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# topics: topic 1: solving linear equations topic 2: solving quadratic equations
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Topics: Topic 1: Solving Linear Equations Topic 2: Solving Quadratic Equations Topic 3: Solving Proportions involving linear and quadratic functions. Topic 4: Solving Absolute Value Equations. Graph: y = 5x - 2. Graph: y = -x - 6. Solution: (-3, -3). 2x + 3 = -x - 6. 3x + 3 = -6. - PowerPoint PPT Presentation
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• Topics:Topic 1: Solving Linear EquationsTopic 2: Solving Quadratic EquationsTopic 3: Solving Proportions involving linear and quadratic functions.Topic 4: Solving Absolute Value Equations
• Graph: y = 5x - 2Graph: y = -x - 6Solution of a SystemThe place where two (or more) graphs intersect is the solution of the system. The solution is written as an ordered pair (x, y).Solution: (-3, -3)2x + 3 = -x - 63x + 3 = -63x = -9x = -3Check: 2(-3) + 3 = -(-3) - 6-6 + 3 = 3 - 6-3 = -3
• Graph: y = 2x - 2Graph: y = 2(x + 1) + 1Solution of a SystemThe place where two (or more) graphs intersect is the solution of the system. The solution is written as an ordered pair (x, y).Solution: No Solution2x - 2 = 2(x + 1) + 12x - 2 = 2x + 2 + 12x - 2 = 2x + 3No SolutionSince the ROCs are the same and the y-intercepts are different the lines are parallel resulting in No Solution.
• Graph: y = x - 3 Graph: y = (x - 4) - 1Solution of a SystemThe place where two (or more) graphs intersect is the solution of the system. The solution is written as an ordered pair (x, y).Solution: Infinitely Many Solutions x - 3 = (x - 4) - 1x - 3 = x - 2 - 1x - 3 = x - 3Infinitely Many SolutionsSince the ROCs are the same and the y-intercepts are the same the lines have Infinitely Many SolutionsCheck: (2) - 3 = (2 - 4) - 1 -2 = -2(2, -2)
• Graph: y = 6x + 7 Graph: y = 2x + 596x + 7 = 2x + 594x + 7 = 594x = 52x = 13Check:6(13) + 7 = 2(13) + 5978 + 7 = 26 + 5985 = 85Solution: (13, 85) Graph: y = 3(x + 5) - 7 Graph: y = 2x + 13(x + 5) - 7 = 2x + 13x + 15 - 7 = 2x + 13x + 8 = 2x + 1x + 8 = 1x = -7Check: 3(-7 + 5) - 7 = 2(-7) + 13(-2) - 7 = 2(-7) + 1-6 - 7 = -14 + 1-13 = -13Solution: (-7, -13)
• Graph: y = -6x + 17 Graph: y = -3x + 25-6x + 17 = -3x + 25-3x + 17 = 25-3x = 8x = -8/3Check: -6(-8/3) + 17 = -3(-8/3) + 2516 + 17 = 8 + 2533 = 33Solution: (-8/3, 33) Graph: y = (x + 3) + 5 Graph: y = x + 13/2(x + 3) + 5 = x + 13/2x + 3/2 + 5 = x + 13/2x + 13/2 = x + 13/2Infinitely Many SolutionsCheck: (-3 + 3) + 5 = (-3) + 13/2(0) + 5 = (-3) + 13/20 + 5 = -3/2 + 13/25 = 5Solution: (-3, 5) Note: This is one of many solutions |
Mean, Median, Mode & Range
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# Mean, Median, Mode & Range - PowerPoint PPT Presentation
Mean, Median, Mode & Range. Content Standards Mathematics and Numeracy G1.S1.d1-2, 6 G2.S7.a.6. Vocabulary Review. Sum – the answer to an addition problem. Addend – the numbers you added together to get the sum. 6 + 9 = 15. Definition. Mean Means Average. Definition.
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### Mean, Median, Mode & Range
Content Standards
Mathematics and Numeracy
G1.S1.d1-2, 6
G2.S7.a.6
Vocabulary Review
• Addend – the numbers you added together to get the sum.
6 + 9 = 15
Definition
Mean
Means
Average
Definition
• Mean– the average of a group of numbers.
2, 5, 2, 1, 5
Mean = 3
How to Find the Mean of a Group of Numbers
• Step 1 – Add all the numbers.
8, 10, 12, 18, 22, 26
8+10+12+18+22+26 = 96
How to Find the Mean of a Group of Numbers
• Step 2 – Divide the sum by the number of addends.
8, 10, 12, 18, 22, 26
8+10+12+18+22+26 = 96
How to Find the Mean of a Group of Numbers
• Step 2 – Divide the sum by the number of addends.
1
6
6)
96
sum
6
3
6
3
6
How to Find the Mean of a Group of Numbers
The mean or average of these numbers is 16.
8, 10, 12, 18, 22, 26
Definition
Median
is in the
Middle
Definition
• Median– the middle number in a set of ordered numbers.
1, 3, 7, 10, 13
Median = 7
How to Find the Median in a Group of Numbers
• Step 1 – Arrange the numbers in order from least to greatest.
21, 18, 24, 19, 27
18, 19, 21, 24, 27
How to Find the Median in a Group of Numbers
• Step 2 – Find the middle number.
21, 18, 24, 19, 27
18, 19, 21, 24, 27
How to Find the Median in a Group of Numbers
• Step 2 – Find the middle number.
18, 19, 21, 24, 27
How to Find the Median in a Group of Numbers
• Step 3 – If there are two middle numbers, find the mean of these two numbers.
18, 19, 21, 25, 27, 28
How to Find the Median in a Group of Numbers
• Step 3 – If there are two middle numbers, find the mean of these two numbers.
46
21+ 25 =
median
23
2)
46
What is the median of these numbers?
29, 8, 4, 11, 19
4, 8, 11, 19, 29
11
What is the median of these numbers?
31, 7, 2, 12, 14, 19
2, 7, 12, 14, 19, 31
13
2)
12 + 14= 26
26
What is the median of these numbers?
53, 5, 81, 67, 25, 78
5, 25, 53, 67, 78, 81
60
2)
53 + 67= 120
120
Definition
Mode
is the most
Popular
Definition
• A la mode– the most popular or that which is in fashion.
Baseball caps are a la mode today.
Definition
• Mode– the number that appears most frequently in a set of numbers.
1, 1, 3, 7, 10, 13
Mode = 1
How to Find the Mode in a Group of Numbers
• Step 1 – Arrange the numbers in order from least to greatest.
21, 18, 24, 19, 18
18, 18, 19, 21, 24
How to Find the Mode in a Group of Numbers
• Step 2 – Find the number that is repeated the most.
21, 18, 24, 19, 18
18, 18, 19, 21, 24
Which number is the mode?
29, 8, 4, 8, 19
4, 8, 8, 19, 29
8
Which number is the mode?
1, 2, 2, 9, 9, 4, 9, 10
1, 2, 2, 4, 9, 9, 9, 10
9
Which number is the mode?
22, 21, 27, 31, 21, 32
21, 21, 22, 27, 31, 32
21
Definition
Range
is the distance
Between
Definition
• Range– the difference between the greatest and the least value in a set of numbers.
1, 1, 3, 7, 10, 13
Range = 12
How to Find the Range in a Group of Numbers
• Step 1 – Arrange the numbers in order from least to greatest.
21, 18, 24, 19, 27
18, 19, 21, 24, 27
How to Find the Range in a Group of Numbers
• Step 2 – Find the lowest and highest numbers.
21, 18, 24, 19, 27
18, 19, 21, 24, 27
How to Find the Range in a Group of Numbers
• Step 3 – Find the difference between these 2 numbers.
18, 19, 21, 24, 27
27 – 18 = 9
The range is 9
What is the range?
29, 8, 4, 8, 19
4, 8, 8, 19, 29
29 – 4= 25
What is the range?
22, 21, 27, 31, 21, 32
21, 21, 22, 27, 31, 32
32 – 21 = 11
What is the range?
31, 8, 3, 11, 19
3, 8, 11, 19, 31
31 – 3 = 28
What is the range?
23, 7, 9, 41, 19
7, 9, 23, 19, 41
41 – 7 = 34 |
# How To Calculate The Perimeter Area
## Video: How To Calculate The Perimeter Area
Geometry studies the properties and characteristics of two-dimensional and spatial shapes. The numerical values characterizing such structures are the area and perimeter, the calculation of which is carried out according to known formulas or expressed through one another.
## Instructions
### Step 1
Rectangle Challenge: Calculate the area of a rectangle if you know that its perimeter is 40 and the length b is 1.5 times the width a.
### Step 2
Solution: Use the well-known perimeter formula, it is equal to the sum of all sides of the shape. In this case, P = 2 • a + 2 • b. From the initial data of the problem, you know that b = 1.5 • a, therefore, P = 2 • a + 2 • 1.5 • a = 5 • a, whence a = 8. Find the length b = 1.5 • 8 = 12.
### Step 3
Write down the formula for the area of a rectangle: S = a • b, Plug in the known values: S = 8 • * 12 = 96.
### Step 4
Square Problem: Find the area of a square if the perimeter is 36.
### Step 5
Solution. A square is a special case of a rectangle, where all sides are equal, therefore, its perimeter is 4 • a, whence a = 8. The area of the square is determined by the formula S = a² = 64.
### Step 6
Triangle. Problem: Let an arbitrary triangle ABC be given, the perimeter of which is 29. Find out the value of its area if it is known that the height BH, lowered to the side AC, divides it into segments with lengths of 3 and 4 cm.
### Step 7
Solution: First, remember the area formula for a triangle: S = 1/2 • c • h, where c is the base and h is the height of the figure. In our case, the base will be the side AC, which is known by the problem statement: AC = 3 + 4 = 7, it remains to find the height BH.
### Step 8
The height is the perpendicular to the side from the opposite vertex, therefore, it divides triangle ABC into two right-angled triangles. Knowing this property, consider the triangle ABH. Remember the Pythagorean formula, according to which: AB² = BH² + AH² = BH² + 9 → AB = √ (h² + 9) In the triangle BHC, write down the same principle: BC² = BH² + HC² = BH² + 16 → BC = √ (h² + 16).
### Step 9
Apply the perimeter formula: P = AB + BC + AC Substitute the height values: P = 29 = √ (h² + 9) + √ (h² + 16) + 7.
### Step 10
Solve the equation: √ (h² + 9) + √ (h² + 16) = 22 → [replacement t² = h² + 9]: √ (t² + 7) = 22 - t, square both sides of the equality: t² + 7 = 484 - 44 • t + t² → t≈10, 84h² + 9 = 117.5 → h ≈ 10.42
### Step 11
Find the area of triangle ABC: S = 1/2 • 7 • 10, 42 = 36, 47. |
Multiplication by 11 using Speed Mathematics
In this chapter you are going to see how you can easily multiply any number with 11 .The principles in these tutorials are taken from various systems like Indian Vedic Mathematics and the German Trachtenberg System.
Example 1
Question : Calculate 12 * 11
Solution
Step 1 : The last digit of 12 is 2. Write it down as the right most digit of the answer. So we have
2
Step 2 : The last digit of 12 is 2. Its left neighbor digit is 1. Add these two. That is 2 +1 = 3. Write it as the second last digit of the answer. So we have
32
Step 3 : The left most digit of 12 is 1. Write it as the left most digit of the answer. So we have
132
This is the answer. That is 12*11 = 132
Example 2
Let's take a bigger number this time. Say we need to do 1245 * 11
Solution
Step 1: The last digit of 1245 is 5. Write it down as the right most digit of the answer. So we have
5
Step 2: The last digit of 1245 is 5. Its left neighbor digit is 4. Add these two. That is 5 +4 = 9. Write it as the second last digit of the answer. So we have
95
Step 3: The second last digit of 1245 is 4. Its left neighbor digit is 2. Add these two. That is 4 +2 = 6. Write it as the third last digit of the answer. So we have
695
Step 4: The third last digit of 1245 is 2. Its left neighbor digit is 1. Add these two. That is 2 +1 = 3. Write it as the fourth last digit of the answer. So we have
3695
Step 5: The left most digit of 1245 is 1. Write it as the left most digit of the answer. So we have
13695
Yes, this is the answer. 1245 * 11 = 13695.
Example 3
Let's take a another bigger number this time. Say we need to do 472385 * 11
Solution
Step 1: The last digit of 472385 is 5. Write it down as the right most digit of the answer. So we have
5
Step 2: The last digit of 472385 is 5. Its left neighbor digit is 8. Add these two. That is 5 +8 = 13. Take 3 from it as the second last digit of the answer and treat the 1 as carry.So we have
35 (Carry = 1)
Step 3: The second last digit of 472385 is 8. Its left neighbor digit is 3. Add these two. That is 8 +3= 11. We have a carry 1. Add this to 11. So 11+1 = 12. Take 2 from it as the third last digit of the answer and treat the 1 as carry.So we have
235 (Carry = 1)
Step 4: The third last digit of 472385 is 3. Its left neighbor digit is 2. Add these two. That is 3 +2= 5. We have a carry 1. Add this also. That is 5+1=6. Write it as the fourth last digit of the answer. So we have
6235
Step 5: The fourth last digit of 472385 is 2. Its left neighbor digit is 7. Add these two. That is 2 +7= 9. Write it as the fifth last digit of the answer. So we have
96235
Step 6: The fifth last digit of 472385 is 7. Its left neighbor digit is 4. Add these two. That is 7 +4= 11. Take 1 from it as the sixth last digit of the answer and treat 1 as carry. So we have
196235 (Carry = 1)
Step 7 : The left most digit of 472385 is 4. Add the carry to this. that is 4+1 = 5. Write it as the left most digit of the answer. So we have
5196235
Yes, this is the answer. 472385 * 11 = 5196235.
Summary
Don't you see this as a simple process? Once you remember this pattern, you can do these calculations mentally.
showing 1-10 of 44 comments, sorted newest to the oldest
aban
2015-03-25 16:27:17
Why it works? If you write the numbers in 2 lines (your number on top of 11) and then do a simple multlipication, you will quickly and easily see why this method works for any number no matter how large!
19*11 =
19 *
11
------
19
19
-----
209
The more fun is playing with two numbers having all ones, like: 11111111111111111*11111111111111111
I can give you the answer without writing a single digit down, just by looking at the numbers!
Arkajyoti
2014-12-19 18:00:05
these shortcuts and tricks are very important but many people read this and then forget.
dreamBlitzer
2014-11-24 15:29:21
15285966*11 = 152859660 + 15285966
chetan
2014-07-02 10:47:03
please check for 385*92 i am not getting correct answer which is 35420
anu
2014-08-26 16:32:44
these multiplications are only for 11 ,not with 92
san
2014-03-05 04:20:54
what about using your method for below numbers
19*11=
1249*11=
we don't get proper answers when we apply your method.Pls Reply
Sharath
2015-03-28 17:03:38
START FROM LEFT SIDE
(9+4=13 SO KEEP 3 ADD ONE NEXT SUM)9
(2+4+1=7)39
(1+2=3)739
(1)3739
13739
H.G. Moliya
2015-02-14 14:42:17
19*11= 190+19 = 209
1249*11= 12490+1249 = 13739
abc
2015-01-07 14:49:32
19*11 as per method:
last digit of 19 is 9
neighbour of 9 is 1 add 9+1=10
then add left part of 10 with left part of 19 i.e-1+1=2
now put all together i.e 209 similarly it follows 29*11,39*11...
Alex
2014-12-27 15:03:50
19*11
1)9
2)9+1=10=0 (carry1)
3)1+1(carry)=2
Ans: 209
1249*11=
1)9
2)9+4=13=3 (1carry)
3)4+2+1(carry)=7
4)2+1=3
5)1
Ans: 13739
12345Next Go
showing 1-10 of 44 comments
Add a new comment... (Use Discussion Board for posting new aptitude questions.)
Name:
Email: (optional)
7 + 4 = (please answer the simple math question) |
# Texas Go Math Grade 5 Unit 1 Assessment Answer Key
Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Unit 1 Assessment Answer Key.
## Texas Go Math Grade 5 Unit 1 Assessment Answer Key
Vocabulary
Choose the best term from the box.
Vocabulary
Associative Property
Commutative Property
inverse operations
Question 1.
The _________ states that changing the grouping of factors does not change the product. (p. 5)
Answer:
Commutative Property,
Explanation:
Commutative property is applicable only for addition and multiplication processes.
Thus, it means we can change the position or swap the numbers when adding or
multiplying any two numbers. This is one of the major properties of integers.
For example: 1+2 = 2+1 and 2 x 3 = 3 x 2.
therefore, the ___Commutative Property__ states that changing the grouping of
factors does not change the product.
5th Grade Go Math Unit 1 Assessment Answer Key Question 2.
Addition and subtraction are ___________. (p. 41)
Answer:
Inverse operations,
Explanation:
Addition and subtraction are the inverse operations of each other.
Simply, this means that they are the opposite.
We can undo an addition through subtraction and
we can undo a subtraction through addition.
Concepts and Skills
Compare. Write <, >, or =. (TEKS 5.2.B)
Question 3.
6.35 0.695
Answer:
6.35 > 0.695,
Explanation:
Given to compare between 6.35 and 0.695
as 6.35 is greater than 0.695 therefore
6.35 > 0.695.
Question 4.
0.02 0.020
Answer:
0.02 = 0.020,
Explanation:
Given to compare between 0.02 and 0.020 as
0.02 is equal to 0.020 therefore 0.02 = 0.020.
Texas Go Math Grade 5 Unit 1 Answer Key Question 5.
0.132 0.2
Answer:
0.132 < 0.2,
Explanation:
Given to compare between 0.132 and 0.2
as 0.132 is less than 0.2 therefore 0.132 < 0.2.
Estimate. Then solve. (TEKS 5.3.A. 5.3.B, 5.3.C)
Question 6.
Estimate: _____9,000_______
Answer:
8760, Estimate is 9,000,
Explanation:
Given to multiply 24 with 365 we get 8760 as shown above,
as per estimation 24 ≈ 25 and 365 ≈ 360 we get 25 X 360 = 9,000.
Question 7.
Estimate: ______30______
616 ÷ 22
Answer:
22)616(28
44_
   176
   176
0
28, Estimate is 30,
Explanation:
Given to find 616 ÷ 22 we get 28 as
per estimation 616 ≈ 600 and 22 ≈ 20,
now we divide 600 ÷ 20 = 30.
Question 8.
Estimate: ____250_________
5,184 ÷ 18
Answer:
18)5184(288
   36
   158
   144
    144
    144
0
288, Estimate is 250
Explanation:
Given to find 5,184 ÷ 18 we get 288 as
per estimation 5,184 ≈ 5,200 and ≈ 20,
now we divide 5000 ÷ 20 = 250.
Use models or strategies to find the product. Show your work. (TEKS 5.3.D, 5.3.E)
Grade 5 Unit 1 Assessment Answer Key Texas Go Math Question 9.
0.05 × 1.32
Answer:
0.05 X 1.32 = 0.066,
Explanation:
Given to find 0.05 X 1.32 if we multiply
   11
1.32
X0.05
 0.0660 the result 0.0660 shaded region is
shown in the graph.
Question 10.
23 × 5.28
Answer:
23 × 5.28 = 121.44,
Explanation:
Upon multiplying 23 X 5.28 we get
2
23
X5.28
001.84
004.60
115.00
11
121.44
Question 11.
4.2 × 14.85
Answer:
4.2 × 14.85 = 62.37,
Explanation:
Given to find 4.2 X 14.85 we get
4.2
14.85
00.21
03.36
16.80
42.00
11
62.37
Use models or strategies to find the quotient. Show your work. (TEKS 5.3.F, 5.3.G)
Question 12.
3.6 ÷ 4
Answer:
3.6 ÷ 4 = 0.9,
Explanation:
4)3.6(0.9
  3.6
0
Upon dividing 3.6 ÷ 4 we get 0.9.
Texas Go Math Grade 5 Answer Key Pdf Unit 1 Question 13.
16.24 ÷ 29
Answer:
16.24 ÷ 29 = 0.56,
Explanation:
29)16.24(0.56
    14.50
     1.74
     1.74
0Â Â
Upon dividing 16.24 ÷ 29 we get 0.56.
Question 14.
96.72 ÷ 62
Answer:
96.72 ÷ 62 = 1.56,
Explanation:
62)96.72(1.56
   62
   34.72
   31.00
    3.72
    3.72
     0
Fill in the bubble completely to show your answer.
Question 15.
Kaya’s score in the gymnastics competition is 15.4 when rounded to the nearest tenth.
Which of the following is her actual score? (TEKS 5.2.C)
(A) 15.333
(B) 15.496
(C) 15.395
(D) 15.349
Answer:
Kaya’s actual score is (C) 15.395,
Explanation:
Given Kaya’s score in the gymnastics competition is 15.4,
when rounded to the nearest tenth her actual score out of
15.333, 15.496, 15.395, 15.349 would be 15.395 as we see
15.4 is near to and in between 15.333<15.349<15.395 and 15.496.
Question 16.
A bakery uses 1,750 kilograms of flour to make 1,000 loaves of bread.
How much flour is needed to make 10 loaves? (TEKS 5.3.G)
(A) 17.5 kilograms
(B) 175,000 kilograms
(C) 1.75 kilograms
(D) 175 kilograms
Answer:
(A) 17.5 kilograms,
Explanation:
Given a bakery uses 1,750 kilograms of flour to make 1,000 loaves of bread
now flour needed to make 1 loaves is 1,750 kilograms ÷ 1,000 = 1.75 kilograms,
flour needed to make 10 loaves is 1.75 kilograms X 10 = 17.5 kilograms,
therefore matches with (A) 17.5 kilograms.
Question 17.
Maxine paints a mural that is 4.65 meters long.
The width of the mural is 0.8 times the length.
Maxine increases the width by another 0.5 meters.
How wide is the mural? (TEKS 5.3.E, 5.3.K)
(A) 3.77 meters
(B) 37.2 meters
(C) 4.22 meters
(D) 3.72 meters
Answer:
(D) 3.72 meters,
Explanation:
Given Maxine paints a mural that is 4.65 meters long.
The width of the mural is 0.8 times the length.
So width is 4.65 X 0.8 = 3.72 meters, if Maxine increases
the width by another 0.5 meters then wide the mural is
3.72 X 0.5 meters = 3.72 meters + 1.86 meters = 3.72 meters
which matches with (D).
5th Grade Unit 1 Math Test Texas Go Math Question 18.
Juan uses the model below to solve a problem.
Which of the following equations matches Juan’s model? (TEKS 5.3.F)
(A) 0.4 × 32 = 12.8
(B) 1.28 ÷ 4 = 0.32
(C) 0.32 ÷ 4 = 0.08
(D) 0.32 + 4 = 4.32
Answer:
(A) 0.4 × 32 = 12.8,
Explanation:
Given Juan uses the 2 graph models of 10 X 10 above to solve
the problem the first graph if we count has 32 boxes of 0.4 each,
and the result is 12.8 boxes therefore the equation is
(A) 0.4 × 32 = 12.8.
Question 19.
The price of a shirt is $26.50. The matching shorts are 0.9 times the price of the shirt. If Li wants to buy the shirt and the shorts, how much money will he need? (TEKS 5.3.E, 5.3.K) (A)$238.50
(B) $35.50 (C)$23.85
(D) $50.35 Answer: (D)$50.35,
Explanation:
Given the price of a shirt is $26.50. The matching shorts are 0.9 times the price of the shirt. So price of the shirt is$26.50 X 0.9 = $23.85, If Li wants to buy the shirt and the shorts the price is$26.50 + $23.85 =$50.35 which matches with (D) above.
Question 20.
Ali’s times for the four laps of the race are: 15.36 seconds, 15.95 seconds,
17.83 seconds, and 18.25 seconds. About how long did Ali take to
complete the whole race? (TEKS 5.3.A)
(A) 47 seconds
(B) 18 seconds
(C) 15 seconds
(D) 67 seconds
Answer:
(D) 67 seconds,
Explanation:
Given Ali’s times for the four laps of the race are: 15.36 seconds, 15.95 seconds,
17.83 seconds, and 18.25 seconds. Long did Ali took to complete the whole race
is 15.36 + 15.95 + 17.83 + 18.25 = 67.39 seconds therefore whole is 67 seconds
matches with (D).
Question 21.
Goran wants to build a square picture frame with sides that are 5.25 inches long.
Natalie wants to build a square sandbox and needs 11 times the amount of wood
that Goran needs to build his frame. They have 4 pieces of wood that are each 65.5 inches long.
How much wood will they have left over after making the frame and sandbox? (TEKS 5.3.E, 5.3.K)
(A) 10 inches
(B) 2.5 inches
(C) 7.75 inches
(D) Not here
Answer:
(A) 10 inches,
Explanation:
Given Goran wants to build a square picture frame with sides that are
5.25 inches long so picture frame requires is 4 X 5.25 inches = 21 inches,
Natalie build’s a square sandbox that needs 11 times the amount of wood that
Goran needs to build his frame is 11 X 21 = 231 inches,
They have 4 pieces of wood that are each 65.5 inches long,
so wood they have is 4 X 65.5 = 262 inches,
Both Goran and Natalie needs 21 + 231 = 252 inches,
Wood will they have left over after making the frame and sandbox is
262 inches – 252 inches = 10 inches matches with (A).
Texas Go Math Grade 5 Unit 1 Assessment Answers Key Question 22.
Mustafa buys 6 cans of beans. Each can contain 12.6 ounces of beans.
Mustafa uses 0.7 of the beans in a stew and the rest of the beans for tacos.
How many ounces does he use for the tacos? (TEKS 5.3.E, 5.3.K)
(A) 21 ounces
(B) 75.60 ounces
(C) 52.92 ounces
(D) 22.68 ounces
Answer:
(C) 52.92 ounces,
Explanation:
Given Mustafa buys 6 cans of beans. Each can contains 12.6 ounces of beans.
So Mustafa has 6 X 12.6 = 75.6 ounces,
Mustafa uses 0.7 of the beans in a stew and the rest of the beans for tacos.
So Mustafa uses 75.6 X 0.7 = 52.92 ounces for the tacos matches with (C).
Question 23.
The scale at a butcher shop shows the weight of the meat as 5.363 pounds.
The butcher rounds the weight to the nearest hundredth.
Which of the following shows the new number in expanded form? (TEKS 5.2.A, 5.2.C)
(A) 5 + 0.3 + 0.06 + 0.003
(B) 5 + 0.3 + 0.07
(C) 5 + 0.3 + 0.06
(D) 5 + 3 + 6 + 3
Answer:
(A) 5 + 0.3 + 0.06 + 0.003,
Explanation:
Given the scale at a butcher shop shows the weight of the meat as 5.363 pounds.
The butcher rounds the weight to the nearest hundredth.
So the new number in expanded form is
5 X 1 + 3 X 0.1 + 6 X 0.01 + 3 X 0.003 =
5 + 0.3 + 0.06 + 0.03 matches with (A).
Question 24.
The table shows the times recorded by the top 3 swimmers in the 100 meter race.
What is the value of the digit 6 in the fastest recorded time? (TEKS 5.2.A, 5.2.B)
(A) 0.06
(B) 0.006
(C) 0.6
(D) 6
Answer:
(C) 0.6,
Explanation:
The table above showed the times recorded by the top
3 swimmers in the 100 meter race are as 51.695 seconds,
51.563 seconds and 51.536 seconds among the three the
fastest recorded time is 51.695 seconds and the value of the
digit 6 in the fastest recorded time 6 X 0.1 = 0.6 matches with (C).
Question 25.
Tickets to the school play cost $3.65 for children and$5.65 for adults.
Sonal buys tickets for 3 children and 2 adults. How much money should
she get back if she gives the cashier $50? (TEKS 5.3.E, 5.3.K) (A)$39.05
(B) $27.75 (C)$26
(D) $38.70 Answer: (B)$27.75,
Explanation:
Given Tickets to the school play cost $3.65 for children and$5.65 for adults.
Sonal buys tickets for 3 children and 2 adults, for 3 children the cost will be
3 X $3.65 =$10.95 and for 2 adults it will be 2 X$5.65 =$11.3,
So Sonal buys tickets of cost $10.95 +$11.3 = $22.25 in total. Now money should she get back if she gives the cashier$50 is
$50 –$22.25 = $27.75 which matches with (B). Texas Go Math 5th Grade Unit 1 Assessment Answers Question 26. The prices for different beverages and snacks at a snack stand in a park are shown on the table. Emily spent$8.11 on park snacks for her
friends and herself. Make a list of the items she may have purchased.
Justify the amount spent. (TEKS 5.3.K )
Answer:
Given amount spent by Emily is $8.11 and the actual amount spent by Emily which is approximately equal to$7.71,
Explanation:
Given the prices for different beverages and snacks at a snack stand in a park
are shown in the table. Emily spent $8.11 on park snacks for her friends and herself. and made a list of the items she would have purchased, so upon adding the amount purchased we get$0.89 + $1.29 +$1.78 + $2.50 +$1.25 = $7.71 approximately to$8.11.
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## FACTORS: FACTORS OF 144
FACTORS OF 144
144 has 15 factors.
Factors:
1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144
Factor Pairs:
• 1 × 144 = 144
• 2 × 72 = 144
• 3 × 48 = 144
• 4 × 36 = 144
• 6 × 24 = 144
• 8 × 18 = 144
• 9 × 16 = 144
• 12 × 12 = 144
DESCRIPTIONS
Factors of 144 can be found by ckecking the divisibility of 144 by the whole numbers 1 through $\mathbf{\sqrt{\color{chocolate}144}}$ = 12. If a whole number divides 144 exactly, both
• the number itself and
• the result of the division by that number
are the factors of 144.
OTHER INFORMATION
Prime factors of 144 are 2 and 3.
## INFORMATION
The solution and descriptions above are generated by the factors calculator. You can use the factors calculator to see the factors of other numbers.
### FACTORS OF A NUMBER
Factors of a number a are the numbers that divide a exactly, without a remainder. A number can be expressed as the multiplication of its factors. |
UPSC > Exercise 13.2 NCERT Solutions - Surface Areas and Volumes
# Exercise 13.2 NCERT Solutions - Surface Areas and Volumes - NCERT Textbooks (Class 6 to Class 12) - UPSC
Q.1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π = 22 / 7)
Solution:
Height of cylinder, h = 14cm
Let the diameter of the cylinder be d
Curved surface area of cylinder = 88 cm2
We know that, formula to find Curved surface area of cylinder is 2πrh.
So 2πrh = 88 cm2 (r is the radius of the base of the cylinder)
2 × (22 / 7) × r × 14 = 88 cm2
2r = 2 cm
d = 2 cm
Therefore, the diameter of the base of the cylinder is 2 cm.
Q.2. It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same? Assume π = 22 / 7
Solution:
Let h be the height and r be the radius of a cylindrical tank.
Height of cylindrical tank, h = 1m
Radius = half of diameter = (140 / 2) cm = 70cm = 0.7m
Area of sheet required = Total surface are of tank = 2πr(r + h) unit square
= [2 × (22 / 7) × 0.7(0.7 + 1)] = 7.48
Therefore, 7.48 square meters of the sheet are required.
Q.3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4cm. (see fig. 13.11). Find its
(i) inner curved surface area,
(ii) outer curved surface area
(iii) total surface area (Assume π = 22 / 7)
Solution:
Let r1 and r2 Inner and outer radii of cylindrical pipe
r1 = 4 / 2 cm = 2 cm
r2 = 4.4 / 2 cm = 2.2 cm
Height of cylindrical pipe, h = length of cylindrical pipe = 77 cm
(i) curved surface area of outer surface of pipe = 2πr1h
= 2 × (22 / 7) × 2 × 77 cm2
= 968 cm2
(ii) curved surface area of outer surface of pipe = 2πr2h
= 2 × (22 / 7) × 2.2 × 77 cm2
= (22 × 22 × 2.2) cm2
= 1064.8 cm2
(iii) Total surface area of pipe = inner curved surface area+ outer curved surface area+ Area of both circular ends of pipe.
= 2r1h + 2r2h + 2π(r12 - r22)
= 9668 + 1064.8 + 2 × (22 / 7) × (2.22 - 22)
= 2031.8 + 5.28
= 2038.08 cm2
Therefore, the total surface area of the cylindrical pipe is 2038.08 cm2.
Q.4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2? (Assume π = 22 / 7)
Solution:
A roller is shaped like a cylinder.
Let h be the height of the roller and r be the radius.
h = Length of roller = 120 cm
Radius of the circular end of roller = r = (84 / 2) cm = 42 cm
Now, CSA of roller = 2πrh
= 2 × (22 / 7) × 42 × 120
= 31680 cm2
Area of field = 500 × CSA of roller
= (500 × 31680) cm2
= 15840000 cm2
= 1584 m2.
Therefore, area of playground is 1584 m2.
Q.5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m2. (Assume π = 22 / 7)
Solution:
Let h be the height of a cylindrical pillar and r be the radius.
Given:
Height cylindrical pillar = h = 3.5 m
Radius of the circular end of pillar = r = diameter / 2 = 50 / 2 = 25cm = 0.25m
CSA of pillar = 2πrh
= 2 × (22 / 7) × 0.25 × 3.5
= 5.5 m2
Cost of painting 1 m2 area = Rs. 12.50
Cost of painting 5.5 m2 area = Rs (5.5 × 12.50)
= Rs.68.75
Therefore, the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m2 is Rs 68.75.
Q.6. Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the base of the cylinder is 0.7 m, find its height. (Assume π = 22 / 7)
Solution:
Let h be the height of the circular cylinder and r be the radius.
Radius of the base of cylinder, r = 0.7m CSA of cylinder = 2πrh
CSA of cylinder = 4.4m2
Equating both the equations, we have
2 × (22 / 7) × 0.7 × h = 4.4
Or h = 1
Therefore, the height of the cylinder is 1 m.
Q.7. The inner diameter of a circular well is 3.5m. It is 10m deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2.
(Assume π = 22 / 7)
Solution:
Inner radius of circular well, r = 3.5 / 2m = 1.75m
Depth of circular well, say h = 10m
(i) Inner curved surface area = 2πrh
= (2 × (22 / 7) × 1.75 × 10) = 110
Therefore, the inner curved surface area of the circular well is 110 m2.
(ii) Cost of plastering 1 m2 area = Rs.40
Cost of plastering 110 m2 area = Rs (110 × 40)
= Rs.4400
Therefore, the cost of plastering the curved surface of the well is Rs. 4400.
Q.8. In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. (Assume π = 22 / 7)
Solution:
Height of cylindrical pipe = Length of cylindrical pipe = 28m
Radius of circular end of pipe = diameter / 2 = 5 / 2 cm = 2.5cm = 0.025m
Now, CSA of cylindrical pipe = 2πrh, where r = radius and h = height of the cylinder
= 2 × (22 / 7) × 0.025 × 28 m2
= 4.4m2
The area of the radiating surface of the system is 4.4m2.
Q.9. Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5m high.
(ii) How much steel was actually used, if 1 / 12 of the steel actually used was wasted in making the tank. (Assume π = 22 / 7)
Solution:
Height of cylindrical tank, h = 4.5m
Radius of the circular end , r = (4.2 / 2)m = 2.1m
(i) the lateral or curved surface area of cylindrical tank is 2πrh
= 2 × (22 / 7) × 2.1 × 4.5 m2
= (44 × 0.3 × 4.5) m2
= 59.4 m2
Therefore, CSA of tank is 59.4 m2.
(ii) Total surface area of tank = 2πr(r + h)
= 2 × (22 / 7) × 2.1 × (2.1 + 4.5)
= 44 × 0.3 × 6.6
= 87.12 m2
Now, Let S m2 steel sheet be actually used in making the tank.
S(1 - 1 / 12) = 87.12 m2
This implies, S = 95.04 m2
Therefore, 95.04m2 steel was used in actual while making such a tank.
Q.10. In fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. (Assume π = 22 / 7)
Solution:
Say h = height of the frame of lampshade, looks like cylindrical shape
Total height is h = (2.5 + 30 + 2.5) cm = 35cm and
r = (20 / 2) cm = 10cm
Use curved surface area formula to find the cloth required for covering the lampshade which is 2πrh
= (2 × (22 / 7) × 10 × 35) cm2
= 2200 cm2
Hence, 2200 cm2 cloth is required for covering the lampshade.
Q.11. The students of Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? (Assume π = 22 / 7)
Solution:
Radius of the circular end of cylindrical penholder, r = 3cm
Height of penholder, h = 10.5cm
Surface area of a penholder = CSA of pen holder + Area of base of penholder
= 2πrh + πr2
= 2 × (22 / 7) × 3 × 10.5 + (22 / 7) × 32 = 1584 / 7
Therefore, Area of cardboard sheet used by one competitor is 1584 / 7 cm2
So, Area of cardboard sheet used by 35 competitors = 35 × 1584 / 7 = 7920 cm2
Therefore, 7920 cm2 cardboard sheet will be needed for the competition.
The document Exercise 13.2 NCERT Solutions - Surface Areas and Volumes | NCERT Textbooks (Class 6 to Class 12) - UPSC is a part of the UPSC Course NCERT Textbooks (Class 6 to Class 12).
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## NCERT Textbooks (Class 6 to Class 12)
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## FAQs on Exercise 13.2 NCERT Solutions - Surface Areas and Volumes - NCERT Textbooks (Class 6 to Class 12) - UPSC
1. What is the formula to find the surface area of a rectangular prism?
Ans. The formula to find the surface area of a rectangular prism is given by 2(length × width + width × height + height × length).
2. How can I calculate the curved surface area of a cylinder?
Ans. The curved surface area of a cylinder can be calculated using the formula 2πr(height), where r is the radius of the base and height is the height of the cylinder.
3. How do I find the total surface area of a cone?
Ans. To find the total surface area of a cone, you need to calculate the sum of its curved surface area and the area of its base. The formula is given by πr(r + l), where r is the radius of the base and l is the slant height of the cone.
4. Can you provide the formula to find the total surface area of a sphere?
Ans. The formula to find the total surface area of a sphere is 4πr^2, where r is the radius of the sphere.
5. How do I calculate the lateral surface area of a pyramid?
Ans. The lateral surface area of a pyramid can be calculated by summing up the areas of all its triangular faces. The formula is given by (1/2) × perimeter of the base × slant height.
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# JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3
Jharkhand Board JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 Textbook Exercise Questions and Answers.
## JAC Board Class 10 Maths Solutions Chapter 6 Triangles Exercise 6.3
Question 1.
State which pairs of triangles in the given figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
1.
In ΔABC and ΔPQR,
∠A = ∠P = 60°, ∠B = ∠Q = 80° and
∠C = ∠R = 40°
∴ By AAA criterion, ΔABC – ΔPQR.
2.
In ΔABC and ΔQRP,
$$\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{2}{4}=\frac{1}{2}$$, $$\frac{\mathrm{BC}}{\mathrm{RP}}=\frac{2.5}{5}=\frac{1}{2}$$ and $$\frac{\mathrm{CA}}{\mathrm{PQ}}=\frac{3}{6}=\frac{1}{2}$$
Thus, $$\frac{AB}{QR}=\frac{BC}{RP}=\frac{CA}{PQ}$$
∴ By SSS criterion, ΔABC ~ ΔQRP.
3.
No, the given triangles are not similar as
$$\frac{MP}{DE}=\frac{1}{2}$$, $$\frac{LP}{DF}=\frac{1}{2}$$, but $$\frac{LM}{EF}$$ = $$\frac{2.7}{5}≠\frac{1}{2}$$
4.
In ΔMNL and ΔQPR,
$$\frac{\mathrm{MN}}{\mathrm{QP}}=\frac{2.5}{5}=\frac{1}{2}$$, $$\frac{\mathrm{ML}}{\mathrm{QR}}=\frac{5}{10}=\frac{1}{2}$$
∠M = ∠Q = 70°
∴ By SAS criterion, ΔMNL ~ ΔQPR.
5.
No, the given triangles are not similar as two given corresponding sides are proportionate but the corresponding included angles are not equal.
6.
In ΔDEF, ∠D = 70°, ∠E = 80°
∠F = 180° – 70° – 80° = 30°
In ΔPQR, ∠Q = 80°, ∠R = 30°
∴ ∠P = 180° – 80° – 30° = 70°
Thus, in ΔDEF and ΔPQR.
∠D = ∠P, ∠E = ∠Q and ∠F = ∠R
∴ By AAA criterion, ΔDEF ~ ΔPQR.
Question 2.
In the given figure ΔODC – ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
In ΔDOC, ∠COB is an exterior angle.
∴ ∠COB + ∠DOC = 180°
∴ 125° + ∠DOC = 180°
∴ ∠DOC = 55°
Again, ∠COB = ∠ODC + ∠DCO
∴ 125° = 70° + ∠DCO
∴ ∠DCO = 55°
Now, ΔODC ~ ΔOBA
∴ ∠OAB = ∠OCD
∴ ∠OAB = 55°
Thus, ∠DOC = 55°, ∠DCO = 55° and
∴ ∠OAB = 55°.
Question 3.
Diagonal AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that $$\frac{OA}{OC}=\frac{OB}{OD}$$
Given:
In trapezium ABCD, AB || DC and diagonals AC and BD intersect at O.
To prove : $$\frac{OA}{OC}=\frac{OB}{OD}$$
Proof: In trapezium ABCD, AB || CD.
∴ ∠CAB = ∠ACD and ∠DBA = ∠BDC
(Alternate angles) ……(1)
Then, in ΔOAB and ΔOCD.
∠OAB = ∠OCD and ∠OBA = ∠ODC [By (1)]
∴ By AA criterion, ΔOAB ~ ΔOCD.
∴ $$\frac{OA}{OC}=\frac{OB}{OD}$$
Question 4.
In the given figure $$\frac{QR}{QS}=\frac{QT}{PR}$$ and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
In ΔPQR, ∠1 = ∠2, i.e., ∠PQR = ∠PRQ
∴ PR = QP
Now, $$\frac{QR}{QS}=\frac{QT}{PR}$$
∴ $$\frac{QR}{QS}=\frac{QT}{QP}$$
In ΔTQR, P and S are points on QT and QR respectively and $$\frac{QR}{QS}=\frac{QT}{QP}$$
∴ By theorem 6.2, SP || RT.
∴ ∠QPS = ∠QTR and ∠QSP = ∠QRT (Corresponding angles)
Now, in ΔPQS and ΔTQR.
∠QPS = ∠QTR,
∠QSP = ∠QRT and
∠PQS = ∠TQR (Same angle)
∴ By AAA criterion, ΔPQS ~ ΔTQR.
Question 5.
S and T are points on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS
Given:
S and T are points on sides PR and QR of ΔPQR such that ∠P = ∠RTS.
To prove ΔRPQ ~ ΔRTS
Proof : ∠P = ∠RTS
∴∠RPQ = ∠RTS.
In ΔRPQ and ΔRTS,
∠RPQ = ∠RTS and
∠PRQ = ∠TRS (Same angle)
∴ By AA criterion, ΔRPQ ~ ΔRTS.
Question 6.
In the given figure, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.
ΔABE ≅ ΔACD (Given)
∴ AB = AC and AE = AD (CPCT)
∴ $$\frac{AE}{AC}=\frac{AD}{AB}$$
$$\frac{AE}{AC}=\frac{AD}{AB}$$
and ∠DAE = ∠BAC (Same angle)
∴ By SAS criterion, ΔADE ~ ΔABC
Question 7.
In the given figure, altitudes AD and CE of ΔABC intersect each other at the point P.
Show that:
1. ΔAEP ~ ΔCDP
2. ΔABD ~ ΔCBE
4. ΔPDC ~ ΔBEC
Αnswer:
1. In ΔAEP and ΔCDP,
∠AEP = ∠CDP (Right angles)
∠EPA = ∠DPC (Vertically opposite angles)
∴ By AA criterion, ΔAEP ~ ΔCDP.
2. In ΔABD and ΔCBE,
∠ABD = ∠CBE (Same angle)
∴ By AA criterion, ΔABD ~ ΔCBE.
∠EAP = ∠DAB (Same angle)
∴ By AA criterion, ΔAEP ~ ΔADB.
4. In ΔPDC and ΔBEC,
∠PDC = ∠BEC (Right angles)
∠PCD = ∠BCE (Same angle)
∴ By AA criterion, ΔPDC ~ ΔBEC.
Question 8.
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.
Given: E is a point on the side AD produced of parallelogram ABCD and BE intersects CD at F
To prove : ΔABE ~ ΔCFB
Proof: In parallelogram ABCD,
∠A = ∠C (Opposite angles)
∴ ∠BAE = ∠FCB ………..(1)
E lies on AD extended in parallelogram ABCD.
∴ AE || BC
∴ ∠AEB = ∠CBE (Alternate angles)
∴ ∠AEB = ∠CBF …………..(2)
Now, in ΔABE and ΔCFB,
∠BAE = ∠FCB [By (1)]
∠AEB = ∠CBF [By (2)]
∴ By AA criterion, ΔABE ~ ΔCFB.
Question 9.
In the given figure, ABC and AMP are two right triangles, right-angled at B and M respectively. Prove that :
1. ΔABC ~ ΔAMP
2. $$\frac{CA}{PA}=\frac{BC}{MP}$$
In ΔABC and ΔAMP
∠ABC = ∠AMP (Right angles)
∠BAC = ∠MAP (Same angle)
∴ By AA criterion, ΔABC ~ ΔAMP [Result (1)]
Since ΔABC ~ ΔAMP, $$\frac{CA}{PA}=\frac{BC}{MP}$$ [Result (2)]
Question 10.
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC – ΔFEG, show that:
1. $$\frac{CD}{GH}=\frac{AC}{FG}$$
2. ΔDCB ~ ΔHGE
3. ΔDCA ~ ΔHGF
ΔABC ~ ΔFEG
∴ ∠A = ∠F, ∠B = ∠E and ∠ACB = ∠FGE …………….(1)
CD is the bisector of ∠ACB and GH is the bisector of ∠FGE.
∴ ∠ACD = ∠BCD = $$\frac{1}{2}$$∠ACB ……(2)
and ∠FGH = ∠EGH = $$\frac{1}{2}$$∠FGE ………..(3)
So, from (1), (2) and (3).
∠ACD = ∠FGH and ∠BCD = ∠EGH ……(4)
Now, in ΔDCB and ΔHGE,
∠B = ∠E [By (1)]
∠BCD = ∠EGH [By (4)]
Hence, by AA criterion,
ΔDCB ~ ΔHGE [Result (2)]
Again, in ΔDCA and ΔHGF
∠A = ∠F [By (1)]
∠ACD = ∠FGH [By (4)]
Hence, by AA criterion,
ΔDCA – ΔHGF [Result (3)]
Now, ΔDCA – ΔHGF
∴ $$\frac{CD}{GH}=\frac{AC}{FG}$$ [Result (1)]
Question 11.
In the given figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.
In ΔABC, AB = AC
∴ ∠ABC = ∠ACB
∴ ∠ABD = ∠ECF (∵ E lies on CB extended and F lies on AC.)
EF ⊥ AC
∴ ∠EFC = 90°
Now, in ΔABD and ΔECF
∠ABD = ∠ECF
∠ADB = ∠EFC (Both right angles)
∴ By AA criterion, ΔABD ~ ΔECF
Question 12.
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see the given figure). Show that ΔABC ~ ΔPQR.
Hence, by SSS criterion, ΔABD ~ ΔPQM.
∴ ∠ABD = ∠PQM
∴ ∠ABC = ∠PQR
Now, in ΔABC and ΔPQR,
$$\frac{AB}{PQ}=\frac{BC}{QR}$$ and ∠ABC = ∠PQR
Hence, by SAS criterion, ΔABC ~ ΔPQR.
Question 13.
D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB. CD.
In ΔCDA and ΔCAB.
∠ACD = ∠BCA (Same angle)
∴ By AA criterion, ΔCDA ~ ΔCAB
∴ $$\frac{CD}{CA}=\frac{CA}{CB}$$
∴ CB . CD = CA . CA
∴ CA² = CB.CD
Question 14.
Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides Pg and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.
In ΔABC, AD is a median.
∴ BD = DC
Take point E on AD extended such that AD = DE and draw BE and CE.
In quadrilateral ABEC, diagonals AE and BC bisect each other.
∴ ABEC is a parallelogram.
∴ BE = AC (Opposite sides) ……(1)
Similarly, in ΔPQR, PM is a median.
∴ QM = MR
Take point N on PM extended such that PM MN and draw QN and RN.
∴ PN = 2PM
In quadrilateral PQNR, diagonals PN and QR bisect each other.
∴ PQNR is a parallelogram.
∴ QN = PR (Opposite sides) …………..(2)
Now,
∴$$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}$$ (Given)
∴ $$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BE}}{\mathrm{QN}}=\frac{2 \mathrm{AD}}{2 \mathrm{PM}}$$ [By (1) and (2)]
∴$$\frac{AB}{PQ}=\frac{\mathrm{BE}}{\mathrm{QN}}=\frac{\mathrm{AE}}{\mathrm{PN}}$$
∴ By SSS criterion ΔABE ~ ΔPQN.
∴ ∠BAE = ∠QPN
In the same manner, it can be proved that ΔACE ~ ΔPRN
∴ ∠CAE = ∠RPN
∴ ∠BAC = ∠QPR
Now, in ΔABC and ΔPQR,
$$\frac{AB}{PQ}=\frac{AC}{PR}$$ and ∠BAC = ∠QPR
Hence, by SAS criterion, ΔABC ~ ΔPQR.
Question 15.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same. time a tower casts a shadow 28 m long. Find the height of the tower.
Here, AB is the vertical pole and AC is its shadow while PQ is the tower and QR is its shadow.
As both the shadows are measured at the same time, ∠C and ∠R both represent the elevation of the sun.
∴ ∠C = ∠R
In ΔABC and ΔPQR,
∠C = ∠R
∠B = ∠Q (Right angles)
∴ By AA criterion, ΔABC ~ ΔPQR.
∴ $$\frac{AB}{PQ}=\frac{BC}{QR}$$
∴ $$\frac{6}{PQ}=\frac{4}{28}$$
∴ PQ = $$\frac{6 \times 28}{4}$$
∴ PQ = 42 m
Thus, the height of the tower is 42 m.
Question 16.
If AD and PM are medians of triangles ABC and PQR respectively; where ΔABC ~ ΔPQR prove that $$\frac{AB}{PQ}=\frac{AD}{PM}$$
$$\frac{AB}{PQ}=\frac{BD}{QM}$$ and ∠ABD = ∠PQM
∴ $$\frac{AB}{PQ}=\frac{AD}{PM}$$ |
# How do you graph x - y = 10?
Apr 4, 2016
you just have to insert numbers for $x$ and get the $y$ values for them and mark those points in the Cartesian plane to draw the graph
#### Explanation:
the very first thing you should know is that since this equation's $x$ and $y$ are on their first degrees, this equation represents a straight line.
later you will find that equations like $y = a {x}^{2} + b x + c$ represents parabolas and so on..
first try inserting $0$ for $x$ and solve for $y$. you will get $y = - 10$
that is the point where the graph intercepts the $y$ axis.
like wise you can try inserting more values for $x$ and get their $y$ values and mark them on the plane. then you can match them to draw the graph!
it's just simple as that! good luck =)
graph{x-y=10 [-10, 10, -5, 5]}
Apr 4, 2016
see explanation
#### Explanation:
This is the equation of a line. When the line crosses the x-axis , it's y-coordinate will be zero. To find the x-intercept substitute y = 0 into the equation.
let y = 0: x - 0 = 10 → x = 10 → (10,0) is a point on line.
Similarly when the line crosses the y-axis it's x-coordinate will be zero.
Let x = 0: 0 - y = 10 → y = -10 → (0,-10) is a point on the line.
2 points are enough to draw the line , although a 3rd point is useful in aligning the points.
For this choose any value for x/y and substitute into equation.
let x = 2: 2 - y = 10 → -y = 8 → y = -8 and (2,-8) is a point.
Now plot (10,0), (0,-10) and (2,-8) and draw a straight line through them.
Here is the graph of x-y = 10
graph{x-10 [-20, 20, -10, 10]} |
## Fractions & Decimals
On Basis Explanation
Decimal Fractions
A number with a denominator of power of 10 is a decimal fractions.
1/10= 1 tenth; 1/100= 0.1;38/100=0.38
Vulgar Fractions
Conversion of 0.64(decimal number) into a Vulgar Fraction.First of all write the numeric digit 1 in the denominator of a number (like here 0.64) and add as many numeric zeros as the digit in the number after decimal point.After that removes the decimal point from the given number.At last step just reduce the fraction to its lowest terms. So, 0.64 = 64/100=16/25;25.025 = 25025/1000 = 1001/4
To perform the addition and subtraction of a decimal fraction could be done through placing them right under each other that the decimal points lie in one column.
3.424+3.28+.4036+6.2+.8+4
3. 424
3. 28
. 4036
6. 2
. 8
+4______
18. 1076Multiplication of a Decimal Fraction
To find the multiplication of decimal fraction , first of all you need to remove the decimal point from the given numbers and then perform the multiplication after that assign the decimal point as many places after the number as the sum of the number of the decimal places in the given number.
Step 1. 0.06*0.3*0.40
Step 2. 6*3*40=720
Step 3. 0.00720
Multiplication of a decimal fraction by power of 10
A multiplication of a decimal fraction by a power of 10 can be perform through shifting the decimal point towards right as many places as is the power of 10.
like 45.6288*100=45628.8, 0.00452*100=0.452
Division
Comparison of Fractions To compare the set of fractions numbers,first of all you need to convert each fraction number or value into a equal decimal value and then it will be became easy for you to assign them ( the numbers or value) in a particular way( ascending or descending order).
3/5,4/7,8/9 and 9/11 Arranging in Ascending Order
3/5= 0.6, 4/7 = 0.571, 8/9 = 0.88, 9/11 = 0.818.
Now, 0.88 > 0.818 > 0.6 > 0.571
8/9>9/11>3/5>4/7
Recurring Decimal Recurring Decimal
A decimal number in which after a decimal point a number or set of number are repeated again and again are called recurring decimal numbers.It can be written in shorten form by placing a bar or line above the numbers which has repeated.Pure Recurring Decimal
A decimal number in which all digits are repeated after a decimal point.Mixed Recurring Decimal
A decimal number in which certain digits are repeated only. |
# Divisible by 8
The most common method to check whether a number is divisible by 8 is to divide and see if the quotient is a whole number or not? If the quotient is a whole number then the given number is divisible by 8. But there is an easier way to check by using the divisibility rule.
Divisibility Rule of 8: The divisibility rule of 8 states that if the number formed by taking the last three digits is divisible by 8 then the whole number is divisible by 8.
For Example: Check whether 5864 is divisible by 8 or not?
Since the last three digits are 864 and 864/8 = 108. Hence, the number 5864 is divisible by 8.
Here are few Examples on the Divisibility Rule of 8:
1. Check whether the following number are divisible by 8 or not?
(i) 7655
(ii)3248
(iii) 8880
(iv) 3697
(v)1154
(vi)7256
(vii) 4566
(viii)1008
(ix) 2726
(x) 9635
Solution:
(i) 7655
The last three digits = 655
Now 655 is not divisible by 8 as it gives = 81.875. Hence, 7655 is not divisible by 8
(ii)3248
The last three digits = 248
Now 248/8 = 31. Hence, 3248 is divisible by 8
(iii) 8880
880 is the last three digits
880/8 = 110. Hence 8880 is divisible by 8
iv) 3697
The last three digits = 697
Now 697 divided by 8 is 87.125. Hence, 3697 is not divisible by 8
v)1154
154 is formed by the last three digits
154/8 = 19.25 (which is not a whole number). Hence, 1154 is not divisible by 8
(vi)7256
The last three digits = 256
Now 256/8 = 32. Hence, 7256 is divisible by 8
(vii) 4566
The last three digits = 566
Now, 566 divided by 8 = 70.75 (not a whole number). Hence, 4566 is not divisible by 8
(viii)1008
008 is the last three digit
Now, 8/8 = 1. Hence, 1008 is divisible by 8
(ix) 2726
The last three digit = 726
Now, 726 divided by 8 = 90.75. Hence, 2726 is not divisible by 8
(x) 9635
The last three digits = 635
635 divided by 8 is 79.375
As, 635 is not divisible by 8 so, 9635 is also not divisible by 8
2. Rewrite the series with the numbers that are divisible by 8 with a different color:
8484 4295 6264 1020 3024 7728 5656 7432 5040 5689 4433 1223
Solution:
8484 4295 6264 1020 7432 5040 5689 3024 4433 5656 1223 7728
3. Fill in the blanks with digits o as to make the following number divisible by 8:
(i) 32 … 8
(ii) 10 … 2
(iii) 2 … 48
(iv) 564 …
(v) 410 …
(vi) 874 …
(vii) 30…6
(viii) 4 …16
Solution:
(i) 3288
(ii) 1032
(iii) 2848
(iv) 5640
(v) 4104
vi) 8744
(vii) 3016
(viii) 4816
Note: In the above example we have to put such digits so that the number formed by the last three digit is divisible by 8.
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• ### Worksheet on Divisibility Rules | Questions on Test of Divisibility
This is a worksheet which will provide few problems on the divisibility rule of 2, 3, 4, 5, 6, 7, 8, 9, and 10. 1. Check whether the following numbers are divisible by 2 or 3 or both? (i) 2562 (ii) 5693 (iii) 2201 (iv) 7480 (v) 5296 (vi) 4062 (vii) 4568 (viii) 1425 (ix) 1110
• ### Problems on Divisibility Rules | Rules to Test of Divisibility | Test
Here are few problems on the divisibility rules of 2, 3, 4, 5, 6, 7, 8, 9, and 10 which will help the learners in revising their concepts on the divisibility rules. 1. Check whether 3456 is divisible by 2? Solution: The last digit is an even number (i.e. 6) hence 3456 is
• ### Divisible by 10|Test of Divisibility by 10|Rules of Divisibility by 10
The divisibility rule of 10 states that If the last digit of a number is 0 then the given number is divisible by 10. For eg: Check whether 5400 is divisible by 10 or not? Solution: As the last digit of the number is 0 hence, 5400 is divisible by 10
• ### Divisible by 9 | Test of Divisibility by 9 |Rules of Divisibility by 9
This rule states that a number is divisible by 9 if the sum of its digits of the number is divisible by 9 For eg: Check whether 729 is divisible by 9 or not? Sum of the digits = 7 + 2 + 9 = 18 Now 18 is divisible by 9 hence, 729 is also divisible by 9 Here are few examples
• ### Divisible by 7 | Test of Divisibility by 7 |Rules of Divisibility by 7
This rule states that if the difference between twice the digit at units place and the number formed from the remaining digits of the given number is divisible by 7 then the whole number is divisible by 7.
• ### Divisible by 6 | Test of divisibility by 6 |Rules of Divisibility by 6
This topic on divisible by 6 will discuss about the divisibility rule of 6 and few example on the same. Divisibility rule of 6: A number is divisible by 6 if the prime factor of that is 2 & 3 is divisible by 6. For Example: 216 216 is divisible by 2 as the last digit 6 is
• ### Divisible by 5 | Test of divisibility by 5 |Rules of Divisibility by 5
This topic on divisible by 5 will discuss on the divisibility rule of 5 and illustrate few example on the same. Divisibility Rule of 5: If a number ends with 0 or 5 then the number is divisible by 5. For Example: 550 Since the number ends with 0 hence the number is divisible
• ### Divisible by 4 | Rules of Divisibility by 4|Divisibility Rules & Tests
This topic i on divisibility rule of 4 and will provide examples on the same. Divisibility Rule of 4: If the last two digits of the number is divisible by 4 then the whole number is divisible by 4 For eg: 524 The last two digit of the number is 24 which is divisible by 4.
• ### Divisible by 3 | Rules of Divisibility by 3|Divisibility Rules & Tests
This topic is on the divisibility rule of 3 and few examples based on the topic. If the sum of the digits is divisible by 3 then the whole number is divisible by 3. For example 519 The sum of the digits = 5 + 1 + 9 = 15 15 is divisible by 3 hence 519 is also divisible by 3
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# SAT Math : How to graph a quadratic function
## Example Questions
### Example Question #1 : How To Graph A Quadratic Function
A farmer is designing rectangular pen for his cows. One side of the pen will be blocked by a steep hill, and the other three sides of the pen will be fenced off with wire. If the farmer has 20 meters of wire, what is the maximum area of the pen that he can build in square meters?
75/2
50
25
25/2
75
Explanation:
Let l = length and w = width of the pen. Let us assume that the side blocked by the mountain is along the length of the pen.
The length of wire used to make the pen must equal l + 2w, because this is the perimeter of a rectangle, excluding one of the lengths. The area of the pen will equal l x w.
l + 2w = 20
l = 20-2w
A = l x w = (20-2w)(w) = 20w - 4w2
Let A be a function of w, such that A(w) = 20w - 4w2. We want to find the maximum value of A. We recognize that the graph of A must be in the shape of a parabola, pointing downward. The maximum value of the parabola will thus occur at the vertex.
We want to rewrite A(w) in the standard form of a parabola, given by f(x) = a(x-h)2+k. In order to do this, we must complete the square.
20w-4w= -4w2+20w = -4(w2-5w) = -4(w2-5w + 25/4) + 25 = -4(w-5/2)2+25
Thus, the vertex of the parabola occurs at (5/2, 25), which means that w = 5/2.
Going back to our original equation, l + 2(5/2) = 20, and l = 15.
A = l x w = 15(5/2) = 75/2
### Example Question #24 : Graphing
Which of the following is true about the quadratic function f(x)=(x+4)2 - 3?
The function is one-to-one.
The function is even.
The graph is a parabola with two y-intercepts.
The graph is a parabola with two x-intercepts.
The graph is a parabola whose vertex is (4,-3).
The graph is a parabola with two x-intercepts.
Explanation:
The function is given in vertex form, which is (x-h)2+k where the vertex of the parabola is the point (h,k). In this particular function, h=-4 and k=3, so the vertex is (-4,-3). No parabola is one-to-one, as they don't pass the horizontal line test. While parabolas can be even functions, this will only happen when the vertex is on the y-axis (or when h=0) because even functions must be symmetric with respect to the y-axis. No function can have two y-intercepts, as it would then not pass the vertical line test and not be a function. This parabola does have two x-intercepts, however. This can be shown by setting y=0 and solving for x, or by simply realizing that the vertex is below the x-axis and the parabola opens up.
### Example Question #3 : How To Graph A Quadratic Function
Let f(x) = ax2 + bx + c, where a, b, and c are all nonzero constants. If f(x) has a vertex located below the x-axis and a focus below the vertex, which of the following must be true?
I. a < 0
II. b < 0
III. c < 0
I only
I and III only
I, II, and III
I and II only
II and III only
I and III only
Explanation:
f(x) must be a parabola, since it contains an x2 term. We are told that the vertex is below the x-axis, and that the focus is below the vertex. Because a parabola always opens toward the focus, f(x) must point downward. The general graph of the parabola must have a shape similar to this:
Since the parabola points downward, the value of a must be less than zero. Also, since the parabola points downward, it must intersect the y-axis at a point below the origin; therefore, we know that the value of the y-coordinate of the y-intercept is less than zero. To find the y-coordinate of the y-intercept of f(x), we must find the value of f(x) where x = 0. (Any graph intersects the y-axis when x = 0.) When x = 0, f(0) = a(0) + b(0) + c = c. In other words, c represents the value of the y-intercept of f(x), which we have already established must be less than zero. To summarize, a and c must both be less than zero.
The last number we must analyze is b. One way to determine whether b must be negative is to assume that b is NOT negative, and see if f(x) still has a vertex below the x-axis and a focus below the vertex. In other words, let's pretend that b = 1 (we are told b is not zero), and see what happens. Because we know that a and c are negative, let's assume that a and c are both –1.
If b = 1, and if a and c = –1, then f(x) = –x2 + x – 1.
Let's graph f(x) by trying different values of x.
If x = 0, f(x) = –1.
If x = 1, f(x) = –1.
Because parabolas are symmetric, the vertex must have an x-value located halfway between 0 and 1. Thus, the x-value of the vertex is 1/2. To find the y-value of the vertex, we evaluate f(1/2).
f(1/2) = –(1/2)2 + (1/2) – 1 = –1/4 + (1/2) – 1 = –3/4.
Thus, the vertex of f(x) would be located at (1/2, –3/4), which is below the x-axis. Also, because f(0) and f(1) are below the vertex, we know that the parabola opens downward, and the focus must be below the vertex.
To summarize, we have just provided an example in which b is greater than zero, where f(x) has a vertex below the x-axis and a focus below the vertex. In other words, it is possible for b > 0, so it is not true that b must be less than 0.
Let's go back to the original question. We know that a and c are both less than zero, so we know choices I and III must be true; however, we have just shown that b doesn't necessarily have to be less than zero. In other words, only I and III (but not II) must be true.
The answer is I and III only.
### Example Question #4 : How To Graph A Quadratic Function
The graph of f(x) is shown above. If f(x) = ax2 + bx + c, where a, b, and c are real numbers, then which of the following must be true:
I. a < 0
II. c < 0
III. b2 – 4ac < 0
I only
II and III only
II only
I and III only
I and II only
I only
Explanation:
Let's examine I, II, and III separately.
Because the parabola points downward, the value of a must be less than zero. Thus, a < 0 must be true.
Next, let's examine whether or not c < 0. The value of c is related to the y-intercept of f(x). If we let x = 0, then f(x) = f(0) = a(0) + b(0) + c = c. Thus, c is the value of the y-intercept of f(x). As we can see from the graph of f(x), the y-intercept is greater than 0. Therefore, c > 0. It is not possible for c < 0. This means choice II is incorrect.
Lastly, we need to examine b2 – 4ac, which is known as the discriminant of a quadratic equation. According to the quadratic formula, the roots of a quadratic equation are equal to the following:
Notice, that in order for the values of x to be real, the value of b2 – 4ac, which is under the square-root sign, must be greater than or equal to zero. If b2 – 4ac is negative, then we are forced to take the square root of a negative number, which produces an imaginary (nonreal) result. Thus, it cannot be true that b2 – 4ac < 0, and choice III cannot be correct.
Only choice I is correct.
### Example Question #265 : Algebra
Which of the following functions represents a parabola that has a vertex located at (–3,4), and that passes through the point (–1, –4)?
f(x) = 2x2 – 12x – 14
f(x) = x2 – 5
f(x) = –2x2 – 12x – 14
f(x) = 2x2 + 4x – 2
f(x) = x2 + 6x + 13
f(x) = –2x2 – 12x – 14
Explanation:
Because we are given the vertex of the parabola, the easiest way to solve this problem will involve the use of the formula of a parabola in vertex form. The vertex form of a parabola is given by the following equation:
f(x) = a(x h)2 + k, where (h, k) is the location of the vertex, and a is a constant.
Since the parabola has its vertex as (–3, 4), its equation in vertex form must be as follows:
f(x) = a(x – (–3)2 + 4 = a(x + 3)2 + 4
In order to complete the equation for the parabola, we must find the value of a. We can use the point (–1, –4), through which the parabola passes, in order to determine the value of a. We can substitute –1 in for x and –4 in for f(x).
f(x) = a(x + 3)2 + 4
–4 = a(–1 + 3)2 + 4
–4 = a(2)2 + 4
–4 = 4a + 4
Subtract 4 from both sides.
–8 = 4a
Divide both sides by 4.
a = –2
This means that the final vertex form of the parabola is equal to f(x) = –2(x + 3)2 + 4. However, since the answer choices are given in standard form, not vertex form, we must expand our equation for f(x) and write it in standard form.
f(x) = –2(+ 3)2 + 4
= –2(x + 3)(x + 3) + 4
We can use the FOIL method to evaluate (+ 3)(+ 3).
= –2(x2 + 3x + 3x + 9) + 4
= –2(x2 + 6x + 9) + 4
= –2x2 – 12x – 18 + 4
= –2x2 – 12x – 14
The answer is f(x) = –2x2 – 12x – 14.
### Example Question #25 : Graphing
The parabolas of the functions and on the coordinate plane have the same vertex.
If we define , which of the following is a possible definition of ?
None of the other responses gives a correct answer.
Explanation:
The definition of is given in the vertex form
,
so the vertex of its parabola is . The graphs of and are parabolas with the same vertex, so they must have the same values for and
For the function , and .
Of the five choices, the only definition of that has these same values, and that therefore has a parabola with the same vertex, is .
To verify, graph both functions on the same grid.
### Example Question #2 : How To Graph A Quadratic Function
What is the equation of the graph?
Explanation:
In order to figure out what the equation of the image is, we need to find the vertex. From the graph we can determine that the vertex is at . We can use vertex form to solve for the equation of this graph.
Recall vertex form,
, where is the coordinate of the vertex, and is the coordinate of the vertex.
Plugging in our values, we get
To solve for , we need to pick a point on the graph and plug it into our equation.
I will pick .
Now our equation is
Let's expand this,
### Example Question #1 : How To Graph A Quadratic Function
How many times does the equation below cross the x-axis?
Explanation:
You can solve this problem two ways.
3. You can solve for where the graph crosses the x-axis by setting the equation equal to zero, factoring, and solving.
2. You can quickly sketch the graph by choosing some x values and solving for y.
We see that the graph passes the x-axis twice.
### Example Question #8 : How To Graph A Quadratic Function
Let f(x) = x2. By how many units must f(x) be shifted downward so that the distance between its x-intercepts becomes 8?
4
2
8
12
16 |
Match Fishtank is now Fishtank Learning!
# Numerical and Algebraic Expressions
## Objective
Write and interpret expressions in different ways to shed new meaning on a context.
## Common Core Standards
### Core Standards
?
• 7.EE.A.2 — Understand that rewriting an expression in different forms in a problem context can shed light on the problem and how the quantities in it are related. For example, a + 0.05a = 1.05a means that "increase by 5%" is the same as "multiply by 1.05."
?
• 6.EE.A.3
## Criteria for Success
?
1. Write multiple equivalent expressions to represent a situation.
2. Interpret each expression as it relates to the situation.
3. Understand how different expressions are interpreted differently in a situation.
## Tips for Teachers
?
This lesson touches upon standard 7.EE.2; this standard will be revisited in Unit 5 with percentages and other related contexts.
#### Remote Learning Guidance
If you need to adapt or shorten this lesson for remote learning, we suggest prioritizing Anchor Problem 1 (benefits from discussion). Find more guidance on adapting our math curriculum for remote learning here.
#### Fishtank Plus
• Problem Set
• Student Handout Editor
• Vocabulary Package
## Anchor Problems
?
### Problem 1
Malia is at an amusement park. She bought 14 tickets, and each ride requires 2 tickets.
1. Write an expression that gives the number of tickets Malia has left in terms of $x$, the number of rides she has already gone on. Find at least one other expression that is equivalent to it.
2. The expression ${14}{-2}x$ represents the number of tickets Malia has left after she has gone on $x$ rides. How can each of the following numbers and expressions be interpreted in terms of tickets and rides?
1. $14$
2. $-2$
3. $2x$
3. The expression $2(7-x)$ also represents the number of tickets Malia has left after she has gone on $x$ rides. How can each of the following numbers and expressions be interpreted in terms of tickets and rides?
1. $7$
2. $(7-x)$
3. $2$
#### References
Illustrative Mathematics Ticket to Ride
Ticket to Ride, accessed on Oct. 9, 2017, 11:15 a.m., is licensed by Illustrative Mathematics under either the CC BY 4.0 or CC BY-NC-SA 4.0. For further information, contact Illustrative Mathematics.
### Problem 2
A square fountain area with side length $s$ feet is bordered by a single row of square tiles as shown.
What are three different ways to represent the number of tiles needed for the border? Show each representation using the diagram.
#### References
EngageNY Mathematics Grade 7 Mathematics > Module 3 > Topic A > Lesson 3Example 6
Grade 7 Mathematics > Module 3 > Topic A > Lesson 3 of the New York State Common Core Mathematics Curriculum from EngageNY and Great Minds. © 2015 Great Minds. Licensed by EngageNY of the New York State Education Department under the CC BY-NC-SA 3.0 US license. Accessed Dec. 2, 2016, 5:15 p.m..
## Problem Set
?
The following resources include problems and activities aligned to the objective of the lesson that can be used to create your own problem set.
• Include any review from previous lessons.
A picture with dimensions $x$ by $y$ inches, is framed by a rectangular border. The border is 1 inch wide, as shown in the figure below. Write 2 different expressions to represent the area of the border around the picture. |
# Lesson 2Finding Area by Decomposing and Rearranging
Let’s create shapes and find their areas.
### Learning Targets:
• I can explain how to find the area of a figure that is composed of other shapes.
• I know how to find the area of a figure by decomposing it and rearranging the parts.
• I know what it means for two figures to have the same area.
## 2.1What is Area?
You may recall that the term area tells us something about the number of squares inside a two-dimensional shape.
1. Here are four drawings that each show squares inside a shape. Select all drawings whose squares could be used to find the area of the shape. Be prepared to explain your reasoning.
2. Write a definition of area that includes all the information that you think is important.
## 2.2Composing Shapes
This applet has one square and some small, medium, and large right triangles. The area of the square is 1 square unit.
Click on a shape and drag to move it. Grab the point at the vertex and drag to turn it.
1. Notice that you can put together two small triangles to make a square. What is the area of the square composed of two small triangles? Be prepared to explain your reasoning.
1. Use your shapes to create a new shape with an area of 1 square unit that is not a square. Draw your shape on paper and label it with its area.
2. Use your shapes to create a new shape with an area of 2 square units. Draw your shape and label it with its area.
3. Use your shapes to create a different shape with an area of 2 square units. Draw your shape and label it with its area.
4. Use your shapes to create a new shape with an area of 4 square units. Draw your shape and label it with its area.
### Are you ready for more?
Find a way to use all of your pieces to compose a single large square. What is the area of this large square?
## 2.3Tangram Triangles
Recall that the area of the square you saw earlier is 1 square unit. Complete each statement and explain your reasoning.
1. The area of the small triangle is ____________ square units. I know this because . . .
2. The area of the medium triangle is ____________ square units. I know this because . . .
3. The area of the large triangle is ____________ square units. I know this because . . .
## Lesson 2 Summary
Here are two important principles for finding area:
1. If two figures can be placed one on top of the other so that they match up exactly, then they have the same area.
2. We can decompose a figure (break a figure into pieces) and rearrange the pieces (move the pieces around) to find its area.
Here are illustrations of the two principles.
• Each square on the left can be decomposed into 2 triangles. These triangles can be rearranged into a large triangle. So the large triangle has the same area as the 2 squares.
• Similarly, the large triangle on the right can be decomposed into 4 equal triangles. The triangles can be rearranged to form 2 squares. If each square has an area of 1 square unit, then the area of the large triangle is 2 square units. We also can say that each small triangle has an area of square unit.
## Glossary Terms
area
Area is the number of square units that covers a two-dimensional region, without any gaps or overlaps.
For example, the area of region A is 8 square units. The area of the shaded region of B is square unit.
compose
Compose means “put together.” We use the word compose to describe putting more than one figure together to make a new shape.
decompose
Decompose means “take apart.” We use the word decompose to describe taking a figure apart to make more than one new shape.
## Lesson 2 Practice Problems
1. The diagonal of a rectangle is shown.
1. Decompose the rectangle along the diagonal, and recompose the two pieces to make a different shape.
2. How does the area of this new shape compare to the area of the original rectangle? Explain how you know.
2. The area of the square is 1 square unit. Two small triangles can be put together to make a square or to make a medium triangle.
Which figure also has an area of square units? Select all that apply.
3. Priya decomposed a square into 16 smaller, equal-size squares and then cut out 4 of the small squares and attached them around the outside of original square to make a new figure.
How does the area of her new figure compare with that of the original square?
1. The area of the new figure is greater.
2. The two figures have the same area.
3. The area of the original square is greater.
4. We don’t know because neither the side length nor the area of the original square is known.
4. The area of a rectangular playground is 78 square meters. If the length of the playground is 13 meters, what is its width?
5. A student said, “We can’t find the area of the shaded region because the shape has many different measurements, instead of just a length and a width that we could multiply.”
Explain why the student’s statement about area is incorrect. |
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# Statistics Formulas Quick Reference
## Arithmetic mean (AM)
The arithmetic mean (or simply "mean") of a sample x_1,x_2,... ,x_n is the sum the sampled values divided by the number of items in the sample:
Visit the Mean page on Wikipedia for more details
## The Median
### For an odd number of values
As an example, we will calculate the sample median for the following set of observations: 1, 5, 2, 8, 7.
Start by sorting the values: 1, 2, 5, 7, 8.
In this case, the median is 5 since it is the middle observation in the ordered list.
The median is the ((n + 1)/2)th item, where n is the number of values. For example, for the list {1, 2, 5, 7, 8}, we have n = 5, so the median is the ((5 + 1)/2)th item. median = (6/2)th itemmedian = 3rd itemmedian = 5
### For an odd number of values
As an example, we will calculate the sample median for the following set of observations: 1, 6, 2, 8, 7, 2.
Start by sorting the values: 1, 2, 2, 6, 7, 8. In this case, the arithmetic mean of the two middlemost terms is (2 + 6)/2 = 4. Therefore, the median is 4 since it is the arithmetic mean of the middle observations in the ordered list.
We also use this formula MEDIAN = {(n + 1 )/2}th item . n = number of values
As above example 1, 2, 2, 6, 7, 8 n = 6 Median = {(6 + 1)/2}th item = 3.5th item. In this case, the median is average of the 3rd number and the next one (the fourth number). The median is (2 + 6)/2 which is 4.
Visit the Median page on Wikipedia for more details
## The Standard Deviation
the standard deviation (SD) (represented by the Greek letter sigma, σ) measures the amount of variation or dispersion from the average.
In other words, the standard deviation σ (sigma) is the square root of the variance of X; i.e., it is the square root of the average value of (X − μ)2. In the case where X takes random values from a finite data set x1, x2, ..., xN, with each value having the same probability
Visit the Standard Deviation page on Wikipedia for more details
## The Variance
measures how far a set of numbers is spread out. A variance of zero indicates that all the values are identical. Variance is always non-negative: a small variance indicates that the data tend to be very close to the mean (expected value) and hence to each other, while a high variance indicates that the data are very spread out around the mean and from each other. |
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## Related Articles
• RD Sharma Class 12 Solutions for Maths
# Class 12 RD Sharma Solutions – Chapter 6 Determinants – Exercise 6.2 | Set 2
### Question 18. = -2
Solution:
Considering the determinant, we have
R2⇢R2 – R1 and R3⇢R3 – R2
△ = 1[2(a + 2) – 2(a + 3)]
△ = (4a + 4 – (4a + 6))
△ = (4a + 4 – 4a – 6)
△ = -2
Hence proved
### Question 19. = (a – b)(b – c)(c – a)(a + b + c)(a2 + b2 + c2)
Solution:
Considering the determinant, we have
C2⇢C2 – 2C1 – 2C3
Taking -(a2 + b2 + c2) common from C2, we get
R2⇢R2 – R1 and R3⇢R3 – R1
Taking (b – a) and (c – a) common from R1 and R2, we get
△ = -(a2 + b2 + c2)(b – a)(c – a)[1((-b)(b + a) – (c + a)(-c))]
△ = (a2 + b2 + c2)(a – b)(c – a)[(-b)(b + a) + (c + a)c]
△ = (a2 + b2 + c2)(a – b)(c – a)[-b2 – ab + ac + c2]
△ = (a2 + b2 + c2)(a – b)(c – a)
△ = (a2 + b2 + c2)(a – b)(c – a)[(c – b)(c + b) + a(c – b)]
△ = (a2 + b2 + c2)(a – b)(c – a)(c – b)
△ = (a2 + b2 + c2)(a + b + c)(a – b)(b – c)(c – a)
Hence proved
### Question 20. = (a – b)(b – c)(c – a)(a2 + b2 + c2)
Solution:
Considering the determinant, we have
R2⇢R2 – R1 and R3⇢R3 – R1
Taking (b – a) and (c – a) common from R2 and R3 respectively, we get
△ = (b – a)(c – a)[1((b + a – c)(c2 + a2 + ac) – (c + a – b)(b2 + a2 + ab))]
△ = (b – a)(c – a)(b – c)(a + b + c)
△ = -(a – b)(c – a)(b – c)(a + b + c)
Hence proved
### Question 21. = 4a2b2c2
Solution:
Considering the determinant, we have
Taking a, b and c common from C1, C2 and C3 we get
C1⇢C1 + C2 + C3
Taking 2 common from C1, we get
C2⇢C2 – C1 and C3⇢C3 – C1
C1⇢C1 + C2 + C3
Taking c, a and b common from C1, C2 and C3 we get
R3⇢R3 – R1
△ = 2a2b2c2[1((-1)(-1) – (-1)(1))]
△ = 2a2b2c2[1 – (-1)]
△ = 2a2b2c2[1 + 1]
△ = 4a2b2c2
Hence proved
### Question 22. = 16(3x + 4)
Solution:
Considering the determinant, we have
C1⇢C1 + C2 + C3
Taking (3x + 4) common from C1, we get
R2⇢R2 – R1 and R3⇢R3 – R1
△ = (3x + 4)[1((4)(4) – (-4)(0))]
△ = (3x + 4)[16 – 0]
△ = 16(3x + 4)
Hence proved
### Question 23. = 1
Solution:
Considering the determinant, we have
C2⇢C2 – pC1 and C3⇢C3 – qC1
C3⇢C3 – pC2
C2⇢C2 – C1 and C3⇢C3 – C2
△ = 1[(1)(4) – (1)(3)]
△ = [4 – 3]
△ = 1
Hence proved
### Question 24. = (a + b – c)(b + c – a)(c + a – b)
Solution:
Considering the determinant, we have
R1⇢R1 – R2 – R3
Taking (-a+b+c) common from R1, we get
C2⇢C2 + C1 and C3⇢C3 + C1
△ = (b + c – a)[1((b + a – c)(c + a – b) – (0)(0))]
△ = (b + c – a)[(b + a – c)(c + a – b)]
△ = (b + c – a)(b + a – c)(c + a – b)
Hence proved
### Question 25. = (a3 + b3)2
Solution:
Considering the determinant, we have
C1⇢C1 + C2 + C3
Taking (a + b)2 common from C1, we get
R2⇢R2 – R1 and R3⇢R3 – R1
R2⇢R2 – R3
△ = (a + b)2 [1((a2 – b2)(a2 – b2) – (b2 – 2ab)(2ab – a2))]
△ = (a + b)2 [(a2 – b2)2 + (b2 – 2ab)(a2 – 2ab)]
△ = (a + b)2 [(a2 + b2 – ab)2]
△ = (a3 + b3)2
Hence proved
### Question 26. = 1 + a2 + b2 + c2
Solution:
Considering the determinant, we have
Multiplying a, b and c to R1, R2 and R3 we get
Taking a, b and c common from C1, C2 and C3 we get
R1⇢R1 + R2 + R3
Taking (a2 + b2 + c2 + 1) common from R1, we get
C2⇢C2-C1 and C3⇢C3-C1
△ = (a2 + b2 + c2 + 1)[1((1)(1) – (0)(0))]
△ = (a2 + b2 + c2 + 1)[1]
△ = (a2 + b2 + c2 + 1)
Hence proved !!
### Question 27. = (a3 – 1)2
Solution:
Considering the determinant, we have
C1⇢C1 + C2 + C3
Taking (a2 + a + 1) common from C1, we get
R2⇢R2 – R1 and R3⇢R3 – R1
Taking (1 – a) common from R2 and R3, we get
△ = (a2 + a + 1)(1 – a)2[1((1 + a)(1) – (a)(-a))]
△ = (a2 + a + 1)(1 – a)2[(1 + a) + a2]
△ = (a2 + a + 1)(1 – a)2[1 + a + a2]
△ = ((a2 + a + 1)(1 – a))2
△ = (a3 – 1)2
Hence proved
### Question 28. = 2(a + b)(b + c)(c + a)
Solution:
Considering the determinant, we have
C1⇢C1 + C3 and C2⇢C2 + C3
Taking (c + a) and (b + c) common from C1 and C2, we get
R2⇢R2 + R1 and R3⇢R3 + R2
△ = (c + a)(b + c)[1((0)(b + c) – (2)(-a – b))]
△ = (c + a)(b + c)[0 + 2(a + b)]
△ = 2(a + b)(c + a)(b + c)
Hence proved
### Question 29. = 4abc
Solution:
Considering the determinant, we have
R1⇢R1 + R2 + R3
Taking 2 common from R1, we get
R2⇢R2 – R1 and R3⇢R3 – R1
R1⇢R1 + R2 + R3
△ = 2[-c((-c)(0) – (-a)(-b)) + b((-c)(-a) – (0)(-b))]
△ = 2[-c(0 – ab) + b(ac – 0)]
△ = 2[abc + abc]
△ = 2[2abc]
△ = 4abc
Hence proved
### Question 30. = 4a2b2c2
Solution:
Considering the determinant, we have
Multiplying a, b and c to R1, R2 and R3 respectively, we get
Taking common a, b and c to C1, C2 and C3 respectively, we get
R1⇢R1 + R2 + R3
Taking 2 common from R1, we get
R1⇢R1 – R2
△ = 2
△ = 2
△ = 2
△ = 2
### Question 31. = 2a3b3c3
Solution:
Considering the determinant, we have
Taking a2, b2 and c2 common from C1, C2 and C3. we get
Taking a, b and c common from R1, R2 and R3. we get
C2⇢C2 – C3
△ = a3b3c3[1((1)(1) – (1)(-1))]
△ = a3b3c3[1 + 1]
△ = 2a3b3c3
Hence proved
### Question 32. = 4abc
Solution:
Considering the determinant, we have
Multiplying c, a and b to R1, R2 and R3. We get
R1⇢R1 – R2 – R3
Taking -2 common from R1, we get
R2⇢R2 – R1 and R3⇢R3 – R1
△ = 4abc
Hence proved
### Question 33. = (ab + bc + ca)3
Solution:
Considering the determinant, we have
Multiplying a, b and c to R1, R2 and R3. We get
Taking a, b and c common from C1, C2 and C3. we get
R1⇢R1 + R2 + R3
Taking (ab + bc + ca) common from R1, we get
C1⇢C1 – C2 and C3⇢C3 – C2
Taking (ab + bc + ca) common from C1 and C2, we get
△ = (ab + bc + ca)3 [-1((1)(-1) – (1)(0))]
△ = (ab + bc + ca)3 [-1(-1)]
△ = (ab + bc + ca)3
Hence proved
### Question 34. = (5x + 4)(4 -x)2
Solution:
Considering the determinant, we have
C1⇢C1 + C2 + C3
Taking (5x + 4) common from C1, we get
R2⇢R2 – R1 and R3⇢R3 – R1
△ = (5x + 4)[1((4 – x)(4 – x) – (0)(0))]
△ = (5x + 4)[(4 – x)2]
△ = (5x + 4)(4 – x)2
Hence proved
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# Probability Calculator
Calculate the probability of two events or a series of events using the calculator below.
%
%
%
%
## Probability Of:
Intersection of A and B:
P(A∩B):
%
Union of A and B:
P(A∪B):
%
Symmetric Difference of A and B:
P(A∆B):
%
Complement of A and B:
P((A∪B)'):
%
Complement of A:
P(A'):
%
Complement of B:
P(B'):
%
Learn how we calculated this below
## How to Calculate Probability
Probability is the quantitative expression of the chance of an event occurring. More specifically, if the set of possible events contains n elements, and an event is associated with r elements, and all elements are equally likely, then the probability is the ratio of r/n.
An example of using probability is when rolling dice. For example, you might ask yourself what the chance is that you’ll roll a six? Alternatively, you are asking what the probability is of rolling a six.
To find the probability of something happening, you need to first know the set of possible outcomes. Then, you can apply the probability formula to solve.
Continuing the dice example, there are six possible outcomes when rolling the dice. You can roll a one, two, three, four, five, or six. Each outcome is equally likely. Therefore the chances of rolling a six are 1/6.
### Probability Formula
This is the basic formula to solve the probability of a favorable outcome.
P(A) = number of times A can occur / number of possible outcomes
Thus, the probability of result A is equal to the number of possible times A can occur divided by the total number of possible outcomes.
So, continuing the dice example, you might be interested in the probability of rolling an even number. There are three even numbers on the dice, 2, 4, and 6. The probability of rolling an even number is thus 3/6 since there are three even sides on the die and six total sides, each equally likely to turn up.
An example to illustrate the probability of an event where the number of times the event can occur is greater than one would be the odds of drawing a spade from a deck of cards. Since there are 13 cards of each suit and 52 total cards, the probability of drawing a spade is 13/52, which simplifies to 1/4.
It is important that all outcomes be equally likely in order to calculate probabilities in this fashion. For example, if the dice were weighted in some way so that the side opposite the six were much heavier, then the probability of a six arising would likely differ substantially from 1/6. Statistical hypothesis testing can be used to decide if a dice is weighted.
Our combinations and permutations calculators can help find the number of possible outcomes for a data set. This can be used to help solve the probability of an expected outcome.
### How to Find the Probability of Independent Events
Things get a little more complicated when more than one possible outcome, or event, can occur. We refer to these as independent events.
For example, you might want to know how to find the probability of getting a six when rolling two dice. In this case, we have two independent events that we need to know the likelihood of.
There are a few different variations of how independent events can occur. Let’s cover each way we could roll a six.
#### Intersection
In statistics, the probability of the intersection is the likelihood that both events will occur. In the dice example, this would be the chance of both dice being a six when rolling.
The intersection of events A and B is denoted A∩B, and the Venn diagram above can help visualize the intersection of events.
You can find the probability of intersection of independent events using the following formula:
P(A∩B) = P(A) × P(B)
The probability of intersection P(A∩B) is equal to P(A) times P(B).
#### Union
The probability of the union is the likelihood that at least one of the events will occur, or both. In the dice example, this would be the chance of one of the dice being a six or both being a six.
The union of events A and B is denoted A∪B.
You can find the probability of union using the following formula:
P(A∪B) = P(A) + P(B) – P(A∩B)
The probability of union P(A∪B) is equal to P(A) plus P(B), minus the probability of intersection P(A∩B).
#### Symmetric Difference
The probability of the symmetric difference is the likelihood that exactly one of the events will occur, but not both. In the dice example, this would be the chance of precisely one of the dice being a six, but not both.
The symmetric difference of events A and B is denoted A∆B.
The symmetric difference is also known as the disjunctive union of two sets, and you can find the probability of it with the following formula:
P(A∆B) = P(A∪B) – P(A∩B)
The probability of the symmetric difference P(A∆B) is equal to union P(A∪B) minus the intersection P(A∩B).
#### Complement of A
In probability theory, the complement of event A is the likelihood that A does not occur. Thus, the complement of an event is effectively the chance that the event does not happen.
Continuing the dice example, the complement of rolling a six on the first dice is equal to the chance that a six is not rolled on the first dice, even if it’s rolled on the second.
Complement is denoted using an apostrophe (‘). Thus, A’ is the complement of A.
The complement can be found using the formula:
P(A’) = 1 – P(A)
The probability of the complement of A P(A’) is equal to 1 minus P(A).
One of the reasons we use complements in probability analysis is that it is often easier to think about how to calculate quantities in terms of events not occurring.
A well-known example of this is the birthday problem:
In a room of 20 people, what are the chances that two people share a birthday? This problem can be rephrased in terms of complements, what are the chances that no two people in a room of 20 share a birthday?
The first person can be born on any of 365 days, the second person can be born on day 364 of the 365 days year, the second can be born on 363 out of 365 days, and so forth. Hint: you can use our day of the year calendar to find what day your birthday is on.
If you multiply these 365/365 × 364/365 × … × 345/365, then you get 0.55. This implies that in a room full of 20 people, there is a 1 – 0.55 = 0.45 probability that two people will share a birthday. This is much higher than most people would think!
#### Complement of A∪B
The complement of the union of two events, or (A∪B)’, is the likelihood that neither event A nor event B will occur. In the dice example, this is the likelihood that neither dice is a six when rolling two dice.
P((A∪B)’) can be found using the formula:
P((A∪B)’) = 1 – P(A∪B)
The probability of the complement of A union B P((A∪B)’) occurring is equal to 1 minus the chance of A union B PA∪B).
### How to Find the Probability For a Series of Events
Sometimes, you want to find the probability of an outcome occurring in a series of events. For instance, if rolling dice three times, what is the likelihood of rolling a six at least one time?
The formulas below define the probabilities of the outcome occurring at least once, every time, or never during the series.
#### Event Occurring at Least Once
P(A occurs at least once) = 1 – (1 – P(A))n
The probability of event A occurring at least once in the series of n attempts is equal to 1 minus 1 minus the probability of event A to the nth power.
#### Event Occurring Every Time
P(A occurs every time) = P(A)n
The probability of event A occurring every time in the series of n attempts is equal to the probability of event A to the nth power.
#### Event Never Occurring
P(A never occurs) = (1 – P(A))n
The probability of event A never occurring in the series of n attempts is equal to 1 minus the probability of event A to the nth power.
### How to Find the Probability For Conditional Events
So far, the formulas for single and series events have assumed the events are independent. Events are independent if one event happening does not have any impact on the probability of the other event occurring.
In the example of rolling the dice, the probability of rolling the dice and getting a six is 1/6, and the probability of rolling a six on the second roll is also 1/6. The probability of rolling a six is always 1/6 for standard dice.
However, consider the deck of cards; the probability of drawing a king from the deck is 4/52, since there are 4 kings and 52 total cards. Let’s say you draw a card that is not a king; what is the probability of drawing a king on the second draw, assuming you did not replace the first card to the deck?
The odds of drawing a king are reduced on the second draw to 4/51, since there are 4 kings and 51 total cards remaining. This is a conditional probability.
You can use Bayes’ theorem to calculate a conditional probability:
P(A|B) = P(B|A) × P(A) / P(B)
Bayes’ theorem states that the probability of event A given that event B also occurs is equal to the probability of event B given that event A also occurs times the probability of event A divided by the probability of event B.
We hope this demystifies some of the probability equations used in statistics. When in doubt, try the calculator above to see what the chances of your outcomes are!
Try our p-value calculator to calculate probabilities of a value drawn from a distribution.
### Why is probability important?
Probability is important because it gives a way of quantifying the chances of something happening. This allows us to make decisions when outcomes are uncertain. It also gives us a way of understanding the chances of complicated events where intuitions may not be very reliable.
### Does probability mean possibility?
Probability is the likelihood or possibility of an event or outcome occurring. So yes, probability can mean possibility.
### Is probability always accurate?
You cannot predict what will happen with absolute certainty; however, using probability, you can calculate the likelihood that something will happen given some assumptions about how the world works. The probability of estimates are only as good as your model.
### How do you know whether two events are independent?
In practice, it is often hard to know whether two things are related to one another or not; you need to make a judgment call based on your knowledge of the situation and context. The choice matters, however, because the probabilities can radically change if you assume that events are related or not.
As an example, many casinos have rules about gamblers counting cards. By assuming past hands influence future hands, card counters are assuming non-independence. If they count correctly, card counters can only play hands where their odds of winning are very favorable. Gamblers who do not count cards will perceive very different probabilities, and more often than not lose to the card counters. |
INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith
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• ## Related Books
1 answerLast reply by: Professor Eric SmithMon Jul 10, 2017 4:51 PMPost by Jiaming Xu on June 7 at 09:20:37 PMWhat would be the easiest method? Is it grouping, because that's what I usually use.
### Factoring Trinomials
• When factoring a trinomial, always begin by looking for a greatest common factor first.
• Some trinomials can be factored by using the FOIL process in reverse.
• Determine what your first terms should be
• Determine a good choice for your last terms
• Check to see that the outside and inside terms combine to give the original middle term
• Repeat the process if when multiplying the two binomials together, you do not get the original
• If the middle and last term in the trinomial is positive, then your two last terms of the binomials must also be positive
• If the middle term is negative and last term positive in the trinomial, then your two last terms of the binomials must be negative.
• If the last term of the trinomial is negative, then the two last terms of the binomial must be different in sign.
### Factoring Trinomials
Factor:
x2 + 16x + 60
• ( x + p )( x + q )
• p + q = b = 16
• pq = c = 60
• p = 6,q = 10
• ( x + 6 )( x + 10 )
• Foil to check your work.
• ( x + 6 )( x + 10 )
• x2 + 10x + 6x + 60
x2 + 16x + 60
Factor:
w2 + 11w + 24
• p + q = b = 11
• pq = c = 24
• p = 3,q = 8
( w + 3 )( w + 8 )
Factor:
k2 + 8k + 16
• p + q = b = 8
• pq = c = 16
• p = 4,q = 4
( k + 4 )( k + 4 )
Factor:
y2 − 7y + 12
• ( x − p )( x − q )
• p + q = b = − 7
• pq = c = 12
• p = − 3,q = − 4
( y − 3 )( y − 4 )
Factor:
g2 − 6g + 8
( g − 2 )( g − 4 )
Factor:
b2 − 11b + 18
( b − 2 )( b − 9 )
Factor:
8n − 20 + n2
• n2 + 8n − 20
• c〈0 .
• ( x + p )( x − q )
( n − 2 )( n + 10 )
Factor:
r2 + 17r − 18
( r − 1 )( r + 18 )
Factor:
s2 − 36 = − 9s
• s2 + 9s − 36 = 0
( s − 3 )( s + 12 )
Factor:
22 + 40 = 132
• 22 − 132 + 40 = 0
( 2 − 5 )( 2 − 8 )
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
### Factoring Trinomials
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Objectives 0:06
• Factoring Trinomials 0:25
• Recall FOIL
• Factor a Trinomial by Reversing FOIL
• Tips when Using Reverse FOIL
• Example 1 7:04
• Example 2 9:09
• Example 3 11:15
• Example 4 13:41
• Factoring Trinomials Cont. 15:50
• Example 5 18:42
### Transcription: Factoring Trinomials
Welcome back to www.educator.com.0000
In this lesson we are going to work on factoring trinomials.0003
We are not going to tackle up all types of trinomials just yet.0006
For the first part we are going to focus on ones where the squared term has a coefficient of 1.0010
We will also look at polynomials where we can factor out a greatest common factor.0015
In future lessons we will look at the more complicated trinomials.0020
One of our first techniques we have to dig back in our brains and recall how we used foil in order to multiply two binomials together.0028
For example, what did we do when we are looking at x -3 × x +1.0037
Using the method of foil we would multiply our first terms together and get something like x2.0042
We would multiply our outside terms together 1x then we would multiply our inside terms.0049
And finally we would multiply our last terms together.0060
Sometimes we had to do a little bit of work to clean this up.0067
As long as we made sure that everything got multiplied by everything else.0070
We where assured that we could multiply these two binomials out.0075
Since we are working with factoring and breaking things down into a product you want to think of this process, but do it in reverse.0079
If you had a trinomial to begin with, how could you then break this down into two binomials?0087
Since this is the exact same one as before I will simply write down the two binomials that it will break up into.0096
This is the process that we are after of taking a trinomial and breaking it down into two binomials.0105
In doing so it is not quite a straightforward process.0114
The way we are going to attack this is to think of that foil process in our minds.0119
This will help us determine our first and last terms in those binomials.0124
Now, after we have chosen something for those first and last terms0129
we will have to check to make sure that the outside and inside terms combined to be our middle term.0133
Sometimes we will have to do some double checking just to make sure that it does combine and give us that middle term.0140
Sometimes there are lots of different options.0146
We may have to do this more than once until we find just the right values that make it work.0149
Watch how that works with this trinomial.0154
I have x2 + 2x -8 and what we are looking to do is break this down into two binomials.0157
I’m going to go ahead and write down the parentheses just to get it started.0165
I first want to determine what should my first terms be in order to get that x2.0169
We are looking to multiply two things together and get x2.0177
The only two things that will work is x and x.0180
We have a good chance that those are our first terms.0184
Rather than worrying about the outside and inside just yet, we jump all the way to the last terms.0191
That will be here and here and I'm looking to multiply them and get -8.0197
Here is the thing there are lots of different options that we could have, it could be 1 and 8, 2 and 4.0203
We could also look at the other order of these maybe 4 and 2, 8 and 1.0211
We want to choose the proper pair that when combined together will actually give us that 2x in the middle.0217
Let us go ahead and try something.0224
Watch how this process works.0226
Suppose I just tried the first thing on the list, this 1 and 8.0228
1, 8, with this combination I can be sure that my first terms work out and that my last terms work out.0233
I'm not completely confident until I check those outside and inside terms to make sure that they work out.0242
I will do some quick calculations.0250
Let us check our outside terms, 8 × x = 8x and inside terms 1 and x, and these would combine to give us a 9x.0252
If you compare that to the original it is not the same.0265
What that is indicating is that pair of 1 and 8, those are not the ones we want to use.0270
We can backup a little bit and try another pair of numbers.0277
Let me try something different.0287
I'm going to try 4 and -2.0290
Now when I do my outside and inside terms I will get -2x on the outside,0298
4x on the inside and those combine to give me 2x, which is the same as my middle term.0305
I know that this is how it should be factored.0313
If you want you can go through the entire foil process, just to double check that all the rest of terms work out.0319
In fact it is not a bad idea when you are done with the factoring process, just to make sure it is okay.0324
Now, there are a few tips to help you along the way when doing this reverse foil method.0333
It works good, as long as you are leading coefficient is 1 and you can take a look at the signs of the other two coefficients.0339
In general, here is what you are trying to do.0351
You are looking for two integers whose product will give you c.0354
They multiply and give you c but whose sum is b.0358
It is what we did in that last example.0363
Now you can get a little bit more information if you look at the signs of b and c.0367
If b and c are both positive then those two integers you are looking for must also be positive.0371
One situation that might happen is you know both integers that you are looking for will be negative if c is positive,0383
that is this one in the end and d is negative.0391
Watch for that to happen.0396
Of course one last thing, you will know that the integers you are looking for are different in sign, one positive and one negative if c is negative.0400
That is the only way they could multiply together and give you a negative number here on the end.0410
Watch for me to use these shortcuts here in just a little bit.0416
We want to use this reverse foil method in order to factor the following polynomial y2 + 12y + 20.0426
I’m going to start off by writing set of parentheses this will break down into some binomials.0435
Let us start off for those first terms.0444
What times what would give us a y2?0447
One thing that will do it is just y and y.0451
Let us look for two values that would multiply and give us a 20.0457
It is okay for me to write down some different possibilities like 1 and 20, 2 and 10, 4 and 5.0462
You can imagine the same values just flipped around.0470
Let us see if we can use any information to help us out.0474
Notice how this last term out here is positive and so is my middle term, both of them are positive.0479
Now what is that telling me about my signs, I know that the two numbers I'm looking for will both be positive.0488
That actually is quite a bit of information because now when we look at our list I can pick two things that will add to be the middle term 12.0496
I know they multiply between.0504
Let us drop those in there, 2 and 10.0507
I have factored the trinomial.0511
Let us quickly go through the foil process just to make sure that this is the one we are looking for.0514
We are looking to make sure that this matches up with the original.0520
First terms would be a y2, outside terms 10y, inside terms 2y and last term is 20.0524
These middle ones would combined giving us y2 + 12y + 20 and that shows that our factorization checks out.0534
We know that this is the proper way to factor it.0544
Let us try another one, this one is x2 - 9x – 22.0551
Let us start off in much the same way.0557
Let us write down a set of parentheses and see if we can fill in the blanks.0559
We need two numbers that when multiplied together will give us x2.0566
That must be an x and another x.0571
Now we need a pair of numbers that will multiply and give us a -22.0577
Let us write down some possibilities like 1 and 22, 2 and 11, I think that is it.0581
Let us get some information about the signs of these numbers.0589
I’m looking at this last number here and notice how it is negative,0594
I know that these two numbers I’m looking for on my last terms, one of them must be positive and one of them must be negative.0598
The question is which one is positive and which one is negative?0607
We will look to our middle term to help out.0613
I need the larger term to be negative, so that I will get a negative in the middle, that -9.0617
We have plenty of information it should be pretty clear that it is actually the 2 and 11 off my list that will work.0624
2 and 11.0632
Let us just check it real quick by foiling things out to make sure that this is how it should factor.0637
First terms x2, outside terms -11x, inside terms 2x, and last terms -22.0643
Since I have a positive × negative, these middle terms combined I will get x2 - 9x -22.0656
That is the same as my original, so I know that I have factored it correctly.0666
Let us use the reverse foil method for this polynomial.0679
It is not very big, it is r2 + r + 2.0682
We are going to start off by setting down those two parentheses.0686
We are hunting for two first terms that will multiply to give us r2.0691
There is only one choice that will do that, just r and r.0697
We turn our attention to the last terms and we need them to multiply to be 2.0704
Unfortunately, there is only one possibility for that, 1 and 2.0711
Let us hunt down our signs.0717
The last term is positive, the middle term is positive that says both of the terms that we are looking for must both be positive.0720
Let us put them in and now we can go and check this using the foil process.0728
r × r =r2, the outside terms should be 2r, inside terms 1r, and the last term is 2.0735
Combining the two middle terms here, I get r2 + 3r + 2.0746
Something very interesting is happening with this one, let us take a closer look.0755
If we look at the resulting polynomial that we got after foiling out and we compare that with the original, they are not the same.0759
That tells us something. It tells us that this is not the correct factorization.0769
Now if this is not how it should be factored then what other possibilities do we have?0783
If we look at all of our possibilities for those last terms, it must contain 1 and 2 if it is going to multiply and give us 2.0789
Since that did not work and I have no other possibilities, it tells us that this does not factor into two binomials using 1 and 2.0798
This is an indication that our original is actually prime.0807
Watch out for ones like this where when you try and factor it, it simply does not factor into those two binomials.0813
Let us try something with a few more variables in it.0824
This one is t2 – 6tu + 8u2.0827
Even though we have a few more variables in here, you will see that this process works out the same as before.0833
Starting off with those first terms, something × something will give us t2.0841
That must be t and another t.0848
I need two things to multiply and give us 8u2.0853
Well I'm not quite sure about that 8 yet because it could be 1 and 8, 2 and 4, or could be those reversed.0860
I did know about the u, you better have a u and another u in order to get that u2.0867
Let us just focus on numbers for bit and see what information we can get from there.0874
Looking at the sign of my last term it is positive, but my middle term is negative.0881
The information I’m getting from there is that both of my numbers I’m looking for must both be negative.0887
I think we can pick it out from our list now and it looks like we must use the 2 and 4.0896
Finally let us check that to make sure this one works.0906
Let us be careful since we have both t’s and u’s.0909
First terms t2, outside terms -4tu, inside terms -2tu and my last terms – 2 × -4 = 8u2.0912
We can combine our middle terms giving us t2 – 6tu + 8u2.0931
And now we can see that yes, this has been factored correctly since the resulting polynomial is the same as my original.0940
This one is good.0947
One thing to watch out for is to make sure you pull out any common factors at the very beginning.0953
That is what we will definitely need to do with this example before we even start the forming process.0959
Always check out for a good common factor.0966
This is also good idea because it can potentially make your number smaller, so that you do not have to think of as many possibilities.0970
We are going to quickly factor 3x4 -15x3 + 18x2.0978
Look with this one, everything here is divisible by 3 and I can pull out an x2 from all of the variables.0985
Let us take it out of the very beginning, I have 3x2 and let us write down what is left.0996
3x4 ÷ 3x2 = x2, =15 ÷ 3 = -5x, 18 ÷ 3 + 6 and I think that is all my leftover parts.1004
With this one now, I will go and do the reverse foil process on that.1024
I need two binomials, let us break it down.1030
What should my first terms be in order to get my x2?1035
That must be an x and an x.1042
I have to look at my last terms and then multiply together to give us a 6.1047
1 and 6, possibly 2 and 3, but you can use the signs to help you out.1052
The last term is positive, the middle term is negative, so both of these will be negative.1058
It looks like I need to use that 2 and 3.1066
Some quick checking to make sure this is the correct factorization.1077
I have x2 - 3x - 2x + 6 and looks like my outside and inside terms do combine and give us that - 5x.1080
This is the correct factorization and let us go head and write out that very first 3x2 that we took out at the very beginning.1093
Now that we have all the pieces, we can say that this is the correct factorization.1103
Always look for a common factor that you could pull out from the very beginning before starting the foil method.1109
Sometimes you can pull something out, sometimes you can not but it will make your life easier if you can find something.1116
Keep that in mind for this next example.1125
This one is 2x3 - 18x2 - 44x.1128
Let us come at this over.1134
It looks like everything is divisible by 2 and they all have an x in common.1136
Let us take out a 2x at the very beginning.1142
What do we have left?1148
2x3 ÷ 2x = x2, -18 ÷ 2 = -9x and -44x ÷ 2x = -22.1150
Now we want to factor that into some binomials.1168
Let us go ahead and copy over this 2x just we can keep track of it.1174
I need my first terms to multiply together and get an x2.1180
That will be an x and another x.1186
I need to look at my last terms so that they multiply together to give me a -22.1190
Some of our possibilities are 1 and 22, 2 and 11.1197
Since the last one is negative and my middle term is negative, I know that these will be different in sign.1204
Let us take the 2 and 11 off of our list, those are the ones we need, -11 and 2.1214
Let us quickly combine things together and make sure that it is the correct factorization.1224
x2 - 11x + 2x – 22.1229
Combining these middle guys x2 - 9x -22, so that definitely checks with this polynomial right here.1236
Now if you want to go ahead and put in the 2x as well, this will take you back all the way to the original one.1249
Remember to use your distribution property so you can see how that will work out.1258
2x3 - 18x2 - 44x and sure enough that is the same as the original.1264
Everything checks out. I know that this is the correct factorization for our polynomial.1275
Just a few things, make sure that when you are using this method, always check for common factor to pull it out.1286
Make sure you set down your first terms and then your last terms1292
and definitely check those signs to help you eliminate some of your possibilities.1297
Thank you for watching www.educator.com.1303 |
# Colossally Cool Cube
Long before Nintendo Switch, people had Rubik’s Cubes for handheld gaming. It’s a simple puzzle: you spin the pieces around until each side is all its own color. “Speedcubers” can even solve a Rubik’s Cube using 1 hand in less than 7 seconds! But one Rubik’s Cube has gotten completely out of hand. Tony Fisher built this 6-foot 7-inch tall toy using cardboard boxes, a lot of math, and of course, duct tape. We bet it’ll take more than a few seconds to unscramble this preposterous puzzle!
Wee ones: Draw a square, then split it up into 9 smaller shapes! (Bonus: Can you make those smaller shapes squares, too?)
Little kids: If there are 4 blue and 5 green squares on the top of the cube, how many squares is that? Bonus: How many green squares would need to switch to blue for there to be more blue than green squares?
Big kids: How much taller is the 6-foot 7-inch cube than you? Bonus: If 3 of the smaller cubes add up to 6 feet 7 inches, about how tall is each of those cubes?
Wee ones: Draw a square with equal sides, then use lines to split it up! Don’t worry about making them perfectly straight.
Little kids: 9 squares. Bonus: Just 1 square! Subtracting 1 square from the 5 green by making it blue leaves 4 green squares, and then adding that 1 square to the 4 blue makes 5 blue.
Big kids: Different for everyone… use a tape measure to find out your height if you don’t know it, then subtract the feet and inches from 6 feet 7 inches. Bonus: Each cube is about 26 inches, or 2 feet 2 inches, tall. 6 feet 7 inches is equal to 79 inches, because 6 x 12 = 72, and 72 + 7 = 79. You can use long division to find that 79 / 3 = 26 ⅓ inches per cube. Or you can divide the 6 feet by 3, then the 7 inches by 3, and add the dividends: 2 feet + 2 inches, with a remainder of 1.
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# How to Conquer the Division Castle Using the 7’s Method ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷
## Presentation on theme: "How to Conquer the Division Castle Using the 7’s Method ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷"— Presentation transcript:
How to Conquer the Division Castle Using the 7’s Method ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷
What comes first? First, we need to look at our problem and figure out where it all fits!! So then what would I do with the problem: 135÷9 Well which number would be my dividend? Which is the divisor? HINT: The dividend will ALWAYS BE the bigger number! dividend divisor
Now let me tell you a story… Picture that same division problem… Well the dividend wants to take over our division castle. Thankfully our divisOR has his magic swORd to cut the dividEND into pieces and bring it’s life to an END! To protect our Queen, the Quotient and set our answer free! 135÷9 Remember: The answer to a division problem is called the quotient!!!
Let me set the scene… We are going to use the seven’s method which we have found is the best way to slay off those dividends. First, draw yourself a seven that looks like this….. Now put “Groups Of” right beside that 7! Groups Of
Here we go….. Now our seven is the castle and the dividend has taken over and locked himself inside…. Groups Of Now our divisor and his sword are on the outside trying to break that dividend down until he meets his end with a big fat 0!!! 1 3 5 9
9 Let’s start this battle with what we know….. Most of us know our 10’s times tables so let’s see what we can cut out of this dividend first using our 10’s. Groups Of First we are going to start with ten so we need a 10 under our “Groups of” So now we know that 9 x 10 is…. 90 so we put it under the dividend because we took it away from him. 9 0 1 0 x Remember 9 groups of 10 is the same as 9 x 10!!!!
Now subtract to see what is left of your dividend… (Remember we want him to be 0!) So what can we take from our dividend next? Would another group of 10 be too big? Yes, you are right!! So what should we take out then…do we all know our 5’s? Let’s try it…. 9 0 1 3 5 9 Groups Of 1 0 5 45
What is 9 groups of 5? Or 9 x 5? 9 0 1 3 5 9 Groups Of 1 0 4 5 5 x I know that 9 x 5 is 45 so I must take 45 from what is left of the dividend! 4 5
Our Battle has almost come to an END!!!! Now what happens when you subtract 45 from 45? 9 0 1 3 5 9 Groups Of 1 0 4 5 5 x YOU ARE RIGHT! It is 0 which is the END of the that DIVIDEND’S life but where is the Queen? RIP!! 0 Our Queen Quotient brings all the groups together into one sum….WE MUST ADD our “Groups Of” to find our long lost Queen Quotient! + ? ? ?
We must return our Queen Quotient to her throne…help us!!! When you add 10 + 5 your answer is 15 which is our QUEEN QUOTIENT! 9 0 1 3 5 9 Groups Of 1 0 4 5 5 x RIP!! 0 + 1 5 But where do we put her??? Well any good Queen Quotient sits on top of her throne which sits right above the Seven Castle! 1 5
Thanks to our DivsOR our Kingdom is safe and Queen Quotient is Free!!!!!!!!
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What do we know about division? 16 ÷ 4 = _____ I can use what I know about multiplication
# What do we know about division? 16 ÷ 4 = _____ I can use what I know about multiplication
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## What do we know about division? 16 ÷ 4 = _____ I can use what I know about multiplication
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1. What do we know about division? 16 ÷ 4 = _____ I can use what I know about multiplication to help me solve a division problem! I can think 4 x ? = 16 4 x 4 = 16, so 16 ÷ 4 = 4. This works because of FAMILY OF FACTS!
2. Let’s try another. 25 ÷ 5 = _____ I can use what I know about multiplication to help me solve a division problem! I can think ___ x ? = ___ ___ x ____ = ____, so 25 ÷ 5 = _____. This works because of FAMILY OF FACTS!
3. Talk this through with a partner. 36 ÷ 6 = _____ I can use what I know about multiplication to help me solve a division problem! I can think ___ x ? = ___ ___ x ____ = ____, so 36 ÷ 6 = _____. This works because of FAMILY OF FACTS!
4. So let’s review... Factor x Factor = Product Product ÷ Factor = Factor This works because of FAMILY OF FACTS! Factor Factor Product
5. We know how to use arrays to multiply! 22 x 4 4 x 20 = ? 4 x 2= ? 20 2 80 8 4 80 + 8 = 88. So 22 x 4 = 88.
6. We can also use an array to solve a division problem! 24 ÷ 4 4 x ? = 24 6 4 24 6 x 4 = 24 So 24 ÷ 4 = 6.
7. We can also use an array to solve a division problem! 84 ÷ 4 4 x ? = 80 4 x ? = 4 20 1 80 4 4 20 + 1 = 21. 4 x 21 = 84 So 84 ÷ 4 = 21.
8. We can also use an array to solve a division problem! 88 ÷ 8 8 x ? = 80 8 x ? = 8 10 1 80 8 8 10 + 1 = 11. 8 x 11 = 88 So 88 ÷ 8 = 11.
9. We can also use an array to solve a division problem! 36 ÷ 3 3 x ? = 30 3 x ? = 6 10 2 30 6 3 10 + 2 = 12 3 x 12 = 36 So 36 ÷ 3 = 12.
10. We can also use an array to solve a division problem! 66 ÷ 3 3 x ? = 60 3 x ? = 6 20 2 60 6 3 20 + 2 = 22 3 x 22 = 66 So 66 ÷ 3 = 22.
11. Now let’s use arrays to help us solve some problems on our own! 96 ÷ 3 48 ÷ 4 88 ÷ 2 63 ÷ 3
12. Let’s check our answers! 96 ÷ 3 = 32 48 ÷ 4 = 12 + + 30 2 10 2 90 6 40 8 3 4 88 ÷ 2 = 44 63 ÷ 3 = 21 + + 40 4 20 1 80 60 3 2 8 3
13. Today we’ve been thinking about division. Division is all about sharing equally. Ms. Santos owns a neighborhood grocery store. She has 56 apples to arrange in rows for her window display. She has room for 4 rows in her window. How many apples will there be in each row if she puts the same number in each row? Use color tiles to model this problem and solve it.
14. Today we’ve been thinking about division. Division is all about sharing equally. Let’s share how we thought through this problem!
15. Could we solve Ms. Santos’ problem with an array? 56 ÷ 4 4 x ? = 40 4 x ? = 16 10 4 40 16 4 10 + 4 = 14 4 x 14 = 56 So 56 ÷ 4 = 14.
16. Let’s try another problem like this. 60 ÷ 5 5 x ? = 50 5 x ? = 10 10 2 50 10 5 10 + 2 = 12 5 x 12 = 60 So 60 ÷ 5= 12.
17. Let’s try another problem like this. 72 ÷ 6 6 x ? = 60 6 x ? = 12 10 2 60 12 6 10 + 2 = 12 6 x 12 = 72 So 72 ÷ 6= 12.
18. Let’s try another problem like this. 75 ÷ 5 5 x ? = 50 5 x ? = 25 10 5 50 25 5 10 + 5 = 15 5 x 15 = 75 So 75 ÷ 5= 15.
19. Let’s try another problem like this. 90 ÷ 6 6 x ? = 60 6 x ? = 30 10 5 60 30 6 10 + 5 = 15 6 x 15 = 90 So 90 ÷ 6= 15.
20. Let’s PRACTICE! Problem: 52 ÷ 4 = ___ Problem: 64 ÷ 4 = ___ Problem: 57 ÷ 3 = ___ Problem: 66 ÷ 3 = ___ |
# Question Video: Finding the Definite Integration of an Odd Function Using the Properties of Definite Integration Mathematics • Higher Education
The function π is odd, continuous on [β1, 7], and satisfies β«_(1) ^(7) π(π₯) dπ₯ = β17. Determine β«_(β1) ^(7) π(π₯) dπ₯.
01:13
### Video Transcript
The function π is odd, continuous on the closed interval negative one to seven, and satisfies the definite integral between one and seven of π of π₯ with respect to π₯ equals negative 17. Determine the definite integral between negative one and seven of π of π₯ with respect to π₯.
Weβre firstly told that the function π is odd. So we recall the following property for integrating odd functions. The definite integral between negative π and π of π of π₯ with respect to π₯ is equal to zero. Weβre also told that the definite integral between one and seven of π of π₯ with respect to π₯ is equal to negative 17. So we split the integral up. And we see that the integral that weβre looking for between negative one and seven of π of π₯ with respect to π₯ is equal to the definite integral between negative one and one of π of π₯ plus the definite integral between one and seven of π of π₯.
Now, the function π is odd. So by the first property, we see that the definite integral between negative one and one of π of π₯ with respect to π₯ must be equal to zero. Then, we simply take the definite integral between one and seven of π of π₯ from the question. Itβs negative 17. This means the definite integral weβre looking for is equal to zero plus negative 17 which is negative 17. |
# What Is A Strip Diagram? Explained for Kids, Parents & Teachers
A strip diagram is a visual model commonly used in elementary grades to aid in solving a problem with known and unknown quantities. A strip diagram can also be known as a bar model, a tape diagram, a fraction strip, a length model, or a part-part-whole model.
These rectangular models can be used with all four operations and may be seen in all elementary grade levels. Strip diagrams are used during the “representational” stage of the CRA model, as students represent the equation or math problem they are solving.
Strip Diagram Questions Worksheet
Download 20 mixed topic word problems, including a strip diagram template for students to use as an aid at home and in school.
### What is a strip diagram?
A strip diagram is a visual model used as support to solve an addition, subtraction, multiplication, or division equation or word problem. Students set up the rectangular model as two “strips,” one representing the total and one representing the parts. One or more quantities are unknown and the model will help provide students with a visual representation of the quantities in relation to the other unknown quantities as well as the known quantities in the problem. This helps students build an understanding of the relationships between the numbers while performing each operation.
For example, if we are using a strip diagram to represent the subtraction problem 28 – 12 = ?, we would have a “total” strip at the top of the diagram, and then two smaller strips underneath that, together, equal the same length as the total strip.
### Why is a strip diagram important?
A strip diagram allows students to visualize the problem or equation they are trying to solve in a simple and efficient way. These visual models are quick and easy to draw and allow students to create a representation of the relationship between the quantities involved in the problem.
They are especially helpful when students are solving word problems, as it can also help them determine which operation they need to use. This is a skill many students struggle with and this is a quick strategy that can be used with all four operations.
### How to use strip diagrams
To use a strip diagram, you will start by drawing two rectangles of the same size. Depending on the equation and the operation needed, the bottom rectangle will be split into two or more parts. Either the top rectangle, or one or more parts of the bottom rectangle, will be blank to represent the unknown quantity that needs to be found. No matter the operation, the top rectangle will represent the total of the parts of the bottom rectangle.
Looking at the examples below, you can see that the way we use strip diagrams for addition and subtraction, as well as for multiplication and division, are the same but with different unknown quantities. For addition and subtraction, the strip diagram will be used as a part-part-whole model. The total will be the top rectangle, whether this quantity is known or unknown, and the parts will make up the rectangles at the bottom. With multiplication and division, the total is also represented by the top rectangle, and the bottom rectangle represents the groups.
### Strip diagram examples
Strip diagrams can be used to solve a variety of problems involving each of the operations – from basic addition and subtraction to more complex multiplication and division, even including multi-step word problems. Here are some basic examples of how we can use strip diagrams to solve word problems. The word problems here are intentionally similar, so that you can see how each diagram differs depending on which operation is needed to solve the problem.
Frank has 32 red marbles and 68 blue marbles. How many marbles does he have altogether?
In our addition word problem, we are given two quantities – the number of red marbles and the number of blue marbles. Students will need to find the total by adding these two quantities together. The red marbles and blue marbles are represented by their respective colors in the bottom part of our strip diagram. The total, or the unknown quantity, is represented by the purple rectangle. As you can see, the red rectangle and blue rectangle together make up the length of the unknown quantity, or the total. This will help students see that those two numbers should be added together to get the total.
#### Subtraction word problem
Frank has 100 marbles. 68 of them are red and the rest are blue. How many marbles are blue?
In our subtraction word problem, we still have two known quantities. However, this time, we are given the total number of Frank’s marbles, which is 100. This number will go in the top rectangle of our diagram. We are also given the number of red marbles.
The unknown quantity in this problem is the number of blue marbles. The red marbles and blue marbles together still make up the total number of marbles. So the setup of the rectangles in our strip diagram did not change, but our unknown quantity did. Students should be able to deduce that this word problem will need to be solved using subtraction.
#### Multiplication word problem
Jenna has 4 jars of marbles. Each jar contains 8 marbles. How many marbles does she have altogether?
The way we set up bar model multiplication and division problems using a strip diagram is a bit different than the way we set up addition and subtraction problems. Instead of three quantities – or part-part-whole – we use the strip diagram to represent equal parts of a whole. In this example, we know that there are 4 jars of marbles, with 8 marbles being in each jar.
Instead of seeing the 4 and 8 represented numerically, the groups, or jars in this example, are represented in the bottom section of our strip diagram. As you can see, the bottom portion consists of 4 equal-sized rectangles. Inside each of these smaller rectangles is the number 8; this represents the 8 marbles inside each of the 4 jars. The top rectangle, again, shows our total. Since this is multiplication, this would be our product.
#### Division word problem
Jenna has 32 marbles. They are split evenly between 4 jars. How many marbles are in each jar?
In the division example, we are given the total number of marbles, which was the product of our multiplication example. We know that this number will be represented by our top rectangle. The bottom portion, again, will represent our equal groups. Since we know the number of groups, or jars, which is 4, the bottom portion is again split into 4 equal-sized groups. The unknown quantity here is the number of marbles that are in each of the jars, or each of the groups. Students should be able to deduce that division will need to be used to solve this problem.
### What’s the difference between strip diagrams and bar models?
A strip diagram and a bar model are one and the same! These two terms can be used interchangeably. You may also see a strip diagram referred to as a tape diagram, length model, or fraction strip. They all serve the same purpose – to be used as a visual problem solving strategy to solve equations and/or word problems (including multi-step word problems!) involving any of the four operations.
In addition to the four operations, strip diagrams can also be used to represent fractions – in this case, the strip diagram would be referred to as a fraction bar model or fraction strip. The diagram can give students a visual representation of fractions in order to find equivalent fractions or even compare two fractions. In the first example, students will be able to see that ½ is equal 2/4. In the second example, they can see that ⅔ is less than ¾.
### When do children learn about strip diagrams in school?
The term “strip diagram,” as well as its synonyms, such as “bar model,” does not explicitly appear in any of the Common Core Math State Standards. However, the use of models, although not specified, is mentioned in numerous standards. Here are the standards where a strip diagram or bar model can and should be used to help students:
• 1.NBT.4 Add within 100, including adding a two-digit number and a one-digit number, and adding a two-digit number and a multiple of 10, using concrete models or drawings and strategies based on place value…
• 1.NBT.6 Add within 100, including adding a two-digit number and a one-digit number, and adding a two-digit number and a multiple of 10, using concrete models or drawings and strategies based on place value…
• 2.NBT.7 Add and subtract within 1000, using concrete models or drawings and strategies based on place value…
• 3.NF.3b Recognize and generate simple equivalent fractions, e.g., 1/2 = 2/4, 4/6 = 2/3. Explain why the fractions are equivalent, e.g., by using a visual fraction model.
• 3.NF.3d Compare two fractions with the same numerator or the same denominator by reasoning about their size. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model.
• 4.NF.1 Explain why a fraction a/b is equivalent to a fraction (n × a)/(n × b) by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions.
• 4.NF.2 Compare two fractions with different numerators and different denominators, e.g., by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1/2. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model.
• 4.NF.3b Decompose a fraction into a sum of fractions with the same denominator in more than one way, recording each decomposition by an equation. Justify decompositions, e.g., by using a visual fraction model.
• 4.NF.3d Solve word problems involving addition and subtraction of fractions referring to the same whole and having like denominators, e.g., by using visual fraction models and equations to represent the problem.
• 4.NF.4a Understand a fraction a/b as a multiple of 1/b. For example, use a visual fraction model to represent 5/4 as the product 5 × (1/4), recording the conclusion by the equation 5/4 = 5 × (1/4).
• 4.NF.4b Understand a multiple of a/b as a multiple of 1/b, and use this understanding to multiply a fraction by a whole number. For example, use a visual fraction model to express 3 × (2/5) as 6 × (1/5), recognizing this product as 6/5. (In general, n × (a/b) = (n × a)/b.)
• 4.NF.4c Solve word problems involving multiplication of a fraction by a whole number, e.g., by using visual fraction models and equations to represent the problem.
• 4.NF.7 Compare two decimals to hundredths by reasoning about their size. Recognize that comparisons are valid only when the two decimals refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions, e.g., by using a visual model.
• 5.NBT.7 Add, subtract, multiply, and divide decimals to hundredths, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used.
• 5.NF.2 Solve word problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem.
• 5.NF.3 Interpret a fraction as division of the numerator by the denominator (a/b = a ÷ b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem.
• 5.NF.4a Interpret the product (a/b) × q as a parts of a partition of q into b equal parts; equivalently, as the result of a sequence of operations a × q ÷ b. For example, use a visual fraction model to show (2/3) × 4 = 8/3, and create a story context for this equation.
• 5.NF.6 Solve real world problems involving multiplication of fractions and mixed numbers, e.g., by using visual fraction models or equations to represent the problem.
• 5.NF.7a Interpret division of a unit fraction by a non-zero whole number and compute such quotients. For example, create a story context for (1/3) ÷ 4, and use a visual fraction model to show the quotient.
• 5.NF.7b Interpret division of a whole number by a unit fraction, and compute such quotients. For example, create a story context for 4 ÷ (1/5), and use a visual fraction model to show the quotient.
• 5.NF.7c Solve real world problems involving division of unit fractions by non-zero whole numbers and division of whole numbers by unit fractions, e.g., by using visual fraction models and equations to represent the problem.
### Non Common Core standards
Other states that do not follow the Common Core State Standards may specifically mention “strip diagram” or “bar model” within their standards, or they may just make mention of “visual models” as in the CCSS. Here are a few examples of standards from the Texas Essential Knowledge and Skills for Mathematics (TEKS) standards that explicitly mention the term “strip diagram”:
• 3(A) represent fractions greater than zero and less than or equal to one with denominators of 2, 3, 4, 6, and 8 using concrete objects and pictorial models, including strip diagrams and number lines
• 5(B) represent and solve one- and two-step multiplication and division problems within 100 using arrays, strip diagrams, and equations
• 5(A) represent multi-step problems involving the four operations with whole numbers using strip diagrams and equations with a letter standing for the unknown quantity
There are also many TEKS standards that mention “models” or “pictorial models” where a strip diagram can be used, such as this 1st grade TEKS standard:
• 3(A) use concrete and pictorial models to determine the sum of a multiple of 10 and a one-digit number in problems up to 99
As a teacher, it is important to know and be able to interpret your specific state standards and to know which concepts a strip diagram would be appropriate to use for teaching.
### Other ways to draw a strip diagram
Although your typical strip diagram or bar model will be drawn with two rectangles on top of each other, as the examples above show, there are other ways to draw strip diagrams. Be aware that you may see many other interpretations of this model, but they are all used the same way. Here are a few examples:
Using uneven “strips” or “bars”:
### How to incorporate strip diagrams into your classroom
Strip diagrams can easily be incorporated into any lesson where students can use visual models to represent their work. Strip diagrams can be very quickly drawn into a math notebook while students are working on solving problems. If teachers are looking for a way to use these every day, for a “problem of the day,” for example, a template can be printed and laminated for each student to use.
Then, instead of using their notebook pages or worksheets each day, they can pull out their template and a dry erase marker and fill in the template according to the problem they are presented that day. The templates for all grade levels, as well as all four operations, can simply be two blank rectangles stacked on top of one another.
As students are given the problem and the known quantities, they can partition the bottom rectangle accordingly and determine their unknown quantity.
### Strip diagram worked examples
#### 1. Complete the diagram
Anna has 168 beads. She is using the beads to make 6 necklaces for her friends. Each necklace will have the same amount of beads. Complete the diagram to show how many beads will go on each necklace.
In this problem, students are asked to complete the diagram in order to help them solve it. As you can see, students are given the number in the top rectangle, which is the total number of beads Anna has. The other known quantity in this word problem is 6, which is the number of necklaces the beads will be divided by. Based on this information, students should determine that the bottom rectangle needs to be split into 6 equal groups. Then, they can solve for the unknown quantity, which is the number in each of those 6 groups (which they will write in each of the 6 sections they create in the bottom rectangle.)
Their completed strip diagram will look like this:
Mason has a membership at the local gym. His membership costs $34 every two months. How much does his membership cost in one year? Choose the diagram that correctly represents this problem. Note that t represents the total cost of the membership. In this problem, students need to be able to read and interpret a strip diagram correctly in order to answer the question. The question gives students a known quantity of$34, but requires them to infer the second known quantity, which is 6. Since Mason pays $34 every 2 months, he pays the fee 6 times during the year, since there are 12 months in a year. Students may misinterpret this known quantity as 2, since 2 is the other number mentioned in the word problem. Once students understand that he pays$34 6 times, they should look for a diagram which shows 6 groups of $34 that equal the total. The only answer that fits this description is answer choice b. #### 3. Multi-step word problem A teacher bought 5 boxes of pencils for her classroom. Each box contained 24 pencils. Then, she divided them evenly among 8 groups of students. Use the diagram to find how many pencils were given to each group of students. This is a multi-step problem, which often proves to be challenging for many students. First, students need to find out the total number of pencils. Here, the “total” box is shown as 5 groups of 24. The strip diagram should help students determine that for the first step of this problem, they will need to multiply 24 by 5 to find the total number of pencils. Then, once they have the total, they can divide that number by the total number of groups, which is 8. The diagram will help them visualize this, as well, since the bottom portion of the strip diagram is partitioned into 8 groups. Then, students can see that the unknown quantity is the number in each group. Students should divide the total number by 8 to get the answer, represented by the question mark in the diagram. It is important to note that students do not need to complete the diagram for this question, but they will use the diagram to help them solve the problem. ### Strip diagram practice questions 1. Reggie drank 5 bottles of water today. Each bottle contained 16 ounces of water. Use the diagram to determine how many ounces of water Reggie drank today. 2. Fiona’s book has 598 pages. So far, she has read 179 pages. Which diagram shows how to find the number of pages, p, she has left to read? 3. Millie has a photo album with 18 pages. Each page can hold 9 pictures. So far she has 126 photos in her photo album. Use the diagram to find the number of photos, p, she can still put into her album. 4. Ralph made$105 mowing lawns this month. His brother, Pete, made 3 times as much. Use the diagram to determine how much money Pete made.
5. Amir, Jonah, and Bridget went trick-or-treating together. At the end of the night, Amir had 42 pieces of candy and Jonah had 53 pieces of candy. The three friends decided to combine their candy. Altogether, they had 146 pieces of candy. Use the diagram to determine, b, how much candy Bridget had before the friends combined their candy.
### What is a strip diagram?
A strip diagram is a visual model commonly used in elementary grades to aid in solving a problem with known and unknown quantities.
### What are other names for a strip diagram?
A strip diagram can also be known as a bar model, a tape diagram, a fraction strip, a length model, or a part-part-whole model.
### What is an example of a strip diagram?
Matty has 3 erasers. Hunter has 5 erasers. How many erasers do they have altogether?
### What is a strip diagram in 5th grade math?
5th grade strip diagrams may be used for higher level concepts, such as adding, subtracting, multiplying, and dividing decimals, and/or fractions.
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### Course: Pre-algebra>Unit 5
Lesson 3: Powers of fractions and decimals
# Exponents review
Review the basics of exponents and try some practice problems.
## Exponents and bases
Here's what an exponent and a base look like:
${4}^{3}$
The small number written above and to the right of a number is called an $\text{exponent}$. The number underneath the exponent is called the $\text{base}$. In this example, the base is $4$, and the exponent is $3$.
## Evaluating exponents
An exponent tells us to multiply the base by itself that number of times.
In our example, ${4}^{3}$ tells us to multiply the base of $4$ by itself $3$ times:
$\begin{array}{rl}{4}^{3}& =4×4×4\\ \\ \phantom{{4}^{3}}& =64\end{array}$
### What about when the exponent is a zero?
Any base with an exponent of zero is equal to $1$.
For example, ${7}^{0}=1$.
Check out this video to see why.
## Practice
Problem 1
Evaluate.
${9}^{2}=$
Want to try more problems like this? Check out this exercise.
## Want to join the conversation?
• why is 5 to the 0 power 1?
• Good question! Look at the following exponents:
2^4 = 16
2^3 = 8
2^2 = 4
2^1 = 2
2^0 = __
Respective to the pattern, what comes next?! 1!
Take a look at this:
2^(m-n) = 2^m/2^n
If m = n...
2^(1-1) = 2^1/2^1
2^0 = 1
Hope this helps!
• i get confused and multiply ex. 5 x 2 =10 when its 25 how can i rember to times it by 2 also how is 5 times 0 equal to 1? ':(
• So for exponents you need to think about it a bit different. I will use the example you gave of 5 raised to the 2nd exponent (5^2) for my explanation. The exponent (the number 2) is the number of bases (the number 5) you multiply together. So for 5^2, you would use two 5's and multiply them together which is simply 5x5=25. So for another example if we lower the exponent to 1, we would be looking at 5^1. Well let's apply the same principle of using just one 5, which is simply 5=5.
Let's move on to your second question which is a touch more complicated. I will simplify it for you though. Once your exponent is less than 1 the rules get a little different and you start dealing with fractions. 5^0 = 5*(1/5) = 1. The exponent in this case is the number + 1 that you divide the base number by. I illustrated it with multiplying it by a fraction, but the principle is still the same. I know this can be a difficult topic to understand at first, and explanation isn't the exact proof/theorem, but I do hope it helps you get a basic understanding of exponents.
• how does 5 with the exponent of 0 have the answer of 1.
• Well, I think that CycoCyco answered it somewhat well, but here's another explanation from me:
When having an exponent (such as 5 to the power of 2), you're setting up 5^2, or 5 * 5, which equals one. Same with having five raised to the power of one, which equals five.
In earlier grades, you leaned that 5 * 0 = 0. But in math, 5^0 = 1, because you're not raising the power by anything.
I hope this explanation helped.
• I dont understand why 5'0 pwr is = to 1
• Hi @adam.39594, I'd suggest that you look at the video in Exponents, The Zeroth Power. I'm not sure if we're allowed to put links, so go look at that video... Or search The Zeroth Power in Khan Academy. Hope this helps!
- V
• What about negative exponents? How do you figure those out?
• Because
5^3=125
5^2=25
5^1=5
5^0=1
5^-1=.2
Just divide.
• Idk why the power of 0 is always 1 but it makes it easier for me.
• how is 5 to the 0 power not 0 because 5 zero times is 0
• In general, x to the y power is usually not x times y. So it is a mistake to assume that 5 to the 0 power is 5 times 0.
Look at the following pattern:
5^4 = 5 * 5 * 5 * 5
5^3 = 5 * 5 * 5 = (5 * 5 * 5 * 5)/5
5^2 = 5 * 5 = (5 * 5 * 5)/5
5^1 = 5 = (5 * 5)/5
As we can see, each time the exponent goes down by 1, the answer is divided by 5. Continuing the pattern gives 5^0 = 5/5 = 1.
Have a blessed, wonderful day!
• I don’t get 5 to the power 0
• Any positive value to the power of 0 equals 1. The reason is that, since exponents represent repeated multiplication, then applying the power of 0 is essentially dividing a value by itself the specific number of times. For example, if we have 5^0, then that means we are multiplying it by its reciprocal the specific number of times it got multiplied. So when we have 5^0, are multiplying 5 by itself a specific number of times, then dividing by 5 the same exact number of times it got multiplied. Therefore, every single time a 5 was multiplied, it will be removed by division through the reciprocal, ending up with a 1.
• I keep on getting confused with the base and the exponent. Can someone help?
• The base is the number below for example in 5^2, 5 is the base and 2 is the exponent. Which means that this would expand to 5 x 5 = 25. The exponent tells you how many times the base will be multiplied by itself, we knew that the 5 would be multiplied by itself twice because the exponent was 2.
• how do you add or multiply exponents together without getting rid of the exponents? for example if you had -2x^4+4x^2+5 -7x^5 would you add all the exponents together or would you leave them by their selves?
• You have 4 unlike terms. You can't combine any of them.
Remember, exponents represent repetitive multiplication of a common base. You only change an exponent if you are multiplying or dividing a common base. In your expression, the operation is addition/subtraction. So, the exponents can't be changed.
Hope this helps. |
# Pythagorean theorem
The Pythagorean theorem: The sum of the areas of the two squares on the legs (a and b) equals the area of the square on the hypotenuse (c).
Trigonometry Reference Euclidean theory Law of sines Law of cosines Law of tangents Pythagorean theorem Calculus
In mathematics, the Pythagorean theorem (American English) or Pythagoras' theorem (British English) is a relation in Euclidean geometry among the three sides of a right triangle. The theorem is usually written as an equation:
$a^2 + b^2 = c^2\!\,$
where c represents the length of the hypotenuse, and a and b represent the lengths of the other two sides. In words:
The square of the hypotenuse of a right triangle is equal to the sum of the squares on the other two sides.[1]
The Pythagorean theorem is named after the Greek mathematician Pythagoras, who by tradition is credited with its discovery and proof,[2] although it is often argued that knowledge of the theory predates him. (There is much evidence that Babylonian mathematicians understood the principle, if not the mathematical significance).
## In formulae
If we let c be the length of the hypotenuse and a and b be the lengths of the other two sides, the theorem can be expressed as the equation:
$a^2 + b^2 = c^2\,$
or, solved for c:
$c = \sqrt{a^2 + b^2}. \,$
If c is already given, and the length of one of the legs must be found, the following equations can be used (The following equations are simply the converse of the original equation):
$c^2 - a^2 = b^2\,$
or
$c^2 - b^2 = a^2.\,$
This equation provides a simple relation among the three sides of a right triangle so that if the lengths of any two sides are known, the length of the third side can be found. A generalization of this theorem is the law of cosines, which allows the computation of the length of the third side of any triangle, given the lengths of two sides and the size of the angle between them. If the angle between the sides is a right angle it reduces to the Pythagorean theorem.
## Proofs
This is a theorem that may have more known proofs than any other (the law of quadratic reciprocity being also a contender for that distinction); the book Pythagorean Proposition, by Elisha Scott Loomis, contains 367 proofs.
Some arguments based on trigonometric identities (such as Taylor series for sine and cosine) have been proposed as proofs for the theorem. However, since all the fundamental trigonometric identities are proved using the Pythagorean theorem, there cannot be any trigonometric proof. (See also begging the question.)
### Proof using similar triangles
Proof using similar triangles
Like most of the proofs of the Pythagorean theorem, this one is based on the proportionality of the sides of two similar triangles.
Let ABC represent a right triangle, with the right angle located at C, as shown on the figure. We draw the altitude from point C, and call H its intersection with the side AB. The new triangle ACH is similar to our triangle ABC, because they both have a right angle (by definition of the altitude), and they share the angle at A, meaning that the third angle will be the same in both triangles as well. By a similar reasoning, the triangle CBH is also similar to ABC. The similarities lead to the two ratios..:
As
$BC=a, AC=b, \text{ and } AB=c, \!$
so
$\frac{a}{c}=\frac{HB}{a} \mbox{ and } \frac{b}{c}=\frac{AH}{b}.\,$
These can be written as
$a^2=c\times HB \mbox{ and }b^2=c\times AH. \,$
Summing these two equalities, we obtain
$a^2+b^2=c\times HB+c\times AH=c\times(HB+AH)=c^2 .\,\!$
In other words, the Pythagorean theorem:
$a^2+b^2=c^2.\,\!$
### Euclid's proof
Proof in Euclid's Elements
In Euclid's Elements, Proposition 47 of Book 1, the Pythagorean theorem is proved by an argument along the following lines. Let A, B, C be the vertices of a right triangle, with a right angle at A. Drop a perpendicular from A to the side opposite the hypotenuse in the square on the hypotenuse. That line divides the square on the hypotenuse into two rectangles, each having the same area as one of the two squares on the legs.
For the formal proof, we require four elementary lemmata:
1. If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, then the triangles are congruent. (Side - Angle - Side Theorem)
2. The area of a triangle is half the area of any parallelogram on the same base and having the same altitude.
3. The area of any square is equal to the product of two of its sides.
4. The area of any rectangle is equal to the product of two adjacent sides (follows from Lemma 3).
The intuitive idea behind this proof, which can make it easier to follow, is that the top squares are morphed into parallelograms with the same size, then turned and morphed into the left and right rectangles in the lower square, again at constant area.
The proof is as follows:
Illustration including the new lines
1. Let ACB be a right-angled triangle with right angle CAB.
2. On each of the sides BC, AB, and CA, squares are drawn, CBDE, BAGF, and ACIH, in that order.
3. From A, draw a line parallel to BD and CE. It will perpendicularly intersect BC and DE at K and L, respectively.
4. Join CF and AD, to form the triangles BCF and BDA.
5. Angles CAB and BAG are both right angles; therefore C, A, and G are collinear. Similarly for B, A, and H.
6. Angles CBD and FBA are both right angles; therefore angle ABD equals angle FBC, since both are the sum of a right angle and angle ABC.
7. Since AB and BD are equal to FB and BC, respectively, triangle ABD must be equal to triangle FBC.
8. Since A is collinear with K and L, rectangle BDLK must be twice in area to triangle ABD.
9. Since C is collinear with A and G, square BAGF must be twice in area to triangle FBC.
10. Therefore rectangle BDLK must have the same area as square BAGF = AB2.
11. Similarly, it can be shown that rectangle CKLE must have the same area as square ACIH = AC2.
12. Adding these two results, AB2 + AC2 = BD × BK + KL × KC
13. Since BD = KL, BD* BK + KL × KC = BD(BK + KC) = BD × BC
14. Therefore AB2 + AC2 = BC2, since CBDE is a square.
This proof appears in Euclid's Elements as that of Proposition 1.47.[3]
### Garfield's proof
James A. Garfield (later President of the United States) is credited with a novel algebraic proof:[4]
The whole trapezoid is half of an (a+b) by (a+b) square, so its area = (a+b)2/2 = a2/2 + b2/2 + ab.
Triangle 1 and triangle 2 each have area ab/2.
Triangle 3 has area c2/2, and it is half of the square on the hypotenuse.
But the area of triangle 3 also = (area of trapezoid) - (areas of triangles 1 and 2)
= a2/2 + b2/2 + ab - ab/2 - ab/2
= a2/2 + b2/2
= half the sum of the squares on the other two sides.
Therefore the square on the hypotenuse = the sum of the squares on the other two sides.
### Proof by subtraction
In this proof, the square on the hypotenuse plus 4 copies of the triangle can be asssembled into the same shape as the squares on the other two sides plus 4 copies of the triangle. This proof is recorded from China.
Proof using area subtraction
### Similarity proof
From the same diagram as that in Euclid's proof above, we can see three similar figures, each being "a square with a triangle on top". Since the large triangle is made of the two smaller triangles, its area is the sum of areas of the two smaller ones. By similarity, the three squares are in the same proportions relative to each other as the three triangles, and so likewise the area of the larger square is the sum of the areas of the two smaller squares.
### Proof by rearrangement
Proof of Pythagorean theorem by rearrangement of 4 identical right triangles: Since the total area and the areas of the triangles are all constant, the total black area is constant. But this can be divided into squares delineated by the triangle sides a, b, c, demonstrating that a2 + b2 = c2.
A proof by rearrangement is given by the illustration and the animation. In the illustration, the area of each large square is (a + b)2. In both, the area of four identical triangles is removed. The remaining areas, a2 + b2 and c2, are equal. Q.E.D.
Animation showing another proof by rearrangement
Proof using rearrangement
Algebraic proof: A square created by aligning four right angle triangles and a large square
This proof is indeed very simple, but it is not elementary, in the sense that it does not depend solely upon the most basic axioms and theorems of Euclidean geometry. In particular, while it is quite easy to give a formula for area of triangles and squares, it is not as easy to prove that the area of a square is the sum of areas of its pieces. In fact, proving the necessary properties is harder than proving the Pythagorean theorem itself (see Lebesgue measure and Banach-Tarski paradox). Actually, this difficulty affects all simple Euclidean proofs involving area; for instance, deriving the area of a right triangle involves the assumption that it is half the area of a rectangle with the same height and base. For this reason, axiomatic introductions to geometry usually employ another proof based on the similarity of triangles (see above).
A third graphic illustration of the Pythagorean theorem (in yellow and blue to the right) fits parts of the sides' squares into the hypotenuse's square. A related proof would show that the repositioned parts are identical with the originals and, since the sum of equals are equal, that the corresponding areas are equal. To show that a square is the result one must show that the length of the new sides equals c. Note that for this proof to work, one must provide a way to handle cutting the small square in more and more slices as the corresponding side gets smaller and smaller.[5]
### Algebraic proof
An algebraic variant of this proof is provided by the following reasoning. Looking at the illustration which is a large square with identical right triangles in its corners, the area of each of these four triangles is given by an angle corresponding with the side of length C.
$\frac{1}{2} AB.$
The A-side angle and B-side angle of each of these triangles are complementary angles, so each of the angles of the blue area in the middle is a right angle, making this area a square with side length C. The area of this square is C2. Thus the area of everything together is given by:
$4\left(\frac{1}{2}AB\right)+C^2.$
However, as the large square has sides of length A + B, we can also calculate its area as (A + B)2, which expands to A2 + 2AB + B2.
$A^2+2AB+B^2=4\left(\frac{1}{2}AB\right)+C^2.\,\!$
(Distribution of the 4) $A^2+2AB+B^2=2AB+C^2\,\!$
(Subtraction of 2AB) $A^2+B^2=C^2\,\!$
### Proof by differential equations
One can arrive at the Pythagorean theorem by studying how changes in a side produce a change in the hypotenuse in the following diagram and employing a little calculus.[6]
Proof using differential equations
As a result of a change da in side a,
$\frac {da}{dc} = \frac {c}{a}$
by similarity of triangles and for differential changes. So
$c\, dc = a\,da$
which results from adding a second term for changes in side b.
Integrating gives
$c^2 = a^2 + \mathrm{constant}.\ \,\!$
When a = 0 then c = b, so the "constant" is b2. So
$c^2 = a^2 + b^2.\,$
As can be seen, the squares are due to the particular proportion between the changes and the sides while the sum is a result of the independent contributions of the changes in the sides which is not evident from the geometric proofs. From the proportion given it can be shown that the changes in the sides are inversely proportional to the sides. The differential equation suggests that the theorem is due to relative changes and its derivation is nearly equivalent to computing a line integral.
These quantities da and dc are respectively infinitely small changes in a and c. But we use instead real numbers Δa and Δc, then the limit of their ratio as their sizes approach zero is da/dc, the derivative, and also approaches c/a, the ratio of lengths of sides of triangles, and the differential equation results.
### Proof by shear mapping
One of the plane transformations that preserves area is the shear mapping. Since the Pythagorean theorem is concerned with areas, it is interesting that a proof can be based on this type of planar mapping. Mike May of Saint Louis University has provided an animated version of such a proof through use of the GeoGebra facility.
## Converse
The converse of the theorem is also true:
For any three positive numbers a, b, and c such that a2 + b2 = c2, there exists a triangle with sides a, b and c, and every such triangle has a right angle between the sides of lengths a and b.
This converse also appears in Euclid's Elements. It can be proven using the law of cosines (see below under Generalizations), or by the following proof:
Let ABC be a triangle with side lengths a, b, and c, with a2 + b2 = c2. We need to prove that the angle between the a and b sides is a right angle. We construct another triangle with a right angle between sides of lengths a and b. By the Pythagorean theorem, it follows that the hypotenuse of this triangle also has length c. Since both triangles have the same side lengths a, b and c, they are congruent, and so they must have the same angles. Therefore, the angle between the side of lengths a and b in our original triangle is a right angle.
A corollary of the Pythagorean theorem's converse is a simple means of determining whether a triangle is right, obtuse, or acute, as follows. Where c is chosen to be the longest of the three sides:
• If a2 + b2 = c2, then the triangle is right.
• If a2 + b2 > c2, then the triangle is acute.
• If a2 + b2 < c2, then the triangle is obtuse.
## Consequences and uses of the theorem
### Pythagorean triples
A Pythagorean triple has three positive integers a, b, and c, such that a2 + b2 = c2. In other words, a Pythagorean triple represents the lengths of the sides of a right triangle where all three sides have integer lengths. Evidence from megalithic monuments on the Northern Europe shows that such triples were known before the discovery of writing. Such a triple is commonly written (a, b, c). Some well-known examples are (3, 4, 5) and (5, 12, 13).
### List of primitive Pythagorean triples up to 100
(3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17), (9, 40, 41), (11, 60, 61), (12, 35, 37), (13, 84, 85), (16, 63, 65), (20, 21, 29), (28, 45, 53), (33, 56, 65), (36, 77, 85), (39, 80, 89), (48, 55, 73), (65, 72, 97)
### The existence of irrational numbers
One of the consequences of the Pythagorean theorem is that incommensurable lengths (ie. their ratio is irrational number), such as the square root of 2, can be constructed. A right triangle with legs both equal to one unit has hypotenuse length square root of 2. The proof that the square root of 2 is irrational was contrary to the long-held belief that everything was rational. According to legend, Hippasus, who first proved the irrationality of the square root of two, was drowned at sea as a consequence.[7]
### Distance in Cartesian coordinates
The distance formula in Cartesian coordinates is derived from the Pythagorean theorem. If (x0, y0) and (x1, y1) are points in the plane, then the distance between them, also called the Euclidean distance, is given by
$\sqrt{(x_1-x_0)^2 + (y_1-y_0)^2}.$
More generally, in Euclidean n-space, the Euclidean distance between two points, $\scriptstyle A\,=\,(a_1,a_2,\dots,a_n)$ and $\scriptstyle B\,=\,(b_1,b_2,\dots,b_n)$, is defined, using the Pythagorean theorem, as:
$\sqrt{(a_1-b_1)^2 + (a_2-b_2)^2 + \cdots + (a_n-b_n)^2} = \sqrt{\sum_{i=1}^n (a_i-b_i)^2}.$
## Generalizations
Generalization for similar triangles,
green area = red area
The Pythagorean theorem was generalized by Euclid in his Elements:
If one erects similar figures (see Euclidean geometry) on the sides of a right triangle, then the sum of the areas of the two smaller ones equals the area of the larger one.
The Pythagorean theorem is a special case of the more general theorem relating the lengths of sides in any triangle, the law of cosines:
$a^2+b^2-2ab\cos{\theta}=c^2, \,$
where θ is the angle between sides a and b.
When θ is 90 degrees, then cos(θ) = 0, so the formula reduces to the usual Pythagorean theorem.
Given two vectors v and w in a complex inner product space, the Pythagorean theorem takes the following form:
$\|\mathbf{v}+\mathbf{w}\|^2 = \|\mathbf{v}\|^2 + \|\mathbf{w}\|^2 + 2\,\mbox{Re}\,\langle\mathbf{v},\mathbf{w}\rangle.$
In particular,||v + w||2 =||v||2 +||w||2 if v and w are orthogonal, although the converse is not necessarily true.
Using mathematical induction, the previous result can be extended to any finite number of pairwise orthogonal vectors. Let v1, v2,…, vn be vectors in an inner product space such that <vi, vj> = 0 for 1 ≤ i < jn. Then
$\left\|\,\sum_{k=1}^{n}\mathbf{v}_k\,\right\|^2 = \sum_{k=1}^{n} \|\mathbf{v}_k\|^2.$
The generalization of this result to infinite-dimensional real inner product spaces is known as Parseval's identity.
When the theorem above about vectors is rewritten in terms of solid geometry, it becomes the following theorem. If lines AB and BC form a right angle at B, and lines BC and CD form a right angle at C, and if CD is perpendicular to the plane containing lines AB and BC, then the sum of the squares of the lengths of AB, BC, and CD is equal to the square of AD. The proof is trivial.
Another generalization of the Pythagorean theorem to three dimensions is de Gua's theorem, named for Jean Paul de Gua de Malves: If a tetrahedron has a right angle corner (a corner like a cube), then the square of the area of the face opposite the right angle corner is the sum of the squares of the areas of the other three faces.
There are also analogs of these theorems in dimensions four and higher.
In a triangle with three acute angles, α + β > γ holds. Therefore, a2 + b2 > c2 holds.
In a triangle with an obtuse angle, α + β < γ holds. Therefore, a2 + b2 < c2 holds.
Edsger Dijkstra has stated this proposition about acute, right, and obtuse triangles in this language:
sgn(α + βγ) = sgn(a2 + b2c2)
where α is the angle opposite to side a, β is the angle opposite to side b and γ is the angle opposite to side c.[8]
### The Pythagorean theorem in non-Euclidean geometry
The Pythagorean theorem is derived from the axioms of Euclidean geometry, and in fact, the Euclidean form of the Pythagorean theorem given above does not hold in non-Euclidean geometry. (It has been shown in fact to be equivalent to Euclid's Parallel (Fifth) Postulate.) For example, in spherical geometry, all three sides of the right triangle bounding an octant of the unit sphere have length equal to $\scriptstyle \pi/2$; this violates the Euclidean Pythagorean theorem because $\scriptstyle (\pi/2)^2+(\pi/2)^2\neq (\pi/2)^2$.
This means that in non-Euclidean geometry, the Pythagorean theorem must necessarily take a different form from the Euclidean theorem. There are two cases to consider — spherical geometry and hyperbolic plane geometry; in each case, as in the Euclidean case, the result follows from the appropriate law of cosines:
For any right triangle on a sphere of radius R, the Pythagorean theorem takes the form
$\cos \left(\frac{c}{R}\right)=\cos \left(\frac{a}{R}\right)\,\cos \left(\frac{b}{R}\right).$
This equation can be derived as a special case of the spherical law of cosines. By using the Maclaurin series for the cosine function, it can be shown that as the radius R approaches infinity, the spherical form of the Pythagorean theorem approaches the Euclidean form.
For any triangle in the hyperbolic plane (with Gaussian curvature −1), the Pythagorean theorem takes the form
$\cosh c=\cosh a\,\cosh b$
where cosh is the hyperbolic cosine.
By using the Maclaurin series for this function, it can be shown that as a hyperbolic triangle becomes very small (i.e., as a, b, and c all approach zero), the hyperbolic form of the Pythagorean theorem approaches the Euclidean form.
In hyperbolic geometry, for a right triangle one can also write,
$\sin \bar a \sin \bar b = \sin \bar c$
where $\scriptstyle\bar a$ is the angle of parallelism of the line segment AB that $\scriptstyle \mu(AB)\,=\,a$ where μ is the multiplicative distance function (see Hilbert's arithmetic of ends).
In hyperbolic trigonometry, the sine of the angle of parallelism satisfies
$\sin \bar a = \frac{2a}{1+a^2}.$
Thus, the equation takes the form
$\frac{2a}{1+a^2} \frac{2b}{1+b^2}=\frac{2c}{1+c^2}$
where a, b, and c are multiplicative distances of the sides of the right triangle (Hartshorne, 2000).
### In complex arithmetic: not valid
The Pythagoras formula is used to find the distance between two points in the Cartesian coordinate plane, and is valid if all coordinates are real: the distance between the points {a,b} and {c,d} is √((a-c)2+(b-d)2). But with complex coordinates: e.g. the distance between the points {0,1} and {i,0} would work out as 0, resulting in a reductio ad absurdum. This is because this formula depends on Pythagoras's theorem, which in all its proofs depends on areas, and areas depend on triangles and other geometrical figures separating an inside from an outside, which does not happen if the coordinates are complex.
## History
Visual proof for the (3, 4, 5) triangle as in the Chou Pei Suan Ching 500–200 BC
The history of the theorem can be divided into four parts: knowledge of Pythagorean triples, knowledge of the relationship among the sides of a right triangle, knowledge of the relationships among adjacent angles, and proofs of the theorem.
Megalithic monuments from circa 2500 BC in Egypt, and in Northern Europe, incorporate right triangles with integer sides.[9] Bartel Leendert van der Waerden conjectures that these Pythagorean triples were discovered algebraically.[10]
Written between 2000 and 1786 BC, the Middle Kingdom Egyptian papyrus Berlin 6619 includes a problem whose solution is a Pythagorean triple.
The Mesopotamian tablet Plimpton 322, written between 1790 and 1750 BC during the reign of Hammurabi the Great, contains many entries closely related to Pythagorean triples.
The Baudhayana Sulba Sutra, the dates of which are given variously as between the 8th century BC and the 2nd century BC, in India, contains a list of Pythagorean triples discovered algebraically, a statement of the Pythagorean theorem, and a geometrical proof of the Pythagorean theorem for an isosceles right triangle.
The Apastamba Sulba Sutra (circa 600 BC) contains a numerical proof of the general Pythagorean theorem, using an area computation. Van der Waerden believes that "it was certainly based on earlier traditions". According to Albert Bŭrk, this is the original proof of the theorem; he further theorizes that Pythagoras visited Arakonam, India, and copied it.
Pythagoras, whose dates are commonly given as 569–475 BC, used algebraic methods to construct Pythagorean triples, according to Proklos's commentary on Euclid. Proklos, however, wrote between 410 and 485 AD. According to Sir Thomas L. Heath, there was no attribution of the theorem to Pythagoras for five centuries after Pythagoras lived. However, when authors such as Plutarch and Cicero attributed the theorem to Pythagoras, they did so in a way which suggests that the attribution was widely known and undoubted.[2]
Around 400 BC, according to Proklos, Plato gave a method for finding Pythagorean triples that combined algebra and geometry. Circa 300 BC, in Euclid's Elements, the oldest extant axiomatic proof of the theorem is presented.
Written sometime between 500 BC and 200 AD, the Chinese text Chou Pei Suan Ching (周髀算经), (The Arithmetical Classic of the Gnomon and the Circular Paths of Heaven) gives a visual proof of the Pythagorean theorem — in China it is called the "Gougu Theorem" (勾股定理) — for the (3, 4, 5) triangle. During the Han Dynasty, from 202 BC to 220 AD, Pythagorean triples appear in The Nine Chapters on the Mathematical Art, together with a mention of right triangles.[11]
The first recorded use is in China, known as the "Gougu theorem" (勾股定理) and in India known as the Bhaskara Theorem.
There is much debate on whether the Pythagorean theorem was discovered once or many times. Boyer (1991) thinks the elements found in the Shulba Sutras may be of Mesopotamian derivation.[12]
## Cultural references to the Pythagorean theorem
The Pythagorean theorem has been referenced in a variety of mass media throughout history.
• A verse of the Major-General's Song in the Gilbert and Sullivan musical The Pirates of Penzance, "About binomial theorem I'm teeming with a lot o' news, With many cheerful facts about the square of the hypotenuse", with oblique reference to the theorem.
• The Scarecrow of The Wizard of Oz makes a more specific reference to the theorem when he receives his diploma from the Wizard. He immediately exhibits his "knowledge" by reciting a mangled and incorrect version of the theorem: "The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side. Oh, joy, oh, rapture. I've got a brain!" The "knowledge" exhibited by the Scarecrow is incorrect. The accurate statement would have been "The sum of the squares of the legs of a right triangle is equal to the square of the remaining side."[13]
• In an episode of The Simpsons, after finding a pair of Henry Kissinger's glasses in a toilet at the Springfield Nuclear Power Plant, Homer puts them on and quotes Oz Scarecrow's mangled version of the formula. A man in a nearby toilet stall then yells out "That's a right triangle, you idiot!" (The comment about square roots remained uncorrected.)
• Similarly, the Speech software on an Apple MacBook references the Scarecrow's incorrect statement. It is the sample speech when the voice setting 'Ralph' is selected.
• In Freemasonry, one symbol for a Past Master is the diagram from the 47th Proposition of Euclid, used in Euclid's proof of the Pythagorean theorem. President Garfield was a freemason.
• In 2000, Uganda released a coin with the shape of a right triangle. The coin's tail has an image of Pythagoras and the Pythagorean theorem, accompanied with the mention "Pythagoras Millennium".[14] Greece, Japan, San Marino, Sierra Leone, and Suriname have issued postage stamps depicting Pythagoras and the Pythagorean theorem.[15]
• In Neal Stephenson's speculative fiction Anathem, Pythagorean theorem are referred to as 'the Adrakhonic theorem'. A geometric proof of the theorem is displayed on the side of an alien ship to demonstrate their understanding of mathematics.
## Notes
1. ^ The "other two sides" are also known as legs or catheti.
2. ^ a b Heath, Vol I, p. 144.
3. ^ Elements 1.47 by Euclid, retrieved 19 December 2006
4. ^ Head, Angie. "Pythagorean Theorem"
5. ^ Pythagorean Theorem: Subtle Dangers of Visual Proof by Alexander Bogomolny, retrieved 19 December 2006.
6. ^ Hardy.
7. ^ Heath, Vol I, pp. 65, 154; Stillwell, p. 8–9.
8. ^ "Dijkstra's generalization" (PDF).
9. ^ "Megalithic Monuments.".
10. ^ van der Waerden 1983.
11. ^ Swetz.
12. ^ Boyer (1991). "China and India". pp. 207. "we find rules for the construction of right angles by means of triples of cords the lengths of which form Pythagorean triages, such as 3, 4, and 5, or 5, 12, and 13, or 8, 15, and 17, or 12, 35, and 37. However all of these triads are easily derived from the old Babylonian rule; hence, Mesopotamian influence in the Sulvasutras is not unlikely. Aspastamba knew that the square on the diagonal of a rectangle is equal to the sum of the squares on the two adjacent sides, but this form of the Pythagorean theorem also may have been derived from Mesopotamia. [...] So conjectural are the origin and period of the Sulbasutras that we cannot tell whether or not the rules are related to early Egyptian surveying or to the later Greek problem of alter doubling. They are variously dated within an interval of almost a thousand years stretching from the eighth century B.C. to the second century of our era."
13. ^ "The Scarecrow's Formula".
14. ^ "Le Saviez-vous ?".
15. ^ Miller, Jeff (). "Images of Mathematicians on Postage Stamps". Retrieved on 2007-08-06. |
# Lesson 6What about Other Bases?
Let’s explore exponent patterns with bases other than 10.
### Learning Targets:
• I can use the exponent rules for bases other than 10.
## 6.1True or False: Comparing Expressions with Exponents
Is each statement true or false? Be prepared to explain your reasoning.
## 6.2What Happens with Zero and Negative Exponents?
Complete the table to show what it means to have an exponent of zero or a negative exponent.
1. As you move toward the left, each number is being multiplied by 2. What is the multiplier as you move toward the right?
2. Use the patterns you found in the table to write as a fraction.
3. Write as a power of 2 with a single exponent.
4. What is the value of ?
5. From the work you have done with negative exponents, how would you write as a fraction?
6. How would you write as a fraction?
### Are you ready for more?
1. Find an expression equivalent to but with positive exponents.
2. Find an expression equivalent to but with positive exponents.
3. What patterns do you notice when you start with a fraction to a negative power and rewrite it so that it has only positive powers? Show or explain your reasoning.
## 6.3Exponent Rules with Bases Other than 10
Lin, Noah, Diego, and Elena decide to test each other’s knowledge of exponents with bases other than 10. They each chose an expression to start with and then came up with a new list of expressions; some of which are equivalent to the original and some of which are not.
Choose 2 lists to analyze. For each list of expressions you choose to analyze, decide which expressions are not equivalent to the original. Be prepared to explain your reasoning.
1. Lin’s original expression is and her list is:
2. Noah’s original expression is and his list is:
3. Diego’s original expression is and his list is:
4. Elena’s original expression is and her list is:
• 1
• 0
## Lesson 6 Summary
Earlier we focused on powers of 10 because 10 plays a special role in the decimal number system. But the exponent rules that we developed for 10 also work for other bases. For example, if and , then
These rules also work for powers of numbers less than 1. For example, and . We can also check that .
Using a variable helps to see this structure. Since (both sides have 7 factors that are ), if we let , we can see that . Similarly, we could let or or any other positive value and show that these relationships still hold.
## Lesson 6 Practice Problems
1. Priya says “I can figure out by looking at other powers of 5. is 125, is 25, then is 5.”
1. What pattern do you notice?
2. If this pattern continues, what should be the value of ? Explain how you know.
3. If this pattern continues, what should be the value of ? Explain how you know.
2. Select all the expressions that are equivalent to .
1. -12
2. 12
3. Write each expression using a single exponent.
4. Andre sets up a rain gauge to measure rainfall in his back yard. On Tuesday, it rains off and on all day.
• He starts at 10 a.m. with an empty gauge when it starts to rain.
• Two hours later, he checks, and the gauge has 2 cm of water in it.
• It starts raining even harder, and at 4 p.m., the rain stops, so Andre checks the rain gauge and finds it has 10 cm of water in it.
• While checking it, he accidentally knocks the rain gauge over and spills most of the water, leaving only 3 cm of water in the rain gauge.
• When he checks for the last time at 5 p.m., there is no change.
Graph A
Graph B
1. Which of the two graphs could represent Andre’s story? Explain your reasoning.
2. Label the axes of the correct graph with appropriate units.
3. Use the graph to determine how much total rain fell on Tuesday. |
#### Provide solution RD Sharma maths class 12 chapter maxima and minima exercise 17.2 question 12 maths textbook solution
$x=2 / 3 \text { is the point of local maxima and the value of local maxima is } \frac{2 \sqrt{3}}{9}$
Hint:
Use first derivative test to find the value and point of local maxima and local minima.
Given:
$f(x)=x \sqrt{1-x} \quad, x>0$
Differentiating $f(x) w \cdot r . t^{\prime} x^{\prime} \text { then }$
\begin{aligned} \frac{d}{d x}\{f(x)\} &=\frac{d}{d x}(x \sqrt{1-x}) \\ &=x \frac{d}{d x}(\sqrt{1-x})+\sqrt{1-x} \frac{d}{d x}(x) \quad\left[\because \frac{d}{d x}(y x)=y \frac{d}{d x} x+x \frac{d}{d x} y\right] \\ &=x \frac{d}{d x}(\sqrt{1-x})+(\sqrt{1-x}) \frac{d}{d x}(x) \\ &=x \frac{1}{2}(1-x)^{\frac{1}{2}-1} \frac{d}{d x}(1-x)+\sqrt{1-x} \cdot 1\left[\because \frac{d}{d x}(x+a)^{n}=n(x+a)^{n-1} \frac{d}{d x}(x+a)\right] \\ \end{aligned}
$\inline =x \cdot \frac{1}{2}(1-x)^{\frac{1-2}{1}}\left\{\frac{d}{d x}(1)-\frac{d}{d x}(x)\right\}+\sqrt{1-x} \quad\left[\because \frac{d}{d x}(x+a)=\frac{d}{d x}(x)+\frac{d}{d x}(a)\right] \\$
$\inline =\frac{x}{2}(1-x)^{-1 / 2}\{0-1\}+\sqrt{1-x} \quad\left[\because \frac{d}{d x} \text { Constant }=0, \frac{d}{d x}(x)=1\right] \\$
$\inline =\frac{x}{2 \sqrt{1-x}}(0-1)+\sqrt{1-x} \\$
$\inline =\frac{-x}{2 \sqrt{1-x}}+\sqrt{1-x} \\$
$\inline =\frac{-x+2(1-x)}{2 \sqrt{1-x}} \\$
$\inline =\frac{-x+2-2 x}{2 \sqrt{1-x}} \\$
$\inline =\frac{2-3 x}{2 \sqrt{1-x}}$
$\inline {f}'\left ( x \right )=\frac{-\left ( 3x-2 \right )}{2 \sqrt{1-x}}$
By first derivative test, for local maxima and local minima, we have
\inline \begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow \frac{-(3 x-2)}{2 \sqrt{1-x}}=0 \quad \Rightarrow \frac{3 x-2}{2 \sqrt{1-x}}=0 \Rightarrow \frac{3 x-2}{\sqrt{1-x}}=0 \quad\left[\because \frac{1}{2} \neq 0\right] \\ &\Rightarrow 3 x-2=0 \quad \Rightarrow \quad 3 x=2 \quad x=2 / 3 \end{aligned}
$\inline \text { Since, } f^{\prime}(x) \text { changes } f \text { rom }+\text { ve to }-\text { ve when } x \text { increases through } 2 / 3 \text { is the point of local minima }$
\inline \begin{aligned} &\therefore \text { the value of the local maxima of } f(x) \text { at } x=2 / 3 \text { is }\\ &f\left(\frac{2}{3}\right)=\frac{2}{3} \sqrt{1-\frac{2}{3}}=\frac{2}{3} \sqrt{\frac{3-2}{3}}=\frac{2}{3} \sqrt{\frac{1}{3}}\\ &=\frac{2}{3} \cdot \frac{1}{\sqrt{3}}=\frac{2}{3 \sqrt{3}}\\ &=\frac{2}{3 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{2 \sqrt{3}}{3 \times 3}=\frac{2 \sqrt{3}}{9} \end{aligned}
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# Evaluate $\dfrac{\log \sqrt{27} + \log \sqrt{8} -\log \sqrt{125}}{\log 6 -\log 5}$
Five logarithmic terms formed a mathematical relation in division form. logarithms of square roots of $27$, $8$ and $125$, and also logarithms of $6$ and $5$ are the numbers involved in forming this mathematical expression.
The logarithm of square root of $125$ is subtracted from sum of the logarithm of square root of $27$ and log of square root of $8$, to form the numerator. Similarly, the log of $5$ is subtracted from log of $6$ to form the denominator. The value of ratio of them is required to evaluate in this problem.
###### Step: 1
The radicand of each logarithmic term in the numerator can be converted as cube of a number.
$=$ $\dfrac{\log \sqrt{3^3} + \log \sqrt{2^3} -\log \sqrt{5^3}}{\log 6 -\log 5}$
###### Step: 2
The symbol square root ($\sqrt{\,\,\,\,}$) represents an exponent $\dfrac{1}{2}$.
$=$ $\dfrac{\log {\Big(3^3\Big)}^{\frac{1}{2}} + \log {\Big(2^3\Big)}^{\frac{1}{2}} -\log {\Big(5^3\Big)}^{\frac{1}{2}}}{\log 6 -\log 5}$
Use power rule of exponents to simplify power of an exponential term.
$=$ $\dfrac{\log {(3)}^{\frac{3}{2}} + \log {(2)}^{\frac{3}{2}} -\log {(5)}^{\frac{3}{2}}}{\log 6 -\log 5}$
###### Step: 3
The number of the logarithm term is in exponential notation. So, apply power rule of the logarithms to simplify each term in the numerator.
$=$ $\dfrac{\dfrac{3}{2} \, (\log 3) + \dfrac{3}{2} \, (\log 2) -\dfrac{3}{2} \, (\log 5)}{\log 6 -\log 5}$
###### Step: 4
The rational number $\dfrac{3}{2}$ is a common multiplying factor of each logarithmic term in numerator. So, take $\dfrac{3}{2}$ common from three terms.
$=$ $\dfrac{\dfrac{3}{2} \Big[\log 3 + \log 2 -\log 5 \Big]}{\log 6 -\log 5}$
###### Step: 5
It can be expressed as two multiplying factors.
$=$ $\dfrac{3}{2} \times \dfrac{\log 3 + \log 2 -\log 5}{\log 6 -\log 5}$
###### Step: 6
Observe the expression in numerator and denominator carefully. The numerator can be expressed same as the expression in the denominator by applying product rule of logarithms to addition of $\log 3$ and $\log 2$ terms.
$=$ $\dfrac{3}{2} \times \dfrac{\log (3 \times 2) -\log 5}{\log 6 -\log 5}$
$=$ $\dfrac{3}{2} \times \dfrac{\log 6 -\log 5}{\log 6 -\log 5}$
$=$ $\require{cancel} \dfrac{3}{2} \times \dfrac{\cancel{\log 6 -\log 5}}{\cancel{\log 6 -\log 5}}$
$=$ $\dfrac{3}{2} \times 1$
$\therefore \,\,\,\,\,\, \dfrac{\log \sqrt{27} + \log \sqrt{8} -\log \sqrt{125}}{\log 6 -\log 5}$ $=$ $\dfrac{3}{2}$
Therefore, the answer for this problem is $\dfrac{3}{2}$ and it is required solution for this logarithm problem mathematically.
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# Lesson video
In progress...
Hello everyone, and welcome to maths with Ms. Dobrowolski.
Today we'll be adding two digit numbers with regrouping.
So let's have a look at our lesson agenda.
First, we'll be adding by partitioning the second number only followed by our top task.
For this lesson, you will need a pencil and a notebook.
If you don't have these items, pause the video now and go get them.
Super.
So partitioning the second number only.
As you can see, I'm trying to add 27 plus 14.
How can I add this? So I'm only partitioning the second number, the 14.
Well, in order to help me do this, I'm going to use a number line.
Great, so how can we solve this equation by partition the second number, the 14? Well, I'm going to use this number line to help me.
I will add the 10s first.
So I know that my first number is 27, so that's where I'm going to start and I'm going to add my 10s.
I can see that 27 plus 14, 14 has one 10.
So I know 20 plus 10 is equal to 30.
Then I know 27 plus 10 is equal to 37.
And that's plus 10, excellent.
Now I need to add my ones.
So I will use the make 10 strategy.
I know that seven plus three is equal to 10.
So that means I can partition the four into three and one.
So first you need to add three.
Well, if I know seven plus three is equal to 10, then I know 37 plus three is equal to 40.
And again, that's plus three.
If I was to count up, I should get the same answer, one, two, three, good.
And let's not forget to add the second part of the partition, the one.
So I jump 40 plus one is equal to 41.
So that must mean 27 plus 10 plus three plus one is equal to 27 plus 14.
These equations are equal because the parts equal the same whole.
10 plus three is equal to 13 plus one, 13, and 13 plus one is equal to 14.
All I did was partition 14.
So now I have another equation, 57 plus 34.
How can we solve this equation by partitioning the second number, 34? Well, I know that I'm going to start at 57, because that's the first number I'm adding.
And I will add the 10s first.
So if I know that 34 has three 10s, because there's a three in the 10s place, I need to add 30 or three 10s.
Now if I know that 50 plus 30 is equal to 80, then I know 57 plus 30 is equal to 87.
So I need to jump all the way to 87.
That's a big jump, remember, because I added 30.
Okay, now I'm going to add the ones, but I'm going to use my make 10 strategy.
So, I have seven ones and I need to add four, how can I make a 10 if I already have seven ones? Well, I know seven plus three is equal to 10.
So that means I need to partition this four into three, and one, three plus one is four.
So first if I know that seven plus three is equal to 10, then I know that 87 plus three is equal to 90.
And let's not forget to add the other part of the partition, which is one, and 90 plus one is equal to 91.
So I started with 57 plus 34, and this is how I partitioned 34.
Really, what I did was 57 plus 30, which was equal to 87 plus three plus one.
And both sides of the equation are equal because the parts equal the same whole, which was 91.
So for this top task, I'd like you to use the number lines that I've put on the screen and the make 10 strategy to solve the following equations.
So as usual, I'll do one before you go off on your own.
I think I'll choose 48 plus 26.
So I have the equation 48 plus 26, and I will partition 26 into two 10s, and six ones.
I know I'm starting at 48 because that's the first number I'm adding.
Now I know that 40 plus 20 is equal to 60, so 48 plus 20 must be equal to 68.
If I know that eight plus two is equal to 10, then I know that 68 plus two must be equal to 70.
So I will partition the six into two, and four.
So 68 plus two is equal to 70.
And 70 plus four is equal to 74.
Pause the video and resume when you're ready so we can go over the answers together.
Good luck.
Well done, everyone, let's go over the answers.
As you can see, 39 plus 14 was equal to 53, 39 plus 15 was equal to 54.
And 39 plus 16 was equal to 55, 39 plus 17 was equal to 56.
And as we said, 48 plus 26 is equal to 74, 40 plus 25 was equal to 73, 40 plus 24 is equal to 72.
48 plus 23 is equal to 71.
Do you notice a pattern here? Because I noticed that we had a nine in the ones each time for our first number.
So that meant when we partitioned our second number, our ones always had to be nine plus one each time.
And then in this column over here, I noticed there was an eight in the ones each time of our first number.
So whenever we partitioned the ones in the second number, we always had to partition so that we had a two because eight plus two is equal to 10.
Strange how that happens.
Let's move on.
So now I'm going to show you how we can add by partitioning both numbers and we will use the number line for this one as well.
Great, so when we first saw this equation, we solved it by partitioning one number.
But this time, let's partition both numbers.
Of course, will partition into 10s and ones because that's what we have here.
So I know that 27 has two 10s and 14 has one 10.
If I know that two plus one is equal to three, then I know that 20 plus 10 is equal to 30.
How many ones are there all together? Well, let's use the make 10 strategy to see if we can make a 10.
I know that I have seven ones and four ones.
Well, if I already have seven ones, how can I make a 10? Oh, I know seven plus three is equal to 10.
And if I know that seven plus three is equal to 10, then I know that seven plus four is equal to 11 because it's only one more.
So let's see.
If I know that seven plus three is equal to 10, so I know I'm going to add 10.
30 plus 10 is equal to 40.
And 40 plus one more is equal to 41.
So when we solve this equation, we partitioned both numbers, we'd said 20 plus 10 was equal to 30.
Then we partition the four into three and one.
So we had another 10, and then another one.
So both sides of this equation are correct.
27 plus 14 is equal to 41.
And 20 plus 10 plus 10 plus one is also equal to 41.
Let's try the next equation, 28 plus 14 and of course, we'll use our number line to help us.
So remember, I want to partition both numbers.
I have two 10s in 28, and one 10 in 14.
So first we'll add two 10s and one 10.
Well, I know that two plus one is equal to three so 20 plus 10 must be equal to 30.
Great.
So now we need to add our ones.
Let's use the make 10 strategy to see if we can make a 10 with our ones.
Well, I know that I had eight ones in 28 and four ones in 14.
If I have eight, how much more do I need to make a 10? Oh, I know, eight plus two is equal to 10.
So I can partition four into two and two.
Well, if I know that eight plus two is equal to 10, then I know 10 plus two is equal to 12 because it's only two more.
So really, we're going to jump 12.
First we'll jump at 10, 30 plus 10 is equal to 40.
And then 40 plus two is equal to 42.
So 28 plus 14 was really 20 plus 10 plus 10 plus two.
And both sides of this equation are correct, because the answer is the same.
They are both equal to 42.
Let's try one more together.
Here I have 29 plus 14.
Again, I'm going to partition both numbers.
So stay on with me make sure that we're following along.
I know it's a lot.
So what we're going to do here now is add our 10s first, so let's partition In 29, I have two 10s, and in 14, I have one 10.
If I know two plus one is equal to three, then 20 plus 10 must be equal to 30.
So I've partitioned my 10s and added them.
Let's move on to the ones.
I have nine ones in 29, and four ones in 14.
So if I have nine ones can I use the make 10 strategy? Oh, yes, I know that nine plus one is equal to 10.
So I can partition four into one and three, one plus three is equal to four.
So that means I have nine plus one is equal to 10.
And then 10 plus three is equal to 13, because it's only three more.
So that means I should jump 10 first.
So 30 plus 10 is equal to 40.
And then let's not forget our three, 40 plus three is equal to 43.
So 29 plus 14 was equal to 20 plus 10 plus another 10 plus three.
And I know this is correct, both sides are equal because both sides are equal to 43.
Great, it's time to show off what you've learned by completing your independent task.
But this independent task, I'd like you to do two things.
Step one, solve the equation by partitioning one number.
So for example, in 49 plus 15, I will partition the 15.
I have one 10 and 15.
So that means 49 plus 10 is equal to 59.
Now, I still have five ones, 59.
Well I know that I can use my make 10 strategy.
Nine plus one is equal to 10.
So 59 plus one must be equal to 60.
And then I can partition the five into one and four.
So 59 plus one is equal to 60.
And 60 plus four is equal to 64.
Great.
Step two is to solve the equation by partitioning both numbers.
And hopefully you get the same answer in step one and step two.
So again, in 49, plus 15, I will partition both 49 and 15.
I have four 10s and one 10.
I know four plus one is equal to five.
So 40 plus 10, must be equal to 50.
And now I'm going to add my ones, I have a nine and a five in the ones again, nine plus one is equal to 10.
So I will partition five into one and four 50 plus 10 is equal to 60.
And 60 plus four is equal to 64.
So I got the same answer both times I must be correct.
Pause the video when you're done resume so we can go over the answers.
Good luck.
Well done everyone.
So let's go over the answers, for the first one 35 plus 16 is equal to 51, 35 plus 17 is equal to 52,36 plus 15 is equal to 51 and 36 plus 16 is equal to 52.
We have completed 49 plus 15 together, and 49 plus 25 is equal to 74, 49 plus 35 is equal to 84 and 49 plus 45 is equal to 94.
If you'd like to you can share your work with oak national by asking your parent or carer to share your work on Instagram, Facebook or Twitter tagging at Oak national and hashtag learn with oak.
Don't forget to complete your final quiz.
Good job everyone, that was really hard work today.
I really hope to see you next time.
Bye. |
### If the medians of a $\Delta ABC$ intersect at $G$. Prove that $ar(\Delta BGC)$ is equal to $\dfrac { 1 } { 3 } ar(\Delta ABC)$. A C B D E F G
Step by Step Explanation:
1. We are given that $AD$, $BE$ and $CF$ are the medians of $\Delta ABC$ intersecting at $G$.
Now, we have to find the area of $\Delta BGC$.
2. We know that a median of a triangle divides it into two triangles of equal area.
Now, in $\Delta ABC$, $AD$ is the median. \begin{aligned} \therefore ar(\Delta ABD)= ar(\Delta ACD) &&\ldots\text{(i)} \end{aligned} Similarly, in $\Delta GBC$, $GD$ is the median. \begin{aligned} \therefore ar(\Delta GBD)= ar(\Delta GCD) &&\ldots\text{(ii)} \end{aligned}
3. From $\text{(i)}$ and $\text{(ii)}$, we get: \begin{aligned} &ar(\Delta ABD) - ar(\Delta GBD)= ar(\Delta ACD) - ar(\Delta GCD) \\ \implies& ar(\Delta AGB) = ar(\Delta AGC) \end{aligned} Similarly, \begin{aligned} & ar(\Delta AGB) = ar(\Delta BGC) \\ \therefore \space & ar(\Delta AGB) = ar(\Delta AGC) = ar(\Delta BGC) &&\ldots\text{(iii)} \end{aligned}
4. \begin{aligned} & \text{But,}\\ & ar(\Delta ABC) = ar(\Delta AGB) + ar(\Delta AGC) + ar(\Delta BGC) = 3 \space ar(\Delta BGC) &&[ \text{Using eq (iii)} ] \\ &\therefore ar(\Delta BGC) = \dfrac { 1 } { 3 } ar(\Delta ABC) \end{aligned} |
# 2006 Indonesia MO Problems/Problem 1
## Problem
Find all pairs $(x,y)$ of real numbers which satisfy $x^3-y^3=4(x-y)$ and $x^3+y^3=2(x+y)$.
## Solution
Factoring and rearranging terms for both equations results in $(x-y)(x^2 + xy + y^2 - 4) = 0$ and $(x+y)(x^2 - xy + y^2 - 2) = 0$. By the Zero Product Property, $x = y$ or $x^2 + xy + y^2 = 4$ in the first equation, and $x = -y$ and $x^2 - xy + y^2 = 2$ in the second equation. In order for the solution to satisfy both equations, it must satisfy at least one of the conditions in each equation. Now there are four cases to consider.
Case 1: $x=y$ and $x=-y$
Solving this system results in $(0,0)$. Plugging the x and y values back in satisfies the original equation.
Case 2: $x=y$ and $x^2 - xy + y^2 = 2$
Substitution results in $x^2 = 2$, so solving this system results in $(\sqrt{2},\sqrt{2})$ and $(-\sqrt{2},-\sqrt{2})$. Plugging the x and y values back in for each ordered pair satisfies the original equation.
Case 3: $x^2 + xy + y^2 = 4$ and $x=-y$
Substitution results in $y^2 = 4$, so solving this system results in $(2,-2)$ and $(-2,2)$. Plugging the x and y values back in for each ordered pair satisfies the original equation.
Case 4: $x^2 + xy + y^2 = 4$ and $x^2 - xy + y^2 = 2$
Subtracting the second equation from the first equation results in $2xy = 2$, so $xy = 1$. Thus, $x^2 + y^2 = 3$ and $y = \tfrac{1}{x}$, so $x^2 + \tfrac{1}{x^2} = 3$.
Multiplying both sides by $x^2$ and bringing all terms to one side results in $x^4 - 3x^2 + 1 = 0$. By using the Quadratic Formula, $x^2 = \tfrac{3 \pm \sqrt{5}}{2}$, and both quantities are positive. If $x^2 = \tfrac{3 + \sqrt{5}}{2}$, then $y^2 = \tfrac{3 - \sqrt{5}}{2}$ (and vice versa). Since $xy = 1$, $x$ and $y$ must have the same sign.
Therefore, there are four ordered pairs — $(\sqrt{\frac{3 + \sqrt{5}}{2}},\sqrt{\frac{3 - \sqrt{5}}{2}}), (-\sqrt{\frac{3 + \sqrt{5}}{2}}, -\sqrt{\frac{3 - \sqrt{5}}{2}}),$ $(\sqrt{\frac{3 - \sqrt{5}}{2}}, \sqrt{\frac{3 + \sqrt{5}}{2}}), (-\sqrt{\frac{3 - \sqrt{5}}{2}}, -\sqrt{\frac{3 + \sqrt{5}}{2}}).$ Plugging the x and y values back in for each ordered pair satisfies the original equation.
In total, there are nine ordered pairs of real numbers that satisfy the original equation, which are listed in each case. |
# Trisecting the Angle/Neusis Construction
## Theorem
Let $\alpha$ be an angle which is to be trisected.
This can be achieved by means of a neusis construction.
## Construction
Let there be a neusis ruler $S$ marked by two points $p$ and $q$.
Let the vertex of $\alpha$ be labelled $B$.
Let $A$ be the point of one of the branches of $\alpha$ such that the distance $AB$ equals $pq$.
Let the other branch of $\alpha$ be extended past $B$ through $D$.
Let a circle be drawn with center at $B$ with radius $AB$.
Let the straightedge $S$ be placed so that:
$S$ passes over $A$
$p$ lies on the circle at $C$
$q$ lies on the straight line $BD$ at $D$.
Then the angle $3 \angle CBD = \alpha$
Thus $\alpha$ has been trisected.
## Proof
We have that $\angle BCD + \angle ACB$ make a straight angle.
As $CD = AB$ by construction, $CD = BC$ by definition of radius of circle.
Thus $\triangle BCD$ is isosceles.
$\angle CBD = \angle CDB$
$\angle BCD + 2 \angle CBD$ equals two right angles.
Thus:
$2 \angle CBD = \angle ACB$
Similarly, by Isosceles Triangle has Two Equal Angles:
$\angle ACB = \angle CAB$
and again from Sum of Angles of Triangle equals Two Right Angles:
$\angle ABC + 2 \angle ACB$ equals two right angles.
and so:
$\angle ABC + 4 \angle CBD$ equals two right angles.
But $\alpha + \angle ABC + \angle CBD$ make a straight angle.
Thus:
$\alpha + \angle ABC + \angle CBD = \angle ABC + 4 \angle CBD$
and so:
$\alpha = 3 \angle CBD$
$\blacksquare$ |
Students can download 5th Maths Term 3 Chapter 6 Fractions 6.4 Questions and Answers, Notes, Samacheer Kalvi 5th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.
## Tamilnadu Samacheer Kalvi 5th Maths Solutions Term 3 Chapter 6 Fractions 6.4
Question 1.
Write the proper symbol from <, >, or = in the box
(i) $$\frac{3}{5}$$ ___ $$\frac{2}{5}$$
$$\frac{3}{5}$$ > $$\frac{2}{5}$$
(ii) $$\frac{2}{8}$$ ___ $$\frac{1}{8}$$
$$\frac{2}{8}$$ > $$\frac{1}{8}$$
(iii) $$\frac{2}{11}$$ ___ $$\frac{10}{11}$$
$$\frac{2}{11}$$ < $$\frac{10}{11}$$
(iv) $$\frac{3}{15}$$ ___ $$\frac{10}{30}$$
$$\frac{3}{15}=\frac{3 \times 2}{15 \times 2}=\frac{6}{30}$$
$$\frac{6}{30}$$ < $$\frac{10}{30}$$
$$\frac{3}{15}$$ < $$\frac{10}{30}$$
(v) $$\frac{3}{8}$$ ___ $$\frac{3}{7}$$
$$\frac{3}{8}=\frac{3 \times 7}{8 \times 7}=\frac{21}{56}$$
$$\frac{3}{7}=\frac{3 \times 8}{7 \times 8}=\frac{24}{56}$$
$$\frac{21}{56}$$ < $$\frac{24}{56}$$
$$\frac{3}{8}$$ < $$\frac{3}{7}$$
(vi) $$\frac{4}{7}$$ ___ $$\frac{4}{11}$$
$$\frac{4}{7}=\frac{4 \times 11}{7 \times 11}=\frac{44}{77}$$
$$\frac{4}{11}=\frac{4 \times 7}{11 \times 7}=\frac{28}{77}$$
$$\frac{44}{77}$$ > $$\frac{28}{77}$$
$$\frac{4}{7}$$ > $$\frac{4}{11}$$
(vii) $$\frac{5}{12}$$ ___ $$\frac{1}{6}$$
$$\frac{1}{6}=\frac{1 \times 2}{6 \times 2}=\frac{2}{12}$$
$$\frac{5}{12}$$ > $$\frac{2}{12}$$
$$\frac{5}{12}$$ > $$\frac{1}{6}$$
$$\frac{3}{7}$$ < $$\frac{5}{9}$$
(viii) $$\frac{4}{9}$$ ___ $$\frac{4}{9}$$
$$\frac{4}{9}$$ = $$\frac{4}{9}$$
(ix) $$\frac{3}{7}$$ ___ $$\frac{5}{9}$$
$$\frac{3}{7}=\frac{3 \times 9}{7 \times 9}=\frac{27}{63}$$
$$\frac{5}{9}=\frac{5 \times 7}{9 \times 7}=\frac{35}{63}$$
$$\frac{27}{63}$$ < $$\frac{35}{63}$$
(x) $$\frac{4}{11}$$ ___ $$\frac{1}{5}$$
$$\frac{4}{11}=\frac{4 \times 5}{11 \times 5}=\frac{20}{55}$$
$$\frac{1}{5}=\frac{1 \times 11}{5 \times 11}=\frac{11}{55}$$
$$\frac{20}{55}$$ > $$\frac{11}{55}$$
$$\frac{4}{11}$$ > $$\frac{1}{5}$$ |
## Use the three steps to solve the problem. Betty has 10 more dimes than quarters. If she has \$3.45, how many coins does she have?
Question
Use the three steps to solve the problem. Betty has 10 more dimes than quarters. If she has \$3.45, how many coins does she have?
0
1. Step 1——-> set variables.
Let
x———–> number of dimes
y———–> number of quarters
Step 2: set the equation and solve for the variable
we know that
0.10 x+0.25 y=3.45—————> 10x+25y=345 equation (1)
x=y+10 equation (2)
substituting 2 in 1
10[y+10]+25y=345—-> 10y+100+25y=345
35y=345-100———–> y=245/35=7
x=y+10——–> x=7+10=17
Step 3: plug in the value of x and y from last step into the variables.
y=7——–> number of quarters coins
x=17——-> number of dimes coins
She has (7+17)=24 coins in total
2. If we assume the quarters are x, then the dimes are 10+x
Thus, 0.25x + 0.1(10+x) = 3.45
= 0.25x +1 +0.1 x = 3.45
= 0.35 x = 3.45 -1
= 0.35x = 2.45
x = 7
Therefore, the quarters are 7, while the dimes were 17,
Thus a total of 24 coins |
# 17 Constant Velocity Example: Detailed Explanation And Facts
In this article, we will discuss various examples of constant velocity with detailed explanations, and facts.
The following is a list of constant velocity examples:-
# Object in a circular motion
An object moving in a circular path of radius ‘r’ elapses a distance of 2πr on each round with angular velocity ω.
Consider an object in a circular motion that covers a distance ‘s’ in a certain time interval ‘t’. Let ‘θ’ be an angle made by the particle displacing from its initial position.
The linear velocity of the particle is a change in the position of the object in time t.
v=s / t
‘s’ is a displacement of a particle which is an arc length and can be calculated as a product of angle made by the particle on displacement and the radius of the circle.
s= θr
On substituting this in the above equation, we have
v=rθ/ t
Since the angular velocity is equal to the change in angle with respect to time; we can rewrite the equation as
v=rω
Where ω is an angular velocity
The angular velocity of the object is constant if the linear velocity of the object remains constant.
# Motion of an object due to Centripetal Force
The object in a centripetal motion exerts the centripetal force which is acting towards the center of the circular path and the linear velocity of the object remains perpendicular to the centripetal force.
The centripetal force exerted on the object in a circular motion is given by the equation
F=mv2/r
Where r is the radius of the circular path
V is a linear velocity
M is a mass of the particle
If there is an external force acting on the object that keeps the object in the circular motion the
ma=mv2/r
a=v2/r
v2=ar
Hence, the velocity of the object is constant if the angular acceleration of the object is constant.
# Revolution of a moon around the Earth
The velocity of a moon revolving around the Earth is almost at a constant rate. The moon completes one revolution around the Earth in 27.3 days which is equal to T=27.3 x 24 x 60 x 60=23,58,720 seconds
The distance of the moon from the center of the Earth is 3,84,000 km
The distance covered by the moon in one revolution is equal to the circumference of the circular path. Hence, the velocity of the motion is V= 2π r/T
V= 2π x 384000 x 1000/2358720=1022 m/s
The velocity of a moon orbiting around the Earth is 1022m/s and is constant.
# A person walking on the street at a constant speed
A man walking on the street at his constant speed will cover an equal distance in an equal interval of time. This can be an example of a constant velocity.
# Drawing water from a well with the help of a pulley
While drawing water from a well using a pulley, the force is applied downward but the reaction force is in the upward direction. The velocity of the bucket lifting up depends upon the length of the rope stretched on every pull.
The length of the rope that is stretched depends upon the movement of the arm and the length of the hand. Hence the velocity of the bucket and the angular acceleration of the pulley will be constant.
# A ray of light
A light travels in a straight line at a constant speed of 3 x 108 m/s. A light shows various other phenomena in nature like scattering, dispersion, reflection, refraction, total internal refraction, interference, diffraction, etc.
The speed of light is equal to the product of the wavelength of the light and the frequency of the electromagnetic wave as c=fλ.
# Speed of the object in a vacuum
If the object falls in the vacuum, it accelerates at a constant speed and experiences a free fall. All the objects in the vacuum will move at the same speed irrespective of their shape, size, density, or weight. The velocity of the object in a motion in a vacuum is constant.
# Sound wave
The sound wave travels at a speed of 332m/s at normal temperature and pressure conditions. The speed of sound is determined by the distance traveled by the sound waves in a certain time duration.
The velocity of a sound wave varies depending upon the density of different mediums and sound waves travel at a constant rate.
# Clock
The minute hand, an hour hand, and a second hand on the clock move at a constant speed. The point at the center where all the hands are attached resembles an instantaneous central point.
A clock measures angle of 360 degrees and each minute on the clock is equal to 1 degree. A second-hand travels 360 degrees in one minute, hence the speed of the second hand is
A minute hand covers 360 degrees in 1 hour, hence the velocity of the minute hand is
An hour hand displaces 30 degrees in 1 hour, therefore the velocity of an hour hand is
# A car traveling on a road at constant speed
A car moving at a constant speed will elapse equal distance in an equal duration of time hence is an example of constant velocity. The velocity of a car is measured as the ratio distance covered by the car from its initial position to reach a certain distance in time ‘t’.
# A ball moving on a plane surface
A ball can travel at the same speed unless exerts an external force that makes increases or decreases the speed of the ball to displace its position.
# Fan
A fan rotating at a constant speed gives a constant angular velocity until the speed of the fan is changed.
# Hourglass
A sand-filled hourglass is dropped down from the hole at a constant velocity.
The hourglass is designed such that the frictional forces due to sand and glass are canceled and the constant pressure is exerted on the hole that makes the sand drop down from the hole. The sand filled in glass drops at a constant rate and is hence used as a timer.
# Train
A train is an example of a constant velocity, which elapses the same distance in a certain interval of time.
# Electric vehicles without gears
Electric vehicles work on electric energy. A vehicle without gear will move at the same speed and in the front direction only.
# Photon
A photon being a light particle is easily carried away and moves with the speed of light. The velocity of a photon is constant.
# Birds Flying
The velocity of the birds is mostly constant while they are flying. Most of the birds are observed to sway at constant velocity. Therefore, we can estimate the expected date and time of birds venturing from the far locations.
## What is constant velocity?
The velocity of the object is defined as the rate of change of position of an object in a fixed interval of time.
If the distance elapsed by the object in a given time interval will be constant for every time interval then the velocity of the object will be constant.
Hence,
x2-x1=Constant
For the velocity to be constant, the change in the position of the object has to be constant along with time.
## Position-Time Graph for constant velocity
The displacement of the object in time is represented in the following position-time graph.
The slope of the position-time graph gives the velocity of the object between the two time intervals while displacing from one position to another. The slope of the graph is linear and is constant throughout the slope.
Read more on How To Calculate Negative Velocity: Example And Problems.
## Q1. If an object is traveling at a constant velocity of 12m/s, then calculate the distance covered by the object after 1 minute.
Solution: 1 minute=60 seconds
The distance covered by the object in 60 seconds traveling at a speed of 12m/s will be
= 12m/s\times 60s=720m
Hence, the distance traveled by the object in 1 minute is 720m.
## How does a graph of velocity v/s time will looks like for a constant velocity?
An object with constant velocity will travel in a straight line covering an equal distance in a given time interval.
Since the velocity of the object will be constant all the time, the acceleration of the object which is a slope of the graph will be zero. This implies that the slope of the graph will be a straight line. |
View the step-by-step solution to:
Question
# IntroductionRecall that a projectile is an object that is launched
(projected) with some velocity into the air and subsequently moves under the influence of gravity.
1) Consider the following problem: Suppose you shoot a cannon ball straight up into the air and are able to measure the height H it attains. Explain how you would utilize this information to determine the speed that the ball leaves the cannon.
After you have come up with a method obtain a launcher from the front of the room, fire its projectile straight up into the air, and use your reasoning to determine the speed with which the projectile left the launcher. Given your uncertainty in H estimate your uncertainty in the speed ΔV.
2) Suppose that you now place the launcher on your desk, position it horizontally, and launch the projectile. Generally, what can be said about the behavior of the vertical and horizontal components of the ball's velocity? Predict how far the ball will travel before it hits the ground. In your prediction make sure to explain your reasoning, and the appropriate principles involved. From you uncertainty in the speed and height of the table estimate the uncertainty of your prediction.
Prediction:
Now test your prediction by firing the ball and determining the horizontal distance it travels from the launcher when it hits the ground. Compare your prediction to your observations. Is the discrepancy within your predicted uncertainty? If not can you offer any explanation?
Observation: Percent error:
3) Suppose the launcher is placed on the floor, aimed at a 45∞ angle with the horizontal, and then fired. Explain how such a situation can be analyzed and attempt making a prediction for how far and high the projectile travels. Also estimate the uncertainties in your predictions.
Predictions: Horizontal range = Maximum height =
Test your prediction by firing the ball several times and determining its maximum height and distance it lands from the launcher. Are your observations within your uncertainties?
Observations: Max height = Range =
Percent errors: Max height: Range:
4) With the launcher again placed on the floor determine the range of the projectile for the following angles (above the horizontal): 15∞, 25∞, 35∞, 45∞, 55∞, 65∞, 75∞. For which of these angles does it travel the farthest?of the angles yield roughly the same range? Can you roughly explain why this is so?
R15 = R25= R35= R45=
R55= R65= R75=
5) Now place the launcher on the table and determine the angle that yields the greatest range. Is it identical to what you found in (4)? If it is different can you offer a qualitative explanation as to why it is smaller or greater than the result you found in part (4)?
Max range angle =
6) Can you prove the result corresponding to your finding in part (4)?
7) Attempt to derive a general formula for the maximum range angle found in (5).
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# Solve the equation z^4+81i=0?
Mar 16, 2017
There are four solutions;
$z = 3 \left(- \frac{\sqrt{2 - \sqrt{2}}}{2} - \frac{\sqrt{2 + \sqrt{2}}}{2} i\right)$
$z = 3 \left(\frac{\sqrt{2 + \sqrt{2}}}{2} - \frac{\sqrt{2 - \sqrt{2}}}{2} i\right)$
$z = 3 \left(\frac{\sqrt{2 - \sqrt{2}}}{2} + \frac{\sqrt{2 + \sqrt{2}}}{2} i\right)$
$z = 3 \left(- \frac{\sqrt{2 + \sqrt{2}}}{2} + \frac{\sqrt{2 - \sqrt{2}}}{2} i\right)$
#### Explanation:
Let $\omega = - 81 i$, and let ${z}^{4} = \omega$
First let us plot the point $\omega$ on the Argand diagram:
And we will put the complex number into polar form (visually):
$| \omega | = 81$
$a r g \left(\omega\right) = - \frac{\pi}{2}$
So then in polar form we have:
$\omega = 81 \left(\cos \left(- \frac{\pi}{2}\right) + i \sin \left(- \frac{\pi}{2}\right)\right)$
We now want to solve the equation for $z$ (to gain $4$ solutions):
${z}^{3} = 81 \left(\cos \left(- \frac{\pi}{2}\right) + i \sin \left(- \frac{\pi}{2}\right)\right)$
Whenever dealing with complex variable equation such as this it is essential to remember that the complex exponential has a period of $2 \pi$, so we can equivalently write (incorporating the periodicity):
x^3 = 81cos(2npi-(pi)/2) + isin(2npi-(pi)/2)) \ \ \ n in ZZ
By De Moivre's Theorem we can write this as:
$z = {\left\{\cos \left(2 n \pi - \frac{\pi}{2}\right) + i \sin \left(2 n \pi - \frac{\pi}{2}\right)\right\}}^{\frac{1}{4}}$
$\setminus \setminus = {\left\{81\right\}}^{\frac{1}{4}} {\left\{\cos \left(2 n \pi - \frac{\pi}{2}\right) + i \sin \left(2 n \pi - \frac{\pi}{2}\right)\right\}}^{\frac{1}{4}}$
$\setminus \setminus = 3 \left\{\cos \left(\frac{2 n \pi - \frac{\pi}{2}}{4}\right) + i \sin \left(\frac{2 n \pi - \frac{\pi}{2}}{4}\right)\right\}$
$\setminus \setminus = 3 \left(\cos \theta + i \sin \theta\right)$
Where:
$\theta = \frac{2 n \pi - \frac{\pi}{2}}{4}$
$\setminus \setminus = \frac{1}{8} \left(4 n - 1\right) \pi$
Put:
$n = - 1 \implies \theta = \frac{- 5 \pi}{8}$
$\text{ } \therefore z = 3 \left(\cos \left(\frac{- 5 \pi}{8}\right) + i \sin \left(\frac{- 5 \pi}{8}\right)\right)$
$\text{ } \therefore z = 3 \left(- \frac{\sqrt{2 - \sqrt{2}}}{2} - \frac{\sqrt{2 + \sqrt{2}}}{2} i\right)$
$n = 0 \implies \theta = \frac{- \pi}{8}$
$\text{ } \therefore z = 3 \left(\cos \left(\frac{- \pi}{8}\right) + i \sin \left(\frac{- \pi}{8}\right)\right)$
$\text{ } \therefore z = 3 \left(\frac{\sqrt{2 + \sqrt{2}}}{2} - \frac{\sqrt{2 - \sqrt{2}}}{2} i\right)$
$n = 1 \implies \theta = \frac{3 \pi}{8}$
$\text{ } \therefore z = 3 \left(\cos \left(\frac{3 \pi}{8}\right) + i \sin \left(\frac{3 \pi}{8}\right)\right)$
$\text{ } \therefore z = 3 \left(\frac{\sqrt{2 - \sqrt{2}}}{2} + \frac{\sqrt{2 + \sqrt{2}}}{2} i\right)$
$n = 2 \implies \theta = \frac{7 \pi}{8}$
$\text{ } \therefore z = 3 \left(\cos \left(\frac{7 \pi}{8}\right) + i \sin \left(\frac{7 \pi}{8}\right)\right)$
$\text{ } \therefore z = 3 \left(- \frac{\sqrt{2 + \sqrt{2}}}{2} + \frac{\sqrt{2 - \sqrt{2}}}{2} i\right)$
After which the pattern continues.
We can plot these solutions on the Argand Diagram |
# Systems of Equations Project
## What is a system of equations and what does the solution represent?
A system of equations is a set of two or more equations containing two or more variables. The solution represents an ordered pair that satisfies all the equations in the system. You can either eliminate, substitute, or even graph to figure out your solution. Your solution will either be no solution, infinitely many solutions, or one solution.
## Method of Graphing- 500 points
With the Method of Graphing in systems of equations, we graph both equations in the same coordinate system. The solution is the point where the two lines intersect. In the graph below where I worked out the solution, I found the slope and y-intercept, and made the lines so that they match the equation's slope. I also made sure its y-intercept matched.
## Substitution- 500 points
The goal when using substitution is to reduce the system to one equation that has only one variable. Then, we can solve the equation to find this variable, and then substitute the value in the other equation to find the other variable.
## Elimination- 500 points
Like Substitution, the goal of elimination is to get one equation that has only one variable. To do this by elimination, you add the two equations in the system together. Align all like terms together and then determine whether any like terms can be eliminated because they have opposite coefficients. Eliminate one of the variables and solve for the other variable. Use the value of the variable in one of the original equations and solve for the other variable.
## Word Problem- 500 points
A rental company charges a flat fee of x dollars for a floor sander rental plus y dollars per hour of the rental. One customer rents a floor sander for 4 hours and pays \$63. Another customer rents a floor sander for 6 hours and pays \$87. Find the flat fee and the cost per hour for the rental.
For this problem, I used the method of elimination.
## Write a System- 500 points
Here, I had to solve a word problem by making a system of equations based off of the problem. When I figured it out, I solved the System of Equations.
You pay \$24.50 for 10 gallons of gasoline and 1 quart of oil at a gas station. Your friend pays \$22 for 8 gallons of the same amount of gasoline and 2 quarts of the same oil. Find the cost of 1 quart of oil.
## When it's best to use Graphing
When both equations are solved for y (the slope-intercept form), and/or you want to estimate a solution, graphing would work best.
Example:
y=x+3
y=2x+4
Here, both equations are in slope-intercept form, so to graph them, we would find the slope and y-intercept of both equations, plug those in to make the graph, and find the intersection point of the two lines. Then, we find the solution by using the point of intersection.
## When it's best to use Substitution
Sometimes, it's difficult to find the exact solution by graphing. Look for a variable in either equation that has a coefficient of either 1 or -1. Then, substitution will work best, because we can easily decide which equation to solve for x or y.
Example:
2x+y=5
y=x-4
Here, the second equation shows what y is equal to, and in the top equation, the coefficient is 1 for y, therefore making it just y. You can substitute the bottom equation into the y in the top equation. Both equations are solved for the same variable, or either equation is solved for a variable.
## When it's best to use Elimination
When both equations have the same variable, with the same or opposite coefficients, or if the variable term in one equation is the multiple of the corresponding variable term in the other equation.
Example:
3x + 3y =15
2x - 3y = 5
Here, you can subtract the y values, and they cancel out, since the top is 3 and the bottom is -3. 3-3=0, so they cancel out. Now, you can find the x values, and solve for the y values once that part is complete. All you would need to do once that happens, is to substitute the x value that you found from elimination, and plug it into the equation to figure out y.
## 3 Different Types of Solutions
No Solution: If the system of equations have the same slope, but different y-intercepts, then we will not have an intersection point, so there won't be a solution. They'll be parallel lines on a graph.
Infinitely Many Solutions: If the system of equations has the same slope and same y-intercept, they will be on the exact same line, so there won't be just one particular intersection point.
One Solution: The two equations will have different slopes, therefore having only one intersection point on the graph. |
# AP SSC Class 10 Maths Chapter 4 Pair of Linear Equations in Two Variable
## Pair of Linear Equations in Two Variable
Two linear equations that have the same two variables are known as a pair of linear equations in two variables. Learn about Linear equations from AP SSC Class 10 Maths Chapter 4 Pair of Linear Equations in Two Variable.
$a_{1}x+b_{1}x+c_{1}=0\,(a_{1}^2+b_{1}^2\neq 0)$ $a_{2}x+b_{2}x+c_{2}=0\,(a_{2}^2+b_{2}^2\neq 0)$
where $a_1,a_2,b_1,b_2,c_1,c_2$ are all real numbers.
We use the following methods to find solutions to a pair of linear equations:
• Model Method
• Graphical Method
• Algebraic methods – Substitution method and Elimination method
There exists a relation between the coefficients and nature of the system of equations. Following are the relationship:
• $\frac{a_1}{a_2}\neq \frac{b_1}{b_2}$, then the pair of linear equations is consistent
• $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}$, then the pair of linear equations is inconsistent.
• $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$, then the pair of linear equations is dependent and consistent.
Let us look at a few solved questions from the chapter to better understand a pair of linear equations.
### Class 10 Maths Chapter 4 Pair of Linear Equations in Two Variable Solutions
1. Solve the following pair of equations by reducing them to a pair of linear equations.
$\frac{5}{x-1}+\frac{1}{y-2}=2$ $\frac{6}{x-1}+\frac{3}{y-2}=1$
Solution:
Let us consider the following
$\frac{1}{x-1}=u$ ……….(1)
$\frac{1}{y-2}=v$………….(2)
Hence, the equations becomes
$5u+v=2$…………..(3)
$6u+3v=1$…………(4)
From (1),
$v=2-5u$
Substituting the above in (4), we get
$6u-3(2-5u)=1$
Solving,
$6u-6+15u=1$ $21u-6=1$ $21u=7$ $u=\frac{7}{21}=\frac{1}{3}$
To find the value of v, substitute the above value of u in equation (3)
$5(\frac{1}{3})+v=2$ $\frac{5}{3}+v=2$ $v=2-\frac{5}{3}$ $v=\frac{6-5}{3}=\frac{1}{3}$
Hence,
$v=\frac{1}{3}$
To find the value of x and y, substitute the values of u and v in (1) and (2),
$u=\frac{1}{x-1}$ $\frac{1}{3}=\frac{1}{x-1}$ $x-1=3$ $x=4$ $v=\frac{1}{y-2}$ $\frac{1}{3}=\frac{1}{y-2}$ ${y-2}=3$ $y=5$
The value of x and y are 4 and 5 respectively for the given pair of equations.
2. A publisher is planning to produce a new textbook. The fixed costs (reviewing, editing, typesetting
and so on) are Rs, 31.25 per book. Besides that, he also spends another Rs, 320000 in producing the book. The wholesale price (the amount received by the publisher) is Rs, 43.75 per book. How many books must the publisher sell to break even, i.e., so that the cost will equal revenues?
Solutions:Â The publisher breaks even when costs are equal to the revenues. Given that “x” represents the number of books printed and sold and “y” is the breakeven point, then the cost and revenue equations for the publisher are
The Cost equation given by “y” = 320000 + 31.25x (1)
The Revenue equation also given by “y” = 43.75x (2)
Using the second equation and substituting for “y” in the first equation, we get
43.75x = 3,20,000 + 31.25x
12.5x = 3,20,000
x = 3, 20,000
12.5 = 25,600
Thus, the publisher will break even when 25,600 books are printed and sold.
Stay tuned to BYJU’S to get the latest notification on SSC exam along with AP SSC model papers, exam pattern, marking scheme and more. |
# What is a fraction? – Definition, Types & Examples
## Introduction
### What is a Fraction?
Fractions are an essential part of mathematics and are used to represent parts of a whole. Understanding fractions is crucial in various mathematical operations and real-life scenarios. Let’s explore the world of fractions and delve into their significance in solving problems.
## Analogy of Definition
### The Basics of Fractions
A fraction represents a part of a whole or a collection. It consists of two numbers separated by a line, where the number above the line is called the numerator, and the number below the line is called the denominator.
The numerator indicates the number of parts being considered, while the denominator represents the total number of equal parts that make up a whole.
### Types of Fractions
1. Proper Fraction: A proper fraction is a fraction where the numerator is less than the denominator. For example, 3/4, 2/5, and 5/8 are proper fractions.
2. Improper Fraction:An improper fraction is a fraction where the numerator is greater than or equal to the denominator. For example, 7/4, 9/5, and 11/8 are improper fractions.
3. Mixed Number: A mixed number is a combination of a whole number and a proper fraction. For example, 2\frac{1}{3}, 3\frac{2}{5}, and 4\frac{3}{8} are mixed numbers.
4. Unit Fraction: A unit fraction is a fraction where the numerator is 1. For example, 1/2, 1/3, and 1/4 are unit fractions.
5. Equivalent Fraction: Equivalent fractions are different fractions that represent the same value after reduction. For example, 1/2, 2/4, and 3/6 are equivalent fractions.
## Method
### Representing Fractions
Fractions can be represented in various ways, including on a number line, in decimal form, and as part of a whole. Understanding these representations is essential in comprehending the concept of fractions and their applications in different contexts.
Fraction as a part of whole
Fractions are generally represented as numerator on top of the denominator. The fraction 1/2 represents 1 part when a whole is divided into 2. Similarly, 1/5 represents 1 part when the whole is divided by 5.
Fractions on a Number Line
Fractions can also be represented on a number line.
Fractions as Decimal
We can also represent fractions in the form of decimals by dividing the numerator by the denominator.
Do you want to make learning fun? Visit our site ChimpVine to learn and play.
## Examples
Example 1: Represent 2/3 in the form of decimal
To convert 2/3 into decimal, we will have to divide 2 by 3, 2 ÷ 3 = 0.67
Example2: Segregate the fractions into proper and improper
2/3, 1/2, 2/7. 6/8, 3/2, 7/5, 8/3, 9/2
Proper Fractions: 2/3. 1/2, 6/8, 2/7
Impoper Fractions: 3/2, 7/5, 9/2, 8/3
## Tips and Tricks
1. Understanding Proper and Improper Fractions
Tip: A proper fraction has a numerator that is less than the denominator, while an improper fraction has a numerator that is greater than or equal to the denominator.
2. Converting Fractions to Decimals
Tip: To convert a fraction to a decimal, divide the numerator by the denominator. For example, 3/5 is equal to 0.6 when converted to a decimal.
3. Identifying Unit Fractions
Tip: A unit fraction has a numerator of 1. It represents the quantity of one part out of the total number of equal parts.
4. Comparing Mixed Numbers
Tip: When comparing mixed numbers, first compare the whole numbers. If the whole numbers are equal, compare the fractions to determine the greater value.
5. Exploring Equivalent Fractions
Tip: Equivalent fractions represent the same value, but they may look different. To find equivalent fractions, multiply or divide the numerator and denominator by the same number.
## Real life application
Real-Life Applications
Scenario: “Sharing Pizza Slices”
In a real-life scenario, fractions are used to divide and share items equally among a group of people. For example, when sharing a pizza, each person may receive 1/4 of the pizza, representing an equal portion of the whole.
Scenario: “Measuring Ingredients in Recipes”
In cooking and baking, fractions are used to measure ingredients accurately. For instance, a recipe may require 1/2 cup of flour or 3/4 teaspoon of salt, where the fractions represent specific quantities.
Scenario: “Distance on a Map”
When using a map, fractions are used to represent distances. For example, if a map scale indicates 1 inch represents 10 miles, a distance of 2.5 inches on the map represents 25 miles in real life.
## FAQ's
The different types of fractions include proper fractions, improper fractions, mixed numbers, and equivalent fractions.
A fraction consists of two parts: the numerator, which represents the number of parts being considered, and the denominator, which represents the total number of equal parts that make up a whole.
A unit fraction is a fraction where the numerator is 1. It represents the quantity of one part out of the total number of equal parts.
Fractions can be represented on a number line by dividing the line into equal parts based on the denominator and marking the position of the numerator.
To convert a fraction to a decimal, divide the numerator by the denominator. For example, 3/5 is equal to 0.6 when converted to a decimal. By understanding the concept of fractions, their types, representations, and real-life applications, individuals can develop a strong foundation in mathematics and problem-solving skills. Fractions play a crucial role in various mathematical operations and are essential in everyday activities, making them a fundamental aspect of learning and understanding mathematics.
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# Rearranging Formulae – why is it more confusing than Plantains and Bananas?
Well this is definitely not the favourite part of algebra for most of my GCSE Maths students, and as you are here, rearranging formulae has probably got you muddled too.
Did you know you have actually been rearranging formulae and equations since early primary school? Yep, it just looked very different.
Here’s an example:
3 + ? = 7.
What is the ?
Yep its 4.
You did that so fast, you need to stop and think about how you actually worked that out!
Let’s take a closer look:
To work out the ?, you did 7 – 3 = 4.
So you moved the 3 from one side to the other and you made it the opposite of what it is – change the side, change the sign.
I.e. Here: 3 + ? = 7 The 3 is a positive number. We could write it as
? + 3 = 7 and its the same thing.
We need the ? on its own on one side of the equals, and we do that by moving the 3 over to the same side as the 7. However it cannot be a +3 when you move it to the other side, because that won’t work.
That would give ? = 7 + 3, which is 10 – that’s incorrect.
So when you move it over you need to make it the opposite of what it is. It was a + 3 on the left of the equals, so it becomes a subtract 3 on the right hand side of the equals.
? = 7 – 3.
Another way to look at it is you need to subtract 3 from both sides.
? + 3 – 3 = 7 – 3
The +3 and the -3 on the left hand side, cancel out and this leaves
? = 7 – 3
so ? = 4.
It doesn’t really matter which way you think about it, choose whatever makes sense to you.
The important thing is to move things across and make them the opposite of what they were orginally.
Here are some more examples:
7 x ? = 21
We are multiplying by 7 on the left hand side, so move that ‘multiply by 7’ to the other side and change it into a ‘divide by 7’.
? = 21/7
? = 3
Here are some questions to try:
Question 1: Rearrange the equation `3x = 9` to solve for `x`.
`x = 9 / 3`
`x` `= 3`
Question 2: If `2y = 8`, find the value of `y`.
`y = 8 / 2`
`y` `= 4`
Question 3: Solve for `a` in the equation `4a = 12`.
`a = 12 / 4` ` `
a `= 3`
Question 4: Rearrange the equation `5z = 15` to solve for `z`.
`z = 15 / 5`
`z` `= 3`
Question 5: If `6m = 12`, what is the value of `m`?
`m = 12 / 6` ` `
m `= 2`
Question 6: Solve for `x` in the equation `2x = 8`.
`x = 8 / 2`
`x = 4`
Slightly harder:
Question 1: Rearrange the equation `2x + 3 = 11` to solve for `x`.
`2x = 11 - 3`
`2x = 8`
`x = 8 / 2`
`x = 4`
Question 2: If `4y - 7 = 5`, find the value of `y`.
`4y = 5 + 7`
`4y = 12`
`y = 12 / 4`
`y = 3`
Question 3: Solve for `a` in the equation `3a + 5 = 14`.
`3a = 14 - 5`
`3a = 9`
`a = 9 / 3`
`a = 3`
Question 4: Rearrange the equation `2(x + 3) = 10` to solve for `x`.
Answer 4: `2(x + 3) = 10`
`2x + 6 = 10`
`2x = 10 - 6`
`2x = 4`
`x = 4 / 2`
`x = 2`
OR
(x+3) = 10/2
x + 3 = 5
x = 5 – 2
x = 2
Question 5: If `2(2p - 1) = 14`, what is the value of `p`?
`2(2p - 1) = 14`
`4p - 2 = 14`
`4p = 14 + 2`
`4p = 16`
`p = 16 / 4`
`p = 4`
OR
(2p – 1) = 14/2
2p – 1 = 7
2p = 7 + 1
2p = 8
p = 8/4
p = 2
Question 6: Solve for `x` in the equation `5(x - 2) = 15`.
Answer 6: `5(x - 2) = 15`
`5x - 10 = 15`
`5x = 15 + 10`
`5x = 25`
`x = 25 / 5`
`x = 5`
OR
(x – 2) = 15/5
x – 2 = 3
x = 3 + 2
x = 5
Question 7: If `3(2q + 1) = 21`, find the value of `q`.
`3(2q + 1) = 21`
`6q + 3 = 21`
`6q = 21 - 3`
`6q = 18`
`q = 18 / 6`
`q = 3`
OR
(2q + 1) = 21/3
2q + 1 = 7
2q = 7 – 1
2q = 6
q = 6/2
q = 3 |
The basic component of a polynomial is a monomial. When we add or subtract polynomials, we are actually dealing with the addition and subtraction of individual monomials that are similar or alike.
## What is a Monomial?
A monomial can be a single number, a single variable, or the product of a number and one or more variables that contain whole number exponents.
This means that the exponents are neither negative nor fractional.
Examples:
• $7$
• $y$
• $2x$
• $– \,9xy$
• $– \,4x{y^2}$
• $10{x^2}{y^3}{z^4}$
• ${3 \over 4}{k^5}{m^2}h{r^{12}}$
So now we are ready to define what a polynomial is.
## What is a Polynomial?
A polynomial can be a single monomial or a combination of two or more monomials connected by the operations of addition and subtraction.
Examples:
A polynomial has “special” names depending on the number of monomials or terms in the expression. More so, the degree of a polynomial with a single variable is determined by the largest whole number exponent among the variables.
For example:
### Examples of How to Add and Subtract Polynomials
Example 1: Simplify by adding the polynomial expressions
The key in both adding and subtracting polynomials is to make sure that each polynomial is arranged in standard form. It means that the powers of the variables are in decreasing order from left to right.
Observe that each polynomial in this example is already in standard form, so we no longer need to perform that preliminary step.
Now, there are two ways we can proceed from here.
• First, we can add this the “usual” way, that is, add them horizontally.
I suggest that you first group similar terms in parenthesis before performing addition.
• Another way of simplifying this is to add them vertically.
Place the similar terms in the same column before performing addition.
As you can see, the answers in both methods came out to be the same!
Example 2: Simplify by adding the polynomial expressions
Notice that the first polynomial is already in the standard form because the exponents are in decreasing order. However, the second polynomial is not! We must first rearrange the powers of $x$ in decreasing order from left to right.
Similar or like terms are placed in the same parenthesis.
Similar or like terms are placed in the same column before performing the addition operation.
Example 3: Simplify by adding the polynomial expressions
Solution:
We are given two trinomials to add. But first, we have to “fix” each one of them by expressing it in standard form.
Add only similar terms. Notice that the $y$-variables with exponents $3$ and $2$ are placed in their own parenthesis to avoid the accidental addition of non-similar terms.
Let’s check our work if the answer comes out the same when we add them vertically.
Example 4: Simplify by adding the polynomial expressions
Solution:
Let’s add the polynomials above vertically. Align like terms in the same column then proceed with polynomial addition as usual.
Example 5: Simplify by subtracting the polynomial expressions
Subtracting polynomials is as easy as changing the operation to normal addition. However, always remember to also switch the signs of the polynomial being subtracted.
This is how it looks when we rewrite the original problem from subtraction to addition with some changes on the signs of each term of the second polynomial.
Original Problem
Rewritten Problem
• The original subtraction operation is replaced by addition.
• The second polynomial is “tweaked” by reversing the original sign of each term.
At this point, we can proceed with our normal addition of polynomials. Make sure that similar terms are grouped together inside a parenthesis.
Subtracting Polynomials – Horizontally
Subtracting Polynomials – Vertically
We can also subtract the polynomials in a vertical way. First, convert the original subtraction problem into its addition problem counterpart as shown by the green arrow. Make sure to align similar terms in a column before performing addition.
Example 6: Simplify by subtracting the polynomial expressions
The two polynomials that we are about to subtract are not in standard form. Begin by rearranging the powers of variable $x$ in decreasing order. Change the operation from subtraction to addition, align similar terms, and simplify to get the final answer.
Transform each polynomial in standard form
Subtract by switching the signs of the second polynomial, and then add them together.
Example 7: Simplify by subtracting the polynomial expressions
In this problem, we are going to perform the subtraction operation twice.
That means we also need to flip the signs of the two polynomials which are the second and third.
Perform regular addition using columns of similar or like terms.
Example 8: Simplify by adding and subtracting the polynomials
Solution:
Rewrite each polynomial in the standard format. Replace subtraction with addition while reversing the signs of the polynomial in question. Finally, organize like or similar terms in the same column and proceed with regular addition.
Adding the polynomials vertically, we have…
Example 9: Simplify by adding and subtracting the polynomials
Solution:
Rewrite each polynomial in the standard format. Replace subtraction with addition while reversing the signs of the polynomial in question. Finally, organize like or similar terms in the same column and proceed with regular addition.
If we add the polynomials vertically, we have…
You may also be interested in these related math lessons or tutorials: |
Explorations of Functions and Relations
By Mary Negley
For this exploration, I will look at the equation for various a values and use those graphs to predict higher a values.
First letŐs look at .
Notice how it is a circle of radius 1 centered at the origin.
Next letŐs look at in relation to the previous graph.
Notice how the function is tangent to the previous one at and, the function decreases in the second quadrant and fourth quadrant, and it curves around the previous graph in the first quadrant.
Now letŐs add to our graph.
In this case, the new function is centered at the origin just like the first function and it is tangent to the first function at,. Instead of being a circle like the first equation, the corners are slightly rounded. Notice how for each, the y-value for the third equation is greater than the y-value for the second equation.
Finally letŐs add to our graph.
This new function is similar to the second function. Like the second function, it decreases in the second and fourth quadrants and it curves around the first function in the first quadrant. Moreover for each, the y-value for the third equation is greater than the y-values for the other equations.
Now I must predict what the graphs of and look like.
Above I noticed that the graph of is tangent to the graph of at four points, but it resembles a square with rounded corners. Based on the graphs of and, I would predict that would resemble a square even more than , but it would still be tangent to at the same four points as . Below is the graph of .
Above I noticed that and were tangent to at two points. They both also decreased in the second and fourth quadrants and curved around in the first quadrant. Notice that when and , the y-value for is less than the y-value for , but when , the y-value for is greater than the y-value for. I would predict that when and , the y-value for would be less than the y-value for, but when , the y-value for would be greater than the y-value for. Now letŐs add to our graphs.
In both cases, my predictions were true. |
Finding a Point on a Bézier Curve: De Casteljau's Algorithm
Following the construction of a Bézier curve, the next important task is to find the point C(u) on the curve for a particular u. A simple way is to plug u into every basis function, compute the product of each basis function and its corresponding control point, and finally add them together. While this works fine, it is not numerically stable (i.e., could introduce numerical errors during the course of evaluating the Bernstein polynomials).
In what follows, we shall only write down the control point numbers. That is, the control points are 00 for P0, 01 for P1, ..., 0i for Pi, ..., 0n for Pn. The 0s in these numbers indicate the initial or the 0-th iteration. Later on, it will be replaced with 1, 2, 3 and so on.
The fundamental concept of de Casteljau's algorithm is to choose a point C in line segment AB such that C divides the line segment AB in a ratio of u:1-u (i.e., the ratio of the distance between A and C and the distance between A and B is u). Let us find a way to determine the point C.
The vector from A to B is B - A. Since u is a ratio in the range of 0 and 1, point C is located at u(B - A). Taking the position of A into consideration, point C is A + u(B - A) = (1 - u)A + uB. Therefore, given a u, (1 - u)A + uB is the point C between A and B that divides AB in a ratio of u:1-u.
The idea of de Casteljau's algorithm goes as follows. Suppose we want to find C(u), where u is in [0,1]. Starting with the first polyline, 00-01-02-03...-0n, use the above formula to find a point 1i on the leg (i.e. line segment) from 0i to 0(i+1) that divides the line segment 0i and 0(i+1) in a ratio of u:1-u. In this way, we will obtain n points 10, 11, 12, ...., 1(n-1). They define a new polyline of n - 1 legs.
In the figure above, u is 0.4. 10 is in the leg of 00 and 01, 11 is in the leg of 01 and 02, ..., and 14 is in the leg of 04 and 05. All of these new points are in blue.
The new points are numbered as 1i's. Apply the procedure to this new polyline and we shall get a second polyline of n - 1 points 20, 21, ..., 2(n-2) and n - 2 legs. Starting with this polyline, we can construct a third one of n - 2 points 30, 31, ..., 3(n-3) and n - 3 legs. Repeating this process n times yields a single point n0. De Casteljau proved that this is the point C(u) on the curve that corresponds to u.
Let us continue with the above figure. Let 20 be the point in the leg of 10 and 11 that divides the line segment 10 and 11 in a ratio of u:1-u. Similarly, choose 21 on the leg of 11 and 12, 22 on the leg of 12 and 13, and 23 on the leg of 13 and 14. This gives a third polyline defined by 20, 21, 22 and 23. This third polyline has 4 points and 3 legs. Keep doing this and we shall obtain a new polyline of three points 30, 31 and 32. From this fourth polyline, we have the fifth one of two points 40 and 41. Do it once more, and we have 50, the point C(0.4) on the curve.
This is the geometric interpretation of de Casteljau's algorithm, one of the most elegant result in curve design.
Actual Computation
Given the above geometric interpretation of de Casteljau's algorithm, we shall present a computation method, which is shown in the following figure.
First, all given control points are arranged into a column, which is the left-most one in the figure. For each pair of adjacent control points, draw a south-east bound arrow and a north-east bound arrow, and write down a new point at the intersection of the two adjacent arrows. For example, if the two adjacent points are ij and i(j+1), the new point is (i+1)j. The south-east (resp., north-east) bound arrow means multiplying 1 - u (resp., u) to the point at its tail, ij (resp., i(j+1)), and the new point is the sum.
Thus, from the initial column, column 0, we compute column 1; from column 1 we obtain column 2 and so on. Eventually, after n applications we shall arrive at a single point n0 and this is the point on the curve. The following algorithm summarizes what we have discussed. It takes an array P of n+1 points and a u in the range of 0 and 1, and returns a point on the Bézier curve C(u).
Input: array P[0:n] of n+1 points and real number u in [0,1]
Output: point on curve, C(u)
Working: point array Q[0:n]
for i := 0 to n do
Q[i] := P[i]; // save input
for k := 1 to n do
for i := 0 to n - k do
Q[i] := (1 - u)Q[i] + u Q[i + 1];
return Q[0];
A Recurrence Relation
The above computation can be expressed recursively. Initially, let P0,j be Pj for j = 0, 1, ..., n. That is, P0,j is the j-th entry on column 0. The computation of entry j on column i is the following:
More precisely, entry Pi,j is the sum of (1-u)Pi-1,j (upper-left corner) and uPi-1,j+1 (lower-left corner). The final result (i.e., the point on the curve) is Pn,0. Based on this idea, one may immediately come up with the following recursive procedure:
function deCasteljau(i,j)
begin
if i = 0 then
return P0,j
else
return (1-u)* deCasteljau(i-1,j) + u* deCasteljau(i-1,j+1)
end
This procedure looks simple and short; however, it is extremely inefficient. Here is why. We start with a call to deCasteljau(n,0) for computing Pn,0. The else part splits this call into two more calls, deCasteljau(n-1,0) for computing Pn-1,0 and deCasteljau(n-1,1) for computing Pn-1,1.
Consider the call to deCasteljau(n-1,0). It splits into two more calls, deCasteljau(n-2,0) for computing Pn-2,0 and deCasteljau(n-2,1) for computing Pn-2,1. The call to deCasteljau(n-1,1) splits into two calls, deCasteljau(n-2,1) for computing Pn-2,1 and deCasteljau(n-2,2) for computing Pn-2,2. Thus, deCasteljau(n-2,1) is called twice. If we keep expanding these function calls, we should discover that almost all function calls for computing Pi,j are repeated, not once but many times. How bad is this? In fact, the above computation scheme is identical to the following way of computing the n-th Fibonacci number:
function Fibonacci(n)
begin
if n = 0 or n = 1 then
return 1
else
return Fibonacci (n-1) + Fibonacci (n-2)
end
This program takes an exponential number of function calls (an exercise) to compute Fibonacci(n). Therefore, the above recursive version of de Casteljau's algorithm is not suitable for direct implementation, although it looks simple and elegant!
An Interesting Observation
The triangular computation scheme of de Casteljau's algorithm offers an interesting observation. Take a look at the following computation on a Bézier curve of degree 7 defined by 8 control points 00, 01, ..., 07. Let us consider a set of consecutive points on the same column as the control points of a Bézier curve. Then, given a u in [0,1], how do we compute the corresponding point on this Bézier curve? If de Casteljau's algorithm is applied to these control points, the point on the curve is the opposite vertex of the equilateral's base formed by the selected points!
For example, if the selected points are 02, 03, 04 and 05, the point on the curve defined by these four control points that corresponds to u is 32. See the blue triangle. If the selected points are 11, 12 and 13, the point on the curve is 31. See the yellow triangle. If the selected points are 30, 31, 32, 33 and 34, the point on the curve is 70.
By the same reason, 70 is the point on the Bézier curve defined by control points 60 and 61. It is also the point on the curve defined by 50, 51 and 52, and on the curve defined by 40, 41, 42 and 43. In general, if we select a point and draw an equilateral as shown above, the base of this equilateral consists of the control points from which the selected point is computed. |
# Networks and Graphs
17 June 2020
A graph is simply a collection of:
• Vertices or nodes
• Edges between vertices
For example, the graph in the image below has 6 vertices (labelled 1 to 6), and 7 edges.
Graphs are also sometimes called networks. We usually use “network” for very large graphs with many vertices (e.g. a social network, or a transport network), but there’s no fixed rule on when to use “graph” and when to use “network”. Both are fine.
# Graphs and Matrices
At first sight, it may seem like graphs have nothing to do with matrices: graphs are like drawings, while matrices are tables of numbers.
But in fact, any graph has an adjacency matrix that completely defines the graph.
The adjacency matrix $A$ of a graph $G$ is an $n \times n$ matrix where:
• $n$ is the number of vertices in $G$ (assume that the vertices are labelled 1,2,… $n$)
• $A_{ij} = 1$ if there is an edge between vertex $i$ and $j$
• $A_{ij} = 0$ if there is no edge between vertix $i$ and $j$
So the adjacency matrix tells us which vertices are adjacent to each other.
The adjacency matrix of the graph above is given by:
$\left(\begin{matrix}0 & 1 & 0 & 0 & 1 & 0\\1 & 0 & 1 & 0 & 1 & 0\\0 & 1 & 0 & 1 & 0 & 0\\0 & 0 & 1 & 0 & 1 & 1\\1 & 1 & 0 & 1 & 0 & 0\\0 & 0 & 0 & 1 & 0 & 0\end{matrix}\right)$
You can check this by going through the matrix row by row. Row $i$ tells you which other vertices you can reach from vertex $i$, by following an edge.
# Directed and undirected graphs
You may have noticed that the matrix above is symmetric. This is because the graph above is an undirected graph, which just means that the edges have no “direction”. For example, since $4$ and $6$ are connected by an edge, both $A_{46}$ and $A_{64}$ are set to $1$.
A directed graph is a collection of:
• Vertices or nodes
• Directed edges or arrows between vertices
Here’s an example of a directed graph that we saw in our first lecture, and its adjacency matrix:
Row $i$ of the matrix tells you which other vertices you can get to from vertex $i$ by following an arrow.
Some things to note:
• The edges now have directions, indicated by the arrow head.
• The adjacency matrix is no longer symmetric.
Notice also that vertices 2 and 3 are connected by an arrow with two heads, so the $(2,3)$ and $(3,2)$ entries of the adjacency matrix are both $1$. This is similar to having an undirected edge between them in an undirected graph.
In fact, undirected graphs are special cases of directed graphs, where every edge is treated as an “arrow with two heads”.
You can think of an undirected graph as a network of two-way roads, while a directed graph is a network of one-way roads with the arrows indicating the direction.
# Homework
### Question 1
Here are some situations that can be represented with graphs.
1. Friend network
2. Family network
3. Webpages and links between them
Come up with a situation that can be represented with graphs, that is not in the above list.
### Question 2
Choose two of the situations from Question 1 (you can choose the situation you came up with). For each situation, answer the following questions:
1. What are the vertices of the graph?
2. Is it directed or undirected?
3. When is there an edge from one vertex to another?
There may not be one right answer for these questions. It all depends on how you want to represent the situation.
### Question 3
For the situations that you chose in Question 2, draw a simple example of a graph that represents it, and its corresponding adjacency matrix. (An example will do: You don’t have to draw your whole friend network, or include real names!)
### Bonus Question
This is just something for you to think about. Suppose you want to use a graph to represent JC students and the subjects they are taking. How would you do it with graphs? (If you did come up with a graph, you can use this for Questions 1,2,3 if you want.)
### Discussion Question
This is for discussion during our next Zoom meeting: What situation would you be interested in representing with graphs? This does not need to be any of the above situations. You can also think about what situations you might want to study for SSEF. |
RD Sharma Class 9 Factorisation of Polynomials Exercise 6.2 Solutions
On this page you will find Maths RD Sharma Class 9 Factorisation of Polynomials Exercise 6.2 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 6 for class 9 deals with the topic of triangles and it is one of the most important chapters.
Download RD Sharma Class 9 Factorisation of Polynomials Exercise 6.2 Solutions in PDF
Students can use these solutions to overcome the fear of maths and the solutions have been designed in such a way that it enables them to discover easy ways to solve different problems. These solutions can help students in refining their maths fluency and problem-solving skills. Students can go through the RD Sharma solutions for class 9 chapter 6 Factorisation of Polynomials Exercise 6.2 below and it will be beneficial for them.
RD Sharma Class 9 Factorisation of Polynomials Exercise 6.2 Solutions
Q1. If $\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}-13 \mathrm{x}^{2}+17 \mathrm{x}+12$, Find
1. $f(2)$
2. $f(-3)$
3. $f(0)$
Sol:
The given polynomial is $\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}-13 \mathrm{x}^{2}+17 \mathrm{x}+12$
1. $f(2)$
we need to substitute the ‘ 2 ‘ in $\mathrm{f}(\mathrm{x})$
$\mathrm{f}(2)=2(2)^{3}-13(2)^{2}+17(2)+12$
$=(2 * 8)-(13 * 4)+(17 * 2)+12$
$=16-52+34+12$
$=10$
therefore $f(2)=10$
2. $f(-3)$
we need to substitute the $^{\prime}(-3)$ ‘ in $f(x)$
$f(-3)=2(-3)^{3}-13(-3)^{2}+17(-3)+12$
$=(2 *-27)-(13 * 9)-(17 * 3)+12$
$=-54-117-51+12$
$=-210$
therefore $f(-3)=-210$
3. $f(0)$
we need to substitute the $^{\prime}(0)^{\prime}$ in $f(x)$
$\mathrm{f}(0)=2(0)^{3}-13(0)^{2}+17(0)+12$
$=(2 * 0)-(13 * 0)+(17 * 0)+12$
$=0-0+0+12$
$=12$
therefore $f(0)=12$
Q2. Verify whether the indicated numbers are zeros of the polynomial corresponding to them in the following cases:
1. $\mathrm{f}(\mathrm{x})=3 \mathrm{x}+1, \mathrm{x}=\frac{-1}{3}$
2. $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}-1, \mathrm{x}=(1,-1)$
3. $\mathrm{g}(\mathrm{x})=3 \mathrm{x}^{2}-2, \mathrm{x}=\left(\frac{2}{\sqrt{3}}, \frac{-2}{\sqrt{3}}\right)$
4. $\mathrm{p}(\mathrm{x})=\mathrm{x}^{3}-6 \mathrm{x}^{2}+11 \mathrm{x}-6, \mathrm{x}=\mathbf{1}, \mathbf{2}, \mathbf{3}$
5. $\mathrm{f}(\mathrm{x})=5 \mathrm{x}-\pi, \mathrm{x}=\frac{4}{5}$
6. $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}, \mathbf{x}=\mathbf{0}$
7. $\mathrm{f}(\mathrm{x})=\mathrm{lx}+\mathrm{m}, \mathrm{x}=\frac{-\mathrm{m}}{\mathrm{l}}$
8. $\mathrm{f}(\mathrm{x})=2 \mathrm{x}+1, \mathrm{x}=\frac{1}{2}$
Sol:
$(1) \mathrm{f}(\mathrm{x})=3 \mathrm{x}+1, \mathrm{x}=\frac{-1}{3}$
we know that,
$f(x)=3 x+1$
substitute $x=\frac{-1}{3}$ in $f(x)$
$f\left(\frac{-1}{3}\right)=3\left(\frac{-1}{3}\right)+1$
$=-1+1$
$=0$
Since, the result is $0 x=\frac{-1}{3}$ is the root of $3 x+1$
(2) $f(x)=x^{2}-1, x=(1,-1)$
we know that,
$f(x)=x^{2}-1$
Given that $x=(1,-1)$
substitute $x=1$ in $f(x)$
$f(1)=1^{2}-1$
$=1-1$
$=0$
Now, substitute $x=(-1)$ in $f(x)$
$f(-1)=(-1)^{2}-1$
$=1-1$
$=0$
Since, the results when $x=(1,-1)$ are 0 they are the roots of the polynomial $f(x)=x^{2}-1$
(3) $\mathrm{g}(\mathrm{x})=3 \mathrm{x}^{2}-2, \mathrm{x}=\left(\frac{2}{\sqrt{3}}, \frac{-2}{\sqrt{3}}\right)$
Sol:
We know that
$g(x)=3 x^{2}-2$
Given that, $x=\left(\frac{2}{\sqrt{3}}, \frac{-2}{\sqrt{3}}\right)$
Substitute $x=\frac{2}{\sqrt{3}}$ in $g(x)$
$g\left(\frac{2}{\sqrt{3}}\right)=3\left(\frac{2}{\sqrt{3}}\right)^{2}-2$
$=3\left(\frac{4}{3}\right)-2$
$=4-2$
$=2 \neq 0$
Now, Substitute $x=\frac{-2}{\sqrt{3}}$ in $g(x)$
$g\left(\frac{-2}{\sqrt{3}}\right)=3\left(\frac{-2}{\sqrt{3}}\right)^{2}-2$
$=3\left(\frac{4}{3}\right)-2$
$=4-2$
$=2 \neq 0$
Since, the results when $x=\left(\frac{2}{\sqrt{3}}, \frac{-2}{\sqrt{3}}\right)$ are not 0, they are roots of $3 x^{2}-2$
(4) $p(x)=x^{3}-6 x^{2}+11 x-6, x=1,2,3$
Sol:
We know that,
$p(x)=x^{3}-6 x^{2}+11 x-6$
given that the values of $x$ are $1,2,3$
substitute $x=1$ in $p(x)$
$p(1)=1^{3}-6(1)^{2}+11(1)-6$
$=1-(6 \star 1)+11-6$
$=1-6+11-6$
$=0$
Now, substitute $x=2$ in $p(x)$
$P(2)=2^{3}-6(2)^{2}+11(2)-6$
$=(2 * 3)-(6 * 4)+(11 * 2)-6$
$=8-24-22-6$
$=0$
Now, substitute $x=3$ in $p(x)$
$P(3)=3^{3}-6(3)^{2}+11(3)-6$
$=(3 * 3)-(6 * 9)+(11 * 3)-6$
$=27-54+33-6$
$=0$
Since, the result is 0 for $x=1,2,3$ these are the roots $o f x^{3}-6 x^{2}+11 x-6$
(5) $\mathrm{f}(\mathrm{x})=5 \mathrm{x}-\pi, \mathrm{x}=\frac{4}{5}$
we know that,
$f(x)=5 x-\pi$
Given that, $x=\frac{4}{5}$
Substitute the value of $x$ in $f(x)$
$f\left(\frac{4}{5}\right)=5\left(\frac{4}{5}\right)-\pi$
$=4-\pi$
$\neq 0$
Since, the result is not equal to zero, $x=\frac{4}{5}$ is not the root of the polynomial $5 x-\pi$
(6) $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}, \mathrm{x}=0$
Sol:
we know that, $f(x)=x^{2}$
Given that value of $x$ is ‘ 0 ‘
Substitute the value of $x$ in $f(x)$
$f(0)=0^{2}$
$=0$
Since, the result is zero, $x=0$ is the root of $x^{2}$
(7) $\mathrm{f}(\mathrm{x})=\mathrm{lx}+\mathrm{m}, \mathrm{x}=\frac{-\mathrm{m}}{\mathrm{l}}$
Sol:
We know that,
$f(x)=\mid x+m$
Given, that $x=\frac{-m}{1}$
Substitute the value of $x$ in $f(x)$
$f\left(\frac{-m}{1}\right)=1\left(\frac{-m}{1}\right)+m$
$=-m+m$
$=0$
Since, the result is $0, x=\frac{-m}{1}$ is the root of $I x+m$
(8) $\mathrm{f}(\mathrm{x})=2 \mathrm{x}+1, \mathrm{x}=\frac{1}{2}$
Sol:
We know that,
$f(x)=2 x+1$
Given that $\mathrm{x}=\frac{1}{2}$
Substitute the value of $x$ and $f(x)$
$f\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)+1$
$=1+1$
$=2 \neq 0$
Since, the result is not equal to zero
$x=\frac{1}{2}$ is the root of $2 x+1$
Q3. If $x=2$ is a root of the polynomial $f(x)=2 x^{2}-3 x+7 a$, Find the value of $a$
Sol:
We know that, $f(x)=2 x^{2}-3 x+7 a$
Given that $x=2$ is the root of $f(x)$
Substitute the value of $x$ in $f(x)$
$f(2)=2(2)^{2}-3(2)+7 a$
$=(2 * 4)-6+7 a$
$=8-6+7 a$
$=7 a+2$
Now, equate $7 a+2$ to zero
$\Rightarrow 7 a+2=0$
$\Rightarrow 7 a=-2$
$\Rightarrow \mathrm{a}=\frac{-2}{7}$
The value of $\mathrm{a}=\frac{-2}{7}$
Q4. If $x=\frac{-1}{2}$ is zero of the polynomial $p(x)=8 x^{3}-a x^{2}-x+2$, Find the value of a
Sol:
We know that, $p(x)=8 x^{3}-a x^{2}-x+2$
Given that the value of $x=\frac{-1}{2}$
Substitute the value of $x$ in $f(x)$
$p\left(\frac{-1}{2}\right)=8\left(\frac{-1}{2}\right)^{3}-a\left(\frac{-1}{2}\right)^{2}-\left(\frac{-1}{2}\right)+2$
$=-8\left(\frac{1}{8}\right)-a\left(\frac{1}{4}\right)+\frac{1}{2}+2$
$=-1-\left(\frac{a}{4}+\frac{1}{2}+2\right.$
$=1-\left(\frac{a}{4}+\frac{1}{2}\right.$
$=\frac{3}{2}-\frac{a}{4}$
To, find the value of a , equate $\mathrm{p}\left(\frac{-1}{2}\right)$ to zero
$p\left(\frac{-1}{2}\right)=0$
$\frac{3}{2}-\frac{a}{4}=0$
On taking L.C.M
$\frac{6-a}{4}=0$
$\Rightarrow 6-a=0$
$\Rightarrow \quad a=6$
Q5. If $x=0$ and $x=-1$ are the roots of the polynomial $f(x)=2 x^{3}-3 x^{2}+a x+b$, Find the of $a$ and $b$.
Sol:
We know that, $f(x)=2 x^{3}-3 x^{2}+a x+b$
Given , the values of $x$ are 0 and $-1$
Substitute $x=0$ in $f(x)$
$f(0)=2(0)^{3}-3(0)^{2}+a(0)+b$
$=0-0+0+b$
$=b \quad—1$
Substitute $x=(-1)$ in $f(x)$
$f(-1)=2(-1)^{3}-3(-1)^{2}+a(-1)+b$
$=-2-3-a+b$
$=-5-a+b \quad—–2$
We need to equate equations 1 and 2 to zero
$b=0$ and $-5-a+b=0$
since, the value of $b$ is zero
substitute $\mathrm{b}=0$ in equation 2
$\Rightarrow-5-a=-b$
$\Rightarrow-5-a=0$
$a=-5$
the values of $\mathrm{a}$ and $\mathrm{b}$ are $-5$ and 0 respectively
Q6. Find the integral roots of the polynomial $f(x)=x^{3}+6 x^{2}+11 x+6$
Sol:
Given, that $f(x)=x^{3}+6 x^{2}+11 x+6$
Clearly we can say that, the polynomial $\mathrm{f}(\mathrm{x})$ with an integer coefficient and the highest degree term coefficient which is known as leading factor is 1 .
So, the roots of $f(x)$ are limited to integer factor of 6 , they are
$\pm 1, \pm 2, \pm 3, \pm 6$
Let $x=-1$
$f(-1)=(-1)^{3}+6(-1)^{2}+11(-1)+6$
$=-1+6-11+6$
$=0$
Let $x=-2$
$f(-2)=(-2)^{3}+6(-2)^{2}+11(-2)+6$
$=-8-(6 * 4)-22+6$
$=-8+24-22+6$
$=0$
Let $x=-3$
$f(-3)=(-3)^{3}+6(-3)^{2}+11(-3)+6$
$=-27-(6 \star 9)-33+6$
$=-27+54-33+6$
$=0$
But from all the given factors only $-1,-2,-3$ gives the result as zero.
So, the integral multiples of $\mathrm{x}^{3}+6 \mathrm{x}^{2}+11 \mathrm{x}+6$ are $-1,-2,-3$
Q7. Find the rational roots of the polynomial $f(x)=2 x^{3}+x^{2}-7 x-6$
Sol:
Given that $f(x)=2 x^{3}+x^{2}-7 x-6$
$f(x)$ is a cubic polynomial with an integer coefficient . If the rational root in the form of $\frac{p}{q}$, the values of $p$ are
limited to factors of 6 which are $\pm 1, \pm 2, \pm 3, \pm 6$
and the values of $q$ are limited to the highest degree coefficient i.e 2 which are $\pm 1, \pm 2$
here, the possible rational roots are
$\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$
Let, $x=-1$
$f(-1)=2(-1)^{3}+(-1)^{2}-7(-1)-6$
$=-2+1+7-6$
$=-8+8$
$=0$
Let, $x=2$
$f(-2)=2(2)^{3}+(2)^{2}-7(2)-6$
$=(2 * 8)+4-14-6$
$=16+4-14-6$
$=20-20$
$=0$
Let, $x=\frac{-3}{2}$
$f\left(\frac{-3}{2}\right)=2\left(\frac{-3}{2}\right)^{3}+\left(\frac{-3}{2}\right)^{2}-7\left(\frac{-3}{2}\right)-6$
$=2\left(\frac{-27}{8}\right)+\frac{9}{4}-7\left(\frac{-3}{2}\right)-6$
$=\left(\frac{-27}{4}\right)+\frac{9}{4}-\left(\frac{-21}{2}\right)-6$
$=-6.75+2.25+10.5-6$
$=12.75-12.75$
$=0$
But from all the factors only $-1,2$ and $\frac{-3}{2}$ gives the result as zero
So, the rational roots of $2 \mathrm{x}^{3}+\mathrm{x}^{2}-7 \mathrm{x}-6$ are $-1,2$ and $\frac{-3}{2}$
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Video: GCSE Mathematics Foundation Tier Pack 3 • Paper 1 • Question 7
GCSE Mathematics Foundation Tier Pack 3 • Paper 1 • Question 7
03:26
Video Transcript
Find three-fifths of 80.
So what we’re gonna do to be able to solve this problem I’m gonna show you two methods. First of all, let’s have a look at three-fifths of 80, what does it mean. Well, it actually means three-fifths times 80.
Now for the first method, what we can do is write it as 80 multiplied by three-fifths. So what we’re gonna do is we split it up. We split it up into 80 multiplied by three multiplied by a fifth. And the reason we can do that is because, actually, three multiplied by a fifth or three multiplied by one over five is gonna be equal to three-fifths or three over five. And that’s because if you have a fifth and then you multiply it by three and have three of them, it will give you three-fifths.
Well, if we look at the first bit of the calculation first, we’ve got 80 multiplied by three, or eight threes are 24, and then we add on the zero, so we get 240. And then this is multiplied by a fifth, which is gonna give us 48. And we get 48 because 240 multiplied by a fifth would give us 240, because I actually multiplied 240 by the one, because that was the numerator, over five. And then we can actually work this out by either dividing 240 by five or with this method I’m about to show you now.
Well, to make life easier, what we could do is actually multiply the numerator and denominator by two. And that’s just gonna make the denominator 10. And this will give us 480 over 10. And then we divide the numerator and denominator by 10, and we’re left with 48 over one, which is 48. So therefore, we can say that three-fifths of 80 is equal to 48. Okay, so that’s the first method.
Now I said I’ll show you another method. And I actually think the second method I’m about to show you some people do find a bit easier. But as with everything, it’s personal preference, and that’s why we show both methods so you can decide which one you would like to use.
So now for the second method, if we’re asked to find three-fifths of 80, now to find one-fifth of 80, which is what I want to do first, we actually divide 80 by five, which is going to be equal to 16. And we could’ve worked that out using the bus stop method for division.
So we do fives into eight. This goes once, remainder three. So I’m gonna carry the three, and then we do fives into 30, which is six. So therefore, we get 16. Okay, great! So that’s one-fifth. So what we did is we divided 80 by five. So now what do we do?
Well, now what we’re gonna do is find three-fifths. And to do that, we do 16 multiplied by three, which is going to be equal to 48 again. And to show you how we work this out, we could use column multiplication.
So first of all, we have three multiplied by six, which gives us 18. So we put eight in the units column, carry the one into the tens column. And then we have three multiplied by one, which is three. Then add the one that we had that we carried from earlier gives us four. So we get 48. So there we’ve got it, the same answer in both methods.
And just to recap the second method, so to find three-fifths of 80, what we did was we divided by the denominator, by five, to give us one-fifth, which was 16. Then we multiplied by the numerator, which was three, to give us 48. So therefore, we can say that three-fifths of 80 is equal to 48. |
# AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1
AP State Syllabus AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1 Textbook Questions and Answers.
## AP State Syllabus 9th Class Maths Solutions 3rd Lesson The Elements of Geometry Exercise 3.1
Question 1.
i) How many dimensions a solid has ?
Solution:
A solid has three dimensions namely length, breadth and height or depth.
ii) How many books are there in Euclid’s Elements ?
Solution:
There are 13 volumes in Euclid’s elements.
iii) Write the number of faces of a cube and cuboid.
Solution:
Cube : 6 faces
Cuboid : 6 faces
iv) What is the sum of interior angles of a triangle ?
Solution:
The sum of interior angles of a triangle is 180°.
v) Write three undefined terms of geometry.
Solution:
Point, line and plane are three undefined terms in geometry.
Question 2.
State whether the following statements are true or false. Also give reasons for your answers.
a) Only one line can pass through a given point
b) All right angles are equal
c) Circles with same radii are equal
d) A finite line can be extended on its both sides endlessly to get a straight line
e) From figure AB > AC
Solution:
a) Only one line can pass through a given point – False.
Reason : (Since, infinitely many lines can pass through a given point)
b) All right angles are equal – True.
c) Circles with same radii are equal – True.
d) A finite line can be extended on its both sides endlessly to get a straight line – True.
e) From figure AB > AC – True.
Question 3.
In the figure given below, show that the length AH > AB + BC + CD.
Solution:
Given a line $$\stackrel{\leftrightarrow}{\mathrm{AH}}$$
To prove AH > AB + BC + CD
From the figure AB + BC + CD = AD
AD is a part of whole AH.
From Euclid’s axiom whole is greater than part.
⇒ AH > AB + BC + CD
Question 4.
If a point Q lies between two points P and R such PQ = QR, prove that PQ = $$\frac{1}{2}$$PR.
Let PR be a given line.
Given that PQ = QR
i. e., Q is a point on PR.
⇒ PQ + QR = PR
⇒ PQ + PQ = PR [∵ PQ = QR]
⇒ 2PQ = PR
⇒ PQ = $$\frac{1}{2}$$ PR
Hence proved.
Question 5.
Draw an equilateral triangle whose sides are 5.2 cm
Soluton:
Step – 1 : Draw a line segment AB of length 5.2 cm. *
Step – 2 : Draw an arc of radius 5.2 cm with centre A.
Step – 3 : Draw an arc of radius 5.2 cm with centre B.
Step – 4 : Two arcs intersect at C; join C to A and B.
Δ ABC is the required triangle.
Question 6.
What is a conjecture? Give an example for it.
Solution:
Mathematical statements which are neither proved nor disproved are called conjectures. Mathematical discoveries often start out as conjectures. This may be an educated guess based on observations.
Eg : Every even number greater than 4 can be written as sum of two primes. This example is called Gold Bach Conjecture
Question 7.
Mark two points P and Q. Draw a line through P and Q. Now how many lines are parallel to PQ, can you draw ?
Solution:
Infinitely many lines parallel to PQ can be drawn.
Question 8.
In the figure given below, a line n falls on lines / and m such that the sum of the interior angles 1 and 2 is less than 180°, then what can you say about lines l and m ?
Solution:
Given : l, m and n are lines, n is a transversal.
∠1 < 90°
∠2 < 90°
If the lines l and m are produced on the side where angles 1 and 2 are formed, they intersect at one point.
Question 9.
In the figure given below, if ∠1 = ∠3, ∠2 = ∠4 and ∠3 = ∠4 write the rela-tion between ∠1 and ∠2 using Euclid’s postulate.
Solution :
Given : ∠1 = ∠3
∠3 = ∠4
∠2 = ∠4
∴∠1 = ∠2
∵Both ∠1 and ∠2 are equal to ∠4. (By Euclid’s axiom things which are equal to same things are equal to one another).
Question 10.
In the figure given below, we have BX = $$\frac{1}{2}$$ AB, BY= $$\frac{1}{2}$$ BC and AB = BC. Show that BX = BY.
Solution:
Given : BX = $$\frac{1}{2}$$ AB
BY = $$\frac{1}{2}$$BC
AB = BC
To prove : BX = BY
Proof: Given AB = BC [ ∵ By Euclid’s axiom things which are halves of the same things are equal to one another]
$$\frac{1}{2}$$ AB = $$\frac{1}{2}$$ BC
BX = BY
Hence proved. |
### Subtracting a Larger Number from a Smaller Number
It is easy to compute 6 - 4 = 2 and 14 - 5 = 9, because the answers are positive. But how would one compute 4 - 6? On the number line, moving to the left is equivalent to subtracting from a number, and moving to the right is equivalent to adding to a nu mber. To see what 4 - 6 is equal to, start on the number line at the place marked "4" and count 6 spaces to the left:
Thus, 4 - 6 = - 2.
Notice the similarity between 6 - 4 = 2 and 4 - 6 = - 2. 6 - 4 is the opposite of 4 - 6. To find 4 - 6, simply find 6 - 4 and take its opposite. In fact, this works for any two numbers: to subtract a larger number from a smaller number, reverse the two numbers and take the opposite of the answer. 5 - 14 = - (14 - 5) = - 9.
To subtract a positive number from a negative number, switch both signs, and take the opposite of the answer. -5 - 7 = - (5 + 7) = - 12 and -8 - 11 = - (8 + 11) = - 19.
### Adding and Subtracting Negative Numbers
As we just saw, adding 6 to a number is the same as moving 6 spaces to the right on the number line. Adding -6 to a number is the same as moving 6 spaces to the left. Thus, adding -6 to a number is the same as subtracting 6 from the number. In general terms, adding a number is the same as subtracting its opposite.
Thus, to add a negative number, subtract its opposite. To subtract a negative number, add its opposite.
Examples:
5 + (- 2) = 5 - 2 = 3
7 - (- 11) = 7 + 11 = 18
-12 - (- 2) = - 12 + 2 = - (12 - 2) = - 10
17 + (- 23) = 17 - 23 = - (23 - 17) = - 6
11 - (- 2) = 11 + 2 = 13
12 - (- 2) + (- 7) = 12 + 2 - 7 = 7 |
## MATHS BITE: 6174
6174 is known as Kaprekar’s Constant. Why is this number important? Perform the following process (called Kaprekar’s Routine):
1. Take any two digit number whose digits are not all identical.
2. Arrange the digits in descending and then ascending order to get two four digit numbers.
3. Subtract the smaller number from the bigger number.
4. Go to step 2 and repeat.
This process will always reach its fixed point 6174 in at most 7 iterations. 6174 is a fixed point as once it has been reached, the process will continue yielding 7641 – 1467 = 6174.
4311-1134=3177
7731-1377=6354
6543-3456=3087
8730-0378=8352
8532-2358=6174
7641-1467=6174
### The Maths Behind it
Each number in the sequence uniquely determines the next number. As there are only finitely many possibilities, eventually the sequence must return to a number it has already hit. This leads to a cycle.
So any starting number will give a sequence that eventually cycles.
There can be many cycles, but for 4 digit numbers in base 10, there happens to be 1 non – trivial cycle, which involves the number 6174.
## MATHS BITE: Kaprekar Numbers
Consider an n-digit number k. Square it, and then add the right n digits to the left n or n-1 digits (by convention, the second part may start with the digit 0, but must be nonzero). If the result is k then it is called a Kaprekar number. They are named after D. R. Kaprekar, a recreational mathematician from India.
We can extend the definition to any base b:
Let X be a non-negative integer and n a positive integer. X is an n-Kaprekar number for base b if there exist non-negative integer A, and positive integer B satisfying:
X2 = Abn + B, where 0 < B < bn
X = A + B
-Wikipedia
### Examples in Base 10
• 297: 2972 = 88209, which can be split into 88 and 209, and 88 + 209 = 297.
• 999: 9992 = 998001, which can be split into 998 and 001, and 998 + 001 = 999.
• In particular, 9, 99, 999… are all Kaprekar numbers.
• More generally, for any base b, there exist infinitely many Kaprekar numbers, including all numbers of the form bn − 1.
• 100: 100 is NOT a Kaprekar number as, although 1002 = 10000 and 100 + 00 = 100, the second part here is zero.
M x
(P.S. Another post on Kaprekar is coming soon!) |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 6.23: Circle Graphs to Display Data
Difficulty Level: At Grade Created by: CK-12
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One day while Taylor was in the candy store, she saw a chart that her Dad had made sitting on the counter.
“What’s this mean?” she asked looking at the chart.
“That is a chart that shows our best sellers. Every other candy sells less than 10%, so I don’t usually include it. These are the top sellers. I keep track of our inventory each month and determine which candies were the top sellers. Then I create a graph of the data,” he explained.
“Where is the graph?”
“I haven’t made it yet.”
“I could do that,” Taylor said smiling.
“Terrific! Go right ahead.”
Taylor was so excited. She could finally put all of her math to work. She knew that a circle graph would be the best way to show the percentages. Here is the chart.
Lollipops - 55%
Licorice - 10%
Chocolates - 20%
Gummy Bears - 15%
Taylor started to work on the circle graph and she thought that she knew what she was doing, but then she got stuck. She couldn’t remember how to change each percentage into a number of degrees.
This is where you come in. It is your turn to help Taylor. Pay attention to this Concept and you will know how to create the circle graph in the end.
### Guidance
Creating a circle graph may seem tricky, but if you think about circle graphs it can become easier to figure out. First, notice that in the graphs at the end of the last Concept, that each percentage was converted to a specific number of degrees.
When you know the number of degrees that a percentage is equal to, you can use a protractor and a circle to draw it in exactly.
To figure this out, we have to figure out each percentage in terms of degrees. How do we do this?
First, we do this by creating a proportion. A percent is out of 100, so we can make a ratio out of any percent.
25% becomes 25100\begin{align*}\frac{25}{100}\end{align*}
15% becomes 15100\begin{align*}\frac{15}{100}\end{align*}
A circle is out of 360\begin{align*}360^ \circ\end{align*}. Since we are trying to figure out the number of degrees, we use a variable over 360 for the second ratio.
Here is a proportion for converting 25% to degrees.
25100=x360\begin{align*}\frac{25}{100} = \frac{x}{360}\end{align*}
Now we cross multiply and solve for the variable x\begin{align*}x\end{align*}. That will be the number of degrees.
100x100xx25%=25(360)=9000=90=90\begin{align*}100x & = 25(360)\\ 100x & = 9000\\ x & =90\\ 25 \% &= 90^\circ\end{align*}
Now if you were going to draw this on a circle graph, you could take a circle and your protractor and measure in a 90\begin{align*}90^\circ\end{align*} angle. That would equal 25% of the graph.
Now let's apply this.
The table shows the number of students in the seventh grade who are studying each foreign language. Make a circle graph that shows the data.
Foreign Language Number of Students Studying Language
Spanish 88
French 48
Italian 16
German 8
Step 1: Find the total number of seventh grade students studying a foreign language. Then find the percent of students studying each language.
88+48+16+8=160\begin{align*}88 + 48 + 16 + 8 = 160\end{align*}
Foreign Language Number of Students Studying Language Percent of Students Studying Language
Spanish 88 88160=1120=55%\begin{align*}\frac{88}{160} = \frac{11}{20} = 55 \%\end{align*}
French 48 48160=310=30%\begin{align*}\frac{48}{160} = \frac{3}{10} = 30 \%\end{align*}
Italian 16 16160=110=10%\begin{align*}\frac{16}{160} = \frac{1}{10} = 10 \%\end{align*}
German 8 8160=120=5%\begin{align*}\frac{8}{160} = \frac{1}{20} = 5 \%\end{align*}
Step 2: Find the measure of the central angle by multiplying 360\begin{align*}360^\circ\end{align*} by the percent.
Foreign Language Number of Students Studying Language Percent of Students Studying Language Degrees in Central Angle
Spanish 88 55% 55% of 360=0.55×360=198\begin{align*}360^\circ = 0.55 \times 360^\circ = 198^\circ\end{align*}
French 48 30% 30% of 360=0.30×360=108\begin{align*}360^\circ = 0.30 \times 360^\circ = 108^\circ\end{align*}
Italian 16 10% 10% of 360=0.10×360=36\begin{align*}360^\circ = 0.10 \times 360^\circ = 36^\circ\end{align*}
German 8 5% 5% of 360=0.05×360=18\begin{align*}360^\circ = 0.05 \times 360^\circ = 18^\circ\end{align*}
Step 3: Draw a circle with a compass. Draw one radius. Use that radius as a side of one central angle. Measure and draw the other central angles using a protractor.
Step 4: Label each sector with a title and percent and give a title to the entire circle graph.
Here is the final graph.
Convert each percent into degrees.
#### Example A
20%
Solution: 72\begin{align*}72^\circ\end{align*}
#### Example B
40%
Solution:144\begin{align*}144^\circ\end{align*}
#### Example C
75%
Solution:270\begin{align*}270^\circ\end{align*}
Here is the original problem once again. Use what you have learned to help Taylor make the circle graph.
One day while Taylor was in the candy store, she saw a chart that her Dad had made sitting on the counter.
“What’s this mean?” she asked looking at the chart.
“That is a chart that shows our best sellers. Every other candy sells less than 10%, so I don’t usually include it. These are the top sellers. I keep track of our inventory each month and determine which candies were the top sellers. Then I create a graph of the data,” he explained.
“Where is the graph?”
“I haven’t made it yet.”
“I could do that,” Taylor said smiling.
“Terrific! Go right ahead.”
Taylor was so excited. She could finally put all of her math to work. She knew that a circle graph would be the best way to show the percentages. Here is the chart.
Lollipops - 55%
Licorice - 10%
Chocolates - 20%
Gummy Bears - 15%
Taylor started to work on the circle graph and she thought that she knew what she was doing, but then she got stuck. She couldn’t remember how to change each percentage into a number of degrees.
First, we need to convert each percentage to a number of degrees. We can do this by multiplying each decimal by 360.
Lollipops .55×360=198\begin{align*}.55 \times 360 = 198^\circ\end{align*}
Licorice .10×360=36\begin{align*}.10 \times 360 = 36^\circ\end{align*}
Chocolates .20×360=72\begin{align*}.20 \times 360 = 72^\circ\end{align*}
Gummy Bears .15×360=54\begin{align*}.15 \times 360 = 54^\circ\end{align*}
Next, Taylor can use a protractor and a circle to create the circle graph. Here is her final work.
### Vocabulary
Here are the vocabulary words in this Concept.
Circle Graph
a visual display of data in a circle. A circle graph is created from percentages with the entire circle representing the whole. The sectors of the circle graph are divided according to degrees which are created out of 360\begin{align*}360^\circ\end{align*}.
Sector
the section of a circle graph. Each section is known as a sector. Each sector can be measured in degrees and given a percentage.
### Guided Practice
Here is the original problem once again.
Convert 30% into degrees.
First, we write a proportion.
30100=x360\begin{align*}\frac{30}{100} = \frac{x}{360}\end{align*}
Next, we cross multiply and solve for the variable.
10x100xx30%=30(360)=10800=108=108\begin{align*}10x &= 30(360)\\ 100x &= 10800\\ x &= 108\\ 30 \% &= 108^\circ\end{align*}
This is our answer.
### Video Review
Here is a video for review.
### Practice
Directions: Use what you have learned to tackle each problem.
1. The table shows the how much money the students in the seventh grade have raised so far for a class trip. Make a circle graph that shows the data.
Fundraiser Amount
Car wash $150 Book sale$175
Bake sale $100 Plant sale$75
2. Make a list of 5 popular ice cream flavors. Then survey your classmates asking them which of the 5 flavors is their favorite ice cream flavor. Use the data to make a circle graph.
3. Use a newspaper to locate a circle graph of some data. Then write five questions about the data.
Directions: Look at each percentage and then use a proportion to find the equivalent number of degrees. You may round your answer when necessary.
4. 12%
5. 25%
6. 28%
7. 42%
8. 19%
9. 80%
10. 90%
11. 34%
12. 15%
13. 5%
14. 10%
15. 78%
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# Lesson 7.5 Dilations.
## Presentation on theme: "Lesson 7.5 Dilations."— Presentation transcript:
Lesson 7.5 Dilations
Vocabulary Dilation Scale Factor Center of Dilation
Dilation A dilation is a transformation that changes the size, but not the shape, of a figure. Remember: translations, reflections, and rotations are transformations that do NOT change the size or shape of a figure. The pupils of your eyes, the black center, works like a camera lens, dilating to let more or less light in, which means it gets smaller or larger.
Scale Factor A scale factor describes how much a figure is enlarged or reduced. A scale factor can be expressed as a decimal, fraction, or percent. A scale factor between 0 and 1 reduces a figure. A scale factor greater than 1 enlarges the figure. Each point in the figure MUST be either enlarged or reduced by the same scale factor or it is not a dilation.
Example 1 Figure ABC has coordinates of (3,3), (1,1), and (4,1). It is dilated by a scale factor of 2. Does the figure reduce or enlarge? To find the new coordinates, multiply each number in the ordered pairs by the scale factor. What are the coordinates for the new figure, A’B’C’?
Example 2 Rectangle WXYZ has the following side measures:
WX = 4 cm, XY = 10 cm, YZ = 4 cm, and ZW = 10 cm What are the measures of W’X’Y’Z’ if it is dilated by a scale factor of ½ ? Does this reduce or enlarge the figure?
Center of Dilation Every dilation has a fixed point that is the center of dilation. To find the center of dilation, draw a line that connects each pair of corresponding vertices. The lines intersect at one point. This point is the center of dilation.
Example 3 Dilate triangle EFG by a scale factor of 1.5 with F as the center of dilation. E (-8, -4), F ( -4, -4), G (-4, -8) When using one of the given points as the center of dilation, that point remains the same, but the other points are multiplied by the scale factor. This means when dilated, E ( ), F (-4, -4), and G ( ). When the origin is used as the center of dilation, each point is multiplied by the scale factor. This means that if the same triangle is dilated by a scale factor of 1.5 as the origin as the center of dilation, the new coordinates would be E( ), F ( ), and G ( ).
Assignment Page 364 #2-14 even and Workbook page 58
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# Week 1 – Math 11 – Arithmetic sequences
to find a value in your sequence ($t_{n}$) you must use this formula…
$t_{n}$$t_{1}$ + (n – 1)(d)
to give an example here is my pattern – 3, 7, 11, 15, 19…
to find $t_{50}$ follow these steps….
the first thing you will need in your formula is $t_{1}$
in this sequence 3 is $t_{1}$ because it is the first number of the sequence, so your formula so far would look like this…
$t_{50}$ = 3 + (n – 1)(d)
next, you will want to subtract 1 from 50, because the next part of our formula is (n – 1). the (n) represents the number in your sequence just as the second number of your sequence is represented by $t_{2}$. so your formula should look like this after you subtracted 1 from 50…
$t_{50}$ = 3 + (49)(d)
the final thing in your sequence is (d). the (d) represents the common difference, the common difference is the rate increasing between numbers. in our sequence, our common difference is 4 because we increase 4 everytime. ex/ 3 +4 = 7 +4 = 11 +4 = 15.
once you know your common difference your formula should look like this…
$t_{50}$ = 3 + (49)(4)
once you calculate your formula it should look something like this…
$t_{50}$ = 3 + (49)(4) = 199
$Part 2$
to find the arithmetic series you need to use this formula…
$S_{n}$$\frac{n}{2}$($t_{1}$$t_{n}$)
first, we will start by adding what we know this far…
$S_{50}$$\frac{50}{2}$(3 + $t_{n}$)
The last thing you need is $t_{n}$. since we are in $S_{50}$ we need to use the awnser of $t_{50}$. Once you add this to your formula it should look like this…
$S_{50}$$\frac{50}{2}$(3 + 199)
When you add this all up you should get this…
$S_{50}$$\frac{50}{2}$(3 + 199) = 5,050
# Week1 – Precal 11 Arithmetic sequence
Arithmetic sequences add the same amount to each time. ex/ 7, 10, 13, 16…
Common difference stands for the amount increasing and is represented by (d). ex/ 7,10,13 d=3
Tn stands for the unknown number you’re trying to find, Tn could be the last number in the sequence or the 50th number.
T is used to represent the numbers in the sequence. ex/
7, 10, 13, 16…
T1,T2,T3,T4
# How dose one build resilience in the face of extream disaster
I think one builds resilience in…
All around the world, there is discrimination in many ways, gender, race, and religion. Sticking together with friends and family can help but also looking at the big picture and pushing through.
# Spoken Word – Procastination
This is the spoken word I wrote
^ ^
| |
The link to the video above
# My Environmental Interactions Post
ill usually start my morning with a hot shower to wake me up. ill also play on my computer after I have made my lunch till I have to go to school. usually I leave the lights of because I forget, but back to topic ill walk to school usually with Liam and arrive there 25 minutes later. at lunch ill bring my lunch bag with my sandwich and a fruit. my sand witch is covered in saran rap witch ill throw in to the garbage. after school ill walk home to the lights I left on, and play on my computer I left on through out the day. through out the night ill either do hockey, witch is off technology, or ill use my computer for home work or games. basically ill be active or on my computer.
How These Effect The Environment – in a bad way
A) when I take a hot shower in the morning I could use cold water which would save water, but not really any one likes to take or even takes cold showers. especially at 7 in the morning
B) I use my computer through out the day, at school, at home, just about every were for more than 40% of the day (when I’m awake). I could cut back on so much teck, but its really hard too. all the information I need to complete just about any assignment also on the computer Is at my finger tips when using this device. I can also entertain my self for hour on end.
C) me leaving the lights on is defiantly a problem. But when I forget to take my fish oil I tend to forget lots and lights are one of the things ill forget on a daily basis. its kind of a huge spree of forgetting.
D) when I go to school and I bring my lunch, ill always have some kind of plastic. once I’ve successfully opened what ever may be inside of the plastic ill throw it out, and it will then be later added to a landfill. it will take decades for it to decompose.
How These Effect The Environment – in a good way
A) when ever I head to school I walk about 95% of the time. walking cuts back on the burning of fossil fuels reducing the amount of green house gasses, which could cause acid rain and melting of the ice caps.
B) another thing I do is breathing, every one breaths so every one dose this. when I exhale and inhale I’m doing something. I’m taking in oxygen and breathing out carbine dioxide. this goes on threw out every 24 hour day till I die. now this could be considered a bad thin due to the large amount of co2 in the atmosphere but I defend it by saying, “what so I just stop breathing ?
C) when I play hockey or volunteer I’m skating. this is physical active instead of being a couch potato.
How These Effect…
Cons:
A) Wasting energy and water.
B) Uses lots of energy
C) uses wasted energy that I don’t need to be using
D) Increases the size of land fills and pollutes.
Pros:
A) when I walk I’m decrasing my carbin footprint helping cut donw on the fosil fuels I would have used if I got a ride.
B) This is something that you cant avoid but it increase the amount of co2 in the atmosphere.
C) I’m more active and it cuts down on the amount of energy I would be using if I stopped doing sports. |
# RECONSTRUCTING A FUNCTION FROM ITS GRADIENT
This section has three parts. In Part 1 we show how to find given its gradient
In Part 2 we show that, although all gradients are expressions of the form
set and , not all such expressions are gradients. In Part 3 we tackle the problem of recognizing which expressions are actually gradients.
## Part 1
Problem1. Find given that
Solution. Since
we have
Integrating with respect to , we have
where is independent of but may depend on . Differentiation with respect to gives
The two equations for can be reconciled only by having
This means that
We could have started by integrating with respect to and then differentiating with respect to . This procedure would have yielded the same result.
Problem 2. Find given that
Solution. Here we have
We will proceed as in the first problem. Integrating with respect to , we have
with independent of . Differentiation with respect to gives
The two equations for can be reconciled only by having
This means that
The function
is the general solution of the vector differential equation
Each particular solution can be obtained by assigning a particular value to the constant of integration .
## Part 2
The next problem shows that not all linear combinations are gradients.
Problem 3. Show that is not a gradient.
Solution. Suppose on the contrary that it is a gradient. Then there exists a function such that
Obviously
and thus
This contradicts the mixed partial derivative theorem: the four partial derivatives under consideration are everywhere continuous and thus, according to the mixed partial derivative theorem, we must have
## Part 3
We come now to the problem of recognizing which linear combinations
are actually gradients. But first we need to review some ideas and establish some new terminology.
An open set (in the plane or in three-space) is said to be connected if any two points of the set can be joined by a polygonal path that lies entirely within the set. An open connected set is called an open region. A curve
is said to be closed if it begins and ends at the same point:
It is said to be simple if it does not intersect itself:
As is intuitively clear, a simple closed curve in the plane separates the plane into two disjoint open connected sets: a bounded inner region consisting of all points surrounded by the curve and an unbounded outer region consisting of all points not surrounded by the curve. (Jordan Curve Theorem.)
Finally, we come to the notion we need in our work with gradients:
Let be an open region of the plane, and let be an arbitrary simple closed curve. is said to be simply connected if is in implies the inner region of is in
THEOREM Let and be functions of two variables, each continuously differentiable on a simply connected open region . The linear combination is a gradient on if and only if
for all .
Proof. A complete proof of this theorem is given in an advanced calculus course. We will prove the result under the additional assumption that has the form of an open rectangle with sides parallel to the coordinate axes. Suppose that is a gradient on this open rectangle , say
Since
we have
Since and have continuous first partials, has continuous second partials. Thus, according to the mixed partials theorem we have
Conversely, suppose that
We must show that is a gradient on . To do this, we choose a point from and form the function
The first integral is independent of . Hence it follows from the fundamental theorem of calculus that
Differentiating with respect to we have
The first term is since once again we are differentiating with respect to the upper limit. In the second term the variable appears under the sign of integration. It can be shown that, since and are continuous,
Anticipating this result, we have
We have now shown that
It follows that is the gradient of on
Example 4. The vector functions
and
are both everywhere defined. The first vector function is the gradient of a function everywhere defined:
The second is not a gradient:
and so
Example 5. The vector function defined on the punctured disc by setting
satisfying
We will see later that is not a gradient on that set. The set is not simply connected so the theorem does not apply.
# Double Integral over a Rectangle
Let be a function continuous on the rectangle
Our object is to define the double integral
Recall that to define the integral
we first introduced some auxiliary notions: partition , upper sum , and lower sum We were then able to define
as the unique number that satisfies the inequality
We will follow exactly the same procedure to define the double integral
First we explain what we mean by a partition of the rectangle . To do this, we begin with a partition
and a partition
The set
is called a partition of . consists of all the grid points
The partition breaks up into nonoverlapping rectangles where for each fixed
On each rectangle , the function takes on a maximum value , and a minimum value The sum of all the products
is called the upper sum for :
The sum of all the products
is called the lower sum for :
Example 1. Consider the function
on the rectangle
As a partition of take
and as a partition of take
The partition then breaks up the initial rectangle into the six rectangles.
On each rectangle , the function takes on its maximum value at the point , the corner farthest from the origin:
Thus
(area of ) +(area of ) + (area of )
+ (area of ) + (area of ) + (area of )
On each rectangle , takes on its minimum value at the point , the corner closest to the origin:
Thus
(area of ) +(area of ) + (area of )
+ (area of ) + (area of ) + (area of )
We return now to the general situation. As in the one-variable case, it can be shown that, if is continuous, then there exists one and only one number that satisfies the inequality
Definition of the double integral over a rectangle
The unique number that satisfies the inequality
is called the double integral of over and is denoted by
# The Double Integral as a Volume
If is nonnegative on the rectangle , the equation
represents a surface that lies above . In this case the double integral
gives the volume of the solid that is bounded below by and bounded above by the surface
To see this, consider a partition of . breaks up into subrectangles ; and thus the entire solid into parts . Since contains a rectangular solid with base and height , we must have
(area of ) volume of .
Since is contained in a rectangular solid with base and height , we must have
volume of (area of ).
In short, for each pair of indices and , we must have
(area of ) volume of (area of ).
Adding up these inequalities, we are forced to conclude that
volume of .
Since is arbitrary, the volume of must be the double integral:
volume of =
The double integral
gives the volume of a solid of constant height erected over . In square units this is just the area of :
# Some Computations
Double integrals are generally computed by techniques that we will take up later. It is possible, however, to evaluate simple double integrals directly from the definition.
Problem 2. Evaluate
where is the rectangle
Solution. Here
We begin with a partition
and a partition
This gives
as an arbitrary partition of On each rectangle , has constant value . This gives and throughout. Thus
Similarly
Since
and was chosen arbitrarily, the inequality must hold for all partitions of . This means that
Remark. If
gives the volume of the rectangular solid of constant height erected over the rectangle .
Problem 3. Evaluate
where
Solution. With with a partition
and a partition
we have
as an arbitrary partition of On each rectangle the function
has a maximum
and a minimum
Thus
and
For each pair of indices and ,
This means that for arbitrary we have
The middle term can be written
The first double sum reduces to
The second double sum reduces to
The sum of these two numbers
satisfies the inequality
The integral is therefore :
Remark. This last integral should not be interpreted as a volume. The expression does not keep a constant sign on
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### Course: Class 6 (Foundation)>Unit 5
Lesson 3: Even and odd numbers
# Intro to even and odd numbers
This video explains the difference between even and odd numbers, and how they interact when added together. Even numbers are multiples of two and can be divided evenly, while odd numbers are not divisible evenly by two. You can easily identify whether a number is even or odd by looking at the ones place.
## Want to join the conversation?
• @ , Khan states that 0 is a multiple of two. Is zero not a multiple of every number?
• Yes, zero is a multiple of every number. That means that you get an even number whenever you multiply by zero, because you get zero.
Be careful though! You cannot say it the other way around and be correct. Two is not a multiple of zero. There is no way to multiply by zero and get two as a result.
• Does even and odd only apply to integers?
• In the context of real numbers, the answer to your question is yes -- the descriptions of "even" and "odd" apply only to integers.
A number n is even if and only if n = 2k, where k is an integer. This definition implies that if n is even then n is an integer, because if k is an integer, then 2k will be an integer.
A number n is odd if and only if n = 2k + 1, where k is an integer. This definition implies that if n is odd then n is an integer, because if k is an integer, then 2k + 1 will be an integer.
Example of an even number --
Consider the number 8
8 = 2*4
4 is an integer. Since 8 can be expressed as 2*4, this means 8 is even.
Example of a number that is not even --
Consider the number 2.2
2.2 = 2*(1.1)
1.1 is not an integer. Since 2.2 cannot be expressed as 2k, where k is an integer, this means 2.2 is not even.
Example of an odd number --
Consider the number -7
-7 = 2*(-4) + 1
-4 is an integer. Since 7 can be expressed as 2*(-4) + 1, this means 7 is odd.
Example of a number that is not odd --
Consider the number 13/10
13/10 = 2*(3/20) + 1
3/20 is not an integer. Since 13/10 cannot be expressed as 2k + 1, where k is an integer, this means 13/10 is not odd.
The concept of parity (the property of being even or odd) can apply to other objects in math as well. For instance, certain functions can be described as odd or even.
• How is 0 an even number?
• Can't an Even + Odd become a even too?
• No, an even number added to an odd number is always an odd number. Try counting out in small numbers by yourself, and you'll notice the result.
• Sal said zero is a multiple of two but, it's also a multiple of 1 so, doesn't that make it odd and even?
• Yes, zero is a multiple of one, but it's considered even because (according to math's principles) all even numbers are two more than the last, and since one is one more than zero, it is not even, so you are half-right,Arbaaz.
• Will 190,398,928 be even?
• Just take a look at the last number. If it is 2, 4, 6, 8, or 0 the number is even. If it is 1, 3, 5, 7, or 9, then the number is odd. So in this case, the number is 8 so the number is even.
• Is 2.3 an even or an odd number? Is 3.2 even?
• Good question! We only talk about even and odd for whole numbers. Numbers like 2.3 and 3.2 are not whole numbers, they're decimals. So, they're neither even nor odd.
• is a negative number even if its multiple by 2 like -68?
• Yes, negative numbers can be even, because they are multiples of 2. Zero is also an even number.
• I have a questions that troubles me. You have made a reasonable case for claiming that the number '0' is even. You present your proof in in a rather mechanical way. On a deeper level however, that seems wrong. the symbol '0' represents 'Nothing'. How can 'nothing' be either even or odd? It is the absence of anything, if you have 0 apples, do you have an even or an odd number of apples? the answer is obvious, you have neither an even nor an odd number of apples, you have NO apples.
On the surface '0' can be classified as an even number but on a conceptual level I don't see how it can be even or odd.
what am I not understanding? Jim
• Well, it isn't really the case like the number 1, where it is neither prime nor composite, but rather because 0 is simply divisible by 2 (the definition of "even") and that it is also in between 2 odd numbers, so it just comes down to logic.
• Are Irrational numbers odd or even? |
# How do you solve and graph 3-2k > -4?
Sep 6, 2015
$k < \frac{7}{2}$
graph{3-2x < -4 [-2.8, 8.297, -2.854, 2.694]}
#### Explanation:
Things you can do with expressions in an inequality which maintain the inequality:
• Add the same amount to each expression
• Subtract the same amount from each expression
• Divide each expression by the same amount provided the amount is greater than zero
• Multiply each expression by the same amount provided the amount is greater than zero
Things you can do with expressions in an inequality which reverse the direction of the inequality sign:
• Exchange the expressions with one another on opposite sides of the inequality (when dealing with a single inequality relation)
• Multiply both sides of the inequality by an amount less than zero
• Divide both sides of the inequality by an amount less than zero
$3 - 2 k < - 4$
Given the above rules, we can subtract $3$ from each expression, to get:
$\textcolor{w h i t e}{\text{XXXX}} - 2 k < - 7$
Then we can divide each expression by $- 2$, to get:
$\textcolor{w h i t e}{\text{XXXX")k > 7/2 color(white)("XXXX}}$(remember dividing by a value less than zero reverses the inequality sign) |
# How do you use the Fundamental Theorem of Calculus to find the derivative of int (u^3) / (1+u^2) du from 2-3x to 5?
Feb 17, 2017
$= \frac{3 {\left(2 - 3 x\right)}^{3}}{1 + {\left(2 - 3 x\right)}^{2}}$
#### Explanation:
The following assumes you mean the derivative wrt $x$.
$\frac{d}{\mathrm{du}} \left({\int}_{a}^{u} f \left(t\right) \setminus \mathrm{dt}\right) = f \left(u\right)$ where $a =$ some constant.
If in fact $u = u \left(x\right)$, we can use the Chain Rule so that:
$\frac{d}{\mathrm{dx}} \left({\int}_{a}^{u \left(x\right)} f \left(t\right) \setminus \mathrm{dt}\right) = f \left(u \left(x\right)\right) \frac{\mathrm{du}}{\mathrm{dx}}$
We can also play with the intervals ..... in summary ${\int}_{a}^{b} = {\int}_{a}^{0} + {\int}_{0}^{b} = {\int}_{0}^{b} - {\int}_{0}^{a}$.... in order to reach this point:
$\frac{d}{\mathrm{dx}} \left({\int}_{{u}_{1} \left(x\right)}^{{u}_{2} \left(x\right)} f \left(t\right) \setminus \mathrm{dt}\right) = f \left({u}_{2} \left(x\right)\right) \frac{{\mathrm{du}}_{2}}{\mathrm{dx}} - f \left({u}_{1} \left(x\right)\right) \frac{{\mathrm{du}}_{1}}{\mathrm{dx}}$
Gasp! With all that, we just need to process this:
$\frac{d}{\mathrm{dx}} \left({\int}_{2 - 3 x}^{5} \frac{{u}^{3}}{1 + {u}^{2}} \mathrm{du}\right)$
$= \frac{{5}^{3}}{1 + {5}^{2}} \frac{d \left(5\right)}{\mathrm{dx}} - \frac{{\left(2 - 3 x\right)}^{3}}{1 + {\left(2 - 3 x\right)}^{2}} \frac{d \left(2 - 3 x\right)}{\mathrm{dx}}$
$= \frac{3 {\left(2 - 3 x\right)}^{3}}{1 + {\left(2 - 3 x\right)}^{2}}$
Maybe you can further simplify that. |
## Saturday, December 16, 2006
### John's percentage growing post
Question #1:
What is a good definition of percent? You should use words symbols, pictures and numerical examples in your definition.
Question #2:
How are three fifths (3/5), 3:2, 60% and 0.6 all the same? Use pictures and words to show you answer.
-if you divide 3 by 5 you will get o.6.
-if you multiply o.6 by 100 you will get 60%.
-to get 3:2 subtract 5 (denominator) to 3 (numerator) you will get 2 as a part.
3 divided by 5=0.6
0.6 multiply by 100=60%
5-3=2 3:2
Question #3
Show 3 different ways to find 35% of 80.
a. 80 divided by 100=0.8
35%x0.8=28
35%/100 = 28/80
b. 35% divided by 100=0.35
0.35x80=28
c.80 divided by 100=0.8
10x0.8= 8
25x0.8= +20
35%/100 = 28/80
Question #4:
Find a link to blogs that deal with percentages. Leave a comment behind and add the link with a review (What the post was talking about.... yes you have to read the post and why others should read the post) Hint In the side bar there are links to other schools. Three of them have done work on percentages!
I found this blog http://linden8m.blogspot.com/ and it's about percentages.
They have great explanations about percentages and good examples too...
Question #5:
The principal announced that 50% of the children in Ms. Stanzi's class met their reading goal for the month and that 55% of the children in Ms Lowrey's class met their reading goal for the month. Ms Stanzi said that a greater number of her students met their reading goal. Could Ms Stanzi be correct? Why or Why not.
Question #6:
Use a hundred grid (unit square) to illustrate the following questions. Once you have explained and illustrated what the question means solve it.
a) 16 is 40% of what number?
16 divided by40=0.4
0.4 multiply by 100=40
40 divided by 100=0.4
40 multiply by o.4=16
16 is the number of 40%.
b.What is 120% of 30?
120 divided by 100=1.2
1.2 multiply by 30=36
john- said...
Setup
5/5
Comments:I labeled it correctly, have the title correct, and put colour for the rest.
Question 1
7/10
Comments:I need more definition for percents, and some pictures too.
Question 2
4/10
Comments:I didn't put a picture for this one.
Question 3
5/10
Comments:I didn't use bubbleshare, and there's no image.
Question 4
4/5
Question 5
Question 6
7/15
Comments:pictures is not that great, and I didn't answer the question correctly.
Overall Mark
32/65
Overall Comment for the Growing Post:I need to put more effort into my growing post, add more pictures and that's it..
kristian said...
Setup
5/5
Comments:I give 5 because you do it nicely
Question 1
7/10
Comments:I give you7 because you need more explanation
Question 2
8/10
Question 3
4.5/10
Comments:I give you a 4.5 bacause you need to use gliffy or bubbleshare on this question
Question 4
5/5
Question 5
0/10
Question 6
11/15
Overall Mark
40.5/65
Overall Comment for the Growing Post....good job!!!!!!!!!!!!
kristian said...
Setup
5/5
Comments:I give 5 because you do it nicely
Question 1
7/10
Comments:I give you7 because you need more explanation
Question 2
8/10
Question 3
4.5/10
Comments:I give you a 4.5 bacause you need to use gliffy or bubbleshare on this question
Question 4
5/5 |
### Taylor Series
#### Definition
A Taylor series is a power series expansion of a function $f(x)$ about a given point. The Taylor series expansion of a differentiable function $f(x)$ about a point $x=c$ is given by:
$\sum_{n=0}^{\infty}\dfrac{f^{(n)}(c)}{n!}(x-c)^n=f(c)+f'(c)(x-c)+\dfrac{f''(c)}{2!}(x-c)^2+\dfrac{f'''(c)}{3!}(x-c)^3+\ldots$
The Taylor series expansion about the point $c=0$ is known as a Maclaurin series expansion.
Taylor series expansions have many applications, including evaluating definite integrals, finding the limit of a function and approximating the value of an expression.
#### Worked Example
###### Example 1
Use the first two terms of a Taylor expansion to approximate $\cos{\left(\dfrac{4\pi}{5}\right)}$.
###### Solution
Here the function to expand is $f(x)=\cos{x}$. Recall that the first two terms of a Taylor expansion about the point $x=c$ are given by $f(x)\approx f(c)+f'(c)(x-c).$
The derivative of $\cos{x}$ with respect to $x$ is $-\sin{x}$, so the first two terms of the Taylor series expansion for $\cos{x}$ are: $f(x)\approx \cos{c}-(x-c)\sin{c}.$
To approximate $\cos{\left(\dfrac{4\pi}{5}\right)}$, a suitable choice for $c$ must be made.
First note that: $\frac{4\pi}{5}=\frac{16\pi}{20}=\frac{(15+1)\pi}{20}=\frac{3\pi}{4}+\frac{\pi}{20}$.
Since $\cos{\left(\dfrac{3\pi}{4}\right)}$ and $\sin{\left(\dfrac{3\pi}{4}\right)}$ are commonly known trigonometric ratios, choose $c=\dfrac{3\pi}{4}$. Substituting $x=\dfrac{4\pi}{5}$ and $c=\dfrac{3\pi}{4}$ into the expansion gives
\begin{align} \cos{\left(\frac{8\pi}{10}\right)} &\approx \cos{\left(\frac{3\pi}{4}\right)}-\left(\frac{8\pi}{10}-\frac{3\pi}{4}\right)\sin{\left(\frac{3\pi}{4}\right)} \\ &= -\frac{1}{\sqrt{2} }-\frac{\pi}{20}\cdot\frac{1}{\sqrt{2} } \\ &= -\frac{1}{\sqrt{2} }\left(1+\frac{\pi}{20}\right). \end{align}
#### Video Examples
###### Taylor Expansion
Prof. Robin Johnson finds the Taylor expansion of $(6+4x)^{\large{^1/_3}}$ about $x=-1$.
##### Applications of Taylor Series
###### Example 1
Prof. Robin Johnson uses Taylor series to approximate $\sqrt{1.1}$.
###### Example 2
Prof. Robin Johnson uses Taylor series to approximate $^3\sqrt{26}$.
###### Example 3
Prof. Robin Johnson uses Taylor series to approximate $\ln{(0.95)}$.
###### Example 4
Prof. Robin Johnson uses Taylor series to approximate $\sin{\left(\dfrac{3\pi}{10}\right)}$.
#### Workbooks
These workbooks produced by HELM are good revision aids, containing key points for revision and many worked examples. |
tgt
## Saturday, 9 August 2014
### CHAPTER 10 - Worked Out Examples
Example: 11
(a) For $|x|\; < \;1$ expand ${(1 - x)^{ - 3}}$ (b) Find the coefficient of ${x^n}$ in the expansion of${(1 + 3x + 6{x^2} + 10{x^3} +\ldots \infty )^{ - n}}$
Solution: 11-(a)
We have
${V_r} = \dfrac{{( - 3)\;(( - 3) - 1)(( - 3) - 2)) \ldots (( - 3) - r + 1)}}{{r!}}$ $= \dfrac{{{{( - 1)}^r}\;(r + 2)!}}{{2(r!)}}$ $= \dfrac{{{{( - 1)}^r}\;(r + 1)\;(r + 2)}}{2}$
Thus,
${V_0} = 1,\;\;{V_1} = - 3,\;{V_2} = 6,\;{V_3} = - 10\ldots$
so that
${(1 - x)^{ - 3}} = 1 + 3x + 6{x^2} + 10{x^3} + \ldots \infty$
Solution: 11-(b)
We use the result of part – $(a)$ in this:
${(1 + 3x + 6{x^2} + 10{x^3}\ldots \infty )^{ - n}} = {\left( {{{(1 - x)}^{ - 3}}} \right)^{ - n}}$ $= {(1 - x)^{3n}}$
The coefficient of ${x^n}$ in this binomial expansion (note: the power is now a positive integer) would be ${( - 1)^n} \cdot {\;^{3n}}{C_n}$.
Example: 12
Find the magnitude of the greatest term in the expansion of ${\left( {1 - 5y} \right)^{ - 2/7}}$for $y = \dfrac{1}{8}$.
Solution: 12
Let us first do the general case: what is the greatest term in the expansion of ${(1 + x)^n}$, where $n$ is an arbitrary rational number. We have,
${T_{r + 1}} = {V_r}\;{x^r}$ and ${T_r} = {V_{r - 1}}\;{x^{r - 1}}$ so that $\dfrac{{{T_{r + 1}}}}{{{T_r}}} = \dfrac{{{V_r}}}{{{V_{r - 1}}}} \cdot x$ $= \dfrac{{n - r + 1}}{r} \cdot x$
Now, let us find the conditions for which this ratio exceeds $1$. We have
$\left| {\;{T_{r + 1}}\;} \right|\; \ge \;\left| {\;{T_r}\;} \right|$ $\Rightarrow \left| {\;\dfrac{{n + 1}}{r} - 1} \right|\; \ge \;\dfrac{1}{{|x|}}$ $\ldots(1)$
For this particular problem, $(1)$ becomes
$\left| {\;\dfrac{{\dfrac{{ - 2}}{7} + 1}}{r} - 1} \right|\; \ge \;\dfrac{1}{{\left| {\dfrac{{ - 5}}{8}} \right|}}$ $\Rightarrow\,\,\,\, \left| {\;\dfrac{5}{{7r}} - 1\;} \right|\; \ge \;\dfrac{8}{5}$ $\Rightarrow \,\,\,\,\dfrac{5}{{7r}}\; \ge \;\dfrac{{13}}{5}$ $\Rightarrow\,\,\,\, r \le \;\dfrac{{25}}{{91}}$ $\Rightarrow\,\,\,\, r=0$
Thus, ${T_{r + 1}} = {T_1}$ is the greatest term, with magnitude $1$ |
Open In App
# The Fox, The Duck and a Circular Pond
A Duck that is being chased by a Fox saves itself by sitting at the centre of a circular pond of radius r. The Duck can only fly from land and not able to fly from the water. Furthermore, Fox cannot swim. The speed of the Fox is four times the speed of the Duck. Assuming that the Duck and Fox are perfectly smart, is it possible for the Duck to ever reach the edge of the pond and fly away to its escape from the ground?
Solution:
• The Duck can’t swim directly away from the Fox because for that the Duck would have to swim a distance r and till that time Fox would have cover half the circumference of the pond i.e., (pi * r). Since the speed of the Fox is four times the speed of the Duck, therefore, he will be able to catch Duck as we know that
`(pi * r) < (4 * r)`
• The solution for this puzzle is that Duck could swim in concentric circles closed to the centre of the pond. Due to this Duck would cover the small circumference of the pond and the Fox will have to cover the larger circumference of the pond.
• Using the above strategy the Duck will make tiny concentric circles around the center and due to this, the Fox will not be able to cope up with the Duck as he has to cover the entire pond.
• As the speed of the Fox is four times the Duck, therefore the Duck must move in a concentric circle of radius r/4 such that the distance cover by Duck and the Fox is same. As long as Duck stays within that concentric circle the Duck will gain some distance over Fox after some time and the Fox will be unable to keep up with the Duck.
• As soon as the Fox is 180 degrees behind the Duck, the Duck would cover the remaining distance (3*r / 4) as distance cover the Fox is less than the distance cover by Duck because
``` (pi * r) > 4 * (3r/4)
```
• At last, The Duck would able to reach the edge of the pond and fly away. |
Question
1. # Write The Expression As The Sine, Cosine, Or Tangent Of An Angle. Sin 57° Cos 13° – Cos 57° Sin 13°
## Introduction
The trigonometric functions of Sine, Cosine, and Tangent are essential tools for understanding how angles measure along the circumference of a circle. But what happens when you’re asked to write an expression using these functions? In this blog post, we’ll look at one such example: Write the expression as the sine, cosine, or tangent of an angle. The expression in question is “sin 57° cos 13° – cos 57° sin 13°”. We’ll walk through how to solve this problem step-by-step and investigate what each part of the equation means. By the end, you’ll have a better understanding of how to use sine, cosine, and tangent to work with expressions.
## The Sine, Cosine, and Tangent of an Angle
In mathematics, the trigonometric functions (also called the circular functions) are functions of an angle. They relate the angles of a triangle to the lengths of its sides. Trigonometric functions are important in many areas, such as engineering, navigation, design, and physics.
The three most common trigonometric functions are sine (sin), cosine (cos), and tangent (tan). These are usually denoted by the letters S, C, and T. The inverse trigonometric functions are usually denoted by Arcsin(sin-1), Arccos(cos-1), and Arctan(tan-1).
To find the value of these trigonometric functions for a given angle, we use a right-angled triangle. The angle is usually denoted by θ (theta).
## The Sin of 57°
The Sin of 57°
We all know that the sin of an angle is the ratio of the opposite side to the hypotenuse. But what if we don’t know the sides? How do we find the sin of 57°?
First, let’s draw a picture. We’ll put the angle at 57° in the middle and label the sides:
Now we can see that, to find the sin of 57°, we just need to find the ratio of Side A to Side B. We can do this using a little trigonometry.
First, we’ll use the cosine function. The cosine of an angle is the ratio of the adjacent side to the hypotenuse. In our case, that’s Side B over Side C. So we can write:
cos(57°) = Side B / Side C
Now we can solve for Side B:
Side B = cos(57°) * Side C ~~ 0.74 * Side C
## The Cosine of 13°
To find the cosine of 13°, we can use the fact that cos(x) = sin(90-x). Therefore, cos(13) = sin(77).
## The Tangent of 57°
The tangent of 57° is the ratio of the side opposite 57° to the side adjacent to 57°. Tangent is abbreviated as tan. Therefore, we can write: |
# For a test of Ho: p=0.5,the z test statistic equals -1.74. Find the p-value for Ha: p<0.5.
The question aims to find out the p-value using the given alternative hypothesis, which is a one-sided hypothesis. Therefore, the p-value will be determined for the left tail test with reference to the standard normal probability table. When the alternative hypothesis states that a certain value for a parameter in the null hypothesis is lesser than the actual value, then left-tail tests are used.
Figure-1 : P-Value and Satistical Significance
Let’s first understand the difference between the Null and Alternative hypotheses. Null hypothesis $H_o$ refers to no association between two parameters of the population, meaning both are the same. Alternative hypothesis $H_a$ is opposite to the null hypothesis and states that there is a difference between two parameters.
## Expert Solution:
In order to calculate the p-value, we will use the standard normal table. According to the given information, the value of the test statistic is given as:$z = -1.74$Null hypothesis $H_o$ is given as:$p = 0.5$Alternative Hypothesis $H_a$ is given as:$p < 0.5$The formula for p-value is given as:$p = P (Z < z)$Where P is the probability:$p = P (Z < -1.74)$The p-value can be calculated by determining the probability less than -1.74 using the standard normal table. Therefore, from the table p-value is given as:$p = 0.0409$
## Alternative Solution:
For the given problem, the p-value will be determined using the standard probability table. Check against the row starting with -1.74 and column with 0.04. The answer obtained will be:$p = P ( Z< -1.74)$$p = 0.0409$Therefore, the p-value for $H_a$ < 0.5 is 0.0409.
## Example:
For a test of $H_o$: $p = 0.5$, the $z$ test statistic equals 1.74. Find the p-value for $H_a: p>0.5$.
Figure-2 : Z-Test Satistic
In this example, the value of test statistic $z$ is 1.74, therefore, it is a right tail test.For calculating the p-value for a right tail test, the formula is given as:$p = 1 – P ( Z > z)$$p = 1 – P ( Z > 1.74)$Now use the standard probability table to find the value.The p-value is given as:$p = 1 – 0.9591$ $p = 0.0409$Therefore, the p-value is 0.0409. |
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# Difference of cubes
Definition.
Difference of cubes of two expressions can be found using the following formula:
a3 - b3 = (a - b)·(a2 + ab + b2)
## Derivation of the formula of difference of cubes
The proof of the formula is very simple. To prove the formula is sufficient to multiply the expression:
(a - b)·(a2 + ab + b2) =
= a3 + a2b + ab2 - ba2 - ab2 - b3 = a3 - b3
## Applying of difference of cubes formula
Difference of cubes formula convenient to use:
• to factorised
• to simplify expressions
## Examples of task
Example 1.
Factorised x3 - 27.
Solution: Apply the difference of cubes formula.
x3 - 27 = x3 - 33 = (x - 3)·(x2 + 3x + 9)
Example 2.
Factorised 8x3 - 27y6.
Solution: Apply the difference of cubes formula.
8x3 - 27y6 = (2x)3 - (3y2)3 =
= (2x - 3y2)·(4x2 + 6xy2 + 9y4)
Example 3.
Simplify the expression 27x3 - 13x - 1.
Solution: Apply the difference of cubes formula in numerator.
27x3 - 13x - 1 = (3x - 1)·(9x2 + 3x +1)3x - 1 = 9x2 + 3x +1
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### 2 3 Subtract Integers
1. 1. Subtracting Real Numbers Chapter 2.3 Subtracting is the same as adding a negative!
2. 2. Warm Up <ul><li>What is the opposite of the following? </li></ul><ul><ul><li>-14 +5.2 -19 19 </li></ul></ul><ul><li>What is the absolute value of the following? </li></ul><ul><ul><li>-14 +5.2 -19 19 </li></ul></ul><ul><li>Find the sum </li></ul><ul><ul><li>-5 + 8= -9 + -12= </li></ul></ul>
3. 3. Warm Up Answers <ul><li>Just change the signs </li></ul><ul><ul><li>+14 -5.2 +19 -19 </li></ul></ul><ul><li>Make them all positive </li></ul><ul><ul><li>+14 +5.2 +19 +19 </li></ul></ul><ul><li>Find the sum </li></ul><ul><ul><li>-5 + 8= +3 -9 + -12= -21 </li></ul></ul>
4. 4. Subtracting Integers is really adding negatives <ul><li>Find the sum 9 + (-6) </li></ul><ul><li>Find the sum 9-(+6)? </li></ul><ul><li>What do you notice about the two number sentences above? </li></ul><ul><li>+3 </li></ul><ul><li>+3 </li></ul>
5. 5. Make a subtraction statement into an addition statement <ul><li>Exchange the signs in the middle. </li></ul><ul><li>4 – (+5) = </li></ul><ul><li>-9 – (+8) = </li></ul><ul><li>-57 – (+25) = </li></ul><ul><li>+87 – (+88)= </li></ul><ul><li>A minus sign IS a negative sign. </li></ul><ul><li>4 + (-5) </li></ul><ul><li>-9 + ( - 8) </li></ul><ul><li>-57 +( - 25) </li></ul><ul><li>+87 + - 88 </li></ul>
6. 6. Two negatives make a positive. <ul><li>4 – (+5) = 4 + (-5) </li></ul><ul><li> Opposite </li></ul><ul><li>statements </li></ul><ul><li>4 – (-5) = 4 + (+5) </li></ul><ul><li>Subtract a negative = add a positive. </li></ul>
7. 7. How do you handle two negative signs in the middle? <ul><li>-9 – (-8) </li></ul><ul><li>Change them both to positive </li></ul><ul><li>-9 - (- 8) </li></ul><ul><li>-9 + (+8)= -1 </li></ul>
8. 8. Try Some. Rewrite each statement <ul><li>-57 – (-25)= </li></ul><ul><li>-17 – (-21)= </li></ul><ul><li>7 – (-13)= </li></ul><ul><li>8 – (-33)= </li></ul><ul><li>-5 – 15 = </li></ul><ul><li>-57 + ( + 25) </li></ul><ul><li>-17 + ( + 21) </li></ul><ul><li>7 + ( + 13) </li></ul><ul><li>8 + ( + 33) </li></ul><ul><li>-5 + - 15 </li></ul>
9. 9. And the answer is? <ul><li>-16 – (-21)= </li></ul><ul><li>-83 – (-17)= </li></ul><ul><li>83 – (-17)= </li></ul><ul><li>47 – (-12)= </li></ul><ul><li>-9 – 22 = </li></ul><ul><li>-16 + (+21)= -5 </li></ul><ul><li>-93 + (+17)= -76 </li></ul><ul><li>83 + (+17)= 100 </li></ul><ul><li>47 + (+12)= 59 </li></ul><ul><li>-9 + - 22 = - 31 </li></ul>
10. 10. pg 71: 23 – 67 odds |
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# Mohit travels 380 km per day. How much will he travel in 120 days?
Last updated date: 13th Jun 2024
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Hint: Here, we need to find the distance Mohit travels in 120 days. We will use the concept of multiplication to get the distance travelled by Mohit in 120 days. Multiplication is the repeated addition of equal groups.
Multiplication helps in adding multiple equal groups quickly. It is denoted by the symbol $\times$. Brackets may also be used to denote multiplication.
For example: Suppose that the number 2 is added repeatedly 15 times, that is $2 + 2 + \ldots \ldots + 2$ 15 times. This can be represented using multiplication as $2 \times 15$.
Now in the question, it is given that the distance travelled by Mohit in 1 day is 380 km.
We can multiply the distance travelled in 1 day by any number $x$, to get the distance travelled by Mohit in $x$ days.
Multiplying the distance travelled in 1 day by 120, we can find the distance travelled by Mohit in 120 days.
Therefore, we get
Distance travelled by Mohit in 120 days $= 380 \times 120$km
Simplifying the expression, we get
Distance travelled by Mohit in 120 days $= 45600$km
Therefore, Mohit travels 45,600 km in 120 days.
Note: We can solve the product $380 \times 120$ more using the algebraic identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$.
Rewriting $380 \times 120$ as a product of sum and difference of two numbers, we get
$380 \times 120 = \left( {250 + 130} \right)\left( {250 - 130} \right)$
Applying the algebraic identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$, we get
$\Rightarrow 380 \times 120 = {250^2} - {130^2}$
Simplifying the expression, we get
$\Rightarrow 380 \times 120 = 62500 - 16900$
Subtracting the terms of the expression, we get
$\Rightarrow 380 \times 120 = 45600$
Thus, distance travelled by Mohit in 120 days $= 45600$km
Therefore, Mohit travels 45,600 km in 120 days. |
# How do you differentiate e^x(x^2-3)?
##### 4 Answers
Oct 26, 2017
$\textcolor{b l u e}{{e}^{x} \left({x}^{2} + 2 x - 3\right)}$
#### Explanation:
You need to use the product rule to differentiate this. The product rule states that:
${f}^{'} \left(a b\right) = a \cdot {f}^{'} \left(b\right) + b \cdot {f}^{'} \left(a\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$
$\frac{\mathrm{dy}}{\mathrm{dx}} \left({x}^{2} - 3\right) = 2 x$
So, put into product rule gives:
${e}^{x} \cdot \left(2 x\right) + \left({x}^{2} - 3\right) \cdot {e}^{x} = \textcolor{b l u e}{{e}^{x} \left({x}^{2} + 2 x - 3\right)}$
Oct 26, 2017
${e}^{x} \left({x}^{2} + 2 x - 3\right)$
#### Explanation:
Use the rule for finding the derivative of the product of two functions.
$\mathmr{if} y = f \left(x\right) g \left(x\right)$, then $\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$
Here, $f \left(x\right) = {e}^{x}$, so $f ' \left(x\right) = {e}^{x}$
$g \left(x\right) = {x}^{2} - 3$, so $g ' \left(x\right) = 2 x$
Put 'em all together: $f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right) =$
${e}^{x} \left({x}^{2} - 3\right) + {e}^{x} 2 x$
...Wolfram tells me I need to factor:
${e}^{x} \left({x}^{2} + 2 x - 3\right)$
GOOD LUCK
Oct 26, 2017
Use the product rule: $f = g \cdot h = f ' = g ' \cdot h + g \cdot h '$, where all are functions of the same variable and are being differentiated with respect to the same variable.
#### Explanation:
I preface this by saying that this at the moment is a product rule question posted in the chain rule section. Thus, I must consider both the possibility that it was posted in the wrong section, and the possibility that it was written incorrectly. Answers for both possibilities are below, please choose the appropriate one.
IF YOU MEANT ${e}^{x} \cdot \left({x}^{2} - 3\right)$
This as written is not a case of the chain rule, but rather of the product rule. The product rule states that given $f \left(x\right) = g \left(x\right) \cdot h \left(x\right) , f ' \left(x\right) = g ' \left(x\right) \cdot h \left(x\right) + g \left(x\right) \cdot h ' \left(x\right)$
Here we have $g \left(x\right) = {e}^{x} , g ' \left(x\right) = {e}^{x} , h \left(x\right) = {x}^{2} - 3 , h ' \left(x\right) = 2 x$. Thus...
$\frac{\mathrm{df}}{\mathrm{dx}} = {e}^{x} \cdot \left({x}^{2} - 3\right) + {e}^{x} \cdot 2 x = {e}^{x} \left({x}^{2} + 2 x - 3\right)$
If you had instead meant ${e}^{{x}^{2} - 3}$, you would use the chain rule.
The chain rule states that for a composition of functions $f \left(g \left(x\right)\right)$ (i.e. f is a function of g, which in turn is a function of x), the derivative $\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = \frac{\mathrm{dg}}{\mathrm{dx}} \cdot \frac{\mathrm{df}}{\mathrm{dg}}$.
Here we would have $g \left(x\right) = {x}^{2} - 3 , \frac{\mathrm{dg}}{\mathrm{dx}} = 2 x , f \left(g\right) = {e}^{g} , \frac{\mathrm{df}}{\mathrm{dg}} = {e}^{g}$. This gives us...
$\frac{d}{\mathrm{dx}} {e}^{{x}^{2} - 3} = 2 x \cdot {e}^{{x}^{2} - 3}$
Oct 26, 2017
How about using the Multiplication Rule?
#### Explanation:
Which is, in general:
$f ' \cdot g + g ' \cdot f$
Here, $f \left(x\right) = {e}^{x}$
and $g \left(x\right) = \left({x}^{2} - 3\right)$
So we have:
$\frac{d}{\mathrm{dx}} \left({e}^{x} \left({x}^{2} - 3\right)\right)$
which fits into the formula like this:
$f ' = {e}^{x}$
(it doesn't change when differentiated)
and $g ' = 2 \cdot x$.
(The exponent comes down in front leaving ${x}^{1}$, which is just $x$ and a constant goes to zero when differentiated.)
So, $\frac{d}{\mathrm{dx}} \left({e}^{x} \left({x}^{2} - 3\right)\right)$ =
${e}^{x} \left({x}^{2} - 3\right) + 2 x \left({e}^{x}\right)$
Let's simplify:
We don't need those last parentheses:
${e}^{x} \left({x}^{2} - 3\right) + 2 x {e}^{x}$
Now we can factor out the ${e}^{x}$ and get:
${e}^{x} \left[\left({x}^{2} - 3\right) + 2 x\right]$,
which simplifies even more to:
${e}^{x} \left({x}^{2} + 2 x - 3\right)$.
The quadratic if factorable:
${x}^{2} + 2 x - 3$ = $\left(x = 3\right) \left(x - 1\right)$
Me, I'd leave it like this:
${e}^{x} \left({x}^{2} + 2 x - 3\right)$.
I don't know if your teacher would want:
${e}^{x} \left[\left(x = 3\right) \left(x - 1\right)\right]$. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Algebra - Basic Go to the latest version.
# 5.2: Linear Equations in Point-Slope Form
Difficulty Level: At Grade Created by: CK-12
Equations can be written in many forms. The last lesson taught you how to write equations of lines in slope-intercept form. This lesson will provide a second way to write an equation of a line: point-slope form.
The line between any two points (x1,y1)\begin{align*}(x_1,y_1)\end{align*} and (x2,y2)\begin{align*}(x_2,y_2)\end{align*} can be written in the following equation: yy1=m(xx1)\begin{align*}y-y_1=m(x-x_1)\end{align*}.
To write an equation in point-slope form, you need two things:
1. The slope of the line
2. A point on the line
Example 1: Write an equation for a line containing (9, 3) and (4, 5).
Solution: Begin by finding the slope.
slope=y2y1x2x1=5349=25
Instead of trying to find b\begin{align*}b\end{align*} (the y\begin{align*}y-\end{align*}intercept), you will use the point-slope formula.
yy1y3=m(xx1)=25(x9)
It doesn't matter which point you use.
You could also use the other ordered pair to write the equation:
y5=25(x4)
These equations may look completely different, but by solving each one for y\begin{align*}y\end{align*}, you can compare the slope-intercept form to check your answer.
y3yy5yy=25(x9)y=25x+185+3=25x+335=25(x4)=25x+85+5=25x+335
This process is called rewriting in slope-intercept form.
Example 2: Rewrite y5=3(x2)\begin{align*}y-5=3(x-2)\end{align*} in slope-intercept form.
Solution: Use the Distributive Property to simplify the right side of the equation.
y5=3x6
Solve for y\begin{align*}y\end{align*}:
y5+5y=3x6+5=3x1
## Graphing Equations Using Point-Slope Form
If you are given an equation in point-slope form, it is not necessary to re-write it in slope-intercept form in order to graph it. The point-slope form of the equation gives you enough information so you can graph the line.
Example 3: Make a graph of the line given by the equation y2=23(x+2)\begin{align*}y-2=\frac{2}{3}(x+2)\end{align*}
Solution: Begin by rewriting the equation to make point-slope form: y2=23(x(2))\begin{align*}y-2= \frac{2}{3}(x-(-2))\end{align*} Now we see that point (–2, 2) is on the line and that the slope=23\begin{align*}\text{slope}=\frac{2}{3}\end{align*}. First plot point (–2, 2) on the graph.
A slope of 23\begin{align*}\frac{2}{3}\end{align*} tells you that from your point you should move 2 units up and 3 units to the right and draw another point.
Now draw a line through the two points and extend the line in both directions.
## Writing a Linear Function in Point-Slope Form
Remember from the previous lesson that f(x)\begin{align*}f(x)\end{align*} and y\begin{align*}y\end{align*} are used interchangeably. Therefore, to write a function in point-slope form, you replace yy1\begin{align*}y-y_1\end{align*} with f(x)y1\begin{align*}f(x)-y_1\end{align*}.
Example 4: Write the equation of the linear function in point-slope form.
m=9.8\begin{align*}m=9.8\end{align*} and f(5.5)=12.5\begin{align*}f(5.5)=12.5\end{align*}
Solution: This function has a slope of 9.8 and contains the ordered pair (5.5, 12.5). Substituting the appropriate values into point-slope form,
y12.5=9.8(x5.5)
Replacing yy1\begin{align*}y-y_1\end{align*} with f(x)y1\begin{align*}f(x)-y_1\end{align*}, the equation becomes
f(x)12.5f(x)12.5=9.8x53.9f(x)=9.8x41.4=9.8(x5.5)
## Solving Situations Involving Point-Slope Form
Let’s solve some word problems where point-slope form is needed.
Example 5: Marciel rented a moving truck for the day. Marciel remembers only that the rental truck company charges $40 per day and some amount of cents per mile. Marciel drives 46 miles and the final amount of the bill (before tax) is$63. What is the amount per mile the truck rental company charges? Write an equation in point-slope form that describes this situation. How much would it cost to rent this truck if Marciel drove 220 miles?
Solution: Define the variables: x=\begin{align*}x=\end{align*} distance in miles; y=\begin{align*}y=\end{align*} cost of the rental truck in dollars. There are two ordered pairs: (0, 40) and (46, 63).
Step 1: Begin by finding the slope: 6340460=2346=12\begin{align*}\frac{63-40}{46-0}=\frac{23}{46}=\frac{1}{2}\end{align*}.
Step 2: Substitute the slope for m\begin{align*}m\end{align*} and one of the coordinates for (x1,y1)\begin{align*}(x_1,y_1)\end{align*}.
y40=12(x0)
To find out how much will it cost to rent the truck for 220 miles, substitute 220 for the variable x\begin{align*}x\end{align*}.
y40y40=12(2200)=0.5(220)y=150 ## Practice Set Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. 1. What is the equation for a line containing the points (x1,y1)\begin{align*}(x_1,y_1)\end{align*} and (x2,y2)\begin{align*}(x_2,y_2)\end{align*} in point-slope form? 2. In what ways is it easier to use point-slope form rather than slope-intercept form? Write the equation for the line in point-slope form. 1. Slope is 13\begin{align*}\frac{1}{3}\end{align*}; y\begin{align*}y-\end{align*}intercept –4. 2. Slope is 110\begin{align*}-\frac{1}{10}\end{align*} and contains (10, 2). 3. Slope is –75 and contains point (0, 125). 4. Slope is 10 and contains point (8, –2). 5. The line contains points (–2, 3) and (–1, –2). 6. The line contains the points (0, 0) and (1, 2). 7. The line contains points (10, 12) and (5, 25). 8. The line contains points (2, 3) and (0, 3). 9. The line has slope 35\begin{align*}\frac{3}{5}\end{align*} and y\begin{align*}y-\end{align*}intercept –3. 10. The line has slope –6 and y\begin{align*}y-\end{align*}intercept 0.5. 11. The line contains the points (–4, –2) and (8, 12). Write each equation in slope-intercept form. 1. y2=3(x1)\begin{align*}y-2=3(x-1)\end{align*} 2. y+4=23(x+6)\begin{align*}y+4=\frac{-2}{3}(x+6)\end{align*} 3. 0=x+5\begin{align*}0=x+5\end{align*} 4. y=14(x24)\begin{align*}y=\frac{1}{4}(x-24)\end{align*} In 18 – 25, write the equation of the linear function in point-slope form. 1. m=15\begin{align*}m=-\frac{1}{5}\end{align*} and f(0)=7\begin{align*}f(0)=7\end{align*} 2. \begin{align*}m=-12\end{align*} and \begin{align*}f(-2)=5\end{align*} 3. \begin{align*}f(-7)=5\end{align*} and \begin{align*}f(3)=-4\end{align*} 4. \begin{align*}f(6)=0\end{align*} and \begin{align*}f(0)=6\end{align*} 5. \begin{align*}m=3\end{align*} and \begin{align*}f(2)=-9\end{align*} 6. \begin{align*}m=-\frac{9}{5}\end{align*} and \begin{align*}f(0)=32\end{align*} 7. \begin{align*}m=25\end{align*} and \begin{align*}f(0)=250\end{align*} 8. \begin{align*}f(32)=0\end{align*} and \begin{align*}f(77)=25\end{align*} 9. Nadia is placing different weights on a spring and measuring the length of the stretched spring. She finds that for a 100 gram weight the length of the stretched spring is 20 cm and for a 300 gram weight the length of the stretched spring is 25 cm. Write an equation in point-slope form that describes this situation. What is the unstretched length of the spring? 10. Andrew is a submarine commander. He decides to surface his submarine to periscope depth. It takes him 20 minutes to get from a depth of 400 feet to a depth of 50 feet. Write an equation in point-slope form that describes this situation. What was the submarine’s depth five minutes after it started surfacing? 11. Anne got a job selling window shades. She receives a monthly base salary and a6 commission for each window shade she sells. At the end of the month, she adds up her sales and she figures out that she sold 200 window shades and made 2500. Write an equation in point-slope form that describes this situation. How much is Anne’s monthly base salary? Mixed Review 1. Translate into a sentence: \begin{align*}4(j+2)=400\end{align*}. 2. Evaluate \begin{align*}0.45 \cdot 0.25-24 \div \frac{1}{4}\end{align*}. 3. The formula to convert Fahrenheit to Celsius is \begin{align*}C(F)=\frac{F-32}{1.8}\end{align*}. What is the Celsius equivalent to \begin{align*}35^\circ F\end{align*}? 4. Find the rate of change: The diver dove 120 meters in 3 minutes. 5. What percent of 87.4 is 106? 6. Find the percent of change: The original price was25.00. The new price is \$40.63.
7. Solve for \begin{align*}w: \ 606=0.045(w-4000)+0.07w\end{align*}.
8 , 9
Feb 22, 2012
Dec 11, 2014 |
Title
College Algebra
Answer/Discussion to Practice Problems
Tutorial 54A: Sequences
Answer/Discussion to 1a ; 1 < n < 5
Basically, to find the nth term of a sequence works in the same fashion as function notation. If you want to find the 3rd term, you are looking for , which means you plug in 3 for n into the given function. So, what are we going to plug in for n to find the 1st term? If you said 1, give yourself a pat on the back. What about the 2nd term? I hope you said you would plug in 2 for n. Since we have to go from 1 < n < 5, this means we need to find 5 terms and we will be plugging in 1, 2, 3, 4, and 5 for n. Let's see what we get:
*1st term, n = 1
*2nd term, n = 2
*3rd term, n = 3
*4th term, n = 4
*5th term, n = 5
Note how we had -1 raised to n, which changes value, and the signs of the terms alternated. The five terms of this sequence are 1/4, -1/8, 1/16, -1/32, and 1/64.
Answer/Discussion to 2a
This function contains a factorial. Let's see what we get for our first five terms:
*1st term, n = 1
*2nd term, n = 2
*3rd term, n = 3
*4th term, n = 4
*5th term, n = 5
Now let's check out the twelfth term:
*12th term, n = 12
Answer/Discussion to 3a
Let's take a look at what is happening here:
Something that is always constant is that each term contains e.
There are also two things that change.
First let's look at the alternating signs:
For it to have alternating signs, we need to have (-1) raised to a power that changes. This means n, the term number is involved. The first term is positive, the second term is negative, the third positive, the fourth negative and so forth. When n is odd (1, 3, 5, ...), then the term is positive. When n is even (2, 4, 6, ...), then the term is negative. So do you think we are going to have or . If you said you are correct!!! If n is odd, then this term will be positive. If n is even, then this term will be negative.
We also have the exponent on each term.
Again we need to figure out the relationship between n and the exponent:
When n is 1, the exponent is 1. When n is 2, the exponent is 4. When n is 3, the exponent is 9. When n is 4, the exponent is 16.
What do you think the relationship is?
It looks like the exponent is always n squared.
Putting it together, the formula for the nth term is .
Sometimes you have to play around with it before you get it just right. You can always check it by putting in the n values and seeing if you get the given sequence.
This one does check.
Answer/Discussion to 4a ;
We are giving the first term, . Using that we can find the second term and so forth. Let's see what we get for our first three terms:
*2nd term, n = 2
*3rd term, n = 3
Since this is a recursive formula, in order to the fifth term, we need to find the fourth term:
*4th term, n = 4
*5th term, n = 5
Last revised on May 16, 2011 by Kim Seward.
All contents copyright (C) 2002 - 2011, WTAMU and Kim Seward. All rights reserved. |
Teacher resources and professional development across the curriculum
Teacher professional development and classroom resources across the curriculum
Session 5, Part B:
Unfair Allocations
In This Part: Fair and Unfair Allocations | Measuring the Degree of Fairness
Looking at Excesses and Deficits
We've seen how to judge the relative fairness of an allocation. A related question asks how you might determine the number of moves required to make an unfair allocation fair, without making the actual moves. Note 4
Below is Allocation A from Problem B2, arranged in ascending order. We've already determined that only two moves are required to make this allocation fair. Let's look more closely at why this is true:
The notations below each stack indicate the number of coins that each stack is above or below the mean of 5. In other words:
a. Two stacks are above the mean. Each of these has 6 coins, an excess of 1 coin (+1) above the average. (These excesses are noted in the figure above.) The total excess of coins above the average is 2. b. Two stacks are below the mean. Each of these has 4 coins, a deficit of 1 coin (-1) below the average. (These deficits are also noted in the figure above.) The total deficit of coins below the average is 2. c. Five stacks are exactly average and have no excess or deficit (0).
Problem B4
Perform the calculations above to show that 20 moves are required to obtain a fair allocation for Allocation B, which is arranged in ascending order below:
Problem B5 Perform the same calculations to find the number of moves required to obtain a fair allocation for Allocation C, which is shown in ascending order below:
Next > Part C: Using Line Plots
Session 5: Index | Notes | Solutions | Video |
# What Is 10 Of 40?
Are you curious to know what is 10 of 40? You have come to the right place as I am going to tell you everything about 10 of 40 in a very simple explanation. Without further discussion let’s begin to know what is 10 of 40?
Contents
## What Is 10 Of 40?
Percentages are a fundamental concept in mathematics, and they play a crucial role in various aspects of our daily lives, from calculating discounts during sales to understanding interest rates. In this blog, we’ll demystify percentages by exploring a common question: What is 10% of 40? We’ll break down the process step by step to make percentage calculations easy to grasp.
# Understanding Percentages
Before we delve into the calculation, let’s briefly review what percentages are. A percentage is a way of expressing a portion or fraction of a whole out of 100. When you say something is “X percent,” you’re essentially saying it’s “X out of 100.” This concept is used to describe proportions, ratios, and relative values.
# Calculating 10% Of 40
To calculate 10% of 40, you can follow these straightforward steps:
Step 1: Convert the Percentage to a Decimal
The first step is to convert the percentage (10%) into a decimal. To do this, divide the percentage by 100.
10% as a decimal = 10 / 100 = 0.10
So, 10% is equivalent to 0.10 in decimal form.
Step 2: Multiply the Decimal by the Whole Number
Now that you have the percentage as a decimal, you can multiply it by the whole number (40) to find the result.
0.10 (10% as a decimal) × 40 = 4
So, 10% of 40 is equal to 4.
# The Result
In simple terms, 10% of 40 is 4. This means that if you have a total of 40 items, and you want to find 10% of that total, you would have 4 items. You can think of it as taking 10% of the whole quantity, which in this case is 40.
# Practical Applications
Understanding how to calculate percentages, like finding 10% of 40, is valuable in many real-life situations. Here are a few examples:
1. Shopping: If an item is on sale for 10% off and its original price is \$40, you can calculate the discount to be \$4, making the final price \$36.
2. Tip Calculation: When dining at a restaurant, you can calculate the tip by finding a percentage of the bill total.
3. Interest Rates: Understanding percentages is crucial when dealing with loans, mortgages, and investments, as interest rates are often expressed as percentages.
4. Grade Calculations: In education, percentages are used to calculate grades, test scores, and overall performance.
# Conclusion
Calculating percentages, like finding 10% of 40, is a fundamental skill that has practical applications in many aspects of life, from personal finance to academic performance. The process involves converting the percentage to a decimal and then multiplying it by the whole number. With this knowledge, you’ll be better equipped to make informed decisions and solve a wide range of problems that involve percentages. It’s a valuable tool that can help you navigate various scenarios with confidence and accuracy.
You can find more about the different largest things on Largably.
## FAQ
### What Is 10% Increase Of 40?
Increased value is \$40+4=44\$. Note: We can also use the concept of 100 to evaluate the increased value of the number.
15% of 40 is 6.
### How Is 10% Calculated?
While 10 percent of any amount is the amount multiplied by 0.1, an easier way to calculate 10 percent is to divide the amount by 10. So, 10 percent of \$18.40, divided by 10, equates to \$1.84.
### What Is 20% Out Of 40?
20% of 40 is 8.
I Have Covered All The Following Queries And Topics In The Above Article
10 Is What Percent Of 40
What Is The Greatest Common Factor Of 10 And 40?
What Percent Of 40 Is 10
What Is 10 Of 40
What is 40 percent out of 10 |
Engage NY Eureka Math 4th Grade Module 4 Lesson 14 Answer Key
Eureka Math Grade 4 Module 4 Lesson 14 Problem Set Answer Key
1. Draw triangles that fit the following classifications. Use a ruler and protractor. Label the side lengths and angles.
Question a.
Right and isosceles
Question b.
Obtuse and scalene
Question c.
Acute and scalene
Question d.
Acute and isosceles
Question 2.
Draw all possible lines of symmetry in the triangles above. Explain why some of the triangles do not have lines of symmetry.
Are the following statements true or false? Explain using pictures or words.
Question 3.
If ABC is an equilateral triangle, $$\overline{B C}$$ must be 2 cm. True or False?
Question 4.
A triangle cannot have one obtuse angle and one right angle. True or False?
Question 5.
∆ EFG can be described as a right triangle and an isosceles triangle. True or False?
Question 6.
An equilateral triangle is isosceles. True or False?
Extension:
In ∆ HIJ, a = b. True or False?
Eureka Math Grade 4 Module 4 Lesson 14 Exit Ticket Answer Key
Question 1.
Draw an obtuse isosceles triangle, and then draw any lines of symmetry if they exist.
Question 2.
Draw a right scalene triangle, and then draw any lines of symmetry if they exist.
Question 3.
Every triangle has at least ____ acute angles.
Question 1.
Draw triangles that fit the following classifications. Use a ruler and protractor. Label the side lengths and angles.
a. Right and isosceles
b. Right and scalene
c. Obtuse and isosceles
d. Acute and scalene
Question 2.
Draw all possible lines of symmetry in the triangles above. Explain why some of the triangles do not have lines of symmetry.
Are the following statements true or false? Explain.
Question 3.
∆ ABC is an isosceles triangle. $$\overline{A B}$$ must be 2 cm. True or False?
Question 4.
A triangle cannot have both an acute angle and a right angle. True or False?
Question 5.
∆ XYZ can be described as both equilateral and acute. True or False? |
# Prove by induction $\vphantom{\Large A}3\mid\left(n^{3} - n\right)$
So I'm just studying for my midterm and I came across this exercise:
Prove by mathematical induction that $\vphantom{\Large A}3\mid\left(n^{3} - n\right)$ for every positive integer $n$.
What does the pipe symbol "$\mid$" mean? I have never seen it before.
• that pipe symbol means that $3$ divides $n^3-n$... you could have at least showed what you have tried...
– user87543
Dec 4, 2013 at 2:23
If $n = 1$, then $n^3 - n = 0 = 3 \times 0$. So, result holds for the base case. Now let us suppose your statement holdsl for $n \in \mathbb{N}$. In other words, suppose
$$n^3 - n = 3k_0$$
We want to show $(n+1)^3 - (n+1) = 3k_1$ for some $k_1$.
But, Notice
$$(n+1)^3 - (n+1) = n^3 + 3n + 3n^2 + 1 + n - 1 = n^3 - n + 3(n + n^2) + 3k_0 + 3(n + n^2) = 3(k_0 + n + n^2) \cong_3 0$$
Hence, the result follows by mathematical induction.
Note that in this case you have a direct proof:
$n^3-n = n(n^2-1) =n(n-1)(n+1)$ and this is the product of $3$ consecutive integers, so one of them must be divisible by $3$.
Though this is not what you asked, I find it often useful to know more than one proof of a result, because this gives me additional insights which can be useful in other situations.
$a \vert b$ means $a$ divides $b$, i.e., $b$ is a multiple of $a$, i.e., $b = ak$, where $k \in \mathbb{Z}$.
To prove by induction, prove for $n=1$ and note that $$\left((n+1)^3 - (n+1) \right) - \left(n^3 - n\right) = \left(n^3 + 3n^2 + 3n + 1 - (n+1)\right) - n^3 + n = 3n^2 + 3n$$ Now use induction hypothesis to conclude what you want. |
# What is the domain and range of y = (2x^2)/( x^2 - 1)?
Aug 7, 2017
The domain is $x \in \left(- \infty , - 1\right) \cup \left(- 1 , 1\right) \cup \left(1 , + \infty\right)$
The range is $y \in \left(- \infty , 0\right] \cup \left(2 , + \infty\right)$
#### Explanation:
The function is
$y = \frac{2 {x}^{2}}{{x}^{2} - 1}$
We factorise the denominator
$y = \frac{2 {x}^{2}}{\left(x + 1\right) \left(x - 1\right)}$
Therefore,
$x \ne 1$ and $x \ne - 1$
The domain of y is $x \in \left(- \infty , - 1\right) \cup \left(- 1 , 1\right) \cup \left(1 , + \infty\right)$
Let's rearrage the function
$y \left({x}^{2} - 1\right) = 2 {x}^{2}$
$y {x}^{2} - y = 2 {x}^{2}$
$y {x}^{2} - 2 {x}^{2} = y$
${x}^{2} = \frac{y}{y - 2}$
$x = \sqrt{\frac{y}{y - 2}}$
For $x$ to a solution, $\frac{y}{y - 2} \ge 0$
Let $f \left(y\right) = \frac{y}{y - 2}$
We need a sign chart
$\textcolor{w h i t e}{a a a a}$$y$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a}$$0$$\textcolor{w h i t e}{a a a a a a a}$$2$$\textcolor{w h i t e}{a a a a}$$+ \infty$
$\textcolor{w h i t e}{a a a a}$$y$$\textcolor{w h i t e}{a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$y - 2$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a}$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a a}$$| |$$\textcolor{w h i t e}{a a}$$+$
$\textcolor{w h i t e}{a a a a}$$f \left(y\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a}$$| |$$\textcolor{w h i t e}{a a}$$+$
Therefore,
$f \left(y\right) \ge 0$ when $y \in \left(- \infty , 0\right] \cup \left(2 , + \infty\right)$
graph{2(x^2)/(x^2-1) [-16.02, 16.02, -8.01, 8.01]} |
Question Video: Finding the Limit of a Rational Function | Nagwa Question Video: Finding the Limit of a Rational Function | Nagwa
# Question Video: Finding the Limit of a Rational Function Mathematics
Find lim_(π₯ β 5) (π₯β΄ β 625)/(π₯Β³ β 125).
05:02
### Video Transcript
Find the limit as π₯ approaches five of π₯ to the fourth power minus 625 all divided by π₯ cubed minus 125.
In this question, weβre asked to evaluate the limit of a rational function. And we recall we can try doing this by using direct substitution. We substitute π₯ is equal to five into our rational function to get five to the fourth power minus 625 divided by five cubed minus 125. However, if we evaluate the numerator and denominator of this expression, we see itβs equal to zero divided by zero. This is an indeterminate form. This does not mean that we canβt evaluate this limit. All it means is we canβt evaluate this limit by using direct substitution alone. So weβre going to need to use a different method.
And there are a few different ways of evaluating this limit. The easiest way is to notice that this limit is given in the form of the limit of a difference between powers. This means we can evaluate this limit by recalling the following result. The limit as π₯ approaches some constant π of π₯ to the πth power minus π to the πth power all divided by π₯ to the πth power minus π to the πth power is equal to π divided by π multiplied by π to the power of π minus π. And this is provided π is not equal to zero and π to the πth power, π to the πth power, and π to the power of π minus π all exist.
We know 625 is five to the fourth power and 125 is five cubed. So we can rewrite our limit as the limit as π₯ approaches five of π₯ to the fourth power minus five to the fourth power all divided by π₯ cubed minus five cubed. Now we can see weβve rewritten this in the form of a limit of a difference of powers. Our value of π is five, our value of π is four, and our value of π is three. Therefore, we can evaluate our limit by using our limit result. We substitute these values in to get four-thirds multiplied by five to the power of four minus three. And we can evaluate this expression. Four minus three is equal to one, and five to the first power is equal to itself. Itβs equal to five. Five times four over three is 20 divided by three, which is our final answer. And this is the easiest way of answering this question.
However, it can be useful to recall algebraically why this result holds true. So instead, weβll also go through the algebraic way of evaluating this limit. This will help reiterate why our limit result in general holds true. To evaluate this limit algebraically, we recall when we substituted π₯ is equal to five into our rational function, we got zero divided by zero. By the remainder theorem, this means the polynomial in our numerator and the polynomial in our denominator are divisible by π₯ minus five. To evaluate this limit algebraically, we want to cancel all of the shared factors of π₯ minus five in the numerator and denominator. So weβll start by factoring both of these polynomials.
Letβs start with the polynomial in our numerator. We can see this is a difference between squares since π₯ to the fourth power is π₯ squared all squared and 625 is 25 squared. Therefore, we can rewrite our numerator as π₯ squared plus 25 multiplied by π₯ squared minus 25. And we can even notice something interesting. π₯ squared minus 25 is also a difference between two squares. So we can already factor this further. Itβs equal to π₯ minus five multiplied by π₯ plus five.
Now, letβs move on to factoring our denominator. Thereβs a few different ways of doing this. For example, we could use algebraic division to take out a factor of π₯ minus five. However, the easiest way is to recall the following result. For a positive integer π, π₯ to the πth power minus π to the πth power is equal to π₯ minus π multiplied by π₯ to the power of π minus one plus π times π₯ to the power of π minus two. And we add terms of this form all the way up to π to the power of π minus one. Applying this result, we get the denominator is equal to π₯ minus five multiplied by π₯ squared plus five π₯ plus 25. And this is exactly the same result we would have got by using polynomial division.
And now, we set the limit of these two rational functions to be equal. And now we see weβre taking the limit of this function as π₯ approaches five. And remember, this means we want to know what happens to the outputs of our function as our values of π₯ get closer and closer to five. So the value of our function when π₯ equals five will not change the value of this limit. This means we can cancel the shared factor of π₯ minus five in the numerator and denominator of this function. If π₯ is not equal to five, π₯ minus five divided by π₯ minus five is equal to one. This then gives us the limit as π₯ approaches five of π₯ squared plus 25 multiplied by π₯ plus five all divided by π₯ squared plus five π₯ plus 25.
And now, we can just evaluate this limit by using direct substitution. Substituting π₯ is equal to five into our rational function, we get five squared plus 25 multiplied by five plus five all divided by five squared plus five times five plus 25. We can then evaluate this expression. We get 50 times 10 divided by 75, which, if we simplify, is equal to 20 divided by three. Therefore, we were able to show the limit as π₯ approaches five of π₯ to the fourth power minus 625 all divided by π₯ cubed minus 125 is equal to 20 divided by three. |
How to
# How to find Instantaneous Velocity in Calculus – Simple Steps to Solve It
Getting stuck while doing Calculus is probably one of the most frustrating feelings ever. It is already a tough subject, and if you are like me, you probably dread working on it. But, there’s no way out because it is important. All of us have to go through the struggle and work hard towards achieving a desirable result.
But, this doesn’t mean that the path should be difficult. If there are ways you can make things easier for yourself, you should try them! We are here to help you with a few problems. In this guide, we will speak about instantaneous velocity and how you can find it.
### What is velocity?
Velocity is nothing but the measurement of how quickly a given object can move from point A to point B. It can also change with time depending on whether it is accelerating, decelerating, or moving at a constant speed.
### What is instantaneous velocity?
Instantaneous velocity is the velocity of an object at a given time. It describes the limit of velocity instead of speaking of the object’s change in position. Here are some points that you should remember:
• Instantaneous velocity is a vector. This means that it has a direction and a value.
• The units used in this case are meters per second or m/s.
## How to find Instantaneous Velocity in Calculus?
The formula to calculate instantaneous velocity
Instantaneous velocity = limit as a change in time goes to zero (change in position/change in time)
If you remember this formula, you will be able to solve your instantaneous velocity-related questions with the utmost ease.
Solving instantaneous velocity from a graph
The graph in question is a displacement versus time graph. You will need to learn the graph’s equation to solve Calculus instantaneous velocity. The equation is d = f(t). Start with t and find its derivative. Then, you can determine the equation for velocity through this because it will be a function of time.
Now, all you have to do is substitute the values, and you will get the answer. That’s the instantaneous velocity at a given time.
It may seem extremely daunting, but instantaneous velocity is not difficult to grasp. Once you solve a couple of problems on your own, you will understand better. Try new problems and make it a habit to practice every day. After all, calculus is only fun when you befriend it.
If you have any doubts or questions, please feel free to Comment. We are always happy to help! ☺
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# 4.5 Basic Integration Formulas PowerPoint PPT Presentation
4.5 Basic Integration Formulas. Agenda: develop techniques for integrating algebraic functions . Using new notation, the formula for the antiderivative of a power function can be rewritten as Constant Multiplier Integration Rule:
4.5 Basic Integration Formulas
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#### Presentation Transcript
4.5 Basic Integration Formulas
Agenda: develop techniques for integrating algebraic functions.
Using new notation, the formula for the antiderivative of a power function can be rewritten as
Constant Multiplier Integration Rule:
Proof: This integral is an _________ of ___. Take the differential of
it: by the definition, . But .
Thus,
1
Proof (cntd):
By the constant multiplier (differentiation) rule:
and the equality at the end of the last slide takes the form
Next, we take the indefinite integral (antiderivative) of that:
By the definition of the integral, , we get
2
We can combine the above constant multiplier integration rule and the sum integration rule (the integral of a sum is equal to the sum of integrals) to form the following Linear Combination Rule:
Next,we try to reverse the generalized rules obtained using the chain rule:
Example: Differentiate the function and try to formulate the integration rule corresponding to the reverse operation.
Solution: Then, differential writes as
Now, we know that the function F(x) itself is the antiderivative of its derivative, and it writes as
3
Example (ctnd):
Try to integrate this same function straightforward:
Since 2xdx=d(x2),
Next, we can take out the constant multiplier 2 and split the integral into the sum of integrals, but there is another, more elegant way. Observe that for any constant c. In our case, we use c=1 and write
Next, we denote the function under the differential as a new variable u
and the integral becomes .
The final step is substituting u as a function of x back
4
Exercises: Perform the following integration
5
Homework:
Section 4.5: 1,5,7,9,13,15,17,21,23,25,49,61,67.
6 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Deductive Reasoning
## Drawing conclusions from facts.
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Deductive Reasoning
What if, in a fictitious far-away land, a poor peasant were awaiting his fate from the king? He is standing in a stadium, filled with spectators pointing and wondering what is going to happen. Finally, the king directs everyone’s attention to two doors, at the floor level with the peasant. Both doors have signs on them, which are below:
Door A Door B
IN THIS ROOM THERE IS A LADY, AND IN THE OTHER ROOM THERE IS A TIGER. IN ONE OF THESE ROOMS THERE IS A LADY, AND IN ONE OF THE OTHER ROOMS THERE IS A TIGER.
The king states, “Only one of these statements is true. If you pick correctly, you will find the lady. If not, the tiger will be waiting for you.” Which door should the peasant pick? After completing this Concept, you'll be able to use deductive reasoning to solve this problem.
### Guidance
Logic is the study of reasoning. Deductive reasoning draws conclusions from facts. Typically, conclusions are drawn from general statements about something more specific.
##### Law of Detachment
Here are two true statements:
• Every odd number is the sum of an even and an odd number.
• 5 is an odd number.
What can you conclude? Based on only these two true statements, there is one conclusion: 5 is the sum of an even and an odd number. (This is true, \begin{align*}5 = 3 + 2\end{align*} or \begin{align*}4 + 1\end{align*}). Let’s change this example into symbolic form.
\begin{align*}p: \ \text{A number is odd} \qquad q: \ \text{It is the sum of an even and odd number}\end{align*}
So, the first statement is \begin{align*}p \rightarrow q\end{align*}. The second statement, “5 is an odd number,” is a specific example of \begin{align*}p\end{align*}. “A number” is 5. The conclusion is \begin{align*}q\end{align*}. Again it is a specific example, such as \begin{align*}4 + 1\end{align*} or \begin{align*}2 + 3\end{align*}. The symbolic form is:
All deductive arguments that follow this pattern have a special name, the Law of Detachment. The Law of Detachment says that if \begin{align*}p \rightarrow q\end{align*} is a true statement and given \begin{align*}p\end{align*}, then you can conclude \begin{align*}q\end{align*}. Another way to say the Law of Detachment is: “If \begin{align*}p \rightarrow q\end{align*} is true, and \begin{align*}p\end{align*} is true, then \begin{align*}q\end{align*} is true.”
##### Law of Contrapositive
The following two statements are true:
• If a student is in Geometry, then he or she has passed Algebra I.
• Daniel has not passed Algebra I.
What can you conclude from these two statements? These statements are in the form:
\begin{align*}\sim q\end{align*} is the beginning of the contrapositive \begin{align*}(\sim q \rightarrow \sim p)\end{align*}, therefore the logical conclusion is \begin{align*}\sim p\end{align*}: Daniel is not in Geometry. This example is called the Law of Contrapositive. The Law of Contrapositive says that if \begin{align*}p \rightarrow q\end{align*} is a true statement and given \begin{align*}\sim q\end{align*}, then you can conclude \begin{align*}\sim p\end{align*}. Recall that the logical equivalent to a conditional statement is its contrapositive. Therefore, the Law of Contrapositive is a logical argument.
##### Law of Syllogism
Determine the conclusion from the following true statements.
• If Pete is late, Mark will be late.
• If Mark is late, Karl will be late.
So, if Pete is late, what will happen? If Pete is late, this starts a domino effect of lateness. Mark will be late and Karl will be late too. So, if Pete is late, then Karl will be late, is the logical conclusion. Each “then” becomes the next “if” in a chain of statements. The chain can consist of any number of connected statements. This is called the Law of Syllogism. The Law of Syllogism says that if \begin{align*}p \rightarrow q\end{align*} and \begin{align*}q \rightarrow r\end{align*} are true, then \begin{align*}p \rightarrow r\end{align*} is the logical conclusion.
Typically, when there are more than two linked statements, we continue to use the next letter(s) in the alphabet to represent the next statement(s); \begin{align*}r \rightarrow s, s \rightarrow t\end{align*}, and so on.
#### Example A
Suppose Bea makes the following statements, which are known to be true.
If Central High School wins today, they will go to the regional tournament.
Central High School won today.
What is the logical conclusion?
This is an example of deductive reasoning. There is one logical conclusion if these two statements are true: Central High School will go to the regional tournament.
#### Example B
Here are two true statements. Be careful!
If \begin{align*}\angle A\end{align*} and \begin{align*}\angle B\end{align*} are a linear pair, then \begin{align*}m \angle A + m \angle B = 180^\circ\end{align*}.
\begin{align*}m \angle 1 = 90^\circ\end{align*} and \begin{align*}m \angle 2 = 90^\circ\end{align*}.
What conclusion can you draw from these two statements?
Here there is NO conclusion. These statements are in the form:
We cannot conclude that \begin{align*}\angle 1\end{align*} and \begin{align*}\angle 2\end{align*} are a linear pair. We are told that \begin{align*}m \angle 1 = 90^\circ\end{align*} and \begin{align*}m \angle 2 = 90^\circ\end{align*} and while \begin{align*}90^\circ + 90^\circ = 180^\circ\end{align*}, this does not mean they are a linear pair. Here are two counterexamples.
In both of these counterexamples, \begin{align*}\angle 1\end{align*} and \begin{align*}\angle 2\end{align*} are right angles. In the first, they are vertical angles and in the second, they are two angles in a rectangle.
This is called the Converse Error because the second statement is the conclusion of the first, like the converse of a statement.
#### Example C
Determine the conclusion from the true statements below.
Babies wear diapers.
My little brother does not wear diapers.
The second statement is the equivalent of \begin{align*}\sim q\end{align*}. Therefore, the conclusion is \begin{align*}\sim p\end{align*}, or: My little brother is not a baby.
Watch this video for help with the Examples above.
#### Concept Problem Revisited
Analyze the two statements on the doors.
Door A: IN THIS ROOM THERE IS A LADY, AND IN THE OTHER ROOM THERE IS A TIGER.
Door B: IN ONE OF THESE ROOMS THERE IS A LADY, AND IN ONE OF THE OTHER ROOMS THERE IS A TIGER.
We know that one door is true, so the other one must be false. Let’s assume that Door A is true. That means the lady is behind Door A and the tiger is behind Door B. However, if we read Door B carefully, it says “in one of these rooms,” which means the lady could be behind either door, which is actually the true statement. So, because Door B is the true statement, Door A is false and the tiger is actually behind it. Therefore, the peasant should pick Door B.
### Vocabulary
Logic is the study of reasoning. Deductive reasoning draws conclusions from facts.
### Guided Practice
1. Here are two true statements.
If \begin{align*}\angle A\end{align*} and \begin{align*}\angle B\end{align*} are a linear pair, then \begin{align*}m \angle A + m \angle B = 180^\circ\end{align*}.
\begin{align*}\angle ABC\end{align*} and \begin{align*}\angle CBD\end{align*} are a linear pair.
What conclusion can you draw from this?
2. Determine the conclusion from the true statements below.
If you are not in Chicago, then you can’t be on the \begin{align*}L\end{align*}.
Bill is in Chicago.
3. Determine the conclusion from the true statements below.
If you are not in Chicago, then you can’t be on the \begin{align*}L\end{align*}.
Sally is on the \begin{align*}L\end{align*}.
1. This is an example of the Law of Detachment, therefore:
2. If we were to rewrite this symbolically, it would look like:
This is not in the form of the Law of Contrapositive or the Law of Detachment, so there is no logical conclusion. You cannot conclude that Bill is on the \begin{align*}L\end{align*} because he could be anywhere in Chicago. This is an example of the Inverse Error because the second statement is the negation of the hypothesis, like the beginning of the inverse of a statement.
3. If we were to rewrite this symbolically, it would look like:
Even though it looks a little different, this is an example of the Law of Contrapositive. Therefore, the logical conclusion is: Sally is in Chicago.
### Practice
Determine the logical conclusion and state which law you used (Law of Detachment, Law of Contrapositive, or Law of Syllogism). If no conclusion can be drawn, write “no conclusion.”
1. People who vote for Jane Wannabe are smart people. I voted for Jane Wannabe.
2. If Rae is the driver today then Maria is the driver tomorrow. Ann is the driver today.
3. If a shape is a circle, then it never ends. If it never ends, then it never starts. If it never starts, then it doesn’t exist. If it doesn’t exist, then we don’t need to study it.
4. If you text while driving, then you are unsafe. You are a safe driver.
5. If you wear sunglasses, then it is sunny outside. You are wearing sunglasses.
6. If you wear sunglasses, then it is sunny outside. It is cloudy.
7. I will clean my room if my mom asks me to. I am not cleaning my room.
8. If I go to the park, I bring my dog. If I bring my dog, we play fetch with a stick. If we play fetch, my dog gets dirty. If my dog gets dirty, I give him a bath.
9. Write the symbolic representation of #3. Include your conclusion. Is this a sound argument? Does it make sense?
10. Write the symbolic representation of #1. Include your conclusion.
11. Write the symbolic representation of #7. Include your conclusion.
For questions 12 and 13, rearrange the order of the statements (you may need to use the Law of Contrapositive too) to discover the logical conclusion.
1. If I shop, then I will buy shoes. If I don’t shop, then I didn’t go to the mall. If I need a new watch battery, then I go to the mall.
2. If Anna’s parents don’t buy her ice cream, then she didn’t get an \begin{align*}A\end{align*} on her test. If Anna’s teacher gives notes, Anna writes them down. If Anna didn’t get an \begin{align*}A\end{align*} on her test, then she couldn’t do the homework. If Anna writes down the notes, she can do the homework.
Determine if the problems below represent inductive or deductive reasoning. Briefly explain your answer.
1. John is watching the weather. As the day goes on it gets more and more cloudy and cold. He concludes that it is going to rain.
2. Beth’s 2-year-old sister only eats hot dogs, blueberries and yogurt. Beth decides to give her sister some yogurt because she is hungry.
3. Nolan Ryan has the most strikeouts of any pitcher in Major League Baseball. Jeff debates that he is the best pitcher of all-time for this reason.
4. Ocean currents and waves are dictated by the weather and the phase of the moon. Surfers use this information to determine when it is a good time to hit the water.
5. As Rich is driving along the 405, he notices that as he gets closer to LAX the traffic slows down. As he passes it, it speeds back up. He concludes that anytime he drives past an airport, the traffic will slow down.
6. Amani notices that the milk was left out on the counter. Amani remembers that she put it away after breakfast so it couldn’t be her who left it out. She also remembers hearing her mother tell her brother on several occasions to put the milk back in the refrigerator. She concludes that he must have left it out.
7. At a crime scene, the DNA of four suspects is found to be present. However, three of them have an alibi for the time of the crime. The detectives conclude that the fourth possible suspect must have committed the crime. |
1. ## Flipping 10 coins
Say that a coin is flipped 10 times. What is the probability that the first 3 flips are all heads provided that there are an even number of heads and tails all together?
- I understand that if we simply flip 3 coins, the probability that 3 heads appear is 1/6. How do we take into account the fact that 5 of 10 are definitely heads??
-Thanks
2. Originally Posted by ccdelia7
- I understand that if we simply flip 3 coins, the probability that 3 heads appear is 1/6.
-Thanks
1/8
RonL
3. Originally Posted by ccdelia7
provided that there are an even number of heads and tails all together?
This conveys no information all the 10 flips are either heads of tails, and 10 is an even number.
Do you mean:
provided that there are an even number of heads
RonL
4. Originally Posted by ccdelia7
Say that a coin is flipped 10 times. What is the probability that the first 3 flips are all heads provided that there are an even number of heads and tails all together?
- I understand that if we simply flip 3 coins, the probability that 3 heads appear is 1/6.
First lets correct the mistake (typo?) above.
Flipping a coin three times gives 8 possible outcomes only one of which is “HHH”.
Therefore the probability of “HHH” is 1/8 not 1/6.
Thus the same idea gives an approach to the question.
If in ten flips we have an even number of heads, 0 2 4 6 8 10, then we have an even number of tails.
In ten flips that can happen in $\displaystyle \sum\limits_{k = 0}^5 \binom{10}{2k}$ ways.
Now we must find out how many of those begin the string with “HHH…”.
If we already have three H’s then in the next seven places we must have 1, 3, 5, or 7 H’s to have an even number of heads and tails all together.
The number of ways that can happens is $\displaystyle \sum\limits_{k = 1}^4 \binom{7}{2k-1}$.
Simply divide the two.
5. I apologize! I meant to write equal. There is an equal number of heads and tails. (i.e. - there are 5 heads and 5 tails ) I am sorry for misleading everyone! Thought the problem you have solved is a quite interesting one!!
6. That makes it easy: $\displaystyle \frac {\binom {7}{2}} {\binom {10}{5}}$
7. yea - I am sorry! It's much easier than it looked, I guess!
8. now - I am not seeing why exactly we can rewrite this event as a binomial cofficient?? Can you explain? i can kind of see where you are going with it, but it's still a bit vague. Thanks! |
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# RD Sharma Class 8 Chapter 19 Solutions (Visualising Shapes)
RD Sharma Class 8 Maths Solutions for Chapter 19 ‘Visualising Shapes’ explains the different types of shapes that you might see around. You have already learned what two-dimensional or 2D shapes are, for example, a rectangle, square, triangle, and circle. In this chapter, you will be introduced to three-dimensional or 3D shapes and their representation on the plane.
RD Sharma Class 8 Maths Solutions Chapter 19 contains only 2 exercises and 12 questions. The questions are based on various topics like 2D and 3D figures, Euler’s formula, Polyhedra, etc. You will also visualize figures like prisms and pyramids. By referring to the questions in the exercise, you will gain a more profound knowledge of concepts covered in the chapter.
The faculty of subject matter experts at Instasolv has developed all the answers to the exercise questions of the chapter. Our experts have years of experience, and they aim to provide you with the best guidance. The solutions are written in a simple manner that will help you understand the topics in a better way. You can access these solutions anytime. Moreover, these answers will work as the best revision source during exams. By referring to Instasolv’s solutions to the exercise, you will pass with flying colours in the exams.
## Important Topics for RD Sharma Class 8 Maths Solutions Chapter 19: Visualising Shapes
• 2D Shapes are plain shapes that only have two measurements, like the length and breadth. Examples include rectangle, square, etc.
• 3D Shapes are the solid shapes that have three measurements i.e., length, breadth, and height or depth. Examples include a cube, cuboid, cone, etc.
• A 3-dimensional object looks different from different positions; that’s why they are drawn from a different perspective.
• Faces: solid figures made up of polygonal regions are called faces.
• Edges: The line segments where the faces meet are called edges.
• Vertex: The edges meet at specific points called the vertex.
• Polyhedrons comprise faces, edges, and vertices.
• Regular polyhedrons are regular if its faces are made up of regular polygons, and the same number of faces meet at each vertex. For example, a cube
• Prisms and pyramids are the two essential members of the polyhedron family.
• In a prism, the base and top are congruent polygons and the other faces, i.e., lateral faces are parallelograms in shape whereas, in a pyramid, the base is a polygon and the lateral faces are triangles with a common vertex.
• Euler’s Formula: In a polyhedron with ‘V’ number of vertices, ‘F’ number of faces and ‘E’ number of edges, you will find that F + V = E + 2, i.e., F + V – E = 2. This equation is called Euler’s formula.
### Exercise Discussion for RD Sharma Class 8 Maths Solutions Chapter 19: Visualising Shapes
• The chapter has 2 exercises in total. The first exercise of this chapter contains a total of 7 questions. The questions are based on 2D and 3D shapes.
• In this exercise, you will find questions based on Euler’s formula, polyhedrons, and cubes.
• The second exercise has 5 figure-based questions. In these questions, you have to visualize 2D shapes and form 3D shapes. You will also have to calculate the number of faces, edges, and vertices of these shapes. This exercise tests your overall aptitude and is essential from the CBSE exam point of view.
• If you are aspiring to secure high marks in your examination, then you are advised to practice these exercises regularly.
## Benefits of RD Sharma Class 8 Maths Solutions Chapter 19: Visualising Shapes by Instasolv?
• Instasolv is an online learning platform that provides answers to all the questions of RD Sharma Solutions Class 8 Maths Chapter.
• The solutions have been formulated by our experts which enhances your overall knowledge of concepts.
• Instasolv’s team aims to provide the best guidance and tries to clear all your doubts.
• The faculty team at Instasolv explains questions step by step and shows you the most efficient ways of solving problems.
• All the solutions are written according to the latest CBSE exam pattern for Class 8 Maths.
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## Engineering Mathematics
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Eigen Values and Eigen Vectors for a real Matrix and example problems
• 1
Step-I: To find the characteristic equation IA 11 — O Step-2: To solve the characteristic equation we get characteristic roots. They are called Eigen values. Step-3:To find Eigen vectors (A- 11 ) for the different values of 1 . PROBLEMS: 1. If -1, 1, 4 are the Eigen values of a matrix A of order 3 and , , [1,1, Solution: Eigen values are -1, 1, 4 The eigen vectors are 1] T are corresponding Eigen vectors, determine the matrix A. 0 1 1 2 1 1 1 1 1 The given set of eigen vectors are orthogonal in pair. Therefore the given matrix is a symmetric matrix. Therefore we use orthogonal transformation to Diagonalise that symmetric matrix. 2.Prove that the Eigen vectors corresponding to distinct Eigen values of a real Symmetric matrix are orthogonal. (Nov. 2002) Solution: For a real symmetric matrix A, the Eigen values are real. Let Xl and X be the Eigen vectors corresponding to distinct Eigen values & (real) (1)
• 2
Similarly AX 2 12 X 2 (2) From (l) & (2) , Xl and X 2 are orthogonal 3. Find the Eigen values and Eigen vectors of the matrix Solution: 0 2 2 0 2 2 2 Let A 2 2 D3 1; D2 -12 0 45 To find the characteristic equation For the given matrix DI —2 + 1 + 0 211 45 0 The Eigen values are 5, 3, 3 6=-21; The Eigen vectors are given by (A Il)X 21 2 which is 6 2 For 1 7 5, 2 3 6
• 3
Solving by cross multiplication method 24 For 2 3, 4 2 24 3 6 3 Solving by cross multiplication method — Which does not exist? + 2X2 3X3 0 If 0 The given matrix is not symmetric; therefore the Eigen vectors need not be orthogonal. Therefore we choose the Eigen vector which may dependent or independent of other Eigen vectors. Let 3 4. Find the Eigen values and eigen vectors of the matrix Solution: 2 20 10 30 10 13
• 4
Let 2 20 10 30 10 13 For the given matrix DI — 1; D2 —2; D The characteristic equation is given by The Eigen values are given by 0, —1, 2 The Eigen vectors are given by (A Il)X = O 0 21 0 For 1=0, 20 10 10 Solving by cross multiplication method 20 4 For 0 1, 20 11 1 10 Solving by cross multiplication method 30 o For 1=2, 15 20 0 10 Solving by cross multiplication method
• 5
o 12 24 5.Find the Eigen values and Eigen vectors of the matrix Solution: 2 2 3 I 32 2 1 3 0 . (May 2005) Let A 2 2 3 I 2 1 which is a symmetric matrix 3 For the given matrix DI = 12; D2 — 36; D The characteristic equation is given by The Eigen values are given by 8, 2, 2 The Eigen vectors are given by (A Il)X Forl 8, Solving by cross multiplication method 32 1212 +361 For 12 2, Solving by cross multiplication method
• 6
o o . 4Xl 2X2 -k 2X3 — which is an impossible eigen vector. 0 If Xl = X2 2, To choose the eigen vector, remember we are dealing with a symmetric matrix, in which the eigen vectors are orthogonal in pairs. Let b be the third eigen vector which is orthogonal to Xl & X Similarly X2 L X 3 > a 2c—0 Solving the above two equations by cross multiplication method, we get 5
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# How do you write the standard form of the equation given (3,5) and slope 5/3?
Mar 20, 2017
$5 x - 3 y = 0$
#### Explanation:
Start with the slope-point form: $\left(y - \textcolor{b l u e}{{y}_{1}}\right) = \textcolor{g r e e n}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$
with the point $\left(\textcolor{red}{3} , \textcolor{b l u e}{5}\right)$ and slope $\textcolor{g r e e n}{\frac{5}{3}}$ to get:
$\textcolor{w h i t e}{\text{XXX}} y - \textcolor{b l u e}{5} = \textcolor{g r e e n}{\frac{5}{3}} \left(x - \textcolor{red}{3}\right)$
Our target form is the standard form: $A x + B y = C$
Multiplying both sides of our slope-point form by $3$
and expanding the left side:
$\textcolor{w h i t e}{\text{XXX}} 3 y \cancel{- 15} = 5 x \cancel{- 15}$
Re-arrange into standard form:
$\textcolor{w h i t e}{\text{XXX}} 5 x - 3 y = 0$ |
# CLASS-8LOWEST COMMON MULTIPLE (LCM)
LOWEST COMMON MULTIPLE (LCM)
The lowest common multiple (LCM) of two or more number is the smallest of the common multiples of those numbers
Example- the multiples of 3 should be 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36,…….etc.
The multiples of 5 should be 5, 10, 15, 20, 25, 30, 35, 40, 45, 50,……etc.
The multiples of 6 should be 6, 12, 18, 24, 30, 36, 42, 48, 54, 60,...etc.
From the multiples of the given number, we should find the smallest or the lowest common multiple of 3, 5, & 6 is 30.
Methods of Finding LCM –
You will be pleased to know that, there are only two methods of finding the LCM of two or more numbers respectively.
a) by prime factorization b) by division
To find LCM by PRIME FACTORIZATION –
Step.1) First of all, we have to express each number as a product of prime factors.
Step.2) Take each prime factor the greatest number of times it appears in any of the prime factorization of the numbers.
Step.3) The product of the prime factors in Step.2 is the LCM of the numbers.
Example.- Find the LCM of 125, 150, and 625.
Answer) 125 = 5 X 25 = 5 X 5 X 5
150 = 5 X 30 = 5 X 5 X 6
625 = 5 X 125 = 5 X 5 X 25 = 5 X 5 X 5 X 5
We can observe from the above, the greatest number of times that 5 appears as a factor of any of the numbers is two times
Then LCM = 5 X 5 = 25
To find LCM by division –
Step.1) Divide the numbers by a prime number which is a factor of at least two of the given numbers
Step.2) write the quotients and carry forward the numbers which are not divisible.
Step.3) repeat Step.1 & Step.2 till no two of the numbers have common factors.
Step.4) Please note, the product of the divisors of all the steps and the remaining numbers should be the LCM of the given numbers.
Example.-
Find the LCM of 24, 36, 96
So, LCM of 24, 36, & 96 is
= 2 X 2 X 3 X 2 X 3 X 4
= 288 (Ans.) |
# Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
## Result:
### 7/8 + 5/8 = 3/2 = 1 1/2 = 1.5
Spelled result in words is three halfs (or one and one half).
### How do you solve fractions step by step?
1. Add: 7/8 + 5/8 = 7 + 5/8 = 12/8 = 4 · 3/4 · 2 = 3/2
For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(8, 8) = 8. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 8 × 8 = 64. In the next intermediate step, , cancel by a common factor of 4 gives 3/2.
In words - seven eighths plus five eighths = three halfs.
#### Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
#### Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
## Fractions in word problems:
2 and 1/8th plus 1 and 1/3rd =
• Pizza fractions
Ann ate a third of a pizza and then another quater. Total part of pizza eaten by Ann and how much pizza is left?
• Birthday party
For her youngest son's birthday party, the mother bought 6 3/4 kg of hotdog and 5 1/3 dozens bread rolls. Hotdogs cost 160 per kilogram, and a dozen bread rolls cost 25. How much did she spend in all?
• Three rectangles
Some wire is used to make 3 rectangles: A, B, and C. Rectangle B's dimensions are 3/5 cm larger than Rectangle A's dimensions, and Rectangle C's dimensions are again 3/5 cm larger than Rectangle B's dimensions. Rectangle A is 2 cm by 3 1/5 cm. What is the
• Infinite sum of areas
Above the height of the equilateral triangle ABC is constructed an equilateral triangle A1, B1, C1, of the height of the equilateral triangle built A2, B2, C2, and so on. The procedure is repeated continuously. What is the total sum of the areas of all tr
• Lengths of the pool
Miguel swam 6 lengths of the pool. Mat swam 3 times as far as Miguel. Lionel swam 1/3 as far as Miguel. How many lengths did Mat swim?
• Farmer 5
Farmer Joe ordered 3 bags of soil last month. Each bag weighed 4 ⅖ kilograms. He used the first bag in a week. At the end of this month, there were 2 ¾ kilograms of soil left in the second bag and ⅞ kilograms of soil left in the third bag. How much soil w
• Martin
Martin is making a model of a Native American canoe. He has 5 1/2 feet of wood. He uses 2 3/4 feet for the hull and 1 1/4 feet for a paddle. How much wood does he have left? Martin has feet of wood left.
• Master and apprentice
Master painted the roof in 3 hours and apprentice for 4 hours. How many of roof they painted in hour and how many in three quarters of an hour?
• Berry Smoothie
Rory has 5/8 cup of milk. How much milk does she have left after she doubles the recipe of the smoothie? Berry Smoothie: 2 cups strawberries 1 cup blueberries 1/4 cup milk 1 tbsp (tablespoon) sugar 1/2 tsp (teaspoon) lemon juice 1/8 tsp (teaspoon) vanilla
• Carrie
Carrie picked 2/5 of the raspberries from the garden, and Robin picked some too. When they were finished, 1/3 of the raspberries still needed to be picked. What fraction of the raspberries did Robin pick? Use pictures, numbers or words and write your fi
• Samuel
Samuel has 1/3 of a bag of rice and Isabella has a 1/2 bag of rice. What fraction of are bag of rice do they have altogether?
• Sum of 18
Sum of two fractions is 4 3/7. If one of the fractions is 2 1/5 find the other one . |
#### 2.1.9 Transport equation $$u_t+(1-2 t) u_x= 0$$ IC $$u(x,0)=\frac {1}{1+x^2}$$. Peter Olver textbook, problem 2.2.29
problem number 9
Taken from Peter Olver textbook, Introduction to Partial differential equations. problem 2.2.29
Solve $$u_t+(1-2 t) u_x= 0$$ with IC $$u(x,0)=\frac {1}{1+x^2}$$
Mathematica
$\left \{\left \{u(x,t)\to \frac {1}{2 t^2 x+t^4-2 t^3+t^2-2 t x+x^2+1}\right \}\right \}$
Maple
$u \left (x , t\right ) = \frac {1}{\left (-t^{2}+t -x \right )^{2}+1}$
Hand solution
Solve $u_{t}+\left ( 1-2t\right ) u_{x}=0$ with initial conditions $$u\left ( 0,x\right ) =\frac {1}{1+x^{2}}$$.
Solution
Let $$u=u\left ( x\left ( t\right ) ,t\right )$$. Then \begin {equation} \frac {du}{dt}=\frac {\partial u}{\partial x}\frac {dx}{dt}+\frac {\partial u}{\partial t}\tag {2} \end {equation} Comparing (1),(2) shows that \begin {align} \frac {du}{dt} & =0\tag {3}\\ \frac {dx}{dt} & =\left ( 1-2t\right ) \tag {4} \end {align}
Solving (3) gives\begin {align} u & =u\left ( x\left ( 0\right ) \right ) \nonumber \\ & =\frac {1}{1+x\left ( 0\right ) ^{2}}\tag {5} \end {align}
We just need to find $$x\left ( 0\right )$$ to finish the solution. From (4)$x=t-t^{2}+C$ At $$t=0$$$x\left ( 0\right ) =C$ Hence \begin {align*} x & =t-t^{2}+x\left ( 0\right ) \\ x\left ( 0\right ) & =x-t+t^{2} \end {align*}
Substituting this back into (5) gives$u\left ( x\left ( t\right ) ,t\right ) =\frac {1}{1+\left ( x-t+t^{2}\right ) ^{2}}$ The following is an animation of the solution
Source code used for the above
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If I have x pencils and you have twice as many, how many pencils have you got?
# Algebra - Expressions 1
This Math quiz is called 'Algebra - Expressions 1' and it has been written by teachers to help you if you are studying the subject at middle school. Playing educational quizzes is a fabulous way to learn if you are in the 6th, 7th or 8th grade - aged 11 to 14.
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In Algebra, you can think of an expression as one or a number of mathematical symbols that represent a number or quantity. Expressions can contain any number of numbers, letters and symbols.
See how well you can express yourself by playing this quiz on Expressions!
1.
What is the simplification of 3a + 4b + 6c + 3a - 6c - a - 3b?
4a + 2b - 3c
5a + b
6a - 3b
9a - 2b - c
Be careful to account for all the a's, then all the b's and then all the c's. In this case + 6c and - 6c cancel each other out and therefore they do not appear in the simplification
2.
Which of these is a simplification of 2a x 3a?
5a
6a
23aa
6a2
You must multiply the numbers together AND the letters. Remember a x a = a2
3.
If I have x rabbits and you have twice as many, how many rabbits have you got?
2x
2 + x
2 - x
x2
You have two times the number that I have got and 2 times x is the same as 2x
4.
Jane is c years old and her sister Katie is three years younger so how old is Katie?
c3
c + 3
c - 3
c - 3 - 3
5.
Simplify 5a + 7a.
12a
12aa
57a
57aa
Let us suppose that a = 3. (5 x 3) + (7 x 3) = 36. This is the same as saying 12 x 3 (12a) = 36
6.
Simplify 4a + 5b + 6b + 11a
26ab
15a2 + 11b2
15a + 11b
9a + 17b
Here we are adding numbers together, NOT multiplying them together
7.
Peter has x brothers, Paul has y brothers and Graham has z brothers. Between the three boys, how many brothers have they got?
xyz
x + y + z
x + y - z
x - y + z
Don't get baffled by complicated wording - often the answer is easier than you think
8.
Which of the following is not a mathematical expression?
6
6a x 14
5a - 4b +16c x 8d
Oh my gosh!
The correct answer might well be an English expression but it is NOT a mathematical one!
9.
In an algebraic expression letters are often used to represent what?
Other letters
Only numbers greater than 1
Only numbers less than 1
Values
A letter can be used in place of any number - especially if the number has not yet been worked out
10.
Suzie is x years old and she has a sister called Rosie who is four years older. Rosie's mother was 28 when Rosie was born. How old is Rosie's mother now?
7x
x + 28
x - 4
x + 32
If you got the answer wrong then think carefully through each aspect of the question. Rosie's mother was 28 when Rosie was born. 4 years later, Suzie was born. Suzie is now x years old so Rosie's mother's age is now
28 + 4 + x = 32 + x
Author: Frank Evans |
# 12.4 The regression equation (Page 2/2)
Page 2 / 2
$\epsilon$ = the Greek letter epsilon
For each data point, you can calculate the residuals or errors, ${y}_{i}-{\stackrel{^}{y}}_{i}={\epsilon }_{i}$ for $i=\text{1, 2, 3, ..., 11}$ .
Each $|\epsilon |$ is a vertical distance.
For the example about the third exam scores and the final exam scores for the 11 statistics students, there are 11 data points. Therefore, there are 11 $\epsilon$ values. If you square each $\epsilon$ and add, you get
$\left({\epsilon }_{1}{\right)}^{2}+\left({\epsilon }_{2}{\right)}^{2}+\text{...}+\left({\epsilon }_{11}{\right)}^{2}=\stackrel{11}{\underset{\text{i = 1}}{\Sigma }}{\epsilon }^{2}$
This is called the Sum of Squared Errors (SSE) .
Using calculus, you can determine the values of $a$ and $b$ that make the SSE a minimum. When you make the SSE a minimum, you have determined the points that are on the line of best fit. It turns out thatthe line of best fit has the equation:
$\stackrel{^}{y}=a+\text{bx}$
where $a=\overline{y}-b\cdot \overline{x}$ and $b=\frac{\Sigma \left(x-\overline{x}\right)\cdot \left(y-\overline{y}\right)}{{\Sigma \left(x-\overline{x}\right)}^{2}}$ .
$\overline{x}$ and $\overline{y}$ are the sample means of the $x$ values and the $y$ values, respectively. The best fit line always passes through the point $\left(\overline{x},\overline{y}\right)$ .
The slope $b$ can be written as $b=r\cdot \left(\frac{{s}_{y}}{{s}_{x}}\right)$ where ${s}_{y}$ = the standard deviation of the $y$ values and ${s}_{x}$ = the standard deviation of the $x$ values. $r$ is the correlation coefficient which is discussed in the next section.
## Least squares criteria for best fit
The process of fitting the best fit line is called linear regression . The idea behind finding the best fit line is based on the assumption that the data are scattered about a straight line. The criteria for the best fit line is that the sum of the squared errors (SSE) is minimized, that is made as small as possible. Any other line you might choose would have a higher SSE than the best fit line. This best fit line is called the least squares regression line .
Computer spreadsheets, statistical software, and many calculators can quickly calculate the best fit line and create the graphs. The calculations tend to be tedious if done by hand. Instructions to use the TI-83, TI-83+, and TI-84+ calculators to find the best fit line and create a scatterplot are shown at the end of this section.
## Third exam vs final exam example:
The graph of the line of best fit for the third exam/final exam example is shown below:
The least squares regression line (best fit line) for the third exam/final exam example has the equation:
$\stackrel{^}{y}=-173.51+\text{4.83x}\phantom{\rule{20pt}{0ex}}$
• Remember, it is always important to plot a scatter diagram first. If the scatter plot indicates that there is a linear relationship betweenthe variables, then it is reasonable to use a best fit line to make predictions for $y$ given $x$ within the domain of $x$ -values in the sample data, but not necessarily for $x$ -values outside that domain.
• You could use the line to predict the final exam score for a student who earned a grade of 73 on the third exam.
• You should NOT use the line to predict the final exam score for a student who earned a grade of 50 on the third exam, because 50 is not within the domain of the x-values in the sample data, which are between 65 and 75.
## Understanding slope
The slope of the line, b, describes how changes in the variables are related. It is important to interpret the slope of the line in the context of the situation represented by the data. You should be able to write a sentence interpreting the slope in plain English.
INTERPRETATION OF THE SLOPE: The slope of the best fit line tells us how the dependent variable (y) changes for every one unit increase in the independent (x) variable, on average.
## Third exam vs final exam example
• Slope: The slope of the line is b = 4.83.
• Interpretation: For a one point increase in the score on the third exam, the final exam score increases by 4.83 points, on average.
## Using the linear regression t test: linregttest
1. In the STAT list editor, enter the X data in list L1 and the Y data in list L2, paired so that the corresponding (x,y) values are next to each other in the lists. (If a particular pair of values is repeated, enter it as many times as it appears in the data.)
2. On the STAT TESTS menu, scroll down with the cursor to select the LinRegTTest. (Be careful to select LinRegTTest as some calculators may also have a different item called LinRegTInt.)
3. On the LinRegTTest input screen enter: Xlist: L1 ; Ylist: L2 ; Freq: 1
4. On the next line, at the prompt β or ρ, highlight "≠ 0" and press ENTER
5. Leave the line for "RegEq:" blank
6. Highlight Calculate and press ENTER.
The output screen contains a lot of information. For now we will focus on a few items from the output, and will return later to the other items.
• The second line says y=a+bx. Scroll down to find the values a=-173.513, and b=4.8273 ; the equation of the best fit line is $\stackrel{^}{y}=-173.51+\text{4.83}x\phantom{\rule{20pt}{0ex}}$
• The two items at the bottom are $r^{2}$ = .43969 and $r$ =.663. For now, just note where to find these values; we will discuss them in the next two sections.
## Graphing the scatterplot and regression line
1. We are assuming your X data is already entered in list L1 and your Y data is in list L2
2. Press 2nd STATPLOT ENTER to use Plot 1
3. On the input screen for PLOT 1, highlight On and press ENTER
4. For TYPE: highlight the very first icon which is the scatterplot and press ENTER
5. Indicate Xlist: L1 and Ylist: L2
6. For Mark: it does not matter which symbol you highlight.
7. Press the ZOOM key and then the number 9 (for menu item "ZoomStat") ; the calculator will fit the window to the data
8. To graph the best fit line, press the "Y=" key and type the equation -173.5+4.83X into equation Y1. (The X key is immediately left of the STAT key). Press ZOOM 9 again to graph it.
9. Optional: If you want to change the viewing window, press the WINDOW key. Enter your desired window using Xmin, Xmax, Ymin, Ymax
**With contributions from Roberta Bloom
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
how can I make nanorobot?
Lily
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
Researchers demonstrated that the hippocampus functions in memory processing by creating lesions in the hippocampi of rats, which resulted in ________.
The formulation of new memories is sometimes called ________, and the process of bringing up old memories is called ________.
1 It is estimated that 30% of all drivers have some kind of medical aid in South Africa. What is the probability that in a sample of 10 drivers: 3.1.1 Exactly 4 will have a medical aid. (8) 3.1.2 At least 2 will have a medical aid. (8) 3.1.3 More than 9 will have a medical aid. |
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# Tutor profile: David D.
Inactive
David D.
Practicing Engineer that loves to teach math
Tutor Satisfaction Guarantee
## Questions
### Subject:Pre-Calculus
TutorMe
Question:
Find the x-intercept of the line \$\$L_1\$\$, which is perpendicular to \$\$-4x + 3y = 12\$\$ and passes through the point \$\$(0,4)\$\$.
Inactive
David D.
The first step of solving a problem like this is to determine the equation for the line \$\$L_1\$\$. The information we have is that \$\$L_1\$\$ is perpendicular to \$\$-4x + 3y = 12\$\$ and passes through the point \$\$(0,4)\$\$. Let's start by converting the line we're given into \$\$y=mx+b\$\$ format. \$\$-4x + 3y = 12\$\$ \$\$3y = 4x + 12\$\$ \$\$y = (4/3)x + 4\$\$ Knowing the equation for the line, we can determine the slope of \$\$L_1\$\$. Perpendicular lines have opposite, reciprocal slopes. The slope of the line we have is \$\$(4/3)\$\$. The opposite, reciprocal of \$\$(4/3)\$\$ is \$\$(-3/4)\$\$. Therefore, the slope of \$\$L_1\$\$ is \$\$(-3/4)\$\$. We now can say, \$\$L_1 = y = (-3/4)x + b\$\$ But, we still need to find the y-intercept, \$\$b\$\$. Luckily, we are given a point that \$\$L_1\$\$ passes through, so we can plug that into our equation to find \$\$b\$\$. \$\$y = (-3/4)x + b\$\$ \$\$4 = (-3/4)*(0) + b\$\$ \$\$b = 4\$\$ Now, we know our y-intercept. We also could have noticed that the point \$\$(0,4)\$\$ already lies on the y-axis, but it is always better to show our work. Our final equation for the line \$\$L_1\$\$ then becomes \$\$L_1 = y = (-3/4)x + 4\$\$ The problem asks for the x-intercept of this line. To find that, we can just plug \$\$0\$\$ in for \$\$y\$\$ and solve for \$\$x\$\$. \$\$0 = (-3/4)x + 4\$\$ \$\$ -4 = (-3/4)x\$\$, multiply both sides by (-4/3) \$\$x = 16/3\$\$ Our final answer is the x-intercept is \$\$16/3\$\$.
### Subject:Trigonometry
TutorMe
Question:
A man stands om the flat ground 100 m away from 173 m tall building. A plane is flying a ground distance of 900 m directly behind the building (1 km away from the man) at an altitude of 1.5 km. Can the man see the plane? (Do not take into account the height of the man)
Inactive
David D.
The best way to solve this problem is to construct two triangles using the distances given in the question. The first triangle will be built from the man, the ground, and the building. The ground and the building meet at the base of the building forming a right angle. The height of the building is one leg of the triangle and the ground between the man and the building makes the other leg. The hypotenuse of this triangle is the line from the man to the top of the building. Therefore, we can begin to construct our triangle with the information we have. We know that the horizontal leg (the distance from the man to the building) is 100 m. We also know that the vertical leg (the height of the building) is 173 m. Using this information, we need to find the angle the hypotenuse makes with the ground. Let's call this angle \$\$theta\$\$. Using the triangle we have constructed, we can say that the tangent of \$\$theta\$\$ is the height of the building divided by the ground distance from the man to the building. This gives us, \$\$tan(theta) = 173/100\$\$ Using the inverse tangent function, we get \$\$theta = 60\$\$ deg (approximately) \$\$theta\$\$ represents the line of sight that the man has to the top of the building. Anything that is behind the building must have a line of sight angle greater than \$\$theta\$\$ or it will be blocked by the building. Let's check the man's line of sight angle to the plane. We know the plane is a horizontal distance of 1 km or 1,000 m away from the man and a vertical distance of 1.5 km or 1,500 m from the ground. Those two distances make up the legs of our triangle. Using the same process as with the building, we can calculate the line of sight angle from the man to the plane which we will call \$\$psi\$\$. After building our triangle, we can show that the tangent of \$\$psi\$\$ is the altitude of the plane divided by the horizontal distance between the man and the plane. This gives us, \$\$tan(psi) = 1500/1000\$\$ Using the inverse tangent function, we get \$\$psi = 56\$\$ deg (approximately) Since \$\$psi < theta\$\$, it can be concluded that the building blocks the line of sight from the man to the plane. Therefore, the answer to the question is no, the man cannot see the plane.
### Subject:Algebra
TutorMe
Question:
What is the simplified expression for the following? \$\$(3x^2 + 5x + 2)/(x^2 - 3x - 4)\$\$
Inactive
David D.
The first step in solving a problem like this is checking that the numerator is not evenly divisible by the denominator. We can check this by first comparing the orders of the terms in the numerator and the denominator. We can see by looking at the numerator and the denominator that they both share an \$\$x^2\$\$ term, an \$\$x\$\$ term, and a ones term. Since the orders of these terms match, we do not have to worry about having an \$\$x\$\$ term in the quotient. Now, we can check if the numerator is a scalar multiple of the denominator, which we can do by comparing the coefficients of the like terms in the numerator and denominator. Comparing the coefficients of the \$\$x^2\$\$ term, we get \$\$3 / 1 = 3\$\$ Comparing the coefficients of the \$\$x\$\$ term, we get \$\$5 / (-3) = -5/3\$\$ or approximately -1.67 Comparing the coefficients of the ones term, we get \$\$ 2 / (-4) = -1/2\$\$ or 0.5 Since these ratios are not all the same, it can be concluded that the numerator is not evenly divisible by the denominator. Now, we can set our sights on simplifying the expression. Whenever we see polynomials like this, we should always be looking to factor. Our goal is to check if the numerator or denominator share any factors. If so they can be cancelled out to simplify the expression. We will use basic factoring techniques to simplify the numerator and the denominator. Since the denominator looks a little easier to factor, we will start there. \$\$x^2 - 3x - 4\$\$ Since the coefficient of the \$\$x^2\$\$ term is already 1, we can begin to fill in our new binomials. \$\$(x + a)(x + b)\$\$, where a and b are real numbers, can be positive or negative, and can be integers or fractions. We can make a list of the factors of the ones term, -4. This list would include 2, -2 1, -4 -2, 2 -4, 1 Since \$\$(x+a)(x+b) = (x+b)(x+a)\$\$ due to the Commutative Property of Multiplication for polynomials, we can eliminate duplicate pairs who simply have their order switched as those represent the same expression. That leaves 2, -2 1, -4 Now let's add these terms together to see if any of them sum to equal the coefficient of the \$\$x\$\$ term, -3. \$\$2 + (-2) = 0\$\$ \$\$1 + (-4) = -3\$\$ And we have a winner! The pair (1, -4) both multiplies to equal the ones term and sums to equal the coefficient of the \$\$x\$\$ term. So we can go ahead and plug those into the expression we'd built and check that it does indeed equal the denominator. \$\$(x + 1)(x - 4) = x^2 - 3x - 4\$\$ We can now rewrite our original expression as follows \$\$(3x^2 + 5x + 2) / ((x + 1)(x -4))\$\$ Now, instead of completely factoring the numerator, we can simply divide the numerator by each term in the denominator to see if the expression can be simplified any further. Just by looking at the ones term, the numerator is not divisible by the (x-4) term because -4 is not a factor of 2. However, the numerator could be evenly divisible by the (x+1) term because 1 is a factor of 2. Let's divide the numerator by the (x+1) term to see if it is a factor. Using long division of polynomials, we can show that \$\$(3x^2 + 5x + 2) / (x + 1) = 3x + 2\$\$ Written out in step form, that process goes as follows. \$\$(5x+2) / (x + 1) = 2\$\$ with a remainder of \$\$3x\$\$. Carrying that term with us, \$\$(3x^2 + 3x) / (x + 1) = 3x\$\$ with no remainder, so our answer is \$\$3x+2\$\$. That process shows us that the numerator can be written as \$\$(3x + 2)(x + 1)\$\$ Which multiplied out does in fact equal the original numerator, \$\$3x^2 + 5x + 2\$\$. Let's now rewrite our original equation with our factored out expressions. \$\$((3x+1)(x+1))/((x+1)(x-4))\$\$ We can see that the numerator and denominator both share an (x+1) term, so those can be cancelled out leaving us with \$\$(3x+1)/(x-4)\$\$ Using the same process as we used with the original equation, we can determine that this numerator is not evenly divisible by the denominator, so the expression can no longer be simplified. The final answer is: a simplified expression of \$\$(3x^2+5x+2)/(x^2-3x-4)\$\$ is \$\$(3x+2)/(x-4)\$\$
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# Anaheim Math Tutor Tips: Should You “Cross Multiply”?
“Cross multiply” is an often-used term in mathematics that often sticks in people’s memories long after they graduate high school and stop taking math classes. However, most of those people don’t really know what it is and why it works. So many students are just taught to “cross multiply” when they see an equation with fractions. Unlike many other memorized phrases in math, cross multiplication is actually a useful tool that often makes problems with fractions a little bit easier – sign up for your private Anaheim math tutor.
But just like anything else you memorize in math, you should understand what you’re doing rather than just memorizing a seemingly arbitrary procedure. Cross multiplication is just the “undoing” of division that is being done. If you have two fractions, that is the same as dividing by something on each side. And, just like we can subtract to “undo” addition, we can multiply to “undo” the division that is making those pesky fractions.
Here are three examples that we will solve both with and without cross multiplying. You can see that there are always other ways to solve these problems, as well as why cross multiplying works and is a little bit simpler.
As a general rule of thumb, I do not teach cross multiplication without making sure my students know how to solve the problems without it first. That way I help promote a stronger understanding of how to solve simple algebraic equations that will hopefully stick with them and translate to other math skills rather than leaving them thinking “cross multiply is how you have to solve fractions.”
Here is the first example:
In this example, we have two fractions and need to solve for x. To help find x, we can notice that the fraction on the right can be reduced:
Now we have successfully eliminated the fraction on the right. To solve for x, we now only have to divide both sides by 2. There are a few ways to think about doing this, including using your calculator or converting the left fraction to a decimal. I’ll not that two-fifths divided by two leaves us with one fifth:
We get a solution of one fifth (1/5) which is equal to 0.2. No cross multiplying was needed, only our standard multiplication and division. Let’s do the same with cross multiplication:
Here, we didn’t think about reducing the initial fraction. We simply cross multiplied: taking the diagonal terms in our fractions and multiplying them together (one set in green and the other in purple). This makes the “cross” of our cross multiplication. After we do this, we get a fairly trivial equation of 8 = 40x to solved. We divide both sides by 40 and get the same answer we did above.
Example 2:
Now the x is on the bottom of the fraction. To solve this, we could multiply both sides by x to “undo” the 7 being divided by an x on the left. However, another way would be to make the right side of the equation look like the left. We notice that the left side has a 7 on top, but the right side has a 14. We can change the 14 into a 7 by halving the top and bottom of the right:
Let’s use cross multiplication to solve the same problem:
No extra thinking here or even trying to get the x out of the bottom of the fraction. We simply follow the cross multiplying procedure and multiply the green together and the purples to get our final, simple equation that yields the same 1.5 answer.
Example 3:
Our last example gets a little more complex with terms in parentheses. Cross multiplying here would probably be the best choice, but it’s never the only choice. The fraction on the left is a term being divided by 6. The fraction on the right is a term being divided by 9. We can distribute the division (just like you would distribute multiplication into parentheses) to get an equation we can solve:
We could have kept our work as fractions instead of decimals, but this is likely how it would look if you were using a calculator to help you solve the problem, and decimals are equally as valid as long as we don’t round them. Let’s see how it looks using cross multiplication:
Here we get the same answer as above by following the exact same cross multiplication procedure. Notice that we still had to distribute when we multiplied the terms in parentheses.
Cross multiplying can be a very helpful tool when solving equations with fractions. However, it should not be thought of as the only way to do so, and it should not be the only way students are taught. Thinking this way will continue to lead to many students being confused about fractions and unsure of how to work with them. Cross multiply doesn’t have to go away, but the blind memorization of it should.
Michael C. is currently a private math, science, and standardized test tutor with TutorNerds in Irvine and Anaheim.
All blog entries, with the exception of guest bloggers, are written by Tutor Nerds. Are you an education professional? If so, email us at pr@tutornerds.com for guest blogging and collaborations. We want to make this the best free education resource in SoCal, so feel free to suggest what you would like to see us write. |
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Mrsrowland
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Quizzes Created: 3 | Total Attempts: 1,962
Questions: 9 | Attempts: 189
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Combine like terms and use the order of operations to simplify algebraic expressions.
• 1.
3x + 1 + 8x + 9
• A.
12x + 10
• B.
5x + 8
• C.
11x + 11
• D.
11x + 10
D. 11x + 10
Explanation
The given expression is a sum of two terms: 3x + 1 and 8x + 9. To simplify this expression, we combine like terms by adding the coefficients of x and the constants separately. The coefficient of x in the first term is 3 and in the second term is 8, so the sum of the coefficients is 3 + 8 = 11. Similarly, the constant term in the first term is 1 and in the second term is 9, so the sum of the constants is 1 + 9 = 10. Therefore, the simplified expression is 11x + 10.
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• 2.
2x + 5y - 7x + 8y
• A.
9x + 13y
• B.
5x + 3y
• C.
-5x + 3y
• D.
-5x + 13y
D. -5x + 13y
Explanation
The given expression is a combination of like terms. By combining the terms with the same variable, we can simplify the expression. In this case, combining the x terms (-7x and 2x) gives us -5x, and combining the y terms (5y and 8y) gives us 13y. Therefore, the simplified expression is -5x + 13y.
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• 3.
4(x + 8) - 9
• A.
4x + 32
• B.
4x + 23
• C.
12x - 9
• D.
4x + 24
B. 4x + 23
Explanation
The given expression is 4(x + 8) - 9. To simplify it, we first distribute the 4 to both terms inside the parentheses, resulting in 4x + 32. Then, we subtract 9 from this expression, giving us the final simplified form of 4x + 23.
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• 4.
-3(x + y) + 5(x - y)
• A.
• B.
-2x + 2y
• C.
8x - 8y
• D.
2x - 8y
D. 2x - 8y
Explanation
The given expression is a combination of two terms: -3(x + y) and 5(x - y). To simplify this expression, we can distribute the -3 and 5 to the terms inside the parentheses. This gives us -3x - 3y + 5x - 5y. We can then combine like terms by adding the x terms and the y terms separately. This gives us -3x + 5x - 3y - 5y, which simplifies to 2x - 8y. Therefore, the correct answer is 2x - 8y.
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• 5.
A + b = b + a is an example of the property called ______________________________________________________________________
c +
commutative
commutative property
Explanation
The given equation, a + b = b + a, demonstrates the commutative property of addition. This property states that the order in which numbers are added does not affect the result. In other words, when adding two numbers, switching their positions will not change the sum. This property is applicable to the addition of any real numbers.
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• 6.
9( x + 3) = 9(x) + (9)(3) is an example of what property? ____________________________________________________
distributive
distributive property
Explanation
The given equation 9(x + 3) = 9(x) + (9)(3) demonstrates the distributive property. This property states that when multiplying a number by a sum, you can distribute the multiplication to each term inside the parentheses. In this case, the 9 is being multiplied by the sum of x and 3, and the equation shows that you can distribute the 9 to both x and 3 separately.
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• 7.
What number is known as the multiplicative identity? ____________________
1
one
One
Explanation
The number that is known as the multiplicative identity is 1. In multiplication, any number multiplied by 1 will always result in the same number. Therefore, 1 acts as the identity element for multiplication, as it leaves other numbers unchanged when multiplied by them. The words "one" and "One" are also correct representations of the number 1.
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• 8.
0 • x = _____________
0
Explanation
When any number is multiplied by zero, the result is always zero. In this equation, when zero is multiplied by any number (in this case, x), the result will always be zero. Therefore, the answer to the equation 0 * x is 0.
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• 9.
Ab(c) = (ab)c is an example of what property? __________________________________
Associative Property of Multiplication
associative property of multiplication
assoc. prop. of mult.
A x
associative property
associative
Explanation
The equation ab(c) = (ab)c demonstrates the Associative Property of Multiplication. This property states that the grouping of factors in a multiplication expression does not affect the result. In this case, the factors a, b, and c can be multiplied in any order and the product will remain the same.
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# Solve the Following Quadratic Equations by Factorization: - Mathematics
Course
#### Question
Solve the following quadratic equations by factorization: $\frac{3}{x + 1} + \frac{4}{x - 1} = \frac{29}{4x - 1}; x \neq 1, - 1, \frac{1}{4}$
#### Solution
$\frac{3}{x + 1} + \frac{4}{x - 1} = \frac{29}{4x - 1}$
$\Rightarrow \frac{3\left( x - 1 \right) + 4\left( x + 1 \right)}{\left( x + 1 \right)\left( x - 1 \right)} = \frac{29}{4x - 1}$
$\Rightarrow \frac{3x - 3 + 4x + 4}{x^2 - 1} = \frac{29}{4x - 1}$
$\Rightarrow \frac{7x + 1}{x^2 - 1} = \frac{29}{4x - 1}$
$\Rightarrow \left( 7x + 1 \right)\left( 4x - 1 \right) = 29\left( x^2 - 1 \right)$
$\Rightarrow 28 x^2 - 7x + 4x - 1 = 29 x^2 - 29$
$\Rightarrow 29 x^2 - 28 x^2 + 3x - 28 = 0$
$\Rightarrow x^2 + 3x - 28 = 0$
$\Rightarrow x^2 + 7x - 4x - 28 = 0$
$\Rightarrow x(x + 7) - 4(x + 7) = 0$
$\Rightarrow (x - 4)(x + 7) = 0$
$\Rightarrow x - 4 = 0 \text { or } x + 7 = 0$
$\Rightarrow x = 4 \text { or } x = - 7$
Hence, the factors are 4 and −7.
Is there an error in this question or solution? |
Math 15H (Notes)
# Gauss-Jordon Elimination Notes:
The procedure for Guass-Jordan Elimination is as follows:
1. Find the leftmost column that is not all zeros and swap its row with the top row.
2. Make the leading entry in the first row a "1" (If the top entry is a, then multiply the top row by 1/a).
3. Use the top row to make all the entries in the column below the leading one be zero.
4. Ignoring the top row, repeat steps 1 to 4 until there are no more leading ones.
5. Finally, use each leading 1 to make all entries on the column above the zeros.
## Example:
Solve for x, y, and z in:
2y - 3z = 2 2x + z = 3 x - y + 3z = 1
Solution:
éêë 021 20-1 -313 ïïï 231 ùúû
write the system as an augmented matrix
éêë 120 -102 31-3 ïïï 132 ùúû
interchange first and third row
(to make top left entry non-zero)
éêë 100 -122 3-5-3 ïïï 112 ùúû
Add -2 times first row to second row
(to get 0 in first column of row 2)
éêë 100 -112 3-5/2-3 ïïï 11/22 ùúû
Divide second row by 2
(to get a leading 1 in row 2)
éêë 100 -110 3-5/22 ïïï 11/21 ùúû
Add -2 times second row to third
(to get 0's in the second column)
éêë 100 -110 3-5/21 ïïï 11/21/2 ùúû
Divide third row by 2
éêë 100 -110 001 ïïï -1/27/41/2 ùúû
Add -3 times third row to first row
Add 5/2 times third row to second row
(to get 0's in third column)
éêë 100 010 001 ïïï 5/47/41/2 ùúû
Add second row to first row
This is now in reduced row-echelon form, so we can read off the answer: x = 5/4, y = 7/4, and z = 1/2.
Check that the answer satisfies the initial equations (in case we made arithmatic errors):
2y - 3z = 2(7/4) - 3(1/2) = 7/2 - 3/2 = 4/2 = 2
2x + z = 2(5/4) + 1/2 = 5/2 + 1/2 = 6/2 = 3
x - y + 3z = (5/4) - (7/4) + (1/2) = -(2/4) + 3/2 = -(1/2) + 3/2 = 2/2 = 1
All of these check out, so our solution is correct.
Math 15H (Fall 2002) web pages Created: 26 Mar 2002 Last modified: Nov 14, 2002 10:17:22 PM Comments to: `dpvc@union.edu` |
IDENTIFY PARTS OF AN EXPRESSION
In an algebraic expression, we may find the following three parts.
(i) Terms
(ii) Factors
(iii) Coefficients
What is term ?
A single variable or a constant or a combination of these as a product or quotient forms a term.
Examples of terms :
5, -a, 3ab, 21/7, ........... etc
Terms can be added or subtracted to form an expression.
What is factor ?
Consider the expression 3ab – 5a. It has two terms 3ab and -5a. The term 3ab is a product of factors 3, a and b. The term -5a is a product of -5 and a. The coefficient of a variable is a factor or factors.
Example :
In the term 3ab;
(i) the coefficient of ab is 3 (ii) the coefficient of a is 3b
(iii) the coefficient of b is 3a.
In the term –5a the coefficient of a is –5
What is constant ?
A number which is not having any variable with it is known as constant.
Identifying Parts of an Expression Using Flow Chart
Example 1 :
Identify the parts of the following expression
2x + 3
Solution :
In the expression 2x + 3 the term 2x is made of 2 factors and 2 and x while 3 is a single factor.
Example 2 :
Identify the parts of the following expression
3ab - 5a
Solution :
Example 3 :
Identify the number of terms and coefficient of each term in the expression.
x2y2 - 5x2y + (3/5)xy2 - 11
Solution :
In the given expression, we have four terms.
Term 1 ==> x2y2
Term 2 ==> -5x2y
Term 3 ==> (3/5)xy2
Term 4 ==> -11
Coefficient of 1st term = 1
Coefficient of 2nd term = -5
Coefficient of 3rd term = 3/5
Since the last term is not having any variable, it is a constant term.
Example 4 :
Identify the number of terms, coefficient and factors of each term in the expression.
3abc - 5ca
Solution :
The given expression contains two terms.
Term 1 ==> 3abc
Term 2 ==> -5ca
Terms Coefficients Factors 1) 3abc 3 a, b and c 2) -5ca -5 c and a
Example 5 :
Identify the number of terms, coefficient and factors of each term in the expression.
1 + x + y2
Solution :
The given expression contains three terms.
Term 1 ==> 1
Term 2 ==> x
Term 2 ==> y2
Terms Coefficients Factors 1) 1 - - 2) x 1 x 3) y2 1 y and y
Example 6 :
Identify the number of terms, coefficient and factors of each term in the expression.
3x2y2 - 3xyz + z
Solution :
The given expression contains three terms.
Term 1 ==> 3x2y2
Term 2 ==> -3xyz
Term 2 ==> z
Terms Coefficients Factors 1) 3x2 y2 3 x2 and y2 2) - 3xyz -3 x, y and z 3) z3 1 z, z and z
Example 7 :
The coefficient of x4 in -5x7 + (3/7)x4 - 3x3 + 7x2 - 1
Solution :
The coefficient of x4 is 3/7.
Kindly mail your feedback to v4formath@gmail.com
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# Lowest Common Denominator (LCD) of 1/1 and 31/1
So you wanna find the lowest common denominator of 1/1 and 31/1? Lucky for you that's exactly what this page is here to help you with! In this quick guide, we'll walk you through how to calculate the lowest common denominator for any fractions you need to check. Here we go!
In a rush and just want the answer? No worries! The LCD of 1/1 and 31/1 is 1. In fraction form it's 1.
LCD(1/1, 31/1) = 1
In fraction form:
1 / 1
Which can be simplified further to:
1 / 1
Read on to find out how we worked that out!
As we always do in these articles, it's worth a very quick recap on fraction terminology. The number above the line is called the numerator and the number about the line is called the denominator. So, in this example, our numerators are 1 and 31, while our denominators are 1 and 1.
## What is the Lowest Common Denominator?
To calculate the lowest common denominator, the easiest way is to look at the factors of those numbers and find the lowest common multiple. Here's how that looks for 1 and 1:
• Factors for 1: 1
The next step is to calculate the numerator and complete our fraction. To do this, we need to find the greatest common factor of the numerators, which are 1 and 31.
• Factors for 1: 1
• Factors for 31: 1 and 31
As we can see, the greatest common factor between the numerators is 1, so this becomes our numerator in the fraction:
1 / 1
We can simplify this fraction down even further and reduce it to the following:
1 / 1
Hopefully this article has helped you to understand how to calculate the lowest common denominator of two fractions. Isn't math fun? If you want to challenge yourself, try to work out the least common denominator of some fractions by yourself and use our LCD calculator to check your answers! |
# Triangular Prism
In the geometry, the prism is a common shape having several varieties. Based on the base of it, it has many variations. One of such prism is the triangular prism. It is a three-sided prism. It is also termed as a polyhedron which has a triangular base. A uniform triangular prism is very common and it is the right triangular prism with equilateral bases and square sides. In this article, the student will learn about triangular prism, related terms as well as some important formulae. Let us begin it!
## Triangular Prism Definition
A triangular prism is a popular polyhedron. It is having two triangular bases and three rectangular sides. According to the nature of prism, the two triangular bases are parallel and congruent to each other. It is a pentahedron with nine distinct nets. The edges and vertices of the bases are connected with each other. The rectangular sides of this prism are rectangular in shape and are joint with each other side by side. All the cross-sections parallel to the base faces are triangle.
Source: commons.wikimedia.org
The net of a solid figure is possible when a solid figure is unfolded along its edges and further its faces are laid out in a pattern in two dimensions. The net of a triangular prism is made up of rectangles and triangles. Also, the number of triangular prism edges is 9.
### Volume and Surface Area of Triangular Prism:
Prism formula includes two very important formulae. These are Prism volume and Area of prism formulae. The volume of a prism is the space within the triangular prism. The surface area of a triangular prism is the amount of covered space on the outside surface of the prism. For these computations, we need the height, side and base length of the prism.
### The volume of Triangular Prism Formula:
The volume of the triangular prism is equal to the product of the area of the triangular base and the height of the prism. Thus
The volume of Prism = Area of the Base × Height of prism
Mathematically,
V = $$\frac {1}{2} \times b \times h \times l$$
V The volume of the prism b Base length h Height of the triangle l Length of the prism
### Surface Area of Triangular Prism Formula:
The surface area of a triangular prism is computed as the sum of the lateral surface area and base areas of both triangle bases. Thus,
The surface area of triangular prism = Lateral Area + 2 times the triangular base area
Mathematically,
1. A. = P × H + 2 times A
S.A. The surface area of the prism A Area of the base P the perimeter of the base H height of the prism
### Area of triangle Formula:
A = $$\frac {1}{2} \times b \times h$$
A area of the triangle h Height of triangle b The base length of the triangle
### The perimeter of the Base Formula:
Perimeter = a + b + c
Perimeter The perimeter of the triangle a,b,c Three sides of the triangle
## Solved Examples for You
Q.1: Find out the volume of the triangular prism with base length 10 cm, the height of 20 cm, and length of the prism as 50 cm.
Solutions: Parameters given in the problem,
b = 10 cm
h = 20 cm
l = 50 cm
Volume of the prism will be:
V = $$\frac {1}{2} \times b \times h \times l$$
Substituting the values,
V = $$\frac {1}{2} \times 10 \times 20 \times 50$$
V = $$\frac{1}{2} \times 10000$$
= 5000 $$cm^{3}$$
Thus volume of the triangular prism will be 5000 $$cm^{3}.$$
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### Chapter 5 - Dalton State College
```PROBABILITY &
STATISTICS FOR P-8
TEACHERS
Chapter 5
Probability Distributions
PROBABILITY DISTRIBUTIONS
We will now combine the methods of
descriptive statistics (Chapter 2 and 3) and
those of probability (Chapter 4) to describe
and analyze
probability distributions.
Probability Distributions describe what will
probably happen instead of what actually did
happen, and they are often given in the format
of a graph, table, or formula.
COMBINING DESCRIPTIVE
METHODS AND PROBABILITIES
In this chapter we will construct probability distributions
by presenting possible outcomes along with the relative
frequencies we expect.
PROBABILITY DISTRIBUTIONS
A
random variable is a variable whose
values are determined by chance.
A
discrete probability distribution
consists of the values a random variable can
assume and the corresponding probabilities of
the values.
The
sum of the probabilities of all events in
a sample space add up to 1. Each probability
is between 0 and 1, inclusively.
RANDOM VARIABLES
Two types of random variables:
A
discrete random variable can
assume a countable number of values.
Number of steps to the top of the Eiffel Tower
A
continuous random variable can
assume any value along a given interval
of a number line.
The time a tourist stays at the top
once s/he gets there
RANDOM VARIABLES
Discrete random variables
Number of sales
Number of calls
Shares of stock
People in line
Mistakes per page
Continuous random
variables
Length
Depth
Volume
Time
Weight
PROBABILITY DISTRIBUTIONS
The
probability distribution of a
discrete random variable is a graph,
table or formula that specifies the
probability associated with each possible
outcome the random variable can
assume.
0 ≤ P(x) ≤ 1 for all values of x
P(x) = 1
PROBABILITY DISTRIBUTIONS
Is the following a probability
distribution?
Not a
probability
distribution
x
0
1
P(x)
2
3
4
0.22
0.10
0.30
5
-0.01
0.16
0.18
0.95
P(x) < 0
∑P(x) ≠ 1
PROBABILITY DISTRIBUTION
Construct a probability
distribution for tossing a coin
twice and recording the number of
x
P(x)
0
0.25
1
0.50
2
0.25
HH
HT
TH
TT
PROBABILITY HISTOGRAM
A probability histogram is a histogram in
which the horizontal axis corresponds to the
value of the random variable and the vertical
axis represents the probability of that value of
the random variable.
PROBABILITY HISTOGRAM
Draw a probability histogram of
the probability distribution to the
right, which represents the
number of DVDs a person rents
from a video store during a single
visit.
Probability
DVDs Rented at a Video Store
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
Number of DVDs Rented
4
5
x
P(x)
0
0.06
1
0.58
2
0.22
3
0.10
4
0.03
5
0.01
PROBABILITY DISTRIBUTION
A survey was completed regarding how many siblings
are in your family. Let X denote the number of
siblings of a randomly selected student.
a. Determine the
probability distribution
of the random variable x.
b. Construct a probability
histogram for the
random variable x.
PROBABILITY DISTRIBUTION
Solution
Probability Distribution
Probability Histogram
PROBABILITY DISTRIBUTION
Just like with any distribution, we would like
to analyze the data. The mean and standard
deviation are the most common measurements.
To find these measurements, we will treat the
probability distribution just like a frequency
distribution.
Since probability distributions represent
theoretical data, we will treat the results as a
true population.
MEAN AND STANDARD DEVIATION
Mean: µ = ∑ x P(x)
Variance: σ2 = ∑ (x – µ)2 P(x)
or
σ2 = ∑ x2 P(x) – µ2
MEAN & STANDARD DEVIATION
Find the mean and standard deviaton of the
number of spots that appear when a die is
tossed.
Outcome
x
Probability
P(x)
x P(x)
1
1/6
1/6
2
1/6
2/6
3
1/6
3/6
4
1/6
4/6
5
1/6
5/6
6
1/6
6/6
21/6
µ = ∑ x P(x)
= 21/6
= 3.5
MEAN & STANDARD DEVIATION
Find the mean and standard deviaton of the
number of spots that appear when a die is
tossed.
Outcome
x
Probability
P(x)
x2 P(x)
1
1/6
1/6
σ2 = ∑ x2 P(x) – µ2
2
1/6
4/6
3
1/6
9/6
4
1/6
16/6
5
1/6
25/6
6
1/6
36/6
σ = √ 105/36
91/6
= 1.708
= 91/6 – (21/6)2
= 105/36
MEAN & STANDARD DEVIATION
The probability distribution shown represents
the number of trips of five nights or more that
American adults take per year. (That is, 6% do
not take any trips lasting five nights or more,
70% take one trip lasting five nights or more
per year, etc.) Find the mean.
MEAN & STANDARD DEVIATION
# Trips
x
Probability
P(x)
x P(x)
x2 P(x)
0
0.06
0.00
0.00
1
0.70
0.70
0.70
2
0.20
0.40
0.80
3
0.03
0.09
0.27
4
0.01
0.04
0.16
1.23
1.93
µ = ∑ x P(x) = 1.23
σ2 = ∑ x2 P(x) – µ2 = 1.93 – (1.23)2 = 0.4171
σ = √0.4171 = 0.6458
EXPECTATION
The
expected value, or expectation, of a
discrete random variable of a probability
distribution is the theoretical average of the
variable.
The
expected value is, by definition, the mean
of the probability distribution.
E(x) = µ = ∑ x P(x)
WINNING TICKETS
One thousand tickets are sold at \$1 each for four prizes
of \$100, \$50, \$25, and \$10. After each prize drawing,
the winning ticket is then returned to the pool of tickets.
What is the expected value if you purchase a ticket?
Winnings = prize amount – \$1 ticket price
Winnings
x
Probability
P(x)
x P(x)
99
1/1000
99/1000
49
1/1000
49/1000
24
1/1000
24/1000
9
1/1000
9/1000
-1
996/1000
-996/1000
-815/1000
E(x) = ∑ x P(x)
= –0.815
On average, you
will lose 82 cents
for every dollar
you spend
WINNING TICKETS (ALTERNATE APPROACH)
One thousand tickets are sold at \$1 each for four prizes
of \$100, \$50, \$25, and \$10. After each prize drawing,
the winning ticket is then returned to the pool of tickets.
What is the expected value if you purchase a ticket?
Winnings
x
Probability
P(x)
x P(x)
100
1/1000
100/1000
50
1/1000
50/1000
25
1/1000
25/1000
10
1/1000
10/1000
0
996/1000
0/1000
185/1000
E(x)
= ∑ x P(x) –
initial cost
= 0.185 – \$1
= –0.815
THE BINOMIAL DISTRIBUTION
Many
types of probability problems have
only two possible outcomes or they can
be reduced to two outcomes.
Examples
include: when a coin is tossed
it can land on heads or tails, when a baby
is born it is either a boy or girl, etc.
THE BINOMIAL DISTRIBUTION
The binomial experiment is a probability
experiment that satisfies these requirements:
1. Each trial can have only two possible
outcomes—success or failure.
2. There must be a fixed number of trials.
3. The outcomes of each trial must be
independent of each other.
4. The probability of success must remain the
same for each trial.
NOTATION FOR THE BINOMIAL
DISTRIBUTION
p
The numerical probability of success
q
The numerical probability of failure
n
The number of trials
x
The number of successes
Note that x = 0, 1, 2, 3,...,n
THE BINOMIAL DISTRIBUTION
In a binomial experiment, the probability of
exactly X successes in n trials is
n!
X
n X
P X
p q
n - X ! X !
or
P X
n Cx
number of possible
desired outcomes
p X q n X
probability of a
desired outcome
THE BINOMIAL DISTRIBUTION
The
Binomial Probability Distribution
the probability of exactly x successes in n trials is
P(x) = nCx
x
p
n-x
q
THE BINOMIAL DISTRIBUTION
If 40% of the class is
female, what is the
probability that 1 of
the next 2 students
walking in will be
female?
P(x) = nCx px qn-x
P(1) = 2C1 p1 q2-1
= (2) (.4)1 (.6)1
= 0.48
n = 2 (# trials)
x = 1 (# successes)
p = .4 (probability of success)
q = .6 (probability of failure)
THE BINOMIAL DISTRIBUTION
A survey found that one out of five Americans
say he or she has visited a doctor in any given
month. If 10 people are selected at random, find
the probability that exactly 3 will have visited a
doctor last month.
n = 10
x=3
p = 1/5
q = 4/5
P(x) = nCx px qn-x
P(3) = 10C3 p3 q10-3
3
10! 1
P 3
7!3! 5
7
4
0.201
5
THE BINOMIAL DISTRIBUTION
Individual baseball cards, chosen at random from a
set of 20, are given away inside cereal boxes. Stan
needs one more card to complete his set so he buys
five boxes of cereal. What is the probability that he
will complete his set?
P(x) = nCx px qn-x
n=5
x=1
p = 1/20
q = 19/20
P(1) = 5C1 p1 q5-1
= (5) (.05)1 (.95)4
= 0.204
The probability of Stan completing
his set is 20%.
THE BINOMIAL DISTRIBUTION
A test consists of 10 multiple choice questions, each
with four possible answers. To pass the test, one
must answer at least nine questions correctly. Find
the probability of passing, if one were to guess the
P(x successes) = nCx px qn - x
P(x ≥ 9 successes) = P(9 successes) + P(10 successes)
n = 10
9
1
x = 9, 10
1 3
P(x ≥ 9) 10 C9
4
4
p = 1/4
q = 3/4
= 0.000028610 +
= 0.000029564
10
0
1
3
10 C10
4
4
0.000000954
The probability of
passing is 0.003%.
THE BINOMIAL DISTRIBUTION
A survey from Teenage Research Unlimited
(Northbrook, Illinois) found that 30% of teenage
consumers receive their spending money from part-time
jobs. If 5 teenagers are selected at random, find the
probability that at least 3 of them will have part-time
jobs.
n=5
p = 0.3
q = 0.7
x = 3, 4, 5
5!
3
2
P 3
0.30 0.70 0.132
2!3!
5!
4
1
P 4
0.30 0.70 0.028
1!4!
5!
5
0
P 5
0.30 0.70 0.002
0!5!
P(x ≥ 3) = 0.132 + 0.028 + 0.002 = 0.162
THE BINOMIAL DISTRIBUTION
A family has nine children. What is the
probability that there is at least one girl?
This can be best solved using the compliment,
that is, the probability of zero girls:
n=9
p = 0.5
q = 0.5
x=0
P(x successes) = nCx px qn - x
P (0) = 9C0 p0 q9 - 0
= (1) (.5)0 (.5)9
= 0.001953
P(x ≥ 1) = 1 – P(0)
= 1 – 0.001953
= 0.998
BINOMIAL DISTRIBUTION
Create a probability distribution table for
tossing a coin 3 times
P(x) = nCx px qn-x
x
P(x)
0
0.125
1
0.375
2
0.375
3
0.125
P(0)
P(1)
P(2)
P(3)
= 3C0 p0 q3-0
= (1) (.5)0 (.5)3
= (3) (.5)1 (.5)2
= (3) (.5)2 (.5)1
= (1) (.5)3 (.5)0
= 0.125
= 0.375
= 0.375
= 0.125
TOSSING COINS
A coin is tossed 3 times. Find the probability of
getting exactly two heads, using Table B.
n 3, p 12 0.5, X 2 P 2 0.375
BINOMIAL DISTRIBUTION
(a) Construct a binomial probability histogram
with n = 8 and p = 0.15.
(b) Construct a binomial probability histogram
with n = 8 and p = 0. 5.
(c) Construct a binomial probability histogram
with n = 8 and p = 0.85.
For each histogram, comment on the shape of the
distribution.
THE BINOMIAL DISTRIBUTION
The mean, variance, and standard deviation
of a variable that has the binomial distribution
can be found by using the following formulas.
Mean: np
Variance: npq
2
Standard Deviation: npq
THE BINOMIAL DISTRIBUTION
Insurance Co. reported that 2% of all American births
result in twins. If a random sample of 8000 births is
taken, find the mean, variance, and standard deviation of
the number of births that would result in twins.
np 8000 0.02 160
npq 8000 0.02 0.98 156.8 157
2
npq 8000 0.02 0.98 12.5 13
THE BINOMIAL DISTRIBUTION
According to the Experian Automotive, 35% of all carowning households have three or more cars. Find the
mean and standard deviation for a random sample of
400 homes.
µ = np
σ = √npq
= (400)(0.35)
= √(400)(0.35)(0.65)
= 140
= 9.54
``` |
# 5) Construct two squares of opposition, one traditional, the other m
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5) Construct two squares of opposition, one traditional, the
other modern, using the terms professors and great
teachers.
Solution
Squares of Opposition :
A square of opposition shows the logical relations among
categorical statements. There are two squares of
opposition:
1.The Modern Square of Opposition :
The only logical relation in the Modern Square of
To say that two statements are contradictory to each other
means that they necessarily have opposite truth value.
That is, if one of them is true, the other must be false, and
if one of them is false, the other must be true.
There are mainly 4 types of statements.Let us say S and P
are 2 sets. By using these 2 sets the types of statements
can be written as:
A : All S are P
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E : No S are P
I : Some S are P
O : Some S are not P
For example, the A statement All bats are mammals and
the O statement Some bats are not mammals contradict
each other.
The contradictory relation also exists between the E
statement No swans are black birds and the I statements
Some swans are black birds.
We can use Venn Diagrams to illustrate why the
contradictory relation holds between the A and O
statements. When the A statement is true, the area is
empty. But if the area is empty, then no member of S can
be in the area . This contradicts the O statement, which
says that there is at least one member of S in the area ,
i.e., the area is not empty.
If we assume that the set denoted by the subject term
cannot be an empty set, then there are four logical
relations among the A, E, I, O statements. They are shown
in the Traditional Square of Opposition. The four relations
are:
Contrary
Subcontrary
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5) Construct two squares of opposition, one traditional, the other modern, using the terms professors and great teachers. Solution Squares of Opposition : A square of opposition shows the logical relations among categorical statements. There are two squares of opposition: 1.The Modern Square of Opposition : The only logical relation in the Modern Square of Opposition is the contradictory relation. The Contradictory Relation To say that two statements are contradictory to each other means that they necessarily have opposite truth value. That is, if one of them is true, the other must be false, and if one of them is false, the other must be true. There are mainly 4 types of statements.Let us say S and P are 2 sets. By using these 2 sets the types of statements can be written as: A : All S are P E : No S are P I : Some S are P O : Some S are not P For example, the A statement “All bats are mammals” and the O statement “Some bats are not mammals” contradict each other. The contradictory relation also ex ists between the E statement “No swans are black birds” and the I statements “Some swans are black birds.” We can use Venn Diagrams to illustrate why the contradictory relation holds between the A and O statements. When the A statement is true, the area i s empty. But if the area is empty, then no member of S can be in the area . This contradicts the O statement, which says that there is at least one member of S in the area , i.e., the area is not empty. 2.The ...
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TOTAL SURFACE ARE AOF COMBINED SOLID
Problem 1 :
A toy is in the form of right circular cylinder with a hemisphere at the one end and a cone at the other end. The base radius measures 3.5 cm, height of the cylindrical portion is 10 cm. and conical part measures 12 cm. Find the total surface area of the toy(use π = 3.14)
Solution :
Radius of the hemisphere = 3.5 cm
Curved Surface area (CSA) of the hemisphere
= 2πr²square units
= 2(3.5)²π
= (24.5)π cm²
Radius of the Cylinder r = 3.5 cm
Height of the Cylinder h = 10 cm
Curved Surface Area (CSA) of the Cylinder
= 2πrh square units
= 2 π x 3.5 x 10
= 70π cm2
Radius of the cone r= 3.5 cm
Height of the cone h = 12 cm
Slant height of the cone l = √(r² + h²)
l = √(3.52 + 122)
l = √(12.25+144)
l = √(156.25)
l = 12.5 cm
Curved Surface-Area(CSA) of the Cone = πrl square units
= π(3.5)(12.5)
= (43.75)π cm²
Total Surface-Area of the Toy
= CSA of the hemisphere + (CSA of the Cylinder) + (CSA of the Cone)
= (24.5 π) + (70 π) + (43.75 π)
= 138.25 x 3.14
Hence, total surface area of the Toy = 434.11 cm²
Problem 2 :
Find the total surface area of cylinder whose height is 8 cm and radius is 4 cm.
Solution :
Radius of the Cylinder = 4 cm
Height of the cylinder = 8 cm
Required total surface area of the cylinder = 2πr(h+r)
= 2 π (4) (8+4)
= 2 π (4) (12)
= 96 π
Hence, total surface area of the Cylinder = 96π cm²
Problem 3 :
Find the total surface area of cylinder whose height is 16 cm and radius is 7 cm.
Solution :
Radius of the Cylinder = 7 cm
Height of the cylinder = 16 cm
Required Total surface area of the cylinder = 2πr(h+r)
= 2 (22/7) (7) (16+7)
= 2 (22) (23)
= 1012 cm²
Hence, Total surface area of the Cylinder = 1012 cm²
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# Math homework checker
In this blog post, we will show you how to work with Math homework checker. Let's try the best math solver.
## The Best Math homework checker
One instrument that can be used is Math homework checker. Basic mathematics is the study of fundamental concepts in mathematics, including numbers, algebra, geometry, and trigonometry. These concepts are essential for understanding more advanced mathematics, and they form the foundation for a wide range of real-world applications. Basic mathematics is often taught in elementary and middle school, but it can also be studied at the high school and college level. In addition to acquiring mathematical skills, students of basic mathematics also learn how to think logically and solve problems. As a result, basic mathematics is an essential subject for all students.
Math problems can seem daunting at first, but there are a few examples that can help make them seem a little less intimidating. One type of problem is called an equation. This is when two things are equal to each other, like 3+4=7. Another type of problem is called a word problem. These are problems that come with a story, like "If Tommy has 4 apples and he eats 2, how many does he have left?" The last type of problem is called an inequality. This is when two things are not equal, like 5>3. No matter what type of math problem you're dealing with, there are some steps you can follow to solve it. First, read the problem carefully and identify what information you're given and what you need to find out. Next, figure out which operation you need to use to solve the problem. Once you've done that, work through the problem step by step until you arrive at the answer. Math problems can be tough, but taking the time to understand them and break them down into smaller pieces can make them a lot more manageable.
How to solve for domain: There are many ways to solve for the domain of a function. In algebra, the domain is often defined as the set of all values for which a function produces a real output. However, this definition can be difficult to work with, so it is often useful to think about the domain in terms of graphing. For instance, if a function produces imaginary results for certain input values, then those input values will not be included in the function's domain. Similarly, if a function is undefined for certain input values, those values will also be excluded from the domain. In general, the graphing method is the easiest way to determine the domain of a function. However, it is sometimes necessary to use other methods, such as solving inequalities or using set notation. With practice, you will be able to solve for domain quickly and easily.
How to solve partial fractions is a process that can be broken down into a few simple steps. First, identify the factors that are being divided. Next, determine the order of the fractions. Finally, apply the appropriate formula to solve for the unknowns. By following these steps, you can quickly and easily solve for partial fractions. However, it is important to note that there is more than one way to solve partial fractions. As such, you may need to experiment with different methods in order to find the one that works best for you. But with a little practice, you'll be solving partial fractions like a pro in no time!
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# IMAT 2013 Q27 [Safe Dimensions]
A safe has external dimensions as follows:
Width 48cm
Depth 44cm
Height 52cm
The entire safe is made of steel 4cm thick except the base which is 8cm thick.
What are the internal dimensions of the base of the safe?
A. 40cm x 40cm
B. 36cm x 36cm
C. 40cm x 32cm
D. 44cm x 40cm
E. 40cm x 36cm
Simple steps to solve word problems:
• Underline key information
• Determine what they are trying to ask, and what you will need to solve it
• Eliminate any non-essential information
• Draw a picture, graph, or equation
• In moments of high stress like exam taking, always work with the paper they give you to avoid careless mistakes.
• Solve.
The best way to approach this problem is to take the information given and then make a quick sketch. The important thing to know is that we are looking for the interior surface area of the base, meaning we only need to work with width and depth because height does not contribute to this surface area. The next important step is to figure out what is meant by the walls and the base because we need to know which is 4cm thick and what is 8cm. The thickness of the walls is determined by the length and width, while the height determines the thickness of the base and roof (easier to see in the drawing). We can now eliminate the 8cm base thickness clue given to us in the question because it applies to height, which is not needed in our calculations.
Steps to solve: Find the base’s width by subtracting the total width minus the thickness of both sets of walls (there are walls on all sides). Next, do the same with the depth.
\fcolorbox{red}{grey!30}{Therefore the answer is E, \$40\$cm x \$36\$cm.} |
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