrm{csc}\left(x-\frac{\pi }{4}\right)$
stretching factor:period:asymptotes:
$f\left(x\right)=5\left(\mathrm{cot}\left(x+\frac{\pi }{2}\right)-3\right)$
For the following exercises, find and graph two periods of the periodic function with the given stretching factor,$\text{\hspace{0.17em}}|A|,\text{\hspace{0.17em}}$period, and phase shift.
A tangent curve,$\text{\hspace{0.17em}}A=1,\text{\hspace{0.17em}}$period of$\text{\hspace{0.17em}}\frac{\pi }{3};\text{\hspace{0.17em}}$and phase shift$\text{\hspace{0.17em}}\left(h,\text{\hspace{0.17em}}k\right)=\left(\frac{\pi }{4},2\right)$
$y=\mathrm{tan}\left(3\left(x-\frac{\pi }{4}\right)\right)+2$
A tangent curve,$\text{\hspace{0.17em}}A=-2,\text{\hspace{0.17em}}$period of$\text{\hspace{0.17em}}\frac{\pi }{4},\text{\hspace{0.17em}}$and phase shift$\text{\hspace{0.17em}}\left(h,\text{\hspace{0.17em}}k\right)=\left(-\frac{\pi }{4},\text{\hspace{0.17em}}-2\right)$
For the following exercises, find an equation for the graph of each function.
$f\left(x\right)=\mathrm{csc}\left(2x\right)$
$f\left(x\right)=\mathrm{csc}\left(4x\right)$
$f\left(x\right)=2\mathrm{csc}x$
$f\left(x\right)=\frac{1}{2}\mathrm{tan}\left(100\pi x\right)$
## Technology
For the following exercises, use a graphing calculator to graph two periods of the given function. Note: most graphing calculators do not have a cosecant button; therefore, you will need to input$\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$as$\text{\hspace{0.17em}}\frac{1}{\mathrm{sin}\text{\hspace{0.17em}}x}.$
$f\left(x\right)=|\mathrm{csc}\left(x\right)|$
$f\left(x\right)=|\mathrm{cot}\left(x\right)|$
$f\left(x\right)={2}^{\mathrm{csc}\left(x\right)}$
$f\left(x\right)=\frac{\mathrm{csc}\left(x\right)}{\mathrm{sec}\left(x\right)}$
Graph$\text{\hspace{0.17em}}f\left(x\right)=1+{\mathrm{sec}}^{2}\left(x\right)-{\mathrm{tan}}^{2}\left(x\right).\text{\hspace{0.17em}}$What is the function shown in the graph?
$f\left(x\right)=\mathrm{sec}\left(0.001x\right)$
$f\left(x\right)=\mathrm{cot}\left(100\pi x\right)$
$f\left(x\right)={\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x$
## Real-World Applications
The function$\text{\hspace{0.17em}}f\left(x\right)=20\mathrm{tan}\left(\frac{\pi }{10}x\right)\text{\hspace{0.17em}}$marks the distance in the movement of a light beam from a police car across a wall for time$\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$in seconds, and distance$\text{\hspace{0.17em}}f\left(x\right),$ in feet.
1. Graph on the interval$\text{\hspace{0.17em}}\left[0,\text{\hspace{0.17em}}5\right].$
2. Find and interpret the stretching factor, period, and asymptote.
3. Evaluate$\text{\hspace{0.17em}}f\left(1\right)\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}f\left(2.5\right)\text{\hspace{0.17em}}$and discuss the function’s values at those inputs.
Standing on the shore of a lake, a fisherman sights a boat far in the distance to his left. Let$\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$measured in radians, be the angle formed by the line of sight to the ship and a line due north from his position. Assume due north is 0 and$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$is measured negative to the left and positive to the right. (See [link].) The boat travels from due west to due east and, ignoring the curvature of the Earth, the distance$\text{\hspace{0.17em}}d\left(x\right),\text{\hspace{0.17em}}$in kilometers, from the fisherman to the boat is given by the function$\text{\hspace{0.17em}}d\left(x\right)=1.5\mathrm{sec}\left(x\right).$
1. What is a reasonable domain for$\text{\hspace{0.17em}}d\left(x\right)?$
2. Graph$\text{\hspace{0.17em}}d\left(x\right)\text{\hspace{0.17em}}$on this domain.
3. Find and discuss the meaning of any vertical asymptotes on the graph of$\text{\hspace{0.17em}}d\left(x\right).$
4. Calculate and interpret$\text{\hspace{0.17em}}d\left(-\frac{\pi }{3}\right).\text{\hspace{0.17em}}$Round to the second decimal place.
5. Calculate and interpret$\text{\hspace{0.17em}}d\left(\frac{\pi }{6}\right).\text{\hspace{0.17em}}$Round to the second decimal place.
6. What is the minimum distance between the fisherman and the boat? When does this occur?
1. $\text{\hspace{0.17em}}\left(-\frac{\pi }{2},\text{\hspace{0.17em}}\frac{\pi }{2}\right);\text{\hspace{0.17em}}$
2. $\text{\hspace{0.17em}}x=-\frac{\pi }{2}\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}x=\frac{\pi }{2};\text{\hspace{0.17em}}$the distance grows without bound as$\text{\hspace{0.17em}}|x|$approaches$\text{\hspace{0.17em}}\frac{\pi }{2}\text{\hspace{0.17em}}$—i.e., at right angles to the line representing due north, the boat would be so far away, the fisherman could not see it;
3. 3; when$\text{\hspace{0.17em}}x=-\frac{\pi }{3},\text{\hspace{0.17em}}$the boat is 3 km away;
4. 1.73; when$\text{\hspace{0.17em}}x=\frac{\pi }{6},\text{\hspace{0.17em}}$the boat is about 1.73 km away;
5. 1.5 km; when$\text{\hspace{0.17em}}x=0\text{\hspace{0.17em}}$
A laser rangefinder is locked on a comet approaching Earth. The distance$\text{\hspace{0.17em}}g\left(x\right),\text{\hspace{0.17em}}$in kilometers, of the comet after$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$days, for$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$in the interval 0 to 30 days, is given by$\text{\hspace{0.17em}}g\left(x\right)=250,000\mathrm{csc}\left(\frac{\pi }{30}x\right).$
1. Graph$\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$on the interval$\text{\hspace{0.17em}}\left[0,\text{\hspace{0.17em}}35\right].$
2. Evaluate$\text{\hspace{0.17em}}g\left(5\right)\text{\hspace{0.17em}}$ and interpret the information.
3. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond?
4. Find and discuss the meaning of any vertical asymptotes.
A video camera is focused on a rocket on a launching pad 2 miles from the camera. The angle of elevation from the ground to the rocket after$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$seconds is$\text{\hspace{0.17em}}\frac{\pi }{120}x.$
1. Write a function expressing the altitude$\text{\hspace{0.17em}}h\left(x\right),\text{\hspace{0.17em}}$in miles, of the rocket above the ground after$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$seconds. Ignore the curvature of the Earth.
2. Graph$\text{\hspace{0.17em}}h\left(x\right)\text{\hspace{0.17em}}$on the interval$\text{\hspace{0.17em}}\left(0,\text{\hspace{0.17em}}60\right).$
3. Evaluate and interpret the values$\text{\hspace{0.17em}}h\left(0\right)\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}h\left(30\right).$
4. What happens to the values of$\text{\hspace{0.17em}}h\left(x\right)\text{\hspace{0.17em}}$as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ approaches 60 seconds? Interpret the meaning of this in terms of the problem.
1. $h\left(x\right)=2\mathrm{tan}\left(\frac{\pi }{120}x\right);$
2. $h\left(0\right)=0:\text{\hspace{0.17em}}$after 0 seconds, the rocket is 0 mi above the ground;$h\left(30\right)=2:\text{\hspace{0.17em}}$after 30 seconds, the rockets is 2 mi high;
3. As$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$approaches 60 seconds, the values of$\text{\hspace{0.17em}}h\left(x\right)\text{\hspace{0.17em}}$grow increasingly large. The distance to the rocket is growing so large that the camera can no longer track it.
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# GRE Quantitative Comparison Tip #5 – Estimation with a Twist
Here’s the whole series of QC tips:
Tip #1: Dealing with Variables
Tip #2: Striving for Equality
Tip #3: Logic over Algebra
Tip #4: Comparing in Parts
Tip #5: Estimation with a Twist
In my last post, we solved the following question:
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Our approach was to first recognize that two of fractions are approximately 1/2 and two are approximately 1/3.
Now if we had used these approximations, we would have been left with:
This would have made us conclude (incorrectly) that the answer is C.
To solve this question using approximation, we applied a twist. We recognized that 213/428 is a little bit less than 1/2, which we denoted as 1/2. We also noticed that 3007/9101 is a little bit less than 1/3, which we denoted as 1/3.
And so on.
With these little twists, we were able to simplify the two columns as:
From here, it was clear that the correct answer is B
Okay, now let’s see if you can apply this approximation with a twist to solve the following question:
Aside
: before you read my solutions, see if you can find additional ways to solve this question.
Okay, first we’ll solve the question using approximation with a twist, and then we’ll solve it using different approaches.
## Approximation with a twist:
First, let’s approximate as follows:
From here, we can drop 4 zeroes from each number to get:
At this point, I’ll apply a nice rule that says:
In other words, the product of two numbers is equal to the product of twice one number and half the other value.
So, in Column A, we’ll double 45+ and halve 64+ to get:
From here, when we compare the products in parts, we can see that Column A must be greater than Column B, so the answer is A.
## Alternative approach #1
As you might have guessed, the two original products are too large to work on a calculator. For example, (641,713)x(451,222)=289,555,023,286 and this number is too large for the GRE’s onscreen calculator. As such, the calculator would display an error message if you tried to perform this calculation.
There are, however, some ways to work around this constraint and still use the calculator.
For example, you could divide each number by 1000 to get:
At this point, the products will still fit into the display of the onscreen calculator and you would clearly see that Column A is greater.
## Alternative approach #2
Another possible approach is to perform the same steps we performed earlier, but stop when we get to:
Originally, we applied that handy rule where we double one number and halve the other. However, we could also use the calculator at this point.
(64)x(45)=2880 and (90)x(32)=2880. This means that (64+)x(45+)=2880+ and (90)x(32)=2880.
So, we get:
Once again, we can see that the answer is A.
### 21 Responses to GRE Quantitative Comparison Tip #5 – Estimation with a Twist
1. Laboni September 19, 2019 at 9:58 pm #
I just love this. Thank you so much
2. Asmita April 9, 2019 at 8:58 am #
I am getting confused at the following point:
Column A:
641,713 x 451,222
Approximated to:
640,000 x 450,000
The approximations are slightly lesser than the original numbers.
Column B:
897,189 x 319,977
Approximated to:
900,000 x 320,000
The approximations are slightly greater than the original numbers.
Therefore, shouldn’t Column A have a negative sign applied and Column B a positive sign.
• Magoosh Test Prep Expert April 15, 2019 at 6:35 pm #
Hi Asmita!
In Column A, the estimates are a little lower than what the actual values are. Therefore, the actual valuewill be higher than the approximation. That’s why it gets a “+”. 🙂
In Column B, the estimates are a little higher than what the actual values are. Therefore, the actual value will be lower than the approximation. That’s why this gets a “-“.
Does that help clear things up? 🙂
3. Julia January 8, 2016 at 5:42 pm #
Hi Brent
I correctly solved both problems before checking for the correct answers by rounding. Is there a weakness to rounding that is unforeseen (by me) that will cause me to get such QC answers incorrect? Thanks!
• Magoosh Test Prep Expert February 1, 2016 at 6:31 am #
Hi Julia,
Rounding can be a powerful tool as long as you make sure the process of rounding doesn’t introduce error, like it would if small differences in rounding affect the final answer or if the answer choices are sufficiently close together that rounding could change your ultimate choice. Since you’re asking this question, though, I think you sound like you’ll be in good shape!
4. Jesley December 9, 2015 at 8:57 pm #
Why or how was a negative sign applied to column b?
• liz July 20, 2017 at 7:46 am #
This means that when this number was estimated, this column actually is a little bit ‘less than’ the written number (like, the saying ‘plus or minus’). That is why column A also has a +. That number is actually little more than the written number, which is an estimate. So, since the answers look to be the same (2880) when you estimate them, you can see that Column A is actually 2880 ‘plus’ and Column B is 2880 ‘minus’, so A is greater.
5. Sahar September 8, 2015 at 6:39 am #
Brent!
Simple rules and your tricks are awesome! 🙂
6. Harsha June 14, 2015 at 9:39 pm #
That’s a terrific way of solving these type of problems. A * B = 2 A * 1/2B. Thanks very much
7. paul nwabuona February 4, 2015 at 2:45 pm #
super cool!!
8. priya November 22, 2014 at 7:33 pm #
awesome! i love it when you teach us shortcut tricks..they are very helpful and fun too!
9. siddharth mehra June 6, 2013 at 2:44 am #
nice tip
10. Tony May 3, 2013 at 4:55 pm #
Brent,
Fantastic system. I’m preparing myself for the GRE …….with you.
So you’re the one who speaks there , on the math part?
Love it
Tony
• Brent Hanneson May 6, 2013 at 11:47 am #
Thanks for the feedback, Tony.
Much appreciated.
Cheers,
Brent
11. Brent Hanneson May 21, 2012 at 5:28 pm #
Great idea – that also works!
Cheers,
Brent
12. Dineshvalmiki May 21, 2012 at 2:48 pm #
Alternative approach:
Column (a) column (b)
641,713 x 451,222 897,189 x 319,977
div (a) with 897,189 and (b) with 641,713
=>
(1/2)+ (1/2)-
• Brent May 21, 2012 at 7:13 pm #
Great idea – that also works!
Cheers,
Brent
13. Erika Klara November 19, 2011 at 5:02 am #
Thank you for your invaluable help in math.
• Brent November 20, 2011 at 9:34 pm #
Thanks Erika.
Cheers,
Brent
14. saranya November 18, 2011 at 8:15 am #
Hi sir, thank you for your tips
• Brent November 18, 2011 at 12:16 pm #
Thanks saranya!
Cheers,
Brent
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# Geometric Sequences
A Geometric Sequence is a sequence of numbers that has the property that the ratio between two consecutive elements is constant, equal to a certain value $$r$$. This value is also known as the common ratio.
Assume that the first term is $$a$$. Then, the next term is $$a r$$, and the next is $$ar^2$$. And so on.
So, in other words, we start with the first term $$a$$, and the next term is always found by multiplying the previous term by $$r$$.
So the first term is $$a_1 = a$$.
The second term is $$a_2 = a r$$.
The third term is $$a_3 = a r^2$$.
Therefore, the general nth term is
$\large a_n = a r^{n-1}$
Summarizing, you need two pieces of information to define a geometric sequence: You need the initial term $$a$$, and the constant ratio $$r$$. Consecutive terms of the sequence are obtained by multiplying the previous term by $$r$$.
### EXAMPLE 1: Example of a geometric sequence
Find the 6th term of a geometric sequence with initial term $$10$$, and $$r = 1/2$$.
Based on the information provided, we have enough information to define the geometric sequence. Indeed, we have the first term $$a = 10$$, and we have the constant ratio $$r = 1/2$$.
The general nth term is
$\large a_n = a r^{n-1}$
so then the 6th term is
$\large \displaystyle a_{10} = a r^{6-1} = 10 \left(\frac{1}{2}\right)^5$ $\large = \frac{10}{32} = \frac{5}{16}$
## Can the Common Ratio be Negative?
Yes, absolutely. The constant ratio $$r$$ can be negative. For example, we can have a geometric sequence with initial term $$a_1 = 1$$ and constant ratio $$r = -2$$. So then, the second term is $$a_2 = 1 \cdot (-2) = -2$$, $$a_3 = (-2) \cdot (-2) = 4$$, and so on.
So, it is the exactly the same rule: in order to the following term, we multiply the previous term by the constant ratio $$r$$, even if the constant ratio is negative.
### EXAMPLE 2
Find the 5th term of a geometric sequence with initial term $$3$$, and $$r = -2$$.
We have enough information to define the geometric sequence, because we have the first term $$a_1 = 3$$, and we have the constant ratio $$r = -2$$.
The general nth term (with negative constant ratio) is
$\large a_n = a r^{n-1} = 3 \cdot (-2)^{n-1}$
so then the 5th term is
$\large \displaystyle a_{5} = a r^{5-1} = 3 \cdot (-2)^{5-1} = 3 \cdot (-2)^4 = 3 \cdot 16 = 48$
### EXAMPLE 3
Consider the sequence 1, 1/2, 1/4, 1/16, ... Is this sequence geometric?
In order for a given sequence to be geometric, the terms need to have a common ratio. In this case, dividing the second term by the first term we get $$(1/2)/1 = 1/2$$.
Then, if we divide the third by the second term: $$(1/4)/(1/2) = 1/2$$. So far so good.
Now, if we divide the fourth by the third term: $$(1/16)/(1/4) = 1/4$$. It fails. It is not a geometric series, because it does not have a common ratio (the ratio is 1/2 for the first two terms, but then it is 1/4, so it is not constant).
Hence, the sequence is NOT a geometric sequence.
## More About the Geometric Sequences
The punchline you need to keep in mind. What is the formula of geometric sequence? Simple
$\large a_n = a r^{n-1}$
where $$a$$ is the initial term and $$r$$ is the constant ratio (or common ratio, as it is also called).
There are a couple of calculators that you may want to use that are related to the concept of geometric sequence, or geometric progression, as it is also called..
• First you can check our infinite geometric series sum calculator, which sums infinite terms of a geometric sequence. This sum will be well defined (converge) if the constant ratio is such that $$|r| < 1$$.
• Also, you will want to use our geometric sequence sum calculator, which computes the sum of terms in a geometric sequence, UP to a certain finite value. This sum is well defined without conditions on the constant ratio $$r$$, provided that we add up to a finite term of the sequence.
### Can a geometric sequence have a common ratio of 1?
Absolutely. The general term for a geometric sequence with a common ratio of 1 is
$\large a_n = a r^{n-1}= a \cdot 1^{n-1} = a$
So, a sequence with common ratio of 1 is a rather boring geometric sequence, with all the terms equal to the first term.
In case you have any suggestion, or if you would like to report a broken solver/calculator, please do not hesitate to contact us. |
# Objectives Write an equation for a circle.
## Presentation on theme: "Objectives Write an equation for a circle."— Presentation transcript:
Objectives Write an equation for a circle. Graph a circle, and identify its center and radius. Vocabulary circle tangent
Notes 1. Write an equation for the circle with center (1, –5) and a radius of 2. Write an equation for the circle with center (–4, 4) and containing the point (–1, 16). 3. Write an equation for the line tangent to the circle x2 + y2 = 17 at the point (4, 1). 4. Which points on the graph shown are within 2 units of the point (0, –2.5)? 2
A circle is the set of points in a plane that are a fixed distance, called the radius, from a fixed point, called the center. Because all of the points on a circle are the same distance from the center of the circle, you can use the Distance Formula to find the equation of a circle.
Example 1: Using the Distance Formula to Write the Equation of a Circle
Write the equation of a circle with center (–3, 4) and radius r = 6. Use the Distance Formula with (x2, y2) = (x, y), (x1, y1) = (–3, 4), and distance equal to the radius, 6. Use the Distance Formula. Substitute. Square both sides.
Notice that r2 and the center are visible in the equation of a circle
Notice that r2 and the center are visible in the equation of a circle. This leads to the formula for a circle with center (h, k) and radius r. If the center of the circle is at the origin, the equation simplifies to x2 + y2 = r2. Helpful Hint
Example 2A: Writing the Equation of a Circle
Write the equation of the circle. the circle with center (0, 6) and radius r = 1 (x – h)2 + (y – k)2 = r2 Equation of a circle (x – 0)2 + (y – 6)2 = 12 Substitute. x2 + (y – 6)2 = 1
Example 2B: Writing the Equation of a Circle
Write the equation of the circle. the circle with center (–4, 11) and containing the point (5, –1) Use the Distance Formula to find the radius. (x + 4)2 + (y – 11)2 = 152 Substitute the values into the equation of a circle. (x + 4)2 + (y – 11)2 = 225
The location of points in relation to a circle can be described by inequalities. The points inside the circle satisfy the inequality (x – h)2 + (x – k)2 < r2. The points outside the circle satisfy the inequality (x – h)2 + (x – k)2 > r2.
Example 3: Consumer Application
Use the map and information given in Example 3 on page 730. Which homes are within 4 miles of a restaurant located at (–1, 1)? The circle has a center (–1, 1) and radius 4. The points insides the circle will satisfy the inequality (x + 1)2 + (y – 1)2 < 42. Points B, C, D and E are within a 4-mile radius . Check Point F(–2, –3) is near the boundary. (–2 + 1)2 + (–3 – 1)2 < 42 (–1)2 + (–4)2 < 42 x < 16 Point F (–2, –3) is not inside the circle.
A tangent is a line in the same plane as the circle that intersects the circle at exactly one point. Recall from geometry that a tangent to a circle is perpendicular to the radius at the point of tangency.
Example 4: Writing the Equation of a Tangent
Write the equation of the line tangent to the circle x2 + y2 = 29 at the point (2, 5). Step 1 Identify the center and radius of the circle. The circle has center of (0, 0) and radius r = . Step 2 Find the slope of the radius at the point of tangency and the slope of the tangent. Use the slope formula.
Use the point-slope formula.
Example 4 Continued Step 3 Find the slope-intercept equation of the tangent by using the point (2, 5) and the slope m = Perpendicular lines have slopes that are opposite reciprocals 2 5 Use the point-slope formula. Substitute (2, 5) (x1 , y1 ) and – for m. 2 5 Rewrite in slope-intercept form.
Example 4 Continued The equation of the line that is tangent to x2 + y2 = 29 at (2, 5) is Graph the circle and the line to check.
Notes 1. Write an equation for the circle with center (1, –5) and a radius of (x – 1)2 + (y + 5)2 = 10 2. Write an equation for the circle with center (–4, 4) and containing the point (–1, 16). (x + 4)2 + (y – 4)2 = 153 3. Write an equation for the line tangent to the circle x2 + y2 = 17 at the point (4, 1). y – 1 = –4(x – 4)
Notes 4. Which points on the graph shown are within 2 units of the point (0, –2.5)? C, F
Circles: Extra Info The following power-point slides contain extra examples and information. Reminder of Lesson Objectives Write an equation for a circle. Graph a circle, and identify its center and radius. 16
Write an equation for a circle.
Graph a circle, and identify its center and radius. Vocabulary circle tangent 17
Check It Out! Example 2 Find the equation of the circle with center (–3, 5) and containing the point (9, 10). Use the Distance Formula to find the radius. Substitute the values into the equation of a circle. (x + 3)2 + (y – 5)2 = 132 (x + 3)2 + (y – 5)2 = 169
Check It Out! Example 4 Write the equation of the line that is tangent to the circle 25 = (x – 1)2 + (y + 2)2, at the point (1, –2). Step 1 Identify the center and radius of the circle. From the equation 25 = (x – 1)2 +(y + 2)2, the circle has center of (1, –2) and radius r = 5.
Check It Out! Example 4 Continued
Step 2 Find the slope of the radius at the point of tangency and the slope of the tangent. Use the slope formula. Substitute (5, –5) for (x2 , y2 ) and (1, –2) for (x1 , y1 ). The slope of the radius is –3 4 Because the slopes of perpendicular lines are negative reciprocals, the slope of the tangent is .
Check It Out! Example 4 Continued
Step 3. Find the slope-intercept equation of the tangent by using the point (5, –5) and the slope Use the point-slope formula. Substitute (5, –5 ) for (x1 , y1 ) and for m. 4 3 Rewrite in slope-intercept form.
Check It Out! Example 4 Continued
The equation of the line that is tangent to 25 = (x – 1)2 + (y + 2)2 at (5, –5) is Check Graph the circle and the line. |
# Manipulation and control 2010 Instructor: Dimitar Dimitrov eMail
```Örebro University
Support notes
Course:
Manipulation and control
2010
Singularity functions
(in this handout we will follow [1])
Main points covered
Instructor: Dimitar Dimitrov
Office: T2228
eMail: [email protected]
Tel: 30 14 82
Question hours: ask appointment by email
www.aass.oru.se/Research/Learning/drdv.html
• Unit-step function
• Unit-ramp function
• Unit-rectangle function
• Unit-triangle function
• Signum function
• Unit-impulse
• Combinations of functions
• Amplitude scaling, time shifting and time scaling
1
2
Unit-step function
2
1.5
1.5
step(t)
1
1
0.5
0.5
0
t
0
on
−0.5
off
on
−0.5
−3
−2
−1
0
1
2
3
2
3
−1
1.5
step(t)
−1.5
1
−2
−2
0
2
4
6
8
10
Figure 1: Example of a sine signal switched on and off. In red is depicted the actual signal,
in dashed blue is a sinusoid. In dashed green is depicted a signal used specify the on and off
times for the sinusoid.
0.5
−0.5
−3
Sines, cosines, and exponentials are all continuous and
differentiable functions at every point in time. In practice,
many important signals are not continuous and differentiable, due to operations like switching a signal on and off
for example.
Question: how to describe the signal in Fig. 1?
t
0
−2
−1
0
1
Figure 2: The unit-step function according to definition (1) (bottom). The top figure depicts
a common way of drawing the unit step. Usually step(t) is denoted simply by u(t).
We define the (continuous time) unit-step function as
1,
step(t) = 12 ,
0,
t>0
t=0
t<0
(1)
Signals of this type can be easily (mathematically) described by multiplying a function that is continuous and
differentiable (in the above case sinusoid) by another function that switches from zero to one or from one to zero at
some finite time.
The unit-step function is useful, because it can mathematically represent a common action in real physical systems - fast switching from one state to another.
3
4
Unit-ramp function
Unit-rectangle function
2.5
2
2
1.5
1.5
rect(t)
ramp(t)
1
1
0.5
0.5
t
0
t
0
−0.5
−0.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
−2
Figure 3: The unit-ramp function (in blue). In green dashed line is depicted the unit-step
function.
Another common type of signal is one that is switched on
at some time and changes linearly afterwards. We define a
unit-ramp function as
ramp(t) =
=
Z
t,
0,
t>0
t≤0
t
step(τ )dτ = t step(t)
(2)
−∞
The area of the rectangular in Fig. 3 is 0.5 (which is the
value of ramp(t) at t = 0.5).
5
−1.5
−1
−0.5
0
0.5
1
1.5
2
Figure 4: The unit-rectangle function.
1,
rect(t) = 21 ,
0,
|t| < 21
|t| = 12
|t| > 12
(3)
The unit-rectangle function is very convenient for describing the switching on, then switching off of a signal.
When rect(t) multiplies another function the result is zero
outside the nonzero range of rect(t) and is equal to the
other function inside the nonzero range of rect(t).
6
Unit-triangle function
1.5
f (t)
step(t + 0.5)
1
2
0.5
1.5
tri(t)
t
0
1
−0.5
0.5
step(t − 0.5)
−1
t
0
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
Figure 5: The unit-rectangle function expressed in terms of step(t + 0.5) − step(t − 0.5).
We can think of the application of rect(t) as “opening of
a gate”, allowing other functions to go through, and then
closing the “gate”.
Note that (as i the case of ramp(t)) rect(t) can be expressed in terms of step(t).
−0.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
Figure 6: The unit-triangle function.
tri(t) =
(
1 − |t|,
0,
|t| < 1
|t| ≥ 1
(5)
The unit-triangle function can be described as
rect(t) = step(t − t0) − step(t − t1),
(t0 < t1)
(4)
where, the notation step(t − t0) means that the unit-step
function is shifted with t0 time units. If t0 > 0, then the
shift would be a delay.
7
tri(t) = ramp(t + 1) − 2ramp(t) + ramp(t − 1),
or
tri(t) = (t + 1)step(t + 1) − 2t step(t) + (t − 1)step(t − 1)
8
Signum function
1.5
Unit-impulse
sgn(t)
1
δ(t + t3 )
0.5
δ(t − t1 )
Aδ(t − t2 )
1
1
1
A
t = t3
t=0
t = t1
t = t2
t
0
−0.5
Figure 8: Graphical representation of impulses. The number on the left side denotes the strength
of the impulse.
−1
−2
δ(t)
−1.5
−1
−0.5
0
0.5
1
1.5
An impulse is commonly “illustrated” as a vertical arrow,
with a number denoting its strength (see Fig. 8).
2
Figure 7: The signum function.
1,
sgn(t) = 0,
−1,
t>0
t=0
t<0
= 2step(t) − 1
(6)
In Fig. 8, the impulse δ(t − t1) is simply δ(t) delayed
with t1 time units. The impulse Aδ(t − t2) is simply δ(t)
delayed with t2 time units and scaled by a factor of A.
We discussed the unit-impulse δ(t) in Handout 7.
For nonzero arguments, the value of the signum function
has a magnitude of one and a sign that is the same as the
sign of its argument.
A Matlab implementation of step(t), ramp(t), rect(t),
and tri(t) is available on the course web-site in file sfunc.m.
In Matlab the signum function is called sign.m.
9
10
Unit-sinc function
Combinations of functions
We already demonstrated that for describing a given
function, we could use sums, differences, products of some
available functions (e.g. step(t)).
4
3
2
For example in order to describe the signal depicted in
Fig. 1 we used
sinc(t)
1
t
0
sin(t)step(t) − sin(t)step(t − 2) + sin(t)step(t − 3) =
sin(t) [step(t) − step(t − 2) + step(t − 3)]
−1
where, step(t) − step(t − 2) + step(t − 3) is depicted in
dashed green line in Fig. 1. Below is given the Matlab code
that generates the figure
t = -3:0.01:10;
y = sin(t);
−2
−3
−5
−4
−3
−2
−1
0
1
2
3
4
5
Figure 9: The unit-sinc function.
The unit-sinc function is related to the unit-rectangle
function (you will use it in your Digital Image Processing
course next year). It is the Fourier transform of the unitrectangle function. sinc(t) is called a unit function, because
its height and area are one. It is defined by
sinc(t) =
Note that sinc(0) = 1;
sin(πt)
πt
(7)
u1 = sfunc(t,1);
u2 = sfunc(t-2,1);
u3 = sfunc(t-3,1);
hold on;
plot(t,y.*(u1-u2+u3),’r’,’LineWidth’,3)
plot(t,y,’b--’)
plot(t,u1-u2+u3,’g--’)
axis([-3 10 -2 2]); grid on;
11
12
Amplitude scaling and shifting
Time scaling
3
5
2
−2g(2(t + 3.5))
)
g( t−3
2
g(t)
4
2g(t + 3.5)
1
3
t
0
g(t − 3)
g(t)
2
−1
1
t
0
−2
−3
−1
−4
−2
−5
−4
−3
−2
−1
0
1
2
3
4
5
−5
−4
−3
−2
−1
0
1
2
3
4
5
Figure 10: Amplitude scaling and time shifting of a signal.
Figure 11: Time scaling.
Scaling the amplitude of a signal g(t) by a constraint A
can be achieved by Ag(t).
Suppose we have an analog tape recording of some music.
When we play the tape in the usual way we hear the music
as performed but if we increase the speed of the tape movement across the sensing head we hear a speeded-up version
of the music. All the frequencies in the original recording
are now higher and the time of the performance is reduced.
If we slow down the tape the opposite effect occurs.
Time shifting of a signal g(t) with t0 time units can be
achieved by g(t − t0). For example let t = {0, 1, 2, 3, 4}
and f (t) = {5, 6, 7, 7, 8},
• f (t − 1) =? for t = 2, t = 5
• 2f (t − 2) =? for t = 4
13
14
Response of LTI system to different excitations.
From Handout 7 we know that if an impulse δ(t) is used
to excite a Linear Time-Invariant (LTI) system, the system’s response is called impulse response and we denoted
it by h(t). Recall the sifting property
h(t) = h(t) ∗ δ(t) =
Z
h(τ )
2
2
1.5
1.5
1
1
0.5
0.5
0
0
−0.5
−0.5
−1
−1
0
∞
−∞
Let us denote the response of an LTI system to a unitstep by hs(t)
hs(t) = h(t) ∗ step(t) =
=
=
∞
Z−∞
∞
Z−∞
t
h(τ )u(t − τ )dτ
h(τ )u(−(τ − t))dτ
τ
3
4
5
0
1
2
1.5
1
1
0.5
0.5
0
0
−0.5
−0.5
−1
−1
2
τ
3
τ
R2
0
2
1
2
4
5
0
1
2
3
4
5
4
5
h(t)
τ
3
Figure 12: Visualization of the convolution h(t) ∗ step(t) for t = 2. Note that in the lower-left
figure, the product h(t)step(−(τ − t)) is depicted with the dashed green line.
(8)
−∞
The effect of convolving h(t) with the unit-step function
is depicted in Fig. 12.
Equation (8) shows that the response of an LTI system
to step(t) is the integral of its impulse response.
Note that we can we can think of step(t) as the integral of
δ(t), hence if an excitation changes to its integral,
the response also changes to its integral.
15
2
1.5
0
h(τ )dτ
1
h(t)step(−(τ − t))
h(τ )δ(t − τ )dτ = h(t)
Z
step(τ − t)
We can also turn these relationships around and say that
since the first derivative is the “inverse” of integration, if
an excitation is changed to its derivative, the
response is also changed to its derivative.
Rt
Recall from equation (2) that ramp(t) = −∞ step(τ )dτ ,
hence the responseof an LTI system to ramp(t) is given by
R t R t
−∞
−∞ h(τ )dτ dτ .
16
References
[1] M. Roberts, “Fundamentals of Signals & Systems,” Mc
Graw Hill, 2008.
17
``` |
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Binomial Distribution
To understand binomial distributions and binomial probability, it helps to understand binomial experiments and some associated notation; so it is discussed below.
Binomial Experiment
A binomial experiment (also known as a Bernoulli trial) is a statistical experiment that has the following properties:
y y y y
The experiment consists of n repeated trials. Each trial can result in just two possible outcomes. We call one of these outcomes a success and the other, a failure. The probability of success, denoted by P, is the same on every trial. The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials.
Consider the following statistical experiment. You flip a coin 2 times and count the number of times the coin lands on heads. This is a binomial experiment because:
y y y y
The experiment consists of repeated trials. We flip a coin 2 times. Each trial can result in just two possible outcomes - heads or tails. The probability of success is constant - 0.5 on every trial. The trials are independent; that is, getting heads on one trial does not affect whether we get heads on other trials.
Notation
y y y y y y
x: The number of successes that result from the binomial experiment. n: The number of trials in the binomial experiment. P: The probability of success on an individual trial. Q: The probability of failure on an individual trial. (This is equal to 1 - P.) b(x; n, P): Binomial probability - the probability that an n-trial binomial experiment results in exactly x successes, when the probability of success on an individual trial is P. nCr: The number of combinations of n things, taken r at a time.
Binomial Distribution
A binomial random variable is the number of successes x in n repeated trials of a binomial experiment. The probability distribution of a binomial random variable is called a binomial distribution (also known as a Bernoulli distribution).
Suppose we flip a coin two times and count the number of heads (successes). The binomial random variable is the number of heads, which can take on values of 0, 1, or 2. The binomial distribution is presented below. Number of heads Probability 0 0.25 1 0.50 2 0.25 The binomial distribution has the following properties:
y y y
The mean of the distribution ( x) is equal to n * P . The variance ( 2x) is n * P * ( 1 - P ). The standard deviation ( x) is sqrt[ n * P * ( 1 - P ) ].
Binomial Probability
The binomial probability refers to the probability that a binomial experiment results in exactly x successes. For example, in the above table, we see that the binomial probability of getting exactly one head in two coin flips is 0.50. Given x, n, and P, we can compute the binomial probability based on the following formula: Binomial Formula. Suppose a binomial experiment consists of n trials and results in x successes. If the probability of success on an individual trial is P, then the binomial probability is: b(x; n, P) = nCx * Px * (1 - P)n - x
Example 1 Suppose a die is tossed 5 times. What is the probability of getting exactly 2 fours? Solution: This is a binomial experiment in which the number of trials is equal to 5, the number of successes is equal to 2, and the probability of success on a single trial is 1/6 or about 0.167. Therefore, the binomial probability is: b(2; 5, 0.167) = 5C2 * (0.167)2 * (0.833)3 b(2; 5, 0.167) = 0.161
Poisson Distribution
Attributes of a Poisson Experiment
A Poisson experiment is a statistical experiment that has the following properties:
y y y y
The experiment results in outcomes that can be classified as successes or failures. The average number of successes ( ) that occurs in a specified region is known. The probability that a success will occur is proportional to the size of the region. The probability that a success will occur in an extremely small region is virtually zero.
Note that the specified region could take many forms. For instance, it could be a length, an area, a volume, a period of time, etc.
Notation
The following notation is helpful, when we talk about the Poisson distribution.
y y y y
e: A constant equal to approximately 2.71828. (Actually, e is the base of the natural logarithm system.) : The mean number of successes that occur in a specified region. x: The actual number of successes that occur in a specified region. P(x; ): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is .
Poisson Distribution
A Poisson random variable is the number of successes that result from a Poisson experiment. The probability distribution of a Poisson random variable is called a Poisson distribution. Given the mean number of successes ( ) that occur in a specified region, we can compute the Poisson probability based on the following formula: Poisson Formula. Suppose we conduct a Poisson experiment, in which the average number of successes within a given region is . Then, the Poisson probability is: P(x; ) = (e- ) ( x) / x! where x is the actual number of successes that result from the experiment, and e is approximately equal to 2.71828. The Poisson distribution has the following properties:
y y
The mean of the distribution is equal to The variance is also equal to .
Example 1 The average number of homes sold by the Acme Realty company is 2 homes per day. What is the probability that exactly 3 homes will be sold tomorrow? Solution: This is a Poisson experiment in which we know the following:
y y y
= 2; since 2 homes are sold per day, on average. x = 3; since we want to find the likelihood that 3 homes will be sold tomorrow. e = 2.71828; since e is a constant equal to approximately 2.71828.
We plug these values into the Poisson formula as follows: P(x; ) = (e- ) ( x) / x! P(3; 2) = (2.71828-2) (23) / 3! P(3; 2) = (0.13534) (8) / 6 P(3; 2) = 0.180 Thus, the probability of selling 3 homes tomorrow is 0.180 .
The Normal Distribution (Bell Curve)
In many natural processes, random variation conforms to a particular probability distribution known as the normal distribution, which is the most commonly observed probability distribution. Mathematicians de Moivre and Laplace used this distribution in the 1700's. In the early 1800's, German mathematician and physicist Karl Gauss used it to analyze astronomical data, and it consequently became known as the Gaussian distribution among the scientific community. The shape of the normal distribution resembles that of a bell, so it sometimes is referred to as the "bell curve", an example of which follows:
Normal Distribution
The above curve is for a data set having a mean of zero. In general, the normal distribution curve is described by the following probability density function:
y y y y
y y
mean standard deviation
If the mean and standard deviation are known, then one essentially knows as much as if one had access to every point in the data set.
The Empirical Rule
The empirical rule is a handy quick estimate of the spread of the data given the mean and standard deviation of a data set that follows the normal distribution.
The empirical rule states that for a normal distribution:
y y y
68% of the data will fall within 1 standard deviation of the mean 95% of the data will fall within 2 standard deviations of the mean Almost all (99.7%) of the data will fall within 3 standard deviations of the mean
Note that these values are approximations. For example, according to the normal curve probability density function, 95% of the data will fall within 1.96 standard deviations of the mean; 2 standard deviations is a convenient approximation.
Normal Distribution and the Central Limit Theorem
The normal distribution is a widely observed distribution. Furthermore, it frequently can be applied to situations in which the data is distributed very differently. This extended applicability is possible because of the central limit theorem, which states that regardless of the distribution of the population, the distribution of the means of random samples approaches a normal distribution for a large sample size. |
RD Sharma 2014 Solutions for Class 7 Math Chapter 14 Lines And Angles are provided here with simple step-by-step explanations. These solutions for Lines And Angles are extremely popular among class 7 students for Math Lines And Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2014 Book of class 7 Math Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2014 Solutions. All RD Sharma 2014 Solutions for class 7 Math are prepared by experts and are 100% accurate.
#### Question 31:
In Fig., three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of x, y, z and u.
$\angle$BOD + $\angle$DOF + $\angle$FOA = 180° (Linear pair)
$\angle$FOA = $\angle$u = $180°-90°-50°=40°$
$\angle \mathrm{FOA}=\angle x=40°$ (Vertically opposite angles)
$\angle \mathrm{BOD}=\angle z=90°$ (Vertically opposite angles)
$\angle \mathrm{EOC}=\angle y=50°$ (Vertically opposite angles)
#### Question 32:
In Fig., find the values of x, y and z.
#### Question 1:
In Fig., line n is a transversal to lines l and m. Identify the following:
(i) Alternate and corresponding angles in Fig. (i).
(ii) Angles alternate to ∠d and ∠g and angles corresponding to angles ∠f and ∠h in Fig. (ii).
(iii) Angle alternate to ∠PQR, angle corresponding to ∠RQF and angle alternate to ∠PQE in Fig. (iii).
(iv) Pairs of interior and exterior angles on the same side of the transversal in Fig. (ii).
(i) Figure (i)
Corresponding angles:
$\angle$EGB and $\angle$GHD
$\angle$HGB and $\angle$FHD
$\angle$EGA and $\angle$GHC
$\angle$AGH and $\angle$CHF
Alternate angles:
$\angle$EGB and $\angle$CHF
$\angle$HGB and $\angle$CHG
$\angle$EGA and $\angle$FHD
$\angle$AGH and $\angle$GHD
(ii) Figure (ii)
Alternate angle to $\angle$d is $\angle$e.
Alternate angle to $\angle$g is $\angle$b.
Also,
Corresponding angle to $\angle$f is $\angle$c.
Corresponding angle to $\angle$h is $\angle$a.
(iii) Figure (iii)
Angle alternate to $\angle$PQR is $\angle$QRA.
Angle corresponding to $\angle$RQF is $\angle$ARB.
Angle alternate to $\angle$POE is $\angle$ARB.
(iv) Figure (ii)
Pair of interior angles are
$\angle$a and $\angle$e
$\angle$d and $\angle$f
Pair of exterior angles are
$\angle$b and $\angle$h
$\angle$c and $\angle$g
#### Question 2:
In Fig., AB and CD are parallel lines intersected by a transversal PQ at L and M respectively. If ∠CMQ = 60°, find all other angles in the figure.
$\angle$ALM = $\angle$CMQ = $60°$ (Corresponding angles)
$\angle$LMD = $\angle$CMQ = $60°$ (Vertically opposite angles)
$\angle$ALM = $\angle$PLB = $60°$ (Vertically opposite angles)
Since
$\angle$CMQ + $\angle$QMD = $180°$ (Linear pair)
QMD = $180°-60°=120°$
$\angle$QMD = $\angle$MLB = $120°$ (Corresponding angles)
$\angle$QMD = $\angle$CML = $120°$ (Vertically opposite angles)
$\angle$MLB = $\angle$ALP = $120°$ (Vertically opposite angles)
#### Question 3:
In Fig., AB and CD are parallel lines intersected by a transversal PQ at L and M respectively. If ∠LMD = 35° find ∠ALM and ∠PLA.
In the given Fig., AB || CD.
#### Question 4:
The line n is transversal to line l and m in Fig. Identify the angle alternate to ∠13, angle corresponding to ∠15, and angle alternate to ∠15.
In this given Fig., line l || m.
Here,
Alternate angle to $\angle$13 is $\angle$7.
Corresponding angle to $\angle$15 is $\angle$7.
Alternate angle to $\angle$15 is $\angle$5.
#### Question 5:
In Fig., line l || m and n is a transversal. If ∠1 = 40°, find all the angles and check that all corresponding angles and alternate angles are equal.
In the given figure, l || m.
Here,
Also,
Thus,
Hence, alternate angles are equal.
#### Question 6:
In Fig., line l || m and a transversal n cuts them at P and Q respectively. If ∠1 = 75°, find all other angles.
In the given figure, l || m, n is a transversal line and ∠1 = 75°.
Thus, we have:
#### Question 7:
In Fig., AB || CD and a transversal PQ cuts them at L and M respectively. If ∠QMD = 100°, find all other angles.
In the given figure, AB || CD, PQ is a transversal line and $\angle$QMD = 100°.
Thus, we have:
$\angle$DMQ + $\angle$QMC = 180° (Linear pair)
$\therefore \angle \mathrm{QMC}=180°-\angle \mathrm{DMQ}=180°-100°=80°$
Thus,
$\angle$DMQ = $\angle$BLM = 100° (Corresponding angles)
$\angle$DMQ = $\angle$CML = 100° (Vertically opposite angles)
$\angle$BLM = $\angle$PLA = 100° (Vertically opposite angles)
Also,
$\angle$CMQ = $\angle$ALM = 80° (Corresponding angles)
$\angle$CMQ = $\angle$DML = 80° (Vertically opposite angles)
$\angle$ALM = $\angle$PLB = 80° (Vertically opposite angles)
#### Question 8:
In Fig., l || m and p || q. Find the values of x, y, z, t.
In the given figure, l || m and p || q.
Thus, we have:
$\angle z=80°$ (Vertically opposite angles)
$\angle z=\angle t=80°$ (Corresponding angles)
$\angle z=\angle y=80°$ (Corresponding angles)
$\angle x=\angle y=80°$ (Corresponding angles)
#### Question 9:
In Fig., line l || m, ∠1 = 120° and ∠2 = 100°, find out ∠3 and ∠4.
In the given figure, ∠1 = 120° and ∠2 =100°.
Since l || m, so
Also,
We know that the sum of all the angles of triangle is 180°.
$\therefore \angle 6+\angle 3+\angle 4=180°\phantom{\rule{0ex}{0ex}}⇒60°+80°+\angle 4=180°\phantom{\rule{0ex}{0ex}}⇒140°+\angle 4=180°\phantom{\rule{0ex}{0ex}}⇒\angle 4=180°-140°=40°$
#### Question 10:
In Fig., line l || m. Find the values of a, b, c, d. Give reasons.
In the given figure, line l || m.
Thus, we have:
#### Question 11:
In Fig., AB || CD and ∠1 and ∠2 are in the ratio 3 : 2. Determine all angles from 1 to 8.
In the given figure, AB || CD and t is a transversal line.
Now, let:
$\angle 1=3x\phantom{\rule{0ex}{0ex}}\angle 2=2x$
Thus, we have:
Now,
#### Question 12:
In Fig., l, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find ∠1, ∠2 and ∠3.
In the given figure, l || m || n and p is a transversal line.
Thus, we have:
#### Question 13:
In Fig., if l || m || n and ∠1 = 60°, find ∠2.
In the given figure, l || m || n and ∠1 = 60°.
Thus, we have:
#### Question 14:
In Fig., if AB || CD and CD || EF, find ∠ACE.
In the given figure, AB || CD and CD || EF.
Extend line CE to E'.
Thus, we have:
#### Question 15:
In Fig., if l || m, n || p and ∠1 = 85°, find ∠2.
In the given figure, l || m, n || p and ∠1 = 85°.
Now, let ∠4 be the adjacent angle of ∠2.
Thus, we have:
$\angle 3+\angle 2=180°$ (Sum of interior angles on the same side of the transversal)
$\therefore \angle 2=180°-\angle 3=180°-85°=95°$
#### Question 16:
In Fig., a transversal n cuts two lines l and m. If ∠1 = 70° and ∠7 = 80°, is l || m?
We know that if the alternate exterior angles of two lines are equal, then the lines are parallel.
In the given figure, are alternate exterior angles, but they are not equal.
Therefore, lines l and m are not parallel.
#### Question 17:
In Fig., a transversal n cuts two lines l and m such that ∠2 = 65° and ∠8 = 65°. Are the lines parallel?
$\angle$2 = $\angle$3 = 65° (Vertically opposite angles)
$\angle$8 = $\angle$6 = 65° (Vertically opposite angles)
∴ $\angle$3 = $\angle$6
l || m (Two lines are parallel if the alternate angles formed with the transversal are equal)
#### Question 18:
In Fig., show that AB || EF.
Extend line CE to E'.
#### Question 19:
In Fig., AB || CD. Find the values of x, y, z.
$\angle x+125°=180°$ (Linear pair)
$\therefore \angle x=180°-125°=55°$
$\angle z=125°$ (Corresponding angles)
$\angle x+\angle z=180°$ (Sum of adjacent interior angles is $180°$)
$\angle x+125°=180°\phantom{\rule{0ex}{0ex}}⇒\angle x=180°-125°=55°$
$\angle x+\angle y=180°$ (Sum of adjacent interior angles is $180°$)
$55°+\angle y=180°\phantom{\rule{0ex}{0ex}}⇒\angle y=180°-55°=125°$
#### Question 20:
In Fig., find out ∠PXR, if PQ || RS.
Draw a line parallel to PQ passing through X.
Here,
(Alternate interior angles)
∵ PQ || RS || XF
∴ $\angle \mathrm{PXR}=\angle \mathrm{PXF}+\angle \mathrm{FXR}=70°+50°=120°$
#### Question 21:
In Fig., we have
(i) ∠MLY = 2 ∠LMQ, find ∠LMQ.
(ii) ∠XLM = (2x − 10)° and ∠LMQ = x + 30°, find x.
(iii) ∠XLM = ∠PML, find ∠ALY
(iv) ∠ALY = (2x − 15)°, and ∠LMQ = (x + 40)°, find x
(i)
(ii)
(iii)
(iv)
#### Question 22:
In Fig., DE || BC. Find the values of x and y.
$\angle$ABC = $\angle$DAB (Alternate interior angles)
$\angle$ACB = $\angle$EAC (Alternate interior angles)
#### Question 23:
In Fig., line AC || line DE and ∠ABD = 32°. Find out the angles x and y if ∠E = 122°.
#### Question 24:
In Fig., side BC of ∆ABC has been produced to D and CE || BA. If ∠ABC = 65°, ∠BAC = 55°, find ∠ACE, ∠ECD and ∠ACD.
$\angle$ABC = $\angle$ECD = 55° (Corresponding angles)
$\angle$BAC = $\angle$ACE = 65° (Alternate interior angles)
Now, $\angle$ACD = $\angle$ACE + $\angle$ECD
⇒ $\angle$ACD = 55° + 65° = 120°
#### Question 25:
In Fig., line CAAB || line CR and line PR || line BD. Find ∠x, ∠y and ∠z.
Since CA ⊥ AB,
$\therefore \angle x=90°$
We know that the sum of all the angles of triangle is 180°.
$\angle$PBC = $\angle$APQ = $70°$ (Corresponding angles)
Since
#### Question 26:
In Fig., PQ || RS. Find the value of x.
#### Question 27:
In Fig., AB || CD and AE || CF; ∠FCG = 90° and ∠BAC = 120°. Find the values of x, y and z.
$\angle$BAC = $\angle$ACG = 120° (Alternate interior angle)
∴ $\angle$ACF + $\angle$FCG = 120°
$\angle$ACF = 120° − 90° = 30°
$\angle$DCA + $\angle$ACG = 180° (Linear pair)
$\angle$x = 180° − 120° = 60°
$\angle$BAC + $\angle$BAE + $\angle$EAC = 360°
$\angle$CAE = 360° − 120° − (60° + 30°) = 150° ($\angle$BAE = $\angle$DCF)
#### Question 28:
In Fig., AB || CD and AC || BD. Find the values of x, y, z.
(i) Since AC || BD and CD || AB, ABCD is a parallelogram.
$\angle$CAB + $\angle$ACD = 180° (Sum of adjacent angles of a parallelogram)
∴ $\angle$ACD = 180° − 65° = 115°
$\angle$CAD = $\angle$CDB = 65° (Opposite angles of a parallelogram)
$\angle$ACD = $\angle$DBA = 115° (Opposite angles of a parallelogram)
(ii) Here,
AC || BD and CD || AB
$\angle$DAC = x = 40° (Alternate interior angle)
$\angle$DAB = y = 35° (Alternate interior angle)
#### Question 29:
In Fig., state which lines are parallel and why?
Let F be the point of intersection of line CD and the line passing through point E.
Since $\angle$ACD and $\angle$CDE are alternate and equal angles, so
$\angle$ACD = 100° = $\angle$CDE
∴ AC || EF
#### Question 30:
In Fig. 87, the corresponding arms of ∠ABC and ∠DEF are parallel. If ∠ABC = 75°, find ∠DEF.
Construction:
Let G be the point of intersection of lines BC and DE.
∵ AB || DE and BC || EF
$\angle \mathrm{ABC}=\angle \mathrm{DGC}=\angle \mathrm{DEF}=75°$ (Corresponding angles)
#### Question 1:
Write down each pair of adjacent angles shown in Fig.
Adjacent angles are the angles that have a common vertex and a common arm.
Following are the adjacent angles in the given figure:
#### Question 2:
In Fig., name all the pairs of adjacent angles.
In figure (i), the adjacent angles are:
In figure (ii), the adjacent angles are:
$\angle$BAD and $\angle$DAC
$\angle$BDA and $\angle$CDA
#### Question 3:
In figure, write down: (i) each linear pair (ii) each pair of vertically opposite angles.
(i) Two adjacent angles are said to form a linear pair of angles if their non-common arms are two opposite rays.
$\angle$1 and $\angle$3
$\angle$1 and $\angle$2
$\angle$4 and $\angle$3
$\angle$4 and $\angle$2
$\angle$5 and $\angle$6
$\angle$5 and $\angle$7
$\angle$6 and $\angle$8
$\angle$7 and $\angle$8
(ii) Two angles formed by two intersecting lines having no common arms are called vertically opposite angles.
$\angle$1 and $\angle$4
$\angle$2 and $\angle$3
$\angle$5 and $\angle$8
$\angle$6 and $\angle$7
#### Question 4:
Are the angles 1 and 2 given in Fig. adjacent angles?
No, because they have no common vertex.
#### Question 5:
Find the complement of each of the following angles:
(i) 35°
(ii) 72°
(iii) 45°
(iv) 85°
Two angles are called complementary angles if the sum of those angles is 90°.
Complementary angles of the following angles are:
#### Question 6:
Find the supplement of each of the following angles:
(i) 70°
(ii) 120°
(iii) 135°
(iv) 90°
Two angles are called supplementary angles if the sum of those angles is 180°.
Supplementary angles of the following angles are:
(i) 180° − 70° = 110°
(ii) 180° − 120° = 60°
(iii) 180° − 135° = 45°
(iv) 180° − 90° = 90°
#### Question 7:
Identify the complementary and supplementary pairs of angles from the following pairs:
(i) 25°, 65°
(ii) 120°, 60°
(iii) 63°, 27°
(iv) 100°, 80°
Since
Therefore, (i) and (iii) are the pairs of complementary angles and (ii) and (iv) are the pairs of supplementary angles.
#### Question 8:
Can two angles be supplementary, if both of them be
(i) obtuse?
(ii) right?
(iii) acute?
(i) No, two obtuse angles cannot be supplementary.
(ii) Yes, two right angles can be supplementary. ($\because \angle 90°+\angle 90°=\angle 180°$)
(iii) No, two acute angles cannot be supplementary.
#### Question 9:
Name the four pairs of supplementary angles shown in Fig.
Following are the supplementary angles:
$\angle$AOC and $\angle$COB
$\angle$BOC and $\angle$DOB
$\angle$BOD and $\angle$DOA
$\angle$AOC and $\angle$DOA
#### Question 10:
In Fig., A, B, C are collinear points and ∠DBA = ∠EBA.
(i) Name two linear pairs
(ii) Name two pairs of supplementary angles.
(i) Linear pairs:
$\angle$ABD and $\angle$DBC
$\angle$ABE and $\angle$EBC
Because every linear pair forms supplementary angles, these angles are:
$\angle$ABD and $\angle$DBC
$\angle$ABE and $\angle$EBC
#### Question 11:
If two supplementary angles have equal measure, what is the measure of each angle?
Let x and y be two supplementary angles that are equal.
$\angle x=\angle y$
According to the question,
$\angle x+\angle y=180°\phantom{\rule{0ex}{0ex}}⇒\angle x+\angle x=180°\phantom{\rule{0ex}{0ex}}⇒2\angle x=180°\phantom{\rule{0ex}{0ex}}⇒\angle x=\frac{180°}{2}=90°\phantom{\rule{0ex}{0ex}}\therefore \angle x=\angle y=90°$
#### Question 12:
If the complement of an angle is 28°, then find the supplement of the angle.
Let x be the complement of the given angle $28°$.
So, supplement of the angle = $180°-62°=118°$
#### Question 13:
In Fig. 19, name each linear pair and each pair of vertically opposite angles:
Two adjacent angles are said to form a linear pair of angles if their non-common arms are two opposite rays.
$\angle$1 and $\angle$2
$\angle$2 and $\angle$3
$\angle$3 and $\angle$4
$\angle$1 and $\angle$4
$\angle$5 and $\angle$6
$\angle$6 and $\angle$7
$\angle$7 and $\angle$8
$\angle$8 and $\angle$5
$\angle$9 and $\angle$10
$\angle$10 and $\angle$11
$\angle$11 and $\angle$12
$\angle$12 and $\angle$9
Two angles formed by two intersecting lines having no common arms are called vertically opposite angles.
$\angle$1 and $\angle$3
$\angle$4 and $\angle$2
$\angle$5 and $\angle$7
$\angle$6 and $\angle$8
$\angle$9 and $\angle$11
$\angle$10 and $\angle$12
#### Question 14:
In Fig., OE is the bisector of ∠BOD. If ∠1 = 70°, find the magnitudes of ∠2, ∠3 and ∠4.
Since OE is the bisector of $\angle$BOD,
#### Question 15:
One of the angles forming a linear pair is a right angle. What can you say about its other angle?
One angle of a linear pair is the right angle, i.e., 90°.
∴ The other angle = 180° 90° = 90°
#### Question 16:
One of the angles forming a linear pair is an obtuse angle. What kind of angle is the other?
If one of the angles of a linear pair is obtuse, then the other angle should be acute; only then can their sum be 180°.
#### Question 17:
One of the angles forming a linear pair is an acute angle. What kind of angle is the other?
In a linear pair, if one angle is acute, then the other angle should be obtuse. Only then their sum can be 180°.
#### Question 18:
Can two acute angles form a linear pair?
No, two acute angles cannot form a linear pair because their sum is always less than 180°.
#### Question 19:
If the supplement of an angle is 65°; then find its complement.
Let be the required angle.
Then, we have:
x + 65° = 180°
$⇒$x = 180° - 65° = 115°
The complement of angle cannot be determined.
#### Question 20:
Find the value of x in each of the following figures.
(i)
Since (Linear pair)
(ii)
(iii)
(iv)
(v)
$2x°+x°+2x°+3x°=180°\phantom{\rule{0ex}{0ex}}⇒8x=180\phantom{\rule{0ex}{0ex}}⇒x=\frac{180}{8}=22.5°$
(vi)
$3x°=105°\phantom{\rule{0ex}{0ex}}⇒x=\frac{105}{3}=35°$
#### Question 21:
In Fig. 22, it being given that ∠1 = 65°, find all other angles.
$\angle 1=\angle 3$ (Vertically opposite angles)
$\therefore \angle 3=65°$
Since $\angle 1+\angle 2=180°$ (Linear pair)
$\therefore \angle 2=180°-65°=115°$
$\angle 2=\angle 4$ (Vertically opposite angles)
$\therefore \angle 4=\angle 2=115°$ and $\angle 3=65°$
#### Question 22:
In Fig., OA and OB are opposite rays:
(i) If x = 25°, what is the value of y?
(ii) If y = 35°, what is the value of x?
$\angle$AOC + $\angle$BOC = 180° (Linear pair)
$⇒\left(2y+5\right)+3x=180°\phantom{\rule{0ex}{0ex}}⇒3x+2y=175°$
(i) If x = 25°, then
$3×25°+2y=175°\phantom{\rule{0ex}{0ex}}⇒75°+2y=175°\phantom{\rule{0ex}{0ex}}⇒2y=175°-75°=100°\phantom{\rule{0ex}{0ex}}⇒y=\frac{100°}{2}=50°$
(ii) If y = 35°, then
$3x+2×35°=175°\phantom{\rule{0ex}{0ex}}⇒3x+70°=175°\phantom{\rule{0ex}{0ex}}⇒3x=175°-70°=105°\phantom{\rule{0ex}{0ex}}⇒x=\frac{105°}{3}=35°$
#### Question 23:
In Fig., write all pairs of adjacent angles and all the linear pairs.
Linear pairs of angles:
#### Question 24:
In Fig. 25, find ∠x. Further find ∠BOC, ∠COD and ∠AOD.
$\angle BOC=x+20°=50°+20°=70°\phantom{\rule{0ex}{0ex}}\angle COD=x=50°\phantom{\rule{0ex}{0ex}}\angle AOD=x+10°=50°+10°=60°\phantom{\rule{0ex}{0ex}}$
#### Question 25:
How many pairs of adjacent angles are formed when two lines intersect in a point?
If two lines intersect at a point, then four adjacent pairs are formed, and those pairs are linear as well.
#### Question 26:
How many pairs of adjacent angles, in all, can you name in Fig.?
There are 10 adjacent pairs in the given figure; they are:
#### Question 27:
In Fig., determine the value of x.
#### Question 28:
In Fig., AOC is a line, find x.
#### Question 29:
In Fig., POS is a line, find x.
$\angle \mathrm{QOP}+\angle \mathrm{QOR}+\angle \mathrm{ROS}=180°$ (Angles on a straight line)
$⇒60°+4x+40°=180°\phantom{\rule{0ex}{0ex}}⇒100°+4x=180°\phantom{\rule{0ex}{0ex}}⇒4x=180°-100°=80°\phantom{\rule{0ex}{0ex}}⇒x=\frac{80°}{4}=20°\phantom{\rule{0ex}{0ex}}$
#### Question 30:
In Fig., lines l1 and l2 intersect at O, forming angles as shown in the figure. If x = 45°, find the values of y, z and u. |
## Allegations and Mixtures Practice Questions – Set 4
Allegations and Mixtures Practice Questions – Set 4
1. There are two mixtures of honey and water in which the ratio of honey and water are as 1 : 3 and 3 : 1 respectively. Two litres are drawn from first mixture and 3 litres from second mixture, are mixed to form another mixture. What is the ratio of honey and water in it?
a) 106 : 69
b) 103 : 72
c) 89 : 86
d) 11 : 9
e) None of these
Answer: D)
Explanation: The percentage of honey in the first mixture = ¼ × 100 = 25%
The percentage of honey in the second mixture = ¾ × 100 = 75%
Given that, 2 litres are drawn from first mixture and 3 litres from second mixture, then
Part of honey in the new mixture = (2 × 25 + 3 × 75)/5 = 275/5 = 55%
Then the ratio of honey and water in the new mixture would be 55 : (100 – 55) = 55 : 45 = 11 : 9
2. Jagtap purchases 30 kg of wheat at the rate of 11.50 per kg and 20 kg of wheat at the rate of Rs. 14.25 per kg. He mixed the two and sold the mixture. Approximately at what price per kg should he sell the mixture to make 30 per cent profit?
a) Rs.16.38
b) Rs.18.20
c) Rs.15.60
d) Rs.14.80
e) None of these
Answer: A)
Explanation: Given, Jagtap purchases 30 kg of wheat at the rate of 11.50 per kg and 20 kg of wheat at the rate of Rs. 14.25 per kg.
Total cost of the mixture = 30 × 11.50 + 20 × 14.25
Total cost of the mixture = Rs. 630
Total kg of mixture = 30 + 20 = 50 kg
Cost of mixture per kg = 630/50 = Rs 12.6 per kg
Given, he sells the mixture to make 30 per cent profit.
Selling price of mixture per kg = 12.6 + 30% of 12.6
Selling price of mixture per kg = RS. 16.38
3. A and B are two alloys of bronze and platinum prepared by mixing metals in the ratio of 4 : 3 and 9 : 5, respectively. If equal quantities of alloys A and B are melted to form a third alloy C, then find the ratio of bronze and platinum in C.
a) 19 : 25
b) 17 : 11
c) 9 : 7
d) 22 : 13
e) 12 : 29
Answer: B)
Explanation: Let x be the quantity of each melted alloy A and B.
Quantity of bronze in A = 4x/7
Quantity of platinum in A = 3x/7
Quantity of bronze in B = 9x/14
Quantity of platinum in B = 5x/14
Total quantity of bronze in alloy C = 4x/7 + 9x/14 = 17x/14
Total quantity of platinum in alloy C = 3x/7 + 5x/14 = 11x/14
Requried ratio = 17x/14 : 11x/14 = 17: 11
4. A liuid P is 2 ½ times as heavy as water and water is 1 3/5 times as heavy as another liquid ‘Q’. The amount of liquid ‘P’ that must be added to 9 litres of the liquid ‘Q’ so that the mixture may weigh as much as an equal volume of water, will be
a) 7 litres
b) 5 1/6 litres
c) 5 litres
d) 2 ¼ litres
e) 3 litres
Answer: D)
Explanation: Let the weight of liquid Q be w kg.
Thus, weight of water = 1 3/5 times as heavy as liquid ‘Q’ = 8w/5
Weight of liquid P = 2 1/2 times as heavy as water = 5/2 × 8w/5=4w
Let the amount of liquid P to be added to 9 litres of Q be x litres.
Volume of mixture of P & Q = (9 + x) litres
Now, according to condition given in the problem, the weight of the mixture should be equal to the weight of equal amount of water.
(4w × x) + (w × 9) = (8w/5) × (9 + x)
4x + 9 = 8/5 × (9+x)
5 × (4x + 9) = 8 × (9 + x)
20x + 45 = 72 + 8x
20x 8x = 72 45
12x = 27
5. In what proportion must water be mixed with milk to gain 33 1/3 per cent by selling the mixture at the cost price?
a) 2 : 3
b) 1 : 3
c) 1 : 4
d) 3 : 4
e) None of these
Answer: B)
Explanation: Let C.P of 1 litre milk = Rs 1
So, S.P of 1 litre mixture = CP of 1L milk = Rs 1
Gain = 33 1/3% = 100/3%
CP of 1L mix = 100/(100 + gain%) × SP
= [100/(100 + 100/3)] × 1 = 100/(400/3) = 3/4
By rule of allegation
CP of 1L water (0) CP of 1L milk (1)
CP of 1L mixture (3/4)
1 – (3/4) = 1/4 (3/4) – 0 = ¾
Quantity of water : quantity of milk = ¼ : ¾ = 1:3
6. A vessel is full of a mixture of spirit and water in which there is found to be 17% of spirit by measure. Ten litres are drawn off and the vessel is filled up with water. The proportion of spirit is now found to be 15 1/9%. How much does the vessel hold?
a) 70 litres
b) 75 litres
c) 80 litres
d) 90 litres
e) 100 litres
Answer: D)
Explanation: Let the initial quantity of mixture be 100x litres
Quantity of spirit in original mixture = 17x litres
Quantity of water in original mixture = 83x litres
10 litres of mixture is drawn off and replaced by 10 litres of water.
Thus, quantity of spirit in new mixture = 17x - (17% of 10 litres) = 17x - 1.7 litres
Quantity of water in new mixture = 83x - (83% of 10 litres) + 10 litres = 83x - 8.3 + 10 litres
= 83x - 1.7 litres
Given: proportion of spirit in new mixture = 15 1/9% = 136/9%
Proportion of water in new mixture = 764/9%
Thus, [(17x – 1.7)/(83x – 1.7)]litres = [(136/9)/(764/9)]
On solving, we get
1700x = 1530
x = 1530/1700 = 0.9 litres
Thus, the initial quantity of mixture will be 100x litres = 100 × 0.9 litres = 90 litres.
7. How much water must be added to a cask containing 40 1/2 litres of spirit worth Rs. 3.92 a litre to reduce the price to Rs. 3.24 a litre?
a) 9 1/3 litre
b) 9 litres
c) 8 ½ litres
d) 8 litres
e) None of these
Answer: C)
Explanation: Quantity of sprit = 40 1/2 litres
Price of sprit = Rs.3.92 per litre (dearer)
Price of water = Rs. 0 (cheaper)
Now if we mix spirit and water, we have to mix in such a proportion that mixture would cost Rs.3.24 per litre
Consider the price at which mixture to be made be mean price i.e. 3.24 per litre.
Proportion in terms of quantity must be given as
Quantity of cheaper/Quantity of dearer = (dearer price – mean price)/(mean price – cheaper price)
=> Quantity of cheaper/(40 1/3) L = (3.92 – 3.24)/3.24 = 0.68/3.24 = 68/324
Quantity of cheaper = 68/324 × 81/2 = 17/2 = 8 ½ L
8. In a mixture of milk and water the proportion of water by weight was 75%. If in the 60 gms mixture 15 gms. Water was added, what would be the percentage of water in the new mixture?
a) 75%
b) 88%
c) 90%
d) 100%
e) None of these
Answer: E)
Explanation: Given in the question, in the mixture of milk and water the proportion of water by weight was 75%.
Initially we have taken 60 gram mixture of milk & water.
Amount of water in this mixture = 75% of 60 gram = 45 gram
Amount of milk in the mixture = (60 45) gram = 15 gram
Externally we have added 15 gram water.
New amount of water = (45 + 15) gram = 60 gram
Now, new volume of mixture = (60 + 15) gram = 75 gram
[Due to addition of 15 gram water]
Percentage of water in new mixture = (Amount of water)/(Amount of Mixture) × 100%
Percentage of water in new mixture = 60/75 × 100% = 80%
9. 80% milk solution is mixed with 30% milk solution to produce 45 liter of 50% milk solution. If the price of 45 liter of 50% milk solution is Rs.1,080 and the price of 30% milk solution is Rs. 20 per liter, then what is the price of 80% milk solution in the final mixture?
a) Rs.640
b) Rs.650
c) Rs.400
d) Rs.250
e) Rs.540
Answer: E)
Explanation: Say there is X liters of 30% milk solution and remaining (45 – X) liters of 80% milk.
Amount of milk in the solution = 0.30X + 0.80(45 – X)
Amount of milk in 45 liters of 50% solution = 0.50 × 45 = 22.5
0.30X + 36 – 0.8X = 22.5
0.5X = 13.5
X = 27 liters
Volume of 80% milk in the mixture = 45 – X = 18 liters.
Price of 30% milk at Rs.20 per liter = 20 × 27 = Rs.540
Price of 80% milk = 1080 – 540 = Rs.540
10. In a 729 litres mixture of milk and water, the ratio of milk to water is 7 : 2. To get a new mixture containing milk and water in the ratio 7 : 3, the amount of water to be added is–
a) 81 litres
b) 71 litres
c) 56 litres
d) 50 litres
e) None of these
Answer: A)
Explanation: Quantity of milk in 729 litres of mixture,
= 7/9 × 729 = 567 litres
Quantity of water = (729 567) litres = 162 litres
Let x litres of water is mixed to get the required ratio of 7 : 3
567/(162 + x) = 7/3 => 7x = 567 => x = 81 liters |
Rd Sharma 2019 Solutions for Class 9 Math Chapter 22 Tabular Representation Of Statistical Data are provided here with simple step-by-step explanations. These solutions for Tabular Representation Of Statistical Data are extremely popular among Class 9 students for Math Tabular Representation Of Statistical Data Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2019 Book of Class 9 Math Chapter 22 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2019 Solutions. All Rd Sharma 2019 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.
Question 1:
What do you understand by the word ''statistics'' in
(i) singular form (ii) plural form?
Answer:
(i) In singular form statistics may be defined as the science of collection, presentation, analysis and interpretation of numerical data.
(ii) In plural form statistics means numerical facts or observations collected with definite purpose.
For examples, the income and expenditure of persons in a particular locality, number of males and females in a particular town are statistics.
Question 2:
Describe some fundamental characteristic of statistics.
Answer:
The plural form statistics has the simplest structure and the singular form statistics has many components. There is only structural difference between singular and plural form statistics. Some of the characteristics of a statistics are
1. Statistics is a collection of observations. So, clearly a single observation cannot form a statistics.
2. Statistics are collected with definite purpose.
3. Statistics are comparable and classified into various types depending on their properties.
4. Statistics are expressed quantitatively and not qualitatively.
Question 3:
What are (i) primary data? (ii) secondary data? Which of the two − the primary or the secondary data − is more reliable and why?
Answer:
(i) When an investigator collects data himself with a definite plan or designs in his (her) mind, it is called primary data.
(ii) Data which are not originally collected rather obtained from published or unpublished sources are known as secondary data.
Question 4:
Why do we group data?
Answer:
To study the features of a collected data, the data must be arranged in a condensed form. There are a number of ways to arrange the data in condensed form, namely,
1. Serial order or alphabetical order
2. Ascending order
3. Descending order
But, if the number of observations is large, then arranging data in ascending or descending or serial order is a tedious job and it does not tell us much except perhaps the minimum(s) and maximum(s) of data. So to make it easily understandable and clear we condense the data into groups or table form.
Question 5:
Explain the meaning of the following terms:
(i) variate
(ii) class-integral
(iii) class-size
(iv) class-mark
(v) frequency
(vi) class limits
(vii) true class limits.
Answer:
(i) A name which takes different values is called variates. For example, the mark obtained by students of class IX in mathematics is variates.
(ii) In a grouped data, the groups are called class-intervals. For example, 0-5, 5-10, 10-15… are class intervals.
(iii) The difference between the true upper limit and the true lower limit of a class is called its class size. For example, the class size of the class-interval 10-15 is
(iv) The mid value of a class is called the class mark. For example, the mid value of the class 10-15 is
(v) The number of observation falling in a particular class is called the frequency of that class or class frequency. For example, if the number of students obtaining marks 60-70 in a particular subject is 60, then the frequency of the class 60-70 is 60.
(vi) Class limits are the boundaries of a class. The left boundary of a class is called the lower limit and the right boundary of a class is called the upper limit. For example, for the class interval 60-70 the lower limit is 60 and the upper limit is 70.
(vii) The class limits of a continuous grouped frequency distribution are called true class limits. For example, 5-10, 10-15, 15-20, 20-25, 25-30 are continuous class intervals, then the true lower and upper limits (class limits) of the class 15-20 are 15 and 20 respectively. If the classes are not continuous, then adjust the class intervals to form continuous grouped class intervals.
Question 6:
The ages of ten students of a group are given below. The ages have been recorded in years and months:
8 - 6, 9 - 0, 8 - 0,4, 9 - 3, 7 - 8, 8 - 11, 8 - 7, 9 - 2, 7 - 10, 8 - 8
(i) What is the lowest age?
(ii) What is the highest age?
(iii) Determine the range?
Answer:
After arranging in ascending order, the ages of the students are 7 years 8 months, 7 years 10 months, 8 years 4 months, 8 years 6 months, 8 years 7 months, 8 years 8 months, 8 years 11 months, 9 years, 9 years 2 months, and 9 years 3 months.
(i) The lowest age is 7 years 8 months.
(ii) The highest age is 9 years 3 months.
(iii) The range of the ages is
Question 7:
The monthly pocket money of six friends is given below:
Rs 45, Rs 30, Rs 50, Rs 25, Rs 45
(a) What is the highest pocket money?
(b) What is the lowest pocket money?
(c) What is the range?
(d) Arrange the amounts of pocket money in ascending order.
Answer:
(i) After arranging in ascending order, the pocket money’s in Rs. are 25, 30, 40, 45, 45 and 50. The highest pocket money is Rs 50.
(ii) The lowest pocket money is Rs 25.
(iii) The range of the pocket money’s is
(iv) The given data (pocket money in Rs.) in ascending order is 25, 30, 40, 45, 45 and 50.
Question 8:
Write the class-size in each of the following:
(a) 0-4, 5-9, 10-14
(b) 10-19, 20-29, 30-39
(c) 100-120, 120-140, 160-180
(d) 0-0.25, 0.25-0, 0.50-0.75
(e) 5-5.01, 5.01-5.02, 5.02-5.03
Answer:
(i) The given classes are 0-4, 5-9 and 10-14. The classes can be written in continuous form as (-0.5)-4.5, 4.5-9.5, and 9.5-14.5. The upper and lower limits of the first class are 4.5 and (-0.5) respectively. Hence, the class size is
(ii) The given classes are 10-19, 20-29, and 30-39. The classes can be written in continuous form as 9.5-19.5, 19.5-29.5, 29.5-39.5. The upper and lower limits of the first class are 19.5 and 9.5 respectively. Hence, the class size is
(iii) The given classes are 100-120, 120-140, 160-180, are in continuous form. The upper and lower limits of the first class are 120 and 100 respectively. Hence, the class size is
(iv) The given classes are 0-0.25, 0.25-0.50, 0.50-0.75, are in continuous form. The upper and lower limits of the first class are 0.25 and 0 respectively. Hence, the class size is
(v) The given classes are 5-5.01, 5.01-5.02, 5.02-5.03, are in continuous form. The upper and lower limits of the first class are 5.01 and 5 respectively. Hence, the class size is
Question 9:
The final marks in mathematics of 30 students are as follows:
53, 61, 48, 60, 78, 68, 55, 100, 67, 90
75, 88, 77, 37, 84, 58, 60, 48, 62, 56
44, 58, 52, 64, 98, 59, 70, 39, 50, 60
(i) Arrange these marks in the ascending order, 30 to 39 one group, 40 to 49 second group, etc.
Now answer the following:
(ii) What is the highest score?
(iii) What is the lowest score?
(iv) What is the range?
(v) If 40 is the pass mark how many have failed?
(vi) How many have scored 75 or more?
(vii) Which observations between 50 and 60 have not actually appeared?
(viii) How many have scored less than 50?
Answer:
(i) The frequency distribution table of grouped frequency such as 30-39, 40-49… is the following:
(ii) The highest score is 100.
(iii) The lowest score is 37.
(iv) The range of the scores is
(v) If 40 is the pass mark, then the number of students failed is 2.
(vi) The number of students scored more than 75 is 8.
(vii) The observations 51, 54 and 57 in between 50-60 have not actually appeared.
(viii) Number of students scored less than 50 is 5.
Question 10:
The weights of new born babies (in kg) in a hospital on a particular day are as follows:
2.3, 2.2, 2.1, 2.7, 2.6, 3.0, 2.5, 2.9, 2.8, 3.1, 2.5, 2.8, 2.7, 2.9, 2.4
(i) Rearrange the weights in descending order.
(ii) Determine the highest weight.
(iii) Determine the lowest weight.
(iv) Determine the range.
(v) How many babies were born on that day?
(vi) How many babies weigh below 2.5 kg?
(vii) How many babies weigh more than 2.8 kg?
(viii) How many babies weigh 2.8 kg?
Answer:
(i) The weights of the new born babies (in kg) in descending order are
3.1, 3.0, 2.9, 2.9, 2.8, 2.8, 2.7, 2.7, 2.6, 2.5, 2.4, 2.4, 2.3, 2.2, 2.1
(ii) The highest weight is 3.1 kg.
(iii) The lowest weight is 2.1 kg.
(iv) The range of the weights is
(v) On that day, the number of babies born is 15.
(vi) Number of babies’ of weight less than 2.5 kg is 4.
(vii) Number of babies’ of weight more than 2.8 kg is 4.
(viii) Number of babies’ of weight 2.8 kg is 2.
Question 11:
The number of runs scored by a cricket player in 25 innings are as follows:
26, 35, 94, 48, 82, 105, 53, 0, 39, 42, 71, 0, 64, 15, 34, 67, 0, 42, 124, 84, 54, 48, 139, 64, 47
(a) Rearrange these runs in ascending order.
(b) Determine the player, is highest score.
(c) How many times did the player not score a run?
(d) How many centuries did he score?
(e) How many times did he score more than 50 runs?
Answer:
(i) After arranging in ascending order, the given data (runs) is 0, 0, 0, 15, 26, 34, 35, 39, 42, 42, 47, 48, 48, 53, 54, 64, 64, 67, 71, 82, 84, 94, 105, 124 and 139.
(ii) The highest score of the player is 139.
(iii) The player did not score any run 3 times.
(iv) He scored 3 centuries.
(v) He scored more than 50 runs 12 times.
Question 12:
The class size of a distribution is 25 and the first class-interval is 200-224. There are seven class-intervals.
(i) Write the class-intervals.
(ii) Write the class-marks of each intervals.
Answer:
Since, there are 7 class intervals and first class is 200–224, therefore all the class intervals are:
(i) 200–224
(ii) 225–249
(iii) 250–274
(iv)275–299
(v)300–324
(vi)325–349
(viii)350–374
We know that the class mark corresponding to each class mark is given by:
Now,
Therefore, the class marks are:
212, 237, 262, 287, 312, 337 and 362
Question 13:
Write the class size and class limits in each of the following:
(i) 104, 114, 124, 134, 144, 154 and 164
(ii) 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, 97 and 102
(iii) 12.5, 17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 47.5
Answer:
(i) Here, the data is distributed uniformly. So the class size h is given by difference between any two consecutive data. So,
If a is the class mark of a class interval and h be the class size, then the lower and upper limits of the class interval are: respectively
Lower limit of first class interval is;
And, upper limit of first class interval is:
Other class limits are:
(ii) Here, the data is distributed uniformly. So the class size h is given by difference between any two consecutive data. So,
If a is the class mark of a class interval and h be the class size, then the lower and upper limits of the class interval are: respectively
Lower limit of first class interval is;
And, upper limit of first class interval is:
Other class limits are:
(iii) Here, the data is distributed uniformly. So the class size h is given by difference between any two consecutive data. So,
If a is the class mark of a class interval and h be the class size, then the lower and upper limits of the class interval are: respectively
Lower limit of first class interval is;
And, upper limit of first class interval is:
Other class limits are:
Question 14:
Following data gives the number of children in 40 families:
1, 2, 6, 5, 1, 5, 1, 3, 2, 6, 2, 3, 4, 2, 0, 0, 4, 4, 3, 2, 2, 0, 0, 1, 2, 2, 4, 3, 2, 1, 0, 5, 1, 2, 4, 3, 4, 1, 6, 2, 2.
Represent it in the form of a frequency distribution.
Answer:
Here, the maximum and minimum values of the variate are 6 and 0 respectively.
So the range = 6 – 0 = 6
Here, we will take class size 1. So we must have 6 classes each of size 1.
Therefore, the frequency distribution in which the lower limit is included and upper limit excluded is:
Question 15:
The marks scored by 40 students of class IX in mathematics are given below:
81, 55, 68, 79, 85, 43, 29, 68, 54, 73, 47, 35, 72, 64, 95, 44, 50, 77, 64, 35, 79, 52, 45, 54, 70, 83, 62, 64, 72, 92, 84, 76, 63, 43, 54, 38, 73, 68, 52, 54
Prepare a frequency distribution with class size of 10 marks.
Answer:
Here, the maximum and minimum values of the variate are 95 and 29 respectively.
So the range = 95 – 29 = 66
Here, we will take class size 10. So we must have 7 classes each of size 10
Lower limit of first class interval is;
And, upper limit of first class interval is:
Therefore, the frequency distribution in which the lower limit is included and upper limit excluded is:
Question 16:
The heights (in cm) of 30 students of class IX are given below:
155, 158, 154, 158, 160, 148, 149, 150, 153, 159, 161, 148, 157, 153, 157, 162, 159, 151, 154, 156, 152, 156, 160, 152, 147, 155, 163, 155, 157, 153
Prepare a frequency distribution table with 160-164 as one of the class intervals.
Answer:
One of the class intervals is 160–164. This means that class size is 4
Here, the maximum and minimum values of the variate are 163 and 147 respectively.
So the range = 163 – 147 = 16
Here, we will take class size 4. So we must have 5 classes each of size 4
Lower limit of first class interval is;
And, upper limit of first class interval is:
Therefore, the frequency distribution in which the lower limit is included and upper limit excluded is:
Question 17:
The monthly wages of 30 workers in a factory are given below:
83.0, 835, 890, 810, 835, 836, 869, 845, 898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 836, 878, 840, 868, 890, 806, 840, 890.
Represent the data in the form of a frequency distribution with class size 10.
Answer:
Here, the maximum and minimum values of the variate are 898 and 804 respectively.
So the range = 898 – 804 = 94
Here, we will take class size 10. So we must have 94/10 i.e. 10 classes each of size 10.
Therefore, the frequency distribution in which the lower limit is included and upper limit excluded is:
Lower limit of first class interval is;
And, upper limit of first class interval is:
Other class limits are:
Question 18:
The daily maximum temperatures (in degree celsius) recorded in a certain city during the month of November are as follows:
25.8, 24.5, 25.6, 20.7, 21.8, 20.5, 20.6, 20.9, 22.3, 22.7, 23.1, 22.8, 22.9, 21.7, 21.3, 20.5, 20.9, 23.1, 22.4, 21.5, 22.7, 22.8, 22.0, 23.9, 24.7, 22.8, 23.8, 24.6, 23.9, 21.1
Represent them as a frequency distribution table with class size 1°C.
Answer:
Here, the maximum and minimum values of the variate are 25.8 and 20.5 respectively.
So the range = 25.8 – 20.5 = 5.3
Here, we will take class size 1.
Lower limit of first class interval is;
And, upper limit of first class interval is:
Therefore, the frequency distribution in which the lower limit is included and upper limit excluded is:
Question 19:
Construct a frequency table with equal class intervals from the following data on the monthly wages (in rupees) of 28 labourers working in a factory, taking one of the class intervals as 210-230 (230 not included):
220, 268, 258, 242, 210, 268, 272, 242, 311, 290, 300, 320, 319, 304, 302, 318, 306, 292, 254, 278, 210, 240, 280, 316, 306, 215, 256, 236.
Answer:
Here, the maximum and minimum values of the variate are 320 and 210 respectively.
So the range = 320 – 210 = 110
Here, we will take class size 20 (As one class interval is 210–230).
Therefore, the frequency distribution in which the lower limit is included and upper limit excluded is:
Question 20:
The blood groups of 30 students of class VIII are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O
Represent this data in the form of a frequency distribution table. Find out which is the most common and which is the rarest blood group among these students.
Answer:
It can be observed that 9 students have their blood group as A, 6 as B, 3 as AB, and 12 as O.
Therefore, the blood group of 30 students of the class can be represented as follows.
Blood group Number of students A 9 B 6 AB 3 O 12 Total 30
It can be observed clearly that the most common blood group and the rarest blood group among these students is O and AB respectively as 12 (maximum number of students) have their blood group as O, and 3 (minimum number of students) have their blood group as AB.
Question 21:
Three coins were tossed 30 times. Each time the number of heads occurring was noted down as follow:
0 1 2 2 1 2 3 1 3 0 1 3 1 1 2 2 0 1 2 1 3 0 0 1 1 2 3 2 2 0
Prepare a frequency distribution table for the data given above.
Answer:
By observing the data given above, the required frequency distribution table can be constructed as follows.
Number of heads Number of times (frequency) 0 6 1 10 2 9 3 5 Total 30
Question 22:
Thirty children were asked about the number of hours they watched T.V. programmes in the previous week. The results were found as follows:
1 6 2 3 5 12 5 8 4 8 10 3 4 12 2 8 15 1 17 6 3 2 8 5 9 6 8 7 14 12
(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5-10.
(ii) How many children watched television for 15 or more hours a week?
Answer:
(i) Our class intervals will be 0 − 5, 5 − 10, 10 −15…..
The grouped frequency distribution table can be constructed as follows.
Hours Number of children 0 − 5 10 5 − 10 13 10 − 15 5 15 − 20 2 Total 30
(ii) The number of children who watched TV for 15 or more hours a week is 2 (i.e., the number of children in class interval 15 − 20).
Question 23:
The daily minimum temperatures in degrees Celsius recorded in a certain Arctic region are as follows:
− 12.5, −10.8, −18.6, −8.4, −10.8, −4.2, −4.8, −6.7, −13.2, −11.8, −2.3, 1.2, 2.6, 0, −2.4, 0, 3.2, 2.7, 3.4, 0, −2.4, −2.4, 0, 3.2, 2.7, 3.4, 0, −2.4, −5.8, −8.9, −14.6, −12.3, −11.5, −7.8, −2.9.
Represent them as frequency distribution table taking −19.9 to − 15 as the first class interval.
Answer:
Since the first class is –19.9 to –15
Therefore, the frequency distribution in which the lower limit is included and upper limit excluded is:
Question 1:
Define cumulative frequency distribution.
Answer:
Cumulative frequency distribution is a table which displays the manner in which cumulative frequencies are distributed over various classes.
In a grouped frequency distribution the cumulative frequency of a class is the total of all frequencies upto and including that particular class. Make sure that, for calculating cumulative frequencies, the classes should be written in ascending order.
For example, the following table gives the cumulative frequency distribution of marks scored by 55 students in a test:
Cumulative frequency distributions are of two types, namely, less than and greater than or more than. For less than cumulative frequency distributions we add up the frequencies from the above and for greater than cumulative frequencies we add up the frequencies from below.
Question 2:
Explain the difference between a frequency distribution and a cumulative frequency distribution.
Answer:
Frequency distribution displays the frequencies of the corresponding class-intervals. But, the cumulative frequency distribution displays the cumulative frequencies of the corresponding classes.
For example, the following table gives the frequency and cumulative frequency distribution of marks scored by 55 students in a test:
In a frequency distribution, the sum of all the frequencies is equal to the total number of observations. But, in the cumulative frequency distribution, the last cumulative frequency is same as the total number of observations. For example, in the above table the sum of all the frequencies is 55, which is same as the total number of students and the last cumulative frequency is 55, which is same as the total number of students.
Question 3:
The marks scored by 55 students in a test are given below:
Marks: 0-5 5-10 10-15 15-20 20-25 25-30 30-35 No. of students 2 6 13 17 11 4 2
Prepare a cumulative frequency table.
Answer:
The marks score by 55 students are given as
The cumulative frequency distribution is the following:
Question 4:
Following are the ages of 360 patients getting medical treatment in a hospital on a day:
Age (in years): 10-20 20-30 30-40 40-50 50-60 60-70 No.of Patients: 90 50 60 80 50 30
Construct a cumulative frequency distribution.
Answer:
The ages in years of 360 patients are given as
The cumulative frequency distribution is the following:
Question 5:
The water bills (in rupees) of 32 houses in a certain street for the period 1.1.98 to 31.3.98 are given below:
56, 43, 32, 38, 56, 24, 68, 85, 52, 47, 35, 58, 63, 74, 27, 84, 69, 35, 44, 75, 55, 30, 54, 65, 45, 67, 95, 72, 43, 65, 35, 59.
Tabulate the data and present the data as a cumulative frequency table using 70-79 as one of the class intervals.
Answer:
The minimum and maximum bills are 24 Rs. and 95 Rs.
The range is.
Given that 70-79 is a class-interval. So, the class size is.
Now calculate
Thus, the number of classes is 8.
The cumulative frequency distribution is the following:
Question 6:
The number of books in different shelves of a library are as follows:
30, 32, 28, 24, 20, 25, 38, 37, 40, 45, 16, 20
19, 24, 27, 30, 32, 34, 35, 42, 27, 28, 19, 34,
38, 39, 42, 29, 24, 27, 22, 29, 31, 19, 27, 25
28, 23, 24, 32, 34, 18, 27, 25, 37, 31, 24, 23,
43, 32, 28, 31, 24, 23, 26, 36, 32, 29, 28, 21
Prepare a cumulative frequencies distribution table using 45-49 as the last class-interval.
Answer:
The minimum and maximum numbers of books in shelves are 16 and 45.
The range is.
Given that 45-49 is the last class-interval. So, the class size is.
Now calculate
Thus, the number of classes is 8.
The cumulative frequency distribution is the following:
Question 7:
Given below are the cumulative frequencies showing the weights of 685 students of a school. Prepare a frequency distribution table.
Weight (in kg) No. of students Below 25 Below 30 Below 35 Below 40 Below 45 Below 50 Below 55 Below 60 Below 65 Below 70 0 24 78 183 294 408 543 621 674 685
Answer:
We make class intervals 0-25, 25-30, 30-35, 35-40, 40-45, 45-50, 50-55, 55-60, 60-65 and 65-70.
We will now find the frequencies of the different class-intervals from the cumulative frequency distribution table.
The frequency of the class 0-25 is 0.
The frequency of the class 25-30 is.
The frequency of the class 30-35 is.
The frequency of the class 35-40 is.
The frequency of the class 40-45 is.
The frequency of the class 45-50 is.
The frequency of the class 50-55 is.
The frequency of the class 55-60 is.
The frequency of the class 60-65 is.
The frequency of the class 65-70 is.
Here is the cumulative frequency distribution table.
Question 8:
The following cumulative frequency distribution table shows the daily electricity consumption (in KW) of 40 factories in an industrial state:
Consumption (in KW) No. of Factories Below 240 Below 270 Below 300 Below 330 Below 360 Below 390 Below 420 1 4 8 24 33 38 40
(i) Represent this as a frequency distribution table.
(ii) Prepare a cumulative frequency table.
Answer:
We make class intervals 0-240, 240-270, 270-300, 300-330, 330-360, 360-390 and 390-420.
(i) We will now find the frequencies of the different class-intervals.
Just proceed reverse to you proceed to generate cumulative frequency
Here is the frequency distribution table.
(ii) We are given the cumulative frequency table and we are asked to prepare new one.
So we will generate in some other fashion. Concentrate on frequency table and prepare the table. Here is the cumulative frequency distribution table.
Question 9:
Given below is a cumulative frequency distribution table showing the ages of people living in a locality:
Age in years No. of persons Above 108 Above 96 Above 84 Above 72 Above 60 Above 48 Above 36 Above 24 Above 12 Above 0 0 1 3 5 20 158 427 809 1026 1124
Prepare a frequency distribution table.
Answer:
We make class intervals 0-12, 12-24, 24-36, 36-48, 48-60, 60-72, 72-84, 84-96 and 96-108.
We will now find the frequencies of the different class-intervals from the cumulative frequency distribution table.
The frequency of the class 0-12 is 1026−1124=98
The frequency of the class 12-24 is1026−809=217
The frequency of the class 24-36 is 809−427=382
The frequency of the class 36-48 is 427−158=269
The frequency of the class 48-60 is 158−20=138
The frequency of the class 60-72 is 20−5=15
The frequency of the class 72-84 is 5−3=2
The frequency of the class 84-96 is 3−1=2
The frequency of the class 96-108 is 1−0=1
Here is the cumulative frequency distribution table.
Question 1:
Mark the correct alternative in each of the following:
Tally marks are used to find
(a) class intervals
(b) range
(c) frequency
(d) upper limits
Answer:
Tally marks are used to find the frequencies.
Hence, the correct choice is (c).
Question 2:
The difference between the highest and lowest values of the observations is called
(a) frequency
(b) mean
(c) range
(d) class-intervals
Answer:
The difference between the highest and lowest values of the observations is called the range. Hence, the correct choice is (c).
Question 3:
The difference between the upper and the lower class limits is called
(a) mid-points
(b) class size
(c) frequency
(d) mean
Answer:
The difference between the upper and the lower class limits is called the class size. Hence, the correct choice is (b).
Question 4:
In the class intervals 10-20, 20-30, 20 is taken in
(a) the interval 10-20
(b) the interval 20-30
(c) both intervals 10-20, 20-30
(d) none of the intervals
Answer:
The given class intervals are 10-20, 20-30. In these class intervals the value 20 is lies in the class interval 20-30. Hence, the correct choice is (b).
Question 5:
In a frequency distribution, the mid-value of a class is 15 and the class intervals is 4. The lower limit of the class is
(a) 10
(b) 12
(c) 13
(d) 14
Answer:
Let l and m respectively be the lower and upper limits of the class. Then the mid-value of the class is and the class-size is.
Therefore, we have two equations
Subtracting the second equation from the first equation, we have
Hence, the lower limit of the class is 13. Thus, the correct choice is (c).
Question 6:
The mid-value of a class interval is 42. If the class size is 10, then the upper and lower limits of the class are:
(a) 47 and 37
(b) 37 and 47
(c) 37.5 and 47.5
(d) 47.5 and 37.5
Answer:
Let l and m respectively be the lower and upper limits of the class. Then the mid-value of the class is and the class-size is.
Given that the mid-value of the class is 42 and the class-size is 10. Therefore, we have two equations
Adding the above two equations, we have
Substituting the value of m in the first equation, we have
Hence, the upper and lower limits of the class are 47 and 37 respectively. Thus, the correct choice is (a).
Question 7:
The number of times a particular item occurs in a given data is called its
(a) variation
(b) frequency
(c) cumulative frequency
(d) class-size
Answer:
The number of times a particular item occurs in a given data is called the frequency of the item. Hence, the correct choice is (b).
Question 8:
The width of each of nine classes in a frequency distribution is 2.5 and the lower class boundary of the lowest class 10.6. Then the upper class boundary of the highest class is
(a) 35.6
(b) 33.1
(c) 30.6
(d) 28.1
Answer:
The number of classes is 9 and the uniform class size is 2.5. The lower limit of the lower class (first class) is 10.6. Therefore, the upper limit of the last class is
Hence, the correct choice is (b).
Question 9:
The following marks were obtained by the students in a test:
81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85, 79, 62
The range of the marks is
(a) 9
(b) 17
(c) 27
(d) 33
Answer:
The marks obtained by the students are 81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85, 79 and 62.
The highest and lowest marks are 95 and 62 respectively. Therefore, the range of marks is
Hence, the correct option is (d).
Question 10:
Tallys are usually marked in a bunch of
(a) 3
(b) 4
(c) 5
(d) 6
Answer:
Tallies are usually marked in a bunch of 4. Hence, the correct option is (b)
Question 11:
Let l be the lower class limit of a class-interval in a frequency distribution and m be the mid point of the class. Then, the upper class limit of the class is
(a) m+$\frac{l+m}{2}$
(b) l+$\frac{m+l}{2}$
(c) 2m − 1
(d) m − 2l
Answer:
Given that, the lower class limit of a class-interval is l and the mid-point of the class is m. Let u be the upper class limit of the class-interval. Therefore, we have
Thus the upper class limit of the class is. Hence, the correct choice is (c).
View NCERT Solutions for all chapters of Class 9 |
# Calculating the Cross-Sectional Area of a Pipe
## We use the inside diameter measurement of a pipe to calculate its cross-sectional area.
In Lesson 3, we learned how to calculate the area of a circle. Now, we'll use that equation to calculate the cross-sectional area of a pipe.
As you know, A = 0.785 d2
Remember that pipes have both inside and outside diameters, which vary by material, pipe wall thickness, and size. To solve the cross-sectional area of a pipe, you'll need the inside diameter measurement.
Then, use the following equation:
A = 0.785 di2
where: A = cross-sectional area of pipe (this is usually expressed in inches)
and di = inside diameter (inches)
Once we know the cross-sectional area, we can calculate the volume and flow of water in a pipe.
# More information related to Calculating the Cross-Sectional Area of a Pipe:
The area of any shape is the amount of space it takes up in one plane of geometry. In simpler terms, the area can be thought of as the footprint left by a shape. Sometimes, as is the case with rectangular shapes, the area is calculated by multiplying the length of the shape by its width. A football field is 120 yards long (including the endzones) by approximately 53 yards wide. The area of a football field is 120 yards x 53 yards = 6,360 square yards. Notice that the units are also multiplied by each other (yards x yards = square yards).
Calculating the area of a triangular object takes our rectangular equation just one step further. Imagine drawing a line from one corner of that same football field to the opposite corner. The two shapes you have created are triangles, each one half of the total area. So, to calculate the area of a triangle, we simply take our same equation for rectangles (Length x Width) and multiply it by 1/2. Just to keep things straight, we often use the word "base" to describe the "width" and "height" to describe "length" when discussing triangles. So, the area of a triangle is: (1/2 base x height).
The area of a circle is a formula that takes a bit of memory. You may recall that the radius is the distance from the center of a circle to its outer edge, which is also the same as 1/2 of the diameter. The area of a circle is calculated as: (Pi x radius2). For example, if we have a circle that has a diameter of 20ft, we would figure out the area by first calculating the radius (1/2 x diameter = 10 feet) and then plugging it into our equation. In this case, the area is Pi (estimated as 3.14) x 10 feet x 10 feet = 314 square feet.
# The Calculating the Cross-Sectional Area of a Pipe slide is featured in the following lessons:
### Water Mains, Pipes & Valves
Duration: minutes
Level: |
# Write a two-way table of observed counts for sex and whether a participant in soccer had a lower-extremity injury or not. Determine a two-way table of expected counts for these data. Show calculations verifying that the value of the chi-square statistic is 2.51.
Write a two-way table of observed counts for sex and whether a participant in soccer had a lower-extremity injury or not. Determine a two-way table of expected counts for these data. Show calculations verifying that the value of the chi-square statistic is 2.51.
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a) Here is the table of observed counts:
$\begin{array}{cccc}\text{Gender / Had injury?}& \text{Yes}& \text{No}& \text{Total}\\ \text{Male}& 153& 1514& 1667\\ \text{Female}& 74& 925& 999\\ \text{Total}& 227& 2439& 2666\end{array}$
The numbers in the ”No” column are simply differences between last column and first column , and the numbers in the last row are sums of columns
b)The expected count for each cell is calculated as
$\text{Expected}=\frac{\text{Total row count}\cdot \text{Total column count}}{\text{Sample size}}$
Here is the table of the expected counts:
$\begin{array}{cccc}\text{Gender / Had injury?}& \text{Yes}& \text{No}& \text{Total}\\ \text{Male}& \frac{1667\cdot 227}{2666}=141.94& \frac{1667\cdot 2439}{2666}=1525.06& 1667\\ \text{Female}& \frac{999\cdot 227}{2666}=85.06& \frac{999\cdot 2439}{2666}=913.94& 999\\ \text{Total}& 227& 2439& 2666\end{array}$
c) The value of chi-square statistics is
${x}^{2}=\sum _{\text{all cells}}\frac{\left(\text{Observed count}-\text{Expected count}{\right)}^{2}}{\text{Expected count}}$
$=\frac{\left(153-141.94{\right)}^{2}}{141.94}+\frac{\left(1514-1525.06{\right)}^{2}}{1525.06}+\frac{\left(74-85.06{\right)}^{2}}{85.06}+\frac{\left(925-913.94{\right)}^{2}}{913.94}$
$=0.8618+0.0802+1.4381+0.1338$
$=2.5139$ |
# Data Interpretation Questions
In mathematics, Data Interpretation refers to analyze, interpret, and make decisions out of a collection of data that has been processed. It is usually presented in various forms like tabular form, bar graphs, line charts, and other similar forms and hence requires an interpretation of some type. In this tutorial, we will learn about data interpretation with the help of some major techniques and examples.
These are the given type of questions based on Data Interpretation
### Questions based on Tabular Form
1. Read the following table carefully and answer the questions.
The given tables show the total population of 6 different cities and the ratio of literate to illiterate population and the percentage of graduates out of the literate population in each city.
Cities Population in thousand Literate to Illiterate Percentage of graduate out of literate
P 24 8:4 10
Q 30 4:6 20
R 36 6:6 25
S 40 7:3 40
T 20 3:2 60
Question 1) Graduate population of city Q and R together is approximately what percent more or less than the graduate population of city P and T together?
Explanation:
Graduate population of the city P and T together
Question 2) The population of city R who are literate but not graduate is how much more than the average graduate population of city S and T together?
Explanation:
The population who are literate but not graduate of city R
So, the required difference = 36000 - 9200 = 28800
Question 3) If the ratio of illiterate males to females in city q is 3:2 and the ratio of graduate males to the female population in city S is 4: 6. Find the ratio of total illiterate male in city Q and graduate female in city S
Explanation:
Illiterate male in city Q
2. The table below shows the number of laptops sold by the 5 sellers and the ratio of Lenovo laptops sold to Sony laptops sold. Read the data carefully and answer the given questions.
Seller Total Laptops Sold Lenovo sold: Sony Sold
P 1800 6 : 3
Q 3000 6: 4
R 3200 3: 5
S 2500 2 : 3
T 1600 4: 4
Question 1) The Lenovo laptop sold by P seller is what percent less than Lenovo laptop sold by Q seller.
Explanation:
Question 2) Find the average number of Sony laptop sold by seller Q, R, and S all together?
Explanation:
Required Average =
Question 3) Find the ratio of Lenovo laptop sold by 'R' to Lenovo laptop sold by 'T '?
Question 4) Sony laptop sold by S and T together is how much less than Sony laptop sold by Q and R together?
Explanation:
Sony laptop sold by S and T together
= 1500 + 800 = 2300
Sony laptop sold by Q and R together
= 1200 + 2000 = 3200
Required difference = 3200 - 2300 = 900
### Questions based on Missing number
1. The table given below shows the number of products sold by 5 shopkeepers on five different days.
Days / Person P Q R S T
Monday 360 220 280 - 290
Tuesday 240 - 350 270 440
Wednesday 580 550 290 460 -
Thursday 380 660 - 680 130
Friday - 420 150 550 300
#### Note: Some data are missing in the above table; calculate the missing data if required.
Question 1) If the total product sold by P and Q, including all five days, is 1800 and 2000 respectively, the product sold by Q on Tuesday is what percent more/less than the product sold by P on Friday?
Explanation:
Product sold by P on Friday = 1800 - (360 + 240 + 580 + 380)
= 1800 - 1560
= 240
Product sold by Q on Tuesday = 3200 - (220 + 550 + 660 + 420)
= 2000 - 1850
= 150
So, the required percentage
Question 2) If the ratio of products sold by Q and R together on Thursday to products sold by R and S together on the same day is 3: 1, find the product sold by R on Thursday?
Explanation:
Let item sold by R on Thursday be a.
According to question
660+a = 3a + 390
2a = 270
a = 135
Question 3) What is the difference in the product sold by P on Monday and Friday. If total products sold by P is 2000?
Explanation:
Product sold by P on Friday = 2000 - 1560 = 440
So, the required difference = 440 - 360 = 80
Question 4) If the average of products sold by S on Monday and Tuesday is 260, then the products sold by S on Monday is what percent of the product sold by T on Friday?
Explanation:
Products sold by S on Monday = 2 260 - 270
= 520 - 270 = 250
Required percentage =
2. The table below shows the population of five cities and the percentage of male, female, and transgender people.
Cities Total Male Female Transgender
P 10000 40% - 32%
Q 12000 - 35% 45 %
R 14000 33 % 37% -
S 16000 30 % - 40%
T 18000 - 36% 44
Question 1) Total number of females in village 'Q' and 'S' together is what percent more than the total number of males in village 'R'?
Explanation:
Total number of females in village 'Q' and 'S' together
= 4500 + 4800
= 9300
The total number of males in village 'R'
Question 2) What will be the difference between the number of males and females in Village 'P'?
Explanation:
Required difference
Question 3) Find the difference between female in village 'S' to males in village 'T' ?
Explanation:
= 4800 - 1600
= 3200
Question 4) Transgender population in village 'P' is what percent of the female population in Village 'Q' ?
Explanation:
Required %
= 45 %
### Questions based on Bar Graph
The bar graph is given below shows the per gram rate of Carbon and Oxygen on various days. Read the data carefully and answer the given question.
Question 1) Peter buy 5 grams of Oxygen on Tuesday and 8 grams of Carbon on Wednesday. Find the total amount paid by him?
Explanation:
Required amount
= 2700 5 + 2400 8
= 13500 + 19200
= 32700
Question 2) Average per gram price of Oxygen on all four days together is how much more than the average per gram price of Carbon on all four days together?
Explanation:
Average per gram price of Oxygen on all the four days
= 2475
Average per gram price of Carbon on all the four days
= 2325
So, the required difference = 2475 - 2325 = 150
Question 3) On Friday, per gram price of both Carbon and Oxygen increased by 20% and 25%, respectively, compared to the price on Thursday. Find the sum of peer gram price of Carbon and Oxygen on Friday?
Explanation:
Required amount
Question 4) What is the ratio of per gram price of Carbon and Oxygen together on Monday to that on Thursday?
Explanation:
Required ratio
Question 5) Jhon bought 4 grams of both Carbon and Oxygen on Wednesday and sold it on Thursday at the given price. Find the total amount lose by him?
Explanation:
= 4 [2600 - 2400 +2200 - 2500]
= 4 [-100]
= 400
2. A Company delivers three different items, namely, food, drinks, and Cosmetics. If the company's total production was the same for all years and % production of three products in the particular year given below, answer the given question.
Question 1) In 2010, the number of food items produced by the company is what percent more or less than cosmetic items produced in the year 2013?
Explanation:
Let's suppose the total production of the company to be K
So, the required percent =
= 39 % less
Question 2) If production in the year 2014 was 1,40,000. Find the difference between the number of food items produced in 2014 and drink items produced in 2011?
Explanation:
Required difference
= 70000 - 28000 = 42000
Question 2) Find the ratio between the number of cosmetic items produced in 2014 and the number of food products produced in 2010.
Explanation:
Let's suppose the total production of the company to be K
So, the required ratio =
Question 3) The difference between food items and drink items produced by the company in 2011 is 20000. Find the average of food and cosmetic items produced by the company in 2010?
Explanation:
Let's suppose total production of the company be K
According to question,
20% of k = 20000
K = 100000
So, the required average =
= 20000 + 50000 / 2
= 70000/2
= 35000
Question 4) Find the total production in 2015 if there is an increase of 20% in total production in 2018 compared to the previous year given that the number of drink items produced in 2012 was 16000?
Explanation:
Let's suppose the production of each previous years be K
According to the question,
40% of k = 16000
K = 40000
So, the total production in 2015
= 48000
### Caselet Data Interpretation
1. The total number of TV sold by a store in January and March is 1800, and the total number of TV sold in January is 40% more than that of in March. The store sold TV of only 4 brands that are LG, Sony, Samsung, and MI. 20% of total TV sold by the store is Sony, and the total number of Sony TV sold by a store in March is 25% less than that of Sony TV sold by a store in January. The number of MI TVs sold by a store in January is 30% more than that of the total Sony TV sold by the store in January. The total number of Samsung TVs sold by a store in January is 20% less than the total LG TV sold by a store in March is 20% more than that of total LG TV sold by a store in January. The total number Of MI and Samsung TVs sold in March is 7:4.
Question 1)
Total Samsung TV sold by a store in January is what percent more than total MI sold by a store in March?
Explanation:
Let total number of LG TV sold in month of January = k
So, the total number of Samsung TV sold in moth Of January = 0.8k
According to question,
K + 0.8k = (1050- 182-236)
= 1.8k = 632
= k = 351
The total number of LG TV sold in January = 351
The total number of Samsung TV sold in January = 0.8 × 351 = 280
Brands January March
Sony 182 137
MI 236 109
Samsung 280 336
LG 351 421
Total 1049 1003
= 156%
Question 2) Find the ratio between total Sony TV sold by a store in January to total MI TV sold by a store in March?
Explanation:
Let total number of LG TV sold in month of January = k
So, the total number of Samsung TV sold in moth Of January = 0.8k
According to question,
K + 0.8k = (1050- 182-236)
= 1.8k = 632
= k = 351
The total number of LG TV sold in January = 351
The total number of Samsung TV sold in January = 0.8 × 351 = 280
Brands January March
Sony 182 137
MI 236 109
Samsung 280 336
LG 351 421
Total 1049 1003
Required ratio = 182/109
Question 3) Find the average number of MI, Samsung, and Sony TV sold by a store in March?
Explanation:
Let total number of LG TV sold in month of January = k
So, the total number of Samsung TV sold in moth Of January = 0.8k
According to question,
K + 0.8k = (1050- 182-236)
= 1.8k = 632
= k = 351
The total number of LG TV sold in January = 351
The total number of Samsung TV sold in January = 0.8 × 351 = 280
Brands January March
Sony 182 137
MI 236 109
Samsung 280 336
LG 351 421
Total 1049 1003
Question 4) Total Samsung TV sold by a store in January is what percent less than total LG TV sold by a store in March?
Explanation:
Let total number of LG TV sold in month of January = k
So, the total number of Samsung TV sold in moth Of January = 0.8k
According to question,
K + 0.8k = (1050- 182-236)
= 1.8k = 632
= k = 351
The total number of LG TV sold in January = 351
The total number of Samsung TV sold in January = 0.8 × 351 = 280
Brands January March
Sony 182 137
MI 236 109
Samsung 280 336
LG 351 421
Total 1049 1003
Required percentage
Question 5) Find the difference between the total LG TV sold by a store in both months and the total Samsung TV sold by a store in both months?
Explanation:
Let total number of LG TV sold in month of January = k
So, the total number of Samsung TV sold in moth Of January = 0.8k
According to question,
K + 0.8k = (1050- 182-236)
= 1.8k = 632
= k = 351
The total number of LG TV sold in January = 351
The total number of Samsung TV sold in January = 0.8 351 = 280
Brands January March
Sony 182 137
MI 236 109
Samsung 280 336
LG 351 421
Total 1049 1003
Required difference = (351 + 421) - (280+336)
= 772 - 616 = 156
2. On a particular day, the total number of people who went to a club for the party is 75% of the total capacity of the club. There is a total four-bar available there, and each has an equal capacity of 400. 25% of people have drunk in bar 1 and 20% remaining drunk in bar 4. The number of people who have drunk in bar 2 is 40% more than of bar 3. The number of males and females who have drunk on that particular day is 3: 5.
Question 1)
If in bar 1 the ratio of the number of males to that of female was 4:3, and in bar 4, the ratio of female to male was 9:7, find the total number of females who have drunk in these two bars is what percent of the total number of females who have drunk on that particular day?
Explanation:
Total capacity of the club = 400 × 4 = 1600
Let's the number of people who have drunk in bar 3 be 100x then number of people in bar 2 be 140x
According to question,
100x + 140x = (1200 - 300 - 180) = 720
So, the total number of people in bar 3 = 100 × 3 = 300 and
the total number of people in bar 2 =
Question 2)
Find the difference between the total number of males who have drunk on that day and the number of people who have drunk in bar 2 and bar 4 together?
Explanation:
Total capacity of the club = 400 × 4 = 1600
Let's the number of people who have drunk in bar 3 be 100x then number of people in bar 2 be 140x
According to question,
100x + 140x = (1200 - 300 - 180) = 720
So, the total number of people in bar 3 = 100 × 3= 300 and
the total number of people in bar 2 = 140 × 3 = 420
Required difference = (420 + 180) - 450
= 600 - 450 = 150
Question 3)
Find the ratio of the average number of people who have drunk in bar 1 and bar 3 to the average of the people who have drunk in bar 2 and bar 4.
Explanation:
Total capacity of the club = 400 × 4 = 1600
Let's the number of people who have drunk in bar 3 be 100x then number of people in bar 2 be 140x
According to question,
100x + 140x = (1200 - 300 - 180) = 720
Question 4) The number of people who have drunk in bar 3 is what percent of the total capacities of Club?
Explanation:
Total capacity of the club = 400 × 4 = 1600
Let's the number of people who have drunk in bar 3 be 100x then number of people in bar 2 be 140x
According to question,
100x + 140x = (1200 - 300 - 180) = 720
Question 5)
The cost of an entry ticket is Rs. 350, but there was a discount of 20% for females and 10% for male people on that day. Find the total income earned by the club on that day?
Explanation:
Total capacity of the club = 400 × 4 = 1600
Let's the number of people who have drunk in bar 3 be 100x then number of people in bar 2 be 140x
According to question,
100x + 140x = (1200 - 300 - 180) = 720
So, the total number of people in bar 3 = 100× 3= 300 and
the total number of people in bar 2 = 140 × 3= 420
On that day,
Total income earned by the club on that day
=(750×280 + 450×315)
= (210000 + 141750)
= Rs. 351,750
3. The total population of Noida sector 3 is 30000, and there are three sectors: P, Q, and R. The ratio of the population of their sectors (P:Q: R) is 7:5:3. The number of males in sector P is 5000 more than that of females in sector R. Ratio of numbers of females in sector P to that if males in sector R is 7:3. The number of females in sector Q is 60% of the number of males in sector P.
Question 1) Find the average number of males in sector P and Q is how much more and less than the number of females in sector P?
Explanation:
Let number of females in sector P and number of males in sector R be 7x and 3x respectively
Number of males in sector P = (14000 - 7x)
Number of females in sector R = (6000 - 3x)
According to question,
(14000 - 7x) - (6000 - 3x) = 5000
=14000 -7x -6000 + 3x = 5000
= 8000 -4x = 5000
X = 750
Number of females in sector P = 7 ×750= 5230
Number of males in sector P = 14000 = 8770
Number of males in sector R = 3×750= 2250
Number of females in sector R = 6000 - 2250 = 3750
Sector Males Females Totals
P 8770 5230 14000
Q 4738 5262 10000
R 2250 3750 6000
= 6754 - 5230 = 1524
Question 2)
The total male population of P and R sectors is what percent of the total population of these two sectors?
Explanation:
Let number of females in sector P and number of males in sector R be 7x and 3x respectively
Number of males in sector P = (14000 - 7x)
Number of females in sector R = (6000 - 3x)
According to question,
(14000 - 7x) - (6000 - 3x) = 5000
=14000 -7x -6000 + 3x = 5000
= 8000 -4x = 5000
X = 750
Number of females in sector P = 7 ×750 = 5230
Number of males in sector P = 14000 -5230 = 8770
Number of males in sector R = 3×750 = 2250
Number of females in sector R = 6000 - 2250 = 3750
Sector Males Females Totals
P 8770 5230 14000
Q 4738 5262 10000
R 2250 3750 6000
Question 3)
Find the ratio of the total number of females in all sectors to that of the total male population in all sectors?
Explanation:
Let number of females in sector P and number of males in sector R be 7x and 3x respectively
Number of males in sector P = (14000 - 7x)
Number of females in sector R = (6000 - 3x)
According to question,
(14000 - 7x) - (6000 - 3x) = 5000
=14000 -7x -6000 + 3x = 5000
= 8000 -4x = 5000
X = 750
Number of females in sector P = 7 ×750 = 5230
Number of males in sector P = 14000 -5230 = 8770
Number of males in sector R = 3×750 = 2250
Number of females in sector R = 6000 - 2250 = 3750
Sector Males Females Totals
P 8770 5230 14000
Q 4738 5262 10000
R 2250 3750 6000
Question 4) The average male population of sectors Q and R is how much more or less than the average female population of these two sectors.
Explanation:
Let number of females in sector P and number of males in sector R be 7x and 3x respectively
Number of males in sector P = (14000 - 7x)
Number of females in sector R = (6000 - 3x)
According to question,
(14000 - 7x) - (6000 - 3x) = 5000
=14000 -7x -6000 + 3x = 5000
= 8000 -4x = 5000
X = 750
Number of females in sector P = 7 ×750 = 5230
Number of males in sector P = 14000 -5230 = 8770
Number of males in sector R = 3×750 = 2250
Number of females in sector R = 6000 - 2250 = 3750
Sector Males Females Totals
P 8770 5230 14000
Q 4738 5262 10000
R 2250 3750 6000
Question 5) 70% and 80% of the female sector P and sector Q respectively are working. Find the working females of Q is how much percent more or less than that of working females of A.
Explanation:
Let number of females in sector P and number of males in sector R be 7x and 3x respectively
Number of males in sector P = (14000 - 7x)
Number of females in sector R = (6000 - 3x)
According to question,
(14000 - 7x) - (6000 - 3x) = 5000
=14000 -7x -6000 + 3x = 5000
= 8000 -4x = 5000
X = 750
Number of females in sector P = 7 ×750 = 5230
Number of males in sector P = 14000 -5230 = 8770
Number of males in sector R = 3×750 = 2250
Number of females in sector R = 6000 - 2250 = 3750
Sector Males Females Totals
P 8770 5230 14000
Q 4738 5262 10000
R 2250 3750 6000
### Questions based on Bar Graph
1. The pie chart given below shows the percentage distribution of five village populations in Two different years (2014 and 2018). Read the given data carefully and answer the questions.
Question 1) Find increment percentage of the population of village T in the year 2014 over the year 2018?
Question 2) Find the ratio between total population of village Q in year 2014 to total population of village Q and R together in the year 2018?
Explanation:
Question 3) Find the difference between population of village P in the year 2018 and Village P in year 2014?
Explanation:
= 600 - 256 = 344
Question 4) Find the average number of populations of village R in both given years?
:
Question 5)
Find the decrement percentage in the population of village S in the year 2014 over the year 2018?
Explanation:
2. The pie chart shown below shows the total income of Jhon in six different months and the percentage distribution in these months. Read the data carefully and answer the given questions.
Question 1) Income of john in March and July together is what percent less than the income of john in August and December together?
Explanation:
Given,
Income of john in the month of March and July together = 12% + 22% = 34%
Income of john in the month of August and December together = 18% + 18% = 36 %
Question 2) Income of John in July and December together is how much more than the income of john in January and March together?
Explanation:
Income of John in July and December together = 22% + 18% = 40%
income of john in January and March together = 16% + 12% = 28%
Question 3) Which month shows the highest percent increment in income as compared to the previous month?
Explanation:
We can see in the pie chart; the July month shows the highest percent increase in income as compared to the previous month, which is equal to
Question 4) Income in November and December together makes how much central angle of the total?
Explanation:
Income in the month of November and December together = 18% + 14 % = 32%
Question 5) Jhon's average income in starting four months (January, March, July, August) in the given six months (January, March, July, August, November, December) is how much less than Jhon's average income in the last four months in the given six months?
Explanation:
Jhon's average income in starting four months (January, March, July, August) =
= 2720
Jhon's average income in starting four months (July, August, November, December)
= 2880
So, the required difference = 2880 - 2720 = 160
### Questions based on Line Graph
1. Read the given line graph carefully and answer the questions. The Line graph shows the percentage of AC sold by six shopkeepers.
Question 1) AC's sold by shopkeeper Q and U together is how much more than AC's sold by shopkeeper P and S together?
Explanation:
AC's sold by shopkeeper Q and U together = (15 + 35)% = 50%
AC's sold by shopkeeper P and S together = (10 + 5)% = 15%
Question 2) AC's sold by shopkeeper T and V together is how much percent more than AC's sold by shopkeeper R and S together?
Explanation:
AC's sold by shopkeeper T and V together = (25+40)% = 65%
than AC's sold by shopkeeper R and S together = (20 + 5)% = 25%
Question 3) V sold only three types of AC's that are W, X, and Y in the ratio 5:6:7. Find the difference between AC's sold by V of type W and Y together and type X?
Explanation:
Total AC's sold by shopkeeper V
Question 4) What is the ratio of the average of AC's sold by shopkeepers T, R, and P together to average AC's sold by shopkeepers P and V together?
Explanation:
AC's sold by shopkeeper T, R and P together = 15+20+10 = 45/3 = 15
AC's sold by shopkeeper P and V together = 40 + 10 = 50/2 = 25
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# Finite set explained
In mathematics, particularly set theory, a finite set is a set that has a finite number of elements. Informally, a finite set is a set which one could in principle count and finish counting. For example,
\{2,4,6,8,10\}
is a finite set with five elements. The number of elements of a finite set is a natural number (possibly zero) and is called the cardinality (or the cardinal number) of the set. A set that is not a finite set is called an infinite set. For example, the set of all positive integers is infinite:
\{1,2,3,\ldots\}.
Finite sets are particularly important in combinatorics, the mathematical study of counting. Many arguments involving finite sets rely on the pigeonhole principle, which states that there cannot exist an injective function from a larger finite set to a smaller finite set.
## Definition and terminology
Formally, a set is called finite if there exists a bijection
f\colonS\to\{1,\ldots,n\}
for some natural number . The number is the set's cardinality, denoted as . The empty set or ∅ is considered finite, with cardinality zero.
If a set is finite, its elements may be written — in many ways — in a sequence:
x1,x2,\ldots,xn(xi\inS, 1\lei\len).
In combinatorics, a finite set with elements is sometimes called an -set and a subset with elements is called a -subset. For example, the set is a 3-set – a finite set with three elements – and is a 2-subset of it.
(Those familiar with the definition of the natural numbers themselves as conventional in set theory, the so-called von Neumann construction, may prefer to use the existence of the bijection
f\colonS\ton
, which is equivalent.)
## Basic properties
Any proper subset of a finite set S is finite and has fewer elements than S itself. As a consequence, there cannot exist a bijection between a finite set S and a proper subset of S. Any set with this property is called Dedekind-finite. Using the standard ZFC axioms for set theory, every Dedekind-finite set is also finite, but this implication cannot be proved in ZF (Zermelo–Fraenkel axioms without the axiom of choice) alone.The axiom of countable choice, a weak version of the axiom of choice, is sufficient to prove this equivalence.
Any injective function between two finite sets of the same cardinality is also a surjective function (a surjection). Similarly, any surjection between two finite sets of the same cardinality is also an injection.
The union of two finite sets is finite, with
|S\cupT|\le|S|+|T|.
In fact, by the inclusion–exclusion principle:
|S\cupT|=|S|+|T|-|S\capT|.
More generally, the union of any finite number of finite sets is finite. The Cartesian product of finite sets is also finite, with:
|S x T|=|S| x |T|.
Similarly, the Cartesian product of finitely many finite sets is finite. A finite set with n elements has 2 distinct subsets. That is, the power set P(S) of a finite set S is finite, with cardinality 2.
Any subset of a finite set is finite. The set of values of a function when applied to elements of a finite set is finite.
All finite sets are countable, but not all countable sets are finite. (Some authors, however, use "countable" to mean "countably infinite", so do not consider finite sets to be countable.)
The free semilattice over a finite set is the set of its non-empty subsets, with the join operation being given by set union.
## Necessary and sufficient conditions for finiteness
In Zermelo–Fraenkel set theory without the axiom of choice (ZF), the following conditions are all equivalent:[1]
1. S is a finite set. That is, S can be placed into a one-to-one correspondence with the set of those natural numbers less than some specific natural number.
2. (Kazimierz Kuratowski) S has all properties which can be proved by mathematical induction beginning with the empty set and adding one new element at a time. (See below for the set-theoretical formulation of Kuratowski finiteness.)
3. (Paul Stäckel) S can be given a total ordering which is well-ordered both forwards and backwards. That is, every non-empty subset of S has both a least and a greatest element in the subset.
4. Every one-to-one function from P(P(S)) into itself is onto. That is, the powerset of the powerset of S is Dedekind-finite (see below).[2]
5. Every surjective function from P(P(S)) onto itself is one-to-one.
6. (Alfred Tarski) Every non-empty family of subsets of S has a minimal element with respect to inclusion.[3] (Equivalently, every non-empty family of subsets of S has a maximal element with respect to inclusion.)
7. S can be well-ordered and any two well-orderings on it are order isomorphic. In other words, the well-orderings on S have exactly one order type.
If the axiom of choice is also assumed (the axiom of countable choice is sufficient[4]), then the following conditions are all equivalent:
1. S is a finite set.
2. (Richard Dedekind) Every one-to-one function from S into itself is onto.
3. Every surjective function from S onto itself is one-to-one.
4. S is empty or every partial ordering of S contains a maximal element.
## Foundational issues
Georg Cantor initiated his theory of sets in order to provide a mathematical treatment of infinite sets. Thus the distinction between the finite and the infinite lies at the core of set theory. Certain foundationalists, the strict finitists, reject the existence of infinite sets and thus recommend a mathematics based solely on finite sets. Mainstream mathematicians consider strict finitism too confining, but acknowledge its relative consistency: the universe of hereditarily finite sets constitutes a model of Zermelo–Fraenkel set theory with the axiom of infinity replaced by its negation.
Even for the majority of mathematicians that embrace infinite sets, in certain important contexts, the formal distinction between the finite and the infinite can remain a delicate matter. The difficulty stems from Gödel's incompleteness theorems. One can interpret the theory of hereditarily finite sets within Peano arithmetic (and certainly also vice versa), so the incompleteness of the theory of Peano arithmetic implies that of the theory of hereditarily finite sets. In particular, there exists a plethora of so-called non-standard models of both theories. A seeming paradox is that there are non-standard models of the theory of hereditarily finite sets which contain infinite sets, but these infinite sets look finite from within the model. (This can happen when the model lacks the sets or functions necessary to witness the infinitude of these sets.) On account of the incompleteness theorems, no first-order predicate, nor even any recursive scheme of first-order predicates, can characterize the standard part of all such models. So, at least from the point of view of first-order logic, one can only hope to describe finiteness approximately.
More generally, informal notions like set, and particularly finite set, may receive interpretations across a range of formal systems varying in their axiomatics and logical apparatus. The best known axiomatic set theories include Zermelo-Fraenkel set theory (ZF), Zermelo-Fraenkel set theory with the Axiom of Choice (ZFC), Von Neumann–Bernays–Gödel set theory (NBG), Non-well-founded set theory, Bertrand Russell's Type theory and all the theories of their various models. One may also choose among classical first-order logic, various higher-order logics and intuitionistic logic.
A formalist might see the meaning of set varying from system to system. Some kinds of Platonists might view particular formal systems as approximating an underlying reality.
## Set-theoretic definitions of finiteness
In contexts where the notion of natural number sits logically prior to any notion of set, one can define a set S as finite if S admits a bijection to some set of natural numbers of the form
\{x|x<n\}
. Mathematicians more typically choose to ground notions of number in set theory, for example they might model natural numbers by the order types of finite well-ordered sets. Such an approach requires a structural definition of finiteness that does not depend on natural numbers.
Various properties that single out the finite sets among all sets in the theory ZFC turn out logically inequivalent in weaker systems such as ZF or intuitionistic set theories. Two definitions feature prominently in the literature, one due to Richard Dedekind, the other to Kazimierz Kuratowski. (Kuratowski's is the definition used above.)
A set S is called Dedekind infinite if there exists an injective, non-surjective function
f:SS
. Such a function exhibits a bijection between S and a proper subset of S, namely the image of f. Given a Dedekind infinite set S, a function f, and an element x that is not in the image of f, we can form an infinite sequence of distinct elements of S, namely
x,f(x),f(f(x)),...
. Conversely, given a sequence in S consisting of distinct elements
x1,x2,x3,...
, we can define a function f such that on elements in the sequence
f(xi)=xi+1
and f behaves like the identity function otherwise. Thus Dedekind infinite sets contain subsets that correspond bijectively with the natural numbers. Dedekind finite naturally means that every injective self-map is also surjective.
Kuratowski finiteness is defined as follows. Given any set S, the binary operation of union endows the powerset P(S) with the structure of a semilattice. Writing K(S) for the sub-semilattice generated by the empty set and the singletons, call set S Kuratowski finite if S itself belongs to K(S).[5] Intuitively, K(S) consists of the finite subsets of S. Crucially, one does not need induction, recursion or a definition of natural numbers to define generated by since one may obtain K(S) simply by taking the intersection of all sub-semilattices containing the empty set and the singletons.
Readers unfamiliar with semilattices and other notions of abstract algebra may prefer an entirely elementary formulation. Kuratowski finite means S lies in the set K(S), constructed as follows. Write M for the set of all subsets X of P(S) such that:
• X contains the empty set;
• For every set T in P(S), if X contains T then X also contains the union of T with any singleton.
Then K(S) may be defined as the intersection of M.
In ZF, Kuratowski finite implies Dedekind finite, but not vice versa. In the parlance of a popular pedagogical formulation, when the axiom of choice fails badly, one may have an infinite family of socks with no way to choose one sock from more than finitely many of the pairs. That would make the set of such socks Dedekind finite: there can be no infinite sequence of socks, because such a sequence would allow a choice of one sock for infinitely many pairs by choosing the first sock in the sequence. However, Kuratowski finiteness would fail for the same set of socks.
### Other concepts of finiteness
In ZF set theory without the axiom of choice, the following concepts of finiteness for a set S are distinct. They are arranged in strictly decreasing order of strength, i.e. if a set S meets a criterion in the list then it meets all of the following criteria. In the absence of the axiom of choice the reverse implications are all unprovable, but if the axiom of choice is assumed then all of these concepts are equivalent.[6] (Note that none of these definitions need the set of finite ordinal numbers to be defined first; they are all pure "set-theoretic" definitions in terms of the equality and membership relations, not involving ω.)
• I-finite. Every non-empty set of subsets of S has a ⊆-maximal element. (This is equivalent to requiring the existence of a ⊆-minimal element. It is also equivalent to the standard numerical concept of finiteness.)
• Ia-finite. For every partition of S into two sets, at least one of the two sets is I-finite. (A set with this property which is not I-finite is called an amorphous set.)
• II-finite. Every non-empty ⊆-monotone set of subsets of S has a ⊆-maximal element.
• III-finite. The power set P(S) is Dedekind finite.
• IV-finite. S is Dedekind finite.
• V-finite. ∣S∣ = 0 or 2 ⋅ ∣S∣ > ∣S|.
• VI-finite. ∣S∣ = 0 or ∣S∣ = 1 or ∣S2 > ∣S∣.
• VII-finite. S is I-finite or not well-orderable.
The forward implications (from strong to weak) are theorems within ZF. Counter-examples to the reverse implications (from weak to strong) in ZF with urelements are found using model theory.[7]
Most of these finiteness definitions and their names are attributed to by . However, definitions I, II, III, IV and V were presented in, together with proofs (or references to proofs) for the forward implications. At that time, model theory was not sufficiently advanced to find the counter-examples.
Each of the properties I-finite thru IV-finite is a notion of smallness in the sense that any subset of a set with such a property will also have the property. This is not true for V-finite thru VII-finite because they may have countably infinite subsets.
## References
• Tarski . Alfred . Alfred Tarski . 1954 . Theorems on the existence of successors of cardinals, and the axiom of choice. Nederl. Akad. Wetensch. Proc. Ser. A., Indagationes Math. . 16. 26–32. 10.1016/S1385-7258(54)50005-3 . 0060555.
• Book: Whitehead. Alfred North. Alfred North Whitehead. Russell. Bertrand. Bertrand Russell. February 2009. 1912. Principia Mathematica. Two. Merchant Books. 978-1-60386-183-0.
## Notes and References
1. Web site: Art of Problem Solving . 2022-09-07 . artofproblemsolving.com.
2. The equivalence of the standard numerical definition of finite sets to the Dedekind-finiteness of the power set of the power set was shown in 1912 by . This Whitehead/Russell theorem is described in more modern language by .
3. , demonstrated that his definition (which is also known as I-finite) is equivalent to Kuratowski's set-theoretical definition, which he then noted is equivalent to the standard numerical definition via the proof by .
4. Book: Handbook of Differential Equations: Ordinary Differential Equations. Canada. A.. Drabek. P.. Fonda. A.. 2005-09-02. Elsevier. 9780080461083. en.
5. The original paper by defined a set S to be finite when
P(S)∖ = ⋂.In other words, S is finite when the set of all non-empty subsets of S is equal to the intersection of all classes X which satisfy:
• all elements of X are non-empty subsets of S,
• the set is an element of X for all x in S,
• X is closed under pairwise unions.
Kuratowski showed that this is equivalent to the numerical definition of a finite set.
6. This list of 8 finiteness concepts is presented with this numbering scheme by both, and, although the details of the presentation of the definitions differ in some respects which do not affect the meanings of the concepts.
7. found counter-examples to each of the reverse implications in Mostowski models. Lévy attributes most of the results to earlier papers by Mostowski and Lindenbaum. |
If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
### Course: Algebra 2 (Eureka Math/EngageNY)>Unit 3
Lesson 13: Topic E: Lesson 29: Arithmetic series
# Arithmetic series intro
Sal explains the formula for the sum of a finite arithmetic series. Created by Sal Khan.
## Want to join the conversation?
• How do you figure out the sum of an alternating sequence like the sum of the first 39 terms of
1 + (-1)^n?
• You can look at it as a sum of two sequences--the first is arithmetic, with initial term a=1 and term difference of d=0. The first term is 1, the 39th ("last") term is 1+0*39=1. Sum of an arithmetic sequence is (first + last)*(#terms/2) = (1+1)*(39/2) = (2)*(39/2) = 39. The second sequence is geometric, with initial term a=-1 and term ratio r=-1. Sum of a geometric series, from another video, is a*(1-r^n)/(1-r)
= (-1)*(1-(-1)^39)/(1-(-1))
= (-1)*(1-(-1))/(1-(-1))
= (-1)*(1+1)/(1+1)
= (-1)*2/2
= -1
You can verify this intuitively by considering that the first term is -1, and then the next 38 terms cancel each other out in pairs (2nd and 3rd cancel, 4th and 5th cancel...38th and 39th cancel).
Sum the two results to get 39 + (-1) = 38.
Not every sequence can be so simply broken down into a sum of other (simple) sequences, but this one can.
• At how is 2+(n-1)=(n+1)? That doesn't make any sense to me.
• 2 + 𝑛 – 1
We can combine the 2 and the -1 to get 2 – 1 = 1. Thus:
2 + 𝑛 – 1 = 𝑛 + 1
Proficiency in this type of algebra is expected for studying arithmetic series. I would recommend looking at the "solving one step equations" playlist and the "combining like terms" video.
• If this is true, why is the sum of all positive integers -1/12?
• It's important to keep in mind that the people who make that series add up to -1/12 are willfully breaking some of the rules of standard mathematics (and usually not mentioning that fact). For the purposes of what they're doing (unless they're just doing it to freak out the newbies), they're not wrong, but it just makes things confusing to people learning the rules of standard math. For what you're learning now, the intuition that the sum of all positive integers should be undefinable because it is infinitely large is the correct attitude.
• how do you write the sigma notation for the sequence 1+5+9+13+17+21 what is n? and what would the general form be?
• nth term: 1 + 4 * ( n - 1 ) = 1 + 4n - 4 = 4n - 3
6
Σ ( 4n - 3 ) = 1 + 5 + 9 + 13 + 17 + 21
n = 1
• I am shortly going to be starting a Calculus course on Khan Academy. Did any of you find Calculus to be easier than Algebra?
• When I completed Unit 9 it listed Mastery Points Unavailable. Why is that?
• Unit 9 and 10 are from an older version of this course and do not yet have Unit Mastery enabled for them.
You can still complete the content like normal, but completing the content in these units won't contribute towards your overall mastery progress in the course.
• So how would you find the sum of an infinite geometric series?
• You cannot add up an infinite number of numbers, but you can take the limit of the sum as n approaches infinity. Sometimes this gives you a number, sometimes it gives you infinity, sometimes it isn't helpful at all. Much of the time, you must be content to know if the sum "converges" or "diverges".
You'll learn about this in calculus. Unfortunately, Sal only has a couple of videos on the basics of this broad subject. Try a Google search for "Infinite Series".
• What does the n stand for?
• "n" is used to represent the term in the sequence. If you need the 3rd term, then n=3. If you need the 10th term, the n=10.
Hope this helps. |
# Blog
## Brainteaser: Space climb
By
,
Difficulty: Fun
Catalina went on a game show and won their biggest prize! Catalina loves a good adventure, so she spent 3/4 of her prize money on a trip into space. She then spent 2/3 of her remaining money on training to climb Mount Everest. Finally, she donated the last remaining \$100,000 to wildlife conservation.
How much money did Catalina win?
## Need a hint?
Start by drawing a long, skinny rectangle to represent Catalina’s total prize money. Next divide your rectangle into four equal parts and shade in three of these parts. The shaded area represents the money she spent on the trip to space. The non-shaded part represents her remaining money.
Now, what would 2/3 of the remaining money look like on your rectangle? We recommend using a different colour to shade in the part she spent training for Everest.
Next ask: what did Catalina do with the remaining, non-shaded fraction of the rectangle?
Catalina won \$1,200,000!
This problem uses lots of fractions. We’ll show you a quick way to solve it at the end but first we’ll work through a visual solution.
Start by drawing a long, skinny rectangle to represent Catalina’s total prize money. Next, follow along with the story and divide your rectangle into 4 equal parts. Then, you can shade in three of these parts to represent the space trip that cost 3/4 of Catalina’s total prize.
The non-shaded fourth of the rectangle is the money Catalina has left over after going to space. We know from the story that she uses 2/3 of this remaining money on training for Everest. So, divide the non-shaded portion into 3 equal parts. Next, use a different coloured pencil to shade in two of these parts to represent the cost of training for Everest.
We know that the last remaining non-shaded portion of the rectangle is equal to \$100,000, Catalina’s wildlife donation. Now, we can work backwards to find the total prize money.
First, looking at our drawing, we can see that the Everest training cost exactly twice the amount donated, or \$200,000. This means that the money remaining after the space trip is exactly 3 times the amount donated, or \$300,000.
Second, we know from the story that the money remaining after the space trip is 1/4 of the total prize money, so Catalina must have won 4 times \$300,000, which is \$1,200,000.
This hints at a faster way to calculate the prize. Start by asking what fraction of the whole is the \$100,000 wildlife donation? Well, it’s 1/3 of a 1/4, which multiplied together makes 1/12. (Use your drawing to convince yourself that the white portion fits exactly 12 times in your big rectangle.) This means \$100,000 is 1/12 of the total prize money. Multiplying \$100,000 by 12 gives us \$1,200,000.
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Prove square root of 3 is Irrational Number Prove square root of 3 is Irrational Number
Here, I am going to tell you the best way of understanding that root of 3 is an Irrational Number. So, we are assuming √3 is a rational number i.e √3=a/b equation (1) Where a and b are integers having no common factor (b≠0). On squaring both side, (√3)2= (a/b) 2 3= a2/b2 equation (2) 3b2=a2 equation (3) where a and b are both odd number and a/b reduce to smallest possible terms. It is not possible that a and b are even because if a and b are even one can always be divided by 2 as we assume a/b is an Rational Numbers. a=2m+1. Assuming a and b are odd b=2n+1 By putting the value of a and b in equation 3: 3(2n+1)2= (2m+1)2 3(4n2+1+4n), = (2m2+1+4m) 12n2+3+12n, =4m2+1+4m 12n2+12n+2, =4m2+4m 6n2+6n+1, =2m2+2m 6n2+6n+1, =2(m2+n).. In this equation all the value of m is always odd and the value of n is always n for all values so this equation has no solution. The values of our assumptions a and b cannot be found so we can say that root of 3 is an Irrational Number. Know More About Constant Law of Limit Worksheets
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What is a Rational Number? The Rational numbers are those numbers which can either be whole numbers or fractions or decimals. Rational numbers can be written as a ratio of two integers in the form 'p/q' where 'p' and 'q' are integers and 'q' is nonzero. A rational number is simply a ratio of two integers, for example1/5 is a rational number (1 divided by 5, or the ratio of 1 to 5). Introduction to Rational numbers Today, I will tell you a story. Once there was a family of Natural numbers where all counting numbers used to live. One day a guest named zero visited the house and requested for a permission to stay there. All were happy; they requested the eldest member of the family Mr. infinite (∞) to grant the permission for 0. Definition of Rational Number In mathematics, Rational Numbers can be defined as a ratio of two integers. Rational numbers can be expressed in form of fraction 'a/b' in which 'a' and 'b' are integers where denominator 'b' not equal to zero. We also have set of rational number and 'Q' is used to represent it. Positive Rational Numbers Positive rational numbers can be expressed as the ratio 'p/q' where, 'p' and 'q' are both positive integers. Any positive rational number can be expressed as a sum of distinct reciprocals of positive integers like: 3/5 = 1/3+2/6+1/18. A positive rational number has infinitely many different such representations called Egyptian. Read More About Constant Laws of Limit Worksheet
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Negative Rational Numbers The numbers which are written in form of a/b where a and b are integers such that b≠0 all comes in the family of Rational Numbers. Positive Rational Numbers are those which have both numerators and denominators as positive or negative. Example: 5/7, 6/5 or (-3/-4) are positive rational numbers. Properties of Rational Numbers The rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers. The properties are called closure properties of rational numbers. Operations on Rational Numbers In mathematics, we generally deal with four types of basic operations called as addition, subtraction, multiplication, and division. We can easily perform these four kinds of operations on different type of Rational Numbers. Rational Numbers on a Number Line A number line is graphical representation of numbers from negative infinity to positive infinity in a single line. Rational numbers can be expressed in fraction (a/b) form, where 'b' is not equation.
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Prove square root of 3 is Irrational Number
Know More About Constant Law of Limit Worksheets It is not possible that a and b are even because if a and b are even one can always be divi...
Prove square root of 3 is Irrational Number
Know More About Constant Law of Limit Worksheets It is not possible that a and b are even because if a and b are even one can always be divi... |
### The notation of a function actually refers to determining the "name" of the function.
It is customary to symbolize a function using letters from the Latin alphabet when the two most common notations are:
• $y$
• $f(x)$
(Of course, similar notations can also be used).
The - inside parentheses expresses that it is an independent variable of the function and the function's dependency ( or ) on it. $x$,$y$,$f$
## Examples with solutions for Notation of a Function
### Exercise #1
Is the given graph a function?
### Step-by-Step Solution
It is important to remember that a function is an equation that assigns to each element in domain X one and only one element in range Y
Let's note that in the graph:
$f(0)=2,f(0)=-2$
In other words, there are two values for the same number.
Therefore, the graph is not a function.
No
### Exercise #2
Which of the following equations corresponds to the function represented in the graph?
### Step-by-Step Solution
Use the formula for finding slope:
$m=\frac{y_2-y_1}{x_2-x_1}$
We take the points:
$(0,-2),(-2,0)$
$m=\frac{-2-0}{0-(-2)}=$
$\frac{-2}{0+2}=$
$\frac{-2}{2}=-1$
We substitute the point and slope into the line equation:
$y=mx+b$
$0=-1\times(-2)+b$
$0=2+b$
We combine like terms:
$0+(-2)=b$
$-2=b$
Therefore, the equation will be:
$y=-x-2$
$y=-x-2$
### Exercise #3
Which of the following equations corresponds to the function represented in the table?
### Step-by-Step Solution
We will use the formula for finding slope:
$m=\frac{y_2-y_1}{x_2-x_1}$
Let's take the points:
$(-1,4),(3,8)$
$m=\frac{8-4}{3-(-1)}=$
$\frac{8-4}{3+1}=$
$\frac{4}{4}=1$
We'll substitute the point and slope into the line equation:
$y=mx+b$
$8=1\times3+b$
$8=3+b$
Let's combine like terms:
$8-3=b$
$5=b$
Therefore, the equation will be:
$y=x+5$
$y=x+5$
### Exercise #4
Is the given graph a function?
No
### Exercise #5
Is the given graph a function?
No
### Exercise #6
Determine whether the following table represents a function
No
### Exercise #7
Determine whether the data in the following table represent a constant function
No
### Exercise #8
Is the given graph a function?
Yes
### Exercise #9
Is the given graph a function?
Yes
### Exercise #10
Is the given graph a function?
Yes
### Exercise #11
Is the given graph a function?
Yes
### Exercise #12
Determine whether the following table represents a function
Yes
### Exercise #13
Determine whether the following table represents a function
Yes
### Exercise #14
Determine whether the following table represents a function
Yes
### Exercise #15
Determine whether the following table represents a function |
# 9th-RATIONAL NUMBERS
55 %
45 %
Entertainment
Published on May 31, 2013
Author: allwynasir
Source: authorstream.com
PowerPoint Presentation: 5 7 2 1 Rational Numbers S.JASMINE SUGIRTHA 9 th CLASS S.JASMINE SUGIRTHA 1 MATHS PowerPoint Presentation: Rational and Irrational Numbers Rational Numbers A rational number is any number that can be expressed as the ratio of two integers. Examples All terminating and repeating decimals can be expressed in this way so they are irrational numbers. a b 4 5 2 2 3 = 8 3 6 = 6 1 2.7 = 27 10 0.625 = 5 8 34.56 = 3456 100 -3 = 3 1 - 0.3 = 1 3 0.27 = 3 11 0.142857 = 1 7 0.7 = 7 10 S.JASMINE SUGIRTHA 2 MATHS Rational: Rational and Irrational Numbers Rational Numbers A rational number is any number that can be expressed as the ratio of two integers. All terminating and repeating decimals can be expressed in this way so they are irrational numbers. a b Show that the terminating decimals below are rational. 0.6 3.8 56.1 3.45 2.157 6 10 38 10 561 10 345 100 2157 1000 Rational S.JASMINE SUGIRTHA 3 MATHS PowerPoint Presentation: Rational and Irrational Numbers Rational Numbers A rational number is any number that can be expressed as the ratio of two integers. All terminating and repeating decimals can be expressed in this way so they are irrational numbers. a b To show that a repeating decimal is rational. Example 1 To show that 0.333… is rational. Let x = 0.333… 10 x = 3.33… 9 x = 3 x = 3/9 x = 1/3 Example 2 To show that 0.4545… is rational. Let x = 0.4545… 100 x = 45.45… 99 x = 45 x = 45/99 x = 5/11 S.JASMINE SUGIRTHA 4 MATHS PowerPoint Presentation: Rational and Irrational Numbers Rational Numbers A rational number is any number that can be expressed as the ratio of two integers. All terminating and repeating decimals can be expressed in this way so they are irrational numbers. a b Question 1 Show that 0.222… is rational. Let x = 0.222… 10 x = 2.22… 9 x = 2 x = 2/9 Question 2 Show that 0.6363… is rational. Let x = 0.6363… 100 x = 63.63… 99 x = 63 x = 63/99 x = 7/11 S.JASMINE SUGIRTHA 5 MATHS PowerPoint Presentation: Rational and Irrational Numbers Rational Numbers A rational number is any number that can be expressed as the ratio of two integers. All terminating and repeating decimals can be expressed in this way so they are irrational numbers. a b 999 x = 273 x = 273/999 9999 x = 1234 x = 1234/9999 Question 3 Show that 0.273is rational. Let x = 0.273 1000 x = 273.273 x = 91/333 Question 4 Show that 0.1234 is rational. Let x = 0.1234 10000 x = 1234.1234 S.JASMINE SUGIRTHA 6 MATHS PowerPoint Presentation: Rational and Irrational Numbers Rational Numbers A rational number is any number that can be expressed as the ratio of two integers. All terminating and repeating decimals can be expressed in this way so they are irrational numbers. a b By looking at the previous examples can you spot a quick method of determining the rational number for any given repeating decimal. 0.1234 1234 9999 0.273 273 999 0.45 45 99 0.3 3 9 S.JASMINE SUGIRTHA 7 MATHS PowerPoint Presentation: Rational and Irrational Numbers Rational Numbers A rational number is any number that can be expressed as the ratio of two integers. All terminating and repeating decimals can be expressed in this way so they are irrational numbers. a b 0.1234 1234 9999 0.273 273 999 0.45 45 99 0.3 3 9 Write the repeating part of the decimal as the numerator and write the denominator as a sequence of 9’s with the same number of digits as the numerator then simplify where necessary. S.JASMINE SUGIRTHA 8 MATHS PowerPoint Presentation: Rational and Irrational Numbers Rational Numbers A rational number is any number that can be expressed as the ratio of two integers. All terminating and repeating decimals can be expressed in this way so they are irrational numbers. a b 1543 9999 628 999 32 99 7 9 0.1543 0.628 0.32 0.7 Write down the rational form for each of the repeating decimals below. S.JASMINE SUGIRTHA 9 MATHS Irrational: Irrational a b Rational and Irrational Numbers Irrational Numbers An irrational number is any number that cannot be expressed as the ratio of two integers. 1 1 Pythagoras The history of irrational numbers begins with a discovery by the Pythagorean School in ancient Greece. A member of the school discovered that the diagonal of a unit square could not be expressed as the ratio of any two whole numbers. The motto of the school was “All is Number” (by which they meant whole numbers). Pythagoras believed in the absoluteness of whole numbers and could not accept the discovery. The member of the group that made it was Hippasus and he was sentenced to death by drowning. S.JASMINE SUGIRTHA 10 MATHS PowerPoint Presentation: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Rational Numbers Irrational Numbers S.JASMINE SUGIRTHA 11 MATHS PowerPoint Presentation: a b Rational and Irrational Numbers Irrational Numbers An irrational number is any number that cannot be expressed as the ratio of two integers. 1 1 Pythagoras Intuition alone may convince you that all points on the “Real Number” line can be constructed from just the infinite set of rational numbers, after all between any two rational numbers we can always find another. It took mathematicians hundreds of years to show that the majority of Real Numbers are in fact irrational. The rationals and irrationals are needed together in order to complete the continuum that is the set of “Real Numbers”. S.JASMINE SUGIRTHA 12 MATHS PowerPoint Presentation: a b Rational and Irrational Numbers Irrational Numbers An irrational number is any number that cannot be expressed as the ratio of two integers. 1 1 Pythagoras Surds are Irrational Numbers We can simplify numbers such as into rational numbers. However, other numbers involving roots such as those shown cannot be reduced to a rational form. Any number of the form which cannot be written as a rational number is called a surd. All irrational numbers are non-terminating , non-repeating decimals. Their decimal expansion form shows no pattern whatsoever. Other irrational numbers include and e , ( Euler’s number ) S.JASMINE SUGIRTHA 13 MATHS PowerPoint Presentation: Rational and Irrational Numbers Multiplication and division of surds. For example: and also for example and S.JASMINE SUGIRTHA 14 MATHS PowerPoint Presentation: Rational and Irrational Numbers Example questions Show that is rational rational Show that is rational rational a b S.JASMINE SUGIRTHA 15 MATHS PowerPoint Presentation: Rational and Irrational Numbers Questions a e State whether each of the following are rational or irrational. b c d f g h irrational rational rational irrational rational rational rational irrational S.JASMINE SUGIRTHA 16 MATHS PowerPoint Presentation: Rational and Irrational Numbers Combining Rationals and Irrationals Addition and subtraction of an integer to an irrational number gives another irrational number, as does multiplication and division. Examples of irrationals S.JASMINE SUGIRTHA 17 MATHS PowerPoint Presentation: Rational and Irrational Numbers Combining Rationals and Irrationals Multiplication and division of an irrational number by another irrational can often lead to a rational number. Examples of Rationals 21 26 8 1 -13 S.JASMINE SUGIRTHA 18 MATHS PowerPoint Presentation: Rational and Irrational Numbers Combining Rationals and Irrationals Determine whether the following are rational or irrational. (a) 0.73 (b) (c) 0.666…. (d) 3.142 (e) (f) (g) (h) (i) (j) (j) (k) (l) irrational rational rational rational irrational irrational irrational rational rational irrational irrational rational rational S.JASMINE SUGIRTHA 19 MATHS END: END THANK U S.JASMINE SUGIRTHA 20 MATHS
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Courses
# Classification of Discontinuity JEE Notes | EduRev
## JEE : Classification of Discontinuity JEE Notes | EduRev
The document Classification of Discontinuity JEE Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.
All you need of JEE at this link: JEE
B. Classification of Discontinuity
Definition :– Let a function f be defined in the neighbourhood of a point c, except perhaps at c itself.
Also let both one–sided limits
Then the point c is called a discontinuity of the first kind in the function f(x).
In more complicated case may not exist because
one or both one-sided limits do not exist. Such condition is called a discontinuity of the second kind.
has a discontinuity of the first kind at x = 0
The function y = |x| /x is defined for all x ∈ R, x ≠0;
but at x = 0 it has a discontinuity of the first kind.
has no limits (neither one-sided nor two-sided) at x = 2 and x = 3 since Therefore x = 2 and x = 3 are discontinuities of the second kind
The function y = ln |x| at the point x = 0 has the limits Consequently, (and also the one-sided limits) do not exist; x = 0 is a discontinuity of the second kind.
It is not true that discontinuities of the second kind only arise when
The situation is more complicated.
Thus, the function y = sin (1/x), has no one-sided limits for x → 0– and x → 0+, and does not tend to infinity as x → 0 There is no limit as x → 0 since the values of the function sin (1/x) do not approach a certain number, but repeat an infinite number of times within the interval from –1 to 1 as x→ 0.
Removable & Irremovable Discontinuity
(a) In case exists but is not equal to f(c) then the function is said to have a removable discontinuity. In this case we can redefine the function such that = f(c) & make it continuous at x = c.
Removable Type Of Discontinuity Can Be Further Classified As :
(i) Missing Point Discontinuity : exists finitely but f(a) is not defined .
e.g. has a missing point discontinuity at x = 1
(ii) Isolated Point Discontinuity : exists & f(a) also exists but
x ≠4 & f (4) = 9 has a break at x = 4.
(b) In case does not exist then it is not possible to make the function continuous by redefining it . Such discontinuities are known as non - removable discontinuity.
Irremovable Type Of Discontinuity Can Be Further Classified As :
(i) Finite discontinuity : e.g. f(x) = x - [x] at all integral x.
(ii) Infinite discontinuity :
(iii) Oscillatory discontinuity : e.g. f(x) = sin1/x at x = 0
In all these cases the value of f(a) of the function at x= a (point of discontinuity) may or may not exist but does not exist.
Remark :
(i) In case of finite discontinuity the non-negative difference between the value of the RHL at x = c & LHL at x = c is called The Jump Of Discontinuity . A function having a finite number of jumps in a given interval I is called a Piece-wise Continuous or Sectionally Continuous function in this interval.
(ii) All Polynomials, Trigonometrical functions, Exponential & Logarithmic functions are continuous in their domains.
(iii) Point functions are to be treated as discontinuous is not continuous at x = 1.
(iv) If f is continuous at x = c & g is continuous at x = f(c) then the composite g[f(x)] is continuous at x = c.
are continuous at x = 0 , hence the composite will also be continuous at x = 0.
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## Mathematics (Maths) Class 12
209 videos|222 docs|124 tests
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The commonly used methods to find the Highest Common Factor (HCF) or the Greatest Common Divisor (GCD) are the prime factorization method and the simultaneous division method. There is another method (less popular) to find HCF which is extremely useful in certain cases but not so useful in the rest.
If the question is, find the HCF of 80 and 216, this is how you can proceed:
Prime factorization:
80 = 2*2*2*2*5
216 = 2*2*2*3*3*3
HCF = 2*2*2 = 8
Simultaneous Division:
After this, as there are no common factors in 10 and 27, we will stop and take 2*2*2 = 8 as HCF.
Division method:
I am illustrating division method step by step.
Step 1: Divide the larger number by the smaller number. So 216 when divided by 80, gives 2 as quotient and 56 as remainder.
Step 2: Divide 80 (divisor of the previous division) by 56. This gives you 1 as quotient, and 24 as remainder.
Step 3: Divide 56 (divisor of the previous division) by 24. This gives you 2 as quotient, and 8 as remainder.
Step 3: Divide 24 (divisor of the previous division) by 8. This gives you 3 as quotient, and 0 remainder. That makes 8 the HCF of 80 and 216.
We will look at another example. Find HCF of 105 and 462.
Step 1: Divide 462 by 105. Quotient is 4 and remainder is 42.
Step 2: Divide 105 by 42. Quotient is 2 and remainder is 21.
Step 3: Divide 42 by 21. Quotient is 2 and remainder is 0. Hence, 21 is the HCF of 105 and 462.
This method is particularly useful when you get numbers that are not easy to factorize. For example, find the HCF of 629 and 851. In the previous examples, we could use the factorization and simultaneous division method as the prime factors were easy to find. But for something like HCF of 629 and 851, this method is super convenient. Give it a try. Answer? 37
Hope you understood this method. If you have any queries, leave them in the comments section. And stay tuned for more questions, concept articles and useful resources. And for all serious CAT aspirants, be a part of our active Facebook community here: CAT preparation with Learningroots
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# Linear Approximation Formula | Linear Interpolation & Regression Formula
## Linear Approximation Formula
For the function of any given value, we have to determine the closest estimation value of a function and it is given by the Linear approximation Formula. The other name for this mathematical concept is tangent line approximation or approximate tangent value of a function.
This is the process to find the equation of a line or closest approximation of a given function for the given value. It is used in Trigonometry or optics. This is an infinitely small like where the curve is almost straight and imitates the function. For the real-valued function F (x), the linear approximation value is given as below –
$\LARGE f(x) = f(x_{0}+f'(x_{0}(x-x_{0})+R_2$
where R2 is the remainder term. The linear approximation, then, is given by
$\LARGE f(x)\approx f(x_{0}+f'(x_{0}(x-x_{0})$
Where,
f(x0) is the value of f(x) at x = x0.
f'(x0) is the derivative value of f(x) at x = x0.
The approximation value is almost equal to the tangent line at a given point a?You could also opt for Euler’s method to find the solution for linear approximation values. In the end, only the closeness of the tangent line matters and tangent values can be calculated with specific formulas in mathematics. The other popular name for linear approximation is linearization.
So, how can you calculate the tangent at a given point a. There are two main things to remember here. First is the slope of a line and second is the particular point (a, b) line passes through. In this case, the equation of the line could be given as –
$\LARGE Y – b = m(X-a)$
Here, values of b and m will not be given automatically but you have to calculate it yourself only with formula derivatives.
### Linear Interpolation Formula
Interpolation is a popular statistical tool in mathematics that is used to calculate the estimated values between two points. Here, we will discuss the formula for the concept. Interpolation is also used in science, businesses, or many other fields too. Every time when you have to predict values between data points, linear interpolation formula is helpful.
Take an example of Tomato plant planted by a curious gardener. He wanted to know how much the plant has grown on the third day? If it is 6mm tall on the third day then its height should increase in a linear pattern in coming days too. It means the growth chart should create a straight line when plotting data on the graph. Here, the linear Approximation formula comes in handy to solve the issue.The Linear Interpolation Formula in mathematics is given as below.
$\large y=y_{1}+\frac{\left(x-x_{1}\right)\left(y-y_{1}\right)}{x_{2}-x_{1}}$
Where,
x1 and y1 are the first coordinates
x2 and y2 are the second coordinates
x is the point to perform the interpolation
y is the interpolated value.
### Linear Regression Formula
Linear regression is the highly common and predictive analysis technique used by the mathematicians or scientists. In this case, there are two variables, one is taken as the explanatory variable, and the other is taken as the dependent variable. With the help of a linear regression model, you can always relate the weights of individuals with heights.
There are a variety of linear regression techniques available in the mathematical world. These are -simple linear regression. Multiple linear regression, Logistic regression, Ordinal regression, Multinomial regression, Discriminant analysis etc. The Formula for linear regression equation in mathematics is given by:
$\large y=a+bx$
a and b are given by the following formulas:
$\large b\left(slope\right)=\frac{n\sum xy-\left(\sum x\right)\left(\sum y\right)}{n\sum x^{2}-\left(\sum x\right)^{2}}$
$\large a\left(intercept\right)=\frac{n\sum y-b\left(\sum x\right)}{n}$
Where,
x and y are two variables on regression line.
b = Slope of the line.
a = y-intercept of the line.
x = Values of first data set.
y = Values of second data set. |
Express each of the following decimals in the form
Question:
Express each of the following decimals in the form $\frac{p}{q}$ :
(i) $0.39$
(ii) $0.750$
(iii) $2.15$
(iv) $7.010$
(v) $9.90$
(vi) $1.0001$
Solution:
(i) Given decimal is $0.39$
Now we have to convert given decimal number into the $\frac{p}{q}$ form
Let $\frac{p}{q}=0.39$
$\Rightarrow \frac{p}{q}=\frac{39}{100}$
(ii) Given decimal is $0.750$
Now we have to convert given decimal number into $\frac{p}{q}$ form
Let $\frac{p}{q}=0.750$
$\Rightarrow \frac{p}{q}=\frac{750}{1000}$
$\Rightarrow \frac{p}{q}=\frac{75}{100}$
$\Rightarrow \frac{p}{q}=\frac{3}{4}$
(iii) Given decimal is $2.15$
Now we have to express the given decimal number into $\frac{p}{q}$ form
Let $\frac{p}{q}=2.15$
$\Rightarrow \frac{p}{q}=\frac{215}{100}$
$\Rightarrow \frac{p}{q}=\frac{43}{20}$
(iv) Given decimal is $7.010$
Now we have to express the given decimal number into $\frac{p}{q}$ form
Let $\frac{p}{q}=7.010$
$\Rightarrow \frac{p}{q}=\frac{7010}{1000}$
$\Rightarrow \frac{p}{q}=\frac{701}{100}$
(v) Given decimal is $9.90$
Now we have to find given decimal number into $\frac{p}{q}$ form
Let $\frac{p}{q}=9.90$
$\Rightarrow \frac{p}{q}=\frac{990}{100}$
$\Rightarrow \frac{p}{q}=\frac{99}{10}$
Hence, $9.90=\frac{99}{10}$
(vi) Given decimal is $1.0001$
Now we have to find given decimal number into $\frac{p}{q}$ form
$\frac{p}{q}=1.0001 \Rightarrow \frac{p}{q}=\frac{10001}{10000}$ |
# Heronian Triangle whose Altitude and Sides are Consecutive Integers
## Theorem
There exists exactly one Heronian triangle one of whose altitudes and its sides are all consecutive integers.
This is the Heronian triangle whose sides are $\tuple {13, 14, 15}$ and which has an altitude $12$.
## Proof
We note that a Heronian triangle whose sides are all consecutive integers is also known as a Fleenor-Heronian triangle.
From Sequence of Fleenor-Heronian Triangles, we have that the smallest such triangles are as follows:
$\tuple {1, 2, 3}$, which has an altitude of $0$
This is the degenerate case where the Heronian triangle is a straight line.
While $0, 1, 2, 3$ is a sequence of $4$ consecutive integers, this is not technically a triangle.
$\tuple {3, 4, 5}$ with area $6$.
It has altitudes $3$, $4$ and $\dfrac {12} 5$.
$\tuple {13, 14, 15}$
This can be constructed by placing the $2$ Pythagorean triangles $\tuple {5, 12, 13}$ and $\tuple {9, 12, 15}$ together along their common side $12$:
Thus the altitude and sides are:
$\tuple {12, 13, 14, 15}$
and this is the Heronian triangle we seek.
It has area $84$.
The next largest Fleenor-Heronian triangle has sides $\tuple {51, 52, 53}$.
Using Heron's Formula, its area is given by:
$\AA = \sqrt {78 \times 25 \times 26 \times 27} = 1170$
Hence its altitudes are:
$45 \frac {45} {51}$, $45$, $44 \frac 8 {53}$
For still larger triangles, the altitudes are never within $1$ unit of the sides:
Consider the triangle with sides $\tuple {a - 1, a, a + 1}$.
Using Heron's Formula, its area is given by:
$\ds \AA$ $=$ $\ds \sqrt {s \paren {s - a + 1} \paren {s - a} \paren {s - a - 1} }$ $\ds$ $=$ $\ds \sqrt {\frac 3 2 a \paren {\frac 1 2 a + 1} \paren {\frac 1 2 a} \paren {\frac 1 2 a - 1} }$ $\ds$ $=$ $\ds \frac a 4 \sqrt {3 \paren {a + 2} \paren {a - 2} }$ $\ds$ $=$ $\ds \frac a 4 \sqrt {3 a^2 - 12}$
Its longest altitude is therefore:
$\ds \frac {2 a} {4 \paren {a - 1} } \sqrt {3 a^2 - 12}$ $<$ $\ds \frac {a^2 \sqrt 3} {2 \paren {a - 1} }$
and we have:
$\ds \frac {a^2 \sqrt 3} {2 \paren {a - 1} }$ $<$ $\ds \paren {a - 1} - 1$ $\ds \leadstoandfrom \ \$ $\ds a^2 \sqrt 3$ $<$ $\ds 2 \paren {a - 1}^2 - 2 \paren {a - 1}$ $\ds \leadstoandfrom \ \$ $\ds 2 a^2 - 4 a + 2 - 2 a + 2 - \sqrt 3 a^2$ $>$ $\ds 0$ $\ds \leadstoandfrom \ \$ $\ds \paren {2 - \sqrt 3} a^2 - 6 a + 4$ $>$ $\ds 0$ $\ds \leadsto \ \$ $\ds a$ $>$ $\ds \frac {6 + \sqrt {6^2 - 4 \times 4 \paren {2 - \sqrt 3} } } {2 \paren {2 - \sqrt 3} }$ Quadratic Formula $\ds$ $\approx$ $\ds 21.7$
This shows that for $a \ge 22$, all altitudes of the triangle is less than $a - 2$.
Hence there are no more examples.
$\blacksquare$ |
# How and when do you use integration by parts?
• 788 views
Integration by parts is a method of integration used when you are attempting to integrate a function which is the product of two functions. If the two products can be expanded there is usually an easier way to integrate them than integration by parts. For example, x2(x - 4) is easier to integrate when expanded to x3 - 4x2.
The general form of the equation for integration by parts is:
∫f(x)g’(x)dx = f(x)g(x) - ∫g(x)f’(x)dx
where f’(x) is the derivative of f(x). It is also commonly seen as:
∫u dv/dx dx = uv - ∫v du/dx dx
where u and v are both function of x.
A good guideline when deciding which function to use as u (or f(x)) is the acronym LIATE:
Logarithmic e.g. ln(x)
Inverse trigonometry e.g. sin-1(x)
Algebraic e.g. x
Trigonometry e.g. sin(x)
Exponential e.g. ex.
Step 1:
Split the integrand (function to be integrated) in to its 2 products.
E.g. ∫xln(x)dx can be split in to x and ln(x).
Step 2:
Decide which function should be u and which should be dv/dx.
E.g. x is algebraic, ln is logarithmic. Logarithmic comes before algebraic in LIATE so u = ln(x) and dv/dx = x.
Step 3:
Find du/dx and v by differentiating and integrating u and dv/dx respectively.
E.g. u = ln(x), du/dx = x-1, dv/dx = x and v = x2/2
Step 4:
Substitute the variables in to the equation for integration by parts.
E.g. ∫xln(x)dx = ln(x)x2/2 - ∫x-1x2/2 dx = ln(x)x2/2 - ∫x/2 dx.
Step 5:
Evaluate the new integral.
E.g ∫xln(x)dx = ln(x)x2/2 - x2/4 + c = x2/4 (2ln(x) - 1) + c where c is a constant of integration.
Step 5 may require you to perform integration by parts again. Also LIATE does not work in every situation. If it does not work, switch the products used for u and dv/dx and try again.
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# Difference Between Union, Intersection and Difference of Sets for JEE Main 2024
Last updated date: 11th Sep 2024
Total views: 106.5k
Views today: 1.06k
## A Dive Into the History of Sets
Sets are the well ordered collection of elements such as numbers, variables, symbols, letters, sentences and so forth. Any object that is distinct, and up to our imagination, can be considered an element of the set. The number of elements inside the set is known as the cardinal number of a set. Generally, a set is represented by a capital letter. For example, L = {35, 89, 104} is a set that contains three elements.
Primarily, sets are divided into many types based on the cardinal number, i.e., based on the number of elements inside the set. If a set consists of only one element, then it is called a Singleton set. If there are no elements, then it is an Empty set. Also, if the number of elements is finite, it could be stated as a Finite set. If not, then it is an Infinite set.
There are still many sets defined on the basis of many other properties. To understand those, let us first discuss the Union, Intersection, and Difference of Sets, which are some of the operations performed on two or more sets to produce a single set.
## What is The Union of Sets?
Let us assume there are two sets L and M and the union of these two sets is represented as \$L\cup M\$.
\$L\cup M\$ is the union of two sets L and M which contain all the elements of L and M.
Note: Elements must be distinct and can not be repeated in the Union Set. Every element is such that it may belong to set L or set M, but it must belong to one of those sets.
The union of sets is commutative, i.e., \$L\cup M=M\cup L\$
Let’s comprehend this concept of the union of sets using an example.
Let L = {b, e, i, m, r, w, z} and M = {2, 5, 7, 9, 11, 13, e, r, x, y, z}
Now, \$L\cup M\$ = {b, e, i, m, r, w, z, 2, 5, 7, 9, 11, 13, x, y}
As you can see here, elements that belong to both L and M are written but they are not repeated. This can be shown in a diagram which needs a Universal set U.
Note: A universal set is a set that consists of all the elements related to the described sets in the question given. Suppose if we are asked to describe a set of integers as the union of two sets, the universal set consists of all the numbers like real, complex, integers, rational or irrational including natural. Out of the numbers which are present in the universal set, we make the union taking only integers out of it.
Image: Union of two sets L and M
In the above image, the blue-coloured circle is the L set and the pink one is the M set. The blue shaded portion is the union of these two sets L and M.
If there are n number of sets namely, L1, L2, L3 and so on up to Ln then the sets formula for the union if sets are given by the equation.
• \$\Sigma_{i=1}^{n}{L_i}=L_{1} \cup L_{2} \cup L_{3}-------L_{n}\$
## Intersection of Sets
Let L and M be the two sets. The intersection set of L and M is demonstrated as
\$L\cap M\$ and this set consist of elements which are part of both L and M.
That is, if an element is a part of the intersection set \$L\cap M\$, then it must belong to both the L set and the M set.
The intersection is also commutative, that is \$L\cap M=M\cap L\$
Suppose L = {12,27,21,35,86,91,97} and M = {2,20,21,47,54,77,86,97}. Now the intersection set, \$L\cap M\$ = {21,86,97} as only these three elements are in common.
The diagram for the intersection of sets is shown below.
Image: Intersection of two sets L and M
In the above image, we can see the part highlighted is the intersection part of the two circles. If there are n numbers of sets, namely L1, L2, L3 and so on up to Ln, then the sets formula for the intersection of these sets is given by
• \$\Sigma_{i=1}^{n}{L_i}=L_{1} \cap L_{2} \cap L_{3}-------L_{n}\$
This shows that the intersection set for the n sets consists of elements that are present in all the n sets.
## Difference Between Sets
If the difference between two sets L and M is L – M, then this set contains the elements which are present in only L but not in M. This is basically removing the elements in the intersection set \$L\cap M\$ from the whole set of L. This gives us the difference between two sets L and M.
Let us understand how this works through a figure.
Image: Difference between sets L and M
As shown here, the highlighted part is the section of only L but not M. This is the difference set, L-M.
Note: The difference set is not commutative, i.e., L – M is not equal to M – L.
Let’s have a look at some of the formulas of sets that are useful in solving the problems.
• \$n(L)-n(L \cap M)=n(L-M)\$
• \$n(L \cup M)=n(L)+n(M)-n(L \cap M)\$
• \$n(L-M)+n(L \cap M)+n(M-L)=n(L \cup M)\$
## Conclusion
Therefore, the union of two sets \$L\cup M\$ can be written as the sum of the difference set L – M and the Intersection set \$L\cap M\$ and the difference set M – L. These formulae are proven and the properties of sets are well defined. The union set and the intersection set satisfy the commutative property, associative property and distributive property.
Hence, the questions like what is an intersection and what is a union of sets get clear through this article. The examples of sets discussed are clear and vivid.
Competitive Exams after 12th Science |
Solving Linear Equations In One Variable
When linear equations consist of only one variable, it becomes easier for us to solve the equations. Such kind of equations represent a vertical (perpendicular to x- axis or parallel to y- axis) or horizontal straight line (perpendicular to y- axis or parallel to x- axis). Thus we get an intersection point on one of the two axes. Solving Linear Equations in One Variable can be thought of finding a real or imaginary value of any unknown variable in the equation.
To understand how we deal with such kind of linear equations, first we have to be sure about that expression consists of one variable only. If we have more than one variable in our equation, we would need that much number of equations to get values of unknown variables. For example, if we have a linear equation given as: 5x = 10. We can directly get value of 'x' by dividing both sides of equation with 5. Thus, x = 2. In case we have an equation of form x + 4 = 2 we would be taking all constant values to right side of equation, leaving the unknown part to the left. In this example we will get x = 2 – 4 = -2.
To graph a linear equation in one variable like in our above examples: for equation 5x = 10 we will be drawing a straight line intersecting x - axis at x = 2 parallel to the y - axis. Similarly, for equation x + 4 = 2 we will be draw a line parallel to y - axis and intersecting x - axis at x = -2. These straight lines will have a constant slope that is m = 0 at every point of intersection on any of the axes.
Solving Literal Equations
In geometry we may come across a situation where we need to solve for some unknown variables other than "usual" one. For instance, formula for area of a square with sides of length “s” is given as A = s * s. Here, solving for “s” for known value of area 'A'. This method of solving a formula for any known variable is known as literal equation solver. So solving...Read More
Addition and multiplication Property of Equality
Equation or equality can be defined as an expression containing equal sign. There are different properties of equality here we will discuss addition and multiplication property of equality.
Addition property of equality: Addition property of equality states that both sides of expression should be equal in an equation.
W...Read More
Solving linear equations
Linear equation is an algebraic equation which may have both constant and variable terms. In other words, Linear equation is actually that expression which consists of equal sign and some variables and constants such as y + 3, y + 3x etc. Variable may be one or more than one. A non - linear equation can be converted into linear equation. Exponential t...Read More
Math Topics |
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Definition of Variable
A variable can be any unknown value, and will be a critical part of our curriculum beginning with algebra all the way up through calculus!
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What if you had a jar filled with dimes and quarters? You know that the total of the coins in the jar is $8.60. How could you write an equation to represent this situation? After completing this Concept, you'll be able to use variables to write equations like this one with unknown quantities. Watch This Guidance No one likes doing the same problem over and over again—that’s why mathematicians invented algebra. Algebra takes the basic principles of math and makes them more general, so we can solve a problem once and then use that solution to solve a group of similar problems. In arithmetic, you’ve dealt with numbers and their arithmetical operations (such as $+, \ -, \ \times, \ \div$ ). In algebra, we use symbols called variables (which are usually letters, such as $x, \ y, \ a, \ b, \ c, \ \ldots$ ) to represent numbers and sometimes processes. For example, we might use the letter $x$ to represent some number we don’t know yet, which we might need to figure out in the course of a problem. Or we might use two letters, like $x$ and $y$ , to show a relationship between two numbers without needing to know what the actual numbers are. The same letters can represent a wide range of possible numbers, and the same letter may represent completely different numbers when used in two different problems. Using variables offers advantages over solving each problem “from scratch.” With variables, we can: • Formulate arithmetical laws such as $a + b = b + a$ for all real numbers $a$ and $b$ . • Refer to “unknown” numbers. For instance: find a number $x$ such that $3x + 1 = 10$ . • Write more compactly about functional relationships such as, “If you sell $x$ tickets, then your profit will be $3x - 10$ dollars, or “ $f(x) = 3x - 10$ ,” where “ $f$ ” is the profit function, and $x$ is the input (i.e. how many tickets you sell). Example A Write an algebraic equation for the perimeter and area of the rectangle below. To find the perimeter, we add the lengths of all 4 sides. We can still do this even if we don’t know the side lengths in numbers, because we can use variables like $l$ and $w$ to represent the unknown length and width. If we start at the top left and work clockwise, and if we use the letter $P$ to represent the perimeter, then we can say: $P = l + w + l + w$ We are adding $2 \ l$ ’s and $2 \ w$ 's, so we can say that: $P = 2 \cdot l + 2 \cdot w$ It's customary in algebra to omit multiplication symbols whenever possible. For example, $11x$ means the same thing as $11 \cdot x$ or $11 \times x$ . We can therefore also write: $P = 2l + 2w$ Area is length multiplied by width . In algebraic terms we get: $A = l \times w \to A = l \cdot w \to A = lw$ Note: $2l + 2w$ by itself is an example of a variable expression ; $P = 2l + 2w$ is an example of an equation . The main difference between expressions and equa tions is the presence of an equa ls sign (=). In the above example, we found the simplest possible ways to express the perimeter and area of a rectangle when we don’t yet know what its length and width actually are. Now, when we encounter a rectangle whose dimensions we do know, we can simply substitute (or plug in ) those values in the above equations. In this chapter, we will encounter many expressions that we can evaluate by plugging in values for the variables involved. Example B Eric has some money in his savings account. How much more money does he need in order to buy a game that costs$98?
Solution:
Let $M$ be the money that Eric still needs and let $S$ be the money that Eric has in his savings account. Then, by subtracting the money he already has from the total money needed, we can figure out how much money he still needs:
$M=98-S.$
Example C
Write an equation for the sum of 3 times some number and 5.
Solution:
Let $S$ be the total sum. Let $n$ be some number. Then 3 times some number is $3\cdot n$ and then the sum of that and 5 is:
$S=3n+5.$
Watch this video for help with the Examples above.
Vocabulary
• We use symbols called variables (which are usually letters, such as $x, \ y, \ a, \ b, \ c, \ \ldots$ ) to represent numbers and sometimes processes.
• $2l + 2w$ by itself is an example of a variable expression ; $P = 2l + 2w$ is an example of an equation . The main difference between expressions and equa tions is the presence of an equa ls sign (=).
Guided Practice
Alex has a certain amount of nickels and dimes in a jar. Write an algebraic equation for how much money she has, in terms of how many nickles and dimes she has.
Solution:
Let $n$ be the number of nickels and $d$ be the number of dimes that Alex has in the jar. Since each nickel is worth $0.05, the amount of money she has in nickels will be: $0.05\cdot n$ Since each dime is worth$0.10, the amount of money she has in dimes will be:
$0.10\cdot d$
This means that the total amount of money $M$ that Alex has will be:
$M=0.05 \cdot n +0.10 \cdot d.$
Simplifying the expressions, we get:
$M=0.05n +0.10d.$
Practice
For 1-4, write the following in a more condensed form by leaving out a multiplication symbol.
1. $2 \times 11x$
2. $1.35 \cdot y$
3. $3 \times \frac { 1 } { 4 }$
4. $\frac { 1 } { 4 } \cdot z$
For 5-10, write an equation for the following situations.
1. The amount of money Andrea has in a jar full of quarters and dimes.
2. The amount of money Michelle has in her coin purse if it only contains quarters, dimes and pennies.
3. The sum of 7 and 6 times some number.
4. 4 less than 20 times some number.
5. The amount of money you will earn if you are paid $10.25 an hour and spend$4.00 round trip to get too and from work.
6. A father earns a \$2000 dividend from an oil investment and distributes it equally amongst his children.
Vocabulary Language: English
Equation
Equation
An equation is a mathematical sentence that describes two equal quantities. Equations contain equals signs.
substitute
substitute
In algebra, to substitute means to replace a variable or term with a specific value.
Variable
Variable
A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.
Variable Expression
Variable Expression
A variable expression is a mathematical phrase that contains at least one variable or unknown quantity.
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# Sepia's question at Yahoo! Answers regarding finding the length of a line segment within a triangle
#### MarkFL
Staff member
Here is the question:
How will you find the length of a line drawn inside a scalene triangle?
ABC is a scalene triangle, in which AB = 3, AC = 4 and BC = 6
D is the line joining A with BC.such that BD : DC : :2 :3
I have posted a link there to this topic so the OP can see my work.
#### MarkFL
Staff member
Re: Sepia's question ay Yahoo! Answers regarding finding the length of a line segment within a trian
Hello Sepia,
Let's first draw a diagram:
We see that:
$$\displaystyle u+v=6$$
and we are given:
$$\displaystyle \frac{u}{v}=\frac{2}{3}\implies u=\frac{2}{3}v$$
Substituting this into the first equation, we then find:
$$\displaystyle \frac{2}{3}v+v=6$$
$$\displaystyle \frac{5}{3}v=6$$
$$\displaystyle v=\frac{18}{5}$$
Now, let's use the Law of Cosines to determine the cosine of the angle $\theta$:
$$\displaystyle 3^2=4^2+6^2-2\cdot4\cdot6\cos(\theta)$$
$$\displaystyle \cos(\theta)=\frac{4^2+6^2-3^2}{2\cdot4\cdot6}=\frac{43}{48}$$
Next, using the Law of Cosines again, we may state:
$$\displaystyle x=\sqrt{4^2+v^2-2\cdot4\cdot v\cos(\theta)}$$
$$\displaystyle x=\sqrt{4^2+\left(\frac{18}{5} \right)^2-2\cdot4\cdot\left(\frac{18}{5} \right)\left(\frac{43}{48} \right)}=\frac{\sqrt{79}}{5}$$ |
# 3.7 Two-dimensional Shapes and Perimeter
## Unit Goals
• Students reason about shapes and their attributes, with a focus on quadrilaterals. They solve problems involving the perimeter and area of shapes.
### Section A Goals
• Reason about shapes and their attributes.
### Section B Goals
• Find the perimeter of two-dimensional shapes, including when all or some side lengths are given.
### Section C Goals
• Solve problems involving perimeter and area, in and out of context.
### Section D Goals
• Apply geometric understanding to solve problems.
### Problem 1
#### Pre-unit
Practicing Standards: 2.G.A.1
1. Which shapes are hexagons? ____________________
2. Draw a quadrilateral and label it Q. Draw a pentagon and label it P.
### Problem 2
#### Pre-unit
Practicing Standards: 3.MD.C.7
1. Finish covering the rectangle with equal-sized squares.
2. Write an addition equation showing how many small squares there are in the rectangle.
3. Write a multiplication equation showing how many small squares there are in the rectangle.
### Problem 3
#### Pre-unit
Practicing Standards: 3.MD.C.7.d
What is the area of the figure?
### Problem 4
#### Pre-unit
Practicing Standards: 2.MD.A.1
Find the length of each rectangle.
### Problem 5
1. Which shapes have 5 angles? ____________________
2. Which shapes have all side lengths the same size? ____________________
### Problem 6
1. Which triangles have one right angle? ____________________
2. Which triangles have two sides that are the same length? ____________________
### Problem 7
One of these quadrilaterals is Diego’s mystery quadrilateral.
List some questions you would ask to figure out Diego’s mystery quadrilateral. How would the answers help you figure out which quadrilateral it is?
### Problem 8
Select all of the quadrilaterals that are rectangles.
A:
B:
C:
D:
E:
### Problem 9
1. Draw a quadrilateral that is not a rhombus and label it A. Explain why your shape is not a rhombus.
2. Draw a quadrilateral that is not a rectangle and label it B. Explain why your shape is not a rectangle.
### Problem 10
#### Exploration
Find one of each of the following shapes in the image.
1. A triangle with a right angle
2. A triangle with 3 equal sides
3. A rhombus
4. A rectangle
5. A square
### Problem 11
#### Exploration
Do any of the triangles have:
1. a right angle?
2. two equal sides?
3. three equal sides?
### Problem 1
Find the perimeter of each shape.
### Problem 2
1. Draw 2 different shapes with perimeter 20 units.
2. What strategy did you use to draw your shapes?
### Problem 3
Find the perimeter of each shape. Explain or show your reasoning.
1.
2.
### Problem 4
All sides of this hexagon have the same length. The perimeter of the hexagon is 96 cm. What are its side lengths? Explain or show your reasoning.
### Problem 5
#### Exploration
The side length of every pattern block is 1 inch.
1. What is the perimeter of the shape? Explain your reasoning.
2. What is the perimeter of the shape if you remove the skinny rhombuses around the boundary?
3. What is the perimeter if you next remove the blue rhombuses? What if you keep removing more shapes?
### Problem 6
#### Exploration
1. Draw some different shapes that you can find the perimeter of. Then find their perimeters.
2. Can you draw a rectangle whose perimeter is odd? Explain or show your reasoning.
3. Can you draw a pentagon or hexagon (or a figure with even more sides) whose perimeter is odd?
### Problem 1
A rectangular card has an area of 60 square centimeters. It is 4 centimeters longer than it is wide. What is the perimeter of the card? Explain or show your reasoning.
### Problem 2
Draw two rectangles with perimeter 20 units on the grid whose areas are different. What are the areas of the rectangles?
### Problem 3
Draw two rectangles on the grid with area 30 square units whose perimeters are different. What are the perimeters of your rectangles?
### Problem 4
#### Exploration
Clare draws a rectangle.
1. She tells you that the perimeter is 36. What rectangle could Clare have drawn?
2. Then she tells you that her rectangle has the biggest area possible. What rectangle could Clare have drawn?
### Problem 5
#### Exploration
Draw a rectangle on the grid but don’t share with your partner. Give your partner clues to help them guess the perimeter and area of your rectangle. Try not to just tell them the side lengths of your rectangle.
### Problem 1
The perimeter of a rectangular park is 444 meters. The park is 175 meters long. How wide is the park? Explain or show your reasoning.
### Problem 2
1. Where do you see rhombuses in the design?
2. Where do you see rectangles in the design?
3. Are there any triangles in the design?
### Problem 3
1. The body of your robot has a perimeter of 64 units. What were some different lengths and widths you could choose that fit on the grid?
2. The head has a perimeter of 36 units. What were some different lengths and widths you could choose for the head that fit on the grid?
### Problem 4
#### Exploration
Han is making a rectangular garden in his backyard. He has 42 feet of fencing and the garden needs to be at least 3 feet wide. What are the possible lengths and widths of the garden Han could make? Explain or show your reasoning. |
Lecture 6
1 / 18
# Lecture 6 - PowerPoint PPT Presentation
Lecture 6. 2.1 Sets 2.2 Set Operations. Definition of Set and Set Theory. Describing Set Membership. Set Builder Notation. Sets and Set Operations. Definition: A set is any collection of distinct things considered as a whole. A set is an unordered collection of objects.
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Lecture 6
2.1 Sets 2.2 Set Operations
Sets and Set Operations
Definition: A set is any collection of distinct things considered as a whole. A set is an unordered collection of objects.
Discuss whether each of the following in a set:
S = {1, 2, 3, 42} V = {x|x is a real number}
T = {1, 1, 2, 3} W = {x|x is not in W}
U = { } Z = {{1,2,3},{2,3,4},{3,4,5}}
P = { { }, { { } }, { { { } } } } Q = {{1,2,3}, {2,3,4},{3,2,1}}
Since the members of a set are in no particular order Q is not a set if its members are sets, but Q is a set if its members are 3-tuples, vectors or some other entity for which membership order is important.
Since f = { } we can rewrite P = { f, {f}, {{f}} } so P is a set containing three elements, namely the empty set, a singleton containing the empty set and a singleton containing a singleton containing the empty set.
If A is a set containing n elements then |A| = n, and is called the cardinality of A.
Given a set S, the power set of S is the set of all subsets of the set S. The power set is S is denoted by P(S).
The ordered n-tuple (a1, a2, . . . , an) is the ordered collection that has a1 as its first element, a2 as its second element . . . and an as its nth element.
Let A and B be sets. The Cartesian product of A and B, denoted by AxB, is the set of all ordered pairs (a,b) where a A and b B. Hence,
AxB = {(a,b)|a A b B}.
Definitions: Set Properties
B
2 4 5
1 1,2 1,4 1,5
3 3,2 3,4 3,5
5 5,2 5,4 5,5
8 8,2 8,4 8,5
A
A = { s, pass, link, stock }
B = { word, port, age, able}
Cartesian Product
A = { 1,3,5,8 }
B = { 2,4,5 }
Set Notation with Quantifiers
For all x, elements of the Reals, x2 is greater than or equal to 0.
There exists an x, element of the Integers, such that x2 equals 1.
For every x, element of the Reals, there exists a y, element of the Reals, such that x times y = 1. (give an exception to show this statement is false)
For every x, element of the Integers, there exists ay, element of the Integers, such that x plus y = 0.
A
B
U
U
U
U
U
A
B
A
B
A
B
A
B
U
U
U
U
A
B
A
B
A
B
A
B
U
U
U
U
A
B
A
B
A
B
A
B
Venn Diagrams
Set Identities
(This is why we had a separate test on first-order logic.)
1 1 1 1 1 1 1 1
1 1 0 1 1 1 0 1
1 0 1 1 1 0 0 1
1 0 0 0 0 0 0 0
0 1 1 1 0 0 0 0
0 1 0 1 0 0 0 0
0 0 1 1 0 0 0 0
0 0 0 0 0 0 0 0
Membership Table
Show that
An Example
Satisfiability Set Enumeration
= |
# Equation of a Circle
## Equation of a Circle
The equation of the circle with center C (h, k) and radius r is (x – h)² + (y – k)² = r².
Proof: Above image represents P (x₁, y₁) is locus of the point, r is radius and C (h, k) is center.
The distance between the CP is r.
let P (x₁, y₁) be a point P lies in the circle PC = r
$$\sqrt{{{({{x}_{1}}-h)}^{2}}+{{({{y}_{1}}-k)}^{2}}}=r$$.
(x₁ – h) ² + (y₁ – k) ² = r²
The locus of P is (x – h)² + (y – k)² = r²
Note:
i) The equation of a circle of the form x² + y² + 2gx + 2fy + c =0
ii) The equation of a circle with center origin (0, 0) and radius r is
(x – 0)² + (y – 0)² = r²
x² + y² = r²
Example 1: Find the equation of the circle with center (1, 4) and radius 5.
Solution: Given that,
C (h, k) = (1, 4), P (x, y) and r = 5
The locus of P is (x-h) ² + (y-k) ² = r²
(x-1) ² + (y-4) ² = 5²
x² – 2x + 1 + y² – 8y + 16 = 25
x² + y² – 2x – 8y – 8 = 0
Required equation of the circle is x² + y² – 2x – 8y – 8 = 0
Example 2: Find the equation of the circle with center origin and radius 9.
Solution: Given that,
C (h, k) = (0, 0), P (x, y) and r = 9
The locus of P is (x-0) ² + (y-0) ² = r²
x² + y² = 9²
x² + y² – 81 = 0. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Congruent Angles and Angle Bisectors
## Bisectors split the angle into two equal halves.
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Practice Congruent Angles and Angle Bisectors
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Congruent Angles and Angle Bisectors
What if you knew that an angle was split exactly in half? How could you use this information to help you solve problems? After completing this Concept, you'll be able to bisect an angle and solve problems related to angle bisectors.
### Watch This
CK-12 Foundation: Chapter1CongruentAnglesandAngleBisectorsA
James Sousa: Angle Bisectors
Watch the first part of this video.
James Sousa: Angle Bisector Exercise 2
### Guidance
When two rays have the same endpoint, an angle is created.
Here, \begin{align*}\overrightarrow{BA}\end{align*} and \begin{align*}\overrightarrow{BC}\end{align*} meet to form an angle. An angle is labeled with an “\begin{align*}\angle\end{align*}” symbol in front of the three letters used to label it. This angle can be labeled \begin{align*}\angle ABC\end{align*} or \begin{align*}\angle CBA\end{align*}. Always put the vertex (the common endpoint of the two rays) in the middle of the three points. It doesn’t matter which side point is written first.
An angle bisector is a ray that divides an angle into two congruent angles, each having a measure exactly half of the original angle. Every angle has exactly one angle bisector.
\begin{align*}\overline{BD}\end{align*} is the angle bisector of \begin{align*}\angle ABC\end{align*}
\begin{align*}\angle ABD & \cong \angle DBC\\ m \angle ABD & = \frac{1}{2} m \angle ABC\end{align*}
Label equal angles with angle markings, as shown below.
##### Investigation: Constructing an Angle Bisector
1. Draw an angle on your paper. Make sure one side is horizontal.
2. Place the pointer on the vertex. Draw an arc that intersects both sides.
3. Move the pointer to the arc intersection with the horizontal side. Make a second arc mark on the interior of the angle. Repeat on the other side. Make sure they intersect.
4. Connect the arc intersections from #3 with the vertex of the angle.
To see an animation of this construction, view http://www.mathsisfun.com/geometry/construct-anglebisect.html.
#### Example A
How many angles are in the picture below? Label each one two different ways.
There are three angles with vertex \begin{align*}U\end{align*}. It might be easier to see them all if we separate them out.
So, the three angles can be labeled, \begin{align*}\angle XUY\end{align*} or \begin{align*} \angle YUX, \ \angle YUZ\end{align*} or \begin{align*} \angle ZUY\end{align*}, and \begin{align*}\angle XUZ\end{align*} or \begin{align*} \angle ZUX\end{align*}.
#### Example B
What is the measure of each angle?
From the picture, we see that the angles are congruent, so the given measures are equal.
\begin{align*}(5x + 7)^\circ & = (3x + 23)^\circ\\ 2x^\circ & = 16^\circ\\ x & = 8^\circ\end{align*}
To find the measure of \begin{align*}\angle ABC\end{align*}, plug in \begin{align*}x = 8^\circ\end{align*} to \begin{align*}(5x + 7)^\circ\end{align*}.
\begin{align*}& (5(8) + 7)^\circ\\ & (40 + 7)^\circ\\ & 47^\circ\end{align*}
Because \begin{align*}m \angle ABC = m \angle XYZ, \ m \angle XYZ = 47^\circ\end{align*} too.
#### Example C
Is \begin{align*}\overline{OP}\end{align*} the angle bisector of \begin{align*} \angle SOT\end{align*}? If \begin{align*}m \angle ROT = 165^\circ\end{align*}, what is \begin{align*}m \angle SOP\end{align*} and \begin{align*}m \angle POT\end{align*}?
Yes, \begin{align*}\overline{OP}\end{align*} is the angle bisector of \begin{align*}\angle SOT\end{align*} according to the markings in the picture. If \begin{align*}m \angle ROT = 165^\circ\end{align*} and \begin{align*}m \angle ROS = 57^\circ\end{align*}, then \begin{align*}m \angle SOT = 165^\circ - 57^\circ = 108^\circ\end{align*}. The \begin{align*}m \angle SOP\end{align*} and \begin{align*}m \angle POT\end{align*} are each half of \begin{align*}108^\circ\end{align*} or \begin{align*}54^\circ\end{align*}.
Watch this video for help with the Examples above.
CK-12 Foundation: Chapter1CongruentAnglesandAngleBisectorsB
### Guided Practice
For exercises 1 and 2, copy the figure below and label it with the following information:
1. \begin{align*}\angle A \cong \angle C\end{align*}
2. \begin{align*}\angle B \cong \angle D\end{align*}
3. Use algebra to determine the value of d:
1. You should have corresponding markings on \begin{align*}\angle A \end{align*} and \begin{align*}\angle C\end{align*}.
2. You should have corresponding markings on \begin{align*}\angle B \end{align*} and \begin{align*}\angle D\end{align*} (that look different from the markings you made in #1).
3. The square marking means it is a \begin{align*}90^\circ\end{align*} angle, so the two angles are congruent. Set up an equation and solve:
\begin{align*}7d-1&=2d +14\\ 5d&=15\\ d&=3\end{align*}
### Explore More
For 1-4, use the following picture to answer the questions.
1. What is the angle bisector of \begin{align*}\angle TPR\end{align*}?
2. What is \begin{align*}m\angle QPR\end{align*}?
3. What is \begin{align*}m\angle TPS\end{align*}?
4. What is \begin{align*}m\angle QPV\end{align*}?
For 5-6, use algebra to determine the value of variable in each problem.
For 7-10, decide if the statement is true or false.
1. Every angle has exactly one angle bisector.
2. Any marking on an angle means that the angle is \begin{align*}90^\circ\end{align*}.
3. An angle bisector divides an angle into three congruent angles.
4. Congruent angles have the same measure.
In Exercises 11-15, use the following information: \begin{align*}Q\end{align*} is in the interior of \begin{align*}\angle ROS\end{align*}. \begin{align*}S\end{align*} is in the interior of \begin{align*}\angle QOP\end{align*}. \begin{align*}P\end{align*} is in the interior of \begin{align*}\angle SOT\end{align*}. \begin{align*}S\end{align*} is in the interior of \begin{align*}\angle ROT\end{align*} and \begin{align*}m \angle ROT = 160^\circ, \ m \angle SOT = 100^\circ,\end{align*} and \begin{align*}m \angle ROQ = m \angle QOS = m \angle POT\end{align*}.
1. Make a sketch.
2. Find \begin{align*}m \angle QOP\end{align*}
3. Find \begin{align*}m \angle QOT\end{align*}
4. Find \begin{align*}m \angle ROQ\end{align*}
5. Find \begin{align*}m \angle SOP\end{align*}
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 1.3.
### My Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
angle bisector
An angle bisector is a ray that splits an angle into two congruent, smaller angles.
Congruent
Congruent figures are identical in size, shape and measure. |
# Mathematics | Predicates and Quantifiers | Set 1
Prerequisite : Introduction to Propositional Logic
Introduction
Consider the following example. We need to convert the following sentence into a mathematical statement using propositional logic only.
"Every person who is 18 years or older, is eligible to vote."
The above statement cannot be adequately expressed using only propositional logic. The problem in trying to do so is that propositional logic is not expressive enough to deal with quantified variables. It would have been easier if the statement were referring to a specific person. But since it is not the case and the statement applies to all people who are 18 years or older, we are stuck.
Therefore we need a more powerful type of logic.
Predicate Logic
Predicate logic is an extension of Propositional logic. It adds the concept of predicates and quantifiers to better capture the meaning of statements that cannot be adequately expressed by propositional logic.
What is a predicate?
Consider the statement, “ is greater than 3″. It has two parts. The first part, the variable , is the subject of the statement. The second part, “is greater than 3”, is the predicate. It refers to a property that the subject of the statement can have.
The statement “ is greater than 3″ can be denoted by where denotes the predicate “is greater than 3” and is the variable.
The predicate can be considered as a function. It tells the truth value of the statement at . Once a value has been assigned to the variable , the statement becomes a proposition and has a truth value.
In general, a statement involving n variables can be denoted by . Here is also referred to as n-place predicate or a n-ary predicate.
• Example 1: Let denote the statement “ > 10″. What are the truth values of and ?
Solution: is equivalent to the statement 11 > 10, which is True.
is equivalent to the statement 5 > 10, which is False.
• Example 2: Let denote the statement ““. What is the truth value of the propositions and ?
Solution: is the statement 1 = 3 + 1, which is False.
is the statement 2 = 1 + 1, which is True.
What are quantifiers?
In predicate logic, predicates are used alongside quantifiers to express the extent to which a predicate is true over a range of elements. Using quantifiers to create such propositions is called quantification.
There are two types of quantification-
1. Universal Quantification- Mathematical statements sometimes assert that a property is true for all the values of a variable in a particular domain, called the domain of discourse. Such a statement is expressed using universal quantification.
The universal quantification of for a particular domain is the proposition that asserts that is true for all values of in this domain. The domain is very important here since it decides the possible values of . The meaning of the universal quantification of changes when the domain is changed. The domain must be specified when a universal quantification is used, as without it, it has no meaning.
Formally,
The universal quantification of is the statement
" for all values of in the domain"
The notation denotes the universal quantification of .
Here is called the universal quantifier.
is read as "for all ".
• Example 1: Let be the statement “ > “. What is the truth value of the statement ?
Solution: As is greater than for any real number, so for all or .
2. Existential Quantification- Some mathematical statements assert that there is an element with a certain property. Such statements are expressed by existential quantification. Existential quantification can be used to form a proposition that is true if and only if is true for at least one value of in the domain.
Formally,
The existential quantification of is the statement
"There exists an element in the domain such that "
The notation denotes the existential quantification of .
Here is called the existential quantifier.
is read as "There is atleast one such such that ".
• Example : Let be the statement “ > 5″. What is the truth value of the statement ?
Solution: is true for all real numbers greater than 5 and false for all real numbers less than 5. So .
To summarise,
Now if we try to convert the statement, given in the beginning of this article, into a mathematical statement using predicate logic, we would get something like-
Here, P(x) is the statement "x is 18 years or older and,
Q(x) is the statement "x is eligible to vote".
Notice that the given statement is not mentioned as a biconditional and yet we used one. This is because Natural language is ambiguous sometimes, and we made an assumption. This assumption was made since it is true that a person can vote if and only if he/she is 18 years or older. Refer Introduction to Propositional Logic for more explanation.
Other Quantifiers –
Although the universal and existential quantifiers are the most important in Mathematics and Computer Science, they are not the only ones. In Fact, there is no limitation on the number of different quantifiers that can be defined, such as “exactly two”, “there are no more than three”, “there are at least 10”, and so on.
Of all the other possible quantifiers, the one that is seen most often is the uniqueness quantifier, denoted by .
The notation states "There exists a unique such that is true".
Quantifiers with restricted domains
As we know that quantifiers are meaningless if the variables they bind do not have a domain. The following abbreviated notation is used to restrict the domain of the variables-
> 0, > 0.
The above statement restricts the domain of , and is a shorthand for writing another proposition, that says , in the statement.
If we try to rewrite this statement using an implication, we would get-
> >
Similarly a statement using Existential quantifier can be restated using a conjuction between the domain restricting proposition and the actual predicate.
1. Restriction of a universal quantification is the same as the
universal quantification of a conditional statement.
2. Restriction of a existentital quantification is the same as the
existential quantification of a conjunction.
Definitions to Note
1. Binding variables- A variable whose occurrence is bound by a quantifier is called
a bound variable. Variables not bound by any quantifiers are called free variables.
2. Scope- The part of the logical expression to which a quantifier is applied is called
the scope of the quantifier.
This topic has been covered in two parts. The second part of this topic is explained in another article – Predicates and Quantifiers – Set 2
References-
First Order Logic – Wikipedia
Quantifiers – Wikipedia
Discrete Mathematics and its Applications, by Kenneth H Rosen
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# 1.01 The real number system
Lesson
Our real numbers system hasn't been around in its current state forever. It was developed slowly over time. The real number system includes rationals, irrationals, integers, whole numbers, and natural numbers.
Classification of numbers is about identifying which set, or sets, a number might belong to. It might be helpful to remember the different types of numbers as a story about filling in the numbers on a number line.
The first numbers we put on the line are the natural numbers.
Natural numbers
The set of natural numbers are the counting numbers, starting from $1$1:
$1,2,3,4,5,6,7,\ldots$1,2,3,4,5,6,7,
Next, we will add $0$0 to our line to show the whole numbers.
Whole numbers
The set of whole numbers are the counting numbers, starting from $0$0:
$0,1,2,3,4,5,6,7,\ldots$0,1,2,3,4,5,6,7,
The left side of this line looks pretty empty. If we add all the negatives we now have a set of numbers called the integers.
Integers
Whole numbers together with negative numbers make up the set of integers:
$\ldots,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,\ldots$,7,6,5,4,3,2,1,0,1,2,3,4,5,6,7,
But are there numbers between the ones we already have marked? The answer is yes - an infinite amount of numbers between every little mark!
What sort of numbers are these? Well, rational numbers are all numbers that indicate whole numbers as well as parts of whole numbers. So fractions, decimals, and percentages are added to our number line to create the set of rational numbers.
Rational numbers
Integers together with all fractions (including repeating or repeating decimals) make up the set of Rational Numbers.
They cannot be listed, but here are some examples:
$\ldots,-8,-7.4,-7,-6,-5.33387,-4,-2,0,\frac{1}{2},75%,1,2,3,3.5656,\ldots$,8,7.4,7,6,5.33387,4,2,0,12,75%,1,2,3,3.5656,
But wait! Our number line is still not quite full. There are still gaps in a few places. These gaps are filled with numbers we call irrational numbers. These are numbers like $\sqrt{21}$21 and $\pi$π:
Now we can revisit our picture of the different sets of numbers in the real number system. Looking at the image below, we can see some examples of numbers that below to each set:
Notice that some number sets are entirely contained within larger number sets. For example, all of the whole numbers like $1,2,3,17,28736,\ldots$1,2,3,17,28736, etc. are also integers. But there are some integers, like $-1,-2,-56,-98324$1,2,56,98324, that are not whole numbers.
Similarly, rational numbers are also real numbers, but the set of real numbers includes all the rational numbers and all the irrational numbers.
#### Practice questions
##### QUestion 1
Height above sea level is expressed as a positive quantity. Which set of numbers is the most appropriate for describing the position of a submarine relative to sea level?
1. Integers
A
Whole numbers
B
Integers
A
Whole numbers
B
##### QUESTION 2
Using the diagram, complete the following statement.
A real number is either:
1. a whole number or an irrational number.
A
an integer or an irrational number.
B
a rational number or an irrational number.
C
an integer or a rational number.
D
a whole number or an irrational number.
A
an integer or an irrational number.
B
a rational number or an irrational number.
C
an integer or a rational number.
D
##### QUESTIOn 3
Using the diagram, classify the number $\sqrt{49}$49.
Select all that apply.
1. $\sqrt{49}$49 is an irrational number.
A
$\sqrt{49}$49 is an integer.
B
$\sqrt{49}$49 is a rational number.
C
$\sqrt{49}$49 is a whole number.
D
$\sqrt{49}$49 is an irrational number.
A
$\sqrt{49}$49 is an integer.
B
$\sqrt{49}$49 is a rational number.
C
$\sqrt{49}$49 is a whole number.
D
### Outcomes
#### 8.NS.1
Give examples of rational and irrational numbers and explain the difference between them. Understand that every number has a decimal expansion; for rational numbers, show that the decimal expansion terminates or repeats, and convert a decimal expansion that repeats into a rational number. |
# Decimal
Let’s learn about the decimal number system.
We'll cover the following
Everyone is familiar with decimal numbers. We use decimal numbers in our calculations everyday. Representing numbers in decimals is just one possible way of writing a number. There are many other writing methods, such as Roman numbers and Chinese characters. This section introduces two other ways of writing numbers: binary and hex. Binary and hex are particularly useful in digital communication. In fact, at the most fundamental level, binary numbers are the basis of all computing. In this section, we’ll make extensive use of the notation for decimal exponentiation. For example, if we write $10^7$ (10 to the power 7), then we mean 10 multiplied by itself seven times:
$10^7 = 10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10$
Note: It is a standard mathematical convention to say that any number raised to the power 0 is 1. For example, $10^0$ = 1.
## Decimal
Let’s understand some of the principles behind decimal numbers.
### Writing a number as a decimal
Consider the number 359. What do 3, 5, and 9 actually represent? Observe that:
$359 = 300 + 50 + 9$
$= (3 \times 100) + (5 \times 10) + (9 \times 1)$
$= (3 \times 10^2 ) + (5 \times 10^1 ) + (9 \times 10^0)$
All we have done in these last three lines is rewrite the number 359 in a form that shows how it can be represented as a sum of multiples of powers of ten. This is why our most familiar numbers are often referred to as decimal numbers: the digits we use to write them to indicate the multiples of powers of ten that are added together to make up the number. Observe two things:
1. The digits of a decimal number can take any of the values from 0 up to 9.
2. Every digit in a decimal number will be multiplied by some power of 10. The powers of 10 start with 0 (the furthest digit to the right) and then increase from right to left.
Decimal numbers are sometimes also referred to as base 10 numbers. This is because the representation of the number in digits is ‘based’ on the powers of 10, as we have just seen. We normally don’t bother indicating that a number is written in base 10, since this tends to be assumed by default. However, because we will be changing numbers from one base into another, we will often need to indicate the base being used. To indicate that the number 359 is a decimal number, we will sometimes write it as $359_{10}$.
As a last remark about decimal (or any number base), any number of zeros can be put in front of a decimal number without changing its value. These are called leading zeros. For example:
$00013 = 013 = 13$
This is consistent with our previous way of expressing decimal numbers, since:
$00013 = (0 \times 10^4 )$
$= (0 \times 10^3 ) + (0 \times 10^2 ) + (1 \times 10^1 ) + (3 \times 10^0)$
$= 0 + 0 + 0 + (1 \times 10^1 ) + (3 \times 10^0) = (1 \times 10^1 ) + (3 \times 10^0) = 13$
Of course, when writing numbers in decimal, we do not normally use any leading zeros.
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## Sunday, August 30, 2015
### AMCAT Quantitative ability Material ( Problems on numbers )
2015 & 2016 Batch Freshers Registration Link
PROBLEMS ON NUMBERS
In this section, questions involving a set of numbers are put in the form of a puzzle. You have to analyze the given conditions, assume the unknown numbers and form equations accordingly, which on solving yield the unknown numbers.
EXAMPLE Problems
Ex.1. A number is as much greater than 36 as is less than 86. Find the number.
Sol. Let the number be x. Then, x - 36 = 86 - x => 2x = 86 + 36 = 122 => x = 61. Hence, the required number is 61.
# Ex. 2. Find a number such that when 15 is subtracted from 7 times the number, the
Result is 10 more than twice the number. (Hotel Management, 2002)
Sol. Let the number be x. Then, 7x - 15 = 2x + 10 => 5x = 25 =>x = 5.
Hence, the required number is 5.
Ex. 3. The sum of a rational number and its reciprocal is 13/6. Find the number.
(S.S.C. 2000)
Sol. Let the number be x.
Then, x + (1/x) = 13/6 => (x2 + 1)/x = 13/6 => 6x2 – 13x + 6 = 0
=> 6x2 – 9x – 4x + 6 = 0 => (3x – 2) (2x – 3) = 0
ð x = 2/3 or x = 3/2
Hence the required number is 2/3 or 3/2.
Ex. 4. The sum of two numbers is 184. If one-third of the one exceeds one-seventh
of the other by 8, find the smaller number.
Sol. Let the numbers be x and (184 - x). Then,
(X/3) - ((184 – x)/7) = 8 => 7x – 3(184 – x) = 168 => 10x = 720 => x = 72.
So, the numbers are 72 and 112. Hence, smaller number = 72.
Ex. 5. The difference of two numbers is 11 and one-fifth of their sum is 9. Find the numbers.
Sol. Let the number be x and y. Then,
x – y = 11 ----(i) and 1/5 (x + y) = 9 => x + y = 45 ----(ii)
Adding (i) and (ii), we get: 2x = 56 or x = 28. Putting x = 28 in (i), we get: y = 17.
Hence, the numbers are 28 and 17.
Ex. 6. If the sum of two numbers is 42 and their product is 437, then find the
absolute difference between the numbers. (S.S.C. 2003)
Sol. Let the numbers be x and y. Then, x + y = 42 and xy = 437
x - y = sqrt[(x + y)2 - 4xy] = sqrt[(42)2 - 4 x 437 ] = sqrt[1764 – 1748] = sqrt[16] = 4.
Required difference = 4.
Ex. 7. The sum of two numbers is 16 and the sum of their squares is 113. Find the
numbers.
Sol. Let the numbers be x and (15 - x).
Then, x2 + (15 - x)2 = 113 => x2 + 225 + X2 - 30x = 113
=> 2x2 - 30x + 112 = 0 => x2 - 15x + 56 = 0
=> (x - 7) (x - 8) = 0 => x = 7 or x = 8.
So, the numbers are 7 and 8.
Ex. 8. The average of four consecutive even numbers is 27. Find the largest of these
numbers.
Sol. Let the four consecutive even numbers be x, x + 2, x + 4 and x + 6.
Then, sum of these numbers = (27 x 4) = 108.
So, x + (x + 2) + (x + 4) + (x + 6) = 108 or 4x = 96 or x = 24.
:. Largest number = (x + 6) = 30.
Ex. 9. The sum of the squares of three consecutive odd numbers is 2531.Find the
numbers.
Sol. Let the numbers be x, x + 2 and x + 4.
Then, X2 + (x + 2)2 + (x + 4)2 = 2531 => 3x2 + 12x - 2511 = 0
=> X2 + 4x - 837 = 0 => (x - 27) (x + 31) = 0 => x = 27.
Hence, the required numbers are 27, 29 and 31.
Ex. 10. Of two numbers, 4 times the smaller one is less then 3 times the 1arger one by 5. If the sum of the numbers is larger than 6 times their difference by 6, find the two numbers.
Sol. Let the numbers be x and y, such that x > y
Then, 3x - 4y = 5 ...(i) and (x + y) - 6 (x - y) = 6 => -5x + 7y = 6 …(ii)
Solving (i) and (ii), we get: x = 59 and y = 43.
Hence, the required numbers are 59 and 43.
Ex. 11. The ratio between a two-digit number and the sum of the digits of that
number is 4 : 1.If the digit in the unit's place is 3 more than the digit in the ten’s place, what is the number?
Sol. Let the ten's digit be x. Then, unit's digit = (x + 3).
Sum of the digits = x + (x + 3) = 2x + 3. Number = l0x + (x + 3) = llx + 3.
11x+3 / 2x + 3 = 4 / 1 => 1lx + 3 = 4 (2x + 3) => 3x = 9 => x = 3.
Hence, required number = 11x + 3 = 36.
Ex. 12. A number consists of two digits. The sum of the digits is 9. If 63 is subtracted
from the number, its digits are interchanged. Find the number.
Sol. Let the ten's digit be x. Then, unit's digit = (9 - x).
Number = l0x + (9 - x) = 9x + 9.
Number obtained by reversing the digits = 10 (9 - x) + x = 90 - 9x.
therefore, (9x + 9) - 63 = 90 - 9x => 18x = 144 => x = 8.
So, ten's digit = 8 and unit's digit = 1.
Hence, the required number is 81.
Ex. 13. A fraction becomes 2/3 when 1 is added to both, its numerator and denominator.
And ,it becomes 1/2 when 1 is subtracted from both the numerator and denominator. Find the fraction.
Sol. Let the required fraction be x/y. Then,
x+1 / y+1 = 2 / 3 => 3x – 2y = - 1 …(i) and x – 1 / y – 1 = 1 / 2
ð 2x – y = 1 …(ii)
Solving (i) and (ii), we get : x = 3 , y = 5
therefore, Required fraction= 3 / 5.
Ex. 14. 50 is divided into two parts such that the sum of their reciprocals is 1/ 12.Find the two parts.
Sol. Let the two parts be x and (50 - x).
Then, 1 / x + 1 / (50 – x) = 1 / 12 => (50 – x + x) / x ( 50 – x) = 1 / 12
=> x2 – 50x + 600 = 0 => (x – 30) ( x – 20) = 0 => x = 30 or x = 20.
So, the parts are 30 and 20.
Ex. 15. If three numbers are added in pairs, the sums equal 10, 19 and 21. Find the
numbers )
Sol. Let the numbers be x, y and z. Then,
x+ y = 10 ...(i) y + z = 19 ...(ii) x + z = 21 …(iii)
Adding (i) ,(ii) and (iii), we get: 2 (x + y + z ) = 50 or (x + y + z) = 25.
Thus, x= (25 - 19) = 6; y = (25 - 21) = 4; z = (25 - 10) = 15.
Hence, the required numbers are 6, 4 and 15. |
### Home > CCG > Chapter Ch10 > Lesson 10.1.5 > Problem10-59
10-59.
In the figure at right, $\overleftrightarrow{ E X }$ is tangent to $⊙O$ at point $X$. $OE=20$ cm and $XE=15$ cm.
1. What is the area of the circle?
According to the definition of the tangent line, what is the measure of $∠EXO$?
How can you use this to find the length of $XO$?
A tangent line is perpendicular to a radius, so the measure of $∠EXO = 90º$, so the triangle is a right triangle. So we can use the Pythagorean Theorem.
$\left(XE\right)^2+\left(XO\right)^2=\left(OE\right)^2\\ \quad \ 15^2+(XO)^2=20^2\\ \qquad \qquad \ \ \ XO^2=175\\ \qquad \qquad \quad \ XO=13.23$
$A = \pi r^2\\ A=\pi (13.23)^2\\ A = 175\pi\text{ cm}^2\\ A = 549.78\text{ cm}^2$
1. What is the area of the sector bounded by $\overline{OX}$ and $\overline{ON}$?
Find the measure of $∠XON$. Use this to find the area of the sector.
$A\approx74.21\text{ cm}^2$
2. Find the area of the region bounded by $\overline{XE}$, $\overline{NE}$, and $\overarc{ N X }$.
Find the area of the triangle formed by $E, X,\text{ and }O$. Subtract the area of the sector from the area of the triangle. |
# What is the factorial number of 9?
## What is the factorial number of 9?
362,880
Factorials of Numbers 1 to 10 Table
n Factorial of a Number n! Value
7 7! 5,040
8 8! 40,320
9 9! 362,880
10 10! 3,628,800
## How do you calculate a factorial?
To find the factorial of a number, multiply the number with the factorial value of the previous number. For example, to know the value of 6! multiply 120 (the factorial of 5) by 6, and get 720.
What is the factorial value of 8?
40320 is a value of 8 factorial .
What does 4 factorial look like?
Factorials are very simple things. They’re just products, indicated by an exclamation mark. For instance, “four factorial” is written as “4!” and means 1×2×3×4 = 24. (“enn factorial”) means the product of all the whole numbers from 1 to n; that is, n!
### What is N factorial equal to?
In more mathematical terms, the factorial of a number (n!) is equal to n(n-1). For example, if you want to calculate the factorial for four, you would write: 4!
### What type of number is 9?
composite number
9 is a composite number, its proper divisors being 1 and 3. It is 3 times 3 and hence the third square number. Nine is a Motzkin number. It is the first composite lucky number, along with the first composite odd number and only single-digit composite odd number.
What is a factorial of 20?
Answer: The factorial of 20 is 2432902008176640000.
What does 7 factorial mean?
Factorial, in mathematics, the product of all positive integers less than or equal to a given positive integer and denoted by that integer and an exclamation point. Thus, factorial seven is written 7!, meaning 1 × 2 × 3 × 4 × 5 × 6 × 7. Factorial zero is defined as equal to 1.
## What is factorial example?
Factorials (!) are products of every whole number from 1 to n. In other words, take the number and multiply through to 1. For example: If n is 3, then 3! is 3 x 2 x 1 = 6.
## How much is 100 factorial?
The number of zeros in 100! will be 24 .
What is a factorial symbol?
The factorial (denoted or represented as n!) for a positive number or integer (which is denoted by n) is the product of all the positive numbers preceding or equivalent to n (the positive integer). The notation for a factorial (n!) was introduced in the early 1800s by Christian Kramp, a French mathematician.
Is 9 an evil number?
Examples. The first evil numbers are: 0, 3, 5, 6, 9, 10, 12, 15, 17, 18, 20, 23, 24, 27, 29, 30, 33, 34, 36, 39 … |
# Differentiate the following functions with respect to x :
Question:
Differentiate the following functions with respect to $x$ :
$\sin ^{-1}\left(\frac{2^{x+1}}{1+4^{x}}\right)$
Solution:
$y=\sin ^{-1}\left\{\frac{2^{x+1}}{1+4^{x}}\right\}$
For function to be defined
$-1 \leq \frac{2^{x+1}}{1+4^{x}} \leq 1$
Since the quantity is positive always
$\Rightarrow 0 \leq \frac{2^{x+1}}{1+4^{x}} \leq 1$
$\Rightarrow 0<2^{x+1} \leq 1+4^{x}$
$\Rightarrow 0<2 \leq 2^{-x}+2^{x}$
This condition is always true, hence function is always defined.
$y=\sin ^{-1}\left\{\frac{2 \times 2^{x}}{1+\left(2^{2}\right)^{x}}\right\}$
Let $2^{x}=\tan \theta$
$y=\sin ^{-1}\left\{\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right\}$
Using $\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$
Now,
$y=\sin ^{-1}(\sin 2 \theta)$
$y=2 \theta$
$y=2 \tan ^{-1}\left(2^{x}\right)$
Differentiating w.r.t $\mathrm{x}$, we get
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(2 \tan ^{-1} 2^{\mathrm{x}}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}}=2 \times \frac{2^{\mathrm{x}} \log 2}{1+\left(2^{\mathrm{x}}\right)^{2}}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2^{\mathrm{x}+1} \log 2}{1+4^{\mathrm{x}}}$ |
# Continued… Obj: draw Box-and-whisker plots representing a set of data Do now: use your calculator to find the mean for 85, 18, 87, 100, 27, 34, 93, 52,
## Presentation on theme: "Continued… Obj: draw Box-and-whisker plots representing a set of data Do now: use your calculator to find the mean for 85, 18, 87, 100, 27, 34, 93, 52,"— Presentation transcript:
continued… Obj: draw Box-and-whisker plots representing a set of data Do now: use your calculator to find the mean for 85, 18, 87, 100, 27, 34, 93, 52, 91, 54, 59, 61, 82, 68, 78
Step 1 - Find the median. 18, 27, 34, 52, 54, 59, 61, 68, 78, 82, 85, 87, 91, 93, 100 68 is the median of this data set. Step 2 – Find the lower quartile. *The lower quartile is the median of the data set to the left of 68. (18, 27, 34, 52, 54, 59, 61,) 68, 78, 82, 85, 87, 91, 93, 100 52 is the lower quartile
Step 3 – Find the upper quartile. *The upper quartile is the median of the data set to the right of 68. 18, 27, 34, 52, 54, 59, 61, 68, (78, 82, 85, 87, 91, 93, 100) 87 is the upper quartile Step 4 – Find the maximum and minimum 18, 27, 34, 52, 54, 59, 61, 68, 78, 82, 85, 87, 91, 93, 100 18 is the minimum and 100 is the maximum.
In order to draw the box-and-whisker plot, Organize the 5 number summary Min – 18 Lower Quartile – 52 Median – 68 Upper Quartile – 87 Max – 100
Notice, the Box includes the lower quartile, median, and upper quartile. The Whiskers extend from the Box to the max and min.
Ex 4: Use the following set of data to create the 5 number summary and a box-and-whisker plot. 3, 7, 11, 11, 15, 21, 23, 39, 41, 45, 50, 61, 87, 99, 220
Median 3, 7, 11, 11, 15, 21, 23, 39, 41, 45, 50, 61, 87, 99, 220 The median is 39
Lower Quartile ( 1 st Quartile ) (3, 7, 11, 11, 15, 21, 23), 39, 41, 45, 50, 61, 87, 99, 220 The lower quartile is 11
Upper Quartile ( 3rd Quartile ) 3, 7, 11, 11, 15, 21, 23, 39, (41, 45, 50, 61, 87, 99, 220) The upper quartile is 61
Maximum 3, 7, 11, 11, 15, 21, 23, 39, 41, 45, 50, 61, 87, 99, 220 The max is 220
Minimum 3, 7, 11, 11, 15, 21, 23, 39, 41, 45, 50, 61, 87, 99, 220 The min is 3
Min - 3 Lower Quartile - 11 Median - 39 Upper Quartile - 61 Max - 220
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7.5: Linear Systems with Multiplication
Difficulty Level: At Grade Created by: CK-12
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What if you were given a system of linear equations like x2y=7\begin{align*}x - 2y = 7\end{align*} and 3x4y=3\begin{align*}3x - 4y = -3\end{align*}? How could you solve for one of the variables by eliminating the other? After completing this Concept, you'll be able to solve a system of linear equations by multiplication and then elimination.
Guidance
So far, we’ve seen that the elimination method works well when the coefficient of one variable happens to be the same (or opposite) in the two equations. But what if the two equations don’t have any coefficients the same?
It turns out that we can still use the elimination method; we just have to make one of the coefficients match. We can accomplish this by multiplying one or both of the equations by a constant.
Here’s a quick review of how to do that. Consider the following questions:
1. If 10 apples cost $5, how much would 30 apples cost? 2. If 3 bananas plus 2 carrots cost$4, how mush would 6 bananas plus 4 carrots cost?
If you look at the first equation, it should be obvious that each apple costs $0.50. So 30 apples should cost$15.00.
The second equation is trickier; it isn’t obvious what the individual price for either bananas or carrots is. Yet we know that the answer to question 2 is 8.00. How? If we look again at question 1, we see that we can write an equation: 10a=5\begin{align*}10a = 5\end{align*} (a\begin{align*}a\end{align*} being the cost of 1 apple). So to find the cost of 30 apples, we could solve for a\begin{align*}a\end{align*} and then multiply by 30—but we could also just multiply both sides of the equation by 3. We would get 30a=15\begin{align*}30a = 15\end{align*}, and that tells us that 30 apples cost15.
And we can do the same thing with the second question. The equation for this situation is 3b+2c=4\begin{align*}3b + 2c = 4\end{align*}, and we can see that we need to solve for (6b+4c)\begin{align*}(6b + 4c)\end{align*}, which is simply 2 times (3b+2c)\begin{align*}(3b + 2c)\end{align*}! So algebraically, we are simply multiplying the entire equation by 2:
2(3b+2c)6b+4c=24=8distribute and multiply:\begin{align*}2(3b + 2c) &= 2 \cdot 4 && distribute \ and \ multiply:\\ 6b + 4c&= 8\end{align*}
So when we multiply an equation, all we are doing is multiplying every term in the equation by a fixed amount.
Solving a Linear System by Multiplying One Equation
If we can multiply every term in an equation by a fixed number (a scalar), that means we can use the addition method on a whole new set of linear systems. We can manipulate the equations in a system to ensure that the coefficients of one of the variables match.
This is easiest to do when the coefficient as a variable in one equation is a multiple of the coefficient in the other equation.
Example A
Solve the system:
7x+4y5x2y=17=11\begin{align*}7x + 4y & = 17\\ 5x - 2y & = 11\end{align*}
Solution
You can easily see that if we multiply the second equation by 2, the coefficients of y\begin{align*}y\end{align*} will be +4 and -4, allowing us to solve the system by addition:
2 times equation 2:
10x4y=22+ (7x+4y)=17 17x=34divide by 17 to get: x=2now add to equation one:\begin{align*}& \qquad \qquad \qquad \ \ \quad 10x - 4y = 22 && now \ add \ to \ equation \ one:\\ & \qquad \qquad \quad \underline{\;\;\;+ \ \ (7x + 4y) = 17\;\;\;\;\;}\\ & \qquad \qquad \qquad \qquad \ \ \ \quad 17x = 34\\ \\ & divide \ by \ 17 \ to \ get: \qquad \ x = 2\end{align*}
Now simply substitute this value for x\begin{align*}x\end{align*} back into equation 1:
72+4y4yy=17=3=0.75since 7×2=14, subtract 14 from both sides:divide by 4:\begin{align*}7 \cdot 2 + 4y &= 17 && since \ 7 \times 2 = 14, \ subtract \ 14 \ from \ both \ sides:\\ 4y &= 3 && divide \ by \ 4:\\ y &= 0.75\end{align*}
Example B
Anne is rowing her boat along a river. Rowing downstream, it takes her 2 minutes to cover 400 yards. Rowing upstream, it takes her 8 minutes to travel the same 400 yards. If she was rowing equally hard in both directions, calculate, in yards per minute, the speed of the river and the speed Anne would travel in calm water.
Solution
Step one: first we convert our problem into equations. We know that distance traveled is equal to speed ×\begin{align*}\times\end{align*} time. We have two unknowns, so we’ll call the speed of the river x\begin{align*}x\end{align*}, and the speed that Anne rows at y\begin{align*}y\end{align*}. When traveling downstream, her total speed is her rowing speed plus the speed of the river, or (x+y)\begin{align*}(x + y)\end{align*}. Going upstream, her speed is hindered by the speed of the river, so her speed upstream is (xy)\begin{align*}(x - y)\end{align*}.
Downstream Equation: 2(x+y)=400\begin{align*}2(x + y) = 400\end{align*}
Upstream Equation: 8(xy)=400\begin{align*}8(x - y) = 400\end{align*}
Distributing gives us the following system:
2x+2y8x8y=400=400\begin{align*}2x + 2y &= 400\\ 8x - 8y &= 400\end{align*}
Right now, we can’t use the method of elimination because none of the coefficients match. But if we multiplied the top equation by 4, the coefficients of y\begin{align*}y\end{align*} would be +8 and -8. Let’s do that:
8x+8y=1,600 + (8x8y)=40016x=2,000\begin{align*}& \quad \qquad \ 8x + 8y = 1,600\\ & \ \underline{\;\;\; + \ \ (8x - 8y) = 400\;\;\;}\\ & \quad \qquad \qquad 16x = 2,000\end{align*}
Now we divide by 16 to obtain x=125\begin{align*}x = 125\end{align*}.
Substitute this value back into the first equation:
2(125+y)125+yy=400=200=75divide both sides by 2:subtract 125 from both sides:\begin{align*}2(125 + y) &= 400 && divide \ both \ sides \ by \ 2:\\ 125 + y &= 200 && subtract \ 125 \ from \ both \ sides:\\ y &= 75\end{align*}
Anne rows at 125 yards per minute, and the river flows at 75 yards per minute.
Solving a Linear System by Multiplying Both Equations
So what do we do if none of the coefficients match and none of them are simple multiples of each other? We do the same thing we do when we’re adding fractions whose denominators aren’t simple multiples of each other. Remember that when we add fractions, we have to find a lowest common denominator—that is, the lowest common multiple of the two denominators—and sometimes we have to rewrite not just one, but both fractions to get them to have a common denominator. Similarly, sometimes we have to multiply both equations by different constants in order to get one of the coefficients to match.
Example C
Andrew and Anne both use the I-Haul truck rental company to move their belongings from home to the dorm rooms on the University of Chicago campus. I-Haul has a charge per day and an additional charge per mile. Andrew travels from San Diego, California, a distance of 2060 miles in five days. Anne travels 880 miles from Norfolk, Virginia, and it takes her three days. If Anne pays $840 and Andrew pays$1845, what does I-Haul charge
a) per day?
b) per mile traveled?
Solution
First, we’ll set up our equations. Again we have 2 unknowns: the daily rate (we’ll call this x\begin{align*}x\end{align*}), and the per-mile rate (we’ll call this y\begin{align*}y\end{align*}).
Anne’s equation: 3x+880y=840\begin{align*}3x + 880y = 840\end{align*}
Andrew’s Equation: 5x+2060y=1845\begin{align*}5x + 2060y = 1845\end{align*}
We can’t just multiply a single equation by an integer number in order to arrive at matching coefficients. But if we look at the coefficients of x\begin{align*}x\end{align*} (as they are easier to deal with than the coefficients of y\begin{align*}y\end{align*}), we see that they both have a common multiple of 15 (in fact 15 is the lowest common multiple). So we can multiply both equations.
Multiply the top equation by 5:
15x+4400y=4200\begin{align*}15x + 4400y = 4200\end{align*}
Multiply the lower equation by 3:
15x+6180y=5535\begin{align*}15x + 6180y = 5535\end{align*}
Subtract:
15x+4400y=4200 (15x+6180y)=5535 1780y=1335Divide by 1780: y=0.75\begin{align*}& \qquad \qquad 15x + 4400y = 4200\\ & \underline{\quad \ - \ \ \ (15x + 6180y) = 5535\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ \ -1780y = -1335\\ \\ & \text{Divide by}\ -1780: \ y = 0.75\end{align*}
Substitute this back into the top equation:
3x+880(0.75)3xx=840=180=60since 880×0.75=660, subtract 660 from both sides:divide both sides by 3\begin{align*}3x + 880(0.75) &= 840 && since \ 880 \times 0.75 = 660, \ subtract \ 660 \ from \ both \ sides:\\ 3x &= 180 && divide \ both \ sides \ by \ 3\\ x &= 60\end{align*}
I-Haul charges $60 per day plus$0.75 per mile.
Watch this video for help with the Examples above.
Vocabulary
• Elimination method: The purpose of the elimination method to solve a system is to cancel, or eliminate, a variable by either adding or subtracting the two equations. Sometimes the equations must be multiplied by scalars first, in order to cancel out a variable.
• Least common multiple: The least common multiple is the smallest value that is divisible by two or more quantities without a remainder.
Guided Practice
Solve the system \begin{align*}\begin{cases} 4x+7y=6\\ 6x+5y=20 \end{cases}\end{align*}.
Solution:
Neither \begin{align*}x\end{align*} nor \begin{align*}y\end{align*} have additive inverse coefficients, but the \begin{align*}x\end{align*}-variables do share a common factor of 2. Thus we can most easily eliminate \begin{align*}x\end{align*}.
In order to make \begin{align*}x\end{align*} have the same coefficient in each equation, we must multiply one equation by the factor not shared in the coefficient of \begin{align*}x\end{align*} in the other equation. We need to multiply the first equation by 3 and the second equation by 2, making one of them negative as well:
\begin{align*}\begin{cases} 3(4x+7y=6)\\ -2(6x+5y=20) \end{cases} & \rightarrow \quad \begin{cases} 12x+21y=18\\ -12x-10y=-40\end{cases}\end{align*}
\begin{align*}\text{Add the two equations.} && 11y&=-22\\ \text{Divide by} \ 11. && y&=-2\end{align*}
To find the \begin{align*}x-\end{align*}value, use the Substitution Property in either equation.
\begin{align*}4x+7(2)&=6\\ 4x+14&=6\\ 4x&=-8\\ x&=-2\end{align*}
The solution to this system is \begin{align*} (-2,-2)\end{align*}.
Practice
Solve the following systems using multiplication.
1. \begin{align*}5x - 10y = 15\!\\ 3x - 2y = 3\end{align*}
2. \begin{align*}5x - y = 10\!\\ 3x - 2y = -1\end{align*}
3. \begin{align*}5x + 7y = 15\!\\ 7x - 3y = 5\end{align*}
4. \begin{align*}9x + 5y = 9\!\\ 12x + 8y = 12.8\end{align*}
5. \begin{align*}4x - 3y = 1\!\\ 3x - 4y = 4\end{align*}
6. \begin{align*}7x - 3y = -3\!\\ 6x + 4y = 3\end{align*}
7. \begin{align*}&x=3y\\ &x-2y=-3\end{align*}
8. \begin{align*}&y=3x+2\\ &y=-2x+7\end{align*}
9. \begin{align*}&5x-5y=5\\ &5x+5y=35\end{align*}
10. \begin{align*}&y=-3x-3\\ &3x-2y+12=0\end{align*}
11. \begin{align*}&3x-4y=3\\ &4y+5x=10\end{align*}
12. \begin{align*}&9x-2y=-4\\ &2x-6y=1\end{align*}
Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Vocabulary Language: English
elimination
The elimination method for solving a system of two equations involves combining the two equations in order to produce one equation in one variable.
Least Common Denominator
The least common denominator or lowest common denominator of two fractions is the smallest number that is a multiple of both of the original denominators.
Least Common Multiple
The least common multiple of two numbers is the smallest number that is a multiple of both of the original numbers.
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# Find the equation of the circle passing through the points (2, 3)
Question:
Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line – 3y – 11 = 0.
Solution:
Let the equation of the required circle be $(x-h)^{2}+(y-k)^{2}=r^{2}$.
Since the circle passes through points (2, 3) and (–1, 1),
$(2-h)^{2}+(3-k)^{2}=r^{2} \ldots(1)$
$(-1-h)^{2}+(1-k)^{2}=r^{2} \ldots(2)$
Since the centre (h, k) of the circle lies on line – 3y – 11 = 0,
– 3k = 11 … (3)
From equations (1) and (2), we obtain
$(2-h)^{2}+(3-k)^{2}=(-1-h)^{2}+(1-k)^{2}$
$\Rightarrow 4-4 h+h^{2}+9-6 k+k^{2}=1+2 h+h^{2}+1-2 k+k^{2}$
$\Rightarrow 4-4 h+9-6 k=1+2 h+1-2 k$
$\Rightarrow 6 h+4 k=11 \ldots(4)$
On solving equations (3) and (4), we obtain $h=\frac{7}{2}$ and $k=\frac{-5}{2}$.
On substituting the values of $h$ and $k$ in equation (1), we obtain
$\left(2-\frac{7}{2}\right)^{2}+\left(3+\frac{5}{2}\right)^{2}=r^{2}$
$\Rightarrow\left(\frac{4-7}{2}\right)^{2}+\left(\frac{6+5}{2}\right)^{2}=r^{2}$
$\Rightarrow\left(\frac{-3}{2}\right)^{2}+\left(\frac{11}{2}\right)^{2}=r^{2}$
$\Rightarrow \frac{9}{4}+\frac{121}{4}=r^{2}$
$\Rightarrow \frac{130}{4}=r^{2}$
Thus, the equation of the required circle is
$\left(x-\frac{7}{2}\right)^{2}+\left(y+\frac{5}{2}\right)^{2}=\frac{130}{4}$
$\left(\frac{2 x-7}{2}\right)^{2}+\left(\frac{2 y+5}{2}\right)^{2}=\frac{130}{4}$
$4 x^{2}-28 x+49+4 y^{2}+20 y+25=130$
$4 x^{2}+4 y^{2}-28 x+20 y-56=0$
$4\left(x^{2}+y^{2}-7 x+5 y-14\right)=0$
$x^{2}+y^{2}-7 x+5 y-14=0$ |
Etymology []. Contrapositive Proof Example Proposition Suppose n 2Z. converse of proposition contrapositive of proposition Contents For the proposition P Q, the proposition Q P is called its converse, and the proposition Q P is called its contrapositive. This latter statement can be proven as follows: suppose that x is not even, then x is odd. (Contrapositive) Let integer n be given. Proof. Converse and Contrapositive Subjects to be Learned. The logical contrapositive of a conditional statement is created by negating the hypothesis and conclusion, then switching them. : a proposition or theorem formed by contradicting both the subject and predicate or both the hypothesis and conclusion of a given proposition or theorem and interchanging them 'if not-B then not-A ' is the contrapositive of 'if A then B ' Although a direct proof can be given, we choose to prove this statement by contraposition. Suppose you have the conditional statement {\color{blue}p} \to {\color{red}q}, we compose the contrapositive statement by interchanging the hypothesis and conclusion of the inverse of the same conditional statement.. This is an example of a case where one has to be careful, the negation is \n ja or n jb." (logic) The inverse of the converse of a given proposition. The positions of p and q of the original statement are switched, and then the opposite of each is considered: $$\sim q \rightarrow \sim p$$. But our main reason for introducing it is that it provides more opportunities to practice writing proofs, both direct and contrapositive. (noun) The proves the contrapositive of the original proposition, Example. For example for the proposition "If it rains, then I get wet", Converse: If I get wet, then it rains. From a proposition, its inverse, its converse, and its contrapositive are derived as follows: Proposition: "If P then … The Contrapositive of a Conditional Statement. The contrapositive of the above statement is: If x is not even, then x 2 is not even.. and contrapositive is the natural choice. 3) The contrapositive statement is a combination of the previous two. contrapositive (plural contrapositives) The inverse of the converse of a given propositionUsage notes []. Contrapositive: If Jennifer does not eat food, then Jennifer is not alive. Try to apply the two step transformation process and write out the proper contrapositive. If 3 - n2, then 3 - n. Proof. We need to nd the contrapositive of the given statement. First we need to negate \n - a and n - b." An example will help to make sense of this new terminology and notation. Definition [~q → ~p] is the contrapositive (contraposition) of the conditional statement [p → q]. Squaring, we have n2 = (3a)2 = 3(3a2) = 3b where b = 3a2. To find the contrapositive, switch and negate both p and q. Let's look at another example. Example 1. What does contrapositive mean? 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Proof a ; b ; n 2Z.If n - ab, then x 2 even. |
### Multiple Representations - Moving To the Common Core
```TO INFINITY AND BEYOND . . .
CATHY SHIDE, CONSULTANT
1
OBJECTIVES
• Integrate the math practices
with word problems
• teachers and students going
• Use different modes of
representation to solve problems
with a focus on Fractions, Ratios,
and Percents
2
“TAPE DIAGRAM”
“A drawing that looks like a
segment of tape, used to illustrate
number relationships. Also known
as a strip diagram, bar model,
fraction strip, or length model.”
Also referenced in “Visual Fraction
Model” definition.
- CCSSM (Glossary) p. 87
3
WHY NOT KEY WORDS?
• Takes away reason for reading
• Does not develop process for problem solving
• Takes away creating equations and finding all the
information
• Takes away from thinking “What else do I know to
solve the problem? What else may I need?”
4
EXPLORE
• Caroline has 46 dolls in her collection. She has 12
more than her cousin, Tammy. How many dolls
does Tammy have?
how much did they have altogether?
• Pedro and Gina started out with an equal number
of coins. Pedro lost 12 coins, and Gina collected
another 36 coins. How many more coins did Gina
have than Pedro?
PROBLEM #1
Cathy and Joan started out
with the same number of coins.
Cathy lost 12 coins and Joan
gained 36. How many more
coins does Joan have than
Cathy?
6
problem, mark the who
or what and unit bars.
accordingly. Cathy’s
loses 12 and Joan gains
36
3. Record the ?, what
students need to find.
Cathy’s coins
Joan’s coins
Cathy’s
coins
Joan’s
coins
12
36
12
Cathy’s
coins
Joan’s
coins
36
?
4. Students solve and try to 12 + 36 = 48
record an equation if
CL + JG = ?
possible.
What Cathy lost plus what Joan gained
equals how many more coins Joan has.
7
Example from the ISAT ER problem from 5th grade.
6.RP.3
Students were creating spirit
necklaces to sell for a fundraiser.
A necklace takes twice as many
4 times as many purple beads as
28 beads. What is the number of
9
7.RP.3
A class had 32 students and
twenty-five percent were
boys. When some new boys
joined the class, the
percentage of boys
increased to 40%. How many
new boys joined the class?
10
7.RP.3
Two students were running for
school president. Student A
Student B. How many
students voted?
11
5.NF.4
The fundraising committee
students sold 5/8 of the pizzas
and took 1/5 of the
remainder for a party. How
many pizzas did the
committee have left to sell?
12
GROUP PROBLEM SOLVING
Work with your colleagues to create:
• A manipulative model with your color tiles
• A tape diagram (bar model) of your
problem
• An equation
• A verbal description of your thought process
• What other questions can be answered
13
WHAT DO YOU KNOW?
up of 3 parts apple juice and
1 part cranberry juice. The
company will use 5 gallon
containers for the cran-apple
mixture.
14
WHAT ARE THE MATH PRACTICES?
•What practices have you
been engaged in?
15
MATH
• USA:
How can I teach my kids to get the answer to this
problem?
Use mathematics they already know. Easy, reliable, works with
bottom half, good for classroom management.
• Japanese:
How can I use this problem to teach the mathematics
of this unit?
Phil Daro, Writer of CCSS in Mathematics, Slide 16,
16
POSING THE PROBLEM
• Whole class: pose problem, make sure
students understand the language, no hints at
solution
• Focus students on the problem situation, not
and ask them to formulate questions that
make situation into a word problem
situation; ramp questions up toward key
mathematics that transfers to other problems
Phil Daro, Writer of CCSS in Mathematics, Slide 80,
17
WHAT PROBLEM TO USE?
•
•
•
•
Problems that draw thinking toward the
mathematics you want to teach. NOT too
routine, right after learning how to solve.
important mathematics students should take
with them? Find a problem that draws
attention to this mathematics
Begin chapter with this problem (from lesson
5 thru 10, or chapter test). This has diagnostic
power. Also shows you where time has to
go.
Also near end of chapter, while still time to
Phil Daro, Writer of CCSS in Mathematics, Slide 81,
respond
18
WHICH IS BIGGER?
• 1/3 or 1/2
3/5 or 6/10
3/4 or 7/8
10/16 or 5/8
2/3 or 6/12
• Katie collected 20 seashells.
Diego collected 3 times as
many seashells as Katie. What
was the total number of
seashells collected?
• Corey paid \$84 for a kite and a
CD player. The CD player cost
6 times as much as the kite.
What is the price difference
between the two items?
• Of the 60 students in the fifth
grade, we know that 3/5 are girls.
We also know that 1/2 of the girls
have blue eyes and one-fourth of
the boys have blue eyes. How
many of the students in the fifth |
Complex Numbers Introduction
Note: this page has been created with the use of AI. Please take caution, and note that the content of this page does not necessarily reflect the opinion of Cratecode.
Mathematics is like a magical world filled with diverse inhabitants – numbers of different shapes, sizes, and properties. One of the most fascinating residents of this world is the complex number. Let's embark on a journey to understand what complex numbers are, their properties, and their role in mathematics.
A Complex World
In the realm of numbers, we have real numbers – the ones we deal with in our day-to-day lives. But there's another type of number lurking in the shadows: imaginary numbers. Imaginary numbers are based on the square root of -1, commonly denoted as i. Now, a complex number is a creature that combines both real and imaginary numbers. It has the form `a + bi`, where `a` is the real part and `bi` is the imaginary part.
Imaginary Numbers
Before diving deeper into complex numbers, let's understand the enigmatic imaginary numbers. They were born out of necessity when mathematicians tried to find square roots of negative numbers. They realized that no real number could satisfy this condition – and thus, the imaginary numbers emerged.
The fundamental imaginary unit i is defined as the square root of -1, so `i^2 = -1`. We can build other imaginary numbers by multiplying i by real numbers. For example, 3i, -5i, and 2.7i are all imaginary numbers.
Complex Numbers: The Hybrid Beings
Complex numbers are like mythical creatures with the body of a real number and the wings of an imaginary number. They combine the powers of both worlds, making them incredibly versatile in mathematics.
Operations with Complex Numbers
Performing operations with complex numbers is like choreographing a dance between real and imaginary parts. Here's how some basic operations work:
• Addition/Subtraction: Add/subtract the real parts and imaginary parts separately. Example: `(3 + 4i) + (2 - 5i) = (3 + 2) + (4i - 5i) = 5 - i`
• Multiplication: Use the distributive property and remember that `i^2 = -1`. Example: `(3 + 4i) * (2 - 5i) = 6 - 15i + 8i - 20i^2 = 6 - 7i + 20 = 26 - 7i`
• Division: Multiply the numerator and denominator by the complex conjugate of the denominator, then simplify. Example: `(3 + 4i) / (2 - 5i) = (3 + 4i) * (2 + 5i) / (2 - 5i) * (2 + 5i) = (26 + 7i) / 29 = (26/29) + (7/29)i`
Complex Conjugate
The complex conjugate of a complex number `a + bi` is the number `a - bi`. It's like a reflection of the original number in the real axis. Complex conjugates play a crucial role in division, as we saw in the example above.
Applications of Complex Numbers
Complex numbers might seem like abstract entities, but they have significant applications in various fields, such as:
• Electrical Engineering: Complex numbers help model alternating current circuits and analyze their behavior.
• Fluid Dynamics: They are used to solve problems involving the flow of fluids in various conditions.
• Quantum Mechanics: Complex numbers are essential in representing quantum states and understanding the behavior of particles at the quantum level.
Conclusion
Complex numbers, the mystical beings of the number world, are intriguing and powerful. They expand our understanding of mathematics, allowing us to explore new dimensions, solve complex problems, and unlock the secrets of the universe. Now that you've been introduced to complex numbers, you're ready to embark on a journey to explore their wonders and applications in various fields. |
Everyday maths 2 (Wales)
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# 5.1 Scale examples
Take a look at the following examples.
## Example 1: Reading scales
Figure 23 Scale with numbered intervals of 50
You can see that this scale is marked up in numbered intervals of 50. However, what does each line in between each numbered interval represent? You can use your judgement to help you to figure out what each small step represents.
Watch this video (https://corbettmaths.com/ 2013/ 04/ 27/ reading-scales/ [Tip: hold Ctrl and click a link to open it in a new tab. (Hide tip)] ) for further information about how to do this.
Alternatively, you can work this out using division. If you count on from 0 to 50 on this scale, there are 5 steps: 50 ÷ 5 = 10, so each step is 10.
• a.The arrow is pointing to the 2nd mark after 50. As the steps are going up in 10s, the arrow is pointing to 70.
• b.The arrow is halfway between the 1st and 2nd step after 150. The first step is 160 and the second is 170 so the arrow is pointing to 165.
## Example 2: Reading scales
Sometimes you will need to read scales where the reading will be a decimal number.
Figure 24 Scale with single numbered intervals going from 3–5
If you look at this scale, it goes up in numbered intervals of 1. From one whole number to the next whole number there are 10 small steps. 1 ÷ 10 = 0.1, so each step is 0.1.
Hint: Have a look at the image to show how to count the number of steps between the numbered markers.
• a.The arrow is pointing to the fourth step after 3 so the arrow is pointing to 3.4.
• b.The arrow is pointing to the eighth step after 4, so the arrow is pointing to 4.8.
FSM_2_CYMRU
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teaching resource
# Counting to 10 Match-Up Activity
• Updated: 05 Nov 2021
Practice counting to 10 with this set of twenty object and number matching cards.
• Editable: Google Slides
• Non-Editable: PDF
• Pages: 6 Pages
• Grades: PK - K
Tag #TeachStarter on Instagram for a chance to be featured!
teaching resource
# Counting to 10 Match-Up Activity
• Updated: 05 Nov 2021
Practice counting to 10 with this set of twenty object and number matching cards.
• Editable: Google Slides
• Non-Editable: PDF
• Pages: 6 Pages
• Grades: PK - K
Practice counting to 10 with this set of twenty object and number matching cards.
When we teach students how to count, they learn to assign a number to whatever they are counting, and that the last number said is the total.
So when you think about it, counting is a lot more than just knowing your numbers
Through this activity, students show they can count 1-10 items in sequence, 1 number per item, demonstrating that the last count indicates how many items were counted regardless of the order.
## How to Play Our Counting to 10 Game
Use this resource in your math center, as a small group activity, or a whole class exercise (see below) to practice counting objects to 10 and pairing each object with its number name.
To play, shuffle the cards and lay them all face down in the center of the playing area. Students will take turns flipping 2 cards to look for matching pairs of objects and digits. The player who finds a match keeps the cards and takes another turn. If the cards aren’t a matching digit and number of objects, turn the cards back over in their spot and move on to the next player. Continue until there are no more cards left in the playing area. The player with the most matches in the end wins!
## Scaffolding + Extension Tips
Challenge students who already understand the concept to choose a card and provide a number that is greater than or less than (directed by the teacher) than the number.
Support students who need help understanding the concepts with manipulatives like counters or connecting cubes, and have them place one on each picture as they count.
## More Ways to Play Counting to 10 Games
Use this resource as independent practice for fast finishers, and for full-class learning opportunities like scoot activities, lesson reviews, formative assessments, and more.
### 🐟 Go Fish (Beginner)
It’s easy to turn these cards into a game of Go Fish! Shuffle the cards and pass out 7 to each player. The remaining cards go in the center of the play area. Taking turns, players ask each other for matches: looking for the object card that matches the number in their hand, or vice versa.
### 🖍️ Draw It! Warm-Up/Exit Game (Intermediate)
This activity can also be used as a formative assessment. Project a number card and have students draw their own set of objects that represent the digit shown.
### 🧑🏫 Show Me! (Advanced)
Give each student a mini dry-erase board and a dry-erase marker. Project an object card and have students count and write the number of objects on their whiteboards. When everyone has written down their answers, say, “Show Me.” Students will flip their boards, allowing you to see who needs extra support with this skill.
Get our 10 Best Scaffolding Strategies here!
## Easily Prepare This Resource for Your Students
Print on cardstock for added durability and longevity. Place all pieces in a folder or large envelope for easy access.
## Before You Download
Use the dropdown icon on the Download button to choose between the PDF or Google Slides version of this resource.
This resource was made in collaboration with Heather Chambers, a teacher in Texas and Teach Starter Collaborator.
You can count on us for more counting activities!
### teaching resource
#### 1-10 Counting Worksheet
A worksheet to practice one-to-one correspondence.
1 pageGrades: PK - K
### teaching resource
#### Counting to 10 Activity
Use this number activity to consolidate your students' knowledge of numbers 1-10.
5 pagesGrades: PK - 1
### teaching resource
#### Ten in the Bed - Counting Activity
A fun, hands-on number rhyme activity to help students sequence numbers and count forward and backward from 1–10.
1 pageGrades: PK - K
### 0 Comments
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# What is reverse operation in math?
## What is reverse operation in math?
An inverse operation reverses the effect of the first operation. For example, if we operated adding two numbers say 5+3 = 8. The inverse operation of this would be the subtraction of these two numbers: 5-3= 2.
## Why do we use inverse operation?
Mathematically, inverse operations are opposite operations. Addition is the opposite of subtraction, division is the opposite of multiplication, and so on. Inverse operations are used to solve simple algebraic equations to more difficult equations that involve exponents, logarithms, and trigonometry.
## What is the inverse of 3x 4?
The inverse function of 3x - 4 is (x+4)/3.
## What is the inverse of 7?
Here, 1⁄7 is called the multiplicative inverse of 7. Similarly, the multiplicative inverse of 13 is 1⁄13. Another word for multiplicative inverse is 'reciprocal'.
## What is the inverse of a 5?
The multiplicative inverse of 5 is 1/5.
## How is multiplication doing the reverse of division?
0:022:41Division (Multiplication in Reverse) - YouTubeYouTube
## Why are multiplication and division inverse operations?
An inverse operation "reverses" another operation. Similarly, multiplication and division are inverses of each other because multiplying and dividing by the same number does not change the original number. For example, 11×5/5 = 11 and 6/2×2 = 6.
## What is the inverse of 3 4?
Multiplicative inverse of a fraction Thus, the multiplicative inverse of 3⁄4 is 4⁄3. The multiplicative inverse or reciprocal of a fraction a⁄b is b⁄a.
## How do you reverse inverse?
How to Invert a Function to Find Its InverseSwitch f(x) and x. When you switch f(x) and x, you get. (Note: To make the notation less clumsy, you can rewrite f(x) as y and then switch x and y.)Change the new f(x) to its proper name — f–1(x). The equation then becomes.Solve for the inverse. This step has three parts:Apr 19, 2017
## What's the inverse of 1?
1 itselfThe multiplicative inverse of 1 is 1 itself.
## What's the inverse of 6?
1/6So the reciprocal of 6 is 1/6 because 6 = 6/1 and 1/6 is the inverse of 6/1.
## Is reverse process of multiplication?
Factoring is the reverse process of multiplication. When factoring, it is always helpful to look for a GCF that can be pulled out of the polynomial expression.
## What is the reverse multiplication?
The division is the reverse process of multiplication. Dividing by a number is equivalent to multiplying by the reciprocal of the number. Another word for multiplicative inverse is 'reciprocal'.
## How do you reverse division?
The inverse of division is multiplication. If you divide by a given number and then multiply by the same number, you will arrive at the same number you started with. Multiplication has the opposite effect to division. For example, here is 10 ÷ 2 = 5.
## Why do we use inverse operations?
Mathematically, inverse operations are opposite operations. Addition is the opposite of subtraction, division is the opposite of multiplication, and so on. Inverse operations are used to solve simple algebraic equations to more difficult equations that involve exponents, logarithms, and trigonometry.
What storage device has a temporary storage?
How long is temporarily suspended PS4?
Can a deer hang in 50 degree weather?
## Does Sam's Club have a one day pass?
The Sam's Club One Day trial pass will get you in the door, however, if you want to buy anything, you will be charged a 10% service fee on your purchases (which may negate the savings of purchasing from Sam's Club in the first place).
## What is Sam's Club member satisfaction warranty?
We offer a 100% satisfaction guarantee on your Membership or your money back. If you are not pleased with any product you buy at the Club, we will replace it, refund your money, or compensate you with a Sam's Club gift card. |
# What is 0.64 Rounded to the Tenths Place?
Rounding numbers is a common mathematical skill that helps to easily make calculations and estimations. Rounding numbers to the tenths place is a relatively common task, and it is a simple process that can be done easily with a few steps. Rounding 0.64 to the tenths place is a useful skill that can help with many calculations.
## Understanding the Tenths Place
In order to understand how to round numbers to the tenths place, it is important to have a clear understanding of the tenths place. The tenths place is the first number after the decimal point and corresponds to the number of tenths in a number. In the number 0.64, the 6 is in the tenths place, which is the first number after the decimal point.
## Rounding to the Tenths Place
Rounding 0.64 to the tenths place requires that you look at the number in the hundredths place, or the second number after the decimal point. In the number 0.64, the 4 is in the hundredths place. If the number in the hundredths place is less than five, the number in the tenths place stays the same, so 0.64 rounded to the tenths place is 0.6. If the number in the hundredths place is five or greater, the number in the tenths place is increased by one, so 0.65 rounded to the tenths place is 0.7.
## Using the Tenths Place to Make Calculations
Rounding numbers to the tenths place can be a useful tool for making calculations. For example, if you are trying to figure out how many tenths of an inch are in a certain measurement, you can round the number to the tenths place and get an approximation of the answer. This is useful for quickly making estimations without having to calculate the exact answer.
##
Rounding numbers to the tenths place is a basic mathematical skill that can be used to quickly make calculations and estimations. Understanding the tenths place and how to round numbers to the tenths place is essential for making calculations, and 0.64 rounded to the tenths place is 0.6. Using this skill can be useful for quickly making estimations and calculations without having to calculate the exact answer. |
# Problem of math
Problem of math is a software program that helps students solve math problems. Our website can solving math problem.
## The Best Problem of math
This Problem of math supplies step-by-step instructions for solving all math troubles. The matrix 3x3 is a common problem in mathematics. In this case, we have a 3-by-3 square of numbers. We want to find the values of A, B and C that solve the equation AB=C. The solution is: A=C/2 B=C/4 C=C/8 When we multiply B by C (or C by -1), we get A. When we divide A by B, we get C. And when we divide C by -1, we get -B. This is a fairly simple way to solve the matrix 3x3. It's also useful to remember if you have any nonlinear equations with matrices, like x^2 + y^2 = 4x+2y. In these cases, you can usually find a solution by finding the roots of the nonlinear equation and plugging it into the matrix equation.
Algebra is a mathematical field that focuses on solving problems using formulas. Algebra equations are expressions that can be used to solve problems. Algebra equations can be written on paper, in equations, or as word problems. One common type of algebra equation is an equation with two unknowns (also called variables). This type of equation could be used to solve the following types of problems: • Finding the length of a string • Finding the volume of a box • Finding the number of steps it takes to climb a certain number of stairs Algebra equations can also be written in other ways, such as as word problems, by using variables and different symbols. For example, “Bill climbed four flights of stairs to get to his apartment” could represent an algebraic equation such as "4x + 3 = 16." In this case, the letter “x” represents one unknown, and “+” represents addition. The letter “=” represents equality, which means that you need to find the value that makes 4x equal to 16. This could be any number from 1 to 4; for example, 1 would make 4x equal 16 and 2 would make 4x equal 8. However, if you were asked to find out how many steps it took Bill to climb four flights of stairs (meaning you didn't know how many steps there
Because math can be so important, it's important to learn how to do it well. That’s where math problems come in. Math problems are used to test your ability to understand math. To solve a math problem, you need to understand what the problem is asking for and be able to calculate that information. If you don’t know how to do this, you might struggle with math problems. However, there are some things you can do to help yourself. First, make sure you know the calculation rules for your grade level. Second, practice by doing simple math problems over and over again until you get them right. Finally, work with a tutor or teacher if you need help. By using these strategies, you can improve your skills and become better at solving math problems!
They are used primarily in science and engineering, although they are also sometimes used for business and economics. They can be used to find the minimum or maximum value of an expression, find a root of a function, find the maximum value of an array, etc. The most common use of a quaratic equation solver is to solve a set of simultaneous linear equations. In this case, the user enters two equations into the program and it will output the solution (either via manual calculation or by generating one of several automatic methods). A quaratic equation solver can also be used to solve any other system of equations with fewer than three variables (for example, it could be used to solve an entire system of four equations). Quaratic equation solvers are very flexible; they can be programmed to perform nearly any type of calculation that can be done with algebraic formulas. They can also be adapted for specific applications; for example, a commercial quaratic equation solver can usually be modified to calculate electricity usage.
Mathematics is a study of symbols and relations that can be applied to anything. It is a very logical and analytical subject, which has many practical applications. Mathematics is not just about memorizing numbers and learning formulas. Mathematics is a language that can be used to describe the world around us. It can be used to describe how a theory works, or how something works physically. Mathematics can also be used to describe what things cost and how much they are worth. Mathematics can even be used to solve crimes! When we have information about an event in our world, we can use mathematics to determine what happened and why it happened. This can help us figure out who did it or how it happened in the first place.
It helps me with everyday math and the fact that there’s barely any ads make it even better. Not only that it gives you math problem answers it explains how to get the solution so you also learn. This app is definitely better than math way, the old app I used for math answers. |
## Section 1.4: Linear Functions and Their Graphs
### Learning Outcomes
• Identify and Interpret the Slope and Vertical Intercept of a Line
• Calculate and interpret the slope of a line.
• Determine the equation of a linear function.
• Graph linear functions.
• Write the equation for a linear function from the graph of a line.
Introduction to Linear Functions
A bamboo forest in China (credit: “JFXie”/Flickr)
Imagine placing a plant in the ground one day and finding that it has doubled its height just a few days later. Although it may seem incredible, this can happen with certain types of bamboo species. These members of the grass family are the fastest-growing plants in the world. One species of bamboo has been observed to grow nearly 1.5 inches every hour. [1] In a twenty-four hour period, this bamboo plant grows about 36 inches, or an incredible 3 feet! A constant rate of change, such as the growth cycle of this bamboo plant, is a linear function.
Recall from Functions and Function Notation that a function is a relation that assigns to every element in the domain exactly one element in the range. Linear functions are a specific type of function that can be used to model many real-world applications, such as plant growth over time. In this chapter, we will explore linear functions, their graphs, and how to relate them to data.
Just as with the growth of a bamboo plant, there are many situations that involve constant change over time. Consider, for example, the first commercial maglev train in the world, the Shanghai MagLev Train. It carries passengers comfortably for a 30-kilometer trip from the airport to the subway station in only eight minutes.[2]
Suppose a maglev train were to travel a long distance, and that the train maintains a constant speed of 83 meters per second for a period of time once it is 250 meters from the station. How can we analyze the train’s distance from the station as a function of time? In this section, we will investigate a kind of function that is useful for this purpose, and use it to investigate real-world situations such as the train’s distance from the station at a given point in time.
## Representing Linear Functions
The function describing the train’s motion is a linear function, which is defined as a function with a constant rate of change, that is, a polynomial of degree 1. There are several ways to represent a linear function, including word form, function notation, tabular form, and graphical form. We will describe the train’s motion as a function using each method.
### Representing a Linear Function in Word Form
Let’s begin by describing the linear function in words. For the train problem we just considered, the following word sentence may be used to describe the function relationship.
• The train’s distance from the station is a function of the time during which the train moves at a constant speed plus its original distance from the station when it began moving at constant speed.
The speed is the rate of change. Recall that a rate of change is a measure of how quickly the dependent variable changes with respect to the independent variable. The rate of change for this example is constant, which means that it is the same for each input value. As the time (input) increases by 1 second, the corresponding distance (output) increases by 83 meters. The train began moving at this constant speed at a distance of 250 meters from the station.
### Representing a Linear Function in Function Notation
Another approach to representing linear functions is by using function notation. One example of function notation is an equation written in the form known as the slope-intercept form of a line, where $x$ is the input value, $m$ is the rate of change, and $b$ is the initial value of the dependent variable.
\begin{align}&\text{Equation form} && y=mx+b \\ &\text{Function notation} && f(x)=mx+b\\& \end{align}
In the example of the train, we might use the notation $D\left(t\right)$ in which the total distance $D$
is a function of the time $t$. The rate, $m$, is 83 meters per second. The initial value of the dependent variable $b$ is the original distance from the station, 250 meters. We can write a generalized equation to represent the motion of the train.
$D(t)=83t+250$
### Representing a Linear Function in Tabular Form
A third method of representing a linear function is through the use of a table. The relationship between the distance from the station and the time is represented in the table in Figure 1. From the table, we can see that the distance changes by 83 meters for every 1 second increase in time.
Figure 1. Tabular representation of the function D showing selected input and output values
### Q & A
Can the input in the previous example be any real number?
No. The input represents time, so while nonnegative rational and irrational numbers are possible, negative real numbers are not possible for this example. The input consists of non-negative real numbers.
### Representing a Linear Function in Graphical Form
Another way to represent linear functions is visually, using a graph. We can use the function relationship from above, $D(t)=83t+250$, to draw a graph, represented in the graph in Figure 2. Notice the graph is a line. When we plot a linear function, the graph is always a line.
The rate of change, which is constant, determines the slant, or slope of the line. The point at which the input value is zero is the vertical intercept, or y-intercept, of the line. We can see from the graph that the y-intercept in the train example we just saw is $\left(0,250\right)$ and represents the distance of the train from the station when it began moving at a constant speed.
Figure 2. The graph of $D(t)=83t+250$. Graphs of linear functions are lines because the rate of change is constant.
Notice that the graph of the train example is restricted, but this is not always the case. Consider the graph of the line $f(x)=2{x}_{}+1$. Ask yourself what numbers can be input to the function, that is, what is the domain of the function? The domain is comprised of all real numbers because any number may be doubled, and then have one added to the product.
### A General Note: Linear Function
A linear function is a function whose graph is a line. Linear functions can be written in the slope-intercept form of a line
$f(x)=mx+b$
where $b$ is the initial or starting value of the function (when input, $x=0$), and $m$ is the constant rate of change, or slope of the function. The y-intercept is at $\left(0,b\right)$.
### Example 1: Using a Linear Function to Find the Pressure on a Diver
The pressure, $P$, in pounds per square inch (PSI) on the diver in Figure 3 depends upon her depth below the water surface, $d$, in feet. This relationship may be modeled by the equation, $P(d)=0.434d+14.696$. Restate this function in words.
Figure 3. (credit: Ilse Reijs and Jan-Noud Hutten)
## Determining Whether a Linear Function is Increasing, Decreasing, or Constant
Figure 4
The linear functions we used in the two previous examples increased over time, but not every linear function does. A linear function may be increasing, decreasing, or constant.
For an increasing function, as with the train example,
the output values increase as the input values increase.
The graph of an increasing function has a positive slope. A line with a positive slope slants upward from left to right as in (a).
For a decreasing function, the slope is negative.
The output values decrease as the input values increase.
A line with a negative slope slants downward from left to right as in (b). If the function is constant, the output values are the same for all input values so the slope is zero. A line with a slope of zero is horizontal as in (c).
### A General Note: Increasing and Decreasing Functions
The slope determines if the function is an increasing linear function, a decreasing linear function, or a constant function.
• $f(x)=mx+b\text{ is an increasing function if }m>0$.
• $f(x)=mx+b\text{ is an decreasing function if }m<0$.
• $f(x)=mx+b\text{ is a constant function if }m=0$.
### Example 2: Deciding whether a Function Is Increasing, Decreasing, or Constant
Some recent studies suggest that a teenager sends an average of 60 texts per day.[3] For each of the following scenarios, find the linear function that describes the relationship between the input value and the output value. Then, determine whether the graph of the function is increasing, decreasing, or constant.
1. The total number of texts a teen sends is considered a function of time in days. The input is the number of days, and output is the total number of texts sent.
2. A teen has a limit of 500 texts per month in his or her data plan. The input is the number of days, and output is the total number of texts remaining for the month.
3. A teen has an unlimited number of texts in his or her data plan for a cost of 50 per month. The input is the number of days, and output is the total cost of texting each month. ## Calculating and Interpreting Slope In the examples we have seen so far, we have had the slope provided for us. However, we often need to calculate the slope given input and output values. Given two values for the input, ${x}_{1}$ and ${x}_{2}$, and two corresponding values for the output, ${y}_{1}$ and ${y}_{2}$ —which can be represented by a set of points, $\left({x}_{1}\text{, }{y}_{1}\right)$ and $\left({x}_{2}\text{, }{y}_{2}\right)$—we can calculate the slope $m$, as follows $m=\frac{\text{change in output (rise)}}{\text{change in input (run)}}=\frac{\Delta y}{\Delta x}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$ where $\Delta y$ is the vertical displacement and $\Delta x$ is the horizontal displacement. Note in function notation two corresponding values for the output ${y}_{1}$ and ${y}_{2}$ for the function $f$, ${y}_{1}=f\left({x}_{1}\right)$ and ${y}_{2}=f\left({x}_{2}\right)$, so we could equivalently write $m=\frac{f\left({x}_{2}\right)-f\left({x}_{1}\right)}{{x}_{2}-{x}_{1}}$ The graph in Figure 5 indicates how the slope of the line between the points, $\left({x}_{1,}{y}_{1}\right)$ and $\left({x}_{2,}{y}_{2}\right)$, is calculated. Recall that the slope measures steepness. The greater the absolute value of the slope, the steeper the line is. Figure 5 The slope of a function is calculated by the change in $y$ divided by the change in $x$. It does not matter which coordinate is used as the $\left({x}_{2}\text{,}{y}_{2}\right)$ and which is the $\left({x}_{1}\text{,}{y}_{1}\right)$, as long as each calculation is started with the elements from the same coordinate pair. ### Q & A Are the units for slope always $\frac{\text{units for the output}}{\text{units for the input}}$ ? Yes. Think of the units as the change of output value for each unit of change in input value. An example of slope could be miles per hour or dollars per day. Notice the units appear as a ratio of units for the output per units for the input. ### A General Note: Calculate Slope The slope, or rate of change, of a function $m$ can be calculated according to the following: $m=\frac{\text{change in output (rise)}}{\text{change in input (run)}}=\frac{\Delta y}{\Delta x}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$ where ${x}_{1}$ and ${x}_{2}$ are input values, ${y}_{1}$ and ${y}_{2}$ are output values. ### How To: Given two points from a linear function, calculate and interpret the slope. 1. Determine the units for output and input values. 2. Calculate the change of output values and change of input values. 3. Interpret the slope as the change in output values per unit of the input value. ### Example 3: Finding the Slope of a Linear Function If $f\left(x\right)$ is a linear function, and $\left(3,-2\right)$ and $\left(8,1\right)$ are points on the line, find the slope. Is this function increasing or decreasing? ### Try It If $f\left(x\right)$ is a linear function, and $\left(2,\text{ }3\right)$ and $\left(0,\text{ }4\right)$ are points on the line, find the slope. Is this function increasing or decreasing? ### Try It ### Example 4: Finding the Population Change from a Linear Function The population of a city increased from 23,400 to 27,800 between 2008 and 2012. Find the change of population per year if we assume the change was constant from 2008 to 2012. ### Try It The population of a small town increased from 1,442 to 1,868 between 2009 and 2012. Find the change of population per year if we assume the change was constant from 2009 to 2012. ### Try It ## Writing the Point-Slope Form of a Linear Equation Up until now, we have been using the slope-intercept form of a linear equation to describe linear functions. Here, we will learn another way to write a linear function, the point-slope form. $y-{y}_{1}=m\left(x-{x}_{1}\right)$ The point-slope form is derived from the slope formula. \begin{align}&{m}=\frac{y-{y}_{1}}{x-{x}_{1}} && \text{assuming }{ x }\ne {x}_{1} \\ &{ m }\left(x-{x}_{1}\right)=\frac{y-{y}_{1}}{x-{x}_{1}}\left(x-{x}_{1}\right) && \text{Multiply both sides by }\left(x-{x}_{1}\right)\\ &{ m }\left(x-{x}_{1}\right)=y-{y}_{1} && \text{Simplify} \\ &y-{y}_{1}={ m }\left(x-{x}_{1}\right) &&\text{Rearrange} \end{align} Keep in mind that the slope-intercept form and the point-slope form can be used to describe the same function. We can move from one form to another using basic algebra. For example, suppose we are given an equation in point-slope form, $y - 4=-\frac{1}{2}\left(x - 6\right)$ . We can convert it to the slope-intercept form as shown. \begin{align}&y - 4=-\frac{1}{2}\left(x - 6\right) \\ &y - 4=-\frac{1}{2}x+3 && \text{Distribute the }-\frac{1}{2}. \\ &y=-\frac{1}{2}x+7 && \text{Add 4 to each side}. \end{align} Therefore, the same line can be described in slope-intercept form as $y=-\frac{1}{2}x+7$. ### A General Note: Point-Slope Form of a Linear Equation The point-slope form of a linear equation takes the form $y-{y}_{1}=m\left(x-{x}_{1}\right)$ where $m$ is the slope, ${x}_{1 }$ and ${y}_{1}$ are the $x$ and $y$ coordinates of a specific point through which the line passes. ## Writing the Equation of a Line Using a Point and the Slope The point-slope form is particularly useful if we know one point and the slope of a line. Suppose, for example, we are told that a line has a slope of 2 and passes through the point $\left(4,1\right)$. We know that $m=2$ and that ${x}_{1}=4$ and ${y}_{1}=1$. We can substitute these values into the general point-slope equation. $\begin{gathered}y-{y}_{1}=m\left(x-{x}_{1}\right)\\ y - 1=2\left(x - 4\right)\end{gathered}$ If we wanted to then rewrite the equation in slope-intercept form, we apply algebraic techniques. \begin{align}&y - 1=2\left(x - 4\right) \\ &y - 1=2x - 8 && \text{Distribute the }2 \\ &y=2x - 7 && \text{Add 1 to each side} \end{align} Both equations, $y - 1=2\left(x - 4\right)$ and $y=2x - 7$, describe the same line. See Figure 6. Figure 6 ### Example 5: Writing Linear Equations Using a Point and the Slope Write the point-slope form of an equation of a line with a slope of 3 that passes through the point $\left(6,-1\right)$. Then rewrite it in the slope-intercept form. ### Try It Write the point-slope form of an equation of a line with a slope of –2 that passes through the point $\left(-2,\text{ }2\right)$. Then rewrite it in the slope-intercept form. ### Try It ## Writing the Equation of a Line Using Two Points The point-slope form of an equation is also useful if we know any two points through which a line passes. Suppose, for example, we know that a line passes through the points $\left(0,\text{ }1\right)$ and $\left(3,\text{ }2\right)$. We can use the coordinates of the two points to find the slope. \begin{align}{m}&=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\ &=\frac{2 - 1}{3 - 0} \\ &=\frac{1}{3} \end{align} Now we can use the slope we found and the coordinates of one of the points to find the equation for the line. Let use (0, 1) for our point. \begin{align}y-{y}_{1}&=m\left(x-{x}_{1}\right)\\ y - 1&=\frac{1}{3}\left(x - 0\right)\end{align} As before, we can use algebra to rewrite the equation in the slope-intercept form. \begin{align}&y - 1=\frac{1}{3}\left(x - 0\right) \\ &y - 1=\frac{1}{3}x && \text{Distribute the }\frac{1}{3} \\ &y=\frac{1}{3}x+1 && \text{Add 1 to each side} \end{align} Both equations describe the line shown in Figure 7. Figure 7 ### Example 6: Writing Linear Equations Using Two Points Write the point-slope form of an equation of a line that passes through the points (5, 1) and (8, 7). Then rewrite it in the slope-intercept form. ### Try It Write the point-slope form of an equation of a line that passes through the points $\left(-1,3\right)$ and $\left(0,0\right)$. Then rewrite it in the slope-intercept form. ## Writing and Interpreting an Equation for a Linear Function Now that we have written equations for linear functions in both the slope-intercept form and the point-slope form, we can choose which method to use based on the information we are given. That information may be provided in the form of a graph, a point and a slope, two points, and so on. Look at the graph of the function f in Figure 8. Figure 8 We are not given the slope of the line, but we can choose any two points on the line to find the slope. Let’s choose (0, 7) and (4, 4). We can use these points to calculate the slope. \begin{align} m&=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}} \\ &=\frac{4 - 7}{4 - 0} \\ &=-\frac{3}{4} \end{align} Now we can substitute the slope and the coordinates of one of the points into the point-slope form. $\begin{gathered}y-{y}_{1}=m\left(x-{x}_{1}\right) \\ y - 4=-\frac{3}{4}\left(x - 4\right) \end{gathered}$ If we want to rewrite the equation in the slope-intercept form, we would find $\begin{gathered}y - 4=-\frac{3}{4}\left(x - 4\right)\hfill \\ y - 4=-\frac{3}{4}x+3\hfill \\ \text{ }y=-\frac{3}{4}x+7\hfill \end{gathered}$ Figure 9 If we wanted to find the slope-intercept form without first writing the point-slope form, we could have recognized that the line crosses the y-axis when the output value is 7. Therefore, b = 7. We now have the initial value b and the slope m so we can substitute m and b into the slope-intercept form of a line. So the function is $f(xt)=-\frac{3}{4}x+7$, and the linear equation would be $y=-\frac{3}{4}x+7$. ### How To: Given the graph of a linear function, write an equation to represent the function. 1. Identify two points on the line. 2. Use the two points to calculate the slope. 3. Determine where the line crosses the y-axis to identify the y-intercept by visual inspection. 4. Substitute the slope and y-intercept into the slope-intercept form of a line equation. ### Example 7: Writing an Equation for a Linear Function Write an equation for a linear function given a graph of f shown in Figure 10. Figure 10 ### Example 8: Writing an Equation for a Linear Function Given Two Points If f is a linear function, with $f\left(3\right)=-2$ , and $f\left(8\right)=1$, find an equation for the function in slope-intercept form. ### Try It If $f\left(x\right)$ is a linear function, with $f\left(2\right)=-11$, and $f\left(4\right)=-25$, find an equation for the function in slope-intercept form. ### Try it ## Graphing Linear Functions As we have seen, the graph of a linear function is a straight line. We were also able to see the points of the function as well as the initial value from a graph. By graphing two functions, then, we can more easily compare their characteristics. There are three basic methods of graphing linear functions. The first is by plotting points and then drawing a line through the points. The second is by using the y-intercept and slope. And the third is by using transformations of the identity function $f(x)=x$. ## Graphing a Function by Plotting Points To find points of a function, we can choose input values, evaluate the function at these input values, and calculate output values. The input values and corresponding output values form coordinate pairs. We then plot the coordinate pairs on a grid. In general, we should evaluate the function at a minimum of two inputs in order to find at least two points on the graph. For example, given the function, $f(x)=2x$, we might use the input values 1 and 2. Evaluating the function for an input value of 1 yields an output value of 2, which is represented by the point (1, 2). Evaluating the function for an input value of 2 yields an output value of 4, which is represented by the point (2, 4). Choosing three points is often advisable because if all three points do not fall on the same line, we know we made an error. ### How To: Given a linear function, graph by plotting points. 1. Choose a minimum of two input values. 2. Evaluate the function at each input value. 3. Use the resulting output values to identify coordinate pairs. 4. Plot the coordinate pairs on a grid. 5. Draw a line through the points. ### Example 9: Graphing by Plotting Points Graph $f\left(x\right)=-\frac{2}{3}x+5$ by plotting points. ### Try It Graph $f\left(x\right)=-\frac{3}{4}x+6$ by plotting points. ## Graphing a Linear Function Using y-intercept and Slope Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. The first characteristic is its y-intercept, which is the point at which the input value is zero. To find the y-intercept, we can set x = 0 in the equation. The other characteristic of the linear function is its slope m, which is a measure of its steepness. Recall that the slope is the rate of change of the function. The slope of a function is equal to the ratio of the change in outputs to the change in inputs. Another way to think about the slope is by dividing the vertical difference, or rise, by the horizontal difference, or run. We encountered both the y-intercept and the slope in Linear Functions. Let’s consider the following function. $f\left(x\right)=\frac{1}{2}x+1$ The slope is $\frac{1}{2}$. Because the slope is positive, we know the graph will slant upward from left to right. The y-intercept is the point on the graph when = 0. The graph crosses the y-axis at (0, 1). Now we know the slope and the y-intercept. We can begin graphing by plotting the point (0,1) We know that the slope is rise over run, $m=\frac{\text{rise}}{\text{run}}$. From our example, we have $m=\frac{1}{2}$, which means that the rise is 1 and the run is 2. So starting from our y-intercept (0, 1), we can rise 1 and then run 2, or run 2 and then rise 1. We repeat until we have a few points, and then we draw a line through the points as shown in Figure 14. Figure 14 ### A General Note: Graphical Interpretation of a Linear Function In the equation $f\left(x\right)=mx+b$ • b is the y-intercept of the graph and indicates the point (0, b) at which the graph crosses the y-axis. • m is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope: $m=\frac{\text{change in output (rise)}}{\text{change in input (run)}}=\frac{\Delta y}{\Delta x}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$ ### Q & A Do all linear functions have y-intercepts? Yes. All linear functions cross the y-axis and therefore have y-intercepts. (Note: A vertical line parallel to the y-axis does not have a y-intercept, but it is not a function.) ### How To: Given the equation for a linear function, graph the function using the y-intercept and slope. 1. Evaluate the function at an input value of zero to find the y-intercept. 2. Identify the slope as the rate of change of the input value. 3. Plot the point represented by the y-intercept. 4. Use $\frac{\text{rise}}{\text{run}}$ to determine at least two more points on the line. 5. Sketch the line that passes through the points. ### Example 10: Graphing by Using the y-intercept and Slope Graph $f\left(x\right)=-\frac{2}{3}x+5$ using the y-intercept and slope. ### Try It Find a point on the graph we drew in Example 10 that has a negative x-value. ### Try It ## Graphing a Linear Function Using Transformations Another option for graphing is to use transformations of the identity function $f(x)=x$ . A function may be transformed by a shift up, down, left, or right. A function may also be transformed using a reflection, stretch, or compression. ### Vertical Stretch or Compression In the equation $f(x)=mx$, the m is acting as the vertical stretch or compression of the identity function. When m is negative, there is also a vertical reflection of the graph. Notice in Figure 16 that multiplying the equation of $f(x)=x$ by m stretches the graph of f by a factor of m units if > 1 and compresses the graph of f by a factor of m units if 0 < < 1. This means the larger the absolute value of m, the steeper the slope. Figure 16. Vertical stretches and compressions and reflections on the function $f\left(x\right)=x$. ### Vertical Shift In $f\left(x\right)=mx+b$, the b acts as the vertical shift, moving the graph up and down without affecting the slope of the line. Notice in Figure 17 that adding a value of b to the equation of $f\left(x\right)=x$ shifts the graph of f a total of b units up if b is positive and |b| units down if b is negative. Figure 17. This graph illustrates vertical shifts of the function $f(x)=x$. Using vertical stretches or compressions along with vertical shifts is another way to look at identifying different types of linear functions. Although this may not be the easiest way to graph this type of function, it is still important to practice each method. ### How To: Given the equation of a linear function, use transformations to graph the linear function in the form $f\left(x\right)=mx+b$. 1. Graph $f\left(x\right)=x$. 2. Vertically stretch or compress the graph by a factor m. 3. Shift the graph up or down b units. ### Example 11: Graphing by Using Transformations Graph $f\left(x\right)=\frac{1}{2}x - 3$ using transformations. ### Try It Graph $f\left(x\right)=4+2x$, using transformations. ### Q & A In Example 3, could we have sketched the graph by reversing the order of the transformations? No. The order of the transformations follows the order of operations. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. This is why we performed the compression first. For example, following the order: Let the input be 2. \begin{align}f(2)&=\frac{1}{2}(2)-3 \\ &=1-3 \\ &=-2 \end{align} ## Writing the Equation for a Function from the Graph of a Line Earlier, we wrote the equation for a linear function from a graph. Now we can extend what we know about graphing linear functions to analyze graphs a little more closely. Begin by taking a look at Figure 21. We can see right away that the graph crosses the y-axis at the point (0, 4) so this is the y-intercept. Figure 21 Then we can calculate the slope by finding the rise and run. We can choose any two points, but let’s look at the point (–2, 0). To get from this point to the y-intercept, we must move up 4 units (rise) and to the right 2 units (run). So the slope must be $m=\frac{\text{rise}}{\text{run}}=\frac{4}{2}=2$ Substituting the slope and y-intercept into the slope-intercept form of a line gives $y=2x+4$ ### How To: Given a graph of linear function, find the equation to describe the function. 1. Identify the y-intercept of an equation. 2. Choose two points to determine the slope. 3. Substitute the y-intercept and slope into the slope-intercept form of a line. ### Example 12: Matching Linear Functions to Their Graphs Match each equation of the linear functions with one of the lines in Figure 22. 1. $f\left(x\right)=2x+3$ 2. $g\left(x\right)=2x - 3$ 3. $h\left(x\right)=-2x+3$ 4. $j\left(x\right)=\frac{1}{2}x+3$ ## Finding the x-intercept of a Line So far, we have been finding the y-intercepts of a function: the point at which the graph of the function crosses the y-axis. A function may also have an x-intercept, which is the x-coordinate of the point where the graph of the function crosses the x-axis. In other words, it is the input value when the output value is zero. To find the x-intercept, set a function f(x) equal to zero and solve for the value of x. For example, consider the function shown. $f\left(x\right)=3x - 6$ Set the function equal to 0 and solve for x. \begin{align}&0=3x - 6 \\ &6=3x \\ &2=x \\ &x=2 \end{align} The graph of the function crosses the x-axis at the point (2, 0). ### Q & A Do all linear functions have x-intercepts? No. However, linear functions of the form = c, where c is a nonzero real number are the only examples of linear functions with no x-intercept. For example, = 5 is a horizontal line 5 units above the x-axis. This function has no x-intercepts. Figure 24 ### A General Note: x-intercept The x-intercept of the function is the point where the graph crosses the x-axis. Points on the x-axis have the form (x,0) so we can find x-intercepts by setting f(x) = 0. For a linear function, we solve the equation mx + = 0 ### Example 13: Finding an x-intercept Find the x-intercept of $f\left(x\right)=\frac{1}{2}x - 3$. ### Try It Find the x-intercept of $f\left(x\right)=\frac{1}{4}x - 4$. ## Describing Horizontal and Vertical Lines There are two special cases of lines on a graph—horizontal and vertical lines. A horizontal line indicates a constant output, or y-value. In Figure 26, we see that the output has a value of 2 for every input value. The change in outputs between any two points, therefore, is 0. In the slope formula, the numerator is 0, so the slope is 0. If we use = 0 in the equation $f(x)=mx+b$, the equation simplifies to $f(x)=b$. In other words, the value of the function is a constant. This graph represents the function $f(x)=2$. Figure 26. A horizontal line representing the function $f(x)=2$. Figure 27 A vertical line indicates a constant input, or x-value. We can see that the input value for every point on the line is 2, but the output value varies. Because this input value is mapped to more than one output value, a vertical line does not represent a function. Notice that between any two points, the change in the input values is zero. In the slope formula, the denominator will be zero, so the slope of a vertical line is undefined. Notice that a vertical line, such as the one in Figure 28, has an x-intercept, but no y-intercept unless it’s the line x = 0. This graph represents the line x = 2. Figure 28. The vertical line, = 2, which does not represent a function. ### A General Note: Horizontal and Vertical Lines Lines can be horizontal or vertical. A horizontal line is a line defined by an equation in the form $f(x)=b$. A vertical line is a line defined by an equation in the form $x=a$. ### Example 14: Writing the Equation of a Horizontal Line Write the equation of the line graphed in Figure 29. Figure 29 ### Example 15: Writing the Equation of a Vertical Line Write the equation of the line graphed in Figure 30. Figure 30 ## Key Equations slope-intercept form of a line $f\left(x\right)=mx+b$ slope $m=\frac{\text{change in output (rise)}}{\text{change in input (run)}}=\frac{\Delta y}{\Delta x}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$ point-slope form of a line $y-{y}_{1}=m\left(x-{x}_{1}\right)$ ## Key Concepts • The ordered pairs given by a linear function represent points on a line. • Linear functions can be represented in words, function notation, tabular form, and graphical form. • The rate of change of a linear function is also known as the slope. • An equation in the slope-intercept form of a line includes the slope and the initial value of the function. • The initial value, or y-intercept, is the output value when the input of a linear function is zero. It is the y-value of the point at which the line crosses the y-axis. • An increasing linear function results in a graph that slants upward from left to right and has a positive slope. • A decreasing linear function results in a graph that slants downward from left to right and has a negative slope. • A constant linear function results in a graph that is a horizontal line. • Analyzing the slope within the context of a problem indicates whether a linear function is increasing, decreasing, or constant. • The slope of a linear function can be calculated by dividing the difference between y-values by the difference in corresponding x-values of any two points on the line. • The slope and initial value can be determined given a graph or any two points on the line. • One type of function notation is the slope-intercept form of an equation. • The point-slope form is useful for finding a linear equation when given the slope of a line and one point. • The point-slope form is also convenient for finding a linear equation when given two points through which a line passes. • The equation for a linear function can be written if the slope m and initial value are known. • A linear function can be used to solve real-world problems. • A linear function can be written from tabular form. • Linear functions may be graphed by plotting points or by using the y-intercept and slope. • Graphs of linear functions may be transformed by using shifts up, down, left, or right, as well as through stretches, compressions, and reflections. • The y-intercept and slope of a line may be used to write the equation of a line. • The x-intercept is the point at which the graph of a linear function crosses the x-axis. • Horizontal lines are written in the form, f(x) = b. • Vertical lines are written in the form, = b. ## Glossary decreasing linear function a function with a negative slope: If $f\left(x\right)=mx+b, \text{then } m<0$. horizontal line a line defined by $f(x)=b$, where b is a real number. The slope of a horizontal line is 0. increasing linear function a function with a positive slope: If $f\left(x\right)=mx+b, \text{then } m>0$. linear function a function with a constant rate of change that is a polynomial of degree 1, and whose graph is a straight line point-slope form the equation for a line that represents a linear function of the form $y-{y}_{1}=m\left(x-{x}_{1}\right)$ slope the ratio of the change in output values to the change in input values; a measure of the steepness of a line slope-intercept form the equation for a line that represents a linear function in the form $f\left(x\right)=mx+b$ vertical line a line defined by x = a, where a is a real number. The slope of a vertical line is undefined. x-intercept the point on the graph of a linear function when the output value is 0; the point at which the graph crosses the horizontal axis y-intercept the value of a function when the input value is zero; also known as initial value ## Section 1.4 Homework Exercises 1. Terry is skiing down a steep hill. Terry’s elevation, E(t), in feet after t seconds is given by $E\left(t\right)=3000-70t$. Write a complete sentence describing Terry’s starting elevation and how it is changing over time. 2. Maria is climbing a mountain. Maria’s elevation, E(t), in feet after t minutes is given by $E\left(t\right)=1200+40t$. Write a complete sentence describing Maria’s starting elevation and how it is changing over time. 3. Jessica is walking home from a friend’s house. After 2 minutes she is 1.4 miles from home. Twelve minutes after leaving, she is 0.9 miles from home. What is her rate in miles per hour? 4. Sonya is currently 10 miles from home and is walking farther away at 2 miles per hour. Write an equation for her distance from home t hours from now. 5. A boat is 100 miles away from the marina, sailing directly toward it at 10 miles per hour. Write an equation for the distance of the boat from the marina after t hours. 6. Timmy goes to the fair with40. Each ride costs \$2. How much money will he have left after riding $n$ rides?
For the following exercises, determine whether the equation of the curve can be written as a linear function.
7. $y=\frac{1}{4}x+6$
8. $y=3x - 5$
9. $y=3{x}^{2}-2$
10. $3x+5y=15$
11. $3{x}^{2}+5y=15$
12. $3x+5{y}^{2}=15$
13. $-2{x}^{2}+3{y}^{2}=6$
14. $-\frac{x - 3}{5}=2y$
For the following exercises, determine whether each function is increasing or decreasing.
15. $f\left(x\right)=4x+3$
16. $g\left(x\right)=5x+6$
17. $a\left(x\right)=5 - 2x$
18. $b\left(x\right)=8 - 3x$
19. $h\left(x\right)=-2x+4$
20. $k\left(x\right)=-4x+1$
21. $j\left(x\right)=\frac{1}{2}x - 3$
22. $p\left(x\right)=\frac{1}{4}x - 5$
23. $n\left(x\right)=-\frac{1}{3}x - 2$
24. $m\left(x\right)=-\frac{3}{8}x+3$
For the following exercises, find the slope of the line that passes through the two given points.
25. $\left(2,\text{ }4\right)$ and $\left(4,\text{ 10}\right)$
26. $\left(1,\text{ 5}\right)$ and $\left(4,\text{ 11}\right)$
27. $\left(-1,\text{4}\right)$ and $\left(5,\text{2}\right)$
28. $\left(8,-2\right)$ and $\left(4,6\right)$
29. $\left(6,\text{ }11\right)$ and $\left(-4, 3\right)$
For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible.
30. $f\left(-5\right)=-4$, and $f\left(5\right)=2$
31. $f\left(-1\right)=4$ and $f\left(5\right)=1$
32. $\left(2,4\right)$ and $\left(4,10\right)$
33. Passes through $\left(1,5\right)$ and $\left(4,11\right)$
34. Passes through $\left(-1,\text{ 4}\right)$ and $\left(5,\text{ 2}\right)$
35. Passes through $\left(-2,\text{ 8}\right)$ and $\left(4,\text{ 6}\right)$
36. x intercept at $\left(-2,\text{ 0}\right)$ and y intercept at $\left(0,-3\right)$
37. x intercept at $\left(-5,\text{ 0}\right)$ and y intercept at $\left(0,\text{ 4}\right)$
For the following exercises, find the slope of the lines graphed.
38.
39.
40.
For the following exercises, write an equation for the lines graphed.
41.
42.
43.
44.
45.
46.
For the following exercises, which of the tables could represent a linear function? For each that could be linear, find a linear equation that models the data.
47.
x 0 5 10 15 g(x) 5 –10 –25 –40
48.
x 0 5 10 15 h(x) 5 30 105 230
49.
x 0 5 10 15 f(x) –5 20 45 70
50.
x 5 10 20 25 k(x) 28 13 58 73
51.
x 0 2 4 6 g(x) 6 –19 –44 –69
52.
x 2 4 6 8 f(x) –4 16 36 56
53.
x 2 4 6 8 f(x) –4 16 36 56
54.
x 0 2 6 8 k(x) 6 31 106 231
55. Find the value of x if a linear function goes through the following points and has the following slope: $\left(x,2\right),\left(-4,6\right),m=3$
56. Find the value of y if a linear function goes through the following points and has the following slope: $\left(10,y\right),\left(25,100\right),m=-5$
57. Find the equation of the line that passes through the following points: $\left(a,\text{ }b\right)$ and $\left(a,\text{ }b+1\right)$
58. Find the equation of the line that passes through the following points: $\left(2a,\text{ }b\right)$ and $\left(a,\text{ }b+1\right)$
59. Find the equation of the line that passes through the following points: $\left(a,\text{ }0\right)$ and $\left(c,\text{ }d\right)$
For the following exercises, match the given linear equation with its graph.
60. $f\left(x\right)=-x - 1$
61. $f\left(x\right)=-2x - 1$
62. $f\left(x\right)=-\frac{1}{2}x - 1$
63. $f\left(x\right)=2$
64. $f\left(x\right)=2+x$
65. $f\left(x\right)=3x+2$
For the following exercises, sketch a line with the given features.
66. An x-intercept of $\left(-\text{4},\text{ 0}\right)$ and y-intercept of $\left(0,\text{ -2}\right)$
67. An x-intercept of $\left(-\text{2},\text{ 0}\right)$ and y-intercept of $\left(0,\text{ 4}\right)$
68. A y-intercept of $\left(0,\text{ 7}\right)$ and slope $-\frac{3}{2}$
69. A y-intercept of $\left(0,\text{ 3}\right)$ and slope $\frac{2}{5}$
70. Passing through the points $\left(-\text{6},\text{ -2}\right)$ and $\left(\text{6},\text{ -6}\right)$
71. Passing through the points $\left(-\text{3},\text{ -4}\right)$ and $\left(\text{3},\text{ 0}\right)$
For the following exercises, sketch the graph of each equation.
72. $f\left(x\right)=-2x - 1$
73. $g\left(x\right)=-3x+2$
74. $h\left(x\right)=\frac{1}{3}x+2$
75. $k\left(x\right)=\frac{2}{3}x - 3$
76. $f\left(t\right)=3+2t$
77. $p\left(t\right)=-2+3t$
78. $x=3$
79. $x=-2$
80. $r\left(x\right)=4$
81. $q\left(x\right)=3$
82. $4x=-9y+36$
83. $\frac{x}{3}-\frac{y}{4}=1$
84. $3x - 5y=15$
85. $3x=15$
86. $3y=12$
87. If $g\left(x\right)$ is the transformation of $f\left(x\right)=x$ after a vertical compression by $\frac{3}{4}$, a shift right by 2, and a shift down by 4
a. Write an equation for $g\left(x\right)$.
b. What is the slope of this line?
c. Find the y-intercept of this line.
88. If $g\left(x\right)$ is the transformation of $f\left(x\right)=x$ after a vertical compression by $\frac{1}{3}$, a shift left by 1, and a shift up by 3
a. Write an equation for $g\left(x\right)$.
b. What is the slope of this line?
c. Find the y-intercept of this line.
For the following exercises, write the equation of the line shown in the graph.
89.
90.
91.
92. |
# Area of a Triangle Formed by a Line Tangent to f(x) and the Axis
Original question
Find the maximum area of a triangle formed in the first quadrant by the x-axis, y-axis and a tangent line to the graph of $f = (x + 2)^{-2}$
So far I've looked here and determined the equation for the height is $$h = \frac{-2}{(x+2)^3}\frac{1}{4}a$$ but when I try setting it to zero in the other answer the x doesn't cancel. I also tried wolframalpha so I'm pretty sure I did not simplify wrong. Any help would be appreciated.
• The area of the triangle is proportional to the product of the $x$- and $y$-intercepts of the tangent line. – amd Nov 16 '17 at 1:41
We must fix a parameter. It's convenient to have that as the $x$ coordinate of the point on the curve where the tangent touches the curve. Let this be $a$.
Slope of the tangent at that point = $\displaystyle -\frac{2}{(a+2)^3}$
Furthermore, the tangent passes through the point $\displaystyle (a,\frac{1}{(a+2)^2})$
So the equation of the tangent line is given by:
$\displaystyle y - \frac{1}{(a+2)^2} = -\frac{2}{(a+2)^3}(x-a)$
Set $y=0$ for $x$-intercept $x_0$ of the tangent line:
$\displaystyle - \frac{1}{(a+2)^2} = -\frac{2}{(a+2)^3}(x_0-a)$
Then set $x=0$ for $y$-intercept $y_0$ of the tangent line:
$\displaystyle y_0 - \frac{1}{(a+2)^2} = -\frac{2}{(a+2)^3}(-a)$
Solve for $x_0$ and $y_0$. The area of the triangle is $\frac 12x_0y_0$. Maximise that expression relative to $a$. |
# Series Resistors and Voltage Divider – Easy Explanation
If the equations are not displayed correctly, please use the desktop view
Contents
A resistor is the most basic element for an electric circuit. This element can be used for converting current from a voltage and vice versa. The resistor is often used to adjust the current and voltage in a circuit. Resistor is also a passive element.
Even a resistor is the most basic element, if a circuit has a complicated combination of several resistors, you may find difficulty to analyze the circuit.
May it be series-connected resistors or parallel-connected resistors, we will learn how to solve them.
Series resistors are multiple resistors connected together in a single path.
Series and parallel resistors can be represented by a single resistance Req. This will help us very well to analyze a circuit.
Keep in mind, resistors are used to limit current in the circuit and linked to Ohm’s law (I=V/R) where higher resistance will make the current lower.
No matter how complex it is, the resistors will follow Ohm’s law and Kirchhoff’s circuit laws.
## Resistors in Series
We can say resistors connected in series if they are connected together in a single wire.
The current has to flow through all the resistors from the first to the end resistor and back to the source terminal.
All the resistors connected in series will have a common current with the same value flowing through all of them. The current that flows through the first resistor must flow through all the rest resistors.
Say, we have a circuit with the terminal A-B as the source terminal and three resistors R1, R2, and R3, respectively as illustrated below,
The mathematical equation is:
## Equivalent Resistance for Series Resistors
After looking at the equations above, we can replace several resistors into a single resistor with “equivalent resistance”.
Say we have two, three, or more resistors connected together in a series connection, their equivalent resistance Req is the sum of all the resistors.
The more we connect resistors into a series circuit, the more resistance we get.
What is the equivalent resistance? We can say:
Equivalent Resistance is a single resistance that represents the resistances of any resistors connected without changing the value of current and voltage in the circuit.
This total resistance is generally known as the Equivalent Resistance and can be defined as; “a single value of resistance that can replace any number of resistors in series without altering the values of the current or the voltage in the circuit “.
Then the equation given for calculating total resistance of the circuit when connecting together resistors in series is given as:
## Series Resistor Equation
Analyzing a series resistor circuit can be done with Kirchhoff’s laws like before.
Consider a single-loop circuit below as the example of series connection.
The two resistors are in series since the same current i flows in both of them.
Applying Ohm’s law to each resistor, we get
Apply KVL to the loop in the clockwise direction, we obtain
Combining two equations above gives
Or
We replace R1 + R2 into Req as its equivalent resistance
and we get
Hence the circuit above can be replaced by the equivalent circuit below. Those two are equivalent because they have the same values of voltages and current at the terminal a-b.
An equivalent circuit above is very useful in simplifying the analysis of a circuit. In general,
The equivalent resistance of any number of resistors connected in series is the sum of the individual resistances.
For N resistors in series then,
Looking back to the circuit above with two resistors, R1 and R2, we can calculate their corresponding voltage.
In order to determine the voltage in each resistor, we substitute
Into
To get
## Series Resistor Combination
Looking from the equivalent resistance equation above, we can simplify some examples here. Note that equivalent resistance for series resistors is the algebraic sum of the individual resistances.
Here we have two resistors with identical resistances. The Req for two resistors is equal to 2R, for three resistors is equal to 3R, and so on.
Here is another example. We have two resistors with different resistances.
The Req for two resistors is equal to
for three resistors is equal to
and so on.
One thing to always remember:
The equivalent resistance Req for series resistors is always greater than the largest resistance of the connected resistor in a circuit.
You can check it by yourself easily.
## Series Resistor Voltage
Even if we have the series resistor voltage equation above, we will learn how to get it and how to use it.
If we have a circuit above and need to know the voltages for each resistor, we need to find the Req first.
Remember to find the Req first if we have multiple resistors connected in a circuit to make the calculation easy.
From series resistance equation we conclude that
Using Ohm’s law, we get current as:
And now we have the current, let us find the voltages for each resistor.
For a note,
The value of the voltage source in a circuit is equal to the sum of the voltage drop or the potential differences of the resistors.
Voltage drop in series combined together is the voltage source applied to the circuit.
Summary,
Using Ohm’s law again:
This proves that
and the value we get from that picture.
## Voltage Divider Circuit
The source voltage v is divided among the resistors in direct proportion to their resistances; the larger the resistance, the larger the voltage drop.
This is called the principle of voltage division and the circuit is called a voltage divider.
From the explanation above we can see that a single 6V voltage source can provide different voltage drops or potential differences across the resistors.
This behavior can make a series resistor circuit act as a voltage divider circuit.
This circuit splits the voltage source across to each resistor proportional to their resistances. The voltage is determined by the resistance of the resistor.
The larger the resistance, the larger the voltage drop and vice versa.
Do you remember what we have learned about Kirchhoff’s voltage law? Kirchhoff’s voltage law (KVL) states that the algebraic sum of all voltages around a closed path (or loop) is zero.
The principle of voltage division is used to divide the voltage source v proportionally to the resistances in the circuit.
The voltage divider example is shown below.
For easier explanation, we will only use two resistors R1 and R2 connected in series. We use 10V voltage source Vi, 4Ω and 6Ω resistors, and put an extra wire to R2 as Vo.
We can use the voltage divider formula to find the Vo. The mathematical equation is:
We can use more than two resistors for voltage divider circuits. But, the voltage for each resistor will be smaller.
Now let us use three resistors to form a voltage divider circuit as shown below.
Hence, the mathematical voltage divider equation for the voltage across the 6Ω is 3V according to:
This proves what we have concluded before, the more resistances we use, the smaller the voltage drop or potential difference across the resistors we get.
In general, if voltage divider has N resistors (R1, R2, …., RN) in series with the voltage source v, the nth resistor will have a voltage drop of
The voltage divider is used to divide a large voltage to a smaller one.
## Series Resistors Summary
After learning a lot of series resistor explanation, here we try to summarize in short explanation:
• Series resistor is a circuit when we connect multiple resistors in a single wire. We connect the end of the first resistor to the head of the second resistor and so on.
• Series resistor connection has the same value of current.
• The voltage drop across each resistor is proportional to the sum of the resistances and follows Ohm’s law (V = I x R).
• Series resistors circuit acts as a voltage divider circuit.
## Series Resistors Example
For better understanding let us review the examples below:
We have a circuit with a 20V voltage source, three resistors with 3Ω, 7Ω, and 10Ω. Find the equivalent resistance Req, the current, and the voltage drop for each resistor in the circuit.
Equivalent resistance Req:
The Req for series resistors is the sum of all the resistances in the circuit.
Current:
To find the current we use Ohm’s law. Hence, the current would be 1 A.
Voltage drop:
Using Ohm’s law again:
## Frequently Asked Question
### How to reduce voltage?
To reduce voltage in an electric circuit, we can simply form a voltage divider circuit with two resistors. To cut voltage in half, we use two resistors with identical resistances connected in series and connect a wire in between them.
### How to divide voltage with resistors?
To divide voltage in half, we can use two resistors with identical resistance in series and place a jumper in between them. For example, two 5Ω resistors will divide 5 V voltage source into 2.5 V.
### What do resistors in series divide?
The electric current through every resistor connected in series will be the same. The voltage drop for each resistor combined together will be equal to the voltage drop in the circuit.
### Why are resistors in series called potential dividers?
Series resistors are able to divide total voltage into several voltage drops for each resistance.
### How do you calculate resistors in a series circuit?
The resistance of resistors in a series circuit is the sum of resistance of each individual resistor.
### How does a voltage divider work in a series circuit?
A voltage divider is made from a pair of resistors where the input voltage is applied to both resistors but the output is only taken from one of them.
### Do voltages add in series?
The voltage increases in series resistors circuit while the voltage is the same for parallel connection.
### What is the voltage divider equation?
Voltage divider equation with two resistors is Vout=(R2/R1+R2)Vin. Where we draw output voltage from the second resistor R2 (bottom resistor).
### What is the rule for resistors in series?
A circuit with series resistors carries the same electric current through all series resistors but their own voltage drop depends on their own resistance. This is where Ohm’s Law (V=IxR) fulfills its role.
### What is a resistor voltage divider?
A voltage divider is basically series resistors with a wire in between them to draw divided output voltage.
### What is the formula to calculate the total resistance in series combination?
The total resistance for series resistors is the sum of all individual resistors in series. The equivalent resistance for series resistors is always higher than the biggest resistance of that series resistors.
### What is the formula of series connection?
The formula for series resistors connection is Req=R1+R2+R3+…+Rn. |
# How do you solve x+3y=-2, 2x-y=10?
Sep 8, 2015
The solutions for the system of equations is
color(blue)(x=4, y=-2
#### Explanation:
$x + 3 y = - 2$, multiplying this equation by $2$
$\textcolor{b l u e}{2 x} + 6 y = - 4$......equation $1$
$\textcolor{b l u e}{2 x} - y = 10$.....equation $2$
Solving by elimination.
Subtracting equation $2$ from $1$ results in elimination of color(blue)(2x
$\cancel{\textcolor{b l u e}{2 x}} + 6 y = - 4$
$- \cancel{\textcolor{b l u e}{2 x}} + y = - 10$
$7 y = - 14$
color(blue)(y=-2
Finding $x$ by substituting $y$ in equation $1$
$x + 3 y = - 2$
$x + 3 \cdot \left(- 2\right) = - 2$
$x - 6 = - 2$
$x = - 2 + 6$
color(blue)(x=4 |
Class 6 Maths Algebra Using Expressions Practically
Using Expressions Practically
The use of expressions comes handy in everyday. For e.g. if we know that the price of one kg onion is Rs.5 less than the price of 1kg potato, then we can use expressions to calculate the price of 1kg onion. Let the price of 1kg potato be x. Then the price of 1kg onion if Rs. (x-5).
Similarly, if Seema is 5 years older than Ritu and the age of Ritu is given to be z years. Then Seema’s age is (z + 5) years.
Examples:
Take Sarita’s present age to be y .
What will be her age 5 years from now?
Her age 5 years from now will be = (y +5) years.
What was her age 3 years back?
Her age 3 years back was = (y – 3) years.
Sarita’s grandfather is 6 times her age. What is the age of her grandfather?
Age of Sarita’s grandfather = (6*y) years
Grandmother is 2 years younger than grandfather. What is grandmother's age?
Age of Sarita’s grandmother = (Age of grandfather – 2) years
= (6*y -2) years
Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father's age?Age of Sarita’s father = (3y + 5) years.
A rectangular box has height h Its length is 5 times the height and breadth is 10 cm less than the length. Express the length and the breadth of the box in terms of the height.
Height of rectangular box = h cm. (given)
Length of rectangular box = 5*h cm
Breadth of rectangular box = (Length – 10) cm
= (5*h -10)cm.
A bus travels at v km per hour. It is going from Daspur to Beespur. After the bus has travelled 5 hours, Beespur is still 20 km away. What is the distance from Daspur to Beespur? Express it using v.
Speed of the bus = v km per hour
Distance travelled in 5 hours = Speed*Time
= Speed (km per hour)* 5 hours
= 5v kms
Distance from Daspur to Beespur =Distance travelled in 5 hours + 20km
= (5v + 20)kms
. |
# Grok all the things
grok (v): to understand (something) intuitively.
# Matrices
๐ถ ย Children (ELI5)
Hey there, young explorer! Are you ready for an exciting journey into the whimsical world of matrices? Let's discover their incredible powers and see how they can help us solve all sorts of number puzzles!
## โจ What Is a Matrix? ๐ค
Imagine you have a bunch of toy cars and you want to organize them in neat rows and columns, just like how they're parked in a parking lot. A matrix is like that parking lot, but instead of cars, it has numbers!
For instance, we could have a matrix like this:
``````2 5 3
7 1 8
4 9 6``````
See how there are three rows and three columns? This matrix is called a 3x3 matrix because it has three rows and three columns! The matrix can be of any size, even as big as a football field, if we wanted it to be!
## ๐ฏ What Can We Do with Matrices? ๐ฒ
Now that we know what a matrix is let's find out what amazing things we can do with them!
Have you ever played with building blocks and wanted to stack them up? That's exactly what we do when we add matrices! If we have two matrices of the same size, we can simply add them together by adding the numbers in the same position. Let's try it with these two matrices:
Matrix A:
``````1 3
4 2``````
Matrix B:
``````9 0
2 6``````
When we add Matrix A and Matrix B, we get:
``````(1+9) (3+0)
(4+2) (2+6)``````
Which gives us:
``````10 3
6 8``````
### 2. Subtracting Matrices ๐
What if we want to take away the numbers instead? We can do that too! Subtracting matrices works a lot like adding them, but instead of adding the numbers in the same position, we subtract them. Let's try it with our Matrix A and Matrix B:
When we subtract Matrix B from Matrix A, we get:
``````(1-9) (3-0)
(4-2) (2-6)``````
Which gives us:
``````-8 3
2 -4``````
And there you have itโwe've just subtracted two matrices!
### 3. Multiplying Matrices ๐ข
Now, let's take our matrix magic to the next level and multiply them! Multiplying matrices might seem tricky at first, but once you get the hang of it, it's really fun!
To multiply two matrices, the number of columns in the first matrix must be equal to the number of rows in the second one. Let's start with these two matrices:
Matrix C:
``````2 4
3 1``````
Matrix D:
``````7 5
6 8``````
Here are the steps to multiply Matrix C and Matrix D:
1. Multiply the elements of the first row of Matrix C with the corresponding elements of the first column of Matrix D, and add them up: `(2 * 7) + (4 * 6) = 14 + 24 = 38`
2. Multiply the elements of the first row of Matrix C with the corresponding elements of the second column of Matrix D, and add them up: `(2 * 5) + (4 * 8) = 10 + 32 = 42`
3. Multiply the elements of the second row of Matrix C with the corresponding elements of the first column of Matrix D, and add them up: `(3 * 7) + (1 * 6) = 21 + 6 = 27`
4. Multiply the elements of the second row of Matrix C with the corresponding elements of the second column of Matrix D, and add them up: `(3 * 5) + (1 * 8) = 15 + 8 = 23`
And our result is:
``````38 42
27 23``````
Woohoo! We've successfully multiplied two matrices!
## ๐ Real-Life Superpowers of Matrices! ๐
You might be wondering, "Where do we even use matrices in real life?" Well, matrices have some fantastic superpowers!
1. Computer Graphics: Matrices are used to create stunning animations in video games and movies so that characters can move, rotate, and scale smoothly!
2. Cryptography: Matrices help in sending secret messages by encoding and decoding them using special matrix operationsโjust like real-life spies!
3. Robotics: Robots use matrices to calculate their movements and understand where they are in a room so they don't bump into your favorite toys!
And that's just the beginning! Matrices are everywhere in Science, Technology, Engineering, and Math (STEM) fields, making our world more amazing every day!
## ๐ Congratulations, Young Explorer! ๐
Well done! Now you know what matrices are and how to add, subtract, and multiply them! You've also discovered the awe-inspiring superpowers of matrices in real life! So go ahead, share your newfound matrix wisdom, and keep exploring the mathematical mysteries of our universe!
Grok.foo is a collection of articles on a variety of technology and programming articles assembled by James Padolsey. Enjoy! And please share! And if you feel like you can donate here so I can create more free content for you. |
# Video: Finding the Composite of Two Functions
Alex Cutbill
Given 𝑓(𝑥) = 3𝑥 − 1 and 𝑔(𝑥) = 𝑥² + 1, which of the following expressions gives (𝑓 ∘ 𝑔)(𝑥)? [A] 3𝑥² + 3 [B] 9𝑥² − 6𝑥 + 3 [C] 3𝑥² +2 [D] 3𝑥² [E] 9𝑥² − 6𝑥 + 2
04:03
### Video Transcript
Given 𝑓 of 𝑥 equals three 𝑥 minus one and 𝑔 of 𝑥 equals 𝑥 squared plus one, which of the following expressions gives 𝑓 of 𝑔 of 𝑥? Is it option A) three 𝑥 squared plus three, option B) nine 𝑥 squared minus six 𝑥 plus three, option C) three 𝑥 squared plus two, option D) three 𝑥 squared, or option E) nine 𝑥 squared minus six 𝑥 plus two.
So what is the thing here which is represented by this 𝑓 and then this small circle and then this 𝑔? Well, it’s a new function which you get by composing the functions 𝑓 and 𝑔. We can explain what this function does by showing its effect on an arbitrary input 𝑥. We have that 𝑓 of 𝑥 is equal to three 𝑥 minus one and 𝑔 of 𝑥 is equal to 𝑥 squared plus one and we’d like to have a similar expression on the right-hand side to show what happens to 𝑥 when 𝑓 of 𝑔 is applied instead of 𝑓 or 𝑔 alone.
How do we do this? Well, we use a definition of function composition which tells us that 𝑓 composed with 𝑔, or 𝑓 of 𝑔, of 𝑥 is equal to 𝑓 of 𝑔 of 𝑥. Before we try to find an expression for 𝑓 of 𝑔 of 𝑥, let’s just see where 𝑓 of 𝑔 takes the number two. And of course, we just replace 𝑥 by two wherever we see it in the above definition. So we get that 𝑓 of 𝑔 of two is equal to 𝑓 of 𝑔 of two. We want to evaluate the right-hand side, and we do that by first finding 𝑔 of two and substituting that it.
What is 𝑔 of two? Well, we can find that by using our expression for 𝑔 of 𝑥. 𝑔 of 𝑥 is 𝑥 squared plus one, so 𝑔 of two is two squared plus one which is five. And we substitute that in. So now we see that 𝑓 of 𝑔 of two is 𝑓 of five. Now I need to find 𝑓 of five. How do we do that? We use the expression that we have for 𝑓 of 𝑥. 𝑓 of 𝑥 is three 𝑥 minus one and so 𝑓 of five is three times five minus one which is 14. So when the function 𝑓 of 𝑔 is given the input two, we get the output 14.
How do you find 𝑓 of 𝑔 of 𝑥 though? It’s going to be very similar to finding 𝑓 of 𝑔 of two except we’re going to be dealing with algebraic expressions instead of numbers. So first again, we evaluated the right-hand side, and we do that by first finding 𝑔 of 𝑥. Well, we have an expression for 𝑔 of 𝑥 in the question, so let’s just substitute that expression in. So we get 𝑓 of 𝑥 squared plus one.
Okay. Now what do we do with that? We use the definition for 𝑓, so 𝑓 of 𝑥 is equal to three 𝑥 minus one. What is 𝑓 of 𝑥 squared plus one then? Well, 𝑓 takes an input and returns three times that input minus one. So the answer is: three times 𝑥 squared plus one minus one. And we can simplify this by first applying the distributive property to multiply out the brackets and then collecting the constant terms to get three 𝑥 squared plus two.
Okay. So now that we have our answer, let’s go through the options, and we can see that C, three 𝑥 squared plus two, is our correct answer. |
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• Level: GCSE
• Subject: Maths
• Word count: 2075
# Investigate Borders - a fencing problem.
Extracts from this document...
Introduction
Borders
Borders
Aim
My aim is to investigate Borders. I will be drawing borders to different squares and finding a formula for each one and finally I will find a Universal Formula.
Introduction
This is what the task tells me:
Here are 2 squares with squares added on each side to make a border, which surrounds the starting squares.
You can then add another border as shown:
Investigate Borders.
Method
First I will find out how many squares needed for the border to a 1x1 square, then 2x1 and so on up to 5x1. Then I will find a formula for the border to each square and also test the formula out to prove that it works. I will predict how many squares needed for the 6th border and find out if my prediction was correct.
Next, I will find out how many squares needed for the border of a 1x2 square, then 2x2, and so on up to 5x2, then 1x3, 2x3, and so on up to 5x3. Again I will find formulas to them and prove that all the formulas work.
Finally, I will put all the formulas together in three groups and find one overall formula for each of the groups. Then, I will get those three formulas and get one Universal formula in the end.
Diagram of Borders of square: 1x1
Table of results for Borders of square: 1x1
Formula
You can always find ‘the nth term’ using the Formula:
‘a’ is simply the value of
Middle
nth term = 4 x 6 + 8 = 32
Common Difference nth Term
Results
My prediction was 32 which is the correct answer, for the number of squares needed for border number 6, and which also proves that the formula is in working order.
Formula to find the number of squares needed for each border (for square 1x2):
Formula = Simplification =
I have already found out the Formula for 1x2 so there’s no need for the Diagrams & Tables, and I have already proved that the Formula works.
Diagram of Borders of square: 2x2
Table of results for Borders of square: 2x2
Formula to find the number of squares needed for each border (for square 2x2):
Common difference = 4
First term = 8
Formula = Simplification =
Experiment
I will try to find the number of squares needed for border number 6 using the formula, I found out, above:
nth term = 4 x 6 + 4 = 28
Common Difference nth Term
Results
My prediction was 28 which is the correct answer, for the number of squares needed for border number 6, and which also proves that the formula is in working order.
Diagram of Borders of square: 3x2
Table of results for Borders of square: 3x2
Formula to find the number of squares needed for each border (for square 3x2):
Common difference = 4
First term = 10
Formula = Simplification =
Experiment
I will try to find the number of squares needed for border number 6 using the formula, I found out, above:
nth term = 4 x 6 + 6 = 30
Common Difference nth Term
Results
Conclusion
5x3 = 4n + 12
Formula for nx3th term:
4 y + ( 2 n + 2 )
e.g. for border number 6:
First replace ‘n’ with 6.
4 y + ( 2 x 6 + 2) = 4 y + 14
Now you can replace the ‘y’ with an ‘n’ to have the formula, to find the number of squares, for border number 6.
4 n + 14
The Universal formula:
The formula for Length x Width = 4 n + 2 L + 2 W – 4 = B (Border)
e.g. for the 6th border of a 5x3 rectangle:
First replace ‘n’ with the border number = 6, ‘L’ with the Length = 5, and ‘W’ with the Width = 3. Then add brackets where necessary.
( 4 x 6 ) + ( 2 x 5 ) + ( 2 x 3) – 4 = B
Then multiply out the brackets:
24 + 10 + 6 – 4 = 36
36 is the correct answer.
Formulas for nx1
1x1 = 4n
2x1 = 4n + 2
3x1 = 4n + 4
4x1 = 4n + 6
5x1 = 4n + 8
Formulas for nx2
1x1 = 4n + 2
2x1 = 4n + 4
3x1 = 4n + 6
4x1 = 4n + 8
5x1 = 4n + 10
Formulas for nx3
1x1 = 4n + 4
2x1 = 4n + 6
3x1 = 4n + 8
4x1 = 4n + 10
5x1 = 4n + 12
Conclusion
In the time available to me, I believe I have researched Borders to the full extent of my ability. I found formulas to squares nx1. I then extended this to squares nx2 and nx3, and I then was able to construct my Universal Formula, which will tell you the number of squares in any border of square nxn, which could be anything from 2x2 to 10x15. I also found that many of my predictions I made along the way turned out to be correct.
I would say that this investigation has been a success. I began this investigation with the aim to find formulas to nx1, nx2 & nx3 and then a Universal Formula and they were achieved.
Created by Syed Islam, 11Q
Syed Islam, 11Q, 3819 – Mr. Abadji
This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section.
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# Related GCSE Number Stairs, Grids and Sequences essays
1. ## Mathematics Coursework: problem solving tasks
3 star(s)
21 = 3n + 3 3 T 4 6 8 10 12 14 = 2n + 2 3 L 4 4 4 4 4 4 = 4 3 + 0 2 4 6 8 10 = 2n - 2 3 Total 8 12 16 20 24 28 = 4n +
2. ## Number Grids and stairs.
Below are the results from the such an exercise on a 10x10, 11x11 and 12x12 numbered grid box Table 4: 10 x 10 Grid 1 2 3. 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21.
1. ## How many squares in a chessboard n x n
+ 36 + 25 + 16 + 9 + 4 +1 =49 + 36 + 25 + 16 + 9 + 4 +1, which gives the total number of squares in this particular square. This result will then be added to the next 8 x 8 square.
2. ## &quot;Multiply the figures in opposite corners of the square and find the difference between ...
Therefore the formula will be more or less the same as the 10 by 10 grid as it will contain (n - 1).In conclusion the formula I have worked out for any square on an 11 by 11 grid is 11(n - 1)
1. ## Investigation of diagonal difference.
a 3 x 2 cutout. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
2. ## Open Box Problem.
This is again because I want to test what the divisor will turn out to be. As you can see from the table on the next page, there is a very significant difference between 10,000/5, which is 2000, and the value of x, which give this open box its maximum volume, which is 2113.
1. ## Mathematics Borders
* If n=1 U1 = an3 + bn2 +cn + d 1 = 1 (1)3 + b(1)2 + c(1) + d 3 = b + c + d = 2 [equation 1] 3 * If n = 2 U2 = 1 (2)3 + b(2)2 + c(2)
2. ## Mathematical Coursework: 3-step stairs
Therefore I will take the pattern number of the total: > 54-6=48 > b= 48 To conclusion my new formula would be: > 6n+48 12cm by 12cm grid 133 134 135 136 137 138 139 140 141 142 143 144 121 122 123 124 125 126 127 128 129 130
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Difference between revisions of "Change Of Coordinate Transformations"
Change of Coordinate Transformations
Field: Other
Image Created By: [[Author:| ]]
Change of Coordinate Transformations
Image displaying different coordinate systems.
Basic Description
A Change Of Coordinate Transformation is a transformation that converts coordinates from one coordinate system to another coordinate system. Transformations such as scaling, rotating, and translating are usually looked upon as changing or manipulating the geometry itself. However, with change of coordinate transformations, it is important to realize that the coordinate representation of the geometry is modified, rather than the geometry itself.
A More Mathematical Explanation
Note: understanding of this explanation requires: *Linear Algebra
Change of coordinate transformations are different for vectors and points. Choose from one of the se [...]
Change of coordinate transformations are different for vectors and points. Choose from one of the sections below to get a better understanding of how the transformations are applied.
Vectors
Consider a coordinate system A, and a vector $\boldsymbol{\vec{p}}$. The coordinates of $\boldsymbol{\vec{p}}$ relative to coordinate system A is $\boldsymbol{\vec{p}}_A = (x, y)$. It is also apparent that
$\boldsymbol{\vec{p}} = x\boldsymbol{\hat{u}} + y\boldsymbol{\hat{v}}$
In which $\boldsymbol{\hat{u}}$ and $\boldsymbol{\hat{v}}$ are unit vectors along the x and y-axes of coordinate system A. Now consider a second coordinate system, B. In coordinate system B,
$\boldsymbol{\vec{p}}_B = x\boldsymbol{\hat{u}}_B + y\boldsymbol{\hat{v}}_B$
More generally, given $\boldsymbol{\vec{p}}_A = (x, y)$ along with $\boldsymbol{\hat{u}}$ and $\boldsymbol{\hat{v}}$, $\boldsymbol{\vec{p}}_B = (x', y')$ may be found using the formula above.
In 3-dimensional space, given $\boldsymbol{\vec{p}}_A = (x, y, z)$ then
$\boldsymbol{\vec{p}}_B = x\boldsymbol{\hat{u}}_B + y\boldsymbol{\hat{v}}_B + z\boldsymbol{\hat{w}}_B$
In which $\boldsymbol{\hat{u}}$, $\boldsymbol{\hat{v}}$, and $\boldsymbol{\hat{w}}$ are unit vectors along the x, y, and z-axes of coordinate system A.
Points
Consider a coordinate system A, and a point $\boldsymbol{q}$. Point $\boldsymbol{q}$ may be expressed as:
$\boldsymbol{q} = x\boldsymbol{\hat{u}} + y\boldsymbol{\hat{v}} + \boldsymbol{O}$
In which $\boldsymbol{\hat{u}}$ and $\boldsymbol{\hat{v}}$ are unit vectors along the x and y-axes of coordinate system A, and $\boldsymbol{O}$ is the origin of coordinate system A. Now consider a second coordinate system, B. In coordinate system B,
$\boldsymbol{q}_B = x\boldsymbol{\hat{u}}_B + y\boldsymbol{\hat{v}}_B + \boldsymbol{O}_B$
More generally, given $\boldsymbol{q}_A = (x, y)$ along with $\boldsymbol{\hat{u}}$, $\boldsymbol{\hat{v}}$, and $\boldsymbol{O}$ relative to coordinate system B, then $\boldsymbol{q}_B = (x', y')$ may be found using the above formula.
In 3-dimensional space, given $\boldsymbol{q}_A = (x, y, z)$ then
$\boldsymbol{q}_B = x\boldsymbol{\hat{u}}_B + y\boldsymbol{\hat{v}}_B + z\boldsymbol{\hat{w}}_B + \boldsymbol{O}_B$
In which $\boldsymbol{\hat{u}}$, $\boldsymbol{\hat{v}}$, and $\boldsymbol{\hat{w}}$ are unit vectors along the x, y, and z-axes of coordinate system A, and $\boldsymbol{O}$ is the origin of coordinate system A.
Matrix Representation
The change of coordinate transformation varies for points and vectors and thus results in two different equations. However, by using a homogeneous coordinates, both cases may be handled with the following equation:
$(x', y', z', w) = x\boldsymbol{u}_B + y\boldsymbol{v}_B + z\boldsymbol{w}_B + w\boldsymbol{O}_B$
When $w = 1$, the equation handles the change of coordinate transformation for points. Finally, when $w = 0$, the equation handles the transformation for vectors. As long as the $w$-coordinate is set correctly, there is no need to keep track of two different equations. |
Equations of Simple Parabolas (Part 2)
# Equations of Simple Parabolas (Part 2)
## Shifting the Parabola
Now, here's some very good news. By using graphical transformations, knowledge of this one simple equation $\,y = ax^2\,$ actually gives full understanding of all parabolas with directrix parallel to the $x$-axis! The results are summarized next.
Shift the parabola (together with its focus and directrix) horizontally by $\,h\,,$ and vertically by $\,k\,.$ This yields the following information:
original equation: $y=ax^2$
shifted equation: $y=a(x-h)^2 + k$
original vertex: $(0,0)$
new vertex: $(h,k)$
original focus: $\displaystyle (0,p) = (0,\frac{1}{4a})$
new focus: $\displaystyle (h,p+k)= (h,\frac{1}{4a}+k)$
original directrix: $y=-p$
new directrix: $y=-p+k$
## Example(Graphing a Shifted Parabola)
In this example, I illustrate the approach that I usually take when asked to give complete information about the parabola $\,y = a(x-h)^2 + k\,.$
Question: Completely describe the graph of the equation $\,y = -3(x+5)^2 + 1\,.$
Solution:
• Concave up or down? Since $\,a=-3\lt 0\,,$ the parabola is concave down (sheds water). Since the focus is always inside a parabola, we know the focus is under the vertex.
• Find the vertex: The vertex of $\,y = a(x-h)^2 + k\,$ is the point $\,(h,k)\,$:
$$\cssId{s34}{y = a( \overset{\text{What value of } \ x\ }{\overset{\text{ makes this zero?}}{\overbrace{ \underset{\text{answer: } \ h}{\underbrace{x-h}}}}})^2\ \ \ + \overset{y\text{-value of the vertex}}{\overbrace{\ \ k\ \ }}}$$
For us, what makes $\,x+5\,$ equal to zero? Answer: $\,-5\,$
Thus, $\,-5\,$ is the $x$-value of the vertex. When $\,x = -5\,,$ the corresponding $y$-value is $\,1\,.$ Thus, the vertex is $\,(-5,1)\,.$
• Find the distance from the vertex to the focus: Using the ‘one over four pee pair’ memory device:
$$\cssId{s42}{p = \frac{1}{4a} = \frac{1}{4(-3)} = -\frac{1}{12}}$$
Thus, the distance from the focus to the vertex is:
$$\cssId{s44}{|p| = \left|-\frac1{12}\right| = \frac{1}{12}}$$
• Find the focus: We already determined that the focus lies below the vertex. So, the focus is:
$$\cssId{s48}{(-5,1-\frac{1}{12}) =(-5,\frac{11}{12})}$$
• Find the equation of the directrix: Every horizontal line has an equation of the form:
$$\cssId{s51}{y = \text{(some #)}}$$
In our example, the focus lies $\,\frac{1}{12}\,$ below the vertex, so the directrix lies $\,\frac{1}{12}\,$ above the vertex. Thus, the equation of the directrix is $\,y=1+\frac{1}{12}\,,$ that is, $\,y=\frac{13}{12}\,.$
• Plot an additional point: Plotting an additional point gives a sense of the ‘width’ of the parabola. For example, when $\,x = -4\,$ we have:
\begin{align} y &= -3(-4+5)^2 + 1\cr &= -3(1) + 1\cr &= -2 \end{align}
In this parabola, the vertex is close to the focus, so the parabola is narrow.
$\,y=-3(x+5)^2+1\,$
For fun, zip up to WolframAlpha and type in any of the following:
vertex of y = -3(x+5)^2 + 1
focus of y = -3(x+5)^2 + 1
directrix of y = -3(x+5)^2 + 1
How easy is that?! |
# Video: Creating Linear Equations with One Variable
The sum of a number and nineteen divided by two is five more than three times the number. Write an equation to represent the statement above. Let π₯ represent the number.
03:14
### Video Transcript
The sum of a number and 19 divided by two is five more than three times the number. Write an equation to represent the statement above. Let π₯ represent the number.
As we are letting π₯ represent the number, then the sum of a number and 19 can be written as π₯ plus 19. We then need to divide this by two. So we have π₯ plus 19 divided by two. We were told that this is five more than three times the number. Three times the number is equal to three π₯ and five more than this is three π₯ plus five. These two expressions are equal to each other.
This means that the equation that represents the statement is π₯ plus 19 divided by two is equal to three π₯ plus five. Whilst we werenβt explicitly asked to in this question, we can solve this equation to calculate the value of the number π₯. Multiplying both sides of the equation by two gives us π₯ plus 19 is equal to two multiplied by three π₯ plus five.
Expanding the bracket or parenthesis gives us six π₯ plus 10 as two multiplied by three π₯ is equal to six π₯ and two multiplied by five is equal to 10. Subtracting π₯ from both sides of this equation gives us 19 is equal to five π₯ plus 10. We can then subtract 10 from both sides of the equation. This leaves us with nine is equal to five π₯ or five π₯ equals nine. Finally, dividing both sides by five gives us a value of π₯ of nine-fifths or nine over five.
Nine-fifths is equivalent to one and four-fifths or 1.8. We can, therefore, see that the value for π₯ was equal to 1.8.
We can substitute this value into the left- and right-hand side of the equation and check that they are equal to each other. 1.8 plus 19 is equal to 20.8 and 20.8 divided by two is equal to 10.4. Therefore, the left-hand side is equal to 10.4. Substituting π₯ equals 1.8 into the right-hand side gives us three multiplied by 1.8 plus five. Three multiplied by 1.8 is equal to 5.4. And adding five to this also gives us 10.4.
We have, therefore, proved that the correct equation was π₯ plus 19 divided by two is equal to three π₯ plus five. And our value for π₯ would be 1.8 or nine-fifths. |
NCERT Exemplar Class 12 Maths Solutions Chapter 4 Determinants
# NCERT Exemplar Class 12 Maths Solutions Chapter 4 Determinants
Edited By Ravindra Pindel | Updated on Sep 15, 2022 04:51 PM IST | #CBSE Class 12th
NCERT exemplar Class 12 Maths solutions chapter 4 Determinants will help in learning the ways to find the determinant of various square matrices, co-factors, the inverse of matrices etc. NCERT exemplar Class 12 Maths chapter 4 solutions are useful for the students to get a deeper and better look at the matrices and how to solve them uniquely. From the scoring point of view, chapter 4 of NCERT Class 12 Maths Solutions can be very crucial for 12 Class students. Students can use NCERT exemplar Class 12 Maths solutions chapter 4 PDF download and study the topic and the solutions offline as well.
Question:1
Using the properties of determinants in evaluate:
$\left|\begin{array}{cc} x^{2}-x+1 & x-1 \\ x+1 & x+1 \end{array}\right|$
$\left|\begin{array}{cc} x^{2}-x+1 & x-1 \\ x+1 & x+1 \end{array}\right|$
\begin{aligned} &\text { If } A=\left|\begin{array}{ll} a & b \\ c & d \end{array}\right| \\ &\text { The value of determinant of A can then be found by: }\\ &|A|=\left|\begin{array}{ll} a & b \\ c & d \end{array}\right|=a d-b c \end{aligned}
Question:2
Using the properties of determinants in evaluate:
$\left|\begin{array}{ccc} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{array}\right|$
$A=\left|\begin{array}{ccc} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{array}\right|$
$C_{1} \rightarrow C_{1}+C_{2}+C_{3}$
$\begin{array}{l} =\left|\begin{array}{ccc} a+x+y+z & y & z \\ a+x+y+z & a+y & z \\ a+x+y+z & y & a+z \end{array}\right| \\\\ =(a+x+y+z)\left|\begin{array}{ccc} 1 & y & z \\ 1 & a+y & z \\ y & a+z \end{array}\right| \end{array}$
$R_{2} \rightarrow R_{2}-R_{1}$
$R_{3} \rightarrow R_{3}-R_{\mathrm{1}}$
$\begin{array}{l} =(a+x+y+z)\left|\begin{array}{ccc} 1 & y & z \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right| \\\\ =(a+x+y+z)\left|\begin{array}{ll} a & 0 \\ 0 & a \end{array}\right|\\\\=a^{2}(a+z+x+y) \end{array}$
Question:3
Using the properties of determinants in evaluate:
$\left|\begin{array}{ccc} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{array}\right|$
$\text{Let } A=\left|\begin{array}{ccc} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{array}\right|$
$=x^{2} y^{2} z^{2}\left|\begin{array}{lll} 0 & x & x \\ y & 0 & y \\ z & z & 0 \end{array}\right|$
$C_{2} \rightarrow C_{2}-C_{3}$
$=x^{2} y^{2} z^{2}\left|\begin{array}{ccc} 0 & 0 & x \\ y & -y & y \\ z & z & 0 \end{array}\right|$
Expand IAI along C3 to get:
$=x^{2} y^{2} z^{2}(x(y z+y z))$
$=x^{2} y^{2} z^{2}(2xy z)$
$=2x^{3} y^{3} z^{3}$
Question:4
Using the properties of determinants in evaluate:
$\left|\begin{array}{ccc} 3 x & -x+y & -x+z \\ x-y & 3 y & z-y \\ x-z & y-z & 3 z \end{array}\right|$
$Let \: \: A=\left|\begin{array}{ccc} 3 x & -x+y & -x+z \\ x-y & 3 y & z-y \\ x-z & y-z & 3 z \end{array}\right|$
Apply -
\\\begin{aligned} &=\left|\begin{array}{ccc} 3 x-x+y-x+z & -x+y & -x+z \\ x-y+3 y+z-y & 3 y & z-y \\ x-z+y-z+3 z & y-z & 3 z \end{array}\right|\\ &=\left|\begin{array}{ccc} x+y+z & -x+y & -x+z \\ x+y+z & 3 y & z-y \\ x+y+z & y-z & 3 z \end{array}\right|\\ &\text { Take }(x+y+z) \text { common from } C_{1}\\ &=(x+y+z)\left|\begin{array}{ccc} 1 & -x+y & -x+z \\ 1 & 3 y & z-y \\ 1 & y-z & 3 z \end{array}\right| \end{aligned}
\\\begin{aligned} &\text { Apply } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1,} \text { you will get }\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 1-1 & 3 \mathrm{y}-(-\mathrm{x}+\mathrm{y}) & \mathrm{z}-\mathrm{y}-(-\mathrm{x}+\mathrm{z}) \\ 1 & \mathrm{y}-\mathrm{z} & 3 \mathrm{z} \end{array}\right|\\\end{aligned}
\\\begin{aligned} &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 0 & 3 \mathrm{y}+\mathrm{x}-\mathrm{y} & \mathrm{z}-\mathrm{y}+\mathrm{x}-\mathrm{z} \\ 1 & \mathrm{y}-\mathrm{z} & 3 \mathrm{z} \end{array}\right|\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y} & \mathrm{x}-\mathrm{y} \\ 1 & \mathrm{y}-\mathrm{z} & 3 \mathrm{z} \end{array}\right|\\ &\text { Now apply, } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y} & \mathrm{x}-\mathrm{y} \\ 1-1 & \mathrm{y}-\mathrm{z}-(-\mathrm{x}+\mathrm{y}) & 3 \mathrm{z}-(-\mathrm{x}+\mathrm{z}) \end{array}\right| \end{aligned}
\\\begin{aligned} &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y} & \mathrm{x}-\mathrm{y} \\ 0 & \mathrm{y}-\mathrm{z}+\mathrm{x}-\mathrm{y} & 3 \mathrm{z}+\mathrm{x}-\mathrm{z} \end{array}\right|\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y} & \mathrm{x}-\mathrm{y} \\ 0 & \mathrm{x}-\mathrm{z} & 2 \mathrm{z}+\mathrm{x} \end{array}\right|\\ &\text { Apply, } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3},\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y}-(-\mathrm{x}+\mathrm{z}) & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y}-(\mathrm{x}-\mathrm{y}) & \mathrm{x}-\mathrm{y} \\ 0 & \mathrm{x}-\mathrm{z}-(2 \mathrm{z}+\mathrm{x}) & 2 \mathrm{z}+\mathrm{x} \end{array}\right|\\\end{aligned}
\\\\\begin{aligned} &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y}+\mathrm{x}-\mathrm{z} & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y}-\mathrm{x}+\mathrm{y} & \mathrm{x}-\mathrm{y} \\ 0 & \mathrm{x}-\mathrm{z}-2 \mathrm{z}-\mathrm{x} & 2 \mathrm{z}+\mathrm{x} \end{array}\right|\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & \mathrm{y}-\mathrm{z} & -\mathrm{x}+\mathrm{z} \\ 0 & 3 \mathrm{y} & \mathrm{x}-\mathrm{y} \\ 0 & -3 \mathrm{z} & 2 \mathrm{z}+\mathrm{x} \end{array}\right| \end{aligned}
Now, expand the determinant along Column 1
Question:5
Using the properties of determinants in evaluate:
$\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|$
$Let\: \: A=\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|$
\\\begin{aligned} &\text { Using } C_{1} \rightarrow C_{1}+C_{2}+C_{3},\\ &=\left|\begin{array}{ccc} x+4+x+x & x & x \\ x+x+4+x & x+4 & x \\ x+x+x+4 & x & x+4 \end{array}\right|\\ &=\left|\begin{array}{ccc} 3 x+4 & x & x \\ 3 x+4 & x+4 & x \\ 3 x+4 & x & x+4 \end{array}\right|\\ &\text { Take }(3 x+4) \text { common form } C \text { , }\\ &=(3 x+4)\left|\begin{array}{ccc} 1 & x & x \\ 1 & x+4 & x \\ 1 & x & x+4 \end{array}\right|\\ &\mathrm{Apply}, \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\\ &=(3 x+4)\left|\begin{array}{ccc} 1 & x & x \\ 0 & 4 & 0 \\ 1-1 & x-x & x+4-x \end{array}\right| \end{aligned}
\\\begin{aligned} &=(3 \mathrm{x}+4)\left|\begin{array}{lll} 1 & \mathrm{x} & \mathrm{x} \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}\right|\\ &\text { Expand along } \mathrm{C}_{1}\\ &=(3 x+4)[1 \times\{(16)-0\}]\\ &=(3 x+4)(16)\\ &=16(3 x+4) \end{aligned}
Question:6
Using the properties of determinants in evaluate:
$\left|\begin{array}{ccc} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|$
$Let\: \: A=\left|\begin{array}{ccc} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|$
\\\begin{aligned} &\text { Apply } R_{1} \rightarrow R_{1}+R_{2}+R_{y}\\ &=\left|\begin{array}{ccc} \mathrm{a}-\mathrm{b}-\mathrm{c}+2 \mathrm{~b}+2 \mathrm{c} & 2 \mathrm{a}+\mathrm{b}-\mathrm{c}-\mathrm{a}+2 \mathrm{c} & 2 \mathrm{a}+2 \mathrm{~b}+\mathrm{c}-\mathrm{a}-\mathrm{b} \\ 2 \mathrm{~b} & \mathrm{~b}-\mathrm{c}-\mathrm{a} & 2 \mathrm{~b} \\ 2 \mathrm{c} & 2 \mathrm{c} & \mathrm{c}-\mathrm{a}-\mathrm{b} \end{array}\right|\\ &=\left|\begin{array}{ccc} \mathrm{a}+\mathrm{b}+\mathrm{c} & \mathrm{a}+\mathrm{b}+\mathrm{c} & \mathrm{a}+\mathrm{b}+\mathrm{c} \\ 2 \mathrm{~b} & \mathrm{~b}-\mathrm{c}-\mathrm{a} & 2 \mathrm{~b} \\ 2 \mathrm{c} & 2 \mathrm{c} & \mathrm{c}-\mathrm{a}-\mathrm{b} \end{array}\right|\\ &\text { Take }(a+b+c) \text { common form the first row }\\ &=(a+b+c)\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|\\ &\text { Apply } C_{2} \rightarrow C_{2}-C_{1} \end{aligned}
$\begin{array}{l} =(a+b+c)\left|\begin{array}{ccc} 1 & 1-1 & 1 \\ 2 b & b-c-a-2 b & 2 b \\ 2 c & 2 c-2 c & c-a-b \end{array}\right| \\ =(a+b+c)\left|\begin{array}{ccc} 1 & 0 & 1 \\ 2 b & -b-c-a & 2 b \\ 2 c & 0 & c-a-b \end{array}\right| \end{array}$
Now apply $\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}$
$\begin{array}{l} =(a+b+c)\left|\begin{array}{ccc} 1 & 0 & 1-1 \\ 2 b & -(a+b+c) & 2 b-2 b \\ 2 c & 0 & c-a-b-2 c \end{array}\right| \\ =(a+b+c)\left|\begin{array}{ccc} 1 & 0 & 0 \\ 2 b & -(a+b+c) & 0 \\ 2 c & 0 & -(a+b+c) \end{array}\right| \end{array}$
Expand the determinant along Row 1
Question:7
Using the properties of determinants in prove that:
$\left|\begin{array}{lll} y^{2} z^{2} & y z & y+z \\ z^{2} x^{2} & z x & z+x \\ x^{2} y^{2} & x y & x+y \end{array}\right|=0$
Taking LHS, $\left|\begin{array}{lll} y^{2} z^{2} & y z & y+z \\ z^{2} x^{2} & z x & z+x \\ x^{2} y^{2} & x y & x+y \end{array}\right|$
\\\begin{aligned} &\text { Multiply and divide } \mathrm{R}_{1} \mathrm{R}_{2}, \mathrm{R}_{3} \text { respectively by } \mathrm{x}, \mathrm{y}, \mathrm{z}\\ &=\frac{1}{x y z}\left|\begin{array}{lll} x y^{2} z^{2} & x y z & x(y+z) \\ y z^{2} x^{2} & y z x & y(z+x) \\ z x^{2} y^{2} & z x y & z(x+y) \end{array}\right|\\ &=\frac{1}{x y z}\left|\begin{array}{lll} x y^{2} z^{2} & x y z & x y+x z \\ x^{2} y z^{2} & x y z & y z+x y \\ x^{2} y^{2} z & x y z & x z+y z \end{array}\right|\\ &\text { Now, take xyz common from the first and second Column }\\ &=\frac{1}{\mathrm{xyz}} \times \mathrm{xyz} \times \mathrm{xyz}\left|\begin{array}{lll} \mathrm{yz} & 1 & \mathrm{xy}+\mathrm{xz} \\ \mathrm{xz} & 1 & \mathrm{yz}+\mathrm{xy} \\ \mathrm{xy} & 1 & \mathrm{xz}+\mathrm{yz} \end{array}\right|\\ &=\operatorname{xyz}\left|\begin{array}{lll} y z & 1 & x y+x z \\ x z & 1 & y z+x y \\ x y & 1 & x z+y z \end{array}\right| \end{aligned}
\\\begin{aligned} &\text { Apply, } \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}+\mathrm{C}_{2}\\ &=\operatorname{xyz}\left|\begin{array}{lll} y z & 1 & x y+x z+y z \\ x z & 1 & y z+x y+x z \\ x y & 1 & x z+y z+x y \end{array}\right|\\ &\text { Take }(\mathrm{xy}+\mathrm{yz}+\mathrm{xz}) \text { common from } \mathrm{C}_{3}\\ &=(\text { xyz })(\mathrm{xy}+\mathrm{yz}+\mathrm{xz})\left|\begin{array}{lll} \mathrm{yz} & 1 & 1 \\ \mathrm{xz} & 1 & 1 \\ \mathrm{xy} & 1 & 1 \end{array}\right| \end{aligned}
Whenever any the values in any two rows or columns of a determinant are identical, the resultant value of that determinant is 0
Hence, $\left|\begin{array}{lll} y^{2} z^{2} & y z & y+z \\ z^{2} x^{2} & z x & z+x \\ x^{2} y^{2} & x y & x+y \end{array}\right|=0$
∴ LHS = RHS
Question:8
Using the properties of determinants in prove that:
$\left|\begin{array}{ccc} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{array}\right|=4 x y z$
LHS given,
$\left|\begin{array}{ccc} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{array}\right|$
\\\begin{aligned} &=\left|\begin{array}{ccc} \mathrm{y}+\mathrm{z}+\mathrm{z}+\mathrm{y} & \mathrm{z}+\mathrm{z}+\mathrm{x}+\mathrm{x} & \mathrm{y}+\mathrm{x}+\mathrm{x}+\mathrm{y} \\ \mathrm{z} & \mathrm{z}+\mathrm{x} & \mathrm{x} \\ \mathrm{y} & \mathrm{x} & \mathrm{x}+\mathrm{y} \end{array}\right|\\ &=\left|\begin{array}{ccc} 2 z+2 y & 2 z+2 x & 2 x+2 y \\ z & z+x & x \\ y & x & x+y \end{array}\right|\\ &2 \text { can be taken common from } \mathrm{R}_{1}\\ &=2\left|\begin{array}{ccc} \mathrm{z}+\mathrm{y} & \mathrm{z}+\mathrm{x} & \mathrm{x}+\mathrm{y} \\ \mathrm{z} & \mathrm{z}+\mathrm{x} & \mathrm{x} \\ \mathrm{y} & \mathrm{x} & \mathrm{x}+\mathrm{y} \end{array}\right|\\ \end{aligned}
\\\begin{aligned} \\&\text { Apply, } R_{1} \rightarrow R_{1}-R_{2}\\ &=2\left|\begin{array}{ccc} \mathrm{z}+\mathrm{y}-\mathrm{z} & \mathrm{z}+\mathrm{x}-(\mathrm{z}+\mathrm{x}) & \mathrm{x}+\mathrm{y}-\mathrm{x} \\ \mathrm{z} & \mathrm{z}+\mathrm{x} & \mathrm{x} \\ \mathrm{y} & \mathrm{x} & \mathrm{x}+\mathrm{y} \end{array}\right| \end{aligned}
\\\begin{aligned} &=2\left|\begin{array}{ccc} y & 0 & y \\ z & z+x & x \\ y & x & x+y \end{array}\right|\\ &\text { Apply, } R_{3} \rightarrow R_{3}-R_{1},\\ &=2\left|\begin{array}{ccc} \mathrm{y} & 0 & \mathrm{y} \\ \mathrm{z} & \mathrm{z}+\mathrm{x} & \mathrm{x} \\ \mathrm{y}-\mathrm{y} & \mathrm{x}-0 & \mathrm{x}+\mathrm{y}-\mathrm{y} \end{array}\right|\\ &=2\left|\begin{array}{ccc} y & 0 & y \\ z & z+x & x \\ 0 & x & x \end{array}\right|\\ &\text { Now, Apply } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3},\\ &=2\left|\begin{array}{ccc} \mathrm{y} & 0 & \mathrm{y} \\ \mathrm{z}-0 & \mathrm{z}+\mathrm{x}-\mathrm{x} & \mathrm{x}-\mathrm{x} \\ 0 & \mathrm{x} & \mathrm{x} \end{array}\right|\\ &=2\left|\begin{array}{lll} y & 0 & y \\ z & z & 0 \\ 0 & x & x \end{array}\right| \end{aligned}
Take y,z,x common from the R1, R2 and R3 respectively
\begin{aligned} &\text { Expand along Column } 1\\ &\begin{array}{r} |A|=a_{11}(-1)^{1+1}\left|\begin{array}{ll} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}\right|+a_{21}(-1)^{2+1}\left|\begin{array}{ll} a_{12} & a_{13} \\ a_{32} & a_{33} \end{array}\right| \\ +a_{31}(-1)^{3+1}\left|\begin{array}{ll} a_{12} & a_{13} \\ a_{22} & a_{23} \end{array}\right| \end{array} \end{aligned}
Hence Proved
Question:9
Using the properties of determinants in prove that:
$\left|\begin{array}{ccc} a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|=(a-1)^{3}$
Take LHS
$\left|\begin{array}{ccc} a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|$
$\\\begin{array}{l} \text { Apply, } R_{1} \rightarrow R_{1}-R_{2} \\ =\left|\begin{array}{ccc} a^{2}+2 a-(2 a+1) & 2 a+1-(a+2) & 1-1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right| \\ \\=\left|\begin{array}{ccc} a^{2}+2 a-2 a-1 & 2 a+1-a-2 & 0 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right| \\ \\=\left|\begin{array}{ccc} a^{2}-1 & a-1 & 0 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right| \end{array}$
\\\begin{aligned} &=\left|\begin{array}{ccc} (a-1)(a+1) & a-1 & 0 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|\left[\because\left(a^{2}-b^{2}\right)=(a-b)(a+b)\right]\\ &\text { Take (a-1) common from R }\\ &=(a-1)\left|\begin{array}{ccc} a+1 & 1 & 0 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|\\ &\text { Apply, } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\\ &=\left|\begin{array}{ccc} a+1 & 1 & 0 \\ 2 a+1-3 & a+2-3 & 1-1 \\ 3 & 3 & 1 \end{array} \mid\right|.\\ &=(a-1)\left|\begin{array}{ccc} a+1 & 1 & 0 \\ 2 a-2 & a-1 & 0 \\ 3 & 3 & 1 \end{array}\right|\\ &=(a-1)\left|\begin{array}{ccc} a+1 & 1 & 0 \\ 2(a-1) & a-1 & 0 \\ 3 & 3 & 1 \end{array}\right| \end{aligned}
Take (a-1) common from $R_2$
\begin{aligned} &=(a-1)^{2}\left|\begin{array}{ccc} a+1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{array}\right|\\ &\text { Expand along } \mathrm{C}_{3} \end{aligned}
$\\\\ = (a -1)\textsuperscript{2} [1 \{ (a + 1) -2 \} ]\\ \\ \vspace{\baselineskip}= (a -1)\textsuperscript{2} [a + 1 -2]\\ \\ \vspace{\baselineskip}= (a -1)\textsuperscript{3}\\ \\ \vspace{\baselineskip}= RHS\\$
Hence,Proved.
Question:10
If A + B + C = 0, then prove that $\begin{array}{|ccc|} 1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{array} =0$
LHS given: $\begin{array}{|ccc|} 1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{array}$
Expand along $R_1$
$\\\begin{array}{l} |\mathrm{A}|=\mathrm{a}_{11}(-1)^{1+1}\left|\begin{array}{cc} \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right|+\mathrm{a}_{12}(-1)^{1+2}\left|\begin{array}{cc} \mathrm{a}_{21} & \mathrm{a}_{23} \\ \mathrm{a}_{31} & \mathrm{a}_{33} \end{array}\right| \\ +\mathrm{a}_{13}(-1)^{1+3}\left|\begin{array}{cc} \mathrm{a}_{21} & \mathrm{a}_{22} \\ \mathrm{a}_{31} & \mathrm{a}_{32} \end{array}\right| \\ =(1)\left|\begin{array}{cc} 1 & \cos \mathrm{A} \\ \cos \mathrm{A} & 1 \end{array}\right|-\cos \mathrm{C}\left|\begin{array}{cc} \cos \mathrm{C} & \cos \mathrm{A} \\ \cos \mathrm{B} & 1 \end{array}\right|+\cos \mathrm{B}\left|\begin{array}{cc} \cos \mathrm{C} & 1 \\ \cos \mathrm{B} & \cos \mathrm{A} \end{array}\right| \end{array}$
Question:11
If the co-ordinates of the vertices of an equilateral triangle with sides of length ‘a’ are $\left (x_1, y_1 \right )$, $\left (x_2, y_2 \right )$, $\left (x_3, y_3 \right )$, then $\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|^{2}=\frac{3 a^{4}}{4}$
Area of a triangle with the given vertices will be:
\\\begin{aligned} &\Rightarrow\left|\begin{array}{lll} \mathrm{x}_{1} & \mathrm{x}_{2} & \mathrm{x}_{3} \\ \mathrm{y}_{1} & \mathrm{y}_{2} & \mathrm{y}_{3} \\ \mathrm{z}_{1} & \mathrm{z}_{2} & \mathrm{z}_{3} \end{array}\right|^{2}=\frac{3}{4} \mathrm{a}^{4}\\ &\text { Hence Proved } \end{aligned}
Question:12
Find the value of θ satisfying $\left|\begin{array}{ccc} 1 & 1 & \sin 3 \theta \\ -4 & 3 & \cos 2 \theta \\ 7 & -7 & -2 \end{array}\right|=0$
Given:
$\left|\begin{array}{ccc} 1 & 1 & \sin 3 \theta \\ -4 & 3 & \cos 2 \theta \\ 7 & -7 & -2 \end{array}\right|=0$
Expand along Row 1
$\begin{array}{l} |\mathrm{A}|=\mathrm{a}_{11}(-1)^{1+1}\left|\begin{array}{cc} \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right|+\mathrm{a}_{12}(-1)^{1+2}\left|\begin{array}{cc} \mathrm{a}_{21} & \mathrm{a}_{23} \\ \mathrm{a}_{31} & \mathrm{a}_{33} \end{array}\right| \\ +\mathrm{a}_{13}(-1)^{1+3}\left|\begin{array}{cc} \mathrm{a}_{21} & \mathrm{a}_{22} \\ \mathrm{a}_{31} & \mathrm{a}_{32} \end{array}\right| \\ =\left|\begin{array}{lll} 1 \end{array}\left|\begin{array}{cc} 3 & \cos 2 \theta \\ -7 & -2 \end{array}\right|-1\left|\begin{array}{cc} -4 & \cos 2 \theta \\ 7 & -2 \end{array}\right|+\sin 3 \theta\left|\begin{array}{cc} -4 & 3 \\ 7 & -7 \end{array}\right|\right. \end{array}$
We know,
$\begin{array}{l} \Rightarrow \mathrm{or} \sin \theta=-\frac{1}{2} \text { or } \sin \theta=\frac{3}{2} \\ \Rightarrow \theta=\mathrm{n} \pi \text { or } \theta=\mathrm{m} \pi+(-1)^{\mathrm{n}}\left(-\frac{\pi}{6}\right) ; \mathrm{m}, \mathrm{n} \in \mathrm{Z} \end{array}$
But it is not possible to have $\sin \theta=\frac{3}{2}$
Question:13
If $\left|\begin{array}{ccc} 4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x \end{array}\right|=0$ then find values of x.
Given:
$\left|\begin{array}{ccc} 4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x \end{array}\right|=0$
\\\begin{aligned} &\text { Apply, } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\\ &\Rightarrow\left|\begin{array}{ccc} 4-x+4+x+4+x & 4+x & 4+x \\ 4+x+4-x+4+x & 4-x & 4+x \\ 4+x+4+x+4-x & 4+x & 4-x \end{array}\right|=0\\ &\text { Take, }(12+\mathrm{x}) \text { common from Row } 1\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 4+x & 4+x \\ 1 & 4-x & 4+x \\ 1 & 4+x & 4-x \end{array}\right|=0\\ &\text { Apply } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}+\mathrm{C}_{3}\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 4+x+4+x & 4+x \\ 1 & 4-x+4+x & 4+x \\ 1 & 4+x+4-x & 4-x \end{array}\right|=0\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 2(4+x) & 4+x \\ 1 & 8 & 4+x \\ 1 & 8 & 4-x \end{array}\right|=0 \end{aligned}
\\\begin{aligned} &\text { Apply, } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 2(4+x) & 4+x \\ 1-1 & 8-8 & 4+x-(4-x) \\ 1 & 8 & 4-x \end{array}\right|=0\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 2(4+x) & 4+x \\ 0 & 0 & 2 x \\ 1 & 8 & 4-x \end{array}\right|=0\\ &\text { Apply } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 2(4+x) & 4+x \\ 0 & 0 & 2 x \\ 1-1 & 8-8-2 x & 4-x-4-x \end{array}\right|=0\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 2(4+x) & 4+x \\ 0 & 0 & 2 x \\ 0 & -2 x & -2 x \end{array}\right|=0 \end{aligned}
Expand along Column 1
Question:14
If a1, a2, a3, ..., ar are in G.P., then prove that the determinant $\left|\begin{array}{ccc} a_{r+1} & a_{r+5} & a_{r+9} \\ a_{r+7} & a_{r+11} & a_{r+15} \\ a_{r+11} & a_{r+17} & a_{r+21} \end{array}\right|$ is independent of r.
A is the first term of G.PR is the common ratio of G.P.
$\\\\ =\mathrm{AR}^{\mathrm{r}} \times \mathrm{AR}^{\mathrm{r}+6} \times \mathrm{AR}^{\mathrm{r}+10}\left|\begin{array}{ccc} 1 & \mathrm{AR}^{4} & \mathrm{AR}^{8} \\ 1 & \mathrm{AR}^{4} & \mathrm{AR}^{8} \\ 1 & \mathrm{AR}^{6} & \mathrm{AR}^{10} \end{array}\right|$
Whenever any the values in any two rows or columns of a determinant are identical, the resultant value of that determinant is 0
Rows 1 and 2 are identical
Question:15
Show that the points (a + 5, a – 4), (a – 2, a + 3) and (a, a) do not lie on a straight line for any value of a.
(a + 5, a – 4), (a – 2, a + 3) and (a, a) is given.
We need to prove that they don’t line in a straight line for any value of a
This can be done by proving the points to be vertices of triangle.
Area of triangle:-
$\\\begin{array}{l} \Delta=\frac{1}{2}\left|\begin{array}{ccc} x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ z_{1} & z_{2} & z_{3} \end{array}\right| \\ =\frac{1}{2}\left|\begin{array}{ccc} a+5 & a-4 & 1 \\ a-2 & a+3 & 1 \\ a & a & 1 \end{array}\right| \\ \text { Apply } R_{2} \rightarrow R_{2}-R_{1} \\ =\frac{1}{2}\left|\begin{array}{ccc} a+5 & a-4 & 1 \\ a-2-a-5 & a+3-a+4 & 1-1 \\ a & a & 1 \end{array}\right| \\\\ =\frac{1}{2}\left| \begin{array}{ccc} a+5 & a-4 & 1 \\ -7 & 7 & 0 \\ a & a & 1 \end{array}\right| \end{array}$
\\\begin{aligned} &\text { Apply, } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\\ &=\frac{1}{2}\left|\begin{array}{ccc} a+5 & a-4 & 1 \\ -7 & 7 & 0 \\ a-a-5 & a-a+4 & 1-1 \end{array}\right|\\ &=\frac{1}{2}\left|\begin{array}{ccc} a+5 & a-4 & 1 \\ -7 & 7 & 0 \\ -5 & 4 & 0 \end{array}\right|\\ &\text { Expand along Column } 3\\ &=\frac{1}{2}[(1)(-28-(7)(-5))]\\ &=\frac{1}{2}(-28+35)=\frac{7}{2} \neq 0 \end{aligned}
This proves that the given points form a triangle therefore do not lie on a straight line.
Question:16
\\\begin{aligned} &\Delta=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1+\cos A & 1+\cos B & 1+\cos C \\ \cos ^{2} A+\cos A & \cos ^{2} B+\cos B & \cos ^{2} C+\cos C \end{array}\right|=0\\ &\text { Apply, } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}\\ &=\left|\begin{array}{ccc} 1 & 1-1 & 1 \\ 1+\cos A & 1+\cos B-1-\cos A & 1+\cos C \\ \cos ^{2} A+\cos A & \cos ^{2} B+\cos B-\cos ^{2} A-\cos A & \cos ^{2} C+\cos C \end{array}\right|=0\\ &=\left|\begin{array}{ccc} 1 & 0 & 1 \\ 1+\cos A & \cos B-\cos A & 1+\cos C \\ \cos ^{2} A+\cos A & \cos ^{2} B+\cos B-\cos ^{2} A-\cos A & \cos ^{2} C+\cos C \end{array}\right|=0\\ &=\left|\begin{array}{ccc} 1 & 0 & 1 \\ 1+\cos A & \cos B-\cos A & 1+\cos C \\ \cos ^{2} A+\cos A & (\cos B-\cos A)(\cos B+\cos A)+(\cos B-\cos A) & \cos ^{2} C+\cos C \end{array}\right|=0 \end{aligned}
\\\begin{aligned} &\begin{array}{|ccc|} 1 & 0 & 1 \\ 1+\cos A & \cos B-\cos A & 1+\cos C \\ \cos ^{2} A+\cos A & (\cos B-\cos A)[(\cos B+\cos A)+1] & \cos ^{2} C+\cos C \end{array} \mid=0\\ &\text { Take, cosB-cosA common from column } 2\\ &\Rightarrow \cos B-\cos A\left|\begin{array}{ccc} 1 & 0 & 1 \\ 1+\cos A & 1 & 1+\cos C \\ \cos ^{2} A+\cos A & (\cos B+\cos A)+1 & \cos ^{2} C+\cos C \end{array}\right|=0\\ &\text { Apply } \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}\\ &\cos B-\cos A\left|\begin{array}{ccc} 1 & 0 & 1-1 \\ 1+\cos A & 1 & 1+\cos C-1-\cos A \\ \cos ^{2} A+\cos A & \cos B+\cos A+1 & \cos ^{2} C+\cos C-\cos ^{2} A-\cos A \end{array}\right|=0 \end{aligned}
\\\begin{aligned} &\Rightarrow \cos B-\cos A\left|\begin{array}{ccc} 1 & 0 & 0 \\ 1+\cos A & 1 & \cos C-\cos A \\ \cos ^{2} A+\cos A & \cos B+\cos A+1 & \cos ^{2} C-\cos ^{2} A+\cos C-\cos A \end{array}\right|=0\\ &\Rightarrow(\cos B-\cos A)\left|\begin{array}{ccc} 1 & 0 & 0 \\ 1+\cos A & 1 & (\cos C-\cos A) \\ \cos ^{2} A+\cos A & \cos B+\cos A+1 & (\cos C-\cos A)(\cos C+\cos A+1) \end{array}\right|=0\\ &\text { Take cosC-cosA common from Column } 3\\ &\Rightarrow(\cos B-\cos A)(\cos C-\cos A)\left|\begin{array}{ccc} 1 & 0 & 0 \\ 1+\cos A & 1 & 1 \\ \cos ^{2} A+\cos A & \cos B+\cos A+1 & \cos C+\cos A+1 \end{array}\right|=0 \end{aligned}
Expand along Row 1
Hence, ΔABC is an isosceles triangle.
Question:17
Find $A^{-1}$ if $A=\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right]$ and show that $A^{-1}=\frac{A^{2}-3 I}{2}$
\\\begin{aligned} &=0-1\{0-1\}+1\{1-0\}\\ &=1+1\\ &=2\\ &\text { To find adj } \mathrm{A} \end{aligned}
$\\\begin{array}{l} a_{11}=\left|\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right|=0-1=-1 \\ a_{12}=-\left|\begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array}\right|=-(0-1)=1 \\ a_{13}=\left|\begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array}\right|=1-0=1 \\ a_{21}=-\left|\begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array}\right|=-(0-1)=1 \\ a_{22}=\left|\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right|=0-1=-1 \end{array}$
$\\\begin{array}{l} \mathrm{a}_{23}=-\left|\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\right|=-(0-1)=1 \\ \mathrm{a}_{31}=\left|\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right|=1-0=1 \\ \mathrm{a}_{32}=-\left|\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\right|=-(0-1)=1 \\ \therefore \mathrm{adj} \mathrm{A}=\left[\begin{array}{ccc} \mathrm{a}_{11} & \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{21} & \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{31} & \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right] \end{array}$
\\\begin{aligned} &\therefore \mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}=\frac{\left[\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]}{2}=\frac{1}{2}\left[\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]\\ &A^{-1}=\frac{A^{2}-31}{2}\\ &\text { Now, we need to prove that }\\ &A^{2}=\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right] \times\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right]=\left[\begin{array}{lll} 0+1+1 & 0+0+1 & 0+1+0 \\ 0+0+1 & 1+0+1 & 1+0+0 \\ 0+1+0 & 1+0+0 & 1+1+0 \end{array}\right]\\ &A^{2}=\left[\begin{array}{lll} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array}\right]\\ &\therefore \frac{A^{2}-3 I}{2}=\frac{1}{2}\left\{\left[\begin{array}{lll} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array}\right]-3\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\right\} \end{aligned}
$\\\begin{array}{l} =\frac{1}{2}\left\{\left[\begin{array}{ccc} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array}\right]-\left[\begin{array}{ccc} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right]\right\} \\ =\frac{1}{2}\left\{\left[\begin{array}{ccc} 2-3 & 1 & 1 \\ 1 & 2-3 & 1 \\ 1 & 1 & 2-3 \end{array}\right]\right\} \\ \left.=\frac{1}{2}\left[\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]\right\} \\ =A^{-1} \end{array}$
Hence Proved
Question:18
If $A=\left[\begin{array}{ccc} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{array}\right]$ find $A^{-1}$. Using $A^{-1}$, solve the system of linear equations $x-2y = 10, 2x -y -z = 8, -2y + z = 7$
Find IAI Expand IAl along Column 1
\\\begin{aligned} &\begin{array}{l} |\mathrm{A}|=\mathrm{a}_{11}(-1)^{1+1}\left|\begin{array}{cc} \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right|+\mathrm{a}_{21}(-1)^{2+1}\left|\begin{array}{cc} \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right| +\mathrm{a}_{31}(-1)^{3+1}\left|\begin{array}{cc} \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{22} & \mathrm{a}_{23} \end{array}\right| \\ |\mathrm{A}|=(1)\left|\begin{array}{cc} -1 & -2 \\ -1 & 1 \end{array}\right|-(-2)\left|\begin{array}{cc} 0 & 0 \\ -1 & 1 \end{array}\right|+0\left|\begin{array}{cc} 2 & 0 \\ -1 & -2 \end{array}\right| \\ =(-1+2)+2(0)+0 \\ =1 \end{array}\\ &\text { To find adj } A \end{aligned}
$\\\begin{array}{l} a_{11}=\left|\begin{array}{cc} -1 & -2 \\ -1 & 1 \end{array}\right|=-1-2=-3 \\ a_{12}=-\left|\begin{array}{cc} -2 & -2 \\ 0 & 1 \end{array}\right|=-(-2+0)=2 \\ a_{13}=\left|\begin{array}{cc} -2 & -1 \\ 0 & -1 \end{array}\right|=2+0=2 \\ a_{21}=-\left|\begin{array}{cc} 2 & 0 \\ -1 & 1 \end{array}\right|=-(2+0)=-2 \\ a_{22}=\left|\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right|=1 \\ a_{23}=-\left|\begin{array}{cc} 1 & 2 \\ 0 & -1 \end{array}\right|=-(-1)=1 \end{array}$
$\\\begin{array}{l} a_{31}=\left|\begin{array}{cc} 2 & 0 \\ -1 & -2 \end{array}\right|=-4 \\ a_{32}=-\left|\begin{array}{cc} 1 & 0 \\ -2 & -2 \end{array}\right|=-(-2)=2 \\ a_{33}=\left|\begin{array}{ccc} 1 & 2 \\ -2 & -1 \end{array}\right|=-1+4=3 \\ \therefore \text { adj } A=\left[\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right]^{T}=\left[\begin{array}{ccc} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{array}\right] \\ \therefore A^{-1}=\frac{\operatorname{adj} A}{|A|}=\frac{\left[\begin{array}{ccc} -3 & -2 & -4 \\ 2 & 1 & 3 \end{array}\right]}{1}=\left[\begin{array}{ccc} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{array}\right] \end{array}$
According to linear equation:
$\\x -2y = 10\\ \\ \vspace{\baselineskip}2x -y -z = 8\\ \\ \vspace{\baselineskip}-2y + z = 7\\ \\$
We know that, AX = B
$A=\left[\begin{array}{ccc} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{array}\right]$
Here,
So, transpose of $\\ A^{-1}$
$\\\begin{array}{l} \Rightarrow X=A^{-1} B \\ \Rightarrow\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{lll} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{array}\right]\left[\begin{array}{l} 10 \\ 8 \\ 7 \end{array}\right] \\ \Rightarrow\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} -30+16+14 \\ -20+8+7 \\ -40+16+21 \end{array}\right] \\ \therefore x=0, y=-5 \text { and } z=-3 \end{array}$
Question:19
Using matrix method, solve the system of equations 3x + 2y – 2z = 3, x + 2y + 3z = 6, 2x – y + z = 2.
$\\Given,\\ \vspace{\baselineskip} 3x + 2y -2z = 3,\\ \\ \vspace{\baselineskip}x + 2y + 3z = 6,\\ \\ \vspace{\baselineskip}2x -y + z = 2\\ \vspace{\baselineskip} We know,\\ \vspace{\baselineskip} AX = B\\$
$\begin{array}{c} |\mathrm{A}|=\mathrm{a}_{11}(-1)^{1+1}\left|\begin{array}{cc} \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right|+\mathrm{a}_{21}(-1)^{2+1}\left|\begin{array}{cc} \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right| +\mathrm{a}_{31}(-1)^{3+1}\left|\begin{array}{cc} \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{22} & \mathrm{a}_{23} \end{array}\right| \\ |\mathrm{A}|=(3)\left|\begin{array}{cc} 2 & 3 \\ -1 & 1 \end{array}\right|-(1)\left|\begin{array}{cc} 2 & -2 \\ -1 & 1 \end{array}\right|+2\left|\begin{array}{cc} 2 & -2 \\ 2 & 3 \end{array}\right| \end{array}$
\\\begin{aligned} &=3(2+3)-1(2-2)+2(6+4)\\ &=3(5)+2(10)\\ &=15+20\\ &=35\\ &\text { To find adj } A\\ &a_{11}=\left|\begin{array}{cc} 2 & 3 \\ -1 & 1 \end{array}\right|=2+3=5\\ &a_{12}=-\left|\begin{array}{ll} 1 & 3 \\ 2 & 1 \end{array}\right|=-(1-6)=5\\ &a_{13}=\left|\begin{array}{cc} 1 & 2 \\ 2 & -1 \end{array}\right|=-1-4=-5 \end{aligned}
$\\\begin{array}{l} a_{21}=-\left|\begin{array}{cc} 2 & -2 \\ -1 & 1 \end{array}\right|=-(2-2)=0 \\ a_{22}=\left|\begin{array}{cc} 3 & -2 \\ 2 & 1 \end{array}\right|=3+4=7 \\ a_{23}=-\left|\begin{array}{cc} 3 & 2 \\ 2 & -1 \end{array}\right|=-(-3-4)=-7 \\ a_{32}=-\left|\begin{array}{cc} 3 & -2 \\ 1 & 3 \end{array}\right|=-(9+2)=-11 \\ a_{33}=\left|\begin{array}{cc} 3 & 2 \\ 1 & 2 \end{array}\right|=6-2=4 \end{array}$
$\\\begin{array}{l} \therefore \text { adj } A=\left[\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right]^{T}=\left[\begin{array}{ccc} 5 & 5 & -5 \\ 0 & 7 & -7 \\ 10 & -11 & 4 \end{array}\right]^{T}=\left[\begin{array}{ccc} 5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4 \end{array}\right] \\ \therefore A^{-1}=\frac{\operatorname{adj} A}{|A|}=\frac{\left[\begin{array}{ccc} 5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4 \end{array}\right]}{35}=\frac{1}{35}\left[\begin{array}{ccc} 5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4 \end{array}\right] \\ \text { Now, } x=A^{-1} B \end{array}$
$\\\begin{array}{l} {\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\frac{1}{35}\left[\begin{array}{ccc} 5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4 \end{array}\right]\left[\begin{array}{l} 3 \\ 6 \\ 2 \end{array}\right]} \\\\ \Rightarrow\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\frac{1}{35}\left[\begin{array}{c} 15+20 \\ 15+42-22 \\ -15+42+8 \end{array}\right] \\ \\\Rightarrow\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\frac{1}{35}\left[\begin{array}{l} 35 \\ 35 \\ 35 \end{array}\right] \\ \\\Rightarrow\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right] \end{array}$
∴ x = 1, y = 1 and z = 1
Question:20
Given $A=\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right], B=\left[\begin{array}{ccc} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right]$ find BA and use this to solve the system of equations $y + 2z = 7, x - y = 3, 2x + 3y + 4z = 17.$
$\\\begin{array}{l} A=\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right] \text { and } B=\left[\begin{array}{ccc} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right] \\ \therefore B A=\left[\begin{array}{ccc} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right]\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right] \\ =\left[\begin{array}{ccc} 2+4 & 2-2 & -4+4 \\ 4-12+8 & 4+6-4 & -8-12+20 \\ -4+4 & 2-2 & -4+10 \end{array}\right] \end{array}$
$\begin{array}{l}\\\ =\left[\begin{array}{lll} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}\right] \\\\ =6\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \end{array}$
$BA = 6I \ldots (i)\\$
Now, given system of equations is:
$\\ \vspace{\baselineskip}y + 2z = 7,\\ \\ \vspace{\baselineskip}x -y = 3,\\ \\ \vspace{\baselineskip}2x + 3y + 4z = 17\\ \\ \vspace{\baselineskip}So,\\$
\\ \begin{aligned} &\left[\begin{array}{ccc} 0 & 1 & 2 \\ 1 & -1 & 0 \\ 2 & 3 & 4 \end{array}\right]\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\left[\begin{array}{c} 7 \\ 3 \\ 17 \end{array}\right]\\ &\text { Apply, } \mathrm{R}_{1} \leftrightarrow \mathrm{R}_{2}\\ &\mathrm{R}_{2} \leftrightarrow \mathrm{R}_{3}\\ &\left[\begin{array}{ccc} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right]\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\left[\begin{array}{c} 3 \\ 17 \\ 7 \end{array}\right]\\ &\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\left[\begin{array}{ccc} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right]^{-1}\left[\begin{array}{c} 3 \\ 17 \\ 7 \end{array}\right]_{\ldots(i i)}\\ &\text { So, } B A=6 I \text { [from eg(i)] } \end{aligned}
$\\\begin{array}{l} \left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right] =\frac{1}{6}\left[\begin{array}{l} 12 \\ -6 \\ 24 \end{array}\right] \\\\ \left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right] ={\left[\begin{array}{c} 2 \\ -1 \\ 4 \end{array}\right]} \\ \therefore \mathrm{x}=2, \mathrm{y}=-1 \text { and } \mathrm{z}=4 \end{array}$
Question:21
If a + b + c ≠ 0 and $\left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|=0$ then prove that a = b = c.
$\\ A=\left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|$
\\\begin{aligned} &\text { Apply } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\\ &=\left|\begin{array}{lll} a+b+c & b & c \\ b+a+c & c & a \\ c+a+b & a & b \end{array}\right|\\ &\text { Take }(a+b+c) \text { common from Column } 1\\ &=(a+b+c)\left|\begin{array}{ccc} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{array}\right| \end{aligned}
Expand along Column 1
$\\\vspace{\baselineskip} = (a + b + c)[(1)(bc -a\textsuperscript{2}) -(1)(b\textsuperscript{2} -ac) + (1)(ba -c\textsuperscript{2})]\\ \\ \vspace{\baselineskip}= (a + b + c)[bc -a\textsuperscript{2} -b\textsuperscript{2} + ac + ab -c\textsuperscript{2}]\\ \\ \vspace{\baselineskip}= (a + b + c)[-(a\textsuperscript{2} + b\textsuperscript{2} + c\textsuperscript{2} -ab -bc -ac)]\\$
$\\ =-\frac{1}{2}(a+b+c)\left(2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 b c-2 a c\right) \\ =-\frac{1}{2}(a+b+c)\left[\left(a^{2}+b^{2}-2 a b\right)+\left(b^{2}+c^{2}-2 b c\right)+\left(c^{2}+a^{2}-2 a c\right)\right] \\ =-\frac{1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] \\ {\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]}$
Given that Δ = 0
$\Rightarrow-\frac{1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]=0$
Hence Proved
Question:22
Prove that $\left|\begin{array}{ccc} \mathrm{bc}-\mathrm{a}^{2} & \mathrm{ca}-\mathrm{b}^{2} & \mathrm{ab}-\mathrm{c}^{2} \\ \mathrm{ca}-\mathrm{b}^{2} & \mathrm{ab}-\mathrm{c}^{2} & \mathrm{bc}-\mathrm{a}^{2} \\ \mathrm{ab}-\mathrm{c}^{2} & \mathrm{bc}-\mathrm{a}^{2} & \mathrm{ca}-\mathrm{b}^{2} \end{array}\right|$ is divisible by a + b + c and find the quotient.
$\left|\begin{array}{ccc} \mathrm{bc}-\mathrm{a}^{2} & \mathrm{ca}-\mathrm{b}^{2} & \mathrm{ab}-\mathrm{c}^{2} \\ \mathrm{ca}-\mathrm{b}^{2} & \mathrm{ab}-\mathrm{c}^{2} & \mathrm{bc}-\mathrm{a}^{2} \\ \mathrm{ab}-\mathrm{c}^{2} & \mathrm{bc}-\mathrm{a}^{2} & \mathrm{ca}-\mathrm{b}^{2} \end{array}\right|$ is given.
Apply, \\\begin{aligned} &=\left|\begin{array}{ccc} \mathrm{b} c-\mathrm{a}^{2}-\mathrm{ca}+\mathrm{b}^{2} & \mathrm{ca}-\mathrm{b}^{2}-\mathrm{ab}+\mathrm{c}^{2} & \mathrm{ab}-\mathrm{c}^{2}-\mathrm{bc}+\mathrm{a}^{2} \\ \mathrm{ca}-\mathrm{b}^{2} & \mathrm{ab}-\mathrm{c}^{2} & \mathrm{bc}-\mathrm{a}^{2} \\ \mathrm{ab}-\mathrm{c}^{2} & \mathrm{bc}-\mathrm{a}^{2} & \mathrm{ca}-\mathrm{b}^{2} \end{array}\right|\\ &=\left|\begin{array}{ccc} (b c-c a)+\left(b^{2}-a^{2}\right) & (c a-a b)+\left(c^{2}-b^{2}\right) & (a b-b c)+\left(a^{2}-c^{2}\right) \\ c a-b^{2} & a b-c^{2} & b c-a^{2} \\ a b-c^{2} & b c-a^{2} & c a-b^{2} \end{array}\right|\\ &=\left|\begin{array}{ccc} c(b-a)+(b-a)(b+a) & a(c-b)+(c-b)(c+b) & b(a-c)+(a-c)(a+c) \\ c a-b^{2} & a b-c^{2} & b c-a^{2} \\ a b-c^{2} & b c-a^{2} & c a-b^{2} \end{array}\right|\\ &=\left|\begin{array}{ccc} (b-a)(c+b+a) & (c-b)(a+c+b) & (a-c)(b+a+c) \\ c a-b^{2} & a b-c^{2} & b c-a^{2} \\ a b-c^{2} & b c-a^{2} & c a-b^{2} \end{array}\right| \end{aligned}
Take (a+b+c) common from Column 1
\\\begin{aligned} &=(a+b+c)\left|\begin{array}{ccc} (b-a) & (c-b) & (a-c) \\ c a-b^{2} & a b-c^{2} & b c-a^{2} \\ a b-c^{2} & b c-a^{2} & c a-b^{2} \end{array}\right|\\ &\text { Apply } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\\ &=(a+b+c)\left|\begin{array}{ccc} (b-a) & (c-b) & (a-c) \\ c a-b^{2}-a b+c^{2} & a b-c^{2}-b c+a^{2} & b c-a^{2}-c a+b^{2} \\ a b-c^{2} & b c-a^{2} & c a-b^{2} \end{array}\right|\\ &=(a+b+c)\left|\begin{array}{ccc} (b-a) & (c-b) & (a-c) \\ (c-b)(a+b+c) & (a-c)(a+b+c) & (b-a)(a+b+c) \\ a b-c^{2} & b c-a^{2} & c a-b^{2} \end{array}\right| \end{aligned}
Take (a+b+c) common from Column 2
\\\begin{aligned} &=(a+b+c)^{2}\left|\begin{array}{ccc} b-a+c-b+a-c & (c-b) & (a-c) \\ c-b+a-c+b-a & (a-c) & (b-a) \\ a b-c^{2}+b c-a^{2}+c a-b^{2} & b c-a^{2} & c a-b^{2} \end{array}\right|\\ &=(a+b+c)^{2}\left|\begin{array}{ccc} 0 & (c-b) & (a-c) \\ 0 & (a-c) & (b-a) \\ a b+b c+c a-\left(a^{2}+b^{2}+c^{2}\right) & b c-a^{2} & c a-b^{2} \end{array}\right| \end{aligned}
Expand along Column 1
$\\\vspace{\baselineskip}=(a + b + c)\textsuperscript{2}[ab + bc + ca -(a\textsuperscript{2} + b\textsuperscript{2} + c\textsuperscript{2})]\textsuperscript{2}\\$
$\\\vspace{\baselineskip}=(a + b + c)(a + b + c)[ab + bc + ca -(a\textsuperscript{2} + b\textsuperscript{2} + c\textsuperscript{2})]^2\\$
The determinant is divisible by (a+b+c) and the quotient is $\\(a + b + c)[ab + bc + ca -(a\textsuperscript{2} + b\textsuperscript{2} + c\textsuperscript{2})]^2\\$
Question:23
If x + y + z = 0, prove that $\left|\begin{array}{lll} x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a \end{array}\right|=x y z\left|\begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array}\right|$
Given LHS,
$\left|\begin{array}{lll} x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a \end{array}\right|$
Expand along Row 1
$\\\\ = xa \{ (za)(ya) -(xc)(xb) \} -(yb) \{ (yc)(ya) -(zb)(xb) \} + (zc) \{ (yc)(xc) -(zb)(za) \} \\ \\ \vspace{\baselineskip}= xa \{ a\textsuperscript{2}yz -x\textsuperscript{2}bc \} -yb \{ y\textsuperscript{2}ac -b\textsuperscript{2}xz \} + zc \{ c\textsuperscript{2}xy -z\textsuperscript{2}ab \} \\ \\ \vspace{\baselineskip}= a\textsuperscript{3}xyz -x\textsuperscript{3}abc -y\textsuperscript{3}abc + b\textsuperscript{3}xyz + c\textsuperscript{3}xyz -z\textsuperscript{3}abc\\ \\ \vspace{\baselineskip}= xyz(a\textsuperscript{3} + b\textsuperscript{3} + c\textsuperscript{3}) -abc(x\textsuperscript{3} + y\textsuperscript{3} + z\textsuperscript{3})\\$
Given x+y+z = 0
$=x y z\left|\begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array}\right|$
Question:24
Given:
$\left|\begin{array}{cc} 2 \mathrm{x} & 5 \\ 8 & \mathrm{x} \end{array}\right|=\left|\begin{array}{cc} 6 & -2 \\ 7 & 3 \end{array}\right|$
\\\begin{aligned} &A=\left|\begin{array}{ll} a & b \\ c & d \end{array}\right|\\ &\text { Then the determinant of } \mathrm{A} \text { is }\\ &|\mathrm{A}|=\left|\begin{array}{ll} \mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d} \end{array}\right|=\mathrm{ad}-\mathrm{b} \mathrm{c} \end{aligned}
Question:25
The value of determinant $\left|\begin{array}{lll} a-b & b+c & a \\ b-c & c+a & b \\ c-a & a+b & c \end{array}\right|$
A. $a^3 + b^3 + c^3$
B. 3 bc
C. $a^3 + b^3 + c^3-3abc$
D. none of these
C)
Given:
$\left|\begin{array}{lll} a-b & b+c & a \\ b-c & c+a & b \\ c-a & a+b & c \end{array}\right|$
Apply C2→ C2 + C3
\\\begin{aligned} &=\left|\begin{array}{lll} a-b & a+b+c & a \\ b-c & c+a+b & b \\ c-a & a+b+c & c \end{array}\right|\\ &\text { Take }(a+b+c) \text { common from Column } 2\\ &=(a+b+c)\left|\begin{array}{lll} a-b & 1 & a \\ b-c & 1 & b \\ c-a & 1 & c \end{array}\right|\\ &\text { Apply } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{3}\\ &=(a+b+c)\left|\begin{array}{lll} a-b-a & 1 & a \\ b-c-b & 1 & b \\ c-a-c & 1 & c \end{array}\right|\\ &=(a+b+c)\left|\begin{array}{ccc} -b & 1 & a \\ -c & 1 & b \\ -a & 1 & c \end{array}\right| \end{aligned}
Expand along Row 1
$\\ \vspace{\baselineskip} = (a + b + c)[(-b) \{ c -b \} -(1) \{ -c\textsuperscript{2} -(-ab) \} + a \{ -c -(-a) \} ]\\ \\ \vspace{\baselineskip}= (a + b + c)(-bc + b\textsuperscript{2} + c\textsuperscript{2} -ab -ac + a\textsuperscript{2})\\ \\ \vspace{\baselineskip}= a(-bc + b\textsuperscript{2} + c\textsuperscript{2} -ab -ac + a\textsuperscript{2}) + b(-bc + b\textsuperscript{2} + c\textsuperscript{2} -ab -ac + a\textsuperscript{2}) + c(-bc + b\textsuperscript{2} + c\textsuperscript{2} -ab -ac + a\textsuperscript{2})\\$
$\\ \\ \vspace{\baselineskip}= -abc + ab\textsuperscript{2} + ac\textsuperscript{2} -a\textsuperscript{2}b -a\textsuperscript{2}c + a\textsuperscript{3} -b\textsuperscript{2}c + b\textsuperscript{3} + bc\textsuperscript{2} -ab\textsuperscript{2} -abc + a\textsuperscript{2}b -bc\textsuperscript{2} + b\textsuperscript{2}c + c\textsuperscript{3} -abc -ac\textsuperscript{2}+ a\textsuperscript{2}c\\ \\ \vspace{\baselineskip}= a\textsuperscript{3} + b\textsuperscript{3} + c\textsuperscript{3} -3abc\\$
Question:26
B)
\\ \begin{aligned} &\text { The area of a triangle with vertices }\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right) \text { is given by }\\ &\Delta=\frac{1}{2}\left|\begin{array}{lll} \mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3} & 1 \end{array}\right|\\ &\Delta=\frac{1}{2}\left|\begin{array}{ccc} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & \mathrm{k} & 1 \end{array}\right|=9\\ &\Delta=\left|\begin{array}{ccc} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & \mathrm{k} & 1 \end{array}\right|=18 \end{aligned}
Expand along Column 2
Question:27
The determinant $\left|\begin{array}{ccc} b^{2}-a b & b-c & b c-a c \\ a b-a^{2} & a-b & b^{2}-a b \\ b c-a c & c-a & a b-a^{2} \end{array}\right|$ equals
A. abc (b–c) (c – a) (a – b)
B. (b–c) (c – a) (a – b)
C. (a + b + c) (b – c) (c – a) (a – b)
D. None of these
D)
Given:
$\left|\begin{array}{ccc} b^{2}-a b & b-c & b c-a c \\ a b-a^{2} & a-b & b^{2}-a b \\ b c-a c & c-a & a b-a^{2} \end{array}\right|$
$\\=\left|\begin{array}{lll}b(b-a) & b-c & c(b-a) \\ a(b-a) & a-b & b(b-a) \\ c(b-a) & c-a & a(b-a)\end{array}\right| \\Take (b-a) common from both Column 1 and 3\\ =(\mathrm{b}-\mathrm{a})(\mathrm{b}-\mathrm{a})\left|\begin{array}{lll}\mathrm{b} & \mathrm{b}-\mathrm{c} & \mathrm{c} \\ \mathrm{a} & \mathrm{a}-\mathrm{b} & \mathrm{b} \\ \mathrm{c} & \mathrm{c}-\mathrm{a} & \mathrm{a}\end{array}\right| \\Apply \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{3}\\ =(\mathrm{b}-\mathrm{a})(\mathrm{b}-\mathrm{a})\left|\begin{array}{lll}\mathrm{b}-\mathrm{c} & \mathrm{b}-\mathrm{c} & \mathrm{c} \\ \mathrm{a}-\mathrm{b} & \mathrm{a}-\mathrm{b} & \mathrm{b} \\ \mathrm{c}-\mathrm{a} & \mathrm{c}-\mathrm{a} & \mathrm{a}\end{array}\right|$
Whenever any two columns or rows in any determinant are equal, its value becomes = 0
Here Column 1 and 2 are identical
Question:28
The number of distinct real roots of $\left|\begin{array}{ccc} \sin x & \cos & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{array}\right|=0$ in the interval $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$ is
A. 0
B. –1
C. 1
D. None of these
C)
Given
$\left|\begin{array}{ccc} \sin x & \cos & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{array}\right|=0$
\begin{aligned} &\text { Apply } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\\ &\Rightarrow\left|\begin{array}{lll} \sin x+\cos x+\cos x & \cos x & \cos x \\ \cos x+\sin x+\cos x & \sin x & \cos x \\ \cos x+\cos x+\sin x & \cos x & \sin x \end{array}\right|=0\\ &\Rightarrow\left|\begin{array}{ccc} 2 \cos x+\sin x & \cos x & \cos x \\ 2 \cos x+\sin x & \sin x & \cos x \\ 2 \cos x+\sin x & \cos x & \sin x \end{array}\right|=0\\ &\text { Take }(2 \cos x+\sin x) \text { common from Column } 1\\ &\Rightarrow 2 \cos x+\sin x\left|\begin{array}{ccc} 1 & \cos x & \cos x \\ 1 & \sin x & \cos x \\ 1 & \cos x & \sin x \end{array}\right|=0\\ &\text { Apply } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\\ &\Rightarrow 2 \cos x+\sin x\left|\begin{array}{ccc} 1 & \cos x & \cos x \\ 1-1 & \sin x-\cos x & \cos x-\cos x \\ 1 & \cos x & \sin x \end{array}\right|=0 \end{aligned}
$\\\Rightarrow 2 \cos x+\sin x\left|\begin{array}{ccc} 1 & \cos x & \cos x \\ 0 & \sin x-\cos x & 0 \\ 1 & \cos x & \sin x \end{array}\right|=0 \\ \text { Apply } R_{3} \rightarrow R_{3}-R_{1} \\ \Rightarrow 2 \cos x+\sin x\left|\begin{array}{ccc} 1 & \cos x & \cos x \\ 0 & \sin x-\cos x & 0 \\ 1-1 & \cos x-\cos x & \sin x-\cos x \end{array}\right|=0 \\$
$\Rightarrow 2 \cos x+\sin x\left|\begin{array}{ccc} 1 & \cos x & \cos x \\ 0 & \sin x-\cos x & 0 \\ 0 & 0 & \sin x-\cos x \end{array}\right|=0$
Expand along Column 1
Only one real and distinct root occurs.
Question:29
B. –1
C. 1
D. None of these
A)
Given:
$\begin{array}{|ccc|} -1 & \cos \mathrm{C} & \cos \mathrm{B} \\ \cos \mathrm{C} & -1 & \cos \mathrm{A} \\ \cos \mathrm{B} & \cos \mathrm{A} & -1 \end{array} \mid$
Expand along Column 1
$\begin{array}{r} |\mathrm{A}|=\mathrm{a}_{11}(-1)^{1+1}\left|\begin{array}{ll} \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right|+\mathrm{a}_{21}(-1)^{2+1}\left|\begin{array}{ll} \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right| +\mathrm{a}_{31}(-1)^{3+1}\left|\begin{array}{ll} \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{22} & \mathrm{a}_{23} \end{array}\right| \end{array}$
$\Delta=(-1)\left|\begin{array}{cc} -1 & \cos A \\ \cos A & -1 \end{array}\right|-\cos C\left|\begin{array}{cc} \cos C & \cos B \\ \cos A & -1 \end{array}\right|+\cos B\left|\begin{array}{cc} \mid \cos C & \cos B \\ -1 & \cos A \end{array}\right|$
$\\ \vspace{\baselineskip}= [(-1) \{ 1 -cos\textsuperscript{2}A \} -cos C \{ -cos C -cos Acos B \} + cos B \{ cos A cos C + cos B \} ]\\ \\ \vspace{\baselineskip}= -1 + cos\textsuperscript{2}A + cos\textsuperscript{2}C + cos A cos B cos C + cos A cos B cos C + cos\textsuperscript{2}B\\ \\ \vspace{\baselineskip}= -1 + cos\textsuperscript{2}A + cos\textsuperscript{2}B + cos\textsuperscript{2}C + 2cos A cos B cos C\\ \vspace{\baselineskip} \text{Using the formula}\\ \vspace{\baselineskip} 1 + cos2A = 2cos\textsuperscript{2}A\\$
\\ \begin{aligned} &=-1+\frac{1+\cos 2 \mathrm{~A}}{2}+\frac{1+\cos 2 \mathrm{~B}}{2}+\frac{1+\cos 2 \mathrm{C}}{2}+2 \cos \mathrm{A} \cos \mathrm{B} \cos \mathrm{C}\\ &\text { Taking L.C.M, we get }\\ &=\frac{-2+1+\cos 2 A+1+\cos 2 B+1+\cos 2 C+4 \cos A \cos B \cos C}{2}\\ &=\frac{1+(\cos 2 A+\cos 2 B)+\cos 2 C+4 \cos C \cos A \cos B}{2}\\ &\text { Now use: } \cos (A+B) \cos (A-B)=2 \cos A \cos B\\ &so, \cos 2 A+\cos 2 B=2 \cos (A+B) \cos (A-B) \end{aligned}
\\ \begin{aligned} &=\frac{1+\cos 2 C+\{2 \cos (A+B) \cos (A-B)\}+4 \cos A \cos B \cos C}{2}\\ &=\frac{1+2 \cos ^{2} C-1+\{2 \cos (A+B) \cos (A-B)\}+4 \cos A \cos B \cos C}{2}\\ &.=\frac{2 \cos ^{2} \mathrm{C}+[2 \cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})\}+4 \cos \mathrm{A} \cos \mathrm{B} \cos C}{2} \ldots (\mathrm{i})\\ &\text { We know that } A, B, C \text { are angles of triangle }\\ &\Rightarrow A+B+C=\pi\\ &\Rightarrow A+B=\pi-C \end{aligned}
$\\ =\frac{2 \cos ^{2} C+\{2 \cos (\pi-C) \cos (A-B)\}+4 \cos A \cos B \cos C}{2} \\ =\frac{2 \cos ^{2} C+\{-2 \cos C \cos (A-B)\}+4 \cos A \cos B \cos C}{2} [\because \cos (\pi-x)=-\cos x] \\ =\frac{-2 \cos C\{\cos (A-B)-\cos C\}+4 \cos A \cos B \cos C}{2}$
Question:30
A)
Given:
$f(t)=\left|\begin{array}{ccc} \cos t & t & 1 \\ 2 \sin t & t & 2 t \\ \sin t & t & t \end{array}\right|$
\\ \begin{aligned} &\text { Divide } \mathrm{R}_{2} \text { and } \mathrm{R}_{3} \text { by } t\\ &f(t)=t^{2}\left|\begin{array}{ccc} \cos t & t & 1 \\ \frac{2 \sin t}{t} & \frac{t}{t} & \frac{2 t}{t} \\ \frac{\operatorname{sint}}{t} & \frac{t}{t} & \frac{t}{t} \end{array}\right|\\ &f(t)=t^{2}\left|\begin{array}{ccc} \cos t & t & 1 \\ \frac{2 \sin t}{ t} & 1 & 2 \\ \frac{\sin t}{t} & 1 & 1 \end{array}\right| \end{aligned}
$\\ \Rightarrow \frac{\mathrm{f}(\mathrm{t})}{\mathrm{t}^{2}}=\frac{\mathrm{t}^{2}}{\mathrm{t}^{2}}\left|\begin{array}{ccc} \cos t & t & 1 \\ \frac{2 \sin t}{ t} & 1 & 2 \\ \frac{\sin t}{t} & 1 & 1 \end{array}\right|$
$\\\Rightarrow \lim _{t \rightarrow 0} \frac{\mathrm{f}(\mathrm{t})}{\mathrm{t}^{2}}=\left|\begin{array}{lll} \lim _{t \rightarrow 0} \cos t & \lim _{t \rightarrow 0} t & \lim _{t \rightarrow 0} 1 \\ \lim _{t \rightarrow 0} \frac{2 \sin t}{t} & \lim _{t \rightarrow 0} 1 & \lim _{t \rightarrow 0} 2 \\ \lim _{t \rightarrow 0} \frac{\sin t}{t} & \lim _{t \rightarrow 0} 1 & \lim _{t \rightarrow 0} 1 \end{array}\right|$
$=\left|\begin{array}{lll} 1 & 0 & 1 \\ 2 & 1 & 2 \\ 1 & 1 & 1 \end{array}\right|\left(\because \lim _{t \rightarrow 0} \frac{\sin t}{t}=1\right)$
Expand along Row 1
$\\ = (1)(1 -2) + (1)(2 -1)\\ \\ \vspace{\baselineskip}= - 1 + 1\\ \vspace{\baselineskip}= 0\\$
Question:31
A)
\\ \begin{aligned} &\text { We have: }\\ &\Delta=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1+\cos \theta & 1 & 1 \end{array}\right|\\ &\text { Apply, } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3} \text { and } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3}\\ &\Rightarrow \Delta=\left|\begin{array}{ccc} 0 & 0 & 1 \\ 0 & \sin \theta & 1 \\ \cos \theta & 0 & 1 \end{array}\right| \end{aligned}
The maximum value of :
Question:32
If $f(x)=\left|\begin{array}{ccc} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{array}\right|$
A. f (a) = 0
B. f (b) = 0
C. f (0) = 0
D. f (1) = 0
C)
We have:
$f(x)=\left|\begin{array}{ccc} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{array}\right|$
\\ \begin{aligned} &\text { If we put } x=a\\ &f(a)=\left|\begin{array}{ccc} 0 & a-a & a-b \\ a+a & 0 & a-c \\ a+b & a+c & 0 \end{array}\right|\\ &=0\left|\begin{array}{cc} 0 & a-c \\ a+c & 0 \end{array}\right|-0\left|\begin{array}{cc} 2 a & a-c \\ a+b & 0 \end{array}\right|+(a-b)\left|\begin{array}{cc} 2 a & 0 \\ a+b & a+c \end{array}\right| \end{aligned}
$\\ \begin{array}{l} f(b)=\left|\begin{array}{ccc} 0 & b-a & b-b \\ b+a & 0 & b-c \\ b+b & b+c & 0 \end{array}\right| \\ =0\left|\begin{array}{cc} 0 & b-c \\ b+c & 0 \end{array}\right|-(b-a)\left|\begin{array}{cc} b+a & b-c \\ 2 b & 0 \end{array}\right|+0\left|\begin{array}{cc} b+a & 0 \\ 2 b & b+c \end{array}\right| \end{array}$
If x = 0 according to the given question:
$\\ \begin{array}{l} f(0)=\left|\begin{array}{ccc} 0 & 0-a & 0-b \\ 0+a & 0 & 0-c \\ 0+b & 0+c & 0 \end{array}\right| \\ =0\left|\begin{array}{cc} 0 & -c \\ c & 0 \end{array}\right|-(-a)\left|\begin{array}{cc} a & -c \\ b & 0 \end{array}\right|+(-b)\left|\begin{array}{ll} a & 0 \\ b & c \end{array}\right| \end{array}$
$\\ \vspace{\baselineskip}= 0 + a [a (0) -(-bc)] -b [ac -b (0)]\\ \\ \vspace{\baselineskip}= a [bc] -b [ac]\\ \\ \vspace{\baselineskip}= abc -abc = 0\\ \vspace{\baselineskip}$
Then the condition is satisfied.
Question:33
If $A=\left[\begin{array}{ccc} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{array}\right]$
then $A^{-1}$ exists if
A. λ = 2
B. λ ≠ 2
C. λ ≠ -2
D. None of these
D)
We have
$A=\left[\begin{array}{ccc} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{array}\right]$
$\therefore \lambda \neq-\frac{8}{5}$
So, $A^{-1}$ exists if and only if $\lambda \neq-\frac{8}{5}$
Question:34
If A and B are invertible matrices, then which of the following is not correct?
A. $adj A = |A|. A^{-1}$
B. $det (A)^{-1} = [det (A)]^{-1}$
C. $(AB)^{-1} = B^{-1} A^{-1}$
D. $(A + B)^{-1} = B^{-1} + A^{-1}$
D)
We know, A and B are invertible matrices
\\ \begin{aligned} &\Rightarrow|\mathrm{A}|^{-1}=\frac{1}{|\mathrm{~A}|}\\ &\therefore \operatorname{det}(A)^{-1}=[\operatorname{det}(A)]^{-1} \ldots(B)\\ &\text { We know that } \frac{|\mathrm{A}|^{-1}}{ }=\frac{\operatorname{adj} \mathrm{A}}{|\vec{A}|}\\ &\Rightarrow \text { adj } A=|A| \cdot A^{-1} \ldots \text { option }(A)\\ &\Rightarrow(\mathrm{A}+\mathrm{B})^{-1}=\frac{1}{|\mathrm{~A}+\mathrm{B}|} \operatorname{adj}(\mathrm{A}+\mathrm{B})\\ &{\text { But }}{\mathrm{B}}^{-1}+\mathrm{A}^{-1}=\frac{1}{|\mathrm{~B}|} \operatorname{adj} \mathrm{B}+\frac{1}{|\mathrm{~A}|} \text { adj } \mathrm{A} \end{aligned}
Question:35
If x, y, z are all different from zero and ,$\left|\begin{array}{ccc} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{array}\right|=0$, then value of $x^{-1} + y^{-1} + z^{-1}$ is
$\\A. x y z\\B. x^{-1} y^{-1} z^{-1}\\C. -x -y -z\\D. -1$
We have
$\left|\begin{array}{ccc} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{array}\right|=0$
$\text { Apply } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{3} \text { and } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3}$
$\left|\begin{array}{ccc} x & 0 & 1 \\ 0 & y & 1 \\ -z & -z & 1+z \end{array}\right|=0$
Expand along Row 1
Divide both side by xyz
$\\ \Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1=0 \\ \therefore \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=x^{-1}+y^{-1}+z^{-1}=-1$
Question:36
The value of the determinant $\left|\begin{array}{ccc} x & x+y & x+2 y \\ x+2 y & x & x+y \\ x+y & x+2 y & x \end{array}\right|$ is
A. $9x^2 (x + y)$
B. $9y^2 (x + y)$
C. $3y^2 (x + y)$
D. $7x^2 (x + y)$
B)
\\\begin{aligned} &\text { Matrix given: }\\ &\left|\begin{array}{ccc} x & x+y & x+2 y \\ x+2 y & x & x+y \\ x+y & x+2 y & x \end{array}\right|\\ &=\mathrm{x}\left|\begin{array}{cc} \mathrm{x} & \mathrm{x}+\mathrm{y} \\ \mathrm{x}+2 \mathrm{y} & \mathrm{x} \end{array}\right|-(\mathrm{x}+\mathrm{y})\left|\begin{array}{cc} \mathrm{x}+2 \mathrm{y} & \mathrm{x}+\mathrm{y} \\ \mathrm{x}+\mathrm{y} & \mathrm{x} \end{array}\right|+(\mathrm{x}+2 \mathrm{y})\left|\begin{array}{cc} \mathrm{x}+2 \mathrm{y} & \mathrm{x} \\ \mathrm{x}+\mathrm{y} & \mathrm{x}+2 \mathrm{y} \end{array}\right| \end{aligned}
$\\\vspace{\baselineskip}= x [x\textsuperscript{2} -(x + y) (x + 2y)] -(x + y) [(x + 2y) (x) -(x + y)\textsuperscript{2}] + (x + 2y) [(x + 2y)\textsuperscript{2} -x (x + y)]\\ \\ \vspace{\baselineskip}= x [x\textsuperscript{2} -x\textsuperscript{2} -3xy -2y\textsuperscript{2}] -(x + y) [x\textsuperscript{2} + 2xy -x\textsuperscript{2} -2xy -y\textsuperscript{2}] + (x + 2y) [x\textsuperscript{2} + 4xy + 4y\textsuperscript{2} -x\textsuperscript{2} -xy]\\ \\ \vspace{\baselineskip}= x [-3xy -2y\textsuperscript{2}] -(x + y) [-y\textsuperscript{2}] + (x + 2y) [3xy + 4y\textsuperscript{2}]\\ \\ \vspace{\baselineskip}= -3x\textsuperscript{2}y -2xy\textsuperscript{2} + xy\textsuperscript{2} + y\textsuperscript{3} +3x\textsuperscript{2}y + 4xy\textsuperscript{2} + 6xy\textsuperscript{2} + 8y\textsuperscript{3}\\ \\ \vspace{\baselineskip}= 9y\textsuperscript{3} + 9xy\textsuperscript{2}\\ \\ \vspace{\baselineskip}= 9y\textsuperscript{2} (x + y)\\$
Question:37
C)
\\ \begin{aligned} &\text { We have: }\\ &\Delta=\left|\begin{array}{ccc} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2 a \end{array}\right|=86\\ &1\left|\begin{array}{cc} \mathrm{a} & -1 \\ 4 & 2 \mathrm{a} \end{array}\right|-(-2)\left|\begin{array}{cc} 2 & -1 \\ 0 & 2 \mathrm{a} \end{array}\right|+5\left|\begin{array}{cc} 2 & \mathrm{a} \\ 0 & 4 \end{array}\right|=86 \end{aligned}
$\\\vspace{\baselineskip} 1 [2a\textsuperscript{2} -(-4)] + 2 [4a -0] + 5 [8 -0] = 86\\ \\ \vspace{\baselineskip} 1 [2a\textsuperscript{2} + 4] + 2 [4a] + 5 [8] = 86\\ \\ \vspace{\baselineskip} 2a\textsuperscript{2} + 4 + 8a + 40 = 86\\ \\ \vspace{\baselineskip} 2a\textsuperscript{2} + 8a + 44 = 86\\ \\ \vspace{\baselineskip}2a\textsuperscript{2} + 8a = 42\\ \\ \vspace{\baselineskip}2 (a\textsuperscript{2} + 4a) = 42\\$
Sum of -7 and 3 = -4
Question:38
Fill in the blanks
If A is a matrix of order 3 × 3, then |3A| = ___.
If A is a matrix of order we know:
Question:39
Fill in the blanks
If A is invertible matrix of order 3 × 3, then |$A^{-1}$|= ____.
If A is invertible matrix of order GivenA = invertible matrix 3x3
$\therefore\left|A^{-1}\right|=\frac{1}{|A|} \\$
Question:40
Fill in the blanks
If x, y, z ∈ R, then the value of determinant $\left|\begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|$ is equal to ___.
\\ \begin{aligned} &\text { Given }\\ &\left|\begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|\\ &\text { Apply } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2}\\ &=\left|\begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2}-\left(2^{x}-2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2}-\left(3^{x}-3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2}-\left(4^{x}-4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|\\ &\text { Apply Formula: }(a+b)^{2}-(a-b)^{2}=4 a b \text { . } \end{aligned}
\\ \begin{aligned} &=\left|\begin{array}{lll} 4\left(2^{x}\right)\left(2^{-x}\right) & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ 4\left(3^{x}\right)\left(3^{-x}\right) & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ 4\left(4^{x}\right)\left(4^{-x}\right) & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|\\ &=\left|\begin{array}{lll} 4 & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ 4 & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ 4 & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|\\ &\text { Column } 1 \text { and } 3 \text { thus become proportional }\\ &\therefore\left|\begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|=0 \end{aligned}
Question:41
Fill in the blanks
If cos 2θ = 0, then $\left|\begin{array}{ccc} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{array}\right|^{2}=$
We know
\\ \begin{aligned} &\therefore \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}} \text { and } \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\\ &\text { Then }\\ &\left|\begin{array}{ccc} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{array}\right|^{2}=\left|\begin{array}{ccc} 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \end{array}\right|^{2}\\ &\Rightarrow\left[0-\frac{1}{\sqrt{2}}\left(\frac{1}{2}\right)+\frac{1}{\sqrt{2}}\left(-\frac{1}{2}\right)\right]^{2}=\left[\frac{-2}{2 \sqrt{2}}\right]^{2}=\left[-\frac{1}{\sqrt{2}}\right]^{2}\\ &\therefore\left|\begin{array}{ccc} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{array}\right|^{2}=\frac{1}{2} \end{aligned}
Question:42
Fill in the blanks
If A is a matrix of order 3 × 3, then $(A^2)^{-1 }$= ____.
For matrix A is of order 3X3
\begin{aligned} \left(A^{2}\right)^{-1} &=(A \cdot A)^{-1} \\ &=A^{-1} \cdot A^{-1} \\ &=\left(A^{-1}\right)^{2} \end{aligned}
Question:43
If A is a matrix of order 3 × 3, then number of minors in determinant of A are
If matrix A is of order 3X3 then
Number of Minors of IAI = 9 as there are 9 elements in a 3x3 matrix
Question:44
Fill in the blanks
The sum of the products of elements of any row with the co-factors of corresponding elements is equal to ___.
If, $A=\left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right|$
We know that the determinant is equal to sum of corresponding co factors of any row or column
Question:45
Fill in the blanks
If x = -9 is a root of $\left|\begin{array}{lll} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{array}\right|=0$ , then other two roots are ___.
\\ \begin{aligned} &\text { We know that }\\ &x=-9 \text { is a root of }\left|\begin{array}{lll} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{array}\right|=0\\ &\Rightarrow \mathrm{x}\left|\begin{array}{ll} \mathrm{x} & 2 \\ 6 & \mathrm{x} \end{array}\right|-3\left|\begin{array}{ll} 2 & 2 \\ 7 & \mathrm{x} \end{array}\right|+7\left|\begin{array}{ll} 2 & \mathrm{x} \\ 7 & 6 \end{array}\right|=0 \end{aligned}
Question:46
Fill in the blanks
$\left|\begin{array}{ccc} 0 & x y z & x-z \\ y-x & 0 & y-z \\ z-x & z-y & 0 \end{array}\right|=$
$(z-x)\left[\begin{array}{cc} \left.1\left|\begin{array}{cc} 0 & \mathrm{y}-\mathrm{z} \\ \mathrm{z}-\mathrm{y} & 0 \end{array}\right|-(\mathrm{xyz})\left|\begin{array}{cc} 1 & \mathrm{y}-\mathrm{z} \\ 1 & 0 \end{array}\right|+(\mathrm{x}-\mathrm{z}) \mid \begin{array}{cc} 1 & 0 \\ 1 & \mathrm{z}-\mathrm{y} \end{array} \|\right] \end{array}\right.$
$\\ \vspace{\baselineskip}= (z -x) [1 [0 -(y -z) (z -y)] - (xyz) [0 - (y - z)] + (x -z) [(z -y) -0]]\\ \\ \vspace{\baselineskip}= (z -x) (z -y) (-y + z -xyz + x -z)\\ \\ \vspace{\baselineskip}= (z -x) (z -y) (x -y -xyz)\\ \\ \vspace{\baselineskip}= (z -x) (y -z) (y -x + xyz)\\ \vspace{\baselineskip}$
Question:49
State True or False for the statements
$(aA)^{-1} = (1/a) A^{-1}$, where a is any real number and A is a square matrix.
For a non singular matrix, aA is invertible such that
$\\ (\mathrm{aA})\left(\frac{1}{\mathrm{a}} \mathrm{A}^{-1}\right)=\left(\mathrm{a} \cdot \frac{1}{\mathrm{a}}\right)\left(\mathrm{AA}^{-1}\right) \\ \text {i.e. } \quad(\mathrm{aA})^{-1}=\frac{1}{\mathrm{a}} \mathrm{A}^{-1} \\$
here a = any non-zero scalar. Here A should be a non-singular matrix which is not given in the statement, thus the statement given in question in false.
Question:50
State True or False for the statements
$|A^{-1}| \neq |A|^{-1}$, where A is non-singular matrix.
We know A is a non singular Matrix
In that case: $AA\textsuperscript{-1} = I.\\ \\$
Thus the statement is false.
Question:53
State True or False for the statements
$\left|\begin{array}{lll} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{array}\right|=0$, where a, b, c are in A.P.
\\ \begin{aligned} &\text { since } a, b, c \text { are in } A P, 2 b=a+c .\\ &\therefore\left|\begin{array}{lll} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{array}\right|=0\\ &A p p l y_{i}, R_{1} \rightarrow R_{1}+R_{3}\\ &\Rightarrow\left|\begin{array}{ccc} 2 x+4 & 2 x+6 & 2 x+a+c \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{array}\right|=0 \end{aligned}
Since 2b = a + c,
$\Rightarrow\left|\begin{array}{ccc} 2(x+2) & 2(x+3) & 2(x+b) \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{array}\right|=0$
We can see that Row 1 and 3 are proportional
Thus determinant = 0
Question:54
State True or False for the statements
$|adj. A| = |A|^2$, where A is a square matrix of order two.
For any square matrix of order n, Here n =2, Thus the statement given in question is false
Question:55
$\left|\begin{array}{ccc} \sin A & \cos A & \sin A+\cos A \\ \sin B & \cos B & \sin B+\cos B \\ \sin C & \cos C & \sin C+\cos C \end{array}\right|$
$\\=\left|\begin{array}{ccc} \sin A & \cos A & \sin A \\ \sin B & \cos B & \sin B \\ \sin C & \cos C & \sin C \end{array}\right|+\left|\begin{array}{ccc} \sin A & \cos A & \cos A \\ \sin B & \cos B & \cos B \\ \sin C & \cos C & \cos C \end{array}\right|$
We can see that columns are identical in both the matrix on Right hand side
Thus Determinant = 0
Statement in question is therefore true
Question:56
State True or False for the statements
If the determinant $\left|\begin{array}{ccc} x+a & p+u & 1+f \\ y+b & q+v & m+g \\ z+c & r+w & n+h \end{array}\right|$ splits into exactly K determinants of order 3, each element of which contains only one term, then the value of K is 8
Split row 2
$\\ \begin{array}{ccc} \left|\begin{array}{ccc} \mathrm{x} & \mathrm{p} & 1 \\ \mathrm{y} & \mathrm{q} & \mathrm{m} \\ \mathrm{z}+\mathrm{c} & \mathrm{r}+\mathrm{w} & \mathrm{n}+\mathrm{h} \end{array}\right|+\left|\begin{array}{ccc} \mathrm{a} & \mathrm{u} & \mathrm{f} \\ \mathrm{y} & \mathrm{q} & \mathrm{m} \\ \mathrm{z}+\mathrm{c} & \mathrm{r}+\mathrm{w} & \mathrm{n}+\mathrm{h} \end{array}\right|+\left|\begin{array}{ccc} \mathrm{x} & \mathrm{p} & \mathrm{l} \\ \mathrm{b} & \mathrm{v} & \mathrm{g} \\ \mathrm{z}+\mathrm{c} & \mathrm{r}+\mathrm{w} & \mathrm{n}+\mathrm{h} \end{array}\right| \\ & +\left|\begin{array}{ccc} \mathrm{a} & \mathrm{u} & \mathrm{f} \\ \mathrm{b} & \mathrm{v} & \mathrm{g} \\ \mathrm{z}+\mathrm{c} & \mathrm{r}+\mathrm{w} & \mathrm{n}+\mathrm{h} \end{array}\right| \end{array}$
We can split all the rows in the same way. Thus the statement given in the question is true.
Question:57
State True or False for the statements
Let $\Delta=\left|\begin{array}{lll} a & p & x \\ b & q & y \\ c & r & z \end{array}\right|=16$ ,then $\Delta_{1}=\left|\begin{array}{lll} p+x & a+x & a+p \\ q+y & b+y & b+q \\ r+z & c+z & c+r \end{array}\right|=32$
$\left|\begin{array}{ccc} 2(p+x+a) & a+x & a+p \\ 2(q+y+b) & b+y & b+q \\ 2(r+z+c) & c+z & c+r \end{array}\right| =32$
2 can be taken common from Column 1
$2\left|\begin{array}{ccc} (p+x+a) & a+x & a+p \\ (q+y+b) & b+y & b+q \\ (r+z+c) & c+z & c+r \end{array}\right| =32$
After that apply C1 → C1 – C2 and C2→ C2 – C3
\\ \begin{aligned} &\left|\begin{array}{lll} p & x-p & a+p \\ q & y-q & b+q \\ r & z-r & c+r \end{array}\right|=16\\ &\left|\begin{array}{lll} p & x & a+p \\ q & y & b+q \\ r & z & c+r \end{array}\right|-\left|\begin{array}{lll} p & p & a+p \\ q & q & b+q \\ r & r & c+r \end{array}\right|=16\\ &\text { Second determinants of column } 2 \text { and } 3 \text { are identical }\\ &\left|\begin{array}{lll} p & x & a+p \\ q & y & b+q \\ r & z & c+r \end{array}\right|-0=16 \end{aligned}
$\\ \left|\begin{array}{lll} p & x & a \\ q & y & b \\ r & z & c \end{array}\right|+\left|\begin{array}{lll} p & x & p \\ q & y & q \\ r & z & r \end{array}\right|=16$
Again, second determinant of column 1 and 3 are identical
$\\ \left|\begin{array}{lll} p & x & a \\ q & y & b \\ r & z & c \end{array}\right|=16$
Hence the statement given in question is true
Question:58
State True or False for the statements
The maximum value of $\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1 & 1 & 1+\cos \theta \end{array}\right|$ is 1/2.
$\\ \begin{array}{l} \Delta=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1 & 1 & 1+\cos \theta \end{array}\right| \\ \text { Apply, } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1} \text { and } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1} \\ \Rightarrow \Delta=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 0 & \sin \theta & 0 \\ 0 & 0 & \cos \theta \end{array}\right| \end{array}$
Since the maximum value of
Thus the given statement is true.
## More About NCERT Exemplar Solutions for Class 12 Maths Chapter 4
Determinants are related to the matrices that are solved in chapter 3 of 12 Class NCERT Maths book. Studying determinants is not just about passing exams, but is about being prepared for higher education in any field of maths, science, economics, etc. In Class 12 Maths NCERT exemplar solutions chapter 4, the students will learn about determinants, their elements, and how to calculate determinants of various square matrices.
### NCERT Exemplar Class 12 Maths Solutions Chapter 4 Determinants Sub-topics covered
The sub-topics that are covered in this chapter of NCERT Class 12 solution are:
• Introduction
• Determinant
• Determinant of a matrix of order one
• Determinant of a matrix of order 2
• Determinant of a matrix of order 3x3
• Properties of determinants
• Area of triangle
• Minors and co-factors
• Adjoint and inverse of a matrix
• Applications of matrices and determinants
• Solution of a system of linear equations using the inverse of matrices
### What will Students Learn in NCERT Exemplar Class 12 Maths Solutions Chapter 4?
• The determinants are a value of scalar nature derived from the square matrix-like 2x2, 3x3, etc. This determinant then helps in understanding the linear equation defined by the matrix.
• Determinants derived help in determining whether the solution of the linear system is unique or not.
• There are many other related topics covered in this Class 12 Maths NCERT exemplar solutions chapter 4, like the area of triangle, co-factors, minors, inverse of a matrix, finding the equation of line, etc.
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## NCERT Exemplar Class 12 Maths Solutions
Chapter 1 Relations and Functions Chapter 2 Inverse Trigonometric Functions Chapter 3 Matrices Chapter 5 Continuity and Differentiability Chapter 6 Applications of Derivatives Chapter 7 Integrals Chapter 8 Applications of Integrals Chapter 9 Differential Equations Chapter 10 Vector Algebra Chapter 11 Three-dimensional Geometry Chapter 12 Linear Programming Chapter 13 Probability
## Benefits of NCERT Exemplar Class 12 Maths Solutions Chapter 4
• We will help in finding the NCERT exemplar Class 12 Maths chapter 4 solutions to the questions that are given in the NCERT book in detail. We don't skip steps or try to solve the questions in a trick or the easy way; instead, our expert guides will help in solving every question in the minutest detail kept in mind.
• The language used for solving the questions is simple and easy to understand. Not only the questions are solved exhaustively; we also keep a marking scheme in mind to help our students score better.
• NCERT Exemplar Class 12 Maths solutions chapter 4 will help in gaining confidence in solving more difficult questions. Detailed steps in easy language will help the students to understand the topics and its intricacies.
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### NCERT Exemplar Class 12 Solutions
NCERT Exemplar Class 12 Chemistry Solutions NCERT Exemplar Class 12 Physics Solutions NCERT Exemplar Class 12 Biology Solutions
### Also, check NCERT Solutions for questions given in the book:
Chapter 1 Relations and Functions Chapter 2 Inverse Trigonometric Functions Chapter 3 Matrices Chapter 4 Determinants Chapter 5 Continuity and Differentiability Chapter 6 Application of Derivatives Chapter 7 Integrals Chapter 8 Application of Integrals Chapter 9 Differential Equations Chapter 10 Vector Algebra Chapter 11 Three Dimensional Geometry Chapter 12 Linear Programming Chapter 13 Probability
#### Must Read NCERT Solution subject wise
##### JEE Main Important Mathematics Formulas
As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters
#### Also Check NCERT Books and NCERT Syllabus here
Yes, you download NCERT exemplar Class 12 Maths solutions chapter 4 pdf by using the webpage to pdf tool available online.
2. What are the important topics of this chapter?
The Properties of Determinants, Adjoint and Inverse of a Matrix and Application of Determinants and Matrices are the more important topics among others as per their weightage.
3. How to study well for boards?
Practice, Practice and Practice. Once you have read the chapters well and made notes, you must practice being fast and precise with answers.
4. How many questions are there in this chapter?
The NCERT exemplar solutions for Class 12 Maths chapter 4 has one exercise with 58 questions for practice.
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### Questions related to CBSE Class 12th
Have a question related to CBSE Class 12th ?
Hi,
The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.
The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.
Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.
Yuvan 01 September,2024
hello mahima,
If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.
hope this helps.
Hello Akash,
If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.
You can get the Previous Year Questions (PYQs) on the official website of the respective board.
I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.
Thank you and wishing you all the best for your bright future.
Hello student,
If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:
• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.
Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.
I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.
A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
Option 1) Option 2) Option 3) Option 4)
A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms−2 :
Option 1) 2.45×10−3 kg Option 2) 6.45×10−3 kg Option 3) 9.89×10−3 kg Option 4) 12.89×10−3 kg
An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
Option 1) Option 2) Option 3) Option 4)
A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
Option 1) Option 2) Option 3) Option 4)
In the reaction,
Option 1) at STP is produced for every mole consumed Option 2) is consumed for ever produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .
How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?
Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2
If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will
Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.
With increase of temperature, which of these changes?
Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.
Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is
Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023
A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is
Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9 |
# math
what is the factorial of a negative number?
The factorial function has singularities at the negative integers.
You can see this as follows. For integers we define:
(n+1)! = (n+1)n!
and we put 0! = 1
So, from 0! you can compute 1! and from that you can compute 2! etc. etc. So, the factorial function is now defined for all positive integers.
Using the same recursive equation you get:
n! = (n+1)!/(n+1)
And you see that substituting -1 for n is problematic :)
Now, you can define the factorial functon for all numbers, except the negative integers as follows.
n! = Integral from zero to infinity of
x^n Exp(-x)
This integral is well defined for all positive n. It satisfies the recursion equation. n doesn't have to be an integer. E.g.:
(1/2)! = 1/2 sqrt(pi)
You can now use the recursion equation to extend the definition of the factorial function to negative numbers. E.g.:
(-1/2)! = (1/2)!/(1/2) = sqrt(pi)
The factorial for a positive integer is as you understand it, ie, 4!= 1*2*3*4
0! is defined as 1
For positive numbers (not integers), we use the gamma function to extend the idea of factorials to all those numbers.
For negative numbers, whole or not, factorial, and gamma functions are undefined.
The factorial for a positive integer is as you understand it, ie, 4!= 1*2*3*4
0! is defined as 1
For positive numbers (not integers), we use the gamma function to extend the idea of factorials to all those numbers.
For negative numbers, whole or not, factorial, and gamma functions are undefined.
I need to add, using recursion, one can find factorials for negative NON Whole numbers, the result will be finite. For Negative whole numbers, ie, -4, it will yield a undefined term (divide by zero is not allowed).
I need to add, using recursion, one can find factorials for negative NON Whole numbers, the result will be finite. For Negative whole numbers, ie, -4, it will yield a undefined term (divide by zero is not allowed).
I disagree with Bob here. The gamma function is only undefined at zero and the negative numbers. It's an analytical function with singularities (poles) at the negative integers.
The integral Exp(-x)x^(n)dx from zero to infinity which defines the factorial function for positive n, is, of course, not defined for negative n. However, one can analytically continue this function to the entire complex plane.
All one needs to do is to repeatedly use the recursion relation:
(x-1)! = x!/x
If x! has been defined as an analytical function between x = 0 and x = 1, then the above recursion defines a function between -1 and 0.
Hi: Factorials are dedfined for non-integer negatives via the gamma function -- a topic beyond the scope of this post. Check out mathworld for a reasonably straight forward read on the topic. The gamma function accomodates for all complex factorials, i.e., z!, where z = a+bi for a,b in R ... excepting the negative integers whose factorials are at complex infinity.
Regards,
Rich B.
1. 👍
2. 👎
3. 👁
## Similar Questions
1. ### Math
The table represents the function f(x). A 2-column table with 7 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2, 3. The second column is labeled f of x with entries negative 9, negative
2. ### Math
Which describes the end behavior of the graph of the function f(x)=-2x^3-5x^2+x? a. f(x) approaches infinity as x approaches negative infinity and f(x) approaches infinity as x approaches infinity b. f(x) approaches negative
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2. ### MATH help
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1. ### Calculus
f is a continuous function with a domain [−3, 9] such that f of x equals 3 for x between negative 3 and 0 including negative 3, equals negative 1 times x plus 3 for x between 0 and 6 inclusive, and equals negative 3 for x
2. ### math HARD
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# How do you translate into mathematical expressions and find the number given The product of two and three less than a number is a negative four?
Jun 30, 2016
#### Answer:
$2 \times \left(x - 3\right) = - 4$
$x = 1$
#### Explanation:
One key word in understanding this sentence is the word "AND".
A product is the answer to a multiplication. When the word 'product' is used, it will always say the product of (one number) AND (another number).
The other key word that we look for is, "IS". This indicates what the answer will be and represents an equal sign in an equation.
Use $x$ as 'the number' referred to
So. translate this English sentence as follows:
The PRODUCT of (two) AND (three less than a number) IS (negative four.)
The PRODUCT of (2) AND (3 less than $x$) IS (-4)
$2 \times \left(x - 3\right) = - 4$
Now that we have an equation we can solve it to find $x$
$2 x - 6 = - 4$
$2 x = - 4 + 6 = 2$
$2 x = 2 \Rightarrow x = 1$
Check: $2 \times \left(1 - 3\right) = 2 \times - 2 = - 4$ |
Sequences, Probability, and Counting Theory
# Geometric Sequences
### Learning Objectives
In this section, you will:
• Find the common ratio for a geometric sequence.
• List the terms of a geometric sequence.
• Use a recursive formula for a geometric sequence.
• Use an explicit formula for a geometric sequence.
Many jobs offer an annual cost-of-living increase to keep salaries consistent with inflation. Suppose, for example, a recent college graduate finds a position as a sales manager earning an annual salary of $26,000. He is promised a 2% cost of living increase each year. His annual salary in any given year can be found by multiplying his salary from the previous year by 102%. His salary will be$26,520 after one year; $27,050.40 after two years;$27,591.41 after three years; and so on. When a salary increases by a constant rate each year, the salary grows by a constant factor. In this section, we will review sequences that grow in this way.
### Finding Common Ratios
The yearly salary values described form a geometric sequence because they change by a constant factor each year. Each term of a geometric sequence increases or decreases by a constant factor called the common ratio. The sequence below is an example of a geometric sequence because each term increases by a constant factor of 6. Multiplying any term of the sequence by the common ratio 6 generates the subsequent term.
### Definition of a Geometric Sequence
A geometric sequence is one in which any term divided by the previous term is a constant. This constant is called the common ratio of the sequence. The common ratio can be found by dividing any term in the sequence by the previous term. If is the initial term of a geometric sequence and is the common ratio, the sequence will be
### How To
Given a set of numbers, determine if they represent a geometric sequence.
1. Divide each term by the previous term.
2. Compare the quotients. If they are the same, a common ratio exists and the sequence is geometric.
### Finding Common Ratios
Is the sequence geometric? If so, find the common ratio.
[hidden-answer a=”fs-id1165137430766″]Divide each term by the previous term to determine whether a common ratio exists.
1. The sequence is geometric because there is a common ratio. The common ratio is 2.
2. The sequence is not geometric because there is not a common ratio.
#### Analysis
The graph of each sequence is shown in (Figure). It seems from the graphs that both (a) and (b) appear have the form of the graph of an exponential function in this viewing window. However, we know that (a) is geometric and so this interpretation holds, but (b) is not.
If you are told that a sequence is geometric, do you have to divide every term by the previous term to find the common ratio?
No. If you know that the sequence is geometric, you can choose any one term in the sequence and divide it by the previous term to find the common ratio.
### Try It
Is the sequence geometric? If so, find the common ratio.
The sequence is not geometric because .
### Try It
Is the sequence geometric? If so, find the common ratio.
The sequence is geometric. The common ratio is .
### Writing Terms of Geometric Sequences
Now that we can identify a geometric sequence, we will learn how to find the terms of a geometric sequence if we are given the first term and the common ratio. The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. For instance, if the first term of a geometric sequence is and the common ratio is we can find subsequent terms by multiplying to get then multiplying the result to get and so on.
The first four terms are
### How To
Given the first term and the common factor, find the first four terms of a geometric sequence.
1. Multiply the initial term, by the common ratio to find the next term,
2. Repeat the process, using to find and then to find until all four terms have been identified.
3. Write the terms separated by commons within brackets.
### Writing the Terms of a Geometric Sequence
List the first four terms of the geometric sequence with and
Multiply by to find Repeat the process, using to find
and so on.
The first four terms are
### Try It
List the first five terms of the geometric sequence with and
### Using Recursive Formulas for Geometric Sequences
A recursive formula allows us to find any term of a geometric sequence by using the previous term. Each term is the product of the common ratio and the previous term. For example, suppose the common ratio is 9. Then each term is nine times the previous term. As with any recursive formula, the initial term must be given.
### Recursive Formula for a Geometric Sequence
The recursive formula for a geometric sequence with common ratio and first term is
### How To
Given the first several terms of a geometric sequence, write its recursive formula.
1. State the initial term.
2. Find the common ratio by dividing any term by the preceding term.
3. Substitute the common ratio into the recursive formula for a geometric sequence.
### Using Recursive Formulas for Geometric Sequences
Write a recursive formula for the following geometric sequence.
The first term is given as 6. The common ratio can be found by dividing the second term by the first term.
Substitute the common ratio into the recursive formula for geometric sequences and define
#### Analysis
The sequence of data points follows an exponential pattern. The common ratio is also the base of an exponential function as shown in (Figure)
Do we have to divide the second term by the first term to find the common ratio?
No. We can divide any term in the sequence by the previous term. It is, however, most common to divide the second term by the first term because it is often the easiest method of finding the common ratio.
### Try It
Write a recursive formula for the following geometric sequence.
### Using Explicit Formulas for Geometric Sequences
Because a geometric sequence is an exponential function whose domain is the set of positive integers, and the common ratio is the base of the function, we can write explicit formulas that allow us to find particular terms.
Let’s take a look at the sequence This is a geometric sequence with a common ratio of 2 and an exponential function with a base of 2. An explicit formula for this sequence is
The graph of the sequence is shown in (Figure).
### Explicit Formula for a Geometric Sequence
The nth term of a geometric sequence is given by the explicit formula:
### Writing Terms of Geometric Sequences Using the Explicit Formula
Given a geometric sequence withandfind
[hidden-answer a=”fs-id1165137408599″]The sequence can be written in terms of the initial term and the common ratio
Find the common ratio using the given fourth term.
Find the second term by multiplying the first term by the common ratio.
#### Analysis
The common ratio is multiplied by the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by multiplying the first term by the common ratio nine times or by multiplying by the common ratio raised to the ninth power.
### Try It
Given a geometric sequence with and , find
### Writing an Explicit Formula for the nth Term of a Geometric Sequence
Write an explicit formula for the term of the following geometric sequence.
[hidden-answer a=”fs-id1165137911060″]The first term is 2. The common ratio can be found by dividing the second term by the first term.
The common ratio is 5. Substitute the common ratio and the first term of the sequence into the formula.
The graph of this sequence in (Figure) shows an exponential pattern.
### Try It
Write an explicit formula for the following geometric sequence.
### Solving Application Problems with Geometric Sequences
In real-world scenarios involving arithmetic sequences, we may need to use an initial term of instead of In these problems, we can alter the explicit formula slightly by using the following formula:
### Solving Application Problems with Geometric Sequences
In 2013, the number of students in a small school is 284. It is estimated that the student population will increase by 4% each year.
1. Write a formula for the student population.
2. Estimate the student population in 2020.
1. The situation can be modeled by a geometric sequence with an initial term of 284. The student population will be 104% of the prior year, so the common ratio is 1.04.
Let be the student population and be the number of years after 2013. Using the explicit formula for a geometric sequence we get
2. We can find the number of years since 2013 by subtracting.
We are looking for the population after 7 years. We can substitute 7 for to estimate the population in 2020.
### Try It
A business starts a new website. Initially the number of hits is 293 due to the curiosity factor. The business estimates the number of hits will increase by 2.6% per week.
1. Write a formula for the number of hits.
2. Estimate the number of hits in 5 weeks.
1. The number of hits will be about 333.
Access these online resources for additional instruction and practice with geometric sequences.
### Key Equations
recursive formula for term of a geometric sequence explicit formula forterm of a geometric sequence
### Key Concepts
• A geometric sequence is a sequence in which the ratio between any two consecutive terms is a constant.
• The constant ratio between two consecutive terms is called the common ratio.
• The common ratio can be found by dividing any term in the sequence by the previous term. See (Figure).
• The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. See (Figure) and (Figure).
• A recursive formula for a geometric sequence with common ratio is given by for .
• As with any recursive formula, the initial term of the sequence must be given. See (Figure).
• An explicit formula for a geometric sequence with common ratio is given by See (Figure).
• In application problems, we sometimes alter the explicit formula slightly to See (Figure).
### Section Exercises
#### Verbal
What is a geometric sequence?
A sequence in which the ratio between any two consecutive terms is constant.
How is the common ratio of a geometric sequence found?
What is the procedure for determining whether a sequence is geometric?
Divide each term in a sequence by the preceding term. If the resulting quotients are equal, then the sequence is geometric.
What is the difference between an arithmetic sequence and a geometric sequence?
Describe how exponential functions and geometric sequences are similar. How are they different?
Both geometric sequences and exponential functions have a constant ratio. However, their domains are not the same. Exponential functions are defined for all real numbers, and geometric sequences are defined only for positive integers. Another difference is that the base of a geometric sequence (the common ratio) can be negative, but the base of an exponential function must be positive.
#### Algebraic
For the following exercises, find the common ratio for the geometric sequence.
The common ratio is
For the following exercises, determine whether the sequence is geometric. If so, find the common ratio.
The sequence is geometric. The common ratio is 2.
The sequence is geometric. The common ratio is
The sequence is geometric. The common ratio is
For the following exercises, write the first five terms of the geometric sequence, given the first term and common ratio.
For the following exercises, write the first five terms of the geometric sequence, given any two terms.
For the following exercises, find the specified term for the geometric sequence, given the first term and common ratio.
The first term is and the common ratio is Find the 5th term.
The first term is 16 and the common ratio is Find the 4th term.
For the following exercises, find the specified term for the geometric sequence, given the first four terms.
Find
Find
For the following exercises, write the first five terms of the geometric sequence.
For the following exercises, write a recursive formula for each geometric sequence.
For the following exercises, write the first five terms of the geometric sequence.
For the following exercises, write an explicit formula for each geometric sequence.
For the following exercises, find the specified term for the geometric sequence given.
Let Find
Let Find
For the following exercises, find the number of terms in the given finite geometric sequence.
There are terms in the sequence.
#### Graphical
For the following exercises, determine whether the graph shown represents a geometric sequence.
The graph does not represent a geometric sequence.
For the following exercises, use the information provided to graph the first five terms of the geometric sequence.
#### Extensions
Use recursive formulas to give two examples of geometric sequences whose 3rd terms are
Use explicit formulas to give two examples of geometric sequences whose 7th terms are
Find the 5th term of the geometric sequence
Find the 7th term of the geometric sequence
At which term does the sequence exceed
The sequence exceeds at the 14th term,
At which term does the sequence begin to have integer values?
For which term does the geometric sequence first have a non-integer value?
is the first non-integer value
Use the recursive formula to write a geometric sequence whose common ratio is an integer. Show the first four terms, and then find the 10th term.
Use the explicit formula to write a geometric sequence whose common ratio is a decimal number between 0 and 1. Show the first 4 terms, and then find the 8th term.
Answers will vary. Example: Explicit formula with a decimal common ratio: First 4 terms:
Is it possible for a sequence to be both arithmetic and geometric? If so, give an example.
### Glossary
common ratio
the ratio between any two consecutive terms in a geometric sequence
geometric sequence
a sequence in which the ratio of a term to a previous term is a constant |
# Distributive Property and combining like terms.. Use the Distributive Property to simplify each expression. 1. 8(m + 5) = 48 2. 5(3x + 9) = 75 3. –2(4.
## Presentation on theme: "Distributive Property and combining like terms.. Use the Distributive Property to simplify each expression. 1. 8(m + 5) = 48 2. 5(3x + 9) = 75 3. –2(4."— Presentation transcript:
Distributive Property and combining like terms.
Use the Distributive Property to simplify each expression. 1. 8(m + 5) = 48 2. 5(3x + 9) = 75 3. –2(4 – 3y) = -50 4. 5(8x – 3) = 100
Term: ◦ a # or a # multiplied by a letter Like Terms: ◦ Terms that contain the same variable (letter) Combine means Add Simplify an Expression: ◦ no parentheses and all like terms have been combined. Equivalent means =
Simplify each algebraic expression by combining like terms. 4x – 7x + 5x (4 – 7 + 5)x 2x2x 8y + (– 6) + 3 – 4y 4y – 3
Simplify: 9m + 3p – 3p – 2m + m 8m + 0p 8m8m
1. Using the Distributive Property 2. Combine all like terms
Simplify each algebraic expression. 4(x + 5) – 3x + 7 4x + 20 – 3x + 7 x + 27 6y – 2(y – 7) + 3 6y – 2y + 14 + 3 4y + 17
Simplify each algebraic expression. 1. 7x – 2x + x 2. –9 + 5y – 2y + 9 3. 18w – 5d + 8d – 21w 4. 4x + 6(x + 2) + 3
If a classmate missed this lesson, how would you describe to them the process of simplifying an expression?
Download ppt "Distributive Property and combining like terms.. Use the Distributive Property to simplify each expression. 1. 8(m + 5) = 48 2. 5(3x + 9) = 75 3. –2(4."
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Module 17 - Riemann Sums and the Definite Integral
Introduction | Lesson 1 | Lesson 2 | Lesson 3 | Self-Test
Lesson 17.2: Left-hand Riemann Sums and the AREA Program
In Lesson 17.1 you used right-hand rectangles to approximate the area of the region bounded by the graph of f(x) = x2, the vertical line x = 1, and the x-axis. In this lesson you will use left-hand Riemann sums to approximate the same area.
The sum of the areas of the rectangles shown above is called a left-hand Riemann sum because the left-hand corner of each rectangle is on the curve.
Defining the Left-Hand Sum Function
Suppose that the area under the curve y = f(x) and above the x-axis between the lines x = a and x = b is approximated using left-hand rectangles. A function similar to the one defined in Lesson 17.1 can be used to find the sum of the areas of the left-hand rectangles. Modifying the function in Y2 can create the new function.
• Open the Y= editor and move the cursor to Y2.
The first left-hand x-coordinate is found by letting k = 0 and the last left-hand x-coordinate is found by letting k = n – 1. This replaces the first right-hand x-coordinate found by letting k = 1 and the last right-hand x-coordinate found by letting k = n.
• Change the initial value and ending value of the sum function from "1, N" to "0, N –1"
The function Y1 = X2 should still be defined from Lesson 17.1. The following exploration uses the same procedure that was used to find the right-hand Riemann sum with four rectangles, except this time the left-hand Riemann sum function will be used to approximate the area under the curve.
• Store 0 in A.
• Store 1 in B.
• Store 4 in N.
• Store (B - A)/N in W.
• Paste Y2 to the Home screen and press to evaluate the left-hand Riemann sum.
The left-hand Riemann sum with 4 rectangles is approximately 0.21875 square units.
Evaluate the left-hand Riemann sum for ten rectangles.
• Store 10 in N.
• Redefine W by recalling and executing the command .
• Evaluate the new sum by entering Y2 on the Home screen again and pressing .
The sum with 10 left-hand rectangles is 0.285 square units.
17.2.1 Evaluate the left-hand Riemann sum using 50 rectangles and using 100 rectangles.
The areas found using left-hand rectangles appear to converge to the same limit as the areas found using right-hand rectangles: 1/3 square unit.
When the number of rectangles approaches infinity, the left-hand Riemann function has the same limit as the right-hand Riemann function. That is, the area under the curve can be found by using either right or left Reimann sum limits as the number of rectangles approaches infinity because they produce the same result. This is true in general for continuous functions. In fact, one can choose any x-value in each subinterval to determine the height of the corresponding rectangle and the limit will still be the same as N goes to .
Many times the midpoint of each subinterval is used to compute the height of each rectangle. Using midpoints usually reduces the error in the approximation of the area under the curve.
The AREA Program
A program created for the TI-83 can be used to illustrate the rectangles that approximate the area under a curve. The program AREA draws the rectangles associated with left, right and
Midpoint Riemann sums are obtained by using the midpoint of each subinterval on the x-axis to determine the height of the corresponding rectangle.
midpoint Riemann sums under a curve and displays the sum of the areas associated with each type. It was used to create the pictures of Riemann sum rectangles earlier in this lesson and in Lesson 17.1. With this program you can create similar visual representations.
• Choose to save the file
• Save the file on your local hard disk in a folder that you can access later
Transferring the Program to the TI-83
• Transfer the program AREA to your TI-83 calculator
• Clear the Home Screen
Using the AREA Program
The following procedure illustrates the approximate area of the region bounded by the graph of f(x) = x2 above the x-axis over the interval [0, 1] by drawing rectangles associated with left, right, and midpoint Riemann sums.
The function X2 should be stored in Y1 in the Y= Editor. Make sure that Y1 is selected.
• Clear the function in Y2.
• Define the viewing window as [0, 1, 1] x [0, 1, 1].
• Press , move the cursor to the line containing AREA and press to paste the command prgmAREA to the Home screen.
• Run the program by pressing .
You should see a prompt for the value of A, the left side of the region.
• Enter 0 and press .
Now you should see a prompt for the value of B, the right side of the region.
• Enter 1 and press .
Next you see a prompt for the number of subintervals, i.e., the number of rectangles.
• Enter 10.
The graph of f(x) = x2 with the left-hand rectangles will be displayed when you press . The pause indicator (four vertical dots) should appear in the upper right-hand corner of the screen.
The screen above illustrates using left-hand rectangles to approximate the area under the curve.
Viewing the Subsequent Screens
Subsequent screens of the program display the approximations obtained using left-hand, right-hand, and midpoint Riemann sums, along with illustrations of the rectangles used in the right-hand and midpoint Riemann sums.
• Display the left-hand Riemann sum by pressing .
The left-hand Riemann sum using 10 rectangles is 0.285 square units.
• Display the right-hand rectangles by pressing .
• Move through the remaining screens by repeatedly pressing .
You should see the right Riemann sum, the illustration of midpoint rectangles, and finally the midpoint Riemann sum.
The sums of the areas of the ten rectangles used to approximate the area of the region using left, right, and midpoint rectangles are 0.285, 0.385, and 0.3325 square units, respectively.
Run the program again several times to illustrate convergence of the Riemann sums. Try 25 rectangles and 50 rectangles. |
# Superposition Theorem
Published
Even with KCL and KVL, as circuits get more complicated, sometimes the setup and the math can become quite complicated. However, using the superposition theorem, in certain situations, you can simplify the circuit by turning off or “suppressing” all the independent power sources except one and solving the circuit, and doing that with all of the power sources, adding up the end result into a single outcome.
That, I imagine, didn’t make much sense. But come back after reading the rest of the tutorial and I bet it’ll be a lot clearer. Let’s try it again. The summation of all of the current produced by the power sources individually is the same as the summation of all of the current produced by the power sources together. It’s linear - each power source linearly affects all other power sources. To simplify things, you can remove all other voltage sources and replace them with a short circuit and remove all other current sources and replace them with an open circuit and solve the circuit. Repeat this with all the power sources, and then sum them up to figure out what the total current will be.
Now that I’ve said the same thing in two different ways, I think an example would be the best way to truly understand it in practice.
## Example #1
Let’s use this simple circuit as an example:
As you can see, there is a voltage source and a current source and a single resistor that we want to know the current through. If you need the superposition theorem to simplify this circuit, I highly recommend you check out our tutorials on KVL and on KCL and do some practice circuits! But, with the voltage source and current source seemingly in contention, I can see how this might cause some confusion. With the superposition theorem, this will be simplified.
First, choose a power source and suppress all other power sources. Again, a suppressed voltage source will become a short circuit and a suppressed current source will become an open circuit. Let’s suppress the current source first.
With the current source now open, we can make two statements based on inspection. With the open circuit in series with the resistor, there will be no current flow through the resistor. Thus, the amount of current through the resistor because of the voltage source is 0. Second, as there’s no current through the resistor, we know that the voltage on BOTH sides of the resistor is 10V.
Now, let’s choose the current source and suppress the voltage source.
As the voltage source becomes a short circuit, we can see that we simply have the current source and the 100 ohm resistor. By inspection, we know that the current, flowing from right to left, is 5A. However, let’s take a moment. To create that 5A, the current source will need to generate 500V (V=IR -> 500V = 5A * 100Ω). So on the right side of the resistor, we’re seeing 500V, and on the left side of the resistor, it is grounded, so it’s at 0V.
Now, the final step of doing the superposition theorem is putting it together.
From the first scenario, we had zero amps going through the resistor and ten volts on both sides of the resistor. From the second, we had 5 amps going through the resistor (right to left), with 500 volts on the right side and 0 volts on the left side. Combining this all together, we get that there are 5 amps going through the resistor (right to left), 510 volts on the right side of the resistor and 10 volts on the left side of the resistor. The common ground between the voltage source and the current source is, as expected, at 0V.
Now, does this make sense intuitively? We know that a current source will create any voltage necessary to output the expected current and we know that a voltage source will create or absorb any current necessary to output the expected voltage. Since the voltage source is raising the voltage on the left hand side, the current source has to generate 510V, so that the voltage across the resistor is 500V, to give the desired 5A. The voltage source on the left, to maintain that 10V, will absorb the 5A. Both power sources have their functional requirements satisfied and the summation of the two sources working separately is the same as the two working at the same time.
Let’s do one more example to see if we can cement your understanding of this concept.
## Example #2
Another circuit that we could almost certainly figure out without using the superposition theorem but since we’re here already, let’s work through it!
First, suppress the 10V voltage source. When it’s suppressed, what do we do with it - open or short the circuit?
Yep, we short it. So, we get the following circuit:
Now we solve this like normal. Do whatever is most comfortable for you, I think I’m going to do this an odd way - not using KCL or KVL but just Ohm’s Law and what we know of series and parallel resistors.
We have 200Ω in series with 300Ω||500Ω (an equivalent resistance of 187.5Ω), so the total current through the series circuits will be:
Which is also the current through the 200Ω resistor. Now that we know the current through that resistor, we can find the voltage across it.
Now we know we have 5 - 2.58V = 2.42V across the other two resistors and can solve for their currents independently.
And, summing 8.1mA and 4.8mA gives us 12.9mA so everything makes sense and is internally consistent! I love it when the math works out, even if it’s simple.
We’ve solved for this source, let’s solve with the 10V source.
I won’t go through the steps this time, but here are the currents through the different resistors - feel free to do it yourself for practice.
200Ω resistor: 16.2mA
300Ω resistor: 22.6mA
500Ω resistor: 6.5mA
Notice that the current through the 200Ω and 500Ω resistors don’t actually add up to the same current as in the 300Ω resistor? That’s just due to rounding errors - don’t worry about it.
BUT! Now we have an interesting dilemma. I didn’t identify the direction the current was flowing in these resistors, but in my calculations, with the two scenarios, the 200Ω and the 500Ω resistors were labelled with the current flowing the opposite way. So we need to decide which direction we will assume the current will flow so we know which of these values are positive and which are negative. In this case, we’ll assume that the current is flowing left to right in the 200Ω resistor, and right to left in the 300Ω resistor.
With that assumption, we get the following values:
First pass:
200Ω resistor: 12.9mA
300Ω resistor: -8.1mA
500Ω resistor: 4.8mA
Second pass:
200Ω resistor: -16.2mA
300Ω resistor: 22.6mA
500Ω resistor: 6.5mA
Summed total:
200Ω resistor: -3.3mA
300Ω resistor: 14.5mA
500Ω resistor: 11.3mA
Surprisingly, the 5V power supply is actually absorbing 3.3mA — not something I was anticipating when randomly putting these values together. There’s not a lot we can do to make a sanity check here, unfortunately, other than it makes sense that current is flowing away from the 10V supply more than the 5V supply and that the currents going into the center node equal zero (again, disregarding rounding issues).
A couple of notes as we wrap things up.
1. Again, this concept only works with linear circuits. Op-amps, microcontrollers, 555 timers, or other non-linear elements will render this invalid.
2. This works with AC circuits as well as DC circuits, as long as the circuit is linear.
3. Any dependent power source must remain in the circuit at all times. You cannot take it out (or suppress it) or leave it in as the sole power source in a circuit.
And that’s it for the superposition theorem. As stated at the beginning of this tutorial: Using the superposition theorem, in certain situations, you can simplify the circuit by suppressing all the independent power sources except one and solving the circuit. Then you do that with all of the power sources, summing the end result into a single outcome. Now that we’ve been through it, that hopefully makes a lot more sense! If you have any questions or comments, leave them here on CircuitBread.com and we’ll try and clarify the issue for you.
Get the latest tools and tutorials, fresh from the toaster. |
# 2016 AIME II Problems/Problem 1
## Problem
Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially.
## Solution 1
Let $r$ be the common ratio, where $r>1$. We then have $ar-9-(a-5)=a(r-1)-4=ar^{2}-25-(ar-9)=ar(r-1)-16$. We now have, letting, subtracting the 2 equations, $ar^{2}+-2ar+a=12$, so we have $3ar=432,$ or $ar=144$, which is how much Betty had. Now we have $144+\dfrac{144}{r}+144r=444$, or $144(r+\dfrac{1}{r})=300$, or $r+\dfrac{1}{r}=\dfrac{25}{12}$, which solving for $r$ gives $r=\dfrac{4}{3}$, since $r>1$, so Alex had $\dfrac{3}{4} \cdot 144=\boxed{108}$ peanuts.
Let $a$ be Alex's peanuts and $k$ the common ratio. Then we have $a(k^2+k+1)=444$. Adding $k$ to both sides and factoring,
$$\frac{444}{a}+k=(k+1)^2$$
For the common difference, $ak=5-(a-5)=ak^2-25-(ak-9)$. Simplifying, $k^2-2k+1=\frac{12}{a}$. Factoring, $$(k-1)^2=\frac{12}{a}$$
$(k+1)^2-(k-1)^2=4k$, so $4k=\frac{444-20}{a}$. Then, $k=\frac{432}{3a}=\frac{432}{a}=\frac{144}{a}$. Substitute k in $(k-1)^2=\frac{12}{a}$ to get $(\frac{144-a}{a})^2=\frac{12}{a}$. Simplifying, expanding, and applying the quadratic formula,
$a=150\pm\frac{\sqrt{300^2-4(144^2)}{2}$ (Error compiling LaTeX. ! File ended while scanning use of \frac .)
Taking out $4^2\cdot3^2$ from under the radical leaves $$a=150\pm6\sqrt{625-576}=108, 192$$ Since Alex's peanut number was the lowest of the trio, and $3*192>444$, Alex initially had \boxed{108} peanutes.
(Solution by BJHHar)
## Solution 3
Let the initial numbers of peanuts Alex, Betty and Charlie had be $a$, $b$, and $c$ respectively. Let the final numbers of peanuts, after eating, be $a'$, $b'$, and $c'$.
We are given that $a + b + c = 444$. Since a total of $5 + 9 + 25 = 39$ peanuts are eaten, we must have $a' + b' + c' = 444 - 39 = 405$. Since $a'$, $b'$, and $c'$ form an arithmetic progression, we have that $a' = b' - x$ and $c' = b' + x$ for some integer $x$. Substituting yields $3b' = 405$ and so $b' = 135$. Since Betty ate $9$ peanuts, it follows that $b = b' + 9 = 144$.
Since $a$, $b$, and $c$ form a geometric progression, we have that $a = \frac{b}{r}$ and $c = br$. Multiplying yields $ac = b^2 = 144^2$. Since $a + c = 444 - b = 300$, it follows that $a = 150 - \lambda$ and $c = 150 + \lambda$ for some integer $\lambda$. Substituting yields $(150-\lambda)(150+\lambda) = 144^2$, which expands and rearranges to $\lambda^2 = 150^2-144^2 = 42^2$. Since $\lambda > 0$, we must have $\lambda = 42$, and so $a = 150 - \lambda = \boxed{108}$. |
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# Module 4 topic 1 2nd
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### Module 4 topic 1 2nd
1. 1. More – Coordinate Plane, Midpoint, and Relations and Functions<br />Module 4 Topic 1 <br />
2. 2. Example 1<br />Remember that the quadrants are I, II, III, or IV.<br />Quadrant I = (positive x-value, positive y-value)<br />Quadrant II = (negative x-value, positive y-value)<br />Quadrant III = (negative x-value, negative y-value)<br />Quadrant IV = (positive x-value, negative y-value) <br />
3. 3. Example 2<br />Similar Problem: Find the coordinates of the midpoint of M(-3, 0) and N(-7, -5).<br />Midpoint is the average of the x-values and the average of the y-values.<br />Midpoint = (x2 + x1) ÷ 2 , (y2 + y1) ÷ 2<br />Midpoint = (-7 + -3) ÷ 2 , (-5 + 0) ÷ 2<br />Midpoint = -10 ÷ 2 , -5 ÷ 2<br />Midpoint = -5, -2.5 so we write it as an ordered pair (-5, -2.5) <br />
4. 4. Example 3<br />Similar Problem: Use the following relation to answer questions 7 - 9.<br />{(2, 4), (7, 8), (-5, 4), (-2, -8), (0, 0)}<br />Is the relation a function?<br />Yes - because the x-values do not repeat! <br />
5. 5. Example 4<br />Similar Problem: Use the following relation to answer questions 7 - 9.<br />{(2, 4), (7, 8), (-5, 4), (-2, -8), (0, 0)}<br />State the domain of the relation.<br />Domain = all x-values of the relation<br />2, 7, -5, -2, 0 but let's put these in order so we have<br />-5, -2, 0, 2, 7 <br />
6. 6. Example 5<br />Similar Problem: Use the following relation to answer questions 7 - 9.<br />{(2, 4), (7, 8), (-5, 4), (-2, -8), (0, 0)}<br />State the range of the relation.<br />Range = all y-values of the relation<br />4, 7, 4, -8, 0 but let's put these in order so we have<br />-8, 0, 4, 4, 7 and let's not repeat any so we have -8, 0 , 4, 7 <br />
7. 7. Example 6<br />Similar Problem: If f(x) = 3x + 6, find f(-4).<br />Plug in -4 for x in the function 3x + 6<br />3(-4) + 6<br />-12 + 6<br />-6 <br /> |
# How do you differentiate y= x^(11cosx)?
##### 1 Answer
May 30, 2018
$\frac{\mathrm{dy}}{\mathrm{dx}} = 11 {x}^{11 \cos \left(x\right)} \left(\cos \frac{x}{x} - \sin \left(x\right) \ln \left(x\right)\right)$
#### Explanation:
$y = {x}^{11 \cos \left(x\right)}$
To deal with tricky exponents like this, let's take the natural logarithm of both sides and remember the rule $\log \left({a}^{b}\right) = b \log \left(a\right)$.
$\ln \left(y\right) = \ln \left({x}^{11 \cos \left(x\right)}\right)$
$\ln \left(y\right) = 11 \cos \left(x\right) \ln \left(x\right)$
Now take the derivative on both sides. On the left, we'll need the chain rule. On the right, we'll use the product rule.
$\frac{1}{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 11 \left(\frac{d}{\mathrm{dx}} \cos \left(x\right)\right) \ln \left(x\right) + 11 \cos \left(x\right) \left(\frac{d}{\mathrm{dx}} \ln \left(x\right)\right)$
$\frac{1}{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 11 \left(- \sin \left(x\right)\right) \ln \left(x\right) + 11 \cos \left(x\right) \left(\frac{1}{x}\right)$
Solving for the derivative:
$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left(\frac{11 \cos \left(x\right)}{x} - 11 \sin \left(x\right) \ln \left(x\right)\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 11 {x}^{11 \cos \left(x\right)} \left(\cos \frac{x}{x} - \sin \left(x\right) \ln \left(x\right)\right)$ |
# How to Teach Zero and Negative Exponents
Would You Rather Listen to the Lesson?
Sometimes in Math we have things come up that need to be proved but our students do not yet have the skill set mathematically to prove the new concept. This is one of those times. However, we can show them the proof they need for zero and negative exponents using patterns.
## Why is a0 = 1?
Usually when we prove things and / or use patterns in math class we want to use the simplest example we can. Which in this case would be a base of 2. However, this is one time I do not use 2 as an example. The reason I don't is because powers of 2 and multiplying / dividing by 2 with low numbers are too closely related in their results and can confuse the students (EX: 2x2=4 and 22 =4).
35 = 243
3= 81
327
3= 9
3= 3
3= 1
Notice that when we divide the answers by the base (3) it gives us the following answers.
243/3 = 81
81/3 = 27
27/3 = 9
9/3 = 3
3/3 = 1
If need be you can now use different bases to show the pattern always works.
(Tip: Don't use 10 for the same reason I suggested not to use 2!)
## Patterns with Negative Bases
You can now continue the pattern by showing them negative bases.
(-3)5 = -243
(-3)= 81
(-3)= -27
(-3)= 9
(-3)= -3
(-3)= 1
Notice that when we divide the answers by the base (-3) it gives us the following answers.
(-243)/(-3) = 81
81/(-3) = -27
(-27)/(-3) = 9
9/(-3) = -3
(-3)/(-3) = 1
## Patterns with Negative Exponents
This patter continues on for negative exponents as well.
33 = 27 32 = 9 31 = 3 30 = 1 3−1= 1/3 3−2 = 1/9 3−3 = 1/27 3−4 = 1/81
So...
1/3 = 1/3
(1/3) / 3 = 1/9
(1/9) / 3 = 1/27
(1/27) / 3 = 1/81
Finally, Make sure to point out that...
1/3 = 1/31
1/9 = 1/32
1/27 = 1/33
1/81 = 1/34
1/243 = 1/35
Now you're ready to move on to Exponents that are Fractions!
Here is your free content for this lesson!
## Zero and Negative Exponents - PDFs
7-1 Guided Notes Teacher Edition - Zero and Negative Exponents (Members Only)
7-1 Lesson Plan - Zero and Negative Exponents (Members Only)
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7-1 Project - Zero and Negative Exponents (Members Only)
## Zero and Negative Exponents - Word Docs & PowerPoints
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# How do you divide (3 + 2i) / (2 - 6i)?
Mar 28, 2016
You must multiply the whole expression by the conjugate of the denominator, just like you did when you were younger when you were to rationalize the denominator.
#### Explanation:
The conjugate forms a difference of squares. Therefore, to find it, you must switch the sign in the middle.
Thus, the conjugate is $2 + 6 i$
$\frac{3 + 2 i}{2 - 6 i} \times \frac{2 + 6 i}{2 + 6 i}$
Now multiply. Don't forget that ${i}^{2} = - 1$
(6 + 4i + 12i - 12)/(4 - (36 xx -1)
Don't forget that you can combine $i$'s with $i$'s but you cannot combine non $i ' s$ with $i ' s$.
$\frac{6 + 16 i - 12}{40}$
$\frac{- 6 + 16 i}{40}$
This is as simplified as possible.
Practice exercises:
1. Evaluate:
a). $\frac{4 + 2 i}{3 - 5 i}$
b). $\frac{4 x + 3 i}{- x + 2 i}$
Good luck! |
Derivation of Work-Energy Theorem
We do various activities in our daily life, sitting and reading a newspaper is also doing some work. According to Physics, any work done that does not involve the displacement of the body is not considered work. Thus, to describe the phenomenon of work energy we will derive the work energy theorem. The work-energy theorem explains the reasons behind the physics of no work. Stay tuned and continue reading this article to get the derivation of the work-energy theorem!
What is Work?
Work is defined as the product of force and displacement. It means when the force acting on an object displaces the object from its original position, then it is said to be work done.
You must also know that moving objects possess kinetic energy. Thus, the work-energy theorem gives the relationship between work and kinetic energy.
Work is the change in kinetic energy.
W= ΔK
Also Read: Class 9 Motion
Work-Energy Theorem
The work-energy theorem is defined as the work done by the net force applied on an object or body is directly proportional to the change in kinetic energy.
The formula of the Work-Energy Theorem is
Kf – Ki = W
Here,
Kf= Final kinetic energy
Ki= Initial Kinetic energy
Kf – Ki is a change in Kinetic energy
W= Work done on an object
Derivation of Work-Energy Theorem
Equations of motion,
v2 = u2 + 2as
Here,
v = final velocity of an object
u = initial velocity of an object
a = constant acceleration
s = displacement of the object
The above equation can also be written as:
v2 – u2 = 2as
Substituting the values of the vector quantities, the equation would be:
v2 – u2 = 2*a*d
Now, multiply both sides of the equation by m/2, we get:
½ mv2 – ½ mu2 = ma*d
According to Newton’s second law of motion, F= ma,
Now, the above equation can be written as:
½ mv2 – ½ mu2 = F*d
We also know that W= F.d and, Kinetic energy = (mv²)/2,
Thus the equation becomes:
Kf – Ki = W
Hence, we have:
ΔK = W
Here ΔK = Kf – Ki
ΔK= Change in kinetic energy
Derivation of Work-Energy Theorem for Variable Force
Here is the proof of the work-energy theorem for variable force:
F= Variable Force
t= Time
Work Done by Force is given by:
W= ∫X1X0 F* dx —equation 1
Here,
x0= initial position
x1 = Final Position
We know that,
Kinetic energy KE= ½ *mv2.
This means that when x0 then K0 and when x1 then K1.
Now, differentiate K w.r.t time (t):
dK/dt= m*v* dv/dt
dK/dt=m*a*dx/dt
dK/dt= F*dx/dt
dK= F*dx
K0K1 dK = ∫x0x1 F* dx
ΔK = W
Hence Proved.
Also Read: Laws of Motion Class 11
Derivation of Work-Energy Theorem for Constant Force
From Newton’s Second Law of Motion: F = ma
Here,
a = acceleration of the object
The velocity of the object increases i.e. v1 to v2 by applying acceleration, and the object gets displaced by a distance d.
Thus the equation becomes:
v22−v12= 2*a*d
We can also write it as:
a = (v22−v12)/2d, or
Now substituting the value of a in F= ma
F = m (v22−v12)/2d, or
Fd = m (v22−v12)/2d, or
Fd = ½ m*v22 – ½ m*v12 —Equation 1
Fd is the work done by the force F to move the object through a distance d.
In equation 1, the quantity
K2 = m*v22/2 is the final Kinetic energy of the object.
K1=m*v12/2 is the initial Kinetic of the object.
Thus equation (1) becomes
W=K2-K1=ΔK- Equation 2
Here,
ΔK = change in Kinetic energy of the object.
From equation 2, it is clear that the work done by a force is equal to the change in kinetic energy.
Relevant Blogs
FAQs
What is the formula of the Work-Energy Theorem?
The Formula for the Work-Energy Theorem is W= Kf – Ki
Here, Kf= Final kinetic energy, Ki= Initial Kinetic energy
Kf – Ki is a change in Kinetic energy
W= Work done on an object
What is the statement of the Work-Energy Theorem?
The work-energy theorem states that the total work done on an object by the application of force is equal to the change in its kinetic energy.
What is the derivation of the work equation?
The derivation of the work equation is W= F*d. Here W= work and it is the product of Force (F) and Displacement (d).
For more information about such informative articles, make sure to check the trending events page of Leverage Edu. |
26
Q:
# The length of a class room floor exceeds its breadth by 25 m. The area of the floor remains unchanged when the length is decreased by 10 m but the breadth is increased by 8 m. The area of the floor is
A) 5100 sq.m B) 4870 sq.m C) 4987 sq.m D) 4442 sq.m
Explanation:
Let the breadth of floor be 'b' m.
Then, length of the floor is 'l = (b + 25)'
Area of the rectangular floor = l x b = (b + 25) × b
According to the question,
(b + 15) (b + 8) = (b + 25) × b
2b = 120
b = 60 m.
l = b + 25 = 60 + 25 = 85 m.
Area of the floor = 85 × 60 = 5100 sq.m.
Q:
The perimeter of a rectangle whose length is 6 m more than its breadth is 84 m. What will be the area of the rectangle?
A) 333 sq.mts B) 330 sq.mts C) 362 sq.mts D) 432 sq.mts
Explanation:
Let the breadth of the rectangle = b mts
Then Length of the rectangle = b + 6 mts
Given perimeter = 84 mts
2(L + B) = 84 mts
2(b+6 + b) = 84
2(2b + 6) = 84
4b + 12 = 84
4b = 84 - 12
4b = 72
b = 18 mts
=> Length = b + 6 = 18 + 6 = 24 mts
Now, required Area of the rectangle = L x B = 24 x 18 = 432 sq. mts
5 353
Q:
How many square units in 13 by 9?
A) 13 B) 9 C) 117 D) 13/9
Explanation:
Number of square units in 13 by 9 is given by the area it forms with length and breadth as 13 & 9
Area = 13 x 9 = 117
Hence, number of square units in 13 by 9 is 117 sq.units.
4 1201
Q:
Square units 13 by 9 of an office area is
A) 97 B) 117 C) 107 D) 127
Explanation:
Square units 13 by 9 of an office means office of length 13 units and breadth 9 units.
Now its area is 13x 9 = 117 square units or units square.
8 2049
Q:
Find the area of the square whose side is equal to the diagonal of a rectangle of length 3 cm and breadth 4 cm.
A) 25 sq.cm B) 16 sq.cm C) 9 sq.cm D) 4 sq.cm
Explanation:
Given length of the rectangle = 3 cm
Breadth of the rectangle = 4 cm
Then, the diagonal of the rectangle
Then, it implies side of square = 5 cm
We know that Area of square = S x S = 5 x 5 = 25 sq.cm.
8 1696
Q:
The sides of a right-angled triangle are 12 cm, 16 cm, 20 cm respectively. A new right angle Δ is made by joining the midpoints of all the sides. This process continues for infinite then calculate the sum of the areas of all the triangles so made.
A) 312 sq.cm B) 128 sq.cm C) 412 sq.cm D) 246 sq.cm
Explanation:
Area = = 96
Sum of Area =
15 2640
Q:
A room is 15 feet long and 12 feet broad. A mat has to be placed on the floor of the room leaving 1.5 feet space from the walls. What will be the cost of the mat at the rate of Rs. 3.50 per squire feet?
A) 378 B) 472.5 C) 496 D) 630
Explanation:
18 2185
Q:
A circular piece of thin wire is converted into a rhombus of side 11 cm. Find the diameter of the circular piece?
A) 28 cm B) 3.5 cm C) 7 cm D) 14 cm
Explanation:
Circular piece is 4 x 11 = 44 cm long,
Then Circumference of circle is given by,
44 = pi x D, where D is the diameter
D = 44 / pi
Take pi = 22 / 7, then
D = 44 / (22/7) = (44 x 7) / 22
D = 14 cm.
18 2401
Q:
The area of a square is equal to the area of a rectangle. The length of the rectangle is 5 cm more than a side of the square and its breadth is 3 cm less than the side of the square. What is the perimeter of the rectangle ?
A) 15 cm B) 18 cm C) 34 cm D) 26 cm
Explanation:
Let the side of the square be 's' cm
length of rectangle = (s+5) cm
(s+5) (s-3) =
$s2$ - 5s - 3s - 15 = $s2$
2s = 15
Perimeter of rectangle = 2(L+B) = 2(s+5 + s–3) = 2(2s + 2)
= 2(15 + 2) = 34 cm |
# How do you integrate int (5x+4)^5 using substitution?
Oct 31, 2016
#### Explanation:
Let $u = 5 x + 4$, then $\mathrm{du} = 5 \mathrm{dx}$ or dx = $\left(\frac{1}{5}\right) \mathrm{du}$
$\int {\left(5 x + 4\right)}^{5} \mathrm{dx} = \left(\frac{1}{5}\right) \int {u}^{5} \mathrm{du} = {u}^{6} / 30 + C$
Reverse the substitution:
$\int {\left(5 x + 4\right)}^{5} \mathrm{dx} = {\left(5 x + 4\right)}^{6} / 30 + C$
Oct 31, 2016
Let $u = 5 x + 4$, then $\mathrm{du} = 5 \mathrm{dx}$ or dx = $\left(\frac{1}{5}\right) \mathrm{du}$
$\int {\left(5 x + 4\right)}^{5} \mathrm{dx} = \left(\frac{1}{5}\right) \int {u}^{5} \mathrm{du} = {u}^{6} / 30 + C$
$\int {\left(5 x + 4\right)}^{5} \mathrm{dx} = {\left(5 x + 4\right)}^{6} / 30 + C$ |
# Multiplication of Fractional Number by a Whole Number
This topic deals with multiplication of a whole number i.e. (0, 1, 2, 3, 4, 5, 6, 7…….) by a fraction i.e. a number in $$\frac{p}{q}$$ form. When we are multiplying a whole number with a fraction or fractional number then first we have to change that whole number into fraction just by placing a 1 in the denominator. As 1 in the denominator has no value hence the actual value of the whole number will remain same if we place 1 in the denominator.
Solved Examples on Multiplication of Fractional Number by a Whole Number:
$$\frac{8}{10}$$ × 5
In this sum 5 is the whole number and 8/10 is the fraction.
Step I: Change the whole number into a fraction by placing 1 in the denominator.
Therefore, $$\frac{8}{10}$$ × $$\frac{5}{1}$$
Step II: Now both the numbers are in fraction and the rule of multiplication of fractional number is that numerator and numerator should be multiplied and similarly denominator and denominator should be multiplied. So, we will do
5 × 8 = 40
And 1 × 10 = 10
Therefore, $$\frac{5}{1}$$ × $$\frac{8}{10}$$ = 40/10
Step III: After multiplying numerator by numerator and denominator by denominator the answer that we will get has to be either changed into lowest term (if possible) and then change into mixed fraction if not a proper fraction or can be left as whole number (if possible).
Therefore, first we will change 40/10 into lowest term
= $$\frac{40 ÷ 10}{10 ÷ 10}$$
= $$\frac{4}{1}$$
= 4 [This can be written as 4 as 1 in the denominator has no value it remains same.]
2. $$\frac{2}{3}$$ of 60
Here, ‘of ‘denotes multiplication,
So, $$\frac{2}{3}$$ × 60
= $$\frac{2}{3}$$ × $$\frac{60}{1}$$; [Changing the whole number 60 into fraction]
= $$\frac{2 × 60}{3 × 1}$$, [Multiplying numerator by numerator and denominator by denominator]
= $$\frac{120}{3}$$
= $$\frac{120 ÷ 3}{3 ÷ 3}$$, [Changing into lowest term]
= $$\frac{40}{1}$$
= 40
3. $$\frac{7}{54}$$ × 3
= $$\frac{7}{54}$$ × $$\frac{3}{1}$$; [Changing the whole number 60 into fraction]
= $$\frac{7 × 3}{54 × 1}$$; [Multiplying numerator by numerator and denominator by denominator]
= $$\frac{21}{54}$$
= $$\frac{21 ÷ 3}{54 ÷ 3}$$; [Changing into lowest term]
= $$\frac{7}{18}$$; [Cannot be changed into mixed fraction as it is a proper fraction]
4. 4$$\frac{3}{15}$$ × 10
= $$\frac{15 × 4 + 3}{15}$$ × 10; [Changing improper fraction into mixed fraction]
= $$\frac{63}{15}$$ × 10
= $$\frac{63}{15}$$ × $$\frac{10}{1}$$; [Changing the whole number 10 into fraction]
= $$\frac{63 × 10}{15 × 1}$$; [Multiplying numerator by numerator and denominator by denominator]
= $$\frac{630}{15}$$
= $$\frac{630 ÷ 15}{15 ÷ 15}$$; [Changing into lowest term]
= $$\frac{42}{1}$$
= 42
5. $$\frac{2}{4}$$ × 15
= $$\frac{2}{4}$$ × $$\frac{15}{1}$$; [Changing the whole number 10 into fraction]
= $$\frac{2 × 15}{4 × 1}$$; [Multiplying numerator by numerator and denominator by denominator]
= $$\frac{30}{4}$$
= $$\frac{30 ÷ 2}{4 ÷ 2}$$; [Changing into lowest term]
= $$\frac{15}{2}$$; [Changing improper fraction into mixed fraction]
= 7$$\frac{1}{2}$$; [Answer is expressed in mixed fraction as it was an improper fraction in the 2nd last step]
Hence, these are the various types of sums where we are multiplying a whole number with a fraction.
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Question 1: A cone of height $15 \ cm$ and diameter $7 \ cm$ is mounted on a hemisphere of same diameter. Determine the volume of the solid thus formed.
Cone: Height $= 15 \ cm$, Diameter $= 7 \ cm$
Hemisphere: Radius $= 3.5 \ cm$
Total volume = volume of the cone + volume of the hemisphere
$= \frac{1}{3} \pi \times (3.5)^2 \times 15 + \frac{1}{2} \times \frac{4}{3} \pi \times (3.5)^3$
$= 192.5 + 89.833 = 282.33 \ cm^3$
$\\$
Question 2: A buoy is made in the form of hemisphere surmounted by a right cone whose circular base coincides with the plane surface of hemisphere. The radius of the base of the cone is $3.5 \ meters$ and its volume is two-third of the hemisphere. Calculate the height of the cone and the surface area of the buoy, correct to two places of decimal.
Cone: Height $= h \ cm$, Diameter $= 7 \ cm$
Hemisphere: Radius $= 3.5 \ cm$
Therefore
$\frac{1}{3} \pi \times (3.5)^2 \times h = \frac{2}{3} \times \frac{1}{2} \times \frac{4}{3} \pi \times (3.5)^3$
$h = 2 \times \frac{4}{3} \times 3.5 = 4.67 \$
Total surface area of the solid $= \pi r l + \frac{1}{2} \times 4 \pi r^2$
$= 3.14 \times (3.5) \times \sqrt{3.5^2+4.67^2} + 2 \times 3.14 \times (3.5)^2$
$= 64.137 + 76.93 = 141.17 \ m^2$
$\\$
Question 3: From a rectangular solid of metal $42 \ cm$ by $30 \ cm$ by $20 \ cm$, a conical cavity of diameter $14 cm and depth$latex 24 \ cm is drilled out. Find:
(i) the surface area of remaining solid,
(ii) the volume of remaining solid,
(iii) the weight of the material drilled out if it weighs $7 gm / cm^3$.
Rectangular solid: $42 \ cm$ by $30 \ cm$ by $20 \ cm$
(i) Surface area of the solid = surface are of the rectangular solid – surface are of the base of the cone +curved surface ares of the cone
$= 2 (lb+bh+ hw) - \pi r^2 + \pi r l$
$= 2 (42 \times 30 + 30 \times 20 + 20 \times 42) - \frac{22}{7} \times (7)^2 + \frac{22}{7} \times 7 \times \sqrt{7^2+24^2}$
$= 5400 - 154 + 550 = 5796 \ cm^2$
(ii) Volume = Volume of the solid – Volume of the cone
$= 42 \times 30 \times 20 - \frac{1}{3} \times \frac{22}{7} \times (7)^2 \times 24$
$= 25200 - 1232 = 23968 \ cm^3$
(iii) Weight of the material drilled $= 1232 \times 7 = 8624 \ gms = 8.624 \ kg$
$\\$
Question 4: A cubical block of side $7 \ cm$ is surmounted by a hemisphere of the largest size. Find the surface area of the resulting solid.
Box: Side $= 7 \ cm$
Hemisphere: Radius $= 3.5 \ cm$
Total surface area = surface area of the box – surface area of one side + surface area of the hemisphere
$= 6 (7)^ - (7)^ + \frac{1}{2} \times 4 \pi \times (3.5)^2$
$= 245 + 77 = 322 \ cm^2$
$\\$
Question 5: A vessel is in the form of an inverted cone. Its height is $8 cm$ and the radius of its top, which is open, is $5 \ cm$. It is filled with water up to the rim. When lead shots each of which is a sphere of radius $0.5 \ cm$ are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Cone: Height $= 8 \ cm$, Radius $= 5 \ cm$
Lead shot: Radius $= 0.5 \ cm$, number of shots $= n$
Therefore: $n \times \frac{4}{3} \pi \times (0.5)^3 = \frac{1}{4} \times \frac{1}{3} \pi \times (5)^2 \times 8$
$\Rightarrow n =$ $\frac{5^2 \times 8}{4 \times 4 \times (0.5)^3}$ $= 100$
$\\$
Question 6: A hemi-spherical bowl has negligible thickness and the length of its circumference is $198 \ cm$. Find the capacity of the bowl.
Circumference $= 198 \ cm$
$\Rightarrow 2 \pi r = 198$
$\Rightarrow r = \frac{198}{2 \pi} = 31.5 \ cm$
Therefore Volume of the bowl $= \frac{1}{2} \times \frac{4}{3} \pi \times (31.5)^3 = 65488.5 \ cm^3$
$\\$
Question 7: Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius $r \ cm$.
Sphere: Radius $= r \ cm$
Cone: Radius $= r$, Height $= r$
The maximum volume of the cone $= \frac{1}{3} \pi \times (r)^2 \times r = \frac{1}{3} \pi r^3$
$\\$
Question 8: The radii of the bases of two solid right circular cones of same height are $r_1$, and $r_2$. respectively. The cones are melted and recast into a solid sphere of radius R. Find the height of each cone in terms of $r_1, r_2$, and $R$.
Cones: Radius $= r_1$, Radius $= r_2$, Height $= h$
Sphere: Radius $= R$
Therefore
$\frac{1}{3} \pi \times (r_1)^2 \times h + \frac{1}{3} \pi \times (r_1)^2 \times h = \frac{4}{3} \pi \times (R)^3$
$\Rightarrow (r_1^2+r_2^2)h = 4R^3$
$\Rightarrow h =$ $\frac{4R^3}{(r_1^2+r_2^2)}$
$\\$
Question 9: A solid metallic hemisphere of diameter $28 \ cm$ is melted and recast into a number of identical solid cones, each of diameter $14 \ cm$ and height $8 \ cm$. Find the number of cones so formed.
Hemisphere: Radius $= 14 \ cm$
Cones: Radius $= 7 \ cm$, Height $= 8 \ cm$, Number of cones $= n$
Therefore $n \times \frac{1}{3} \pi \times (7)^2 \times 8 = \frac{2}{3} \pi \times (14)^3$
$\Rightarrow n =$ $\frac{2 \times 14^3}{7^2 \times 8}$ $= 14$
$\\$
Question 10: A cone and a hemisphere have the same base and the same height. Find the ratio between their volumes.
Cone: Radius $= r$, Height $= h$
Hemisphere: Radius $= r$
Ratio of their volumes $=$ $\frac{Volume \ of \ Cone}{Volume \ of \ Hemisphere}$
$=$ $\frac{\frac{1}{3} \pi \times (r)^2 \times h}{ \frac{4}{3} \pi \times (r)^3} = \frac{1}{2}$
$\\$ |
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Review of basic maths
1. 1. Review of Basic MathematicsWhole NumbersAddition and SubtractionAddition is indicated by +. Subtraction is indicated by −. Commutative is a specialmathematical name we give to certain operations. It means that we can do theoperation in any order. Addition is commutative because we know that 2 + 4 means the same thing as 4 + 2Subtraction is not commutative since 21 − 6 is not the same as 6 − 21For this reason addition can be done in any order. Generally we calculate from leftto right across the page. We can sometimes rearrange the order of a sum involvingboth additions and subtractions but when we do this we must remember to keep thenumber with the sign immediately preceeding it. For example 3+5+3+4+7+3 = 3+3+3+4+5+7 6 + 7 − 10 + 2 − 1 − 2 = 6 + 7 + 2 − 10 − 1 − 2Multiplication and DivisionMultiplication is indicated by × or ∗. Sometimes, as long as there will be no confusionwe use juxtaposition to indicate multiplication, particular when we use letters torepresent quantities. For example 3a means 3 ×a and xy means x ×y. Multiplicationis a commutative operation i.e. we can reverse the order e.g. 3 × 4 = 4 × 3. Division is indicated by the symbols ÷, / or − . Division is not commutativesince 4 ÷ 2 = 2 ÷ 4. Multiplication and division should be done from left to rightacross the page. although you can rearrange the order of an expression involvingonly multiplications. For example 3×4×5 = 5×4×3 ab3a = 3aab 1
2. 2. IndicesLike any profession or discipline mathematics has developed short hand notation fora variety of operations. A common shorthand notation that you may be familiarwith is the use of indices or powers. The indice indicates how many times a numbershould be multiplied by itself, so for instance 103 = 10 × 10 × 10 = 1000 3 X4 = X × X × X × X 4A negative index indicates that the power should be on the bottom of a fraction witha 1 on the top. So we have 1 1 1 10−2 = 2 = = 10 10 × 10 100 1 1 y−3 = 3 = y y ×y ×yAdditional Rules Associated with Multiplicaton and DivisionZeroAny quantity multiplied by zero is zero, and since we can perform multiplication inany order zero times any quantity is zero. So 8×0 = 0×8 = 0 21 21 0× = ×0=0 456 456 3×a×0 = 0×a×3 = 0Zero divided by any quantity is zero. However the operation of dividing by zero isNOT DEFINED. So 0 = 0 0 ÷ 20 = 0 0/2a = 0 8But 35 and 16b ÷ 0 are not defined 0OneMultiplying any number by 1 leaves it unchanged a×1= a = 1×a 5×1 = 5 = 1×5 478 478 ×1= 478 1098 = 1× 1098 1098 2
3. 3. Negative NumbersAddition and SubtractionLook at the diagram below which we will refer to as the number line. When wesubtract a positive number (or add a negative number) we move the pointer to theleft to get a smaller number. Represented below is the sum 3 − 5 = −2. .............................. .............................. ........ ........ ...... ...... ... .. ...... ..... ..... ..... .... .... . .... .... .... .... .... .. .... .... ... ... ... ... ... ... . ... ... . ... ... . - -4 -3 -2 -1 0 1 2 3 4 5 Similarly, when we add a positive number (or subtract a negative number) wemove the pointer to the right to get a larger number. The diagram below represents−3 + 7 = 4. ..................................... ..................................... ............. ............ ......... ........ ........ ........ ....... ...... ....... ...... ...... ...... .. . ...... ...... ..... ..... .. ...... ..... ..... ..... ..... .... .... .... . . .. .... .... . j ..... ..... .. . - -4 -3 -2 -1 0 1 2 3 4 5 Notice that subtracting a positive number and adding a negative number resultin the same action, a move to the left on the number line. Adding a positive numberand subtracting a negative number also result in the same action, a move to the righton the number line. So 10 − 4 gives the same result as 10 + −4 and 2 + 6 gives the same result as 2 − −6Multiplication and DivisionMultiplication and division of negative numbers is almost exactly the same as multi-plication of the counting numbers. The only difference is what happens to the sign.It is easy to see that 4 × −5 = −20 and logical to then assume that −4 × 5 = −20But what about −4 × −5? This is equal to 20. So we have the following: (+a) × (+b) = +ab e.g. 3×2 = 6 (+a) × (−b) = −ab e.g. 3 × −2 = −6 (−a) × (+b) = −ab e.g. −3 × 2 = −6 (−a) × (−b) = +ab e.g. −3 × −2 = 6 3
4. 4. Similarly a (+a) ÷ (+b) = + e.g. 8÷4 = 2 b a (+a) ÷ (−b) = − e.g. 8 ÷ −4 = −2 b a (−a) ÷ (+b) = − e.g. −8 ÷ 4 = −2 b a (−a) ÷ (−b) = + e.g. −8 ÷ −4 = 2 bGeneralising the above result we have: • Multiplying or dividing two quantities with the Same sign will give a Positive answer. • Multiplying or dividing two quantities with Unike signs will give a Negative answer.Powers of 10Our number system increases by powers of 10 as we move to the left and decreasesby powers of 10 as we move to the right. Given the number 123 456.78, the 1 tellsus how many 100,000’s there are, the 4 indicates the number of 100’s and the 7indicates the number of tenths. Similarly when we multiply two numbers togethersuch as 200 × 3000 we can use our knowledge of the place value system to representthis as (2 × 100) × (3 × 1000). Since the order is not important in multiplication (i.e.2 × 3 × 4 = 4 × 3 × 2 ) we can rewrite this as: 200 × 3000 = (2 × 100) × (3 × 1000) = (2 × 3) × (100 × 1000) = 6 × 100000 = 600000Some calculator informationSometimes calculators give us the answer to questions in something called scientificnotation. Scientific notation is a way of writing very big or very small numbers. Itconsists of writing the number in two parts, the first part is a number between 1 and10 and the second part is a power of 10. So 2 million (2 000 000) would be writtenas 2 × 106. Similarly 0.00345 would be written as 3.45 × 10−3. Our calculators willrevert to using scientific notation if the answer calculated is very small or very large.Try this exercise 1 ÷ 987654321 4
5. 5. Notice that the answer given by most of your calculators (there may be small varia-tions here depending on the model calculator you have) is 1 ÷ 987654321 = 1.0125−09What this actually means is that the answer to the sum is 1.0125 × 10−9 in otherwords 0.0000000010125. Keep a watch out in that top right hand corner for this kindof notation.Distributive LawSuppose you have four children and each child requires a pencil case (\$2.50), a ruler(\$1.25), an exercise book (\$2.25) and a set of coloured pencils (\$12) for school. One way we can calculate the cost is by multiplying each item by 4 and addingthe result: (4 × \$2.50) + (4 × \$1.25) + (4 × \$2.25) + (4 × \$12) = \$10 + \$5 + \$9 + \$48 = \$72This took quite a bit of effort so perhaps there might be a simpler method. Why notwork out how much it will cost for one child and then multiply by 4? (\$2.50 + \$1.25 + \$2.25 + \$12) × 4 = \$18 × 4 = \$72The fact that 4 × (2.50 + 1.25 + 2.25 + 12) = 4 × 2.5 + 4 × 1.25 + 4 × 2.25 + 4 × 12is called the distributive law in mathematics. When we do a calculation involvingbrackets we generally do the calculation inside the brackets first. When that isnot possible, for example if you have an algebraic expression then you can use thedistributive law to expand the expression. Sometimes we leave out the times symbolwhen multiplying a bracket by a quantity, this is another example of mathematicalshorthand, so the above could be written 4(2.5 + 1.25 + 2.25 + 12)Order of OperationsConsider the following situation. You have worked from 2 p.m. to 9 p.m. on a majorproposal. As you were required to complete it by the following day you can claimovertime. The rates from 9 a.m. to 5 p.m. are \$25 per hour and from 5 p.m. tomidnight rise to \$37.50 per hour. Mathematically, this can be expressed as 3 × \$25 + 4 × \$37.50. How much do youthink you earned? Which attempt below is most reasonable? 5
6. 6. Attempt 1 Attempt 2 3 × 25 + 4 × 37.50 3 × 25 + 4 × 37.50 = 75 + 4 × 37.50 = (3 × 25) + (4 × 37.50) = 79 × 37.50 = 75 + 150 = 2962.50 = 225 Mathematical expressions can be read by anyone regardless of their spoken lan-guage. In order to avoid confusion certain conventions (or accepted methods) mustbe followed. One of these is the order in which we carry out arithmetic. Brackets: Evaluate the expression inside the brackets first eg. (3 + 5) = 8 Indices: Evaluate the expressions raised to a power eg. (3 + 5)2 = 82 = 64 Division: Divide or multiply in order from left to right. Multiplication: eg. 6 × 5 ÷ 3 = 30 ÷ 3 = 10 Addition: Add or subtract in order from left to right. Subtraction: eg. 4 + 5 − 2 + 4 = 9 − 2 + 4 = 7 + 4 = 11 This can easily be remembered by the acronym BIDMAS.Some more calculator informationModern calculators have been programmed to observe the order of operations andalso come equipped with a bracket function. You just enter the brackets in theappropriate spots as you enter the calculation. Some older calculators with a bracketfunction will tell you how many brackets you have opened, they do this by havingon the display (01 or [01.Calculations involving DecimalsZeroes to the right of a decimal point with no digits following have no value. Thus 10.4 10.40 10.400000are all equal in value. There is however a concept of significant figures in the sciencesand when we look at the numbers above they are viewed slightly differently. Thishas to do with the significance of the figures, i.e. to what level of precision do wemeasure something. In that case 10.4 is telling us that we measured to the nearesttenth and 10.400000 is telling us that the measurement is to the nearest millionth. 6
7. 7. RoundingWhen we round decimals to a certain number of decimal places we are replacing thefigure we have with the one that is closest to it with that number of decimal places.An example: Round 1.25687 to 2 decimal places 1. Firstly look at the decimal place after the one you want to round to (in our example this would be the third decimal place) 2. If the number in the next decimal place is a 6,7,8 or 9, then you will be rounding up, so you add 1 to the number in the place you are interested in and you have rounded. In our example the number in the third place is a 6 so we round up. We change the 5 in the second place to a 6 and our rounded number is 1.26 3. If the number in the place after the one we are interested in is a 0,1,2,3 or 4 we round down, i.e. we just write the number out as it is to the required number of places. 4. If the number in the place after the one we are interested in is a 5, then we need to look at what follows it. Cover the number from the beginning to the place you are interested in, for example, suppose we are rounding 2.47568 to three decimal places we look at just the 568 and we ask is that closer to 500 or 600. Since its closer to 600 we get a rounded number of 2.476 5. If only a 5 follows the place we are interested in then different disciplines have different conventions for the rounding. You can either round up or down since 5 is exactly half way between 0 and 10.Addition and Subtraction of DecimalsTo add or subtract decimal numbers you should always remember to line up thedecimal point and then add or subtract normally. For example: 25 + 25.5 + .025 + 2.25 and 36.25 − 6.475 25.000 + 36.250 − 25.500 6.475 0.025 29.775 2.250 52.775 7
8. 8. Multiplication of DecimalsTwo decimals are multiplied together in the same way as two whole numbers are.Firstly, carry out the multiplication ignoring the decimal points. Then count up thenumber of digits after the decimal point in both numbers and add them together. Fi-nally return the decimal point to the appropriate position in the product by countingback from the right hand digit. For example Evaluate 2.6 × 0.005First multiply 26 by 5: 26 × 5 = 130 Now, count up the digits after the decimal point 1 + 3 = 4 Finally return the decimal point to the product four places back from the righthand digit, adding zeroes in front if necessary 4 . 0130Another way of looking at this is 1 1 2.6 × 0.005 = (26 × 1 10 ) × (5 × 1 1000 ) = (26 × 5) × × 10 1000 = (130 × 10000 ) = 1 130 10000 = 0.130As above we have 2.6 × .005 = 0.0130Division of DecimalsWhen dividing numbers with decimals by whole numbers, the only essential rule is toplace the decimal point in the answer exactly where it occurs in the decimal number.For example: 0.238 ÷ 7 = 0.034 0.034 7)0.238When the divisor (the number you are dividing by is also a decimal, the essentialrule is: Move the decimal point in the divisor enough places to the right so that itbecomes a whole number. Now move the decimal point in the dividend (the numberbeing divided) the same number of places to the right (adding zeroes if necessary).Finally carry out the division as in the example above. For example: 0.24 ÷ 0.3 = 2.4 ÷ 3 = 0.8 25 ÷ 0.05 = 2500 ÷ 5 = 500 8
9. 9. Calculations Involving FractionsChanging Mixed Numbers to Improper Fractions3 4 can be expressed as 19 since there are 15 fifths in 3 wholes plus 4 fifths. This can 5 5be worked out by multiplying the whole number by the denominator and adding thenumerator.Changing Improper Fractions to Mixed NumbersImproper fractions are fractions in which the numerator is larger than the denomi-nator. For example 12 . You can use the fraction key on your calculator to convert 5these to mixed numbers or you can divide by hand. For example: 12 ÷ 5 = 2 remainder 2The remainder can be expressed as the fraction 2 . So 5 12 5 = 22. 5Multiplying FractionsBefore performing any operations on fractions always represent it in the form a even bif this makes it an improper fraction. Once all fractions are in this form simplymultiply the numerators together and then multiply the denominators together. Forexample: 3 1 18 1 × 18 18 9 4 One half of 3 = × = = = =1 5 2 5 2×5 10 5 5Division of FractionsBefore performing any operations on fractions always represent it in the form a beven if this makes it an improper fraction. To divide by a fraction is the same asmultiplying by its reciprocal. The reciprocal of a fraction a is the fraction a . So b bthe reciprocal of 3 is 4 , the reciprocal of 1 is 2 which is just 2 and the reciprocal of 4 3 2 15 is 1 . So to divide by 2 is the same as to multiply by 3 . For example: 5 3 2 4 2 4 3 4×3 12 6 ÷ = × = = = 7 3 7 2 7×2 14 7Addition and Subtraction of FractionsTo add and subtract fractions you must be dealing with the same kind of fractions.In other words to add or subtract fractions you need to convert the fractions so thatboth fractions have the same denominator (i.e. the same number on the bottom). Two fractions are said to be equivalent if they represent the same part of onewhole. So 2 is equivalent to 4 since they represent the same amount. To find 3 6 9
10. 10. equivalent fractions you multiply the top and the bottom of the fraction by the sameamount. For example: 2 2×2 4 12 6 = = = = 7 7×2 14 42 21are all equivalent fractions. To add or subtract fractions you need to find equivalentfractions to each part of the sum so that all the fractions in the sum have the samedenominator and then you simply add or subtract the numerators (i.e. the topnumbers). So 2 1 2×3 1×7 6 7 6+7 13 + = + = + = = 7 3 7×3 3×7 21 21 21 21 3 1 3×3 1×4 9 4 9−4 5 − = − = − = = 4 3 4×3 3×4 12 12 12 12PercentagesPercentages are fractions with a denominator of 100. Often there will not be 100things or 100 people out of which to express a fraction or a percentage. When thisis the case you will need to find an equivalent fraction out of 100 by multiplying by100% which is the same as multiplying by 1. 1 1×5 5 1 1 100 = = = 5% or = × 100% = % = 5% 20 20 × 5 100 20 20 20 3 3 × 10 30 3 3 300 = = = 30% or = × 100% = % = 30% 10 10 × 10 100 10 10 10 2 2 × 20 40 2 2 200 = = = 40% or = × 100% = % = 40% 5 5 × 20 100 5 5 5 Therefore you can either find out how many times the denominator goes into 100and multiply the top and bottom of the fraction by this number or multiply thenumerator by 100, then divide through by the denominator. When you are asked to find a percentage you must find the fraction first. Forinstance if there are 15 men, 13 women and 22 children in a group, the fractionof men in the group is 15 since there are 50 = 15 + 13 + 22 people in the group 50altogether. You can either multiply by 100 % (if you have a calculator) or find theequivalent fraction out of 100. In both cases you will reach an answer of 30%. 10
11. 11. Interchanging Fractions, Decimals and PercentagesChanging Fractions to DecimalsUsing your calculator divide the top number (numerator) by the bottom number(denominator) to express a fraction as a decimal. If you do not have a calculator wecan do it manually. Be sure to put a decimal point after the numerator and add afew zeros. 3 can be expressed as 3.00 ÷ 4 = 0.75 4Changing Fractions to PercentagesExpress the fraction as a decimal (as shown above) then multiply the result by 100%.This can be done easily by shifting the decimal point two places to the right. 3 = 3.000 ÷ 8 = 0.375 = 37.5% 8When the denominator of the fraction to be converted goes evenly into 100, thepercentage can be found using equivalent fractions. 3 = 5 100 3 20 60 × = = 60% 5 20 100Converting Percentages to DecimalsIn order to convert a percentage into a decimal, divide be 100. This can be doneeasily by shifting the decimal point two places to the left. If there is no decimal pointbe sure to place a decimal point to the right of the whole number. 45.8% = 0.458 and 0.5% = 0.005 and 7% = 7.0% = 0.07Ordering Fractions, Decimals and PercentagesIn order to compare numbers they must all be presented in the same form withthe same number of decimal places. It is usually easiest to convert everything todecimals. Once you have done this, write the numbers underneath each other liningup the decimal points. Fill in any blanks with zeros. Compare the whole numberside first. If there is a match, then compare the fractional side. For example: Expressthe following numbers in order from smallest to largest. 3 46 0.5, 5.3%, 0.54, 47%, , , 3 5 100 11
12. 12. 1. Convert all the numbers to decimals: 0.5, 0.053, 0.54, 0.47, 0.6, 0.46, 3 2. Line the numbers up 0.5 0.053 0.54 0.47 0.6 0.46 3 3. Fill in all of the blanks with zeros 0.500 0.053 0.540 0.470 0.600 0.460 3.000 4. Compare the whole numbers. If there is a match, compare the fractional side. Since six of the numbers begin with a 0. we must compare the right hand side. Clearly 53 460 470 500 540 600 5. Express the numbers from smallest to largest 46 3 5.3%, , 47%, 0.5, 0.54, , 3 100 5RatioAny fraction can also be expressed as a ratio. 15 can be written as 15:50. However 50be careful of the wording of these types of questions. The ratio of men to the totalnumber in the group is 15:50 but the ratio of men to women is 15:35. 12
13. 13. Some common mathematical notationHere are a few of the shorthand symbols used in mathematics Sign Meaning Sign Meaning + add − minus or subtract × multiply ÷ divide = equals 2 square greater than less than ≥ greater than or equal to ≤ less than or equal to ≈ approximately equal to = not equal to √ square root ∴ therefore Another convention in maths is to leave out the multiplication sign if it will notlead to confusion, so, for example 2a means 2 × a and 4(1 + 2) means 4 × (1 + 2).Introductory AlgebraSubstitutionWhen we use algebra in maths it is often as a convenient form of shorthand. We wantto express a relationship that always holds true for certain things and so we representthese things by letters. For example the length of the perimeter of a rectangle is 2times the length plus 2 times the height. We can express this as 2l + 2h = P wherel represents the length, h represents the height and P represents the perimeter. Substitution is the process of putting numbers in for the letter representatives ofquantities. So, if the length of a rectangle is 4 and the height is 7 then l = 4 andh = 7 and P = 2 × 4 + 2 × 7 = 8 + 14 = 22. We have substituted l = 4 and h = 7into the equation. Another example: Let x = 5, y = 7 and find z when z = 3y + 4x2Then substituting we have z = 3 × 7 + 4 × (5)2 = 3 × 7 + 4 × 25 = 21 + 100 = 121Solving Algebraic EquationsWhen we are asked to solve an algebraic equation for z, say, we are being asked toget z on one side of the equation by itself and the rest of the information on the otherside. (Remember that the relationship still holds when both sides of the equation 13
14. 14. are multiplied or divided by the same thing or when the same quantity is added toor subtracted from each side). In this example we are going to solve for y. 3y + 4 = 19 3y + 4 − 4 = 19 − 4 (subtract 4 from each side of the equation) 3y = 15 3y ÷ 3 = 15 ÷ 3 (divide each side of the equation by 3) y = 5PlottingTo plot the following data on the average height of boys and girls at different ages x : Age (years) 2 3 4 5 6 7 8 9 y : Height (cms) 85 95 100 110 115 122 127 132 Firstly draw two straight lines at right angles to one another. It is customary tohave one horizontal line and the other vertical. The x variable is generally plottedalong the horizontal axis and the y variable along the vertical axis. The horizon-tal axis is usually the independant variable and the vertical axis is the dependantvariable. Next, you need to select a suitable scale for use on each of the two axes. Thescales on the axes can be different from each other. Look first at the x values. In theabove we are told that age is the x variable. Find the smallest and the largest value.The scale chosen has to be such that it will accommadate both of these values. Inthe above we have to fit in values for x between 2 and 9. Similarly for the y variablewe need to fit in values between 85 and 132. 160 120 Height (cms) 80 40 0 2 3 4 5 6 7 8 9 Age (years) 14
15. 15. Alternatively a break in a scale can be used if all the values to be shown lie withina small range. 140 120 This symbol shows Height (cms) that the scale has been ’broken’ i.e. 100 some values have been missed out .................... 80 .. ............................. ....................................... ................ - ... .. ... 0 .. 2 3 4 5 6 7 8 9 Age (years) The axes and scale chosen above will satisfy these requirements. At age 2 yearsthe child has a height of 85 cm. To plot this we go along the x-axis to where the 2is indicated and then go vertically to opposit where the 85 is on the y-axis. Similarly for a child of 3 years and a height of 95 cm. The shorthand wayof indicating these points is (2, 85) and (3, 95). They are plotted together in thediagram below. 140 120 Height (cms) 100 t t 80 . .... ............. . ... 0 . 2 3 4 5 6 7 8 9 Age (years) Plotting the remaining and labelling the display we get: Height of Children at Different Ages 140 t t 120 t t t Height (cms) t 100 t t 80 . .... . ............. .. ... 0 . 2 3 4 5 6 7 8 9 Age (years) 15
16. 16. Sometimes it is preferable to have the axes intersect at zero for both variables. Ifwe do that with the above data then we get the following graph. 160 t t t t 120 t t t Height (cms) t 80 40 0 1 2 3 4 5 6 7 8 9 Age (years) 16 |
# Prove that sec*2 theeta + cosec*2 theeta =sec*2 theeta x cosec*2 theeta?
Mar 1, 2018
Proved in explanation...
#### Explanation:
${\sec}^{2} \theta + \cos e {c}^{2} \theta$ can be written as
$\frac{1}{\cos} ^ 2 \theta + \frac{1}{\sin} ^ 2 \theta$
Taking lcm
$\frac{{\sin}^{2} \theta + {\cos}^{2} \theta}{{\sin}^{2} \theta \cdot {\cos}^{2} \theta}$
$\textcolor{b l u e}{\text{recall} {\sin}^{2} A + C o {s}^{2} A = 1}$
$\frac{1}{{\sin}^{2} \theta \cdot {\cos}^{2} \theta}$
$\frac{1}{\sin} ^ 2 \theta \cdot \frac{1}{\cos} ^ 2 \theta$
cosec^2 theta×sec^2 theta
Hence proved
Mar 1, 2018
See below.
#### Explanation:
${\sec}^{2} \left(\theta\right) + {\csc}^{2} \left(\theta\right) = {\sec}^{2} \left(\theta\right) {\csc}^{2} \left(\theta\right)$
Identities:
1) $\textcolor{red}{\boldsymbol{{\sec}^{2} \left(x\right) = \frac{1}{\cos} ^ 2 \left(x\right)}}$
2) $\textcolor{red}{\boldsymbol{{\csc}^{2} \left(x\right) = \frac{1}{\sin} ^ 2 \left(x\right)}}$
3) $\textcolor{red}{\boldsymbol{{\sin}^{2} x + {\cos}^{2} x = 1}}$
$L H S$
${\sec}^{2} \left(\theta\right) + {\csc}^{2} \left(\theta\right)$
Using identities 1 and 2:
$\frac{1}{\cos} ^ 2 \left(\theta\right) + \frac{1}{\sin} ^ 2 \left(\theta\right)$
$\frac{{\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right)}{{\cos}^{2} \left(\theta\right) {\sin}^{2} \left(\theta\right)}$
Using identity 3
$\frac{1}{{\cos}^{2} \left(\theta\right) {\sin}^{2} \left(\theta\right)}$
By identities 1 and 2:
${\sec}^{2} \left(\theta\right) {\csc}^{2} \left(\theta\right)$
$L H S \equiv R H S$ |
# Algebra help factoring
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## The Best Algebra help factoring
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How do you simplify -5+2(8/13)-6times2 using order of operations?
Jun 2, 2017
$- \frac{205}{13}$ or $- 15 \frac{10}{13}$
Explanation:
The order of operations can be remembered by the mnemonic device PEMDAS (often P lease E xcuse M y D ear A unt S ally). This acronym refers to the order of operations, which is:
1. Parenthesis
2. Exponents
3. Multiplication
4. Division
6. Subtraction
You are given
$- 5 + 2 \left(\frac{8}{13}\right) - 6 \times 2$
Parenthesis
You are given parenthesis, but there are no expressions to simplify. You can leave this as it is.
NOTE: You might be tempted to plug $8 \div 13$ in your calculator. Unfortunately, this gives you a fraction which repeats forever as $0. \overline{615384}$. This will give an answer which is not exact and so should be avoided in this problem.
Exponents
There are no exponents in this problem. Move on to the next letter.
Multiplication
You can now multiply terms
$- 5 + \frac{2 \times 8}{13} - 12 = - 5 + \frac{16}{13} - 12$
Division
If the fraction could be reduced, reducing factions counts as division. However, because the fraction is already in lowest terms, you can move on to the next step.
There are no two numbers you can add. Move on.
Subtraction
$- 5 + \frac{16}{13} - 12 = - \frac{5 \times 13}{13} + \frac{16}{13} - 12 = - \frac{65}{13} + \frac{16}{13} - 12$
$= - \frac{49}{13} - 12 = - \frac{49}{13} - \frac{12 \times 13}{13} = - \frac{49}{13} - \frac{156}{13}$
$= - \frac{205}{13}$
Jun 2, 2017
$= - 15 \frac{10}{13}$
Explanation:
Count the number of terms first, before you do any calculations.
Simplify each term to a single answer.
Within each term you do operations in the following order:
• brackets first
• then powers and roots
• multiply and divide
Each term will simplify to one answer - add or subtract at the end.
$\textcolor{red}{- 5} \textcolor{b l u e}{+ 2 \left(\frac{8}{13}\right)} \textcolor{g r e e n}{- 6 \times 2} \text{ } \leftarrow$ there are 3 terms
$= \textcolor{red}{- 5} \textcolor{b l u e}{+ \frac{16}{13}} \textcolor{g r e e n}{- 12} \text{ } \leftarrow$ add 3 terms
$= \textcolor{red}{- 5} \textcolor{b l u e}{+ 1 \frac{3}{13}} \textcolor{g r e e n}{- 12}$
$= 1 \frac{3}{13} - 17$
$= - 15 \frac{10}{13}$ |
Illustrative Mathematics Grade 8, Unit 8, Lesson 13: Cube Roots
Learning Targets:
• When I have a cube root, I can reason about which two whole numbers it is between.
Related Pages
Illustrative Math
Lesson 13: Cube Roots
Let’s compare cube roots.
Illustrative Math Unit 8.8, Lesson 13 (printable worksheets)
Lesson 13 Summary
Remember that square roots of whole numbers are defined as side lengths of squares. For example, √17 is the side length of a square whose area is 17. We define cube roots similarly, but using cubes instead of squares. The number ∛17, pronounced “the cube root of 17,” is the edge length of a cube which has a volume of 17.
We can approximate the values of cube roots by observing the whole numbers around it and remembering the relationship between cube roots and cubes. For example, ∛20 is between 2 and 3 since 23 = 8 and 33 = 27, and 20 is between 8 and 27. Similarly, since 100 is between 43 and 53, we know ∛100 is between 4 and 5. Many calculators have a cube root function which can be used to approximate the value of a cube root more precisely. Using our numbers from before, a calculator will show that ∛20 ≈ 2.7144 and that ∛20 ≈ 4.6416.
Also like square roots, most cube roots of whole numbers are irrational. The only time the cube root of a number is a whole number is when the original number is a perfect cube.
Lesson 13.1 True or False: Cubed
Decide if each statement is true or false.
Lesson 13.2 Cube Root Values
What two whole numbers does each cube root lie between? Be prepared to explain your reasoning.
Lesson 13.3 Card Sort: Rooted in the Number Line
The numbers x, y, and z are positive, and:
1. Plot x, y, and z on the number line. Be prepared to share your reasoning with the class.
2. Plot -∛2 on the number line.
Diego knows that 82 = 64 and that 43 = 64. He says that this means the following are all true:
• √64 = 8
• ∛64 = 4
• √-64 = -8
• ∛-64 = -4
√64 = 8 is true because 82 = 64
∛64 = 4 is true because 43 = 64
√-64 = -8 is not true because (-8)2 = 64 and not -64
∛-64 = -4 is true because (-4)3 = -64
Lesson 13 Practice Problems
1. Find the positive solution to each equation. If the solution is irrational, write the solution using square root or cube root notation.
2. For each cube root, find the two whole numbers that it lies between.
3. Order the following values from least to greatest:
4. Find the value of each variable, to the nearest tenth.
5. A standard city block in Manhattan is a rectangle measuring 80 m by 270 m. A resident wants to get from one corner of a block to the opposite corner of a block that contains a park. She wonders about the difference between cutting across the diagonal through the park compared to going around the park, along the streets. How much shorter would her walk be going through the park? Round your answer to the nearest meter.
The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
## 7.1 Sampling bowl activity
### 7.1.1 What proportion of this bowl’s balls are red?
Take a look at the bowl in Figure 7.1. It has a certain number of red and a certain number of white balls all of equal size. Furthermore, it appears the bowl has been mixed beforehand, as there does not seem to be any coherent pattern to the spatial distribution of the red and white balls.
Let’s now ask ourselves, what proportion of this bowl’s balls are red?
One way to answer this question would be to perform an exhaustive count: remove each ball individually, count the number of red balls and the number of white balls, and divide the number of red balls by the total number of balls. However, this would be a long and tedious process.
### 7.1.2 Using the shovel once
Instead of performing an exhaustive count, let’s insert a shovel into the bowl as seen in Figure 7.2. Using the shovel, let’s remove $$5 \cdot 10 = 50$$ balls, as seen in Figure 7.3.
Observe that 17 of the balls are red and thus 0.34 = 34% of the shovel’s balls are red. We can view the proportion of balls that are red in this shovel as a guess of the proportion of balls that are red in the entire bowl. While not as exact as doing an exhaustive count of all the balls in the bowl, our guess of 34% took much less time and energy to make.
However, say, we started this activity over from the beginning. In other words, we replace the 50 balls back into the bowl and start over. Would we remove exactly 17 red balls again? In other words, would our guess at the proportion of the bowl’s balls that are red be exactly 34% again? Maybe?
What if we repeated this activity several times following the process shown in Figure 7.4? Would we obtain exactly 17 red balls each time? In other words, would our guess at the proportion of the bowl’s balls that are red be exactly 34% every time? Surely not. Let’s repeat this exercise several times with the help of 33 groups of friends to understand how the value differs with repetition.
### 7.1.3 Using the shovel 33 times
Each of our 33 groups of friends will do the following:
• Use the shovel to remove 50 balls each.
• Count the number of red balls and thus compute the proportion of the 50 balls that are red.
• Return the balls into the bowl.
• Mix the contents of the bowl a little to not let a previous group’s results influence the next group’s.
Each of our 33 groups of friends make note of their proportion of red balls from their sample collected. Each group then marks their proportion of their 50 balls that were red in the appropriate bin in a hand-drawn histogram as seen in Figure 7.5.
Recall from Section 2.5 that histograms allow us to visualize the distribution of a numerical variable. In particular, where the center of the values falls and how the values vary. A partially completed histogram of the first 10 out of 33 groups of friends’ results can be seen in Figure 7.6.
Observe the following in the histogram in Figure 7.6:
• At the low end, one group removed 50 balls from the bowl with proportion red between 0.20 and 0.25.
• At the high end, another group removed 50 balls from the bowl with proportion between 0.45 and 0.5 red.
• However, the most frequently occurring proportions were between 0.30 and 0.35 red, right in the middle of the distribution.
• The shape of this distribution is somewhat bell-shaped.
Let’s construct this same hand-drawn histogram in R using your data visualization skills that you honed in Chapter 2. We saved our 33 groups of friends’ results in the tactile_prop_red data frame included in the moderndive package. Run the following to display the first 10 of 33 rows:
tactile_prop_red
# A tibble: 33 x 4
group replicate red_balls prop_red
<chr> <int> <int> <dbl>
1 Ilyas, Yohan 1 21 0.42
2 Morgan, Terrance 2 17 0.34
3 Martin, Thomas 3 21 0.42
4 Clark, Frank 4 21 0.42
5 Riddhi, Karina 5 18 0.36
6 Andrew, Tyler 6 19 0.38
7 Julia 7 19 0.38
8 Rachel, Lauren 8 11 0.22
9 Daniel, Caroline 9 15 0.3
10 Josh, Maeve 10 17 0.34
# … with 23 more rows
Observe for each group that we have their names, the number of red_balls they obtained, and the corresponding proportion out of 50 balls that were red named prop_red. We also have a replicate variable enumerating each of the 33 groups. We chose this name because each row can be viewed as one instance of a replicated (in other words repeated) activity: using the shovel to remove 50 balls and computing the proportion of those balls that are red.
Let’s visualize the distribution of these 33 proportions using geom_histogram() with binwidth = 0.05 in Figure 7.7. This is a computerized and complete version of the partially completed hand-drawn histogram you saw in Figure 7.6. Note that setting boundary = 0.4 indicates that we want a binning scheme such that one of the bins’ boundary is at 0.4. This helps us to more closely align this histogram with the hand-drawn histogram in Figure 7.6.
ggplot(tactile_prop_red, aes(x = prop_red)) +
geom_histogram(binwidth = 0.05, boundary = 0.4, color = "white") +
labs(x = "Proportion of 50 balls that were red",
title = "Distribution of 33 proportions red")
### 7.1.4 What did we just do?
What we just demonstrated in this activity is the statistical concept of sampling. We would like to know the proportion of the bowl’s balls that are red. Because the bowl has a large number of balls, performing an exhaustive count of the red and white balls would be time-consuming. We thus extracted a sample of 50 balls using the shovel to make an estimate. Using this sample of 50 balls, we estimated the proportion of the bowl’s balls that are red to be 34%.
Moreover, because we mixed the balls before each use of the shovel, the samples were randomly drawn. Because each sample was drawn at random, the samples were different from each other. Because the samples were different from each other, we obtained the different proportions red observed in Figure 7.7. This is known as the concept of sampling variation.
The purpose of this sampling activity was to develop an understanding of two key concepts relating to sampling:
1. Understanding the effect of sampling variation.
2. Understanding the effect of sample size on sampling variation.
In Section 7.2, we’ll mimic the hands-on sampling activity we just performed on a computer. This will allow us not only to repeat the sampling exercise much more than 33 times, but it will also allow us to use shovels with different numbers of slots than just 50.
Afterwards, we’ll present you with definitions, terminology, and notation related to sampling in Section 7.3. As in many disciplines, such necessary background knowledge may seem inaccessible and even confusing at first. However, as with many difficult topics, if you truly understand the underlying concepts and practice, practice, practice, you’ll be able to master them.
To tie the contents of this chapter to the real world, we’ll present an example of one of the most recognizable uses of sampling: polls. In Section 7.4 we’ll look at a particular case study: a 2013 poll on then U.S. President Barack Obama’s popularity among young Americans, conducted by Kennedy School’s Institute of Politics at Harvard University. To close this chapter, we’ll generalize the “sampling from a bowl” exercise to other sampling scenarios and present a theoretical result known as the Central Limit Theorem.
Learning check
(LC7.1) Why was it important to mix the bowl before we sampled the balls?
(LC7.2) Why is it that our 33 groups of friends did not all have the same numbers of balls that were red out of 50, and hence different proportions red? |
# Resolve each of the following quadratic trinomial into factor:
Question:
Resolve each of the following quadratic trinomial into factor:
6x2 − 5xy − 6y2
Solution:
The given expression is $6 \mathrm{x}^{2}-5 \mathrm{xy}-6 \mathrm{y}^{2}$. (Coefficient of $\mathrm{x}^{2}=6$, coefficient of $\mathrm{x}=-5 \mathrm{y}$ and constant term $=-6 \mathrm{y}^{2}$ )
We will split the coefficient of $\mathrm{x}$ into two parts such that their sum is $-5 \mathrm{y}$ and their product equals the product of the coefficient of $\mathrm{x}^{2}$ and the constant term, i.e., $6 \times\left(-6 \mathrm{y}^{2}\right)=-36 \mathrm{y}^{2} .$
Now,
$(-9 y)+4 y=-5 y$
and
$(-9 y) \times 4 y=-36 y^{2}$
Replacing the middle term $-5 x y$ by $-9 x y+4 x y$, we get:
$6 x^{2}-5 x y-6 y^{2}=6 x^{2}-9 x y+4 x y-6 y^{2}$
$=\left(6 x^{2}-9 x y\right)+\left(4 x y-6 y^{2}\right)$
$=3 x(2 x-3 y)+2 y(2 x-3 y)$
$=(3 x+2 y)(2 x-3 y)$ |
# 3.2 Domain and range (Page 8/11)
Page 8 / 11
## Verbal
Why does the domain differ for different functions?
The domain of a function depends upon what values of the independent variable make the function undefined or imaginary.
How do we determine the domain of a function defined by an equation?
Explain why the domain of $\text{\hspace{0.17em}}f\left(x\right)=\sqrt[3]{x}\text{\hspace{0.17em}}$ is different from the domain of $\text{\hspace{0.17em}}f\left(x\right)=\sqrt[]{x}.$
There is no restriction on $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}f\left(x\right)=\sqrt[3]{x}\text{\hspace{0.17em}}$ because you can take the cube root of any real number. So the domain is all real numbers, $\text{\hspace{0.17em}}\left(-\infty ,\infty \right).\text{\hspace{0.17em}}$ When dealing with the set of real numbers, you cannot take the square root of negative numbers. So $\text{\hspace{0.17em}}x$ -values are restricted for $\text{\hspace{0.17em}}f\left(x\right)=\sqrt[]{x}\text{\hspace{0.17em}}$ to nonnegative numbers and the domain is $\text{\hspace{0.17em}}\left[0,\infty \right).$
When describing sets of numbers using interval notation, when do you use a parenthesis and when do you use a bracket?
How do you graph a piecewise function?
Graph each formula of the piecewise function over its corresponding domain. Use the same scale for the $\text{\hspace{0.17em}}x$ -axis and $\text{\hspace{0.17em}}y$ -axis for each graph. Indicate inclusive endpoints with a solid circle and exclusive endpoints with an open circle. Use an arrow to indicate $\text{\hspace{0.17em}}-\infty \text{\hspace{0.17em}}$ or Combine the graphs to find the graph of the piecewise function.
## Algebraic
For the following exercises, find the domain of each function using interval notation.
$f\left(x\right)=-2x\left(x-1\right)\left(x-2\right)$
$f\left(x\right)=5-2{x}^{2}$
$\left(-\infty ,\infty \right)$
$f\left(x\right)=3\sqrt{x-2}$
$f\left(x\right)=3-\sqrt{6-2x}$
$\left(-\infty ,3\right]$
$f\left(x\right)=\sqrt{4-3x}$
$\begin{array}{l}\\ f\left(x\right)=\sqrt[]{{x}^{2}+4}\end{array}$
$\left(-\infty ,\infty \right)$
$f\left(x\right)=\sqrt[3]{1-2x}$
$f\left(x\right)=\sqrt[3]{x-1}$
$\left(-\infty ,\infty \right)$
$f\left(x\right)=\frac{9}{x-6}$
$f\left(x\right)=\frac{3x+1}{4x+2}$
$\left(-\infty ,-\frac{1}{2}\right)\cup \left(-\frac{1}{2},\infty \right)$
$f\left(x\right)=\frac{\sqrt{x+4}}{x-4}$
$f\left(x\right)=\frac{x-3}{{x}^{2}+9x-22}$
$\left(-\infty ,-11\right)\cup \left(-11,2\right)\cup \left(2,\infty \right)$
$f\left(x\right)=\frac{1}{{x}^{2}-x-6}$
$f\left(x\right)=\frac{2{x}^{3}-250}{{x}^{2}-2x-15}$
$\left(-\infty ,-3\right)\cup \left(-3,5\right)\cup \left(5,\infty \right)$
$\frac{5}{\sqrt{x-3}}$
$\frac{2x+1}{\sqrt{5-x}}$
$\left(-\infty ,5\right)$
$f\left(x\right)=\frac{\sqrt{x-4}}{\sqrt{x-6}}$
$f\left(x\right)=\frac{\sqrt{x-6}}{\sqrt{x-4}}$
$\left[6,\infty \right)$
$f\left(x\right)=\frac{x}{x}$
$f\left(x\right)=\frac{{x}^{2}-9x}{{x}^{2}-81}$
$\left(-\infty ,-9\right)\cup \left(-9,9\right)\cup \left(9,\infty \right)$
Find the domain of the function $\text{\hspace{0.17em}}f\left(x\right)=\sqrt{2{x}^{3}-50x}\text{\hspace{0.17em}}$ by:
1. using algebra.
2. graphing the function in the radicand and determining intervals on the x -axis for which the radicand is nonnegative.
## Graphical
For the following exercises, write the domain and range of each function using interval notation.
domain: $\text{\hspace{0.17em}}\left(2,8\right],\text{\hspace{0.17em}}$ range $\text{\hspace{0.17em}}\left[6,8\right)\text{\hspace{0.17em}}$
domain: range:
domain: range: $\text{\hspace{0.17em}}\left[0,2\right]$
domain: $\text{\hspace{0.17em}}\left(-\infty ,1\right],\text{\hspace{0.17em}}$ range: $\text{\hspace{0.17em}}\left[0,\infty \right)\text{\hspace{0.17em}}$
domain: $\text{\hspace{0.17em}}\left[-6,-\frac{1}{6}\right]\cup \left[\frac{1}{6},6\right];\text{\hspace{0.17em}}$ range: $\text{\hspace{0.17em}}\left[-6,-\frac{1}{6}\right]\cup \left[\frac{1}{6},6\right]\text{\hspace{0.17em}}$
domain: range: $\text{\hspace{0.17em}}\left[0,\infty \right)\text{\hspace{0.17em}}$
For the following exercises, sketch a graph of the piecewise function. Write the domain in interval notation.
$f\left(x\right)=\left\{\begin{array}{lll}x+1\hfill & \text{if}\hfill & x<-2\hfill \\ -2x-3\hfill & \text{if}\hfill & x\ge -2\hfill \end{array}$
$f\left(x\right)=\left\{\begin{array}{lll}2x-1\hfill & \text{if}\hfill & x<1\hfill \\ 1+x\hfill & \text{if}\hfill & x\ge 1\hfill \end{array}$
domain: $\text{\hspace{0.17em}}\left(-\infty ,\infty \right)$
$f\left(x\right)=\left\{\begin{array}{c}x+1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x<0\\ x-1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x>0\end{array}$
$f\left(x\right)=\left\{\begin{array}{ccc}3& \text{if}& x<0\\ \sqrt{x}& \text{if}& x\ge 0\end{array}$
domain: $\text{\hspace{0.17em}}\left(-\infty ,\infty \right)$
$f\left(x\right)=\left\{\begin{array}{r}\hfill \begin{array}{r}\hfill {x}^{2}\\ \hfill x+2\end{array}\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{l}\text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x<0\hfill \\ \text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\ge 0\hfill \end{array}$
domain: $\text{\hspace{0.17em}}\left(-\infty ,\infty \right)$
$f\left(x\right)=\left\{\begin{array}{ccc}x+1& \text{if}& x<1\\ {x}^{3}& \text{if}& x\ge 1\end{array}$
$f\left(x\right)=\left\{\begin{array}{c}|x|\\ 1\end{array}\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x<2\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\ge 2\hfill \end{array}$
domain: $\text{\hspace{0.17em}}\left(-\infty ,\infty \right)$
## Numeric
For the following exercises, given each function $f,$ evaluate $f\left(-3\right),\text{\hspace{0.17em}}f\left(-2\right),\text{\hspace{0.17em}}f\left(-1\right),$ and $f\left(0\right).$
$f\left(x\right)=\left\{\begin{array}{lll}x+1\hfill & \text{if}\hfill & x<-2\hfill \\ -2x-3\hfill & \text{if}\hfill & x\ge -2\hfill \end{array}$
$\begin{array}{cccc}f\left(-3\right)=1;& f\left(-2\right)=0;& f\left(-1\right)=0;& f\left(0\right)=0\end{array}$
For the following exercises, given each function $\text{\hspace{0.17em}}f,\text{\hspace{0.17em}}$ evaluate $f\left(-1\right),\text{\hspace{0.17em}}f\left(0\right),\text{\hspace{0.17em}}f\left(2\right),\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(4\right).$
$f\left(x\right)=\left\{\begin{array}{lll}7x+3\hfill & \text{if}\hfill & x<0\hfill \\ 7x+6\hfill & \text{if}\hfill & x\ge 0\hfill \end{array}$
$\begin{array}{cccc}f\left(-1\right)=-4;& f\left(0\right)=6;& f\left(2\right)=20;& f\left(4\right)=34\end{array}$
$f\left(x\right)=\left\{\begin{array}{ccc}{x}^{2}-2& \text{if}& x<2\\ 4+|x-5|& \text{if}& x\ge 2\end{array}$
$f\left(x\right)=\left\{\begin{array}{ccc}5x& \text{if}& x<0\\ 3& \text{if}& 0\le x\le 3\\ {x}^{2}& \text{if}& x>3\end{array}$
$\begin{array}{cccc}f\left(-1\right)=-5;& f\left(0\right)=3;& f\left(2\right)=3;& f\left(4\right)=16\end{array}$
For the following exercises, write the domain for the piecewise function in interval notation.
domain: $\text{\hspace{0.17em}}\left(-\infty ,1\right)\cup \left(1,\infty \right)$
$f\left(x\right)=\left\{\begin{array}{c}2x-3\\ -3{x}^{2}\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{c}\text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x<0\\ \text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\ge 2\end{array}$
## Technology
Graph $\text{\hspace{0.17em}}y=\frac{1}{{x}^{2}}\text{\hspace{0.17em}}$ on the viewing window $\text{\hspace{0.17em}}\left[-0.5,-0.1\right]\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left[0.1,0.5\right].\text{\hspace{0.17em}}$ Determine the corresponding range for the viewing window. Show the graphs.
window: $\text{\hspace{0.17em}}\left[-0.5,-0.1\right];\text{\hspace{0.17em}}$ range:
window: range:
Graph $\text{\hspace{0.17em}}y=\frac{1}{x}\text{\hspace{0.17em}}$ on the viewing window $\text{\hspace{0.17em}}\left[-0.5,-0.1\right]\text{\hspace{0.17em}}$ and Determine the corresponding range for the viewing window. Show the graphs.
## Extension
Suppose the range of a function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ is What is the range of $\text{\hspace{0.17em}}|f\left(x\right)|?$
Create a function in which the range is all nonnegative real numbers.
Create a function in which the domain is $\text{\hspace{0.17em}}x>2.$
Many answers. One function is $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{\sqrt{x-2}}.$
## Real-world applications
The height $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ of a projectile is a function of the time $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ it is in the air. The height in feet for $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ seconds is given by the function $h\left(t\right)=-16{t}^{2}+96t.$ What is the domain of the function? What does the domain mean in the context of the problem?
The domain is it takes 6 seconds for the projectile to leave the ground and return to the ground
The cost in dollars of making $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ items is given by the function $\text{\hspace{0.17em}}C\left(x\right)=10x+500.$
1. The fixed cost is determined when zero items are produced. Find the fixed cost for this item.
2. What is the cost of making 25 items?
3. Suppose the maximum cost allowed is \$1500. What are the domain and range of the cost function, $\text{\hspace{0.17em}}C\left(x\right)?$
#### Questions & Answers
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
Shivam Reply
give me the waec 2019 questions
Aaron Reply
the polar co-ordinate of the point (-1, -1)
Sumit Reply
prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x
Rockstar Reply
tanh`(x-iy) =A+iB, find A and B
Pankaj Reply
B=Ai-itan(hx-hiy)
Rukmini
what is the addition of 101011 with 101010
Branded Reply
If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
Jack
extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero
archana Reply
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
Kc Reply
1+cos²A/cos²A=2cosec²A-1
Ramesh Reply
test for convergence the series 1+x/2+2!/9x3
success Reply
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
Lhorren Reply
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
jancy Reply
answer
Ajith
exponential series
Naveen
what is subgroup
Purshotam Reply
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
Macmillan Reply
e power cos hyperbolic (x+iy)
Vinay Reply
10y
Michael
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# Table of 9
If you have already memorized the table of 8 and you want to learn how to multiply by nineBelow you have compiled some very valuable material that will facilitate your learning without hesitation.
The multiplication table of nine is one of the most complicated but with our tricks, it will be much easier. Don't miss it.
## How to learn the table of 9
Imagine that you have the table of 9 in front of you and you only need to know the result of multiplying nine by each number from 1 to 10. To solve it, you have to write two columns of numbers as follows:
• First column: write from 0 to 9.
• Second column: write from 9 to 0.
If you've done it right, it should look something like this:
• 9 x 1 = 0 9
• 9 x 2 = 1 8
• 9 x 3 = 2 7
• 9 x 4 = 3 6
• 9 x 5 = 4 5
• 9 x 6 = 5 4
• 9 x 7 = 6 3
• 9 x 8 = 7 2
• 9 x 9 = 8 1
• 9 x 10 = 9 0
With this simple technique, you can write the multiplication table of 9 in a matter of a few seconds.
## Multiplication table of 9
If you need a much more complete multiplication table of 9 and that collects a greater number of multiplications, below you have one that goes from 0 to the number 100.
You can save or print this table to have it always at hand.
9 x 0 = 0 9 x 20 = 180 9 x 40 = 360 9 x 60 = 540 9 x 80 = 720 9 x 1 = 9 9 x 21 = 189 9 x 41 = 369 9 x 61 = 549 9 x 81 = 729 9 x 2 = 18 9 x 22 = 198 9 x 42 = 378 9 x 62 = 558 9 x 82 = 738 9 x 3 = 27 9 x 23 = 207 9 x 43 = 387 9 x 63 = 567 9 x 83 = 747 9 x 4 = 36 9 x 24 = 216 9 x 44 = 396 9 x 64 = 576 9 x 84 = 756 9 x 5 = 45 9 x 25 = 225 9 x 45 = 405 9 x 65 = 585 9 x 85 = 765 9 x 6 = 54 9 x 26 = 234 9 x 46 = 414 9 x 66 = 594 9 x 86 = 774 9 x 7 = 63 9 x 27 = 243 9 x 47 = 423 9 x 67 = 603 9 x 87 = 783 9 x 8 = 72 9 x 28 = 252 9 x 48 = 432 9 x 68 = 612 9 x 88 = 792 9 x 9 = 81 9 x 29 = 261 9 x 49 = 441 9 x 69 = 621 9 x 89 = 801 9 x 10 = 90 9 x 30 = 270 9 x 50 = 450 9 x 70 = 630 9 x 90 = 810 9 x 11 = 99 9 x 31 = 279 9 x 51 = 459 9 x 71 = 639 9 x 91 = 819 9 x 12 = 108 9 x 32 = 288 9 x 52 = 468 9 x 72 = 648 9 x 92 = 828 9 x 13 = 117 9 x 33 = 297 9 x 53 = 477 9 x 73 = 657 9 x 93 = 837 9 x 14 = 126 9 x 34 = 306 9 x 54 = 486 9 x 74 = 666 9 x 94 = 846 9 x 15 = 135 9 x 35 = 315 9 x 55 = 495 9 x 75 = 675 9 x 95 = 855 9 x 16 = 144 9 x 36 = 324 9 x 56 = 504 9 x 76 = 684 9 x 96 = 864 9 x 17 = 153 9 x 37 = 333 9 x 57 = 513 9 x 77 = 693 9 x 97 = 873 9 x 18 = 162 9 x 38 = 342 9 x 58 = 522 9 x 78 = 702 9 x 98 = 882 9 x 19 = 171 9 x 39 = 351 9 x 59 = 531 9 x 79 = 711 9 x 99 = 891 9 x 100 = 900
## Image of the table of 9
If you prefer a image of the 9 times table to save it more easily, then you have the same as the previous one but in photo format so you can do with it whatever you want.
## Song of the table of 9
We finish the section dedicated to the multiplication table for the number nine with a song designed for the little ones of the house.
Learn the multiplication tables it is always easier if there is a song that makes it easier to remember the results, and with the one you have in this section, they are sure to learn it much better. |
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Lesson 7: Intro to inequalities with variables
# Graphing inequalities review
Review graphing inequalities with variables on number lines, and then try some practice problems.
## Inequalities
Inequalities show the relation between two expressions that are not equal.
Below are some examples of inequalities:
$9>7$
$x<5$
### Inequalities symbols
SymbolMeaning
$>$Greater than
$\underset{―}{>}$Greater than or equal to
$<$Less than
$\underset{―}{<}$Less than or equal to
## Graphing inequalities with variables
We can use a number line to show the possible solutions to an inequality.
Example 1: $x>4$
An inequality like $x>4$ tells us that $x$ can be any value greater than $4$.
We can show this on a number line by putting an open circle on $4$ and shading the numbers that are greater than $4$.
Example 2:
If we have either the $\underset{―}{>}$ or $\underset{―}{<}$ symbol in our inequality, we shade in the circle to show that the variable may be equal to that number.
For example, is graphed as follows:
This number line shows that $y$ is either equal to $3$ or less than $3$.
## Practice
Problem 1
Choose the inequality that represents the following graph.
Want to try more problems like this? Check out these exercises:
Inequality from graph
Plotting inequalities
## Want to join the conversation?
• ok so umm we have to find the inequalities by know what's less than and greater than right?
• Yes, does this mean you get the two confused? If you make the sign with your left hand (<), left is less. If you make the sign with your right hand (>), Tony the Tiger says right is Grrrrrrrrreater.
• The questions were easy, but didn't really make me think. And if i do get those my teacher should only assign like 3
• Most of the questions were easy however my teacher loves to assign few but very tricky questions.
• i dont understand why circles need to be full or hollow
• Full circle means we include the number:
X ≥ 3 This means X can be 3 OR greater
Hollow circle means we do NOT include the number:
X > 3 This means X can ONLY be greater than 3
• what is this usefull for my life? exept for not loosing maths but WHY
• We never know. Don't waste your time with those questions. You do not study to pass the maths. You study to make your brain stronger and faster. When the future problem comes, you will be able to decide if inequalities will be needed or not and will solve then.
• Whats a good way to memorize which way the line on the number line goes? I keep forgetting.
• If the variable is on the left, the inequality tells you which way to draw the line. For example:
x<6
Notice the inequality is pointing to the left, so your line goes to the left.
x>6
Notice the inequality points to the right, the line goes to the right.
Alternatively, you need to know that the smaller numbers are on the left of the number line and the larger numbers are to the right. A common tip used to help students remember what each inequality symbol means it to think of the symbol as the mouth of a hungry alligator. Its open mouth will always face the larger value.
6>x
Notice, the alligator wants to eat the 6 so it is larger than x. So, x must be numbers smaller than 6 and your line would get drawn to the left.
6<x
Notice, the alligator wants to eat the X, so the x is larger than 6 and your line needs to be to the right of the 6.
Hope this helps.
• I dont understand some of the questions
• some of them are to trick you reread them
• 2x-7y=30 what is x
• I don't know if it's actually possible to find the value of x in this equation without knowing what y is, but it might be helpful to start by getting x by itself
2x - 7y = 30 add 7y to both sides
2x = 30 + 7y divide both sides by 2
x = (30 + 7y)/2
but wait, what if we replace x with (30+7y)/2, because they're equal
We'd get
2((30+7y)/2)-7y=30
30+7y-7y=30
30=30...
wait never-mind. That's technically correct, but also entirely useless. I still don't know what x is.
If anyone knows if it's possible to find x with only this information, then let me know. |
# 2.02 Factoring with sum and difference of cubes
Lesson
Previously, factoring binomials only covered the difference of squares. Can a binomial be factored if the terms are cubic?
### Sum and difference of cubes
Polynomials may be factored in various ways, including, but not limited to grouping or applying general patterns such as difference of squares, sum and difference of cubes, and perfect square trinomials. Algebra I covered all of the main factoring methods except one: sum and difference of cubes
A polynomial in the form $a^3+b^3$a3+b3 is a sum of cubes and in the form $a^3-b^3$a3b3 is a difference of cubes.
Let's start by looking at the general forms of these rules.
General forms for factoring binomials with cubic terms
Sum of two cubes: $a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)$a3+b3=(a+b)(a2ab+b2)
Difference of two cubes: $a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$a3b3=(ab)(a2+ab+b2)
Careful!
The expression $a^3+b^3$a3+b3 is not the same as $\left(a+b\right)^3$(a+b)3
$2^3+5^3$23+53 $\ne$≠ $\left(2+5\right)^3$(2+5)3 $8+125$8+125 $\ne$≠ $7^3$73 $133$133 $\ne$≠ $343$343
Now let's look at some examples and see this process in action.
#### Worked examples
##### Question 1
Factor: $x^3+125$x3+125
Think: Since the expression is a sum of two terms and each are perfect cubes ($x$xand $5$5, respectively), then apply:
$a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)$a3+b3=(a+b)(a2ab+b2)
Do: Using the perfect cubes, substitute the values into formula for $a$a and $b$b
$a^3+b^3$a3+b3 $=$= $\left(a+b\right)\left(a^2-ab+b^2\right)$(a+b)(a2−ab+b2) $x^3+5^3$x3+53 $=$= $\left(x+5\right)\left(x^2-x\times5+5^2\right)$(x+5)(x2−x×5+52) $x^3+125$x3+125 $=$= $\left(x+5\right)\left(x^2-5x+25\right)$(x+5)(x2−5x+25)
##### Question 2
Factor: $27x^3-1$27x31
Think: Since the expression is a difference of two terms and each are perfect cubes ($3x$3x and $1$1, respectively), then apply:
$a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$a3b3=(ab)(a2+ab+b2)
Do: Using the perfect cubes, substitute the values into formula for $a$a and $b$b
$a^3-b^3$a3−b3 $=$= $\left(a-b\right)\left(a^2+ab+b^2\right)$(a−b)(a2+ab+b2) $\left(3x\right)^3-1^3$(3x)3−13 $=$= $\left(3x-1\right)\left(\left(3x\right)^2+3x\times1+1^2\right)$(3x−1)((3x)2+3x×1+12) $27x^3-1$27x3−1 $=$= $\left(3x-1\right)\left(9x^2+3x+1\right)$(3x−1)(9x2+3x+1)
#### Practice questions
##### Question 3
Factor $x^3+512$x3+512.
##### Question 4
Factor $4m^3-32n^3$4m332n3.
### Outcomes
#### MGSE9-12.A.SSE.2
Use the structure of an expression to rewrite it in different equivalent forms. For example,see x^4 - y^4 as (x^2)^2 - (y^2)^2, thus recognizing it as a difference of squares that can be factored as (x^2 - y^2)(x^2 + y^2). |
# Simple Equations – Chapter 4/ CBSE Class 7 Maths
Simple Equations – Chapter 4/ CBSE Class 7 Mathematics is about the Model Questions that you can expect for Yearly Examination. Here you can find out practice problems for Class 7 Mathematics.
Simple Equations – Chapter 4/Extra Questions/ CBSE Class 7 Mathematics
Model Questions/Important Questions
Fill in the blanks:
1. If 3a = 42, then value of a is ————–
2. If 2x – 4 = 6, then value of x is ———-
3. The solution of the equation 3y + 3 = 12 is ————-
4. ‘Three times a number y subtracted from 12 is 5’ can be written as ———-
5. If 5p + 2 = 12, then the value of p is ————-
6. Solve 5p – 3 = 27
7. Solve 2(m + 4) = 8
8. Manu’s father’s age is 4 years more than 3 times Manu’s age. Find Manu’s age if his father is 55 years old.
9. The sum of 2 times of a number and 7 is 35. Find the number?
10. Solve:
a) 3p = 0
b) 3y + 4 = 25
11. Find the value of y for which the expressions y – 15 and 2y + 1 are equal.
12. If k + 3 = 8, find the value of 3k – 7?
13. 10 times a number is 3 less than 4 times the same number. Find the number?
14. Write the algebraic equation and solve; 3 more than 5 times a number is 18.
15. Solve
a) 3(p + 4) = 5
b) 2y – 5 = 5
1. a = 14
2. x = 5
3. y = 3
4. 12 – 3y = 5
5. p = 2
6. 5p – 3 = 27
5p = 27 + 3 = 30
p = 30/5 = 6
7. 2(m + 4) = 8
m + 4 = 8/2 = 4
m = 4 – 4 = 0
8. Let Manu’s age be x.
Given 3x + 4 = 55
3x = 55 – 4 = 51
x = 51/3 = 17
9. Let the number be y.
Given 2y + 7 = 35
2y = 35 – 7 = 28
y = 28/2 = 14
10. a) 3p = 0
p = 0/3 = 0
b) 3y + 4 = 25
3y = 25 – 4 = 21
y = 21/3 =7
11. Given y – 15 = 2y + 1
y – 2y = 1 + 15
-y = 16
y = -16
12. If k + 3 = 8, k = 8 – 3 = 5
Then 3k – 7 =3 x 5 – 7 = 15 – 7 = 8
13. The equation is 10x = 4x – 3
10x – 4x = -3
6x = -3
x = -3/6 = -1/2
14. Let the number be z.
The required equation is 3 + 5z = 18
15. a) 3( p + 4) = 5
3p + 12 = 5
3p = 5 – 12 = -7
p = -7/3
b) 2y – 5 = 5
2y = 5 + 5 = 10
y = 10/2 = 5 |
# 2021 AIME I Problems/Problem 5
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.
## Solution 1
Let the terms be $a-b$, $a$, and $a+b$. Then we want $(a-b)^2+a^2+(a+b)^2=ab^2$, or $3a^2+2b^2=ab^2$. Rearranging, we get $b^2=\frac{3a^2}{a-2}$. Simplifying further, $b^2=3a+6+\frac{12}{a-2}$. Looking at this second equation, since the right side must be an integer, $a-2$ must equal $\pm1, 2, 3, 4, 6, 12$. Looking at the first equation, we see $a>2$ since $b^2$ is positive. This means we must test $a=3, 4, 5, 6, 8, 14$. After testing these, we see that only $a=5$ and $a=14$ work which give $b=5$ and $b=7$ respectively. Thus the answer is $10+21=\boxed{031}$. ~JHawk0224
Note: If you don't understand the simplification of $b^2=\frac{3a^2}{a-2}$, you an actually just use synthetic division and arrive at the same place ~ Anonymous
## Solution 2
Let the common difference be $d$ and let the middle term be $x$. Then, we have that the sequence is $$x-d,~x,~x+d.$$ This means that the sum of the squares of the 3 terms of the sequence is $$(x-d)^2+x^2+(x+d)^2=x^2-2xd+d^2+x^2+x^2+2xd+d^2=3x^2+2d^2.$$ We know that this must be equal to $xd^2,$ so we can write that $$3x^2+2d^2=xd^2,$$ and it follows that $$3x^2-xd^2+2d^2=3x^2-\left(d^2\right)x+2d^2=0.$$
Now, we can treat $d$ as a constant and use the quadratic formula to get $$x=\frac{d^2\pm \sqrt{d^4-4(3)(2d^2)}}{6}.$$ We can factor pull $d^2$ out of the square root to get $$x=\frac{d^2\pm d\sqrt{d^2-24}}{6}.$$ Here, it is easy to test values of $d$. We find that $d=5$ and $d=7$ are the only positive integer values of $d$ that make $\sqrt{d^2-24}$ a positive integer.$^{*}$ $d=5$ gives $x=5$ and $x=\frac{10}{3}$, but we can ignore the latter. $d=7$ gives $x=14$, as well as a fraction which we can ignore.
Since $d=5,~x=5$ and $d=7, x=14$ are the only two solutions and we want the sum of the third terms, our answer is $(5+5)+(7+14)=10+21=\boxed{031}$. -BorealBear, minor edit by Kinglogic
$^*$To prove this, let $\sqrt{d^2-24} = k$, then $d^2-k^2=24$ which is $(d+k)(d-k)=24,$ then remembering that $d$ and $k$ are integers see if you can figure it out. -PureSwag
## Solution 3
Proceed as in solution 2, until we reach $$3x^2+2d^2=xd^2,$$ Write
$d^2=\frac{3x^2}{x-2}$, it follows that $x-2=3k^2$ for some (positive) integer k and $k \mid x$.
Taking both sides modulo $k$, $-2 \equiv 0 \pmod{k}$, so $k \mid 2 \rightarrow k=1,2$.
When $k=1$, we have $x=5$ and $d=5$. When $k=2$, we have $x=14$ and $d=7$. Summing the two cases, we have $10+21=\framebox{031}$.
-Ross Gao
## Solution 4 (Combining Solution 1 and Solution 3)
As in Solution 1, write the three integers in the sequence as $a-d$, $a$, and $a+d$.
Then the sum of the squares of the three integers is $(a-d)^2+a^2+(a+d)^2 = 3a^2+2d^2$.
Setting this equal to the middle term times the common difference squared, which is $ad^2$,
and solving for $d^2$ we get:
$3a^2+2d^2 = ad^2 \implies ad^2-2d^2 = 3a^2 \implies d^2(a-2) = 3a^2 \implies d^2 = \frac{3a^2}{a-2}$
The numerator has to be positive, so the denominator has to be positive too for the sequence
to be strictly increasing; that is, $a>2$.
For $\frac{3a^2}{a-2}$ to be a perfect square, $\frac{3}{a-2}$ must be a perfect square as well.
This means that $a-2$ is divisible by 3, and whatever left over is a perfect square.
We can express this as an equation: let the perfect square left over be $n^2$. Then:
$3n^2 = a-2$. Now when you divide the numerator and denominator by 3, you are left with
$d^2 = \frac{a^2}{n^2} \implies d = \frac{a}{n}$. Because the sequence is of integers, d must also be an
integer, which means that $n$ must divide $a$.
Taking the above equation we can solve for $a$: $3n^2 = a-2 \implies a = 3n^2+2$.
This means that $3n^2+2$ is divisible by $n$. $3n^2$ is automatically divisible by $n$, so
$2$ must be divisible by $n$. Then $n$ must be either of $\{1,2\}$. Plugging back into the equation,
$n = 1 \implies a = 5 \implies d = 5$, so $a+d = 5+5 = 10$.
$n = 2 \implies a = 14 \implies d = 7$, so $a+d = 14+7 = 21$.
Finally, $10+21 = \boxed{031}$
-KingRavi
## Solution 5
Following from previous solutions, we derive $3x^2+2a^2=xa^2.$ We divide both sides to get $3\left(\frac{x}{a}\right)^2+2=x.$ Since $x$ is an integer, $\frac{x}{a}$ must also be an integer, so we have $x=pa$, for some factor $p$. We then get $3p^2+2=pa.$ We then take this to modulo $p$, getting $2\equiv 0 \pmod p.$ The only possibilities for $p$ are therefore 1 and 2. We plug these into $3p^2+2=pa$, for $a=5$ and $x=5$, giving us the sequence $0,5,10$, or $2a=14$ and $x=14$, for the sequence $7,14,21.$ $10+21 = \boxed{031}.$
-RYang2
## See Also
2021 AIME I (Problems • Answer Key • Resources) Preceded byProblem 4 Followed byProblem 6 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# 1.4 Polynomials
Page 1 / 15
In this section students will:
• Identify the degree and leading coefficient of polynomials.
• Multiply polynomials.
• Use FOIL to multiply binomials.
• Perform operations with polynomials of several variables.
Earl is building a doghouse, whose front is in the shape of a square topped with a triangle. There will be a rectangular door through which the dog can enter and exit the house. Earl wants to find the area of the front of the doghouse so that he can purchase the correct amount of paint. Using the measurements of the front of the house, shown in [link] , we can create an expression that combines several variable terms, allowing us to solve this problem and others like it.
First find the area of the square in square feet.
$\begin{array}{ccc}\hfill A& =& {s}^{2}\hfill \\ & =& {\left(2x\right)}^{2}\hfill \\ & =& 4{x}^{2}\hfill \end{array}$
Then find the area of the triangle in square feet.
Next find the area of the rectangular door in square feet.
$\begin{array}{ccc}\hfill A& =& lw\hfill \\ & =& x\cdot 1\hfill \\ \hfill & =& x\hfill \end{array}$
The area of the front of the doghouse can be found by adding the areas of the square and the triangle, and then subtracting the area of the rectangle. When we do this, we get $\text{\hspace{0.17em}}4{x}^{2}+\frac{3}{2}x-x\text{\hspace{0.17em}}{\text{ft}}^{2},$ or $\text{\hspace{0.17em}}4{x}^{2}+\frac{1}{2}x\text{\hspace{0.17em}}$ ft 2 .
In this section, we will examine expressions such as this one, which combine several variable terms.
## Identifying the degree and leading coefficient of polynomials
The formula just found is an example of a polynomial , which is a sum of or difference of terms, each consisting of a variable raised to a nonnegative integer power. A number multiplied by a variable raised to an exponent, such as $\text{\hspace{0.17em}}384\pi ,$ is known as a coefficient . Coefficients can be positive, negative, or zero, and can be whole numbers, decimals, or fractions. Each product $\text{\hspace{0.17em}}{a}_{i}{x}^{i},$ such as $\text{\hspace{0.17em}}384\pi w,$ is a term of a polynomial . If a term does not contain a variable, it is called a constant .
A polynomial containing only one term, such as $\text{\hspace{0.17em}}5{x}^{4},$ is called a monomial . A polynomial containing two terms, such as $\text{\hspace{0.17em}}2x-9,$ is called a binomial . A polynomial containing three terms, such as $\text{\hspace{0.17em}}-3{x}^{2}+8x-7,$ is called a trinomial .
We can find the degree of a polynomial by identifying the highest power of the variable that occurs in the polynomial. The term with the highest degree is called the leading term because it is usually written first. The coefficient of the leading term is called the leading coefficient . When a polynomial is written so that the powers are descending, we say that it is in standard form.
## Polynomials
A polynomial is an expression that can be written in the form
${a}_{n}{x}^{n}+...+{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}$
Each real number a i is called a coefficient . The number $\text{\hspace{0.17em}}{a}_{0}\text{\hspace{0.17em}}$ that is not multiplied by a variable is called a constant . Each product $\text{\hspace{0.17em}}{a}_{i}{x}^{i}\text{\hspace{0.17em}}$ is a term of a polynomial . The highest power of the variable that occurs in the polynomial is called the degree of a polynomial. The leading term is the term with the highest power, and its coefficient is called the leading coefficient .
Given a polynomial expression, identify the degree and leading coefficient .
1. Find the highest power of x to determine the degree.
2. Identify the term containing the highest power of x to find the leading term.
3. Identify the coefficient of the leading term.
what is linear equation with one unknown 2x+5=3
-4
Joel
x=-4
Joel
x=-1
Joan
I was wrong. I didn't move all constants to the right of the equation.
Joel
x=-1
Cristian
what is the VA Ha D R X int Y int of f(x) =x²+4x+4/x+2 f(x) =x³-1/x-1
can I get help with this?
Wayne
Are they two separate problems or are the two functions a system?
Wilson
Also, is the first x squared in "x+4x+4"
Wilson
x^2+4x+4?
Wilson
thank you
Wilson
Wilson
f(x)=x square-root 2 +2x+1 how to solve this value
Wilson
what is algebra
The product of two is 32. Find a function that represents the sum of their squares.
Paul
if theta =30degree so COS2 theta = 1- 10 square theta upon 1 + tan squared theta
how to compute this 1. g(1-x) 2. f(x-2) 3. g (-x-/5) 4. f (x)- g (x)
hi
John
hi
Grace
what sup friend
John
not much For functions, there are two conditions for a function to be the inverse function: 1--- g(f(x)) = x for all x in the domain of f 2---f(g(x)) = x for all x in the domain of g Notice in both cases you will get back to the element that you started with, namely, x.
Grace
sin theta=3/4.prove that sec square theta barabar 1 + tan square theta by cosec square theta minus cos square theta
acha se dhek ke bata sin theta ke value
Ajay
sin theta ke ja gha sin square theta hoga
Ajay
I want to know trigonometry but I can't understand it anyone who can help
Yh
Idowu
which part of trig?
Nyemba
functions
Siyabonga
trigonometry
Ganapathi
differentiation doubhts
Ganapathi
hi
Ganapathi
hello
Brittany
Prove that 4sin50-3tan 50=1
False statement so you cannot prove it
Wilson
f(x)= 1 x f(x)=1x is shifted down 4 units and to the right 3 units.
f (x) = −3x + 5 and g (x) = x − 5 /−3
Sebit
what are real numbers
I want to know partial fraction Decomposition.
classes of function in mathematics
divide y2_8y2+5y2/y2
wish i knew calculus to understand what's going on 🙂
@dashawn ... in simple terms, a derivative is the tangent line of the function. which gives the rate of change at that instant. to calculate. given f(x)==ax^n. then f'(x)=n*ax^n-1 . hope that help.
Christopher
thanks bro
Dashawn
maybe when i start calculus in a few months i won't be that lost 😎
Dashawn |
Courses
# Short Answer Type Questions- Statistics Class 10 Notes | EduRev
## Class 10 : Short Answer Type Questions- Statistics Class 10 Notes | EduRev
The document Short Answer Type Questions- Statistics Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.
All you need of Class 10 at this link: Class 10
Q1. Find the mode of the data:
Marks 0-10 10-20 20-30 30-40 40-50 No. of students 3 12 32 20 6
Sol. Here, modal class is 20–30
f1 = 32, f2 = 20 and f0 = 12
Since, the lower limit of the modal class
l = 20
[∵ h = 10]
Q2. The percentage marks obtained by 100 students in an examination are given below:
Marks 30-35 35-40 40-45 45-50 50-55 55-60 60-65 Frequency 10 16 18 23 18 8 7
Find the median from the above data.
Sol. We have:
Marks Frequency cf 30-35 10 10 + 0 = 10 35-40 16 16 + 10 = 26 40-45 18 18 + 26 = 44 45-50 23 23 + 44 = 67 50-55 18 18 + 67 = 85 55-60 8 8 + 85 = 93 60-65 7 7 + 93 = 100
Here,
∴ The median class is 45−50, such that
l = 45, cf = 44, f = 23 and h = 5
Q3. Write a frequency distribution table for the following data:
Marks Above 0 Above 10 Above 20 Above 30 Above 40 Above 50 No. of students 30 28 21 15 10 0
Sol. Since,
30 − 28 = 2
28 − 21 = 7
21 − 15 = 6
15 − 10 = 5
10 − 0 =10
The required frequency distribution is:
Marks Number of students 0-10 2 10-20 7 20-30 6 30-40 5 40-50 10 Total 30
Q4. Find the median of the following data:
Class interval 0-20 20-40 40-60 60-80 80-100 100-120 Frequency 7 8 12 10 8 5
Sol.
Class Interval Frequency Cumulativefrequency 0-20 7 7 20-40 8 15 40-60 12 27 60-80 10 37 80-100 8 45 100-120 5 50 Total 50
∵ Median class is 40–60
∴ l = 40, f = 12, CF = 15 and h = 20
Since,
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