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This blog is managed by Song Hock Chye, author of Improve Your Thinking Skills in Maths (P1-P3 series), which is published and distributed by EPH.
## Sunday, February 21, 2010
### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q18
Ace Drama Company sold some tickets for a children's performance. It sold the same number of \$8 and \$5 tickets in Week 1 and collected a total of \$1664. In Week 2, it sold 96 \$8 and \$5 tickets. If the company collected \$632 more from the sale of \$8 tickets than the \$5 tickets in the two weeks, how many \$8 tickets were sold altogether?
Solution
Week 1
\$8 tickets --> 8 units
\$5 tickets --> 5 units
Total 13 units --> \$1664
1 unit --> \$1664 divided by 13 = \$128
Value of tickets sold in Week 1
\$8 tickets --> 8 x \$128 = \$1024
\$5 tickets --> 5 x \$128 = \$640
Difference between \$8 tickets and \$5 tickets in Week 1
\$1024 - \$640 = \$384
Difference between \$8 tickets and \$5 tickets in Week 2
\$632 - \$384 = \$248
If 96 tickets in Week 2 were equally sold between \$5 tickets and \$8 tickets, there will be 48 tickets each.
\$5 tickets --> 48 x \$5 = \$240
\$8 tickets --> 48 x \$8 = \$384
Difference between \$8 tickets and \$5 tickets in Week 2 if equal number of \$5 tickets and \$8 sold, would be
\$384 - \$240 = \$144
But the difference is \$248 and not \$144. Instead, we have,
\$248 - \$144 = \$104 (more)
For every \$8 sold instead of \$5, there would be an increase of (\$8 + \$5) = \$13
\$104 (more) divided by \$13 = 8 (tickets more)
48 tickets + 8 tickets (more) = 56 \$8 tickets sold on 2nd day.
1st Week --> \$1024 divided by \$8 = 128 (tickets)
Total number of \$8 tickets
--> 128 + 56 = 184
## Friday, February 19, 2010
### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q17
John had some red and blue marbles in a box. The sum of 1/4 of the red marbles and 2/5 of the blue marbles in the box is 64. The sum of 3/4 of the red marbles and 4/5 of the blue marble in the box is 144.
a) How many red marbles are there in the box?
b) How many blue marbles are there in the box?
Solution
Red (all marbles) --> R R R R
Blue (all marbles) --> B B B B B
1/4 or red marbles and 2/5 or blue marbles -- > 64
R + B B --> 64
3/4 of red marble and 4/5 of blue marbles --> 144
R R R + B B B B --> 144
R + B B --> 64 (multiply all by 2)
R R + B B B B --> 128
We now have
R R R + B B B B --> 144
R R + B B B B --> 128
R --> 144 - 128 = 16
a)
Number of Red Marbles
R R R R --> 4 x 16 = 64
Answer: 64 red marbles
b)
R + B B --> 64
16 + B B --> 64
B B --> 64 - 16 = 48
B --> 48 divided by 2 = 24
Number of Blue Marbles
B B B B B --> 5 x 24 = 120
Answer: 120 blue marbles
### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q16
The ratio of the number of boys to the number of girls in School A is 4:1. the ratio of the number of boys to the number of girls in School B is 2:3. School A had twice as many pupils as School B.
a) What is the ratio of boys in School A to the number of girls in School B?
b) After 30 girls left School A to join School B, the ratio of the number of boys to the number of girls in School B is now 5:8. How many girls are there in School B now?
Solution
* School A is multiplied by 2 to give a total of 10 units. This is because School B has 5 units. School A has twice the number of pupils as School B.
(a)
Boys from School A --> 8 units
Girls from School B --> 3 units
Ratio of number of boys in School A to number of girls in School B
--> 8 : 3
Answer: 8 : 3
b)
School B
* Before is multiplied by 5 and After is multiplied by 2 to make the boys have a common unit of 10 for both Before and After, because there was no transfer of boys.
Girls increased by 1 unit after the transfer
1 unit--> 30
Number of girls in School B after transfer
16 units --> 16 x 30 = 480
### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q15
Three men, A, B and C, worked together to paint a wall. If the painting was done by one man, the time taken to complete the wall for A, B and C would have been 6 hours, 8 hours and 12 hours respectively. A and B had painted for 3 hours after which A rested. B and C then continued with the painting. What would be the total number of hours taken to complete the wall? (Give your answer as a mixed number.)
Solution
First 3 hours
A painted for 3 h --> 3/6 = 1/2 (of the wall)*
B painted for 3 h --> 3/8 = 3/8 (of the wall)**
1/2 + 3/8 = 7/8 (of the wall was painted in first 3 hours)
* A takes 6h to paint the whole wall, therefore, in 3h, 3/6
** B takes 8h to paint the whole wall, therefore, in 3h, 3/8
A rests, B and C continue to paint remaining 1/8 of wall
Ratio of hours taken to paint whole wall
B : C
8 : 12
2 : 3
B takes less time than C, therefore B would have painted more of the wall
Ratio of amount of wall painted
B : C
3 : 2 --> Total units is 5
Time taken for B to paint whole wall
5/5 of wall --> 8h
Therefore, 3/5 of wall --> (3/5) x 8h = (24/5)h
Since only 1/8 of wall remains
1/8 of wall left --> (1/8) x (24/5)h = 3/5 h
Total time taken
--> 3h + 3/5 h = 3 and 3/5 hours
Answer: 3 and 3/5 hours
### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q14
Mr Lim paid \$134.40 for some jackfruits and pomeloes. The cost of the pomelo was 0.8 that of a jackfruit. A pomelo cost \$5.60. If all the pomeloes cost \$22.40 more than the jackfruits, how many fruits did he buy?
Solution
2 units --> \$134.40 - \$22.40 = \$112
1 unit --> \$112 divided by 2 = \$56
Cost of Pomeloes
--> \$56 + \$22.40 = \$78.40
Number of Pomeloes bought
--> \$78.40 divided \$5.60 (per pomelo)
= 14 pomeloes
Cost of of 1 pomelo is 0.8 of 1 jackfruit
Pomelo --> 8 units
Jackfruit--> 10 units
(Pomelo) 8 units --> \$5.60
1 unit --> \$5.60 divided by 8 = \$0.70
(Jackfruit) 10 units --> 10 x \$0.70 = \$7
Number of Jackfruits bought
--> Total cost - cost of pomeloes
\$134.40 - \$78.40 = \$56
\$56 divided by \$7 (per jackfruit) = 8 (jackfruits)
Total number of fruits
14 (pomeloes) + 8 (jackfruits) = 22 fruits
### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q13
The ratio of Jane's allowance to Olivia's allowance was 4:3. After Jane and Olivia were given \$15 and \$8 respectively, the ratio of Jane's allowance to Olivia's allowance became 3:2. How much allowance did Jane have at first?
Solution
Jane
4 units (at first) + \$15 (given) --> 3 parts
Olivia
3 units (at first) + \$8 --> 2 parts
2 parts --> 3 units + \$8
1 part --> 3/2 units + \$4
3 parts --> 3 x 3/2 units + 3 x \$4
= 9/2 units + \$12
(Jane) 3 parts --> 4 units + \$15
(Olivia) 3 parts --> 9/2 units + \$12
(Olivia) 9/2 units + \$12 --> (Jane) 4 units + \$15
9/2 units + \$12 --> 8/2 units + \$15
9/2 units - 8/2 units --> \$15 - \$12
1/2 unit --> \$3
1 unit --> \$6
Jane at first
4 units --> 4 x \$6 = \$24
## Wednesday, February 17, 2010
### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q12
Rahim's age is 2/9 of his grandfather's. His grandfather will be 100 years old in 19 years' time. In how many years' time will Rahim's age be 1/4 of his grandfather?
Solution
Now
Rahim --> 2 units
Grandfather --> 9 units
19 years' time (Grandfather)
9 units + 19 --> 100
9 units --> 100 - 19 = 81
1 unit --> 81 divided by 9 = 9
Now
Rahim --> 2 x 9 years = 18 years
Grandfather --> 9 x 9 years = 81 years
Difference between Grandfather and Rahim
--> 81 years - 18 years = 63 years
Future (when Rahim 1/4 of Grandfather's age)
Rahim --> 1 unit
Grandfahter --> 4 units
But Rahim is 63 years younger
4 units - 1 unit --> 63 years
3 units --> 63 years
1 unit --> 63 years divided by 3 = 21 years
Rahim now --> 18 years
Rahim in future --> 21 years
21 years - 18 years = 3 years
### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q11
There were some marbles in a box. Sofie took out 2/5 of the marbles and put in 6 more. Then John took out 1/6 of the remaining marbles and put in 5 more. there were 25 marbles left. How many marbles were in the box at first?
Solution
6 units - 1 unit + 6 - 1 + 5 --> 25 (left in box)
5 units + 10 --> 25
5 units --> 25 - 10 = 15
1 unit --> 15 divided by 5 = 3
Marbles at first
--> 10 units x 3
= 30
(At first, there were 10 units and not 5 units because the 5 units have been cut into halves, giving a total of 10 smaller units)
### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q10
In the figure shown below, SUVX is a square. STU is an equilateral triangle and TXW is a straight line.
a) Find the value of Angle STX.
b) Find the value of Angle WVX.
Solution
a)
Line TZ passes through V, while line TY is passes through the centre of Line SU.
Angle STX is 1/4 of Angle STU.
Angle STU is 60 degrees (Triangle STU is equilateral)
Angle STX --> (1/4) x 60 degrees = 15 degrees
b)
Angle SXT = 15 degrees (Triangle STX is isosceles)
Angle TXV --> (90 - 15) degrees = 75 degrees
Angle WXV --> (180 - 75) degrees = 105 degrees
Angle WVX
--> (180 - 105) degrees divided by 2 = 37.5 degrees
(Triangle WVX is isosceles)
### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q9
Nigel had a total of 227 durians and pears. He sold half of the durians and bought another 40 pears. As a result, he had an equal number of durians and pears.
a) How many durians were there at first?
b) How many pears were there at first?
Solution
(a)
3 units --> 267
1 unit --> 267 divided by 3 = 89
(Durians) 2 units --> 2 x 89 = 178
(b)
(Pears) 1 unit - 40
--> 89 - 40 = 49
### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q8
Wilson and Yi Lin had \$71 altogether. Yi Lin and Patrick had \$105 altogether. Wilsom hand 3/5 of the money that Patrick had. How much money did Yi Lin have?
Solution
Yi Lin + Patrick
? + 5 units --> \$105
Yi Lin + Wilson
? + 3 units --> \$71
5 units - 3 units --> \$105 - \$71
2 units --> \$34
1 unit --> \$34 divided by 2 = \$17
Yi Lin + Wilson
? + 3 units --> \$71
? + (3 x \$17) --> \$71
? + \$51 --> \$71
? --> \$71 - \$51 = \$20
### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q6
(a) Arif is 2x years old. His father is 4 times as old as he. His mother is 3 years younger than his father. What is their total age in terms of x?
(b) If x is = 4, find their total age.
Solution
(a)
Arif --> 2x
Father --> 4 x 2x = 8x
Mother --> 8x - 3
Total --> 2x + 8x + 8x - 3 = 18x - 3
Answer: (18x - 3) years
(b)
18(4) - 3
= 72 - 3
= 69
### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q5
A container with Bottle A placed in it has a mass of 4.27 kg. An identical container with Bottle B placed in it has a mass of 6.58 kg. The mass of Bottle A is 1/3 that of Bottle B. What is the mass of the Bottle A? Give your answer correct to 2 decimal places.
Solution
Bottle A --> 1 unit
Bottle B --> 3 unit
Container + Bottle A (1 unit) --> 4.27 kg
Container + Bottle B (3 units) --> 6.58 kg
3 units - 1 unit --> 6.58 kg - 4.27 kg
2 units --> 2.31 kg
1 unit --> 2.31 kg divided by 2 units = 1.155 kg
1.155 kg --> 1.16 kg correct to 2 decimal places
### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q4
Box A contains only one-dollar coins. Box B contains only fifty-cent coins and Box C contains only twenty-cent coins. Box A has 5 times as many coins as Box C. Box B contains 12 coins fewer than Box A. Box C contains half the number of coins in Box B. How much money is there in Box B?
Solution
2 units --> 5 units - 12
3 units --> 12
1 unit --> 12 divided by 3 = 4
(Box B)
5 units - 12
-->(5 x 4) - 12
= 20 - 12
= 8 (coins)
8 coins x \$0.50 = \$4
### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q3
The ratio of the number of Chloe's stickers to the number of Faith's stickers is 3:5. The ratio of the number of Faith's stickers to Melissa's stickers is 6:7. If Melissa has 204 stickers more than Chloe, how many stickers do they have altogether?
Solution
Melissa - Chloe --> 204
35 units - 18 units --> 204
17 units --> 204
1 unit --> 204 divided by 17 = 12
All stickers
Chloe + Faith + Melissa
18 + 30 + 35 = 83
83 units --> 83 x 12 = 996
### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q1
LMNO is a square. PQN and PLQ are isosceles triangles. Angle QNM is 23 degrees. Find Angle NPQ.
Solution
Angle PNO --> 23 deg (mirror image of Angle MNQ)
Angle OPN --> 90 deg - 23 deg = 67 deg
Angle LPQ --> 45 deg (Triangle LQP is isosceles and Angle LPQ is a rt angle)
Angle NPQ --> (180 - 45 - 67) deg = 68 deg
## Wednesday, February 10, 2010
### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q18
Ai Tong School organized a 2-day camp. On the first day, the number of boys was 600 more than the girls. On the second day, the number of boys decreased by 10% but the number of girls increased by 10%. If there were 2540 children on the second day, how many children were there on the first day?
Solution
2nd Day
Boys --> 0.9 unit + 540
Girls --> 1.1 unit
Total --> 2540
0.9 unit + 540 + 1.1 unit --> 2540
2 units + 540 --> 2540
2 units --> 2540 - 540 = 2000
1 unit --> 2000 divided by 2 = 1000
1st Day
2 units + 600
(2 x 1000) + 600
= 2000 + 600
= 2600
### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q17
75% of the children in the stadium were girls. After 52 girls and 4 boys left, the remaining children formed groups of 8. In each group, there were 3 boys. How many children were there in the stadium at first?
Solution
Note - "Before" for girls is 3 parts, while "Before" for boys is 1 part. This is to represent 75% girls and 25% boys, before 52 girls and 4 boys left.
From the table above
4 units --> 40
1 unit --> 40 divided by 4 = 10
Total boys (after)
3 units --> 3 x 10 = 30
Total girls (after)
5 units --> 5 x 10 = 50
Total children (after)
30 + 50 = 80
Total children at first
80 + 52 + 4 = 136
### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q16
The number of pupils in Team A to the number of pupils in Team B is in the ratio of 7:6. If 45 pupils are transferred from Team A to Team B, the ratio will become 2:3. How many pupils are there altogether?
Solution
Note - "Before" is multiplied by 5, while "After" is multiplied by 13 so as to have an equal number of total units of 65 each. The total number of units for "Before" and "After" are the same because there was no increase or decrease in the total number of pupils before or after the transfer.
Team A had a decrease of
35 units – 26 units --> 45 (pupils)
9 units --> 45
1 unit --> 45 divided by 9 = 5
Total number of pupils
Team A + Team B
26 units + 39 units --> 65 units
65 units --> 5 x 65 = 325
## Tuesday, February 09, 2010
### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q15
At a party, only 4/9 of the invited guests came. The ratio of the number of women to the number of men present was 3 : 4. If 80 more men turned up for the party, the number of men would be twice the number of women. How many guests were invited?
Solution
Before 80 more men turned up
Women : Men
3 : 4
After 80 more men turned up
Women : Men
3 : 6
There was an increase of 2 units for men
2 units --> 80 (men)
1 unit --> 80 divided by 2 = 40
Total present at the party before 80 more men turned up
7 units --> 280
4/9 of invited guests --> 280
1/9 --> 280 divided by 4 = 70
All invited guests
9/9 --> 9 x 70 = 630
### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q14
For every 200 books Johnson sells, he earns \$8. He will receive an addition of \$20 for every 3000 books sold. How many books must he sell to earn \$700?
Solution
1 group of 200 books --> \$8
3000 books divided by 200 books --> 15 groups of 200 books
15 groups of 200 books
--> 15 x \$8 + \$20 commission
= \$120 + \$20
= \$140
To earn \$700,
\$700 divided by \$140 --> 5 groups of \$140
1 group of \$140
--> 15 groups of 200 books
5 groups of \$140 --> 5 x 15 groups of 200 books
=75 groups of 200 books
75 groups of 200 books
--> 75 x 200 books
= 15 000 books
Answer: 15 000 books
### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q13
The length of a rectangle is thrice as long as its width. Its perimeter is 16p cm.
a) Find the length of the rectangle in terms of p.
b) Find the area of the rectangle if p = 4.
Solution
Perimeter
--> 3 units + 1 unit + 3 units + 1 unit
= 8 units
8 units --> 16p cm
1 unit --> 16p cm divided 8 = 2p cm
a)
Length
3 units --> 3 x 2p cm = 6p cm
Answer: 192 square cm
### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q12
20 similar pails of water can fill 5/12 of a container. Another 8 similar pails and 105 similar bowls of water are needed to fill the container to its brim. How many such bowls of water are needed to fill the empty container completely?
Solution
20 pails --> 5/12 of container (x2)
8 pails + 105 bowls --> remaining 7/12 of container (x5)
40 pails --> 10/12 of container (after x2)
40 pails + 525 bowls --> 35/12 containers (after (x5)
From the above we can see that 525 bowls can fill up:-
525 bowls --> 35/12 - 10/12 = 25/12 containers
25/12 --> 525 bowls
1/12 --> 525 divided 25 = 21 bowls
(Full container) 12/12 --> 12 x 21 = 252
### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q11
James spends 20% of his monthly income on transport, 30% of it on food and 10% of the remainder on clothes. He saves the rest of his income. If his monthly savings is \$900, find his monthly income.
Solution
(Saves)
90% of 50% --> \$900
(90/100) x (50/100) x 100% --> \$900
45% --> \$900
1%--> \$900 divided by 45 = \$20
(Monthly Income)
100% --> 100 x \$20 = \$2000
### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q10
The diagrams below show tiling patterns. Each tile is a square of side 1 cm.
What is the perimeter of Pattern 10?
Solution
(Add 6 cm for every new pattern)
Pattern 1 --> 10 cm
Pattern 2 --> 16 cm
Pattern 3 --> 22 cm
Pattern 4 --> 28 cm
Pattern 5 --> 34 cm
Pattern 6 --> 40 cm
Pattern 7 --> 46 cm
Pattern 8 --> 52 cm
Pattern 9 --> 58 cm
Pattern 10 --> 64 cm
## Friday, February 05, 2010
### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q9
A snail fell into a well that is 300 cm deep.
In the first hour, it climbed 80 cm up the well.
In the second hour, it climbed 70 cm up the well.
Each hour, it managed to climb 10 cm less than the hour before.
How many hours did it take to climb out of the well?
Solution
1st h --> 80 cm
2nd h --> 70 cm
3rd h --> 60 cm
4th h --> 50 cm
5th h --> 40 cm
(80 + 70 + 60 + 50 + 40) cm = 300 cm
--> 5 hours
### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q8
The rectangle ACEG is divided into 4 parts. BCEF is a square. Each part has a different area. Find the area X.
Area of X
--> 4 cm x 3 cm = 12 square cm
Answer: 12 square cm
### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q7
The table below shows the rates for water consumption.
a) Find the amount paid for 50 cubic metres of water used.
b) If 7% GST is imposed on the total amount, how much is the GST correct to the nearest 10-cent?
Water Consumption Rates
First 20 cubic metres ----- \$1.33 per cubic metre
Next 20 cubic metres ----- \$1.46 per cubic metre
Additional amount above 40 cubic metres ----- \$1.73 cubic meter
Solution
a)
1st 20 cubic metres --> 20 x \$1.33 = \$26.60
2nd 20 cubic metres --> 20 x \$1.26 = \$29.20
Next 10 cubic metres --> 10 x \$1.73 = \$17.30
Total Amount
--> \$26.60 + \$29.60 + \$17.10 = \$73.30
b)
7 % x \$73.10 = \$5.117
Approximately --> \$5.10 (to nearest 10-cent)
## Thursday, February 04, 2010
### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q6
Zen had a piece of rectangular paper. He cut away 1/9 of the breadth and 2/5 of the length. Then he measured the remaining piece of rectangular paper. He found that the breadth to be 16 cm and the length to be 15 cm. Find the area of the original piece of paper.
Solution
8/9 --> 16 cm
1/9 --> 16 cm divided by 8 = 2 cm
9/9 --> 9 x 2 cm = 18 cm
Length
3/5 --> 15 cm
1/5 --> 15 cm divided by 3 = 5 cm
5/5 --> 5 x 5 cm = 25 cm
Original area --> 25 cm x 18 cm = 450 square cm
Answer: 450 square cm
### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q5
The figure below is made up of 2 squares, A and B. The ratio of Square A to Square B is 1:4. The shaded part is 1/5 of Square A. What fraction of the whole figure is shaded?
Solution
Shaded part --> 1 unit
Area of Square A --> 5 units (shaded area is 1/5 of Area of Square A)
Note that Square A has 4 units of unshaded and 1 unit of shaded area.
Area of Square B --> 4 x 5 units = 20 units (Area of Square B is 4x that of Square A)
Note that Square B has 19 units of unshaded and 1 unit of shaded area.
Total Area
--> 1 unit (shaded) + 4 units (unshaded Square A) + 19 units (unshaded Square B)
= 24 units
Fraction of shaded area to area of whole figure
--> 1/24
### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q4
In a right-angled triangle, the two sides which form the right angle are 16 cm by 12 cm respectively. How many such triangles are needed to form the smallest square?
Horizontal --> 4 x 12 cm = 48 cm
Vertical --> 3 x 16 cm = 48 cm
The above figure (not drawn to scale) is therefore a square.
The smallest 1 unit rectangle has 2 triangles.
There are 3 rows and 4 columns of the smallest 1 unit rectangle.
3 x 4 = 12 smallest unit of rectangles
2 triangles x 12 = 24 triangles
## Wednesday, February 03, 2010
### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q3
The ratio of Paul's age to Sue's age is 4:3. In 8 years' time the sum of their ages will be 86. What is Paul's age now?
Solution
7 units + 8 + 8 --> 86
7 units --> 86 - 8 - 8 = 70
7 units --> 70
1 unit --> 70 divided by 7 = 10
Paul's age now
4 units --> 4 x 10 = 40
Answer: 40 years old
### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q2
Each skirt costs \$m and 3 blouses cost \$62.50. Find the cost of 2 skirts and 6 blouses.
Solution
Skirt --> \$m
2 skirts --> 2 x \$m = \$2m
3 blouses --> \$62.50
6 blouses --> 2 x \$62.50 = \$125
Total --> \$2m + \$125m
= \$(2m + 125) (Answer)
### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q1
Express the perimeter of the figure below in terms of t.
Solution
Perimeter
--> (6 + t + 5 + 7 +11 + 7 + t) cm
= (2t + 36) cm (Answer) |
# Advice 1: How to find the sum of the roots of the equation
Determining the amount of roots of the equation - one of the necessary steps in solving quadratic equations (equations of the form ax2 + bx + c = 0, where the coefficients a, b, and c are arbitrary numbers, with a ≠ 0) using the theorem of vieta.
Instruction
1
Write the quadratic equation in the form ax2 + bx + c = 0
Example:
Original equation: 12 + x2= 8x
Correct the equation: x2 - 8x + 12 = 0
2
Apply the vieta theorem, according to which the sum of the roots of the equation will be equal to the number "b", taken with the opposite sign, and their work - a "c".
Example:
In this equation b=-8, c=12, respectively:
x1+x2=8
x1∗x2=12
3
Learn positive and negative numbers are roots of equations. If the product and the sum of the roots is positive number, each of the roots is positive. If the product of the roots is positive and the sum of the roots is negative, then both root – negative. If the product of the roots is negative, the roots one root is the " + " sign and another sign "-" In this case, you must use an additional rule: "If the sum of the roots is a positive number greater than the module root is also positive and if the sum of the roots is negative - the larger the module root is negative."
Example:
In this equation the sum and product of positive numbers: 8 and 12, then both the root - the positive number.
4
Solve this system of equations by choosing from the roots. Will be convenient to begin with the selection of the multipliers, and then, to check, substitute each pair of multipliers into the second equation and test whether the sum of these roots solution.
Example:
x1∗x2=12
Suitable pairs of roots are respectively: 12 and 1, 6 and 2, 4 and 3
Check out the pair using equation x1+x2=8. Pair
12 + 1 ≠ 8
6 + 2 = 8
4 + 3 ≠ 8
Accordingly, the roots of an equation are numbers 6 and 8.
Note
In this example, was a variant of the quadratic equation where a=1. To the same way to solve a full quadratic equation, where a&ne 1, it is necessary to make an auxiliary equation, giving "a" to the unit.
Use this method for solving equations in order to quickly find the roots. Also it will help if you need to solve the equation in the mind, without resorting to the records.
# Advice 2: How to find the root of the equation
The equation is called equality of the form f(x,y,...)=g(x,y,..) where f and g are functions of one or several variables. To find the root of an equation means finding the set of arguments at which this equality occurs.
You will need
• Knowledge in mathematical analysis.
Instruction
1
Let's say you have an equation of the form: x+2=x/5. For starters, move all components of this equality from the right side to the left, changing the sign of the component on the opposite. In the right part of this equation will be zero, i.e., we get the following: x+2-x/5 = 0.
2
We give similar terms. Will receive the following: 4x/5 + 2 = 0.
3
Further, from the obtained equation, we find the unknown addend, which in this case is X. the resulting value of the unknown variable will be the solution to the original equation. In this case, we get the following: x = -2,5.
Note
The solutions may be extra roots. They will not be the solution to the original equation, even if you do all correctly decided. Be sure to check all of the solution.
The obtained values of the unknown always check. It's easy to do, substituting the obtained value in the original equation. If the equality is true, then the decision is correct.
# Advice 3: How to find the negative root of the equation
If the substitution number in the equation is true equality, such a number is called a root. The roots can be positive, negative or zero. Among the many roots of the equation distinguish a maximum and a minimum.
Instruction
1
Find all the roots of equations, among them select negative, if any. For example, the quadratic equation 2x2-3x+1=0. Apply the formula for finding roots of a quadratic equation: x(1,2)=[3±√(9-8)]/2=[3±√1]/2=[3±1]/2, then x1=2, x2=1. It is easy to see that the negative is not among them.
2
To find the roots of quadratic equations you can also use the theorem of vieta. According to this theorem x1+x1=-b, x1∙x2=c, where b and c are respectively the coefficients of the equation x2+bx+c=0. Using this theorem, you can calculate the discriminant b2-4ac, which in some cases can significantly simplify the task.
3
If in a square equation the coefficient of x is even, can be used not primary, and short formulas for finding roots. If the basic formula looks like x(1,2)=[-b±√(b2-4ac)]/2a, then in abbreviated form it is written as: x(1,2)=[-b/2±√(b2/4-ac)]/a. If in a square equation, no intercept, simply make x the brackets. And sometimes the left part of the folds to complete the square: x2+2x+1=(x+1)2.
4
Are the types of equations that do not give a single number, a whole lot of solutions. For example, the trigonometric equation. So, the answer to the equation 2sin2(2x)+5sin(2x)-3=0 is x=π/4+πk, where k is an integer. That is, if you substitute any integer values of the parameter k the argument x will satisfy the given equation.
5
In trigonometric problems, you may need to find all the negative roots, or the maximum of the negative. In solving such problems, apply logical reasoning or the method of mathematical induction. Several substitute integer values for k into the expression x=π/4+πk and see how it behaves argument. By the way, the largest negative root in the previous equation will be x=-3π/4 for k=1.
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# Probability and Statistics (Test 4)
## Problem Solving And Reasoning : Mathematics Or Quantitative Aptitude
| Home | | Problem Solving And Reasoning | | Mathematics Or Quantitative Aptitude | | Probability and Statistics |
Probability and Statistics
| Probability and Statistics |
Q.1
Direction: Study the following information carefully and answer the questions that follow:
A committee of five members is to be formed out of 3 trainees, 4 professors and 6 engineers.
In how many ways this can be done if at least 3 trainees to have to be included in a committee
A. 25
B. 95
C. 65
D. 45
E. None of these
Answer : Option D
Explaination / Solution:
Number of Ways when if at least 3 trainees include in committee =
3C3*10C2 = 1*45 --> 45
Workspace
Report
Q.2
Direction: Study the following information carefully and answer the questions that follow:
A committee of five members is to be formed out of 3 trainees, 4 professors and 6 engineers.
In how many ways this can be done if 2 trainees and 3 engineer to have to be included in a committee
A. 32
B. 60
C. 45
D. 90
E. None of these
Answer : Option B
Explaination / Solution:
If 2 trainees and 3 engineers include in committee
3C2*6C3 = 3*20 --> 60
Workspace
Report
Q.3
Direction: Study the following information carefully and answer the questions that follow:
A committee of five members is to be formed out of 3 trainees, 4 professors and 6 engineers.
In how many ways this can be done if at least 2 trainees to have to be included in a committee
A. 340
B. 820
C. 560
D. 405
E. None of these
Answer : Option D
Explaination / Solution:
If at least 2 trainees include in committee
3C2*10C3 + 3C3*10C2 = 3*120+1*45
= 405
Workspace
Report
Q.4
Direction: Study the following information carefully and answer the questions that follow:
A committee of five members is to be formed out of 3 trainees, 4 professors and 6 engineers.
In how many ways this can be done if 1 trainees and 4 engineers be included in a committee
A. 45
B. 32
C. 60
D. 36
E. None of these
Answer : Option A
Explaination / Solution:
If 1 trainees and 4 engineers include in committee
3C1*6C4 = 3*15 --> 45
Workspace
Report
Q.5
Direction: Study the following information carefully and answer the questions that follow:
A committee of five members is to be formed out of 3 trainees, 4 professors and 6 engineers.
In how many ways this can be done if 3 engineers and 2 professors or 2 trainees and 3 professor be included in a committee
A. 140
B. 160
C. 132
D. 155
E. None of these
Answer : Option C
Explaination / Solution:
If 3 engineers and 2 professors or 2 trainees and 3 professors include in committee
6C3*4C2 + 3C2*4C3 = 20*6 + 3*4 = 120 + 12 = 132
Workspace
Report
Q.6
Direction: Study the given information carefully and answer the questions that follow—
A store contains 3 red, 4 blue, 4 green and 4 whites’ bats.
If two bats are picked at random, what is the probability that both are white?
A. 1/3
B. 3/35
C. 3/5
D. 2/35
E. None of these
Answer : Option D
Explaination / Solution:
Probabilities if both are White
4C2/15C2 = 6/105 → 2/35
Workspace
Report
Q.7
If four bats are picked at random, what is the probability that two are blue and two is green?
A. 12/455
B. 35/355
C. 18/455
D. 18/35
E. None of these
Answer : Option A
Explaination / Solution:
Probability if two are Blue and two are Green =
[(4C2*4C2)/15C4] = (6*6)/1365 → 12/455
Workspace
Report
Q.8
If three bats are picked at random, what is the probability that at least one is green?
A. 2/5
B. 60/65
C. 44/455
D. 58/91
E. None of these
Answer : Option D
Explaination / Solution:
Probability if at least one is green
[1-(11C3/15C3)] = [1-(165/455)] → 58/91
Workspace
Report
Q.9
If two bats are picked at random, what is the probability that either both are red or both are white?
A. 3/5
B. 3/35
C. 14/25
D. 4/105
E. None of these
Answer : Option B
Explaination / Solution:
Probabilities if both either are Red or either are White
(3C2 + 4C2)/15C2 = (3 + 6)/105 → 3/35
Workspace
Report
Q.10
If two bats are picked at random, what is the probability that none is white?
A. 8/21
B. 11/21
C. 13/55
D. 11/22
E. None of these
Answer : Option B
Explaination / Solution:
Probabilities if none are white
[11C2/15C2] = (55)/(105) = 11/21
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Report |
## Monday, September 28, 2015
### Draw Polygons with a Given Perimeter using Pro-Bot
Can you program the Pro-Bot to draw polygons & compound figures with a given perimeter?
This is an assignment that I designed for our Grade 3 students. It relates to the Common Core Math Standards: Geometric measurement: recognize perimeterFor this exercise, I highly recommend using graph paper, as it provides a helpful medium for the kids to work out the math problems. Provide at least one sheet per child to work out the problem and then additional sheets as required for the groups to draw the figures using Pro-Bot. Here is a link to a graph paper with 1 cm grid in PDF format; you can make copies for the students to draw on using the Pro-Bot.
### Perimeter of a Figure
A perimeter is a path that surrounds a two-dimensional shape. The term may be used for either the path or its length. It can be thought of as the length of the outline of a shape. (Wiki)
How do you calculate the perimeter of a given figure? You add the length of all the sides of that figure that form its outline. Say you are given a square with sides 3 cm each. The perimeter of the square is 3 + 3 + 3 + 3 = 12 cm. Similarly, a 5 cm x 6 cm rectangle. has a perimeter of 5 + 6 + 5 + 6 = 22 cm.
Can we do the reverse too? Given the perimeter, can we come up with the design for a figure with that perimeter?
Let's look at an example. Given a perimeter of 12 cm, how many polygons can we draw? We can draw multiple polygons, all with the exact same perimeter of 12 cm. In the figure below, you can see:
• a square 3 cm x 3 cm,
• a rectangle 5 cm x 1 cm,
• a rectangle 4 cm x 2 cm,
• a hexagon with sides 3 cm, 1 cm, 1 cm, 1 cm, 4 cm, 2 cm
Can you think of more polygons with a perimeter of 12 cm?
Let's now look at a scenario that shows the practical application of the concept of perimeters. And then program the Pro-Bot to draw a few polygons with a given perimeter. |
iGCSE (2021 Edition)
2.05 Parallel and perpendicular lines
Lesson
Parallel lines
Parallel lines are lines that have the same gradient.
Two parallel lines that never cross and don't have any points in common.
Equations of parallel lines
Let's look at how we can identify parallel lines given their equations.
Equation Form Characteristic of parallel lines Examples
$y=mx+c$y=mx+c Parallel lines have the same $m$m value.
$y=2x-1$y=2x1
$y=4+2x$y=4+2x
For every straight line $y=mx+c$y=mx+c, there exist infinitely many lines parallel to it.
Here is the line $y=x$y=x
Here are two more lines in the same family of parallel lines: $y=x+3$y=x+3 and $y=x-6$y=x6.
Notice that they have the same gradient ($m$m-value) but different $x$x and $y$y-intercepts ($c$c-values).
Lines parallel to the axes
Horizontal lines
Horizontal lines are lines where the $y$y-value is always the same.
Let's look at the coordinates for $A$A, $B$B and $C$C on this line.
$A=\left(-8,4\right)$A=(8,4)
$B=\left(-2,4\right)$B=(2,4)
$C=\left(7,4\right)$C=(7,4)
All the $y$y-coordinates are the same, $y=4$y=4
This means that regardless of the $x$x-value the $y$y value is always $4$4.
The equation of this line is $y=4$y=4
So if the equation of a straight line is $y=c$y=c, then it will be a horizontal line passing through the point $\left(0,c\right)$(0,c).
The $x$x-axis itself is a horizontal line. The equation of the $x$x-axis is $y=0$y=0.
Horizontal lines
All horizontal lines are parallel to the $x$x-axis and are of the form $y=c$y=c.
They have a gradient of $0$0.
Vertical lines
Vertical lines are lines where the $x$x-value is always the same.
Let's look at the coordinates for $A$A, $B$B and $C$C on this line.
$A=\left(-3,8\right)$A=(3,8)
$B=\left(-3,3\right)$B=(3,3)
$C=\left(-3,-3\right)$C=(3,3)
All the $x$x-coordinates are the same, $x=-3$x=3
This means that regardless of the $y$y-value the $x$x-value is always $-3$3.
The equation of this line is $x=-3$x=3
So if an equation of a straight line is $x=c$x=c, then it will be a vertical line passing through the point $\left(c,0\right)$(c,0).
The $y$y-axis itself is a vertical line. The equation of the $y$y-axis is $x=0$x=0.
Vertical lines
All vertical lines are parallel to the $y$y-axis and are of the form $x=c$x=c
Perpendicular lines
Lines that meet at right angles ($90^\circ$90°) are called perpendicular lines.
Play with this applet creating pairs of perpendicular lines.
Fill in this table as you go.
Gradient of line 1 $m_1$m1 $m_2$m2 $m_1\times m_2$m1×m2
What do you notice about the product of the gradients of lines $1$1 and $2$2? (The pair of perpendicular lines)
You will have discovered the perpendicular lines have gradients whose product is equal to $-1$1.
We say that $m_1$m1 is the negative reciprocal of $m_2$m2.
Negative reciprocal is a complex sounding term, but it just means two numbers that have opposite signs and are reciprocals of each other.
Here are some examples of negative reciprocals:
$2$2 and $-\frac{1}{2}$12
$\frac{3}{4}$34 and $-\frac{4}{3}$43
$-10$10 and $\frac{1}{10}$110
Perpendicular lines
• Two lines are perpendicular if their gradients are negative reciprocals of each other.
• To test if lines are perpendicular multiply the gradients together. If the result is $-1$1 then the lines are perpendicular.
Practice questions
Question 1
Which lines are parallel to $y=-3x+2$y=3x+2?
Select all correct options.
1. $y=3x$y=3x
A
$y=-\frac{2x}{3}+8$y=2x3+8
B
$-3y-x=5$3yx=5
C
$y=-10-3x$y=103x
D
$y+3x=7$y+3x=7
E
$y=3x$y=3x
A
$y=-\frac{2x}{3}+8$y=2x3+8
B
$-3y-x=5$3yx=5
C
$y=-10-3x$y=103x
D
$y+3x=7$y+3x=7
E
Question 2
Write down the equation of a line that is parallel to the $x$x-axis and passes through $\left(-10,2\right)$(10,2).
Question 3
Consider the following points on the number plane:
$A$A $\left(2,-1\right)$(2,1)
$B$B $\left(4,-7\right)$(4,7)
$C$C $\left(-3,1\right)$(3,1)
$D$D $\left(-6,10\right)$(6,10)
1. First, calculate the gradient of the line $AB$AB.
2. Now, find the gradient of the line $CD$CD.
3. Is the line $CD$CD parallel to $AB$AB?
Yes
A
No
B
Yes
A
No
B
Question 4
Assess whether the points $A$A, $B$B and $C$C are collinear.
1. If $A$A and $B$B have the coordinates $\left(-4,3\right)$(4,3) and $\left(-2,7\right)$(2,7) respectively, evaluate the gradient of $AB$AB.
2. If $C$C has the coordinates $\left(-7,-3\right)$(7,3), evaluate the gradient of $BC$BC.
3. Based on these two gradients, are $A$A, $B$B and $C$C collinear?
Yes
A
No
B
Yes
A
No
B
Question 5
A line goes through A$\left(3,2\right)$(3,2) and B$\left(-2,4\right)$(2,4):
1. Find the gradient of the given line.
2. Find the equation of the line that has a $y$y-intercept of $1$1 and is parallel to the line that goes through $A$A$\left(3,2\right)$(3,2) and $B$B$\left(-2,4\right)$(2,4).
Question 6
Find the equation of a line that is perpendicular to $y=-\frac{x}{2}+5$y=x2+5, and goes through the point $\left(0,6\right)$(0,6).
Outcomes
0606C8.4A
Know and use the condition for two lines to be parallel or perpendicular. |
# Part 6: Planar Transformations
In this guide, we explain what translations, reflections and rotations of multiple of 90˚ are with detailed diagrams. We'll also give you some checkpoint questions to test your knowledge and solutions.
Planar transformations produce interesting patterns which can be useful in everyday life.
For example, mirrors reflect objects allowing cars to see around corners, and roundabouts rotate cars allowing them to pass through intersections.
If we are provided with a coordinate system such as the cartesian plane, we can study these transformations in mathematical detail.
## NSW Syllabus Outcomes
Stage 4 NSW Syllabus Syllabus Explanation Describe translations, reflections in an axis, and rotations of multiples of $$90°$$ on the Cartesian plane using coordinates (ACMMG181) Use the notation to name the ‘image‘ resulting from a transformation of a point on the Cartesian planePlot and determine the coordinates for resulting from translating one or more timesPlot and determine the coordinates for resulting from reflecting in either the $$x-axis$$ or $$y-axis$$Plot and determine the coordinates for resulting from rotating by a multiple of $$90°$$ about the originRecognise that a combination of translations and/or reflections and/or rotations can produce the same result as another combination of translations and/or reflections and/or rotations
## Assumed knowledge for Planar transformations
Students should have developed spatial awareness and understand how to plot points on a cartesian plane.
## Planar transformation
The three types of planar transformations covered are translations, rotations and reflections.
Translating points means moving them along the plane by a certain, consistent amount.
For example, the following shape and its image are a translation:
To determine a mathematical rule for a translation, we should consider corresponding points on the original shape and its image.
Here, we can consider $$A$$ and $$A’$$:
• $$\ A$$ is $$(2,2)$$ and $$A’$$ is $$(6,4)$$.
We can see that the $$x-coordinate$$ has changed from $$2$$ to $$6$$ and the $$y-coordinate$$ has changed from $$2$$ to $$4$$.
So we have translated $$4$$ across and $$2$$ up.
Hence, the rule for this translation is $$A(x,y) \rightarrow A'(x + 4, \ y + 2)$$
Using this translation, if we are given an original point such as $$B(5,3)$$, we can find $$B’$$ by applying the same translation:
• $$B’$$ is $$(5+4,3+2)$$ which is $$(9,5)$$.
Rotating points involves moving the points by an angle relative to a given point.
For example, the following shape has been rotated about $$(0,0)$$ by $$90$$ degrees anticlockwise:
Again, to find the mathematical rule, let’s consider the coordinates of corresponding points.
From $$B(5,3)$$ to $$B’(-3,5)$$, we have swapped $$x$$ with $$-y$$ and $$y$$ with $$x$$.
So we can express our rule as $$(x,y) \rightarrow (-y,x)$$.
Using this planar transformation, we can find the coordinates of another point, such as $$C(5,1)$$, whose image will be $$C(5,1) \rightarrow C’ (-1,5)$$.
Reflecting points involves copying the points across a line so that their distance to the line is preserved.
For example, the following shape has been reflected about the $$x-axis$$:
To determine a mathematical rule for a reflection, we should consider corresponding points on the original shape and its image.
If we consider $$C$$ and $$C’$$:
• $$C$$ is $$(5,1)$$ and $$C’$$ is $$(-5,1)$$.
We can see that the $$x-coordinate$$ has changed from $$5$$ to $$-1$$ and the $$y-coordinate$$ remains unchanged.
Hence we can express the rule for this translation as $$C(x,y) \rightarrow C'(-x,y)$$.
Using this transformation, we can find the coordinates of other points, such as $$D(3,1) \rightarrow D'(-3,1)$$.
## Checkpoint Questions
1. Translate the following set of points $$3$$ units up and $$2$$ units left. Then, determine a mathematical rule for the given translation.
2. Rotate the following set of points by $$90$$ degrees clockwise around the origin. Then, determine a mathematical rule for the given translation.
3. Reflect the following set of points across the $$y-axis$$. Then, determine a mathematical rule for the given translation.
## Solutions
1. $$P(x,y) \rightarrow P'(x-2, \ y+3)$$
2. $$P(x,y) \rightarrow P'(y, -x)$$
3. $$P(x,y) \rightarrow P(x, -y)$$
## Summary of Planar Transformations
Translation, rotation and reflection are three common planar translations of objects on a coordinate plane.
In this article, we demonstrated how you can define mathematical rules for each of these transformations to apply them to different points.
© Matrix Education and www.matrix.edu.au, 2023. Unauthorised use and/or duplication of this material without express and written permission from this site’s author and/or owner is strictly prohibited. Excerpts and links may be used, provided that full and clear credit is given to Matrix Education and www.matrix.edu.au with appropriate and specific direction to the original content. |
4/9 Rounded to the Nearest Hundredth
Here we will show you step-by-step how to get the solution to: "What is 4/9 rounded to the nearest hundredth?"
First, we convert the fraction to a decimal number by dividing the numerator by the denominator:
4 / 9 = 0.444
There are two parts to the decimal number above:
Integer Part: 0
Fractional Part: 444
Now, we will make the Fractional Part just two digits (nearest hundredth) by using our rounding rules.*
In this case, Rule I applies, so 4/9 (or 0.444) rounded to the nearest hundredth in decimal format is:
0.44
Next, we will make 4/9 rounded to the nearest hundredth in fraction format. Since you can divide our decimal format answer above by 1 and keep the same value, you can make it like this:
0.44 = 0.44/1
Then, we multiply the numerator and denominator by 100 to get rid of the decimal point:
(0.44 x 100) / (1 x 100) = 44/100
That's it. 4/9 rounded to the nearest hundredth is displayed below (simplified if necessary):
11/25
Fraction Rounded to the Nearest Hundredth
"What is 4/9 rounded to the nearest hundredth?" is not the only question we can answer. Enter another fraction here.
/
* Rounding Rules
Rule I) If the last digit in the fractional part is less than 5, then simply remove the last the digit of the fractional part.
Rule II) If the last digit in the fractional part is 5 or more and the second digit in the fractional part is less than 9, then add 1 to the second digit of the fractional part and remove the third digit.
Rule III) If the last digit in the fractional part is 5 or more and the second digit in the fractional part is 9, and the first digit in the fractional part is less than 9, then add 1 to the first digit in the fractional part and make the second digit in the fractional part 0. Then remove the third digit.
Rule IV) If the last digit in the fractional part is 5 or more and the second digit in the fractional part is 9, and the first digit in the fractional part is 9, then add 1 to the integer part and make the fractional part 00.
What is 4/10 rounded to the nearest hundredth?
Here is the next fraction on our list that we have rounded to the nearest hundredth. |
# Warm Up Solve each equation for x. 1. y = x + 3 2. y = 3x – 4
## Presentation on theme: "Warm Up Solve each equation for x. 1. y = x + 3 2. y = 3x – 4"— Presentation transcript:
Warm Up Solve each equation for x. 1. y = x + 3 2. y = 3x – 4 Simplify each expression. x = y – 3 3. 2(x – 5) 2x – 10 4. 12 – 3(x + 1) 9 – 3x
Warm Up Continued Evaluate each expression for the given value of x. 5. x + 8 for x = 6 6. 3(x – 7) for x =10 12 9
Objective Solve linear equations in two variables by substitution.
Sometimes it is difficult to identify the exact solution to a system by graphing. In this case, you can use a method called substitution. The goal when using substitution is to reduce the system to one equation that has only one variable. Then you can solve this equation by the methods taught in Chapter 2.
Solving Systems of Equations by Substitution
Step 2 Step 3 Step 4 Step 5 Solve for one variable in at least one equation, if necessary. Step 1 Substitute the resulting expression into the other equation. Solve that equation to get the value of the first variable. Substitute that value into one of the original equations and solve. Write the values from steps 3 and 4 as an ordered pair, (x, y), and check.
Example 1A: Solving a System of Linear Equations by Substitution
Solve the system by substitution. y = 3x y = x – 2 Step 1 y = 3x Both equations are solved for y. y = x – 2 Step 2 y = x – 2 3x = x – 2 Substitute 3x for y in the second equation. Step 3 –x –x 2x = –2 2x = –2 x = –1 Solve for x. Subtract x from both sides and then divide by 2.
Example 1A Continued Solve the system by substitution.
Write one of the original equations. Step 4 y = 3x y = 3(–1) y = –3 Substitute –1 for x. Write the solution as an ordered pair. Step 5 (–1, –3) Check Substitute (–1, –3) into both equations in the system. y = 3x –3 3(–1) –3 –3 y = x – 2 –3 –1 – 2 –3 –3
Example 1B: Solving a System of Linear Equations by Substitution
Solve the system by substitution. y = x + 1 4x + y = 6 The first equation is solved for y. Step 1 y = x + 1 Step 2 4x + y = 6 4x + (x + 1) = 6 Substitute x + 1 for y in the second equation. 5x + 1 = 6 Simplify. Solve for x. Step 3 –1 –1 5x = 5 x = 1 5x = 5 Subtract 1 from both sides. Divide both sides by 5.
Example1B Continued Solve the system by substitution.
Write one of the original equations. Step 4 y = x + 1 y = 1 + 1 y = 2 Substitute 1 for x. Write the solution as an ordered pair. Step 5 (1, 2) Check Substitute (1, 2) into both equations in the system. y = x + 1 2 2 4x + y = 6 4(1) 6 6
Example 1C: Solving a System of Linear Equations by Substitution
Solve the system by substitution. x + 2y = –1 x – y = 5 Step 1 x + 2y = –1 Solve the first equation for x by subtracting 2y from both sides. −2y −2y x = –2y – 1 Step 2 x – y = 5 (–2y – 1) – y = 5 Substitute –2y – 1 for x in the second equation. –3y – 1 = 5 Simplify.
Example 1C Continued Step 3 –3y – 1 = 5 Solve for y. +1 +1 –3y = 6 Add 1 to both sides. –3y = 6 –3 –3 y = –2 Divide both sides by –3. Step 4 x – y = 5 Write one of the original equations. x – (–2) = 5 x + 2 = 5 Substitute –2 for y. –2 –2 x = 3 Subtract 2 from both sides. Write the solution as an ordered pair. Step 5 (3, –2)
Check It Out! Example 1a Solve the system by substitution. y = x + 3 y = 2x + 5 Step 1 y = x + 3 y = 2x + 5 Both equations are solved for y. Step 2 2x + 5 = x + 3 y = x + 3 Substitute 2x + 5 for y in the first equation. –x – 5 –x – 5 x = –2 Step 3 2x + 5 = x + 3 Solve for x. Subtract x and 5 from both sides.
Check It Out! Example 1a Continued
Solve the system by substitution. Write one of the original equations. Step 4 y = x + 3 y = –2 + 3 y = 1 Substitute –2 for x. Step 5 (–2, 1) Write the solution as an ordered pair.
Check It Out! Example 1b Solve the system by substitution. x = 2y – 4 x + 8y = 16 Step 1 x = 2y – 4 The first equation is solved for x. (2y – 4) + 8y = 16 x + 8y = 16 Step 2 Substitute 2y – 4 for x in the second equation. Step 3 10y – 4 = 16 Simplify. Then solve for y. 10y = 20 Add 4 to both sides. 10y = Divide both sides by 10. y = 2
Check It Out! Example 1b Continued
Solve the system by substitution. Step 4 x + 8y = 16 Write one of the original equations. x + 8(2) = 16 Substitute 2 for y. x + 16 = 16 Simplify. x = 0 – 16 –16 Subtract 16 from both sides. Write the solution as an ordered pair. Step 5 (0, 2)
Check It Out! Example 1c Solve the system by substitution. 2x + y = –4 x + y = –7 Solve the second equation for x by subtracting y from each side. Step 1 x + y = –7 – y – y x = –y – 7 2(–y – 7) + y = –4 x = –y – 7 Step 2 Substitute –y – 7 for x in the first equation. 2(–y – 7) + y = –4 Distribute 2. –2y – 14 + y = –4
Check It Out! Example 1c Continued
Solve the system by substitution. Step 3 –2y – 14 + y = –4 Combine like terms. –y – 14 = –4 –y = 10 Add 14 to each side. y = –10 Step 4 x + y = –7 Write one of the original equations. x + (–10) = –7 Substitute –10 for y. x – 10 = – 7
Check It Out! Example 1c Continued
Solve the system by substitution. Step 5 x – 10 = –7 Add 10 to both sides. x = 3 Step 6 (3, –10) Write the solution as an ordered pair.
Sometimes you substitute an expression for a variable that has a coefficient. When solving for the second variable in this situation, you can use the Distributive Property.
When you solve one equation for a variable, you must substitute the value or expression into the other original equation, not the one that had just been solved. Caution
Example 2: Using the Distributive Property
y + 6x = 11 Solve by substitution. 3x + 2y = –5 Solve the first equation for y by subtracting 6x from each side. Step 1 y + 6x = 11 – 6x – 6x y = –6x + 11 3x + 2(–6x + 11) = –5 3x + 2y = –5 Step 2 Substitute –6x + 11 for y in the second equation. 3x + 2(–6x + 11) = –5 Distribute 2 to the expression in parenthesis.
Example 2 Continued y + 6x = 11 Solve by substitution. 3x + 2y = –5 Step 3 3x + 2(–6x) + 2(11) = –5 Simplify. Solve for x. 3x – 12x + 22 = –5 –9x + 22 = –5 –9x = –27 – 22 –22 Subtract 22 from both sides. –9x = –27 – –9 Divide both sides by –9. x = 3
Example 2 Continued y + 6x = 11 Solve by substitution. 3x + 2y = –5 Write one of the original equations. Step 4 y + 6x = 11 y + 6(3) = 11 Substitute 3 for x. y + 18 = 11 Simplify. –18 –18 y = –7 Subtract 18 from each side. Step 5 (3, –7) Write the solution as an ordered pair.
Check It Out! Example 2 –2x + y = 8 Solve by substitution. 3x + 2y = 9 Step 1 –2x + y = 8 Solve the first equation for y by adding 2x to each side. + 2x x y = 2x + 8 3x + 2(2x + 8) = 9 3x + 2y = 9 Step 2 Substitute 2x + 8 for y in the second equation. 3x + 2(2x + 8) = 9 Distribute 2 to the expression in parenthesis.
Check It Out! Example 2 Continued
–2x + y = 8 Solve by substitution. 3x + 2y = 9 Step 3 3x + 2(2x) + 2(8) = 9 Simplify. Solve for x. 3x + 4x = 9 7x = 9 7x = –7 –16 –16 Subtract 16 from both sides. 7x = –7 Divide both sides by 7. x = –1
Check It Out! Example 2 Continued
–2x + y = 8 Solve by substitution. 3x + 2y = 9 Write one of the original equations. Step 4 –2x + y = 8 –2(–1) + y = 8 Substitute –1 for x. y + 2 = 8 Simplify. –2 –2 y = 6 Subtract 2 from each side. Step 5 (–1, 6) Write the solution as an ordered pair.
Lesson Quiz: Part I Solve each system by substitution. 1. 2. 3. y = 2x (–2, –4) x = 6y – 11 (1, 2) 3x – 2y = –1 –3x + y = –1 x – y = 4
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# Min, Max and the Five-Number Summary Part B: The Median and the Three-Number Summary (35 Minutes)
In This Part: The Median
Another useful summary measure for a collection of data is the median. As you learned in Session 2, the median is the middle data value in an ordered list. Here’s one way to find the median of our ordered noodles.
First, place your 11 noodles in order from shortest to longest on a new piece of paper or cardboard. Your arrangement should look something like this:
Next, remove two noodles at a time, one from each end, and put them to the side:
Continue this process until only one noodle remains. This noodle is the median. Label it “Med”:
Notice that the median divides the set of 11 noodles into two groups of equal size — the five noodles shorter than the median and the five noodles longer than the median. Another way to say this is that there are just as many noodles before the median as there are after the median.
Problem B1
If you could see only the median noodle, what would you know about the other noodles?
What would knowing the median tell you about each of the first five (the shortest five) noodles? What would it tell you about each of the last five (the longest five) noodles?
Problem B2
If you could see only the median noodle, describe some information you would not know about the other noodles.
In This Part: The Three-Noodle Summary
Now remove all the noodles except Min, Med, and Max.
We’ll call this display the “Three-Noodle Summary.
Problem B3
If you could see Min, Med, and Max, what would you know about the other noodles? Be specific about how this compares to Problem A3 (where you only knew Min and Max) and Problem B1 (where you only knew Med).
Problem B4
Describe some information you still wouldn’t know about the other noodles from the Three-Noodle Summary.
In This Part: The Three-Number Summary
Now let’s convert the Three-Noodle Summary to the Three-Number Summary. If they’re not already there, place the three noodles — Min, Med, and Max — in order on the horizontal axis.
Next add a vertical number line, and mark the lengths of the three noodles. (Left)
Remove the noodles, and you’re left with the Three-Number Summary. (Right)
Problem B5
If we call the length of the fourth noodle N4, how does N4 compare to Min, Med, and Max? What wouldn’t you know about N4 if you only knew Min, Med, and Max?
In This Part: Even Data Sets
In the previous example, it wasn’t hard to find the median because there were 11 noodles — an odd number. For an odd number of noodles, the median is the noodle in the middle. But how do we find the median for an even number of noodles?
Add a 12th noodle, with a different length from the other 11 noodles, to the original collection. Arrange the noodles in order from shortest to longest.
Problem B6
Using the method of removing pairs of noodles (the longest and the shortest), try to determine the median noodle length. What happens?
This time, there won’t be one remaining noodle in the middle — there will be two! If you remove this middle pair, you’ll have no noodles left.
Therefore, you’ll need to draw a line midway between the two remaining noodles to play the role of the median. The length of this line should be halfway between the lengths of the two middle noodles:
Move the middle pair aside, and you can see your new median:
Notice that this median still divides the set of noodles into two groups of the same size — the six noodles shorter than the median and the six noodles longer than the median:
The major difference is that, this time, the median is not one of the original noodles; it was computed to divide the set into two equal parts.
Note: It is a common mistake to include this median in your data set when you’ve added it in this way. This median, however, is not part of your data set.
Video Segment
In this video segment, participants discuss the process of finding the median of a data set with an even number of values (in this case n = 20). Watch this video segment to review the process you used in Problem B6 or if you would like further explanation.
Note: The data set used by the onscreen participants is different from the one provided above.
Problem B7
If you could see only the median of a set of 12, what would you know about the other noodles?
You can convert the Three-Noodle Summary for these 12 noodles to the Three-Number Summary in the same way you did it for the set of 11 noodles:
Add a vertical number line, and mark the lengths of the three noodles:
Remove the noodles, and you’re left with the Three-Number Summary:
In This Part: Review
As we have seen with the noodle examples, the median divides ordered numeric data into two groups, each with the same number of data values.
If you only know the Three-Number Summary (Min, Med, and Max) for a set of data, you can still glean quite a bit of information about the data. You know that all the data values are between Min and Max, and you know that Med divides the data into two groups of equal size. One group contains data values to the left of Med, and the other group contains data values to the right of Med. You also know that the group of values to the left of the median must be lower than (or equal to) the median in value, and that the group of values to the right of the median must be greater than (or equal to) the median in value.
### Solutions
Problem B1
You would know that there must be exactly five noodles shorter than the median noodle and five noodles longer than the median noodle.
Problem B2
You would not know the actual values of any of the other noodles: The five shorter noodles could be extremely short, the five longer noodles could be many feet long, they could all be fairly close in size to the median, etc. You would also not know or be able to estimate the maximum or minimum length of the other noodles.
Problem B3
You would know that all of the noodles are between Min and Max, and you can divide the noodles into two equal groups: five that are shorter than Med (including Min) and five that are longer than Med (including Max). This information gives you two specific intervals that contain an equal number of noodles, and all of the noodles are contained in these intervals. This is different from Problem A3, where you knew nothing about the size of the noodles between Min and Max, and from Problem B1, where you knew nothing about the upper and lower boundaries of your data set.
Problem B4
You still wouldn’t know the lengths of the noodles in the two intervals between Min and Med, or between Med and Max. These noodles could be very close to Med, very close to the extreme values, evenly spread within the intervals, or something else entirely. There is no way to know without more information.
Problem B5
You would know that N4 must be larger than Min, smaller than Med, and smaller than Max. This is true because N6 is the median, and N4 must be smaller than N6. You still wouldn’t know N4’s actual value or whether N4 was closer to Min or to Med. (A common mistake is to claim that N4 must be closer to Med than it is to Min. This is not necessarily true, since the values of N2 through N5 can be anywhere in the interval between Min and Med; for example, they could all be very close to Min.) |
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Add the integers: -5 + 3
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When adding integers first decided what kind of numbers you have: two positives, two negatives, or one of each. In this problem, you have one of each. When you have one of each you subtract the two numbers and take the sign of the larger number. So 5 - 3 is 2. Since 5 is bigger than 3 we have more negatives so our answer is -2.
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Add the fractions: 1/2 + 1/4.
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In order to add fractions you first have to find a common denominator. The denominators here are 2 and 4. Think of a number that both 2 and 4 go into. 4 is the easiest. To change the 2 to a 4 you must multiply the denominator and the numerator by 2 because whatever you do to the top you have to do to the bottom. So 1/2 becomes 2/4. Now add 2/4 and 1/4 by adding the numerators. So 1 + 2 makes 3. The denominator remains the same. So the answer is 3/4. Since there are no common factors for 3 or 4 besides 1 we cannot reduce this fraction.
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Solve the equation for x: 2x + 5x - 6 = 6x + 4
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3-2: Solving Systems of Equations using Substitution
3-2: Solving Systems of Equations using Substitution
3-2: Solving Systems of Equations using Substitution
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Presentation Transcript
1. 3-2: Solving Systems of Equations using Substitution
2. Solving Systems of Equations using Substitution Steps: 1. Solve one equation for one variable (y= ; x= ; a=) 2. Substitute the expression from step one into the other equation. 3. Simplify and solve the equation. 4. Substitute back into either original equation to find the value of the other variable. 5. Check the solution in both equations of the system.
3. Example #1: y = 4x 3x + y = -21 Step 1:Solve one equation for one variable. y = 4x(This equation is already solved for y.) Step 2: Substitute the expression from step one into the other equation. 3x + y = -21 3x + 4x = -21 Step 3: Simplify and solve the equation. 7x = -21 x = -3
4. y = 4x 3x + y = -21 Step 4: Substitute back into either original equation to find the value of the other variable. 3x + y = -21 3(-3) + y = -21 -9 + y = -21 y = -12 Solution to the system is (-3, -12).
5. y = 4x 3x + y = -21 Step 5: Check the solution in both equations. Solution to the system is (-3,-12). 3x + y = -21 3(-3) + (-12) = -21 -9 + (-12) = -21 -21= -21 y = 4x -12 = 4(-3) -12 = -12
6. Example #2: x + y = 10 5x – y = 2 Step 1: Solve one equation for one variable. x + y = 10 y = -x +10 Step 2: Substitute the expression from step one into the other equation. 5x - y = 2 5x -(-x +10) = 2
7. x + y = 10 5x – y = 2 Step 3: Simplify and solve the equation. 5x -(-x + 10) = 2 5x + x -10 = 2 6x -10 = 2 6x = 12 x = 2
8. x + y = 10 5x – y = 2 Step 4: Substitute back into either original equation to find the value of the other variable. x + y = 10 2 + y = 10 y = 8 Solution to the system is (2,8).
9. x + y = 10 5x – y = 2 Step 5: Check the solution in both equations. Solution to the system is (2, 8). 5x – y = 2 5(2) - (8) = 2 10 – 8 = 2 2 = 2 x + y =10 2 + 8 =10 10 =10 |
# Limits via Algebra
Most of the time, it's more precise (and a lot faster) to find limits using algebra.
When finding a limit of the form , where f(x) is just one nice algebraic expression, the first thing to do is plug a into the function to see if it exists. If f(a) exists, then that's the answer. This won't work for piecewise functions, since there could be gaps separating the pieces of the function.
### Sample Problem
Find .
The first thing we do is plug in 3 and see if we find a value.
That's a perfectly good fraction, therefore
Here's an example of how we might not find a solution.
### Sample Problem
Find .
The first thing we do is plug in 0 and see if we find a number.
= ?
The fraction is undefined, since we can't divide by zero.
### Sample Problem
Find .
The first thing we do is see if we can plug in 2.
When we're asked to find the limit of a quotient, if we plug in a number and find there's a good chance we can do something about it: we simplify the quotient and try again.
### Sample Problem
Find .
We already tried to plug in 2 and that didn't work. We'll simplify the fraction by factoring the polynomials.
x2x – 2 = (x + 1)(x – 2)
and
x2 – 5x + 6 = (x – 2)(x – 3).
Now we can see why we got before: 2 is a root of the polynomial in the numerator, and also a root of the polynomial in the denominator. Here's the point where we can do something useful: cancel the term (x – 2) from the numerator and denominator:
.
Finally, put in 2 again:
We can now say
= 2.
Why does this work? Let
The numerator of f(x) can be factored as
which equals (x – 1) for every value of x except -1 (when x = -1, the function is undefined).
We'll say that again, because it's important. After factoring f, we see that we can think of it as
.
If we graph this, we find the line x – 1 with a "hole" in the graph at x = -1 since f(-1) is undefined:
It's like the function f(x) is trying to be x – 1, but failing at one spot (poor function!). Here's the good news: since f(x) is trying to be x – 1, we can find the limit by instead finding
. |
# Division Of Rational Numbers Worksheet
A Reasonable Figures Worksheet can help your child become a little more acquainted with the principles associated with this proportion of integers. In this worksheet, students will be able to remedy 12 diverse difficulties related to rational expressions. They are going to learn how to multiply two or more figures, group them in sets, and figure out their goods. They will also exercise simplifying realistic expressions. As soon as they have mastered these methods, this worksheet might be a valuable device for furthering their studies. Division Of Rational Numbers Worksheet.
## Reasonable Figures are a proportion of integers
The two main varieties of phone numbers: irrational and rational. Reasonable figures are considered complete figures, in contrast to irrational figures usually do not replicate, and also have an unlimited quantity of numbers. Irrational amounts are non-zero, non-terminating decimals, and square roots which are not ideal squares. These types of numbers are not used often in everyday life, but they are often used in math applications.
To define a rational amount, you need to understand just what a reasonable quantity is. An integer is really a total amount, plus a rational variety is really a proportion of two integers. The percentage of two integers is definitely the amount on top separated through the amount at the base. If two integers are two and five, this would be an integer, for example. However, there are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They can be manufactured in to a portion
A rational number has a numerator and denominator which are not absolutely no. Consequently they are often depicted as a portion. Along with their integer numerators and denominators, reasonable figures can furthermore have a unfavorable worth. The unfavorable importance should be positioned on the left of as well as its absolute value is its length from absolutely nothing. To simplify this example, we are going to claim that .0333333 can be a small percentage that can be composed being a 1/3.
Together with bad integers, a realistic variety may also be produced in a small percentage. By way of example, /18,572 is really a reasonable variety, when -1/ is just not. Any portion composed of integers is rational, as long as the denominator is not going to include a and can be written for an integer. Furthermore, a decimal that leads to a point is also a rational number.
## They are sensation
In spite of their brand, reasonable figures don’t make a lot perception. In math, they can be solitary organizations by using a special span on the quantity line. This means that once we add up one thing, we could order the size and style by its proportion to the original volume. This retains true regardless if there are endless logical figures in between two particular phone numbers. In other words, numbers should make sense only if they are ordered. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
In real life, if we want to know the length of a string of pearls, we can use a rational number. To obtain the duration of a pearl, for instance, we might count up its breadth. Just one pearl is 15 kilos, which is a reasonable amount. In addition, a pound’s weight is equal to 10 kilos. Therefore, we must be able to split a lb by 10, without be worried about the size of a single pearl.
## They can be expressed being a decimal
If you’ve ever tried to convert a number to its decimal form, you’ve most likely seen a problem that involves a repeated fraction. A decimal number could be composed as being a a number of of two integers, so four times five is the same as seven. An identical problem involves the repetitive fraction 2/1, and each side needs to be divided up by 99 to obtain the proper answer. But how can you make your conversion process? Here are a few illustrations.
A reasonable number can also be designed in various forms, which includes fractions along with a decimal. A great way to symbolize a realistic variety in the decimal is usually to split it into its fractional equal. There are three ways to split a realistic amount, and every one of these ways brings its decimal counterpart. One of these ways is always to separate it into its fractional equivalent, and that’s what’s called a terminating decimal. |
# 17. EXERCISES
The exercises are numbered according to the numbers of the relevant sections of the manual.
Exercise 2.1 Mean value and variance
In this exercise we use part of the length-frequency data of the coral trout (Plectropomus leopardus) presented in Fig. 3.4.0.2, namely those in the length interval 23-29 cm. These fish are assumed to belong to one cohort. The length-frequencies are presented in Fig. 17.2.1.
Read the frequencies, F(j) from Fig. 17.2.1 and complete the worksheet. Calculate mean, variance and standard deviation.
Worksheet 2.1
j L(j) - L(j) + dL F(j) (j) F(j) * (j) (j) - F(j) * ((j) - )2 1 - -2.968 2 - -2.468 3 - -1.968 4 - -1.468 5 - -0.968 6 - -0.468 7 - 0.032 8 26.5-27.0 6 26.75 160.50 0.532 1.698 9 27.0-27.5 2 54.50 1.032 2.130 10 27.5-28.0 2 55. 50 1.532 4.694 11 28.5-29.0 2 56.50 2.032 8.258 12 1 28.75 2.532 6.411 sums S F(j) 31 = s2 = s =
Fig. 17.2.1 Length-frequency sample
Exercise 2.2 The normal distribution
This exercise consists of fitting a normal distribution to the length-frequency sample of Exercise 2.1, by using the expression:
(Eq. 2.2.1)
for a sufficient number of x-values allowing you to draw the bell-shaped curve.
For your convenience introduce the auxiliary symbols:
so that the formula above can be written
Since A and B do not depend on L and as they are going to be used many times, it is convenient to calculate them separately before-hand.
1) Calculate A and B
B = -1/(2s2) =
2) Calculate Fc(x) for the following values of x:
Worksheet 2.2
x Fc(x) x Fc(x) 22.0 26.0 22.5 26.5 23.0 27.0 23.5 27.5 24.0 28.0 24.5 28.5 25.0 29.0 25.5 29.5
3) Draw the bell-shaped curve on Fig. 17.2.1
Exercise 2.3 Confidence limits
Calculate the 95% confidence interval for the mean value estimated in Exercise 2.1.
Exercise 2.4 Ordinary linear regression analysis
It is often observed that the more boats participate in a fishery the lower the catch per boat will be. This is not surprising when one considers the fish stock as a limited resource which all boats have to share. In Chapter 9 we shall deal with the fisheries theory behind this model.
The data shown below in the worksheet are from the Pakistan shrimp fishery (Van Zalinge and Sparre, 1986).
1) Draw the scatter diagram.
2) Calculate intercept and slope (use the worksheet).
3) Draw the regression line in the scatter diagram.
4) Calculate the 95% confidence limits of a and b.
Worksheet 2.4
number of boats catch per boat per year year i x(i) x(i)2 y(i) y(i)2 x(i) * y(i) 1971 1 456 43.5 19836.0 1972 2 536 44.6 23905.6 1973 3 554 38.4 21273.6 1974 4 675 23.8 16065.0 1975 5 702 25.2 17690.4 1976 6 730 532900 30.5 930.25 1977 7 750 562500 27.4 750.76 1978 8 918 842724 21.1 445.21 1979 9 928 861184 26.1 681.21 1980 10 897 804609 28.9 835.21 Total 7146 309.5 211099.5 = = = = sx = sy = slope: intercept: = variance of b: sb = variance of a: sa = Student's distribution: tn-2 = confidence limits of b and a: b - sb * tn-2, b + sb * tn-2 = [________________,________________] a - sa * tn-2, a + sa * tn-2 = [________________,________________]
Exercise 2.5 The correlation coefficient
Refer to Exercise 2.4. Does the correlation coefficient make sense in the example of catch per boat regressed on number of boats? Consider which of the variables is the natural candidate as independent variable. Can we (in principle) decide in advance on the values of one of them?
Irrespective of your findings in the first part of the exercise carry out the calculation of the 95% confidence limits of r.
Exercise 2.6 Linear transformations of normal distributions, used as a tool to separate two overlapping normal distributions (the Bhattacharya method)
Fig. 17.2.6A shows a frequency distribution which is the result of two overlapping normal distributions "a" and "b". We assume that the length-frequencies presented in Fig. 17.2.6B are also a combination of two normal distributions. The aim of the exercise is to separate these two components. The total sample size is 398. Assume that each component has 50% of the total or 199. Further assume that the frequencies at the left somewhat below the top are fully representative for component "a", while those at the bottom of the right side are fully representative for component "b".
Fig. 17.2.6A Combined distribution of two overlapping normal distributions
Fig. 17.2.6B Length-frequency sample (assumed to consist of two normal distributions
1) Complete Worksheet 2.6a.
2) Plot D ln F(z) = y' against x + dL/2 = z and decide which points lie on straight lines with negative slopes (see Fig. 2.6.5).
3) On the basis of the plot select the points to be used for the linear regressions. (Avoid the area of overlap and points based on very few observations). Do the two linear regressions and determine a and b.
4) Calculate , s2 = -1/b and s = Ö s2 for each component.
5) Draw the two plots which represent each distribution in linear form.
6) We now want to convert the straight lines into the corresponding theoretical (calculated) normal distributions. Using Eq. 2.2.1 calculate Fc(x) for both normal distributions for a sufficient number of x-values to allow you to draw the two bell-shaped curves superimposed on Fig. 17.2.6B. Assume n = 199 for both components. (Use the same method as presented in Exercise 2.2). Complete Worksheet 2.6b.
Worksheet 2.6a
interval x F(x) ln F(x) D ln F(z) z = x + dL/2 4-5 4.5 2 0.693 0.916 5 5-6 5.5 5 1.609 0.875 6 6-7 6.5 12 7 7-8 7.5 24 8-9 8.5 35 9-10 9.5 42 10-11 10. 5 42 11-12 11.5 46 12-13 12.5 56 13-14 13.5 58 14-15 14.5 45 15-16 15.5 22 3.091 -1.145 16 16-17 16.5 7 1.946 -1.253 17 17-18 17.5 2 0.693
Worksheet 2.6b
First component Second component B = B = x Fc(x) first Fc(x) second x Fc(x) first Fc(x) second 1.5 11.5 2.5 12.5 3.5 13.5 4.5 14.5 5.5 15.5 6.5 16.5 7.5 17.5 8.5 18.5 9.5 19.5 10.5 20.5
Exercise 3.1 The von Bertalanffy growth equation
The growth parameters of the Malabar blood snapper (Lutjanus malabaricus) in the Arafura Sea were reported by Edwards (1985) as:
K = 0.168 per year
L¥ = 70.7 cm (standard length)
t0 = 0.418 years
Edwards also estimated the standard length/weight relationship for Lutjanus malabaricus:
w = 0.041 * L2.842 (weight in g and standard length in cm)
as well as the relationship between standard length (S.L.) and total length (T.L.):
T.L. = 0.21 + 1.18 * S.L.
Complete the worksheet and draw the following three curves:
1) Standard length as a function of age
2) Total length as a function of age
3) Weight as a function of age
Worksheet 3.1
age standard length total length body weight age standard length total length body weight years cm cm g years cm cm g 0.5 8 1.0 9 1.5 10 2 12 3 14 4 16 5 (do not use ages above 16 in the graph) 6 7 20 50
Exercise 3.1.2 The weight-based von Bertalanffy growth equation
Pauly (1980) determined the following parameters for the pony fish or slipmouth (Leiognathus splendens) from western Indonesia:
L¥ = 14 cm
q = 0.02332
K = 1.0 per year
t0 = -0.2 year
Complete the worksheet and draw the length and the weight-converted von Bertalanffy growth curves.
Worksheet 3.1.2
age t length L(t) weight w(t) age t length L(t) weight w(t) 0 0.9 0.1 1.0 0.2 1.2 0.3 1.4 0.4 1.6 0.5 1.8 0.6 2.0 0.7 2.5 0.8 3.0
Exercise 3.2.1 Data from age readings and length compositions (age/length key)
Consider Table 3.2.1.1 (age/length key) and suppose we caught a total of 2400 fish of the species in question during the cruise from which this age/length key was obtained and that only 439 specimens of Table 3.2.1.1 were aged. The remaining fish were all measured for length. To reduce the computational work of the exercise only a part (386 fish) of this length-frequency sample is used. This part is shown in the worksheet.
Estimate how many of these 386 fish belonged to each of the four cohorts listed in Table 3.2.1.1, by completing the worksheet.
Worksheet 3.2.1
cohort 1982 S 1981 A 1981 S 1980 A 1982 S 1981 A 1981 S 1980 A length interval key number in length sample numbers per cohort 35-36 0.800 0.200 0 0 53 42.4 10.6 0 0 36-37 0.636 0.273 0.091 0 61 38.8 16.7 5.6 0 37-38 49 38-39 52 39-40 70 40-41 52 41-42 0.222 0.444 0.222 0.111 49 10.9 21.8 10.0 5.4 total 386 187.2. 133.8
Exercise 3.3.1 The Gulland and Holt plot
Randall (1962) tagged, released and recaptured ocean surgeon fish (Acanthurus bahianus) near the Virgin Islands. Data of 11 of the recaptured fish are shown in the worksheet, in the form of their length at release (column B) and at recapture (column C) and the length of the time between release and recapture (column D).
1) Estimate K and L for the ocean surgeon fish using the Gulland and Holt plot.
2) Calculate the 95% confidence limits of the estimate of K.
Worksheet 3.3.1
A B C D E F fish no. L(t) L(t + D t) D t cm cm days cm/year cm (y) (x) 1 9.7 10.2 53 2 10.5 10.9 33 3 10.9 11.8 108 4 11.1 12.0 102 5 12.4 15.5 272 6 12.8 13.6 48 7 14.0 14.3 53 8 16.1 16.4 73 9 16.3 16.5 63 10 17.0 17.2 106 11 17.7 18.0 111 a (intercept) = b (slope) = K = L¥ = sb = tn-2 = confidence interval of K =
Exercise 3.3.2 The Ford-Walford plot and Chapman's method
Postel (1955) reports the following length/age relationship for Atlantic yellowfin tuna (Thunnus albacares) off Senegal:
age(years) fork length(cm) 1 35 2 55 3 75 4 90 5 105 6 115
Estimate K and L¥ using the Ford-Walford plot and Chapman's method.
Worksheet 3.3.2
Plot FORD-WALFORD CHAPMAN t L(t) (x) L(t + D t) (y) L(t) (x) L(t + D t) - L(t) (y) 1 2 3 4 5 a (intercept) b (slope) tn-2 confidence limits of b K L¥
Exercise 3.3.3 The von Bertalanffy plot
Cassie (1954) presented the length-frequency sample of 256 seabreams (Chrysophrys auratus) shown in the figure. He resolved this sample into normally distributed components (similar to Fig. 3.2.2.2) using the Cassie method (cf. Section 3.4.3) and found the following mean lengths for four age groups (cf. Fig. 17.3.3.3):
A B C D age group mean length(inches) D L/D t 0 3.22 2.11 4.28 1 5.33 2.29 6.48 2 7.62 2.12 8.68 3 9.74
Note: a Gulland and Holt plot gives (cf. Columns C and D): K = -0.002 and L¥ = -950 inches, which makes no sense whatsoever.
1) Estimate K from the von Bertalanffy plot.
2) Why does it not make sense to ask you to estimate t0?
Exercise 3.4.1 Bhattacharya's method
Weber and Jothy (1977) presented the length-frequency sample of 1069 threadfin breams (Nemipterus nematophorus) shown in Fig. 17.3.4.1A. These fish were caught during a survey from 29 March to 1 May 1972, in the South China Sea bordering Sarawak. The lengths measured are total lengths from the snout to the tip of the lower lobe of the caudal fin.
Figs. 17.3.4.1B and 17.3.4.1C show the Bhattacharya plots for the data in Fig. 17.3.4.1A, where B is based on the original data in 5 mm length intervals and C on the same data regrouped in 1 cm intervals. You should proceed with Fig. C for two reasons: 1) because it appears easier to see a structure in Fig. C than in Fig. B and 2) because the number of calculations is much lower.
1) Resolve the length-frequency sample (1 cm groups, Fig. C) into normally distributed components and estimate thereby mean length and standard deviations for each component. Use the four worksheets and plot the regression lines.
2) Estimate L¥ and K using a Gulland and Holt plot. Draw the plot.
3) Do you think the analysis could have been improved by using Fig. B (5 mm length groups) instead of Fig. C (1 cm groups)?
Worksheet 3.4.1a
A B C D E F G H I length interval(cm) N1+ ln N1+ D ln N1+(y) L(x) D ln N1 ln N1 N1 N2+ 5.75-6.75 1 0 - - - - 1 0 6.75-7.75 26 3.258 (3.258) 6.75 1.262 - 26 0 7.75-8.75 42# 3.738# 0.480 7.75 0.354 3.738# 42# 0 8.75-9.75 19 2.944 -0.793 8.75 -0.554 3.183 19 0 9.75-10.75 5 9.75 10.75-11.75 15 10.75 11.75-12.75 41 11.75 12.75-13.75 125 12.75 13.75-14.75 135 13.75 14.75-15.75 102 14.75 15.75-16.75 131 15.75 16.75-17.75 106 16.75 17.75-18.75 86 17.75 18.75-19.75 59 18.75 19.75-20.75 43 19.75 20.75-21.75 45 20.75 21.75-22.75 56 21.75 22.75-23.75 20 22.75 23.75-24.75 8 23.75 24.75-25.75 3 24.75 25.75-26.75 1 25.75 Total 1069 a (intercept) = b (slope) =
Worksheet 3.4.1b
A B C D E F G H I interval N2+ ln N2+ D ln N2+ L D ln N2 ln N2 N2 N3+ 5.75-6.75 6.75-7.75 6.75 7.75-8.75 7. 75 8.75-9.75 8.75 9.75-10.75 9.75 10.75-11.75 10.75 11.75-12.75 11.75 12.75-13.75 12.75 13.75-14.75 13.75 14.75-15.75 14.75 15.75-16.75 15.75 16.75-17.75 16.75 17.75-18.75 17.75 18.75-19.75 18.75 19.75-20.75 19.75 20.75-21.75 20.75 21.75-22.75 21.75 22.75-23.75 22.75 23.75-24.75 23.75 24.75-25.75 24.75 25.75-26.75 25.75 Total a (intercept) = b (slope) =
Worksheet 3.4.1c
A B C D E P G H I interval N3+ ln N3+ D ln N3+ L D ln N3 ln N3 N3 N4+ 5.75-6.75 6.75-7.75 6.75 7.75-8.75 7.75 8.75-9.75 8.75 9.75-10.75 9.75 10.75-11.75 10.75 11.75-12.75 11.75 12.75-13.75 12.75 13.75-14.75 13.75 14.75-15.75 14.75 15.75-16.75 15.75 16.75-17.75 16.75 17.75-18.75 17.75 18.75-19.75 18.75 19.75-20.75 19.75 20.75-21.75 20.75 21.75-22.75 21.75 22.75-23.75 22.75 23.75-24.75 23.75 24.75-25.75 24.75 25.75-26.75 25.75 Total a (intercept) = b (slope) =
Worksheet 3.4.1d
A B C D E F G H I interval N4+ ln N4+ D ln N4+ L D ln N4 ln N4 N4 N5+ 5.75-6.75 - 6.75-7.75 6.75 7.75-8.75 7.75 8.75-9.75 8.75 9.75-10.75 9.75 10.75-11.75 10.75 11.75-12.75 11.75 12.75-13.75 12.75 13.75-14.75 13.75 14.75-15.75 14.75 15.75-16.75 15.75 16.75-17.75 16.75 17.75-18.75 17.75 18.75-19.75 18.75 19.75-20.75 19.75 20.75-21.75 20.75 21.75-22.75 21.75 22.75-23.75 22.75 23.75-24.75 23.75 24.75-25.75 24.75 25.75-26.75 25.75 Total a (intercept) = b (slope) = =
Exercise 3.4.2 Modal progression analysis
Fig. 17.3.4.2A shows a time series over twelve months of ponyfish (Leiognathus splendens) from Manila Bay, Philippines, 1957-58. (Data from Tiews and Caces-Borja, 1965; figure redrawn from Ingles and Pauly, 1984). The numbers at the right hand side of the bar diagram indicate the sample sizes, while the height of the bars represents the percentages of the total number per length group.
Fig. 17.3.4.2B shows a time series of six samples of mackerel, (Rastrelliger kanagurta) from Palawan, Philippines, 1965. (Data from Research Division, BFAR, Manila; figure redrawn from Ingles and Pauly, 1984).
1) Fit by eye growth curves to these two time series, trying to follow the modal progression (as was done in Fig. 3.4.2.6). Start by fitting a straight line and then add some curvature to it, but do not be too particular about it. (Actually one should have carried out a Bhattacharya or similar analysis for each sample, but because of the amount of work involved in that approach, we take the easier, but less dependable, eye-fit. This exercise aims at illustrating only the principles of modal progression analysis - not the exact procedure).
2) Read from the eye-fitted growth curves pairs of (t, L) = (time of sampling, length), and use the Gulland and Holt plot to estimate K and L¥ . Assume that the samples were taken on the first day of the month. Read for Leiognathus splendens only the length for the samples indicated by "*" in Fig. A, as the figure is too small for a precise reading of each month. Use the worksheet.
3) Use the von Bertalanffy plot to estimate t0.
Worksheet 3.4.2
A. Leiognathus splendens:
GULLAND AND HOLT PLOT VON BERTALANFFY PLOT time of sampling L(t) D L/D t t - ln (1 - L/L¥ ) 1 June 1 Sep. 1 Dec. 1 March a (intercept) (slope, -K or K) L¥ = - a/b = t0 = - a/b = L(t) = ___________ [1 - exp (- _______ (t - _________ ))]
Fig. 17.3.4.2A Time series of length-frequencies of ponyfish. Data source: Tiews and Caces-Borja, 1965
B. Rastrelliger kanagurta:
GULLAND AND HOLT PLOT VON BERTALANFFY PLOT time of sampling L(t) D L/D t t - ln (1 - L/L¥ ) 1 Feb 1 March 1 May 1 June 1 July 1 August a (intercept) (slope, -K or K) L¥ = - a/b = t0 = - a/b = L(t) = ___________ [1 - exp (- _______ (t - _________ ))]
Fig. 17.3.4.2B Time series of length-frequencies of Indian mackerel. Data source: BFAR, Manila
Exercise 3.5.1 ELEFAN I
This exercise aims at explaining the details of the length-frequency restructuring process. Fig. 17.3.5.1A shows a (hypothetical) length-frequency sample, where the line shows the moving average. The worksheet table shows the calculation procedure and some results. Further explanations are given below for each step of the procedure.
1) Fill in the missing figures in the worksheet table.
2) Draw the bar diagram of the restructured data on the worksheet figure (B).
Worksheet 3.5.1
RESTRUCTURING OF LENGTH FREQUENCY SAMPLE STEP1 STEP2 STEP3 STEP4a STEP4b STEP5 STEP6 mid-lengthL orig. freq.FRQ (L) MA (L) FRQ/MA zeroes de-emphasized points highest positive points 5 4 4.6 a) 0.870 - 0.197 h) 2 -0.197 -0.109 p) 10 13 4.6 2 0.966 k) 0.966 s) 15 6 4.8 b) 1.250 e) 1 0.123 l) 0.123 20 0 4.0 0 -1.000 1 0 25 1 0.714 -0.341 i) 3 -0.341 -0.188 30 0 0.4 0 -1.000 2 35 0 1.0 c) 0 f) 1 -1.000 40 1 1.000 -0.077 2 -0.077 45 3 1.770 j) 2 1.062 m) 1.062 50 1 1 -0.127 q) 55 0 0 -1.000 1 -1.000 0 r) 60 1 0.4 d) 3 0.523 n) S = SP = (S /12) = M = 1.083 g) SN = ASP = - SP/SN = R = 0.552 o)
Fig. 17.3.5.1A Hypothetical length-frequency sample. Line indicates moving average over 5 neighbours
Step 1: Calculate the moving average, MA(L) over 5 neighbours.
Examples: (see Fig. 17.3.5.1 A and worksheet table)
MA (5) = (0 + 0 + 4 + 13 + 6)/5 = 4.6 a)
(two zeroes added at start of the sample)
MA (15) = (4 + 13 + 6 + 0 + 1)/5 = 4.8 b)
MA (35) = (1 + 0 + 0 + 1 + 3)/5 = 1.0 c)
MA (60) = (1 + 0 + 1 + 0 + 0)/5 = 0.4 d)
Step 2: Divide the original frequencies, FRQ(L), by the moving average (MA) and calculate their mean value, M:
Examples:
6/4.8 = 1.25 e)
0/1 = 0 f)
(12 = number of length intervals)
Step 3: Divide FRQ/MA by M and subtract 1
Examples:
0.870/1.083 - 1 = -0.197 h)
0.714/1.083 - 1 = -0.341 i)
3.000/1.083 - 1 = 1.770 j)
Step 4a: Count numbers of "zero neighbours" among the four neighbours (two zeroes added to each end of the sample).
Step 4b: De-emphasize positive isolated values: For each "zero-neighbour" the isolated point is reduced by 20%:
and if there are "zero-neighbours" then multiply this value by [1 - 0.2 * (no. of zeroes)]
Examples:
1.610 * (1 - 0.2 * 2) = 0.966 k)
0.154 * (1 - 0.2 * 1) = 0.123 l)
1.770 * (1 - 0.2 * 2) = 1.062 m)
1.308 * (1 - 0.2 * 3) = 0.523 n)
Note: In the most recent version (Gayanilo, Soriano and Pauly, 1988) the de-emphasizing has been made more pronounced by using the factor:
Step 4c: Calculate sum, SP, of positive (restructured) FRQs and calculate sum, SN, of negative (restructured) FRQs and calculate the ratio R = - SP/SN
Example:
SP = 0.966 + 0.123 + 1.062 + 0.523 = 2.674
SN = -0.197 - 1 - 0.340 - 1 - 1 - 0.076 - 0.230 - 1 = -4.845
R = - SP/SN = 2.674/4.845 = 0.552 o)
then multiply this value by R. Values > 0 are not changed.
Examples:
-0.197 * 0.552 = -0.109 p)
-0.231 * 0.552 = -0.123 q)
FRQ (55) = 0 r)
Plot the points in the diagram (Fig. 17.3.5.1B).
Step 6: Calculate ASP (available sum of peaks). Identify the highest point in each sequence of intervals with positive points (a "sequence" may consist of a single interval)
Examples:
0.966 is the highest point in the positive sequence 10-15 cm s)
1.062 is the highest point in the positive sequence 45-45 cm
0.523 is the highest point in the positive sequence 60-60 cm
ASP = 0.966 + 1.062 + 0.523 = 2.551
Fig. 17.3.5.1B Diagram for plotting points obtained after Step 5 (see text)
Exercise 3.5.1a ELEFAN I, continued
This exercise aims at illustrating the importance of the choice of the size of the length interval (cf. Exercise 3.4.1).
Fig. 17.3.5.1C1 shows a length-frequency sample (from Macdonald and Pitcher, 1979) of 523 pike from Heming Lake, Canada, grouped in 2 cm length intervals. There are five cohorts, determined on the basis of age reading of scales with the mean lengths shown in the following table:
ageyears mean lengthcm standard deviationcm 1 23.3 2.44 2 33.1 3.00 3 41.3 4.27 4 51.2 5.08 5 61.3 7.07
These data put us in a position to test ELEFAN I.
Fig. 17.3.5.1C2 shows the normally distributed components derived from scale readings, and Fig. C3 shows the restructured data.
Except for the largest fish ELEFAN I manages to place the ASPs (indicated by arrows) close to where the "true" mean lengths of the cohorts are, but like all other methods ELEFAN I has difficulties in handling the largest (oldest) fish.
Repeat the restructuring using Worksheet 3.5.1a on the basis of 4 cm intervals (see worksheet figure) instead of 2 cm intervals. Compare the results with those presented in Figs. 17.3.5.1C1 and C2.
Fig. 17.3.5.1D Regrouped length-frequency data, 4 cm length intervals (see Fig. 17.3.5.1C)
Worksheet 3.5.1a
RESTRUCTURING OF LENGTH FREQUENCY SAMPLE STEP1 STEP2 STEP3 STEP4a STEP4b STEP5 STEP6 mid-lengthL orig. freq.FRQ(L) MA(L) FRQ/MA zeroes de-emphasized points highest positive points 20 14 24 32 28 45 32 109 36 115 40 78 44 45 48 29 52 23 56 11 60 12 64 5 68 2 72 1 76 2 S = SP = (S /15) = M = SN = ASP = -SP/SN = R =
Fig. 17.3.5.1E Diagram for plotting points obtained after Step 5 using data from Fig. 17.3.5.1D
Exercise 4.2 The dynamics of a cohort (exponential decay model with variable Z)
Consider a cohort of a demersal fish species recruiting at an age t, which is arbitrarily put to zero. Recruitment is N (0) = 10000.
1) Calculate, using the worksheet, for the first ten half year periods the number of survivors at the beginning of each period and the numbers caught when mortality rates are as shown below:
age group(years) natural mortality fishing mortality Comments t1 - t2 M F 0.0-0.5 2.0 0.0 Cohort still on the nursery ground and exposed to heavy predation due to small size 0.5-1.0 1.5 0.0 1.0-1.5 0.5 0.2 Cohort under migration to fishing ground. Some fish escape through meshes 1.5-2.0 0.3 0.4 2.0-2.5 0.3 0.6 Cohort under full exploitation 2.5-3.0 0.3 0.6 3.0-3.5 0.3 0.6 3.5-4.0 0.3 0.6 Predation pressure reduced 4.0-4.5 0.3 0.6 4.5-5.0 0.3 0.6 Recruitment: N (0) = 10000
2) Give a graphical presentation of the results.
Worksheet 4.2
t1 - t2 M F Z e-0.5Z N(t1) N(t2) N(t1) - N(t2) F/Z C(t1, t2) 0.0-0.5 2.0 0.0 0.5-1.0 1.5 0.0 1.0-1.5 0.5 0.2 1.5-2.0 0.3 0.4 2.0-2.5 0.3 0.6 2.5-3.0 0.3 0.6 3.0-3.5 0.3 0.6 3.5-4.0 0.3 0.6 4.0-4.5 0.3 0.6 4.5-5.0 0.3 0.6
Exercise 4.2a The dynamics of a cohort (the formula for average number of survivors, Eq. 4.2.9)
Calculate the average number of survivors during the last 3 years for the cohort dealt with in Exercise 4.2 using the exact expression (Eq. 4.2.9) and the approximation demonstrated in Fig. 4.2.3, i.e. calculate N(2.0, 5.0).
Exercise 4.3 Estimation of Z from CPUE data
Assume that in Table 3.2.1.2 the numbers observed are the numbers caught of each cohort per hour trawling on 15 October 1983.
Estimate the total mortality for the stock under the assumption of constant recruitment, using Eq. 4.3.0.3:
Worksheet 4.3
cohort 1982 A 1982 S 1981 A 1981 S 1980 A 1) age t2 1.14 1.64 2.14 2.64 3.14 CPUE 111 67 40 24 15 cohort age t1 CPUE 1983 S 0.64 182 1982 A 1.14 111 ------ 1982 S 1.64 67 ------ ------ 1981 A 2.14 40 ------ ------ ------ 1981 S 2.64 24 ------ ------ ------ ------ 1) A = autumn, S = spring
Exercise 4.4.3 The linearized catch curve based on age composition data
Use the data presented in Table 4.4.3.1 of North Sea whiting (1974-1980).
Estimate Z from the catches of the 1974-cohort after plotting the catch curve. Calculate the confidence limits of the estimate of Z.
Worksheet 4.4.3
age(years)t yeary C(y, t, t+1) ln C(y, t, t+1) remarks (x) (y) 0 1 2 3 4 5 6 7 1981 - - slope: b = sb2 = [(sy/sx)2 - b2]/(n-2) = sb = sb * tn-2 = ________________ z = _______ ± _______
Exercise 4.4.5 The linearized catch curve based on length composition data
Length-frequency data from Ziegler (1979) for the threadfin bream (Nemipterus japonicus) from Manila Bay are given in the worksheet below, L¥ = 29.2 cm, K = 0.607 per year.
1) Carry out the length-converted catch curve analysis, using the worksheet.
2) Draw the catch curve.
3) Calculate the confidence limits for each estimate of Z.
Worksheet 4.4.5
L1 - L2 C (L1, L2) t(L1) D t z(slope) remarks a) b) c) (y) 7-8 11 not used, not under full exploitation 8-9 69 9-10 187 10-11 133 ? 11-12 114 ? 12-13 261 ? 13-14 386 ? 14-15 445 ? 15-16 535 ? 16-17 407 ? 17-18 428 ? 18-19 338 ? 19-20 184 ? 20-21 73 ? 21-22 37 ? 22-23 21 ? 23-24 19 ? 24-25 8 ? 25-26 7 too close to L¥ 26-27 2
Formulas to be used:
a) Eq. 3.3.3.2
b) Eq. 4.4.5.1
c) Eq. 4.4.5.2
Details of the regression analyses:
length group slope number of obs. Student's distrib. variance of slope stand. dev. of slope confidence limits of Z L1 - L2 Z n tn-2 sb2 sb Z ± tn-2 * sb
Exercise 4.4.6 The cumulated catch curve based on length composition data (Jones and van Zalinge method)
Length-frequency data from Ziegler (1979) for the threadfin bream (Nemipterus japonicus) from Manila Bay are given in the worksheet below,
L¥ = 29.2 cm, K = 0.607 per year.
1) Determine Z/K by the Jones and van Zalinge method, using the worksheet. (Start cumulation at largest length group).
2) Plot the "catch curve".
3) Calculate the 95% confidence limits for each estimate of Z (worksheet).
Worksheet 4.4.6
L1 - L2 C(L1, L2) S C (L1, L¥ ) cumulated ln S C (L1, L¥ ) ln (L¥ - L1) Z/K remarks (y) (x) (slope) 7-8 11 not used, not under full exploitation 8-9 69 9-10 187 10-11 133 ? 11-12 114 ? 12-13 261 ? 13-14 386 ? 14-15 445 ? 15-16 535 ? 16-17 407 ? 17-18 428 ? 18-19 338 ? 19-20 184 ? 20-21 73 ? 21-22 37 ? 22-23 21 ? 23-24 19 ? 24-25 8 ? 25-26 7 too close to L¥ 26-27 2
Details of the regression analyses
length group slope* K number of obs. Student's distrib. variance of slope stand. dev. of slope confidence limits of Z L1 - L2 Z n tn-2 sb2 sb Z ± K * tn-2 * sb
Exercise 4.4.6a The Jones and van Zalinge method applied to shrimp
Carapace length-frequency data for female shrimp (Penaeus semisulcatus) from Kuwait waters, 1974-1975, from Jones and van Zalinge (1981), are presented in the worksheet below. L¥ = 47.5 mm (carapace length). Input data are total landings in millions of shrimps per year by the Kuwait industrial shrimp fishery.
Note: In this case the length intervals have different sizes, because the length groups have been derived from commercial size groups, which are given in number of tails per pound (1 kg = 2.2 pounds).
1) Determine Z/K by the Jones and van Zalinge method using the worksheet.
2) Plot the "catch curve".
3) Calculate the 95 % confidence limits for each estimate of Z/K.
Worksheet 4.4.6a
carapace lengthmm numbers landed/year(millions) cumulated numbers/year(millions) remarks L1 - L2 C(L1, L2) S C(L1, L¥ ) ln S C(L1, L¥ ) ln (L¥ - L1) Z/K (y) (x) (slope) 11.18-18.55 2.81 18.55-22.15 1.30 22.15-25.27 2.96 25.27-27.58 3.18 27.58-29.06 2.00 29.06-30.87 1.89 30.87-33.16 1.78 33.16-36.19 0.98 36.19-40.50 0.63 40.50-47.50 0.63
Details of the regression analyses:
lower length slope number of obs. Student's distrib. variance of slope stand. dev. of slope confidence limits of slope L1 Z/K n tn-2 sb2 sb Z/K ± tn-2 * sb
Exercise 4.5.1 Beverton and Holt's Z-equation based on length data (applied to shrimp)
The same data as for Exercise 4.4.6a (from Jones and van Zalinge, 1981) on Penaeus semisulcatus are given in the worksheet below. L¥ = 47.5 mm (carapace length).
Estimate Z/K using Beverton and Holt's Z-equation (Eq. 4.5.1.1) and the worksheet (start cumulations at largest length group).
Worksheet 4.5.1
A B C D E F G H carapace length groupmm numbers landed/year(millions) cumulated catch mid-length *) *) *) *) L' (L1) - L2 C(L1, L2) S C(L1, L¥ ) Z/K 11.18-18.55 2.81 18.55-22.15 1.30 22.15-25.27 2.96 25.27-27.58 3.18 27.58-29.06 2.00 29.06-30.87 1.89 30.87-33.16 1.78 33.16-36.19 0.98 36.19-40.50 0.63 40.50-47.50 0.63 *) Column E: catch per length group * mid lengthColumn F: cumulation of column EColumn G: column F divided by column C
Exercise 4.5.4 The Powell-Wetherall method
Fork-length distribution (in %) of the blue-striped grunt (Haemulon sciurus) caught in traps at the Port Royal reefs off Jamaica during surveys in 1969-1973, are given in the worksheet below (from Munro, 1983, Table 10.35 p. 137).
1) Complete the worksheet, from the bottom.
2) Make the Powell-Wetherall plot and decide on the points to be included in the regression analysis.
3) Estimate Z/K and L (in fork-length).
4) What are the basic assumptions underlying the method?
Worksheet 4.5.4
A B C D *) E *) F *) G *) H *) L1 - L2(L' = L1) C(L1, L2) (% catch) S C(L',¥)(% cumulated) (x) (y) 14-15 1.8 14.5 15-16 3.4 15.5 16-17 5.8 16.5 17-18 8.4 17.5 18-19 9.1 18.5 19-20 10.2 19.5 20-21 14.3 20.5 21-22 13.7 21.5 22-23 10.0 22.5 23-24 6.3 23.5 24-25 6.4 24.5 25-26 5.3 25.5 26-27 3.3 26.5 27-28 1.8 27.5 28-29 0.3 28.5 *) Column D: sum column B (from the bottom)Column E: column B * column CColumn F: sum column E (from bottom)Column G: divide column F by column DColumn H: column G - column A (L' = L1)
Exercise 4.6 Plot of Z on effort (estimation of M and q)
For the trawl fishery in the Gulf of Thailand the effort (in millions of trawling hours) and the mean lengths of bulls eye (Priacanthus tayenus) over the years 1966-1974 were taken from Boonyubol and Hongskul (1978) and South China Sea Fisheries Development Programme (1978) and presented in the worksheet below (L¥ = 29.0 cm, K = 1.2 per year, Lc = 7.6 cm).
1) Calculate Z, using the worksheet.
2) Plot Z against effort and determine M (intercept) and q (slope).
3) Calculate the 95% confidence limits for the estimates of M and q.
Use the following two sets of input data (years):
a) The years 1966-1970
b) The years 1966-1974 and comment on the results.
Worksheet 4.6
year effort a) mean length cm 1966 2.08 15.7 1.97 1967 2.80 15.5 1968 3.50 16.1 1969 3.60 14.9 1970 3.80 14.4 1071 no data 1972 no data 1973 9.94 12.8 1974 6.06 12.8 a) in millions of trawling hours
Exercise 5.2 Age-based cohort analysis (Pope's cohort analysis)
Catch data by age group of the North Sea whiting (from ICES, 1981a) are presented in Tables 5.1.1 and 4.4.3.1.
1) Calculate fishing mortalities for the 1974 cohort (catch numbers given in Table 5.1.1 and M = 0.2 per year) by Pope's cohort analysis under the two different assumptions on the F for the oldest age group:
F6 = 1.0 per year
F6 = 2.0 per year
2) Plot F against age for the two cases above as well as for the case of Table 5.1.1, where
F6 = 0.5 per year
3) Discuss the significance of the choice of the terminal F (F6). Which of the three alternatives do you prefer? (Base your decision on the solution to Exercise 4.4.3, which deals with the same data set).
Exercise 5.3 Jones' length-based cohort analysis
As in Exercises 4.4.6a and 4.5.1 we use the landings of female Penaeus semisulcatus of the 74/75-cohort from Kuwait waters (from Jones and van Zalinge, 1981). These data were derived from the total number of processed prawns in each of ten market categories (cf. Worksheet 5.3).
1) Using Worksheet 5.3 and the formulas given below, estimate fishing mortalities and stock numbers by means of Jones' length-based cohort analysis, using the parameters:
K = 2.6 per year
M = 3.9 per year
L¥ = 47.5 mm (carapace length)
2) Give your opinion on our choice of terminal F/Z (= 0.1).
3) Is the cohort analysis a dependable method in this case? (The value of M is a "guesstimate").
Worksheet 5.3
length group nat. mort. factor number caught(mill.) number of survivors exploitation rate fishing mort. total mort. g) a) b) c) d) e) L1 - L2 H(L1, L2) C(L1, L2) N(L1) F/Z F Z 11.18-18.55 2.81 18.55-22.15 1.30 22.1.5-25.27 2.96 25.27-27.58 3.18 27.58-29.06 2.00 29.06-30.87 1.89 30.87-33.16 1.78 33.16-36.19 0.98 36.19-40.50 0.63 40.50-47.50 0.63 f) a) b) N(L1) = [N(L2) * H(L1, L2) + C(L1, L2)] * H(L1, L2) c) F/Z = C(L1, L2)/[N(L1) - N(L2)] d) F = M * (F/Z)/(1 - F/Z) e) Z = F + M f) N(last L1) = C(last L1, L¥ )/(F/Z) g) carapace lengths in mm corresponding to the market categories (in units of number of tails per pound):
no/lb: 400 110 70 50 40 35 30 25 20 <15 L1: 11.18 18.55 22.15 25.27 27.58 29.06 30.87 33.16 36.19 40.5 L2: 18.55 22.15 25.27 27.58 29.06 30.87 33.16 36.19 40.5 47.5
Exercise 6.1 A mathematical model for the selection ogive
Draw a selection curve using the parameters:
L50% = 13.6 cm and L75% = 14.6 cm
Use the logistic curve SL = 1/[1 + exp(S1 - S2 * L)]
Exercise 6.5 Estimation of the selection ogive from a catch curve
Data on catch by length group of Upeneus vittatus were taken from Table 4.4.5.1. K = 0.59 per year, L¥ = 23.1 cm, t0 = -0.08 year
1) Estimate the logistic curve St = 1/[1 + exp(T1 - T2 * t)]
2) Estimate L50% = L¥ * [1 - exp(K * (t0 - t50%))] and L75%
3) Evaluate the choice of first length interval given in Table 4.4.5.1.
Worksheet 6.5
A B C D E F G H I length groupL1 - L2 ta) D t C(L1, L2) ln (C/D t)b) St obs.c) ln (1/S - 1)d) est.e) remarks (x) (y) 6-7 0.56 0.102 3 3.38 (not used) 7-8 0.67 0.109 143 7.18 8-9 0.78 0.116 271 7.76 9-10 0.90 0.125 318 7.86 10-11 1.03 0.134 416 8.04 11-12 1.17 0.146 488 8.11 12-13 1.32 0.160 614 8.25 13-14 1.49 0.177 613f) 8.15 used for the analysis to estimate Z (see Table 4.4.5.1) 14-15 1.67 0.197 493 f) 7.83 15-16 1.88 0.223 278 f) 7.13 16-17 2.12 0.257 93 f) 5.89 17-18 2.40 0.303 73 f) 5.48 18-19 2.74 0.370 7 f) 2.94 19-20 3.15 0.473 2 f) 1.44 20-21 3.70 0.659 2 1.11 21-22 4.53 1.094 0 - 22-23 6.19 4.094 1 -1.40 23-24 - - 1 - a) t [(L1 + L2)/2], age corresponding to interval mid-length b) ln(C/D t), dependent variable in catch curve regression analysis c) S(t) obs. = C/[D t * exp(a - Z * t)], observed selection ogive Z = 4.19 and a = 14.8 (from Table 4.4.5.1) d) ln(1/S - 1), dependent variable in regression e) S(t) est. = 1/[1 + exp(T1 - T2 * t)], theoretical (estimated) selection ogive f) points used in the catch curve analysis (cf. Table 4.4.5.1)
Exercise 6.7 Using a selection curve to adjust catch samples
1) Adjust the length-frequencies for Upeneus vittatus (from the data given in Table 4.4.5.1) using the results of Exercise 6.5:
L50% = 13.6 cm and L75% = 14.6 cm
S1 =
S2 =
SL =
2) Draw a histogram of the original and the adjusted frequencies excluding the raised (estimated unbiased) frequencies which you think are not safely estimated.
Worksheet 6.7
length groupL1 - L2 midpoint observed biased sample selection ogiveSL estimated unbiased sample 6-7 3 7-8 143 8-9 271 9-10 318 10-11 416 11-12 488 12-13 614 13-14 613 14-15 493 15-16 278 16-17 93 17-18 73 18-19 7 19-20 2 20-21 2 21-22 0 22-23 1 23-24 1
Exercise 7.2 Stratified random sampling versus simple random sampling and proportional sampling
This exercise illustrates the gain in precision obtained from stratification. Use Table 7.2.2.
1) Estimate the variance of the mean landing Y from three different sampling methods, when the total sample size is n = 20, using the worksheets.
a) Simple random sampling
b) Proportional sampling: a sample of 20% from each stratum
Worksheet 7.2 for a) and b)
stratum j s(j) s(j)2 N(j) 1 large 2 medium 3 small total as defined by Eq. 2.1.3. a) Simple random sampling b) Proportional sampling
Worksheet 7.2 for c)
stratum s(j) * N(j) 1 large 2 medium 3 small total 1.00 n = 20 c) Optimum stratified sampling
2) Calculate the standard deviations and compare the allocations per stratum.
random proportional optimum allocation per stratum 1 large 2 medium 3 small
Exercise 8.3 The yield per recruit model of Beverton and Holt (yield per recruit, biomass per recruit as a function of F)
Pauly (1980) determined the following parameters for Leiognathus splendens (cf. Exercise 3.1.2). W¥ = 64 g, K = 1.0 per year, t0 = -0.2 year, Tr = 0.2 year, M = 1.8 per year.
1) Draw the Y/R and the B/R curves, for three different values of Tc: Tc = Tr = 0.2 year, Tc = 0.3 year and Tc = 1.0 year.
Worksheet 8.3
Tc = Tr = 0.2 Tc = 0.3 Tc = 1.0 F Y/R B/R Y/R B/R Y/R B/R 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.5 4.0 4.5 5.0 100.0
2) Try to explain why MSY increases when Tc increases (without the use of mathematics). Is the above statement a general rule, i.e. does it hold for any increase of Tc?
3) Read the (approximate) values of FMSY and MSY/R from the worksheet. Comment on your findings under the assumption that the present level of F is 1.0.
Exercise 8.4 Beverton and Holt's relative yield per recruit concept
For the swordfish (Xiphias gladius) off Florida, Berkeley and Houde (1980) determined the parameters:
L¥ = 309 cm, K = 0.0949 per year and M = 0.18 per year
Draw the relative yield per recruit curve, (Y/R') as a function of E, for two different values of the 50% retention length:
Lc = 118 cm and Lc = 150 cm.
Worksheet 8.4
Lc = 118 cm Lc = 150 cm E (Y/R)' (Y/R)' (F) 0 0 0.1 0.020 0.2 0.045 0.3 0.077 0.4 0.120 0.5 M = 0.180 0.6 0.270 0.7 0.42 0.8 0.72 0.9 1.62 1.0 ¥
Exercise 8.6 A predictive age-based model (Thompson and Bell analysis)
In the (hypothetical example) given in the table below a fish stock is exploited by two different gears, viz. beach seines and gill nets. These gears account for the total catch from the stock. A sampling programme for estimation of total numbers caught by age group and by gear has been running for the years 1975-1985.
Based on the total numbers caught a VPA has been made and the estimated F values for the last data year (1985) have been separated into a beach seine component, FB and a gill net component FG (cf. Eq. 8.6.1). The average recruitment (number of 0-group fish) for the years 1975 to 1985 has been estimated from VPA to be 1000000 fish. The natural mortalities are assumed to take the age-specific values. These data are presented in part a of the worksheet.
Use Worksheet 8.6a to solve the following problems:
1) Under the assumption that fishing mortality remains the same as in 1985 and that the recruitment is of average size, predict (based on the assumption of equilibrium):
1.1) The number of survivors (stock numbers) by age group.
1.2) Numbers caught by age group for each gear.
1.3) Yield of each gear.
Use Worksheet 8.6b to solve the following problems:
2) Under the assumption that the gill net effort remains the same as in 1985 but that the beach seine fishery is closed (and that the recruitment is of average size) predict as 1.1, 1.2 and 1.3 above.
3) Would you, based on the results of 1) and 2) recommend a closure of the beach seine fishery?
Worksheet 8.6
a. No change in fishing effort:
age group mean weight (g) beach seine mortality gill net mortality natural mortality total mortality stock number beach seine catch gill net catch beach seine yield gill net yield total yield t w FB FG M Z '000 CB CG YB YG YB + YG 0 8 0.05 0.00 2.00 1000 1 283 0.40 0.00 0.80 2 1155 0.10 0.19 0.30 3 2406 0.01 0.59 0.20 4 3764 0.00 0.33 0.20 5 5046 0.00 0.09 0.20 6 6164 0.00 0.02 0.20 7 7090 0.00 0.00 0.20 total Z = FB + FG + M N(t + 1) = N(t) * exp(-Z) CB = FB * N * (1 - exp(-Z))/Z CG = FG * N * (1 - exp(-Z))/Z
b. Closure of the beach seine fishery:
age group mean weight (g) beach seine mortality gill net mortality natural mortality total mortality stock number beach seine catch gill net catch beach seine yield gill net yield total yield t w FB FG M Z '000 CB CG YB YG YB + YG 0 8 1 283 2 1155 3 2406 4 3764 5 5046 6 6164 7 7090 total
Exercise 8.7 A predictive length-based model (Thompson and Bell analysis)
For this exercise a hypothetical example is used:
M = 0.3 per year, K = 0.3 per year, L¥ = 60.0 cm
Recruitment, N(10, 15) = 1000
length class fishing mortality mean body weight g price per kg natural mortality factor L1 - L2 F (L1, L2) (L1, L2) H (L2, L2) a) 10-15 0.03 19.5 1.0 1.05409 15-20 0.20 53.6 1.0 1.06066 20-25 0.40 113.9 1.5 1.06904 25-30 0.70 207.9 1.5 1.08012 30-35 0.70 343.3 2.0 1.09544 35-40 0.70 527.3 2.0 1.11803 40-L¥ 0.70 767.7 2.0 - a) H(L1, L2) = ((L¥ - L1)/(L¥ - L2))M/2K
Do the length-converted Thompson and Bell analysis on the example.
Worksheet 8.7
length classL1-L2 P(L1, L2) N(L1)a) N(L2)a) mean biomass*D tb) catchC(L1, L2)c) yield(L1, L2)d) value(L1, L2)e) 10-15 0.03 1000 15-20 0.20 20-25 0.40 25-30 0.70 30-35 0.70 35-40 0.70 40-L¥ 0.70 f) Total _____ a) N(L1) of a length group is equivalent to the N(L2) of the previous length groupN(L2) = N(L1) * [1/H(L1, L2) - E(L1, L2)]/[H(L1, L2) - E(L1, L2)]where E(L1, L2) = F(L1, L2)/Z(L1. L2) where Nmean(L1, L2) * Dt = [N(L1) - N(L2)]/Z(L1, L2) c) C(L1, L2) = F(L1, L2) * Nmean(L1, L2) * D t e) value(L1, L2) = yield(L1, L2) * price(L1, L2)
Exercise 8.7a A predictive length-based model (yield curve, Thompson and Bell analysis)
1) Do the same exercise as in Exercise 8.7 but under the assumption of a 100% increase in fishing effort (Worksheet 8.7a).
Worksheet 8.7a
length classL1-L2 F(L1, L2) N(L1)a) N(L2)a) mean biomass*D tb) catchC(L1, L2)c) yield(L1, L2)d) value(L1, L2)e) 10-15 1000 15-20 20-25 25-30 30-35 35-40 40-L¥ f) Total _____ a) N(L1) of a length group is equivalent to the N(L2) of the previous length groupN(L2) = N(L1) * [1/H(L1, L2) - E(L1, L2)]/[H(L1, L2) - E(L1, L2)]where E(L1, L2) = F(L1, L2)/Z(L1. L2) where Nmean(L1, L2) * Dt = [N(L1) - N(L2)]/Z(L1, L2) c) C(L1, L2) = F(L1, L2) * Nmean(L1, L2) * D t e) value(L1, L2) = yield(L1, L2) * price(L1, L2)
2) Use the result of 1) combined with the solution to Exercise 8.7 and the results given in the table below to draw the yield, the mean biomass and the value curves.
F-factor yield mean biomass value x * D t 0.0 0.00 1445.41 0.00 0.2 116.38 865.89 226.11 0.4 154.48 585.63 296.49 0.6 165.12 426.42 312.70 0.8 164.75 326.87 307.56 1.0 1.2 153.25 213.94 277.35 1.4 146.23 180.15 260.38 1.6 139.37 154.84 244.14 1.8 132.95 135.40 229.10 2.0 MSY = 165.8 at X = 0.69 biomass at MSY = 378.8 MSE = 312.9 at X = 0.61 biomass at MSE = 405.7
Exercise 9.1 The Schaefer model and the Fox model
In Worksheet 9.1 are given total catch and total effort in standard boat days for the years 1969 through 1978 for the shrimp fishery in the Arafura Sea. Catches are mainly composed of the five species Penaeus merguiensis, Penaeus semisulcatus, Penaeus monodon, Metapenaeus ensis and Parapenaeopsis sculptilis (from Naamin and Noer, 1980).
1) Calculate Y/f (kg per boat day) and ln (Y/f) and plot them against effort.
2) Estimate MSY and fMSY by the Schaefer model.
3) Estimate MSY and fMSY by the Fox model.
4) Plot yield against effort and draw the yield curves estimated by the two methods.
Worksheet 9.1
year yield (tonnes) headless effort f(i) boat days Schaefer Y/f kg/boat day Fox ln (Y/f) ln(kg/boat day) i Y(i) (x) (y) (y) 1969 546.7 1224 1970 812.4 2202 1971 2493.3 6684 1972 4358.6 12418 1973 6891.5 16019 1974 6532.0 21552 1975 4737.1 24570 1976 5567.4 29441 1977 5687.7 28575 1978 5984.0 30172 mean values standard deviations intercept (Schaefer: a, Fox: c) *) slope (Schaefer: b. Fox: d) *) *) a, b replaced by c, d for the Fox model
continuation of Worksheet 9.1
Schaefer Fox variance of slopesb2 = [(sy/sx)2 - b2]/(10-2) standard deviation of slope, sbconfidence limits of slope,upper limit, b + tn-2 * sblower limit, b - tn-2 * sb variance of intercept standard deviation of interceptStudent's distribution, tn-2confidence limits of interceptupper limit, a + tn-2* salower limit, a - tn-2 * sa MSY - a2/(4b) = -(1/d) * exp(c - 1) = fMSY - a/(2b) = - 1/d =
Worksheet 9.1a (for drawing the yield curves)
fboat days Schaeferyield (tonnes) Foxyield (tonnes) 5000 10000 15000 20000 25000 fMSY 30000 35000 fMSY 40000 45000
Exercise 13.8 The swept area method, precision of the estimate of biomass, estimation of MSY and optimal allocation of hauls
The data for this exercise were taken from report no. 8 of Project KEN/74/023: "Offshore trawling survey", which deals with the stock assessment of Kenyan demersal resources from surveys in the period 1979-81. The data used here are a modified set on the catch of the small-spotted grunt, Pomadasys opercularis. The data are given as catch in weight per unit time (Cw/t) in kg per hour trawling for 23 hauls covering two strata (in Worksheet 13.8). The vessel speed, current speed, both in knots (nautical mile per hour) and trawl wing spread (hr * X2) are also given.
1) Apply Eq. 13.5.3 to calculate the distance, D, covered per hour and Eq. 13.5.1 to calculate the area swept per hour, a, for each haul. Calculate the yield, Cw, per unit of area for each haul using Eq. 13.6.2 (data in the worksheet, 1 nautical mile (nm) = 1852 m).
2) Calculate for each stratum the estimate of mean catch per unit area Ca and the confidence limits of the estimates (using Eq. 2.3.1). Calculate using Eqs. 13.7.5 and 13.6.3 an estimate of the mean biomass for the total area, when A1 = 24 square nautical miles (sq.nm), A2 = 53 sq.nm and X1 (catchability) is assigned the value 0.5.
3) Estimate MSY using Gulland's formula, with M = Z = 0.6 per year (i.e. we assume a virgin stock).
4) Construct a graph showing the maximum relative error for the mean catch per area against the number of hauls for each of the two strata. We define (cf. Section 7.1, Fig. 7.1.1)
where s is the standard deviation of the estimate of the catch in weight per unit area:
5) Assume that you have financial resources to make 200 hauls. Allocate these 200 hauls between the two strata for optimum stratified sampling (cf. Section 7.2).
Worksheet 13.8
STRATUM 1:
A B C D E F G H I J HAUL CPUE VESSEL CURRENT TRAWL AREA CPUA no.i Cw/hkg/h speedVSknots coursedir Vdegrees speedCSknots directiondir Cdegrees spreadhr * X2m distanceDnm sweptasq.nm Cw/a = Cakg/sq.nm 1 7.0 2.8 220 0.5 90 18 2 7.0 3.0 210 0.5 180 16 3 5.0 3.0 200 0.3 135 17 4 4.0 3.0 180 0.4 230 18 5 1.0 3.0 90 0.5 270 17 6 4.0 3.0 45 0.4 160 18 7 9.0 3.5 25 0.4 200 18 8 0.0 3.0 210 0.3 300 18 9 0.0 3.5 0 0.4 0 18 10 14.0 2.8 45 0.6 0 18 11 8.0 3.0 120 0.3 300 18 STRATUM 2: 12 42.0 4.0 30 0.5 160 17 13 98.0 3.3 215 0.4 90 17 14 223.0 3.9 30 0.0 0 17 15 59.0 3.8 35 0.3 180 17 16 32.0 3.5 210 0.5 270 17 17 6.0 2.8 210 0.5 330 17 18 66.0 3.8 45 0.5 30 17 19 60.0 4.0 30 0.5 180 18 20 48.0 4.0 210 0.5 180 18 21 52.0 3.8 20 0.4 180 18 22 48.0 4.0 30 0.5 190 18 23 18.0 3.0 210 0.3 190 18
Confidence limits of stratum number of hauls standard deviation Student's distr. confidence limits for n s s/Ö n tn-1 1 2
Worksheet 13.8a (for plotting graph maximum relative error)
number of hauls Student's distribution stratum 1 stratum 2 n tn-1 e a) e a) 5 2.78 10 2.26 20 2.09 50 2.01 100 1.98 200 1.97
Worksheet 13.8b (optimum allocation)
stratum standard deviation of Ca area of stratum s A A * s A * s/S A * s 200 * A * s/S A * s 1 2 Total |
Paul's Online Math Notes
[Notes]
Calculus I - Notes
Limits Previous Chapter Next Chapter Applications of Derivatives The Definition of the Derivative Previous Section Next Section Differentiation Formulas
## Interpretations of the Derivative
Before moving on to the section where we learn how to compute derivatives by avoiding the limits we were evaluating in the previous section we need to take a quick look at some of the interpretations of the derivative. All of these interpretations arise from recalling how our definition of the derivative came about. The definition came about by noticing that all the problems that we worked in the first section in the chapter on limits required us to evaluate the same limit.
Rate of Change
The first interpretation of a derivative is rate of change. This was not the first problem that we looked at in the limit chapter, but it is the most important interpretation of the derivative. If represents a quantity at any x then the derivative represents the instantaneous rate of change of at .
Example 1 Suppose that the amount of water in a holding tank at t minutes is given by . Determine each of the following. (a) Is the volume of water in the tank increasing or decreasing at minute? [Solution] (b) Is the volume of water in the tank increasing or decreasing at minutes? [Solution] (c) Is the volume of water in the tank changing faster at or minutes? [Solution] (d) Is the volume of water in the tank ever not changing? If so, when? [Solution] Solution In the solution to this example we will use both notations for the derivative just to get you familiar with the different notations. We are going to need the rate of change of the volume to answer these questions. This means that we will need the derivative of this function since that will give us a formula for the rate of change at any time t. Now, notice that the function giving the volume of water in the tank is the same function that we saw in Example 1 in the last section except the letters have changed. The change in letters between the function in this example versus the function in the example from the last section won’t affect the work and so we can just use the answer from that example with an appropriate change in letters. The derivative is. Recall from our work in the first limits section that we determined that if the rate of change was positive then the quantity was increasing and if the rate of change was negative then the quantity was decreasing. We can now work the problem. (a) Is the volume of water in the tank increasing or decreasing at minute? In this case all that we need is the rate of change of the volume at or, So, at the rate of change is negative and so the volume must be decreasing at this time. (b) Is the volume of water in the tank increasing or decreasing at minutes? Again, we will need the rate of change at . In this case the rate of change is positive and so the volume must be increasing at . (c) Is the volume of water in the tank changing faster at or minutes? To answer this question all that we look at is the size of the rate of change and we don’t worry about the sign of the rate of change. All that we need to know here is that the larger the number the faster the rate of change. So, in this case the volume is changing faster at than at . (d) Is the volume of water in the tank ever not changing? If so, when? The volume will not be changing if it has a rate of change of zero. In order to have a rate of change of zero this means that the derivative must be zero. So, to answer this question we will then need to solve This is easy enough to do. So at the volume isn’t changing. Note that all this is saying is that for a brief instant the volume isn’t changing. It doesn’t say that at this point the volume will quit changing permanently. If we go back to our answers from parts (a) and (b) we can get an idea about what is going on. At the volume is decreasing and at the volume is increasing. So at some point in time the volume needs to switch from decreasing to increasing. That time is . This is the time in which the volume goes from decreasing to increasing and so for the briefest instant in time the volume will quit changing as it changes from decreasing to increasing.
Note that one of the more common mistakes that students make in these kinds of problems is to try and determine increasing/decreasing from the function values rather than the derivatives. In this case if we took the function values at , and we would get,
Clearly as we go from to the volume has decreased. This might lead us to decide that AT the volume is decreasing. However, we just can’t say that. All we can say is that between and the volume has decreased at some point in time. The only way to know what is happening right at is to compute and look at its sign to determine increasing/decreasing. In this case is negative and so the volume really is decreasing at .
Now, if we’d plugged into the function rather than the derivative we would have gotten the correct answer for even though our reasoning would have been wrong. It’s important to not let this give you the idea that this will always be the case. It just happened to work out in the case of .
To see that this won’t always work let’s now look at . If we plug and into the volume we can see that again as we go from to the volume has decreased. Again, however all this says is that the volume HAS decreased somewhere between and . It does NOT say that the volume is decreasing at . The only way to know what is going on right at is to compute and in this case is positive and so the volume is actually increasing at .
So, be careful. When asked to determine if a function is increasing or decreasing at a point make sure and look at the derivative. It is the only sure way to get the correct answer. We are not looking to determine is the function has increased/decreased by the time we reach a particular point. We are looking to determine if the function is increasing/decreasing at that point in question.
Slope of Tangent Line
This is the next major interpretation of the derivative. The slope of the tangent line to at is . The tangent line then is given by,
Example 2 Find the tangent line to the following function at . Solution We first need the derivative of the function and we found that in Example 3 in the last section. The derivative is, Now all that we need is the function value and derivative (for the slope) at . The tangent line is then,
Velocity
Recall that this can be thought of as a special case of the rate of change interpretation. If the position of an object is given by after t units of time the velocity of the object at is given by .
Example 3 Suppose that the position of an object after t hours is given by, Answer both of the following about this object. (a) Is the object moving to the right or the left at hours? [Solution] (b) Does the object ever stop moving? [Solution] Solution Once again we need the derivative and we found that in Example 2 in the last section. The derivative is, (a) Is the object moving to the right or the left at hours? To determine if the object is moving to the right (velocity is positive) or left (velocity is negative) we need the derivative at . So the velocity at is positive and so the object is moving to the right at . (b) Does the object ever stop moving? The object will stop moving if the velocity is ever zero. However, note that the only way a rational expression will ever be zero is if the numerator is zero. Since the numerator of the derivative (and hence the speed) is a constant it can’t be zero. Therefore, the object will never stop moving. In fact, we can say a little more here. The object will always be moving to the right since the velocity is always positive.
We’ve seen three major interpretations of the derivative here. You will need to remember these, especially the rate of change, as they will show up continually throughout this course.
Before we leave this section let’s work one more example that encompasses some of the ideas discussed here and is just a nice example to work.
Example 4 Below is the sketch of a function . Sketch the graph of the derivative of this function, . Solution At first glance this seems to an all but impossible task. However, if you have some basic knowledge of the interpretations of the derivative you can get a sketch of the derivative. It will not be a perfect sketch for the most part, but you should be able to get most of the basic features of the derivative in the sketch. Let’s start off with the following sketch of the function with a couple of additions. Notice that at , , and the tangent line to the function is horizontal. This means that the slope of the tangent line must be zero. Now, we know that the slope of the tangent line at a particular point is also the value of the derivative of the function at that point. Therefore, we now know that, This is a good starting point for us. It gives us a few points on the graph of the derivative. It also breaks the domain of the function up into regions where the function is increasing and decreasing. We know, from our discussions above, that if the function is increasing at a point then the derivative must be positive at that point. Likewise, we know that if the function is decreasing at a point then the derivative must be negative at that point. We can now give the following information about the derivative. Remember that we are giving the signs of the derivatives here and these are solely a function of whether the function is increasing or decreasing. The sign of the function itself is completely immaterial here and will not in any way effect the sign of the derivative. This may still seem like we don’t have enough information to get a sketch, but we can get a little bit more information about the derivative from the graph of the function. In the range we know that the derivative must be negative, however we can also see that the derivative needs to be increasing in this range. It is negative here until we reach and at this point the derivative must be zero. The only way for the derivative to be negative to the left of and zero at is for the derivative to increase as we increase x towards . Now, in the range we know that the derivative must be zero at the endpoints and positive in between the two endpoints. Directly to the right of the derivative must also be increasing (because it starts at zero and then goes positive therefore it must be increasing). So, the derivative in this range must start out increasing and must eventually get back to zero at . So, at some point in this interval the derivative must start decreasing before it reaches . Now, we have to be careful here because this is just general behavior here at the two endpoints. We won’t know where the derivative goes from increasing to decreasing and it may well change between increasing and decreasing several times before we reach . All we can really say is that immediately to the right of the derivative will be increasing and immediately to the left of the derivative will be decreasing. Next, for the ranges and we know the derivative will be zero at the endpoints and negative in between. Also, following the type of reasoning given above we can see in each of these ranges that the derivative will be decreasing just to the right of the left hand endpoint and increasing just to the left of the right hand endpoint. Finally, in the last region we know that the derivative is zero at and positive to the right of . Once again, following the reasoning above, the derivative must also be increasing in this range. Putting all of this material together (and always taking the simplest choices for increasing and/or decreasing information) gives us the following sketch for the derivative. Note that this was done with the actual derivative and so is in fact accurate. Any sketch you do will probably not look quite the same. The “humps” in each of the regions may be at different places and/or different heights for example. Also, note that we left off the vertical scale because given the information that we’ve got at this point there was no real way to know this information. That doesn’t mean however that we can’t get some ideas of specific points on the derivative other than where we know the derivative to be zero. To see this let’s check out the following graph of the function (not the derivative, but the function). At and we’ve sketched in a couple of tangent lines. We can use the basic rise/run slope concept to estimate the value of the derivative at these points. Let’s start at . We’ve got two points on the line here. We can see that each seem to be about one-quarter of the way off the grid line. So, taking that into account and the fact that we go through one complete grid we can see that the slope of the tangent line, and hence the derivative, is approximately -1.5. At it looks like (with some heavy estimation) that the second point is about 6.5 grids above the first point and so the slope of the tangent line here, and hence the derivative, is approximately 6.5. Here is the sketch of the derivative with the vertical scale included and from this we can see that in fact our estimates are pretty close to reality. Note that this idea of estimating values of derivatives can be a tricky process and does require a fair amount of (possible bad) approximations so while it can be used, you need to be careful with it.
We’ll close out this section by noting that while we’re not going to include an example here we could also use the graph of the derivative to give us a sketch of the function itself. In fact, in the next chapter where we discuss some applications of the derivative we will be looking using information the derivative gives us to sketch the graph of a function.
The Definition of the Derivative Previous Section Next Section Differentiation Formulas Limits Previous Chapter Next Chapter Applications of Derivatives
[Notes]
© 2003 - 2016 Paul Dawkins |
"The Basics of Proof Writing: Defining, Identifying, and Solving with Theorems" In this lesson, you will learn what a proof is and how to construct a proof for a given hypothesis and conclusion. You will also become familiar with
Slide1this lesson defines what a proof is and shows how to write aproof for a given hypothesis and conclusions. You will be able to identify the postulates, axioms, and theorems that justify the statements in a
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About "The Basics of Proof Writing: Defining, Identifying, and Solving with Theorems" In this lesson, you will learn what a proof is and how to construct a proof for a given hypothesis and conclusion. You will also become familiar with
PowerPoint presentation about '"The Basics of Proof Writing: Defining, Identifying, and Solving with Theorems" In this lesson, you will learn what a proof is and how to construct a proof for a given hypothesis and conclusion. You will also become familiar with'. This presentation describes the topic on Slide1this lesson defines what a proof is and shows how to write aproof for a given hypothesis and conclusions. You will be able to identify the postulates, axioms, and theorems that justify the statements in a. The key topics included in this slideshow are . Download this presentation absolutely free.
Presentation Transcript
Slide1this lesson defines what a proof is and shows how to write aproof for a given hypothesis and conclusions.
Slide2You will be able to identify the postulates, axioms, and theorems that justify the statements in a proof. You will be able to use a theorem to solve problems.
Slide3Theorem: A cat has nine tails. Proof: No cat has eight tails. A cat has one tail more than no cat. Therefore, a cat has nine tails.
Slide4Proof – A series of true statements leading to a desired conclusion Theorem – A statement that can be proven true Given – Specified Prove – To show that a conclusion is true
Slide5If Angles are vertical angles, then their measures are equal. To start a proof, clearly state what is given and is to prove . Given or hypothesis: “angles are vertical” Conclusion: “their measures are equal”
Slide6Next, draw a picture of the given. a b c l d m a and d are vertical angles c and b are vertical angles
Slide7To Prove: m a = m d m c = m b Work in 2 columns. You are now ready.
Slide8Statement Lines l & m intersect to form vertical angles a & d m a + m b = 180 o m d + m b = 180 o Proof 1. Given 2. a & b are adjacent on m and are supplementary 3. b and d are adjacent on l and are supplementary
Slide9m a + m b = m b + m d m a = m d 4) Axiom I, substitution and steps 2 & 3 5) Axiom 3, if equals are subtracted from equals, the differences are equal |
# How do you calculate a 20% slope?
## How do you calculate a 20% slope?
Calculating the Slope Percentage Slope percentage is calculated in much the same way as the gradient. Convert the rise and run to the same units and then divide the rise by the run. Multiply this number by 100 and you have the percentage slope.
## What is a 20% grade?
Slope
18 1 32.5
19 1 34.4
20 1 36.4
What is the formula for calculating gradient?
To calculate the gradient of a straight line we choose two points on the line itself. The difference in height (y co-ordinates) ÷ The difference in width (x co-ordinates). If the answer is a positive value then the line is uphill in direction. If the answer is a negative value then the line is downhill in direction.
### What is a 15% slope?
Example: a road with 15% slope has an angle of 8.53°.
### How do you find the gradient percentage?
Percent of slope is determined by dividing the amount of elevation change by the amount of horizontal distance covered (sometimes referred to as “the rise divided by the run”), and then multiplying the result by 100.
What is a 20 incline?
0-10% = moderate incline 10-15% = slightly steep incline 15-20 = pretty steep incline 20-25% = steep incline 25-30+% = very steep incline.
#### What is a 20 30 in percentage?
Now we can see that our fraction is 66.666666666667/100, which means that 20/30 as a percentage is 66.6667%.
#### What is the C in Y MX C?
The equation y=mx+c is the general equation of any straight line where m is the gradient of the line (how steep the line is) and c is the y -intercept (the point in which the line crosses the y -axis).
2. FORMULA: Gradient = VI (Difference in height)
3. HE (Horizontal distance)
4. MAGNETIC DECLINATION AND MAGNETIC BEARING.
5. Mag N True N.
## What is a 30 degree slope?
30 degrees is equivalent to a 58% grade which is another way to describe the magnitude of a slope.
## How to work out the ratio gradient of a slope?
To work out the gradient of an existing slope you will need the distance of the slope (run) and the height of the slope (rise). The method that follows may not be the most mathematical way to work out the ratio gradient of a slope – but it is really simple and has always worked for me! Remember to convert units so they are the same.
How do you calculate the percent of a steep hill?
When driving, have you ever noticed that the warning sign indicating a steep hill has a percentage on it? This Percent Slope Calculator can calculate that slope percentage! To get the percent of a slope, simply enter the Rise (height of the hill) and the Run (width of the hill) below and press “Calculate Slope”. How to Calculate Percent Slope
### What is an example of slope as Grade?
Example – Slope as Grade. Slope as grade for an elevation of 1 m over a distance of 2 m can be calculated as. Sgrade (%) = (1 m)/ (2 m) = 50 (%) |
Introduction
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# Introduction
## Introduction
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##### Presentation Transcript
1. Introduction Geometric figures can be graphed in the coordinate plane, as well as manipulated. However, before sliding and reflecting figures, the definitions of some important concepts must be discussed. Each of the manipulations that will be discussed will move points along a parallel line, a perpendicular line, or a circular arc. In this lesson, each of these paths and their components will be introduced. 5.1.1: Defining Terms
2. Key Concepts A point is not something with dimension; a point is a “somewhere.” A point is an exact position or location in a given plane. In the coordinate plane, these locations are referred to with an ordered pair (x, y), which tells us where the point is horizontally and vertically. The symbol A (x, y) is used to represent point A at the location (x, y). 5.1.1: Defining Terms
3. Key Concepts, continued A line requires two points to be defined. A line is the set of points between two reference points and the infinite number of points that continue beyond those two points in either direction. A line is infinite, without beginning or end. This is shown in the diagram below with the use of arrows. The symbol is used to represent line AB. 5.1.1: Defining Terms
4. Key Concepts, continued You can find the linear distance between two points on a given line. Distance along a line is written as d(PQ) where P and Q are points on a line. Like a line, a ray is defined by two points; however, a ray has only one endpoint. The symbol is used to represent ray AB. 5.1.1: Defining Terms
5. Key Concepts, continued Similarly, a line segment is also defined by two points, but both of those points are endpoints. A line segment can be measured because it has two endpoints and finite length. Line segments are used to form geometric figures. The symbol is used to represent line segment AB. 5.1.1: Defining Terms
6. Key Concepts, continued An angle is formed where two line segments or rays share an endpoint, or where a line intersects with another line, ray, or line segment. The difference in direction of the parts is called the angle. Angles can be measured in degrees or radians. The symbol is use to represent angle A. A represents the vertex of the angle. Sometimes it is necessary to use three letters to avoid confusion. In the diagram below, can be used to represent the same angle, . Notice that A is the vertex of the angle and it will always be listed in between the points on the angle’s rays. 5.1.1: Defining Terms
7. Key Concepts, continued An acute angle measures less than 90° but greater than 0°. An obtuse angle measures greater than 90° but less than 180°. A right angle measures exactly 90°. Two relationships between lines that will help us define transformations are parallel and perpendicular. Parallel lines are two lines that have unique points and never cross. If parallel lines share one point, then they will share every point; in other words, a line is parallel to itself. 5.1.1: Defining Terms
8. Key Concepts, continued Perpendicular lines meet at a right angle (90°), creating four right angles. 5.1.1: Defining Terms
9. Key Concepts, continued A circle is the set of points on a plane at a certain distance, or radius, from a single point, the center. Notice that a radius is a line segment. Therefore, if we draw any two radii of a circle, we create an angle where the two radii share a common endpoint, the center of the circle. 5.1.1: Defining Terms
10. Key Concepts, continued Creating an angle inside a circle allows us to define a circular arc, the set of points along the circle between the endpoints of the radii that are not shared. The arc length, or distance along a circular arc,is dependent on the length of the radius and the angle that creates the arc—the greater the radius or angle, the longer the arc. 5.1.1: Defining Terms
11. Common Errors/Misconceptions mislabeling angles or not including enough points to specify an angle misusing terms and notations finding the length of incorrect arcs 5.1.1: Defining Terms
12. Guided Practice Example 4 Given the following: Are and parallel? Are and parallel? Explain. 5.1.1: Defining Terms
13. Guided Practice: Example 4, continued and intersect at the same angle and . will never cross . Therefore, is parallel to . 5.1.1: Defining Terms
14. Guided Practice: Example 4, continued and intersect at the same angle, but . As you move from Z to Yon , you move closer to, and will eventually intersect, . Therefore, is not parallel to . ✔ 5.1.1: Defining Terms
15. Guided Practice: Example 4, continued 5.1.1: Defining Terms
16. Guided Practice Example 5 Refer to the figures below. Given , is the set of points with center B a circle? Given , is the set of points with center Y a circle? 5.1.1: Defining Terms
17. Guided Practice: Example 5, continued The set of points with center B is a circle because all points are equidistant from the center, B. 5.1.1: Defining Terms
18. Guided Practice: Example 5, continued The set of points with center Y is not a circle because the points vary in distance from the center, Y. ✔ 5.1.1: Defining Terms
19. Guided Practice: Example 5, continued 5.1.1: Defining Terms
20. Introduction The word transform means “to change.” In geometry, a transformation changes the position, shape, or size of a figure on a coordinate plane. The original figure, called a preimage, is changed or moved, and the resulting figure is called an image. We will be focusing on three different transformations: translations, reflections, and rotations. These transformations are all examples of isometry, meaning the new image is congruent to the preimage.
21. In this lesson, we will learn to describe transformations as functions on points in the coordinate plane. the potential inputs for a transformation function f in the coordinate plane will be a real number coordinate pair, (x, y), and each output will be a real number coordinate pair, f(x, y) the x and y values will change. Example: f(x, y) = (x + 1, y + 2) means for any ordered pair (x, y) add 1 to the x coordinate and add 2 to the y coordinate
22. Finally, transformations are generally applied to a set of points such as a line, triangle, square or other figure. In geometry, these figures are described by points, P, rather than coordinates (x, y), and transformation functions are often given the letters R, S, or T We will see T(x, y) written T(P) or P', known as “P prime.” A transformation T on a point P is a function where T(P) is P'.
23. Example 1 Given the point P(5, 3) and T(x, y)=(x + 2, y + 2), what are the coordinates of T(P)?
24. Guided Practice: Example 1, continued Identify the point given. We are givenP(5, 3). 5.1.2: Transformations As Functions
25. Guided Practice: Example 1, continued Identify the transformation. We are given T(P)=(x + 2, y + 2). 5.1.2: Transformations As Functions
26. Guided Practice: Example 1, continued Calculate the new coordinate. T(P)=(x + 2, y + 2) (5 + 2, 3 + 2) (7, 5) T(P) = (7, 5) ✔ 5.1.2: Transformations As Functions
27. We can also define transformations using a subscript notation. T(x, y)=(x + 2, y - 1) can be defined as T2, -1 meaning add two to the x value and subtract one from the y value. Example 2 Plot the points A’, B’, C’ using the translation T2, -1 B B’ A A’ A(1, 2) => B(4, 4) => C(3, 0) => A’(3, 1) B’(6, 3) C’(5, -1) C C’
28. Guided Practice Example 3 Given the transformation of a translation T5, –3, and the points P (–2, 1) and Q (4, 1), show that the transformation of a translation is isometric by calculating the distances, or lengths, of and . 5.1.2: Transformations As Functions
29. Guided Practice: Example 3, continued Plot the points of the preimage. 5.1.2: Transformations As Functions
30. Guided Practice: Example 3, continued Transform the points. T5, –3(x, y) = (x + 5, y – 3) 5.1.2: Transformations As Functions
31. Guided Practice: Example 3, continued Plot the image points. 5.1.2: Transformations As Functions
32. Guided Practice: Example 3, continued Calculate the distance, d, of each segment from the preimage and the image and compare them. Since the line segments are horizontal, count the number of units the segment spans to determine the distance. d(PQ) = 5 The distances of the segments are the same. The translation of the segment is isometric. ✔ 5.1.2: Transformations As Functions
33. Guided Practice: Example 3, continued 5.1.2: Transformations As Functions |
# Multiplying a constant and a linear monomial
A constant is a quantity which does not change. It is a quantity whose value is fixed and not variable for example the numbers 3, 8, 21…π, etc. are constants.
A monomial is a number, or a variable or the product of a number and one or more variables. For example, -5, abc/6, x... are monomials.
A linear monomial is an expression which has only one term and whose highest degree is one. It cannot contain any addition or subtraction signs or any negative exponents.
Multiplying a constant like 5 with a linear monomial like x
gives the result as follows 5 × x = 5x
Simplify the expression shown:
−13 × 7z
### Solution
Step 1:
The constant is −13 and the linear monomial is 7z
Step 2:
Simplifying
−13 × 7z = −91z
So, −13 × 7z = −91z
Simplify the expression shown:
$\left ( \frac{-5}{11} \right ) \times 9$mn
### Solution
Step 1:
The constant is $\left ( \frac{-5}{11} \right )$ and the linear monomial is 9mn
Step 2:
Simplifying
$\left ( \frac{-5}{11} \right ) \times 9mn = \left( \frac{−45mn}{11} \right )$
So, $\left (\frac{−5}{11} \right) \times 9mn = \left( \frac{−45mn}{11} \right)$
Simplify the expression shown:
$\left ( \frac{9}{12} \right) \times (3p)$
### Solution
Step 1:
The constant is $\left ( \frac{9}{12} \right)$ and the linear monomial is 3p
Step 2:
Simplifying
$\left ( \frac{9}{12} \right) \times (3p) = \left( \frac{9p}{4} \right)$
So, $\left ( \frac{9}{12} \right) \times (3p) = \left( \frac{9p}{4} \right)$ |
# Circle A has a radius of 3 and a center of (5 ,9 ). Circle B has a radius of 4 and a center of (1 ,2 ). If circle B is translated by <3 ,2 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?
Feb 9, 2018
${R}_{A} + {R}_{B}$ $\textcolor{g r e e n}{\left(7\right)}$ $>$ $\vec{{O}_{A} {O}_{B}}$ $\textcolor{b l u e}{\left(4.123\right)}$, the two circles A & B overlap.
#### Explanation:
Circle A ${O}_{A} \left(5 , 9\right) , {R}_{A} = 3$, Circle B ${O}_{B} \left(1 , 2\right) , {R}_{B} = 4$
${O}_{B}$ $t r a n s l a t e d$ $b y$ $\left(3 , 2\right)$
New ${O}_{B} = \left(\begin{matrix}1 + 3 \\ 2 + 3\end{matrix}\right) \implies \left(\begin{matrix}4 \\ 5\end{matrix}\right)$
Distance $\vec{{O}_{A} {O}_{B}} = \sqrt{{\left(4 - 5\right)}^{2} + {\left(5 - 9\right)}^{2}} = \sqrt{17} \approx 4.123$
Sum of radii ${R}_{A} + {R}_{B} = 3 + 4 = 7$
Since ${R}_{A} + {R}_{B}$ $\textcolor{g r e e n}{\left(7\right)}$ $>$ $\vec{{O}_{A} {O}_{B}}$ $\textcolor{b l u e}{\left(4.123\right)}$, the two circles A & B overlap. |
# How do you solve 1/2x > 22?
Apr 10, 2018
$x > 44$
#### Explanation:
Treat the inequality sign as an equals sign so you can isolate the variable. Multiply both sides by $2$ so $x$ is isolated.
$\frac{1}{2} \cdot 2 \cdot x > 22 \cdot 2$
$x > 44$
Apr 10, 2018
$x > 44$
#### Explanation:
$\text{multiply both sides by 2}$
$\cancel{2} \times \frac{x}{\cancel{2}} > 2 \times 22$
$\Rightarrow x > 44 \text{ is the solution}$ |
# How do you complete the square to solve 4x^2 - 7x - 2 = 0?
May 23, 2015
In this way:
$4 {x}^{2} - 7 x - 2 = 0 \Rightarrow 4 \left({x}^{2} - \frac{7}{4} x\right) - 2 = 0 \Rightarrow$
$4 \left({x}^{2} - \frac{7}{4} x + \frac{49}{64} - \frac{49}{64}\right) - 2 = 0 \Rightarrow$
$4 \left({x}^{2} - \frac{7}{4} x + \frac{49}{64}\right) - \frac{49}{16} - 2 = 0 \Rightarrow$
$4 {\left(x - \frac{7}{8}\right)}^{2} = \frac{49}{16} + 2 \Rightarrow 4 {\left(x - \frac{7}{8}\right)}^{2} = \frac{49 + 32}{16} \Rightarrow$
${\left(x - \frac{7}{8}\right)}^{2} = \frac{81}{16} \cdot \frac{1}{4} \Rightarrow {\left(x - \frac{7}{8}\right)}^{2} = \frac{81}{64} \Rightarrow$
$x - \frac{7}{8} = \pm \frac{9}{8} \Rightarrow x = \frac{7}{8} \pm \frac{9}{8} \Rightarrow$
${x}_{1} = \frac{7}{8} - \frac{9}{8} = - \frac{2}{8} = - \frac{1}{4}$
${x}_{2} = \frac{7}{8} + \frac{9}{8} = \frac{16}{8} = 2$. |
# How do you differentiate f(x)= (4 x^2 + 5x -8 )/ (x- 1 ) using the quotient rule?
Jan 13, 2017
$f ' \left(x\right) = \frac{4 {x}^{2} - 8 x + 3}{x - 1} ^ 2$
#### Explanation:
differentiate using the $\textcolor{b l u e}{\text{quotient rule}}$
$\text{Given " f(x)=(g(x))/(h(x))" then}$
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
$\text{here } g \left(x\right) = 4 {x}^{2} + 5 x - 8 \Rightarrow g ' \left(x\right) = 8 x + 5$
$\text{and } h \left(x\right) = x - 1 \Rightarrow h ' \left(x\right) = 1$
$\Rightarrow f ' \left(x\right) = \frac{\left(x - 1\right) \left(8 x + 5\right) - \left(4 {x}^{2} + 5 x - 8\right) .1}{x - 1} ^ 2$
$= \frac{8 {x}^{2} - 3 x - 5 - 4 {x}^{2} - 5 x + 8}{x - 1} ^ 2$
$= \frac{4 {x}^{2} - 8 x + 3}{x - 1} ^ 2$ |
# Help with math
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## The Best Help with math
Help with math is a mathematical instrument that assists to solve math equations. Solving an equation is all about finding the value of the variable that makes the equation true. There are a few different steps that you can follow to solve an equation, but the process essentially boils down to two things: using inverse operations to isolate the variable, and then using algebraic methods to find the value of the variable. Let's take a look at an example to see how this works in practice. Suppose we want to solve the equation 2x+3=11. First, we would use inverse operations to isolate the variable by subtracting 3 from both sides of the equation. This would give us 2x=8. Next, we would use algebraic methods to solve for x by dividing both sides of the equation by 2. This would give us x=4. So, the solution to our equation is x=4. By following these steps, you can solve any equation you come across. Just remember to take your time and triple check your work!
Math can be a difficult subject for many people. understand. That's where a math solver website can come in handy. These websites can provide you with the answers to Math problems, as well as step-by-step solutions to show you how they solved the problem. This can be a valuable resource when you're stuck on a Math problem and can't seem to solve it on your own. In addition, many Math solver websites also offer forums where you can ask Math questions and get help from other users. So if you're struggling with Math, don't hesitate to use a math solver website to get the help you need.
Once the equation has been factored, you can solve each factor by setting it equal to zero and using the quadratic formula. Another method for solving the square is to complete the square. This involves adding a constant to both sides of the equation so that one side is a perfect square. Once this is done, you can take the square root of both sides and solve for the variable. Finally, you can use graphing to solve the square. To do this, you will need to plot the points associated with the equation and then find the intersection of the two lines. Whichever method you choose, solving the square can be a simple process as long as you have a strong understanding of algebra.
Solving for an exponent can be tricky, but there are a few tips that can help. First, make sure to identify the base and the exponent. The base is the number that is being multiplied, and the exponent is the number of times that it is being multiplied. For example, in the equation 8 2, the base is 8 and the exponent is 2. Once you have identified the base and exponent, you can begin to solve for the exponent. To do this, take the logarithm of both sides of the equation. This will allow you to move the exponent from one side of the equation to the other. For example, if you take the logarithm of both sides of 8 2 = 64, you getlog(8 2) = log(64). Solving this equation for x gives you x = 2log(8), which means that 8 2 = 64. In other words, when solving for an exponent, you can take the logarithm of both sides of the equation to simplify it.
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## Arc Lengths of Curves
### Learning Outcomes
• Determine the length of a curve, $y=f(x),$ between two points
• Determine the length of a curve, $x=g(y),$ between two points
## Arc Length of the Curve $y$ = $f$($x$)
In previous applications of integration, we required the function $f(x)$ to be integrable, or at most continuous. However, for calculating arc length we have a more stringent requirement for $f(x).$ Here, we require $f(x)$ to be differentiable, and furthermore we require its derivative, ${f}^{\prime }(x),$ to be continuous. Functions like this, which have continuous derivatives, are called smooth. (This property comes up again in later chapters.)
Let $f(x)$ be a smooth function defined over $\left[a,b\right].$ We want to calculate the length of the curve from the point $(a,f(a))$ to the point $(b,f(b)).$ We start by using line segments to approximate the length of the curve. For $i=0,1,2\text{,…},n,$ let $P=\left\{{x}_{i}\right\}$ be a regular partition of $\left[a,b\right].$ Then, for $i=1,2\text{,…},n,$ construct a line segment from the point $({x}_{i-1},f({x}_{i-1}))$ to the point $({x}_{i},f({x}_{i})).$ Although it might seem logical to use either horizontal or vertical line segments, we want our line segments to approximate the curve as closely as possible. Figure 1 depicts this construct for $n=5.$
Figure 1. We can approximate the length of a curve by adding line segments.
To help us find the length of each line segment, we look at the change in vertical distance as well as the change in horizontal distance over each interval. Because we have used a regular partition, the change in horizontal distance over each interval is given by $\text{Δ}x.$ The change in vertical distance varies from interval to interval, though, so we use $\text{Δ}{y}_{i}=f({x}_{i})-f({x}_{i-1})$ to represent the change in vertical distance over the interval $\left[{x}_{i-1},{x}_{i}\right],$ as shown in Figure 2. Note that some (or all) $\text{Δ}{y}_{i}$ may be negative.
Figure 2. A representative line segment approximates the curve over the interval $\left[{x}_{i-1},{x}_{i}\right].$
By the Pythagorean theorem, the length of the line segment is $\sqrt{{(\text{Δ}x)}^{2}+{(\text{Δ}{y}_{i})}^{2}}.$ We can also write this as $\text{Δ}x\sqrt{1+{((\text{Δ}{y}_{i})\text{/}(\text{Δ}x))}^{2}}.$ Now, by the Mean Value Theorem, there is a point ${x}_{i}^{*}\in \left[{x}_{i-1},{x}_{i}\right]$ such that ${f}^{\prime }({x}_{i}^{*})=(\text{Δ}{y}_{i})\text{/}(\text{Δ}x).$ Then the length of the line segment is given by $\text{Δ}x\sqrt{1+{\left[{f}^{\prime }({x}_{i}^{*})\right]}^{2}}.$ Adding up the lengths of all the line segments, we get
$\text{Arc Length}\approx \underset{i=1}{\overset{n}{\text{∑}}}\sqrt{1+{\left[{f}^{\prime }({x}_{i}^{*})\right]}^{2}}\text{Δ}x.$
This is a Riemann sum. Taking the limit as $n\to \infty ,$ we have
$\text{Arc Length}=\underset{n\to \infty }{\text{lim}}\underset{i=1}{\overset{n}{\text{∑}}}\sqrt{1+{\left[{f}^{\prime }({x}_{i}^{*})\right]}^{2}}\text{Δ}x={\displaystyle\int }_{a}^{b}\sqrt{1+{\left[{f}^{\prime }(x)\right]}^{2}}dx.$
We summarize these findings in the following theorem.
### Arc Length for $y$ = $f$($x$)
Let $f(x)$ be a smooth function over the interval $\left[a,b\right].$ Then the arc length of the portion of the graph of $f(x)$ from the point $(a,f(a))$ to the point $(b,f(b))$ is given by
$\text{Arc Length}={\displaystyle\int }_{a}^{b}\sqrt{1+{\left[{f}^{\prime }(x)\right]}^{2}}dx.$
Note that we are integrating an expression involving ${f}^{\prime }(x),$ so we need to be sure ${f}^{\prime }(x)$ is integrable. This is why we require $f(x)$ to be smooth. The following example shows how to apply the theorem.
### Example: Calculating the Arc Length of a Function of $x$
Let $f(x)=2{x}^{3\text{/}2}.$ Calculate the arc length of the graph of $f(x)$ over the interval $\left[0,1\right].$ Round the answer to three decimal places.
### Try It
Let $f(x)=\left(\dfrac{4}{3}\right){x}^{3\text{/}2}.$ Calculate the arc length of the graph of $f(x)$ over the interval $\left[0,1\right].$ Round the answer to three decimal places.
Watch the following video to see the worked solution to the above Try It.
Although it is nice to have a formula for calculating arc length, this particular theorem can generate expressions that are difficult to integrate. We study some techniques for integration in Introduction to Techniques of Integration in the second volume of this text. In some cases, we may have to use a computer or calculator to approximate the value of the integral.
### Example: Using a Computer or Calculator to Determine the Arc Length of a Function of $x$
Let $f(x)={x}^{2}.$ Calculate the arc length of the graph of $f(x)$ over the interval $\left[1,3\right].$
### Try It
Let $f(x)= \sin x.$ Calculate the arc length of the graph of $f(x)$ over the interval $\left[0,\pi \right].$ Use a computer or calculator to approximate the value of the integral.
Watch the following video to see the worked solution to the above Try It.
## Arc Length of the Curve $x$ = $g$($y$)
We have just seen how to approximate the length of a curve with line segments. If we want to find the arc length of the graph of a function of $y,$ we can repeat the same process, except we partition the $y\text{-axis}$ instead of the $x\text{-axis}.$ Figure 3 shows a representative line segment.
Figure 3. A representative line segment over the interval $\left[{y}_{i-1},{y}_{i}\right].$
Then the length of the line segment is $\sqrt{{(\text{Δ}y)}^{2}+{(\text{Δ}{x}_{i})}^{2}},$ which can also be written as $\text{Δ}y\sqrt{1+{((\text{Δ}{x}_{i})\text{/}(\text{Δ}y))}^{2}}.$ If we now follow the same development we did earlier, we get a formula for arc length of a function $x=g(y).$
### Arc Length for $x$ = $g$($y$)
Let $g(y)$ be a smooth function over an interval $\left[c,d\right].$ Then, the arc length of the graph of $g(y)$ from the point $(c,g(c))$ to the point $(d,g(d))$ is given by
$\text{Arc Length}={\displaystyle\int }_{c}^{d}\sqrt{1+{\left[{g}^{\prime }(y)\right]}^{2}}dy$
### Example: Calculating the Arc Length of a Function of $y$
Let $g(y)=3{y}^{3}.$ Calculate the arc length of the graph of $g(y)$ over the interval $\left[1,2\right].$
### Try It
Let $g(y)=\frac{1}{y}.$ Calculate the arc length of the graph of $g(y)$ over the interval $\left[1,4\right].$ Use a computer or calculator to approximate the value of the integral. |
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# If $\overrightarrow a ,\overrightarrow {{\text{ }}b} ,\overrightarrow {{\text{ }}c}$ are the unit vectors such that $\overrightarrow a \cdot \overrightarrow b = 0 = \overrightarrow a \cdot \overrightarrow b$ and the angle between $\overrightarrow b$ and $\overrightarrow c$ is $\dfrac{\pi }{3}$. Then find the value of $\left| {\overrightarrow a \times \overrightarrow b - \overrightarrow a \times \left. {\overrightarrow c } \right|} \right.$
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Hint: In this question we have to find the magnitude of the given vector using the given condition using unit vectors and angle. For that first we are going to find the unit vector and dot product value and then using a given angle we can find the required cross product value of the given vector.
From the question, $\overrightarrow a ,\overrightarrow b ,\overrightarrow c$ are unit vector
Then the magnitude value of unit vector is $1$
$\therefore \left| {\overrightarrow a } \right| = 1$, $\left| {\overrightarrow b } \right| = 1$,$\left| {\overrightarrow c } \right| = 1$
We know that if the dot product of two vector is 0 then the two vectors are said to be perpendicular,
$\therefore \overrightarrow a .\overrightarrow b = 0$ then $\overrightarrow a$ is perpendicular to $\overrightarrow b$
$\therefore \overrightarrow a .\overrightarrow c = 0$ then $\overrightarrow a$ is perpendicular to $\overrightarrow c$
So, $\overrightarrow a$ is perpendicular to both $\overrightarrow b$ and $\overrightarrow c$ then $\overrightarrow b$ and $\overrightarrow c$ are parallel, then we obtain that $\overrightarrow a$ is parallel to $\overrightarrow b \times \overrightarrow c$.
That is, $\overrightarrow a$ is parallel to $\overrightarrow b \times \overrightarrow c$
Thus, we obtain $\left| {\overrightarrow a \times \overrightarrow b - \overrightarrow a \times \left. {\overrightarrow c } \right|} \right.$
Now take $a$ common, we get
$\Rightarrow \left| {\overrightarrow a \times (\overrightarrow b - \left. {\overrightarrow c )} \right|} \right. - - - - - - - - - - - (1)$
We know that if $\overrightarrow b$ and $\overrightarrow c$ are parallel, then
$\Rightarrow \overrightarrow b = \lambda \overrightarrow c$
Now substitute $\overrightarrow b$ value in equation (1), we get
$\Rightarrow \left| {\overrightarrow a \times (\lambda \overrightarrow c - \left. {\overrightarrow c )} \right|} \right.$
Now taking $\overrightarrow c$ common, we get
$\Rightarrow \left| {\overrightarrow a \times (\lambda - \left. {1)\overrightarrow c } \right|} \right.$
We can separate ${\text{(}}\lambda {\text{ - 1)}}$ from the modulus, we get
$\Rightarrow {\text{(}}\lambda {\text{ - 1)}}\left| {\overrightarrow a \times \left. {\overrightarrow c } \right|} \right. - - - - - - - - - - - - (2)$
We know that,
$\Rightarrow \left| {\overrightarrow a \times \left. {\overrightarrow c } \right|} \right. = \left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\sin \theta - - - - - - - - - - (3)$
Given that the angle between $\overrightarrow a$ and $\overrightarrow c$ is $\dfrac{\pi }{3}$ because $\overrightarrow a$ and $\overrightarrow c$ is perpendicular and substitute the angle in equation (3), we get
$\Rightarrow \left| {\overrightarrow a \times \left. {\overrightarrow c } \right|} \right. = \left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\sin \dfrac{\pi }{2}$
Since $\left| {\overrightarrow a } \right| = 1$, $\left| {\overrightarrow c } \right| = 1$, then
$\Rightarrow \left| {\overrightarrow a \times \left. {\overrightarrow c } \right|} \right. = 1 \cdot 1.\sin {90^ \circ }$
$\Rightarrow \left| {\overrightarrow a \times \left. {\overrightarrow c } \right|} \right. = \sin {60^ \circ } - - - - - - - - - - (4)$
We know that, $\sin {90^ \circ } = 1$ substitute $\sin {60^ \circ }$ value in equation (4), we get
$\Rightarrow \left| {\overrightarrow a \times \left. {\overrightarrow c } \right|} \right. = 1 - - - - - - - - - - (5)$
Now, Substitute equation (5) in (2)
$\Rightarrow \left( {\lambda {\text{ - 1}}} \right)\left| {\overrightarrow a \times \left. {\overrightarrow c } \right|} \right.$
$\Rightarrow \left( {\lambda - 1} \right) \cdot 1$
$\Rightarrow \left( {\lambda - 1} \right) - - - - - - - - - - - - - - - (6)$
Similarly, now take $\overrightarrow a$ is parallel to $\overrightarrow b \times \overrightarrow c$,
We know that if $\overrightarrow b$ and $\overrightarrow c$ are parallel, then
$\Rightarrow \overrightarrow b = \lambda \overrightarrow c$
Similarly $\overrightarrow a$ is parallel to $\overrightarrow b \times \overrightarrow c$
$\Rightarrow \overrightarrow a = \lambda (\overrightarrow b \times \overrightarrow c )$
$\Rightarrow \left| {\overrightarrow a } \right| = \lambda \left| {(\overrightarrow b \times \overrightarrow c )} \right|$
We know that $\overrightarrow a ,{\text{ }}\overrightarrow b ,{\text{ }}\overrightarrow c$ are unit vectors
$\Rightarrow \left| {\overrightarrow a } \right| = \lambda \left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\sin {60^ \circ }$
Given that $\left| {\overrightarrow a } \right| = 1$, $\left| {\overrightarrow b } \right| = 1$, $\left| {\overrightarrow c } \right| = 1$and $\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}$
By substituting, we get
$\Rightarrow 1 = \lambda \cdot 1 \cdot 1 \cdot \dfrac{{\sqrt 3 }}{2}$
Now we get $\lambda$ value,
$\Rightarrow \lambda = \dfrac{2}{{\sqrt 3 }}$
Now substitute $\lambda$ value in equation (6), we get
$\Rightarrow \left( {\lambda - 1} \right) - - - - - - - - - - - - - - - (6)$
$\Rightarrow \left( {\dfrac{2}{{\sqrt 3 }} - 1} \right)$
Multiply and divide by $\sqrt 3$ we get,
$\Rightarrow \left( {\dfrac{2}{{\sqrt 3 }} - \dfrac{{\sqrt 3 }}{{\sqrt 3 }}} \right)$
Denominator for the both terms are equal, we get,
$\Rightarrow \left( {\dfrac{{2 - \sqrt 3 }}{{\sqrt 3 }}} \right)$
$\therefore$ Thus the value $\left| {\overrightarrow a \times \overrightarrow b - \overrightarrow a \times \left. {\overrightarrow c } \right|} \right.$ is $\dfrac{{2 - \sqrt 3 }}{{\sqrt 3 }}$
Note: This kind of problem has to be derived in a substitution way because the question gives ideas to solve in the manner of dot product as well as cross product. The basic vector calculation and algebraic calculation take place. And also we need to know some trigonometric ratio values and fraction addition. |
# 1984 AHSME Problems/Problem 15
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem 15
If $\sin{2x}\sin{3x}=\cos{2x}\cos{3x}$, then one value for $x$ is
$\mathrm{(A) \ }18^\circ \qquad \mathrm{(B) \ }30^\circ \qquad \mathrm{(C) \ } 36^\circ \qquad \mathrm{(D) \ }45^\circ \qquad \mathrm{(E) \ } 60^\circ$
## Solution
We divide both sides of the equation by $\cos{2x}\times\cos{3x}$ to get $\frac{\sin{2x}\times\sin{3x}}{\cos{2x}\times\cos{3x}}=1$, or $\tan{2x}\times\tan{3x}=1$.
This looks a lot like the formula relating the slopes of two perpendicular lines, which is $m_1\times m_2=-1$, where $m_1$ and $m_2$ are the slopes. It's made even more relatable by the fact that the tangent of an angle can be defined by the slope of the line that makes that angle with the x-axis.
We can make this look even more like the slope formula by multiplying both sides by $-1$:
$\tan{2x}\times-\tan{3x}=-1$, and using the trigonometric identity $-\tan{x}=\tan{-x}$, we have $\tan{2x}\times\tan{-3x}=-1$.
Now it's time for a diagram:
$[asy] unitsize(2.54cm); draw(unitcircle); draw((0,-1.25)--(0,1.25)); draw((-1.25,0)--(1.25,0)); draw((0,0)--(cos(2pi/10),sin(2pi/10))); draw((0,0)--(cos(-3pi/10),sin(-3pi/10))); label("\tan{-3x}",(cos(-3pi/10),sin(-3pi/10)),SE); label("\tan{2x}",(cos(2pi/10),sin(2pi/10)),NE); label("2x",(.125,.03),ENE); label("3x",(.125,-.06),ESE); [/asy]$ Since the product of the two slopes, $\tan{2x}$ and $\tan{-3x}$, is $-1$, the lines are perpendicular, and the angle between them is $\frac{\pi}{2}$. The angle between them is also $2x+3x=5x$, so $5x=\frac{\pi}{2}$ and $x=\frac{\pi}{10}$, or $18^\circ, \boxed{\text{A}}$.
NOTE: To show that $18^\circ$ is not the only solution, we can also set $5x$ equal to another angle measure congruent to $\frac{\pi}{2}$, such as $\frac{5\pi}{2}$, yielding another solution as $\frac{\pi}{2}=90^\circ$, which clearly is a solution to the equation.
## Solution 2
We can simply try the answers. We quickly see that $18^\circ$ works, since $\sin(36^\circ)\sin(54^\circ)=\cos(54^\circ)\cos(36^\circ)=\cos(36^\circ)\cos(54^\circ)$ by the identity $\cos(x)=\sin(90^\circ-x)$
## Solution 3
Start by subtracting $\sin{2x}\sin{3x}$ from both sides to get $\cos{2x}\cos{3x}-\sin{2x}\sin{3x}=0$. We recognize that this is of the form $\cos{a}\cos{b}-\sin{a}\sin{b}=\cos{(a+b)}$, so $\cos{2x}\cos{3x}-\sin{2x}\sin{3x}=\cos{(2x+3x)}=0$. $\cos{90^\circ}=0$, so $x=\boxed{18^\circ}$.
~purplepenguin2
1984 AHSME (Problems • Answer Key • Resources) Preceded byProblem 14 Followed byProblem 16 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions |
# Is 7579 prime or composite?
Here is the answer to questions like: Is 7579 prime or composite? or is 7579 a prime or a composite number?
Use the Prime Factorization tool below to discover if any given number is prime or composite and in this case calculate the its prime factors. See also in this web page a Prime Factorization Chart with all primes from 1 to 1000.
### Prime Factorization Calculator
Enter the integer number you want to get its prime factors:
Ex.: 2, 3, 4, 11, 10225, etc.
Prime factorization result:
The number 7579 is a composite number so, it is possible to factorize it. In other words, 7579 can be divided by 1, by itself and at least by 11, 13 and 53. A composite number is a positive integer that has at least one positive divisor other than one or the number itself. In other words, a composite number is any integer greater than one that is not a prime number.
The prime factorization of 7579 = 11•13•53.
The prime factors of 7579 are 11, 13 and 53.
### Factor tree or prime decomposition for 7579
As 7579 is a composite number, we can draw its factor tree:
## What is prime factorization?
### Definition of prime factorization
The prime factorization is the decomposition of a composite number into a product of prime factors that, if multiplied, recreate the original number. Factors by definition are the numbers that multiply to create another number. A prime number is an integer greater than one which is divided only by one and by itself. For example, the only divisors of 7 are 1 and 7, so 7 is a prime number, while the number 72 has divisors deived from 23•32 like 2, 3, 4, 6, 8, 12, 24 ... and 72 itself, making 72 not a prime number. Note the the only "prime" factors of 72 are 2 and 3 which are prime numbers.
## Prime factorization example 1
Let's find the prime factorization of 72.
### Solution 1
Start with the smallest prime number that divides into 72, in this case 2. We can write 72 as:
72 = 2 x 36
Now find the smallest prime number that divides into 36. Again we can use 2, and write the 36 as 2 x 18, to give.
72 = 2 x 2 x 18
18 also divides by 2 (18 = 2 x 9), so we have:
72 = 2 x 2 x 2 x 9
9 divides by 3 (9 = 3 x 3), so we have:
72 = 2 x 2 x 2 x 3 x 3
2, 2, 2, 3 and 3 are all prime numbers, so we have our answer.
In short, we would write the solution as:
72 = 2 x 36
72 = 2 x 2 x 18
72 = 2 x 2 x 2 x 9
72 = 2 x 2 x 2 x 3 x 3
72 = 23 x 32 (prime factorization exponential form)
### Solution 2
Using a factor tree:
• Procedure:
• Find 2 factors of the number;
• Look at the 2 factors and determine if at least one of them is not prime;
• If it is not a prime factor it;
• Repeat this process until all factors are prime.
See how to factor the number 72:
72 / \ 2 36 / \ 2 18 / \ 2 9 / \ 3 3 72 is not prime --> divide by 2 36 is not prime --> divide by 2 18 is not prime --> divide by 2 9 is not prime --> divide by 3 3 and 3 are prime --> stop
Taking the left-hand numbers and the right-most number of the last row (dividers) an multiplying then, we have
72 = 2 x 2 x 2 x 3 x 3
72 = 23 x 32 (prime factorization exponential form)
Note that these dividers are the prime factors. They are also called the leaves of the factor tree.
## Prime factorization example 2
See how to factor the number 588:
588 / \ 2 294 / \ 2 147 / \ 3 49 / \ 7 7 588 is not prime --> divide by 2 294 is not prime --> divide by 2 147 is not prime --> divide by 3 49 is not prime --> divide by 7 7 and 7 are prime --> stop
Taking the left-hand numbers and the right-most number of the last row (dividers) an multiplying then, we have
588 = 2 x 2 x 3 x 7 x 7
588 = 22 x 3 x 72 (prime factorization exponential form)
## Prime Factorization Chart 1-1000
nPrime
Factorization
2 =2
3 =3
4 =2•2
5 =5
6 =2•3
7 =7
8 =2•2•2
9 =3•3
10 =2•5
11 =11
12 =2•2•3
13 =13
14 =2•7
15 =3•5
16 =2•2•2•2
17 =17
18 =2•3•3
19 =19
20 =2•2•5
21 =3•7
22 =2•11
23 =23
24 =2•2•2•3
25 =5•5
26 =2•13
27 =3•3•3
28 =2•2•7
29 =29
30 =2•3•5
31 =31
32 =2•2•2•2•2
33 =3•11
34 =2•17
35 =5•7
36 =2•2•3•3
37 =37
38 =2•19
39 =3•13
40 =2•2•2•5
41 =41
42 =2•3•7
43 =43
44 =2•2•11
45 =3•3•5
46 =2•23
47 =47
48 =2•2•2•2•3
49 =7•7
50 =2•5•5
51 =3•17
52 =2•2•13
53 =53
54 =2•3•3•3
55 =5•11
56 =2•2•2•7
57 =3•19
58 =2•29
59 =59
60 =2•2•3•5
61 =61
62 =2•31
63 =3•3•7
64 =2•2•2•2•2•2
65 =5•13
66 =2•3•11
67 =67
68 =2•2•17
69 =3•23
70 =2•5•7
71 =71
72 =2•2•2•3•3
73 =73
74 =2•37
75 =3•5•5
76 =2•2•19
77 =7•11
78 =2•3•13
79 =79
80 =2•2•2•2•5
81 =3•3•3•3
82 =2•41
83 =83
84 =2•2•3•7
85 =5•17
86 =2•43
87 =3•29
88 =2•2•2•11
89 =89
90 =2•3•3•5
91 =7•13
92 =2•2•23
93 =3•31
94 =2•47
95 =5•19
96 =2•2•2•2•2•3
97 =97
98 =2•7•7
99 =3•3•11
100 =2•2•5•5
101 =101
102 =2•3•17
103 =103
104 =2•2•2•13
105 =3•5•7
106 =2•53
107 =107
108 =2•2•3•3•3
109 =109
110 =2•5•11
111 =3•37
112 =2•2•2•2•7
113 =113
114 =2•3•19
115 =5•23
116 =2•2•29
117 =3•3•13
118 =2•59
119 =7•17
120 =2•2•2•3•5
121 =11•11
122 =2•61
123 =3•41
124 =2•2•31
125 =5•5•5
126 =2•3•3•7
127 =127
128 =2•2•2•2•2•2•2
129 =3•43
130 =2•5•13
131 =131
132 =2•2•3•11
133 =7•19
134 =2•67
135 =3•3•3•5
136 =2•2•2•17
137 =137
138 =2•3•23
139 =139
140 =2•2•5•7
141 =3•47
142 =2•71
143 =11•13
144 =2•2•2•2•3•3
145 =5•29
146 =2•73
147 =3•7•7
148 =2•2•37
149 =149
150 =2•3•5•5
151 =151
152 =2•2•2•19
153 =3•3•17
154 =2•7•11
155 =5•31
156 =2•2•3•13
157 =157
158 =2•79
159 =3•53
160 =2•2•2•2•2•5
161 =7•23
162 =2•3•3•3•3
163 =163
164 =2•2•41
165 =3•5•11
166 =2•83
167 =167
168 =2•2•2•3•7
169 =13•13
170 =2•5•17
171 =3•3•19
172 =2•2•43
173 =173
174 =2•3•29
175 =5•5•7
176 =2•2•2•2•11
177 =3•59
178 =2•89
179 =179
180 =2•2•3•3•5
181 =181
182 =2•7•13
183 =3•61
184 =2•2•2•23
185 =5•37
186 =2•3•31
187 =11•17
188 =2•2•47
189 =3•3•3•7
190 =2•5•19
191 =191
192 =2•2•2•2•2•2•3
193 =193
194 =2•97
195 =3•5•13
196 =2•2•7•7
197 =197
198 =2•3•3•11
199 =199
200 =2•2•2•5•5
201 =3•67
202 =2•101
203 =7•29
204 =2•2•3•17
205 =5•41
206 =2•103
207 =3•3•23
208 =2•2•2•2•13
209 =11•19
210 =2•3•5•7
211 =211
212 =2•2•53
213 =3•71
214 =2•107
215 =5•43
216 =2•2•2•3•3•3
217 =7•31
218 =2•109
219 =3•73
220 =2•2•5•11
221 =13•17
222 =2•3•37
223 =223
224 =2•2•2•2•2•7
225 =3•3•5•5
226 =2•113
227 =227
228 =2•2•3•19
229 =229
230 =2•5•23
231 =3•7•11
232 =2•2•2•29
233 =233
234 =2•3•3•13
235 =5•47
236 =2•2•59
237 =3•79
238 =2•7•17
239 =239
240 =2•2•2•2•3•5
241 =241
242 =2•11•11
243 =3•3•3•3•3
244 =2•2•61
245 =5•7•7
246 =2•3•41
247 =13•19
248 =2•2•2•31
249 =3•83
250 =2•5•5•5
nPrime
Factorization
251 =251
252 =2•2•3•3•7
253 =11•23
254 =2•127
255 =3•5•17
256 =2•2•2•2•2•2•2•2
257 =257
258 =2•3•43
259 =7•37
260 =2•2•5•13
261 =3•3•29
262 =2•131
263 =263
264 =2•2•2•3•11
265 =5•53
266 =2•7•19
267 =3•89
268 =2•2•67
269 =269
270 =2•3•3•3•5
271 =271
272 =2•2•2•2•17
273 =3•7•13
274 =2•137
275 =5•5•11
276 =2•2•3•23
277 =277
278 =2•139
279 =3•3•31
280 =2•2•2•5•7
281 =281
282 =2•3•47
283 =283
284 =2•2•71
285 =3•5•19
286 =2•11•13
287 =7•41
288 =2•2•2•2•2•3•3
289 =17•17
290 =2•5•29
291 =3•97
292 =2•2•73
293 =293
294 =2•3•7•7
295 =5•59
296 =2•2•2•37
297 =3•3•3•11
298 =2•149
299 =13•23
300 =2•2•3•5•5
301 =7•43
302 =2•151
303 =3•101
304 =2•2•2•2•19
305 =5•61
306 =2•3•3•17
307 =307
308 =2•2•7•11
309 =3•103
310 =2•5•31
311 =311
312 =2•2•2•3•13
313 =313
314 =2•157
315 =3•3•5•7
316 =2•2•79
317 =317
318 =2•3•53
319 =11•29
320 =2•2•2•2•2•2•5
321 =3•107
322 =2•7•23
323 =17•19
324 =2•2•3•3•3•3
325 =5•5•13
326 =2•163
327 =3•109
328 =2•2•2•41
329 =7•47
330 =2•3•5•11
331 =331
332 =2•2•83
333 =3•3•37
334 =2•167
335 =5•67
336 =2•2•2•2•3•7
337 =337
338 =2•13•13
339 =3•113
340 =2•2•5•17
341 =11•31
342 =2•3•3•19
343 =7•7•7
344 =2•2•2•43
345 =3•5•23
346 =2•173
347 =347
348 =2•2•3•29
349 =349
350 =2•5•5•7
351 =3•3•3•13
352 =2•2•2•2•2•11
353 =353
354 =2•3•59
355 =5•71
356 =2•2•89
357 =3•7•17
358 =2•179
359 =359
360 =2•2•2•3•3•5
361 =19•19
362 =2•181
363 =3•11•11
364 =2•2•7•13
365 =5•73
366 =2•3•61
367 =367
368 =2•2•2•2•23
369 =3•3•41
370 =2•5•37
371 =7•53
372 =2•2•3•31
373 =373
374 =2•11•17
375 =3•5•5•5
376 =2•2•2•47
377 =13•29
378 =2•3•3•3•7
379 =379
380 =2•2•5•19
381 =3•127
382 =2•191
383 =383
384 =2•2•2•2•2•2•2•3
385 =5•7•11
386 =2•193
387 =3•3•43
388 =2•2•97
389 =389
390 =2•3•5•13
391 =17•23
392 =2•2•2•7•7
393 =3•131
394 =2•197
395 =5•79
396 =2•2•3•3•11
397 =397
398 =2•199
399 =3•7•19
400 =2•2•2•2•5•5
401 =401
402 =2•3•67
403 =13•31
404 =2•2•101
405 =3•3•3•3•5
406 =2•7•29
407 =11•37
408 =2•2•2•3•17
409 =409
410 =2•5•41
411 =3•137
412 =2•2•103
413 =7•59
414 =2•3•3•23
415 =5•83
416 =2•2•2•2•2•13
417 =3•139
418 =2•11•19
419 =419
420 =2•2•3•5•7
421 =421
422 =2•211
423 =3•3•47
424 =2•2•2•53
425 =5•5•17
426 =2•3•71
427 =7•61
428 =2•2•107
429 =3•11•13
430 =2•5•43
431 =431
432 =2•2•2•2•3•3•3
433 =433
434 =2•7•31
435 =3•5•29
436 =2•2•109
437 =19•23
438 =2•3•73
439 =439
440 =2•2•2•5•11
441 =3•3•7•7
442 =2•13•17
443 =443
444 =2•2•3•37
445 =5•89
446 =2•223
447 =3•149
448 =2•2•2•2•2•2•7
449 =449
450 =2•3•3•5•5
451 =11•41
452 =2•2•113
453 =3•151
454 =2•227
455 =5•7•13
456 =2•2•2•3•19
457 =457
458 =2•229
459 =3•3•3•17
460 =2•2•5•23
461 =461
462 =2•3•7•11
463 =463
464 =2•2•2•2•29
465 =3•5•31
466 =2•233
467 =467
468 =2•2•3•3•13
469 =7•67
470 =2•5•47
471 =3•157
472 =2•2•2•59
473 =11•43
474 =2•3•79
475 =5•5•19
476 =2•2•7•17
477 =3•3•53
478 =2•239
479 =479
480 =2•2•2•2•2•3•5
481 =13•37
482 =2•241
483 =3•7•23
484 =2•2•11•11
485 =5•97
486 =2•3•3•3•3•3
487 =487
488 =2•2•2•61
489 =3•163
490 =2•5•7•7
491 =491
492 =2•2•3•41
493 =17•29
494 =2•13•19
495 =3•3•5•11
496 =2•2•2•2•31
497 =7•71
498 =2•3•83
499 =499
500 =2•2•5•5•5
nPrime
Factorization
501 =3•167
502 =2•251
503 =503
504 =2•2•2•3•3•7
505 =5•101
506 =2•11•23
507 =3•13•13
508 =2•2•127
509 =509
510 =2•3•5•17
511 =7•73
512 =2•2•2•2•2•2•2•2•2
513 =3•3•3•19
514 =2•257
515 =5•103
516 =2•2•3•43
517 =11•47
518 =2•7•37
519 =3•173
520 =2•2•2•5•13
521 =521
522 =2•3•3•29
523 =523
524 =2•2•131
525 =3•5•5•7
526 =2•263
527 =17•31
528 =2•2•2•2•3•11
529 =23•23
530 =2•5•53
531 =3•3•59
532 =2•2•7•19
533 =13•41
534 =2•3•89
535 =5•107
536 =2•2•2•67
537 =3•179
538 =2•269
539 =7•7•11
540 =2•2•3•3•3•5
541 =541
542 =2•271
543 =3•181
544 =2•2•2•2•2•17
545 =5•109
546 =2•3•7•13
547 =547
548 =2•2•137
549 =3•3•61
550 =2•5•5•11
551 =19•29
552 =2•2•2•3•23
553 =7•79
554 =2•277
555 =3•5•37
556 =2•2•139
557 =557
558 =2•3•3•31
559 =13•43
560 =2•2•2•2•5•7
561 =3•11•17
562 =2•281
563 =563
564 =2•2•3•47
565 =5•113
566 =2•283
567 =3•3•3•3•7
568 =2•2•2•71
569 =569
570 =2•3•5•19
571 =571
572 =2•2•11•13
573 =3•191
574 =2•7•41
575 =5•5•23
576 =2•2•2•2•2•2•3•3
577 =577
578 =2•17•17
579 =3•193
580 =2•2•5•29
581 =7•83
582 =2•3•97
583 =11•53
584 =2•2•2•73
585 =3•3•5•13
586 =2•293
587 =587
588 =2•2•3•7•7
589 =19•31
590 =2•5•59
591 =3•197
592 =2•2•2•2•37
593 =593
594 =2•3•3•3•11
595 =5•7•17
596 =2•2•149
597 =3•199
598 =2•13•23
599 =599
600 =2•2•2•3•5•5
601 =601
602 =2•7•43
603 =3•3•67
604 =2•2•151
605 =5•11•11
606 =2•3•101
607 =607
608 =2•2•2•2•2•19
609 =3•7•29
610 =2•5•61
611 =13•47
612 =2•2•3•3•17
613 =613
614 =2•307
615 =3•5•41
616 =2•2•2•7•11
617 =617
618 =2•3•103
619 =619
620 =2•2•5•31
621 =3•3•3•23
622 =2•311
623 =7•89
624 =2•2•2•2•3•13
625 =5•5•5•5
626 =2•313
627 =3•11•19
628 =2•2•157
629 =17•37
630 =2•3•3•5•7
631 =631
632 =2•2•2•79
633 =3•211
634 =2•317
635 =5•127
636 =2•2•3•53
637 =7•7•13
638 =2•11•29
639 =3•3•71
640 =2•2•2•2•2•2•2•5
641 =641
642 =2•3•107
643 =643
644 =2•2•7•23
645 =3•5•43
646 =2•17•19
647 =647
648 =2•2•2•3•3•3•3
649 =11•59
650 =2•5•5•13
651 =3•7•31
652 =2•2•163
653 =653
654 =2•3•109
655 =5•131
656 =2•2•2•2•41
657 =3•3•73
658 =2•7•47
659 =659
660 =2•2•3•5•11
661 =661
662 =2•331
663 =3•13•17
664 =2•2•2•83
665 =5•7•19
666 =2•3•3•37
667 =23•29
668 =2•2•167
669 =3•223
670 =2•5•67
671 =11•61
672 =2•2•2•2•2•3•7
673 =673
674 =2•337
675 =3•3•3•5•5
676 =2•2•13•13
677 =677
678 =2•3•113
679 =7•97
680 =2•2•2•5•17
681 =3•227
682 =2•11•31
683 =683
684 =2•2•3•3•19
685 =5•137
686 =2•7•7•7
687 =3•229
688 =2•2•2•2•43
689 =13•53
690 =2•3•5•23
691 =691
692 =2•2•173
693 =3•3•7•11
694 =2•347
695 =5•139
696 =2•2•2•3•29
697 =17•41
698 =2•349
699 =3•233
700 =2•2•5•5•7
701 =701
702 =2•3•3•3•13
703 =19•37
704 =2•2•2•2•2•2•11
705 =3•5•47
706 =2•353
707 =7•101
708 =2•2•3•59
709 =709
710 =2•5•71
711 =3•3•79
712 =2•2•2•89
713 =23•31
714 =2•3•7•17
715 =5•11•13
716 =2•2•179
717 =3•239
718 =2•359
719 =719
720 =2•2•2•2•3•3•5
721 =7•103
722 =2•19•19
723 =3•241
724 =2•2•181
725 =5•5•29
726 =2•3•11•11
727 =727
728 =2•2•2•7•13
729 =3•3•3•3•3•3
730 =2•5•73
731 =17•43
732 =2•2•3•61
733 =733
734 =2•367
735 =3•5•7•7
736 =2•2•2•2•2•23
737 =11•67
738 =2•3•3•41
739 =739
740 =2•2•5•37
741 =3•13•19
742 =2•7•53
743 =743
744 =2•2•2•3•31
745 =5•149
746 =2•373
747 =3•3•83
748 =2•2•11•17
749 =7•107
750 =2•3•5•5•5
nPrime
Factorization
751 =751
752 =2•2•2•2•47
753 =3•251
754 =2•13•29
755 =5•151
756 =2•2•3•3•3•7
757 =757
758 =2•379
759 =3•11•23
760 =2•2•2•5•19
761 =761
762 =2•3•127
763 =7•109
764 =2•2•191
765 =3•3•5•17
766 =2•383
767 =13•59
768 =2•2•2•2•2•2•2•2•3
769 =769
770 =2•5•7•11
771 =3•257
772 =2•2•193
773 =773
774 =2•3•3•43
775 =5•5•31
776 =2•2•2•97
777 =3•7•37
778 =2•389
779 =19•41
780 =2•2•3•5•13
781 =11•71
782 =2•17•23
783 =3•3•3•29
784 =2•2•2•2•7•7
785 =5•157
786 =2•3•131
787 =787
788 =2•2•197
789 =3•263
790 =2•5•79
791 =7•113
792 =2•2•2•3•3•11
793 =13•61
794 =2•397
795 =3•5•53
796 =2•2•199
797 =797
798 =2•3•7•19
799 =17•47
800 =2•2•2•2•2•5•5
801 =3•3•89
802 =2•401
803 =11•73
804 =2•2•3•67
805 =5•7•23
806 =2•13•31
807 =3•269
808 =2•2•2•101
809 =809
810 =2•3•3•3•3•5
811 =811
812 =2•2•7•29
813 =3•271
814 =2•11•37
815 =5•163
816 =2•2•2•2•3•17
817 =19•43
818 =2•409
819 =3•3•7•13
820 =2•2•5•41
821 =821
822 =2•3•137
823 =823
824 =2•2•2•103
825 =3•5•5•11
826 =2•7•59
827 =827
828 =2•2•3•3•23
829 =829
830 =2•5•83
831 =3•277
832 =2•2•2•2•2•2•13
833 =7•7•17
834 =2•3•139
835 =5•167
836 =2•2•11•19
837 =3•3•3•31
838 =2•419
839 =839
840 =2•2•2•3•5•7
841 =29•29
842 =2•421
843 =3•281
844 =2•2•211
845 =5•13•13
846 =2•3•3•47
847 =7•11•11
848 =2•2•2•2•53
849 =3•283
850 =2•5•5•17
851 =23•37
852 =2•2•3•71
853 =853
854 =2•7•61
855 =3•3•5•19
856 =2•2•2•107
857 =857
858 =2•3•11•13
859 =859
860 =2•2•5•43
861 =3•7•41
862 =2•431
863 =863
864 =2•2•2•2•2•3•3•3
865 =5•173
866 =2•433
867 =3•17•17
868 =2•2•7•31
869 =11•79
870 =2•3•5•29
871 =13•67
872 =2•2•2•109
873 =3•3•97
874 =2•19•23
875 =5•5•5•7
876 =2•2•3•73
877 =877
878 =2•439
879 =3•293
880 =2•2•2•2•5•11
881 =881
882 =2•3•3•7•7
883 =883
884 =2•2•13•17
885 =3•5•59
886 =2•443
887 =887
888 =2•2•2•3•37
889 =7•127
890 =2•5•89
891 =3•3•3•3•11
892 =2•2•223
893 =19•47
894 =2•3•149
895 =5•179
896 =2•2•2•2•2•2•2•7
897 =3•13•23
898 =2•449
899 =29•31
900 =2•2•3•3•5•5
901 =17•53
902 =2•11•41
903 =3•7•43
904 =2•2•2•113
905 =5•181
906 =2•3•151
907 =907
908 =2•2•227
909 =3•3•101
910 =2•5•7•13
911 =911
912 =2•2•2•2•3•19
913 =11•83
914 =2•457
915 =3•5•61
916 =2•2•229
917 =7•131
918 =2•3•3•3•17
919 =919
920 =2•2•2•5•23
921 =3•307
922 =2•461
923 =13•71
924 =2•2•3•7•11
925 =5•5•37
926 =2•463
927 =3•3•103
928 =2•2•2•2•2•29
929 =929
930 =2•3•5•31
931 =7•7•19
932 =2•2•233
933 =3•311
934 =2•467
935 =5•11•17
936 =2•2•2•3•3•13
937 =937
938 =2•7•67
939 =3•313
940 =2•2•5•47
941 =941
942 =2•3•157
943 =23•41
944 =2•2•2•2•59
945 =3•3•3•5•7
946 =2•11•43
947 =947
948 =2•2•3•79
949 =13•73
950 =2•5•5•19
951 =3•317
952 =2•2•2•7•17
953 =953
954 =2•3•3•53
955 =5•191
956 =2•2•239
957 =3•11•29
958 =2•479
959 =7•137
960 =2•2•2•2•2•2•3•5
961 =31•31
962 =2•13•37
963 =3•3•107
964 =2•2•241
965 =5•193
966 =2•3•7•23
967 =967
968 =2•2•2•11•11
969 =3•17•19
970 =2•5•97
971 =971
972 =2•2•3•3•3•3•3
973 =7•139
974 =2•487
975 =3•5•5•13
976 =2•2•2•2•61
977 =977
978 =2•3•163
979 =11•89
980 =2•2•5•7•7
981 =3•3•109
982 =2•491
983 =983
984 =2•2•2•3•41
985 =5•197
986 =2•17•29
987 =3•7•47
988 =2•2•13•19
989 =23•43
990 =2•3•3•5•11
991 =991
992 =2•2•2•2•2•31
993 =3•331
994 =2•7•71
995 =5•199
996 =2•2•3•83
997 =997
998 =2•499
999 =3•3•3•37
1000 =2•2•2•5•5•5 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Directed Line Segments
## Definition and representation of vectors.
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Directed Line Segments
While babysitting your nephew one day you have a discussion regarding math. As a second grader, he is learning to add and subtract quantities. He asks you about what you're doing in math class, and you explain to him that you have just been introduced to vectors. Once you explain to him that a vector is a mathematical quantity that has both magnitude and direction, his question is both simple and yet brilliant: "Why do you need vectors? Can't everything just be described with numbers that don't have direction?"
### Directed Line Segments
A vector is represented diagrammatically by a directed line segment or arrow. A directed line segment has both magnitude and direction. Magnitude refers to the length of the directed line segment and is usually based on a scale. The vector quantity represented, such as influence of the wind or water current may be completely invisible.
A 25 mph wind is blowing from the northwest. If 1 cm=5 mph\begin{align*}1\ cm = 5\ mph\end{align*}, then the vector would look like this:
An object affected by this wind would travel in a southeast direction at 25 mph.
A vector is said to be in standard position if its initial point is at the origin. The initial point is where the vector begins and the terminal point is where it ends. The axes are arbitrary. They just give a place to draw the vector.
If we know the coordinates of a vector’s initial point and terminal point, we can use these coordinates to find the magnitude and direction of the vector.
All vectors have magnitude. This measures the total distance moved, total velocity, force or acceleration. “Distance” here applies to the magnitude of the vector even though the vector is a measure of velocity, force, or acceleration. In order to find the magnitude of a vector, we use the distance formula. A vector can have a negative magnitude. A force acting on a block pushing it at 20 lbs north can be also written as vector acting on the block from the south with a magnitude of -20 lbs. Such negative magnitudes can be confusing; making a diagram helps. The -20 lbs south can be re-written as +20 lbs north without changing the vector. Magnitude is also called the absolute value of a vector.
#### Finding the Length of a Vector
If a vector starts at the origin and has a terminal point with coordinates (3,5), find the length of the vector.
If we know the coordinates of the initial point and the terminal point, we can find the magnitude by using the distance formula. Initial point (0,0) and terminal point (3,5).
|v|=(30)2+(50)2=9+25=5.8\begin{align*}|\vec{v}| = \sqrt{(3 - 0)^2 + (5 - 0)^2} = \sqrt{9 + 25} = 5.8\end{align*} The magnitude of v\begin{align*}\vec{v}\end{align*} is 5.8.
If we don’t know the coordinates of the vector, we must use a ruler and the given scale to find the magnitude. Also notice the notation of a vector, which is usually a lower case letter (typically u,v\begin{align*}u, v\end{align*}, or w\begin{align*}w\end{align*}) in italics, with an arrow over it, which indicates direction. If a vector is in standard position, we can use trigonometric ratios such as sine, cosine and tangent to find the direction of that vector.
#### Find the Direction of a Vector
If a vector is in standard position and its terminal point has coordinates of (12, 9) what is the direction?
The horizontal distance is 12 while the vertical distance is 9. We can use the tangent function since we know the opposite and adjacent sides of our triangle.
tanθtan1912=912=36.9\begin{align*}\tan \theta & = \frac{9}{12} \\ \tan^{-1} \frac{9}{12} & = 36.9^\circ\end{align*}
So, the direction of the vector is 36.9\begin{align*}36.9^\circ\end{align*}.
If the vector isn’t in standard position and we don’t know the coordinates of the terminal point, we must a protractor to find the direction.
Two vectors are equal if they have the same magnitude and direction. Look at the figures below for a visual understanding of equal vectors.
#### Determining if Two Vectors are Equal
a\begin{align*}\vec{a}\end{align*} is in standard position with terminal point (-4, 12)
b\begin{align*}\vec{b}\end{align*} has an initial point of (7, -6) and terminal point (3, 6)
You need to determine if both the magnitude and the direction are the same.
Magnitude:|a||b|Direction:ab=(0(4))2+(012)2=16+144=160=410=(73)2+(66)2=16+144=160=410tanθ=124θ=108.43tanθ=6673=124θ=108.43\begin{align*}\text{Magnitude}:\quad |\vec{a}| & = \sqrt{(0-(-4))^2 + (0 - 12)^2} = \sqrt{16+144} = \sqrt{160} = 4\sqrt{10} \\ |\vec{b}| & = \sqrt{(7-3)^2 + (-6-6)^2} = \sqrt{16+144} = \sqrt{160} = 4 \sqrt{10} \\ \text{Direction}:\quad \vec{a} & \rightarrow \tan \theta = \frac{12}{-4} \rightarrow \theta = 108.43^\circ \\ \vec{b} & \rightarrow \tan \theta = \frac{-6-6}{7-3} = \frac{-12}{4} \rightarrow \theta = 108.43^\circ\end{align*}
Because the magnitude and the direction are the same, we can conclude that the two vectors are equal.
### Examples
#### Example 1
Your nephew's thinking is quite good. Many things in the world can be described by numbers, without the use of direction. However, math needs to "line up" with reality. If something doesn't work with just numbers, there needs to be a new type of mathematical quantity to describe the behavior completely. For example, consider two cars that are both moving at 25 miles per hour. Will they collide?
You can see that there isn't enough information to answer the question. You don't know which way the cars are going. If the two cars are going in the same direction, then they won't collide. If, however, they are going directly at each other, then they will certainly collide. In order to describe the behavior of the cars completely, a quantity is needed that is not just the magnitude of the car's motion, but also the direction - which is why vectors are needed.
#### Example 2
Given the initial and terminal coordinates below, find the magnitude and direction of the vector that results.
initial ( 2, 4) terminal (8, 6)
|a|=(28)2+(46)2=6.3\begin{align*}|\vec{a}| = \sqrt{(2-8)^2 + (4-6)^2} = 6.3\end{align*}, direction =tan1(4628)=18.4\begin{align*}=\tan^{-1} \left (\frac{4-6}{2-8} \right ) = 18.4^\circ\end{align*}
#### Example 3
Given the initial and terminal coordinates below, find the magnitude and direction of the vector that results.
initial (5, -2) terminal (3, 1)
\begin{align*}|\vec{a}| = \sqrt{(5-3)^2 + (-2-1)^2} = 3.6\end{align*}, direction \begin{align*}=\tan^{-1} \left (\frac{-2-1}{5-3} \right ) = 123.7^\circ\end{align*}. Note that when you use your calculator to solve for \begin{align*}\tan^{-1} (\frac{-2-1}{5-3})\end{align*}, you will get \begin{align*}-56.3^\circ\end{align*}. The calculator produces this answer because the range of the calculator’s \begin{align*}y = \tan^{-1} x\end{align*} function is limited to \begin{align*}-90^\circ < y < 90^\circ\end{align*}. You need to sketch a draft of the vector to see that its direction when placed in standard position is into the second quadrant (and not the fourth quadrant), and so the correct angle is calculated by moving the angle into the second quadrant through the equation \begin{align*}-56.3^\circ + 180^\circ=123.7^\circ.\end{align*}
#### Example 4
Assume \begin{align*}\vec{a}\end{align*} is in standard position. For the terminal point (12, 18), find the magnitude and direction of the vector.
\begin{align*}|\vec{a}| = \sqrt{12^2 + 18^2} = 21.6\end{align*}, direction \begin{align*}=\tan^{-1} \left (\frac{18}{12} \right ) = 56.3^\circ \end{align*}
### Review
1. What is the difference between the magnitude and direction of a vector?
2. How can you determine the magnitude of a vector if you know its initial point and terminal point?
3. How can you determine the direction of a vector if you know its initial point and terminal point?
4. How can you determine whether or not two vectors are equal?
5. If a vector starts at the origin and has a terminal point with coordinates (2, 7), find the magnitude of the vector.
6. If a vector is in standard position and its terminal point has coordinates of (3, 9), what is the direction of the vector?
7. If a vector has an initial point at (1, 6) and has a terminal point at (5, 9), find the magnitude of the vector.
8. If a vector has an initial point at (1, 4) and has a terminal point at (8, 7), what is the direction of the vector?
Given the initial and terminal coordinates below, find the magnitude and direction of the vector that results.
1. initial (4, -1); terminal (5, 3)
2. initial (2, -3); terminal (4, 5)
3. initial (3, 2); terminal (0, 3)
4. initial (-2, 5); terminal (2, 1)
Determine if the two vectors are equal.
1. \begin{align*}\vec{a}\end{align*} is in standard position with terminal point (1, 5) and \begin{align*}\vec{b}\end{align*} has an initial point (3, -2) and terminal point (4, 2).
2. \begin{align*}\vec{c}\end{align*} has an initial point (-3, 1) and terminal point (1, 2) and \begin{align*}\vec{d}\end{align*} has an initial point (3, 5) and terminal point (7, 6).
3. \begin{align*}\vec{e}\end{align*} is in standard position with terminal point (2, 3) and \begin{align*}\vec{f}\end{align*} has an initial point (1, -6) and terminal point (3, -9).
To see the Review answers, open this PDF file and look for section 5.13.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Directed Line Segment A directed line segment is a portion of a line that has both a magnitude and direction.
Magnitude The magnitude of a line segment or vector is the length of the line segment or vector.
Vector A vector is a mathematical quantity that has both a magnitude and a direction. |
# How to Calculate Density - Worked Example Problem
## Finding the Ratio Between Mass and Volume
Density is the measurement of the amount of mass per unit of volume. In order to calculate density, you need to know the mass and volume of the item. The formula for density is:
density = mass/volume
The mass is usually the easy part while finding volume can be tricky. Simple shaped objects are usually given in homework problems such as using a cube, brick or sphere. For a simple shape, use a formula to find volume. For irregular shapes, the easiest solution is to measure volume displaced by placing the object in a liquid.
This example problem shows the steps needed to calculate the density of an object and a liquid when given the mass and volume.
### Key Takeaways: How to Calculate Density
• Density is how much matter is contained within a volume. A dense object weighs more than a less dense object that is the same size. An object less dense than water will float on it; one with greater density will sink.
• The density equation is density equals mass per unit volume or D = M / V.
• The key to solving for density is to report the proper mass and volume units. If you are asked to give density in different units from the mass and volume, you will need to convert them.
Question 1: What is the density of a cube of sugar weighing 11.2 grams measuring 2 cm on a side?
Step 1: Find the mass and volume of the sugar cube.
Mass = 11.2 grams
Volume = cube with 2 cm sides.
Volume of a cube = (length of side)3
Volume = (2 cm)3
Volume = 8 cm3
Step 2: Plug your variables into the density formula.
density = mass/volume
density = 11.2 grams/8 cm3
density = 1.4 grams/cm3
Answer 1: The sugar cube has a density of 1.4 grams/cm3.
Question 2: A solution of water and salt contains 25 grams of salt in 250 mL of water. What is the density of the salt water? (Use density of water = 1 g/mL)
Step 1: Find the mass and volume of the salt water.
This time, there are two masses. The mass of the salt and the mass of the water are both needed to find the mass of the salt water. The mass of the salt is given, but the only the volume of water is given. We've also been given the density of water, so we can calculate the mass of the water.
densitywater = masswater/volumewater
solve for masswater,
masswater = densitywater·volumewater
masswater = 1 g/mL · 250 mL
masswater = 250 grams
Now we have enough to find the mass of the salt water.
masstotal = masssalt + masswater
masstotal = 25 g + 250 g
masstotal = 275 g
Volume of the salt water is 250 mL.
Step 2: Plug your values into the density formula.
density = mass/volume
density = 275 g/250 mL
density = 1.1 g/mL
Answer 2: The salt water has a density of 1.1 grams/mL.
## Finding Volume by Displacement
If you're given a regular solid object, you can measure its dimensions and calculate its volume. Unfortunately, the volume of few objects in the real world can be measured this easily! Sometimes you need to calculate volume by displacement.
How do you measure displacement? Say you have a metal toy soldier. You can tell it is heavy enough to sink in water, but you can't use a ruler to measure its dimensions. To measure the toy's volume, fill a graduated cylinder about half way with water. Record the volume. Add the toy. Make sure to displace any air bubbles that may stick to it. Record the new volume measurement. The volume of the toy soldier is the final volume minus the initial volume. You can measure the mass of the (dry) toy and then calculate density.
## Tips for Density Calculations
In some cases, the mass will be given to you. If not, you'll need to obtain it yourself by weighing the object. When obtaining mass, be aware of how accurate and precise the measurement will be. The same goes for measuring volume. Obviously, you'll get a more precise measurement using a graduated cylinder than using a beaker, however, you may not need such a close measurement. The significant figures reported in the density calculation are those of your least precise measurement. So, if your mass is 22 kg, reporting a volume measurement to the nearest microliter is unnecessary.
Another important concept to keep in mind is whether your answer makes sense. If an object seems heavy for its size, it should have a high density value. How high? Keep in mind the density of water is about 1 g/cm³. Objects less dense than this float in water, while those that are more dense sink in water. If an object sinks in water, your density value better be greater than 1!
## More Homework Help
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# Lines Angles and Transversals: Definition with Examples
Sep 9, 2022
### Key Concepts
• Identify angles created by parallel lines cut by a transversal
• Find unknown angle measures
• Use algebra to find unknown angle measures
## Lines Angles and Transversals
1. What is meant by similar figures?
1. Which symbols is used to indicate similarity?
1. What is the sequence of transformations for the image given?
1. Similar figures are two figures having the same shape. The objects which are of exactly the same shape and size are known as congruent objects.
1. This symbol is used for similarity.
1. Reflect over the x-axis, then translate (x+6, y)
## Angles, Lines, and Transversals
#### Angle:
Angles are formed when two lines intersect at a point.
#### Line:
A line is a one-dimensional figure, which has length but no width. A line is made of a set of points which is extended in opposite directions infinitely.
A transversal is a line that passes through two lines in the same plane at two distinct points.
Transversal:
### Identify Angles Created by Parallel Lines Cut by a Transversal
#### Parallel lines:
Parallel lines are the lines that do not intersect or meet each other at any point in a plane.
#### Transversal:
When any two parallel lines are cut by a transversal, many pairs of angles are formed.
There is a relationship that exists between these pairs of angles.
While some of them are congruent, the others are supplementary.
### Example:
#### Identify angles created by parallel lines cut by a transversal.
Sol:
Parallel Lines Cut by a Transversal
From the figure corresponding angles formed by the intersection of the transversal are:
∠1 and ∠5
∠2 and ∠6
∠3 and ∠7
∠4 and ∠8
The pair of corresponding angles are equal in measure, that is,
∠1 = ∠5, ∠2 = ∠6, ∠3 = ∠7, and ∠4 = ∠8
### Alternate interior angles:
Alternate interior angles are formed on the inside of two parallel lines which are intersected by a transversal. In the figure given above, there are two pairs of alternate interior angles.
∠3 and ∠6
∠4 and ∠5
The pair of alternate interior angles are equal in measure, that is, ∠3 = ∠6, and ∠4 = ∠5
### Alternate exterior angles:
When two parallel lines are cut by a transversal, the pairs of angles formed on either side of the transversal are named as alternate exterior angles. From the figure given above, there are two pairs of alternate exterior angles.
∠1 and ∠8
∠2 and ∠7
The pair of alternate exterior angles are equal in measure, that is, ∠1 = ∠8, and ∠2 = ∠7
### Consecutive interior angles:
When two parallel lines are cut by a transversal, the pairs of angles formed on the inside of one side of the transversal are called consecutive interior angles or co-interior angles. From the figure, there are two pairs of consecutive interior angles.
∠4 and ∠6
∠3 and ∠5
Unlike the other pairs given above, the pair of consecutive interior angles are supplementary, that is, ∠4 + ∠6 = 180°, and ∠3 + ∠5 = 180°.
### Find Unknown Angle Measures
Example:
What is the measures of m ∠6? Explain.
Sol:
Use what you know about the angles created when parallel lines are cut by transversal.
m ∠6 + 59° = 180°
m ∠6 = 180° – 59°
m ∠6 = 121°
### Use Algebra to Find Unknown Angle Measures
Algebra:
An algebraic expression in algebra is formed using integer constants, variables, and basic arithmetic operations of addition (+), subtraction (-), multiplication (×), and division (/).
Below image is the example for algebraic expression.
Example:
In the figure given below, let the lines l1 and l2 be parallel and t is transversal. Find the value of x.
Sol:
From the given figure,
∠ (2x + 20) ° and ∠ (3x – 10) ° are corresponding angles.
So, they are equal.
Then, we have
(2x + 20) ° = (3x – 10)°
2x + 20 = 3x – 10
Subtract 2x from each side.
20 = x – 10
30 = x
## Exercise
1. Given the following two parallel lines that have been cut by a transversal. ∠1 and which angle make up alternate interior angles?
1. Find the missing angle measures.
1. Find the unknown angle.
1. Find the value of x in the following figure.
1. What type of angle pair is ∠1 and ∠3?
1. Solve for x.
1. Find x.
1. Given the following two parallel lines that have been cut by a transversal.
Which two angles would be alternate exterior?
1. For the given figure, can you conclude that r∥s? Explain.
1. Solve for x.
### What we have learned:
• Understand Angles, Lines and Transversals
• Identify Angles Created by Parallel Lines Cut by a Transversal
• Find Unknown Angle Measures
• Use Algebra to Find Unknown Angle Measures
Concept Map:
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […] |
### The Pythagorean Theorum
Wednesday, February 18, 2009
We were given an assignment about Pythagoras, a Greek mathematician who supposedly lived a very long time ago. We were also given a list of important vocabulary that could be of use to us.
The legs are the shorter part of a right angle triangle, while the hypotenuse is the longer part of a right angle triangle.
R.A.T. is an abbreviated form of Right Angle Triangle, which is a triangle that has a 90° angle.
The Greeks were an ancient civilization who greatly influencial in philosophy, educational systems, etc. etc.
A theorum is a statement proved that it is right judging by previous evidence.
The three pictures are linked together because it has to with the Pythagorean Theorum. The Pythagorean therum tells us the relationship between the legs and hyptonuse of a right angle triangle, that is why there is a picture of a right angle triangle in the post (we know it is a right angle because of the square at the corner). There is a picture of a square because in the Pythagorean Theorum you make multiple right angles, resulting in a square. The last picture is of Pythagoras himself, or rather a sculpture of him. It was his theorum, so that is why he is there.
You are able to explain the artifacts using the vocabulary because it all has to do with the theorum. You use right angles the entire time because you are trying to find the relationship between the legs of a right angle triangle.
The shapes you use are right angle triangles, and squares. The formula is A²+B²=C². The "guy" is obviously Pythagoras, and we care in Grade 8 Math because he supposedly created this formula. He was uber intelligent Greek mathematician who had mad skills in that particular subject, such as many others.
Below is a picture of how to find a missing length of an R.A.T.
Below is a picture of a word problem.
Now here are pictures of my Pythagoras Booklet.
Word Problem 2
a) The diagonal length of the square is 4.2 cm2
-- each square is 3x3 so you use a2+b2=c2.
9+9=c2
18=c2
18 [square root] = 4.2
4.2= c
b) the diagonal of the checker board is 22.627 cm2
- theres 64 squares, and 64 divided by 4 is 16. 16 is the side length. So from there you use A2+b2=C2. Then you get 22.627cm2
Here are my Pythagoras Videos. |
Question Video: Finding the Quotient of a Complex Number Raised to a Power in Polar Form | Nagwa Question Video: Finding the Quotient of a Complex Number Raised to a Power in Polar Form | Nagwa
# Question Video: Finding the Quotient of a Complex Number Raised to a Power in Polar Form Mathematics • Third Year of Secondary School
## Join Nagwa Classes
Given that πβ = cos (2π/3) + π sin (2π/3) and πβ = cos (π/6) + π sin (π/6), find (πββ΅/πβΒ²).
03:42
### Video Transcript
Given that π sub one is equal to the cos of two π by three plus π sin of two π by three and π sub two is equal to the cos of π by six plus π sin of π by six, find π sub one to the fifth power divided by π sub two squared.
In this question, weβre given two complex numbers, π sub one and π sub two, in polar form. We need to use this to determine π sub one to the fifth power divided by π sub two squared. And thereβs many different ways we can approach answering this question. Since we need to find π sub one to the fifth power and π sub two squared and both of these are given in polar form, letβs start by recalling de Moivreβs theorem.
One version of de Moivreβs theorem tells us for any integer value of π and real number π, cos π plus π sin π all raised to the πth power is equal to cos of ππ plus π sin of ππ. And we can use this to evaluate π sub one to the fifth power and π sub two squared.
Letβs start with π sub one raised to the fifth power. This is the cos of two π by three plus π sin of two π by three all raised to the fifth power. And our exponent is an integer value of five. De Moivreβs theorem then tells us we can evaluate this by multiplying each of the arguments by five. And since five times two π by three is 10π by three, this gives us the cos of 10π by three plus π sin of 10π by three. We can also use this to find π sub two all squared. Thatβs the cos of π by six plus π sin of π by six all squared. This time, our exponent is two, which is once again an integer. And de Moivreβs theorem tells us we can evaluate this by multiplying the argument by two. And since two times π by six is π by three, this gives us the cos of π by three plus π sin of π by three. This then gives us that π sub one raised to the fifth power divided by π sub two squared is equal to the cos of 10π by three plus π sin of 10π by three all divided by the cos of π by three plus π sin of π by three.
This is then the quotient of two complex numbers given in polar form. And we can evaluate this by recalling how we find the quotient of two complex numbers given in polar form. If the complex number in our denominator is nonzero, then π times the cos of π plus π sin of π all divided by π times the cos of π plus π sin of π is equal to π over π multiplied by the cos of π minus π plus π sin of π minus π. In other words, we find the quotient of their moduli and we find the difference in their arguments.
And in the quotient of the two complex numbers weβre given, we can see thereβs no coefficient. In other words, their moduli are both one. So we only need to find the difference in the arguments. We get cos of 10π by three minus π by three plus π sin of 10π by three minus π by three. And 10π by three minus π by three is equal to nine π by three, which simplifies to give us three π. And remember, we can add and subtract integer multiples of two π from the argument. This gives us an argument of π, which simplifies to give us cos of π plus π sin of π.
And itβs worth noting we can simplify this further. The cos of π is negative one and the sin of π is zero. So this simplifies to give us negative one. However, weβll leave our answer in polar form because this is the form π sub one and π sub two is given in.
Therefore, we were able to show if π sub one is cos of two π by three plus π sin of two π by three and π sub two is cos of π by six plus π sin of π by six, then π sub one to the fifth power divided by π sub two squared in polar form is cos of π plus π sin of π.
## Join Nagwa Classes
Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!
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# The Addition and Resolution of Vectors: The Force Table TLSAMP 2005.
## Presentation on theme: "The Addition and Resolution of Vectors: The Force Table TLSAMP 2005."— Presentation transcript:
The Addition and Resolution of Vectors: The Force Table TLSAMP 2005
Group Members Jazmon Malone Christian Mallet Kristen Williams Janna Lipford Rhonda Laird
Force At Work
Group Two
Objective To analyze different methods of vector addition.
Hypothesis We believe that the sum of two vectors is not equal to the sum of their magnitude.
Group Two At Work
Materials Three Weights Calibrator Thread Force Table Scissors Calculator Pulleys Washer
Methods Collected all materials Make all pulleys parallel to the force table Next you will cut the thread and attach it to each weight Attach each weight to the washer After attaching each weight to the washer, place one weight on 30 degrees and the other weight on 120 degrees. With the remainder weight you will attempt to balance the washer. When the washer is balanced you will record the degree and the weight that it took to balance the other two weights
Methods (Continued) The weight is the magnitude of the resultant force, and degree is 180 degrees different from the direction of the resultant force Input your results into the following formula Record your results on a table. Repeat this experiment three more times. Compare the results and come to a conclusion
Data F1F2 ANGLE 1 IN DEGREES ANGLE 1 IN RADIANS ANGLE 2 IN DEGREES ANGLE 2 IN RADIANS 50 300.52361202.0944 50 200.34911001.7453 50 00.0000901.5708 50 100.1745951.6581 50 0.87271602.7925 50 1001.7453300.5236 50 1152.0071400.6981 50 350.61091452.5307
Data (Continued) cos(a1)cos(a2)sum(Fx)sin(a1)sin(a2)sum(Fy) 0.866025-0.518.301270190.50.86602540468.30127019 0.939693-0.1736538.302222160.3420201430.98480775366.34139482 16.13E-175001 0.984808-0.0871644.882600510.1736481780.99619469858.49214379 0.642788-0.93969-14.845250550.7660444430.34202014355.40322932 -0.173650.86602534.618861310.9848077530.574.24038765 -0.422620.76604417.171309070.9063077870.6427876177.45476984 0.819152-0.8191500.573576436 57.35764364
Data (Continued) R Balancing ForceTHETA IN RADIANS THETA IN DEGREES 70.71068701.30899693975 76.60444801.04719755160 70.71068450.78539816345 73.7277352.50.91629785753 57.3576475-1.308996939-75 81.915264.9951.13446401465 79.3353377.51.3526301778 57.35764000
Data Continued cos(a1)cos(a2)sum(Fx)sin(a1)sin(a2)sum(Fy) 0.866025-0.518.301270190.50.86602540468.30127019 0.939693-0.1736538.302222160.3420201430.98480775366.34139482 16.13E-175001 0.984808-0.0871644.882600510.1736481780.99619469858.49214379 0.642788-0.93969-14.845250550.7660444430.34202014355.40322932 -0.173650.86602534.618861310.9848077530.574.24038765 -0.422620.76604417.171309070.9063077870.6427876177.45476984 0.819152-0.8191500.573576436 57.35764364
Data Continued Theoretical Force Empirical Force 70.7106870 76.6044480 70.7106845 73.7277352.5 57.3576475 81.915264.995 79.3353377.5 57.357640
Hypothesis Testing Ho: Theoretical Force = Empirical Force Ha: Theoretical Force is not equal to the Empirical Force By using the Paired t-Test we found that the p-value is equal to By using the Paired t-Test we found that the p-value is equal to 0.157964475 The p-value is less than 0.05, so we failed to reject the null hypothesis.
Data Continued t-Test: Paired Two Sample for Means Theoretical ForceBalancing Force Mean70.9649199958.124375 Variance85.69359314701.3294674 Observations88 Pearson Correlation0.528446146 Hypothesized Mean Difference0 df7 t Stat1.580675693 P(T<=t) one-tail0.078982238 t Critical one-tail1.894577508 P(T<=t) two-tail0.157964475 t Critical two-tail2.36462256
Group Two with the Force Table
Conclusion Our Empirical Result supports the theoretical result. The theoretical result is the sum of the forces and magnitude. Our Empirical Result supports the theoretical result. The theoretical result is the sum of the forces and magnitude. |
# NCERT Solutions For Class 6 Maths Fractions Exercise 7.4
## NCERT Solutions For Class 6 Maths Fractions Exercise 7.4
NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.4
Exercise 7.4
Question 1.
Write shaded portion as fraction. Arrange them in ascending and descending order using correct sign ‘<‘, ‘=’, ‘>’ between the fractions.
(c) Show $\frac { 2 }{ 4 }$, $\frac { 4 }{ 6 }$, $\frac { 8 }{ 6 }$and $\frac { 6 }{ 6 }$on the number line. Put appropriate signs between the fractions given.
Solution:
(a) Total number of divisions = 8
(i) Number of shaded parts = 3
∴ Fraction = $\frac { 3 }{ 8 }$
(ii) Total number of divisions = 8
Number of shaded parts = 6
∴ Fraction = $\frac { 6 }{ 8 }$
(iii) Total number of divisions = 8
Number of shaded parts = 4
∴ Fraction = $\frac { 4 }{ 8 }$
(iv) Total number of divisions = 8
Number of shaded part = 1
∴ Fraction = $\frac { 1 }{ 8 }$
Now the fractions are:
$\frac { 3 }{ 8 }$, $\frac { 6 }{ 8 }$, $\frac { 4 }{ 8 }$and $\frac { 1 }{ 8 }$with same denominator.
(b)(i) Total number of divisions = 9
Number of shaded parts = 8
∴ Fraction = $\frac { 8 }{ 9 }$
(ii) Total number of divisions = 9
Number of shaded parts = 4
∴ Fraction = $\frac { 4 }{ 9 }$
(iii) Total number of divisions = 9
Number of shaded parts = 3
∴ Fraction = $\frac { 3 }{ 9 }$
(iv) Total number of divisions = 9
Number of shaded parts = 6
∴ Fraction = $\frac { 6 }{ 9 }$
∴ Fractions are $\frac { 8 }{ 9 }$, $\frac { 4 }{ 9 }$, $\frac { 3 }{ 9 }$, $\frac { 6 }{ 9 }$with same denominator.
Question 2.
Compare the fractions and put an appropriate sign.
Solution:
Here, denominators of the two fractions are same and 3 < 5.
Here, numerators of the fractions are same and 7 > 4.
Here, denominators of the two fractions are same and 4 < 5.
Here, numerators of the two fractions are same and 5 < 7.
Question 3.
Make five more such pairs and put appropriate signs.
Solution:
Question 4.
Look at the figures and write ’<’, or ’>’ ’=’ between the given pairs of fractions.
Make five more such problems and solve them with your friends
Solution:
Make five more such problems yourself and solve them with your friends.
Question 5.
How quickly can you do this? Fill appropriate sign. ‘<‘, ‘=’, ‘>’.
Solution:
Question 6.
The following fractions represent just three different numbers. Separate them into three groups of equivalent fractions, by changing each one to its simplest form.
Solution:
Now grouping the above fractions into equivalent fractions, we have
Question 7.
Find answers to the following. Write and indicate how you solved them.
Solution:
By cross-multiplying, we get
5 x 5 = 25 and 4 x 9 = 36
Since 25 ≠ 36
By cross-multiplying, we get
9 x 9 = 81 and 16 x 5 =80
Since 81 ≠ 80
By cross-multiplying, we get
4 x 20 = 80 and 5 x 16 = 80
Since 80 = 80
By cross-multiplying, we get
1 x 30 = 30 and 4 x 15 = 60
Question 8.
Ila read 25 pages of a book containing 100 pages.
Lalita read $\frac { 2 }{ 5 }$of the same book. Who read less?
Solution:
Ila reads 25 pages out of 100 pages.
Lalita reads $\frac { 2 }{ 5 }$of the same book.
Comparing $\frac { 1 }{ 4 }$and $\frac { 2 }{ 5 }$, we get
1 x 5 = 5 and 2 x 4 = 8
Since 5 < 8
$\frac { 1 }{ 4 }$< $\frac { 2 }{ 5 }$
Question 9.
Rafiq exercised for $\frac { 3 }{ 6 }$of an hour, while Rohit exercised for $\frac { 3 }{ 4 }$of an hour. Who exercised for a longer time?
Solution:
Rafiq exercised for $\frac { 3 }{ 6 }$of an hour.
Rohit exercised for $\frac { 3 }{ 4 }$of an hour.
Comparing $\frac { 3 }{ 6 }$and $\frac { 3 }{ 4 }$, we get
3 x 4 = 12 and 3 x 6 = 18
Since 12 < 18
$\frac { 3 }{ 4 }$> $\frac { 3 }{ 6 }$
Hence Rohit exercised for longer time.
Question 10.
In a class A of 25 students, 20 passed in first class, in another class B of 30 students, 24 passed in first class. In which class was a greater fraction of students getting first class?
Solution:
In class A, 20 students passed in first class out of 25 students.
∴ Fraction of students getting first class
In class B, 24 students passed in first class out of 30 students.
∴ Fraction of students getting first class
Comparing the two fractions, we get $\frac { 4 }{ 5 }$= $\frac { 4 }{ 5 }$
Hence, both the class A and B have the same fractions.
+ |
# Lesson 3
Representing Exponential Growth
Let’s explore exponential growth.
### 3.1: Math Talk: Exponent Rules
Rewrite each expression as a power of 2.
$$2^3 \boldcdot 2^4$$
$$2^5 \boldcdot 2$$
$$2^{10} \div 2^7$$
$$2^9 \div 2$$
### 3.2: What Does $x^0$ Mean?
1. Complete the table. Take advantage of any patterns you notice.
$$x$$ $$3^x$$ 4 3 2 1 0 81 27
2. Here are some equations. Find the solution to each equation using what you know about exponent rules. Be prepared to explain your reasoning.
1. $$9^?\boldcdot 9^7 = 9^7$$
2. $$\dfrac {9^{12}}{9^?}= 9^{12}$$
3. What is the value of $$5^0$$? What about $$2^0$$?
We know, for example, that $$(2+3)+5=2+(3+5)$$ and $$2\boldcdot (3\boldcdot 5)=(2\boldcdot 3)\boldcdot 5$$. The grouping with parentheses does not affect the value of the expression.
Is this true for exponents? That is, are the numbers $$2^{(3^5)}$$ and $$(2^3)^5$$ equal? If not, which is bigger? Which of the two would you choose as the meaning of the expression $$2^{3^5}$$ written without parentheses?
### 3.3: Multiplying Microbes
1. In a biology lab, 500 bacteria reproduce by splitting. Every hour, on the hour, each bacterium splits into two bacteria.
1. Write an expression to show how to find the number of bacteria after each hour listed in the table.
2. Write an equation relating $$n$$, the number of bacteria, to $$t$$, the number of hours.
3. Use your equation to find $$n$$ when $$t$$ is 0. What does this value of $$n$$ mean in this situation?
hour number of bacteria
0 500
1
2
3
6
t
2. In a different biology lab, a population of single-cell parasites also reproduces hourly. An equation which gives the number of parasites, $$p$$, after $$t$$ hours is $$p = 100 \boldcdot 3^t.$$ Explain what the numbers 100 and 3 mean in this situation.
### 3.4: Graphing the Microbes
1. Refer back to your work in the table of the previous task. Use that information and the given coordinate planes to graph the following:
a. Graph $$(t,n)$$ when $$t$$ is 0, 1, 2, 3, and 4.
b. Graph $$(t,p)$$ when $$t$$ is 0, 1, 2, 3, and 4. (If you get stuck, you can create a table.)
2. On the graph of $$n$$, where can you see each number that appears in the equation?
3. On the graph of $$p$$, where can you see each number that appears in the equation?
### Summary
In relationships where the change is exponential, a quantity is repeatedly multiplied by the same amount. The multiplier is called the growth factor.
Suppose a population of cells starts at 500 and triples every day. The number of cells each day can be calculated as follows:
number of days number of cells
0 500
1 1,500 (or $$500 \boldcdot 3$$)
2 4,500 (or $$500 \boldcdot 3\boldcdot 3$$, or $$500 \boldcdot 3^2$$)
3 13,500 (or $$500 \boldcdot 3\boldcdot 3 \boldcdot 3$$, or $$500 \boldcdot 3^3$$)
$$d$$ $$500 \boldcdot 3^d$$
We can see that the number of cells ($$p$$) is changing exponentially, and that $$p$$ can be found by multiplying 500 by 3 as many times as the number of days ($$d$$) since the 500 cells were observed. The growth factor is 3. To model this situation, we can write this equation: $$\displaystyle p = 500 \boldcdot 3^d$$.
The equation can be used to find the population on any day, including day 0, when the population was first measured. On day 0, the population is $$500 \boldcdot 3^0$$. Since $$3^0 = 1$$, this is $$500 \boldcdot 1$$ or 500.
Here is a graph of the daily cell population. The point $$(0,500)$$ on the graph means that on day 0, the population starts at 500.
Each point is 3 times higher on the graph than the previous point. $$(1,1500)$$ is 3 times higher than $$(0,500)$$, and $$(2,4500)$$ is 3 times higher than $$(1,1500)$$.
### Glossary Entries
• growth factor
In an exponential function, the output is multiplied by the same factor every time the input increases by one. The multiplier is called the growth factor. |
Math Real functions Polynomial functions
# Polynomial functions
Polynomial functions have an equation in the form:
$f(x)=a_n\cdot x^n+a_{n-1}\cdot x^{n-1}+...$ $+a_1\cdot x+a_0$
i
### Hint
Polynomial functions are a special type of rational functions $\frac{g(x)}{h(x)}$ with the denominator $h(x) = 1$.
The function term is called polynomial and the exponent $n$ indicates the degree of the polynomial.
The following are typical function graphs for nth degree functions:
$n=1$ linear functions* $f(x)=x$
$n=2$ quadratic functions* $f(x)=x^2$
$n=3$ cubic functions* $f(x)=x^3$
* In the examples, all functions are also power functions, but linear, quadratic and cubic functions are only power functions if there is no other summand behind them. For example, these are not power functions: $f(x)=x+5$ and $f(x)=4x^2-x+3$
!
### Note
Power functions of the form $f(x)=ax^n$ are only polynomial functions if $n\in\mathbb{N}$.
Power functions with $n\in\mathbb{Z}$ and $n<0$ (e.g. $x^{-3}$) are rational functions.
### Examples
Other examples of polynomial functions are:
• $f(x)=10x-5$
• $f(x)=x^3+x$
• $f(x)=x^4+10x^3+2x^2+4$
!
### Remember
The graph of an nth degree function has at most
• $n$ zeros
• $(n-1)$ extrema
• $(n-2)$ inflection points
### Example
The pictured 3rd degree function has:
• three zeros ($x_0$)
• two extrema ($H$ = maximum and $T$ = minimum)
• one inflection point ($W$) |
Class 6 Maths Fractions Adding or Subtracting Unlike Fractions
• These fractions have different denominators.
• Firstly, the fractions are converted into equivalent fractions with a common denominator.
• To do so, the LCM of denominators is calculated.
• The fractions are converted into like fractions with a common denominator.
• The common denominator is retained
Subtraction of Unlike Fractions
• These fractions have different denominators.
• Firstly, the fractions are converted into equivalent fractions with a common denominator.
• To do so, the LCM of denominators is calculated.
• The fractions are converted into like fractions with a common denominator.
• The common denominator is retained
• The numerators are subtracted.
Problem: Solve 2/3 + 1/7
Solution:
These fractions have different denominators.
Firstly, the fractions are converted into equivalent fractions with a common denominator.
To do so, the LCM of denominators is calculated.
So, the fractions are :
= 2*7/3*7 and 1*3/7*3
= 14/21 and 3/21
Adding the numerators we get :
= 14+3/21
= 17/21
Problem: A piece of wire 7/8 metre long broke into two pieces. One piece was ¼ metre long. How long is the other piece?
Solution:
Total length of the wire = 7/8 metre
Length of one piece of the wire = ¼ metre
Length of the remaining piece of the wire = 7/8 – ¼
Since these are unlike fractions, they are converted into equivalent fractions with common denominator.
The LCM of 8 and 4 is 8. 7/8 already has the denominator of the value 8. ¼ can be converted into 2/8 by multiplying both the numerator and denominator by 2.
Now, we get :
7/8 – ¼
= 7/8 – 2/8
= (7-2)/8
=5/8 metre
Hence the length of the remaining piece of wire is 5/8 metre
. |
Prime Factor
Curriculum Goal
Junior: Number Sense
• Represent composite numbers as a product of their prime factors, including through the use of factor trees
Context
• Students use factor trees and multiplication facts to identify prime factors of composite numbers.
• Students should be familiar with factors, prime factors, prime numbers, and composite numbers
• This activity is inspired by the board game “Prime Climb”.
Materials
• Prime Climb Prompt Sheet (Appendix A)
• Worksheet (Appendix B)
• Prime Climb Challenge Slides (Appendix C)
• Pencil crayons or markers (red, orange, green, blue and purple)
Lesson
Introduction
• Before starting the activity provide students with an overview of the terms they will be engaging with. Ask students: What are factors? What are prime numbers? What are prime factors? What are composite numbers?
Term Definition Example Factor A number you multiply with another number to get a product Factors of 6: 1, 2, 3, and 6 1 × 6, 6 × 1, 2 × 3, and 3 × 2 Prime Numbers Has two factors Can only be divided by one and the number itself 2, 3, 5, 7, 11, etc. E.g., 5 is divisible by 1 and 5 Prime Factors Are numbers you multiple with another number to get a product that can only be divided by one and the number itself Prime Factors of 6: 2 and 3 The numbers 2 and 3 can both only be divided by one and the number itself making them prime 2 × 3 and 3 x 2 Composite Numbers Has more than two factors Can be divided by all its factors 4, 6, 8, 9, etc. E.g., 6 is divisible by 1,2,3, and 6
Lesson
• Project the Prime Climb prompt sheet (Appendix A). Tell students: Look at circles one to 20 and write down what you notice and wonder about the different colours and number of sections in each circle. What do you think the different colours represent? What do you think the sections in each circle represent?
• As students share what they notice, record their ideas so the class can see them. Pay particular attention to any ideas on prime numbers, composite numbers, and prime factors.
• Ensure students understand the relationship between the colours/patterns in the circles.
• Each circle has a white center indicating that each number is divisible by 1
• Prime numbers are one solid colour and only have one section (e.g., 1, 2, 3, 5, 7, 11, 13, 17, 19)
• Composite numbers have two or more sections (e.g., 4, 6, 8, 9, 10, etc.)
• Ask students to share the knowledge they already have: Who can share a prime number? Who can share a composite number? What are the factors? How do you know?
• Continue by projecting Appendix B. Work on the first challenge collaboratively.
• Tell students that in the left column, they will identify factors before creating a factor tree, which is used to determine the prime factors of a number greater than one. Emphasize that a factor tree is not complete until the composite numbers are broken down to their prime factors. Remind students that there may be more than one way to get to the answer.
• Ask students: What are the factors of the number 36? What does the factor tree look like? What are the prime factors? Is the number prime or composite? How do you know? What does the factor circle look like?
• Prompt students to start with the number 36. Tell students: 1 × 36 = 36. What are two other numbers that we can multiple together to get a product or total of 36?
• If students are unsure of how to start this process ask them: Two multiped by what number equals 36? When the number 18 is shared draw two lines like the ones in blue above and write 2 at the bottom of one and 18 at the bottom of the other.
• Ask students: Are either of these numbers prime? Why or why not? Students should share that two is a prime number as it can only be divided by one and itself.
• Instruct students to circle the two as it is a prime number.
• Repeat the same instructions until you reach two prime numbers.
• Tell students to write all the prime numbers horizontally which should be circled at the bottom of the box where it says “Prime factors:” with a space left in between each value.
• Continue by asking students: What do these numbers represent? Allow a couple of students to share they ideas before telling students to multiple these four terms together. Students should find that when the prime factors of a number are multiplied together the product is that number.
• Tell students to add multiplication symbols in the spaces they left between the prime factors.
• Proceed by asking students: Is 36 a prime or composite number? Students should share that 36 is a composite number as it has more than two factors.
• Ask students: Since we know the prime factors of the number 36 are 2, 2, 3, and 3, how many sections will our circle have? Students should share that this circle will have four sections as the number 36 has four prime factors.
• Continue by asking: Now that we know our factor circle has four sections what colour will these sections be? Students should share that since two of the prime factors were multiples of two, two of the sections will be orange and since two of the prime factors were multiples of three two of the sections will be green.
• Project Appendix A to allow students to compare their answer to the factor circle provided.
• Advance to the next challenge when it seems appropriate.
Conclusion
• Tell students that they will now select two number in between twenty and sixty to carry out the same procedures they completed for the former challenges.
When students have independently grasped the idea of the challenges section students into group of 4 to determine the factor tree and circle for numbers between 60 and 99.
Look Fors
• Are students able to differ between factors and multiples?
• Are students able to use mental math multiplication to identify prime factors in prime and composite numbers?
• What language are students using to describe the patterns they notice in the numbers? |
`18/(x^2-3x)-6/(x-3)=5/x` Solve the equation by using the LCD. Check for extraneous solutions.
`18/(x^2-3x)-6/(x-3)=5/x`
LCD is `x^2-3x=x(x-3)`
Multiply all the terms of the equation by LCD and simplify,
`x(x-3)(18/(x^2-3x))-x(x-3)(6/(x-3))=x(x-3)(5/x)`
`18-6x=(x-3)5`
`18-6x=5x-15`
`-6x-5x=-15-18`
`-11x=-33`
`x=(-33)/(-11)`
`x=3`
Let's check the solution by plugging in the original equation,
`18/(3^2-3*3)-6/(3-3)=5/3`
`18/0-6/0=5/3`
Since the solution x=3, yields a denominator of zero, so it's an extraneous solution and the original equation has no solution.
Approved by eNotes Editorial Team
LCD is an acronym for least common denominator. It is the product of distinct factors on the denominator side. Basically, find LCD is the same as finding the LCM (least common multiple) of the denominators.
For the given equation `18/(x^2-3x)-6/(x-3)=5/x` , the denominators are `(x^2-3x)` , `(x-3)` , and `x` . Note: The factored form of the denominator `x^2-3x` is ` x(x-3)` .
Based on the list of factors, The distinct factors are `x` and `(x-3).`
Thus,` LCD= x*(x-3) ` or `(x^2-3x)` .
To simplify the equation, we multiply each term by the LCD.
`18/(x^2-3x)*(x^2-3x)-6/(x-3)*x*(x-3) =5/x*x*(x-3) `
Cancel out common factors to get rid of the denominators.
`18 -6*x =5*(x-3)`
Apply distribution property.
`18-6x=5x-15`
Subtract 18 from both sides.
`18-6x-18=5x-15-18`
`-6x= 5x -33`
Subtract `5x ` from both sides of the equation.
`-6x-5x= 5x -33-5x`
`-11x=-33`
Divide both sides by `-11` .
`(-11x)/(-11)= (-33)/(-11)`
`x=3`
To check for extraneous solution, plug-in `x=3` on `18/(x^2-3x)-6/(x-3)=5/x` .
`18/(3^2-3*3)-6/(3-3)=?5/3`
`18/(9-9)-6/0=?5/3`
`18/0-6/0=?5/3`
undefined -undefined=?`5/3 ` FALSE.
Note: Any value divided by `0` results to undefined value.
An undefined result implies the x value is an extraneous solution.
Therefore, the `x=3` is an extraneous solution.
There is no real solution for the given equation `18/(x^2-3x)-6/(x-3)=5/x` .
Approved by eNotes Editorial Team |
# Matchstick Challenge: Can You Solve 3*9=5 Equation By Moving 1 Matchstick?
Puzzle lovers and problem solvers often enjoy tackling brain teasers that challenge their thinking. One classic puzzle involves matchsticks and math. We have a famous one for you: Can you make the equation 3*9=5 correct by moving just one matchstick? Let’s explore this intriguing challenge.
### The Challenge
We start with the equation 3*9=5. It looks like a multiplication problem, but there’s a twist – you can’t change the numbers. You’re only allowed to move one matchstick to make it right.
### The Puzzle-Solving Process
To solve this puzzle, you must think creatively and consider unconventional ways to use the matchsticks. Here’s how to do it step by step:
1. Assess the Initial Equation:
Take a good look at 3*9=5. You’ll see it doesn’t make sense in math. Normally, 3 times 9 equals 27, not 5.
2. Identify the Matchstick to Move:
Now, pick the matchstick you want to move carefully. Since you can only move one, you need to choose wisely.
3. Rearrange the Matchstick:
To fix this equation, you have to change the “3” into a different number. Move one matchstick from the “3” to another spot to create a new number.
4. Create the Correct Equation:
After moving the matchstick, you should have a new equation that makes sense and follows the math rules. Your goal is to change 3*9=5 into a mathematically correct equation.
### The Clever Solution
The solution to this matchstick puzzle is clever. Move the horizontal matchstick from the “5” and place it vertically against the “3.” This creates the symbol “≠,” which means “not equal to.” Now, the equation reads as 3*9≠5, which is true because 3 times 9 is not equal to 5.
### Matchstick Puzzles Challenge Our Minds
Puzzles like this are excellent for sharpening problem-solving skills and sparking creative thinking. They make you think differently and explore unique ways to find answers. While the original equation may seem impossible to fix, moving just one matchstick can turn it into a proper math statement.
### More Brain-Teasing Fun:
Puzzles like the 3*9=5 challenge remind us that math and logic can be both entertaining and puzzling. The next time you come across a matchstick puzzle or any brain teaser, approach it with an open mind and a willingness to think creatively. It’s the process of solving these puzzles that enhances our thinking and brings joy to our minds. So, have fun with math!
### Can You Solve 3*9=5 with One Matchstick Move? It’s Possible!
For puzzle enthusiasts and problem solvers, tackling brain-teasers that challenge creative thinking and logical reasoning is a favorite pastime. One classic puzzle involves manipulating matchsticks to create a new equation by altering an existing one. In this article, we present a popular matchstick puzzle: Can you solve the equation 3*9=5 by moving just one matchstick? Let’s dive into this intriguing conundrum and explore the solution.
### The Challenge
The initial equation we’re presented with is 3*9=5. At first glance, it appears to be a straightforward multiplication problem, but there’s a catch: you can’t change the numbers; you can only move one matchstick to make the equation correct.
### The Puzzle-Solving Process
To solve this puzzle, you need to think outside the box and consider unconventional ways to manipulate the matchsticks. Here’s a step-by-step guide to solving the equation:
1. Assess the Initial Equation:
Begin by carefully examining the equation 3*9=5. You’ll quickly realize that it doesn’t make sense mathematically. In standard arithmetic, 3 multiplied by 9 equals 27, not 5.
2. Identify the Matchstick to Move:
Next, identify which matchstick you want to move. Since you can only move one, it’s crucial to choose the right matchstick to achieve the desired outcome.
3. Rearrange the Matchstick:
To turn this equation into a mathematically accurate one, you’ll need to change the “3” into a different number. Move one of the matchsticks from the “3” to another location to create a new numeral.
4. Create the Correct Equation:
Once you’ve moved the matchstick, you should now have a new equation that makes sense and adheres to standard mathematical rules. Your goal is to transform 3*9=5 into a mathematically accurate equation.
### The Clever Solution
The solution to this matchstick puzzle is quite clever. By moving the horizontal matchstick from the “5” and placing it vertically against the “3,” you can create the mathematical symbol for “not equal to.” The equation then reads as 3*9≠5, which is a valid and true statement because 3 multiplied by 9 does not equal 5.
### Matchstick Puzzles Challenge Your Brain
Puzzles like the 3*9=5 challenge are excellent exercises for honing your problem-solving skills and stimulating creative thinking. They require you to look beyond the obvious and consider unconventional approaches to arrive at a solution. While the initial equation may seem impossible to fix, moving just one matchstick can turn it into a valid mathematical statement.
### More Brain-Teasing Fun
Puzzles like this one remind us that mathematics and logic can be both fun and challenging. So, the next time you encounter a matchstick puzzle or any other brain teaser, remember to approach it with an open mind and a willingness to think creatively. After all, it’s the journey of solving these puzzles that truly enriches our cognitive abilities and brings joy to our minds. |
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# Percentages, simple and compound interest; time, distance and speed
Percentages, simple and compound interest; time, distance and speed
Percentages, simple and compound interest; time, distance and speed
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### Percentages, simple and compound interest; time, distance and speed
1. 1. PERCENTAGES, SIMPLE AND COMPOUND INTEREST, TIME, DISTANCE AND SPEED NUMBERS
2. 2. PERCENTAGES Introduction on how to calculate a percentage. https://www.youtube.com/watch?v=rR95Cbcjz us&t=237s
3. 3. FOR EXAMPLE Type1: What is the 23% of \$500? So we write 23% as a fraction: 23 100 Then we multiply times \$500: 23 100 × \$500 = \$115 Type 2: If 17% of the total price is \$1445, what is the total price? \$1445 is 17% of some number So we call X = 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑡ℎ𝑎𝑡 1445 𝑖𝑠 17% 𝑜𝑓 17% × X = 0.17 × X = 1445 now we solve for X 𝑋 = \$1445 0.17 = \$8500
4. 4. FOR EXAMPLE Type 2: (continued) 18% of the students at a school play tennis. If the school has 216 tennis players, then how many students are there? 216 is 18% of some number So we call: 𝑋 = 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑡ℎ𝑎𝑡 216 𝑖𝑠 18% 𝑜𝑓 18% × 𝑋 = 0.18 × 𝑋 = 216 Solve for X 𝑋 = 216 0.18 = 1200 students Type 3: What percentage is 85 of 250? Now we will use the following formula: Percentage = 𝑎𝑚𝑜𝑢𝑛𝑡 𝑡𝑜𝑡𝑎𝑙 × 100% So we have: 𝑎𝑚𝑜𝑢𝑛𝑡=85 𝑡𝑜𝑡𝑎𝑙=250 × 100% = 34%
5. 5. PERCENTAGE CHANGE – EXAMPLES In a sale the Price of a shirt was reduced from \$40 to \$32. Find the percentage decrease. This is a similar Type 3 question, so we will be using the following formula: 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑐ℎ𝑎𝑛𝑔𝑒 = 𝐶ℎ𝑎𝑛𝑔𝑒 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 × 100% In this case, the original is \$40 and the change in the Price is \$40-\$32=\$8 So 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑐ℎ𝑎𝑛𝑔𝑒 = 8 32 × 100% = 20% Type 4: A car worth \$5000 loses 15% of its value in a year. What is it worth after one year? If it loses 15% of its value after a year, it means that after one year the car is going to cost 85% of the value that cost at the beginning of the year. It means that it is going to cost 85% of \$5000: 85% × \$5000 = 0.85 × \$5000 = \$4250
6. 6. PERCENTAGE CHANGE - EXAMPLES Type 4: (continued) A footbal club’s average attendance was 41,200. The following year the attendance rose by 4%. What was its new average attendance? If the attendance rose by 4% it means that 4% of 41,200 will be added to the following year: 4% of 41,200 = 0.04x41200 = 1648 So the new average attendance will be: 41200+1648=42848 Another way: since it increased by 4%, the new average attendance will be 104% of 41200 1.04x41200 = 42848 The value of a house in 2007 was \$300000. What was its value in 2008 after a fall of 8%? Since there was a fall of 8%, it means that the house now costs 92% of the cost in 2007 92% of \$300000 = 0.92x\$300000 = \$276000 Another way: since there was a fall of 8%, 0.08x\$300000 = \$24000 so the new cost will be: \$300000 - \$24000 = \$276000
7. 7. SIMPLE AND COMPOUND INTEREST • When you invest money in a bank offering simple interest you only get interest on the original principal. • I = PRT (I = interest earned, P is the investment, R is the percentage rate and T is the time) Simple interest • When you invest money in a bank offering compound interest, the interest you get each year is added to your principal and the next year’s interest is paid on the increased amount in your account. Compound interest If you sabe money in a savings scheme (for example, with a bank), the initial amount you invest is called the PRINCIPAL and you recieve interest on your money.
8. 8. FOR EXAMPLE – SIMPLE INTEREST Joseph invests \$500 at 4% per annum simple interest. How much will he have at the end of 5 years? (per annum=per year) Interest = 4% of \$500 = 0.04 x \$500 = \$20 per year Total interest in 5 years = \$20 x 5 = \$100 Joseph will have: \$500 + \$100 = \$600 Aruna invests \$480 at 5% simple interest per annum. What is the investment worth after: a) 7 months A whole year’s interest = 5% of \$480 = 0.05 x \$480 = \$24 7 months interest = \$24 x 7 12 = \$14 After 7 months, Aruna’s investment will be worth: \$480 + \$14 = \$494 b) 3 years 3 years interest = 3 x \$24 = \$72 After 3 years Aruna’s investment will be worth: \$480 + \$72 = \$552
9. 9. FOR EXAMPLE – COMPOUND INTEREST Ali invests \$200 at 3% compound interest. What amount will he have after 3 years? Year 1: 3% of \$200 = 0.03 x \$200 = \$6 \$200 + \$6 = \$206 at the start of year 2 Year 2: 3% of \$206 = 0.03 x \$206 = \$6.18 \$206 + \$6.18 = \$212.18 at the start of year 3 Year 3: 3% of \$212.18 = \$6.37 (to 2 decimal places) \$212.18 + \$6.37 = \$218.55 at the end of year 3 Rose borrows \$800 at 7% compound interest for 2 years. How much does she owe at the end of the 2 years? 7% of \$800 = 0.07 x \$800 = \$56 after 1 year, Rose owes \$856 7% of \$856 = 0.07 x \$856 = \$59.92 after 2 years Rose owes \$915.92
10. 10. FOR EXAMPLE – COMPOUND INTEREST (CONTINUED) A car loses 30% of its value each year. If it cost \$18000 then, how much is it worth after 3 years? One way: to calculate the value of the car after each year \$18000 x 0.3 = \$5400 the value after year 1: \$18000 - \$5400 = \$12600 \$12600 x 0.3 = \$3780 the value after year 2: \$12600 - \$3780 = \$8820 \$8820 x 0.3 = \$2646 the value after year 3: \$8820 - \$2646 = \$6174 Another way: NOTE: At the end of each year the car is worth 70% (0.7) of its value at the start of that year. \$18000 x 0.7 = \$12600 the value after year 1 \$12600 x 0.7 = \$8820 the value after year 2 \$18000 × 0.73 = \$6174 \$8820 x 0.7 = \$6174 the value after year 3
11. 11. WORKING WITH TIME Units used for measuring time: 1 year = 365 days = 8,760 hours = 525,600 minutes = 31,536,000 seconds Since 1 day has 24 hours & 1 hour has 60 minute & 1 minute has 60 seconds. This numbers are the ones you have to use to go from one unit to another. There are two ways to express time: 24-hr or 12-hr. (The 12-hr has AM and PM) Remember: when doing the calculations on your calculator, the decimal isn’t the minute!!! For example: 5.3 hours ≠ 5 hours and 3 minutes or 5 hours and 30 minutes, it means 5 hours and 18 minutes. You can try with www.springfrog.com/converter/decimal-time.htm
12. 12. FOR EXAMPLE A girl arrived at a rehearsal at 1935 and left 2129. For how long was she at the rehearsal? 1935 to 2135 is 2 hours, but she stayed till 2129 (6 minutes less than 2 hours) she stayed for 1 hour and 54 minutes A plane leaves Shanghai at 1055 local time and arrives at London the same day at 1550 local time. The website says that the flight takes 12h 55min. How many hours is the time in London behind the time in Shanghai? First: I want to know at what time from Shanghai did the plane arrive to London (A que hora de Shanghai llego el avión a Londres) 1055 + 12h 55min = 2350 so the plane arrives at 2350 (Shanghai time). This is 1550 London time, so 2350 – 1550 = 8hrs so London is 8 hours behind Shanghai.
13. 13. FOR EXAMPLE (CONTINUED) Derek wanted to get from Twickenham to Wembley. A website gave him the information shown in the table: (a) How long did the journey take in total? (b) At which station did Derek wait for 8 minutes? (a) He departed at 1158 and arrived at 1302. 1158 + 1hr = 1258 + 4 minutes = 1302 So the journey took 1 hour and 4 minutes (b) 1158 – 1203 = 5min; 1211 – 1203 = 8min; 1230 – 1211 = 9min; 1235 – 1230 = 5min; 1242 – 1235 = 7min; 1248 – 1242 = 6min; 1302 – 1248 =4min. Derek waited for 8 minutes at the Richmond Rail Station. Depart Twickenham Rail Station 1158 Arrive Richmond Rail Station 1203 Depart Richmond Rail Station 1211 Arrive Willesden Junction Underground Station 1230 Depart Willesden Junction Underground Station 1235 Arrive Wembley Central Station 1242 Depart Wembley Central Station 1248 Arrive Wembley Stadium 1302
14. 14. TIME, DISTANCE AND SPEED 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑠𝑝𝑒𝑒𝑑(𝑆) = 𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑒𝑑 (𝐷) 𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 (𝑇) Example: A car travels at 189km at an average speed of 60km/hr, how long does the journey take? S = D/T T = D/S 𝑇 = 189𝑘𝑚 60 𝑘𝑚 ℎ𝑟 = 3.15ℎ𝑟𝑠 = 3ℎ𝑟 9𝑚𝑖𝑛 |
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Simplify the following using the laws of logarithms:
(a) $\quad \log _9 27$
(b) $\quad \log _3 2 \cdot \log _2 5 \cdot \log _5 9$
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(a) There are two methods for simplifying this expression:
Method 1
Change to base 3
\begin{aligned} & \log _9 27 \\ & =\frac{\log _3 27}{\log _3 9} \\ & =\frac{\log _3 3^3}{\log _3 3^2} \\ & =\frac{3 \log _3 3}{2 \log _3 3} \\ & =\frac{3}{2} \end{aligned}
Method 2
Change to base 10
\begin{aligned} & \log _9 27 \\ & =\frac{\log _{10} 27}{\log _{10} 9} \\ & =\frac{\log 27}{\log 9} \\ & =\frac{\log 3^3}{\log 3^2} \\ & =\frac{3 \log 3}{2 \log 3}=\frac{3}{2} \end{aligned}
(b)
\begin{aligned} & \log _3 2 \cdot \log _2 5 \cdot \log _5 9 \\ & =\frac{\log 2}{\log 3} \cdot \frac{\log 5}{\log 2} \cdot \frac{\log 9}{\log 5} \\ & =\frac{\log 2}{\log 3} \cdot \frac{\log 5}{\log 2} \cdot \frac{\log 3^2}{\log 5} \\ & =\frac{\log 2}{\log 3} \cdot \frac{\log 5}{\log 2} \cdot \frac{2 \log 3}{\log 5} \\ & =2 \end{aligned}
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Today’s Lesson
Today’s Lesson. Using Tree Diagrams to Find the Probabilities of Events. Warm-up Activity. Let’s review adding and subtracting integers. 0 1 2 3 4 5 6 7 8 9. –9 –8 –7 –6 –5 –4 –3 –2 –1. On a number line, which direction do you move for a positive number?. Right.
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1. Today’s Lesson Using Tree Diagrams to Find the Probabilities of Events
2. Warm-up Activity Let’s review adding and subtracting integers.
3. 0 1 2 3 4 5 6 7 8 9 –9 –8 –7 –6 –5 –4 –3 –2 –1 On a number line, which direction do you move for a positive number? Right
4. On a number line, which direction do you move for a negative number? 0 1 2 3 4 5 6 7 8 9 –9 –8 –7 –6 –5 –4 –3 –2 –1 Left
5. 0 1 2 3 4 5 6 7 8 9 –9 –8 –7 –6 –5 –4 –3 –2 –1 Use the number line to solve the problems below. a. 7 – 10 = b. –8 + 11 = c. –3 – 6 = d. –5 + 9 = –3 3 –9 4
6. Solve the following problems. If needed, use a number line. a. –12 + 18 = b. 15 – 19 = c. –13 + 19 = d. –7 –13 = 6 – 4 6 –20
7. Whole-Class Skills Lesson Today we are going to conduct a number of experiments and list all possible outcomes using tree diagrams to find the probabilities of events.
8. Use a tree diagram to list all the possible outcomes. Hayley can choose from black or tan pants and a white, yellow, or purple shirt.
9. T B W Y P W Y P (TW) (TY) (TP) (BW) (BY) (BP)
10. B T W Y P W Y P (TW) (TY) (TP) (BW) (BY) (BP) List all the possible outcomes. {TW, TY, TP, BW, BY, BP}
11. B T W Y P W Y P (TW) (TY) (TP) (BW) (BY) (BP) How many possible outcomes are there? 6
12. 1 6 Possible Outcomes: {BW, BV, BP, TW, TY, TP} What is the probability of wearing tan pants and white shirt? There is only 1 outcome of tan pants and white shirt out of 6 possible outcomes.
13. 1 3 = {TW, TY, TP} 2 6 Possible Outcomes: {BW, BV, BP, TW, TY, TP} What is the probability of wearing tan pants?
14. List all the possible outcomes. At a city fair, Mitchell wants a candy apple. At a booth, he could choose from a red apple (R) or green apple (G), strawberry candy coating (S) or caramel coating (C), and peanut (P) or walnuts (W). Possible Outcomes {RSP, RSW, RCP, RCW, GSP, GSW, GCP, GCW}
15. Possible Outcomes {RSP, RSW, RCP, RCW, GSP, GSW, GCP, GCW} How many possible outcomes? 8
16. 2 1 = 8 4 {RSW, GSW} Possible Outcomes {RSP, RSW, RCP, RCW, GSP, GSW, GCP, GCW} What is the probability of having strawberry candy coating and walnuts?
17. 6 3 = 8 4 {RSP, RSW, RCP, RCW, GSW, GCW} Possible Outcomes {RSP, RSW, RCP, RCW, GSP, GSW, GCP, GCW} What is the probability of having a red apple or peanuts?
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# RD Sharma solutions for Class 11 Mathematics Textbook chapter 32 - Statistics [Latest edition]
## Chapter 32: Statistics
Exercise 32.1Exercise 32.2Exercise 32.3Exercise 32.4Exercise 32.5Exercise 32.6Exercise 32.7Others
Exercise 32.1 [Page 6]
### RD Sharma solutions for Class 11 Mathematics Textbook Chapter 32 Statistics Exercise 32.1 [Page 6]
Exercise 32.1 | Q 1.1 | Page 6
Calculate the mean deviation about the median of the observation:
3011, 2780, 3020, 2354, 3541, 4150, 5000
Exercise 32.1 | Q 1.2 | Page 6
Calculate the mean deviation about the median of the observation:
38, 70, 48, 34, 42, 55, 63, 46, 54, 44
Exercise 32.1 | Q 1.3 | Page 6
Calculate the mean deviation about the median of the observation:
34, 66, 30, 38, 44, 50, 40, 60, 42, 51
Exercise 32.1 | Q 1.4 | Page 6
Calculate the mean deviation about the median of the observation:
22, 24, 30, 27, 29, 31, 25, 28, 41, 42
Exercise 32.1 | Q 1.5 | Page 6
Calculate the mean deviation about the median of the observation:
38, 70, 48, 34, 63, 42, 55, 44, 53, 47
Exercise 32.1 | Q 2.1 | Page 6
Calculate the mean deviation from the mean for the data:
4, 7, 8, 9, 10, 12, 13, 17
Exercise 32.1 | Q 2.2 | Page 6
Calculate the mean deviation from the mean for the data:
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Exercise 32.1 | Q 2.3 | Page 6
Calculate the mean deviation from the mean for the data:
38, 70, 48, 40, 42, 55, 63, 46, 54, 44a
Exercise 32.1 | Q 2.4 | Page 6
Calculate the mean deviation from the mean for the data:
(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Exercise 32.1 | Q 3 | Page 6
Calculate the mean deviation of the following income groups of five and seven members from their medians:
IIncome in Rs. IIIncome in Rs. 40004200440046004800 300400042004400460048005800
Exercise 32.1 | Q 4.1 | Page 6
The lengths (in cm) of 10 rods in a shop are given below:
40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2
Find mean deviation from median
Exercise 32.1 | Q 4.2 | Page 6
The lengths (in cm) of 10 rods in a shop are given below:
40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2
Find mean deviation from the mean also.
Exercise 32.1 | Q 5.1 | Page 6
In 34, 66, 30, 38, 44, 50, 40, 60, 42, 51 find the number of observations lying between
$\bar{ X }$ − M.D. and
$\bar{ X }$ + M.D, where M.D. is the mean deviation from the mean.
Exercise 32.1 | Q 5.2 | Page 6
In 22, 24, 30, 27, 29, 31, 25, 28, 41, 42 find the number of observations lying between
$\bar { X }$ − M.D. and
$\bar { X }$ + M.D, where M.D. is the mean deviation from the mean.
Exercise 32.1 | Q 5.3 | Page 6
In 38, 70, 48, 34, 63, 42, 55, 44, 53, 47 find the number of observations lying between
$\bar { X }$ − M.D. and
$\bar { X }$ + M.D, where M.D. is the mean deviation from the mean.
Exercise 32.2 [Page 11]
### RD Sharma solutions for Class 11 Mathematics Textbook Chapter 32 Statistics Exercise 32.2 [Page 11]
Exercise 32.2 | Q 1 | Page 11
Calculate the mean deviation from the median of the following frequency distribution:
Heights in inches 58 59 60 61 62 63 64 65 66 No. of students 15 20 32 35 35 22 20 10 8
Exercise 32.2 | Q 2 | Page 11
The number of telephone calls received at an exchange in 245 successive one-minute intervals are shown in the following frequency distribution:
Number of calls 0 1 2 3 4 5 6 7 Frequency 14 21 25 43 51 40 39 12
Compute the mean deviation about median.
Exercise 32.2 | Q 3 | Page 11
Calculate the mean deviation about the median of the following frequency distribution:
xi 5 7 9 11 13 15 17 fi 2 4 6 8 10 12 8
Exercise 32.2 | Q 4.1 | Page 11
Find the mean deviation from the mean for the data:
xi 5 7 9 10 12 15 fi 8 6 2 2 2 6
Exercise 32.2 | Q 4.2 | Page 11
Find the mean deviation from the mean for the data:
xi 5 10 15 20 25 fi 7 4 6 3 5
Exercise 32.2 | Q 4.3 | Page 11
Find the mean deviation from the mean for the data:
xi 10 30 50 70 90 fi 4 24 28 16 8
Exercise 32.2 | Q 4.4 | Page 11
Find the mean deviation from the mean for the data:
Size 20 21 22 23 24 Frequency 6 4 5 1 4
Exercise 32.2 | Q 4.5 | Page 11
Find the mean deviation from the mean for the data:
Size 1 3 5 7 9 11 13 15 Frequency 3 3 4 14 7 4 3 4
Exercise 32.2 | Q 5.1 | Page 11
Find the mean deviation from the median for the data:
xi 15 21 27 30 35 fi 3 5 6 7 8
Exercise 32.2 | Q 5.2 | Page 11
Find the mean deviation from the median for the data:
xi 74 89 42 54 91 94 35 fi 20 12 2 4 5 3 4
Exercise 32.3 [Pages 16 - 17]
### RD Sharma solutions for Class 11 Mathematics Textbook Chapter 32 Statistics Exercise 32.3 [Pages 16 - 17]
Exercise 32.3 | Q 1 | Page 16
Compute the mean deviation from the median of the following distribution:
Class 0-10 10-20 20-30 30-40 40-50 Frequency 5 10 20 5 10
Exercise 32.3 | Q 2.1 | Page 16
Find the mean deviation from the mean for the data:
Classes 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800 Frequencies 4 8 9 10 7 5 4 3
Exercise 32.3 | Q 2.2 | Page 16
Find the mean deviation from the mean for the data:
Classes 95-105 105-115 115-125 125-135 135-145 145-155 Frequencies 9 13 16 26 30 12
Exercise 32.3 | Q 2.3 | Page 16
Find the mean deviation from the mean for the data:
Classes 0-10 10-20 20-30 30-40 40-50 50-60 Frequencies 6 8 14 16 4 2
Exercise 32.3 | Q 3 | Page 16
Compute mean deviation from mean of the following distribution:
Mark 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 No. of students 8 10 15 25 20 18 9 5
Exercise 32.3 | Q 4 | Page 16
The age distribution of 100 life-insurance policy holders is as follows:
Age (on nearest birth day) 17-19.5 20-25.5 26-35.5 36-40.5 41-50.5 51-55.5 56-60.5 61-70.5 No. of persons 5 16 12 26 14 12 6 5
Calculate the mean deviation from the median age
Exercise 32.3 | Q 5 | Page 16
Find the mean deviation from the mean and from median of the following distribution:
Marks 0-10 10-20 20-30 30-40 40-50 No. of students 5 8 15 16 6
Exercise 32.3 | Q 6 | Page 16
Calculate mean deviation about median age for the age distribution of 100 persons given below:
Age: 16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55 Number of persons 5 6 12 14 26 12 16 9
Exercise 32.3 | Q 7 | Page 17
Calculate the mean deviation about the mean for the following frequency distribution:
Class interval: 0–4 4–8 8–12 12–16 16–20 Frequency 4 6 8 5 2
Exercise 32.3 | Q 8 | Page 17
Calculate mean deviation from the median of the following data:
Class interval: 0–6 6–12 12–18 18–24 24–30 Frequency 4 5 3 6 2
Exercise 32.4 [Page 28]
### RD Sharma solutions for Class 11 Mathematics Textbook Chapter 32 Statistics Exercise 32.4 [Page 28]
Exercise 32.4 | Q 1.1 | Page 28
Find the mean, variance and standard deviation for the data:
2, 4, 5, 6, 8, 17.
Exercise 32.4 | Q 1.2 | Page 28
Find the mean, variance and standard deviation for the data:
6, 7, 10, 12, 13, 4, 8, 12.
Exercise 32.4 | Q 1.3 | Page 28
Find the mean, variance and standard deviation for the data:
227, 235, 255, 269, 292, 299, 312, 321, 333, 348.
Exercise 32.4 | Q 1.4 | Page 28
Find the mean, variance and standard deviation for the data 15, 22, 27, 11, 9, 21, 14, 9.
Exercise 32.4 | Q 2 | Page 28
The variance of 20 observations is 5. If each observation is multiplied by 2, find the variance of the resulting observations.
Exercise 32.4 | Q 3 | Page 28
The variance of 15 observations is 4. If each observation is increased by 9, find the variance of the resulting observations.
Exercise 32.4 | Q 4 | Page 28
The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations.
Exercise 32.4 | Q 5 | Page 28
The mean and standard deviation of 6 observations are 8 and 4 respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.
Exercise 32.4 | Q 6 | Page 28
The mean and variance of 8 observations are 9 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.
Exercise 32.4 | Q 7 | Page 28
For a group of 200 candidates, the mean and standard deviations of scores were found to be 40 and 15 respectively. Later on it was discovered that the scores of 43 and 35 were misread as 34 and 53 respectively. Find the correct mean and standard deviation.
Exercise 32.4 | Q 8 | Page 28
The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation?
Exercise 32.4 | Q 9 | Page 28
The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
(i) If wrong item is omitted
(ii) if it is replaced by 12.
Exercise 32.4 | Q 10 | Page 28
The mean and standard deviation of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations were omitted.
Exercise 32.4 | Q 11 | Page 28
Show that the two formulae for the standard deviation of ungrouped data
$\sigma = \sqrt{\frac{1}{n} \sum \left( x_i - X \right)^2_{}}$ and
$\sigma' = \sqrt{\frac{1}{n} \sum x_i^2 - X^2_{}}$ are equivalent, where $X = \frac{1}{n}\sum_{} x_i$
Exercise 32.5 [Pages 37 - 38]
### RD Sharma solutions for Class 11 Mathematics Textbook Chapter 32 Statistics Exercise 32.5 [Pages 37 - 38]
Exercise 32.5 | Q 1 | Page 37
Find the standard deviation for the following distribution:
x : 4.5 14.5 24.5 34.5 44.5 54.5 64.5 f : 1 5 12 22 17 9 4
Exercise 32.5 | Q 2 | Page 38
Table below shows the frequency f with which 'x' alpha particles were radiated from a diskette
x : 0 1 2 3 4 5 6 7 8 9 10 11 12 f : 51 203 383 525 532 408 273 139 43 27 10 4 2
Calculate the mean and variance.
Exercise 32.5 | Q 3 | Page 38
Find the mean, mode, S.D. and coefficient of skewness for the following data:
Year render: 10 20 30 40 50 60 No. of persons (cumulative): 15 32 51 78 97 109
Exercise 32.5 | Q 4 | Page 38
Find the standard deviation for the following data:
x : 3 8 13 18 23 f : 7 10 15 10 6
Exercise 32.6 [Pages 41 - 42]
### RD Sharma solutions for Class 11 Mathematics Textbook Chapter 32 Statistics Exercise 32.6 [Pages 41 - 42]
Exercise 32.6 | Q 1 | Page 41
Calculate the mean and S.D. for the following data:
Expenditure in Rs: 0-10 10-20 20-30 30-40 40-50 Frequency: 14 13 27 21 15
Exercise 32.6 | Q 2 | Page 41
Calculate the standard deviation for the following data:
Class: 0-30 30-60 60-90 90-120 120-150 150-180 180-210 Frequency: 9 17 43 82 81 44 24
Exercise 32.6 | Q 3 | Page 42
Calculate the A.M. and S.D. for the following distribution:
Class: 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Frequency: 18 16 15 12 10 5 2 1
Exercise 32.6 | Q 4 | Page 42
A student obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was later found that one observation was wrongly copied as 50, the correct figure being 40. Find the correct mean and S.D.
Exercise 32.6 | Q 5 | Page 42
Calculate the mean, median and standard deviation of the following distribution:
Class-interval: 31-35 36-40 41-45 46-50 51-55 56-60 61-65 66-70 Frequency: 2 3 8 12 16 5 2 3
Exercise 32.6 | Q 6 | Page 42
Find the mean and variance of frequency distribution given below:
xi: 1 ≤ x < 3 3 ≤ x < 5 5 ≤ x < 7 7 ≤ x < 10 fi: 6 4 5 1
Exercise 32.6 | Q 7 | Page 42
The weight of coffee in 70 jars is shown in the following table:
Weight (in grams): 200–201 201–202 202–203 203–204 204–205 205–206 Frequency: 13 27 18 10 1 1
Determine the variance and standard deviation of the above distribution.
Exercise 32.6 | Q 8 | Page 42
Mean and standard deviation of 100 observations were found to be 40 and 10 respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.
Exercise 32.6 | Q 9 | Page 42
While calculating the mean and variance of 10 readings, a student wrongly used the reading of 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.
Exercise 32.6 | Q 10
Calculate the mean, variance and standard deviation of the following frequency distribution.
Class: 1–10 10–20 20–30 30–40 40–50 50–60 Frequency: 11 29 18 4 5 3
Exercise 32.7 [Pages 47 - 49]
### RD Sharma solutions for Class 11 Mathematics Textbook Chapter 32 Statistics Exercise 32.7 [Pages 47 - 49]
Exercise 32.7 | Q 1 | Page 47
Two plants A and B of a factory show following results about the number of workers and the wages paid to them
Plant A Plant B No. of workers 5000 6000 Average monthly wages Rs 2500 Rs 2500 Variance of distribution of wages 81 100
In which plant A or B is there greater variability in individual wages?
Exercise 32.7 | Q 2 | Page 47
The means and standard deviations of heights ans weights of 50 students of a class are as follows:
Weights Heights Mean 63.2 kg 63.2 inch Standard deviation 5.6 kg 11.5 inch
Which shows more variability, heights or weights?
Exercise 32.7 | Q 3 | Page 48
Coefficient of variation of two distributions are 60% and 70% and their standard deviations are 21 and 16 respectively. What are their arithmetic means?
Exercise 32.7 | Q 4 | Page 48
Calculate coefficient of variation from the following data:
Income (in Rs): 1000-1700 1700-2400 2400-3100 3100-3800 3800-4500 4500-5200 No. of families: 12 18 20 25 35 10
Exercise 32.7 | Q 5 | Page 48
An analysis of the weekly wages paid to workers in two firms A and B, belonging to the same industry gives the following results:
Firm A Firm B No. of wage earners 586 648 Average weekly wages Rs 52.5 Rs. 47.5 Variance of the 100 121 distribution of wages
(i) Which firm A or B pays out larger amount as weekly wages?
(ii) Which firm A or B has greater variability in individual wages?
Exercise 32.7 | Q 6 | Page 48
The following are some particulars of the distribution of weights of boys and girls in a class:
Number Boys Girls 100 50 Mean weight 60 kg 45 kg Variance 9 4
Which of the distributions is more variable?
Exercise 32.7 | Q 7 | Page 48
The mean and standard deviation of marks obtained by 50 students of a class in three subjects, mathematics, physics and chemistry are given below:
Subject Mathematics Physics Chemistry Mean 42 32 40.9 Standard Deviation 12 15 20
Which of the three subjects shows the highest variability in marks and which shows the lowest?
Exercise 32.7 | Q 8 | Page 48
From the data given below state which group is more variable, G1 or G2?
Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Group G1 9 17 32 33 40 10 9 Group G2 10 20 30 25 43 15 7
Exercise 32.7 | Q 9 | Page 48
Find the coefficient of variation for the following data:
Size (in cms): 10-15 15-20 20-25 25-30 30-35 35-40 No. of items: 2 8 20 35 20 15
Exercise 32.7 | Q 10 | Page 48
From the prices of shares X and Y given below: find out which is more stable in value:
X: 35 54 52 53 56 58 52 50 51 49 Y: 108 107 105 105 106 107 104 103 104 101
Exercise 32.7 | Q 11 | Page 49
Life of bulbs produced by two factories A and B are given below:
Length of life(in hours): 550–650 650–750 750–850 850–950 950–1050 Factory A:(Number of bulbs) 10 22 52 20 16 Factory B:(Number of bulbs) 8 60 24 16 12
The bulbs of which factory are more consistent from the point of view of length of life?
Exercise 32.7 | Q 12 | Page 49
Following are the marks obtained,out of 100 by two students Ravi and Hashina in 10 tests:
Ravi: 25 50 45 30 70 42 36 48 35 60 Hashina: 10 70 50 20 95 55 42 60 48 80
Who is more intelligent and who is more consistent?
[Page 49]
### RD Sharma solutions for Class 11 Mathematics Textbook Chapter 32 Statistics [Page 49]
Q 1 | Page 49
Write the variance of first n natural numbers.
Q 2 | Page 49
If the sum of the squares of deviations for 10 observations taken from their mean is 2.5, then write the value of standard deviation.
Q 3 | Page 49
If x1x2, ..., xn are n values of a variable X and y1y2, ..., yn are n values of variable Y such that yi = axi + bi = 1, 2, ..., n, then write Var(Y) in terms of Var(X).
Q 4 | Page 49
If X and Y are two variates connected by the relation
$Y = \frac{aX + b}{c}$ and Var (X) = σ2, then write the expression for the standard deviation of Y.
Q 5 | Page 49
In a series of 20 observations, 10 observations are each equal to k and each of the remaining half is equal to − k. If the standard deviation of the observations is 2, then write the value of k.
Q 6 | Page 49
If each observation of a raw data whose standard deviation is σ is multiplied by a, then write the S.D. of the new set of observations.
Q 7 | Page 49
If a variable X takes values 0, 1, 2,..., n with frequencies nC0nC1nC2 , ... , nCn, then write variance X.
[Pages 50 - 52]
### RD Sharma solutions for Class 11 Mathematics Textbook Chapter 32 Statistics [Pages 50 - 52]
Q 1 | Page 50
For a frequency distribution mean deviation from mean is computed by
• M.D. = $\frac{\Sigma f}{\Sigma f \left| d \right|}$
• M.D. = $\frac{\Sigma d}{\Sigma f}$
• M.D. = $\frac{\Sigma f d}{\Sigma f}$
• M.D. = $\frac{\Sigma f \left| d \right|}{\Sigma f}$
Q 2 | Page 50
For a frequency distribution standard deviation is computed by applying the formula
• $\sigma = \sqrt{\frac{\Sigma f d^2}{\Sigma f} - \left( \frac{\Sigma f d}{\Sigma f} \right)^2}$
• $\sigma = \sqrt{\left( \frac{\Sigma f d}{\Sigma f} \right)^2 - \frac{\Sigma f d^2}{\Sigma f}}$
• $\sigma = \sqrt{\frac{\Sigma f d^2}{\Sigma f} - \frac{\Sigma fd}{\Sigma f}}$
• $\sqrt{\left( \frac{\Sigma fd}{\Sigma f} \right)^2 - \frac{\Sigma f d^2}{\Sigma f}}$
Q 3 | Page 50
If v is the variance and σ is the standard deviation, then
• $v = \frac{1}{\sigma^2}$
• $v = \frac{1}{\sigma}$
• v = σ2
• v2 = σ
Q 4 | Page 50
The mean deviation from the median is
• equal to that measured from another value
• maximum if all observations are positive
• greater than that measured from any other value.
• less than that measured from any other value.
Q 5 | Page 50
If n = 10, $X = 12$ and $\Sigma x_i^2 = 1530$ , then the coefficient of variation is
• 36%
• 41%
• 25%
• none of these
Q 6 | Page 50
The standard deviation of the data:
x: 1 a a2 .... an f: nC0 nC1 nC2 .... nCn
is
• $\left( \frac{1 + a^2}{2} \right)^n - \left( \frac{1 + a}{2} \right)^n$
• $\left( \frac{1 + a^2}{2} \right)^{2n} - \left( \frac{1 + a}{2} \right)^n$
• $\left( \frac{1 + a}{2} \right)^{2n} - \left( \frac{1 + a^2}{2} \right)^n$
• none of these
Q 7 | Page 50
The mean deviation of the series aa + da + 2d, ..., a + 2n from its mean is
• $\frac{(n + 1) d}{2n + 1}$
• $\frac{nd}{2n + 1}$
• $\frac{n (n + 1) d}{2n + 1}$
• $\frac{(2n + 1) d}{n (n + 1)}$
Q 8 | Page 50
A batsman scores runs in 10 innings as 38, 70, 48, 34, 42, 55, 63, 46, 54 and 44. The mean deviation about mean is
• 8.6
• 6.4
• 10.6
• 7.6
• None of these
Q 9 | Page 51
The mean deviation of the numbers 3, 4, 5, 6, 7 from the mean is
• 25
• 5
• 1.2
• 0
Q 10 | Page 51
The sum of the squares deviations for 10 observations taken from their mean 50 is 250. The coefficient of variation is
• 10 %
• 40 %
• 50 %
• none of these
Q 11 | Page 51
Let x1x2, ..., xn be values taken by a variable X and y1y2, ..., yn be the values taken by a variable Y such that yi = axi + bi = 1, 2,..., n. Then,
• Var (Y) = a2 Var (X)
• Var (X) = a2 Var (Y)
• Var (X) = Var (X) + b
• none of these
Q 12 | Page 51
If the standard deviation of a variable X is σ, then the standard deviation of variable $\frac{a X + b}{c}$ is
• a σ
• $\frac{a}{c}\sigma$
• $\left| \frac{a}{c} \right| \sigma$
• $\frac{a\sigma + b}{c}$
Q 13 | Page 51
If the S.D. of a set of observations is 8 and if each observation is divided by −2, the S.D. of the new set of observations will be
• −4
• −8
• 8
• 4
Q 14 | Page 51
If two variates X and Y are connected by the relation $Y = \frac{a X + b}{c}$ , where abc are constants such that ac < 0, then
• $\sigma_Y = \frac{a}{c} \sigma_X$
• $\sigma_Y = - \frac{a}{c} \sigma_X$
• $\sigma_Y = \frac{a}{c} \sigma_X + b$
• none of these
Q 15 | Page 51
If for a sample of size 60, we have the following information $\sum_{} x_i^2 = 18000$ and $\sum_{} x_i = 960$ , then the variance is
• 6.63
• 16
• 22
• 44
Q 16 | Page 51
Let abcdbe the observations with mean m and standard deviation s. The standard deviation of the observations a + kb + kc + kd + ke + k is
• s
• ks
• s + k
• $\frac{s}{k}$
Q 17 | Page 51
The standard deviation of first 10 natural numbers is
• 5.5
• 3.87
• 2.97
• 2.87
Q 18 | Page 51
Consider the first 10 positive integers. If we multiply each number by −1 and then add 1 to each number, the variance of the numbers so obtained is
• 8.25
• 6.5
• 3.87
• 2.87
Q 19 | Page 51
Consider the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. If 1 is added to each number, the variance of the numbers so obtained is
• 6.5
• 2.87
• 3.87
• 8.25
Q 20 | Page 51
The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is
• 50,000
• 250,000
• 252500
• 255000
Q 21 | Page 51
Let x1x2, ..., xn be n observations. Let $y_i = a x_i + b$ for i = 1, 2, 3, ..., n, where a and b are constants. If the mean of $x_i 's$ is 48 and their standard deviation is 12, the mean of $y_i 's$ is 55 and standard deviation of $y_i 's$ is 15, the values of a and are
• a = 1.25, b = −5
• a = −1.25, b = 5
• a = 2.5, b = −5
• a = 2.5, b = 5
Q 22 | Page 51
The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is
• 2
• 2.57
• 3
• 3.57
Q 23 | Page 52
The mean deviation for n observations $x_1 , x_2 , . . . , x_n$ from their mean $\bar{X}$ is given by
• $\sum^n_{i = 1} \left( x_i - X \right)$
• $\frac{1}{n} \sum^n_{i = 1} \left( x_i - X \right)$
• $\sum^n_{i = 1} \left( x_i - X \right)^2$
• $\frac{1}{n} \sum^n_{i = 1} \left( x_i - X \right)^2$
Q 24 | Page 52
Let $x_1 , x_2 , . . . , x_n$ be n observations and $X$ be their arithmetic mean. The standard deviation is given by
• $\sum^n_{i = 1} \left( x_i - X \right)^2$
• $\frac{1}{n}\sum^n_{i = 1}\left( x_i - X \right)^2$
• $\sqrt{\frac{1}{n} \sum^n_{i = 1} \left( x_i - X \right)^2}$
• $\sqrt{\frac{1}{n} \sum^n_{i = 1} x_i^2 - X^2}$
Q 25 | Page 52
The standard deviation of the observations 6, 5, 9, 13, 12, 8, 10 is
• $\sqrt{6}$
• $\frac{52}{7}$
• $\sqrt{\frac{52}{7}}$
## Chapter 32: Statistics
Exercise 32.1Exercise 32.2Exercise 32.3Exercise 32.4Exercise 32.5Exercise 32.6Exercise 32.7Others
## RD Sharma solutions for Class 11 Mathematics Textbook chapter 32 - Statistics
RD Sharma solutions for Class 11 Mathematics Textbook chapter 32 (Statistics) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Class 11 Mathematics Textbook solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. RD Sharma textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Class 11 Mathematics Textbook chapter 32 Statistics are Central Tendency - Median, Measures of Dispersion, Concept of Range, Mean Deviation, Introduction of Variance and Standard Deviation, Standard Deviation, Standard Deviation of a Discrete Frequency Distribution, Standard Deviation of a Continuous Frequency Distribution, Shortcut Method to Find Variance and Standard Deviation, Introduction of Analysis of Frequency Distributions, Comparison of Two Frequency Distributions with Same Mean, Statistics Concept, Central Tendency - Mean, Measures of Dispersion - Quartile Deviation, Standard Deviation - by Short Cut Method, Concept of Mode.
Using RD Sharma Class 11 solutions Statistics exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in RD Sharma Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 11 prefer RD Sharma Textbook Solutions to score more in exam.
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# Difference between revisions of "2001 AIME I Problems/Problem 6"
## Problem
A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
## Solutions
### Solution 1
Recast the problem entirely as a block-walking problem. Call the respective dice $a, b, c, d$. In the diagram below, the lowest $y$-coordinate at each of $a$, $b$, $c$, and $d$ corresponds to the value of the roll.
The red path corresponds to the sequence of rolls $2, 3, 5, 5$. This establishes a one-to-one correspondence between valid dice roll sequences and block walking paths.
The solution to this problem is therefore $\dfrac{\binom{9}{4}}{6^4} = {\dfrac{7}{72}}$. So the answer is $\boxed{079}$.
### Solution 2
If we take any combination of four numbers, there is only one way to order them in a non-decreasing order. It suffices now to find the number of combinations for four numbers from $\{1,2,3,4,5,6\}$. We can visualize this as taking the four dice and splitting them into 6 slots (each slot representing one of {1,2,3,4,5,6}), or dividing them amongst 5 separators. Thus, there are ${9\choose4} = 126$ outcomes of four dice. The solution is therefore $\frac{126}{6^4} = \frac{7}{72}$, and $7 + 72 = \boxed{079}$.
### Solution 3
Call the dice rolls $a, b, c, d$. The difference between the $a$ and $d$ distinguishes the number of possible rolls there are.
• If $a - d = 0$, then the values of $b,\ c$ are set, and so there are $6$ values for $a,\ d$.
• If $a - d = 1$, then there are ${3\choose2} = 3$ ways to arrange for values of $b,\ c$, but only $5$ values for $a,\ d$.
• If $a - d = 2$, then there are ${4\choose2} = 6$ ways to arrange $b, c$, and there are only $6 - 2 = 4$ values for $a, d$.
Continuing, we see that the sum is equal to $\sum_{i = 0}^{5}{i+2\choose2}(6 - i) = 6 + 15 + 24 + 30 + 30 + 21 = 126$. The requested probability is $\frac{126}{6^4} = \frac{7}{72}$.
### Solution 4
The dice rolls can be in the form
ABCD
AABC
AABB
AAAB
AAAA
where A, B, C, D are some possible value of the dice rolls. (These forms are not keeping track of whether or not the dice are in ascending order, just the possible outcomes.)
1. Now, for the first case, there are ${6\choose4} = 15$ ways for this. We do not have to consider the order because the combination counts only one of the permutations; we can say that it counts the correct (ascending order) permutation.
2. Second case: ${6\choose3} = 20$ ways to pick 3 numbers, ${3\choose1}$ ways to pick 1 of those 3 to duplicate. A total of 60 for this case.
3. Third case: ${6\choose2}=15$ ways to pick 2 numbers. We will duplicate both, so nothing else in this case matters.
4. Fourth case: ${6\choose2} = 15$ ways to pick 2 numbers. We pick one to duplicate with ${2\choose1} = 2$, so there are a total of 30 in this case.
5. Fifth case: ${6\choose1} = 6$; all get duplicated so nothing else matters.
There are a total of $6^4$ possible dice rolls.
Thus,
$\frac{m}{n} = \frac{15 + 60 + 15 + 30 + 6}{6^4} = \frac{126}{6^4} = \frac{7}{72}$
### Solution 5
Consider the number of possible dice roll combinations which work after $1$ roll, after $2$ rolls, and so on. There is 6 possible rolls for the first dice. If the number rolled is a 1, then there are 6 further values that are possible for the second dice; if the number rolled is a 2, then there are 5 further values that are possible for the second dice, and so on.
Suppose we generalize this as a function, say $f(l,r)$ return the number of possible combinations after $l$ rolls and $r$ being the beginning value of the first roll. It becomes clear that from above, $f(1,r) = 1$; every value of $l$ after that is equal to the sum of the number of combinations of $l - 1$ rolls that have a starting value of at least $r$. If we slowly count through and add up all the possible combinations we get $\frac{7}{72}$ possibilities.
### Solution 6
In a manner similar to the above solution, instead consider breaking it down into two sets of two dice rolls. The first subset must have a maximum value which is $\le$ the minimum value of the second subset.
• If the first subset ends in a 1, there is $1$ such subset and there are $6 + 5 + 4 + 3 + 2 + 1 = \frac{6}{2}(6 + 1) = 21$ ways of making the second subset.
• If the first subset ends in a 2, there is $2$ such subsets and there are $5 + \ldots + 1 = \frac{5}{2}(5 + 1) = 15$ ways of making the second subset.
Thus, the number of combinations is $\sum_{i=1}^6 i \cdot \left(\frac{(7 - i)(8 - i)}{2}\right) = 21 + 30 + 30 + 24 + 15 + 6 = 126$, and the probability again is $\frac7{72}$, giving $m+n=\boxed{079}$.
### Solution 7 Recursion Formula
If you're too tired to think about any of the above smart transformations of the problem, a recursion formula can be a robust way to the correct answer.
We just need to work out the valid cases for each roll. Denote by $N_{k}(n)$ the number of valid cases in the $k+1$-th roll when the number $n$ is rolled, for $k=1,2,3$ and $1 <= n <= 6$. Then we have the following recursion formula: $$N_{1}(n) = n$$ $$N_{2}(n) = N_{1}(1) + N_{1}(2) + ... + N_{1}(n)=N_{2}(n-1)+N_{1}(n)$$ $$N_{3}(n) = N_{2}(1) + N_{2}(2) + ... + N_{2}(n)=N_{3}(n-1)+N_{2}(n)$$ The logic is that, if $n$ is rolled, then the number of valid cases is the subtotal of valid cases in the preceding roll for the outcome of 1 to $n$. The recursion can be easily calculated by hand when $N_k(n)$ are put in columns side by side, given the fact that te numbers are smaller than 100. Finally, the total number of cases is $$N = \sum_{n=1}^{6} N_{3}(n)$$ and $$P = \frac{N}{6^4}$$
### Solution 8
We use expected value.
The first dice averages to $3.5$. So, we start off with $\frac{7}{2}$. Now, we have a $\frac{1}{2}$ chance.
Next, we consider the number that we picked. There is a total of $1$ for the probabilities when we pick $1$, $2$, or $3$. So, we get $\frac{1}{3}$. Finally, we have $1$ and $2$. This gives $\frac{\frac{1}{3}}{2}=\frac{1}{6}$. Multiplying, we get $\frac{7}{72} \implies \boxed{079}$. ~asdf334 (I actually don't know if this is a valid sol, it seems sketchy to me lol)
### Solution 9: Observation of each case
Lets try casework and observe the cases. Notice that if the last roll is a $1$, then the only dice rolls may be $1-1-1-1$, which is only $1$ possibility. Observe that if the last roll is $2$, then there are $4 = 1 + 3$ possibilities. When the last roll is a $3$, there are $10 = 1 + 3 + 6$ possibilities. Notice when the last roll is $n$, the number of cases is the sum of the first $n$ positive triangular numbers. This is easily provable when observing the numbers of possibilities after assigning a value to the last and second to last rolls. So there are a total of $1 + 1 + 3 + 1 + 3 + 6 + 1 + 3 + 6 + 10 + 1 + 3 + 6 + 10 + 15 + 1 + 3 + 6 + 10 + 15 + 21 = 126$ possibilities. So the probability is $\frac{126}{6^4} = \frac{7}{72}$ and $7 + 72 = \boxed{079}$. ~skyscraper
### Solution 10: Distributions
This is equivalent to picking a four-element sequence of $\{1, 2, 3, 4, 5, 6\}$ with repetition. Notice that once the sequence is picked, there is one and only one way to order these so that they form a sequence of rolls satisfying our conditions.
Now count the number of such four-element sequences, let $a$ be the number of $1$s in the sequence, $b$ be the number of $2$s, $c$ $3$s, $d$ $4$s, $e$ $5$s, and $f$ $6$s. Now we see that we must have $$a + b + c + d + e + f = 4$$ with $a, b, c, d, e, f$ being nonnegative integers since there are a total of $4$ numbers picked. The number of solutions to this is $\dbinom{9}{4},$ so our total number is equal to $\dfrac{\binom{9}{4}}{6^4} = \dfrac{7}{72},$ making our answer $\boxed{079}.$
~Ilikeapos
## Solution 11: The proof no one will read
We can construct a bijection using stars and bars as follows: Consider the sequence of $6$ balls $OOOOOO$. Now place $4$ bars in between them. For instance, the configuration $O||O|OO|OO$ represents the rolls $1,1,2,2$. Now you might think, "Oh now it's easy. $\binom{10}{4}$." However, what if we have $|O|OOO|O|O$? Maybe we could remove the last ball. But then we have $5$ balls. How do we rid of this ambiguity? Well we could give it a value. If there is nothing to the right of a bar, give it the value of $6$. Now the answer is simple. $\binom{5+4}{4}=\binom{9}{4}$. Then we have $6^4$ different rolls, and now, the answer is simply $\frac{\binom{9}{4}}{6^4}$, which gives the answer of $\boxed{079}$. (Remember to always include the $0$ in this competition for $1$ or $2$ digit numbers).
~th1nq3r
## See also
2001 AIME I (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 7 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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# Congruence in Overlapping Triangles
Instructor: Melanie Olczak
Melanie has taught high school Mathematics courses for the past ten years and has a master's degree in Mathematics Education.
This lesson will provide an introduction to how to determine if triangles that overlap are congruent to each other. Definitions of congruent triangles and the five ways to prove triangles congruent are explained.
## Congruent Triangles
Two triangles are congruent if they are exactly the same size and shape, which means they have the same angle measures and the same side lengths. If we know that all the sides and all the angles are congruent in two triangles, then we know that the two triangles are congruent. To help show that two triangles are congruent and to understand which angles and which sides are congruent, we use tick marks to show corresponding sides and angles.
We don't necessarily need to know that all the sides and angles are congruent to prove two triangles are congruent.
## Ways To Prove Triangles are Congruent
Now, let's explore the different ways you can tell if two triangles are congruent and look at examples demonstrating these methods.
#### Side-Side-Side (SSS)
If we know that three sides of one triangle are congruent to three sides of another triangle, then the two triangles are congruent.
#### Side-Angle-Side (SAS)
If we know that two sides and the angle between them are congruent to the two sides and the angle between them in another triangle, then the two triangles are congruent. It is important that the angle be the included angle, which means that it is between the two sides.
#### Angle-Side-Angle (ASA)
If we know that two angles and the side between them are congruent to the two angles and the side between them in another triangle, then the two triangles are congruent. In this case, the side is the included side, because it is between the two angles.
#### Angle-Angle-Side (AAS)
If we know that two angles and a side are congruent to the two angles and a side in another triangle, then the two triangles are congruent. The side here is the non-included side, which means it is not the side between the two angles.
#### Hypotenuse-Leg (HL)
If the hypotenuse and a leg of a right triangle are congruent to the hypotenuse and leg of another right triangle, then the two triangles are congruent.
Ways that do not work: There is no way to prove two triangles congruent if you have two sides and an angle that is not between the two sides. You also cannot prove two triangles congruent just by knowing all three angles are congruent.
## Congruence in Triangles with Overlapping Parts
#### Example 1
In this example, the two triangles are next to each other and they share a common side, AC. Since AC is the same side in both triangles, we know that it must be congruent to itself. This property is called the reflexive property of congruence, which says that any segment or angle is congruent to itself.
Since the triangles have three sides congruent to each other, then these two triangles are congruence by SSS.
#### Example 2
Triangle ABC and triangle ADC share a common side, AC. Again, by the reflexive property, we know that side AC is congruent to itself.
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# 2005 AMC 12A Problems/Problem 19
## Problem
A faulty car odometer proceeds from digit 3 to digit 5, always skipping the digit 4, regardless of position. If the odometer now reads 002005, how many miles has the car actually traveled? $(\mathrm {A}) \ 1404 \qquad (\mathrm {B}) \ 1462 \qquad (\mathrm {C})\ 1604 \qquad (\mathrm {D}) \ 1605 \qquad (\mathrm {E})\ 1804$
## Solutions
### Solution 1
We find the number of numbers with a $4$ and subtract from $2005$. Quick counting tells us that there are $200$ numbers with a 4 in the hundreds place, $200$ numbers with a 4 in the tens place, and $201$ numbers with a 4 in the units place (counting $2004$). Now we apply the Principle of Inclusion-Exclusion. There are $20$ numbers with a 4 in the hundreds and in the tens, and $20$ for both the other two intersections. The intersection of all three sets is just $2$. So we get:
$2005-(200+200+201-20-20-20+2) = 1462 \Longrightarrow \mathrm{(B)}$
### Solution 2
Alternatively, consider that counting without the number $4$ is almost equivalent to counting in base $9$; only, in base $9$, the number $9$ is not counted. Since $4$ is skipped, the symbol $5$ represents $4$ miles of travel, and we have traveled $2004_9$ miles. By basic conversion, $2004_9=9^3(2)+9^0(4)=729(2)+1(4)=1458+4=\boxed{1462}$
### Solution 3
Since any numbers containing one or more $4$s were skipped, we need only to find the numbers that don't contain a $4$ at all. First we consider $1$ - $1999$. Single digits are simply four digit numbers with a zero in all but the ones place (this concept applies to double and triple digits numbers as well). From $1$ - $1999$, we have $2$ possibilities for the thousands place, and $9$ possibilities for the hundreds, tens, and ones places. This is $2 \cdot 9 \cdot 9 \cdot 9-1$ possibilities (because $0000$ doesn't count) or $1457$ numbers. From $2000$ - $2005$ there are $6$ numbers, $5$ of which don't contain a $4$. Therefore the total is $1457 + 5$, or $1462$ $\Rightarrow$ $\boxed{\text{B}}$.
### Solution 4
We seek to find the amount of numbers that contain at least one $4,$ and subtract this number from $2005.$
We can simply apply casework to this problem.
The amount of numbers with at least one $4$ that are one or two digit numbers are $4,14,24,34,40-49,54,\cdots,94$ which gives $19$ numbers.
The amount of three digit numbers with at least one $4$ is $8*19+100=252.$
The amount of four digit numbers with at least one $4$ is $252+1+19=272$
This, our answer is $2005-19-252-272=1462,$ or $\boxed{B}.$
~coolmath2017
### Solution 5(Super fast)
This is very analogous to base $9$. But, in base $9$, we don't have a $9$. So, this means that these are equal except for that base 9 will be one more than the operation here. $2005_9 = 5+0+0+1458 = 1463$. $1463 - 1 = 1462$
Therefore, our answer is $\boxed {1462}$
~Arcticturn |
# 22. Complex Numbers Solutions Part 3
1. Since $$P, A, B$$ are collinear
$\therefore \begin{vmatrix}z & \overline{z} & 1\\a & \overline{a} & 1\\ b & \overline{b} & 1\end{vmatrix} = 0 \Rightarrow z(\overline{a} - \overline{b}) - \overline{z}(a - b) + (a\overline{b} - \overline{a}b) = 0$
Similarly, $$P, C, D$$ are collinear
$\Rightarrow z(\overline{c} - \overline{d}) - \overline{z}(c - d) + (c\overline{d} - \overline{c}d) = 0$
Multiplying the first equation with $$c - d$$ and second with $$a - b$$ and subtracting resulting equations we get
$z(\overline{a} - \overline{b})(c - d) - \overline{z}(\overline{c} - \overline{d})(a - b) = (c\overline{d} - \overline{c}d)(a - b) - (a\overline{b} - \overline{a}b)(c - d)$
Given that $$a, b, c, d$$ all lie on the circle $$|z| = r \therefore |z|^2 = r^2 \Rightarrow z\overline{z} = r^2 \Rightarrow \overline{z} = \frac{r^2}{z}$$
Substituting accordingly in previous equation
$z\left(\frac{r^2}{a} - \frac{r^2}{b}\right)(c - d) - z\left(\frac{r^2}{c} - \frac{r^2}{d}\right)(a - b) = \left(\frac{cr^2}{d} - \frac{dr^2}{c}\right)(a - b) - \left(\frac{ar^2}{b} - \frac{br^2}{a}\right)(c - d)$
Solving this will yield desired result.
2. We have,
$\frac{z + 1}{z - 2} = \frac{3 + t + i\sqrt{3 - t^2}}{1 + t + i\sqrt{3 - t^2}} \therefore \left|\frac{z + 1}{z - 1}\right|^2 = \frac{(3 + t)^2 + 3 - t^2}{(1 + t)^2 + 3 - t^2} = \frac{6(t + 2)}{2(t + 2)} = 3$
Thus modulus of required fraction is independent of $$t$$.
Also, $$z = x + iy = 2 + t + i\sqrt{3 - t^2} \Rightarrow y = 3 - (x - 2)^2$$ or $$(x - 2)^2 + y^2 = 3.$$
Thus locus of complex number represents a circle with center at $$(2, 0)$$ having radius 3.
3. Given,
$\begin{vmatrix}a & b & c\\b & c & a\\c & a & b\end{vmatrix} = 0 \Rightarrow a^3 + b^3 + c^3 - 3abc = 0 \Rightarrow (a + b + c)[a^2 + b^2 + c^2 - ab -bc - ca] = 0 \Rightarrow (a - b)^2 + (b - c)^2 + (c - a)^2 = 0 \Rightarrow a = b = c [\because a + b + c \ne 0 \because z_1 \ne = 0]$
The circle made by these is shown below:
Now OA = OB = OC where O is the origin and A, B and C are the points representing $$z_1, z_2$$ and $$z_3$$ respectively.
$$\therefore$$ O is the circumcenter of $$\triangle ABC.$$
Now
$arg\left(\frac{z_3}{z_2}\right) = \angle BOC = 2\angle BAC = 2arg\left(\frac{z_3 - z_1}{z_2 - z_1}\right) = arg\left(\frac{z_3 - z_1}{z_2 - z_1}\right)^2$
Hence, proven.
4. $z_2 = \frac{OQ}{OP}z_1e^{i\theta} = cos\theta z_1e^{i\theta}$
and
$z_3 = \frac{OR}{OP}z_1e^{i2\theta} = cos2\theta z_1e^{i2\theta}$
Hence,
$z_2^2cos2\theta = z_1z_3cos^2\theta.$
5. Given circles are $$|z| = 1$$ or $$x^2 + y^2 - 1 = 0$$ and $$|z - 1| = 4$$ or $$x^2 + y^2 -2x - 15 = 0.$$
Let the circles cut by these two circles orthogonally is
$x^2 + y^2 + 2gx + 2fy +c = 0$
Now since first two circles cut this third one orthogonally
$\therefore 2g.0 + 2f.0 = c - 1 \Rightarrow c = 1$
and
$\therefore 2g(-1) + 2f.0 = c - 15 \Rightarrow g = 7$
Therefore the required circles are
$x^2 + y^2 + 14x + 2fy + 1 = 0 |z + 7 + if| = \sqrt{48 + f^2}$
6. Given $$|z + 3| = t^2 - 2t + 6$$ represents a circle with center (-3, 0) and radius $$t^2 -2t + 6.$$ The inequality $$|z - 3\sqrt{3}i| < t^2$$ means $$z$$ lies in the interior of circle having center at $$(0, 3\sqrt{3})$$ having radius $$t^2.$$
Let A is center of first circle and B is center of second circle. Clearly when both the circles are disjoint or touching then no solution is possible.
Further solution is left as an exercise.
7. Let $$z = x + iy$$
$\frac{az + b}{cz + d} = \frac{ax + iay + b}{cx + icy + d} = \frac{[(ax + b) + iay][(cx + d) - icy]}{(cx + d)^2 + d^2} Im\left(\frac{az + b}{cz + d}\right) = \frac{ay(cx + d) - cy(ax + b)}{(cx + d)^2 + d^2} \Rightarrow \frac{ady - bcy}{(cx + d)^2 + d^2}$
Now since $$ad > bc$$ sign is same as $$y$$ i.e. positive. Hence, proven.
8. Given,
$z_1 = \frac{i(z_2 + 1)}{z_2 - 1} \Rightarrow x_1 + iy_1 = \frac{i(x_2 + iy_2 + 1)}{(x_2 - 1) + iy_2} = \frac{[-y_2 + i(x_2 + 1)][(x_2 - 1) + iy_2]}{(x_2 - 1)^2 + y_2^2}$
Now equating for real part and then evaluating the desired equation will yield the result.
9. $$sin25\theta + icos25\theta$$ This question is left as an exercise.
10. Let $$z = x + iy.$$ Now we have
$z^2 + |z| = x^2 - y^2 + i2xy + \sqrt{x^2 - y^2} = 0$
Equating imaginary parts we have $$2xy = 0$$ which means either $$x = 0$$ or $$y = 0.$$ Let $$y = 0$$ then we have
$x^2 + \sqrt{x^2} = 0$
Since $$x$$ is real the only possible solution is $$x = 0$$. So $$z = 0.$$
If $$x = 0$$ then we have
$y^2 + \sqrt{-y^2} = 0 y^4 + y^2 = 0 y^2 = -1 \Rightarrow y = \pm i$
Thus we have $$z = \pm i.$$
11. Problem no. 111 to 118 have been left as exercises to the reader.
1. Given,
$|1 - \overline{z_1}z_2|^2 - |z_1 - z_2|^2 = (1 - \overline{z_1}z_2)(1 - z_1\overline{z_2}) - (z_1 - z_2)(\overline{z_1} - \overline{z_2}) [\because |z|^2 = z\overline{z}] = (1 - \overline{z_1}z_2 - z_1\overline{z_2} + |z_1|^2|z_2|^2) - (|z_1|^2 - \overline{z_1}z_2 - z_1\overline{z_2} + |z_2|^2) = (1 - |z_1|^2 - |z_2|^2 + |z_1|^2|z_2|^2) = (1 - |z_1|^2)(1 - |z_2|^2)$
2. Consider two complex numbers $$z_1 = a_1 + ib_1$$ and $$z_2 = a_2 + ib_2.$$ Now we have to prove $$|z_1 + z_2| \le |z_1| + |z_2|$$ which can be further extended to prove the result.
$\sqrt{(a_1 + a_2)^2 + (b_2 + b_2)^2} \le \sqrt{a_1^2 + b_1^2} + \sqrt{a_2^2 + b_2^2}$
Squaring both sides and simplifying we get
$a_1a_2 + b_1b_2 \le \sqrt{(a_1^2 + b_1^2)(a_2^2 + b_2^2)} \Rightarrow (a_1a_2 + b_1b_2)^2 - (a_1^2 + b_1^2)(a_2^2 + b_2^2) \le 0 \Rightarrow -(a_1b_2 - a_2b_1)^2 \le 0$
which is true. Hence, proven.
3. We have,
$\left|\frac{\overline{z_1} - 2\overline{z_2}}{2 - z_1\overline{z_2}}\right| = 1 \Rightarrow |\overline{z_1} - 2\overline{z_2}|^2 = |2 - z_1\overline{z_2}|^2 \Rightarrow (\overline{z_1} - 2\overline{z_2})(z_1 - 2z_2) = (2 - z_1\overline{z_2})(2 - \overline{z_1}z_2) \Rightarrow |z_1|^2 - 2z_1\overline{z_2} - 2\overline{z_1}z_2 + 4|z_2|^2 = 4 - 2z_1\overline{z_2} - 2\overline{z_1}z_2 + |z_1|^2|z_2|^2 \Rightarrow |z_1|^2|z_2|^2 - 4|z_2|^2 - |z_1|^2 - 4 = 0 \Rightarrow \because |z_1| \ne 1 |z_2| = 2$
4. It can be solves similarly as 121 and is left as an exercise.
5. We have,
$\left|\frac{z_1 + z_2}{2} + \sqrt{z_1z_2}\right| + \left|\frac{z_1 + z_2}{2} - \sqrt{z_1z_2}\right| = \frac{1}{2}\left|(\sqrt{z_1} + \sqrt{z_2})^2\right| + \frac{1}{2}\left|(\sqrt{z_1} - \sqrt{z_2})^2\right| = |z_1| + |z_2|$
6. From problem no. 54 it follows that $$|a + \sqrt{a^2 - b^2}| + |a - \sqrt{a^2 - b^2}| = |a + b| + |a - b|.$$
Substituting $$a = \beta$$ and $$b = \sqrt{\alpha\gamma}$$ we have
$|\beta + \sqrt{\alpha\gamma}| + |\beta - \sqrt{\alpha\gamma}| = |\alpha|\left(|\frac{\beta}{\alpha} + \sqrt{\frac{\gamma}{\alpha}}| + |\frac{\beta}{\alpha} - \sqrt{\frac{\gamma}{\alpha}}|\right) = |alpha|\left(|-z_1 - z_2 + \sqrt{z_1z_2}| + |-z_1 - z_2 - \sqrt{z_1z_2}|\right) = |\alpha|(|z_1| + |z_2|)$
7. We have,
$|a| = 1 \Rightarrow |a|^2 = 1 \Rightarrow a\overline{a} = 1 \Rightarrow \overline{a} = \frac{1}{a}$
From this we can write
$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \overline{a} + \overline{b} + \overline{c} = 3$
8. Given,
$|z + 4| \le 3 \Rightarrow |z + 1 + 3| \le 3 \Rightarrow |z + 1| + 3 \le 3 [\because |z_1 + z_2| \le |z_1| + |z_2|] |z + 1| \le 0$
Following similarly
$|z + 4| = |z + 1 + 3| \ge |z + 1| - 3 \Rightarrow |z + 1| \ge 6$
So least value is 0 and greatest value is 6.
9. Let $$z_1 = r_1cos\theta_1 + isin\theta_1$$ and $$z_2 = r_2cos\theta_2 + isin\theta_2.$$ Now it can be easily shown that
$4|z_1 + z_2|^2 - (|z_1| + |z_2|)^2\left(\frac{z_1}{|z_2|} + \frac{z_2}{|z_2|}\right)^2 \ge 0$
10. Given equation is $$z^2 + az + b = 0.$$ Let $$p, q$$ are two of its roots. Then we have $$p + q = -a$$ and $$pq = b.$$ Taking modulus of both we have $$|p + q| = |a|$$ and $$|pq| = b.$$ Now it is required that $$|p| = |q| = 1.$$ Therefore we have $$|p + q| \le |p| + |q| = 2 \therefore |a| \le 2.$$ Similarly $$|b| = |pq| = |p||q| = 1.$$ Since $$p, q$$ have unit modulii we can have them as $$p = cos\theta_1 + isin\theta_1$$ and $$q = cos\theta_2 + isin\theta_2.$$
$pq = (cos\theta_1 + isin\theta_1)(cos\theta_2 + isin\theta_2) = cos(\theta_1 + \theta_2) + isin(\theta_1 + \theta_2) \therefore arg(b) = arg(pq) = \theta_1 + \theta_2 p + q = cos\theta_1 + isin\theta_1 + cos\theta_2 + isin\theta_2 = cos^2{\frac{\theta_1}{2}} + i^2sin^2{\frac{\theta_1}{2}} + i2sin{\frac{\theta_1}{2}}cos{\frac{\theta_1}{2}} + cos^2{\frac{\theta_2}{2}} + i^2sin^2{\frac{\theta_2}{2}} + i2sin{\frac{\theta_2}{2}}cos{\frac{\theta_2}{2}} = cos\frac{\theta_1 + \theta_2}{2} + isin\frac{\theta_1 + \theta_2}{2} \therefore arg(a) = arg(p + q) = \frac{\theta_1 + \theta_2}{2} \therefore argb = 2arga$
11. Let $$a = x + iy.$$ First we consider first two inequalities
$|z| \le |Re(z)| + |Im(z)| \Rightarrow \sqrt{x^2 + y^2} \le x + y$
Sqauting we have
$x^2 + y^2 \le x^2 + y^2 + 2xy \Rightarrow 2xy \ge 0$
which is true. Now we consider last two inequalities
$|Re(z)| + |Im(z)| \le \sqrt{2}|z| \Rightarrow x + y \le \sqrt{2(x^2 + y^2)}$
Squaring we get
$x^2 + y^2 + 2xy \le 2(x^2 + y^2) \Rightarrow (x - y)^2 \ge 0$
which is also true. Hence, proven.
12. Translating the given equation we have
$|z|^2 - 2|z| - 4 \ge 0$
The greatest root of this equation is $$\sqrt{5} + 1.$$ Hence proven.
13. Since $$\alpha, \beta, \gamma, \delta$$ are root of the equation.
$(x - \alpha)(x - \beta)(x - \gamma)(x - \delta) = ax^4 + bx^3 + cx^2 + dx + e$
Substituting $$x = i$$ we get following
$(i - \alpha)(i - \beta)(i - \gamma)(i - \delta) = ai^4 + bi^3 + ci^2 + di + e \Rightarrow (1 + i\alpha)(1 + i\beta)(1 + i\gamma)(1 + i\delta) = a - ib - c + id + e$
Taking modulus and squaring we get our desired result.
14. This is similar to 131 and is left as an exercise.
15. Let $$|z_1| = |z_2| = |z_3| = R.$$ $$\therefore$$ origin is the circumcenter of triangle. Since triangle is also equilateral circumcenter and origin coincide. Therefore, origin is also centroid. Thus
$\frac{z_1 + z_2 + z_3}{3} = 0 \Rightarrow z_1 + z_2 + z_3 = 0$
16. Similar to 133 it can be proven that it is an equilateral triangle. Now since $$|z_1| = |z_2| = |z_3| = 1$$ therefore it is an equilateral triangle inscribed in an unit circle.
17. Circumcenter of an equilateral triangle is given by $$z_0 = \frac{z_1 + z_2 + z_3}{3}$$ which is same as centroid. Now since this triangle is equilateral
$\sum z_1^2 = \sum z_1z_2 (\sum z_1)^2 = \sum z_1^2 + 2\sum z_1z_2 = 3\sum z_1^2$
Also,
$z_0 = \frac{\sum z_1}{3} \Rightarrow \sum z_1 = 3z_0 \Rightarrow 3\sum z_1^2 = 9z_0^2 \Rightarrow \sum z_1^2 = 3z_0^2$
18. Since $$z_1, z_2$$ and origin form an equilateral triangle we have
$z_1^2 + z_2^2 + 0^2 - z_1z_2 - z_2*0 - z_1*0 = 0$
Hence, proven.
19. From 136 $$z_1, z_2$$ and origin will form a triangle if $$z_1^2 + z_2^2 - z_1z_2 = 0.$$ Therefore,
$(z_1 + z_2)^2 = 3z_1z_2 \Rightarrow a^2 = 3b.$
20. Centroid of the triangle is given by $$\frac{z_1 + z_2 + z_3}{3}$$ i.e. $$\frac{-3\alpha}{3}$$ i.e. $$-\alpha.$$ Triangle will be equilateral if
$z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1 \Rightarrow (z_1 + z_2 + z_3)^2 = 3(z_1z_2 + z_2z_3 + z_3z_1) \Rightarrow 9\alpha^2 = 9\beta \Rightarrow \alpha^2 = \beta$
21. Given,
$z_2 = \frac{z_1 + z_3}{2}$
Clearly, from section formula we can deduce that $$z_2$$ divides line segment joining $$z_1$$ and $$z_3$$ in two equal segments hence the complex numbers are collinear.
22. $$z_3$$ will divide line segment joining $$z_1$$ and $$z_2$$ either externally or internally. Now section formula can be used to prove remaining.
23. We have,
$\frac{z - i}{ z + i}$
as a purely imaginary quantity. Let $$z = x + iy.$$
$\frac{[x + i(y - 1)][x - i(y + 1)])}{x^2 + (y + 1)^2}$
Equating real part to 0 we have
$\Rightarrow x^2 + y^2 - 1 = 0$
Therefore locus of z represents the circle $$x^2 + y^2 = 1.$$
24. $$z$$ represents the ring between the concentric circles whose center is at (3, 4i) having radii 1 and 2.
25. Let $$z = x+ iy.$$ Now we have
$\sqrt{(x - 1)^2 + y^2} + \sqrt{(x + 1)^2 + y^2} \le 4$
Let $$L + M = 4$$
$L^2 - M^2 = -4x \therefore L^2 - M^2 = -x \therefore 4L^2 = (4 - x)^2 4(x^2 + y^2 - 2x + 1) = 16 + x^2 - 8x 3x^2 + 4y^2 = 12 \frac{x^2}{4} + \frac{y^2}{3} = 1$
Hence it represent the above ellipse.
26. Let $$z = x + iy$$ then we have
$x = t + 5 \text{ and } y = \sqrt{4 -t^2} \Rightarrow (x - 5)^2 = t^2 \text{ and } y^2 = 4 -t^2$
Adding we get, $$(x - 5)^2 + y^2 = 4$$ which represents a circle with radius at (5, 0) with radius 2.
27. Given $$\frac{z^2}{z - 1}$$ is real i.e. its imaginary part is zero.
$\frac{(x^2 - y^2 + i2xy)((x - 1) - iy)}{(x - 1)^2 + y^2}$
Equating imaginary part to 0 we have
$x^2 + y^2 - 2x = 0 \therefore (x - 1)^2 + y^ = 1$
which represents a circle having center at (1, 0) and radius unity.
28. Given, $$|z^2 + (-1)| = |z|^2 + |(-1)| \Rightarrow \frac{z^2}{-1}$$ is non-negative real number. Thus $$z$$ is purely imaginary number. Thus locus of z is a straight line.
Question 147 to 149 are left as exercises.
1. Given,
$\log_{\sqrt{3}}\frac{|z|^2 - |z| + 1}{2 + |z|} < 2 \Rightarrow \frac{|z|^2 - |z| + 1}{2 + |z|} < (\sqrt{3})^2 \Rightarrow |z|^2 - 4|z| - 5 < 0 \Rightarrow |z| < 5$ |
# Lecture14 - Lecture 14 Factoring Algorithms(Taken from...
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Lecture 14: Factoring Algorithms (Taken from Stinson section 4.8 and Elementary Number Theory , Rosen) Fermat Factorization Note that if n (odd) n = ab, then we can rewrite n = s 2 - t 2 , for some value of s and t, since n = (s-t)(s+t), where s = (a+b)/2 and t=(a-b)/2. Both s and t are integers since a and b must both be odd. Now, to use the Fermat factorization, let n = s 2 - t 2 , then we have that t 2 = s 2 - n. Simply start plugging in successive values for s starting with the minimal s for which s 2 > n, until you find a solution for t. Example: n=6077, start s=78. 78 2 - 6077 = 7 79 2 - 6077 = 164 80 2 - 6077 = 323 81 2 - 6077 = 484 = 22 2 , so 6077 = 81 2 - 22 2 = (81 - 22)(81 + 22) = 59x103 Pollard Rho Method Given that n=pq, the basic idea behind this method is to find to integers x and y such that p | (x - y) but n does NOT divide into x - y. Once you find these integers, computing gcd(n, x-y) will yield a non-trivial factor of n. The manner in which the Pollard Rho Method suggests finding x and y is as follows: x 0 = 2 (or any random value mod n) x k+1 f(x k ) mod n f should be a polynomial that produces a large set of numbers before it repeats. A function that works well in practice is f(x) = x 2 + 1. Even when we produce this sequence of values, it would seem as if we should check the difference between each pair of numbers produced. This is O(m 2 ) GCD calls if m values
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• Level: GCSE
• Subject: Maths
• Word count: 2856
# The Gradient Function Maths Investigation
Extracts from this document...
Introduction
Introduction
A lot of graphs produce lines that are curves. Some of the curves are steep, and some are not.
In this investigation I will be looking to work out a formula, which will work out the gradient of any curve.
There are three methods for working out the gradient of a curve, all of them using a tangent:
• The Tangent Method
• The Increment method
• General Proof
This is the graph of y=x². I will find out the gradient of this curve, by using the three methods I mentioned above. I will use the point x=2 for this graph.
Tangent Method: The tangent method is already shown on the graph. I have drawn a line which touches the edge of the curve, when x=2. Once I have drawn the line I then turn it into a triangle, and look at the change in y, and the change in x. Finally I use the formula:
GRADIENT= VERTICAL (CHANGE IN Y AXIS)
HORIZONTAL (CHANGE IN X AXIS)
Therefore the gradient of the curve equals:
The only problem with the tangent method, is that it is only an estimation.
Increment Method: The increment method is where you plot two points (P and Q) on the curve, and the draw a line to join them. This
Middle
5
This is the graph of y=x5. I will continue to investigate the gradients of the graphs y=xn.
Increment Method: x=3
P(3,35) and Q(4,45)
=45-35
4-3
=781
P(3,35) and Q(3.75,3.755)
=3.755-35
3.75-3
=664.8
P(3,35) and Q(3.5,3.55)
=3.55-35
3.5-3
=564.4
P(3,35) and Q(3.25,3.255)
=3.255-35
3.25-3
=478.4
P(3,35) and Q(3.01,3.015)
=3.015-35
3.01-3
=407.7
P(3,35) and Q(3.001,3.0015)
=3.0015-35
3.001-3
=405.3
P(3,35) and Q(3.0001,3.00015)
=3.00015-35
3.0001-3
=405.03
From these results we can conclude that the gradient on an y=x, when x=3 is 405.
Instead of using general proof to work out the equation for this graph, I will use another method, that I have taught myself. It is called Binomial Expansion.
A binomial expansion is the result of a binomial expression, like (x+h) being raised to a power.
The simplest binomial expansion is (x+1), but I will show you the expansion of (x+h)5. So the expansion is:
1 x5 + 5 x h + 10 x h + 10 x h + 5 x h + 1 h
x5 + 5x4h + 10x3h2 + 10x2h2 + 5xh4 + h5
We can now, add the remainder of the formula and simplify this down, like in general proof:
x5+5x4h+10x3h2+10x2h2+5xh4+h5-x5
h
=h(5x4+10x3h+10x2h+5xh3+h4)
h
=(5x4+10x3h+10x2h+5xh3+h4)
=5x4
Therefore when:
Conclusion
n-1, into two.
For example, we shall take the graph, 2x²+3x³:
As we can see, two terms involved, therefore if we were to use our formula naxn-1, it would not work, as there is more than one term. Therefore we have to add something to the formula, to include the second term. So when there are formulas that have more than one term involved, the formula is:
naxn-1+laxn-1
Here the change is, the inclusion of laxn-1. This of course is the second term. The formula would, change depending on the amount of terms involved.
So therefore form this formula I can work out the gradient of the graph y=2x²+3x³.
Let’s first take the graph 2x². I, previously got the formula 4x. The second graph 3x³, we haven’t looked at before, but with our formula, we can see that it is equal 9x².
This means that to find the gradient of the curve y=2x²+3x³, we use the formula 4x+9x².
Therefore the gradient of the curve when:
x=1 is 4+9=13
x=2 is 8+36=44
x=3 is 12+81=93
x=4 is 16+144=160
If I were to work out a similar graph to, y=2x²+3x³, using differentiation, I would use exactly the same method. I would take each term in turn, and work out their respective formula, thus enabling me to put this into the general formula.
This student written piece of work is one of many that can be found in our GCSE Gradient Function section.
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# What is 201/74 as a decimal?
## Solution and how to convert 201 / 74 into a decimal
201 / 74 = 2.716
Fraction conversions explained:
• 201 divided by 74
• Numerator: 201
• Denominator: 74
• Decimal: 2.716
• Percentage: 2.716%
Converting 201/74 to 2.716 starts with defining whether or not the number should be represented by a fraction, decimal, or even a percentage. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. But in some cases, fractions make more sense, i.e., cooking or baking and in other situations decimals make more sense as in leaving a tip or purchasing an item on sale. If we need to convert a fraction quickly, let's find out how and when we should.
201 / 74 as a percentage 201 / 74 as a fraction 201 / 74 as a decimal
2.716% - Convert percentages 201 / 74 201 / 74 = 2.716
## 201/74 is 201 divided by 74
The first step of teaching our students how to convert to and from decimals and fractions is understanding what the fraction is telling is. 201 is being divided into 74. Think of this as our directions and now we just need to be able to assemble the project! Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 201 divided by 74. We must divide 201 into 74 to find out how many whole parts it will have plus representing the remainder in decimal form. This is our equation:
### Numerator: 201
• Numerators are the top number of the fraction which represent the parts of the equation. Any value greater than fifty will be more difficult to covert to a decimal. The bad news is that it's an odd number which makes it harder to covert in your head. Large two-digit conversions are tough. Especially without a calculator. Let's look at the fraction's denominator 74.
### Denominator: 74
• Denominators are located at the bottom of the fraction, representing the total number of parts. Larger values over fifty like 74 makes conversion to decimals tougher. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Ultimately, don't be afraid of double-digit denominators. Now let's dive into how we convert into decimal format.
## How to convert 201/74 to 2.716
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 74 \enclose{longdiv}{ 201 }$$
To solve, we will use left-to-right long division. This method allows us to solve for pieces of the equation rather than trying to do it all at once.
### Step 2: Solve for how many whole groups you can divide 74 into 201
$$\require{enclose} 00.2 \\ 74 \enclose{longdiv}{ 201.0 }$$
How many whole groups of 74 can you pull from 2010? 148 Multiple this number by our furthest left number, 74, (remember, left-to-right long division) to get our first number to our conversion.
### Step 3: Subtract the remainder
$$\require{enclose} 00.2 \\ 74 \enclose{longdiv}{ 201.0 } \\ \underline{ 148 \phantom{00} } \\ 1862 \phantom{0}$$
If there is no remainder, you’re done! If you have a remainder over 74, go back. Your solution will need a bit of adjustment. If you have a number less than 74, continue!
### Step 4: Repeat step 3 until you have no remainder
Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals is a necessity. They each bring clarity to numbers and values of every day life. Same goes for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But each represent values in everyday life! Without them, we’re stuck rounding and guessing. Here are real life examples:
### When you should convert 201/74 into a decimal
Tax - Taxes are also in decimal form. Example. \$0.06 on every dollar spent. This is also represented in percentages.
### When to convert 2.716 to 201/74 as a fraction
Progress - If we were writing an essay and the teacher asked how close we are to done. We wouldn't say .5 of the way there. We'd say we're half-way there. A fraction here would be more clear and direct.
### Practice Decimal Conversion with your Classroom
• If 201/74 = 2.716 what would it be as a percentage?
• What is 1 + 201/74 in decimal form?
• What is 1 - 201/74 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 2.716 + 1/2?
### Convert more fractions to decimals
From 201 Numerator From 74 Denominator What is 201/64 as a decimal? What is 191/74 as a decimal? What is 201/65 as a decimal? What is 192/74 as a decimal? What is 201/66 as a decimal? What is 193/74 as a decimal? What is 201/67 as a decimal? What is 194/74 as a decimal? What is 201/68 as a decimal? What is 195/74 as a decimal? What is 201/69 as a decimal? What is 196/74 as a decimal? What is 201/70 as a decimal? What is 197/74 as a decimal? What is 201/71 as a decimal? What is 198/74 as a decimal? What is 201/72 as a decimal? What is 199/74 as a decimal? What is 201/73 as a decimal? What is 200/74 as a decimal? What is 201/74 as a decimal? What is 201/74 as a decimal? What is 201/75 as a decimal? What is 202/74 as a decimal? What is 201/76 as a decimal? What is 203/74 as a decimal? What is 201/77 as a decimal? What is 204/74 as a decimal? What is 201/78 as a decimal? What is 205/74 as a decimal? What is 201/79 as a decimal? What is 206/74 as a decimal? What is 201/80 as a decimal? What is 207/74 as a decimal? What is 201/81 as a decimal? What is 208/74 as a decimal? What is 201/82 as a decimal? What is 209/74 as a decimal? What is 201/83 as a decimal? What is 210/74 as a decimal? What is 201/84 as a decimal? What is 211/74 as a decimal?
### Convert similar fractions to percentages
From 201 Numerator From 74 Denominator 202/74 as a percentage 201/75 as a percentage 203/74 as a percentage 201/76 as a percentage 204/74 as a percentage 201/77 as a percentage 205/74 as a percentage 201/78 as a percentage 206/74 as a percentage 201/79 as a percentage 207/74 as a percentage 201/80 as a percentage 208/74 as a percentage 201/81 as a percentage 209/74 as a percentage 201/82 as a percentage 210/74 as a percentage 201/83 as a percentage 211/74 as a percentage 201/84 as a percentage |
# SAT: Math Practice Quiz! Test
Approved & Edited by ProProfs Editorial Team
The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes.
| By Mmmaxwell
M
Mmmaxwell
Community Contributor
Quizzes Created: 32 | Total Attempts: 53,189
Questions: 16 | Attempts: 700
Settings
.
• 1.
### If (1/3)y + 9 = 0, then y =
• A.
-27
• B.
-9
• C.
-3
• D.
3
• E.
27
A. -27
Explanation
To find the value of y, we can start by isolating the variable. We can begin by subtracting 9 from both sides of the equation: (1/3)y = -9. Next, we can multiply both sides by 3 to eliminate the fraction: y = -27. Therefore, the correct answer is -27.
Rate this question:
• 2.
### In the figure above, P, Q, and R lie on the same line. P is the center of the larger circle, and Q is the center of the smaller circle. If the radius of the larger circle is 4, what is the radius of the smaller circle?
• A.
1
• B.
2
• C.
4
• D.
8
• E.
16
B. 2
Explanation
Since P is the center of the larger circle and Q is the center of the smaller circle, the line connecting P and Q is the line passing through the centers of both circles. Since P, Q, and R lie on the same line, R must also lie on this line. Therefore, the line connecting P and R is also the line passing through the centers of both circles. Since the radius of the larger circle is 4, the distance between P and R is 8. Since the radius of a circle is half of its diameter, the radius of the smaller circle is half of 8, which is 4. Therefore, the radius of the smaller circle is 2.
Rate this question:
• 3.
### Roy planted corn on 1/5 of his land. If he planted 45 acres of corn, how many acres of land does he have?
• A.
90
• B.
112 1/2
• C.
135
• D.
225
• E.
337 1/2
D. 225
Explanation
If Roy planted 45 acres of corn, which represents 1/5 of his land, we can determine the total number of acres he has by multiplying 45 by 5. This calculation gives us 225 acres, which is the correct answer.
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• 4.
### 6, 10, 18, 34, 66 The first number in the list above is 6. Which of the following gives a rule for finding each successive number in the list?
• A.
Add 4 to the preceding number.
• B.
Take 1/2 of the preceding number and then add 7 to that result.
• C.
Double the preceding number and then subtract 2 from that result.
• D.
Subtract 2 from the preceding number and then double that result.
• E.
Triple the preceding number and then subtract 8 from that result.
C. Double the preceding number and then subtract 2 from that result.
Explanation
Each number in the list is obtained by doubling the preceding number and then subtracting 2 from that result. Starting with 6, doubling it gives 12, and then subtracting 2 gives 10. Continuing this pattern, we get 18, 34, and 66.
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• 5.
### The two semicircles in the figure above have centers R and S, respectively. If RS = 12, what is the total length of the darkened curve?
• A.
8Ï€
• B.
9Ï€
• C.
12Ï€
• D.
15Ï€
• E.
16Ï€
C. 12Ï€
Explanation
The total length of the darkened curve can be found by adding the circumference of both semicircles. Since the radius of each semicircle is not given, we cannot determine the exact length. However, we do know that RS = 12, which means the distance between the centers of the semicircles is 12. Therefore, the total length of the darkened curve is equal to the circumference of one semicircle plus the circumference of the other semicircle, which is equal to 2Ï€r + 2Ï€r = 4Ï€r. Since the radius is not given, we cannot determine the exact length, but it will be equal to 12Ï€.
Rate this question:
• 6.
### If h and k are positive numbers and h + k = 7, then (7 - k)/h =
• A.
1
• B.
0
• C.
-1
• D.
H
• E.
K - 1
A. 1
Explanation
Given that h and k are positive numbers and h + k = 7, we can substitute the value of h + k into the expression (7 - k)/h. This gives us (7 - k)/h = (7 - k)/(7 - h). Simplifying this expression further, we get (7 - k)/(7 - h) = 1. Therefore, the correct answer is 1.
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• 7.
### The table above shows the populations of two countries and their population densities. The number of square miles in the area of Country B is approximately how much greater than the number of square miles in the area of Country A ?
• A.
200
• B.
3,600
• C.
5,000
• D.
8,000
• E.
950,000,000
D. 8,000
Explanation
Country B has a population density of 200 people per square mile. The number of square miles in the area of Country B is approximately 8,000 greater than the number of square miles in the area of Country A.
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• 8.
### If x2 = x + 6, which of the following must be true?
• A.
X = 6
• B.
X< 3
• C.
X > 0
• D.
X^2 < x
• E.
X^2 > x
E. X^2 > x
Explanation
If x^2 = x + 6, then rearranging the equation we get x^2 - x - 6 = 0. Factoring this quadratic equation, we have (x - 3)(x + 2) = 0. Therefore, the solutions for x are x = 3 and x = -2. By substituting these values into the inequality x^2 > x, we find that x = 3 satisfies the inequality, but x = -2 does not. Hence, x^2 > x must be true.
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• 9.
### Let the functions f be defined by f(x) = 5x - 2a, where a is a constant. If f(10) + f(5) = 55, what is the value of a?
• A.
-5
• B.
0
• C.
5
• D.
10
• E.
20
C. 5
Explanation
The given question states that the sum of f(10) and f(5) is equal to 55. Substituting the values of x into the function, we get 5(10) - 2a + 5(5) - 2a = 55. Simplifying this equation, we have 50 - 2a + 25 - 2a = 55. Combining like terms, we get 75 - 4a = 55. Solving for a, we subtract 75 from both sides and divide by -4, giving us a = 5.
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• 10.
### A number is called “even-odd” if it is halfway between an even integer and an odd integer. If x is an even-odd number, which of the following must be true? I. 2x is an integer. II. 2x is even-odd. III. x is halfway between two even integers.
• A.
I only
• B.
II only
• C.
I and II only
• D.
II and III only
• E.
I, II, and III
A. I only
Explanation
If x is an even-odd number, it means that x is halfway between an even integer and an odd integer. Since an even integer multiplied by 2 is always an integer, statement I must be true. However, statement II cannot be determined because it is not specified whether multiplying x by 2 will still result in a number that is halfway between an even and odd integer. Statement III cannot be determined either because it is not specified whether x is exactly halfway between two even integers or if it is closer to one of them.
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• 11.
### If m is a positive integer, which of the following is NOT equal to (24)m ?
• A.
2^(4m)
• B.
4^(2m)
• C.
2^m(2^(3m))
• D.
4^m(2^m)
• E.
16^m
D. 4^m(2^m)
Explanation
The expression (24)m can be simplified to 2^(4m), which means that it is equal to 2 raised to the power of 4m. Looking at the answer choices, all of them involve raising 2 to some power except for 4^m(2^m). This expression involves multiplying 4^m and 2^m together, which is not equivalent to 2^(4m). Therefore, 4^m(2^m) is NOT equal to (24)m.
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• 12.
### In the figure above, l || m. Which of the following must equal 180?
• A.
K + n + r
• B.
K + p + s
• C.
N + p + s
• D.
P + n + t
• E.
R + s + t
B. K + p + s
Explanation
Since l || m, we can conclude that the angles formed by the transversal line are congruent. Therefore, the sum of the angles k, p, and s must equal 180 degrees.
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• 13.
### How many different ordered pairs (x, y) are there such that x is an even integer, where 4 ≤ x ≤ 10, and y is an integer, where 4 < y < 10 ?
• A.
8
• B.
12
• C.
20
• D.
30
• E.
36
C. 20
Explanation
There are 4 even integers between 4 and 10, which are 4, 6, 8, and 10. There are 5 integers between 4 and 10, which are 5, 6, 7, 8, and 9. Therefore, there are 4 even integers for each integer between 4 and 10, resulting in a total of 4 * 5 = 20 different ordered pairs (x, y).
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• 14.
### N(t) = 500(0.81)t The function above can be used to model the population of a certain endangered species of animal. If n(t) gives the number of the species living t decades after the year 1900, which of the following is true about the population of the species from 1900 to 1920 ?
• A.
• B.
• C.
• D.
• E.
C. It decreased by about 180.
Explanation
The function n(t) = 500(0.81)t represents the population of the species t decades after the year 1900. Since we are interested in the population from 1900 to 1920, we can substitute t = 2 into the equation.
n(2) = 500(0.81)2
n(2) = 500(0.6561)
n(2) ≈ 328.05
Therefore, the population decreased by about 180 individuals from 1900 to 1920.
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• 15.
### A sphere of radius r inside a cube touches each one of the six sides of the cube. What is the volume of the cube, in terms of r ?
• A.
R^3
• B.
2r^3
• C.
4r^3
• D.
(4/3)r^3
• E.
8r^3
E. 8r^3
Explanation
The cube has a side length equal to twice the radius of the sphere. Since the sphere touches each side of the cube, the diagonal of the cube is equal to the diameter of the sphere. Using the Pythagorean theorem, we can find that the diagonal of the cube is equal to 2√3 times the side length of the cube. Therefore, the side length of the cube is equal to r√3. The volume of the cube is then (r√3)^3 = 27r^3. Simplifying, we get the answer of 8r^3.
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• 16.
### According to the graph above, in which year was the ratio of the number of students enrolled at School B to the number of students enrolled at School A the greatest?
• A.
1990
• B.
1991
• C.
1992
• D.
1993
• E.
1994
E. 1994
Explanation
Based on the graph, the ratio of the number of students enrolled at School B to the number of students enrolled at School A is highest in the year 1994.
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Related Topics |
# Sets Chapter 2 - PowerPoint PPT Presentation
1 / 12
Sets Chapter 2. 1. Sets: A Problem Solving Tool 2.1. 2. Sets. A set is a well-defined collection of objects, called elements or members of the set. Let A = {1 , 2, 3, 4 }, which means that “ A is the set containing the elements 1 , 2, 3 , and 4.”
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Sets Chapter 2
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1
## Sets: A Problem Solving Tool2.1
2
Sets
A set is a well-defined collection of objects, called elements or members of the set.
Let A = {1, 2, 3, 4}, which means that “A is the set containing the elements 1, 2, 3, and 4.”
4 A “4 is in A” or “4 is an element of the set A”
6 A“6 is not an element of A”
3
• Let X = {a, b, x, y}. Fill in the blank with or to make
each statement correct.
• a___X
• x___X
• A___X
4
Sets of Numbers Commonly used in Mathematics
N = {x | x is a natural number} = {1, 2, 3, …}
W = {x | x is a whole number} = {0,1, 2, 3, …}
I = {x | x is an integer} = {…, 2, 1, 0,1, 2, 3, …}
Q = {x | x is a rational number} = {x | x, is of the form
a/b where a, b are integers and b 0}
Q = {x | x is a rational number} = {x | x has a terminating or repeating decimal pattern}
S = {x | x is a irrational number} = {x | x is not rational}
R = {x | x is a real number} = {x | x is the union of the rational and irrational numbers}
5
Describing Sets
• 1. Verbal or written description
• Roster Method or Listing
• Set-builder notation
The Empty Set
The symbol { } or Ø represents the empty, or null set.
NOTE:
{Ø} is not the null set. It is the set containing the null set.
6
Describe the following sets using the roster method and set builder notation.
• The set of counting numbers less than 8.
• The set of counting numbers between 3 and 8.
• The set of counting numbers between 6 and 7.
• The set of counting numbers greater than 3.
a. {1, 2, 3, 4, 5, 6, 7}
b. {4, 5, 6, 7}
c. { } or φ
d. {4, 5, 6,…}
7
Equality of Sets
Two sets A and B are equal, A = B, if they have the same elements not necessarily listed in the same order.
State whether the sets A and B are equal.
• A={2n + 1|n is a counting number} B={2n – 1|n is a counting number}
• A = {1,1,2,2,3} , B = {2,1,3}
1. A ≠ B , not equal
2. A = B , equal
8
Subsets
The set A is a subset of B, A ⊆ B, if every element of A is also an element of B.
Proper Subsets
A set A is said to be a proper subset of B, A ⊂ B, if
A ⊆ B but A B.
The Universal Set
The universal set,U, is the set of all elements under discussion.
9
List all the subsets and indicate which are the proper subsets of the given set.
• U = {a, b}
• U = {1, 2, 3}
• Ø, {a}, {b}, and {a, b}. The first three are proper subsets.
Ø
{1}, {2}, {3}
{1, 2}, {1, 3}, {2, 3}
{1, 2, 3}
The first seven are proper subsets.
Note: by definition Ø is a subset of any set.
10
Number of Subsets of a Set
A set of n elements has 2n subsets.
Refer to the problems on the previous slide:
22 = 4 and 23 = 8.
Number of Proper Subsets of a Set
A set of n elements has 2n – 1 proper subsets.
Refer to the problems on the previous slide:
22 1 = 4 1 = 3 and 23 1 = 8 1 = 7.
11
### Example:
Given: A = { 12, 19, 26,…, 68} Find
the number of subsets
the number of proper subsets
To solve this problem you need to determine how many elements are in the set.
Notice the elements in this set are evenly spaced and form what is called an arithmetic sequence. The next term is found by adding a common difference of seven in this example. The total number of elements can be found without having to list all of the elements in this set using the formula An = A1 + (n – 1)d.
An is the last term in the sequence, A1 is the first term, d is the common difference and n is the number of terms in the sequence. Substituting the numbers into the equation and solving for n yields the number of terms in this set.
a. Number of subsets = 29 = 512
b. Number of proper subsets = 29– 1 = 512 – 1 = 511
68 = 12 + (n – 1)(7)
–12 –12
56 = (n – 1)(7)
7 7
8 = n – 1
+1 + 1
9 = n
12
END |
380 minus 1 percent
This is where you will learn how to calculate three hundred eighty minus one percent (380 minus 1 percent). We will first explain and illustrate with pictures so you get a complete understanding of what 380 minus 1 percent means, and then we will give you the formula at the very end.
We start by showing you the image below of a dark blue box that contains 380 of something.
380
(100%)
1 percent means 1 per hundred, so for each hundred in 380, you want to subtract 1. Thus, you divide 380 by 100 and then multiply the quotient by 1 to find out how much to subtract. Here is the math to calculate how much we should subtract:
(380 ÷ 100) × 1
= 3.8
We made a pink square that we put on top of the image shown above to illustrate how much 1 percent is of the total 380:
The dark blue not covered up by the pink is 380 minus 1 percent. Thus, we simply subtract the 3.8 from 380 to get the answer:
380 - 3.8
= 376.2
The explanation and illustrations above are the educational way of calculating 380 minus 1 percent. You can also, of course, use formulas to calculate 380 minus 1%.
Below we show you two formulas that you can use to calculate 380 minus 1 percent and similar problems in the future.
Formula 1
Number - ((Number × Percent/100))
380 - ((380 × 1/100))
380 - 3.8
= 376.2
Formula 2
Number × (1 - (Percent/100))
380 × (1 - (1/100))
380 × 0.99
= 376.2
Number Minus Percent
Go here if you need to calculate any other number minus any other percent.
381 minus 1 percent
Here is the next percent tutorial on our list that may be of interest. |
### Improve Your Mind With Some Simple Math Games
Mathematics has long been recognized for it's mental benefits.
Working with numbers improves your concentration, memory, focus, problem solving skills, and general clarity of thought.
To enjoy these benefits, you don't have to indulge in any complicated formulas. All you need is a few minutes daily practice playing some simple math games.
And before you rush out to buy the latest Xbox console and software, realize that numbers are all around you...
Look at the clock on your computer. Usually it's located in the lower right-hand corner of your screen (or use any clock to tell the time).
The 24-hour format works best. On my computer right now, the time is 15:38.
There are all kinds of creative games you can play with this. Here are ten to get you started:
#1 Add the single digits together from left to right:
1 + 5 + 3 + 8 = (say "6... 9... 17")
#2 Add the single digits together from right to left:
8 + 3 + 5 + 1 = (say "11... 16.. 17")
#3 Add the inner and outer digits together, then add the resulting pairs together:
1 + 8 = 9
5 + 3 = 8
and so 9 + 8 = 17
#4 Add the single digits on either side of the colon, and multiply the results:
1 + 5 = 6 and 3 + 8 = 11
6 x 11 = 66
#5 Subtract the single digits on either side of the colon, and multiply the results:
1 from 5 = 4
3 from 8 = 5
4 x 5 = 20
#6 Multiply the single digits on either side of the colon, and multiply the results:
1 x 5 = 5 and 3 x 8 = 24
5 x 24 = 120
#7 Add the two-digit numbers either side of the colon:
Add 15 to 38 to get 53
#8 Subtract the two-digit numbers either side of the colon:
Subtract 15 from 38 to get 23
#9 Divide the two-digit numbers either side of the colon:
Divide 15 into 38 to get 2 remainder 8
#10 Feeling brave? Multiply the two-digit numbers either side of the colon:
Multiply 15 by 38 to get... 570
You can repeat the above exercises as many times a day as you like.
Try them anytime you have a spare minute, like when you're placed on hold in a telephone queue.
You may not turn into a mathematical genius, but you'll certainly keep your brain in gear!
Murdo Macleod is co-author of the popular "Fun With Figures" mental math course, which shows anyone aged between 8 and 80 the easy way to do impressive mental calculations. Visit the website today for more details at: http://www.FunWithFigures.com
Murdo Macleod |
# How Are Linear Equations Used in Everyday Life?
By Jessica Smith; Updated June 02, 2017
Linear equations use one or more variables where one variable is dependent on the other. Almost any situation where there is an unknown quantity can be represented by a linear equation, like figuring out income over time, calculating mileage rates, or predicting profit. Many people use linear equations every day, even if they do the calculations in their head without drawing a line graph.
### Variable Costs
Imagine that you are taking a taxi while on vacation. You know that the taxi service charges \$9 to pick your family up from your hotel and another \$0.15 per mile for the trip. Without knowing how many miles it will be to each destination, you can set up a linear equation that can be used to find the cost of any taxi trip you take on your trip. By using "x" to represent the number of miles to your destination and "y" to represent the cost of that taxi ride, the linear equation would be: y = 0.15x + 9.
### Rates
Linear equations can be a useful tool for comparing rates of pay. For example, if one company offers to pay you \$450 per week and the other offers \$10 per hour, and both ask you to work 40 hours per week, which company is offering the better rate of pay? A linear equation can help you figure it out! The first company's offer is expressed as 450 = 40x. The second company's offer is expressed as y = 10(40). After comparing the two offers, the equations tell you that the first company is offering the better rate of pay at \$11.25 per hour.
### Budgeting
A party planner has a limited budget for an upcoming event. She'll need to figure out how much it will cost her client to rent a space and pay per person for meals. If the cost of the rental space is \$780 and the price per person for food is \$9.75, a linear equation can be constructed to show the total cost, expressed as y, for any number of people in attendance, or x. The linear equation would be written as y = 9.75x + 780. With this equation, the party planner can substitute any number of party guests and give her client the actual cost of the event with the food and rental costs included.
### Making Predictions
One of the most helpful ways to apply linear equations in everyday life is to make predictions about what will happen in the future. If a bake sale committee spends \$200 in initial start up costs and then earns \$150 per month in sales, the linear equation y = 150x - 200 can be used to predict cumulative profits from month to month. For instance, after six months, the committee can expect to have netted \$700 because (150 x 6) - 200 = \$700. While real world factors certainly impact how accurate predictions are, they can be a good indication of what to expect in the future. Linear equations are a tool that make this possible. |
# Common Core: High School - Number and Quantity : Complex Number Form: CCSS.Math.Content.HSN-CN.A.1
## Example Questions
← Previous 1
### Example Question #1 : Complex Number Form: Ccss.Math.Content.Hsn Cn.A.1
Simplifying the following.
Explanation:
This question tests one's ability to perform arithmetic operations on complex numbers. Questions like this introduces and builds on the concept of complex numbers. Recall that a complex number by definition contains a negative square. In mathematical terms this is expressed as follows.
Performing arithmetic operations on complex numbers relies on the understanding of the various algebraic operations and properties (distributive, associative, and commutative properties) as well as the imaginary, complex number .
For the purpose of Common Core Standards, "know there is a complex number such that , and every complex number has a form with and are reals", falls within the Cluster A of "perform arithmetic operations with complex numbers" (CCSS.MATH.CONTENT.HSF.CN.A).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Perform multiplication between the two terms.
Recall that multiplication between two radicand terms (terms under the square root sign), can be combined as one using the communicative property with multiplication.
Step 2: Factor the two term in the expression.
Step 3: Pull out common terms that exists in the radicand.
Remember that when a number appears under the square root sign, one of the numbers can be brought out front and the other one is canceled out.
Step 4: Use the identity that .
### Example Question #2 : Complex Number Form: Ccss.Math.Content.Hsn Cn.A.1
Express as a pure imaginary number.
Explanation:
A pure imaginary number is expressed as , where is a positive real number and represents the imaginary unit.
### Example Question #3 : Complex Number Form: Ccss.Math.Content.Hsn Cn.A.1
Express as a pure imaginary number.
Explanation:
A pure imaginary number is expressed as , where is a positive real number and represents the imaginary unit.
Simplify:
Explanation:
Simplify:
Explanation:
Simplify:
Explanation:
### Example Question #7 : Complex Number Form: Ccss.Math.Content.Hsn Cn.A.1
Simplify
Explanation:
The powers of are:
This pattern continues for every successive four power of . Thus:
### Example Question #8 : Complex Number Form: Ccss.Math.Content.Hsn Cn.A.1
Simplify:
Explanation:
The powers of are:
This pattern continues for every successive four power of . Thus:
### Example Question #9 : Complex Number Form: Ccss.Math.Content.Hsn Cn.A.1
Simplify:
Explanation:
The powers of are:
This pattern continues for every successive four power of . Thus:
### Example Question #10 : Complex Number Form: Ccss.Math.Content.Hsn Cn.A.1
Simplify:
Explanation:
The powers of are:
This pattern continues for every successive four power of . Thus:
To simplify to a larger power, simply break it into terms, as these simplify to 1.
← Previous 1 |
FutureStarr
1 Percent of 20
1 Percent of 20
The average tech startup pivots 6. 4 times before achieving sustainable growth. But for the small percentage of startups that actually reach (or already have) sustainable growth, that number drops to just 1 in 20.
Multiply
. A real-world example could be: there are two girls in a group of five children. What's the percentage of girls? In other words, we want to know what's the ratio of girls to all children. It's 2 out of 5, or 2/5. We call the first number (2) a numerator and the second number (5) a denominator because this is a fraction. To calculate the percentage, multiply this fraction by 100 and add a percent sign.
As your maths skills develop, you can begin to see other ways of arriving at the same answer. The laptop example above is quite straightforward and with practise, you can use your mental maths skills to think about this problem in a different way to make it easier. In this case, you are trying to find 20%, so instead of finding 1% and then multiplying it by 20, you can find 10% and then simply double it. We know that 10% is the same as 1/10th and we can divide a number by 10 by moving the decimal place one place to left (removing a zero from 500). Therefore 10% of £500 is £50 and 20% is £100. (Source: www.skillsyouneed.com)
Divide
Do you have problems with simplifying fractions? The best way to solve this is by finding the GCF (Greatest Common Factor) of the numerator and denominator and divide both of them by GCF. You might find our GCF and LCM calculator to be convenient here. It searches all the factors of both numbers and then shows the greatest common one. As the name suggests, it also estimates the LCM which stands for the Least Common Multiple.
Although Ancient Romans used Roman numerals I, V, X, L, and so on, calculations were often performed in fractions that were divided by 100. It was equivalent to the computing of percentages that we know today. Computations with a denominator of 100 became more standard after the introduction of the decimal system. Many medieval arithmetic texts applied this method to describe finances, e.g., interest rates. However, the percent sign % we know today only became popular a little while ago, in the 20th century, after years of constant evolution. (Source: www.omnicalculator.com)
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# A Rectangle’S Length is 5 Cm Less than Twice Its Width. If the Length is Decreased by 5 Cm and Width is Increased by 2 Cm - Mathematics
Sum
A rectangle’s length is 5 cm less than twice its width. If the length is decreased by 5 cm and width is increased by 2 cm; the perimeter of the resulting rectangle will be 74 cm. Find the length and width of the origi¬nal rectangle.
#### Solution
Let width of the original rectangle = x cm
Length of the original rectangle = (2x – 5)cm
Now, new length of the rectangle = 2x – 5 – 5 = (2x – 10) cm
New width of the rectangle = (x + 2) cm
New perimeter = 2[Length+Width] = 2[2x – 10 + x + 2] = 2[3x – 8] = (6x – 16) cm
Given; new perimeter = 74 cm
6x – 16 = 74
⇒ 6x = 74 + 16
⇒ 6x = 90
⇒ x = 15
Length of the original rectangle = 2x – 5 = 2 x 15 – 5 = 30 – 5 = 25 cm
Width of the original rectangle = x = 15 cm
Concept: Solving Linear Inequations
Is there an error in this question or solution?
Chapter 14: Linear Equations in one Variable - Exercise 14 (B) [Page 169]
#### APPEARS IN
Selina Concise Mathematics Class 8 ICSE
Chapter 14 Linear Equations in one Variable
Exercise 14 (B) | Q 11 | Page 169
Share |
# How do you simplify 2^2 \cdot 2^4 \cdot 2^6?
Mar 3, 2018
See a solution process below:
#### Explanation:
Use this rule for exponents to simplify the expression:
${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$
${2}^{\textcolor{red}{2}} \times {2}^{\textcolor{b l u e}{4}} \times {2}^{\textcolor{g r e e n}{6}} \implies {2}^{\textcolor{red}{2} + \textcolor{b l u e}{4} + \textcolor{g r e e n}{6}} \implies {2}^{12}$
${2}^{12}$ in its simplest form is 4096
Mar 3, 2018
${2}^{12}$
#### Explanation:
It's easier to think about the problem by writing it out like this:
$\left(2 \cdot 2\right) \cdot \left(2 \cdot 2 \cdot 2 \cdot 2\right) \cdot \left(2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2\right)$
Count the number of twos,
${2}^{12}$
while it helps to think about it that way, the easiest way to solve it is to just add the exponents.
${2}^{2} \cdot {2}^{4} \cdot {2}^{6}$
$2 + 4 = 6$
$6 + 6 = 12$
${2}^{12}$
I hope this helps :) |
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Transcript
Measures of
Dispersion
Week 3
What is dispersion?
• Dispersion is how the data is spread out,
or dispersed from the mean.
• The smaller the dispersion values, the
more consistent the data.
• The larger the dispersion values, the more
spread out the data values are. This
means that the data is not as consistent.
Consider these sets of data:
• Grades from Test # 1 =
- 81,83,83,82,86,81,87,80,81,86
• Grades from Test # 2 =
- 95,74,65,90,87,97,60,81,99,76
• What differences do you see between the
two sets?
• What are the Mean scores? Ranges?
• Do you believe these grades tell a story?
Important Symbols to remember
= mean
• X = an individual value
• N = Population size
• n = sample population size
• i = 1st data value in population
Variance
• The average of the squares of
each difference of a data value
and the mean.
Standard Deviation
• is the measure of the average distance
between individual data points and their
mean.
• It is the square root of the variance.
• The lower case Greek letter sigma is used
to denote standard deviation.
How to Calculate Standard
Deviation
• Given the data set {5, 6, 8, 9}, calculate
the standard deviation.
• Step 1: find the mean of the data set
Sum of items x1 x2 x3 ... xn
Mean
Count
n
5 6 8 9 28
Mean
7
4
4
How to Calculate Standard
Deviation
• Step 2: Find the difference between each
data point and the mean.
5 7 2
6 7 1
87 1
97 2
How to Calculate Standard
Deviation
• Step 3: Square the difference between
each data point and the mean.
2 4
2
1 1
2
1 1
2
2 4
2
How to Calculate Standard
Deviation
• Step 4: Sum the squares of the
differences between each data point and
the mean.
4 1 1 4 10
How to Calculate Standard
Deviation
• Step 5: Take the square root of the sum
of the squares of the differences divided
by the total number of data points;
10
10 3.16227766
1.58113883
4
2
2
* The average distance between individual data
points and the mean is 1.58113883 units from 7
Standard Deviation
• Formula of what we just did:
n
1
2
( Xi X )
n i 1
• For sample S.D. use 1/(n-1)
When to use Pop. vs. Sample
• When we have the actual entire population
(for example our class, 29 students), we
would use the Population formula.
• If the problem tells us to use a particular
formula; Pop. v. Samp.
• If we are working with less entire
population of a much larger group, we will
use the sample formula.
• (Which is one taken away from the pop. total)
Why is this useful?
• It provides clues as to how representative
the mean is of the individual data points.
• For example, consider the following two
data sets with the same means, but
different standard deviations.
Bowler # 1
{98, 99, 101, 102}
X 100
1.58113883
Bowler # 2
{30 ,51, 149, 169}
X 100
78.10889834
The mean with the standard deviation provides a better description of the data set.
TI-83 to Calculate Standard Deviation.
• Step 1. Press STAT,EDIT,1:EDIT
• Step 2. Enter your data in the L1 column,
pressing enter after every data entry.
• Step 3. Press STAT, CALC,1-Var stats
• Step 4. Scroll down to the lower case
symbol for the Greek letter sigma
• calculator help.
Let’s try one more by hand:
(1) Find the population standard deviation
for the following Stats class test grades:
78, 84, 88, 92, 68, 82, 92, 72, 88, 86, 76, 90
(a) How many grades fall within one SD of the mean?
(b) What percent fall within one SD of the mean?
* Now check it with the calculator!
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November 24, 2022
The information can look overwhelming at first. However, offer yourself some grace and room so there’s no hurry or strain while working through these problems. To be efficient at quadratic equations like a pro, you will need a good sense of humor, patience, and good understanding.
Now, let’s begin learning!
## What Is the Quadratic Equation?
At its core, a quadratic equation is a mathematical formula that portrays distinct scenarios in which the rate of deviation is quadratic or proportional to the square of few variable.
Though it may look similar to an abstract concept, it is simply an algebraic equation expressed like a linear equation. It ordinarily has two solutions and utilizes complex roots to solve them, one positive root and one negative, employing the quadratic equation. Unraveling both the roots should equal zero.
### Definition of a Quadratic Equation
First, bear in mind that a quadratic expression is a polynomial equation that comprises of a quadratic function. It is a second-degree equation, and its conventional form is:
ax2 + bx + c
Where “a,” “b,” and “c” are variables. We can utilize this equation to figure out x if we replace these terms into the quadratic equation! (We’ll go through it later.)
Ever quadratic equations can be scripted like this, which results in working them out simply, relatively speaking.
### Example of a quadratic equation
Let’s compare the following equation to the previous formula:
x2 + 5x + 6 = 0
As we can observe, there are 2 variables and an independent term, and one of the variables is squared. Thus, compared to the quadratic equation, we can confidently state this is a quadratic equation.
Generally, you can find these kinds of equations when measuring a parabola, that is a U-shaped curve that can be plotted on an XY axis with the data that a quadratic equation gives us.
Now that we understand what quadratic equations are and what they look like, let’s move forward to figuring them out.
## How to Work on a Quadratic Equation Using the Quadratic Formula
Although quadratic equations might seem very intricate when starting, they can be broken down into multiple easy steps utilizing a straightforward formula. The formula for solving quadratic equations includes setting the equal terms and utilizing basic algebraic operations like multiplication and division to obtain two answers.
Once all functions have been performed, we can solve for the values of the variable. The answer take us one step closer to discover answer to our actual question.
### Steps to Working on a Quadratic Equation Utilizing the Quadratic Formula
Let’s promptly place in the common quadratic equation again so we don’t forget what it looks like
ax2 + bx + c=0
Before working on anything, keep in mind to separate the variables on one side of the equation. Here are the 3 steps to solve a quadratic equation.
#### Step 1: Note the equation in standard mode.
If there are variables on both sides of the equation, total all equivalent terms on one side, so the left-hand side of the equation is equivalent to zero, just like the conventional model of a quadratic equation.
#### Step 2: Factor the equation if workable
The standard equation you will end up with should be factored, usually using the perfect square method. If it isn’t possible, put the terms in the quadratic formula, which will be your closest friend for solving quadratic equations. The quadratic formula looks something like this:
x=-bb2-4ac2a
All the terms correspond to the same terms in a conventional form of a quadratic equation. You’ll be utilizing this significantly, so it pays to remember it.
#### Step 3: Apply the zero product rule and work out the linear equation to discard possibilities.
Now once you possess two terms resulting in zero, solve them to obtain two answers for x. We get two results because the answer for a square root can either be negative or positive.
### Example 1
2x2 + 4x - x2 = 5
Now, let’s fragment down this equation. First, clarify and place it in the standard form.
x2 + 4x - 5 = 0
Next, let's identify the terms. If we contrast these to a standard quadratic equation, we will get the coefficients of x as follows:
a=1
b=4
c=-5
To solve quadratic equations, let's replace this into the quadratic formula and solve for “+/-” to involve both square root.
x=-bb2-4ac2a
x=-442-(4*1*-5)2*1
We work on the second-degree equation to get:
x=-416+202
x=-4362
Now, let’s clarify the square root to obtain two linear equations and work out:
x=-4+62 x=-4-62
x = 1 x = -5
Next, you have your answers! You can review your work by using these terms with the first equation.
12 + (4*1) - 5 = 0
1 + 4 - 5 = 0
Or
-52 + (4*-5) - 5 = 0
25 - 20 - 5 = 0
### Example 2
Let's try one more example.
3x2 + 13x = 10
Let’s begin, put it in the standard form so it results in 0.
3x2 + 13x - 10 = 0
To solve this, we will put in the values like this:
a = 3
b = 13
c = -10
Solve for x employing the quadratic formula!
x=-bb2-4ac2a
x=-13132-(4*3x-10)2*3
Let’s streamline this as much as feasible by figuring it out exactly like we did in the previous example. Work out all simple equations step by step.
x=-13169-(-120)6
x=-132896
You can figure out x by taking the negative and positive square roots.
x=-13+176 x=-13-176
x=46 x=-306
x=23 x=-5
3*(2/3)2 + (13*2/3) - 10 = 0
4/3 + 26/3 - 10 = 0
30/3 - 10 = 0
10 - 10 = 0
Or
3*-52 + (13*-5) - 10 = 0
75 - 65 - 10 =0
And that's it! You will solve quadratic equations like nobody’s business with some patience and practice!
Given this synopsis of quadratic equations and their fundamental formula, students can now take on this complex topic with confidence. By starting with this straightforward explanation, children gain a strong understanding before taking on further complicated ideas later in their studies. |
# Ratio and Proportion Worksheet
Ratio and proportions are essential for kids when it comes to learning mathematics. Also, the ratio and proportion worksheet allow them to understand measurements effectively. They help introduce the idea of ratios to children, which they will later use when learning about algebra. The ratio compares 2 quantities of a similar kind, represented as a:b or a/b. When 2 or more ratios are equal, they are in a proportion. Ratio and proportion worksheets pdf with answers help teach kids to apply these concepts in everyday lives such as distances, weights, time and heights. Ratio and proportion is a vital foundations for other mathematical concepts, and by practising ratio and proportion worksheet pdf, kids can master this concept.
Suggested Article: Fractions Worksheets
## What is Ratio?
A ratio compares 2 quantities of a similar kind and represents how much of a quantity is in other quantities. It is classified into 2 types, namely, part to part ratio and part to whole ratio. A part to part ratio represents how 2 groups are associated. For instance, the ratio of girls and boys in a sports team is 20:15. A part to whole ratio represents the association between a particular group to an entire lot. For instance, out of 20 people, 10 love chocolates. Hence, the part to whole ratio is 10:20, which implies that 10 people out of 20 people love chocolates. To help children learn this concept, introduce them to BYJU’S ratio and proportion worksheet.
Also, refer to Prime Factorization Worksheets.
## What is Proportion?
A proportion is a comparison between two numerals. It is also referred to as a share, part or a number that is contemplated compared to the whole. When 2 ratios are equal, they are said to be in proportion. Proportions are represented by :: or =. The two types of proportions are direct and inverse proportion. Also, provide ratio and proportion worksheets pdf with answers to kids, so that they can practise.
Suggested Article: Ratio and Proportion Worksheets
### Properties of Proportion
A proportion is a situation in which 2 ratios are equal. These are the properties of proportion.
• Addendo – If p : q = x : y, then the value of the ratio is p + q : x + y
• Subtrahendo – If p : q = x : y, then value of the ratio is p – q : x – y
• Dividendo – If p : q = x : y, then p – q : q = x – y : y
• Componendo – If p : q = x : y, then p + q : q = x + y : y
• Alternendo – If p : q = x : y, then p : x = q: y
• Invertendo – If p : q = x : y, then q : p = y : x
• Componendo and dividendo – If p : q = x : y, then p + q : p – q = x + y : x – y
## A List of Worksheets of Ratio and Proportion
The ratio and proportion worksheet helps them grasp the concept of ratio and proportions, which is vital for other maths topics, too. This worksheet of ratio and proportion also helps them apply ratios to real-life situations. Ratio and proportion worksheet pdf teaches the little learners how to work with measurements, understand equivalencies and use fractions. Learning about ratio and proportions can be fun. By the time kids reach Classes 1, 2 and 3, provide them with BYJU’S ratio and proportion word problems worksheet with answers pdf. Provided below is a comprehensive list of ratio and proportion worksheet for the little ones to practise.
### Ratio and Proportion Word Problems Worksheet with Answers Pdf
For more Kids Learning activities similar to Factors Worksheets visit BYJU’S website.
## Frequently Asked Questions on Ratio and Proportion Worksheet
Q1
### Define ratio in maths.
A ratio compares 2 quantities of a similar kind and represents how much of a quantity is in other quantities. The ratio is classified into 2 types, namely part to part ratio and part to whole ratio. A part to part ratio represents how 2 groups are associated. For instance, the ratio of girls and boys in a sports team is 20:15.
Q2
### Define proportion in maths.
A proportion is a comparison between two numerals. It is also referred to as a share, part or a number that is contemplated compared to the whole. When 2 ratios are equal, they are said to be in proportion. Proportions are represented by :: or =. Types of proportions are direct and inverse proportion.
Q3
### What are the properties of proportion?
The properties of proportion are addendo, subtrahendo, dividendo, componendo, alternendo, invertendo, componendo and dividendo. |
# Hardest maths questions – find a^4 + b^4 + c^4
Okay this question is from the 2006
competition and only 1% of students got this one right, although this one isn’t
multiple-choice so that does make it harder. All right, if you haven’t tried
this one before then pause the video here and have a go. All right so,
simultaneous equations… two main methods to deal with these: one is substitution and
the other one is elimination. Now I did actually try this one out with
substitution but it quickly got very messy; you don’t want to use substitution
for this one. Instead we’re just going to sort of take each equation… I’m going to
call this equation one, equation two and three, and we’ll just see what we can get
out of this. So the first thing that I thought of was, what if we squared both
sides of equation one? Because then we’re going to get some “a squared”s and “b
squared”s and maybe that’s going to be able to be combined with equation two
somehow. So let’s work out equation one squared. So this may not exactly be the
same as elimination method that you’re familiar with but it’s basically
the same sort of thing. Okay so square both sides of this. On the
left, if you think of it like… if you’re having trouble expanding it, then
maybe write it out like this first. So we’ve got a squared from there and then
you’ll have an ab, then there’ll also be an ab from those and, you know, there’s gonna
be bc and bc again and so on, so you’re gonna get a squared + b squared +
c squared and then there’s two of each of the cross terms, I guess. Okay well that’s really good because
then we can subtract equation two from that and that’s going to get rid of that,
so I might just cross that out and change that to six on the right, so if we
just subtract that from both sides. And then let’s divide both sides by 2, so
that gives us ab + bc + ac is 3, so that might be useful. Now
another thing we could do is we could multiply equation 1 by equation 2
because that’s going to give us like an a cubed and b cubed and so on, so we might
be able to combine that with equation 3 somehow. So equation 1 times equation 2:
on the left we’ve got a + b + c times a² + b² + c² .
That looked like a 4. And that equals that times that. So when we expand this
we’re going to get like the a cubed from there and the b cubed and so on and then
we’re also gonna get ab² plus ac² from there, and we’re gonna get a ba²,
whoops ba², plus bc² and a ca² and cb². So that’s a bit annoying, but at least
now we can subtract equation 3 from that. So if I just subtract equation 3,
get rid of that, and subtract 22 from there, so we get that that is 18, so that
might be useful later on. Now what else we could do is with this equation we
might be able to do something with that. I might call that equation 4, and then we
could try something like equation 1 times equation 4 and see what happens.
With this one I just sort of tried a lot of things and not everything was
helpful, but most of it was. One of the things I tried that wasn’t helpful I
think was I tried multiplying equation 1 by equation 3. There might be a way to do it
with that but I didn’t… it didn’t lead to anything for me. Let’s just try out
everything we can think of. So equation 1 times equation 4, so on the left we’ve
got the a + b + c times that and then on the right we’ve got the 4
times 3, so 12. So that is going to make a²b plus…
we’ll have an abc plus a²c plus ab² plus b²c
plus another abc plus another abc from there, plus bc² and
ac² from there, and I’m running out of space again. Okay so we’ve got 3abc from
those and then plus the rest of this stuff, which is actually the same as what
we had here, right? Because we’ve got the a²b which was there and a²c
which was there and so on, and we know that all of that is 18, so
we’ve got 3abc + 18=12, so that’s good because we can work out abc.
If we shift that over we get a negative 6 and then divide by 3 and we get
negative 2, so that might come in handy. Another thing that we could do is we
could square equation 4, see what we can come up with with that. So equation 4
squared, so that squared. You’re gonna get like the square of each one and then
plus the cross terms, and there’s gonna be two of each, like because it’s pretty
much the same as squaring equation 1, if you remember that had the squares and
then 2 times the cross terms. So… oops, where’s equation 4 gone? So by the cross terms I mean like ab
times bc and so on. So ab²c and we’re gonna get a²bc and abc².
Okay and then on the right we’ve got 9. Okay. Now with this bit you
might notice that we can factorise that because we’ve got abc in each term, so
that’s abc times a + b + c. Well we know a + b + c,
that was 4, and we know abc is -2, so we’ve got 4 times -2,
that bit’s -8, and then times that 2 is -16,
so like that whole thing is negative 16. If we shift that over we’ll get 25, so
we know that all of that is 25. Okay and then another thing that we could do
because we’re looking for a to the fourth so I sort of thought of two ways
that we might be able to get a to the fourth. We could do equation 1 times
equation 3; that didn’t work out too well. Well maybe it does, I just couldn’t
see a way to make it work out. The other way that you could do it is you can
square equation 2 because that’s going to give you an a to the fourth and so on,
so let’s try that. So that will give you a⁴ + b⁴ + c⁴ and again
2 times the cross terms, so a²c², b²c² and a²b², and that was
10 so 10 squared is 100. Okay well now we’ve got that that is 25,
so that whole thing is 50, we can shift that across, and we get that that is 50
(because 100 – 50) so that is the answer to that one. Okay so I went back and had
another go at substitution method and worked it out this time. So I don’t want
to spend too long on this but let me just quickly go through what I did. So I
took that equation and solved it for c and then subbed it into equation 2, and played
around with that and got to there, and then subbed it in to equation 3 as well, and that
was annoying to expand that cubed but yeah you can do that. Some stuff cancels.
And then what I did was instead of solving this for b which would be kind
of like the standard way to do substitution. You can do that
and then you get like… it’s a quadratic in b and so it’s like plus/minus square
roots of stuff and that’s where I gave up before because it was a bit too messy.
But instead of doing that what I did this time was I decided to solve this
for b² and then sub that in there and also in there and see what happens.
And it turns out that a lot of stuff cancels and you just end up with this
cubic. Now if you just try a few values like often you’ll get solutions for, you know,
if you try a=1 and a=-1 and 2 and -2 and so on,
you might find some solutions. So I tried a=2 and that worked; that gave 0. So
what you can do then is divide that by a minus 2 because you know that that’s a
factor of it, so if you divide that by a minus 2 you get a² – 2a – 1
so either you’ve got a=2 or this equals 0, so solve that just using
quadratic formula you would get a equals 1 plus or minus root 2. So because of the
symmetry of the situation, where you’ve got like a and b and c are essentially
interchangeable, like if you swap a and b the system of equations are still
like the same thing. That means that these solutions for a are also solutions
for b and c, so it might be like a equals b… sorry a=2, b=1 + root 2,
and c=1 – root 2. So then to work out a⁴ + b⁴ + c⁴, you just raise all those things to the power 4. So I
did that. If you expand those then a lot of stuff cancels, like you’ve got the
+4√2 there and the -4√2 there. The stuff that’s left
after you’ve cancelled… you’ve got 1 and 12 and 4 that makes 17 and then
you’ve also got 17 there and the 2⁴ is 16, so you add all those
up and you get 50. So that’s another way that you can do it. The solutions
from the booklet actually did it sort of a way that was in between those two,
because what they did was they started out the same as what I was doing before, with
like elimination basically, but once they worked out abc and ab + bc + ac,
they used that to find the coefficients of the cubic. So they went like straight
from that to this point, and then did the same thing from there. So there’s like
three different methods so far that I know of that you can use to solve this.
Okay let me know if you have any questions and stay tuned for the next
video. You might also like to check out the inheritance question that I did for
the junior students because that was a senior question as well and only 8%
of the seniors got that one right. |
How do you solve the rational equation (4-8x)/(1-x+4)=8/(x+1)?
Jun 4, 2018
$\frac{4 - 8 x}{1 - x + 4} = \frac{8}{x + 1}$
$\frac{4 - 8 x}{5 - x} = \frac{8}{x + 1}$
Cross multiply to remove the fractions
$\left(4 - 8 x\right) \left(x + 1\right) = 8 \left(5 - x\right)$
Expand the brackets
$4 x + 4 - 8 {x}^{2} - 8 x = 40 - 8 x$
$4 - 4 x - 8 {x}^{2} = 40 - 8 x$
Add $8 {x}^{2}$ to both sides
$4 - 4 x = 8 {x}^{2} - 8 x + 40$
add $4 x$ to both sides
4=8x^2-4x+40#
subtract 4 from both sides
$8 {x}^{2} - 4 x + 36 = 0$
Divide both sides by 4
$2 {x}^{2} - x + 9 = 0$
$x = \frac{1 \setminus \pm \sqrt{1 - 4 \times 2 \times 9}}{2 \times 2}$
$x = \frac{1 \setminus \pm \sqrt{- 71}}{4}$ |
# A card is drawn from a pack of 52 cards. The probability of getting an ace is 1/52. True or False?
By BYJU'S Exam Prep
Updated on: September 25th, 2023
The statement that if a card is drawn from a pack of 52 cards, the probability of getting an ace is 1/52 is false. There are a total of 52 cards in a pack and in each pack, there are 4 aces. Hence, the probability of getting an ace in a deck of cards will be 1/13. A probability is always calculated in the range of 0 to 1. In this range, 0 denotes an event that is impossible and 1 is an event that is possible.
## The Probability of Getting an Ace from a Deck of Cards
By definition, a probability is a chance of an event happening or not happening. The probability of an event is proportional to the total number of outcomes divided by number of favorable outcomes. The formula to calculate the probability of an event is –
P(E) = Total number of outcomes/Number of favorable outcomes
### Solution
Now, to understand the probability of drawing an ace in a deck of cards, we will consider the following.
Total number of cards in a deck = 52
Total number of aces in a deck of cards = 4
Now, that we have clarity on the number of cards, we can find the probability by following the formula mentioned above.
Probability of getting an ace in a deck of cards = 52/4 = 1/13
Thus, the statement that the probability of getting an ace in a deck of cards is 1/52 is false.
Summary:
## A card is drawn from a pack of 52 cards. The probability of getting an ace is 1/52. True or False?
The statement is false. In a deck of cards, there are 52 cards and the total number of aces is 4. Hence, the probability will be 4/52 or 1/13 when taking the total number of outcomes and the number of favorable outcomes into consideration.
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# An Inequality in Triangle
For $a,b,c\gt 0$ the sides of a triangle
$W(a,b,c)=a^3b+b^3c+c^3a-a^2b^2-b^2c^2-c^2a^2\ge 0$
and $W(a, b, c) = 0$ if and only if $a = b = c.$
### Proof
Since $a,$ $b$ and $c$ are sides of a triangle, we have $a + b\ge c,$ $b + c \ge a$ and $c + a\ge b.$ Introducing change of variables $a = x + y,$ $b = y + z,$ $c = z + x,$ using that $x, y,z\ge 0,$ and substituting in the equation, we obtain
$W=x^3y+y^3z+z^3x-xyz(x+y+z).$
The main tool in the proof will be the weighted arithmetic mean-geometric means inequality:
$\displaystyle \bigg(\frac{w_{1}F_{1}+w_{2}F_{2}+w_{3}F_{3}}{w_{1}+w_{2}+w_{3}}\bigg)^{w}\ge F^{w_{1}}_{1}F^{w_{2}}_{2}F^{w_{3}}_{3}.$
where $w=w_{1}+w_{2}+w_{3}$ and equality occurs only if $F_{1}=F_{2}=F_{3},$ for positive weights.
The first step is to establish
$\displaystyle \frac{w_{1}x^3y+w_{2}y^3z+w_{3}z^3x}{w_{1}+w_{2}+w_{3}}\ge x^2yz$
using appropriate weights:
$\begin{cases} 3w_{1}+w_{3}&=2w_{1}+2w_{2}+2w_{3}\\ w_{1}+3w_{2}&=w_{1}+w_{2}+w_{3}\\ w_{2}+3w_{3}&=w_{1}+w_{2}+w_{3}. \end{cases}$
Hence, for $w_{1}=4,$ $w_{2}=1,$ $w_{3}=2,$ we obtain
$\displaystyle \frac{4}{7}x^3y+\frac{1}{7}y^3z+\frac{2}{7}z^3x\ge x^2yz.$
Permuting the variables twice, we obtain two more inequalities. Adding all three inequalities, we obtain the desired one. The equality occurs only if $x^3 y = y^3z = z^3x,$ which is equivalent to $x = y = z,$ provided $xyz\ne 0.$ Note that if at least one variable vanishes, say $x = 0,$ and equality occurs, then $y^3z = 0$ and either $a = x + y = 0$ or $c = z + x = 0,$ contradicting our assumption.
### Acknowledgment
The inequality above has been published by Maxim Arnold and Vadim Zharnitsky in The American Mathematical Monthly (April 2015, 378-379) as an ancillary lemma in a bigger article. Arnold and Zharnitsky found the inequality useful. I lifted it from their article, being charmed with the proof. |
# How do you simplify 36/sqrt15?
##### 2 Answers
Jul 4, 2017
$\frac{36}{\sqrt{15}} = \frac{12 \sqrt{15}}{5}$
#### Explanation:
$\frac{36}{\sqrt{15}}$
= $\frac{36}{\sqrt{15}} \times \frac{\sqrt{15}}{\sqrt{15}}$
= $\frac{36 \times \sqrt{15}}{\sqrt{15} \times \sqrt{15}}$
= $\frac{36 \times \sqrt{15}}{15}$
= $\frac{{\cancel{36}}^{12} \times \sqrt{15}}{{\cancel{15}}^{5}}$
= $\frac{12 \sqrt{15}}{5}$
Jul 4, 2017
$\frac{12 \sqrt{15}}{5}$
#### Explanation:
To simplify a fraction with a radical, we want to make sure that there is no radical on the bottom. In order to do this, we can multiply the fraction by $\frac{\sqrt{15}}{\sqrt{15}}$"
$\frac{36}{\sqrt{15}} \cdot \frac{\sqrt{15}}{\sqrt{15}}$
$= \frac{36 \cdot \sqrt{15}}{\sqrt{15} \cdot \sqrt{15}}$
$= \frac{36 \sqrt{15}}{15}$
We have one last step to simplify this expression. Notice that the top and bottom both share a factor of 3 now. So, we need to divide both the top and bottom by 3:
$\frac{36 \sqrt{15}}{15} \div \frac{3}{3}$
$= \frac{36 \sqrt{15} \div 3}{15 \div 3}$
$= \frac{12 \sqrt{15}}{5}$
Final Answer |
## University Calculus: Early Transcendentals (3rd Edition)
Published by Pearson
# Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 338: 46
#### Answer
$$\int^{3}_{0}\frac{ydy}{\sqrt{5y+1}}=\frac{36}{25}$$
#### Work Step by Step
$$A=\int^{3}_{0}\frac{ydy}{\sqrt{5y+1}}$$ We set $u=\sqrt{5y+1}$, which means $u^2=5y+1$ and $y=\frac{u^2-1}{5}$ $$du=\frac{(5y+1)'}{2\sqrt{5y+1}}dy=\frac{5}{2\sqrt{5y+1}}dy$$ $$\frac{dy}{\sqrt{5y+1}}=\frac{2}{5}du$$ - For $y=3$, we have $u=\sqrt{5\times3+1}=4$ - For $y=0$, we have $u=\sqrt{5\times0+1}=1$ Therefore, $$A=\frac{2}{5}\int^{4}_{1}\frac{u^2-1}{5}du=\frac{2}{25}\int^{4}_{1}(u^2-1)du$$ $$A=\frac{2}{25}\Big(\frac{u^3}{3}-u\Big)\Big]^{4}_{1}$$ $$A=\frac{2}{25}\Big((\frac{64}{3}-4)-(\frac{1}{3}-1)\Big)$$ $$A=\frac{2}{25}\Big(\frac{52}{3}-(-\frac{2}{3})\Big)$$ $$A=\frac{2}{25}\times\frac{54}{3}=\frac{36}{25}$$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. |
# SAT Math : How to find the slope of parallel lines
## Example Questions
### Example Question #101 : Lines
Consider line c to be y= -4x - 7. Which is the reflection of line c across the x-axis?
y = 4x – 7
y = –4x + 7
y = 4x + 7
y = (1/4)x + 7
y = 4x + 7
Explanation:
A line reflected across the x-axis will have the negative value of the slope and intercept. This leaves y= 4x + 7.
### Example Question #1 : Parallel Lines
If line J passes through the points (1, 3) and (2, 4) and line K passes through (0, x) and (10, 3), what would be the value of x in order for lines J and K to be parallel?
–13
13
1
7
–7
–7
Explanation:
Find the slope of line J, (4 – 3)/(2 – 1) = 1
Now use this slope in for the equation of line K of the form y = mx + b for the other point (10, 3)
3 = 10 + b → b = –7
So for the point (0, X) → X = 0 – 7,
so x = –7 when these two lines are parallel.
### Example Question #102 : Lines
Line is represented by the equation .
If line passes through the points and , and if is parallel to , then what is the value of ?
Explanation:
We are told that lines l and m are parallel. This means that the slope of line m must be the same as the slope of l. Line l is written in the standard form of Ax + By = C, so its slope is equal to –A/B, or –2/–3, which equals 2/3. Therefore, the slope of line m must also be 2/3.
We are told that line m passes through the points (1, 4) and (2, a). The slope between these two points must equal 2/3. We can use the formula for the slope between two points and then set this equal to 2/3.
slope = (a – 4)/(2 – 1) = a – 4 = 2/3
a – 4 = 2/3
Multiply both sides by 3:
3(a – 4) = 2
3a – 12 = 2
3a = 14
a = 14/3
### Example Question #1 : How To Find The Slope Of Parallel Lines
What is the slope of a line parallel to the line: -15x + 5y = 30 ?
3
-15
1/3
30
3
Explanation:
First, put the equation in slope-intercept form: y = 3x + 6. From there we can see the slope of this line is 3 and since the slope of any line parallel to another line is the same, the slope will also be 3.
### Example Question #11 : Lines
What is the slope of any line parallel to –6x + 5y = 12?
6/5
12
5/6
12/5
6
6/5
Explanation:
This problem requires an understanding of the makeup of an equation of a line. This problem gives an equation of a line in y = mx + b form, but we will need to algebraically manipulate the equation to determine its slope. Once we have determined the slope of the line given we can determine the slope of any line parallel to it, becasue parallel lines have identical slopes. By dividing both sides of the equation by 5, we are able to obtain an equation for this line that is in a more recognizable y = mx + b form. The equation of the line then becomes y = 6/5x + 12/5, we can see that the slope of this line is 6/5.
### Example Question #11 : Parallel Lines
What is the slope of a line that is parallel to the line 11x + 4y - 2 = 9 – 4x ?
Explanation:
We rearrange the line to express it in slope intercept form.
Any line parallel to this original line will have the same slope.
### Example Question #1 : How To Find The Slope Of Parallel Lines
In the standard (x, y) coordinate plane, what is the slope of a line parallel to the line with equation ?
Explanation:
Parallel lines will have equal slopes. To solve, we simply need to rearrange the given equation into slope-intercept form to find its slope.
The slope of the given line is . Any lines that run parallel to the given line will also have a slope of .
### Example Question #11 : Lines
What is the slope of a line that is parallel to the line ?
Explanation:
Parallel lines have the same slope. The question requires you to find the slope of the given function. The best way to do this is to put the equation in slope-intercept form (y = mx + b) by solving for y.
First subtract 6x on both sides to get 3y = –6x + 12.
Then divide each term by 3 to get y = –2x + 4.
In the form y = mx + b, m represents the slope. So the coefficient of the x term is the slope, and –2 is the correct answer.
### Example Question #1 : How To Find The Slope Of Parallel Lines
Solve each problem and decide which is the best of the choices given.
What is the slope of a line parallel to the following equation of a line?
Explanation:
Parallel lines have the same slope. So to find the slope of the given equation by getting the equation in slope-intercept form where is the slope of the line.
First, subtract both sides by giving you,
.
A parallel line would have a slope of .
### Example Question #2 : How To Find The Slope Of Parallel Lines
Which of the following lines is parallel to ? |
Here you will learn how to solve quadratic equation by factorisation with examples.
Let’s begin –
Step 1. Splitting of middle term :
(i) If the product of a and c = +ac
then we have to choose two factors ac whose sum is equal to b.
(ii) If the product of a and c = -ac
then we have to choose two factors of ac whose difference is equal to b.
Step 2 : Let the factors of $$ax^2 + bx + c$$ be (dx + e) and (fx + g)
$$\implies$$ (dx + e) (fx + g) = 0
Either dx + e = or fx + g = 0
$$\implies$$ x = -$$e\over d$$ or x = -$$g\over f$$
Example : Solve the equation : $$2x^2 – 11x + 12$$ = 0.
Solution : We have, $$2x^2 – 11x + 12$$ = 0
$$2x^2 – 8x – 3x + 12$$ = 0
2x (x – 4) – 3 (x – 4) = 0
(x – 4) (2x – 3) = 0
$$\implies$$ x – 4 = 0 or 2x – 4 = 0
$$\implies$$ x = 4 or x = $$3\over 2$$
Hence, x = 4 and x = $$3\over 2$$ are the roots of the given equation.
Example : Solve the equation : $$3x^2 – 14x – 5$$ = 0.
Solution : We have, $$3x^2 – 14x – 5$$ = 0
$$3x^2 – 15x + x – 5$$ = 0
3x (x – 5) + 1 (x – 5) = 0
(x – 5) (3x + 1) = 0
$$\implies$$ x – 5 = 0 or 3x + 1 = 0
$$\implies$$ x = 5 or x = -$$1\over 3$$
Hence, x = 5 and x = -$$1\over 3$$ are the roots of the given equation. |
# 2018 AMC 10A Problems/Problem 21
## Problem
Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$-plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$
## Solution 1
Substituting $y=x^2-a$ into $x^2+y^2=a^2$, we get $$x^2+(x^2-a)^2=a^2 \implies x^2+x^4-2ax^2=0 \implies x^2(x^2-(2a-1))=0$$ Since this is a quartic, there are 4 total roots (counting multiplicity). We see that $x=0$ always at least one intersection at $(0,-a)$ (and is in fact a double root).
The other two intersection points have $x$ coordinates $\pm\sqrt{2a-1}$. We must have $2a-1> 0,$ otherwise we are in the case where the parabola lies entirely above the circle (tangent to it at the point $(0,a)$). This only results in a single intersection point in the real coordinate plane. Thus, we see $\boxed{\textbf{(E) }a>\frac12}$.
(projecteulerlover)
## Solution 2
$[asy] Label f; f.p=fontsize(6); xaxis(-2,2,Ticks(f, 0.2)); yaxis(-2,2,Ticks(f, 0.2)); real g(real x) { return x^2-1; } draw(graph(g, 1.7, -1.7)); real h(real x) { return sqrt(1-x^2); } draw(graph(h, 1, -1)); real j(real x) { return -sqrt(1-x^2); } draw(graph(j, 1, -1)); [/asy]$
Looking at a graph, it is obvious that the two curves intersect at (0, -a). We also see that if the parabola goes 'in' the circle. Then, by going out of it (as it will), it will intersect five times, an impossibility. Thus we only look for cases where the parabola becomes externally tangent to the circle. We have $x^2 - a = -\sqrt{a^2 - x^2}$. Squaring both sides and solving yields $x^4 - (2a - 1)x^2 = 0$. Since x = 0 is already accounted for, we only need to find 1 solution for $x^2 = 2a - 1$, where the right hand side portion is obviously increasing. Since a = 1/2 begets x = 0 (an overcount), we have $\boxed{\textbf{(E) }a>\frac12}$ is the right answer.
Solution by JohnHankock
## Solution 3
This describes a unit parabola, with a circle centered on the axis of symmetry and tangent to the vertex. As the curvature of the unit parabola at the vertex is 2, the radius of the circle that matches it has a radius of $\frac{1}{2}$. This circle is tangent to an infinitesimally close pair of points, one on each side. Therefore, it is tangent to only 1 point. When a larger circle is used, it is tangent to 3 points because the points on either side are now separated from the vertex. Therefore, $\boxed{\textbf{(E) }a>\frac12}$ is correct.
$QED \blacksquare$
## Solution 4
Notice, the equations are of that of a circle of radius a centered at the origin and a parabola translated down by a units. They always intersect at the point $(0, a)$, and they have symmetry across the y-axis, thus, for them to intersect at exactly 3 points, it suffices to find the y solution.
First, rewrite the second equation to $y=x^2-a\implies x^2=y+a$ And substitute into the first equation: $y+a+y^2=a^2$ Since we're only interested in seeing the interval in which a can exist, we find the discriminant: $1-4a+4a^2$. This value must not be less than 0 (It is the square root part of the quadratic formula). To find when it is 0, we find the roots: $$4a^2-4a+1=0 \implies a=\frac{4\pm\sqrt{16-16}}{8}=\frac{1}{2}$$ Since $\lim_{a\to \infty}(4a^2-4a+1)=\infty$, our range is $\boxed{\textbf{(E) }a>\frac12}$.
Solution by ktong
## Solution 5 (Cheating with Answer Choices)
Simply plug in $a = 0, \frac{1}{2}, \frac{1}{4}, 1$ and sole the systems. (This shouldn't take too long.) And then realize that only $a=1$ yields three real solutions for $x$, so we are done and the answer is $\boxed{\textbf{(E) }a>\frac12}$.
~ ccx09
## Solution 6 (Calculus Needed)
In order to solve for the values of $a$, we need to just count multiplicities of the roots when the equations are set equal to each other: in other words, take the derivative. We know that $\sqrt{x^2 - a^2} = x^2 - a$. Now, we take square of both sides, and rearrange to obtain $x^4 - (2a + 1)x^2 + 2a^2 = 0$. Now, we make take the second derivative of the equation to obtain $6x^2 - (4a + 2) = 0$. Now, we must take discriminant. Since we need the roots of that equation to be real and not repetitive (otherwise they would not intersect each other at three points), the discriminant must be greater than zero. Thus,
$b^2 - 4ac > 0 \rightarrow 0 - 4(6)(4a + 2) > 0 \rightarrow a > \frac{1}{2}$ The answer is $\boxed{\textbf{(E) }a>\frac12}$ and we are done.
~awesome1st |
# 15-Apr-15Created by Mr. Lafferty1 Statistics www.mathsrevision.com Mode, Mean, Median and Range Semi-Interquartile Range ( SIQR ) Nat 5 Quartiles Boxplots.
## Presentation on theme: "15-Apr-15Created by Mr. Lafferty1 Statistics www.mathsrevision.com Mode, Mean, Median and Range Semi-Interquartile Range ( SIQR ) Nat 5 Quartiles Boxplots."— Presentation transcript:
15-Apr-15Created by Mr. Lafferty1 Statistics www.mathsrevision.com Mode, Mean, Median and Range Semi-Interquartile Range ( SIQR ) Nat 5 Quartiles Boxplots – Five Figure Summary Full Standard Deviation Sample Standard Deviation Exam questions
Nat 5 15-Apr-15Created by Mr Lafferty Maths Dept Starter Questions www.mathsrevision.com xoxo 42 o
Nat 5 15-Apr-15Created by Mr Lafferty Maths Dept Statistics Learning Intention Success Criteria 1.Understand the terms 1.Understand the terms mean, range, median and mode. 1.We are revising the terms mean, median, mode and range. 2.To be able to calculate mean, range, mode and median. www.mathsrevision.com Averages
15-Apr-15Created by Mr Lafferty Maths Dept Finding the mode The mode or modal value in a set of data is the data value that appears the most often. For example, the number of goals scored by the local football team in the last ten games is: What is the modal score? Is it possible to have more than one modal value? Is it possible to have no modal value? Yes 2,1,2,0, 2,3,1,2,1.2,1,2,0, 2,3,1,2,1. 2. Statistics www.mathsrevision.com Nat 5
15-Apr-15Created by Mr Lafferty Maths Dept The mean The mean is the most commonly used average. To calculate the mean of a set of values we add together the values and divide by the total number of values. For example, the mean of 3, 6, 7, 9 and 9 is Mean = Sum of values Number of values Statistics www.mathsrevision.com Nat 5
15-Apr-15Created by Mr Lafferty Maths Dept Finding the median The median is the middle value of a set of numbers arranged in order. For example, Find the median of 10,7,9,12,7,8,6, Write the values in order: 6,7, 8,9,10,12. The median is the middle value. Statistics www.mathsrevision.com Nat 5
15-Apr-15Created by Mr Lafferty Maths Dept Finding the median When there is an even number of values, there will be two values in the middle. For example, Find the median of 56, 42, 47, 51, 65 and 43. The values in order are: There are two middle values, 47 and 51. 42,43,47,51,56,65. 47 + 51 2 = 98 2 = 49 Statistics www.mathsrevision.com Nat 5
15-Apr-15Created by Mr Lafferty Maths Dept Finding the range The range of a set of data is a measure of how the data is spread across the distribution. To find the range we subtract the lowest value in the set from the highest value. Range = Highest value – Lowest value When the range is large; the values vary widely in size. When the range is small; the values are similar in size. Statistics www.mathsrevision.com Nat 5
15-Apr-15Created by Mr Lafferty Maths Dept Here are the high jump scores for two girls in metres. Joanna1.621.411.351.201.15 Kirsty1.591.451.411.301.30 Find the range for each girl’s results and use this to find out who is consistently better. Joanna’s range = 1.62 – 1.15 = 0.47 Kirsty’s range = 1.59 – 1.30 = 0.29 The range Statistics www.mathsrevision.com Nat 5 Kirsty is consistently better !
15-Apr-15Created by Mr. Lafferty Maths Dept. www.mathsrevision.com No of Bulbs (c) Freq.(f) Example : This table shows the number of light bulbs used in people’s living rooms 7 5 1 20 2 5 1 2 3 4 5 Totals Frequency Tables Working Out the Mean Adding a third column to this table will help us find the total number of bulbs and the ‘Mean’. 7 x 1 = 7 5 x 3 = 15 1 x 5 = 5 2 x 4 = 8 5 x 2 = 10 45 (f) x (B) Nat 5
15-Apr-15Created by Mr Lafferty Maths Dept Now try N5 TJ Ex 11.1 Ch11 (page 104) www.mathsrevision.com Statistics Nat 5 Averages
15-Apr-15Created by Mr. Lafferty12 www.mathsrevision.com Lesson Starter Q1. Q2.Calculate sin 90 o Q3.Factorise 5y 2 – 10y Q4. A circle is divided into 10 equal pieces. Find the arc length of one piece of the circle if the radius is 5cm. Nat 5
15-Apr-15Created by Mr. Lafferty Maths Dept. www.mathsrevision.com Learning Intention Success Criteria 1.We are learning about Quartiles. 1.Understand the term Quartile. Quartiles Nat 5 2.Be able to calculate the Quartiles for a set of data.
15-Apr-15Created by Mr Lafferty Maths Dept Statistics www.mathsrevision.com Quartiles :Splits a dataset into 4 equal lengths. Nat 5 Quartiles 25% Q 1 Q 2 Q 3 25%50%75% Median
15-Apr-15Created by Mr Lafferty Maths Dept Statistics www.mathsrevision.com Nat 5 Quartiles Note : Dividing the number of values in the dataset by 4 and looking at the remainder helps to identify quartiles. R1 means to can simply pick out Q 2 (Median) R2 means to can simply pick out Q 1 and Q 3 R3 means to can simply pick out Q 1, Q 2 and Q 3 R0 means you need calculate them all
Quartiles 15-Apr-15Created by Mr Lafferty Maths Dept Statistics www.mathsrevision.com Example 1 :For a list of 9 numbers find the SIQR 3, 3, 7, 8, 10, 9, 1, 5, 9 2 numbers2 numbers2 numbers 2 numbers Q1Q2Q3 The quartiles are Q 1 :the 2 nd and 3 rd numbers Q 2 :the 5 th number Q 3 :the 7 th and 8 th number. Nat 5 1 No. 3 7 9 9 ÷ 4 = 2R1 1 3 3 5 7 8 9 9 10 Semi-interquartile Range (SIQR) = ( Q 3 – Q 1 ) ÷ 2 = ( 9 – 3 ) ÷ 2 = 3
Quartiles 15-Apr-15Created by Mr Lafferty Maths Dept Statistics www.mathsrevision.com Example 3 :For the ordered list find the SIQR. 3, 6, 2, 10, 12, 3, 4 1 number1 number1 number1 number Q1Q2Q3 The quartiles are Q 1 :the 2 nd number Q 2 :the 4 th number Q 3 :the 6 th number. 7 ÷ 4 = 1R3 Nat 5 3 4 10 Semi-interquartile Range (SIQR) = ( Q 3 – Q 1 ) ÷ 2 = ( 10 – 3 ) ÷ 2 = 3.5 2 3 3 4 6 10 12
Nat 5 15-Apr-15Created by Mr Lafferty Maths Dept Now try N5 TJ Ex 11.2 Ch11 (page 106) www.mathsrevision.com Statistics Averages
15-Apr-15Created by Mr. Lafferty19 www.mathsrevision.com Lesson Starter In pairs you have 3 minutes to explain the various steps of factorising. Nat 5
15-Apr-15Created by Mr. Lafferty Maths Dept. www.mathsrevision.com Learning Intention Success Criteria 1.We are learning about Semi-Interquartile Range. 1.Understand the term Semi- Interquartile Range. Semi-Interquartile Range Nat 5 2.Be able to calculate the SIQR.
Nat 5 The range is not a good measure of spread because one extreme, (very high or very low value can have a big effect). Another measure of spread is called the Semi - Interquartile Range and is generally a better measure of spread because it is not affected by extreme values. Inter-Quartile Range www.mathsrevision.com
Upper Quartile = 10 Q3Q3 Lower Quartile = 4 Q1Q1 Median = 8 Q2Q2 3, 4, 4, 6, 8, 8, 8, 9, 10, 10, 15, Finding the Semi-Interquartile range. 6, 3, 9, 8, 4, 10, 8, 4, 15, 8, 10 Order the data Inter- Quartile Range = (10 - 4)/2 = 3 Example 1: Find the median and quartiles for the data below.
12, 6, 4, 9, 8, 4, 9, 8, 5, 9, 8, 10 4, 4, 5, 6, 8, 8, 8, 9, 9, 9, 10, 12 Order the data Inter- Quartile Range = (9 - 5½) = 1¾ Example 2: Find the median and quartiles for the data below. Lower Quartile = 5½ Q1Q1 Upper Quartile = 9 Q3Q3 Median = 8 Q2Q2 Finding the Semi-Interquartile range.
Nat 5 15-Apr-15Created by Mr Lafferty Maths Dept Now try N5 TJ Ex 11.3 Ch11 (page 108) www.mathsrevision.com Statistics
15-Apr-15Created by Mr. Lafferty25 www.mathsrevision.com Lesson Starter In pairs you have 3 minutes to come up with questions on Straight Line Theory ( Remember you needed to know the answers to the questions ) Nat 5
15-Apr-15Created by Mr. Lafferty Maths Dept. www.mathsrevision.com Learning Intention Success Criteria 1.We are learning about Boxplots and five figure summary. 1.Calculate five figure summary. Boxplots ( 5 figure Summary) Nat 5 2.Be able to construct a boxplot.
456789101112 Median Lower Quartile Upper Quartile Lowest Value Highest Value Box Whisker 130140150160170180190 Boys Girls cm Box and Whisker Diagrams. Box plots are useful for comparing two or more sets of data like that shown below for heights of boys and girls in a class. Anatomy of a Box and Whisker Diagram.
Lower Quartile = 5½ Q1Q1 Upper Quartile = 9 Q3Q3 Median = 8 Q2Q2 456789101112 4, 4, 5, 6, 8, 8, 8, 9, 9, 9, 10, 12 Example 1: Draw a Box plot for the data below Drawing a Box Plot.
Upper Quartile = 10 Q3Q3 Lower Quartile = 4 Q1Q1 Median = 8 Q2Q2 3, 4, 4, 6, 8, 8, 8, 9, 10, 10, 15, Example 2: Draw a Box plot for the data below Drawing a Box Plot. 34567891011 1213 1415
Upper Quartile = 180 QuQu Lower Quartile = 158 QLQL Median = 171 Q2Q2 Question: Stuart recorded the heights in cm of boys in his class as shown below. Draw a box plot for this data. Drawing a Box Plot. 137, 148, 155, 158, 165, 166, 166, 171, 171, 173, 175, 180, 184, 186, 186 130140150160170180190cm
2. The boys are taller on average. Question: Gemma recorded the heights in cm of girls in the same class and constructed a box plot from the data. The box plots for both boys and girls are shown below. Use the box plots to choose some correct statements comparing heights of boys and girls in the class. Justify your answers. Drawing a Box Plot. 130140150160170180190 Boys Girls cm 1. The girls are taller on average. 3. The girls show less variability in height. 4. The boys show less variability in height. 5. The smallest person is a girl 6. The tallest person is a boy
Nat 5 15-Apr-15Created by Mr Lafferty Maths Dept Now try N5 TJ Ex 11.4 Ch11 (page 109) www.mathsrevision.com Statistics
15-Apr-15Created by Mr. Lafferty Maths Dept. Starter Questions Starter Questions www.mathsrevision.com Nat 5
15-Apr-15Created by Mr. Lafferty Maths Dept. www.mathsrevision.com Learning Intention Success Criteria 1.Know the term Standard Deviation. 1. We are learning the term Standard Deviation for a collection of data. 2.Calculate the Standard Deviation for a collection of data. Nat 5 Standard Deviation For a FULL set of Data
Nat 5 15-Apr-15Created by Mr. Lafferty Maths Dept. www.mathsrevision.com Standard Deviation For a FULL set of Data The range measures spread. Unfortunately any big change in either the largest value or smallest score will mean a big change in the range, even though only one number may have changed. The semi-interquartile range is less sensitive to a single number changing but again it is only really based on two of the score.
Nat 5 15-Apr-15Created by Mr. Lafferty Maths Dept. www.mathsrevision.com Standard Deviation For a FULL set of Data A measure of spread which uses all the data is the Standard Deviation The deviation of a score is how much the score differs from the mean.
Nat 5 ScoreDeviation(Deviation) 2 70 72 75 78 80 Totals375 Example 1 :Find the standard deviation of these five scores 70, 72, 75, 78, 80. Standard Deviation For a FULL set of Data Step 1 : Find the mean 375 ÷ 5 = 75 Step 3 : (Deviation) 2 15-Apr-15Created by Mr. Lafferty Maths Dept. www.mathsrevision.com -5 -3 0 3 5 0 25 9 0 9 25 68 Step 2 : Score - Mean Step 4 : Mean square deviation 68 ÷ 5 = 13.6 Step 5 : Take the square root of step 4 √13.6 = 3.7 Standard Deviation is 3.7 (to 1d.p.)
Nat 5 Example 2 :Find the standard deviation of these six amounts of money £12, £18, £27, £36, £37, £50. Standard Deviation For a FULL set of Data Step 1 : Find the mean 180 ÷ 6 = 30 15-Apr-15Created by Mr. Lafferty Maths Dept. www.mathsrevision.com Step 2 : Score - Mean Step 3 : (Deviation) 2 Step 4 : Mean square deviation 962 ÷ 6 = 160.33 ScoreDeviation(Deviation) 2 12 18 27 36 37 50 Totals180 -18 -12 -3 6 7 20 324 144 9 36 49 400 0 962 Step 5 : Take the square root of step 4 √160.33 = 12.7 (to 1d.p.) Standard Deviation is £12.70
Nat 5 15-Apr-15Created by Mr. Lafferty Maths Dept. www.mathsrevision.com Standard Deviation For a FULL set of Data When Standard Deviation is LOW it means the data values are close to the MEAN. When Standard Deviation is HIGH it means the data values are spread out from the MEAN. MeanMean
Nat 5 15-Apr-15Created by Mr. Lafferty Maths Dept. Now try N5 TJ Ex 11.5 Q1 & Q2 Ch11 (page 111) www.mathsrevision.com Standard Deviation For a FULL set of Data
15-Apr-15Created by Mr. Lafferty Maths Dept. Starter Questions Starter Questions www.mathsrevision.com Nat 5 In pairs you have 6 mins to write down everything you know about the circle theory. Come up with a circle type of question you could be asked at National 5 Level.
15-Apr-15Created by Mr. Lafferty Maths Dept. www.mathsrevision.com Learning Intention Success Criteria 1. We are learning how to calculate the Sample Standard deviation for a sample of data. Standard Deviation For a Sample of Data Standard deviation Nat 5 1.Know the term Sample Standard Deviation. 2.Calculate the Sample Standard Deviation for a collection of data.
Nat 5 15-Apr-15Created by Mr. Lafferty Maths Dept. www.mathsrevision.com Standard Deviation For a Sample of Data In real life situations it is normal to work with a sample of data ( survey / questionnaire ). We can use two formulae to calculate the sample deviation. s = standard deviation n = number in sample ∑ = The sum of x = sample mean We will use this version because it is easier to use in practice !
Nat 5 Example 1a : Eight athletes have heart rates 70, 72, 73, 74, 75, 76, 76 and 76. 15-Apr-15Created by Mr. Lafferty Maths Dept. www.mathsrevision.com Standard Deviation For a Sample of Data Heart rate (x)x2x2 70 72 73 74 75 76 Totals 4900 5184 5329 5476 5625 5776 5776 5776 ∑x 2 = 43842 ∑x = 592 Step 2 : Square all the values and find the total Step 3 : Use formula to calculate sample deviation Step 1 : Sum all the values Q1a. Calculate the mean : 592 ÷ 8 = 74 Q1a. Calculate the sample deviation
Nat 5 Created by Mr. Lafferty Maths Dept. Heart rate (x)x2x2 80 81 83 90 94 96 100 Totals 6400 6561 6889 8100 8836 9216 9216 10000 Example 1b : Eight office staff train as athletes. Their Pulse rates are 80, 81, 83, 90, 94, 96, 96 and 100 BPM 15-Apr-15 www.mathsrevision.com Standard Deviation For a Sample of Data ∑x = 720 Q1b(ii) Calculate the sample deviation Q1b(i) Calculate the mean : 720 ÷ 8 = 90 ∑x 2 = 65218
Nat 5 15-Apr-15Created by Mr. Lafferty Maths Dept. www.mathsrevision.com Standard Deviation For a Sample of Data Q1b(iii) Who are fitter the athletes or staff. Compare means Athletes are fitter Staff Athletes Q1b(iv) What does the deviation tell us. Staff data is more spread out.
Nat 5 15-Apr-15Created by Mr. Lafferty Maths Dept. Now try N5 TJ Ex 11.5 Q3 onwards Ch11 (page 113) www.mathsrevision.com Standard Deviation For a FULL set of Data
Calculate the mean and standard deviation
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# Using the Definition of a Derivative…
Q: Use the definition of a derivative to take the derivative of f(x) = 6sqrt(x)
f ‘ (x) = lim (as h goes to 0) of [f(x+h) – f(x)] / h
So, using our given equation: f(x) = 6sqrt(x), we have:
f(x+h) = 6sqrt(x+h)
f(x) = 6sqrt(x)
So, the derivative is:
Step 1: The set-up) f ‘ (x) = limit as h goes to 0 of [6sqrt(x+h) – 6sqrt(x)]/h
Now, we need to do some simplifying before plugging in 0 for h.
The general strategy to use when you see square roots is to multiply both top and bottom of the fraction by the conjugate, which is (6sqrt(x+h) + 6sqrt(x):
Step 2: Simplifying) [6sqrt(x+h) – 6sqrt(x)] * (6sqrt(x+h) + 6sqrt(x) ) / h * (6sqrt(x+h) + 6sqrt(x) )
Simplify this more and nice things happen (distribute / FOIL the top):
36(x+h) – 36(x) / [h *(6sqrt(x+h) + 6sqrt(x))]
Simplify numerator more:
(36x + 36h – 36x) / [h *(6sqrt(x+h) + 6sqrt(x))]
36h / [h *(6sqrt(x+h) + 6sqrt(x))]
Cancel out “h”
36 / (6sqrt(x+h) + 6sqrt(x))]
Now, take the limit as h goes to 0 (plug in 0 for h):
36 / (6sqrt(x+0) + 6sqrt(x))] =
36 / (6sqrt(x) + 6sqrt(x))
36 / 12 sqrt(x) |
## The linear nature of graphs
An example
Let's start with an example. The picture below shows the graph of a function we've seen already: Notice that at , the graph passes through (1,1). Now let's pretend that we can look at the graph under a microscope and that we can continually increase the magnification of the microscope. Clicking on the "Zoom" button below increases the magnification by a factor of two.
The important thing is that the closer we zoom in, the more the graph looks like a straight line through the point (1,1). We call this line the tangent line to the graph through (1,1).
Questions:
What is the slope of the tangent line? Since we know that the tangent line passes through the point (1,1), find the equation of the tangent line. (Here is a review of straight lines and their equations should you feel you need it.) Can you think of a function whose graph would not look like a straight line if you zoomed in like this? After you think about this, have a look at this example. Is the absolute value function differentiable at x = 1?
Important Jargon
We say that a function is differentiable at x if it looks more and more like a straight line near the point x. For instance, the function is differentiable at while the absolute value function is not differentiable at If a function is differentiable at a point x, we call the slope of its tangent line the derivative of at x. We will write this as . For instance, the function is differentiable at x = 1 and the derivative is
Conclusions
Maybe now you're wondering what the point is behind all of this. That's such a good question that we're going to answer it many times and in many ways throughout this course. You see, straight lines are the graphs of functions which have the form and these are exceptionally easy functions to work with. For instance, you only need to know two pieces of information -- a slope and a point through which the graph passes -- to know everything there is to know about such a function. A principal theme in this course is to use information which can be easily extracted from the tangent line to deduce information about the original graph, which could be difficult to deduce more directly. For instance, in our example above, we know that the slope of the tangent line is 1/2. Since this is positive, we know that the tangent line is rising as we move to the right. What is important is that the graph shares this property with the tangent line; that is, a small increase in the x-coordinate will increase the y-coordinate of the point on the graph. Later in the term, we will see the power of this kind of reasoning. In this section, we have built a sequence of snapshots of the graph each of which look more like a straight line. However, if we made very careful measurements, we would find that the graphs in the snapshots are still bent just a little bit. However, we can still imagine what the straight line would look like. In a way, it's like trying to get all the sand out of your car after a day at the beach. At first, there's lots of sand, but after you clean and clean, your car gets closer and closer to being spotless. And while there's always a few specks of sand hiding someplace so that you can never actually have a perfectly clean car, it's all right to think that your car is perfectly clean.
Confused?
If you find this confusing, you shouldn't worry too much (at least for now). These kinds of questions bamboozled some of history's greatest thinkers for several thousand years: ancient mathematicians asked questions like these, but it wasn't until the 1600's that people started to formulate useful answers to them. What we are really talking about is the fundamental concept of Calculus which we'll soon learn to call a limit . In the next few sections, we'll look at the derivative from a few new angles and find some nifty ways to compute with them. Gradually, the picture will become clear. |
# One Tailed Test or Two in Hypothesis Testing; One Tailed Distribution Area
Contents (Click to slip to that section):
## One tailed test or two in Hypothesis Testing: Overview
In hypothesis testing, you are asked to decide if a claim is true or not. For example, if someone says “all Floridian’s have a 50% increased chance of melanoma”, it’s up to you to decide if this claim holds merit. One of the first steps is to look up a z-score, and in order to do that, you need to know if it’s a one tailed test or two. You can figure this out in just a couple of steps.
## One tailed test or two in Hypothesis Testing: Steps
If you’re lucky enough to be given a picture, you’ll be able to tell if your test is one-tailed or two-tailed by comparing it to the image above. However, most of the time you’re given questions, not pictures. So it’s a matter of deciphering the problem and picking out the important piece of information. You’re basically looking for keywords like equals, more than, or less than.
Example question #1: A government official claims that the dropout rate for local schools is 25%. Last year, 190 out of 603 students dropped out. Is there enough evidence to reject the government official’s claim?
Example question #2: A government official claims that the dropout rate for local schools is less than 25%. Last year, 190 out of 603 students dropped out. Is there enough evidence to reject the government official’s claim?
Example question #3: A government official claims that the dropout rate for local schools is greater than 25%. Last year, 190 out of 603 students dropped out. Is there enough evidence to reject the government official’s claim?
Step 2: Rephrase the claim in the question with an equation.
1. In example question #1, Drop out rate = 25%
2. In example question #2, Drop out rate < 25%
3. In example question #3, Drop out rate > 25%.
Step 3: If step 2 has an equals sign in it, this is a two-tailed test. If it has > or < it is a one-tailed test.
Like the explanation? Check out the Statistics How To Handbook, which has hundreds of easy to understand definitions and examples, just like this one!
## One Tailed Test or Two: Onto some more technical stuff
The above should have given you a brief overview of the differences between one-tailed tests and two-tailed tests. For the very beginning of your stats class, that’s probably all the information you need to get by. But once you hit ANOVA and regression analysis, things get a little more challenging.
## 1. Alpha levels
Alpha levels (sometimes just called “significance levels”) are used in hypothesis tests; it is the probability of making the wrong decision when the null hypothesis is true. A one-tailed test has the entire 5% of the alpha level in one tail (in either the left, or the right tail). A two-tailed test splits your alpha level in half (as in the image to the left).
Let’s say you’re working with the standard alpha level of 0.5 (5%). A two tailed test will have half of this (2.5%) in each tail. Very simply, the hypothesis test might go like this:
1. A null hypothesis might state that the mean = x. You’re testing if the mean is way above this or way below.
2. You run a t-test, which churns out a t-statistic.
3. If this test statistic falls in the top 2.5% or bottom 2.5% of its probability distribution (in this case, the t-distribution), you would reject the null hypothesis.
The “cut off” areas created by your alpha levels are called rejection regions. It’s where you would reject the null hypothesis, if your test statistic happens to fall into one of those rejection areas. The terms “one tailed” and “two tailed” can more precisely be defined as referring to where your rejection regions are located.
## 2. Power
A one-tailed test is where you are only interested in one direction. If a mean is x, you might want to know if a set of results is more than x or less than x. A one-tailed test is more powerful than a two-tailed test, as you aren’t considering an effect in the opposite direction.
## 3. When Should You Use a One-Tailed Test?
In the above examples, you were given specific wording like “greater than” or “less than.” Sometimes you, the researcher, do not have this information and you have to choose the test.
For example, you develop a drug which you think is just as effective as a drug already on the market (it also happens to be cheaper). You could run a two-tailed test (to test that it is more effective and to also check that it is less effective). But you don’t really care about it being more effective, just that it isn’t any less effective (after all, your drug is cheaper). You can run a one-tailed test to check that your drug is at least as effective as the existing drug.
On the other hand, it would be inappropriate (and perhaps, unethical) to run a one-tailed test for this scenario in the opposite direction (i.e. to show the drug is more effective). This sounds reasonable until you consider there may be certain circumstances where the drug is less effective. If you fail to test for that, your research will be useless.
Consider both directions when deciding if you should run a one tailed test or two. If you can skip one tail and it’s not irresponsible or unethical to do so, then you can run a one-tailed test.
## One tailed Test or Two: How to find the area of a one-tailed distribution: Steps
There are a few ways to find the area under a one tailed distribution curve. The easiest, by far, is looking up the value in a table like the z-table. A z-table gives you percentages, which represent the area under a curve. For example, a table value of 0.5000 is 50% of the area and 0.2000 is 20% of the area.
If you are looking for other area problems*, see the normal distribution curve index. The index lists seven possible types of area, including two tailed, one tailed, and areas to the left and right of z.
*You can also calculate areas with integral calculus. See The Area Problem.
Note: In order to use a z-table, you need to split your z-value up into decimal places (e.g. tenths and hundredths). For example, if you are asked to find the area in a one tailed distribution with a z-value of 0.21, split this into tenths (0.2) and hundredths (0.01).
## One tailed distribution: Steps for finding the area in a z-table
Step 1: Look up your z-score in the z-table. Looking up the value means finding the intersection of your two decimals (see note above). For example, if you are asked to find the area in a one tailed distribution to the left of z = -0.46, look up 0.46 in the table (note: ignore negative values. If you have a negative value, use its absolute value). The table below shows that the value in the intersection for 0.46 is .1772. This figure was obtained by looking up 0.4 in the left hand column and 0.06 in the top row.
z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359
0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753
0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141
0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517
0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879
0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224
Step 2: Take the area you just found in step 2 and add .500. That’s because the area in the right-hand z-table is the area between the mean and the z-score. You want the entire area up to that point, so:
.5000 + .1772 = .6772.
Step 3: Subtract from 1 to get the tail area:
1 – .6772 = 0.3228.
That’s it!
## One Tailed Test or Two: References
Gonick, L. (1993). The Cartoon Guide to Statistics. HarperPerennial.
Heath, D. (2002). An Introduction to Experimental Design and Statistics for Biology. CRC Press.
IDRE: FAQ: What are the differences between one-tailed and two-tailed tests? Retrieved May 27, 2018 from: https://stats.idre.ucla.edu/other/mult-pkg/faq/general/faq-what-are-the-differences-between-one-tailed-and-two-tailed-tests/ |
NSW Syllabuses
# Mathematics K–10 - Stage 2 - Number and Algebra Addition and Subtraction
### Outcomes
#### A student:
• MA2-1WM
uses appropriate terminology to describe, and symbols to represent, mathematical ideas
• MA2-2WM
selects and uses appropriate mental or written strategies, or technology, to solve problems
• MA2-3WM
checks the accuracy of a statement and explains the reasoning used
• MA2-5NA
uses mental and written strategies for addition and subtraction involving two-, three-, four- and five-digit numbers
### Content
• Students:
• Recall addition facts for single-digit numbers and related subtraction facts to develop increasingly efficient mental strategies for computation (ACMNA055)
• add three or more single-digit numbers
• model and apply the associative property of addition to aid mental computation, eg 2 + 3 + 8 = 2 + 8 + 3 = 10 + 3 = 13
• apply known single-digit addition and subtraction facts to mental strategies for addition and subtraction of two-, three- and four-digit numbers, including:
• the jump strategy on an empty number line, eg 823 + 56: 823 + 50 = 873, 873 + 6 = 879
• the split strategy, eg 23 + 35: 20 + 30 + 3 + 5 = 58
• the compensation strategy, eg 63 + 29: 63 + 30 = 93, subtract 1 to obtain 92
• using patterns to extend number facts, eg 500 – 200: 5 – 2 = 3, so 500 – 200 = 300
• bridging the decades, eg 34 + 26: 34 + 6 = 40, 40 + 20 = 60
• changing the order of addends to form multiples of 10, eg 16 + 8 + 4: add 16 to 4 first
• using place value to partition numbers, eg 2500 + 670: 2500 + 600 + 70 = 3170
• partitioning numbers in non-standard forms, eg 500 + 670: 670 = 500 + 170, so 500 + 670 = 500 + 500 + 170, which is 1000 + 170 = 1170
• choose and apply efficient strategies for addition and subtraction (Problem Solving)
• discuss and compare different methods of addition and subtraction (Communicating)
• use concrete materials to model the addition and subtraction of two or more numbers, with and without trading, and record the method used
• select, use and record a variety of mental strategies to solve addition and subtraction problems, including word problems, with numbers of up to four digits
• give a reasonable estimate for a problem, explain how the estimate was obtained, and check the solution (Communicating, Reasoning)
• use the equals sign to record equivalent number sentences involving addition and subtraction and so to mean 'is the same as', rather than to mean to perform an operation, eg 32 – 13 = 30 – 11
• check given number sentences to determine if they are true or false and explain why,
eg 'Is 39 – 12 = 15 + 11 true? Why or why not?' (Communicating, Reasoning)
• Recognise and explain the connection between addition and subtraction (ACMNA054)
• demonstrate how addition and subtraction are inverse operations
• explain and check solutions to problems, including by using the inverse operation
• Represent money values in multiple ways and count the change required for simple transactions to the nearest five cents (ACMNA059)
• calculate equivalent amounts of money using different denominations, eg 70 cents can be made up of three 20-cent coins and a 10-cent coin, or two 20-cent coins and three 10-cent coins, etc
• perform simple calculations with money, including finding change, and round to the nearest five cents
• calculate mentally to give change
### Background Information
An inverse operation is an operation that reverses the effect of the original operation. Addition and subtraction are inverse operations; multiplication and division are inverse operations.
In Stage 2, it is important that students apply and extend their repertoire of mental strategies for addition and subtraction. The use of concrete materials to model the addition and subtraction of two or more numbers, with and without trading, is intended to provide a foundation for the introduction of the formal algorithm in Addition and Subtraction 2.
One-cent and two-cent coins were withdrawn by the Australian Government in 1990. Prices can still be expressed in one-cent increments, but the final bill is rounded to the nearest five cents (except for electronic transactions), eg
$5.36,$5.37 round to $5.35$5.38, $5.39,$5.41, $5.42 round to$5.40
$5.43,$5.44 round to \$5.45.
### Language
Students should be able to communicate using the following language: plus, add, addition, minus, the difference between, subtract, subtraction, equals, is equal to, is the same as, number sentence, empty number line, strategy, digit, estimate, round to.
Students need to understand the different uses for the = sign, eg 4 + 1 = 5, where the = sign indicates that the right side of the number sentence contains 'the answer' and should be read to mean 'equals', compared to a statement of equality such as 4 + 1 = 3 + 2, where the = sign should be read to mean 'is the same as'.
### National Numeracy Learning Progression links to this Mathematics outcome
When working towards the outcome MA2‑5NA the sub-elements (and levels) of Quantifying numbers (QuN9-QuN10), Additive strategies (AdS6-AdS8), Understanding money (UnM4-UnM7) and Number patterns and algebraic thinking (NPA4-NPA5) describe observable behaviours that can aid teachers in making evidence-based decisions about student development and future learning.
The progression sub-elements and indicators can be viewed by accessing the National Numeracy Learning Progression.
### Outcomes
#### A student:
• MA2-1WM
uses appropriate terminology to describe, and symbols to represent, mathematical ideas
• MA2-2WM
selects and uses appropriate mental or written strategies, or technology, to solve problems
• MA2-3WM
checks the accuracy of a statement and explains the reasoning used
• MA2-5NA
uses mental and written strategies for addition and subtraction involving two-, three-, four- and five-digit numbers
### Content
• Students:
• Apply place value to partition, rearrange and regroup numbers to at least tens of thousands to assist calculations and solve problems (ACMNA073)
• select, use and record a variety of mental strategies to solve addition and subtraction problems, including word problems, with numbers of up to and including five digits, eg 159 + 23: 'I added 20 to 159 to get 179, then I added 3 more to get 182', or use an empty number line:
• pose simple addition and subtraction problems and apply appropriate strategies to solve them (Communicating, Problem Solving)
• use a formal written algorithm to record addition and subtraction calculations involving two-, three-, four- and five-digit numbers, eg
• solve problems involving purchases and the calculation of change to the nearest five cents, with and without the use of digital technologies (ACMNA080)
• solve addition and subtraction problems involving money, with and without the use of digital technologies
• use a variety of strategies to solve unfamiliar problems involving money (Communicating, Problem Solving)
• reflect on their chosen method of solution for a money problem, considering whether it can be improved (Communicating, Reasoning)
• calculate change and round to the nearest five cents
• use estimation to check the reasonableness of solutions to addition and subtraction problems, including those involving money
### Background Information
Students should be encouraged to estimate answers before attempting to solve problems in concrete or symbolic form. There is still a need to emphasise mental computation, even though students can now use a formal written method.
When developing a formal written algorithm, it will be necessary to sequence the examples to cover the range of possibilities, which include questions without trading, questions with trading in one or more places, and questions with one or more zeros in the first number.
This example shows a suitable layout for the decomposition method:
### Language
Students should be able to communicate using the following language: plus, add, addition, minus, the difference between, subtract, subtraction, equals, is equal to, empty number line, strategy, digit, estimate, round to, change (noun, in transactions of money).
Word problems requiring subtraction usually fall into two types − either 'take away' or 'comparison'.
Take away – How many remain after some are removed?
eg 'I have 30 apples in a box and give away 12. How many apples do I have left in the box?'
Comparison – How many more need to be added to a group? What is the difference between two groups?
eg 'I have 18 apples. How many more apples do I need to have 30 apples in total?', 'Mary has 30 apples and I have 12 apples. How many more apples than me does Mary have?'
Students need to be able to translate from these different language contexts into a subtraction calculation.
The word 'difference' has a specific meaning in a subtraction context. Difficulties could arise for some students with phrasing in relation to subtraction problems, eg '10 take away 9' should give a response different from that for '10 was taken away from 9'.
### National Numeracy Learning Progression links to this Mathematics outcome
When working towards the outcome MA2‑5NA the sub-elements (and levels) of Quantifying numbers (QuN10-QuN11), Additive strategies (AdS7-AdS8), Understanding money (UnM4-UnM7) and Number patterns and algebraic thinking (NPA5) describe observable behaviours that can aid teachers in making evidence-based decisions about student development and future learning.
The progression sub-elements and indicators can be viewed by accessing the National Numeracy Learning Progression. |
# Basic Geometry: Types of AnglesPosted on : 12-07-2018 Posted by : Admin
## Introduction
• Angle is the shape formed by the union of two rays diverting from a common endpoint.
• This common endpoint from which the two lines divert is called as vertex.
• The two straight sides of the angle are called its legs or arms.
• The area between the rays that make up an angle is known as interior of the angle. Interior of the angle extends away from the vertex to infinity.
• The remaining place on the plane which does not belong to interior is called as exterior of the angle.
• Two angles are said to be congruent if their measurements are equal. It is denoted by ≅.
## Bisector of an angle
Bisector of an angle is a ray which divides the interior of the angle into two equal measures. In other words, it is a line which cuts an angle into two equal angles. Bisector of an angle is also known as angle bisector. Angle bisector passes through the vertex of the angle. An angle bisector can be external or internal.
Internal angle bisector: In the given figure below, the ray OX bisects the internal angle ∠XYZ. Now ∠XYO and ∠OYZ are congruent.
Therefore ∠XYO ≅ ∠OYZ
Here OX is the internal angle bisector.
Exterior angle bisector: In the given figure below, the ray OM bisects the external angle ∠X1YZ. Now ∠ X1YO and ∠OYZ are congruent.
Therefore ∠X1YO ≅ ∠OYZ.
Here OM is the external angle bisector.
## Types of angles
The following is the detailed study of few important types of angles which are useful in geometry.
1. Acute angle: The angle whose value lies between 0 to 90° is called acute angle.
2. Right angle: The angle whose value is 90° is called right angle.
3. Obtuse angle: The angle whose value lies between 90 to 180° is called obtuse angle.
4. Straight angle: The angle whose value is 180° is called straight angle.
5. Reflex angle: The angle whose value lies between 180 to 360° is called reflex angle.
6. Full angle: The angle whose value is 360° is called complete or full angle. This angle is also known as full rotation angle.
For example let us consider two parallel lines X1 and X2. Now let another line AB intersect both the parallel lines X1 and X2. The following are the facts that can be observed from these lines.
Corresponding angles are equal
∠1=∠5
∠4=∠8
∠2=∠6
∠3=∠7
Alternate interior angles are equal
∠3=∠5
∠4=∠6
Vertically opposite angles are equal
∠2=∠4
∠1=∠3
∠5=∠7
∠6=∠8
## Pairs of angles
Two angles are called as adjacent angles, when they have a common side and common vertex. Their interiors are separate and these angles do not overlap. In the given example figure, ∠AOB and ∠BOC are adjacent angles.
## Linear pair
Two angles are called a linear pair, when they have a common side and their other two sides are opposite rays. The angles that form a linear pair are always adjacent. The sum of angles in a linear pair is 180°. In the given example figure, ∠AOC and ∠COB form a linear pair. Because ∠AOC + ∠COB= 180°
## Complimentary angle
If the sum of two angles is 90°, then they are called complimentary angles. Each angle is complement of the other angle. If ∠AOC + ∠COB= 90°, then ∠AOC and ∠COB are complementary angles.
## Supplementary angle
If the sum of two angles is 180°, then they are called supplementary angles. Each angle is supplement of the other angle. The angles of linear pair are supplementary angles.
∠AOB + ∠CED= 120° + 60° = 180°
Therefore these angles are supplementary
## Vertically opposite angles
Two angles are said to be vertically opposite angles if their sides form two pairs of opposite rays. In other words, when two lines intersect, two pairs of vertically opposite angles are formed. These are the angles opposite to each other when two lines cross each other. For example, let us consider two lines AB and DC intersecting at point ‘O’.
From this we can deduce the following,
Vertically opposite angles are similar or congruent in nature,
∠AOC and ∠DOB are vertically opposite (∠AOC ≅ ∠DOB)
∠AOD and ∠COB are also vertically opposite (∠AOD ≅ ∠COB)
Sum of all angles at the intersection point is 360°, ∠AOC+∠DOB+∠AOD+∠COB= 360°
## Corresponding angles
When two lines are intersected by a transversal, then they form four pairs of corresponding angles. For example, let us consider two lines EF and GH being intersected at points B and C by the transversal line AD.
From this we can deduce the following,
The four pairs of corresponding angles are,
∠ABE, ∠BCG
∠EBC, ∠GCD
∠ABF, ∠BCH
∠FBC, ∠HCD
When two parallel lines are intersected by a transversal line, the pair of corresponding angles so formed is congruent.
Hence is EF is parallel to GH and AD is the transversal line then,
∠ABE ≅ ∠BCG
∠EBC ≅ ∠GCD
∠ABF ≅ ∠BCH
∠FBC ≅ ∠HCD
Conversely, if the transversal line intersects two lines and if one pair of corresponding angles is congruent then two lines are parallel. Hence, when one pair of corresponding angles is congruent, then all the pairs of corresponding angles are congruent.
## Alternate angles
When two parallel lines are intersected by a transversal line, they form two pairs of alternate angles.
∠EBC, ∠BCH
∠FBC, ∠GCB
When two parallel lines are intersected by a transversal line, the pairs of alternate angles so formed are congruent.
∠EBC ≅ ∠BCH
∠FBC ≅ ∠GCB
Conversely, if the transversal line intersects two lines and if one pair of alternate angles is congruent then two lines are parallel. Hence, when one pair of alternate angles is congruent, then all the pairs of alternate angles are congruent.
## Interior angles
When two lines are intersected by a transversal line, they form two pairs of interior angles. For example, the two pairs of interior angles in the given figure are ∠EBC, ∠GCB and ∠FBC, ∠BCH
When two parallel lines are intersected by a transversal line, the pairs of interior angles so formed are supplementary.
∠EBC+∠GCB=180°
∠FBC+∠BCH=180°
Conversely, if the transversal line intersects two lines and if one pair of interior angles is supplementary then two lines are parallel. Hence, when one pair of interior angles is supplementary, the other pair is also supplementary. And all the pairs of alternate and corresponding angles are congruent.
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Strand: NUMBER AND OPERATIONS - FRACTIONS (4.NF)
Extend understanding of equivalence and ordering of fractions (Standards 4.NF.1-2). Build fractions from unit fractions by applying and extending previous understandings of operations on whole numbers (Standards 4.NF.3-4). Understand decimal notation to the hundredths and compare decimal fractions with denominators of 10 and 100 (Standards 4.NF.5- 7). Denominators for fourth grade are limited to 2, 3, 4, 5, 6, 8, 10, 12, and 100.
This problem simply asks students to add the tenths and hundredths together.
• Addition of Fractions Using a Visual Model
Adding two fractions with unlike denominators is the focus of this video lesson. Students will learn how to use a visual model to work with these fractions. NOTE: You have to create a Free PBS Account to view this web page, but it is easy to do and worth the effort.
• Comparing Fractions Using Benchmarks Game
The goal of this task is to determine appropriate benchmarks for fractions with a focus on providing explanations that demonstrate deep conceptual understanding.
• Comparing Fractions with Lines
This lesson will help students simplify fractions, compare and order them on a number line, and estimate their value.
• Comparing Sums of Unit Fractions
The purpose of this task is to help develop students' understanding of addition of fractions; it is intended as an instructional task.
• Comparing Two Different Pizzas
Students are given the example of two whole pizzas, one large and one medium. They must determine what fraction of the two pizzas has been eaten if someone eats 2 slices of the medium pizza. They must explain their reasoning and draw a picture to illustrate their solution.
• Cynthia's Perfect Punch
The purpose of this task is for students to estimate and compute sums of mixed numbers in the context of a student making a punch recipe.
• Dimes and Pennies
This task asks: "A dime is 1/10 of a dollar and a penny is 1/100 of a dollar. What fraction of a dollar is 6 dimes and 3 pennies? Write your answer in both fraction and decimal form."
• Doubling Numerators and Denominators
The purpose of this task is to assess whether students understand the meaning of the numerator and the denominator in a fraction. This task is not appropriate for a high-stakes summative assessment, but it could be very helpful in gauging students' flexibility with the meaning of fractions.
• Equivalent Fractions Finder
Students doing this lesson will visually experiment with the relationship between the value of fractions and areas within a square or a circle. In doing so they will practice simplifying fractions and compare fractions.
• Expanded Fractions and Decimals
The purpose of this task is for students to show they understand the connection between fraction and decimal notation by writing the same numbers both ways.
• Explaining Fraction Equivalence with Pictures
The purpose of this task is to provide students with an opportunity to explain fraction equivalence through visual models in a particular example.
• Extending Multiplication From Whole Numbers to Fractions
This task can be used at the beginning of a fourth grade unit on multiplication with fractions to evaluate what students know about multiplication.
• Fraction Conversion 2 (with percents)
When completing this lesson students will understand how to convert fractions, decimals, and percentages.
• Fraction Equivalence
Students are given this task: "Explain why 6/10=60.100. Draw a picture to illustrate your explanation."
• Fraction King
This is a lesson plan to help students compare two fractions and understand the addition and subtraction of fractions.
• Fractions and Rectangles
The primary goal of this task is for students to use pictures to explain the equivalence between 3/12 and 1/4.
In order to assist educators with the implementation of the Common Core, the New York State Education Department provides curricular modules in Pre-K-Grade 12 English Language Arts and Mathematics that schools and districts can adopt or adapt for local purposes.
• Grade 4 Mathematics Module 5: Fraction Equivalence, Ordering, and Operations
In this 40-day module, students build on their Grade 3 work with unit fractions as they explore fraction equivalence and extend this understanding to mixed numbers. This leads to the comparison of fractions and mixed numbers and the representation of both in a variety of models. Benchmark fractions play an important part in students ability to generalize and reason about relative fraction and mixed number sizes. Students then have the opportunity to apply what they know to be true for whole number operations to the new concepts of fraction and mixed number operations.
• Grade 4 Mathematics Module 6: Decimal Fractions
This 20-day module gives students their first opportunity to explore decimal numbers via their relationship to decimal fractions, expressing a given quantity in both fraction and decimal forms. Utilizing the understanding of fractions developed throughout Module 5, students apply the same reasoning to decimal numbers, building a solid foundation for Grade 5 work with decimal operations.
• Grade 4 Unit 3: Fraction Equivalents (Georgia Standards)
In this unit students will understand representations of simple equivalent fractions and compare fractions with different numerators and different denominators.
• Grade 4 Unit 4: Operations with Fractions (Georgia Standards)
In this unit students will identify visual and written representations of fractions, understand representations of simple equivalent fractions, understand the concept of mixed numbers with common denominators to 12, add and subtract fractions with common denominators, add and subtract mixed numbers with common denominators and convert mixed numbers to improper fractions and improper fractions to mixed fractions.
• Grade 4 Unit 5: Fractions and Decimals (Georgia Standards)
In this unit, students will express fractions with denominators of 10 and 100 as decimals, understand the relationship between decimals and the base ten system, understand decimal notation for fractions, use fractions with denominators of 10 and 100 interchangeably with decimals and express a fraction with a denominator 10 as an equivalent fraction with a denominator 100.
• Histograms and Bar Graphs
This lesson will introduce students to histograms and bar graphs and help them understand the difference between them.
• How Many Tenths and Hundredths?
In this task students are given a list of equations such as "1 tenth + 4 hundredths = ______________ hundredths" and asked to finish the equations to make true statements.
• IXL Game: Add and Subtract Fractions
This game helps fourth graders understand how to compare two fractions with different numerators and different denominators. This is just one of many online games that supports the Utah Math core. Note: The IXL site requires subscription for unlimited use.
• Listing fractions in increasing size
This activity asks students to order a group of fractions from smallest to largest and explain their reasoning.
• Making 22 Seventeenths in Different Ways
This is a straightforward task related to adding fractions with the same denominator. The main purpose is to emphasize that there are many ways to decompose a fraction as a sum of fractions, similar to decompositions of whole numbers that students should have seen in earlier grades. It is suitable for assessment or, with opportunity for classroom discussion, it could also be useful in instruction.
• Money in the piggy bank
Students are given the scenario of a student opening a piggy bank and finding an array of coins. They must determine the number of coins, what fraction is made up of dimes and the total value of the coins.
• Multiply Fractions Jeopardy
This game can be played by one or two players as they solve multiplication problems with fractions.
• Number and Operations - Fractions (4.NF) - Fourth Grade Core Guide
The Utah State Board of Education (USBE) and educators around the state of Utah developed these guides for Fourth Grade Mathematics - Number and Operations - Fractions (4.NF)
• Peaches
This task provides a context where it is appropriate for students to subtract fractions with a common denominator; it could be used for either assessment or instructional purposes.
• Plastic Building Blocks
The purpose of this task is to have students add mixed numbers with like denominators when given a problem in the context of two students building a castle out of building blocks.
• Rational Number Project
This portal leads to 28 lesson plans designed to help students understand the four operations with fractions.
• Running Laps
The purpose of this task is for students to compare two fractions that arise in a context. Because the fractions are equal, students need to be able to explain how they know that.
• Single Fraction Finder
In this lesson students will practice dividing a whole into a fraction and visualize fractions on a number line.
• Subtraction of Fractions Using a Visual Model
Students will learn how to subtract fractions with unlike denominators by using a visual model. NOTE: You have to create a Free PBS Account to view this web page, but it is easy to do and worth the effort.
• Sugar in Six Cans of Soda
In this task students are given the statement "For a certain brand of orange soda, each can contains 4/15 cup of sugar." They are asked to calculate how many cups of sugar are in six cans and then draw a picture representing the answer.
• Using Benchmarks to Compare Fractions
This task is intended primarily for instruction. The goal is to provide examples for comparing two fractions, 1/5 and 2/7 in this case, by finding a benchmark fraction which lies in between the two.
• Using Place Value
Each part of this task highlights a slightly different aspect of place value as it relates to decimal notation. More than simply being comfortable with decimal notation, the point is for students to be able to move fluidly between and among the different ways that a single value can be represented and to understand the relative size of the numbers in each place.
• What Fraction of this Shape is Red?
This Teaching Channel video shows how students can explore part and whole by creating pattern block designs. (4 minutes)
• Writing a Mixed Number as an Equivalent Fraction
This task relates to writing a mixed number as an equivalent fraction. The purpose of this task is to help students understand and articulate the reasons for the steps in the usual algorithm for converting a mixed number into an equivalent fraction,
http://www.uen.org - in partnership with Utah State Board of Education (USBE) and Utah System of Higher Education (USHE). Send questions or comments to USBE Specialists - Trish French or Molly Basham and see the Mathematics - Elementary website. For general questions about Utah's Core Standards contact the Director - Jennifer Throndsen.
These materials have been produced by and for the teachers of the State of Utah. Copies of these materials may be freely reproduced for teacher and classroom use. When distributing these materials, credit should be given to Utah State Board of Education. These materials may not be published, in whole or part, or in any other format, without the written permission of the Utah State Board of Education, 250 East 500 South, PO Box 144200, Salt Lake City, Utah 84114-4200. |
## Data Handling
EXERCISE 3.4
CBSE Class 7 Maths
Q1: Tell whether the following is certain to happen, impossible can happen but not certain.
(i) You are older today than yesterday.
(ii) A tossed coin will land heads up.
(iii) A die when tossed shall land up with 8 on top.
(iv) The next traffic light seen will be green.
(v) Tomorrow will be a cloudy day.
Answer:
(i) It is certain to happen.
(ii) It can happen but not certain.
(iii)It is impossible.
(iv) It can happen but not certain.
(v) It can happen but not certain.
Q2: There are 6 marbles in a box with numbers from 1 to 6 marked on each of them.
(i) What is the probability of drawing a marble with number 2?
(ii) What is the probability of drawing a marble with number 5?
Answer:
Total marbles from 1 to 6 marked in a box = 6
(i) The probability of drawing a marble with number 2.
P (drawing one marble) = ⅙
(ii) The probability of drawing a marble with number 5.
P (drawing one marble) = ⅙
Q3: A coin is flipped to decide which team starts the game. What is the probability that your team will start?
Answer:
A coin has two possible outcomes Head and Tail.
Probability of getting Head or Tail is equal.
P (Starting game) = ½
Other Questions:
Q4: A 5₹ coin is tossed, what are the possible outcomes?
Answer: Two outcomes i.e. either Head or Tail.
Q5: A box of 600 electric bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. What is the probability that it is a non defective bulb?
Answer: Total electric bulbs = 600
Defective bulbs = 12
Non Defective bulbs = 600 - 12 = 588
P (picking up non-defective bulb) = Non Defective bulbs ÷ Total bulbs = 588 ÷ 600 = 0.98
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# ZingPath: Perimeter and Area of Polygonal Regions
Searching for
## Perimeter and Area of Polygonal Regions
Learn in a way your textbook can't show you.
Explore the full path to learning Perimeter and Area of Polygonal Regions
### Lesson Focus
#### The Relationship Between Perimeter and Area
Pre-Algebra
Determine the smallest or largest possible perimeter given a fixed area, and determine the smallest or largest possible area given a fixed perimeter.
### Now You Know
After completing this tutorial, you will be able to complete the following:
• Distinguish between perimeter and area.
• Determine the largest or smallest perimeter, given the area.
• Determine the largest or smallest area, given the perimeter.
### Everything You'll Have Covered
Area
Area is the measurement of interior space of a two-dimensional shape, or the surface of three-dimensional figure; it is the amount of surface a shape covers. Area is measured in square units. The area of the rectangle can be calculated using the following formula: A = l × w, where l represents the length and w represents the width.
For the shape above, the length equals 4 units and the width equals 3 units. A = l × w
A = 4 × 3 = 12 square units
Perimeter
Perimeter is the length of the boundary of the shape. It can be calculated adding the measurements of each side. For a rectangle, it can be calculated using the following formula: P = l + w + l + w, or P = 2l + 2w.
For the shape above, the length equals 4 units and the width equals 3 units.P = 4 + 3 + 4 + 3 P = (2 × 4) + (2 × 3) = 14 units
Area and perimeter of irregular shapes
If irregular shapes are drawn on grid paper, as in the Activity Object, you can use the grid to help you calculate the area and perimeter.
• For area, simply count all the unit squares that make up the shape:There are 12 unit squares that make up this shape, so the area equals 12 square units.
• For perimeter, count the sides of each unite square that makes up the shape: There are 24 units that make up the perimeter of shape. Therefore, the perimeter equals 24 units.
### Tutorial Details
Approximate Time 20 Minutes Pre-requisite Concepts Students should know the definitions of area and perimeter. Course Pre-Algebra Type of Tutorial Concept Development Key Vocabulary area, perimeter, shape |
# 4.01 Mean
Lesson
The mean is described as the average of the numbers in a data set. It is defined as the sum of the scores divided by the number of scores.
The symbol for the mean of a sample is $\overline{x}$x, whilst the population mean is represented by the symbol $\mu$μ (Greek letter 'mu'). We typically don't have data for every member of the population, so we usually don't know $\mu$μ exactly, but we can estimate it by using the sample mean, $\overline{x}$x, from a well designed survey.
If certain scores are repeated, such as when information is given in a frequency table then we can find the total sum of all scores by multiplying each unique score by its frequency, then adding them all up.
We summarise the calculation of the mean below.
Mean
The mean of a set of data is calculated by:
$\text{Mean}=\frac{\text{Total sum of all scores}}{\text{Number of scores}}$Mean=Total sum of all scoresNumber of scores
If certain scores are repeated, then:
$\text{Total sum of all scores}=\text{sum of}\ \left(\text{Unique score}\times\text{Frequency}\right)$Total sum of all scores=sum of (Unique score×Frequency)
Now let's look at a few examples of calculating the mean of different data sets.
#### Worked examples
##### Example 1
Find the mean from the data in the stem plot below.
Stem Leaf $2$2 $3$3 $8$8 $3$3 $1$1 $1$1 $1$1 $4$4 $0$0 $3$3 $5$5 $0$0 $3$3 $8$8 $8$8 $6$6 $2$2 $2$2 $9$9 $7$7 $1$1 $8$8 $8$8 $3$3 $9$9 $0$0 $0$0 $1$1
Think: We can find the mean by adding up all of the scores, then dividing the total by the number of scores.
Do:
$\text{Mean}$Mean $=$= $\frac{\text{Total of all scores}}{\text{Number of scores}}$Total of all scoresNumber of scores $=$= $\frac{23+28+3\times31+40+43+50+53+2\times58+2\times62+69+71+78+83+2\times90+91}{20}$23+28+3×31+40+43+50+53+2×58+2×62+69+71+78+83+2×90+9120 $=$= $\frac{1142}{20}$114220 $=$= $57.1$57.1
##### Example 2
A statistician has organised a set of data into the frequency table shown.
Score ($x$x) Frequency ($f$f)
$44$44 $8$8
$46$46 $10$10
$48$48 $6$6
$50$50 $18$18
$52$52 $5$5
(a) Complete the frequency distribution table by adding a column showing the total sum for each unique score.
Think: For each unique score ($x$x-value), multiply it by the number of times that score appears. In other words, multiply the unique score by its frequency $\left(f\right)$(f) to find the total sum for that score.
Do: So for a score of $44$44, which occurred $8$8 times, the total score is $44\times8=352$44×8=352. Completing the entire column, we get the following table.
Score ($x$x) Frequency ($f$f) $fx$fx
$44$44 $8$8 $352$352
$46$46 $10$10 $460$460
$48$48 $6$6 $288$288
$50$50 $18$18 $900$900
$52$52 $5$5 $260$260
Totals $47$47 $2260$2260
(b) Calculate the mean of this data set. Round your answer to two decimal places.
Think: We calculate the mean by dividing the sum of the scores (that is, the sum of all the $fx$fx's) by the number of scores (the total frequency).
Do:
$\text{Mean}$Mean $=$= $\frac{\text{Total of all scores}}{\text{Number of scores}}$Total of all scoresNumber of scores $=$= $\frac{2260}{47}$226047 $=$= $48.09$48.09 ($2$2 d.p.)
Using technology
Throughout this chapter and in particular for moderate to large data sets, you should use appropriate technology such as a calculator with a statistics program or your computer.
Tips:
• Familiarise yourself with the program and the types of calculations and graphs it is capable of creating.
• Ensure settings are correct for the data given, this is particularly important when changing between data that is in a simple list to data that is in a frequency table.
• Take note of the different symbols used for the different calculations we will encounter.
#### Practice questions
##### Question 1
Find the mean of the following scores:
$8$8, $15$15, $6$6, $27$27, $3$3.
##### Question 2
In each game of the season, a basketball team recorded the number of 'three-point shots' they scored. The results for the season are represented in the given dot plot.
1. What was the total number of points scored from three-point shots during the season?
2. Considering the total number of points, what was the mean number of points scored each game? Round to 2 decimal places if necessary.
3. What was the mean number of three point shots per game this season? Leave your answer to two decimal places if necessary.
##### Question 3
The mean of $4$4 scores is $21$21. If three of the scores are $17$17, $3$3 and $8$8, find the $4$4th score (call it $x$x).
### Outcomes
#### 3.3.1.2
calculate measures of central tendency, the mean and the median from a dataset |
# Fraction Rules
Pronunciation: /ˈfræk.ʃən rulz/ Explain
Figure 1: Fraction
Fraction rules are a set of algebraic rules for working with fractions. A fraction has a numerator and a denominator. A fraction represents a division operation. The numerator is the dividend. The denominator is the divisor.
Fraction Rules
OperationEquationsExamplesDescription
Adding two fractions[2]To add fractions, transform each fraction so they have a common denominator. Add the numerators and use the common denominator as the denominator. Reduce the fraction. See Operations on Fractions: Addition and Subtraction.
Subtracting two fractionsTo subtract fractions, transform each fraction so they have a common denominator. Subtract the numerators and use the common denominator as the denominator. Reduce the fraction. See Operations on Fractions: Addition and Subtraction.
Multiplying two fractions[2]To multiply fractions, multiply the numerators and multiply the denominators. Reduce the fraction. See Operations on Fractions: Multiplication.
Multiplying a fraction and a whole number.To multiply a fraction and a whole number, multiply the numerator by the whole number. The denominator remains unchanged. Reduce the fraction if possible.
Dividing two fractions[2]To divide fractions, flip the divisor upside down then multiply by the dividend. Reduce the fraction. See Operations on Fractions: Division.
Dividing a fraction by an integer.To divide a fraction by a whole number, convert the whole number to a fraction, the divide the fractions.
Raising a fraction to a power.See Operations on Fractions: Exponentiation.
Converting a mixed number to an improper fraction.To convert a mixed number to an improper fraction, multiply the whole part by the denominator and add the product to the numerator. The denominator remains unchanged. See How to Convert a Mixed Number to a Fraction.
Converting an improper fraction to a mixed number.To convert an improper fraction to a mixed number, divide the numerator by the denominator using a remainder. The mixed number is the quotient plus the remainder divided by the denominator. See How to Convert a Fraction to a Mixed Number..
Zero numerator.Applying the property of multiplying by zero, a zero numerator with a zero denominator is zero. See Property of Multiplying by 0.
Zero denominator.Since division by zero is undefined, a zero denominator makes the fraction undefined.
One minus sign.Since , apply the associative property of multiplication to get
Two minus signs.Since , apply the associative property of multiplication to get
If a fraction has the same nonzero numerator and denominator, the fraction's value is 1.Anything except 0 divided by itself is 1.
Any integer can be made into a fraction.Since , apply the property of multiplying by 1: . See Property of Multiplying by 1.
Reducing fractions.Given two arbitrary values a and b, and values c, d, and e such that a = c · d and b = c · e, . See Reducing Fractions.
Building fractions.Given a fraction a / b and a number d that is a multiple of d, find e such that b · e = d, then a / b = (a · e) / (b · e).
Operations on complex fractions.Simplify the complex fractions, then use the rules for simple fractions.To manipulate a complex fraction, convert it to a simple fraction, then follow the rules for simple fractions. See Complex Fraction.
Converting a decimal number to a fraction.To convert a decimal to a fraction, change the decimal to a whole number and divide it by 10n where n is the number of digits after the decimal point.
Converting a percentage to a fraction.To convert a percentage to a fraction, use the percentage as the numerator, 100 as the denominator, then simplify.
Comparing fractions with like denominators.To compare fractions with like denominators, compare the numerators. The relationship between the fractions is the same as the relationship between the denominators.
Comparing fractions with unlike denominators.To compare fractions with unlike denominators, either convert them to a decimal or transform them to a common denominator, then compare them.
Table 1
### References
1. McAdams, David E.. All Math Words Dictionary, fraction rules. 2nd Classroom edition 20150108-4799968. pg 82. Life is a Story Problem LLC. January 8, 2015. Buy the book
2. Fine, Henry B., Ph. D.. Number-System of Algebra Treated Theoretically and Historically. 2nd edition. pp 12-15. www.archive.org. D. C. Heath & Co., Boston, U.S.A.. 1907. Last Accessed 7/11/2018. http://www.archive.org/stream/thenumbersystemo17920gut/17920-pdf#page/n21/mode/1up/search/fraction. Buy the book
3. Oberg, Erik. Arithmetic Simplified. pp 21-31. www.archive.org. Industrial Press. 1914. Last Accessed 7/11/2018. http://www.archive.org/stream/arithmeticsimpli00oberrich#page/21/mode/1up/search/fraction. Buy the book
4. Oberg, Erik. Elementary Algebra. pg 23. www.archive.org. Industrial Press. 1914. Last Accessed 7/11/2018. http://www.archive.org/stream/elementaryalgebr00oberrich#page/n26/mode/1up/search/fraction. Buy the book
5. Bettinger, Alvin K. and Englund, John A.. Algebra and Trigonometry. pp 9-11,36-40. www.archive.org. International Textbook Company. January 1963. Last Accessed 7/11/2018. http://www.archive.org/stream/algebraandtrigon033520mbp#page/n18/mode/1up. Buy the book
McAdams, David E. Fraction Rules. 4/21/2019. All Math Words Encyclopedia. Life is a Story Problem LLC. https://www.allmathwords.org/en/f/fractionrules.html.
### Revision History
4/21/2019: Modified equations and expression to match the new format. (McAdams, David E.)
12/21/2018: Reviewed and corrected IPA pronunication. (McAdams, David E.)
8/28/2018: Corrected spelling. (McAdams, David E.) |
How to Identify the Four Conic Sections in Graph Form - dummies
# How to Identify the Four Conic Sections in Graph Form
Each conic section has its own standard form of an equation with x- and y-variables that you can graph on the coordinate plane. You can write the equation of a conic section if you are given key points on the graph. You can alter the shape of each of these graphs in various ways, but the general graph shapes still remain true to the type of curve that they are.
Cutting the right cone with a plane to get conic sections.
This figure illustrates how a plane intersects the cone (the top and bottom half are considered two halves of one cone) to create the conic sections, and the following list explains the figure.
• Circle: A circle is the set of all points that are a given distance (the radius, r) from a given point (the center). To get a circle from the right cone, the plane slices occurs horizontally through either the top or bottom half of the cone.
• Parabola: A parabola is a curve in which every point is equidistant from one point (the focus) and a line (the directrix). It looks a lot like the letter U, although it may be upside down or sideways. To form a parabola, the plane slices through parallel to the side of the right cone).
• Ellipse: An ellipse is the set of all points where the sum of the distances from two points (the foci) is constant. You may be more familiar with another term for ellipse, oval. In order to get an ellipse from the right cone, the plane must slice through the cone at a shallow enough angle where it is slicing through only one-half of the cone. (Note: if the plane slices horizontally through the cone, a circle is created. A circle is considered a special type of ellipse.)
• Hyperbola: A hyperbola is the set of points where the difference of the distances between two points is constant. The shape of the hyperbola is difficult to describe without a picture, but it looks visually like two parabolas (although they’re very different mathematically) mirroring one another with some space between them. To get a hyperbola, the plane must slice through the right cone and a steep enough angle where it is slicing through both halves of the cone.
Most of the time, sketching a conic is not enough. Each conic section has its own set of information that you usually have to give to supplement the graph. You have to indicate where the center, vertices, major and minor axes, and foci are located. Often, this information is more important than the graph itself. Besides, knowing all this valuable info helps you sketch the graph more accurately than you could without it. |
# The Euler Equation
The Euler (pron. oiler) Equation must surely rate as one of the most elegant, beautiful and awe-inspiring formula in maths. Its consequences are diverse and shocking when you consider its simplicity.
The Euler Equation is often given by the following: etheta i = cos theta + i sin theta
### Proof of the Euler Equation
Let us consider the function y = cos x + i sin x. Continuing to treat i like any other number, we have, by differentiation:
dy/dx = -sin x + i cos x = i(cos x + i sin x)
Hence dy/dx = iy
=> i dx/dy = 1/y
=> ix = ln y + c
But when x = 0, y = 1. So c = 0.
=> ix = ln y
=> y = eix
So cos x + i sin x = eix
It may be objected that we have nowhere defined the meaning of a number such as ez when z is complex; but the reader should not be deterred by such inhibitions! Indeed, the above paragraph may be regarded as providing, if not a definition, at least a reasonable exposition of the meaning of eix, making it consistent with the familiar processes of mathematics.
### Consequences of the Euler Equation
One of the most fundemental consequences of the Euler Equation is shown by taking theta to equal pi radians. When this is done then the equation (once rearranged slightly) gives the following:
epi i + 1 = 0
This unites the five most important numbers in maths; pi, i, e, 1 and 0 into one relation and so the Euler Equation is taken as one of the most important points of unification. From this equation one gains a glimmer of how the whole of maths fits together.
This glimmer is reinforced if one takes the natural log of both sides of the equation (after subtracting one from both sides). This allows us to define the natural log of the negative numbers in the complex plane as follows:
ln -1 = i pi
A far reaching result which defines a whole family of hyperbolic (or modular) forms (strongly related to hyperbolic functions) based around two mutually perpendicular complex planes (represented commonly by Argand Diagrams) sharing no axii. These were linked to another, seemingly unrelated, part of maths known as eliptic curves by the Taniyama-Shimura theorum which (as the first part of the Langland's Programme) became part of the quest for a Grand Unified Mathematics and forms the basis for some of the most important maths today. Indeed, the famed proof of Fermat's Last Theorum by Andrew Wiles was in fact also the proof of Taniyama-Shimura and so became dually celebrated as a triumph over an amateur tease-artist and as the stabilisation of an increasingly shaky foundation of an entire branch of maths.
Not bad for such a simple equation. |
# 1.7: Inverse Functions - Mathematics
Learning Objectives
• Verify inverse functions.
• Determine the domain and range of an inverse function, and restrict the domain of a function to make it one-to-one.
• Find or evaluate the inverse of a function.
• Use the graph of a one-to-one function to graph its inverse function on the same axes.
A reversible heat pump is a climate-control system that is an air conditioner and a heater in a single device. Operated in one direction, it pumps heat out of a house to provide cooling. Operating in reverse, it pumps heat into the building from the outside, even in cool weather, to provide heating. As a heater, a heat pump is several times more efficient than conventional electrical resistance heating.
If some physical machines can run in two directions, we might ask whether some of the function “machines” we have been studying can also run backwards. Figure (PageIndex{1}) provides a visual representation of this question. In this section, we will consider the reverse nature of functions.
Figure (PageIndex{1}): Can a function “machine” operate in reverse?
## Verifying That Two Functions Are Inverse Functions
Suppose a fashion designer traveling to Milan for a fashion show wants to know what the temperature will be. He is not familiar with the Celsius scale. To get an idea of how temperature measurements are related, he asks his assistant, Betty, to convert 75 degrees Fahrenheit to degrees Celsius. She finds the formula
[C=dfrac{5}{9}(F−32)]
and substitutes 75 for (F) to calculate
[dfrac{5}{9}(75−32)approx24^{circ}]
Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, he sends his assistant the week’s weather forecast from Figure (PageIndex{2}) for Milan, and asks her to convert all of the temperatures to degrees Fahrenheit.
At first, Betty considers using the formula she has already found to complete the conversions. After all, she knows her algebra, and can easily solve the equation for (F) after substituting a value for (C). For example, to convert 26 degrees Celsius, she could write
[egin{align} 26&=dfrac{5}{9}(F-32) 26⋅dfrac{9}{5}&=F−32 F&=26⋅dfrac{9}{5}+32approx79end{align}]
After considering this option for a moment, however, she realizes that solving the equation for each of the temperatures will be awfully tedious. She realizes that since evaluation is easier than solving, it would be much more convenient to have a different formula, one that takes the Celsius temperature and outputs the Fahrenheit temperature.
The formula for which Betty is searching corresponds to the idea of an inverse function, which is a function for which the input of the original function becomes the output of the inverse function and the output of the original function becomes the input of the inverse function.
Given a function (f(x)), we represent its inverse as (f^{−1}(x)), read as “(f) inverse of (x).” The raised −1 is part of the notation. It is not an exponent; it does not imply a power of −1 . In other words, (f^{−1}(x)) does not mean (frac{1}{f(x)}) because (frac{1}{f(x)}) is the reciprocal of (f) and not the inverse.
The “exponent-like” notation comes from an analogy between function composition and multiplication: just as (a^{−1}a=1) (1 is the identity element for multiplication) for any nonzero number (a), so (f^{−1}{circ}f) equals the identity function, that is,
[(f^{−1}{circ}f)(x)=f^{−1}(f(x))=f^{−1}(y)=x]
This holds for all (x) in the domain of (f). Informally, this means that inverse functions “undo” each other. However, just as zero does not have a reciprocal, some functions do not have inverses.
Given a function (f(x)), we can verify whether some other function (g(x)) is the inverse of (f(x)) by checking whether either (g(f(x))=x) or (f(g(x))=x) is true. We can test whichever equation is more convenient to work with because they are logically equivalent (that is, if one is true, then so is the other.)
For example, (y=4x) and (y=frac{1}{4}x) are inverse functions.
[(f^{−1}{circ}f)(x)=f^{-1}(4x)=dfrac{1}{4}(4x)=x]
and
[(f{circ}f^{−1})(x)=fBig(dfrac{1}{4}xBig)=4Big(dfrac{1}{4}xBig)=x]
A few coordinate pairs from the graph of the function (y=4x) are ((−2, −8)), ((0, 0)), and ((2, 8)). A few coordinate pairs from the graph of the function (y=frac{1}{4}x) are ((−8, −2)), ((0, 0)), and ((8, 2)). If we interchange the input and output of each coordinate pair of a function, the interchanged coordinate pairs would appear on the graph of the inverse function.
Definition: Inverse Function
For any one-to-one function (f(x)=y), a function (f^{−1}(x)) is an inverse function of (f) if (f^{−1}(y)=x). This can also be written as (f^{−1}(f(x))=x) for all (x) in the domain of (f). It also follows that (f(f^{−1}(x))=x) for all (x) in the domain of (f^{−1}) if (f^{−1}) is the inverse of (f).
The notation (f^{−1}) is read “(f) inverse.” Like any other function, we can use any variable name as the input for (f^{−1}), so we will often write (f^{−1}(x)), which we read as “(f) inverse of (x).” Keep in mind that
[f^{−1}(x) eqdfrac{1}{f(x)}]
and not all functions have inverses.
Example (PageIndex{1}): Identifying an Inverse Function for a Given Input-Output Pair
If for a particular one-to-one function (f(2)=4) and (f(5)=12), what are the corresponding input and output values for the inverse function?
Solution
The inverse function reverses the input and output quantities, so if
[f(2)=4, ext{ then } f^{-1}(4)=2 ; f(5)=12, ext{ then }f^{-1}(12)=5].
Alternatively, if we want to name the inverse function (g), then (g(4)=2) and (g(12)=5).
Analysis
Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed. See Table (PageIndex{1}).
Table (PageIndex{1})
((x,f(x)))((x,g(x)))
((2,4))((4,2))
((5,12))((12,5))
Exercise (PageIndex{1})
Given that (h^{-1}(6)=2), what are the corresponding input and output values of the original function (h)?
(h(2)=6)
How To: Given two functions (f(x)) and (g(x)), test whether the functions are inverses of each other.
1. Determine whether (f(g(x))=x) or (g(f(x))=x).
2. If both statements are true, then (g=f^{-1}) and (f=g^{-1}). If either statement is false, then both are false, and (g{ eq}f^{-1}) and (f{ eq}g^{-1}).
Example (PageIndex{2}): Testing Inverse Relationships Algebraically
If (f(x)=frac{1}{x+2}) and (g(x)=frac{1}{x}−2), is (g=f^{-1})?
Solution
[egin{align} g(f(x))&=dfrac{1}{(frac{1}{x+2})−2} &=x+2−2 &=x end{align}]
so
[g=f^{-1} ext{ and } f=g^{-1}]
This is enough to answer yes to the question, but we can also verify the other formula.
[egin{align} f(g(x))&=dfrac{1}{frac{1}{x}-2+2} &= dfrac{1}{frac{1}{x}} &=x end{align}]
Analysis
Notice the inverse operations are in reverse order of the operations from the original function.
Exercise (PageIndex{2})
If (f(x)=x^3−4) and (g(x)=sqrt[3]{x+4}), is (g=f^{-1})?
Yes
Example (PageIndex{3}): Determining Inverse Relationships for Power Functions
If (f(x)=x^3) (the cube function) and (g(x)=frac{1}{3}x), is (g=f^{-1})?
Solution
[f(g(x))=dfrac{x^3}{27}{ eq}x]
No, the functions are not inverses.
Analysis
The correct inverse to the cube is, of course, the cube root (sqrt[3]{x}=x^{frac{1}{3}}), that is, the one-third is an exponent, not a multiplier.
Exercise (PageIndex{3})
If (f(x)=(x−1)^3) and (g(x)=sqrt[3]{x}+1), is (g=f^{-1})?
Yes
## Finding Domain and Range of Inverse Functions
The outputs of the function (f) are the inputs to (f^{-1}), so the range of (f) is also the domain of (f^{-1}). Likewise, because the inputs to (f) are the outputs of (f^{-1}), the domain of (f) is the range of (f^{-1}). We can visualize the situation as in Figure (PageIndex{3}).
Figure (PageIndex{3}): Domain and range of a function and its inverse.
When a function has no inverse function, it is possible to create a new function where that new function on a limited domain does have an inverse function. For example, the inverse of (f(x)=sqrt{x}) is (f^{-1}(x)=x^2), because a square “undoes” a square root; but the square is only the inverse of the square root on the domain (left[0,infty ight)), since that is the range of (f(x)=sqrt{x}).
We can look at this problem from the other side, starting with the square (toolkit quadratic) function (f(x)=x^2). If we want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and –3. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the “inverse” is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an inverse function. In order for a function to have an inverse, it must be a one-to-one function.
In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function (f(x)=x^2) with its range limited to (left[0,infty ight)), which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).
If (f(x)=(x−1)^2) on ([1,∞)), then the inverse function is (f^{-1}(x)=sqrt{x}+1).
• The domain of (f) = range of (f^{-1} = left[1,infty ight)).
• The domain of (f^{-1}) = range of (f = left[0,infty ight)).
Is it possible for a function to have more than one inverse?
No. If two supposedly different functions, say, (g) and h, both meet the definition of being inverses of another function (f), then you can prove that (g=h). We have just seen that some functions only have inverses if we restrict the domain of the original function. In these cases, there may be more than one way to restrict the domain, leading to different inverses. However, on any one domain, the original function still has only one unique inverse.
Note: Domain and Range of Inverse Functions
The range of a function (f(x)) is the domain of the inverse function (f^{-1}(x)).
The domain of (f(x)) is the range of (f^{-1}(x)).
How To: Given a function, find the domain and range of its inverse.
1. If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse.
2. If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function.
Example (PageIndex{4}): Finding the Inverses of Toolkit Functions
Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The toolkit functions are reviewed in Table (PageIndex{2}). We restrict the domain in such a fashion that the function assumes all y-values exactly once.
Table (PageIndex{2})
(f(x)=c)(f(x)=x)(f(x)=x^2)(f(x)=x^3)(f(x)=frac{1}{x})
Reciprocal squaredCube RootSquare RootAbsolute Value
(f(x)=frac{1}{x^2})(f(x)=sqrt[3]{x})(f(x)=sqrt{x})(f(x)=|x|)
Solution
The constant function is not one-to-one, and there is no domain (except a single point) on which it could be one-to-one, so the constant function has no meaningful inverse.
The absolute value function can be restricted to the domain (left[0,infty ight)),where it is equal to the identity function.
The reciprocal-squared function can be restricted to the domain ((0,infty)).
Analysis
We can see that these functions (if unrestricted) are not one-to-one by looking at their graphs, shown in Figure (PageIndex{4}). They both would fail the horizontal line test. However, if a function is restricted to a certain domain so that it passes the horizontal line test, then in that restricted domain, it can have an inverse.
Figure (PageIndex{4}): (a) Absolute value (b) Reciprocal squared
(PageIndex{4}): The domain of function (f) is ((1,infty)) and the range of function (f) is ((−infty,−2)). Find the domain and range of the inverse function.
Solution
The domain of function (f^{-1}) is ((−infty,−2)) and the range of function (f^{-1}) is ((1,infty)).
## Finding and Evaluating Inverse Functions
Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases.
### Inverting Tabular Functions
Suppose we want to find the inverse of a function represented in table form. Remember that the domain of a function is the range of the inverse and the range of the function is the domain of the inverse. So we need to interchange the domain and range.
Each row (or column) of inputs becomes the row (or column) of outputs for the inverse function. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function.
Example (PageIndex{5}): Interpreting the Inverse of a Tabular Function
A function (f(t)) is given in Table (PageIndex{3}), showing distance in miles that a car has traveled in (t) minutes. Find and interpret (f^{-1}(70))
(t) (minutes) (f(t)) (miles) 30 50 70 90 20 40 60 70
The inverse function takes an output of (f) and returns an input for (f). So in the expression (f^{-1}(70)), 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function (f), 90 minutes, so (f^{-1}(70)=90). The interpretation of this is that, to drive 70 miles, it took 90 minutes.
Alternatively, recall that the definition of the inverse was that if (f(a)=b), then (f^{-1}(b)=a). By this definition, if we are given (f^{-1}(70)=a), then we are looking for a value (a) so that (f(a)=70). In this case, we are looking for a (t) so that (f(t)=70), which is when (t=90).
Exercise (PageIndex{5})
Using Table (PageIndex{4}), find and interpret (a) (f(60)),and (b) (f^{-1}(60)).
(t) (minutes) (f(t)) (miles) 30 50 60 70 90 20 40 50 60 70
(f(60)=50). In 60 minutes, 50 miles are traveled.
(f^{-1}(60)=70). To travel 60 miles, it will take 70 minutes.
### Evaluating the Inverse of a Function, Given a Graph of the Original Function
We saw in Functions and Function Notation that the domain of a function can be read by observing the horizontal extent of its graph. We find the domain of the inverse function by observing the vertical extent of the graph of the original function, because this corresponds to the horizontal extent of the inverse function. Similarly, we find the range of the inverse function by observing the horizontal extent of the graph of the original function, as this is the vertical extent of the inverse function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function’s graph.
Given the graph of a function, evaluate its inverse at specific points.
1. Find the desired input on the y-axis of the given graph.
2. Read the inverse function’s output from the x-axis of the given graph.
Example (PageIndex{6}): Evaluating a Function and Its Inverse from a Graph at Specific Points
A function (g(x)) is given in Figure (PageIndex{5}). Find (g(3)) and (g^{-1}(3)).
.
Solution
To evaluate (g(3)), we find 3 on the x-axis and find the corresponding output value on the y-axis. The point ((3,1)) tells us that (g(3)=1).
To evaluate (g^{-1}(3)), recall that by definition (g^{-1}(3)) means the value of (x) for which (g(x)=3). By looking for the output value 3 on the vertical axis, we find the point ((5,3)) on the graph, which means (g(5)=3), so by definition, (g^{-1}(3)=5.) See Figure (PageIndex{6}).
Exercise (PageIndex{6})
Using the graph in Figure (PageIndex{6}), (a) find (g^{-1}(1)),and (b) estimate (g^{-1}(4)).
3
5.6
### Finding Inverses of Functions Represented by Formulas
Sometimes we will need to know an inverse function for all elements of its domain, not just a few. If the original function is given as a formula— for example, (y) as a function of (x)— we can often find the inverse function by solving to obtain (x) as a function of (y).
How To: Given a function represented by a formula, find the inverse.
1. Make sure (f) is a one-to-one function.
2. Solve for (x)
3. Interchange (x) and (y).
Example (PageIndex{7}): Inverting the Fahrenheit-to-Celsius Function
Find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature.
[C=dfrac{5}{9}(F−32)]
Solution
[egin{align} C&=frac{5}{9}(F-32) C{cdot}frac{9}{5}&=F−32 F&=frac{9}{5}C+32end{align}]
By solving in general, we have uncovered the inverse function. If
[C=h(F)=dfrac{5}{9}(F−32)],
then
[F=h^{-1}(C)=dfrac{9}{5}C+32.]
In this case, we introduced a function (h) to represent the conversion because the input and output variables are descriptive, and writing (C^{-1}) could get confusing.
Exercise (PageIndex{7})
Solve for (x) in terms of (y) given (y=frac{1}{3}(x−5))
(x=3y+5)
Example (PageIndex{8}): Solving to Find an Inverse Function
Find the inverse of the function (f(x)=frac{2}{x−3}+4).
Solution
[egin{align} y&=dfrac{2}{x−3+4} & ext{Set up an equation.} y−4&=dfrac{2}{x−3} & ext{Subtract 4 from both sides.} x−3&=dfrac{2}{y−4} & ext{Multiply both sides by x−3 and divide by y−4.} x&=dfrac{2}{y−4}+3 & ext{Add 3 to both sides.} end{align}]
So (f^{-1}(y)=frac{2}{y−4}+3) or (f^{-1}(x)=frac{2}{x−4}+3).
Analysis
The domain and range of (f) exclude the values 3 and 4, respectively. (f) and (f^{-1}) are equal at two points but are not the same function, as we can see by creating Table (PageIndex{5}).
(x) (f(x)) 1 2 5 (f^{-1}(y)) 3 2 5 (y)
Example (PageIndex{9}): Solving to Find an Inverse with Radicals
Find the inverse of the function (f(x)=2+sqrt{x−4}).
Solution
[ egin{align} y&=2+sqrt{x-4} (y-2)^2&=x-4 x&=(y-2)^2+4 end{align}]
So (f^{-1}(x)=(x−2)^2+4).
The domain of (f) is (left[4,infty ight)). Notice that the range of (f) is (left[2,infty ight)), so this means that the domain of the inverse function (f^{-1}) is also (left[2,infty ight))
Analysis
The formula we found for (f^{-1}(x)) looks like it would be valid for all real (x). However, (f^{-1}) itself must have an inverse (namely, (f) ) so we have to restrict the domain of (f^{-1}) to (left[2,infty ight)) in order to make (f^{-1}) a one-to-one function. This domain of (f^{-1}) is exactly the range of (f).
Exercise (PageIndex{8})
What is the inverse of the function (f(x)=2-sqrt{x})? State the domains of both the function and the inverse function.
(f^{-1}(x)=(2−x)^2); domain of (f): (left[0,infty ight)); domain of (f^{-1}): (left(−infty,2 ight])
## Finding Inverse Functions and Their Graphs
Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function (f(x)=x^2) restricted to the domain (left[0,infty ight)), on which this function is one-to-one, and graph it as in Figure (PageIndex{7}).
Figure (PageIndex{7}): Quadratic function with domain restricted to ([0, infty)).
Restricting the domain to (left[0,infty ight)) makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain.
We already know that the inverse of the toolkit quadratic function is the square root function, that is, (f^{-1}(x)=sqrt{x}). What happens if we graph both (f) and (f^{-1}) on the same set of axes, using the x-axis for the input to both (f) and (f^{-1})?
We notice a distinct relationship: The graph of (f^{-1}(x)) is the graph of (f(x)) reflected about the diagonal line (y=x), which we will call the identity line, shown in Figure (PageIndex{8}).
.
Figure (PageIndex{8}): Square and square-root functions on the non-negative domain
This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes.
Example (PageIndex{10}): Finding the Inverse of a Function Using Reflection about the Identity Line
Given the graph of (f(x)) in Figure (PageIndex{9}), sketch a graph of (f^{-1}(x)).
This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of ((0,infty)) and range of ((−infty,infty)), so the inverse will have a domain of ((−infty,infty)) and range of ((0,infty)).
If we reflect this graph over the line (y=x), the point ((1,0)) reflects to ((0,1)) and the point ((4,2)) reflects to ((2,4)). Sketching the inverse on the same axes as the original graph gives Figure (PageIndex{10}).
Exercise (PageIndex{1})
Draw graphs of the functions (f) and (f^{-1}) from Example (PageIndex{8}).
Is there any function that is equal to its own inverse?
Yes. If (f=f^{-1}), then (f(f(x))=x), and we can think of several functions that have this property. The identity function
does, and so does the reciprocal function, because
[dfrac{1}{frac{1}{x}}=x]
Any function (f(x)=c−x), where (c) is a constant, is also equal to its own inverse.
## Key Concepts
• If (g(x)) is the inverse of (f(x)), then (g(f(x))=f(g(x))=x).
• Each of the toolkit functions has an inverse.
• For a function to have an inverse, it must be one-to-one (pass the horizontal line test).
• A function that is not one-to-one over its entire domain may be one-to-one on part of its domain.
• For a tabular function, exchange the input and output rows to obtain the inverse.
• The inverse of a function can be determined at specific points on its graph.
• To find the inverse of a formula, solve the equation (y=f(x)) for (x) as a function of (y). Then exchange the labels (x) and (y).
• The graph of an inverse function is the reflection of the graph of the original function across the line (y=x).
## 1.7 Integrals Resulting in Inverse Trigonometric Functions
In this section we focus on integrals that result in inverse trigonometric functions. We have worked with these functions before. Recall from Functions and Graphs that trigonometric functions are not one-to-one unless the domains are restricted. When working with inverses of trigonometric functions, we always need to be careful to take these restrictions into account. Also in Derivatives, we developed formulas for derivatives of inverse trigonometric functions. The formulas developed there give rise directly to integration formulas involving inverse trigonometric functions.
### Integrals that Result in Inverse Sine Functions
Let us begin this last section of the chapter with the three formulas. Along with these formulas, we use substitution to evaluate the integrals. We prove the formula for the inverse sine integral.
### Rule: Integration Formulas Resulting in Inverse Trigonometric Functions
The following integration formulas yield inverse trigonometric functions. Assume a > 0 a > 0 :
## 1.7: Inverse Functions - Mathematics
In the last example from the previous section we looked at the two functions (fleft( x ight) = 3x - 2) and (gleft( x ight) = frac <3>+ frac<2><3>) and saw that
[left( ight)left( x ight) = left( ight)left( x ight) = x]
and as noted in that section this means that these are very special functions. Let’s see just what makes them so special. Consider the following evaluations.
In the first case we plugged (x = - 1) into (fleft( x ight)) and got a value of -5. We then turned around and plugged (x = - 5) into (gleft( x ight)) and got a value of -1, the number that we started off with.
In the second case we did something similar. Here we plugged (x = 2) into (gleft( x ight)) and got a value of(frac<4><3>), we turned around and plugged this into (fleft( x ight)) and got a value of 2, which is again the number that we started with.
Note that we really are doing some function composition here. The first case is really,
[left( ight)left( < - 1> ight) = gleft[ ight)> ight] = gleft[ < - 5> ight] = - 1]
and the second case is really,
Note as well that these both agree with the formula for the compositions that we found in the previous section. We get back out of the function evaluation the number that we originally plugged into the composition.
So, just what is going on here? In some way we can think of these two functions as undoing what the other did to a number. In the first case we plugged (x = - 1) into (fleft( x ight)) and then plugged the result from this function evaluation back into (gleft( x ight)) and in some way (gleft( x ight)) undid what (fleft( x ight)) had done to (x = - 1) and gave us back the original (x) that we started with.
Function pairs that exhibit this behavior are called inverse functions. Before formally defining inverse functions and the notation that we’re going to use for them we need to get a definition out of the way.
A function is called one-to-one if no two values of (x) produce the same (y). This is a fairly simple definition of one-to-one but it takes an example of a function that isn’t one-to-one to show just what it means. Before doing that however we should note that this definition of one-to-one is not really the mathematically correct definition of one-to-one. It is identical to the mathematically correct definition it just doesn’t use all the notation from the formal definition.
Now, let’s see an example of a function that isn’t one-to-one. The function (fleft( x ight) = ) is not one-to-one because both (fleft( < - 2> ight) = 4) and (fleft( 2 ight) = 4). In other words, there are two different values of (x) that produce the same value of (y). Note that we can turn (fleft( x ight) = ) into a one-to-one function if we restrict ourselves to (0 le x < infty ). This can sometimes be done with functions.
Showing that a function is one-to-one is often a tedious and difficult process. For the most part we are going to assume that the functions that we’re going to be dealing with in this section are one-to-one. We did need to talk about one-to-one functions however since only one-to-one functions can be inverse functions.
Now, let’s formally define just what inverse functions are.
#### Inverse Functions
Given two one-to-one functions (fleft( x ight)) and (gleft( x ight)) if
then we say that (fleft( x ight)) and (gleft( x ight)) are inverses of each other. More specifically we will say that (gleft( x ight)) is the inverse of (fleft( x ight)) and denote it by
[gleft( x ight) = >left( x ight)]
Likewise, we could also say that (fleft( x ight)) is the inverse of (gleft( x ight)) and denote it by
[fleft( x ight) = >left( x ight)]
The notation that we use really depends upon the problem. In most cases either is acceptable.
For the two functions that we started off this section with we could write either of the following two sets of notation.
Now, be careful with the notation for inverses. The “-1” is NOT an exponent despite the fact that it sure does look like one! When dealing with inverse functions we’ve got to remember that
This is one of the more common mistakes that students make when first studying inverse functions.
The process for finding the inverse of a function is a fairly simple one although there is a couple of steps that can on occasion be somewhat messy. Here is the process
#### Finding the Inverse of a Function
Given the function (fleft( x ight)) we want to find the inverse function, (>left( x ight)).
1. First, replace (fleft( x ight)) with (y). This is done to make the rest of the process easier.
2. Replace every (x) with a (y) and replace every (y) with an (x).
3. Solve the equation from Step 2 for (y). This is the step where mistakes are most often made so be careful with this step.
4. Replace (y) with (>left( x ight)). In other words, we’ve managed to find the inverse at this point!
5. Verify your work by checking that (left( >> ight)left( x ight) = x) and (left( <> circ f> ight)left( x ight) = x) are both true. This work can sometimes be messy making it easy to make mistakes so again be careful.
That’s the process. Most of the steps are not all that bad but as mentioned in the process there are a couple of steps that we really need to be careful with.
In the verification step we technically really do need to check that both (left( >> ight)left( x ight) = x) and (left( <> circ f> ight)left( x ight) = x) are true. For all the functions that we are going to be looking at in this section if one is true then the other will also be true. However, there are functions (they are far beyond the scope of this course however) for which it is possible for only one of these to be true. This is brought up because in all the problems here we will be just checking one of them. We just need to always remember that technically we should check both.
Now, we already know what the inverse to this function is as we’ve already done some work with it. However, it would be nice to actually start with this since we know what we should get. This will work as a nice verification of the process.
So, let’s get started. We’ll first replace (fleft( x ight)) with (y).
Next, replace all (x)’s with (y) and all y’s with (x).
Finally replace (y) with (>left( x ight)).
Now, we need to verify the results. We already took care of this in the previous section, however, we really should follow the process so we’ll do that here. It doesn’t matter which of the two that we check we just need to check one of them. This time we’ll check that (left( >> ight)left( x ight) = x) is true.
Now the fact that we’re now using (gleft( x ight)) instead of (fleft( x ight)) doesn’t change how the process works. Here are the first few steps.
Now, to solve for (y) we will need to first square both sides and then proceed as normal.
Finally let’s verify and this time we’ll use the other one just so we can say that we’ve gotten both down somewhere in an example.
So, we did the work correctly and we do indeed have the inverse.
Before we move on we should also acknowledge the restrictions of (x ge 0) that we gave in the problem statement but never apparently did anything with. Note that this restriction is required to make sure that the inverse, (>left( x ight)) given above is in fact one-to-one.
Without this restriction the inverse would not be one-to-one as is easily seen by a couple of quick evaluations.
Therefore, the restriction is required in order to make sure the inverse is one-to-one.
The next example can be a little messy so be careful with the work here.
The first couple of steps are pretty much the same as the previous examples so here they are,
Now, be careful with the solution step. With this kind of problem it is very easy to make a mistake here.
[eginxleft( <2y - 5> ight) & = y + 4 2xy - 5x & = y + 4 2xy - y & = 4 + 5x left( <2x - 1> ight)y & = 4 + 5x y & = frac<<4 + 5x>><<2x - 1>>end]
So, if we’ve done all of our work correctly the inverse should be,
Finally, we’ll need to do the verification. This is also a fairly messy process and it doesn’t really matter which one we work with.
Okay, this is a mess. Let’s simplify things up a little bit by multiplying the numerator and denominator by (2x - 1).
Wow. That was a lot of work, but it all worked out in the end. We did all of our work correctly and we do in fact have the inverse.
There is one final topic that we need to address quickly before we leave this section. There is an interesting relationship between the graph of a function and its inverse.
Here is the graph of the function and inverse from the first two examples. We’ll not deal with the final example since that is a function that we haven’t really talked about graphing yet.
In both cases we can see that the graph of the inverse is a reflection of the actual function about the line (y = x). This will always be the case with the graphs of a function and its inverse.
An inverse function is a function which does the “reverse” of a given function. More formally, if (f) is a function with domain (X), then (^<-1>) is its inverse function if and only if (^<-1>left(fleft(x ight) ight)=x) for every (x in X).
A function must be a one-to-one relation if its inverse is to be a function. If a function (f) has an inverse function (f^<-1>), then (f) is said to be invertible.
Given the function (f(x)), we determine the inverse (f^<-1>(x)) by:
• interchanging (x) and (y) in the equation
• making (y) the subject of the equation
• expressing the new equation in function notation.
Note: if the inverse is not a function then it cannot be written in function notation. For example, the inverse of (f(x) = 3x^2) cannot be written as (f^<-1>(x) = pm sqrt<3>x>) as it is not a function. We write the inverse as (y = pm sqrt<3>x>) and conclude that (f) is not invertible.
If we represent the function (f) and the inverse function (^<-1>) graphically, the two graphs are reflected about the line (y=x). Any point on the line (y = x) has (x)- and (y)-coordinates with the same numerical value, for example ((-3-3)) and (left( frac<4><5> frac<4> <5> ight)). Therefore interchanging the (x)- and (y)-values makes no difference.
This diagram shows an exponential function (black graph) and its inverse (blue graph) reflected about the line (y = x) (grey line).
Important: for (^<-1>), the superscript (- ext<1>) is not an exponent. It is the notation for indicating the inverse of a function. Do not confuse this with exponents, such as (left( frac<1> <2> ight)^<-1>) or (3 + x^<-1>).
Be careful not to confuse the inverse of a function and the reciprocal of a function:
## Inverse Functions Contents: This page corresponds to § 1.7 (p. 150) of the text. Suggested Problems from Text p.158 #1-4, 5, 8, 9, 12, 13, 15, 18, 21, 22, 27, 31, 34, 37, 46, 48, 51, 71, 74, 83 Definition of Inverse Function
Before defining the inverse of a function we need to have the right mental image of function.
Consider the function f(x) = 2x + 1. We know how to evaluate f at 3, f(3) = 2*3 + 1 = 7. In this section it helps to think of f as transforming a 3 into a 7, and f transforms a 5 into an 11, etc.
Now that we think of f as "acting on" numbers and transforming them, we can define the inverse of f as the function that "undoes" what f did. In other words, the inverse of f needs to take 7 back to 3, and take -3 back to -2, etc.
Let g(x) = (x - 1)/2. Then g(7) = 3, g(-3) = -2, and g(11) = 5, so g seems to be undoing what f did, at least for these three values. To prove that g is the inverse of f we must show that this is true for any value of x in the domain of f. In other words, g must take f(x) back to x for all values of x in the domain of f. So, g(f(x)) = x must hold for all x in the domain of f. The way to check this condition is to see that the formula for g(f(x)) simplifies to x.
g(f(x)) = g(2x + 1) = (2x + 1 -1)/2 = 2x/2 = x.
This simplification shows that if we choose any number and let f act it, then applying g to the result recovers our original number. We also need to see that this process works in reverse, or that f also undoes what g does.
f(g(x)) = f((x - 1)/2) = 2(x - 1)/2 + 1 = x - 1 + 1 = x.
Letting f -1 denote the inverse of f, we have just shown that g = f -1 .
Let f and g be two functions. If
f(g(x)) = x and g(f(x)) = x,
then g is the inverse of f and f is the inverse of g.
(a) Open the Java Calculator and enter the formulas for f and g. Note that you take a cube root by raising to the (1/3), and you do need to enter the exponent as (1/3), and not a decimal approximation. So the text for the g box will be
(x - 2)^(1/3)
Use the calculator to evaluate f(g(4)) and g(f(-3)). g is the inverse of f, but due to round off error, the calculator may not return the exact value that you start with. Try f(g(-2)). The answers will vary for different computers. However, on our test machine f(g(4)) returned 4 g(f(-3)) returned 3 but, f(g(-2)) returned -1.9999999999999991, which is pretty close to -2.
The calculator can give us a good indication that g is the inverse of f, but we cannot check all possible values of x.
(b) Prove that g is the inverse of f by simplifying the formulas for f(g(x) and g(f(x)).
## Graphs of Inverse Functions
We have seen examples of reflections in the plane. The reflection of a point (a,b) about the x-axis is (a,-b), and the reflection of (a,b) about the y-axis is (-a,b). Now we want to reflect about the line y = x.
The reflection of the point (a,b) about the line y = x is the point (b,a) .
Let f(x) = x 3 + 2. Then f(2) = 10 and the point (2,10) is on the graph of f. The inverse of f must take 10 back to 2, i.e. f -1 (10)=2, so the point (10,2) is on the graph of f -1 . The point (10,2) is the reflection in the line y = x of the point (2,10). The same argument can be made for all points on the graphs of f and f -1 .
The graph of f -1 is the reflection about the line y = x of the graph of f.
## Existence of an Inverse
Some functions do not have inverse functions. For example, consider f(x) = x 2 . There are two numbers that f takes to 4, f(2) = 4 and f(-2) = 4. If f had an inverse, then the fact that f(2) = 4 would imply that the inverse of f takes 4 back to 2. On the other hand, since f(-2) = 4, the inverse of f would have to take 4 to -2. Therefore, there is no function that is the inverse of f.
Look at the same problem in terms of graphs. If f had an inverse, then its graph would be the reflection of the graph of f about the line y = x. The graph of f and its reflection about y = x are drawn below.
Note that the reflected graph does not pass the vertical line test, so it is not the graph of a function.
This generalizes as follows: A function f has an inverse if and only if when its graph is reflected about the line y = x, the result is the graph of a function (passes the vertical line test). But this can be simplified. We can tell before we reflect the graph whether or not any vertical line will intersect more than once by looking at how horizontal lines intersect the original graph!
### Horizontal Line Test
Let f be a function.
If any horizontal line intersects the graph of f more than once, then f does not have an inverse.
If no horizontal line intersects the graph of f more than once, then f does have an inverse.
The property of having an inverse is very important in mathematics, and it has a name.
Definition : A function f is one-to-one if and only if f has an inverse.
The following definition is equivalent, and it is the one most commonly given for one-to-one.
Alternate Definition : A function f is one-to-one if, for every a and b in its domain, f(a) = f(b) implies a = b.
Graph the following functions and determine whether or not they have inverses.
(a) f(x) = (x - 3) x 2 . Answer
(b) f(x) = x 3 + 3x 2 +3x. Answer
(c) f(x) = x ^(1/3) ( the cube root of x). Answer
## Finding Inverses
Example 1. First consider a simple example f(x) = 3x + 2 .
The graph of f is a line with slope 3, so it passes the horizontal line test and does have an inverse.
There are two steps required to evaluate f at a number x. First we multiply x by 3, then we add 2.
Thinking of the inverse function as undoing what f did, we must undo these steps in reverse order.
The steps required to evaluate f -1 are to first undo the adding of 2 by subtracting 2. Then we undo multiplication by 3 by dividing by 3.
Therefore, f -1 (x) = (x - 2)/3.
Steps for finding the inverse of a function f.
1. Replace f(x) by y in the equation describing the function.
2. Interchange x and y. In other words, replace every x by a y and vice versa.
3. Solve for y.
4. Replace y by f -1 (x).
Step 1 y = 6 - x/2. Step 2 x = 6 - y/2. Step 3 x = 6 - y/2. y = 12 - 2x. Step 4 f -1 (x) = 12 - 2x.
Step 2 often confuses students. We could omit step 2, and solve for x instead of y, but then we would end up with a formula in y instead of x. The formula would be the same, but the variable would be different. To avoid this we simply interchange the roles of x and y before we solve.
This is the function we worked with in Exercise 1. From its graph (shown above) we see that it does have an inverse. (In fact, its inverse was given in Exercise 1.)
Step 1 y = x 3 + 2. Step 2 x = y 3 + 2. Step 3 x - 2 = y 3 .
(x - 2)^(1/3) = y. Step 4 f -1 (x) = (x - 2)^(1/3).
Graph f(x) = 1 - 2x 3 to see that it does have an inverse. Find f -1 (x). Answer
## MAT 112 Ancient and Contemporary Mathematics
Sometimes it is possible to “undo” the effect of a function. In these cases, we can define a function that reverses the effects of another function. A function that we can “undo” is called invertible.
In the video in Figure 7.5.1 we introduce inverse functions and give examples. For details and examples, see our treatment of inverse
We give the formal definition of an invertible function and of the inverse of an invertible function.
###### Definition 7.5.2 .
Let (A) and (B) be non-empty sets. We say that a function (f:A o B) is if for every (bin B) there is exactly one (ain A) such that (f(a)=b ext<.>) The of an invertible function (f:A o B ext<,>) denoted by (f^<-1> ext<,>) is the function (f^<-1>:B o A) that assigns to each element (b in B) the unique element (a in A) such that (f(a) = b ext<.>)
In other words, a function (f:A o B) is invertible if every (bin B) has exactly one preimage (ain A ext<.>) So if (f(a) = b ext<,>) then (f^<-1>(b) = a ext<.>)
Confirm your understanding of the definition by completing it in the exercise.
###### Checkpoint 7.5.3 . Definition of invertible function.
We investigate whether the two functions (mathrm ) and ( ext) are invertible.
###### Example 7.5.5 . Are (mathrm ) and ( ext) invertible ?
We use the functions (mathrm colon N o I) and (mathrm colon I o G) from Example 7.1.5 and Example 7.1.7 given by the tables in Figure 7.1.4 and Figure 7.1.6 respectively.
The function (mathrm colon N o I) where (I) is the set of student identification numbers and (N) is the set of student names is invertible as long as every student has a different name. The function (mathrm^<-1>: I o N) is the function that tells us the student's name for a given identification number. With the table in Figure 7.1.4 we get
Recall the function grade from Example 7.1.7. The function (mathrm colon I o G) where (I) is the set of identification numbers and (G) is the set of grades is not invertible since many students may earn the same grade in a class. Both the students with the identification number 1007 and 1008 earn an A in MAT 112. We have (mathrm ( ext < 1007 >) = ext) and ( ext ( ext < 1008 >) = ext ext<,>) and we would not be able to uniquely define (mathrm^<-1>(A) ext<.>)
In Figure 7.5.6 we give an example of an invertible function from (_5) to (_5) and its inverse. The function (e) in Figure 7.5.9 illustrates that for an invertible function the domain and codomain do not have to be the same.
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## Math: How to Find the Inverse of a Function
The inverse function of a function f is mostly denoted as f -1 . A function f has an input variable x and gives then an output f(x). The inverse of a function f does exactly the opposite. Instead it uses as input f(x) and then as output it gives the x that when you would fill it in in f will give you f(x). To be more clear:
If f(x) = y then f -1 (y) = x. So the output of the inverse is indeed the value that you should fill in in f to get y. So f(f -1 (x)) = x.
Not every function has an inverse. A function that does have an inverse is called invertible. Only if f is bijective an inverse of f will exist. But what does this mean?
The easy explanation of a function that is bijective is a function that is both injective and surjective. However, for most of you this will not make it any clearer.
A function is injective if there are no two inputs that map to the same output. Or said differently: every output is reached by at most one input.
An example of a function that is not injective is f(x) = x 2 if we take as domain all real numbers. If we fill in -2 and 2 both give the same output, namely 4. So x 2 is not injective and therefore also not bijective and hence it won&apost have an inverse.
A function is surjective if every possible number in the range is reached, so in our case if every real number can be reached. So f(x)= x 2 is also not surjective if you take as range all real numbers, since for example -2 cannot be reached since a square is always positive.
So while you might think that the inverse of f(x) = x 2 would be f -1 (y) = sqrt(y) this is only true when we treat f as a function from the nonnegative numbers to the nonnegative numbers, since only then it is a bijection.
This does show that the inverse of a function is unique, meaning that every function has only one inverse.
## More Questions with Solutions
Use the table below to find the following if possible:
1) g -1 (0) , b) g -1 (-10) , c) g -1 (- 5) , d) g -1 (-7) , e) g -1 (3)
Solution
a) According to the the definition of the inverse function:
a = g -1 (0) if and only if g(a) = 0
Which means that a is the value of x such g(x) = 0.
Using the table above for x = 11, g(x) = 0. Hence a = 11 and therefore g -1 (0) = 11
b) a = g - 1 (- 5) if and only if g(a) = - 5
The value of x for which g(x) = - 5 is equal to 0 and therefore g -1 ( - 5) = 0
c) a = g -1 (-10) if and only if g(a) = - 10
There is no value of x for which g(x) = -10 and therefore g -1 (-10) is undefined.
d) a = g -1 (- 7) if and only if g(a) = - 7
There no value of x for which g(x) = - 7 and therefore g -1 (- 7) is undefined.
e) a = g -1 (3) if and only if g(a) = 3
The value of x for which g(x) = 3 is equal to - 2 and therefore g -1 (3) = - 2
## Questions on Inverse Functions with Solutions and Answers
Analytical and graphing methods are used to solve maths problems and questions related to inverse functions. Detailed solutions are also presented. Several questions involve the use of the property that the graphs of a function and the graph of its inverse are reflection of each other on the line y = x.
1) Sketch the graph of the inverse of f in the same system of axes.
2) Find the inverse of and check your answer using some points.
Solution
1) Locate few points on the graph of f. Here is a list of points whose coordinates (a , b) can easily be determined from the graph:
(1 , 1) , (0 , -1) , (-1 , -3)
On the graph of the inverse function, the above points will have coordinates (b , a) as follows:
(1 , 1) , (-1 , 0) , (-3 , -1)
Plot the above points and sketch the graph of the inverse of f so that the two graphs are reflection of each other on the line y = x as shown below.
1) Sketch the inverse of f in the same graph.
2) Find the inverse of and check your answer using some points.
Solution
1) Locate few points on the graph of f. A possible list of points whose coordinates (a , b) is as follows:
(1.5 , 0) , (2 , 1) , (6 , 3)
On the graph of the inverse function, the above points will have coordinates (b , a) as follows:
(0 , 1.5) , (1 , 2) , (3 , 6)
Plot the above points and sketch the graph of the inverse of f so that the two graphs are reflection of each other on the line y = x as shown below.
2) Write the given function f(x) = √(2 x - 3) as an equation in two unknowns.
y = √(2 x - 3)
Solve the above for x. First square both sides
2 x - 3 = y 2
2 x = y 2 + 3
x = (y 2 + 3) / 2
Interchange x and y and write the equation of the inverse function f -1 and write the domain of the inverse.
y = (x 2 + 3) / 2
f -1 (x) = (x 2 + 3) / 2 , x ≥ 0 (domain which is the range of f from its graph above)
We now verify that the points (0 , 1.5) , (1 , 2) and (3 , 6) used to sketch the graph of the inverse function are on the graph of f -1 .
f -1 (0) = (0 2 + 3) / 2 = 1.5
f -1 (1) = (1 2 + 3) / 2 = 2
f -1 (3) = (3 2 + 3) / 2 = 6
Solution
1) Use the graph to find points on the graph of f. A possible list of points whose coordinates (a , b) is as follows:
(0 , 3) , (2 , -1) , (5 , - 3)
On the graph of the inverse function, the above points will have coordinates (b , a) as follows:
(3 , 0) , (- 1 , 2) , (- 3 , 5)
Plot the above points and sketch the graph of the inverse of f so that the two graphs are reflection of each other on the line y = x as shown below.
2) We now determine f -1 (x). For -3 ≤ x ≤ - 1 , f -1 (x) has a linear expression with slope m1 through the points (- 1 , 2) , (- 3 , 5) given by
m1 = (5 - 2) / (-3 - (-1)) = - 3 / 2
For -3 ≤ x ≤ - 1, f -1 (x) is given by:
f -1 (x) = - (3 / 2)(x - (-1)) + 2 = - (3 / 2)(x + 1) + 2
For - 1 < x ≤ 3 , f -1 (x) has a linear expression with slope through the points (- 1 , 2) , (3 , 0) given by
m2 = (0 - 2) / (3 - (-1)) = - 1 / 2
For - 1 < x ≤ 3, f -1 (x) is given by:
f -1 (x) = - (1 / 2)(x - (-1)) + 2 = - (1 / 2)(x + 1) + 2
1) What is the domain and range of f?
2) Sketch the graph of f -1 .
3) Find f -1 (x) (include domain).
Solution
1) f(x) is defined as a real number if the radicand 2 / x - 1 is greater than or equal to 0. Hence we need to solve the inequality:
2 / x - 1 ≥ 0
(2 - x) / x ≥ 0
The expression on the left of the inequality changes sign at the zeros of the numerator and denominator which are x = 2 and x = 0. See table below.
Domain: (0 , 2]
Range: (-∞ , 0]
2) Points on the graph of f
(2 , 0) , (1 , -1)
The above points on the graph of the inverse function, will have coordinates (b , a) as follows:
(0 , 2) , (- 1 , 1)
Plot the above points and sketch the graph of the inverse of f so that the two graphs are reflection of each other on the line y = x as shown below.
3) Write f(x) as an equation in y and x.
( y = -sqrt-1> )
Solve the above equation for x. Square both sides of the above equation
( y^2 = dfrac<2>-1 )
( dfrac<2> = y^2 + 1 )
( x = dfrac<2> )
Interchange x and y and write the inverse function
( y = dfrac<2> )
( f^<-1>(x) = dfrac<2> )
Domain and range of f -1 are the range and domain of f . Hence
Domain of f -1 : (-∞ , 0]
Range of f -1 : (0 , 2]
Solution
For each graph, select points whose coordinates are easy to determine. Use these points and also the reflection of the graph of function f and its inverse on the line y = x to skectch to sketch the inverse functions as shown below |
Unitary Matrix: Definition, Properties, Formula & Examples
Unitary Matrix is a non-singular matrix with complex numbers. The product of conjugation of unitary matrix and transpose of unitary matrix results in the Identity matrix. Students can find detailed information regarding the unitary matrix likewise definition, formula, properties, and examples from here. Thus have a look at this page to learn in detail about the unitary matrix.
Unitary Matrix Definition
A unitary matrix is a matrix whose inverse is equal to the conjugate transpose. Conjugate transpose is referred to as the Hermitian matrix. The unitary matrices may also be non-square matrices but the matrices have orthonormal columns. The unitary matrix is denoted by U.
Unitary Matrix Formula
Formulas of a unitary matrix are
U*U = UU* = I
where I is an identity matrix
Properties of Unitary Matrix
1. The two complex vectors are x and y, this x and y are multiplied by U that is, (Ux, Uy) = (x, y).
2. U is normal (U*U = UU*)
3. U is diagonalizable, that is, U is unitarily similar to a diagonal matrix, as a consequence of the spectral theorem.
Therefore U is a decomposition of the form U=VDV*
where V is unitary
D is diagonal and unitary.
4. |det(U)| = 1
5. Its eigenspaces are orthogonal.
6. U can be written as U = e power iH, where
e indicates the matrix exponential.
i is the imaginary unit.
H is a Hermitian matrix.
Unitary Matrix Solved Problems
Example 1.
Find the matrix $$A =\left[\begin{matrix} 1/√2 & 1/√2 \cr 1/√2 i & -1/√2 i \cr \end{matrix} \right]$$ is a unitary matrix or not.
Solution:
Given that the matrix is
$$A =\left[\begin{matrix} 1/√2 & 1/√2 \cr 1/√2 i & -1/√2 i \cr \end{matrix} \right]$$
The conjugate transpose of a matrix A is A*
A*$$=\left[\begin{matrix} 1/√2 & -1/√2 i \cr 1/√2 & 1/√2 i \cr \end{matrix} \right]$$
A A* $$=\left[\begin{matrix} ½ – ½ i² & ½ + ½ i² \cr ½ + ½ i² & ½ – ½ i² \cr \end{matrix} \right]$$
A A* $$=\left[\begin{matrix} 1 & 0 \cr 0 & 1 \cr \end{matrix} \right]$$
Therefore A A* = I
Therefore the given matrix is unitary
Example 2.
Show that the matrix $$A =\left[\begin{matrix} 1+i & 1-i \cr 1-i & 1+i \cr \end{matrix} \right]$$ × ½ is Unitary or not
Solution:
Given that the matrix is
$$A =\left[\begin{matrix} 1+i & 1-i \cr 1-i & 1+i \cr \end{matrix} \right]$$ × ½
The Conjugate transpose of a matrix A is A*.
$$A* =\left[\begin{matrix} 1-i & 1+i \cr 1+i & 1-i \cr \end{matrix} \right]$$ × ½
$$A A* =\left[\begin{matrix} 4 & 0 \cr 0 & 4 \cr \end{matrix} \right]$$ × ¼.
Therefore the given matrix is Unitary
Example 3.
If the matrix A = ½ $$=\left[\begin{matrix} 1 & -i & -1+i \cr i & 1 & 1+i \cr 1+i & -1+i 0 \cr \end{matrix} \right]$$ Is unitary or not.
Solution:
Given that the matrix is
A = ½ $$\left[\begin{matrix} 1 & -i & -1+i \cr i & 1 & 1+i \cr 1+I -1+I 0 \cr \end{matrix} \right]$$
The Conjugate transpose of a matrix A = A*
A* = ½ $$=\left[\begin{matrix} 1 & -i & -1-i \cr i & 1 & 1+i \cr 1+i & -1+i & 0 \cr \end{matrix} \right]$$
Then A A*
A A* = 1/4 $$= \left[\begin{matrix} 1 & 0 & 0 \cr 0 & 1 & 0\cr 0 & 0 & 1 \cr \end{matrix} \right]$$
Hence the given matrix is Unitary.
Example 4.
Find the matrix A = ⅓ $$=\left[\begin{matrix} 2+i & 2i \cr 2i & 2-i \cr \end{matrix} \right]$$ is Unitary or not.
Solution:
Given that the matrix is
A = ⅓ $$=\left[\begin{matrix} 2+i & 2i \cr 2i & 2-i \cr \end{matrix} \right]$$
The Conjugate transpose of the matrix A is A*
A* = ⅓$$=\left[\begin{matrix} 2-i & -2i \cr -2i & 2+i \cr \end{matrix} \right]$$
A A* = ⅓ $$=\left[\begin{matrix} 2+i & 2i \cr 2i & 2-i \cr \end{matrix} \right]$$ × ⅓ $$=\left[\begin{matrix} 2-i & -2i \cr -2i & 2+i \cr \end{matrix} \right]$$
A A* = 1/9 $$=\left[\begin{matrix} 9 & 0 \cr 0 & 9 \cr \end{matrix} \right]$$
Hence the given matrix is Unitary.
Example 5.
If the matrix $$A =\left[\begin{matrix} 0 & 1+2i \cr -1+2i & 0 \cr \end{matrix} \right]$$ then shows that (I-A)(I+A) inverse is unitary or not.
Solution:
Given that the matrix is
$$A =\left[\begin{matrix} 0 & 1+2i \cr -1+2i & 0 \cr \end{matrix} \right]$$
Let $$I =\left[\begin{matrix} 1 & 0 \cr 0 & 1 \cr \end{matrix} \right]$$
$$I+A =\left[\begin{matrix} 1 & 1+2i \cr -1+2i & 1 \cr \end{matrix} \right]$$
$$I-A =\left[\begin{matrix} 1 & -1+2i \cr 1-2i & 1 \cr \end{matrix} \right]$$
I-A inverse = 1/|I+A| $$=\left[\begin{matrix} 0 & 1+2i \cr -1+2i & 0 \cr \end{matrix} \right]$$
(I-A) inverse = ⅙ $$A =\left[\begin{matrix} 0 & 1+2i \cr -1+2i & 0 \cr \end{matrix} \right]$$
(I+A)(I-A) inverse = 1/36 $$=\left[\begin{matrix} 36 & 0 \cr 0 & 36 \cr \end{matrix} \right]$$
Hence the given matrix is Unitary.
Related Topics:
FAQs on Unitary Matrix
1. What is the Order of a Unitary Matrix?
The order of the unitary matrix is n × n which means the same number of rows and the same number of columns. It is a square matrix.
2. What are the Properties of the Unitary Matrix?
The properties of a unitary matrix are
• A unitary matrix is a nonsingular matrix and an Invertible matrix.
• The product of two unitary matrices is a unitary matrix.
• The addition and subtraction of two unitary matrices is also a unitary matrix.
3. What is a Unitary Matrix?
A unitary matrix is a square matrix of complex figures. And the product of the antipode of a unitary matrix, and the conjugate transpose of a unitary matrix is equal to the identity matrix. U power H = U power-1. |
# HBSE 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.5
Haryana State Board HBSE 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.5 Textbook Exercise Questions and Answers.
## Haryana Board 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Exercise 9.5
Question 1.
Use a suitable identity to get each of the following products :
(i) (x + 3) (x + 3)
(ii) (2y + 5) (2y + 5)
(iii) (2a – 7) (2a – 7)
(iv) (3a – $$\frac{1}{2}$$)(3a – $$\frac{1}{2}$$)
(v) (1.1m – 0.4) (1.1m + 0.4)
(vi) (a2 + b2) (-a2 + b2)
(vii) (6x- 7) (6x + 7)
(viii) (-a + c) (-a + c)
(ix)
(x) (7a – 9b) (7a – 9b).
Solution:
(i) ∵ (x + 9) (x + b)
= x2 + (a + b) x + ab
∴ (x + 3) (x + 3) = x2 + (3 + 3)x +3 × 3
= x2 + 6x + 9.
(ii) ∵ (a + b)2 = a2 + 2ab + b2
∴ (2y + 5) (2y + 5) = (2y + 5)2
= (2y)2 + 2 × 2y × 5 + (5)2
= 4y2 + 20y + 25.
(iii) (2a – 7) (2a – 7)
= (2a-7)2
[∵ (a – b)2 = a2 – 2ab + b2]
= (2a)2 – 2 × 2a × 7 + (7)2
= 4a2 – 28a+ 49.
(iv) (3a – $$\frac{1}{2}$$)(3a – $$\frac{1}{2}$$) = (3a – $$\frac{1}{2}$$)2
= (3a)2 – 2 × 3a × $$\frac{1}{2}$$ + ($$\frac{1}{2}$$)2
= 9a2 – 3a + $$\frac{1}{4}$$
(v) (1.1m -0.4) (1.1m + 0.4)
∵ (a + b) (a – b) = a2 – b2
∴ (1.1m + 0.4) (1.1m – 0.4)
= (1.1m)2 – (0.4)2
= 1.21m2 – 0.16.
(vi) (a2 + b2) (-a2 + b2) = (b2 + a2) (b2 – a2)
= (b2)2 – (a2)2 – b4 – a4.
(vii) (6x – 7) (6x + 7) = (6x)2 – (7)2
= 36x2 – 49.
(viii) (c -a) (c- a) = (c – a)2
= c2 – 2ca, + a2
(x) (7a – 9b) (7a – 9b)
= (7a – 9b)2
= (7a)2 – 2 x 7a x 9b + (9b)2
= 49a2 – 126ab + 81b2.
Question 2.
Use the identity (x + a) (x + b) = x2 + (a + b) x + ab to find the following products :
(i) (x + 3) (x + 7)
(ii) (4x + 5) (4x + 1)
(iii) (4x – 5)(4x – 1)
(iv) (4x + 5)(4x – 1)
(v) (2x + 5y) (2x + 3y)
(vi) (2a2 + 9) (2a2 + 5)
(vii) (xyz – 4) (xyz – 2).
Solution:
(i) (x + 3) (x + 7)
= x2 + (3 + 7)x + 3 × 7
= x2 + 10x + 21.
(iii) (4x + 5) (4x + 1)
= (4x)2 + (5 + 1)x + 5 × 1
= 16x2 + 6x + 5.
(iii) (4x – 5) (4x – 1)
= (4x)2 + [-5 + (-1)]x + (-5x – 1)
= 16x2 – 6x + 5.
(iv) (4x + 5) (4x – 1)
= (4x)2 + [5 + (-1)]x + [5 × (-1)]
= 16x2 + 4x – 5.
(v) (2x + 5y) (2x + 3y)
= (2x)2 + (5y + 3y)x + 5y × 3y
= 4x2 + 8xy + 15y2.
(vi) (2a2 + 9) (2a2 + 5)
= (2a2)2 + (9 + 5)2a2 + 9 × 5
= 4a4 + 28a2 + 45.
(vii) (xyz – 4) (xyz – 2)
= (xyz)2 + [(-4) + (-2)] xyz + (-4) x (-2)
= x2y2z2 – 6xyz + 8.
Question 3.
Find the following squares by using the identities :
(i) (b – 7)2
(ii) (xy + 3z)2
(iii) (6x2 – 5y)2
(iv)
(v) (0.4p – 0.5q)2
(vi) (2xy + 5y)2
Solution:
(i) b – 7)2 = b2 – 2 x b x 7 + 72
[∵ (a – b)2 = a2 – 2ab + b2]
= b2 – 14 b + 49.
(ii) (xy + 3z)2 = (xy)2 + 2 × xy × 3z + (3z)2
[∵ (a + b)2 = a2 + 2ab + b2]
= x2y2 + 6xyz2 + 9z2.
(iii) (6x2 – 5y)2 = (6x2)2 – 2 × 6x2 × 5y + (5y)2
= 36x4 – 60x2y + 25y2.
(iv)
(v) (0.4p – 0.5q)2 = (0.4p)2 – 2 × 0.4p × 0.5q + (0.5q)2
= 0.16p2 – 0.40pq + 0.25q2.
(vi) (2xy + 5y)2 = (2xy)2 + 2 × 2xy × 5y + (5y)2
= 4x2y2 + 20xy2 + 25y2.
Question 4.
Simplify :
(i) (a2 – b2)2
(ii) (2x + 5)2 – (2x – 5)2
(iii) (7m – 8n)2 + (7m + 8n)2
(iv) (4m + 5n)2 + (5m + 4n)2
(v) (2.5p – 1.5q)2 – (1.5p – 2.5q)2
(vi) (ab + bc)2 – 2ab2c
(vii) (m2 – n2m)2 + 2m3n2.
Solution:
(i) (a2 – b2)2 = (a2)2 – 2 × a2 × b2 + (b2)2
= a4 – 2a2b2 + b4.
(ii) (2x + 5)2 – (2x – 5)2
= [(2x + 5) + (2x – 5)] [(2x + 5) – (2x – 5)]
= (2x + 5 + 2x – 5) (2x + 5 – 2x + 5)
= 4x × 10 = 40x.
(iii) (7m – 8n)2 + (7m + 8n)2
= (7m)2 – 2 × 7m × 8n + (8n)2 + (7m)2 + 2 × 7m × 8n + (8n)2
= 2 × (7m)2 + 2 × (8n)2.
= 2 × 49m2 + 2 × 64n2
= 98m2 + 128n2.
(iv) (4m + 5n)2 + (5m + 4n)2
= (4m)2 + 2 × 4m × 5n + (5n)2 + (5m)2 + 2 × 5m × 4n + (4n)2
= 16m2 + 40mn + 25n2 + 25m2 + 40mn + 16n2
= 41m2 + 41n2 + 80mn
(v) (2.5p – 1.5q)2 – (1.5p – 2.5q)2
= [(2.5p)2 – 2 × 2.5p × 1.5q + (1.5q)2] – [(1.5p)2 – 2 × 1.5p × 2.5q + (2.5q)2]
= 6.25p2 – 0.75pq + 2.25q2 – 2.25p2 + 0.75pq + 6.25q2
= 12.50p2.
(vi) (ab + bc)2 – 2ab2c
= (ab)2 + 2 × ab × bc + (bc)2 – 2ab2c
= a2b2 + 2ab2c + b2c2 – 2ab2c
= a2b2 + b2c2.
(vii) (m2 – n2m)2 + 2m3n2
= (m2)2 – 2 × m2 × n2m + (n2m)2 + 2m3n2
= m4 – 2m3n2 + n4m2 + 2m3n2 = m4 + n4m2.
Question 5.
Show that:
(i) (3x + 7)2 – 84x = (3x – 7)2
(ii) (9p – 5q)2 + 180pq = (9p + 5q)2
(iii) (4pq + 3q)2 – (4pq – 3q)2 = 48pq2
(iv) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0.
Solution:
(i) (3x + 7)2 – 84= (3x – 7)2
L.H.S. = (32)2 + 2 × 3x × 7 + (7)2 – 84x
= 9x2 + 42x + 49 – 84x
= 9x2 – 42x + 49 = (3x – 7)2
= R.H.S. Hence proved.
(ii) (9p – 5q)2 + 180pg = (9q + 5q)2
L.H.S. = (9p)2 + 2 × 9p × 5q + (5q)2 + 180pg
= (9p)2 + 2 × 9pq × 5q + (5q)2
= (9p + 5q)2
= R.H.S. Hence proved.
(iii) (4pq + 3q)2 – (4pq – 3q)2 = 48pq2
L.H.S. = [(4pg + 3q) + (4pq – 3q)]
[(4pq + 3q) – (4pq – 3q)]
= (4pq + 3q + 4pq – 3g)
(4pq + 3q – 4pq + 3q)
= 8pq × 6q = 48pq2
= R.H.S. Hence proved.
(iv) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
L.H.S. = a2 – b2 + b2 – c2 + c2 – a2 = 0
= R.H.S Hence proved.
Question 6.
Using identities, evaluate.
(i) (71)2
(ii) (99)2
(iii) (102)2
(iv) (998)2
(v) (5.2)2
(vi) 297 × 303
(vii) 78 × 82
(viii) (8.9)2
(ix) 10.5 × 9.5
Solution:
(i) (70 + 1)2 = (70)2 + 2 × 70 × 1 + (1)2
= 4900 + 140 + 1
= 5041.
(ii) (99)2 = (100 – 1)2
= (100)2 – 2 × 100 × 1 + (1)2
= 10000 – 200 + 1 = 9801
(iii) (102)2 = (100 + 2)2
= (100)2 + 2 × 100 × 2 + (2)2
= 10000 + 400 + 4 = 10402.
(iv) (998)2 = (1000 – 2)2
= (1000)2 – 2 × 1000 × 2 + (2)2
= 1000000 – 4000 + 4
= 996004.
(v) (5.2)2 = (5 + 0.2)2
= (5)2 + 2 × 5 × 0.2 + (0.2)2
= 25 + 2.0 + 0.04 = 27.04.
(vi) 297 × 303 = (300 – 3) (300 + 3)
= (300)2 – (3)2
= 90000 – 9
= 89991
(vii) 78 × 82 = (80 – 2) (80 + 2)
= (80)2 – (2)2
= 6400 – 4
= 6396.
(viii) (8.9)2 = (9 – 0.1)2
= (9)2 – 2 × 9 × 0.1 + (0.1)2
= 81 – 1.8 + 0.01
= 79.21.
(ix) (10.5) × (9.5) = (10 + 0.5) (10 – 0.5)
= (10)2 – (0.5)2
= 100 – 0.25
= 99.75.
Question 7.
Using a2 – b2 = (a + b) (a – b), find
(i) (51)2 – (49)2
(ii) (1.02)2 – (0.98)2
(iii) (153)2 – (147)2
(iv) (12.1)2 – (7.9)2
Solution:
(i) (51 + 49) (51 – 49) = 100 × 2 = 200.
(ii) (1.02)2 – (0.98)2
= (1.02 + 0.98) (1.02 – 0.98)
= 2.00 × 0.04 = 0.08.
(iii) (153)2 – (147)2
= (153 + 147) (153 – 147)
= 200 × 6 = 1200.
(iv) (12.1)2 – (7.9)2 = (12.1 + 7.9) (12.1 – 7.9)
= 20.0 × 4.2 = 84.0 = 84.
Question 8.
Using
(x + a) (x + b) = x2 + (a + b) x + ab, find
(i) 103 × 104
(ii) 5.1 × 5.2
(iii) 103 × 98
(iv) 9.7 × 9.8.
Solution:
(i) (100 + 3) (100 + 4)
= (100)2 + (3 + 4) × 100 + 3 × 4
= 10000 + 700 + 12 = 10712.
(ii) (5 + 0.1) (5 + 0.2)
= 52 + (0.1 + 0.2) × 5 + (0.1 × 0.2)
= 25 + 0.02 × 5 + 0.02
= 25 + 0.10 + 0.02
= 25.12.
(iii) (100 + 3) (100 – 2)
= (100)2 + [3 + (-2)] × 100 + 3 × (-2)
= 10000 + 100 – 6
= 10094.
(iv) 9.7 × 9.8 = (10 – 0.3) (10 – 0.2)
= (10)2 + (-0.3 + (-o.2)] × 10 + (-0.3) × (-0.2)
= 100 – 5 – 0.06
= 94.94. |
# Connecticut - Grade 1 - Math - Geometry - Composite Shapes - 1.G.2
### Description
Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.1
• State - Connecticut
• Standard ID - 1.G.2
• Subjects - Math Common Core
• Math
• Geometry
## More Connecticut Topics
Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
Given a two-digit number, mentally find 10 more or 10 less than the number, without having to count; explain the reasoning used.
Express the length of an object as a whole number of length units, by laying multiple copies of a shorter object (the length unit) end to end; understand that the length measurement of an object is the number of same-size length units that span it with no gaps or overlaps. Limit to contexts where the object being measured is spanned by a whole number of length units with no gaps or overlaps.
Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: A. 10 can be thought of as a bundle of ten ones — called a “ten.” B. The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. C. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones).
Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares. |
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# Bijective Functions
You may have used QR codes for various purposes before. Each QR code contains some information in them and is used to uniquely identify an item or service. Every QR code uniquely identifies one and only one such item/service. Mathematically, the mapping between the QR code and the object that it identifies is an example of a bijective function.
In this section, we will look at the bijective function and understand it in the different forms of function.
## Defining a Bijective Function
Suppose we have two sets, $$A$$ and $$B$$, and a function $$f$$ points from $$A$$ to $$B$$ $$(f:A\to B)$$. If every element in codomain $$B$$ is pointed to by at least one element in domain $$A$$, the function is called a bijective function.
A function $$f:A\to B$$ is bijective if, for every $$y$$ in $$B$$, there is exactly one $$x$$ in $$A$$ such that $$f(x)=y$$.
A bijective function is both injective (one-one function) and surjective (onto function) in nature.
If every element of the range is mapped to exactly one element from the domain is called the injective function. That is, no element of the domain points to more than one element of the range.
In a surjective function, every element of the co-domain is an image of at least one element of the domain.
Injective and Surjective functions, StudySmarter Originals
So, it logically follows that if a function is both injective and surjective in nature, it means that every element of the domain has a unique image in the co-domain, such that all elements of the co-domain are also part of the range (have a corresponding element in the domain). Such a function is called a bijective function.
You can consider a bijective function to be a perfect one-to-one correspondence. Every element in the domain has exactly one corresponding image in the co-domain, and vice-versa.
Bijective function, StudySmarter Originals
Note that the onto function is not bijective, as it needs to be a one-one function to be bijective. Let's take a look at the difference between these two to understand it better.
## Difference between Bijective and Surjective functions
We will see the difference between bijective and surjective functions in the following table.
Bijective function Surjective function A function $$(f:A\to B)$$ is bijective if, for every $$y$$ in $$B$$, there is exactly one $$x$$ in $$A$$ such that $$f(x)=y$$. A function $$(f:A\to B)$$ is surjective if for every $$y$$ in $$B$$ there is at least one $$x$$ in $$A$$ such that $$f(x)=y$$. A bijective function is both one-one and onto function. A surjective function is onto function. The domain and co-domain have an equal number of elements. A co-domain can be an image for more than one element of the domain. Bijective graphs have exactly one horizontal line intersection in the graph. Surjective graphs have at least one horizontal line intersection in the graph. Example - $$f:\mathbb{R}\to \mathbb{R}, f(x)=2x$$ Example - $$f:\mathbb{R}\to \mathbb{R}, f(x)=x^{3}-3x$$
## Composition of Bijective functions
Consider the functions $$f:A\to B , g:B\to C$$. Then the composition of the function $$(g\circ f)(x)=g(f(x))$$ from function $$A$$ to $$C$$. The composition of the bijective function is derived from the composition of injective and surjective functions.
The function $$f:A\to B , g:B\to C$$ are injective function, then the composition $$g\circ f$$ is also injective. Similarly, for the two surjective functions $$f$$ and $$g$$, their composition $$g\circ f$$ is also surjective.
Suppose both $$f:A\to B$$ and $$g:B\to C$$ are bijective. This implies that both $$f$$ and $$g$$ are both injective and surjective as well. The composition of the functions $$g\circ f$$ is both injective and surjective. Hence, the composition of function $$g\circ f$$ is bijective.
Note that if $$g\circ f$$ is bijective, then it can only be possible that $$f$$ is injective and $$g$$ is surjective.
Bijective composition, StudySmarter Originals
## Bijective function graph
We can determine a bijective function based on the plotted graph too. To identify a bijective function graph, we consider a horizontal line test based on injective and surjective functions. For a function to be bijective both the test for injective and surjective should be satisfied.
### Horizontal line test
This test is used to check the injective, surjective, and bijective functions. We determine the type of function based on the number of intersection points with the horizontal line and the given graph.
To check this, draw horizontal lines from different points. If each horizontal line intersects the graph at most one point then, it is an injective function. If the function is surjective, then a horizontal line should intersect at at least one point. So, when checking for bijective function, there should be exactly one intersecting point with a horizontal line.
## Bijective function examples
Show bijection for the function $$f:\mathbb{R}\to \mathbb{R}, f(x)=x$$.
Solution:
Consider the function $$f(x)=x$$, where the domain and co-domain are the set of all real numbers.
All values in the co-domain correspond to a unique value in the domain. Thus, the function is bijective in nature.
It is injective because every value of $$x$$ leads to a different value of $$y$$. It is surjective because any possible real number $$r$$ can have a corresponding value $$x$$ such that $$f(x)=r$$.
When a bijective function is drawn on a graph, a horizontal line parallel to the X-axis must intersect the graph at exactly one point (horizontal line test).
The following graph demonstrates this for the function $$f(x)=x$$.
Graph for function $$f(x)=x$$, StudySmarter Originals
Verify if the function $$f:\mathbb{R}\to \mathbb{R}, f(x)=x^{2}$$ is bijective or not.
Solution:
Here for the given function, the range of the function only includes values $$\ge 0$$. But the co-domain includes all negative real numbers too. And the members of the co-domain can be images of multiple members of the domain, for example $$f(2)=f(-2)=4$$. Hence, the function $$f(x)=x^{2}$$ is not injective. So, $$f(x)=x^{2}$$ is not bijective.
When we draw the function on a graph, we can notice how it fails the horizontal line test as it intersects at two different points.
Non-bijective function graph for $$f(x)=x^{2}$$, StudySmarter Originals
Is the function $$f(x)=2x$$ bijective? Also, show for which domain and co-domain.
Solution:
When we set the domain and co-domain of the function to the set of all real numbers, it was a bijective function.
Hence, for $$f:\mathbb{R}\to \mathbb{R}, f(x)=2x$$ is bijective.
However, if we restrict the domain and co-domain of the function to the set of all natural numbers, this no longer remains a bijective function. Since the range would include all even numbers but exclude all odd numbers, but they remain part of the co-domain. For example, it is impossible to get $$f(x)=3$$, for any natural number value of $$x$$. Thus, the function is not surjective, and consequently not bijective.
So, $$f:\mathbb{N}\to \mathbb{N}, f(x)=2x$$$f:\mathrm{ℕ}\to \mathrm{ℕ},f\left(x\right)=2x$ is not bijective.
## Bijective Functions - Key takeaways
• A bijective function is both injective and surjective in nature.
• A function $$f:A\to B$$ is bijective if, for every $$y$$ in $$B$$, there is exactly one $$x$$ in $$A$$ such that $$f(x)=y$$.
• A bijective function is one-one and onto function, but an onto function is not a bijective function.
• The composition of bijective functions is again a bijective function.
• For a bijective function, there should be exactly one intersecting point with a horizontal line.
#### Flashcards in Bijective Functions 3
###### Learn with 3 Bijective Functions flashcards in the free StudySmarter app
We have 14,000 flashcards about Dynamic Landscapes.
What is a bijective function?
A function that is both injective and surjective is called a bijective function.
How to prove a function is bijective?
To prove that a function is bijective, first prove that it is injective and then prove that it is surjective.
Are continuous functions bijective?
All bijective functions are continuous but not all continuous functions are bijective.
Are all linear functions bijective?
All linear continuous functions are bijective. For example f(x)=2x.
What is bijective function with example?
The function f(x)=x is an example of a bijective function as it is both injective and surjective,
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# Question Video: Solving Word Problems by Subtracting Numbers up to 999
A shop has 953 bottles of drink. 230 were sold. How many are left?
02:05
### Video Transcript
A shop has 953 bottles of drink. 230 were sold. How many are left?
When we solve word problems, we always start by highlighting the key numbers and phrases. A shop had 953 bottles of drink. 230 are sold. We need to calculate how many bottles are left.
Now we can write the equation; 953 bottles, 230 are sold. Next we need to figure out which operation to use, plus or minus. The shop had 953 bottles and then sold some.
That’s correct! The operation we need to use is minus because 230 bottles were already sold. Once we’ve subtracted 230 bottles from the 953, that will tell us how many bottles are left.
Now we can calculate the answer using column subtraction, 953 minus 230. Starting in the ones column, three minus zero is three. Moving across to the 10s column, five minus three is two.
Finally, we’re moving to the 100s column: nine minus two is seven. 953 minus 230 equals 723. The shop had 953 bottles of drink. They sold 230, which means that 723 were left. We calculated our answer using column subtraction. |
Question Video: Finding the Surface Area of a Sphere Given Its Diameter | Nagwa Question Video: Finding the Surface Area of a Sphere Given Its Diameter | Nagwa
# Question Video: Finding the Surface Area of a Sphere Given Its Diameter Mathematics • Second Year of Preparatory School
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Find the surface area of a sphere whose diameter is 12.6 cm. Use 𝜋 = 22/7.
03:23
### Video Transcript
Find the surface area of a sphere whose diameter is 12.6 centimeters. Use 𝜋 equals 22 over seven.
The formula we need for calculating the surface area of a sphere is this, four 𝜋𝑟 squared, where 𝑟 is the radius of the sphere. In this question though, we haven’t been given the radius; we’ve been given the length of the sphere’s diameter. That’s no great problem, though, because we know the relationship that exists between the radius and diameter of the sphere. The radius is half the length of the diameter. So if the diameter is 12.6 centimeters, the radius is half of this. That’s 6.3 centimeters. So we can substitute this value for the radius directly into our formula for the surface area, giving four 𝜋 multiplied by 6.3 squared. And remember, it’s only the radius we’re squaring.
Now, the question actually asks us to use 22 over seven as an approximation for 𝜋. So this suggests we haven’t got access to a calculator in this question. Our surface area becomes four multiplied by 22 over seven multiplied by 6.3 squared. And let’s see how we could work this out without a calculator. First, we can write 6.3 squared as 6.3 multiplied by 6.3. Now, we should spot that seven is a factor of 63. So we can divide 6.3 by seven relatively easily. Using a short division or bus-stop method, firstly, there are no sevens in six, so we put a zero and carry the six. And then there are nine sevens in 63. So 6.3 divided by seven is 0.9. So our calculation becomes four multiplied by 22 multiplied by 0.9 multiplied by 6.3.
We can work this calculation out in pairs. Four multiplied by 22, first of all, is 88. And to work out 0.9 multiplied by 6.3, we can first work out 63 times nine, which is 567, and then recall that we need to divide by 10 twice in order to give the answer to the decimal calculation. So dividing 567 by 100 gives 5.67. This also makes sense from an estimation point of view. We’re multiplying 6.3 by something a little less than one, so the answer we get should be a little less than 6.3, and 5.67 is reasonable.
Finally, we can work out 5.67 multiplied by 88 by first working out 567 multiplied by 88 which is 49896 and then dividing this value by 100, which gives 498.96. As the units for the diameter were centimeters, the units for the surface area will be square centimeters. And so we have our answer to the problem. Using 22 over seven as an approximation for 𝜋, we found that the surface area of the sphere whose diameter is 12.6 centimeters is 498.96 square centimeters. Remember, the key point in this question was that we needed to calculate the radius of the sphere first before we could calculate its surface area.
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# How do you find the slope and intercept to graph y=3x+4?
Apr 1, 2018
$b = 4 , m = 3$
#### Explanation:
The intercept and slope are already given. This equation is in the form $y = m x + b$, where $b$ is the y-intercept (0,4) and $m$ is the slope, $3$.
Apr 1, 2018
Slope: $3$
Y-Intercept: $\left(4 , 0\right)$
X-Intercept: $\left(0 , - \frac{4}{3}\right)$
#### Explanation:
You should be armed with the knowledge what the slope-intercept form is. It is $y = m x + b$. $m$ stands for the slope, and $b$ stands for the y-intercept. So, seeing from the equation, $y = 3 x + 4$, 3 is the slope and $\left(4 , 0\right)$ is the y-intercept. To find the x-intercept, you have to plug 0 in for $y$ and just solve for $x$. Then, you should have $0 = 3 x + 4$. You subtract $4$ to isolate the variable. Lastly, you divide $3$ to get $- \frac{4}{3}$. The x-intercept is $\left(0 , - \frac{4}{3}\right)$ |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Five Number Summary
Quartiles or medians and the minimum and maximum values.
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Practice Five Number Summary
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Five Number Summary
When given a long list of numbers, it is useful to summarize the data. One way to summarize the data is to give the lowest number, the highest number and the middle number. In addition to these three numbers it is also useful to give the median of the lower half of the data and the median of the upper half of the data. These five numbers give a very concise summary of the data.
What is the five number summary of the following data?
0, 0, 1, 2, 63, 61, 27, 13
Guidance
Suppose you have ordered data with observations. The rank of each observation is shown by its index.
In data sets that are large enough, you can divide the numbers into four parts called quartiles. The quartiles of interest are the first quartile, , the second quartile, , and the third quartile . The second quartile, , is defined to be the median of the data. The first quartile, , is defined to be the median of the lower half of the data. The third quartile, , is similarly defined to be the median of the upper half of the data.
These three numbers in addition to the minimum and maximum values are the five number summary. Note that there are variations of the five number summary that you can study in a statistics course.
Example A
Compute the five number summary for the following data.
2, 7, 17, 19, 25, 26, 26, 32
Solution: There are 8 observations total.
• Lowest value (minimum) : 2
• (Note that this is the median of the first half of the data - 2, 7, 17, 19)
• (Note that this is the median of the full set of data)
• (Note that this is the median of the second half of the data - 25, 26, 26, 32)
• Upper value (maximum) : 32
Example B
Compute the five number summary for the following data:
4, 8, 11, 11, 12, 14, 16, 20, 21, 25
Solution: There are 10 observations total.
• Lowest value (minimum) : 4
• (Note that this is the median of the first half of the data - 4, 8, 11, 11, 12)
• (Note that this is the median of the full set of data)
• (Note that this is the median of the second half of the data - 14, 16, 20, 21, 25)
• Upper value (maximum) : 25
Example C
Compute the five number summary for the following data:
3, 7, 10, 14, 19, 19, 23, 27, 29
Solution: There are 9 observations total. To calculate and , you should include the median in both the lower half and upper half calculations.
• Lowest value (minimum) : 3
• (this is the median of 3, 7, 10, 14, 19)
• (this is the median of 19, 19, 23, 27, 29)
• Upper value (maximum) : 29
Concept Problem Revisited
To compute the five number summary, it helps to order the data.
0, 0, 1, 2, 13, 27, 61, 63
• Since there are 8 observations, the median is the average of the and observations:
• The lowest observation is 0.
• The highest observation is 63.
• The middle of the lower half is
• The middle of the upper half is
The five number summary is 0, 0.5, 7.5, 44, 63
Vocabulary
The rank of an observation is the number of observations that are less than or equal to the value of that observation.
Data is divided into four parts by the first quartile , second quartile and third quartile . The second quartile is also known as the median.
Guided Practice
1. Create a set of data that meets the following five number summary:
{2, 5, 9, 18, 20}
2. Compute the five number summary for the following data:
1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 6, 7, 8, 9, 10, 15
3. Compute the five number summary for the following data:
1, 4, 96, 356, 2557, 9881, 14420, 20100
1. Suppose there are 8 data points. The lowest point must be 2 and the highest point must be 20. The middle two points must average to be 9 so they could be 8 and 10. The second and third points must average to be 5 so they could be 4 and 6. The sixth and seventh points need to average to be 18 so they could be 18 and 18. Here is one possible answer:
2, 4, 6, 8, 10, 18, 18, 20
2. There are 20 observations.
• Lower : 1
• Upper : 15
3. There are 8 observations.
• Lower : 1
• Upper: 20100
Practice
Compute the five number summary for each of the following sets of data:
1. 0.16, 0.08, 0.27, 0.20, 0.22, 0.32, 0.25, 0.18, 0.28, 0.27
2. 77, 79, 80, 86, 87, 87, 94, 99
3. 79, 53, 82, 91, 87, 98, 80, 93
4. 91, 85, 76, 86, 96, 51, 68, 92, 85, 72, 66, 88, 93, 82, 84
5. 335, 233, 185, 392, 235, 518, 281, 208, 318
6. 38, 33, 41, 37, 54, 39, 38, 71, 49, 48, 42, 38
7. 3, 7, 8, 5, 12, 14, 21, 13, 18
8. 6, 22, 11, 25, 16, 26, 28, 37, 37, 38, 33, 40, 34, 39, 23, 11, 48, 49, 8, 26, 18, 17, 27, 14
9. 9, 10, 12, 13, 10, 14, 8, 10, 12, 6, 8, 11, 12, 12, 9, 11, 10, 15, 10, 8, 8, 12, 10, 14, 10, 9, 7, 5, 11, 15, 8, 9, 17, 12, 12, 13, 7, 14, 6, 17, 11, 15, 10, 13, 9, 7, 12, 13, 10, 12
10. 49, 57, 53, 54, 49, 67, 51, 57, 56, 59, 57, 50, 49, 52, 53, 50, 58
11. 18, 20, 24, 21, 5, 23, 19, 22
12. 900, 840, 880, 880, 800, 860, 720, 720, 620, 860, 970, 950, 890, 810, 810, 820, 800, 770, 850, 740, 900, 1070, 930, 850, 950, 980, 980, 880, 960, 940, 960, 940, 880, 800, 850, 880, 760, 740, 750, 760, 890, 840, 780, 810, 760, 810, 790, 810, 820, 850
13. 13, 15, 19, 14, 26, 17, 12, 42, 18
14. 25, 33, 55, 32, 17, 19, 15, 18, 21
15. 149, 123, 126, 122, 129, 120
To view the Explore More answers, open this PDF file and look for section 15.3.
Vocabulary Language: English
first quartile
first quartile
The first quartile, also known as $Q_1$, is the median of the lower half of the data.
five number summary
five number summary
The five number summary of a set of data is the minimum, first quartile, second quartile, third quartile, and maximum.
Lower quartile
Lower quartile
The lower quartile, also known as $Q_1$, is the median of the lower half of the data.
Maximum
Maximum
The largest number in a data set.
Median
Median
The median of a data set is the middle value of an organized data set.
Minimum
Minimum
The minimum is the smallest value in a data set.
Quartile
Quartile
A quartile is each of four equal groups that a data set can be divided into.
rank
rank
The rank of an observation is the number of observations that are less than or equal to the value of that observation.
second quartile
second quartile
The second quartile, also known as $Q_2$, is the median of the data.
third quartile
third quartile
The third quartile, also known as $Q_3$, is the median of the upper half of the data.
Upper Quartile
Upper Quartile
The upper quartile, also known as $Q_3$, is the median of the upper half of the data. |
# From Parallelogram to Rhombus
🏆Practice from a parallelogram to a rhombus
You will be able to determine that the parallelogram is a rhombus if at least one of the following conditions is met:
1. If in the parallelogram there is a pair of adjacent equal sides - it is a rhombus.
2. If in the parallelogram the diagonals bisect each other, forming angles of $90^o$ degrees, that is, they are perpendicular - it is a rhombus.
3. If in the parallelogram one of the diagonals is the bisector - it is a rhombus.
## Test yourself on from a parallelogram to a rhombus!
Given the parallelogram:
Is this parallelogram a rhombus?
## From Parallelogram to Rhombus
How will you realize that the parallelogram in front of you is a rhombus?
We are here to teach you 3 criteria by which you can demonstrate, quickly and simply, that you have a rhombus in front of you.
First let's remember the definition of a rhombus.
The definition of a rhombus says the following: it is a parallelogram that has a pair of sides (or edges) adjacent that are equal.
### First criterion - Adjacent sides equal
If the parallelogram has a pair of equal adjacent sides, it is a rhombus.
You can use this theorem to show that it is a rhombus without having to prove it.
However, for you to understand the logic behind it, we will demonstrate this theorem below.
Data:
Parallelogram $ABCD$
$AB=BC$
We have to prove that: $ABCD$ is a rhombus
Solution:
Since we have that $ABCD$ is a parallelogram, we deduce that:
$AB=DC$
$AD=BC$ because, in the parallelogram, each pair of opposite sides are also equivalent.
Let's observe our data $BC=AB$
Now, we can determine that all sides of the parallelogram are equivalent according to the transitive relation.
We obtain:
$AB=DC=BC=AD$
Indeed, $ABCD$ is a quadrilateral with all its sides equal, therefore, it is a rhombus.
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### Second criterion - Perpendicular diagonals
If in the parallelogram the diagonals bisect each other forming angles of $90^o$ degrees, that is, they are perpendicular, it is a rhombus.
You can use this theorem to show that it is a rhombus without having to prove it.
However, for you to understand the logic behind it, we will demonstrate this theorem below.
Data:
Parallelogram $ABCD$
$AC⊥BD$
It must be demonstrated that: $ABCD$ is a rhombus
Solution:
Based on the first criterion, we already know that, it is enough for us to see that in the parallelogram there are two adjacent equivalent sides for us to know that it is a rhombus.
We can see that there is in this parallelogram a pair of adjacent equivalent sides if we use the congruence of triangles $ABE$ and $BEC$.
We will place them on top of each other:
Side $AE=AE$ common side, of the same length.
Angle $∢AEB=∢BEC$ These angles measure 90º since we know that the diagonals that create them are perpendicular.
Side $AE=CE$ in the parallelogram, the diagonals intersect
From this it follows that: $⊿ABE=⊿BEC$ according to SAS
And, consequently,
we will be able to determine that: $AB=BC$
According to the congruence, corresponding sides are equivalent.
We will notice that $AB$ and $BC$ are adjacent equivalent sides in the parallelogram and, therefore, we will be able to determine that $ABCD$ is a rhombus since a parallelogram with a pair of adjacent equivalent sides is a rhombus.
### Third Criterion - Diagonal equal to bisector
If in the parallelogram one of the diagonals is the bisector - it is a rhombus.
You can use this theorem to show that it is a rhombus without having to prove it.
However, for you to understand the logic behind it, we will prove this theorem below.
Data:
Parallelogram $ABCD$
$∢c1=∢c2$
We have to prove that: $ABCD$ is a rhombus
Solution:
Let's remember, a parallelogram that has a pair of equal adjacent sides is a rhombus.
Let's start:
From the data we have we can deduce $AB∥DC$ since in a parallelogram, the opposite sides are also parallel.
Then: $∢BAC=∢C1$, alternate angles between parallel lines are equivalent.
According to the transitive relation, $∢BAC=∢C2$.
Now we can deduce that: $AB=BC$
In the triangle, opposite equivalent angles there are sides equal to each other.
Now we can determine that $ABCD$ is a rhombus.
We found in the parallelogram a pair of equal adjacent sides, therefore, it is a rhombus.
Suggestion:
To remember the three theorems that prove that a parallelogram is a rhombus, try to remember the three key terms: sides, diagonals, and angles.
Great!
Now you know all the criteria to prove that a parallelogram is a rhombus.
## Examples and exercises with solutions from parallelogram to rhombus
### Exercise #1
Given the parallelogram:
Is this parallelogram a rhombus?
### Step-by-Step Solution
The definition of a rhombus is "a quadrilateral with all sides equal"
Therefore, the square in the diagram is indeed a rhombus
True
### Exercise #2
Look at the parallelogram below:
The diagonals form 90 degrees at the center of the parallelogram.
Is this parallelogram a rhombus?
### Step-by-Step Solution
The parallelogram whose diagonals are perpendicular to each other (meaning the angle between them is $90\degree$) is a rhombus, therefore the given parallelogram is a rhombus.
Yes.
### Exercise #3
Given the parallelogram:
Is this parallelogram a rhombus?
Not true
### Exercise #4
Given the parallelogram:
Is this parallelogram a rhombus?
Not true
### Exercise #5
Given the parallelogram:
Is this parallelogram a rhombus? |
# How To Solve & Graph Quadratic Inequalities (6 Cases)
Solving and graphing quadratic equations is one thing, but quadratic inequalities add another layer of work to be done. However, if you have a method, you can solve these problems without much trouble.
So, how do you solve & graph quadratic inequalities? To solve a quadratic inequality, first convert it to standard form. Next, identify the case you are in, and find the zeros of the quadratic. Then, graph the corresponding parabola from the quadratic. Finally, shade the appropriate region on the graph based on signs and the inequality symbol.
Of course, some cases are easier to solve and graph than others, but there is a way to do every problem.
In this article, we’ll take a closer look at the steps to solve and graph quadratic inequalities. We’ll pay special attention to the possible cases for the solution sets and what to do in each situation.
Let’s get started.
## How To Solve & Graph Quadratic Inequalities
There are 7 steps to take if you want to solve and graph a quadratic inequality. Doing them in this order ensures that you won’t miss anything along the way:
• 1. Convert the quadratic inequality to standard form with a > 0.
• 2. Find the value of vy (the y-coordinate of the vertex of the parabola).
• 3. Use vy to identify the case for the solutions.
• 4. Find the zeros of the quadratic.
• 5. Write the solution set for the quadratic inequality.
• 6. Graph the parabola corresponding to the quadratic.
• 7. Shade the appropriate region on the graph, based on the inequality sign.
First, we’ll explain each of these steps in more detail. Then, we’ll look at some examples to make things clear.
### 1. Convert The Quadratic Inequality To Standard Form With a > 0
Standard form just means that one side of the inequality is zero. This would only require subtraction.
However, we want to take the extra step of making a > 0 (that is, a positive coefficient for x2). Sometimes this will be faster with multiplication by -1, which will require us to switch the direction of the inequality symbol.
For example, let’s say we have the quadratic inequality -2x2 + 3x + 4 > 5. To convert to standard form with a > 0:
• -2x2 + 3x + 4 > 5 [original inequality]
• -2x2 + 3x + 4 – 5 > 5 – 5 [subtract 5 from both sides]
• -2x2 + 3x – 1 > 0
• -1(-2x2 + 3x – 1) < 0 [multiply both sides by -1, and switch direction of inequality symbol]
• 2x2 – 3x + 1 < 0
Now our quadratic inequality is in standard form, with one side equal to zero. We also have a > 0 (since a = 2), which makes our work easier in future steps.
### 2. Find the Value of Vy (The y-coordinate Of The Vertex Of The Parabola)
Remember that for a quadratic f(x) = ax2 + bx + c, we can graph the corresponding parabola. This parabola has a vertex with coordinates (vx, vy), where:
• vx = -b / 2a
• vy = (4ac – b2) / 4a
So, to find vy (the y coordinate of the vertex), all we need to do is plug a, b, and c into vy, the formula for the y coordinate of the vertex of a parabola.
For example, let’s go back to the inequality we converted to standard form earlier:
• 2x2 – 3x + 1 < 0
Here, we have a = 2, b = -3, and c = 1.
Substituting these into the vy formula gives us:
• vy = (4ac – b2) / 4a
• vy = (4(2)(1) – (-3)2) / 4(2)
• vy = (8 – 9) / 8
• vy = -1 / 8
So, the y coordinate of the vertex of the parabola is -1/8.
### 3. Use vy To Identify The Case For The Solutions
There are 6 cases for the solutions of the quadratic inequality, depending on the sign of vy and the direction of the inequality symbol:
#### Case 1: vy > 0
In this case, the y-coordinate of the vertex is positive (it is above the x-axis). Since a > 0, the parabola opens upward, which means the parabola is always above the x-axis, and the quadratic f(x) is always positive.
The following 4 subcases are possible for vy > 0:
• f(x) > 0: every x value in the domain is a solution.
• f(x) >= 0: every x value in the domain is a solution.
• f(x) < 0: there is no solution.
• f(x) <= 0: there is no solution.
#### Case 2: vy = 0
In this case, the y-coordinate of the vertex is positive (it lies on the x-axis). Since a > 0, the parabola opens upward, which means the parabola is always on or above the x-axis, and the quadratic f(x) is always positive or zero.
The following 4 subcases are possible for vy = 0:
• f(x) > 0: every x value in the domain is a solution, except for vx, the x-coordinate of the vertex.
• f(x) >= 0: every x value in the domain is a solution.
• f(x) < 0: there is no solution.
• f(x) <= 0: the only solution is vx, the x-coordinate of the vertex.
#### Case 3: vy < 0
In this case, the y-coordinate of the vertex is negative (it is below the x-axis). Since a > 0, the parabola opens upward, which means the parabola is sometimes above the x-axis, sometimes below the x-axis, and sometimes intersecting the x-axis (this is where we find the zeros).
So, f(x) can take values that are positive, negative, or zero. In fact, f(x) has two zeros in this case, r and s (r < s).
The following 4 subcases are possible for vy < 0:
• f(x) > 0: some x values in the domain are solutions: x < r and x > s.
• f(x) >= 0: some x values in the domain are solutions: x <= r and x >= s.
• f(x) < 0: some x values in the domain are solutions: r < x < s.
• f(x) <= 0: some x values in the domain are solutions: r <= x <= s.
Continuing with our example from earlier, we see that vy = -1/8 puts us in Case 3, vy < 0. We also have f(x) < 0, which is the 3rd subcase.
This means that some x values in the domain are solutions: the x values between the zeros of the quadratic.
The following table summarizes the cases for vy, the sign of the inequality, and the solutions of the quadratic inequality.
### 4. Find The Zeros Of The Quadratic
You can find the zeros of a quadratic in multiple ways. One way is to factor the quadratic and set the factors equal to zero.
Another way is to use the quadratic formula. This is the most versatile method, since it will give us the zeros of any quadratic (no matter how difficult it is to factor!)
Once we have the zeros of the quadratic (call them r and s), we can write out the solution sets for the inequality and then proceed to graph the parabola and shade the appropriate regions.
In our example from earlier, the quadratic is f(x) = 2x2 – 3x + 1, with a = 2, b = -3, and c = 1.
Using the quadratic formula (pictured below), we find:
r = 1/2 and s = 1 are the solutions for the quadratic (note that r < s).
### 5. Write The Solution Set For The Quadratic Inequality
Now that we have the case and the zeros, we can write the solution set for the quadratic inequality.
Following along with the example from earlier, remember that we were in Case 3 (vy < 0) and subcase 3 (f(x) < 0).
This means that our solution set is r < x < s, or ½ < x < 1. We can also write this set as (1/2, 1).
Remember that an open parenthesis means that the endpoint is not included in the set.
### 6. Graph The Parabola Corresponding To The Quadratic
Now we graph the parabola using the following type of curve:
• a solid curve for inclusive inequalities (>= or <= symbols)
• a dashed curve for strict inequalities (> or < symbols)
Going back to our example from before, let’s say we want to graph y < f(x). In that case, we will use a dashed line to graph the parabola.
We know that the zeros are r = 1/2 and s = 1. We also know that the vertex is at the point (vx, vy), or (3/4, -1/8).
If we plot these three points to start our graph, we can sketch in the shape of a parabola that intersects all three points.
The graph of f(x) = 2x2 – 3x + 1 is shown below (drawn as a dashed curve).
### 7. Shade The Appropriate Region Of The Graph, Based On The Inequality Sign
To shade the appropriate region, we shade above or below the curve depending on the sign of the inequality:
• For y > f(x) > 0 or y >= f(x), we shade above the parabola curve.
• For y < f(x) or y <= f(x), we shade below the parabola curve.
Going back to our example from earlier, if we want to graph y < f(x), we will want to shade below the curve.
## Example: Solve & Graph A Quadratic Inequality
Now let’s walk through the steps to solve and graph the quadratic inequality 2x2 + 6x – 12 > 8. We’ll take it step by step.
### Step 1 is to convert to standard form with a > 0.
We can just subtract 8 from both sides to get:
• 2x2 + 6x – 20 > 0
### Step 2 is to find vy = (4ac – b2) / 4a:
vy = (4ac – b2) / 4a
vy = (4(2)(-20) – 62) / 4(2)
vy = (-160 – 36) / 8
vy = -24.5
### Step 3 is to use vy to identify the case for the solutions.
Since vy < 0, we are in Case 3. Since f(x) > 0, we are in subcase 1. The solutions are x < r and x > s.
### Step 4 is to find the zeros of the quadratic.
After using the quadratic formula with a = 2, b = 6, and c = -20, we get r = -5 and s = 2.
### Step 5 is to write the solution set for the quadratic inequality.
We have x < r and x > s, or x < -5 and x > 2.
### Step 6 is to graph the parabola corresponding to the quadratic function f(x) = 2x2 + 6x – 12.
We use a dashed curve, since we have a strict inequality (the > symbol).
### Step 7 is to shade the appropriate region.
If we have the inequality y > f(x), then we want to shade above the curve.
## Conclusion
Now you know how to solve and graph quadratic inequalities. Hopefully this method makes things a little easier to understand and a little less overwhelming. |
# 3.3 Rates of change and behavior of graphs (Page 5/15)
Page 5 / 15
Access this online resource for additional instruction and practice with rates of change.
## Key equations
Average rate of change $\frac{\Delta y}{\Delta x}=\frac{f\left({x}_{2}\right)-f\left({x}_{1}\right)}{{x}_{2}-{x}_{1}}$
## Key concepts
• A rate of change relates a change in an output quantity to a change in an input quantity. The average rate of change is determined using only the beginning and ending data. See [link] .
• Identifying points that mark the interval on a graph can be used to find the average rate of change. See [link] .
• Comparing pairs of input and output values in a table can also be used to find the average rate of change. See [link] .
• An average rate of change can also be computed by determining the function values at the endpoints of an interval described by a formula. See [link] and [link] .
• The average rate of change can sometimes be determined as an expression. See [link] .
• A function is increasing where its rate of change is positive and decreasing where its rate of change is negative. See [link] .
• A local maximum is where a function changes from increasing to decreasing and has an output value larger (more positive or less negative) than output values at neighboring input values.
• A local minimum is where the function changes from decreasing to increasing (as the input increases) and has an output value smaller (more negative or less positive) than output values at neighboring input values.
• Minima and maxima are also called extrema.
• We can find local extrema from a graph. See [link] and [link] .
• The highest and lowest points on a graph indicate the maxima and minima. See [link] .
## Verbal
Can the average rate of change of a function be constant?
Yes, the average rate of change of all linear functions is constant.
If a function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ is increasing on $\text{\hspace{0.17em}}\left(a,b\right)\text{\hspace{0.17em}}$ and decreasing on $\text{\hspace{0.17em}}\left(b,c\right),\text{\hspace{0.17em}}$ then what can be said about the local extremum of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ on $\text{\hspace{0.17em}}\left(a,c\right)?\text{\hspace{0.17em}}$
How are the absolute maximum and minimum similar to and different from the local extrema?
The absolute maximum and minimum relate to the entire graph, whereas the local extrema relate only to a specific region around an open interval.
How does the graph of the absolute value function compare to the graph of the quadratic function, $\text{\hspace{0.17em}}y={x}^{2},\text{\hspace{0.17em}}$ in terms of increasing and decreasing intervals?
## Algebraic
For the following exercises, find the average rate of change of each function on the interval specified for real numbers $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}h$ in simplest form.
$f\left(x\right)=4{x}^{2}-7\text{\hspace{0.17em}}$ on
$4\left(b+1\right)$
$g\left(x\right)=2{x}^{2}-9\text{\hspace{0.17em}}$ on
$p\left(x\right)=3x+4\text{\hspace{0.17em}}$ on
3
$k\left(x\right)=4x-2\text{\hspace{0.17em}}$ on
$f\left(x\right)=2{x}^{2}+1\text{\hspace{0.17em}}$ on $\text{\hspace{0.17em}}\left[x,x+h\right]$
$4x+2h$
$g\left(x\right)=3{x}^{2}-2\text{\hspace{0.17em}}$ on $\text{\hspace{0.17em}}\left[x,x+h\right]$
$a\left(t\right)=\frac{1}{t+4}\text{\hspace{0.17em}}$ on $\text{\hspace{0.17em}}\left[9,9+h\right]$
$\frac{-1}{13\left(13+h\right)}$
$b\left(x\right)=\frac{1}{x+3}\text{\hspace{0.17em}}$ on $\text{\hspace{0.17em}}\left[1,1+h\right]$
$j\left(x\right)=3{x}^{3}\text{\hspace{0.17em}}$ on $\text{\hspace{0.17em}}\left[1,1+h\right]$
$3{h}^{2}+9h+9$
$r\left(t\right)=4{t}^{3}\text{\hspace{0.17em}}$ on $\text{\hspace{0.17em}}\left[2,2+h\right]$
$\frac{f\left(x+h\right)-f\left(x\right)}{h}\text{\hspace{0.17em}}$ given $\text{\hspace{0.17em}}f\left(x\right)=2{x}^{2}-3x\text{\hspace{0.17em}}$ on $\text{\hspace{0.17em}}\left[x,x+h\right]$
$4x+2h-3$
## Graphical
For the following exercises, consider the graph of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ shown in [link] .
Estimate the average rate of change from $\text{\hspace{0.17em}}x=1\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}x=4.$
Estimate the average rate of change from $\text{\hspace{0.17em}}x=2\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}x=5.$
$\frac{4}{3}$
For the following exercises, use the graph of each function to estimate the intervals on which the function is increasing or decreasing.
increasing on $\text{\hspace{0.17em}}\left(-\infty ,-2.5\right)\cup \left(1,\infty \right),\text{\hspace{0.17em}}$ decreasing on
increasing on $\text{\hspace{0.17em}}\left(-\infty ,1\right)\cup \left(3,4\right),\text{\hspace{0.17em}}$ decreasing on $\text{\hspace{0.17em}}\left(1,3\right)\cup \left(4,\infty \right)$
For the following exercises, consider the graph shown in [link] .
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
e power cos hyperbolic (x+iy)
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
Dua
Yes
ahmed
Thank you
Dua
give me treganamentry question
Solve 2cos x + 3sin x = 0.5 |
## Number System Questions and Answers Part-1
1. When we reverse the digits of the number 13, the increases by 18. How many other two digit numbers increases by 18 when their digits are reversed?
a) 5
b) 6
c) 7
d) 8
Explanation: Let the numbers are in the form of (10x + y), so when the digits of the number are reversed the number becomes (10y + x)
According to question,
(10y + x) - (10x + y) = 18
9(y - x) = 18
→ y - x = 2
So, the possible pairs of (x, y) are (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9)
we need the number other than 13.
There are 6 possible numbers i.e. 24, 35, 46, 57, 68, 79
So, total numbers of possible numbers are 6
2. Find the LCM and HCF of 2.5, 0.5 and 0.175.
a) 17.5
b) 5
c) 7.5
d) 2.5
Explanation:
\eqalign{ & 2.5 = \frac{{25}}{{10}}, \cr & 0.5 = \frac{5}{{10}}, \cr & 0.175 = \frac{{175}}{{1000}}, \cr}
Now,
LCM of two or more fractions is given by:
\eqalign{ & \frac{{{\text{LCM}}\,{\text{of}}\,{\text{Numerators}}}}{{{\text{HCF}}\,{\text{of}}\,{\text{Denominators}}}} \cr & \frac{{{\text{LCM}}\,{\text{of}}\,25,\,5,\,175}}{{{\text{HCF}}\,{\text{of}}\,10,\,10,\,1000}} \cr & = \frac{{175}}{{10}} \cr & = 17.5 \cr}
3. If A381 is divisible by 11, find the value of the smallest natural number A?
a) 9
b) 7
c) 6
d) 5
Explanation: A number is divisible by 11 if the difference of the sum of the digits in the odd places and sum of the digits in even place is zero or divisible by 11.
Hence, (A + 8) - (3 + 1) = 0 or multiple of 11.
To get the difference 0 or multiple of 11, we need 7 at the place of A.
So, sum of odd place - sum of even place
= 15 - 4 = 11. And this is divisible by 11.
4. The greatest number which will divides: 4003, 4126 and 4249, leaving the same remainder in each case:
a) 43
b) 41
c) 45
d) None of these
Explanation: Rule- Greatest number with which if we divide P, Q, R and it leaves same remainder in each case. Number is of form = HCF of (P - Q), (P - R)
Therefore, HCF of (4126 - 4003), (4249 - 4003) = HCF of 123, 246 = 41.
5. LCM of two numbers is 936. If their HCF is 4 and one of the numbers is 72, the other is
a) 62
b) 42
c) 52
d) None of these
Explanation: Let two numbers be N1 and N2.
Now,
HCF of the numbers × LCM of the numbers = Multiplication of the numbers.
936 × 4 = 72 × N1
N1 = $$\frac{{936 \times 4}}{{72}}$$
N1 = 52
6. Two alarm clocks ring their alarms at regular intervals of 50 seconds and 48 seconds. If they first beep together at 12 noon, at what time will they beep again for first time?
a) 12:10 PM
b) 12:12 PM
c) 12:11 PM
d) 12:20 PM
Explanation: They will ring together after,
LCM of 48 and 50 secs.
48 = 2 × 2 × 2 × 2 × 3;
50 = 2 × 5 × 5
LCM = 2 × 2 × 2 × 2 × 3 × 5 × 5 = 1200 secs
= 20 min.
They will beep together at 12:20
7. The sum of the digits of two-digit number is 10, while when the digits are reversed, the number decrease by 54. Find the changed number.
a) 28
b) 19
c) 37
d) 46
Explanation: Let number be (10x + y)
According to question,
(10x + y) - (10y + x) = 54
10x - 10y + y - x = 54
9x - 9y = 54
x - y = 6 -------(i)
Sum of digits,
(x + y) = 10 ------- (ii)
(i) - (ii)
So, x - y - x - y = 6 - 10
-2y = -4
y = 2 and, x = 8
Then, the required number is
= (10y + x)
= 10 × 2 + 8
= 28
8. Find the HCF of (3125-1) and (335-1).
a) 35 - 1
b) 312 - 1
c) 34 - 1
d) None of these
Explanation: Rule - The HCF of (am - 1) and (an - 1) is given by (aHCF of m, n - 1)
Thus for this question the answer is (35 - 1)
Since, 5 is the HCF of 35 and 125
9. The HCF of 2472, 1284 and a third number 'N' is 12. If their LCM is 23 × 32 × 5 × 103 × 107, then the number 'N' is:
a) 22 × 32 × 7
b) 22 × 33 × 10
c) 22 × 32 × 5
d) None of these
Explanation: HCF of the numbers × LCM of the numbers = Multiplication of the numbers.
(12) × (23 × 32 × 5 × 103 × 107) = 2472 × 1284 × N
Hence,
N = $$\frac{{\{ \left( {12} \right) \times \left( {{2^3} \times {3^2} \times 5 \times 103 \times 107} \right)\} }}{{2472 \times 1284}}$$
N = 3 × 5
10. The greatest number, which when subtracted from 5834, gives a number exactly divisible by each of 20, 28, 32, 35 is.
a) 1120
b) 5600
c) 4714
d) 5200 |
# Eureka Math Algebra 1 Module 2 Lesson 8 Answer Key
## Engage NY Eureka Math Algebra 1 Module 2 Lesson 8 Answer Key
### Eureka Math Algebra 1 Module 2 Lesson 8 Exploratory Challenge Answer Key
Exploratory Challenge 1: Country Data
A science museum has a Traveling Around the World exhibit. Using 3D technology, participants can make a virtual tour of cities and towns around the world. Students at Waldo High School registered with the museum to participate in a virtual tour of Kenya, visiting the capital city of Nairobi and several small towns. Before they take the tour, however, their mathematics class decided to study Kenya using demographic data from 2010 provided by the United States Census Bureau. They also obtained data for the United States from 2010 to compare to data for Kenya.
The following histograms represent the age distributions of the two countries.
Review with students what each interval of ages represents. For example, the first interval represents people
whose ages are 0≤x<5. Pose the following questions:
→ What percent of people in Kenya are younger than 5?
→ What ages are represented by the intervals along the horizontal axis?
→ If x represents age, then the first interval would be 0≤x<5, the second interval would be 5≤x<10, etc.
→ What does the first bar (0≤x<5) mean in the U.S. histogram?
→ The percent of people in the U.S. who are younger than 5
Exploratory Challenge 2:
Learning More About the Countries Using Box Plots and Histograms
A random sample of 200 people from Kenya in 2010 was discussed in previous lessons. A random sample of 200 people from the United States is also available for study. Box plots constructed using the ages of the people in these two samples are shown below.
Then, discuss:
→ What information is displayed in a box plot?
→ Median, minimum, and maximum values, quartiles, and IQR are displayed.
→ What does the (*) represent on the box plot for Kenya?
→ It represents extreme values or outliers.
→ Can we find this same information in the histograms from Example 1?
→ No. Median, Q1, Q2, and minimum and maximum are not clear from a histogram. It could only be used to estimate these values.
→ Remind students that the histogram represents the entire population of Kenya, whereas the box plot only represents a sample of 200 people.
### Eureka Math Algebra 1 Module 2 Lesson 8 Exercise Answer Key
Exercises 1–8
Exercise 1.
How do the shapes of the two histograms differ?
The bars in the Kenya histogram slowly decline; the distribution is skewed with a tail to the right. The bars in the U.S. histogram are even for a while and then show a more rapid decline.
Exercise 2.
Approximately what percent of people in Kenya are between the ages of 0 and 10 years?
Approximately 32% (17% are ages 0 to 5 years, and 15% are ages 5 to 10 years.)
Exercise 3.
Approximately what percent of people in the United States are between the ages of 0 and 10 years?
Approximately 13%
Exercise 4.
Approximately what percent of people in Kenya are 60 years or older?
Approximately 5%
Exercise 5.
Approximately what percent of people in the United States are 60 years or older?
Approximately 20%
Exercise 6.
The population of Kenya in 2010 was approximately 41 million people. What is the approximate number of people in Kenya between the ages of 0 and 10 years?
32% of 41 million people is approximately 13,120,000 people.
Exercise 7.
The population of the United States in 2010 was approximately 309 million people. What is the approximate number of people in the United States between the ages of 0 and 10 years?
13% of 309 million people is approximately 40,170,000 people.
Exercise 8.
The Waldo High School students started planning for their virtual visit of the neighborhoods in Nairobi and several towns in Kenya. Do you think they will see many teenagers? Will they see many senior citizens who are 70 or older? Explain your answer based on the histogram.
Adding a portion of the percent of people in the 10 to 14 years old group and the percent of people 15 to 19 years old approximates the estimate of the percent of teenagers. About 15% represents teenagers. Students are likely to see teenagers as this is a relatively large percent of the population. According to the histogram, approximately 3% of the population in Kenya is 70 or older. As a result, students are unlikely to see many senior citizens 70 or older.
Exercises 9–16
Exercise 9.
Adrian, a senior at Waldo High School, stated that the box plots indicate that the United States has a lot of older people compared to Kenya. Would you agree? How would you describe the difference in the ages of people in these two countries based on the above box plots?
Yes. The population of the United States has a much greater percent of people in the older age intervals.
Exercise 10.
Estimate the median age of a person in Kenya and the median age of a person in the United States using the box plots.
The median Kenyan age is slightly less than 20 years, while the median U.S. age is slightly less than 40 years.
Exercise 11.
Using the box plot, 25% of the people in the United States are younger than what age? How did you determine that age?
25% are younger than approximately 18 years. I used the value of Q1, or the first quartile.
Exercise 12.
Using the box plot, approximately what percent of people in Kenya are younger than 18 years old?
Approximately 50% of the people in Kenya are less than 18 years old.
Exercise 13.
Could you have estimated the mean age of a person from Kenya using the box plot? Explain your answer.
No, the box plot does not provide an estimate of the mean age.
Exercise 14.
The mean age of people in the United States is approximately 38 years. Using the histogram, estimate the percent of people in the United States who are younger than the mean age in the United States.
Approximately 50% of the U.S. population is less than 38 years old.
Exercise 15.
If the median age is used to describe a typical person in Kenya, what percent of people in Kenya are younger than the median age? Is the mean or median age a better description of a typical person in Kenya? Explain your answer.
50% of the people in Kenya are less than the median age. The median is a better indicator of a typical age because the distribution is skewed.
Exercise 16.
What is the IQR of the ages in the sample from the United States? What is the IQR of the ages in the sample from Kenya? If the IQRs are used to compare countries, what does a smaller IQR indicate about a country? Use Kenya and the United States to explain your answer.
The IQR for the United States is 58 years – 18 years, or 40 years; the IQR for Kenya is 36 years – 7 years,
or 31 years. A smaller IQR indicates that more of the sample is around the median age, which you can see from looking at the histogram.
### Eureka Math Algebra 1 Module 2 Lesson 8 Problem Set Answer Key
The following box plot summarizes ages for a random sample from a made-up country named Math Country.
Question 1.
Make up your own sample of forty ages that could be represented by the box plot for Math Country. Use a dot plot to represent the ages of the forty people in Math Country.
Many possible dot plots would be correct. Analyze individually. Ten of the ages need to be between 0 and 25 years old, ten of the ages need to be between 25 and 40 years old, ten of the ages need to be between 40 and 70 years old, and ten of the ages need to between 70 and 90 years old.
Question 2.
Is the sample of forty ages represented in your dot plot of Math Country the only sample that could be represented by the box plot? Explain your answer.
There are many possible dot plots that might be represented by this box plot. Any data set with the same 5-number summary would result in this same box plot.
Question 3.
The following is a dot plot of sixty ages from a random sample of people from Japan in 2010.
Draw a box plot over this dot plot.
The following is the box plot of the ages of the sample of people from Japan:
Question 4.
Based on your box plot, would the median age of people in Japan be closer to the median age of people in Kenya or the United States? Justify your answer.
The median age of Japan would be closer to the median age of the United States than to the median age of Kenya. The box plot indicates that the median age of Japan is approximately 45 years old. This median age is even greater than the median age of the United States.
Question 5.
What does the box plot of this sample from Japan indicate about the possible differences in the age distributions of people from Japan and Kenya?
A much greater percent of the people in Japan are in the older age groups than is the case for Kenya.
### Eureka Math Algebra 1 Module 2 Lesson 8 Exit Ticket Answer Key
Question 1.
Using the histograms of the population distributions of the United States and Kenya in 2010, approximately what percent of the people in the United States were between 15 and 50 years old? Approximately what percent of the people in Kenya were between 15 and 50 years old? |
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