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# Divisibility Tests by 8 and 12
We will discuss here about the rules of divisibility tests by 8 and 12 with the help of different types of problems.
1. If ‘a’ is a positive perfect square integer, then a(a - 1) is always divisible by
(a) 12
(b) multiple of 12
(c) 12 - x
(d) 24
Solution:
‘a’ is a positive perfect square integer.
Let, a = x2
Now, a (a – 1) = x2(x2 – 1)
Therefore, a(a – 1) is always divisible by 12
Answer: (a)
Note: x2(x2 – 1) is always divisible by 12 for any positive integral values of x.
2. If m and n are two digits of the number 653mn such that this number is divisible by 80, then (m + n) is equal to
(a) 2
(b) 3
(c) 4
(d) 6
Solution:
653xy is divisible by 80
Therefore, the values of y must be 0.
Now, 53x must be divisible by 8.
Therefore, the value of x = 6
Thus, the required sum of (x + y) = (6 + 0) = 6
Answer: (d)
Note: The number formed by last three digits when divisible by 8, then the number is divisible by 8.
3. The sum of first 45 natural numbers will be divisible by
(a) 21
(b) 23
(c) 44
(d) 46
Solution:
Number of natural numbers (n) is 45
Therefore, Sum of numbers divisible by 45 and 46 ÷ 2 = 23
Therefore, according to the given options the required number is 23.
Answer: (b)
Note: Sum of ‘n’ terms of natural numbers is always divisible by {n or n/2 or (n + 1) or (n + 1)/2} and also by the factors of n or (n + 1)
4. How many digits from the unit’s digit must be divisible by 32, to make the complete number is divisible by 32?
(a) 2
(b) 4
(c) 5
(d) None of these
Solution:
32 = 25
Therefore, required number of digits is 5
Answer: (c)
Note: Power of ‘2’ and ‘5’ indicate the number of digits from the unit’s digit to decide whether the number is divisible by what number.
5. If 4a3 + 984 = 13b7, which is divisible by 11, then find the value of (a + b)
(a) 8
(b) 9
(c) 10
(d) 11
Solution:
13b7 is divisible by 11
Therefore, (3 + 7) – (1 + b) = 0
Or, 10 – 1 + b = 0
Therefore, b = 9
Now, 4a3 + 984 = 1397
Thus, a = 9 – 8 = 1
Therefore, required values of (a + b) = (1 + 9) = 10
Answer: (c)
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## 8.1 Radicals
### Learning Objectives
1. Find square roots.
2. Find cube roots.
3. Find nth roots.
4. Simplify expressions using the product and quotient rules for radicals.
## Square Roots
The square rootThe number that, when multiplied by itself, yields the original number. of a number is that number that when multiplied by itself yields the original number. For example, 4 is a square root of 16, because $4 2 =16$. Since $( −4 ) 2 =16$, we can say that −4 is a square root of 16 as well. Every positive real number has two square roots, one positive and one negative. For this reason, we use the radical sign to denote the principal (nonnegative) square rootThe positive square root of a real number, denoted with the symbol . and a negative sign in front of the radical to denote the negative square root.
Zero is the only real number with one square root.
If the radicandThe expression a within a radical sign, $an$., the number inside the radical sign, is nonnegative and can be factored as the square of another nonnegative number, then the square root of the number is apparent. In this case, we have the following property:
Example 1: Find the square root.
a. $36$
b. $144$
c. $0.04$
d. $19$
Solution:
a. $36=62=6$
b. $144=122=12$
c. $0.04=(0.2)2=0.2$
d. $19=( 1 3)2=13$
Example 2: Find the negative square root.
a. $−4$
b. $−1$
Solution:
a. $−4=−22=−2$
b. $−1=−12=−1$
The radicand may not always be a perfect square. If a positive integer is not a perfect square, then its square root will be irrational. For example, $2$ is an irrational number and can be approximated on most calculators using the square root button.
Next, consider the square root of a negative number. To determine the square root of −9, you must find a number that when squared results in −9:
However, any real number squared always results in a positive number:
The square root of a negative number is currently left undefined. For now, we will state that $−9$ is not a real a number.
## Cube Roots
The cube rootThe number that, when used as a factor with itself three times, yields the original number; it is denoted with the symbol . of a number is that number that when multiplied by itself three times yields the original number. Furthermore, we denote a cube root using the symbol , where 3 is called the indexThe positive integer n in the notation that is used to indicate an nth root.. For example,
The product of three equal factors will be positive if the factor is positive and negative if the factor is negative. For this reason, any real number will have only one real cube root. Hence the technicalities associated with the principal root do not apply. For example,
In general, given any real number a, we have the following property:
When simplifying cube roots, look for factors that are perfect cubes.
Example 3: Find the cube root.
a. $273$
b. $643$
c. $03$
d. $183$
Solution:
a. $273=333=3$
b. $643=433=4$
c. $03=033=0$
d. $183=( 1 2)33=12$
Example 4: Find the cube root.
a. $−83$
b. $−13$
c. $−1273$
Solution:
a. $−83=(−2)33=−2$
b. $−13=(−1)33=−1$
c. $−1273=(− 1 3)33=−13$
It may be the case that the radicand is not a perfect cube. If an integer is not a perfect cube, then its cube root will be irrational. For example, $23$ is an irrational number which can be approximated on most calculators using the root button. Depending on the calculator, we typically type in the index prior to pushing the button and then the radicand as follows:
Therefore, we have
## nth Roots
For any integer $n≥2$, we define the nth rootThe number that, when raised to the nth power, yields the original number. of a positive real number as that number that when raised to the nth power yields the original number. Given any nonnegative real number a, we have the following property:
Here $n$ is called the index and $an$ is called the radicand. Furthermore, we can refer to the entire expression $an$ as a radicalUsed when referring to an expression of the form $an$.. When the index is an integer greater than 3, we say “fourth root”, “fifth root”, and so on. The nth root of any number is apparent if we can write the radicand with an exponent equal to the index.
Example 5: Find the nth root.
a. $814$
b. $325$
c. $17$
d. $1164$
Solution:
a. $814=344=3$
b. $325=255=2$
c. $17=177=1$
d. $1164=( 1 2)44=12$
If the index is $n=2$, then the radical indicates a square root and it is customary to write the radical without the index, as illustrated below:
We have already taken care to define the principal square root of a number. At this point, we extend this idea to nth roots when n is even. For example, 3 is a fourth root of 81, because $34=81$. And since $(−3)4=81$, we can say that −3 is a fourth root of 81 as well. Hence we use the radical sign to denote the principal (nonnegative) nth rootThe positive nth root when n is even. when n is even. In this case, for any real number a, we use the following property:
For example,
The negative nth root, when n is even, will be denoted using a negative sign in front of the radical .
We have seen that the square root of a negative number is not real because any real number, when squared, will result in a positive number. In fact, a similar problem arises for any even index:
Here the fourth root of −81 is not a real number because the fourth power of any real number is always positive.
Example 6: Simplify.
a. $−164$
b. $−164$
Solution:
a. The radicand is negative and the index is even. Therefore, there is no real number that when raised to the fourth power is −16.
b. Here the radicand is positive. Furthermore, $16=24$, and we can simplify as follows:
When n is odd, the same problems do not occur. The product of an odd number of positive factors is positive and the product of an odd number of negative factors is negative. Hence when the index n is odd, there is only one real nth root for any real number a. And we have the following property:
Example 7: Find the nth root.
a. $−325$
b. $−17$
Solution:
a. $−325=(−2)55=−2$
b. $−17=(−1)77=−1$
Try this! Find the fourth root: $6254$.
Answer: 5
### Video Solution
(click to see video)
Summary: When n is odd, the nth root is positive or negative depending on the sign of the radicand.
When n is even, the nth root is positive or not real depending on the sign of the radicand.
## Simplifying Using the Product and Quotient Rule for Radicals
It will not always be the case that the radicand is a perfect power of the given index. If not, we use the following two properties to simplify them. If a and b represent positive real numbers, then we have
Product rule for radicals$a⋅bn=an⋅bn$, where a and b represent positive real numbers.: $a⋅bn=an⋅bn$ Quotient rule for radicals$abn=anbn$, where a and b represent positive real numbers.: $abn=anbn$
A radical is simplifiedA radical where the radicand does not consist of any factor that can be written as a perfect power of the index. if it does not contain any factor that can be written as a perfect power of the index.
Example 8: Simplify: $12$.
Solution: Here 12 can be written as 4 ⋅ 3, where 4 is a perfect square.
We can verify our answer on a calculator:
Also, it is worth noting that
Answer: $23$
Example 9: Simplify: $135$.
Solution: Begin by finding the largest perfect square factor of 135.
Therefore,
Answer: $315$
Example 10: Simplify: $50121$.
Solution: Begin by finding the prime factorizations of both 50 and 121. This will enable us to easily determine the largest perfect square factors.
Therefore,
Answer: $5211$
Example 11: Simplify: $1623$.
Solution: Use the prime factorization of 162 to find the largest perfect cube factor:
Replace the radicand with this factorization and then apply the product rule for radicals.
We can verify our answer on a calculator.
Answer: $3 63$
Try this! Simplify: $2 963$.
Answer: $4 123$
### Video Solution
(click to see video)
Example 12: Simplify: $−965$.
Solution: Here we note that the index is odd and the radicand is negative; hence the result will be negative. We can factor the radicand as follows:
Then simplify:
Answer: $−2 35$
Example 13: Simplify: $−8643$.
Solution: In this case, consider the equivalent fraction with $−8=(−2)3$ in the numerator and then simplify.
Answer: −1/2
Try this! Simplify $−1083$.
Answer: $−3 43$
### Video Solution
(click to see video)
### Key Takeaways
• The square root of a number is that number that when multiplied by itself yields the original number. When the radicand a is positive, $a2=a$. When the radicand is negative, the result is not a real number.
• The cube root of a number is that number that when used as a factor with itself three times yields the original number. The cube root may be positive or negative depending on the sign of the radicand. Therefore, for any real number a, we have the property $a33=a$.
• When working with nth roots, n determines the definition that applies. We use $ann=a$ when n is odd and $ann=|a|$ when n is even. When n is even, the negative nth root is denoted with a negative sign in front of the radical sign.
• To simplify square roots, look for the largest perfect square factor of the radicand and then apply the product or quotient rule for radicals.
• To simplify cube roots, look for the largest perfect cube factor of the radicand and then apply the product or quotient rule for radicals.
• To simplify nth roots, look for the factors that have a power that is equal to the index n and then apply the product or quotient rule for radicals. Typically, the process is streamlined if you work with the prime factorization of the radicand.
### Topic Exercises
Part A: Radicals
Simplify.
1. $81$
2. $100$
3. $64$
4. $121$
5. $0$
6. $1$
7. $0.25$
8. $0.01$
9. $1.21$
10. $2.25$
11. $14$
12. $136$
13. $2516$
14. $925$
15. $−25$
16. $−9$
17. $−36$
18. $−81$
19. $−100$
20. $−1$
21. $273$
22. $1253$
23. $643$
24. $83$
25. $183$
26. $1643$
27. $8273$
28. $641253$
29. $0.0013$
30. $1,0003$
31. $−13$
32. $−83$
33. $−273$
34. $−643$
35. $−183$
36. $−27643$
37. $−8273$
38. $−11253$
39. $814$
40. $6254$
41. $164$
42. $10,0004$
43. $325$
44. $15$
45. $2435$
46. $100,0005$
47. $−164$
48. $−16$
49. $−325$
50. $−15$
51. $−1$
52. $−164$
53. $−5 −273$
54. $−2 −83$
55. $5 −1,0003$
56. $3 −2435$
57. $10 −164$
58. $2 −646$
59. $325$
60. $64$
61. $2 273$
62. $8 2435$
63. $−7 83$
64. $−4 6254$
65. $6 100,0005$
66. $5 1287$
Part B: Simplifying Radicals
Simplify.
67. $32$
68. $250$
69. $80$
70. $150$
71. $160$
72. $60$
73. $175$
74. $216$
75. $5112$
76. $10135$
77. $5049$
78. $−2120$
79. $−3162$
80. $89$
81. $45121$
82. $9681$
83. $543$
84. $243$
85. $483$
86. $813$
87. $403$
88. $1203$
89. $1623$
90. $5003$
91. $541253$
92. $403433$
93. $5 −483$
94. $2 −1083$
95. $8 964$
96. $7 1624$
97. $1605$
98. $4865$
99. $2242435$
100. $5325$
Simplify. Give the exact answer and the approximate answer rounded to the nearest hundredth.
101. $8$
102. $200$
103. $45$
104. $72$
105. $34$
106. $59$
107. $3225$
108. $4849$
109. $803$
110. $3203$
111. $483$
112. $2703$
Rewrite the following as a radical expression with coefficient 1.
113. $215$
114. $37$
115. $510$
116. $103$
117. $2 73$
118. $3 63$
119. $2 54$
120. $3 24$
121. The formula for the area A of a square is $A=s2$. If the area is 18 square units, then what is the length of each side?
122. Calculate the length of a side of a square with an area of 60 square centimeters.
123. The formula for the volume V of a cube is $V=s3$. If the volume of a cube is 112 cubic units, then what is the length of each side?
124. Calculate the length of a side of a cube with a volume of 54 cubic centimeters.
Part C: Discussion Board
125. Explain why there are two square roots for any nonzero real number.
126. Explain why there is only one cube root for any real number.
127. What is the square root of 1, and what is the cube root of 1? Explain why.
128. Explain why $−1$ is not a real number and why $−13$ is a real number.
### Answers
1: 9
3: 8
5: 0
7: 0.5
9: 1.1
11: 1/2
13: 5/4
15: Not a real number
17: −6
19: −10
21: 3
23: 4
25: 1/2
27: 2/3
29: 0.1
31: −1
33: −3
35: −1/2
37: −2/3
39: 3
41: 2
43: 2
45: 3
47: −2
49: −2
51: Not a real number
53: 15
55: −50
57: Not a real number
59: 15
61: 6
63: −14
65: 60
67: $42$
69: $45$
71: $410$
73: $57$
75: $207$
77: $527$
79: $−272$
81: $3511$
83: $3 23$
85: $2 63$
87: $2 53$
89: $3 63$
91: $3 235$
93: $−10 63$
95: $16 64$
97: $2 55$
99: $2 753$
101: $22≈2.83$
103: $35≈6.71$
105: $32≈0.87$
107: $425≈1.13$
109: $2 103≈4.31$
111: $2 63≈3.63$
113: $60$
115: $250$
117: $563$
119: $804$
121: $32$ units
123: $2 143$ units |
Elementary Geometry for College Students (5th Edition)
First, we need to prove $\triangle FED\cong\triangle GED$ by SSS. Then we can deduce $\angle DEF\cong\angle DEG$. Then prove $\angle DEF$ and $\angle DEG$ must be right $\angle$s. That means $\overline{DE}\bot\overline{FG}$
*PLANNING: First, we need to prove $\triangle FED\cong\triangle GED$. Then we can deduce $\angle DEF\cong\angle DEG$. So $\angle DEF$ and $\angle DEG$ must be right $\angle$s. That means $\overline{DE}\bot\overline{FG}$ 1. $E$ is the midpoint of $\overline{FG}$ (Given) 2. $\overline{EF}\cong\overline{EG}$ (The midpoint of a line divides it into 2 congruent lines) 3. $\overline{DF}\cong\overline{DG}$ (Given) 4. $\overline{DE}\cong\overline{DE}$ (Identity) So now we have all 3 sides of $\triangle FED$ are congruent with 3 corresponding sides of $\triangle GED$. Therefore, 5. $\triangle FED\cong\triangle GED$ (SSS) 6. $\angle DEF\cong\angle DEG$ (CPCTC) However, we see that $\angle DEF+\angle DEG=\angle FEG=180^o$ (since $\overline{FG}$ is a line) Therefore, the value of each angle must be $90^o$. So, 7. $\angle DEF$ and $\angle DEG$ are both right $\angle$s. 8. $\overline{DE}\bot\overline{FG}$ (if a line intersects another one and creates 2 right angles, then those 2 lines are perpendicular with each other) |
Red Marbles, Blue Marbles
By | April 16, 2010
Problem: you have two jars, 50 red marbles, 50 blue marbles. you need to place all the marbles into the jars such that when you blindly pick one marble out of one jar, you maximize the chances that it will be red. (when picking, you’ll first randomly pick a jar, and then randomly pick a marble out of that jar) you can arrange the marbles however you like, but each marble must be in a jar.
Solution
Chance! chance is easy if you know how to do the formula. we know that we have two choices to make. first we’ll pick a jar, and each jar will have a 1/2 chance of being picked. then we’ll pick a marble, and depending how we stack the marbles, we’ll have a (# of red marbles in jar)/(# of total marbles in jar) chance of getting a red one.
for example, say we put all the red marbles into jar A and all the blue ones into jar B. then our chances for picking a red one are:
1/2 chance we pick jar A * 50/50 chance we pick a red marble
1/2 chance we pick jar B * 0/50 chance we pick a red marble
do the math and you get 1/2 chance for a red marble from jar A and a 0/2 chance for a red marble from jar B. add ‘em up and you get the result = 1/2 chance for picking a red marble.
think about it for awhile and see if you can figure out the right combination. we had a 50/50 (guaranteed) chance in picking a red marble from jar A, but we didn’t have to have 50 red marbles in there to guarantee those fantastic odds, did we? we could’ve just left 1 red marble in there and the odds are still 1/1. then we can take all those other marbles and throw them in jar B to help the odds out there.
let’s look at those chances:
1/2 we pick jar A * 1/1 we pick a red marble
1/2 we pick jar B * 49/99 we pick a red marble
do the math and add them up to get 1/2 + 49/198 = 148/198, which is almost 3/4.
we can prove these are the best odds in a somewhat non-formal way as follows. our goal is to maximize the odds of picking a red marble. therefore we can subdivide this goal into maximizing the odds of picking a red marble in jar A and maximizing the odds of picking a red marble in jar B. if we do that, then we will have achieved our goal. it is true that by placing more red marbles into a jar we will increase the chances of picking a red marble. it is also true that by reducing the number of blue marbles in a jar we will increase the odds also. we’ve maximized the odds in jar A since 1/1 is the maximum odds by reducing the number of blue marbles to 0 (the minimum). we’ve also maximized the number of red marbles in jar B. if we added any more red marbles to jar B we would have to take them out of jar A which reduce the odds there to 0 (very bad). if we took any more blue ones out of jar B we would have to put them in jar A which reduce the odds there by 50% (very bad).
it wasn’t really a good proof, but QED anyway 😛 |
## Inverse of Matrix Worksheets
In this page inverse of matrix worksheets we are going to see practice questions of the topic matrix.
1) Reversal law for inverse
If A and B are any two non singular matrices of the same order,then AB is also non singular and (AB)⁻¹ = B⁻¹ A⁻¹ the inverse of a product is the product of the inverses taken in the reverse order.
2) Reversal law of Transposes
If A and B are matrices comfortable to multiplication,then (AB)^T = B^T A^T
3) Inverse law
For any non singular matrix A. (A^T)⁻¹ = (A⁻¹)^T
These are the properties in the topic inverse of a matrix.
Definition:
If A is a non-singular matrix,there exists an inverse which is given by
Inverse of matrix Practice questions
1) Find the inverse of the following matrix
2 1 1 1 1 1 1 -1 2
inverse of matrix worksheet
Solution
2) Find the inverse of the following matrix
1 2 1 2 -1 2 1 1 -2
inverse of matrix worksheets
Solution
3) Find the inverse of the following matrix
6 2 3 3 1 1 10 3 4
Solution
4) Find the inverse of the following matrix
2 5 7 1 1 1 2 1 -1
Solution
5) Find the inverse of the following matrix
3 1 -1 2 -1 2 2 1 -2
Solution
Quote on Mathematics
“Mathematics, without this we can do nothing in our life. Each and everything around us is math.
Math is not only solving problems and finding solutions and it is also doing many things in our day to day life. They are:
It divides sorrow and multiplies forgiveness and love.
Some people would not be able accept that the subject Math is easy to understand. That is because; they are unable to realize how the life is complicated. The problems in the subject Math are easier to solve than the problems in our real life. When we people are able to solve all the problems in the complicated life, why can we not solve the simple math problems?
Many people think that the subject math is always complicated and it exists to make things from simple to complicate. But the real existence of the subject math is to make things from complicate to simple.”
2. Matrix Inverse Calculator - 2x2 Matrix
3. Matrix Inverse Calculator - 3x3 Matrix
4. Matrix Inverse Calculator - 4x4 Matrix
5. Cramer's Rule Calculator - 3x3 Matrix
6. Matrix Addition Calculator - 3x3 Matrix
7. Matrix Subtraction Calculator - 3x3 Matrix
8. Matrix Multiplication Calculator - 2x2 Matrix
9. Matrix Multiplication Calculator - 3x3 Matrix
10. Matrix Determinant Calculator - 3x3 & 2x2 Matrix
11. Matrix Addition Calculator - 2x2 Matrix
12. Matrix Subtraction Calculator- 2x2 Matrix
13. Matrix Addition Calculator - 4x4 Matrix
14. Matrix Subtraction Calculator- 4x4 Matrix
15. Matrix Multiplication Calculator - 4x4 Matrix
16. Matrix Determinant Calculator - 4x4 matrix
17. Squared Matrix Calculator
18. Transpose Matrix Calculator |
# Matrix multiplication is pretty tough- so i will cover that in class. In the meantime , compute the following if A=begin{bmatrix}2&1&1 -1&-1&4 end{bmatrix} , B=begin{bmatrix}0 & 2 -4 & 12&-3 end{bmatrix} , C=begin{bmatrix}6 & -1 3 & 0-2&5 end{bmatrix} , D=begin{bmatrix}2 & -3&4 -3& 1&-2 end{bmatrix} If the operation is not possible , write NOT POSSIBLE and be able to explain why a)A+B b)B+C c)2A
Question
Matrices
Matrix multiplication is pretty tough- so i will cover that in class. In the meantime , compute the following if
$$A=\begin{bmatrix}2&1&1 \\-1&-1&4 \end{bmatrix} , B=\begin{bmatrix}0 & 2 \\-4 & 1\\2&-3 \end{bmatrix} , C=\begin{bmatrix}6 & -1 \\3 & 0\\-2&5 \end{bmatrix} , D=\begin{bmatrix}2 & -3&4 \\-3& 1&-2 \end{bmatrix}$$
If the operation is not possible , write NOT POSSIBLE and be able to explain why
a)A+B
b)B+C
c)2A
2021-01-05
Step 1
$$\text{Given : } A=\begin{bmatrix}2&1&1 \\-1&-1&4 \end{bmatrix} , B=\begin{bmatrix}0 & 2 \\-4 & 1\\2&-3 \end{bmatrix} , C=\begin{bmatrix}6 & -1 \\3 & 0\\-2&5 \end{bmatrix} , D=\begin{bmatrix}2 & -3&4 \\-3& 1&-2 \end{bmatrix}$$
Solution for question a:
To compute A+B:
Note that two matrices may be added if and only if they have the same dimension, that is, they must have the same number of rows and columns.
Here, Dimension of $$A=2 \times 3$$
And, Dimension of $$B=3 \times 2$$
Matrix A and B do not have the same dimension. Hence, matrix A and B cannot be added.
Therefore it is not possible to perform A+B.
Step 2
Solution for question b:
Here,
Dimension of matrix $$B=3 \times 2$$
Dimension of matrix $$C=3 \times 2$$
Both matrices B and C have the same dimension. Hence, matrix B and C can be added.
Further,
$$B+C=\begin{bmatrix}0 & 2 \\-4 & 1\\2&-3 \end{bmatrix}+\begin{bmatrix}6 & -1 \\3 & 0\\-2&5 \end{bmatrix}$$
$$=\begin{bmatrix}0+6 & 2+(-1) \\-4+3 & 1+0\\2+(-2)&(-3)+5 \end{bmatrix}$$
$$=\begin{bmatrix}6 &1 \\-1 &1\\0&2 \end{bmatrix}$$
Therefore,
$$B+C=\begin{bmatrix}6 &1 \\-1 &1\\0&2 \end{bmatrix}$$
Step 3
Solution for question c:
To compute 2A multiply each entry of the matrix A by 2.
$$2A=2\begin{bmatrix}2&1&1 \\-1&-1&4 \end{bmatrix}$$
$$=\begin{bmatrix}4&2&2 \\-2&-2&8 \end{bmatrix}$$
therefore,
$$2A=\begin{bmatrix}4&2&2 \\-2&-2&8 \end{bmatrix}$$
### Relevant Questions
A random sample of $$n_1 = 14$$ winter days in Denver gave a sample mean pollution index $$x_1 = 43$$.
Previous studies show that $$\sigma_1 = 19$$.
For Englewood (a suburb of Denver), a random sample of $$n_2 = 12$$ winter days gave a sample mean pollution index of $$x_2 = 37$$.
Previous studies show that $$\sigma_2 = 13$$.
Assume the pollution index is normally distributed in both Englewood and Denver.
(a) State the null and alternate hypotheses.
$$H_0:\mu_1=\mu_2.\mu_1>\mu_2$$
$$H_0:\mu_1<\mu_2.\mu_1=\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1<\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1\neq\mu_2$$
(b) What sampling distribution will you use? What assumptions are you making? NKS The Student's t. We assume that both population distributions are approximately normal with known standard deviations.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference $$\mu_1 - \mu_2$$. Round your answer to two decimal places.) NKS (d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(e) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \alpha?
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. (g) Find a 99% confidence interval for
$$\mu_1 - \mu_2$$.
(Round your answers to two decimal places.)
lower limit
upper limit
(h) Explain the meaning of the confidence interval in the context of the problem.
Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.
Refer to the following matrices.
$$A=\begin{bmatrix}2 & -3&7&-4 \\-11 & 2&6&7 \\6 & 0&2&7 \\5 & 1&5&-8 \end{bmatrix} B=\begin{bmatrix}3 & -1&2 \\0 & 1&4 \\3 & 2&1 \\-1 & 0&8 \end{bmatrix} , C=\begin{bmatrix}1& 0&3 &4&5 \end{bmatrix} , D =\begin{bmatrix}1\\ 3\\-2 \\0 \end{bmatrix}$$
Identify the row matrix. Matrix C is a row matrix.
$$A=\begin{bmatrix}2& 1&1 \\-1 & -1&4 \end{bmatrix} B=\begin{bmatrix}0& 2 \\-4 & 1\\2 & -3 \end{bmatrix} C=\begin{bmatrix}6& -1 \\3 & 0\\-2 & 5 \end{bmatrix} D=\begin{bmatrix}2& -3&4 \\-3 & 1&-2 \end{bmatrix}$$
a)$$A-3D$$
b)$$B+\frac{1}{2}$$
c) $$C+ \frac{1}{2}B$$
(a),(b),(c) need to be solved
The bulk density of soil is defined as the mass of dry solidsper unit bulk volume. A high bulk density implies a compact soilwith few pores. Bulk density is an important factor in influencing root development, seedling emergence, and aeration. Let X denotethe bulk density of Pima clay loam. Studies show that X is normally distributed with $$\displaystyle\mu={1.5}$$ and $$\displaystyle\sigma={0.2}\frac{{g}}{{c}}{m}^{{3}}$$.
(a) What is thedensity for X? Sketch a graph of the density function. Indicate onthis graph the probability that X lies between 1.1 and 1.9. Findthis probability.
(b) Find the probability that arandomly selected sample of Pima clay loam will have bulk densityless than $$\displaystyle{0.9}\frac{{g}}{{c}}{m}^{{3}}$$.
(c) Would you be surprised if a randomly selected sample of this type of soil has a bulkdensity in excess of $$\displaystyle{2.0}\frac{{g}}{{c}}{m}^{{3}}$$? Explain, based on theprobability of this occurring.
(d) What point has the property that only 10% of the soil samples have bulk density this high orhigher?
(e) What is the moment generating function for X?
Solve for X in the equation, given
$$3X + 2A = B$$
$$A=\begin{bmatrix}-4 & 0 \\1 & -5\\-3&2 \end{bmatrix} \text{ and } B=\begin{bmatrix}1 & 2 \\ -2 & 1 \\ 4&4 \end{bmatrix}$$
The product of matrix B and C is matrix D
$$\begin{bmatrix}2 & -1&4 \\g & 0&3\\2&h&0 \end{bmatrix} \times \begin{bmatrix}-1 & 5 \\4&f\\-3&1 \end{bmatrix}=\begin{bmatrix}i & 24 \\-16&-4\\4&e \end{bmatrix}$$
3.From the expression above, what should be the value of e?
4.From the expression above, what should be the value of g?
5.From the expression above, what should be the value of f?
If $$A=\begin{bmatrix}1 & 1 \\3 & 4 \end{bmatrix} , B=\begin{bmatrix}2 \\1 \end{bmatrix} ,C=\begin{bmatrix}-7 & 1 \\0 & 4 \end{bmatrix},D=\begin{bmatrix}3 & 2 & 1 \end{bmatrix} \text{ and } E=\begin{bmatrix}2 & 3&4 \\1 & 2&-1 \end{bmatrix}$$
Find , if possible,
a) A+B , C-A and D-E b)AB, BA , CA , AC , DA , DB , BD , EB , BE and AE c) 7C , -3D and KE
$$A=\begin{bmatrix}4 & 0 \\-3 & 5 \\ 0 & 1 \end{bmatrix} B=\begin{bmatrix}5 & 1 \\-2 & -2 \end{bmatrix} C=\begin{bmatrix}1 & -1 \\-1 & 1 \end{bmatrix}$$
$$A=\begin{bmatrix}1 & -2&1 \\0 & 2&-1\\2&1&1 \end{bmatrix}$$
$$B=\begin{bmatrix}2 & 1&-1 \\1 & -1&0\\2&-1&1 \end{bmatrix}$$
(a)$$\begin{pmatrix}3 & 2&|&8 \\1 & 5&|&7 \end{pmatrix}$$
(b)$$\begin{pmatrix}5 & -2&1&|&3 \\2 & 3&-4&|&0 \end{pmatrix}$$
(c)$$\begin{pmatrix}2 & 1&4&|&-1 \\4 & -2&3&|&4 \\5 & 2&6&|&-1 \end{pmatrix}$$
(d)$$\begin{pmatrix}4 & -3&1&2&|&4 \\3 & 1&-5&6&|&5 \\1 & 1&2&4&|&8\\5 & 1&3&-2&|&7 \end{pmatrix}$$ |
The top number says how many slices we have. The bottom number says how many equal slices the whole pizza was cut into.
You are watching: What is between 1/4 and 1/2
## Equivalent Fractions
Some fractions may look different, but are really the same, for example:
4/8 = 2/4 = 1/2 (Four-Eighths) (Two-Quarters) (One-Half) = =
It is usually best to show an answer using the simplest fraction ( 1/2 in this case ). That is called Simplifying, or Reducing the Fraction
## Numerator / Denominator
We call the top number the Numerator, it is the number of parts we have.We call the bottom number the Denominator, it is the number of parts the whole is divided into.
See more: Where In A Prokaryotic Cell Do The Reactions Of Glycolysis Occur?
NumeratorDenominator
You just have to remember those names! (If you forget just think "Down"-ominator)
It is easy to add fractions with the same denominator (same bottom number):
1/4 + 1/4 = 2/4 = 1/2 (One-Quarter) (One-Quarter) (Two-Quarters) (One-Half) + = =
One-quarter plus one-quarter equals two-quarters, equals one-half
Another example:
5/8 + 1/8 = 6/8 = 3/4 + = =
Five-eighths plus one-eighth equals six-eighths, equals three-quarters
## Adding Fractions with Different Denominators
But what about when the denominators (the bottom numbers) are not the same?
3/8 + 1/4 = ? + =
Three-eighths plus one-quarter equals ... what?
We must somehow make the denominators the same.
In this case it is easy, because we know that 1/4 is the same as 2/8 :
3/8 + 2/8 = 5/8 + =
Three-eighths plus two-eighths equals five-eighths
There are two popular methods to make the denominators the same:
(They both work nicely, use the one you prefer.)
## Other Things We Can Do With Fractions
We can also:
Visit the Fractions Index to find out even more.
Fractions Index Equivalent Fractions Adding Fractions Subtracting Fractions Multiplying Fractions Dividing Fractions Greatest Common Factor Least Common Multiple |
Informative line
### Kinetic Energy And Work Energy Theorem
Learn kinetic energy of a rotating body, Find the angular velocity of the rod when it becomes vertical and the work done by gravity when the rod turned by an angle ?.
# Fixed Axis
• Rotation about a fixed axis :
In rotational motion, the particles forming the rigid body, move in parallel planes along circles centered on the same fixed axis, called the axis of rotation.
• Axis of rotation :
Figure shows a rigid body of arbitrary shape in rotation about a fixed axis, called the axis of rotation or the rotation axis.
#### A disk of radius R rotates in such a manner that its center moves in a vertical circle of radius $$\dfrac{R}{2}$$ and its velocity is always perpendicular to the plane of the disk. Then, the axis of rotation is
A
B
C
D
×
As the center moves perpendicular to the plane of the disk, its center lies at O and the chord AB, passing through O.
Therefore, the line perpendicular to the planes of circular motion i.e., in the plane of disk is the axis of rotation. Hence, option (C) is correct.
In option (A), distance between center of mass and axis of rotation is $$\dfrac{R}{3}$$ . So, it can't rotate in a circle of radius $$\dfrac{R}{2}$$ .
Hence, option (A) is incorrect.
In option (B), the distance between axis of rotation and center of mass is R. So, it can't rotate in a circle of radius $$\dfrac{R}{2}$$ .
Hence, option (B) is incorrect.
In option (D), the distance between axis of rotation and center mass is zero. So, it can't rotate in a vertical circle of radius $$\dfrac{R}{2}$$ .
Hence, option (D) is incorrect.
### A disk of radius R rotates in such a manner that its center moves in a vertical circle of radius $$\dfrac{R}{2}$$ and its velocity is always perpendicular to the plane of the disk. Then, the axis of rotation is
A
B
C
D
Option C is Correct
# Kinetic Energy of a Rotating Body
• The kinetic energy of a particle of mass m, moving with velocity v, is given by $$K = \dfrac{1}{2} m v^2$$.
• In linear motion, the linear variables i.e., displacement (s), velocity (v) and acceleration (a) are same for all particles of a body.
• In a pure rotational motion, these linear variables have different values for different particles of the body. Thus, these variables cannot be associated with the rotating body.
• For a body which is in pure rotational motion the angular variables i.e., angular displacement, angular velocity and angular acceleration have same value for all particles of a body.
• Using this, the kinetic energy of a rigid body can be calculated by adding kinetic energy of all the particles
$$K E= \int \dfrac{1}{2} (dm) v^2$$
Since, v = r $$\omega$$
$$K E= \int \dfrac{1}{2} (dm) \;r^2_{{p}/{o}} \;\omega^2$$
$$K E= \dfrac{\omega^2}{2} \int(dm) \;r^2_{{p}{/o}}$$
Note :- $$\omega$$ of all the particles on a rigid body is same.
• Moment of inertia of a body about an axis passing through O is given by :
$$I_0 = \int dm (r_ {{p}/{o}})^2$$
• Therefore, Kinetic energy, $$K = \dfrac{1}{2} I_0 \;\omega^2$$
• Here, $$r_ {{p}/{o}}$$ is the perpendicular distance of the points from an axis passing through O.
• $$I_0$$ is the moment of inertia of the body about the axis passing through O i.e., axis of rotation.
#### A rod AB of mass (M) 6kg and length (L) 2m is rotating about A with an angular velocity ($$\omega$$) 2 rad/sec. Find its kinetic energy.
A 14 J
B 20 J
C 16 J
D 18 J
×
Moment of inertia of rod about A
$$I_ {{rod}/{A}} = \dfrac{ML^2}{3}$$
Kinetic energy , $$K E = \dfrac {1}{2} \; \dfrac{ML^2}{3} \omega^2$$
$$= \dfrac {1}{2}\times\left( \dfrac{6(2)^2\times(2)^2}{3}\right)$$
= 16 Joules
### A rod AB of mass (M) 6kg and length (L) 2m is rotating about A with an angular velocity ($$\omega$$) 2 rad/sec. Find its kinetic energy.
A
14 J
.
B
20 J
C
16 J
D
18 J
Option C is Correct
#### A rigid body having moment of inertia $$I$$, about the given axis of mass m, hinged at O is kept in a position such that the line joining center of mass to O is horizontal. Given position of center of mass from O is $$X_{cm}$$. Find the angular velocity under gravity when CO turns by an angle $$\theta$$.
A $$\sqrt {\dfrac{2mg X_{cm} sin\,\theta} {I}}$$
B $$\sqrt {\dfrac{mg X_{cm} sin\,\theta}{I}}$$
C $$\sqrt {\dfrac{2mg X_{cm}}{ I}}$$
D $$\sqrt {\dfrac{2mg X_{cm} sin\theta}{3 I}}$$
×
Work done by gravity,
$$W_g = mg h _{cm}$$
where $$h _{cm}$$ is displacement of center of mass in vertical direction.
So, $$h _{cm} = X_{cm}sin\,\theta$$
$$\because$$ $$W_g = mg \;X_{cm}sin\,\theta$$
$$K.E=mg X_{cm} sin\,\theta$$............(1)
Also, $$K.E=\dfrac{1}{2} I\omega^2$$................(2)
$$I$$ is moment of inertia of the body about the given axis of rotation.
From (1) and (2)
$$mg X_{cm} sin\,\theta = \dfrac{1}{2} I \omega^2$$
$$\omega =\sqrt {\dfrac{2mg X_{cm} sin\,\theta}{I}}$$
### A rigid body having moment of inertia $$I$$, about the given axis of mass m, hinged at O is kept in a position such that the line joining center of mass to O is horizontal. Given position of center of mass from O is $$X_{cm}$$. Find the angular velocity under gravity when CO turns by an angle $$\theta$$.
A
$$\sqrt {\dfrac{2mg X_{cm} sin\,\theta} {I}}$$
.
B
$$\sqrt {\dfrac{mg X_{cm} sin\,\theta}{I}}$$
C
$$\sqrt {\dfrac{2mg X_{cm}}{ I}}$$
D
$$\sqrt {\dfrac{2mg X_{cm} sin\theta}{3 I}}$$
Option A is Correct
#### A uniform rod AB of mass $$m_1$$, and length $$\ell$$ is hinged at a point O such that AO is $$\dfrac{\ell}{3}$$. If a point mass m2 is attached at point A. Then, find the angular velocity of the system, when it is released from horizontal position and turned by an angle $$\dfrac {\pi}{6}$$. Take downward direction to be positive.
A $$\sqrt {3(m_1-2m_2)g \over 2(m_1 + m_2)\ell}$$
B $$\sqrt {2(m_1 + m_2){\dfrac{g}{\ell}}}$$
C $$\sqrt {\dfrac{3}{2} \dfrac{g}{\ell}}$$
D $$\sqrt {(m_1-2m_2) \over \ell g}$$
×
Center of mass, C of the rod is at a distance $$\dfrac{\ell}{6}$$ from O when it turns by an angle $$\dfrac{\pi}{6}$$, where $$h_{cm}$$ is the vertical displacement of center of mass.
Vertical displacement of center of mass of rod
$${(h_{cm})_{rod}} = \dfrac{\ell}{6} sin \dfrac{\pi}{6}$$
$$= \dfrac{\ell}{12}$$ (downwards)
Vertical displacement of center of mass of particle
$${(h_{cm})_{particle}} = \dfrac{\ell}{3} sin \dfrac{\pi}{6}$$
$$= -\dfrac{\ell}{6}$$ (upwards)
Total work done under gravity
$$W_{g} = m_{1} g{(h_{cm})}_{rod} + {m_2}g {(h_{cm})_{particle}}$$
$$W_{g}= m_{1}g\dfrac{\ell}{12} - m_{2}g\dfrac{\ell}{6}$$
Then, by work - energy theorem
$$\Delta KE = W_g$$
$$K E_{f} - KE_{i}= (m_{1}g\dfrac{\ell}{12} - m_{2}g\dfrac{\ell}{6})$$
$$KE_{f} = (\dfrac{m_{1}}{2}-m_{2})g\dfrac{\ell}{6}$$
$$KE_{f} = ({m_{1}}- 2m_{2})g\dfrac{\ell}{12}$$..................(1)
Moment of inertia of a system containing two or more bodies about an axis, is sum of $$I$$ of the bodies about the same given axis.
Thus, $$I = I_{rod} + I_{particle}$$
$$I = \big[ \dfrac {m_{1}\ell^2}{12}+ m_1(\dfrac{\ell}{2})^2\big]+ m_{2} (\dfrac{\ell}{3})^2$$
$$I = \dfrac{m_1\ell^2}{9} +\dfrac{m_2\ell^2}{9} = (m_{1} + m_{2}) \dfrac{\ell^2}{9}$$
Therefore, $$KE = \dfrac{1}{2} I \omega^2 = \dfrac{1}{2} (m_{1}+m_{2}) \dfrac{\ell^{2}}{9}\omega^2$$ .......................(2)
From (1) and (2),
$$= (m_{1}-2m_{2})g\dfrac{\ell}{12} = \dfrac{1}{2}(m_{1}+m_{2})\dfrac{\ell}{9} \omega^2$$
$$\omega= \sqrt {3(m_1-2m_2)g \over 2(m_1 + m_2)\ell}$$
### A uniform rod AB of mass $$m_1$$, and length $$\ell$$ is hinged at a point O such that AO is $$\dfrac{\ell}{3}$$. If a point mass m2 is attached at point A. Then, find the angular velocity of the system, when it is released from horizontal position and turned by an angle $$\dfrac {\pi}{6}$$. Take downward direction to be positive.
A
$$\sqrt {3(m_1-2m_2)g \over 2(m_1 + m_2)\ell}$$
.
B
$$\sqrt {2(m_1 + m_2){\dfrac{g}{\ell}}}$$
C
$$\sqrt {\dfrac{3}{2} \dfrac{g}{\ell}}$$
D
$$\sqrt {(m_1-2m_2) \over \ell g}$$
Option A is Correct
# Work Done by Gravity, when a Rod Turns by an Angle $$\theta$$
• Consider a rod of mass m and length $$\ell$$, which is hinged at one end and kept in horizontal position initially, as shown in figure.
• When the rod turned by an angle $$\theta$$, then its position can be shown as in figure.
• The center of mass (C) of a uniform rod lies at a distance $$\dfrac{L}{2}$$ from one end.
• The vertical displacement of center of mass is, $$h_ {cm} = \dfrac{\ell}{2} sin\theta$$
• The work done by gravity is $$W_g = m g h_{cm}$$
$$W_ g = mg\dfrac{\ell}{2} sin\theta$$
#### A uniform rod of mass m, length $$\ell$$ is hinged at one end and kept in horizontal position initially, as shown. Find the work done by gravity when the rod turned by an angle $$\theta$$.
A $$mg\ell sin\theta$$
B $$mg\ell cos\theta$$
C $$mg \dfrac{\ell}{2} cos\theta$$
D $$mg \dfrac{\ell}{2} sin\theta$$
×
Center of mass (C) of a uniform rod lies at a distance $$\dfrac{L}{2}$$ from one end.
Vertical displacement of center of mass, $$h_ {cm} = \dfrac{\ell}{2} sin\theta$$
Work done by gravity, $$W_g = m g h_{cm}$$
$$= mg\dfrac{\ell}{2} sin\theta$$
### A uniform rod of mass m, length $$\ell$$ is hinged at one end and kept in horizontal position initially, as shown. Find the work done by gravity when the rod turned by an angle $$\theta$$.
A
$$mg\ell sin\theta$$
.
B
$$mg\ell cos\theta$$
C
$$mg \dfrac{\ell}{2} cos\theta$$
D
$$mg \dfrac{\ell}{2} sin\theta$$
Option D is Correct
# Angular Velocity and Kinetic Energy of Rigid Body when it Rotates under Gravity
• Consider a uniform rod of mass m and length $$\ell$$, which is hinged at one end and kept in horizontal position initially as shown in figure.
• When it becomes vertical under the action of gravity. then the position of rod will be as shown in figure.
• The work done by gravity is $$W_ g = mgh_{cm}$$, where $$h_{cm}$$ is displacement of center of mass in vertical direction.
• By work energy-theorem,
Change in kinetic energy = Total work done
$$K E_f \;– \;KE _i =W_g$$
$$K E_f =W_g$$ [ Since, $$K E _i =0$$ ]
$$K E =mg \dfrac{\ell}{2}$$ ...........(1)
Also, $$K E = \dfrac{1}{2} I \omega^2$$ ..............( 2)
where $$I$$ is the moment of inertia of the body about the given axis of rotation.
From (1) and (2), $$\omega$$ can be calculated.
#### A uniform rod of mass m and length $$\ell$$ is hinged at one end and kept in horizontal position initially, as shown. Find the angular velocity of the rod when it becomes vertical. Consider downward motion to be positive.
A $$\sqrt {\dfrac{2g} {\ell}}$$
B $$\sqrt {\dfrac{g} {\ell}}$$
C $$\sqrt {\dfrac{3g}{\ell}}$$
D $$\sqrt {2g{\ell}}$$
×
Work done by gravity,
$$W_g = mg h _{cm}$$
where $$h _{cm}$$ is displacement of center of mass in vertical direction.
By work–Energy theorem,
change in kinetic energy = Total work done
$$K E_f \;– \;KE _i =W_g$$
Since, $$KE _i =0$$
Therefore, $$K E_f =W_g$$
Kinetic Energy $$K E =mg \dfrac{\ell}{2}$$............(1)
Also, $$K E = \dfrac{1}{2} I\omega^2$$................(2)
$$I$$ is moment of inertia of the body about the given axis of rotation.
From (1) and (2)
$$mg \dfrac{\ell}{2} = \dfrac{1}{2} I\omega^2$$
$$mg \dfrac{\ell}{2} = \dfrac{1}{2}\dfrac{m\ell^2}{3} \omega^2$$ $$[\because\,\,I = \dfrac{m\ell^2}{3}]$$
$$\omega = \sqrt {\dfrac{3g}{\ell}}$$
### A uniform rod of mass m and length $$\ell$$ is hinged at one end and kept in horizontal position initially, as shown. Find the angular velocity of the rod when it becomes vertical. Consider downward motion to be positive.
A
$$\sqrt {\dfrac{2g} {\ell}}$$
.
B
$$\sqrt {\dfrac{g} {\ell}}$$
C
$$\sqrt {\dfrac{3g}{\ell}}$$
D
$$\sqrt {2g{\ell}}$$
Option C is Correct |
Introduction on how to measure angles learning:
A ruler in the geometry is used to measure the lines and the units used in the rulers are inches or centimeters .In geometry we measure the angles using a tool called the protractor. A protractor uses units called degrees to measure the angles. The protractor measures the angle from left to right. Now we are going to see how to measure the angles and its learning.
## How to measure angles learning : types of angles
There are four types of angles. They are acute angles, obtuse angles, right angles and straight angles.
Learning right angles:
The right angles which measures 90 degrees.
Learning straight angles:
The straight angles which measures 180 degrees and it is a straight line.
Learning acute angles:
An acute angle is an angle which measures less than 90 degrees.
Learning obtuse angles:
Obtuse angle is nothing but the measure will be more than 90 degrees and less than 180 degrees.
Learning Supplement angles:
The sum of two angles up to 180 degree is called as the supplement angle.
## How to measure angles learning : examples
Now we are going to see about the learning of angles and how to measure with some examples.
Ex 1:
In the given rectangle, what is the type of angle present in the given figure and measure the angles present in it?
Sol:
In the rectangle all the angles are right angles because it measures 90 degree each in all corners. It consists of four 90 degrees.
Ex 2:
How to find the type of angle present in the triangle, if the measures are 34°, 60°, and 86°?
Sol:
In a triangle the sum of the measure of three angles are given as,
34° + 60° + 86° = 180°
Thus the angles present in this are acute type of angles.
Ex 3:
Find the supplement of the angle of 47°
Sol:
The supplementary angle of 47° = 180° – 47° = 133°.
The supplementary angle of 133° = 180° – 47° = 133°.
The angles 47° and 133° are supplementary angles.
Ex 4:
An angle is 20° more than its complement. What is its measure?
Sol:
Let us take the unknown degree as x°
Its complement = 90°-x°
90°-x°+20°= x°
2x° = 110°
X° = 55°
Thus the measure of the angle is 55 degree. |
# Lesson 1
Introducing Ratios and Ratio Language
## 1.1: What Kind and How Many? (5 minutes)
### Warm-up
In this warm-up, students compare figures and sort them into categories.
### Launch
Display the image for all to see. Give students 1 minute of quiet think time followed by 2 minutes of partner discussion.
### Student Facing
Think of different ways you could sort these figures. What categories could you use? How many groups would you have?
### Student Response
For access, consult one of our IM Certified Partners.
### Anticipated Misconceptions
If students struggle to create their own categories, prompt them to consider a specific attribute of the figures, such as the size, color, or shape.
### Activity Synthesis
Record all the ways students answered the question for all to see. Ask a student to explain how they sorted the figures. Ask if anyone saw it a different way until all the different ways of seeing the shapes have been shared. Emphasize that the important thing is to describe the way they sorted them clearly enough that everyone agrees that it is a reasonable way to sort them. Tell students we will be looking at different ways of seeing the same set of objects in the next activity.
## 1.2: The Teacher’s Collection (10 minutes)
### Activity
This activity introduces students to ratio language and notation through examples based on a collection of everyday objects. Students learn that a ratio is an association between quantities, and that this association can be expressed in multiple ways.
After discussing examples of ratio language and notation for one way of categorizing the objects in the collection, students write ratios to describe the quantities for another way of categorizing objects in the collection.
As students work, circulate and identify those who:
• Create different categories from the given collection.
• Create categories whose quantities can be rearranged into smaller groups (e.g. 6 A’s and 4 B’s can be expressed as “for every 3 A’s there are 2 B’s”).
• Express the same ratio in opposite order or by using different words (e.g. “the ratio of A to B is 7 to 3,” and “for every 7 A’s there are 3 B’s”).
Have a collection of objects ready to display for the launch. Make sure there are different ways the collection can be sorted. For example, the dinosaurs below can be categorized by color (green, orange, and purple), by the number of legs they stand on (standing on 4 legs or on 2 legs), or by the features along their backs (crest, white stripe, or nothing).
Familiar classroom objects such as binder clips or pattern blocks can also be used to form collections. This picture shows a collection of binder clips that could be categorized by size (small, medium, and large) or by color (black, green, and blue) .
### Launch
Display a collection of objects for all to see. Give students 2 minutes of quiet think time to come up with as many different categories for sorting the collection as they can think of. Record students’ categories for all to see. Sort the collection into one of the student-suggested categories and count the number of items in each. Record the number of objects in each category and display for all to see. For example:
category A: green category B: orange category C: purple
3 2 4
Explain that we can talk about the quantities in the different categories using something called ratios. Tell students: “A ratio is an association between two or more quantities.” We use a colon, or the word “to,” between two values we are associating.
Share the following examples (adapt them to suit your collection) and display them for all to see. Keep the examples visible for the duration of the lesson.
• The ratio of purple to orange dinosaurs is 4 to 2.
• The ratio of purple to orange dinosaurs is $$4:2$$.
• The ratio of orange to purple dinosaurs is 2 to 4.
• The ratio of orange to purple dinosaurs is $$2:4$$
Explain that we can also associate two quantities using the phrase “for every $$a$$ of these, there are $$b$$ of those.” Add the following examples to the display.
• For every 3 green dinosaurs there are 4 purple dinosaurs.
• There are 4 purple dinosaurs for every 2 orange dinosaurs.
Finally, find two categories whose items can be rearranged into smaller groups, e.g. 4 purple dinosaurs to 2 orange dinosaurs. Point out that in some cases we can associate the same categories using different numbers. Share the following example and add it to the display.
• For every 2 purple dinosaurs, there is 1 orange dinosaur.
Have students write two or three sentences to describe ratios between the categories they suggested.
Representation: Develop Language and Symbols. Create a display of important terms and vocabulary. Invite students to suggest language or diagrams to include that will support their understanding of: ratio and category.
Supports accessibility for: Conceptual processing; Language
### Student Facing
1. Think of a way to sort your teacher’s collection into two or three categories. Count the items in each category, and record the information in the table.
2. Write at least two sentences that describe ratios in the collection. Remember, there are many ways to write a ratio:
• The ratio of one category to another category is ________ to ________.
• The ratio of one category to another category is ________ : ________.
• There are _______ of one category for every _______ of another category.
### Student Response
For access, consult one of our IM Certified Partners.
### Anticipated Misconceptions
Students may write ratios with no descriptive words. $$8:2$$ is a good start, but part of writing a ratio is stating what those numbers mean. Draw students’ attention to the sentence stems in the task statement; encourage them to use those words.
### Activity Synthesis
Invite several students to share their categories and sentences. Display them for all to see, attending to correct ratio language. Be sure to include students who express the same categories in reverse order, in different words, or with a different set of numbers (which students will later call an equivalent ratio). Leave several sentences displayed for students to see and use as a reference while working on the next task.
## 1.3: The Student’s Collection (20 minutes)
### Activity
In this activity, students write ratios to describe objects in their own collection. They create a display of objects and circulate to look at their classmates’ work. Students see that there are several ways to write ratios to describe the same situation.
### Launch
Invite students to share what types of items are in their personal collections. If students did not bring in a collection, pair them with another student, or provide them with an extra collection that you have brought in for that purpose.
Provide access to tools for creating a visual display. Tell students they will pause their work before creating a visual display to get their sentences approved.
Representing, Conversing: MLR7 Compare and Connect. Use this routine to help students consider audience when preparing to display their work. Display the list of items that should be included on the display and ask students, “what kinds of details could you include on your display to help a reader understand the ratios you’ve used to describe the objects in your collection?” Record ideas and display for all to see. Examples of these types of details or annotations include: the order in which representations are organized on the display, attaching written notes or details to certain representations, using specific vocabulary or phrases, or using color or arrows to show connections between representations. If time allows, ask students to describe specific examples of additional details that other groups used that helped them to interpret and understand their displays.
Design Principle(s): Maximize meta-awareness; Optimize output
### Student Facing
1. Sort your collection into three categories. You can experiment with different ways of arranging these categories. Then, count the items in each category, and record the information in the table.
2. Write at least two sentences that describe ratios in the collection. Remember, there are many ways to write a ratio:
• The ratio of one category to another category is ________ to ________.
• The ratio of one category to another category is ________ : ________.
• There are _______ of one category for every _______ of another category.
3. Make a visual display of your items that clearly shows one of your statements. Be prepared to share your display with the class.
### Student Response
For access, consult one of our IM Certified Partners.
### Student Facing
#### Are you ready for more?
1. Use two colors to shade the rectangle so there are 2 square units of one color for every 1 square unit of the other color.
2. The rectangle you just colored has an area of 24 square units. Draw a different shape that does not have an area of 24 square units, but that can also be shaded with two colors in a $$2:1$$ ratio. Shade your new shape using two colors.
### Student Response
For access, consult one of our IM Certified Partners.
### Anticipated Misconceptions
Watch for students simply writing a numerical ratio, such as $$3:7$$, without any descriptive words. Draw their attention to the sentence stems in the task statement.
### Activity Synthesis
Once students have had enough time to create their displays, circulate through each display and listen to how students describe their ratios.
As students present their displays, point out the various ways that students chose to showcase their work, including different ways to say the same ratio. Ask students who used two sets of numbers to describe the same categories (e.g., 8 to 2 and “4 for every 1”) to demonstrate the two ways of grouping the objects.
## Lesson Synthesis
### Lesson Synthesis
This lesson is all about how to use ratio language and notation to describe an association between two or more quantities. Wrap up the lesson by drawing a diagram for all to see of, say, 6 squares and 3 circles.
Say, “One way to write this ratio is, there are 6 squares for every 3 circles. What are some other ways to write this ratio?” Some correct options might be:
• The ratio of squares to circles is $$6:3$$.
• The ratio of circles to squares is 3 to 6.
• There are 2 squares for every 1 circle.
Display this diagram and the associated sentences the class comes up with somewhere in the classroom so students can refer back to the correct ratio and rate language during subsequent lessons.
Consider posing some more general questions, such as:
• Explain what a ratio is in your own words.
• What things must you pay attention to when writing a ratio?
• What are some words and phrases that are used to write a ratio?
## 1.4: Cool-down - A Collection of Animals (5 minutes)
### Cool-Down
For access, consult one of our IM Certified Partners.
## Student Lesson Summary
### Student Facing
A ratio is an association between two or more quantities. There are many ways to describe a situation in terms of ratios. For example, look at this collection:
Here are some correct ways to describe the collection:
• The ratio of squares to circles is $$6:3$$.
• The ratio of circles to squares is 3 to 6.
Notice that the shapes can be arranged in equal groups, which allow us to describe the shapes using other numbers.
• There are 2 squares for every 1 circle.
• There is 1 circle for every 2 squares. |
4
Q:
# There are 5 consecutive odd numbers. If the difference between square of the average of first two odd number and the of the average last two odd numbers is 396, what is the smallest odd number?
A) 29 B) 27 C) 31 D) 33
Explanation:
Let the five consecutive odd numbers be x-4, x-2, x, x+2, x+4
According to the question,
Difference between square of the average of first two odd number and the of the average last
two odd numbers is 396
i.e, x+3 and x-3
Hence, the smallest odd number is 33 - 4 = 29.
Q:
What is the common ratio of the following geometric sequence?
4, 2, 1, 0.5, 0.25, 0.125,...
A) 0.5 B) -1 C) 1.5 D) -0.5
Explanation:
In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.
Herein the given sequence, 4, 2, 1, 0.5, 0.25, 0.125,...
Common Ratio r = 2/4 = 1/2 = 0.5/1 = 0.25/0.5 = 0.125/0.25 == 0.5.
0 46
Q:
What is the place value of 6 in 64?
A) 6 B) 60 C) 64 D) 10
Explanation:
We know that,
● Each digit has a fixed position called its place.
● Each digit has a value depending on its place called the place value of the digit.
● The face value of a digit for any place in the given number is the value of the digit itself
● Place value of a digit = (face value of the digit) × (value of the place).
Hence, the place value of 6 in 64 = 6 x 10 = 60.
5 483
Q:
Which one of the following is not a prime number?
A) 91 B) 71 C) 41 D) 31
Explanation:
Prime Numbers :: Numbers which are divisible by only 1 and itself are Prime Numbers.
It's answer will be 91.
Because 91 can be divisible by 7,13,91,1.
It is quite clear that prime number should be divisible only by itself and by 1.
5 465
Q:
What are the Multiples of 6 and the Common Multiples of 4 and 6?
A) 12, 34, 42 B) 12, 18, 36, C) 6, 4, 14 D) 4, 8, 16
Answer & Explanation Answer: B) 12, 18, 36,
Explanation:
The Multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60.
The Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60.
The Common Multiples of 4 & 6 upto 60 numbers are given as 12, 24, 36, 48, 60.
8 464
Q:
12 students of class IX took part in GK quiz. If the number of boys is 4 more than the number of girls. Find the number of boys and girls took part in the quiz ?
A) 7, 5 B) 9, 3 C) 8, 4 D) 6, 6
Answer & Explanation Answer: C) 8, 4
Explanation:
Let the number of girls = x
=> x + x + 4 = 12
=> 2x = 8
=> x = 4
=> Number of girls = 4
=> Number of boys = 4 + 4 = 8
14 570
Q:
A number is 2 more than thrice the second number and 4 times the smaller number is 5 more than the greater number. What is greater number ?
A) 7 B) 23 C) 9 D) 21
Explanation:
Let the greater and smaller number be p and q respectively.
4q = p + 5 ------ (I)
p = 3q+2 ------- (II)
From equation (I) and (II)
q = 7
p = 23
8 730
Q:
What will be added to the unit digits of the number 86236 so that number will be divisible by 9 ?
A) 1 B) 2 C) 3 D) 0
Explanation:
A number is divisible by 9 only if the sum of the digits of the number is divisible by 9.
Here 86236 = 8 + 6 + 2 + 3 + 6 = 25
We must add 2 to 25 to become 27 which is divisible by 9.
4 1070
Q:
How many numbers up to 101 and 300 are divisible by 11 ?
A) 18 B) 20 C) 19 D) 17 |
# Differentiation Rules - Finding the Derivative of a Constant Times a Function
In this post I'm going to explain another one of the differentiation rules for working with derivatives. This time, I will show you how to find the derivative of a constant times a function.
In case you have missed them, I am creating a series of posts that explain some basic concepts in differential calculus. So far, in my first lesson I explained how to find the derivative of a constant function, and then I followed that with a post about the power rule for derivatives. So far, these are some of the basic rules for calculating derivatives, and it is a good idea to become very familiar with them so that you can apply them all as required on more elaborate problems later on.
This derivatives rule is a very simple one, but that doesn't make it any less important. Consider that you have any function f(x), and it is multiplied by a constant. Assuming that f'(x) in fact exists, f'(x) times a constant is equal to the constant times this derivative. That may sound a little wordy, so maybe this equation will be a little clearer:
So, all you really need to be concerned with is calculating the derivative f'(x). Then, multiplying by the constant is a simple calculation that you can add at the end.
If you're curious about what this rule means, here is one way of looking at this. If you have an equation such as y = f(x), if you consider multiplying this function by a constant, what you are doing is essentially stretching the graph vertically. So, comparatively, y = 2f(x) is a graph that is twice the height. Therefore, if you then consider the slope along this function, since you have doubled the rise but not the run (the slope calculation), you have a doubled slope at every point. And since the derivative of a function represents the slope of the line at a point, you can then see how this rule all comes together. Basically, if you stretch the function by a constant factor, you can simply multiply the slope (the derivative) by this factor as well.
As I said, there isn't much to remember about this particular derivative rule, but it is very important to know. It will often need to be considered in addition to other rules that I have/will outline in this series. One example would be to calculate the derivative of something like 4x3. In this case, you'd need to draw upon this rule, as well as the power rule from my last post. There are a few other basic differentiation rules like this one that I will cover in my next posts. Learn all of these rules well, and you'll have no problem differentiating complicated functions!
One final comment - if you thought this post was helpful in any way, then please do me a favour and click on the Like and +1 buttons found on this page! Thanks for your support and for visiting. Be sure to come back if you need help with any other maths concepts. |
Let us try the effect of repeating several times over the operation of differentiating a function (see chapter 3). Begin with a concrete case.
Let $$y = x^5$$. \begin{aligned} &\text{First differentiation, } &&=5x^4. && \\ &\text{Second differentiation, } &&5 \times 4x^3 &&= 20x^3. \\ &\text{Third differentiation, } &&5 \times 4 \times 3x^2 &&= 60x^2. \\ &\text{Fourth differentiation, } &&5 \times 4 \times 3 \times 2x &&= 120x. \\ &\text{Fifth differentiation, } &&5 \times 4 \times 3 \times 2 \times 1 &&= 120. \\ &\text{Sixth differentiation, } && &&= 0.\end{aligned}
There is a certain notation, with which we are already acquainted (see chapter 3), used by some writers, that is very convenient. This is to employ the general symbol $$f(x)$$ for any function of $$x$$. Here the symbol $$f(~)$$ is read as “function of,” without saying what particular function is meant. So the statement $$y=f(x)$$ merely tells us that $$y$$ is a function of $$x$$, it may be $$x^2$$ or $$ax^n$$, or $$\cos x$$ or any other complicated function of $$x$$.
The corresponding symbol for the differential coefficient is $$f'(x)$$, which is simpler to write than $$\dfrac{dy}{dx}$$. This is called the “derived function” of $$x$$.
Suppose we differentiate over again, we shall get the “second derived function” or second differential coefficient, which is denoted by $$f^{\prime\prime}(x)$$; and so on.
Now let us generalize.
Let $y = f(x) = x^n$.
\begin{aligned} &\text{First differentiation, } &&f'(x)=nx^{n-1}. \\ &\text{Second differentiation, } &&f^{\prime\prime}(x)=n(n-1)x^{n-2}. \\ &\text{Third differentiation, }&& f^{\prime\prime\prime}(x)=5 n(n-1)(n-2)x^{n-3}. \\ &\text{Fourth differentiation, } &&f^{\prime\prime\prime\prime}(x)=n(n-1)(n-2)(n-3)x^{n-4}.\\ & &&etc.\end{aligned}
But this is not the only way of indicating successive differentiations. For,
\begin{aligned} &\text{if the original function be} &&y=f(x) \\ &\text{once differentiating gives} && \dfrac{dy}{dx}= f'(x) \\ &\text{twice differentiating gives} && \dfrac{d\dfrac{dy}{dx}}{dx}= f^{\prime\prime}(x)\end{aligned}
and this is more conveniently written as $$\dfrac{d^2y}{(dx)^2}$$, or more usually $$\dfrac{d^2y}{dx^2}$$. Similarly, we may write as the result of thrice differentiating, $$\dfrac{d^3y}{dx^3} = f^{\prime\prime\prime}(x)$$.
### Examples.
Example 1
Now let us try $$y = f(x) = 7x^4 + 3.5x^3 – \frac{1}{2}x^2 + x – 2$$.
Solution
\begin{aligned} \frac{dy}{dx} &= f'(x) = 28x^3 + 10.5x^2 – x + 1, \\ \frac{d^2y}{dx^2} &= f^{\prime\prime}(x) = 84x^2 + 21x – 1, \\ \frac{d^3y}{dx^3} &= f^{\prime\prime\prime}(x) = 168x + 21, \\ \frac{d^4y}{dx^4} &= f^{\prime\prime\prime\prime}(x) = 168, \\ \frac{d^5y}{dx^5} &= f^{\prime\prime\prime\prime\prime}(x) = 0.\end{aligned}
Example 2
In a similar manner if $$y = \phi(x) = 3x(x^2 – 4)$$,
Solution
\begin{aligned} \phi'(x) &= \frac{dy}{dx} = 3\bigl[x \times 2x + (x^2 – 4) \times 1\bigr] = 3(3x^2 – 4), \\ \phi^{\prime\prime}(x) &= \frac{d^2y}{dx^2} = 3 \times 6x = 18x, \\ \phi^{\prime\prime\prime}(x) &= \frac{d^3y}{dx^3} = 18, \\ \phi^{\prime\prime\prime\prime}(x) &= \frac{d^4y}{dx^4} = 0.\end{aligned}
## Exercises IV.
Find $$\dfrac{dy}{dx}$$ and $$\dfrac{d^2y}{dx^2}$$ for the following expressions:
$$(1) y = 17x + 12x^2$$ $$(2) y = \dfrac{x^2 + a}{x + a}$$
$$(3) y = 1 + \dfrac{x}{1} + \dfrac{x^2}{1\times2} + \dfrac{x^3}{1\times2\times3} + \dfrac{x^4}{1\times2\times3\times4}$$.
$$(4)$$ Find the 2nd and 3rd derived functions in the Exercises (chapter 6), No. 1 to No. 7, and in the Examples given (chapter 6), No. 1 to No. 7.
Solution
$$(1) 17 + 24x$$; $$24$$. $$(2) \dfrac{x^2 + 2ax – a}{(x + a)^2}$$; $$\dfrac{2a(a + 1)}{(x + a)^3}$$
$$(3) 1 + x + \dfrac{x^2}{1 \times 2} + \dfrac{x^3}{\times 2 \times 3}$$; $$1 + x + \dfrac{x^2}{1 \times 2}$$.
$$(4)$$ (Exercises, chapter 6):
• $$(1)$$ (a) $$\dfrac{d^2 y}{dx^2} = \dfrac{d^3 y}{dx^3} = 1 + x + \frac{1}{2}x^2 + \frac{1}{6} x^3 + \ldots$$
(b) $$2a$$, $$0$$. (c) $$2$$, $$0$$. (d) $$6x + 6a$$, $$6$$.
• $$(2)$$ $$-b$$, $$0$$. $$(3)$$ $$2$$, $$0$$.
• $$(4)$$ $$\begin{gathered}[t] 56440x^3 – 196212x^2 – 4488x + 8192. \\ 169320x^2 – 392424x – 4488. \end{gathered}$$
• $$(6) 2$$, $$0$$. $$(6)$$ $$371.80453x$$, $$371.80453$$.
• $$(7) \dfrac{30}{(3x + 2)^3}$$, $$-\dfrac{270}{(3x + 2)^4}$$.
(Examples, chapter 6):
• $$(1) \dfrac{6a}{b^2} x$$, $$\dfrac{6a}{b^2}$$. $$(2) \dfrac{3a \sqrt{b}} {2 \sqrt{x}} – \dfrac{6b \sqrt[3]{a}}{x^3}$$, $$\dfrac{18b \sqrt[3]{a}}{x^4} – \dfrac{3a \sqrt{b}}{4 \sqrt{x^3}}$$
$$(3) \dfrac{2}{\sqrt[3]{\theta^8}} – \dfrac{1.056}{\sqrt[5]{\theta^{11}}}$$, $$\dfrac{2.3232}{\sqrt[5]{\theta^{16}}} – \dfrac{16}{3 \sqrt[3]{\theta^{11}}}$$.
$$(4)$$ $$\begin{gathered}[t] 810t^4 – 648t^3 + 479.52t^2 – 139.968t + 26.64. \\ 3240t^3 – 1944t^2 + 959.04t – 139.968. \end{gathered}$$
$$(5) 12x + 2$$, $$12$$.
$$(6) 6x^2 – 9x$$, $$12x – 9$$.
$$(7)$$ \begin{aligned}[t] &\dfrac{3}{4} \left(\dfrac{1}{\sqrt{\theta}} + \dfrac{1}{\sqrt{\theta^5}}\right) +\dfrac{1}{4} \left(\dfrac{15}{\sqrt{\theta^7}} – \dfrac{1}{\sqrt{\theta^3}}\right). \\ &\dfrac{3}{8} \left(\dfrac{1}{\sqrt{\theta^5}} – \dfrac{1}{\sqrt{\theta^3}}\right) -\dfrac{15}{8}\left(\dfrac{7}{\sqrt{\theta^9}} + \dfrac{1}{\sqrt{\theta^7}}\right). \end{aligned} |
# 2019 AIME I Problems/Problem 6
## Problem 6
In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$.
## Solution 1 (Trig)
Let $\angle MKN=\alpha$ and $\angle LNK=\beta$. Note $\angle KLP=\beta$.
Then, $KP=28\sin\beta=8\cos\alpha$. Furthermore, $KN=\frac{65}{\sin\alpha}=\frac{28}{\sin\beta} \Rightarrow 65\sin\beta=28\sin\alpha$.
Dividing the equations gives $$\frac{65}{28}=\frac{28\sin\alpha}{8\cos\alpha}=\frac{7}{2}\tan\alpha\Rightarrow \tan\alpha=\frac{65}{98}$$
Thus, $MK=\frac{MN}{\tan\alpha}=98$, so $MO=MK-KO=\boxed{090}$.
## Solution 2 (Similar triangles)
$[asy] size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); [/asy]$
First, let $P$ be the intersection of $LO$ and $KN$ as shown above. Note that $m\angle KPL = 90^{\circ}$ as given in the problem. Since $\angle KPL \cong \angle KLN$ and $\angle PKL \cong \angle LKN$, $\triangle PKL \sim \triangle LKN$ by AA similarity. Similarly, $\triangle KMN \sim \triangle KPO$. Using these similarities we see that $$\frac{KP}{KL} = \frac{KL}{KN}$$ $$KP = \frac{KL^2}{KN} = \frac{28^2}{KN} = \frac{784}{KN}$$ and $$\frac{KP}{KO} = \frac{KM}{KN}$$ $$KP = \frac{KO \cdot KM}{KN} = \frac{8\cdot KM}{KN}$$ Combining the two equations, we get $$\frac{8\cdot KM}{KN} = \frac{784}{KN}$$ $$8 \cdot KM = 28^2$$ $$KM = 98$$ Since $KM = KO + MO$, we get $MO = 98 -8 = \boxed{090}$.
## Solution 3 (Similar triangles, orthocenters)
Extend $KL$ and $NM$ past $L$ and $M$ respectively to meet at $P$. Let $H$ be the intersection of diagonals $KM$ and $LN$ (this is the orthocenter of $\triangle KNP$).
As $\triangle KOL \sim \triangle KHP$ (as $LO \parallel PH$, using the fact that $H$ is the orthocenter), we may let $OH = 8k$ and $LP = 28k$.
Then using similarity with triangles $\triangle KLH$ and $\triangle KMP$ we have
$$\frac{28}{8+8k} = \frac{8+8k+HM}{28+28k}$$
Cross-multiplying and dividing by $4+4k$ gives $2(8+8k+HM) = 28 \cdot 7 = 196$ so $MO = 8k + HM = \frac{196}{2} - 8 = \boxed{090}$. (Solution by scrabbler94)
## Solution 4 (Algebraic Bashing)
First, let $P$ be the intersection of $LO$ and $KN$. We can use the right triangles in the problem to create equations. Let $a=NP, b=PK, c=NO, d=OM, e=OP, f=PC,$ and $g=NC.$ We are trying to find $d.$ We can find $7$ equations. They are $$4225+d^2=c^2,$$ $$4225+d^2+16d+64=a^2+2ab+b^2,$$ $$a^2+e^2=c^2,$$ $$b^2+e^2=64,$$ $$b^2+e^2+2ef+f^2=784,$$ $$a^2+e^2+2ef+f^2=g^2,$$ and $$g^2+784=a^2+2ab+b^2.$$ We can subtract the fifth equation from the sixth equation to get $a^2-b^2=g^2-784.$ We can subtract the fourth equation from the third equation to get $a^2-b^2=c^2-64.$ Combining these equations gives $c^2-64=g^2-784$ so $g^2=c^2+720.$ Substituting this into the seventh equation gives $c^2+1504=a^2+2ab+b^2.$ Substituting this into the second equation gives $4225+d^2+16d+64=c^2+1504$. Subtracting the first equation from this gives $16d+64=1504.$ Solving this equation, we find that $d=\boxed{090}.$ (Solution by DottedCaculator)
## Solution 5 (5-second PoP)
$[asy] size(8cm); pair K, L, M, NN, X, O; K=(-sqrt(98^2+65^2)/2, 0); NN=(sqrt(98^2+65^2)/2, 0); L=sqrt(98^2+65^2)/2*dir(180-2*aSin(28/sqrt(98^2+65^2))); M=sqrt(98^2+65^2)/2*dir(2*aSin(65/sqrt(98^2+65^2))); X=foot(L, K, NN); O=extension(L, X, K, M); draw(K -- L -- M -- NN -- K -- M); draw(L -- NN); draw(arc((K+NN)/2, NN, K)); draw(L -- X, dashed); draw(arc((O+NN)/2, NN, X), dashed); draw(rightanglemark(K, L, NN, 100)); draw(rightanglemark(K, M, NN, 100)); draw(rightanglemark(L, X, NN, 100)); dot("K", K, SW); dot("L", L, unit(L)); dot("M", M, unit(M)); dot("N", NN, SE); dot("X", X, S); [/asy]$ Notice that $KLMN$ is inscribed in the circle with diameter $\overline{KN}$ and $XOMN$ is inscribed in the circle with diameter $\overline{ON}$. Furthermore, $(XLN)$ is tangent to $\overline{KL}$. Then, $$KO\cdot KM=KX\cdot KN=KL^2\implies KM=\frac{28^2}{8}=98,$$and $MO=KM-KO=\boxed{090}$.
(Solution by TheUltimate123)
## Solution 6 (Alternative PoP)
$[asy] size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); [/asy]$
Call the base of the altitude from $L$ to $NK$ point $P$. Let $PO=x$. Now, we have that $KP=\sqrt{64-x^2}$ by the Pythagorean Theorem. Once again by Pythagorean, $LO=\sqrt{720+x^2}-x$. Using Power of a Point, we have
$$(KO)(OM)=(LO)(OQ)$$ ($Q$ is the intersection of $OL$ with the circle $\neq L$)
$$8(MO)=(\sqrt{720+x^2}+x)(\sqrt{720+x^2}-x)$$
$$8(MO)=720$$
$$MO=\boxed{090}$$.
(Solution by RootThreeOverTwo) |
# What is 38/50 as a percentage?
How to show 38/50 as a percent? Here are the easiest ways to do so!
The usage of percentages is very vital in mathematical expressions and problems. Hence you need to have a clear concept about the working principles and the methods of the percentage that are required for solving a sum.
Here we will learn about how to change any fraction to its percentage form. Now we shall take the simple example of 38 of 50 and explore the methods in which a fraction can be converted to its percentage state.
What is 38/50 as a percent? The answer is 38/50 as a percentage is 76%
## Basic concepts of converting a fraction to percentage
1) A fraction is made of two parts, the numerator, and the denominator. The numerator refers to the numbers or the digits that are present above the line of division and the denominator is the digit or the value that is present below the line of division.
2) Here in the given statement 38 is the numerator and 50 is the denominator.
3) A percentage is described as expressing any number in terms of hundred. It is the method by which we learn to compare or understand their greatness or smallness by comparing it with the whole value of 100. In other words, we can say that a percentage is defined as a fraction of 100.
4) So if we write 50% then we mean 50/100 and if we say 60% then it means 60/100.
5) When we divide a fraction, the numerator becomes the dividend which gets divided in the process and the denominator becomes the divisor which is used to divide the dividend. Here 38 is the dividend and 50 is the divisor when we need to divide the given fraction 38/50.
## Calculation to show 38/50 as a percent
Two different methods are involved in the conversion of a fraction to its percentage form.
### Method #1
Here you are going to learn about the first method where we have to balance the fraction and change the denominator to 100.
Step 1
When you need to change the given denominator 50 to a hundred, you have to divide 100 by 50.
#### 100÷50=2
$\frac{100}{50}=2$
Step 2
Now you may use this answer to multiply with both the numerator and the denominator to maintain a balance in the fraction 38/50.
Hence,
#### (38×2)÷(50×2)=76/100
$\frac{(38\times2)}{(50\times2)}=\frac{76}{100}$
Thus the required results for the conversion of the given fraction 38/50 to percentage is 76%.
#### 38/50 as a percent = 76%
Also read: What is 13/15 as a percentage?
### Method #2
Here we will convert the given fraction to decimals and then change it into a percentage.
Step 1
Here we will first change the given fraction 38/50 to decimals and to do so you need to divide the numerator by the denominator where the numerator 38 becomes the dividend and 50 the denominator becomes the divisor.
Thus,
#### 38÷50= 0.76
Step 2
Now you need to multiply the obtained decimal value by 100 so that you can find the results of the given fraction 38/50 into a percentage.
So,
#### 0.76×100= 76%
Hence we can see that through the two different methods of conversion of a fraction to a percentage, the results are always the same in the case of conversion 38/50 to percentage.
#### 38/50 as a percent = 76%
Both the methods are extremely easy and you do not need much effort to change your given fraction to a percentage. However, you do need to practice both the methods well so that you remember them and can solve the sums on your own when presented with one.
Also read: What is 1/3 as a percentage? |
# Lesson 6
Areas in Histograms
### Lesson Narrative
The mathematical purpose of this lesson is to apply the concepts of mean and standard deviation to data modeled using a normal distribution and to use the area under a normal curve to estimate percentiles. The work of this lesson connects to previous work because students were introduced to the normal distribution as a way to model distributions that are approximately symmetric and bell-shaped. The work of this lesson connects to upcoming work because students will use the normal distribution to estimate the proportion of data values falling within given intervals. When students make connections between histograms, normal distributions, the mean, and the standard deviation to estimate population proportions, students are reasoning abstractly and quantitatively (MP2).
Technology isn’t required for this lesson, but there are opportunities for students to choose to use appropriate technology to solve problems. We recommend making technology available.
### Learning Goals
Teacher Facing
• Describe (orally) the relationship between areas in a histogram and proportions of data.
• Generalize (orally and in writing) that normally distributed data always have the same proportion of values in an interval with endpoints described by the mean and standard deviation.
### Student Facing
• Let’s find proportions of data in certain intervals.
### Required Preparation
A copy of the blackline master should be made available for each group of 2 students for the activity Story Submissions
### Student Facing
• I can calculate a proportion of a set of data that matches a shaded area in a histogram.
• I recognize the patterns of proportions that occur in distributions that are approximately normal in shape.
Building Towards
### Glossary Entries
• normal distribution
A specific distribution in statistics whose graph is symmetric and bell-shaped, has an area of 1 between the $$x$$-axis and the graph, and has the $$x$$-axis as a horizontal asymptote.
• relative frequency histogram
A histogram where the height of each bar is the fraction of the entire data set that falls into the corresponding interval (that is, it is the relative frequency with which the data values fall into that interval).
### Print Formatted Materials
For access, consult one of our IM Certified Partners. |
Question Video: Recognizing That on a Distance–Time Graph a Steeper Gradient Means a Greater Speed | Nagwa Question Video: Recognizing That on a Distance–Time Graph a Steeper Gradient Means a Greater Speed | Nagwa
# Question Video: Recognizing That on a Distance–Time Graph a Steeper Gradient Means a Greater Speed Science • Third Year of Preparatory School
The distance–time graph shows an object that changes from moving at one uniform speed to moving at a different uniform speed. Which color line shows the greater speed?
02:08
### Video Transcript
The distance–time graph shows an object that changes from moving at one uniform speed to moving at a different uniform speed. Which color line shows the greater speed? (A) The blue line, (B) the red line, (C) both lines show the same speed.
In this question, we are given a distance–time graph that we’re told shows the movement of an object that changes from moving at one uniform speed to moving at a different uniform speed. We are asked to look at this graph and figure out which of the colored lines represents the movement that has the greater speed. In order to figure this out, we will need to recall how to read a distance–time graph and how to use such a graph to find an object’s speed.
A distance–time graph, like its name suggests, is a graph measuring the distance that an object has traveled and the amount of time that it has traveled for. We can see that we have distance plotted on the vertical axis and time on the horizontal axis. Recall that the speed of an object is defined as the distance traveled by that object divided by the time it traveled for. This means we could find the speed of each of this object’s movements if we had the values for this graph. But even without numerical values, we can still compare the two line sections representing the two sections of movement. In particular, we are interested in the slope of the lines.
The slope of a line is defined as the change in the vertical coordinate divided by the change in the horizontal coordinate. Because the speed of an object is equal to the distance traveled divided by the time taken and because the vertical coordinate on a distance–time graph is distance and the horizontal coordinate is time, then the slope of a line on a distance–time graph gives us the speed of the object. A larger change in distance over a given change of time will correspond to a steeper slope and a greater speed.
So let’s look at the blue and red lines on this graph and compare them. Looking at the slopes of the lines, we can see that the red line is steeper than the blue line. This means that the red line represents the movement with the greater speed. Therefore, option (B), the red line, is the correct answer.
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# Mastering Method Math: A Step-by-Step Guide
Title: Mastering Method Math: A Step-by-Step Guide
Subtitle: Unlocking the Secrets of Math with Method Math
Introduction
Method Math is a powerful tool for learning and understanding math. It is a problem-solving approach that is based on the idea that mathematics is a language that can be understood with the proper techniques. With Method Math, students learn to break down problems into smaller, manageable pieces and then use logic and reasoning to solve them. This approach makes math easier to understand and can help students gain a deeper understanding of the subject. In this guide, we will explore the basics of Method Math and provide a step-by-step guide for mastering it.
Body
The first step in mastering Method Math is to become familiar with the basic concepts. This includes understanding the terms, symbols, and operations used in math. It is also important to understand the different types of equations and how they are used. Once you have a basic understanding of the concepts, you can begin to practice solving problems.
The next step is to learn how to break down problems into smaller pieces. This involves looking at the problem and identifying the key information that needs to be solved. Once this is done, you can begin to use the techniques of Method Math to solve the problem. This includes using equations, diagrams, and other visual aids to help you understand the problem.
Examples
Let’s look at an example of how Method Math can be used to solve a problem. Suppose you have the equation 2x + 3 = 7. To solve this problem using Method Math, you would first identify the key information. In this case, the key information is that the equation is an equation with two variables, x and 7. You would then use the techniques of Method Math to solve for x. This would involve using equations and diagrams to help you understand the problem and identify the solution.
For example, you could draw a diagram of the equation and use it to solve for x. This would involve drawing a line from the left side of the equation to the right side and then finding the point where the line intersects with the x-axis. This point is the solution to the equation, which is x = 2.
FAQ Section
Q: What is Method Math?
A: Method Math is a problem-solving approach that is based on the idea that mathematics is a language that can be understood with the proper techniques. It involves breaking down problems into smaller pieces and then using logic and reasoning to solve them.
Q: How can Method Math help me understand math?
A: Method Math can help you gain a deeper understanding of math by breaking down problems into smaller, manageable pieces. This makes math easier to understand and can help you gain a better grasp of the subject.
Q: What is the first step in mastering Method Math?
A: The first step in mastering Method Math is to become familiar with the basic concepts of math, including the terms, symbols, and operations used. Once you have a basic understanding of the concepts, you can begin to practice solving problems.
Summary
In summary, Method Math is a powerful tool for learning and understanding math. It is based on the idea that mathematics is a language that can be understood with the proper techniques. With Method Math, students learn to break down problems into smaller, manageable pieces and then use logic and reasoning to solve them. This guide has provided a step-by-step guide for mastering Method Math, including becoming familiar with the basic concepts, learning how to break down problems into smaller pieces, and using equations and diagrams to help understand the problem.
Conclusion
Method Math is a powerful tool for learning and understanding math. With the proper techniques and practice, students can gain a deeper understanding of the subject and become more confident in their math abilities. This guide has provided a step-by-step guide for mastering Method Math, so that students can unlock the secrets of math and become successful in their math studies.
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# Single Variable Multiplication Equations
## Solve one - step equations using multiplication.
0%
Progress
Practice Single Variable Multiplication Equations
Progress
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Single Variable Multiplication Equations
Mr. Ricky’s Biology class is going to the botanical gardens to study the different kinds of flowers and plants that are there. They have raised 68 dollars for tickets so far. The student rate for each ticket is 5 dollars. Mr. Ricky asked the students to figure out how many tickets they can buy. How many tickets can they afford with 68 dollars?
In this concept, you will learn to solve single variable multiplication equations.
### Guidance
In the algebraic equation below, the variable z\begin{align*}z\end{align*} represents a number.
z×2=8
What number does z\begin{align*}z\end{align*} represent?
You can find out by asking yourself, “What number, when multiplied by 2, equals 8?”
Since 4×2=8\begin{align*}4 \times 2 = 8\end{align*}z\begin{align*}z\end{align*} must be equal to 4.
To solve a more complex equation, such as z×7=105\begin{align*}z\times 7 = 105\end{align*}, you should use another strategy for solving the equation.
To solve an equation in which a variable is multiplied by a number, you can use the inverse operation of multiplication-division. To isolate the variable you divide both sides of the equation by that number to find the value of the variable.
You can divide both sides of the equation by the same number and not change the equality because of the Division Property of Equality, which states:
If a=b\begin{align*}a=b\end{align*} and c0\begin{align*}c \neq 0\end{align*}, then ac=bc\begin{align*}\frac{a}{c} = \frac{b}{c}\end{align*}.
This means that if you divide one side of an equation by a nonzero number, c\begin{align*}c\end{align*}, you must divide the other side of the equation by that same number, c\begin{align*}c\end{align*}, to keep the values on both sides equal.
Let’s look at an example.
Solve for z\begin{align*}z\end{align*}.
z×7=105
In this equation, z\begin{align*}z\end{align*} is multiplied by 7. So, to isolate the variable z\begin{align*}z\end{align*}, you can divide both sides of the equation by 7.
First, using the division property of equality, divide both sides of the equation by 7.
z×7z×77==1051057
Next, separate the fraction z×77\begin{align*}\frac{z \times 7}{7}\end{align*}, and simplify.
z×77z×1z===151515
The answer is z=15\begin{align*}z = 15\end{align*}.
Here is another example.
Solve for r\begin{align*}r\end{align*}.
8r=128
In this equation, -8 is multiplied by r\begin{align*}r\end{align*}. So, using the division property of equality, you can divide both sides of the equation by -8 to solve for r\begin{align*}r\end{align*}.
First, divide both sides of the equation by -8.
8r8r8==1281288
Next, separate the fraction 8r8\begin{align*}\frac{-8r}{-8}\end{align*}, and simplify.
8r81rr===12881616
The answer is r=16\begin{align*}r = -16\end{align*}.
### Guided Practice
Sarvenaz earns $8 for each hour she works. She earned a total of$168 last week.
1. Write an equation to represent h\begin{align*}h\end{align*}, the number of hours she worked last week.
2. Determine how many hours Sarvenaz worked last week.
First, complete part a.
Let h\begin{align*}h\end{align*} be the number of hours Sarvenaz worked. She earns $8 for each hour she works, so you multiply the number of hours she worked by$8 to find the total amount she earned. Write a multiplication equation.
8h=168
Next, work on part b.
Solve the equation 8h=168\begin{align*}8h = 168\end{align*} to find h\begin{align*}h\end{align*}, the number of hours she worked last week.
First, use the division property of equality to divide both sides of the equations by 8.
8h8h8==1681688
Next, separate the fraction 8h8\begin{align*}\frac{8h}{8}\end{align*} and simplify.
88h1hh===212121
The answer is Sarvenaz works 21 hours last week.
### Examples
Solve each equation.
#### Example 1
4x=12
First, use the division property of equality, and divide both sides of the equation by -4.
4x4x4==12124
Next, separate the fraction 4x4\begin{align*}\frac{-4x}{-4}\end{align*} and simplify.
44x1xx===333
The answer is x=3\begin{align*}x = -3\end{align*}.
#### Example 2
8a=64
First, use the division property of equality and divide both sides of the equation by 8.
8a8a8==64648
Next, separate the fraction 8a8\begin{align*}\frac{8a}{8}\end{align*} and simplify.
88a1aa===888
The answer is a=8\begin{align*}a = 8\end{align*}.
#### Example 3
9b=81
First, use the division property of equality and divide both sides of the equation by 9.
9b9b9==81819
Next, separate the fraction 9b9\begin{align*}\frac{9b}{9}\end{align*} and simplify.
99b1bb===999
The answer is b=9\begin{align*}b = 9\end{align*}.
Remember Mr. Ricky’s Biology class?
The class has raised 68 dollars for their trip and is wondering how many 5 dollar tickets they can buy.
First, write an equation to represent this information. Let \begin{align*}d\end{align*}, be the number of tickets they can buy. You can say that, \begin{align*}d\end{align*} times the price of each ticket, 5 dollars, is 68 dollars.
Next, use the division property of equality and divide both sides of the equation by 5.
Then, separate the fraction \begin{align*}\frac{5d}{5}\end{align*} and simplify.
Next, interpret the result.
The students can buy 13.6 tickets. You can’t buy .6 of a ticket. So, the students can only buy 13 tickets with 3 dollars left over.
### Explore More
Solve each single variable multiplication equation for the missing value.
1. \begin{align*}4x = 16\end{align*}
2. \begin{align*}6x = 72\end{align*}
3. \begin{align*}-6x = 72\end{align*}
4. \begin{align*}-3y = 24\end{align*}
5. \begin{align*}-3y =-24\end{align*}
6. \begin{align*}-5x = -45\end{align*}
7. \begin{align*}-1.4x = 2.8\end{align*}
8. \begin{align*}3.5a = 7\end{align*}
9. \begin{align*}7a =-49\end{align*}
10. \begin{align*}14b = -42\end{align*}
11. \begin{align*}24b = -48\end{align*}
12. \begin{align*}-24b = -48\end{align*}
13. \begin{align*}34b =-102\end{align*}
14. \begin{align*}84x = 252\end{align*}
15. \begin{align*}-84x = -252\end{align*}
### Vocabulary Language: English
Inverse Operation
Inverse Operation
Inverse operations are operations that "undo" each other. Multiplication is the inverse operation of division. Addition is the inverse operation of subtraction.
Product
Product
The product is the result after two amounts have been multiplied.
Quotient
Quotient
The quotient is the result after two amounts have been divided. |
# Magic Sum Puzzles
Place the numbers 1 to 5 in the circles above, one number per circle, so that the sum of numbers on each of the two diagonal lines is 9.
What number goes in the middle?
# Magic Sum Puzzles
We'll spend the next few questions considering the puzzle below:
Place all the numbers from 1 to 6 (one in each circle) so that the sum of numbers on each of the three marked lines comes out to be 12.
It can help to break the puzzle up into regions and compare them; one method is shown below. When we calculate all the sums described in the puzzle, what is the most important difference between the circles marked in red and those marked in green?
# Magic Sum Puzzles
If $$R$$ is the sum of numbers in the inner red section and $$G$$ is the sum of numbers in the outer green section, which equation is true? (Remember we're using the numbers 1 through 6.)
# Magic Sum Puzzles
ORIGINAL QUESTION: Place all the numbers from 1 to 6 (one in each circle) so that the sum of numbers on each of the three marked lines comes out to be 12.
Before trying to solve the puzzle, let's think of the sums together. There are three sums—one on each line and each equals 12—so the total of the sums is $$12 \times 3 = 36 .$$
Which equation must be true?
# Magic Sum Puzzles
Based on the previous problems, we now know that $$R + G = 21$$ and $$2R + G = 36 .$$ We can use algebra to combine these statements and figure out what $$R$$ and $$G$$ are.
What is $$R?$$ (Hint: Subtract the first equation from the second.)
# Magic Sum Puzzles
Place all the numbers from 1 to 6 (one in each circle) so that the sum of numbers on each of the three marked lines comes out to be 12.
We have now learned that $$R$$ sums to 15. What number must go in the top left circle if 4 is in the top right?
# Magic Sum Puzzles
Fill the grid above with the numbers 1 through 6 (one in each square). There will be five sums: two made by adding up the numbers in each row, and three by adding up the numbers in each column.
There is a way to fill the grid so four of the sums are the same and one will be different. What is the sum that is different?
$$($$Hint: $$1 + 2 + 3 + 4 + 5 + 6 = 21 .)$$
# Magic Sum Puzzles
Let's change up the goal slightly!
Suppose you want to fill the puzzle above with the numbers 1 through 6 (one each circle) so that every set of three circles next to each other add up to the same sum.
Is this possible to do?
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# Proofs in Mathematics PowerPoint PPT Presentation
Proofs in Mathematics. In this section, we discuss the proofs of a number of interesting results in mathematics. The main goal is to see how mathematical proofs are constructed. But first, we need some terminology.
Proofs in Mathematics
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## Proofs in Mathematics
In this section, we discuss the proofs of a number of interesting results in mathematics.
The main goal is to see how mathematical proofs are constructed. But first, we need some terminology.
The natural numbers are the positive integers or whole numbers, 1, 2, 3, 4, 5, ….
The integers are the set consisting of the natural numbers, zero (0), and the negatives of the natural numbers:
… , -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, ….
The rational numbers are all the numbers which are quotients of integers (except that we do not allow any quotients in which the denominator is zero).
The word “rational" comes from “ratio”.
We begin with a famous problem. It is said that when the ancient Greeks discovered this fact, they sacrificed two hundred oxen to celebrate:
The square-root of 2 is “irrational”, i.e., not a rational number.
Proof: Suppose √2 is a rational number, that is, it is possible
to write √2 = a/b, for some natural numbers a and b. As usual,
we can assume that this fraction a/b is in lowest terms, which amounts to saying that a and b have no common divisors other than 1.
Squaring both sides of the equation gives
2 =
or 2 b2 = a2
= 4k2
b2 = 2k2
Now the left side of this equation is an even number, so the right side must also be even. But this forces a to be even, because a2 is odd if a is odd.
But if a is even, i.e., is divisible by 2, then a2 must be divisible by 4. After all, if a = 2k, then a2 = (2k)2 = 2k(2k) = 4k2 , which is clearly divisible by 4.
Now if 2 b2 is divisible by 4, then b2 must be divisible by 2 and, as before, we conclude that b is even. But this means a and b are both even,which contradicts our assumption that they have no common divisors.
The fact that we arrived at a contradiction means it must have been wrong to make the assumptions we did. In this case, this means it was wrong to assume that the square-root of 2 is rational.
We conclude that it is irrational, completing the proof.
QED
This is an example of what mathematicians call “proof by contradiction”. You assume that the result you want is false and from that derive a contradiction. Then you can conclude that the result must be true.
### Hilbert's Hotel. David Hilbert (1862 — 1943) was a celebrated German mathematician.
A number of important concepts in mathematics are named after him,
but there is also a more frivolous example, known as Hilbert's Hotel.
This establishment has an infinite number of rooms, numbered 1, 2, 3, 4, 5, . . . .
### One day, there is a convention, and every room is full, but then a large group of additional guests arrives.
“No problem", says the clerk. He simply asks each guest to move to the room with double his or her current number.
All the guests move to the even-numbered rooms, freeing up all the odd-numbered rooms for the newcomers.
This story can be summarized by saying that there are just as many even numbers as there are numbers altogether. This seems peculiar to us, because usually if we take part of something, we expect it to be smaller.
The point, of course, is that that logic does not apply to infinite sets.
### A set is called “countable" if it has the same number of elements as the natural numbers 1, 2, 3, 4, 5, . . . .
We have just seen that the even numbers constitute a countable set.
The set of integers . . . , -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, . . . is another (the negative numbers
. . . , 5, -4, -3, -2, -1 clearly correspond to the natural numbers, but then they also correspond to the even numbers).
We need to convince ourselves that the remaining numbers 0, 1, 2, 3, 4, 5, . . . . correspond to the odd numbers.
### But this is not hard: if we double each number and add 1, we get 1, 3, 5, 7, 9, 11, . . . ).
A little more surprising is the set of rational numbers a/b . We shall now see that the
rationals are countable.
Actually, since we have seen that combining two countable sets gives a countable set (like the even and odd numbers combining to make all the natural numbers), it is enough to see that the set of positive rational numbers is countable.
### We begin by making a table.
numerator
Denominator
Eliminate any duplications.
numerator
Denominator
Then enumerate the positive rational numbers in a zigzag pattern.
We find that there are countably many positive rational numbers.
### A number n > 1 is “prime" if the only natural numbers which divide it are 1 and n.
The first few prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
But 4 and 6 are not prime, since each is divisible by 2. Likewise, 15 is divisible by 3 and 5.
It is natural to wonder if the prime numbers come to an end, or if there are infinitely many.
We need to recall the principle of “prime factorization”, which says that any number N can be written as a product of prime numbers: N = p1p2p3 … pm;where p1, p2, p3, . . . , pm are prime numbers, possibly with repetitions.
Moreover, this factorization is unique, apart from the order in which the factors are written.
So, for example,
15 = 3 . 5, 8 = 2 . 2 . 2, 36 = 2 . 2 . 3 . 3.
Now suppose that there were only finitely many prime numbers, and label themp1, p2, p3, . . . , pm . Consider the numberM = (p1. p2. p3. . . . . pm ) + 1.
Clearly M is not divisible by any of the primes p1, p2, p3, . . . , pm , because if you divide M by any of them, there is always a remainder of 1.
But since we are assuming these are the only primes, this contradicts the principle of prime factorization.
This contradiction proves that there must be infinitely many primes.QED
Johann Carl Friedrich Gauss (1777 –1855) |
# How to solve a differential equation of this form?
Please consider a differential equation of the form: $$(ax + by + c) dx + (ex + fy + g) dy = 0$$ For the special case of $$c = g = 0$$, then this equation is homogenous and I know how to solve it. Normally, I would solve this equation by setting up the following system of equations: \begin{align*} ax + by + c &= 0 \\ ex + fy + g &= 0 \end{align*} Assuming that this set of equations of a unique solution, I know how to solve differential equation of this form. However, I how do I solve a differential equation of this form when the above system of differential equations does not have a unique solution?
I am thinking the correct substitution is: $$z = ax + by$$ which gives me: \begin{align*} dz &= a\, dx \\ dz &= b\, dy \end{align*} Do I have this right?
This result is adapted from "The Fluxional Calculus. An Elementary Treatise" by Thomas Jephson, 1830.
Given
$$\tag 1 (ax + by + c) dx + (ex + fy + g) dy = 0$$
To approach solving this, we need to be complete by doing case work.
Case 1:
Assume $$ax+by+c = v$$ and $$ex +fy+g = w$$.
Taking derivatives
$$a dx + b dy = dv \\ e dx + f dy = dw$$
Using elimination
$$dx = \dfrac{f dv - b dw}{af - be} \\ dy = \dfrac{a dw - e dv}{af - be}$$
Substituting these into $$(1)$$ and multiplying by $$(af - be)$$, assuming $$(af - be) \ne 0$$
$$v(f dv - b dw) + w(a dw - e dv) = 0$$
This can be written as
$$(f v -e w)dv + (a w - b v) dw = 0$$
This is a homogeneous case.
Case 2:
Assume $$x = v + m$$ and $$y = w + n$$, then $$(1)$$ becomes
$$(a v + b w + a m + b n + c)dv + (e v + f w + e m + f n + g)dw = 0$$
Suppose
$$(a m + b n + c) = 0 \\ (e m + fn + g) = 0$$
From this, find the required values for $$m$$ and $$n$$.
Now this becomes another homogenous case.
Case 3:
If $$b = e$$ in $$(1)$$, we have an Exact Equation and can just integrate to get the result.
Case 4:
If $$af = be$$, then we are dividing by zero and $$m = n = \infty$$ and the exactness fails.
However, this case may be reduced to
$$(a x + b y) dx + (a x + b y)dy + c dx + g dy = 0$$
If we substitute $$v = ax + by$$ and the resulting $$dy$$, we will have a Separable Equation.
You might also like to review this handy flowchart. |
# How do you solve abs(8 - 3x) = 11?
Jul 3, 2015
The answer is $x = - 1$ or $x = \frac{19}{3}$.
#### Explanation:
First of all we must remember the definition of the absolute value, which is done by cases:
If $x > 0 \implies \left\mid x \right\mid = x$
If $x < 0 \implies \left\mid x \right\mid = - x$
Applying this to our question, we obtain the following:
If $\left(8 - 3 x\right) > 0 \implies \left\mid 8 - 3 x \right\mid = 8 - 3 x$
Then, $\left\mid 8 - 3 x \right\mid = 11 \implies 8 - 3 x = 11 \implies - 3 x = 3 \implies x = - 1$
If $\left(8 - 3 x\right) < 0 \implies \left\mid 8 - 3 x \right\mid = - \left(8 - 3 x\right) = 3 x - 8$
Then, $\left\mid 8 - 3 x \right\mid = 11 \implies 3 x - 8 = 11 \implies 3 x = 19 \implies x = \frac{19}{3}$ |
# Q-Intro: Inversion about the Mean
In this introduction article, we discuss and try to understand about projection, outer product and how to achieve the Inversion about the mean using Unitary Gates.
Let’s explore inversion about mean with the following example:
Consider a sequence of integers: 53, 38, 17, 23, 79.
The average of this numbers is a = 42
The average is the number such that the sum of the lengths of the lines above the average is the same as the sum of the lengths of the lines below.
Suppose we wanted to change the sequence so that each element of the original sequence above the average would be the same distance from the average but below. Furthermore, each element of the original sequence below the average would be the same distance from the average but above.
In other words, we are trying to inverting each element around the average.
For example, the first number, 53 is
a - 53 = 42 - 53 =-11 units away from the average.
We must add a = 42 to -11 and get a + (a - 53) = 31.
The second element of the original sequence, 38, is
a - 38 = 42 - 38 = 4 units below the average.
We must add a = 42 to 4 and get a + (a - 38) = 46
The third element of the original sequence, 17, is
a - 17 = 42 - 17 = 25 units below the average.
We must add a = 42 to 25 and get a + (a - 17) = 67
The fourth element of the original sequence, 23, is
a - 23 = 42 - 23 = 19 units below the average.
We must add a = 42 to 19 and get a + (a - 23) = 61
The fifth element of the original sequence, 79, is
a - 79 = 42 - 79 = -37 units below the average.
We must add a = 42 to -37 and get a + (a - 23) = 5
In general, we shall change each element to v
The above sequence(as shown in fig 1) becomes 31, 46, 67, 61, and 5 as shown in fig 2. Notice that the average of this sequence remains 42.
what we have seen is invert around the mean, the below points become above points and above points become below points.
Let us write this in terms of matrices. Rather than writing the numbers as a sequence,
consider a vector V as shown below
and consider the matrix A(all-1’s) divided by no. of elements as shown below
where 5 is the number of elements and let’s see what happens when perform the operation A * V as shown below
after performing the multiplication with 5×5 by 5×1 gives us 5×1 as shown below
53 + 38 + 17 + 23 + 79 = 210 and 210/5 = 42 the resultant as shown in the below fig.
It is easy to see that A is a matrix that finds the average of a sequence/Vector V.
In terms of matrices, the formula v’ = -v + 2a becomes V’ = -V + 2AV = (- I + 2A)V.
Let’s calculate -I + 2A as shown below
after performing the matrix addition, the result is as shown below
And, as expected, (-I + 2A)V = V’ ,
in our case,
after performing multiplication the resultant is as shown below
after simplifying
which gives us
Let us generalize, rather than dealing with five numbers, let us deal with 2^n numbers. Given n qubits, there are 2^n possible states. A state is a 2^n vector. Consider the following 2^n-by-2^n matrix:
and -I+2A can be as shown below
we can identify that A² = A
Multiplying a state by −I + 2A will invert amplitudes about the mean. which we have observed above with an example
We must show that −I + 2A is a unitary matrix.
First, observe that the ad joint of −I + 2A is itself. Then, using the properties of matrix multiplication and realizing that matrices act very much like polynomials,
we have (−I + 2A) * (−I + 2A) = +I − 2A − 2A + 4 A² = I − 4A + 4A²
from A² = A we can write as
I − 4A + 4A = I
Note: In linear algebra and functional analysis, a projection is a linear transformation P from a vector space to itself such that P² = P. That is, whenever P is applied twice to any vector, it gives the same result as if it were applied once (i.e. P is idempotent).
A projection on a vector space V is a linear operator P: V → V such that P² = P
Projection matrix:
In the finite-dimensional case, a square matrix P is called a projection matrix if it is equal to its square, i.e. P² = P
A square matrix P is called an orthogonal projection matrix if P²= P = P*
Outer Product:
In linear algebra, the outer product of two vectors is a matrix. If the two vectors have dimensions n and m, then their outer product is an n × m matrix.
Given two vectors of size m × 1 and n × 1 as shown below
their outer product, denoted u ⊗ v , is defined as the m × n matrix A obtained by multiplying each element of u by each element of v as shown below
After performing multiplication, we get matrix A as shown below
For complex vectors, it is often useful to take the conjugate transpose of v, denoted (v†) as shown below
In Bra-ket notation, as an example we can write A as shown below
similarly, The conjugate transpose (also called Hermitian conjugate) of a ket is the corresponding bra as shown below
from the fig outer2 as mentioned above we can write the outer product in terms of bra-ket notation as shown below
From the below fig braket 4, we can observe that the outer product is similar to the matrix A as mentioned in fig outer3 (above) we can replace matrix A with |u⟩⟨v|
Coming back to our original discussion Inversion about the mean
in the expression -I+2A, here A is an all-1's matrix with a division number of elements.
the matrix to be all-1’s it shall have “u = v” or in other wards |u⟩⟨u|
let see the equivalent notation in quantum case:
The operator Us = 2|s⟩⟨s| -I is a reflection through |s⟩
How to write this operator in our known quantum gates?
Let’s explore little deep about -I + 2A, specially A
we know A is all-1's matrix divide by number of elements or qubits
we write this as follows : A= J/N where J is a all-1's matrix(commonly used in physics) and N = 2^n.
then let’s define |ψ⟩ as shown below
and the complex conjugate as shown below
the outer product of these bra ket as shown below
the result is all-1’s matrix as we expected so we can achieve the matrix J using the states |0⟩ and|1⟩
then finally A= J/N, here N = 2 for the example we consider, following is the way we achieve J/2
ket plus state as shown below
bra plus state as shown below
the outer product of plus state as shown below, we gives us J/2 as expected
in case of N qubits J/N can be achieved as shown below
Let’s further simplify, with so far we achieved as shown below
we validate, the above expression can be written as shown below in the following
by solving what HIH means as shown below
After multiplying the first two matrices from right to left as shown below
after performing the multiplication the result is an Identity matrix as shown below
from this result we can observe that HIH = I, replacing what we observed result as shown below
In case of n qubits as shown below
which rises to the following circuit as shown below, -I + 2A inversion about the mean
Let’s further understanding which gate is equivalent to 2|0⟩⟨0|-I
simplifying gives the expression is equivalent to Pauli Z matrix
what is the result of H(2|0⟩⟨0|-I)H?
2|0⟩⟨0|-I = pauli -z = Z gate
We see, what exactly HZH do!
from right to left solving as shown below
which gives as follows
further simplifying
and the result of the HZH gives us X as shown below
HZH = 𝜎𝑥 which means HZH performs the action of bit flip as we expected.
by performing the same in circuit composer we see as below. apply superposition to the circuit
when add Z gate which gives as HZ as shown below
and finally when we add H gate again to collapse the superposition and draw the measurement as shown below
in the case of multi qubit as shown below
it performs bit flip, |0⟩ to |1⟩ and |1⟩ to |0⟩, it is performing the projection as we discussed.
--
--
## More from Anonymousket
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# Mixed fractions calculator
The calculator performs basic and advanced operations with mixed numbers, fractions, integers, decimals. Mixed fractions are also called mixed numbers. A mixed fraction is a whole number and a proper fraction combined, i.e. one and three-quarters. The calculator evaluates the expression or solves the equation with step-by-step calculation progress information. Solve problems with two or more mixed numbers fractions in one expression.
The calculator performs basic and advanced operations with mixed numbers, fractions, integers, decimals. Mixed fractions are also called mixed numbers. A mixed fraction is a whole number and a proper fraction combined, i.e. one and three-quarters. The calculator evaluates the expression or solves the equation with step-by-step calculation progress information. Solve problems with two or more mixed numbers fractions in one expression.
## Result:
### 1/4 + 1/6 = 5/12 ≅ 0.4166667
Spelled result in words is five twelfths.
### Calculation steps
1. Add: 1/4 + 1/6 = 1 · 3/4 · 3 + 1 · 2/6 · 2 = 3/12 + 2/12 = 3 + 2/12 = 5/12
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(4, 6) = 12. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 4 × 6 = 24. In the next intermediate step, the fraction result cannot be further simplified by canceling.
In words - one quarter plus one sixth = five twelfths.
#### Examples:
sum of two mixed numbers: 1 3/4 + 2 3/8
addition of three mixed numbers: 1 3/8 + 6 11/13 + 5 7/8
addition of two mixed numbers: 2 1/2 + 4 2/3
subtracting two mixed numbers: 7 1/2 - 5 3/4
multiplication of mixed numbers: 3 3/4 * 2 2/5
comparing mixed numbers: 3 1/4 2 1/3
changing improper fraction to mixed number: 9/4
What is 3/4 as a mixed number: 3/4
subtracting mixed number and fraction: 1 3/5 - 5/6
sum mixed number and an improper fraction: 1 3/5 + 11/5
## Mixed fractions in word problems:
Add this two mixed numbers: 1 5/6 + 2 2/11=
Add two mixed fractions: 2 4/6 + 1 3/6
2 and 1 8th plus 1 and 1 3rd =
Why does 1 3/4 + 2 9/10 equal 4.65? How do you solve this?
3 3/4 + 2 3/5 + 5 1/2 Show your solution.
What is 4 1/2+2/7-213/14?
• Clock mathematics
If it is now 7:38 pm, what time will it be in 30,033,996,480 minutes from now?
• Two pizzas
Jacobs mom bought two whole pizzas. He ate 2/10 of the pizza and his dad ate 1 1/5. How much is left.
• Marbles
Dave had 40 marbles. Junjun has 2 1/5 more than Dave’s marbles. How many marbles do they have altogether?
• The book 4
Mr. Kinion read 3 3/4 chapters in his book on Monday. He then read 2 4/6 more chapters on Tuesday. How many chapters has he read so far?
• A textile
A textile store sold a bolt of denim (what jeans are made out of). In one day, the following number of yards were purchased from the one bolt: 5 2/3, 7, 4 2/3, 8 5/8, 9 3/5, 10 ½, and 8. How many yards were sold?
• Painting classrooms
Martha solicited 3 2/3 liters of paint for the brigade. Their city mayor gave their school another 2 2/5 liters of paint. If each classroom needs 1 3/7 liters, how many classrooms can be painted?
• Ingrid
Ingrid watched television for 2 1/4 hours and Devon watched for 3 2/3 hours. For how many hours did they watch television in total?
• A 14.5-gallon
A 14.5-gallon gasoline tank is 3/4 full. How many gallons will it take to fill the tank? Write your answer as a mixed number.
• Arelli
Arelli had 20 minutes to do a three-problem quiz. She spent 9 7/10 minutes on question A and 3 2/5 minutes on question B. How much time did she have left for question C? Solve on paper to find the answer as a fraction. |
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# Differentiation
## Derivatives
Derivative is the rate of change in the slope of the curve of the function.
Derivative can be finding using the formula,
At some places derivatives are also represented by fꞌ(x).
## Differentiation
The process of finding a derivative is called Differentiation. Normally a dependent variable is expressed in terms of independent variables by means of an equation. Now when we find the differential coefficient of the dependent variable with respect to the independent variable, what we do is we try to find out another equation by which the change in the dependent variable (for any infinitesimal change in independent variable) is relatable to the independent variable, whatever be the value of the independent variable.
Hence, derivative is a measure of how a function changes as its input changes. Informally, the derivative is the ratio of the infinitesimal change of the output over the infinitesimal change of the input producing that change of output. Geometrically, for a real valued function of a single real variable, the derivative at a point equals the slope of the tangent line to the graph of the function at that point. The process of finding a derivative is called differentiation.
In fact, differentiation and integration are the two fundamental operations in single-variable calculus. The figure given below illustrates the exact difference between integration and differentiation:
## Notation
There are a number of ways of writing the derivative of a function. Though all these are same but it is essential to know them so as to avoid any kind of confusion:
Suppose we are finding the derivative of x2, then its derivative may be written as:
• If y = x2, dy/dx = 2x
• d/dx (x2) = 2x
This says that the derivative of x2 with respect to x is 2x.
• If f(x) = x2, f'(x) = 2x
This says that is f(x) = x2, the derivative of f(x) is 2x.
## Geometrical Interpretation of Differentiation
Let y = f(x) be a function, and let P(a, f(a)) and Q(a+h, f(a+h)) be two points on the graph of the function that are close to each other. This graph is shown below:
Joining the points P and Q with a straight line gives us the secant on the graph of the function. And in the ?PQR, the gradient of the line is given by
m = {f(a+h) – f(a)}/(a+h-a) = {f(a+h) – f(a)} / h
In limiting process, i.e. as Q approaches P, h becomes really small, almost close to zero.
Hence, m = (change in y)/(change in x)
= {f(a+h) – f(a)} / h
lim h→0 (?y/?x) = {f(a+h) – f(a)} / h
As Q → P, the chord/secant PQ tends to be a tangent at P for the curve y = f(x). Thus the limiting becomes the slope of the tangent at P for y = f(x).
We denote lim h→0 (?y/?x) = dy/dx, and
lim h→0 [f(a+h) – f(a)]/ h = f’(a)
Therefore dy/dx = f’ (a) = m
Now, we have already seen that ‘m’ is the slope or gradient of the tangent at P for f(x).
f’(a) = tan θ = slope of f(x) at x = a.
## Rules of differentiation
The differentiation of functions is carried out in accordance with some rules. These differentiation rules have been listed with the help of the following chart:
All these rules will be discussed in detail in the coming sections. Here, we shall give a brief outline of these rules:
(1) Constant rule:
If the function f is a constant, then its derivative is zero. Moreover if a function is multiplied by a constant then its derivative is given by
{cf(x)}' = cf'(X)
(2) Sum Rule:
{f(x) ± g(x)}' = f'(x) ± g'(x)
(αf + βg)' = αf' + βg' , for all functions f and g and real numbers α and β.
(3) Product Rule:
(fg)' = f'g + g'f, for all functions f and g.
(4) Quotient Rule:
(f/g)' = (f'g – fg')/g2, for all functions f and g such that g ≠ 0.
(5) Chain Rule:
If f(x) = h(g(x)), then f'(x) = h'(g(x)). g'(x)
So while using the chain rule remember the following points: (i) Express the original function as a simpler function of u, where u is a function of x. (ii) Differentiate the two functions you now have. Multiply the derivatives together, leaving your answer in terms of the original function (i.e. in x's rather than u's).
## Logarithmic Differentiation
To differentiate some special functions using logarithm is called Logarithm Differentiation. When it is difficult to differentiate the function then we use the differentiation using logarithms.
Logarithm Differentiation starts with taking the natural logarithm that is, logarithm to the base e on the both sides.
In order to compute the derivative of:
• A function which is the product or quotient of a number of functions
y = g1(x)g2(x)g3(x) ….. or [(g1(x) g2(x) g3(x) ….) / (h1(x) h2(x) h3(x) ….)
• A function of the form [f(x)]g(x) where f and g are both derivable, it is better to take the logarithm of the function first and then differentiate.
• Or, we can write it as y = (f(x))g(x) = e g(x).ln f(x)?
Example
Differentiate y = xx
Solution:
y = xx
We will take the logarithm to both sides
In y = In xx
= x In x
Now we will differentiate both sides by using chain rule to the left-hand side as y represents a function of x and use the product rule on the right-hand side.
Multiply both sides with y
yꞌ = y (1 + In x)
= xx(1 + In x)
?We list below certain differentiation formulae for finding the derivative of functions:
Given below are some more formulae of some logarithmic and trigonometric functions:
d/dx (xx) = xx (1 + ln x) d/dx (logax) = 1/ x ln a d/dx (loga f(x)) = 1/ f(x) ln a . d/dx f(x) d/dx (sin x) = (sin x)' = cos x (cos x)' = -sin x (tan x)' = sec2 x (sec x)' = sec x tan x (cosec x)' = -cosec x cot x (cot x)' = -cosec2x
## Steps of Logarithm Differentiation
Logarithm Differentiation is the shortcut which helps in differentiating without using the product rule.
If we have to differentiate the above function then we don’t need to multiply it at all
Step 1: We can take the natural logarithm of both sides.
Step 2: use the product rule of the log.
loga(m × n) = logam + logan
So it will be like this
Step 3: Differentiate both sides using the chain rule of differentiation.
Step 4: Multiply both sides with f(x).
We now discuss some of the differentiation examples followed by some differentiation questions:
### Illustration 1:
What is the gradient of the curve y = 2x3 at the point (3, 54)?
### Solution:
The gradient of the curve is given by its derivative so the question actually requires you to compute the derivative.
The derivative is dy/dx = 6x2
When x = 3, dy/dx = 6 × 9 = 54.
Hence, the required gradient is 54.
________________________________________________________________________________
### Illustration 2:
Find the derivative of
### Solution:
We will use the quotient rule to find the derivative.
Now we know that
……….(1)
…….(2)
Hence, combining (1) and (2) we get the required answer.
_______________________________________________________________________________
### Illustration 3:
Find the derivative of f(x) = x4 + sin (x2) – ln (x)ex + 7.
### Solution:
Using the formulae, we get the derivative as
___________________________________________________________________________-
### Illustration 4:
Find the derivative of (x2 + 5x – 3)/ 3 x½.
### Solution:
This looks hard, but it isn't. The trick is to simplify the expression first: do the division (divide each term on the numerator by 3x½.
Doing this we obtain
(1/3)x3/2 + (5/3)x½ - x (using the laws of indices).
So differentiating term by term: ½ x½ + (5/6)x + ½x-3/2.
Q1. Both the derivative and the integral of this function are same
(a) the function is ex.
(b) the function is e2x.
(c) the function is 2ex.
(d) the function is ln x.
Q2. If f(x) = h(g(x)), then f'(x) =
(a) h'(g’(x)). g(x)
(b) h(g’(x)). g'(x)
(c) h'(g(x)). g(x)
(d) h'(g(x)). g'(x)
Q3. The derivative of logax is
(a) 1/x ln ax
(b) 1/x ln ax
(c) 1/x ln ax
(d) 1/x ln ax
Q4. The derivative of a constant function is
(a) zero
(b) the function itself.
(c) constant
(d) none of these
Q5. the derivative of the function [f(x)]n is
(a) n f(x)n (f’(x))
(b) n f’(x)n-1 (f(x))
(c) n f(x)n-1 (f’(x))
(d) f(x)n-1 (f’(x))
Q1. Q2. Q3. Q4. Q5. (a) (d) (b) (a) (c)
## Related Resources
To read more, Buy study materials of Methods of Differentiation comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
Logarithmic Differentiation
### Course Features
• Video Lectures
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• Previous Year Papers
• Mind Map
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Towards a just, equitable, humane and sustainable society
# Parts and wholes : Types of Fraction
0
R. Gomathy
## Represented in Different Way, But it should give same meaning CBSE, Maths, Grade – V
Prior Knowledge:
• An understanding about fractions - numerator and denominator.
• Knows to pictorially represent proper fractions.
Objective:
• To develop knowledge about the types of fraction - proper, improper and mixed fractions.
• Convert improper fraction to mixed fraction and represent it pictorially.
• Develop conceptual understanding about improper and mixed fraction.
• Solve simple problems related to mixed and improper fractions.
## ENGAGE:
Students were asked to pictorially represent as many fractional numbers as they can in five minutes. They then exchanged notebooks and corrected each other’s answers and pointed out if they found any mismatch between the fractional number and its pictorial representation. They discussed and arrive at the rig ht answers. Students understanding about fraction was assessed. I used this opportunity to clarify their misconceptions.
## EXPLORE:
Students were asked to pictorially represent the given fractions.
During this activity, students were easily able to represent the proper fractions. But their first response to improper fractions was that it is not possible to pictorially represent them.
## EXPLAIN:
These are some of the picture drawn by students.
Students tried to represent the improper fractions based on their understanding. One student explained her picture saying that since 4 is in the denominator, the box needs to be divided into four parts. And 5 being the numerator, she drew five such boxes.
Some students struggled with this concept. So they were asked to explain their rationale while representing proper fractions like ¼, 2/4, ¾, 4/4. I asked them a few questions to check their understanding levels.
1. Why did you divide the square into four parts? What does that mean?
2. What does ¼ stand for?
3. What is each part called?
Students responses were that since 4 is in the denominator it was divided it into 4 equal parts. The shaded region or selected parts denote the numerator.
Since they had understood the concept of proper fractions, they could be taught improper fractions now. So I asked them to draw the pictures for all the proper fractions that were written on the board.
Here I asked a few more questions based on the pictures. I asked them how many parts are shaded in 1/4? - One. How many portions are shaded in the second picture? – Two. Likewise, in the rest of the pictures three and four portions are shaded. So, In order to get 5/4 how many such portions would have to be shaded? - 5 portions should be shaded and students shaded 5 portions.
I explained that the fully shaded portion is called a whole and is represented as one and the other is ¼ . So 5/4 = 1¼. Then the children pictorially represented the other fractions 6/4, 7/4 and 8/4 and also wrote them as mixed
fractions.
## ELABORATE:
I have drawn a few fraction diagram in board. students were asked to write the corresponding mixed fraction and improper fraction for the given diagram.
Assessment:
Students’ ability to count the whole and the parts and write it as mixed fraction and their ability to convert a given mixed fraction into improper fraction was assessed.
Activity 2:
In this activity, I have given a few improper fraction number. Students have to shade the picture and write it as mixed fraction in their note book.
EVALUATE
A game was conducted to evaluate the children. I drew many fractions in chits and put them in a bowl and passed them around. Music was played while they passed the bowl. When the music stopped, the student who had the bowl, had to pick a chit and say 3 points about that fractional number or diagram.
For example: the chits had both numbers and diagrams. Like 4/5, 5/4, 6/10, 9/7
If a chit had 5/4 written on it, students said that:
5/4 is an improper fraction.
It has 5 ¼ parts in it.
It has 1 whole and a ¼ part.
Activity 2: Students had to write a few improper fractions and draw the picture of it and write it as mixed fraction.
In the same way students have to write a mixed fraction, draw picture for it and write the improper for it.
Assessment:
Through the these activities students’ knowledge and understanding about the concept was evaluated. Some of the common mistakes made by students are discussed here. Instead of counting only the shaded portions, they counted the entire region. So 2¼ was written as 9/12 instead of 9/4. Students could write improper fractions but could not write it as multiple fractions. They were given practice until they were able to get it right. |
# 6.5 Transformation of graphs (Page 2/5)
Page 2 / 5
$af\left(x\right)+d;\phantom{\rule{1em}{0ex}}a,d\in R$
These changes are called external or post-composition modifications.
## Arithmetic operations
Addition and subtraction to independent variable x is represented as :
$x+c;\phantom{\rule{1em}{0ex}}c\in R$
The notation represents addition operation when c is positive and subtraction when c is negative. In particular, we should underline that notation “bx+c” does not represent addition to independent variable. Rather it represents addition/ subtraction to “bx”. We shall develop proper algorithm to handle such operations subsequently. Similarly, addition and subtraction operation on function is represented as :
$\mathrm{f\left(x\right)}+d;\phantom{\rule{1em}{0ex}}d\in R$
Again, “af(x) + d” is addition/ subtraction to “af(x)” not to “f(x)”.
## Product/division operations
Product and division operations are defined with a positive constant for both independent variable and function. It is because negation i.e. multiplication or division with -1 is a separate operation from the point of graphical effect. In the case of product operation, the magnitude of constants (a or b) is greater than 1 such that resulting value is greater than the original value.
$bx;\phantom{\rule{1em}{0ex}}|b|>1\phantom{\rule{1em}{0ex}}\text{for independent variable}$ $af\left(x\right);\phantom{\rule{1em}{0ex}}|a|>1\phantom{\rule{1em}{0ex}}\text{for function}$
The division operation is eqivalent to product operation when value of multiplier is less than 1. In this case, magnitude of constants (a or b) is less than 1 such that resulting value is less than the original value.
$bx;\phantom{\rule{1em}{0ex}}0<|b|<1\phantom{\rule{1em}{0ex}}\text{for independent variable}$ $af\left(x\right);\phantom{\rule{1em}{0ex}}0<|a|<1\phantom{\rule{1em}{0ex}}\text{for function}$
## Negation
Negation means multiplication or division by -1.
## Effect of arithmetic operations
Addition/ subtraction operation on independent variable results in shifting of core graph along x-axis i..e horizontally. Similarly, product/division operations results in scaling (shrinking or stretching) of core graph horizontally. The change in graphs due to negation is reflected as mirroring (across y–axis) horizontally. Clearly, modifications resulting from modification to input modifies core graph horizontally. Another important aspect of these modification is that changes takes place opposite to that of operation on independent variable. For example, when “2” is added to independent variable, then core graph shifts left which is opposite to the direction of increasing x. A multiplication by 2 shrinks the graph horizontally by a factor 2, whereas division by 2 stretches the graph by a factor of 2.
On the other hand, modification in the output of function is reflected in change in graphs along y-axis i.e. vertically. Effects such as shifting, scaling (shrinking or stretching) or mirroring across x-axis takes place in vertical direction. Also, the effect of modification in output is in the direction of modification as against effects due to modifications to input. A multiplication of function by a positive constant greater than 1, for example, stretches the graph in y-direction as expected. These aspects will be clear as we study each of the modifications mentioned here.
## Forms of representation
There is a bit of ambiguity about the nature of constants in symbolic representation of transformation. Consider the representation,
A stone propelled from a catapult with a speed of 50ms-1 attains a height of 100m. Calculate the time of flight, calculate the angle of projection, calculate the range attained
58asagravitasnal firce
Amar
water boil at 100 and why
what is upper limit of speed
what temperature is 0 k
Riya
0k is the lower limit of the themordynamic scale which is equalt to -273 In celcius scale
Mustapha
How MKS system is the subset of SI system?
which colour has the shortest wavelength in the white light spectrum
if x=a-b, a=5.8cm b=3.22 cm find percentage error in x
x=5.8-3.22 x=2.58
what is the definition of resolution of forces
what is energy?
Ability of doing work is called energy energy neither be create nor destryoed but change in one form to an other form
Abdul
motion
Mustapha
highlights of atomic physics
Benjamin
can anyone tell who founded equations of motion !?
n=a+b/T² find the linear express
أوك
عباس
Quiklyyy
Moment of inertia of a bar in terms of perpendicular axis theorem
How should i know when to add/subtract the velocities and when to use the Pythagoras theorem?
Centre of mass of two uniform rods of same length but made of different materials and kept at L-shape meeting point is origin of coordinate |
# How coin counters work
## Instructions
Draw 15 circles in a row on the board. Ask the students to count them. It is ideal if some of the students get a total other than 15, because this will show how easy it is to lose track when counting by ones. Try to elicit ideas for better ways of handling this task.
If possible, give 15-40 counters to each student. If this is not possible for logistical or budgetary reasons, just have them use drawn circles. Ask the students to arrange 15 counters or circles in an orderly way. Guide them to the idea of making a 3x5 or 5x3 grid, but first let them try to figure out on their own that if they try to make 2 or 4 rows, they won't be even. When they are done, emphasize the fact that some students made a 3x5 grid, and some made a 5x3.
Ask the students to add up their rows. Depending on how they arranged their counters, some will add 3+3+3+3+3, and some will add 5+5+5. Emphasize that in each case, it adds up to 15. Discuss why this is the case, and whether it will always be the case for a similar problem. Ask a 3x5 student if the 5x3 student next to him/her was "wrong," and vice-versa.
Discuss the fact that adding 3 five times was a bit time consuming. Ask what would happen if we had 99 rows of 3. Hopefully the students will recognize that it would take all day if we needed to add 3 to itself 99 times, and will be happy to learn that there is a faster way. Most students love the prospect of a shortcut.
Explain that we have an operation called multiplication, which is nothing more than repeated addition. Instead of adding 3 five times, we can multiply 3 times 5. Elicit the fact that this represents three rows of five counters each. For the students who arranged their grids the other way, elicit that they added five rows of three counters each. The main thing is to stress that either way is the same thing. You could try explaining that this is called the commutative property. It doesn't matter we group the counters in a 3x5 or a 5x3 grid. It's still 15 counters.
At this point, introduce the students
to a 6x6 multiplication table. Don't just "fill it in." Work with the counters or circles on the board to compute the value that belongs in each cell of the table. Continue to emphasize the concept of commutativity, meaning that once we have 6x4, we automatically have 4x6. Emphasize that once we compute a cell of the table, we don't have to "reinvent the wheel" every time we need to do that computation.
Convince the students that they simply must memorize the multiplication table, and understand where all the values come from. Teach them that the answer to a multiplication problem is called the product, and that they will hear this word constantly throughout high school. Emphasize that if they cannot do simple multiplication problems in their head quickly and easily, they will have tons of trouble in later math. Students need to understand that multiplication is not just something given to them to keep them busy.
Explore the concepts of multiplying by 0 and 1. Why do the answers work out the way they do? Explain to students that a typical multiplication table allows them to find products up through 12x12, but that they will later learn a procedure which will allow them to multiply any two numbers, including very large ones. Discuss the fact that a table larger than 12x12 is not very practical.
Begin to introduce students to word problems that are solved using multiplication. A typical problem is, "There are 5 children at a party, and each child will be given 6 candies. How many candies are needed?" Discuss the fact that the keyword "each" implies multiplication in this case. Ask students for some other real-world scenarios where multiplication would be helpful.
Note that in an ideal world, multiplication should not be taught until students are fully comfortable with addition, but this is often not possible. Multiplication should not be taught "by rote," meaning that students need to understand what multiplication really is. It is meaningless to give students a page of practice exercises which they will solve without any thought at all, but will simply solve by way of a multiplication table or a calculator. With that said, once students fully understand the concept of multiplication, and when to use it, you can reinforce basic multiplication facts by having them use flashcards, perhaps in pairs or small groups. Good luck!
Source: ehow.com
Category: Bank |
Home Education Fraction, definitions, types, and examples.
# Fraction, definitions, types, and examples.
0
whenever the word fraction is mentioned, it simply means part of something or part of a whole. for instance, if a quantity is divided into four equal parts, then we can say that each part is one-fourth of the quantity. which is often written as $\frac{1}{4}$
A fraction like $\frac{2}{7}$ of mangoes simply means you divide the mangoes into seven equal parts and take two of those seventh
When you have numbers like $\frac{1}{8}$ and $\frac{1}{7}$ they are called vulgar fractions or common fractions. mathematically we write fractions as $\frac{A}{B}$. The top number is called the numerator and the down number is called the denominator.
## Types of Fractions
There are only three types of fractions, this includes common fractions, decimal fractions, and percentages.
Common fraction: when we say fraction, we often refer to a common fraction. examples are $\frac{1}{2}, \frac{4}{7}$.
Numbers in a form of 0.78, 0.15 are called decimals fractions. It is often called decimals. We write them with a decimal point.
Now, numbers that are written in this form 78%, 25% are called percentages. We write them with percentages or the symbol %.
## Types of common fraction
There are five types of common fractions, which includes:
. Proper fractions: this is a type of fraction whose numerator is less than the denominator. Examples $\frac{15}{17}, \frac{3}{7}, \frac{9}{21}$
. Improper fraction: this is a type of fraction whose numerator is bigger than the denominator. Examples $\frac{8}{3}, \frac{18}{17}, \frac{156}{26}$
. Like fractions: when two or more fractions often have the same denominators then they are called Like fractions. Examples $\frac{3}{8}, \frac{5}{8}, \frac{1}{8}$
. Unlike fractions: when two or more fractions have different denominators, then they are called, unlike fractions. Examples $\frac{1}{7}$ and $\frac{6}{8}$, $\frac{3}{9}$ and $\frac{7}{15}$.
. Mixed numbers: These fractions often contain a whole number part and a fraction part. some people at times called them mixed fractions. Examples 2$\frac{7}{29}$, 1$\frac{9}{15}$
## Equivalent fraction
when two or more fractions look different but have the same value then it is called equivalent fractions. If you have a fraction like $\frac{3}{4}$. To find the equivalent fractions of this fraction, multiply both the numerator and denominator by the same number. For instance
$\frac{3\times2}{4\times2} = \frac{6}{8}\\$
$\frac{3\times3}{4\times3} = \frac{9}{12}$
this, therefore, means that $\frac{3}{4}$, $\frac{6}{8}$, and $\frac{9}{12}$ are equivalent fractions
Examples: Find out which of the following pairs of fractions are equivalent
a. $\frac{4}{5}$ and $\frac{16}{20}$
b. $\frac{3}{5}$ and $\frac{33}{55}$
c. $\frac{6}{7}$ and $\frac{24}{28}$
## solutions
a. $\frac{4}{5} = \frac{4\times4}{5\times4} = \frac{16}{20}$
Hence $\frac{4}{5}$ and $\frac{16}{20}$ are equivalent fractions
READ: Extend the provision of free internet connectivity to technical universities. vice-chancellor
b. $\frac{3}{5} = \frac{3\times11}{5\times11}= \frac{33}{55}$
Hence $\frac{3}{5}$ and $\frac{33}{55}$ are equivalent fractions.
c. $\frac{6}{7} = \frac{6\times4}{7\times4} = \frac{24}{28}$
Hence $\frac{6}{7}$ and $\frac{24}{28}$ are equivalent fractions |
## Week 4 in math 10
Fractional Exponents
This week I learnt how to fractional exponents into radicals in order to evaluate it. Here are the steps I used.
First you take your base and place it inside the radicant. Then the bottom half of the fractional exponent (denominator) becomes the index and the top half (numerator) becomes the exponent on the inside of the radicant. You would then find the prime factorization of the base to find the root. You leave the exponent on the inside alone. After that you take the root of the base you now attach the exponent from the inside and place it on the root you found. The last step is to evaluate all that.
And if you were to convert a radical into a single power you would take the radicant which would turn into your base, the index would then become your denominator and the exponent inside would become your numerator.
## Week 3 in math 10
Exponents
This week in math 10 we started a new unit on exponents. And one thing that really stuck out for me is using the laws to simplify equations. It also helps to know what to do when you chose the law.
Laws:
Multiplication law – When you 2 or more variables with the same base and are sitting side by side, but different exponents all you need to do is keep the same base and add all the exponents together. (ex. $5^3$ x $5^5$ = $5^8$) (3+5=8)
Division law – When you 2 or more variables with the same base and have one on top of the other but have different exponents all you need to do is keep the base and subtract the exponents and whichever side has the bigger exponent is where you place what’s left of the exponents. (ex. $2^9$ ÷ $2^3$ = $2^6$) (9-3=6)
Power of a power law – When you have an equation with brackets and have an exponent on the inside and the outside of the brackets all you need to do is times the outside exponent by the inside exponents. (ex. ($5^3$)$^2$ = $5^6$) (3×2=6)
Zero exponent law – When you have a variable the zero as the exponent it always equals 1, because there are no copies of the variable so you times the exponent by the coefficient. (ex. $7^0$ = 1)
After knowing how to use these laws it now becomes easier to simplify equations.
## Week 2 in math 10
Week 2 in math 10
Something that I learned this week is how to convert entire radicals into mixed radicals and vice-versa. In the beginning, I was a bit confused but after doing the homework and asking my classmates for help I began to understand.
This is the process I use to convert the entire radical. First, I would use prime factorization on the radicand, then from there I would look at the index, for example, in $\sqrt[4]{48}$ the index is 4, meaning that after finding the prime factorization of 48 you would look for the same prime numbers and place them in groups of 4.(Depending on the index given that indicates how many prime numbers are placed in one group.) The prime numbers that are in a group are placed on the outside of the radical which becomes the coefficient, and all the remainders that were not grouped are either multiplied together or left alone. If the remainders are multiple different groups of prime numbers you multiply them, but if there is only one group of the same prime number there’s nothing to multiply it by so you leave it as it is. The remainders stay on the inside of the radical.
Ex. |
# You asked: Are the diagonals of a kite congruent?
Contents
The diagonals are not congruent, but they are always perpendicular. In other words, the diagonals of a kite will always intersect at right angles.
## Are the sides of a kite congruent?
Kite properties include (1) two pairs of consecutive, congruent sides, (2) congruent non-vertex angles and (3) perpendicular diagonals. Other important polygon properties to be familiar with include trapezoid properties, parallelogram properties, rhombus properties, and rectangle and square properties.
## Is a kite shape congruent?
A kite is a quadrilateral shape with two pairs of adjacent (touching), congruent (equal-length) sides.
## Is it true that the diagonals of a kite intersect perpendicularly?
The Diagonals of a Kite are Perpendicular to Each Other
We have already shown that the diagonal that connects the two corners formed by the sides that are equal bisects the angles at those corners. So it is now easy to show another property of the diagonals of kites- they are perpendicular to each other.
## How do you find the diagonals of a kite?
Explanation: In order to solve this problem, first observe that the red diagonal line divides the kite into two triangles that each have side lengths of and. Notice, the hypotenuse of the interior triangle is the red diagonal. Therefore, use the Pythagorean theorem: , where the length of the red diagonal.
## How many pairs of congruent sides does a kite have?
A kite is a quadrilateral with exactly two pairs of adjacent congruent sides. (This definition excludes rhombi.
## How do you prove a kite is congruent?
If two disjoint pairs of consecutive sides of a quadrilateral are congruent, then it’s a kite (reverse of the kite definition). If one of the diagonals of a quadrilateral is the perpendicular bisector of the other, then it’s a kite (converse of a property).
## Which pair of angles are congruent in kite case?
In a kite, one pair of opposite angles are congruent.
## Do the diagonals of a kite bisect the opposite angles?
THEOREM: If a quadrilateral is a kite, it has one diagonal that bisects a pair of opposite angles.
## Do the diagonals of a kite bisect vertex angles?
A quadrilateral is a kite if: one diagonal bisects the vertex angles through which it passes, or. one diagonal is the perpendicular bisector of the other diagonal.
## What do you mean by congruent?
: having the same size and shape congruent triangles.
## Are the triangles of a kite congruent?
The angles between the congruent sides are called vertex angles. The other angles are called non-vertex angles. If we draw the diagonal through the vertex angles, we would have two congruent triangles. Theorem: The non-vertex angles of a kite are congruent.
## What is true about diagonals of a kite?
The diagonals of a kite are perpendicular to each other. The longer diagonal of the kite bisects the shorter diagonal. The area of a kite is equal to half of the product of the length of its diagonals. The perimeter of a kite is equal to the sum of the length of all of its sides.
## What are congruent diagonals?
Properties of a square
The diagonals are congruent. The diagonals are perpendicular to and bisect each other. A square is a special type of parallelogram whose all angles and sides are equal. Also, a parallelogram becomes a square when the diagonals are equal and right bisectors of each other.
## Why are the diagonals of a kite perpendicular?
The intersection of the diagonals of a kite form 90 degree (right) angles. This means that they are perpendicular. The longer diagonal of a kite bisects the shorter one. This means that the longer diagonal cuts the shorter one in half. |
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# Evaluate the following integral:
Question:
Evaluate the following integral:
$\int \frac{\sin 2 x}{(1+\sin x)(2+\sin x)} d x$
Solution:
The denominator is factorized, so let separate the fraction through partial fraction, hence let
$\frac{\sin 2 x}{(1+\sin x)(2+\sin x)}=\frac{A}{(1+\sin x)}+\frac{B}{2+\sin x} \ldots \ldots$ (i)
$\Rightarrow \frac{\sin 2 x}{(1+\sin x)(2+\sin x)}=\frac{A(2+\sin x)+B(1+\sin x)}{(1+\sin x)(2+\sin x)}$
$\Rightarrow \sin 2 x=A(2+\sin x)+B(1+\sin x)=2 A+A \sin x+B+B \sin x$
$\Rightarrow 2 \sin x \cos x=\sin x(A+B)+(2 A+B) \ldots \ldots$ (ii)
We need to solve for $A$ and $B$.
We will equate similar terms, we get.
$2 A+B=0 \Rightarrow B=-2 A$
And $A+B=2 \cos x$
Substituting the value of $B$, we get
$A-2 A=2 \cos x \Rightarrow A=-2 \cos x$
Hence $B=-2 A=-2(-2 \cos x)$
$\Rightarrow B=4 \cos x$
We put the values of $A$ and $B$ values back into our partial fractions in equation (i) and replace this as the integrand. We get
$\int\left[\frac{\sin 2 x}{(1+\sin x)(2+\sin x)}\right] d x$
$\Rightarrow \int\left[\frac{A}{(1+\sin x)}+\frac{B}{2+\sin x}\right] d x$
$\Rightarrow \int\left[\frac{-2 \cos x}{(1+\sin x)}+\frac{4 \cos x}{2+\sin x}\right] d x$
Split up the integral,
$\Rightarrow-\int \frac{2 \cos x}{(1+\sin x)} d x+\int \frac{4 \cos x}{2+\sin x} d x$
Let substitute
$u=\sin x \Rightarrow d u=\cos x d x$
so the above equation becomes,
$\Rightarrow-2 \int \frac{1}{(1+u)} d u+4 \int \frac{1}{2+u} d u$
Now substitute
$v=1+u \Rightarrow d v=d u$
$z=2+u \Rightarrow d z=d u$
So above equation becomes,
$\Rightarrow-2 \int \frac{1}{(v)} d v+4 \int \frac{1}{z} d z$
On integrating we get
$\Rightarrow-2 \log |v|+4 \log |z|+C$
Substituting back, we get
$\Rightarrow 4 \log |2+u|-2 \log |1+u|+C$
$\Rightarrow 4 \log |2+\sin x|-2 \log |1+\sin x|+C$
Applying logarithm rule, we get
$\Rightarrow \log \left|(2+\sin x)^{4}\right|-\log \left|(1+\sin x)^{2}\right|+C$
$\Rightarrow \log \left|\frac{(2+\sin x)^{4}}{(1+\sin x)^{2}}\right|+C$
Note: the absolute value signs account for the domain of the natural log function $(x>0)$.
Hence,
$\int \frac{\sin 2 x}{(1+\sin x)(2+\sin x)} d x=\log \left|\frac{(2+\sin x)^{4}}{(1+\sin x)^{2}}\right|+C$ |
# 3.1 Signed numbers
Page 1 / 1
This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. The basic operations with real numbers are presented in this chapter. The concept of absolute value is discussed both geometrically and symbolically. The geometric presentation offers a visual understanding of the meaning of |x|. The symbolic presentation includes a literal explanation of how to use the definition. Negative exponents are developed, using reciprocals and the rules of exponents the student has already learned. Scientific notation is also included, using unique and real-life examples.Objectives of this module: be familiar with positive and negative numbers and with the concept of opposites.
## Overview
• Positive and Negative Numbers
• Opposites
## Positive and negative numbers
When we studied the number line in Section [link] we noted that
Each point on the number line corresponds to a real number, and each real number is located at a unique point on the number line.
## Positive and negative numbers
Each real number has a sign inherently associated with it. A real number is said to be a positive number if it is located to the right of 0 on the number line. It is a negative number if it is located to the left of 0 on the number line.
## The notation of signed numbers
A number is denoted as positive if it is directly preceded by a $"+"$ sign or no sign at all.
A number is denoted as negative if it is directly preceded by a $"-"$ sign.
The $"+"$ and $"-"$ signs now have two meanings:
$+$ can denote the operation of addition or a positive number.
$-$ can denote the operation of subtraction or a negative number.
## Read the $"-"$ Sign as "negative"
To avoid any confusion between "sign" and "operation," it is preferable to read the sign of a number as "positive" or "negative."
## Sample set a
$-8$ should be read as "negative eight" rather than "minus eight."
$4+\left(-2\right)$ should be read as "four plus negative two" rather than "four plus minus two."
$-6+\left(-3\right)$ should be read as "negative six plus negative three" rather than "minus six plusminus three."
$-15-\left(-6\right)$ should be read as "negative fifteen minus negative six" rather than "minus fifteenminus minus six."
$-5+7$ should be read as "negative five plus seven" rather than "minus five plus seven."
$0-2$ should be read as "zero minus two."
## Practice set a
Write each expression in words.
$4+10$
four plus ten
$7+\left(-4\right)$
seven plus negative four
$-9+2$
negative nine plus two
$-16-\left(+8\right)$
negative sixteen minus positive eight
$-1-\left(-9\right)$
negative one minus negative nine
$0+\left(-7\right)$
zero plus negative seven
## Opposites
On the number line, each real number has an image on the opposite side of 0. For this reason we say that each real number has an opposite. Opposites are the same distance from zero but have opposite signs.
The opposite of a real number is denoted by placing a negative sign directly in front of the number. Thus, if $a$ is any real number, then $-a$ is its opposite. Notice that the letter $a$ is a variable. Thus, $"a"$ need not be positive, and $"-a"$ need not be negative.
If $a$ is a real number, $-a$ is opposite $a$ on the number line and $a$ is opposite $-a$ on the number line.
$-\left(-a\right)$ is opposite $-a$ on the number line. This implies that $-\left(-a\right)=a$ .
This property of opposites suggests the double-negative property for real numbers.
## The double-negative property
If $a$ is a real number, then
$-\left(-a\right)=a$
## Sample set b
If $a=3$ , then $-a=-3$ and $-\left(-a\right)=-\left(-3\right)=3$ .
If $a=-4$ , then $-a=-\left(-4\right)=4$ and $-\left(-a\right)=a=-4$ .
## Practice set b
Find the opposite of each real number.
8
$-8$
17
$-17$
$-6$
6
$-15$
15
$-\left(-1\right)$
$-1$ , since $-\left(-1\right)=1$
$-\left[-\left(-7\right)\right]$
7
Suppose that $a$ is a positive number. What type of number is $-a$ ?
If $a$ is positive, $-a$ is negative.
Suppose that $a$ is a negative number. What type of number is $-a$ ?
If $a$ is negative, $-a$ is positive.
Suppose we do not know the sign of the number $m$ . Can we say that $-m$ is positive, negative, or that we do notknow ?
We must say that we do not know.
## Exercises
A number is denoted as positive if it is directly preceded by ____________________ .
a plus sign or no sign at all
A number is denoted as negative if it is directly preceded by ____________________ .
For the following problems, how should the real numbers be read ? (Write in words.)
$-5$
a negative five
$-3$
12
twelve
10
$-\left(-4\right)$
negative negative four
$-\left(-1\right)$
For the following problems, write the expressions in words.
$5+7$
five plus seven
$2+6$
$11+\left(-2\right)$
eleven plus negative two
$1+\left(-5\right)$
$6-\left(-8\right)$
six minus negative eight
$0-\left(-15\right)$
Rewrite the following problems in a simpler form.
$-\left(-8\right)$
$-\left(-8\right)=8$
$-\left(-5\right)$
$-\left(-2\right)$
2
$-\left(-9\right)$
$-\left(-1\right)$
1
$-\left(-4\right)$
$-\left[-\left(-3\right)\right]$
$-3$
$-\left[-\left(-10\right)\right]$
$-\left[-\left(-6\right)\right]$
$-6$
$-\left[-\left(-15\right)\right]$
$-\left\{-\left[-\left(-26\right)\right]\right\}$
26
$-\left\{-\left[-\left(-11\right)\right]\right\}$
$-\left\{-\left[-\left(-31\right)\right]\right\}$
31
$-\left\{-\left[-\left(-14\right)\right]\right\}$
$-\left[-\left(12\right)\right]$
12
$-\left[-\left(2\right)\right]$
$-\left[-\left(17\right)\right]$
17
$-\left[-\left(42\right)\right]$
$5-\left(-2\right)$
$5-\left(-2\right)=5+2=7$
$6-\left(-14\right)$
$10-\left(-6\right)$
16
$18-\left(-12\right)$
$31-\left(-1\right)$
32
$54-\left(-18\right)$
$6-\left(-3\right)-\left(-4\right)$
13
$2-\left(-1\right)-\left(-8\right)$
$15-\left(-6\right)-\left(-5\right)$
26
$24-\left(-8\right)-\left(-13\right)$
## Exercises for review
( [link] ) There is only one real number for which ${\left(5a\right)}^{2}=5{a}^{2}$ . What is the number?
0
( [link] ) Simplify $\left(3xy\right)\left(2{x}^{2}{y}^{3}\right)\left(4{x}^{2}{y}^{4}\right)$ .
( [link] ) Simplify ${x}^{n+3}\cdot {x}^{5}$ .
${x}^{n+8}$
( [link] ) Simplify ${\left({a}^{3}{b}^{2}{c}^{4}\right)}^{4}$ .
( [link] ) Simplify ${\left(\frac{4{a}^{2}b}{3x{y}^{3}}\right)}^{2}$ .
$\frac{16{a}^{4}{b}^{2}}{9{x}^{2}{y}^{6}}$
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
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
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
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
how did you get the value of 2000N.What calculations are needed to arrive at it
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Stage 4 - Stage 5
# Graphing Absolute Value Functions
Lesson
The graphs of the absolute value function applied to other functions often have sharp turns at points where the original function changes from positive to negative. That is, there may be points at which the absolute value function of a function is not differentiable. If you haven't yet started calculus, this just means that there may be points that are not smooth, in fact in absolute value functions these points often appear 'pointed'.
For example, the graph of $y(x)=|x^2-1|$y(x)=|x21| has the shape shown in the following sketch.
## Critical Points
In considering the behaviour of a function that has an absolute value component, we will often need to investigate how the function changes at certain critical points. This is done with a view to writing separate expressions, without the absolute value symbols, that represent the function over particular intervals in its domain of definition.
The function $f(x)=\frac{|x^2-7x+10|}{x-2}$f(x)=|x27x+10|x2 is undefined at $x=2$x=2 since the denominator would be zero there. We observe that the denominator is positive for $x>2$x>2 and negative for $x<2$x<2.
The numerator, however, is always non-negative. So, the function is negative when $x<2$x<2 but non-negative when $x>2$x>2.
The numerator can be written as the product $|(x-2)(x-5)|$|(x2)(x5)| or equivalently, $|x-2||x-5|$|x2||x5|. So, we see that critical changes occur at $x=2$x=2 and at $x=5$x=5.
When $x<2$x<2, the numerator is positive, $(x-2)(x-5)>0$(x2)(x5)>0; when $22<x<5, the numerator is negative,$(x-2)(x-5)<0$(x2)(x5)<0 and when$x>5$x>5$(x-2)(x-5)>0$(x2)(x5)>0. The numerator has the value zero at$x=5$x=5. Thus, we can write three different function definitions, applicable over the parts of the domain. When$x<2$x<2,$f(x)=\frac{x^2-7x+10}{x-2}=x-5$f(x)=x27x+10x2=x5. (The function is negative and increasing.) When$22<x5$f(x)=-\frac{x^2-7x+10}{x-2}=-x+5$f(x)=x27x+10x2=x+5. (The function is positive and decreasing or zero.)
When $x>5$x>5$f(x)=\frac{x^2-7x+10}{x-2}=x-5$f(x)=x27x+10x2=x5. (The function is positive and increasing.)
## Beginning to sketch
As with other functions, we can make a sketch representing the graph of the function by constructing a table of values and then plotting the points from the table. We understand the graph of a function to be the set of all points that satisfy the function definition. There are usually infinitely many such points and so, we must make a selection from them in order to make a sketch approximating the graph.
The sketch will be accurate if enough points are included from the table or if certain special points, crucial to the shape of the graph, are chosen. For absolute value functions, it is useful to include the critical points at which abrupt changes of gradient occur. That is, we look for the points at which the expression inside the absolute value brackets changes from negative to positive.
As well as finding the critical points, we need to use what is known about sketching graphs of functions that are not absolute value functions. If the function is linear, for example, we need another point to define each line segment or we need to know a gradient. Similarly, we may need the zeros and turning points of polynomial functions, and in other cases, identifying asymptotes could be important.
#### Example 1
The function given by $y(x)=3|x-2|+1$y(x)=3|x2|+1, is to be explored. As part of the investigation, we wish to sketch the graph.
We see that there is a critical value, $x=2$x=2, and that $y(2)=1$y(2)=1. When $x\le2$x2, the function is equivalent to $y(x)=3(2-x)+1$y(x)=3(2x)+1 or more simply, $y(x)=-3x+7$y(x)=3x+7. For this part of the graph, we could choose another value, $x=0$x=0 and obtain the point $(0,7)$(0,7).
When $x>2$x>2, the function is equivalent to $y(x)=3(x-2)+1$y(x)=3(x2)+1 or more simply, $y(x)=3x-5$y(x)=3x5. Another point on this part of the graph is the point $(3,4)$(3,4).
These three points are sufficient to determine the graph.
The following example is more complicated but it illustrates some of the techniques that may be needed.
#### Example 2
The function $f$f given by $f(x)=|x^2+3x|-2x+1$f(x)=|x2+3x|2x+1, has critical points where the expression $x^2+3x$x2+3x is zero. That is, when $x=0$x=0 and when $x=-3$x=3
When $x>0$x>0, and when $x<-3$x<3$f(x)$f(x) can be written $f_1(x)=x^2+x+1$f1(x)=x2+x+1. Between these values, when $-3\le x\le0$3x0, $f(x)$f(x) takes the form $f_2(x)=-x^2-5x+1$f2(x)=x25x+1.
Thus, the critical points are $(-3,7)$(3,7) and $(0,1)$(0,1).
Now, $f_1(x)=x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}$f1(x)=x2+x+1=(x+12)2+34 by completing the square. From this, it can be seen that $f_1$f1 cannot be less than $\frac{3}{4}$34 and this minimum occurs when $x=-\frac{1}{2}$x=12. Although this point is not in the domain of $f_1$f1 and therefore does not belong to the graph, it will be helpful in determining the shape of the sketch. We note that $f_1$f1 has no real zeros and its graph must be entirely above the $x$x-axis.
Similarly, we can rewrite $f_2=-x^2-5x+1$f2=x25x+1 as $f_2(x)=-\left(\left(x+\frac{5}{2}\right)^2-\frac{29}{4}\right)$f2(x)=((x+52)2294). This has a maximum of $\frac{29}{4}$294 at $x=-\frac{5}{2}$x=52
From the quadratic formula or otherwise, we determine that $f_2$f2 has zeros at $\frac{5\pm\sqrt{29}}{-2}$5±292. That is, at approximately $x=-5.2$x=5.2 and $x=0.2$x=0.2.
This collection of six points is enough to give a good indication of the shape of the graph of $f$f.
#### Worked Examples
##### Question 1
Consider the function $y=\left|x\right|$y=|x|.
1. Complete the table.
$x$x $y$y $-2$−2 $-1$−1 $0$0 $1$1 $2$2 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
2. Hence sketch a graph of the function.
3. What are the coordinates of the vertex?
4. What is the domain?
$x\le0$x0
A
$-\infty<x< B$x\ge0$x0 C$x>0$x>0 D 5. What is the range?$y\le0$y0 A$-\infty<y<
B
$y\ge0$y0
C
$y>0$y>0
D
##### Question 2
Graph $y=-\frac{1}{4}\left|x+2\right|$y=14|x+2| $+$+ $4$4.
If the graph of $y=\left|x\right|$y=|x| is compressed vertically by a factor of $\frac{1}{4}$14, reflected across the $x$x-axis and translated $3$3 unit(s) left and $3$3 unit(s) up, what is the equation of the new graph? |
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### April 4, 2014
1. 1. Holt Algebra 1 9-3 Graphing Quadratic Functions TGIF April 4, 2014 Today: Review from Yesterday Getting to Know the Quadratic Function: (how the b value changes the parabola) Class Work
2. 2. Holt Algebra 1 9-3 Graphing Quadratic Functions Important Concepts from Yesterday: 1. There is no single "right way" to graph a quadratic function. In fact the goal is for you to understand and use a variety of methods, so that you can choose the best (easiest) method for a given problem. 2. The axis of symmetry is an important part of parabolas and can save you much time and effort if you understand its properties. Because a parabola is symmetrical, each point is the same number of units away from the axis of symmetry. Helpful Hint 2.5 You must have at least 5 points to graph the parabola.
3. 3. Holt Algebra 1 9-3 Graphing Quadratic Functions 4. The vertex is an (x,y) coordinate on the AOS, and is either the minimum or maximum y value Important Concepts from Yesterday: 5. The vertex is an (x,y) coordinate on the AOS, and is either the minimum or maximum y value 6. All parabolas begin from the parent function (y = x2), and are moved around the coordinate plane from changes in the a, b, and c values. 3. The axis of symmetry has a single coordinate (x) and represents the exact center of the parabola. 7. A quadratic equation with no solutions will not cross the x-axis at any point. It can still be graphed using other methods.
4. 4. Holt Algebra 1 9-3 Graphing Quadratic Functions Review from Yesterday: How the a and c values affect the quadratic function y = ax2 + bx + c Start with the parent function, which is... First, how does a change in a affect the parabola y = x2 Effects of the a, b, & c values
5. 5. Holt Algebra 1 9-3 Graphing Quadratic FunctionsEffects of the a, b, & c values 1. The greater the value of 'a', the narrower and steeper the graph. 2. A positive 'a' value results in parabola which turns up and has a vertex minimum. 3. A negative 'a' value results in parabola which turns down and has a vertex maximum.
6. 6. Holt Algebra 1 9-3 Graphing Quadratic FunctionsEffects of the a, b, & c values How does a change in c affect the parabola? The value of c is also used to find the y-intercept. Set the 'x' values = 0, and find the intercept. We would expect the value of 'c' in this graph to be.......
7. 7. Holt Algebra 1 9-3 Graphing Quadratic Functions Effects of the a, b, & c values How changes in 'b' affect the parabola: Why does a positive b value (see aqua, b = 2) result in a shift 2 units to the left?
8. 8. Holt Algebra 1 9-3 Graphing Quadratic Functions Graph the quadratic function. y = x2 + 4x + 4 Step 2 Find the axis of symmetry, Step 1: Try to picture what the graph will look like before you start. Use the a,b, and c values to determine your prediction Step 3: Determine the best method(s) to solve that particular function. Step 4 : Plot at least 5 points, then connect the dots to complete the parabola. then find 'y' to complete the coordinates for the vertex
9. 9. Holt Algebra 1 9-3 Graphing Quadratic Functions Step 2 Find the axis of symmetry and the vertex. y = x2 + 4x + 2 Substitute for x to find the y coordinate The x-coordinate of the vertex is... The y-coordinate is Find at least 4 more points, then graph. This is also the AOS
10. 10. Holt Algebra 1 9-3 Graphing Quadratic Functions Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points. Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve. y = x2 + 4x + 2 Example 2 Continued
11. 11. Holt Algebra 1 9-3 Graphing Quadratic Functions Class Work: Graphing Quadratic Functions
12. 12. Holt Algebra 1 9-3 Graphing Quadratic Functions |
# Question Video: Finding the Missing Term in the Vector Forces Expression Mathematics
Complete the following: In the given figure, π
β = (π Γ β¦)/(sin (πβ + πβ)).
03:33
### Video Transcript
Complete the following. In the given figure, π
sub one is equal to π multiplied by what divided by the sin of π sub one plus π sub two.
The diagram shows a parallelogram of forces, where the sides of the parallelogram are equal to the magnitudes of vectors π
sub one and π
sub two and the diagonal of the parallelogram is the resultant of these forces π. In order to find an expression for π
sub one, we can begin by finding the missing angles in triangle ππ΄πΆ. Angles π΅ππΆ and ππΆπ΄ are alternate angles, meaning that angle ππΆπ΄ is equal to π sub two.
We know that angles in a triangle sum to 180 degrees. This means that the measure of angle ππ΄πΆ is equal to 180 degrees minus π sub one plus π sub two. We can now use the sine rule or law of sines to find an expression for π
sub one in terms of π together with π sub one and π sub two. The law of sines states that π over sin π΄ is equal to π over sin π΅, which is equal to π over sin πΆ, where lowercase π, π, and π are the lengths of the three sides of our triangle and capital π΄, π΅, and πΆ are the measures of the angles opposite the corresponding sides.
In triangle ππ΄πΆ, side length ππ΄ is opposite the angle at πΆ. This gives us π
sub one over sin of π sub two. The resultant π is opposite the angle equal to 180 degrees minus π sub one plus π sub two. Substituting this into the law of sines gives us π over sin of 180 degrees minus π sub one plus π sub two. Recalling the symmetry of the sine function, we know that the sin of 180 degrees minus πΌ is equal to sin πΌ. This means that the right-hand side of our equation can be rewritten as π over sin of π sub one plus π sub two. In order to make π
sub one the subject of our equation, we can multiply through by sin of π sub two. π
sub one is therefore equal to π multiplied by sin of π sub two divided by sin of π sub one plus π sub two.
And we can therefore conclude that the missing part of the equation in the question is sin of π sub two. |
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# Solve the differential equation:$\left( {x{y^2} + x} \right)dx + \left( {{x^2}y + y} \right)dy = 0$.
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Hint: Separate the terms with $x$ variable on one side and terms with $y$ variable on other side. And then solve the equation integrating both sides.
The given differential equation is $\left( {x{y^2} + x} \right)dx + \left( {{x^2}y + y} \right)dy = 0$. This can be simplified as:
$\Rightarrow x\left( {{y^2} + 1} \right)dx = - y\left( {{x^2} + 1} \right)dy, \\ \Rightarrow \dfrac{x}{{\left( {{x^2} + 1} \right)}}dx = - \dfrac{y}{{\left( {{y^2} + 1} \right)}}dy, \\ \Rightarrow \dfrac{{2x}}{{\left( {{x^2} + 1} \right)}}dx = - \dfrac{{2y}}{{\left( {{y^2} + 1} \right)}}dy \\$
Integrating both sides, we’ll get:
$\Rightarrow \int {\dfrac{{2x}}{{\left( {{x^2} + 1} \right)}}dx} = - \int {\dfrac{{2y}}{{\left( {{y^2} + 1} \right)}}dy,}$
We know that $\int {\dfrac{{2x}}{{\left( {{x^2} + 1} \right)}}dx} = \log \left| {{x^2} + 1} \right| + C$, Using this in the above equation, we’ll get:
$\Rightarrow \log \left| {{x^2} + 1} \right| = - \log \left| {{y^2} + 1} \right| + C, \\ \Rightarrow \log \left| {{x^2} + 1} \right| + \log \left| {{y^2} + 1} \right| = C, \\ \Rightarrow \log \left( {\left| {{x^2} + 1} \right|\left| {{y^2} + 1} \right|} \right) = C, \\ \Rightarrow \left| {\left( {{x^2} + 1} \right)\left( {{y^2} + 1} \right)} \right| = {e^C}, \\ \Rightarrow \left( {{x^2} + 1} \right)\left( {{y^2} + 1} \right) = \pm {e^C} \\$
Thus the solution of the differential equation is $\left( {{x^2} + 1} \right)\left( {{y^2} + 1} \right) = \pm {e^C}$
Note: The method used in solving the above differential equation is called variable separation method i.e. keeping the terms containing the same variable on one side and terms having other variables on the other side. And then integrating on both the sides. |
# Distinct balls into distinct boxes with a minimal number of balls in each box
Find the number of ways to distribute $8$ distinct balls into $3$ distinct boxes if each box must hold at least $2$ balls.
The stars and bars approach would not work because the balls are non-identical. Stirling Numbers of the second kind would also not work directly, because each box doesn't just have to be non-empty; they must contain at least $2$ balls each.
I tried the following approach: Since each box must contain at least $3$ balls each, then the $8$ distinct balls must be divided into distinct partitions of sizes $2,2,4$ or $2,3,3$.
For the first case, we choose $2$ of the $8$ balls for the first partition, $2$ of the $6$ remaining balls for the second partition, and $4$ of the $4$ remaining balls for the third partition. Since the partitions are distinct, we multiply by $3!$ to give:
$$\binom{8}{2}\binom{6}{2}\binom{4}{4}3!$$
Apply the same approach, the number of ways for the second case is:
$$\binom{8}{2}\binom{6}{3}\binom{3}{3}3!$$
Hence, the total number of ways is
$$\binom{8}{2}\binom{6}{2}\binom{4}{4}3! +\binom{8}{2}\binom{6}{3}\binom{3}{3}3!$$
Is my approach valid? Also, is it possible to transform the question to make use of Stirling numbers of the second kind?
$8=4+2+2=2+4+2=2+2+4$ gives $3\times\frac{8!}{4!2!2!}$ possibilities.
$8=2+3+3=3+2+3=3+3+2$ gives $3\times\frac{8!}{2!3!3!}$ possibilities.
So not $3!$ but $3$ as extra factor to avoid double counting.
edit
Label the boxes: box1, box2 and box3.
Number of possibilities to end up with $4$ balls in box1: $\binom{8}{4}\binom{4}{2}\binom{2}{2}=\frac{8!}{4!2!2!}=420$.
The same for $4$ balls in box2 and the same for $4$ ballsi in box3. The events are disjoint, so this amounts in $3\times 420$ possibilities for ending up with $4$ balls in one of the boxes.
Number of possibilities to end up with $2$ balls in box1 and $3$ balls in the other boxes: $\binom{8}{2}\binom{6}{3}\binom{3}{3}=\frac{8!}{2!3!3!}=560$.
The same for $2$ balls in box2 and $3$ in box1 and box3. Also the same for $2$ balls in box3 and $3$ in box1 and box2. The events are disjoint, so this amounts in $3\times 560$ possibilities for ending up with $2$ balls in one of the boxes and $3$ in the other boxes.
• This seems to work for smaller cases! ... but could you elaborate more? If $\{A, B\}, \{C,D\}, \{E,F,G,H\}$ is a possible partitioning, then why not multiply by $3!$ to permute the partitions? Aug 16 '14 at 8:56
• I must think it over and do not exclude that I made a mistake. After $5$ minutes I will delete. Later I hope to come back. Aug 16 '14 at 9:05
• $\left\{ \left\{ A,B\right\} ,\left\{ C,D\right\} ,\left\{ E,F,G,H\right\} \right\}$ and $\left\{ \left\{ C,D\right\} ,\left\{ A,B\right\} ,\left\{ E,F,G,H\right\} \right\}$ are allready different choices. However $\left\{ \left\{ C,D\right\} ,\left\{ A,B\right\} ,\left\{ E,F,G,H\right\} \right\}$ also appears as rearrangement of $\left\{ \left\{ A,B\right\} ,\left\{ C,D\right\} ,\left\{ E,F,G,H\right\} \right\}$. So it is counted double. Aug 16 '14 at 10:50
Associate a different colour to each of the three boxes; then you are looking for the number of ways to colour the balls such that the frequency vectors of colours used is one of $$(2,2,4), (2,3,3), (2,4,2), (3,2,3), (3,3,2), (4,2,2)$$ The number for each of these colourings is given by a trinomial coefficient, and by symmtry one can simply compute $$3\binom8{2,2,4}+3\binom8{2,3,3} = 3\times(420+560) = 2940.$$
So basically your approach is correct, but your rather unclearly motivated multiplication by $3!=6$ is not right; rather, each pattern should be multiplied by the number of its distinct permutations, which happens to be$~3$ here both for $(2,2,4)$ and for $(2,3,3)$. |
# Basics of Set Theory – Part 2( Finite,Infinite & Equal Sets)
Finite and Infinite Sets
Let A = {1, 2, 3, 4, 5}, B = {a, b, c, d, e, g}
A set having definit number of elements is called a finite set and a set having indefinite number of elements is called infinite set. Empty set is a finite set.
Set A = {M, A, H, E, S, H} is finite set with 5 elements and Set B = { 2, 3, 5, 7, 11, 13} is finite set with 6 elements. Number of elements of a set means the number of distinct elements of the set and is denoted as n (S). If n (S)
is a natural number, then S is non-empty finite set.
Consider a Set C of all odd numbers. We see that the number of elements of this
set is not finite since there are infinite number of odd numbers. We say that the set
of odd numbers is an infinite set. The sets A and B given above are finite sets
and n(A) = 5, n(B) = 6 .
Example 6 State which of the following sets are finite or infinite :
(i) {x : x Î N and (x – 1) (x –2) = 0} N is set of natural numbers.
(ii) {x : x Î N and x2 = 4}
(iii) {x : x Î N and 2x –1 = 0}
(iv) {x : x Î N and x is prime}
(v) {x : x Î N and x is odd}
Solution (i) Given set = {1, 2}. Hence, it is finite.
(ii) Given set = {2}. Hence, it is finite.
(iii) Given set = f. Hence, it is finite.
(iv) The given set is the set of all prime numbers and since set of prime
numbers is infinite. Hence the given set is infinite
(v) Since there are infinite number of odd numbers, hence, the given set is
infinite.
Equal Sets
If each and every element of one set is same as that of the other set then both sets are equal.
We consider the following examples :
(i) Let A = {1, 2, 3, 4} and B = {3, 1, 4, 2}. Then A = B.
(ii) Let A be the set of prime numbers less than 6 and P the set of prime factors
of 30. Then A and P are equal, since 2, 3 and 5 are the only prime factors of
30 and also these are less than 6.
A set does not change if one or more elements of the set are repeated.
For example, the sets A = {1, 2, 3} and B = {2, 2, 1, 3, 3} are equal, since each
element of A is in B and vice-versa. That is why we generally do not repeat any
element in describing a set.
Example 7 Find the pairs of equal sets, if any, give reasons:
A = {0}, B = {x : x > 15 and x < 5},
C = {x : x – 5 = 0 }, D = {x: x2 = 25},
E = {x : x is an integral positive root of the equation x2 – 2x –15 = 0}.
Solution Since 0 Î A and 0 does not belong to any of the sets B, C, D and E, it
follows that, A ≠ B, A ≠ C, A ≠ D, A ≠≠ E.
Since B = f but none of the other sets are empty. Therefore B ≠ C, B ≠ D
and B ≠ E. Also C = {5} but –5 Î D, hence C ≠ D.
Since E = {5}, C = E. Further, D = {–5, 5} and E = {5}, we find that, D ≠ E.
Thus, the only pair of equal sets is C and E.
Example 8 Which of the following pairs of sets are equal? Justify your answer.
(i) X, the set of letters in “ALLOY” and B, the set of letters in “LOYAL”.
(ii) A = {n : n Î Z and n2 ≤ 4} where Z is set of all integers
and B = {x : x Î R and x2 – 3x + 2 = 0}.
Solution (i) We have, X = {A, L, L, O, Y}, B = {L, O, Y, A, L}. Then X and B are
equal sets as repetition of elements in a set do not change a set. Thus,
X = {A, L, O, Y} = B
(ii) A = {–2, –1, 0, 1, 2}, B = {1, 2}. Since 0 Î A and 0 Ï B, A and B are not equal sets.
EXERCISE
1. Which of the following are examples of the null set
(i) Set of odd natural numbers divisible by 2
(ii) Set of even prime numbers
(iii) { x : x is a natural numbers, x < 5 and x > 7 }
(iv) { y : y is a point common to any two parallel lines}
2. Which of the following sets are finite or infinite
(i) The set of months of a year
(ii) {1, 2, 3, . . .}
(iii) {1, 2, 3, . . .99, 100}
(iv) The set of positive integers greater than 100
(v) The set of prime numbers less than 99
3. State whether each of the following set is finite or infinite:
(i) The set of lines which are parallel to the x-axis
(ii) The set of letters in the English alphabet
(iii) The set of numbers which are multiple of 5
(iv) The set of animals living on the earth
(v) The set of circles passing through the origin (0,0)
4. In the following, state whether A = B or not:
(i) A = { a, b, c, d } B = { d, c, b, a }
(ii) A = { 4, 8, 12, 16 } B = { 8, 4, 16, 18}
(iii) A = {2, 4, 6, 8, 10} B = { x : x is positive even integer and x ≤ 10}
(iv) A = { x : x is a multiple of 10}, B = { 10, 15, 20, 25, 30, . . . }
5. Are the following pair of sets equal ? Give reasons.
(i) A = {2, 3}, B = {x : x is solution of x2 + 5x + 6 = 0}
(ii) A = { x : x is a letter in the word FOLLOW}
B = { y : y is a letter in the word WOLF}
6. From the sets given below, select equal sets :
A = { 2, 4, 8, 12}, B = { 1, 2, 3, 4}, C = { 4, 8, 12, 14}, D = { 3, 1, 4, 2}
E = {–1, 1}, F = { 0, a}, G = {1, –1}, H = { 0, 1}
Updated: June 6, 2014 — 3:24 pm |
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# Sample Solution Paper 2 - Term- 1 Mathematics, Class 6 Class 6 Notes | EduRev
## Class 6 : Sample Solution Paper 2 - Term- 1 Mathematics, Class 6 Class 6 Notes | EduRev
``` Page 1
CBSE VI | Mathematics
Sample Paper 2 – Solution
CBSE Board
Class VI Mathematics
Term I
Sample Paper 2 – Solution
Time: 2 ½ hours Total Marks: 80
Section A
Arrange the numbers in place-value chart:
Cr T L L T Th Th H T O
4 7 8 9 6 3 0 4
4 7 8 9 6 3 4 0
Clearly both numbers have 8 digits.
At crores, ten lakhs, lakhs, ten thousands, thousands and hundeds place both have
the same digits i.e. 4, 7, 8, 9, 6, 3 respectively.
But at tens place, first number has 0 and second number has 4.
Clearly, 0 < 4
Hence
47896304 < 47896340
To add 0 and 4 on number line, move 4 steps to the right of 0.
Estimate the product by rounding off 52 to its nearest tens and 188 to its nearest
hundreds.
52 can be rounded off to its nearest tens as 50 and 188 can be rounded off to its
nearest hundreds as 200.
So, the required estimation of the product is 50 × 200 = 10000
Since,
Therefore, 36 = 2 × 2 × 3 × 3
Page 2
CBSE VI | Mathematics
Sample Paper 2 – Solution
CBSE Board
Class VI Mathematics
Term I
Sample Paper 2 – Solution
Time: 2 ½ hours Total Marks: 80
Section A
Arrange the numbers in place-value chart:
Cr T L L T Th Th H T O
4 7 8 9 6 3 0 4
4 7 8 9 6 3 4 0
Clearly both numbers have 8 digits.
At crores, ten lakhs, lakhs, ten thousands, thousands and hundeds place both have
the same digits i.e. 4, 7, 8, 9, 6, 3 respectively.
But at tens place, first number has 0 and second number has 4.
Clearly, 0 < 4
Hence
47896304 < 47896340
To add 0 and 4 on number line, move 4 steps to the right of 0.
Estimate the product by rounding off 52 to its nearest tens and 188 to its nearest
hundreds.
52 can be rounded off to its nearest tens as 50 and 188 can be rounded off to its
nearest hundreds as 200.
So, the required estimation of the product is 50 × 200 = 10000
Since,
Therefore, 36 = 2 × 2 × 3 × 3
CBSE VI | Mathematics
Sample Paper 2 – Solution
13 + (12 – 6 × 3) = 13 + (12 – 18) = 13 – 6 = 7
NO and PQ can be extended indefinitely on both sides, so they are lines. On extending,
it can be seen that they would meet at a point. Hence, they are intersecting lines.
The number just before 1000000 is one less than 1000000.
The required number = 1000000 – 1 = 999999.
On a number line, –110 lies next to –111 on the right.
Therefore, the successor of –111 is –110.
The given fraction is
Dividing the numerator and denominator by 3, we get
Thus,
is equivalent to .
The numbers 138 and 432 are divisible by both 2 and 3 and hence by 6.
The number 653 is neither divisible by 3 nor by 2 and hence not by 6.
Now, consider the number 531.
Since, the sum of the digits of the number 531 is divisible by 3, so 531 is divisible by 3.
But it is not an even number, so it is not divisible by 2.
Thus, 531 is divisible by 3 but not by 6.
(i) ? A, ? B (ii) ? B, ? C (iii) ? C, ? D (iv) ? D, ? A.
Page 3
CBSE VI | Mathematics
Sample Paper 2 – Solution
CBSE Board
Class VI Mathematics
Term I
Sample Paper 2 – Solution
Time: 2 ½ hours Total Marks: 80
Section A
Arrange the numbers in place-value chart:
Cr T L L T Th Th H T O
4 7 8 9 6 3 0 4
4 7 8 9 6 3 4 0
Clearly both numbers have 8 digits.
At crores, ten lakhs, lakhs, ten thousands, thousands and hundeds place both have
the same digits i.e. 4, 7, 8, 9, 6, 3 respectively.
But at tens place, first number has 0 and second number has 4.
Clearly, 0 < 4
Hence
47896304 < 47896340
To add 0 and 4 on number line, move 4 steps to the right of 0.
Estimate the product by rounding off 52 to its nearest tens and 188 to its nearest
hundreds.
52 can be rounded off to its nearest tens as 50 and 188 can be rounded off to its
nearest hundreds as 200.
So, the required estimation of the product is 50 × 200 = 10000
Since,
Therefore, 36 = 2 × 2 × 3 × 3
CBSE VI | Mathematics
Sample Paper 2 – Solution
13 + (12 – 6 × 3) = 13 + (12 – 18) = 13 – 6 = 7
NO and PQ can be extended indefinitely on both sides, so they are lines. On extending,
it can be seen that they would meet at a point. Hence, they are intersecting lines.
The number just before 1000000 is one less than 1000000.
The required number = 1000000 – 1 = 999999.
On a number line, –110 lies next to –111 on the right.
Therefore, the successor of –111 is –110.
The given fraction is
Dividing the numerator and denominator by 3, we get
Thus,
is equivalent to .
The numbers 138 and 432 are divisible by both 2 and 3 and hence by 6.
The number 653 is neither divisible by 3 nor by 2 and hence not by 6.
Now, consider the number 531.
Since, the sum of the digits of the number 531 is divisible by 3, so 531 is divisible by 3.
But it is not an even number, so it is not divisible by 2.
Thus, 531 is divisible by 3 but not by 6.
(i) ? A, ? B (ii) ? B, ? C (iii) ? C, ? D (iv) ? D, ? A.
CBSE VI | Mathematics
Sample Paper 2 – Solution
Section B
13.
T Cr Cr T L L T Th Th H T O
(i)
7 0 7 0 7 5
(ii)
5 3 6 1 8 4 9 3
i. Seven lakh seven thousand seventy five.
ii. Five crore thirty-six lakh eighteen thousand four hundred ninety three.
14.
1. Two lines in the same plane which never intersect are called parallel lines.
2. Parallel lines remain the same distance apart over their entire length.
15.
i. Going 6 m to the West
ii. A withdrawal of Rs 100
iii. 10 km below sea level
iv. Spending Rs 500
16. Population of the village = 13295
Increase in population= Average growth - 1 = 399.
Population in the successive year = 13295 + 399 = 13694
17.
Prime factorisation of 455 is 5 × 7 × 13
Therefore, the dimensions of the cuboid are 5 cm, 7 cm, 13 cm.
Page 4
CBSE VI | Mathematics
Sample Paper 2 – Solution
CBSE Board
Class VI Mathematics
Term I
Sample Paper 2 – Solution
Time: 2 ½ hours Total Marks: 80
Section A
Arrange the numbers in place-value chart:
Cr T L L T Th Th H T O
4 7 8 9 6 3 0 4
4 7 8 9 6 3 4 0
Clearly both numbers have 8 digits.
At crores, ten lakhs, lakhs, ten thousands, thousands and hundeds place both have
the same digits i.e. 4, 7, 8, 9, 6, 3 respectively.
But at tens place, first number has 0 and second number has 4.
Clearly, 0 < 4
Hence
47896304 < 47896340
To add 0 and 4 on number line, move 4 steps to the right of 0.
Estimate the product by rounding off 52 to its nearest tens and 188 to its nearest
hundreds.
52 can be rounded off to its nearest tens as 50 and 188 can be rounded off to its
nearest hundreds as 200.
So, the required estimation of the product is 50 × 200 = 10000
Since,
Therefore, 36 = 2 × 2 × 3 × 3
CBSE VI | Mathematics
Sample Paper 2 – Solution
13 + (12 – 6 × 3) = 13 + (12 – 18) = 13 – 6 = 7
NO and PQ can be extended indefinitely on both sides, so they are lines. On extending,
it can be seen that they would meet at a point. Hence, they are intersecting lines.
The number just before 1000000 is one less than 1000000.
The required number = 1000000 – 1 = 999999.
On a number line, –110 lies next to –111 on the right.
Therefore, the successor of –111 is –110.
The given fraction is
Dividing the numerator and denominator by 3, we get
Thus,
is equivalent to .
The numbers 138 and 432 are divisible by both 2 and 3 and hence by 6.
The number 653 is neither divisible by 3 nor by 2 and hence not by 6.
Now, consider the number 531.
Since, the sum of the digits of the number 531 is divisible by 3, so 531 is divisible by 3.
But it is not an even number, so it is not divisible by 2.
Thus, 531 is divisible by 3 but not by 6.
(i) ? A, ? B (ii) ? B, ? C (iii) ? C, ? D (iv) ? D, ? A.
CBSE VI | Mathematics
Sample Paper 2 – Solution
Section B
13.
T Cr Cr T L L T Th Th H T O
(i)
7 0 7 0 7 5
(ii)
5 3 6 1 8 4 9 3
i. Seven lakh seven thousand seventy five.
ii. Five crore thirty-six lakh eighteen thousand four hundred ninety three.
14.
1. Two lines in the same plane which never intersect are called parallel lines.
2. Parallel lines remain the same distance apart over their entire length.
15.
i. Going 6 m to the West
ii. A withdrawal of Rs 100
iii. 10 km below sea level
iv. Spending Rs 500
16. Population of the village = 13295
Increase in population= Average growth - 1 = 399.
Population in the successive year = 13295 + 399 = 13694
17.
Prime factorisation of 455 is 5 × 7 × 13
Therefore, the dimensions of the cuboid are 5 cm, 7 cm, 13 cm.
CBSE VI | Mathematics
Sample Paper 2 – Solution
18. The opposite sides of a parallelogram are parallel and equal.
Therefore, LM = NO
19. 90, 91, 92, 93, 94, 95, 96 are the required numbers.
20. Number of circles in step 1 = 3 = 1 × 2 + 1
Number of circles in step 2 = 5 = 2 × 2 + 1
Thus, we can observe that the number of circles is obtained by multiplying the step
number by 2 and then adding 1.
Therefore, number of circles in the 100
th
step = (100 × 2) + 1 = 201
21.
20570 = 2 × 5 × 11 × 11 × 17
22. Anna is 7 feet above sea level.
She jumps 3 feet down and walks another 2 feet down. Total distance travelled
downwards = 3 + 2 = 5 feet.
23. (–13) + (–19) + (+15) + (–10)
= –13 – 19 + 15 – 10
= –13 – 19 – 10 + 15
= –42 + 15
= –27
24. 21397 can be estimated as 21000
27807 can be estimated as 28000
42305 can be estimated as 42000
On adding, we get 21000 + 28000 + 42000 = 91000
Page 5
CBSE VI | Mathematics
Sample Paper 2 – Solution
CBSE Board
Class VI Mathematics
Term I
Sample Paper 2 – Solution
Time: 2 ½ hours Total Marks: 80
Section A
Arrange the numbers in place-value chart:
Cr T L L T Th Th H T O
4 7 8 9 6 3 0 4
4 7 8 9 6 3 4 0
Clearly both numbers have 8 digits.
At crores, ten lakhs, lakhs, ten thousands, thousands and hundeds place both have
the same digits i.e. 4, 7, 8, 9, 6, 3 respectively.
But at tens place, first number has 0 and second number has 4.
Clearly, 0 < 4
Hence
47896304 < 47896340
To add 0 and 4 on number line, move 4 steps to the right of 0.
Estimate the product by rounding off 52 to its nearest tens and 188 to its nearest
hundreds.
52 can be rounded off to its nearest tens as 50 and 188 can be rounded off to its
nearest hundreds as 200.
So, the required estimation of the product is 50 × 200 = 10000
Since,
Therefore, 36 = 2 × 2 × 3 × 3
CBSE VI | Mathematics
Sample Paper 2 – Solution
13 + (12 – 6 × 3) = 13 + (12 – 18) = 13 – 6 = 7
NO and PQ can be extended indefinitely on both sides, so they are lines. On extending,
it can be seen that they would meet at a point. Hence, they are intersecting lines.
The number just before 1000000 is one less than 1000000.
The required number = 1000000 – 1 = 999999.
On a number line, –110 lies next to –111 on the right.
Therefore, the successor of –111 is –110.
The given fraction is
Dividing the numerator and denominator by 3, we get
Thus,
is equivalent to .
The numbers 138 and 432 are divisible by both 2 and 3 and hence by 6.
The number 653 is neither divisible by 3 nor by 2 and hence not by 6.
Now, consider the number 531.
Since, the sum of the digits of the number 531 is divisible by 3, so 531 is divisible by 3.
But it is not an even number, so it is not divisible by 2.
Thus, 531 is divisible by 3 but not by 6.
(i) ? A, ? B (ii) ? B, ? C (iii) ? C, ? D (iv) ? D, ? A.
CBSE VI | Mathematics
Sample Paper 2 – Solution
Section B
13.
T Cr Cr T L L T Th Th H T O
(i)
7 0 7 0 7 5
(ii)
5 3 6 1 8 4 9 3
i. Seven lakh seven thousand seventy five.
ii. Five crore thirty-six lakh eighteen thousand four hundred ninety three.
14.
1. Two lines in the same plane which never intersect are called parallel lines.
2. Parallel lines remain the same distance apart over their entire length.
15.
i. Going 6 m to the West
ii. A withdrawal of Rs 100
iii. 10 km below sea level
iv. Spending Rs 500
16. Population of the village = 13295
Increase in population= Average growth - 1 = 399.
Population in the successive year = 13295 + 399 = 13694
17.
Prime factorisation of 455 is 5 × 7 × 13
Therefore, the dimensions of the cuboid are 5 cm, 7 cm, 13 cm.
CBSE VI | Mathematics
Sample Paper 2 – Solution
18. The opposite sides of a parallelogram are parallel and equal.
Therefore, LM = NO
19. 90, 91, 92, 93, 94, 95, 96 are the required numbers.
20. Number of circles in step 1 = 3 = 1 × 2 + 1
Number of circles in step 2 = 5 = 2 × 2 + 1
Thus, we can observe that the number of circles is obtained by multiplying the step
number by 2 and then adding 1.
Therefore, number of circles in the 100
th
step = (100 × 2) + 1 = 201
21.
20570 = 2 × 5 × 11 × 11 × 17
22. Anna is 7 feet above sea level.
She jumps 3 feet down and walks another 2 feet down. Total distance travelled
downwards = 3 + 2 = 5 feet.
23. (–13) + (–19) + (+15) + (–10)
= –13 – 19 + 15 – 10
= –13 – 19 – 10 + 15
= –42 + 15
= –27
24. 21397 can be estimated as 21000
27807 can be estimated as 28000
42305 can be estimated as 42000
On adding, we get 21000 + 28000 + 42000 = 91000
CBSE VI | Mathematics
Sample Paper 2 – Solution
Section C
25. C stands for 100
D stands for 500
V stands for 5
I stands for 1
X stands for 10
M stands for 1000
In ascending order, the numbers can be arranged as
1 < 5 < 10 < 100 < 500 < 1000
Thus, the given roman numerals can be arranged in ascending order as
I, V, X, C, D, M
26. First we find the LCM of 48, 60, 72.
48 = 2 × 2 × 2 × 2 × 3
60 = 2 × 2 × 3 × 5
72 = 2 × 2 × 2 × 3 × 3
LCM = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720
Hence, they will meet after = 2 rounds.
27. The given fractions are .
LCM of 2, 3, 6, 9 = (2 x 3 x 3) = 18
So, we convert each of the given fractions into an equivalent fraction with 18 as the
denominator.
Thus, we have:
Hence, the like fractions are
```
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## Mathematics (Maths) Class 6
191 videos|230 docs|43 tests
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Prove the following trigonometric identities.
Question:
Prove the following trigonometric identities.
(i) $\frac{1+\cos \theta+\sin \theta}{1+\cos \theta-\sin \theta}=\frac{1+\sin \theta}{\cos \theta}$
(ii) $\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta}$
(iii) $\frac{\cos \theta-\sin \theta+1}{\cos \theta+\sin \theta-1}=\operatorname{cosec} \theta+\cot \theta$
(iv) $(\sin \theta+\cos \theta)(\tan \theta+\cot \theta)=\sec \theta+\operatorname{cosec} \theta$
Solution:
(i) We have to prove the following identity-
$\frac{1+\cos \theta+\sin \theta}{1+\cos \theta-\sin \theta}=\frac{1+\sin \theta}{\cos \theta}$
Consider the LHS.
$\frac{1+\cos \theta+\sin \theta}{1+\cos \theta-\sin \theta}$
$=\left(\frac{1+\cos \theta+\sin \theta}{1+\cos \theta-\sin \theta}\right)\left(\frac{1+\cos \theta+\sin \theta}{1+\cos \theta+\sin \theta}\right)$
$=\frac{(1+\cos \theta+\sin \theta)^{2}}{(1+\cos \theta)^{2}-\sin ^{2} \theta}$
$=\frac{2+2(\cos \theta+\sin \theta+\sin \theta \cos \theta)}{2 \cos ^{2} \theta+2 \cos \theta}$
$=\frac{2(1+\cos \theta)(1+\sin \theta)}{2 \cos \theta(1+\cos \theta)}$
$=\frac{1+\sin \theta}{\cos \theta}$
= RHS
Hence proved.
(ii) We have to prove the following identity-
$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta}$
Consider the LHS.
$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}$
$=\left(\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}\right)\left(\frac{\sin \theta+\cos \theta+1}{\sin \theta+\cos \theta+1}\right)$
$=\frac{(\sin \theta+1)^{2}-\cos ^{2} \theta}{(\sin \theta+\cos \theta)^{2}-1}$
$=\frac{2 \sin ^{2} \theta+2 \sin \theta}{2 \sin \theta \cos \theta}$
$=\frac{2 \sin \theta(1+\sin \theta)}{2 \sin \theta \cos \theta}$
$=\frac{1+\sin \theta}{\cos \theta}$
$=\left(\frac{1+\sin \theta}{\cos \theta}\right)\left(\frac{1-\sin \theta}{1-\sin \theta}\right)$
$=\frac{\cos \theta}{1-\sin \theta}$
$=\frac{1}{\sec \theta-\tan \theta}$ (Divide numerator and denominator by $\cos \theta$ )
RHS
Hence proved.
(iii) We have to prove the following identity-
$\frac{\cos \theta-\sin \theta+1}{\cos \theta+\sin \theta-1}=\operatorname{cosec} \theta+\cot \theta$
Consider the LHS.
$\frac{\cos \theta-\sin \theta+1}{\cos \theta+\sin \theta-1}$
$=\frac{\cos \theta-\sin \theta+1}{\cos \theta+\sin \theta-1} \times \frac{\cos \theta+\sin \theta+1}{\cos \theta+\sin \theta+1}$
$=\frac{(\cos \theta+1)^{2}-(\sin \theta)^{2}}{(\cos \theta+\sin \theta)^{2}-(1)^{2}}$
$=\frac{\cos ^{2} \theta+1+2 \cos \theta-\sin ^{2} \theta}{\cos ^{2} \theta+\sin ^{2} \theta+2 \cos \theta \sin \theta-1}$
$=\frac{\cos ^{2} \theta+1+2 \cos \theta-\left(1-\cos ^{2} \theta\right)}{1+2 \cos \theta \sin \theta-1}$
$=\frac{2 \cos ^{2} \theta+2 \cos \theta}{2 \cos \theta \sin \theta}$
$=\frac{2 \cos \theta(\cos \theta+1)}{2 \cos \theta \sin \theta}$
$=\frac{\cos \theta+1}{\sin \theta}$
$=\frac{\cos \theta}{\sin \theta}+\frac{1}{\sin \theta}$
$=\cot \theta+\operatorname{cosec} \theta$
= RHS
Hence proved.
(iv)
Consider the LHS.
$(\sin \theta+\cos \theta)(\tan \theta+\cot \theta)$
$=(\sin \theta+\cos \theta)\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)$
$=(\sin \theta+\cos \theta)\left(\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \times \cos \theta}\right)$
$=\frac{\sin \theta+\cos \theta}{\sin \theta \times \cos \theta} \quad\left[\sin ^{2} \theta+\cos ^{2} \theta=1\right]$
$=\frac{1}{\cos \theta}+\frac{1}{\sin \theta}$
$=\sec \theta+\operatorname{cosec} \theta$
= RHS
Hence proved. |
6.1.5 Random Vectors
When dealing with multiple random variables, it is sometimes useful to use vector and matrix notations. This makes the formulas more compact and lets us use facts from linear algebra. In this section, we briefly explore this avenue. The reader should be familiar with matrix algebra before reading this section. When we have $n$ random variables $X_1$, $X_2$, ..., $X_n$ we can put them in a (column) vector $\textbf{X}$: $$\nonumber \textbf{X} = \begin{bmatrix} X_1 \\%[5pt] X_2 \\%[5pt] . \\[-10pt] . \\[-10pt] . \\[5pt] X_n \end{bmatrix}.$$ We call $\textbf{X}$ a random vector. Here $\textbf{X}$ is an $n$-dimensional vector because it consists of $n$ random variables. In this book, we usually use bold capital letters such as $\mathbf{X,Y}$ and $\mathbf{Z}$ to represent a random vector. To show a possible value of a random vector we usually use bold lowercase letters such as x, y and z. Thus, we can write the CDF of the random vector $\textbf{X}$ as \begin{align} F_{\mathbf{\large{X}}}(\textbf{x})&=F_{X_{\large{1}}, X_{\large{2}}, ..., X_{\large{n}}}(x_1, x_2, ..., x_n) \\ &=P(X_1 \leq x_1, X_2 \leq x_2, ..., X_n \leq x_n). \end{align} If the $X_i$'s are jointly continuous, the PDF of X can be written as \begin{align} f_{\mathbf{\large{X}}}(\textbf{x})=f_{X_{\large{1}}, X_{\large{2}}, ..., X_{\large{n}}}(x_1, x_2,..., x_n). \end{align}
Expectation:
The expected value vector or the mean vector of the random vector X is defined as \begin{align} \nonumber E\mathbf{X} = \begin{bmatrix} EX_1 \\%[5pt] EX_2 \\%[5pt] . \\[-10pt] . \\[-10pt] . \\[5pt] EX_n \end{bmatrix}. \end{align} Similarly, a random matrix is a matrix whose elements are random variables. In particular, we can have an $m$ by $n$ random matrix M as \begin{align} \nonumber \mathbf{M} = \begin{bmatrix} X_{11} & X_{12} & ... & X_{1n} \\%[5pt] X_{21} & X_{22} & ... & X_{2n} \\%[5pt] . & . & . & .\\[-10pt] . & . & . & . \\[-10pt] . & . & . & . \\[5pt] X_{m1} & X_{m2} & ... & X_{mn} \end{bmatrix}. \end{align} We sometimes write this as M$=[X_{ij}]$, which means that $X_{ij}$ is the element in the $i$th row and $j$th column of M. The mean matrix of M is given by \begin{align} \nonumber E\mathbf{M} = \begin{bmatrix} EX_{11} & EX_{12} & ... & EX_{1n} \\%[5pt] EX_{21} & EX_{22} & ... & EX_{2n} \\%[5pt] . & . & . & .\\[-10pt] . & . & . & . \\[-10pt] . & . & . & . \\[5pt] EX_{m1} & EX_{m2} & ... & EX_{mn} \end{bmatrix}. \end{align} Linearity of expectation is also valid for random vectors and matrices. In particular, let X be an $n$-dimensional random vector and the random vector Y be defined as \begin{align} \mathbf{Y}=\mathbf{AX+b}, \end{align} where A is a fixed (non-random) $m$ by $n$ matrix and b is a fixed $m$-dimensional vector. Then we have \begin{align} E\mathbf{Y}=\mathbf{AEX+b}. \end{align} Also, if $\mathbf{X_1, X_2, \cdots, X_k}$ are $n$-dimensional random vectors, then we have \begin{align} \mathbf{E[X_1+X_2+ \cdots +X_k]=EX_1+EX_2+ \cdots +EX_k}. \end{align}
Correlation and Covariance Matrix
For a random vector $\mathbf{X}$, we define the correlation matrix, $\mathbf{R_X}$, as \begin{align} \nonumber \mathbf{R_X=E[XX^{T}]} = E \begin{bmatrix} X_1^2 & X_1 X_2 & ... & X_1 X_n \\%[5pt] X_2 X_1 & X_2^2 & ... & X_2 X_n \\%[5pt] . & . & . & .\\[-10pt] . & . & . & . \\[-10pt] . & . & . & . \\[5pt] X_n X_1 & X_n X_2 & ... & X_n^2 \end{bmatrix} = \begin{bmatrix} EX_1^2 & E[X_1 X_2] & ... & E[X_1 X_n] \\%[5pt] EX_2 X_1 & E[X_2^2] & ... & E[X_2 X_n] \\%[5pt] . & . & . & .\\[-10pt] . & . & . & . \\[-10pt] . & . & . & . \\[5pt] E[X_n X_1] & E[X_n X_2] & ... & E[X_n^2] \end{bmatrix}, \end{align} where $^T$ shows matrix transposition.
The covariance matrix, $C_X$, is defined as \begin{align} \nonumber \mathbf{C_X}&=\mathbf{E[(X-EX)(X-EX)^{T}]} \\ \nonumber \\ \nonumber &= E \begin{bmatrix} (X_1-EX_1)^2 & (X_1-EX_1)(X_2-EX_2) & ... & (X_1-EX_1)(X_n-EX_n) \\%[5pt] (X_2-EX_2)(X_1-EX_1) & (X_2-EX_2)^2 & ... & (X_2-EX_2) (X_n-EX_n) \\%[5pt] . & . & . & .\\[-10pt] . & . & . & . \\[-10pt] . & . & . & . \\[5pt] (X_n-EX_n) (X_1-EX_1) & (X_n-EX_n) (X_2-EX_2) & ... & (X_n-EX_n)^2 \end{bmatrix} \\
\nonumber &= \begin{bmatrix} \textrm{Var}(X_1) & \textrm{Cov}(X_1,X_2)& ... & \textrm{Cov}(X_1,X_n) \\%[5pt] \textrm{Cov}(X_2,X_1) & \textrm{Var}(X_2) & ... & \textrm{Cov}(X_2,X_n) \\%[5pt] . & . & . & .\\[-10pt] . & . & . & . \\[-10pt] . & . & . & . \\[5pt] \textrm{Cov}(X_n,X_1) & \textrm{Cov}(X_n X_2) & ... & \textrm{Var}(X_n) \end{bmatrix}. \end{align} The covariance matrix is a generalization of the variance of a random variable. Remember that for a random variable, we have $\textrm{Var}(X)=EX^2-(EX)^2$. The following example extends this formula to random vectors.
Example
For a random vector X, show \begin{align} \mathbf{C_X}= \mathbf{R_X-EX EX^T}. \end{align}
• Solution
• We have \begin{align} \nonumber \mathbf{C_X}&=\mathbf{E[(X-EX)(X-EX)^{T}]} \\ \nonumber &=\mathbf{E[(X-EX)(X^T-EX^T)]} \\ \nonumber &=\mathbf{E[X X^T]-EX EX^T-EX EX^T+EX EX^T} \hspace{10pt} \textrm{(by linearity of expectation)} \\ \nonumber &=\mathbf{R_X-EX EX^T}. \end{align}
\begin{align} &\textrm{Correlation matrix of X: } \\ \nonumber & \hspace{80pt} \mathbf{R_X=E[X X^{T}]} \\ &\textrm{Covariance matrix of X: } \\ \nonumber & \hspace{80pt} \mathbf{C_X=E[(X-EX)(X-EX)^{T}]=R_X-EX EX^T} \end{align}
Example
Let X be an $n$-dimensional random vector and the random vector Y be defined as \begin{align}%\label{} \mathbf{Y=A X+b}, \end{align} where A is a fixed $m$ by $n$ matrix and b is a fixed $m$-dimensional vector. Show that \begin{align} \mathbf{C_Y=A C_X A^T}. \end{align}
• Solution
• Note that by linearity of expectation, we have \begin{align} \mathbf{EY=A EX+b}. \end{align} By definition, we have \begin{align} \mathbf{C_Y}&=\mathbf{E[(Y-EY)(Y-EY)^{T}]}\\ &=\mathbf{E[(A X+b-A EX-b)(AX+b-A EX-b)^T]}\\ &=E\mathbf{[A(X-EX)(X-EX)^T A^T ]}\\ &=\mathbf{AE[(X-EX)(X-EX)^T] A^T} &\textrm{(by linearity of expectation)}\\ &= \mathbf{A C_X A^T}. \end{align}
Example
Let $X$ and $Y$ be two jointly continuous random variables with joint PDF $$\nonumber f_{X,Y}(x,y) = \left\{ \begin{array}{l l} \frac{3}{2}x^2+y & \quad 0<x,y<1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right.$$ and let the random vector U be defined as $$\nonumber \mathbf{U} = \begin{bmatrix} X \\%[5pt] Y \end{bmatrix}.$$ Find the correlation and covariance matrices of U.
• Solution
• We first obtain the marginal PDFs of $X$ and $Y$. Note that $R_X=R_Y=(0,1)$. We have for $x \in R_X$ \begin{align} f_X(x)&=\int_{0}^{1} \frac{3}{2}x^2+y \hspace{5pt}dy\\ &=\frac{3}{2}x^2+\frac{1}{2}, \hspace{10pt} \textrm{for $0<x<1$}. \end{align} Similarly, for $y \in R_Y$, we have \begin{align} f_Y(y)&=\int_{0}^{1} \frac{3}{2}x^2+y \hspace{5pt}dx\\ &=y+\frac{1}{2}, \hspace{10pt} \textrm{for $0<y<1$}. \end{align} From these, we obtain $EX=\frac{5}{8}$, $EX^2=\frac{7}{15}$, $EY=\frac{7}{12}$, and $EY^2=\frac{5}{12}$. We also need $EXY$. By LOTUS, we can write \begin{align} EXY&=\int_{0}^{1} \int_{0}^{1} xy \left(\frac{3}{2}x^2+y\right) dxdy\\ &= \int_{0}^{1} \frac{3}{8}y+\frac{1}{2}y^2 dy\\ &=\frac{17}{48}. \end{align} From this, we also obtain \begin{align} \textrm{Cov}(X,Y)&=EXY-EXEY\\ &=\frac{17}{48}-\frac{5}{8}.\frac{7}{12}\\ &=-\frac{1}{96}. \end{align} The correlation matrix $R_U$ is given by $$\nonumber \mathbf{R_U=E[UU^{T}]} = \begin{bmatrix} EX^2 & EXY \\%[5pt] EYX & EY^2 \end{bmatrix} = \begin{bmatrix} \frac{7}{15} & \frac{17}{48} \\[5pt] \frac{17}{48} & \frac{5}{12} \end{bmatrix}.$$ The covariance matrix $\mathbf{C_U}$ is given by $$\nonumber \mathbf{C_U} = \begin{bmatrix} \textrm{Var}(X) & \textrm{Cov}(X,Y) \\%[5pt] \textrm{Cov}(Y,X) & \textrm{Var}(Y) \end{bmatrix} = \begin{bmatrix} \frac{73}{960} & -\frac{1}{96} \\[5pt] -\frac{1}{96} & \frac{11}{144} \end{bmatrix}.$$
Properties of the Covariance Matrix:
The covariance matrix is the generalization of the variance to random vectors. It is an important matrix and is used extensively. Let's take a moment and discuss its properties. Here, we use concepts from linear algebra such as eigenvalues and positive definiteness. First note that, for any random vector X, the covariance matrix $\mathbf{C_X}$ is a symmetric matrix. This is because if $\mathbf{C_X}=[c_{ij}]$, then \begin{align}%\label{} c_{ij}=\textrm{Cov}(X_i,X_j)=\textrm{Cov}(X_j,X_i)=c_{ji}. \end{align} Thus, the covariance matrix has all the nice properties of symmetric matrices. In particular, $\mathbf{C_X}$ can be diagonalized and all the eigenvalues of $\mathbf{C_X}$ are real. Here, we assume X is a real random vector, i.e., the $X_i$'s can only take real values. A special important property of the covariance matrix is that it is positive semi-definite (PSD). Remember from linear algebra that a symmetric matrix M is positive semi-definite (PSD) if, for all vectors b, we have \begin{align} \mathbf{b^T M b} \geq 0. \end{align} Also, M is said to be positive definite (PD), if for all vectors b$\neq 0$, we have \begin{align} \mathbf{b^T M b} > 0. \end{align} By the above definitions, we note that every PD matrix is also PSD, but the converse is not generally true. Here, we show that covariance matrices are always PSD.
Theorem . Let $\mathbf{X}$ be a random vector with $n$ elements. Then, its covariance matrix $\mathbf{C_X}$ is positive semi-definite(PSD).
Proof. Let b be any fixed vector with $n$ elements. Define the random variable $Y$ as \begin{align} Y=\mathbf{b^T (X-EX)}. \end{align} We have \begin{align} 0 &\leq EY^2\\ &=E(YY^T)\\ &=\mathbf{b^T} E\big[\mathbf{(X-EX)(X-EX)^T}\big]\mathbf{b} \\ &=\mathbf{b^T C_X b}. \end{align} Note that the eigenvalues of a PSD matrix are always larger than or equal to zero. If all the eigenvalues are strictly larger than zero, then the matrix is positive definite. From linear algebra, we know that a real symmetric matrix is positive definite if and only if all its eigenvalues are positive. Since $\mathbf{C_X}$ is a real symmetric matrix, we can state the following theorem.
Theorem . Let $\mathbf{X}$ be a random vector with $n$ elements. Then its covariance matrix $\mathbf{C_X}$ is positive definite (PD), if and only if all its eigenvalues are larger than zero. Equivalently, $\mathbf{C_X}$ is positive definite (PD), if and only if $\det(\mathbf{C_X})>0$.
Note that the second part of the theorem is implied by the first part. This is because the determinant of a matrix is the product of its eigenvalues, and we already know that all eigenvalues of $\mathbf{C_X}$ are larger than or equal to zero.
Example
Let $X$ and $Y$ be two independent $Uniform(0,1)$ random variables. Let the random vectors U and V be defined as $$\nonumber \mathbf{U} = \begin{bmatrix} X \\%[5pt] X+Y \end{bmatrix}, \hspace{5pt} \mathbf{V} = \begin{bmatrix} X \\%[5pt] Y \\ X+Y \end{bmatrix}.$$ Determine whether $\mathbf{C_U}$ and $\mathbf{C_V}$ are positive definite.
• Solution
• Let us first find $\mathbf{C_U}$. We have $$\nonumber \mathbf{C_U} = \begin{bmatrix} \textrm{Var}(X) & \textrm{Cov}(X,X+Y) \\%[5pt] \textrm{Cov}(X+Y,X) & \textrm{Var}(X+Y) \end{bmatrix}.$$ Since $X$ and $Y$ are independent $Uniform(0,1)$ random variables, we have \begin{align}%\label{} \textrm{Var}(X)=\textrm{Var}(&Y)=\frac{1}{12},\\ \textrm{Cov}(X,X+Y)&=\textrm{Cov}(X,X)+\textrm{Cov}(X,Y)\\ &=\frac{1}{12}+0=\frac{1}{12},\\ \textrm{Var}(X+Y)=&\textrm{Var}(X)+\textrm{Var}(Y)=\frac{1}{6}. \end{align} Thus, $$\nonumber \mathbf{C_U} = \begin{bmatrix} \frac{1}{12} & \frac{1}{12} \\[5pt] \frac{1}{12} & \frac{1}{6} \end{bmatrix}.$$ So we conclude \begin{align}%\label{} \det(\mathbf{C_U})&=\frac{1}{12} . \frac{1}{6} - \frac{1}{12} . \frac{1}{12}\\ &=\frac{1}{144}>0. \end{align} Therefore, $\mathbf{C_U}$ is positive definite. For $\mathbf{C_V}$, we have \begin{align} \nonumber \mathbf{C_V} &= \begin{bmatrix} \textrm{Var}(X) & \textrm{Cov}(X,Y) & \textrm{Cov}(X,X+Y) \\%[5pt] \textrm{Cov}(Y,X) & \textrm{Var}(Y) &\textrm{Cov}(Y,X+Y) \\ \textrm{Cov}(X+Y,X) &\textrm{Cov}(X+Y,Y) & \textrm{Var}(X+Y) \end{bmatrix}\\ &= \begin{bmatrix} \frac{1}{12} & 0 & \frac{1}{12} \\[5pt] 0 & \frac{1}{12} & \frac{1}{12} \\[5pt] \frac{1}{12} & \frac{1}{12} & \frac{1}{6} \end{bmatrix}. \end{align} So we conclude \begin{align}%\label{} \det(\mathbf{C_V})&=\frac{1}{12} \left(\frac{1}{12} \cdot \frac{1}{6} - \frac{1}{12} \cdot \frac{1}{12} \right)-0+ \frac{1}{12} \left(0 - \frac{1}{12} \cdot \frac{1}{12} \right)\\ &=0. \end{align} Thus, $\mathbf{C_V}$ is not positive definite (we already know that it is positive semi-definite).
Finally, if we have two random vectors, $\mathbf{X}$ and $\mathbf{Y}$, we can define the cross correlation matrix of $\mathbf{X}$ and $\mathbf{Y}$ as \begin{align} \nonumber \mathbf{R_{XY}=E[X Y^T]}. \end{align} Also, the cross covariance matrix of X and Y is \begin{align} \nonumber \mathbf{C_{XY}=E[(X-EX)(Y-EY)^T]}. \end{align}
Functions of Random Vectors: The Method of Transformations
A function of a random vector is a random vector. Thus, the methods that we discussed regarding functions of two random variables can be used to find distributions of functions of random vectors. For example, we can state a more general form of Theorem 5.1 (method of transformations). Let us first explain the method and then see some examples on how to use it. Let $\textbf{X}$ be an $n$-dimensional random vector with joint PDF $f_{\mathbf{X}}(\mathbf{x})$. Let $G:\mathbb{R}^n \mapsto \mathbb{R}^n$ be a continuous and invertible function with continuous partial derivatives and let $H=G^{-1}$. Suppose that the random vector $\mathbf{Y}$ is given by $\mathbf{Y}=G(\mathbf{X})$ and thus $\mathbf{X}=G^{-1}(\mathbf{Y})=H(\mathbf{Y})$. That is, $$\nonumber \mathbf{X} = \begin{bmatrix} X_1 \\%[5pt] X_2 \\%[5pt] . \\[-10pt] . \\[-10pt] . \\[5pt] X_n \end{bmatrix} =\begin{bmatrix} H_1(Y_1,Y_2,...,Y_n) \\%[5pt] H_2(Y_1,Y_2,...,Y_n) \\%[5pt] . \\[-10pt] . \\[-10pt] . \\[5pt] H_n(Y_1,Y_2,...,Y_n) \end{bmatrix}.$$ Then, the PDF of Y, $f_{Y_1,Y_2,...,Y_n}(y_1,y_2,...,y_n)$, is given by
\begin{align} \nonumber f_{\mathbf{Y}}(\mathbf{y})=f_{\mathbf{X}}\big(H(\mathbf{y})\big) |J| \end{align}
where $J$ is the Jacobian of $H$ defined by \begin{align} \nonumber J= \det \begin{bmatrix} \frac{\partial H_1}{\partial y_1} & \frac{\partial H_1}{\partial y_2} & ... & \frac{\partial H_1}{\partial y_n} \\[5pt] \frac{\partial H_2}{\partial y_1} & \frac{\partial H_2}{\partial y_2} & ... & \frac{\partial H_2}{\partial y_n} \\ \vdots &\vdots &\vdots &\vdots \\ \frac{\partial H_n}{\partial y_1} & \frac{\partial H_n}{\partial y_2} &... & \frac{\partial H_n}{\partial y_n} \\ \end{bmatrix}, \end{align} and evaluated at $(y_1,y_2,...,y_n)$.
Example
Let X be an $n$-dimensional random vector. Let A be a fixed (non-random) invertible $n$ by $n$ matrix, and b be a fixed $n$-dimensional vector. Define the random vector Y as \begin{align}%\label{} \mathbf{Y=AX+b}. \end{align} Find the PDF of $\mathbf{Y}$ in terms of PDF of $\mathbf{X}$.
• Solution
• Since $A$ is invertible, we can write \begin{align}%\label{} X=A^{-1}(Y-b). \end{align} We can also check that \begin{align} \nonumber J= \det (A^{-1})=\frac{1}{\det(A)}. \end{align} Thus, we conclude that \begin{align} f_{Y}(y)=\frac{1}{|\det(A)|}f_{X}\big(A^{-1}(y-b) \big). \end{align}
Normal (Gaussian) Random Vectors:
We discussed two jointly normal random variables previously in Section 5.3.2. In particular, two random variables $X$ and $Y$ are said to be bivariate normal or jointly normal, if $aX+bY$ has normal distribution for all $a,b \in \mathbb{R}$. We can extend this definition to $n$ jointly normal random variables.
Random variables $X_1$, $X_2$,..., $X_n$ are said to be jointly normal if, for all $a_1$,$a_2$,..., $a_n$ $\in \mathbb{R}$, the random variable \begin{align}%\label{} a_1X_1+a_2X_2+...+a_nX_n \end{align} is a normal random variable.
As before, we agree that the constant zero is a normal random variable with zero mean and variance, i.e., $N(0,0)$. When we have several jointly normal random variables, we often put them in a vector. The resulting random vector is a called a normal (Gaussian) random vector.
A random vector $$\nonumber \textbf{X} = \begin{bmatrix} X_1 \\%[5pt] X_2 \\%[5pt] . \\[-10pt] . \\[-10pt] . \\[5pt] X_n \end{bmatrix}$$ is said to be normal or Gaussian if the random variables $X_1$, $X_2$,..., $X_n$ are jointly normal.
To find the general form for the PDF of a Gaussian random vector it is convenient to start from the simplest case where the $X_i$'s are independent and identically distributed (i.i.d.), $X_i \sim N(0,1)$. In this case, we know how to find the joint PDF. It is simply the product of the individual (marginal) PDFs. Let's call such a random vector the standard normal random vector. So, let $$\nonumber \textbf{Z} = \begin{bmatrix} Z_1 \\%[5pt] Z_2 \\%[5pt] . \\[-10pt] . \\[-10pt] . \\[5pt] Z_n \end{bmatrix},$$ where $Z_i$'s are i.i.d. and $Z_i \sim N(0,1)$. Then, we have \begin{align}%\label{} f_{\mathbf{Z}}(\mathbf{z})&=f_{Z_1, Z_2, ..., Z_n}(z_1, z_2, ..., z_n)\\ &=\prod_{i=1}^{n} f_{Z_i}(z_i)\\ &=\frac{1}{(2\pi)^{\frac{n}{2}}} \exp \left\{-\frac{1}{2} \sum_{i=1}^{n} z_i^2 \right\}\\ &=\frac{1}{(2\pi)^{\frac{n}{2}}} \exp \left\{-\frac{1}{2} \mathbf{z}^T\mathbf{z} \right\}. \end{align}
For a standard normal random vector Z, where $Z_i$'s are i.i.d. and $Z_i \sim N(0,1)$, the PDF is given by \begin{align}%\label{} f_{\mathbf{Z}}(\mathbf{z})=\frac{1}{(2\pi)^{\frac{n}{2}}} \exp \left\{-\frac{1}{2} \mathbf{z}^T\mathbf{z} \right\}. \end{align}
Now, we need to extend this formula to a general normal random vector X with mean $\mathbf{m}$ and covariance matrix C. This is very similar to when we defined general normal random variables from the standard normal random variable. We remember that if $Z \sim N(0,1)$, then the random variable $X=\sigma Z+ \mu$ has $N(\mu, \sigma^2)$ distribution. We would like to do the same thing for normal random vectors.
Assume that I have a normal random vector X with mean $\mathbf{m}$ and covariance matrix C. We write $\mathbf{X} \sim N(\mathbf{m},\mathbf{C})$. Further, assume that $\mathbf{C}$ is a positive definite matrix.(The positive definiteness assumption here does not create any limitations. We already know that $\mathbf{C}$ is positive semi-definite (Theorem 6.2), so $\det(\textbf{C}) \geq 0$. We also know that $\mathbf{C}$ is positive definite if and only if $\det(\textbf{C})>0$ (Theorem 6.3). So here, we are only excluding the case $\det(\textbf{C})=0$. If $\det(\textbf{C})=0$, then you can show that you can write some $X_i$'s as a linear combination of others, so indeed we can remove them from the vector without losing any information.) Then from linear algebra we know that there exists an $n$ by $n$ matrix Q such that \begin{align}%\label{} &\textbf{Q}\textbf{Q}^T=\textbf{I} \quad (\textbf{I} \textrm{ is the identity matrix}),\\ &\textbf{C}= \textbf{Q} \textbf{D} \textbf{Q}^T, \end{align} where D is a diagonal matrix $$\nonumber D = \begin{bmatrix} d_{11} & 0 & ... & 0 \\%[5pt] 0 & d_{22} & ... & 0 \\%[5pt] . & . & . & .\\[-10pt] . & . & . & . \\[-10pt] . & . & . & . \\[5pt] 0 & 0 & ... & d_{nn} \end{bmatrix}.$$ The positive definiteness assumption guarantees that all $d_{ii}$'s are positive. Let's define $$\nonumber D^{\frac{1}{2}} = \begin{bmatrix} \sqrt{d_{11}} & 0 & ... & 0 \\%[5pt] 0 & \sqrt{d_{22}} & ... & 0 \\%[5pt] . & . & . & .\\[-10pt] . & . & . & . \\[-10pt] . & . & . & . \\[5pt] 0 & 0 & ... & \sqrt{d_{nn}} \end{bmatrix}.$$ We have $D^{\frac{1}{2}} D^{\frac{1}{2}}=\mathbf{D}$ and $D^{\frac{1}{2}}={D^{\frac{1}{2}}}^{T}$. Also define \begin{align}%\label{} \mathbf{A}= Q D^{\frac{1}{2}} Q^T. \end{align} Then, \begin{align}%\label{} \mathbf{A} \mathbf{A}^{T}= \mathbf{A}^{T}\mathbf{A}=\mathbf{C}. \end{align} Now we are ready to define the transformation that converts a standard Gaussian vector to $\mathbf{X} \sim N(\mathbf{m},\mathbf{C})$. Let Z be a standard Gaussian vector, i.e., $\mathbf{Z} \sim N(\mathbf{0},\mathbf{I})$. Define \begin{align}%\label{} \mathbf{X}=\mathbf{A}\mathbf{Z}+\mathbf{m}. \end{align} We claim that $\mathbf{X} \sim N(\mathbf{m},\mathbf{C})$. To see this, first note that $\mathbf{X}$ is a normal random vector. The reason is that any linear combination of components of $\mathbf{X}$ is indeed a linear combination of components of $\mathbf{Z}$ plus a constant. Thus, every linear combination of components of $\mathbf{X}$ is a normal random variable. It remains to show that $E\mathbf{X}=\mathbf{m}$ and $\mathbf{C_X}=\mathbf{C}$. First note that by linearity of expectation we have \begin{align}%\label{} E\mathbf{X} &= E \left[\mathbf{A}\mathbf{Z}+\mathbf{m}\right]\\ &= \mathbf{A} E[\mathbf{Z}] +\mathbf{m}\\ &= \mathbf{m}. \end{align} Also, by Example 6.12 we have \begin{align}%\label{} C_X&=A C_Z A^T\\ &=\mathbf{A} \mathbf{A}^{T} & (\textrm{since } C_Z=\mathbf{I})\\ &=\mathbf{C}. \end{align} Thus, we have shown that X is a random vector with mean $\mathbf{m}$ and covariance matrix C. Now we can use Example 6.15 to find the PDF of X. We have \begin{align} f_{\mathbf{X}}(\mathbf{x})&=\frac{1}{|\det(\mathbf{A})|}f_{\mathbf{Z}}\big(\mathbf{A}^{-1}(\mathbf{x}-\mathbf{m}) \big) \\ &=\frac{1}{(2\pi)^{\large{\frac{n}{2}}} |\det(\mathbf{A})|} \exp \left\{-\frac{1}{2} (\mathbf{A}^{-1}(\mathbf{x}-\mathbf{m}))^T(\mathbf{A}^{-1}(\mathbf{x}-\mathbf{m})) \right\}\\ &=\frac{1}{(2\pi)^{\large{\frac{n}{2}}} \sqrt{\det(\mathbf{C})}} \exp \left\{-\frac{1}{2} (\mathbf{x}-\mathbf{m})^T\mathbf{A}^{-T} \mathbf{A}^{-1}(\mathbf{x}-\mathbf{m}) \right\}\\ &=\frac{1}{(2\pi)^{\large{\frac{n}{2}}} \sqrt{\det(\mathbf{C})}} \exp \left\{-\frac{1}{2} (\mathbf{x}-\mathbf{m})^T \mathbf{C}^{-1}(\mathbf{x}-\mathbf{m}) \right\}. \end{align}
For a normal random vector $\mathbf{X}$ with mean $\mathbf{m}$ and covariance matrix $\mathbf{C}$, the PDF is given by \begin{align}\label{eq:pdf-normal-vec} f_{\mathbf{X}}(\mathbf{x})=\frac{1}{(2\pi)^{\large{\frac{n}{2}}} \sqrt{\det\textbf{C}}} \exp \left\{-\frac{1}{2} (\textbf{x}-\textbf{m})^T \mathbf{C}^{-1}(\textbf{x}-\textbf{m}) \right\} \hspace{20pt} (6.1) \end{align}
Example
Let $X$ and $Y$ be two jointly normal random variables with $X \sim N(\mu_X,\sigma_X)$, $Y \sim N(\mu_Y,\sigma_Y)$, and $\rho(X,Y)=\rho$. Show that the above PDF formula for PDF of $\begin{bmatrix} X \\%[5pt] Y \end{bmatrix}$ is the same as $f_{X,Y}(x,y)$ given in Definition 5.4 in Section 5.3.2. That is, \begin{align} \label{eq:bivariate-normal} f_{XY}(x,y)&=\frac{1}{2 \pi \sigma_X \sigma_Y \sqrt{1-\rho^2}} \cdot \\ & \exp \left\{-\frac{1}{2 (1-\rho^2)}\bigg[\bigg(\frac{x-\mu_X}{\sigma_X}\bigg)^2 +\bigg(\frac{y-\mu_Y}{\sigma_Y}\bigg)^2-2\rho \frac{(x-\mu_X)(y-\mu_Y)}{\sigma_X \sigma_Y} \bigg] \right\}. \end{align}
• Solution
• Both formulas are in the form $ae^{-\frac{1}{2}b}$. Thus, it suffices to show that they have the same $a$ and $b$. Here we have $$m = \begin{bmatrix} \mu_X \\%[5pt] \mu_Y \end{bmatrix}.$$ We also have $$\nonumber C = \begin{bmatrix} \textrm{Var}(X) & \textrm{Cov}(X,Y) \\%[5pt] \textrm{Cov}(Y,X) & \textrm{Var}(Y) \end{bmatrix}=\begin{bmatrix} \sigma^2_X & \rho \sigma_X \sigma_Y \\%[5pt] \rho \sigma_X \sigma_Y & \sigma^2_Y \end{bmatrix}.$$ From this, we obtain \begin{align}%\label{} \det \mathbf{C}= \sigma^2_X \sigma^2_Y(1-\rho^2). \end{align} Thus, in both formulas for PDF $a$ is given by \begin{align}%\label{} a= \frac{1}{2 \pi \sigma_X \sigma_Y\sqrt{1-\rho^2}}. \end{align} Next, we check $b$. We have $$\nonumber C^{-1} = \frac{1}{\sigma^2_X \sigma^2_Y(1-\rho^2)}\begin{bmatrix} \sigma^2_Y & -\rho \sigma_X \sigma_Y \\%[5pt] -\rho \sigma_X \sigma_Y & \sigma^2_X \end{bmatrix}.$$ Now by matrix multiplication we obtain \begin{align}%\label{} (x-m)^T \mathbf{C}^{-1}(x-m) &= \\ &\hspace{-40pt}=\frac{1}{\sigma^2_X \sigma^2_Y(1-\rho^2)} \begin{bmatrix} x-\mu_X\\ y-\mu_Y \end{bmatrix}^{T} \begin{bmatrix} \sigma^2_Y & -\rho \sigma_X \sigma_Y \\%[5pt] -\rho \sigma_X \sigma_Y & \sigma^2_X \end{bmatrix} \begin{bmatrix} x-\mu_X\\ y-\mu_Y \end{bmatrix}\\ &\hspace{-40pt}=-\frac{1}{2 (1-\rho^2)}\left[\left(\frac{x-\mu_X}{\sigma_X}\right)^2 +\left(\frac{y-\mu_Y}{\sigma_Y}\right)^2-2\rho \frac{(x-\mu_X)(y-\mu_Y)}{\sigma_X \sigma_Y} \right], \end{align} which agrees with the formula in Definition 5.4.
Remember that two jointly normal random variables $X$ and $Y$ are independent if and only if they are uncorrelated. We can extend this to multiple jointly normal random variables. Thus, if you have a normal random vector whose components are uncorrelated, you can conclude that the components are independent. To show this, note that if the $X_i$'s are uncorrelated, then the covariance matrix $\mathbf{C}_\mathbf{X}$ is diagonal, so its inverse $\mathbf{C}^{-1}_\mathbf{X}$ is also diagonal. You can see that in this case the PDF (Equation 6.1) becomes the products of marginal PDFs.
If $\mathbf{X}=[X_1,X_2,...,X_n]^T$ is a normal random vector, and we know $\textrm{Cov}(X_i,X_j)=0$ for all $i \neq j$, then $X_1$,$X_2$, ..., $X_n$ are independent.
Another important result is that if $\mathbf{X}=[X_1,X_2,...,X_n]^T$ is a normal random vector then $\mathbf{Y}=\mathbf{A}\mathbf{X}+\mathbf{b}$ is also a random vector because any linear combination of components of $\mathbf{Y}$ is also a linear combination of components of $\mathbf{X}$ plus a constant value.
If $\mathbf{X}=[X_1,X_2,...,X_n]^T$ is a normal random vector, $\mathbf{X} \sim N(\mathbf{m},\mathbf{C})$, $\mathbf{A}$ is an $m$ by $n$ fixed matrix, and $\mathbf{b}$ is an $m$-dimensional fixed vector, then the random vector $\mathbf{Y}=\mathbf{A}\mathbf{X}+\mathbf{b}$ is a normal random vector with mean $\mathbf{A}E\mathbf{X}+\mathbf{b}$ and covariance matrix $\mathbf{A} \mathbf{C} \mathbf{A}^T$. \begin{align}%\label{} \mathbf{Y} \sim N(\mathbf{A}E\mathbf{X}+\mathbf{b},\mathbf{A} \mathbf{C} \mathbf{A}^T) \end{align}
The print version of the book is available through Amazon here. |
## Wednesday, December 9, 2015
### High School Math Solutions – Trigonometry Calculator, Trig Identities
In a previous post, we talked about trig simplification. Trig identities are very similar to this concept. An identity is the equality of two expressions. When given a trig identity, you have to prove that both sides of the equation are equivalent. This requires using some skills from trig simplification and learning some new skills. One big difference between trig identities and trig simplification is that you are able to work both sides of the equation to show both sides are equivalent. Just like trig simplification, there is no exact recipe on how to prove trig identities. Memorizing trig identities will make proving trig identities 100 times easier.
Trig identities to memorize:
I’ll go over some tips to help make proving trig identities a little easier, and then we will go through an example step by step, so you can understand the thought process when proving trig identities.
Tips:
1. Don’t work on both sides of the equation at the same time
2. Start on the more complicated side
3. Try converting everything into cosine and sine
4. Try working on the other side if you get stuck
5. Memorize the identities
6. If you get frustrated, take a break and look at it again with fresh eyes
1. We will start on the more complicated side (right side) and convert everything to sine and cosine.
=\frac{1-\frac{\sin^2(x)}{\cos^2(x)}}{2\frac{\sin(x)}{\cos(x)}}
2. It looks a little messy, so let’s simplify it.
=\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}\cdot\frac{\cos(x)}{2\sin(x)}
=\frac{\cos^2(x)-\sin^2(x)}{2\cos(x)\sin(x)}
3. I’m stumped as to what to do next, so we’ll start working on the left side and convert everything to sine and cosine.
=\frac{\cos(2x)}{\sin(2x)}
4. This looks familiar. Let’s use the double angle identities. We know what identity to use for cos(2x) based on what the right side of the equation looks like.
=\frac{\cos^2(x)-\sin^2(x)}{2\ cos(x)\sin(x)}
5. The left side now matches the right side. We’re done!
Proving trig identities take a lot of practice. Let’s look at two more examples. |
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# Area and Perimeter of Triangles
## Area is half the base times the height while the perimeter is the sum of the sides.
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Practice Area and Perimeter of Triangles
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Area or Perimeter of Triangles and Quadrilaterals
For both a rhombus and a kite, area can be found if you know the lengths of the diagonals. What is special about the diagonals of these shapes?
#### Watch This
Watch the portions of this video having to do with triangles and quadrilaterals.
http://www.youtube.com/watch?v=ZASBmoylCPc James Sousa: Area and Perimeter Formulas
#### Guidance
Perimeter is the distance around a shape. To find the perimeter of any two dimensional shape, find the sum of the lengths of all the sides.
Area is the number of square units it takes to cover a two dimensional shape. The most basic shape to find the area of is a rectangle. The area of a rectangle is base times height.
\begin{align*}Area_{Rectangle}=bh\end{align*}
By dissection, a parallelogram can be turned into a rectangle.
Therefore, the area of a parallelogram is also base times height.
\begin{align*}Area_{Parallelogram}=bh\end{align*}
You can think of any triangle as half a parallelogram. If you rotate a triangle \begin{align*}180^\circ\end{align*} about the midpoint of one of its sides, the original triangle and the new triangle will be a parallelogram.
Therefore, the area of a triangle is base times height divided by two. Remember that any of the three sides can be the base. Also remember that the height must be perpendicular to the base and extend to the highest point of the triangle.
\begin{align*}{Area}_{Triangle}=\frac{bh}{2}=\frac{1}{2}bh\end{align*}
These and other area formulas that are good to know are shown below.
Shape Area Formula Picture Rectangle \begin{align*}A=bh\end{align*} Parallelogram \begin{align*}A=bh\end{align*} Triangle \begin{align*}A=\frac{bh}{2}\end{align*} Trapezoid \begin{align*}A=\frac{(b_1+b_2)h}{2}\end{align*} Rhombus \begin{align*}A=bh\end{align*} or \begin{align*}A=\frac{d_1d_2}{2}\end{align*} Kite \begin{align*}A=\frac{d_1d_2}{2}\end{align*} Square \begin{align*}A=s^2\end{align*}
Example A
Derive the formula for the area of a trapezoid.
Solution: A trapezoid can be thought of as half a parallelogram by rotating it \begin{align*}180^\circ\end{align*} about the midpoint of one of its non-parallel sides.
The base of this parallelogram is \begin{align*}b_1+b_2\end{align*} and the height is \begin{align*}h\end{align*}. The area of the parallelogram is \begin{align*}(b_1+b_2)h\end{align*}. Therefore, the area of the trapezoid is \begin{align*}\frac{(b_1+b_2)h}{2}\end{align*}.
Example B
Find the area of the trapezoid.
Solution: \begin{align*}b_1=6\end{align*}, \begin{align*}b_2=10\end{align*}, \begin{align*}h=8\end{align*}.
\begin{align*}Area=\frac{(6+10)8}{2}=\frac{(16)8}{2}=64 \ units^2\end{align*}
Example C
Find the area and perimeter of a square with side length 10 inches.
Solution: Perimeter is the distance around the shape. A square has four congruent sides, so each side is 10 inches. The perimeter is 40 in.
Area is the number of square units it takes to cover the shape. The area is \begin{align*}10^2=100 \ in^2\end{align*}.
Concept Problem Revisited
The diagonals of both rhombuses and kites are perpendicular. This means that when the shapes are broken down into triangles, one diagonal can be the base of the triangle and a portion of the other diagonal will be the height.
#### Vocabulary
Perimeter is the distance around a shape.
Area is the number of square units it takes to cover a two dimensional shape.
#### Guided Practice
1. The diagonals of a rhombus bisect each other. This means that they cut each other in half. The diagonals are also perpendicular. Derive the area formula for a rhombus \begin{align*}A=\frac{d_1d_2}{2}\end{align*}.
2. True or false: To find the area of a parallelogram you just multiply the lengths of two adjacent sides.
3. Explain why the formula \begin{align*}A=bh\end{align*} works to find the area of a rhombus.
1. The diagonals divide the rhombus into four triangles. For each of the triangles, the base and height are \begin{align*}\frac{d_1}{2}\end{align*} and \begin{align*}\frac{d_2}{2}\end{align*}. Therefore, the area of each triangle is \begin{align*}\frac{\left(\frac{d_1}{2} \right) \left(\frac{d_2}{2} \right)}{2}=\frac{d_1d_2}{8}\end{align*}. The rhombus is made up of four triangles with the same area. The area of the rhombus is \begin{align*}4 \left(\frac{d_1d_2}{8} \right)=\frac{d_1d_2}{2}\end{align*}.
2. False. The base and height need to be perpendicular. In a generic parallelogram, two adjacent sides will not be perpendicular.
3. \begin{align*}A=bh\end{align*} is the formula for the area for any parallelogram. Since a rhombus is a parallelogram, this area formula works for a rhombus.
#### Practice
1. Find the area and perimeter of a rectangle with a length of 12 inches and a width of 15 inches.
2. Find the area and perimeter of a right triangle with legs 3 cm and 4 cm and hypotenuse 5 cm.
3. Find the area of a trapezoid with bases 4 cm and 12 cm and height 9 cm.
4. The perimeter of a rectangle is 150 cm. The length is 4 times more than the width. What is the length of the rectangle?
5. The area of a triangle is \begin{align*}30 \ cm^2 \end{align*}. The base is twice as long as the height. What is the height of the triangle?
6. The perimeter of a right triangle is 24 in. The area is \begin{align*}24 \ in^2\end{align*}. The hypotenuse is 4 inches longer than the base. What are the lengths of the sides of the triangle?
7. Why is it necessary to use square units (such as \begin{align*}in^2\end{align*}) when referring to area?
8. Why don't you use square units when referring to perimeter?
9. When finding the area of a trapezoid, does it matter which sides you label as \begin{align*}b_1\end{align*} and \begin{align*}b_2\end{align*}?
10. Find the area of a square with a diagonal of 12 in.
11. Explain in your own words why you divide by two in the formula for the area of a triangle.
12. Explain in your own words why you divide by two in the formula for the area of a trapezoid.
13. Find the area of a kite with diagonals 10 cm and 18 cm.
14. Derive the formula \begin{align*}A=s^2\end{align*} for the area of a square.
15. The diagonals of a kite are perpendicular and one diagonal bisects the other diagonal. Derive the formula \begin{align*}A=\frac{d_1d_2}{2}\end{align*} for the area of a kite.
### Vocabulary Language: English
Area
Area
Area is the space within the perimeter of a two-dimensional figure.
Perimeter
Perimeter
Perimeter is the distance around a two-dimensional figure.
Perpendicular
Perpendicular
Perpendicular lines are lines that intersect at a $90^{\circ}$ angle. The product of the slopes of two perpendicular lines is -1.
Right Angle
Right Angle
A right angle is an angle equal to 90 degrees.
Right Triangle
Right Triangle
A right triangle is a triangle with one 90 degree angle.
Area of a Parallelogram
Area of a Parallelogram
The area of a parallelogram is equal to the base multiplied by the height: A = bh. The height of a parallelogram is always perpendicular to the base (the sides are not the height).
Area of a Triangle
Area of a Triangle
The area of a triangle is half the area of a parallelogram. Hence the formula: $A = \frac{1}{2}bh \text{ or } A = \frac{bh}{2}$. |
Multiplying fractions sounds daunting! Do you multiply the denominator or the numerator? Which is which?
Learning how to multiply fractions by whole numbers is a useful skill that you can use in your everyday life, including baking, sports, and crafting!
## Numerators and Denominators
Before we learn how to multiply fractions, we need to know the two parts of a fraction.
### Numerator
The number above the line is called the numerator.The numerator is the number of parts of the whole amount.
### Denominator
The number below the line is called the denominator.The denominator is the number of equal parts the whole is divided into.
Sometimes, you'll see a fraction written with a slash instead of a horizontal line — for example: 3/4. The number that falls to the left is equivalent to the numerator, and the number that falls on the right is the denominator.
In the example above, the numerator is 1, and the denominator is 2.
## Multiply whole numbers
Now, we'll get into multiplying whole numbers. It's easier than you think!
When multiplying fractions by a whole number, make your whole number the numerator. Then, put 1 as the denominator. A fraction is simply telling you to divide. 4/1 is equivalent to 4.
Let's try an example. You want to multiply 2/5 by 4:
Image created by the author via WebWhiteboard
All you have to do is multiply across. Multiply the numerators and multiply the denominators. Your answer would be 8/5.
#### Quiz
It takes 3/4 of a cup of flour to bake one muffin. You want to bake 7 muffins but you only have a cup that measures a quarter (1/4) of a cup. How many quarter cups of flour will it take to bake these muffins in total?
## Improper fractions and mixed numbers
When you multiply fractions by a whole number, you end up with an improper fraction. An improper fraction means that the numerator is greater than the denominator. You can convert improper fractions to mixed numbers by dividing the numerator by the denominator and leaving the remainder as a fraction.
Let's look at our example from the previous step:
Image created by the author via WebWhiteboard
5 goes into 8 once, and we're left with 3. So 8/5 written as a mixed number would be 1 and 3/5.
Image created by the author via WebWhiteboard
#### Quiz
If a football game played 5 quarters instead of the typical 4, how would the total number of quarters played be written as a mixed number?
## Take Action
Now you can easily multiply whole numbers by fractions. Check out these other helpful resources related to fractions! |
# Factors of 38: Prime Factorization, Methods, Tree, and Examples
The factors of 38 are any integers smaller than 38 which divides 38 without leaving any remainder. In other words, it can be any integer smaller than 38, which multiplies with another integer to give us the result of 38.
Figure 1 – All possible Factors of 38
Before deepening this topic, we shall first clarify the basic concepts for better understanding. One of them is the concept of prime and composite numbers.A prime number is a number that only has two factors, one and the number itself. Other numbers having more than two factors are categorized as composite numbers. Calculating factors for prime numbers is comparatively easy.The second point to remember is that a factor number can be positive or negative but cannot be in decimals or fractions. A number has to be a whole number to be a factor.Any number, except for 1, will have two factors at least. One of them is 1, and the second one will be the number itself. This is why one will always be the common factor between the factors of any two numbers.Now that the basic concepts of factors are clear, it will be easy for you to learn the factors of 38 through this article. Further in this article, we will learn to calculate factors of 38 through multiplication and division. We will also learn about the methods of prime factorization and factor trees. For the quick application of these concepts, solved examples will be provided at the end of this article.
## What Are the Factors of 38?
The factors of 38 are 1, 2, 19, and the number 38 itself. These are the only numbers to divide 38 fully without leaving any remainder. Any other integers smaller than 38 would not be able to give us the same results. This data also proves that 38 is a composite number as it has more than two primary factors. In total, there are only four factors of 38.
## How To Calculate the Factors of 38?
To find out the factors of any number, two methods can be used; the first one is multiplication and the second one is division. Both methods are straightforward and reliable for calculating the correct answers.You only have to be cautious about considering every single number on your list for working. You might lose a score if the number you missed is a factor for your given number. Let’s start with the division method. For this, we will first half the given number. This will save us time as integers larger than half of the given number can never be the factor of that number except for the number itself. We will consider all integers between 1 and that number and start dividing the given number by those integers. The integer that can divide the given number fully, without leaving any remainder, will be a factor along with the quotient for that number. Let’s look at the example of 38 to understand this concept better:To get the half of 38, we will divide it by 2. This will give us 19. Now we will start to divide 38 by numbers between 1 and 19. Upon dividing 38 by 2, we get 19 as the quotient, and nothing is left as the remainder. Now both 2 and 19 (the divisor and the quotient) are considered as the factors of 38. When we repeat this process with other numbers, none of them can divide 38 fully. So our factors of 38 will be 1, 2, 19, and 38.For calculating factors by the multiplication method, we will try multiplying different integers together. Our goal here is to get 38 as the product of multiplication. If any pair of the numbers make up to the product ‘38’, both integers will be considered as factors of 38. For better understanding, have a look at these examples:
1 x 38 = 38
2 x 19 = 38
Since the result retrieved is 38, both the numbers will be considered product factors.Hence it is proved by both of these methods that 1, 2, 19, and 38 are the only factors of 38.
## Factors of 38 by Prime Factorization
Prime factorization can be called a decomposition method as a number is decomposed into its factors until prime factors are not achieved. It is done to know the prime factors of an integer.For this method, we will start by dividing the given number with the smallest number, except for 1, which divides our given number fully. Like in the case of 38, we will start by dividing 38 by 2, and our answer will be 19.For 38, our work will stop here as the results achieved were prime numbers. So we know that the prime factors of 38 are 2 and 19. To represent this answer, we can use the notation 2 x 19.You can take a look at the Prime Factorization of 38 below:
Figure 2 – Prime Factorization of 38
## Factor Tree of 38
A factor tree is one method to represent a number’s factors. Apart from representing prime factorization in notations form, a factor tree can also be used. In a factor tree, the given number for the method is written on the top. It is divided by a small number first, and the answers are written below by extending the branches. This process continues until prime numbers are not achieved. The tree ends with prime numbers at the bottom.For the factor tree of 38, we will write 38 on top. Then we will start dividing 38 by smaller divisible numbers first. Remember that we won’t divide it by 1. Another smaller divisible number is 2, so we will divide 38 by 2. We will get 19 as our answer. Now we will extend one line below 38 and write one below it; then, we will open another line similarly and write 19 below it. Since both numbers are prime numbers, we have reached the end of the factor tree. So the factor tree of 38 will end here. A factor tree of 38 will be attached by the end of this article. You can look at it to comprehend this knowledge with the actual representation of the factor tree. The diagram of the Factor Tree of 38 is attached below:
Figure 3 – Factor Tree of 38
## Factors of 38 in Pairs
Any two numbers which multiply to give you a product of any given number form a factor pair of that number. These numbers can be positive or negative, but they cannot be decimals or fractions.Earlier in this article, when we learned to calculate factors through the division method, we found that both the divisor and quotient are considered factors. Now for this concept, we will pair both of those numbers together to form a factor pair of that given number.The first step in finding the factor pairs is to know whether the given number is a prime number or a composite number. If it is a prime number, both actors will be paired together to make a factor pair.On the other hand, if the number is composite, you will have to start by multiplying its factors with each other to find a pair that gives you the product of that given number. This can be a little time-consuming for numbers with many factors as they will have many factor pairs. For finding factor pairs of 38, we will make a list of its factors first, as shown below:
Factor list = 1, 2, 19, 38
Looking at the factor list, we can tell it’s a composite number. Now we will multiply these numbers with each other to find out which numbers give us the product 38.
1 x 2 = 2
1 x 38 = 38
2 x 19 = 38
2 x 38 = 76
38 x 19 = 722
Now that we have made a list of every possible product, we can figure out that there are only two pairs that give us the result 38. Only those two pairs, (1, 38) and (2, 19), will be considered as our answer.
## Factors of 38 Solved Examples
Now that we have gained enough knowledge about factors of 38 let’s look at a few solved examples to memorize the concepts.
### Example 1
Make a list of even factors of 38.
### Solution
We will look into the list of factors of 38, and then we will identify even numbers for our answer. The factors of 38 are 1, 2, 19, and 38, so our solutions will be 2 and 38.
### Example 2
How to determine that 38 is a composite number?
### Solution
Factors of 38 are 1, 2, 19, and 38. This proves it’s a composite number, as prime numbers only have two factors.
### Example 3
List common factors between factors 2 and 38.
### Solution
Factors of 2: 1 and 2
Factors of 38: 1, 2, 19, and 38.
Common factors: 1 and 2.All images/mathematical drawings are made using GeoGebra. |
# Class 7 Maths NCERT Solutions for Chapter – 3 Data Handling EX – 3.4
## Data Handling
Question 1.
Tell whether the following is certain to happen, impossible, can happen but not certain.
1. You are older today than yesterday.
2. A tossed coin will land heads up.
3. A die when tossed shall land up with 8 on top.
4. The next traffic light seen will be green.
5. Tomorrow will be a cloudy day.
Solution:
1. Certain to happen
2. Can happen but not certain
3. Impossible
4. Can happen but not certain
5. Can happen but not certain
Question 2.
There are 6 marbles in a box with numbers from 1 to 6 marked on each of them.
(i)
What is the probability of drawing a marble with number 2 ?
(ii) What is the probability of drawing a marble with number 5 ?
Solution:
Out of 6 marbles, one can be drawn in 6 ways. So, total number of events = 6
(i) The marble with number 2 can be obtained only in one way.
∴ Required probability = $\frac { 1 }{ 6 }$
(ii) The marble with number 5 can be obtained only in one way.
∴ Required probability = $\frac { 1 }{ 6 }$
Question 3.
A coin is flipped to decide which team starts the game. What is the probability that your team will start?
Solution:
On tossing of a coin, the possible outcomes are head (H) or tail (T).
Required probability = $\frac { 1 }{ 2 }$ |
## College Algebra (11th Edition)
$\dfrac{4}{a^2}$
$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $\dfrac{4a^5(a^{-1})^3}{(a^{-2})^{-2}} .$ $\bf{\text{Solution Details:}}$ Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{4a^5a^{-1(3)}}{a^{-2(-2)}} \\\\= \dfrac{4a^5a^{-3}}{a^{4}} .\end{array} Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{4a^{5+(-3)}}{a^{4}} \\\\= \dfrac{4a^{5-3}}{a^{4}} \\\\= \dfrac{4a^{2}}{a^{4}} .\end{array} Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} 4a^{2-4} \\\\= 4a^{-2} .\end{array} Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{4}{a^2} .\end{array} |
# What Is A Way To Wright 100,000 Using Exponents (2023)
Mathematics Middle School
We have to wright 100,000 using exponents. The exponent of number shows how many times you use the number in multiplication.
For example: 64 = 8², because: 64 = 8 · 8or: 64 = 4³, because 64 = 4 · 4 · 4 In this case:100,000 = 10 · 10 · 10 · 10 · 10Therefore:100,000 = 10^5 ( or "10 to the fifth power" )Answer: 10^5.
Step-by-step explanation: since 10x 5 equals 100,000
## Related Questions
Will uses 3/4 cup of olive oil to make 3 batches of saled dressing.how much oil does will use for one batch of saled dressing
1/4 cup of oilif 3/4 gives you 3 batches of salad dressing just subtract the top number by 2 and that will give you one 1/4 of a cup of oil
Will will use 1/4 cup of olive oil for one batch of salad dressing. If there is 3/4 cups used for 3 batches, one batch is equal to 1/4.
at memorial hospital 9 of the 12 babies born on Tuesday were boys In simplest form What fraction of the babies born on Tuesday were boys
9/12 = (9/3) / (12/3)= 3/4.
3/4 of the babies born on Tuesday were boys~
Fraction
Step-by-step explanation:
Given : At memorial hospital 9 of the 12 babies born on Tuesday were boys
To find : What fraction of the babies born on Tuesday were boys
Solution : As we know total number of babies born in memorial hospital = 12
Out of 12 babies number of boys born on Tuesday = 9
So the fraction of babies born on Tuesday were boys
Therefore, The fraction or simplest form of fraction
Alma just received a 6% raise in salary. Before the raise, she was making \$52,000 per year. How much more will Alma earn next year?
The rise in Alma's salary is:
(6%)* \$52,000= \$3,120
\$52,000+ \$3,120= \$55,120
6% of \$52000 is \$3,120
Adding that to what she already makes means she will make \$55,120 next year.
List the following from least to greatest 75% 1/2 0.375 5/8
.375 1/2 5/8 75% --that's the order, hope it helped:)
Rectangular Washington County measures 15.9 miles by 9.1 miles. What is the county's area?
Formula for Area= L x W (length x Width). 15.9 X 9.1= 144.69
Step-by-step explanation:
ricardo has 2 cases of video games with the same number of game in each cases . he give 4 games to his brothers . Ricardo has 10 games left. How many video games were in each case ?
12 because you split the 4 from the two boxes.
Jenna skis 2 1/3 miles down the mountain. Her instructor skis 1 1/2 times as far. Does Jenna ski a shorter, greater, or the same distance as her instructor?
She skis a shorter amount than her instructor because it is 5 1/2 times more than her instructor
carter drank 15.75 gallons of water in 4 weeks.he drank the same amount of water each day.about how many days altogether will it take him to drink 20 gallons?
To answer this first lets see how many gallons of water he drinks per week.
15.75÷ 4 =3.9375
Now lets divide to see how much he drinks each day.
3.9375÷ 7 =0.5625
20÷0.5625 =35.5555...
So, it will take Carter about 35 and a half days to finish 20 gallons.
Hope I helped ya!!
Rachel is putting 3,902 bagels into boxes. If she puts 8 bagels in each box, how many boxes can she fill?? Write the answer as a mixed number
3902/8=487.75
So, the answer is 487 and 3/4
U have to divide 3902 and 8
How many fractions are equivalent to 4/5
There are an infinite number of them.
-- Think of any number.
-- Multiply 4 by your number. Write it on top.
-- Multiply 5 by your number. Write it on the bottom.
-- You have a new fraction that's equivalent to 4/5 .
How many different numbers can you think of ?
That's how many new fractions you can create
that are ALL equivalent to 4/5 .
There is an infinite amount of numbers equivalent to 4/5 such as:
8/10,12/15.16/20,20/25
Which exponential function goes through the points (1,8) and (4,64)? A) f(x)=4(2)^x
B) f(x)=2(4)^x
C) f(x)=4(2)^-x
D) f(x)=2(4)-x
Plug in (4,64) into each answer choice.
A) 4(2)^x
64=4(2)^4
64 = 4(16)
64 = 64
B) f(x) = 2(4)^x
64 = 2(4)^4
64 = 2(256)
64≠ 512
C) f(x) = 4(2)^{-x}
64 = 4(2)^{-4}
64 = 4(-0.0625)
64≠ 0.25
D) f(x) = 2(4)^{-x}
64 = 2(4)^{-4}
64 = 2(0.00390625)
64≠0.0078125
a bottle contains 3.5liter of water. A second bottle contains 3,750 millimeters of water .how many more millimeters are in. the larger bottle than in the smaller bottle
Well, if you convert 3.5 liters to 3,500 milliliters, all you do from there is do
3,750-3,500 =250 milliliters
3.5 L= 3,500 mL
3,750 mL- 3,500 mL= 250 mL
The larger bottle has 250 mL more than the smaller one has~
How many pounds are equivalent to 40 ounces
The equivalent to 40 pounds are 2.5pounds.
an elevator travels 117 feet in 6.5 seconds. what is the elevators speed as a unit rate? explain answer.
The elevator need to climb 117 feet and it reaches 117 feet in 6.5 seconds only. Now, let’s solve for the unit rate of the elevator. What is it’s speed. => 117 feet is the distance that the elevator needs to climb. => 6.5 seconds is the now. To find the speed let’s follow that formula. S = d/t => 117 feet / 6.5 seconds => 18 feet per seconds Unit rate of the elevator speed = 18 feet/s
Each month for 7 months, Samuel mows 3 lawns. How many more lawns does he need to mow before he has mowed 29 lawns. I think it is 8 more lawns and not 7.
7x3=21; 29-21=8.
Am I wrong?
Answer: No, He needs to to mow 7 more lawns before he has mowed 29 lawns.
Step-by-step explanation:
Given: The number of lawns by Samuel in each month =- 3
The number of months he mowed the lawns = 7
Then, the number of lawns mowed by Samuel =
Let x be the number of lawns he needs to mow before he has mowed 29 lawns, then x
What I did is 7×3 and got 21 then counted up in till I got to 29 and I got 6 not 8.
stickers are made with the same ratio of width to length. a sticker 2 inches has a length of 4 inches. complete the table
x (Widht) y (Length)
1 2
2 4
3 6
4 8
5 10
Step-by-step explanation:
We can simplify the ratio 2 : 4 by dividing both terms by the greatest common factor, in this case by 2
2/2=1
4/2=2
So, with this information we can find the length multiply by 2 the width
y(length)=2*x
2=4,3=6,4=8,5=10
inches will be x and length will be y you multiply x by 2 to get y
Minxia counted 40 green cars and 20 silver cars in the parking lot. If the number of green cars stays the same, how many more silver card would need to be added so the ratio of green cars to silver cars is 1 to 3?
The number of silver cars that need to be added will be 100.
What are ratio and proportion?
A ratio is a group of sequentially ordered numbers a and b expressed as a/b, where b is never equal to zero. When two objects are equal, a statement is said to be proportional.
Minxia counted 40 green vehicles and 20 silver vehicles in the parking garage. On the off chance that the quantity of green vehicles remains something similar.
Let 'x' be the number of silver cars that need to be added.
If the ratio of green cars to silver cars is 1: 3. Then the number of silver cars that need to be added will be given as,
40 / (x + 20) = 1 / 3
Simplify the equation, then we have
40 / (x + 20) = 1 / 3
(x + 20) = 40 x 3
x + 20 = 120
x = 100
The number of silver cars that need to be added will be 100.
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What I did is that: I multiply 40 by 3 because 40 is 1/3 and I have to find the answer for 3/3 (Confusing, sorry I don't know how to explain it correctly). I got 120, subtract 20 from 120 beacause ther is 20 silver cards. The answer is 100
A cartoon lasts 18 minutes.How many seconds is the cartoon?
The Cartoon lasts 1080 seconds.
what is Unitary Method?
The unitary technique involves first determining the value of a single unit, followed by the value of the necessary number of units.
For example, Let's say Ram spends 36 Rs. for a dozen (12) bananas.
12 bananas will set you back 36 Rs. 1 banana costs 36 x 12 = 3 Rupees.
As a result, one banana costs three rupees. Let's say we need to calculate the price of 15 bananas.
This may be done as follows: 15 bananas cost 3 rupees each; 15 units cost 45 rupees.
Given:
A cartoon lasts 18 minutes.
We know, 1 minute = 60 seconds
So, 18 minutes= 60 x 18
= 1080 seconds
Hence, the cartoon lasts 1080 seconds.
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18 Minutes
So 1 minute equals to 60 seconds.
So 60 times 18 equals 1,080.
Hope this helped and if you need more help then just message me instead :)
Alyssa divides a granola bar into halves. How many equal parts are there?
Answer: If Alyssa divides a granola bar into halves then there are two equal parts.
Step-by-step explanation:
We know that to divide a thing in half means to divide or split it into two equal parts.
Therefore if Alyssa divides a granola bar into halves then it means that she divides a granola bar in two equal parts .
It means there are two equal parts of the granola bar.
Hence, the number of equal parts of granola bar when divided into halves = 2
1 split into two equal parts so the answer is 2
Suppose you have a coupon for a 20% discount. You buy a game that costs \$38. The sales tax rate is 5.5%. Sales tax applies to the cost after the discount. What is the total cost of the game?
\$32.072 ≈\$32.07
20% of \$38 is equal to \$7.60
\$38 - \$7.60 = \$30.40
5.5% of \$30.40 = 1.672
\$30.40 + \$1.672 = \$32.072
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# 5.3 The Normal Distribution
## LEARNING OBJECTIVES
• Describe properties of the normal distribution.
• Apply the Empirical Rule for normal distributions.
The normal distribution is the most important of all the distributions. It is widely used and even more widely abused. Its graph is a symmetric, bell-shaped curve. You see the bell curve in almost all disciplines, including psychology, business, economics, the sciences, nursing, and, of course, mathematics. Some of your instructors may use the normal distribution to help determine your grade. Most IQ scores are normally distributed. Often real-estate prices fit a normal distribution. The normal distribution is extremely important, but it cannot be applied to everything in the real world.
Properties of the normal distribution include:
• The curve of a normal distribution is symmetric and bell-shaped.
• The center of a normal distribution is at the mean $\mu$.
• In a normal distribution, the mean, the median, and the mode are equal.
• The curve is symmetric about a vertical line drawn through the mean.
• The tails of a normal distribution extend to infinity in both directions along the $x$-axis.
• The standard deviation, $\sigma$, of a normal distribution determines how wide or narrow the curve is.
• The total area under the curve of a normal distribution equals 1.
A normal distribution is completely determined by its mean $\mu$ and its standard deviation $\sigma$, which means there are an infinite number of normal distributions. The mean $\mu$ determines the center of the distribution—a change in the value of $\mu$ causes the graph to shift to the left or right. The standard deviation $\sigma$ determines the shape of the bell. Because the area under the curve must equal one, a change in the standard deviation $\sigma$ causes a change in the shape of the curve—the curve becomes fatter or skinnier depending on the value of $\sigma$.
The normal distribution is a continuous probability distribution. As we saw in the previous section, the area under the curve of the normal distribution equals the probability that the corresponding normal random variable takes on a value with in a given interval. That is, the probability that the normal random variable is in between $a$ and $b$ equals the area under the normal curve in between $x=a$ and $x=b$.
Watch this video: ck12.org normal distribution problems: Quantitative sense of normal distributions | Khan Academy by Khan Academy [10:52]
## The Empirical Rule
For a normal distribution with mean $\mu$ and standard deviation $\sigma$, then the Empirical Rule says the following:
• About $68\%$ of the values lie between $–1 \times \sigma$ and $+1 \times \sigma$ of the mean $\mu$.
• In other words, about $68\%$ of the data fall with one standard deviation of the mean.
• About $95\%$ of the values lie between $–2 \times \sigma$ and $+2 \times \sigma$ of the mean $\mu$.
• In other words, about $95\%$ of the data fall with two standard deviation of the mean.
• About $99.7\%$ of the values lie between $–3 \times \sigma$ and $+3 \times \sigma$ of the mean $\mu$.
• In other words, about $99.7\%$ of the data fall with three standard deviation of the mean
The empirical rule is also known as the $68-95-99.7$ rule.
## EXAMPLE
Suppose a normal distribution has a mean $50$ and a standard deviation $6$.
About $68\%$ of the values lie between $–1 \times \sigma = -1 \times 6 = –6$ and $1 \times \sigma = 1 \times 6 = 6$ of the mean $50$. The values $50 – 6 = 44$ and $50 + 6 = 56$ are within one standard deviation of the mean $50$. So $68\%$ of the values in this distribution are between 44 and 56.
About $95\%$ of the values lie between $–2 \times \sigma = -2 \times 6 = –12$ and $2 \times \sigma = 2 \times 6 = 12$ of the mean $50$. The values $50 – 12 = 38$ and $50 + 12 = 62$ are within two standard deviations of the mean $50$. So $95\%$ of the values in the distribution are between 38 and 62.
About $99.7\%$ of the $x$ values lie between $–3 \times \sigma = -3 \times 6 = –18$ and $3 \times \sigma = 3 \times 6 = 18$ of the mean $50$. The values $50 – 18 = 32$ and $50 + 18= 68$ are within three standard deviations of the mean $50$. So $99.7\%$ of the values in the distribution are between 32 and 68.
## TRY IT
Suppose a normal distribution has a mean $25$ and a standard deviation $5$. Between what values of does $68\%$ of the data lie?
Click to see Solution
• Between $25+(-1) \times 5=20$ and $25+1 \times 5=30$.
## EXAMPLE
From 1984 to 1985, the height of 15 to 18-year-old males from Chile follows a normal distribution with mean $172.36$cm and standard deviation $6.34$cm.
1. About $68\%$ of the heights of 15 to 18-year old males in Chile from 1984 to 1985 lie between what two values?
2. About $95\%$ of the heights of 15 to 18-year old males in Chile from 1984 to 1985 lie between what two values?
3. About $99.7\%$ of the heights of 15 to 18-year old males in Chile from 1984 to 1985 lie between what two values?
Solution:
1. $\displaystyle{\mu+(-1)\times \sigma=172.36+(-1)\times 6.34=166.02}$ and $\displaystyle{\mu+1 \times \sigma=172.36+1 \times 6.34=178.70}$
2. $\displaystyle{\mu+(-2)\times \sigma=172.36+(-2)\times 6.34=159.68}$ and $\displaystyle{\mu+2 \times \sigma=172.36+2 \times 6.34=185.04}$
3. $\displaystyle{\mu+(-3)\times \sigma=172.36+(-1)\times 6.34=153.34}$ and $\displaystyle{\mu+3 \times \sigma=172.36+2 \times 6.34=191.36}$
## TRY IT
The scores on a college entrance exam have an approximate normal distribution with a mean $52$ points and a standard deviation of $11$ points.
1. About $68\%$ of the exam scores lie between what two values?
2. About $95\%$ of the exam scores lie between what two values?
3. About $99.7\%$ of the exam scores lie between what two values?
Click to see Solution
1. About $68\%$ of the scores lie between the values $52+(-1) \times 11=41$ and $52+1 \times 11=63$.
2. About $95\%$ of the values lie between the values $52+(-2) \times 11=30$ and $52+2 \times 11=74$.
3. About $99.7\%$ of the values lie between the values $52+(-3) \times 11=19$ and $52+3 \times 11=85$.
Watch this video: Empirical Rule| Probability and Statistics | Khan Academy by Khan Academy [10:25]
## Concept Review
The normal distribution is the most frequently used distribution in statistics. The graph of a normal distribution is a symmetric, bell-shaped curve centered at the mean of the distribution. The probability that a normal random variable takes on a value in inside an interval equals the area under the corresponding normal distribution curve.
For a normal distribution, the empirical rule states that 68% of the data falls within one standard deviation of the mean, 95% of the data falls within two standard deviations, and 99.7% of the data falls within three standard deviations of the mean. |
# Week 10 – precalculus 11
Dividing radicals is not as easy as multiplying, though it does include multiplying. We will need to use distributive property and potentially adding to rationlize the denominator. We cannot divide by a radical so we must make the denominator a whole number. We normally multiply the denominator by itself and multiply that with the top as well but if we have a binomial we must multiply the top and bottom by the conjugate of the denominator. A conjugate is the same as the denominator but the plus or minus sign will be the opposite of the denominator. To see how this works we must show it. The goal is to simplify the expression as much as possible. We already know how to multiply radicals, this is just combining those skills with some adding.
First we will start with an expression like $\frac {4\sqrt {2} - 6\sqrt {5}}{2\sqrt {3}}$
Because it is a monomial denominator we will multiply the top and bottom by $2\sqrt{3}$
$\frac {8\sqrt {6} - 12\sqrt {15}}{4\sqrt {9}}$
The denominator now should become 3×4
$\frac {8\sqrt {6} - 12\sqrt {15}} {12}$
Because we cannot simply the numerators values or add them together we will simply divide them both by a common factor as with the denominator. Giving us:
$\frac {2\sqrt {6} - 3\sqrt {15}} {3}$
Let’s say that we are given a expression with a binomial denominator.
We are given $\frac {5\sqrt {3} + 4\sqrt {2}}{2\sqrt {3} - \sqrt 2}$
To solve this we will have to multply the top and bottom by the conjugate of the denominator.
We are given $\frac {(5\sqrt {3} + 4\sqrt {2}) (2\sqrt {3}+ \sqrt {2})}{(2\sqrt {3} - \sqrt 2) (2\sqrt {3}+ \sqrt {2})}$
Then we will foil/muliply
$\frac {10\sqrt {9} + 5\sqrt {6} + 8\sqrt {6} + 4\sqrt {4}}{4\sqrt {9} + 2\sqrt {6} - 2\sqrt {6} - \sqrt {4}}$
Many of these we can either add together or turn into whole numbers. We will also add together the whole numbers. Two of the radicals on the denominator cancel out.
$\frac {38 + 13\sqrt {6}}{10}$
At this point, the numbers cannot be divided by a common number. This is the simplified form. This it how you use multplication to rationalize a denominator and simplify a radical expression in a division setting. |
Smartick is an online platform for children to master math in only 15 minutes a day
Jun16
# How to Solve Double Digit Division
###### In today’s post we are going to explain how to solve double digit division.
Before beginning to learn how to solve double digit division, it is important that you become familiar with these terms, because we will use them later.
Dividend: the number that is being divided.
Divisor: the number by which the dividend is divided.
Quotient: the result of division.
Remainder: the amount that is left over after division.
Once you have seen this, you know where to place each number in the division. Now, we have to follow these steps:
1. Take the first digits of the dividend, the same number of digits that the divisor has. If the number taken from the dividend is smaller than the divisor, you need to take the next digit of the dividend.
2. Divide the first number of the dividend (or the two first numbers if the previous step took another digit) by the first digit of the divisor. Write the result of this division in the space of the quotient.
3. Multiply the digit of the quotient by the divisor, write the result beneath the dividend and subtract it. If you cannot, because the dividend is smaller, you will have to choose a smaller number in the quotient until it can subtract.
4. After subtraction, drop the next digit of the dividend and repeat from step 2 until there are no more remaining numbers in the dividend.
That’s the concept, but we are going to go through it with an example.
We are going to solve the following double digit division:
1. Take the first digits of the dividend: in this case 57. But as 57 is smaller than 73, you have to take one more digit: 573.
1. To divide 573 by 73, we take the first two digits of the dividend: 57 and divide them by the first digit of the divisor:
57 ÷ 7 = 8
1. Write the 8 in the quotient and multiply it by the divisor:
8 x 73 = 584
But 584 is bigger than 573; therefore, 8 “does not fit”. You have to choose the preceding number and multiply again:
7 x 73 = 511
511 is smaller than the dividend; therefore 7 “does fit”. We write 511 beneath the digits of the dividend and then divide and subtract:
1. Drop the next digit of the dividend, which is 8. Now, you have to divide 628 by 73. Repeat the previous steps:
Divide the first two digits of the dividend by the first digit of the divisor and write it in the space of the quotient:
62 ÷ 7 = 8
Multiply that digit by the divisor:
8 x 73 = 584
584 is less than 628; therefore, we can subtract:
628 – 584 = 44
The result of this division is 78 and a remainder of 44.
I hope that you have learned with this post how to do double-digit division.
Do not hesitate to leave your comments!
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• maryssaMay 19 2020, 9:34 AM
I forgot how to do it
• ArpilApr 02 2020, 4:46 AM
Thank you so much for this trick
• Bhargab Jyoti GoswamiFeb 26 2020, 12:05 AM
Thanks
Actually I forgot that
• EasonJun 16 2019, 1:27 PM
Very useful but a bit too complicated for me to remember.
• Gaurav GuptaMay 22 2019, 12:48 AM
Lovely,
Thanks
• SachithaJan 05 2019, 5:50 AM
Understand
• Vanita SomvanshiApr 10 2018, 8:55 PM
impressive !
• KareenApr 09 2018, 7:32 PM
I love this demonstration. Thanks very much. |
# Lesson 5: The Graph Scale
Chapter 3: Transformations of
Graphs and Data
Lesson 5: The Graph Scale-Change
Theorem
Mrs. Parziale
Vocabulary:
•
Vertical stretch: A scale change that makes the original graph taller or shorter
•
Horizontal stretch: a scale change that makes the original graph wider or
skinnier.
•
Scale change: a stretch or shrink applied to the graph vertically or horizontally
•
Vertical scale change: The value that changes the vertical values of the graph.
•
Horizontal scale change: The value that changes the horizontal values of the
graph.
•
Size change: When the same vertical and horizontal scale change occurs.
Example 1:
Consider the graph of y x 4 x
(a) Complete the table and graph on the grid:
3
x y
-2
-1
0
1
2
y
5
4
3
2
1
–5
–4
–3 –2 –1
–1
–2
–3
–4
–5
1
22
33
44
55
xx
(b) Replace (y) with
y
3
.
y
9
1. Solve the new equation for y and
graph it on the same grid at right.
8
7
6
5
4
2. What happens to the ycoordinates?
3
2
1
–6
3. This is called a vertical stretch of
magnitude 3 .
–5
–4
–3
–2
–1
–1
–2
–3
–4
–5
–6
4. Under what scale change is the
new figure a vertical scale change
of the original?
–7
–8
–9
1
2
3
4
5
6
x
(c) Replace (x) with
x
y
.
9
2
1. Solve the new equation for
y and graph it.
2. What happens to the xcoordinates?
3. This is called a horizontal
stretch of magnitude 2 .
4. Under what scale change is
the new figure a horizontal
scale change of the
original?
8
7
6
5
4
3
2
1
–6
–5
–4
–3
–2
–1
–1
–2
–3
–4
–5
–6
–7
–8
–9
1
2
3
4
5
6
x
y
(d) Let f ( x ) x 4 x .
Find an equation for
g(x), the image of f(x)
under
9
3
8
7
6
5
4
3
2
1
–6
–5
–4
–3
–2
–1
–1
–2
S ( x , y ) (2 x , 3 y )
–3
–4
–5
–6
–7
–8
What is happening to
each part of the graph?
–9
1
2
3
4
5
6
x
S ( x , y ) (2 x , 3 y )
• How is the x changed?
• Change: horizontal
stretch two times wider.
y
9
8
7
6
5
4
3
2
1
–9 –8 –7 –6 –5 –4 –3 –2 –1
–1
–2
• How is the y changed?
• Change: vertical stretch
three times the original.
–3
–4
–5
–6
–7
–8
–9
1
2
3
4
5
6
7
8
9 x
Graph Scale-Change Theorem
In a relation described by a sentence in (x) and (y),
the following two processes yield the same
graph:
(1) replace (x) by
x
a
and (y) by
y
b
in the sentence
( x , y ) ( ax , by )
(2) apply the scale change __________________
to
the graph of the original relation.
size change
Note: If a = b, then you have performed a __________
If a = negative, the graph has been reflected (flipped) over the y-axis
If b = negative, the graph has been reflected over the x-axis
So, What’s the Equation?
(d) Find an equation for g(x), the image of f(x)
under S ( x , y ) (2 x , 3 y )
y
9
8
7
f ( x) y x 4 x
3
6
5
4
3
x
y
-2
0
x
y
2
1
–9 –8 –7
–6
–5
–4
–3
–2
–1
–1
–2
-1
3
0
0
1
-3
–3
–4
–5
–6
–7
–8
–9
2
0
1
2
3
4
5
6
7
8
9
xx
Example 2:
• Consider y x . Find an equation for the
x
function under S ( x , y ) , 2 y
3
• Describe what happens to all of the x values:
• Describe what happens to all of the y values:
x
S ( x, y ) , 2 y
3
• Find the equation for the transformed image
by
– Replace (x) with ______________
– Replace (y) with ______________
– Now make the new equation (remember to
simplify to y= form):
Graph It!
y
9
y x
8
7
6
5
4
y 2 3x
3
2
1
–9
–8
–7
–6
–5
–4
–3
–2
–1
–1
–2
–3
–4
–5
–6
–7
–8
–9
1
2
3
4
5
6
7
8
9
x
Closure - Example 3:
The graph to the right is
x
y = f(x). Draw 3 y f ( ) .
4
• What should happen to all
of the x values?
• What should happen to all
of the y values? |
# Hyperbolic Cosine Calculator
The hyperbolic cosine function has important applications in mathematics, physics, and engineering. In this calculator, we will explore hyperbolic cosine, provide examples of hyperbolic cosine calculations, explain how to use an online hyperbolic cosine calculator, and discuss its real-life applications.
Number : Hyperbolic Cosine :
## Similar Calculators:
Hyperbolic cosine, often denoted as cosh(x), is a mathematical function that is part of the hyperbolic trigonometric functions. It is defined as:
$$\cosh(x) = \frac{e^x + e^{-x}}{2}$$
## Understanding the Hyperbolic Cosine Calculator
The hyperbolic cosine calculator is a tool that simplifies the process of finding the cosh(x) value for a given input value of x. It operates based on the mathematical definition of hyperbolic cosine.
### Examples of Hyperbolic Cosine Calculations
#### Example 1: Calculating Hyperbolic Cosine of a Value
Let’s start with a basic example. Suppose we want to find the hyperbolic cosine of $$x = 2$$. Using the hyperbolic cosine calculator:
$$\cosh(2) = \frac{e^2 + e^{-2}}{2} \approx 3.7622$$
The hyperbolic cosine of 2 is approximately 3.7622.
#### Example 2: Hyperbolic Cosine of a Negative Value
Hyperbolic cosine calculations can handle negative values as well. Let’s find the hyperbolic cosine of $$x = -1$$:
$$\cosh(-1) = \frac{e^{-1} + e^{1}}{2} \approx 1.5431$$
The hyperbolic cosine of -1 is approximately 1.5431.
#### Example 3: Calculating Hyperbolic Cosine Using an Identity
Hyperbolic cosine values can also be calculated using the identity:
$$\cosh(x) = \sqrt{\frac{e^{2x} + 1}{2}}$$
Let’s calculate $$\cosh(3)$$ using this identity:
$$\cosh(3) = \sqrt{\frac{e^{2 \cdot 3} + 1}{2}} \approx 10.0677$$
The hyperbolic cosine of 3 is approximately 10.0677.
## Solution Explanation
The examples provided demonstrate how to use the hyperbolic cosine calculator to find cosh(x) values for different inputs, including positive and negative values. Understanding hyperbolic cosine is essential in various mathematical and scientific applications.
## How to Use an Online Hyperbolic Cosine Calculator
Using an online hyperbolic cosine calculator is straightforward. Here’s a general guide:
1. Input the Value: Enter the value $$x$$ for which you want to calculate the hyperbolic cosine.
2. Click Calculate: Press the “Calculate” button to obtain the cosh(x) value.
3. View the Result: The calculator will display the value corresponding to the input value.
Online hyperbolic cosine calculators provide quick and accurate results, making hyperbolic cosine calculations convenient.
## FAQs
### Q1: What is Hyperbolic Cosine?
Hyperbolic cosine ($$\cosh(x)$$) is a mathematical function that is part of the hyperbolic trigonometric functions. It is defined as $$\cosh(x) = \frac{e^x + e^{-x}}{2}$$.
### Q2: What Are the Properties of Hyperbolic Cosine?
Hyperbolic cosine has properties similar to cosine. It is an even function, and its range is the set of all positive real numbers. Additionally, $$\cosh(0) = 1$$.
### Q3: How Do I Calculate Hyperbolic Cosine Manually?
Hyperbolic cosine can be calculated manually using its mathematical definition or the identity $$\cosh(x) = \sqrt{\frac{e^{2x} + 1}{2}}$$.
### Q4: What Are the Limits of Hyperbolic Cosine Calculations?
Hyperbolic cosine calculations can handle a wide range of values, but they may become impractical for extremely large or small inputs. Additionally, cosh(x) is always positive.
### Q5: How is Hyperbolic Cosine Used in Real Life?
Hyperbolic cosine has applications in fields like physics and engineering, particularly in problems related to heat conduction, waveforms, and modeling physical systems. |
## Simplifying Expressions
In algebra, letters are used to stand for numbers.
The letters are often called variables or pronumerals.
It is necessary to be able to add, subtract, multipy and divide algebraic terms.
### Definitions
An algebraic term can be made up of coefficient, variables and exponents. Like terms have the same variables and exponents. e.g. { 2a, 4a, -6a } are like terms.a is common to all. { x, 3a2, 6 } are unlike terms. There is no common term. An algebraic expression is a group of terms. e.g. 4x + 3y is an expression.
### Addition and Subtraction
An expression involving addition and subtraction can be simplified only if it contains like terms.
The like terms are collected together and then added or subtracted.
Examples Answers Simplify: (a) 5a + 6a − 2a (a) 5a + 6a − 2a = 9a (b) 18pq − 10pq + 4pq (b) 18pq − 10pq + 4pq = 12pq (c) 3c + 4d2 + 5c − d2 (c) 3c + 4d2 + 5c − d2 = 8c + 3d2 (d) 4x2 + 10x2 − 3x2 4x2 + 10x2 − 3x2 = 11x2
### Multiplying
3xy means "3 multiplied by x multiplied by y ".
In terms of this type the number comes first and the variables are usually placed in alphabetical order.
e.g. 3q × 4p = 12pq
5c × 3e × 2d = 30cde
Note that numbers are multiplied first and letters (variables) are placed in alphabetical order.
### Dividing
8xy4x means " 8xy divided by 4x ".
This type of expression is simplified by cancelling.
e.g. 2y
Note Cancelling can only be done between numbers or letters on the top line and numbers or letters on the bottom line.
### Words into Symbols
Many maths problems are written in words and these words often have to be changed into algebraic expressions.
Words Algebra 3 is added to x x + 3 5 is subtracted from y y − 5 z is multiplied by 4 4z w is divided by 10 w⁄10 The product of x and y xy The difference between a and 5 a − 5 The sum of 7 and w 7 + w p is muliplied by q and 5 is added pq + 5 6 multipied by c is equal to 30 6c = 30 |
# Tessellations.
## Presentation on theme: "Tessellations."— Presentation transcript:
Tessellations
Tessellation A tessellation or a tiling is a way to cover a floor with shapes so that there is no overlapping or gaps. Remember the last jigsaw puzzle piece you put together? Well, that was a tessellation. The shapes were just really weird.
Examples Brick walls are tessellations. The rectangular face of each brick is a tile on the wall. Chess and checkers are played on a tiling. Each colored square on the board is a tile, and the board is an example of a periodic tiling.
Examples Mother nature is a great producer of tilings. The honeycomb of a beehive is a periodic tiling by hexagons. Each piece of dried mud in a mudflat is a tile. This tiling doesn't have a regular, repeating pattern. Every tile has a different shape. In contrast, in our other examples there was just one shape.
Alhambra The Alhambra, a Moor palace in Granada, Spain, is one of today’s finest examples of the mathematical art of 13th century Islamic artists.
Tesselmania Motivated by what he experienced at Alhambra, Maurits Cornelis Escher created many tilings.
Regular tiling To talk about the differences and similarities of tilings it comes in handy to know some of the terminology and rules. We’ll start with the simplest type of tiling, called a regular tiling. It has three rules: The tessellation must cover a plane with no gaps or overlaps. The tiles must be copies of one regular polygon. Each vertex must join another vertex. Can we tessellate using these game rules? Let’s see.
Regular tiling Tessellations with squares, the regular quadrilateral, can obviously tile a plane. Note what happens at each vertex. The interior angle of each square is 90º. If we sum the angles around a vertex, we get 90º + 90º + 90º + 90º = 360º. How many squares to make 1 complete rotation?
Regular tiling Which other regular polygons do you think can tile the plane?
Triangles Triangles? Yep!
How many triangles to make 1 complete rotation? The interior angle of every equilateral triangle is 60º. If we sum the angles around a vertex, we get 60º + 60º + 60º + 60º + 60º + 60º = 360º again!.
Pentagons Will pentagons work?
The interior angle of a pentagon is 108º, and 108º + 108º + 108º = 324º.
Hexagons Hexagons? The interior angle is 120º, and 120º + 120º + 120º = 360º. How many hexagons to make 1 complete rotation?
Heptagons Heptagons? Octagons?
Not without getting overlaps. In fact, all polygons with more than six sides will overlap.
Regular tiling So, the only regular polygons that tessellate the plane are triangles, squares and hexagons. That was an easy game. Let’s make it a bit more rewarding.
Semiregular tiling A semiregular tiling has the same game rules except that now we can use more than one type of regular polygon. Here is an example made from a square, hexagon, and dodecahedron: To name a tessellation, work your way around one vertex counting the number of sides of the polygons that form the vertex. Go around the vertex such that the smallest possible numbers appear first.
Semiregular tiling Here is another example made from three triangles and two squares: There are only 8 semiregular tessellations, and we’ve now seen two of them: the and the Your in-class construction will help you find the remaining 6 semiregular tessellations.
Demiregular tiling The 3 regular tessellations (by equilateral triangles, by squares, and by regular hexagons) and the 8 semiregular tessellations you just found are called 1-uniform tilings because all the vertices are identical. If the arrangement at each vertex in a tessellation of regular polygons is not the same, then the tessellation is called a demiregular tessellation. If there are two different types of vertices, the tiling is called 2-uniform. If there are three different types of vertices, the tiling is called 3-uniform.
Examples There are 20 different 2-uniform tessellations of regular polygons. / / /
Summary Regular Tessellation Semiregular Tessellation
Only one regular polygon used to tile Semiregular Tessellation Uses more than one regular polygon Has the same pattern of polygons AT EVERY VERTEX Demiregular Tessellation Has DIFFERENT patterns of polygons used at vertices Must name all different patterns.
Name the Tessellation SemiRegular 4.6.12
Regular? SemiRegular? DemiRegular? SemiRegular
Name the Tessellation Demiregular 3.12.12/3.4.3.12
Regular? SemiRegular? DemiRegular? Demiregular /
Name the Tessellation Demiregular 3.3.3.3.3.3/3.3.4.12
Regular? SemiRegular? DemiRegular? Demiregular /
Name the Tessellation DemiRegular 3.6.3.6/3.3.6.6
Regular? SemiRegular? DemiRegular? DemiRegular /
Name the Tessellation SemiRegular 3.3.4.3.4
Regular? SemiRegular? DemiRegular? SemiRegular
A Tessellation Review: The Basics…
REGULAR POLYGONS… have 3 or more sides. have 3 or more angles.
all sides are equal. all angles are equal.
What Is A Tessellation?
A REGULAR TESSELLATION is…
a tessellation made up of congruent regular polygons. Regular polygons are polygons that are the same size and shape. Regular means that the sides are all the same length.
What famous artist uses tessellations in his work?
This is a piece by the artist, M.C.Escher. Can you guess the title??? LIZARDS!!!
A REGULAR TESSELLATION is…
a tessellation made up of congruent regular polygons. REMEMBER… Regular polygons are polygons that are the same size and shape. Regular means that the sides are all the same length.
Extra! Extra! Other Tessellation PowerPoint Information
Student Tessellation Webquest |
# Percentages 17/10/2015 1 17 October 2015. Contents Converting between Fractions Decimals and Percentages Finding a Percentage Profit & Loss Reverse Percentages.
## Presentation on theme: "Percentages 17/10/2015 1 17 October 2015. Contents Converting between Fractions Decimals and Percentages Finding a Percentage Profit & Loss Reverse Percentages."— Presentation transcript:
Percentages 17/10/2015 1 17 October 2015
Contents Converting between Fractions Decimals and Percentages Finding a Percentage Profit & Loss Reverse Percentages Writing as a Percentage 17/10/2015 2
Converting between F, D & P Converting a percentage to a fraction 40% means 40 out of every 100 Don’t forget to cancel down, if possible
Converting between F, D & P Converting a percentage to a decimal 67% means i.e. 67 100 Remember by 100 ? HTU 67 6 7 So, 67% = 0.67
Converting between F, D & P Converting a decimal to a percentage Reverse the process i.e. X by 100 HTU 043 4 3 So, 0.43 = 43%
Converting between F, D & P Converting a fraction to a percentage Convert to a decimal first then to a percentage i.e. Then, change to a percentage, x by 100 So, = 0.6 = 60%
Finding a Percentage - Remember, 10% = To find we by 10 Also, 1% = And, to find we by 100 Without a Calculator
Finding a Percentage - Use these facts to find any percentage i.e. Find 32% of \$240 10% = \$24 and 1% = \$2.40 So, 30% = 3 x \$24 = \$72 and 2% = 2 x \$2.40 = \$4.80 32% = \$72.00 + 4.80 = \$76.80 Without a Calculator
Finding a Percentage - 55% of 120 children at the theatre were boys, how many were boys ? 10% = 12 and 1% = 1.2 So, 50% = 5 x 12 = 60 and 5% = 5 x 1.2 = 6 55% = 60 + 6 = 66 boys (5% can also be found by using ½ of 10%) Without a Calculator
Finding a Percentage - Change percentage to a decimal first eg. Find 28% of 690 28% = 0.28 and “of” means multiply So 28% of 690 is 0.28 x 690 Type into your calculator Answer = 193.2 With a Calculator
Finding a Percentage - Another example, find 17.5% of \$250 So, 0.175 x 250 Type into calculator Answer = \$43.75 Find 32.5% of 1200 … 0.325 x 1200 = With a Calculator 390
Profit & Loss 2 types of question Type 1 - A car was bought for \$1200 and was later sold at a 15% profit, how much was it sold for ? Find 15% and then add it on to \$1200 If it were sold for a 24% loss Find 24% and then take it off the \$1200
Profit & Loss Type 2 – A car was bought for \$1200 and later sold for \$1500, what is the percentage profit ? Use the format To create a fraction Cancel to simplest form and then change to a percentage Actual Profit (or Loss) Original Amount
Profit & Loss A car was bought for \$1200 and later sold for \$1400, what is the percentage profit ? Actual Profit Original Amount
Profit & Loss A cycle was bought for \$600 and later sold for \$450, what is the percentage loss ? Actual Loss Original Amount
Reverse Percentages The original amount is always 100% A reduction of 20% means the new price is 80% of original An increase of 15% means the new price is 115% of original Use the calculator method to find original amount
Reverse Percentages eg. In a 25% sale a sofa costs \$480, how much did it cost before the sale ? 25% reduction means 75% of original i.e. 100% - 25% = 75% So, £480 0.75 = \$640 Price before Sale ? Price after Sale £480 ÷ 0.75 x 0.75
Reverse Percentages eg. Following a 10% increase petrol now costs \$1.20 per litre, how much did it cost before the increase ? 10% increase means 110% of original So, \$1.20 1.10 = \$1.09 per litre Price before increase ? New Price \$1.20 ÷ 1.10 x 1.10
Writing as a Percentage One quantity as a percentage of another eg. Aylish scored 32 out of 50 in a science test and 48 out of 80 in maths Write as a fraction first, then cancel down ScienceMaths = 64%= 60%
Writing as a Percentage What percentage of cars are Green ? 22 out of 122 were green, so Change to a decimal Then convert to a percentage Car Park Survey ColourFrequency Green22 Silver43 Black57 =18%
Session Summary Converting between Fractions Decimals and Percentages Finding a Percentage Profit & Loss Reverse Percentages Writing as a Percentage Next week - Ratio 17/10/2015 21
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# High School Geometry Practice - Measurement
## CCSS Geometry Problems
This quiz is designed in accordance with the following Common Core State (CCS) standards:
Question Common Core Standard # 1 G.GMD.2 + 2 G.GMD.3 3 G.GMD.3 4 G.GMD.3 5 G.GMD.3 6 G.GMD.3 7 G.GMD.3 8 G.GMD.4 9 G.GMD.4
indicates a modeling standard.
+ indicates a college and career ready standard.
All questions of this quiz illustrate the difficulty of the high school G.MD standards required for all students, including the college and career ready students.
Question 1: The rectangular solid S and the prism P in the figure below have the same height. The base of S is a square with the sides of length a and the base of P is a right triangle with catheti of lengths a and 2a. If the volume of P is 200cm3, what is the volume of S?
Question 2: The standard ice hockey rink is rectangular with rounded corners, as shown in the figure below. Calculate the volume of ice from the rink, if the height of the ice is one tenth of a foot.
Question 3: An ice cream cone can be modeled by a semi sphere on top of a cone, as shown in the picture below. Calculate the volume of the ice cream cone as a function of r, if the height of the cone is h = 4r.
Questions 4 and 5: A sphere with a radius of 5cm is inscribed in a cylinder.
The ratio between the volume of the cylinder and the volume of the sphere is .
A cross-section by a horizontal plane at a height of 2 cm from the bottom of the cylinder intersects both solids. What is the area of the portion of the cross section inside the cylinder but outside the sphere?
Question 6: In the figure below, a sphere is inscribed in a right circular cone. What is the height of the cone if R = 4 inches and r = 2 inches?
Question 7: Given the triangle ABC in the figure below, sides AB and AC have the following lengths: AB = 3 and AC = 6. What is the volume of the shape formed by rotating the triangle around AB?
Question 8: The rectangle ABCD in the figure below has sides of lengths a and b. The volume of the shape formed by rotating the rectangle around AB is V1 and the volume of the shape formed by rotating the rectangle around BC is V2.
Calculate the ratio V1/V2.
Press the Submit button to see the results. |
# How do you show a function is discontinuous at a point?
## How do you show a function is discontinuous at a point?
To show from the (u03b5, u03b4)-definition of continuity that a function is discontinuous at a point x0, we need to negate the statement: For every u03b5 x26gt; 0 there exists u03b4 x26gt; 0 such that |x u2212 x0| x26lt; u03b4 implies |f(x) u2212 f(x0)| x26lt; u03b5. Its negative is the following (check that you understand this!): There exists an u03b5 x26gt; 0 such that for
## How do you know if a function is discontinuous?
If the function factors and the bottom term cancels, the discontinuity at the x-value for which the denominator was zero is removable, so the graph has a hole in it. After canceling, it leaves you with x 7. Therefore x + 3 0 (or x 3) is a removable discontinuity the graph has a hole, like you see in Figure a.
## What is an example of a discontinuous function?
A discontinuous function is a function that has a discontinuity at one or more values mainly because of the denominator of a function is being zero at that points. For example, if the denominator is (x-1), the function will have a discontinuity at x1.
## How do you know if a graph is discontinuous?
If the function factors and the bottom term cancels, the discontinuity at the x-value for which the denominator was zero is removable, so the graph has a hole in it. After canceling, it leaves you with x 7. Therefore x + 3 0 (or x 3) is a removable discontinuity the graph has a hole, like you see in Figure a.
## What are the 3 types of discontinuous functions?
There are three types of discontinuities: Removable, Jump and Infinite.
## What kind of functions are discontinuous?
A discontinuous function is the opposite. It is a function that is not a continuous curve, meaning that it has points that are isolated from each other on a graph. When you put your pencil down to draw a discontinuous function, you must lift your pencil up at least one point before it is complete.
## How do you know if a function is continuous or discontinuous?
Explanation: Start by factoring the numerator and denominator of the function. A point of discontinuity occurs when a number is both a zero of the numerator and denominator. Since is a zero for both the numerator and denominator, there is a point of discontinuity there.
## How do you know if a function is discontinuous on a graph?
Definition. A function f is continuous at a if limxaf(x)f(a). We say f is discontinuous at a if f is not continuous at a.
## How do you tell if a graph is continuous or discontinuous?
A function being continuous at a point means that the two-sided limit at that point exists and is equal to the function’s value. Point/removable discontinuity is when the two-sided limit exists, but isn’t equal to the function’s value.
## What makes a graph discontinuous?
Discontinuous functions are functions that are not a continuous curve – there is a hole or jump in the graph. It is an area where the graph cannot continue without being transported somewhere else. There are many types of continuities.
## What does discontinuous look like on a graph?
The point, or removable, discontinuity is only for a single value of x, and it looks like single points that are separated from the rest of a function on a graph. A jump discontinuity is where the value of f(x) jumps at a particular point.
## What are the types of discontinuous functions?
There are two types of discontinuities: removable and non-removable. Then there are two types of non-removable discontinuities: jump or infinite discontinuities. Removable discontinuities are also known as holes. They occur when factors can be algebraically removed or canceled from rational functions.
## What are the 4 types of discontinuity?
There are four types of discontinuities you have to know: jump, point, essential, and removable.
## What are the 3 conditions for a function to be continuous?
Key Concepts. For a function to be continuous at a point, it must be defined at that point, its limit must exist at the point, and the value of the function at that point must equal the value of the limit at that point.
## How can you classify discontinuities?
Discontinuities can be classified as jump, infinite, removable, endpoint, or mixed. Removable discontinuities are characterized by the fact that the limit exists. Removable discontinuities can be fixed by re-defining the function.
## What types of functions are discontinuous?
Summary
• There are two types of discontinuities: removable and non-removable.
• Removable discontinuities are also known as holes.
• Jump discontinuities occur when a function has two ends that don’t meet, even if the hole is filled in at one of the ends. |
# Free Simple Equations 01 Practice Test - 7th grade
If 50 is subtracted from a number , the result is 4. Find the number.
A.
54
B.
46
C.
25
D.
45
#### SOLUTION
Solution : A
Let us say 'x' is the number . Given that 50 is subtracted from it and the result obtained is 4.
So mathematically, x50=4 is the equation.
Thus, x=50+4=54
54 is the number.
Krishna and Karan are on the see-saw. Krishna's weight is 60 kg and Karan's is 80 kg. Who among the following kids can help Krishna balance the see-saw. (All of them can sit at extreme ends only).
A. Peter who has weight of 21 kg
B. Alison who has weight of 23 kg
C.
Pandu who has weight of 20 kg
D.
Mahi who has weight of 10 kg
#### SOLUTION
Solution : C
It is given that Krishna's weight is 60 kg and Karan's weight is 80 kg. To balance the see-saw Krishna needs another person.
Let the required weight needed to balance the see-saw be x kg
Then, 60+x=80 x=8060=20
So, Krishna should take help of a kid who weighs exactly 20 kg. Thus, Pandu would be the correct selection.
In a quiz program,the correct answer gets you 2 marks. For each wrong answer, one mark is deducted from the total score. There are 50 questions to be answered. If Preeti scored 76 marks for the quiz, how many of her questions were right?
A. 26
B. 13
C. 42
D. 38
#### SOLUTION
Solution : C
As the number of questions answered is unknown, let us assume that unknown variable as 'x'.
If the number of questions answered correctly are x, then the number of questions answered incorrectly will be (50x).
For each correct answer, she gets 2 marks. So the marks scored for x correctly answered questions are 2x. Similarly, the marks deducted for (50-x) incorrectly answered questions are (50x)×1=50x
Now, the total marks scored by Preeti will be
2x- (50 -x) = 2x -50 +x = 3x -50 (For each wrong answer, one mark is deducted from the total score)
The final equation is, 3x -50 = 76
3x = 50 +76 = 126
x=1263=42
So, the total number of questions answered correctly by Preeti are 42.
Step 1: 3x+2=17
Step 2: 3x=172
This process of shifting the number from one side of the equation to the other side is called as ___.
A.
translation
B.
root of equation
C.
transposition
D.
correlation
#### SOLUTION
Solution : C
Transposition is nothing but balancing the other side of the equality when we make any change on one side.
Here, in the equation 3x+2=17, when we add 2 on both sides, we get 3x+22=172 i.e., 3x=172.
Instead of following the above process, 2 is shifted to the other side of the equation.
Therefore, we get 3x=172
This process is known as transposition.
Karan and Kundan are two brothers. Presently Karan's age is four times the age of Kundan. After 6 years Karan's age will be twice as that of Kundan. What is the present age of Kundan?
__
#### SOLUTION
Solution :
Let the present age of Kundan be x.
So, Karan's present age is 4x.
After 6 years,
Kundan's age = (x+6)
Karan's age = (4x+6)
Also, Karan's age would be twice of Kundan.
So, (4x +6) = 2 (x +6)
4x +6 = 2x +12
x = 3
One pleasant morning, a number 'a' goes for a jog. Another number 'b' passes by 'a' and comments "I am four more than your half”. 'a' gets offended and starts shouting at 'b'. Now, both of them got into a big fight about their value and soon a big crowd gathered. A wise man, Krishna comes, looks at both the numbers, and says, "If you look closely, you are both equals.” 'a' and 'b' realised that the wise man was correct. What are the numbers 'a' and 'b'?
A. 8,8
B. 8,4
C. 4,4
D. 8,10
#### SOLUTION
Solution : A
From the given information, b=4+a2
The wise man Krishna said that both a and b are equal.
So, a=b
Hence subsituting in a = b in the first equation , we have
a=4+a2
aa2=4
a2=4
a=8
Since a = b b = 8
Bittu tells Kitty " I have 5 chocolates more than seven times the number of chocolates you have”, if Kitty has 5 chocolates, how many chocolates does Bittu have ?
___
#### SOLUTION
Solution :
The number of chocolates Bittu has
= 5+(7×number of chocolates Kitty has)=5+(7×5)
=40
My granddaughter is 10 years old in 2015. I am 8 more than 6 times her age. I was born in the year
#### SOLUTION
Solution :
Grand daughter's age is 10 years now. So her grandpa's age is
8+6×10=8+60=68
Grandpa is 68 years old by 2015. So he was born in 1947.
There is always an equality sign in an equation.
A.
True
B.
False
#### SOLUTION
Solution : A
An expression having equality sign is called an equation.
2a15=10 and 10=15+2a are not the same equations.
A.
True
B.
False
#### SOLUTION
Solution : B
An equation remains same even if the LHS and RHS are interchanged. |
# Olympiad Test: Direction Sense Test -1
## 10 Questions MCQ Test Mathematical Olympiad Class 8 | Olympiad Test: Direction Sense Test -1
Description
Attempt Olympiad Test: Direction Sense Test -1 | 10 questions in 20 minutes | Mock test for Class 8 preparation | Free important questions MCQ to study Mathematical Olympiad Class 8 for Class 8 Exam | Download free PDF with solutions
QUESTION: 1
### Nitesh faces towards North. Turning to his right he walks 20 meters. He then turns to his left and walks 20 meters, then he moves 30 m to his right then turns to his right again and walks 45 meters. At last he turns to his right and moves 35 meters. In which direction is he now from the starting position?
Solution:
Starting position of Nitesh is A and his final position is F which is in the South-East direction from starting point.
QUESTION: 2
### Ranjan is looking for Ratan. He went 20m in the east before turning to his right. He went 20m before turning to his right again to look Ratan at Mohan’s position 30m from this point. Ratan was not there. From that point he went 100m to his north before meeting Ratan. How did Ranjan meet Ratan from the starting point?
Solution:
In the above Fig. A is Ranjan’s position and E is Ratan’s position. The required distance = AE
QUESTION: 3
### Dinesh walks 10m in front and 10m to the right then turning to his left three times he walks 5m, 15m, 15 m respectively. How far is he from his starting position?
Solution:
Dinesh starts from A his final position is at F.
AF = DE – GD = 15 – 10 = 5 m
QUESTION: 4
Pritam walks 15m towards south then turning to his right he walks 30m then turning to his left he walks 20m. Again he turns to his left and walks 30m. How far is he from his starting position?
Solution:
Pritam’s starting point is at A.
Final position is E.
AE = A B + BE
= AB + CD
= 15 + 20
= 35 m
QUESTION: 5
Ankit is facing East. He turns 100° in the clockwise direction and then 145° in the anticlockwise direction. In which direction is he facing now?
Solution:
Ankit’s final position is OC, which is in North-East.
QUESTION: 6
Nishi walked 20m towards North, took a left turn and walked 10m. She again took a left turn and walked 20m. How far and in which direction is she from starting point?
Solution:
Starting point = A
End point = D
D is 10 m and in west direction from A.
QUESTION: 7
Milan walks 10m towards North. From there he walks 6m towards south. Then he walks 3m East. How far and in which direction is he with reference to the starting point?
Solution:
Starting point = A
End point = D
AD is North-east of starting point.
QUESTION: 8
Nitu leave from her house, she first walks 20 m in North-west direction and then 20 m in South west direction. Next she walks 20 m in south east direction. At last she turns towards her house. In which direction is she moving?
Solution:
Starting point = A
Ending point = D
DA is the final position that is North-east.
QUESTION: 9
Arman is facing South. Arman turns right and walk 20m. Then he turns right again and walk 10m. Then he tums left and walk 10m and then turning right walk 20m. Then he turn right again and walk some distance. In which direction is he from the initial point?
Solution:
Starting point = A, Ending point = F
The point F is in the North-east direction.
QUESTION: 10
Mohit walked 30 m towards East, took a right turn and walked 50m then he took a left turn and walked 30m. In which direction is he now from initial point?
Solution:
Starting point = A
Ending point = D
D is in the South-east direction.
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# Lesson 4
How Many Groups? (Part 1)
Let’s play with blocks and diagrams to think about division with fractions.
### 4.1: Equal-sized Groups
Write a multiplication equation and a division equation for each sentence or diagram.
1. Eight $5 bills are worth$40.
2. There are 9 thirds in 3 ones.
### 4.2: Reasoning with Pattern Blocks
Use the pattern blocks in the applet to answer the questions. (If you need help aligning the pieces, you can turn on the grid.)
1. If a hexagon represents 1 whole, what fraction do each of the following shapes represent? Be prepared to show or explain your reasoning.
1. 1 triangle
2. 1 rhombus
3. 1 trapezoid
4. 4 triangles
5. 3 rhombuses
6. 2 hexagons
7. 1 hexagon and 1 trapezoid
2. Here are Elena’s diagrams for $$2 \boldcdot \frac12 = 1$$ and $$6 \boldcdot \frac13 = 2$$. Do you think these diagrams represent the equations? Explain or show your reasoning.
3. Use pattern blocks to represent each multiplication equation. Remember that a hexagon represents 1 whole.
1. $$3 \boldcdot \frac 16=\frac12$$
2. $$2 \boldcdot \frac 32=3$$
4. Answer the questions. If you get stuck, consider using pattern blocks.
1. How many $$\frac 12$$s are in 4?
2. How many $$\frac23$$s are in 2?
3. How many $$\frac16$$s are in $$1\frac12$$?
### Summary
Some problems that involve equal-sized groups also involve fractions. Here is an example: “How many $$\frac16$$ are in 2?” We can express this question with multiplication and division equations. $$\displaystyle {?} \boldcdot \frac16 = 2$$ $$\displaystyle 2 \div \frac16 = {?}$$
Pattern-block diagrams can help us make sense of such problems. Here is a set of pattern blocks.
If the hexagon represents 1 whole, then a triangle must represent $$\frac16$$, because 6 triangles make 1 hexagon. We can use the triangle to represent the $$\frac 16$$ in the problem.
Twelve triangles make 2 hexagons, which means there are 12 groups of $$\frac16$$ in 2.
If we write the 12 in the place of the “?” in the original equations, we have: $$\displaystyle 12 \boldcdot \frac16 = 2$$
$$\displaystyle 2 \div \frac16 = 12$$ |
# How do you explain odd numbers to children?
## How do you explain odd numbers to children?
Children in Key Stage 1 need to learn about odd and even numbers. An even number is a number that can be divided into two equal groups. An odd number is a number that cannot be divided into two equal groups.
## What is the importance of mathematics in your profession?
For example, they need math to treat patients, understand medical tests, and operate medical equipment. In fact, math is a basic skill in most occupations. Therefore, math skills are very important to your teen’s future and job success.
## Why is it important to learn odd and even numbers?
Identifying even and odd numbers is an important skill that children need to help them understand our number system and aid in their preparation to group whole number operations. It will also help prepare them to learn division, prime numbers and even square roots.
## Which is the odd number?
Odd numbers are whole numbers that cannot be divided exactly into pairs. Odd numbers, when divided by 2, leave a remainder of 1. 1, 3, 5, 7, 9, 11, 13, 15 … are sequential odd numbers. Odd numbers have the digits 1, 3, 5, 7 or 9 in their ones place.
## Why do odd numbers look better?
An odd number of details is more effective at capturing your gaze. Odd numbers force your eyes to move around the grouping–and by extension, the room. That forced movement is the heart of visual interest. It’s for that reason that a set of three is more appealing and memorable than something paired off in two’s.
## Why is physic important?
Physics plays an important role in many of the liberal arts disciplines and contributes to society’s understanding of such areas as energy, weather, medical science, and space exploration. A rigorous major provides students with the preparation required for graduate study in physics.
## Why do I like even numbers?
King and Janiszewski argue that our brains process even numbers more easily than odd numbers, and that this increased fluency translates as liking for the product. Even numbers are more easily processed, they say, because they appear more frequently in the times tables.
## What are odd numbers from 1 to 100?
The odd numbers from 1 to 100 are: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99.
## How do you teach even numbers to kindergarten?
Here are five easy ways to teach odd and even numbers!
1. Line the Children Up in Pairs. I think it is easiest for the kids to understand what an odd number is FIRST by lining them up by twos.
2. Explain the Concept and Sing It.
3. Review It Daily with the Calendar.
4. Sing the Count by Twos Song, and Write Those Numbers!
## Is 95 odd or even?
95 is an odd number. |
# 7/30 simplified
#### Understand the Problem
The question is asking to simplify the fraction 7/30. The approach involves finding the greatest common divisor (GCD) of the numerator and denominator and dividing both by that number.
The simplified fraction is $\frac{7}{30}$.
The simplified fraction is $\frac{7}{30}$.
#### Steps to Solve
1. Identify the GCD of 7 and 30
To simplify the fraction, we first find the greatest common divisor (GCD) of the numerator (7) and the denominator (30).
The prime factorization of 7 is just 7 (since it's a prime number), and the prime factorization of 30 is $2 \times 3 \times 5$.
Since there are no common factors, the GCD of 7 and 30 is 1.
1. Divide by the GCD
Next, we simplify the fraction by dividing both the numerator and the denominator by the GCD.
[ \frac{7}{30} = \frac{7 \div 1}{30 \div 1} = \frac{7}{30} ]
1. State the simplified fraction
Since the GCD is 1, the fraction is already in its simplest form.
Thus, the simplified fraction remains:
[ \frac{7}{30} ]
The simplified fraction is $\frac{7}{30}$.
The fraction $\frac{7}{30}$ is in its simplest form as 7 and 30 have no common factors other than 1. This means that it cannot be simplified further. |
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# Year 5
### Using Our Maths A
#### Following Units 1-5
These activities bring together the different mathematical skills that your child should now know. By combining these skills in puzzles and activities your child will need to reason and think of solutions, as well as complete calculations. This will help your child to really understand the skills they are learning.
Talk through each group of questions and ask your child how they are going to solve the challenges.
It’s good to get stuck sometimes! When this happens discuss the challenge in the following order:
• What do we need to find out to solve the challenge?
• What do we know by looking at the question?
• Want could we do first?
• How would that help us?
Is there anything else we could do?
#### Ordering tricky numbers
1. Put these numbers in order from the smallest to the largest. Then pick two numbers that are next to each other and explain what makes one bigger than the other. Do this again for other pairs. Being able to explain your thinking is just as important and getting the numbers in the right order.
303, 3303, 30300, 3300, 333, 33, 3003
2856, 8256, 6825, 5628, 5682, 2658, 6852, 8562
2. What is the value of each digit in red? Explain how you know.
243.607
805.01
200.5
609.99
1356.82
303000
5432641
8328688
3. Calculate these on paper and then check them with a calculator.
2.86 + 3.09 – 1.06
264.11 – 86.09 + 66.66
3056 – 89.99
58.61 + 359 – 26.04
88.8 + 99.9 – 66.6
4. Count up in steps of 10000
38,——–,———-,———,———-,———-.
5. Count up in steps of 10,000
860,—————–, —————,—————-,—————,————-.
6. Count down in steps of 100
1462,———-,———–,————–,————-,————-.
7. Count up in steps of 100,000
890,————–,—————,—————,—————–,—————.
8. Count down in steps of 1000
26,412,——————–,—————–,—————–,—————-,————-. |
## Intersection of concentric, unit area sphere and square
Where do a concentric sphere and square, both of area 1, intersect?
The shortest distance from the center on this square is 1/2 units. The radius of the sphere with a surface area of 1 is inferior to 1/2, meaning that the square and the sphere meat at every point on the circle of the same radius which is concentric and coplanar to the square.
Mentor Blog Entries: 9 Do you mean square and circle? Or Sphere and cube? Area Or Volumn?
## Intersection of concentric, unit area sphere and square
Sorry, it was getting late. I meant to say
Where do a concentric circle and square, both of area 1, intersect?
Well, where in terms of what? Interval of points on the circle? If so, assuming the sides of the square are parallel to the axes, we have to calculate on which interval does the distance between the center (here the origin) and the edges of the square is equal or greater than $$\frac{1}{\sqrt{\pi}}$$.
To recap and revise: A square of 4 units area is concentric to a circle of 4 units area. Given that their common center falls on an origin of Cartesian coordinates and the square's corners are located at (1, 1) (1, -1) (-1,-1) (-1, 1), what coordinates describe the points at which the square and circle overlap?
Well, first, what's the radius of the circle? You can get that from the area. Now, draw yourself a picture where you estimate where the intersections are (you can do this now, since you know the radius of the circle). Each intersection point will be on a side of the square, so that will give you one of the coordinates of the point. To get the other one, draw the line from the origin to the point and the line from the point to the other axis (the one whose coordinate you're missing). You should now be able to get the second coordinate for the point. The other seven points' coordinates should then be obvious by inspection, given the symmetry of the problem.
Recognitions: Gold Member Science Advisor Staff Emeritus A circle of area 1 has radius $1/(2\pi)$ and so a circle of area 1 with center at the origin of a coordinate system can be represented by the equation $x^2+ y^2= 1/4\pi^2$. A square of area 1 has sides of length 1 and so a square of area 1 with center ata the origin of a coordinate system can be represented as the are contained within the lines x= 1/2, y= 1/2, x= -1/2, y= -1/2. Where does $x^2+ y^2= 1/4\pi^2$ intersect each of those? Loren, I would have thought you would have been able to do a problem like this easily!
Thanks for your confidence, Walls of Ivy. Indeed I must think twice before assuming something significant. Arrgh! Next time I will refer to paper and pencil before jumping to confusions.
HallsofIvy is clearly correct, but I find a geometric solution easier than an algebraic one. I would simply draw the right triangle with the origin at one end of the hypotenuse and the intersection point in question at the other end. One leg of the triangle has length 1 and the hypotenuse has length R, the radius of the circle. Good ol' Pythagoras gives us the other leg, which is the other coordinate of the intersection point. Different strokes for different folks - there's nothing better about this method than the one HallsofIvy offered; it's just personal preference. Of course, you end up solving the same equations, you just get there a different way.
Quote by HallsofIvy A circle of area 1 has radius $1/(2\pi)$ and so a circle of area 1 with center at the origin of a coordinate system can be represented by the equation $x^2+ y^2= 1/4\pi^2$. A square of area 1 has sides of length 1 and so a square of area 1 with center ata the origin of a coordinate system can be represented as the are contained within the lines x= 1/2, y= 1/2, x= -1/2, y= -1/2. Where does $x^2+ y^2= 1/4\pi^2$ intersect each of those? Loren, I would have thought you would have been able to do a problem like this easily!
You got distracted a bit there, you were thinking of circumference instead of area. The radius is $$\frac{1}{\sqrt{\pi}}$$. |
# Inequalities for SQA National 5 Maths
1. Basics of inequalities
2. Linear inequalities
3. Direction of inequalities
While in an equation the "=" sign indicates that the sides are identical, inequalities are used when they aren't. Different inequalities represent different relationships between the two sides. The definitions of the different symbols that can be used in inequalities are given in the diagram.
Linear inequalities can be solved just as linear equations using inverse operations, with the only difference being that the inequality sign is kept throughout rather than the "=".
When solving inequalities, if it is required to multiply or divide by a negative number, the direction of the inequality sign must be reversed. When there is an expression in terms of the variable that is being solved in the denominator, multiply every term by the square of the expression, so that there is no need to reverse the inequality sign, as this will guarantee that the number that the terms are being multiplied by is positive.
# 1
If y > 11, what are the possible values of y? Assume that y is an integer.
y > 11 means that y is greater than 11; therefore, the possible values of y are 12, 13, 14, 15, 16, … .
12, 13, 14, 15, 16, …
# 2
A square of side 2x + 5 has a perimeter less than 30. Find the range of possible values of x.
Since the perimeter of the square is less than 30, 4(2x + 5) < 30.
Expanding the left side gives 8x + 20 < 30.
Subtracting both sides by 20 gives 8x < 10.
Dividing both sides by 8 gives x < 1.25.
Since the sides of the square cannot be negative, 2x + 5 > 0.
Subtracting 5 from both sides gives 2x > −5.
Dividing both sides by 2 gives x > −5/2.
Putting the two inequalities together gives −5/2 < x < 5/4.
−2.5 < x < 1.25
# 3
Solve 3(2x − 7) < x − 1 and show the range of possible values of x on a number line.
image
# 4
Solve 22x + 8 > 24.
Dividing all terms by 2 gives 11x + 4 > 12.
Subtracting both sides by 4 gives 11x > 8.
Dividing both sides by 11 gives x > 8/11.
x > 8/11
# 5
Given that z is an integer and z ≥ −6, state the possible values of z.
z ≥ −6 means that z is greater than or equal to −6, therefore the possible values of z are −6, −5, −4, −3, −2, … .
−6, −5, −4, −3, −2, …
End of page |
Splash Screen.
Presentation on theme: "Splash Screen."— Presentation transcript:
Splash Screen
Five-Minute Check (over Lesson 3–5) Main Idea
Example 1: Multiply Decimals Example 2: Multiply Decimals Example 3: Annex Zeros in the Product Example 4: Annex Zeros in the Product Example 5: Multiply by 10, 100, or 1,000 Lesson Menu
Estimate and find the product of decimals and whole numbers.
Main Idea/Vocabulary
Since the estimate is 76, place the decimal point after the 5.
Multiply Decimals Find 18.9 × 4. Method 1 Use estimation. Round 18.9 to 19. 18.9 × 4 → 19 × 4 or 76 3 3 18.9 × 4 Since the estimate is 76, place the decimal point after the 5. 7 5 . 6 Example 1
Method 2 Count decimal places.
Multiply Decimals Method 2 Count decimal places. 3 3 18.9 × 4 one decimal place 7 5 . 6 Count one decimal place from the right. Answer: 75.6 Example 1
Find 12.6 × 8. A B C D. 1,008 A B C D Example 1
Since the estimate is 3.5, place the decimal point after the 3.
Multiply Decimals Find 0.56 × 7. Method 1 Use estimation. Round 0.56 to 0.5. 0.56 × 7 → 0.5 × 7 or 3.5 4 0.56 × 7 Since the estimate is 3.5, place the decimal point after the 3. 3 9 . 2 Example 2
Method 2 Count decimal places.
Multiply Decimals Method 2 Count decimal places. 4 0.56 × 7 two decimal places 3 9 . 2 Count two decimal places from the right. Answer: 3.92 Example 2
Find 0.83 × 4. A B. 3.32 C. 33.2 D. 332 A B C D Example 2
Annex Zeros in the Product
Find 3 × 1 0.016 × 3 three decimal places 0. 4 8 Annex a zero on the left of 48 to make three decimal places. Answer: Example 3
Find 4 × A B. 0.92 C. 9.2 D. 92 A B C D Example 3
Annex Zeros in the Product
ALGEBRA Evaluate 5g if g = 5g = 5 × Replace g with × four decimal places 0.0455 Annex a zero to make four decimal places. Answer: Example 4
ALGEBRA Evaluate 3x if x = 0.0062.
C. 1.86 D. 18.6 A B C D Example 4
Method 1 Use paper and pencil.
Multiply by 10, 100, or 1,000 MENTAL MATH Find 3.25 × 100. Method 1 Use paper and pencil. 100 × 3.25 two decimal places + 325.00 two decimal places Answer: Example 5
Multiply by 10, 100, or 1,000 Method 2 Use mental math.
Move the decimal point to the right the same number of zeros that are in 100, or 2 places. 3.25 × 100 = 3.25 Answer: 325 Example 5
MENTAL MATH Find 2.4 × 1,000. A. 24 B. 240 C. 2,400 D. 24,000 A B C D
Example 5
End of the Lesson End of the Lesson
Five-Minute Check (over Lesson 3–5) Image Bank Math Tools
Modeling Decimals Dividing Decimals Modeling Decimals Resources
Find 8.7 + 11.5. 19.2 20.1 20.2 21.2 (over Lesson 3-5) A B C D
Five Minute Check 1
Find 17.15 + 12.96. 29.11 29.95 30.06 30.11 (over Lesson 3-5) A B C D
Five Minute Check 2
Find 7.5 – 4.1. 3.4 3.6 4.4 5.6 (over Lesson 3-5) A B C D
Five Minute Check 3
Find 34.64 – 27.18. 7.22 7.46 7.72 8.56 (over Lesson 3-5) A B C D
Five Minute Check 4
Evaluate the expression a – b if a = 105.8 and b = 95.9.
(over Lesson 3-5) Evaluate the expression a – b if a = and b = 95.9. 8.9 9.7 9.9 10.3 A B C D Five Minute Check 5
(over Lesson 3-5) Based on the information in the table, how much greater was Michael Young’s batting average than Jose Reyes’ batting average? .058 .055 .041 .028 A B C D Five Minute Check 5
End of Custom Shows |
Lesson Explainer: The Domain and the Range of a Radical Function | Nagwa Lesson Explainer: The Domain and the Range of a Radical Function | Nagwa
# Lesson Explainer: The Domain and the Range of a Radical Function Mathematics • Second Year of Secondary School
## Join Nagwa Classes
In this explainer, we will learn how to find the domain and the range of a radical function either from its graph or from its defining rule.
In particular, we will focus on the domain and range of functions involving the square and the cube roots.
Let us begin by recalling the definitions of domain and range of a function.
### Theorem: Domain and Range of a Function
The domain of a function is a set of all possible values such that the expression is defined.
The range of a function is a set of all possible values the expression can take, when is any number from the domain of the function.
For example, consider the function . If there is a real number satisfying , it must be the case that . We know that the square of any real number is not negative; thus must be nonnegative. This tells us that the domain of the square root function is , which is expressed in the interval notation as .
The graph of the square root function is shown below.
In the figure above, the square root function is graphed over the interval . Although the graph of the function appears to become flatter for larger values of , it continues to increase without a bound. We can observe this effect by graphing the square root function over a larger interval, .
We can see that the -value of the graph keeps increasing for larger values of . In fact, given any large positive number , we know that . Also, we know that . Hence, the range of the square root function is .
### Theorem: Domain and Range of the Square Root Function
The domain and range of the square root function, , is .
More generally, the domain of a composite square root function can be identified by finding the values of satisfying .
Let us consider a few examples where we identify the domain and the range of functions involving the square root.
### Example 1: Finding the Domain of Root Functions
Consider the function .
1. Find the domain of .
2. Find the range of .
Part 1
We recall that the square root cannot take a negative number as an argument. Hence, the domain of the given function is found by setting the expression inside the square root to be greater than or equal to zero. In other words,
This leads to , which is in interval notation.
The domain of is .
Part 2
The range of a function is the set of all possible function values. We know that the range of the square root function is . In other words, for any number in the interval , we can find some number that satisfies . This means that the number satisfies
Hence, any number in the interval is a possible function value of .
The range of is .
Let us consider another example for obtaining the domain of a composite square root function.
### Example 2: Finding the Domain of Root Functions
Find the domain of the function .
Recall that the domain of a function is the set of -values satisfying
In this example, . Hence, the domain of this function is the set of -values such that
Rearranging this inequality leads to , which is written as in interval notation.
The domain of is .
In the next example, we will determine the correct graph of a composite square root function by considering its domain and range.
### Example 3: Identifying the Graph of a Radical Function
Which of the following is the graph of ?
Let us use the domain and the range of to identify the graph. We know that the domain of is the set of -values satisfying
In this example, . Hence, the domain of this function is the set of -values such that
Rearranging this inequality leads to , which is written as in interval notation. Hence, the domain of is .
Recall that the range of the square root function is . We have observed that where . Since the range of is all real numbers, must have the same range as . This leads to the range of , which is .
Let us identify which one of the graphs represents a function whose domain is and whose range is . From each given graph, the domain of the function corresponding to the graph is the part of the horizontal axis where the graph exists. Also, the range of the function is the part of the vertical axis where the graph exists. We obtain the domain and range of each function using its graph.
1. Domain: , range:
2. Domain: , range:
3. Domain: , range:
4. Domain: , range:
5. Domain: , range:
Therefore, the only possible match for is D.
We might recall that when working with rational functions, we must be careful to restrict the domain to ensure that the function on the denominator of the expression cannot be equal to zero. Let us demonstrate how to find the domain of a function which is a ratio of two composite square root functions.
### Example 4: Finding the Domain of Rational Functions
Determine the domain of the function .
In the given function, we observe two types of restrictions to the domain.
• Square root: there are two expressions and . We recall that the square root function cannot take a negative number.
• Denominator: this is the expression . We recall that the denominator of a fraction cannot equal zero.
We begin by considering the restrictions imposed by the square roots. For to be well defined, we need . This leads to , which is the interval .
For to be well defined, we need , which leads to , or .
Finally, let us consider the denominator. Since the denominator cannot equal zero, we need to exclude the case when . Squaring both sides of this equation gives , leading to . Hence, we need to impose .
The domain of this function is the set of -values satisfying all three conditions:
To see how these three restrictions interact, let us draw a number line with these restrictions.
In the diagram above, the purple highlight represents the interval , the green highlight represents the interval , and the red X represents the restriction . We can draw an interval that satisfies the intersection of these restrictions simultaneously.
Hence, the domain of is .
### Example 5: Finding the Domain of Rational Functions
Find the domain of .
In the given function, we observe two types of restrictions to the domain.
• Square root: there are the three expressions , , and . We recall that the square root function cannot take a negative number.
• Denominator: this is the expression . We recall that the denominator of a fraction cannot equal zero.
We begin by considering the restrictions imposed by the square roots. For to be well defined, we need . This leads to , which is the interval .
Similarly, the expressions and lead to the intervals and respectively.
Next, let us consider the denominator. Since the denominator cannot be equal to zero, we need to exclude the case when
We can add to both sides of this equation to obtain
Hence, we need to impose .
We have found four domain restrictions for :
Let us draw a number line to visualize how these restrictions interact.
In the diagram above, the purple highlight represents the interval , the green highlight represents the interval , the blue highlight represents , and the red X represents the restriction . We can draw the set satisfying all four restrictions simultaneously.
Hence, the domain of is .
In previous examples, we consider the domain and range of square root functions with linear expressions, . In the next example, we will find the domain and the range of a composite square root function where the expression inside the square root involves the absolute value function.
### Example 6: Finding the Domain of Rational Functions
Consider the function .
1. Find the domain of .
2. Find the range of .
Part 1
Let us find the domain of the given function. We know that the domain of is the set of -values satisfying
In the function , the expression is under the square root. Hence, we need to find the values of such that
To solve an inequality with an absolute value, we begin by isolating the absolute value on one side of the inequality. Adding to both sides of the inequality, we obtain
We can consider this in two separate situations. First, consider the case that is nonnegative. Since the absolute value does not change a nonnegative number, the inequality is the same as
Second, consider the case when is negative. Since the absolute value takes away a negative sign, needs to be at least to satisfy the inequality . In other words, . Together, is the same as
The domain of is .
Part 2
Let us find the range of . The range of a function is the set of all possible values of the function. We can obtain the range of a function by considering what are the largest and the smallest possible values of the function. Since any function value is a square root of a number, we know that it cannot be negative. Thus, 0 would be the smallest possible function value if it is possible. For this, we need the expression under the square root to be equal to zero. This leads to
This is possible when or . This means that the smallest possible value of is 0.
To find the largest possible value of , we note that is a nonnegative number. Since the expression inside the square root is , the function would have the largest value when , which happens when . In this case, the function value is . So, the largest possible value of is 2.
To finally conclude that the range of this function is , we need to know that all values between 0 and 2 are possible. If is any value between 0 and 2, let us find a number such that . We have
Squaring both sides of the equation,
Rearranging the equation,
As we have done previously, we can split this equation into the two cases, depending on the sign of . But, since we just need to find one possible value , let us just take . In this case, we have which leads to . Let us check if . Substituting into the function , we obtain
Since , we know that , so we can ignore the absolute value in . Continuing from above:
Since , . Hence, we have shown that, for any , we have that satisfies .
The range of is .
We have considered many examples on domain and range of functions involving the square root. Unlike the square root, the cube root function does not impose any restrictions on the domain or the range. Following is the graph of the cube root function, .
Unlike the square root function, we note that the function extends to the left and the right side of the -axis, indicating that the cube root can take any real numbers. We also note that the -values tend to positive or negative infinity as we move to the left or the right. This indicates that the range of the cube root function is all real numbers.
### Theorem: Domain and Range of the Cube Root Function
The domain and range of the cube root function, , are all real numbers. This is denoted as or .
In the next example, we will identify the domain of a cube root function where the expression under the cube root is linear, .
### Example 7: Finding the Domain of a Cubic Root Function
Determine the domain of the function .
Recall that the domain and the range of the cube root function is . In other words, the cube root function does not impose any domain restriction. Since the expression, , under the cube root does not have any domain restriction, there are no restrictions to possible -values for this function.
Hence, the domain of is all real numbers, .
In our final example, we will find the domain and the range of a function that involves both the square and the cube roots.
### Example 8: Finding the Domain of Rational Functions
Consider the function .
1. Find the domain of .
2. Find the range of .
Part 1
Let us find the domain of . We recall the domain restrictions for the square and the cube roots.
• The square root function, , has the domain .
• The cube root function, , has no domain restrictions. The domain of the cube root function is all real numbers, or .
Since the cube root does not impose any domain restriction, we only need to consider the restriction from the square root expression . Setting the expression to be nonnegative,
Rearranging this inequality leads to .
The domain of is .
Part 2
Let us find the range of . First, consider the expression . Recall that the range of the square root function is . Note that can be written as
Since the range of is all real numbers, the range of must be the same as the range of . Thus, the range of is . This tells us that the expression will output nonnegative values. Then, possible values for can be written as
The largest possible value of this happens when , which gives us . Since the largest function value is 5 and we know that the cube root function, , tends to to the left of -axis, we would like to conclude that the range of the function is . Let us carefully justify this conclusion.
If is some number satisfying , we need to show that
We can raise both sides of the equation to the cubic power to obtain
Since , we have , which means . This tells us that is a possible value of the function as long as .
The range of is .
Let us finish by recapping a few important concepts from this explainer.
### Key Points
• The domain and range of the square root function is .
More generally, the domain of a composite square root function can be identified by finding the values of satisfying .
• The domain and range of the cube root function, , are all real numbers. This is denoted as or .
• If the range of the function is all real numbers, then the range of or is or respectively.
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• Realistic Exam Questions |
# How do you calculate your axial strength?
## How do you calculate your axial strength?
The simplest formula for axial stress is force divided by cross-sectional area.
## Is torque directly proportional to force?
Torque is directly proportional to the radius of rotation. Ans. True. Since torque is the product of force and radius of rotation, increasing this radius will also increase the resulting torque.
How is axial force measured?
The axial force is measured by systematically decreasing the axial force from the nominal one.
### How do you calculate an axial load?
Measure the total horizontal distance traversed by the load (e.g., the total distance in the horizontal direction of a cable holding up a sign). Measure the total vertical distance traversed by the load. Divide the distance in the vertical direction by the distance in the horizontal direction.
### How to calculate axial force in a bolt?
The relation between applied torque and axial force – or load – in a bolt can be calculated as. T = K F d (1) where. T = wrench torque (Nm, in lb) K = constant that depends on the bolt material and size. d = nominal bolt diameter (m, in) F = axial bolt force (N, lb) Typical values for K with mild-steel bolts in range 1/4″ to 1″:
How do you calculate torque on a bolt?
Bolt Torque Calculator Calculate required bolt torque The relation between applied torque and axial force – or load – in a bolt can be calculated in this general equation as T = K F d (1 – l/100) (1)
#### What is the relation between torque and axial force?
The relation between applied torque and axial force – or load – in a bolt can be calculated in this general equation as T = K F d (1 – l/100) (1)
#### How to calculate the clamping force of a bolt?
Calculation of Clamping force from bolt torque 1 T = Torque (in-lb) 2 K = Constant to account for friction (0.15 – 0.2 for these units) 3 D = Bolt diameter (inches) 4 P = Clamping Force (lb) More
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# 35 7 7 Skills Practice Geometric Sequences As Exponential Functions
## Introduction
Geometric sequences and exponential functions are fundamental concepts in mathematics. They are used in various fields, including finance, physics, and computer science. Understanding and being able to practice these skills is crucial for students and professionals alike. In this article, we will explore seven skills that will help you master geometric sequences as exponential functions.
### Skill 1: Identifying Geometric Sequences
The first skill you need to practice is identifying geometric sequences. A geometric sequence is a sequence of numbers where each term is found by multiplying the previous term by a constant ratio. To identify a geometric sequence, look for a consistent ratio between consecutive terms. For example, in the sequence 2, 6, 18, 54, the ratio between consecutive terms is 3.
### Skill 2: Finding the Common Ratio
Once you have identified a geometric sequence, the next skill to practice is finding the common ratio. The common ratio is the constant value by which each term is multiplied to obtain the next term. To find the common ratio, divide any term in the sequence by its preceding term. For example, in the sequence 2, 6, 18, 54, the common ratio is 3.
### Skill 3: Writing the Recursive Formula
Another important skill is writing the recursive formula for a geometric sequence. The recursive formula defines each term in the sequence in terms of the preceding term(s). To write the recursive formula, use the common ratio and the initial term of the sequence. For example, in the sequence 2, 6, 18, 54, the recursive formula is given by an = 3 * an-1, where an represents the n-th term.
### Skill 4: Writing the Explicit Formula
In addition to the recursive formula, it is also important to practice writing the explicit formula for a geometric sequence. The explicit formula allows you to directly calculate any term in the sequence without having to compute the preceding terms. To write the explicit formula, use the initial term and the common ratio. For example, in the sequence 2, 6, 18, 54, the explicit formula is given by an = 2 * 3n-1.
### Skill 5: Finding the Sum of a Geometric Sequence
Next, you should practice finding the sum of a geometric sequence. The sum of a geometric sequence is the total value obtained by adding up all the terms in the sequence. To find the sum, use the formula Sn = a(1 - rn)/(1 - r), where Sn represents the sum of the first n terms, a is the initial term, and r is the common ratio.
### Skill 6: Recognizing Exponential Functions
Understanding exponential functions is another essential skill when practicing geometric sequences. An exponential function is a function where the independent variable appears as an exponent. Exponential functions are often used to model growth or decay processes. To recognize an exponential function, look for a constant base raised to a variable exponent. For example, f(x) = 2x is an exponential function.
### Skill 7: Converting Geometric Sequences to Exponential Functions
The final skill you should practice is converting geometric sequences to exponential functions. This skill allows you to represent a geometric sequence as an exponential function. To convert a geometric sequence to an exponential function, use the explicit formula and rewrite it in function notation. For example, the geometric sequence 2, 6, 18, 54 can be represented as the exponential function f(n) = 2 * 3n-1.
## Conclusion
Mastering the skills of practicing geometric sequences as exponential functions is essential for anyone working with mathematical concepts. By practicing identifying geometric sequences, finding the common ratio, writing the recursive and explicit formulas, finding the sum of a geometric sequence, recognizing exponential functions, and converting geometric sequences to exponential functions, you will gain a solid foundation in these fundamental mathematical concepts. Keep practicing these skills, and you will become more confident in your ability to work with geometric sequences and exponential functions. |
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# Division of Fractions and Whole Numbers
You may find our Fractions - Parts of Numbers page useful to read in conjunction with this article.
## Division of Fractions
Division is the most challenging of the four mathematical operations, right? Thankfully, not when it comes to dealing with fractions.
The key to knowing how to divide fractions is understanding that the operation of multiplying is the opposite of division. If we can make a multiplication sum then it's not too hard - as shown in the previous walkthrough.
To turn a division sum into a multiplication sum, we need to take the fraction which we are dividing by and turn it on its head. Once you've done this (effectively making it the 'opposite' of the number you started with) then you can multiply instead of dividing, and get the same answer!
Let's explore an example:
By turning the number we're dividing by upside down, while turning the division sign into a multiplication sign, we're keeping the answer the same. Effectively, we're finding the 'opposite of the opposite', which is the same as we started with.
If we carry out the simple multiplication sums, we get:
Now we have an improper fraction; the way we make it into a mixed number (the regular way of expressing an answer) is to see how many times the bottom number goes into the top. The answer is 2, with a remainder of 2. The answer is therefore:
Don't be afraid of getting numbers which are larger than the ones you start with. If both numbers are normal fractions then you will get a larger answer than the initial figures.
## Division of Fractions and Whole Numbers
Dividing fractions is not as intimidating as it appears - see the walkthrough - but when you get questions which involve a whole or mixed number as well it may seem harder. In fact, there's only an extra step to take before carrying out the same operations as before. If the mixed number or whole number is turned into a fraction, then you can treat it as a division sum as per the previous example.
Let's see how it works:
What is a half divided by two and a quarter?
The theory says that if you turn the second fraction upside-down then you can simply multiply the digits and get an answer. The problem is that our second figure is not a fraction but it can become one easily enough. With reference to the mixed numbers walkthrough, we need to change the two and a quarter into just quarters. There are four quarters in a whole one, so two and a quarter is the same as nine quarters.
The process is just following what we've done before - change the mixed number and then change the division sum to a multiplication. The final step is to simplify the fraction and put it in its lowest form.
What do you do if you encounter a number which is whole, and doesn't have a denominator? Well, 4 is the same as 4 รท 1 so can be expressed as 4 over 1. Any number can be expressed as itself over the number one. Do this and you have a fraction with which you can work. |
We had lots of solutions sent in for this activity. Brookfield Junior School obviously got very involved and sent in many solutions. Here is an example of two of them.
Firstly, from Bill:
The formula to Ip dip is n≥8 when n stands for the number of friends so that means if the number of friends is eight or over you should be in the eighth position. To solve the first bit the position is the remainder of the number of the friends divided by eight. so the formula is p=8/n= the remainder.
Secondly from Tomas:
If you had two people you would start in second position.
If you had three people you would start in second position.
If you had four people you would start in fourth position.
If you had five people you would start in third position.
If you had six people you would start in second position.
If you had seven people you would start in first position.
If you had eight people you would start in second position and so on.
From Mossley Primary School we had solutions sent in from Kati-leigh, Luke, Chiara, Amelia and Emily. Here is one example:
3: right friend clockwise
4: second to right clockwise
5: third to right clockwise
6: left clockwise
7: yourself
8: right clockwise
9: second to right clockwise
10: third to right clockwise
11: third to left
Amelia and Bea from Barton C E V A Primary School sent in the following
If there were 8 or more players, then always go for the 8th place.
If there were fewer than 8 players then you would find how many players there are and see what number adds to the amount of players to get to 8.You may need to use multiplication facts with this:
7 players: 1x7= 7 7 + 1 = 8 You go in 1st place
6 players: 1x6= 6 6 + 2 = 8 You go in 2nd place
5 players: 1x5= 5 5 + 3 = 8 You go in 3rd place
4 players: 1x4= 4 4 + 4 = 8 You go in 4th place
3 players: 2x3= 6 6 + 2 = 8 You go in 2nd place
2 players: 3x2 = 6 6 +2 = 8 You go in 2nd place
This is because there are 8 words in the rhyme.
Thank you, all of you, for the ideas you sent in, in order to solve this challenge. |
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# Perpendicular Lines in the Coordinate Plane
## Lines with opposite sign, reciprocal slopes that intersect at right angles.
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Perpendicular Lines in the Coordinate Plane
What if you were given two perpendicular lines in the coordinate plane? What could you say about their slopes? After completing this Concept, you'll be able to answer this question. You'll also find the equations of perpendicular lines and determine if two lines are perpendicular based on their slopes.
### Guidance
Perpendicular lines are two lines that intersect at a \begin{align*}90^\circ\end{align*}, or right, angle. In the coordinate plane, that would look like this:
If we take a closer look at these two lines, the slope of one is -4 and the other is \begin{align*}\frac{1}{4}\end{align*}.
This can be generalized to any pair of perpendicular lines in the coordinate plane. The slopes of perpendicular lines are opposite reciprocals of each other.
#### Example A
Find the equation of the line that is perpendicular to \begin{align*}y=-\frac{1}{3}x+4\end{align*} and passes through (9, -5).
First, the slope is the opposite reciprocal of \begin{align*}-\frac{1}{3}\end{align*}. So, \begin{align*}m = 3\end{align*}. Plug in 9 for \begin{align*}x\end{align*} and -5 for \begin{align*}y\end{align*} to solve for the new \begin{align*}y-\end{align*}intercept \begin{align*}(b)\end{align*}.
\begin{align*}-5 & = 3(9)+b\\ -5 & = 27+b\\ -32 & = b\end{align*}
Therefore, the equation of the perpendicular line is \begin{align*}y=3x-32\end{align*}.
#### Example B
Graph \begin{align*}3x-4y=8\end{align*} and \begin{align*}4x+3y=15\end{align*}. Determine if they are perpendicular.
First, we have to change each equation into slope-intercept form. In other words, we need to solve each equation for \begin{align*}y\end{align*}.
\begin{align*}3x-4y & = 8 && 4x+3y = 15\\ -4y & = -3x+8 && 3y = -4x + 15\\ y & = \frac{3}{4}x-2 && y = -\frac{4}{3}x+5\end{align*}
Now that the lines are in slope-intercept form (also called \begin{align*}y-\end{align*}intercept form), we can tell they are perpendicular because their slopes are opposite reciprocals.
#### Example C
Find the equation of the line that is perpendicular to the line \begin{align*}y=2x+7\end{align*} and goes through the point (2, -2).
The perpendicular line goes through (2, -2), but the slope is \begin{align*}-\frac{1}{2}\end{align*} because we need to take the opposite reciprocal of \begin{align*}2\end{align*}.
\begin{align*}y & = -\frac{1}{2}x+b\\ -2 & = -\frac{1}{2}(2) + b\\ -2 & = -1+b\\ -1 & =b\end{align*}
The equation is \begin{align*}y = -\frac{1}{2}x-1\end{align*}.
### Guided Practice
Find the slope of the lines that are perpendicular to the lines below.
1. \begin{align*}y=2x+3\end{align*}
2. \begin{align*}y = -\frac{2}{3}x-5\end{align*}
3. \begin{align*}y=x+2\end{align*}
1. \begin{align*}m = 2\end{align*}, so \begin{align*}m_\perp\end{align*} is the opposite reciprocal, \begin{align*}m_\perp = -\frac{1}{2}\end{align*}.
2. \begin{align*}m = -\frac{2}{3}\end{align*}, take the reciprocal and change the sign, \begin{align*}m_\perp=\frac{3}{2}\end{align*}.
3. Because there is no number in front of \begin{align*}x\end{align*}, the slope is 1. The reciprocal of 1 is 1, so the only thing to do is make it negative, \begin{align*}m_\perp= -1\end{align*}.
### Practice
Determine if each pair of lines are perpendicular. Then, graph each pair on the same set of axes.
1. \begin{align*}y=-2x+3\end{align*} and \begin{align*}y = \frac{1}{2}x+3\end{align*}
2. \begin{align*}y=-3x+1\end{align*} and \begin{align*}y=3x-1\end{align*}
3. \begin{align*}2x-3y=6\end{align*} and \begin{align*}3x+2y=6\end{align*}
4. \begin{align*}x-3y=-3\end{align*} and \begin{align*}x+3y=9\end{align*}
Determine the equation of the line that is perpendicular to the given line, through the given point.
1. \begin{align*}y=x-1; \ (-6, 2)\end{align*}
2. \begin{align*}y=3x+4; \ (9, -7)\end{align*}
3. \begin{align*}5x-2y= 6; \ (5, 5)\end{align*}
4. \begin{align*}y = 4; \ (-1, 3)\end{align*}
Find the equations of the two lines in each graph below. Then, determine if the two lines are perpendicular.
For the line and point below, find a perpendicular line through the given point.
### Vocabulary Language: English Spanish
perpendicular
perpendicular
Two lines that intersect at a $90^\circ$, or right, angle. The slopes of perpendicular lines are opposite reciprocals of each other. |
### Applications of DIfferentiation Part V
Connected Rates of Change
Example 1
A curve has the equation $y = (5x-2)(\sqrt{2x + 2})$
a) Find $\frac{dy}{dx}$ in the form of $\frac{ ax }{ \sqrt{2x + 2} }$
Using Product rule
\begin{aligned} \frac{dy}{dx} &= (5x-2)\frac{1}{2}( 2x + 2)^{-\frac{1}{2}}( 2 ) + 5\sqrt{2x + 2} \\ &= \frac{ 5x - 2 }{ ( 2x + 2)^{\frac{1}{2} } } + 5\sqrt{2x + 2} \\ &= \frac{ 5x - 2 + 2x + 2 }{ \sqrt{2x + 2} } \\ &= \frac{ 7x }{ \sqrt{2x + 2} } \\ \end{aligned}
b) Hence find the rate of change of x when x = 7 when y is change at a rate of 4 units per second
Given $x = 7 \qquad \frac{dy}{dt} = 4$
\begin{aligned} \frac{dy}{dt} &= \frac{dy}{dx} \times \frac{dx}{dt} \\ 4 &= \frac{ 7x }{ \sqrt{2x + 2} } \times \frac{dx}{dt} \\ 4 &= \frac{ 49 }{ \sqrt{14 + 2} } \times \frac{dx}{dt} \\ \frac{dx}{dt} &= \frac{16}{49} \\ \end{aligned} |
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Lesson 16: Derivatives of Logarithmic and Exponential Functions
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We show the the derivative of the exponential function is itself! And the derivative of the natural logarithm function is the reciprocal function. We also show how logarithms can make complicated differentiation problems easier.
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Lesson 16: Derivatives of Logarithmic and Exponential Functions
1. 1. Section 3.3 Derivatives of Exponential and Logarithmic Functions V63.0121.027, Calculus I October 22, 2009 . . Image credit: heipei . . . . . .
2. 2. Outline Derivative of the natural exponential function Exponential Growth Derivative of the natural logarithm function Derivatives of other exponentials and logarithms Other exponentials Other logarithms Logarithmic Differentiation The power rule for irrational powers . . . . . .
3. 3. Derivatives of Exponential Functions Fact If f(x) = ax , then f′ (x) = f′ (0)ax . . . . . . .
4. 4. Derivatives of Exponential Functions Fact If f(x) = ax , then f′ (x) = f′ (0)ax . Proof. Follow your nose: f(x + h) − f(x) a x+ h − a x f′ (x) = lim = lim h→0 h h→0 h a x a h − ax a h−1 = lim = ax · lim = ax · f′ (0). h→0 h h→0 h . . . . . .
5. 5. Derivatives of Exponential Functions Fact If f(x) = ax , then f′ (x) = f′ (0)ax . Proof. Follow your nose: f(x + h) − f(x) a x+ h − a x f′ (x) = lim = lim h→0 h h→0 h a x a h − ax a h−1 = lim = ax · lim = ax · f′ (0). h→0 h h→0 h To reiterate: the derivative of an exponential function is a constant times that function. Much different from polynomials! . . . . . .
6. 6. The funny limit in the case of e Remember the definition of e: ( ) 1 n e = lim 1 + = lim (1 + h)1/h n→∞ n h→0 Question eh − 1 What is lim ? h→0 h . . . . . .
7. 7. The funny limit in the case of e Remember the definition of e: ( ) 1 n e = lim 1 + = lim (1 + h)1/h n→∞ n h→0 Question eh − 1 What is lim ? h→0 h Answer If h is small enough, e ≈ (1 + h)1/h . So [ ]h eh − 1 (1 + h)1/h − 1 (1 + h) − 1 h ≈ = = =1 h h h h . . . . . .
8. 8. The funny limit in the case of e Remember the definition of e: ( ) 1 n e = lim 1 + = lim (1 + h)1/h n→∞ n h→0 Question eh − 1 What is lim ? h→0 h Answer If h is small enough, e ≈ (1 + h)1/h . So [ ]h eh − 1 (1 + h)1/h − 1 (1 + h) − 1 h ≈ = = =1 h h h h eh − 1 So in the limit we get equality: lim =1 h→0 h . . . . . .
9. 9. Derivative of the natural exponential function From ( ) d x ah − 1 eh − 1 a = lim ax and lim =1 dx h→0 h h→0 h we get: Theorem d x e = ex dx . . . . . .
10. 10. Exponential Growth Commonly misused term to say something grows exponentially It means the rate of change (derivative) is proportional to the current value Examples: Natural population growth, compounded interest, social networks . . . . . .
11. 11. Examples Examples Find these derivatives: e3x 2 ex x 2 ex . . . . . .
12. 12. Examples Examples Find these derivatives: e3x 2 ex x 2 ex Solution d 3x e = 3e3x dx . . . . . .
13. 13. Examples Examples Find these derivatives: e3x 2 ex x 2 ex Solution d 3x e = 3e3x dx d x2 2 d 2 e = ex (x2 ) = 2xex dx dx . . . . . .
14. 14. Examples Examples Find these derivatives: e3x 2 ex x 2 ex Solution d 3x e = 3e3x dx d x2 2 d 2 e = ex (x2 ) = 2xex dx dx d 2 x x e = 2xex + x2 ex dx . . . . . .
15. 15. Outline Derivative of the natural exponential function Exponential Growth Derivative of the natural logarithm function Derivatives of other exponentials and logarithms Other exponentials Other logarithms Logarithmic Differentiation The power rule for irrational powers . . . . . .
16. 16. Derivative of the natural logarithm function Let y = ln x. Then x = ey so . . . . . .
17. 17. Derivative of the natural logarithm function Let y = ln x. Then x = ey so dy ey =1 dx . . . . . .
18. 18. Derivative of the natural logarithm function Let y = ln x. Then x = ey so dy ey =1 dx dy 1 1 =⇒ = y = dx e x . . . . . .
19. 19. Derivative of the natural logarithm function Let y = ln x. Then x = ey so dy ey =1 dx dy 1 1 =⇒ = y = dx e x So: Fact d 1 ln x = dx x . . . . . .
20. 20. Derivative of the natural logarithm function Let y = ln x. Then y . x = ey so dy ey =1 dx .n x l dy 1 1 =⇒ = y = dx e x . x . So: Fact d 1 ln x = dx x . . . . . .
21. 21. Derivative of the natural logarithm function Let y = ln x. Then y . x = ey so dy ey =1 dx .n x l dy 1 1 1 =⇒ = y = . dx e x x . x . So: Fact d 1 ln x = dx x . . . . . .
22. 22. The Tower of Powers y y′ The derivative of a x3 3x2 power function is a power function of one x2 2x1 lower power x1 1x0 x0 0 ? ? x−1 −1x−2 x−2 −2x−3 . . . . . .
23. 23. The Tower of Powers y y′ The derivative of a x3 3x2 power function is a power function of one x2 2x1 lower power x1 1x0 Each power function is x 0 0 the derivative of another power function, except ? x −1 x−1 x−1 −1x−2 x−2 −2x−3 . . . . . .
24. 24. The Tower of Powers y y′ The derivative of a x3 3x2 power function is a power function of one x2 2x1 lower power x1 1x0 Each power function is x 0 0 the derivative of another power function, except ln x x −1 x−1 x−1 −1x−2 ln x fills in this gap precisely. x−2 −2x−3 . . . . . .
25. 25. Outline Derivative of the natural exponential function Exponential Growth Derivative of the natural logarithm function Derivatives of other exponentials and logarithms Other exponentials Other logarithms Logarithmic Differentiation The power rule for irrational powers . . . . . .
26. 26. Other logarithms Example d x Use implicit differentiation to find a. dx . . . . . .
27. 27. Other logarithms Example d x Use implicit differentiation to find a. dx Solution Let y = ax , so ln y = ln ax = x ln a . . . . . .
28. 28. Other logarithms Example d x Use implicit differentiation to find a. dx Solution Let y = ax , so ln y = ln ax = x ln a Differentiate implicitly: 1 dy dy = ln a =⇒ = (ln a)y = (ln a)ax y dx dx . . . . . .
29. 29. Other logarithms Example d x Use implicit differentiation to find a. dx Solution Let y = ax , so ln y = ln ax = x ln a Differentiate implicitly: 1 dy dy = ln a =⇒ = (ln a)y = (ln a)ax y dx dx Before we showed y′ = y′ (0)y, so now we know that 2h − 1 3h − 1 ln 2 = lim ≈ 0.693 ln 3 = lim ≈ 1.10 h→0 h h→0 h . . . . . .
30. 30. Other logarithms Example d Find loga x. dx . . . . . .
31. 31. Other logarithms Example d Find loga x. dx Solution Let y = loga x, so ay = x. . . . . . .
32. 32. Other logarithms Example d Find loga x. dx Solution Let y = loga x, so ay = x. Now differentiate implicitly: dy dy 1 1 (ln a)ay = 1 =⇒ = y = dx dx a ln a x ln a . . . . . .
33. 33. Other logarithms Example d Find loga x. dx Solution Let y = loga x, so ay = x. Now differentiate implicitly: dy dy 1 1 (ln a)ay = 1 =⇒ = y = dx dx a ln a x ln a Another way to see this is to take the natural logarithm: ln x ay = x =⇒ y ln a = ln x =⇒ y = ln a dy 1 1 So = . dx ln a x . . . . . .
34. 34. More examples Example d Find log2 (x2 + 1) dx . . . . . .
35. 35. More examples Example d Find log2 (x2 + 1) dx Answer dy 1 1 2x = 2+1 (2x) = dx ln 2 x (ln 2)(x2 + 1) . . . . . .
36. 36. Outline Derivative of the natural exponential function Exponential Growth Derivative of the natural logarithm function Derivatives of other exponentials and logarithms Other exponentials Other logarithms Logarithmic Differentiation The power rule for irrational powers . . . . . .
37. 37. A nasty derivative Example √ (x2 + 1) x + 3 Let y = . Find y′ . x−1 . . . . . .
38. 38. A nasty derivative Example √ (x2 + 1) x + 3 Let y = . Find y′ . x−1 Solution We use the quotient rule, and the product rule in the numerator: [ √ ] √ ′ (x − 1) 2x x + 3 + (x2 + 1) 1 (x + 3)−1/2 − (x2 + 1) x + 3(1) 2 y = (x − 1)2 √ √ 2x x + 3 (x2 + 1) (x 2 + 1 ) x + 3 = + √ − (x − 1 ) 2 x + 3(x − 1) (x − 1)2 . . . . . .
39. 39. Another way √ (x 2 + 1 ) x + 3 y= x−1 1 ln y = ln(x2 + 1) + ln(x + 3) − ln(x − 1) 2 1 dy 2x 1 1 = 2 + − y dx x + 1 2(x + 3) x − 1 So ( ) dy 2x 1 1 = + − y dx x2 + 1 2(x + 3) x − 1 ( ) √ 2x 1 1 (x2 + 1) x + 3 = + − x2 + 1 2(x + 3) x − 1 x−1 . . . . . .
40. 40. Compare and contrast Using the product, quotient, and power rules: √ √ ′ 2x x + 3 (x2 + 1) (x2 + 1) x + 3 y = + √ − (x − 1) 2 x + 3(x − 1) (x − 1)2 Using logarithmic differentiation: ( ) √ ′ 2x 1 1 (x2 + 1) x + 3 y = + − x2 + 1 2(x + 3) x − 1 x−1 . . . . . .
41. 41. Compare and contrast Using the product, quotient, and power rules: √ √ ′ 2x x + 3 (x2 + 1) (x2 + 1) x + 3 y = + √ − (x − 1) 2 x + 3(x − 1) (x − 1)2 Using logarithmic differentiation: ( ) √ ′ 2x 1 1 (x2 + 1) x + 3 y = + − x2 + 1 2(x + 3) x − 1 x−1 Are these the same? . . . . . .
42. 42. Compare and contrast Using the product, quotient, and power rules: √ √ ′ 2x x + 3 (x2 + 1) (x2 + 1) x + 3 y = + √ − (x − 1) 2 x + 3(x − 1) (x − 1)2 Using logarithmic differentiation: ( ) √ ′ 2x 1 1 (x2 + 1) x + 3 y = + − x2 + 1 2(x + 3) x − 1 x−1 Are these the same? Which do you like better? . . . . . .
43. 43. Compare and contrast Using the product, quotient, and power rules: √ √ ′ 2x x + 3 (x2 + 1) (x2 + 1) x + 3 y = + √ − (x − 1) 2 x + 3(x − 1) (x − 1)2 Using logarithmic differentiation: ( ) √ ′ 2x 1 1 (x2 + 1) x + 3 y = + − x2 + 1 2(x + 3) x − 1 x−1 Are these the same? Which do you like better? What kinds of expressions are well-suited for logarithmic differentiation? . . . . . .
44. 44. Derivatives of powers Let y = xx . Which of these is true? (A) Since y is a power function, y′ = x · xx−1 = xx . (B) Since y is an exponential function, y′ = (ln x) · xx (C) Neither . . . . . .
45. 45. Derivatives of powers Let y = xx . Which of these is true? (A) Since y is a power function, y′ = x · xx−1 = xx . (B) Since y is an exponential function, y′ = (ln x) · xx (C) Neither . . . . . .
46. 46. It’s neither! Or both? If y = xx , then ln y = x ln x 1 dy 1 = x · + ln x = 1 + ln x y dx x dy = xx + (ln x)xx dx Each of these terms is one of the wrong answers! . . . . . .
47. 47. Derivative of arbitrary powers Fact (The power rule) Let y = xr . Then y′ = rxr−1 . . . . . . .
48. 48. Derivative of arbitrary powers Fact (The power rule) Let y = xr . Then y′ = rxr−1 . Proof. y = xr =⇒ ln y = r ln x Now differentiate: 1 dy r = y dx x dy y =⇒ = r = rxr−1 dx x . . . . . . |
# How do you color in Desmos?
It is just a click away. In Desmos, you can draw, color, and even write notes and equations in a browser. You can use this app to draw pictures, make your own shapes, or even complete math problems. Whether you are just a beginner or an advanced student, Desmos has you covered.
Contents
## How do you make a square on Desmos?
A square is a polygon that is both equilateral and isosceles right. A square with these properties is called a regular rectangle. The length and width are both equal. To create a square, draw all four sides of your rectangle, then change the size until the length and width are the same. If you do not want to enter values, you can click the small square on the right-hand side of the Desmos canvas window and hit the “Create square” button.
## Correspondingly, how do you fill in color on Desmos?
Color fills are used to fill a space inside or outside of the canvas – as I mentioned, you choose a color that best fills it. It is easy to get color in just a few clicks. How do I color on a line on Desmos?
Colors are used to color areas inside or outside of the canvas. Simply click and drag a color over the area inside the canvas area.
## How do you solve system of equations?
Solve: You can use the following approach to solve a system of linear equations. Write each equation in the form AX=B. Divide the first row of the equation by the second row of the equation. So 2AX = 1/3 in order to convert 3A into a row you multiply a 1 and 3X to get 3A/3X=1. The solution is 3X.
First, you will find the equation of a line that is parallel and intersects the vertical line in one point. Then you will substitute the coordinates of that intersection (a point). This gives a point on the equation. Then you have to factor the resulting equation.
## What part do you shade for inequalities?
Shading or shading (also known as shading and shading) is a technique for suggesting the direction and intensity of light coming from a specific point of view in a painting.
## How do you find the Directrix of a parabola?
The equation of a parabola is y=mx+b. Substituting the values of x, y, and m to identify the equation: y=1,000x + 15. If the equation of a parabola is given by one of the following forms f(x)=ax2+bx+c or f(y)=ay2+by+c, then the equation is a parabola. The origin of the coordinate axis can now be found using the equation y=-b/a+2=0.
## What are the four inequality symbols?
Four inequality signs are often used in school, college, and professional mathematics: “<" ">“.
## How do you make a quadrilateral on Desmos?
To convert a quadrilateral to a quadrilateral in Desmos, just move the mouse cursor to either one of the four corners; The area between the two corners is the quadrilateral that your cursor is hovering over. Click in the quadrilateral to draw it.
## It’s the best choice to select a sphere to learn pencil shading.How do you find the center and radius of a circle?
When you want to find either the x or y coordinate of a point on the circle by the radius, you divide the distance by the radius. For example, you can find the x coordinate of a point on the circle defined by the equation r2 – r2 = – 2r by dividing -2 by r.
## How do you graph a shape on a graphing calculator?
To graph the shape on your calculator, you do the following steps: Place your calculator’s y-axis cursor over the shape to select the vertical scale. Use the numeric keypad to move the cursor above or below the curve or curve on the y-axis. Note: You can’t graph more than one shape at a time.
## How do you make a parabola sideways?
To give you some sense of how a parabola is formed, think of a line with two points. Let’s call these points (0,0) and (x, y). This gives you two points. From these two points, you can draw a line to the left that makes a parabola.
## Beside above, how do you shade a circle?
Shading a circle does not add a solid shape like square, rectangle or an object. One way to shade a circle would be to use a rectangle and trace three straight vertical lines. Use this vertical line to trace across the top of the white area just below the top horizontal line. |
# Difference Between Variable and Parameter
Many people get confused when differentiating between a variable and parameter. Although these two terms have a number of similarities, they are also some key differences between them. Both variables and parameters are commonly used in the field of algebra, mathematics and engineering. Variables and parameters are used often to solve complex engineering problems.
Essentially, a variable can be any number that is dependent on another entity. If the value of that entity changes, the variable will also increase or decrease. Parameter on the other hand, acts a bridge between two or more variables to form a meaningful relationship. From physics to mathematics, engineering to statistical analysis, variables and parameters help us relate equations and quantities. The main difference between these two terms is that while variables keep changing, the parameter remains constant.
There are two types of variables including continuous variables and discrete variables. Mathematical and engineering problems can be solved easily be identifying and categorizing the variables in any given equation.
### Instructions
• 1
Variable
A variable is a number of a quantity that changes with change in entity it is related to.
Consider a sample of a car moving on a road. The speed of the car can be calculated by using the formula given below.
S = D/T
Where S is the speed of the car
D is the distance covered
And T is the time taken to cover that distance
Now, in the equation above, distance and time are constants whereas speed or S is the variable. Any change recorded in the distance or time will result in a change in the speed variable. If the car covered 20 meters in 60 seconds then the speed of the car will be 20/60, which is equal to 0.33 m/sec. However, if the distance is increased to 30 meters in 60 seconds then the speed of the vehicle will be increased to 30/60 or 0.5 m/sec, showing the speed is a variable quantity, directly related to the distance and time.
Image courtesy: euler.rene-grothmann.de
• 2
Parameter:
Parameter is an entity that acts as a bridge between two or more variables in any given equation.
Parameters may have different dimension than variables. In the equation given below, x and y are the variables.
X2 + y2 = 1
Solving the above equation gives,
X = cos (w)
Y = sin (w)
The values of parameter can be substituted in the equations above to find the solution.
Image courtesy: euler.rene-grothmann.de |
# Question #c3471
Jan 31, 2017
$\textcolor{red}{5} x - \textcolor{b l u e}{3} y = \textcolor{g r e e n}{0}$
#### Explanation:
First, we can obtain an equation for the line using the point-slope formula. The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$
Where $\textcolor{b l u e}{m}$ is the slope and $\textcolor{red}{\left(\left({x}_{1} , {y}_{1}\right)\right)}$ is a point the line passes through.
Substituting the values from the problem gives:
$\left(y - \textcolor{red}{5}\right) = \textcolor{b l u e}{\frac{5}{3}} \left(x - \textcolor{red}{3}\right)$
The standard form of a linear equation is:
$\textcolor{red}{A} x + \textcolor{b l u e}{B} y = \textcolor{g r e e n}{C}$
where, if at all possible, $\textcolor{red}{A}$, $\textcolor{b l u e}{B}$, and $\textcolor{g r e e n}{C}$are integers, and A is non-negative, and, A, B, and C have no common factors other than 1
We can begin solving for this form by multiplying each side of the equation by $\textcolor{p u r p \le}{3}$ to eliminate the fraction:
$\textcolor{p u r p \le}{3} \left(y - \textcolor{red}{5}\right) = \textcolor{p u r p \le}{3} \times \textcolor{b l u e}{\frac{5}{3}} \left(x - \textcolor{red}{3}\right)$
$3 y - 15 = \cancel{\textcolor{p u r p \le}{3}} \times \textcolor{b l u e}{\frac{5}{\cancel{3}}} \left(x - \textcolor{red}{3}\right)$
$3 y - 15 = 5 x - 15$
Next we can add $\textcolor{red}{15}$ and subtract $\textcolor{b l u e}{5 x}$ to each side of the equation to isolate the constant on the right side of the equation while maintaining the balance of the equation.
$- \textcolor{b l u e}{5 x} + 3 y - 15 + \textcolor{red}{15} = - \textcolor{b l u e}{5 x} + 5 x - 15 + \textcolor{red}{15}$
$- 5 x + 3 y - 0 = 0 - 0$
$- 5 x + 3 y = 0$
Now, we can multiple each side of the equation by $\textcolor{red}{- 1}$ to make the coefficient of $x$ positive or non-negative.
$\textcolor{red}{- 1} \left(- 5 x + 3 y\right) = \textcolor{red}{- 1} \times 0$
$5 x - 3 y = 0$ |
# How do you differentiate f(x) = (e^x-4x)/(2x-e^x) using the quotient rule?
Jan 23, 2017
The answer is $= \frac{2 {e}^{x} \left(1 - x\right)}{2 x - {e}^{x}} ^ 2$
#### Explanation:
The quotient rule is
$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{{v}^{2}}$
Here,
$u = {e}^{x} - 4 x$, $\implies$, $u ' = {e}^{x} - 4$
$v = 2 x - {e}^{x}$, $\implies$, $v ' = 2 - {e}^{x}$
Therefore,
$f ' \left(x\right) = \frac{\left({e}^{x} - 4\right) \left(2 x - {e}^{x}\right) - \left({e}^{x} - 4 x\right) \left(2 - {e}^{x}\right)}{2 x - {e}^{x}} ^ 2$
$= \frac{2 x {e}^{x} - {\cancel{e}}^{2 x} - \cancel{8} x + 4 {e}^{x} - 2 {e}^{x} + {\cancel{e}}^{2 x} + \cancel{8} x - 4 x {e}^{x}}{2 x - {e}^{x}} ^ 2$
$= \frac{2 {e}^{x} - 2 x {e}^{x}}{2 x - {e}^{x}} ^ 2$
$= \frac{2 {e}^{x} \left(1 - x\right)}{2 x - {e}^{x}} ^ 2$ |
Ex 1.1
Chapter 1 Class 12 Relation and Functions
Serial order wise
Get live Maths 1-on-1 Classs - Class 6 to 12
### Transcript
Ex 1.1, 9 (Introduction) Show that each of the relation R in the set A = {x ∈ Z: 0 ≤ x ≤ 12} , given by (i) R = { (a, b):|a – b| is a multiple of 4} is an equivalence relation. Find the set of all elements related to 1 in each case. Modulus function |1| = 1 |2| = 2 |0| = 0 |−1| = 1 |−3| = 3 Ex 1.1, 9 Show that each of the relation R in the set A = {x ∈ Z: 0 ≤ x ≤ 12} , given by (i) R = { (a, b):|a – b| is a multiple of 4} is an equivalence relation. Find the set of all elements related to 1 in each case. A = {x ∈ Z: 0 ≤ x ≤ 12} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} R = {(a,b) : |a – b| is a multiple of 4} Note: Multiples of 4 are 0, 4, 8, 12 i.e. |a – b| can be 0, 4, 8, 12 only Check reflexive Since |a – a| = |0| = 0 & 0 is a multiple of 4 So, |a – a| is a multiple of 4 ∴ (a, a) ∈ R, ∴ R is reflexive. Check symmetric We know that |a – b| = |b – a| Hence, if |a – b| is a multiple of 4, then |b – a| is also a multiple of 4 Hence, If (a, b) ∈ R, then (b, a) ∈ R ∴ R is symmetric |a – b| = |–(b – a)| = |b – a| Check transitive If |a – b| is a multiple of 4, then (a – b) is a multiple of 4 Similarly, if |b – c| is a multiple of 4 , then (b – c) is a multiple of 4 Now, Sum of multiple of 4 is also a multiple of 4 a – b + b – c is a multiple of 4 ⇒ a – c is a multiple of 4 Hence, |a – c| is a multiple of 4 ∴ If |a – b| &|b – c| is a multiple of 4 , then |a – c| is a multiple of 4 i.e. If (a, b) ∈ R & (b, c) ∈ R , then (a, c) ∈ R ∴ R is transitive Rough 8 & 12 are multiple of 4 8 + 12 = 20 is also a multiple of 4 Since R is reflexive, symmetric and transitive, it is equivalence relation We need to find set of elements related to 1 R = {(a, b) : |a – b| is a multiple of 4} & A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} Here a = 1, The set of elements related to 1 are {1, 5, 9} |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# 9.6: Factoring Special Products
Difficulty Level: At Grade Created by: CK-12
When we learned how to multiply binomials, we talked about two special products: the Sum and Difference Formula and the Square of a Binomial Formula. In this lesson, we will learn how to recognize and factor these special products.
## Factoring the Difference of Two Squares
We use the Sum and Difference Formula to factor a difference of two squares. A difference of two squares can be a quadratic polynomial in this form: a2b2\begin{align*}a^2-b^2\end{align*}. Both terms in the polynomial are perfect squares. In a case like this, the polynomial factors into the sum and difference of the square root of each term.
a2b2=(a+b)(ab)
In these problems, the key is figuring out what the a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} terms are. Let’s do some examples of this type.
Example 1: Factor the difference of squares.
(a) x29\begin{align*}x^2-9\end{align*}
(b) x2y21\begin{align*}x^2y^2-1\end{align*}
Solution:
(a) Rewrite as x29\begin{align*}x^2-9\end{align*} as x232\begin{align*}x^2-3^2\end{align*}. Now it is obvious that it is a difference of squares.
We substitute the values of a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} for the Sum and Difference Formula:
(x+3)(x3)
The answer is x29=(x+3)(x3)\begin{align*}x^2-9=(x+3)(x-3)\end{align*}.
(b) Rewrite as x2y21\begin{align*}x^2y^2-1\end{align*} as (xy)212\begin{align*}(xy)^2-1^2\end{align*}. This factors as (xy+1)(xy1)\begin{align*}(xy+1)(xy-1)\end{align*}.
## Factoring Perfect Square Trinomials
A perfect square trinomial has the form
a2+2ab+b2ora22ab+b2
The factored form of a perfect square trinomial has the form
And(a+b)2 if a2+2(ab)+b2(ab)2 if a22(ab)+b2
In these problems, the key is figuring out what the a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} terms are. Let’s do some examples of this type.
Example: x2+8x+16\begin{align*}x^2+8x+16\end{align*}
Solution: Check that the first term and the last term are perfect squares.
x2+8x+16asx2+8x+42.
Check that the middle term is twice the product of the square roots of the first and the last terms. This is true also since we can rewrite them.
x2+8x+16asx2+24x+42
This means we can factor x2+8x+16\begin{align*}x^2+8x+16\end{align*} as (x+4)2\begin{align*}(x+4)^2\end{align*}.
Example 2: Factor x24x+4\begin{align*}x^2-4x+4\end{align*}.
Solution: Rewrite x24x+4\begin{align*}x^2-4x+4\end{align*} as x2+2(2)x+(2)2\begin{align*}x^2+2 \cdot (-2) \cdot x+(-2)^2\end{align*}.
We notice that this is a perfect square trinomial and we can factor it as: (x2)2\begin{align*}(x-2)^2\end{align*}.
## Solving Polynomial Equations Involving Special Products
We have learned how to factor quadratic polynomials that are helpful in solving polynomial equations like ax2+bx+c=0\begin{align*}ax^2+bx+c=0\end{align*}. Remember that to solve polynomials in expanded form, we use the following steps:
Step 1: Rewrite the equation in standard form such that: Polynomial expression = 0.
Step 2: Factor the polynomial completely.
Step 3: Use the Zero Product Property to set each factor equal to zero.
Step 4: Solve each equation from step 3.
Example 3: Solve the following polynomial equations.
x2+7x+6=0
Solution: No need to rewrite because it is already in the correct form.
Factor: We write 6 as a product of the following numbers:
6x2+7x+6=6×1=0andfactors as6+1=7(x+1)(x+6)=0
Set each factor equal to zero:
x+1=0orx+6=0
Solve:
x=1orx=6
Check: Substitute each solution back into the original equation.
(1)2+7(1)+6(6)2+7(6)+6=1+(7)+6=0=36+(42)+6=0
## 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.
Factor the following perfect square trinomials.
1. x2+8x+16\begin{align*}x^2+8x+16\end{align*}
2. x218x+81\begin{align*}x^2-18x+81\end{align*}
3. x2+24x144\begin{align*}-x^2+24x-144\end{align*}
4. x2+14x+49\begin{align*}x^2+14x+49\end{align*}
5. 4x24x+1\begin{align*}4x^2-4x+1\end{align*}
6. 25x2+60x+36\begin{align*}25x^2+60x+36\end{align*}
7. 4x212xy+9y2\begin{align*}4x^2-12xy+9y^2\end{align*}
8. x4+22x2+121\begin{align*}x^4+22x^2+121\end{align*}
Factor the following difference of squares.
1. x24\begin{align*}x^2-4\end{align*}
2. x236\begin{align*}x^2-36\end{align*}
3. x2+100\begin{align*}-x^2+100\end{align*}
4. x2400\begin{align*}x^2-400\end{align*}
5. 9x24\begin{align*}9x^2-4\end{align*}
6. 25x249\begin{align*}25x^2-49\end{align*}
7. 36x2+25\begin{align*}-36x^2+25\end{align*}
8. 16x281y2\begin{align*}16x^2-81y^2\end{align*}
Solve the following quadratic equations using factoring.
1. x211x+30=0\begin{align*}x^2-11x+30=0\end{align*}
2. x2+4x=21\begin{align*}x^2+4x=21\end{align*}
3. x2+49=14x\begin{align*}x^2+49=14x\end{align*}
4. x264=0\begin{align*}x^2-64=0\end{align*}
5. x224x+144=0\begin{align*}x^2-24x+144=0\end{align*}
6. 4x225=0\begin{align*}4x^2-25=0\end{align*}
7. x2+26x=169\begin{align*}x^2+26x=-169\end{align*}
8. x216x60=0\begin{align*}-x^2-16x-60=0\end{align*}
Mixed Review
1. Find the value for k\begin{align*}k\end{align*} that creates an infinite number of solutions to the system {3x+7y=1kx14y=2\begin{align*}\begin{cases} 3x+7y=1\\ kx-14y=-2 \end{cases}\end{align*}.
2. A restaurant has two kinds of rice, three choices of mein, and four kinds of sauce. How many plate combinations can be created if you choose one of each?
3. Graph y5=13(x+4)\begin{align*}y-5= \frac{1}{3}(x+4)\end{align*}. Identify its slope.
4. \$600 was deposited into an account earning 8% interest compounded annually.
1. Write the exponential model to represent this situation.
2. How much money will be in the account after six years?
5. Divide 489÷315\begin{align*}4 \frac{8}{9} \div -3\frac{1}{5}\end{align*}.
6. Identify an integer than is even and not a natural number.
8 , 9
Feb 22, 2012
Oct 19, 2015 |
# A factory makes tennis rackets and cricket bats.
Question:
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsman’s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftsman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time.
(ii) What number of rackets and bats must be made if the factory is to work at full capacity?
(ii) If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the maximum profit of the factory when it works at full capacity.
Solution:
(i) Let the number of rackets and the number of bats to be made be x and y respectively.
The machine time is not available for more than 42 hours.
$\therefore 1.5 x+3 y \leq 42$ (1)
The craftsman’s time is not available for more than 24 hours.
$\therefore 3 x+y \leq 24$ (2)
The factory is to work at full capacity. Therefore,
$1.5 x+3 y=42$
$3 x+y=24$
On solving these equations, we obtain
x = 4 and y = 12
Thus, 4 rackets and 12 bats must be made.
(i) The given information can be complied in a table as follows.
$\therefore 1.5 x+3 y \leq 42$
$3 x+y \leq 24$
$x, y \geq 0$
The profit on a racket is Rs 20 and on a bat is Rs 10.
$\therefore Z=20 x+10 y$
$3 x+y \leq 24$
$x, y \geq 0$
The profit on a racket is Rs 20 and on a bat is Rs 10.
$\therefore Z=20 x+10 y$
The mathematical formulation of the given problem is
Maximize $Z=20 x+10 y$ ...(1)
subject to the constraints,
$1.5 x+3 y \leq 42 \ldots$ (2)
$3 x+y \leq 24 \ldots$ (3)
$x, y \geq 0 \ldots$ (4)
The feasible region determined by the system of constraints is as follows.
The corner points are A (8, 0), B (4, 12), C (0, 14), and O (0, 0).
The values of Z at these corner points are as follows.
Thus, the maximum profit of the factory when it works to its full capacity is Rs 200. |
# Lesson 17
Two Related Quantities, Part 1
### Lesson Narrative
This lesson is the first of two that apply new understanding of algebraic expressions and equations to represent relationships between two quantities. Students use and make connections between tables, graphs, and equations that represent these relationships.
In this lesson, students revisit and extend their understanding of equivalent ratios. A familiar scenario of mixing paints in a given ratio provides the context for writing equations that represents the relationship between two quantities. Students then create a table of values that shows how changes in one quantity affect changes in the other, and graph the points from the table in the coordinate plane. They are invited to notice that these points lie on a line. Students will study proportional relationships in more depth in grade 7.
Students learn that relationships between two quantities can be described by two different but related equations with one quantity, the dependent variable, affected by changes in the other quantity, the independent variable. When people engage in mathematical modeling, which variable is considered independent and which is considered dependent is often the choice of the modeler (though sometimes the situation suggests choosing one way over the other). The context in this lesson was intentionally chosen because the context does not suggest a preference about which quantity is chosen as the independent variable.
Teacher Notes for IM 6–8 Accelerated
Due to the sequence in IM 6–8 Math Accelerated, the work with proportional relationships referenced in the lesson narrative is in the next unit and not in Grade 7 as stated.
### Learning Goals
Teacher Facing
• Compare and contrast (orally) graphs and equations that represent a relationship between the same quantities but have the independent and dependent variables switched.
• Comprehend the terms “independent variable” and “dependent variable” (in spoken and written language).
• Create a table, graph, and equation to represent the relationship between quantities in a set of equivalent ratios.
### Student Facing
Let’s use equations and graphs to describe relationships with ratios.
### Student Facing
• I can create tables and graphs that show the relationship between two amounts in a given ratio.
• I can write an equation with variables that shows the relationship between two amounts in a given ratio.
### Glossary Entries
• dependent variable
The dependent variable is the result of a calculation.
For example, a boat travels at a constant speed of 25 miles per hour. The equation $$d=25t$$ describes the relationship between the boat's distance and time. The dependent variable is the distance traveled, because $$d$$ is the result of multiplying 25 by $$t$$.
• independent variable
The independent variable is used to calculate the value of another variable.
For example, a boat travels at a constant speed of 25 miles per hour. The equation $$d=25t$$ describes the relationship between the boat's distance and time. The independent variable is time, because $$t$$ is multiplied by 25 to get $$d$$.
### Print Formatted Materials
Student Task Statements pdf docx Cumulative Practice Problem Set pdf docx Cool Down Log In Teacher Guide Log In Teacher Presentation Materials pdf docx |
geometric puzzles requiring rotation
Geometric problems that require rotation are difficult in general.
Problem 1
Given is an equilateral triangle ABC with inside a point P such that:
PC = 3
PA = 4
PB = 5
To calculate: the length a of the triangle.
The solution is reached by a 60 degrees counterclockwise rotation of triangle BPC around center B.
After rotation P becomes P', C' = A.
Triangle PP'B is equilateral as well so PP'=5.
Now look at triangle PC'P'.
The sides have lengths 3,4,5 which makes it right-angled (Pythagoras lemma).
After rotation the sides AP and AP' are penpendicular.
So, before rotation their intersection had an angle of 90 + 60 = 150 degrees.(LAPC)
Apply the cosine rule in triangle APC to calculate side AC = a:
a2 = 32 + 42 - 2.3.4.cos(1500)
a2 = 25 - 24.(-0,866)
a = 6,766
successive approximation, a numerical method.
The program is written in Delphi.
Problem 2
Given is square ABCD with point P inside such that:
AP = 1
BP = 2
CP = 3
Unknown is the side of the square.
Two rotations are required:
1.
C is the center. Rotate triangle CPB 90 degrees clockwise.
2.
A is the center. Rotate triangle ABP 90 degrees counterclockwise.
In above figure P'DP'' is a straight line because P'D is a 90 degrees clockwise rotation of BP
and P''D is a 90 degrees counterclockwise rotation of BP.
Triangle PCP' is isosceles and right-angled so calculation of PP' is possible.
The same is true for triangle APP'' and side PP''.
In triangle P''PP' now PD is median and the lengths of the sides are known.
Using the formula for medians (a special case of Stewart's lemma) PD may be calculated.
note: "zwaartelijn" is Dutch for median.
Finally:
What length has the side of the square?
Hint:
1. use the cosine rule or
2. realize that pentagon APCP'P'' has the same area as square ABCD
Problem 3
Given is right angled isosceles triangle ABC (AB = AC).
The 3,4,5 lengths remind of the Pythagoras lemma.
Rotate ABC clockwise 90 degrees using A as the center of rotation.
So:
C becomes C'= B
D --> D'
E --> E'
see picture below: |
# The Golden Ratio
## Diagram 1:-
Perfection: We all have different definitions of achievement. For example, the meanings of perfect grades, perfect looks, complete personalities, etc., depends on individual preferences, moods, and situations. In other words, perfection is in the eyes of the beholder and therefore subjective.
However, for Mathematicians perfection means perfect numbers, maybe, as defined by the Golden Ratio. The golden ratio is often thought of as the ideal ratio between different measurements of an object. It may be difficult to visualize but let us discuss using Diagram 1.
All the segmented regions within Diagram 1 follow the golden ratio; i.e., that sides a/b=(a+b)/a which equals ~1.6180… This may be difficult to understand from the get-go so I will simplify it this diagram 1. This diagram is made up of one square (aa) and one rectangle (ab). The dimensions of the Square aa are 34 & 34, and the aspects of the rectangle ab are 34 & 21. If you take the side a(34) and divide it by the side b(21), you get 1.619 the golden ratio. If you make the sum of the sides a and b (34+21) and divide it by the side a (55/34), you get 1.618, the golden ratio!
This diagram is made up of one square (aa) and one rectangle (ab). The dimensions of the Square aa are 34 & 34, and the sizes of the rectangle ab are 34 & 21. If you take the side a(34) and divide it by the side b(21), you get 1.619 the golden ratio. If you take the sum of the sides a and b (34+21) and divide it by the side a (55/34), you get 1.618, the golden ratio!
## The golden ratio is a ratio we derived from observing the proportions of natural objects. For example:
This shell a de novo natural product, not a human-made object, exhibits this golden ratio. The power of observation made us incorporate the golden ratio in many of our designs (human-made objects). The Greek civilization dating back to 500 BCE started incorporating this ratio in their structures. Since then, the golden ratio has been used in a variety of art forms including photography. This is the power of observation. Inspired enough, you can explore more about the golden ratio and its origin using the following resources. |
# 4.6 Exponential and logarithmic equations (Page 4/8)
Page 4 / 8
## Using the definition of a logarithm to solve logarithmic equations
For any algebraic expression $\text{\hspace{0.17em}}S\text{\hspace{0.17em}}$ and real numbers $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}c,$ where
## Using algebra to solve a logarithmic equation
Solve $\text{\hspace{0.17em}}2\mathrm{ln}x+3=7.$
Solve $\text{\hspace{0.17em}}6+\mathrm{ln}x=10.$
$x={e}^{4}$
## Using algebra before and after using the definition of the natural logarithm
Solve $\text{\hspace{0.17em}}2\mathrm{ln}\left(6x\right)=7.$
Solve $\text{\hspace{0.17em}}2\mathrm{ln}\left(x+1\right)=10.$
$x={e}^{5}-1$
## Using a graph to understand the solution to a logarithmic equation
Solve $\text{\hspace{0.17em}}\mathrm{ln}x=3.$
[link] represents the graph of the equation. On the graph, the x -coordinate of the point at which the two graphs intersect is close to 20. In other words $\text{\hspace{0.17em}}{e}^{3}\approx 20.\text{\hspace{0.17em}}$ A calculator gives a better approximation: $\text{\hspace{0.17em}}{e}^{3}\approx 20.0855.$
Use a graphing calculator to estimate the approximate solution to the logarithmic equation $\text{\hspace{0.17em}}{2}^{x}=1000\text{\hspace{0.17em}}$ to 2 decimal places.
$x\approx 9.97$
## Using the one-to-one property of logarithms to solve logarithmic equations
As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers $\text{\hspace{0.17em}}x>0,$ $S>0,$ $T>0\text{\hspace{0.17em}}$ and any positive real number $\text{\hspace{0.17em}}b,$ where $\text{\hspace{0.17em}}b\ne 1,$
For example,
So, if $\text{\hspace{0.17em}}x-1=8,$ then we can solve for $\text{\hspace{0.17em}}x,$ and we get $\text{\hspace{0.17em}}x=9.\text{\hspace{0.17em}}$ To check, we can substitute $\text{\hspace{0.17em}}x=9\text{\hspace{0.17em}}$ into the original equation: $\text{\hspace{0.17em}}{\mathrm{log}}_{2}\left(9-1\right)={\mathrm{log}}_{2}\left(8\right)=3.\text{\hspace{0.17em}}$ In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.
For example, consider the equation $\text{\hspace{0.17em}}\mathrm{log}\left(3x-2\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(x+4\right).\text{\hspace{0.17em}}$ To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for $\text{\hspace{0.17em}}x:$
x=-b+_Гb2-(4ac) ______________ 2a
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the  that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake |
# 4.4: Polynomial Division
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##### Learning Objectives
• Divide polynomials using long division
• Divide polynomials using synthetic division
• Apply the factor and remainder theorems to polynomials
Try these questions prior to beginning this section to help determine if you are set up for success:
1. Simplify the expression:
1. $$\quad (4x^2+2x+1)(x-4)+1$$
2. $$\quad \dfrac{15z^4-18z^3+24z^2-9z}{3z}$$
1. $$\quad 4x^3-14x^2-7x-3$$
2. $$\quad 5z^3-6z^2+8z-3$$
If you missed this problem or feel you could use more practice, review [ 2.6: Multiplying Polynomial Expressions and 2.13: Simplifying, Multiplying, and Dividing Rational Expressions]
Suppose we wish to find the x-intercepts of $$f(x) = x^3 + 4x^2-5x-14$$. Setting $$f(x)=0$$ results in the polynomial equation $$x^3 + 4x^2-5x-14=0$$. Despite all of the factoring techniques we learned in Intermediate Algebra courses, this equation foils us at every turn since these factoring techniques do no work. If we graph $$f$$ using a graphing calculator window [-5,5]x[-16,16], we get
The graph suggests that the function has three x-intercepts, one of which is at $$x=2$$. It's easy to show that $$f(2) = 0$$, but the other two x-intercepts seem to be less friendly. Even though we could use the 'Zero' command to find decimal approximations for these x-intercepts, we seek a method to find the exact values of the other x-intercepts. Based on our experience in the previous section, if there is an x-intercept at $$x=2$$, it seems that there should be a factor of $$(x-2)$$ for $$f(x)$$. In other words, we should expect that $$x^3 + 4x^2-5x-14=(x-2) \, q(x)$$, where $$q(x)$$ is some other polynomial. How could we find such a $$q(x)$$, if it even exists? The answer comes from our old friend, division.
We are familiar with long division from ordinary arithmetic. We begin by dividing into the digits of the dividend that have the greatest place value. We divide, multiply, subtract, include the digit in the next place value position, and repeat. For example, let’s divide 178 by 3 using long division.
Long Division
Step 1: $$5 \times 3=15$$ and $$17-15=2$$ Step 2: Bring down the 8 Step 3: $$9 \times 3=27$$ and $$28-27=1$$ Answer: $$59 R 1$$ or $$59 \frac{1}{3}$$
Another way to look at the solution is as a sum of parts. This should look familiar, since it is the same method used to check division in elementary arithmetic.
\begin{align*} \text{dividend}&=(\text{divisor}{\cdot}\text{quotient})+\text{remainder} \\ 178&=(3{\cdot}59)+1 \\ &=177+1 \\ &=178\end{align*}
We can also state that $$\frac{178}{3}=59+\frac{1}{3}$$ which is the algebraic form of the mixed number $$59\frac{1}{3}$$
We call this the Division Algorithm.
## Long Division of Polynomials
Now, let's revisit the the polynomial $$f(x) = x^3 + 4x^2-5x-14$$. In the last section we saw that we could write a polynomial as a product of factors, each corresponding to a x-intercept. If we know that there is an x-intercept at $$x = 2$$ for $$f(x)$$, then we might guess that the polynomial could be factored as $$x^{3} +4x^{2} -5x-14=(x-2)$$ (something). To find that "something," we can use polynomial division.
##### Example $$\PageIndex{1}$$
Divide $$x^{3} +4x^{2} -5x-14$$ by $$x-2$$.
Solution
Start by writing the problem out in long division form
$$x - 2 \longdiv { x ^ { 3 } + 4 x ^ { 2 } - 5 x - 1 4 }$$
Now we divide the leading terms: $$x^{3} \div x=x^{2}$$. It is best to align it above the same-powered term in the dividend. Now, multiply that $$x^{2}$$ by $$x-2$$ and write the result below the dividend.
Again, divide the leading term of the remainder by the leading term of the divisor. $$6x^{2} \div x=6x$$. We add this to the result, multiply 6x by $$x-2$$, and subtract.
This tells us $$x^{3} +4x^{2} -5x-14$$ divided by $$x-2$$ is $$x^{2} +6x+7$$, with a remainder of zero. This also means that we can factor $$x^{3} +4x^{2} -5x-14$$ as $$\left(x-2\right)\left(x^{2} +6x+7\right)$$.
Now, if we want to find the other x-intercepts, we now solve $$x^2+6x+7=0$$. Since this doesn't factor nicely we can complete the square or use the Quadratic Formula to get $$x = -3 \pm \sqrt{2}$$. The point of this section is to focus on how to perform division. Before looking at more examples, let's note what we can expect when we divide polynomials.
##### The Division Algorithm for Polynomials
Suppose $$d(x)$$ and $$p(x)$$ are nonzero polynomials where the degree of $$p$$ is greater than or equal to the degree of $$d$$. There exist two unique polynomials, $$q(x)$$ and $$r(x)$$, such that $p(x) = d(x) \, q(x) + r(x),\nonumber$
or $\frac{p(x)}{d(x)}=q(x)+\frac{r(x)}{d(x)} \nonumber$
where either $$r(x) = 0$$ or the degree of $$r$$ is strictly less than the degree of $$d$$.
The polynomial $$p$$ is called the dividend; $$d$$ is the divisor; $$q$$ is the quotient; $$r$$ is the remainder. If $$r(x)=0$$ then $$d$$ is called a factor of $$p$$.
Because of the division, the remainder will either be zero, or a polynomial of lower degree than d(x). Because of this, if we divide a polynomial by a term of the form $$x-c$$, then the remainder will be zero or a constant.
If $$p(x)=(x-c)q(x)+r$$, then $$p(c)=(c-c)q(c)+r=0+r=r$$, which establishes the Remainder Theorem.
##### The Remainder Theorem
If $$p(x)$$ is a polynomial of degree 1 or greater and c is a real number, then when p(x) is divided by $$x-c$$, the remainder is $$p(c)$$.
If $$x-c$$ is a factor of the polynomial $$p$$, then $$p(x)=(x-c)q(x)$$ for some polynomial $$q$$. Then $$p(c)=(c-c)q(c)=0$$, showing $$c$$ is a zero of the polynomial. This shouldn’t surprise us - we already knew that if the polynomial factors it reveals the roots.
If $$p(c)=0$$, then the remainder theorem tells us that if p is divided by $$x-c$$, then the remainder will be zero, which means $$x-c$$ is a factor of $$p$$.
The solutions to an equation $$p(x)=0$$ are called zeros of $$p(x)$$. Zeros can be real numbers or complex numbers. As we have seen with quadratics, real zeros corresponding to x-coordinates of x-intercepts and complex zeros do not.
##### The Factor Theorem
If $$p(x)$$ is a nonzero polynomial, then the real number $$c$$ is a zero of $$p(x)$$ if and only if $$x-c$$ is a factor of $$p(x)$$.
##### Example $$\PageIndex{2}$$
Divide $$5x^2+3x−2$$ by $$x+1$$.
Solution
The quotient is $$5x−2$$. The remainder is 0. We write the result as
$\dfrac{5x^2+3x−2}{x+1}=5x−2\nonumber$
Analysis
This division problem had a remainder of 0. This tells us that $$(x+1)$$ is a factor of $$5x^2+3x−2$$, meaning $$5x^2+3x−2=(x+1)(5x−2)$$, and that $$x=-1$$ is a zero of $$p(x)=5x^2+3x−2$$.
##### Example $$\PageIndex{3}$$
Divide $$6x^3+11x^2−31x+15$$ by $$3x−2$$.
Solution
There is a remainder of 1. We can express the result as:
$\dfrac{6x^3+11x^2−31x+15}{3x−2}=2x^2+5x−7+\dfrac{1}{3x−2}\nonumber$
Analysis
We can check our work by using the Division Algorithm to rewrite the solution, then multiply.
$(3x−2)(2x^2+5x−7)+1=6x^3+11x^2−31x+15\nonumber$
Notice, as we write our result,
• the dividend is $$6x^3+11x^2−31x+15$$
• the divisor is $$3x−2$$
• the quotient is $$2x^2+5x−7$$
• the remainder is $$1$$. Since the remainder is $$1$$ we know that $$3x−2$$ is not a factor of $$6x^3+11x^2−31x+15$$.
## Synthetic Division
Since dividing by $$x-c$$ is a way to check if a number, $$c$$, is a zero of the polynomial, it would be nice to have a faster way to divide by $$x-c$$ than having to use long division every time. Happily, quicker ways have been discovered.
Let’s look back at the long division we did in Example 1 and try to streamline it. First, let’s distribute the negatives in the subtraction steps:
Next, observe that the terms $$-x^{3}$$, $$-6x^{2}$$, and $$-7x$$ are the exact opposite of the terms above them. Also note that the terms we ‘bring down’ (namely the $$\mathrm{-}$$5x and $$\mathrm{-}$$14) aren’t really necessary to recopy since we know they will cancel out when we subtract, so let's omit them, too:
Now, let’s move things up a bit and, for reasons which will become clear in a moment, copy the $$x^{3}$$ into the last row.
Note that by arranging things in this manner, each term in the last row is obtained by adding the two terms above it. Notice also that the quotient polynomial can be obtained by dividing each of the first three terms in the last row by $$x$$ and adding the results. If you take the time to work back through the original division problem, you will find that this is exactly the way we determined the quotient polynomial.
This means that we no longer need to write the quotient polynomial down, nor the $$x$$ in the divisor, to determine our answer.
We’ve streamlined things quite a bit so far, but we can still do more. Let’s take a moment to remind ourselves where the $$2x^{2}$$, $$12x$$ and 14 came from in the second row. Each of these terms was obtained by multiplying the terms in the quotient, $$x^{2}$$, 6x and 7, respectively, by the -2 in $$x - 2$$, then by -1 when we changed the subtraction to addition. Multiplying by -2 then by -1 is the same as multiplying by 2, so we replace the -2 in the divisor by 2. Furthermore, the coefficients of the quotient polynomial match the coefficients of the first three terms in the last row, so we now take the plunge and write only the coefficients of the terms to get
We have constructed a synthetic division tableau for this polynomial division problem. Let’s re-work our division problem using this tableau to see how it greatly streamlines the division process. To divide $$x^{3} +4x^{2} -5x-14$$ by $$x-2$$, we write 2 in the place of the divisor and the coefficients of $$x^{3} +4x^{2} -5x-14$$ in for the dividend. Then "bring down" the first coefficient of the dividend.
Next, take the 2 from the divisor and multiply by the 1 that was "brought down" to get 2. Write this underneath the 4, then add to get 6.
Now take the 2 from the divisor times the 6 to get 12, and add it to the -5 to get 7.
Finally, take the 2 in the divisor times the 7 to get 14, and add it to the -14 to get 0.
The first three numbers in the last row of our tableau are the coefficients of the quotient polynomial. Remember, we started with a third degree polynomial and divided by a first degree polynomial, so the quotient is a second degree polynomial. Hence the quotient is $$x^{2} +6x+7$$. The number in the box is the remainder. Synthetic division is our tool of choice for dividing polynomials by divisors of the form $$x - c$$. It is important to note that it works only for these kinds of divisors. Also take note that when a polynomial (of degree at least 1) is divided by $$x - c$$, the result will be a polynomial of exactly one less degree. Finally, it is worth the time to trace each step in synthetic division back to its corresponding step in long division.
##### Example $$\PageIndex{4}$$
Use synthetic division to divide $$5x^{3} -2x^{2} +1$$ by $$x-3$$.
Solution
When setting up the synthetic division tableau, we need to enter 0 for the coefficient of $$x$$ in the dividend. Doing so gives
Since the dividend was a third degree polynomial, the quotient is a quadratic polynomial with coefficients 5, 13 and 39. Our quotient is $$q(x)=5x^{2} +13x+39$$ and the remainder is $$r(x) = 118$$. We can express this as $\frac{5x^{3} -2x^{2} +1}{x-3}=5x^{2} +13x+39+\frac{118}{x-3}\nonumber$
We also know that $$x-3$$ is not a factor of $$5x^{3} -2x^{2} +1$$ since the remainder is not zero.
##### Example $$\PageIndex{5}$$
Divide $$x^{3} +8$$ by $$x+2$$.
Solution
For this division, we rewrite $$x+2$$ as $$x-\left(-2\right)$$ and proceed as before.
The quotient is $$x^{2} -2x+4$$ and the remainder is zero. Since the remainder is zero, $$x+2$$ is a factor of $$x^{3} +8$$.
$x^{3} +8=(x+2)\left(x^{2} -2x+4\right)\nonumber$
We can write our answer as $\dfrac{x^{3} +8}{x+2}=x^{2} -2x+4\nonumber$
Note in the last example that there are placeholders for the missing $$x^2$$ and $$x$$ terms. When using synthetic division you must put place holders for the missing terms for the division to work properly.
##### You Try $$\PageIndex{1}$$
Divide $$4x^{4} -8x^{2} -5x$$ by $$x-3$$ using synthetic division.
$$4x^4 - 8x^2 - 5x$$ divided by $$x -3$$ is $$4x^3 + 12x^2 + 28x + 79$$ with remainder 237
Using this process allows us to find the real zeros of polynomials, presuming we can figure out at least one root. We’ll explore how to do that in the next section.
##### Example $$\PageIndex{6}$$
The polynomial $$p(x)=4x^{4} -4x^{3} -11x^{2} +12x-3$$ has a horizontal intercept at $$x=\dfrac{1}{2}$$ with multiplicity 2. Find the other intercepts of $$p(x)$$.
Solution
Since $$x=\dfrac{1}{2}$$ is an intercept with multiplicity 2, then $$x-\dfrac{1}{2}$$ is a factor twice. Use synthetic division to divide by $$x-\dfrac{1}{2}$$ twice.
From the first division, we get $$4x^{4} -4x^{3} -11x^{2} +12x-3=\left(x-\dfrac{1}{2} \right)\left(4x^{3} -2x^{2} -x-6\right)$$ The second division tells us
$4x^{4} -4x^{3} -11x^{2} +12x-3=\left(x-\dfrac{1}{2} \right)\left(x-\dfrac{1}{2} \right)\left(4x^{2} -12\right)\nonumber$
To find the remaining intercepts, we set $$4x^{2} -12=0$$ and get $$x=\pm \sqrt{3}$$.
Note this also means $$4x^{4} -4x^{3} -11x^{2} +12x-3=4\left(x-\dfrac{1}{2} \right)\left(x-\dfrac{1}{2} \right)\left(x-\sqrt{3} \right)\left(x+\sqrt{3} \right)$$.
Let's summarize what we've learned about zeros and polynomial functions:
##### Connections Between Zeros, Factors, and Graphs of Polynomial Functions
Suppose $$p$$ is a polynomial function of degree $$n \geq 1$$. The following statements are equivalent:
• The real number $$c$$ is a zero of $$p$$
• $$p(c) = 0$$
• $$x = c$$ is a solution to the polynomial equation $$p(x) = 0$$
• $$(x - c)$$ is a factor of $$p(x)$$
• The point $$(c, 0)$$ is an $$x$$-intercept of the graph of $$y = p(x)$$
This page titled 4.4: Polynomial Division is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Katherine Skelton. |
## Latest Post
Mixed numbers are numbers with extra than two elements. Numbers may be labeled based on the variety of things they have got. If a range has the simplest elements – 1 and the quantity itself, then it is a top variety. However, most numbers have extra factors and are known as composite numbers. On this web page, we can find out about the distinction between high and composite numbers, smallest composite numbers, and extraordinary composite numbers. The ultimate one is exciting because there are numerous ordinary high numbers in preference to 2, which is the handiest even top range. Click here https://cricfor.com/
## What Are Composite Numbers?
A range that is divisible via quite a number apart from 1 and the quantity itself is referred to as a composite variety. For instance, four is a range of this is divisible by using 1, 2, and four, so it’s far a composite number.
## Composite Variety Definition
Mixed numbers may be described as the ones herbal numbers that have greater than two elements. Let us study extra about composite numbers with examples.
## Examples Of Entire Numbers
4, 6, eight, 9, and 10 are the first few composite numbers. Let us take 6 and eight. In the above example, 6 and 8 are called composite numbers because they’ve greater than 2 factors. Let us continue to recognize the important residences of composite numbers.
Get to know more about various subjects 80 inches in cm
## How To Locate Composite Variety?
To find a composite variety, we discover the factors of the given wide variety. If quite a number has greater than two factors, it is complex. The high-quality way to find out a composite quantity is to do a divisibility test. The divisibility test facilitates us to determine whether several is a high quantity or a composite wide variety. Divisibility method that several is divisible (without the rest) through quite a number apart from 1 and itself.
To do this, we need to check to peer if the variety may be divided by using those commonplace elements: 2, 3, five, 7, eleven, and thirteen. If the given range is even, then the number is divisible through 2. Begin check. The quantity ends with 0 or 5, then check it with five. If the quantity can not be divided by way of any of those numbers, then the variety is a prime number. For example, sixty-eight is divisible by 2, this means that it has elements other than 1 and sixty-eight, consequently, we can say that 68 is a composite variety.
## Homes Of Composite Numbers
A composite variety is a nice integer that may be fashioned using multiplying two smaller high-quality integers. Note the residences of the composite range listed under:
All composite numbers are exactly divisible by smaller numbers which can be high or mixed.
Every composite variety is made from two or more high numbers.
Let us test the properties of the complex range 72 to understand the concept higher.
The parent shows that multiplying those wonderful integers gives us a composite quantity.
## Styles Of Combined Numbers
The primary sorts of composite numbers in arithmetic are bizarre composite numbers and even composite numbers.
## Abnormal Combined Variety
All peculiar numbers which aren’t prime are atypical composite numbers. For example, 9, 15, 21, 25, and 27 are bizarre composite numbers. Consider the numbers 1, 2, 3, 4, 9, 10, eleven, 12, and 15. Here 9 and 15 are abnormal composite numbers because those two bizarre numbers have more than 2 elements and they satisfy the circumstance of composite numbers.
## Even Blended Numbers
All even numbers which are not prime are even prime numbers. For instance, four, 6, 8, 10, 12, 14, and sixteen are even numbers. Consider the numbers 1, 2, three, four, 9, 10, eleven, 12, and 15 once more. Here 4, 10, and 12 are even composite numbers because those even numbers have more than 2 factors and fulfill the situation of composite numbers.
## Smallest Composite Quantity
The smallest composite range is four. A composite number is defined as a number that has a divisor apart from 1 and the number itself. Now, as we start counting: 1, 2, three, 4, five, 6, …. Similarly, we see that 1 isn’t a composite range because its only divisor is 1. 2 is not a composite range as it has the handiest two divisors, i.E. 1 and the range itself 2. Three isn’t a composite quantity as it has the most effective two divisors, i.E. 1 and the quantity itself 3. However, whilst we come to the number 4, we recognize that its divisors are 1, 2, and four. Hence, variety 4 satisfies the criteria of a composite quantity. Hence 4 is the smallest composite quantity.
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# Upper Quartile(KS2, Year 4)
homesitemapstatisticsthe upper quartile
The upper quartile is the middle number between the median and the highest number. It is the middle number of the upper half of a set of numbers. The upper quartile is denoted Q3.
## Understanding the Upper Quartile and the Quartiles
The upper quartile is the highest of the three quartiles. The three quartiles divide a set of numbers, that are in numerical order, into four equal groups:
• The middle quartile is also known as the median. It is the middle number in the set. It divides the set in two halves: a lower half and an upper half.
• The upper quartile is the middle number of the upper half. It divides the upper half in two.
## Finding the Upper Quartile
There are different methods for finding the upper quartile, which give different values.
• The Moore and McCabe method excludes the median from the upper half. The upper quartile is the middle number of the upper half:
• The Tukey method includes the median in the upper half. The upper quarter is the middle number of the upper half:
Note: When the upper half has an even number of numbers, the middle number is halfway between the middle two numbers. It is the mean of the middle two numbers:
Q3 = (8 + 9) ÷ 2 = 8.5
• The Mendenhall and Sincich method uses a formula to find the position of the upper quartile. The formula is shown below:
In this formula, n is how many numbers there are in the set. In our example, n is 11.
Position of Q3 = 3(n + 1) ÷ 4 = 3 × (11 + 1) ÷ 4
Position of Q3 = 36 ÷ 4
Position of Q3 = 9th
The upper quartile is the 9th number in the set:
## What's In a Name?
Quartile comes from the Latin word "quartus", meaning 'a fourth'. It comes from the same root as 'quarter'.
## Put Your Numbers in Order
The quartiles of a set of numbers divide the numbers into four equal groups when the numbers are in order. Imagine you were asked to find the upper quartile of the numbers below. Don't be tempted to jump right in.
3 5 4 1 2
Put the numbers in order and then find the upper quartile:
1 2 3 4 5
This page was written by Stephen Clarke.
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# Math Expressions Grade 2 Unit 1 Lesson 18 Answer Key More Complex Compare Problems
## Math Expressions Common Core Grade 2 Unit 1 Lesson 18 Answer Key More Complex Compare Problems
Draw comparison bars. Write an equation. Solve the problem. Show your work.
Question 1.
Morgan sees 15 birds on a bird-watching trip. She sees 6 more birds than Shari. How many birds does Shari see?
Shari saw 9 birds
Explanation:
Morgan sees 15 birds
Finally she saw 6 more birds than Shari
So,15 – 6 = 9
Question 2.
There are 5 fewer trucks than cars in the parking lot. If there are 8 trucks, how many cars are there?
13 cars
Explanation:
Total trucks = 8
5 fewer trucks than cars
So, total cars 8 + 5 = 13
Question 3.
Anh makes I 2 quilts. Krista makes 7 fewer quilts than Anh. How many quilts does Krista make?
5 quilts
Explanation:
Anh makes = 12 quilts
Krista makes =7 fewer quilts
12 – 7 = 5
Question 4.
There are 8 fewer tigers than lions at the zoo. There are 8 tigers at the zoo. How many lions does the zoo have?
16 lions
Explanation:
8 tigersĀ are less than lions in zoo
Total tigers are 8
So, total lions 8 + 8 = 16
Find the unknown addend (unknown partner).
Question 1.
3 + __________ = 12
14 – __________ = 8
15 – 6 = __________
3 + 9 = 12
14 – 6 = 8
15 – 6 = 9
Explanation:
In the Math mountain, the top part of the mountain to called Total or Sum and it consists of two parts called Partners or Addends
The addition of two whole numbers results in the total amount or sum of those values combined.
12 – 3 = 9
14 – 8 = 6
15 – 6 =9
Question 2.
4 + __________ = 13
15 – __________ = 7
14 – 7 = __________
13 – 4 = 9
15 – 7 = 8
14 – 7 = 7
Explanation:
In the Math mountain, the top part of the mountain to called Total or Sum and it consists of two parts called Partners or Addends
The addition of two whole numbers results in the total amount or sum of those values combined.
13 – 4 = 9
15 – 7 = 8
14 – 7 = 7
Solve the word problems. Show your work.
Question 3.
There are 13 dancers in the front row. 7 dancers are in the back row. How many fewer dancers are in the back row than are in the front row?
6 dancers
Explanation:
Number of dancers in front row = 12
Number of dancers in back row = 7
Still back row need 6 more dancers than front row
13 – 7 = 6
Question 4.
There are 8 birds in the red cage. The blue cage has 4 more birds than the red cage. How many birds are in the blue cage?
12 birds
Explanation:
Number of birds in red cage = 8
Number of birds in blue cage = 4 more than red cage
Total birds in blue cage = 8 + 4 =12
Question 5.
Stretch Your Thinking When would you use a drawing of comparison bars for a word problem? |
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# System of Equation Word Problems
Please see the attached document for better formatting:
_____________________________________________________
Review examples 2, 3, and 4 in section 8.4 of the text. How does the author determine what the first equation should be? What about the second equation?
How are these examples similar? How are they different? Post a similar to example 2,3 and 4.
Post for classmate to solve.
2.) Translate. The first row of the table and the fourth sentence of the problem tell us that a total of 63 pupae was received. Thus we have one equation:
p+q=63
Since each pupa of morpho granadensis costs \$4.15 and p pupae were received, 4.15p is the cost for the morpho granadensis species. Similarly, 1.50q is the cost of the battus polydamus species. From the third row of the table and the information in the statement of the problem, we get a second equation:
4.15p+1.50q=147.50
We can multiply by 100 on both sides of this equation in order to clear the decimals. This gives us the following system of equations as a translation:
p+q=63 (1)
415p+150q=14,750 (2)
3.) Solve. We decide to use the elimination method to solve the system. We eliminate q by multiplying equation (1) by -150 and adding it to equation (2).
-150p-150q = -9450 Multiplying equation(1) by -150
415p+150q = 14750
P = 20-------------------solving for
To find q, we substitute 20 for p in equation (1) and solve for q:
p+q=63
20+q=63
q=43
We obtain (20,43), or p=20,q=43.
4.) Check We check in the original problem. Remember that p is the number of pupae of morpho granadensis and q is the number of pupae of battus polydamus.
Number of pupae; p+q=20+43=63
Cost of morpho granadensis: \$4.15p=4.15(20)= \$83.00
Cost of battus polydamus: \$1.50q=1.50(43)=64.50
Total:= \$147.50
#### Solution Summary
This solution includes step by step solutions of solving and checking solutions of word problems on system of equations.
\$2.19 |
# Conjugate Complex Numbers
Definition of conjugate complex numbers: In any two complex numbers, if only the sign of the imaginary part differ then, they are known as complex conjugate of each other.
Conjugate of a complex number z = a + ib, denoted by $$\bar{z}$$, is defined as
$$\bar{z}$$ = a - ib i.e., $$\overline{a + ib}$$ = a - ib.
For example,
(i) Conjugate of z$$_{1}$$ = 5 + 4i is $$\bar{z_{1}}$$ = 5 - 4i
(ii) Conjugate of z$$_{2}$$ = - 8 - i is $$\bar{z_{2}}$$ = - 8 + i
(iii) conjugate of z$$_{3}$$ = 9i is $$\bar{z_{3}}$$ = - 9i.
(iv) $$\overline{6 + 7i}$$ = 6 - 7i, $$\overline{6 - 7i}$$ = 6 + 7i
(v) $$\overline{-6 - 13i}$$ = -6 + 13i, $$\overline{-6 + 13i}$$ = -6 - 13i
Properties of conjugate of a complex number:
If z, z$$_{1}$$ and z$$_{2}$$ are complex number, then
(i) $$\bar{(\bar{z})}$$ = z
Or, If $$\bar{z}$$ be the conjugate of z then $$\bar{\bar{z}}$$ = z.
Proof:
Let z = a + ib where x and y are real and i = √-1. Then by definition, (conjugate of z) = $$\bar{z}$$ = a - ib.
Therefore, (conjugate of $$\bar{z}$$) = $$\bar{\bar{z}}$$ = a + ib = z. Proved.
(ii) $$\bar{z_{1} + z_{2}}$$ = $$\bar{z_{1}}$$ + $$\bar{z_{2}}$$
Proof:
If z$$_{1}$$ = a + ib and z$$_{2}$$ = c + id then $$\bar{z_{1}}$$ = a - ib and $$\bar{z_{2}}$$ = c - id
Now, z$$_{1}$$ + z$$_{2}$$ = a + ib + c + id = a + c + i(b + d)
Therefore, $$\overline{z_{1} + z_{2}}$$ = a + c - i(b + d) = a - ib + c - id = $$\bar{z_{1}}$$ + $$\bar{z_{2}}$$
(iii) $$\overline{z_{1} - z_{2}}$$ = $$\bar{z_{1}}$$ - $$\bar{z_{2}}$$
Proof:
If z$$_{1}$$ = a + ib and z$$_{2}$$ = c + id then $$\bar{z_{1}}$$ = a - ib and $$\bar{z_{2}}$$ = c - id
Now, z$$_{1}$$ - z$$_{2}$$ = a + ib - c - id = a - c + i(b - d)
Therefore, $$\overline{z_{1} - z_{2}}$$ = a - c - i(b - d)= a - ib - c + id = (a - ib) - (c - id) = $$\bar{z_{1}}$$ - $$\bar{z_{2}}$$
(iv) $$\overline{z_{1}z_{2}}$$ = $$\bar{z_{1}}$$$$\bar{z_{2}}$$
Proof:
If z$$_{1}$$ = a + ib and z$$_{2}$$ = c + id then
$$\overline{z_{1}z_{2}}$$ = $$\overline{(a + ib)(c + id)}$$ = $$\overline{(ac - bd) + i(ad + bc)}$$ = (ac - bd) - i(ad + bc)
Also, $$\bar{z_{1}}$$$$\bar{z_{2}}$$ = (a – ib)(c – id) = (ac – bd) – i(ad + bc)
Therefore, $$\overline{z_{1}z_{2}}$$ = $$\bar{z_{1}}$$$$\bar{z_{2}}$$ proved.
(v) $$\overline{(\frac{z_{1}}{z_{2}}}) = \frac{\bar{z_{1}}}{\bar{z_{2}}}$$, provided z$$_{2}$$ ≠ 0
Proof:
According to the problem
z$$_{2}$$ ≠ 0 ⇒ $$\bar{z_{2}}$$ ≠ 0
Let, $$(\frac{z_{1}}{z_{2}})$$ = z$$_{3}$$
z$$_{1}$$ = z$$_{2}$$ z$$_{3}$$
⇒ $$\bar{z_{1}}$$ = $$\bar{z_{2} z_{3}}$$
⇒ $$\frac{\bar{z_{1}}}{\bar{z_{2}}}$$ = $$\bar{z_{3}}$$
⇒ $$\overline{(\frac{z_{1}}{z_{2}}}) = \frac{\bar{z_{1}}}{\bar{z_{2}}}$$, [Since z$$_{3}$$ = $$(\frac{z_{1}}{z_{2}})$$] Proved.
(vi) |$$\bar{z}$$| = |z|
Proof:
Let z = a + ib then $$\bar{z}$$ = a - ib
Therefore, |$$\bar{z}$$| = $$\sqrt{a^{2} + (-b)^{2}}$$ = $$\sqrt{a^{2} + b^{2}}$$ = |z| Proved.
(vii) z$$\bar{z}$$ = |z|$$^{2}$$
Proof:
Let z = a + ib, then $$\bar{z}$$ = a - ib
Therefore, z$$\bar{z}$$ = (a + ib)(a - ib)
= a$$^{2}$$ – (ib)$$^{2}$$
= a$$^{2}$$ – i$$^{2}$$b$$^{2}$$
= a$$^{2}$$ + b$$^{2}$$, since i$$^{2}$$ = -1
= $$(\sqrt{a^{2} + b^{2}})^{2}$$
= |z|$$^{2}$$. Proved.
(viii) z$$^{-1}$$ = $$\frac{\bar{z}}{|z|^{2}}$$, provided z ≠ 0
Proof:
Let z = a + ib ≠ 0, then |z| ≠ 0.
Therefore, z$$\bar{z}$$ = (a + ib)(a – ib) = a$$^{2}$$ + b$$^{2}$$ = |z|$$^{2}$$
⇒ $$\frac{z\bar{z}}{|z|^{2}}$$ = 1
⇒ $$\frac{\bar{z}}{|z|^{2}}$$ = $$\frac{1}{z}$$ = z$$^{-1}$$
Therefore, z$$^{-1}$$ = $$\frac{\bar{z}}{|z|^{2}}$$, provided z ≠ 0. Proved.
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# Basic Functions
Polynomials
Exponential Functions
Trigonometric Functions
lim
x 0
sin x
x
Trigonometric Identities
The Number e
Index
FAQ
Polynomials
Definition
P a0 a1x a2 x 2
an x n
## The polynomial P is of degree n.
A number x for which P(x)=0 is called a root of the
polynomial P.
Theorem
Index
## A polynomial of degree n has at most n real roots.
Polynomials may have no real roots, but a polynomial of
an odd degree has always at least one real root.
FAQ
## Graphs of Linear Polynomials
Graphs of linear polynomials y = ax + b are straight lines. The coefficient
a determines the angle at which the line intersects the x axis.
## Graphs of the linear
polynomials:
1. y = 2x+1 (the red line)
## 2. y = -3x+2 (the black line)
3. y = -3x + 3 (the blue line)
Index
FAQ
## Graphs of Higher Degree Polynomials
The behaviour of a polynomial P a0 a1x
## If an 0 and n is odd, then as x also P x .
Likewise: as x also P x .
Problem
## The picture on the right shows the
graphs and all roots of a 4th degree
polynomial and of a 5th degree
polynomial. Which is which?
Solution
## The blue curve must be the graph of
the 4th degree polynomial because
of its behavior as x grows or gets
smaller.
Index
FAQ
## Measuring of Angles (1)
Angles are formed by two half-lines starting
from a common vertex. One of the half-lines
is the starting side of the angle, the other one
is the ending side. In this picture the starting
side of the angle is blue, and the red line is
the ending side.
Angles are measured by drawing a circle
of radius 1 and with center at the vertex
of the angle. The size, in radians, of the
angle in question is the length of the
black arc of this circle as indicated in the
picture.
In the above we have assumed that the angle is
oriented in such a way that when walking along
the black arc from the starting side to the
ending side, then the vertex is on our left.
Index
FAQ
## Measuring of Angles (2)
The first picture on the right shows a
positive angle.
## The angle becomes negative if the orientation
gets reversed. This is illustrated in the second
picture.
## This definition implies that angles are always
between -2 and 2. By allowing angles to
rotate more than once around the vertex, one
generalizes the concept of angles to angles
greater than 2 or smaller than - 2.
Index
FAQ
## Trigonometric Functions (1)
Consider positive angles , as indicated in the pictures.
sin
Definition
## The quantities sin and cos are defined
by placing the angle at the origin with starting
side on the positive x -axis. The intersection point
of the end side and the circle with radius 1 and with
center at the origin is cos ,sin .
cos cos .
Index
sin
## This definition applies for positive angles.
We extend that to the negative angles by
setting
sin sin and
cos
cos
FAQ
sin2 cos2 1
sin
definition.
Definition
tan
sin
cos
cot
cos
sin
cos
Graphs of:
1.
and
2.
Index
FAQ
## Trigonometric Functions (3)
The size of an angle is measured as the length
of the arc, indicated in the picture, on a circle
of radius 1 with center at the vertex.
## On the other hand, sin() is the length of the red
line segment in the picture.
Lemma
sin
## For positive angles , sin .
The above inequality is obvious by the above picture. For negative angles
the inequality is reversed.
Index
FAQ
## Trigonometric Functions (4)
Trigonometric functions sin and cos are
everywhere continuous, and lim sin 0 and lim cos 1.
0
sin tan .
Hence
sin
sin
Lemma
Index
1
.
cos
lim
sin
sin
FAQ
tan
Examples
Problem 1
Solution
Compute lim
sin 2 x
x 0
Rewrite
sin 2 x
x
sin 2 x
2
.
2x
sin 2x
x 0
Hence
Index
sin 2x
x
2x
1.
sin 2x
2
2.
x 0
2x
FAQ
Examples
Problem 2
Compute lim
x 0
Rewrite
Solution
sin sin x
x
sin sin x
x
sin x
sin sin x
x 0
sin x
1.This follows
Hence
sin sin x
Index
sin x
1.
x 0
## Mika Seppl: Basic Functions
FAQ
Trigonometric Identities 1
Defining Identities
1
csc
sin
tan
1
sec
cos
sin
cos
cot
1
cot
tan
cos
sin
Derived Identities
sin sin
cos =cos
## sin 2 sin cos 2 cos
sin2 +cos2 =1
sin x y sin x cos y cos x sin y
cos x y cos x cos y sin x sin y
Index
## Mika Seppl: Basic Functions
FAQ
Trigonometric Identities 2
Derived Identities (contd)
## sin x y sin x cos y cos x sin y
cos x y cos x cos y sin x sin y
tan x y
tan x tan y
1 tan x tan y
tan x y
tan x tan y
1 tan x tan y
cos 2x 2cos2 x 1
cos 2x 1 2sin2 x
cos x
2
Index
1 cos 2 x
2
sin x
2
## Mika Seppl: Basic Functions
1 cos 2 x
2
FAQ
Exponential Functions
Exponential functions are functions of the form
f x ax.
Assuming that a 0, a x is a well defined expression for all x .
## The picture on the right shows the graphs of the
functions:
x
1
1) y , the red curve
2
2) y 1x , the black line
x
3
3) y , the blue curve, and
2
x
5
4) y , the green curve.
2
Index
## Mika Seppl: Basic Functions
FAQ
The Number e
From the picture it appears obvious that,
as the parameter a grows, also the slope
of the tangent, at x 0, of the graph of the
a=5/2
a=1/2
a=3/2
function a x grows.
a=1
Definition
## The mathematical constant e is defined
as the unique number e for which the slope
of the tangent of the graph of e x at x 0
is 1.
e2.718281828
Index
## The slope of a tangent
line is the tangent of the
angle at which the
tangent line intersects
the x-axis.
FAQ |
# Find the slope of a line. slope rise run Main Idea/Vocabulary.
## Presentation on theme: "Find the slope of a line. slope rise run Main Idea/Vocabulary."— Presentation transcript:
Find the slope of a line. slope rise run Main Idea/Vocabulary
ACCESS RAMPS The access ramp from the sidewalk to the door of a hotel rises 8 inches for every horizontal change of 96 inches. What is the slope of the access ramp? Answer: Example 1
ACCESS RAMPS The access ramp from the sidewalk to the door of an office building rises 14 inches for every horizontal change of 210 inches. What is the slope of the access ramp? A. B. C. D. A B C D Example 1
Find Slope Using a Graph
Find the slope of the line. Choose two points on the line. The vertical change is –3 units while the horizontal change is 2 units. Example 2
Find Slope Using a Graph
Find the slope of the line.
A. B. C. D. A B C D Example 2
Find Slope Using a Table
The points given in the table lie on a line. Find the slope of the line. Then graph the line. +2 +3 Example 3
Find Slope Using a Table
The points given in the table lie on a line. Find the slope of the line. Then graph the line.
C. D. A B C D Example 3
KC
Find Slope Using Coordinates
Find the slope of the line that passes through A(3, 3) and B(2, 0). Definition of slope (x1, y1) = (3, 3), (x2, y2) = (2,0) Simplify. Answer: The slope is 3. Example 4
Find Slope Using Coordinates
Check When going from left to right, the graph of the line slants upward. This is correct for positive slope. Example 4
Find the slope of the line that passes through A(4, 3) and B(1, 0).
C. 2 D. 5 A B C D Example 4
Find Slope Using Coordinates
Find the slope of the line that passes through X(–2, 3) and Y(3, 0). Definition of slope (x1, y1) = (–2, 3), (x2, y2) = (3,0) Simplify. Answer: Example 5
Find Slope Using Coordinates
Check When going from left to right, the graph of the line slants downward. This is correct for a negative slope. Example 5
Find the slope of the line that passes through X(–3, 3) and Y(1, 0).
B. C. D. A B C D Example 5
Which of the following is a graph of the function y = x – 2?
(over Lesson 9-3) Which of the following is a graph of the function y = x – 2? A. B. C. D. A B C D Five Minute Check 1
Which of the following is a graph of the function y = 4x?
(over Lesson 9-3) Which of the following is a graph of the function y = 4x? A. B. C. D. A B C D Five Minute Check 2
(over Lesson 9-3) A local repairman charges \$20 per visit plus \$15 per hour of work. Which choice shows an equation for the repairman’s rates and a graph of the equation? A. B. C. D. A B C D Five Minute Check 3
Which function is graphed in the figure shown?
(over Lesson 9-3) Which function is graphed in the figure shown? A. y = x – 1 B. y = 2x – 1 C. y = x D. y = 3x – 3 A B C D Five Minute Check 4
End of Custom Shows
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### Still have math questions?
Arithmetic
Question
3. Multiply or divide. Draw a model to explain your thinking. $$4 \div \frac { 1 } { 3 }$$$$\frac { 1 } { 2 } \div 3$$
$$4 \div \frac { 1 } { 3 } = 12$$
Steps
$$4 \div \frac { 1 } { 3 }$$
Convert element to fraction: $$4 = \frac { 4 } { 1 }$$
$$= \frac { 4 } { 1 } \div \frac { 1 } { 3 }$$
Apply the fraction rule: $$\frac { a } { b } \div \frac { c } { d } = \frac { a } { b } \times \frac { d } { c }$$
$$= \frac { 4 } { 1 } \times \frac { 3 } { 1 }$$
Apply rule $$\frac { a } { 1 } = a$$
$$= 4 \times 3$$
Multiply the numbers: $$4 \times 3 = 12$$
$$= 12$$
$$\frac { 1 } { 2 } \div 3 = \frac { 1 } { 6 }$$
Steps
$$\frac { 1 } { 2 } \div 3$$
Convert element to fraction: $$3 = \frac { 3 } { 1 }$$
$$= \frac { 1 } { 2 } \div \frac { 3 } { 1 }$$
Apply the fraction rule: $$\frac { a } { b } \div \frac { c } { d } = \frac { a } { b } \times \frac { d } { c }$$
$$= \frac { 1 } { 2 } \times \frac { 1 } { 3 }$$
Multiply fractions: $$\frac { a } { b } \times \frac { c } { d } = \frac { a \times c } { b \times d }$$
$$= \frac { 1 \times 1 } { 2 \times 3 }$$
Multiply the numbers: $$1 \times 1 = 1$$ $$= \frac { 1 } { 2 \times 3 }$$
Multiply the numbers: $$2 \times 3 = 6$$
$$= \frac { 1 } { 6 }$$
Solution
View full explanation on CameraMath App. |
# Understanding Linear Algebra
## Section2.3The span of a set of vectors
Matrix multiplication allows us to rewrite a linear system in the form $$A\xvec = \bvec\text{.}$$ Besides being a more compact way of expressing a linear system, this form allows us to think about linear systems geometrically since matrix multiplication is defined in terms of linear combinations of vectors.
We now return to our two fundamental questions, rephrased here in terms of matrix multiplication.
• Existence: Is there a solution to the equation $$A\xvec=\bvec\text{?}$$
• Uniqueness: If there is a solution to the equation $$A\xvec=\bvec\text{,}$$ is it unique?
In this section, we focus on the existence question and see how it leads to the concept of the span of a set of vectors.
### Preview Activity2.3.1.The existence of solutions.
1. If the equation $$A\xvec = \bvec$$ is inconsistent, what can we say about the pivot positions of the augmented matrix $$\left[\begin{array}{r|r} A \amp \bvec \end{array}\right]\text{?}$$
2. Consider the matrix $$A$$
\begin{equation*} A = \left[ \begin{array}{rrr} 1 \amp 0 \amp -2 \\ -2 \amp 2 \amp 2 \\ 1 \amp 1 \amp -3 \end{array}\right]\text{.} \end{equation*}
If $$\bvec=\threevec{2}{2}{5}\text{,}$$ is the equation $$A\xvec = \bvec$$ consistent? If so, find a solution.
3. If $$\bvec=\threevec{2}{2}{6}\text{,}$$ is the equation $$A\xvec = \bvec$$ consistent? If so, find a solution.
4. Identify the pivot positions of $$A\text{.}$$
5. For our two choices of the vector $$\bvec\text{,}$$ one equation $$A\xvec = \bvec$$ has a solution and the other does not. What feature of the pivot positions of the matrix $$A$$ tells us to expect this?
### Subsection2.3.1The span of a set of vectors
In the preview activity, we considered a $$3\times3$$ matrix $$A$$ and found that the equation $$A\xvec = \bvec$$ has a solution for some vectors $$\bvec$$ in $$\real^3$$ and has no solution for others. We will introduce a concept called span that describes the vectors $$\bvec$$ for which there is a solution.
We can write an $$m\times n$$ matrix $$A$$ in terms of its columns
\begin{equation*} A = \left[\begin{array}{rrrr} \vvec_1\amp\vvec_2\amp\cdots\amp\vvec_n \end{array}\right]\text{.} \end{equation*}
Remember that Proposition 2.2.4 says that the equation $$A\xvec = \bvec$$ is consistent if and only if we can express $$\bvec$$ as a linear combination of $$\vvec_1,\vvec_2,\ldots,\vvec_n\text{.}$$
#### Definition2.3.1.
The span of a set of vectors $$\vvec_1,\vvec_2,\ldots,\vvec_n$$ is the set of all linear combinations that can be formed from the vectors.
Alternatively, if $$A = \begin{bmatrix} \vvec_1 \amp \vvec_2 \amp \cdots \amp \vvec_n \end{bmatrix}\text{,}$$ then the span of the vectors consists of all vectors $$\bvec$$ for which the equation $$A\xvec = \bvec$$ is consistent.
#### Example2.3.2.
Considering the set of vectors $$\vvec=\twovec{-2}{1}$$ and $$\wvec = \twovec{8}{-4}\text{,}$$ we see that the vector
\begin{equation*} \bvec = 3\vvec + \wvec = \twovec2{-1} \end{equation*}
is one vector in the span of the vectors $$\vvec$$ and $$\wvec$$ because it is a linear combination of $$\vvec$$ and $$\wvec\text{.}$$
To determine whether the vector $$\bvec=\twovec{5}{2}$$ is in the span of $$\vvec$$ and $$\wvec\text{,}$$ we form the matrix
\begin{equation*} A = \begin{bmatrix} \vvec \amp \wvec \end{bmatrix} = \begin{bmatrix} -2 \amp 8 \\ 1 \amp -4 \\ \end{bmatrix} \end{equation*}
and consider the equation $$A\xvec=\bvec\text{.}$$ We have
\begin{equation*} \left[ \begin{array}{rr|r} -2 \amp 8 \amp 5 \\ 1 \amp -4 \amp 2 \\ \end{array} \right] \sim \left[ \begin{array}{rr|r} 1 \amp -4 \amp 0 \\ 0 \amp 0 \amp 1 \\ \end{array} \right], \end{equation*}
which shows that the equation $$A\xvec = \bvec$$ is inconsistent. Therefore, $$\bvec=\twovec52$$ is one vector that is not in the span of $$\vvec$$ and $$\wvec\text{.}$$
#### Activity2.3.2.
Let’s look at two examples to develop some intuition for the concept of span.
1. First, we will consider the set of vectors
\begin{equation*} \vvec = \twovec{1}{2}, ~~~\wvec = \twovec{-2}{-4}\text{.} \end{equation*}
1. What vector is the linear combination of $$\vvec$$ and $$\wvec$$ with weights:
• $$c = 2$$ and $$d=0\text{?}$$
• $$c = 1$$ and $$d=1\text{?}$$
• $$c = 0$$ and $$d=-1\text{?}$$
2. Can the vector $$\twovec{2}{4}$$ be expressed as a linear combination of $$\vvec$$ and $$\wvec\text{?}$$ Is the vector $$\twovec{2}{4}$$ in the span of $$\vvec$$ and $$\wvec\text{?}$$
3. Can the vector $$\twovec{3}{0}$$ be expressed as a linear combination of $$\vvec$$ and $$\wvec\text{?}$$ Is the vector $$\twovec{3}{0}$$ in the span of $$\vvec$$ and $$\wvec\text{?}$$
4. Describe the set of vectors in the span of $$\vvec$$ and $$\wvec\text{.}$$
5. For what vectors $$\bvec$$ does the equation
\begin{equation*} \left[\begin{array}{rr} 1 \amp -2 \\ 2 \amp -4 \end{array}\right] \xvec = \bvec \end{equation*}
have a solution?
2. We will now look at an example where
\begin{equation*} \vvec = \twovec{2}{1}, ~~~\wvec = \twovec{1}{2}\text{.} \end{equation*}
1. What vector is the linear combination of $$\vvec$$ and $$\wvec$$ with weights:
• $$c = 2$$ and $$d=0\text{?}$$
• $$c = 1$$ and $$d=1\text{?}$$
• $$c = 0$$ and $$d=-1\text{?}$$
2. Can the vector $$\twovec{-2}{2}$$ be expressed as a linear combination of $$\vvec$$ and $$\wvec\text{?}$$ Is the vector $$\twovec{-2}{2}$$ in the span of $$\vvec$$ and $$\wvec\text{?}$$
3. Can the vector $$\twovec{3}{0}$$ be expressed as a linear combination of $$\vvec$$ and $$\wvec\text{?}$$ Is the vector $$\twovec{3}{0}$$ in the span of $$\vvec$$ and $$\wvec\text{?}$$
4. Describe the set of vectors in the span of $$\vvec$$ and $$\wvec\text{.}$$
5. For what vectors $$\bvec$$ does the equation
\begin{equation*} \left[\begin{array}{rr} 2 \amp 1 \\ 1 \amp 2 \end{array}\right] \xvec = \bvec \end{equation*}
have a solution?
This activity aims to convey the geometric meaning of span. Remember that we can think of a linear combination of the two vectors $$\vvec$$ and $$\wvec$$ as a recipe for walking in the plane $$\real^2\text{.}$$ We first move a prescribed amount in the direction of $$\vvec$$ and then a prescribed amount in the direction of $$\wvec\text{.}$$ The span consists of all the places we can walk to.
#### Example2.3.5.
Let’s consider the vectors $$\vvec=\twovec20$$ and $$\wvec=\twovec{-1}1$$ as shown in Figure 2.3.6.
The figure shows us that $$\bvec = \vvec + 2\wvec = \twovec02$$ is a linear combination of $$\vvec$$ and $$\wvec\text{.}$$ Indeed, we can verify this algebraically by constructing the linear system
\begin{equation*} \begin{bmatrix}\vvec \amp \wvec \end{bmatrix} ~ \xvec = \twovec02, \end{equation*}
whose corresponding augmented matrix has the reduced row echelon form
\begin{equation*} \left[ \begin{array}{rr|r} 2 \amp -1 \amp 0 \\ 0 \amp 1 \amp 2 \\ \end{array} \right] \sim \left[ \begin{array}{rr|r} 1 \amp 0 \amp 1 \\ 0 \amp 1 \amp 2 \\ \end{array} \right]. \end{equation*}
Because this system is consistent, we know that $$\bvec=\twovec02$$ is in the span of $$\vvec$$ and $$\wvec\text{.}$$
In fact, we can say more. Notice that the coefficient matrix
\begin{equation*} \begin{bmatrix} 2 \amp -1 \\ 0 \amp 1 \\ \end{bmatrix} \sim \begin{bmatrix} 1 \amp 0 \\ 0 \amp 1 \\ \end{bmatrix} \end{equation*}
has a pivot position in every row. This means that for any other vector $$\bvec\text{,}$$ the augmented matrix corresponding to the equation $$\begin{bmatrix}\vvec \amp \wvec \end{bmatrix} ~\xvec = \bvec$$ cannot have a pivot position in its rightmost column:
\begin{equation*} \left[ \begin{array}{rr|r} 2 \amp -1 \amp * \\ 0 \amp 1 \amp * \\ \end{array} \right] \sim \left[ \begin{array}{rr|r} 1 \amp 0 \amp * \\ 0 \amp 1 \amp * \\ \end{array} \right]. \end{equation*}
Therefore, the equation $$\begin{bmatrix}\vvec \amp \wvec \end{bmatrix} ~\xvec = \bvec$$ is consistent for every two-dimensional vector $$\bvec\text{,}$$ which tells us that every two-dimensional vector is in the span of $$\vvec$$ and $$\wvec\text{.}$$ In this case, we say that the span of $$\vvec$$ and $$\wvec$$ is $$\real^2\text{.}$$
The intuitive meaning is that we can walk to any point in the plane by moving an appropriate distance in the $$\vvec$$ and $$\wvec$$ directions.
#### Example2.3.7.
Now let’s consider the vectors $$\vvec=\twovec{-1}1$$ and $$\wvec=\twovec2{-2}$$ as shown in Figure 2.3.8.
From the figure, we expect that $$\bvec = \twovec02$$ is not a linear combination of $$\vvec$$ and $$\wvec\text{.}$$ Once again, we can verify this algebraically by constructing the linear system
\begin{equation*} \begin{bmatrix}\vvec \amp \wvec \end{bmatrix} ~ \xvec = \twovec02. \end{equation*}
The augmented matrix has the reduced row echelon form
\begin{equation*} \left[ \begin{array}{rr|r} -1 \amp 2 \amp 0 \\ 1 \amp -2 \amp 2 \\ \end{array} \right] \sim \left[ \begin{array}{rr|r} 1 \amp -2 \amp 0 \\ 0 \amp 0 \amp 1 \\ \end{array} \right], \end{equation*}
from which we see that the system is inconsistent. Therefore, $$\bvec=\twovec02$$ is not in the span of $$\vvec$$ and $$\wvec\text{.}$$
We should expect this behavior from the coefficient matrix
\begin{equation*} \begin{bmatrix} -1 \amp 2 \\ 1 \amp -2 \\ \end{bmatrix} \sim \begin{bmatrix} 1 \amp -2 \\ 0 \amp 0 \\ \end{bmatrix}. \end{equation*}
Because the second row of the coefficient matrix does not have a pivot position, it is possible for a linear system $$\begin{bmatrix}\vvec \amp \wvec \end{bmatrix} ~\xvec = \bvec$$ to have a pivot position in its rightmost column:
\begin{equation*} \left[ \begin{array}{rr|r} -1 \amp 2 \amp * \\ 1 \amp -2 \amp * \\ \end{array} \right] \sim \left[ \begin{array}{rr|r} 1 \amp -2 \amp 0 \\ 0 \amp 0 \amp 1 \\ \end{array} \right]. \end{equation*}
If we notice that $$\wvec = -2\vvec\text{,}$$ we see that any linear combination of $$\vvec$$ and $$\wvec\text{,}$$
\begin{equation*} c\vvec + d\wvec = c\vvec -2d\vvec = (c-2d)\vvec, \end{equation*}
is actually a scalar multiple of $$\vvec\text{.}$$ Therefore, the span of $$\vvec$$ and $$\wvec$$ is the line defined by the vector $$\vvec\text{.}$$ Intuitively, this means that we can only walk to points on this line using these two vectors.
#### Notation2.3.9.
We will denote the span of the set of vectors $$\vvec_1,\vvec_2,\ldots,\vvec_n$$ by $$\laspan{\vvec_1,\vvec_2,\ldots,\vvec_n}\text{.}$$
In Example 2.3.5, we saw that $$\laspan{\vvec,\wvec} = \real^2\text{.}$$ However, for the vectors in Example 2.3.7, we saw that $$\laspan{\vvec,\wvec}$$ is simply a line.
### Subsection2.3.2Pivot positions and span
A set of vectors $$\vvec_1,\vvec_2,\ldots,\vvec_n$$ naturally defines a matrix $$A = \begin{bmatrix}\vvec_1\amp\vvec_2\amp\cdots\vvec_n\end{bmatrix}$$ whose columns are the given vectors. As we’ve seen, a vector $$\bvec$$ is in $$\laspan{\vvec_1,\vvec_2,\ldots,\vvec_n}$$ precisely when the linear system $$A\xvec=\bvec$$ is consistent.
The previous examples point to the fact that the span is related to the pivot positions of $$A\text{.}$$ While Section 2.4 and Section 3.5 develop this idea more fully, we will now examine the possibilities in $$\real^3\text{.}$$
#### Activity2.3.3.
In this activity, we will look at the span of sets of vectors in $$\real^3\text{.}$$
1. Suppose $$\vvec=\threevec{1}{2}{1}\text{.}$$ Give a geometric description of $$\laspan{\vvec}$$ and a rough sketch of $$\vvec$$ and its span in Figure 2.3.10.
2. Now consider the two vectors
\begin{equation*} \evec_1 = \threevec{1}{0}{0},~~~ \evec_2 = \threevec{0}{1}{0}\text{.} \end{equation*}
Sketch the vectors below. Then give a geometric description of $$\laspan{\evec_1,\evec_2}$$ and a rough sketch of the span in Figure 2.3.11.
3. Let’s now look at this situation algebraically by writing write $$\bvec = \threevec{b_1}{b_2}{b_3}\text{.}$$ Determine the conditions on $$b_1\text{,}$$ $$b_2\text{,}$$ and $$b_3$$ so that $$\bvec$$ is in $$\laspan{\evec_1,\evec_2}$$ by considering the linear system
\begin{equation*} \left[\begin{array}{rr} \evec_1 \amp \evec_2 \\ \end{array}\right] ~\xvec = \bvec \end{equation*}
or
\begin{equation*} \left[\begin{array}{rr} 1 \amp 0 \\ 0 \amp 1 \\ 0 \amp 0 \\ \end{array}\right] \xvec = \threevec{b_1}{b_2}{b_3}\text{.} \end{equation*}
Explain how this relates to your sketch of $$\laspan{\evec_1,\evec_2}\text{.}$$
4. Consider the vectors
\begin{equation*} \vvec_1 = \threevec{1}{1}{-1},~~ \vvec_2 = \threevec{0}{2}{1}. \end{equation*}
1. Is the vector $$\bvec=\threevec{1}{-2}{4}$$ in $$\laspan{\vvec_1,\vvec_2}\text{?}$$
2. Is the vector $$\bvec=\threevec{-2}{0}{3}$$ in $$\laspan{\vvec_1,\vvec_2}\text{?}$$
3. Give a geometric description of $$\laspan{\vvec_1,\vvec_2}\text{.}$$
5. Consider the vectors
\begin{equation*} \vvec_1 = \threevec{1}{1}{-1}, \vvec_2 = \threevec{0}{2}{1}, \vvec_3 = \threevec{1}{-2}{4}\text{.} \end{equation*}
Form the matrix $$\left[\begin{array}{rrrr} \vvec_1 \amp \vvec_2 \amp \vvec_3 \end{array}\right]$$ and find its reduced row echelon form.
What does this tell you about $$\laspan{\vvec_1,\vvec_2,\vvec_3}\text{?}$$
6. If the span of a set of vectors $$\vvec_1,\vvec_2,\ldots,\vvec_n$$ is $$\real^3\text{,}$$ what can you say about the pivot positions of the matrix $$\left[\begin{array}{rrrr} \vvec_1\amp\vvec_2\amp\ldots\amp\vvec_n \end{array}\right]\text{?}$$
7. What is the smallest number of vectors such that $$\laspan{\vvec_1,\vvec_2,\ldots,\vvec_n} = \real^3\text{?}$$
The types of sets that appear as the span of a set of vectors in $$\real^3$$ are relatively simple.
• First, with a single nonzero vector, all linear combinations are simply scalar multiples of that vector so that the span of this vector is a line, as shown in Figure 2.3.12.
Notice that the matrix formed by this vector has one pivot position. For example,
\begin{equation*} \threevec{-2}{3}{1} \sim \threevec{1}{0}{0}\text{.} \end{equation*}
• The span of two vectors in $$\real^3$$ that do not lie on the same line will be a plane, as seen in Figure 2.3.13.
For example, the vectors
\begin{equation*} \vvec_1=\threevec{-2}{3}{1},~~~ \vvec_2=\threevec{1}{-1}{3} \end{equation*}
\begin{equation*} \left[\begin{array}{rr} -2 \amp 1 \\ 3 \amp -1 \\ 1 \amp 3 \\ \end{array}\right] \sim \left[\begin{array}{rr} 1 \amp 0 \\ 0 \amp 1 \\ 0 \amp 0 \\ \end{array}\right] \end{equation*}
with two pivot positions.
• Finally, a set of three vectors, such as
\begin{equation*} \vvec_1=\threevec12{-1},~~~ \vvec_2=\threevec201,~~~ \vvec_3=\threevec{-2}20 \end{equation*}
may form a matrix having three pivot positions
\begin{equation*} \left[\begin{array}{rrr} \vvec_1 \amp \vvec_2 \amp \vvec_3 \end{array}\right] = \left[\begin{array}{rrr} 1 \amp 2 \amp -2 \\ 2 \amp 0 \amp 2 \\ -1 \amp 1 \amp 0 \\ \end{array}\right] \sim \left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \\ \end{array}\right], \end{equation*}
one in every row. When this happens, no matter how we augment this matrix, it is impossible to obtain a pivot position in the rightmost column:
\begin{equation*} \left[\begin{array}{rrr|r} 1 \amp 2 \amp -2 \amp *\\ 2 \amp 0 \amp 2 \amp * \\ -1 \amp 1 \amp 0 \amp * \\ \end{array}\right] \sim \left[\begin{array}{rrr|r} 1 \amp 0 \amp 0 \amp *\\ 0 \amp 1 \amp 0 \amp * \\ 0 \amp 0 \amp 1 \amp * \\ \end{array}\right]. \end{equation*}
Therefore, any linear system $$\begin{bmatrix}\vvec_1\amp\vvec_2\amp\vvec_3\end{bmatrix} ~\xvec = \bvec$$ is consistent, which tells us that $$\laspan{\vvec_1,\vvec_2,\vvec_3} = \real^3\text{.}$$
To summarize, we looked at the pivot positions in a matrix whose columns are the three-dimensional vectors $$\vvec_1,\vvec_2,\ldots,\vvec_n\text{.}$$ We found that with
• one pivot position, the span was a line.
• two pivot positions, the span was a plane.
• three pivot positions, the span was $$\real^3\text{.}$$
Though we will return to these ideas later, for now take note of the fact that the span of a set of vectors in $$\real^3$$ is a relatively simple, familiar geometric object.
The reasoning that led us to conclude that the span of a set of vectors is $$\real^3$$ when the associated matrix has a pivot position in every row applies more generally.
This tells us something important about the number of vectors needed to span $$\real^m\text{.}$$ Suppose we have $$n$$ vectors $$\vvec_1,\vvec_2,\ldots,\vvec_n$$ that span $$\real^m\text{.}$$ The proposition tells us that the matrix $$A = \left[\begin{array}{rrrr} \vvec_1\amp\vvec_2\amp\ldots\amp\vvec_n \end{array}\right]$$ has a pivot position in every row, such as in this reduced row echelon matrix.
\begin{equation*} \left[\begin{array}{rrrrrr} 1 \amp 0 \amp * \amp 0 \amp * \amp 0 \\ 0 \amp 1 \amp * \amp 0 \amp * \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \amp * \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 1 \\ \end{array}\right]. \end{equation*}
Since a matrix can have at most one pivot position in a column, there must be at least as many columns as there are rows, which implies that $$n\geq m\text{.}$$ For instance, if we have a set of vectors that span $$\real^{632}\text{,}$$ there must be at least 632 vectors in the set.
We have thought about a linear combination of a set of vectors $$\vvec_1,\vvec_2,\ldots,\vvec_n$$ as the result of walking a certain distance in the direction of $$\vvec_1\text{,}$$ followed by walking a certain distance in the direction of $$\vvec_2\text{,}$$ and so on. If $$\laspan{\vvec_1,\vvec_2,\ldots,\vvec_n} = \real^m\text{,}$$ this means that we can walk to every point in $$\real^m$$ using the directions $$\vvec_1,\vvec_2,\ldots,\vvec_n\text{.}$$ Intuitively, this proposition is telling us that we need at least $$m$$ directions to have the flexibility needed to reach every point in $$\real^m\text{.}$$
#### Terminology.
Because span is a concept that is connected to a set of vectors, we say, “The span of the set of vectors $$\vvec_1, \vvec_2, \ldots, \vvec_n$$ is ....” While it may be tempting to say, “The span of the matrix $$A$$ is ...,” we should instead say “The span of the columns of the matrix $$A$$ is ....”
### Subsection2.3.3Summary
We defined the span of a set of vectors and developed some intuition for this concept through a series of examples.
• The span of a set of vectors $$\vvec_1,\vvec_2,\ldots,\vvec_n$$ is the set of linear combinations of the vectors. We denote the span by $$\laspan{\vvec_1,\vvec_2,\ldots,\vvec_n}\text{.}$$
• A vector $$\bvec$$ is in $$\laspan{\vvec_1,\vvec_2,\ldots,\vvec_n}$$ if and only if the linear system
\begin{equation*} \left[\begin{array}{rrrr} \vvec_1\amp\vvec_2\amp\ldots\vvec_n \end{array}\right] ~\xvec = \bvec \end{equation*}
is consistent.
• If the $$m\times n$$ matrix
\begin{equation*} \left[\begin{array}{rrrr} \vvec_1\amp\vvec_2\amp\ldots\vvec_n \end{array}\right] \end{equation*}
has a pivot position in every row, then the span of these vectors is $$\real^m\text{;}$$ that is,
\begin{equation*} \laspan{\vvec_1,\vvec_2,\ldots,\vvec_n} = \real^m. \end{equation*}
• Any set of vectors that spans $$\real^m$$ must have at least $$m$$ vectors.
### Exercises2.3.4Exercises
#### 1.
In this exercise, we will consider the span of some sets of two- and three-dimensional vectors.
1. Consider the vectors
\begin{equation*} \vvec_1 = \twovec{1}{-2}, \vvec_2 = \twovec{4}{3}\text{.} \end{equation*}
1. Is $$\bvec = \twovec{2}{1}$$ in $$\laspan{\vvec_1,\vvec_2}\text{?}$$
2. Give a geometric description of $$\laspan{\vvec_1,\vvec_2}\text{.}$$
2. Consider the vectors
\begin{equation*} \vvec_1=\threevec{2}{1}{3}, \vvec_2=\threevec{-2}{0}{2}, \vvec_3=\threevec{6}{1}{-1}\text{.} \end{equation*}
1. Is the vector $$\bvec=\threevec{-10}{-1}{5}$$ in $$\laspan{\vvec_1,\vvec_2,\vvec_3}\text{?}$$
2. Is the vector $$\vvec_3$$ in $$\laspan{\vvec_1,\vvec_2,\vvec_3}\text{?}$$
3. Is the vector $$\bvec=\threevec{3}{3}{-1}$$ in $$\laspan{\vvec_1,\vvec_2,\vvec_3}\text{?}$$
4. Give a geometric description of $$\laspan{\vvec_1,\vvec_2,\vvec_3}\text{.}$$
#### 2.
Provide a justification for your response to the following questions.
1. Suppose you have a set of vectors $$\vvec_1,\vvec_2,\ldots,\vvec_n\text{.}$$ Can you guarantee that $$\zerovec$$ is in $$\laspan{\vvec_1\,\vvec_2,\ldots,\vvec_n}\text{?}$$
2. Suppose that $$A$$ is an $$m \times n$$ matrix. Can you guarantee that the equation $$A\xvec = \zerovec$$ is consistent?
3. What is $$\laspan{\zerovec,\zerovec,\ldots,\zerovec}\text{?}$$
#### 3.
For both parts of this exercise, give a geometric description of sets of the vectors $$\bvec$$ and include a sketch.
1. For which vectors $$\bvec$$ in $$\real^2$$ is the equation
\begin{equation*} \left[\begin{array}{rr} 3 \amp -6 \\ -2 \amp 4 \\ \end{array}\right] \xvec = \bvec \end{equation*}
consistent?
2. For which vectors $$\bvec$$ in $$\real^2$$ is the equation
\begin{equation*} \left[\begin{array}{rr} 3 \amp -6 \\ -2 \amp 2 \\ \end{array}\right] \xvec = \bvec \end{equation*}
consistent?
#### 4.
Consider the following matrices:
\begin{equation*} A = \left[\begin{array}{rrrr} 3 \amp 0 \amp -1 \amp 1 \\ 1 \amp -1 \amp 3 \amp 7 \\ 3 \amp -2 \amp 1 \amp 5 \\ -1 \amp 2 \amp 2 \amp 3 \\ \end{array}\right],~~~ B = \left[\begin{array}{rrrr} 3 \amp 0 \amp -1 \amp 4 \\ 1 \amp -1 \amp 3 \amp -1 \\ 3 \amp -2 \amp 1 \amp 3 \\ -1 \amp 2 \amp 2 \amp 1 \\ \end{array}\right]\text{.} \end{equation*}
Do the columns of $$A$$ span $$\real^4\text{?}$$ Do the columns of $$B$$ span $$\real^4\text{?}$$
#### 5.
Determine whether the following statements are true or false and provide a justification for your response. Throughout, we will assume that the matrix $$A$$ has columns $$\vvec_1,\vvec_2,\ldots,\vvec_n\text{;}$$ that is,
\begin{equation*} A = \left[\begin{array}{rrrr} \vvec_1\amp\vvec_2\amp\ldots\amp\vvec_n \end{array}\right]\text{.} \end{equation*}
1. If the equation $$A\xvec = \bvec$$ is consistent, then $$\bvec$$ is in $$\laspan{\vvec_1,\vvec_2,\ldots,\vvec_n}\text{.}$$
2. The equation $$A\xvec = \vvec_1$$ is consistent.
3. If $$\vvec_1\text{,}$$ $$\vvec_2\text{,}$$ $$\vvec_3\text{,}$$ and $$\vvec_4$$ are vectors in $$\real^3\text{,}$$ then their span is $$\real^3\text{.}$$
4. If $$\bvec$$ is a linear combination of $$\vvec_1, \vvec_2,\ldots,\vvec_n\text{,}$$ then $$\bvec$$ is in $$\laspan{\vvec_1,\vvec_2,\ldots,\vvec_n}\text{.}$$
5. If $$A$$ is an $$8032\times 427$$ matrix, then the span of the columns of $$A$$ is a set of vectors in $$\real^{427}\text{.}$$
#### 6.
This exercise asks you to construct some matrices whose columns span a given set.
1. Construct a $$3\times3$$ matrix whose columns span $$\real^3\text{.}$$
2. Construct a $$3\times3$$ matrix whose columns span a plane in $$\real^3\text{.}$$
3. Construct a $$3\times3$$ matrix whose columns span a line in $$\real^3\text{.}$$
#### 7.
Provide a justification for your response to the following questions.
1. Suppose that we have vectors in $$\real^8\text{,}$$ $$\vvec_1,\vvec_2,\ldots,\vvec_{10}\text{,}$$ whose span is $$\real^8\text{.}$$ Can every vector $$\bvec$$ in $$\real^8$$ be written as a linear combination of $$\vvec_1,\vvec_2,\ldots,\vvec_{10}\text{?}$$
2. Suppose that we have vectors in $$\real^8\text{,}$$ $$\vvec_1,\vvec_2,\ldots,\vvec_{10}\text{,}$$ whose span is $$\real^8\text{.}$$ Can every vector $$\bvec$$ in $$\real^8$$ be written uniquely as a linear combination of $$\vvec_1,\vvec_2,\ldots,\vvec_{10}\text{?}$$
3. Do the vectors
\begin{equation*} \evec_1=\threevec{1}{0}{0}, \evec_2=\threevec{0}{1}{0}, \evec_3=\threevec{0}{0}{1} \end{equation*}
span $$\real^3\text{?}$$
4. Suppose that $$\vvec_1,\vvec_2,\ldots,\vvec_n$$ span $$\real^{438}\text{.}$$ What can you guarantee about the value of $$n\text{?}$$
5. Can 17 vectors in $$\real^{20}$$ span $$\real^{20}\text{?}$$
#### 8.
The following observation will be helpful in this exercise. If we want to find a solution to the equation $$AB\xvec = \bvec\text{,}$$ we could first find a solution to the equation $$A\yvec = \bvec$$ and then find a solution to the equation $$B\xvec = \yvec\text{.}$$
Suppose that $$A$$ is a $$3\times 4$$ matrix whose columns span $$\real^3$$ and $$B$$ is a $$4\times 5$$ matrix. In this case, we can form the product $$AB\text{.}$$
1. What is the shape of the product $$AB\text{?}$$
2. Can you guarantee that the columns of $$AB$$ span $$\real^3\text{?}$$
3. If you know additionally that the span of the columns of $$B$$ is $$\real^4\text{,}$$ can you guarantee that the columns of $$AB$$ span $$\real^3\text{?}$$
#### 9.
Suppose that $$A$$ is a $$12\times12$$ matrix and that, for some vector $$\bvec\text{,}$$ the equation $$A\xvec=\bvec$$ has a unique solution.
1. What can you say about the pivot positions of $$A\text{?}$$
2. What can you say about the span of the columns of $$A\text{?}$$
3. If $$\cvec$$ is some other vector in $$\real^{12}\text{,}$$ what can you conclude about the equation $$A\xvec = \cvec\text{?}$$
4. What can you about the solution space to the equation $$A\xvec =\zerovec\text{?}$$
#### 10.
Suppose that
\begin{equation*} \vvec_1 = \fourvec{3}{1}{3}{-1}, \vvec_2 = \fourvec{0}{-1}{-2}{2}, \vvec_3 = \fourvec{-3}{-3}{-7}{5}\text{.} \end{equation*}
1. Is $$\vvec_3$$ a linear combination of $$\vvec_1$$ and $$\vvec_2\text{?}$$ If so, find weights such that $$\vvec_3 = a\vvec_1+b\vvec_2\text{.}$$
2. Show that a linear combination
\begin{equation*} a\vvec_1 + b\vvec_2 + c\vvec_3 \end{equation*}
can be rewritten as a linear combination of $$\vvec_1$$ and $$\vvec_2\text{.}$$
3. Explain why $$\laspan{\vvec_1,\vvec_2,\vvec_3} = \laspan{\vvec_1,\vvec_2}\text{.}$$
#### 11.
As defined in this section, the span of a set of vectors is generated by taking all possible linear combinations of those vectors. This exercise will demonstrate the fact that the span can also be realized as the solution space to a linear system.
We will consider the vectors
\begin{equation*} \vvec_1=\threevec{1}{0}{-2}, \vvec_2=\threevec{2}{1}{0}, \vvec_3=\threevec{1}{1}{2} \end{equation*}
1. Is every vector in $$\real^3$$ in $$\laspan{\vvec_1,\vvec_2,\vvec_3}\text{?}$$ If not, describe the span.
2. To describe $$\laspan{\vvec_1,\vvec_2,\vvec_3}$$ as the solution space of a linear system, we will write
\begin{equation*} \bvec=\threevec{a}{b}{c}\text{.} \end{equation*}
If $$\bvec$$ is in $$\laspan{\vvec_1,\vvec_2,\vvec_3}\text{,}$$ then the linear system corresponding to the augmented matrix
\begin{equation*} \left[\begin{array}{rrr|r} 1 \amp 2 \amp 1 \amp a \\ 0 \amp 1 \amp 1 \amp b \\ -2\amp 0 \amp 2 \amp c \\ \end{array}\right] \end{equation*}
must be consistent. This means that a pivot cannot occur in the rightmost column. Perform row operations to put this augmented matrix into a triangular form. Now identify an equation in $$a\text{,}$$ $$b\text{,}$$ and $$c$$ that tells us when there is no pivot in the rightmost column. The solution space to this equation describes $$\laspan{\vvec_1,\vvec_2,\vvec_3}\text{.}$$
3. In this example, the matrix formed by the vectors $$\left[\begin{array}{rrr} \vvec_1\amp\vvec_2\amp\vvec_2 \\ \end{array}\right]$$ has two pivot positions. Suppose we were to consider another example in which this matrix had had only one pivot position. How would this have changed the linear system describing $$\laspan{\vvec_1,\vvec_2,\vvec_3}\text{?}$$ |
# Question Video: Factorizing the Sum of Two Cubes Mathematics
The expression π₯Β³ + 27 has two factors. One factor is π₯ + 3. What is the other factor?
02:06
### Video Transcript
The expression π₯ cubed plus 27 has two factors. One factor is π₯ plus three. What is the other factor?
π₯ cubed plus 27 is something called a sum of cubes, where π cubed plus π cubed is equal to π plus π times π squared minus ππ plus π squared. So our π₯ cubed plus 27, we know that one of the factors is π₯ plus three. And that matches with the π plus π. So letβs go back to our original expression and make sure that π should be equal to π₯ and π should be equal to three.
So if π cubed is equal to π₯ cubed and π cubed is equal to 27, we can solve for π and π by taking the cube root of both sides of the equation, leaving π to be equal to π₯, because the cube root of π₯ cubed is π₯. And now to solve for π, we take the cube root of both sides of the equation. And we find that π is equal to three, because the cube root of 27 is three. And this is because 27 is equal to nine times three. And nine is equal to three times three. So three threes is equal to three cubed. So the cube root of three cubed is three.
So just as we said, π is equal to π₯ and π is equal to three. So to find the other factor, we simply need to plug π and π into this expression. π squared will be equal to π₯ squared minus π times π, so minus π₯ times three, plus π squared.
So letβs go ahead and simplify this. And we get that itβs equal to π₯ squared minus three π₯ plus nine. This is what the other factor would be. So hereβs what we started with. Here was one of the factors. And we solved for the missing factor. So once again, the missing factor is equal to π₯ squared minus three π₯ plus nine.
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# EXPONENTIAL GROWTH AND DECAY WORD PROBLEMS
## About "Exponential growth and decay word problems"
Exponential growth and decay word problems :
To solve exponential growth and decay word problems, we have to be aware of exponential growth and decay functions.
Let us consider the following two examples.
When we invest some money in a bank, it grows year by year, because of the interest paid by the bank.
We buy a car and use it for some years. When it becomes too old, we would like to sell it.
In the first example, we will be keen to know the final value (Amount invested + Interest) of our deposit. To know the final value of the deposit, we have to use growth function.
In the second example, we will be eager to know the sale value of the car (Purchased price - depreciation). Here we have to use decay function.
In this way, growth and decay functions are being used in our life.
## Exponential growth and Decay functions - Formulas
Many real world phenomena are being modeled by functions which describe how things grow or decay as time passes.
Let us see the functions which use to estimate and growth and decay.
Formula 1 :
The formula given below is related to compound interest formula and represents the case where interest is being compounded continuously.
That is, at any instant the balance is changing at a rate that equals "r" times the current balance.
We use this formula, when it is given "exponential growth/or decay"
A ---> Ending amount
---> Beginning amount
---> Growth/Decay rate
---> Time
Note :
If it is decay function, the value of "r" will be negative.
Formula 2 :
The formula given below is compound interest formula and represents the case where interest is being compounded annually or the growth is being compounded once the term is completed.
A ---> Ending amount
---> Beginning amount
---> Growth/Decay rate
---> No. of years / Time
Note :
If it is decay function, the value of "r" will be negative.
Formula 3 :
The formula given below is related to geometric progression. Here, the initial amount will grow/decay at the constant ratio "b".
A ---> Ending amount
---> Beginning amount
---> Growth/Decay ratio
---> No. of years / terms
Note :
If it is growth function, we will have "r" > 1
If it is decay function, we will have 0 < r < 1
## Exponential growth and decay word problems - Examples
To have better understanding on "Exponential growth and decay word problems", let us look at some examples.
Example 1 :
David owns a chain of fast food restaurants that operated 200 stores in 1999. If the rate of increase is 8% annually, how many stores does the restaurant operate in 2007 ?
Solution :
Number of years between 1999 and 2007 is
n = 2007 - 1999 = 8
No. of stores in the year 2007 = P(1+r)
Here, P = 200, r = 8% or 0.08, n = 8
No. of stores in the year 2007 = 200(1+0.08)
No. of stores in the year 2007 = 200(1.08)
No. of stores in the year 2007 = 200(1+0.08)
No. of stores in the year 2007 = 200(1.8509)
No. of stores in the year 2007 = 370.18
Hence, the number of stores in the year 2007 is 370 (approximately)
Example 2 :
You invest \$2500 in bank which pays 10% interest per year compounded continuously. What will be the value of the investment after 10 years ?
Solution :
We have to use the formula given below to know the value of the investment after 3 years.
Here,
A = Final value of the deposit
P = 2500, r = 10% or 0.1, t = 10, e = 2.71828 and also
rt = 0.1x10 = 1
A = 2500(2.71828)¹
A = 6795.70
Hence, the value of the investment after 10 years is \$6795.70
Example 3 :
Suppose a radio active substance decays at a rate of 3.5% per hour. What percent of substance will be left after 6 hours ?
Solution :
Since the initial amount of substance is not given and the problem is based on percentage, we have to assume that the initial amount of substance is 100.
We have to use the formula given below to find the percent of substance after 6 hours.
Here,
A = Amount of substance after 6 hours.
P = 100, r = -3.5% or -0.035, t = 6
(Here, the value of "r" is taken in negative sign. because the substance decays)
A = 100(1-0.035)
A = 100(0.935)
A = 100(0.8075)
A = 80.75
Since the initial amount of substance is assumed as 100, the percent of substance left after 6 hours is 80.75%
Example 4 :
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture initially, how many bacteria will be present at the end of 8th hour?
Solution :
Note that the number of bacteria present in the culture doubles at the end of successive hours.
Since it grows at the constant ratio "2", the growth is based is on geometric progression.
We have to use the formula given below to find the no. of bacteria present at the end of 8th hour.
Here,
A = No. of bacteria at the end of 8th hour
a = 30, b = 2 and x = 8
A = 30x2
A = 30x256
A = 7680
Hence, the number of bacteria at the end of 8th hour is 7680.
Example 5 :
A sum of money placed at compound interest doubles itself in 3 years. If interest is being compounded annually, in how many years will it amount to four times itself ?
Solution :
Let "P" be the amount invested initially.
From the given information, P becomes 2P in 3 years.
Since the investment is in compound interest, for the 4th year, the principal will be 2P.
And 2P becomes 4P (it doubles itself) in the next 3 years.
Therefore, at the end of 6 years accumulated value will be 4P.
Hence, the amount deposited will amount to 4 times itself in 6 years.
After having gone through the stuff given above, we hope that the students would have understood "Exponential growth and decay word problems". |
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