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# Unit conversion multiplication
This is going to sound like a really dumb question. At first I thought it was just my brain failing to function, but I can't work my mind around it. I'm using a problem to demonstrate the conceptual issue, but what I care about is why things work however they may.
Let's say we have a rectangle $5~\text{cm} \times 10~\text{cm}$. What is the area in meters?
If we say it is an area of $50~\text{cm}$, our answer should be $.05~\text{m}^2$.
On the other hand, if we were to convert to meters first, $.05~\text{m} \cdot .1~\text{m} =.005~\text{m}^2$.
• Do you mean a rectangle with side lengths $5~\text{cm}$ and $10~\text{cm}$. Squares have congruent sides. – N. F. Taussig Mar 6 '16 at 3:08
There are $100~\text{cm}$ in a meter. A square meter is the area of a square with side length $1~\text{m} = 100~\text{cm}$. Hence, $$1~\text{m}^2 = (1~\text{m})(1~\text{m}) = (100~\text{cm})(100~\text{cm}) = 10000~\text{cm}^2$$ Thus, to convert an area given in $\text{cm}^2$ to an area given in $\text{m}^2$, we must multiply by the conversion factor $$\frac{1~\text{m}^2}{10000~\text{cm}^2}$$ Hence, in your example, the area in square meters of a rectangle with area $50~\text{cm}^2$ is $$50~\text{cm}^2 \cdot \frac{1~\text{m}^2}{10000~\text{cm}^2} = 0.005~\text{m}^2$$
i think you're underestimating how many $cm^2$ there are in a $m^2$.
50 $cm^2 * \frac{1 m^2}{10,000 cm^2} = .005 m^2$ |
# 7.10 Compare fractions
Lesson
## Ideas
Being able to identify how many equal parts are in a fraction model will help us compare fractions in this lesson. Let's try this problem to review.
### Examples
#### Example 1
Here is a shape divided into parts, use it to complete the statements.
a
This shape has equal parts.
Worked Solution
Create a strategy
Count the number of smaller parts.
Apply the idea
There are 3 small triangles in the shape.
This shape has 3 equal parts.
b
Each part is \dfrac{⬚}{⬚} of the whole.
Worked Solution
Create a strategy
One shaded part would look like this:
We can write this fraction as:
Apply the idea
There is 1 shaded part out of 3 total parts.
Each part is \dfrac{1}{3} of the whole.
Idea summary
We can write a fraction from a fraction model as:
## Compare fractions
This video will show us how to compare fractions using models.
### Examples
#### Example 2
Which fraction is larger?
A
B
Worked Solution
Create a strategy
All of the pieces are equal in size, so we need to count which shape has more shaded pieces.
Apply the idea
Option A has 1 shaded piece while option B has 2. So option B is the larger fraction.
Idea summary
To compare fractions using fraction models with the same size pieces, count the number of pieces shaded.
## Complements to one whole
Let's look at finding complements to 1 whole.
### Examples
#### Example 3
If I have 1 third, how many more thirds do I need to make a whole?
Worked Solution
Create a strategy
Count the number of parts you need to shade to cover the whole shape.
Apply the idea
There are 2 more parts to be shaded to cover the whole.\text{Number of thirds} = 2
Idea summary
When comparing fractions, if the denominators are the same, then we can compare the numerators. The denominator also tells us how many parts make up one whole.
### Outcomes
#### MA2-7NA
represents, models and compares commonly used fractions and decimals |
Chapter 5.1 - PowerPoint PPT Presentation
1 / 20
Chapter 5.1. Bisectors of Triangles. Concept. Use the Perpendicular Bisector Theorems. A. Find BC . BC = AC Perpendicular Bisector Theorem BC = 8.5Substitution. Answer: 8.5. Use the Perpendicular Bisector Theorems. B. Find XY . Answer: 6.
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Chapter 5.1
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Chapter 5.1
Bisectors of Triangles
Concept
Use the Perpendicular Bisector Theorems
A. Find BC.
BC= ACPerpendicular Bisector Theorem
BC= 8.5Substitution
Use the Perpendicular Bisector Theorems
B. Find XY.
Use the Perpendicular Bisector Theorems
C. Find PQ.
A. Find NO.
A.4.6
B.9.2
C.18.4
D.36.8
B. Find TU.
A.2
B.4
C.8
D.16
C. Find EH.
A.8
B.12
C.16
D.20
Concept
Use the Angle Bisector Theorems
A. Find DB.
DB= DCAngle Bisector Theorem
DB= 5Substitution
Use the Angle Bisector Theorems
B. Find mWYZ.
Use the Angle Bisector Theorems
C. Find QS.
Answer: So, QS = 4(3) – 1 or 11.
A. Find the measure of SR.
A.22
B.5.5
C.11
D.2.25
B. Find the measure of HFI.
A.28
B.30
C.15
D.30
C. Find the measure of UV.
A.7
B.14
C.19
D.25
Concept
Use the Incenter Theorem
A. Find ST if S is the incenter of ΔMNP.
By the Incenter Theorem, since S is equidistant from the sides of ΔMNP,ST = SU.
Find ST by using the Pythagorean Theorem.
Use the Incenter Theorem
B. Find mSPU if S is the incenter of ΔMNP.
A. Find the measure of GF if D is the incenter of ΔACF.
A.12
B.144
C.8
D.65 |
# Search by Topic
#### Resources tagged with Limits similar to Integral Equation:
Filter by: Content type:
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### There are 19 results
Broad Topics > Pre-Calculus and Calculus > Limits
### Fractional Calculus II
##### Stage: 5
Here explore some ideas of how the definitions and methods of calculus change if you integrate or differentiate n times when n is not a whole number.
### Fractional Calculus I
##### Stage: 5
You can differentiate and integrate n times but what if n is not a whole number? This generalisation of calculus was introduced and discussed on askNRICH by some school students.
### Fractional Calculus III
##### Stage: 5
Fractional calculus is a generalisation of ordinary calculus where you can differentiate n times when n is not a whole number.
### Triangle Incircle Iteration
##### Stage: 4 Challenge Level:
Keep constructing triangles in the incircle of the previous triangle. What happens?
### Golden Eggs
##### Stage: 5 Challenge Level:
Find a connection between the shape of a special ellipse and an infinite string of nested square roots.
### There's a Limit
##### Stage: 4 and 5 Challenge Level:
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?
### Reciprocal Triangles
##### Stage: 5 Challenge Level:
Prove that the sum of the reciprocals of the first n triangular numbers gets closer and closer to 2 as n grows.
### Squaring the Circle and Circling the Square
##### Stage: 4 Challenge Level:
If you continue the pattern, can you predict what each of the following areas will be? Try to explain your prediction.
### Discrete Trends
##### Stage: 5 Challenge Level:
Find the maximum value of n to the power 1/n and prove that it is a maximum.
### Exponential Trend
##### Stage: 5 Challenge Level:
Find all the turning points of y=x^{1/x} for x>0 and decide whether each is a maximum or minimum. Give a sketch of the graph.
### Squareflake
##### Stage: 5 Challenge Level:
A finite area inside and infinite skin! You can paint the interior of this fractal with a small tin of paint but you could never get enough paint to paint the edge.
### Spokes
##### Stage: 5 Challenge Level:
Draw three equal line segments in a unit circle to divide the circle into four parts of equal area.
### Witch of Agnesi
##### Stage: 5 Challenge Level:
Sketch the members of the family of graphs given by y = a^3/(x^2+a^2) for a=1, 2 and 3.
### Over the Pole
##### Stage: 5 Challenge Level:
Two places are diametrically opposite each other on the same line of latitude. Compare the distances between them travelling along the line of latitude and travelling over the nearest pole.
### Production Equation
##### Stage: 5 Challenge Level:
Each week a company produces X units and sells p per cent of its stock. How should the company plan its warehouse space?
### Rain or Shine
##### Stage: 5 Challenge Level:
Predict future weather using the probability that tomorrow is wet given today is wet and the probability that tomorrow is wet given that today is dry.
### Converging Product
##### Stage: 5 Challenge Level:
In the limit you get the sum of an infinite geometric series. What about an infinite product (1+x)(1+x^2)(1+x^4)... ?
### Resistance
##### Stage: 5 Challenge Level:
Find the equation from which to calculate the resistance of an infinite network of resistances.
### Golden Fractions
##### Stage: 5 Challenge Level:
Find the link between a sequence of continued fractions and the ratio of succesive Fibonacci numbers. |
M203J Worksheet [1] The teenage pregnancy rate in 2006 was 72 teenage girls in every 1000 (http://www.guttmacher.org/pubs/USTPtrends.pdf). Home pregnancy kits (for women who collect and test their own samples) was found to have an overall sensitivity of 75% and a specificity of around 65% (http://www.medicine.ox.ac.uk/bandolier/band64/b64-7.html). (a) If a pregnant teen uses one of these home pregnancy kits, what's the chance that it will correctly say she is pregnant? This is the sensitivity; i.e., 75%. (b) If a pregnant teen uses one of these home pregnancy kits, what's the chance of a false negative (i.e., that it will incorrectly say she is not pregnant)? This is 100% - 75% = 25%. (c) If a teen who is not pregnant uses one of these home pregnancy kits, what's the chance that it will correctly say she is not pregnant? This is the specificity; i.e., 65%. (d) If a teen who is not pregnant uses one of these home pregnancy kits, what's the chance of a false positive (i.e., that it will incorrectly say she is pregnant)? This is 100% - 65% = 35%. (e) If a randomly chosen teen uses one of these home pregnancy kits and gets a negative result, what's the chance it is a false negative (i.e., what's the chance she is pregnant in spite of the test saying she's not)? Consider a random sample of 10,000 teen girls: Negative Test Positive Test Not pregnant 6032 = 65% of 9280 3248 = 9280-6032 9280=10000-720 Pregnant 180 = 720-540 540 = 75% of 720 720=(72/1000)10000 Total=10000 There are 180 false negatives out of 6032 + 180 = 6212 negatives all together, so the chance is 180/6212 = 2.9%. I.e., there's only a 3% chance the test is wrong. [2] If you flip a fair coin 100 times and find the fraction p^ of heads, what's the chance that p^ ≥ 0.6 (i.e., what is the chance that you get at least 60 heads)? Use what we've learned about normal distributions, by assuming that p^ is normally distributed (in this case p = 0.5 since we're assuming the coin is fair). [This is the same problem as choosing a sample of size n = 100 from a normally distributed population with p = 0.50, and asking what percentage of samples have p^ ≥ 60%.] The formula says σ = √(p(1-p)/n) = √(0.5(1-0.5)/100) = 5%, and x = 60% gives z = 2 (i.e., 60% is two standard deviations above the mean p = 50%). We know 95% of samples have p^ within 2σ of p = 50%, so 100% - 95% = 5% will have p^ outside this range, with half of them at or above 2σ; i.e., there's a 2.5% chance that p^ will be 60% or more. [3] Monday's Omaha World-Herald, October 25, 2010, published a poll based on sampling 607 registered voters in Omaha (the polling was done October 17-21). It found that 44% planned to vote for Lee Terry for Congress, 39% planned to vote for Terry's opponent, Tom White, and 17% were undecided or planned not to vote. The claimed margin of error was ± 4%. (a) What margin of error do you find from the formula assuming p = 0.44? Answer: ± 4% (b) What margin of error do you find from the formula assuming p = 0.39? Answer: ± 3.96% (c) What e would you need in order to have a 99.7% chance that p is in the range p^ ± e% ? Answer: e = 3s = 6% (d) Note that with a margin of error of 4%, it's possible that White is ahead of Terry. If we want to try to tell who's really ahead we might want a smaller margin of error. What sample size would you need to have a margin of error of ±2% for Terry's p^ = 44% ? s is half of the margin of error, so .01 = s = √(p^(1-p^)/n). Solving for n gives n = p^(1-p^)/s^2 = 0.44*0.56/0.01^2 = 2464. Note however, that it really doesn't help much to reduce the error of margin here; there are so many undecided voters that, even if you knew how everyone else stood, you still wouldn't be very sure of the outcome of the election. |
# Equation And Problem Solving
Worked-out word problems on linear equations with solutions explained step-by-step in different types of examples. Solution: Then the other number = x 9Let the number be x. Therefore, x 4 = 2(x - 5 4) ⇒ x 4 = 2(x - 1) ⇒ x 4 = 2x - 2⇒ x 4 = 2x - 2⇒ x - 2x = -2 - 4⇒ -x = -6⇒ x = 6Therefore, Aaron’s present age = x - 5 = 6 - 5 = 1Therefore, present age of Ron = 6 years and present age of Aaron = 1 year.5. Then the other multiple of 5 will be x 5 and their sum = 55Therefore, x x 5 = 55⇒ 2x 5 = 55⇒ 2x = 55 - 5⇒ 2x = 50⇒ x = 50/2 ⇒ x = 25 Therefore, the multiples of 5, i.e., x 5 = 25 5 = 30Therefore, the two consecutive multiples of 5 whose sum is 55 are 25 and 30. The difference in the measures of two complementary angles is 12°. ⇒ 3x/5 - x/2 = 4⇒ (6x - 5x)/10 = 4⇒ x/10 = 4⇒ x = 40The required number is 40.There are several problems which involve relations among known and unknown numbers and can be put in the form of equations. Sum of two numbers = 25According to question, x x 9 = 25⇒ 2x 9 = 25⇒ 2x = 25 - 9 (transposing 9 to the R. S changes to -9) ⇒ 2x = 16⇒ 2x/2 = 16/2 (divide by 2 on both the sides) ⇒ x = 8Therefore, x 9 = 8 9 = 17Therefore, the two numbers are 8 and 17.2. A number is divided into two parts, such that one part is 10 more than the other. Try to follow the methods of solving word problems on linear equations and then observe the detailed instruction on the application of equations to solve the problems.
Tags: Example Dissertation TitlesTricks To Make An Essay LongerMaster'S Thesis Proposal OutlineCreative And Professional WritingAnthropology Dissertation AbstractTcd Thesis SubmissionPicnic At Hanging Rock EssaysSad Incident Essay
According to the question; Ron will be twice as old as Aaron. Complement of x = 90 - x Given their difference = 12°Therefore, (90 - x) - x = 12°⇒ 90 - 2x = 12⇒ -2x = 12 - 90⇒ -2x = -78⇒ 2x/2 = 78/2⇒ x = 39Therefore, 90 - x = 90 - 39 = 51 Therefore, the two complementary angles are 39° and 51°9. If the table costs \$40 more than the chair, find the cost of the table and the chair. Solution: Let the number be x, then 3/5 ᵗʰ of the number = 3x/5Also, 1/2 of the number = x/2 According to the question, 3/5 ᵗʰ of the number is 4 more than 1/2 of the number.
Solution: Let the breadth of the rectangle be x, Then the length of the rectangle = 2x Perimeter of the rectangle = 72Therefore, according to the question2(x 2x) = 72⇒ 2 × 3x = 72⇒ 6x = 72 ⇒ x = 72/6⇒ x = 12We know, length of the rectangle = 2x = 2 × 12 = 24Therefore, length of the rectangle is 24 m and breadth of the rectangle is 12 m. Then Aaron’s present age = x - 5After 4 years Ron’s age = x 4, Aaron’s age x - 5 4. Then the cost of the table = \$ 40 x The cost of 3 chairs = 3 × x = 3x and the cost of 2 tables 2(40 x) Total cost of 2 tables and 3 chairs = \$705Therefore, 2(40 x) 3x = 70580 2x 3x = 70580 5x = 7055x = 705 - 805x = 625/5x = 125 and 40 x = 40 125 = 165Therefore, the cost of each chair is \$125 and that of each table is \$165. If 3/5 ᵗʰ of a number is 4 more than 1/2 the number, then what is the number?
In "real life", these problems can be incredibly complex.
This is one reason why linear algebra (the study of linear systems and related concepts) is its own branch of mathematics.
How long would it take to paint the house if they worked together?
## Essays On Judicial Review - Equation And Problem Solving
Step 2: Solve the equation created in the first step.Solving equations can be tough, especially if you've forgotten or have trouble understanding the tools at your disposal.One of those tools is the subtraction property of equality, and it lets you subtract the same number from both sides of an equation. Solving equations can be tough, especially if you've forgotten or have trouble understanding the tools at your disposal. This tutorial shows you how to take a word problem and translate it into a mathematical equation involving fractions.Also, real-world problems such as tipping in a restaurant, finding the price of a sale item, and buying enough paint for a room all involve using formulas.Introduce a problem to students that requires them to use a formula to solve the problem.The following problem would be best solved using a formula: Students can use the formula F = 1.8C 32 to find the solution.Using a Formula is a problem-solving strategy that can be used for problems that involve converting units or measuring geometric objects.Using a Formula is a problem-solving strategy that students can use to find answers to math problems involving geometry, percents, measurement, or algebra.To solve these problems, students must choose the appropriate formula and substitute data in the correct places of a formula.For example: Math problems requiring formulas can be simple, with few criteria needed to solve them, or they can be multidimensional, requiring charts or tables to organize students' thinking.Including more than one formula in a problem, or having multiple correct answers to a problem will help stretch this strategy.
## Comments Equation And Problem Solving
• ###### Kinematic Equations and Problem-Solving - The Physics.
In the above equations, the symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands.…
• ###### Word Problems Equations And Inequalities Siyavula
To solve word problems we need to write a set of equations that represent the problem mathematically. The solution of the equations is then the solution to the.…
• ###### Word Problems on Linear Equations Equations in One.
Worked-out word problems on linear equations with solutions explained step-by-step in different. Steps involved in solving a linear equation word problem…
• ###### Sample Math 101 Test Problems - WSU Math Department
Sample problems are under the links in the "Sample Problems" column and the corresponding review material is under the. Solving Linear Equations.…
• ###### Algebra - Linear Equations Practice Problems
Apr 25, 2018. Here is a set of practice problems to accompany the Linear Equations section of the Solving Equations and Inequalities chapter of the notes for.…
• ###### The 5 Steps of Problem Solving - Quick and Dirty Tips
Dec 30, 2016. Do you ever find yourself stuck on math problems before you even get. to turn a word problem into an algebraic equation and then solve it.…
• ###### Step-by-Step Math Problem Solver
QuickMath allows students to get instant solutions to all kinds of math problems, from algebra and equation solving right through to calculus and matrices.… |
Math Challenge: Computing the Average Rotational Speed of Earth
Or of any celestial body. Here I discuss a solution that can be explained to high school students, to get them interested in mathematics, statistics and probabilities. A few interesting related problems further enhance the pedagogical value of this discussion.
I stumbled upon this kind of problems when learning advanced mathematics in my postgraduate studies, in a course entitled stochastic geometry. Just formulating the problem required advanced knowledge of sophisticated measure theory and stochastic spatial processes in some special mathematical spaces. The lessons were very theoretical, focusing on highly abstract and generic mathematical settings. This removed much of the fun of this otherwise very exciting field full of mysterious and remarkable results.
Here I take a radically opposite path, proving that these concepts can be introduced to high school students, without rigorous definitions, without even a basic knowledge of the concept of probability or random variable. The purpose is to get them challenged and interested, maybe hoping a few of them will become data scientists or mathematicians.
Average Rotational Speed of Earth
So here is the “high school” solution to this problem. Here we are talking about the speed for a location on the surface of Earth. The rotation is around Earth’s polar axis.
Earth being a sphere, it has plenty of symmetries, and all we need to do is to compute the rotational speed of one point for each latitude, weight these points appropriately, and compute a weighted average across the latitudes. Since there are more points located at the equator than at the 45 degrees latitude, one would expect the weight at the equator to be the largest.
Let x be the angle specifying the latitude: from 0 at the equator, to p/2 at the North pole. The solution will be a weighted average with the following features:
• x being a continuous variable, we are dealing with an integral rather than a sum
• Lower bound is x = 0, upper bound is x = p/2
• Let w(x) be the weight attached to x, and L(x) be the length of the circle, parallel to the equator, and containing x
Let’s also define:
• R = radius of Earth
• T = one day (our time unit)
At the equator, the speed is of course 2pR/T. At x, the speed is L(x)/T. The average speed is
v = (1/T) x Int{ L(x) w(x) } / Int{ w(x) }
where the integral Int is between 0 and p/2.
Here, an heuristic but rudimentary argument (I call it intuition) shows that w(x) is equal to L(x). The computation of L(x) is straightforward, requiring basic trigonometry only:
L(x) = 2pR cos x.
Combining all the pieces together, we obtain the following formula for the average rotational speed of Earth:
For writing the above mathematical formula, I used an online mathematical editor known as HostMath, entered the formula, took a screenshot and saved it as an image, then inserted it in this article as an image.
The computation of the integrals involved is also straightforward (high school grade) and can even be performed online using the WolphramAlpha tool. Below is a screenshot:
Interesting related facts and problems
Now that we have computed the formula, it is easy to find that
• A plane flies faster than the Earth is rotating, when circling close to the North pole, but not as fast as Earth when circling near the equator. At which latitude are both speeds identical?
• At which latitude does a point on Earth’s surface rotate (along Earth’s axis) at the speed of sound?
• Pick up a random location on Earth’s surface. What is the expected distance covered in one day, circling around Earth’s polar axis? Solution: This is the same problem but formulated in a different way.
• The average speed is equal to p/4 times the speed at the equator.
• Potential classroom experiment: Pick up n locations randomly distributed on Earth’s surface, compute the speed at each location, then average these speeds. Does the average speed converge to the theoretical result as n tends to infinity? This experiment can be done using Monte-Carlo simulations on a computer. However, to generate random points on a sphere, read this tutorial. The problem is more complicated than it seems at first glance.
Top DSC Resources |
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Top
Octagon Calculator
Top
Octagon Calculator is an online tool used to find area and perimeter of the octagon. Octagon Calculator is valid only for regular octagon.
Octagon is one of the shapes of polygon. It is a closed plane shape and also enclosed by straight sides. The eight straight sides are also same. Octagon has eight same angles. In octagon, there are eight vertices and eight edges. The interior angle is 135 degree in the octagon shape.
## Step by Step Calculation
Step 1 :
Octagon Formula's
Area of the Octagon
A = 2 * a2 * ( 1 + $\sqrt{2}$ )
Where a = side length
Perimeter of the Octagon
P = 8 * a
Step 2 :
Put the values in the above formula's and solve it further.
## Example Problems
1. ### Find the area and perimeter of the octagon for the value of side length is 7 cm.
Step 1 :
Given: The length of side of the octagon is 7 cm
Area of the Octagon
A = 2 * a2 * ( 1 + $\sqrt{2}$ )
Where a = side length
Perimeter of the Octagon
P = 8 * a
Step 2 :
Put the values in the above formula's and solve it further.
Area of the Octagon
A = 2 * 72 * ( 1 + $\sqrt{2}$ )
A = 2 * 49 * (1 + 1.4142)
A = 236.59 cm2
Perimeter of the Octagon
P = 8 * a
P = 8 * 7 cm
P = 56 cm
Area of the Octagon = 236.59 cm2
Perimeter of the octagon = 56 cm
2. ### Find the area and perimeter of the octagon for the value of side length is 11 m.
Step 1 :
Given: The length of side of the octagon is 11 m
Area of the Octagon
A = 2 * a2 * ( 1 + $\sqrt{2}$ )
Where a = side length
Perimeter of the Octagon
P = 8 * a
Step 2 :
Put the values in the above formula's and solve it further.
Area of the Octagon
A = 2 * 112 * ( 1 + $\sqrt{2}$ )
A = 2 * 121 * (1 + 1.4142)
A = 584.23 m2
Perimeter of the Octagon
P = 8 * a
P = 8 * 11 m
P = 88 m |
# Word Problems Involving Different Operations
Related Topics:
More Lessons for Grade 5
Math Worksheets
Examples, solutions, videos, worksheets, stories, and songs to help Grade 5 students learn how to solve Word Problems Involving Different Operations.
Assorted Basic Word Problems
Examples:
1. A class has twenty-seven students. Fourteen students move to another class. How many students are now in class?
2. A store had 24 brownies. The brownies were sold to 8 customers, and each customer bought the same amount. How many brownies did each customer buy?
3. Jane has 5 rows of dimes, and each row has 8 dimes in it. How many dimes does she have in total?
4. Steve has 25 marbles. He finds 27 more. How many marbles does he have now? Create story contexts for numerical expressions and tape diagrams, and solve real word problems
Example:
Chase volunteers at an animal shelter after school, feeding and playing with the cats.
A. If he can make 5 servings of cat food from a third of a kilogram of food, how much does one serving weigh?
B. If Chase wants to give the same serving size to each of the 20 cats, how many kilograms of food will he need?
Solve word problems using tape diagrams and fraction by fraction multiplication
Example:
Anthony bought an 8-ffot board. He cut off 3/4 of the board to build a shelf, and gave the rest to his brother for an art project. How many inches long was the piece Anthony gave to his brother?
Multiply any whole number by a fraction using tape diagrams.
Examples:
1. There are 48 students going on a field trip. One-fourth are girls. How many boys are going on the trip?
2. Three angles are labeled below with arcs. The smallest angle is 3/8 as large as the 160° angle. Find the value of angle a.
3. Abbie spent 5/8 of her money and saved the rest. If she spent \$45, how much money did she have at first?
4. Mrs. Harrison used 16 ounces of dark chocolate while baking. She used 2/5 of the chocolate to make some frosting and used the rest to make brownies. How much more chocolate did Mrs. Harrison use in the brownies than in the frosting?
5. A straight angle is split into two smaller angles as shown. The smaller angle's measure is 1/6 that of a straight angle. What is the value of angle a?
6. Annabel and Eric made 17 ounces of pizza dough. They used 5/8 of the dough to make a pizza and used the rest to make calzones. What is the difference between the amount of dough they used to make pizza, and the amount of dough the used to make calzonee?
7. The New York Rangers hockey team won 3/4 of their games last season. If they lost 21 games, how many games did they paly in the entire season?
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 find any asymptotes of f(x)=(x+3)/(x^2-4)?
Oct 16, 2016
There are two vertical asymptotes at $x = - 2$ and $x = 2$
There is a horizontal asymptote at 0 as $x \to \infty$
There is a horizontal asymptote at 0 as $x \to - \infty$
#### Explanation:
$f \left(x\right) = \frac{x + 3}{{x}^{2} - 4}$
$\therefore f \left(x\right) = \frac{x + 3}{{x}^{2} - {2}^{2}}$
$\therefore f \left(x\right) = \frac{x + 3}{\left(x + 2\right) \left(x - 2\right)}$
Vertical Asymptotes
These occur when the denominator is zero
$\therefore \left(x + 2\right) \left(x - 2\right) = 0$
$x = \pm 2$#
Horizontal Asymptotes
We also need to examine the behaviour as $x \to \pm \infty$
Now for large $x$,
$x + 3 \approx x$ and $x + 2 \approx x$ and $x - 3 \approx x$
We have, $f \left(x\right) = \frac{x + 3}{\left(x + 2\right) \left(x - 2\right)}$
and so, for large $x$, $f \left(x\right) \approx \frac{x}{x} ^ 2 \approx \frac{1}{x}$
so, as $x \to - \infty \implies f \left(x\right) \to {0}^{-}$
and, as $x \to \infty \implies f \left(x\right) \to {0}^{+}$
Summary
There are two vertical asymptotes at $x = - 2$ and $x = 2$
There is a horizontal asymptote at 0 as $x \to \infty$
There is a horizontal asymptote at 0 as $x \to - \infty$
The graph validates this;
graph{(x+3)/((x+2)(x-2)) [-10, 10, -5, 5]} |
# Mathematical Proof The Eagle In The USPS Logo Is FAST!
The logo for the United States Postal Service is a mean-looking eagle. But a true fluid dynamics geek might look at it and realize that eagle is moving so fast it’s causing a shock wave. But just how fast is it moving? [Andrew Higgins] asked and answered this question, posting his analysis of the logo’s supersonic travel. He claims it’s Mach 4.9, but, how do we know? Science!
It turns out if something is going fast enough, you can tell just how fast with a simple picture! We’ve all seen pictures of jets breaking the sound barrier, this gives us information about the jet’s speed.
## How does it work?
Think about it like this: sound moves at roughly 330 m/s on Earth at sea level. If an object moves through air at that velocity, the air disturbances are transmitted as sound waves. If it’s moving faster than sound, those waves get distributed downstream, behind the moving object. The distance of these waves behind the moving object is dependent on the object’s speed.
This creates a line of these interactions known as a “Mach line.” Find the angle difference of the Mach line and the direction of travel and you have the “Mach angle” (denoted by α or µ).
There is a simple formula for determining the speed of an object using the Mach angle, the speed of sound (a), and an object’s velocity (v): sin(µ) = a / v. The ratio of to a is known as the Mach number, (M). If an object is going exactly the speed of sound, it’s going Mach 1 (because v = a).
Since Mach number (M) is v / a, we can plug it into the formula from above as 1 / M and use [Andrew]’s calculation shown in the image at the top of the article for a Mach angle (µ) of ~11.7°:
$\bf \sin ( \mu ) = \frac{1}{M} \\ \\ M = \frac{1}{\sin(\mu)} \\ \\ M = \frac{1}{\sin(11.7)} \\ \\ M = \frac{1}{0.202787295357} \\ \\ M = 4.9312753949380048$
The real question is, did the USPS chose Mach 4.93 as a hint to some secret government postal project? Or, was it simply a 1993 logo designer’s attempt to “capture the ethos of a modern era which continues today”?
## 33 thoughts on “Mathematical Proof The Eagle In The USPS Logo Is FAST!”
1. Dave says:
To put it into terms of an actual speed, you would need to know the density of the medium in which it is moving.
What’s the density of red tape?
1. MCP says:
Infinite!
1. Alex Rossie says:
Excellent got a real life guffaw out of me.
For AusPost the logo should be a brick through your window.
2. Comedicles says:
In air / gas it depends only on temperature. There is no density involved. It is about the speed of the molecules and their elastic collisions, which is temperature.
2. starhawk says:
I hate to break the news to you, Danny, but April First has already come and gone…
3. tomás zerolo says:
Density and… elasticity and things.
What if it’s just surface waves on water?
But then you’d have to state speed in knots, wouldn’t you?
1. Queeg says:
Knots are pretty much like miles per hour, just more expensive.
4. Paulie says:
Is that faster than an unladen swallow?
1. blackikarus says:
African or European swallow?
2. RW ver 0.0.1 says:
African or European?
5. RoboMonkey says:
You’re using coconuts!
6. Sam says:
I prefer W. A. S. T. E.
…Ugh. They took the angle using horizontal as reference, not the direction of travel. C-.
1. RW ver 0.0.1 says:
There’s a surprising amount of college level treatment of aerodynamics that asserts that anything acting vertically is lift, anything acting horizontally in one direction is drag, anything acting in the opposite direction to drag is thrust, and all that acts vertically downward is weight. Everything at 90 degrees to each other, always. I mean if you do it thoroughly it all comes out in the vectors but it inhibits intuitive feel for what is going on, and will lead you to booboos that you might otherwise catch, especially if applying it to other than aeroplanes.
I myself, NASA and any self respecting physicist prefer this at least..
https://www.grc.nasa.gov/WWW/K-12/airplane/climb.html
2. Slacker says:
If we assume the top and bottom shocks are symmetrical and the eagle is moving at an upward angle then we can just use half the angle between the shocks for the calculation.
Measuring the two shocks I get an angle of 33 degrees, so that gives α = 16.5, and a resulting speed of about mach 3.52
8. Osgeld says:
doesn’t matter how fast it is when every other package goes to the wrong fuggin city
1. Yaakov says:
Where do you live???
9. 8bitwiz says:
Where the hell was this 2 1/2 weeks ago when it would have been perfect?
10. Rick says:
Uh … I believe this submission was sent via USPS … it’s actually early! /s
11. The logo is about the only thing that is fast about the USPS. Everything is a “day late and a dollar short” at my apartment. Birthday cards often show up a week later! Since I have gone almost completely online with my bills, I might as well have a trashcan with my apartment number on it instead of a mailbox because that’s all I get anymore…
1. Wernicke says:
I generally seem to get things the few things that I get through USPS pretty much on time, but for some reason, all the birthday cards that I send usually are about a week late. Hmmm… could I be doing something wrong?
1. USPS worker says:
Birthday Cards take longer to inspect and see if they could steel the money inside.
12. “MATHEMATICAL PROOF THE EAGLE IN THE USPS LOGO IS FAST!” is frankly a bit incorrect. (Both factually and grammatically. (should be “mathematics proves the eagle in”))
The formulas for calculating mach speeds are actually derived from empirical observations.
The fact that a gas stops behaving like a fluid when crossing the speed of sound makes it rather simple to express the angle of the resulting shock wave in proportion to the speed of the object relative to the speed of sound of the gas.
But if a gas were to have other ways of interaction that traveled faster than the speed of sound, then the shock wave would behave differently. Now, most gases do have other ways to interact that are in fact faster than the speed of sound (primarily thermal radiation, changes in gravity due to changes in density, etc) but since most of these other interactions are frankly rather negligible in comparison to the particles just getting pushed about as if they were marbles, then we can largely just consider the later case in our approximations.
Meaning that we have made the problem into a simple geometrical expression of the air getting pushed aside at the speed of sound, while our object moves forth at a speed greater than the speed of sound. This in turn gives us a simple triangle. Though it is only an approximation. One that in this case happens to be very close reality relatively speaking.
But the accuracy of this approximation would diminish if gases had more significant modes of interactions above the speed of sound.
Converting the mach speed to kt, mph, km/h, or m/s would require more empirically derived constants to calculate these values. It can’t be done with mathematics alone to be fair.
Though, some people say that mathematics can prove/explain anything. Or that for example logic is simply a subfield of mathematics. (something that is fairly trivial to disprove.)
A mathematical proof on the other hand doesn’t just toss away part of a problem just because it is negligible.
(in physics and engineering on the other hand, tossing out negligible effects is fairly common and fully accepted.)
13. echodelta says:
Though it’s in the Constitution, there are rumblings of a roost and wing clipping of the speeding eagle. The solvency of the pension is the stone around it’s neck. Paper ads (in the screen age) are it’s mainstay, mine filled a trashcan under the mailbox that was full after a year from USPS. That’s a lot of nest building material.
1. Gregg Eshelman says:
The government mandated the USPS has to squirrel away a massive amount of money for their retired workers’ pension fund. Far more money than could possibly be disbursed even if every former postal employee lived to 100.
Last I looked it up, the fund has grown to over \$10 billion. The USPS needs a whole new fleet of delivery trucks and could definitely use some more employees, especially when they rarely, if ever, have all the registers staffed at the busiest times.
The fleet of Grumman LLV’s are 26 to 33 years old and the ones that haven’t gone completely to pieces have been rebuilt at least once.
14. bruce says:
I was once stopped behind a postal truck…and the logo looked like a duck with its head turned back to the left over its white wings……and I could never see the eagle in the logo again
1. Thopter says:
It takes a bit of squinting, but I think I can see your duck.
15. Big Bird says:
But is it really the eagle that is fast? Is it traveling solely under it’s own power? Or did they launch it somehow?
What this really represents is a late 80s through early 90s project where the USPS was experimenting with delivering food right to people’s door. You see firing a bird through a specialized cannon was a way to pluck, cook and deliver the bird all in one step. Had they succeeded such efficiency would have revolutionized the food industry.
Alas, they never quite figured out how to land the food in a way that was safe, clean and appetizing and the project ran out of funding. However, no one really knows what goes on at their Isla Nublar research facitlity. Given recent events it seems likely they would revisit this project.
1. Yaakov says:
Luv it. One can only hope! 🐔
16. Justin says:
What a terribly inefficient way to calculate the Mach number!
Gross approximation or not, if we assume the sine and arctangent (or, ahem, “inverse of tangent”, as some people like to call it) are a reasonable pair of formulas for obtaining a workable Mach number… there is a better way to obtain the same result!
Enter Pythagoras!
You see, arctangent takes “rise over run” and gives us the angle. We have a “run” of 4.83, for a “rise” of 1. Fair enough; we get angle “alpha”.
But then this Mach formula, as illustrated, turns that angle “alpha” right back into a ratio of right-angle triangle sides!
Sine takes an angle and gives us “opposite over hypotenuse”. Well, the side “opposite” of “alpha” has a value of 1, so we get 1/h, where h is the hypotenuse.
But look! Mach number is 1/sine(alpha), or 1/(1/h)… which is… h! The hypotenuse! The Mach number is the hypotenuse!
So a simple application of the Pythagorean theorem can give us the Mach number, DIRECTLY, as long as the “rise” of the slope is normalized to 1! …Which in this diagram, it already is.
The calculation? M = sqrt(1 + 4.83^2)
Why is this better? Because it involves only three operations: a square, a square root, and dirt-simple addition of +1.
Computers and microprocessors can square a number very quickly and efficiently, and there are efficient binary algorithms especially tuned for calculating square roots. Oh and incrementing a number by 1 is so blazingly fast on a microprocessor, it barely even needs a mention.
Compare with trigonometric calculations, which can only be done either by enormously space-inefficient floating-point look-up tables, or by iterating through enormously time-inefficient Taylor/Maclaurin polynomials until the desired level of precision is achieved.
If you’re going to program a microprocessor to do such calculations (which seems fitting, given the typical audiences of HACKADAY)… then for the love of god and all that is holy, please, Please, PLEASE give some thought to the mathematics involved, and optimize your algorithms as best as possible for your application.
17. This goes slightly over my dumb head, but I’m pretty sure they chose Mach 4.93.
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## Methods to Determine the Break-Even Point
### Learning Outcomes
• Determine the break-even point using the equation method, the formula method, and in dollar sales and sales units
So the Minnesota Kayak Company has these awesome new kayaks they are going to introduce to the market. They are a new company and need help in determining pricing, costs and how many kayaks they will need to sell in a month to break even. They are looking to you to help them determine if the selling price and costs will help them to reach their goals. They give you the following information to work with:
Price per kayak $500 variable costs per kayak$225 Contribution margin per kayak $275 Fixed costs/month$7,700
With this information, it is your task to find the breakeven point using the three different methods. Let’s look first at the equation method:
The equation method utilizes the profit equation introduced earlier.
$\text{Profit}=\text{Selling price}-\text{Variable Expenses}-\text{Fixed Expenses}$
Also, let’s revisit the contribution margin concept and some shortcuts.
• $\text{Contribution margin}=\text{Selling price}-\text{Variable expenses}$
• Profit= P
• Contribution Margin= CM
• Quantity= Q
• Fixed Expenses = F
• Variable Expenses = V
So in our kayak example we are looking for a break even point meaning that the profit = $0 We can then put together our break even point utilizing the equation method as follows: $\begin{array}{rcl}\0&=&\text{Unit CM}\times\text{Q}-\text{F}\\\0&=&\275\times\text{Q}-\7,700\\\7,700&=&\275\times\text{Q}\\\dfrac{\7,700}{275}&=&\text{Q}\\28&=&\text{Q}\end{array}$ Minnesota Kayak needs to sell 28 kayaks at$500 each to break even.
The formula method gets to the same answer in a different way. It is kind of a shortcut to the equation method:
$\begin{array}{rcl}\text{Unit Sales to Break Even}&=&\dfrac{\text{F}}{\text{Unit CM}}\\\dfrac{\7,700}{\275}&=&28\text{ kayaks}\end{array}$
So regardless of the method used, you get to the same result! |
GreeneMath.com - Special Products Test
Special Products Test
When we multiply with polynomials, some problem types repeat themselves often. We generally refer to these as special products. It is very beneficial to memorize the formulas for all of the special products. This will allow us to conduct the multiplication in such a scenario very quickly.
Test Objectives:
•Demonstrate the ability to quickly find the product of a binomial squared
•Demonstrate the ability to quickly find the product of the sum and difference of the same two terms
•Demonstrate the ability to quickly find the product of a binomial cubed
Special Products Test:
#1:
Instructions: Find each product.
a) (x + 4)2
b) (4x - 8)(4x + 8)
Watch the Step by Step Video Solution
|
View the Written Solution
#2:
Instructions: Find each product.
a) (2u - 7)(2u + 7)
b) (5x2 + 7y2)(5x2 - 7y2)
Watch the Step by Step Video Solution
|
View the Written Solution
#3:
Instructions: Find each product.
a) (6 + 4p2)2
b) (10y2 + 8x)2
Watch the Step by Step Video Solution
|
View the Written Solution
#4:
Instructions: Find each product.
a) (7n3 - 10m)2
b) (-10x2 - 10y2)2
Watch the Step by Step Video Solution
|
View the Written Solution
#5:
Instructions: Find each product.
a) (5x2 + 3y3)3
b) (9x3z5 - 2y4)3
Watch the Step by Step Video Solution
|
View the Written Solution
Written Solutions:
#1:
Solution:
a) x2 + 8x + 16
b) 16x2 - 64
Watch the Step by Step Video Solution
#2:
Solution:
a) 4u2 - 49v2
b) 25x4 - 49y4
Watch the Step by Step Video Solution
#3:
Solution:
a) 16p4 + 48p2 + 36
b) 100y4 + 160xy2 + 64x2
Watch the Step by Step Video Solution
#4:
Solution:
a) 49n6 - 140n3m + 100m2
b) 100x4 + 200x2y2 + 100y4
Watch the Step by Step Video Solution
#5:
Solution:
a) 125x6 + 225x4y3 + 135x2y6 + 27y9
b) 729x9z15 - 486x6y4z10 + 108x3y8z5 - 8y12
Watch the Step by Step Video Solution |
# State True or False. If $a$,$b$,$c$ are in A.P., then $b + c$,$c + a$,$a + b$ are also in A.P. $A$. $True$B.False Answer Verified Hint: The above problem is related to arithmetic progression. First, we should be able to define an arithmetic sequence. After that its properties can be used to solve the above problem. Arithmetic Progression is a sequence of numbers such that the difference of any two consecutive numbers is constant. General {n^{th}} term of an A.P. is given by: {a_n} = a + \left( {n - 1} \right)d Where a is the first term of the A.P. sequence and d is known as the common difference. Given in the problem, a,b,c are in A.P. \Rightarrow First term = a \Rightarrow Second term = a + d = b \Rightarrow Third term = a + 2d = c \Rightarrow b - a = c - b = d ……………………………….. (1) Here d is the common difference. Since we need to check whether b + c,c + a,a + b are in A.P. or not. We need to find the common difference of the consecutive terms. \Rightarrow c + a - \left( {b + c} \right) = c + a - b - c = a - b Using equation (1) in above, we get c + a - \left( {b + c} \right) = a - b = - d ………………………….(2) Similarly, a + b - \left( {c + a} \right) = a + b - c - a = b - c \Rightarrow a + b - \left( {c + a} \right) = b - c = - d ………………...(3) From (2) and (3) , the common difference of the consecutive terms is equal. \Rightarrow$b + c$,$c + a$,$a + b$ are in A.P.
Hence option $(A)$ is the correct answer. |
# Math for nine-year-olds: fold, punch and cut for symmetry!
Today I had the pleasure to visit my daughter’s fourth-grade classroom for some fun mathematical activities. The topic was Symmetry! I planned some paper-folding activities, involving hole-punching and cutting, aiming to display the dynamism that is present in the concept of symmetry. Symmetry occurs when a figure can leap up, transforming itself through space, and land again exactly upon itself in different ways. I sought to have the students experience this dynamic action not only in their minds, as they imagined various symmetries for the figures we considered, but also physically, as they folded a paper along a line of symmetry, checking that this fold brought the figure exactly to itself.
The exercises were good plain fun, and some of the kids wanted to do them again and again, the very same ones. Here is the pdf file for the entire project.
To get started, we began with a very simple case, a square with two dots on it, which the girls folded and then punched through with a hole punch, so that one punch would punch through both holes at once.
Next, we handled a few patterns that required two folds. I told the kids to look for the lines of symmetry and fold on them. Fold the paper in such a way that with ONE punch, you can make exactly the whole pattern.
Don’t worry if the holes you punch do not line up exactly perfectly with the dots — if the holes are all touching their intended target and there are the right number of holes, then I told the kids it is great!
The three-fold patterns are a bit more challenging, but almost all of the kids were able to do it. They did need some help punching through, however, since it sometimes requires some strength to punch through many layers.
With these further patterns, some of the folds don’t completely cover the paper. So double-check and make sure that you won’t end up with unwanted extra holes!
The second half of the project involved several cutting challenges. To begin, let’s suppose that you wanted to cut a square out of the middle of a piece of paper. How would you do it? Perhaps you might want to poke your scissors through and then cut around the square in four cuts. But can you do it in just ONE cut? Yes! If you fold the paper on the diagonals of the square, then you can make one quick snip to cut out exactly the square, leaving the frame completely intact.
Similarly, one can cut out many other shapes in just one cut. The rectangle and triangle are a little trickier than you might think at first, since at a middle stage of folding, you’ll find that you end up folding a shorter line onto a longer one, but the point is that this is completely fine, since one cut will go through both. Give it a try!
Here are a few harder shapes, requiring more folds.
It is an amazing mathematical fact, the fold-and-cut theorem, that ANY shape consisting of finitely many straight line segments can be made with just one cut after folding. Here are a few challenges, which many of the fourth-graders were able to do during my visit.
What a lot of fun the visit was! I shall look forward to returning to the school next time.
In case anyone is interested, I have made available a pdf file of this project. I would like to thank Mike Lawler, whose tweet put me onto the topic. And see also the awesome Numberphile video on the fold-and-cut theorem, featuring mathematician Katie Steckles.
See more of my Math for Kids!
## 27 thoughts on “Math for nine-year-olds: fold, punch and cut for symmetry!”
1. Next year, when you teach them about forcing, you can use this to teach them about symmetric extensions! 🙂
• Can you cut out every desired submodel with one cut?
• I’ve likened symmetric models to buyer’s remorse: you begin with a model, and you keep regretting the generic you’ve chosen at the “generic filters store”. At the end you are left with a bunch of sets which make a model of ZF, but not necessarily ZFC.
I don’t know if it’s possible to cut it in a single cut. But surely the template of a symmetric model is the part which does not “really” depend on the choice of generic, and thus impertinent to this choice, and only those names which are interpreted the same way by “most” generics.
Maybe you should wait with this until they are 11 years old… 🙂
2. This is awesome! The exploration and the video. Can’t wait to show my students. They will dive right into this. Thanks for sharing!
3. The opposite ideas also are interesting to explore: mark fold lines on the paper and indicate either a location for a hole to be punched or a line to be cut. Can they work out what will result?
• Yes, I had actually done that with a few of the students, and it works well. You don’t have to mark the fold lines, but just actually fold the paper, and then unfold it, and mark one hole. Where will the other holes be? The students draws all the holes, and then actually does it, to check how well they did.
4. Thanks this gavey family something to do at ihop while we were waiting for pancakes.
• Pancakes and math for dinner! Sounds great!
5. GREAT !
You might remember I have translated in french some times ago some of your posts
* graph coloring, chromatic numbers, and Eulerian paths and circuits : http://toysfab.com/2014/07/coloriages-de-graphes-et-nombres-chromatiques/
* graph theory for kids! : http://toysfab.com/2016/06/la-theorie-des-graphes-pour-enfants/
My kids and people around me have unsurprisingly highly appreciated these 2 articles, I’d like to keep nurturing them…
Congratulations for this work and kind regards !
FXF
6. I love this activity. I find myself wondering if it would be enhanced or destroyed by incorporating patty paper into the exploration. That is, putting these designs onto a translucent paper before folding will emphasize the overlay, and probably add clarity to some of the symmetries. It’s not clear to me if that would also take some of the fun out of the project; I suspect it wouldn’t.
• Nice idea! Why not give it a try, and then report back here how it worked? We’d love to hear about it.
7. Pingback: Symmetry – G∞AT
8. I have been googling math club ideas for my kids’ school (3rd-, 4th- and 5th-graders) for months and finding your blog is analogous to hitting the jackpot. I did some Nim last week and will borrow your Nim ideas for this week. Symmetry next. Thank you!
• Glad to hear it! Let me know how things go, and good luck to your young mathematicians.
• Yes, indeed. I had said this in the post (toward the end), with the same link. |
# Find the value of $\bigstar$: Puzzle 5 - Every little Symbol
This puzzle replaces all numbers (and operations) with other symbols.
Your job, as the title suggests, is to find what value fits in the place of $\bigstar$.
All symbols abide to the following rules:
• Each numerical symbol represents integers and only integers. This means fractions and irrational numbers like $\sqrt2$ are not allowed. However, negative numbers and zero are allowed.
• Any symbol that is NOT numerical must be one of the following operations: $\{+,-,\times,\text{^}\}$. Notice how all operation are binary operations. This means that all operation symbols must have a number on their left and on their right. Use that fact to your advantage!
1. Each symbol represents a unique number/operation. This means that for any two symbols $\alpha$ and $\beta$ which are in the same puzzle, $\alpha\neq\beta$.
2. The following equations are satisfied (this is the heart of the puzzle): $$\text{I. }a\,@\,b=a\,@\,c \\ \space \\ \text{II. }a\,\#\,a=d \\ \space \\ \text{III. }d\,\#\,d=c \\ \space \\ \text{IV. }b=c\,\\,d \\ \space \\ \text{V. }e\,\\,e\,\%\,e\,\\,b=c\,\%\,d\,@\,c \\ \space \\ \text{VI. }\bigstar=e\,\\,d\,\%\,a$$
### What is a Solution?
A solution is a value for $\bigstar$, such that, for the group of numerical symbols in the puzzle $S_1$ and for the operational symbols in the puzzle $S_2$ there exists a one-to-one function $f:S_1\to\Bbb Z$ and another one-to-one function $g:S_2\to\{+,-,\times,\text{^}\}$ which, after replacing all provided symbols using these functions, satisfies all given equations.
### What is a Correct Answer?
An answer is considered correct if you can prove that a certain value for $\bigstar$ is a solution. This can be done easily by getting a function from every symbol in the puzzle to the correct values (that is, find an example for $f:S_1\to\Bbb Z$ and $g:S_2\to\{+,-,\times,\text{^}\}$).
An answer will be accepted if it is the first correct answer to also prove that the solution is the only solution. In other words, there is no other possible value for $\bigstar$.
Good luck!
Previous puzzles in the series:
Next Puzzle
• NOTE: once an answer is accepted, the next one in the series will be posted shortly after. Side Note: to get $\bigstar$ use \bigstar in MathJax, and to get $\text^$ use \text^. This could come in handy when writing an answer to this. – NODO55 Jan 18 '18 at 7:29
• Would it be possible for you give this series more descriptive titles? Perhaps use something that's different about this puzzle e.g. the missing operators. – boboquack Jan 18 '18 at 8:49
• Then what do you do when there isn't anything too special about the puzzle, or if the special part of the puzzle is practically a giveaway? – NODO55 Jan 18 '18 at 9:11
• Side note to solution: apparently I completely missed a possible value for a... but surprisingly it worked out in the end, so good job solvers! – NODO55 Jan 18 '18 at 9:37
## 2 Answers
$\bigstar$ is:
$11$
With
$@ = \text^$
$\# = +$
$\$ = \times\% = -a = 1b = 8c = 4d = 2e = 6$## Explanation:$\text{I. }a\,@\,b=a\,@\,c$There are 2 ways this could be true with all distinct symbols: Possibility 1:$@$is$\times$operator, and$a$is 0.$@ = \timesa = 0$Possibility 2:$@$is$\text^$operator, and$a$can be either$0$or$\pm1$.$@ = \text^a = 1\space or\space 0\space or -1\text{II. }a\,\#\,a=d\text{III. }d\,\#\,d=c$From this operation, we can see that$a$cannot be 0 since no matter the operation, there will be a duplicate between$a$,$c$, and$d$. As such, only possibility 2 is valid:$@ = \text^a = 1\space or\space -1$Building from that,$\#$is not multiplication since$d$and$c$would be$\pm1$too, and there would be at least a duplicate between$a$,$b$and$c$.$\#$is not subtraction either, cause then$d$would be 0, and$c$would be equal to$d$. Therefore:$\# = +a = 1\space or -1d = 2\space or -2c = 4\space or -4\text{IV. }b=c\,\$\,d$
if $\$$is -, b would be the same as d. Therefore: \ = \times b = 8 By elimination: \% = - \text{V. }e\,\\,e\,\%\,e\,\\,b=c\,\%\,d\,@\,c e * e - e * b = c - d ^ c e^{2} - 8e = (\pm4) - 2^{(\pm4)} If a, b and c are negative: e^{2} - 8e = -4 - (-2^{-4}) e^2 - 8e = -4 + 1/16 No integer solution for e, so this is invalid. Thus a, b and c are positive: e^{2} - 8e = 4 - 2^4 e(e-8) = -12 e = \{2, 6\} Since d is already 2, e cannot be 2 too. Therefore: e = 6 \text{VI. }\bigstar=e\,\\,d\,\%\,a 6 * 2 - 1 = 11 This should leave no other possibilities, since we've exhausted all possibilities for a-e and the symbols. • with Eq.1, why cant @ be * ? – NODO55 Jan 18 '18 at 8:06 • Ah, my bad. It is possible for @ to be * if a = 0, but eq.2 quickly shows that if a = 0, d and c will be, too. Thus 0 is out of the question. I'll rework my explanation accordingly. – votbear Jan 18 '18 at 8:09 • Another note: why cant \# be *? a=-1 seems to work with Eq.2 – NODO55 Jan 18 '18 at 8:12 • Yep, just mentioned that case too in my edit. I think that made it easier to read/follow too, thank you for the tips! – votbear Jan 18 '18 at 8:16 • You've got a slight problem left, if a=0 and # = ^, then 0^0=1, so you need to go one step further and say that in that case, you would have d = 1 and c = 1^1 = 1 which is a contradiction. – Florian Bourse Jan 18 '18 at 9:53 a=1, b=8, c=4, d=2, e=6, \bigstar=11. Because From \text{I. }, @ cannot be + or -. If @ is \times, a is 0 and from \text{II. }, d will also be 0. so @ is \text{^}, so a\text{^}(b-c)=1 (we already know that a is not 0), so a=\pm1. \text{#} cannot be -, otherwise from \text{III. }, c=0, and from \text{IV. }, if \ is \times, b=0, if \$$ is not$\times$,$b=d$, neither of these cases are acceptable. if$\text{#}$is$\times$, then from$\text{II. }$,$d=1$and$a=-1$, and from$\text{I. }$,$b$and$c$are of same parity. So from$\text{IV. }$,$\$$can neither be + or -. there are nothing left to use for \$$.
Hence $\text{#}$ must be $+$, and from $\text{II. }$, $d=2a$, from $\text{III. }$, $c=2d=4a$
From $\text{IV. }$, $\$$cannot be - otherwise c=d. So \$$ is$\times$and from$\text{IV. }$,$b=cd=8a^2$. Now$\%$is minus. Plug everything into$\text{V. }e\times e-e\times b=c-d\text{^}ce(e-8a^2)=4a-(2a)^{4a}$We know from the very beginning that$a=\pm 1$, So LHS is clearly an integer, and if$a=-1$the RHS will not be an integer. Hence$a=1, b=8, c=4, d=2$, plug them to the RHS of$\text{V. }$to get$-12$. Solve a quadratic equation the get$e=2$or$e=6$.$d=2$, so$e=6$. Now plug everything to the final formula to get$\bigstar=6\times2-1=11\$
• Still struggling with type setting. Not sure what I have done wrong but can you take a look? @NODO55 – Weijun Zhou Jan 18 '18 at 8:44
• You need to add 2 spaces at the end of the sentence for it to register as a newline. – votbear Jan 18 '18 at 8:50
• Thank you for helping me out! You saved my afternoon. @Votbear – Weijun Zhou Jan 18 '18 at 8:52
• Seems fine to me. – NODO55 Jan 18 '18 at 9:35 |
# The Decimal Representation Of 200
## Introduction
In this article, we will delve into the decimal representation of the number 200. For those who may not be familiar with decimal numbers, they are a system of numbers that use a base of ten and include ten digits (0-9). This system is widely used in everyday life, from counting money to measuring time.
## What is 200?
200 is a whole number that falls under the category of even numbers. It is the product of multiplying 100 by 2. In mathematical terms, it is represented by the numeral 200.
## Decimal Representation of 200
The decimal representation of 200 is simply the number 200 followed by a decimal point and a zero. In other words, it is 200.0. This might seem trivial, but it is important to understand the decimal representation of numbers as it is used in various mathematical calculations.
## Converting 200 to a Fraction
To convert 200 to a fraction, we simply place it over a denominator of 1, which gives us 200/1. This fraction can be simplified by dividing both the numerator and denominator by a common factor, which is 1. The simplified fraction is therefore 200/1, which is equivalent to the whole number 200.
## 200 as a Percentage
To express 200 as a percentage, we multiply it by 100. This gives us 20,000%. This might seem like a large number, but it is important to remember that percentages are used to represent a portion of a whole, and in this case, the whole is 200.
## 200 as a Decimal Percentage
A decimal percentage is a percentage expressed as a decimal number. To express 200 as a decimal percentage, we simply divide it by 100. This gives us 2.00, which is equivalent to 200%.
## 200 in Scientific Notation
Scientific notation is a way of writing very large or very small numbers using powers of ten. To express 200 in scientific notation, we write it as 2 x 10^2. This means that 200 is equal to 2 multiplied by 10 raised to the power of 2.
## 200 in Roman Numerals
Roman numerals are a system of numbers that originated in ancient Rome. To represent 200 in Roman numerals, we use the letter CC, which stands for 100 + 100.
## 200 in Binary
Binary is a system of numbers that uses only two digits – 0 and 1. To represent 200 in binary, we need to convert it to its binary equivalent. This is done by dividing 200 by 2 repeatedly until we get a quotient of 0. The remainders of these divisions are then read from bottom to top to get the binary equivalent. In this case, the binary equivalent of 200 is 11001000.
## 200 in Octal
Octal is a system of numbers that uses a base of eight and includes eight digits (0-7). To represent 200 in octal, we need to convert it to its octal equivalent. This is done by dividing 200 by 8 repeatedly until we get a quotient of 0. The remainders of these divisions are then read from bottom to top to get the octal equivalent. In this case, the octal equivalent of 200 is 310.
Hexadecimal is a system of numbers that uses a base of sixteen and includes sixteen digits (0-9 and A-F). To represent 200 in hexadecimal, we need to convert it to its hexadecimal equivalent. This is done by dividing 200 by 16 repeatedly until we get a quotient of 0. The remainders of these divisions are then read from bottom to top to get the hexadecimal equivalent. In this case, the hexadecimal equivalent of 200 is C8.
## 200 in Different Number Systems
We have seen that 200 can be represented in different number systems, including decimal, binary, octal, and hexadecimal. Each of these systems has its own unique properties and uses. For example, binary is used in computer programming, octal is used in Unix file permissions, and hexadecimal is used in color codes.
## Properties of 200
200 is an even number, which means that it is divisible by 2 without leaving a remainder. It is also a composite number, which means that it has more than two factors. The factors of 200 are 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, and 200.
## Applications of 200
The number 200 has various applications in different fields. For example, it is commonly used in measurements of time, such as 200 seconds, 200 minutes, and 200 hours. It is also used in measurements of money, such as 200 dollars, 200 euros, and 200 pounds.
## Conclusion
In conclusion, we have explored the decimal representation of the number 200 and seen how it can be represented in different number systems and formats. We have also looked at the properties and applications of 200. Understanding the decimal representation of numbers is crucial in many mathematical calculations, and knowing the various representations of a number can be useful in different fields. |
# Illustrative Mathematics Unit 6.3, Lesson 15: Finding This Percent of That
Learning Targets:
• I can solve different problems like “What is 40% of 60?” by dividing and multiplying.
Related Pages
Illustrative Math
### Lesson 15: Finding This Percent of That
Let’s solve percentage problems like a pro.
Illustrative Math Unit 6.3, Lesson 15 (printable worksheets)
### Lesson 15 Summary
The following diagram shows how to find the percent of a number.
### Lesson 15.1 Number Talk: Decimals
Find the value of each expression mentally. (0.23) · 100
50 ÷ 100
145 · 1/100
7 ÷ 100
Scroll down the page for the answer to the “Are you ready for more?” section.
### Lesson 15.2 Audience Size
A school held several evening activities last month—a music concert, a basketball game, a drama play, and literacy night. The music concert was attended by 250 people. How many people came to each of the other activities?
1. Attendance at a basketball game was 30% of attendance at the concert.
2. Attendance at the drama play was 140% of attendance at the concert.
3. Attendance at literacy night was 44% of attendance at the concert.
#### Are you ready for more?
50% of the people who attended the drama play also attended the music concert. What percentage of the people who attended the music concert also attended the drama play?
From lesson 5.2, we know that 350 people attended the drama play.
50% of 350 = (0.5) · 350 = 175 people attended both the drama play and music concert.
From lesson 5.2, we know that 250 people attended the music concert.
The percentage of the people who attended the music concert also attended the drama play would be 175 ÷ 250 · 100% = 70%
#### Lesson 15.3 - Everything is On Sale
During a sale, every item in a store is 80% of its regular price.
1. If the regular price of a T-shirt is \$10, what is its sale price?
2. The regular prices of five items are shown here. Find the sale price of each item.
item 1 item 2 item 3 item 4 item 5 regular price \$1 \$4 \$10 \$55 \$120 sale price
3. You found 80% of many values. Was there a process you repeated over and over to find the sale prices? If so, describe it. 4. Select all the expressions that could be used to find 80% of x. Be prepared to explain your reasoning.
#### Lesson 15 Practice Problems
1. a. To find 40% of 75, Priya calculates ⅖ · 75. Does her calculation give the correct value for 40% of 75? Explain or show how you know.
b. If x represents a number, does ⅖ · 75 always represent 40% of that number? Explain your reasoning.
2. Han spent 75 minutes practicing the piano over the weekend. For each question, explain or show your reasoning.
a. Priya practiced the violin for 152% as much time as Han practiced the piano. How long did she practice?
b. Tyler practiced the clarinet for 64% as much time as Han practiced the piano. How long did he practice?
3. Last Sunday 1,575 people visited the amusement park. 56% of the visitors were adults, 16% were teenagers, and 28% were children ages 12 and under. Find the number of adults, teenagers, and children that visited the park.
4. Order from greatest to least:
• 55% of 180
• 300% of 26
• 12% of 700
1. Complete each statement.
a. 20% of 60 is ________
b. 25% of ________ is 6
c. ________% of 100 is 14
d. 50% of 90 is ________
e. 10% of ________ is 7
f. 30% of 70 is ________
2. A shopper needs 24 sandwich rolls. The store sells identical rolls in 2 differently sized packages. They sell a six-pack for \$5.28 and a four-pack for \$3.40. Should the shopper buy 4 six-packs or 6 four-packs? Explain your reasoning.
3. On a field trip, there are 3 chaperones for every 20 students. There are 92 people on the trip. Answer these questions. If you get stuck, consider using a tape diagram.
a. How many chaperones are there?
b. How many children are there?
The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
# Coded Inequalities MCQ Quiz - Objective Question with Answer for Coded Inequalities - Download Free PDF
Last updated on Oct 20, 2023
Coded inequalities are kind of questions in which each symbol corresponds to a specific sign. Here, we need to replace all the symbols with their respective signs and then check which of the conclusions follows the inequality given. Coded inequalities are kind of an upgraded version of simple inequalities where we need to replace all the symbols with signs first and then compare the relations. There is nothing new except the way these questions are asked. The underlying concepts of simple inequalities and coded inequalities are same. The questions given below will give students an idea about how these questions are asked in real examinations.
## Latest Coded Inequalities MCQ Objective Questions
#### Comprehension:
Direction: Study the given information and answer which conclusion definitely follows:
P @ Q means P is neither less nor equal to Q.
P # Q means P is neither greater nor equal to Q.
P \$ Q means P is neither less nor greater than Q.
P & Q means P is not greater than Q.
P % Q means P is not less than Q.
Statement: Y # X & Z # N; X % R \$ T @ W
Conclusions:
I) N @ W
II) T % Y
III) Z % R
1. Only conclusion I is true.
2. Only conclusion II is true.
3. Both conclusion I and III is true.
4. All are true
5. None is true
Option 3 : Both conclusion I and III is true.
#### Coded Inequalities Question 1 Detailed Solution
Given:
P is @ # \$ & % to Q > < = ≤ ≥
Now,
Statements: Y # X & Z # N; X % R \$ T @ W
Y < X ≤ Z < N; X ≥ R = T > W
Y < X ≥ R = T > W
N > Z ≥ X ≥ R = T > W
Conclusions:
I) N @ W → N > W → It is given that N > Z X ≥ R = T > W. As we can see that N is greater than W, this conclusion follows.
II) T % Y → T Y → It is given that Y < X ≥ R = T. As there is no direct relation between Y and T, this conclusion also doesn't follow.
III) Z % R → Z R → It is given that Z X ≥ R. As we can see that Z is greater than or equal to R, this conclusion also follows.
Hence, the correct answer is Only conclusion I and III is true.
#### Comprehension:
Direction: Study the given information and answer which conclusion definitely follows:
P @ Q means P is neither less nor equal to Q.
P # Q means P is neither greater nor equal to Q.
P \$ Q means P is neither less nor greater than Q.
P & Q means P is not greater than Q.
P % Q means P is not less than Q.
Statements: M @ L \$ Y \$ C # V & O; S @ N @ Y & G
Conclusions:
I) V # G
II) S @ C
III) O % M
1. Only conclusion I is true.
2. Only conclusion II is true.
3. Both conclusion I and II is true.
4. All are true
5. None is true
Option 2 : Only conclusion II is true.
#### Coded Inequalities Question 2 Detailed Solution
Given:
P is @ # \$ & % to Q > < = ≤ ≥
Now,
Statements: M @ L \$ Y \$ C # V & O; S @ N @ Y & G
M > L = Y = C < V ≤ O; S > N > Y ≤ G
M > L = Y = C ≤ G
M > L = Y = C < N < S
S > N > Y = C < V ≤ O
G ≥ Y = C < V ≤ O
Conclusions:
I) V # G → V < G → It is given that G ≥ Y = C < V. As there is no direct relation between V and G, this conclusion doesn't follow.
II) S @ C → S > C → It is given that S > N > Y = C. As we can see that S is greater than C, this conclusion follows.
III) O % M → O M → It is given that M > L = Y = C < V ≤ O. As there is no direct relation between O and M, this conclusion also doesn't follow.
Hence, the correct answer is Only conclusion II is true.
#### Comprehension:
Direction: Study the given information and answer which conclusion definitely follows:
P @ Q means P is neither less nor equal to Q.
P # Q means P is neither greater nor equal to Q.
P \$ Q means P is neither less nor greater than Q.
P & Q means P is not greater than Q.
P % Q means P is not less than Q.
Statements: T # N @ M & H # D; M \$ L # Z; N % V @ C
Conclusions:
I) N % Z
II) H @ C
III) V # D
1. Only conclusion I is true.
2. Only conclusion II is true.
3. Both conclusion I and II is true.
4. All are true
5. None is true
Option 5 : None is true
#### Coded Inequalities Question 3 Detailed Solution
Given:
P is @ # \$ & % to Q > < = ≤ ≥
Now,
Statements: T # N @ M & H # D; M \$ L # Z; N % V @ C
T < N > M ≤ H < D; M = L < Z; N ≥ V > C
T < N > M = L < Z
T < N ≥ V > C
H > D ≥ M = L < Z
C < V ≤ N > M ≤ H < D
Conclusions:
I) N % Z → N ≥ Z → It is given that N > M = L < Z. As there is no direct relation between N and Z, this conclusion doesn't follow.
II) H @ C → H > C → It is given that C < V ≤ N > M ≤ H. As there is no direct relation between H and C, this conclusion also doesn't follow.
III) V # D → V < D → It is given that V ≤ N > M ≤ H < D. As there is no direct relation between V and D, this conclusion also doesn't follow.
Hence, the correct answer is None is true.
#### Coded Inequalities Question 4:
The symbols a, f, #, ⋆ and ∞ are used with the following meanings :
R a S implies that R is greater than S.
R f S implies that R is lesser than S.
R # S implies that R is equal to S.
R ⋆ S implies that R is either greater than or equal to S.
R ∞ S implies that R is either lesser than or equal to S.
Based on the above statements, answer the following :
Statements :
M a N, O f M, M ∞ P
Conclusions :
I. M ⋆ P
II. O f P
1. Conclusion I is true but II is false.
2. Conclusion II is true but I is false.
3. Both the conclusions are true.
4. Neither I nor II is true.
Option 2 : Conclusion II is true but I is false.
#### Coded Inequalities Question 4 Detailed Solution
Statements :
M a N means M > N
O f M means O < M
M ∞ P means M ≤ P
Conclusions :
I. M ⋆ P → means M ≥ P . This conclusion cannot be drawn from the above statements. Hence conclusion I is false.
II. O f P → means O < P. Since it is given that O < M and M ≤ P, we can conclude that O < P. Hence conclusion II is true.
Hence option 2 :Conclusion II is true but I is false is correct.
#### Coded Inequalities Question 5:
Direction: In the following question assuming the given statement to be true, find which of the conclusion among given conclusions is/are definitely true and then give your answers accordingly.
‘A & B’ means ‘A is either smaller than or equal to B.’
‘A @ B’ means ‘A is neither smaller than nor equal to B.’
‘A % B’ means ‘A is either greater than or equal to B.’
‘A # B’ means ‘A is neither greater than nor equal to B.’
‘A + B’ means ‘A is neither greater than nor smaller than B.’
Statements: A % D @ G + H # C; B & F @ C + E
Conclusions:
I. A @ E
II. G # F
1. Only I follow
2. Only II follow
3. Both I and II follows
4. Neither I nor II follow
5. Either I or II follow
Option 2 : Only II follow
#### Coded Inequalities Question 5 Detailed Solution
As per the information given,
A is Symbol & @ % # + Meaning ≤ > ≥ < = to B
Given Statement: A % D @ G + H # C; B & F @ C + E
On converting: A ≥ D > G = H < C; B ≤ F > C = E
On combining: A ≥ D > G = H < C = E < F ≥ B
Conclusions:
I. A @ E → A > E → False (as A ≥ D > G = H < C = E).
II. G # F → G < F → True (as G = H < C = E < F).
Hence, only II follow.
## Top Coded Inequalities MCQ Objective Questions
#### Comprehension:
Direction: In the following questions, the symbols *, #, %, & and \$ are used with the following meaning as illustrated below:
‘X * Y’ means ‘X is neither less than nor greater than Y’.
‘X # Y’ means ‘X is either greater than or equal to Y’.
‘X % Y’ means ‘X is less than Y’.
‘X & Y’ means ‘X is neither less than nor equal to Y’.
‘X \$ Y’ means ‘X is not greater than Y’.
Now in each of the following questions assuming the given statements to be True, find which of the conclusion/s given below them is/are definitely True?
Statements:
A % B, C & D, F * E # C, D % A
Conclusions:
I. D % B
II. E & A
III. F & D
1. Only Conclusion I is True.
2. Only Conclusion II is True.
3. Only Conclusion III is True.
4. Both Conclusions I and III are True.
5. Both Conclusions II and III are True.
Option 4 : Both Conclusions I and III are True.
#### Coded Inequalities Question 6 Detailed Solution
According to the given information,
X is Symbol * # % & \$ Meaning = ≥ < > ≤ to Y
Given Statements: A % B, C & D, F * E # C, D % A
On converting: A < B, C > D, F = E ≥ C, D < A
On combining: F = E ≥ C > D < A < B
Conclusions:
I. D % B → D < B → True (as D < A < B→ D < B)
II. E & A → E > A → False (as E ≥ C > D < A → E > D < A → thus clear relation between E and A can’t be determined)
III. F & D → F > D → True (as F = E ≥ C > D → F ≥ C > D → F > D)
Hence, both conclusions I and III are true.
#### Comprehension:
Direction: In the following questions, the symbols *, #, %, & and \$ are used with the following meaning as illustrated below:
‘X * Y’ means ‘X is neither less than nor greater than Y’.
‘X # Y’ means ‘X is either greater than or equal to Y’.
‘X % Y’ means ‘X is less than Y’.
‘X & Y’ means ‘X is neither less than nor equal to Y’.
‘X \$ Y’ means ‘X is not greater than Y’.
Now in each of the following questions assuming the given statements to be True, find which of the conclusion/s given below them is/are definitely True?
Statements
Q & P, Y * U, D % P, Y # D
Conclusions:
I. U * D
II. D \$ U
III. D & U
1. Only Conclusion I is True.
2. Only Conclusion II is True.
3. Both Conclusions I and II are True.
4. Either Conclusion I or II is True.
5. Either Conclusion I or III is True.
Option 2 : Only Conclusion II is True.
#### Coded Inequalities Question 7 Detailed Solution
According to the given information,
X is Symbol * # % & \$ Meaning = ≥ < > ≤ to Y
Statement: Q & P, Y * U, D % P, Y # D
On converting: Q > P, Y = U, D < P, Y ≥ D
On combining: Q > P > D ≤ Y = U
Conclusions:
I. U * D → U = D → False (as D ≤ Y = U → D ≤ U)
II. D \$ U → D U → True (as D ≤ Y = U → D ≤ U)
III. D & U → D > U → False (as D ≤ Y = U → D ≤ U)
Hence, Only conclusion II is true.
#### Coded Inequalities Question 8
In the following questions, the symbols \$, #, @, % and * illustrate the following meanings.
• P \$ Q - P is not smaller than Q
• P # Q - P is neither greater than nor equal to Q.
• P @ Q - P is neither smaller than nor equal to Q.
• P % Q - P not greater than Q
• P * Q - P is neither greater than nor smaller than Q
Statements:
K # L, L % M, M * N, N # O
Conclusions:
I. K # M
II. K * M
III. L % O
1. I only
2. Either I or II only
3. III only
4. All I, II and III
Option 1 : I only
#### Coded Inequalities Question 8 Detailed Solution
P is \$ # @ % * ≥ < > ≤ = To Q
K # L → K < L
L % M → L ≤ M
M * N → M = N
N # O → N < O
Combining all the statements together, we get:
K < L ≤ M = N < O
I. K # M → K < M → True (As, K < L ≤ M → K < M)
II. K * M → K = M → False (As, K < L ≤ M → K < M)
III. L % O → L ≤ O → False (As, L ≤ M = N < O → L < O)
Hence, ‘I only’ is the correct answer.
#### Coded Inequalities Question 9
Directions: In the following question assuming the given statements to be true, find which of the conclusion among given conclusions is/are definitely true and then give your answers accordingly.
Statements: H > F ≤ O ≤ L; F ≥ V < D
Conclusions:
I. L ≥ V
II. O > D
1. Only I is true
2. Only II is true
3. Both I and II are true
4. Either I or II is true
5. Neither I nor II is true
Option 1 : Only I is true
#### Coded Inequalities Question 9 Detailed Solution
Given statements: H > F ≤ O ≤ L; F ≥ V < D
Combined Statement: H > F, L ≥ O ≥ F ≥ V < D
Conclusions:
I. L ≥ V → True ( as L ≥ O ≥ F ≥ V → L ≥ V )
II. O > D → False ( as O ≥ F ≥ V < D → thus clear relation between O and D cannot be determined)
Hence, only I is true.
#### Comprehension:
Directions: In these types of questions the symbols \$, *, @, ©, # are used with the following meanings as illustrated below:
‘P\$Q’ means ‘P is smaller than Q’
‘P©Q’ means P is greater than Q’
‘P#Q’ means P is equal to Q’
‘P@Q’ means P is either equal to or greater than Q
‘P*Q’ means P is either equal to or less than Q
Now in each of the following questions assuming the given statements to be True, find which of the conclusion/s given below them is/are definitely True?
Statements:
B © C # D @ X; E * X; D @ Z
Conclusions:
II. Z @ B
1. if only Conclusion I is true.
2. if only Conclusion II is true.
3. if either Conclusion I or II is true.
4. if neither Conclusion I or II is true.
5. if both conclusion I and II are true.
Option 1 : if only Conclusion I is true.
#### Coded Inequalities Question 10 Detailed Solution
P is Symbol # © \$ * @ Meaning = > < ≤ ≥ to Q
Statements: B © C # D @ X; E * X; D @ Z
On converting: B > C = D ≥ X ≥ E; B > C = D ≥ Z
Conclusion:
I. B > E → TRUE (B > C = D ≥ X ≥ E)
II. Z ≥ B → False (B > C = D ≥ Z)
Hence Only Conclusion I is true.
#### Coded Inequalities Question 11
Directions: In the following question assuming the given statements to be true, find which of the conclusion among given conclusions is/are definitely true and then give your answers accordingly.
Statements: E > R < T, V > Z ≥ K = T
Conclusions:
I. V < E
II. R < Z
1. Only I is True
2. Only II is True
3. Both I and II are True
4. Either I and II is True
5. Neither I and II is True
Option 2 : Only II is True
#### Coded Inequalities Question 11 Detailed Solution
Given statements: E > R < T, V > Z ≥ K = T
Combined Statement: V > Z ≥ K = T > R < E
Conclusions:
I. V < E → False ( as V > Z ≥ K = T > R < E → thus clear relation between V and E cannot be determined)
II. R < Z → True ( as Z ≥ K = T > R → R < T = K ≤ Z → R < Z)
Hence, only II is true.
#### Coded Inequalities Question 12
Read the given statement and conclusions carefully. Decide which of the given conclusions is/are true based on the given statement.
Statement:
Z > H ≥ Y > R < D ≥ F > X
Conclusions:
I. Z ≥ X
II. F H
1. Neither conclusion I nor II is true
2. Only conclusion II is true
3. Only conclusion I is true
4. Both conclusions I and II are true
Option 1 : Neither conclusion I nor II is true
#### Coded Inequalities Question 12 Detailed Solution
Given: Z > H ≥ Y > R < D ≥ F > X
The logic followed here is:
There is no relationship between Z and X, F and H. This is obtained from given inequality.
Conclusion I: Z ≥ X → False (as Z > H ≥ Y > R < D ≥ F > X, no clear relation between Z and X can be formed)
Conclusion II: F > → False ( as H ≥ Y > R < D ≥ F, no clear relation between F and H can be formed)
Hence, Neither conclusion I nor II is true is the correct answer.
#### Coded Inequalities Question 13
Directions: In the following question, the symbols δ, @, ©, % and ⋆ are used with the following meaning as illustrated below.
‘A © B’ means ‘A is not smaller than B’.‘A % B’ means ‘A is neither smaller than nor equal to B ’.‘A ⋆ B’ means ‘A is neither greater than nor equal to B’.‘A δ B’ means ‘A is not greater than B’.‘A @ B’ means ‘A is neither greater than nor smaller than B’.
Now in each of the following questions assuming the given statements to be true, find which of the four conclusions I, II, II and IV given below them is / are definitely true and give your answer accordingly.
Statements:
Conclusions:
I. B % D
II. C @ A
III. D @ A
IV. B ⋆ D
1. Only I and III are true
2. Only III and IV are true
3. Only III is true
4. Only IV is true
Option 4 : Only IV is true
#### Coded Inequalities Question 13 Detailed Solution
Decode the information as:
A is Symbol δ ⋆ © % @ Meaning ≤ < ≥ > = Than B
After decoding the statements using the above table:
A ≥ B, C ≥ B, D > C
Combined decoded statement is:
D > C ≥ B ≤ A
Conclusions:
I. B > D → False (as D > C ≥ B → D > B)
II. C = A → False (as C ≥ B ≤ A → No direct relation found between A and C)
III. D = A → False (as C ≥ B ≤ A → No direct relation found between A and D)
IV. B < D → True (as D > C ≥ B → D > B)
Therefore, only IV is true.
#### Coded Inequalities Question 14
If A * Z means A is greater than Z
A # Z means A equal to Z
A + Z means A is less than or equal to Z
A − Z means A is neither greater than nor equal to Z,
then which conclusion(s) is/are correct according to the given expression?
Z # P + T − U * C # M
Conclusions:
I. Z < U
II. M > U
1. Neither I nor II
2. Only I
3. Both I and II
4. Only II
Option 2 : Only I
#### Coded Inequalities Question 14 Detailed Solution
Given:
A * Z means A is greater than Z
A # Z means A equal to Z
A + Z means A is less than or equal to Z
A − Z means A is neither greater than nor equal to Z,
Let us first decode the given symbols and then draw a family tree.
A is Sign * # + - Meaning > = ≤ < of Z
Given expression: Z # P + T − U * C # M
Now break the code Z = P T < U > C = M
Conclusions:
I. Z < U → True (as Z = P T < U gives; as Z T and T < U so Z < U; True).
II. M > U → False (as U > C = M gives; as U > M so M > U; False).
Hence, the correct answer is “Only I”.
#### Coded Inequalities Question 15
In the question two statements are given, followed by two conclusions, I and II. You have to consider the statements to be true even if it seems to be at variance from commonly known facts. You have to decide which of the given conclusions, if any, follows from the given statements.
Statements:
P = U < M < K ≤ I > N
Conclusions:
I. N ≥ K
II. I > P
1. If only conclusion II is true
2. If either conclusion I or II is true
3. If only conclusion I is true
4. If neither conclusion I nor II is true
Option 1 : If only conclusion II is true
#### Coded Inequalities Question 15 Detailed Solution
Given statement is,
P = U < M < K ≤ I > N
Conclusions:
I. N ≥ K, this is false as there is no relation between N and K
II: I > P, this is true.
Hence, “only conclusion II” is true. |
# Step by Step- RCC roof Slab steel calculation- Numerical Example
In this article, I have described step by step RCC roof Slab steel calculation. So read it carefully one by one to get a clear understanding.
Slab is a structural member which enable people to move from one floor and one place to another place in a flat. With out slab one can not take benefit of any structure like, building, culvert, bridge. etc.
## What is RCC Slab?
Simply, the slab in which steel bars are used is known as RCC Slab. The full form of RCC is reinforced concrete cement. Where Reinforcement is called a steel rod.
The slab should be strong enough to bear the required load according to design. The slab should be provided steel bars to withheld tensile force, so this slab is called the RCC slab.
I have given an example for the calculation of the quantity steel rod in slab below.
Let us take an example of a slab.
## Numecal for RCC roof Slab steel calculation
Q) The length and breadth of slab is 10 meter and 7 meter respectively.
Let, The shorter span has provided the main rod and longer span has provided distribution rod.
Beside this the slab has following information,
For main bars
Length of on rod = 7 meter
Diameter = 12 mm
Center to Center gap (c/c) = 5 inch
For Distribution bars
Length of on rod = 8 meter
Diameter = 10 mm
Center to Center gap (c/c) = 5 inch
Solution:-
Remember that, slab does not have bent up bar. It means the slab only has straight bars. But, don’t worry you can calculate that also in the same process.
Bent up bar is also called crank bars which are in sloped form at the edge of the slab. you can see in this picture below.
Now you have to see the calculation carefully to understand step by step.
#### Step:- 1
Calculate no. of Steel bar
For the main bar,
= L”/(c/c) + 1
= (8/0.127) +1
= 63.99
You can provide 64 bars
Now, For Distribution Bar
= L/(c/c) + 1
= (7/0.127)+ 1
=56.11
You can Provide 57 bars
#### Step:-2
Calculate total length of steel bars
For main bars,
= 64 * Length of one Main Rod
= 64* 7
= 448 m
Simillarly,
For distribution bars ,
= 57* Length of one Distribution bars
= 57* 8
= 456 m
#### Step:-3
Calculate weight of Steel bar
##### For main bar,
={(D^2)/162} * L
= {(12^2)/162} * 448
= 398.2 kg
Keep in mind that unit of Diameter (D) must be in millimeter and Length (L) must be in meter. And final result will be in Kg.
##### For Distribution bar
W = {(D^2)/162} * L
= {(10^2)/162} * 456
= 281.48 kg
Hence, in this way you can calculate weight or quantity of steel rod in a slab. |
# Math camera online
Math camera online can support pupils to understand the material and improve their grades. Our website will give you answers to homework.
## The Best Math camera online
Here, we debate how Math camera online can help students learn Algebra. Once this has been accomplished, the resulting equation can be solved for the remaining variable. In some cases, it may not be possible to use elimination to solve a system of linear equations. However, by understanding how to use this method, it is usually possible to simplify a system of equations so that it can be solved using other methods.
Factor calculators are a great tool for anyone who wants to quickly figure out the factors of a number. Although there are various ways to calculate factors, a factor calculator can be especially helpful if you're working with large numbers or if you need to find all of the factors of a number. Generally, you simply enter the number that you want to factor into the calculator and then hit the "calculate" button. The calculator will then display all of the factors of that number. In addition, some factor calculators will also show you any prime factors that may be present. Factor calculators can be found online or in some math textbooks.
How to solve factorials? There are a couple different ways to do this. The most common way is to use the factorial symbol. This symbol looks like an exclamation point. To use it, you write the number that you want to find the factorial of and then put the symbol after it. For example, if you wanted to find the factorial of five, you would write 5!. The other way to solve for factorials is to use multiplication. To do this, you would take the number that you want to find the factorial of and multiply it by every number below it until you reach one. Using the same example from before, if you wanted to find the factorial of five using multiplication, you would take 5 and multiply it by 4, 3, 2, 1. This would give you the answer of 120. So, these are two different ways that you can solve for factorials!
In mathematics, the domain of a function is the set of all input values for which the function produces a result. For example, the domain of the function f(x) = x2 is all real numbers except for negative numbers, because the square of a negative number is undefined. To find the domain of a function, one must first identify all of the possible input values. Then, one must determine which input values will produce an undefined result. The set of all input values that produce a defined result is the domain of the function. In some cases, it may be possible to solve for the domain algebraically. For example, if f(x) = 1/x, then the domain is all real numbers except for 0, because division by 0 is undefined. However, in other cases it may not be possible to solve for the domain algebraically. In such cases, one can use graphing to approximate thedomain.
Another method is to use exponential equations. Exponential equations are equivalent to log equations, so they can be manipulated in the same way. By using these methods, you can solve natural log equations with relative ease.
This is a great app that is unlike no other because it goes above and beyond any other app. One feature I love about this app is that it not only provides you with the answer but it gives in depth details to help explain why the solution is true. Without it I'd probably struggle a lot in math. Thank you, developers! Keep up the great work!
Xanthia Moore
The app is really good, really good! The only problem that I see is that when I am using the calculator instead of having a period, it has a comma, and for me it is a little confusing. It would be nice to have an option of having both of them though.
Felicity Peterson
Solve precalculus math problems Solve math problems show work Dividing polynomials solver System of equations by elimination solver College math equations and answers |
Perimeters and Areas of Similar Polygons
# Perimeters and Areas of Similar Polygons
You may want to review: Similarity, Ratios, and Proportions
Suppose that a triangle with sides $\,a\,,$ $\,b\,$ and $\,c\,$ has been scaled by a positive number $\,s\,$ to get a similar triangle with corresponding sides $\,A\,,$ $\,B\,,$ and $\,C\,.$
Thus, $\,A = sa\,$ and $\,B = sb\,$ and $\,C = sc\,.$
Let's investigate the relationship between the perimeters of these two triangles:
\begin{align} &\cssId{s8}{\text{perimeter of original triangle} = a + b + c}\cr\cr &\cssId{s9}{\text{perimeter of scaled triangle}}\cr &\qquad \cssId{s10}{= A + B + C}\cr &\qquad \cssId{s11}{= sa + sb + sc}\cr &\qquad \cssId{s12}{= s(a + b + c)}\cr &\qquad \cssId{s13}{= s(\text{perimeter of original triangle})}\cr \end{align}
Thus, the perimeter ends up being scaled by the same factor that scales the sides!
This can be re-phrased in terms of ratios. Notice that:
\begin{align} &\cssId{s17}{\frac{\text{scaled perimeter}}{\text{original perimeter}}}\cr\cr &\qquad \cssId{s18}{= \frac{A+B+C}{a+b+c}}\cr\cr &\qquad \cssId{s19}{= \frac{sa+sb+sc}{a+b+c}}\cr\cr &\qquad \cssId{s20}{= \frac{s(a+b+c)}{a+b+c}}\cr\cr &\qquad \cssId{s21}{= s}\cr\cr &\qquad \cssId{s22}{= \text{scaling factor}} \end{align}
and
$$\begin{gather} \cssId{s24}{\frac{\text{scaled side}}{\text{original side}}} \cssId{s25}{= \frac{A}{a}} \cssId{s26}{= \frac{sa}{a}} \cssId{s27}{= s}\cr\cr \cssId{s28}{\frac{\text{scaled side}}{\text{original side}} = \frac{B}{b} = \frac{sb}{b} = s}\cr\cr \cssId{s29}{\frac{\text{scaled side}}{\text{original side}} = \frac{C}{c} = \frac{sc}{c} = s} \end{gather}$$
So, the ratio of the perimeters is the same as the ratio of corresponding sides.
A similar calculation shows that this result is indeed true for polygons in general:
THEOREM Perimeters of Similar Polygons
The ratio of the perimeters of two similar polygons is equal to the ratio of the corresponding sides.
That is:
\begin{align} &\cssId{s36}{\frac{\text{perimeter of scaled polygon}}{\text{perimeter of original polygon}}}\cr\cr &\qquad \cssId{s37}{= \text{the scaling factor}}\cr\cr &\qquad \cssId{s38}{= \frac{\text{scaled side}}{\text{original side}}} \end{align}
Area, on the other hand, behaves quite differently.
Suppose you have a square of side $\,a\,$ that has been scaled by $\,s\,$ to get a square of side $\,A\,.$ Thus, $\,A = sa\,.$
Let's investigate the relationship between the areas of these two squares:
\begin{align} &\cssId{s45}{\text{area of original square}} \cssId{s46}{= a\cdot a} \cssId{s47}{= a^2}\cr\cr &\cssId{s48}{\text{area of scaled square}}\cr &\qquad \cssId{s49}{= A\cdot A}\cr &\qquad \cssId{s50}{= (sa)\cdot(sa)}\cr &\qquad \cssId{s51}{= s^2\cdot a^2}\cr &\qquad \cssId{s52}{= s^2(\text{area of original square})} \end{align}
Thus, the area ends up being scaled by the square of the factor that scales the sides!
Again, this can be re-phrased in terms of ratios. Notice that:
\begin{align} &\cssId{s56}{\frac{\text{scaled square area}}{\text{original square area}}}\cr\cr &\qquad \cssId{s57}{= \frac{A\cdot A}{a \cdot a}}\cr\cr &\qquad \cssId{s58}{= \frac{(sa)\cdot (sa)}{a\cdot a}}\cr\cr &\qquad \cssId{s59}{= \frac{s^2a^2}{a^2}}\cr\cr &\qquad \cssId{s60}{= s^2}\cr\cr &\qquad \cssId{s61}{= \text{the square of the ratio of corresponding sides}} \end{align}
This idea leads to the following theorem:
THEOREM Areas of Similar Polygons
The ratio of the areas of two similar polygons is equal to the square of the ratio of the corresponding sides.
## Examples
Question: Suppose that a triangle has sides of lengths $\,x\,,$ $\,y\,,$ and $\,z\,.$ The sides of the triangle are scaled by a factor of $\,5\,$ to get a similar triangle. What is a formula for the perimeter of the new triangle?
Solution: The perimeter is scaled by the same factor as the sides. The original perimeter is $\,x+y+z\,.$ The new (scaled) perimeter is $\,5(x+y+z)\,.$
Question: Suppose that the sides of a polygon are scaled by a factor of $\,3\,$ to get a similar polygon. This new polygon has area $\,198\,.$ What is the area of the original polygon?
Solution: The area is scaled by the square of the factor that scales the sides. Let $\,A\,$ denote the (unknown) original area. We are told that the new (scaled) area is $\,198\,,$ and the scaling factor is $\,3\,.$
So, $\,A\,$ gets scaled by $\,3^2\,$ to give $\,198\,$; i.e., $\,3^2A = 198\,.$
Solving for $\,A\,$ gives $\,A = \frac{198}{3^2} = 22\,,$ so the area of the original polygon is $\,22\,.$ |
Definitions
An set of vectors is an orthogonal set if each pair of vectors is orthogonal to eachother such that
(1)
\begin{align} \begin{Bmatrix} u_1,...,u_n \end{Bmatrix}\\ u_1 \cdot u_2 = 0, u_1 \cdot u_2 = 0, ... , u_{n-1\text{}} \cdot u_n = 0 \end{align}
In addition, if there is a subspace S formed by an orthogonal set, that set is linearly independent and is thus a basis for S. Therefore, if there exists a vector $\vec{y}$ is S, then $\vec{y}$ can be formed by the orthogonal set
(2)
\begin{align} \vec{y} = c_1\vec{u_1}+c_2\vec{u_2}+...+c_n\vec{u_n} \end{align}
We can therefore multiply each term by $\vec{u_1}$ for instance, and since all other u's are perpendicular to $u_1$, all other terms drop out except $\vec{y} \cdot \vec{u_1} = c_1 \vec{u_1}^2$. This can be simplified and $c_1$ can be found using
(3)
\begin{align} \frac{\vec{y} \cdot \vec{u_1}}{\vec{u_1}^2} = c_1 \end{align}
This is true for all c's so we now have the tools to find any vector in S by a linear combination of the orthogonal basis.
## Orthogonal Projection
Now, if we're looking to express $\vec{y}$ as a sum of two vectors, one which is a multiple of $\vec{u}$ and one perpendicular to $\vec{u}$, we must apply some of what we just learned. We know the scalar $c_1$ to create a vector in S that is a component of $\vec{y}$ is $\frac{\vec{y} \cdot \vec{u_1}}{\vec{u_1}^2}$. Using this to scale $\vec{u_1}$ which we'll call $\hat{y}$, we find
(4)
\begin{align} \vec{y} = \hat{y} + \vec{v} \end{align}
where $\vec{v}$ is the perpendicular vector such that
(5)
\begin{align} \vec{y} - \hat{y} = \vec{v} \\ \vec{v} \cdot \hat{y} = 0 \end{align}
Orthonormal Sets
An Orthonormal set is a set of orthogonal vectors whose magnitudes are all 1.
To normalize a vector, just divide each component by the magnitude of the vector so…
(6)
\begin{align} \hat{x} = \frac{\vec{x}}{|\vec{x}|} \end{align}
Then the orthonormal set has the property $U^TU = I$. Thus, $U^T = U^{-1}$.
page revision: 0, last edited: 02 May 2015 17:45 |
# Find Four Consecutive Integers Whose Sum is 138 by 3 Ways
Result:
Verify:
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The sum of four consecutive integers is 138, what are these integers? I can easily tell you that the answer are 33, 34, 35 and 36. You must be interested in how to find these 4 consecutive integers whose sum is 138. There are three methods here, let us introduce them one by one below.
## How to find four consecutive integers whose sum is 138?
### 1. Hypothetical method
Assuming that N is used to represent the first integer, so the second integer is N + 1, the third integer is N + 2, the 4th integer is N + 3.
Therefore, the sum of 4 consecutive integers is 138, which can be expressed by the equation
N + (N + 1) + (N + 2) + (N + 3) = 138
Solve this equation
N + (N + 1) + (N + 2) + (N + 3) = 138
N + N + 1 + N + 2 + N + 3 = 138
4 * N + 6 = 138
4 * N = 138 – 6
4 * N = 132
N = 132 / 4
N = 33
Now, we can get that 33 is the first integer of 4 consecutive integers whose sum is 138. So, the second integer is 34, the third integer is 35, the 4th integer is 36.
The answer came out, the sum of 4 consecutive integers is 138, these three integers are 33, 34, 35 and 36.
Verify: 33 + 34 + 35 + 36 = 138. Correct!
This is the most common method, let’s look at the second method, which is my favorite method.
### 2. Formula method
According to the consecutive integers calculator based on sum, we can know that the sum of M consecutive integers is S, the first integer formula is
First(n) = S / M – (M – 1) / 2
M represents the number of consecutive integers.
S stands for sum.
Back to the problem we want to solve: find four consecutive integers whose sum is 138. In here, M = 4, S = 138. Replace them in the formula to calculate the first integer
First(n) = 138 / 4 – (4 – 1) / 2 = 34.5 – 3 / 2 = 34.5 – 1.5 = 33
The first integer is 33, so, it can be easily calculated that the second integer is 34, the third integer is 35, the 4th integer is 36.
Thus, the answer are also 33, 34, 35 and 36. Same as the first method. Then look at the third method, which is the simplest one.
### 3. Average method
The principle is very simple, because we want to calculate the consecutive integers, so after calculating the average, we can find the integers near the average.
The sum of 4 consecutive integers is 138, so, the average of these 4 consecutive integers is 138 / 4 = 34.5. The integers around 34.5 are 32, 33, 34, 35, 36, 37, 38. Now the problem is simple, find 4 consecutive integers from 32 to 38 and their average is 34.5. The answer are 33, 34, 35 and 36.
So the sum of 4 consecutive integers is 138, these integers are 33, 34, 35, 36. The results are consistent with the above two methods. Is it very simple?
## Problems can be sloved by this answer
Now, we have found out these 4 consecutive integers whose sum is 138, some problems can be easily solved.
• 1. What are four consecutive integers whose sum is 138?
The answer are 33, 34, 35 and 36.
• 2. What is the smallest of four consecutive integers whose sum is 138?
The smallest number is 33.
• 3. What is the greatest of four consecutive integers whose sum is 138?
The greatest number is 36.
• 4. What is the average of four consecutive integers whose sum is 138?
The average of these 4 consecutive integers is (33 + 34 + 35 + 36) / 4 = 138 / 4 = 34.5.
• 5. What is the product of four consecutive integers whose sum is 138?
The sum of 4 consecutive integers is 138, the product of them is 33 * 34 * 35 * 36 = 1413720.
• 6. What is the sum of square of four consecutive integers whose sum is 138?
The sum of 4 consecutive integers is 138, the sum of their squares is 332 + 342 + 352 + 362 = 4766.
## Summarize
On this page, In addition to introducing three methods how to find four consecutive integers whose sum is 138, it also provides a calculator that calculates four consecutive integers based on the sum. If you encounter a similar problem next time, you can directly use this calculator to calculate the answer, which is very convenient.
Of course, if the problem you encounter is more complicated, such as: the number of consecutive integers is not 4, or you need to calculate consecutive odd integers or even integers. You can use our other more advanced sum-based consecutive integers calculator, where you can specify the number of consecutive integers and select consecutive integers type: natural integers, odd integers or even integers. I believe it can help you.
Well, that’s it, the above are three solutions to find four consecutive integers whose sum is 138. Personally, I prefer the second method, because it can accurately calculate the first integer and the calculation process is very simple. So how about you? Please leave a message and tell me which method you like? |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 2.7: Square Roots and Real Numbers
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
At the end of this lesson, students will be able to:
• Find square roots.
• Approximate square roots.
• Identify irrational numbers.
• Classify real numbers.
• Graph and order real numbers.
## Vocabulary
Terms introduced in this lesson:
square root
principal square root
positive square root
perfect squares
prime factors
rational number
irrational number
approximation
non-repeating decimals
simplest form
integer
sub-intervals
## Teaching Strategies and Tips
Contrary to what students think, there is no discrepancy between there being two possible values for b\begin{align*}b\end{align*}, given a positive x\begin{align*}x\end{align*}, such that b2=x\begin{align*}b^2 = x\end{align*} and only one value for x\begin{align*}\sqrt{x}\end{align*}. The positive number b\begin{align*}b\end{align*} is the principal square root. It is the value of the function.
Point out that ¯¯¯¯¯¯¯\begin{align*}\surd^{\overline{\;\;\;}}\end{align*} and 2\begin{align*}\sqrt[2]{}\end{align*} mean the same thing. The 2\begin{align*}“2”\end{align*} is understood.
In Example 1:
• Use factor trees to break down the radicands into as many perfect squares as possible.
• Primes which appear an even number of times constitute a perfect square.
• Any unpaired factors are left under the radical sign.
• The convention is to leave any irreducible radical in the form: (square root)(irreducible part)\begin{align*}\mathrm{(square\ root)} \sqrt{(\mathrm{irreducible \ part})}\end{align*}
Lots of practice early on makes it easier for students when the radicals become more complicated and involve variable expressions.
Use Example 3 to multiply, divide, and simplify radical expressions. State which rules were used in classroom examples.
In Example 4, students use a calculator and round their answer to three decimal places. Review rounding decimals.
Motivate irrational numbers in Example 5.
• Irrational numbers complete the set of real numbers.
• They cannot be expressed as ratio of two integers.
• They have an unending (non-terminating), seemingly random (non-repeating) decimal expansion.
• Some irrational numbers: π,1π,2,3,any prime\begin{align*}\pi, \frac{1} {\pi}, \sqrt{2}, \sqrt{3}, \sqrt{\mathrm{any \ prime}}\end{align*}
In Example 5, students identify the given numbers. A number is rational if it
• can be expressed as a fraction of integers.
• has a finite decimal expansion.
• has a repeating block of digits in its decimal expansion.
Have students check their calculator displays for repeating blocks of digits, not just patterns. For example, the two numbers below have obvious patterns, but no repeating blocks, and therefore are irrational:
• 0.010010001000010000010000001\begin{align*}0.010010001000010000010000001 \ldots \end{align*}
• 0.12345678910111213141516171819202122\begin{align*}0.12345678910111213141516171819202122 \ldots\end{align*}
Remind students that integers can be written as fractions with a 1\begin{align*}1\end{align*} in the denominator. See Example 6a and 6b. The strategy in Example 6e is to simplify first.
The strategy in Example 8 is to use a calculator to find the decimal expansions of the numbers rounded to as many places as is needed to identify each. Since all the numbers are between 3.1\begin{align*}3.1\end{align*} and 3.2\begin{align*}3.2\end{align*}, going out to three decimal places is sufficient.
## Error Troubleshooting
In Problem 3 of the Review Questions:
• Students should find repeating blocks of digits before claiming that a number is rational.
• It is not sufficient to claim that the “unpredictable” decimals of a number on a calculator display belong to an irrational number. The decimal expansion of 2/19\begin{align*}2/19\end{align*} has 18\begin{align*}18\end{align*} seemingly random digits in its repeating block; therefore, the rational number 2/19\begin{align*}2/19\end{align*} would go “unchecked” on an ordinary calculator display because it could not show enough digits.
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Breaking up the second number is a mental math strategy for addition. Some students may find this method more efficient than left-to-right addition.
This strategy involves breaking up the second number in an equation into more manageable parts. Like many other mental math strategies, this strategy encourages students to think flexibly and to manipulate numbers in different ways. This is the big goal of mental math!
As you look at the examples given, you’ll notice that this strategy reinforces place value understanding, as students are breaking the second number into its expanded form.
EXAMPLES
Let’s take a look at how to perform this strategy. Whenever you introduce a new strategy in your classroom, be sure to use small, easy to work with numbers. This will ensure that students can focus on the strategy itself rather than struggling with big numbers while trying to master a new strategy.
In this example, we will add 14+12. We will break the 12 into a 10 and a 2.
Now we add. First we add 14+10 to make 24, and then add the remaining 2 to make 26.
Let’s try another example. Here we will solve 35+46. First we break the 46 into a 40 and a 6.
We will add 35+40 to make 75, and then add the remaining 6 to make 81.
Breaking up the second number can be used with more digits as well. Let’s try a 3-digit plus 3-digit equation. Here we will add 124+345. We can first break up the 345 into a 300, a 40, and a 5.
Now we begin by adding 124+300 to make 424. Then we will add the 40 to make 464. Lastly we will add the remaining 5 to make 469.
FLEXIBLE THINKING
One of the greatest aspects of mental math is that there is not a series of steps to memorize. Really we just want our students to understand what the numbers mean and be able to manipulate them in a way that works for each individual student.
Suppose we have the equation 213+214.
One student might choose to break up the second number and add 213+200+10+4.
Another student might choose to break up the second number into only two parts and add 200+210+4.
A third student might choose to add these numbers using left to right addition.
There is no ONE right way. When students know the values of the digits, and understand what the numbers mean, we open up the options for how an equation can be solved.
This is one strategy that you definitely will want to incorporate into your math instruction. It can be used in many different ways and you will notice that your students begin using this sort of thinking for other math concepts as well.
NEXT STEPS:
• If you would like full support for teaching addition strategies in your classroom, check out The Addition Station HEREYou’ll find this strategy in the Addition Stations for the upper grades. These Math Stations are self-paced, student-centered stations for the basic math strategies. Students move through the levels at their own pace, ensuring that they are always challenged, and working to their full potential. |
# What is the implicit derivative of 25=sin(xy)/x-3xy?
Nov 19, 2015
Long time since tried this but I am having a go!
$\textcolor{b l u e}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sin \left(x y\right) - y x \cos \left(x y\right)}{{x}^{2} \cos \left(x y\right) - 3 {x}^{3}} + \frac{3 y}{\cos \left(x y\right) - 3 x}}$
$\textcolor{red}{\text{You will need to check this!!}}$
#### Explanation:
$\frac{d}{\mathrm{dx}} \left(\frac{\sin \left(x y\right)}{x} - 3 x y\right) = \frac{d}{\mathrm{dx}} \left(25\right)$
$\frac{d}{\mathrm{dx}} \left(\frac{\sin \left(x y\right)}{x}\right) - \frac{d}{\mathrm{dx}} \left(3 x y\right) = 0$
Using std forms $\frac{d}{\mathrm{dx}} \left(u v\right) = v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}$
$\textcolor{w h i t e}{\times \times \times \times \times} \frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let $\textcolor{w h i t e}{\times} w = u v \to x y$
then ${\textcolor{g r e e n}{\frac{\mathrm{dw}}{\mathrm{dx}} \to \frac{\mathrm{dy}}{\mathrm{dx}} = y + x \frac{\mathrm{dy}}{\mathrm{dx}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(1\right)}}_{\text{confirmed}}$
Let $t = \sin \left(x y\right) = \sin \left(w\right)$
then ${\textcolor{g r e e n}{\frac{\mathrm{dt}}{\mathrm{dx}} \to \frac{\mathrm{dy}}{\mathrm{dx}} = \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) \cos \left(x y\right) \ldots \ldots \ldots \ldots . \left(2\right)}}_{\text{confirmed}}$
color(green)(d/dx(3xy) = 3(y+x(dy)/dx) "from (1) "................(3))_"confirmed"
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider: $\frac{d}{\mathrm{dx}} \left(\frac{\sin \left(x y\right)}{x}\right) \to \frac{u}{v}$
$\frac{x \left\{y \cos \left(x y\right) + x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \cos \left(x y\right)\right\} - \sin \left(x y\right)}{x} ^ 2 \ldots \ldots {.}_{\textcolor{red}{\text{corrected y-> ycos(xy)}}}$
${\textcolor{g r e e n}{\frac{y \cos \left(x y\right)}{x} + \frac{\mathrm{dy}}{\mathrm{dx}} \cos \left(x y\right) - \sin \frac{x y}{x} ^ 2. \ldots \ldots . . \left(4\right)}}_{\textcolor{red}{\left(\text{corrected} y \to y \cos \left(x y\right)\right)}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Combining (3) and (4)}}$
$\left\{\frac{y}{x} \cos \left(x y\right) + \frac{\mathrm{dy}}{\mathrm{dx}} \cos \left(x y\right) - \sin \frac{x y}{x} ^ 2\right\} - \left\{3 y + 3 x \frac{\mathrm{dy}}{\mathrm{dx}}\right\} = {0}_{\text{corrected } y \to y \cos \left(x y\right)}$
Collecting like terms $\textcolor{red}{\text{(rebuilt calculations)}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} \cos \left(x y\right) - 3 x \frac{\mathrm{dy}}{\mathrm{dx}} \textcolor{red}{=} \frac{\sin \left(x y\right)}{x} ^ 2 - \frac{y}{x} \cos \left(x y\right) + 3 {y}_{\textcolor{red}{\text{ corrected } y \to y \cos \left(x y\right)}}$
$\textcolor{w h i t e}{x}$
$\frac{\mathrm{dy}}{\mathrm{dx}} \textcolor{red}{=} \frac{1}{\cos \left(x y\right) - 3 x} \left(\frac{\sin \left(x y\right)}{x} ^ 2 - \frac{y}{x} \cos \left(x y\right) + 3 y\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} \textcolor{red}{=} \frac{1}{\cos \left(x y\right) - 3 x} \left(\frac{\sin \left(x y\right) - y x \cos \left(x y\right)}{x} ^ 2 + 3 y\right)$
$\textcolor{b l u e}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sin \left(x y\right) - y x \cos \left(x y\right)}{{x}^{2} \cos \left(x y\right) - 3 {x}^{3}} + \frac{3 y}{\cos \left(x y\right) - 3 x}}$
Nov 19, 2015
By working step-wise taking derivatives of individual components you should end up with
(dy)/(dx) = (xycos(xy)-sin(xy)-3x^2y)/(3x^3-x^2cos(xy)
#### Explanation:
This is complex so check carefully before assuming what follows is valid:
Breaking the equation up into small pieces and working on the derivatives for each:
Part 1: The $\textcolor{red}{x y}$ component of $25 = \sin \frac{\textcolor{red}{x y}}{x} - 3 x y$
Finding the derivative of $x y$ is of the form for finding the derivative of $u v$ and we know $\frac{d \left(u v\right)}{\mathrm{dx}} = v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}$
so
$\textcolor{w h i t e}{\text{XXX}} \frac{d \left(x y\right)}{\mathrm{dx}} = y \frac{\cancel{\mathrm{dx}}}{\cancel{\mathrm{dx}}} + x \frac{\mathrm{dy}}{\mathrm{dx}}$
$\textcolor{w h i t e}{\text{XXXXXX}} = y + x \frac{\mathrm{dy}}{\mathrm{dx}}$
Part 2: The $\textcolor{red}{\sin \left(x y\right)}$ component of $25 = \frac{\textcolor{red}{\sin \left(x y\right)}}{x} - 3 x y$
Finding the derivative of $\sin \left(x y\right)$ is of the form for finding the derivative of $f \left(g \left(u\right)\right)$ and we know
$\textcolor{w h i t e}{\text{XXX}} \frac{\mathrm{df} \left(g \left(x\right)\right)}{\mathrm{dx}} = \frac{\mathrm{df} \left(g \left(x\right)\right)}{d \left(g \left(x\right)\right)} \cdot \frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}}$
with $f \left(x\right) = \sin \left(x\right)$ and $g \left(x\right) = x y$
so
$\textcolor{w h i t e}{\text{XXX}} \frac{\mathrm{ds} \in \left(x y\right)}{\mathrm{dx}} = \cos \left(x y\right) \cdot \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right)$
Part 3: The $\textcolor{red}{\sin \frac{x y}{x}}$ component of $25 = \textcolor{red}{\sin \frac{x y}{x}} - 3 x y$
Finding the derivative of $\sin \frac{x y}{x}$ is of the form for finding the derivative of $\frac{u}{v}$ and we know
$\textcolor{w h i t e}{\text{XXX}} d \frac{\frac{u}{v}}{\mathrm{dx}} = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{{v}^{2}}$
with $v = x$ and $u = \sin \left(x y\right)$
so
$\textcolor{w h i t e}{\text{XXX}} \frac{d \frac{\sin \left(x y\right)}{x}}{\mathrm{dx}} = \frac{x \cdot \left(\cos \left(x y\right) \cdot \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right)\right) - \sin \left(x y\right) \frac{\cancel{\mathrm{dx}}}{\cancel{\mathrm{dx}}}}{{x}^{2}}$
$\textcolor{w h i t e}{\text{XXXXXXXX}} = \frac{\cos \left(x y\right) \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right)}{x} - \frac{\sin \left(x y\right)}{{x}^{2}}$
Part 4: The $\textcolor{red}{x y}$ component of $25 = \sin \frac{x y}{x} - 3 \textcolor{red}{x y}$
from Part 1 we already have
$\textcolor{w h i t e}{\text{XXX}}$(d(xy))/(dx) = y+x(dy)/(dx)
Putting the pieces together
Take the derivative of both sides
$d \frac{25}{\mathrm{dx}} = \frac{d \left(\sin \frac{x}{x} - 3 x y\right)}{\mathrm{dx}}$
$0 = \frac{\mathrm{ds} \in \frac{x}{x}}{\mathrm{dx}} - 3 \left(\frac{d \left(x y\right)}{\mathrm{dx}}\right)$
$\frac{\cos \left(x y\right) \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right)}{x} - \frac{\sin \left(x y\right)}{{x}^{2}} - 3 \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$
$\frac{y \cos \left(x y\right)}{x} + \cos \left(x y\right) \frac{\mathrm{dy}}{\mathrm{dx}} - \sin \frac{x y}{{x}^{2}} - 3 y - 3 x \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\frac{y \cos \left(x y\right)}{x} - \sin \frac{x y}{{x}^{2}} - 3 y = \left(3 x - \cos \left(x y\right)\right) \frac{\mathrm{dy}}{\mathrm{dx}}$
(dy)/(dx) = (xycos(xy)-sin(xy)-3x^2y)/(3x^3-x^2cos(xy)# |
# RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment
Here you can get solutions of RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment. These Solutions are part of RS Aggarwal Solutions Class 10. we have given RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment download pdf.
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## RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment
### Exercise 18
Question 1:
Radius = [latex]frac { Diameter }{ 2 } =frac { 35 }{ 2 } cm [/latex]
Circumference of circle = 2πr = [latex]left( 2times frac { 22 }{ 7 } times frac { 35 }{ 2 } right) cm [/latex] = 110 cm
∴ Area of circle = πr2 = [latex]left( frac { 22 }{ 7 } times frac { 35 }{ 2 } times frac { 35 }{ 2 } right) [/latex] cm2
= 962.5 cm2
More Resources
Question 2:
Circumference of circle = 2πr = 39.6 cm
Question 3:
Area of circle = πr2 = 301.84
Circumference of circle = 2πr = [latex]left( 2times frac { 22 }{ 7 } times 9.8 right) [/latex] = 61.6 cm
Question 4:
Let radius of circle be r
Then, diameter = 2 r
circumference – Diameter = 16.8
Circumference of circle = 2πr = [latex]left( 2times frac { 22 }{ 7 } times 3.92 right) [/latex] cm = 24.64 cm
Question 5:
Let the radius of circle be r cm
Then, circumference – radius = 37 cm
Question 6:
Area of square = (side)2 = 484 cm2
⇒ side = [latex]sqrt { 484 } cm [/latex] = 22 cm
Perimeter of square = 4 × side = 4 × 22 = 88 cm
Circumference of circle = Perimeter of square
Question 7:
Area of equilateral = [latex]frac { sqrt { 3 } }{ 4 } { a }^{ 2 } [/latex] = 121√3
Perimeter of equilateral triangle = 3a = (3 × 22) cm
= 66 cm
Circumference of circle = Perimeter of circle
2πr = 66
⇒ [latex]left( 2times frac { 22 }{ 7 } times r right) [/latex] cm = 66
⇒ r = 10.5 cm
Area of circle = πr2 = [latex]left( frac { 22 }{ 7 } times 10.5times 10.5 right) [/latex] cm2
= 346.5 cm2
Question 8:
Let the radius of park be r meter
Question 9:
Let the radii of circles be x cm and (7 – x) cm
Circumference of the circles are 26 cm and 18 cm
Question 10:
Area of first circle = πr2 = 962.5 cm2
Area of second circle = πR2 = 1386 cm2
Width of ring R – r = (21 – 17.5) cm = 3.5 cm
Question 11:
Area of outer circle = π[latex]r_1^2 [/latex] = [latex]left( frac { 22 }{ 7 } times 23times 23 right) [/latex] cm2
= 1662.5
Area of inner circle = π[latex]r_2^2 [/latex] = [latex]left( frac { 22 }{ 7 } times 12times 12 right) [/latex] cm2
= 452.2 cm2
Area of ring = Outer area – inner area
= (1662.5 – 452.5) cm2 = 1210 cm2
Question 12:
Inner radius of the circular park = 17 m
Width of the path = 8 m
Outer radius of the circular park = (17 + 8)m = 25 m
Area of path = π[(25)2-(17)2] = cm2
Area = 1056 m2
Question 13:
Let the inner and outer radii of the circular tacks be r meter and R meter respectively. Then
Inner circumference = 440 meter
Since the track is 14 m wide every where.
Therefore,
Outer radius R = r + 14m = (70 + 14) m = 84 m
Outer circumference = 2πR
= [latex]left( 2times frac { 22 }{ 7 } times 84 right)m [/latex] = 528 m
Rate of fencing = Rs. 5 per meter
Total cost of fencing = Rs. (528 × 5) = Rs. 2640
Area of circular ring = πR2 – πr2
Cost of levelling = Rs 0.25 per m2
Cost of levelling the track = Rs(6776 × 0.25) = Rs. 1694
Question 14:
Let r m and R m be the radii of inner circle and outer boundaries respectively.
Then, 2r = 352 and 2R = 396
Width of the track = (R – r) m
Area the track = π(R2 – r) = π (R+r)(R-r)
Question 15:
Area of rectangle = (120 × 90)
= 10800 m2
Area of circular lawn = [Area of rectangle – Area of park excluding circular lawn]
= [10800 – 2950] m2 = 7850 m2
Area of circular lawn = 7850 m2
⇒ πr2 = 7850 m2
Hence, radius of the circular lawn = 50 m
Question 16:
Area of the shaded region = (area of circle with OA as diameter) + (area of semicircle ∆DBC) – (area of ∆BCD)
Area of circle with OA as diameter = πr2
OB = 7 cm, CD = AB = 14 cm
Area of semicircle ∆DBC =
= 72
Question 17:
Diameter of bigger circle = AC = 54 cm
Radius of bigger circle = [latex]frac { AC }{ 2 } [/latex]
= [latex](frac { 54 }{ 2 }) [/latex] cm = 27 cm
Diameter AB of smaller circle = AC – BC = 54-10 = 44 cm
Radius of smaller circle = [latex]frac { 44 }{ 2 } [/latex] cm = 22 cm
Area of bigger circle = πR2 = [latex]left( frac { 22 }{ 7 } times 27times 27 right) [/latex] cm2
= 2291. 14 cm2
Area of smaller circle = πr2 = [latex]left( frac { 22 }{ 7 } times 22times 22 right) [/latex] cm2
= 1521. 11 cm2
Area of shaded region = area of bigger circle – area of smaller circle
= (2291. 14 – 1521. 11) cm2 = 770 cm2
Question 18:
PS = 12 cm
PQ = QR = RS = 4 cm, QS = 8 cm
Perimeter = arc PTS + arc PBQ + arc QES
Area of shaded region = (area of the semicircle PBQ) + (area of semicircle PTS)-(Area of semicircle QES)
Question 19:
Length of the inner curved portion
= (400 – 2 × 90) m
= 220 m
Let the radius of each inner curved part be r
Inner radius = 35 m, outer radius = (35 + 14) = 49 m
Area of the track = (area of 2 rectangles each 90 m × 14 m) + (area of circular ring with R = 49 m, r = 35 m
Length of outer boundary of the track
Question 20:
OP = OR = OQ = r
Let OQ and PR intersect at S
We know the diagonals of a rhombus bisect each other at right angle.
Therefore we have
Question 21:
Diameter of the inscribed circle = Side of the square = 10 cm
Radius of the inscribed circle = 5 cm
Diameter of the circumscribed circle
= Diagonal of the square
= (√2×10) cm
Radius of circumscribed circle = 5√2 cm
(i) Area of inscribed circle = [latex]left( frac { 22 }{ 7 } times 5times 5 right) [/latex] = 78.57 cm2
(ii) Area of the circumscribed circle
Question 22:
Let the radius of circle be r cm
Then diagonal of square = diameter of circle = 2r cm
Area of the circle = πr2 cm2
Question 23:
Let the radius of circle be r cm
Let each side of the triangle be a cm
And height be h cm
Question 24:
Radius of the wheel = 42 cm
Circumference of wheel = 2πr = [latex]left( 2times frac { 22 }{ 7 } times 42 right) [/latex] = 264 cm
Distance travelled = 19.8 km = 1980000 cm
Number of revolutions = [latex]frac { 1980000 }{ 264 } [/latex] = 7500
Question 25:
Radius of wheel = 2.1 m
Circumference of wheel = 2πr = [latex]left( 2times frac { 22 }{ 7 } times 2.1 right) [/latex] = 13.2 m
Distance covered in one revolution = 13.2 m
Distance covered in 75 revolutions = (13.2 × 75) m = 990 m
= [latex]frac { 990 }{ 1000 } [/latex] km
Distance a covered in 1 minute = [latex]frac { 99 }{ 100 } [/latex] km
Distance covered in 1 hour = [latex]frac { 99 }{ 100 } times 60[/latex] km = 59.4 km
Question 26:
Distance covered by the wheel in 1 revolution
The circumference of the wheel = 198 cm
Let the diameter of the wheel be d cm
Hence diameter of the wheel is 63 cm
Question 27:
Radius of the wheel = r = [latex]frac { 60 }{ 2 } [/latex] = 30 cm
Circumference of the wheel = 2πr = [latex]left( 2times frac { 22 }{ 7 } times 30 right) [/latex] = [latex]frac { 1320 }{ 7 } [/latex] cm
Distance covered in 140 revolution
Distance covered in one hour = [latex]frac { 264 }{ 1000 } times 60[/latex] = 15.84 km
Question 28:
Distance covered by a wheel in 1minute
Circumference of a wheel = 2πr = [latex]left( 2times frac { 22 }{ 7 } times 70 right) [/latex] = 440 cm
Number of revolution in 1 min = [latex]frac { 121000 }{ 440 } [/latex] = 275
Question 29:
Area of quadrant = [latex]frac { 1 }{ 4 } [/latex] πr2
Circumference of circle = 2πr = 22
Question 30:
Area which the horse can graze = Area of the quadrant of radius 21 m
Area ungrazed = [(70×52) – 346.5] m2
= 3293.5 m2
Question 31:
Each angle of equilateral triangle is 60°
Area that the horse cannot graze is 36.68 m2
Question 32:
Each side of the square is 14 cm
Then, area of square = (14 × 14) cm2
= 196 cm2
Thus, radius of each circle 7 cm
Required area = area of square ABCD – 4 (area of sector with r = 7 cm, θ= 90°)
Area of the shaded region = 42 cm2
Question 33:
Area of square = (4 × 4) cm2
= 16 cm2
Area of four quadrant corners
Radius of inner circle = 2/2 = 1 cm
Area of circle at the center = πr2 = (3.14 × 1 × 1) cm2
= 3.14 cm2
Area of shaded region = [area of square – area of four corner quadrants – area of circle at the centre]
= [16 – 3.14 – 3.14] cm= 9.72 cm2
Question 34:
Area of rectangle = (20 × 15) m= 300 m2
Area of 4 corners as quadrants of circle
Area of remaining part = (area of rectangle – area of four quadrants of circles)
= (300 – 38.5) m2 = 261.5 m2
Question 35:
Ungrazed area
Question 36:
Shaded area = (area of quadrant) – (area of DAOD)
Question 37:
Area of flower bed = (area of quadrant OPQ) – (area of the quadrant ORS)
Question 38:
Let A, B, C be the centres of these circles. Joint AB, BC, CA
Required area=(area of ∆ABC with each side a = 12 cm) – 3(area of sector with r = 6, θ = 60°)
The area enclosed = 5.76 cm2
Question 39:
Let A, B, C be the centers of these circles. Join AB, BC, CA
Required area= (area of ∆ABC with each side 2) – 3[area of sector with r = a cm, θ = 60°]
Question 40:
Let A, B, C, D be the centres of these circles
Join AB, BC, CD and DA
Side of square = 10 cm
Area of square ABCD
= (10 × 10) cm2
= 100 cm2
Area of each sector =
= 19.625 cm2
Required area = [area of sq. ABCD – 4(area of each sector)]
= (100 – 4 × 19.625) cm2
= (100 – 78.5) = 21.5 cm2
Question 41:
Required area = [area of square – areas of quadrants of circles]
Let the side = 2a unit and radius = a units
Area of square = (side × side) = (2a × 2a) sq. units = 4a2 sq.units
Question 42:
Let the side of square = a m
Area of square = (a × a) cm = a2m2
Side of square = 40 m
Therefore, radius of semi circle = 20 m
Area of semi circle =
= 628 m2
Area of four semi circles = (4 × 628) m= 2512 m2
Cost of turfing the plot of of area 1 m2 = Rs. 1.25
Cost of turfing the plot of area 2512 m= Rs. (1.25 × 2512)
= Rs. 3140
Question 43:
Area of rectangular lawn in the middle
= (50 × 35) = 1750 m2
Radius of semi circles = [latex]frac { 35 }{ 2 } [/latex] = 17.5 m
Area of lawn = (area of rectangle + area of semi circle)
= (1750 + 962.5) m2 = 2712.5 m2
Question 44:
Area of plot which cow can graze when r = 16 m is πr2
= [latex]left( frac { 22 }{ 7 } times 10.5times 10.5 right) [/latex]
= 804.5 m2
Area of plot which cow can graze when radius is increased to 23 m
= [latex]left( frac { 22 }{ 7 } times 10.5times 10.5 right) [/latex]
= 1662.57 m2
Additional ground = Area covered by increased rope – old area
= (1662.57 – 804.5)m2 = 858 m2
Question 45:
Given: ABC is right angled at A with AB = 6 cm and AC = 8 cm
Let us join OA, OB and OC
ar(∆AOC) + ar(∆OAB) + ar(∆BOC) = ar(∆ABC)
Question 46:
Question 47:
Area of region ABCDEFA = area of square ABDE + area of semi circle BCD – area of ∆AFE
Question 48:
Side of the square ABCD = 14 cm
Area of square ABCD = 14 × 14 = 196 cm2
Radius of each circle = [latex]frac { 14 }{ 4 } [/latex] = 3.5 cm
Area of the circles = 4 × area of one circle
Area of shaded region = Area of square – area of 4 circles
= 196 – 154 = 42 cm2
Question 49:
Diameter AC = 2.8 + 1.4
= 4.2 cm
Radius r1 = [latex]frac { 4.2 }{ 2 } [/latex] = 2.1 cm
Length of semi-circle ADC = πr= π × 2.1 = 2.1 π cm
Diameter AB = 2.8 cm
Radius r2 = 1.4 cm
Length of semi- circle AEB = πr2= π × 1.4 = 1.4 π cm
Diameter BC = 1.4 cm
Radius r3 = [latex]frac { 1.4 }{ 2 } [/latex] = 0.7 cm
Length of semi – circle BFC = π × 0.7 = 0.7 π cm
Perimeter of shaded region = 2.1 + 1.4 + 0.7 = 4.2 π cm
= 4.2 × [latex]frac { 22 }{ 7 } [/latex] = 13.2 cm
Question 50:
Area of shaded region = Area of ∆ABC + Area of semi-circle APB + Area of semi circle AQC – Area of semicircle BAC
Further in ∆ABC, ∠A = 90°
Adding (1), (2), (3) and subtracting (4)
Question 51:
In ∆PQR, ∠P = 90°, PQ = 24 cm, PR = 7 cm
Area of semicircle
Area of ∆PQR = [latex]frac { 1 }{ 2 } [/latex] × 7 × 24 cm2 = 84 cm2
Shaded area = 245.31 – 84 = 161.31 cm2
Question 52:
ABCDEF is a hexagon.
∠AOB = 60°, Radius = 35 cm
Area of sector AOB
Area of ∆AOB =
= 530.425 cm2
Area of segment APB = (641.083 – 530.425) cm= 110.658 cm2
Area of design (shaded area) = 6 × 110.658 cm2 = 663.948 cm2
= 663.95 cm2
Question 53:
In ∆ABC, ∠A = 90°, AB = 6cm, BC = 10 cm
Area of ∆ABC =
Let r be the radius of circle of centre O
Question 54:
Area of equilateral triangle ABC = 49√3 cm2
Let a be its side
Area of sector BDF =
Area of sector BDF = Area of sector CDE = Area of sector AEF
Sum of area of all the sectors
= [latex]frac { 77 }{ 3 } [/latex] × 3 cm2 = 77 cm2
Shaded area = Area of ∆ABC – sum of area of all sectors
= 49√3 – 77 = (84.77 – 77.00) cm2
= 77.7 cm2
Question 55:
In ∆ABC, ∠B = 90°, AB = 48 cm, BC = 14 cm
Area of semi-circle APC
Area of quadrant BDC with radius 14 cm
Shaded area = Area of ∆ABC + Area of semi-circle APC – Area of quadrant BDC
= ( 336+982.14-154 ) cm2
= ( 1318.14-154 ) cm= 1164.14 cm2
Question 56:
Radius of quadrant ABED = 16 cm
Its area =
Area of ∆ABD = [latex]left( frac { 1 }{ 2 } times 16times 16 right) [/latex] cm2
= 128 cm2
Area of segment DEB
Area of segment DFB = [latex]frac { 512 }{ 7 } [/latex] cm2
Total area of segments = 2 × [latex]frac { 512 }{ 7 } [/latex] cm2 = [latex]frac { 1024 }{ 7 } [/latex] cm2
Shaded area = Area of square ABCD – Total area of segments
Question 57:
Radius of circular table cover = 70 cm
Area of the circular cover =
Shaded area = Area of circle – Area of ∆ABC
= (15400 – 6365.1)
Question 58:
Area of the sector of circle =
r = 14 cm and θ = 45°
Question 59:
Length of the arc
Length of arc = ( 17.5 × [latex]frac { 22 }{ 7 } [/latex] ) cm = 55 cm
Area of the sector =
= ( [latex]frac { 22 }{ 7 } [/latex] × 183.75 ) cm2 = 577.5 cm2
Question 60:
Length of arc of circle = 44 cm
Radius of circle = 17.5 cm
Area of sector =
= ( 22 × 17.5) cm2 = 385 cm2
Question 61:
Let sector of circle is OAB
Perimeter of a sector of circle =31 cm
OA + OB + length of arc AB = 31 cm
6.5 + 6.5 + arc AB = 31 cm
arc AB = 31 – 13
= 18 cm
Question 62:
Area of the sector of circle =
Radius = 10.5 cm
Question 63:
Length of the pendulum = radius of sector = r cm
Question 64:
Length of arc =
Circumference of circle = 2πr
Area of circle =
= 962.5 cm2
Question 65:
Circumference of circle = 2πr
Question 66:
Angle described by the minute hand in 60 minutes θ = 360°
Angle described by minute hand in 20 minutes
Required area swept by the minute hand in 20 minutes
= Area of the sector(with r = 15 cm and θ = 120°)
Question 67:
θ = 56° and let radius is r cm
Area of sector =
Hence radius = 6cm
Question 68:
Question 69:
In 2 days, the short hand will complete 4 rounds
∴ Distance travelled by its tip in 2 days
=4(circumference of the circle with r = 4 cm)
= (4 × 2 × 4) cm = 32 cm
In 2 days, the long hand will complete 48 rounds
∴ length moved by its tip
= 48(circumference of the circle with r = 6cm)
= (48 × 2 × 6) cm = 576 cm
∴ Sum of the lengths moved
= (32 + 576) = 608 cm
= (608 × 3.14) cm = 1909.12 cm
Question 70:
∆OAB is equilateral.
So, ∠AOB = 60°
Length of arc BDA = (2π × 12 – arc ACB) cm
= (24π – 4π) cm = (20π) cm
= (20 × 3.14) cm = 62.8 cm
Area of the minor segment ACBA
Question 71:
Let AB be the chord of circle of centre O and radius = 6 cm such that ∠AOB = 90°
Area of sector = OACBO
Area of ∆AOB =
Area of minor segment ACBA
= (area of sector OACBO) – (area of ∆OAB)
= (28.29 – 18) cm2 = 10.29 cm2
Area of major segment BDAB
Question 72:
Let OA = 5√2 cm , OB = 5√2 cm
And AB = 10 cm
Area of ∆AOB =
= 25 cm2
Area of minor segment = (area of sector OACBO) – (area of ∆OAB)
= ( 39.25 – 25 ) cm2 = 14.25 cm2
Question 73:
Area of sector OACBO
Area of minor segment ACBA
Area of major segment BADB
Question 74:
Let AB be the chord of circle of centre O and radius = 30 cm such that AOB = 60°
Area of the sector OACBO
Area of ∆OAB =
Area of the minor segment ACBA
= (area of the sector OACBO) – (area of the ∆OAB)
=(471 – 389.25) cm2 = 81.75 cm2
Area of the major segment BADB
= (area of circle) – (area of the minor segment)
= [(3.14 × 30 × 30) – 81.75)] cm2 = 2744.25 cm2
Question 75:
Let the major arc be x cm long
Then, length of the minor arc = [latex]frac { 1 }{ 5 } [/latex] x cm
Circumference =
Question 76:
Radius of the front wheel = 40 cm = [latex]frac { 2 }{ 5 } [/latex] m
Circumference of the front wheel =
Distance moved by it in 800 revolution
Circumference of rear wheel = (2π × 1)m = (2π) m
Required number of revolutions =
Complete RS Aggarwal Solutions Class 10
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# Video: Addition on a Number Line
Tim Burnham
Learn how to carry out addition of two numbers by finding the first number on a number line, then counting along the same number of steps as the second number to arrive at the sum.
03:53
### Video Transcript
In this video, we’re going to look at how to count on a number line to help us to add numbers together.
First, let’s add six and three.
Look at this addition sentence. Remember that this is the addition symbol; it tells us to add two numbers. So we have to add six and three. And remember that this symbol means equal; the number in the green box will be equal to six add three.
Now, let’s use the number line to find our sum. First, we start with six. So, let’s find six on the number line and put a counter on it. Now that the counter is on six, we have to add three; let’s jump three times. We can imagine a frog picking up the counter and helping us to make our three jumps: one, two, three.
So from six, we took one, two, three jumps. So we started on six, we took three jumps, and we landed on nine. So the sum of six and three is nine. Now our addition sentence says six add three is equal to nine. And now you know how to add using a number line. Let’s try one more question.
What is seven plus four?
Remember that plus is another word for add. So, we’ve got to add seven and four; let’s write these numbers in our addition sentence. Seven plus four equals what? We’ve got to find the total. To add seven and four, first we have to find seven on the number line. And we put our counter on the number seven. Now we have to add four.
That means we have to make four jumps on the number line, and this time that frog isn’t going to help us: one, two, three, four. So we started at seven, we added four, and we ended up at eleven. This means that eleven is equal to seven plus four.
Okay, let’s just go through the steps that we need when using a number line to add. First, find one of the numbers on the number line. Second, jump the other number of steps to count on: one, two, three, four, five. And thirdly, the number that you land on is the sum. And in this case, that was eleven. All you have to do now is find a friendly frog to help you jump along a number line to add your numbers together. Good luck. |
# y = ⅔f(x) Vertical Stretches about an Axis.
## Presentation on theme: "y = ⅔f(x) Vertical Stretches about an Axis."— Presentation transcript:
y = ⅔f(x) Vertical Stretches about an Axis.
The graph of y = f(x) is stretched vertically about the x-axis by a factor of ⅔ . Horizontal Stretches about Axis. y = f(bx) y = f(3x) The graph of y = f(x) has been stretched horizontally by factor of 2. Math 30-1
y + 4 = f(3x – 12) y + 4 = f(3(x – 4)) Factor Stretch Translate
** You must factor out the stretch if there is also a translation** y + 4 = f(3x – 12) Factor Stretch Translate y + 4 = f(3(x – 4)) Math 30-1
The graph of the function y = f(x) is transformed to produce the graph of the function y = g(x)
An equation for g(x) in terms of f(x) is A. B. C. D. Math 30-1
1.2B Exploring Reflections
Math 30-1
Reflect the Graph of y = f(x) in the x-axis
f(x) = | x | (-6, 6) (6, 6) Family of Functions f(x) = | x | (-2, 2) (2, 2) f(x) = -| x | (2, -2) (-2, -2) f(x) = -| x | (-6, -6) (6, -6) Math 30-1
Reflections in the x-axis affect the y-coordinate.
Notice (x, y) → (x, -y). The variable y is replaced by (-y) in the ordered pair and in the function equation . y = f(x) → -y = f(x) or y = -f(x) The original equation y = |x| maps to -y = |x| or y = -|x| The invariant points are on the line of reflection, the x-axis. The invariant points would be the x-intercepts Math 30-1
Reflecting y = f(x) Graph y = -f(x) y = f(x) y = -f(x)
Domain and Range Math 30-1
Notice domain and range.
Graphing a Reflection of y = f(x) in the y-axis (7, 7) (-7, 7) (-5, 5) (5, 5) (7, 3) (-7, 3) Notice domain and range. Math 30-1
Notice (x, y) → (-x, y). The variable x is replaced by (-x).
Reflections in the y-axis affect the x-coordinate. Notice (x, y) → (-x, y). The variable x is replaced by (-x). y = f(x) → y = f(-x). The original equation x - 5 = 0.5(y - 5)2 becomes (-x) - 5 = 0.5(y - 5)2 The invariant points are on the line of reflection, the y-axis. The invariant points would be the y-intercepts. Math 30-1
If the point (6, -1) is on the graph of f(x), what would be the
Given the graph of f(x), graph the image function g(x) that would be the graph of f(-x). f(x) g(x) If the point (6, -1) is on the graph of f(x), what would be the corresponding point on the graph of g(x). (-6, -1) Math 30-1
g(x) = f(-x) g(x) = -f(x) f(x) = g(-x) f(x) = -g(x)
Which transformation is true regarding the two graphs g(x) = f(-x) g(x) = -f(x) f(x) = g(-x) f(x) = -g(x) Math 30-1
Transformations of Functions
Given that the point (2, 6) is on the graph of f(x), state the corresponding point after the following transformations of f(x). a) f(x - 3) - 4 b) -f(x + 2) - 1 c) f(-x + 2) + 3 f(-(x - 2)) + 3 (5, 2) (0, -7) (0, 9) Math 30-1
True or False The graph of y = -f(-x - 2) – 3 is a horizontal translation of the graph of y = f(x) 2 units right. Multiple Choice Which of the following transformations on y = f(x) would have the y-intercepts as invariant points? - y = f(x) B. y = f(-x) C. y = f(x + 2) D. y = 2f(x) Order of Transformations: Factor Reflections Stretches Translations Math 30-1
Transformation Summary
Type of Transformation Replace what in equation? With What? Resulting Effect Resulting Equations Vertical Translation y y - k Graph moves k units up. y-k = f(x) y + k Graph moves k units down. y + k = f(x) Horizontal Translation x x - h Graph moves h units right. y = f(x - h) x + h Graph moves h units left. y = f(x + h) Vertical Reflection -y Graph is reflected in the x-axis y = -f(x) Horizontal Reflection -x Graph is reflected in the y-axis y = f(-x)
Assignment Page 28 1,3, 5c,d, 7b,d, 10, 15 Math 30-1 |
Kindergarten Math
Resources to help teach and assess Common Core Kindergarten Math
Kindergarten Resource Blog
Engage NY (Eureka) Video Lessons and Homework Solutions
Engage NY (Eureka) Homework and Video Examples
Printable Math Activities by Standard
In Kindergarten, instructional time should focus on two critical areas: (1) representing and comparing whole numbers, initially with sets of objects; (2) describing shapes and space. More learning time in Kindergarten should be devoted to number than to other topics.
Students use numbers, including written numerals, to represent quantities and to solve quantitative problems, such as counting objects in a set; counting out a given number of objects; comparing sets or numerals; and modeling simple joining and separating situations with sets of objects, or eventually with equations such as 5 + 2 = 7 and 7 – 2 = 5. (Kindergarten students should see addition and subtraction equations, and student writing of equations in kindergarten is encouraged, but it is not required.) Students choose, combine, and apply effective strategies for answering quantitative questions, including quickly recognizing the cardinalities of small sets of objects, counting and producing sets of given sizes, counting the number of objects in combined sets, or counting the number of objects that remain in a set after some are taken away.
Students describe their physical world using geometric ideas (e.g., shape, orientation, spatial relations) and vocabulary. They identify, name, and describe basic two-dimensional shapes, such as squares, triangles, circles, rectangles, and hexagons, presented in a variety of ways (e.g., with different sizes and orientations), as well as three-dimensional shapes such as cubes, cones, cylinders, and spheres. They use basic shapes and spatial reasoning to model objects in their environment and to construct more complex shapes.
Counting & Cardinality
Know number names and the count sequence.
CCSS.MATH.CONTENT.K.CC.A.1
Count to 100 by ones and by tens.
CCSS.MATH.CONTENT.K.CC.A.2
Count forward beginning from a given number within the known sequence (instead of having to begin at 1).
CCSS.MATH.CONTENT.K.CC.A.3
Write numbers from 0 to 20. Represent a number of objects with a written numeral 0-20 (with 0 representing a count of no objects).
Count to tell the number of objects.
CCSS.MATH.CONTENT.K.CC.B.4
Understand the relationship between numbers and quantities; connect counting to cardinality.
CCSS.MATH.CONTENT.K.CC.B.4.A
When counting objects, say the number names in the standard order, pairing each object with one and only one number name and each number name with one and only one object.
CCSS.MATH.CONTENT.K.CC.B.4.B
Understand that the last number name said tells the number of objects counted. The number of objects is the same regardless of their arrangement or the order in which they were counted.
CCSS.MATH.CONTENT.K.CC.B.4.C
Understand that each successive number name refers to a quantity that is one larger.
CCSS.MATH.CONTENT.K.CC.B.5
Count to answer “how many?” questions about as many as 20 things arranged in a line, a rectangular array, or a circle, or as many as 10 things in a scattered configuration; given a number from 1-20, count out that many objects.
Compare numbers.
CCSS.MATH.CONTENT.K.CC.C.6
Identify whether the number of objects in one group is greater than, less than, or equal to the number of objects in another group, e.g., by using matching and counting strategies. (Include groups with up to ten objects.)
CCSS.MATH.CONTENT.K.CC.C.7
Compare two numbers between 1 and 10 presented as written numerals.
Operations & Algebraic Thinking
Understand addition as putting together and adding to, and understand subtraction as taking apart and taking from.
CCSS.MATH.CONTENT.K.OA.A.1
Represent addition and subtraction with objects, fingers, mental images, drawings, sounds (e.g., claps), acting out situations, verbal explanations, expressions, or equations.
CCSS.MATH.CONTENT.K.OA.A.2
Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem.
CCSS.MATH.CONTENT.K.OA.A.3
Decompose numbers less than or equal to 10 into pairs in more than one way, e.g., by using objects or drawings, and record each decomposition by a drawing or equation (e.g., 5 = 2 + 3 and 5 = 4 + 1).
CCSS.MATH.CONTENT.K.OA.A.4
For any number from 1 to 9, find the number that makes 10 when added to the given number, e.g., by using objects or drawings, and record the answer with a drawing or equation.
CCSS.MATH.CONTENT.K.OA.A.5
Fluently add and subtract within 5.
Number & Operations in Base Ten
CCSS.MATH.CONTENT.K.NBT.A.1
Work with numbers 11-19 to gain foundations for place value.
CCSS.MATH.CONTENT.K.NBT.A.1
Compose and decompose numbers from 11 to 19 into ten ones and some further ones, e.g., by using objects or drawings, and record each composition or decomposition by a drawing or equation (such as 18 = 10 + 8); understand that these numbers are composed of ten ones and one, two, three, four, five, six, seven, eight, or nine ones.
Measurement & Data
Describe and compare measurable attributes.
CCSS.MATH.CONTENT.K.MD.A.1
Describe measurable attributes of objects, such as length or weight. Describe several measurable attributes of a single object.
CCSS.MATH.CONTENT.K.MD.A.2
Directly compare two objects with a measurable attribute in common, to see which object has “more of”/”less of” the attribute, and describe the difference. For example, directly compare the heights of two children and describe one child as taller/shorter.
Classify objects and count the number of objects in each category.
CCSS.MATH.CONTENT.K.MD.B.3
Classify objects into given categories; count the numbers of objects in each category and sort the categories by count. (Include groups with up to ten objects.)
Geometry
Identify and describe shapes.
CCSS.MATH.CONTENT.K.G.A.1
Describe objects in the environment using names of shapes, and describe the relative positions of these objects using terms such as abovebelowbesidein front ofbehind, and next to.
CCSS.MATH.CONTENT.K.G.A.2
Correctly name shapes regardless of their orientations or overall size.
CCSS.MATH.CONTENT.K.G.A.3
Identify shapes as two-dimensional (lying in a plane, “flat”) or three-dimensional (“solid”).
Analyze, compare, create, and compose shapes.
CCSS.MATH.CONTENT.K.G.B.4
Analyze and compare two- and three-dimensional shapes, in different sizes and orientations, using informal language to describe their similarities, differences, parts (e.g., number of sides and vertices/”corners”) and other attributes (e.g., having sides of equal length).
CCSS.MATH.CONTENT.K.G.B.5
Model shapes in the world by building shapes from components (e.g., sticks and clay balls) and drawing shapes.
CCSS.MATH.CONTENT.K.G.B.6
Compose simple shapes to form larger shapes. For example, “Can you join these two triangles with full sides touching to make a rectangle?” |
# 6.2 Construct and Interpret Binomial Distributions
## Presentation on theme: "6.2 Construct and Interpret Binomial Distributions"— Presentation transcript:
6.2 Construct and Interpret Binomial Distributions
Page 388 What is a probability distribution? What is a binomial distribution? What is a binomial experiment?
Probability Distribution
A probability distribution is a function that gives the probability of each possible value of a random variable. The sum of all the probabilities in a probability distribution must equal one. Probability Distribution for Rolling a Die x 1 2 3 4 5 6 P(x) 1 6
Let X be a random variable that represents the sum when two six-sided dice are rolled. Make a table and a histogram showing the probability distribution for X. SOLUTION The possible values of X are the integers from 2 to 12. The table shows how many outcomes of rolling two dice produce each value of X. Divide the number of outcomes for X by 36 to find P(X).
What is the most likely sum when rolling two six-sided dice?
SOLUTION The most likely sum when rolling two six-sided dice is the value of X for which P(X) is greatest. This probability is greatest for X = 7. So, the most likely sum when rolling the two dice is 7.
What is the probability that the sum of the two dice is at least 10?
The probability that the sum of the two dice is at least 10 is: P(X > 10 ) = P(X = 10) + P(X = 11) + P(X = 12) = 3 36 + 2 1 = 6 36 = 1 6 0.167
6.2 Assignment day 1 Page 391, 1-9
6.2 Construct and Interpret Binomial Distributions day 2
Page 388
Binomial Distribution
A binomial distribution shows the probabilities of the outcomes of a binomial experiment. Binomial Experiments: There are 𝑛 independent trials. Each trial has only two possible outcomes: success and failure. The probability of success is the same for each trial. This probability is denoted by 𝑝. The probability of failure is given by 1−𝑝. For a binomial experiment, the probability of exactly 𝑘 successes in n trials is: P(𝑘 successes)=𝑛𝐶𝑘 𝑝 𝑘 (1−𝑝) 𝑛−𝑘
Sports Surveys According to a survey, about 41% of U.S. households have a soccer ball. Suppose you ask 6 randomly chosen U.S. households whether they have a soccer ball. Draw a histogram of the binomial distribution for your survey. SOLUTION The probability that a randomly selected household has a soccer ball is p = Because you survey 6 households, n = 6.
P(k = 0) = 6C0(0.41)0(0.59)6 0.042 P(k = 1) = 6C1(0.41)1(0.59)5 0.176 P(k = 2) = 6C2(0.41)2(0.59)4 0.306 P(k = 3) = 6C3(0.41)3(0.59)3 0.283 P(k = 4) = 6C3(0.41)4(0.59)2 0.148 P(k = 5) = 6C5(0.41)5(0.59)1 0.041 P(k = 6) = 6C6(0.41)6(0.59)0 0.005
A histogram of the distribution is shown.
What is the most likely outcome of the survey? SOLUTION The most likely outcome of the survey is the value of k for which P(k) is greatest. This probability is greatest for k = 2. So, the most likely outcome is that 2 of the 6 households have a soccer ball.
What is the probability that at most 2 households have a soccer ball?
SOLUTION The probability that at most 2 households have a soccer ball is: P(k < 2) = P(k = 2) + P(k = 1) + P(k = 0) 0.524 ANSWER So, the probability is about 52%.
Classifying Distributions
Probability Distributions can be classified in two ways: Symmetric if a vertical line can be drawn that divides the histogram into two parts that are mirror images. Skewed if the distribution is not symmetric.
Describe the shape of the binomial distribution that shows the probability of exactly k successes in 8 trials if (a) p = 0.5 SOLUTION Symmetric; the left half is a mirror image of the right half.
Describe the shape of the binomial distribution that shows the probability of exactly 𝑘 successes in 8 trials if (b) 𝑝 = 0.9. SOLUTION Skewed; the distribution is not symmetric about any vertical line.
How can you figure this out without a histogram?
A binomial experiment consists of 5 trials with probability p of success on each trial. Describe the shape of the binomial distribution that shows the probability of exactly k successes if (a) p = 0.4 and (b) p = 0.5 ANSWER a. The distribution is skewed since it is not symmetric about a vertical line. b. The distribution is symmetric since it is symmetric about a vertical line. How can you figure this out without a histogram? As p gets farther from 0.5, its shape becomes more skewed.
What is a binomial distribution?
A probability distribution is a function that gives the probabilities of each possible outcome for a random variable. The sum of all probabilities in a probability distribution must equal 1. For a binomial experiment, the probability of exactly 𝑘 successes in 𝑛 trials can be calculated from 𝑛𝐶𝑘 𝑝 𝑘 (1−𝑝) 𝑛−𝑘 where n is the number of independent trials and 𝑝 is the probability of success. A binomial distribution is a probability distribution that shows the probabilities of the outcomes of a binomial experiment. To construct a binomial distribution, list each of the possible outcomes. Then calculate the probability of each of the outcomes.
6.2 Assignment day 2 Page 391, 13-15, 21-23, 30-38, 43-45 |
# SAT Math : How to find range
## Example Questions
← Previous 1
### Example Question #1 : Range
Find the range of the following sequence:
1, 3, 4, 6, 11, 23
6
5
8
22
22
Explanation:
To find the range, subtract the smallest number from the largest number. In this case we get 23 - 1 = 22 so the range is 22.
### Example Question #81 : Statistics
What is the range of the data set below?
103, 132, 241, 251, 623, 174, 132, 170, 843, 375, 120, 641, 384, 222, 833
103
545
473
833
740
740
Explanation:
The range of a set of data is found by subtracting the smallest item from the largest. In this case, 843 - 103 = 740.
### Example Question #1 : Range
Find the range of the data set:
25, 83 , 51, 13, 37, 21, 52, 58
13
42.5
44
70
83
70
Explanation:
44 is the median of the data. 42.5 is the arithmetic mean. 13 is the minimum value. 83 is the maximum value.
To find the range, subtract the minimum value from the maximum value:
83 - 13 = 70
### Example Question #2 : Range
F, G, H are the only three numbers in a sequence in which each number is twice the number before it. The three number have an arithmetic mean of 21. What is the value of H?
36
22
63
30
24
36
Explanation:
We know that H = 2G = 4F, and G = 2F and that (F + G + H)/3=21. If we substitute each term with its F equivalent we get:
(F + 2F + 4F)/3 = 21
7F/3 = 21
7F = 63
F = 9
If we substitute F = 9 into H = 4F, we get H = 36.
The correct answer is 36.
### Example Question #151 : Data Analysis
A set of numbers consists of eleven consecutive even integers. If the median of the set is 62, what is the range of the set?
20
62
40
10
52
Explanation:
We don't know what the numbers are yet, so let's just call them a1, a2, a3, a4....a11. Let's assume a1 is the smallest, and a11 is the greatest.
If we were going to find the median of our set, we would need to line them all up, from least to greatest.
a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11
The median is the middle number, which in this case about be a6, because there are five numbers before it and five numbers after it.
We know that the median is 62, so that means that there must be five numbers before 62 and five numbers after. Because the numbers are all consecutive even integers, the set must look like this:
52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72
To find the range of this set, we must find the difference between the largest and smallest numbers. The largest is 72, and the smallest is 52, so the range is 72 - 52 = 20.
The answer is 20.
### Example Question #1 : How To Find Range
The median of a set of eleven consecutive odd integers is 39. What is the range of the set?
39
20
10
29
30
20
Explanation:
Let's represent the eleven integers as A, B, C, D, E, F, G, H, I, J, and K. Let's assume that if we lined them up from least to greatest, the set would be as follows:
A, B, C, D, E, F, G, H, I, J, K
The median in this set is the middle number. In this case, the middle number will always be F, because there are five numbers before it and five numbers after it.
We are told that the median is 39. Thus F = 39, and our set looks like this:
A, B, C, D, E, 39, G, H, I, J, K
Since we know that all the numbers are consecutive odd integers, the set must consist of the five odd numbers before 39 and the five odd numbers after 39. In other words, the set is this:
29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49
The question asks for the range of the set, which is the difference between the largest and smallest numbers. In this case, the range is 49 – 29 = 20.
The answer is 20.
### Example Question #1 : How To Find Range
Find the range.
Explanation:
First, arrange the numbers from highest to lowest, then subtract the highest number from the lowest number.
### Example Question #1 : Range
You have the following data set:
What is the range of the data set?
Explanation:
The range can be obtained by subtracting the lowest number in the set from the highest number in the set.
The lowest number in the set is 2 and the highest number is 66.
Therefore, is the range of the data set.
### Example Question #1 : How To Find Range
The following numbers are selected from a set of data:
Which of the following could NOT be the range of the data set?
Explanation:
The range of the data is the difference between the largest number and the smallest number. The range of the given data is ; therefore, the range of the entire data set must be equal to or greater than . So, it could not be .
### Example Question #5 : Range
What is the range of the following data set? |
29 Linear combinations of vectors
A linear combination is a sum of basic elements, each of which has been scaled. For instance, in Block 1 we looked at linear combinations of functions such as $g(t) = A + B e^{kt}\ .$ This combines the basic functions $$\text{one}(t)$$ and $$e^{kt}$$ scaled respectively by $$A$$ and $$B$$.
Like functions, vectors can be scaled and added. In one sense, this is just a matter of arithmetic. However, new concepts become accessible using the geometrical interpretation of vectors. This chapter builds gradually to one such concept: “slices” of an embedding space called subspaces. Many questions about constructing approximations—How good can the approximation be? How to make it better?—are clarified by seeing the possibilities for approximation as the elements of a subspace.
29.1 Scaling vectors
To scale a vector means to change its length without altering its direction. For instance, scaling by a negative number flips the vector tip-for-tail. Figure 29.1 shows two vectors $$\vec{v}$$ and $$\vec{w}$$ together with several scaled versions.
Arithmetically, scaling a vector is accomplished by multiplying each component of the vector by the scalar, e.g.
$\vec{u} = \left[\begin{array}{r}1.5\\-1\end{array}\right]\ \ \ \ 2\vec{u} = \left[\begin{array}{r}3\\-2\end{array}\right]\ \ \ \ -\frac{1}{2}\vec{u} = \left[\begin{array}{r}-0.75\\0.5\end{array}\right]\ \ \ \$
Geometrically, however, a vector corresponds to one step in a journey. For example, a vector scaled by 2.5 is a journey of two-and-a-half steps; scaling by -10 means traveling backward ten steps.
The subspace associated with a single vector is the set of all possible journeys that scaling a vector can accomplish. Visually, this corresponds to all the points on an infinitely long line defined by two points: the tip and the tail of the vector.
To add two vectors, place them tip to tail (without changing the direction). The sum is the new vector running from the tail of the first one to the tip of the second. (Figure 29.2)
Adding vectors in this way takes advantage of the rootlessness of a vector. So long as we keep the direction and length the same, we can move a vector to any convenient place. For adding vectors, the convenient arrangement is to place the tail of the second vector at the tip of the first. The result—the blue pencil in Figure 29.2—runs from the first (yellow) pencil’s tail to the second (green) pencil’s tip.
Arithmetically, vector addition is simply a matter of applying addition component-by-component. For instance, consider adding two vectors $$\vec{v}$$ and $$\vec{w}$$:
$\underbrace{\left[\begin{array}{r}1.5\\-1\\2\\6\end{array}\right]}_\vec{v} + \underbrace{\left[\begin{array}{r}2\\4\\-2\\-3.2\end{array}\right]}_\vec{w} = \underbrace{\left[\begin{array}{r}3.5\\3\\0\\2.8\end{array}\right]}_{\vec{v} + \vec{w}}$
Adding vectors makes sense only when they inhabit the same embedding space. In other words, the vectors must have the same number of components.
Arithmetic subtraction of one vector from another is a simple componentwise operation. For example:
$\underbrace{\left[\begin{array}{r}1.5\\-1\\2\\6\end{array}\right]}_\vec{v} {\Large -} \underbrace{\left[\begin{array}{r}2\\4\\-2\\-3.2\end{array}\right]}_\vec{w} = \underbrace{\left[\begin{array}{r}-0.5\\-5\\4\\9.2\end{array}\right]}_{\vec{v} - \vec{w}}\ .$
From a geometrical point of view, many people like to think of $$\vec{v} - \vec{w}$$ in terms of placing the two vectors tail to tail as in Figure 29.3. Read the result as the vector running from the tip of $$\vec{v}$$ to the tip of $$\vec{w}$$. In Figure 29.3, the yellow vector is $$\vec{v}$$ and the blue vector is $$\vec{w}$$. The result of the subtraction is the green vector.
29.3 Linear combinations
Thinking of a scaled vector as a “step” of a given length in a given direction leads us to conceive of a linear combination of vectors as step-by-step instructions for a journey. A central use for the formalism of vectors is to guide our thinking and our algorithms for figuring out how best to get from one “place” to another. We have used quotation marks around “place” because we are not necessarily referring to a physical destination. We will get to what else we might mean by “place” later in this Block.
As a fanciful example of getting to a “place,” consider a treasure hunt. Here are the instructions to get there:
1. On June 1, go to the flagpole before sunrise.
2. At 6:32, walk 213 paces away from the Sun.
3. At 12:19, walk 126 paces toward the Sun.
The Sun’s position varies over the day. Consequently, the direction of the Sun on June 1 at 6:32 is different than at 12:19. (Figure 29.4)
The treasure-hunt directions are in the form of a linear combination of vectors. So far, we know the direction of each vector. Imagine that the length is one stride or pace. (Admittedly, not a scientific unit of length.) Scaling $$\color{magenta}{\text{the magenta vector}}$$ by -213 and $$\color{blue}{\text{the blue vector}}$$ by 126, then adding the two scaled vectors gives a vector that takes you from the flagpole to the treasure.
A stickler for details might point out that the “direction of the sun” has an upward component. Common sense dictates that the walk is in the direction of the Sun as projected onto Earth’s surface. Chapter 30 deals with projections of vectors.
29.4 Functions as vectors
Since calculus is about functions, one naturally expects a chapter on vectors in a calculus textbook to make a connection between functions and vectors. Recall from Section 7.5 (entitled “Functions and data”) the idea of representing a function as a table of inputs and the corresponding outputs.
Here is such a table with some of our pattern-book functions.
t one(t) identity(t) exp(t) sin(t) pnorm(t)
0.0 1 0.0 1.000000 0.0000000 0.5000000
0.1 1 0.1 1.105171 0.0998334 0.5398278
0.2 1 0.2 1.221403 0.1986693 0.5792597
0.3 1 0.3 1.349859 0.2955202 0.6179114
0.4 1 0.4 1.491825 0.3894183 0.6554217
... 51 rows in total ...
4.6 1 4.6 99.48432 -0.9936910 0.9999979
4.7 1 4.7 109.94717 -0.9999233 0.9999987
4.8 1 4.8 121.51042 -0.9961646 0.9999992
4.9 1 4.9 134.28978 -0.9824526 0.9999995
5.0 1 5.0 148.41316 -0.9589243 0.9999997
In the tabular representation of the pattern-book functions, each function is a column of numbers—a vector.
Functions that we construct by linear combination are, in this vector format, just a linear combination of the vectors. For instance, the function $$g(t) \equiv 3 - 2 t$$ is $$3\cdot \text{one}(t) - 2 \cdot \text{identity}(t)$$
t one(t) identity(t) g(t)
0.0 1 0.0 3.0
0.1 1 0.1 2.8
0.2 1 0.2 2.6
0.3 1 0.3 2.4
0.4 1 0.4 2.2
... 51 rows in total ...
4.6 1 4.6 -6.2
4.7 1 4.7 -6.4
4.8 1 4.8 -6.6
4.9 1 4.9 -6.8
5.0 1 5.0 -7.0
The table above is a collection of four vectors: $$\vec{\mathtt t}$$, $$\vec{\mathtt{ one(t)}}$$, $$\vec{\mathtt{identity(t)}}$$, and $$\vec{\mathtt{g(t)}}$$. Each of those vectors has 51 components. In math-speak, we can say that the vectors are “embedded in a 51-dimensional space.”
In the table, the function output is tabulated only for select, discrete values of the input. Such discreteness is not a fundamental problem. Interpolation techniques such as that described in Section 7.5 enable evaluation of the function for a continuous input.
29.5 Matrices and linear combinations
A collection of vectors, such as the one displayed in the previous table, is called a matrix. Each vector in a matrix must have the same number of components.
As mathematical notation, we will use bold-faced, capital letters to stand for matrices, for example, $$\mathit{M}$$. The symbol $$\rightleftharpoons$$ is a reminder that a matrix can contain multiple vectors, just as the symbol $$\rightharpoonup$$ in $$\vec{v}$$ reminds us that the name “$$v$$” refers to a vector. (It is typical in mathematical writing to use single letters to refer to vectors instead of the descriptive, multi-letter names used in data frames.)
In the conventions for data, we give a name to each data frame column so that we can refer to it individually. In the conventions used in vector mathematics, single letters refer to the individual vectors.
As a case in point, let’s look at a matrix $$\mathit{M}$$ containing the two vectors which we’ve previously called $$\vec{\mathtt{one(t)}}$$ and $$\vec{\mathtt{identity(t)}}$$: $$.$$ The linear combination which we might previous have called $$3\cdot \vec{\mathtt{t}} - 2\,\vec{\mathtt{identity(t)}}$$ can be thought of as
$\left[\overbrace{\begin{array}{r} 1\\ 1 \\ 1 \\ 1 \\ \vdots &\\ 1 \\ 1 \end{array}}^{3 \times} \stackrel{\begin{array}{r} \\ \\ \\ \\ \\ \\ \\ \\ \end{array}}{\Large + \ } \overbrace{\begin{array}{r} 0\\ 0.1 \\ 0.2 \\ 0.3 \\ \vdots\\ 4.9 \\ 5.0 \end{array}}^{-2 \times}\right] = \left[\begin{array}{r} \\ \\ 3\\ 2.8\\2.6\\2.4\\\vdots\\-6.8\\-7.0\\ \\ \\ \end{array}\right]\ ,$ but this is not conventional notation. Instead, we would write this more concisely as
In symbolic form, the linear combination of the columns of $$\mathit{M}$$ using respectively the scalars in $$\vec{w}$$ is simply $$\mathit{M} \, \vec{w}$$. The construction of such linear combinations is called matrix multiplication.
Naturally, the operation only makes sense if there are as many components to $$\vec{w}$$ as columns in $$\mathit{M}$$.
“Matrix multiplication” might better have been called “$$\mathit{M}$$ linearly combined by $$\vec{w}$$.” Nevertheless, “matrix multiplication” is the standard term for such linear combinations.
In R, make vectors with the rbind() command, short for “bind rows,” as in
rbind(2, 5, -3)
## [,1]
## [1,] 2
## [2,] 5
## [3,] -3
Note that the vector components appear as successive arguments to the rbind() function.
Collect multiple vectors into a matrix with the cbind() command, short for “bind columns.” The arguments to cbind() will typically be vectors created by rbind(). For instance, the matrix
$\mathit{A} \equiv \left[\vec{u}\ \ \vec{v}\right]\ \ \text{where}\ \ \vec{u} \equiv \left[\begin{array}{r}2\\5\\-3\end{array}\right]\ \ \text{and}\ \ \vec{v} \equiv \left[\begin{array}{r}1\\-4\\0\end{array}\right]$
can be constructed in R with these commands.
u <- rbind(2, 5, -3)
v <- rbind(1, -4, 0)
A <- cbind(u, v)
A
## [,1] [,2]
## [1,] 2 1
## [2,] 5 -4
## [3,] -3 0
To compute the linear combination $$3 \vec{u} + 1 \vec{v}$$, that is, $$\mathit{A} \cdot \left[\begin{array}{r}3\\1\end{array}\right]$$ you use the matrix multiplication operator %*%. For instance, the following defines a vector $\vec{x} \equiv \left[\begin{array}{r}3\\1\end{array}\right]$ to do the job in a way that is easy to read:
x <- rbind(3, 1)
A %*% x
## [,1]
## [1,] 7
## [2,] 11
## [3,] -9
It is a mistake to use * instead of %*% for matrix multiplication. Remember that * is for componentwise multiplication, which is different from matrix multiplication. Componentwise multiplication with vectors and matrices will usually give an error message as with:
A * x
## Error in A * x: non-conformable arrays
The phrase “non-conformable arrays” is R-speak for “I do not know how to do componentwise multiplication with two incompatibly shaped objects.
29.6 Sub-spaces
Previously, we have said that a vector with $$n$$ components is “embedded” in an $$n$$-dimensional space. Think of an embedding space as a kind of club with restricted membership. For instance, a vector with two elements is properly a member of the 2-dimensional club, but a vector with more or fewer than two elements cannot have a place in the two-dimensional club. Similarly, there are clubs for 3-component vectors, 4-component vectors, and so on.
The clubhouse itself is a kind of space, the space in which any and all of the vectors that are eligible for membership can be embedded.
Now imagine the clubhouse arranged into meeting rooms. Each meeting room is just part of the clubhouse space. Which part? That depends on a set of vectors who sponsor the meeting. For instance, in the ten-dimensional clubhouse, a few members, let’s say $$\color{blue}{\vec{u}}$$ and $$\color{magenta}{\vec{v}}$$ decide to sponsor a meeting. That meeting room, part of the whole clubhouse space, is called a subspace.
A subspace has its own rules for admission. Vectors belong to the subspace only if they are a linear combination of the sponsoring members. The sponsoring members define the subspace, but the subspace itself consists of an infinity of vectors: all possible vectors that amount to a linear combination of the sponsors.
As an example, consider the clubhouse that is open to any and all vectors with three components. The diagram in Figure 29.5 shows the clubhouse with just two members present, $$\color{blue}{\vec{u}}$$ and $$\color{magenta}{\vec{v}}$$.
Any vector can individually sponsor a subspace. In Figure 29.5 the subspace sponsored by $$\color{blue}{\vec{u}}$$ is the extended line through $$\color{blue}{\vec{u}}$$, that is, all the possible scaled versions of $$\color{blue}{\vec{u}}$$. Similarly, the subspace sponsored by $$\color{magenta}{\vec{v}}$$ is the extended line through $$\color{magenta}{\vec{v}}$$. Each of these subspaces is one-dimensional.
Multiple vectors can sponsor a subspace. The subspace sponsored by both $$\color{blue}{\vec{u}}$$ and $$\color{magenta}{\vec{v}}$$ contains all the vectors that can be constructed as linear combinations of $$\color{blue}{\vec{u}}$$ and $$\color{magenta}{\vec{v}}$$; the gray subspace in Figure 29.6.
On the other hand, the subspace sponsored by $$\color{magenta}{\vec{v}}$$ and $$\color{blue}{\vec{u}}$$ is not the entire clubhouse. $$\color{magenta}{\vec{v}}$$ and $$\color{blue}{\vec{u}}$$ lie in a common plane, but not all the vectors in the 3-dimensional clubhouse lied in that plane. In fact, if you rotate Figure 29.6 to “look down the barrel” of either $$\color{magenta}{\vec{v}}$$ or $$\color{blue}{\vec{u}}$$, the plane will entirely disappear from view. A subspace is an infinitesimal slice of the embedding space.
“Sponsored a subspace” is metaphorical. In technical language, we speak of the subspace spanned by a set of vectors in the same embedding space. Usually, we refer to a “set of vectors” as a matrix. For instance, letting $\mathit{M} \equiv \left[{\Large \strut}\color{blue}{\vec{u}}\ \ \color{magenta}{\vec{v}}\right]\ ,$ the gray plane in Figure 29.6 is the subspace spanned by $$\mathit{M}$$ or, more concisely, $$span(\mathit{M})$$.
For a more concrete, everyday representation of the subspace spanned by two vectors, a worthwhile experiment is to pick up two pencils pointing in different directions. Place the eraser ends together, pinched between thumb and forefinger. Point the whole rigid assembly in any desired direction; the angle between them will remain the same.
Place a card on top of the pencils, slipping it between pressed fingers to hold it tightly in place. The card is another kind of geometrical object: a planar surface. The orientation of two vectors together determines the orientation of the surface. This simple fact will be significant later on.
One can replace the pencils with line segments drawn on the card underneath each pencil. Now, the angle is readily measurable in two dimensions. The angle between two vectors in three dimensions is the same as the angle drawn on the two-dimension surface that rests on the vectors.
Notice that one can also lay a card along a single vector. What is different here is that the card can be rolled around the pencil while still staying in contact; there are many different orientations for such a card even while the vector stays fixed. So a single fixed vector does not uniquely determine the orientation of the planar surface in which the vector can reside. It takes two vectors to determine a unique planar surface.
29.7 Exercises
Problem with NA NA |
How To Calculate Slope
## What are the steps to find slope?
How To Calculate Slope
1. What is Slope?
2. 4 Steps to Calculate Slope.
3. Step 1: Find two points anywhere on the line.
4. Step 2: Count the rise (the units up or down) to get from one point to the next.
5. Step 3: Determine the run (the units left to right) to get from one point to the next.
6. Step 4: If possible, simplify the fraction.
## What is the standard form of 450?
For instance, standard form of a number ‘450’ in general is ‘450’, but we can also rewrite this number in another form popularly known as scientific notation as, ‘4.50 * 102’. Hence a standard form is considered to be the format, mostly preferred in common.
## How do you write in expanded form?
Expanded form or expanded notation is a way of writing numbers to see the math value of individual digits. When numbers are separated into individual place values and decimal places they can also form a mathematical expression. 5,325 in expanded notation form is 5,000 + 300 + 20 + 5 = 5,325.
## How do you write a point slope equation?
Point-slope is the general form y-y₁=m(x-x₁) for linear equations. It emphasizes the slope of the line and a point on the line (that is not the y-intercept). Watch this video to learn more about it and see some examples.
## What is the slope formula for 2 points?
Use the slope formula to find the slope of a line given the coordinates of two points on the line. The slope formula is m=(y2-y1)/(x2-x1), or the change in the y values over the change in the x values. The coordinates of the first point represent x1 and y1. The coordinates of the second points are x2, y2.
## How do you write 90 in expanded form?
1. Expanded Form. When you write a number in expanded form, you write a number in the form of an addition statement that shows place value.
2. Expanded Form.
3. 65 = 60 + 5.
4. 56 = 50 + 6.
5. 91 = 90 + 1.
6. 24 = 20 + 4.
7. 76 = 70 + 6.
8. 37 = 30 + 7.
## What is an example of point slope form?
Examples of Applying the Concept of Point-Slope Form of a Line. The slope is given as m = 3 m = 3 m=3, and the point (2,5) has coordinates of x 1 = 2 {x_1} = 2 x1=2 and y 1 = 5 {y_1} = 5 y1=5. Now plug the known values into the slope-intercept form to get the final answer.
## How do you write an equation given a point?
Find the Equation of a Line Given That You Know a Point on the Line And Its Slope. The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept. If you a point that a line passes through, and its slope, this page will show you how to find the equation of the line.
## What is the slope of 5?
Slope is equal to rise over run. Linda, The equation for a line is y=mx+b where m is the slope and b is the y intercept (where x=0) if we look at the line you were given y=5 and compare it to the general equation we find b=5 and m=0. So the answer is 0.
## What is the formula for slope and y-intercept?
The equation of any straight line, called a linear equation, can be written as: y = mx + b, where m is the slope of the line and b is the y-intercept. The y-intercept of this line is the value of y at the point where the line crosses the y axis.
## What is standard form in algebra?
The standard form for linear equations in two variables is Ax+By=C. For example, 2x+3y=5 is a linear equation in standard form. When an equation is given in this form, it’s pretty easy to find both intercepts (x and y). This form is also very useful when solving systems of two linear equations.
## What is 59.69 Written in expanded form?
In expanded form, we write a number as the sum of the place values it has. Thus, 59.69 = 50 + 9 + 0.60 + 0.09.
## Do you simplify slope?
If you set the rise over the run, you get a fraction that describes the slope. Sometimes this fraction can be further simplified by dividing the numerator and the denominator by their greatest common factor. Divide the numerator by this number. If the result is zero, the simplified slope of the line is also zero.
## What is the formula of two point form?
Since we know two points on the line, we use the two-point form to find its equation. The final equation is in the slope-intercept form, y=mx+b y = m x + b . Find the equation of a straight line whose x-intercept is a and y-intercept is b , as shown in the following figure. |
# 2x + y = 3. Since the second equation is precisely the same as the first equation, it is enough to find x and y satisfying the system
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## Transcription
1 1. Systems of linear equations We are interested in the solutions to systems of linear equations. A linear equation is of the form 3x 5y + 2z + w = 3. The key thing is that we don t multiply the variables together nor do we raise powers, nor takes logs or introduce sine and cosines. A system of linear equations is of the form 3x 5y + 2z = 3 2x + y + 5z = 4. This is a system of two linear equations in three variables. The first equation is a system consisting of one linear equation in four variables. In this class we will be more interested in the nature of the solutions rather than the exact solutions themselves. So let us try to form a picture of what to expect. Suppose we start with an easy case. A system of two linear equations in two variables: x 2y = 1 2x + y = 3. It is easy to check that this has the unique solution x = y = 1. To see that this is not the only qualitative behaviour, suppose we consider the system 2x + y = 3 2x + y = 3. Since the second equation is precisely the same as the first equation, it is enough to find x and y satisfying the system 2x + y = 3. In other words the solution set is infinite. Any point of the line given by the equation 2x + y = 3 will do. The problem is that the second equation does not impose independent conditions. Note that we can disguise (to a certain extent) this problem by writing down the system 2x + y = 3 4x + 2y = 6. All we did was take the second equation and multiply it by 2. But we are not fooled, we still realise that the second equation imposes no more conditions than the first. It is not independent of the first equation. 1
2 We can tweak this example to get a system 2x + y = 3 2x + y = 2. Now this system has no solutions whatsoever. If 2x + y = 3 then 2x + y 2. The key point is that these three examples exhaust the qualitative behaviour: One solution. Infinitely many solutions. No solutions. In other words if there are at least two solutions then there are infinitely many solutions. In particular it turns out that the answer is never that there are two solutions. We say that a linear system is inconsistent if there are no solutions and otherwise we also say that the system is consistent. It is instructive to think how this comes out geometrically. Suppose there are two variables x and y. Then we can represent the solutions to a system of linear equations in x and y as a set of points in the plane. Points in the plane are represented by a pair of points (x, y) and we will refer to these points as vectors. The set of all such points is called R 2, R 2 = { (x, y) x, y R }. Okay, so what are the possibilities for the solution set? Well suppose we have a system of one equation. The solution set is a line. Suppose we have a system of two equations. If the equations impose independent conditions, then we get a single point, in fact the intersection of the two lines represented by each equation. Are there other possible solution sets? Yes, we already saw we might get no solutions. Either take a system of two equations with no solutions (in fact parallel lines) or a system of three equations, where we get three different points when we intersect each pair of lines. If we have three equations can we still get a solution? Yes suppose all three lines are concurrent (pass through the same point). It is interesting to see how this works algebraically. Suppose we consider the system 2x + y = 3 x 2y = 1 3x y = 2. Then the vector (x, y) = (1, 1) still satisfies all three equations. Note that the third equation is nothing more than the sum of the first two equations. As soon as one realises this fact, it is clear that the third 2
3 equation fails to impose independent conditions, that is the third equation depends on the first two. Note one further possibility. The following represents a system of three linear equations in two variables, with a line of solutions: x + y = 1 2x + 2y = 2 3x + 3y = 3. These equations are very far from being independent. In fact no matter how many equations there are, it is still possible (but more and more unlikely) that there are solutions. Finally let me point out some unusual possibilities. If we have a system of no equations, then the solution set is the whole of R 2. In fact 0x + 0y = 0, represents a single equation whose solution set is R 2. token 0x + 0y = 1, By the same is a linear equation with no solutions. In summary the solution set to a system of linear equations in two variables exhibits one of four different qualitative possibilities: (1) No solutions. (2) One solution. (3) A line of solutions. (4) The whole plane R 2. (Strictly speaking, so far we have only shown that these possibilities occur, we have not shown that these are the only possibilities.). In other words the case when there are infinitely many solutions can be be further refined into the last two cases. Now let us delve deeper into characterising the possible behaviour of the solutions. Let us consider a system of three linear equations in three unknowns. Can we list all geometric possibilities? One possibility is we get a point. For example x = 0 y = 0 z = 0. The first equation cuts down the space of solutions to a plane, the plane { (0, y, z) y, z R }. 3
4 The second equation represents another plane which intersects the first plane in a line { (0, 0, z) z R }. The last equation defines a plane which intersects this line in the origin, (0, 0, 0). Or the first two equations could give a line and the third equation might also contain the same line, x = 0 y = 0 x + y = 0. The last equation is a sum of the first two equations. But we could just as well consider x = 0 y = 0 3x 2y = 0. Again the problem is that the third equation is not independent from the first two equations. There are many other possibilities. Two planes could be parallel, in which case there are no solutions. It is time for some general notation and definitions. A vector (r 1, r 2,..., r n ) R n, is a solution to a linear equation if a 1 x 1 + a 2 x a n x n = b. a 1 r 1 + a 2 r a n r n = b. A system of m linear equations in n variables has the form a 11 x 1 + a 12 x a 1n x n = b 1 a 21 x 1 + a 22 x a 2n x n = b =. a m1 x 1 + a m2 x a mn x n = b m. A solution to a system of linear equations is a vector (r 1, r 2,..., r n ) R n which satisfies all m equations simultaneously. The solution set is the set of all solutions. Okay let us now turn to the boring bit. Given a system of linear equations, how does one solve the system? Again, let us start with a 4
5 simple example, x 2y = 1 2x + y = 3. There are many ways to solve this system, but let me focus on a particular method which generalises well to larger systems. Use the first equation to eliminate x from the second equation. To do this, take the first equation, multiply it by 2 and add it to the second equation (we prefer to think of multiplying by a negative number and adding rather than subtracting). We get a new system of equations, x 2y = 1 5y = 5. Since the second equation does not involve x at all, it is straightforward to use the second equation to get y = 1. Now use that value and substitute it into the first equation to determine x, x 2 = 1, so that x = 1. This method is called Gaussian elimination. The last step, where we go from the bottom to the top and recursively solve for the variables is called backwards substitution. Let us do the same thing with a system of three equations in three variables: x 2y z = 4 2x 3y + z = 10 x + 5y + 11z = 3. We start with the first equation and use it to eliminate the appearance of x from the second equation. To do this, take the first equation, multiply it by 2 and add it to the second equation. x 2y z = 4 y + 3z = 2 x + 5y + 11z = 3. Now let us eliminate the appearance of x from the third equation. To do this add the first equation to the third equation. x 2y z = 4 y + 3z = 2 3y + 10z = 7. 5
6 The next step is to eliminate y from the third equation by using the second equation. To do this, take the second equation, multiply it by 3 and add it to the third equation, x 2y z = 4 y + 3z = 2 z = 1. This completes the elimination step. Notice the characteristic upside down staircase shape of the equations. Now we do backwards substitution. Since the last equation does not involve either x or y, we read off the value for z from the last equation, to get z = 1. Now take this value for z and use the second equation, which does not involve x to solve for y, y + 3 = 2. This gives y = 1. Finally substitute the values for x and y to determine x, x = 4, to get x = 3. The solution is (x, y, z) = (3, 1, 1) R 3. It is easy to check that this is indeed a solution. Strictly speaking we should also check that there are no other solutions. We will come to this point later. Let us introduce some notation, which for the time being we will think of as just being a convenience. Instead of carrying around the variables x, y, z etc, let us just put the information into a table. We represent the system x 2y z = 4 2x 3y + z = 10 x + 5y + 11z = 3. by the augmented matrix B = The coefficient matrix is A =
7 A is a matrix with three rows and three columns, denoted 3 3. Note that the rows represent equations and the columns variables. Let b = b is also known as a column vector. Clearly b, as a matrix, is 3 1. Then the augmented matrix is formed from the two matrices A and b ( A b ) If we let v = x y, z then we may also write down the system of equations very compactly as Av = b. It is possible to understand the previous method of solving linear equations in terms of the augmented matrix only. We start with, The first step is to take the first row, multiply it by 2 and add it to the second row, This puts a zero in the second row, first column. The next step is to take the first row, multiply it by 1 and add it to the third row, This puts a zero in the third row, first column. The next step is to take the second row, multiply it by 3 and add it to the third row, This puts a zero in the third row, second column. At this stage, we think of this augmented matrix as representing a system of equations and use the old method of back substitution to solve this system. 7
8 Let us look at a slightly more complicated example. Suppose that we start with a system of four equations in four unknowns, x + 3y 2z w = 1 6x 15y + 9z + 9w = 9 x z + 4w = 5 4x + 10y 5z 2w = 3. We first replace this by the augmented matrix, The first step is to use the one in the first row, first column to eliminate the 6, 1 and 4 from the first column. To do this we add 6, 1, 4 times the first row to the second, third and fourth row, to get The next step is to multiply the second row by 1/3 to get a one in the second row, second column, Now we eliminate the 3 and the 2 in the second column. To do this we add 3 and 2 times the second row to the third and fourth row, The next step is to get a one in the third row, third column. We cannot do this by rescaling the third row. We can do it by swapping the third and fourth rows,
9 Consider what happens when we try to solve the resulting linear equations by back substitution. The last equation reads 0x + 0y + 0z + 0w = 2. There are no values for x, y, z and w which work. The original system has no solutions. Now suppose we start with the system x + 3y 2z w = 1 6x 15y + 9z + 9w = 9 x z + 4w = 4 4x + 10y 5z 2w = 5. The only thing we have changed is 5 to 4 = 5 1. If we follow the same steps as before, we get down to the same matrix, except that the last entry is 0 = 1 1 (remember at some point we swapped two rows) The last equation x + 0y + 0z + 0w = 0,. places no restriction on x, y, z and w. The previous equation reads, z + 4w = 1. Using this equation, we can solve for z in terms of w, to get z = 1 4w. We can use this value for z in the second equation to determine y in terms of w, y (1 4w) + w = 1, to get y = 2 5w. Finally, we can use this value for y and z in the first equation to solve for x, to get We get the family of solutions, x + 3(2 5w) 2(1 4w) w = 1, x = 5 8w. (x, y, z, w) = ( 5 + 8w, 2 5w, 1 4w, w). Note that no matter the value of w, we get a solution for the original system of linear equations. For example if we pick w = 0, we get the solution (x, y, z, w) = ( 5, 2, 1, 0), 9
10 but if we choose the value w = 1, we get the solution (x, y, z, w) = (3, 3, 3, 1). In particular this system has infinitely many solutions. As before the reason for this is because the original system of equations was not independent. In fact even the first three equations are not independent. We can think of this as being the same thing as the first three rows are not independent. The third row minus four times the first row is the same as the second row plus the first row (we can see this by following the steps of the Gaussian elimination). This is the same as saying the third row is equal to the second row plus three times the first row. It is then automatic that any solution to the first equation and the second equations is a solution to the third equation. Note also that we can represent the solutions in a slightly different way, (x, y, z, w) = ( 5, 2, 1, 0) + w(8, 5, 4, 1). 10
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# Cartesian Product of Sets
## Ordered Pairs
In sets, the order of elements is not important. For example, the sets {2, 3} and {3, 2} are equal to each other. However, there are many instances in mathematics where the order of elements is essential. So, for example, the pairs of numbers with coordinates (2, 3) and (3, 2) represent different points on the plane. This leads to the concept of ordered pairs.
An ordered pair is defined as a set of two objects together with an order associated with them. Ordered pairs are usually written in parentheses (as opposed to curly braces, which are used for writing sets).
In the ordered pair $$\left( {a,b} \right),$$ the element $$a$$ is called the first entry or first component, and $$b$$ is called the second entry or second component of the pair.
Two ordered pairs $$\left( {a,b} \right)$$ and $$\left( {c,d} \right)$$ are equal if and only if $$a = c$$ and $$b = d.$$ In general,
$\left( {a,b} \right) \ne \left( {b,a} \right).$
The equality $$\left( {a,b} \right) = \left( {b,a} \right)$$ is possible only if $$a = b.$$
## Tuples
The concept of ordered pair can be extended to more than two elements. An ordered $$n-$$tuple is a set of $$n$$ objects together with an order associated with them. Tuples are usually denoted by $$\left( {{a_1},{a_2}, \ldots, {a_n}} \right).$$ The element $${a_i}$$ $$\left({i = 1,2, \ldots, n}\right)$$ is called the $$i\text{th}$$ entry or component, and $$n$$ is called the length of the tuple.
Similarly to ordered pairs, the order in which elements appear in a tuple is important. Two tuples of the same length $$\left( {{a_1},{a_2}, \ldots, {a_n}} \right)$$ and $$\left( {{b_1},{b_2}, \ldots, {b_n}} \right)$$ are said to be equal if and only if $${a_i} = {b_i}$$ for all $${i = 1,2, \ldots, n}.$$ So the following tuples are not equal to each other:
$\left( {1,2,3,4,5} \right) \ne \left( {3,2,1,5,4} \right).$
Unlike sets, tuples may contain a certain element more than once:
$\left( {1,2,3,2,1,1,1} \right).$
Ordered pairs are sometimes referred as $$2-$$tuples.
## Cartesian Product of Two Sets
Suppose that $$A$$ and $$B$$ are non-empty sets. The Cartesian product of two sets $$A$$ and $$B,$$ denoted $$A \times B,$$ is the set of all possible ordered pairs $$\left( {a,b} \right),$$ where $$a \in A$$ and $$b \in B:$$
$A \times B = \left\{ {\left( {a,b} \right) \mid a \in A \text{ and } b \in B} \right\}.$
The Cartesian product is also known as the cross product.
The figure below shows the Cartesian product of the sets $$A = \left\{ {1,2,3} \right\}$$ and $$B = \left\{ {x,y} \right\}.$$
It consists of $$6$$ ordered pairs:
$A \times B = \left\{ {\left( {1,x} \right),\left( {2,x} \right),\left( {3,x} \right),\left( {1,y} \right),\left( {2,y} \right),\left( {3,y} \right)} \right\}.$
Similarly, we can find the Cartesian product $$B \times A:$$
$B \times A = \left\{ {\left( {x,1} \right),\left( {y,1} \right),\left( {x,2} \right),\left( {y,2} \right),\left( {x,3} \right),\left( {y,3} \right)} \right\}.$
As you can see from this example, the Cartesian products $$A \times B$$ and $$B \times A$$ do not contain exactly the same ordered pairs. So, in general, $$A \times B \ne B \times A.$$
If $$A = B,$$ then $$A \times B$$ is called the Cartesian square of the set $$A$$ and is denoted by $$A^2:$$
${A^2} = \left\{ {\left( {a,b} \right) \mid a \in A \text{ and } b \in A} \right\}.$
## Cartesian Product of Several Sets
Cartesian products may also be defined on more than two sets.
Let $${A_1}, \ldots ,{A_n}$$ be $$n$$ non-empty sets. The Cartesian product $${A_1} \times \ldots \times {A_n}$$ is defined as the set of all possible ordered $$n-$$tuples $$\left({{a_1}, \ldots ,{a_n}}\right),$$ where $${a_i} \in {A_i}$$ and $${i = 1,\ldots, n}.$$
If $${A_1} = \ldots = {A_n} = A,$$ then $${A_1} \times \ldots \times {A_n}$$ is called the $$n\text{th}$$ Cartesian power of the set $$A$$ and is denoted by $${A^n}.$$
## Some Properties of Cartesian Product
1. The Cartesian product is non-commutative:
$A \times B \ne B \times A$
2. $$A \times B = B \times A,$$ if only $$A = B.$$
3. $${A \times B = \varnothing},$$ if either $$A = \varnothing$$ or $$B = \varnothing.$$
4. The Cartesian product is non-associative:
$\left( {A \times B} \right) \times C \ne A \times \left( {B \times C} \right)$
5. Distributive property over set intersection:
$A \times \left( {B \cap C} \right) = \left( {A \times B} \right) \cap \left( {A \times C} \right)$
6. Distributive property over set union:
$A \times \left( {B \cup C} \right) = \left( {A \times B} \right) \cup \left( {A \times C} \right)$
7. Distributive property over set difference:
$A \times \left( {B \backslash C} \right) = \left( {A \times B} \right) \backslash \left( {A \times C} \right)$
8. If $$A \subseteq B,$$ then $$A \times C \subseteq B \times C$$ for any set $$C.$$
## Cardinality of Cartesian Product
The сardinality of a Cartesian product of two sets is equal to the product of the cardinalities of the sets:
$\left| {A \times B} \right| = \left| {B \times A} \right| = \left| A \right| \times \left| B \right|.$
Similarly,
$\left| {{A_1} \times \ldots \times {A_n}} \right| = \left| {{A_1}} \right| \times \ldots \times \left| {{A_n}} \right|.$
See solved problems on Page 2. |
Students can go through AP Board 8th Class Maths Notes Chapter 11 Algebraic Expressions to understand and remember the concepts easily.
## AP State Board Syllabus 8th Class Maths Notes Chapter 11 Algebraic Expressions
→ There are number of situations in which we need to multiply algebraic expressions.
→ A monomial multiplied by a monomial always gives a monomial.
→ While multiplying a polynomial by a monomial, we multiply every term in the polynomial by the monomial.
→ In carrying out the multiplication of an algebraic expression with another algebraic expression (monomial/ binomial/ trinomial etc.) we multiply term by term i.e. every term of the expression is multiplied by every term in the another expression.
→ An identity is an equation, which is true for-all values of the variables in the equation. On the other hand, an equation is true only for certain values of its variables. An equation is not an identity.
→ The following are identities:
I. (a + b)2 = a2 + 2ab + b2
II. (a – b)2 = a2 – 2ab + b2
III. (a + b) (a -b) = a2 – b2
IV. (x + a) (x + b) = x2 + (a + b)x + ab
→ The above four identities are useful in carrying out squares and products of algebraic expressions. They also allow easy alternative methods to calculate products of numbers and so on.
Note:
We know that
(+) × (+) = +
(+) × (-) = –
(-) × (+) = –
(-) × (-) = + |
# SUBTRACTING FRACTIONS WITH LIKE DENOMINATORS
The following steps will be useful to subtract fractions with like denominators.
Step 1 :
When two fractions with the same denominator are subtracted, take the denominator once.
Step 2 :
Now, subtract the numerators and simplify, if required.
Example 1 :
Find the value of :
1/5 - 2/5
Solution :
1/5 - 2/5
The given two fractions have the same denominator. That is 5.
So, take the denominator once and subtract the numerators.
= (1 - 2) / 5
= -1 / 5
Therefore,
1/5 - 2/5 = -1/5
Example 2 :
Subtract 2/14 from 5 / 14.
Solution :
5/14 - 2/14
The given two fractions have the same denominator. That is 14.
So, take the denominator once and subtract the numerators.
= (5 - 2) / 14
= 3 / 14
Therefore,
5/14 - 2/14 = 3/14
Example 3 :
Subtract 47/125 from 87/125.
Solution :
87/125 - 47/125
The given two fractions have the same denominator. That is 125.
So, take the denominator once and subtract the numerators.
= (87 - 47) / 125
= 40 / 125
Simplify.
= 8 / 25
Therefore,
87/125 - 47/125 = 8/25
Example 4 :
Find the value of :
1/5 + 2/5 - 2/5
Solution :
1/5 + 2/5 - 2/5
The given fractions have the same denominator. That is 5.
So, take the denominator once and simplify the numerators.
= (1 + 2 - 2) / 5
= 1 / 5
Therefore,
1/5 + 2/5 + 2/5 = 1/5
Example 5 :
Find the value of :
2/13 - 7/13 + 12/13 - 5/13
Solution :
2/13 - 7/13 + 12/13 - 5/13
The given fractions have the same denominator. That is 13.
So, take the denominator once and simplify the numerators.
= (2 - 7 + 12 - 5) / 13
= 2 / 13
Therefore,
2/13 - 7/13 + 12/13 - 5/13 = 2/13
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# 9.7 Higher roots (Page 4/8)
Page 4 / 8
Simplify: $\sqrt[5]{3x}+\sqrt[5]{3x}$ $3\sqrt[3]{9}-\sqrt[3]{9}$ .
$2\sqrt[5]{3x}$ $2\sqrt[3]{9}$
Simplify: $\sqrt[4]{10y}+\sqrt[4]{10y}$ $5\sqrt[6]{32}-3\sqrt[6]{32}$ .
$2\sqrt[4]{10y}$ $2\sqrt[6]{32}$
When an expression does not appear to have like radicals, we will simplify each radical first. Sometimes this leads to an expression with like radicals.
Simplify: $\sqrt[3]{54}-\sqrt[3]{16}$ $\sqrt[4]{48}+\sqrt[4]{243}$ .
## Solution
1. $\begin{array}{ccc}& & \sqrt[3]{54}-\sqrt[3]{16}\hfill \\ \\ \\ \text{Rewrite each radicand using perfect cube factors.}\hfill & & \sqrt[3]{27}·\sqrt[3]{2}-\sqrt[3]{8}·\sqrt[3]{2}\hfill \\ \\ \\ \text{Rewrite the perfect cubes.}\hfill & & \sqrt[3]{{\left(3\right)}^{3}}\phantom{\rule{0.2em}{0ex}}\sqrt[3]{2}-\sqrt[3]{{\left(2\right)}^{3}}\phantom{\rule{0.2em}{0ex}}\sqrt[3]{2}\hfill \\ \\ \\ \text{Simplify the radicals where possible.}\hfill & & 3\sqrt[3]{2}-2\sqrt[3]{2}\hfill \\ \\ \\ \text{Combine like radicals.}\hfill & & \sqrt[3]{2}\hfill \end{array}$
2. $\begin{array}{ccc}& & \phantom{\rule{2em}{0ex}}\sqrt[4]{48}+\sqrt[4]{243}\hfill \\ \\ \\ \text{Rewrite using perfect fourth power factors.}\hfill & & \phantom{\rule{2em}{0ex}}\sqrt[4]{16}·\sqrt[4]{3}+\sqrt[4]{81}·\sqrt[4]{3}\hfill \\ \\ \\ \text{Rewrite the perfect fourth powers.}\hfill & & \phantom{\rule{2em}{0ex}}\sqrt[4]{{\left(2\right)}^{4}}\phantom{\rule{0.2em}{0ex}}\sqrt[4]{3}+\sqrt[4]{{\left(3\right)}^{4}}\phantom{\rule{0.2em}{0ex}}\sqrt[4]{3}\hfill \\ \\ \\ \text{Simplify the radicals where possible.}\hfill & & \phantom{\rule{2em}{0ex}}2\sqrt[4]{3}+3\sqrt[4]{3}\hfill \\ \\ \\ \text{Combine like radicals.}\hfill & & \phantom{\rule{2em}{0ex}}5\sqrt[4]{3}\hfill \end{array}$
Simplify: $\sqrt[3]{192}-\sqrt[3]{81}$ $\sqrt[4]{32}+\sqrt[4]{512}$ .
$\sqrt[3]{3}$ $6\sqrt[4]{2}$
Simplify: $\sqrt[3]{108}-\sqrt[3]{250}$ $\sqrt[5]{64}+\sqrt[5]{486}$ .
$\text{−}\sqrt[3]{2}$ $5\sqrt[5]{2}$
Simplify: $\sqrt[3]{24{x}^{4}}-\sqrt[3]{-81{x}^{7}}$ $\sqrt[4]{162{y}^{9}}+\sqrt[4]{516{y}^{5}}$ .
## Solution
1. $\begin{array}{ccc}& & \phantom{\rule{4em}{0ex}}\sqrt[3]{24{x}^{4}}-\sqrt[3]{-81{x}^{7}}\hfill \\ \\ \\ \text{Rewrite each radicand using perfect cube factors.}\hfill & & \phantom{\rule{4em}{0ex}}\sqrt[3]{8{x}^{3}}·\sqrt[3]{3x}-\sqrt[3]{-27{x}^{6}}·\sqrt[3]{3x}\hfill \\ \\ \\ \text{Rewrite the perfect cubes.}\hfill & & \phantom{\rule{4em}{0ex}}\sqrt[3]{{\left(2x\right)}^{3}}\phantom{\rule{0.2em}{0ex}}\sqrt[3]{3x}-\sqrt[3]{{\left(-3{x}^{2}\right)}^{3}}\phantom{\rule{0.2em}{0ex}}\sqrt[3]{3x}\hfill \\ \\ \\ \text{Simplify the radicals where possible.}\hfill & & \phantom{\rule{4em}{0ex}}2x\sqrt[3]{3x}-\left(-3{x}^{2}\sqrt[3]{3x}\right)\hfill \end{array}$
2. $\begin{array}{ccc}& & \sqrt[4]{162{y}^{9}}+\sqrt[4]{516{y}^{5}}\hfill \\ \\ \\ \text{Rewrite each radicand using perfect fourth power factors.}\hfill & & \sqrt[4]{81{y}^{8}}·\sqrt[4]{2y}+\sqrt[4]{256{y}^{4}}·\sqrt[4]{2y}\hfill \\ \\ \\ \text{Rewrite the perfect fourth powers.}\hfill & & \sqrt[4]{{\left(3{y}^{2}\right)}^{4}}·\sqrt[4]{2y}+\sqrt[4]{{\left(4y\right)}^{4}}·\sqrt[4]{2y}\hfill \\ \\ \\ \text{Simplify the radicals where possible.}\hfill & & 3{y}^{2}\sqrt[4]{2y}+4|y|\sqrt[4]{2y}\hfill \end{array}$
Simplify: $\sqrt[3]{32{y}^{5}}-\sqrt[3]{-108{y}^{8}}$ $\sqrt[4]{243{r}^{11}}+\sqrt[4]{768{r}^{10}}$ .
$2y\sqrt[3]{4{y}^{2}}+3{y}^{2}\sqrt[3]{4{y}^{2}}$ $3{r}^{2}\sqrt[4]{3{r}^{3}}+4{r}^{2}\sqrt[4]{3{r}^{2}}$
Simplify: $\sqrt[3]{40{z}^{7}}-\sqrt[3]{-135{z}^{4}}$ $\sqrt[4]{80{s}^{13}}+\sqrt[4]{1280{s}^{6}}$ .
$2{z}^{2}\sqrt[3]{5z}+3z\sqrt[3]{5z}$ $2|{s}^{3}|\sqrt[4]{5s}+4|s|\sqrt[4]{5s}$
Access these online resources for additional instruction and practice with simplifying higher roots.
## Key concepts
• Properties of
• $\sqrt[n]{a}$ when $n$ is an even number and
• $a\ge 0$ , then $\sqrt[n]{a}$ is a real number
• $a<0$ , then $\sqrt[n]{a}$ is not a real number
• When $n$ is an odd number, $\sqrt[n]{a}$ is a real number for all values of a .
• For any integer $n\ge 2$ , when n is odd $\sqrt[n]{{a}^{n}}=a$
• For any integer $n\ge 2$ , when n is even $\sqrt[n]{{a}^{n}}=|a|$
• $\sqrt[n]{a}$ is considered simplified if a has no factors of ${m}^{n}$ .
• Product Property of n th Roots
$\sqrt[n]{ab}=\sqrt[n]{a}·\sqrt[n]{b}\phantom{\rule{0.5em}{0ex}}\text{and}\phantom{\rule{0.5em}{0ex}}\sqrt[n]{a}·\sqrt[n]{b}=\sqrt[n]{ab}$
• Quotient Property of n th Roots
$\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}\phantom{\rule{0.5em}{0ex}}\text{and}\phantom{\rule{0.5em}{0ex}}\frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}}$
• To combine like radicals, simply add or subtract the coefficients while keeping the radical the same.
## Practice makes perfect
Simplify Expressions with Higher Roots
In the following exercises, simplify.
$\sqrt[3]{216}$
$\sqrt[4]{256}$
$\sqrt[5]{32}$
$\sqrt[3]{27}$
$\sqrt[4]{16}$
$\sqrt[5]{243}$
$3$ $2$ $3$
$\sqrt[3]{512}$
$\sqrt[4]{81}$
$\sqrt[5]{1}$
$\sqrt[3]{125}$
$\sqrt[4]{1296}$
$\sqrt[5]{1024}$
$5$ $6$ $4$
$\sqrt[3]{-8}$
$\sqrt[4]{-81}$
$\sqrt[5]{-32}$
$\sqrt[3]{-64}$
$\sqrt[4]{-16}$
$\sqrt[5]{-243}$
$-4$ $\text{not real}$ $-3$
$\sqrt[3]{-125}$
$\sqrt[4]{-1296}$
$\sqrt[5]{-1024}$
$\sqrt[3]{-512}$
$\sqrt[4]{-81}$
$\sqrt[5]{-1}$
$-8$ not a real number $-1$
$\sqrt[5]{{u}^{5}}$
$\sqrt[8]{{v}^{8}}$
$\sqrt[3]{{a}^{3}}$
$a$ $|b|$
$\sqrt[4]{{y}^{4}}$
$\sqrt[7]{{m}^{7}}$
$\sqrt[8]{{k}^{8}}$
$\sqrt[6]{{p}^{6}}$
$|k|$ $|p|$
$\sqrt[3]{{x}^{9}}$
$\sqrt[4]{{y}^{12}}$
$\sqrt[5]{{a}^{10}}$
$\sqrt[3]{{b}^{27}}$
${a}^{2}$ ${b}^{9}$
$\sqrt[4]{{m}^{8}}$
$\sqrt[5]{{n}^{20}}$
$\sqrt[6]{{r}^{12}}$
$\sqrt[3]{{s}^{30}}$
${r}^{2}$ ${s}^{10}$
$\sqrt[4]{16{x}^{8}}$
$\sqrt[6]{64{y}^{12}}$
$\sqrt[3]{-8{c}^{9}}$
$\sqrt[3]{125{d}^{15}}$
$-2{c}^{3}$ $5{d}^{5}$
$\sqrt[3]{216{a}^{6}}$
$\sqrt[5]{32{b}^{20}}$
$\sqrt[7]{128{r}^{14}}$
$\sqrt[4]{81{s}^{24}}$
$2{r}^{2}$ $3{s}^{6}$
Use the Product Property to Simplify Expressions with Higher Roots
In the following exercises, simplify.
$\sqrt[3]{{r}^{5}}$ $\sqrt[4]{{s}^{10}}$
$\sqrt[5]{{u}^{7}}$ $\sqrt[6]{{v}^{11}}$
$u\sqrt[5]{{u}^{2}}$ $v\sqrt[6]{{v}^{5}}$
$\sqrt[4]{{m}^{5}}$ $\sqrt[8]{{n}^{10}}$
$\sqrt[5]{{p}^{8}}$ $\sqrt[3]{{q}^{8}}$
$p\sqrt[5]{{p}^{3}}$ ${q}^{2}\sqrt[3]{{q}^{2}}$
$\sqrt[4]{32}$ $\sqrt[5]{64}$
$\sqrt[3]{625}$ $\sqrt[6]{128}$
$5\sqrt[3]{5}$ $2\sqrt[6]{2}$
$\sqrt[5]{64}$ $\sqrt[3]{256}$
$\sqrt[4]{3125}$ $\sqrt[3]{81}$
$5\sqrt[4]{5}$ $3\sqrt[3]{3}$
$\sqrt[3]{108{x}^{5}}$ $\sqrt[4]{48{y}^{6}}$
$\sqrt[5]{96{a}^{7}}$ $\sqrt[3]{375{b}^{4}}$
$2a\sqrt[5]{3{a}^{2}}$ $5b\sqrt[3]{3b}$
$\sqrt[4]{405{m}^{10}}$ $\sqrt[5]{160{n}^{8}}$
$\sqrt[3]{512{p}^{5}}$ $\sqrt[4]{324{q}^{7}}$
$8p\sqrt[3]{{p}^{2}}$ $3q\sqrt[4]{4{q}^{3}}$
$\sqrt[3]{-864}$ $\sqrt[4]{-256}$
$\sqrt[5]{-486}$ $\sqrt[6]{-64}$
$-3\sqrt[5]{2}$ $\text{not real}$
$\sqrt[5]{-32}$ $\sqrt[8]{-1}$
$\sqrt[3]{-8}$ $\sqrt[4]{-16}$
$-2$ $\text{not real}$
Use the Quotient Property to Simplify Expressions with Higher Roots
In the following exercises, simplify.
$\sqrt[3]{\frac{{p}^{11}}{{p}^{2}}}$ $\sqrt[4]{\frac{{q}^{17}}{{q}^{13}}}$
$\sqrt[5]{\frac{{d}^{12}}{{d}^{7}}}$ $\sqrt[8]{\frac{{m}^{12}}{{m}^{4}}}$
$d$ $|m|$
$\sqrt[5]{\frac{{u}^{21}}{{u}^{11}}}$ $\sqrt[6]{\frac{{v}^{30}}{{v}^{12}}}$
$\sqrt[3]{\frac{{r}^{14}}{{r}^{5}}}$ $\sqrt[4]{\frac{{c}^{21}}{{c}^{9}}}$
${r}^{2}$ $|{c}^{3}|$
$\frac{\sqrt[4]{64}}{\sqrt[4]{2}}$ $\frac{\sqrt[5]{128{x}^{8}}}{\sqrt[5]{2{x}^{2}}}$
$\frac{\sqrt[3]{-625}}{\sqrt[3]{5}}$ $\frac{\sqrt[4]{80{m}^{7}}}{\sqrt[4]{5m}}$
$-5$ $4m\sqrt[4]{{m}^{2}}$
$\sqrt[3]{\frac{1050}{2}}$ $\sqrt[4]{\frac{486{y}^{9}}{2{y}^{3}}}$
$\sqrt[3]{\frac{162}{6}}$ $\sqrt[4]{\frac{160{r}^{10}}{5{r}^{3}}}$
$3\sqrt[3]{6}$ $2|r|\sqrt[4]{2{r}^{3}}$
$\sqrt[3]{\frac{54{a}^{8}}{{b}^{3}}}$ $\sqrt[4]{\frac{64{c}^{5}}{{d}^{2}}}$
$\sqrt[5]{\frac{96{r}^{11}}{{s}^{3}}}$ $\sqrt[6]{\frac{128{u}^{7}}{{v}^{3}}}$
$\frac{2{r}^{2}\sqrt[5]{3r}}{{s}^{3}}$ $\frac{2{u}^{3}\sqrt[6]{2uv3}}{v}$
$\sqrt[3]{\frac{81{s}^{8}}{{t}^{3}}}$ $\sqrt[4]{\frac{64{p}^{15}}{{q}^{12}}}$
$\sqrt[3]{\frac{625{u}^{10}}{{v}^{3}}}$ $\sqrt[4]{\frac{729{c}^{21}}{{d}^{8}}}$
$\frac{5{u}^{3}\sqrt[3]{5u}}{v}$ $\frac{3{c}^{5}\sqrt[4]{9c}}{{d}^{2}}$
In the following exercises, simplify.
$\sqrt[7]{8p}+\sqrt[7]{8p}$
$3\sqrt[3]{25}-\sqrt[3]{25}$
$\sqrt[3]{15q}+\sqrt[3]{15q}$
$2\sqrt[4]{27}-6\sqrt[4]{27}$
$2\sqrt[3]{15q}$ $-4\sqrt[4]{27}$
$3\sqrt[5]{9x}+7\sqrt[5]{9x}$
$8\sqrt[7]{3q}-2\sqrt[7]{3q}$
$\sqrt[3]{81}-\sqrt[3]{192}$
$\sqrt[4]{512}-\sqrt[4]{32}$
$\sqrt[3]{250}-\sqrt[3]{54}$
$\sqrt[4]{243}-\sqrt[4]{1875}$
$5\sqrt[3]{5}-3\sqrt[3]{2}$ $-2\sqrt[4]{3}$
$\sqrt[3]{128}+\sqrt[3]{250}$
$\sqrt[5]{729}+\sqrt[5]{96}$
$\sqrt[4]{243}+\sqrt[4]{1250}$
$\sqrt[3]{2000}+\sqrt[3]{54}$
$3\sqrt[4]{3}+5\sqrt[4]{2}$ $13\sqrt[3]{2}$
$\sqrt[3]{64{a}^{10}}-\sqrt[3]{-216{a}^{12}}$
$\sqrt[4]{486{u}^{7}}+\sqrt[4]{768{u}^{3}}$
$\sqrt[3]{80{b}^{5}}-\sqrt[3]{-270{b}^{3}}$
$\sqrt[4]{160{v}^{10}}-\sqrt[4]{1280{v}^{3}}$
$2b\sqrt[3]{10{b}^{2}}+3b\sqrt[3]{10}$ $2{v}^{2}\sqrt[4]{10{v}^{2}}-4\sqrt[4]{5{v}^{3}}$
Mixed Practice
In the following exercises, simplify.
$\sqrt[4]{16}$
$\sqrt[6]{64}$
$2$
$\sqrt[3]{{a}^{3}}$
$|b|$
$\sqrt[3]{-8{c}^{9}}$
$\sqrt[3]{125{d}^{15}}$
$5{d}^{5}$
$\sqrt[3]{{r}^{5}}$
$\sqrt[4]{{s}^{10}}$
${s}^{2}\sqrt[4]{{s}^{2}}$
$\sqrt[3]{108{x}^{5}}$
$\sqrt[4]{48{y}^{6}}$
$2y\sqrt[4]{3{y}^{2}}$
$\sqrt[5]{-486}$
$\sqrt[6]{-64}$
$\text{not real}$
$\frac{\sqrt[4]{64}}{\sqrt[4]{2}}$
$\frac{\sqrt[5]{128{x}^{8}}}{\sqrt[5]{2{x}^{2}}}$
$2x\sqrt[5]{2x}$
$\sqrt[5]{\frac{96{r}^{11}}{{s}^{3}}}$
$\sqrt[6]{\frac{128{u}^{7}}{{v}^{3}}}$
$\frac{2{u}^{3}\sqrt[6]{2uv3}}{v}$
$\sqrt[3]{81}-\sqrt[3]{192}$
$\sqrt[4]{512}-\sqrt[4]{32}$
$4\sqrt[4]{2}$
$\sqrt[3]{64{a}^{10}}-\sqrt[3]{-216{a}^{12}}$
$\sqrt[4]{486{u}^{7}}+\sqrt[4]{768{u}^{3}}$
$3u\sqrt[4]{6{u}^{3}}+4\sqrt[4]{3{u}^{3}}$
## Everyday math
Population growth The expression $10·{x}^{n}$ models the growth of a mold population after $n$ generations. There were 10 spores at the start, and each had $x$ offspring. So $10·{x}^{n}$ is the number of offspring at the fifth generation. At the fifth generation there were 10,240 offspring. Simplify the expression $\sqrt[5]{\frac{10,240}{10}}$ to determine the number of offspring of each spore.
Spread of a virus The expression $3·{x}^{n}$ models the spread of a virus after $n$ cycles. There were three people originally infected with the virus, and each of them infected $x$ people. So $3·{x}^{4}$ is the number of people infected on the fourth cycle. At the fourth cycle 1875 people were infected. Simplify the expression $\sqrt[4]{\frac{1875}{3}}$ to determine the number of people each person infected.
$5$
## Writing exercises
Explain how you know that $\sqrt[5]{{x}^{10}}={x}^{2}$ .
Explain why $\sqrt[4]{-64}$ is not a real number but $\sqrt[3]{-64}$ is.
## Self check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
What does this checklist tell you about your mastery of this section? What steps will you take to improve?
how did you get the value of 2000N.What calculations are needed to arrive at it
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Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hours longer than Tamara to go 80 miles, how fast can Samantha ride her bike? Got questions? Get instant answers now!
how do u solve that question
Seera
Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hours longer than Tamara to go 80 miles, how fast can Samantha ride her bike?
Seera
Speed=distance ÷ time
Tremayne
x-3y =1; 3x-2y+4=0 graph
Brandon has a cup of quarters and dimes with a total of 5.55$. The number of quarters is five less than three times the number of dimes ashley Reply app is wrong how can 350 be divisible by 3. Raheem Reply June needs 48 gallons of punch for a party and has two different coolers to carry it in. The bigger cooler is five times as large as the smaller cooler. How many gallons can each cooler hold? Susanna Reply Susanna if the first cooler holds five times the gallons then the other cooler. The big cooler holda 40 gallons and the 2nd will hold 8 gallons is that correct? Georgie @Susanna that person is correct if you divide 40 by 8 you can see it's 5 it's simple Ashley @Geogie my bad that was meant for u Ashley Hi everyone, I'm glad to be connected with you all. from France. Lorris Reply I'm getting "math processing error" on math problems. Anyone know why? Ray Reply Can you all help me I don't get any of this Jade Reply 4^×=9 Alberto Reply Did anyone else have trouble getting in quiz link for linear inequalities? Sireka Reply operation of trinomial Justin Reply y=2×+9 Jacob Reply Keshad gets paid$2,400 per month plus 6% of his sales. His brother earns $3,300 per month. For what amount of total sales will Keshad’s monthly pay be higher than his brother’s monthly pay? Hector Reply Mayra has$124 in her checking account. She writes a check for $152. What is the New Balance in her checking account? REVOLUTION Reply -28$
ashley
-\$28
Stephanie |
# Simplify the expression (4x+8)+(-6x). Explain how the associative and commutative properties were used to solve the expression?
Jul 14, 2018
$- 2 \left(x - 4\right)$
#### Explanation:
1. use the distribution property to change $+ \times \left(- 6 x\right)$ to -6x ( + xx - = -)
2. remove the parenthesis giving 4x +8
( there are several ways to proceed from here my choice is)
3. use the commutative property to move the +8 and - 6x
$+ 4 x + 8 - 6 x = 4 x - 6 x + 8$ commutative property
4. Use the associative property to group + 4x -6x
$+ 4 x - 6 x + 8 = \left(+ 4 x - 6 x\right) + 8$ associative property
5. Use algebraic addition to solve for $\left(+ 4 x - 6 x\right)$
$\left(+ 4 x - 6 x\right) + 8 = - 2 x + 8$
6. Use the reverse distribution principal to simply and remove common terms
$- 2 \times \left\{\frac{- 2 x}{- 2} + \frac{8}{- 2}\right\} = - 2 \left(x - 4\right)$ |
# How do you find the area of a scalene triangle?
## How do you find the area of a scalene triangle?
We can find the scalene triangle’s area when the length of its two sides and the included angle are given.
1. When two sides b and c and included angle A is known, the area of the triangle is, Area = (1/2) bc × sin A.
2. When sides a and c and included angle B is known, the area of the triangle is, Area = (1/2) ac × sin B.
## Can we use Heron’s formula in scalene triangle?
Heron’s formula is used to find the area of a triangle when we know the length of all its sides. It is also termed as Hero’s Formula. We can Heron’s formula to find different types of triangles, such as scalene, isosceles and equilateral triangles.
What is a formula of scalene triangle?
The Formula for Scalene Triangle Area of Triangle = \frac{1}{2} \times b \times h, where b is the base and h is the height. The formula for Perimeter of any Triangle is, P = a+ b + c where a, b, c are the length of the sides.
### What is the Heron area formula?
Heron’s formula, formula credited to Heron of Alexandria (c. 62 ce) for finding the area of a triangle in terms of the lengths of its sides. In symbols, if a, b, and c are the lengths of the sides: Area = Square root of√s(s – a)(s – b)(s – c) where s is half the perimeter, or (a + b + c)/2.
### What is the base of a scalene triangle?
In a scalene triangle, the side that we consider to be the bottom side, or the side opposite the top vertex, is called the base of the triangle, and we can find the length of the base of a scalene triangle using its area and its height.
What is the area of ISO scalene triangle?
List of Formulas to Find Isosceles Triangle Area
Formulas to Find Area of Isosceles Triangle
Using base and Height A = ½ × b × h
Using all three sides A = ½[√(a2 − b2 ⁄4) × b]
Using the length of 2 sides and an angle between them A = ½ × b × c × sin(α)
#### How do you find the Semiperimeter of a triangle?
In other words, if the lengths of all the sides of a triangle are given, we can calculate the semi perimeter by adding the sides and dividing the sum by 2. The formula that is used to find the semi perimeter of triangle is, Semi perimeter = (a + b + c)/2, where ‘a’, ‘b’, ‘c’ are the three sides of the triangle.
#### What is Stalin triangle?
A scalene triangle is a triangle in which all three sides have different lengths. Also the angles of a scalene triangle have different measures. Some right triangles can be a scalene triangle when the other two angles or the legs are not congruent.
How do you find the semiperimeter of a triangle?
## How do you solve a scalene right triangle?
The formula for the perimeter of a scalene triangle is equal to the sum of the length of sides of a triangle and it is given as: The Perimeter of a Scalene Triangle = a + b + c units. |
# Surface Area of a Right Prism
How to find the surface area of a right prism: definition, formula, examples, and their solutions.
## Prism
A prism is a 3D figure
that has
a pair of polygon [bases] (brown)
and [lateral faces]
which are adjacent to both bases.
The [bases] are congruent and parallel.
The [lateral faces] are the faces
that are not the bases.
## Right Prism
A right prism is a prism
whose lateral faces are all rectangles
and are perpendicular to the bases.
## Formula
A = 2B + Ph
A: Surface area of a right prism
B: Base area
P: Perimeter of the base
h: height of the prism
2B: Sum of the base areas
Ph: Lateral area
## Example 1
Set the bottom face as the base.
The base is a rectangle.
Its sides are 7 and 5.
So B = 35.
Area of a rectangle
The sides of the base are 7 and 5.
And there are two pairs of 7 and 5.
So the perimeter of the base is
P = (7 + 5)⋅2.
7 + 5 = 12
12⋅2 = 24
So P = 24.
It says
the given figure is a right prism.
So h = 8.
B = 35
P = 24
h = 8
So A = 2⋅35 + 24⋅8.
2⋅35 = 70
24⋅8 = 192
70 + 192 = 262
So A = 262.
## Example 2
Set the front face as the base.
The base is an equilateral triangle.
Its sides are all 4.
So B = (√3/4)⋅42.
Area of an equilateral triangle
Cancel the denominator 4
and reduce 42 to 4.
Then B = 4√3.
The sides of the base are all 4.
And there are three congruent sides.
So the perimeter of the base is
P = 4⋅3.
4⋅3 = 12
So P = 12.
It says
the given figure is a right prism.
So h = 7.
B = 4√3
P = 12
h = 7
So A = 2⋅4√3 + 12⋅7.
2⋅4√3 = 8√3
12⋅7 = 84
Arrange the terms. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
10.4: Area of Circles
Difficulty Level: At Grade Created by: CK-12
Introduction
The Quilting Survey
Jillian loves quilting. At first, she thought that she would love it because it is something that she could do with her grandmother, but now she is sure that she actually really loves the quilting itself. Jillian loves creating something with her hands and seeing the finished project.
Jillian asked her grandmother how long she had been quilting and her grandmother told her that she had been quilting for a long time, long before Jillian was even born. Jillian began to wonder how many other people in the world quilt.
During a trip to the library, Jillian used the computer to do some research. She found that the number of people quilting in the United States has increased significantly from 2006 to 2009. Quilters.com completed a survey and here are their results.
In 2006, 21.3 million people were quilting. That means that about 13% of all Americans were making quilts.
In 2009, 27 million people were quilting. That means that about 17% of all Americans were quilting.
That is an increase of 4% in just three years! It might not seem like much, but it is a significant increase!
Jillian wants to show her grandmother the results of the survey. She has decided to create a picture of the data to show how the information has changed. To do this, she is going to create a circle graph.
She will create two circle graphs, one for 2006 and one for 2009.
Do you know how to do this? During this lesson you will continue learning about circles. At the end of the lesson, you will see how Jillian used a circle to display the quilting data.
What You Will Learn
• Find areas of circles given radius or diameter.
• Find radius or diameter of circles given area.
• Find areas of combined figures involving parts of circles.
• Display real-world data using circle graphs.
Teaching Time
I. Find Areas of Circles Given Radius or Diameter
In the last Concept you learned about the parts of a circle and about finding the circumference of a circle. This Concept is going to focus on the inside of the circle. The inside of a circle is called the area of the circle.
What is the area of a circle?
Remember back to working with quadrilaterals? The area of the quadrilateral is the surface or space inside the quadrilateral. Well, the area of a circle is the same thing. It is the area inside the circle that we are measuring.
In this picture, the area of the circle is yellow.
How do you measure the area of a circle?
To figure out the area of a circle, we are going to need a couple of different measurements. The first one is pi. We will need to use the numerical value for pi, or 3.14, to represent the ratio between the diameter and the circumference.
The next measurement we need to use is the radius. Remember that the radius is the distance from the center of a circle to the edge, or is 1/2 of the circle's diameter.
We calculate the area of a circle by multiplying the radius squared (multiplied by itself) by pi (3.14).
Here is the formula.
\begin{align*}A = \pi r^2\end{align*}
Here is an example.
Example
Find the area of the circle.
Next, we use our formula and the given information.
\begin{align*}A & = \pi r^2\\ A & = (3.14)(3^2)\\ A & = (3.14)(9)\\ A & = 28.26 \ mm^2\end{align*}
What about if you have been given the diameter and not the radius?
If you have been given the diameter and not the radius, you can still figure out the area of the circle. You start by dividing the measure of the diameter in half since the radius is one-half the measure of the diameter.
Then you use the formula and solve for area.
Let’s look at an example.
Example
Notice that the diameter is 6 inches. We can divide this in half to find the radius.
6 \begin{align*}\div\end{align*} 2 \begin{align*}=\end{align*} 3 inches \begin{align*}=\end{align*} radius
Next, we substitute the given values into the formula and solve for the area of the circle.
\begin{align*}A & = \pi r^2\\ A & = (3.14)(3^2)\\ A & = (3.14)(9)\\ A & = 28.26 \ in^2\end{align*}
Now it’s time for you to try a few on your own. Find the area of the circle using the given radius or diameter.
1. \begin{align*}r = 4 \ cm\end{align*}
2. \begin{align*}r = 8 \ cm\end{align*}
3. \begin{align*}d = 4 \ in\end{align*}
Take a few minutes to check your work. Did you remember to label your answer using square inches or centimeters?
II. Find the Radius or Diameter of Circles Given Area
Now that you know how to find the area of a circle given a radius or diameter, we can work backwards and use the area to find the radius or the diameter.
It is time to use your detective skills again!!
Now, let’s look at an example.
Example
The area of the circle is \begin{align*}153.86 \ in^2\end{align*}, what is the radius? What is the diameter?
This problem asks for you to figure out two different things. First, let’s find the radius and then we can use that measure to figure out the diameter.
Let’s begin by using the formula for finding the area of a circle.
\begin{align*}A & = \pi r^ 2\\ 153.86 & = (3.14)r^2\end{align*}
We substituted in the given information. We know the area, and we know that the measure for pi is 3.14. Next, we can divide the area by pi. This will help to get us one step closer to figuring out the radius.
\begin{align*}{3.14 \overline{ ) {153.86}}}\end{align*}
Remember, when you divide decimals, we move the decimal two places in the divisor and the dividend. Here is our new division problem.
\begin{align*}{314 \overline{ ) {15386}}}\end{align*}
Yes. It is a large number to divide, but don’t let that stop you. Just work through it step by step and you will be able to find the correct answer!
\begin{align*}& \overset{\qquad \ \quad 49}{314 \overline{ ) {15386}}}\\ & \quad \underline{-1256 \ \ }\\ & \qquad \ 2826\\ & \quad \ \ \underline{-2826}\\ & \qquad \qquad 0\end{align*}
So far, our answer is 49, but that is not the radius.
\begin{align*}49 = r^2\end{align*}
We need to figure out which number times itself is equal to 49.
Now we know that the radius is 7 because 7 \begin{align*}\times\end{align*} 7 \begin{align*}=\end{align*} 49.
What is the diameter?
The measure of the radius is one-half the measure of the diameter. If the radius is 7, then the diameter is double that.
7 \begin{align*}\times\end{align*} 2 \begin{align*}=\end{align*} 14
The diameter is 14 inches.
Warning! Working backwards is tricky! Be sure that you take your time when working through problems!
Try a few of these on your own.
1. If the area of a circle is \begin{align*}12.56 \ cm^2\end{align*}. What is the radius? What is the diameter?
2. If the area of a circle is \begin{align*}113.04 \ m^2\end{align*}. What is the radius? What is the diameter?
III. Find Areas of Combined Figures Involving Parts of Circles
Sometimes, there will be figures that aren’t quadrilaterals and they aren’t circles either, they are combined figures. A combined figure is a figure that is made up of more than one type of polygon. You can still figure out the area of combined figures, but you will have to think about how to do it!!
Let’s look at an example.
Example
What is the area of the figure?
To solve this problem, you first have to look at which figures have been combined. Here you have one-half of a circle and a rectangle.
We will need to figure out the area of the rectangle, the area of half of the circle and then add the two areas.
This will give us the area of the combined figure.
\begin{align*}A = lw\end{align*}
The length of the rectangle is 6 inches. The width of the rectangle is 3 inches.
\begin{align*}A & = (6)(3)\\ A & = 18 \ in^2\end{align*}
Next, we find the area of the circle. We can start by noticing that the length of the rectangle is also the diameter of the circle. The diameter of the circle is 6 inches. We can start by figuring out the area of one whole circle and then divide that in half for the area of half of the circle.
If the diameter of the circle is 6 inches, then the radius is 3 inches. Remember that the radius is one-half of the diameter.
\begin{align*}A & = \pi r^2\\ A & = (3.14)(3^2)\\ A & = 28.26 \ in^2\end{align*}
This is the area of the whole circle. Our figure only has half of a circle, so we divide this in half.
28.26 \begin{align*}\div\end{align*} 2 \begin{align*}=\end{align*} 14.13 in
Now we combine the area of the rectangle with the area of the half circle. This will equal the area of the entire figure.
18 + 14.13 = 32.13
The area of the figure is \begin{align*}32.13 \ in^2\end{align*}.
Try one of these on your own. Remember, separate the figure and find the area of the parts, then combine the areas.
Take a few minutes to check your work with a friend, then move on to the next section.
IV. Display Real-World Data Using Circle Graphs
In our introduction problem, Jillian wants to display her quilting data in a circle graph. We can use circle graphs to display real-world data. In fact, we do it all the time.
What is a circle graph?
A circle graph is a visual way to display data using circles and parts of a circle.
A circle graph uses a circle to indicate 100%. The entire circle represents 100% and each section of a circle represents some part out of 100.
You can see here that this circle graph is divided into five sections. Each section represents a part of a whole.
Back in Chapter Two, we looked at the spending habits of a teenager. Here you can see that 50% of his money went into savings. 40% of his money was spent on food and that 10% of his money went to baseball cards.
When Jillian creates her circle graphs, she will be able to create ones that show how quilting has grown from 2006 to 2009. Let’s go and revisit that introductory problem now.
Real Life Example Completed
The Quilting Survey
Here is the original problem once again. Reread it and underline any important information.
Jillian loves quilting. At first, she thought that she would love it because it is something that she could do with her grandmother, but now she is sure that she actually really loves the quilting itself. Jillian loves creating something with her hands and seeing the finished project.
Jillian asked her grandmother how long she had been quilting and her grandmother told her that she had been quilting for a long time, long before Jillian was even born. Jillian began to wonder how many other people in the world quilt.
During a trip to the library, Jillian used the computer to do some research. She found that the number of people quilting in the United States has increased significantly from 2006 to 2009. Quilters.com completed a survey and here are their results.
In 2006, 21.3 million people were quilting. That means that about 13% of all Americans were making quilts.
In 2009, 27 million people were quilting. That means that about 17% of all Americans were quilting.
That is an increase of 4% in just three years! It might not seem like much, but it is a significant increase!
Jillian wants to show her grandmother the results of the survey. She has decided to create a picture of the data to show how the information has changed. To do this, she is going to create a circle graph.
She will create two circle graphs, one for 2006 and one for 2009.
Let’s look at Jillian’s data.
The first circle graph will show that 13% out of 100% of people were quilting in 2006. Here is the circle graph.
The second circle graph shows that in 2009, the number of people quilting increased to 17% out of 100%.
Now Jillian has two circle graphs that she can share with her grandmother.
Vocabulary
Here are the vocabulary words that are found in this lesson.
Area
the surface or space of the figure inside the perimeter.
the measure of the distance halfway across a circle.
Diameter
the measure of the distance across a circle
Squaring
uses the exponent 2 to show that a number is being multiplied by itself. \begin{align*}3^2 = 3 \times 3\end{align*}
Pi
the ratio of the diameter to the circumference. The numerical value of pi is 3.14.
Technology Integration
Other Videos:
1. http://www.mathplayground.com/mv_area_circles.html – This is a basic blackboard video on finding the area of a circle by Brightstorm.
Time to Practice
Directions: Find the area of the following circles given the radius.
1. \begin{align*}r = 4 \ in\end{align*}
2. \begin{align*}r = 5 \ cm\end{align*}
3. \begin{align*}r = 8 \ in\end{align*}
4. \begin{align*}r = 2 \ cm\end{align*}
5. \begin{align*}r = 7 \ m\end{align*}
6. \begin{align*}r = 9 \ in\end{align*}
7. \begin{align*}r = 10 \ ft\end{align*}
8. \begin{align*}r = 11 \ cm\end{align*}
9. \begin{align*}r = 20 \ ft\end{align*}
10. \begin{align*}r = 30 \ miles\end{align*}
Directions: Find the area of the following circles given the diameter.
11. \begin{align*}d = 10 \ in\end{align*}
12. \begin{align*}d = 12 \ m\end{align*}
13. \begin{align*}d = 14 \ cm\end{align*}
14. \begin{align*}d = 16 \ ft\end{align*}
15. \begin{align*}d = 18 \ in\end{align*}
16. \begin{align*}d = 22 \ ft\end{align*}
17. \begin{align*}d = 24 \ cm\end{align*}
18. \begin{align*}d = 28 \ m\end{align*}
19. \begin{align*}d = 30 \ m\end{align*}
20. \begin{align*}d = 36 \ ft\end{align*}
Directions: Find the area of each of the following.
21.
22.
23.
24.
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## Types of Graphs-
Before you go through this article, make sure that you have gone through the previous article on various Types of Graphs in Graph Theory.
We have discussed-
• A graph is a collection of vertices connected to each other through a set of edges.
• The study of graphs is known as Graph Theory.
## Euler Graph-
An Euler graph may be defined as-
Any connected graph is called as an Euler Graph if and only if all its vertices are of even degree.ORAn Euler Graph is a connected graph that contains an Euler Circuit.
### Euler Graph Example-
The following graph is an example of an Euler graph-
Here,
• This graph is a connected graph and all its vertices are of even degree.
• Therefore, it is an Euler graph.
Alternatively, the above graph contains an Euler circuit BACEDCB, so it is an Euler graph.
## Euler Path-
Euler path is also known as Euler Trail or Euler Walk.
• If there exists a Trail in the connected graph that contains all the edges of the graph, then that trail is called as an Euler trail.
OR
• If there exists a walk in the connected graph that visits every edge of the graph exactly once with or without repeating the vertices, then such a walk is called as an Euler walk.
### NOTE
A graph will contain an Euler path if and only if it contains at most two vertices of odd degree.
### Euler Path Examples-
Examples of Euler path are as follows-
## Euler Circuit-
Euler circuit is also known as Euler Cycle or Euler Tour.
• If there exists a Circuit in the connected graph that contains all the edges of the graph, then that circuit is called as an Euler circuit.
OR
• If there exists a walk in the connected graph that starts and ends at the same vertex and visits every edge of the graph exactly once with or without repeating the vertices, then such a walk is called as an Euler circuit.
OR
• An Euler trail that starts and ends at the same vertex is called as an Euler circuit.
OR
• A closed Euler trail is called as an Euler circuit.
### NOTE
A graph will contain an Euler circuit if and only if all its vertices are of even degree.
### Euler Circuit Examples-
Examples of Euler circuit are as follows-
## Semi-Euler Graph-
If a connected graph contains an Euler trail but does not contain an Euler circuit, then such a graph is called as a semi-Euler graph.
Thus, for a graph to be a semi-Euler graph, following two conditions must be satisfied-
• Graph must be connected.
• Graph must contain an Euler trail.
### Example-
Here,
• This graph contains an Euler trail BCDBAD.
• But it does not contain an Euler circuit.
• Therefore, it is a semi-Euler graph.
## Important Notes-
### Note-01:
To check whether any graph is an Euler graph or not, any one of the following two ways may be used-
• If the graph is connected and contains an Euler circuit, then it is an Euler graph.
• If all the vertices of the graph are of even degree, then it is an Euler graph.
### Note-02:
To check whether any graph contains an Euler circuit or not,
• Just make sure that all its vertices are of even degree.
• If all its vertices are of even degree, then graph contains an Euler circuit otherwise not.
### Note-03:
To check whether any graph is a semi-Euler graph or not,
• Just make sure that it is connected and contains an Euler trail.
• If the graph is connected and contains an Euler trail, then graph is a semi-Euler graph otherwise not.
### Note-04:
To check whether any graph contains an Euler trail or not,
• Just make sure that the number of vertices in the graph with odd degree are not more than 2.
• If the number of vertices with odd degree are at most 2, then graph contains an Euler trail otherwise not.
### Note-05:
• A graph will definitely contain an Euler trail if it contains an Euler circuit.
• A graph may or may not contain an Euler circuit if it contains an Euler trail.
### Note-06:
• An Euler graph is definitely be a semi-Euler graph.
• But a semi-Euler graph may or may not be an Euler graph.
## Problems-
Which of the following is / are Euler Graphs?
## Solutions-
If all the vertices of a graph are of even degree, then graph is an Euler Graph otherwise not.
Using the above rule, we have-
A) It is an Euler graph.
B) It is not an Euler graph.
C) It is not an Euler graph.
D) It is not an Euler graph.
E) It is an Euler graph.
F) It is not an Euler graph.
To gain better understanding about Euler Graphs in Graph Theory,
Watch this Video Lecture
Next Article- Hamiltonian Graph
Get more notes and other study material of Graph Theory.
Watch video lectures by visiting our YouTube channel LearnVidFun.
## Chromatic Number-
Before you go through this article, make sure that you have gone through the previous article on Chromatic Number.
We gave discussed-
• Graph Coloring is a process of assigning colors to the vertices of a graph.
• It ensures that no two adjacent vertices of the graph are colored with the same color.
• Chromatic Number is the minimum number of colors required to properly color any graph.
In this article, we will discuss how to find Chromatic Number of any graph.
## Graph Coloring Algorithm-
• There exists no efficient algorithm for coloring a graph with minimum number of colors.
• Graph Coloring is a NP complete problem.
However, a following greedy algorithm is known for finding the chromatic number of any given graph.
## Greedy Algorithm-
### Step-01:
Color first vertex with the first color.
### Step-02:
Now, consider the remaining (V-1) vertices one by one and do the following-
• Color the currently picked vertex with the lowest numbered color if it has not been used to color any of its adjacent vertices.
• If it has been used, then choose the next least numbered color.
• If all the previously used colors have been used, then assign a new color to the currently picked vertex.
## Drawbacks of Greedy Algorithm-
There are following drawbacks of the above Greedy Algorithm-
• The above algorithm does not always use minimum number of colors.
• The number of colors used sometimes depend on the order in which the vertices are processed.
Also Read- Types of Graphs in Graph Theory
## Problem-01:
Find chromatic number of the following graph-
## Solution-
Applying Greedy Algorithm, we have-
Vertex a b c d e f Color C1 C2 C1 C2 C1 C2
From here,
• Minimum number of colors used to color the given graph are 2.
• Therefore, Chromatic Number of the given graph = 2.
The given graph may be properly colored using 2 colors as shown below-
## Problem-02:
Find chromatic number of the following graph-
## Solution-
Applying Greedy Algorithm, we have-
Vertex a b c d e f Color C1 C2 C2 C3 C3 C1
From here,
• Minimum number of colors used to color the given graph are 3.
• Therefore, Chromatic Number of the given graph = 3.
The given graph may be properly colored using 3 colors as shown below-
## Problem-03:
Find chromatic number of the following graph-
## Solution-
Applying Greedy Algorithm, we have-
Vertex a b c d e f g Color C1 C2 C1 C3 C2 C3 C4
From here,
• Minimum number of colors used to color the given graph are 4.
• Therefore, Chromatic Number of the given graph = 4.
The given graph may be properly colored using 4 colors as shown below-
## Problem-04:
Find chromatic number of the following graph-
## Solution-
Applying Greedy Algorithm, we have-
Vertex a b c d e f Color C1 C2 C3 C1 C2 C3
From here,
• Minimum number of colors used to color the given graph are 3.
• Therefore, Chromatic Number of the given graph = 3.
The given graph may be properly colored using 3 colors as shown below-
## Problem-05:
Find chromatic number of the following graph-
## Solution-
Applying Greedy Algorithm,
• Minimum number of colors required to color the given graph are 3.
• Therefore, Chromatic Number of the given graph = 3.
The given graph may be properly colored using 3 colors as shown below-
To gain better understanding about How to Find Chromatic Number,
Watch this Video Lecture
Get more notes and other study material of Graph Theory.
Watch video lectures by visiting our YouTube channel LearnVidFun.
## Graph Coloring-
Graph Coloring is a process of assigning colors to the vertices of a graphsuch that no two adjacent vertices of it are assigned the same color.
• Graph Coloring is also called as Vertex Coloring.
• It ensures that there exists no edge in the graph whose end vertices are colored with the same color.
• Such a graph is called as a Properly colored graph.
### Graph Coloring Example-
The following graph is an example of a properly colored graph-
In this graph,
• No two adjacent vertices are colored with the same color.
• Therefore, it is a properly colored graph.
### Graph Coloring Applications-
Some important applications of graph coloring are as follows-
• Map Coloring
• Preparing Time Table
• Assignment
• Conflict Resolution
• Sudoku
## Chromatic Number-
Chromatic Number is the minimum number of colors required to properly color any graph.ORChromatic Number is the minimum number of colors required to color any graphsuch that no two adjacent vertices of it are assigned the same color.
### Chromatic Number Example-
Consider the following graph-
In this graph,
• No two adjacent vertices are colored with the same color.
• Minimum number of colors required to properly color the vertices = 3.
• Therefore, Chromatic number of this graph = 3.
• We can not properly color this graph with less than 3 colors.
Also Read- Types of Graphs in Graph Theory
## Chromatic Number Of Graphs-
Chromatic Number of some common types of graphs are as follows-
## 1. Cycle Graph-
• A simple graph of ‘n’ vertices (n>=3) and ‘n’ edges forming a cycle of length ‘n’ is called as a cycle graph.
• In a cycle graph, all the vertices are of degree 2.
### Chromatic Number
• If number of vertices in cycle graph is even, then its chromatic number = 2.
• If number of vertices in cycle graph is odd, then its chromatic number = 3.
## 2. Planar Graphs-
A Planar Graph is a graph that can be drawn in a plane such that none of its edges cross each other.
### Chromatic Number
Chromatic Number of any Planar Graph
= Less than or equal to 4
### Examples-
• All the above cycle graphs are also planar graphs.
• Chromatic number of each graph is less than or equal to 4.
## 3. Complete Graphs-
• A complete graph is a graph in which every two distinct vertices are joined by exactly one edge.
• In a complete graph, each vertex is connected with every other vertex.
• So to properly it, as many different colors are needed as there are number of vertices in the given graph.
### Chromatic Number
Chromatic Number of any Complete Graph
= Number of vertices in that Complete Graph
## 4. Bipartite Graphs-
• A Bipartite Graph consists of two sets of vertices X and Y.
• The edges only join vertices in X to vertices in Y, not vertices within a set.
### Chromatic Number
Chromatic Number of any Bipartite Graph
= 2
## 5. Trees-
• A Tree is a special type of connected graph in which there are no circuits.
• Every tree is a bipartite graph.
• So, chromatic number of a tree with any number of vertices = 2.
### Chromatic Number
Chromatic Number of any tree
= 2
### Examples-
To gain better understanding about Graph Coloring & Chromatic Number,
Watch this Video Lecture
Next Article- Graph Coloring Algorithm
Get more notes and other study material of Graph Theory.
Watch video lectures by visiting our YouTube channel LearnVidFun.
## Types of Graphs-
Before you go through this article, make sure that you have gone through the previous article on various Types of Graphs in Graph Theory.
We have discussed-
• A graph is a collection of vertices connected to each other through a set of edges.
• The study of graphs is known as Graph Theory.
## Planar Graph-
A planar graph may be defined as-
In graph theory,Planar graph is a graph that can be drawn in a plane such that none of its edges cross each other.
### Planar Graph Example-
The following graph is an example of a planar graph-
Here,
• In this graph, no two edges cross each other.
• Therefore, it is a planar graph.
## Regions of Plane-
The planar representation of the graph splits the plane into connected areas called as Regions of the plane.
Each region has some degree associated with it given as-
• Degree of Interior region = Number of edges enclosing that region
• Degree of Exterior region = Number of edges exposed to that region
### Example-
Consider the following planar graph-
Here, this planar graph splits the plane into 4 regions- R1, R2, R3 and R4 where-
• Degree (R1) = 3
• Degree (R2) = 3
• Degree (R3) = 3
• Degree (R4) = 5
## Planar Graph Chromatic Number-
• Chromatic Number of any planar graph is always less than or equal to 4.
• Thus, any planar graph always requires maximum 4 colors for coloring its vertices.
## Planar Graph Properties-
### Property-01:
In any planar graph, Sum of degrees of all the vertices = 2 x Total number of edges in the graph
### Property-02:
In any planar graph, Sum of degrees of all the regions = 2 x Total number of edges in the graph
### Case-01:
In any planar graph, if degree of each region is K, then-
K x |R| = 2 x |E|
### Case-02:
In any planar graph, if degree of each region is at least K (>=K), then-
K x |R| <= 2 x |E|
### Case-03:
In any planar graph, if degree of each region is at most K (<=K), then-
K x |R| >= 2 x |E|
### Property-03:
If G is a connected planar simple graph with ‘e’ edges, ‘v’ vertices and ‘r’ number of regions in the planar representation of G, then-
r = e – v + 2
This is known as Euler’s Formula.
It remains same in all the planar representations of the graph.
### Property-04:
If G is a planar graph with k components, then-
r = e – v + (k + 1)
## Problem-01:
Let G be a connected planar simple graph with 25 vertices and 60 edges. Find the number of regions in G.
## Solution-
Given-
• Number of vertices (v) = 25
• Number of edges (e) = 60
By Euler’s formula, we know r = e – v + 2.
Substituting the values, we get-
Number of regions (r)
= 60 – 25 + 2
= 37
Thus, Total number of regions in G = 37.
## Problem-02:
Let G be a planar graph with 10 vertices, 3 components and 9 edges. Find the number of regions in G.
## Solution-
Given-
• Number of vertices (v) = 10
• Number of edges (e) = 9
• Number of components (k) = 3
By Euler’s formula, we know r = e – v + (k+1).
Substituting the values, we get-
Number of regions (r)
= 9 – 10 + (3+1)
= -1 + 4
= 3
Thus, Total number of regions in G = 3.
## Problem-03:
Let G be a connected planar simple graph with 20 vertices and degree of each vertex is 3. Find the number of regions in G.
## Solution-
Given-
• Number of vertices (v) = 20
• Degree of each vertex (d) = 3
### Calculating Total Number Of Edges (e)-
By sum of degrees of vertices theorem, we have-
Sum of degrees of all the vertices = 2 x Total number of edges
Number of vertices x Degree of each vertex = 2 x Total number of edges
20 x 3 = 2 x e
∴ e = 30
Thus, Total number of edges in G = 30.
### Calculating Total Number Of Regions (r)-
By Euler’s formula, we know r = e – v + 2.
Substituting the values, we get-
Number of regions (r)
= 30 – 20 + 2
= 12
Thus, Total number of regions in G = 12.
## Problem-04:
Let G be a connected planar simple graph with 35 regions, degree of each region is 6. Find the number of vertices in G.
## Solution-
Given-
• Number of regions (n) = 35
• Degree of each region (d) = 6
### Calculating Total Number Of Edges (e)-
By sum of degrees of regions theorem, we have-
Sum of degrees of all the regions = 2 x Total number of edges
Number of regions x Degree of each region = 2 x Total number of edges
35 x 6 = 2 x e
∴ e = 105
Thus, Total number of edges in G = 105.
### Calculating Total Number Of Vertices (v)-
By Euler’s formula, we know r = e – v + 2.
Substituting the values, we get-
35 = 105 – v + 2
∴ v = 72
Thus, Total number of vertices in G = 72.
## Problem-05:
Let G be a connected planar graph with 12 vertices, 30 edges and degree of each region is k. Find the value of k.
## Solution-
Given-
• Number of vertices (v) = 12
• Number of edges (e) = 30
• Degree of each region (d) = k
### Calculating Total Number Of Regions (r)-
By Euler’s formula, we know r = e – v + 2.
Substituting the values, we get-
Number of regions (r)
= 30 – 12 + 2
= 20
Thus, Total number of regions in G = 20.
### Calculating Value Of k-
By sum of degrees of regions theorem, we have-
Sum of degrees of all the regions = 2 x Total number of edges
Number of regions x Degree of each region = 2 x Total number of edges
20 x k = 2 x 30
∴ k = 3
Thus, Degree of each region in G = 3.
## Problem-06:
What is the maximum number of regions possible in a simple planar graph with 10 edges?
## Solution-
In a simple planar graph, degree of each region is >= 3.
So, we have 3 x |R| <= 2 x |E|.
Substituting the value |E| = 10, we get-
3 x |R| <= 2 x 10
|R| <= 6.67
|R| <= 6
Thus, Maximum number of regions in G = 6.
## Problem-07:
What is the minimum number of edges necessary in a simple planar graph with 15 regions?
## Solution-
In a simple planar graph, degree of each region is >= 3.
So, we have 3 x |R| <= 2 x |E|.
Substituting the value |R| = 15, we get-
3 x 15 <= 2 x |E|
|E| >= 22.5
|E| >= 23
Thus, Minimum number of edges required in G = 23.
To gain better understanding about Planar Graphs in Graph Theory,
Watch this Video Lecture
Next Article- Euler Graph
Get more notes and other study material of Graph Theory.
Watch video lectures by visiting our YouTube channel LearnVidFun.
## Walk in Graph Theory-
In graph theory,
• A walk is defined as a finite length alternating sequence of vertices and edges.
• The total number of edges covered in a walk is called as Length of the Walk.
### Walk in Graph Theory Example-
Consider the following graph-
In this graph, few examples of walk are-
• a , b , c , e , d (Length = 4)
• d , b , a , c , e , d , e , c (Length = 7)
• e , c , b , a , c , e , d (Length = 6)
## Open Walk in Graph Theory-
In graph theory, a walk is called as an Open walk if-
• Length of the walk is greater than zero
• And the vertices at which the walk starts and ends are different.
## Closed Walk in Graph Theory-
In graph theory, a walk is called as a Closed walk if-
• Length of the walk is greater than zero
• And the vertices at which the walk starts and ends are same.
### NOTE
It is important to note the following points-
• If length of the walk = 0, then it is called as a Trivial Walk.
• Both vertices and edges can repeat in a walk whether it is an open walk or a closed walk.
## Path in Graph Theory-
In graph theory, a path is defined as an open walk in which-
• Neither vertices (except possibly the starting and ending vertices) are allowed to repeat.
• Nor edges are allowed to repeat.
## Cycle in Graph Theory-
In graph theory, a cycle is defined as a closed walk in which-
• Neither vertices (except possibly the starting and ending vertices) are allowed to repeat.
• Nor edges are allowed to repeat.
OR
In graph theory, a closed path is called as a cycle.
## Trail in Graph Theory-
In graph theory, a trail is defined as an open walk in which-
• Vertices may repeat.
• But edges are not allowed to repeat.
## Circuit in Graph Theory-
In graph theory, a circuit is defined as a closed walk in which-
• Vertices may repeat.
• But edges are not allowed to repeat.
OR
In graph theory, a closed trail is called as a circuit.
### NOTE
It is important to note the following points-
• Every path is a trail but every trail need not be a path.
• Every cycle is a circuit but every circuit need not be a cycle.
• For directed graphs, we put term “directed” in front of all the terms defined above.
## Important Chart-
The following chart summarizes the above definitions and is helpful in remembering them-
Also Read- Types of Graphs in Graph Theory
## Problem-01:
Consider the following graph-
Decide which of the following sequences of vertices determine walks.
For those that are walks, decide whether it is a circuit, a path, a cycle or a trail.
1. a , b , g , f , c , b
2. b , g , f , c , b , g , a
3. c , e , f , c
4. c , e , f , c , e
5. a , b , f , a
6. f , d , e , c , b
1. Trail
2. Walk
3. Cycle
4. Walk
5. Not a walk
6. Path
## Problem-02:
Consider the following graph-
Consider the following sequences of vertices and answer the questions that follow-
1. x , v , y , w , v
2. x , u , x , u , x
3. x , u , v , y , x
4. x , v , y , w , v , u , x
1. Which of the above given sequences are directed walks?
2. What are the lengths of directed walks?
3. Which directed walks are also directed paths?
4. Which directed walks are also directed cycles?
## Solution-
### Part-01:
• Only (A) and (B) are the directed walks.
• (C) is not a directed walk since there exists no arc from vertex u to vertex v.
• (D) is not a directed walk since there exists no arc from vertex v to vertex u.
### Part-02:
Both the directed walks (A) and (B) have length = 4.
### Part-03:
• Neither (A) nor (B) are directed paths.
• This is because vertices repeat in both of them.
• Vertex v repeats in Walk (A) and vertex u repeats in walk (B).
### Part-04:
• Neither of them are directed cycles.
• Walk (A) does not represent a directed cycle because its starting and ending vertices are not same.
• Walk (B) does not represent a directed cycle because it repeats vertices/edges.
## Problem-03:
Consider the following graph-
Observe the given sequences and predict the nature of walk in each case-
1. v1e1v2e2v3e2v2
2. v4e7v1e1v2e2v3e3v4e4v5
3. v1e1v2e2v3e3v4e4v5
4. v1e1v2e2v3e3v4e7v1
5. v6e5v5e4v4e3v3e2v2e1v1e7v4e6v6
## Solution-
1. Open walk
2. Trail (Not a path because vertex v4 is repeated)
3. Path
4. Cycle
5. Circuit (Not a cycle because vertex v4 is repeated)
To gain better understanding about Walk in Graph Theory,
Watch this Video Lecture
Next Article- Graph Coloring
Get more notes and other study material of Graph Theory.
Watch video lectures by visiting our YouTube channel LearnVidFun. |
# Wilson's theorem, infinite monkey's Theorem
Source: Internet
Author: User
Wilson's theorem, infinite monkey's Theorem
Wilson's theorem proves that if p is a prime number, p can be divisible by (p-1 )! + 1
First: First test the strange Value
If p = 2, true;
If p = 3, true;
Then we will start to study the situation where p> = 5
Second: let's prove several conclusions first.
Set A = {2, 3, 4 ,......, P-2}
Note B = {a, 2a, 3a ,......, (P-1)}
[There will be no number of modulo p with the same remainder in B .......................................... Conclusion 1
Proof: Set b1a, b2a, B, b2, b1, [1, P-1] to different numbers.
Assume that b1a has b2a (mod p)
Then | b1-b2 | a limit 0 (mod p)
However | b1-b2 | ε [1, P-2]
So | b1-b2 | a, B
However, it is clear that no number in B can be divisible by p, so the question cannot be set.
So Conclusion 1 proves
Because B has a certain number of elements, the remainder of non-p-multiples modulo p is also only a certain number of elements. Therefore, the remainder of modulo p in B forms a set of C = {1, 2, 3 ,..., P-1 }]
In B, the number of the remainder 1 divided by p is ba, B, A, and B! = ....................... Conclusion 2
Proof: If B = 1, ba = a, because a is A, so! = 1, not true
If B is P-1, p-a (mod p) is a p-a (p). Because A is a, p-a is [2, P-2], not true.
If B = a, then a2 contains 1 (mod p), so (a + 1) (A-1) + mod p), Because p is a prime number, so (a + 1) and (A-1) have and only one is a multiple of p. because a in A, so (a + 1) in [P-1], (A-1) in [1, P-3], so they can not be a multiple of p
Conclusion 2: Evidence]
[Conclusion 2: If a is different, B is different .............................. Conclusion 3
Proof: Set a1! = A2, but all belong to A, and ba1 ba2 has 1 (mod p)
Yi Zhi ba1, ba2 ε B, but according to conclusion 1, it is not true
Conclusion 3: Evidence]
Third: If we have proved so many conclusions, this theorem has basically surfaced.
Conclusion 1 tells us that in B, each a has a unique bits, [1, P-1], so that AB branch 1 (mod p)
Conclusion 3 sublimation Conclusion 1: The range of B is reduced to B, [2, P-2], and a = B. that is to say, in [2, P-2], each number can find a unique inverse element (Note: p is obviously an even number, therefore, all integers in this range can be divided into a pair of children), and the reverse element belongs to [2, P-2].
P-1 )! + 1 second (p-1) + 1 second 0 (mod p)
Related Keywords:
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# How do you find the derivative of y = x^(cos x)?
##### 1 Answer
Feb 18, 2017
$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{\cos} x \left(\frac{\cos x}{x} - \ln x \sin x\right)$
#### Explanation:
Take the natural log of both sides and move the cos in front of the natural log.
$\ln y = \left(\cos x\right) \ln x$
Use implicit differentiation. You have to take the product rule of the right side.
$\left(\frac{1}{y}\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \left(\cos x\right) \left(\frac{1}{x}\right) + \left(\ln x\right) \left(- \sin x\right)$
$\left(\frac{1}{y}\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{\cos x}{x} - \ln x \sin x$
Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left(\frac{\cos x}{x} - \ln x \sin x\right)$
Plug in ${x}^{\cos} x$ for y
$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{\cos} x \left(\frac{\cos x}{x} - \ln x \sin x\right)$ |
##### Geometry: 1,001 Practice Problems For Dummies (+ Free Online Practice)
In geometry, when you rotate an image, the sign of the degree of rotation tells you the direction in which the image is rotating. A positive degree measurement means you're rotating counterclockwise, whereas a negative degree measurement means you're rotating clockwise.
The following practice questions test your knowledge of rotations by asking you to rotate an octagon, and then to rotate coordinates of an image about the origin.
## Practice questions
1. The figure shows a regular octagon with l, m, and p as lines of symmetry.
Find the image of the given point after the following rotation:
2. Use your knowledge of rotations to answer this question: What are the coordinates of the image of (4, –6) after a clockwise rotation of 90 degrees about the origin?
1. O
When doing the composition of transformations, you perform the transformation closest to the point first. Therefore, you perform the transformations from right to left.
Rotate Point A counterclockwise 405 degrees first. Rotating 405 degrees means you're rotating more than a full revolution:
This means you're really just rotating the point 45 degrees counterclockwise:
Take the new point and reflect it over line l:
Take the new point and rotate it 135 degrees clockwise:
2. (–6, –4)
All the rules for rotations are written so that when you're rotating counterclockwise, a full revolution is 360 degrees. Rotating 90 degrees clockwise is the same as rotating 270 degrees counterclockwise. Rotating 270 degrees counterclockwise about the origin is the same as reflecting over the line y = x and then reflecting over the x-axis. This means that the point (x, y) will become the point (y, –x). In this question, (4, –6) will become (–6, –4). |
# Equation solving
In mathematics, equation solving is the problem of finding what values (numbers, functions, sets...) fulfill a condition stated as an equality (an equation). Usually, this condition involves expressions with variables (or unknowns), which are to be substituted by values in order for the equality to hold. More precisely, an equation involves some free variables.
In one general case, we have a situation such as
f(x0,...,xn)=c, c constant
which has a set of solutions S in the form
{(a0,...,an)∈Tn|f(a0,...,an)=c}
with T the domain of the function. Note that the set of solutions can be empty (there are no solutions), singleton (there is exactly 1 solution), finite (there are only n number of solutions), or infinite (there are always solutions).
For example, an expression such as
3x+2y=21z
can be solved by first modifying the equation in some way as to preserve the equality, such as subtracting both sides by 21z to obtain
3x+2y-21z=0
Now, it occurs that in solving this equation, that there is not just one solution to this equation, but a infinite set of solutions, which can be written
{(x, y, z)|3x+2y-21z=0}.
One particular solution is x = 20/3, y = 11, z = 2. In fact, this particular set of solutions describe a plane in three dimensions, which passes through the point (20/3, 11, 2).
Contents
## Solution sets
If the solution set is empty, then there are no such ai such that
f(x0,...,xn)=c
becomes true.
For example, let us examine the classic one-variable case, given a function
[itex]f : \mathbb{R}^+ \rightarrow \mathbb{R} ; x \longmapsto x^2[itex]
consider the equation
f(x) = -1
The solution set is {}, in that no positive real number solves this equation. However note that in attempting to find solutions for this equation, if we modify the function's definition - more specifically, the function's domain, we can find solutions to this equation. So, if we were instead to define
[itex]g : \mathbb{C} \rightarrow \mathbb{C} ; x \longmapsto x^2[itex]
g(x) = -1
has a solution set {i, -i}, where i is the imaginary unit. This equation has exactly two solutions.
We have already seen that certain solutions sets can describe surfaces. For example, in studying elementary mathematics, one knows that the solution set of an equation in the form ax=b with a,b real-valued constants, this forms a line in the vector space R2. However, it may not always be easy to graphically depict solutions sets - for example, the solution set to an equation in the form ax+by+cz+dw=k (with a, b, c, d, and k real-valued constants) is a hyperplane.
## Methods of solution
In simple cases, it is rather easy to solve an equation provided certain conditions are met. However, in more complicated cases, exact symbolic forms for solutions are often difficult to obtain or cumbersome to manipulate with, and an approximate numerical solution may be in fact sufficient for use.
### Inverse functions
In the simple case of a function of one variable, say, h(x), we can solve an equation of the form
h(x)=c, c constant
by considering what is known as the inverse function of h.
If h : A -> B, the inverse function, denoted h-1, defined as h-1 : B -> A is a function such that h-1(h(x)) = h(h-1(x)) = x.
Now, if we apply the inverse function to both sides of
h(x)=c, c constant
we obtain
h-1(h(x))=h-1(c)
x = h-1(c)
and we have found the solution to the equation. However, depending on the function, the inverse may be difficult to be defined, or may not be a function on all of the set B (only on some subset), and have many values at some point.
### Numerical methods
With more complicated equations, simple methods to solve equations can fail. In certain circumstances, a root-finding algorithm can be used to find a numerical solution to an equation, which within some applications can be entirely sufficient to solve some problem.
#### Taylor series
One well-studied area of mathematics involves examining whether we can create some simple function to approximate a more complex equation near a given point. In fact, polynomials in one or several variables can be used to approximate functions in this way - these are known as Taylor series.
## Solving other equations
It is important to note that we can create even more complex equations, involving differential operators, matrices, and so on. The underlying principle of solving equations by finding a value which satisfies the equation is maintained, but with vastly differing methodologies used to find them.de:Lösen von Gleichungen nl:oplossen van vergelijkingen
• Art and Cultures
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Space and Astronomy |
## Engage NY Eureka Math 1st Grade Module 1 Lesson 18 Answer Key
### Eureka Math Grade 1 Module 1 Lesson 18 Problem Set Answer Key
Question 1.
Add. Color the balloons that match the number in the boy’s mind. Find expressions that are equal. Connect them below with = to make true number sentences.
a.
b.
Question 2.
Are these number sentences true?
If it is false, rewrite the number sentence to make it true.
a. 3 + 1 = 2 + 2
____________
3 + 1 = 2 + 2
4 = 4
b. 9 + 1 = 1 + 2
____________
9 + 1 = 1 + 2
9 + 1 = 1 + 2
10 not equal to 3
correct expression is
9 + 1 = 1 + 9
10 = 10
c. 2 + 3 = 1 + 4
____________
2 + 3 = 1 + 4
5 = 5
d. 5 + 1 = 4 + 2
____________
5 + 1 = 4 + 2
6 = 6
e. 4 + 3 = 3 + 5
____________
4 + 3 = 3 + 5
7 not equal to 8
correct expression is
4 + 3 = 3 + 4
7 = 7
f. 0 + 10 = 2 + 8
____________
0 + 10 = 2 + 8
10 = 10
g. 6 + 3 = 4 + 5
____________
6 + 3 = 4 + 5
9 = 9
h. 3 + 7 = 2 + 6
____________
3 + 7 = 2 + 6
10 not equal to 8
correct expression is
3 + 7 = 7 + 3
10 = 10
Question 3.
Write a number in the expression and solve.
a. 1 + __ = 3 + 2
1 + 4 = 3 + 2
b. __ + 4 = 2 + 5
1 + 4 = 2 + 5
c. __ + 5 = 6 + __
2+ 5 = 6 + 1
d. 7 + __ = 8 + __
7 + 1= 8 + 0
### Eureka Math Grade 1 Module 1 Lesson 18 Exit Ticket Answer Key
Find two ways to fix each number sentence to make it true.
a.
b.
### Eureka Math Grade 1 Module 1 Lesson 18 Homework Answer Key
Question 1.
The pictures below are not equal. Make the pictures equal, and write a true number sentence.
Question 2.
Circle the true number sentences, and rewrite the false sentences to make them true.
a.
b.
c.
d.
e.
f.
g.
h.
i.
Question 3.
Find the missing part to make the number of sentences true.
a. 8 + 0 = __ + 4
8 + 0 = 4 + 4
b. 7 + 2 = 9 + __
7 + 2 = 9 + 0
c. 5 + 2 = 4 + __ |
# Unit 5 Review Stations
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## Objective
SWBAT: • Create and describe a ratio between two quantities • Calculate a unit rate • Use rate reasoning to solve problems • Use ratios to convert measurements • Find a percent of a quantity as a rate per 100
#### Big Idea
If the stores have equivalent prices, what do you know about n? Students review for the Unit 5 test by completing stations.
## Do Now
5 minutes
See my Do Now in my Strategy folder that explains my beginning of class routines.
Often, I create do nows that have problems that connect to the task that students will be working on that day. Here I want students to think about what they have learned throughout the unit. I have students share out with the class.
## Review Stations
30 minutes
Notes:
• Before this lesson I use quiz data to Create Homogeneous Groups.
• I will start off working together with these students to focus on topics in the review that they need help on. I may omit some problems for these students so that they can focus on quality, instead of quantity.
There are many ways to do stations. Today I Create Homogeneous Groups. I identify which stations groups need to focus on first. I set out the copies of the stations throughout the room. I Post a Key for each station somewhere in the classroom. I review expectations for the day.
Students are engaging in MP1: Make sense of problems and persevere in solving them, MP2: Reason abstractly and quantitatively, and MP6: Attend to precision. Groups can move at their own pace, each time they complete a station, they check in with me. I quickly scan the work. If they are on track, I send them to check their answers at the key. If I see repeated mistakes I circle the problems and have them return to work on them. Once they have successfully completed a station, they check it off on their To Do List and move on to another station.
At the end of the work time, I hand around several staplers for students to staple their completed station worksheets together.
## Closure
15 minutes
For Closure I show them the two rates. I explain that Hidden Sweets charges \$0.50 per ounce of candy and Sugar Heaven charges \$ n for each pound of candy. I ask students what they know about n if both stores have equivalent prices. Students participate in a Think Write Pair ShareI call on students to share out their ideas. I want students to remember that there are 16 ounces in one pound. If they forget that fact, I tell them. Therefore, Sugar Heaven must charge \$8 per pound. Students are engaging in MP7: Look for and make use of structure.
Then I present a second situation. I tell students that Sugar Heaven’s price is 10% cheaper than Hidden Sweet’s prices. What must n equal if this is the case? Again students participate in a Think Write Pair Share. I call on students to share out their ideas. Some students may take the equivalent rate of \$8 per pound and take 10% off. Other students may start with the cost of \$0.50 per ounce and take 10% off and then find the price per pound.
Students pass in their completed station work. For Homework, I tell students that they need to complete the stations that they did not finish. If some students have not completed a large amount, I will give them an extra day or two to complete the work. |
# 6th Class Mathematics Commercial Mathematics COMMERCIAL MATHEMATICS
COMMERCIAL MATHEMATICS
Category : 6th Class
Learning Objective
• To understand the term percentage and value of percentage.
• To understand the terms cost price, selling price, profit and loss.
• To learn how to calculate profit, loss, profit percent, loss percent, selling price cost price.
• To understand the terms simple interest and amount.
• To learn how to calculate simple interest and amount.
PERCENTAGE
The term percent means "for every hundred". A fraction whose denominator is 100 is called percentage and the numerator of the fraction is called the rate percent. Thus, when we say a man made a profit of 20 percent we mean to say that he gained Rs 20 for every hundred rupees he invested in the business, i.e., 20/100 rupees for each Rupee. The abbreviation of percent is p.c. and it is generally denoted by %.
VALUE OF PERCENTAGE
Value of percentage always depends on the quantity to which it refers consider the statement:
“65% of the students in this class are boys". From the context, it is understood that boys form 65% of the total number of students in the class. To know the value of 65%, the value of the total number of student should be mown.
If the total number of students is 200, then, the number of boys $=\frac{200\times 65}{100}=130;$ It can also be written as $(200)\times (0.65)=130$.
Note that the expressions 6%, 63%, 72%, 155% etc. do not have any value intrinsic to themselves. Their values depend on the quantities to which they refer.
COMPARING PERCENTAGES
Which of the three 25%, 5% and 125% is the largest?
If should be remembered that no comparison can be made about the above percentages, because they do not have intrinsic values. If 25% refers to 25% of 10,000 then its value is 0.25 x 10,000 = 2,500 and if 75% of 100, its value is 0.75 x 100 =75.
And so we can conclude that 25% of 10,000 > 75% of 100.
Note: Percentages can be compared only when the quantities they refer to are known.
IMPORTANT RESULTS
1. To express a percentage as a fraction divide it by $100\Rightarrow$ a percentage = 1/100.
Example:
Express the following as fraction
(a) 25%
(b) $33\frac{1}{3}%$
Sol.
(a) $25%\,=\frac{25}{10}\left[ \text{since}\,\text{ }\!\!%\!\!\text{ }\,\text{means}\,\frac{1}{100} \right]=\frac{1}{4}$
(b) $33\frac{1}{3}%\,=\frac{100}{3}%\,=\frac{100}{3\times 100}=\frac{1}{3}$
2. To express a fraction as a percent multiply it by $100\,\frac{a}{b}\,=\left[ \left( \frac{a}{b}\times 100 \right) \right]\,%$
Example:
Express $\frac{1}{8}$ as a percentage.
Sol.
$\frac{1}{8}=\frac{1}{8}\times 100%$
$=\frac{100}{8}%=\frac{25}{2}%\,=12\frac{1}{2}%$
3. To express percentage as a decimal we remove the symbol% and shift the decimal point by two places the left.
Example:
Express $6\frac{1}{2}%$ as a decimal.
Sol.
$6\frac{1}{2}%=\frac{13}{2}%\,=6.5%\,=\frac{65}{100}=0.065$
4. To express decimal as a percentage we shift the decimal point by two places to the right and write number obtained with the symbol % or simply we multiply the decimal with 100.
Example:
0.345 as a percentage.
Sol. $0.345\times 100%$
= 34.5%
If A is R% of a given number N, then $N=\frac{A\times 100}{R}$
Example:
25% of a number is 80. What is the number?
Sol.
Let the number be X.
According to the given condition
$\frac{25}{100}\,\times \,X=80\Rightarrow \,X=\frac{80\times 100}{25}=320$.
PROFIT AND LOSS
COST PRICE (CP)
The price for which an article is bought is called its cost price.
SELLING PRICE (SP)
The price at which an article is sold is called its selling price.
PROFIT (GAIN)
The difference between the selling price and cost price is called the profit. For profit, selling price should be greater than cost price.
LOSS
The difference between the cost price and the selling price is called the loss. When cost price is greater than the selling price there is a loss.
Wit and loss is generally represented as a percent of the cost price, unless otherwise stated.
FORMULAE
• Profit = S.P – C.P, if S.P > C.P
• Loss C.P – S.P, if C.P > S.P
• Profit%$\frac{\text{Profit}\,\times 100}{C.P}$
• $Loss%\frac{Loss\,\times 100}{C.P}$
• $S.P=\left( \frac{100+\text{Profit}%}{100} \right)\times C.P$
• $S.P=\left( \frac{100+Loss%}{100} \right)\times C.P$
• $C.P=\left( \frac{100}{100+\text{Profit}%} \right)\times S.P$
• $C.P=\left( \frac{100}{100-\text{Profit}%} \right)\,\times S.P$
Example:
A man buys a radio for Rs 600 and sells it at a gain of 25%, what will be the selling price for him?
Sol.
C.P = Rs 600 and Gain (Profit) % = 25%
$S.P=\left( \frac{100+\text{Profit}%}{100} \right)\times C.P$
$=\left( \frac{100+25}{100} \right)$
$=\frac{125}{100}\times 600$
= Rs 70
Example:
Find the cost price of good, sells at Rs 160, having Loss% = 20%
Sol.
S.P = Rs 160% = 20%
$C.P=\frac{S.P\times 100}{100-Loss%}$
$=\frac{160\times 100}{80}$
= Rs 200
Cost price of goods = Rs 200.
SIMPLE INTEREST
Suppose Vishal borrows money from a bank for his higher studies, then at the end of the specified period, he has to pay the money he borrowed and some additional money for the privilege of having used the bank money
Now, we can define the following terms:
The money borrowed is called the Principal ‘p’;
The additional money paid is called the Interest 'SI;
The total money paid is called the Amount 'A’
Formula
1. Amount = Principal + Interests A =P+S.I.
2. Simple Interest
$=\frac{\text{Principal}\,\text{ }\!\!\times\!\!\text{ Time}\,\text{ }\!\!\times\!\!\text{ Rate}}{\text{100}}\,\Rightarrow \,S.I.=\frac{P\times T\times R}{100}$
3. Principal
$=\frac{\text{100 }\!\!\times\!\!\text{ Simple}\,\text{Interest}}{\text{Time }\!\!\times\!\!\text{ Rate}}\,\Rightarrow \,P=\frac{100\times S.I.}{T\times R}$
4. Rate $=\frac{\text{100 }\!\!\times\!\!\text{ Simple}\,\text{Interest}}{\text{Principle}\,\times \text{Time}}\,\Rightarrow \,R=\frac{100\times S.I.}{P\times T}$
5. Time $=\frac{\text{100 }\!\!\times\!\!\text{ Simple}\,\text{Interest}}{\text{Principle}\,\text{ }\!\!\times\!\!\text{ Rate}}\,\Rightarrow \,T=\frac{100\times S.I.}{P\times R}$
Example
Find (i) the interest, (ii) the amount on Rs 5000 at 5% per annum S.I. for 4 years.
Sol.
(i) Given P = Rs 5000, T = 4 years, R = 5%
We know $S.I.=\frac{P\times T\times R}{100}$$\therefore \,\,S.I.\,=\frac{5000\times 4\times 5}{100}$
Hence, S.I. =Rs 1000.
(ii) Amount = P + S.I.
$\therefore$ Amount = Rs 5000 + Rs 1000 = Rs 6000.
Example
What sum of money will amount to Rs 1400 at the rate of 8% per annum S.I. in 5 years?
Sol. Let the principal be Rs 100
100x5x8
$\therefore$ S.I. on Rs 100 for 5 years at 8% $=\frac{100\times 5\times 8}{100}=5\times 8=\text{Rs}40$
$\therefore$ Amount = P + S.I.
=Rs (100+40)=Rs 140.
If the amount is Rs 140, then Principal = Rs 100
If the amount is Rs 1400, then Principal
$=\frac{100\times 1400}{140}$
$\therefore$ Principal = Rs 1000.
Example
In what time will Rs 450 amount to Rs 540 at 5% per annum S.I.?
Sol. Given A = Rs 540, P = Rs 450, R = 5%, T = Rs
$\therefore$ S.I. = A - P = Rs 540 - Rs 450 == Rs 90
$\therefore$ $T=\frac{100\times S.I.}{P\times R}$
$\therefore$ $T=\frac{100\times 90}{450\times 5}=4$
$\therefore$ Time = 4 years.
Example
In how many years will a sum of money double itself at 10% per annum S.I.?
Sol. Let the principle be Rs 'x’
$\therefore$ Amount == 2x
$\therefore$ S.I. = A – P
$\therefore$ S.I. $=2x-x=x$
$T=\frac{100\times S.I.}{P\times R}$
$\therefore$ $T=\frac{100\times x}{x\times 10}\,=10$
$\therefore$ Time = 10 years.
Note:
1. If a sum of money doubles, means S.I. = 2P - P = P
2. If a sum of money triple, means S.I. = 3P - P = 2P
Example
If Rs 800 amounts to Rs 960 in 4 years, find the rate percent per annum S.I.
Sol. Given: P = Rs 800, A, = Rs 960
$\therefore$ S.I. = Rs 960 – Rs 800 = Rs 160
$R=\frac{100\times S.I.}{P\times T}\,=\frac{100\times 160}{800\times }=5$
$\therefore$ Rate == 5%.
AVERAGE OR MEAN
If ${{X}_{1}},{{X}_{2}},\,{{X}_{3}}....{{X}_{n}}$ be n observations then their arithmetic mean is given by:
$\overline{X}=\frac{{{X}_{1}}+{{X}_{2}}+......{{X}_{n}}}{n}$
Example
Find the mean height if the heights of 5 persons are 144cm, 153 cm, 150cm, 158 cm and 155 cm respectively
Sol.
Mean Height $=\frac{144+153+150+158+155}{5}$
$=\frac{760}{5}\,=152\,cm$
Example
Find the mean of the first 10 odd numbers.
Sol.
First ten odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
So mean
$(\overline{x})=\frac{1+3+5+9+11+13+15+17+19}{10}$
$=\frac{100}{10}=10$
SPEED
The speed of a body is defined as the distance covered by it in unit time.
Speed $=\frac{\text{Distance}}{\text{Time}}$ distance is constant
Time $=\frac{\text{Distance}}{\text{Speed}};$ time is constant
Distance = Speed x Time; Speed is constant
UNITS OF MEASUREMENT
• Time is measure in seconds (sec), minutes (min) or hours (hr)
• Distance is usually measured in metres (m), kilometres (Km), miles, yards or feet.
• Speed is usually measured in metre/sec (m/s), kilometre/hour (Km/hr), miles/hr.
AVERAGE SPEED
Total distance travelled
Average speed $=\frac{\text{Total}\,\text{distance}\,\text{travelled}}{\text{Total}\,\text{time}\,\text{take}}$
Example
Find the Speed of train travels 90 km in 2 hours.
Sol.
$Speed\,=\frac{\text{Distance}}{\text{Time}}$
Speed $=\frac{90}{2}$
Speed = 45 Km/hr
Example
Find the average speed of a person if he goes 5 km from point A to B and then goes 3 km from point B to C, he takes 2 hours to cover the distance from point A to C.
Sol.
Total distance covered by man = (5 + 3) km = 8 km.
Average speed
$=\frac{\text{Total}\,\text{distance}}{\text{Total}\,\text{time}\,\text{take}}=\frac{8}{2\,}=4\,km/hr$
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# How do you solve 6n - 4 = n + 11?
##### 1 Answer
Jan 18, 2017
See the entire solution process below:
#### Explanation:
First, add and subtract the necessary terms from each side of the equation to isolate the $n$ terms on one side of the equation and the constants on the other side of the equation while keeping the equation balanced:
$6 n - 4 + \textcolor{red}{4} - \textcolor{b l u e}{n} = n + 11 + \textcolor{red}{4} - \textcolor{b l u e}{n}$
$6 n - \textcolor{b l u e}{n} - 4 + \textcolor{red}{4} = n - \textcolor{b l u e}{n} + 11 + \textcolor{red}{4}$
$\left(6 - 1\right) n - 0 = 0 + 15$
$5 n = 15$
Next, divide each side of the equation by $\textcolor{red}{5}$ to solve for $n$ while keeping the equation balanced:
$\frac{5 n}{\textcolor{red}{5}} = \frac{15}{\textcolor{red}{5}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} n}{\cancel{\textcolor{red}{5}}} = 3$
$n = 3$ |
# JEE Main & Advanced Mathematics Functions Some Important Definitions
## Some Important Definitions
Category : JEE Main & Advanced
(1) Real numbers : Real numbers are those which are either rational or irrational. The set of real numbers is denoted by $R$.
• (2) Related quantities : When two quantities are such that the change in one is accompanied by the change in other, e., if the value of one quantity depends upon the other, then they are called related quantities.
(3) Variable: A variable is a symbol which can assume any value out of a given set of values.
(i) Independent variable : A variable which can take any arbitrary value, is called independent variable.
(ii) Dependent variable : A variable whose value depends upon the independent variable is called dependent variable.
(4) Constant : A constant is a symbol which does not change its value, i.e., retains the same value throughout a set of mathematical operation. These are generally denoted by $a,\,\,b,\,\,c$ etc. There are two types of constant, absolute constant and arbitrary constant.
(5) Absolute value : The absolute value of a number $x,$ denoted by $|x|,$ is a number that satisfies the conditions
$|x|=\left\{ \begin{matrix} -x \\ \,\,0 \\ \,\,\,x \\ \end{matrix}\,\,\,\begin{matrix} \text{if} \\ \text{if} \\ \text{if} \\ \end{matrix}\,\,\begin{matrix} x<0 \\ x=0 \\ x>0 \\ \end{matrix}. \right.$ We also define $|x|,$as follows, $|x|=$ maximum $\{x,\,\,-x\}$ or $|x|=\sqrt{{{x}^{2}}}$.
(6) Fractional part : We know that $x\ge [x].$ The difference between the number $'x'$ and it’s integral value $'[x]'$ is called the fractional part of $x$ and is symbolically denoted as $\{x\}$. Thus, $\{x\}=x-[x]$e.g., if $x=4.92$ then $[x]=4$ and $\{x\}=0.92$.
Fractional part of any number is always non-negative and less than one.
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### Section 4.11 : Linear Approximations
For problems 1 – 4 find a linear approximation to the function at the given point.
1. $$f\left( x \right) = \cos \left( {2x} \right)$$ at $$x = \pi$$
2. $$h\left( z \right) = \ln \left( {{z^2} + 5} \right)$$ at $$z = 2$$
3. $$g\left( x \right) = 2 - 9x - 3{x^2} - {x^3}$$ at $$x = - 1$$
4. $$g\left( t \right) = {{\bf{e}}^{\sin \left( t \right)}}$$ at $$t = - 4$$
5. Find the linear approximation to $$h\left( y \right) = \sin \left( {y + 1} \right)$$ at $$y = 0$$. Use the linear approximation to approximate the value of $$\sin \left( 2 \right)$$ and $$\sin \left( {15} \right)$$. Compare the approximated values to the exact values.
6. Find the linear approximation to $$R\left( t \right) = \sqrt[5]{t}$$ at $$t = 32$$. Use the linear approximation to approximate the value of $$\sqrt[5]{{31}}$$ and $$\sqrt[5]{3}$$. Compare the approximated values to the exact values.
7. Find the linear approximation to $$h\left( x \right) = {{\bf{e}}^{1 - x}}$$ at $$x = 1$$. Use the linear approximation to approximate the value of $${\bf{e}}$$ and $${{\bf{e}}^{ - 4}}$$. Compare the approximated values to the exact values.
For problems 8 – 10 estimate the given value using a linear approximation and without using any kind of computational aid.
1. $$\ln \left( {1.1} \right)$$
2. $$\sqrt {8.9}$$
3. $$\sec \left( {0.1} \right)$$ |
# Fraction calculator
This fraction calculator performs basic and advanced fraction operations, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. The calculator helps in finding value from multiple fractions operations. Solve problems with two, three, or more fractions and numbers in one expression.
## The result:
### 41/8 + 21/16 = 99/16 = 6 3/16 = 6.1875
Spelled result in words is ninety-nine sixteenths (or six and three sixteenths).
### How do we solve fractions step by step?
1. Conversion a mixed number 4 1/8 to a improper fraction: 4 1/8 = 4 1/8 = 4 · 8 + 1/8 = 32 + 1/8 = 33/8
To find a new numerator:
a) Multiply the whole number 4 by the denominator 8. Whole number 4 equally 4 * 8/8 = 32/8
b) Add the answer from the previous step 32 to the numerator 1. New numerator is 32 + 1 = 33
c) Write a previous answer (new numerator 33) over the denominator 8.
Four and one eighth is thirty-three eighths.
2. Conversion a mixed number 2 1/16 to a improper fraction: 2 1/16 = 2 1/16 = 2 · 16 + 1/16 = 32 + 1/16 = 33/16
To find a new numerator:
a) Multiply the whole number 2 by the denominator 16. Whole number 2 equally 2 * 16/16 = 32/16
b) Add the answer from the previous step 32 to the numerator 1. New numerator is 32 + 1 = 33
c) Write a previous answer (new numerator 33) over the denominator 16.
Two and one sixteenth is thirty-three sixteenths.
3. Add: 33/8 + 33/16 = 33 · 2/8 · 2 + 33/16 = 66/16 + 33/16 = 66 + 33/16 = 99/16
It is suitable to adjust both fractions to a common (equal, identical) denominator for adding, subtracting, and comparing fractions. The common denominator you can calculate as the least common multiple of both denominators - LCM(8, 16) = 16. It is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 8 × 16 = 128. In the following intermediate step, it cannot further simplify the fraction result by canceling.
In other words - thirty-three eighths plus thirty-three sixteenths is ninety-nine sixteenths.
#### Rules for expressions with fractions:
Fractions - use a forward slash to divide the numerator by the denominator, i.e., for five-hundredths, enter 5/100. If you use mixed numbers, leave a space between the whole and fraction parts.
Mixed numerals (mixed numbers or fractions) keep one space between the integer and
fraction and use a forward slash to input fractions i.e., 1 2/3 . An example of a negative mixed fraction: -5 1/2.
Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1/2 : 1/3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
### Math Symbols
SymbolSymbol nameSymbol MeaningExample
+plus signaddition 1/2 + 1/3
-minus signsubtraction 1 1/2 - 2/3
*asteriskmultiplication 2/3 * 3/4
×times signmultiplication 2/3 × 5/6
:division signdivision 1/2 : 3
/division slashdivision 1/3 / 5
:coloncomplex fraction 1/2 : 1/3
^caretexponentiation / power 1/4^3
()parenthesescalculate expression inside first-3/5 - (-1/4)
The calculator follows well-known rules for the order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
MDAS - Multiplication and Division have the same precedence over Addition and Subtraction. The MDAS rule is the order of operations part of the PEMDAS rule.
Be careful; always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) have the same priority and must evaluate from left to right. |
# NCERT Solutions For Class 9th Maths Chapter 5 : Introduction to Euclid’s Geometry
CBSE NCERT Solutions For Class 9th Maths Chapter 5 : Introduction to Euclid’s Geometry. NCERT Solutins For Class 9 Mathematics. Exercise 5.1, Exercise 5.2
### NCERT Solutions for Class IX Maths: Chapter 5 – Introduction to Euclid’s Geometry
Page No:85
Exercise 5.1
1. Which of the following statements are true and which are false? Give reasons for your
(i) Only one line can pass through a single point.
(ii) There are an infinite number of lines which pass through two distinct points.
(iii) A terminated line can be produced indefinitely on both the sides.
(iv) If two circles are equal, then their radii are equal.
(v) In Fig. 5.9, if AB = PQ and PQ = XY, then AB = XY.
(i) False. There can be infinite line drawn passing through a single point.
(ii) False. Only one line can be drawn which passes through two distinct points.
(iii) True. A terminated line can be produced indefinitely on both the sides.
In geometry, a line can be extended in both direction. A line means infinite long length.
(iv) True. If two circles are equal, then their radii are equal.
By superposition, we will find that the centre and circumference of the both circles coincide. Hence, their radius must be equal.
(v) True. By Euclid’s first axiom things which are equal to the same thing, are equal to one another.
2. Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they, and how might you define them?
(i) parallel lines (ii) perpendicular lines (iii) line segment
(iv) radius of a circle (v) square
Yes, other terms need to be defined first which are:
Plane: A plane is flat surface on which geometric figures are drawn.
Point: A point is a dot drawn on a plane surface and is dimensionless.
Line: A line is collection of points which can extends in both direction and has only length not breadth.
(i) Parallel lines: When two or more never intersect each other in a plane and perpendicular distance between them is always constant then they are said to be parallel lines.
(ii) Perpendicular lines: When two lines intersect each other at right angle in a plane then they are said to be perpendicular to each other.
(iii) Line segment: A line segment is a part of a line with two end points and cannot be extended further.
(iv) Radius of circle: The fixed distance between the centre and the circumference of the circle is called the radius of the circle.
(v) Square: A square is a quadrilateral in which all the four sides are equal and each internal angle is a right angle.
3. Consider two ‘postulates’ given below:
(i) Given any two distinct points A and B, there exists a third point C which is in between A and B.
(ii) There exist at least three points that are not on the same line.
Do these postulates contain any undefined terms? Are these postulates consistent? Do they follow from Euclid’s postulates? Explain.
Undefined terms in the postulates:
→ Many points lie in a plane. But here it is not given about the position of the point C whether it lies on the line segment joining AB or not.
→ Also, there is no information about the plane whether the points are in same plane or not.
Yes, these postulates are consistent when we deal with these two situation:
(i) Point C is lying in between and on the line segment joining A and B.
(ii) Point C not lies on the line segment joining A and B.
Page No: 86
4. If a point C lies between two points A and B such that AC = BC, then prove that AC = 1/2 AB. Explain by drawing the figure.
Here, AC = BC
AC + AC = BC + AC
also, BC +AC = AB (as it coincides with line segment AB)
∴ 2 AC = AB (If equals are added to equals, the wholes are equal.)
⇒ AC = 1/2 AB.
5. In Question 4, point C is called a mid-point of line segment AB. Prove that every line segment has one and only one mid-point.
Let A and B be the line segment and points P and Q be two different mid points of AB.
Now,
∴ P and Q are midpoints of AB.
Therefore AP=PB and also AQ = QB.
also, PB + AP = AB (as it coincides with line segment AB)
Similarly, QB + AQ = AB.
Now,
AP + AP = PB + AP (If equals are added to equals, the wholes are equal.)
⇒ 2 AP = AB — (i)
Similarly,
2 AQ = AB — (ii)
From (i) and (ii)
2 AP = 2 AQ (Things which are equal to the same thing are equal to one another.)
⇒ AP = AQ (Things which are double of the same things are equal to one another.)
Thus, P and Q are the same points. This contradicts the fact that P and Q are two different mid points of AB. Thus, it is proved hat every line segment has one and only one mid-point.
6. In Fig. 5.10, if AC = BD, then prove that AB = CD.
Given, AC = BD
From the figure,
AC = AB + BC
BD = BC + CD
⇒ AB + BC = BC + CD
According to Euclid’s axiom, when equals are subtracted from equals, remainders are also equal.
Subtracting BC both sides,
AB + BC – BC = BC + CD – BC
AB = CD
7. Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that the question is not about the fifth postulate.)
Axiom 5 : The whole is always greater than the part.
Take an example of a cake. When it is whole it will measures 2 pound but when we took out a part from it and measures its weigh it will came out lower than the previous one. So, the fifth axiom of Euclid is true for all the universal things. That is why it is considered a ‘universal truth’.
Page No: 88
Exercise 5.2
1. How would you rewrite Euclid’s fifth postulate so that it would be easier to understand?
The fifth postulates is about parallel lines.
When two or more never intersect each other in a plane and perpendicular distance between them is always constant then they are said to be parallel lines.
Two facts of the postulates:
(i) If P doesn’t lie on l then we can draw a line through P which will be parallel to the line l.
(ii) There will be only one line can be drawn through P which is parallel to the line l.
2. Does Euclid’s fifth postulate imply the existence of parallel lines? Explain. |
# Chapter 6 - Systems of Equations and Inequalities - 6-1 Solving Systems by Graphing - Practice and Problem-Solving Exercises - Page 363: 10
After graphing the functions we find that the two intercept at $(2,4)$. The two functions are consistent and independent because they are two different functions that share one intersection point and have different slopes
#### Work Step by Step
#$1$ The standard notation for linear functions is $y=mx + b$, where m=slope of the function (rise over run) and b=the y-intercept of the function. #$2$ For Equation 1, $y= 2x$, the $m$ value is $2$, which means the slope is $2$, in other words this means for every increase of 1 in the positive x-direction the y-value will increase by $2$ in the positive y-direction. There is no $b$ value so the y-intercept is $(0,0)$. $y= 2x; slope=2 and y-intercept= (0,0)$ #$3$ Equation 2, $y= -2+8$, the $m$ value, or slope, is $-2$, which means for every increase of 1 in the positive x-direction the y-value will decrease by 2 There is $b$ value so the y-intercept is (0,8) $y= -2x+8; slope= -2 and y-intercept= (0,8)$ #$4$ When we graph the $2$ functions, $Equation 1$ corresponds with $f(x)=2x$ in the $blue color$ and we can see that it has a slope of $2$ and its y-intercept is at $(0,0)$. $Equation 2$ corresponds with $g(x)=8-2x$ in the $red color$ and it has a slope of $-2$ and its y-intercept is at $(0, 8)$. We can see in the graph that the two functions intersect at $(2, 4)$. This means that the two functions equal one another at this point. #$5$ We can double check our result by making the functions equal one another and solving for x. After doing this we find that the two functions equal one another when $x = 2$. By plugging this back into the functions we can check this again because when $x = 2$ both functions are $y = 4$. Solution- $2x = -2x + 8$ $+2x .. +2x$ $4x = 8$ $4x\div4=8\div4$ $x = 2$
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# 2.3 Set algebra
## 2.3.1 Commutativity and associativity
This section is about the laws that union, intersection, difference, and complement obey.
###### Theorem 2.3.1.
For all sets $A$ and $B$,
• $A\cap B=B\cap A$, and
• $A\cup B=B\cup A$.
These are called the commutativity properties for intersection and union. These results might seem obvious, but we will write out the proofs carefully because the method of using logical equivalences will be applied to more complex set identities later.
###### Proof.
By definition,
$\displaystyle A\cap B$ $\displaystyle=\{x:x\in A\wedge x\in B\}$ $\displaystyle B\cap A$ $\displaystyle=\{x:x\in B\wedge x\in A\}.$
Theorem 1.5.1 tells us that for any two WFFs $\phi$ and $\psi$, the formulas $(\phi\wedge\psi)$ and $(\psi\wedge\phi)$ are logically equivalent: one is true if and only if the other is true. So
$x\in A\wedge x\in B$
is true if and only if
$x\in B\wedge x\in A$
is true. This shows that for any $x$ we have $x\in A\cap B$ if and only if $x\in B\cap A$, so by definition of set equality, $A\cap B=B\cap A$.
The argument for $\cup$ is the same, except that we use the logical equivalence $(\phi\vee\psi)\equiv(\psi\vee\psi)$. ∎
What this proof shows is that if you have a set $X$ defined in set builder notation using a logical formula
$X=\{x:P(x)\}$
then it is equal to any other set defined using a logically equivalent formula.
###### Theorem 2.3.2.
For all sets $A$, $B$, $C$ we have
• $A\cap(B\cap C)=(A\cap B)\cap C$ and
• $A\cup(B\cup C)=(A\cup B)\cup C$.
This is the associativity property for $\cap$ and $\cup$.
###### Proof.
Like the proof of the last theorem, these equalities follow from the associativity properties for $\wedge$ and $\vee$ we saw in Theorem 1.5.1. ∎
Associativity tells us means there’s no ambiguity in writing
$A\cup B\cup C\text{ or }A\cap B\cap C$
without any brackets to indicate which union or intersection should be done first. Compare this with
$1+2+3.$
There’s no need for brackets because it doesn’t matter whether you do 1+2 first then add 3, or whether you add 1 to the result of 2 + 3. On the other hand $1+2\times 3$ or $1-2-3$ require either brackets or a convention on which operation to do first. Similarly $A\cup(B\cap C)$ is different to $(A\cup B)\cap C$ in general, so the brackets here are obligatory, and $A\setminus(B\setminus C)$ is different to $(A\setminus B)\setminus C$.
## 2.3.2 The distributive laws
Because we defined unions and intersections using logical conditions on set elements, they should obey laws that come from the results we proved about $\wedge$ and $\vee$.
###### Theorem 2.3.3.
For any sets $A$, $B$, $C$
1. 1.
$A\cup(B\cap C)=(A\cup B)\cap(A\cup C)$ and
2. 2.
$A\cap(B\cup C)=(A\cap B)\cup(A\cap C)$.
###### Proof.
Consider the first of these identities. The left hand side consists of all things $x$ such that
$x\in A\vee(x\in B\wedge x\in C).$ (2.2)
The right hand side consists of all things $x$ such that
$(x\in A\vee x\in B)\wedge(x\in A\vee x\in C)$ (2.3)
By Theorem 1.6.1, for any WFFs $\phi,\psi,\theta$ we have $\phi\vee(\psi\wedge\theta)\equiv(\phi\vee\psi)\wedge(\phi\vee\theta)$. Thus any $x$ makes (2.2) true if and only if it makes (2.3) true. So $x$ belongs to the first set if and only if it belongs to the second, therefore the two sets are equal.
The second identity can be proved similarly. ∎ |
# NCERT Solution (Ex 1.1) - Chapter 1: Rational Number, Maths, Class 8 Notes - Class 8
## Class 8: NCERT Solution (Ex 1.1) - Chapter 1: Rational Number, Maths, Class 8 Notes - Class 8
The document NCERT Solution (Ex 1.1) - Chapter 1: Rational Number, Maths, Class 8 Notes - Class 8 is a part of Class 8 category.
All you need of Class 8 at this link: Class 8
Exercise 1.1 (Rational Number)
Question 1:
Using appropriate properties find:
(i)
[Using associative property]
[Using associative property]
(II)
[Using associative property]
[Using distributive property]
Question 2:
Write the additive inverse of each of the following:
(i) 2/8
(ii) -5/9
(iii) -6/-5
(iv) 2/-9
(v) 19/-6
We know that additive inverse of a rational number
(i) Additive inverse of 2/8 is -2/8
(ii) Additive inverse of -5/9 is 5/9
(iii) Additive inverse of -6/-5 is -6/5
(iv) Additive inverse of 2/-9 is 2/9
(v) Additive inverse of19/-6 is 19/6
Question 3:
Verify that -(-x)= x for:
(i)x= 11/15
(ii)x= 13/17
(i) Putting x= 11/15 in -(-x) =x
Hence, verified.
(ii)
Hence, verified.
Question 4:
Find the multiplicative inverse of the following:
We know that multiplicative inverse of a rational number
(i) Multiplicative inverse of - 13 is -1/13
(ii) Multiplicative inverse of -13/19 is -19/13
(iii) Multiplicative inverse of 1/5 is 5
(iv) Multiplicative inverse of
(v) Multiplicative inverse of
(vi) Multiplicative inverse of -1 is 1/-1
Question 5:
Name the property under multiplication used in each of the following:
(i) 1 is the multiplicative identity.
(ii) Commutative property.
(iii) Multiplicative Inverse property.
Question 6:
Multiply 6/13 by the reciprocal of -7/
16
The reciprocal of -7/16 is -16/7
According to the question,
Question 7:
Tell what property allows you to compute
By using associative property of multiplication, a x (b x c) = (a x b) x c.
Question 8:
Is 8/9 the multiplicative inverse of
Why or why not?
Since multiplicative inverse of a rational number a is
Therefore,
But its product must be positive 1.
Therefore, 8/9 is not the multiplicative inverse of
Question 9:
Is 0.3 the multiplicative inverse of
Why or why not?
Since multiplicative inverse of a rational number a is
Therefore,
Therefore, Yes 0.3 is the multiplicative inverse of
Question 10:
Write:
(i) The rational number that does not have a reciprocal.
(ii) The rational numbers that are equal to their reciprocals.
(iii) The rational number that is equal to its negative.
(i) 0
(ii) 1 and -1
(iii) 0
The document NCERT Solution (Ex 1.1) - Chapter 1: Rational Number, Maths, Class 8 Notes - Class 8 is a part of Class 8 category.
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# Understanding Numbers, counting, and basic addition and subtraction
Everyone utilizes mathematics daily, making it an essential ability. Math is used in everything from keeping track of time to calculating bills. Understanding numbers, counting, and fundamental operations like addition and subtraction are the first steps on the trip. Let's examine these ideas one by one.
Start Your Child's Math Journey Now!
## The Number Concept
Quantities are symbolized by numbers. They provide us with a means of measurement and comparison, which aids in our understanding of the world. For example, the number "3" indicates how many apples you have when you say you have three. A fundamental understanding of numbers serves as the basis for all mathematical ideas.
Important Points:
• Quantities are represented by numbers.
• One can measure, compare, and count using numbers.
• The number system has an unlimited range beginning at 0.
## Counting Is the First Math Concept
Finding the number of items in a group is done by counting. It is the most fundamental arithmetic ability that kids pick up, typically via counting fingers, toys, or even steps.
Important Points:
• Commence counting with 1 and work your way up (1, 2, 3, etc.).
• Comprehending the notion of "how many" is facilitated by counting.
• It's crucial to count in numerical order, without omitting any.
### Activities for Counting Practice:
• Count the books on the shelf or the passing automobiles in your immediate surroundings.
• To encourage learning, use counting games or number charts.
• To better comprehend subtraction, count backward.
Combining two or more quantities to determine the total is the process of addition. You can calculate your total amount when you add all the numbers.
Important Points:
• The plus (+) symbol is used to denote addition.
• The sum is the outcome of addition.
• You can use items or your fingers to count little numbers when adding them.
For instance, if you were given three apples and you already had two, you would have five apples. This is equivalent to 2 + 3 = 5.
• Play games where you have to combine various objects.
• Utilizing a number line, practice adding numbers.
## Simple Subtraction Removing Amounts
The process of subtracting one quantity from another to get the remaining amount is known as subtraction. It is the antithesis of adding and a necessary life skill.
Important Points:
• The symbol for subtraction is negative (-).
• The difference is what remains after subtraction.
• Objects or a number line can also be used to illustrate subtraction.
For instance, if you have five apples and you give away two of them, you will still have three. This is equivalent to 5 - 2 = 3.
### Exercises for Subtraction Practice:
• Utilize commonplace situations, such as splitting food, to master subtraction.
• Use toys or money to solve basic subtraction problems.
• Use a number line to practice subtraction and help you visualize the process.
## The Connection Between Subtraction and Addition
Subtraction and addition have a strong relationship. Gaining proficiency in one aids in mastering the other. For instance, it would be simple to determine that 5 - 3 = 2 if you already knew that 2 + 3 = 5.
Important Points:
• The operations of addition and subtraction are inverse.
• Understanding addition facts facilitates comprehension of subtraction.
• To gain confidence, practice solving similar addition and subtraction problems.
## Adding Interest to Math with Games and Activities
Math education doesn't have to be tedious. Understanding numbers, counting, addition, and subtraction may be made more interesting by incorporating entertaining games and activities.
### Suggestions for Having Fun While Learning:
• Play counting board games, such as "Chutes and Ladders."
• To practice addition and subtraction, use math applications or online games.
• To make arithmetic more relevant, incorporate your child's favorite characters into narrative puzzles.
The foundation of mathematics is counting, number comprehension, and simple addition and subtraction. Through the implementation of captivating exercises, kids may cultivate a robust foundation in mathematics. Recall that math is all around us; embrace it and enjoy your education!
Q.1: What are numbers, and why are they important?
Ans: Numbers are symbols that represent quantities. They are essential for counting, measuring, comparing, and performing various mathematical operations. Numbers help us understand and interact with the world around us.
Q.2: How do I teach my child to count?
Ans: Start by counting everyday objects like toys, books, or snacks. Encourage your child to count out loud and point to each item as they count. Using counting songs, games, and number charts can also make learning fun and engaging.
Q.3: What is the difference between counting and addition?
Ans: Counting is the process of determining the quantity of items in a group, usually by listing numbers in sequence (1, 2, 3, etc.). Addition, on the other hand, involves combining two or more quantities to find a total. For example, counting how many apples you have is different from adding more apples to the group.
Q.4: How can I help my child understand basic addition? |
# 242-535 ADA: 15. Basic Math1 Objective o a reminder about dot product and cross product, and their use for analyzing line segments (e.g. do two segments.
## Presentation on theme: "242-535 ADA: 15. Basic Math1 Objective o a reminder about dot product and cross product, and their use for analyzing line segments (e.g. do two segments."— Presentation transcript:
242-535 ADA: 15. Basic Math1 Objective o a reminder about dot product and cross product, and their use for analyzing line segments (e.g. do two segments intersect and the distance of a point to a line) Algorithm Design and Analysis (ADA) 242-535, Semester 1 2014-2015 15. Some Basic Maths Used in CG
242-535 ADA: 15. Basic Math2 1.Simple Geometry 2.Dot (Scalar) Product 3.Cross (Vector) Product 4.The CCW() Function 5.Line Segment Intersection 6.Distance Between a Point and a Line 7.Area of a Convex PolygonOverview
242-535 ADA: 15. Basic Math3 Points o (x, y) Cartesian coordinates o (r, Θ) Polar coordinates Line Segments o Positioned using their end points Lines o 2-tuple (m, c) using y = mx + c o 3-tuple (a, b, c) using ax + by = c o Two points P 0, P 1 on the line using P(t) = (1-t)P 0 + t P 1 Polygons o Ordered list of points 1. Simple Geometry
Line Segments & Vectors p = (x, y ) 1 2 O = (0, 0)x y 1 2 Points (vectors): p, p, p p = p p 1 2 1 2 2 1 p p = (x x, y y ) 1 2 Line segment: p p = p p 2 1 1 2
242-535 ADA: 15. Basic Math5 Given the two vectors a = and b = then the dot product is a · b = a1*b1 + a2*b2 + a3*b3 Example a =, b = a · b = 0*2 + -3*3 + 7*1 = 9 – 7 = 2 2. Dot (Scalar) Product
Dot Product as Geometry a b = |a| |b| cos θ The dot product is the product of the magnitudes of the two vectors a and b and the cosine of the angle between them. The angle between a and b is: Θ = cos -1 ((a b) / |a||b|) The angle between a and b is: Θ = cos -1 ((a b) / |a||b|)
242-535 ADA: 15. Basic Math7 Angle Example
242-535 ADA: 15. Basic Math8 It can be used to get the projection (length) of vector a onto vector b. Uses of Dot Product
242-535 ADA: 15. Basic Math9 3. Cross (Vector) Product
242-535 ADA: 15. Basic Math10 Cross Product Example
242-535 ADA: 15. Basic Math11 The cross product a × b is defined as a vector c that is perpendicular to both a and b, with a direction given by the right-hand rule (the unit vector n ) and a magnitude equal to the area of the parallelogram that the vectors span. Cross Product as Geometry the area of the parallelogram. a b = |a| |b | sin θ n
242-535 ADA: 15. Basic Math12 Angle Example
242-535 ADA: 15. Basic Math13 You can use the cross product of the vectors a and b to generate a perpendicular vector, which is the normal to the plane defined by a and b. Normals are very useful when calculating things like lighting in 3D computer games, and up/down in FPSs. Uses of Cross Product
242-535 ADA: 15. Basic Math14 Suppose we have three vectors a, b, c which form a 3D figure like so: Geometric Properties b a c
242-535 ADA: 15. Basic Math15 Area of the parallelogram of the front face of the object is: area = | a x b | Volume of the parallelepiped (the entire object) is: volume = | a · ( b x c ) | If the volume == 0 then it must mean that all three vectors lie in the same plane o in terms of the box diagram, it means that |a| is 0 units above the b x c face Area and Volume
242-535 ADA: 15. Basic Math17 A basic geometric function: o Returns 1 if the points are in counter-clockwise order (ccw) o Returns -1 if the points are in clockwise order (cw) o Returns 0 if the points are collinear CCW() utilises cross product. 4. The CCW() Function 2 0 1 ccw(p0, p1, p2) returns 1
Turning of Consecutive Segments Counterclockwise Clockwise No turn (collinear) p p p 0 1 2 p p p 0 1 2 pp p 0 12 p p p p > 0 0 1 0 2 p p p p < 0 0 1 0 2 p p p p = 0 0 1 0 2 Segments p p and p p. Move from p to p then to p. 0 1 1 2 0 1 2
242-535 ADA: 15. Basic Math19 5. Line Segment Intersection p4 p3 p1 p2 Boundary cases must be considered.
Using Cross Products p p 1 2 p p 3 4 Two line segments intersect iff each of the two pairs of cross products below have different signs (or one cross product in the pair is 0). (p p ) ( p p ) and ( p p ) ( p p ) 1 4 3 4 2 4 3 4 3 2 1 2 4 2 1 2 p 1 p 2 p 3 p 4 3 1 // the line through p p // intersects p p. 4 23 // the line through p p // intersects p p. 4 12
242-535 ADA: 15. Basic Math22 SEGMENTS-INTERSECT(p 1, p 2, p 3, p 4 ) 1 d 1 = CCW(p 3, p 4, p 1 ) 2 d 2 = CCW(p 3, p 4, p 2 ) 3 d 3 = CCW(p 1, p 2, p 3 ) 4 d 4 = CCW(p 1, p 2, p 4 ) 5 if ((d 1 > 0 and d 2 0)) && ((d 3 > 0 and d 4 0)) 6 return true 7 elseif d 1 = 0 and ON-SEGMENT(p 3, p 4, p 1 ) 8 return true 9 elseif d 2 = 0 and ON-SEGMENT(p 3, p 4, p 2 ) 10 return true 11 elseif d 3 = 0 and ON-SEGMENT(p 1, p 2, p 3 ) 12 return true 13 elseif d 4 = 0 and ON-SEGMENT(p 1, p 2, p 4 ) 14 return true 15 else return falseCode lines straddle check boundary cases
242-535 ADA: 15. Basic Math23 CCW(p i, p j, p k ) // called DIRECTION in CLRS return (p k - p i ) × (p j - p i ) ON-SEGMENT(p i, p j, p k ) 1 if min(x i, x j ) ≤ x k ≤ max(x i, x j ) && min(y i, y j ) ≤ y k ≤ max(y i, y j ) 2 return true 3 else return false
242-535 ADA: 15. Basic Math24 Intersection Cases
242-535 ADA: 15. Basic Math26 Given a point and a line or line segment, determine the shortest distance between them. 6. Distance Between a Point and a Line P3 P1 P2 h The equation of a line defined through two points P1 (x1,y1) and P2 (x2,y2) is P = P1 + u (P2 - P1) using the dot product
242-535 ADA: 15. Basic Math27 The point P3 (x3, y3) is closest to the line at the tangent to the line which passes through P3, that is, the dot product of the tangent and line is 0, thus ( P3 - P ) ( P2 - P1 ) = 0 Substituting the equation of the line gives: [ P3 - P1 - u( P2 - P1 )] ( P2 - P1 ) = 0 Solving this gives the value of u: Uses a (b+c) = (a b) + (a c)
242-535 ADA: 15. Basic Math28 Substituting this into the equation of the line gives the point of intersection (x, y) of the tangent as: x = x1 + u (x2 - x1) y = y1 + u (y2 - y1) The distance therefore between the point P3 and the line is the distance between (x, y) and P3.
242-535 ADA: 15. Basic Math29 double distToSegment(Point2D p1, Point2D p2, Point2D p3) { double xDelta = p2.getX() - p1.getX(); double yDelta = p2.getY() - p1.getY(); if ((xDelta == 0) && (yDelta == 0)) return -1; // p1 and p2 are the same point double u = ((p3.getX() - p1.getX()) * xDelta + (p3.getY() - p1.getY()) * yDelta) / (xDelta * xDelta + yDelta * yDelta); Point2D closePt; // the (x,y) of the previous slide if (u < 0) closePt = p1; else if (u > 1) closePt = p2; else closePt = new Point2D.Double(p1.getX() + u * xDelta, p1.getY() + u * yDelta); return closePt.distance(p3); }Code
242-535 ADA: 15. Basic Math30 The coordinates (x1, y1), (x2, y2), (x3, y3),..., (xn, yn) of a convex polygon are arranged as shown. They must be in counter-clockwise order around the polygon, beginning and ending at the same point. 7. Area of a Convex Polygon (x 1, y 1 ) (x n, y n ) (x 2, y 2 ) … …
242-535 ADA: 15. Basic Math31 Example Can be proved using induction on area defined in terms of triangles.
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# Samacheer Kalvi 12th Business Maths Solutions Chapter 3 Integral Calculus II Ex 3.1
Students can download 12th Business Maths Chapter 3 Integral Calculus II Ex 3.1 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.
## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 3 Integral Calculus II Ex 3.1
Question 1.
Using Integration, find the area of the region bounded the line 2y + x = 8, the x-axis and the lines x = 2, x = 4
Solution:
The given lines are 2y + x = 8, x-axis, x = 2, x = 4
Question 2.
Find the area bounded by the lines y – 2x – 4 = 0, y = 1, y = 3 and the y-axis.
Solution:
Given lines are y – 2x – 4 = 0, y = 1, y = 3, y-axis
y – 2x = 4 = 0 ⇒ x = $$\frac{y-4}{2}$$
We observe that the required area lies to the left to the y-axis
Question 3.
Calculate the area bounded by the parabola y2 = 4ax and its latus rectum.
Solution:
The latus rectum of the parabola is the line x = a through its focus (a, 0) perpendicular to the x-axis.
Equation of parabola
y² = 4ax
y = $$\sqrt { 4ax}$$ ⇒ y = 2√a √x
Required area = 2[Area in the first quadrant between the limits x = 0 and x = a]
Question 4.
Find the area bounded by the line y = x, the x-axis and the ordinates x = 1, x = 2.
Solution:
Given lines are y = x, x-axis, x = 1, x = 2
Question 5.
Using integration, find the area of the region bounded by the line y – 1 = x, the x axis and the ordinates x = -2, x = 3
Solution:
Given lines are y – 1 = x, x-axis, x = -2, x = 3
Question 6.
Find the area of the region lying in the first quadrant bounded by the region y = 4x2, x = 0, y = 0 and y = 4.
Solution:
The given parabola is y = 4x2
$$x^{2}=\frac{y}{4}$$
comparing with the standared form x2 = 4ay
$$4 a=\frac{1}{4} \Rightarrow a=\frac{1}{16}$$
The parabola is symmetric about y-axis
We require the area in the first quadrant.
Question 7.
Find the area bounded by the curve y = x2 and the line y = 4
Solution:
Given the parabola is y = x2 and line y = 4
The parabola is symmetrical about the y-axis.
So required area = 2 [Area in the first quadrant between limits y = 0 and y = 4] |
# 微分方程学代写|DIFFERENTIAL EQUATIONS MATH221 University of Liverpool Assignment
Assignment-daixieTM为您提供利物浦大学University of Liverpool DIFFERENTIAL EQUATIONS MATH221微分方程学代写代考辅导服务！
## Instructions:
Differential equations are indeed fundamental to many areas of mathematics and science, including physics, engineering, economics, and biology, just to name a few. They allow us to model and analyze complex systems and phenomena that are too difficult or impossible to understand using algebraic or geometric techniques alone.
The five parts of the module you mentioned seem to cover a broad range of topics in differential equations, from basic first-order ODEs to more advanced PDEs. It’s great that the module emphasizes both theory and applications, as both are important for understanding and using differential equations effectively.
Solving differential equations can be a difficult and sometimes frustrating task, as there are often many different methods and techniques that can be used, and the solutions can be complex and difficult to interpret. However, with practice and patience, it’s possible to become proficient in solving and analyzing these equations.
Overall, it sounds like MATH201 will be a valuable and challenging module for students who are interested in pursuing further studies in mathematics, science, or engineering. Good luck to all who take it!
In (a)-(c) we consider the autonomous equation $\dot{x}=2 x-3 x^2+x^3$.
(a) Sketch the phase line of this equation.
(a) The equation is $\dot{x}=x(x-1)(x-2)$. The phase line has three equilibria $x=0,1,2$. For $x<0$, the arrow points down. For $02$, the arrow points up.
(b) Sketch the graphs of some solutions. Be sure to include at least one solution with values in each interval above, below, and between the critical points.
(b) The horizontal axis is $t$ and the vertical axis is $x$. There are three constant solutions $x(t) \equiv 0,1,2$. Their graphs are horizontal. Below $x=0$, all solutions are decreasisng and they tend to $-\infty$.
Between $x=0$ and $x=1$, all solutions are increasing and they approach $x=1$. Between $x=1$ and $x=2$, all solutions are decreasing and they approach $x=1$. Above $x=2$, all solutions are increasing and they tend to $+\infty$.
(c) Some solutions have points of inflection. What are the possible values of $x(a)$ if a nonconstant solution $x(t)$ has a point of inflection at $t=a$ ?
(c) A point of inflection $(a, x(a))$ is where $\ddot{x}$ changes sign. In particular, $\ddot{x}(a)$ must be zero. Differentiating the given equation with respect to $t$, we have
$$\ddot{x}=2 \dot{x}-6 x \dot{x}+3 x^2 \dot{x}=\dot{x}\left(2-6 x+3 x^2\right) \text {. }$$
If $x(t)$ is not a constant solution, $\dot{x}(a) \neq 0$ so that $x(a)$ must satisfy
$$2-6 x(a)+3 x(a)^2=0 \quad \Leftrightarrow \quad x(a)=1 \pm \frac{1}{\sqrt{3}} .$$ |
# Arithmetic Progression
## NCERT Exercise 5.3
### Part 5
Question 7: Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution: We have; n = 22, d = 7 and a22 = 149
We know;
a_n = a + (n – 1)d
Or, 149 = a + 21 xx 7
Or, a = 149 – 147 = 2
The sum can be calculated as follows:
S=n/2[2a+(n-1)d]
=(22)/(2)(2+149)
=11xx151=1661
Question 8: Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution: We have; a2 = 14, a3 = 18 and n = 51
Here;
a_3 – a_2 = 18 – 14 = 4
Hence, d = 4
Or, a_2 – a = 4
Or, 14 – a = 4
Or, a = 10
Now, sum can be calculated as follows:
S=n/2[2a+(n-1)d]
=(51)/(2)(2xx10+50xx4)
=51(10+50xx2)
=51xx110=5610
Thus, sum = 5610
Question 9: If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Solution: Here;
S_7=7/2(2a+6d)=49
Or, 7(a+3d)=49
Or, a+3d=49÷7=7
This is the 4th term
Similarly;
S_(17)=(17)/(2)(2a+16d)=289
Or, 17(a+8d)=289
Or, a+8d=289÷17=17
This is the 9th term
Subtracting 4th term from 9th term, we get;
a + 8d – a – 3d = 17 – 7
Or, 5d = 10
Or, d = 2
Using the value of d in equation for fourth term, we can find a as follows:
a + 3d =7
Or, a + 6 = 7
Or, a = 1
Using the values of a and d; we can find the sum of first n terms as follows:
S=n/2[2a+(n-1)d]
=n/2[2xx1+(n-1)2]
=n(1+n-1)=n^2
Thus, sum of n terms of this AP = n2 |
##### Notes
Daily Math Practice provides a week's worth of problems ideal for morning work, homework, or review. Week 8 supports Common Core Mathematics Standards 1.OA.A.1, 1.OA.B.3, 1.OA.B.4, 1.OA.C.5, and 1.OA.C.6.
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# Grade 1 - Daily Math Practice - Week 8 (Grade 1)
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DAILY MATH PRACTICE - WEEK 8 DAY 1 DAY 2 Three facts of a fact family are shown. Write the missing fact.15 - 6 = 96 + 9 = 1515 - 9 = 6 8 + = 1616 - 8 = Circle the greater number. Count down to subtract.$ul({:(,18),(-,6):})$ Circle the greater number. Count down to subtract.$ul({:(,13),(-,4):})$ Subtract. $ul({:(,2),(-,1):})$ Subtract. $ul({:(,5),(-,3):})$ Grace picks 20 pumpkins on Monday. She picks 9 on Tuesday. How many more pumpkins did Grace pick on Monday than Tuesday? A jug holds 16 cups of cider. Nari pours cups of cider for her friends. There are 6 cups left in the jug. How many cups of cider did Nari pour?
DAILY MATH PRACTICE - WEEK 8 DAY 3 DAY 4 Three facts of a fact family are shown. Write the missing fact.8 + 4 = 1212 - 8 = 44 + 8 = 12 7 + = 1111 - 7 = Circle the greater number. Count down to subtract.$ul({:(,10),(-,5):})$ Circle the greater number. Count down to subtract.$ul({:(,14),(-,6):})$ Subtract. $ul({:(,4),(-,2):})$ Subtract. $ul({:(,3),(-,2):})$ There are 20 pencils in a box. Alex sharpens 8. How many pencils still need to be sharpened? There 14 pairs of boots on a shelf. 5 pairs are green. The rest are red. How many pairs of red boots are on the shelf?
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Question
# If $\cos ec\theta = \dfrac{{p + q}}{{p - q}}$, then $\cot \left( {\dfrac{\pi }{4} + \dfrac{\theta }{2}} \right) =$A) $\sqrt {\dfrac{q}{p}}$B) $\sqrt {\dfrac{p}{q}}$C) $pq$D) $\sqrt {pq}$
Hint: Here we will first use the following identity:-
$\cos ec\theta = \dfrac{1}{{\sin \theta }}$ then we will use componendo dividendo and then finally use the following identities to get the desired answer.
$\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} \\ \cot \theta = \dfrac{1}{{\tan \theta }} \\$
The given equation is:-
$\cos ec\theta = \dfrac{{p + q}}{{p - q}}$
Now applying the following identity
$\cos ec\theta = \dfrac{1}{{\sin \theta }}$
We get:-
$\dfrac{1}{{\sin \theta }} = \dfrac{{p + q}}{{p - q}}$
Now applying componendo and dividendo we get:-
$\dfrac{{1 + \sin \theta }}{{1 - \sin \theta }} = \dfrac{{p + q + p - q}}{{p + q - \left( {p - q} \right)}}$
Solving it further we get:-
$\dfrac{{1 + \sin \theta }}{{1 - \sin \theta }} = \dfrac{{2p}}{{p + q - p + q}} \\ \Rightarrow \dfrac{{1 + \sin \theta }}{{1 - \sin \theta }} = \dfrac{{2p}}{{2q}} \\ \Rightarrow \dfrac{{1 + \sin \theta }}{{1 - \sin \theta }} = \dfrac{p}{q} \\$
Now we know that:-
$\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$
Hence substituting the value we get:-
$\dfrac{{1 + 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}{{1 - 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}} = \dfrac{p}{q}$……………………………….(1)
Now we know that:-
${\sin ^2}\theta + {\cos ^2}\theta = 1$
Hence, ${\sin ^2}\dfrac{\theta }{2} + {\cos ^2}\dfrac{\theta }{2} = 1$
Hence substituting this value in equation1 we get:-
$\dfrac{{{{\sin }^2}\dfrac{\theta }{2} + {{\cos }^2}\dfrac{\theta }{2} + 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}{{{{\sin }^2}\dfrac{\theta }{2} + {{\cos }^2}\dfrac{\theta }{2} - 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}} = \dfrac{p}{q}$
Now using the following identities:-
${\left( {A + B} \right)^2} = {A^2} + {B^2} + 2AB \\ {\left( {A - B} \right)^2} = {A^2} + {B^2} - 2AB \\$
We get:-
$\dfrac{{{{\left( {\sin \dfrac{\theta }{2} + \cos \dfrac{\theta }{2}} \right)}^2}}}{{{{\left( {\sin \dfrac{\theta }{2} - \cos \dfrac{\theta }{2}} \right)}^2}}} = \dfrac{p}{q}$
Solving it further we get:-
$\dfrac{{{{\left( {\sin \dfrac{\theta }{2} + \cos \dfrac{\theta }{2}} \right)}^2}}}{{{{\left( {\cos \dfrac{\theta }{2} - \sin \dfrac{\theta }{2}} \right)}^2}}} = \dfrac{p}{q}$
Now taking ${\cos ^2}\dfrac{\theta }{2}$ common from numerator and denominator we get:-
${\left( {\dfrac{{\dfrac{{\cos \dfrac{\theta }{2}}}{{\cos \dfrac{\theta }{2}}} + \dfrac{{\sin \dfrac{\theta }{2}}}{{\cos \dfrac{\theta }{2}}}}}{{\dfrac{{\cos \dfrac{\theta }{2}}}{{\cos \dfrac{\theta }{2}}} - \dfrac{{\sin \dfrac{\theta }{2}}}{{\cos \dfrac{\theta }{2}}}}}} \right)^2} = \dfrac{p}{q}$
Now we know that:
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
Hence, $\tan \dfrac{\theta }{2} = \dfrac{{\sin \dfrac{\theta }{2}}}{{\cos \dfrac{\theta }{2}}}$
Therefore substituting the values we get:-
${\left( {\dfrac{{1 + \tan \dfrac{\theta }{2}}}{{1 - \tan \dfrac{\theta }{2}}}} \right)^2} = \dfrac{p}{q}$
Now we know that
$\tan \dfrac{\pi }{4} = 1$
Hence substituting the value we get:-
${\left( {\dfrac{{\tan \dfrac{\pi }{4} + \tan \dfrac{\theta }{2}}}{{\tan \dfrac{\pi }{4} - \tan \dfrac{\theta }{2}}}} \right)^2} = \dfrac{p}{q}$
Now we know that
$\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
Hence applying this identity in above equation we get:-
${\left[ {\tan \left( {\dfrac{\pi }{4} + \dfrac{\theta }{2}} \right)} \right]^2} = \dfrac{p}{q}$
Now taking square root of both the sides we get:-
$\sqrt {{{\left[ {\tan \left( {\dfrac{\pi }{4} + \dfrac{\theta }{2}} \right)} \right]}^2}} = \sqrt {\dfrac{p}{q}}$
Simplifying it further we get:-
$\tan \left( {\dfrac{\pi }{4} + \dfrac{\theta }{2}} \right) = \sqrt {\dfrac{p}{q}}$……………………….(2)
Now we know that
$\cot \theta = \dfrac{1}{{\tan \theta }}$
Hence we will take reciprocal of equation 2 and then apply this identity to get desired value:-
$\dfrac{1}{{\tan \left( {\dfrac{\pi }{4} + \dfrac{\theta }{2}} \right)}} = \sqrt {\dfrac{q}{p}}$
Now applying the identity we get:-
$\cot \left( {\dfrac{\pi }{4} + \dfrac{\theta }{2}} \right) = \sqrt {\dfrac{q}{p}}$
Therefore option B is the correct option.
Note: Students might make mistake in making the squares of the quantities using the identities:
${\left( {A + B} \right)^2} = {A^2} + {B^2} + 2AB \\ {\left( {A - B} \right)^2} = {A^2} + {B^2} - 2AB \\$
In such questions we need to use the given information and then then transform it into required form to get the desired answer. |
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 1.2: Real Numbers - Algebra Essentials
[ "article:topic", "exponent", "identity property of addition", "identity property of multiplication", "inverse property of addition", "distributive property", "Exponential Notation", "authorname:openstax", "algebraic expression", "associative property of addition", "associative property of multiplication", "base", "commutative property of addition", "constant", "equation", "formula", "integers", "inverse property of multiplication", "irrational numbers", "natural numbers", "order of operations", "rational numbers", "real number line", "real numbers", "variable", "whole numbers" ]
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
Skills to Develop
• Classify a real number as a natural, whole, integer, rational, or irrational number.
• Perform calculations using order of operations.
• Use the following properties of real numbers: commutative, associative, distributive, inverse, and identity.
• Evaluate algebraic expressions.
• Simplify algebraic expressions.
It is often said that mathematics is the language of science. If this is true, then the language of mathematics is numbers. The earliest use of numbers occurred 100 centuries ago in the Middle East to count, or enumerate items. Farmers, cattlemen, and tradesmen used tokens, stones, or markers to signify a single quantity—a sheaf of grain, a head of livestock, or a fixed length of cloth, for example. Doing so made commerce possible, leading to improved communications and the spread of civilization.
Three to four thousand years ago, Egyptians introduced fractions. They first used them to show reciprocals. Later, they used them to represent the amount when a quantity was divided into equal parts.
But what if there were no cattle to trade or an entire crop of grain was lost in a flood? How could someone indicate the existence of nothing? From earliest times, people had thought of a “base state” while counting and used various symbols to represent this null condition. However, it was not until about the fifth century A.D. in India that zero was added to the number system and used as a numeral in calculations.
Clearly, there was also a need for numbers to represent loss or debt. In India, in the seventh century A.D., negative numbers were used as solutions to mathematical equations and commercial debts. The opposites of the counting numbers expanded the number system even further.
Because of the evolution of the number system, we can now perform complex calculations using these and other categories of real numbers. In this section, we will explore sets of numbers, calculations with different kinds of numbers, and the use of numbers in expressions.
### Classifying a Real Number
The numbers we use for counting, or enumerating items, are the natural numbers: 1, 2, 3, 4, 5, and so on. We describe them in set notation as {1,2,3,...}where the ellipsis (…) indicates that the numbers continue to infinity. The natural numbers are, of course, also called the counting numbers. Any time we enumerate the members of a team, count the coins in a collection, or tally the trees in a grove, we are using the set of natural numbers. The set of whole numbers is the set of natural numbers plus zero: {0,1,2,3,...}.
The set of integers adds the opposites of the natural numbers to the set of whole numbers: {...,−3,−2,−1,0,1,2,3,...}.It is useful to note that the set of integers is made up of three distinct subsets: negative integers, zero, and positive integers. In this sense, the positive integers are just the natural numbers. Another way to think about it is that the natural numbers are a subset of the integers.
negative integers zero positive integers
…,−3,−2,−1, 0, 1,2,3,
The set of rational numbers is written as {mnm and n are integers and n0}.Notice from the definition that rational numbers are fractions (or quotients) containing integers in both the numerator and the denominator, and the denominator is never 0. We can also see that every natural number, whole number, and integer is a rational number with a denominator of 1.
Because they are fractions, any rational number can also be expressed in decimal form. Any rational number can be represented as either:
1. a terminating decimal:$$\frac{15}{8}$$ =1.875,or
2. a repeating decimal:$$\frac{4}{11}$$ =0.36363636=0.$$\overline{36}$$
We use a line drawn over the repeating block of numbers instead of writing the group multiple times.
Example
Writing Integers as Rational Numbers
Write each of the following as a rational number.
1. 7
2. 0
3. –8
Solution
Write a fraction with the integer in the numerator and 1 in the denominator.
1. $$7= \frac{7}{1}$$
2. $$0= \frac{0}{1}$$
3. $$−8= \frac{8}{1}$$
Exercise
Write each of the following as a rational number.
1. 11
2. 3
3. –4
Solution
1. $$\frac{11}{1}$$
2. $$\frac{3}{1}$$
3. $$\frac{4}{1}$$
Example
Identifying Rational Numbers
Write each of the following rational numbers as either a terminating or repeating decimal.
1. $$\frac{5}{7}$$
2. $$\frac{15}{5}$$
3. $$\frac{13}{25}$$
Solution
Write each fraction as a decimal by dividing the numerator by the denominator.
1. $$\frac{5}{7}$$ =−0.$$\overline{714285}$$ ,a repeating decimal
2. $$\frac{15}{5}$$ =3(or 3.0), a terminating decimal
3. $$\frac{13}{25}$$ =0.52, a terminating decimal
Exercise
Write each of the following rational numbers as either a terminating or repeating decimal.
1. $$\frac{68}{17}$$
2. $$\frac{8}{13}$$
3. $$\frac{13}{25}$$
Solution
1. 4 (or 4.0), terminating;
2. 0.$$\overline{615384}$$ ,repeating;
3. –0.85, terminating
#### Irrational Numbers
At some point in the ancient past, someone discovered that not all numbers are rational numbers. A builder, for instance, may have found that the diagonal of a square with unit sides was not 2 or even 32 ,but was something else. Or a garment maker might have observed that the ratio of the circumference to the diameter of a roll of cloth was a little bit more than 3, but still not a rational number. Such numbers are said to be irrational because they cannot be written as fractions. These numbers make up the set of irrational numbers. Irrational numbers cannot be expressed as a fraction of two integers. It is impossible to describe this set of numbers by a single rule except to say that a number is irrational if it is not rational. So we write this as shown.
{h|h is not a rational number}
Example
Differentiating Rational and Irrational Numbers
Determine whether each of the following numbers is rational or irrational. If it is rational, determine whether it is a terminating or repeating decimal.
1. $$\sqrt{25}$$
2. $$\frac{33}{9}$$
3. $$\sqrt{11}$$
4. $$\frac{17}{34}$$
5. 0.3033033303333
Solution
1. $$\sqrt{25}$$ :This can be simplified as $$\sqrt{25}$$= 5.Therefore,$$\sqrt{25}$$is rational.
2. $$\frac{33}{9}$$ :Because it is a fraction,$$\frac{33}{9}$$is a rational number. Next, simplify and divide. $$\frac{33}{9}$$=$$\cancel{frac{33}{9}}$$So,339is rational and a repeating decimal.
3. $$\sqrt{11}$$ :This cannot be simplified any further. Therefore,$$\sqrt{11}$$is an irrational number.
4. $$\frac{17}{34}$$:Because it is a fraction,$$\frac{17}{34}$$is a rational number. Simplify and divide. 1734=17<svg style="width: 1.61em; height: 1.313em;"><g fill="none" stroke="black" stroke-width="1.4175"><line x1="1" x2="25.7775" y1="20.468700000000002" y2="1.4175"/></g></svg>134<svg style="width: 1.61em; height: 1.313em;"><g fill="none" stroke="black" stroke-width="1.4175"><line x1="1" x2="25.7775" y1="20.5695" y2="1.4175"/></g></svg>2=12=0.5So,$$\frac{17}{34}$$is rational and a terminating decimal.
5. 0.3033033303333is not a terminating decimal. Also note that there is no repeating pattern because the group of 3s increases each time. Therefore it is neither a terminating nor a repeating decimal and, hence, not a rational number. It is an irrational number.
Exercise
Determine whether each of the following numbers is rational or irrational. If it is rational, determine whether it is a terminating or repeating decimal.
1. $$\frac{7}{77}$$
2. $$\sqrt{81}$$
3. 4.27027002700027
4. $$\frac{91}{13}$$
5. $$\sqrt{39}$$
Solution
1. rational and repeating;
2. rational and terminating;
3. irrational;
4. rational and repeating;
5. irrational
#### Real Numbers
Given any number n, we know that n is either rational or irrational. It cannot be both. The sets of rational and irrational numbers together make up the set of real numbers. As we saw with integers, the real numbers can be divided into three subsets: negative real numbers, zero, and positive real numbers. Each subset includes fractions, decimals, and irrational numbers according to their algebraic sign (+ or –). Zero is considered neither positive nor negative.
The real numbers can be visualized on a horizontal number line with an arbitrary point chosen as 0, with negative numbers to the left of 0 and positive numbers to the right of 0. A fixed unit distance is then used to mark off each integer (or other basic value) on either side of 0. Any real number corresponds to a unique position on the number line.The converse is also true: Each location on the number line corresponds to exactly one real number. This is known as a one-to-one correspondence. We refer to this as the real number line as shown in Figure(1.2.1)
Figure 1.2.1 The real number line
Example
Classifying Real Numbers
Classify each number as either positive or negative and as either rational or irrational. Does the number lie to the left or the right of 0 on the number line?
1. -$$\frac{10}{3}$$
2. -$$\sqrt{5}$$
3. −6π
4. 0.615384615384
Solution
1. -$$\frac{10}{3}$$is negative and rational. It lies to the left of 0 on the number line.
2. -$$\sqrt{5}$$is positive and irrational. It lies to the right of 0.
3. $$-\sqrt{289}=-\sqrt{17^2}=−17$$is negative and rational. It lies to the left of 0.
4. −6πis negative and irrational. It lies to the left of 0.
5. 0.615384615384is a repeating decimal so it is rational and positive. It lies to the right of 0.
Exercise
Classify each number as either positive or negative and as either rational or irrational. Does the number lie to the left or the right of 0 on the number line?
1. $$\sqrt{73}$$
2. −11.411411411
3. $$\frac{47}{19}$$
4. -$$\frac{\sqrt{5}}{2}$$
5. 6.210735
Solution
1. positive, irrational; right
2. negative, rational; left
3. positive, rational; right
4. negative, irrational; left
5. positive, rational; right
#### Sets of Numbers as Subsets
Beginning with the natural numbers, we have expanded each set to form a larger set, meaning that there is a subset relationship between the sets of numbers we have encountered so far. These relationships become more obvious when seen as a diagram, such as Figure(1.2.2).
Figure(1.2.2) :Sets of numbers
N: the set of natural numbers
W: the set of whole numbers
I: the set of integers
Q: the set of rational numbers
Q´: the set of irrational numbers
Note
SETS OF NUMBERS
The set of natural numbers includes the numbers used for counting: {1,2,3,...}.
The set of whole numbers is the set of natural numbers plus zero: {0,1,2,3,...}.
The set of integers adds the negative natural numbers to the set of whole numbers: {...,−3,−2,−1,0,1,2,3,...}.
The set of rational numbers includes fractions written as {mn∣∣m and n are integers and n0}.
The set of irrational numbers is the set of numbers that are not rational, are nonrepeating, and are nonterminating: {h∣∣h is not a rational number}
Example
Differentiating the Sets of Numbers
Classify each number as being a natural number (N), whole number (W), integer (I), rational number (Q), and/or irrational number (Q′).
1. $$\sqrt{36}$$
2. $$\frac{8}{3}$$
3. $$\sqrt{73}$$
4. −6
5. 3.2121121112
Solution
N W I Q Q′
a.$$\sqrt{36}$$ =6 X X X X
b.$$\frac{8}{3}$$ =2. $$\overline{6}$$ X
c.$$\sqrt{73}$$ X
d. –6 X X
e. 3.2121121112... X
Exercise
Classify each number as being a natural number (N), whole number (W), integer (I), rational number (Q), and/or irrational number (Q′).
1. -$$\frac{35}{7}$$
2. 0
3. $$\sqrt{169}$$
4. $$\sqrt{24}$$
5. 4.763763763
Solution
N W I Q Q'
a.-$$\frac{35}{7}$$ X X
b. 0 X X X
c.$$\sqrt{169}$$ X X X X
d.$$\sqrt{24}$$ X
e. 4.763763763... X
Exercise
Classify each number as being a natural number (N), whole number (W), integer (I), rational number (Q), and/or irrational number (Q′).
1. -$$\frac{35}{7}$$
2. 0
3. $$\sqrt{169}$$
4. $$\sqrt{24}$$
5. 4.763763763
Solution:
N W I Q Q'
a.-$$\frac{35}{7}$$ X X
b. 0 X X X
c.$$\sqrt{169}$$ X X X X
d.$$\sqrt{24}$$ X
e. 4.763763763... X
### Performing Calculations Using the Order of Operations
When we multiply a number by itself, we square it or raise it to a power of 2. For example,$$4^2 =4\times4=16$$.We can raise any number to any power. In general, the exponential notation an means that the number or variable a is used as a factor n times.
$a^n=a⋅a⋅a⋅…⋅a$ n factors
In this notation, an is read as the nth power of a , where a is called the base and n is called the exponent. A term in exponential notation may be part of a mathematical expression, which is a combination of numbers and operations. For example, 24+6⋅23−42 is a mathematical expression.
To evaluate a mathematical expression, we perform the various operations. However, we do not perform them in any random order. We use the order of operations. This is a sequence of rules for evaluating such expressions.
Recall that in mathematics we use parentheses ( ), brackets [ ], and braces { } to group numbers and expressions so that anything appearing within the symbols is treated as a unit. Additionally, fraction bars, radicals, and absolute value bars are treated as grouping symbols. When evaluating a mathematical expression, begin by simplifying expressions within grouping symbols.
The next step is to address any exponents or radicals. Afterward, perform multiplication and division from left to right and finally addition and subtraction from left to right.
Let’s take a look at the expression provided.
$24+6 \times \dfrac{2}{3} − 4^2$
There are no grouping symbols, so we move on to exponents or radicals. The number 4 is raised to a power of 2, so simplify 42 as 16.
$24+6 \times \dfrac{2}{3} − 4^2$
$24+6 \times \dfrac{2}{3} − 16$
Next, perform multiplication or division, left to right.
$24+6 \times \dfrac{2}{3} − 16$
$24+4-16$
Lastly, perform addition or subtraction, left to right.
$24+4−16$
$28−16$
$12$
Therefore,$24+6 \times \dfrac{2}{3} − 4^2 =12$
For some complicated expressions, several passes through the order of operations will be needed. For instance, there may be a radical expression inside parentheses that must be simplified before the parentheses are evaluated. Following the order of operations ensures that anyone simplifying the same mathematical expression will get the same result.
Note
ORDER OF OPERATIONS
Operations in mathematical expressions must be evaluated in a systematic order, which can be simplified using the acronym PEMDAS:
P(arentheses)
E(xponents)
M(ultiplication) and D(ivision)
A(ddition) and S(ubtraction)
HOW TO: Given a mathematical expression, simplify it using the order of operations.
1. Simplify any expressions within grouping symbols.
2. Simplify any expressions containing exponents or radicals.
3. Perform any multiplication and division in order, from left to right.
4. Perform any addition and subtraction in order, from left to right.
Example
Using the Order of Operations
Use the order of operations to evaluate each of the following expressions.
1. $$(3+2)^2+4 \times(6+2)$$
2. $$\dfrac{5^2-4}{7}- \sqrt{11-2}$$
3. $$6−|5−8|+3\times(4−1)$$
4. $$\dfrac{14-3 \times2}{2 \times5-3^2}$$
5. $$7\times(5\times3)−2\times[(6−3)−4^2]+1$$
Solution
a. $$(3\times2)^2-4\times(6+2)=(6)^2-4\times(8)$$ Simplify parentheses
$$=36-4\times8$$ Simplify exponent
$$=36-32$$ Simplify multiplication
$$=4$$ Simplify subtraction
b. $$\dfrac{5^2-4}{7}- \sqrt{11-2}= \dfrac{5^2-4}{7}-\sqrt{9}$$ Simplify grouping symbols (radical)
$$=\dfrac{5^2-4}{7}-3$$ Simplify radical
$$=\dfrac{25-4}{7}-3$$ Simplify exponent
$$=\dfrac{21}{7}-3$$ Simplify subtraction in numerator
$$=3-3$$ Simplify division
$$=0$$ Simplify subtraction
Note that in the first step, the radical is treated as a grouping symbol, like parentheses. Also, in the third step, the fraction bar is considered a grouping symbol so the numerator is considered to be grouped.
c. $$6−|5−8|+3\times(4−1)=6-|-3|+3\times3$$ Simplify inside grouping symbols
$$=6-3+3\times3$$ Simplify absolute value
$$=6-3+9$$ Simplify multiplication
$$=3+9$$ Simplify subtraction
$$=12$$ Simplify addition
d. $$\dfrac{14-3 \times2}{2 \times5-3^2}=\dfrac{14-3 \times2}{2 \times5-9}$$ Simplify exponent
$$=\dfrac{14-6}{10-9}$$ Simplify products
$$=\dfrac{8}{1}$$ Simplify differences
$$=8$$ Simplify quotient
In this example, the fraction bar separates the numerator and denominator, which we simplify separately until the last step.
e. $$7\times(5\times3)−2\times[(6−3)−4^2]+1=7\times(15)-2\times[(3)-4^2]+1$$ Simplify inside parentheses
$$=7\times(15)-2\times(3-16)+1$$ Simplify exponent
$$=7\times(15)-2\times(-13)+1$$ Subtract
$$=105+26+1$$ Multiply
$$=132$$ Add
Exercise
Use the order of operations to evaluate each of the following expressions.
a. $$\sqrt{5^2-4^2}+7\times(5-4)^2$$
b. $$1+\dfrac{7\times5-8\times4}{9-6}$$
c. $$|1.8-4.3|+0.4\times\sqrt{15+10}$$
d. $$\dfrac{1}{2}\times[5\times3^2-7^2]+\dfrac{1}{3}\times9^2$$
e. $$[(3-8^2)-4]-(3-8)$$
Solution
a. 10
b. 2
c. 4.5
d. 25
e. 26
### Using Properties of Real Numbers
For some activities we perform, the order of certain operations does not matter, but the order of other operations does. For example, it does not make a difference if we put on the right shoe before the left or vice-versa. However, it does matter whether we put on shoes or socks first. The same thing is true for operations in mathematics.
#### Commutative Properties
The commutative property of addition states that numbers may be added in any order without affecting the sum.
$a+b=b+a$
We can better see this relationship when using real numbers.
$$(−2)+7=5$$ and $$7+(−2)=5$$
Similarly, the commutative property of multiplication states that numbers may be multiplied in any order without affecting the product.
$a\times b=b\times a$
Again, consider an example with real numbers.
$(−11)\times(−4)=44$and$(−4)\times(−11)=44$
It is important to note that neither subtraction nor division is commutative. For example, $$17−5$$ is not the same as $$5−17$$. Similarly, $$20÷5≠5÷20$$.
#### Associative Properties
The associative property of multiplication tells us that it does not matter how we group numbers when multiplying. We can move the grouping symbols to make the calculation easier, and the product remains the same.
$a(bc)=(ab)c$
Consider this example.
$$(3\times4)\times5=60$$and$$3\times(4\times5)=60$$
The associative property of addition tells us that numbers may be grouped differently without affecting the sum.
$a+(b+c)=(a+b)+c$
This property can be especially helpful when dealing with negative integers. Consider this example.
$$[15+(−9)]+23=29$$and$$15+[(−9)+23]=29$$
Are subtraction and division associative? Review these examples.
$8-(3-15)\overset{?}{=}(8-3)-15$
$8-(-12)\overset{?}{=}5-15$
$20≠20-10$
$64÷(8÷4)\overset{?}{=}(64÷8)÷4$
$64÷2\overset{?}{=}8÷4$
$32≠2$
As we can see, neither subtraction nor division is associative.
#### Distributive Property
The distributive property states that the product of a factor times a sum is the sum of the factor times each term in the sum.
$a\times(b+c)=a\times b+a\times c$
This property combines both addition and multiplication (and is the only property to do so). Let us consider an example.
Note that 4 is outside the grouping symbols, so we distribute the 4 by multiplying it by 12, multiplying it by –7, and adding the products.
Example
To be more precise when describing this property, we say that multiplication distributes over addition. The reverse is not true, as we can see in this example.
$6+(3\times5)\overset{?}{=}(6+3)\times(6\times5)$
$6+(15)\overset{?}{=}(9)\times(11)$
$21≠99$
Multiplication does not distribute over subtraction, and division distributes over neither addition nor subtraction.
A special case of the distributive property occurs when a sum of terms is subtracted.
$a−b=a+(−b)$
For example, consider the difference $$12−(5+3)$$ . We can rewrite the difference of the two terms $$12$$ and $$(5+3)$$ by turning the subtraction expression into addition of the opposite. So instead of subtracting $$(5+3)$$ , we add the opposite.
$12+(−1)\times(5+3)$
Now, distribute −1 and simplify the result.
$12−(5+3)=12+(−1)\times(5+3)$
$=12+[(-1)\times5+(-1)\times3]$
$=12+(-8)$
$=4$
This seems like a lot of trouble for a simple sum, but it illustrates a powerful result that will be useful once we introduce algebraic terms. To subtract a sum of terms, change the sign of each term and add the results. With this in mind, we can rewrite the last example.
$12−(5+3)=12+(-5-3)$
$=12-8$
$=4$
#### Identity Properties
The identity property of addition states that there is a unique number, called the additive identity (0) that, when added to a number, results in the original number.
$a+0=a$
The identity property of multiplication states that there is a unique number, called the multiplicative identity (1) that, when multiplied by a number, results in the original number.
$a\times 1=a$
For example, we have $$(−6)+0=−6$$ and$$23\times1=23$$ . There are no exceptions for these properties; they work for every real number, including 0 and 1.
#### Inverse Properties
The inverse property of addition states that, for every real number a, there is a unique number, called the additive inverse (or opposite), denoted−a, that, when added to the original number, results in the additive identity, 0.
$a+(−a)=0$
For example, if $$a =−8$$, the additive inverse is $$8$$, since $$(−8)+8=0$$.
The inverse property of multiplication holds for all real numbers except 0 because the reciprocal of 0 is not defined. The property states that, for every real number a, there is a unique number, called the multiplicative inverse (or reciprocal), denoted 1a , that, when multiplied by the original number, results in the multiplicative identity, 1.
$a\times \dfrac{1}{a}=1$
For example, if $$a =−\dfrac{2}{3}$$, the reciprocal, denoted $$\dfrac{1}{a}$$ , is $$-\dfrac{3}{2}$$ because
$a⋅\dfrac{1}{a}=(−\dfrac{2}{3})\times(−\dfrac{3}{2})=1$
Note
PROPERTIES OF REAL NUMBERS
The following properties hold for real numbers a, b, and c.
Addition Multiplication Commutative Property $$a+b=b+a$$ $$a\times b=b\times a$$ Associative Property $$a+(b+c)=(a+b)+c$$ $$a(bc)=(ab)c$$ Distributive Property $$a\times (b+c)=a\times b+a\times c$$ Identity Property There exists a unique real number called the additive identity, 0, such that, for any real number a $$a+0=a$$ There exists a unique real number called the multiplicative identity, 1, such that, for any real number a $$a\times 1=a$$ Inverse Property Every real number a has an additive inverse, or opposite, denoted –a, such that $$a+(−a)=0$$ Every nonzero real number a has a multiplicative inverse, or reciprocal, denoted 1a , such that $$a\times (\dfrac{1}{a})=1$$
Example
Using Properties of Real Numbers
Use the properties of real numbers to rewrite and simplify each expression. State which properties apply.
a. $$3\times 6+3\times 4$$
b. $$(5+8)+(−8)$$
c. $$6−(15+9)$$
d. $$\dfrac{4}{7}\times(\dfrac{2}{3}\times \dfrac{7}{4})$$
e. $$100\times[0.75+(−2.38)]$$
Solution:
a. $$3\times6+3\times4=3\times(6+4)$$ Distributive property
$$=3\times10$$ Simplify
$$=30$$ Simplify
b. $$(5+8)+(-8)=5+[8+(-8)]$$ Associative property of addition
$$=5+0$$ Inverse property of addition
$$=5$$ Identity property of addition
c. $$6-(15+9)=6+[(-15)+(-9)]$$ Distributive property
$$=6+(-24)$$ Simplify
$$=-18$$ Simplify
d. $$\dfrac{4}{7}\times(\dfrac{2}{3}\times\dfrac{7}{4})=\dfrac{4}{7}\times(\dfrac{7}{4}\times\dfrac{2}{3})$$ Commutative property of multiplication
$$=(\dfrac{4}{7}\times\dfrac{7}{4})\times\dfrac{2}{3}$$ Associative property of multiplication
$$=1\times\dfrac{2}{3}$$ Inverse property of multiplication
$$=\dfrac{2}{3}$$ Identity property of multiplication
e. $$100\times[0.75+(-2.38)]=100\times0.75+100\times(-2.38)$$ Distributive property
$$=75+(-238)$$ Simplify
$$=-163$$ Simplify
Exercise
Use the properties of real numbers to rewrite and simplify each expression. State which properties apply.
a. $$(-\dfrac{23}{5})\times[11\times(-\dfrac{5}{23})]$$
b. $$5\times(6.2+0.4)$$
c. $$18-(7-15)$$
d. $$\dfrac{17}{18}+[\dfrac{4}{9}+(-\dfrac{17}{18})]$$
e. $$6\times(-3)+6\times3$$
Solution:
a. $$11$$, commutative property of multiplication, associative property of multiplication, inverse property of multiplication, identity property of multiplication;
b. $$33$$, distributive property;
c. $$26$$, distributive property;
d. $$\dfrac{4}{9}$$, commutative property of addition, associative property of addition, inverse property of addition, identity property of addition;
e. $$0$$, distributive property, inverse property of addition, identity property of addition
#### Evaluating Algebraic Expressions
So far, the mathematical expressions we have seen have involved real numbers only. In mathematics, we may see expressions such as $$x +5$$, $$\dfrac{4}{3}\pi r^3$$, or $$\sqrt{2m^3 n^2}$$. In the expression $$x +5$$, $$5$$ is called a constant because it does not vary and x is called a variable because it does. (In naming the variable, ignore any exponents or radicals containing the variable.) An algebraic expression is a collection of constants and variables joined together by the algebraic operations of addition, subtraction, multiplication, and division.
We have already seen some real number examples of exponential notation, a shorthand method of writing products of the same factor. When variables are used, the constants and variables are treated the same way.
$(-3)^5 =(-3)\times(-3)\times(-3)\times(-3)\times(-3)$ $x^5=x\times x\times x\times x\times x$
$(2\times7)^3=(2\times7)\times(2\times7)\times(2\times7)$
$(yz)^3=(yz)\times(yz)\times(yz)$
In each case, the exponent tells us how many factors of the base to use, whether the base consists of constants or variables.
Any variable in an algebraic expression may take on or be assigned different values. When that happens, the value of the algebraic expression changes. To evaluate an algebraic expression means to determine the value of the expression for a given value of each variable in the expression. Replace each variable in the expression with the given value, then simplify the resulting expression using the order of operations. If the algebraic expression contains more than one variable, replace each variable with its assigned value and simplify the expression as before.
Example
Describing Algebraic Expressions
List the constants and variables for each algebraic expression.
a. $$x + 5$$
b. $$\dfrac{4}{3}\pi r^3$$
c. $$\sqrt{2m^3 n^2}$$
Constants Variables
a. $$x + 5$$ $$5$$ $$x$$
b. $$\dfrac{4}{3}\pi r^3$$ $$\dfrac{4}{3}$$, $$\pi$$ $$r$$
c. $$\sqrt{2m^3 n^2}$$ $$2$$ $$m$$,$$n$$
Exercise
List the constants and variables for each algebraic expression.
a. $$2\pi r(r+h)$$
b. $$2(L + W)$$
c. $$4y^3+y$$
Constants Variables
a. $$2\pi r(r+h)$$ $$2$$,$$\pi$$ $$r$$,$$h$$
b. $$2(L + W)$$ $$2$$ $$L$$, $$W$$
c. $$4y^3+y$$ $$4$$ $$y$$
Example
Evaluating an Algebraic Expression at Different Values
Evaluate the expression$$2x7$$for each value for $$x$$.
a. $$x=0$$
b. $$x=1$$
c. $$x=12$$
d. $$x=−4$$
Solution:
a. Substitute $$0$$ for x.
$2x-7=2(0)-7$
$=0-7$
$=-7$
b. Substitute $$1$$ for x.
$2x-7=2(1)-7$
$=2-7$
$=-5$
c. Substitute $$\dfrac{1}{2}) for x. $2x-7=2(\dfrac{1}{2})-7$ $=1-7$ $=-6$ d. Substitute \(−4$$ for x.
$2x-7=2(-4)-7$
$=-8-7$
$=-15$
Exercise
Evaluate the expression $$11−3y$$ for each value for $$y$$.
a. $$y=2$$
b. $$y=0$$
c. $$y=\dfrac{2}{3}$$
d. $$y=−5$$
Solution:
a. $$5$$;
b. $$11$$;
c. $$9$$;
d. $$26$$
Example
Evaluating Algebraic Expressions
Evaluate each expression for the given values.
a. $$x+5$$ for $$x=-5$$
b. $$\dfrac{t}{2t-1}$$ for $$t=10$$
c. $$\dfrac{4}{3}\pi r^3$$ for $$r=5$$
d. $$a+ab+b$$ for $$a=11$$ , $$b=-8$$
e. $$\sqrt{2m^3 n^2}$$ for $$m=2$$ , $$n=3$$
Solution
a. Substitute−5for x .
$x+5=(-5)+5$
$=0$
b. Substitute 10 for t .
$\dfrac{t}{2t-1}=\dfrac{(10)}{2(10)-1}$
$\dfrac{10}{20-1}$
$\dfrac{10}{19}$
c. Substitute 5 for .
$\dfrac{4}{3} \pi r^3=\dfrac{4}{3}\pi (5)^3$
$=\dfrac{4}{3}\pi (125)$
$=\dfrac{500}{3}\pi$
d. Substitute 11 for and –8 for .
$a+ab+b=(11)+(11)(-8)+(-8)$
$=11-88-8$
$=-85$
e. Substitute 2 for mand 3 for n.
$\sqrt{2m^3 n^2}=\sqrt{2(2)^3 (3)^2}$
$=\sqrt{2(8)(9)}$
$=\sqrt{144}$
$=12$
Exercise
Evaluate each expression for the given values.
a. $$\dfrac{y+3}{y-3}$$ for $$y=5$$
b. $$7-2t$$ for $$t=-2$$
c. $$\dfrac{1}{3}\pi r^2$$ for $$r=11$$
d. $$(p^2 q)^3$$ for $$p=-2$$, $$q=3$$
e. $$4(m-n)-5(n-m)$$ for $$m=\dfrac{2}{3}$$ $$n=\dfrac{1}{3}$$
Solution
a. $$4$$
b. $$11$$
c. $$\dfrac{121}{3}\pi$$
d. $$1728$$
e. $$3$$
#### Formulas
An equation is a mathematical statement indicating that two expressions are equal. The expressions can be numerical or algebraic. The equation is not inherently true or false, but only a proposition. The values that make the equation true, the solutions, are found using the properties of real numbers and other results. For example, the equation $$2x +1= 7$$ has the unique solution of 3 because when we substitute 3 for x in the equation, we obtain the true statement $$2(3)+1=7$$.
A formula is an equation expressing a relationship between constant and variable quantities. Very often, the equation is a means of finding the value of one quantity (often a single variable) in terms of another or other quantities. One of the most common examples is the formula for finding the area A of a circle in terms of the radius r of the circle: $$A= \pi r^2$$ . For any value of r , the area A can be found by evaluating the expression $$\pi r^2$$ .
Example
Using a Formula
A right circular cylinder with radius r and height h has the surface area S (in square units) given by the formula $$S=2\pi r(r+h)$$. See Figure 1.2.3. Find the surface area of a cylinder with radius 6 in. and height 9 in. Leave the answer in terms of $$\pi$$ .
Figure 1.2.3
Evaluate the expression $$2\pi r(r+h)$$ for $$r=6$$ and $$h=9$$.
$S=2\pi r(r+h)$
$=2\pi (6)[(6)+(9)]$
$=2\pi(6)(15)$
$=180\pi$
The surface area is $$180\pi$$ square inches
Exercise
A photograph with length L and width W is placed in a matte of width 8 centimeters (cm). The area of the matte (in square centimeters, or $$cm^2$$ is found to be A=(L+16)(W+16)LW.See Figure 1.2.4 Find the area of a matte for a photograph with length 32 cm and width 24 cm.
Figure 1.2.4
Solution:
$$1152cm^2$$
#### Simplifying Algebraic Expressions
Sometimes we can simplify an algebraic expression to make it easier to evaluate or to use in some other way. To do so, we use the properties of real numbers. We can use the same properties in formulas because they contain algebraic expressions.
Example
Simplifying Algebraic Expressions
Simplify each algebraic expression.
a. $$3x-2y+x-3y-7$$
b. $$2r-5(3-r)+4$$
c. $$(4t-\dfrac{5}{4}s)-(\dfrac{2}{3}t+2s)$$
d. $$2mn-5m+3mn+n$$
Solution:
a.$$3x-2y+x-3y-7=3x+x-2y-3y-7$$ Commutative property of addition
$$=4x-5y-7$$ Simplify
b. $$2r-5(3-r)+4=2r-15+5r+4$$ Distributive property
$$=2r+5y-15+4$$ Commutative property of addition
$$=7r-11$$ Simplify
c. $$(4t-\dfrac{5}{4}s)-(\dfrac{2}{3}t+2s)=4t-\dfrac{5}{4}s-\dfrac{2}{3}t-2s$$ Distributive property
$$=4t-\dfrac{2}{3}t-\dfrac{5}{4}s-2s$$ Commutative property of addition
$$=\dfrac{10}{3}t-\dfrac{13}{4}s$$ Simplify
d. $$2mn-5m+3mn+n=2mn+3mn-5m+n$$ Commutative property of addition
$$=5mn-5m+n$$ Simplify
Exercise
Simplify each algebraic expression.
a. $$\dfrac{2}{3}y−2(\dfrac{4}{3}y+z)$$
b. $$\dfrac{5}{t}−2−\dfrac{3}{t}+1$$
c. $$4p(q−1)+q(1−p)$$
d. $$9r−(s+2r)+(6−s)$$
Solution:
a. $$−2y−2z$$ or $$−2(y+z)$$
b. $$\dfrac{2}{t}−1$$
c. $$3pq−4p+q$$
d. $$7r−2s+6$$
Example
Simplifying a Formula
A rectangle with length L and width W has a perimeter P given by $$P =L+W+L+W$$ . Simplify this expression.
$$P =L+W+L+W$$
$$P =L+L+W+W$$ Commutative property of addition
$$P =2L+2W$$ Simplify
$$P=2(L+W)$$ Distributive property
Exercise
If the amount P is deposited into an account paying simple interest r for time t , the total value of the deposit A is given by $$A =P+Prt$$ . Simplify the expression. (This formula will be explored in more detail later in the course.)
Solution:
$$A=P(1+rt)$$
Access these online resources for additional instruction and practice with real numbers.
### Key Concepts
• Rational numbers may be written as fractions or terminating or repeating decimals. See Example and Example.
• Determine whether a number is rational or irrational by writing it as a decimal. See Example.
• The rational numbers and irrational numbers make up the set of real numbers. See Example. A number can be classified as natural, whole, integer, rational, or irrational. See Example.
• The order of operations is used to evaluate expressions. See Example.
• The real numbers under the operations of addition and multiplication obey basic rules, known as the properties of real numbers. These are the commutative properties, the associative properties, the distributive property, the identity properties, and the inverse properties. See Example.
• Algebraic expressions are composed of constants and variables that are combined using addition, subtraction, multiplication, and division. See Example. They take on a numerical value when evaluated by replacing variables with constants. See Example,Example, and Example
• Formulas are equations in which one quantity is represented in terms of other quantities. They may be simplified or evaluated as any mathematical expression. See Example and Example.
## Verbal
Is2an example of a rational terminating, rational repeating, or irrational number? Tell why it fits that category.
irrational number. The square root of two does not terminate, and it does not repeat a pattern. It cannot be written as a quotient of two integers, so it is irrational.
What is the order of operations? What acronym is used to describe the order of operations, and what does it stand for?
What do the Associative Properties allow us to do when following the order of operations? Explain your answer.
The Associative Properties state that the sum or product of multiple numbers can be grouped differently without affecting the result. This is because the same operation is performed (either addition or subtraction), so the terms can be re-ordered.
## Numeric
For the following exercises, simplify the given expression.
10+2×(53)
6÷2(81÷32)
−6
18+(68)3
−2×[16÷(84)2]2
−2
46+2×7
3(58)
−9
4+610÷2
12÷(36÷9)+6
9
(4+5)2÷3
312×2+19
4
2+8×7÷4
5+(6+4)11
4
918÷32
14×3÷76
0
9(3+11)×2
6+2×21
9
64÷(8+4×2)
9+4(22)
25
(12÷3×3)2
25÷527
−6
(157)×(37)
2×49(−1)
17
4225×15
12(31)÷6
4
## Algebraic
For the following exercises, solve for the variable.
8(x+3)=64
4y+8=2y
−4
(11a+3)18a=−4
4z2z(1+4)=36
−6
4y(72)2=−200
(2x)2+1=−3
±1
8(2+4)15b=b
2(11c4)=36
2
4(31)x=4
14(8w42)=0
2
For the following exercises, simplify the expression.
4x+x(137)
2y(4)2y11
−14y11
a23(64)12a÷6
8b4b(3)+1
−4b+1
5l÷3l×(96)
7z3+z×62
43z3
4×3+18x÷912
9(y+8)27
9y+45
(96t4)2
6+12b3×6b
−6b+6
18y2(1+7y)
(49)2×27x
16x3
8(3m)+1(8)
9x+4x(2+3)4(2x+3x)
9x
524(3x)
## Real-World Applications
For the following exercises, consider this scenario: Fred earns $40 mowing lawns. He spends$10 on mp3s, puts half of what is left in a savings account, and gets another $5 for washing his neighbor’s car. Write the expression that represents the number of dollars Fred keeps (and does not put in his savings account). Remember the order of operations. 12(4010)+5 How much money does Fred keep? For the following exercises, solve the given problem. According to the U.S. Mint, the diameter of a quarter is 0.955 inches. The circumference of the quarter would be the diameter multiplied byπ.Is the circumference of a quarter a whole number, a rational number, or an irrational number? irrational number Jessica and her roommate, Adriana, have decided to share a change jar for joint expenses. Jessica put her loose change in the jar first, and then Adriana put her change in the jar. We know that it does not matter in which order the change was added to the jar. What property of addition describes this fact? For the following exercises, consider this scenario: There is a mound ofgpounds of gravel in a quarry. Throughout the day, 400 pounds of gravel is added to the mound. Two orders of 600 pounds are sold and the gravel is removed from the mound. At the end of the day, the mound has 1,200 pounds of gravel. Write the equation that describes the situation. g+4002(600)=1200 Solve for g. For the following exercise, solve the given problem. Ramon runs the marketing department at his company. His department gets a budget every year, and every year, he must spend the entire budget without going over. If he spends less than the budget, then his department gets a smaller budget the following year. At the beginning of this year, Ramon got$2.5 million for the annual marketing budget. He must spend the budget such that2,500,000x=0.What property of addition tells us what the value of x must be?
## Technology
For the following exercises, use a graphing calculator to solve for x. Round the answers to the nearest hundredth.
0.5(12.3)248x=35
(0.250.75)2x7.2=9.9
68.4
## Extensions
If a whole number is not a natural number, what must the number be?
Determine whether the statement is true or false: The multiplicative inverse of a rational number is also rational.
true
Determine whether the statement is true or false: The product of a rational and irrational number is always irrational.
Determine whether the simplified expression is rational or irrational:−184(5)(−1).
irrational
Determine whether the simplified expression is rational or irrational:−16+4(5)+5.
The division of two whole numbers will always result in what type of number?
rational
What property of real numbers would simplify the following expression:4+7(x1)?
## Glossary
algebraic expression
constants and variables combined using addition, subtraction, multiplication, and division
the sum of three numbers may be grouped differently without affecting the result; in symbols,a+(b+c)=(a+b)+c
associative property of multiplication
the product of three numbers may be grouped differently without affecting the result; in symbols,a(bc)=(ab)c
base
in exponential notation, the expression that is being multiplied
two numbers may be added in either order without affecting the result; in symbols,a+b=b+a
commutative property of multiplication
two numbers may be multiplied in any order without affecting the result; in symbols,ab=ba
constant
a quantity that does not change value
distributive property
the product of a factor times a sum is the sum of the factor times each term in the sum; in symbols,a(b+c)=ab+ac
equation
a mathematical statement indicating that two expressions are equal
exponent
in exponential notation, the raised number or variable that indicates how many times the base is being multiplied
exponential notation
a shorthand method of writing products of the same factor
formula
an equation expressing a relationship between constant and variable quantities
there is a unique number, called the additive identity, 0, which, when added to a number, results in the original number; in symbols,a+0=a
identity property of multiplication
there is a unique number, called the multiplicative identity, 1, which, when multiplied by a number, results in the original number; in symbols,a1=a
integers
the set consisting of the natural numbers, their opposites, and 0:{,−3,−2,−1,0,1,2,3,…}
for every real numbera,there is a unique number, called the additive inverse (or opposite), denoteda,which, when added to the original number, results in the additive identity, 0; in symbols,a+(a)=0
inverse property of multiplication
for every non-zero real numbera,there is a unique number, called the multiplicative inverse (or reciprocal), denoted1a,which, when multiplied by the original number, results in the multiplicative identity, 1; in symbols,a1a=1
irrational numbers
the set of all numbers that are not rational; they cannot be written as either a terminating or repeating decimal; they cannot be expressed as a fraction of two integers
natural numbers
the set of counting numbers:{1,2,3,…}
order of operations
a set of rules governing how mathematical expressions are to be evaluated, assigning priorities to operations
rational numbers
the set of all numbers of the formmn,wheremandnare integers andn0.Any rational number may be written as a fraction or a terminating or repeating decimal.
real number line
a horizontal line used to represent the real numbers. An arbitrary fixed point is chosen to represent 0; positive numbers lie to the right of 0 and negative numbers to the left.
real numbers
the sets of rational numbers and irrational numbers taken together
variable
a quantity that may change value
whole numbers
the set consisting of 0 plus the natural numbers:{0,1,2,3,…} |
# What is the ans please?
Mar 5, 2018
The correct answer is $\text{option A}$
#### Explanation:
The inequality is
$\frac{1}{4 x - 3} \le x$
Therefore,
$x - \frac{1}{4 x - 3} \ge 0$
Putting on the same denominator
$\frac{x \left(4 x - 3\right) - 1}{4 x - 3} \ge 0$
$\frac{4 {x}^{2} - 3 x - 1}{4 x - 3} \ge 0$
The correct answer is $\text{option A}$
Mar 5, 2018
$0 \le \frac{4 {x}^{2} - 3 x - 1}{4 x - 3}$ (part a)
#### Explanation:
given
$\textcolor{g r e e n}{\implies \frac{1}{4 x - 3} \le x}$
Subtract $\textcolor{b l u e}{\frac{1}{4 x - 3}}$ from both sides.
$\frac{1}{4 x - 3} - \textcolor{red}{\frac{1}{4 x - 3}} \le x - \textcolor{red}{\frac{1}{4 x - 3}}$
$0 \le \frac{x \left(4 x - 3\right) - 1}{4 x - 3}$
$\textcolor{f i r e b r i c k}{0 \le \frac{4 {x}^{2} - 3 x - 1}{4 x - 3}}$ or part a
Mar 6, 2018
I will use this online graphing calculator to show that the correct answer is A
#### Explanation:
First I will show you $\textcolor{red}{\frac{1}{4 x - 3} \le x}$
Please observe that it is saying that y is any value that you like but domain of x is $- \frac{1}{4} \le x < \frac{3}{4}$ and $x \ge 1$.
I will save the graph of A for after I have shown that selections B through E are NOT the solutions:
Here is Selection B $\textcolor{b l u e}{\frac{4 {x}^{2} - 3 x - 1}{4 x - 3} \le 0}$:
Please observe that it is saying that y is any value that you like but domain of x is $- \frac{1}{4} \le x$ and $\frac{3}{4} < x \le 1$; this clearly disagrees with the graph of the original equation.
Here is Selection C $\textcolor{g r e e n}{4 {x}^{2} - 3 x - 1 \ge 0}$
Please observe that it is saying that y is any value that you like but domain of x is $- \frac{1}{4} \le x$ and $x \ge 1$; this clearly disagrees with the graph of the original equation.
Here is Selection D $\textcolor{p u r p \le}{4 {x}^{2} - 3 x - 1 \le 0}$
Please observe that it is saying that y is any value that you like but domain of x is $- \frac{1}{4} \le x \le 1$; this clearly disagrees with the graph of the original equation.
I cannot use the inequality function for selection E $\textcolor{\mathmr{and} a n \ge}{4 {x}^{2} + 3 x + 1 \le 0}$, because $\textcolor{\mathmr{and} a n \ge}{y = 4 {x}^{2} + 3 x + 1}$ is never less than 0 and, therefore, the domain of the inequality is the null set but I will show you the parabola:
Finally, the correct answer Selection A $\frac{4 {x}^{2} - 3 x - 1}{4 x - 3} \ge 0$
Please observe that the graph of the original inequality is identical to this graph with the same domain, $- \frac{1}{4} \le x < \frac{3}{4}$ and $x \ge 1$. |
# Math Tricks and Fermat’s Little Theorem
So you think you’re a math whiz. You storm into parties armed with math’s most flamboyant tricks. You can recite the digits of π and e to 50 digits—whether in base 10 or 12. You can calculate squares with ease, since you’ve mastered the difference of squares x2 – y2 = (x + y)(x – y). In tackling 572, simply notice that 572 – 72 = (57 + 7)(57 – 7) = 64*50 = 3200. Adding 72 to both sides gives 572 = 3249.
Image by Hashir Milhan from
Wikimedia Commons under
Creative Commons.
You can also approximate square roots using the truncated Taylor series x ≈ c + (x – c2)/(2c) where c2 is the closest perfect square less than or equal to x. So √17 ≈ 4 + (17 – 16)/(2*4) = 4.125, whereas √17 = 4.123105 . . ..
But do you know what number theory is? It’s not taught in high school, and everyone’s repertoire of math tricks needs some number theory. Mastering modular arithmetic—the first step in number theory—will make you the life of the party. Calculating 83 mod 7 just means find the remainder after dividing by 7: 83 = 11*7 + 6, so 83 ≡ 6 mod 7. But it’s actually easier since 83 = 7*12 + (-1), so 83 ≡ -1 ≡ 6 mod 7. Modular arithmetic reveals the secrets of divisibility. Everybody knows the trick to see whether 3 divides a number; you just add the digits and check if 3 divides that number. But the reasoning is obvious when you write m = 10nan + 10n-1an-1 + . . . + 10a1 + a0 where the ai are the digits of m. Each 10k has a remainder of 1 modulo 3 so man + an-1 + . . . + a1 + a0 mod 3. Using this method generates tricks for other integers.
For instance, if 13 divides m, then 13 divides a0 – 3a1 – 4a2a3 + 3a4 + 4a5 + a6 – 3a7 – 4a+ . . . and the pattern continues. This is because
10 ≡ -3 mod 13,
102 ≡ 10*10 ≡ (-3)(-3) ≡ 9 ≡ -4 mod 13,
103 ≡ 102*10 ≡ (-4)(-3) ≡ 12 ≡ -1 mod 13,
104 ≡ 103*10 ≡ (-1)*(-3) ≡ 3 mod 13,
105 ≡ 104*10 ≡ 3*(-3) ≡ -9 ≡ 4 mod 13,
and 106 ≡ 105*10 ≡ 4*(-3) ≡ -12 ≡ 1 mod 13.
From 106 and onwards the pattern repeats. In fact, calculating 10n mod k for successive n will reveal the divisibility rule for k.
Then comes Fermat’s little theorem, the key to solving seemingly impossible calculations.
Fermat. Image from Wikimedia
Commons. Under public domain.
The theorem states for a prime p and integer a that aa mod p. If p doesn’t divide a, then ap -1 ≡ 1 mod p. I’ll illustrate the power of this little result in a computation. Let’s find 2371 mod 5. We’ll be using 24 ≡ 1 mod 5, which we get from Fermat’s little theorem. Now 2371 = 236823 =(24)9223, so by the theorem, 2371 = (24)9223 ≡ 19223 ≡ 1*23 ≡ 8 ≡ 3 mod 5. Exploiting Fermat’s little theorem can impress your friends, but try to avoid questions. Computing residues modulo a composite number—calculating b mod n for a composite number n—may require paper and ruin the magic.
Leonhard Euler proved a more general version of Fermat’s little theorem; it’s called the Euler-Fermat theorem. This theorem isn’t for parties; explaining it to the non-mathematically inclined will always require paper and some time. Nonetheless, it will impress at dinner if you have a napkin and pen.
Understanding this theorem requires Euler’s totient function φ(n).
Euler. From Wikimedia
Commons. Under public domain.
The number φ(n) for some n is the number of positive integers coprime with n that are less than or equal to n. Two numbers a and b are coprime if their greatest common factor is one. Hence 14 and 3 are coprime because their biggest shared factor is 1, but 21 and 14 aren’t coprime because they have a common divisor of 7. Moreover, φ(14) = 6 because 14 has six numbers less than or equal to it that are coprime with it: 1, 3, 5, 9, 11, and 13. Notice that if p is prime, φ(p) = p – 1 because every number less than p is coprime with p.
Now the Euler-Fermat theorem states that aφ(n) ≡ 1 mod n, which looks similar to ap -1 ≡ 1 mod p for a prime p. In fact, if = φ(p) = p – 1 for a prime p, the theorem reduces to Fermat’s little theorem.
Fermat’s little theorem has another generalization, Lagrange’s theorem. Joseph-Louis Lagrange was Euler’s student. Lagrange’s theorem generalizes both the previous theorems and doesn’t even require numbers. But due to the required background in group theory, I won’t go over the theorem. You can find links to more information on Lagrange’s theorem below.
Remember, a math whiz doesn’t need props like a magician does. Hook your audience with some modular arithmetic, and reel the people in with Fermat’s little theorem. If you want to get complicated, the most you’ll need is a pen and some paper.
Sources and cool stuff:
Modular arithmetic: http://nrich.maths.org/4350
Proof of Fermat’s little theorem: https://primes.utm.edu/notes/proofs/FermatsLittleTheorem.html
Euler-Fermat theorem and its proof: http://www.artofproblemsolving.com/Wiki/index.php/Euler%27s_Totient_Theorem
Lagrange’s theorem (only for the brave): http://cims.nyu.edu/~kiryl/teaching/aa/les102903.pdf
Number theory textbooks: Gordan Savin’s Numbers, Groups, and Cryptography and George E. Andrews’s Number Theory
Interesting sources of math tricks and problems: Paul Zeitz’s The Art, Craft of Problem Solving and The USSR Olympiad Problem Book, and What is Mathematics by Richard Courant and Herbert Robbins |
PHY160-Solution-Ch1.1-F07
# PHY160-Solution-Ch1.1-F07 - 2 INTRODUCTION AND MATHEMATICAL...
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2 INTRODUCTION AND MATHEMATICAL CONCEPTS CHAPTER 1 INTRODUCTION AND MATHEMATICAL CONCEPTS PROBLEMS ____________________________________________________________________________________________ 1. REASONING AND SOLUTION We use the fact that 0.200 g = 1 carat and that, under the conditions stated, 1000 g has a weight of 2.205 lb to construct the two conversion factors: (0.200 g)/(1 carat) = 1 and (2.205 lb)/(1000 g) = 1 . Then, 3106 carats ( 29 0.200 g 1 carat 2.205 lb 1000 g 1.37 lb = 3. REASONING a. To convert the speed from miles per hour (mi/h) to kilometers per hour (km/h), we need to convert miles to kilometers. This conversion is achieved by using the relation 1.609 km = 1 mi (see the page facing the inside of the front cover of the text). b. To convert the speed from miles per hour (mi/h) to meters per second (m/s), we must convert miles to meters and hours to seconds. This is accomplished by using the conversions 1 mi = 1609 m and 1 h = 3600. SOLUTION a. Multiplying the speed of 34.0 mi/h by a factor of unity, (1.609 km)/(1 mi) = 1, we find the speed of the bicyclists is ( 29 mi mi Speed = 34.0 1 34.0 h = 1.609km h 1 mi km 54.7 h = b. Multiplying the speed of 34.0 mi/h by two factors of unity, (1609 m)/(1 mi) = 1 and (1 h)/(3600 s) = 1, the speed of the bicyclists is ( 29 ( 29 mi mi Speed = 34.0 1 1 34.0 h = h 1609m 1 mi 1 h m 15.2 3600s s = _____________________________________________________________________________ _ 5. REASONING When converting between units, we write down the units explicitly in the calculations and treat them like any algebraic quantity. We construct the appropriate conversion factor (equal to unity) so that the final result has the desired units. SOLUTION a. Since 1.0 × 10 3 grams = 1.0 kilogram, it follows that the appropriate conversion factor is (1.0 × 10 3 g)/(1.0 kg) = 1 . Therefore,
Chapter 1 Problems 3 6 5 10 kg - ( 29 3 1.0 10 g 1.0 kg 1 3 5 10 g - = b. Since 1.0 × 10 3 milligrams = 1.0 gram, 3 5 10 g - ( 29 3 1.0 10 mg 1.0 g 1 5 mg = c. Since 1.0 × 10 6 micrograms = 1.0 gram, 3 5 10 g - ( 29 6 1.0 10 g 1.0 g μ 3 5 10 g = ____________________________________________________________________________________________ 7. REASONING AND SOLUTION a. F = [M][L]/[T] 2 ; ma
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# How do you solve 2x+7y=10 and x-2y=15?
Aug 5, 2015
$\left\{\begin{matrix}x = \frac{125}{11} \\ y = - \frac{20}{11}\end{matrix}\right.$
#### Explanation:
You could solve this sytem of equations by multiplication.
More specifically, you can multiply the second equation by $- 2$ to get
$- 2 \cdot \left(x - 2 y\right) = - 2 \cdot 15$
$- 2 x + 4 y = - 30$
The two equations now look like this
$\left\{\begin{matrix}2 x + 7 y = 10 \\ - 2 x + 4 y = - 30\end{matrix}\right.$
Next, add the left side and the right side of the equations separately to cancel out the $x$-term
$\textcolor{red}{\cancel{\textcolor{b l a c k}{2 x}}} + 7 y - \textcolor{red}{\cancel{\textcolor{b l a c k}{2 x}}} + 4 y = 10 + \left(- 30\right)$
$11 y = - 20 \implies y = \textcolor{g r e e n}{- \frac{20}{11}}$
Now take this value of $y$ and use it in the first equation to find the value of $x$
$2 x + 7 \cdot \frac{- 20}{11} = 10$
$2 x = 10 + \frac{140}{11}$
$2 x = \frac{250}{11} \implies x = \frac{250}{11} \cdot \frac{1}{2} = \textcolor{g r e e n}{\frac{125}{11}}$ |
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# Data Organization and Interpretation for Praxis II ParaPro Test Prep Study Guide (page 4)
By
Updated on Jul 5, 2011
### Circle Graphs
Circle graphs are often used to show what percent of a whole is taken up by different components of that whole. This type of graph is representative of a total amount, and is usually divided into percentages. Each section of the chart represents a portion of the whole, and all of these sections added together will equal 100%. The following chart shows the three styles of model homes in a new development, and what percentage of each there is.
The chart shows the different styles of model homes. The categories add up to 100% (25 + 30 + 45 = 100). To find the percentage of estate homes, you can look at the pie chart and see that 45% of the homes are done in the estate style.
If you know the total number of items in a circle graph, you can calculate how many there are of each component. You just need to multiply the percent by the total.
Example
There are 500 homes in the new development. How many of them are chateaus?
To calculate the number of components (chateaus) out of the total (500), you need to multiply the percent times the total.
25% × 500 = 0.25 × 500 = 125
There are 125 chateaus in the development.
### Line Graph
A line graph is a graph used to show a change over time. The line moves from left to right to show how the data changes over a time period. If a line is slanted up, it represents an increase, whereas a line sloping down represents a decrease. A flat line indicates no change.
In the line graph below, the number of delinquent payments is charted for the first quarter of the year. Each week, the number of customers with outstanding bills is added together and recorded.
There is an increase in delinquency for the first two weeks, and then the level is maintained for an additional two weeks. There is a steep decrease after week five (initially) until the ninth week, where it levels off again—but this time at 0. The 11th week shows a radical increase followed by a little jump up at week 12, and then a decrease at week 13. It is also interesting to see that the first and last weeks have identical values.
This line graph shows an obvious trend because the points go up as the line moves to the right. That means there is a positive trend. A question on the ParaPro Assessment may ask you to make a prediction based on the trend. Each year, the profits of the company go up by about \$3 million. Therefore, if you were asked to predict the profits of the company in 2010, you could add \$3 million to the profits in 2009. An appropriate prediction for the company's total profits in 2010, based on the trend in the graph, would be \$21 million.
### Mean, Median, and Mode
It is important to understand trends in data. To do that, look at where the center of the data lies. There are a number of ways to find the center of a set of data.
### Mean
Average usually refers to the arithmetic mean (usually just called the mean). To find the mean of a set of numbers, add all of the numbers together and divide by the quantity of numbers in the set.
average = (sum of set) ÷ (quantity of set)
Example
Find the average of 9, 4, 7, 6, and 4.
(Divide by 5 because there are five numbers in the set.) The mean, or average, of the set is 6.
### Median
Another center of data is the median, which is the number in the center, if you arrange all the data in ascending or descending order. To find the median of a set of numbers, arrange the numbers in ascending or descending order and find the middle value. If the set contains an odd number of elements, then choose the middle value. If the set contains an even number of elements, simply average the two middle values.
Example
Find the median of the number set: 1, 5, 4, 7, 2.
First arrange the set in order—1, 2, 4, 5, 7—and then find the middle value. Because there are five values, the middle value is the third one: 4. The median is 4.
Example
Find the median of the number set: 1, 6, 3, 7, 2, 8.
First arrange the set in order—1, 2, 3, 6, 7, 8—and then find the middle values, 3 and 6.
Find the average of the numbers 3 and 6:
= 4.5. The median is 4.5.
### Mode
The mode of a set of numbers is the number that appears the greatest number of times.
Example
Find the mode for the following data set: 1, 2, 5, 9, 4, 2, 9, 6, 9, 7.
For the number set 1, 2, 5, 9, 4, 2, 9, 6, 9, 7, the number 9 is the mode, because it appears the most frequently.
The practice quiz for this study guide can be found at:
Math for Praxis II ParaPro Test Prep Practice Problems |
# Sophomores constitute approximately what percent of the total
Sophomores constitute approximately what percent of the total enrollment?
1. \$23%\$
2. \$20%\$
3. \$18%\$
4. \$15%\$
5. \$8%\$
So, you were trying to be a good test taker and practice for the GRE with PowerPrep online. Buuuut then you had some questions about the quant section—specifically question 16 of Section 6 of Practice Test 1. Those questions testing our knowledge of Graphical Methods for Describing Data can be kind of tricky, but never fear, PrepScholar has got your back!
## Survey the Question
Let’s search the problem for clues as to what it will be testing, as this will help shift our minds to think about what type of math knowledge we’ll use to solve this question. Pay attention to any words that sound math-specific and anything special about what the numbers look like, and mark them on our paper.
Looks like we’ll need to analyze two tables containing data in this question, so it likely will draw on what we’ve learned about Graphical Methods for Describing Data. Let’s keep what we’ve learned about this skill at the tip of our minds as we approach this question.
## What Do We Know?
Let’s carefully read through the question and make a list of the things that we know.
1. We want to find the percent of sophomores out of all of the students
## Develop a Plan
We want to know the percent of all of the students who are sophomores. We know that if we had the number of sophomores and the total number of students, then we could just divide these two values, multiply by \$100%\$ and have our answer.
The total number of students is given to us in the first table: \$1{,}400\$. Sophomore can be males or females, so we can get the number of sophomores by adding together the male sophomores and the females sophomores. Looks like we have a well-developed plan, so let’s finish solving this question.
## Solve the Question
Adding together the male and female sophomores, we get:
\$\Number \of \Sophomores\$ \$=\$ \$\Male \Sophomores + \Female \Sophomores\$ \$ \$ \$ \$ \$\Number \of \Sophomores\$ \$=\$ \$215+109\$ \$ \$ \$ \$ \$\Number \of \Sophomores\$ \$=\$ \$324\$
Now we just need to divide by the total number of students to get the percent that are sophomores
\$\Percent \of \Sophomores\$ \$=\$ \${\Number \of \Sophomores}/{\Total \Number \of \Students}·100%\$ \$ \$ \$ \$ \$\Percent \of \Sophomores\$ \$=\$ \$324/{1{,}400}·100%\$ \$ \$ \$ \$ \$\Percent \of \Sophomores\$ \$=\$ \$23%\$
Well, that wasn’t so bad after all. That problem never stood a chance, not with our knowledge about how percentages work. The correct answer is A, \$23%\$.
## What Did We Learn
To find percents, we just do the number we’re interested in divided by the total number. Here, we just had to be careful to make sure that we added together the number of male and female sophomores to get the total number of sophomores. We can be sure that the devious GRE test writers put in wrong answer choices to punish us if we forgot to add these together. Great work by us being patient and taking the time to read the full question to avoid this type of careless error.
Want more expert GRE prep? Sign up for the five-day free trial of our PrepScholar GRE Online Prep Program to access your personalized study plan with 90 interactive lessons and over 1600 GRE questions.
Have questions? Leave a comment or send us an email at [email protected]. |
eozoischgc
2021-12-27
Evaluate the following integrals.
$\int x{10}^{x}dx$
Jillian Edgerton
Step 1
Given: $I=\int x{\left(10\right)}^{x}dx$
for evaluating given integral, we will use integral by parts theorem
according to this theorem
$\int {f}^{\prime }\left(x\right)g\left(x\right)dx=g\left(x\right)\int {f}^{\prime }\left(x\right)dx-\int \left[\left({g}^{\prime }\left(x\right)\right)\int {f}^{\prime }\left(x\right)dx\right]dx+c$
Step 2
so,
$I=\int \left(x\right)\left({10}^{x}\right)dx$
$=x\int \left({10}^{x}\right)dx-\int \left[1\int {10}^{x}dx\right]dx$
$\left(\because \int {a}^{x}dx=\frac{{a}^{x}}{\mathrm{ln}a}+c\right)$
$=x\left(\frac{{10}^{x}}{\mathrm{ln}10}\right)-\int \left(\frac{{10}^{x}}{\mathrm{ln}10}\right)dx+c$
$=\frac{x\left({10}^{x}\right)}{\mathrm{ln}10}-\frac{1}{\mathrm{ln}10}\left(\frac{{10}^{x}}{\mathrm{ln}10}\right)+c$
hence, given integral is $\frac{x\left({10}^{x}\right)}{\mathrm{ln}10}-\frac{{10}^{x}}{{\left(\mathrm{ln}10\right)}^{2}}+c$.
Andrew Reyes
$\int x\cdot {10}^{x}dx$
Integration piece by piece: $\int f{g}^{\prime }=fg-\int {f}^{\prime }g$
$f=x,{g}^{\prime }={10}^{x}$
$=\frac{x\cdot {10}^{x}}{\mathrm{ln}\left(10\right)}-\int \frac{{10}^{x}}{\mathrm{ln}\left(10\right)}dx$
$\int \frac{{10}^{x}}{\mathrm{ln}\left(10\right)}dx$
Let's apply linearity:
$=\frac{1}{\mathrm{ln}\left(10\right)}\int {10}^{x}dx$
$\int {10}^{x}dx$
Integral of exponential function:
$\int {a}^{x}dx=\frac{{a}^{x}}{\mathrm{ln}\left(a\right)}$ at a=10:
$=\frac{{10}^{x}}{\mathrm{ln}\left(10\right)}$
$\frac{1}{\mathrm{ln}\left(10\right)}\int {10}^{x}dx$
$=\frac{{10}^{x}}{{\mathrm{ln}}^{2}\left(10\right)}$
$\frac{x\cdot {10}^{x}}{\mathrm{ln}\left(10\right)}-\int \frac{{10}^{x}}{\mathrm{ln}\left(10\right)}dx$
$=\frac{x\cdot {10}^{x}}{\mathrm{ln}\left(10\right)}-\frac{{10}^{x}}{{\mathrm{ln}}^{2}\left(10\right)}$
$\int x{10}^{x}dx$
$=\frac{x\cdot {10}^{x}}{\mathrm{ln}\left(10\right)}-\frac{{10}^{x}}{{\mathrm{ln}}^{2}\left(10\right)}+C$
Let's rewrite / simplify:
$=\frac{\left(\mathrm{ln}\left(10\right)x-1\right)\cdot {10}^{x}}{{\mathrm{ln}}^{2}\left(10\right)}+C$
Vasquez
$\int x×{10}^{x}dx$
Prepare for integration by parts
u=x
$dv={10}^{x}dx$
du=dx
$v=\frac{{10}^{x}}{\mathrm{ln}\left(10\right)}$
$x×\frac{{10}^{x}}{\mathrm{ln}\left(10\right)}-\int \frac{{10}^{x}}{\mathrm{ln}\left(10\right)}dx$
$x×\frac{{10}^{x}}{\mathrm{ln}\left(10\right)}-\frac{1}{\mathrm{ln}\left(10\right)}×\int {10}^{x}dx$
$x×\frac{{10}^{x}}{\mathrm{ln}\left(10\right)}-\frac{1}{\mathrm{ln}\left(10\right)}×\frac{{10}^{x}}{\mathrm{ln}\left(10\right)}$
Simplify
$\frac{x×{10}^{x}}{\mathrm{ln}\left(10\right)}-\frac{{10}^{x}}{\mathrm{ln}\left(10{\right)}^{2}}$
$\frac{x×{10}^{x}}{\mathrm{ln}\left(10\right)}-\frac{{10}^{x}}{\mathrm{ln}\left(10{\right)}^{2}}+C$ |
# 15.9
(1)
## 15.9 Change of Variables in
(2)
### Change of Variables in Multiple Integrals
In one-dimensional calculus we often use a change of
variable (a substitution) to simplify an integral. By reversing the roles of x and u, we can write
where x = g(u) and a = g(c), b = g(d). Another way of writing Formula 1 is as follows:
(3)
3
### Change of Variables in Multiple Integrals
A change of variables can also be useful in double integrals.
We have already seen one example of this: conversion to polar coordinates. The new variables r and θ are related to the old variables x and y by the equations
x = r cos θ y = r sin θ
and the change of variables formula can be written as f(x, y) dA = f(r cos θ, r sin θ) r dr dθ
where S is the region in the rθ-plane that corresponds to the region R in the xy-plane.
(4)
### Change of Variables in Multiple Integrals
More generally, we consider a change of variables that is given by a transformation T from the uv-plane to the
xy-plane:
T(u, v) = (x, y)
where x and y are related to u and v by the equations x = g(u, v) y = h(u, v)
or, as we sometimes write,
x = x(u, v) y = y(u, v)
(5)
5
### Change of Variables in Multiple Integrals
We usually assume that T is a C1 transformation, which means that g and h have continuous first-order partial
derivatives.
A transformation T is really just a function whose domain and range are both subsets of
If T(u1, v1) = (x1, y1), then the point (x1, y1) is called the image of the point (u1, v1).
If no two points have the same image, T is called one-to-one.
(6)
### Change of Variables in Multiple Integrals
Figure 1 shows the effect of a transformation T on a region S in the uv-plane.
T transforms S into a region R in the xy-plane called the image of S, consisting of the images of all points in S.
(7)
7
### Change of Variables in Multiple Integrals
If T is a one-to-one transformation, then it has an inverse transformation T–1 from the xy-plane to the uv-plane and it may be possible to solve Equations 3 for u and v in terms of x and y:
u = G(x, y) v = H(x, y)
(8)
### Example 1
A transformation is defined by the equations x = u2 – v2 y = 2uv Find the image of the square
S = {(u, v) | 0 ≤ u ≤ 1, 0 ≤ v ≤ 1}.
Solution:
The transformation maps the boundary of S into the boundary of the image.
So we begin by finding the images of the sides of S.
(9)
9
### Example 1 – Solution
The first side, S1, is given by v = 0 (0 ≤ u ≤ 1).
(See Figure 2.)
Figure 2
cont’d
(10)
### Example 1 – Solution
From the given equations we have x = u2, y = 0, and so 0 ≤ x ≤ 1.
Thus S1 is mapped into the line segment from (0, 0) to (1, 0) in the xy-plane.
The second side, S2, is u = 1 (0 ≤ v ≤ 1) and, putting u = 1 in the given equations, we get
x = 1 – v2 y = 2v
cont’d
(11)
11
### Example 1 – Solution
Eliminating v, we obtain
x = 1 – 0 ≤ x ≤ 1
which is part of a parabola.
Similarly, S3 is given by v = 1 (0 ≤ u ≤ 1), whose image is the parabolic arc
x = – 1 –1 ≤ x ≤ 0
cont’d
(12)
### Example 1 – Solution
Finally, S4 is given by u = 0 (0 ≤ v ≤ 1) whose image is x = –v2, y = 0, that is, –1 ≤ x ≤ 0. (Notice that as we move around the square in the counterclockwise direction, we also move around the parabolic region
in the counterclockwise direction.)
The image of S is the region R (shown in Figure 2) bounded by the x-axis and the parabolas
given by Equations 4 and 5.
cont’d
(13)
13
### Change of Variables in Multiple Integrals
Now let’s see how a change of variables affects a double integral. We start with a small rectangle S in the uv-plane whose lower left corner is the point (u0, v0) and whose
dimensions are ∆u and ∆v. (See Figure 3.)
Figure 3
(14)
### Change of Variables in Multiple Integrals
The image of S is a region R in the xy-plane, one of whose boundary points is (x0, y0) = T(u0, v0).
The vector
r(u, v) = g(u, v) i + h(u, v) j
is the position vector of the image of the point (u, v).
The equation of the lower side of S is v = v0, whose image curve is given by the vector function r(u, v ).
(15)
15
### Change of Variables in Multiple Integrals
The tangent vector at (x0, y0) to this image curve is ru = gu(u0, v0) i + hu(u0, v0) j
Similarly, the tangent vector at (x0, y0) to the image curve of the left side of S (namely, u = u0) is
rv = gv(u0, v0) i + hv(u0, v0) j
(16)
### Change of Variables in Multiple Integrals
We can approximate the image region R = T(S) by a parallelogram determined by the secant vectors
a = r(u0 + ∆u, v0) – r(u0, v0) b = r(u0, v0 + ∆v) – r(u0, v0) shown in Figure 4.
(17)
17
### Change of Variables in Multiple Integrals
But
and so r(u0 + ∆u, v0) – r(u0, v0) ≈ ∆u ru Similarly r(u0, v0 + ∆v) – r(u0, v0) ≈ ∆v rv
This means that we can approximate R by a parallelogram determined by the vectors ∆u ru and ∆v rv. (See Figure 5.)
Figure 5
(18)
### Change of Variables in Multiple Integrals
Therefore we can approximate the area of R by the area of this parallelogram, which is
| (∆u ru) × (∆v rv)| = | ru × rv| ∆u ∆v
Computing the cross product, we obtain
(19)
19
### Change of Variables in Multiple Integrals
The determinant that arises in this calculation is called the Jacobian of the transformation and is given a special
notation.
With this notation we can use Equation 6 to give an approximation to the area ∆A of R:
where the Jacobian is evaluated at (u0, v0).
(20)
### Change of Variables in Multiple Integrals
Next we divide a region S in the uv-plane into rectangles Sij and call their images in the xy-plane Rij. (See Figure 6.)
(21)
21
### Change of Variables in Multiple Integrals
Applying the approximation (8) to each Rij, we approximate the double integral of f over R as follows:
where the Jacobian is evaluated at (ui, vj). Notice that this double sum is a Riemann sum for the integral
(22)
### Change of Variables in Multiple Integrals
The foregoing argument suggests that the following theorem is true.
Theorem 9 says that we change from an integral in x and y to an integral in u and v by expressing x and y in terms of u and v and writing
(23)
23
### Change of Variables in Multiple Integrals
Notice the similarity between Theorem 9 and the one-dimensional formula in Equation 2.
Instead of the derivative dx/du, we have the absolute value of the Jacobian, that is, |∂(x, y)/∂(u, v)|.
As a first illustration of Theorem 9, we show that the
formula for integration in polar coordinates is just a special case.
(24)
### Change of Variables in Multiple Integrals
Here the transformation T from the rθ-plane to the xy-plane is given by
x = g(r, θ) = r cos θ y = h(r, θ) = r sin θ
and the geometry of the transformation is shown in Figure 7.
T maps an ordinary rectangle in the rθ -plane to a polar rectangle in the xy-plane.
(25)
25
### Change of Variables in Multiple Integrals
The Jacobian of T is
Thus Theorem 9 gives
(26)
(27)
27
### Triple Integrals
There is a similar change of variables formula for triple integrals.
Let T be a transformation that maps a region S in
uvw-space onto a region R in xyz-space by means of the equations
x = g(u, v, w) y = h(u, v, w) z = k(u, v, w)
(28)
### Triple Integrals
The Jacobian of T is the following 3 × 3 determinant:
Under hypotheses similar to those in Theorem 9, we have the following formula for triple integrals:
(29)
29
### Example 4
Use Formula 13 to derive the formula for triple integration in spherical coordinates.
Solution:
Here the change of variables is given by
x = ρ sin φ cos θ y = ρ sin φ sin θ z = ρ cos φ
We compute the Jacobian as follows:
(30)
### Example 4 – Solution
= cos φ (–ρ2 sin φ cos φ sin2 θ – ρ2 sin φ cos φ cos2 θ) – ρ sin φ (ρ sin2 φ cos2 θ + ρ sin2 φ sin2 θ)
= –ρ2 sin φ cos2 φ – ρ2 sin φ sin2 φ
= –ρ2 sin φ
cont’d
(31)
31
### Example 4 – Solution
Therefore
and Formula 13 gives
f(x, y, z) dV
= f(ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) ρ2 sin φ dρ dθ dφ
cont’d
Updating...
## References
Related subjects : |
# How to apply l'Hopital to calculate the limit of the function (x^2-x-6)/(x+2) x-->-2
giorgiana1976 | Student
To calculate the limit, we'll substitute x by -2 in the expression of the function.
lim f(x) = lim ( 4 + 2 -6 )/( -2 + 2 )
lim ( x^2 - x -6 )/( x+2 ) = (6-6)/(-2+2) = 0/0
We notice that we've obtained an indetermination case, 0/0 type.
We'll apply l'Hospital's rule, by differentiating separately the numerator and denominator.
We'll differentiate the numerator:
( x^2 - x -6)' = 2x - 1
We'll differentiate the denominator:
(x+2)' = 1
We'll re-write the limit:
lim ( x^2 - x -6 )/( x+ 2 ) = lim (2x-1)
We'll substitute x by -2:
lim (2x-1) = 2*(-2) - 1 = -4-1
The limit of the function, when x->-2, is: lim (x^2-x-6)/(x+2) = -5.
tonys538 | Student
In a limit `lim_(x->a) (f(x))/(g(x))` , if `(f(a))/(g(a)) = 0/0` or `(+-oo)/(+-oo)` , it is possible to apply l'Hospoital's rule and substitute `(f(x))/(g(x))` with `((f(x))')/((g(x))')`
In the given problem, the limit `lim_(x->-2)(x^2-x-6)/(x+2)` is required.
If we substitute x = -2, the result is `(4 + 2 - 6)/(-2 + 2) = 0/0` . This is an indeterminate form and we can apply l'Hospital's rule. Replace the numerator and denominator by their derivatives.
This gives:
`lim_(x->-2) (2x - 1)/(1)`
Substituting x = -2 gives (2*-2 -1)/1 = -5
The required limit is `lim_(x->-2)(x^2-x-6)/(x+2) = -5` |
## 4 spoons and 3 forks cost \$15 4 spoons and 1 fork cost \$13 how many do the forks cost?
Question
4 spoons and 3 forks cost \$15 4 spoons and 1 fork cost \$13 how many do the forks cost?
0
1. Let’s say that s=spoons and f=forks.
According to this, we can say that:
{4s+3f=15
{4s+f=13
This is a system.
We can then use elimination to solve and isolate the variable:
(4s-3f=15)
– (4s+f=13)
We can then subtract and conclude that:
-4f=2
-f=1/2
f=-1/2 |
# Binomial Distribution
## What is the 'Binomial Distribution'
The binomial distribution is a probability distribution that summarizes the likelihood that a value will take one of two independent values under a given set of parameters or assumptions. The underlying assumptions of the binomial distribution are that there is only one outcome for each trial, that each trial has the same probability of success and that each trial is mutually exclusive, or independent of each other.
## BREAKING DOWN 'Binomial Distribution'
A binomial distribution summarizes the number of trials, or observations, when each trial has the same probability of attaining one particular value. The binomial distribution determines the probability of observing a specified number of successful outcomes in a specified number of trials. The expected value, or mean, of a binomial distribution is calculated by multiplying the number of trials by the probability of successes. For example, the expected value of the number of heads in 100 trials is 50, or (100 * 0.5).
## Bernoulli Trial
The binomial distribution is the sum of a series of multiple independent and identically distributed Bernoulli trials. In a Bernoulli trial, the experiment is said to be random and could only have two possible outcomes: success or failure. For example, flipping a coin is considered to be a Bernoulli trial; each trial can only take one of two values (heads or tails), each success has the same probability (the probability of flipping a head is 0.5) and the results of one trial do not influence the results of another.
## Binomial Distribution Example
The binomial distribution is calculated by multiplying the probability of success raised to the power of the number of successes and the probability of failure raised to the power of the difference between the number of successes and number of trials. Then, multiply the product by the combination between the number of trials and the number of successes.
For example, assume that casino created a new game in which participants are able to place bets on the number of heads or tails in a specified number of coin flips. Assume a participant wants to place a \$10 bet that there would be exactly six heads in 20 coin flips. The participant wants to calculate the probability of this occurring, and therefore, he uses the calculation for the binomial distribution. The probability was calculated as: (20! / (6! * (20 - 6)!)) * (0.50)^(6) * (1 - 0.50) ^ (20 - 6). Consequently, the probability of exactly six heads occurring in 20 coin flips is 0.037, or 3.7%. The expected value was 10 heads in this case, wo the participant made a poor bet. |
Sine, Cosine and Tangent ratios of a triangle. How to write the trig ratios of right triangles
# Sine, Cosine, Tangent Ratios
Practice writing the ratios
### Video Tutorial How to write sohcahtoa ratios, given side lengths
#### What do we mean by 'ratio' of sides?
Sine, cosine and tangent of an angle represent the ratios that are always true for given angles. Remember these ratios only apply to right triangles.
The 3 triangles pictured below illustrate this.
Diagram 1
Although the side lengths are different , each one has a 37° angle, and as you can see, the sine of 37 is always the same!
In other words, $$sin(\red { 37^{\circ} } )$$ is always $$\red {.6 }$$ .
(Note: I rounded to the nearest tenth) That's, of course, why we can use a calculator to find these sine, cosine and tangent ratios.
### Practice Problems
##### Problem 1
Step 1
In the diagram 2, what side is adjacent to $$\angle L$$?
$$\overline{ML}$$.
Step 2
What side is the hypotenuse?
$$\overline{ LN}$$
Step 3
Calculate $$cos(\angle L)$$.
Use sochcahtoa to help remember the ratio.
$cos(\angle \red L) = \frac{adjacent}{hypot} \\ cos(\angle \red L) = \frac{8}{10} \\ = .8$
Step 4
Calculate $$cos(\angle N)$$ (a different angle from prior question, look carefully at letters).
$cos(\angle \red N) = \frac{6}{10} \\ = .6$
##### Problem 2
What is the sine ratio of $$\angle C$$ ?
Remember the sine ratio
$sin(\angle C) =\frac{\text{opp } }{ hypot} \\ sin(\angle C) = \frac{6}{10} \\ sin(\angle C) = \frac{6}{10} \\ sin(\angle C) = .6$
##### Problem 3
What is the cosine ratio of $$\angle C$$ in $$\triangle ABC$$ ?
Remember the cosine ratio
$cos(\angle C) =\frac{\text{adj } }{ hypot} \\ cos(C) = \frac{4}{5} \\ cos(C) = .8$
##### Problem 4
What is the tangent ratio of $$\angle A$$ ?
Use sochcahtoa to help remember the ratio.
$tan(A) = \frac{opposite}{adjacent} \\ = \frac{24}{7} \\= 3.42857142$
##### Problem 5
Find the sine, cosine and tangent of $$\angle R$$.
Use sochcahtoa to help remember the ratios.
$$\text{ for } \angle \red R \\ \boxed { Sine } \\ sin(\red R )= \frac{opp}{hyp} \\ sin(\red R )= \frac{12}{13} \\ sin(\red R )= .923 \\ \boxed{ cosine } \\ cos(\red R)= \frac{adj}{hyp} \\ cos(\red R)= \frac{9}{13} = \\ cos(\red R).69 \\ \boxed{ tangent } \\ tan(\red R)= \frac{opp}{adj} \\ tan(\red R) = \frac{12}{9} \\ tan(\red R)= 1.3$$
##### Problem 6
What is $$sin(\angle X)$$?
$$sin(\red X ) = \frac{opp}{hyp} \\ sin(\red X ) = \frac{24}{25} \\ sin(\red X ) = .96$$
##### Problem 7
What is $$cos(\angle X)$$?
$$sin(\red X) = \frac{adj}{hyp} \\ sin(\red X) = \frac{7}{25} \\ sin(\red X) = .28$$
##### Problem 8
Calculate : $$\text{ a) } sin(\angle H) \\ \text{ b) }cos(\angle H) \\ \text{ c) } tan(\angle H)$$
$$sin(\angle \red H) = \frac{3}{5} \\ sin( \angle \red H) = .6 \\ cos(\angle \red H) =\frac{ 4}{5} \\ cos (\angle \red H)= .8 \\ tan(\angle \red H) = \frac{3}{4} \\ tan(\angle \red H)= .75$$
### Practice Problems II
##### Problem 9
Which angle below has a tangent ratio of $$\frac{3}{4}$$?
Diagram A
Diagram B
Use sochcahtoa to help remember the ratio for tangent.
$$tangent = \frac{opp}{adj}$$
So which angle has a tangent that is equivalent to $$\frac{3}{4}$$?
You only have 2 options. Either $$\angle L$$ or $$\angle K$$. So let's calculate the tangent for each angle.
$tan( \angle K) = \frac{12}{9} \\ \frac{12}{9} \red {\ne} \frac 3 4$
$tan( \angle L) = \frac{9}{12} \\ \frac{9}{12} = \frac 3 4$
##### Problem 10
Which angle below has a cosine of $$\frac{3}{5}$$?
Diagram A
Diagram B
Use sochcahtoa to help remember the ratio for tangent.
$$cosine = \frac{adj}{hyp}$$
So which angle has a cosine that is equivalent to $$\frac 3 5$$?
You only have 2 options. Either $$\angle L$$ or $$\angle K$$. So let's calculate the tangent for each angle.
$cos( \angle K) = \frac{9}{15} \\ \frac{9}{15} \red {\ne} \frac 3 4$
$cos(L) = \frac{12}{15} \\ \frac{12}{15} = \frac 3 5$
##### Problem 11
Which angle below has a tangent $$\approx$$ .29167?
This is a little trickier because you are given the ratio as a decimal; however, you only have two options. Either $$\angle A$$ or $$\angle C$$.
$$tan(\angle A) = \frac{48}{14} \\ tan(\angle A)= 3.42857$$
$$tan(\angle C) = \frac{14}{48} \\ tan(\angle C) = .29167$$
##### Problem 12
Which angle below has a tangent of 2.4?
Again, there are two options (angles R or P), but since your ratio is greater than 1 you might quickly be able to notice that it must be R.
$$tan(\angle P) = \frac{5}{12} \\ tan(\angle P) = 0. 41666$$
$$tan(\angle R) = \frac{12}{5} \\ tan(\angle R) = 2.4$$
### Word Problems
##### Problem 13
Step 1
It's easiest to do a word problem like this one, by first drawing the triangle and labelling the sides. We know the opposite side of $$\angle K$$ and we know the hypotenuse.
To get the tangent ratio we need to know the length of the adjacent side.
How can we find the length of the adjacent side?
Step 2
Use the Pythagorean theorem!
$$a^2 + b^2 = c^2 \\ 3^2 + b^2 = 5^2 \\ b^2 = 5^2- 3^2 = 25-9 = 16 \\ b = 4$$
Now, use the tangent ratio!
Step 3
Use the Pythagorean theorem!
$$tan(k) = \frac{opposite}{adjacent} \\ tan(k) = \frac{3}{4}$$
##### Problem 14
Step 1
Draw this triangle and label the sides:
Remember that the cosine ratio = $$\frac{adjacent}{hypotenuse}$$.
How can we find the length of the opposite side?
(Remember that sine involves the opposite so we need to find that somehow).
Step 2
Use the Pythagorean theorem!
$$a^2 + b^2 = c^2 \\ 7^2 + b^2 = 25^2 \\ b^2 = 25^2- 7^2 = 625 - 49 = 576 \\ b = \sqrt{576} =24$$
Now, use the sine ratio!
Step 3
$$sin(b) = \frac{opposite}{hypotenuse}$$
top | Practice I | Practice II | Word Problems | #Challenge Problem
##### Challenge Problem
Be careful!
In the triangle below, which angle has a sine ratio of 2.6?
There is no angle that has a sine of 2.6.
Remember that the maximum value of the sine ratio is 1.
The same, is true, by the way of cosine ratio (max value is 1). |
Journey into cryptography
# Modular arithmetic
## An Introduction to Modular Math
When we divide two integers we will have an equation that looks like the following:
\dfrac{A}{B} = Q \text{ remainder } R
A is the dividend
B is the divisor
Q is the quotient
R is the remainder
Sometimes, we are only interested in what the remainder is when we divide A by B.
For these cases there is an operator called the modulo operator (abbreviated as mod).
Using the same A, B, Q, and R as above, we would have: A \text{ mod } B = R
We would say this as A modulo B is congruent to R. Where B is referred to as the modulus.
For example:
\begin{eqnarray} \dfrac{13}{5} &=& 2 \text{ remainder } \bf{3} \\ 13 \text{ mod } 5 &=& \bf{3} \\ \end{eqnarray}
## Visualize modulus with clocks
Observe what happens when we increment numbers by one and then divide them by 3.
\begin{eqnarray} \frac{0}{3} &=& 0 \text { remainder } \bf{0} \\ \frac{1}{3} &=& 0 \text { remainder } \bf{1} \\ \frac{2}{3} &=& 0 \text { remainder } \bf{2} \\ \frac{3}{3} &=& 1 \text { remainder } \bf{0} \\ \frac{4}{3} &=& 1 \text { remainder } \bf{1} \\ \frac{5}{3} &=& 1 \text { remainder } \bf{2} \\ \frac{6}{3} &=& 2 \text { remainder } \bf{0} \end{eqnarray}
The remainders start at 0 and increases by 1 each time, until the number reaches one less than the number we are dividing by. After that, the sequence repeats.
By noticing this, we can visualize the modulo operator by using circles.
We write 0 at the top of a circle and continuing clockwise writing integers 1, 2, ... up to one less than the modulus.
For example, a clock with the 12 replaced by a 0 would be the circle for a modulus of 12.
To find the result of A \text{ mod } B we can follow these steps:
1. Construct this clock for size B
2. Start at 0 and move around the clock A steps
3. Wherever we land is our solution.
(If the number is positive we step clockwise, if it's negative we step counter-clockwise.)
## Examples
### 8 \text{ mod } 4 = ?
With a modulus of 4 we make a clock with numbers 0, 1, 2, 3.
We start at 0 and go through 8 numbers in a clockwise sequence 1, 2, 3, 0, 1, 2, 3, 0.
We ended up at 0 so 8 \text{ mod } 4 = \bf{0}.
### 7 \text{ mod } 2 = ?
With a modulus of 2 we make a clock with numbers 0, 1.
We start at 0 and go through 7 numbers in a clockwise sequence 1, 0, 1, 0, 1, 0, 1.
We ended up at 1 so 7 \text{ mod } 2 = \bf{1}.
### -5 \text{ mod } 3 = ?
With a modulus of 3 we we make a clock with numbers 0, 1, 2.
We start at 0 and go through 5 numbers in counter-clockwise sequence (5 is negative) 2, 1, 0, 2, 1.
We ended up at 1 so -5 \text{ mod } 3 = \bf{1}.
## Conclusion
If we have A \text{ mod } B and we increase A by a multiple of \bf{B}, we will end up in the same spot, i.e.
A \text{ mod } B = (A + K \cdot B) \text{ mod } B for any integer \bf{K}.
For example:
\begin{eqnarray} 3 \text{ mod } 10 = 3 \\ 13 \text{ mod } 10 = 3 \\ 23 \text{ mod } 10 = 3 \\ 33 \text{ mod } 10 = 3 \\ \end{eqnarray}
## Notes to the Reader
### mod in programming languages and calculators
Many programming languages, and calculators, have a mod operator, typically represented with the % symbol. If you calculate the result of a negative number, some languages will give you a negative result.
e.g.
-5 % 3 = -2.In a future article we will explain, why this happens, and what it means.
### Congruence Modulo
You may see an expression like:
A \equiv B\ (\text{mod } C)
This says that A is congruent to B modulo C. It is similar to the expressions we used here, but not quite the same.
In the next article we will explain, what it means, and how it is related to the expressions above. |
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# How do you find the exact value of the six trigonometric functions of ${{45}^{\circ }}$?
Last updated date: 13th Jun 2024
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Hint: We first express all the six trigonometric functions. We divide them in primary ratios and their inverse ratios. We also find all possible relations between those ratios. Then we take the angle values of ${{0}^{\circ }}$ for all the six trigonometric functions.
Complete step by step solution:
We first complete the list of all the six trigonometric functions.
The main three trigonometric ratio functions are $\sin \theta ,\cos \theta ,\tan \theta$. The inverse of these three functions is $\csc \theta ,\sec \theta ,\cot \theta$. Also, we can express $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$.
Therefore, the relations are $\csc \theta =\dfrac{1}{\sin \theta },\sec \theta =\dfrac{1}{\cos \theta },\cot \theta =\dfrac{1}{\tan \theta }$.
We can also express these ratios with respect to a specific angle $\theta$ of a right-angle triangle and use the sides of that triangle to find the value of the ratio.
A right-angle triangle has three sides and they are base, height, hypotenuse. We express the ratios in $\sin \theta =\dfrac{\text{height}}{\text{hypotenuse}},\cos \theta =\dfrac{\text{base}}{\text{hypotenuse}},\tan \theta =\dfrac{\text{height}}{\text{base}}$.
Similarly, $\csc \theta =\dfrac{\text{hypotenuse}}{\text{height}},\sec \theta =\dfrac{\text{hypotenuse}}{\text{base}},\cot \theta =\dfrac{\text{base}}{\text{height}}$.
Now we express the values of these ratios for the conventional angles of ${{45}^{\circ }}$.
Ratios angles (in degree) values $\sin \theta$ ${{45}^{\circ }}$ $\dfrac{1}{\sqrt{2}}$ $\cos \theta$ ${{45}^{\circ }}$ $\dfrac{1}{\sqrt{2}}$ $\tan \theta$ ${{45}^{\circ }}$ 1 $\csc \theta$ ${{45}^{\circ }}$ $\sqrt{2}$ $\sec \theta$ ${{45}^{\circ }}$ $\sqrt{2}$ $\cot \theta$ ${{45}^{\circ }}$ 1
Note: We need to remember that in mathematics, the trigonometric functions are real functions which relate an angle of a right-angled triangle to ratios of two side lengths. They are widely used in all sciences that are related to geometry, such as navigation, solid mechanics, celestial mechanics, geodesy, and many others. |
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• Level: GCSE
• Subject: Maths
• Word count: 1042
# Mathematics GCSE - Hidden Faces.
Extracts from this document...
Introduction
Shloimi Werjuka Mathematics Coursework
Mathematics GCSE
Hidden Faces
In order to find the number of hidden faces when eight cubes are placed on a table, in a row, I counted the total amount of faces (68), which added up to 48. I then counted the amount of visible faces (26) and subtracted it off the total amount of faces (48-26). This added up to 22 hidden sides.
I then had to investigate the number of hidden faces for other rows of cubes. I started by drawing out the outcomes for the first nine rows of cubes (below):
I decided to show this information in a table (below):
I decided to show this information on a graph (below):
From this information I have noticed that the number of hidden faces are going up by three each time. In order to find the number of hidden faces for other rows of cubes, it is necessary to have a rule.
Row 2
Row 3
Row 1
Instead of trying to find the number of hidden faces I looked at the visible faces and I took that away from the total amount of faces.
Middle
I can see that 3n is 36 and then I will minus 2. So 36-2 = 16, which is correct, so I now know that the formula is correct.
Another way of working out the nth term is to use the graph. Using the formula y=m+c. The gradient is 3/1=3 and the line passes the y-axis at –2. So the formula is y=3-2. So the nth term would be 3n-2.
Moving on, I now need to work out how to calculate the number of hidden faces when there are more than one row of cubes, I therefore need to find patterns to find the formulas for different number of rows. So I drew the first three diagrams for each number of rows (below):
From these diagrams I can see a few patterns. I can see that with one row, when another cube is placed at the end another three hidden faces are added. With this information I can work out the rest of the column with one row. I noticed that with two rows when another two cubes are placed at the end then another eight hidden faces are added.
Conclusion
256, this equals 240. The number of shown faces is (42) 2 (the front and back) + (52) 2 (both sides) + (45) 1 (only top, bottom is hidden), this adds up to 56. So 240 –56 =184 hidden faces.
Now I will show the above as a formula that will work for all cuboids.
The total amount of faces is (width height length number of faces per cube) WHL6 = 6LWH. The number of visible faces = (2WH) (front and back) + (2LH) (both sides) + (2 WL) (top), this equals 2WH+2LH+WL. So the formula for calculating the number of hidden faces of any cuboid is: 6LWH-(2WH+2LH+WL)
= 6LWH-2WH-2LH-WL.
In order to make sure this is the correct formula, I will check it on another cuboid and see if it works.
H = 3
W = 6
L = 4
I will now work out the number of hidden faces for this cuboid using the formula above. 6LWH-2WH-2LH-WL = 6463-(263)-(243)-(64) = 348 hidden faces. I will now check that 348 is the correct answer by working it out a different way. I counted the visible sides, and then subtracted that from the total amount of faces and I ended up with 348. So from this I can see that the formula is correct.
Y:\svn\trunk\engine\docs\working\acumen6\64067.doc page of
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# Related GCSE Hidden Faces and Cubes essays
1. ## An investigation to look at shapes made up of other shapes (starting with triangles, ...
I am certain that the reason my universal formula ends (Y-2) and not just Y is because when taking the number of sides of a shape into account, you only want to know the ones that are not touching any others.
2. ## Maths-hidden faces
For example, by using the dimensions of a cuboid, I was able to work out the shown faces of a 30-cubed cuboid, (length=5, width=3, height=2): (3 x 5)+ 2(2x3) + 2(5x2) = 47 (W x L)+ 2(H x W) + 2(L x H)
1. ## Cubes and Cuboids Investigation.
No. of cubes with 0 painted faces (Y) 2*3*4 0 2*3*5 0 3*4*5 6 'font-size:14.0pt; '>Y= (A-2)*(B-2)*(C-2) 'font-size:14.0pt; '>Again we can say, as none of the cubes with no painted faces are on the outside of the cuboid, you have to minus 1 from each side of the length, therefore A-2, B-2 and C-2 are going to be in the formula.
2. ## I am doing an investigation to look at shapes made up of other shapes ...
The number of sides of a shape may well be incorporated as well, so I have also put these into the table. Shape composed of: � Max. Perimeter � No. of sides � 10 Triangles 12 3 10 Squares 22 4 10 Hexagons 42 6 One thing that catches my eye straight away is the maximum perimeter.
1. ## Maths Investigation -Painted Cubes
(a - 2)(b - 2)(c - 2) (7 - 2)(8 - 2)(9 - 2) 5 x 6 x 7 = 210 Now to see whether my formulas would work on any sized cubes/cuboids, I decided to test them on a cuboid sized 3 x 5 x 9...
2. ## shapes investigation coursework
X=(14+4-2)/(3-2) � X=16/1 � X=16 C And where P=24, D=5 and the shape is S... X=(24+10-2)/(4-2) � X=32/2 � X=16 C And where P=34, D=4 and the shape is H... X=(34+8-2)/(6-2) � X=40/4 � X=10 C This shows that my universal formula works correctly, and also that my predictions
1. ## volumes of open ended prisms
of the prism to get the volume. To prove that this formula works I will input values for the equilateral triangle and confirm that this formula will give me the volume: A = 8cm, B = 8cm, C = 8cm, Y = 32cm.
2. ## GCSE Maths Coursework - Shapes Investigation
In triangles I did not have an existing formula to work with, so this does not apply to them. Assuming I do find a formula in terms of T=, Q= or H=, I will be able to simply rearrange them to give D= and P=.
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# An ascending gradient of 1 in 100 meets a descending gradient of 1 in 50. The length of summit curve required to provide over turning sight distance of 500 m will be:
Free Practice With Testbook Mock Tests
## Options:
1. 938.25 m
2. 781.25 m
3. 470.25 m
4. 170.25 m
### Correct Answer: Option 2 (Solution Below)
This question was previously asked in
TN TRB Civil 2012 Paper
## Solution:
Concept:
Length of summit curve is given by,
If we assume L > OSD
$${\rm{L}} = \frac{{{\rm{N}}{{\rm{S}}^2}}}{{2{{\left[ {\sqrt {{{\rm{h}}_1}} + \sqrt {{{\rm{h}}_2}} } \right]}^2}}}$$
If we assume L < OSD
$${\rm{L}} = 2{\rm{S}} - \frac{{2{{\left[ {\sqrt {{{\rm{h}}_1}} + \sqrt {{{\rm{h}}_2}} } \right]}^2}}}{{\rm{N}}}$$
Where, N = Deviation angle & S = Stoping distance
Calculation:
Given: Upward gradient = 1 in 100, Downward gradient = 1 in 50, Stopping sight distance = 500 m, and $${\rm{N}} = \frac{1}{{100}} - \left( { - \frac{1}{{50}}} \right) = 0.03$$
For overturning sight distance, h1 = 1.2 m & h2 = 1.2 m
Case 1: If L > OSD
$${\rm{L}} = \frac{{0.03 \times {{500}^2}}}{{2{{\left[ {\sqrt {1.2} + \sqrt {1.2} } \right]}^2}}} = 781.25{\rm{\;m}}$$
Case 2: If L < OSD
$${\rm{L}} = 2 \times 500 - \frac{{2{{\left[ {\sqrt {1.2} + \sqrt {1.2} } \right]}^2}}}{{0.03}} = 680{\rm{\;m}}$$
Thus the assumption is not correct.
∴ Length of summit curve is 781.25 m. |
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# Cách tìm môđun của số phức cực hay, chi tiết – Toán lớp 12
## Introduction
Welcome to Kienthucykhoa.com! In this article, we will dive into the topic of finding the modulus of complex numbers. Whether you are a student or just interested in mathematics, understanding this concept is essential. So let’s get started!
## Modulus of Complex Numbers
The modulus of a complex number, denoted as |z|, is the distance between the origin and the point z in the complex plane. It represents the magnitude or absolute value of the complex number.
### Method
To find the modulus of a complex number z, follow these steps:
1. Square the real part of z and the imaginary part of z.
2. Add these squared values together.
3. Take the square root of the sum obtained.
For any complex number z ∈ C, we have the following result:
|z| = √(x^2 + y^2)
## Illustrative Examples
Now, let’s explore some examples to grasp the concept better.
### Example 1:
Find the complex numbers z that satisfy the following:
A. z1 = -1 + i; z2 = 1 – i
B. z1 = 1 + i; z2 = -1 – i
C. z1 = -1 + i; z2 = -1 – i
D. z1 = 1 + i; z2 = 1 – i
Solution:
We need to find the values of x and y that satisfy x^2 + y^2 = 2.
Therefore, we can choose option D.
### Example 2:
Given the complex number z = 2 – 3i, calculate |z|.
Solution:
The modulus of z can be calculated as follows:
|z| = √(2^2 + (-3)^2)
Choosing option C, we have |z| = √13.
### Example 3:
Consider two complex numbers z1 = 1 + 3i and z2 = 2 – i. Find P = |z1 + z2|.
Solution:
We can find the modulus of the sum of z1 and z2:
z1 + z2 = (1 + 3i) + (2 – i) = 3 + 2i
|z1 + z2| = √((3^2) + (2^2))
Choosing option D, we get P = √13.
## Conclusion
In this article, we explored how to find the modulus of complex numbers. Remember, the modulus represents the magnitude or absolute value of a complex number. By applying the steps mentioned earlier, you can easily calculate the modulus. So the next time you encounter a complex number, you will be confident in determining its modulus.
For more comprehensive lessons and exercises on complex numbers and other mathematical topics, visit Kienthucykhoa.com.
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## Lesson: Using Inverse Operations to Solve Equations Introducing the Concept
Although students have been introduced to the concept of using inverse operations to solve equations, they may not have understood that they are actually performing the same operation on both sides of the equation. By modeling this concept, students will move from solving equations through the use of related facts to solving by using inverse operations.
Materials: pan balance, 50 connecting cubes
Prerequisite Skills and Concepts: Students should know how to write and evaluate algebraic expressions and how to write and solve addition and subtraction equations using inverse operations. They should also know basic multiplication and division facts.
Begin by reviewing the difference between an expression and an equation.
• Say: We've learned two ways to solve an equation. One way is by using a function table. The other way is by using inverse operations.
• Ask: What is the inverse of multiplication? (division) What is the inverse of division? (multiplication) What makes multiplication and division inverse operations?
Elicit that multiplication and division undo each another. Encourage students to provide examples in word or numerical form.
• Say: That's right. Multiplication and division are opposites. When you multiply, you combine equal-sized groups. When you divide, you break apart equal-sized groups.
• Ask: How would you use inverse operations to solve 6m = 18?
Write the equation on the chalkboard. Students may say that since 6 times a number is 18, 18 ÷ 6 must be equal to that number. So, since 18 ÷ 6 = 3, m = 3.
• Say: Let's model this equation by placing cubes on a pan balance. Think of the middle of the pan balance as the equals sign in the equation. Both sides of the pan balance must be level, just like both sides of an equation must be equal.
On one side of the balance place 6 cube trains that are each 3 cubes long. On the other side, count out 18 cubes. Explain what you are doing as you place the cubes in the pans. Then point out that the pans are now balanced.
• Say: Let's see what happens if I multiply this side of the equation by 2. Point to the side with the 6 cube trains that are each 3 cubes long.
• Ask: Right now we have 6 groups of 3 on this side. If I multiply the number of groups by 2, how many groups of 3 will I have?
Students should say 12. Add 6 more cube trains that are each 3 cubes long to that side of the balance.
• Ask: The pans are no longer balanced. What do you think I can do to the other pan to get them to balance? Point to the pan containing the 18 cubes. Elicit that you will need to double the number of cubes in the second pan to get the pans to balance. Add 18 cubes to the second pan.
• Say: We can see that the pans are once again balanced after we multiplied both sides by two. Let's make sure this works with the numerical version of the equation.
Model multiplying both sides of the equation by two on the board and then solve for m.
6m x 2 = 18 x 2 12m = 36 Think: 36 ÷ 12 = 3 m = 3
• Ask: What was the value of the variable in the first equation? (3) What was the value of the variable after both sides of the equation were multiplied by 2? (3)
• Say: Now let's divide both sides of this new equation by 3 to see if we get the same value for m. We have 12 groups of 3 on this side of the equation.
• Ask: If we divide the number of groups by 3, how many groups will we have left?
Students may say 4. Remove 8 cube trains. Repeat with the other side and remove 24 cubes. Discuss the fact that the pans are once again balanced. On the chalkboard, divide both sides of 12m = 36 by 3. Then solve. Discuss the fact that the value of m is still 3. Repeat this procedure, first adding a number to each side of 4m = 12, and then subtracting a number from each side of the new equation to show that the value of the variable remains the same.
• Ask: What happens to the value of a variable when you perform the same operation on both sides of an equation?
Students may say that the value remains the same.
• Say: That's right. When you perform the same operation on both sides of the equation, the value of the variable does not change. This is why we can use inverse operations to solve an equation. Look at our original equation, 6m = 18. To find the value of m, we can divide both sides of the equation by 6. Work through the problem on the board as you talk the students through it.
• Say: Six divided by six is one, leaving one m on the left side of the equation. Eighteen divided by six is three. So m equals three.
You may wish to have students work in pairs to further practice this concept using counters. |
# Free math solving apps
In this blog post, we will show you how to work with Free math solving apps. Let's try the best math solver.
## The Best Free math solving apps
There is Free math solving apps that can make the process much easier. The definite integral is the mathematical way of calculating the area under a curve. It is used in calculus and physics to describe areas under curves, areas under surfaces, or volumes. One way to solve definite integrals is by using a trapezoidal rule (sometimes called a triangle rule). This rule is used to approximate the area under a curve by drawing trapezoids of varying sizes and then adding their areas. The first step is to find the height and width of the trapezoid you want. This can be done by drawing a vertical line down the middle of the trapezoid, and then marking off 3 equal segments along both sides. Next, draw an arc connecting the top points of the rectangle, and then mark off 2 equal segments along both sides. Finally, connect the bottom points of the rectangle and mark off 1 equal segment along both sides. The total area is then simply the sum of these 4 areas. Another way to solve definite integrals is by using integration by parts (also known as partial fractions). This method involves finding an expression for an integral that uses only one-half of it—for example, finding f(x) = x2 + 5x + 6 where x = 2/3. Then you can use this expression in place of all terms except for f(x) on both sides of the equation to get . This method sometimes gives more accurate
A 3x3 matrix is made up of three 3x3 matrices. To solve a 3x3 matrix, we can start by finding the least-squares solution to the following equation: With this method, a computer is used to find the linear combination of all of the three coefficients that minimizes the sum of squares.
There are many equation solvers in the market, but not all of them will give you the answer you’re looking for. So what distinguishes the best from the rest? Here are a few things to look out for: accuracy, ease of use and compatibility. A good equation solver should be able to calculate results quickly and accurately, so it’s important to make sure that it’s compatible with your device. It should also be easy to understand, so if you encounter any hiccups along the way, you can get to where you need to go without getting frustrated. The best three equation solver is one that can solve equations in real-time while also being easy to use and compatible with most devices.
It can be solved by using the elimination method. The elimination method is a process that involves removing all the foods that are causing symptoms. It’s important to remove all possible triggers, as well as foods, drinks, and medications that you suspect could be causing your symptoms. After eliminating these possible triggers, you can start reintroducing one new food at a time until you find the one that causes symptoms. You should avoid eating this one food for a few days and then reintroduce it. If you can eat it comfortably and don’t feel any symptoms, then continue with the reintroduction process until you find the trigger that causes symptoms. Once you find this trigger, avoid eating it and use other measures, such as probiotics, to help relieve your symptoms.
If you've ever taken a math class, you've probably had to do some complicated math problems. These can be tricky at first to solve, but there are a few tricks you can use to make them a little bit easier. Try looking for patterns in the numbers or use your knowledge of basic math to figure out the answer. If the question is too hard, try to break it down into smaller pieces and solve each part separately. Once you understand how each part works, you'll be able to put them together to come up with the final answer. If you're feeling challenged by a problem, don't give up right away. Think about how you might be able to simplify it. For example, if there are two sets of numbers and you know one set is larger than the other, it might be easier to just add one number until they match. You can also look at other possible solutions and see if there's something that might work better for your situation.
It a great app that many of my math and algebra teachers have recommended to me. The app goes through each way you can solve the problem step by step including graphs and different solutions. You can also edit any solution you believe is wrong and can be solve another way.
Mariah Green
This is by far the most helpful app I have ever seen, anyone who is having trouble with math or needs any extra help I would indefinitely recommend this app to you. Almost no adds at all and can understand even my sister's handwriting. Very helpful for those in high level math as it shows you all the working in extreme detail and if you still don't get the question, you can just tap on the step you don't understand and the app will explain in extreme detail.
Aaliyah Murphy
Solving problems algebraically answers How to do math equations Precalculus online help How to solve logarithmic functions Help with math homework app Triangle solver law of sines |
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# Law of Cosines
## Relationship between the three sides and an angle for non-right triangles.
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Practice Law of Cosines
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Law of Cosines
The Law of Cosines is a generalized Pythagorean Theorem that allows you to solve for the missing sides and angles of a triangle even if it is not a right triangle. Suppose you have a triangle with sides 11, 12 and 13. What is the measure of the angle opposite the 11?
#### Guidance
The Law of Cosines is:
$c^2=a^2+b^2-2ab \cdot \cos C$
It is important to understand the proof:
You know four facts from the picture:
$a=a_1+a_2 \qquad \ (1)$
$b^2=a_1^2+h^2 \qquad (2)$
$c^2=a_2^2+h^2 \qquad (3)$
$\cos C=\frac{a_1}{b} \qquad \quad (4)$
Once you verify for yourself that you agree with each of these facts, check algebraically that these next two facts must be true.
$a_2=a-a_1 \qquad \ \ (5, \text{ from } 1)$
$a_1=b \cdot \cos C \qquad (6, \text{ from } 4)$
Now the Law of Cosines is ready to be proved using substitution, FOIL, more substitution and rewriting to get the order of terms right.
$c^2&=a_2^2+h^2 && (3 \text{ again})\\c^2&=(a-a_1)^2+h^2 && (\text{substitute using }5)\\c^2&=a^2-2a \cdot a_1+a_1^2+h^2 && (\text{FOIL})\\c^2&=a^2-2a \cdot b \cdot \cos C+a_1^2+h^2 && (\text{substitute using }6)\\c^2&=a^2-2a \cdot b \cdot \cos C+b^2 && (\text{substitute using }2)\\c^2&=a^2+b^2-2ab \cdot \cos C && (\text{rearrange terms})$
There are only two types of problems in which it is appropriate to use the Law of Cosines. The first is when you are given all three sides of a triangle and asked to find an unknown angle. This is called SSS like in geometry. The second situation where you will use the Law of Cosines is when you are given two sides and the included angle and you need to find the third side. This is called SAS.
Example A
Determine the measure of angle $D$ .
Solution: It is necessary to set up the Law of Cosines equation very carefully with $D$ corresponding to the opposite side of 230. The letters are not $ABC$ like in the proof, but those letters can always be changed to match the problem as long as the angle in the cosine corresponds to the side used in the left side of the equation.
$c^2&=a^2+b^2-2ab \cdot \cos C\\230^2&=120^2+150^2-2 \cdot 120 \cdot 150 \cdot \cos D\\230^2-120^2-150^2&=-2 \cdot 120 \cdot 150 \cdot \cos D\\\frac{230^2-120^2-150^2}{-2 \cdot 120 \cdot 150}&=\cos D\\D&=\cos^{-1} \left ( \frac{230^2-120^2-150^2}{-2 \cdot 120 \cdot 150} \right ) \approx 116.4^\circ \approx 2.03 \ radians$
Example B
Determine the length of side $p$ .
Solution:
$c^2&=a^2+b^2-2ab \cdot \cos C\\p^2&=212^2+388^2-2 \cdot 212 \cdot 388 \cdot \cos 82^\circ\\p^2 & \approx 194192.02 \ldots\\p & \approx 440.7$
Example C
Determine the degree measure of angle $N$ .
Solution: This problem must be done in two parts. First apply the Law of Cosines to determine the length of side $m$ . This is a SAS situation like Example B. Once you have all three sides you will be in the SSS situation like in Example A and can apply the Law of Cosines again to find the unknown angle $N$ .
$c^2&=a^2+b^2-2ab \cdot \cos C\\m^2&=38^2+40^2-2 \cdot 38 \cdot 40 \cdot \cos 93^\circ\\m^2& \approx 3203.1\ldots\\m& \approx 56.59 \ldots$
Now that you have all three sides you can apply the Law of Cosines again to find the unknown angle $N$ . Remember to match angle $N$ with the corresponding side length of 38 inches. It is also best to store $m$ into your calculator and use the unrounded number in your future calculations.
$c^2&=a^2+b^2-2ab \cdot \cos C\\38^2&=40^2+(56.59\ldots)^2-2 \cdot 40 \cdot (56.59\ldots) \cdot \cos N\\38^2-40^2-(56.59\ldots)^2&=-2 \cdot 40 \cdot (56.59\ldots) \cdot \cos N\\\frac{38^2-40^2-(56.59\ldots)^2}{-2 \cdot 40 \cdot (56.59\ldots)}&=\cos N\\N&=\cos^{-1} \left ( \frac{38^2-40^2-(56.59\ldots)^2}{-2 \cdot 40 \cdot (56.59\ldots) } \right ) \approx 42.1^\circ$
Concept Problem Revisited
A triangle that has sides 11, 12 and 13 is not going to be a right triangle. In order to solve for the missing angle you need to use the Law of Cosines because this is a SSS situation.
$c^2&=a^2+b^2-2ab \cdot \cos C\\11^2&=12^2+(13)^2-2 \cdot 12 \cdot 13 \cdot \cos C\\C&=\cos^{-1}\left ( \frac{11^2-12^2-13^2}{-2 \cdot 12 \cdot 13} \right ) \approx 52.02\ldots^\circ$
#### Vocabulary
The Law of Cosines is a generalized Pythagorean Theorem that allows you to solve for the missing sides and angles of a triangle even if it is not a right triangle.
SSS refers to Side, Side, Side and refers to a property of congruent triangles in geometry. In this case it refers to the fact that all three sides are known in the problem.
SAS refers to Side, Angle, Side and refers to a property of congruent triangles in geometry. In this case it refers to the fact that the known quantities of a triangle are two sides and the included angle.
Included angle is the angle between two sides.
#### Guided Practice
1. Determine the length of side $r$ .
2. Determine the measure of angle $T$ in degrees.
3. Determine the measure of angle $S$ in radians.
1. $r^2 = 36^2 + 42^2 - 2 \cdot 36 \cdot 42 \cdot \cos 63$
$r = 41.07 \ldots$
2. $36^2 = (41.07 \ldots)^2+42^2-2 \cdot (41.07 \ldots) \cdot 42 \cdot \cos T$
$T \approx 51.34 \ldots^\circ$
3. You could repeat the process from the previous question, or use the knowledge that the three angles in a triangle add up to 180.
$& 63+51.34 \ldots +S=180\\& S \approx 65.65^\circ \cdot \frac{\pi}{180^\circ} \approx 1.145 \ldots radians$
#### Practice
For all problems, find angles in degrees rounded to one decimal place.
In $\Delta ABC, a=12, b=15,$ and $c=20$ .
1. Find the measure of angle $A$ .
2. Find the measure of angle $B$ .
3. Find the measure of angle $C$ .
4. Find the measure of angle $C$ in a different way.
In $\Delta DEF, d=20, e=10,$ and $f=16$ .
5. Find the measure of angle $D$ .
6. Find the measure of angle $E$ .
7. Find the measure of angle $F$ .
In $\Delta GHI,g=19, \angle H=55^\circ,$ and $i=12$ .
8. Find the length of $h$ .
9. Find the measure of angle $G$ .
10. Find the measure of angle $I$ .
11. Explain why the Law of Cosines is connected to the Pythagorean Theorem.
12. What are the two types of problems where you might use the Law of Cosines?
Determine whether or not each triangle is possible.
13. $a=5, b=6, c=15$
14. $a=1, b=5, c=4$
15. $a=5, b=6, c=10$
### Vocabulary Language: English
Included Angle
Included Angle
The included angle in a triangle is the angle between two known sides.
SAS
SAS
SAS means side, angle, side, and refers to the fact that two sides and the included angle of a triangle are known.
SSS
SSS
SSS means side, side, side and refers to the fact that all three sides of a triangle are known in a problem. |
# What is Taylor Series
Suppose you need to calculate $\sin{138}$ and don’t have a calculator at hand. How’d you do that? The way out is to approximate your function with something more convenient to work with, for example, polynomials: $x, x^2, x^3$ and so on. In this section, we’re going to discuss Taylor series which is an expansion of function into infinite sum of power functions. The series is called in honor of English mathematician Brook Taylor, though it was known before Taylor’s works. Taylor series is applied for approximation of function by polynomials. Such approach allows to replace initial more or less complicated function with the sum of simpler ones. Let’s get started.
Suppose we want to approximate some function $f(x)$ at the vicinity of some point $x_0$.
We need a function which will resemble behavior of the given function $f(x)$ at some neighborhood of the point $x_0$. Surely, at least we can take a constant: $f_0=f(x_0)$ which equals $f(x)$ at the point $x_0$.
Thus, $f_{approx}=f(x_0)$ is a horizontal straight line as you can see. But what about slope of $f(x)$ (in other words, the first derivative) at the point $x_0$? Obviously, $f_{approx}(x)$ doesn’t approximate that because $f’_{approx}=0$ at any point. Initial function $f(x)$ is a curve, while our approximation $f_{approx}$ is just a horizontal line. Not a great approximation indeed. By the way, it’s called zero degree approximation because $f_0$ is a zero degree polynomial function. To construct function which will approximate $f(x)$ along with its first derivative at the point $x_0$ we should add something to $f_0$. And this addend should be chosen so that it’ll be equal to zero at the point $x_0$. Let’s add the following term:
$$f_1(x)=(x-x_0)f’(x_0)$$
Factor $(x-x0)$ provides that our new updated expression holds the value of the initial function at the point $x_0$:
$$f_1(x_0)=(x_0-x_0)f’(x_0)=0$$
Video version of this tutorial is available on our youtube channel:
Thus, we obtain:
$$f_{approx}(x)=f_0+f_1=f(x_0)+(x-x_0)f’(x_0)$$
This is called the first degree approximation because $f_0+f_1$ is the first degree polynomial. As we can see, now the following holds for approximation function $f_{approx}$:
$$f_{approx}(x_0)=f(x_0)$$
$$f’_{approx}(x_0)=f’(x_0)$$
$$f’’_{approx}(x)=0$$
This time we’ve also obtained a straight line, but its slope at the point x_0 is the same as the slope of initial function $f(x)$.
But still our approximation is not very good. Let’s continue. We want now to add something to $f_{approx}$ so that it could approximate the second derivative of $f(x)$. Consider the following term:
$$f_2(x)=\frac{(x-x_0)^2}{2}f’’(x_0)$$
As you can see, along with $(x-x_0)^2$ there appears factor $\frac{1}{2}$. It’s because when we differentiate square, appears $2$ and so $\frac{1}{2}\cdot 2=1$ and we get rid of these integers.
$$f_{approx}(x)= f_0+f_1+f_2=f(x_0)+\frac{(x-x_0)^2}{2}f’’(x_0)$$
We’ve obtained a parabolic (quadratic) function, and that’s why it’s called the second degree approximation. As you may notice, each time we add terms the following condition hold: every new term turns into zero at $x_0$ and also gives zero at derivatives except the highest one it approximates. Particularly, $f_2$ approximates the second derivative. So $f_2$ turns into zero at the point $x_0$, also $f_2’(x_0)=0$. But $f’’_2(x_0)=f’’(x_0)$. The next derivation turns it into zero again.
For future needs we can represent obtained approximation as follows:
$$f_{approx}(x)=f(x_0 )+\frac{f'(x_0 )}{1!} (x-x_0 )+\frac{f^{\prime \prime}(x_0 )}{2!} (x-x_0 )^2$$
Thus, we’ve constructed approximation of the initial function so that it resembles $f(x)$ along with its first and second derivatives at the point $x_0$. Obtained approximation is a parabola.
As we can see, it touches $f(x)$ better than previous straight line. In the same manner we can proceed and construct approximation that would resemble function $f(x)$ along with derivatives of any order in the vicinity of $x_0$. We’d obtain series of polynomial functions $(x-x_0)^n$. The more terms we add the better approximation we get.
Let function $f(x)$ be differentiable in some vicinity of the point $x=x_0$. Series
$$\sum _{k=0}^{\infty} {\frac {f^{(k)}(x_0 )} {k!} (x-x_0 )^k}=f(x_0 )+\frac{f'(x_0 )}{1!} (x-x_0 )+\frac{f^{\prime \prime} (x_0 )}{2!} (x-x_0 )^2+⋯$$
approximates function $f(x)$ in the vicinity of the point $x_0$. This series is called Taylor series of the function $f(x)$ at the point $x-0$.
In case $x_0=0$, the series is written as follows:
$$\sum _{k=0}^{\infty} {\frac {f^{(k)}(0)} {k!}x^k}=f(0)+\frac{f'(0)}{1!}x+\frac{f^{\prime \prime} (0)}{2!}x^2+⋯$$
This series is called MacLaurin series.
Not for any function Taylor series converges. The idea is that in certain cases, not all the time, you can rewrite your function as an infinite sum of other functions. And you only do it in a certain tiny neighborhood of the fixed point of your choice. The following theorem takes place.
Let function $f(x)$ have $n+1$ derivative at some vicinity of the point $x=x_0$. Then, if function can be expanded into series due to powers of $(x-x_0)$ , this expansion is unique and is expressed by the following formula:
\begin{aligned}f(x)=f(x_0 )+\frac{f'(x_0 )}{1!} (x-x_0 )+\frac{f^{\prime \prime} (x_0 )}{2!} (x-x_0 )^2+⋯+R_{n+1}(x)=&\\ \sum _{k=0}^{n} {\frac {f^{(k)}(x_0 )} {k!} (x-x_0 )^k}+…+R_{n+1}(x)\end{aligned}
where $R_{n+1}(x)$ is a remainder term. It can be represented in different ways. Remainder term should be placed if we consider finite number of terms, because in such case our function is approximated only due to certain degree of derivative and no further. This means that approximation and function itself differ and therefrom this remainder term appears. Remainder term, thus, indicates difference between function and its approximation by Taylor series.
But what about $\sin{138}$ we started with, you may ask. In the next section we’ll show you how to obtain Taylor series for common functions and explain how to apply it further in homework tasks. Do math!
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Show first derivative
Show second derivative
Show intervals where $$f(x)$$ is increasing
Show intervals where $$f(x)$$ is decreasing
Show intervals where $$f(x)$$ is concave up
Show intervals where $$f(x)$$ is concave down
## About The Shape of a Function
A function $$f(x)$$ is said to be increasing on an interval $$I$$ if, for all $$x_0$$ and $$x_1$$ in $$I$$ with $$x_0\lt x_1$$, we have $$f(x_0) \lt f(x_1)$$. This means that as you trace the graph from left to right on such an interval, your finger would move up. Similarly, a function is decreasing on an interval $$I$$ if, for all $$x_0$$ and $$x_1$$ in $$I$$ with $$x_0\lt x_1$$, we have $$f(x_0) \gt f(x_1)$$.
A function $$f(x)$$ is said to be concave up on an interval $$I$$ if its first derivative is increasing on $$I$$. Graphically, this means the function is curved and forming a bowl shape. Similarly, a function is concave down when its first derivative is decreasing. When a function is concave down, it is curved and forming an upside down bowl (open umbrella) shape.
The graph of a function $$f(x)$$ is closely related to the graphs of its first and second derivatives:
• When the graph of the function $$f(x)$$ is increasing, the value of $$f'(x)$$ is positive, so the graph of $$f'(x)$$ will lie above the $$x$$-axis.
• When the graph of the function $$f(x)$$ is decreasing, the value of $$f'(x)$$ is negative, so the graph of $$f'(x)$$ will lie below the $$x$$-axis.
• When the slope of the line tangent to $$f(x)$$ is 0, the value of $$f'(x)$$ is 0, so the graph of $$f'(x)$$ will have an $$x$$-intercept.
• When the graph of the function $$f(x)$$ is concave up, the value of $$f''(x)$$ is positive, so the graph of $$f''(x)$$ will lie above the $$x$$-axis.
• When the graph of the function $$f(x)$$ is concave down, the value of $$f''(x)$$ is negative, so the graph of $$f''(x)$$ will lie below the $$x$$-axis.
• When the graph of $$f(x)$$ has an inflection point, the value of $$f''(x)$$ is 0, so the graph of $$f''(x)$$ will have an $$x$$-intercept.
• If $$f(x)$$ is a polynomial with degree $$n$$, then $$f'(x)$$ will be a polynomial of degree $$n-1$$ and $$f''(x)$$ will be a polynomial of degree $$n-2$$.
## Using the Applet
The applet will generate the graph of a random polynomial function. Use the controls to highlight the intervals of the graph where it is increasing, decreasing, concave up, and concave down, and to display the first and second derivatives. |
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# MEAN vs. MEDIAN
Mean & Median is one of the most innate parts of statistical mathematics. The branch deals with assembly & analysis of data in different and diverse ways for intended calculation. Mean along with median serve these purposes distinctly.
## What is Mean & Median?
Mean is stated as an average of some numbers. In simple terms, the mean is the most probable result of the combination of given data. Median states the middle-most data set sorted in specific ways with an equal number of higher and smaller numbers or data.
One of the significant differences between mean vs. median is outliers; it has no impact on the median but may have a considerable effect on the mean number. When calculating the mean, each of the numbers affects the final number whereas outliers will have a minimum or no impact on the final numbers
### How To Calculate Mean?
Steps required to calculate the Mean
Step 1: Add all the quantities to get a total number
Step 2: Count the number in a series
Step 3: Now, divide the total found in Step 1 by the number of Step 2.
You will be able to find the mean of quantities and numbers by following the above steps.
In Microsoft Excel, you can also use the formula =AVERAGE(N1, N2, N3……)
### Three kinds of Mean explained below:
• Arithmetic Mean: – It is calculated as the sum of the terms divided by the number of terms.
• For a, b, c being terms and n, i.e., the number of terms being three
Arithmetic mean given as (a+b+c)/3
• Geometric Mean: – It is calculated as the product of terms with the power of inverse of the number of terms.
• For a, b, c being terms and n i.e., number of terms being three
Geometric mean is given as ∛(a x b x c)
• Harmonic Mean: – It is calculated as the number of terms divided by the summation of inverse the of each term.
• For a, b, c being terms and n i.e., number of terms being three
Harmonic mean defined as 3/(1/a+1/b+1/c)
#### How To Calculate Median?
The following steps will guide you to find the median of the number series.
Step 1: Arrange the numbers in descending to ascending order.
Step 2: Count the number of units in a series.
Step 3: Check if the number in Step 2 is Even or Odd
Step 4: If the Step -3 is Even, take the two digits in the center and find the average and if it is Odd, then take the middle number as the Median
For example, the series of numbers is 91, 94, 93, 89, 95, 88, 97, 87, and 100.
Step-1: 87,88, 89, 91, 93, 94,95,97, 100
Step -2: Total units in the series is 9
Step-3: Odd
Step-4: Center of the series is 5th number, median is 93
In Microsoft Excel, you can also use the formula =MEDIAN(N1, N2, N3……)
finance/mean-vs-median |
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### Course: Arithmetic>Unit 18
Lesson 10: Adding & subtracting negative fractions
# Adding fractions with different signs
Use a number line to add fractions with different signs. Created by Sal Khan and Monterey Institute for Technology and Education.
## Want to join the conversation?
• Well if you want to do it without a number line then you just have to do the math. What I mean is that you have to first know how to add and subtract fractions and you have to learn how to learn how to add and subtract mixed fractions, also you have to learn how to add and subtract integers. So yeah it takes a long time to do this kind of math to learn how solve a problem without a number line. Understand?
• Seems a bit much for one equation, is there any simpler/faster way to do it?
• Yes, you can do just do the math without the visualization alone, the number line is just for better understanding. It is a bit complicated, but it is the fastest and most simple way to handle the question if you understand the math to it.
3 1/8 + 3/4 +(-2 1/6) = x
Let's solve for x.
``x = 3 1/8 + 3/4 - 2 1/6 [delete the ()]x = 3 1/8 + 6/8 - 2 1/6 [adjust the fractions for addition]x = 25/8 + 6/8 - 2 1/6 [convert into improper fraction]x = 31/8 - 2 1/6 [do the addition]x = 31/8 - 13/6 [convert into improper fraction]x = 93/24 - 52/24 [adjust the fractions for subtraction]x = 41/24``
or...
``x = 1 17/24``
• When I worked through this problem, I left them as mixed numbers and didn’t convert them to improper fractions. I came to the same answer (41/24 or 1&17/24). Did you convert them for a reason other than preference? Thanks
• Yes, the "short way" of doing this problem is just working on the fractions of the mixed numbers (1/8+3/4-1/6) and subtracting the numbers (3-2). The reason this method works is because, in reality, 1&17/24 is the same as 1+17/24 (and this goes for all mixed numbers). Sal converted the mixed numbers mostly because of preference, but it is easier to convert to improper fractions when the problem would result in a negative fraction.
For example:
3&1/8 - 3/4 -2&1/6
See, now the fractions added up would result in a negative fraction (-19/24) and then you would still have to subtract 19/24 from 1. So, you would have to convert 1 to 24/24 and subtract. By converting all the mixed numbers into improper fractions, the sum of all the improper fractions is the final answer.
In conclusion, both methods work (Sal's method is the one taught in schools, normally) and, depending on the question (whether the fractions of the mixed numbers come out negative), one can work faster than the other.
• this is so confusing and there is so much to remember
• Its alright could you explain what you are having trouble understanding?
• crazy
• I tried to follow the lesson and I came to -11/6 instead of -13/6.
Could we get a clarification on this or I missed a step?
Look forward to her from you guys! example 6. -2 = -12/6 + 1/6= -11/6
• 6-2 is not the same as -12/6+1/6 but -12/6+1/6 is -11/6
• While in the video type the word awesome, and the time bar turns rainbow, it is really cool.
• Ain't no way
(1 vote)
• Is it the same principle but with fractions?🤔
• what are the steps
(1 vote)
• Well, the first step would be to simplify the equation. For example, if I had the problem
-6 3/7 - -(2 7/9) - 8 3/10
then I could simplify it to:
-6 3/7 + 2 2/5 - 8 3/10
Then, you add the fractions together, and the first step for that is to find the common denominator. So, the common denominator of 7, 5, and 10, (the denominators), would be 70, and change the denominators to 70, and multiply the numerators by the number you multiplied the denominator by.
Here's an example, using our above equation;
-6 3/7 + 2 2/5 - 8 3/10
So, we change the denominators to 70, so our equation would be:
-6 3/70 + 2 2/50 - 8 3/70
But now our numerator wouldn't look right, so we have to multiply the numerators by the number we multiplied the denominator by.
So, for -6 3/70 to go from 7 to 70, we would have to multiply by 10, so we multiply 3 by to to get 30, so from:
-6 3/7 to
-6 30/70
The actual value of the fraction stays the same, though this format makes it easier to add/subtract.
So, if we repeat this process, our equation will be
-6 30/70 + 2 28/70 - 8 21/70
Now we do the actual addition. Let's focus on the whole numbers first.
So -6 + 2 - 8
What is 6 to the left of 2? -4! And what is 8 to the left of -4? -12! You can use a number line to help you out. So, the sum of all the whole numbers is 12.
Now let's do the fractions:
-30/70 + 28/70 - 21/70 = -23/70
Make sure you include all the minuses and plusses that were in the equation.
The last step is to add the two products.
-12 + -23/70 = -12 23/70
You then simplify, but this specific fraction is already simplified, so our final answer was:
-12 23/70
I know this looks long and like a bore to read through, but I still hope this helps you! Have a good day!😁 |
# Common Core: High School - Geometry : Similarity, Right Triangles, & Trigonometry
## Example Questions
← Previous 1 3 4 5 6 7 8 9 10 11 12
### Example Question #1 : Similarity, Right Triangles, & Trigonometry
If the red figure is an object and the blue figure is an object after dilation, what is the scale factor?
Explanation:
The best way to solve for the scale factor is to find the same vertex from each object, and divide their components.
Let's use and
Let's divide the x-coordinates together.
Since we are going from the smaller object to the larger object, we know that our scale factor is going to be greater than one.
So our final answer is going to be.
### Example Question #1 : Dilations Given Center And Scale Factor: Ccss.Math.Content.Hsg Srt.A.1
If the red figure is an object and the blue figure is an object after dilation, what is the scale factor?
Explanation:
The best way to solve for the scale factor is to find the same vertex from each object, and divide their components.
Let's use
and
Let's divide the x-coordinates together.
Since we are going from the smaller object to the larger object, we know that our scale factor is going to be greater than one.
So our final answer is going to be.
### Example Question #111 : High School: Geometry
If the red figure is an object and the blue figure is an object after dilation, what is the scale factor?
Explanation:
The best way to solve for the scale factor is to find the same vertex from each object, and divide their components.
Let's use and
Let's divide the x-coordinates together.
Since we are going from the smaller object to the larger object, we know that our scale factor is going to be greater than one.
So our final answer is going to be.
### Example Question #1 : Dilations Given Center And Scale Factor: Ccss.Math.Content.Hsg Srt.A.1
If the red figure is an object and the blue figure is an object after dilation, what is the scale factor?
Explanation:
The best way to solve for the scale factor is to find the same vertex from each object, and divide their components.
Let's use and
Let's divide the x-coordinates together.
Since we are going from the smaller object to the larger object, we know that our scale factor is going to be greater than one.
So our final answer is going to be.
### Example Question #3 : Similarity, Right Triangles, & Trigonometry
If the blue figure is an object and the red is an object after dilation, what is the scale factor?
Explanation:
The best way to solve for the scale factor is to find the same vertex from each object, and divide their components.
Let's use and
Let's divide the x-coordinates together.
Since we are going from the larger object to the smaller object, we know that our scale factor is going to be less than one.
So our final answer is going to be.
### Example Question #5 : Similarity, Right Triangles, & Trigonometry
If the blue figure is an object and the red is an object after dilation, what is the scale factor?
Explanation:
The best way to solve for the scale factor is to find the same vertex from each object, and divide their components.
Let's use and
Let's divide the x-coordinates together.
Since we are going from the larger object to the smaller object, we know that our scale factor is going to be less than one.
So our final answer is going to be.
### Example Question #1 : Dilations Given Center And Scale Factor: Ccss.Math.Content.Hsg Srt.A.1
If the blue figure is an object and the red is an object after dilation, what is the scale factor?
Explanation:
The best way to solve for the scale factor is to find the same vertex from each object, and divide their components.
Let's use and
Let's divide the x-coordinates together.
Since we are going from the larger object to the smaller object, we know that our scale factor is going to be less than one.
So our final answer is going to be.
### Example Question #2 : Dilations Given Center And Scale Factor: Ccss.Math.Content.Hsg Srt.A.1
If the red figure is an object and the blue figure is an object after dilation, what is the scale factor?
Explanation:
The best way to solve for the scale factor is to find the same vertex from each object, and divide their components.
Let's use and
Let's divide the x-coordinates together.
Since we are going from the smaller object to the larger object, we know that our scale factor is going to be greater than one.
So our final answer is going to be.
### Example Question #3 : Dilations Given Center And Scale Factor: Ccss.Math.Content.Hsg Srt.A.1
If the blue figure is an object and the red is an object after dilation, what is the scale factor?
Explanation:
The best way to solve for the scale factor is to find the same vertex from each object, and divide their components.
Let's use and
Let's divide the x-coordinates together.
Since we are going from the larger object to the smaller object, we know that our scale factor is going to be less than one.
So our final answer is going to be.
### Example Question #1 : Dilations Given Center And Scale Factor: Ccss.Math.Content.Hsg Srt.A.1
If the red figure is an object and the blue figure is an object after dilation, what is the scale factor?
Explanation:
The best way to solve for the scale factor is to find the same vertex from each object, and divide their components.
Let's use and
Let's divide the x-coordinates together.
Since we are going from the smaller object to the larger object, we know that our scale factor is going to be greater than one.
So our final answer is going to be.
← Previous 1 3 4 5 6 7 8 9 10 11 12 |
# A Simpler Solution to an Optimization Problem.
The Problem:
My Solution: Let me model this situation in the first quadrant of the $$x$$-$$y$$ plane. Let us take the line $$y=mx+c$$ as the ladder (moving in the first quadrant). This is how the ladder moving in the corridor looks:
The $$y$$-intercept of the line is $$c$$ and the $$x$$-intercept is $$\left(-\frac{c}{m}\right)$$. Since the ladder is moving in the first quadrant, we have $$m<0 \text{ and } c>0$$ Let $$L$$ be the length of the ladder. Then, $$L^2=c^2+\left(\frac{c}{m}\right)^2$$
or
$$c=-\frac{mL}{\sqrt{1+m^2}} \text{ (Negative sign because m<0)}$$
and therefore, the equation of the ladder becomes $$y=mx-\frac{mL}{\sqrt{1+m^2}}$$
As the ladder is transported through the corridor, the distance between the ladder and the corner becomes smaller up to a minimum and then start increasing again. The goal here is to choose $$L$$ such that the ladder just touches the corner as it clears the corridor. This $$L$$ would be the max length of the ladder.
Let the gap along the $$y$$-direction between the corner and the ladder be $$s$$. Let the co-ordinates of the corner be $$(a,b)$$. Then, $$s=b-ma+\frac{mL}{\sqrt{1+m^2}}$$
For a given ladder, this distance hits a minimum value for a certain value of $$m$$. Let's find that out $$\frac{ds}{dm}=-a+\frac{L}{\sqrt{1+m^2}}-\frac{Lm^2}{(1+m^2)^\frac{3}{2}}=0$$
Solving which we obtain the value of $$m$$ which makes $$s$$ hit its minimum value $$m=-\left\{\left(\frac{L}{a}\right)^\frac{2}{3}-1\right\}^\frac{1}{2} \text{ (Negative because m<0)}$$
Therefore, $$s_{\text{min}}=b+a\left\{\left(\frac{L}{a}\right)^\frac{2}{3}-1\right\}^\frac{1}{2}+L\left\{1-\left(\frac{a}{L}\right)^\frac{2}{3}\right\}$$
The value of $$s_{\text{min}}$$ in the case of this problem is zero. Therefore, solving the above equation, we get $$L=a\left\{\left(\frac{b}{a}\right)^\frac{2}{3}+1\right\}^\frac{3}{2}=\left(b^\frac{2}{3}+a^\frac{2}{3}\right)^\frac{3}{2}$$
Plugging, $$a=8$$ and $$b=6$$ as asked in the question we get $$L=19.7313 \approx 20 \text{ feet}$$
Is this correct? If yes, can we make an easier model to solve this problem? (Only in the context of calculus)
Your answer is correct. Note that the question says to round $$\color{red}{down}$$ to the nearest foot.
Consider
You have two right-angled triangles, and they share the same angle $$\theta$$ since the horizontal black and green lines are parallel. Then you can shown that the length of the ladder is given by
$$L(\theta)=\frac{6}{\sin(\theta)}+\frac{8}{\cos(\theta)}$$
for $$0<\theta<\frac{\pi}{2}$$. Then
$$L'(\theta)=0\implies \tan^3(\theta)=\frac{3}{4}\implies \theta \approx 0.73752$$ $$\implies L\approx 19.7313$$
• This model assumes the ladder to be in contact with the corner at all times. But the ladder touches the corner only momentarily. Sep 18 at 16:56
• This is the length of the longest ladder you can carry horizontally around the corner of the corridor. Sep 18 at 19:35
Your animation of the sliding ladder is alright. The answer given here cannot get any simpler. There is corner contact only at a single instant.
Easier modeling is by using symbols with trig. Start with making a sketch of the corner.
$$\text{For easy trig typing using shorthand}\;$$ $$s = \sin \phi, c = \cos \phi \text{, slant ladder length = }$$
$$L=\frac {b}{s}+ \frac{a}{c} \tag 1$$
Differentiate with respect to $$\phi$$
$$\frac{-bc}{s^2}+\frac{as}{c^2}=0 \tag2$$
Simplify to find $$\tan \phi$$
$$\frac{s}{c}=\tan \phi= {\left( \frac {b}{a}\right)}^ {\frac13} \tag 3$$
Construct right triangle with Pythagorean thm ( even if the dimension does not tally to linear dimension ) resolving $$(s,c)$$ to conveniently plug their values into (1) and then simplify
$$L={\left( a^{\frac23}+ b^{\frac23}\right)}^ {\frac32} \tag 4$$
Actually it is an Astroid envelope of the sliding ladder having equation
$$x^{\frac23}+ y^{\frac23}= L^{\frac23}$$
on to which you juxtaposed a sharp corner from right side.
The 3 lines ( two blue,one red) have the same length 19.7313 units. Only the red line touches the corner $$(8.6)$$ . The blue lines do not touch the corner, there is considerable gap/clearance.
Now plug in numerical values for symbols
$$a=6,\; b=8,\;L \approx 19.7313 \;ft, \; \phi= 42.2568^{\circ}; \tag 5$$
If corridor widths are changed you know what to do.
• This model assumes the ladder to be in contact with the corner at all times. But the ladder touches the corner only momentarily. Sep 18 at 16:57
• Not at all. The model does not assume contact with corner always but assumes contact only with the wall and floor along which the ladder slides. This is ensured in the formulation. Your misconception can be hopefully cleared by sketch just now added. To drive home this point I mentioned already about the Astroid that touches the corner at a single point. All other positions of the ladder keeps it clear from re-entrant corner without any contact or interference as shown. Sep 18 at 20:58
$$x = {b\over \sin \varphi}$$ $$\ell- x = {a\over \cos \varphi}$$ so $$\boxed{\ell = {b\over \sin \varphi}+ {a\over \cos \varphi}}$$
Now take the derivative of it with respect to $$\varphi$$
• This model assumes the ladder to be in contact with the corner at all times. But the ladder touches the corner only momentarily. Sep 18 at 16:57
• So?..............................
– Aqua
Sep 18 at 17:08 |
# Solving Examples of Trigonometry
Trigonometry is one of the branch in mathematics, it is having so many examples and applications. Trigonometry is mainly used in civil engineering,it is very useful in engineering field. it is having angles, using the angles we can use in construction field.Its applications is used in so many math divisions.
Here we are discussing the examples of trigonometry
Basic formulas in trigonometry:
## Solved examples of trigonometry
Example1:
Show that cos 3A = 4cos^3A - 3cosA
Proof :
cos(3A) = cos(2A+A)
We know formula cos(A+B) = cosAcosB - sinAsinB
Cos ( 2A + A ) = cos(2A)cos(A) - sin(2A)sinA
cos(2A) = cos(A+A)
= cos²A - sin²A
= cos²A - (1-cos²A)
= 2cos²A - 1
Hence, Cos 3A = (2cos²A-1)cosA - 2sin²AcosA
= 2cos³A - cosA - 2(1-cos²A)cosA
= 2cos³A - cosA - 2cosA + 2cos³A
= 4cos³A - 3cosA
So LHS= RHS
Example 2:
Find the exact value of 1. tan 30 2. cos 60 and 3. cos 45
Solution:
Here we using right angle triangle properties
A 30-60-90 right triangle has sides that are in the ratio of
1 : √3 : 2
And, a 45-45-90 right triangle has sides that are
1 : 1 : √2
From these ratios
tan 30 º = (√3)/3
cos 60 º = 1/2
cos 45 º = (√2)/2
Example3:
Prove that cos(60+A)cos(60-A)-sin(60+A)sin(60-A)= 0.5
Solution:
cos(60+A)cos(60-A)-sin(60+A)sin(60-A)
= cos(60+A+60-A)
= cos(120)
=Cos(180-60 )
= cos60
= 1/2
= 0.5
So LHS= RHS
Example 4:
What is the exact value of cos150
Solution:
cos 150 = cos (180 - 30)
We know formula cos (A - B) = cos A cos B + sin A sin B
cos 150 = cos 180 cos 30 + sin 180 sin 30
cos 150 = (-1) cos 30 + (0) sin 30
cos 150 = - cos 30
= -`sqrt(3)/2` |
# Prove $\blacktriangle ABC$ is a right triangle
Am I missing something here? My homework says to prove that the given triangle is a right triangle, but it does not appear to be a right triangle mathematically.
Let $A =(-3, 2)$, $B=(1, 0)$, and $C=(4,6)$.
Prove that $\blacktriangle ABC$ is a right triangle.
I have tried the slope method and it didn't work, I also tried the distance (Pythagorean) method and that didn't work either.
• Work out the squares of the side lengths and show $AB^2+BC^2=AC^2$ - no square roots required! Commented Sep 23, 2015 at 21:52
• It works with both methods. Make a drawing first to see what angle is right.
– A.Γ.
Commented Sep 23, 2015 at 21:54
Let's try the Pythagorean theorem:
Side $AB = [1- -3,0-2] = [4,-2]$, with length $\sqrt{20}$
Side $BC = [4- 1,6-0] = [3,6]$, with length $\sqrt{45}$
Side $AC = [4- -3,6-2] = [7,4]$, with length $\sqrt{65}$
$|AB|^2 + |BC|^2 = (\sqrt{20})^2+(\sqrt{45})^2 = 20 + 45 = 65 = (\sqrt{65})^2 = |AC|^2$
So the triangle must be right
• I would call it by the name Law of cosines rather than Pythagorean theorem (which for me is just the other direction), but it is okay. Commented Sep 23, 2015 at 22:11
I tried to solve it using the slope concept and indeed $A$, $B$ and $C$ are endpoints of a right triangle.
The slope of the line connecting $AB$ is $\frac{-1}{2}$.
The slope of the line connecting $AC$ is $\frac{4}{7}$.
Lastly, the slope of the line connecting $BC$ is 2.
Now since the product of the slope of line connecting $AB$ and the line connecting $BC$ is $\frac{-1}{2}\cdot 2=-1$, the line connecting $AB$ and the line connecting $BC$ are perpendicular by including line connecting $AC$ one can generate a triangle. This triangle a right triangle since the line connecting $AB$ and the line connecting $BC$ are perpendicular .
Translate the triangle by $(-1,0)$. The new coordinates of the vertices become: $$A(-4,2),\quad B(0,0),\quad C(3,6)$$ and since the dot product between $A$ and $C$ is zero, $\widehat{ABC}$ is a right angle.
Add points $D=(-3,0)$ and $E=(4,0)$ and show that $\Delta ABD \sim\Delta BCE$. Then show that $\measuredangle ABC=90^\circ$.
Do you see why the slopes of perpendicular lines are negative reciprocals of each other?
If you calculate again the distances you will see that $$AC^2=AB^2+BC^2$$ |
# Prealgebra for Two-Year Colleges/Appendix (procedures)/Lowest common multiple
To find the Lowest Common Multiple (LCM) of several numbers, we first express each number as a product of its prime factors.
For example, if we wish to find the LCM of 60, 12 and 102 we write
$\begin{matrix} 60=2^2 \cdot 3 \cdot 5 \\ 12=2^2 \cdot 3 \\ 102=2 \cdot 3 \cdot 17 \end{matrix}$
The product of the highest power of each different factor appearing is the LCM.
For example in this case, $2^2\cdot3\cdot5\cdot17=1020$. You can see that 1020 is a multiple of 12, 60 and 102 ... the lowest common multiple of all three numbers.
Another example: What is the LCM of 36, 45, and 27?
Solution: Factorise each of the numbers
$\begin{matrix} 36=2^2\cdot3^2\\ 45=5\cdot3^2 \\ 27=3^3 \end{matrix}$
The product of the highest power of each different factor appearing is the LCM, i.e;
$2^2\cdot5\cdot3^3=540$
#### Properties of the LCMEdit
If the LCM of the numbers is found and 1 is subtracted from the LCM then the remainder when divided by each of the numbers whose LCM is found would have a remainder that is 1 less than the divisor. For example if the LCM of 2 numbers 10 and 9 is 90. Then 90-1=89 and 89 divided by 10 leaves a remainder of 9 and the same number divided by 9 leaves a remainder of 8. |
Question
# To solve:\displaystyle{\left(\begin{matrix}{x}-{2}{y}={2}\\{2}{x}+{3}{y}={11}\\{y}-{4}{z}=-{7}\end{matrix}\right)}
Alternate coordinate systems
To solve:
$$\displaystyle{\left(\begin{matrix}{x}-{2}{y}={2}\\{2}{x}+{3}{y}={11}\\{y}-{4}{z}=-{7}\end{matrix}\right)}$$
2020-10-22
Calculation:
We have to solve
$$\displaystyle{x}-{2}{y}={2}\ldots{\left({1}\right)}$$
$$\displaystyle{2}{x}+{3}{y}={11}\ldots{\left({2}\right)}$$
$$\displaystyle{y}-{4}{z}=-{7}\ldots{\left({3}\right)}$$
We have to choose equation 1 and equation 2 and eliminate the same variable x.
Multiply both sides of first equation by -2 and we have to add equation 1 and 2, we get
$$\displaystyle{\left({1}\right)}\times-{2}\to-{2}{x}+{4}{y}=-{4}$$
$$\displaystyle{\left({2}\right)}\to{2}{x}+{3}{y}={11}$$
$$\displaystyle{7}{y}={7}$$
$$\displaystyle{y}={1}$$
Now we have to choose a different pair of equations and eliminate the same variable.
Substitude $$\displaystyle{y}={1}$$ in equation 1, we get
$$\displaystyle{x}-{2}{y}={2}$$
$$\displaystyle{x}-{2}{\left({1}\right)}={2}$$
$$\displaystyle{x}-{2}={2}$$
$$\displaystyle{x}={2}+{2}$$
$$\displaystyle{x}={4}$$
Substitute $$\displaystyle{y}={1}$$ in equation 3, we get
$$\displaystyle{y}-{4}{z}=-{7}$$
$$\displaystyle{1}-{4}{z}=-{7}$$
$$\displaystyle-{4}{z}=-{7}-{1}$$
$$\displaystyle-{4}{z}=-{8}$$
Divide both sides by -4
$$\displaystyle\frac{{-{4}{z}}}{{-{4}}}=\frac{{-{8}}}{{-{4}}}$$
$$\displaystyle{z}={2}$$
The solution set for the system is {(4, 1, 2)}
To check:
Substitute $$\displaystyle{x}={4},{y}={1}$$ in equation 1
$$\displaystyle{x}-{2}{y}={2}$$
$$\displaystyle{4}-{2}{\left({1}\right)}={2}$$
$$\displaystyle{4}-{2}={2}$$
$$\displaystyle{2}={2}$$[True]
Substitute $$\displaystyle{x}={4},{y}={1}$$ in equation 2
$$\displaystyle{2}{x}+{3}{y}={11}$$
$$\displaystyle{2}{\left({4}\right)}+{3}{\left({1}\right)}={11}$$
$$\displaystyle{8}+{3}={11}$$
$$\displaystyle{11}={11}$$ [True]
Substitute $$\displaystyle{y}={1},{z}={2}$$ in equation 3
$$\displaystyle{y}-{4}{z}=-{7}$$
$$\displaystyle{1}-{4}{\left({2}\right)}=-{7}$$
$$\displaystyle{1}-{8}=-{7}$$
$$\displaystyle-{7}=-{7}$$ [True]
The solution set $$\displaystyle={\left\lbrace{\left({4},{1},{2}\right)}\right\rbrace}$$
Conclusion:
The solution set for the system is {(4, 1, 2)} |
Square Root
# What is the square root of 27? How to find the square root of 27?
Written by Prerit Jain
Updated on: 12 Aug 2023
### What is the square root of 27? How to find the square root of 27?
The square root of 27 is 5.1961. The square root of 27 is denoted using √27 or 271/2. In the radical form, the square root of 27 is represented as 3√3.
The square root of a number is determined when a quantity of a number is produced when multiplied by itself or a factor of a number when multiplied by itself gives the original number.
In simple terms, when a value is multiplied by itself it gives the original number. For a better understanding,
the number 15 when multiplied by itself (15×15) = 225. The square root for (√225) is 15.
Square root of √27 = √27±3√3
The square root of 27 in decimal form,
√27= 5.1961
The square root of 27 in exponent form,
271/2= 5.1961
## What is the square root of 27?
The square root of a number when multiplied itself gives the original number. The square root of 27 is expressed as √27
√27 = √(Number) × √(Number)
√27 = (5.1961×5.1961)
√27 = √(5.1961)2
Now, remove the square on the right-hand side we get.
√27 = 5.1961
Thus, by multiplying 5.1961 we get the number 27.
## The square root of 27 in radical form
The number 27 is a perfect cube number and hence can be expressed in radical form.
√27 = √3×√3×√3
√27 = 3×√32
√27 = 3√3
## The calculation for the square root of 27
To find the square root of any number first the given number is usually checked to determine whether they are a perfect square number or not. The square root of the number is found using the long division method.
For the numbers with a perfect square, it is easy to find the square roots and it is a bit tough for non-square values. Numbers like 4,9, 16,25, etc., are perfect squares. Numbers such as 2,3,7 and 18 are not perfect square numbers.
27 is not a rational number or a perfect square number.
Prime factorization cannot be used to find the square root of 27. Hence, the long division method is used to find the square root of 27.
## Methods to find the square root
There are three methods to find the square root of a number
• Prime factorization method
• Long division method
• Repeated subtraction
These methods described above do not apply to finding the square root of any number.
### Prime factorization method
To find the square root of a number using the prime factorization method first knows the prime factors of the numbers. Let us take n as a prime number, by grouping their squares we get n2 now multiplying them we get the square root of the number.
• Step 1: Find the prime factor for the given number
√27 = 3×3×3
• Step 2: Pair the prime factors
√27 = 3×32
Take square on both sides
√27=√3×√32
You can cancel the square with the square root we get
√27 = 3√3
• Step 3: Multiply the factors
The √3 = 1.732
√27 = 3×1.732
√27 = 5.196
### Long division method
Long division is one of the easiest methods to find the square root of any number. It was the preferred method to be used for the non-perfect square numbers. Find the integer that can divide the number and proceed with the long division.
Steps to find the square root of 27:
• Step 1: Find the smallest integer that can divide the number.
• Step 2: Keep following the long division using divisor and dividend.
• Step 3: When the particular number of satisfaction is reached the quotient is the square root of the number.
### Find the square root of 27
• Step 1: Find the smallest integer that can divide the number. 27 is not a perfect root number and 5 is the nearest number that can divide it.
• Step 2: Keep following the long division using divisor and dividend.
• Step 3: When the particular number of satisfaction is reached the quotient is the square root of the number. Hence, the √ 27 = 5.1961
## Solved Examples
Example 1: Find the square root of 225 using the prime factorization method
Solution:
Prime factor of 225 = 3×3×5×5
Now, multiply the squares 3×5 = 15
Hence, the square root of 225 is
√225 = 15.
Example 2: What is the square root of 27 using the long division method?
Solution:
The √27 = 5.1961
Example 3: Solve (√3×√27)+9
Solution:
The number 27 is a perfect cube number and can be written as 3√3
(√3×√27)+9 = (3√3×√3)+9
(√3×√27)+9 = 3×3+9
(√3×√27)+9 = 9+9
(√3×√27)+9 = 18.
Example 4: What is an irrational number?
Solution:
A number is said to be an irrational number if it cannot be expressed in the form of a ratio or fractions.
Examples of irrational numbers √2, √3, √5, or √27.
Example 5: Solve 16√27
Solution:
The √27 = 5.1961
16√27 = 16×5.1961
16√27 = 83.1376
## FAQs on the square root of 27
The 27 = 5.1961
### Which method is used to find the √27?
The standard method that is used to find the square root of any number is the long division method.
### How to write the square root of 27 in exponential form?
(27)1/2 or 270.5 is the exponential way to write 27.
### How to write the square root of 27 in radical form?
The radical form of writing square root is 33
### Can I find the square root of 27 using other methods?
Apart from the long division method, prime factorization and repeated subtraction are two methods that are used to find the square root of the number. Repeated subtraction is not applicable to irrational numbers.
### Is √27 Is it an irrational number?
Yes, 27 is an irrational number since the number is not equal to zero. The square root of 27 cannot be expressed in ratios or fractions.
### What are the methods to find the square root of a number?
There are three methods to find the square root of a number
Prime factorization method
Long division method
Repeated subtraction
Written by by
Prerit Jain
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# Solve quadratic equation by completing square
We have a equation: $$2x^2+7x+3$$
I tried to find the vertex of the parabola by this formula: $$a(x-h)^2+k$$
but I could not get it.
I got this: but it is not right.
$$2(x^2+ \frac{7}{2}x+\frac{49}{4})+3-\frac{49}{2}$$
What's the problem? Is it possible to solve it by completing square?
• Why do you want to find the vertex by strictly completing the square ? – Rebellos Nov 24 '18 at 11:13
• Because my book is given this formula. – Ehsan Zehtabchi Nov 24 '18 at 11:16
• The formula makes no sense. You don't need a negative sign there. – Rebellos Nov 24 '18 at 11:18
Completing the square, it is :
$$2x^2 + 7x + 3 = 2\bigg(x^2+ \frac{7}{2}x + \frac{49}{16}\bigg) - \frac{25}{8} = 2\bigg(x + \frac{7}{4}\bigg)^2 - \frac{25}{8}$$
Thus, the vertex can be found at the point where the "completed square term" equals zero :
$$x+\frac{7}{4} = 0 \Leftrightarrow x = -\frac{7}{4}$$
This is truly the vertex point, as it can be cross-validated by the fact that a vertex of a parabola can be found at :
$$x_v = -\frac{b}{2a} \Rightarrow x_v = -\frac{7}{2\cdot 2}$$
Now, the other coordinate of the vertex is the remaining constant term on the right, thus $$-25/8$$.
This means that the given vertex coordinates, are :
$$(h,k) = \bigg(-\frac{7}{4},-\frac{25}{8}\bigg)$$
It's possible.
\begin{align*} 2x^2 + 7x + 3 & = 2\left(x^2 + \frac{7}{2} x \right) + 3 = 2\left(\left(x + \frac{7}{4}\right)^2 - \left(\frac{7}{4}\right)^2 \right) + 3 \\ & = 2 \left(x + \frac{7}{4}\right)^2 - \frac{49}{8} + 3 = 2 \left(x + \frac{7}{4}\right)^2 - \frac{25}{8} \end{align*} So we have the vertex at $$(h,k) = \left(- \frac{7}{4}, - \frac{25}{8}\right)$$.
In completing the square, you want to follow a plan like below.
$$ax^2+bx+c = 0$$
$$x^2+\frac{b}{a}x+\frac{c}{a} = 0 \implies x^2+\frac{b}{a}x = -\frac{c}{a}$$
You want to reach a quadratic trinomial on the LHS so you can solve it. This is done by adding $$\big(\frac{b}{2a}\big)^2$$ to both sides.
$$x^2+\frac{b}{a}x+\bigg(\frac{b}{2a}\bigg)^2 = -\frac{c}{a}+\frac{b^2}{4a^2}$$
$$\bigg(x+\frac{b}{2a}\bigg)^2 = \frac{b^2-4ac}{4a^2}$$
After which you can take the square root of both sides and solve for $$x$$. Can you use this general plan to solve the equation you’ve given? |
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