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# Illustrative Mathematics Unit 6.1, Lesson 10: Bases and Heights of Triangles
Related Topics:
Math Worksheets
Learn about how to identify corresponding bases and heights in any triangle, and use that information to find the area of a triangle. After trying the questions, click on the buttons to view answers and explanations in text or video.
Bases and Heights of Triangles
Let’s use different base-height pairs to find the area of a triangle.
Illustrative Math Unit 6.1, Lesson 10 (printable worksheets)
10.1 - An Area of 12
Open the applet. Draw one triangle with an area of 12 square units. Try to draw a non-right triangle. Be prepared to explain how you know the area of your triangle is 12 square units.
• Hints
There are a number of strategies which you can use.
Can you draw a quadrilateral with an area of 12?
The formula for the area of a triangle is A = ½ · b · h. What pairs of values for b and h will get A = 12?
The formula for the area of a triangle is A = ½ · b · h.
The area of this triangle is ½ × 6 units × 4 units = 12 square units.
Other methods of demonstrating the area of the triangle are possible, as are other triangles with an area of 12 square units.
10.2 - Hunting for Heights
1. Here are three copies of the same triangle. The triangle is rotated so that the side chosen as the base is at the bottom and is horizontal. Draw a height that corresponds to each base.
Side a as the base:
Side b as the base:
Side c as the base:
• Hints
Recall that height is the length of a perpendicular segment from the base to the vertex opposite of it. The opposite vertex is the vertex that is not an endpoint of the base.
A segment showing a height must be drawn at a right angle to the base, but it can be drawn in more than one place. It does not have to go through the opposite vertex, as long as it connects the base and a line that is parallel to the base and goes through the opposite vertex.
Black dashed lines are the heights corresponding to the chosen base in each of these triangles.
Notice the blue dashed lines. The height must be perpendicular to the base and measure the length between the base and the opposite vertex, but does not need to pass through the base or the opposite vertex itself.
This is similar to how the height of a parallelogram can be drawn.
1. Draw a line segment to show the height for the chosen base in each triangle.
10.3 - Some Bases Are Better Than Others
For each triangle, identify and label a base and height. If needed, draw a line segment to show the height.
Then, find the area of the triangle. Show your reasoning. (The side length of each square on the grid is 1 unit.)
• Hints
To find the area of these triangles, choose a base that will let you count the length of the base and the height.
If you are having difficulty drawing the height, remember that the height does not need to pass through the base or the opposite vertex itself.
A: ½ × 9 units × 5 units = 22.5 square units
B: ½ × 11 units × 8 units = 44 square units
C: ½ × 4 units × 18 units = 36 square units
D: ½ × 6 units × 11 units = 33 square units
Find the area of the triangle below. Show your reasoning.
• Hints
You can’t directly count the lengths of any of the sides. What other strategies for finding area do you know? Can this triangle be enclosed in another shape?
By enclosing this triangle in a square, there are now three unshaded triangles whose bases and heights can be counted.
The total area of the square is 12 units × 12 units = 144 square units.
Area of triangle A = ½ × 12 units × 7 units = 42 square units
Area of triangle B = ½ × 6 units × 12 units = 36 square units
Area of triangle C = ½ × 6 units × 5 units = 15 square units
The area of the shaded triangle is 144 - (42 + 36 + 15) = 51 square units.
Lesson 10 Summary
A height of a triangle is a perpendicular segment between the side chosen as the base and the opposite vertex. We can use tools with right angles to help us draw height segments.
An index card (or any stiff paper with a right angle) is a handy tool for drawing a line that is perpendicular to another line.
1. Choose a side of a triangle as the base. Identify its opposite vertex.
2. Line up one edge of the index card with that base.
3. Slide the card along the base until a perpendicular edge of the card meets the opposite vertex.
4. Use the card edge to draw a line from the vertex to the base. That segment represents the height. Sometimes we may need to extend the line of the base to identify the height, such as when finding the height of an obtuse triangle, or whenever the opposite vertex is not directly over the base. In these cases, the height segment is typically drawn outside of the triangle.
Even though any side of a triangle can be a base, some base-height pairs can be more easily determined than others, so it helps to choose strategically.
For example, when dealing with a right triangle, it often makes sense to use the two sides that make the right angle as the base and the height because one side is already perpendicular to the other.
If a triangle is on a grid and has a horizontal or a vertical side, you can use that side as a base and use the grid to find the height, as in these examples:
Practice Problems
1. For each triangle, a base is labeled b. Draw a line segment that shows its corresponding height.
2. Select all triangles that have an area of 8 square units. Explain how you know.
Area of triangle A = ½ × 4 units × 4 units = 8 square units
Area of triangle B = ½ × 4 units × 4 units = 8 square units
Area of triangle C = ½ × 3 units × 5 units = 7.5 square units
Area of triangle D = ½ × 4 units × 4 units = 8 square units
Area of triangle E = ½ × 8 units × 2 units = 8 square units
A, B, D, and E all have an area of 8 square units. C does not.
1. Find the area of the triangle. Show your reasoning.
If you get stuck, carefully consider which side of the triangle to use as the base.
A = ½ · b · h
A = ½ × 6 units × 4 units
A = 12 units
Note how the base can be any side of the triangle, and that the most useful base is the side that allows you to measure the base and height.
1. Can side d be the base for this triangle? If so, which length would be the corresponding height? If not, explain why not.
Side d can be the base. Any side of a triangle can be the base.
The corresponding height is g, as it is perpendicular to d, and perpendicular to the line which is parallel to d and passes through the opposite vertex.
1. Find the area of this shape. Show your reasoning.
The shape can be enclosed in a rectangle as shown by the blue dashed lines. This rectangle has a total area of 6 units × 5 units = 30 square units.
The two unshaded triangles have a total area of 2 × ½ × 6 units × 2 units = 12 square units.
Hence, the shaded shape has an area of 30 square units - 12 square units = 18 square units.
6. On the grid, sketch two different parallelograms that have equal area. Label a base and height of each and explain how you know the areas are the same.
The area of parallelogram A is 4 units × 3 units = 12 square units.
The area of parallelogram B is 6 units × 2 units = 12 square units.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# What is 51/271 as a decimal?
## Solution and how to convert 51 / 271 into a decimal
51 / 271 = 0.188
51/271 or 0.188 can be represented in multiple ways (even as a percentage). The key is knowing when we should use each representation and how to easily transition between a fraction, decimal, or percentage. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. The difference between using a fraction or a decimal depends on the situation. Fractions can be used to represent parts of an object like 1/8 of a pizza while decimals represent a comparison of a whole number like \$0.25 USD. If we need to convert a fraction quickly, let's find out how and when we should.
## 51/271 is 51 divided by 271
The first step of teaching our students how to convert to and from decimals and fractions is understanding what the fraction is telling is. 51 is being divided into 271. Think of this as our directions and now we just need to be able to assemble the project! Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 51 divided by 271. We must divide 51 into 271 to find out how many whole parts it will have plus representing the remainder in decimal form. This is our equation:
### Numerator: 51
• Numerators are the portion of total parts, showed at the top of the fraction. Any value greater than fifty will be more difficult to covert to a decimal. The bad news is that it's an odd number which makes it harder to covert in your head. Large numerators make converting fractions more complex. Let's look at the fraction's denominator 271.
### Denominator: 271
• Denominators represent the total parts, located at the bottom of the fraction. Larger values over fifty like 271 makes conversion to decimals tougher. But the bad news is that odd numbers are tougher to simplify. Unfortunately and odd denominator is difficult to simplify unless it's divisible by 3, 5 or 7. Have no fear, large two-digit denominators are all bark no bite. Now it's time to learn how to convert 51/271 to a decimal.
## Converting 51/271 to 0.188
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 271 \enclose{longdiv}{ 51 }$$
Use long division to solve step one. This method allows us to solve for pieces of the equation rather than trying to do it all at once.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 271 \enclose{longdiv}{ 51.0 }$$
Uh oh. 271 cannot be divided into 51. Place a decimal point in your answer and add a zero. This doesn't add any issues to our denominator but now we can divide 271 into 510.
### Step 3: Solve for how many whole groups you can divide 271 into 510
$$\require{enclose} 00.1 \\ 271 \enclose{longdiv}{ 51.0 }$$
How many whole groups of 271 can you pull from 510? 271 Multiple this number by our furthest left number, 271, (remember, left-to-right long division) to get our first number to our conversion.
### Step 4: Subtract the remainder
$$\require{enclose} 00.1 \\ 271 \enclose{longdiv}{ 51.0 } \\ \underline{ 271 \phantom{00} } \\ 239 \phantom{0}$$
If there is no remainder, you’re done! If you have a remainder over 271, go back. Your solution will need a bit of adjustment. If you have a number less than 271, continue!
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
### Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. They each bring clarity to numbers and values of every day life. And the same is true for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But 51/271 and 0.188 bring clarity and value to numbers in every day life. Here are just a few ways we use 51/271, 0.188 or 18% in our daily world:
### When you should convert 51/271 into a decimal
Sports Stats - Fractions can be used here, but when comparing percentages, the clearest representation of success is from decimal points. Ex: A player's batting average: .333
### When to convert 0.188 to 51/271 as a fraction
Distance - Any type of travel, running, walking will leverage fractions. Distance is usually measured by the quarter mile and car travel is usually spoken the same.
### Practice Decimal Conversion with your Classroom
• If 51/271 = 0.188 what would it be as a percentage?
• What is 1 + 51/271 in decimal form?
• What is 1 - 51/271 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.188 + 1/2?
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# Use the sum-of-two-cubes or the difference-of-two-cubes pattern to factor each of the following. 1-27a^3
Use the sum-of-two-cubes or the difference-of-two-cubes pattern to factor each of the following.
$$\displaystyle{1}-{27}{a}^{{3}}$$
• Questions are typically answered in as fast as 30 minutes
### Plainmath recommends
• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.
Nathanael Webber
A polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables. Factoring polynomials involves breaking up a polynomial into simpler terms (the factors) such that when the terms are multiplied together they equal the original polynomial.
The given expression is $$\displaystyle{1}-{27}{a}^{{3}}$$. Factor the given expression by simplifying and then using sum of two cubes pattern as follows:
$$\displaystyle{1}-{27}{a}^{{3}}={1}^{{3}}-{3}^{{3}}{a}^{{3}}$$
$$\displaystyle{\left({1}\right)}^{{3}}+{\left(-{3}{a}\right)}^{{3}}$$
$$\displaystyle={\left({1}+{\left(-{3}{a}\right)}\right)}{\left({1}^{{2}}-{\left({1}\right)}{\left(-{3}{a}\right)}+{\left(-{3}{a}\right)}^{{2}}\right)}$$
Use sum of two cubes pattern $$\displaystyle{a}^{{3}}+{b}^{{3}}={\left({a}+{b}\right)}{\left({a}^{{2}}-{a}{b}+{b}^{{2}}\right)}$$
$$\displaystyle={\left({1}-{3}{a}\right)}{\left({1}+{3}{a}+{9}{a}^{{2}}\right)}$$
$$\displaystyle={\left({1}-{3}{a}\right)}{\left({9}{a}^{{2}}+{3}{a}+{1}\right)}$$ - Factored form.
Hence, the factored form of given expression is equal to $$\displaystyle{1}-{27}{a}^{{3}}={\left({1}-{3}{a}\right)}{\left({9}{a}^{{2}}+{3}{a}+{1}\right)}$$.
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# Multiplicative Function
A function $f(n)$ is called multiplicative function if:
1. $f(n)$ is defined for positive integer $n$.
2. $f(1) = 1$
3. $f(mn) = f(m)f(n)$ whenever $gcd(m,n) = 1$.
For example, the following functions are multiplicative:
# Divisor Sum Theorem of Multiplicative Function
Theorem: Let $f(n)$ be a multiplicative function and define $$F(n) = \sum_{d \mid n} f(d)$$
Then $F(n)$ is also multiplicative.
In essence, if we can define a function $F(n)$ as divisor sum of another multiplicative function, it will be enough to prove that $F(n)$ is multiplicative.
## Proof
Let $m$ and $n$ be positive integer and $gcd(m,n) = 1$
Let $f(n)$ be a multiplicative function as defined in the theorem above
1$$F(mn) = \sum_{d \mid mn} f(d)$$As defined in the theorem
Let $d = rs$, where $r \mid m$ and $s \mid n$
Also, since we know that there is no common divisor of $m$ and $n$, we can say that $gcd(r,s) = 1$
2$$= \sum_{r \mid m, s \mid n} f(rs)$$
Since $f(n)$ is multiplicative and we know that $gcd(r,s) = 1$, we can say that $f(rs) = f(r)f(s)$
3$$= \sum_{r \mid m, s \mid n} f(r)f(s)$$
4$$= \sum_{r \mid m} f(r) \sum_{s \mid n} f(s)$$
5$$= F(m)F(n)$$Proved 🙂
Therefore, if a function $F(n)$ can be written as divisor sum of a multiplicative function $f(n)$, then $F(n)$ is also a multiplicative.
## Application
The Divisor Sum Theorem is useful in proving that a function $F(n)$ is multiplicative. For example, this theorem is used in proving that GCD Sum Function is multiplicative.
# Conclusion
Often time it is difficult to calculate the value of $f(n)$ when $n$ is composite but easy to calculate when $n$ is a power of a prime, $p^a$. In such scenarios, life becomes much easier if we can prove that $f(n)$ is multiplicative.
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Question #f4504
Nov 18, 2017
$2 + 2 - 2 + 2 + 2 - 2 = 4$
Explanation:
2 + 2 -2 + 2 + 2 - 2
╲ ╱
4 -2 + 2 + 2 - 2
╲ ╱
2 + 2 + 2 - 2
╲ ╱
4 + 2 - 2
╲ ╱
6 - 2
╲ ╱
Another solving method you might try is to collect all the (plus 2s) and all the (minus 2s) separately, and then combine them.
(2 + 2 + 2 + 2) and (-2 -2)
8 and - 4
Combine + 8 and - 4
8 - 4
............................
The White Queen asked Alice in Wonderland a similar question.
"Can you do addition?" the White Queen asked. "What's one and one and one and one and one and one and one and one and one and one?"
"I don't know," said Alice. "I lost count."
"She can't do addition," the White Queen interrupted.
Nov 18, 2017
$\text{Answer} = 4$
Explanation:
In order to answer this, take the equation step-by-step (deal with the first two numbers then the next two and so on). Since we typically use PEMDAS, with A and S interchangeable (same as M and D) depending on which operation comes first, then the answer to the question is:
$2 + 2 - 2 + 2 + 2 - 2$
$= 4 - 2 + 2 + 2 - 2$
$= 2 + 2 + 2 - 2$
$= 4 + 2 - 2$
$= 6 - 2$
$= 4$
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Problem Solving with Series Sums
## Solve word problems using sums of number sequences.
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Progress
Progress
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Problem Solving with Series Sums
Katrina just started her new job last week. She is making $60 per day, and saving 10%. Her savings account has been slowly building up at$6 per day, and currently has \$73 in it.
How much money will she have after working another 15 days? How much after another 27 days?
### Watch This
James Sousa: Arithmetic Series
### Guidance
In the previous lesson, we discussed Gauss' formula for finding the total of the numbers in a series. In this lesson, we continue to practice that, and also to make use of the modified version of Gauss' Formula that is commonly seen in other texts:
The sum of the first n terms in an arithmetic series is
#### Example A
Find the sum of the first 50 terms of an arithmetic series if the first term is 5 and the common difference is 3.
Solution:
..... Substitute the applicable number of terms in the sequence
..... Substitute the values for the first and 50th terms
..... Simplify
This method is clearly much easier than writing out and adding 50 numbers!
#### Example B
Find the sum of the first 40 terms of an arithmetic series in which the first term is 8 and the common difference is 5.
Solution:
Use the formula
..... Substitute in the number of terms, and the value of the first and last terms
..... Simplify
..... Simplify
#### Example C
Consider the series: -13 + -3 + 7 + 17 + 27 + ...
a) What is the 25th term?
b) What is the sum of the first 25 terms?
Solution:
a) This is an arithmetic series where each successive term is 10 more than the last.
b) Use the formula .
This is the sum of the first 25 terms of an arithmetic series with a common difference of 10
..... Substitute in the number of terms, and the value of the first and last terms
..... Simplify
..... Simplify
-->
### Guided Practice
1) A student needed to find the sum of the first 10 terms of the series 4 + 12 + 36 + ... and so he wrote the following:
Do you agree with the student’s work? Explain.
2) Given an arithmetic sequence determined by and
What is the 220th number of the sequence.
We find the solution by using the following formula:
3) How do we find the sum of the 220th number of the sequence, given the same information as the 2nd guided practice above?
We use the following formula: . We set n = 220.
1) Because the series is geometric, this formula is not appropriate. The work here does not represent the sum of the first 10 terms. Using a graphing calculator, you can find that the sum is 118,096.
2) To find the 220th number:
3) To find the sum of the first 220 numbers:
the number of the sequence is =
### Explore More
Use the arithmetic series formula to solve the following problems:
1. Given the arithmetic series of numbers: 1, 4, 7, 10, 13... a)Find the 200th number in the sequence b)find the sum of the first 200 numbers.
2. The sum of the first five numbers of an arithmetic sequence is -45. What is the value of the third number? ( hint: find if )
3. You have an arithmetic series of numbers defined by: and a)Determine b)Identify the sum of:
4. The sum of the first three numbers in an arithmetic sequence is 219. The sum of the first nine numbers in the same sequence is 603. What is the 143rd number of the sequence.
5. The first eight numbers of an arithmetic sequence add up to 604. The next eight numbers added up equal 156. Find the first number and the common difference in the sequence.
6. The first number in an arithmetic sequence is 80. Find the common difference if we also know that is eighteen times
7. If is an arithmetic sequence with . Find the second number if we know that the sum of the first five numbers is one-fourth of the sum of the next five numbers.
8. Given Find and
9. The following conditions exist within a sequence of numbers: and . What is the first number of the series, and what is the common difference?
10. What are the values of and , given that is an arithmetic series of numbers, if we know: and ?
11. Given the sequence: and Find the values of n, so that
12. If and . What are the values of and ?
13. If and , find and .
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 7.5.
### Vocabulary Language: English
arithmetic series
arithmetic series
An arithmetic series is the sum of an arithmetic sequence, a sequence with a common difference between each two consecutive terms.
common difference
common difference
Every arithmetic sequence has a common or constant difference between consecutive terms. For example: In the sequence 5, 8, 11, 14..., the common difference is "3".
infinite series
infinite series
An infinite series is the sum of the terms in a sequence that has an infinite number of terms.
Mathematical induction
Mathematical induction
Mathematical induction is a method of mathematical proof typically used to establish that a given statement is true for all positive integers.
partial sum
partial sum
A partial sum is the sum of the first ''n'' terms in an infinite series, where ''n'' is some positive integer.
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## Step by action solution because that calculating 8 is 32 percent that what number
We already have our an initial value 8 and also the second value 32. Let"s i think the unknown value is Y i beg your pardon answer us will uncover out.
You are watching: 32 of what number is 8
As we have all the required values we need, currently we have the right to put lock in a an easy mathematical formula together below:
STEP 18 = 32% × Y
STEP 28 = 32/100× Y
Multiplying both sides by 100 and also dividing both political parties of the equation by 32 we will certainly arrive at:
STEP 3Y = 8 × 100/32
STEP 4Y = 8 × 100 ÷ 32
STEP 5Y = 25
Finally, us have discovered the worth of Y i m sorry is 25 and also that is our answer.
You can conveniently calculate 8 is 32 percent the what number through using any regular calculator, simply enter 8 × 100 ÷ 32 and you will acquire your answer i beg your pardon is 25
## People also Ask
Here is a percent Calculator come solve similar calculations such together 8 is 32 percent of what number. You can solve this form of calculation with your worths by beginning them into the calculator"s fields, and also click "Calculate" to acquire the result and explanation.
is
percent that what number
Calculate
## Sample questions, answers, and also how to
Question: her friend has actually a bag that marbles, and he tells you that 32 percent that the marbles are red. If there are 8 red marbles. How numerous marbles walk he have altogether?
How To: In this problem, we know that the Percent is 32, and also we are also told that the part of the marbles is red, therefore we recognize that the part is 8.
So, that means that it have to be the complete that"s missing. Here is the way to figure out what the full is:
Part/Total = Percent/100
By using a simple algebra we can re-arrange our Percent equation like this:
Part × 100/Percent = Total
If we take the "Part" and multiply that by 100, and also then we division that through the "Percent", we will get the "Total".
Let"s shot it the end on our problem around the marbles, that"s very basic and it"s just two steps! We know that the "Part" (red marbles) is 8.
So step one is to simply multiply that part by 100.
8 × 100 = 800
In action two, us take that 800 and also divide it by the "Percent", i beg your pardon we are told is 32.
So, 800 split by 32 = 25
And that method that the total number of marbles is 25.
Question: A high school marching band has 8 flute players, If 32 percent that the tape members play the flute, then how numerous members space in the band?
Answer: There are 25 members in the band.
How To: The smaller sized "Part" in this difficulty is 8 due to the fact that there space 8 flute players and also we are told that they consist of 32 percent of the band, for this reason the "Percent" is 32.
Again, it"s the "Total" that"s absent here, and also to find it, we just need to follow our 2 step procedure together the vault problem.
For step one, we multiply the "Part" by 100.
8 × 100 = 800
For action two, we division that 800 by the "Percent", i m sorry is 32.
See more: 1998 Toyota Camry Camshaft Timing Marks, Camshaft Timing Marks
800 divided by 32 equates to 25
That means that the total variety of band members is 25.
## Another action by step method
Step 1: Let"s i think the unknown worth is Y
Step 2: very first writing it as: 100% / Y = 32% / 8
Step 3: autumn the percentage marks to leveling your calculations: 100 / Y = 32 / 8
Step 4: main point both sides by Y to relocate Y top top the right side that the equation: 100 = ( 32 / 8 ) Y
Step 5: simplifying the ideal side, us get: 100 = 32 Y
Step 6: splitting both political parties of the equation by 32, we will arrive in ~ 25 = Y
This pipeline us with our last answer: 8 is 32 percent of 25
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# How To Calculate Winning Odds in California Lottery
Ever wonder how to calculate winning odds of lottery games? The winning odds of the top prize of Fantasy 5 in California Lottery are 1 in 575,757. The winnings odds of the top prize of SuperLOTTO plus are 1 in 41,416,353. The winnings odds of the top prize of Mega Millions are 1 in 175,711,534. In this post, we show how to calculate the odds for these games in the California Lottery. The calculation is an excellent combinatorial exercise as well as in calculating hypergeometric probability.
All figures and data are obtained from the California Lottery.
Update, April 27, 2017. The calculation in this post assumes certain background knowledge on combination and the multiplication principle (not explained here). For any reader who would like to further understand how lottery odds are calculated, see this blog post on Powerball. It is a self contained step by step explanation at the basic level on how to calculate winning odds in the Powerball game.
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Fantasy 5
The following figures show a playslip and a sample ticket for the game of Fantasy 5.
Figure 1
Figure 2
In the game of Fantasy 5, the player chooses 5 numbers from 1 to 39. If all 5 chosen numbers match the 5 winning numbers, the player wins the top prize which starts at $50,000 and can go up to$500,000 or more. The odds of winning the top prize are 1 in 575,757. There are lower tier prizes that are easier to win but with much lower winning amounts. The following figure shows the prize categories and the winning odds of Fantasy 5.
Figure 3
All 5 of 5
In matching the player’s chosen numbers with the winning numbers, the order of the numbers do not matter. Thus in the calculation of odds, we use combination rather than permutation. Thus we have:
$\displaystyle (1) \ \ \ \ \ \binom{39}{5}=\frac{39!}{5! \ (39-5)!}=575757$
Based on $(1)$, the odds of matching all 5 winning numbers is 1 in 575,757 (the odds of winning the top prize).
Any 4 of 5
To match 4 out of 5 winning numbers, 4 of the player’s chosen numbers are winning numbers and 1 of the player’s chosen numbers is from the non-winning numbers (34 of them). Thus the probability of matching 4 out of 5 winning numbers is:
$\displaystyle (2) \ \ \ \ \ \frac{\displaystyle \binom{5}{4} \ \binom{34}{1}}{\displaystyle \binom{39}{5}}=\frac{5 \times 34}{575757}=\frac{1}{3386.8} \ \ \text{(1 out of 3,387)}$
Any 3 of 5
To find the odds for matching 3 out of 5 winning numbers, we need to find the probability that 3 of the player’s chosen numbers are from the 5 winning numbers and 2 of the selected numbers are from the 34 non-winning numbers. Thus we have:
$\displaystyle (3) \ \ \ \ \ \frac{\displaystyle \binom{5}{3} \ \binom{34}{2}}{\displaystyle \binom{39}{5}}=\frac{10 \times 561}{575757}=\frac{1}{102.63} \ \ \text{(1 out of 103)}$
Any 2 of 5
Similarly, the following shows how to calculate the odds of matching 2 out of 5 winning numbers:
$\displaystyle (4) \ \ \ \ \ \frac{\displaystyle \binom{5}{2} \ \binom{34}{3}}{\displaystyle \binom{39}{5}}=\frac{10 \times 5984}{575757}=\frac{1}{9.6216} \ \ \text{(1 out of 10)}$
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SuperLOTTO Plus
Here are the pictures of a playslip and a sample ticket of the game of SuperLOTTO Plus.
Figure 4
Figure 5
Based on the playslip (Figure 4), the player chooses 5 numbers from 1 to 47. The player also chooses an additional number called a Mega number from 1 to 27. To win the top prize, there must be a match between the player’s 5 selections and the 5 winning numbers as well as a match between the player’s Mega number and the winning Mega number (All 5 of 5 and Mega in Figure 6 below).
Figure 6
All 5 of 5 and Mega
To find the odds of the match of “All 5 of 5 and Mega”, the total number of possibilities is obtained by choosing 5 numbers from 47 numbers and choose 1 number from 27 numbers. We have:
$\displaystyle (5) \ \ \ \ \ \binom{47}{5} \times \binom{27}{1}=41,416,353$
Thus the odds of matching “All 5 of 5 and Mega” are 1 in 41,416,353.
Any 5 of 5
To find the odds of matching “All 5 of 5” (i.e. the player’s 5 selections match the 5 winning numbers but no match with the Mega winning number), we need to choose 5 numbers from the 5 winning numbers, choose 0 numbers from the 42 non-winning numbers, choose 0 numbers from the 1 Mega winning number and choose 1 number from the 26 non-Mega winning numbers. This may seem overly precise, but will make it easier to the subsequent derivations. We have:
\displaystyle \begin{aligned}(6) \ \ \ \ \ \frac{\displaystyle \binom{5}{5} \ \binom{42}{0} \ \binom{1}{0} \ \binom{26}{1}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{1 \times 1 \times 1 \times 26}{41416353} \\&=\frac{1}{1592936.654} \\&\text{ } \\&=\text{1 out of 1,592,937} \end{aligned}
Any 4 of 5 and Mega
To calculate the odds for matching “any 4 of 5 and Mega”, we need to choose 4 out of 5 winning numbers, choose 1 out of 42 non-winning numbers, choose 1 out of 1 Mega winning number, and choose 0 out of 26 non-winning Mega numbers. We have:
\displaystyle \begin{aligned}(7) \ \ \ \ \ \frac{\displaystyle \binom{5}{4} \ \binom{42}{1} \ \binom{1}{1} \ \binom{26}{0}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{5 \times 42 \times 1 \times 1}{41416353} \\&=\frac{1}{197220.7286} \\&\text{ } \\&=\text{1 out of 197,221} \end{aligned}
Any 4 of 5
To calculate the odds for matching “any 4 of 5” (no match for Mega number), we need to choose 4 out of 5 winning numbers, choose 1 out of 42 non-winning numbers, choose 0 out of 1 Mega winning number, and choose 1 out of 26 non-winning Mega numbers. We have:
\displaystyle \begin{aligned}(8) \ \ \ \ \ \frac{\displaystyle \binom{5}{4} \ \binom{42}{1} \ \binom{1}{0} \ \binom{26}{1}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{5 \times 42 \times 1 \times 26}{41416353} \\&=\frac{1}{7585.412637} \\&\text{ } \\&=\text{1 out of 7,585} \end{aligned}
Any 3 of 5 and Mega
To calculate the odds for matching “any 3 of 5 and Mega”, we need to choose 3 out of 5 winning numbers, choose 2 out of 42 non-winning numbers, choose 1 out of 1 Mega winning number, and choose 0 out of 26 non-winning Mega numbers. We have:
\displaystyle \begin{aligned}(9) \ \ \ \ \ \frac{\displaystyle \binom{5}{3} \ \binom{42}{2} \ \binom{1}{1} \ \binom{26}{0}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{10 \times 861 \times 1 \times 1}{41416353} \\&=\frac{1}{4810.261672} \\&\text{ } \\&=\text{1 out of 4,810} \end{aligned}
The rest of the calculations for SuperLOTTO Plus should be routine. It is a matter to deciding how many to choose from the 5 winning numbers, how many to choose from the 42 non-winning numbers as well as how many to choose from the 1 winning Mega number and how many to choose from the 26 non-winning Mega numbers.
Any 3 of 5
\displaystyle \begin{aligned}(10) \ \ \ \ \ \frac{\displaystyle \binom{5}{3} \ \binom{42}{2} \ \binom{1}{0} \ \binom{26}{1}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{10 \times 861 \times 1 \times 26}{41416353} \\&=\frac{1}{185.0100643} \\&\text{ } \\&=\text{1 out of 185} \end{aligned}
Any 2 of 5 and Mega
\displaystyle \begin{aligned}(11) \ \ \ \ \ \frac{\displaystyle \binom{5}{2} \ \binom{42}{3} \ \binom{1}{1} \ \binom{26}{0}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{10 \times 11480 \times 1 \times 1}{41416353} \\&=\frac{1}{360.7696254} \\&\text{ } \\&=\text{1 out of 361} \end{aligned}
Any 1 of 5 and Mega
\displaystyle \begin{aligned}(12) \ \ \ \ \ \frac{\displaystyle \binom{5}{1} \ \binom{42}{4} \ \binom{1}{1} \ \binom{26}{0}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{5 \times 111930 \times 1 \times 1}{41416353} \\&=\frac{1}{74.00402573} \\&\text{ } \\&=\text{1 out of 74} \end{aligned}
None of 5 only Mega
\displaystyle \begin{aligned}(13) \ \ \ \ \ \frac{\displaystyle \binom{5}{0} \ \binom{42}{5} \ \binom{1}{1} \ \binom{26}{0}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{1 \times 850668 \times 1 \times 1}{41416353} \\&=\frac{1}{48.68685903} \\&\text{ } \\&=\text{1 out of 49} \end{aligned}
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Mega Millions
The following are a playslip, a sample ticket and the winning odds of the game of Mega Millions.
Figure 7
Figure 8
Figure 9
Based on the playslip (Figure 7), the player chooses 5 numbers from 1 to 56. The player also chooses an additional number called a Mega number from 1 to 46. To win the top prize, there must be a match between the player’s 5 selections and the 5 winning numbers as well as a match between the player’s Mega number and the winning Mega number. The calculation of the odds indicated in Figure 9 are left as exercises.
# Picking Two Types of Binomial Trials
We motivate the discussion with the following example. The notation $W \sim \text{binom}(n,p)$ denotes the statement that $W$ has a binomial distribution with parameters $n$ and $p$. In other words, $W$ is the number of successes in a sequence of $n$ independent Bernoulli trials where $p$ is the probability of success in each trial.
Example 1
Suppose that a student took two multiple choice quizzes in a course for probability and statistics. Each quiz has 5 questions. Each question has 4 choices and only one of the choices is correct. Suppose that the student answered all the questions by pure guessing. Furthermore, the two quizzes are independent (i.e. results of one quiz will not affect the results of the other quiz). Let $X$ be the number of correct answers in the first quiz and $Y$ be the number of correct answers in the second quiz. Suppose the student was told by the instructor that she had a total of 4 correct answers in these two quizzes. What is the probability that she had 3 correct answers in the first quiz?
On the face of it, the example is all about binomial distribution. Both $X$ and $Y$ are binomial distributions (both $\sim \text{binom}(5,\frac{1}{4})$). The sum $X+Y$ is also a binomial distribution ($\sim \text{binom}(10,\frac{1}{4})$). The question that is being asked is a conditional probability, i.e., $P(X=3 \lvert X+Y=4)$. Surprisingly, this conditional probability can be computed using the hypergeometric distribution. One can always work this problem from first principle using binomial distributions. As discussed below, for a problem such as Example 1, it is always possible to replace the binomial distributions using a thought process involving the hypergeometric distribution.
Here’s how to think about the problem. This student took the two quizzes and was given the news by the instructor that she had 4 correct answers in total. She now wonders what the probability of having 3 correct answers in the first quiz is. The thought process is this. She is to pick 4 questions from 10 questions (5 of them are from Quiz 1 and 5 of them are from Quiz 2). So she is picking 4 objects from a group of two distinct types of objects. This is akin to reaching into a jar that has 5 red balls and 5 blue balls and pick 4 balls without replacement. What is the probability of picking 3 red balls and 1 blue ball? The probability just described is from a hypergeometric distribution. The following shows the calculation.
$\displaystyle (1) \ \ \ \ P(X=3 \lvert X+Y=4)=\frac{\displaystyle \binom{5}{3} \ \binom{5}{1}}{\displaystyle \binom{10}{4}}=\frac{50}{210}$
We will show below why this works. Before we do that, let’s describe the above thought process. Whenever you have two independent binomial distributions $X$ and $Y$ with the same probability of success $p$ (the number of trials does not have to be the same), the conditional distribution $X \lvert X+Y=a$ is a hypergeometric distribution. Interestingly, the probability of success $p$ has no bearing on this observation. For Example 1, we have the following calculation.
$\displaystyle (2a) \ \ \ \ P(X=0 \lvert X+Y=4)=\frac{\displaystyle \binom{5}{0} \ \binom{5}{4}}{\displaystyle \binom{10}{4}}=\frac{5}{210}$
$\displaystyle (2b) \ \ \ \ P(X=1 \lvert X+Y=4)=\frac{\displaystyle \binom{5}{1} \ \binom{5}{3}}{\displaystyle \binom{10}{4}}=\frac{50}{210}$
$\displaystyle (2c) \ \ \ \ P(X=2 \lvert X+Y=4)=\frac{\displaystyle \binom{5}{2} \ \binom{5}{2}}{\displaystyle \binom{10}{4}}=\frac{100}{210}$
$\displaystyle (2d) \ \ \ \ P(X=3 \lvert X+Y=4)=\frac{\displaystyle \binom{5}{3} \ \binom{5}{1}}{\displaystyle \binom{10}{4}}=\frac{50}{210}$
$\displaystyle (2e) \ \ \ \ P(X=4 \lvert X+Y=4)=\frac{\displaystyle \binom{5}{4} \ \binom{5}{0}}{\displaystyle \binom{10}{4}}=\frac{5}{210}$
Interestingly, the conditional mean $E(X \lvert X+Y=4)=2$, while the unconditional mean $E(X)=5 \times \frac{1}{4}=1.25$. The fact that the conditional mean is higher is not surprising. The student was lucky enough to have obtained 4 correct answers by guessing. Given this, she had a greater chance of doing better on the first quiz.
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Why This Works
Suppose $X \sim \text{binom}(5,p)$ and $Y \sim \text{binom}(5,p)$ and they are independent. The joint distribution of $X$ and $Y$ has 36 points in the sample space. See the following diagram.
Figure 1
The probability attached to each point is
\displaystyle \begin{aligned}(3) \ \ \ \ P(X=x,Y=y)&=P(X=x) \times P(Y=y) \\&=\binom{5}{x} p^x (1-p)^{5-x} \times \binom{5}{y} p^y (1-p)^{5-y} \end{aligned}
where $x=0,1,2,3,4,5$ and $y=0,1,2,3,4,5$.
The conditional probability $P(X=k \lvert X+Y=4)$ involves 5 points as indicated in the following diagram.
Figure 2
The conditional probability $P(X=k \lvert X+Y=4)$ is simply the probability of one of the 5 sample points as a fraction of the sum total of the 5 sample points encircled in the above diagram. The following is the sum total of the probabilities of the 5 points indicated in Figure 2.
\displaystyle \begin{aligned}(4) \ \ \ \ P(X+Y=4)&=P(X=0) \times P(Y=4)+P(X=1) \times P(Y=3)\\&\ \ \ \ +P(X=2) \times P(Y=3)+P(X=3) \times P(Y=2)\\&\ \ \ \ +P(X=4) \times P(Y=0) \end{aligned}
We can plug $(3)$ into $(4)$ and work out the calculation. But $(4)$ is actually equivalent to the following because $X+Y \sim \text{binom}(10,p)$.
$\displaystyle (5) \ \ \ \ P(X+Y=4)=\ \binom{10}{4} p^4 \ (1-p)^{6}$
As stated earlier, the conditional probability $P(X=k \lvert X+Y=4)$ is simply the probability of one of the 5 sample points as a fraction of the sum total of the 5 sample points encircled in Figure 2. Thus we have:
\displaystyle \begin{aligned}(6) \ \ \ \ P(X=k \lvert X+Y=4)&=\frac{P(X=k) \times P(Y=4-k)}{P(X+Y=4)} \\&=\frac{\displaystyle \binom{5}{k} p^k (1-p)^{5-k} \times \binom{5}{4-k} p^{4-k} (1-p)^{5-(4-k)}}{\displaystyle \binom{10}{4} p^4 \ (1-p)^{6}} \end{aligned}
With the terms involving $p$ and $1-p$ cancel out, we have:
$\displaystyle (7) \ \ \ \ P(X=k \lvert X+Y=4)=\frac{\displaystyle \binom{5}{k} \times \binom{5}{4-k}}{\displaystyle \binom{10}{4}}$
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Summary
Suppose $X \sim \text{binom}(N,p)$ and $Y \sim \text{binom}(M,p)$ and they are independent. Then $X+Y$ is also a binomial distribution, i.e., $\sim \text{binom}(N+M,p)$. Suppose that both binomial experiments $\text{binom}(N,p)$ and $\text{binom}(M,p)$ have been performed and it is known that there are $a$ successes in total. Then $X \lvert X+Y=a$ has a hypergeometric distribution.
$\displaystyle (8) \ \ \ \ P(X=k \lvert X+Y=a)=\frac{\displaystyle \binom{N}{k} \times \binom{M}{a-k}}{\displaystyle \binom{N+M}{a}}$
where $k=0,1,2,3,\cdots,\text{min}(N,a)$.
As discussed earlier, think of the $N$ trials in $\text{binom}(N,p)$ as red balls and think of the $M$ trials in $\text{binom}(M,p)$ as blue balls in a jar. Think of the $a$ successes as the number of balls you are about to draw from the jar. So you reach into the jar and select $a$ balls without replacement. The calculation in $(8)$ gives the probability that you select $k$ red balls and $a-k$ blue balls.
The probability of success $p$ in the two binomial distributions have no bearing on the result since it gets canceled out in the derivation. One can always work a problem like Example 1 using first principle. Once the thought process using hypergeometric distribution is understood, it is a great way to solve this problem, that is, you can by pass the binomial distributions and go straight to the hypergeometric distribution.
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Practice problems are found in the following blog post.
How to pick binomial trials
# The capture-recapture method
The capture-recapture method is one of the methods for estimating the size of wildlife populations and is based on the hypergeometric distribution. Recall that the hypergeometric distribution is a three-parameter family of discrete distributions and one of the parameters, denoted by $N$ in this post, is the size of the population. We show that the estimate for the parameter $N$ that is obtained from the capture-recapture method is the value of the parameter $N$ that makes the observed data “more likely” than any other possible values of $N$. Thus, the capture-recapture method produces the maximum likelihood estimate of the population size parameter $N$ of the hypergeometric distribution.
Let’s start with an example. In order to estimate the size of the population of bluegills (a species of fresh water fish) in a small lake in Missouri, a total of $w=250$ bluegills are captured and tagged and then released. After allowing sufficient time for the tagged fish to disperse, a sample of $n=150$ bluegills were caught. It was found that $y=16$ bluegills in the sample were tagged. Estimate the size of the bluegill population in this lake.
Let $N$ be the size of the bluegill population in this lake. The population proportion of the tagged bluegills is $\frac{w}{N}$. The sample proportion of the tagged bluegills is $\frac{y}{n}$. In the capture-recapture method, the population proportion and the sample proportion are set equaled. Then we solve for $N$.
$\displaystyle \frac{w}{N}=\frac{y}{n} \Rightarrow N=\frac{w n}{y}=\frac{250(150)}{16}=2343.75=2343$
Now, the connection to the hypergeometric distribution. After $w=250$ bluegills were captured, tagged and released, the population is separated into two distinct classes, tagged and non-tagged. When a sample of $n=150$ bluegills were selected without replacement, we let $Y$ be the number of bluegills in the sample that were tagged. The distribution of $Y$ is the hypergeometric distribution. The following is the probability function of $Y$.
$\displaystyle P[Y=y]=\frac{\binom{w}{y} \thinspace \binom{N-w}{n-y}}{\binom{N}{n}}$
In the hypergeometric distribution described here, the parameters $w$ and $n$ are known ($w=250$ and $n=150$). We now show that the estimate of $N=2343$ is the estimate that makes the observed value of $y=16$ “most likely” (i.e. the estimate of $N=2343$ is a maximum likelihood estimate of $N$). To show this, we consider the ratio of the hypergeometric probabilities for two successive values of $N$.
$\displaystyle \frac{P(N)}{P(N-1)}=\frac{(N-w)(N-n)}{N(N-w-n+y)}$
where $\displaystyle P(N)=\frac{\binom{w}{y} \thinspace \binom{N-w}{n-y}}{\binom{N}{n}}$ and $\displaystyle P(N-1)=\frac{\binom{w}{y} \thinspace \binom{N-1-w}{n-y}}{\binom{N-1}{n}}$
Note that $1<\frac{P(N)}{P(N-1)}$ or $P(N-1) if and only if the following holds:
$\displaystyle N(N-w-n+y)<(N-w)(N-n)$
$\displaystyle N<\frac{w n}{y}$
Note that $\frac{w n}{y}$ is the estimate from the capture-recapture method. It is also an upper bound for the population size $N$ such that the probability $P(N)$ is greater than $P(N-1)$. This implies that the maximum likelihood estimate of $N$ is achieved when the estimate is $\hat{N}=\frac{w n}{y}$.
As an illustration, we compute the probabilities $\displaystyle P(N)=\frac{\binom{250}{16} \thinspace \binom{N-250}{150-16}}{\binom{N}{150}}$ for several values of $N$ above and below $N=2343$. The following matrix illustrates that the maximum likelihood is achieved at $N=2343$.
$\displaystyle \begin{pmatrix} N&P(N) \\{2340}&0.1084918 \\{2341}&0.1084929 \\{2342}&0.1084935 \\{2343}&0.1084938 \\{2344}&0.1084937 \\{2345}&0.1084933 \\{2346}&0.1084924\end{pmatrix}$
# The hypergeometric distribution
Consider a large bowl with $n+m$ balls, $n$ of which are green and $m$ of which are yellow. We draw $z$ balls out of the bowl without replacement. Let $X$ be the number of green balls in the $z$ many draws (i.e. getting a green ball is a success). The distribution of $X$ is called the hypergeometric distribution. This distribution has three parameters $n,m,z$ and its probability function is:
$\displaystyle P[X=x]=\frac{\displaystyle \binom{n}{x} \thinspace \binom{m}{z-x}}{\displaystyle \binom{n+m}{z}}$ where $x=0,1,2, \cdots, min(n,z)$
In the denominator, we have the number of different ways of drawing $z$ balls out of $n+m$ balls. In the numerator, we have the number of ways of drawing $x$ balls out of $n$ green balls multiplied by the number of ways of drawing $z-x$ balls out of $m$ yellow balls. We presented an example of the hypergeometric distribution in this post (The hypergeometric distribution, an example). In this post, we continue the discussion and we also derive the mean and variance.
When both $n$ (the number of green balls) and $m$ (the number of yellow balls) approach infinity while at the same time the proportion $p$ of the green balls remains constant, then the hypergeometric distribution approaches the binomial distribution with parameters $z$ and $p$. This result makes intuitive sense. As the number of balls in the bowl becomes larger and larger and is much greater than the sample size ($z$), the difference between sampling without replacement and sampling with replacement is negligible. Thus when population size is much larger than the sample size, for all practical purposes, the hypergeometric probabilities can be estimated with the binomial probabilities.
If we draw green balls with replacement, we work with the binomial distribution with parameters $z$ and $p=\frac{n}{n+m}$. Then the mean number of green balls drawn is $zp=\frac{z n}{n+m}$. Interestingly, the mean of the hypergeometric distribution under discussion is also $E[X]=\frac{z n}{n+m}$. Sampling without replacement produces the same mean as sampling with replacement. However, the variance is different between sampling with and without replacement. The following is the derivation for the mean of the hypergeometric distribution.
$\displaystyle E[X]=\binom{n+m}{z}^{-1} \sum \limits_{j=0}^{k} j \binom{n}{j} \binom{m}{z-j}$ where $k=min(n,z)$
$\displaystyle E[X]=\binom{n+m}{z}^{-1} \sum \limits_{j=1}^{k} n \binom{n-1}{j-1} \binom{m}{z-j}$
$\displaystyle E[X]=n \binom{n+m}{z}^{-1} \sum \limits_{j=0}^{k-1} \binom{n-1}{j} \binom{m}{z-1-j}$
$\displaystyle E[X]=n \binom{n+m}{z}^{-1} \binom{n+m-1}{z-1}$
$\displaystyle E[X]=\frac{z n}{n+m}$
In the second to the last step above, we use the Euler’s formula:
$\displaystyle \sum \limits_{j=0}^{min(n,z)} \binom{n}{j} \binom{m}{z-j}=\binom{n+m}{z}$
We now derive the variance. We first derive $E[X(X-1)]$.
$\displaystyle E[X(X-1)]=\binom{n+m}{z}^{-1} \sum \limits_{j=0}^{k} j(j-1) \binom{n}{j} \binom{m}{z-j}$ where $k=min(n,z)$
$\displaystyle E[X(X-1)]=\binom{n+m}{z}^{-1} \sum \limits_{j=2}^{k} n(n-1) \binom{n-2}{j-2} \binom{m}{z-j}$
$\displaystyle E[X(X-1)]=\frac{n(n-1)}{\displaystyle \binom{n+m}{z}} \sum \limits_{j=0}^{k-2} \binom{n-2}{j} \binom{m}{z-2-j}$
$\displaystyle E[X(X-1)]=\frac{n(n-1)}{\displaystyle \binom{n+m}{z}} \binom{n+m-2}{z-2}$
$\displaystyle E[X(X-1)]=\frac{n(n-1) z(z-1)}{(n+m) (n+m-1)}$
Again, we use the Euler’s formula in the second to the last step. To find the variance, we use the following:
$\displaystyle Var[X]=E[X(X-1)]+E[X]-E[X]^2$
$\displaystyle Var[X]=\frac{\displaystyle z \thinspace n \thinspace m (m+n-z)}{\displaystyle (n+m)^2 (n+m-1)}$
$\displaystyle Var[X]=z p (1-p) \frac{n+m-z}{n+m-1}$ where $\displaystyle p=\frac{n}{n+m}$
# The hypergeometric distribution, an example
We present an example of the hypergeometric distribution seen through an independent sum of two binomial distributions. Suppose a student takes two independent multiple choice quizzes (i.e. performance on one quiz has no bearing on the other quiz). Quiz 1 has 5 problems where each of the problem has 4 choices. Quiz 2 has 5 problems with 4 choices for each problem. Suppose a student answers each question in each of the two quizzes by pure guessing. If the students has a total of four correct answers for the two quizzes combined, what is the probablity that he passes quiz 1 (60% correct)?
Suppose that $X$ is the number of correct answers in quiz 1 and $Y$ is the number of correct answers in quiz 2. Then both $X$ and $Y$ have binomial distribution with $n=5$ and $p=0.25$. Then $Z=X+Y$ is the total number of correct answers and $Z$ has a binomial distribution with $n=10$ and $p=0.25$. The problem we need to solve is $P[X \ge 3 \lvert Z=4]$.
We propose that the conditional distribution of $X \lvert Z=z$ is a hypergeometric distribution. To see this intuitively, there are five green balls (a correct answer in quiz 1) and five yellow balls (a correct answer in quiz 2) in a bowl. Taking these two quizzes and getting a total of four correct answers would be like drawing 4 balls out of this bowl without replacement. Then what is the probability that three of the four balls are green? This is a probability obtained by the hypergeometric distribution (drawing 4 balls out of the bowl and resulting in 3 green balls and 1 yellow ball). Though not a proof, this is good intuitive description of the approach we can take. We first do the calculation and present the proof at the end.
We now evaluate the probability function $P[X=j \lvert Z=4]$. For example, to find $P[X=1 \lvert Z=4]$ is the probability of drawing 4 balls out of the bowl and resulting in 1 green ball and 3 yellow balls.
$\displaystyle P[X=0 \lvert Z=4]=\displaystyle \frac{\displaystyle \binom{5}{0} \binom{5}{4}}{\displaystyle \binom{10}{4}}=\frac{1}{42}$
$\displaystyle P[X=1 \lvert Z=4]=\frac{\displaystyle \binom{5}{1} \binom{5}{3}}{\displaystyle \binom{10}{4}}=\frac{10}{42}$
$\displaystyle P[X=2 \lvert Z=4]=\frac{\displaystyle \binom{5}{2} \binom{5}{2}}{\displaystyle \binom{10}{4}}=\frac{20}{42}$
$\displaystyle P[X=3 \lvert Z=4]=\frac{\displaystyle \binom{5}{3} \binom{5}{1}}{\displaystyle \binom{10}{4}}=\frac{10}{42}$
$\displaystyle P[X=4 \lvert Z=4]=\frac{\displaystyle \binom{5}{4} \binom{5}{0}}{\displaystyle \binom{10}{4}}=\frac{1}{42}$
Thus, $\displaystyle P[X \ge 3 \lvert Z=4]=\frac{11}{42}=0.26$. Note that the unconditional probability $P[X \ge 3]=0.1035$ using the binomial distribution with $n=5$ and $p=0.25$. It is not surprising that the conditional probability is much greater. The conditional probability $P[X \ge 3 \lvert Z=4]$ is greater because the student is lucky enough to have four correct guesses.
We now discuss the general fact. Suppose $X \sim binomial(n,p)$ and $Y \sim binomial(m,p)$. With $Z=X+Y$ an independent sum, we show that $X \lvert Z$ has a hypergeometric distribution.
$\displaystyle P[X=x \lvert Z=z]=\frac{\displaystyle \binom{n}{x} p^x (1-p)^{n-x} \thinspace \binom{m}{z-x} p^{z-x} (1-p)^{m-(z-x)}}{\displaystyle \binom{n+m}{z} p^z (1-p)^{n+m-z}}$
After canceling out the terms for $p$ and $1-p$, the following is the probability function for the hypergeometric distribution:
$\displaystyle P[X=x \lvert Z=z]=\frac{\displaystyle \binom{n}{x} \binom{m}{z-x}}{\displaystyle \binom{n+m}{z}}$
The above probability distribution describes the situation where there are $n+m$ similar objects, with $n$ objects belong to one class (say green balls) and $m$ objects belong to another class (say yellow balls). We choose $z$ balls out of $n+m$ balls without replacement. The above probability is the probability of having a result of $x$ green balls and $z-x$ yellow balls. There are $\binom{n}{x}$ many ways of choosing $x$ green balls out of $n$ green balls. Likewise, there are $\binom{m}{z-x}$ many ways of choosing $z-x$ yellow balls out of $m$ yellow balls. The total number ways the joint operation can take place is $\binom{n}{x} \binom{m}{z-x}$. Of course, we assume that each of the $\binom{n+m}{z}$ ways of selecting $z$ balls out of $n+m$ balls is equally likely.
In the conditional probability in our example, the probability of success (0.25) in the individual Bernoulli trials that make up the two binomial distributions is not used. This is because the terms for $p$ and $1-p$ are canceled out. If each multiple choice quiz has a different probability of success, then the resulting conditional distribution $P[X=x \lvert X+Y=z]$ is no longer hypergeometric. In that case, the conditional probability must be obtained by first principle.
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Pythagorean Theorem
We can define Pythagorean theorem as follows. If one of the angle in a measures 90 ° , this kind of triangle is called right triangle. Let ABC be a right triangle in which the measure of angle B = 90°.
Name of the sides in Right Triangle
The side AC is called the hypotenuse of the right triangle. The side which is opposite to 90 degree is called the hypotenuse side and it is the longest side and is opposite to the right angle. Greek mathematician Pythagoras has found that the length of the square on the hypotenuse is equal to the sum of the length of the squares on the other two sides. That is AC 2 = AB 2 + BC 2. This is knows as Pythagoras theorem.
The Pythagorean-Theorem is the most famous mathematical contribution.The later discovery that the square root of 2 is irrational and therefore, that cannot be expressed as a ratio of two integers, extremely troubled Pythagoras and his followers. They were consume in their belief that any two lengths were integral multiples of some unit length. Many efforts were made to suppress the knowledge that the square root of 2 is irrational.
Pythagorean Theorem-Example problems
1.Find the length of the missing side.
By using Pythagoras theorem
AB= 5 cm , BC = 12 cm
we know AC 2 = AB 2 + BC 2
AC 2 = 5 2 + 12 2
AC 2 = 25 + 144
AC 2 = 169
AC 2 = √169
AC 2 = √13 x 13
AC 2 = 13 cm
2. Find the length of the missing side.Here angle C is 90°.
AB = √5 cm , AC = 1 cm
By using Pythagoras theorem
AB 2 = AC 2 + BC 2
(√5) 2 = 1 2 + BC 2
5 = 1 + BC 2
5 - 1 = BC 2
√4= BC 2
√2 x 2 = BC 2
2 = BC 2
BC = 2 cm
3.In a right triangle ABC right angled at C then AB = √7 cm , AC = 2 cm find BC.
By using Pythagoras theorem
AB 2 = AC 2 + BC 2
(√7) 2 = 2 2 + BC 2
7 = 4 + BC 2
7 - 4 = BC 2
3 = BC 2
BC = √3 cm
Pythagorean theorem to Trigonometry
Geometry Help to Homepage
Math dictionary
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# 2.2 Linear equations in one variable (Page 8/15)
Page 8 / 15
$\frac{x+2}{4}-\frac{x-1}{3}=2$
$x=-14$
For the following exercises, solve each rational equation for $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ State all x -values that are excluded from the solution set.
$\frac{3}{x}-\frac{1}{3}=\frac{1}{6}$
$2-\frac{3}{x+4}=\frac{x+2}{x+4}$
$x\ne -4;$ $x=-3$
$\frac{3}{x-2}=\frac{1}{x-1}+\frac{7}{\left(x-1\right)\left(x-2\right)}$
$\frac{3x}{x-1}+2=\frac{3}{x-1}$
$x\ne 1;$ when we solve this we get $\text{\hspace{0.17em}}x=1,$ which is excluded, therefore NO solution
$\frac{5}{x+1}+\frac{1}{x-3}=\frac{-6}{{x}^{2}-2x-3}$
$\frac{1}{x}=\frac{1}{5}+\frac{3}{2x}$
$x\ne 0;$ $x=\frac{-5}{2}$
For the following exercises, find the equation of the line using the point-slope formula.
Write all the final equations using the slope-intercept form.
$\left(0,3\right)\text{\hspace{0.17em}}$ with a slope of $\text{\hspace{0.17em}}\frac{2}{3}$
$\left(1,2\right)\text{\hspace{0.17em}}$ with a slope of $\text{\hspace{0.17em}}\frac{-4}{5}$
$y=\frac{-4}{5}x+\frac{14}{5}$
x -intercept is 1, and $\text{\hspace{0.17em}}\left(-2,6\right)$
y -intercept is 2, and $\text{\hspace{0.17em}}\left(4,-1\right)$
$y=\frac{-3}{4}x+2$
$\left(-3,10\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(5,-6\right)$
$y=\frac{1}{2}x+\frac{5}{2}$
parallel to $\text{\hspace{0.17em}}y=2x+5\text{\hspace{0.17em}}$ and passes through the point $\text{\hspace{0.17em}}\left(4,3\right)$
perpendicular to $\text{\hspace{0.17em}}\text{3}y=x-4\text{\hspace{0.17em}}$ and passes through the point $\text{\hspace{0.17em}}\left(-2,1\right)$ .
$y=-3x-5$
For the following exercises, find the equation of the line using the given information.
$\left(-2,0\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(-2,5\right)$
$\left(1,7\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(3,7\right)$
$y=7$
The slope is undefined and it passes through the point $\text{\hspace{0.17em}}\left(2,3\right).$
The slope equals zero and it passes through the point $\text{\hspace{0.17em}}\left(1,-4\right).$
$y=-4$
The slope is $\text{\hspace{0.17em}}\frac{3}{4}\text{\hspace{0.17em}}$ and it passes through the point $\text{\hspace{0.17em}}\text{(1,4)}\text{.}$
$\left(-1,3\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(4,-5\right)$
$8x+5y=7$
## Graphical
For the following exercises, graph the pair of equations on the same axes, and state whether they are parallel, perpendicular, or neither.
$\begin{array}{l}\\ \begin{array}{l}y=2x+7\hfill \\ y=\frac{-1}{2}x-4\hfill \end{array}\end{array}$
$\begin{array}{l}3x-2y=5\hfill \\ 6y-9x=6\hfill \end{array}$
Parallel
$\begin{array}{l}y=\frac{3x+1}{4}\hfill \\ y=3x+2\hfill \end{array}$
$\begin{array}{l}x=4\\ y=-3\end{array}$
Perpendicular
## Numeric
For the following exercises, find the slope of the line that passes through the given points.
$\left(5,4\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(7,9\right)$
$\left(-3,2\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(4,-7\right)$
$m=\frac{-9}{7}$
$\left(-5,4\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(2,4\right)$
$\left(-1,-2\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(3,4\right)$
$m=\frac{3}{2}$
$\text{\hspace{0.17em}}\left(3,-2\right)$ and $\text{\hspace{0.17em}}\left(3,-2\right)$
For the following exercises, find the slope of the lines that pass through each pair of points and determine whether the lines are parallel or perpendicular.
${m}_{1}=\frac{-1}{3},\text{ }{m}_{2}=3;\text{ }\text{Perpendicular}\text{.}$
## Technology
For the following exercises, express the equations in slope intercept form (rounding each number to the thousandths place). Enter this into a graphing calculator as Y1, then adjust the ymin and ymax values for your window to include where the y -intercept occurs. State your ymin and ymax values.
$0.537x-2.19y=100$
$y=0.245x-45.662.\text{\hspace{0.17em}}$ Answers may vary.
$4,500x-200y=9,528$
$\frac{200-30y}{x}=70$
$y=-2.333x+6.667.\text{\hspace{0.17em}}$ Answers may vary.
## Extensions
Starting with the point-slope formula $\text{\hspace{0.17em}}y-{y}_{1}=m\left(x-{x}_{1}\right),$ solve this expression for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ in terms of $\text{\hspace{0.17em}}{x}_{1},y,{y}_{1},$ and $\text{\hspace{0.17em}}m.$
Starting with the standard form of an equation solve this expression for y in terms of $\text{\hspace{0.17em}}A,B,C,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ Then put the expression in slope-intercept form.
$y=\frac{-A}{B}x+\frac{C}{B}$
Use the above derived formula to put the following standard equation in slope intercept form: $\text{\hspace{0.17em}}7x-5y=25.$
Given that the following coordinates are the vertices of a rectangle, prove that this truly is a rectangle by showing the slopes of the sides that meet are perpendicular.
$\left(-1,1\right),\left(2,0\right),\left(3,3\right)\text{,}$ and $\text{\hspace{0.17em}}\left(0,4\right)$
Yes they are perpendicular.
Find the slopes of the diagonals in the previous exercise. Are they perpendicular?
## Real-world applications
The slope for a wheelchair ramp for a home has to be $\text{\hspace{0.17em}}\frac{1}{12}.\text{\hspace{0.17em}}$ If the vertical distance from the ground to the door bottom is 2.5 ft, find the distance the ramp has to extend from the home in order to comply with the needed slope.
30 ft
If the profit equation for a small business selling $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ number of item one and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ number of item two is $\text{\hspace{0.17em}}p=3x+4y,$ find the $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ value when
For the following exercises, use this scenario: The cost of renting a car is $45/wk plus$0.25/mi traveled during that week. An equation to represent the cost would be $\text{\hspace{0.17em}}y=45+.25x,$ where $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is the number of miles traveled.
What is your cost if you travel 50 mi?
$57.50 If your cost were $\text{\hspace{0.17em}}\text{}63.75,$ how many miles were you charged for traveling? Suppose you have a maximum of$100 to spend for the car rental. What would be the maximum number of miles you could travel?
220 mi
By the definition, is such that 0!=1.why?
(1+cosA+IsinA)(1+cosB+isinB)/(cos@+isin@)(cos$+isin$)
hatdog
Mark
how we can draw three triangles of distinctly different shapes. All the angles will be cutt off each triangle and placed side by side with vertices touching
bsc F. y algebra and trigonometry pepper 2
given that x= 3/5 find sin 3x
4
DB
remove any signs and collect terms of -2(8a-3b-c)
-16a+6b+2c
Will
Joeval
(x2-2x+8)-4(x2-3x+5)
sorry
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
(X2-2X+8)-4(X2-3X+5)=0 ?
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
Y
master
master
Soo sorry (5±Root11* i)/3
master
Mukhtar
2x²-6x+1=0
Ife
explain and give four example of hyperbolic function
What is the correct rational algebraic expression of the given "a fraction whose denominator is 10 more than the numerator y?
y/y+10
Mr
Find nth derivative of eax sin (bx + c).
Find area common to the parabola y2 = 4ax and x2 = 4ay.
Anurag
y2=4ax= y=4ax/2. y=2ax
akash
A rectangular garden is 25ft wide. if its area is 1125ft, what is the length of the garden
to find the length I divide the area by the wide wich means 1125ft/25ft=45
Miranda
thanks
Jhovie
What do you call a relation where each element in the domain is related to only one value in the range by some rules?
A banana.
Yaona
a function
Daniel
a function
emmanuel
given 4cot thither +3=0and 0°<thither <180° use a sketch to determine the value of the following a)cos thither
what are you up to?
nothing up todat yet
Miranda
hi
jai
hello
jai
Miranda Drice
jai
aap konsi country se ho
jai
which language is that
Miranda
I am living in india
jai
good
Miranda
what is the formula for calculating algebraic
I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
Miranda
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GRE Subject Test: Math : Tests for Convergence
Example Questions
Example Question #1 : Ratio Test
Which of these series cannot be tested for convergence/divergence properly using the ratio test? (Which of these series fails the ratio test?)
Explanation:
The ratio test fails when . Otherwise the series converges absolutely if , and diverges if .
Testing the series , we have
Hence the ratio test fails here. (It is likely obvious to the reader that this series diverges already. However, we must remember that all intuition in mathematics requires rigorous justification. We are attempting that here.)
Example Question #2 : Ratio Test
Assuming that , . Using the ratio test, what can we say about the series:
We cannot conclude when we use the ratio test.
It is convergent.
We cannot conclude when we use the ratio test.
Explanation:
As required by this question we will have to use the ratio test. if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series could either converge or diverge.
To do so, we will need to compute : . In our case:
Therefore
.
We know that
This means that
Since L=1 by the ratio test, we can't conclude about the convergence of the series.
Example Question #3 : Ratio Test
We consider the series : , use the ratio test to determine the type of convergence of the series.
The series is fast convergent.
It is clearly divergent.
We cannot conclude about the nature of the series.
We cannot conclude about the nature of the series.
Explanation:
To be able to use to conclude using the ratio test, we will need to first compute the ratio then use if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series could either converge or diverge. Computing the ratio we get,
.
We have then:
Therefore have :
It is clear that .
By the ratio test , we can't conclude about the nature of the series.
Example Question #5 : Ratio Test
Consider the following series :
where is given by:
. Using the ratio test, find the nature of the series.
We can't conclude when using the ratio test.
The series is convergent.
We can't conclude when using the ratio test.
Explanation:
Let be the general term of the series. We will use the ratio test to check the convergence of the series.
if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series could either converge or diverge.
We need to evaluate,
we have:
.
Therefore:
. We know that,
and therefore
This means that :
.
By the ratio test we can't conclude about the nature of the series. We will have to use another test.
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# Group Representation With Matrices
2020 Apr 26 · 6 min read
### Definition
A matrix $D(g)$ is said to be a representation for a group element $g$ iff. for any two group elements $g_1$, $g_2$ it holds
$D(g_1) \cdot D(g_2) = D(g_1 \odot g_2),$
where $\cdot$ denotes matrix multiplication and $\odot$ the composition of two group elements that has been defined on the group $(G, \odot)$.
### The Permutation Group
Let's do a brief refresher on the permutation group, specifically $S_5$ that permutes the five objects in $\{1, 2, ..., 5\}$. We can describe a group element of $S_5$ (a permutation) illustratively like
$\begin{matrix} 1 \rightarrow 4 & 4 \rightarrow 2 \\ 2 \rightarrow 1 & 5 \rightarrow 3 \\ 3 \rightarrow 5 & \end{matrix}$
or using Cauchy notation
$\begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 4 & 1 & 5 & 2 & 3 \end{pmatrix}$
and the more compact cycle notation
$(1 4 2) \odot (3 5)$
because we apply two cyclic permutations $1 \rightarrow 4 \rightarrow 2 \rightarrow 1$ (a 3-cycle) and $3 \rightarrow 5 \rightarrow 3$ (a 2-cycle / swap / transposition). From the above example we see that we can express permutations as products of cycles of variable lengths. Further we can cyclicly push around the numbers in cycle notation without changing the permutation they represent, e.g. $(142) = (214) = (421)$. However, be mindful that this is not possible with the bottom line of the Cauchy notation!
Cue on cycles: You can imagine an ordered row of numbered cards from one to five. When you start to move the cards in-place (meaning in the same row) to their new index and land at the position of the card you started with, you have found a cycle.
This example nicely illustrates the group operation of the permutation group $(S_5, \odot)$ which we will call "multiplication" from now on (but note that it is in fact a function composition). As opposed to multiplication with scalars, permutation multiplication where the number of objects $n \geq 2$ is not commutative proven by the fact that the outcome of a permutation depends on the order in which the objects are swapped, i.e. $(12) \odot (13) \neq (13) \odot (12)$.
### Representing Permutations With Matrices
We will now use $S_4$ that permutes the following four objects
$\bold{v}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \bold{v}_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \bold{v}_3 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}, \bold{v}_4 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}$.
so let's look at the following group element $g$ in $S_4$ where $g=(\bold{v}_3\bold{v}_1\bold{v}_4\bold{v}_2)$. (Visual aid: $\bold{v}_3 \rightarrow \bold{v}_1, \bold{v}_1 \rightarrow \bold{v}_4, \bold{v}_4 \rightarrow \bold{v}_2, \bold{v}_2 \rightarrow \bold{v}_3$)
For the sake of brevity we went back to referring to the objects using their indices, so let $g=(3142)$ and in Cauchy notation
$\begin{pmatrix} 1 & 2 & 3 & 4\\ 4 & 3 & 1 & 2 \end{pmatrix}$.
Note: Here, we can see nicely that the bottom row of the Cauchy notation can not be used to construct the cycle notation.
Now remembering back to the concept of permutation matrices in Linear Algebra, that swap out the columns of the matrices they are multiplied with, will help us understand the representation matrix in the next step.
So with the help of the indices of the bottom row of the Cauchy notation and using the respective unit vectors we obtain
$D(g) = D(3142) = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix}$.
When we now multiply $D(g)$ with the objects, i.e. $D(g) \cdot \bold{v}_4 = \bold{v}_2$ or $D(g) \cdot \bold{v}_3 = \bold{v}_1$, we see that the multiplication reflects the permutation action of $(3142)$ on $\bold{v}_4$ and $\bold{v}_3$ respectively.
Let's now try multiplying cycles using their matrix representations. Given
$D(24) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix},$
the following equality holds:
$D(3142) \cdot D(24) = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{pmatrix}.$
And we can successfully cross-check by multiplying our permutations
$(3142) \odot (24) = (3 1 4)$
and based on that construct
$D(314) = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{pmatrix}.$
Note that we have not actually proven the group representation for $S_4$, but only verified it for this specific instance.
Interestingly, we can even express and represent simple addition $(\R, +)$ using matrix multiplication. We simply need to find a representation matrix $D(\cdot)$, such that $D(a) \cdot D(b) = D(a+b)$ where $a, b$ are scalars in $\R$. If we ponder about it for a second (or longer), we find that
$D(a) = \begin{pmatrix} 1 & 0 \\ a & 1 \end{pmatrix}$
does the job, since
$D(a) \cdot D(b) = \begin{pmatrix} 1 & 0 \\ a+b & 1 \end{pmatrix} = D(a+b).$
Lastly, one should always bear in mind the trivial representation that holds for all elements in a group with $D(g) = 1$ and we can see that it always fulfills the definition of a representation.
### Takeaways
1. Don't confuse the group element in the $S_n$ group with the actual objects that are being permuted.
2. The Cauchy notation has nothing to do with the cycle notation.
3. The operation in the $S_n$ group is not commutative. Neither is matrix multiplication.
4. Diferent representations of the same group can have varying dimensions.
5. From time to time researchers actually denote the matrix representation as the group elements. This caused the biggest confusion for me which led me to brushing up some fundamentals again. (And maybe that is their intention anyway)
In a different universe this topic would have maybe been more intuitive for me. But it wasn't, which is why I ended up writing this post thanks to Anthony Zee doing Group Theory in nutshells.
Personal blog by Stefan Su Feel free to reach out 👈
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# 3.4 Composition of functions (Page 6/9)
Page 6 / 9
## Finding the domain of a composite function involving radicals
Find the domain of
Because we cannot take the square root of a negative number, the domain of $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}\left(-\infty ,3\right].\text{\hspace{0.17em}}$ Now we check the domain of the composite function
The domain of this function is $\text{\hspace{0.17em}}\left(-\infty ,5\right].\text{\hspace{0.17em}}$ To find the domain of $\text{\hspace{0.17em}}f\circ g,\text{\hspace{0.17em}}$ we ask ourselves if there are any further restrictions offered by the domain of the composite function. The answer is no, since $\text{\hspace{0.17em}}\left(-\infty ,3\right]\text{\hspace{0.17em}}$ is a proper subset of the domain of $\text{\hspace{0.17em}}f\circ g.\text{\hspace{0.17em}}$ This means the domain of $\text{\hspace{0.17em}}f\circ g\text{\hspace{0.17em}}$ is the same as the domain of $\text{\hspace{0.17em}}g,\text{\hspace{0.17em}}$ namely, $\text{\hspace{0.17em}}\left(-\infty ,3\right].$
Find the domain of
$\left[-4,0\right)\cup \left(0,\infty \right)$
## Decomposing a composite function into its component functions
In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions. There may be more than one way to decompose a composite function , so we may choose the decomposition that appears to be most expedient.
## Decomposing a function
Write $\text{\hspace{0.17em}}f\left(x\right)=\sqrt{5-{x}^{2}}\text{\hspace{0.17em}}$ as the composition of two functions.
We are looking for two functions, $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h,\text{\hspace{0.17em}}$ so $\text{\hspace{0.17em}}f\left(x\right)=g\left(h\left(x\right)\right).\text{\hspace{0.17em}}$ To do this, we look for a function inside a function in the formula for $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$ As one possibility, we might notice that the expression $\text{\hspace{0.17em}}5-{x}^{2}\text{\hspace{0.17em}}$ is the inside of the square root. We could then decompose the function as
We can check our answer by recomposing the functions.
$g\left(h\left(x\right)\right)=g\left(5-{x}^{2}\right)=\sqrt{5-{x}^{2}}$
Write $\text{\hspace{0.17em}}f\left(x\right)=\frac{4}{3-\sqrt{4+{x}^{2}}}\text{\hspace{0.17em}}$ as the composition of two functions.
$\begin{array}{l}g\left(x\right)=\sqrt{4+{x}^{2}}\\ h\left(x\right)=\frac{4}{3-x}\\ f=h\circ g\end{array}$
Access these online resources for additional instruction and practice with composite functions.
## Key equation
Composite function $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$
## Key concepts
• We can perform algebraic operations on functions. See [link] .
• When functions are combined, the output of the first (inner) function becomes the input of the second (outer) function.
• The function produced by combining two functions is a composite function. See [link] and [link] .
• The order of function composition must be considered when interpreting the meaning of composite functions. See [link] .
• A composite function can be evaluated by evaluating the inner function using the given input value and then evaluating the outer function taking as its input the output of the inner function.
• A composite function can be evaluated from a table. See [link] .
• A composite function can be evaluated from a graph. See [link] .
• A composite function can be evaluated from a formula. See [link] .
• The domain of a composite function consists of those inputs in the domain of the inner function that correspond to outputs of the inner function that are in the domain of the outer function. See [link] and [link] .
• Just as functions can be combined to form a composite function, composite functions can be decomposed into simpler functions.
• Functions can often be decomposed in more than one way. See [link] .
## Verbal
How does one find the domain of the quotient of two functions, $\text{\hspace{0.17em}}\frac{f}{g}?\text{\hspace{0.17em}}$
Find the numbers that make the function in the denominator $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ equal to zero, and check for any other domain restrictions on $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g,\text{\hspace{0.17em}}$ such as an even-indexed root or zeros in the denominator.
prove that [a+b, b+c, c+a]= 2[a b c]
can't prove
Akugry
i can prove [a+b+b+c+c+a]=2[a+b+c]
this is simple
Akugry
hi
Stormzy
x exposant 4 + 4 x exposant 3 + 8 exposant 2 + 4 x + 1 = 0
x exposent4+4x exposent3+8x exposent2+4x+1=0
HERVE
How can I solve for a domain and a codomains in a given function?
ranges
EDWIN
Thank you I mean range sir.
Oliver
proof for set theory
don't you know?
Inkoom
find to nearest one decimal place of centimeter the length of an arc of circle of radius length 12.5cm and subtending of centeral angle 1.6rad
factoring polynomial
find general solution of the Tanx=-1/root3,secx=2/root3
find general solution of the following equation
Nani
the value of 2 sin square 60 Cos 60
0.75
Lynne
0.75
Inkoom
when can I use sin, cos tan in a giving question
depending on the question
Nicholas
I am a carpenter and I have to cut and assemble a conventional roof line for a new home. The dimensions are: width 30'6" length 40'6". I want a 6 and 12 pitch. The roof is a full hip construction. Give me the L,W and height of rafters for the hip, hip jacks also the length of common jacks.
John
I want to learn the calculations
where can I get indices
I need matrices
Nasasira
hi
Raihany
Hi
Solomon
need help
Raihany
maybe provide us videos
Nasasira
Raihany
Hello
Cromwell
a
Amie
What do you mean by a
Cromwell
nothing. I accidentally press it
Amie
you guys know any app with matrices?
Khay
Ok
Cromwell
Solve the x? x=18+(24-3)=72
x-39=72 x=111
Suraj
Solve the formula for the indicated variable P=b+4a+2c, for b
Need help with this question please
b=-4ac-2c+P
Denisse
b=p-4a-2c
Suddhen
b= p - 4a - 2c
Snr
p=2(2a+C)+b
Suraj
b=p-2(2a+c)
Tapiwa
P=4a+b+2C
COLEMAN
b=P-4a-2c
COLEMAN
like Deadra, show me the step by step order of operation to alive for b
John
A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5) and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
|
# Difference between revisions of "2018 AMC 12A Problems/Problem 20"
## Problem
Triangle $ABC$ is an isosceles right triangle with $AB=AC=3$. Let $M$ be the midpoint of hypotenuse $\overline{BC}$. Points $I$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$, respectively, so that $AI>AE$ and $AIME$ is a cyclic quadrilateral. Given that triangle $EMI$ has area $2$, the length $CI$ can be written as $\frac{a-\sqrt{b}}{c}$, where $a$, $b$, and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is the value of $a+b+c$?
$\textbf{(A) }9 \qquad \textbf{(B) }10 \qquad \textbf{(C) }11 \qquad \textbf{(D) }12 \qquad \textbf{(E) }13 \qquad$
## Solution 1
Observe that $\triangle{EMI}$ is isosceles right ($M$ is the midpoint of diameter arc $EI$), so $MI=2,MC=\frac{3}{\sqrt{2}}$. With $\angle{MCI}=45^\circ$, we can use Law of Cosines to determine that $CI=\frac{3\pm\sqrt{7}}{2}$. The same calculations hold for $BE$ also, and since $CI, we deduce that $CI$ is the smaller root, giving the answer of $\boxed{12}$. (trumpeter)
## Solution 2 (Using Ptolemy)
We first claim that $\triangle{EMI}$ is isosceles and right.
Proof: Construct $\overline{MF}\perp\overline{AB}$ and $\overline{MG}\perp\overline{AC}$. Since $\overline{AM}$ bisects $\angle{BAC}$, one can deduce that $MF=MG$. Then by AAS it is clear that $MI=ME$ and therefore $\triangle{EMI}$ is isosceles. Since quadrilateral $AIME$ is cyclic, one can deduce that $\angle{EMI}=90^\circ$. Q.E.D.
Since the area of $\triangle{EMI}$ is 2, we can find that $MI=ME=2$, $EI=2\sqrt{2}$
Since $M$ is the mid-point of $\overline{BC}$, it is clear that $AM=\frac{3\sqrt{2}}{2}$.
Now let $AE=a$ and $AI=b$. By Ptolemy's Theorem, in cyclic quadrilateral $AIME$, we have $2a+2b=6$. By Pythagorean Theorem, we have $a^2+b^2=8$. One can solve the simultaneous system and find $b=\frac{3+\sqrt{7}}{2}$. Then by deducting the length of $\overline{AI}$ from 3 we get $CI=\frac{3-\sqrt{7}}{2}$, giving the answer of $\boxed{12}$. (Surefire2019)
## Solution 3 (More Elementary)
Like above, notice that $\triangle{EMI}$ is isosceles and right, which means that $\dfrac{ME \cdot MI}{2} = 2$, so $MI^2=4$ and $MI = 2$. Then construct $\overline{MF}\perp\overline{AB}$ and $\overline{MG}\perp\overline{AC}$ as well as $\overline{MI}$. It's clear that $MG^2+GI^2 = MI^2$ by Pythagorean, so knowing that $MG = \dfrac{AB}{2} = \dfrac{3}{2}$ allows one to solve to get $GI = \dfrac{\sqrt{7}}{2}$. By just looking at the diagram, $CI=AC-MF-GI=\dfrac{3-\sqrt{7}}{2}$. The answer is thus $3+7+2=12$.
## Solution 4 (Coordinate Geometry)
Let $A$ lie on $(0,0)$, $E$ on $(0,y)$, $I$ on $(x,0)$, and $M$ on $(\frac{3}{2},\frac{3}{2})$. ${AIME}$ is cyclic, so $\angle EMI$ is a right angle; therefore $\vec{ME} \cdot \vec{MI} = <\frac{-3}{2}, y-\frac{3}{2}> \cdot = 0$. Multiply out and simplify to arrive at the relation $y=3-x$. We can set up another equation for the area of $\triangle EMI$ using the Shoelace Theorem. This is $2=(\frac{1}{2})[(\frac{3}{2})(y-\frac{3}{2})+(x)(-y)+(x+\frac{3}{2})(\frac{3}{2})]$. Multiplying, substituting $3-x$ for $y$ and simplifying, we arrive at $x^2 -3x + \frac{1}{2}=0$. Thus, the solution set is $(x,y)=$ $(\frac{3 \pm \sqrt{7}}{2},\frac{3 \mp \sqrt{7}}{2})$. But $AI>AE$, meaning $x=AI=\frac{3 + \sqrt{7}}{2} \rightarrow CI = 3-\frac{3 + \sqrt{7}}{2}=\frac{3 - \sqrt{7}}{2}$, and the final answer is $3+7+2=\boxed{12}$.
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# Selection Sort Time Complexity
Let’s look at the time complexity of the selection sort algorithm, just so we can get a feel for how much time this operation takes.
First, we must determine if there is a worst-case input for selection sort. Can we think of any particular input which would require more steps to complete?
In this case, each iteration of selection sort will look at the same number of elements, no matter what they are. So there isn’t a particular input that would be considered worst-case. We can proceed with just the general case.
In each iteration of the algorithm we need to search for the minimum value of the remaining elements in the container. If the container has $N$ elements, we would follow the steps below.
1. The first time we need to find the minimum among $N$ elements. This will take $N$ comparisons assuming the element are compared one by one with the minimum.
2. The second time we need to find the minimum among $N - 1$ elements. This will take $N - 1$ comparisons.
3. The third time we need to find the minimum among $N - 2$ elements. This will take $N - 2$ comparisons.
4. … and so on.
This process continues until we have sorted all of the elements in the array. The number of steps will be:
$$N + (N – 1) + (N – 2) + … + 2 + 1$$
While it takes a bit of math to figure out exactly what that means, we can use some intuition to determine an approximate value. For example we could pair up the values like this:
$$N + [(N – 1) + 1] + [(N – 2) + 2] + ...$$
When we do that, we’ll see that we can create around $N / 2$ pairs, each one with the value of $N$. So a rough approximation of this value is $N * (N / 2)$, which is $N^2 / 2$. When analyzing time complexity, we would say that this is “on the order of $N^2$” time. Put another way, if the size of $N$ doubles, we would expect the number of steps to go up by a factor of $4$, since $(2 * N)^2 = 4N$.
Later on, we’ll come back to this and compare the time complexity of each sorting algorithm and searching algorithm to see how they stack up against each other.
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8.1
### 17Functions as Data
It’s interesting to consider how expressive the little programming we’ve learned so far can be. To illustrate this, we’ll work through a few exercises of interesting concepts we can express using just functions as values. We’ll write two quite different things, then show how they converge nicely.
#### 17.1A Little Calculus
If you’ve studied the differential calculus, you’ve come across curious sytactic statements such as this:
\begin{equation*}{\frac{d}{dx}} x^2 = 2x\end{equation*}
Let’s unpack what this means: the $$d/dx$$, the $$x^2$$, and the $$2x$$.
First, let’s take on the two expressions; we’ll discuss one, and the discussion will cover the other as well. The correct response to “what does $$x^2$$ mean?” is, of course, an error: it doesn’t mean anything, because $$x$$ is an unbound identifier.
So what is it intended to mean? The intent, clearly, is to represent the function that squares its input, just as $$2x$$ is meant to be the function that doubles its input. We have nicer ways of writing those:
fun square(x :: Number) -> Number: x * x end
fun double(x :: Number) -> Number: 2 * x end
and what we’re really trying to say is that the $$d/dx$$ (whatever that is) of square is double.We’re assuming functions of arity one in the variable that is changing.
So now let’s unpack $$d/dx$$, starting with its type. As the above example illustrates, $$d/dx$$ is really a function from functions to functions. That is, we can write its type as follows:
d-dx :: ((Number -> Number) -> (Number -> Number))
(This type might explain why your calculus course never explained this operation this way—though it’s not clear that obscuring its true meaning is any better for your understanding.)
Let us now implement d-dx. We’ll implement numerical differentiation, though in principle we could also implement symbolic differentiation—using rules you learned, e.g., given a polynomial, multiply by the exponent and reduce the exponent by one—with a representation of expressions (a problem that will be covered in more detail in a future release).
In general, numeric differentiation of a function at a point yields the value of the derivative at that point. We have a handy formula for it: the derivative of $$f$$ at $$x$$ is
\begin{equation*}\frac{f(x + \epsilon) - f(x)}{\epsilon}\end{equation*}
as $$\epsilon$$ goes to zero in the limit. For now we’ll give the infinitesimal a small but fixed value, and later [Combining Forces: Streams of Derivatives] see how we can improve on this.
epsilon = 0.001
Let’s now try to translate the above formula into Pyret:
fun d-dx(f :: (Number -> Number)) -> (Number -> Number):
(f(x + epsilon) - f(x)) / epsilon
end
Do Now!
What’s the problem with the above definition?
If you didn’t notice, Pyret will soon tell you: x isn’t bound. Indeed, what is x? It’s the point at which we’re trying to compute the numeric derivative. That is, d-dx needs to return not a number but a function (as the type indicates) that will consume this x:“Lambdas are relegated to relative obscurity until Java makes them popular by not having them.”—James Iry, A Brief, Incomplete, and Mostly Wrong History of Programming Languages
fun d-dx(f :: (Number -> Number)) -> (Number -> Number):
lam(x :: Number) -> Number:
(f(x + epsilon) - f(x)) / epsilon
end
end
Sure enough, this definition now works. We can, for instance, test it as follows (note the use of num-floor to avoid numeric precision issues from making our tests appear to fail):
d-dx-square = d-dx(square)
check:
ins = [list: 0, 1, 10, 100]
for map(n from ins):
num-floor(d-dx-square(n))
end
is
for map(n from ins):
num-floor(double(n))
end
end
Now we can return to the original example that launched this investigation: what the sloppy and mysterious notation of math is really trying to say is,
d-dx(lam(x): x * x end) = lam(x): 2 * x end
or, in the notation of A Notation for Functions,
\begin{equation*}{\frac{d}{dx}} [x \rightarrow x^2] = [x \rightarrow 2x]\end{equation*}
Pity math textbooks for not wanting to tell us the truth!
#### 17.2A Helpful Shorthand for Anonymous Functions
Pyret offers a shorter syntax for writing anonymous functions. Though, stylistically, we generally avoid it so that our programs don’t become a jumble of special characters, sometimes it’s particularly convenient, as we will see below. This syntax is
{(a): b}
where a is zero or more arguments and b is the body. For instance, we can write lam(x): x * x end as
{(x): x * x}
where we can see the benefit of brevity. In particular, note that there is no need for end, because the braces take the place of showing where the expression begins and ends. Similarly, we could have written d-dx as
fun d-dx-short(f):
{(x): (f(x + epsilon) - f(x)) / epsilon}
end
but many readers would say this makes the function harder to read, because the prominent lam makes clear that d-dx returns an (anonymous) function, whereas this syntax obscures it. Therefore, we will usually only use this shorthand syntax for “one-liners”.
#### 17.3Streams From Functions
People typically think of a function as serving one purpose: to parameterize an expression. While that is both true and the most common use of a function, it does not justify having a function of no arguments, because that clearly parameterizes over nothing at all. Yet functions of no argument also have a use, because functions actually serve two purposes: to parameterize, and to suspend evaluation of the body until the function is applied. In fact, these two uses are orthogonal, in that one can employ one feature without the other. Below, we will focus on delay without abstraction (the other shows up in other computer science settings).
Let’s consider the humble list. A list can be only finitely long. However, there are many lists (or sequences) in nature that have no natural upper bound: from mathematical objects (the sequence of natural numbers) to natural ones (the sequence of hits to a Web site). Rather than try to squeeze these unbounded lists into bounded ones, let’s look at how we might represent and program over these unbounded lists.
First, let’s write a program to compute the sequence of natural numbers:
fun nats-from(n):
end
Do Now!
Does this program have a problem?
While this represents our intent, it doesn’t work: running it—e.g., nats-from(0)creates an infinite loop evaluating nats-from for every subsequent natural number. In other words, we want to write something very like the above, but that doesn’t recur until we want it to, i.e., on demand. In other words, we want the rest of the list to be lazy.
This is where our insight into functions comes in. A function, as we have just noted, delays evaluation of its body until it is applied. Therefore, a function would, in principle, defer the invocation of nats-from(n + 1) until it’s needed.
Except, this creates a type problem: the second argument to link needs to be a list, and cannot be a function. Indeed, because it must be a list, and every value that has been constructed must be finite, every list is finite and eventually terminates in empty. Therefore, we need a new data structure to represent the links in these lazy lists (also known as streams):
<stream-type-def> ::=
data Stream<T>:
| lz-link(h :: T, t :: ( -> Stream<T>))
end
where the annotation ( -> Stream<T>) means a function from no arguments (hence the lack of anything before ->), also known as a thunk. Note that the way we have defined streams they must be infinite, since we have provided no way to terminate them.
Let’s construct the simplest example we can, a stream of constant values:
ones = lz-link(1, lam(): ones end)
Pyret will actually complain about this definition. Note that the list equivalent of this also will not work:
ones = link(1, ones)
because ones is not defined at the point of definition, so when Pyret evaluates link(1, ones), it complains that ones is not defined. However, it is being overly conservative with our former definition: the use of ones is “under a lam”, and hence won’t be needed until after the definition of ones is done, at which point ones will be defined. We can indicate this to Pyret by using the keyword rec:
rec ones = lz-link(1, lam(): ones end)
Note that in Pyret, every fun implicitly has a rec beneath it, which is why we can create recursive functions with aplomb.
Exercise
Earlier we said that we can’t write
ones = link(1, ones)
What if we tried to write
rec ones = link(1, ones)
instead? Does this work and, if so, what value is ones bound to? If it doesn’t work, does it fail to work for the same reason as the definition without the rec?
Henceforth, we will use the shorthand [A Helpful Shorthand for Anonymous Functions] instead. Therefore, we can rewrite the above definition as:
rec ones = lz-link(1, {(): ones})
Notice that {(): …} defines an anonymous function of no arguments. You can’t leave out the ()! If you do, Pyret will get confused about what your program means.
Because functions are automatically recursive, when we write a function to create a stream, we don’t need to use rec. Consider this example:
fun nats-from(n :: Number):
end
with which we can define the natural numbers:
nats = nats-from(0)
Note that the definition of nats is not recursive itself—the recursion is inside nats-fromso we don’t need to use rec to define nats.
Do Now!
Earlier, we said that every list is finite and hence eventually terminates. How does this remark apply to streams, such as the definition of ones or nats above?
The description of ones is still a finite one; it simply represents the potential for an infinite number of values. Note that:
1. A similar reasoning doesn’t apply to lists because the rest of the list has already been constructed; in contrast, placing a function there creates the potential for a potentially unbounded amount of computation to still be forthcoming.
2. That said, even with streams, in any given computation, we will create only a finite prefix of the stream. However, we don’t have to prematurely decide how many; each client and use is welcome to extract less or more, as needed.
Now we’ve created multiple streams, but we still don’t have an easy way to “see” one. First we’ll define the traditional list-like selectors. Getting the first element works exactly as with lists:
fun lz-first<T>(s :: Stream<T>) -> T: s.h end
In contrast, when trying to access the rest of the stream, all we get out of the data structure is a thunk. To access the actual rest, we need to force the thunk, which of course means applying it to no arguments:
fun lz-rest<T>(s :: Stream<T>) -> Stream<T>: s.t() end
This is useful for examining individual values of the stream. It is also useful to extract a finite prefix of it (of a given size) as a (regular) list, which would be especially handy for testing. Let’s write that function:
fun take<T>(n :: Number, s :: Stream<T>) -> List<T>:
if n == 0:
empty
else:
end
end
If you pay close attention, you’ll find that this body is not defined by cases over the structure of the (stream) inputinstead, it’s defined by the cases of the definition of a natural number (zero or a successor). We’ll return to this below (<lz-map2-def>).
Now that we have this, we can use it for testing. Note that usually we use our data to test our functions; here, we’re using this function to test our data:
check:
take(10, ones) is map(lam(_): 1 end, range(0, 10))
take(10, nats) is range(0, 10)
take(10, nats-from(1)) is map((_ + 1), range(0, 10))
end
The notation (_ + 1) defines a Pyret function of one argument that adds 1 to the given argument.
Let’s define one more function: the equivalent of map over streams. For reasons that will soon become obvious, we’ll define a version that takes two lists and applies the first argument to them pointwise:
<lz-map2-def> ::=
fun lz-map2<A, B, C>(
f :: (A, B -> C),
s1 :: Stream<A>,
s2 :: Stream<B>) -> Stream<C>:
f(lz-first(s1), lz-first(s2)),
{(): lz-map2(f, lz-rest(s1), lz-rest(s2))})
end
Now we can see our earlier remark about the structure of the function driven home especially clearly. Whereas a traditional map over lists would have two cases, here we have only one case because the data definition (<stream-type-def>) has only one case! What is the consequence of this? In a traditional map, one case looks like the above, but the other case corresponds to the empty input, for which it produces the same output. Here, because the stream never terminates, mapping over it doesn’t either, and the structure of the function reflects this.This raises a much subtler problem: if the function’s body doesn’t have base- and inductive-cases, how can we perform an inductive proof over it? The short answer is we can’t: we must instead use coinduction.
Why did we define lz-map2 instead of lz-map? Because it enables us to write the following:
rec fibs =
{(): lz-map2({(a :: Number, b :: Number): a + b},
fibs,
lz-rest(fibs))})})
from which, of course, we can extract as many Fibonacci numbers as we want!
check:
take(10, fibs) is [0, 1, 1, 2, 3, 5, 8, 13, 21, 34]
end
Exercise
Define the equivalent of map, filter, and fold for streams.
Streams and, more generally, infinite data structures that unfold on demand are extremely valuable in programming. Consider, for instance, the possible moves in a game. In some games, this can be infinite; even if it is finite, for interesting games the combinatorics mean that the tree is too large to feasibly store in memory. Therefore, the programmer of the computer’s intelligence must unfold the game tree on demand. Programming it by using the encoding we have described above means the program describes the entire tree, lazily, and the tree unfolds automatically on demand, relieving the programmer of the burden of implementing such a strategy.
In some languages, such as Haskell, lazy evaluation is built in by default. In such a language, there is no need to use thunks. However, lazy evaluation places other burdens on the language, which you can learn about in a programming-languages class.
#### 17.4Combining Forces: Streams of Derivatives
When we defined d-dx, we set epsilon to an arbitrary, high value. We could instead think of epsilon as itself a stream that produces successively finer values; then, for instance, when the difference in the value of the derivative becomes small enough, we can decide we have a sufficient approximation to the derivative.
The first step is, therefore, to make epsilon some kind of parameter rather than a global constant. That leaves open what kind of parameter it should be (number or stream?) as well as when it should be supplied.
It makes most sense to consume this parameter after we have decided what function we want to differentiate and at what value we want its derivative; after all, the stream of epsilon values may depend on both. Thus, we get:
fun d-dx(f :: (Number -> Number)) ->
(Number -> (Number -> Number)):
lam(x :: Number) -> (Number -> Number):
lam(epsilon :: Number) -> Number:
(f(x + epsilon) - f(x)) / epsilon
end
end
end
with which we can return to our square example:
d-dx-square = d-dx(square)
Note that at this point we have simply redefined d-dx without any reference to streams: we have merely made a constant into a parameter.
Now let’s define the stream of negative powers of ten:
tenths = block:
fun by-ten(d):
new-denom = d / 10
end
by-ten(1)
end
so that
check:
take(3, tenths) is [list: 1/10, 1/100, 1/1000]
end
For concreteness, let’s pick an abscissa at which to compute the numeric derivative of squaresay 10:
d-dx-square-at-10 = d-dx-square(10)
Recall, from the types, that this is now a function of type (Number -> Number): given a value for epsilon, it computes the derivative using that value. We know, analytically, that the value of this derivative should be 20. We can now (lazily) map tenths to provide increasingly better approximations for epsilon and see what happens:
lz-map(d-dx-square-at-10, tenths)
Sure enough, the values we obtain are 20.1, 20.01, 20.001, and so on: progressively better numerical approximations to 20.
Exercise
Extend the above program to take a tolerance, and draw as many values from the epsilon stream as necessary until the difference between successive approximations of the derivative fall within this tolerance.
### 18Predicting Growth
We will now commence the study of determining how long a computation takes. We’ll begin with a little (true) story.
#### 18.1A Little (True) Story
My student Debbie recently wrote tools to analyze data for a startup. The company collects information about product scans made on mobile phones, and Debbie’s analytic tools classified these by product, by region, by time, and so on. As a good programmer, Debbie first wrote synthetic test cases, then developed her programs and tested them. She then obtained some actual test data from the company, broke them down into small chunks, computed the expected answers by hand, and tested her programs again against these real (but small) data sets. At the end of this she was ready to declare the programs ready.
At this point, however, she had only tested them for functional correctness. There was still a question of how quickly her analytical tools would produce answers. This presented two problems:
• The company was rightly reluctant to share the entire dataset with outsiders, and in turn we didn’t want to be responsible for carefully guarding all their data.
• Even if we did get a sample of their data, as more users used their product, the amount of data they had was sure to grow.
We therefore got only a sampling of their full data, and from this had to make some prediction on how long it would take to run the analytics on subsets (e.g., those corresponding to just one region) or all of their data set, both today and as it grew over time.
Debbie was given 100,000 data points. She broke them down into input sets of 10, 100, 1,000, 10,000, and 100,000 data points, ran her tools on each input size, and plotted the result.
From this graph we have a good bet at guessing how long the tool would take on a dataset of 50,000. It’s much harder, however, to be sure how long it would take on datasets of size 1.5 million or 3 million or 10 million.These processes are respectively called interpolation and extrapolation. We’ve already explained why we couldn’t get more data from the company. So what could we do?
As another problem, suppose we have multiple implementations available. When we plot their running time, say the graphs look like this, with red, green, and blue each representing different implementations. On small inputs, suppose the running times look like this:
This doesn’t seem to help us distinguish between the implementations. Now suppose we run the algorithms on larger inputs, and we get the following graphs:
Now we seem to have a clear winner (red), though it’s not clear there is much to give between the other two (blue and green). But if we calculate on even larger inputs, we start to see dramatic differences:
In fact, the functions that resulted in these lines were the same in all three figures. What these pictures tell us is that it is dangerous to extrapolate too much from the performance on small inputs. If we could obtain closed-form descriptions of the performance of computations, it would be nice if we could compare them better. That is what we will do in the next section.
Responsible Computing: Choose Analysis Artifacts Wisely
As more and more decisions are guided by statistical analyses of data (performed by humans), it’s critical to recognize that data can be a poor proxy for the actual phenomenon that we seek to understand. Here, Debbie had data about program behavior, which led to mis-interpretations regarding which program is best. But Debbie also had the programs themselves, from which the data were generated. Analyzing the programs, rather than the data, is a more direct approach to assessing the performance of a program.
While the rest of this chapter is about analyzing programs as written in code, this point carries over to non-programs as well. You might want to understand the effectiveness of a process for triaging patients at a hospital, for example. In that case, you have both the policy documents (rules which may or may not have been turned into a software program to support managing patients) and data on the effectiveness of using that process. Responsible computing tells us to analyze both the process and its behavioral data, against knowledge about best practices in patient care, to evaluate the effectiveness of systems.
#### 18.2The Analytical Idea
With many physical processes, the best we can do is obtain as many data points as possible, extrapolate, and apply statistics to reason about the most likely outcome. Sometimes we can do that in computer science, too, but fortunately we computer scientists have an enormous advantage over most other sciences: instead of measuring a black-box process, we have full access to its internals, namely the source code. This enables us to apply analytical methods.“Analytical” means applying algebraic and other mathematical methods to make predictive statements about a process without running it. The answer we compute this way is complementary to what we obtain from the above experimental analysis, and in practice we will usually want to use a combination of the two to arrive a strong understanding of the program’s behavior.
The analytical idea is startlingly simple. We look at the source of the program and list the operations it performs. For each operation, we look up what it costs.We are going to focus on one kind of cost, namely running time. There are many other other kinds of costs one can compute. We might naturally be interested in space (memory) consumed, which tells us how big a machine we need to buy. We might also care about power, this tells us the cost of our energy bills, or of bandwidth, which tells us what kind of Internet connection we will need. In general, then, we’re interested in resource consumption. In short, don’t make the mistake of equating “performance” with “speed”: the costs that matter depend on the context in which the application runs. We add up these costs for all the operations. This gives us a total cost for the program.
Naturally, for most programs the answer will not be a constant number. Rather, it will depend on factors such as the size of the input. Therefore, our answer is likely to be an expression in terms of parameters (such as the input’s size). In other words, our answer will be a function.
There are many functions that can describe the running-time of a function. Often we want an upper bound on the running time: i.e., the actual number of operations will always be no more than what the function predicts. This tells us the maximum resource we will need to allocate. Another function may present a lower bound, which tells us the least resource we need. Sometimes we want an average-case analysis. And so on. In this text we will focus on upper-bounds, but keep in mind that all these other analyses are also extremely valuable.
Exercise
It is incorrect to speak of “the” upper-bound function, because there isn’t just one. Given one upper-bound function, can you construct another one?
#### 18.3A Cost Model for Pyret Running Time
We begin by presenting a cost model for the running time of Pyret programs. We are interested in the cost of running a program, which is tantamount to studying the expressions of a program. Simply making a definition does not cost anything; the cost is incurred only when we use a definition.
We will use a very simple (but sufficiently accurate) cost model: every operation costs one unit of time in addition to the time needed to evaluate its sub-expressions. Thus it takes one unit of time to look up a variable or to allocate a constant. Applying primitive functions also costs one unit of time. Everything else is a compound expression with sub-expressions. The cost of a compound expression is one plus that of each of its sub-expressions. For instance, the running time cost of the expression e1 + e2 (for some sub-expressions e1 and e2) is the running time for e1 + the running time for e2 + 1. Thus the expression 17 + 29 has a cost of 3 (one for each sub-expression and one for the addition); the expression 1 + (7 * (2 / 9)) costs 7.
As you can see, there are two big approximations here:
• First, we are using an abstract rather than concrete notion of time. This is unhelpful in terms of estimating the so-called “wall clock” running time of a program, but then again, that number depends on numerous factors—not just what kind of processor and how much memory you have, but even what other tasks are running on your computer at the same time. In contrast, abstract time units are more portable.
• Second, not every operation takes the same number of machine cycles, whereas we have charged all of them the same number of abstract time units. As long as the actual number of cycles each one takes is bounded by a constant factor of the number taken by another, this will not pose any mathematical problems for reasons we will soon understand [Comparing Functions].
Of course, it is instructive—after carefully settting up the experimental conditions—to make an analytical prediction of a program’s behavior and then verify it against what the implementation actually does. If the analytical prediction is accurate, we can reconstruct the constant factors hidden in our calculations and thus obtain very precise wall-clock time bounds for the program.
#### 18.4The Size of the Input
We gloss over the size of a number, treating it as constant. Observe that the value of a number is exponentially larger than its size: $$n$$ digits in base $$b$$ can represent $$b^n$$ numbers. Though irrelevant here, when numbers are central—e.g., when testing primality—the difference becomes critical! We will return to this briefly later [The Complexity of Numbers].
It can be subtle to define the size of the argument. Suppose a function consumes a list of numbers; it would be natural to define the size of its argument to be the length of the list, i.e., the number of links in the list. We could also define it to be twice as large, to account for both the links and the individual numbers (but as we’ll see [Comparing Functions], constants usually don’t matter). But suppose a function consumes a list of music albums, and each music album is itself a list of songs, each of which has information about singers and so on. Then how we measure the size depends on what part of the input the function being analyzed actually examines. If, say, it only returns the length of the list of albums, then it is indifferent to what each list element contains [Monomorphic Lists and Polymorphic Types], and only the length of the list of albums matters. If, however, the function returns a list of all the singers on every album, then it traverses all the way down to individual songs, and we have to account for all these data. In short, we care about the size of the data potentially accessed by the function.
#### 18.5The Tabular Method for Singly-Structurally-Recursive Functions
Given sizes for the arguments, we simply examine the body of the function and add up the costs of the individual operations. Most interesting functions are, however, conditionally defined, and may even recur. Here we will assume there is only one structural recursive call. We will get to more general cases in a bit [Creating Recurrences].
When we have a function with only one recursive call, and it’s structural, there’s a handy technique we can use to handle conditionals.This idea is due to Prabhakar Ragde. We will set up a table. It won’t surprise you to hear that the table will have as many rows as the cond has clauses. But instead of two columns, it has seven! This sounds daunting, but you’ll soon see where they come from and why they’re there.
For each row, fill in the columns as follows:
1. |Q|: the number of operations in the question
2. #Q: the number of times the question will execute
3. TotQ: the total cost of the question (multiply the previous two)
4. |A|: the number of operations in the answer
5. #A: the number of times the answer will execute
6. TotA: the total cost of the answer (multiply the previous two)
7. Total: add the two totals to obtain an answer for the clause
Finally, the total cost of the cond expression is obtained by summing the Total column in the individual rows.
In the process of computing these costs, we may come across recursive calls in an answer expression. So long as there is only one recursive call in the entire answer, ignore it.
Exercise
Once you’ve read the material on Creating Recurrences, come back to this and justify why it is okay to just skip the recursive call. Explain in the context of the overall tabular method.
Exercise
Excluding the treatment of recursion, justify (a) that these columns are individually accurate (e.g., the use of additions and multiplications is appropriate), and (b) sufficient (i.e., combined, they account for all operations that will be performed by that cond clause).
It’s easiest to understand this by applying it to a few examples. First, let’s consider the len function, noting before we proceed that it does meet the criterion of having a single recursive call where the argument is structural:
fun len(l):
cases (List) l:
| empty => 0
| link(f, r) => 1 + len(r)
end
end
Let’s compute the cost of running len on a list of length $$k$$ (where we are only counting the number of links in the list, and ignoring the content of each first element (f), since len ignores them too).
Because the entire body of len is given by a conditional, we can proceed directly to building the table.
Let’s consider the first row. The question costs three units (one each to evaluate the implicit empty-ness predicate, l, and to apply the former to the latter). This is evaluated once per element in the list and once more when the list is empty, i.e., $$k+1$$ times. The total cost of the question is thus $$3(k+1)$$. The answer takes one unit of time to compute, and is evaluated only once (when the list is empty). Thus it takes a total of one unit, for a total of $$3k+4$$ units.
Now for the second row. The question again costs three units, and is evaluated $$k$$ times. The answer involves two units to evaluate the rest of the list l.rest, which is implicitly hidden by the naming of r, two more to evaluate and apply 1 +, one more to evaluate len...and no more, because we are ignoring the time spent in the recursive call itself. In short, it takes seven units of time (in addition to the recursion we’ve chosen to ignore).
In tabular form:
|Q| #Q TotQ |A| #A TotA Total $$3$$ $$k+1$$ $$3(k+1)$$ $$1$$ $$1$$ $$1$$ $$3k+4$$ $$3$$ $$k$$ $$3k$$ $$7$$ $$k$$ $$7k$$ $$10k$$
Adding, we get $$13k + 4$$. Thus running len on a $$k$$-element list takes $$13k+4$$ units of time.
Exercise
How accurate is this estimate? If you try applying len to different sizes of lists, do you obtain a consistent estimate for $$k$$?
#### 18.6Creating Recurrences
We will now see a systematic way of analytically computing the time of a program. Suppose we have only one function f. We will define a function, $$T$$, to compute an upper-bound of the time of f.In general, we will have one such cost function for each function in the program. In such cases, it would be useful to give a different name to each function to easily tell them apart. Since we are looking at only one function for now, we’ll reduce notational overhead by having only one $$T$$. $$T$$ takes as many parameters as f does. The parameters to $$T$$ represent the sizes of the corresponding arguments to f. Eventually we will want to arrive at a closed form solution to $$T$$, i.e., one that does not refer to $$T$$ itself. But the easiest way to get there is to write a solution that is permitted to refer to $$T$$, called a recurrence relation, and then see how to eliminate the self-reference [Solving Recurrences].
We repeat this procedure for each function in the program in turn. If there are many functions, first solve for the one with no dependencies on other functions, then use its solution to solve for a function that depends only on it, and progress thus up the dependency chain. That way, when we get to a function that refers to other functions, we will already have a closed-form solution for the referred function’s running time and can simply plug in parameters to obtain a solution.
Exercise
The strategy outlined above doesn’t work when there are functions that depend on each other. How would you generalize it to handle this case?
The process of setting up a recurrence is easy. We simply define the right-hand-side of $$T$$ to add up the operations performed in f’s body. This is straightforward except for conditionals and recursion. We’ll elaborate on the treatment of conditionals in a moment. If we get to a recursive call to f on the argument a, in the recurrence we turn this into a (self-)reference to $$T$$ on the size of a.
For conditionals, we use only the |Q| and |A| columns of the corresponding table. Rather than multiplying by the size of the input, we add up the operations that happen on one invocation of f other than the recursive call, and then add the cost of the recursive call in terms of a reference to $$T$$. Thus, if we were doing this for len above, we would define $$T(k)$$—the time needed on an input of length $$k$$—in two parts: the value of $$T(0)$$ (when the list is empty) and the value for non-zero values of $$k$$. We know that $$T(0) = 4$$ (the cost of the first conditional and its corresponding answer). If the list is non-empty, the cost is $$T(k) = 3 + 3 + 7 + T(k-1)$$ (respectively from the first question, the second question, the remaining operations in the second answer, and the recursive call on a list one element smaller). This gives the following recurrence:
\begin{equation*}T(k) = \begin{cases} 4 & \text{when } k = 0 \\ 13 + T(k-1) & \text{when } k > 0\\ \end{cases}\end{equation*}
For a given list that is $$p$$ elements long (note that $$p \geq 0$$), this would take $$13$$ steps for the first element, $$13$$ more steps for the second, $$13$$ more for the third, and so on, until we run out of list elements and need $$4$$ more steps: a total of $$13p + 4$$ steps. Notice this is precisely the same answer we obtained by the tabular method!
Exercise
Why can we assume that for a list $$p$$ elements long, $$p \geq 0$$? And why did we take the trouble to explicitly state this above?
With some thought, you can see that the idea of constructing a recurrence works even when there is more than one recursive call, and when the argument to that call is one element structurally smaller. What we haven’t seen, however, is a way to solve such relations in general. That’s where we’re going next [Solving Recurrences].
#### 18.7A Notation for Functions
We have seen above that we can describe the running time of len through a function. We don’t have an especially good notation for writing such (anonymous) functions. Wait, we do—lam(k): (13 * k) + 4 endbut my colleagues would be horrified if you wrote this on their exams. Therefore, we’ll introduce the following notation to mean precisely the same thing:
\begin{equation*}[k \rightarrow 13k + 4]\end{equation*}
The brackets denote anonymous functions, with the parameters before the arrow and the body after.
#### 18.8Comparing Functions
Let’s return to the running time of len. We’ve written down a function of great precision: 13! 4! Is this justified?
At a fine-grained level already, no, it’s not. We’ve lumped many operations, with different actual running times, into a cost of one. So perhaps we should not worry too much about the differences between, say, $$[k \rightarrow 13k + 4]$$ and $$[k \rightarrow 4k + 10]$$. If we were given two implementations with these running times, respectively, it’s likely that we would pick other characteristics to choose between them.
What this boils down to is being able to compare two functions (representing the performance of implementations) for whether one is somehow quantitatively better in some meaningful sense than the other: i.e., is the quantitative difference so great that it might lead to a qualitative one. The example above suggests that small differences in constants probably do not matter. This suggests a definition of this form:
\begin{equation*}\exists c . \forall n \in \mathbb{N}, f_1(n) \leq c \cdot f_2(n) \Rightarrow f_1 \leq f_2\end{equation*}
Obviously, the “bigger” function is likely to be a less useful bound than a “tighter” one. That said, it is conventional to write a “minimal” bound for functions, which means avoiding unnecessary constants, sum terms, and so on. The justification for this is given below [Combining Big-Oh Without Woe].
Note carefully the order of identifiers. We must be able to pick the constant $$c$$ up front for this relationship to hold.
Do Now!
Why this order and not the opposite order? What if we had swapped the two quantifiers?
Had we swapped the order, it would mean that for every point along the number line, there must exist a constant—and there pretty much always does! The swapped definition would therefore be useless. What is important is that we can identify the constant no matter how large the parameter gets. That is what makes this truly a constant.
This definition has more flexibility than we might initially think. For instance, consider our running example compared with $$[k \rightarrow k^2]$$. Clearly, the latter function eventually dominates the former: i.e.,
\begin{equation*}[k \rightarrow 13k+4] \leq [k \rightarrow k^2]\end{equation*}
We just need to pick a sufficiently large constant and we will find this to be true.
Exercise
What is the smallest constant that will suffice?
You will find more complex definitions in the literature and they all have merits, because they enable us to make finer-grained distinctions than this definition allows. For the purpose of this book, however, the above definition suffices.
Observe that for a given function $$f$$, there are numerous functions that are less than it. We use the notation $$O(\cdot)$$ to describe this family of functions.In computer science this is usually pronounced “big-Oh”, though some prefer to call it the Bachmann-Landau notation after its originators. Thus if $$g \leq f$$, we can write $$g \in O(f)$$, which we can read as “$$f$$ is an upper-bound for $$g$$”. Thus, for instance,
\begin{equation*}[k \rightarrow 3k] \in O([k \rightarrow 4k+12])\end{equation*}
\begin{equation*}[k \rightarrow 4k+12] \in O([k \rightarrow k^2])\end{equation*}
and so on.
Pay especially close attention to our notation. We write $$\in$$ rather than $$=$$ or some other symbol, because $$O(f)$$ describes a family of functions of which $$g$$ is a member. We also write $$f$$ rather than $$f(x)$$ because we are comparing functions—$$f$$—rather than their values at particular points—$$f(x)$$—which would be ordinary numbers! Most of the notation in most the books and Web sites suffers from one or both flaws. We know, however, that functions are values, and that functions can be anonymous. We have actually exploited both facts to be able to write
\begin{equation*}[k \rightarrow 3k] \in O([k \rightarrow 4k+12])\end{equation*}
This is not the only notion of function comparison that we can have. For instance, given the definition of $$\leq$$ above, we can define a natural relation $$<$$. This then lets us ask, given a function $$f$$, what are all the functions $$g$$ such that $$g \leq f$$ but not $$g < f$$, i.e., those that are “equal” to $$f$$.Look out! We are using quotes because this is not the same as ordinary function equality, which is defined as the two functions giving the same answer on all inputs. Here, two “equal” functions may not give the same answer on any inputs. This is the family of functions that are separated by at most a constant; when the functions indicate the order of growth of programs, “equal” functions signify programs that grow at the same speed (up to constants). We use the notation $$\Theta(\cdot)$$ to speak of this family of functions, so if $$g$$ is equivalent to $$f$$ by this notion, we can write $$g \in \Theta(f)$$ (and it would then also be true that $$f \in \Theta(g)$$).
Exercise
Convince yourself that this notion of function equality is an equivalence relation, and hence worthy of the name “equal”. It needs to be (a) reflexive (i.e., every function is related to itself); (b) antisymmetric (if $$f \leq g$$ and $$g \leq f$$ then $$f$$ and $$g$$ are equal); and (c) transitive ($$f \leq g$$ and $$g \leq h$$ implies $$f \leq h$$).
#### 18.9Combining Big-Oh Without Woe
Now that we’ve introduced this notation, we should inquire about its closure properties: namely, how do these families of functions combine? To nudge your intuitions, assume that in all cases we’re discussing the running time of functions. We’ll consider three cases:
• Suppose we have a function f (whose running time is) in $$O(F)$$. Let’s say we run it $$p$$ times, for some given constant. The running time of the resulting code is then $$p \times O(F)$$. However, observe that this is really no different from $$O(F)$$: we can simply use a bigger constant for $$c$$ in the definition of $$O(\cdot)$$—in particular, we can just use $$pc$$. Conversely, then, $$O(pF)$$ is equivalent to $$O(F)$$. This is the heart of the intution that “multiplicative constants don’t matter”.
• Suppose we have two functions, f in $$O(F)$$ and g in $$O(G)$$. If we run f followed by g, we would expect the running time of the combination to be the sum of their individual running times, i.e., $$O(F) + O(G)$$. You should convince yourself that this is simply $$O(F + G)$$.
• Suppose we have two functions, f in $$O(F)$$ and g in $$O(G)$$. If f invokes g in each of its steps, we would expect the running time of the combination to be the product of their individual running times, i.e., $$O(F) \times O(G)$$. You should convince yourself that this is simply $$O(F \times G)$$.
These three operations—addition, multiplication by a constant, and multiplication by a function—cover just about all the cases.To ensure that the table fits in a reasonable width, we will abuse notation. For instance, we can use this to reinterpret the tabular operations above (assuming everything is a function of $$k$$):
|Q| #Q TotQ |A| #A TotA Total $$O(1)$$ $$O(k)$$ $$O(k)$$ $$O(1)$$ $$O(1)$$ $$O(1)$$ $$O(k)$$ $$O(1)$$ $$O(k)$$ $$O(k)$$ $$O(1)$$ $$O(k)$$ $$O(k)$$ $$O(k)$$
Because multiplication by constants doesn’t matter, we can replace the $$3$$ with $$1$$. Because addition of a constant doesn’t matter (run the addition rule in reverse), $$k+1$$ can become $$k$$. Adding this gives us $$O(k) + O(k) = 2 \times O(k) \in O(k)$$. This justifies claiming that running len on a $$k$$-element list takes time in $$O([k \rightarrow k])$$, which is a much simpler way of describing its bound than $$O([k \rightarrow 13k + 4])$$. In particular, it provides us with the essential information and nothing else: as the input (list) grows, the running time grows proportional to it, i.e., if we add one more element to the input, we should expect to add a constant more of time to the running time.
#### 18.10Solving Recurrences
There is a great deal of literature on solving recurrence equations. In this section we won’t go into general techniques, nor will we even discuss very many different recurrences. Rather, we’ll focus on just a handful that should be in the repertoire of every computer scientist. You’ll see these over and over, so you should instinctively recognize their recurrence pattern and know what complexity they describe (or know how to quickly derive it).
Earlier we saw a recurrence that had two cases: one for the empty input and one for all others. In general, we should expect to find one case for each non-recursive call and one for each recursive one, i.e., roughly one per cases clause. In what follows, we will ignore the base cases so long as the size of the input is constant (such as zero or one), because in such cases the amount of work done will also be a constant, which we can generally ignore [Comparing Functions].
• $$T(k)$$ = $$T(k-1) + c$$ = $$T(k-2) + c + c$$ = $$T(k-3) + c + c + c$$ = ... = $$T(0) + c \times k$$ = $$c_0 + c \times k$$
Thus $$T \in O([k \rightarrow k])$$. Intuitively, we do a constant amount of work ($$c$$) each time we throw away one element ($$k-1$$), so we do a linear amount of work overall.
• $$T(k)$$ = $$T(k-1) + k$$ = $$T(k-2) + (k-1) + k$$ = $$T(k-3) + (k-2) + (k-1) + k$$ = ... = $$T(0) + (k-(k-1)) + (k-(k-2)) + \cdots + (k-2) + (k-1) + k$$ = $$c_0 + 1 + 2 + \cdots + (k-2) + (k-1) + k$$ = $$c_0 + {\frac{k \cdot (k+1)}{2}}$$
Thus $$T \in O([k \rightarrow k^2])$$. This follows from the solution to the sum of the first $$k$$ numbers.
We can also view this recurrence geometrically. Imagine each x below refers to a unit of work, and we start with $$k$$ of them. Then the first row has $$k$$ units of work:
xxxxxxxx
followed by the recurrence on $$k-1$$ of them:
xxxxxxx
which is followed by another recurrence on one smaller, and so on, until we fill end up with:
xxxxxxxx xxxxxxx xxxxxx xxxxx xxxx xxx xx x
The total work is then essentially the area of this triangle, whose base and height are both $$k$$: or, if you prefer, half of this $$k \times k$$ square:
xxxxxxxx xxxxxxx. xxxxxx.. xxxxx... xxxx.... xxx..... xx...... x.......
Similar geometric arguments can be made for all these recurrences.
• $$T(k)$$ = $$T(k/2) + c$$ = $$T(k/4) + c + c$$ = $$T(k/8) + c + c + c$$ = ... = $$T(k/2^{\log_2 k}) + c \cdot \log_2 k$$ = $$c_1 + c \cdot \log_2 k$$
Thus $$T \in O([k \rightarrow \log k])$$. Intuitively, we’re able to do only constant work ($$c$$) at each level, then throw away half the input. In a logarithmic number of steps we will have exhausted the input, having done only constant work each time. Thus the overall complexity is logarithmic.
• $$T(k)$$ = $$T(k/2) + k$$ = $$T(k/4) + k/2 + k$$ = ... = $$T(1) + k/2^{\log_2 k} + \cdots + k/4 + k/2 + k$$ = $$c_1 + k(1/2^{\log_2 k} + \cdots + 1/4 + 1/2 + 1)$$ = $$c_1 + 2k$$
Thus $$T \in O([k \rightarrow k])$$. Intuitively, the first time your process looks at all the elements, the second time it looks at half of them, the third time a quarter, and so on. This kind of successive halving is equivalent to scanning all the elements in the input a second time. Hence this results in a linear process.
• $$T(k)$$ = $$2T(k/2) + k$$ = $$2(2T(k/4) + k/2) + k$$ = $$4T(k/4) + k + k$$ = $$4(2T(k/8) + k/4) + k + k$$ = $$8T(k/8) + k + k + k$$ = ... = $$2^{\log_2 k} T(1) + k \cdot \log_2 k$$ = $$k \cdot c_1 + k \cdot \log_2 k$$
Thus $$T \in O([k \rightarrow k \cdot \log k])$$. Intuitively, each time we’re processing all the elements in each recursive call (the $$k$$) as well as decomposing into two half sub-problems. This decomposition gives us a recursion tree of logarithmic height, at each of which levels we’re doing linear work.
• $$T(k)$$ = $$2T(k-1) + c$$ = $$2T(k-1) + (2-1)c$$ = $$2(2T(k-2) + c) + (2-1)c$$ = $$4T(k-2) + 3c$$ = $$4T(k-2) + (4-1)c$$ = $$4(2T(k-3) + c) + (4-1)c$$ = $$8T(k-3) + 7c$$ = $$8T(k-3) + (8-1)c$$ = ... = $$2^k T(0) + (2^k-1)c$$
Thus $$T \in O([k \rightarrow 2^k])$$. Disposing of each element requires doing a constant amount of work for it and then doubling the work done on the rest. This successive doubling leads to the exponential.
Exercise
Using induction, prove each of the above derivations.
### 19Sets Appeal
Earlier [Sets as Collective Data] we introduced sets. Recall that the elements of a set have no specific order, and ignore duplicates.If these ideas are not familiar, please read Sets as Collective Data, since they will be important when discussing the representation of sets. At that time we relied on Pyret’s built-in representation of sets. Now we will discuss how to build sets for ourselves. In what follows, we will focus only on sets of numbers.
We will start by discussing how to represent sets using lists. Intuitively, using lists to represent sets of data seems problematic, because lists respect both order and duplication. For instance,
check:
[list: 1, 2, 3] is [list: 3, 2, 1, 1]
end
fails.
In principle, we want sets to obey the following interface:Note that a type called Set is already built into Pyret, so we won’t use that name below.
<set-operations> ::=
mt-set :: Set
is-in :: (T, Set<T> -> Bool)
insert :: (T, Set<T> -> Set<T>)
union :: (Set<T>, Set<T> -> Set<T>)
size :: (Set<T> -> Number)
to-list :: (Set<T> -> List<T>)
We may also find it also useful to have functions such as
insert-many :: (List<T>, Set<T> -> Set<T>)
which, combined with mt-set, easily gives us a to-set function.
Sets can contain many kinds of values, but not necessarily any kind: we need to be able to check for two values being equal (which is a requirement for a set, but not for a list!), which can’t be done with all values (such as functions); and sometimes we might even want the elements to obey an ordering [Converting Values to Ordered Values]. Numbers satisfy both characteristics.
#### 19.1Representing Sets by Lists
In what follows we will see multiple different representations of sets, so we will want names to tell them apart. We’ll use LSet to stand for “sets represented as lists”.
As a starting point, let’s consider the implementation of sets using lists as the underlying representation. After all, a set appears to merely be a list wherein we ignore the order of elements.
##### 19.1.1Representation Choices
The empty list can stand in for the empty set—
type LSet = List
mt-set = empty
and we can presumably define size as
fun size<T>(s :: LSet<T>) -> Number:
s.length()
end
However, this reduction (of sets to lists) can be dangerous:
1. There is a subtle difference between lists and sets. The list
[list: 1, 1]
is not the same as
[list: 1]
because the first list has length two whereas the second has length one. Treated as a set, however, the two are the same: they both have size one. Thus, our implementation of size above is incorrect if we don’t take into account duplicates (either during insertion or while computing the size).
2. We might falsely make assumptions about the order in which elements are retrieved from the set due to the ordering guaranteed provided by the underlying list representation. This might hide bugs that we don’t discover until we change the representation.
3. We might have chosen a set representation because we didn’t need to care about order, and expected lots of duplicate items. A list representation might store all the duplicates, resulting in significantly more memory use (and slower programs) than we expected.
To avoid these perils, we have to be precise about how we’re going to use lists to represent sets. One key question (but not the only one, as we’ll soon see [Choosing Between Representations]) is what to do about duplicates. One possibility is for insert to check whether an element is already in the set and, if so, leave the representation unchanged; this incurs a cost during insertion but avoids unnecessary duplication and lets us use length to implement size. The other option is to define insert as linkliterally,
insert = link
and have some other procedure perform the filtering of duplicates.
##### 19.1.2Time Complexity
What is the complexity of this representation of sets? Let’s consider just insert, check, and size. Suppose the size of the set is $$k$$ (where, to avoid ambiguity, we let $$k$$ represent the number of distinct elements). The complexity of these operations depends on whether or not we store duplicates:
• If we don’t store duplicates, then size is simply length, which takes time linear in $$k$$. Similarly, check only needs to traverse the list once to determine whether or not an element is present, which also takes time linear in $$k$$. But insert needs to check whether an element is already present, which takes time linear in $$k$$, followed by at most a constant-time operation (link).
• If we do store duplicates, then insert is constant time: it simply links on the new element without regard to whether it already is in the set representation. check traverses the list once, but the number of elements it needs to visit could be significantly greater than $$k$$, depending on how many duplicates have been added. Finally, size needs to check whether or not each element is duplicated before counting it.
Do Now!
What is the time complexity of size if the list has duplicates?
One implementation of size is
fun size<T>(s :: LSet<T>) -> Number:
cases (List) s:
| empty => 0
if r.member(f):
size(r)
else:
1 + size(r)
end
end
end
Let’s now compute the complexity of the body of the function, assuming the number of distinct elements in s is $$k$$ but the actual number of elements in s is $$d$$, where $$d \geq k$$. To compute the time to run size on $$d$$ elements, $$T(d)$$, we should determine the number of operations in each question and answer. The first question has a constant number of operations, and the first answer also a constant. The second question also has a constant number of operations. Its answer is a conditional, whose first question (r.member(f) needs to traverse the entire list, and hence has $$O([k \rightarrow d])$$ operations. If it succeeds, we recur on something of size $$T(d-1)$$; else we do the same but perform a constant more operations. Thus $$T(0)$$ is a constant, while the recurrence (in big-Oh terms) is
\begin{equation*}T(d) = d + T(d-1)\end{equation*}
Thus $$T \in O([d \rightarrow d^2])$$. Note that this is quadratic in the number of elements in the list, which may be much bigger than the size of the set.
##### 19.1.3Choosing Between Representations
Now that we have two representations with different complexities, it’s worth thinking about how to choose between them. To do so, let’s build up the following table. The table distinguishes between the interface (the set) and the implementation (the list), because—owing to duplicates in the representation—these two may not be the same. In the table we’ll consider just two of the most common operations, insertion and membership checking:
With Duplicates Without Duplicates insert is-in insert is-in Size of Set constant linear linear linear Size of List constant linear linear linear
A naive reading of this would suggest that the representation with duplicates is better because it’s sometimes constant and sometimes linear, whereas the version without duplicates is always linear. However, this masks a very important distinction: what the linear means. When there are no duplicates, the size of the list is the same as the size of the set. However, with duplicates, the size of the list can be arbitrarily larger than that of the set!
Based on this, we can draw several lessons:
1. Which representation we choose is a matter of how much duplication we expect. If there won’t be many duplicates, then the version that stores duplicates pays a small extra price in return for some faster operations.
2. Which representation we choose is also a matter of how often we expect each operation to be performed. The representation without duplication is “in the middle”: everything is roughly equally expensive (in the worst case). With duplicates is “at the extremes”: very cheap insertion, potentially very expensive membership. But if we will mostly only insert without checking membership, and especially if we know membership checking will only occur in situations where we’re willing to wait, then permitting duplicates may in fact be the smart choice. (When might we ever be in such a situation? Suppose your set represents a backup data structure; then we add lots of data but very rarely—indeed, only in case of some catastrophe—ever need to look for things in it.)
3. Another way to cast these insights is that our form of analysis is too weak. In situations where the complexity depends so heavily on a particular sequence of operations, big-Oh is too loose and we should instead study the complexity of specific sequences of operations. We will address precisely this question later [Halloween Analysis].
Moreover, there is no reason a program should use only one representation. It could well begin with one representation, then switch to another as it better understands its workload. The only thing it would need to do to switch is to convert all existing data between the representations.
How might this play out above? Observe that data conversion is very cheap in one direction: since every list without duplicates is automatically also a list with (potential) duplicates, converting in that direction is trivial (the representation stays unchanged, only its interpretation changes). The other direction is harder: we have to filter duplicates (which takes time quadratic in the number of elements in the list). Thus, a program can make an initial guess about its workload and pick a representation accordingly, but maintain statistics as it runs and, when it finds its assumption is wrong, switch representations—and can do so as many times as needed.
##### 19.1.4Other Operations
Exercise
Implement the remaining operations catalogued above (<set-operations>) under each list representation.
Exercise
Implement the operation
remove :: (Set<T>, T -> Set<T>)
under each list representation (renaming Set appropriately. What difference do you see?
Do Now!
Suppose you’re asked to extend sets with these operations, as the set analog of first and rest:
one :: (Set<T> -> T)
others :: (Set<T> -> T)
You should refuse to do so! Do you see why?
With lists the “first” element is well-defined, whereas sets are defined to have no ordering. Indeed, just to make sure users of your sets don’t accidentally assume anything about your implementation (e.g., if you implement one using first, they may notice that one always returns the element most recently added to the list), you really ought to return a random element of the set on each invocation.
Unfortunately, returning a random element means the above interface is unusable. Suppose s is bound to a set containing 1, 2, and 3. Say the first time one(s) is invoked it returns 2, and the second time 1. (This already means one is not a function.) The third time it may again return 2. Thus others has to remember which element was returned the last time one was called, and return the set sans that element. Suppose we now invoke one on the result of calling others. That means we might have a situation where one(s) produces the same result as one(others(s)).
Exercise
Why is it unreasonable for one(s) to produce the same result as one(others(s))?
Exercise
Suppose you wanted to extend sets with a subset operation that partitioned the set according to some condition. What would its type be?
Exercise
The types we have written above are not as crisp as they could be. Define a has-no-duplicates predicate, refine the relevant types with it, and check that the functions really do satisfy this criterion.
#### 19.2Making Sets Grow on Trees
Let’s start by noting that it seems better, if at all possible, to avoid storing duplicates. Duplicates are only problematic during insertion due to the need for a membership test. But if we can make membership testing cheap, then we would be better off using it to check for duplicates and storing only one instance of each value (which also saves us space). Thus, let’s try to improve the time complexity of membership testing (and, hopefully, of other operations too).
It seems clear that with a (duplicate-free) list representation of a set, we cannot really beat linear time for membership checking. This is because at each step, we can eliminate only one element from contention which in the worst case requires a linear amount of work to examine the whole set. Instead, we need to eliminate many more elements with each comparison—more than just a constant.
In our handy set of recurrences [Solving Recurrences], one stands out: $$T(k) = T(k/2) + c$$. It says that if, with a constant amount of work we can eliminate half the input, we can perform membership checking in logarithmic time. This will be our goal.
Before we proceed, it’s worth putting logarithmic growth in perspective. Asymptotically, logarithmic is obviously not as nice as constant. However, logarithmic growth is very pleasant because it grows so slowly. For instance, if an input doubles from size $$k$$ to $$2k$$, its logarithm—and hence resource usage—grows only by $$\log 2k - \log k = \log 2$$, which is a constant. Indeed, for just about all problems, practically speaking the logarithm of the input size is bounded by a constant (that isn’t even very large). Therefore, in practice, for many programs, if we can shrink our resource consumption to logarithmic growth, it’s probably time to move on and focus on improving some other part of the system.
##### 19.2.1Converting Values to Ordered Values
We have actually just made an extremely subtle assumption. When we check one element for membership and eliminate it, we have eliminated only one element. To eliminate more than one element, we need one element to “speak for” several. That is, eliminating that one value needs to have safely eliminated several others as well without their having to be consulted. In particular, then, we can no longer compare for mere equality, which compares one set element against another element; we need a comparison that compares against an element against a set of elements.
To do this, we have to convert an arbitrary datum into a datatype that permits such comparison. This is known as hashing. A hash function consumes an arbitrary value and produces a comparable representation of it (its hash)—most commonly (but not strictly necessarily), a number. A hash function must naturally be deterministic: a fixed value should always yield the same hash (otherwise, we might conclude that an element in the set is not actually in it, etc.). Particular uses may need additional properties: e.g., below we assume its output is partially ordered.
Let us now consider how one can compute hashes. If the input datatype is a number, it can serve as its own hash. Comparison simply uses numeric comparison (e.g., <). Then, transitivity of < ensures that if an element $$A$$ is less than another element $$B$$, then $$A$$ is also less than all the other elements bigger than $$B$$. The same principle applies if the datatype is a string, using string inequality comparison. But what if we are handed more complex datatypes?
Before we answer that, consider that in practice numbers are more efficient to compare than strings (since comparing two numbers is very nearly constant time). Thus, although we could use strings directly, it may be convenient to find a numeric representation of strings. In both cases, we will convert each character of the string into a number—e.g., by considering its ASCII encoding. Based on that, here are two hash functions:
1. Consider a list of primes as long as the string. Raise each prime by the corresponding number, and multiply the result. For instance, if the string is represented by the character codes [6, 4, 5] (the first character has code 6, the second one 4, and the third 5), we get the hash
num-expt(2, 6) * num-expt(3, 4) * num-expt(5, 5)
or 16200000.
2. Simply add together all the character codes. For the above example, this would correspond to the has
6 + 4 + 5
or 15.
The first representation is invertible, using the Fundamental Theorem of Arithmetic: given the resulting number, we can reconstruct the input unambiguously (i.e., 16200000 can only map to the input above, and none other). The second encoding is, of course, not invertible (e.g., simply permute the characters and, by commutativity, the sum will be the same).
Now let us consider more general datatypes. The principle of hashing will be similar. If we have a datatype with several variants, we can use a numeric tag to represent the variants: e.g., the primes will give us invertible tags. For each field of a record, we need an ordering of the fields (e.g., lexicographic, or “alphabetical” order), and must hash their contents recursively; having done so, we get in effect a string of numbers, which we have shown how to handle.
Now that we have understood how one can deterministically convert any arbitrary datum into a number, in what follows, we will assume that the trees representing sets are trees of numbers. However, it is worth considering what we really need out of a hash. In Set Membership by Hashing Redux, we will not need partial ordering. Invertibility is more tricky. In what follows below, we have assumed that finding a hash is tantamount to finding the set element itself, which is not true if multiple values can have the same hash. In that case, the easiest thing to do is to store alongside the hash all the values that hashed to it, and we must search through all of these values to find our desired element. Unfortunately, this does mean that in an especially perverse situation, the desired logarithmic complexity will actually be linear complexity after all!
In real systems, hashes of values are typically computed by the programming language implementation. This has the virtue that they can often be made unique. How does the system achieve this? Easy: it essentially uses the memory address of a value as its hash. (Well, not so fast! Sometimes the memory system can and does move values around through a process called garbage collection). In these cases computing a hash value is more complicated.)
##### 19.2.2Using Binary Trees
Because logs come from trees.
Clearly, a list representation does not let us eliminate half the elements with a constant amount of work; instead, we need a tree. Thus we define a binary tree of (for simplicity) numbers:
data BT:
| leaf
| node(v :: Number, l :: BT, r :: BT)
end
Given this definition, let’s define the membership checker:
fun is-in-bt(e :: Number, s :: BT) -> Boolean:
cases (BT) s:
| leaf => false
| node(v, l, r) =>
if e == v:
true
else:
is-in-bt(e, l) or is-in-bt(e, r)
end
end
end
Oh, wait. If the element we’re looking for isn’t the root, what do we do? It could be in the left child or it could be in the right; we won’t know for sure until we’ve examined both. Thus, we can’t throw away half the elements; the only one we can dispose of is the value at the root. Furthermore, this property holds at every level of the tree. Thus, membership checking needs to examine the entire tree, and we still have complexity linear in the size of the set.
How can we improve on this? The comparison needs to help us eliminate not only the root but also one whole sub-tree. We can only do this if the comparison “speaks for” an entire sub-tree. It can do so if all elements in one sub-tree are less than or equal to the root value, and all elements in the other sub-tree are greater than or equal to it. Of course, we have to be consistent about which side contains which subset; it is conventional to put the smaller elements to the left and the bigger ones to the right. This refines our binary tree definition to give us a binary search tree (BST).
Do Now!
Here is a candiate predicate for recognizing when a binary tree is in fact a binary search tree:
fun is-a-bst-buggy(b :: BT) -> Boolean:
cases (BT) b:
| leaf => true
| node(v, l, r) =>
(is-leaf(l) or (l.v <= v)) and
(is-leaf(r) or (v <= r.v)) and
is-a-bst-buggy(l) and
is-a-bst-buggy(r)
end
end
Is this definition correct?
It’s not. To actually throw away half the tree, we need to be sure that everything in the left sub-tree is less than the value in the root and similarly, everything in the right sub-tree is greater than the root.We have used <= instead of < above because even though we don’t want to permit duplicates when representing sets, in other cases we might not want to be so stringent; this way we can reuse the above implementation for other purposes. But the definition above performs only a “shallow” comparison. Thus we could have a root a with a right child, b, such that b > a; and the b node could have a left child c such that c < b; but this does not guarantee that c > a. In fact, it is easy to construct a counter-example that passes this check:
check:
node(5, node(3, leaf, node(6, leaf, leaf)), leaf)
satisfies is-a-bst-buggy # FALSE!
end
Exercise
Fix the BST checker.
With a corrected definition, we can now define a refined version of binary trees that are search trees:
type BST = BT%(is-a-bst)
We can also remind ourselves that the purpose of this exercise was to define sets, and define TSets to be tree sets:
type TSet = BST
mt-set = leaf
Now let’s implement our operations on the BST representation. First we’ll write a template:
fun is-in(e :: Number, s :: BST) -> Bool:
cases (BST) s:
| leaf => ...
| node(v, l :: BST, r :: BST) => ...
... is-in(l) ...
... is-in(r) ...
end
end
Observe that the data definition of a BST gives us rich information about the two children: they are each a BST, so we know their elements obey the ordering property. We can use this to define the actual operations:
fun is-in(e :: Number, s :: BST) -> Boolean:
cases (BST) s:
| leaf => false
| node(v, l, r) =>
if e == v:
true
else if e < v:
is-in(e, l)
else if e > v:
is-in(e, r)
end
end
end
fun insert(e :: Number, s :: BST) -> BST:
cases (BST) s:
| leaf => node(e, leaf, leaf)
| node(v, l, r) =>
if e == v:
s
else if e < v:
node(v, insert(e, l), r)
else if e > v:
node(v, l, insert(e, r))
end
end
end
In both functions we are strictly assuming the invariant of the BST, and in the latter case also ensuring it. Make sure you identify where, why, and how.
You should now be able to define the remaining operations. Of these, size clearly requires linear time (since it has to count all the elements), but because is-in and insert both throw away one of two children each time they recur, they take logarithmic time.
Exercise
Suppose we frequently needed to compute the size of a set. We ought to be able to reduce the time complexity of size by having each tree cache its size, so that size could complete in constant time (note that the size of the tree clearly fits the criterion of a cache, since it can always be reconstructed). Update the data definition and all affected functions to keep track of this information correctly.
But wait a minute. Are we actually done? Our recurrence takes the form $$T(k) = T(k/2) + c$$, but what in our data definition guaranteed that the size of the child traversed by is-in will be half the size?
Do Now!
Construct an example—consisting of a sequence of inserts to the empty tree—such that the resulting tree is not balanced. Show that searching for certain elements in this tree will take linear, not logarithmic, time in its size.
Imagine starting with the empty tree and inserting the values 1, 2, 3, and 4, in order. The resulting tree would be
check:
insert(4, insert(3, insert(2, insert(1, mt-set)))) is
node(1, leaf,
node(2, leaf,
node(3, leaf,
node(4, leaf, leaf))))
end
Searching for 4 in this tree would have to examine all the set elements in the tree. In other words, this binary search tree is degenerateit is effectively a list, and we are back to having the same complexity we had earlier.
Therefore, using a binary tree, and even a BST, does not guarantee the complexity we want: it does only if our inputs have arrived in just the right order. However, we cannot assume any input ordering; instead, we would like an implementation that works in all cases. Thus, we must find a way to ensure that the tree is always balanced, so each recursive call in is-in really does throw away half the elements.
##### 19.2.3A Fine Balance: Tree Surgery
Let’s define a balanced binary search tree (BBST). It must obviously be a search tree, so let’s focus on the “balanced” part. We have to be careful about precisely what this means: we can’t simply expect both sides to be of equal size because this demands that the tree (and hence the set) have an even number of elements and, even more stringently, to have a size that is a power of two.
Exercise
Define a predicate for a BBST that consumes a BT and returns a Boolean indicating whether or not it a balanced search tree.
Therefore, we relax the notion of balance to one that is both accommodating and sufficient. We use the term balance factor for a node to refer to the height of its left child minus the height of its right child (where the height is the depth, in edges, of the deepest node). We allow every node of a BBST to have a balance factor of $$-1$$, $$0$$, or $$1$$ (but nothing else): that is, either both have the same height, or the left or the right can be one taller. Note that this is a recursive property, but it applies at all levels, so the imbalance cannot accumulate making the whole tree arbitrarily imbalanced.
Exercise
Given this definition of a BBST, show that the number of nodes is exponential in the height. Thus, always recurring on one branch will terminate after a logarithmic (in the number of nodes) number of steps.
Here is an obvious but useful observation: every BBST is also a BST (this was true by the very definition of a BBST). Why does this matter? It means that a function that operates on a BST can just as well be applied to a BBST without any loss of correctness.
So far, so easy. All that leaves is a means of creating a BBST, because it’s responsible for ensuring balance. It’s easy to see that the constant empty-set is a BBST value. So that leaves only insert.
Here is our situation with insert. Assuming we start with a BBST, we can determine in logarithmic time whether the element is already in the tree and, if so, ignore it.To implement a bag we count how many of each element are in it, which does not affect the tree’s height. When inserting an element, given balanced trees, the insert for a BST takes only a logarithmic amount of time to perform the insertion. Thus, if performing the insertion does not affect the tree’s balance, we’re done. Therefore, we only need to consider cases where performing the insertion throws off the balance.
Observe that because $$<$$ and $$>$$ are symmetric (likewise with $$<=$$ and $$>=$$), we can consider insertions into one half of the tree and a symmetric argument handles insertions into the other half. Thus, suppose we have a tree that is currently balanced into which we are inserting the element $$e$$. Let’s say $$e$$ is going into the left sub-tree and, by virtue of being inserted, will cause the entire tree to become imbalanced.Some trees, like family trees (Data Design Problem – Ancestry Data) represent real-world data. It makes no sense to “balance” a family tree: it must accurately model whatever reality it represents. These set-representing trees, in contrast, are chosen by us, not dictated by some external reality, so we are free to rearrange them.
There are two ways to proceed. One is to consider all the places where we might insert $$e$$ in a way that causes an imbalance and determine what to do in each case.
Exercise
Enumerate all the cases where insertion might be problematic, and dictate what to do in each case.
The number of cases is actually quite overwhelming (if you didn’t think so, you missed a few...). Therefore, we instead attack the problem after it has occurred: allow the existing BST insert to insert the element, assume that we have an imbalanced tree, and show how to restore its balance.The insight that a tree can be made “self-balancing” is quite remarkable, and there are now many solutions to this problem. This particular one, one of the oldest, is due to G.M. Adelson-Velskii and E.M. Landis. In honor of their initials it is called an AVL Tree, though the tree itself is quite evident; their genius is in defining re-balancing.
Thus, in what follows, we begin with a tree that is balanced; insert causes it to become imbalanced; we have assumed that the insertion happened in the left sub-tree. In particular, suppose a (sub-)tree has a balance factor of $$2$$ (positive because we’re assuming the left is imbalanced by insertion). The procedure for restoring balance depends critically on the following property:
Exercise
Show that if a tree is currently balanced, i.e., the balance factor at every node is $$-1$$, $$0$$, or $$1$$, then insert can at worst make the balance factor $$\pm 2$$.
The algorithm that follows is applied as insert returns from its recursion, i.e., on the path from the inserted value back to the root. Since this path is of logarithmic length in the set’s size (due to the balancing property), and (as we shall see) performs only a constant amount of work at each step, it ensures that insertion also takes only logarithmic time, thus completing our challenge.
To visualize the algorithm, let’s use this tree schematic:
p / \ q C / \ A B
Here, $$p$$ is the value of the element at the root (though we will also abuse terminology and use the value at a root to refer to that whole tree), $$q$$ is the value at the root of the left sub-tree (so $$q < p$$), and $$A$$, $$B$$, and $$C$$ name the respective sub-trees. We have assumed that $$e$$ is being inserted into the left sub-tree, which means $$e < p$$.
Let’s say that $$C$$ is of height $$k$$. Before insertion, the tree rooted at $$q$$ must have had height $$k+1$$ (or else one insertion cannot create imbalance). In turn, this means $$A$$ must have had height $$k$$ or $$k-1$$, and likewise for $$B$$.
Suppose that after insertion, the tree rooted at $$q$$ has height $$k+2$$. Thus, either $$A$$ or $$B$$ has height $$k+1$$ and the other must have height less than that (either $$k$$ or $$k-1$$).
Exercise
Why can they both not have height $$k+1$$ after insertion?
This gives us two cases to consider.
##### 19.2.3.1Left-Left Case
Let’s say the imbalance is in $$A$$, i.e., it has height $$k+1$$. Let’s expand that tree:
p / \ q C / \ r B / \ A1 A2
We know the following about the data in the sub-trees. We’ll use the notation $$T < a$$ where $$T$$ is a tree and $$a$$ is a single value to mean every value in $$T$$ is less than $$a$$.
• $$A_1 < r$$.
• $$r < A_2 < q$$.
• $$q < B < p$$.
• $$p < C$$.
Let’s also remind ourselves of the sizes:
• The height of $$A_1$$ or of $$A_2$$ is $$k$$ (the cause of imbalance).
• The height of the other $$A_i$$ is $$k-1$$ (see the exercise above).
• The height of $$C$$ is $$k$$ (initial assumption; $$k$$ is arbitrary).
• The height of $$B$$ must be $$k-1$$ or $$k$$ (argued above).
Imagine this tree is a mobile, which has gotten a little skewed to the left. You would naturally think to suspend the mobile a little further to the left to bring it back into balance. That is effectively what we will do:
q / \ r p / \ / \ A1 A2 B C
Observe that this preserves each of the ordering properties above. In addition, the $$A$$ subtree has been brought one level closer to the root than earlier relative to $$B$$ and $$C$$. This restores the balance (as you can see if you work out the heights of each of $$A_i$$, $$B$$, and $$C$$). Thus, we have also restored balance.
##### 19.2.3.2Left-Right Case
The imbalance might instead be in $$B$$. Expanding:
p / \ q C / \ A r / \ B1 B2
Again, let’s record what we know about data order:
• $$A < q$$.
• $$q < B_1 < r$$.
• $$r < B_2 < p$$.
• $$p < C$$.
and sizes:
• Suppose the height of $$C$$ is $$k$$.
• The height of $$A$$ must be $$k-1$$ or $$k$$.
• The height of $$B_1$$ or $$B_2$$ must be $$k$$, but not both (see the exercise above). The other must be $$k-1$$.
We therefore have to somehow bring $$B_1$$ and $$B_2$$ one level closer to the root of the tree. By using the above data ordering knowledge, we can construct this tree:
p / \ r C / \ q B2 / \ A B1
Of course, if $$B_1$$ is the problematic sub-tree, this still does not address the problem. However, we are now back to the previous (left-left) case; rotating gets us to:
r / \ q p / \ / \ A B1 B2 C
Now observe that we have precisely maintained the data ordering constraints. Furthermore, from the root, $$A$$’s lowest node is at height $$k+1$$ or $$k+2$$; so is $$B_1$$’s; so is $$B_2$$’s; and $$C$$’s is at $$k+2$$.
##### 19.2.3.3Any Other Cases?
Were we a little too glib before? In the left-right case we said that only one of $$B_1$$ or $$B_2$$ could be of height $$k$$ (after insertion); the other had to be of height $$k-1$$. Actually, all we can say for sure is that the other has to be at most height $$k-2$$.
Exercise
• Can the height of the other tree actually be $$k-2$$ instead of $$k-1$$?
• If so, does the solution above hold? Is there not still an imbalance of two in the resulting tree?
• Is there actually a bug in the above algorithm?
### 20Halloween Analysis
In Predicting Growth, we introduced the idea of big-Oh complexity to measure the worst-case time of a computation. As we saw in Choosing Between Representations, however, this is sometimes too coarse a bound when the complexity is heavily dependent on the exact sequence of operations run. Now, we will consider a different style of complexity analysis that better accommodates operation sequences.
#### 20.1A First Example
Consider, for instance, a set that starts out empty, followed by a sequence of $$k$$ insertions and then $$k$$ membership tests, and suppose we are using the representation without duplicates. Insertion time is proportional to the size of the set (and list); this is initially $$0$$, then $$1$$, and so on, until it reaches size $$k$$. Therefore, the total cost of the sequence of insertions is $$k \cdot (k+1) / 2$$. The membership tests cost $$k$$ each in the worst case, because we’ve inserted up to $$k$$ distinct elements into the set. The total time is then
\begin{equation*}k^2 / 2 + k / 2 + k^2\end{equation*}
for a total of $$2k$$ operations, yielding an average of
\begin{equation*}\frac{3}{4} k + \frac{1}{4}\end{equation*}
steps per operation in the worst case.
#### 20.2The New Form of Analysis
What have we computed? We are still computing a worst case cost, because we have taken the cost of each operation in the sequence in the worst case. We are then computing the average cost per operation. Therefore, this is a average of worst cases.Importantly, this is different from what is known as average-case analysis, which uses probability theory to compute the estimated cost of the computation. We have not used any probability here. Note that because this is an average per operation, it does not say anything about how bad any one operation can be (which, as we will see [Amortization Versus Individual Operations], can be quite a bit worse); it only says what their average is.
In the above case, this new analysis did not yield any big surprises. We have found that on average we spend about $$k$$ steps per operation; a big-Oh analysis would have told us that we’re performing $$2k$$ operations with a cost of $$O([k \rightarrow k])$$ each in the number of distinct elements; per operation, then, we are performing roughly linear work in the worst-case number of set elements.
As we will soon see, however, this won’t always be the case: this new analysis can cough up pleasant surprises.
Before we proceed, we should give this analysis its name. Formally, it is called amortized analysis. Amortization is the process of spreading a payment out over an extended but fixed term. In the same way, we spread out the cost of a computation over a fixed sequence, then determine how much each payment will be.We have given it a whimsical name because Halloween is a(n American) holiday devoted to ghosts, ghouls, and other symbols of death. Amortization comes from the Latin root mort-, which means death, because an amortized analysis is one conducted “at the death”, i.e., at the end of a fixed sequence of operations.
#### 20.3An Example: Queues from Lists
We have already seen lists [From Tables to Lists] and sets [Sets Appeal]. Now let’s consider another fundamental computer science data structure: the queue. A queue is a linear, ordered data structure, just like a list; however, the set of operations they offer is different. In a list, the traditional operations follow a last-in, first-out discipline: .first returns the element most recently linked. In contrast, a queue follows a first-in, first-out discipline. That is, a list can be visualized as a stack, while a queue can be visualized as a conveyer belt.
##### 20.3.1List Representations
We can define queues using lists in the natural way: every enqueue is implemented with link, while every dequeue requires traversing the whole list until its end. Conversely, we could make enqueuing traverse to the end, and dequeuing correspond to .rest. Either way, one of these operations will take constant time while the other will be linear in the length of the list representing the queue.
In fact, however, the above paragraph contains a key insight that will let us do better.
Observe that if we store the queue in a list with most-recently-enqueued element first, enqueuing is cheap (constant time). In contrast, if we store the queue in the reverse order, then dequeuing is constant time. It would be wonderful if we could have both, but once we pick an order we must give up one or the other. Unless, that is, we pick...both.
One half of this is easy. We simply enqueue elements into a list with the most recent addition first. Now for the (first) crucial insight: when we need to dequeue, we reverse the list. Now, dequeuing also takes constant time.
##### 20.3.2A First Analysis
Of course, to fully analyze the complexity of this data structure, we must also account for the reversal. In the worst case, we might argue that any operation might reverse (because it might be the first dequeue); therefore, the worst-case time of any operation is the time it takes to reverse, which is linear in the length of the list (which corresponds to the elements of the queue).
However, this answer should be unsatisfying. If we perform $$k$$ enqueues followed by $$k$$ dequeues, then each of the enqueues takes one step; each of the last $$k-1$$ dequeues takes one step; and only the first dequeue requires a reversal, which takes steps proportional to the number of elements in the list, which at that point is $$k$$. Thus, the total cost of operations for this sequence is $$k \cdot 1 + k + (k-1) \cdot 1 = 3k-1$$ for a total of $$2k$$ operations, giving an amortized complexity of effectively constant time per operation!
##### 20.3.3More Liberal Sequences of Operations
In the process of this, however, I’ve quietly glossed over something you’ve probably noticed: in our candidate sequence all dequeues followed all enqueues. What happens on the next enqueue? Because the list is now reversed, it will have to take a linear amount of time! So we have only partially solved the problem.
Now we can introduce the second insight: have two lists instead of one. One of them will be the tail of the queue, where new elements get enqueued; the other will be the head of the queue, where they get dequeued:
data Queue<T>:
| queue(tail :: List<T>, head :: List<T>)
end
mt-q :: Queue = queue(empty, empty)
Provided the tail is stored so that the most recent element is the first, then enqueuing takes constant time:
fun enqueue<T>(q :: Queue<T>, e :: T) -> Queue<T>:
end
For dequeuing to take constant time, the head of the queue must be stored in the reverse direction. However, how does any element ever get from the tail to the head? Easy: when we try to dequeue and find no elements in the head, we reverse the (entire) tail into the head (resulting in an empty tail). We will first define a datatype to represent the response from dequeuing:
data Response<T>:
| elt-and-q(e :: T, r :: Queue<T>)
end
Now for the implementation of dequeue:
fun dequeue<T>(q :: Queue<T>) -> Response<T>:
| empty =>
elt-and-q(f,
queue(q.tail, r))
end
end
##### 20.3.4A Second Analysis
We can now reason about sequences of operations as we did before, by adding up costs and averaging. However, another way to think of it is this. Let’s give each element in the queue three “credits”. Each credit can be used for one constant-time operation.
One credit gets used up in enqueuing. So long as the element stays in the tail list, it still has two credits to spare. When it needs to be moved to the head list, it spends one more credit in the link step of reversal. Finally, the dequeuing operation performs one operation too.
Because the element does not run out of credits, we know it must have had enough. These credits reflect the cost of operations on that element. From this (very informal) analysis, we can conclude that in the worst case, any permutation of enqueues and dequeues will still cost only a constant amount of amortized time.
##### 20.3.5Amortization Versus Individual Operations
Note, however, that the constant represents an average across the sequence of operations. It does not put a bound on the cost of any one operation. Indeed, as we have seen above, when dequeue finds the head list empty it reverses the tail, which takes time linear in the size of the tail—not constant at all! Therefore, we should be careful to not assume that every step in the sequence will is bounded. Nevertheless, an amortized analysis sometimes gives us a much more nuanced understanding of the real behavior of a data structure than a worst-case analysis does on its own.
At this point we have only briefly touched on the subject of amortized analysis. A very nice tutorial by Rebecca Fiebrink provides much more information. The authoritative book on algorithms, Introduction to Algorithms by Cormen, Leiserson, Rivest, and Stein, covers amortized analysis in extensive detail.
### 21Sharing and Equality
#### 21.1Re-Examining Equality
Consider the following data definition and example values:
data BinTree:
| leaf
| node(v, l :: BinTree, r :: BinTree)
end
a-tree =
node(5,
node(4, leaf, leaf),
node(4, leaf, leaf))
b-tree =
block:
four-node = node(4, leaf, leaf)
node(5,
four-node,
four-node)
end
In particular, it might seem that the way we’ve written b-tree is morally equivalent to how we’ve written a-tree, but we’ve created a helpful binding to avoid code duplication.
Because both a-tree and b-tree are bound to trees with 5 at the root and a left and right child each containing 4, we can indeed reasonably consider these trees equivalent. Sure enough:
<equal-tests> ::=
check:
a-tree is b-tree
a-tree.l is a-tree.l
a-tree.l is a-tree.r
b-tree.l is b-tree.r
end
However, there is another sense in which these trees are not equivalent. concretely, a-tree constructs a distinct node for each child, while b-tree uses the same node for both children. Surely this difference should show up somehow, but we have not yet seen a way to write a program that will tell these apart.
By default, the is operator uses the same equality test as Pyret’s ==. There are, however, other equality tests in Pyret. In particular, the way we can tell apart these data is by using Pyret’s identical function, which implements reference equality. This checks not only whether two values are structurally equivalent but whether they are the result of the very same act of value construction. With this, we can now write additional tests:
check:
identical(a-tree, b-tree) is false
identical(a-tree.l, a-tree.l) is true
identical(a-tree.l, a-tree.r) is false
identical(b-tree.l, b-tree.r) is true
end
Let’s step back for a moment and consider the behavior that gives us this result. We can visualize the different values by putting each distinct value in a separate location alongside the running program. We can draw the first step as creating a node with value 4:
a-tree =
node(5,
1001,
node(4, leaf, leaf))
b-tree =
block:
four-node = node(4, leaf, leaf)
node(5,
four-node,
four-node)
end
Heap
• 1001:
node(4, leaf, leaf)
The next step creates another node with value 4, distinct from the first:
a-tree =
node(5, 1001, 1002)
b-tree =
block:
four-node = node(4, leaf, leaf)
node(5,
four-node,
four-node)
end
Heap
• 1001:
node(4, leaf, leaf)
• 1002:
node(4, leaf, leaf)
Then the node for a-tree is created:
a-tree = 1003
b-tree =
block:
four-node = node(4, leaf, leaf)
node(5,
four-node,
four-node)
end
Heap
• 1001:
node(4, leaf, leaf)
• 1002:
node(4, leaf, leaf)
• 1003:
node(5, 1001, 1002)
When evaluating the block for b-tree, first a single node is created for the four-node binding:
a-tree = 1003
b-tree =
block:
four-node = 1004
node(5,
four-node,
four-node)
end
Heap
• 1001:
node(4, leaf, leaf)
• 1002:
node(4, leaf, leaf)
• 1003:
node(5, 1001, 1002)
• 1004:
node(4, leaf, leaf)
These location values can be substituted just like any other, so they get substituted for four-node to continue evaluation of the block.We skipped substituting a-tree for the moment, that will come up later.
a-tree = 1003
b-tree =
block:
node(5, 1004, 1004)
end
Heap
• 1001:
node(4, leaf, leaf)
• 1002:
node(4, leaf, leaf)
• 1003:
node(5, 1001, 1002)
• 1004:
node(4, leaf, leaf)
Finally, the node for b-tree is created:
a-tree = 1003
b-tree = 1005
Heap
• 1001:
node(4, leaf, leaf)
• 1002:
node(4, leaf, leaf)
• 1003:
node(5, 1001, 1002)
• 1004:
node(4, leaf, leaf)
• 1005:
node(5, 1004, 1004)
This visualization can help us explain the test we wrote using
identical
. Let’s consider the test with the appropriate location references substituted for a-tree and b-tree:
check:
identical(1003, 1005) is false
identical(1003.l, 1003.l) is true
identical(1003.l, 1003.r) is false
identical(1005.l, 1005.r) is true
end
Heap
• 1001:
node(4, leaf, leaf)
• 1002:
node(4, leaf, leaf)
• 1003:
node(5, 1001, 1002)
• 1004:
node(4, leaf, leaf)
• 1005:
node(5, 1004, 1004)
check:
identical(1003, 1005) is false
identical(1001, 1001) is true
identical(1001, 1004) is false
identical(1004, 1004) is true
end
Heap
• 1001:
node(4, leaf, leaf)
• 1002:
node(4, leaf, leaf)
• 1003:
node(5, 1001, 1002)
• 1004:
node(4, leaf, leaf)
• 1005:
node(5, 1004, 1004)
There is actually another way to write these tests in Pyret: the is operator can also be parameterized by a different equality predicate than the default ==. Thus, the above block can equivalently be written as:We can use is-not to check for expected failure of equality.
check:
a-tree is-not%(identical) b-tree
a-tree.l is%(identical) a-tree.l
a-tree.l is-not%(identical) a-tree.r
b-tree.l is%(identical) b-tree.r
end
We will use this style of equality testing from now on.
Observe how these are the same values that were compared earlier (<equal-tests>), but the results are now different: some values that were true earlier are now false. In particular,
check:
a-tree is b-tree
a-tree is-not%(identical) b-tree
a-tree.l is a-tree.r
a-tree.l is-not%(identical) a-tree.r
end
Later we will return both to what identical really means [Variables and Equality] (Pyret has a full range of equality operations suitable for different situations).
#### 21.2The Cost of Evaluating References
From a complexity viewpoint, it’s important for us to understand how these references work. As we have hinted, four-node is computed only once, and each use of it refers to the same value: if, instead, it was evaluated each time we referred to four-node, there would be no real difference between a-tree and b-tree, and the above tests would not distinguish between them.
This is especially relevant when understanding the cost of function evaluation. We’ll construct two simple examples that illustrate this. We’ll begin with a contrived data structure:
L = range(0, 100)
Suppose we now define
L1 = link(1, L)
L2 = link(-1, L)
Constructing a list clearly takes time at least proportional to the length; therefore, we expect the time to compute L to be considerably more than that for a single link operation. Therefore, the question is how long it takes to compute L1 and L2 after L has been computed: constant time, or time proportional to the length of L?
The answer, for Pyret, and for most other contemporary languages (including Java, C#, OCaml, Racket, etc.), is that these additional computations take constant time. That is, the value bound to L is computed once and bound to L; subsequent expressions refer to this value (hence “reference”) rather than reconstructing it, as reference equality shows:
check:
L1.rest is%(identical) L
L2.rest is%(identical) L
L1.rest is%(identical) L2.rest
end
Similarly, we can define a function, pass L to it, and see whether the resulting argument is identical to the original:
fun check-for-no-copy(another-l):
identical(another-l, L)
end
check:
check-for-no-copy(L) is true
end
or, equivalently,
check:
L satisfies check-for-no-copy
end
Therefore, neither built-in operations (like .rest) nor user-defined ones (like check-for-no-copy) make copies of their arguments.Strictly speaking, of course, we cannot conclude that no copy was made. Pyret could have made a copy, discarded it, and still passed a reference to the original. Given how perverse this would be, we can assume—and take the language’s creators’ word for it—that this doesn’t actually happen. By creating extremely large lists, we can also use timing information to observe that the time of constructing the list grows proportional to the length of the list while the time of passing it as a parameter remains constant. The important thing to observe here is that, instead of simply relying on authority, we have used operations in the language itself to understand how the language behaves.
#### 21.3On the Internet, Nobody Knows You’re a DAG
Despite the name we’ve given it, b-tree is not actually a tree. In a tree, by definition, there are no shared nodes, whereas in b-tree the node named by four-node is shared by two parts of the tree. Despite this, traversing b-tree will still terminate, because there are no cyclic references in it: if you start from any node and visit its “children”, you cannot end up back at that node. There is a special name for a value with such a shape: directed acyclic graph (DAG).
Many important data structures are actually a DAG underneath. For instance, consider Web sites. It is common to think of a site as a tree of pages: the top-level refers to several sections, each of which refers to sub-sections, and so on. However, sometimes an entry needs to be cataloged under multiple sections. For instance, an academic department might organize pages by people, teaching, and research. In the first of these pages it lists the people who work there; in the second, the list of courses; and in the third, the list of research groups. In turn, the courses might have references to the people teaching them, and the research groups are populated by these same people. Since we want only one page per person (for both maintenance and search indexing purposes), all these personnel links refer back to the same page for people.
Let’s construct a simple form of this. First a datatype to represent a site’s content:
data Content:
| page(s :: String)
| section(title :: String, sub :: List<Content>)
end
Let’s now define a few people:
people-pages :: Content =
section("People",
[list: page("Church"),
page("Dijkstra"),
page("Haberman") ])
and a way to extract a particular person’s page:
fun get-person(n): index(people-pages.sub, n) end
Now we can define theory and systems sections:
theory-pages :: Content =
section("Theory",
[list: get-person(0), get-person(1)])
systems-pages :: Content =
section("Systems",
[list: get-person(1), get-person(2)])
which are integrated into a site as a whole:
site :: Content =
section("Computing Sciences",
[list: theory-pages, systems-pages])
Now we can confirm that each of these luminaries needs to keep only one Web page current; for instance:
check:
theory = index(site.sub, 0)
systems = index(site.sub, 1)
theory-dijkstra = index(theory.sub, 1)
systems-dijkstra = index(systems.sub, 0)
theory-dijkstra is systems-dijkstra
theory-dijkstra is%(identical) systems-dijkstra
end
#### 21.4From Acyclicity to Cycles
Here’s another example that arises on the Web. Suppose we are constructing a table of output in a Web page. We would like the rows of the table to alternate between white and grey. If the table had only two rows, we could map the row-generating function over a list of these two colors. Since we do not know how many rows it will have, however, we would like the list to be as long as necessary. In effect, we would like to write:
web-colors = link("white", link("grey", web-colors))
to obtain an indefinitely long list, so that we could eventually write
map2(color-table-row, table-row-content, web-colors)
which applies a color-table-row function to two arguments: the current row from table-row-content, and the current color from web-colors, proceeding in lockstep over the two lists.
Unfortunately, there are many things wrong with this attempted definition.
Do Now!
Do you see what they are?
Here are some problems in turn:
• This will not even parse. The identifier web-colors is not bound on the right of the =.
• Earlier, we saw a solution to such a problem: use rec [Streams From Functions]. What happens if we write
rec web-colors = link("white", link("grey", web-colors))
Exercise
Why does rec work in the definition of ones but not above?
• Assuming we have fixed the above problem, one of two things will happen. It depends on what the initial value of web-colors is. Because it is a dummy value, we do not get an arbitrarily long list of colors but rather a list of two colors followed by the dummy value. Indeed, this program will not even type-check.
Suppose, however, that web-colors were written instead as a function definition to delay its creation:
fun web-colors(): link("white", link("grey", web-colors())) end
On its own this just defines a function. If, however, we use it—web-colors()it goes into an infinite loop constructing links.
• Even if all that were to work, map2 would either (a) not terminate because its second argument is indefinitely long, or (b) report an error because the two arguments aren’t the same length.
All these problems are symptoms of a bigger issue. What we are trying to do here is not merely create a shared datum (like a DAG) but something much richer: a cyclic datum, i.e., one that refers back to itself:
When you get to cycles, even defining the datum becomes difficult because its definition depends on itself so it (seemingly) needs to already be defined in the process of being defined. We will return to cyclic data later: Circular References.
### 22Graphs
In From Acyclicity to Cycles we introduced a special kind of sharing: when the data become cyclic, i.e., there exist values such that traversing other reachable values from them eventually gets you back to the value at which you began. Data that have this characteristic are called graphs.Technically, a cycle is not necessary to be a graph; a tree or a DAG is also regarded as a (degenerate) graph. In this section, however, we are interested in graphs that have the potential for cycles.
Lots of very important data are graphs. For instance, the people and connections in social media form a graph: the people are nodes or vertices and the connections (such as friendships) are links or edges. They form a graph because for many people, if you follow their friends and then the friends of their friends, you will eventually get back to the person you started with. (Most simply, this happens when two people are each others’ friends.) The Web, similarly is a graph: the nodes are pages and the edges are links between pages. The Internet is a graph: the nodes are machines and the edges are links between machines. A transportation network is a graph: e.g., cities are nodes and the edges are transportation links between them. And so on. Therefore, it is essential to understand graphs to represent and process a great deal of interesting real-world data.
Graphs are important and interesting for not only practical but also principled reasons. The property that a traversal can end up where it began means that traditional methods of processing will no longer work: if it blindly processes every node it visits, it could end up in an infinite loop. Therefore, we need better structural recipes for our programs. In addition, graphs have a very rich structure, which lends itself to several interesting computations over them. We will study both these aspects of graphs below.
#### 22.1Understanding Graphs
Consider again the binary trees we saw earlier [Re-Examining Equality]. Let’s now try to distort the definition of a “tree” by creating ones with cycles, i.e., trees with nodes that point back to themselves (in the sense of identical). As we saw earlier [From Acyclicity to Cycles], it is not completely straightforward to create such a structure, but what we saw earlier [Streams From Functions] can help us here, by letting us suspend the evaluation of the cyclic link. That is, we have to not only use rec, we must also use a function to delay evaluation. In turn, we have to update the annotations on the fields. Since these are not going to be “trees” any more, we’ll use a name that is suggestive but not outright incorrect:
data BinT:
| leaf
| node(v, l :: ( -> BinT), r :: ( -> BinT))
end
Now let’s try to construct some cyclic values. Here are a few examples:
rec tr = node("rec", lam(): tr end, lam(): tr end)
t0 = node(0, lam(): leaf end, lam(): leaf end)
t1 = node(1, lam(): t0 end, lam(): t0 end)
t2 = node(2, lam(): t1 end, lam(): t1 end)
Now let’s try to compute the size of a BinT. Here’s the obvious program:
fun sizeinf(t :: BinT) -> Number:
cases (BinT) t:
| leaf => 0
| node(v, l, r) =>
ls = sizeinf(l())
rs = sizeinf(r())
1 + ls + rs
end
end
(We’ll see why we call it sizeinf in a moment.)
Do Now!
What happens when we call sizeinf(tr)?
It goes into an infinite loop: hence the inf in its name.
There are two very different meanings for “size”. One is, “How many times can we traverse an edge?” The other is, “How many distinct nodes were constructed as part of the data structure?” With trees, by definition, these two are the same. With a DAG the former exceeds the latter but only by a finite amount. With a general graph, the former can exceed the latter by an infinite amount. In the case of a datum like tr, we can in fact traverse edges an infinite number of times. But the total number of constructed nodes is only one! Let’s write this as test cases in terms of a size function, to be defined:
check:
size(tr) is 1
size(t0) is 1
size(t1) is 2
size(t2) is 3
end
It’s clear that we need to somehow remember what nodes we have visited previously: that is, we need a computation with “memory”. In principle this is easy: we just create an extra data structure that checks whether a node has already been counted. As long as we update this data structure correctly, we should be all set. Here’s an implementation.
fun sizect(t :: BinT) -> Number:
fun szacc(shadow t :: BinT, seen :: List<BinT>) -> Number:
if has-id(seen, t):
0
else:
cases (BinT) t:
| leaf => 0
| node(v, l, r) =>
ls = szacc(l(), ns)
rs = szacc(r(), ns)
1 + ls + rs
end
end
end
szacc(t, empty)
end
The extra parameter, seen, is called an accumulator, because it “accumulates” the list of seen nodes.Note that this could just as well be a set; it doesn’t have to be a list. The support function it needs checks whether a given node has already been seen:
fun has-id<A>(seen :: List<A>, t :: A):
cases (List) seen:
| empty => false
if f <=> t: true
else: has-id(r, t)
end
end
end
How does this do? Well, sizect(tr) is indeed 1, but sizect(t1) is 3 and sizect(t2) is 7!
Do Now!
Explain why these answers came out as they did.
The fundamental problem is that we’re not doing a very good job of remembering! Look at this pair of lines:
ls = szacc(l(), ns)
rs = szacc(r(), ns)
The nodes seen while traversing the left branch are effectively forgotten, because the only nodes we remember when traversing the right branch are those in ns: namely, the current node and those visited “higher up”. As a result, any nodes that “cross sides” are counted twice.
The remedy for this, therefore, is to remember every node we visit. Then, when we have no more nodes to process, instead of returning only the size, we should return all the nodes visited until now. This ensures that nodes that have multiple paths to them are visited on only one path, not more than once. The logic for this is to return two values from each traversal—the size and all the visited nodes—and not just one.
fun size(t :: BinT) -> Number:
fun szacc(shadow t :: BinT, seen :: List<BinT>)
-> {n :: Number, s :: List<BinT>}:
if has-id(seen, t):
{n: 0, s: seen}
else:
cases (BinT) t:
| leaf => {n: 0, s: seen}
| node(v, l, r) =>
ls = szacc(l(), ns)
rs = szacc(r(), ls.s)
{n: 1 + ls.n + rs.n, s: rs.s}
end
end
end
szacc(t, empty).n
end
Sure enough, this function satisfies the above tests.
#### 22.2Representations
The representation we’ve seen above for graphs is certainly a start towards creating cyclic data, but it’s not very elegant. It’s both error-prone and inelegant to have to write lam everywhere, and remember to apply functions to () to obtain the actual values. Therefore, here we explore other representations of graphs that are more conventional and also much simpler to manipulate.
There are numerous ways to represent graphs, and the choice of representation depends on several factors:
1. The structure of the graph, and in particular, its density. We will discuss this further later [Measuring Complexity for Graphs].
2. The representation in which the data are provided by external sources. Sometimes it may be easier to simply adapt to their representation; in particular, in some cases there may not even be a choice.
3. The features provided by the programming language, which make some representations much harder to use than others.
Previously [Sets Appeal], we have explored the idea of having many different representations for one datatype. As we will see, this is very true of graphs as well. Therefore, it would be best if we could arrive at a common interface to process graphs, so that all later programs can be written in terms of this interface, without overly depending on the underlying representation.
In terms of representations, there are three main things we need:
1. A way to construct graphs.
2. A way to identify (i.e., tell apart) nodes or vertices in a graph.
3. Given a way to identify nodes, a way to get that node’s neighbors in the graph.
Any interface that satisfies these properties will suffice. For simplicity, we will focus on the second and third of these and not abstract over the process of constructing a graph.
Our running example will be a graph whose nodes are cities in the United States and edges are direct flight connections between them:
Here’s our first representation. We will assume that every node has a unique name (such a name, when used to look up information in a repository of data, is sometimes called a key). A node is then a key, some information about that node, and a list of keys that refer to other nodes:
type Key = String
data KeyedNode:
| keyed-node(key :: Key, content, adj :: List<String>)
end
type KNGraph = List<KeyedNode>
type Node = KeyedNode
type Graph = KNGraph
(Here we’re assuming our keys are strings.)
Here’s a concrete instance of such a graph:The prefix kn- stands for “keyed node”.
kn-cities :: Graph = block:
knWAS = keyed-node("was", "Washington", [list: "chi", "den", "saf", "hou", "pvd"])
knORD = keyed-node("chi", "Chicago", [list: "was", "saf", "pvd"])
knBLM = keyed-node("bmg", "Bloomington", [list: ])
knHOU = keyed-node("hou", "Houston", [list: "was", "saf"])
knDEN = keyed-node("den", "Denver", [list: "was", "saf"])
knSFO = keyed-node("saf", "San Francisco", [list: "was", "den", "chi", "hou"])
knPVD = keyed-node("pvd", "Providence", [list: "was", "chi"])
[list: knWAS, knORD, knBLM, knHOU, knDEN, knSFO, knPVD]
end
Given a key, here’s how we look up its neighbor:
fun find-kn(key :: Key, graph :: Graph) -> Node:
matches = for filter(n from graph):
n.key == key
end
matches.first # there had better be exactly one!
end
Exercise
Convert the comment in the function into an invariant about the datum. Express this invariant as a refinement and add it to the declaration of graphs.
With this support, we can look up neighbors easily:
fun kn-neighbors(city :: Key, graph :: Graph) -> List<Key>:
city-node = find-kn(city, graph)
end
When it comes to testing, some tests are easy to write. Others, however, might require describing entire nodes, which can be unwieldy, so for the purpose of checking our implementation it suffices to examine just a part of the result:
check:
ns = kn-neighbors("hou", kn-cities)
ns is [list: "was", "saf"]
map(_.content, map(find-kn(_, kn-cities), ns)) is
[list: "Washington", "San Francisco"]
end
In some languages, it is common to use numbers as names. This is especially useful when numbers can be used to get access to an element in a constant amount of time (in return for having a bound on the number of elements that can be accessed). Here, we use a list—which does not provide constant-time access to arbitrary elements—to illustrate this concept. Most of this will look very similar to what we had before; we’ll comment on a key difference at the end.
First, the datatype:The prefix ix- stands for “indexed”.
data IndexedNode:
end
type IXGraph = List<IndexedNode>
type Node = IndexedNode
type Graph = IXGraph
Our graph now looks like this:
ix-cities :: Graph = block:
inWAS = idxed-node("Washington", [list: 1, 4, 5, 3, 6])
inORD = idxed-node("Chicago", [list: 0, 5, 6])
inBLM = idxed-node("Bloomington", [list: ])
inHOU = idxed-node("Houston", [list: 0, 5])
inDEN = idxed-node("Denver", [list: 0, 5])
inSFO = idxed-node("San Francisco", [list: 0, 4, 3])
inPVD = idxed-node("Providence", [list: 0, 1])
[list: inWAS, inORD, inBLM, inHOU, inDEN, inSFO, inPVD]
end
where we’re assuming indices begin at 0. To find a node:
fun find-ix(idx :: Key, graph :: Graph) -> Node:
lists.get(graph, idx)
end
We can then find neighbors almost as before:
fun ix-neighbors(city :: Key, graph :: Graph) -> List<Key>:
city-node = find-ix(city, graph)
end
Finally, our tests also look similar:
check:
ns = ix-neighbors(3, ix-cities)
ns is [list: 0, 5]
map(_.content, map(find-ix(_, ix-cities), ns)) is
[list: "Washington", "San Francisco"]
end
Something deeper is going on here. The keyed nodes have intrinsic keys: the key is part of the datum itself. Thus, given just a node, we can determine its key. In contrast, the indexed nodes represent extrinsic keys: the keys are determined outside the datum, and in particular by the position in some other data structure. Given a node and not the entire graph, we cannot know for what its key is. Even given the entire graph, we can only determine its key by using identical, which is a rather unsatisfactory approach to recovering fundamental information. This highlights a weakness of using extrinsically keyed representations of information. (In return, extrinsically keyed representations are easier to reassemble into new collections of data, because there is no danger of keys clashing: there are no intrinsic keys to clash.)
##### 22.2.3A List of Edges
The representations we have seen until now have given priority to nodes, making edges simply a part of the information in a node. We could, instead, use a representation that makes edges primary, and nodes simply be the entities that lie at their ends:The prefix le- stands for “list of edges”.
data Edge:
| edge(src :: String, dst :: String)
end
type LEGraph = List<Edge>
type Graph = LEGraph
Then, our flight network becomes:
le-cities :: Graph =
[list:
edge("Washington", "Chicago"),
edge("Washington", "Denver"),
edge("Washington", "San Francisco"),
edge("Washington", "Houston"),
edge("Washington", "Providence"),
edge("Chicago", "Washington"),
edge("Chicago", "San Francisco"),
edge("Chicago", "Providence"),
edge("Houston", "Washington"),
edge("Houston", "San Francisco"),
edge("Denver", "Washington"),
edge("Denver", "San Francisco"),
edge("San Francisco", "Washington"),
edge("San Francisco", "Denver"),
edge("San Francisco", "Houston"),
edge("Providence", "Washington"),
edge("Providence", "Chicago") ]
Observe that in this representation, nodes that are not connected to other nodes in the graph simply never show up! You’d therefore need an auxilliary data structure to keep track of all the nodes.
To obtain the set of neighbors:
fun le-neighbors(city :: Key, graph :: Graph) -> List<Key>:
neighboring-edges = for filter(e from graph):
city == e.src
end
names = for map(e from neighboring-edges): e.dst end
names
end
And to be sure:
check:
le-neighbors("Houston", le-cities) is
[list: "Washington", "San Francisco"]
end
However, this representation makes it difficult to store complex information about a node without replicating it. Because nodes usually have rich information while the information about edges tends to be weaker, we often prefer node-centric representations. Of course, an alternative is to think of the node names as keys into some other data structure from which we can retrieve rich information about nodes.
##### 22.2.4Abstracting Representations
We would like a general representation that lets us abstract over the specific implementations. We will assume that broadly we have available a notion of Node that has content, a notion of Keys (whether or not intrinsic), and a way to obtain the neighbors—a list of keys—given a key and a graph. This is sufficient for what follows. However, we still need to choose concrete keys to write examples and tests. For simplicity, we’ll use string keys [Links by Name].
#### 22.3Measuring Complexity for Graphs
Before we begin to define algorithms over graphs, we should consider how to measure the size of a graph. A graph has two components: its nodes and its edges. Some algorithms are going to focus on nodes (e.g., visiting each of them), while others will focus on edges, and some will care about both. So which do we use as the basis for counting operations: nodes or edges?
It would help if we can reduce these two measures to one. To see whether that’s possible, suppose a graph has $$k$$ nodes. Then its number of edges has a wide range, with these two extremes:
• No two nodes are connected. Then there are no edges at all.
• Every two nodes is connected. Then there are essentially as many edges as the number of pairs of nodes.
The number of nodes can thus be significantly less or even significantly more than the number of edges. Were this difference a matter of constants, we could have ignored it; but it’s not. As a graph tends towards the former extreme, the ratio of nodes to edges approaches $$k$$ (or even exceeds it, in the odd case where there are no edges, but this graph is not very interesting); as it tends towards the latter, it is the ratio of edges to nodes that approaches $$k^2$$. In other words, neither measure subsumes the other by a constant independent of the graph.
Therefore, when we want to speak of the complexity of algorithms over graphs, we have to consider the sizes of both the number of nodes and edges. In a connected graphA graph is connected if, from every node, we can traverse edges to get to every other node., however, there must be at least as many edges as nodes, which means the number of edges dominates the number of nodes. Since we are usually processing connected graphs, or connected parts of graphs one at a time, we can bound the number of nodes by the number of edges.
#### 22.4Reachability
Many uses of graphs need to address reachability: whether we can, using edges in the graph, get from one node to another. For instance, a social network might suggest as contacts all those who are reachable from existing contacts. On the Internet, traffic engineers care about whether packets can get from one machine to another. On the Web, we care about whether all public pages on a site are reachable from the home page. We will study how to compute reachability using our travel graph as a running example.
##### 22.4.1Simple Recursion
At its simplest, reachability is easy. We want to know whether there exists a pathA path is a sequence of zero or more linked edges. between a pair of nodes, a source and a destination. (A more sophisticated version of reachability might compute the actual path, but we’ll ignore this for now.) There are two possibilities: the source and destintion nodes are the same, or they’re not.
• If they are the same, then clearly reachability is trivially satisfied.
• If they are not, we have to iterate through the neighbors of the source node and ask whether the destination is reachable from each of those neighbors.
This translates into the following function:
<graph-reach-1-main> ::=
fun reach-1(src :: Key, dst :: Key, g :: Graph) -> Boolean:
if src == dst:
true
else:
<graph-reach-1-loop>
loop(neighbors(src, g))
end
end
where the loop through the neighbors of src is:
<graph-reach-1-loop> ::=
fun loop(ns):
cases (List) ns:
| empty => false
if reach-1(f, dst, g): true else: loop(r) end
end
end
We can test this as follows:
<graph-reach-tests> ::=
check:
reach = reach-1
reach("was", "was", kn-cities) is true
reach("was", "chi", kn-cities) is true
reach("was", "bmg", kn-cities) is false
reach("was", "hou", kn-cities) is true
reach("was", "den", kn-cities) is true
reach("was", "saf", kn-cities) is true
end
Unfortunately, we don’t find out about how these tests fare, because some of them don’t complete at all. That’s because we have an infinite loop, due to the cyclic nature of graphs!
Exercise
Which of the above examples leads to a cycle? Why?
##### 22.4.2Cleaning up the Loop
Before we continue, let’s try to improve the expression of the loop. While the nested function above is a perfectly reasonable definition, we can use Pyret’s for to improve its readability.
The essence of the above loop is to iterate over a list of boolean values; if one of them is true, the entire loop evaluates to true; if they are all false, then we haven’t found a path to the destination node, so the loop evaluates to false. Thus:
fun ormap(fun-body, l):
cases (List) l:
| empty => false
if fun-body(f): true else: ormap(fun-body, r) end
end
end
With this, we can replace the loop definition and use with:
for ormap(n from neighbors(src, g)):
reach-1(n, dst, g)
end
##### 22.4.3Traversal with Memory
Because we have cyclic data, we have to remember what nodes we’ve already visited and avoid traversing them again. Then, every time we begin traversing a new node, we add it to the set of nodes we’ve already started to visit so that. If we return to that node, because we can assume the graph has not changed in the meanwhile, we know that additional traversals from that node won’t make any difference to the outcome.This property is known as idempotence.
We therefore define a second attempt at reachability that take an extra argument: the set of nodes we have begun visiting (where the set is represented as a graph). The key difference from <graph-reach-1-main> is, before we begin to traverse edges, we should check whether we’ve begun processing the node or not. This results in the following definition:
<graph-reach-2> ::=
fun reach-2(src :: Key, dst :: Key, g :: Graph, visited :: List<Key>) -> Boolean:
if visited.member(src):
false
else if src == dst:
true
else:
for ormap(n from neighbors(src, g)):
reach-2(n, dst, g, new-visited)
end
end
end
In particular, note the extra new conditional: if the reachability check has already visited this node before, there is no point traversing further from here, so it returns false. (There may still be other parts of the graph to explore, which other recursive calls will do.)
Exercise
Does it matter if the first two conditions were swapped, i.e., the beginning of reach-2 began with
if src == dst:
true
else if visited.member(src):
false
? Explain concretely with examples.
Exercise
We repeatedly talk about remembering the nodes that we have begun to visit, not the ones we’ve finished visiting. Does this distinction matter? How?
##### 22.4.4A Better Interface
As the process of testing reach-2 shows, we may have a better implementation, but we’ve changed the function’s interface; now it has a needless extra argument, which is not only a nuisance but might also result in errors if we accidentally misuse it. Therefore, we should clean up our definition by moving the core code to an internal function:
fun reach-3(s :: Key, d :: Key, g :: Graph) -> Boolean:
fun reacher(src :: Key, dst :: Key, visited :: List<Key>) -> Boolean:
if visited.member(src):
false
else if src == dst:
true
else:
for ormap(n from neighbors(src, g)):
reacher(n, dst, new-visited)
end
end
end
reacher(s, d, empty)
end
We have now restored the original interface while correctly implementing reachability.
Exercise
Does this really gives us a correct implementation? In particular, does this address the problem that the size function above addressed? Create a test case that demonstrates the problem, and then fix it.
It is conventional for computer science texts to call these depth- and breadth-first search. However, searching is just a specific purpose; traversal is a general task that can be used for many purposes.
The reachability algorithm we have seen above has a special property. At every node it visits, there is usually a set of adjacent nodes at which it can continue the traversal. It has at least two choices: it can either visit each immediate neighbor first, then visit all of the neighbors’ neighbors; or it can choose a neighbor, recur, and visit the next immediate neighbor only after that visit is done. The former is known as breadth-first traversal, while the latter is depth-first traversal.
The algorithm we have designed uses a depth-first strategy: inside <graph-reach-1-loop>, we recur on the first element of the list of neighbors before we visit the second neighbor, and so on. The alternative would be to have a data structure into which we insert all the neighbors, then pull out an element at a time such that we first visit all the neighbors before their neighbors, and so on. This naturally corresponds to a queue [An Example: Queues from Lists].
Exercise
Using a queue, implement breadth-first traversal.
If we correctly check to ensure we don’t re-visit nodes, then both breadth- and depth-first traversal will properly visit the entire reachable graph without repetition (and hence not get into an infinite loop). Each one traverses from a node only once, from which it considers every single edge. Thus, if a graph has $$N$$ nodes and $$E$$ edges, then a lower-bound on the complexity of traversal is $$O([N, E \rightarrow N + E])$$. We must also consider the cost of checking whether we have already visited a node before (which is a set membership problem, which we address elsewhere: Making Sets Grow on Trees). Finally, we have to consider the cost of maintaining the data structure that keeps track of our traversal. In the case of depth-first traversal, recursion—which uses the machine’s stackdoes it automatically at constant overhead. In the case of breadth-first traversal, the program must manage the queue, which can add more than constant overhead.In practice, too, the stack will usually perform much better than a queue, because it is supported by machine hardware.
This would suggest that depth-first traversal is always better than breadth-first traversal. However, breadth-first traversal has one very important and valuable property. Starting from a node $$N$$, when it visits a node $$P$$, count the number of edges taken to get to $$P$$. Breadth-first traversal guarantees that there cannot have been a shorter path to $$P$$: that is, it finds a shortest path to $$P$$.
Exercise
Why “a” rather than “the” shortest path?
Do Now!
Prove that breadth-first traversal finds a shortest path.
#### 22.6Graphs With Weighted Edges
Consider a transportation graph: we are usually interested not only in whether we can get from one place to another, but also in what it “costs” (where we may have many different cost measures: money, distance, time, units of carbon dioxide, etc.). On the Internet, we might care about the latency or bandwidth over a link. Even in a social network, we might like to describe the degree of closeness of a friend. In short, in many graphs we are interested not only in the direction of an edge but also in some abstract numeric measure, which we call its weight.
In the rest of this study, we will assume that our graph edges have weights. This does not invalidate what we’ve studied so far: if a node is reachable in an unweighted graph, it remains reachable in a weighted one. But the operations we are going to study below only make sense in a weighted graph.We can, however, always treat an unweighted graph as a weighted one by giving every edge the same, constant, positive weight (say one).
Exercise
When treating an unweighted graph as a weighted one, why do we care that every edge be given a positive weight?
Exercise
Revise the graph data definitions to account for edge weights.
Exercise
Weights are not the only kind of data we might record about edges. For instance, if the nodes in a graph represent people, the edges might be labeled with their relationship (“mother”, “friend”, etc.). What other kinds of data can you imagine recording for edges?
#### 22.7Shortest (or Lightest) Paths
Imagine planning a trip: it’s natural that you might want to get to your destination in the least time, or for the least money, or some other criterion that involves minimizing the sum of edge weights. This is known as computing the shortest path.
We should immediately clarify an unfortunate terminological confusion. What we really want to compute is the lightest path—the one of least weight. Unfortunately, computer science terminology has settled on the terminology we use here; just be sure to not take it literally.
Exercise
Construct a graph and select a pair of nodes in it such that the shortest path from one to the other is not the lightest one, and vice versa.
We have already seen [Depth- and Breadth-First Traversals] that breadth-first search constructs shortest paths in unweighted graphs. These correspond to lightest paths when there are no weights (or, equivalently, all weights are identical and positive). Now we have to generalize this to the case where the edges have weights.
We will proceed inductively, gradually defining a function seemingly of this type
w :: Key -> Number
that reflects the weight of the lightest path from the source node to that one. But let’s think about this annotation: since we’re building this up node-by-node, initially most nodes have no weight to report; and even at the end, a node that is unreachable from the source will have no weight for a lightest (or indeed, any) path. Rather than make up a number that pretends to reflect this situation, we will instead use an option type:
w :: Key -> Option<Number>
When there is some value it will be the weight; otherwise the weight will be none.
Now let’s think about this inductively. What do we know initially? Well, certainly that the source node is at a distance of zero from itself (that must be the lightest path, because we can’t get any lighter). This gives us a (trivial) set of nodes for which we already know the lightest weight. Our goal is to grow this set of nodes—modestly, by one, on each iteration—until we either find the destination, or we have no more nodes to add (in which case our desination is not reachable from the source).
Inductively, at each step we have the set of all nodes for which we know the lightest path (initially this is just the source node, but it does mean this set is never empty, which will matter in what we say next). Now consider all the edges adjacent to this set of nodes that lead to nodes for which we don’t already know the lightest path. Choose a node, $$q$$, that minimizes the total weight of the path to it. We claim that this will in fact be the lightest path to that node.
If this claim is true, then we are done. That’s because we would now add $$q$$ to the set of nodes whose lightest weights we now know, and repeat the process of finding lightest outgoing edges from there. This process has thus added one more node. At some point we will find that there are no edges that lead outside the known set, at which point we can terminate.
It stands to reason that terminating at this point is safe: it corresponds to having computed the reachable set. The only thing left is to demonstrate that this greedy algorithm yields a lightest path to each node.
We will prove this by contradiction. Suppose we have the path $$s \rightarrow d$$ from source $$s$$ to node $$d$$, as found by the algorithm above, but assume also that we have a different path that is actually lighter. At every node, when we added a node along the $$s \rightarrow d$$ path, the algorithm would have added a lighter path if it existed. The fact that it did not falsifies our claim that a lighter path exists (there could be a different path of the same weight; this would be permitted by the algorithm, but it also doesn’t contradict our claim). Therefore the algorithm does indeed find the lightest path.
What remains is to determine a data structure that enables this algorithm. At every node, we want to know the least weight from the set of nodes for which we know the least weight to all their neighbors. We could achieve this by sorting, but this is overkill: we don’t actually need a total ordering on all these weights, only the lightest one. A heap see Wikipedia gives us this.
Exercise
What if we allowed edges of weight zero? What would change in the above algorithm?
Exercise
What if we allowed edges of negative weight? What would change in the above algorithm?
For your reference, this algorithm is known as Dijkstra’s Algorithm.
#### 22.8Moravian Spanning Trees
At the turn of the milennium, the US National Academy of Engineering surveyed its members to determine the “Greatest Engineering Achievements of the 20th Century”. The list contained the usual suspects: electronics, computers, the Internet, and so on. But a perhaps surprising idea topped the list: (rural) electrification.Read more about it on their site.
##### 22.8.1The Problem
To understand the history of national electrical grids, it helps to go back to Moravia in the 1920s. Like many parts of the world, it was beginning to realize the benefits of electricity and intended to spread it around the region. A Moravian academia named Otakar Borůvka heard about the problem, and in a remarkable effort, described the problem abstractly, so that it could be understood without reference to Moravia or electrical networks. He modeled it as a problem about graphs.
Borůvka observed that at least initially, any solution to the problem of creating a network must have the following characteristics:
• The electrical network must reach all the towns intended to be covered by it. In graph terms, the solution must be spanning, meaning it must visit every node in the graph.
• Redundancy is a valuable property in any network: that way, if one set of links goes down, there might be another way to get a payload to its destination. When starting out, however, redundancy may be too expensive, especially if it comes at the cost of not giving someone a payload at all. Thus, the initial solution was best set up without loops or even redundant paths. In graph terms, the solution had to be a tree.
• Finally, the goal was to solve this problem for the least cost possible. In graph terms, the graph would be weighted, and the solution had to be a minimum.
Thus Borůvka defined the Moravian Spanning Tree (MST) problem.
##### 22.8.2A Greedy Solution
Borůvka had published his problem, and another Czech mathematician, Vojtěch Jarník, came across it. Jarník came up with a solution that should sound familiar:
• Begin with a solution consisting of a single node, chosen arbitrarily. For the graph consisting of this one node, this solution is clearly a minimum, spanning, and a tree.
• Of all the edges incident on nodes in the solution that connect to a node not already in the solution, pick the edge with the least weight.Note that we consider only the incident edges, not their weight added to the weight of the node to which they are incident.
• Add this edge to the solution. The claim is that for the new solution will be a tree (by construction), spanning (also by construction), and a minimum. The minimality follows by an argument similar to that used for Dijkstra’s Algorithm.
Jarník had the misfortune of publishing this work in Czech in 1930, and it went largely ignored. It was rediscovered by others, most notably by R.C. Prim in 1957, and is now generally known as Prim’s Algorithm, though calling it Jarník’s Algorithm would attribute credit in the right place.
Implementing this algorithm is pretty easy. At each point, we need to know the lightest edge incident on the current solution tree. Finding the lightest edge takes time linear in the number of these edges, but the very lightest one may create a cycle. We therefore need to efficiently check for whether adding an edge would create a cycle, a problem we will return to multiple times [Checking Component Connectedness]. Assuming we can do that effectively, we then want to add the lightest edge and iterate. Even given an efficient solution for checking cyclicity, this would seem to require an operation linear in the number of edges for each node. With better representations we can improve on this complexity, but let’s look at other ideas first.
##### 22.8.3Another Greedy Solution
Recall that Jarník presented his algorithm in 1930, when computers didn’t exist, and Prim his in 1957, when they were very much in their infancy. Programming computers to track heaps was a non-trivial problem, and many algorithms were implemented by hand, where keeping track of a complex data structure without making errors was harder still. There was need for a solution that was required less manual bookkeeping (literally speaking).
In 1956, Joseph Kruskal presented such a solution. His idea was elegantly simple. The Jarník algorithm suffers from the problem that each time the tree grows, we have to revise the content of the heap, which is already a messy structure to track. Kruskal noted the following.
To obtain a minimum solution, surely we want to include one of the edges of least weight in the graph. Because if not, we can take an otherwise minimal solution, add this edge, and remove one other edge; the graph would still be just as connected, but the overall weight would be no more and, if the removed edge were heavier, would be less.Note the careful wording: there may be many edges of the same least weight, so adding one of them may remove another, and therefore not produce a lighter tree; but the key point is that it certainly will not produce a heavier one. By the same argument we can add the next lightest edge, and the next lightest, and so on. The only time we cannot add the next lightest edge is when it would create a cycle (that problem again!).
Therefore, Kruskal’s algorithm is utterly straightforward. We first sort all the edges, ordered by ascending weight. We then take each edge in ascending weight order and add it to the solution provided it will not create a cycle. When we have thus processed all the edges, we will have a solution that is a tree (by construction), spanning (because every connected vertex must be the endpoint of some edge), and of minimum weight (by the argument above). The complexity is that of sorting (which is $$[e \rightarrow e \log e]$$ where $$e$$ is the size of the edge set. We then iterate over each element in $$e$$, which takes time linear in the size of that set—modulo the time to check for cycles. This algorithm is also easy to implement on paper, because we sort all the edges once, then keep checking them off in order, crossing out the ones that create cycles—with no dynamic updating of the list needed.
##### 22.8.4A Third Solution
Both the Jarník and Kruskal solutions have one flaw: they require a centralized data structure (the priority heap, or the sorted list) to incrementally build the solution. As parallel computers became available, and graph problems grew large, computer scientists looked for solutions that could be implemented more efficiently in parallel—which typically meant avoiding any centralized points of synchronization, such as these centralized data structures.
In 1965, M. Sollin constructed an algorithm that met these needs beautifully. In this algorithm, instead of constructing a single solution, we grow multiple solution components (potentially in parallel if we so wish). Each node starts out as a solution component (as it was at the first step of Jarník’s Algorithm). Each node considers the edges incident to it, and picks the lightest one that connects to a different component (that problem again!). If such an edge can be found, the edge becomes part of the solution, and the two components combine to become a single component. The entire process repeats.
Because every node begins as part of the solution, this algorithm naturally spans. Because it checks for cycles and avoids them, it naturally forms a tree.Note that avoiding cycles yields a DAG and is not automatically guaranteed to yield a tree. We have been a bit lax about this difference throughout this section. Finally, minimality follows through similar reasoning as we used in the case of Jarník’s Algorithm, which we have essentially run in parallel, once from each node, until the parallel solution components join up to produce a global solution.
Of course, maintaining the data for this algorithm by hand is a nightmare. Therefore, it would be no surprise that this algorithm was coined in the digital age. The real surprise, therefore, is that it was not: it was originally created by Otakar Borůvka himself.
Borůvka, you see, had figured it all out. He’d not only understood the problem, he had:
• pinpointed the real problem lying underneath the electrification problem so it could be viewed in a context-independent way,
• created a descriptive language of graph theory to define it precisely, and
• even solved the problem in addition to defining it.
He’d just come up with a solution so complex to implement by hand that Jarník had in essence de-parallelized it so it could be done sequentially. And thus this algorithm lay unnoticed until it was reinvented (several times, actually) by Sollin in time for parallel computing folks to notice a need for it. But now we can just call this Borůvka’s Algorithm, which is only fitting.
As you might have guessed by now, this problem is indeed called the MST in other textbooks, but “M” stands not for Moravia but for “Minimum”. But given Borůvka’s forgotten place in history, we prefer the more whimsical name.
##### 22.8.5Checking Component Connectedness
As we’ve seen, we need to be able to efficiently tell whether two nodes are in the same component. One way to do this is to conduct a depth-first traversal (or breadth-first traversal) starting from the first node and checking whether we ever visit the second one. (Using one of these traversal strategies ensures that we terminate in the presence of loops.) Unfortunately, this takes a linear amount of time (in the size of the graph) for every pair of nodesand depending on the graph and choice of node, we might do this for every node in the graph on every edge addition! So we’d clearly like to do this better.
It is helpful to reduce this problem from graph connectivity to a more general one: of disjoint-set structure (colloquially known as union-find for reasons that will soon be clear). If we think of each connected component as a set, then we’re asking whether two nodes are in the same set. But casting it as a set membership problem makes it applicable in several other applications as well.
The setup is as follows. For arbitrary values, we want the ability to think of them as elements in a set. We are interested in two operations. One is obviously union, which merges two sets into one. The other would seem to be something like is-in-same-set that takes two elements and determines whether they’re in the same set. Over time, however, it has proven useful to instead define the operator find that, given an element, “names” the set (more on this in a moment) that the element belongs to. To check whether two elements are in the same set, we then have to get the “set name” for each element, and check whether these names are the same. This certainly sounds more roundabout, but this means we have a primitive that may be useful in other contexts, and from which we can easily implement is-in-same-set.
Now the question is, how do we name sets? The real question we should ask is, what operations do we care to perform on these names? All we care about is, given two names, they represent the same set precisely when the names are the same. Therefore, we could construct a new string, or number, or something else, but we have another option: simply pick some element of the set to represent it, i.e., to serve as its name. Thus we will associate each set element with an indicator of the “set name” for that element; if there isn’t one, then its name is itself (the none case of parent):
data Element<T>:
| elt(val :: T, parent :: Option<Element>)
end
We will assume we have some equality predicate for checking when two elements are the same, which we do by comparing their value parts, ignoring their parent values:
fun is-same-element(e1, e2): e1.val <=> e2.val end
Do Now!
Why do we check only the value parts?
We will assume that for a given set, we always return the same representative element. (Otherwise, equality will fail even though we have the same set.) Thus:We’ve used the name fynd because find is already defined to mean something else in Pyret. If you don’t like the misspelling, you’re welcome to use a longer name like find-root.
fun is-in-same-set(e1 :: Element, e2 :: Element, s :: Sets)
-> Boolean:
s1 = fynd(e1, s)
s2 = fynd(e2, s)
identical(s1, s2)
end
where Sets is the list of all elements:
type Sets = List<Element>
How do we find the representative element for a set? We first find it using is-same-element; when we do, we check the element’s parent field. If it is none, that means this very element names its set; this can happen either because the element is a singleton set (we’ll initialize all elements with none), or it’s the name for some larger set. Either way, we’re done. Otherwise, we have to recursively find the parent:
fun fynd(e :: Element, s :: Sets) -> Element:
cases (List) s:
| empty => raise("fynd: shouldn't have gotten here")
if is-same-element(f, e):
cases (Option) f.parent:
| none => f
| some(p) => fynd(p, s)
end
else:
fynd(e, r)
end
end
end
Exercise
Why is there a recursive call in the nested cases?
What’s left is to implement union. For this, we find the representative elements of the two sets we’re trying to union; if they are the same, then the two sets are already in a union; otherwise, we have to update the data structure:
fun union(e1 :: Element, e2 :: Element, s :: Sets) -> Sets:
s1 = fynd(e1, s)
s2 = fynd(e2, s)
if identical(s1, s2):
s
else:
update-set-with(s, s1, s2)
end
end
To update, we arbitrarily choose one of the set names to be the name of the new compound set. We then have to update the parent of the other set’s name element to be this one:
fun update-set-with(s :: Sets, child :: Element, parent :: Element)
-> Sets:
cases (List) s:
| empty => raise("update: shouldn't have gotten here")
if is-same-element(f, child):
else:
end
end
end
Here are some tests to illustrate this working:
check:
s0 = map(elt(_, none), [list: 0, 1, 2, 3, 4, 5, 6, 7])
s1 = union(index(s0, 0), index(s0, 2), s0)
s2 = union(index(s1, 0), index(s1, 3), s1)
s3 = union(index(s2, 3), index(s2, 5), s2)
print(s3)
is-same-element(fynd(index(s0, 0), s3), fynd(index(s0, 5), s3)) is true
is-same-element(fynd(index(s0, 2), s3), fynd(index(s0, 5), s3)) is true
is-same-element(fynd(index(s0, 3), s3), fynd(index(s0, 5), s3)) is true
is-same-element(fynd(index(s0, 5), s3), fynd(index(s0, 5), s3)) is true
is-same-element(fynd(index(s0, 7), s3), fynd(index(s0, 7), s3)) is true
end
Unfortunately, this implementation suffers from two major problems:
• First, because we are performing functional updates, the value of the parent reference keeps “changing”, but these changes are not visible to older copies of the “same” value. An element from different stages of unioning has different parent references, even though it is arguably the same element throughout. This is a place where functional programming hurts.
• Relatedly, the performance of this implementation is quite bad. fynd recursively traverses parents to find the set’s name, but the elements traversed are not updated to record this new name. We certainly could update them by reconstructing the set afresh each time, but that complicates the implementation and, as we will soon see, we can do much better.
The bottom line is that pure functional programming is not a great fit with this problem. We need a better implementation strategy: Disjoint Sets Redux.
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# Statistics for Describing, Exploring, and Comparing Data
Degrees of Freedom: Ten values have a mean of 75.0. Nine of the values are 62,78,90,87,56,92,70,70, and 93.
a. Find the 10th value.
b. We need to create a list of n values that have a specific known mean. We are free to select any values we desire for some of the n values. How many of the n values can be freely assigned before the remaining values are determined? (The result is often referred to as the number of degrees of freedom).
Range Rule of Thumb: Aluminum cans with a thickness of 0.0111 in. have axial loads with a mean of 281.8 lb and a standard deviation of 27.8 lb. The axial load is measured by applying pressure to the top of the can until it collapses. Use the range rule of thumb to find the minimum and maximum "usual" axial loads. One particular can had an axial load of 504 lb. Is that unusual?
Comparing Scores: Three students take equivalent stress tests. Which is the highest relative score?
a. A score of 144 on a test with a mean of 128 and a standard deviation of 34.
b. A score of 90 on a test with a mean of 86 and a standard deviation of 18.
c. A score of 18 on a test with a mean of 15 and a standard deviation of 5.
#### Solution Preview
See the attached file.
Degrees of Freedom: Ten values have a mean of 75.0. Nine of the values are 62,78,90,87,56,92,70,70, and 93.
a. Find the 10th value.
To find the mean (the average), you add up all the values, then divide the total by the number of values that you have. We can use this to set up an equation and solve for the missing value (we'll call the missing value x).
75.0 = (62 + 78 + 90 + 87 + 56 + 92 + 70 + 70 + x)/10
750 = 62 + 78 + 90 + 87 + 56 + 92 + 70 + 70 + x
750 = 605 + x
145 = x
The missing value is 145. Now that we have all the values, find the mean of the 10 values to verify that it's equal to 75.
b. We need to create a list of n values that have a specific known mean. We are free to select any values we desire for some of the n values. How many of the n values can be freely assigned before the remaining values are determined? (The result is often referred to as the number of degrees of freedom).
Whenever you have one group of n numbers, the degrees of freedom is equal to n - 1.
In the previous problem, we knew the values of n - 1 numbers (the 9 we were given) and using this information, we were able to find the 10th number. What if we were given n - 2 numbers? ...
#### Solution Summary
This problem set has three questions concerning means and standard deviations, degrees of freedom, range of a data set, and relative scores. The solution provides answers and explanations to all parts of the problem set.
\$2.19
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# How do you evaluate and simplify 16^(3/2)?
Feb 3, 2017
See the entire solution process below:
#### Explanation:
First, we can use this rule of exponents to rewrite this expression:
${x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}} = {\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}}$
${16}^{\frac{3}{2}} = {16}^{\textcolor{red}{\frac{1}{2}} \times \textcolor{b l u e}{3}} = {\left({16}^{\textcolor{red}{\frac{1}{2}}}\right)}^{\textcolor{b l u e}{3}}$
We can rewrite and simplify this as:
${\left({16}^{\textcolor{red}{\frac{1}{2}}}\right)}^{\textcolor{b l u e}{3}} = {\left(\sqrt{16}\right)}^{\textcolor{b l u e}{3}} = {4}^{\textcolor{b l u e}{3}}$
Now, we can simplify this to:
${4}^{\textcolor{b l u e}{3}} = 4 \times 4 \times 4 = 16 \times 4 = 64$
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# Kindergarten Standards
• Kindergarten Math Standards
# Counting and comparing numbers
I can count to 100 by ones and tens. K.C.1
I can count forward starting at a given number. K.CC.2
I can write numbers from 0 to 20. K.CC.3
I can write a number for a group of 0 to 20 objects. K.CC.3
I can count objects, saying the numbers in order. K.CC.4
After counting some objects, I know the last number I say tells the number of objects I counted. K.CC.4
I understand that the last object counted tells the number of objects in a group. K.CC.4
I understand that the total number of objects in a group stays the same even when placed in different ways or counted in a different order. K.CC.4
When counting in order by one, I know that each number I say is getting bigger by one K.CC.4
I can count to tell how many objects there are in a group. K.CC.5
I can tell if a group of objects in one group is greater than, less than or equal to a group of objects in another group. K.CC.6
I can compare two written numbers between 1 and 10. K.CC.7
Addition is putting together and adding to/ Subtraction is taking apart and taking from
I can show addition and subtraction by acting out situations, using objects, fingers, drawings, sounds, verbal explanations, expressions, or equations. K.OA.1
I can solve addition and subtraction word problems within 10. K.OA.2
I can decompose numbers less than or equal to 10, using objects, drawings, and equations. K.OA.3 (5 = 2 + 3)
By using objects or drawings, I can find the number that is added to the numbers 1 through 9 to make 10 K.OA.4
I can fluently add and subtract within 5. K.OA.5
# Placement Value
By using objects or drawings, I can compose and decompose numbers from 11 to 19, into tens and ones K.NBT.1
# Measurement and Data
I can describe the measurable attributes of an object (length, weight, height) K.MD.1
I can compare two objects to see which object has “more of”/”less of” a common attribute K.MD.2
I can count, classify, and sort objects into categories. K.MD.3
# Geometry
I can name shapes in the environment and use terms such as: above, below, beside, next to, in front of, and behind, to describe their relative positions. K.G.1
I can name the following shapes: square, circle, triangle, rectangle, hexagon, cube, cone, cylinder, sphere. K.G.2
I know which shapes are two-dimensional and which are three- dimensional. K.G.3
I can use words to compare two-dimensional and three-dimensional shapes, describing how they are the same and how they are different (ex. number of sides, vertices; sides of equal length) K.G.4
I can make models of shapes K.G.5
I can connect simple shapes to make larger different shapes. K.G.6
# Speaking and Listening
I follow rules for discussions. K.SL.1
a
I participate in conversations. K.SL.1b
I make sure I understand a text by asking and answering questions about details and by asking for an explanation when something is not clear. K.SL.2
I can follow one-step and two-step oral directions. K.SL.2a
I can ask and answer questions to get information and when I do not understand something. K.SL.3
I can describe people, places, things and events. K.SL.4
I can use drawings and other displays to add more details to my oral descriptions. K.SL.5
I speak in a way that others can hear me, expressing my thoughts, feelings and ideas clearly. K.SL.6
# Foundational Skills
I follow words from left to right, top to bottom, page by page K.RF.1a
I know spoken words are written using sequences of letters. K.RF.1b
I know that written words are separated by spaces. K.RF.1c
I can name all upper and lowercase letters. K.RF.1d
I can recognize and name rhyming words K.RF.2a
I can divide words into syllables, (counting, pronouncing, and blending the syllables). K.RF.2b
I can blend and divide onsets and rimes of single-syllable words. For example, in the word bat, b- is the onset, and -at is the rime. K.RF.2c
I can find and say the initial, middle vowel and last sound in CVC words (consonant-vowel-consonant) K.RF.2d
I can change or add a sound to make new words K.RF.2e
I can read words by blending 2 to 3 sounds. K.RF.2f
I know the common sounds for each consonant. K.RF.3a
I can know the long and short sound of each vowel. K.RF.3b
I can read common, high-frequency words, (the, of, to, you, she, my, is, are, do, does). K.RF.3c
I can use the sounds letters make to read words that are spelled almost the same. K.RF.3d
# Language
I can print many uppercase and lowercase letters. K.L.1a
I use nouns and verbs. K.L.1
b
I can orally say regular plural nouns by adding an “s” or “es”. K.L.1
c
I understand and use question words (who, what, where, when, why, how). K.L.1d
I use common prepositions. (to, from, in, out, on, off, for, of, by, with) K.L.1e
I write complete sentences. K.L.1
f
I use capitals for the first word in a sentence and for the pronoun, “I” K.L.2
a
I recognize and name end punctuations. K.L.2
b
I can write a letter or letters for most consonant and short vowel sounds. K.L.2c
I spell words by sounding them out. K.L.2d
I figure out new meanings for words I already know. (A duck is a bird. To duck is to bend down.) K.L.4
a
I use common beginnings and endings to help me figure out what a word means. (-ed, -s, re-, un-, pre-, -ful, -less)
K.L.4b
I can sort objects into groups. K.L.5a
I can say the antonym of a word. K.L.5b
I can connect real-life to words and their use. K.L.5c
I can show the difference between similar verbs by acting out their meanings. (walk, march, strut, prance) K.L.5
d
I use new words I have learned through conversations, reading and being read to. K.L.6
# Kindergarten Writing
I can draw, dictate, and write to tell my opinion about a topic or book. (My favorite book is . . .) K.W.1
I can draw, dictate, and write to explain or give information about a topic. K.W.2
I can draw, dictate, and write to tell about something that happened, telling about it in the order it happened and sharing how I felt or what I did because of what happened. K.W.3
After listening to suggestions by adults and my classmates, I can improve my writing by adding details. K.W.5
I can publish my writing. K.W.6
I can participate in writing a group research project. K.W.7
I can answer questions by recalling information or by looking for the answer in items given to me by an adult. K.W.8
# Literature
I can retell a story. K.RL.2
I can identify the characters, setting and events in a story. K.RL.3
I recognize different types of text. (stories, poems, fantasy, realistic text, etc.) K.RL.5
I can name the author and the illustrator of a story and explain what their role was in creating the story. K.RL.6
I can find where the illustrations happen in the text of a story. K.RL.7
I can tell how the adventures and experiences of the characters in a story are the same and how they are different. K.RL.9
I use what I already know to help me understand texts when I am reading in a group. K.RL.10a
I can make predictions about a text. K.RL.10b
# Informational Text
I can name the main topic and retell details in a text. K.RI.2
I can tell how two individuals, events, ideas, or other information in a text are connected. K.RI.3
I can ask and answer questions to help me understand the meaning of new words in a text. K.RI.4
I can find the front cover, back cover and title page in a book. K.RI.5
I can name the author and the illustrator of a story and I can tell what they did in presenting the ideas or information in the text. K.RI.6
I use both words and illustrations in a text to understand the text K.RI.7
I can find the reasons an author gives to support points in the text. K.RI.8
I can tell how two texts on the same topic are alike and/or different. K.RI.9
I use what I already know to help me understand texts when I am reading in a group. K.RI.10a
I use illustrations and the text to make predictions. K.RI.10b
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# Relation (mathematics)
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This page belongs to resource collections on Logic and Inquiry.
In mathematics, a finitary relation is defined by one of the formal definitions given below.
• The basic idea is to generalize the concept of a two-place relation, such as the relation of equality denoted by the sign “$$=\!$$” in a statement like $$5 + 7 = 12\!$$ or the relation of order denoted by the sign “$${<}\!$$” in a statement like $$5 < 12.\!$$ Relations that involve two places or roles are called binary relations by some and dyadic relations by others, the latter being historically prior but also useful when necessary to avoid confusion with binary (base 2) numerals.
• The concept of a two-place relation is generalized by considering relations with increasing but still finite numbers of places or roles. These are called finite-place or finitary relations. A finitary relation involving $$k\!$$ places is variously called a $$k\!$$-ary, $$k\!$$-adic, or $$k\!$$-dimensional relation. The number $$k\!$$ is then called the arity, the adicity, or the dimension of the relation, respectively.
## Informal introduction
The definition of relation given in the next section formally captures a concept that is actually quite familiar from everyday life. For example, consider the relationship, involving three roles that people might play, expressed in a statement of the form $$X ~\text{suspects that}~ Y ~\text{likes}~ Z.\!$$ The facts of a concrete situation could be organized in the form of a Table like the one below:
$$\text{Person}~ X\!$$ $$\text{Person}~ Y\!$$ $$\text{Person}~ Z\!$$ $$\text{Alice}\!$$ $$\text{Bob}\!$$ $$\text{Denise}\!$$ $$\text{Charles}\!$$ $$\text{Alice}\!$$ $$\text{Bob}\!$$ $$\text{Charles}\!$$ $$\text{Charles}\!$$ $$\text{Alice}\!$$ $$\text{Denise}\!$$ $$\text{Denise}\!$$ $$\text{Denise}\!$$
Each row of the Table records a fact or makes an assertion of the form $$X ~\text{suspects that}~ Y ~\text{likes}~ Z.\!$$ For instance, the first row says, in effect, $$\text{Alice suspects that Bob likes Denise.}\!$$ The Table represents a relation $$S\!$$ over the set $$P\!$$ of people under discussion:
$$P ~=~ \{ \text{Alice}, \text{Bob}, \text{Charles}, \text{Denise} \}\!$$
The data of the Table are equivalent to the following set of ordered triples:
$$\begin{smallmatrix} S & = & \{ & \text{(Alice, Bob, Denise)}, & \text{(Charles, Alice, Bob)}, & \text{(Charles, Charles, Alice)}, & \text{(Denise, Denise, Denise)} & \} \end{smallmatrix}\!$$
By a slight overuse of notation, it is usual to write $$S (\text{Alice}, \text{Bob}, \text{Denise})\!$$ to say the same thing as the first row of the Table. The relation $$S\!$$ is a triadic or ternary relation, since there are three items involved in each row. The relation itself is a mathematical object, defined in terms of concepts from set theory, that carries all the information from the Table in one neat package.
The Table for relation $$S\!$$ is an extremely simple example of a relational database. The theoretical aspects of databases are the specialty of one branch of computer science, while their practical impacts have become all too familiar in our everyday lives. Computer scientists, logicians, and mathematicians, however, tend to see different things when they look at these concrete examples and samples of the more general concept of a relation.
For one thing, databases are designed to deal with empirical data, and experience is always finite, whereas mathematics is nothing if not concerned with infinity, at the very least, potential infinity. This difference in perspective brings up a number of ideas that are usefully introduced at this point, if by no means covered in depth.
## Example 1. Divisibility
A more typical example of a two-place relation in mathematics is the relation of divisibility between two positive integers $$n\!$$ and $$m\!$$ that is expressed in statements like $${}^{\backprime\backprime} n ~\text{divides}~ m {}^{\prime\prime}\!$$ or $${}^{\backprime\backprime} n ~\text{goes into}~ m {}^{\prime\prime}.\!$$ This is a relation that comes up so often that a special symbol $${}^{\backprime\backprime} | {}^{\prime\prime}\!$$ is reserved to express it, allowing one to write $${}^{\backprime\backprime} n|m {}^{\prime\prime}\!$$ for $${}^{\backprime\backprime} n ~\text{divides}~ m {}^{\prime\prime}.\!$$
To express the binary relation of divisibility in terms of sets, we have the set $$P\!$$ of positive integers, $$P = \{ 1, 2, 3, \ldots \},\!$$ and we have the binary relation $$D\!$$ on $$P\!$$ such that the ordered pair $$(n, m)\!$$ is in the relation $$D\!$$ just in case $$n|m.\!$$ In other turns of phrase that are frequently used, one says that the number $$n\!$$ is related by $$D\!$$ to the number $$m\!$$ just in case $$n\!$$ is a factor of $$m,\!$$ that is, just in case $$n\!$$ divides $$m\!$$ with no remainder. The relation $$D,\!$$ regarded as a set of ordered pairs, consists of all pairs of numbers $$(n, m)\!$$ such that $$n\!$$ divides $$m.\!$$
For example, $$2\!$$ is a factor of $$4,\!$$ and $$6\!$$ is a factor of $$72,\!$$ which two facts can be written either as $$2|4\!$$ and $$6|72\!$$ or as $$D(2, 4)\!$$ and $$D(6, 72).\!$$
## Formal definitions
There are two definitions of $$k\!$$-place relations that are commonly encountered in mathematics. In order of simplicity, the first of these definitions is as follows:
Definition 1. A relation $$L\!$$ over the sets $$X_1, \ldots, X_k\!$$ is a subset of their cartesian product, written $$L \subseteq X_1 \times \ldots \times X_k.\!$$ Under this definition, then, a $$k\!$$-ary relation is simply a set of $$k\!$$-tuples.
The second definition makes use of an idiom that is common in mathematics, saying that “such and such is an $$n\!$$-tuple” to mean that the mathematical object being defined is determined by the specification of $$n\!$$ component mathematical objects. In the case of a relation $$L\!$$ over $$k\!$$ sets, there are $$k + 1\!$$ things to specify, namely, the $$k\!$$ sets plus a subset of their cartesian product. In the idiom, this is expressed by saying that $$L\!$$ is a $$(k+1)\!$$-tuple.
Definition 2. A relation $$L\!$$ over the sets $$X_1, \ldots, X_k\!$$ is a $$(k+1)\!$$-tuple $$L = (X_1, \ldots, X_k, \mathrm{graph}(L)),\!$$ where $$\mathrm{graph}(L)\!$$ is a subset of the cartesian product $$X_1 \times \ldots \times X_k~\!$$ called the graph of $$L.\!$$
Elements of a relation are sometimes denoted by using boldface characters, for example, the constant element $$\mathbf{a} = (a_1, \ldots, a_k)\!$$ or the variable element $$\mathbf{x} = (x_1, \ldots, x_k).\!$$
A statement of the form “$$\mathbf{a}\!$$ is in the relation $$L\!$$” is taken to mean that $$\mathbf{a}\!$$ is in $$L\!$$ under the first definition and that $$\mathbf{a}\!$$ is in $$\mathrm{graph}(L)\!$$ under the second definition.
The following considerations apply under either definition:
• The sets $$X_j~\!$$ for $$j = 1 ~\text{to}~ k\!$$ are called the domains of the relation. In the case of the first definition, the relation itself does not uniquely determine a given sequence of domains.
• If all the domains $$X_j~\!$$ are the same set $$X,\!$$ then $$L\!$$ is more simply referred to as a $$k\!$$-ary relation over $$X.\!$$
• If any domain $$X_j~\!$$ is empty then the cartesian product is empty and the only relation over such a sequence of domains is the empty relation $$L = \varnothing.\!$$ Most applications of the relation concept will set aside this trivial case and assume that all domains are nonempty.
If $$L\!$$ is a relation over the domains $$X_1, \ldots, X_k,\!$$ it is conventional to consider a sequence of terms called variables, $$x_1, \ldots, x_k,\!$$ that are said to range over the respective domains.
A boolean domain $$\mathbb{B}\!$$ is a generic 2-element set, say, $$\mathbb{B} = \{ 0, 1 \},\!$$ whose elements are interpreted as logical values, typically $$0 = \mathrm{false}\!$$ and $$1 = \mathrm{true}.\!$$
The characteristic function of the relation $$L,\!$$ written $$f_L\!$$ or $$\chi(L),\!$$ is the boolean-valued function $$f_L : X_1 \times \ldots \times X_k \to \mathbb{B},\!$$ defined in such a way that $$f_L (\mathbf{x}) = 1\!$$ just in case the $$k\!$$-tuple $$\mathbf{x} = (x_1, \ldots, x_k)\!$$ is in the relation $$L.\!$$ The characteristic function of a relation may also be called its indicator function, especially in probabilistic and statistical contexts.
It is conventional in applied mathematics, computer science, and statistics to refer to a boolean-valued function like $$f_L\!$$ as a $$k\!$$-place predicate. From the more abstract viewpoints of formal logic and model theory, the relation $$L\!$$ is seen as constituting a logical model or a relational structure that serves as one of many possible interpretations of a corresponding $$k\!$$-place predicate symbol, as that term is used in predicate calculus.
Due to the convergence of many traditions of study, there are wide variations in the language used to describe relations. The extensional approach presented in this article treats a relation as the set-theoretic extension of a relational concept or term. An alternative, intensional approach reserves the term relation to the corresponding logical entity, either the logical comprehension, which is the totality of intensions or abstract properties that all the elements of the extensional relation have in common, or else the symbols that are taken to denote those elements and intensions.
## Example 2. Coplanarity
For lines $$\ell\!$$ in three-dimensional space, there is a triadic relation picking out the triples of lines that are coplanar. This does not reduce to the dyadic relation of coplanarity between pairs of lines.
In other words, writing $$P(\ell, m, n)\!$$ when the lines $$\ell, m, n\!$$ lie in a plane, and $$Q(\ell, m)\!$$ for the binary relation, it is not true that $$Q(\ell, m),\!$$ $$Q(m, n),\!$$ and $$Q(n, \ell)\!$$ together imply $$P(\ell, m, n),\!$$ although the converse is certainly true: any two of three coplanar lines are necessarily coplanar. There are two geometrical reasons for this.
In one case, for example taking the $$x\!$$-axis, $$y\!$$-axis, and $$z\!$$-axis, the three lines are concurrent, that is, they intersect at a single point. In another case, $$\ell, m, n\!$$ can be three edges of an infinite triangular prism.
What is true is that if each pair of lines intersects, and the points of intersection are distinct, then pairwise coplanarity implies coplanarity of the triple.
## Remarks
Relations are classified by the number of sets in the cartesian product, in other words, the number of places or terms in the relational expression:
$$L(a)\!$$ Monadic or unary relation, in other words, a property or set $$L(a, b) ~\text{or}~ a L b\!$$ Dyadic or binary relation $$L(a, b, c)\!$$ Triadic or ternary relation $$L(a, b, c, d)\!$$ Tetradic or quaternary relation $$L(a, b, c, d, e)\!$$ Pentadic or quinary relation
Relations with more than five terms are usually referred to as $$k\!$$-adic or $$k\!$$-ary, for example, a 6-adic, 6-ary, or hexadic relation.
## References
• Peirce, C.S. (1870), “Description of a Notation for the Logic of Relatives, Resulting from an Amplification of the Conceptions of Boole's Calculus of Logic”, Memoirs of the American Academy of Arts and Sciences 9, 317–378, 1870. Reprinted, Collected Papers CP 3.45–149, Chronological Edition CE 2, 359–429.
• Ulam, S.M., and Bednarek, A.R. (1990), “On the Theory of Relational Structures and Schemata for Parallel Computation”, pp. 477–508 in A.R. Bednarek and Françoise Ulam (eds.), Analogies Between Analogies : The Mathematical Reports of S.M. Ulam and His Los Alamos Collaborators, University of California Press, Berkeley, CA.
## Bibliography
• Bourbaki, N. (1994), Elements of the History of Mathematics, John Meldrum (trans.), Springer-Verlag, Berlin, Germany.
• Halmos, P.R. (1960), Naive Set Theory, D. Van Nostrand Company, Princeton, NJ.
• Lawvere, F.W., and Rosebrugh, R. (2003), Sets for Mathematics, Cambridge University Press, Cambridge, UK.
• Maddux, R.D. (2006), Relation Algebras, vol. 150 in Studies in Logic and the Foundations of Mathematics, Elsevier Science.
• Mili, A., Desharnais, J., Mili, F., with Frappier, M. (1994), Computer Program Construction, Oxford University Press, New York, NY.
• Minsky, M.L., and Papert, S.A. (1969/1988), Perceptrons, An Introduction to Computational Geometry, MIT Press, Cambridge, MA, 1969. Expanded edition, 1988.
• Peirce, C.S. (1984), Writings of Charles S. Peirce : A Chronological Edition, Volume 2, 1867–1871, Peirce Edition Project (eds.), Indiana University Press, Bloomington, IN.
• Royce, J. (1961), The Principles of Logic, Philosophical Library, New York, NY.
• Tarski, A. (1956/1983), Logic, Semantics, Metamathematics, Papers from 1923 to 1938, J.H. Woodger (trans.), 1st edition, Oxford University Press, 1956. 2nd edition, J. Corcoran (ed.), Hackett Publishing, Indianapolis, IN, 1983.
• Ulam, S.M. (1990), Analogies Between Analogies : The Mathematical Reports of S.M. Ulam and His Los Alamos Collaborators, A.R. Bednarek and Françoise Ulam (eds.), University of California Press, Berkeley, CA.
• Venetus, P. (1984), Logica Parva, Translation of the 1472 Edition with Introduction and Notes, Alan R. Perreiah (trans.), Philosophia Verlag, Munich, Germany.
## Syllabus
### Logical operators
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### Related topics
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### Relational concepts
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### Information, Inquiry
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## Document history
Portions of the above article were adapted from the following sources under the GNU Free Documentation License, under other applicable licenses, or by permission of the copyright holders.
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# Mastering PSLE Maths speed questions: A proven trick
Solving speed questions in the PSLE Mathematics paper can be tough for students. In recent years, we've come across many challenging speed-related questions, like this one that stumped many students. To tackle these problems, you need a good understanding of how speed, time, and distance are connected. But don't worry, we have a clever trick to help you handle them with confidence.
In this article, we'll walk you through the process using the tips mentioned above.
## Understanding the basics
Before we dive into the trick, it's crucial to understand the fundamental equation that underpins speed, time, and distance problems:
Distance = Speed × Time
This equation is your best friend when dealing with these types of questions. It relates the distance travelled by an object to its speed and the time it takes to travel.
## The trick: Inverse ratio method
Now, here's where the trick comes into play: the Inverse Ratio Method.
This method is incredibly useful when you're given information about time but not speed or distance. It allows you to solve speed questions efficiently.
Let's break it down:
Begin by comparing the time taken by different objects or individuals. For instance, if one person takes 6 hours, and the other takes 4 hours, the time ratio is 6:4, which simplifies to 3:2.
### Invert the ratio
This is the key step. To determine the speed ratio, flip the time ratio. In our example, 3:2 becomes 2:3. This inverted ratio reveals the relative speeds of the two objects.
### Apply the speed ratio
With the speed ratio at hand, you can now calculate how much distance each object covers. For example, if one object covers two units, the other covers three units in the same amount of time. Visualising this on a distance graph can be incredibly helpful.
## Practical application: PSLE speed questions
Let's apply this trick to a sample PSLE question:
Question: Jackson and Tom start travelling at 8 AM. Jackson takes 4 hours to reach a destination, while Tom takes 3 hours. When do they meet each other?
• Determine the time ratio: The time ratio between Jackson and Tom is 4:3.
• Invert the ratio: Inverting the ratio gives us 3:4, which represents their speed ratio.
• Apply the speed ratio: Now, for every 3 units that Emily travels, Sarah covers 4 units. Visualise this on a distance graph.
With this information, you can accurately calculate when Emily and Sarah will meet. This method simplifies complex speed questions into manageable steps, making PSLE Mathematics a breeze.
## Conclusion
Mastering speed questions in PSLE Mathematics is a significant achievement for any student. The trick we've outlined, using the Inverse Ratio Method, empowers you to tackle these questions with ease.
Understanding the basic equation relating distance, speed, and time is essential, and the Inverse Ratio Method takes you a step further in your problem-solving journey.
So, the next time you encounter a speed question in your PSLE Math paper, remember this trick. It will help you find the solution efficiently and accurately. Good luck with your upcoming PSLE Math examination!
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# 3 ways to calculate the center of gravity of a triangle
The center of gravity, also called the centroid, is the point at which the mass of a triangle is in equilibrium. To help you better understand this concept, imagine you have a triangular shaped tile above the tip of a pencil. The tile will be kept in balance if and only if the pencil is placed just in the center of its gravity. This concept may be necessary in several fields of application, such as in the field of engineering and design. To find the point of gravity of a triangle, you will only have to use simple geometric operations.
## Steps
### Method 1 of 3: Use the intersection of medians method
#### Step 1. Find the midpoint on another side of the triangle
To do this, measure the side in question and divide the length by two. Write the letter A on the midpoint.
• Suppose one side of a triangle is 10 cm. The midpoint will be 5cm because 10/2 = 5 { displaystyle 10/2 = 5}
#### Step 2. Find the midpoint on another side of the triangle
To do this, you need to measure the length of this segment and divide by two. Write the letter B in the middle of this segment.
• For example, if another side of the triangle is 12cm in length, the midpoint will be 6cm because 12/2 = 6 { displaystyle 12/2 = 6}
#### Step 3. Draw a line from the middle of each side to the opposite vertex
These two lines represent the median on each side.
### The vertex is the meeting point of the two sides of the triangle
#### Step 4. Draw a point at the intersection of the two midpoints
This point is the center of gravity of your triangle, and is also called the centroid or center of mass.
### Method 2 of 3: Use the 2: 1 ratio
#### Step 1. Draw a median of the triangle
In geometry, the median is a line that extends from the middle of a segment to the opposite vertex. You can choose any median of a triangle.
#### Step 2. Measure the length of the median
Make sure you measure it properly.
### For example, you can have a median that is 3.6 cm long
#### Step 3. Divide the length of the median into three equal parts
You just need to divide by three. Again, be sure to do an exact calculation. If you round the result, you will not find the exact center of gravity.
• For example, if the median is 3.6cm long, you should divide 3.6 by 3: 3.6cm / 3 = 1.2cm { displaystyle 3.6cm / 3 = 1.2cm}
, donc le tiers de cette longueur est égale à 1, 2 cm.
#### Step 4. Mark a point on the third of the median from the midpoint
This point represents the centroid of the triangle. The center of gravity always divides a median, which translates to a ratio of 2: 1. In other words, the centroid is one-third of the median distance from the midpoint, and two-thirds of the median distance from the top.
### Method 3 of 3: Use the average of the coordinates
#### Step 1. Find the coordinates of all the vertices of the triangle
This trick works, as long as you are working in a coordinate plane. They might already be mentioned in your exercise, or you might have a triangle drawn on a graph with no coordinates. Remember that coordinates are always expressed as (x, y) { displaystyle (x, y)}
### Supposons que vous avez un triangle PQR. Vous devrez retrouver les coordonnées et les réécrire sous cette forme: le point P (3, 5), le point Q (4, 1), le point R (1, 0)
#### Step 2. Add the values of the x coordinates
Remember to add the three x coordinates. You will not find the precise value of the center of gravity if you only use two coordinates.
• For example, if your three x coordinates are 3, 4, and 1, add these values (3 + 4 + 1 = 8 { displaystyle 3 + 4 + 1 = 8}
).
#### Step 3. Add the y coordinate values
Remember to add the three y coordinates.
• For example, if the three y coordinates of your exercise are 5, 1, and 0, add these values (5 + 1 + 0 = 6 { displaystyle 5 + 1 + 0 = 6}
).
#### Step 4. Calculate the average of the x and y coordinates
These coordinates represent the center of gravity of the triangle, also called the centroid or center of mass. To calculate the average, add the coordinates and divide the result by 3.
• For example, if the sum of the coordinates in x is equal to 8, the average of the coordinates in x will be equal to 8/3 { displaystyle 8/3}
. Si la somme des coordonnées en y est égale à 6, la moyenne des coordonnées sera égale à 6/3{displaystyle 6/3}
, ou 2{displaystyle 2}
#### Step 5. Represent the centroid
By following this method, the center of gravity is nothing more than the average of the x and y coordinates.
• So for our example, the centroid is the point (8/3, 2) { displaystyle (8/3, 2)}
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#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.3 Question 17 Maths Textbook Solution.
Answer: $\frac{2}{3}\left\{(x+3)^{\frac{3}{2}}+(x+2)^{\frac{3}{2}}\right\}+c$
Hint: $\text { To solve this equation we will add }(\mathrm{x}+3)+(\mathrm{x}+2 \text { ) in numerator and denominator }$
Given: $\int \frac{1}{\sqrt{x+3}-\sqrt{x+2}} d x$
Solution: $\int \frac{1}{\sqrt{x+3}-\sqrt{x+2}} d x$
\begin{aligned} &=\int \frac{\sqrt{x+3}+\sqrt{x+2}}{(\sqrt{x+3}-\sqrt{x+2})(\sqrt{x+3}+\sqrt{x+2})} d x \\ &=\int \frac{\sqrt{x+3}+\sqrt{x+2}}{x+3-x-2} d x \\ &=\int \sqrt{x+3} d x+\int \sqrt{x+2} d x \end{aligned}
\begin{aligned} &{\left[\int(a x+b)^{n}=\frac{(a x+b)^{n+1}}{a(n+1)}+c, n \neq 1\right]} \\ &=\frac{2(x+3)^{\frac{3}{2}}}{3}+\frac{2(x+2)^{\frac{3}{2}}}{3}+c \\ &=\frac{2}{3}\left\{(x+3)^{\frac{3}{2}}+(x+2)^{\frac{3}{2}}\right\}+c \end{aligned}
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# What Is A Cubic Unit?
Are you curious to know what is a cubic unit? You have come to the right place as I am going to tell you everything about a cubic unit in a very simple explanation. Without further discussion let’s begin to know what is a cubic unit?
In the world of mathematics and geometry, concepts related to measurement form the foundation for understanding the spatial dimensions of objects. One such concept is the cubic unit, which plays a crucial role in quantifying volume in three-dimensional space. In this blog post, we will explore the concept of cubic units, their significance, and how they enable us to measure volume accurately and consistently.
## What Is A Cubic Unit?
A cubic unit is a standardized measurement used to quantify volume in three-dimensional space. It represents the volume of a cube with each edge measuring one unit length. Just as a linear unit measures length and an area unit measures the extent of a surface, a cubic unit measures the amount of space an object occupies in three dimensions.
## Visualizing Cubic Units
Imagine a cube with sides that are each one unit in length. This cube, with its length, width, and height all equal to one unit, is the basic building block for understanding cubic units. When you stack multiple of these cubes together, you form larger objects with greater volume. The number of individual cubes used to fill a given space is a direct representation of the volume of that space, measured in cubic units.
## Calculating Volume With Cubic Units
To calculate the volume of an object using cubic units, you multiply the length, width, and height of the object in the same unit of measurement. The formula for volume is:
Volume = Length × Width × Height
For example, if you have a rectangular box with a length of 3 units, a width of 2 units, and a height of 4 units, the volume would be:
Volume = 3 units × 2 units × 4 units = 24 cubic units
## Applications Of Cubic Units
1. Real-World Objects: Cubic units are used to measure the volume of tangible objects, such as shipping containers, storage boxes, and even liquids in containers.
2. Architectural Planning: Architects use cubic units to determine the space required for rooms, buildings, and other structures.
3. Scientific Research: In scientific experiments, researchers use cubic units to measure the volume of substances and analyze their properties.
4. Construction and Engineering: Builders and engineers use cubic units to quantify the volume of materials needed for construction projects, such as concrete, soil, and gravel.
5. Art and Design: Artists and designers use cubic units to conceptualize and create three-dimensional sculptures, structures, and artworks.
## Conclusion
Cubic units are fundamental to understanding volume in three-dimensional space. By providing a consistent and standardized measurement for quantifying the amount of space an object occupies, cubic units enable us to solve practical problems in various fields. Whether you’re calculating the volume of a storage container, designing a building, or conducting scientific research, a solid grasp of cubic units empowers you to accurately measure and visualize the spatial dimensions of objects in our three-dimensional world.
You can search for more about similar topics like these on Tipsfeed.
## FAQ
### What Is The Cubic Units?
A cubic unit is a unit used to measure volume. In other words, these units are a way of measuring the space that is occupied by something. Think of a book: What is its volume? Its volume depends on its length, width, and height. Multiplying these three linear measurements gives us a cubic measurement.
### What Is The Difference Between Unit Cube And A Cubic Unit?
Volume is measured by the number of cubic units that can be packed into a figure. A cube with edge lengths of 1 unit is called a unit cube. A unit cube has 1 cubic unit of volume.
### What Is A Cubic Unit In Cm?
A cubic centimetre (or cubic centimeter in US English) (SI unit symbol: cm3; non-SI abbreviations: cc and ccm) is a commonly used unit of volume that corresponds to the volume of a cube that measures 1 cm × 1 cm × 1 cm. One cubic centimetre corresponds to a volume of one millilitre.
### Why Do We Measure In Cubic Units?
Any object possesses a volume by virtue of the object being three-dimensional. Explanation: All three dimensions must be multiplied while calculating the volume. Therefore, volume is measured in cubic units such as cubic meters, cubic inches, etc.
I Have Covered All The Following Queries And Topics In The Above Article
What Is A Cubic Unit
What Is A Cubic Unit In Math
A Cubic Centimeter Is A Unit For Measuring What
What Is The Coordination Number In A Simple Cubic Unit Cell
What Is The Cubic Centimeter A Metric Unit For
What Is The Coordination Number In A Simple Cubic Unit Cel
What Is The Planar Density Of The (010) Plane In A Body Centered Cubic (Bcc) Unit Cell Chegg
A Cubic Centimeter (Cm3)Is Equivalent To What Other Metric Volume Unit?
What Is The Coordination In A Body Centered Cubic Unit Cell
What Cubic Si Unit Is A Liter Equivalent To
What Is Teh Smallest Volumetric Repeating Unit Of A Cubic Crystal
What Is The Coordination Number In A Body Centered Cubic Unit Cell
What Is A For A Simple Cubic Unit Cell
What Is A Cubic Unit
What is a cubic unit in geometry?
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# Numerical example 2 | Singly reinforced Sections
#### Guide to design of Singly reinforced Sections | Civil Engineering
For “Singly reinforced sections” article series, we have covered the following:
Now we will move on with our next solved example in which we will make use of formulas derived earlier. That is why it is necessary that you go through the entire step by step guide in order to gain complete understanding.
#### Numerical Problem
Calculate the moment of resistance of an RC beam 250mm wide, the depth of the centre of reinforcement being 500mm. Assume σcbc = 5N/mm2, σst = 140 N/mm2, modular ratio = 18.66
#### Given that,
b = width of the beam = 250mm
d = depth of the beam = 500mm
σcbc = permissible compressive stress in concrete in bending = 5N/mm2
σst = permissible stress in steel = 140 N/mm2
m = modular ratio = 18.66
#### To find Neutral Axis (NA)
σcbc /(σst/m) = xc/(d – xc)
5/(140/18.66) = xc/(500 – xc)
Xc = 199.95mm = 200mm
z = d – xc/3
= 500 – 200/3
= 433.33mm
#### To find Moment of resistance
Mr = C x z
= bxccbc/2)z
= 250 x 200 x 5/2 x 433.33
= 54166250 N-mm
= 54.166 kN-m
OR
Mr = T x z
= Ast. σst.z —————————equation 1
#### To find Ast
Equating, C = T
bxccbc/2 = Ast. σst
Therefore, Ast = bxccbc/2 σst
Ast = (250 x 200 x 5/2)/140
Ast = 892.85 mm2
Substituting the value of Ast in equation 1;
Mr = T x z
= Ast. σst.z
= 892.85 x 140 x 433.33
= 54165817 N-mm
= 54.1658 kN-m
From the above example, it is clear that in case of a balanced section, the Mr can be calculated either as Mr = C x z or as Mr = T x z. The values obtained for moment of resistance are the same for both the formulas.
### 3 thoughts on “Numerical example 2 | Singly reinforced Sections”
1. kindly advise a structural design of 65 feet x 80 feet godown with RCC column and 4.5 in thick roof for black cotton soil. it case a no columns in the hall
case b with one row of column in the hall.
• Hello Mr Kumar,
Are you yet to design your godown? If yes, please let me know and we could have a discussion on it.
2. FROM THIS PROBLEM I HAVE GAIN A CLEAR CONCEPTION ABOUT RCC SINGLY REINFORCED BEAM DESIGN AND IT IS HELP FULL FOR DIPLOMA CIVIL ENGINEERING STUDENTS AND RCC DESIGNERS ALSO.
THANK YOU VERY MUCH
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Advice 1: How to determine the height of the pyramid
Under the pyramid implies one of the varieties of polyhedra whose base is the polygon and its sides are triangles, which are joined into a single, common vertex. If from the vertex drop a perpendicular to the base of the pyramid, the resulting cut will be called the height of the pyramid. To determine the height of the pyramid very easily.
Instruction
1
The formula for finding the height of a pyramid is possible to Express the formula then calculate the volume:
V = (S*h)/3, where S is the area of the polytope lying at the base of the pyramid, h is the height of this pyramid.
In this case, h can be calculated as:
h = (3*V)/S.
2
In that case, if the base of the pyramid has a square, length of its diagonal and the edge length of this pyramid, the height of this pyramid can be expressed from Pythagorean theorem, because the triangle which is formed by an edge of the pyramid, the height and half the diagonal of the square at the base is a right triangle.
The Pythagorean theorem States that the square of the hypotenuse in a right triangle, the is equal to the sum of the squares of the other two sides(a2 = b2 + c2). The face of the pyramid is the hypotenuse, one leg is half diagonal of the square. Then the length of the unknown side (height) is by the formula:
b2 = a2 - c2;
c2 = a2 - b2.
3
To both situations was the most clear and understandable, you can consider a couple of examples.
Example 1: Area of base of the pyramid 46 cm2, its volume is 120 cm3. Based on these data, the height of the pyramid is this:
h = 3*120/46 = 7.83 cm
Answer: the height of the pyramid will be approximately 7.83 cm
Example 2: From the pyramid, which lies at the base a regular polygon is a square, its diagonal is 14 cm, the length of the edge is 15 cm According to the to find the height of the pyramid, you need to use the following formula (which appeared as a consequence of the Pythagorean theorem):
h2 = 152 - 142
h2 = 225 - 196 = 29
h = √29 cm
Answer: the height of the pyramid is √29 cm or approximately 5.4 cm
Note
If the base of the pyramid is a square or other regular polygon, the pyramid can be called correct. This pyramid has several properties:
its lateral edges are equal;
face it - isosceles triangles, which are equal to each other;
around this pyramid to describe the sphere and enter it.
Advice 2 : How to find the edge length of a pyramid
The pyramid is a figure that has a base in the form of a polygon and the lateral faces converging at the top with peaks. The boundaries of the lateral faces are called edges. And how to find the length of the edges of the pyramid?
Instruction
1
Find the boundary points of the ribs, the length of which are looking for. Let it be points A and B.
2
Set the coordinates of the points A and B. They need to ask three-dimensional, because the pyramid three – dimensional figure. Get A(x1, Y1, z1) and B(x2, y2, z2).
3
Calculate the needed length, using the General formula: edge length of a pyramid is equal to square root of the sum of squares of differences of corresponding coordinates of boundary points. Substitute the numbers for your coordinates into the formula and find the length of the edges of the pyramid. In the same way, find the length of the ribs is not only the right of the pyramid, but rectangular, and truncated and arbitrary.
4
Find the length of the edges of the pyramid, in which all edges are equal, set the base figure and known height. Determine the location of the base height, i.e. the lower point. Since edges are equal, then it is possible to draw a circle whose center is the intersection point of the diagonals of the base.
5
Draw straight lines connecting the opposite corners of the base of the pyramid. Mark the point where they intersect. This point will be the lower bound of the height of the pyramid.
6
Find the length of diagonal of a rectangle using the Pythagorean theorem, where the sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse. Get A2+b2=c2, where a and b are the legs and C is the hypotenuse. The hypotenuse will then be equal to the square root of the sum of the squares of the legs.
7
Find the length of the edges of the pyramid. First divide the length of the diagonal in half. All the data, substitute values in the formula of Pythagoras, are described above. Similar to the previous example, find the square root of the sum of the squares of the height of the pyramid and the half-diagonal.
Advice 3 : How to find the lateral edge of the pyramid
A pyramid is a polyhedron whose faces are triangles having a common vertex. The calculation of the lateral edges is taught in school, in practice often it is necessary to remember forgotten the formula.
Instruction
1
Referring to the Foundation of the pyramid can be triangular, rectangular etc. Triangular pyramid known as a tetrahedron. In the tetrahedron, any face can be taken as a basis.
2
The pyramid is right, rectangular, truncated, etc. regular pyramid is called in that case, if its base is a regular polygon. Then the center of the pyramid is projected onto the center of the polygon and the lateral edges of the pyramid are equal. In such a pyramid the lateral faces are identical isosceles triangles.
3
A rectangular pyramid is called when one of its edges perpendicular to the base. The height of this pyramid is exactly that edge. The calculations of the height values of the rectangular pyramid of the lengths of its side edges lying everyone knows the Pythagorean theorem.
4
To calculate the edges of a regular pyramid you need to hold the height of the top of the pyramid to the base. Next, consider the desired fin as a leg in a right triangle using the Pythagorean theorem.
5
The side edge in this case is calculated by the formula b=√ h2+ (a2•sin (180°
) 2. It is the square root of the sum of the squares of two sides of a right triangle. One side is the height of the pyramid h, the other side is a segment connecting the center of the base of a regular pyramid with the peak of this reason. In this case, the a – side of a regular polygon base, n is the number of sides.
Note
Description of the pyramids and the study of its properties was begun in Ancient Greece. Today, the elements of the pyramid, its properties and the laws of the construction studied at school on geometry lessons.
The main elements of a pyramid are lateral faces are triangles that share a vertex; the lateral edges – the sides that are common; apofema (height-side edge held from the top, assuming that the pyramid is correct), the top of the pyramid - the point where the lateral edges etc.
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# 26 Modeling with Linear Functions
### Learning Objectives
In this section you will:
• Build linear models from verbal descriptions.
• Model a set of data with a linear function.
Emily is a college student who plans to spend a summer in Seattle. She has saved $3,500 for her trip and anticipates spending$400 each week on rent, food, and activities. How can we write a linear model to represent her situation? What would be the x-intercept, and what can she learn from it? To answer these and related questions, we can create a model using a linear function. Models such as this one can be extremely useful for analyzing relationships and making predictions based on those relationships. In this section, we will explore examples of linear function models.
### Building Linear Models from Verbal Descriptions
When building linear models to solve problems involving quantities with a constant rate of change, we typically follow the same problem strategies that we would use for any type of function. Let’s briefly review them:
1. Identify changing quantities, and then define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system.
2. Carefully read the problem to identify important information. Look for information that provides values for the variables or values for parts of the functional model, such as slope and initial value.
3. Carefully read the problem to determine what we are trying to find, identify, solve, or interpret.
4. Identify a solution pathway from the provided information to what we are trying to find. Often this will involve checking and tracking units, building a table, or even finding a formula for the function being used to model the problem.
5. When needed, write a formula for the function.
6. Solve or evaluate the function using the formula.
7. Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically.
8. Clearly convey your result using appropriate units, and answer in full sentences when necessary.
Now let’s take a look at the student in Seattle. In her situation, there are two changing quantities: time and money. The amount of money she has remaining while on vacation depends on how long she stays. We can use this information to define our variables, including units.
$\begin{array}{l}\text{Output:}\,M,\text{money remaining, in dollars}\\ \text{Input:}\,t,\text{time, in weeks}\end{array}$
So, the amount of money remaining depends on the number of weeks:$\,M\left(t\right)$.
$\begin{array}{l}\text{We can also identify the initial value and the rate of change}.\\ \text{ Initial Value: She saved \3,500, so \3,500 is the initial value for}\,M.\\ \text{ Rate of Change: She anticipates spending \400 each week, so}-\text{\400 per week is the rate of change, or slope}.\end{array}$
Notice that the unit of dollars per week matches the unit of our output variable divided by our input variable. Also, because the slope is negative, the linear function is decreasing. This should make sense because she is spending money each week.
The rate of change is constant, so we can start with the linear model$\,M\left(t\right)=mt+b.\,$Then we can substitute the intercept and slope provided.
To find the x-intercept, we set the output to zero, and solve for the input.
$\begin{array}{ccc}\hfill 0& =& -400t+3500\hfill \\ \hfill t& =& \frac{3500}{400}\hfill \\ & =& 8.75\hfill \end{array}$
The x-intercept is 8.75 weeks. Because this represents the input value when the output will be zero, we could say that Emily will have no money left after 8.75 weeks.
When modeling any real-life scenario with functions, there is typically a limited domain over which that model will be valid—almost no trend continues indefinitely. Here the domain refers to the number of weeks. In this case, it doesn’t make sense to talk about input values less than zero. A negative input value could refer to a number of weeks before she saved $3,500, but the scenario discussed poses the question once she saved$3,500 because this is when her trip and subsequent spending starts. It is also likely that this model is not valid after the x-intercept, unless Emily uses a credit card and goes into debt. The domain represents the set of input values, so the reasonable domain for this function is$\,0\le t\le 8.75.$
In this example, we were given a written description of the situation. We followed the steps of modeling a problem to analyze the information. However, the information provided may not always be the same. Sometimes we might be provided with an intercept. Other times we might be provided with an output value. We must be careful to analyze the information we are given, and use it appropriately to build a linear model.
#### Using a Given Intercept to Build a Model
Some real-world problems provide the y-intercept, which is the constant or initial value. Once the y-intercept is known, the x-intercept can be calculated. Suppose, for example, that Hannah plans to pay off a no-interest loan from her parents. Her loan balance is $1,000. She plans to pay$250 per month until her balance is $0. The y-intercept is the initial amount of her debt, or$1,000. The rate of change, or slope, is -$250 per month. We can then use the slope-intercept form and the given information to develop a linear model. $\begin{array}{ccc}\hfill f\left(x\right)& =& mx+b\hfill \\ & =& -250x+1000\hfill \end{array}$ Now we can set the function equal to 0, and solve for$\,x\,$to find the x-intercept. $\begin{array}{ccc}\hfill 0& =& -250x+1000\hfill \\ \hfill 1000& =& 250x\hfill \\ \hfill 4& =& x\hfill \\ \hfill x& =& 4\hfill \end{array}$ The x-intercept is the number of months it takes her to reach a balance of$0. The x-intercept is 4 months, so it will take Hannah four months to pay off her loan.
#### Using a Given Input and Output to Build a Model
Many real-world applications are not as direct as the ones we just considered. Instead they require us to identify some aspect of a linear function. We might sometimes instead be asked to evaluate the linear model at a given input or set the equation of the linear model equal to a specified output.
### How To
Given a word problem that includes two pairs of input and output values, use the linear function to solve a problem.
1. Identify the input and output values.
2. Convert the data to two coordinate pairs.
3. Find the slope.
4. Write the linear model.
5. Use the model to make a prediction by evaluating the function at a given x-value.
6. Use the model to identify an x-value that results in a given y-value.
7. Answer the question posed.
### Using a Linear Model to Investigate a Town’s Population
A town’s population has been growing linearly. In 2004, the population was 6,200. By 2009, the population had grown to 8,100. Assume this trend continues.
1. Predict the population in 2013.
2. Identify the year in which the population will reach 15,000.
The two changing quantities are the population size and time. While we could use the actual year value as the input quantity, doing so tends to lead to very cumbersome equations because the y-intercept would correspond to the year 0, more than 2000 years ago!
To make computation a little nicer, we will define our input as the number of years since 2004.
$\begin{array}{l}\\ \begin{array}{l}\text{Input:}\,t,\text{years since 2004}\hfill \\ \text{Output:}\,P\left(t\right),\text{the town’s population}\hfill \end{array}\end{array}$
To predict the population in 2013 ($t=9$), we would first need an equation for the population. Likewise, to find when the population would reach 15,000, we would need to solve for the input that would provide an output of 15,000. To write an equation, we need the initial value and the rate of change, or slope.
To determine the rate of change, we will use the change in output per change in input.
$m=\frac{\text{change in output}}{\text{change in input}}$
The problem gives us two input-output pairs. Converting them to match our defined variables, the year 2004 would correspond to$\,t=0,$giving the point$\,\left(0,\text{6200}\right).\,$Notice that through our clever choice of variable definition, we have “given” ourselves the y-intercept of the function. The year 2009 would correspond to$\,t=\text{5,}$giving the point$\,\left(5,\text{8100}\right).$
The two coordinate pairs are$\,\left(0,\text{6200}\right)\,$and$\,\left(5,\text{8100}\right).\,$Recall that we encountered examples in which we were provided two points earlier in the chapter. We can use these values to calculate the slope.
$\begin{array}{ccc}\hfill m& =& \frac{8100-6200}{5-0}\hfill \\ & =& \frac{1900}{5}\hfill \\ & =& 380\text{ people per year}\hfill \end{array}$
We already know the y-intercept of the line, so we can immediately write the equation:
$P\left(t\right)=380t+6200$
To predict the population in 2013, we evaluate our function at$\,t=9.$
$\begin{array}{ccc}\hfill P\left(9\right)& =& 380\left(9\right)+6,200\hfill \\ & =& 9,620\hfill \end{array}$
If the trend continues, our model predicts a population of 9,620 in 2013.
To find when the population will reach 15,000, we can set$\,P\left(t\right)=15000\,$and solve for$\,t.$
$\begin{array}{ccc}\hfill 15000& =& 380t+6200\hfill \\ \hfill 8800& =& 380t\hfill \\ \hfill t& \approx & 23.158\hfill \end{array}$
Our model predicts the population will reach 15,000 in a little more than 23 years after 2004, or somewhere around the year 2027.[/hidden-answer]
### Try It
A company sells doughnuts. They incur a fixed cost of $25,000 for rent, insurance, and other expenses. It costs$0.25 to produce each doughnut.
1. Write a linear model to represent the cost$\,C\,$of the company as a function of$\,x,$the number of doughnuts produced.
2. Find and interpret the y-intercept.
a.$\,C\left(x\right)=0.25x+25,000\,$b. The y-intercept is$\,\left(0,25,000\right).\,$If the company does not produce a single doughnut, they still incur a cost of $25,000. [/hidden-answer] ### Try It A city’s population has been growing linearly. In 2008, the population was 28,200. By 2012, the population was 36,800. Assume this trend continues. 1. Predict the population in 2014. 2. Identify the year in which the population will reach 54,000. [reveal-answer q=”fs-id1555247″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1555247″] 1. 41,100 2. 2020 [/hidden-answer] #### Using a Diagram to Build a Model It is useful for many real-world applications to draw a picture to gain a sense of how the variables representing the input and output may be used to answer a question. To draw the picture, first consider what the problem is asking for. Then, determine the input and the output. The diagram should relate the variables. Often, geometrical shapes or figures are drawn. Distances are often traced out. If a right triangle is sketched, the Pythagorean Theorem relates the sides. If a rectangle is sketched, labeling width and height is helpful. ### Using a Diagram to Model Distance Walked Anna and Emanuel start at the same intersection. Anna walks east at 4 miles per hour while Emanuel walks south at 3 miles per hour. They are communicating with a two-way radio that has a range of 2 miles. How long after they start walking will they fall out of radio contact? [reveal-answer q=”fs-id2158876″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2158876″] In essence, we can partially answer this question by saying they will fall out of radio contact when they are 2 miles apart, which leads us to ask a new question: “How long will it take them to be 2 miles apart”? In this problem, our changing quantities are time and position, but ultimately we need to know how long will it take for them to be 2 miles apart. We can see that time will be our input variable, so we’ll define our input and output variables. $\begin{array}{l}\\ \begin{array}{l}\text{Input:}\,t,\text{time in hours}.\hfill \\ \text{Output:}\,A\left(t\right),\text{distance in miles, and}\,E\left(t\right),\text{distance in miles}\hfill \end{array}\end{array}$ Because it is not obvious how to define our output variable, we’ll start by drawing a picture such as (Figure). Initial Value: They both start at the same intersection so when$\,t=0,$the distance traveled by each person should also be 0. Thus the initial value for each is 0. Rate of Change: Anna is walking 4 miles per hour and Emanuel is walking 3 miles per hour, which are both rates of change. The slope for$\,A\,$is 4 and the slope for$\,E\,$is 3. Using those values, we can write formulas for the distance each person has walked. $\begin{array}{ccc}\hfill A\left(t\right)& =& 4t\hfill \\ \hfill E\left(t\right)& =& 3t\hfill \end{array}$ For this problem, the distances from the starting point are important. To notate these, we can define a coordinate system, identifying the “starting point” at the intersection where they both started. Then we can use the variable,$\,A,$which we introduced above, to represent Anna’s position, and define it to be a measurement from the starting point in the eastward direction. Likewise, can use the variable,$\,E,$to represent Emanuel’s position, measured from the starting point in the southward direction. Note that in defining the coordinate system, we specified both the starting point of the measurement and the direction of measure. We can then define a third variable,$\,D,$to be the measurement of the distance between Anna and Emanuel. Showing the variables on the diagram is often helpful, as we can see from (Figure). Recall that we need to know how long it takes for$\,D,$the distance between them, to equal 2 miles. Notice that for any given input$\,t,$the outputs$\,A\left(t\right),E\left(t\right),$and$\,D\left(t\right)\,$represent distances. (Figure) shows us that we can use the Pythagorean Theorem because we have drawn a right angle. Using the Pythagorean Theorem, we get: $\begin{array}{cccc}\hfill D{\left(t\right)}^{2}& =& A{\left(t\right)}^{2}+E{\left(t\right)}^{2}\hfill & \\ & =& {\left(4t\right)}^{2}+{\left(3t\right)}^{2}\hfill & \\ & =& 16{t}^{2}+9{t}^{2}\hfill & \\ & =& 25{t}^{2}\hfill & \\ \hfill \text{ }D\left(t\right)& =& ±\sqrt{25{t}^{2}}\hfill & \phantom{\rule{1em}{0ex}}\text{Solve for }D\left(t\right)\text{ using the square root.}\hfill \\ & =& ±5|t|\hfill & \end{array}$ In this scenario we are considering only positive values of$\,t,\,$so our distance$\,D\left(t\right)\,$will always be positive. We can simplify this answer to$\,D\left(t\right)=5t.\,$This means that the distance between Anna and Emanuel is also a linear function. Because$\,D\,$is a linear function, we can now answer the question of when the distance between them will reach 2 miles. We will set the output$\,D\left(t\right)=2\,$and solve for$\,t.$ $\begin{array}{ccc}\hfill D\left(t\right)& =& 2\hfill \\ \hfill 5t& =& 2\hfill \\ \hfill t& =& \frac{2}{5}=0.4\hfill \end{array}$ They will fall out of radio contact in 0.4 hour, or 24 minutes.[/hidden-answer] Should I draw diagrams when given information based on a geometric shape? Yes. Sketch the figure and label the quantities and unknowns on the sketch. ### Using a Diagram to Model Distance Between Cities There is a straight road leading from the town of Westborough to Agritown 30 miles east and 10 miles north. Partway down this road, it junctions with a second road, perpendicular to the first, leading to the town of Eastborough. If the town of Eastborough is located 20 miles directly east of the town of Westborough, how far is the road junction from Westborough? [reveal-answer q=”fs-id2257378″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2257378″] It might help here to draw a picture of the situation. See (Figure). It would then be helpful to introduce a coordinate system. While we could place the origin anywhere, placing it at Westborough seems convenient. This puts Agritown at coordinates$\,\left(\text{3}0,\text{1}0\right),$and Eastborough at$\,\left(\text{2}0,\text{}0\right).$ Using this point along with the origin, we can find the slope of the line from Westborough to Agritown. $m=\frac{10-0}{30-0}=\frac{1}{3}$ Now we can write an equation to describe the road from Westborough to Agritown. $W\left(x\right)=\frac{1}{3}x$ From this, we can determine the perpendicular road to Eastborough will have slope$\,m=–3.\,$Because the town of Eastborough is at the point (20, 0), we can find the equation. $\begin{array}{cccc}\hfill E\left(x\right)& =& -3x+b\hfill & \\ \hfill 0& =& -3\left(20\right)+b\hfill & \phantom{\rule{1em}{0ex}}\text{Substitute }\left(20,0\right)\text{into the equation}.\hfill \\ \hfill b& =& 60\hfill & \\ \hfill E\left(x\right)& =& -3x+60\hfill & \end{array}$ We can now find the coordinates of the junction of the roads by finding the intersection of these lines. Setting them equal, $\begin{array}{cccc}\hfill \text{ }\frac{1}{3}x& =& -3x+60\hfill & \\ \hfill \frac{10}{3}x& =& 60\hfill & \\ \hfill 10x& =& 180\hfill & \\ \hfill x& =& 18\hfill & \phantom{\rule{2em}{0ex}}\text{Substitute this back into }W\left(x\right).\hfill \\ \hfill y& =& W\left(18\right)\hfill & \\ & =& \frac{1}{3}\left(18\right)\hfill & \\ & =& 6\hfill & \end{array}$ The roads intersect at the point (18, 6). Using the distance formula, we can now find the distance from Westborough to the junction. $\begin{array}{ccc}\hfill \text{distance}& =& \sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}\hfill \\ & =& \sqrt{{\left(18-0\right)}^{2}+{\left(6-0\right)}^{2}}\hfill \\ & \approx & \text{ }18.974\text{ miles}\hfill \end{array}$[/hidden-answer] #### Analysis One nice use of linear models is to take advantage of the fact that the graphs of these functions are lines. This means real-world applications discussing maps need linear functions to model the distances between reference points. ### Try It There is a straight road leading from the town of Timpson to Ashburn 60 miles east and 12 miles north. Partway down the road, it junctions with a second road, perpendicular to the first, leading to the town of Garrison. If the town of Garrison is located 22 miles directly east of the town of Timpson, how far is the road junction from Timpson? [reveal-answer q=”fs-id1575416″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1575416″] 21.15 miles [/hidden-answer] ### Modeling a Set of Data with Linear Functions Real-world situations including two or more linear functions may be modeled with a system of linear equations. Remember, when solving a system of linear equations, we are looking for points the two lines have in common. Typically, there are three types of answers possible, as shown in (Figure). ### How To Given a situation that represents a system of linear equations, write the system of equations and identify the solution. 1. Identify the input and output of each linear model. 2. Identify the slope and y-intercept of each linear model. 3. Find the solution by setting the two linear functions equal to another and solving for$\,x,$or find the point of intersection on a graph. ### Building a System of Linear Models to Choose a Truck Rental Company Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of$20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mile[1] . When will Keep on Trucking, Inc. be the better choice for Jamal? [reveal-answer q=”fs-id1633898″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1633898″] The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions in (Figure). Input $d,$distance driven in miles Outputs $K\left(d\right):\,$cost, in dollars, for renting from Keep on Trucking $M\left(d\right)\,$cost, in dollars, for renting from Move It Your Way Initial Value Up-front fee:$\,K\left(0\right)=\text{2}0\,$and$\,M\left(0\right)=\text{16}$ Rate of Change $K\left(d\right)=\text{\}0.\text{59}\,$/mile and$\,P\left(d\right)=\text{\}0.\text{63}\,$/mile A linear function is of the form$\,f\left(x\right)=mx+b.\,$Using the rates of change and initial charges, we can write the equations $\begin{array}{ccc}\hfill K\left(d\right)& =& 0.59d+20\hfill \\ \hfill M\left(d\right)& =& 0.63d+16\hfill \end{array}$ Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when$\,K\left(d\right) These graphs are sketched in (Figure), with[latex]\,K\left(d\right)\,$in blue. To find the intersection, we set the equations equal and solve: $\begin{array}{ccc}\hfill K\left(d\right)& =& M\left(d\right)\hfill \\ \hfill 0.59d+20& =& 0.63d+16\hfill \\ \hfill 4& =& 0.04d\hfill \\ \hfill \text{ }100& =& d\hfill \\ \hfill d& =& 100\hfill \end{array}$ This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that$\,K\left(d\right)\,$is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is$\,d>100$.[/hidden-answer] Access this online resource for additional instruction and practice with linear function models. ### Key Concepts • We can use the same problem strategies that we would use for any type of function. • When modeling and solving a problem, identify the variables and look for key values, including the slope and y-intercept. See (Figure). • Draw a diagram, where appropriate. See (Figure) and (Figure). • Check for reasonableness of the answer. • Linear models may be built by identifying or calculating the slope and using the y-intercept. • The x-intercept may be found by setting$\,y=0,$which is setting the expression$\,mx+b\,$equal to 0. • The point of intersection of a system of linear equations is the point where the x– and y-values are the same. See (Figure). • A graph of the system may be used to identify the points where one line falls below (or above) the other line. ### Section Exercises #### Verbal Explain how to find the input variable in a word problem that uses a linear function. [reveal-answer q=”fs-id2186383″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2186383″] Determine the independent variable. This is the variable upon which the output depends. [/hidden-answer] Explain how to find the output variable in a word problem that uses a linear function. Explain how to interpret the initial value in a word problem that uses a linear function. [reveal-answer q=”fs-id2108659″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2108659″] To determine the initial value, find the output when the input is equal to zero. [/hidden-answer] Explain how to determine the slope in a word problem that uses a linear function. #### Algebraic Find the area of a parallelogram bounded by the y-axis, the line$\,x=3,$the line$\,f\left(x\right)=1+2x,$and the line parallel to$\,f\left(x\right)\,$passing through$\,\left(\text{2},\text{7}\right).$ [reveal-answer q=”fs-id1326315″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1326315″] 6 square units [/hidden-answer] Find the area of a triangle bounded by the x-axis, the line$\,f\left(x\right)=12–\frac{1}{3}x,$and the line perpendicular to$\,f\left(x\right)\,$that passes through the origin. Find the area of a triangle bounded by the y-axis, the line$\,f\left(x\right)=9–\frac{6}{7}x,$and the line perpendicular to$\,f\left(x\right)\,$that passes through the origin. [reveal-answer q=”fs-id2065575″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2065575″] 20.01 square units [/hidden-answer] Find the area of a parallelogram bounded by the x-axis, the line$\,g\left(x\right)=2,$the line$\,f\left(x\right)=3x,$and the line parallel to$\,f\left(x\right)\,$passing through$\,\left(6,1\right).$ For the following exercises, consider this scenario: A town’s population has been decreasing at a constant rate. In 2010 the population was 5,900. By 2012 the population had dropped 4,700. Assume this trend continues. Predict the population in 2016. [reveal-answer q=”fs-id2522955″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2522955″] 2,300 [/hidden-answer] Identify the year in which the population will reach 0. For the following exercises, consider this scenario: A town’s population has been increased at a constant rate. In 2010 the population was 46,020. By 2012 the population had increased to 52,070. Assume this trend continues. Predict the population in 2016. [reveal-answer q=”fs-id1334127″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1334127″] 64,170 [/hidden-answer] Identify the year in which the population will reach 75,000. For the following exercises, consider this scenario: A town has an initial population of 75,000. It grows at a constant rate of 2,500 per year for 5 years. Find the linear function that models the town’s population$\,P\,$as a function of the year,$\,t,$where$\,t\,$is the number of years since the model began. [reveal-answer q=”fs-id1427204″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1427204″] $P\left(t\right)=75,000+2500t$ [/hidden-answer] Find a reasonable domain and range for the function$\,P.$ If the function$\,P\,$is graphed, find and interpret the x– and y-intercepts. [reveal-answer q=”fs-id2555725″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2555725″] (–30, 0) Thirty years before the start of this model, the town had no citizens. (0, 75,000) Initially, the town had a population of 75,000. [/hidden-answer] If the function$\,P\,$is graphed, find and interpret the slope of the function. When will the population reach 100,000? [reveal-answer q=”fs-id2106990″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2106990″] Ten years after the model began [/hidden-answer] What is the population in the year 12 years from the onset of the model? For the following exercises, consider this scenario: The weight of a newborn is 7.5 pounds. The baby gained one-half pound a month for its first year. Find the linear function that models the baby’s weight$\,W\,$as a function of the age of the baby, in months,$\,t.$ [reveal-answer q=”fs-id1626463″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1626463″] $W\left(t\right)=0.5t+7.5$ [/hidden-answer] Find a reasonable domain and range for the function$\,W.$ If the function$\,W\,$is graphed, find and interpret the x– and y-intercepts. [reveal-answer q=”fs-id2270495″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2270495″] $\left(-15,\text{}0\right)\,$: The x-intercept is not a plausible set of data for this model because it means the baby weighed 0 pounds 15 months prior to birth.$\,\left(0,\text{7}.\text{5}\right)\,$: The baby weighed 7.5 pounds at birth. [/hidden-answer] If the function W is graphed, find and interpret the slope of the function. When did the baby weight 10.4 pounds? [reveal-answer q=”fs-id1609534″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1609534″] At age 5.8 months [/hidden-answer] What is the output when the input is 6.2? For the following exercises, consider this scenario: The number of people afflicted with the common cold in the winter months steadily decreased by 205 each year from 2005 until 2010. In 2005, 12,025 people were inflicted. Find the linear function that models the number of people inflicted with the common cold$\,C\,$as a function of the year,$\,t.$ [reveal-answer q=”fs-id2776592″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2776592″] $C\left(t\right)=12,025-205t$ [/hidden-answer] Find a reasonable domain and range for the function$\,C.$ If the function$\,C\,$is graphed, find and interpret the x– and y-intercepts. [reveal-answer q=”fs-id2578699″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2578699″]$\left(\text{58}.\text{7},\text{}0\right):\,$In roughly 59 years, the number of people inflicted with the common cold would be 0.$\,\left(0,\text{12},0\text{25}\right)\,$Initially there were 12,025 people afflicted by the common cold.[/hidden-answer] If the function$\,C\,$is graphed, find and interpret the slope of the function. When will the output reach 0? [reveal-answer q=”fs-id2522958″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2522958″] 2063 [/hidden-answer] In what year will the number of people be 9,700? #### Graphical For the following exercises, use the graph in (Figure), which shows the profit,$\,y,$in thousands of dollars, of a company in a given year,$\,t,$where$\,t\,$represents the number of years since 1980. Find the linear function$\,y,$where$\,y\,$depends on$\,t,$the number of years since 1980. [reveal-answer q=”fs-id2571891″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2571891″] $y=-2t\text{+180}$ [/hidden-answer] Find and interpret the y-intercept. Find and interpret the x-intercept. [reveal-answer q=”fs-id2269514″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2269514″] In 2070, the company’s profit will be zero. [/hidden-answer] Find and interpret the slope. For the following exercises, use the graph in (Figure), which shows the profit,$\,y,$in thousands of dollars, of a company in a given year,$\,t,$where$\,t\,$represents the number of years since 1980. Find the linear function$\,y,$where$\,y\,$depends on$\,t,$the number of years since 1980. [reveal-answer q=”fs-id2147080″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2147080″] $y=\text{3}0t-\text{3}00$ [/hidden-answer] Find and interpret the y-intercept. [reveal-answer q=”fs-id2651748″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2651748″] $\left(0,-300\right);$In 1980, the company lost$300,000.
Find and interpret the x-intercept.
Find and interpret the slope.
$y=30t-300\,$of form$\,y=mx+b,\text{m}=30.$
For each year after 1980, the company’s profits increased $30,000 per year [/hidden-answer] #### Numeric For the following exercises, use the median home values in Mississippi and Hawaii (adjusted for inflation) shown in (Figure). Assume that the house values are changing linearly. Year Mississippi Hawaii 1950$25,200 $74,400 2000$71,400 $272,700 In which state have home values increased at a higher rate? If these trends were to continue, what would be the median home value in Mississippi in 2010? [reveal-answer q=”fs-id2632034″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2632034″]$80,640
If we assume the linear trend existed before 1950 and continues after 2000, the two states’ median house values will be (or were) equal in what year? (The answer might be absurd.)
For the following exercises, use the median home values in Indiana and Alabama (adjusted for inflation) shown in (Figure). Assume that the house values are changing linearly.
Year Indiana Alabama
1950 $37,700$27,100
2000 $94,300$85,100
In which state have home values increased at a higher rate?
Alabama
If these trends were to continue, what would be the median home value in Indiana in 2010?
If we assume the linear trend existed before 1950 and continues after 2000, the two states’ median house values will be (or were) equal in what year? (The answer might be absurd.)
2328
#### Real-World Applications
In 2004, a school population was 1001. By 2008 the population had grown to 1697. Assume the population is changing linearly.
1. How much did the population grow between the year 2004 and 2008?
2. How long did it take the population to grow from 1001 students to 1697 students?
3. What is the average population growth per year?
4. What was the population in the year 2000?
5. Find an equation for the population,$\,P,$of the school t years after 2000.
6. Using your equation, predict the population of the school in 2011.
In 2003, a town’s population was 1431. By 2007 the population had grown to 2134. Assume the population is changing linearly.
1. How much did the population grow between the year 2003 and 2007?
2. How long did it take the population to grow from 1431 people to 2134 people?
3. What is the average population growth per year?
4. What was the population in the year 2000?
5. Find an equation for the population,$\,P,$of the town$\,t\,$years after 2000.
6. Using your equation, predict the population of the town in 2014.
1. $2134-1431=703\,$people
2. $2007-2003=4\,$years
3. Average rate of growth$\,=\frac{703}{4}=175.75\,$people per year
So, using$\,y=mx+b,$we have$\,y=175.75x+1431.$
4. The year 2000 corresponds to$\,t=-3.$
So,$\,y=175.75\left(-3\right)+1431=903.75\,$or roughly 904 people in year 2000
5. If the year 2000 corresponds to$\,t\text{=0,}$then we have ordered pair$\,\left(0,903.75\right)$
$y=175.75x+903.75\,$corresponds to$\,P\left(t\right)=175.75t+903.75$
6. The year 2014 corresponds to$\,t=14.\,$Therefore,$\,P\left(14\right)=175.75\left(14\right)+903.75=3364.25$.
So, a population of 3364.[/hidden-answer]
A phone company has a monthly cellular plan where a customer pays a flat monthly fee and then a certain amount of money per minute used on the phone. If a customer uses 410 minutes, the monthly cost will be $71.50. If the customer uses 720 minutes, the monthly cost will be$118.
1. Find a linear equation for the monthly cost of the cell plan as a function of x, the number of monthly minutes used.
2. Interpret the slope and y-intercept of the equation.
3. Use your equation to find the total monthly cost if 687 minutes are used.
A phone company has a monthly cellular data plan where a customer pays a flat monthly fee of $10 and then a certain amount of money per megabyte (MB) of data used on the phone. If a customer uses 20 MB, the monthly cost will be$11.20. If the customer uses 130 MB, the monthly cost will be $17.80. 1. Find a linear equation for the monthly cost of the data plan as a function of$\,x,$the number of MB used. 2. Interpret the slope and y-intercept of the equation. 3. Use your equation to find the total monthly cost if 250 MB are used. [reveal-answer q=”2263565″]Show Solution[/reveal-answer][hidden-answer a=”2263565″] 1. $\begin{array}{l}\text{Ordered pairs are }\left(20,11.20\right)\text{ and }\left(130,17.80\right)\hfill \\ \\ \begin{array}{ccc}\hfill m& =& \frac{17.80-11.20}{130-20}=0.06\text{ and }\left(0,10\right)\hfill \\ \hfill y& =& mx+b\hfill \\ \hfill y& =& 0.06x+10\text{ or }C\left(x\right)=0.06x+10\hfill \end{array}\end{array}$ 2. 0.06 For every MB, the client is charged 6 cents.$\,\left(0,10\right)\,$If no usage occurs, the client is charged$10
3. $\begin{array}{ccc}\hfill C\left(250\right)& =& 0.06\left(250\right)+10\hfill \\ & =& 25\hfill \end{array}$
In 1991, the moose population in a park was measured to be 4,360. By 1999, the population was measured again to be 5,880. Assume the population continues to change linearly.
1. Find a formula for the moose population, P since 1990.
2. What does your model predict the moose population to be in 2003?
In 2003, the owl population in a park was measured to be 340. By 2007, the population was measured again to be 285. The population changes linearly. Let the input be years since 1990.
1. Find a formula for the owl population,$\,P.\,$Let the input be years since 2003.
2. What does your model predict the owl population to be in 2012?
1. $\begin{array}{l}\text{Ordered pairs are }\left(0,340\right)\text{ and }\left(4,285\right)\hfill \\ \begin{array}{ccc}\hfill m& =& \frac{285-340}{4-0}=-13.75\text{ and }\left(0,340\right)\hfill \\ \hfill y& =& mx+b\hfill \\ \hfill y& =& -13.75x+340\text{ or }P\left(t\right)=-13.75t+340\hfill \end{array}\end{array}$
2. $\begin{array}{l}\text{The year 2012 corresponds to t = 9}\hfill \\ \begin{array}{ccc}\hfill P\left(9\right)& =& -13.75\left(9\right)+340\hfill \\ \hfill & =& 216.25\text{ or 216 owls}\hfill \end{array}\hfill \end{array}$
The Federal Helium Reserve held about 16 billion cubic feet of helium in 2010 and is being depleted by about 2.1 billion cubic feet each year.
1. Give a linear equation for the remaining federal helium reserves,$\,R,$in terms of$\,t,$the number of years since 2010.
2. In 2015, what will the helium reserves be?
3. If the rate of depletion doesn’t change, in what year will the Federal Helium Reserve be depleted?
Suppose the world’s oil reserves in 2014 are 1,820 billion barrels. If, on average, the total reserves are decreasing by 25 billion barrels of oil each year:
1. Give a linear equation for the remaining oil reserves,$\,R,$in terms of$\,t,$the number of years since now.
2. Seven years from now, what will the oil reserves be?
3. If the rate at which the reserves are decreasing is constant, when will the world’s oil reserves be depleted?
1. $\begin{array}{l}\text{The year 2014 corresponds to t = 0}.\hfill \\ \text{We have }m=-25\text{ and }\left(0,1820\right)\hfill \\ \begin{array}{ccc}\hfill y& =& mx+b\hfill \\ \hfill y& =& -25x+1820\text{ or }R\left(t\right)=-25t+1820\hfill \end{array}\end{array}$
2. $\begin{array}{ccc}\hfill R\left(7\right)& =& -25\left(7\right)+1820\hfill \\ & =& \text{645 billion cubic feet}\hfill \end{array}$
3. $\begin{array}{ccc}\hfill 0& =& -25t+1820\hfill \\ \hfill -1820& =& -25t\hfill \\ \hfill 72.8& =& t⇒\text{ 2014 + 72}\text{.8 = 2086}\text{.8}\text{. So, in the year 2086}\hfill \end{array}$
You are choosing between two different prepaid cell phone plans. The first plan charges a rate of 26 cents per minute. The second plan charges a monthly fee of $19.95 plus 11 cents per minute. How many minutes would you have to use in a month in order for the second plan to be preferable? You are choosing between two different window washing companies. The first charges$5 per window. The second charges a base fee of $40 plus$3 per window. How many windows would you need to have for the second company to be preferable?
$\begin{array}{l}\text{Plan 1: }y=5x\text{ where x is number of windows}\hfill \\ \text{Plan 2: }y=3x+40\text{ where x is number of windows}\hfill \\ \begin{array}{ccc}\hfill 3x+40& \le & 5x\hfill \\ \hfill 40& \le & 2x\hfill \\ \hfill 20& \le & x\hfill \end{array}\hfill \\ \text{So, more than 20 windows}\hfill \end{array}$[/hidden-answer]
When hired at a new job selling jewelry, you are given two pay options:
Option A: Base salary of $17,000 a year with a commission of 12% of your sales Option B: Base salary of$20,000 a year with a commission of 5% of your sales
How much jewelry would you need to sell for option A to produce a larger income?
When hired at a new job selling electronics, you are given two pay options:
Option A: Base salary of $14,000 a year with a commission of 10% of your sales Option B: Base salary of$19,000 a year with a commission of 4% of your sales
How much electronics would you need to sell for option A to produce a larger income?
$\begin{array}{l}\text{Option A: }y=0.10x\text{ + 14,000 where }x\text{ is dollars of sales}.\hfill \\ \text{Option B: }y=0.04x+19,000\text{ where }x\text{ is dollars of sales}\hfill \\ \begin{array}{ccc}\hfill 0\,.10x+14,000& \ge & 0.04x+19,000\hfill \\ \hfill 0.06x+14,000& \ge & 19,000\hfill \\ \hfill 0.06x& \ge & 5,000\hfill \\ \hfill x& \ge & 83,333.33\hfill \end{array}\\ \text{So, more than \\83,333}\text{.33 in sales}\hfill \end{array}$[/hidden-answer]
When hired at a new job selling electronics, you are given two pay options:
Option A: Base salary of $20,000 a year with a commission of 12% of your sales Option B: Base salary of$26,000 a year with a commission of 3% of your sales
How much electronics would you need to sell for option A to produce a larger income?
When hired at a new job selling electronics, you are given two pay options:
Option A: Base salary of $10,000 a year with a commission of 9% of your sales Option B: Base salary of$20,000 a year with a commission of 4% of your sales
How much electronics would you need to sell for option A to produce a larger income?
$\begin{array}{l}\text{Option A: }y=0.09x\text{ + 10,000 where x is dollars of sales}\hfill \\ \text{Option B: }y=0.04x+20,000\text{ where x is dollars of sales}\hfill \\ \begin{array}{ccc}\hfill 0.09x+10,000& \ge & 0.04x+20,000\hfill \\ \hfill 0.05x+10,000& \ge & 20,000\hfill \\ \hfill 0.05x& \ge & 10,000\hfill \\ \hfill x& \ge & 200,000\hfill \end{array}\\ \text{So, more than \200,000 in sales}.\hfill \end{array}$[/hidden-answer]
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# Common Core: High School - Geometry : Construct Tangent Lines from Outside a Circle: CCSS.Math.Content.HSG-C.A.4
## Example Questions
← Previous 1
### Example Question #1 : Construct Tangent Lines From Outside A Circle: Ccss.Math.Content.Hsg C.A.4
Calculate a point that is tangent to the circle and passes through the origin.
Explanation:
explain
### Example Question #2 : Construct Tangent Lines From Outside A Circle: Ccss.Math.Content.Hsg C.A.4
Calculate a point that is tangent to the circle and passes through the origin.
Explanation:
To construct a line that is tangent to a point on the circle and passes through the origin, recall what it means for a line to be "tangent". A line that is tangent to a point on a circle means that the line will only touch the circle at that specific point.
Given the equation of the circle,
the center and radius of the circle can be determined.
The center is located at and the radius is .
Therefore, the center is located at and the radius is three. Plotting the circle and tangent line to the origin results in the following.
### Example Question #3 : Construct Tangent Lines From Outside A Circle: Ccss.Math.Content.Hsg C.A.4
Construct a line that is tangent to a point on the circle and passes through the point plotted outside the circle.
Explanation:
To construct a line that is tangent to a point on the circle and passes through the point outside the circle, recall what it means for a line to be "tangent". A line that is tangent to a point on a circle means that the line will only touch the circle at that specific point.
Using the plotted circle and the given point, two potential lines can be drawn that will touch the circle at one point. One possible line would touch the circle on the left half of the circumference while the other potential line would touch the circle on the right half.
Constructing a potential tangent line, a point can be plotted on the circle as follows.
From here, connect the given point outside the circle to the point on the circle with a straight line. Thus resulting in a tangent line to the circle.
### Example Question #4 : Construct Tangent Lines From Outside A Circle: Ccss.Math.Content.Hsg C.A.4
Determine whether the statement is true or false.
The line is tangent to the circle.
True
False
True
Explanation:
To construct a line that is tangent to a point on the circle, recall what it means for a line to be "tangent". A line that is tangent to a point on a circle means that the line will only touch the circle at that specific point.
Looking at the image, it is seen that the line only touches the circle once therefore, the line is tangent to the circle. Thus this statement is true.
### Example Question #4 : Construct Tangent Lines From Outside A Circle: Ccss.Math.Content.Hsg C.A.4
Construct a line that is tangent to a point on the circle and passes through the point plotted outside the circle.
Explanation:
To construct a line that is tangent to a point on the circle and passes through the point outside the circle, recall what it means for a line to be "tangent". A line that is tangent to a point on a circle means that the line will only touch the circle at that specific point.
Using the plotted circle and the given point, two potential lines can be drawn that will touch the circle at one point. One possible line would touch the circle on the left half of the circumference while the other potential line would touch the circle on the right half.
Constructing a potential tangent line, a point can be plotted on the circle as follows.
From here, connect the given point outside the circle to the point on the circle with a straight line. Thus resulting in a tangent line to the circle.
### Example Question #4 : Construct Tangent Lines From Outside A Circle: Ccss.Math.Content.Hsg C.A.4
Determine whether the statement is true or false.
Given a circle, a tangent line to the circle can be constructed if it intersects the circle at two points.
False
True
False
Explanation:
To construct a line that is tangent to a point on the circle, recall what it means for a line to be "tangent". A line that is tangent to a point on a circle means that the line will only touch the circle at that specific point.
Therefore, by definition the statement is false.
### Example Question #3 : Construct Tangent Lines From Outside A Circle: Ccss.Math.Content.Hsg C.A.4
Determine whether the statement is true or false.
The line is tangent to the circle.
False
True
True
Explanation:
To construct a line that is tangent to a point on the circle, recall what it means for a line to be "tangent". A line that is tangent to a point on a circle means that the line will only touch the circle at that specific point.
Therefore, looking at the graph
it is seen that the line only intersects the circle once. Thus, the statement "The line is tangent to the circle." is true.
### Example Question #7 : Construct Tangent Lines From Outside A Circle: Ccss.Math.Content.Hsg C.A.4
Calculate a point that is tangent to the circle and passes through the point .
Explanation:
To construct a line that is tangent to a point on the circle and passes through the point , recall what it means for a line to be "tangent". A line that is tangent to a point on a circle means that the line will only touch the circle at that specific point.
Given the equation of the circle,
the center and radius of the circle can be determined.
The center is located at and the radius is .
Therefore, the center is located at and the radius is three. Plotting the circle and tangent line to the point results in the following.
Therefore, the point on the circle that creates a tangent line, .
### Example Question #4 : Construct Tangent Lines From Outside A Circle: Ccss.Math.Content.Hsg C.A.4
Calculate a point that is tangent to the circle and passes through the point .
Explanation:
To construct a line that is tangent to a point on the circle and passes through the point , recall what it means for a line to be "tangent". A line that is tangent to a point on a circle means that the line will only touch the circle at that specific point.
Given the equation of the circle,
the center and radius of the circle can be determined.
The center is located at and the radius is .
Therefore, the center is located at and the radius is three. Plotting the circle and tangent line to the point results in the following.
Therefore, the point on the circle that creates a tangent line with the point given is .
### Example Question #5 : Construct Tangent Lines From Outside A Circle: Ccss.Math.Content.Hsg C.A.4
Construct a line that is tangent to the circle.
None of the images depict a tangent line.
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# GED Math : Operations with Negative Numbers
## Example Questions
### Example Question #155 : Basic Operations
Simplify:
Explanation:
Evaluate by using the order of operations. Double negatives result into positive.
Evaluate the sum.
### Example Question #321 : Ged Math
Solve:
Explanation:
Simplify the signs first. Double negatives result in a positive.
### Example Question #321 : Ged Math
Simplify the following expression:
Explanation:
Simplify the following expression:
We are dealing with a complex series of negative numbers. We have to combine these accurately to get the correct answer.
Let's deal with the first two terms. When subtracting from a negative number, we are essentially adding a negative to an already negative number. Therefore,
Which can be combined to get:
Now, we need to deal with the third term. Subtracting a negative is the same as adding a postive (recall that two negative signs make a positive).
So, our final answer is 10
### Example Question #321 : Ged Math
Evaluate:
Explanation:
A negative number raised to an even-numbered power is equal to the absolute value of the number taken to that power. Therefore,
This number is equal to 4 taken as a factor four times, so
### Example Question #323 : Ged Math
What is the value of the expression if and ?
Explanation:
To find the value of , plug in the given x and y values into the appropriate places.
For and ,
Now, follow the order of operations to simplify. Tackle the exponent first.
Next, multiply.
Finally, subtract.
### Example Question #153 : Basic Operations
Explanation:
Adding negatives is the same as subtracting. This is because we are adding negative numbers to our positive number, which in turn will make it smaller than when we originally started.
Let's take and add our to it, making sure to count down because this is a negative we are working with.
.
We have counted down from our original number, . We can see that is what we'll get when we add .
### Example Question #41 : Operations With Negative Numbers
Solve this equation:
Explanation:
Subtracting a negative can sound a little confusing, but it is rather simple.
When subtracting a negative, you are adding it to your positive number. If that sounds confusing, then think of Leave, Change, Opposite.
We shall use our equation to demonstrate this technique. We have minus . We're going to leave our alone, change our minus into a plus, and opposite our into a .
Our equation now should be because we changed our sign and our subtraction. If this still confuses you I suggest you use a calculator and input and you should get the same answer, which is .
### Example Question #41 : Operations With Negative Numbers
Solve the equation:
Explanation:
When multiplying with a negative, you need to understand when to bring the negative sign over in the final answer. You only have the number be negative if the one of the numbers is negative and the other is positive. If both numbers are negative, then they will cancel each other and be positive.
For our example we can see that we only have one negative number and a positive number, and respectively. So our answer has to be negative.
Multiply the two numbers as you would, making sure that your end result has a negative in it.
### Example Question #42 : Operations With Negative Numbers
Solve the equation:
Explanation:
When multiplying with a negative, you need to understand when to bring the negative sign over in the final answer. You only have the number be negative if the one of the numbers is negative and the other is positive. If both numbers are negative, then they will cancel each other and be positive.
For our example, we can see that both of our numbers are negative, and . This means that our final answer will be positive, as both of the negative signs will cancel each other out.
Multiply the two numbers together as you would, making sure to leave out the negative in your final answer.
### Example Question #328 : Ged Math
Solve the equation:
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New Math – Made Easy A New Approach to Expressing Exponentiation and Logarithms by August Klein < [email protected]
It seems to me that the way we go about teaching logarithms is all wrong. Or to put it a little less starkly, I think there is a better way to explain, define, and implement logarithms, roots, and exponents.
To go back to , is just repetitive . It is a shortcut, or shorthand, for repeated addition. For example, we compact 3+3+3+3+3 into 3x5. To undo the 3x5=15, was created. Thus to get back the multiplicand in 3x5=15 we divide by 5, getting 3. Likewise, to get back the multiplier we divide by the multiplicand 3, giving us the multiplier 5
To go back one step further, we do the same in addition. Take the sum 5+6=11. 5 is the augend and 6 is the addend. To undo addition, was created. So if we know that the addend is 6, we subtract 6 (from the sum, 11) to find the augend, 5.
Similarly exponentiation is just repeated multiplication. Thus 5x5x5 becomes 53. (Here the exponent is 3 and the is 5). Now suppose we have the 125 (the expansion) and we know that this was created from the exponent 3, but we do not know the base. How do we find it? What do we do to x 3 = 125 to find the value of the x? For this, roots were created. Thus, the 3rd root of 125 yields 5, the base. 3√x3 = 3√125 I x = 5 Now to undo multiplication, we only need one , division. That is because multiplication is commutative. 5x6 = 6x5. However, exponentiation is not commutative. 54 ≠ 4 5 Taking roots will get back the base. But we need another operator to get back the exponent. Thus logs. To undo 5 x = 125 we apply logs. Thus, log to the base 5 of 125 is 3. x log 55 = log 5125 II x = 3 Thus, logs are the way of undoing an expression “ax “ and finding the x.
So, instead of teaching that logs are an aid to multiplying log ab = log a + log b ab = antilog (log a + log b) or teaching that they are an area under a hyperbola, approach them simply as the way of undoing exponentiation - finding the exponent!
Thus, logs can be introduced in school at the same time as finding roots, and roots in general. Logs are really no more complicated than finding roots.
However, the nomenclature we use is cumbersome. Instead, I use the symbol “ ↑” for exponentiation, “ ↓” for the log , and “ √” for the root function.
Thus, 53 is 5 ↑3 log 5 125 is 125 ↓5 3√125 is 3 √125 Both systems of notation look essentially the same, except that in the case of logs, the base now comes at the end, which is really more logical. * 5 ↑3 = 125 125 ↓5 = 3 3 √125= 5 In mathematical terms, the base (5) raised to the power (3) equals the expansion (125). I like to read this “5 up 3 equals 125”. Likewise, I read 125 ↓5 as “125 down 5” or “125 log 5” and 3 √125 as “3 root 125” or 3 rd root 125
This nomenclature has a number of advantages, not the least of which, it is uncluttered, and can all be written on one line, without subscripts and superscripts, certainly an advantage with computers and in programing.
Also, and Most Important , it makes the solving of equations more transparent . Take b ↑p = s ( base ↑ power = expan Sion)
To solve for b, the base, we take the p th root of both sides ( p √ ) – analagous to I above p√(b ↑p) = p √s or b=p √s (1)
To solve for p, the exponent or the power, we take the log b of both sides ( ↓b ) – as in II (b ↑p) ↓b = s ↓b or p=s ↓b (2) Now let’s look at the 2 new expressions introduced in (1) and (2) above, p√s and s ↓b. Take s ↓b = p (expansion ↓ base = power) or log to base b of the expansion = p
To solve for s, we apply “b ↑” to both sides of the equation. b↑(s ↓b) = b ↑p or s =b↑p (3)
To get back the b, we apply “ √s” to both sides of the equation. (s ↓b) √s = p √s or b =p √s (4)
Take p √s = b **
To solve for p, we apply “s ↓” to both sides of the equation s↓(p √s) = s ↓b or p =s ↓b (5)
To solve for s, we apply “ ↑p” to both sides of the equation p√s) ↑p = b ↑p or s =b↑p (6)
* And necessary to allow (2) thru (5) above to work, with unknown in the middle . See Addendum I ** p, when it occurs in front of the √ sign, is frequently called the index 2
To summarize, then, there are 3 operators - ↑, ↓, and √. These lead to 3 possible equations. 1) b↑p = s 2) s↓b = p 3) p√s = b In each of these 3 operations, we may wish to solve for the 1 st term on the left side of the equation, or we may want to solve for the 2 nd term on the left side of the equation. This then leads to 6 (possible) solutions. These are:
1) b ↑p = s →→ p√(b ↑p) = p √s or b = p √s (solving for the base) →→ (b ↑p) ↓b = s ↓b or p = s ↓b (solving for the power)
2) s ↓b = p →→ b↑(s ↓b) = b ↑p or s = b ↑p →→ (s ↓b) √s = p √s or b = p √s
3) p √s = b →→ s↓(p √s) = s ↓b or p = s ↓b →→ (p √s) ↑p = b ↑p or s = b ↑p
The recipe, then, for solving any of the 3 equations, mechanically, is just follow these 2 rules (1) Use the operators √, ↑, ↓, and √ in just that order. ** (2) Arrange it so that the symbol you want to solve for is in the middle.
Example: to undo b ↑p = s To solve for p 1) We want p in the middle . Take the next operator after ↑, (ie:↓), and place it after (b ↑p) (b ↑p) ↓ _ = s↓_ 2) Reuse the b after the ↓, so that the p is in the middle, between 2 b’s (b ↑p) ↓b = s ↓b or p=s ↓b
To solve for b 1) We want b in the middle . Take the operator before ↑ (ie: √) and place it before (b↑p) _ √(b ↓p) = _ √s 2) Place p before the √, so that we take the p th root of (b ↑p) and of s. p √(b ↑p) = p √s or b = p √s
So now we can readily solve any of the 3 equations x ↑y = z x ↓y = z x √y = z for either x or y.
** Note that in equations 1) thru 6) on page 2, the operators always occur in the √↑↓√ and in 1)to 3) above and 1)to 6) below they always occur in this same sequence, √↑ , or ↑↓ , or ↓√ 3
Again to summarize;
↑↓√ Form . Conventional Form * p (b ↑p) ↓b = p log b b = p (2) log b s (s ↓b) √s = b √ s = b (4) (p √s) ↑p = s (p√ s) p = s (6)
p√(b ↑p) = b P√( b P) = b (1) log s b↑(s ↓b) = s b b = s (3)
s↓(p √s) = p log p√s s = p (5)
All of the 6 equations on the left above are associative – that is, they do not need parentheses.
The ↑↓√ system could be especially helpful and convenient when using the hand held calculator. Mine is a TI-89. It has only an ^ key (equivalent to the “ ↑” in the ↑↓√ system), and an “ln” key. To get a log function such as 125 ↓5, I have to remember that log b a=(log a)/(log c b)** and so enter ln(125)/ln(5) − (9 steps on the TI-89). If the calculator (or computer) had a ↓ key, I could enter directly the 125 ↓5, punch enter, and get 3 (3 steps)
Likewise, if the calculator had a “ √” key, I could solve directly 3√125 – (3 steps), instead of entering 125^(1÷3) – (7 steps) ***
Slightly more complicated , such as (3 √4096) ↓4 vs. ln(4096^(1÷3))÷ln(4) ** (7 steps) (15 steps) would be vastly simpler to enter
* This is the numbering used on page 2 to identify & introduce these same 6 formulas ** Formula 4) in Appendix - a ↓b=(a ↓e)/(b ↓e) ***Formula 1) in Appendix
4
CONCLUSION: The ↑↓√ system has a number of advantages.
1) Everything is written on one line, without subscripts and superscripts, facilitating programing, and writing papers such as this, and adding to legibility
2) It could facilitate entry of calculations into the hand held calculator (as well as the computer)
3) It would make the teaching of logarithms, roots, and exponents easier, clearer, and more readily understandable
4) The reading of an expression is simplified. Instead of saying, for 5 3, “5 to the 3 rd power”, we can say – “5 up 3”. For log 10 125, instead of “log to the base 10 of 125”, say “125 down 10”, and instead of “the 3 rd root of 125”, say “3 root 125” (this is similar to saying “three times five” for “three multiplied by five”)
5) It would make the use and implementation of logarithms, etc. by students more logical , and more mechanical. I know the idea of “mechanical” and “by recipe” may bother some mathematicians, but I know this system has made my hand calculations much easier and less prone to mistakes. The system is so logical, and beautifully symmetrical (see the left hand column at the top of page 4).
Most important – all 3 operations, ↑, ↓, and √, have the same structure , and we use the same basic procedures for unraveling each of them.
Gus Klein 160 La Questa Way Woodside, CA 94062 (650) 851-8520
5
These solutions depend on the following facts (a ↑b) ↓a = b and a↑(b ↓a) = b (a ↓b) √a = b a↓(b √a) = b (a √b) ↑a = b a√(b ↑a) = b
Note that operators always occur in the order ↓↑ , ↓√ , or √↑ Note also that the unknown to be solved for always occurs in the middle, keeping everything symmetrical and easy to apply. Note additionally that, unlike log bs=p, this whole system depends on the base occurring after the expansion - s↓b = p. Otherwise the system falls down
Thus we can easily and quickly solve any of the 3 equations. x↑y = z (1) x↓y = z (2) x√y = z (3) for either x or y
For example: To solve (3) for x
Take: x√2187 = 3 I would have difficulty solving this by conventional means But to solve by the “ ↑↓√ “ method is easy
x √2187=3 2187 ↓x√2187=2187 ↓3 x = ln 2187÷ln 3 = 7 (if Calculator had a “ ↓” key, could skip this step)
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OK, to solve x√2187=3 by conventional means:
x√2187 = 3 2187 1/x = 3 ln (2187 1/x ) = ln 3 (1/x) ln 2187 = ln 3 1/x = ln 3 ÷ ln 2187 x = ln 2187 ÷ ln 3 x = 7
Another example: To solve (2) for y By conventional means Take 1024 ↓ y = 5 log y1024 = 5 log 5 1024 ↓y√1024 = 5√1024 y y 1024 = y y = 1024 ↑(1/5) = 4 * 1024 = y 5 5 √1024 = 5√y5 5 √1024 = y y = 1024 ^ (1/5) = 4
You don’t really have to memorize all those formulas in the Appendix *
*And if the calculator had a “ ↓” key and a “ √” key, you wouldn’t have to use any of them.
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Put another way: here is a different illustration of the simplification brought about by my system (the ↑↓√ system). The 3 formulas in the left hand column below, and the 3 formulas in the right hand column, are all completely covered, in my system, by the 3 simple formulas in the center column.
b log b log a a = b a↑b↓a = b (A) a a =b log a b b √a= b a↓b√a = b (B) log a( √a)=b a a a a ( √b) = b a√b↑a = b (C) √b = b
Formulas in the middle column are associative, and do not need parentheses. Left hand column is equivalent to placing parens around the first 3 characters in center column (ie: (a ↑b) ↓a = b). Right hand col. is the same as placing parens around the 3rd , 4th & 5 th characters (ie: a ↑(b ↓a)=b). However, the center column without parens serves to replace either the left or right hand column.
The point here is that the 3 simple equations in the center column completely replace all 6 of the complicated formulas in the right and left hand columns. To get the 3 simple equations, just space the letters a,b,a between the operators ↑↓ , ↓√ , or √↑ .
Why is all this significant? The 6 formulas in the 2 outside columns are necessary to solve equations involving logs, roots, and exponents, with 2 knowns and one unknown.
For example: Take : log x 2187 = 7. Solution: log x2187 √2187 = 7√2187 Or * x log x2187 = x 7 x = 2187 (1÷7) 2187 = x 7 x = 3 2187 (1÷7) = x x =3
But you can readily solve this by my method. Just follow the 2 rules on page 3, as follows:
Problem: 2187 ↓x =7 Solution: 2187 ↓x√2187=7 √2187 Applying formula (B) above x=3
Simple, straightforward, and (almost) foolproof
*Alternate Solution – This is how I would solve it, conventionally
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During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.
# 2019 AMC 8 Problems/Problem 3
## Problem 3
Which of the following is the correct order of the fractions $\frac{15}{11},\frac{19}{15},$ and $\frac{17}{13},$ from least to greatest?
$\textbf{(A) }\frac{15}{11}< \frac{17}{13}< \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}< \frac{19}{15}<\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}<\frac{19}{15}<\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}<\frac{15}{11}<\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}<\frac{17}{13}<\frac{15}{11}$
## Solution 1
Each one is in the form $\frac{x+4}{x}$ so we are really comparing $\frac{4}{11}, \frac{4}{15},$ and $\frac{4}{13}$ where you can see $\frac{4}{11}>\frac{4}{13}>\frac{4}{15}$ so the answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$.
## Solution 2
We take a common denominator: $$\frac{15}{11},\frac{19}{15}, \frac{17}{13} = \frac{15\cdot 15 \cdot 13}{11\cdot 15 \cdot 13},\frac{19 \cdot 11 \cdot 13}{15\cdot 11 \cdot 13}, \frac{17 \cdot 11 \cdot 15}{13\cdot 11 \cdot 15} = \frac{2925}{2145},\frac{2717}{2145},\frac{2805}{2145}.$$
Since $2717<2805<2925$ it follows that the answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$.
-xMidnightFirex
~ dolphin7 - I took your idea and made it an explanation.
## Solution 3
When $\frac{x}{y}>1$ and $z>0$, $\frac{x+z}{y+z}<\frac{x}{y}$. Hence, the answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$. ~ ryjs
This is also similar to Problem 20 on the AMC 2012.
## Solution 4(probably won't use this solution)
We use our insane mental calculator to find out that $\frac{15}{11} \approx 1.36$, $\frac{19}{15} \approx 1.27$, and $\frac{17}{13} \approx 1.31$. Thus, our answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$.
~~ by an insane math guy
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## How do you work out 48360 divided by 3100?
Table of Contents
48,360 / 3,100 = 4836 / 310.
## What does 6 divided by 4 look like?
Answer: 6 divided by 4 is 1.5 Let us understand this with the long division shown below. Explanation: 6 divided by 4 can be written as 6/4 = 3/2.
What is the answer when 6 is divided by 13?
Using a calculator, if you typed in 6 divided by 13, you’d get 0.4615. You could also express 6/13 as a mixed fraction: 0 6/13.
### What is a 2 divided by 6?
1/ 3
Answer: The value of 2 divided by 6 as a fraction is 1/ 3.
### How do you solve divide?
There are a few main steps to solving a long division problem: divide, multiply, subtract, bringing the number down, and repeating the process.
1. Step One: Set up the Equation.
2. Step Two: Divide.
3. Step Three: Multiply.
4. Step Four: Subtract.
5. Step Five: Pull Down the Next Number.
6. Step Six: Repeat.
How do you know which number to divide first?
Quick Tips
1. We should always start to divide the numbers from first place of the given number.
2. The order of dividing some number is thousands, hundreds, tens… and so on…
3. When we divide any number, divisor must be smaller than the number of dividend.
## How do you write 4 divided by 6?
4 divided by 6, denoted 4 ÷ 6, is equal to 2/3. In decimal form, the answer is the repeating decimal .
## Does 6 divided by 4 have a remainder?
The number 6 divided by 4 is 1 with a remainder of 2 (6 / 4 = 1 R. 2). This is also written as 6 /4 = 1.5.
How do you work out 6 divided by 15?
6 divided by 15 is 2/5 as a fraction or 0.4 as a decimal.
### How do you solve 6 divided by 11?
The first number, 6, is called the dividend….Extra calculations for you
1. Using a calculator, if you typed in 6 divided by 11, you’d get 0.5455.
2. You could also express 6/11 as a mixed fraction: 0 6/11.
### How do you solve 6 divided by 2?
Using a calculator, if you typed in 6 divided by 2, you’d get 3. You could also express 6/2 as a mixed fraction: 3 0/2. If you look at the mixed fraction 3 0/2, you’ll see that the numerator is the same as the remainder (0), the denominator is our original divisor (2), and the whole number is our final answer (3).
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# Laws of Exponents PowerPoint PPT Presentation
Laws of Exponents. 7-2 through 7-4. What we call expanded notation. What we call expanded notation. What we call expanded notation. Putting it all together…. Putting it all together… =3 3 3 3 3 3. Putting it all together… =3 3 3 3 3 3
Laws of Exponents
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## Laws of Exponents
7-2 through 7-4
• What we call expanded notation
• What we call expanded notation
• What we call expanded notation
• Putting it all together…
• Putting it all together…
• =3 3 3 3 33
• Putting it all together…
• =3 3 3 3 33
• All told, how many 3’s?
• 6, so our final answer, in exponential notation, is…
• =3 3 3 3 33
• All told, how many 3’s?
• 6, so our final answer, in exponential notation, is…
• =
• All told, how many 3’s?
• Is there a quicker way?
• =
• All told, how many 3’s?
• Is there a quicker way?
• =
• All told, how many 3’s?
Absolutely; we can gain the same answer by
• Is there a quicker way?
• =
• All told, how many 3’s?
### Giving us our 1st law:
• Is there a quicker way?
• =
• All told, how many 3’s?
### Giving us our 1st law:
-When we multiply powers with the same base, we add their individual exponents.
### Make sure to differentiate between…
Exponents and Coefficients
### Make sure to differentiate between…
Exponents and Coefficients
Students will often get mixed up and apply the wrong operation to the problem.
4(9) = 36
4(9) = 36
4(9) = 36
4(9) = 36
=
=
=
6(9) = 54
=
6(9) = 54
### First steps into…
Scientific Notation!!!
1 < |a| < 10
1 < |a| < 10
n is an integer.
256,000
0.0041
256,000
0.0041
256,000
0.0041
### Keep in mind:
• a is negative when the original number is negative.
### Keep in mind:
• a is negative when the original number is negative.
• With small decimals, the absolute value of the exponent is equal to the # of zeroes, if you include a lead zero before the decimal place.
### Real world application:
• At 20 Celsius, one of water has a mass of about 9.98 grams.
### Real world application:
• At 20 Celsius, one of water has a mass of about 9.98 grams. Each gram of water contains about 3.34 molecules of water
### Real world application:
• At 20 Celsius, one of water has a mass of about 9.98 grams. Each gram of water contains about 3.34 molecules of water. How many molecules of water are contained in a swimming pool containing 200 of water?
O =
O =
O =
O =
O =
O =
O =
POWER LAW!!!
### Next Law
POWER LAW!!!
Raising an exponent inside parentheses to another exponent.
POWER LAW!!!
Example:
POWER LAW!!!
Example:
POWER LAW!!!
Example:
POWER LAW!!!
Example: =
Example: =
Example: =
### Expanded!
Which I can obtain much faster by…
Example: =
### Multiplication
Which I can obtain much faster by…
Example: =
### Power Law:
When raising an exponent to another exponent, we multiply the individual exponents
=
=
=
=
=
### Gentle reminder…
Coefficients are raised to the exponent
### Application:
Raising a product to a power
### Application:
Raising a product to a power
=
=
=
=
### Real world…
• The expression gives the kinetic energy, in joules, of an object of mass of m kg traveling at a speed of v meters per second.
### Real world…
• What is the kinetic energy of an experimental unmanned jet with a mass of kg traveling at a speed of about m/s?
=
=
=
=
=
½ (
=
½ (
joules
### Last Law…
• Moving in the opposite direction
### Again, with expanded notation:
What are a great deal of the 4’s going to do to each other?
### Cancel Out!
What are a great deal of the 4’s going to do to each other?
### Cancel Out!
What are a great deal of the 4’s going to do to each other?
### Cancel Out!
What are we left with?
### Cancel Out!
What are we left with?
### Cancel Out!
What are we left with?
Which is equivalent to…
### Cancel Out!
What are we left with?
Which is equivalent to…
### What’s the “magic” math way of turn 5 and 3 into 2?
What are we left with?
Which is equivalent to…
### Division!!!
What are we left with?
Which is equivalent to…
### Division Property of Exponents…
When dividing powers with the same base, we subtract their exponents.
ONE RESTRICTION:
### Division Property of Exponents…
ONE RESTRICTION: a 0
### Does everyone know why?
ONE RESTRICTION: a 0
### Quick hits
• When the larger exponent is in the denominator,
• When the larger exponent is in the denominator,
you can subtract
the top from the
bottom and put
### Does anyone need to see an example?
• When the larger exponent is in the denominator,
you can subtract
the top from the
bottom and put
### Scientific Notation & Population Density
Population Density is defined as:
### Scientific Notation & Population Density
Population Density is defined as:
### Example
• If the country of Wazup has a population of 3.5 • and a land size of 7 •,what is its population density?
### Example
• If the country of Wazup has a population of 3.5 • and a land size of 7 •,what is its population density?
### Last Application
• Raising an exponent to a power:
### Last Application
• Raising an exponent to a power:
• b 0
### Last Application
• When raising a fraction to an exponent
### Last Application
• When raising a fraction to an exponent
raise both the numerator and denominator to the exponent.
### Choices, choices…
• Simplify the following expression:
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Calculate the iterated integral 8 8 2 y y2 cosx dx dy 0
Calculating the iterated integral 8 8 2 y y2 cosx dx dy 0 can be a daunting task. But with a little creativity, it can be done!
1. Introduction
In this guide, we will show how to calculate the iterated integral 8 8 2 y y2 cosx dx dy 0. We will first evaluate the inner integral, then the outer one.
Assuming y does not equal 0, we can set up the definite integral for the innermost integral as follows:
$$\int_0^{2\pi} \cos(x) \, dx = \sin(2\pi) – \sin(0) = 0$$
We can then solve for the next integral:
$$\int_0^8 \left(\int_0^{2\pi} \cos(x) \, dx\right) y^2 \, dy = \int_0^8 0y^2 \, dy = 0$$
2. What is an iterated integral?
An iterated integral is an integral where the order of integration matters. In other words, it is not enough to simply know the value of the inner integral and the value of the outer integral; one must also know in what order to integrate them. This can be represented using a double integral sign, like this:
In this case, we would first integrate with respect to y from 0 to 8, and then with respect to x from 2 to 8. (Remember, the innermost variable goes first.) The result would be:
2cos(x)dxdy=8(1-cos(8))
Keep in mind that this is different from a regular (or single) integral, which would be written like this:
In this case, the order of integration doesn’t matter because we’re only integrating once. The result would be:
2cos(x)dxdy=4(1-cos(8))
3. How do you calculate an iterated integral?
An iterated integral is an integral where the integrand (the function being integrated) is itself a function of more than one variable. In other words, it’s an integral where you have to do more than one integral to get the final answer.
To calculate an iterated integral, you need to first determine the order of integration. This is usually done by looking at the limits of integration; whoever is “innermost” will be integrated first. In the example above, we would integrate with respect to y first, then x. To do this, we would need to break up the region of integration into two parts:
The inner part, where y is between 8 and 2 𝑦2 cos(𝑥), and
The outer part, where y is between 8 and 2 𝑦2 cos(𝑥).
We would then need to calculate two integrals:
∫82 𝑦2 cos(𝑥)𝑑𝑦 and ∫28 𝑦2 cos(𝑥)𝑑𝑦
4. What is the meaning of the iterated integral 8 8 2 y y2 cosx dx dy 0 ?
This iterated integral represents the double integral of cosx with respect to y from 0 to 8, and then with respect to x from 0 to 8.
5. How can you use iterated integrals to solve problems?
You can use iterated integrals to solve problems by breaking them down into smaller parts. This makes it easier to see how the different parts of the problem interact with each other. It also makes it easier to find the best solution to the problem.
6. What are the benefits of using iterated integrals?
There are several benefits of using iterated integrals:
1. You can calculate integrals over more complicated regions than with regular integrals.
2. You can often calculate iterated integrals faster than regular integrals.
3. Iterated integrals can be easier to visualize than regular integrals.
7. Are there any drawbacks to using iterated integrals?
There are several drawbacks to using iterated integrals. First, they can be computationally intensive, making them impractical for large data sets. Second, they can be sensitive to outliers, meaning that a few unusual data points can skew the results. Finally, they don’t always converge to a single answer, which can make interpretation difficult.
8. Conclusion
After calculating the iterated integral, we find that it equals 8. This means that the double integral of 8yy2*cos(x) over the domain D is equal to 8.
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Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :
# Pentagon
A pentagon is a geometrical shape, which has five sides and five angles. Here, “Penta” denotes five and “gon” denotes angle. The pentagon is one of the types of polygons. The sum of all the interior angles for a regular pentagon is 540 degrees.
In Geometry, we study about different types of shapes. The two-dimensional shape that is composed of straight lines and interior angles is known as a polygon. The examples of polygons are:
• Triangle (three-sided polygon)
• Quadrilateral (four-sided polygon)
• Pentagon (five-sided polygon)
• Hexagon (six-sided polygon)
• Heptagon (seven-sided polygon)
• Octagon (eight-sided polygon) and so on.
In this article, you can learn about the five-sided polygon called “pentagon” with proper definition, shape, sides, properties along with its perimeter and area of a pentagon formula in detail.
## Pentagon Definition
A pentagon is a polygon with 5 sides and 5 angles. The word “pentagon” is made up of two words, namely Penta and Gonia, which means five angles. All the sides of a pentagon meet with each other end to end to form a shape. Therefore,
The number of sides of a pentagon = 5
## Pentagon shape
Like other polygons such as triangle, quadrilaterals, square, rectangle, etc., the pentagon is also a polygon that contains five sides and five angles.
Depending on the sides, angles, and vertices, there are different types of pentagon shapes, such as
• Regular and Irregular pentagon
• Convex and Concave pentagon
### Regular and Irregular Pentagon
If a pentagon is regular, then all the sides are equal in length, and five angles are of equal measures. If the pentagon does not have equal side length and angle measure, then it is known as an irregular pentagon.
### Regular Pentagon
As defined above, a regular pentagon contains five congruent sides. Thus, we can easily find the perimeter of this type of pentagon. This can be observed from the below given figure.
Here, all the five sides are equal so the perimeter of a regular pentagon is five times the length of any one of its sides.
Also, we can divide a regular pentagon into five similar triangles as shown below.
Thus, we can say that the area of a regular pentagon is equal to the 5 times area of a triangle with sides the same as the pentagon.
### Convex and Concave Pentagon
If all the vertices of a pentagon are pointing outwards, it is known as a convex pentagon. If a pentagon has at least one vertex pointing inside, then the pentagon is known as a concave pentagon.
## Pentagon Properties
Some properties of the pentagon are as follows:
• In the pentagon, the sum of the interior angles is equal to 540°.
• If all the sides are equal and all the angles are of equal measure, then it is a regular pentagon. Otherwise, it is irregular.
• In the regular pentagon, each interior angle measures 108°, and each exterior angle measures 72°.
• An equilateral pentagon has 5 equal sides.
• The sum of the interior angles of a rectangular pentagon is 540°.
## Area of a Pentagon
For a regular pentagon with side and apothem length, then the formula to find the area of a pentagon is given as
Area of a Pentagon, A = (5/2) ×Side Length ×Apothem square units
If only the side length of a pentagon is given, then
Area = 5s2 / (4 tan 36°) Square units
If only the radius of a pentagon is given, then
Area =(5/2)r2 sin 72° Square units
## Perimeter of Pentagon
Since all the sides “a” of a regular pentagon are of equal measure, then the perimeter or circumference of a pentagon is written as,
The perimeter of a pentagon, P = 5a units
## Pentagon Solved problem
Question: Find the area and perimeter of a regular pentagon whose side is 5 cm and apothem length is 6 cm.
Solution:
Given:
The side of a pentagon, a = 5 cm
Apothem Length = 6 cm
We know that
The area of a pentagon, A = (5/2) ×Side Length ×Apothem square units
Substitute side = 5 cm, Apothem = 6 cm in formula,
A = (5/2) × 5 × 6
A = 5 × 5 × 3
A = 75
Therefore, the area of a pentagon is 75 cm2
The perimeter of a pentagon, P = 5a units
P = 5(5)
P = 25 cm
Hence, the perimeter of a pentagon is 25 cm.
Based on the properties of pentagons, there are other types of pentagons that exist in geometry. They are:
### 1. Equilateral pentagon
A polygon with five sides of equal length is called an equilateral pentagon. However, all the five internal angles of a pentagon can take a range of sets of values. They are thus allowing it to form a family of pentagons. Therefore, the regular pentagon is unique up to similarity. Because it is an equilateral and equiangular (since its five angles are equal) pentagon.
### 2. Cyclic pentagon
If all the vertices of a pentagon lie on the circumference of a circle, then it is called a cyclic pentagon. The regular pentagon is the best example of a cyclic pentagon. The area of a cyclic pentagon can be represented as one fourth the square root of one of the roots of a septic equation. Here, the coefficients of the equation are functions of the sides of the pentagon. This applies to both regular and irregular pentagons.
Line of Symmetry of a Pentagon:
When coming to the line symmetry, every polygon has a certain number of lines of symmetry. For example, a square has 4 lines of symmetry. In the same way, a regular pentagon has 5 lines of symmetry.
Stay tuned with BYJU’S – The Learning App to learn all the interesting Maths concepts and also explore videos to learn with ease.
## Frequently Asked Questions on Pentagon – FAQs
### How many sides does a pentagon have?
As the name specified “penta” means 5 and “gon” means angle. Thus, a pentagon shape has 5 sides and 5 angles.
### What are the types of pentagons?
The different types of pentagons are:
Simple pentagon
Complex pentagon
Regular pentagon
Irregular pentagon
Concave pentagon
Convex pentagon
Equilateral pentagon
Cyclic pentagon
### What is an irregular pentagon shape?
If the pentagon does not have equal side length and angle measure, then it is known as an irregular pentagon.
### What are the 12 kinds of polygons?
In geometry, the 12 kinds of polygons are:
Triangle ( three-sided polygon)
Quadrilateral ( four-sided polygon)
Pentagon ( five-sided polygon)
Hexagon ( six-sided polygon)
Heptagon ( seven-sided polygon)
Octagon ( eight-sided polygon)
Nonagon (nine-sided polygon)
Decagon (ten-sided polygon)
Icosagon (twenty-sided polygon)
Triacontagon (thirty-sided polygon)
Hectagon (100-sided polygon)
Chiliagon (1000-sided polygon)
### What is a 12 sided shape called?
A 12 sided shape or the twelve-sided polygon is called a Dodecagon. The interior angle of a regular dodecagon is 150 degrees.
### What is a 99 sided shape called?
In geometry, an enneacontagon or enenecontagon or 90-gon is a ninety-sided polygon. Thus, the 99 sided shape is called nonacontakainonagon or enneacontanonagon. Here, enneaconta is the prefix for the number of sides numbered from 90 to 99.
Related Links Triangle Heptagon Octagon Square
Test your Knowledge on Pentagon
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# Difference between revisions of "2002 AIME II Problems/Problem 15"
## Problem
Circles $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ intersect at two points, one of which is $(9,6)$, and the product of the radii is $68$. The x-axis and the line $y = mx$, where $m > 0$, are tangent to both circles. It is given that $m$ can be written in the form $a\sqrt {b}/c$, where $a$, $b$, and $c$ are positive integers, $b$ is not divisible by the square of any prime, and $a$ and $c$ are relatively prime. Find $a + b + c$.
## Solution 1
Let the smaller angle between the $x$-axis and the line $y=mx$ be $\theta$. Note that the centers of the two circles lie on the angle bisector of the angle between the $x$-axis and the line $y=mx$. Also note that if $(x,y)$ is on said angle bisector, we have that $\frac{y}{x}=\tan{\frac{\theta}{2}}$. Let $\tan{\frac{\theta}{2}}=m_1$, for convenience. Therefore if $(x,y)$ is on the angle bisector, then $x=\frac{y}{m_1}$. Now let the centers of the two relevant circles be $(a/m_1 , a)$ and $(b/m_1 , b)$ for some positive reals $a$ and $b$. These two circles are tangent to the $x$-axis, so the radii of the circles are $a$ and $b$ respectively. We know that the point $(9,6)$ is a point on both circles, so we have that
$$(9-\frac{a}{m_1})^2+(6-a)^2=a^2$$
$$(9-\frac{b}{m_1})^2+(6-b)^2=b^2$$
Expanding these and manipulating terms gives
$$\frac{1}{m_1^2}a^2-[(18/m_1)+12]a+117=0$$
$$\frac{1}{m_1^2}b^2-[(18/m_1)+12]b+117=0$$
It follows that $a$ and $b$ are the roots of the quadratic
$$\frac{1}{m_1^2}x^2-[(18/m_1)+12]x+117=0$$
It follows from Vieta's Formulas that the product of the roots of this quadratic is $117m_1^2$, but we were also given that the product of the radii was 68. Therefore $68=117m_1^2$, or $m_1^2=\frac{68}{117}$. Note that the half-angle formula for tangents is
$$\tan{\frac{\theta}{2}}=\sqrt{\frac{1-\cos{\theta}}{1+\cos{\theta}}}$$
Therefore
$$\frac{68}{117}=\frac{1-\cos{\theta}}{1+\cos{\theta}}$$
Solving for $\cos{\theta}$ gives that $\cos{\theta}=\frac{49}{185}$. It then follows that $\sin{\theta}=\sqrt{1-\cos^2{\theta}}=\frac{12\sqrt{221}}{185}$.
It then follows that $m=\tan{\theta}=\frac{12\sqrt{221}}{49}$. Therefore $a=12$, $b=221$, and $c=49$. The desired answer is then $12+221+49=\boxed{282}$.
## Solution 2 (Alcumus)
Let $r_1$ and $r_2$ be the radii of the circles. Then the centers of the circles are of the form $(kr_1,r_1)$ and $(kr_2,r_2)$ for the same constant $k,$ since the two centers are collinear with the origin. Since $(9,6)$ lies on both circles, $$(kr - 9)^2 + (r - 6)^2 = r^2,$$where $r$ represents either radius. Expanding, we get $$k^2 r^2 - (18k + 12) r + 117 = 0.$$We are told the product of the circles is 68, so by Vieta's formulas, $\frac{117}{k^2} = 68.$ Hence, $k^2 = \frac{117}{68},$ and $k = \sqrt{\frac{117}{68}}.$
[asy] unitsize(0.25 cm);
pair[] O; real[] r; pair P;
r[1] = 4.096; r[2] = 16.6; O[1] = (r[1]/(2/3*sqrt(17/13)),r[1]); O[2] = (r[2]/(2/3*sqrt(17/13)),r[2]); P = reflect(O[1],O[2])*(9,6);
draw(Circle(O[1],r[1])); //draw(Circle(O[2],r[2])); draw(arc(O[2],r[2],130,300)); draw((0,0)--(8,12*sqrt(221)/49*8)); draw((0,0)--(30,0)); draw((0,0)--O[1]--(O[1].x,0)); draw(O[1]--(O[1] + reflect((0,0),(10,12*sqrt(221)/49*10))*(O[1]))/2);
label("$y = mx$", (8,12*sqrt(221)/49*8), N);
dot("$(9,6)$", (9,6), NE); dot("$(kr,r)$", O[1], N); dot(P,red); [/asy]
Since the circle is tangent to the line $y = mx,$ the distance from the center $(kr,r)$ to the line is $r.$ We can write $y = mx$ as $y - mx = 0,$ so from the distance formula, $$\frac{|r - krm|}{\sqrt{1 + m^2}} = r.$$Squaring both sides, we get $$\frac{(r - krm)^2}{1 + m^2} = r^2,$$so $(r - krm)^2 = r^2 (1 + m^2).$ Since $r \neq 0,$ we can divide both sides by 0, to get $$(1 - km)^2 = 1 + m^2.$$Then $1 - 2km + k^2 m^2 = 1 + m^2,$ so $m^2 (1 - k^2) + 2km = 0.$ Since $m \neq 0,$ $$m(1 - k^2) + 2k = 0.$$Hence, $$m = \frac{2k}{k^2 - 1} = \frac{2 \sqrt{\frac{117}{68}}}{\frac{117}{68} - 1} = \boxed{\frac{12 \sqrt{221}}{49}}.$$
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How to Solve Zero and Negative Exponents – HowFlux
# How to Solve Zero and Negative Exponents
An exponent can be a positive number, a negative number and even a value of “zero”. If you are provided with exponents in fraction or a value of p/q for example y4/y3, you can solve it up as y4-3= y1=y. As I told you earlier that an exponent can be either a value of positive number, negative number or even zero; it is never a sure show probability that the examiner will only put forward a value in form p/q in front of you and instead he may even put forward zero or negative exponents in front of you. In order to be able to solve such exponent types, the following tips can be bought in use:-
## How to Solve Zero and Negative Exponents
#### 1. Solving the zero exponents:-
We always take the value for a zero exponent as 1 and, I can explain why. Suppose there is an exponential equation with zero value before us. i.e. y4/y4 and we know that when we solve this equation, we will get the answer as y4-4=y0 but y4 always equals to y*y*y*y where *sign stands for multiply and thus when we divide y*y*y*y by y*y*y*y, we get the value as 1 as something divided by something always equals to one and this is why we always take the value for zero exponents as one and here x is never equal to zero.
#### 2. Solving the Negative Exponents:-
Suppose a negative exponent is given to you for example 8-1, you can easily solve it by taking it as 1/8. If instead you would have been given a value 10-1, it would have resulted as 1/10. Actually, while solving the negative exponents, the negative value of power gets positive when we apply it downside as a denominator and a positive value downside always becomes negative when we take it upside applying it with the numerator.
#### 3. A Sample Solution:-
Suppose someone asks you a question, “Simplify: (84)0 * 4-1 where * stands for multiply. The initial part of this question includes a zero value of exponent while the final value includes a negative exponent and we can apply both the rules mentioned in above two points here in this equation to solve it. i.e. (84)0 * 4-1 = 1= 1.1/4=1/4 answer.
#### 4. Ending the Catharsis of Values:-
I make Mathematics as simple here for you while solving the exponential values as here you just have to keep two things in your mind. The first one is to subtract the powers when you get them with the same exponent in the form of p/q and second is to take value of zero exponents as “1” and value of negative exponent as the inverse value of the same number.
#### 5. Keep in Mind the Identities:-
The teacher will keep on telling you some identities based on the same topic when he teaches the chapter of exponential values. Just cram these values or write them on a spare piece of paper while solving the questions and you will soon be able to solve even the most complicated questions of the same chapter.
### About the Author: Editorial Team
Hello from your friendly Editorial Team at HowFlux.com. We thank you for visiting our website and hope you find our articles both fun and educational. We try to cover a wide range of topics and have over 3,000 articles posted for you to enjoy. Thank you again for being a loyal reader!
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# Moment of Inertia and Rotational Energy
Recall one of Newton’s Law’s in the context of rotational motion that we introduced last time:
$\sum \vec{\tau}=\vec{\tau}_{\text{net}}\propto\vec{\alpha}$
That is, the same of all the torques on an object is proportional to the angular acceleration of the object. In this lessen, we will determine that the proportionality constant is exactly.
## Relating Torque and Angular Acceleration
Similar to how we approached introducing concepts in rotational motion in the past, we will start off with a well-known result in linear motion and use it to derive results in rotational motion. Consider Newton’s 2nd Law, which tells us how to relate the acceleration of an object to the net force acting on that object:
$\sum \vec{F}=m\vec{a}$
where $\vec{a}$ is the tangential acceleration of the rotating object, and therefore, we know that the net force $\sum\vec{F}$ is also acting in the tangential direction, since from this equation, $\sum\vec{F}$ and $\vec{a}$ are parallel to one another.
We can take the cross product of both sides of this equation with the vector $\vec{r}$, the radial position vector from the center of rotation:
$\vec{r}\times\sum\vec{F}=m\vec{r}\times\vec{a}$
Notice that by the definition of torque, the left hand side simplifies to
$\sum \vec{\tau}=m\vec{r}\times\vec{a}$
$\vec{r}$ is in the $\hat{r}$ direction and $\vec{a}$ is in the tangential $\hat{\theta}$ direction, so we can easily evaluate the cross product on the right hand side:
$\sum \vec{\tau}=mra\hat{z}$
Furthermore, note that by the definition of angular acceleration, we know that $a=r\alpha$, where $a$ is the magnitude of the tangential acceleration and $\alpha$ is the magnitude of the angular acceleration. Therefore, we can substitute to find
$\sum\vec{\tau}=mr(r\alpha)\hat{z}=mr^2\alpha\hat{z}$
We also know that angular acceleration is in the $\hat{z}$ direction, meaning that $\vec{\alpha}=a\hat{z}$. Furthermore, we can also define a quantity called the moment of inertia $I=mr^2$, which we will discuss in detail below. Therefore, substituting this into the right hand side of this equation,
$$\boxed{\sum\vec{\tau}=I\vec{\alpha}, \text{ where }I=mr^2}$$
This is Newton’s Law for rotational motion: the net torque on a rotating object is equal to the object’s moment of inertia multiplied by the angular acceleration vector.
## Moment of Inertia
We have introduced the concept of the moment of inertia, so let’s dive into what exactly this quantity is.
$m$ is simply the mass of the object. The $r$ contribution to the moment of inertia $I=mr^2$ is the distance of the object from the center of rotation. Based on this definition, we can tell that like most other rotational motion quantities, the moment of inertia of an object will change depending on where the origin/center of rotation is defined. Therefore, in approaching a rotational problem, it is important to keep the convention of the location of the center of rotation constant throughout the problem.
One important note is that often times, there are multiple objects that make up the entirety of the rotating system, so the moment of inertia needs to account for all of these rotating object contributions. Luckily for us, the moment of inertia of the system is equal to the sum of all of the individual moment of inertias, making the calculation quite easy:
$I=\sum mr^2=m_1r_1^2+m_2r_2^2+\cdots$
Of course, if the object has a continuous distribution of mass, we can use an integral form of this equation:
$I=\int dm\text{ }r^2$
As we can see from this equation an important property to take notice of is that
The moment of inertia $I$ only depends on the geometry of the object and where the center of rotation is defined.
Let’s go over a few examples of how to calculate the moment of inertia of different objects and/or groups of objects. We will illustrate four different examples below:
1. Rotating Baton illustrates how to calculate the moment of inertia of a finite, discrete set up, using the equation $I=\sum mr^2$.
2. Rotating Uniform Rod illustrates how to calculate the moment of inertia of a homogeneous mass distribution using the equation $I=\int dm\text{ }r^2$ in one dimension. A basic knowledge of calculus and integration in one dimension is necessary.
3. Thin Cylindrical Shell illustrates how to calculate the moment of inertia of a homogeneous mass distribution using the equation $I=\int dm\text{ }r^2$ in two dimensions. A knowledge of multivariable calculus, multi-dimensional integration, and cylindrical coordinate systems is necessary.
4. Rotating Solid Sphere illustrates how to calculate the moment of inertia of a homogeneous mass distribution using the equation $I=\int dm\text{ }r^2$ in two dimensions. A knowledge of multivariable calculus, multi-dimensional integration, and cylindrical coordinate systems is necessary.
### Example 1: Rotating Baton
Imagine that we have a rigid rod of length $\ell$ with two masses, each of mass $m$, attached to either end. We can assume that the rod itself is very light in comparison to the two end masses, such that the mass of the rod itself can be neglected.
If the center of rotation is in the center of the baton, then the rod rotation will look something like
Using the definition of the moment of inertia above, we can calculate it as
$I=\sum mr^2=m\left(\frac{\ell}{2}\right)^2+m\left(\frac{\ell}{2}\right)^2=\frac{m\ell^2}{2}$
If the center of rotation is on the edge of the baton, then the moment of inertia can be calculated as
Again, applying the definition of the moment inertia again,
$I=\sum mr^2=m(0)^2+m(\ell)^2=m\ell^2$
$I=\frac{m\ell^2}{2}$ if the center of rotation is at the center of the baton.
$I=m\ell^2$ if the center of rotation is at the edge of the baton.
### Example 2: Rotating Uniform Rod
Now, let us assume that we have a uniform, homogenous rod with a total mass $M$ that is evenly distributed through the entirety of the length of the one-dimensional rod.
First, let’s assume that the center of rotation is at the center of the rod.
In this case, the moment of inertia can be calculated using the integral formula $I=\int dm r^2$. In order to calculate $I$ using this formula, we always need to start first by writing the mass in terms of the mass density and the spatial variable(s). In this case, the uniform mass density is $\rho=\frac{\text{total mass}}{\text{total length}}=\frac{M}{\ell}$. Therefore, we know that
$dm=\frac{M}{\ell}dr$
Plugging this into the integral formula and integrating along the entirety of the length of the rod from $r=-\ell/2$ to $r=\ell/2$,
$I=\int dm\text{ }r^2=\frac{M}{\ell}\int_{-\ell/2}^{\ell/2}dr r^2=\frac{M}{3\ell}\left(\frac{\ell^3}{8}-\frac{-\ell^3}{8}\right)=\frac{M}{3\ell}\frac{\ell^3}{4}=\frac{1}{12}M\ell^2$
If the center of rotation is instead on the edge of the rod, then our rotating rod instead looks like
In this case, the moment of inertia can be calculated using a similar method, but this time integrating from $r=0$ to $r=\ell$ since our origin/center of rotation has been shifted. Using the fact that $dm=\frac{M}{\ell}dr$ is still true,
$I=\int dm\text{ }r^2=\frac{M}{\ell}\int_{0}^{\ell}dr r^2=\frac{M}{3\ell}\left(\ell^3-0\right)=\frac{1}{3}M\ell^2$
$I=\frac{1}{12}M\ell^2$ if the center of rotation is at the center of the rod.
$I=\frac{1}{3}M\ell^2$ if the center of rotation is at the edge of the rod.
### Example 3: Thin Cylindrical Shell
Consider a thin, hollow 2-D cylindrical shell rotating about its central axis.
The area shaded in blue represents a small differential area element with total area $da=Rd\theta dz$, which can be derived using the differential spatial elements in cylindrical coordinates. The mass density of this shell is
$\rho=\frac{\text{total mass}}{\text{total area}}=\frac{M}{2\pi R\ell}$
where $\ell$ is the height of the cylinder and $R$ is the radius of the cylinder. Using the same method to relate the differential mass and spatial elements, we have
$dm=\rho da=\frac{M}{2\pi R\ell}Rd\theta dz=\frac{Md\theta dz}{2\pi \ell}$
Therefore, using the definition of the moment of inertia and integrating along the entirety of the circumference in the $\hat{\theta}$ direction and up the cylinder in the $\hat{z}$ direction,
$I=\int dm\text{ }r^2=\frac{MR^2}{2\pi \ell}\int_0^\ell dz\int_0^{2\pi}d\theta=\frac{MR^2}{2\pi \ell}(\ell)(2\pi)$
Simplifying this expression gives us the moment of inertia of this rotating object.
$$I=MR^2$$
### Example 4: Rotating Solid Sphere
Finally, let’s consider a solid, three-dimensional (meaning not hollow) sphere rotating on its central axis:
The small volume differential element of the sphere is given by $dV=rdrd\theta dz$, which can be derived using the differential spatial elements in cylindrical coordinates. The mass density of this shell is
$\rho=\frac{\text{total mass}}{\text{total volume}}=\frac{M}{\frac{4}{3}\pi R^3}=\frac{3M}{4\pi R^3}$
where $R$ is the radius of the solid sphere. Using the same method to relate the differential mass and spatial elements as in the earlier example problems, we have
$dm=\rho dV=\frac{3M}{4\pi R^3}rdr\text{ }d\theta\text{ }dz$
Using the definition of the moment of inertia and integrating along the entirety of the circumference in the $\hat{\theta}$ direction, along the radial direction in the $\hat{r}$ direction, and “up” the sphere in the $\hat{z}$ direction, we have that the total moment of inertia is by
$I=\int dm\text{ }r^2=\frac{3M}{4\pi R^3}\int_0^{2\pi}d\theta\int_{-R}^{R}dz\int_0^{r_{\text{max}}}dr r^3$
Here, $r_{\text{max}}$ is the maximum distance from the central rotation axis, and is a function of $z$.
From this diagram and using the Pythagorean theorem, we can conclude that
$z^2+r_{\text{max}}^2=R^2\longrightarrow r_{\text{max}}=\sqrt{R^2-z^2}$
Therefore, using this expression for $r_{\text{max}}$ as a function of $z$, our expression for the moment of inertia can be rewritten as
$I=\frac{3M}{4\pi R^3}\int_0^{2\pi}d\theta\int_{-R}^{R}dz\left(\frac{1}{4}r_{\text{max}}^4\right)=\frac{3M}{16 R^3}\int_0^{2\pi}d\theta\int_{-R}^{R}dz(R^2-z^2)^2$
It can be shown using standard integration techniques (i.e. expanding the integrand using the binomial theorem and then using the power rule) that $\int_{-R}^Rdz(R^2-z^2)^2=\frac{16R^5}{15}$. Therefore,
$I=\frac{3M}{16 R^3}\int_0^{2\pi}d\theta\frac{16R^5}{15}=\frac{3M}{16 R^3}(2\pi)\frac{16R^5}{15}$
Simplifying the expression on the right hand side gives
$$I=\frac{2}{5}MR^2$$
### Moments of Inertia of Other Common Geometries
Using similar processes as the ones shown in examples 3 and 4 above, the moments of inertia of other common geometric objects can be derived mathematically. For brevity, we will link the Wikipedia page to the results here. I don’t necessarily recommend you to memorize the moments, but it may help to keep this page handy.
## Rotational Energy
Previously, we discussed that objects moving with some velocity $v$ inherently have a kinetic energy given by
$U=\frac{1}{2}mv^2$
As you might expect, an object rotating about some axis with an angular speed $\omega$ also has an inherent energy, which we refer to as rotational energy. While rotational energy is not the same thing as kinetic energy, it turns out that we can derive an expression of it using the expression for kinetic energy. If we take the velocity $v$ as the tangential velocity of a rotating object, then we know from before that $v=r\omega$, where $\omega$ is the magnitude of the angular velocity, $v$ is the tangential speed, and $r$ is the distance of the object from the rotating center. Substituting this into the expression from above gives
$U=\frac{1}{2}m(r\omega)^2=\frac{1}{2}(mr^2)\omega^2=\frac{1}{2}I\omega^2$
using the definition of the moment of inertia from above. Therefore, we can conclude that the rotational energy of a rotating object is given by
$$U=\frac{1}{2}I\omega^2$$
### Example
The rotational energy of a rotating object is simply just another form of energy, and fits into our current working paradigm of different types of energies we have discussed. This means that it also contributes to and obeys the conservation of energy of objects. As an example, let’s consider the following problem:
A solid, three-dimensional ball of mass $m$ and radius $r$ starts from rest at a height $h$ and rolls down a slope, as shown in the figure below.
What is the linear speed $v$ of the ball when it leaves the incline? Assume that the ball rolls without slipping.
We can use conservation of energy to solve this problem. At the top of the inclined plane, the ball is not rotating or moving, and has gravitational potential energy $mgh$. At the bottom of the inclined plane, the ball has no gravitational potential energy, and but has both kinetic and rotational energy. Since there is no friction and total energy is conserved, we can conclude that
$mgh=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2$
From example 4 from above, we know that the moment of inertia of a solid, three-dimensional sphere is $I=\frac{2}{5}mr^2$, so
$mgh=\frac{1}{2}mv^2+\frac{1}{2}\frac{2}{5}mr^2\omega^2=\frac{1}{2}mv^2+\frac{1}{5}m(r\omega)^2$
Using the fact that the tangential speed $v=r\omega$,
$mgh=\frac{1}{2}mv^2+\frac{1}{5}mv^2=\frac{7}{10}mv^2 \longrightarrow v^2=\frac{10}{7}gh$
Therefore, we can conclude that the velocity of the ball at the bottom of the inclined plane is
$$v=\sqrt{\frac{10}{7}gh}$$
## Exercises
### Problem 1
A disk, a hoop, and a solid sphere are released at the same time at the top of an inclined plane. They are all uniform and roll without slipping. In what order do they reach the bottom of the plane?
### Problem 2
Consider a solid cylinder rolling about its axis.
1. The moment of inertia of a solid cylinder about its axis is $I=\frac{1}{2}MR^2$, where $R$ is the radius of the cylinder and $M$ is the mass of the cylinder. If you were able to understand examples 3 and 4 from above and have experience with multivariable calculus and multi-dimensional integrals, show that the moment of inertia of this object is indeed $I=\frac{1}{2}MR^2$.
2. If this cylinder rolls without slipping, what is the ratio of its rotational energy to its kinetic energy?
### Problem 3
This problem is adapted from Caltech’s Ph1a course.
A mass $m$ is connected to a vertical revolving axle by two massless strings of length $L$, each making an angle of $45^{\circ}$ with the axle as shown.
Both the axle and mass are revolving with large angular velocity $\omega$. You may neglect the radius of the axle.
1. Draw a free body diagram for the forces acting on the mass $m$.
2. Find the tensions $T_1$ and $T_2$ in the two strings. Give your answer in terms of $\omega, m, L, g$.
3. What is the minimum angular velocity $\omega_{\text{min}}$ such that both strings remain taut? Give your answer in terms of $\omega, m, L, g$.
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# 5.2 Drag forces
Page 1 / 6
• Express mathematically the drag force.
• Discuss the applications of drag force.
• Define terminal velocity.
• Determine the terminal velocity given mass.
Another interesting force in everyday life is the force of drag on an object when it is moving in a fluid (either a gas or a liquid). You feel the drag force when you move your hand through water. You might also feel it if you move your hand during a strong wind. The faster you move your hand, the harder it is to move. You feel a smaller drag force when you tilt your hand so only the side goes through the air—you have decreased the area of your hand that faces the direction of motion. Like friction, the drag force always opposes the motion of an object. Unlike simple friction, the drag force is proportional to some function of the velocity of the object in that fluid. This functionality is complicated and depends upon the shape of the object, its size, its velocity, and the fluid it is in. For most large objects such as bicyclists, cars, and baseballs not moving too slowly, the magnitude of the drag force ${F}_{\text{D}}$ is found to be proportional to the square of the speed of the object. We can write this relationship mathematically as ${F}_{\text{D}}\propto \phantom{\rule{0.15em}{0ex}}{v}^{2}$ . When taking into account other factors, this relationship becomes
${F}_{\text{D}}=\frac{1}{2}C\rho {\text{Av}}^{2}\text{,}$
where $C$ is the drag coefficient, $A$ is the area of the object facing the fluid, and $\rho$ is the density of the fluid. (Recall that density is mass per unit volume.) This equation can also be written in a more generalized fashion as ${F}_{\text{D}}={\text{bv}}^{2}$ , where $b$ is a constant equivalent to $0\text{.5}\mathrm{C\rho A}$ . We have set the exponent for these equations as 2 because, when an object is moving at high velocity through air, the magnitude of the drag force is proportional to the square of the speed. As we shall see in a few pages on fluid dynamics, for small particles moving at low speeds in a fluid, the exponent is equal to 1.
## Drag force
Drag force ${F}_{\text{D}}$ is found to be proportional to the square of the speed of the object. Mathematically
${F}_{\text{D}}\propto \phantom{\rule{0.25em}{0ex}}{v}^{2}$
${F}_{\text{D}}=\frac{1}{2}\mathrm{C\rho }{\text{Av}}^{2}\text{,}$
where $C$ is the drag coefficient, $A$ is the area of the object facing the fluid, and $\rho$ is the density of the fluid.
Athletes as well as car designers seek to reduce the drag force to lower their race times. (See [link] ). “Aerodynamic” shaping of an automobile can reduce the drag force and so increase a car’s gas mileage.
The value of the drag coefficient, $C$ , is determined empirically, usually with the use of a wind tunnel. (See [link] ).
The drag coefficient can depend upon velocity, but we will assume that it is a constant here. [link] lists some typical drag coefficients for a variety of objects. Notice that the drag coefficient is a dimensionless quantity. At highway speeds, over 50% of the power of a car is used to overcome air drag. The most fuel-efficient cruising speed is about 70–80 km/h (about 45–50 mi/h). For this reason, during the 1970s oil crisis in the United States, maximum speeds on highways were set at about 90 km/h (55 mi/h).
explain how a body becomes electrically charged based on the presence of charged particles
induction
babar
induction
DEMGUE
definitely by induction
Raymond
induction
Raymond
induction
Shah
induction
Korodhso
please why does a needle sinks in water
DEMGUE
induction
Korodhso
induction
Auwal
what are the calculations of Newton's third law of motiow
what is dark matter
(in some cosmological theories) non-luminous material which is postulated to exist in space and which could take either of two forms: weakly interacting particles ( cold dark matter ) or high-energy randomly moving particles created soon after the Big Bang ( hot dark matter ).
Usman
if the mass of a trolley is 0.1kg. calculate the weight of plasticine that is needed to compensate friction. (take g=10m/s and u=0.2)
what is a galaxy
what isflow rate of volume
flow rate is the volume of fluid which passes per unit time;
Rev
flow rate or discharge represnts the flow passing in unit volume per unit time
bhat
When two charges q1 and q2 are 6 and 5 coulomb what is ratio of force
When reducing the mass of a racing bike, the greatest benefit is realized from reducing the mass of the tires and wheel rims. Why does this allow a racer to achieve greater accelerations than would an identical reduction in the mass of the bicycle’s frame?
nehemiah
why is it proportional
i don't know
y
nehemiah
what are the relationship between distance and displacement
They are interchangeable.
Shii
Distance is scalar, displacement is vector because it must involve a direction as well as a magnitude. distance is the measurement of where you are and where you were displacement is a measurement of the change in position
Shii
Thanks a lot
Usman
I'm beginner in physics so I can't reason why v=u+at change to v2=u2+2as and vice versa
Usman
what is kinematics
praveen
kinematics is study of motion without considering the causes of the motion
Theo
The study of motion without considering the cause 0f it
Usman
why electrons close to the nucleus have less energy and why do electrons far from the nucleus have more energy
Theo
thank you frds
praveen
plz what is the third law of thermodynamics
third law of thermodynamics states that at 0k the particles will collalse its also known as death of universe it was framed at that time when it waa nt posible to reach 0k but it was proved wrong
bhat
I have not try that experiment but I think it will magnet....
Hey Rev. it will
Jeff
I do think so, it will
Chidera
yes it will
lasisi
If a magnet is in a pool of water, would it be able to have a magnetic field?.
yes Stella it would
Jeff
formula for electric current
Fokoua
what are you given?
Kudzy
what is current
Fokoua
I=q/t
saifullahi
Current is the flow of electric charge per unit time.
saifullahi
What are semi conductors
saifullahi
materials that allows charge to flow at varying conditions, temperature for instance.
Mokua
these are materials which have electrical conductivity greater than the insulators but less than metal, in these materials energy band Gap is very narrow as compared to insulators
Sunil
materials that allows charge to flow at varying conditions, temperature for instance.
Obasi
wao so awesome
Fokoua
At what point in the oscillation of beam will a body leave it?
Atambiri
what is gravitational force
what is meant by the term law
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# How do you find the GCF for the following list 80, 36?
Oct 21, 2016
The GCF of $80$ and $36$ is $4$
#### Explanation:
One method for finding the GCF of two numbers goes as follows:
• Divide the larger number by the smaller to give a quotient and remainder.
• If the remainder is $0$ then the smaller number is the GCF.
• Otherwise, repeat with the smaller number and the remainder.
In our example we find:
$\frac{80}{36} = 2 \text{ }$ with remainder $8$
$\frac{36}{8} = 4 \text{ }$ with remainder $4$
$\frac{8}{4} = 2 \text{ }$ with remainder $0$
So the GCF of $80$ and $36$ is $4$
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# What is Standard Error? Know About It Today
Today, we will be discussing one of the common mathematical terms that is used very much in modern mathematics. The name of the term is Standard Error of Mean. It is primarily used to measure variability. Using the Standard Error, we can have an estimation regarding the sample’s standard deviation.
Using Standard Error, one can obtain the accuracy as well as the consistency and efficiency of a sample. Using it, we can understand how accurately a sampling distribution is related to the population.
The application of Standard Error is wide. It can use in statistics as well in economics. Researchers who involve in the calculation of regression as well as perform the hypothesis testing also use them in great amounts. Inferential statistics is also one place where Standard Error is used. Following are various terms that co-relate with Standard Error.
• The Median of standard error
• Standard error of the regression coefficient
• Standard error of Mean. The standard error of mean is generally denoted by SEM.
• A variance of Standard error.
Now, we will be going to look at the Standard error of the Mean. (SEM)
## Standard Error of Mean Formula
Below shown is the Standard error formula.
x-=n
Here, x represents the Standard Deviation of the population while n represents the size of the sample. It generally coins as the number of observations conducted in the sample.
There is another formula through which the calculation also does. This formula uses when the population standard deviation can ignore. In place of it, the sample standard deviation can take into consideration. Note that this formula can only apply when the samples known to us exist as statistically independent.
x-=sn
Here, n represents the size of the sample. It generally coins as the number of observations conducted in the sample. Meanwhile, s represents the Standard Deviation of the sample.
## How important Standard Error of Mean is?
Source: teachingchannel.com
At the point when an example of various observations extract from a population and the sample mean to determine, it fills in as a gauge of the population mean. Generally, the Sample mean that has been calculated might vary from the actual mean of the population.
It will help the analyst’s examination to recognize the degree of variation that has been caused. It is the place where the standard error of the mean becomes an integral factor.
There might be the existence of a case in which multiple samples retract from the population, in such case, the standard error of the mean is the standard deviation of various means of the sample from the population mean.
In any case, numerous examples may not generally be accessible to the analyst. Luckily, the standard error of the mean is widely available and can be determined from a solitary example itself.
The method to calculate the same is as follows. The standard deviation of the various observations of the sample is divided by the square root of the size of the sample.
## What Is the Relationship Between Sem and the Size of the Sample?
Source: medium.com
In general times, it is quite evident that as the size of the sample increases, it represents more amount of substance from the population. This way, it becomes closer and closer to the actual reality of the population.
Let us take an example that will make you understand the concept with ease. Consider the scenario of 100 students that are taking part in a school examination. In this examination, let us take our sample from the students. In Sample A, we will pick 20 students randomly and in sample B, we will pick 70 students from the group of students.
Sample B will be more precise for the calculation of the average marks of the student as it has a greater number of students inside it. That means that the standard error of the mean of Sample A will be high compared to B.
There will be less chance of error to occur in sample B compared to sample A. For ideal cases, the sampling error will tend to move towards zero, if the number of participants or the observations is increasing. More the number of observations, the less the standard error percentage.
Also, from the formula, we can note that the Standard error of the mean is inversely proportional to the size of the sample. Thus, if the sample size increase by 16 times, then it is evident from the formula that the error will decrease 4 times.
## The Conclusion:
This was all about the standard error and standard error of the mean. The standard error is an important concept for research, for analysis as well as for educational purposes of the students. Using Standard error, we can determine the amount of variation that has been caused.
We hope that the article has served its purpose and whenever you need to look for the terms, you can find us right here! For all financial aid and guidance, keep making your way to FinanceShed.
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# Class 9 Maths MCQ – Parallel Lines and Transversal – 3
This set of Class 9 Maths Chapter 6 Multiple Choice Questions & Answers (MCQs) focuses on “Parallel Lines and Transversal – 3”.
1. Which of the following pairs are parallel?
a) AB and CD
b) AB and PQ
c) EF and CD
d) EF and PQ
Explanation: ∠AZY + ∠ZYC = 62° + 110° = 162°
Since AB and CD are two lines with PQ as transversal, ∠AZY and ∠ZYC are interior angles on the same side of transversal and their sum is not equal to 180°. So, Line AB is not parallel to CD.
Now, ∠EXY = ∠XYZ = 110°
EF and PQ are two lines with CD as transversal, ∠EXY and ∠XYZ are alternate interior angles and are equal. So, Line EF is parallel to PQ.
2. Find the value of ∠x if line l || m and line n || p.
a) 90°
b) 75°
c) 110°
d) 120°
Explanation:
∠1 = 75° (Vertically Opposite Angles)
Line n || p ⇒ ∠1 = ∠2 = 75° (Alternate Interior Angles)
Also, Line l || m ⇒ ∠2 = x° (Corresponding Angles)
⇒ ∠x = 75°.
3. Find the value of k if ∠1 = 150 – k and ∠2 = 60 + k and line l || m.
a) 80°
b) 75°
c) 45°
d) 150°
Explanation: Since line l || m, ∠1 = ∠2 (Corresponding Angles)
⇒ 150 – k = 60 + k
⇒ 2k = 150 – 60
⇒ k = 45°.
4. Find the value of y if line l || m.
a) 60°
b) 75°
c) 55°
d) 80°
Explanation:
∠1 + ∠(60 + y) = 180° (Linear Pair)
⇒ ∠1 = 180° – 60° – y
⇒ ∠1 = 120° – y
Since line l || m, ∠1 = ∠y = 75° (Alternate Interior Angles)
⇒ 120° – y = y
⇒ 2y = 120°
⇒ y = 60°.
5. Find the value of x, y and z if y : z = 2 : 3 and AB || CD.
a) 40°, 40°, 40°
b) 56°, 70°, 62°
c) 40°, 56°, 84°
d) 32°, 50°, 50°
Explanation: Since line AB || CD, ∠APQ = ∠APC (Alternate Interior Angles)
⇒ ∠x = 40°
Now, y : z = 2 : 3 ⇒ y = 2k, z = 3k ————— (1)
Also, ∠AQR + ∠QRC = 180° (Sum of Interior Angles on the same side of transversal is 180°)
⇒ 40° + y + z = 180°
⇒ 40° + 2k + 3k = 180° (From equation 1)
⇒ 5k = 140°
⇒ k = 28°
So, y = 2k = 2 x 28 = 56°
and z = 3k = 3 x 28 = 84°
6. Find the value of ∠PQR if AB || CD || EF.
a) 70°
b) 120°
c) 110°
d) 80°
Explanation: Line AB || CD || EF
Now, ∠CRS = ∠RSF = 20° (Alternate Interior Angles, CD || EF)
Also, 120° + ∠QRC = 180° (Sum of Interior Angles on the same side of transversal is 180°, AB || CD)
⇒ ∠QRC = 180° – 120° = 60°
Now, ∠QRS = ∠CRS + ∠QRC
⇒ ∠QRS = 20° + 60°
⇒ ∠QRS = 80°.
7. Find the value of ∠QRT if PQ || SR and SR || UT.
a) 10°
b) 20°
c) 110°
d) 120°
Explanation: Extend Line SR till A
Line PQ|| RS ⇒ ∠PQR = ∠QRS = 50° (Alternate Interior Angles)
Line PQ|| RS ⇒ ∠UTR = ∠TRA = 140° (Alternate Interior Angles)
Also, ∠QRS + ∠QRA = 180° (Sum of Interior Angles on the same side of transversal is 180°)
⇒ ∠QRA = 180° – 50° = 130°
Now, ∠QRT = ∠TRA – ∠QRA
⇒ ∠QRT = 140° – 130°
⇒ ∠QRT = 10°.
Sanfoundry Global Education & Learning Series – Mathematics – Class 9.
To practice all chapters and topics of class 9 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.
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# Area of trapezium formula
Learn about the trapezium, area of trapezium and its properties. This is one of the quadrilaterals that we have in geometry. We can call the trapezium as convex quadrilateral because all its angles will be less than 180°.
Now we will study the definition of the trapezium and we will see the shape of the trapezium. After that, we will find out the formula for finding the area of trapezium.
## Definition of trapezium:
In a quadrilateral, if any one pair of opposite sides are parallel and other pair of opposite sides are non parallel then the quadrilateral is called a trapezium.
## The Formula for finding area of trapezium:
Let PQRS be a trapezium.
Here PQ and RS are parallel sides and PS and RQ are non parallel sides. The perpendicular distance between parallel sides gives the height of trapezium. Let us represent height by the letter “h”.
Therefore, the area of trapezium is given by
A = ½ ( sum of the parallel sides) x height.
= ½ ( PQ + RS ) x h
## Properties of trapezium:
→As PQ and RS are parallel sides so, ∠P + ∠S is supplementary and also ∠Q + ∠R is supplementary.
→A line parallel to parallel sides of a trapezium will be dividing non parallel sides of trapezium proportionally.
## Isosceles trapezium:
In a given trapezium if non parallel sides are equal then that trapezium is called an isosceles trapezium.
The sides PQ and RS are parallel and the non parallel sides PS and RQ are equal. So the above given trapezium is called isosceles trapezium.
### Example:
Find the height of trapezium if the area is 540 m2 and the two parallel sides are 13m and 5m respectively.
Solution:
Given
Area = 540 m2
Parallel sides = 13m and 5m
We know that
Area = ½ (sum of the parallel sides) x height
540 = ½ ( 13 + 5 ) x h
540 x 2 = 18 x h
1080 = 18 x h
∴ h = 1080 / 18 = 60 m
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# Get the Right Triangle Questions Right on the GRE – Part I
There will always be some questions related to the right-angled triangle in every GRE test. Hence, you must have the right triangle related concepts on tips!
### The Basics
The Pythagorean Theorem is a basic concept which states that in a right triangle, the square of the hypotenuse equals the sum of the squares of the length of the other two sides of the triangle. Diagrammatically, in the following right triangle,
AB2 = BC2 + AC2; also written as c2 = a2 + b2.
Conventionally, the side opposite to angle C is represented by c, the side opposite to angle A is represented by a, and the side opposite to angle B is represented by b.
Another concept that emerges from this theorem is the concept of Pythagorean Triples. This essentially means that a triangle will be a right-angled triangle if the sum of one side is equal to the sum of squares of the other two sides.
Related Blog: How to Score 320+ in GRE in 30 days?
### Pythagorean Triples
Any triangle with the sides in the ratio 3:4:5 will be a right triangle because of 32 + 42 = 52.
3, 4, and 5 is a Pythagorean Triple. So are all the multiples of 3, 4, and 5 — 6, 8, and 10; 9, 12, and 15; and so on.
Even though there are infinite Pythagorean Triples, one has only to remember the significant ones, which are:
3, 4, and 5
6, 8, and 10
5, 12, and 13
Once in a while, you may come across 10, 24, and 26 or some higher multiples of 3, 4, and 5. However, this doesn’t mean that in a right triangle the sides must always have integer values! Instead, Pythagorean Triples are just integer solutions to the equation c2 = a2 + b2.
Two other types of right triangles are prevalent on the GRE. These are 45-45-90 triangles and 30-60-90 triangles.
### 30-60-90 Triangles
If in a right triangle, the three angles measure 30o, 60o, and 90o, the corresponding (opposite) sides will be in the ratio 1: √3: 2.
An important thing to note that a 30-60-90 right triangle is half of an equilateral triangle. There are a few corollaries to this.
1. The area of an equilateral triangle with side a is given by
2. In any right triangle, if one leg is half the hypotenuse, the triangle has got to be a 30-60-90 triangle, and that leg will be the one opposite to the 30o angle.
### Right Angled Triangle Sides 45-45-90
If in a right triangle, the three angles measure 45o, 45o, and 90o, the corresponding (opposite) sides will be in the ratio 1: 1: √2.
These triangles are also referred to as isosceles right triangles.
The fact that an isosceles right triangle is half of a square leads to other relationships such as:
1. In a square, the diagonal is √two times any side of the square and
2. The area of a square is half the diagonal’s square (½ d2 – where d is the length of the diagonal)
The above two triangles along with the Pythagorean Triples are known as special right triangles.
### Special Right Triangles and GRE
It is seen that the GRE includes several Pythagorean Triples related questions.. Here are a few examples.
Try to solve the questions yourself, and then look at the notes that follow.
Here is another one.
No. 1: In the circle above, AC is a diameter, B is a point on the circumference, and AB < BC. If AC = 20 and the area of the triangle is 96, what is the length of AB?
Note: The angle in a semi-circle is a right angle. Therefore, the triangle ABC is a right triangle right angled at B. Many of us may be tempted to write the relationships = 96 and AB2 + BC2 = 202 and proceed to solve the equations but wait a moment. If AC (the hypotenuse) = 20, which Pythagorean Triple it corresponds to? 3, 4, and 5 multiplied by 4. That is 12, 16, and 20. Why don’t we try it out before we start solving the set of equations? × 12 × 16 = 96. Voila! You have got it! AB = 12. GRE’s obsession with Pythagorean Triples works to our advantage.
Related Blog: GRE Text Completion: Tips & Strategies
Here is another weird way or rather a surreptitious way in which GRE may test your alertness about Pythagorean Triples.
Figure not drawn to scale
No. 2: The figure above has two circles with the shaded region entrapped between them. If the radius of the larger circle is 13 and the area of the shaded region is 144π. What is the radius of the smaller circle?
Note: If you can figure out that 144 is 122, you do not have to do anything other than recalling the Pythagorean Triple 5, 12, and 13.
52π (Area of the smaller circular region) + 122π (Area of the shaded region) = 132π (Area of the more prominent circular region). The answer is 5.
No. 3: In a coordinate plane, what is the distance between the points A (3, 3) and B (8, −9)?
Note: As soon as we see this question, we may be tempted to use the distance formula. Wait a moment. As you look at the positive difference between the values of the x-coordinates and the positive difference between the values of the y-coordinates, you find that the differences are 5 and 12, respectively. Therefore, the line AB is the hypotenuse of a right triangle with legs measuring 5 and 12, and, therefore, the distance has got to be 13. If you are conscious of GRE’s obsession with the Pythagorean Triples, you are better off!
After you get a knack of the types of questions GRE includes concerning Pythagorean Triples, you can easily utilize this as your advantage. Knowledge of such nuances helps you ace the GRE preparation; in our classes at Manya–The Princeton Review, we impart just that knowledge.
By the way, what do you see in the numbers below?
345681091215121620152025512131024263456810912151216201520255121310242634568109121512162015202551213102426
Liked this article? We would love to see you spread some love by sharing this article, as it may help many who are looking to study in other countries.
Learn top experienced tips to ace the GRE: Download our FREE, Complete Study Guide to the GRE!
## Improve Your Vocabulary with GRE WordsApp
If you are preparing for GRE and struggling with your Vocabulary then Manya GRE WordsApp is the ideal choice for you. Manya GRE WordsApp is a simple and efficient way to improve your vocabulary for GRE Exam. This app will make it simple to memorize words and to improve your GRE vocabulary in bite-sized pieces.
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# 10n+3(4n+2)+5 is divisible by (n∈N)
A
7
B
5
C
9
D
7
Video Solution
Text Solution
Generated By DoubtnutGPT
## To prove that 10n+3(4n+2)+5 is divisible by 9 for all n∈N, we will use the principle of mathematical induction.Step 1: Base CaseWe start by checking the base case, n=1.P(1)=101+3(41+2)+5Calculating this gives:P(1)=10+3(43)+5=10+3(64)+5=10+192+5=207Now we check if 207 is divisible by 9:207÷9=23(exact division)Thus, P(1) is true.Step 2: Inductive HypothesisAssume that the statement is true for some integer k, i.e.,P(k)=10k+3(4k+2)+5 is divisible by 9.This means there exists an integer λ such that:10k+3(4k+2)+5=9λ.Step 3: Inductive StepWe need to show that P(k+1) is also true:P(k+1)=10k+1+3(4(k+1)+2)+5.This can be rewritten as:P(k+1)=10⋅10k+3(4k+3)+5.Using the property of exponents, we can express 4k+3 as:4k+3=4k⋅43=64⋅4k.Thus,P(k+1)=10⋅10k+3(64⋅4k)+5.This simplifies to:P(k+1)=10⋅10k+192⋅4k+5.Step 4: Substitute the Inductive HypothesisFrom the inductive hypothesis, we know:10k+3(4k+2)+5=9λ.We can express 3(4k+2) as 3(4k⋅16):P(k)=10k+48⋅4k+5=9λ.Now, substituting this back into our equation for P(k+1):P(k+1)=10⋅10k+192⋅4k+5=10(9λ−48⋅4k−5)+192⋅4k+5.This leads to:P(k+1)=90λ+192⋅4k+5−10⋅5.=90λ+192⋅4k−50+5.=90λ+192⋅4k−45.Step 5: Check Divisibility by 9Now we need to check if P(k+1) is divisible by 9:P(k+1)=90λ+192⋅4k−45.Since 90λ is divisible by 9 and 192⋅4k can be expressed as 9m (where m is an integer), we can conclude that:P(k+1) is divisible by 9.ConclusionBy the principle of mathematical induction, since P(1) is true and P(k)⇒P(k+1) is true, we conclude that:10n+3(4n+2)+5 is divisible by 9 for all n∈N.
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Updated on:7/8/2024
### Knowledge Check
• Question 1 - Select One
## If 10n+3.4n+2+k is divisible by 9,∀n∈N then the least positive integral value of k is
A(a) 1
B(b) 3
C(c) 5
D(d) 7
• Question 2 - Select One
## For all n∈N,2n+1+32n−1 is divisible by: (i) 5 (ii) 7 (iii) 14 (iv) 135
A5
B7
C14
D135
Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc
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## Thursday, August 11, 2016
### Resistors in Series and Parallel Problems and Solutions
Resistors in Series
Resistors in series means end to end connection. When they are connected in series, we can find that the current will be the same across all of them and the supplied voltage across them is distributed proportional to the resistance.
We can find the effective resistance of the system when number of resistors are connected in series and also find out the voltage drop across each resistors as shown below. The effective resistance increases in series combination.
This concept is explained here in this video for your reference.
Resistors in parallel
When similar end of all resistors are connected together and the same with the other ends, the connection is called parallel connection. In parallel connection, the voltage across all the elements is same and the current across them is distributed such that it is inversely proportional to resistance.
We can find the effective resistance and the current in each element as shown below. The effective resistance in parallel is less than even the small value of the circuit. The effective resistance decreases in parallel combination.
In a video lesson over you tube parallel resistence resultant is found as shown in the link below.
Variation of resistance Problems and Solutions
We need to find the effective resistance between the two points as shown in the given picture. We can simply solve the problem by bisecting the entire circuit into two identical parts. The two parts are in series with each other and symmetrical. If we are able to find the resistance of one part, by adding the same value to that, we can find the total resistance of the system given.
As we have bisected the resistor into two parts, its length so its resistance also becomes half. By identifying the resistors in series and parallel and measuring their effective resistance, we can find the total resistance of the circuit as shown below.
Problem and Solution
This problem is about percentage change in the resistance of a wire when there is a change in its length alone. No information is given in the problem about its area. As the volume of the wire remains constant, we need to write area in terms of length and volume. Thus we can prove that resistance is directly proportional to the square of the length of the wire.
The problem is solved as shown below.
Problem and solution
This problem is about variation of resistance with mass of the wire. There is information in the problem about its length but not about area. We can change the area interms of mass and solve the problem as shown in the diagram below.
Problem and solution
This problem is about current passing in a wire when multiple wires are connected in parallel. The total current across the combination is given to us and resistance of each wire is given. We know that if resistors are connected in parallel, the voltage across them is same. Hence the current flow is reciprocal to resistance and the problem can be solved as shown below.
Problem and solution
This problem is about finding a voltage across a resistor when multiple resistors are connected in the circuit as shown in the diagram.
By identifying the elements and currents across them, we can solve the problem as shown below.
Problem and solution
If two parts of a circle separated by an angle are two wires having different resistance, we need to measure the effective resistance of the circuit. The problem is solved as shown below.
Problem and solution
This problem is about to find the effective resistance of the system where infinite resistors are connected as shown in the diagram. We can identify the symmetry in the ladder and we can say that the circuit is the combination of similar symmetrical parts. Except one part, we can assume that all other parts together are having some resistance and even with the other remaining ladder, the answer still remains same. Simply because of infinite ladders, adding or removing one ladder is not going to make a big difference to the entire system significantly.
Thus we can solve the problem as shown below.
Related Post
Capacitors in Series and parallel with Problems and Solutions
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# EQUATIONS AND LINEAR FUNCTIONS PRACTICE EOC
Problem 1 :
If 8x = -4(x + 3), then x = ?
A) -1 B) 1 C) 3/4 D) 1/4
Solution :
8x = -4(x + 3)
8x = -4x - 12
8x + 4x = -12
12x = -12
x = -12/12
x = -1
Problem 2 :
Solve for x :
9x2 - c = d
Solution :
9x2 - c = d
9x2 = d + c
Dividing by 9 on both sides.
x2 = (d + c)/9
x = √[(d + c)/9]
x = √(d + c)/3
So, option A is correct.
Problem 3 :
Which inequality is represented by the graph below.
Solution :
Since it is the falling line, rise = 2 and run = 1
Slope = rise / run = 2
y-intercept = 1
y = 2x + 1
Choosing a point below the line (-1, 0),
0 = 2(-1) + 1
0 = -2 + 1
0 = -1
The line given is a dotted line, we should use dotted line.
y > 2x + 1
Problem 4 :
Jared can run 520 yards in one minute. How fast does he run in feet per second?
A) 12 B) 26 C) 1560 D) 16
Solution :
He can run 520 yards in one minute
3 ft = 1 yard
520 yard = 520(3)
= 1560 ft
In one minute, he can run 1560 ft.
1 minute = 60 seconds
60 second = 1560 ft
1 second = 1560/60
1 second = 26 ft
Problem 5 :
There are three consecutive integers such that the sum of the two smallest integers is 17 less than three times the largest. What is the smallest integer?
A) 5 B) 7 C) 12 D) 6
Solution :
Let x, x + 1 and x + 2 be three consecutive integers.
x + x + 1 = 3(x + 2) - 17
2x + 1 = 3x + 6 - 17
2x - 3x = -17 + 6 - 1
-x = -12
x = 12
So, the smallest integer is 12.
Problem 6 :
Which expression is equivalent
Solution :
Problem 7 :
Which graph below displays the equation 3x - 4y = 28
Solution :
3x - 4y = 28
Finding x and y -intercepts, we get
x-intercept :Put y = 03x - 4(0) = 283x = 28x = 28/3x = 9.3 y-intercept :Put x = 03(0) - 4y = 28-4y = 28y = -28/4y = -7
In option A, the line has x-intercept is positive and y-intercept is negative.
So, option A is correct.
Problem 8 :
Compare the slope of f(x) = -2x + 3 and the slope of the chart of g(x) below.
What is the positive difference between the slopes of f(x) and g(x)?
A) 1 B) 5 C) 8 D) 17
Solution :
Slope of g(x) from the table,
(2, -8) and (4, -2)
Slope (m) = [-2 - (-8)] / (4 - 2)
= (-2 + 8) / 2
= 6/2
= 3
f(x) = -2x + 3
Comparing with y = mx + b
Slope of f(x) = -2, slope of g(x) = 3. slope of g(x) is greater.
f(x) - g(x) = -2 - 3 ==> -5
Problem 9 :
Gregory teaches martial arts. He charges a one-time processing fee of \$5.00 and the cost of the classes is shown below. Let x represent the number of classes and y represent the cost of classes.
Based on this information, what will it cost to take 10 classes?
A) \$123 B) \$128 C) \$118 D) \$153
Solution :
(1, 15) and (2, 27)
Rate of change = (27 - 15) / (2 - 1)
= 12/1
= 12
y-intercept (initial cost) = 5
y = 12x + 5
Cost for 10 classes :
y = 12(10) + 5
y = 120 + 5
y = 125
Problem 10 :
Jerami is going to deposit an equal amount of money into a checking account each month until he has saved \$2,000. The amount of money, y, in the account after x months can be modeled by the equation y = 35x + 250.
What does the slope of the graph of the equation represent?
[A] The amount of money deposited monthly
[B] The amount of money originally in the account
[C] The number of months it would take to earn \$250
[D] The number of months it would take to reach \$2,000
Solution :
Slope means rate of change, the amount he saves monthly. So, option A is correct.
Problem 11 :
Find the range of the function represented in the graph.
A) The range consists of values from -5 to 3.
B) The range consists of values from -4 to 6.
C) The range consists of values from -5 to 6.
D) The range consists of values from -4 to 3.
Solution :
By observing the graph, the graph spread vertically in between -5 to 3.
Problem 12 :
Which equation represents the line passing through the points (3, 2) and (–9, 6)?
A) x – 3y = 9 B) x + 3y = 9
C) 3x y = -9 D) 3x + y = 9
Solution :
(3, 2) and (–9, 6)
Slope (m) = (6 - 2) / (-9 - 3)
= -4/12
= -1/3
Equation of the line :
y - y1 = m(x - x1)
y - 2 = (-1/3) (x - 3)
3(y - 2) = -1(x - 3)
3y - 6 = -x + 3
x + 3y = 3 + 6
x + 3y = 9
Problem 13 :
Which of the following represents the linear equation
3(x + 2) = 12 – 2y
in standard form?
A) y = -3/2x + 3 B) y = 3/2x - 3 C) 3x – 2y = 10
D) 3x + 2y = 6
Solution :
3(x + 2) = 12 – 2y
3x + 6 = 12 - 2y
3x + 2y = 12 - 6
3x + 2y =6
2y = 6 - 3x
y = -3x/2 + 6/2
y = -3x/2 + 3
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# How to Calculate Force From Velocity
Written by pauline gill
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The first equation taught in dynamics is F= ma which is "force equals mass times acceleration." This equation describes the force that must be exerted on an object of known weight or mass as it is accelerated or decelerated. If a racing bicycle with a rider travelling at 20 miles per hour must stop within a certain distance, you can calculate how much force will be applied to the caliper brake on the rim of the rear wheel. You can also prove that doubling velocity quadruples (squares) the force required to stop.
Skill level:
Easy
### Things you need
• Calculator
• Compendium of physics formulas
## Instructions
1. 1
Define the velocity to force application. In this example, the bicycle with its rider weighs 95.3 Kilogram. The rider notes a white stop line that is 30 feet in front of him when he applies the brake. Since you already know the velocity, you now have enough information to calculate the required braking force.
2. 2
Solve for time T, which will enable you to calculate acceleration, or in this case, deceleration. The average velocity over the 30 feet is 20 mph divided by two, or 10 mph, which is 14.66-feet-per-second. If the 30 feet are covered at an average speed of 14.66-feet-per-second, then it takes 2.045 seconds to stop.
3. 3
Solve for acceleration using the 2.045 seconds to cover 30 feet. Since the distance calculation is D = v(0) x T +1/2 (a) T^2, the first term can be ignored since all distance covered is accounted for by the deceleration to zero. Therefore, 30 feet equals ½ a xT^2, which is 30 = ½ a x 2.045 ^2 or 30 = 1/2 a x 4.18. Re-arranging, a = 30 x 2/4.18 = 14.35 feet per second/sec.
4. 4
Solve for force using the F=ma basic equation. Force F = 210 x 14.35 feet per second/sec / 32.2 feet per second/sec (acceleration of gravity) or 42.4 Kilogram of force consistently applied by the brake to the rim for 2.045 seconds to stop the bike. This is probably right at the practical limit of this bicycle's capability to stop.
5. 5
Prove that doubling the velocity quadruples the required force. A 40 mile per hour velocity would result in a stopping time of 1.023 seconds, one-half of 2.045 seconds in the first instance. The D = ½ x a x T^2 term would work out to an acceleration of a = 30 x 2/1.046, or 57.36 feet per second/sec. F = ma would therefore work out to F = 170 Kilogram, very unreasonable for a caliper brake on a skinny racing tire. This foolish rider would never stop from 40 mph in the 30-foot distance, and they would streak right past the stop sign.
#### Tips and warnings
• Always remember that stopping force quadruples as velocity doubles.
• Accelerating quickly to a given velocity uses more force and far more fuel than smooth acceleration.
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# Patterns with Time
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Sometimes you will be given a pattern of time, either written or in pictures. In such cases, you have to examine them and figure out what the pattern is.
Here are the steps you need to follow:
• Determine how much time has been added or subtracted between the first two times.
• Double check to make sure the same amount of time passes between the 2nd and 3rd time. If it does, you have found the pattern.
• Add or subtract the same amount of time to continue the pattern.
Let's take a look at some examples:
## Example 1
The deliveries occur at these times: 4:10 PM, 4:30 PM, 4:50 PM. If this pattern continues, when will the next delivery occur?
## Solution :
First we need to see how much time has passed between the first two deliveries:
Let's subtract the minutes:
30 - 10 = 20, so 20 minutes have passed.
We now need to see if this pattern stays true.
Did 20 minutes pass between 4:30 and 4:50?
50 - 30 = 20,
Yes, the pattern is 20 minutes.
We must find the next time by adding 20 minutes to 4:50 PM, but we can't just add 20 to 50 because we'd get 4:70 and that time doesn't exist.
Let's start by thinking how much time will pass until 5:00. It would take 10 minutes (because there are 60 minutes in an hour and it's already 4:50).
That leaves how much more time?
20 minutes - 10 minutes = 10 minutes.
What is 10 minutes past 5:00?
It is 5:10 PM
Yes, it makes sense that 5:10 PM is 20 minutes after 4:50 PM.
## Example 2
Determine what time will come next in the pattern given below:
## Solution :
First we need to tell the time on each clock:
6:15, 7:30, 8:45, 10:00
Now we need to determine how much time passed in between the first two clocks.
1 hour past 6:15 is 7:15
How many minutes are there between 7:15 and 7:30?
It is 15 because 30 - 15 = 15.
So, 1 hour and 15 minutes have passed.
Let's make sure this is the right pattern by checking the next set.
Is there 1 hour and 15 minutes in between 7:30 and 8:45?
Yes, so this is the right pattern.
We could check if the pattern is true for 8:45 and 10:00 (which it is), but we don't have to because if it's true for two sets, it will continue to be true for the rest of the pattern.
We now need to find the next time by adding 1 hour 15 minutes to 10:00.
1 hour past 10:00 is 11:00 and then 15 minutes later is 11:15.
(We don't know whether or not it is AM or PM)
Does the answer make sense? Yes, it does because 11:15 is 1 hour 15 minutes past 10:00.
## Example 3
Eric is timing his older brother as he runs around the track. The times after his first three laps were 4:10 PM, 4:12 PM, and 4:14 PM. If he can keep up this pace and the pattern continues, what time will it be after he finishes 8 laps in total?
## Solution :
Let's start by finding the pattern.
There are 2 minutes in between 4:10 and 4:12,
There are 2 minutes in between 4:12 and 4:14,
So we know this is our pattern.
We can easily continue this pattern, but we have to find out what will be the time after 8 laps.
So let's expand this pattern to 8 laps:
4:10, 4:12, 4:14, 4:16, 4:18, 4:20, 4:22, 4:24
Answer: After 8 laps the time will be 4:24 PM.
Does this answer make sense? Yes, this is a reasonable time for how fast he is running.
## Patterns with Time
Determining patterns with time:
• Determine how much time has been added or subtracted between the first two times.
• Double check to make sure the same amount of time passes between the 2nd and 3rd time. If it does, you have found the pattern.
• Add or subtract the same amount of time to continue the pattern.
## Similar Lessons
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# How do you find 3%?
Another example of converting a decimal to a percentage is 0.03 x 100 = 3% or 3 percent. However, if you are required to convert 3/20 to a percentage, you should divide 3 by 20 = 0.15. Then multiply 0.15 by 100 = 15% or 15 percent.
## How do you figure 2/3 of someone’s salary?
Change two-thirds to a decimal and then multiply the decimal and your number. To convert 2/3 to decimal, divide the numerator by the denominator: 2 / 3 = 0.66666 7, which you can round to 0.67. For example, to find 2/3 of 21: 0.67 * 21 = 14.07.
## What’s a 45 out of 50?
Now we can see that our fraction is 90/100, which means that 45/50 as a percentage is 90%. And there you have it!
## What is a 36 out of 50?
Now we can see that our fraction is 72/100, which means that 36/50 as a percentage is 72%. And there you have it!
## What is a 3rd of 300?
30100 of 300 = 30100 × 300. Therefore, the answer is 90. If you are using a calculator, simply enter 30÷100×300 which will give you 90 as the answer.
## What is 1/3 in a whole number?
Since, the digit after decimal is less than 5, it will be rounded down to 0. Hence, 1/3 as a whole number will be 0.
## What is a 48 out of 50?
What is this? Now we can see that our fraction is 96/100, which means that 48/50 as a percentage is 96%.
## What is a 49 out of 50?
Now we can see that our fraction is 98/100, which means that 49/50 as a percentage is 98%. And there you have it!
## What is a 50% grade?
What are grade letters? For example “A” could be used to represent grades of 80% and above, “B” to represent grades between 70 and 80%, “C” to represent grades between 50 and 70%, and so on. Alternatively, you could have “Pass” for grades above 50% and “Fail” for grades below 50%.
## How do you find 1/3 of a number?
Thirds are calculated by dividing by 3. For example: One third of 24 =1/3 of 24 = 24/3 = 8. One third of 33 =1/3 of 33 = 33/3 = 11.
## What is a 3rd of 6?
One third of 6 equals 2. If 6 is divided into three equal parts, or thirds, one of those parts is equal to 2.
## How do you write 2 3 on a calculator?
Change two-thirds to a decimal and then multiply the decimal and your number. To convert 2/3 to decimal, divide the numerator by the denominator: 2 / 3 = 0.66666 7, which you can round to 0.67. For example, to find 2/3 of 21: 0.67 * 21 = 14.07.
## How much is a third of a pizza?
Each slice represents one-third of our pizza. So how many thirds make up the whole pizza? One-third, two-thirds, three-thirds. Three out of three or three-thirds make up one whole.
## How do you get 2/3 of a number?
To find 2/3 of a whole number we have to multiply the number by 2 and divide it by 3. To find two-thirds of 18, multiply 2/3 x 18/1 to get 36/3. 36/3 is again simplified as 12.
## How do you write 2/3 as a fraction?
Answer: 4/6, 6/9, 8/12, 10/15 are equivalent to 2/3. All those fractions obtained by multiplying both the numerator and denominator of 2/3 by the same number are equivalent to 2/3. All equivalent fractions get reduced to the same fraction in their simplest form.
## What is a 2.39 GPA?
A 2.3 GPA, or Grade Point Average, is equivalent to a C+ letter grade on a 4.0 GPA scale, and a percentage grade of 77–79.
## What grade is a 4 6?
Answer: 4 out of 6 is represented as 66.67% in the percentage form. Let us find 4 out of 6 as a percentage.
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A9_soln
# A9_soln - Math 235 Assignment 9 Solutions 1 Let ~u =(1 i 2...
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Unformatted text preview: Math 235 Assignment 9 Solutions 1. Let ~u = (1 + i, 2- i ) and ~v = (3- i, 3 + i ). Use the standard inner product on C n to calculate < ~u,~v > , < ~v,~u > , and k ~v k . Solution: We have < ~u,~v > = (1 + i, 2- i ) · (3- i, 3 + i ) = (1 + i )(3 + i ) + (2- i )(3- i ) = 2 + 4 i + 5- 5 i = 7- i < ~v,~u > = (3- i, 3 + i ) · (1 + i, 2- i ) = (3- i )(1- i ) + (3 + i )(2 + i ) = 2- 4 i + 5 + 5 i = 7 + i k ~v k = p < ~v,~v > = q (3- i, 3 + i ) · (3- i, 3 + i ) = p (3- i )(3 + i ) + (3 + i )(3- i ) = √ 20 2. Determine which of the following matrices is unitary. a) A = (1 + i ) / √ 7- 5 / √ 35 (1 + 2 i ) / √ 7 (3 + i ) / √ 35 . Solution: Observe that Solution: Let ~u = ((1 + i ) / √ 7 , (1 + 2 i ) / √ 7), and ~v = (- 5 / √ 35 , (3 + i ) / √ 35). Then we have < ~u,~u > = 1 7 [(1 + i )(1- i ) + (1 + 2 i )(1- 2 i )] = 1 7 [2 + 5] = 1 < ~v,~v > = 1 35 [(- 5)(- 5) + (3 + i )(3- i )] = 1 35 [25 + 10] = 1 < ~u,~v > = 1 √ 7 √ 35 [(1 + i )(- 5) + (1 + 2 i )(3- i )] = 1 √ 7 √ 35 [- 5- 5 i + 5 + 5 i ] = 0 Hence, { ~u,~v } is an orthonormal basis for C 2 and so...
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## This note was uploaded on 04/18/2010 for the course MATH 235 taught by Professor Celmin during the Spring '08 term at Waterloo.
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A9_soln - Math 235 Assignment 9 Solutions 1 Let ~u =(1 i 2...
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# What is 51/216 as a decimal?
## Solution and how to convert 51 / 216 into a decimal
51 / 216 = 0.236
Converting 51/216 to 0.236 starts with defining whether or not the number should be represented by a fraction, decimal, or even a percentage. Fractions and decimals represent parts of a whole, sometimes representing numbers less than 1. Choosing which to use starts with the real life scenario. Fractions are clearer representation of objects (half of a cake, 1/3 of our time) while decimals represent comparison numbers a better (.333 batting average, pricing: \$1.50 USD). After deciding on which representation is best, let's dive into how we can convert fractions to decimals.
## 51/216 is 51 divided by 216
The first step of teaching our students how to convert to and from decimals and fractions is understanding what the fraction is telling is. 51 is being divided into 216. Think of this as our directions and now we just need to be able to assemble the project! The numerator is the top number in a fraction. The denominator is the bottom number. This is our equation! We must divide 51 into 216 to find out how many whole parts it will have plus representing the remainder in decimal form. Here's 51/216 as our equation:
### Numerator: 51
• Numerators are the top number of the fraction which represent the parts of the equation. Any value greater than fifty will be more difficult to covert to a decimal. The bad news is that it's an odd number which makes it harder to covert in your head. Values closer to one-hundred make converting to fractions more complex. Let's look at the fraction's denominator 216.
### Denominator: 216
• Denominators are located at the bottom of the fraction, representing the total number of parts. Larger values over fifty like 216 makes conversion to decimals tougher. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Overall, two-digit denominators are no problem with long division. Next, let's go over how to convert a 51/216 to 0.236.
## Converting 51/216 to 0.236
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 216 \enclose{longdiv}{ 51 }$$
To solve, we will use left-to-right long division. This method allows us to solve for pieces of the equation rather than trying to do it all at once.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 216 \enclose{longdiv}{ 51.0 }$$
We've hit our first challenge. 51 cannot be divided into 216! Place a decimal point in your answer and add a zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 216 into 51 + 0 or 510.
### Step 3: Solve for how many whole groups you can divide 216 into 510
$$\require{enclose} 00.2 \\ 216 \enclose{longdiv}{ 51.0 }$$
How many whole groups of 216 can you pull from 510? 432 Multiple this number by our furthest left number, 216, (remember, left-to-right long division) to get our first number to our conversion.
### Step 4: Subtract the remainder
$$\require{enclose} 00.2 \\ 216 \enclose{longdiv}{ 51.0 } \\ \underline{ 432 \phantom{00} } \\ 78 \phantom{0}$$
If there is no remainder, you’re done! If you still have numbers left over, continue to the next step.
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals is a necessity. They each bring clarity to numbers and values of every day life. Same goes for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But they all represent how numbers show us value in the real world. Here are examples of when we should use each.
### When you should convert 51/216 into a decimal
Speed - Let's say you're playing baseball and a Major League scout picks up a radar gun to see how fast you throw. Your MPH will not be 90 and 51/216 MPH. The radar will read: 90.23 MPH. This simplifies the value.
### When to convert 0.236 to 51/216 as a fraction
Pizza Math - Let's say you're at a birthday party and would like some pizza. You aren't going to ask for 1/4 of the pie. You're going to ask for 2 slices which usually means 2 of 8 or 2/8s (simplified to 1/4).
### Practice Decimal Conversion with your Classroom
• If 51/216 = 0.236 what would it be as a percentage?
• What is 1 + 51/216 in decimal form?
• What is 1 - 51/216 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.236 + 1/2?
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Browse Questions
# Solve the following differential equations : $xdy-ydx=\sqrt{x^2+y^2}dx$
Toolbox:
• A differential equation of the form $\large\frac{dy}{dx }$$= F(x,y) is said to be homogenous if F(x,y) is a homogenous function of degree zero. • To solve this type of equations substitute y = vx and \large\frac{dy}{dx }$$= v + x\large\frac{dv}{dx}$
Step 1:
We can rewrite the given equation as $\large\frac{dy}{dx }=\large\frac{ \sqrt {x^2 - y^2} + y}{x}$
$F(x,y) =\large\frac{ \sqrt {x^2+y^2} +y}{x}$
$F(kx,ky) = \large\frac{\sqrt{ k^2x^2+ k^2y^2} +ky}{kx} =$$k^0.F(x,y) Hence this is a homogenous equation with degree zero. Step 2: Using the information in the tool box let us substitute y = vx and \large\frac{dy}{x} =$$ v + x\large\frac{dv}{dx}$
$v + x\large\frac{dv}{dx} =\frac{ \sqrt{ x^2 + v^2x^2} + vx}{x}$
taking $x$ as the common factorand cancelling we get,
$v + x\large\frac{dv}{dx} $$= \sqrt { 1+v^2)}+v Cancelling v on both sides we get, x\large\frac{dv}{dx}$$= \sqrt{ (1+v^2)}$
Seperating the variables we get,
$\large\frac{dv}{\sqrt{1+v^2}} =\frac{ dx}{x}$
Step 3:
Integrating on both sides we get,
$\log|v+(\sqrt{ 1+v^2})| = \log x + \log C$
writing the value of $v =\large\frac{ y}{x}$
$\log(\large\frac{y}{x}) $$+ \sqrt{ 1+(y^2/x^2) }= \log Cx \log\large\frac{y + \sqrt{(x^2+y^2)}}{x }$$= \log Cx$
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# Divisibility Rules of Numbers
## Divisibility Rules
These divisibility rules will help in a faster calculation, thus saving a lot of time; especially in the exams.
### Divisibility Rules for 2
• If a number ends with either 0 or even digit, it is divisible by 2.
• Examples- 10, 32, 44, 56, 188 etc.
### Divisibility Rules for 3
• When the sum of the digits of a number is divisible by 3, then the number is divisible by 3.
• Example- 4563 (Here the sum of the digits i.e., 4+5+6+3=18 is divisible by 3 so the number must be divisible by 3).
### Divisibility Rules for 4
• When the number made by the last two digits of the given number is divisible by 4.
• Also, if the number having two or more zeros at the end.
• Examples- 400, 8500, 754000, 564, 33832 etc.
### Divisibility Rules for 5
• Number ending with either 0 or 5 is divisible by 5
• Examples- 55, 15, 50, 65, 85 etc.
### Divisibility Rules for 6
• The number must be divisible by 2 and 3 both.
• Examples- 6, 12, 66, 72 etc.
### Divisibility Rules for 7
• When the difference between twice the digit at One’s Place and the number formed by other digits is either zero or a multiple of 7.
• Example- 658 is divisible by 7 i.e., (658 — 65 – 2*8 = 49 is divisible by 7).
### Divisibility Rules for 8
• If the number made by the last three digits of a given number is divisible by 8.
• Also, if the number is having three or more zeros at the end.
• Examples- 9256, 65000, 895740000 etc.
### Divisibility Rule for 9
• When the sum of all digits of the given number is divisible by 9.
• Example- 85869 (8+5+8+6+9=36 is divisible by 9).
### Divisibility Rule for 10
• A number ending with zero is divisible by 10
• Examples- 10, 50, 5600, 450, 8886540 etc.
### Divisibility Rules for 11
• If the sums of digits at odd and even places are equal or differ by a number divisible by 11.
• Example- 2865432
Sum of digits at odd places 2+6+4+3=15
Sum of digits at even places 8+5+2= 15
Both are equal, so divisible by 11
• Example- 217382
Sum of digits at odd places 2+7+8=17
Sum of digits at even places 1+3+2= 6
Differ by 11 (17-6), so divisible by 11
### Divisibility Rules for 12
• If a number is divisible by both 4 and 3.
• Examples- 144, 3600, 2064 etc.
### Divisibility Rules for 14
• If the number is divisible by both 7 and 2.
• Example- 98, 15694 etc.
### Divisibility Rules for 15
• If the given number is divisible by 5 and 3 is also divisible by 15.
• Examples- 183555, 135 etc.
### Divisibility Rule for 16
• A number is divisible by 16 if the number made by its last 4 digits is divisible by 16.
• Example- 126304 etc.
### Divisibility Rule for 18
• If a number is even and is divisible by 9, then it is divisible by 18.
• Examples- 108, 4356 etc.
### Divisibility Rule for 25
• If last two digits are either zero or divisible by 25.
• Examples- 75, 50, 13550, 1275 etc.
### Divisibility Rule for 125
• If the last three digits are divisible by 125.
Examples- 630125; Here last three digits 125 is divisible by 125, so the number 630125 is also divisible by 125.
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For an integer n greater than 1, n! denotes t...
### Related Test
For an integer n greater than 1, n! denotes the product of all integers from 1 to n, inclusive. If x and y are two distinct positive integers such that y > x, what are the values of x and y?
(1) The ratio of the lowest common multiple and the highest common factor of y! and x! is 20:1.
(2) The lowest common multiple and the highest common factor of y! and x! have 3 and 2 prime factors respectively
• a)
Statement (1) ALONE is sufficient, but statement (2) alone is
• b)
Statement (2) ALONE is sufficient, but statement (1) alone is
• c)
BOTH statements (1) and (2) TOGETHER are sufficient to
• d)
EACH statement ALONE is sufficient to answer the question
• e)
Statements (1) and (2) TOGETHER are NOT sufficient to
problem are needed.
Step 1 & 2: Understand Question and Draw Inference
• n! = 1*2*3……*n
• x, y are distinct integers > o
• y > x
To Find: Values of x and y
Step 3 : Analyze Statement 1 independent
1. The ratio of the lowest common multiple and the highest common factor of y! and x! is 20:1.
• x! = 1*2*3…..*x
• y! = 1*2*3….x*(x+1)…….* y ( as y > x)
• So, we can write: y! = x! * a, where a is a positive integer
• As y! is a multiple of x!, LCM( x!, y!) = y!
• Similarly, as x! is a factor of y!, GCD(x!, y!) = x!
So,
Hence, 20 needs to be expressed as a product of 1 or more consecutive positive
integers. Following cases are possible:
• As 20 = 22 * 5, following cases are possible:
• 20 = 4 * 5
• So, (x+1) * y = 20. Therefore, x + 1 = 4, i.e. x = 3 and y = 5
• The other possible case is when (x+1) = y = 20, i.e. y = 20 and x = 19. So,
As we do not have unique values of x and y, the statement is insufficient to answer
Step 4 : Analyze Statement 2 independent
2. The lowest common multiple and the highest common factor of y! and x!
have 3 and 2 prime factors respectively.
• we know that LCM(x!, y!) = y! = 1*2*3*4*5…y (we deduced this in the analysis of statement-1)
• As y! has 3 prime factors only, they have to be the smallest 3 prime factors, i.e. {2, 3 , 5}. So, 5! ≤ y! < 7! as 7! will have 4 prime factors {2, 3, 5, 7}
• 5 ≤ y < 7. So, y = 5 or 6
• We also know that GCD(x!, y!) = x! = 1*2*3… x (we deduced this in the analysis of statement-1)
• As x! has 2 prime factors only, they have to be the smallest 2 prime factors , i.e. {2, 3}. So, 3!≤ x! < 5!, as 5! will have 3 prime factors {2, 3, 5}
• 3≤ x < 5 So, x = {3, 4}
As we do not have unique values of x and y, insufficient to answer
Step 5: Analyze Both Statements Together (if needed)
1. From statement-1, we inferred that y = {20, 5} and x = {19, 3}
2. From statement-2, we inferred that y = { 5, 6} and x = { 3,4 }
Combining both the statements, we have y = 5 and x = 3.
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Lesson Plans and Worksheets for Grade 3
Lesson Plans and Worksheets for all Grades
Videos, examples, solutions, and lessons to help Grade 3 students learn to identify arithmetic patterns (including patterns in the addition table or multiplication table), and explain them using properties of operations. For example, observe that 4 times a number is always even, and explain why 4 times a number can be decomposed into two equal addends.
Common Core: 3.OA.9
### Suggested Learning Target
• I can identify and describe arithmetic patterns in number charts, addition tables, and multiplication tables.
• I can explain arithmetic patterns using properties of operations.
Identify and explain arithmetic patterns using properties of operations (CCSS: 3.OA.9)
Identifying Patterns: 3.OA.9
How to identify an arithmetic pattern?
Explain why the pattern makes sense.
Identify arithmetic patterns
How to identify arithmetic patterns (including patterns in the addition table or multiplication table), and explain them using properties of operations. For example, observe that 4 times a number is always even, and explain why 4 times a number can be decomposed into two equal addends.
Examples:
1. There are 8 rafts floating down the river. Each raft is carrying 4 people. Find the total number of people in the rafts.
2. A farmer is collecting eggs from the hen house. On Monday, the farmer collects 5 eggs. On Tuesday, he collects 12 eggs. On Wednesday, the farmer collects 19 eggs. If this pattern continues, how many eggs does he collect on Saturday?
3. A bottling plant packages one box of bottles water each minute. Each number below represents the total number of bottles packaged after 1 more minute. How many bottles are in each box. Complete the pattern.
4. Anna has 15 shells. She gives away 6 shells to her sister. She wants to line up the remaining shells into 3 equal rows. How many shells will be in each row? For Teachers
Discover Number Patterns With Skip Counting
Grade 3 / Math / Number Sense
CCSS: Math.3.OA.D.9 Math.Practice.MP3 Math.Practice.MP7
Lesson Objective
Look for patterns when counting by 200s
3.oa.9
The Common Core State Standards (CCSS) videos are designed to support states, schools, and teachers in the implementation of the CCSS. Each video is an audiovisual resource that focuses on one or more specific standards and usually includes examples/illustrations geared to enhancing understanding. The intent of each content-focused video is to clarify the meaning of the individual standard rather than to be a guide on how to teach each standard although the examples can be adapted for instructional use.
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# How do you factor 15x^2 - x – 2 by grouping?
Jun 16, 2016
$15 {x}^{2} - x - 2 = \left(3 x + 1\right) \left(5 x - 2\right)$
#### Explanation:
To factorize $15 {x}^{2} - x - 2$ by grouping, we have to first split coefficient of middle term $- 1$ in two parts so that their product is the product of coefficient of other two i.e. $15 \times \left(- 2\right) = - 30$.
It is apparent that these could be $- 6$ and $5$. Hence,
$15 {x}^{2} - x - 2$
= $15 {x}^{2} - 6 x + 5 x - 2$
Now we group them as follows
$3 x \left(5 x - 2\right) + 1 \left(5 x - 2\right)$
= $\left(3 x + 1\right) \left(5 x - 2\right)$
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# Computing the length of a path
The rectangular grounds at Hillingham University is going to have a new path built which takes the curved shape of $y=2.2x^2$, starting from the south-west corner - (taken as the origin in a graph) over to the point on the path where $x=3.7$. The constructors of this path want to know the path's length.
What is the exact length of this path?
-
The length of a curve is given by
$$\int_a^b \sqrt{1+(f'(x))^2}\ dx.$$
Let $f(x)= y$, $f'(x) = \frac{dy}{dx}$, $a=0$ and $b=3.7$, and compute the integral in a straightforward manner.
Where does this come from?
Consider estimating the length of a segment of the curve by forming a right triangle whose vertices are $(x,f(x))$, $(x+\Delta x, f(x))$ and $(x+\Delta x, f(x+\Delta x))$.
Let $\Delta y = f(x+\Delta x)-f(x)$.
Then, the length of the hypotenuse, $s$, is given by the Pythagorean Theorem $s = (\Delta x)^2 + (\Delta y)^2$.
Taking the limit as $\Delta x \to 0$, using the notation of infinitesimals, we see that $\Delta y \to dy$ and $\Delta x \to dx$.
To get the total length of the curve, we add up all these infinitesimal lengths:
$$\int_a^b ds.$$
But from the Pythagorean theorem we have that $ds^2 = dx^2+dy^2$, so $\frac{ds^2}{dx^2} = 1+\frac{dy^2}{dx^2} = 1+\left(f'(x)\right)^2$.
Finally, multiplying $dx^2$ back over and taking the square root, we have
$$ds = \sqrt{1+(f'(x))^2}\ dx.$$
The length of the curve is found by adding up all these infinitesimal hypotenuses, so
$$L(f(x)) = \int_a^b ds = \int_a^b \sqrt{1+(f'(x))^2}\ dx.$$
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is this a standard result? how do you know this? – Anona anon Feb 20 '13 at 17:37
also, why does the function have to be differentiated first? – Anona anon Feb 20 '13 at 17:39
This is a standard result that comes from basic calculus. I'll expand the answer. – Arkamis Feb 20 '13 at 17:39
would you integrate this using trig substitution ? if so how? thank you – Anona anon Feb 21 '13 at 15:41
Well, $f(x) = 2.2x^2$, so $f'(x) = 4.4x$, so you integrate $\left[1+4.4x\right]^{1/2}$ through basic techniques. Substitution is a way; alternatively, you can recognize that multiplying the integral by 4.4 inside, and 1/4.4 outside gives you $\frac{1}{4.4}\int_0^{3.7} u^{1/2}\ du.$ – Arkamis Feb 21 '13 at 16:42
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# In the equation $x\cos(\theta) + y\sin(\theta) = z$ how do I solve in terms of $\theta$?
In the equation $$x\cos(\theta) + y\sin(\theta) = z,$$ how do I solve in terms of $\theta$? i.e $\theta = \dots$.
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Linear equations in $\sin \theta$ and $\cos \theta$ can be solved by a resolvent quadratic equation. One method is to write the $\sin \theta$ and $\cos \theta$ functions in terms of the $\tan (\theta/2)$. – Américo Tavares Apr 15 '11 at 15:44
There are various possible strategies. I will mention one approach. Of course if there is a solution, there are infinitely many, since we can add $2\pi$ to any solution and get another solution.
Let's change notation a little. We are interested in the equation $$a\cos\theta + b\sin\theta=q$$
Rewrite this equation as $$\frac{a}{\sqrt{a^2+b^2}}\cos\theta+ \frac{b}{\sqrt{a^2+b^2}}\sin\theta=\frac{q}{\sqrt{a^2+b^2}}$$ Now let $\phi$ be the angle whose sine is $a/\sqrt{a^2+b^2}$ and whose cosine is $b/\sqrt{a^2+b^2}$. By a formula that I hope is familiar (sine of a sum of angles), the equation can be rewritten as $$\sin(\phi+\theta)=\frac{q}{\sqrt{a^2+b^2}}$$ Look at the right-hand side. If its absolute value is greater than $1$, there will be no (real) solution. Otherwise, for simplicity, call the right-hand side $w$. Then we can write that $\phi+\theta=\arcsin w$ or $\phi+\theta=\pi-\arcsin w$. Now remember that whatever solutions you get through this process, anything obtained by adding $2n\pi$, where $n$ is an integer, to a solution, is also a solution.
The reason I went in detail through this approach is that in Physics, it is often important to express $a\cos\theta+b\sin\theta$ in the form $c\sin(\phi+\theta)$ that we used to solve the equation. There are other approaches.
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Almost right, but you slipped at the end: $\sin(\phi+\theta)=\sin(w)$ means that $\phi+\theta$ belongs to $\arcsin(w)+2\pi\mathbb{Z}$ or to $\pi-\arcsin(w)+2\pi\mathbb{Z}$. – Did Apr 15 '11 at 16:14
You did not get what I wrote: unless $|w|=1$, the set of solutions of the equation $\sin(\phi+\theta)=w$ cannot be written as the set of $\theta_0+2n\pi$, for a well chosen $\theta_0$ and for every integer $n$. So your solution as written is false. – Did Apr 15 '11 at 16:36
@Didier Pau: Sorry, I missed part of the content of your comment. I have edited the answer to take account of what you wrote. – André Nicolas Apr 15 '11 at 16:37
This expands my comment above. As I wrote here "certain trigonometric equations such as the linear equations in $\sin x$ and $\cos x$ can be solved by a resolvent quadratic equation. One method is to write the $\sin x$ and $\cos x$ functions in terms of (...) $\tan$ of the half-angle".
Applying this method, since
$$\cos \theta =\frac{1-\tan ^{2}\frac{\theta }{2}}{1+\tan ^{2}\frac{\theta }{2% }}\qquad\text{and}\qquad\sin \theta =\frac{2\tan \frac{\theta }{2}}{1+\tan ^{2}\frac{\theta }{2}},$$
$$x\frac{1-\tan ^{2}\frac{\theta }{2}}{1+\tan ^{2}\frac{\theta }{2}}+y\frac{% 2\tan \frac{\theta }{2}}{1+\tan ^{2}\frac{\theta }{2}}=z.$$
Let $u=\tan \frac{\theta }{2}$. Then we can write it as
$$\left( x+z\right) u^{2}-2yu+z-x=0,$$
which has the solutions
$$u=\tan \frac{\theta }{2}=\frac{1}{2\left( x+z\right) }\left( 2y\pm2\sqrt{% y^{2}+x^{2}-z^{2}}\right).$$
Thus
$$\theta =2\arctan \left( \frac{1}{ x+z }\left( y\pm % \sqrt{y^{2}+x^{2}-z^{2}}\right) \right) +2n\pi,\qquad n\in\mathbb{Z} .$$
This method is valid iff $\theta \neq (2k+1)\pi$, with $k\in\mathbb{Z}$.
A different technique is to use an auxiliary angle.
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Thank you! I had a math teacher that insisted Math was a sport. – user9624 Apr 15 '11 at 16:58
You are welcome! I adapted this technique from my 1967 Trigonometry text book (J. Calado, Compêndio de Trigonometria), where it is presented and worked out in detail. I deleted my explanation of the auxiliary angle technique (answered before by user6312). – Américo Tavares Apr 15 '11 at 21:13
Let us introduce $c=\cos \theta$. Then your equation reads $$x c + s y \sqrt{1-c^2} =z$$ where $s=\pm 1$ is related to the quadrant of $\theta$ (+1 in the first and second quadrant and -1 in the third and forth). Subtracting $x c$ from both sides then squaring the equation yields $$y^2 (1-c^2) = (z- xc)^2 = z^2 -2 zx c + x^2 c^2.$$ This is a quadratic equation with the solutions $$c_\pm = \frac{x z \pm y \sqrt{x^2+ y^2 - z^2}}{x^2 + y^2}.$$ In order that $c_\pm$ are real, we need $z^2 \leq x^2+y^2$. As we have squared the equation, we have to check whether $c_\pm$ solves the original equation. Indeed, $c_\pm$ solves the original equation with $s_\pm=\text{sgn}[(z-xc_\pm)/y]$. Therefore, we have the two solutions (mod $2\pi$) $$\theta = s_\pm \arccos(c_\pm).$$
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Hmmm... I doubt that. First, both solutions $c_+$ and $c_-$ of the quadratic equation are admissible. Second, the set of solutions $\theta$ is either empty or of the form $a\pm b+2\pi\mathbb{Z}$. – Did Apr 15 '11 at 16:10
@Didier Piau: thank you, I corrected the mistake. Regarding the $+2\pi \mathbb{Z}$, as $\theta$ is an angle everything is mod $2\pi$. – Fabian Apr 15 '11 at 16:18
Did you check that $|c_\pm|\le1$? You know, me saying so should not be enough... And I am not sure I follow your dealing with the signs. – Did Apr 15 '11 at 16:31
@Didier Piau: yes I check it (I actually started writing the edited post before I read your comment...) – Fabian Apr 15 '11 at 16:33
@Didier Piau: the signs are easy. When I square the equation, I have a √ on one side and $(z-xc)/sy$ on the other. The latter should be positive for the equality to hold. – Fabian Apr 15 '11 at 16:36
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# Lesson 22
Solving Rational Equations
• Let’s think about how to solve rational equations strategically.
### Problem 1
Identify all values of $$x$$ that make the equation true.
1. $$\frac{2x+1}{x}=\frac{1}{x-2}$$
2. $$\frac{1}{x+2}=\frac{2}{x-1}$$
3. $$\frac{x+3}{1-x} = \frac{x+1}{x+2}$$
4. $$\frac{x+2}{x+8}= \frac{1}{x+2}$$
### Problem 2
Kiran is solving $$\frac{2x-3}{x-1} = \frac{2}{x(x-1)}$$ for $$x$$, and he uses these steps:
\begin{align} \frac{2x-3}{x-1} &= \frac{2}{x(x-1)}\\ (x-1)\left(\frac{2x-3}{x-1} \right) &= x(x-1) \left( \frac{2}{x(x-1)} \right)\\ 2x-3 &= 2\\ 2x &= 5 \\ x &= 2.5 \\ \end{align}
He checks his answer and finds that it isn't a solution to the original equation, so he writes “no solutions.” Unfortunately, Kiran made a mistake while solving. Find his error and calculate the actual solution(s).
### Problem 3
Identify all values of $$x$$ that make the equation true.
1. $$x=\frac{25}{x}$$
2. $$x+2= \frac{6x-3}{x}$$
3. $$\frac{x}{x^2} = \frac{3}{x}$$
4. $$\frac{6x^2+18x}{2x^3} = \frac{5}{x}$$
### Problem 4
Is this the graph of $$g(x)=\text-x^4(x+3)$$ or $$h(x)=x^4(x+3)$$? Explain how you know.
(From Unit 2, Lesson 10.)
### Problem 5
Rewrite the rational function $$g(x) = \frac{x-9}{x}$$ in the form $$g(x) = c + \frac{r}{x}$$, where $$c$$ and $$r$$ are constants.
(From Unit 2, Lesson 18.)
### Problem 6
Elena has a boat that would go 9 miles per hour in still water. She travels downstream for a certain distance and then back upstream to where she started. Elena notices that it takes her 4 hours to travel upstream and 2 hours to travel downstream. The river’s speed is $$r$$ miles per hour. Write an expression that will help her solve for $$r$$.
(From Unit 2, Lesson 21.)
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# Calculating the Value of e
There are several ways to calculate the value of e. Let's look at the historical development.
## Using a Binomial Expansion
If n is very large (approaches infinity) the value of (1+1/n)^napproaches e.
This is not an efficient way to find e. Even if we go out to n = 100,000, our value is only correct to the 4th decimal place.
e~~[(1+1/n)^n]_(n=100000) =2.718268237
Here's the graph demonstrating this expansion:
e = 2.71828...
y=(1+1/n)^n
Graph of y=(1+1/n)^n, showing the limit as n->+-oo is e.
## Another Expansion
As n becomes very small, (1+n)^(1"/"n) approaches the value of e.
We can obtain reasonable accuracy with a very small value of n.
e~~[(1+n)^(1"/"n)]_(n=0.000000001) =2.718281827
Let's see the graph of the situation.
e = 2.71828...
y=(1+n)^(1//n)
Graph of y=(1+n)^(1"/"n), showing the y-intecept (the limit as x->0) is e.
(There is actually a "hole" at n = 0. Can you understand why?)
## Newton's Series Expansion for e
The series expansion for e is
e^x=1+x+1/2x^2+1/6x^3+ 1/24x^4+...
Replacing x with 1, we have:
e=1+1+1/2(1)^2+ 1/6(1)^3+ 1/24(1)^4+...
We can write this as:
e=sum_(n=0)^oo(1/(n!))
This series converges to give us the answer correct to 9 decimal places using 12 steps:
e~~sum_(n=0)^12(1/(n!))=2.718281828
## Brothers' Formulae
Recently, new formulae have been developed by Brothers (2004) which make the calculation of e very efficient.
e=sum_(n=0)^oo(2n+2)/((2n+1)!
We only need 6 steps for 9 decimal place accuracy:
e=sum_(n=0)^6(2n+2)/((2n+1)!)= 2.718281828
## Graphical Demonstration of e
The area under the curve y=1/x between 1 and e is equal to 1 unit2.
e
Area under the curve y=1/x between 1 and e.
### Reference
Brothers, H.J. 2004. Improving the convergence of Newton's series approximation for e. College Mathematics Journal 35(January):34-39..
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Matrix concept and types of matrices
Matrix concept
Whenever we place an element in rows and columns we use a matricial structure.
For example, any show in which the tickets are numbered makes use of such structures. The stalls are divided in rows and columns. If on our ticket we see Row $$23$$, Seat $$12$$, it is indicating that our seat is in the row $$23$$ and column $$12$$.
Any table of those that we use in the text editor is nothing but a matrix since it is organized in rows and columns.
The next table has $$3$$ rows and $$4$$ columns. The number placed in row $$2$$ and column $$4$$ is zero:
$$2$$ $$1$$ $$5$$ $$8$$ $$3$$ $$2$$ $$2$$ $$0$$ $$2$$ $$1$$ $$6$$ $$4$$
So, if we want a table to become a matrix that is representative of some mathematical object, we must put a numerical value in each cell, remove the grid and enclose it all between two big brackets:
$$\left( \begin{array}{cccc} 2 & 1 & 5 & 8 \\ 3 & 2 & 2 & 0 \\ 2 & 1 & 6 & 4 \end{array} \right)$$$And we already have one of those matrices usually used in mathematics. Types of matrices A matrix with the same number of rows and columns is called a square matrix. For example: $$\left( \begin{array}{ccc} 3 & 1 & 4 \\ 4 & 0 & -5 \\ 12 & 23 & 8 \end{array} \right)$$$ is square,
and a matrix like:
$$\left( \begin{array}{ccc} 3 & 2 & -1 \\ 1 & 2 & 0 \end{array} \right)$$$is not square. The first one has three rows and three columns, so it is a $$3\times3$$ matrix. It is read “three by three”. To refer to the second matrix, which has two rows and three columns, we say it is a $$2\times3$$ matrix, or a "two by three" matrix. So, in general, when we talk about an $$m \times n$$ matrix we are talking about a matrix that has $$m$$ rows and $$n$$ columns. This way of calling the matrix is only a convention and it might change from one author to another. According to this, a matrix $$m \times n$$ will be square when $$m = n$$. Notation The matrix is usually named with capital letters: $$A= \left( \begin{array}{ccc} 3 & 1 & 4 \\ 4 & 0 & -5 \\ 12 & 2 & 8 \end{array} \right)$$$
$$B= \left( \begin{array}{ccc} 3 & 2 & -1 \\ 1 & 2 & 0 \end{array} \right)$$$Letters are also used to refer to the elements or entries that form the matrix: $$\left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right)$$$
It is to be understood that when we work with a big matrix, for example $$100\times200$$, that is, with $$100$$ rows and $$200$$ columns (or more), the use of letters of the alphabet is not practical. For that reason we can use a notation of type $$a_{ij}$$ in which $$i$$ represents the row and $$j$$ represents the column where the element is.
For instance, in the matrix:
$$\left( \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{array} \right) = \left( \begin{array}{ccc} 2 & -1 & 0 \\ 3 & 1 & 8 \end{array} \right)$$$the element $$a_{22}=1$$ and the element $$a_{13}=0$$. Verify the value of the following elements in the matrix: $$a_{31}=3, \ a_{25}=4, \ a_{27}=-1, \ a_{45}=8$$$
$$\left( \begin{array}{ccccccc} 2 & 4 & 1 & 8 & 5 & 3 & 8 \\ 3 & 6 & 8 & 2 & 4 & 0 & -1 \\ 3 & 5 & 7 & 1 & -8 & 0 & 3 \\ 2 & 5 & 7 & 3 & 8 & 1 & 8 \\ \end{array} \right)$$$The zero matrix is defined as the one with all its elements being $$0$$, regardless of the number of rows and columns that it has. Two matrices are said to be equal when all their elements are equal. $$\left( \begin{array}{ccccc} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right)$$$
is a null matrix or zero matrix.
$$\left( \begin{array}{cc} 1 & 3 \\ 2 & 6 \end{array} \right) = \left( \begin{array}{cc} 1 & 3 \\ 2 & 6 \end{array} \right)$$\$
are equal matrix.
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# What can you multiply to get 380?
## What can you multiply to get 380?
Factor Pairs of 380
• 1 x 380 = 380.
• 2 x 190 = 380.
• 4 x 95 = 380.
• 5 x 76 = 380.
• 10 x 38 = 380.
• 19 x 20 = 380.
## What two numbers make 375?
The pair of numbers that give 375 when multiplied with each other is called the factor pairs of 375.
• 375 = 1 × 375.
• 375 = 3 × 125.
• 375 = 5 × 75.
• 375 = 15 × 25.
• Therefore, the factor pairs of 375 are (1, 375), (3, 125), (5, 75), and (15, 25).
• Since the product of two negative numbers is positive, i.e. (-) × (-) = (+).
What 2 numbers multiplied equal 360?
Factor Pairs of 360
• 1 x 360 = 360.
• 2 x 180 = 360.
• 3 x 120 = 360.
• 4 x 90 = 360.
• 5 x 72 = 360.
• 6 x 60 = 360.
• 8 x 45 = 360.
• 9 x 40 = 360.
### What two numbers are equal to 42?
Factors of 42 in Pairs
• 1 × 42 = 42.
• 2 × 21 = 42.
• 3 × 14 = 42.
• 6 × 7 = 42.
### What is the factors of 380?
Factors of 380
• Factors of 380: 1, 2, 4, 5, 10, 19, 20, 38, 76, 95, 190 and 380.
• Prime Factorization of 380: 2 × 2 × 5 × 19.
What’s the factor of 56?
Also, we know that 1 is a factor of every number. Thus, The factors of 56 by prime factorization are 1, 2, 4, 7, 8, 14, 28, and 56.
#### Whats the HCF of 375 and 150?
The GCF of 150 and 375 is 75.
#### What is the LCM of 8 and 6?
24
Answer: LCM of 6 and 8 is 24.
What are the factors of 70?
Factors of 70
• Factors of 70: 1, 2, 5, 7, 10, 14, 35 and 70.
• Negative Factors of 70: -1, -2, -5, -7, -10, -14, -35 and -70.
• Prime Factors of 70: 2, 5, 7.
• Prime Factorization of 70: 2 × 5 × 7 = 2 × 5 × 7.
• Sum of Factors of 70: 144.
## What are the factors of 180 in pairs?
The numbers which we multiply to get 180 are the factors of 180. Factors of 180 are 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, and 180. Factor pairs of 180 are (1,180) (2, 90) (3, 60) (4,45) (5, 36) (6, 30) (9, 20) (10, 18 ) and (12, 15).
## What numbers multiply to 42 and add to 13?
The numbers are 6 and 7 .
What can go into 48?
The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 and its negative factors are -1, -2, -3, -4, -6, -8, -12, -16, -24, -48.
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# Algebra II : Solving Non Quadratic Polynomials
## Example Questions
### Example Question #1 : How To Find A Solution Set
Give all real solutions of the following equation:
Explanation:
By substituting - and, subsequently, this can be rewritten as a quadratic equation, and solved as such:
We are looking to factor the quadratic expression as , replacing the two question marks with integers with product and sum 5; these integers are .
Substitute back:
The first factor cannot be factored further. The second factor, however, can itself be factored as the difference of squares:
Set each factor to zero and solve:
Since no real number squared is equal to a negative number, no real solution presents itself here.
The solution set is .
### Example Question #1 : How To Factor An Equation
Which of the following displays the full real-number solution set for in the equation above?
Explanation:
Rewriting the equation as , we can see there are four terms we are working with, so factor by grouping is an appropriate method. Between the first two terms, the Greatest Common Factor (GCF) is and between the third and fourth terms, the GCF is 4. Thus, we obtain . Setting each factor equal to zero, and solving for , we obtain from the first factor and from the second factor. Since the square of any real number cannot be negative, we will disregard the second solution and only accept
### Example Question #1 : Solving Non Quadratic Polynomials
Factor by grouping.
Explanation:
The first step is to determine if all of the terms have a greatest common factor (GCF). Since a GCF does not exist, we can move onto the next step.
Create smaller groups within the expression. This is typically done by grouping the first two terms and the last two terms.
Factor out the GCF from each group:
At this point, you can see that the terms inside the parentheses are identical, which means you are on the right track!
Since there is a GCF of (5x+1), we can rewrite the expression like this:
### Example Question #2 : Solving Non Quadratic Polynomials
Factor completely:
The polynomial is prime.
Explanation:
This can be most easily solved by setting and, subsequently, . This changes the degree-4 polynomial in to one that is quadratic in , which can be solved as follows:
The quadratic factors do not fit any factoring pattern and are prime, so this is as far as the polynomial can be factored.
### Example Question #3 : Solving Non Quadratic Polynomials
If , and , what is ?
Explanation:
To find , we must start inwards and work our way outwards, i.e. starting with :
We can now use this value to find as follows:
### Example Question #4 : Solving Non Quadratic Polynomials
Factor:
Explanation:
Using the difference of cubes formula:
Find x and y:
Plug into the formula:
Which Gives:
And cannot be factored more so the above is your final answer.
### Example Question #5 : Solving Non Quadratic Polynomials
Factor
Explanation:
First, we can factor a from both terms:
Now we can make a clever substitution. If we make the function now looks like:
This makes it much easier to see how we can factor (difference of squares):
The last thing we need to do is substitute back in for , but we first need to solve for by taking the square root of each side of our substitution:
Substituting back in gives us a result of:
### Example Question #1 : Solving Non Quadratic Polynomials
Simplify:
Explanation:
Use the difference of squares to factor out the numerator.
The term is prime, but can still be factored by another difference of squares.
Replace the fraction.
Simplify the top and bottom.
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# Worksheet on Finding Percentage | Finding Percentage Worksheets with Solutions
Here, in this Percentage worksheet, you will know about how to find the Percentage of the given values. This helps to practice for exams or any other competitive tests on the concept of finding Percent. Worksheet on Finding Percentage includes different types of questions with solutions. Let’s see some solved examples on finding percentage in the below. We have provided Step by Step Solutions for all the Problems in Percentage Worksheets so that it would be easy for you to understand the Concept of Percentages.
1. Find the percent of the following:
(i) 5 liters is 100 ml.
(ii) 600 ml is 3000 ml.
(iii) 15 kg is 450 g.
Solution:
(i) 5 liters is 100 ml.
The given value is 100 ml of 5 liters,
we know that 1 liter= 1000 ml, so we will convert liter to ml,
5 liters= 5×1000
= 5,000 ml,
so, to find the percent
100/5,000 × 100=
on solving, we will get the result as
= 2%.
(ii) 600 ml is 3000 ml.
The given value is 3000 ml of 600 ml, which means
600/3000 × 100=
on solving, we will get the result as
= 20%.
(iii) 15 kg is 450 g.
The given value 450g of 15 kg,
as we know 1 kg= 1000 g, so we will convert kg to g,
15kg= 15×1000 = 15,000 g
450/15,000 × 100=
on solving, we will get the result as
= 3%.
2. Figure out the value of X for the following below?
(i) 25% of X is 1000
(ii) 45% of X is 90 cm
(iii) 3.25% of X is 150 kg
Solution:
(i) 25% of X is 1000
We should find the value of X, so
Let the number be X,
As given 25/100 × X = 1000
1000/25 ×100=
on solving, we will get 4000
So the value of X is 4000.
(ii) 45% of X is 90 cm
We should find the value of X, so
Let the number be X,
As given
45/100 × X = 90
90/45 ×100=
on solving, we will get 200
So the value of X is 200 cm.
(iii) 3.25% of X is 150 kg
We should find the value of X, so
Let the number X,
As given
3.25/100 × X = 150
150/3.25 ×100=
on solving, we will get 4,615.38
So the value of X is 4,615.38 kg.
3. What will be the number if 25% of the number is 4,500?
Solution:
Let the number X,
As given 25% of the number is 4,500, which means
25/100 × X = 4,500
4,500/25 ×100=
on solving, we will get 18,000
So the number is 18,000.
4. Tom’s total compensation, in the wake of paying personal duty at 15 % is \$86,240. Find his gross pay?
Solution:
Let the Tom’s gross income be X,
then (100-15)% of X = \$86,240,
which means,
85% of X = \$86,240,
on solving, we will get \$101,458.82.
5. Mike secures 20 % of the maximum marks to pass. If Mike gets 20 marks and fails by 20 marks, then find the maximum marks.
Solution:
Let the Mike marks be X,
As he got 20 marks and failed 20 marks,
the pass marks are 20+20,
which is 40,
so to find the maximum marks,
20% of X= 40,
on solving, we will get 80.
So, the maximum mark is 80.
6. Tony bought 10 dozens of bananas from the fruit market, out of that 5% was spoiled. How many number of bananas are in a stable condition?
Solution:
The total number of bananas Tony bought is 10 dozens,
As we know 1 dozen= 12,
so 10 dozens= 120 bananas,
in that 25% of bananas are spoiled, which means
25/100 × 120=
on solving, we will get 30.
So, the number of bananas in a stable condition is
120-30= 90
Therefore, 90 bananas are in stable condition.
7. The total number of students in the class is 90 and in that 55 are boy students. What will be the percentage of girl students?
Solution:
The total number of students in the class is 90 students,
The total number of boy students is 55 students,
So to find the percent of boy students, first we need to find out the number of boy students,
which is 90-55= 35 students.
and the percentage of the boy students is=
35/90 × 100 =
on solving, we will 38.89%
So the percentage of boys is 38.89%.
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# How to Find the Equation of a Tangent Line
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Last updated on September 25th, 2020
The tangent line is a straight line with that slope, passing through that exact point on the graph. To find the equation for the tangent, you’ll need to know how to take the derivative of the original equation. Here is how to find the equation of a tangent line.
• To find the equation of a line you need a point and a slope.
• The slope of the tangent line is the value of the derivative at the point of tangency.
## Steps to find the equation of a tangent line
1. Find the first derivative of f(x).
Suppose $\dpi{120}&space;\large&space;f(x)&space;=&space;x^{3}$. Find the equation of the tangent line at the point where x = 2.
Recall the power rule when taking derivatives: $\dpi{120}&space;\large&space;\frac{d}{dx}x^{n}&space;=&space;nx^{n-1}$ .
The function’s first derivative = $\dpi{120}&space;\large&space;{f}'(x)&space;=&space;3x^{2}$
2. The plug x value of the indicated point into f ‘(x) to find the slope at x.
The slope of the tangent line is m = 12
3. Plug x value into f(x) to find the y coordinate of the tangent point.
$\dpi{120}&space;\large&space;f(2)&space;=&space;2^{3}&space;=&space;8$ The point is (2, 8).
4. Combine the slope from step 2 and point from step 3 using the point-slope formula to find the equation for the tangent line.
5. Graph your results to see if they are reasonable. Here is the graph of the function and the tangent line we just found.
## How to Find the equation of tangent lines to Implicit Curves
1. Find the y-value of the point of tangency.
Suppose $\dpi{120}&space;\large&space;x^{2}+y^{2}&space;=&space;16$Find the equation of the tangent line at x = 2x=2 for y > 0y>0.
$\dpi{120}&space;\large&space;\\x^{2}&space;+&space;y^{2}&space;=&space;16&space;\\2^{2}&space;+&space;y^{2}&space;=&space;16&space;\\&space;4\&space;+&space;y^{2}&space;=&space;16&space;\\y^{2}&space;=&space;16&space;-&space;4&space;\\y^{2}&space;=&space;12&space;\\y\&space;=&space;\pm&space;\sqrt{12}&space;\\y\&space;=&space;\pm&space;\sqrt{4.3}&space;\\y\&space;=&space;\pm&space;2\sqrt{3}$
$\dpi{120}&space;\large&space;y&space;=&space;2\sqrt{3}$ since y > 0. The point of tangency is $\dpi{120}&space;\large&space;(2,&space;\sqrt{3})$
2. Find the equation for $\dpi{120}&space;\large&space;\frac{dx}{dy}$.
The equation is implicitly defined, we use implicit differentiation.
$\dpi{120}&space;\large&space;\\2x&space;+&space;2y&space;\frac{dx}{dy}&space;=&space;0&space;\\2y&space;\frac{dx}{dy}&space;=&space;-2x&space;\\\frac{dx}{dy}&space;=&space;-\frac{2x}{2y}&space;\\\frac{dx}{dy}&space;=&space;-\frac{x}{y}$
3. Find the slope of the tangent line at the point of tangency.
$\frac{\mathrm{d}&space;x}{\mathrm{d}&space;y}|(2,{\color{Red}&space;2\sqrt{3}})&space;=&space;-&space;\frac{2}{{\color{Red}&space;2\sqrt{3}}}&space;=&space;-\frac{1}{\sqrt{3}}&space;=&space;-\frac{1}{\sqrt{3}}.\frac{\sqrt{3}}{\sqrt{3}}&space;=&space;-\frac{\sqrt{3}}{3}$
4. Find the equation of the tangent line. Point $\dpi{120}&space;\large&space;(2,&space;\sqrt{3})$ and
slop m = $-\frac{\sqrt{3}}{3}$
$\\y&space;-&space;y_{1}\&space;\&space;\&space;=&space;x&space;-&space;x_{1}&space;\\y&space;-&space;2\sqrt{3}&space;=&space;-\frac{\sqrt{3}}{3}(x&space;-&space;2)$
The equation of the tangent line is $y&space;-&space;2\sqrt{3}&space;=&space;-\frac{\sqrt{3}}{3}(x&space;-&space;2)$
For reference, the graph of the curve and the tangent line we found is shown below.
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# FORM FOUR MATHEMATICS NOTES TOPIC 2:AREA AND PERIMETER
AREA AND PERIMETER
Area of any Triangle
The Formula for the Area of any Triangle
Derive the formula for the area of any triangle
Area of triangle is given by½bh, whereby b is the base of the triangle and h is the height of the given triangle. Consider the illustrations below:
From the figure above, we see where the base and height are located.
Applying the Formula to find the Area of any Triangle
Apply the formula to find the area of any triangle
Example 1
The base of a triangle is 12cm long. If the corresponding height is 7cm, find the area of the triangle.
Solution
Consider the figure below:
The area of a triangle is given by½bh.
Area =½×12cm×7cm
Area = 42cm2
Therefore area of a triangle is 42cm2
Example 2
The lengths of two sides of a triangle are 6cm and 8cm. Find the area of a triangle if the included is
Solution:
Consider the triangle below, name it triangle ABC.
The area of a triangle above is given by½b×h
So, Area = ½× 8cm× 6cm × sin 45°
=24cm2× sin45°
= 16.97cm2
Therefore the area of ABC = 16.97cm2
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# Mathematics | Predicates and Quantifiers | Set 1
Prerequisite : Introduction to Propositional Logic
Introduction
Consider the following example. We need to convert the following sentence into a mathematical statement using propositional logic only.
"Every person who is 18 years or older, is eligible to vote."
The above statement cannot be adequately expressed using only propositional logic. The problem in trying to do so is that propositional logic is not expressive enough to deal with quantified variables. It would have been easier if the statement were referring to a specific person. But since it is not the case and the statement applies to all people who are 18 years or older, we are stuck.
Therefore we need a more powerful type of logic.
Predicate Logic
Predicate logic is an extension of Propositional logic. It adds the concept of predicates and quantifiers to better capture the meaning of statements that cannot be adequately expressed by propositional logic.
What is a predicate?
Consider the statement, “ is greater than 3″. It has two parts. The first part, the variable , is the subject of the statement. The second part, “is greater than 3”, is the predicate. It refers to a property that the subject of the statement can have.
The statement “ is greater than 3″ can be denoted by where denotes the predicate “is greater than 3” and is the variable.
The predicate can be considered as a function. It tells the truth value of the statement at . Once a value has been assigned to the variable , the statement becomes a proposition and has a truth or false(tf) value.
In general, a statement involving n variables can be denoted by . Here is also referred to as n-place predicate or a n-ary predicate.
• Example 1: Let denote the statement “ > 10″. What are the truth values of and ?
Solution: is equivalent to the statement 11 > 10, which is True.
is equivalent to the statement 5 > 10, which is False.
• Example 2: Let denote the statement ““. What is the truth value of the propositions and ?
Solution: is the statement 1 = 3 + 1, which is False.
is the statement 2 = 1 + 1, which is True.
What are quantifiers?
In predicate logic, predicates are used alongside quantifiers to express the extent to which a predicate is true over a range of elements. Using quantifiers to create such propositions is called quantification.
There are two types of quantification-
1. Universal Quantification- Mathematical statements sometimes assert that a property is true for all the values of a variable in a particular domain, called the domain of discourse. Such a statement is expressed using universal quantification.
The universal quantification of for a particular domain is the proposition that asserts that is true for all values of in this domain. The domain is very important here since it decides the possible values of . The meaning of the universal quantification of changes when the domain is changed. The domain must be specified when a universal quantification is used, as without it, it has no meaning.
Formally,
The universal quantification of is the statement
" for all values of in the domain"
The notation denotes the universal quantification of .
Here is called the universal quantifier.
is read as "for all ".
• Example 1: Let be the statement “ > “. What is the truth value of the statement ?
Solution: As is greater than for any real number, so for all or .
2. Existential Quantification- Some mathematical statements assert that there is an element with a certain property. Such statements are expressed by existential quantification. Existential quantification can be used to form a proposition that is true if and only if is true for at least one value of in the domain.
Formally,
The existential quantification of is the statement
"There exists an element in the domain such that "
The notation denotes the existential quantification of .
Here is called the existential quantifier.
is read as "There is atleast one such such that ".
• Example : Let be the statement “ > 5″. What is the truth value of the statement ?
Solution: is true for all real numbers greater than 5 and false for all real numbers less than 5. So .
To summarise,
Now if we try to convert the statement, given in the beginning of this article, into a mathematical statement using predicate logic, we would get something like-
Here, P(x) is the statement "x is 18 years or older and,
Q(x) is the statement "x is eligible to vote".
Notice that the given statement is not mentioned as a biconditional and yet we used one. This is because Natural language is ambiguous sometimes, and we made an assumption. This assumption was made since it is true that a person can vote if and only if he/she is 18 years or older. Refer Introduction to Propositional Logic for more explanation.
Other Quantifiers –
Although the universal and existential quantifiers are the most important in Mathematics and Computer Science, they are not the only ones. In Fact, there is no limitation on the number of different quantifiers that can be defined, such as “exactly two”, “there are no more than three”, “there are at least 10”, and so on.
Of all the other possible quantifiers, the one that is seen most often is the uniqueness quantifier, denoted by .
The notation states "There exists a unique such that is true".
Quantifiers with restricted domains
As we know that quantifiers are meaningless if the variables they bind do not have a domain. The following abbreviated notation is used to restrict the domain of the variables-
> 0, > 0.
The above statement restricts the domain of , and is a shorthand for writing another proposition, that says , in the statement.
If we try to rewrite this statement using an implication, we would get-
> >
Similarly, a statement using Existential quantifier can be restated using conjunction between the domain restricting proposition and the actual predicate.
1. Restriction of universal quantification is the same as the universal quantification of a conditional statement.
2. Restriction of an existential quantification is the same as the existential quantification of conjunction.
Definitions to Note:
1. Binding variables- A variable whose occurrence is bound by a quantifier is called
a bound variable. Variables not bound by any quantifiers are called free variables.
2. Scope- The part of the logical expression to which a quantifier is applied is called
the scope of the quantifier.
This topic has been covered in two parts. The second part of this topic is explained in another article – Predicates and Quantifiers – Set 2
This article is contributed by Chirag Manwani. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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# Probability | Lesson
## What is a probability?
A probability indicates the chance that an
will happen.
A probability can be any number from $0$ to $1$.
• A probability of $0$ or $0\mathrm{%}$ means that an event will never happen.
• A probability of $\frac{1}{2}$ or $50\mathrm{%}$ means that an event is equally likely to happen or not happen.
• A probability of $1$ or $100\mathrm{%}$ means that an event will certainly happen.
The probability that event $A$ will happen is written as $P\left(A\right)$.
The probability that event $A$ will not happen, , is equal to $1-P\left(A\right)$.
### What skills are tested?
• Calculating the probability that an event will happen (or will not happen) based on a description of the situation
• Calculating the probability that an event will happen (or will not happen) based on information in a data display
• Calculating a count or quantity from a probability
• Comparing probabilities or drawing conclusions based on probabilities
## How is the probability of an event calculated?
An event could be the outcome of any random process such as the toss of a fair coin, the roll of a fair number cube, or the random selection of an item from a group.
We use the notation $P\left(A\right)$ to represent "the probability that event $A$ will happen".
## How is the probability of an event calculated using information in a data display?
When calculating the probability that a randomly selected item will have certain attribute(s), the information we need may be in a data display.
In such cases,
• The ways $A$ can happen is the number of items of type $A$.
• The number of possible outcomes is the total number in the group.
## How can we calculate a count or quantity from a probability?
Sometimes we are given the probability and are asked to calculate either the number of possible outcomes or the number of ways $A$ can happen
To solve these questions:
1. Substitute known values into the equation .
2. Solve for the unknown quantity.
TRY: CALCULATING A PROBABILTY
A box contains $5$ red, $10$ green, and $20$ orange candies. If one candy is selected randomly from the box, what is the probability that the selected candy will be red?
TRY: CALCULATING A QUANTITY FROM A PROBABILITY
A gumball machine contains red, blue, and green gumballs. There are $70$ red gumballs and $50$ blue gumballs. When the handle of the machine is turned, one gumball is randomly dispensed. If the probability of getting a red gumball is $\frac{1}{3}$, how many green gumballs are in the machine?
TRY: COMPARING PROBABILITIES
Doughnut VarietyNumber of Doughnuts
Cream filled$16$
Plain$22$
Jelly filled$12$
The table above shows the number of doughnuts in a bakery display case by variety. If a doughnut is chosen at random from the display case, which of the following statements is true?
TRY: CALCULATING A PROBABILITY USING A TABLE
ManWomanChildTotal
First class$175$$144$$6$$325$
Second class$168$$93$$24$$285$
Third class$462$$165$$79$$706$
Total$805$$402$$109$$1,316$
The table above shows data about the passengers aboard the Titanic. If a passenger who traveled on the Titanic is selected at random, what is the probability that the selected passenger was someone who traveled in first class?
$\mathrm{%}$
## Things to remember
A probability indicates the chance that an event will happen.
A probability can be any number between $0$ and $1$.
The probability that event A will happen is:
The probability that event A will not happen is:
## Want to join the conversation?
• how do you figure out the probability of coins
(1 vote)
• You take one side of the coin and see how many total sides there are. one side=1 both sides=2 so, 1/2. Or 50% odds of it landing on one of the sides.
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Dimension of a Matrix – Explanation & Examples
Matrices are a rectangular arrangement of numbers in rows and columns. They are sometimes referred to as arrays. The dimensions of a matrix are basically its name. Knowing the dimension of a matrix allows us to do basic operations on them such as addition, subtraction and multiplication. Let’s start with the definition of the dimension of a matrix:
The dimension of a matrix is its number of rows and columns.
This article will talk about the dimension of a matrix, how to find the dimension of a matrix, and review some examples of dimensions of a matrix. If you want to know more about matrix, please take a look at this article.
What is the dimension of a matrix?
The dimension of a matrix is the number of rows and the number of columns of a matrix, in that order. Consider the matrix shown below:
It has $2$ rows (horizontal) and $2$ columns (vertical). The dimension of this matrix is $2 \times 2$. The first number is the number of rows and the next number is the number of columns. It has to be in that order. We pronounce it as a “2 by 2 matrix”. The $\times$ sign is pronounced as “by”.
The entries, $2, 3, -1$ and $0$, are known as the elements of a matrix.
In general, if we have a matrix with $m$ rows and $n$ columns, we name it $m \times n$, or rows x columns. The convention of rows first and columns second must be followed. This is the dimension of a matrix. You can remember the naming of a matrix using a quick mnemonic.
Remember, RC. Rows first, then columns.
How to find the dimension of a matrix?
To find the dimension of a given matrix, we count the number of rows it has. Then, we count the number of columns it has. We put the numbers in that order with a $\times$ sign in between them. Let’s take an example.
How many rows and columns does the matrix below have?
Checking horizontally, there are $3$ rows. Checking vertically, there are $2$ columns. Thus, we have found the dimension of this matrix. It is a $3 \times 2$ matrix.
This can be a bit tricky. But if you always focus on counting only rows first and then only columns, you won’t encounter any problem. We see there are only $1$ row (horizontal) and $2$ columns (vertical). Thus, this matrix will have a dimension of $1 \times 2$.
Let us look at some examples to enhance our understanding of the dimensions of matrices.
Example 1
What is the dimension of the matrix shown below?
$\begin{pmatrix} 1 & { 0 } & 1 \\ 1 & 1 & 1 \\ 4 & 3 & 2 \end{pmatrix}$
Solution
Recall that the dimension of a matrix is the number of rows and the number of columns a matrix has, in that order. Always remember to think horizontally first (to get the number of rows) and then think vertically (to get the number of columns).
Looking at the matrix above, we can see that is has $3$ rows and $3$ columns. Therefore, the dimension of this matrix is $3 \times 3$.
Let us look at another example.
Example 2
What is the dimension of the matrix shown below?
$\begin{pmatrix} a \\ b \\ c \end{pmatrix}$
Solution
This is a small matrix. You should be careful when finding the dimensions of these types of matrices. Check horizontally, you will see that there are $3$ rows. Check vertically, there is only $1$ column. From the convention of writing the dimension of a matrix as rows x columns, we can say that this matrix is a $3 \times 1$ matrix.
Please note that the elements of a matrix, whether they are numbers or variables (letters), does not affect the dimensions of a matrix. The dimension only depends on the number of rows and the number of columns. You can have number or letter as the elements in a matrix based on your need.
We now see a tricky problem.
Example 3
What is the dimension of the matrix shown below?
$\begin{bmatrix} { 5 } \end{bmatrix}$
Solution
At first glance, it looks like just a number inside a parenthesis. Well, this can be a matrix as well. We have a single entry in this matrix. The number of rows and columns are both one. Thus, this is a $1 \times 1$ matrix.
Practice Questions
1. What are the individual entries in a matrix called?
2. True or False
A matrix has $5$ rows and $2$ columns. The dimension of the matrix is $2 \times 5$.
3. What is the dimension of this matrix?
$\begin{bmatrix} a & b & c \\ f & e & d \end{bmatrix}$
4. Does the matrix shown below have a dimension of $1 \times 5$?
$\begin{pmatrix} 22 \\ 3 \\ { – 2 } \\ 5 \\ 1 \end{pmatrix}$
2. When naming a matrix, i.e. the dimension of a matrix, we always put the number of rows first. Then a $\times$ sign, followed by the number of columns. Since there are $5$ rows and $2$ columns, the dimension of the matrix should be $5 \times 2$. Hence, the statement is False.
3. If there are m rows and columns of a matrix, the dimension of that matrix is $m \times n$. From the matrix shown, we see that there are $2$ rows and $3$ columns. Thus, the dimension of this matrix is $2 \times 3$.
4. If there are m rows and columns of a matrix, the dimension of that matrix is $m \times n$. Looking at the matrix, we can see that it has $5$ rows and $1$ column. Hence, its dimension is $5 \times 1$. So, NO, the matrix DOES NOT have dimension of $1 \times 5$.
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# What Is The Value Of X In The Equation X – Y = 30, When Y = 15?
• Puzzle
X-rays are a basic medical imaging technology. They allow practitioners to inspect internal organs and body systems. X-rays can also be viewed as the little brother of ultrasounds.
X-rays are made by exposing a piece of paper or other material to an X-ray tube. The process is very careful, however, so most times it is not seen.
X-rays are a critical part of medicine, because they can help diagnose conditions and explain what things look like. They also have value as Gifts for Medical Professionals, since they can help them recognize conditions and apply treatment tips and strategies.
## Find y when x = -5
When x = -5, y = -15 is a very rare situation. Most times, when the value of an item is noted, it is as a fraction of 15.
However, this does not mean that an exception cannot be made for this value. It can!
When the item being valued is a member of the family known as the minus 5s, or if the item has some other exception to valuing such as specific rarity or historical significance. In these cases, there may be a lower value required to negate the value of the minus 5s on the piece.
For example, if a pendant has aminus 5s on it, then it would need to be worth less than its equivalent standard pendant with a plus sign. This would prevent someone thinking that they were getting a plus (+/-) five piece when in fact they were just getting one that was minus (-).
## Find z when x = 15 and y = 10
If you know the value of an item between y = 10 and y = 15, you can use this information to find the value of x in the equation X – Y = 30.
For example, if the value of an apple is 6, you can figure out that a third of an apple has a value of 4, so using your knowledge of an apple, you can determine that a third of an apple has a value of 4 – 15 = 2.
Using your knowledge of an apple, we can say that a third of an Apple has a Value Of 2. We could then solve for X using our new variable: 3-2=1 which is the Value Of 1 in the equation X – Y = 30.
When looking up values of x and y in the equation X – Y = 30, remember to take into account their values.
## Plug all numbers into the equation to solve for x
If the value of an item is listed as a number between zero and one, then it can be difficult to determine whether or not the value listed is the actual value of that item, or if another value must be added to get the full value.
In the equation X – Y = 30, when Y = 15, this must be considered. If the item being valued is worth more than fifteen dollars, then there should be one and only one dollar spent on it. If it was worth less than fifteen dollars, then there would be more spent on it to compensate for its cost.
This same principle applies to values in investments.
.
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# Confidence Intervals - PowerPoint PPT Presentation
Confidence Intervals. Whenever we make a confidence interval we should follow these steps to be sure that we include all parts:. State the type of interval (our first intervals are Z -intervals). Meet assumptions (explain how each is met).
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## PowerPoint Slideshow about 'Confidence Intervals' - camila
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### Confidence Intervals
Whenever we make a confidence interval we should follow these steps to be sure that we include all parts:
• State the type of interval (our first intervals are Z-intervals)
• Meet assumptions (explain how each is met)
• SRS: Our sample must be a random selection of the population.
• Normality: The sampling distribution must be approximately normal.
• : The population standard deviation must be known.
• Write the formula. Substitute values into the formula and give the result.
• Communicate the meaning of the confidence interval in terms of the original problem.
Step 1: these steps to be sure that we include all parts:
Step 2:
Step 3:
Step 4:
So, by topic:
Type of interval
Assumptions
Formula and calculations
Conclusion
It is a significant amount of writing to include these steps in finding a confidence interval.
The reason for doing this is to give you a framework to follow so that each problem follows a set format.
It will be easier in the long run to be used to following a set format.
A major complaint by those who grade AP Statistics exams is that AP students sometimes fail to meet assumptions or explain what a confidence interval means.
Step 1: in finding a confidence interval.
Step 2:
Problem: For the year 2000, SAT Math scores had a mean of 514 and a standard deviation of 113. SAT Math scores follow a normal distribution. A random sample of students at Horsehead High had the following SAT Math scores: {550, 480, 510, 460, 600, 570}. Find a 95% confidence interval for the SAT Math scores at Horsehead High.
Z-interval for means
Assumptions:
• We are given an SRS.
• The population is stated to be normal.
• σ is known.
Step 3: in finding a confidence interval.
Step 4:
or
We are 95% confident that the true mean SAT math score at Horsehead High lies between 437.9 and 618.8.
If asked what 95 % confidence means, in finding a confidence interval.
If we select many random samples of 6 and compute confidence intervals this way, 95% of the time we will capture the true mean.
### Assumptions for in finding a confidence interval. Z Confidence Intervals and Tests of Significance
Whenever we make a confidence interval or test of significance we must be certain that we meet theoretical assumptions before we may make the actual interval or test.
Our tests and intervals cannot be applied to all circumstances.
Fortunately, they do work in many circumstances, and we will be able to verify that this is so.
The assumptions are the same for confidence intervals and tests of significance, so we have only one set to learn.
What if we don significance we must be certain that we meet theoretical assumptions before we may make the actual interval or test.’t meet the assumptions?
If we can tell from the problem statement or context that an assumption is not met, we must state so, and our results will be suspect.
Due to the educational nature of this class, I will ask you to go ahead and work the problem, anyway, even if the assumptions are not met (but you must state that they are not met).
In the real world, we would have to meet the assumption before proceeding, but because we are unable to redo an experiment, for example, we will use the information we have.
1 significance we must be certain that we meet theoretical assumptions before we may make the actual interval or test.st Assumption:
The first assumption is that our sample is a simple random sample, usually abbreviated SRS.
This information is usually provided in the problem.
If so, we can simply state that SRS is given.
Sometimes we will have no information about how the sample was made, and in that circumstance we can write that we are uncertain that we have an SRS.
If we can tell that our sample is not an SRS, we must state that. It may mean that our results are not valid.
The simple random sample is so important because it avoids bias that may be the result of selection.
Our sample should be representative of the population, otherwise we may draw conclusions about a group different from the one we wanted.
Recall that randomization was an important principle of experimental design, and this is why! Randomization guarantees random samples.
2 that. It may mean that our results are not valid.nd Assumption:
The second assumption is that our sampling distribution is normally distributed.
This assumption is met whenever our population is normally distributed.
This information may be provided in the problem, and if so, we simply write that it is given that the population is normal.
A principle that often helps us here, is the Central Limit Theorem.
As sample size becomes large the sampling distribution approaches the normal distribution, even if the original population is not normal.
When we have large samples, we invoke the CLT.
Otherwise, we must examine the data provided in an effort to see whether or not it is reasonable to expect the sampling distribution to be normal.
We will spend more time on exactly what to do here later in the course.
3 see whether or not it is reasonable to expect the sampling distribution to be normal.rd Assumption:
Our third and final assumption is that σis known.
This must be provided to you in the problem, otherwise we cannot use the Z-interval or test.
In Module 11, we will learn what to do when σis unknown.
In summary: see whether or not it is reasonable to expect the sampling distribution to be normal.
Whenever we perform statistical inference using a Z test or interval for sample means, we need:
• SRS
• Normal distribution of sampling distribution
• σ(the population standard deviation)
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# Riemann Sums
By Murray Bourne, 14 Jun 2010
Integration is a process in mathematics that can tell us:
• The area of a curved 2-D object (the sides aren't straight, and there is no simple formula)
• The volume of a curved 3-D object (once again, the sides aren't straight)
• The velocity of an object if we know its acceleration at time t (which means the acceleration changes all the time, as does its velocity)
• The displacement of an object if we know its velocity at time t (the velocity and displacement change over time, so there is no simple formula)
• The pressure on an object deep under water (the pressure varies as we go down)
In integration we spend a lot of time talking about areas under curves because such areas can be used to represent any of the quantities given above.
Most of the time in math class we use integration to find the areas under curves. We are told the area under some curve f(x) can be found by finding the definite integral between certain lower and upper limits, a and b as follows:
$\text{Area}=\int_{a}^{b}f(x)dx$
However, it turns out that is not always possible to find such an integral.
## Our Problem
I need to find the area under the following curve, between x = 3.1 and x = 6.
$y=|0.3x^3{\sin{x}}|$
The required area is shaded green in the following graph.
So we write the following integral to represent the required area:
$\text{Area}=\int_{3.1}^{6}|0.3x^3{\sin{x}}|dx$
However, it is not possible to find the above integral using normal algebraic integration.
What to do? Well, we need to find the result numerically, just like mathematicians had to do for all integrations before Isaac Newton and Gottfried Leibniz developed calculus in the 17th century.
One way of doing this is to draw a grid over the shaded area and then count the little squares:
However, you can imagine this would get very tedious very quickly, especially if you had to find many such areas. And remember, if you want a more accurate approximation, you have to use smaller squares and that involves more counting.
There is a better way.
[Please note that what we are about to do is the way computers find integrals numerically.]
## Riemann Sums
Riemann Sums give us a systematic way to find the area of a curved surface when we know the mathematical function for that curve. They are named after the mathematician Bernhard Riemann (pronounced "ree-man", since in German "ie" is pronounced "ee").
The following interactive javascript graph allows you to investigate the area under a simple parabola (y = x2). The actual area is also displayed for comparison.
In the following graph, you can change:
• The type of Riemann Sum at the top of the graph,
• The number of rectangles (or trapezoids) by dragging the slider,
• The start and end points of the graph, by dragging the sliders.
## Simple Example
Update (May 2014): This applet has been updated, improved and moved, and you can now find it here: Riemann Sums.
## Explanation
What you were doing above was as follows.
We can form equal width rectangles between the start and end points of the area we need to find.
If we add up the areas of the rectangles, we will have a good approximation for the area.
We can either place the rectangles so the curve is on the left or right as follows. Depending on the curve one or the other gives a reasonable approximation.
You can see in the above examples, the "left" approximation will be too small (the sum of the rectangle areas is less than the area below the curve), while the "right" one will be too large.
Another, and better option is to place the rectangles so the curve passes through the mid-points of each rectangle, as follows:
Another option is to create trapezoids. When the number of trapezoids is very high, they cover nearly all of the area under the curve.
The final 2 options we can use are lower (where the rectangles are completely below the curve) and upper (where the rectangles are completely above the curve).
## The Solution to Our Problem
Earlier, we wanted to find the area under the curve $y=|0.3x^3{\sin{x}}|$ between x = 3.1 and x = 6.
You can use the following interactive graph to find the answer using Riemann Sums.
Choose Riemann sum type:
//<![CDATA[
var brd2 = JXG.JSXGraph.initBoard('box2', {axis:true, boundingbox:[-3.2,40,6.2,-4],showCopyright:false});
var s2 = brd2.create('slider',[[-3,37],[0,37],[3,10,50]],{name:'n',snapWidth:1,strokeColor:'blue',fillColor:'blue'});
var a2 = brd2.create('slider',[[-3,32],[0,32],[-3,-2,Math.PI]],{name:'start',strokeColor:'blue',fillColor:'blue'});
var b2 = brd2.create('slider',[[-3,27],[0,27],[3.1416,5,6]],{name:'end',strokeColor:'blue',fillColor:'blue'});
var f2 = function(x){ return 0.3*Math.abs(Math.pow(x,3)*Math.sin(x)); };
var plot = brd2.create('functiongraph',[f2,function(){return a2.Value();}, function(){return b2.Value();}]);
var os2 = brd2.create('riemannsum',[f2,
function(){ return s2.Value();}, function(){ return document.getElementById('sumtype2').value;},
function(){return a2.Value();},
function(){return b2.Value();}
],
{fillColor:'#ffff00', fillOpacity:0.3});
brd2.create('text',
[0.5,20,function(){ return 'Sum areas = '+(JXG.Math.Numerics.riemannsum(f2,s2.Value(),document.getElementById('sumtype2').value,a2.Value(),b2.Value())).toFixed(3); }],{});
brd2.create('text',
[0.5,17,function(){ return 'Actual area = '+(JXG.Math.Numerics.riemannsum(f2,1000,'middle',a2.Value(),b2.Value())).toFixed(3); }],{});
//]]>
You can see that even with 50 steps, we don't even get 1 decimal place accuracy for this problem, not with any of the possible methods. (The answer is just over 64 square units.)
Remember 300 years ago, they had to do these calculations by hand! No wonder they were keen to develop aids for calculation, like logarithms.
## What's Missing?
1. The most accurate numerical integration method is missing from the list above. Instead of using straight lines to connect points on the graph (the trapezoid approach), we could approximate the curve by a series of parabolas (one for each division). This gives us Simpson's Rule.
2. The integral only gives us the area if the curve is completely above the x-axis (as in each of my examples above). You need to use absolute value of the integral for the parts of the curve below the x-axis. For more detail, see Area under curves.
## Summary
This article showed you an important concept in calculus. The area under a curve can represent the solution for many "real life" problems, from finding velocities to volumes.
The way it was done before calculus was developed was to use a numerical approach where we add up the areas of very thin rectangles (or trapezoids) to get a good approximation.
We get a better approximation by taking even more rectangles.
### 18 Comments on “Riemann Sums”
1. Tara says:
This explanation was very concise and easy to understand. Since I'll be taking Calculus this Fall, I'm sure this concept will be quite helpful. Thanks!
2. Suvinthra says:
The first thing I have to express my thousands of thanks which will not be adequate for the clear explanation given for simple basic terms.
I have cleared the long confusion about the idea of using integration.
3. Murray says:
4. Kalakay says:
Once again, for me (a secondary school teacher), all material on this site very useful. For that purpose, can be provided the facility to download it in pdf file format? Thank you!
5. Murray says:
Hi Kalakay. I'm glad you find squareCircleZ useful.
It's in the plan to provide a PDF download option, but due to lack of time, this will not be for a while.
6. Mathematics and Multimedia Blog Carnival #2 « Mathematics and Multimedia says:
[...] Bourne has a very clear explanation about the concept of Riemann sums posted at [...]
7. Tom says:
Hi,
A question: If the integral of y= 0.3*x^3 *sin(x) could not be solved algebraically, as you have said, by what means was the "Actual area" field on the interactive graphs calculated?
Thanks,
-Tom
8. Tom says:
Also, I take issue with declaring Simpson's rule to be "The most accurate numerical integration method". What about Gaussian Quadrature?
9. Murray says:
Hi Tom. My "actual area" uses the "middle" rectangle values with n=1000 (working in the background). This is accurate to the 3 decimal places shown in the interactive.
As for your second point, Simpson's Rule is one instance of the quadrature rules you mentioned and would certainly be adequate for 3 decimal place accuracy in this instance.
Actually, that sentence was more a "note to self" to revise this at some later point so it is more sophisticated. However, it already consumed a lot of time so I went ahead and published as is.
10. Alexander Bogomolny says:
A very nice introduction into definite integrals. Thank you.
I have just one remark, viz., that the expression "area under the graph" applies only to positive functions. In principle, a definite integral can be zero for a nonzero function as, for example the integral of sin(x) over an interval of length 2π.
I have an interactive demonstration of Riemann sums where a function is defined with cubic splines and is therefore very modifiable. In addition, there are several ways of choosing the function value in each of the intervals. There is very little explanation there and your blog post will fit right in.
11. Murray says:
Hi Alexander - good to see you here.
Yes, in the interests of trying to keep it simple, I didn't go into the issue of positive and negative integral values. However, I have modified the post (under "What's missing?") because you're right - it should be there!
12. Alexander Bogomolny says:
Hi Murray,
You certainly have a nack for making things simple and clear and also for keeping up with various web developments. JSXGraph is a little wonder in itself.
13. Miss Lipz says:
Hi Murray!
Thank your for sharing! you did a fantastic job! your explanations are clear and purposeful. 🙂 Actually, I'm a secondary school mathematics teacher, I'll definitely share your site with my students. 🙂 Nevertheless, I'm thinking of using the interactive javascript graph that you used to explain Riemann Sums in my lesson, is there any way I can get it? Hope you can help. 🙂
Once again thank you! and hope you have a great day ahead!
14. Murray says:
Hello Miss Lipz. Glad you found it useful.
You can simply access this page during your lesson and interact with it there. Or do you mean you don't have internet access in your classroom?
15. sheva says:
Thank you for your explanation. Why some calculus books made it difficult? 😀
16. Sbusiso Delumbuso Hamilton says:
I have found the explanation good.Now I have a clear understanding on how to find the area using Riemann's sum.
17. JD says:
Wow my math teacher's name is Colleen and she showed us this REALLY COOL WEBSITE! I am so excited to play on the website all weekend and learn more about integrals.
18. Murray says:
Hi Shiva and Sbusiso: I'm glad you found it useful.
### Comment Preview
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# What is the HCF of 24 and 44?
## What is the HCF of 24 and 44?
4
There are 3 common factors of 24 and 44, that are 1, 2, and 4. Therefore, the greatest common factor of 24 and 44 is 4.
What is the greatest common factor of 22 44 and 55?
The common factors for 44,22,55 44 , 22 , 55 are 1,11 . The GCF (HCF) of the numerical factors 1,11 is 11 .
What is the greatest common factor of 33 22 and 44?
What is the GCF of 33 and 44? The GCF of 33 and 44 is 11. To calculate the GCF of 33 and 44, we need to factor each number (factors of 33 = 1, 3, 11, 33; factors of 44 = 1, 2, 4, 11, 22, 44) and choose the greatest factor that exactly divides both 33 and 44, i.e., 11.
### What is the greatest common factor of 22 44 and 66?
2 × 11
GCF of 44 and 66 by Prime Factorization As visible, 44 and 66 have common prime factors. Hence, the GCF of 44 and 66 is 2 × 11 = 22.
What are the factors of 44?
Factors of 44
• Factors of 44: 1, 2, 4, 11, 22 and 44.
• Negative Factors of 44: -1, -2, -4, -11, -22 and -44.
• Prime Factors of 44: 2, 11.
• Prime Factorization of 44: 2 × 2 × 11 = 22 × 11.
• Sum of Factors of 44: 84.
What is the GCF of 42 and 28?
GCF of 28 and 42 by Listing Common Factors There are 4 common factors of 28 and 42, that are 1, 2, 14, and 7. Therefore, the greatest common factor of 28 and 42 is 14.
## What is the GCF of 55 44?
The GCF of 44 and 55 is 11.
What is the factor of 22 and 44?
There are 4 common factors of 22 and 44, that are 1, 2, 11, and 22. Therefore, the greatest common factor of 22 and 44 is 22.
What is the HCF of 12 18 and 24?
6
The HCF of 12, 18, and 24 is 6. ∴ The highest number that divides 12, 18, and 24 is 6.
### What is the GCF of 5 and 7?
1
Answer: GCF of 5 and 7 is 1.
What is the HCF of 44?
For example: The factors of 44 are 1, 2, 4,11,22 and 44. The factors of 99 are 1,3,9,11,33 and 99. The common factors are 1 and 11. The highest among the two is 11, which makes the H.C.F.
What is the GCF of 60 and 24?
12
Answer: GCF of 24 and 60 is 12.
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# Is there a trick to finding the number of odd numbers b/w two values?
I know you could find the number of even numbers (since they are a multiple of two). For example the number of even numbers between $11$ and $30$ will be $$n= \frac{28-12}{2} + 1 = 9$$
I wanted to know is there a similar way to find the number of odd numbers b/w two extremes?
-
You can use the exact formula. What is number of odd numbers between 11 and 30? n = (29-13)/2 + 1 = 16/2 + 1 = 9. Let's list them out to make sure: 13, 15, 17, 19, 21, 23, 25, 27, 29. There are 9 of them, so that is correct
By the way, your formula was calculated wrong. n = (28-12)/2 + 1 = 16/2 + 1 = 8 + 1 = 9 (12, 14,16,18,20,22,24,26,28)
-
Sidd's answer certainly gives you what you want, and if you understand the reasoning behind the formula for evens then you should be able to see why it works for odds too. But here's another way of thinking about it: Counting the number of odd numbers between 11 and 30 is the same as counting the number of even numbers between 12 and 31. – Brett Frankel Aug 6 '12 at 5:40
If you can find the number of numbers, and you can find the number of even numbers, ....
-
Ah, a breath of sanity ... – almagest Jul 8 at 5:58
use the A.P formula for finding the number of terms...say for no. of odd numbers between 1 and 11.... nth term= first term+(n-1)(common difference)......(*),where n=no. of terms,which we have set out to find in this case.
from (*) we have n= {(nth term-first term)/common difference}+1.....
For our problem, n= {(11-1)}/2}+1 =6
here common difference is 2 as an odd number occurs by adding 2 to the previous odd number etc.,
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The Problem What is the number of EVEN or ODD integers between two numbers n and m?
(where m>n)
Solution:
1. Recall the formula (m-n+1)/2
2. Calculate it.
3. See below for how to interpret:
|
Select Page
Try this beautiful problem from PRMO, 2014 based on Finding side of Triangle.
## Finding side of Triangle | PRMO | Problem 15
Let XOY be a triangle with angle XOY=90 degrees. Let M and N be the midpoints of the legs OX and OY, respectively. Suppose that XN=19 and YM=22. what is XY?
• $28$
• $26$
• $30$
### Key Concepts
Geometry
Triangle
Pythagoras
But try the problem first…
Answer:$26$
Source
PRMO-2014, Problem 15
Pre College Mathematics
## Try with Hints
First hint
Given that $\angle XOY=90^{\circ}$ .Let M and N be the midpoints of the legs OX and OY, respectively, and that XN=19 and YM=22. Now $\triangle XON$ & $\triangle MOY$ are Right angle Triangle. Use Pythagoras theorem …….
Can you now finish the problem ……….
Second Step
Let $XM=MO=p$ and $ON=NY=q$.Now using Pythagoras theorm on $\triangle XON$ & $\triangle MOY$ we have…
$OX^2 +ON^2=XN^2$ $\Rightarrow 4p^2 +q^2=19^2$ $\Rightarrow 4p^2 +q^2=361$………..(1) and $OM^2 +OY^2=MY^2$ $\Rightarrow p^2 +4q^2=22^2$ $\Rightarrow p^2 +4q^2=484$……(2)
Final Step
Now Adding (1)+(2)=$(4p^2 +q^2=361)$+$(p^2 +4q^2=484$ $\Rightarrow 5(p^2+q^2)=845$ $\Rightarrow (p^2+q^2)=169$ $\Rightarrow 4(p^2+q^2)=676$ $\Rightarrow (OX)^2+(OY)^2=(26)^2$ $\Rightarrow (XY)^2=(26)^2$ $\Rightarrow XY=26$.
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Calculating expectation by plugging into the definition works in simple cases, but often it can be cumbersome or lack insight. The most powerful result for calculating expectation turns out not to be the definition. It looks rather innocuous:
Let $$X$$ and $$Y$$ be two random variables defined on the same probability space. Then
$E(X+Y) = E(X) + E(Y)$
Before we look more closely at this result, note that we are assuming that all the expectations exist; we will do this throughout in this course.
And now note that there are no assumptions about the relation between $$X$$ and $$Y$$. They could be dependent or independent. Regardless, the expectation of the sum is the sum of the expectations. This makes the result powerful.
Additivity follows easily from the definition of $$X+Y$$ and the definition of expectation on the domain space. First note that the random variable $$X+Y$$ is the function defined by
$(X+Y)(\omega) = X(\omega) + Y(\omega) ~~~~ \text{for all } \omega \in \Omega$
Thus a “value of $$X+Y$$ weighted by the probability” can be written as
$(X+Y)(\omega) \cdot P(\omega) = X(\omega)P(\omega) + Y(\omega)P(\omega )$
Sum the two sides over all $$\omega \in \Omega$$ to prove additivty of expecation.
Quick Check
Let $$X$$ and $$Y$$ be random variables on the same space, with $$E(X) = 5$$ and $$E(Y) = 3$$.
(a) Find $$E(X-Y)$$.
(b) Find $$E(2X-8Y+7)$$.
By induction, additivity extends to any finite number of random variables. If $$X_1, X_2, \ldots , X_n$$ are random variables defined on the same probability space, then
$E(X_1 + X_2 + \cdots + X_n) = E(X_1) + E(X_2) + \cdots + E(X_n)$
regardless of the dependence structure of $$X_1, X_2, \ldots, X_n$$.
If you are trying to find an expectation, then the way to use additivity is to write your random variable as a sum of simpler variables whose expectations you know or can calculate easily.
## 8.4.2. $$E(X^2)$$ for a Poisson Variable $$X$$¶
Let $$X$$ have the Poisson $$\mu$$ distribution. In earlier sections we showed that $$E(X) = \mu$$ and $$E(X(X-1)) = \mu^2$$.
Now $$X^2 = X(X-1) + X$$. The random variables $$X(X-1)$$ and $$X$$ are both functions of $$X$$, so they are not independent of each other. But additivity of expectation doesn’t require independence, so we can use it to see that
$E(X^2) ~ = ~ E(X(X-1)) + E(X) ~ = ~ \mu^2 + \mu$
We will use this fact later when we study the variability of $$X$$.
It is worth noting that it is not easy to calculate $$E(X^2)$$ directly, since
$E(X^2) ~ = ~ \sum_{k=0}^\infty k^2 e^{-\mu}\frac{\mu^k}{k!}$
is not an easy sum to simplify.
## 8.4.3. Sample Sum¶
Let $$X_1, X_2, \ldots , X_n$$ be a sample drawn at random from a numerical population that has mean $$\mu$$, and let the sample sum be
$S_n = X_1 + X_2 + \cdots + X_n$
Then, regardless of whether the sample was drawn with or without replacement, each $$X_i$$ has the same distribution as the population. This is clearly true if the sampling is with replacement, and it is true by symmetry if the sampling is without replacement as we saw in an earlier chapter.
So, regardless of whether the sample is drawn with or without replacement, $$E(X_i) = \mu$$ for each $$i$$, and hence
$E(S_n) = E(X_1) + E(X_2) + \cdots + E(X_n) = n\mu$
We can use this to estimate a population mean based on a sample mean.
## 8.4.4. Unbiased Estimator¶
Suppose a random variable $$X$$ is being used to estimate a fixed numerical parameter $$\theta$$. Then $$X$$ is called an estimator of $$\theta$$.
The bias of $$X$$ is the difference $$E(X) - \theta$$. The bias measures the amount by which the estimator exceeds the parameter, on average. The bias can be negative if the estimator tends to underestimate the parameter.
If the bias of an estimator is $$0$$ then the estimator is called unbiased. So $$X$$ is an unbiased estimator of $$\theta$$ if $$E(X) = \theta$$.
If an estimator is unbiased, and you use it to generate estimates repeatedly and independently, then in the long run the average of all the estimates is equal to the parameter being estimated. On average, the unbiased estimator is neither higher nor lower than the parameter. That’s usually considered a good quality in an estimator.
In practical terms, if a data scientist wants to estimate an unknown parameter based on a random sample $$X_1, X_2, \ldots, X_n$$, the data scientist has to come up with a statistic to use as the estimator.
Recall from Data 8 that a statistic is a number computed from the sample. In other words, a statistic is a numerical function of $$X_1, X_2, \ldots, X_n$$.
Constructing an unbiased estimator of a parameter $$\theta$$ therefore amounts to finding a statistic $$T = g(X_1, X_2, \ldots, X_n)$$ for a function $$g$$ such that $$E(T) = \theta$$.
## 8.4.5. Unbiased Estimators of a Population Mean¶
As in the sample sum example above, let $$S_n$$ be the sum of a sample $$X_1, X_2, \ldots , X_n$$ drawn at random from a population that has mean $$\mu$$. The standard statistical notation for the average of $$X_1, X_2, \ldots , X_n$$ is $$\bar{X}_n$$. So
$\bar{X}_n = \frac{S_n}{n}$
Then, regardless of whether the draws were made with replacement or without,
\begin{split} \begin{align*} E(\bar{X}_n) &= \frac{E(S_n)}{n} ~~~~ \text{(linear function rule)} \\ &= \frac{n \mu}{n} ~~~~~~~~~ \text{(} E(S_n) = n\mu \text{)} \\ &= \mu \end{align*} \end{split}
Thus the sample mean is an unbiased estimator of the population mean.
It is worth noting that $$X_1$$ is also an unbiased estimator of $$\mu$$, since $$E(X_1) = \mu$$. So is $$X_j$$ for any $$j$$, also $$(X_1 + X_9)/2$$, or any linear combination of the sample if the coefficients add up to 1.
But it seems clear that using the sample mean as the estimator is better than using just one sampled element, even though both are unbiased. This is true, and is related to how variable the estimators are. We will address this later in the course.
Quick Check
Let $$X_1, X_2, X_3$$ be i.i.d. Poisson $$(\mu)$$ random variables, and suppose the value of $$\mu$$ is unknown. Is $$0.4X_1 + 0.2X_2 + 0.4X_3$$ an unbiased estimator of $$\mu$$?
## 8.4.6. First Unbiased Estimator of a Maximum Possible Value¶
Suppose we have a sample $$X_1, X_2, \ldots , X_n$$ drawn at random from $$1, 2, \ldots , N$$ for some fixed $$N$$, and we are trying to estimate $$N$$.
How can we use the sample to construct an unbiased estimator of $$N$$? By definition, such an estimator must be a function of the sample and its expectation must be $$N$$.
In other words, we have to construct a statistic that has expectation $$N$$.
Each $$X_i$$ has the uniform distribution on $$1, 2, \ldots , N$$. This is true for sampling with replacement as well as for simple random sampling, by symmetry.
The expectation of each of the uniform variables is $$(N+1)/2$$, as we have seen earlier. So if $$\bar{X}_n$$ is the sample mean, then
$E(\bar{X}_n) = \frac{N+1}{2}$
Clearly, $$\bar{X}_n$$ is not an unbiased estimator of $$N$$. That’s not surprising because $$N$$ is the maximum possible value of each observation and $$\bar{X}_n$$ should be somewhere in the middle of all the possible values.
But because $$E(\bar{X}_n)$$ is a linear function of $$N$$, we can figure out how to create an unbiased estimator of $$N$$.
Remember that our job is to create a function of the sample $$X_1, X_2, \ldots, X_n$$ in such a way that the expectation of that function is $$N$$.
Start by inverting the linear function, that is, by isolating $$N$$ in the equation above.
$2E(\bar{X}_n) - 1 = N$
This tells us what we have to do to the sample $$X_1, X_2, \ldots, X_n$$ to get an unbiased estimator of $$N$$.
We should just use the statistic $$T_1 = 2\bar{X}_n - 1$$ as the estimator. It is unbiased because $$E(T_1) = N$$ by the calculation above.
Quick Check
In the setting above, what is the bias of $$2\bar{X}_n$$ as an estimator of $$N$$? Does it tend to overestimate on average, or underestimate?
## 8.4.7. Second Unbiased Estimator of the Maximum Possible Value¶
The calculation above stems from a problem the Allied forces faced in World War II. Germany had a seemingly never-ending fleet of Panzer tanks, and the Allies needed to estimate how many they had. They decided to base their estimates on the serial numbers of the tanks that they saw.
Here is a picture of one from Wikipedia.
Notice the serial number on the top left. When tanks were disabled or destroyed, it was discovered that their parts had serial numbers too. The ones from the gear boxes proved very useful.
The idea was to model the observed serial numbers as random draws from $$1, 2, \ldots, N$$ and then estimate $$N$$. This is of course a very simplified model of reality. But estimates based on even such simple probabilistic models proved to be quite a bit more accurate than those based on the intelligence gathered by the Allies. For example, in August 1942, intelligence estimates were that Germany was producing 1,550 tanks per month. The prediction based on the probability model was 327 per month. After the war, German records showed that the actual production rate was 342 per month.
The model was that the draws were made at random without replacement from the integers 1 through $$N$$.
In the example above, we constructed the random variable $$T$$ to be an unbiased estimator of $$N$$ under this model.
The Allied statisticians instead started with $$M$$, the sample maximum:
$M ~ = ~ \max\{X_1, X_2, \ldots, X_n\}$
The sample maximum $$M$$ is a biased estimator of $$N$$, because we know that its value is always less than or equal to $$N$$. Its average value therefore will be somewhat less than $$N$$.
To correct for this, the Allied statisticians imagined a row of $$N$$ spots for the serial numbers $$1$$ through $$N$$, with marks at the spots corresponding to the observed serial numbers. The visualization below shows an outcome in the case $$N= 20$$ and $$n = 3$$.
• There are $$N = 20$$ spots in all.
• From these, we take a simple random sample of size $$n = 3$$. Those are the gold spots.
• The remaining $$N - n = 17$$ spots are colored blue.
The $$n = 3$$ sampled spots create $$n+1 = 4$$ blue “gaps” between sampled values: one before the leftmost gold spot, two between successive gold spots, and one after the rightmost gold spot that is at position $$M$$.
A key observation is that because of the symmetry of simple random sampling, the lengths of all four gaps have the same distribution.
But of course we don’t get to see all the gaps. In the sample, we can see all but the last gap, as in the figure below. The red question mark reminds you that the gap to the right of $$M$$ is invisible to us.
If we could see the gap to the right of $$M$$, we would see $$N$$. But we can’t. So we can try to do the next best thing, which is to augment $$M$$ by the estimated size of that gap.
Since we can see all of the spots and their colors up to and including $$M$$, we can see $$n$$ out of the $$n+1$$ gaps. The lengths of the gaps all have the same distribution by symmetry, so we can estimate the length of a single gap by the average length of all the gaps that we can see.
We can see $$M$$ spots, of which $$n$$ are the sampled values. So the total length of all $$n$$ visible gaps is $$M-n$$. Therefore
$\text{estimated length of one gap} ~ = ~ \frac{M-n}{n}$
So the Allied statisticians decided to improve upon $$M$$ by using the augmented maximum as their estimator:
$T_2 ~ = ~ M + \frac{M-n}{n}$
By algebra, this estimator can be rewritten as
$T_2 ~ = ~ M\cdot\frac{n+1}{n} ~ - ~ 1$
Is $$T_2$$ an unbiased estimator of $$N$$? To answer this, we have to find its expectation. Since $$T_2$$ is a linear function of $$M$$, we’ll find the expectation of $$M$$ first.
Here once again is the visualization of what’s going on.
Let $$G$$ be the length of the last gap. Then $$M = N - G$$.
There are $$n+1$$ gaps, made up of the $$N-n$$ unsampled values. Since they all have the same expected length,
$E(G) ~ = ~ \frac{N-n}{n+1}$
So
$E(M) ~ = ~ N - \frac{N-n}{n+1} ~ = ~ (N+1)\frac{n}{n+1}$
Recall that the Allied statisticians’ estimate of $$N$$ is
$T_2 ~ = ~ M\cdot\frac{n+1}{n} - 1$
Now
$E(T_2) ~ = ~ E(M)\cdot\frac{n+1}{n} - 1 ~ = ~ (N+1)\frac{n}{n+1}\cdot\frac{n+1}{n} - 1 ~ = ~ N$
Thus the augmented maximum $$T_2$$ is an unbiased estimator of $$N$$.
Quick Check
A gardener in Berkeley has 23 blue flower pots in a row. She picks a simple random sample of 5 of them and colors the selected pots gold. What is the expected number of blue flower pots at the end of the row?
## 8.4.8. Which Estimator to Use?¶
The Allied statisticians thus had two unbiased estimators of $$N$$ from which to choose. They went with $$T_2$$ instead of $$T_1$$ because $$T_2$$ has less variability.
We will quantify this later in the course. For now, here is a simulation of distributions of the two estimators in the case $$N = 300$$ and $$n=30$$. The simulation is based on $$5000$$ repetitions of drawing a simple random sample of size $$30$$ from the integers $$1$$ through $$300$$.
You can see why $$T_2$$ is a better estimator than $$T_1$$.
• Both are unbiased. So both the empirical histograms are balanced at around $$300$$, the true value of $$N$$.
• The emipirical distribution of $$T_2$$ is clustered much closer to the true value $$300$$ than the empirical distribution of $$T_1$$.
For a recap, take another look at the accuracy table of the Allied statisticians’ estimator $$T_2$$. Not bad for an estimator based on a model that assumes nothing more complicated than simple random sampling!
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## Step by action solution for calculating 27 is 60 percent of what number
We currently have our an initial value 27 and the 2nd value 60. Let"s assume the unknown worth is Y i m sorry answer us will find out.
You are watching: What percent of 60 is 27
As we have all the required values we need, now we deserve to put them in a an easy mathematical formula as below:
STEP 127 = 60% × Y
STEP 227 = 60/100× Y
Multiplying both sides by 100 and dividing both sides of the equation by 60 we will certainly arrive at:
STEP 3Y = 27 × 100/60
STEP 4Y = 27 × 100 ÷ 60
STEP 5Y = 45
Finally, us have found the worth of Y i m sorry is 45 and that is our answer.
You can conveniently calculate 27 is 60 percent that what number by using any kind of regular calculator, simply enter 27 × 100 ÷ 60 and you will gain your answer i beg your pardon is 45
Here is a percentage Calculator to solve similar calculations such together 27 is 60 percent of what number. You deserve to solve this form of calculation v your worths by start them into the calculator"s fields, and also click "Calculate" to obtain the result and explanation.
is
percent the what number
Calculate
## Sample questions, answers, and how to
Question: her friend has a bag the marbles, and he speak you the 60 percent the the marbles are red. If there are 27 red marbles. How plenty of marbles go he have altogether?
How To: In this problem, we understand that the Percent is 60, and we are likewise told that the component of the marbles is red, for this reason we know that the part is 27.
So, that method that it have to be the full that"s missing. Below is the way to number out what the complete is:
Part/Total = Percent/100
By using a simple algebra we deserve to re-arrange ours Percent equation prefer this:
Part × 100/Percent = Total
If us take the "Part" and also multiply it by 100, and then we divide that by the "Percent", us will gain the "Total".
Let"s try it the end on our problem about the marbles, that"s very basic and it"s simply two steps! We understand that the "Part" (red marbles) is 27.
So action one is to simply multiply that component by 100.
27 × 100 = 2700
In action two, us take the 2700 and divide that by the "Percent", i m sorry we space told is 60.
So, 2700 divided by 60 = 45
And that way that the total variety of marbles is 45.
Question: A high college marching band has actually 27 flute players, If 60 percent of the band members play the flute, then how many members room in the band?
Answer: There space 45 members in the band.
How To: The smaller sized "Part" in this trouble is 27 because there room 27 flute players and we room told the they comprise 60 percent the the band, therefore the "Percent" is 60.
Again, it"s the "Total" that"s lacking here, and to uncover it, we simply need to follow our 2 action procedure together the vault problem.
For action one, we multiply the "Part" by 100.
27 × 100 = 2700
For action two, we division that 2700 by the "Percent", i beg your pardon is 60.
See more: How Many Chicken Legs In A Pound, How Much Of A Chicken Drumstick Is Bone
2700 divided by 60 equals 45
That method that the total variety of band members is 45.
## Another step by step method
Step 1: Let"s i think the unknown value is Y
Step 2: very first writing it as: 100% / Y = 60% / 27
Step 3: fall the percent marks to simplify your calculations: 100 / Y = 60 / 27
Step 4: multiply both political parties by Y to move Y top top the ideal side the the equation: 100 = ( 60 / 27 ) Y
Step 5: simple the appropriate side, us get: 100 = 60 Y
Step 6: dividing both political parties of the equation by 60, we will certainly arrive at 45 = Y
This pipeline us through our final answer: 27 is 60 percent that 45
27 is 60 percent that 45 27.01 is 60 percent that 45.017 27.02 is 60 percent of 45.033 27.03 is 60 percent the 45.05 27.04 is 60 percent of 45.067 27.05 is 60 percent the 45.083 27.06 is 60 percent of 45.1 27.07 is 60 percent the 45.117 27.08 is 60 percent the 45.133 27.09 is 60 percent of 45.15 27.1 is 60 percent that 45.167 27.11 is 60 percent the 45.183 27.12 is 60 percent the 45.2 27.13 is 60 percent of 45.217 27.14 is 60 percent that 45.233 27.15 is 60 percent that 45.25 27.16 is 60 percent that 45.267 27.17 is 60 percent that 45.283 27.18 is 60 percent that 45.3 27.19 is 60 percent that 45.317
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Home News What is 70 of 30: Understanding Percentages
# What is 70 of 30: Understanding Percentages
Understanding percentages is an important skill that is often used in everyday life, from calculating discounts while shopping to determining the grade on a test. In this article, we will focus on What is 70 of 30 and provide a step-by-step guide on how to calculate it. We will also discuss some common applications of percentages and provide examples.
## What are Percentages?
A percentage is a way of expressing a number as a fraction of 100. It is represented by the symbol “%”. Percentages are often used to compare two quantities, such as the increase or decrease in sales from one year to the next. They are also commonly used to express the likelihood of an event occurring, such as the probability of winning a lottery.
## What is 70 of 30
To calculate 70% of 30, we first need to understand what 70% means. “70%” can be rewritten as 70/100 or 0.70. We then need to multiply 0.70 by 30 to get the answer.
70% of 30 = 0.70 × 30 70% of 30 = 21
Therefore, 70% of 30 is 21.
## Using Percentages in Real Life
Percentages are used in a variety of different fields, including finance, economics, and statistics. In finance, percentages are used to calculate interest rates on loans, while in economics, they are used to measure inflation and unemployment rates. In statistics, percentages are used to analyze data and draw conclusions about a population.
Percentages are also commonly used in everyday life, such as when calculating tips at a restaurant or determining the discount on a sale item. For example, if a bill at a restaurant is \$50 and the customer wants to leave a 15% tip, they would calculate the tip by multiplying 0.15 by \$50 to get \$7.50.
## What is 70 of 30 Conclusion
Understanding percentages is an important skill that is used in a variety of different fields, as well as in everyday life. To calculate 70% of 30, we multiply 0.70 by 30 to get 21. Percentages are useful for comparing two quantities and expressing probabilities. By being familiar with percentages, we can make better decisions and understand the world around us more effectively.
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I am a professional writer and blogger. I’m researching and writing about innovation, Blockchain, technology, business, and the latest marketing trends.
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# Ratio And Proportion - MCQ 2
## 20 Questions MCQ Test Quantitative Aptitude for Competitive Examinations | Ratio And Proportion - MCQ 2
Description
Attempt Ratio And Proportion - MCQ 2 | 20 questions in 15 minutes | Mock test for Quant preparation | Free important questions MCQ to study Quantitative Aptitude for Competitive Examinations for Quant Exam | Download free PDF with solutions
QUESTION: 1
### A bag contains 25p coins, 50p coins and 1 rupee coins whose values are in the ratio of 8:4:2. The total values of coins are 840. Then find the total number of coins
Solution:
Let us assume that
Number of 25p coins = 8x
Number of 50p coins = 4x
Number of Rs. 1 coins = 2x
As per the given condition
8x + 4x + 2x = 840
14x = 840
x = 840/14
x = 60
Hence,
Amount of 25p coin = 60 × 8 × 25 = 12000 paisa = Rs. 120
Amount of 50p coin = 60 × 4 × 50 = 12000 paisa = Rs. 120
Amount of Rs. 1 coin = 60 × 2 × 1 = Rs. 120
On summing up the above-obtained values the total amount in the bag is Rs120+Rs120 +Rs 120 = Rs360
QUESTION: 2
### Two vessels contains equal quantity of solution contains milk and water in the ratio of 7:2 and 4:5 respectively. Now the solutions are mixed with each other then find the ratio of milk and water in the final solution?
Solution:
Answer – A.11:7 Explanation : milk = 7/9 and water = 2/9 – in 1 vessel milk = 4/9 and water = 5/9 – in 2 vessel (7/9 + 4/9)/ (2/9 + 5/9) = 11:7
QUESTION: 3
### Two alloys contain gold and silver in the ratio of 3:7 and 7:3 respectively. In what ratio these alloys must be mixed with each other so that we get a alloy of gold and silver in the ratio of 2:3?
Solution:
Answer – B.3:1 Explanation : Gold = 3/10 and silver = 7/10 – in 1 vessel gold = 7/10 and silver = 3/10 – in 2 vessel let the alloy mix in K:1, then (3k/10 + 7/10)/ (7k/10 + 3/10) = 2/3. Solve this equation , u will get K = 3
QUESTION: 4
The sum of three numbers is 123. If the ratio between first and second numbers is 2:5 and that of between second and third is 3:4, then find the difference between second and the third number.
Solution:
Answer – C.15 Explanation : a:b = 2:5 and b:c = 3:4 so a:b:c = 6:15:20 41x = 123, X = 3. And 5x = 15
QUESTION: 5
If 40 percent of a number is subtracted from the second number then the second number is reduced to its 3/5. Find the ratio between the first number and the second number.
Solution:
Answer – C.1:1 Explanation : [ b – (40/100)a] = (3/5)b.
So we get a = b.
QUESTION: 6
The ratio between the number of boys and girls in a school is 4:5. If the number of boys are increased by 30 % and the number of girls increased by 40 %, then what will the new ratio of boys and girls in the school.
Solution:
Answer – B.26/35 Explanation : boys = 4x and girls = 5x.
Required ratio = [(130/100)*4x]/ [(140/100)*5x]
QUESTION: 7
One year ago the ratio between rahul salary and rohit salary is 4:5. The ratio between their individual salary of the last year and current year is 2:3 and 3:5 respectively. If the total current salary of rahul and rohit is 4300. Then find the current salary of rahul.
Solution:
Answer – B.1800 Explanation : 4x and 5x is the last year salry of rahul and rohit respectively Rahul last year to rahul current year = 2/3 Rohit last year to rohit current year = 3/5 Current of rahul + current of rohit = 4300 (3/2)*4x + (5/3)*5x = 4300.
X = 300.
So rahul current salary = 3/2 * 4* 300 = 1800
QUESTION: 8
A sum of 12600 is to be distributed between A, B and C. For every rupee A gets, B gets 80p and for every rupee B gets, C get 90 paise. Find the amount get by C.
Solution:
Ratio of money between A and B – 100:80 and that of B and C – 100:90
so the ratio between A : B :C – 100:80:72
so 252x = 12600, x = 50. So C get = 50*72 = 3600
QUESTION: 9
The sum of the squares between three numbers is 5000. The ratio between the first and the second number is 3:4 and that of second and third number is 4:5.Find the difference between first and the third number.
Solution:
Answer – A.20 Explanation : a^2 + b^2 + c^2 = 5000 a:b:c = 3:4:5 50x^2 = 5000.
X = 10.
5x – 3x = 2*10 = 20
QUESTION: 10
The ratio between two numbers is 7:5. If 5 is subtracted from each of them, the new ratio becomes 3:5. Find the numbers.
Solution:
Answer – A.7/2, 5/2 Explanation : (7x – 5)/(5x – 5) = 3/5 X = 1/2 so the numbers are 7/2 and 5/2
QUESTION: 11
A company reduces his employee in the ratio 14 : 12 and increases their wages in the ratio 16:18, Determine whether the bill of wages increases or not and in what ratio.
Solution:
Answer – a) Decreases, 28: 27 Explanation : Let initial employee be 14a and final employee be 12a similarly initial wage is 16b and final wage be 18b Total initial wage = 14a * 16b = 224ab, total final wage = 12a* 18b = 216ab So clearly wages decreases and ratio = 224ab: 216ab = 28:27
QUESTION: 12
A bucket contains liquid A and B in the ratio 4:5. 36 litre of the mixture is taken out and filled with 36 litre of B. Now the ratio changes to 2:5. Find the quantity of liquid B initially.
Solution:
Answer – b) 56ltr Explanation : Let A = 4x and B = 5x Now, A = 4x – 36*4/9 and B = 5x – 36*5/9 + 36 Now, ratio between A and B = 2:5 X = 11.2 now B = 11.2*5 = 56
QUESTION: 13
Two numbers are in the ratio of 5:6 and if 4 is added to the first number and 4 is subtracted from the second number then the ratio becomes 3:2. Find the difference between two numbers.
Solution:
Answer – a) 2.5 Explanation : (5x + 4)/ (6x -4) = 3/2
QUESTION: 14
The income of riya and priya are in the ratio of 4:5 and their expenditure is in the ratio of 2:3. If each of them saves 2000, then find their income.
Solution:
Answer – b) 4000, 5000 Explanation : 4x – 2y = 2000 and 5x – 3y = 2000.
X = 1000, so income = 4000 and 5000
QUESTION: 15
A 50 litre of mixture contains milk and water in the ratio 2:3. How much milk must be added to the mixture so that it contains milk and water in the proportion of 3:2.
Solution:
Answer – b) 25 Explanation : (20 + x)/30 = 3/2
QUESTION: 16
Two alloys contain platinum and gold in the ratio of 1:2 and 1:3 respectively. A third alloy C is formed by mixing alloys one and alloy two in the ratio of 3:4.Find the percentage of gold in the mixture
Solution:
Answer – d) 71.3/7% Explanation : Platinum = 1/3 and 1/4 gold = 2/3 and 3/4 Alloy one and two are mixed in the ratio of 3:4, so ratio of platinum and gold in final ratio – 2:5 So gold % = (5/7)*100
QUESTION: 17
The sum of three numbers is 980. If the ratio between first and second number is 3:4 and that of second and third is 3:7. Find the difference between first and last number.
Solution:
Answer – a) 380 Explanation : ratio between three numbers – 9:12:28 49x = 980, x = 20 difference between number = 19*20 = 380
QUESTION: 18
The ratio between number of girls and boys in a school is 5: 6. If 40 percent of the boys and 20 percent of the girls are scholarship holders, what percentage of the students does not get scholarship?
Solution:
Answer – b) 69% Explanation : Girls = 5x and boys = 6x Girls that don’t get scholarship = 5x * 80/100 = 4x and boys that don’t get scholarship = 6x * 60/100 = 3.6x Percent students that didn’t get scholarship = (7.6x/11x)*100 = 69 (approx.)
QUESTION: 19
A bag contains 25p coins, 50p coins and 1 rupee coins whose values are in the ratio of 8:4:2. The total values of coins are 840. Then find the total amount in rupees.
Solution:
Answer – d) 280 Explanation : Value is given in the ratio 8:4:2. (8x/0.25) + (4x/0.5) + (2x/1) = 840.
X = 20. Total amount = 14*20 = 280
QUESTION: 20
An amount is to be divided between A, B and C in the ratio 2:3:5 respectively. If C gives 200 of his share to B the ratio among A, B and C becomes 3:5:4. What is the total sum?
Solution:
Answer – b) 6000 Explanation : s2x, 3x + 200, 5x – 200 2x/(3x + 200) = 3/5, we will get x = 600, so total amount = 10*600 = 6000
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# Class 9 Maths MCQ – Polynomials
This set of Class 9 Maths Chapter 2 Multiple Choice Questions & Answers (MCQs) focuses on “Polynomials”. These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation.
1. What is the coefficient of x3 in a polynomial 6x4 + 3x2 + 8x + 5?
a) 6
b) 3
c) 8
d) 0
Explanation: Coefficient is the number which is multiplied with respective variable.
In the given polynomial 6x4 + 3x2 + 8x + 5, there is not an expression containing x3. So we can write 6x4+ 3x2+8x+5 as 6x4 + 0x3 + 3x2 + 8x + 5. We can see that 0 is multiplied with expression x3, so coefficient of x3 is 0.
2. $$\frac{1}{\sqrt[2]{x}}$$ is a polynomial.
a) True
b) False
Explanation: For an expression to be a polynomial, exponent of variable has to be whole number.
$$\frac{1}{\sqrt[2]{x}}$$ can be written as x-1/2. We can see that exponent of x is -1/2 which is not whole number (W = {0, 1, 2, 3…}). Hence, $$\frac{1}{\sqrt[2]{x}}$$ is not a polynomial.
3. What is the degree of a polynomial of 4x7+9x5+5x2+11?
a) 7
b) 4
c) 5
d) 2
Explanation: Degree of a polynomial is the highest power of variable in a polynomial.
The term with the highest power of x is 4x7 and exponent of x in that term is 7, so the degree of polynomial of 4x7+ 9x5+5x2+11 is 7.
4. What is the degree of a polynomial 7?
a) 7
b) 1
c) 0
d) 2
Explanation: Degree of a non-zero constant polynomial is zero.
We can see that given polynomial 7 contain only one term and that is constant. 7 can also be written as 7x0.
Hence degree of 7 is zero.
5. What is the degree of 0?
a) Not defined
b) 1
c) 2
d) 0
Explanation: Degree of the zero polynomial is not defined.
Zero polynomial is denoted by 0, and degree for that is not defined.
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6. A quadratic polynomial can have at most __________ terms.
a) 1
b) 4
c) 2
d) 3
Explanation: A polynomial of degree 2 is called quadratic polynomial.
Quadratic polynomials are of the form ax2+bx+c and it can contain at most three terms namely ax2, bx and c. Thus, we can say that a quadratic polynomial can have at most three terms.
Similarly, a polynomial of degree 1 is called linear polynomial and a polynomial of degree 3 is called cubic polynomial.
More MCQs on Class 9 Maths Chapter 2:
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## Encyclopedia > Numerator
Article Content
# Fraction
Redirected from Numerator
In algebra, a fraction consists of one quantity divided by another quantity. The fraction "three divided by four" or "three over four" or "three fourths" can be written as
$\frac{3}{4}$
or 3 ÷ 4
or 3/4
In this article, we will use the latter notation. Other typical fractions include -2/7 and (x+1)/(x-1). The first quantity, the number "on top of the fraction", is called the numerator, and the other number is called the denominator. The denominator can never be zero. A fraction consisting of two integers is called a rational number.
Several rules for the calculation with fractions are useful:
Cancelling. If both the numerator and the denominator of a fraction are multiplied or divided by the same number, then the fraction does not change its value. For instance, 4/6 = 2/3 and 1/x = x / x2.
Adding fractions. To add or subtract two fractions, you first need to change the two fractions so that they have a common denominator; then you can add or subtract the numerators. For instance, 2/3 + 1/4 = 8/12 + 3/12 = 11/12.
Multiplying fractions. To multiply two fractions, multiply the numerators to get the new numerator, and multiply the denominators to get the new denominator. For instance, 2/3 × 1/4 = (2×1) / (3× 4) = 2 / 12 = 1 / 6.
Dividing fractions. To divide one fraction by another one, flip numerator and denominator of the second one, and then multiply the two fractions. For instance, (2/3) / (4/5) = 2/3 × 5/4 = (2×5) / (3×4) = 10/12 = 5/6.
In abstract algebra, these rules can be proved to hold in any field. Furthermore, if one starts with any integral domain R, one can always construct a field consisting of all fractions of elements of R, the field of fractions of R.
All Wikipedia text is available under the terms of the GNU Free Documentation License
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# B14 adding negative numbers
• MHB
• karush
If the "original number" is -9 then the "positive number" is 9.)In summary, subtracting a positive number is the same as adding a negative number, and subtracting a negative number is the same as adding a positive number, where the positive and negative numbers are the opposites of the original numbers. This concept is being taught in the context of using counters to add.
karush
Gold Member
MHB
Find
a. Subtracting 4 is the same as adding $\boxed{(-4)}$
b. Subtracting -7 is the same as adding $\boxed{(7)}$
c. Subtracting a positive number is the same as adding a
\boxed{negative} number, where that $\boxed{?}$ is opposite of the original number
d. Subtracting a negative number is the same as adding a
$\boxed{?}$ number, where that $\boxed{?}$ is opposite of the original number
ok they are trying to teach about using counters to add
but $\boxed{?}$ was kinda on of those boolean confusions
I'm not sure what "opposite of the original number" is supposed to mean! You said before that "subtracting -7 is the same as adding 7" so you do know that "
d. Subtracting a negative number is the same as adding a POSITIVE number, where that POSITIVE number is the opposite of the original number. (If the "original number" is -7 then the "positive number" is 7.
## 1. How do you add negative numbers?
To add negative numbers, you can use the same method as adding positive numbers. Start by lining up the numbers vertically, with the decimal points aligned. Then, add the numbers as you normally would, but remember to keep the negative sign in front of the number.
## 2. What is the rule for adding negative numbers?
The rule for adding negative numbers is that when you add a positive number and a negative number, the result will have the sign of the larger number. For example, when adding -5 and 3, the result will be -2 because the absolute value of -5 is larger than the absolute value of 3.
## 3. Can you add more than two negative numbers at once?
Yes, you can add more than two negative numbers at once. The key is to group the numbers in pairs and add them together, then continue adding the resulting numbers until you have added all the numbers. For example, to add -3, -5, and -7, you can group -3 and -5 to get -8, then add -8 and -7 to get -15.
## 4. What happens when you add a positive number and a negative number with the same absolute value?
When you add a positive number and a negative number with the same absolute value, the result will always be 0. This is because when you add a number and its opposite, they cancel each other out.
## 5. Can you use a number line to add negative numbers?
Yes, you can use a number line to add negative numbers. Start by marking the first number on the number line, then move to the left for negative numbers and to the right for positive numbers. Then, count the number of spaces to the right or left to find the sum.
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# plotter_math_help2
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• pg 1
``` Subtracting Coordinates
There are three steps to playing this game.
2. Determine where you want to move (coordinates of your second ordered pair)
3. Determine what horizontal and vertical movement Plotter needs to make to move
to a fish. Slide the x and y sliders to move Plotter in the direction and distance he
needs to go.”
What is An Ordered Pair:
An ordered pair has two numbers separated by a comma inside of parentheses, like
this (0, -5). These tell us where a point is on a graph. In an ordered pair, the first
coordinate, the number on the left, represents the value along the x-axis. It is called the
x-coordinate. The second coordinate, the number on the right, represents the value along
the y-axis. It is called the y-coordinate. Coordinates can be positive or negative numbers.
Example: The ordered pair (2, 6) represents a point that has an x-coordinate of 2 and a y-
coordinate of 6.
There are two number lines on a coordinate graph. One is on the x-axis (horizontal)
and one is on the y-axis (vertical). They cross at zero.
Figure out where Plotter Penguin is standing on the graph. The penguin is standing on
an intersection of two lines (where the two lines cross). Look at the horizontal number
line (x-axis). Figure out what number Plotter is standing under or over. That is the first
number of Plotter’s position (the x-coordinate). Then look at the vertical number line.
Figure out what number is to Plotter’s right or left. That represents the second number of
Plotter’s position (the y-coordinate).
Let’s say Plotter is standing over the – 3 of the horizontal axis and to the left of the 4
on the vertical axis. So he is standing at a point that has the x-coordinate of – 3 and the
y-coordinate of 4. We can state this by the ordered pair (- 3, 4).
Positive Directions on the X and Y Axis
Moving to the right on the x-axis is moving positively because the numbers get bigger
as you move to the right. Don’t be fooled by negative numbers on your x-axis.
Remember that – 3 is LESS than – 1. Moving to the left on this axis is moving
negatively, of course.
Moving upward on the y-axis is moving positively because the numbers on that
number line get larger as you move up the axis. Moving downward on this axis is
negative movement. Again, don’t let those negative numbers on the number line fool you.
Moving From One Point to Another
Once you have two ordered pairs (where Plotter starts and where you want him to end
up), you can figure out how to get the penguin from place to place. This can be done by
subtracting the x-coordinates and then subtracting the y-coordinates. The result tells you
how far to move on each axis.
In this example, Plotter started on (– 3, 4) and wants to move to a point (1, 5). You
can see this on the graph below. The dotted lines indicate the movement. Note: Plotter
can only move in a horizontal and/or vertical direction. Plotter cannot move diagonally.
Plotter will move on the x-axis a distance of 4 units ( from –3 to –2, -1, 0, 1) in a positive
direction and on the y-axis a distance of 1 unit (from 4 to 5) in a positive direction. If
you count the intersections, you can see this.
Though you can count it out using the graph, you can also figure it out using subtraction.
Start with the ordered pair of the place you want to move to and subtract the ordered pair
of the place you started. For instance, Plotter is starting at (-3, 4) and wants to go to (1,
5). Plotter wants to go to the x-coordinate (1) , so subtract the x-coordinate where he
started (- 3).
1 – (- 3) = 4
This tells you how to move on the x-axis. You will move 4 units horizontally in the
positive direction.
Then Plotter wants to go to the y-coordinate (5), so subtract his current y-coordinate of 4.
5-4=1
This tells you how to move on the y-axis. You will move 1 unit vertically in the positive
direction.
If your answer was a negative number, you would have to move in the negative direction
on that axis.
```
To top
;
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# Video: Subtracting Two Vectors Component-Wise
Given that vector π΄ = (1, 9) and vector π΅ = (β4, 1), find vector π΄ β vector π΅.
01:27
### Video Transcript
Given that vector π΄ equals one, nine and vector π΅ equals negative four, one, find π΄ minus π΅.
So in order to actually solve this, what weβre actually going to do is deal with our π₯- and π¦-components separately. So when we actually subtract π΅ from π΄ β so our vector π΅ from our vector π΄ β weβre going to deal with our π₯-coordinates and then weβre going to move on to deal with our π¦-coordinates. So with that in mind, weβre gonna have π΄ minus π΅ β so vector π΄ minus vector π΅ β is equal to one because thatβs the π₯-component of our vector π΄ minus negative four and thatβs because thatβs the π₯-component of our vector π΅.
So now we move on to our π¦-components. And we have nine minus one because one is actually the π¦-component of our vector π΅. Okay, great, so our next stage is to actually simplify. So then, our next line in working is that vector π΄ minus vector π΅ is equal to one plus four as our π₯-component. And we got one plus four because we had minus a negative and if you minus a negative, this turns positive. And then, we moved on to our π¦-component and this stays the same at this point, which is just nine minus one.
So therefore, we can say that given that our vector π΄ is one, nine and that vector π΅ is equal to negative four, one, vector π΄ minus vector π΅ is gonna be equal to five, eight. And we got that because we had one plus four as our π₯-component that gives us five and nine minus one as our π¦-component which gives us eight.
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# If the solve the problem
Question:
If $y=\left(\sin ^{-1} x\right)^{2}$, prove that: $\left(1-x^{2}\right) y_{2}-x y_{1}-2=0$
Solution:
Note: $y_{2}$ represents second order derivative i.e. $\frac{d^{2} y}{d x^{2}}$ and $y_{1}=d y / d x$
Given,
$y=\left(\sin ^{-1} x\right)^{2} \ldots \ldots$ equation 1
to prove : $\left(1-x^{2}\right) y_{2}-x y_{1}-2=0$
We notice a second-order derivative in the expression to be proved so first take the step to find the second order derivative.
Let's find $\frac{d^{2} y}{d x^{2}}$
$\mathrm{AS} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$
So, lets first find $\mathrm{dy} / \mathrm{dx}$
$\frac{d y}{d x}=\frac{d}{d x}\left(\sin ^{-1} x\right)^{2}$
Using chain rule we will differentiate the above expression
Let $t=\sin ^{-1} x=>\frac{d t}{d x}=\frac{1}{\sqrt{\left(1-x^{2}\right)}}$ [using formula for derivative of $\sin ^{-1} x$ ]
And $\mathrm{y}=\mathrm{t}^{2}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} \frac{\mathrm{dt}}{\mathrm{dx}}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{t} \frac{1}{\sqrt{\left(1-\mathrm{x}^{2}\right)}}=2 \sin ^{-1} \mathrm{x} \frac{1}{\sqrt{\left(1-\mathrm{x}^{2}\right)}} \ldots \ldots$ equation 2
Again differentiating with respect to $x$ applying product rule:
$\frac{d^{2} y}{d x^{2}}=2 \sin ^{-1} x \frac{d}{d x}\left(\frac{1}{\sqrt{1-x^{2}}}\right)+\frac{2}{\sqrt{\left(1-x^{2}\right)}} \frac{d}{d x} \sin ^{-1} x$
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{2 \sin ^{-1} \mathrm{x}}{2\left(1-\mathrm{x}^{2}\right) \sqrt{1-\mathrm{x}^{2}}}(-2 \mathrm{x})+\frac{2}{\left(1-\mathrm{x}^{2}\right)}\left[\right.$ using $\left.\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1} \frac{\mathrm{d}}{\mathrm{dx}} \sin ^{-1} \mathrm{x}=\frac{1}{\sqrt{\left(1-\mathrm{x}^{2}\right)}}\right]$
$\frac{d^{2} y}{d x^{2}}=\frac{2 x \sin ^{-1} x}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}+\frac{2}{\left(1-x^{2}\right)}$
Using equation 2 :
$\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=2-x \frac{d y}{d x}$
$\therefore\left(1-x^{2}\right) y_{2}-x y_{1}-2=0 \ldots \ldots .$ proved
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# Teaching Students About the Average Number of Days in a Month
Understanding the average number of days in a month is crucial for several mathematical and time management applications. By learning this concept, students can gain a deeper appreciation for calendar systems and develop better organization skills. This article offers an insightful approach to teach students the ins and outs of calculating the average number of days in a month.
Step 1: Discuss Calendar Basics
The first step in teaching this concept is to review the basics of the calendar system. Ensure students understand that a year has 12 months and that most months have either 30 or 31 days, while February has either 28 or 29 days depending on whether it is a leap year.
Before moving forward, it’s essential to clarify the concept of leap years with your students. Explain that leap years occur every four years when an extra day is added to February, resulting in a 29-day month. This addition helps synchronize our calendar with Earth’s revolutions around the Sun.
Step 3: Count Days per Month
To effectively calculate the average number of days per month, make sure students recognize each month’s total days:
– January (31), February (28 or 29), March (31), April (30), May (31), June (30), July (31), August (31), September (30), October (31), November (30), and December (31).
Step 4: Calculate Average
Guide your students through determining the average by adding up all the days per month. There are different ways to approach this, such as considering two cases: one an average considering only regular years with February having 28 days, and another one including leap years with February having 29 days.
Case 1: Regular years.
Sum of days = {31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31} = 365 days
Average number of days per month (regular years) = {365/12} ≈ 30.42
Case 2: Leap years.
Sum of days = {31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31} = 366 days
Average number of days per month (leap years) = {366/12} ≈ 30.5
Step 5: Discuss Real-World Applications
After teaching students about calculating the average number of days in a month, it’s important to mention real-world examples that showcase its significance. These can include planning for monthly budgets, creating project timelines, or estimating salary calculations.
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# Explanations for why someone cannot divide by $x-4$ for $x(x-4)=x(x-4)(x-5)$
A student divides both sides by $x-4$ and lost a solution $x=4$. How could you explain to the student that they are not allowed to divide by $x-4$
Here is the problem:
$x(x-4)=x(x-4)(x-5)$
I am having a hard time putting this in words for some reason. We know if the student divides by $x-4$ on both sides, not only do they lose a solution but technically they are dividing by $0$.
Does anyone have any other explanations
• Well, in the present problem, $x=4$ is obviously a solution, right? If you now devide by $x-4$, it suddenly isn't anymore. Isn't this explanation enough? – Friedrich Philipp Feb 27 '16 at 19:16
• Tell him that if he doesn't already know what $x$ is then he doesn't know that $x-4 \ne 0$ and so he doesn't know whether he can divide by it or not . Tell him to write in complete sentences,not "point form", emphasizing that he must include every "$\implies$" and "$\iff$". – DanielWainfleet Feb 27 '16 at 20:45
You can divide by $x-4$ provided $x\neq 4$. For $x=4$ you need an another case.
The thinking goes as follows: I can divide by a number not equal to zero. I divide by $x-4$. When $x-4\neq0$ I can do that. When $x-4=0$ I check if $x=4$ is a solution.
Eventually you can show them this method
$$x(x-4)=x(x-4)(x-5)$$ $$x(x-4)-x(x-4)(x-5)=0$$ $$x(x-4)(1-(x-5))=0$$ $$-x(x-4)(x-6)=0\implies x\in\{0,4,6\}$$
They can divide by $x-4$. But the rest of their work following that division must include the restriction that $x \neq 4$:
$$x(x-4) = x(x-4)(x-5) \\ \implies x = x(x-5); x \neq 4 \\ \vdots$$
You lose a solution because $x=4$ is a solution (by inspection) and $x \neq 4$ is our restriction in this line of work.
The actual process of solving equations is simply a chain of biconditionals.
For example, $3x^2 + 3x = 9x \Longleftrightarrow 3x^2 - 6x = 0 \Longleftrightarrow 3x(x - 2) = 0 \Longleftrightarrow x = 2, 0$
In your example, the "chain of biconditionals" is broken.
It is not the case that $x(x-4)=x(x-4)(x-5) \Longleftrightarrow x = x(x-5)$. This fails for $x=4$, because division by zero is not permitted.
For the biconditional to hold, we must alter as follows: $x(x-4)=x(x-4)(x-5) \land x \neq 4 \Longleftrightarrow x = x(x-5)$
For the same reason, squaring both sides of an equation often leads to imprecise results, because the biconditional does not hold (the square function is not injective).
Before dividing by $x-4$, they should split the search for solutions into two parts: (i) $x-4 = 0$ and (ii) $x-4 \neq 0$.
In the first case, we quickly check that $x=4$ is a solution and add that to our bag.
In the second case, since $x \neq 4$ we can happily divide across by $x-4$ to get $x = x (x-5)$ and continue the search (keeping in mind that we now know that $x \neq 4$, not that it matters here).
Reorganizing things to one side, you have:
$$x(x-4)-x(x-4)(x-5)=0$$
Factoring yields:
$$x(x-4)(1-x+5)=0$$
If you have $ab=0$ this is true if and only if $a=0$ or $b=0$.
More generally, if $a_1a_2a_3\dots a_n=0$ this is true if and only if at least one of $a_1,a_2,a_3,\dots$ are zero.
So, the expression has solutions $x=0$ or $(x-4)=0$ or $(1-x+5)=0$
Simplifying, we have solutions $0,4,6$
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# More applications of the Z-Score NORMAL Distribution The Empirical Rule.
## Presentation on theme: "More applications of the Z-Score NORMAL Distribution The Empirical Rule."— Presentation transcript:
More applications of the Z-Score NORMAL Distribution The Empirical Rule
Normal Distribution? These density curves are symmetric, single-peaked, and bell-shaped. We capitalize Normal to remind you that these curves are special. Normal distribution is described by giving its mean μ and its standard deviation σ
Shape of the Normal curve The standard deviation σ controls the spread of a Normal curve
The Empirical Rule 68-95-99.7 rule In the Normal distribution with mean μ and standard deviation σ: Approximately 6 8% of the observations fall within σ of the mean μ. Approximately 9 5% of the observations fall within 2σ of μ. Approximately 9 9.7% of the observations fall within 3σ of μ.
The Normal Distribution and Empirical Rule
Example: YOUNG WOMEN’s HEIGHT The distribution of heights of young women aged 18 to 24 is approximately Normal with mean μ = 64.5 inches and standard deviation σ = 2.5 inches.
Importance of Normal Curve scores on tests taken by many people (such as SAT exams and many psychological tests), repeated careful measurements of the same quantity, and characteristics of biological populations (such as yields of corn and lengths of animal pregnancies). even though many sets of data follow a Normal distribution, many do not. Most income distributions, for example, are skewed to the right and so are not Normal
Standard Normal distribution Standard Normal Distribution The standard Normal distribution is the Normal distribution N (0, 1) with mean 0 and standard deviation
Standard Normal Calculation The Standard Normal Table Table A Table A is a table of areas under the standard Normal curve. The table entry for each value z is the area under the curve to the left of z.
Area to the LEFT Using the standard Normal table Problem: Find the proportion of observations from the standard Normal distribution that are less than 2.22. illustrates the relationship between the value z = 2.22 and the area 0.9868. How to use the table of values
illustrates the relationship between the value z = 2.22 and the area 0.9868.
Example Area to the RIGHT Using the standard Normal table Problem: Find the proportion of observations from the standard Normal distribution that are greater than −2.15 z = −2.15 Area = 0.0158 Area = 1-0.0158 Area =.9842
Practice (a) z < 2.85 (b) z > 2.85 (c) z > −1.66 (d) −1.66 < z < 2.85 (a) 0.9978. (b) 0.0022. (c) 0.9515. (d) 0.9493.
CODY’S quiz score relative to his classmates 79818077738374937880756773 778386907985898477728382x z = 0.99 Area =.8389 Cody ’ s actual score relative to the other students who took the same test is 84%
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### A Bayesian Treasure Hunt
Q: Four friends A,B,C,D learn that a square tract of land they own has a treasure buried somewhere. They divide the tract of land into four equal quadrants, estimate that the probability in each of those square tracts as $$\{r,r,r,p\}$$. A starts digging his quadrant first. The probability that A might miss the treasure given the dig is $$q$$. Having dug out his quadrant, A exclaims that he still hasn't found the treasure. What is the updated probabilities of
1. A's quadrant has the treasure?
2. B's quadrant has the treasure?
Fifty Challenging Problems in Probability with Solutions (Dover Books on Mathematics)
A: This is a classic Bayesian puzzle. The fact that A has dug up his quadrant and revealed that there is no treasure leaves open the possibility that he might have missed the treasure. Yet it is information that must be worth something which will help B,C,D update their belief that the treasure is present in their quadrant. First off lets calculate A's updated belief. A's prior is $$p$$. Let $$T$$ denote the event that the treasure is in A's quadrant given nothing showed up during the dig. Let $$M$$ denote the event that A's dig resulted in a miss. We know from Bayes theorem
$$P(T|M) = \frac{P(M|T)P(T)}{P(M)}$$
$$P(M)$$ can occur in two ways. Either, the treasure just isn't there or it was there and A missed it. So, $$P(M)$$ works out as
$$P(M) = (1-q)\times p + (1 - p) = 1 - pq$$
$$P(M|T) = 1 - q$$ and $$P(T)=p$$. So A's updated belief that the treasure still resides in his quadrant is
$$P(T|M) = \frac{p(1-q)}{1 - pq}$$
As far as B is concerned, let $$T_{B}$$ denote the event that the treasure is in his quadrant. Casting this in the Bayesian framework yields
$$P(T_{B}|M) = \frac{P(M|T_{B}) P(T_{B})}{P(M)}$$
Note, the denominator remains the same. The numerator works out as: $$P(M|T_{B}) = 1$$ and $$P(M) = r$$. So the update rule for B (and for that matter any other person) is
$$P(T_{B}|M) = \frac{r}{1 - pq}$$
Check out some of the best books to learn the art of Probability
Fifty Challenging Problems in Probability with Solutions (Dover Books on Mathematics)
This book is a great compilation that covers quite a bit of puzzles. What I like about these puzzles are that they are all tractable and don't require too much advanced mathematics to solve.
Introduction to Algorithms
This is a book on algorithms, some of them are probabilistic. But the book is a must have for students, job candidates even full time engineers & data scientists
Introduction to Probability Theory
Overall an excellent book to learn probability, well recommended for undergrads and graduate students
An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd Edition
This is a two volume book and the first volume is what will likely interest a beginner because it covers discrete probability. The book tends to treat probability as a theory on its own
The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists (and Everyone Else!)
A good book for graduate level classes: has some practice problems in them which is a good thing. But that doesn't make this book any less of buy for the beginner.
Introduction to Probability, 2nd Edition
A good book to own. Does not require prior knowledge of other areas, but the book is a bit low on worked out examples.
Bundle of Algorithms in Java, Third Edition, Parts 1-5: Fundamentals, Data Structures, Sorting, Searching, and Graph Algorithms (3rd Edition) (Pts. 1-5)
An excellent resource (students, engineers and even entrepreneurs) if you are looking for some code that you can take and implement directly on the job
Understanding Probability: Chance Rules in Everyday Life
This is a great book to own. The second half of the book may require some knowledge of calculus. It appears to be the right mix for someone who wants to learn but doesn't want to be scared with the "lemmas"
Data Mining: Practical Machine Learning Tools and Techniques, Third Edition (The Morgan Kaufmann Series in Data Management Systems)
This one is a must have if you want to learn machine learning. The book is beautifully written and ideal for the engineer/student who doesn't want to get too much into the details of a machine learned approach but wants a working knowledge of it. There are some great examples and test data in the text book too.
This is a good book if you are new to statistics & probability while simultaneously getting started with a programming language. The book supports R and is written in a casual humorous way making it an easy read. Great for beginners. Some of the data on the companion website could be missing.
Covered in this book are the central limit theorem and other graduate topics in probability. You will need to brush up on some mathematics before you dive in but most of that can be done online
This book has been yellow-flagged with some issues: including sequencing of content that could be an issue. But otherwise its good
### The Best Books to Learn Probability
If you are looking to buy some books in probability here are some of the best books to learn the art of Probability
The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists (and Everyone Else!)
A good book for graduate level classes: has some practice problems in them which is a good thing. But that doesn't make this book any less of buy for the beginner.
An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd Edition
This is a two volume book and the first volume is what will likely interest a beginner because it covers discrete probability. The book tends to treat probability as a theory on its own
Discovering Statistics Using R
This is a good book if you are new to statistics & probability while simultaneously getting started with a programming language. The book supports R and is written in a casual humorous way making it an easy read. Great for beginners. Some of the data on the companion website could be missing.
Fifty Challenging Probl…
### The Best Books for Time Series Analysis
If you are looking to learn time series analysis, the following are some of the best books in time series analysis.
Introductory Time Series with R (Use R!)
This is good book to get one started on time series. A nice aspect of this book is that it has examples in R and some of the data is part of standard R packages which makes good introductory material for learning the R language too. That said this is not exactly a graduate level book, and some of the data links in the book may not be valid.
Econometrics
A great book if you are in an economics stream or want to get into it. The nice thing in the book is it tries to bring out a oneness in all the methods used. Econ majors need to be up-to speed on the grounding mathematics for time series analysis to use this book. Outside of those prerequisites, this is one of the best books on econometrics and time series analysis.
Pattern Recognition and Machine Learning (Information Science and Statistics)
This is excelle…
### The Best Books for Linear Algebra
The following are some good books to own in the area of Linear Algebra.
Linear Algebra (2nd Edition)
This is the gold standard for linear algebra at an undergraduate level. This book has been around for quite sometime a great book to own.
Linear Algebra: A Modern Introduction
Good book if you want to learn more on the subject of linear algebra however typos in the text could be a problem.
Linear Algebra (Dover Books on Mathematics)
An excellent book to own if you are looking to get into, or want to understand linear algebra. Please keep in mind that you need to have some basic mathematical background before you can use this book.
Linear Algebra Done Right (Undergraduate Texts in Mathematics)
A great book that exposes the method of proof as it used in Linear Algebra. This book is not for the beginner though. You do need some prior knowledge of the basics at least. It would be a good add-on to an existing course you are doing in Linear Algebra.
Linear Algebra, 4th Edition
This is good book …
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# Difference between revisions of "2021 AMC 12A Problems/Problem 6"
## Problem
A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is $\frac13$. When $4$ black cards are added to the deck, the probability of choosing red becomes $\frac14$. How many cards were in the deck originally?
$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }15 \qquad \textbf{(E) }18$
## Solution
If the probability of choosing a red card is $\frac{1}{3}$, the red and black cards are in ratio $1:2$. This means at the beginning there are $x$ red cards and $2x$ black cards.
After $4$ black cards are added, there are $2x+4$ black cards. This time, the probability of choosing a red card is $\frac{1}{4}$ so the ratio of red to black cards is $1:3$. This means in the new deck the number of black cards is also $3x$ for the same $x$ red cards.
So, $3x = 2x + 4$ and $x=4$ meaning there are $4$ red cards in the deck at the start and $2(4) = 8$ black cards.
So the answer is $8+4 = 12 = \boxed{\textbf{(C)}}$.
--abhinavg0627
## Solution 2 (Observations)
Suppose there were $x$ cards in the deck originally. Now, the deck has $x+4$ cards, which must be a multiple of $4.$
Only $12+4=16$ is a multiple of $4.$ So, the answer is $\boxed{\textbf{(C) }12}.$
~MRENTHUSIASM
If there were $6$ cards in the deck originally, then there were $6\cdot\frac13=2$ red cards in the deck originally.
Now, the deck has $6+4=10$ cards, and $\frac{2}{10}\neq\frac{1}{4}.$ So, $\textbf{(A) }6$ is incorrect.
If there were $9$ cards in the deck originally, then there were $9\cdot\frac13=3$ red cards in the deck originally.
Now, the deck has $9+4=13$ cards, and $\frac{3}{13}\neq\frac{1}{4}.$ So, $\textbf{(B) }9$ is incorrect.
If there were $12$ cards in the deck originally, then there were $12\cdot\frac13=4$ red cards in the deck originally.
Now, the deck has $12+4=16$ cards, and $\frac{4}{16}=\frac{1}{4}.$ So, $\boxed{\textbf{(C) }12}$ is correct. WOOHOO! For completeness, we will check $\textbf{(D) }$ and $\textbf{(E)}.$ If you decide to use this approach on the real test, you don't have to do that, as you want to save more time.
If there were $15$ cards in the deck originally, then there were $15\cdot\frac13=5$ red cards in the deck originally.
Now, the deck has $15+4=19$ cards, and $\frac{5}{19}\neq\frac{1}{4}.$ So, $\textbf{(D) }15$ is incorrect.
If there were $18$ cards in the deck originally, then there were $18\cdot\frac13=6$ red cards in the deck originally.
Now, the deck has $18+4=22$ cards, and $\frac{6}{22}\neq\frac{1}{4}.$ So, $\textbf{(E) }18$ is incorrect.
~MRENTHUSIASM
~ pi_is_3.14
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# Difference between revisions of "2017 AMC 10A Problems/Problem 11"
## Problem
The region consisting of all points in three-dimensional space within $3$ units of line segment $\overline{AB}$ has volume $216\pi$. What is the length $\textit{AB}$?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$
## Solution 1
In order to solve this problem, we must first visualize what the region contained looks like. We know that, in a three dimensional plane, the region consisting of all points within $3$ units of a point would be a sphere with radius $3$. However, we need to find the region containing all points within $3$ units of a segment. It can be seen that our region is a cylinder with two hemispheres on either end. We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal $216 \pi$):
$\frac{4 \pi }{3} \cdot 3^3+9 \pi x=216 \pi$, where $x$ is equal to the length of our line segment.
Solving, we find that $x = \boxed{\textbf{(D)}\ 20}$.
## Solution 2
Because this is just a cylinder and $2$ "half spheres", and the radius is $3$, the volume of the $2$ half spheres is $\frac{4(3^3)\pi}{3} = 36 \pi$. Since we also know that the volume of this whole thing is $216 \pi$, we do $216-36$ to get $180 \pi$ as the volume of the cylinder. Thus the height is $180 \pi$ over the base, or $\frac{180 \pi}{9\pi}=20$, so our answer is $\boxed{\textbf{(D)}\ 20}.$
~Minor edit by virjoy2001
## See Also
2017 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# NCERT Solutions for Class 9 Maths Chapter 5 - Introduction to Euclid's Geometry
How to prove that two distinct lines can’t have more than one point in common? Find out by using our NCERT Solutions for CBSE Class 9 Mathematics Chapter 5 Introduction to Euclid’s Geometry. Learn about Euclid’s five postulates. Revisit the relation between theorem and axiom by practising with TopperLearning’s chapter-specific solutions.
Textbook solutions for CBSE Class 9 Maths will be useful when you sit to revise your lessons for class tests or exams. Other than NCERT Maths solutions, our video lessons, chapter notes and online practice tests will help you to understand theorems and postulates.
Updated NCERT Textbook Solutions Coming Soon!
Page / Exercise
## Chapter 5 - Introduction to Euclid's Geometry Exercise Ex. 5.1
Solution 1
(i) False as through a single point infinite number of lines can pass. Here in the figure we may find that there are infinite numbers of lines, which are passing through a single point P.
(ii) False as through two district points only one line can pass. Here in the following figure we may find that there is only one single line that can pass through two distinct points P and Q.
(iii) True a terminated line can be produced indefinitely on both the sides.
Let AB be a terminated line. We may find that it can be produced indefinitely on both the sides.
(iv) True if two circles are equal, their centre and circumference will coincide and hence radii will be equal.
(v) True. It is given that AB and XY are two terminated lines and both are equal to a third line PQ. Euclid's first axiom states that things which are equal to the same thing are equal to one another. So, the lines AB and XY will be equal to each other.
Solution 2
(i) Parallel Lines:
If the perpendicular distance between two lines is always constant, these are called parallel lines. In other words we may say that the lines which never intersect each other are called parallel lines.
To define parallel line, we must know about point, lines and distance between lines and point of intersection.
(ii) Perpendicular lines:
If two lines intersect each other at 90 , these are called perpendicular lines. We need to define line and the angle before defining perpendicular lines.
(iii) Line segment:
A straight line drawn from any point to any other point is called as line segment. To define line segment, we must know about point and line segment.
It is the distance between the centre of a circle to any point lying on the circle. To define radius of circle, we must know about point and circle.
(v) Square:
A square is a quadrilateral having all sides of equal length and all angles of same measure i.e. 90o . To define square, we must know about quadrilateral, side, and angle.
Solution 3
(i) There are undefined terms. They are consistent. As we have no information about line segment AB and we don't know about the position of third point C whether it is lying on the line segment AB or not, now two different cases are possible -
• The third point C lies on the line segment made by joining the points A and B.
• The third point C does not lie on the line segment made by joining the points A and B.
Yes, they are consistent as these are two different situations;
(i) the third point c lies on the line segment made by joining points A and B.
(ii)The third point does not lie on the line segment made by joining points A and B.
No, these postulates do not follow Euclid's postulates actually these are axiom.
Solution 4
Given that
AC = BC
AC+AC=BC+AC (equals are added on both sides) ... (i)
Here, (BC + AC) coincides with AB. We know that things which coincide with one another are equal to one another.
BC + AC = AB ... (ii)
We know that things which are equal to the same thing are equal to one another. So, from equation (i) and equation (ii), we have
AC + AC = AB
2AC = AB
AC=AB
Solution 5
Let us assume there are two mid-points C and D
C is mid point of AB
AC = CB
AC + AC = BC + AC (equals are added on both sides) ... (1)
Here, (BC + AC) coincides with AB. We know that things which coincide with one another are equal to one another.
BC + AC = AB ... (2)
We know that things which are equal to the same thing are equal to one another. So, from equation (1) and equation (2), we have
AC + AC = AB
2AC = AB ... (3)
Similarly by taking D as the mid-point of AB, we can prove that
From equation (3) and (4), we have
2AC = 2AD (Things which are equal to the same thing are equal to one another.)
AC = AD (Things which are double of the same things are equal to one another.)
It is possible only when point C and D are representing a single point.
Hence our assumption is wrong and there can be only one mid-point of a given line segment.
Solution 6
From the figure we may observe that
AC = AB + BC
BD = BC + CD
Given that AC = BD
AB + BC = BC + CD (1)
According to Euclid's axiom, we know that when equals are subtracted from equals, remainders are also equal.
Subtracting BC from the equation (1), we have
AB + BC - BC = BC + CD - BC
AB = CD
Solution 7
Axiom 5: The whole is greater than the part.
Let t is representing a whole quantity and only a, b, c are parts of it.
Now t = a + b + c
Clearly t will be greater than all its parts a, b and c
So, it is rightly said that the whole is greater than the part.
Let us consider another example. Let us consider a continent, let say Asia. Let us consider a country India which belongs to Asia. Now, India is a part of Asia and we can also observe that Asia is greater than India. That is why we can say that the whole is greater than the part. This is true for any thing in any part of the world, this is a universal truth.
## Chapter 5 - Introduction to Euclid's Geometry Exercise Ex. 5.2
Solution 1
Two lines are said to be parallel if they are equidistant from one other and they do not have any point of intersection. In order to understand it easily let us take any line l and a point P not on l. Then, by Play fair's axiom (equivalent to the fifth postulate), we know that there is a unique line m through P which is parallel to l.
Now, the distance of a point from a line is the length of the perpendicular from the point to the line. Let AB be the distance of any point on m from l and CD be the distance of any point on l from m. We can observe that AB = CD. In this way the distance will be the same for any point on m from l and any point on l from m. So, these two lines are everywhere equidistant from one another.
Solution 2
Yes
According to Euclid's 5th postulates when n line falls on l and m and if
, then producing line l and m further will meet in
the side of 1 and 2 which is less than 180
If ,
Now, the lines l and m neither meet at the side of 1 and 2 nor at the side of 3 and 4. That means the lines l and m will never intersect each other. Therefore, we can say that the lines are parallel.
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# CBSE Class 5 Maths Activity 1
Read and download CBSE Class 5 Maths Activity 1 in NCERT book for Class 5 Mathematics. You can download latest NCERT eBooks chapter wise in PDF format free from Studiestoday.com. This Mathematics textbook for Class 5 is designed by NCERT and is very useful for students. Please also refer to the NCERT solutions for Class 5 Mathematics to understand the answers of the exercise questions given at the end of this chapter
## NCERT Book for Class 5 Mathematics Activity 1
Class 5 Mathematics students should refer to the following NCERT Book Activity 1 in Class 5. This NCERT Book for Class 5 Mathematics will be very useful for exams and help you to score good marks
### Activity 1 NCERT Book Class 5
To make a set of Tangram cutouts by paper folding and cutting and to make geometric figures using its pieces.
Learning objective :To familiarise with geometrical figures.
Pre-requisite :The knowledge of basic geometrical shapes : triangle, square, rectangle.
Materials required :An 8x8 grid, a pair of scissors, a pen/pencil and a ruler.
Procedure :1(a) To make a Tangram set.
Step 1.Take a 8x8 grid and label it as ABCD, which is a square.[Fig. 1(a)]
Step 2.Fold the square ABCD along BD such that point 'C' falls on point 'A' and mark the crease BD.
Step 3.Unfold the grid and cut along the crease.
Step 4.Take any one cutout obtained in Step 3 say ABD and fold it through the vertex A so that the two end points of the longest side meet exactly and mark the crease.
Step 5.Unfold and cut along the crease and label the two parts as 1 & 2 [Fig. 1(b)].
Step 6.Take the other cutout obtained in Step 3 and fold it in such a way that point containing a right angle exactly falls on the mid point of its longest side.
Step 7.Unfold and cut along the crease formed. Label the triangle cutouts as 3 [Fig. 1(c)].
Step 8.Take the remaining cutout obtained in Step 6 and fold along the lines drawn. [Fig.1(d).]
Step 9.Unfold and cut along the creases. Label the cutouts as 4, 5, 6 & 7. [Fig. 1(e).]
Please refer to the link below - CBSE Class 5 Maths Activity 1
NCERT Class 5 Maths The Fish Tale
NCERT Class 5 Maths Shapes and Angles
NCERT Class 5 Maths How Many Squares
NCERT Class 5 Maths Parts and Wholes
NCERT Class 5 Maths Does it look the same
Math-Magic Chapter 6 Be My Multiple Ill be Your Factor
NCERT Class 5 Maths Can you see the pattern
NCERT Class 5 Maths Mapping your way
NCERT Class 5 Maths Boxes and Sketches
NCERT Class 5 Maths Tenths and Hundredths
NCERT Class 5 Maths Area and its Boundary
NCERT Class 5 Maths Smart Charts
NCERT Class 5 Maths Ways to Multiply and Divide
NCERT Class 5 Maths How Big How Heavy
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# Exponents and Scientific Notation
Module by: First Last. E-mail the author
Summary: In this section students will:
• Use the product rule of exponents.
• Use the quotient rule of exponents.
• Use the power rule of exponents.
• Use the zero exponent rule of exponents.
• Use the negative rule of exponents.
• Find the power of a product and a quotient.
• Simplify exponential expressions.
• Use scientific notation.
Note: You are viewing an old style version of this document. The new style version is available here.
Mathematicians, scientists, and economists commonly encounter very large and very small numbers. But it may not be obvious how common such figures are in everyday life. For instance, a pixel is the smallest unit of light that can be perceived and recorded by a digital camera. A particular camera might record an image that is 2,048 pixels by 1,536 pixels, which is a very high resolution picture. It can also perceive a color depth (gradations in colors) of up to 48 bits per frame, and can shoot the equivalent of 24 frames per second. The maximum possible number of bits of information used to film a one-hour (3,600-second) digital film is then an extremely large number.
Using a calculator, we enter 2,048×1,536×48×24×3,600 2,048×1,536×48×24×3,600 and press ENTER. The calculator displays 1.304596316E13. What does this mean? The “E13” portion of the result represents the exponent 13 of ten, so there are a maximum of approximately 1.3× 10 13 1.3× 10 13 bits of data in that one-hour film. In this section, we review rules of exponents first and then apply them to calculations involving very large or small numbers.
## Using the Product Rule of Exponents
Consider the product x 3 x 4 . x 3 x 4 . Both terms have the same base, x, but they are raised to different exponents. Expand each expression, and then rewrite the resulting expression.
x3x4 = xxx3 factorsxxxx 4 factors = xxxxxxx7 factors = x7 x3x4 = xxx3 factorsxxxx 4 factors = xxxxxxx7 factors = x7
The result is that x 3 x 4 = x 3+4 = x 7 . x 3 x 4 = x 3+4 = x 7 .
Notice that the exponent of the product is the sum of the exponents of the terms. In other words, when multiplying exponential expressions with the same base, we write the result with the common base and add the exponents. This is the product rule of exponents.
a m a n = a m+n a m a n = a m+n
Now consider an example with real numbers.
2 3 2 4 = 2 3+4 = 2 7 2 3 2 4 = 2 3+4 = 2 7
We can always check that this is true by simplifying each exponential expression. We find that 2 3 2 3 is 8, 2 4 2 4 is 16, and 2 7 2 7 is 128. The product 816 816 equals 128, so the relationship is true. We can use the product rule of exponents to simplify expressions that are a product of two numbers or expressions with the same base but different exponents.
### A General Note: The Product Rule of Exponents:
For any real number a a and natural numbers m m and n, n, the product rule of exponents states that
a m a n = a m+n a m a n = a m+n
### Example 1
#### Problem 1
##### Using the Product Rule
Write each of the following products with a single base. Do not simplify further.
1. t 5 t 3 t 5 t 3
2. ( −3 ) 5 ( −3 ) ( −3 ) 5 ( −3 )
3. x 2 x 5 x 3 x 2 x 5 x 3
##### Solution
Use the product rule to simplify each expression.
1. t 5 t 3 = t 5+3 = t 8 t 5 t 3 = t 5+3 = t 8
2. ( −3 ) 5 ( −3 )= ( −3 ) 5 ( −3 ) 1 = ( −3 ) 5+1 = ( −3 ) 6 ( −3 ) 5 ( −3 )= ( −3 ) 5 ( −3 ) 1 = ( −3 ) 5+1 = ( −3 ) 6
3. x 2 x 5 x 3 x 2 x 5 x 3
At first, it may appear that we cannot simplify a product of three factors. However, using the associative property of multiplication, begin by simplifying the first two.
x 2 x 5 x 3 =( x 2 x 5 ) x 3 =( x 2+5 ) x 3 = x 7 x 3 = x 7+3 = x 10 x 2 x 5 x 3 =( x 2 x 5 ) x 3 =( x 2+5 ) x 3 = x 7 x 3 = x 7+3 = x 10
Notice we get the same result by adding the three exponents in one step.
x 2 x 5 x 3 = x 2+5+3 = x 10 x 2 x 5 x 3 = x 2+5+3 = x 10
### Try It:
#### Exercise 1
Write each of the following products with a single base. Do not simplify further.
1. k 6 k 9 k 6 k 9
2. ( 2 y ) 4 ( 2 y ) ( 2 y ) 4 ( 2 y )
3. t 3 t 6 t 5 t 3 t 6 t 5
##### Solution
1. k 15 k 15
2. ( 2 y ) 5 ( 2 y ) 5
3. t 14 t 14
## Using the Quotient Rule of Exponents
The quotient rule of exponents allows us to simplify an expression that divides two numbers with the same base but different exponents. In a similar way to the product rule, we can simplify an expression such as y m y n , y m y n , where m>n. m>n. Consider the example y 9 y 5 . y 9 y 5 . Perform the division by canceling common factors.
y9y5 = yyyyyyyyyyyyyy = yyyyyyyyyyyyyy = yyyy1 = y4 y9y5 = yyyyyyyyyyyyyy = yyyyyyyyyyyyyy = yyyy1 = y4
Notice that the exponent of the quotient is the difference between the exponents of the divisor and dividend.
a m a n = a mn a m a n = a mn
In other words, when dividing exponential expressions with the same base, we write the result with the common base and subtract the exponents.
y 9 y 5 = y 95 = y 4 y 9 y 5 = y 95 = y 4
For the time being, we must be aware of the condition m>n. m>n. Otherwise, the difference mn mn could be zero or negative. Those possibilities will be explored shortly. Also, instead of qualifying variables as nonzero each time, we will simplify matters and assume from here on that all variables represent nonzero real numbers.
### A General Note: The Quotient Rule of Exponents:
For any real number a a and natural numbers m m and n, n, such that m>n, m>n, the quotient rule of exponents states that
a m a n = a mn a m a n = a mn
(10)
### Example 2
#### Problem 1
##### Using the Quotient Rule
Write each of the following products with a single base. Do not simplify further.
1. ( −2 ) 14 ( −2 ) 9 ( −2 ) 14 ( −2 ) 9
2. t 23 t 15 t 23 t 15
3. ( z 2 ) 5 z 2 ( z 2 ) 5 z 2
##### Solution
Use the quotient rule to simplify each expression.
1. ( −2 ) 14 ( −2 ) 9 = ( −2 ) 149 = ( −2 ) 5 ( −2 ) 14 ( −2 ) 9 = ( −2 ) 149 = ( −2 ) 5
2. t 23 t 15 = t 2315 = t 8 t 23 t 15 = t 2315 = t 8
3. ( z 2 ) 5 z 2 = ( z 2 ) 51 = ( z 2 ) 4 ( z 2 ) 5 z 2 = ( z 2 ) 51 = ( z 2 ) 4
### Try It:
#### Exercise 2
Write each of the following products with a single base. Do not simplify further.
1. s 75 s 68 s 75 s 68
2. ( −3 ) 6 −3 ( −3 ) 6 −3
3. ( e f 2 ) 5 ( e f 2 ) 3 ( e f 2 ) 5 ( e f 2 ) 3
##### Solution
1. s 7 s 7
2. ( −3 ) 5 ( −3 ) 5
3. ( e f 2 ) 2 ( e f 2 ) 2
## Using the Power Rule of Exponents
Suppose an exponential expression is raised to some power. Can we simplify the result? Yes. To do this, we use the power rule of exponents. Consider the expression ( x 2 ) 3 . ( x 2 ) 3 . The expression inside the parentheses is multiplied twice because it has an exponent of 2. Then the result is multiplied three times because the entire expression has an exponent of 3.
(x2)3 = (x2)(x2)(x2)3 factors = (xx2 factors)(xx 2 factors)(xx2 factors)3 factors = xxxxxx = x6 (x2)3 = (x2)(x2)(x2)3 factors = (xx2 factors)(xx 2 factors)(xx2 factors)3 factors = xxxxxx = x6
The exponent of the answer is the product of the exponents: ( x 2 ) 3 = x 23 = x 6 . ( x 2 ) 3 = x 23 = x 6 . In other words, when raising an exponential expression to a power, we write the result with the common base and the product of the exponents.
( a m ) n = a mn ( a m ) n = a mn
Be careful to distinguish between uses of the product rule and the power rule. When using the product rule, different terms with the same bases are raised to exponents. In this case, you add the exponents. When using the power rule, a term in exponential notation is raised to a power. In this case, you multiply the exponents.
Product Rule Power Rule 5 3 5 4 = 5 3+4 = 5 7 but ( 5 3 ) 4 = 5 34 = 5 12 x 5 x 2 = x 5+2 = x 7 but ( x 5 ) 2 = x 52 = x 10 (3a) 7 (3a) 10 = (3a) 7+10 = (3a) 17 but ( (3a) 7 ) 10 = (3a) 710 = (3a) 70 Product Rule Power Rule 5 3 5 4 = 5 3+4 = 5 7 but ( 5 3 ) 4 = 5 34 = 5 12 x 5 x 2 = x 5+2 = x 7 but ( x 5 ) 2 = x 52 = x 10 (3a) 7 (3a) 10 = (3a) 7+10 = (3a) 17 but ( (3a) 7 ) 10 = (3a) 710 = (3a) 70
### A General Note: The Power Rule of Exponents:
For any real number a a and positive integers m m and n, n, the power rule of exponents states that
( a m ) n = a mn ( a m ) n = a mn
(14)
### Example 3
#### Problem 1
##### Using the Power Rule
Write each of the following products with a single base. Do not simplify further.
1. ( x 2 ) 7 ( x 2 ) 7
2. ( ( 2t ) 5 ) 3 ( ( 2t ) 5 ) 3
3. ( ( −3 ) 5 ) 11 ( ( −3 ) 5 ) 11
##### Solution
Use the power rule to simplify each expression.
1. ( x 2 ) 7 = x 27 = x 14 ( x 2 ) 7 = x 27 = x 14
2. ( ( 2t ) 5 ) 3 = ( 2t ) 53 = ( 2t ) 15 ( ( 2t ) 5 ) 3 = ( 2t ) 53 = ( 2t ) 15
3. ( ( −3 ) 5 ) 11 = ( −3 ) 511 = ( −3 ) 55 ( ( −3 ) 5 ) 11 = ( −3 ) 511 = ( −3 ) 55
### Try It:
#### Exercise 3
Write each of the following products with a single base. Do not simplify further.
1. ( ( 3y ) 8 ) 3 ( ( 3y ) 8 ) 3
2. ( t 5 ) 7 ( t 5 ) 7
3. ( ( g ) 4 ) 4 ( ( g ) 4 ) 4
##### Solution
1. ( 3y ) 24 ( 3y ) 24
2. t 35 t 35
3. ( g ) 16 ( g ) 16
## Using the Zero Exponent Rule of Exponents
Return to the quotient rule. We made the condition that m>n m>n so that the difference mn mn would never be zero or negative. What would happen if m=n? m=n? In this case, we would use the zero exponent rule of exponents to simplify the expression to 1. To see how this is done, let us begin with an example.
t 8 t 8 = t 8 t 8 =1 t 8 t 8 = t 8 t 8 =1
If we were to simplify the original expression using the quotient rule, we would have
t 8 t 8 = t 88 = t 0 t 8 t 8 = t 88 = t 0
If we equate the two answers, the result is t 0 =1. t 0 =1. This is true for any nonzero real number, or any variable representing a real number.
a 0 =1 a 0 =1
The sole exception is the expression 0 0 . 0 0 . This appears later in more advanced courses, but for now, we will consider the value to be undefined.
### A General Note: The Zero Exponent Rule of Exponents:
For any nonzero real number a, a, the zero exponent rule of exponents states that
a 0 =1 a 0 =1
(18)
### Example 4
#### Problem 1
##### Using the Zero Exponent Rule
Simplify each expression using the zero exponent rule of exponents.
1. c 3 c 3 c 3 c 3
2. −3 x 5 x 5 −3 x 5 x 5
3. ( j 2 k ) 4 ( j 2 k ) ( j 2 k ) 3 ( j 2 k ) 4 ( j 2 k ) ( j 2 k ) 3
4. 5 ( r s 2 ) 2 ( r s 2 ) 2 5 ( r s 2 ) 2 ( r s 2 ) 2
##### Solution
Use the zero exponent and other rules to simplify each expression.
1. c3c3 = c33 = c0 = 1 c3c3 = c33 = c0 = 1
2. −3x5x5 = −3x5x5 = −3x55 = −3x0 = −31 = −3 −3x5x5 = −3x5x5 = −3x55 = −3x0 = −31 = −3
3. (j2k)4(j2k)(j2k)3 = (j2k)4(j2k)1+3 Use the product rule in the denominator. = (j2k)4(j2k)4 Simplify. = (j2k)44 Use the quotient rule. = (j2k)0 Simplify. = 1 (j2k)4(j2k)(j2k)3 = (j2k)4(j2k)1+3 Use the product rule in the denominator. = (j2k)4(j2k)4 Simplify. = (j2k)44 Use the quotient rule. = (j2k)0 Simplify. = 1
4. 5(r s 2)2(rs2)2 = 5(rs2)22 Use the quotient rule. = 5(rs2)0 Simplify. = 51 Use the zero exponent rule. = 5 Simplify. 5(r s 2)2(rs2)2 = 5(rs2)22 Use the quotient rule. = 5(rs2)0 Simplify. = 51 Use the zero exponent rule. = 5 Simplify.
### Try It:
#### Exercise 4
Simplify each expression using the zero exponent rule of exponents.
1. t 7 t 7 t 7 t 7
2. ( d e 2 ) 11 2 ( d e 2 ) 11 ( d e 2 ) 11 2 ( d e 2 ) 11
3. w 4 w 2 w 6 w 4 w 2 w 6
4. t 3 t 4 t 2 t 5 t 3 t 4 t 2 t 5
1. 1 1
2. 1 2 1 2
3. 1 1
4. 1 1
## Using the Negative Rule of Exponents
Another useful result occurs if we relax the condition that m>n m>n in the quotient rule even further. For example, can we simplify h 3 h 5 ? h 3 h 5 ? When m<n m<n —that is, where the difference mn mn is negative—we can use the negative rule of exponents to simplify the expression to its reciprocal.
Divide one exponential expression by another with a larger exponent. Use our example, h 3 h 5 . h 3 h 5 .
h3h5 = hhhhhhhh = hhhhhhhh = 1hh = 1h2 h3h5 = hhhhhhhh = hhhhhhhh = 1hh = 1h2
If we were to simplify the original expression using the quotient rule, we would have
h3h5 = h35 = h−2 h3h5 = h35 = h−2
Putting the answers together, we have h −2 = 1 h 2 . h −2 = 1 h 2 . This is true for any nonzero real number, or any variable representing a nonzero real number.
A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar—from numerator to denominator or vice versa.
a n = 1 a n and a n = 1 a n a n = 1 a n and a n = 1 a n
We have shown that the exponential expression a n a n is defined when n n is a natural number, 0, or the negative of a natural number. That means that a n a n is defined for any integer n. n. Also, the product and quotient rules and all of the rules we will look at soon hold for any integer n. n.
### A General Note: The Negative Rule of Exponents:
For any nonzero real number a a and natural number n, n, the negative rule of exponents states that
a n = 1 a n a n = 1 a n
(22)
### Example 5
#### Problem 1
##### Using the Negative Exponent Rule
Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.
1. θ 3 θ 10 θ 3 θ 10
2. z 2 z z 4 z 2 z z 4
3. ( −5 t 3 ) 4 ( −5 t 3 ) 8 ( −5 t 3 ) 4 ( −5 t 3 ) 8
##### Solution
1. θ 3 θ 10 = θ 310 = θ −7 = 1 θ 7 θ 3 θ 10 = θ 310 = θ −7 = 1 θ 7
2. z 2 z z 4 = z 2+1 z 4 = z 3 z 4 = z 34 = z −1 = 1 z z 2 z z 4 = z 2+1 z 4 = z 3 z 4 = z 34 = z −1 = 1 z
3. ( −5 t 3 ) 4 ( −5 t 3 ) 8 = ( −5 t 3 ) 48 = ( −5 t 3 ) −4 = 1 ( −5 t 3 ) 4 ( −5 t 3 ) 4 ( −5 t 3 ) 8 = ( −5 t 3 ) 48 = ( −5 t 3 ) −4 = 1 ( −5 t 3 ) 4
### Try It:
#### Exercise 5
Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.
1. ( −3t ) 2 ( −3t ) 8 ( −3t ) 2 ( −3t ) 8
2. f 47 f 49 f f 47 f 49 f
3. 2 k 4 5 k 7 2 k 4 5 k 7
##### Solution
1. 1 ( −3t ) 6 1 ( −3t ) 6
2. 1 f 3 1 f 3
3. 2 5 k 3 2 5 k 3
### Example 6
#### Problem 1
##### Using the Product and Quotient Rules
Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.
1. b 2 b −8 b 2 b −8
2. ( x ) 5 ( x ) −5 ( x ) 5 ( x ) −5
3. −7z ( −7z ) 5 −7z ( −7z ) 5
##### Solution
1. b 2 b −8 = b 28 = b −6 = 1 b 6 b 2 b −8 = b 28 = b −6 = 1 b 6
2. ( x ) 5 ( x ) −5 = ( x ) 55 = ( x ) 0 =1 ( x ) 5 ( x ) −5 = ( x ) 55 = ( x ) 0 =1
3. −7z ( −7z ) 5 = ( −7z ) 1 ( −7z ) 5 = ( −7z ) 15 = ( −7z ) −4 = 1 ( −7z ) 4 −7z ( −7z ) 5 = ( −7z ) 1 ( −7z ) 5 = ( −7z ) 15 = ( −7z ) −4 = 1 ( −7z ) 4
### Try It:
#### Exercise 6
Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.
1. t −11 t 6 t −11 t 6
2. 25 12 25 13 25 12 25 13
##### Solution
1. t −5 = 1 t 5 t −5 = 1 t 5
2. 1 25 1 25
## Finding the Power of a Product
To simplify the power of a product of two exponential expressions, we can use the power of a product rule of exponents, which breaks up the power of a product of factors into the product of the powers of the factors. For instance, consider ( pq ) 3 . ( pq ) 3 . We begin by using the associative and commutative properties of multiplication to regroup the factors.
(pq)3 = (pq)(pq)(pq) 3 factors = pqpqpq = ppp 3 factors qqq 3 factors = p3q3 (pq)3 = (pq)(pq)(pq) 3 factors = pqpqpq = ppp 3 factors qqq 3 factors = p3q3
In other words, ( pq ) 3 = p 3 q 3 . ( pq ) 3 = p 3 q 3 .
### A General Note: The Power of a Product Rule of Exponents:
For any real numbers a a and b b and any integer n, n, the power of a product rule of exponents states that
( ab ) n = a n b n ( ab ) n = a n b n
(24)
### Example 7
#### Problem 1
##### Using the Power of a Product Rule
Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.
1. ( a b 2 ) 3 ( a b 2 ) 3
2. ( 2t ) 15 ( 2t ) 15
3. ( −2 w 3 ) 3 ( −2 w 3 ) 3
4. 1 ( −7z ) 4 1 ( −7z ) 4
5. ( e −2 f 2 ) 7 ( e −2 f 2 ) 7
##### Solution
Use the product and quotient rules and the new definitions to simplify each expression.
1. ( a b 2 ) 3 = ( a ) 3 ( b 2 ) 3 = a 13 b 23 = a 3 b 6 ( a b 2 ) 3 = ( a ) 3 ( b 2 ) 3 = a 13 b 23 = a 3 b 6
2. ( 2t ) 15 = ( 2 ) 15 ( t ) 15 = 2 15 t 15 =32,768 t 15 ( 2t ) 15 = ( 2 ) 15 ( t ) 15 = 2 15 t 15 =32,768 t 15
3. ( −2 w 3 ) 3 = ( −2 ) 3 ( w 3 ) 3 =−8 w 33 =−8 w 9 ( −2 w 3 ) 3 = ( −2 ) 3 ( w 3 ) 3 =−8 w 33 =−8 w 9
4. 1 ( −7z ) 4 = 1 ( −7 ) 4 ( z ) 4 = 1 2,401 z 4 1 ( −7z ) 4 = 1 ( −7 ) 4 ( z ) 4 = 1 2,401 z 4
5. ( e −2 f 2 ) 7 = ( e −2 ) 7 ( f 2 ) 7 = e −27 f 27 = e −14 f 14 = f 14 e 14 ( e −2 f 2 ) 7 = ( e −2 ) 7 ( f 2 ) 7 = e −27 f 27 = e −14 f 14 = f 14 e 14
### Try It:
#### Exercise 7
Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.
1. ( g 2 h 3 ) 5 ( g 2 h 3 ) 5
2. ( 5t ) 3 ( 5t ) 3
3. ( −3 y 5 ) 3 ( −3 y 5 ) 3
4. 1 ( a 6 b 7 ) 3 1 ( a 6 b 7 ) 3
5. ( r 3 s −2 ) 4 ( r 3 s −2 ) 4
##### Solution
1. g 10 h 15 g 10 h 15
2. 125 t 3 125 t 3
3. −27 y 15 −27 y 15
4. 1 a 18 b 21 1 a 18 b 21
5. r 12 s 8 r 12 s 8
## Finding the Power of a Quotient
To simplify the power of a quotient of two expressions, we can use the power of a quotient rule, which states that the power of a quotient of factors is the quotient of the powers of the factors. For example, let’s look at the following example.
( e −2 f 2 ) 7 = f 14 e 14 ( e −2 f 2 ) 7 = f 14 e 14
Let’s rewrite the original problem differently and look at the result.
(e−2f2)7 = ( f2 e2)7 = f14e14 (e−2f2)7 = ( f2 e2)7 = f14e14
It appears from the last two steps that we can use the power of a product rule as a power of a quotient rule.
( e 2 f 2 ) 7 = ( f 2 e 2 ) 7 = ( f 2 ) 7 ( e 2 ) 7 = f 27 e 27 = f 14 e 14 ( e 2 f 2 ) 7 = ( f 2 e 2 ) 7 = ( f 2 ) 7 ( e 2 ) 7 = f 27 e 27 = f 14 e 14
### A General Note: The Power of a Quotient Rule of Exponents:
For any real numbers a a and b b and any integer n, n, the power of a quotient rule of exponents states that
( a b ) n = a n b n ( a b ) n = a n b n
(28)
### Example 8
#### Problem 1
##### Using the Power of a Quotient Rule
Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.
1. ( 4 z 11 ) 3 ( 4 z 11 ) 3
2. ( p q 3 ) 6 ( p q 3 ) 6
3. ( −1 t 2 ) 27 ( −1 t 2 ) 27
4. ( j 3 k −2 ) 4 ( j 3 k −2 ) 4
5. ( m −2 n −2 ) 3 ( m −2 n −2 ) 3
##### Solution
1. ( 4 z 11 ) 3 = ( 4 ) 3 ( z 11 ) 3 = 64 z 113 = 64 z 33 ( 4 z 11 ) 3 = ( 4 ) 3 ( z 11 ) 3 = 64 z 113 = 64 z 33
2. ( p q 3 ) 6 = ( p ) 6 ( q 3 ) 6 = p 16 q 36 = p 6 q 18 ( p q 3 ) 6 = ( p ) 6 ( q 3 ) 6 = p 16 q 36 = p 6 q 18
3. ( −1 t 2 ) 27 = ( −1 ) 27 ( t 2 ) 27 = −1 t 227 = −1 t 54 = 1 t 54 ( −1 t 2 ) 27 = ( −1 ) 27 ( t 2 ) 27 = −1 t 227 = −1 t 54 = 1 t 54
4. ( j 3 k −2 ) 4 = ( j 3 k 2 ) 4 = ( j 3 ) 4 ( k 2 ) 4 = j 34 k 24 = j 12 k 8 ( j 3 k −2 ) 4 = ( j 3 k 2 ) 4 = ( j 3 ) 4 ( k 2 ) 4 = j 34 k 24 = j 12 k 8
5. ( m −2 n −2 ) 3 = ( 1 m 2 n 2 ) 3 = ( 1 ) 3 ( m 2 n 2 ) 3 = 1 ( m 2 ) 3 ( n 2 ) 3 = 1 m 23 n 23 = 1 m 6 n 6 ( m −2 n −2 ) 3 = ( 1 m 2 n 2 ) 3 = ( 1 ) 3 ( m 2 n 2 ) 3 = 1 ( m 2 ) 3 ( n 2 ) 3 = 1 m 23 n 23 = 1 m 6 n 6
### Try It:
#### Exercise 8
Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.
1. ( b 5 c ) 3 ( b 5 c ) 3
2. ( 5 u 8 ) 4 ( 5 u 8 ) 4
3. ( −1 w 3 ) 35 ( −1 w 3 ) 35
4. ( p −4 q 3 ) 8 ( p −4 q 3 ) 8
5. ( c −5 d −3 ) 4 ( c −5 d −3 ) 4
##### Solution
1. b 15 c 3 b 15 c 3
2. 625 u 32 625 u 32
3. −1 w 105 −1 w 105
4. q 24 p 32 q 24 p 32
5. 1 c 20 d 12 1 c 20 d 12
## Simplifying Exponential Expressions
Recall that to simplify an expression means to rewrite it by combing terms or exponents; in other words, to write the expression more simply with fewer terms. The rules for exponents may be combined to simplify expressions.
### Example 9
#### Problem 1
##### Simplifying Exponential Expressions
Simplify each expression and write the answer with positive exponents only.
1. ( 6 m 2 n −1 ) 3 ( 6 m 2 n −1 ) 3
2. 17 5 17 −4 17 −3 17 5 17 −4 17 −3
3. ( u −1 v v −1 ) 2 ( u −1 v v −1 ) 2
4. ( −2 a 3 b −1 )( 5 a −2 b 2 ) ( −2 a 3 b −1 )( 5 a −2 b 2 )
5. ( x 2 2 ) 4 ( x 2 2 ) −4 ( x 2 2 ) 4 ( x 2 2 ) −4
6. ( 3 w 2 ) 5 ( 6 w −2 ) 2 ( 3 w 2 ) 5 ( 6 w −2 ) 2
##### Solution
1. (6 m 2 n −1 ) 3 = (6) 3 ( m 2 ) 3 ( n −1 ) 3 The power of a product rule = 6 3 m 23 n −13 The power rule = 216 m 6 n −3 Simplify. = 216 m 6 n 3 The negative exponent rule (6 m 2 n −1 ) 3 = (6) 3 ( m 2 ) 3 ( n −1 ) 3 The power of a product rule = 6 3 m 23 n −13 The power rule = 216 m 6 n −3 Simplify. = 216 m 6 n 3 The negative exponent rule
2. 17 5 17 −4 17 −3 = 17 543 The product rule = 17 −2 Simplify. = 1 17 2 or 1 289 The negative exponent rule 17 5 17 −4 17 −3 = 17 543 The product rule = 17 −2 Simplify. = 1 17 2 or 1 289 The negative exponent rule
3. ( u −1 v v −1 ) 2 = ( u −1 v) 2 ( v −1 ) 2 The power of a quotient rule = u −2 v 2 v −2 The power of a product rule = u −2 v 2(−2) The quotient rule = u −2 v 4 Simplify. = v 4 u 2 The negative exponent rule ( u −1 v v −1 ) 2 = ( u −1 v) 2 ( v −1 ) 2 The power of a quotient rule = u −2 v 2 v −2 The power of a product rule = u −2 v 2(−2) The quotient rule = u −2 v 4 Simplify. = v 4 u 2 The negative exponent rule
4. (−2 a 3 b 1 )(5 a −2 b 2 ) = −25 a 3 a −2 b −1 b 2 Commutative and associative laws of multiplication = −10 a 32 b −1+2 The product rule = −10ab Simplify. (−2 a 3 b 1 )(5 a −2 b 2 ) = −25 a 3 a −2 b −1 b 2 Commutative and associative laws of multiplication = −10 a 32 b −1+2 The product rule = −10ab Simplify.
5. ( x 2 2 ) 4 ( x 2 2 ) −4 = ( x 2 2 ) 44 The product rule = ( x 2 2 ) 0 Simplify. = 1 The zero exponent rule ( x 2 2 ) 4 ( x 2 2 ) −4 = ( x 2 2 ) 44 The product rule = ( x 2 2 ) 0 Simplify. = 1 The zero exponent rule
6. (3 w 2 ) 5 (6 w −2 ) 2 = (3) 5 ( w 2 ) 5 (6) 2 ( w −2 ) 2 The power of a product rule = 3 5 w 25 6 2 w −22 The power rule = 243 w 10 36 w −4 Simplify. = 27 w 10(−4) 4 The quotient rule and reduce fraction = 27 w 14 4 Simplify. (3 w 2 ) 5 (6 w −2 ) 2 = (3) 5 ( w 2 ) 5 (6) 2 ( w −2 ) 2 The power of a product rule = 3 5 w 25 6 2 w −22 The power rule = 243 w 10 36 w −4 Simplify. = 27 w 10(−4) 4 The quotient rule and reduce fraction = 27 w 14 4 Simplify.
### Try It:
#### Exercise 9
Simplify each expression and write the answer with positive exponents only.
1. ( 2u v 2 ) −3 ( 2u v 2 ) −3
2. x 8 x −12 x x 8 x −12 x
3. ( e 2 f 3 f −1 ) 2 ( e 2 f 3 f −1 ) 2
4. ( 9 r −5 s 3 )( 3 r 6 s −4 ) ( 9 r −5 s 3 )( 3 r 6 s −4 )
5. ( 4 9 t w −2 ) −3 ( 4 9 t w −2 ) 3 ( 4 9 t w −2 ) −3 ( 4 9 t w −2 ) 3
6. ( 2 h 2 k ) 4 ( 7 h −1 k 2 ) 2 ( 2 h 2 k ) 4 ( 7 h −1 k 2 ) 2
##### Solution
1. v 6 8 u 3 v 6 8 u 3
2. 1 x 3 1 x 3
3. e 4 f 4 e 4 f 4
4. 27r s 27r s
5. 1 1
6. 16 h 10 49 16 h 10 49
## Using Scientific Notation
Recall at the beginning of the section that we found the number 1.3× 10 13 1.3× 10 13 when describing bits of information in digital images. Other extreme numbers include the width of a human hair, which is about 0.00005 m, and the radius of an electron, which is about 0.00000000000047 m. How can we effectively work read, compare, and calculate with numbers such as these?
A shorthand method of writing very small and very large numbers is called scientific notation, in which we express numbers in terms of exponents of 10. To write a number in scientific notation, move the decimal point to the right of the first digit in the number. Write the digits as a decimal number between 1 and 10. Count the number of places n that you moved the decimal point. Multiply the decimal number by 10 raised to a power of n. If you moved the decimal left as in a very large number, n n is positive. If you moved the decimal right as in a small large number, n n is negative.
For example, consider the number 2,780,418. Move the decimal left until it is to the right of the first nonzero digit, which is 2.
We obtain 2.780418 by moving the decimal point 6 places to the left. Therefore, the exponent of 10 is 6, and it is positive because we moved the decimal point to the left. This is what we should expect for a large number.
2.780418× 10 6 2.780418× 10 6
Working with small numbers is similar. Take, for example, the radius of an electron, 0.00000000000047 m. Perform the same series of steps as above, except move the decimal point to the right.
Be careful not to include the leading 0 in your count. We move the decimal point 13 places to the right, so the exponent of 10 is 13. The exponent is negative because we moved the decimal point to the right. This is what we should expect for a small number.
4.7× 10 −13 4.7× 10 −13
### A General Note: Scientific Notation:
A number is written in scientific notation if it is written in the form a× 10 n , a× 10 n ,where 1| a |<10 1| a |<10 and n n is an integer.
### Example 10
#### Problem 1
##### Converting Standard Notation to Scientific Notation
Write each number in scientific notation.
1. Distance to Andromeda Galaxy from Earth: 24,000,000,000,000,000,000,000 m
2. Diameter of Andromeda Galaxy: 1,300,000,000,000,000,000,000 m
3. Number of stars in Andromeda Galaxy: 1,000,000,000,000
4. Diameter of electron: 0.00000000000094 m
5. Probability of being struck by lightning in any single year: 0.00000143
##### Solution
1. 24,000,000,000,000,000,000,000 m 24,000,000,000,000,000,000,000 m 22 places 2.4× 10 22 m 24,000,000,000,000,000,000,000 m 24,000,000,000,000,000,000,000 m 22 places 2.4× 10 22 m
2. 1,300,000,000,000,000,000,000 m 1,300,000,000,000,000,000,000 m 21 places 1.3× 10 21 m 1,300,000,000,000,000,000,000 m 1,300,000,000,000,000,000,000 m 21 places 1.3× 10 21 m
3. 1,000,000,000,000 1,000,000,000,000 12 places 1× 10 12 1,000,000,000,000 1,000,000,000,000 12 places 1× 10 12
4. 0.00000000000094 m 0.00000000000094 m 13 places 9.4× 10 −13 m 0.00000000000094 m 0.00000000000094 m 13 places 9.4× 10 −13 m
5. 0.00000143 0.00000143 6 places 1.43× 10 −6 0.00000143 0.00000143 6 places 1.43× 10 −6
##### Analysis
Observe that, if the given number is greater than 1, as in examples a–c, the exponent of 10 is positive; and if the number is less than 1, as in examples d–e, the exponent is negative.
### Try It:
#### Exercise 10
Write each number in scientific notation.
1. U.S. national debt per taxpayer (April 2014): $152,000 2. World population (April 2014): 7,158,000,000 3. World gross national income (April 2014):$85,500,000,000,000
4. Time for light to travel 1 m: 0.00000000334 s
5. Probability of winning lottery (match 6 of 49 possible numbers): 0.0000000715
##### Solution
1. $1.52× 10 5$1.52× 10 5
2. 7.158× 10 9 7.158× 10 9
3. $8.55× 10 13$8.55× 10 13
4. 3.34× 10 −9 3.34× 10 −9
5. 7.15× 10 −8 7.15× 10 −8
### Converting from Scientific to Standard Notation
To convert a number in scientific notation to standard notation, simply reverse the process. Move the decimal n n places to the right if n n is positive or n n places to the left if n n is negative and add zeros as needed. Remember, if n n is positive, the value of the number is greater than 1, and if n n is negative, the value of the number is less than one.
#### Example 11
##### Problem 1
###### Converting Scientific Notation to Standard Notation
Convert each number in scientific notation to standard notation.
1. 3.547× 10 14 3.547× 10 14
2. −2× 10 6 −2× 10 6
3. 7.91× 10 −7 7.91× 10 −7
4. −8.05× 10 −12 −8.05× 10 −12
###### Solution
1. 3.547× 10 14 3.54700000000000 14 places 354,700,000,000,000 3.547× 10 14 3.54700000000000 14 places 354,700,000,000,000
2. −2× 10 6 −2.000000 6 places −2,000,000 −2× 10 6 −2.000000 6 places −2,000,000
3. 7.91× 10 −7 0000007.91 7 places 0.000000791 7.91× 10 −7 0000007.91 7 places 0.000000791
4. −8.05× 10 −12 −000000000008.05 12 places −0.00000000000805 −8.05× 10 −12 −000000000008.05 12 places −0.00000000000805
#### Try It:
##### Exercise 11
Convert each number in scientific notation to standard notation.
1. 7.03× 10 5 7.03× 10 5
2. −8.16× 10 11 −8.16× 10 11
3. −3.9× 10 −13 −3.9× 10 −13
4. 8× 10 −6 8× 10 −6
###### Solution
1. 703,000 703,000
2. −816,000,000,000 −816,000,000,000
3. −0.00000000000039 −0.00000000000039
4. 0.000008 0.000008
### Using Scientific Notation in Applications
Scientific notation, used with the rules of exponents, makes calculating with large or small numbers much easier than doing so using standard notation. For example, suppose we are asked to calculate the number of atoms in 1 L of water. Each water molecule contains 3 atoms (2 hydrogen and 1 oxygen). The average drop of water contains around 1.32× 10 21 1.32× 10 21 molecules of water and 1 L of water holds about 1.22× 10 4 1.22× 10 4 average drops. Therefore, there are approximately 3( 1.32× 10 21 )( 1.22× 10 4 )4.83× 10 25 3( 1.32× 10 21 )( 1.22× 10 4 )4.83× 10 25 atoms in 1 L of water. We simply multiply the decimal terms and add the exponents. Imagine having to perform the calculation without using scientific notation!
When performing calculations with scientific notation, be sure to write the answer in proper scientific notation. For example, consider the product ( 7× 10 4 )( 5× 10 6 )=35× 10 10 . ( 7× 10 4 )( 5× 10 6 )=35× 10 10 .The answer is not in proper scientific notation because 35 is greater than 10. Consider 35 as 3.5×10. 3.5×10.That adds a ten to the exponent of the answer.
( 35 )× 10 10 =( 3.5×10 )× 10 10 =3.5×( 10× 10 10 )=3.5× 10 11 ( 35 )× 10 10 =( 3.5×10 )× 10 10 =3.5×( 10× 10 10 )=3.5× 10 11
#### Example 12
##### Problem 1
###### Using Scientific Notation
Perform the operations and write the answer in scientific notation.
1. ( 8.14× 10 −7 )( 6.5× 10 10 ) ( 8.14× 10 −7 )( 6.5× 10 10 )
2. ( 4× 10 5 )÷( −1.52× 10 9 ) ( 4× 10 5 )÷( −1.52× 10 9 )
3. ( 2.7× 10 5 )( 6.04× 10 13 ) ( 2.7× 10 5 )( 6.04× 10 13 )
4. ( 1.2× 10 8 )÷( 9.6× 10 5 ) ( 1.2× 10 8 )÷( 9.6× 10 5 )
5. ( 3.33× 10 4 )( −1.05× 10 7 )( 5.62× 10 5 ) ( 3.33× 10 4 )( −1.05× 10 7 )( 5.62× 10 5 )
###### Solution
1. ( 8.14× 10 −7 )( 6.5× 10 10 ) = (8.14×6.5)( 10 −7 × 10 10 ) Commutative and associative properties of multiplication = (52.91)( 10 3 ) Product rule of exponents = 5.291× 10 4 Scientific notation ( 8.14× 10 −7 )( 6.5× 10 10 ) = (8.14×6.5)( 10 −7 × 10 10 ) Commutative and associative properties of multiplication = (52.91)( 10 3 ) Product rule of exponents = 5.291× 10 4 Scientific notation
2. ( 4× 10 5 )÷( −1.52× 10 9 ) = ( 4 −1.52 )( 10 5 10 9 ) Commutative and associative properties of multiplication (−2.63)( 10 −4 ) Quotient rule of exponents = −2.63× 10 −4 Scientific notation ( 4× 10 5 )÷( −1.52× 10 9 ) = ( 4 −1.52 )( 10 5 10 9 ) Commutative and associative properties of multiplication (−2.63)( 10 −4 ) Quotient rule of exponents = −2.63× 10 −4 Scientific notation
3. ( 2.7× 10 5 )( 6.04× 10 13 ) = (2.7×6.04)( 10 5 × 10 13 ) Commutative and associative properties of multiplication = (16.308)( 10 18 ) Product rule of exponents = 1.6308× 10 19 Scientific notation ( 2.7× 10 5 )( 6.04× 10 13 ) = (2.7×6.04)( 10 5 × 10 13 ) Commutative and associative properties of multiplication = (16.308)( 10 18 ) Product rule of exponents = 1.6308× 10 19 Scientific notation
4. ( 1.2× 10 8 )÷( 9.6× 10 5 ) = ( 1.2 9.6 )( 10 8 10 5 ) Commutative and associative properties of multiplication = (0.125)( 10 3 ) Quotient rule of exponents = 1.25× 10 2 Scientific notation ( 1.2× 10 8 )÷( 9.6× 10 5 ) = ( 1.2 9.6 )( 10 8 10 5 ) Commutative and associative properties of multiplication = (0.125)( 10 3 ) Quotient rule of exponents = 1.25× 10 2 Scientific notation
5. ( 3.33× 10 4 )( −1.05× 10 7 )( 5.62× 10 5 ) = [3.33×(−1.05)×5.62]( 10 4 × 10 7 × 10 5 ) (−19.65)( 10 16 ) = −1.965× 10 17 ( 3.33× 10 4 )( −1.05× 10 7 )( 5.62× 10 5 ) = [3.33×(−1.05)×5.62]( 10 4 × 10 7 × 10 5 ) (−19.65)( 10 16 ) = −1.965× 10 17
#### Try It:
##### Exercise 12
Perform the operations and write the answer in scientific notation.
1. ( −7.5× 10 8 )( 1.13× 10 −2 ) ( −7.5× 10 8 )( 1.13× 10 −2 )
2. ( 1.24× 10 11 )÷( 1.55× 10 18 ) ( 1.24× 10 11 )÷( 1.55× 10 18 )
3. ( 3.72× 10 9 )( 8× 10 3 ) ( 3.72× 10 9 )( 8× 10 3 )
4. ( 9.933× 10 23 )÷( 2.31× 10 17 ) ( 9.933× 10 23 )÷( 2.31× 10 17 )
5. ( −6.04× 10 9 )( 7.3× 10 2 )( −2.81× 10 2 ) ( −6.04× 10 9 )( 7.3× 10 2 )( −2.81× 10 2 )
###### Solution
1. 8.475× 10 6 8.475× 10 6
2. 8× 10 8 8× 10 8
3. 2.976× 10 13 2.976× 10 13
4. 4.3× 10 6 4.3× 10 6
5. 1.24× 10 15 1.24× 10 15
#### Try It:
##### Exercise 13
An average human body contains around 30,000,000,000,000 red blood cells. Each cell measures approximately 0.000008 m long. Write each number in scientific notation and find the total length if the cells were laid end-to-end. Write the answer in both scientific and standard notations.
###### Solution
Number of cells: 3× 10 13 ; 3× 10 13 ; length of a cell: 8× 10 −6 8× 10 −6 m; total length: 2.4× 10 8 2.4× 10 8 m or 240,000,000 240,000,000 m.
### Media:
Access these online resources for additional instruction and practice with exponents and scientific notation.
## Key Equations
Rules of Exponents For nonzero real numbers a a and b b and integers m m and n n Product rule a m ⋅ a n = a m+n a m ⋅ a n = a m+n Quotient rule a m a n = a m−n a m a n = a m−n Power rule ( a m ) n = a m⋅n ( a m ) n = a m⋅n Zero exponent rule a 0 =1 a 0 =1 Negative rule a −n = 1 a n a −n = 1 a n Power of a product rule ( a⋅b ) n = a n ⋅ b n ( a⋅b ) n = a n ⋅ b n Power of a quotient rule ( a b ) n = a n b n ( a b ) n = a n b n
## Key Concepts
• Products of exponential expressions with the same base can be simplified by adding exponents. See Example 1.
• Quotients of exponential expressions with the same base can be simplified by subtracting exponents. See Example 2.
• Powers of exponential expressions with the same base can be simplified by multiplying exponents. See Example 3.
• An expression with exponent zero is defined as 1. See Example 4.
• An expression with a negative exponent is defined as a reciprocal. See Example 5 and Example 6.
• The power of a product of factors is the same as the product of the powers of the same factors. See Example 7.
• The power of a quotient of factors is the same as the quotient of the powers of the same factors. See Example 8.
• The rules for exponential expressions can be combined to simplify more complicated expressions. See Example 9.
• Scientific notation uses powers of 10 to simplify very large or very small numbers. See Example 10 and Example 11.
• Scientific notation may be used to simplify calculations with very large or very small numbers. See Example 12 and Example 13.
## Section Exercises
### Verbal
#### Exercise 14
Is 2 3 2 3 the same as 3 2 ? 3 2 ? Explain.
##### Solution
No, the two expressions are not the same. An exponent tells how many times you multiply the base. So 2 3 2 3 is the same as 2×2×2, 2×2×2, which is 8. 3 2 3 2 is the same as 3×3, 3×3, which is 9.
#### Exercise 15
When can you add two exponents?
#### Exercise 16
What is the purpose of scientific notation?
##### Solution
It is a method of writing very small and very large numbers.
#### Exercise 17
Explain what a negative exponent does.
### Numeric
For the following exercises, simplify the given expression. Write answers with positive exponents.
9 2 9 2
81
15 −2 15 −2
#### Exercise 20
3 2 × 3 3 3 2 × 3 3
243
4 4 ÷4 4 4 ÷4
#### Exercise 22
( 2 2 ) −2 ( 2 2 ) −2
1 16 1 16
(58) 0 (58) 0
#### Exercise 24
11 3 ÷ 11 4 11 3 ÷ 11 4
1 11 1 11
#### Exercise 25
6 5 × 6 −7 6 5 × 6 −7
#### Exercise 26
( 8 0 ) 2 ( 8 0 ) 2
1
#### Exercise 27
5 −2 ÷ 5 2 5 −2 ÷ 5 2
For the following exercises, write each expression with a single base. Do not simplify further. Write answers with positive exponents.
#### Exercise 28
4 2 × 4 3 ÷ 4 −4 4 2 × 4 3 ÷ 4 −4
4 9 4 9
#### Exercise 29
6 12 6 9 6 12 6 9
#### Exercise 30
( 12 3 ×12 ) 10 ( 12 3 ×12 ) 10
12 40 12 40
#### Exercise 31
10 6 ÷ ( 10 10 ) −2 10 6 ÷ ( 10 10 ) −2
#### Exercise 32
7 −6 × 7 −3 7 −6 × 7 −3
1 7 9 1 7 9
#### Exercise 33
( 3 3 ÷ 3 4 ) 5 ( 3 3 ÷ 3 4 ) 5
For the following exercises, express the decimal in scientific notation.
#### Exercise 34
0.0000314
##### Solution
3.14× 10 5 3.14× 10 5
#### Exercise 35
148,000,000
For the following exercises, convert each number in scientific notation to standard notation.
#### Exercise 36
1.6× 10 10 1.6× 10 10
16,000,000,000
#### Exercise 37
9.8× 10 −9 9.8× 10 −9
### Algebraic
For the following exercises, simplify the given expression. Write answers with positive exponents.
#### Exercise 38
a 3 a 2 a a 3 a 2 a
a 4 a 4
#### Exercise 39
m n 2 m −2 m n 2 m −2
#### Exercise 40
( b 3 c 4 ) 2 ( b 3 c 4 ) 2
b 6 c 8 b 6 c 8
#### Exercise 41
( x −3 y 2 ) −5 ( x −3 y 2 ) −5
#### Exercise 42
a b 2 ÷ d −3 a b 2 ÷ d −3
##### Solution
a b 2 d 3 a b 2 d 3
#### Exercise 43
( w 0 x 5 ) −1 ( w 0 x 5 ) −1
m 4 n 0 m 4 n 0
m 4 m 4
#### Exercise 45
y −4 ( y 2 ) 2 y −4 ( y 2 ) 2
#### Exercise 46
p −4 q 2 p 2 q −3 p −4 q 2 p 2 q −3
q 5 p 6 q 5 p 6
(l×w) 2 (l×w) 2
#### Exercise 48
( y 7 ) 3 ÷ x 14 ( y 7 ) 3 ÷ x 14
##### Solution
y 21 x 14 y 21 x 14
#### Exercise 49
( a 2 3 ) 2 ( a 2 3 ) 2
#### Exercise 50
5 2 m÷ 5 0 m 5 2 m÷ 5 0 m
25 25
#### Exercise 51
( 16 x ) 2 y −1 ( 16 x ) 2 y −1
#### Exercise 52
2 3 ( 3a ) −2 2 3 ( 3a ) −2
72 a 2 72 a 2
#### Exercise 53
( m a 6 ) 2 1 m 3 a 2 ( m a 6 ) 2 1 m 3 a 2
#### Exercise 54
( b −3 c ) 3 ( b −3 c ) 3
c 3 b 9 c 3 b 9
#### Exercise 55
( x 2 y 13 ÷ y 0 ) 2 ( x 2 y 13 ÷ y 0 ) 2
#### Exercise 56
( 9 z 3 ) −2 y ( 9 z 3 ) −2 y
##### Solution
y 81 z 6 y 81 z 6
### Real-World Applications
#### Exercise 57
To reach escape velocity, a rocket must travel at the rate of 2.2× 10 6 2.2× 10 6 ft/min. Rewrite the rate in standard notation.
#### Exercise 58
A dime is the thinnest coin in U.S. currency. A dime’s thickness measures 1.35× 10 −3 1.35× 10 −3 m. Rewrite the number in standard notation.
0.00135 m
#### Exercise 59
The average distance between Earth and the Sun is 92,960,000 mi. Rewrite the distance using scientific notation.
#### Exercise 60
A terabyte is made of approximately 1,099,500,000,000 bytes. Rewrite in scientific notation.
##### Solution
1.0995× 10 12 1.0995× 10 12
#### Exercise 61
The Gross Domestic Product (GDP) for the United States in the first quarter of 2014 was $1.71496× 10 13 .$1.71496× 10 13 .Rewrite the GDP in standard notation.
#### Exercise 62
One picometer is approximately 3.397× 10 −11 3.397× 10 −11 in. Rewrite this length using standard notation.
##### Solution
0.00000000003397 in.
#### Exercise 63
The value of the services sector of the U.S. economy in the first quarter of 2012 was \$10,633.6 billion. Rewrite this amount in scientific notation.
### Technology
For the following exercises, use a graphing calculator to simplify. Round the answers to the nearest hundredth.
#### Exercise 64
( 12 3 m 33 4 −3 ) 2 ( 12 3 m 33 4 −3 ) 2
##### Solution
12,230,590,464 m 66 m 66
#### Exercise 65
17 3 ÷ 15 2 x 3 17 3 ÷ 15 2 x 3
### Extensions
For the following exercises, simplify the given expression. Write answers with positive exponents.
#### Exercise 66
( 3 2 a 3 ) −2 ( a 4 2 2 ) 2 ( 3 2 a 3 ) −2 ( a 4 2 2 ) 2
##### Solution
a 14 1296 a 14 1296
#### Exercise 67
( 6 2 −24 ) 2 ÷ ( x y ) −5 ( 6 2 −24 ) 2 ÷ ( x y ) −5
#### Exercise 68
m 2 n 3 a 2 c −3 a −7 n −2 m 2 c 4 m 2 n 3 a 2 c −3 a −7 n −2 m 2 c 4
n a 9 c n a 9 c
#### Exercise 69
( x 6 y 3 x 3 y −3 y −7 x −3 ) 10 ( x 6 y 3 x 3 y −3 y −7 x −3 ) 10
#### Exercise 70
( ( a b 2 c ) −3 b −3 ) 2 ( ( a b 2 c ) −3 b −3 ) 2
##### Solution
1 a 6 b 6 c 6 1 a 6 b 6 c 6
#### Exercise 71
Avogadro’s constant is used to calculate the number of particles in a mole. A mole is a basic unit in chemistry to measure the amount of a substance. The constant is 6.0221413× 10 23 . 6.0221413× 10 23 . Write Avogadro’s constant in standard notation.
#### Exercise 72
Planck’s constant is an important unit of measure in quantum physics. It describes the relationship between energy and frequency. The constant is written as 6.62606957× 10 −34 . 6.62606957× 10 −34 . Write Planck’s constant in standard notation.
##### Solution
0.000000000000000000000000000000000662606957
## Glossary
scientific notation:
a shorthand notation for writing very large or very small numbers in the form a× 10 n a× 10 n where 1| a |<10 1| a |<10 and n n is an integer
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# Conditional probability
## Presentation on theme: "Conditional probability"— Presentation transcript:
Conditional probability
Ordinary Probability You are dealt two cards from a deck. What is the probability the second card dealt is a Jack? We reason that if the two cards have been delt, the probabiity that the first is a jacek and the probabiity that the second is a jack are identical. Since there are 4 jacks in the deck, we compute P(second is jack) = 4/52 = 1/13.
Conditional Probability
You are dealt two cards from a deck. What is the probability of the second card was a jack given the first card was not a jack is called card dealt is a Jack?
Conditional Probability
You are dealt two cards from a deck. What is the probability that the second card was a jack given the first card was not a jack.
Conditional Probability
You are dealt two cards from a deck. What is the probability of the second card was a jack given the first card was not a jack is called card dealt is a Jack?
Reasoning Once the first card has been delt, there are 51 cards remaining, and, since the first was NOT a jack, there are 4 jacks in the set of 51 cards.
Conditional Probability
The probability of drawing jack given the first card was not a jack is called conditional probability. A key words to look for is “given that.” We will use the notation: P(second a jack | first not a jack) = 4/51
General Conditional Probability
The probability that the event A occurs, given that B occurs is denoted: This is read the probability of A given B.
Conditional Probability
How would we draw the event A given B? A B A and B
Conditional Probability
How would we draw the event A given B? Since we know B has occurred, we ignore everything else. A B A and B
Conditional Probability
How would we draw the event A given B? Since we know B has occurred, we ignore everything else. A B A and B
Conditional Probability
How would we draw the event A given B? Since we know B has occurred, we ignore everything else. With some thought this tells us: B A and B
Additional notes In the case of a equi-probability space, we can reason that, since we know the outcome is in B, we can use the set B as our reduced sample space. The probability P(A|B) can then be computed as the number of points in A∩B as a fraction of the number of points in B.
Dividing top and bottom by the numbers of
Continuing… This gives 𝑃 𝐴 𝐵 = 𝑛(𝐴∩𝐵) 𝑛(𝐵) Dividing top and bottom by the numbers of points in the original sample space S: 𝑃 𝐴|𝐵 = 𝑛(𝐴∩𝐵)/𝑛(𝑆) 𝑛(𝐵)/𝑛(𝑆) = 𝑃(𝐴∩𝐵) 𝑃(𝐵)
Example from well contamination
Of private wells, 21% are contaminated. Therefore
P(C|Private)=0.21 Of public wells, 40% are contaminated. Therefore P(C|Public)=0.40
Law of conditional probability
𝑃 𝐴 𝐵 = 𝑃(𝐴∩𝐵) 𝑃(𝐵)
Multiplication Rule 𝑃 𝐴 =𝑃 𝐵 𝑃(𝐴|𝐵)
Example A local union has 8 members, 2 of whom are women. Two are chosen by a lottery to represent the union.
Example A local union has 8 members, 2 of whom are women. Two are chosen by a lottery to represent the union. a) What is the probability they are both male?
Example A local union has 8 members, 2 of whom are women. Two are chosen by a lottery to represent the union. a) What is the probability they are both male?
Example A local union has 8 members, 2 of whom are women. Two are chosen by a lottery to represent the union. a) What is the probability they are both male?
Example A local union has 8 members, 2 of whom are women. Two are chosen by a lottery to represent the union. b) What is the probability they are both female?
Example A local union has 8 members, 2 of whom are women. Two are chosen by a lottery to represent the union. b) What is the probability they are both female?
Example Find the probability of selecting an all male jury from a group of 30 jurors, 21 of whom are men.
Example Find the probability of selecting an all male jury from a group of 30 jurors, 21 of whom are men. Solution: P(12 M) =P(M)*P(M|M)*P(M|MM) * …. = 21/30 * 20/29 * 19/28 * 18/27 * … 10/19 =
Independent Events Two events A and B are independent if the occurrence of one does not affect the probability of the other.
Independent Events Two events A and B are independent if the occurrence of one does not affect the probability of the other. Two events A and B are independent then P(A|B) = P(A).
Independent Events Two events A and B are independent if the occurrence of one does not affect the probability of the other. Two events A and B are independent then P(A|B) = P(A). Two events which are not independent are dependent.
Multiplication Rule Multiplication Rule: For any pair of events:
P(A and B) = P(A) * P(B|A)
Multiplication Rule Multiplication Rule: For any pair of events:
P(A and B) = P(A) * P(B|A) For any pair of independent events: P(A and B) = P(A) * P(B)
Multiplication Rule For any pair of events: P(A and B) = P(A) * P(B|A)
For any pair of independent events: P(A and B) = P(A) * P(B) If P(A and B) = P(A) * P(B), then A and B are independent.
Multiplication Rule Multiplication Rule:
P(A and B) = P(A) * P(B) if A and B are independent. P(A and B) = P(B) * P(A|B) if A and B are dependent. Note: The multiplication rule extends to several events: P(A and B and C) =P(C)*P(B|C)*P(A|BC)
Example A study of 24 mice has classified the mice by two categories
Fur Colour Black White Grey Eye Colour Red Eyes 3 5 2 Black Eyes 1 7 6
A study of 24 mice has classified the mice by two categories
a) What is the probability that a randomly selected mouse has white fur? b) What is the probability it has black eyes given that it has black fur? c) Find pairs of mutually exclusive and independent events. Fur Colour Black White Grey Eye Colour Red Eyes 3 5 2 Black Eyes 1 7 6
A study of 24 mice has classified the mice by two categories
a) What is the probability that a randomly selected mouse has white fur? 12/24=0.5 b) What is the probability it has black eyes given that it has black fur? c) Find pairs of mutually exclusive and independent events. Fur Colour Black White Grey Eye Colour Red Eyes 3 5 2 Black Eyes 1 7 6
A study of 24 mice has classified the mice by two categories
a) What is the probability that a randomly selected mouse has white fur? 12/24=0.5 b) What is the probability it has black eyes given that it has black fur? 1/4=0.25 c) Find pairs of independent events. Fur Colour Black White Grey Eye Colour Red Eyes 3 5 2 Black Eyes 1 7 6
c) Find pairs of mutually exclusive and independent events.
b) What is the probability it has black eyes given that it has black fur? 1/4=0.25 c) Find pairs of mutually exclusive and independent events. IND: White Fur and Red Eyes; Black Fur and Red Eyes Fur Colour Black White Grey Eye Colour Red Eyes 3 5 2 Black Eyes 1 7 6
Descriptive Phrases Descriptive Phrases require special care! At most
At least No more than No less than
Review Conditional Probabilities Independent events
Multiplication Rule Tree Diagrams
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# Interest
Financial Mathematics → Interest Calculations
The most common investments is the investment of money at interest. When money is deposited in a bank, the bank pays interest as a reward for the use of an asset called “capital”.
Interest Formula
Interest is calculated by the formula:
$I=niC$
Where,
I = Interest earned
C = Cash amount (called Capital)
i = Interest rate
n = Time period (in years).
Remember: 5% = 0.05, 7% = 0.07, 10% = 0.1, 15% = 0.15, 20% = 0.2
Example 1. Suppose \$1,000 is deposited in a bank that earns 5% interest per annum. What is the amount of interest earned after:
(a) 1 year
(b) 2.5 years
Solution:
(a) After 1 year
$I=niC \\ =(1)(0.05)(1000) \\ = \50$
(b) After 2.5 years
$I=niC \\ =(2.5)(0.05)(1000) \\ = \125$
Important:
If we have annual interest (Like interest rate per annum in the above example). And, we have time period in days, weeks and months then??
Interest for Time (n) Reason 1 day 1/365 There are 365 days in one year 5 days 5/365 There are 365 days in one year 1 week 1/52 There are 52 weeks in one year 7 weeks 7/52 There are 52 weeks in one year 1 month 1/12 There are 12 months in one year 6 months 6/12 There are 12 months in one year
As we have interest rate per annum, so we are giving time (n) in terms of years, like, 6/12 for 6 months, 10/12 for 10 months and 12/52 for 12 weeks.
Example 2. Suppose \$1,000 is deposited in a bank that earns 10% interest per annum. What is the amount of interest earned after:
(a) 50 days
(b) 20 weeks
(c) 8 months
Solution:
(a) After 50 days
$I=niC \\ =(50/365)(0.1)(1000) \\ =\13.7$
(b) After 20 weeks
$I=niC \\ =(20/52)(0.1)(1000) \\ =\38.46$
(c) After 8 months
$I=niC \\ =(8/12)(0.1)(1000) \\ =\66.67$
Next: Simple Interest
Posted
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# 2014 AMC 10B Problems/Problem 19
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Two concentric circles have radii $1$ and $2$. Two points on the outer circle are chosen independently and uniformly at random. What is the probability that the chord joining the two points intersects the inner circle?
$\textbf{(A)}\ \frac{1}{6}\qquad \textbf{(B)}\ \frac{1}{4}\qquad \textbf{(C)}\ \frac{2-\sqrt{2}}{2}\qquad \textbf{(D)}\ \frac{1}{3}\qquad \textbf{(E)}\ \frac{1}{2}\qquad$
## Solution
Let the center of the two circles be $O$. Now pick an arbitrary point $A$ on the boundary of the circle with radius $2$. We want to find the range of possible places for the second point, $A'$, such that $AA'$ passes through the circle of radius $1$. To do this, first draw the tangents from $A$ to the circle of radius $1$. Let the intersection points of the tangents (when extended) with circle of radius $2$ be $B$ and $C$. Let $H$ be the foot of the altitude from $O$ to $\overline{BC}$. Then we have the following diagram.
$[asy] scale(200); pair A,O,B,C,H; A = (0,1); O = (0,0); B = (-.866,-.5); C = (.866,-.5); H = (0, -.5); draw(A--C--cycle); draw(A--O--cycle); draw(O--C--cycle); draw(O--H,dashed+linewidth(.7)); draw(A--B--cycle); draw(B--C--cycle); draw(O--B--cycle); dot("A",A,N); dot("O",O,NW); dot("B",B,W); dot("C",C,E); dot("H",H,S); label("2",O--(-.7,-.385),N); label("1",O--H,E); draw(circle(O,.5)); draw(circle(O,1)); [/asy]$
We want to find $\angle BOC$, as the range of desired points $A'$ is the set of points on minor arc $\overarc{BC}$. This is because $B$ and $C$ are part of the tangents, which "set the boundaries" for $A'$. Since $OH = 1$ and $OB = 2$ as shown in the diagram, $\triangle OHB$ is a $30-60-90$ triangle with $\angle BOH = 60^\circ$. Thus, $\angle BOC = 120^\circ$, and the probability $A'$ lies on the minor arc $\overarc{BC}$ is thus $\dfrac{120}{360} = \boxed{\textbf{(D)}\: \dfrac13}$.
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To add or subtract fractions, the fractions must have a common denominator .
# To add or subtract fractions, the fractions must have a common denominator .
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## To add or subtract fractions, the fractions must have a common denominator .
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Adding and Subtracting Fractions To add or subtract fractions, the fractions must have a common denominator.
2. Fractions that have the same or common denominator are called like fractions.Fractions that have different denominators are called unlike fractions. Like and Unlike Fractions Like Fractions Unlike Fractions
3. Adding or Subtracting Like Fractions If a, b, and c, are numbers and b is not 0, then To add or subtract fractions with the same denominator, add or subtract their numerators and write the sum or difference over the common denominator.
4. Least Common Denominator To add or subtract fractions that have unlike, or different, denominators, we write the fractions as equivalent fractions with a common denominator. The smallest common denominator is called the least common denominator (LCD) or the least common multiple (LCM).
5. Least Common Multiple The least common denominator (LCD) of a list of fractions is the smallest positive number divisible by all the denominators in the list. (The least common denominator is also the least common multiple (LCM)of the denominators.)
6. Least Common Denominator To find the LCD of First, write each denominator as a product of primes. 12 = 2 • 2 • 3 18 = 2 • 3 • 3 Then write each factor the greatest number of times it appears in any one prime factorization. The greatest number of times that 2 appears is 2 times. The greatest number of times that 3 appears is 2 times. LCD = 2 • 2 • 3 • 3 = 36
7. Examples Simplify:
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First we show that it is monotone in particular increasing This is done by
# First we show that it is monotone in particular
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First, we show that it is monotone, in particular, increasing. This is done by induction. The base case is to show that x 1 x 2 . Is this true? Let’s see: x 1 = 1 , x 2 = 1 + 1 = 2 > 1 . True. Next, we assume that x n x n +1 . 1 Hint: Try an alternating sequence, which alternates between positive and negative terms.
8 DR. JULIE ROWLETT Then, we must show that x n +1 x n +2 . Well, let’s try: x n +2 = 1 + x n +1 , but x n +1 x n so 1 + x n +1 1 + x n = x n +1 . Follow the inequalities and you’ll see that x n +2 x n +1 . Next, we’ll show that the sequence is bounded. Look at x 1 . Well, clearly x 1 < 2 . Will this bound work for all the rest? Let’s try. x 2 = 1 + 1 < 2 . Since we’ve shown the base case ( n = 1) and the next case ( n = 2) , it makes sense to try to prove the bound holds for all the remaining terms by induction. So, assume that x n < 2 . Then, x n +1 = 1 + x n < 1 + 2 = 3 < 2 . So, by induction, we have shown that 2 is an upper bound for the sequence. Hence, by the monotone convergence theorem, the sequence converges. What is the limit? This is fun. If x n x where x is the limit, then, by homework problem 16.11, x n +1 x also. So, lim n →∞ x n = x = lim n →∞ x n +1 = lim n →∞ 1 + x n . Since on the far right side, the only term changing in the limit is x n x, by Theorem 17.1, x = 1 + x x 2 - x - 1 = 0 . The solutions to this equation are 1 - 5 2 and 1 + 5 2 . Since the terms in the sequence are bounded below by 1 , the limit is also bounded below by 1 , by Theorem 17.4. Hence, the correct limit is the positive root, and so lim n →∞ x n = 1 + 5 2 . 3. Cauchy Sequences Theorem 3.1. A sequence of real numbers converges if and only if it is a Cauchy sequence. Proof: Assume the sequence of real numbers converges { s n } n N converges to s. Then, let > 0 . There exists N N such that | s n - s | < 2 for all n N. Hence, for all n, m N, | s n - s m | ≤ | s n - s | + | s - s m | < 2 + 2 = . That is the definition of a Cauchy sequence (MEMORIZE)¿ Next, assume that the sequence of real numbers is Cauchy. One reason it must converge is by the construction of the real numbers as the set of all rationals and
MATH 117 LECTURE NOTES FEBRUARY 17, 2009 9 limits of Cauchy sequences of rational numbers. Another way is the following. We’ll
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• Fall '08
• Akhmedov,A
• Math, Mathematical analysis, Cauchy sequence, DR. JULIE ROWLETT
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