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Introductory Algebra for College Students (7th Edition)
$$x = 7700$$ To check if we have the correct solution, let us plug $7700$ into the original equation for $x$. If both sides of the equation equal one another, then the solution is correct: $$0.6(7700 + 300) = 0.65(7700) - 205$$ Let's evaluate what is inside the parentheses first: $$0.6(8000) = 0.65(7700) - 205$$ Multiply first: $$4800 = 5005 - 205$$ Lastly, we do the subtraction: $$4800 = 4800$$ Both sides of the equation are equal, so we know our solution is correct.
To solve this equation, we work step-by-step to isolate the variable on one side and the constant on the other. To do this, we use the order of operations, represented by the acronym PEMDAS. This means that we evaluate parentheses first, then exponents and radicals, then multiplication and division, and, finally, addition and subtraction. In this problem, we use distributive property first to get rid of all the parentheses: $$0.6(x) + 0.6(300) = 0.65x - 205$$ We now multiply out the terms: $$0.6x + 180 = 0.65x - 205$$ Subtract $0.6x$ from both sides of the equation to isolate the variable to one side of the equation: $$180 = 0.05x - 205$$ Add $205$ to both sides of the equation to isolate the constants to the other side of the equation: $$385 = 0.05x$$ Divide both sides of the equation by $0.05$ to solve for $x$: $$x = 7700$$ To check if we have the correct solution, let us plug $7700$ into the original equation for $x$. If both sides of the equation equal one another, then the solution is correct: $$0.6(7700 + 300) = 0.65(7700) - 205$$ Let's evaluate what is inside the parentheses first: $$0.6(8000) = 0.65(7700) - 205$$ Multiply first: $$4800 = 5005 - 205$$ Lastly, we do the subtraction: $$4800 = 4800$$ Both sides of the equation are equal, so we know our solution is correct. |
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# 2.1: Calculating Basic Probabilities
Difficulty Level: At Grade Created by: Bruce DeWItt
### Learning Objectives
• Understand how to calculate and write a probability
• Understand what constitutes chance behavior
• Understand the concept of the Law of Large Numbers
Probabilities give us an idea of how likely it is for a certain event to happen. For example, when a coin is flipped, the chance that it comes up heads is 50%. Probabilities can be expressed as decimals, fractions, percents, or ratios. We could have said the probability of flipping heads is , 0.5, 12\begin{align*}\frac{1}{2}\end{align*}, 50% or 1:2. Each of these conveys the idea that we should expect to get a heads half of the time. Probabilities only give us an idea of what to expect in the long run. However, they do not tell us what will happen in the short term.
Suppose we flip a coin 10 times in a row and get heads each time. The next coin flip is still a random event because while we cannot tell for certain what the next flip will be, we can be certain that about 50% of all tosses over a long set of tosses will be heads. Some people think that we are on a roll so we are more likely to get another heads. Others will say that getting tails is more likely because we are due to get tails. The truth is that we cannot tell what will happen on the next flip. The only thing we know for certain is that there is a 50% chance that the coin will be heads on its next flip. If we continue to flip this same coin hundreds of times, we would expect the percent of heads to get closer and closer to 50%.
Chance Behavior is not predictable in the short term, however, it has long term predictability. The Law of Large Numbers tells us that despite the results on a small number of flips, we will eventually get closer to the theoretical probability. The outcomes in any random event will always get close to the theoretical probability if the event is repeated a large number of times. We might roll a die 4 times in a row and get a 6 each time, however, if we rolled this die hundreds of times, the percent of time that we get a 6 will get closer and closer to the theoretical probability of 16\begin{align*}\frac{1}{6}\end{align*}.
When calculating a probability, we divide the number of favorable outcomes (outcomes we are interested in) by the total number of outcomes. In other words, the probability that outcome 'A' occurs is found by the formulaP(A)=#offavorableoutcomestotal#ofoutcomes\begin{align*}P(A)=\frac{\#\;of\;favorable\;outcomes}{total\; \#\;of\;outcomes}\end{align*}.
Consider a standard deck of 52 playing cards.
If we asked the question "What is the probability of being dealt a face card (jack, queen, or king)?", we would need to count how many cards are face cards and then divide by the total number of cards, 52.
In this situation there are 12 face cards and 52 cards overall so our probability of getting a face card is 1252=3130.23\begin{align*}\frac{12}{52}=\frac{3}{13}\approx0.23\end{align*}.
In probability, there are outcomes that are sure to happen and there are outcomes that are impossible. If we are once again dealing with a standard 52 card deck, the chance of being dealt either a red card or a black card if one card is dealt is 100%. The chance of being dealt a blue card is 0% since there are no blue cards in a standard deck. All random events have probabilities between 0 and 1. In addition, the sum total of the probabilities for all possible outcomes in the sample space is equal to 1. In other words, if an event occurs, there is a 100% chance that one of the possible outcomes will happen. The list below summarizes these rules.
a) The probability of a sure thing is 1.
b) The probability of an impossible outcome is 0.
c) The sum of the probabilities of all possible outcomes is 1.
d) The probability for any random event must be somewhere from 0 to 1.
As shown earlier, we notate the probability of event 'A' happening as P(A). For example, the probability of rolling a three on a six-sided die can be written P(3)=16\begin{align*}P(3)=\frac{1}{6}\end{align*}. Sometimes we are interested in the probability of an event not occurring. This is called the complement of the event. We can write the probability of the complement of event 'A' happening as P(~A), P(not A), or P(Ac)\begin{align*}P(A^c)\end{align*}. The formula for the complement of an event is P(notA)=1P(A)\begin{align*}P(not\;A)=1-P(A)\end{align*}. On our die rolling question, P(~3)=1P(3)=116=56\begin{align*}= 1-P(3)=1-\frac{1}{6}=\frac{5}{6}\end{align*}. In other words, there is a 56\begin{align*}\frac{5}{6}\end{align*} chance of the dice not landing on a 3. It is important to notice that the probability of an event happening and the probability of its complement always add up to 1.
#### Example 1
Which of the following situations are random events?
i) A student looks through their closet to decide what shirt to wear to school.
ii) A student labels each of their 6 pairs of shoes 1 through 6 and then rolls a single die to decide which pair to wear.
iii) The state legislature decides to increase funding to schools by 3%.
iv) A professional golfer makes a hole-in-one on a 200 yard hole.
#### Solution
Situations i) and ii) are not random events. In both cases, there are additional factors that are influencing the decision. The day of the week or the temperature outside might influence your shirt choice and how much money the state legislature happens to have might influence funding.
Both situations iii) and iv) are random events because while we can't predict what will happen in this particular instance, we can make long term predictions. We can predict the percent of the time the student might end up with the shoes labeled #2 and we can predict the percent of the time that the golfer will make a hole-in-one based upon previous performance.
#### Example 2
In the game of pool, there are a total of 15 balls. Balls numbered 1-8 are solid and balls 9-15 are striped. There are two pool balls of each color, for example, there are two yellow pool balls. One of those are solid and one of those are striped. The only exception to this is that there is only 1 black pool ball, the eight ball, and it is solid.
Suppose the pool balls were put in a bag and a single pool ball is pulled out of the bag. What is the probability that the ball:
a) is yellow?
b) is striped?
c) has a number on it that is greater than 10?
d) is not striped?
#### Solution
a) P(Y)=2150.13\begin{align*}P(Y)=\frac{2}{15}\approx0.13\end{align*}
b) P(Striped)=7150.47\begin{align*}P(Striped)=\frac{7}{15}\approx0.47\end{align*}
c) P(>10)=515=130.33\begin{align*}P(>10)=\frac{5}{15}=\frac{1}{3}\approx0.33\end{align*}
d) P(~Striped)=1P(Striped)=1715=8150.53\begin{align*}1-P({Striped})=1-\frac{7}{15}=\frac{8}{15}\approx0.53\end{align*}
In addition to these types of questions, we can also calculate probabilities by incorporating our counting methods from Chapter 1. Recall that the probability of an event occurring is the number of favorable outcomes divided by the number of total possible outcomes.
#### Example 3
A jury of 12 people is to be selected from a group of 12 men and 8 women. What is the probability that the jury has at least 6 women on it?
#### Solution
The total number of outcomes possible is based upon selecting 12 members from a pool of 20. Since order will not matter, there are 20C12=125,970\begin{align*}_{20} C _{12}=125,970 \end{align*} ways to pick a jury of 12. We now want to have at least 6 women on the jury. This means we could have 6 women and 6 men or 7 women and 5 men or 8 women and 4 men on the jury. Mathematically, this would be 8C6×12C6+8C7×12C5+8C8×12C4=28×924+8×792+1×495=25,872+6,336+495=32,703\begin{align*}_8 C _6\times_{12} C_6+_8C_7\times_{12}C_5+_8C_8\times_{12}C_4=28\times924+8\times792+1\times495=25,872+6,336+495=32,703\end{align*}. There are 32,703 ways to have at least 6 women on the jury out of a possible total of 125,970 different juries or 32,703125,9700.26\begin{align*}\frac{32,703}{125,970}\approx0.26\end{align*}. There is about a 26% chance that the jury will have at least 6 women on it.
Sometimes, data is organized in a Venn diagram, as shown in Example 4 on the following page. We will examine these in greater depth in section 2.3 but for now, it is important to understand that a Venn diagram is an organizational tool that makes it easier to interpret a situation and answer basic probability questions.
#### Example 4
A class of 30 students is surveyed to see whether or not they had a science class and/or a math class this trimester. There are 18 students that have a math class, 14 students who have a science class, and 4 students who have neither. It also turns out that this includes 6 students who currently have both classes. The results of the survey are shown in the Venn diagram below.
a) How many total students are taking a math class this trimester?
b) What is the probability that a randomly selected student is taking a math class this trimester?
c) What is the probability that a randomly selected student is taking both a math and science class this trimester?
d) What is the probability that a randomly selected student is not taking either a math or science class this trimester?
#### Solution
a) There are 12 kids who only have a math class and 6 kids who have both a math a science class this trimester for a total of 18 kids.
b) P(Math)=1830=35=0.6\begin{align*}P(Math)=\frac{18}{30}=\frac{3}{5}=0.6\end{align*}
c) P(Math&Science)=630=15=0.2\begin{align*}P(Math \And Science)=\frac{6}{30}=\frac{1}{5}=0.2\end{align*}
d) P(NoMathorScience)=430=2150.13\begin{align*}P(No Math or Science)=\frac{4}{30}=\frac{2}{15}\approx 0.13\end{align*}
### Problem Set 2.1
#### Exercises
For problems 1-5, express your answer both as a fraction (reduce if possible) and as a decimal to the nearest hundredth.
1) Suppose a single card is dealt from a standard deck of 52 cards. Find the probability that the card is:
a) a red card.
b) a face card.
c) an ace.
d) a three.
e) a club.
f) the three of clubs.
g) a black king.
2) A bag contains some jelly beans. There are a total of 6 red jelly beans, 4 green jelly beans, 2 black jelly beans, 5 yellow jelly beans, and 3 orange jelly beans in the bag. Suppose one jelly bean is drawn from the bag.
a) Find P(purple).
b) Find P(yellow).
c) Find P(~red).
3) A single 6-sided die is rolled one time. Find the probability that the result is:
a) a three
b) a seven
c) an even number
d) a prime number
e) a number equal to or greater than 5.
4) The game Scattegories® uses a 20-sided die. It has all the letters of the alphabet on it except Q, U, V, X, Y, and Z. Find each probability below if the die is rolled one time.
a) P(Vowel)
b) P(~Vowel)
c) P(Q)
d) P(Qc)
e) P(a letter alphabetically after Q)
5) The month of October in a 2011 calendar has 31 days with October 1st being a Saturday as shown in the calendar on the following page. Suppose a day is randomly selected. Find each probability.
a) P(weekend)
b) P(not a weekend)
c) P(October 31st)
d) P(October 32nd)
e) P(~October 31st)
f) P(an odd-numbered day)
6) A roulette wheel contains 38 slots. When the wheel is spun, a ball is dropped onto the wheel and the ball will stop on one of the slots. There are 18 black slots, 18 red slots, and 2 green slots. Suppose the ball on a roulette wheel has landed on red four times in a row. What is the chance that the ball will drop on red on the next spin?
7) A coin has been flipped 10 times. Suppose that it has come up heads on only 2 out of those ten times.
a) What percent of the time has the coin come up heads?
b) Suppose we flip the coin 90 more times and 45 of those 90 flips come up heads. Of the 100 flips completed so far, what percent of the time has the coin come up heads?
c) Suppose we continue to flip the coin an additional 900 times and that 450 of those 900 flips come up heads. Of the 1000 flips completed, what percent of the time has the coin come up heads?
d) As we flipped the coin more and more, the percentage of heads got closer and closer to 50% despite the fact that only 2 of the first 10 flips were heads. What rule does this illustrate?
8) Two 6-sided dice are rolled and we keep track of the total on the two dice.
a) Make a 6 by 6 grid showing the different totals that you can get when rolling the two dice.
b) What is the probability that you get doubles?
c) What is the probability that you get a total of 7?
d) What is the probability that you get a total of at least 8?
9) The high school concert choir has 7 boys and 15 girls. The teacher needs to pick three soloists for the next concert but all of the members are so good she decides to randomly select the three students for the solos.
a) In how many ways can the teacher select the 3 students?
b) What is the probability that all three students selected are girls?
c) What is the probability that at least one boy is selected?
10) A test begins with 5 multiple choice questions with four options on each question. It then has 5 true/false questions.
a) How many answer keys are possible?
b) What is the probability of getting every question correct if a student guesses on each question. Leave your answer as a fraction.
11) A lawn and garden store is moving locations and needs to move its riding lawn mowers to the new store. They have 8 mowers with 36-inch decks, 15 mowers with 42-inch decks, and 6 mowers with 48-inch decks that need to be moved. The trailer they are using can move a total of 8 mowers on each load so several trips will have to be made.
a) In how many ways can 8 mowers be randomly selected for the first load?
b) What is the probability that all the mowers with 48-inch decks get selected for the first load? Leave your answer as a reduced fraction.
c) What is the probability that the first load has exactly two 36-inch deck mowers, four 42-inch deck mowers, and two 48-inch deck mowers?
#### Review Exercises
12) In how many ways can three students be selected for a committee if there are 11 students from which to select?
13) A hockey player needs new skates, a new helmet, and a new stick. Hockey Central has 5 brands of skates, 6 brands of helmets, and 8 brands of sticks. In how many different ways can the player select one of each item?
14) Two standard 6-sided dice are rolled and the results from the two dice are added together. Build a grid to determine which outcome is most likely to occur.
15) On a TV game show, three contestants must each pick a box which they believe contains the grand prize based upon clues given about each box. In how many different ways can this be done if there are 10 boxes from which to choose?
### Notes/Highlights Having trouble? Report an issue.
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# Solving exponential equation | Exponential and logarithmic functions | Algebra II | Khan Academy
Let’s say we have the function y equal to 5 by (2 to the power of t). And someone comes and tells you: “Look, this seems like an interesting feature, but I’m curious – I like the number 1111, I’m curious at what point, for what t value my y will be equal to 1111. ” I advise you to pause the video and think about it on your own. At what t value will this y be equal to or approximately equal to 1111. If necessary, you can even use a calculator. I accept that he tried. Let’s solve this together now. We wonder when 5 over (2 to the power of t) is equal to 1111. Let’s write this down. When 5 by (2 of degree t) is equal to 1111? When we do something algebraic, it is always useful to see if we can isolate the variable we are trying to find – we are trying to find out what t value this will do equal to this here. A good first step would be to try let’s remove this 5 from the left side, so let’s divide the left side by 5. If I want this to continue to be equality, I have to do the same thing with both sides. We get 2 to the power of t is equal to 1111/5. How to find t? Which function is the opposite of the indicative function? This will be the logarithm. If I say that a of degree b is equal to c, this means that the logarithm of c at base a is equal to b. a of degree b is equal to c. The logarithm of c at the base a tells us to what extent it should to raise a to get c. I have to raise a to power b to get c. a of degree b is equal to c. These two statements are equivalent. Let us make a logarithm at base 2 on both sides of this equation. On the left side you have a logarithm of (2 to the power of t) at base 2. On the right side you have a logarithm of 1111/5 at base 2. Why is this useful here? This is how much we need to raise 2, to get 2 to the power of t. To get 2 to the power of t, we need to raise 2 to the power of t. This thing is here it is simply simplified to t. This is simplified to t. On the right side we have a logarithm at base 2, we have it all here. I’ll just write it down – t is equal to the logarithm of 1111/5 at base 2. This is an expression that gives us the value of t, but the next question is how to calculate how much it is. If you take out your calculator, you will quickly notice that there is no logarithm button at base 2, so how do we calculate it? Here we must apply a very useful property of degrees. If we have a logarithm of whatever at base 2 … Let me write it like this, if we have a logarithm of c at base a, we can calculate this as the logarithm of c on any basis, on a logarithm of a at the base this same thing. As this unknown must be the same. Our calculator is useful because it has a “log” button. you just push it, and that’s the logarithm at base 10. If you press “In”, this is a natural logarithm or a logarithm at the base \’e\’. I prefer to use a logarithm at base 10, so it will be the same thing as a logarithm of 1111/5 at base 10 on a logarithm of 2 at base 10. We can take out our calculator and we can use a logarithm on the base ‘e’ if we want – this will be a natural logarithm, but i will just use the “log” button. This is a logarithm of 1111/5 – this is the part here. This is the logarithm at base 10 – this is indicated by the “log” button. Divided by a logarithm of 2 at base 10 – this gives us 7 with many digits after the decimal point, but is approximately equal to 7,796. This is approximately equal to 7,796 so that when t is approximately equal to that, y will be equal to 1111. |
# Introduction to Mensuration
## What is Mensuration?
Mensuration is an area of Geometry that is concerned with finding the area and volume.
So, we often divide Mensuration in two branches:
• 2 D Mensuration - finding area of 2 D figures, such as a triangle, circle etc.
• 3 D Mensuration - finding volume and surface area of 3 D figures, such as a cube, cone, sphere etc.
We have already covered the 2 D Mensuration part in the Geometry module. Here we will focus solely on 3 D Mensuration.
Here’s a bird’s eye view of all the 3 D solid figures that we are going to deal with:
Now let us understand the terms like Volume, Lateral Surface Area, etc.
## Important Terms in Mensuration
### What is Volume?
Volume of a 3-D (three dimensional) object is the amount of space occupied by it.
To find the volume of any solid, we generally just multiply the area of its base with its height.
We use cubic units to indicate the volume of a solid, e.g. cubic centimeter ($cm^3$), cubic millimeter ($mm^3$), etc.
To measure or indicate the volume of a liquid, we generally use the unit of litre, millilitre, etc.
#### Conversion between units of Volume
1 cubic millimeter ($mm^3$) = 1 mm × 1 mm × 1 mm
1 cubic centimeter ($cm^3$) = 1 cm × 1 cm × 1 cm = 10 mm × 10 mm × 10 mm = 1000 $mm^3$
1 cubic decimeter ($dm^3$) = 1 dm × 1 dm × 1 dm = 10 cm × 10 cm × 10 cm = 1000 $cm^3$
1 cubic meter ($m^3$) = 1 m × 1 m × 1 m = 10 dm × 10 dm × 10 dm = 1000 $dm^3$
1 cubic millimeter = 1 microliter = $10^{-6}$ litre
1 cubic centimeter = 1 millilitre = $10^{-3}$ litre
1 cubic decimeter = 1 litre
1 cubic meter = 1 kilolitre = $10^3$ liters
### What is Lateral Surface Area?
The lateral surface area of a solid object is the area of whole of the object, except the area of its base and top.
To find the lateral surface area of any solid, we generally just multiply the perimeter of its base with its height.
### What is Total Surface Area?
As the name suggests, total surface area of a solid object is the area of whole of the object (including its base and top).
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Time and Work Questions
Time and Work is the most common section in SSC and State PSC Exam Question Papers.
If you are seriously preparing for competitive examinations, you cannot complete your preparation without covering the questions in Time and Work.
If you go through the previous SSC and PSC question papers, you can find at least one question from this category.
Though the questions may look tough, you can answer every question that falls under the category of Time and Work with preparation.
This tutorial will help you learn everything about Time and Work problems. We will teach you all Time and Work formulas to help you solve the questions immediately.
The first part of this guide covers the concepts. The second part covers frequently asked SSC and PSC questions in the Work and Time category.
Those who are preparing for various government jobs through SSC, PSC, and UPSC may find this guide helpful.
Time and Work - Important Formulas
• If a man completes his work in 'n' days, work done by him in one day is 1/n.
• If the work done by a man in one day is 1/n, he can finish the complete work in n days.
• Time and work are always inversely proportional to each other.
• If M1 persons can do W1 work in D1 days and M2 persons can do W2 work in D2 days, the general formula to find the total work or total time is :
M1 * D1 * W2 = M2 * D2 * W1
The above concepts are important to solve problems in Time and Work aptitude questions. Now let us check how to solve these questions using shortcuts.
Time & Work - Solved Examples
1. A can finish a specific work in 10 days and B can do the same in 20 days. Find the time required to complete the work by both A and B together.
Ans:
Work done in a day by A = 1/(10)
Work done by B in a day = 1/(20)
Work done in a day by both A and B together = 1/(10) + 1/(20) = (2 + 1)/(20) =3/20
Apply the first law explained in this tutorial. Time and work are inversely proportional to each other.
=> reverse of 3/20 => 20/3 days
2. 40 persons make 60 bricks in 3 hours. If 8 persons leave the work, how many bricks will be made by the remaining persons in the next 6 hours?
Ans:
Here we need to use the formula M1 * D1 * W2 = M2 * D2 * W1
Here M1 = 40 M2= 32 W1 = 60 D1= 3 D2= 6 W2 = ?
```
M2 * D2 * W1
w2 = ------------------
M1 * D1
```
=> W2 = (32 * 6 * 60)/ (40 *3) = 96 bricks
3. If X can do a piece of work in 25 days, and Y can finish it in 20 days. They worked together for 5 days and then X left. How many days Y will take to complete the work?
Ans:
Work was done by X in a day = 1/25
Work was done by Y in a day =1/20
Work done by both X and Y in 5 days = 5 * (1/25 + 1/20) = 9/20
Remaining work = 1 - (9/20) =11/20
Y finish 1/20 work in a day. So to finish 11/20 work, Y needs 11 more days. |
# For prakeet only | Mathematics homework help
In this discussion, you will simplify and compare equivalent expressions written both in radical form and with rational (fractional) exponents. Read the following instructions in order and view the example to complete this discussion:
#### How many pages is this assigment?
• Find the rational exponent problems assigned to you in the table below.
If the last letter of your first name is On pages 576 – 577, do the following problems A or L 42 and 101 B or K 96 and 60 C or J 46 and 104 D or I 94 and 62 E or H 52 and 102 F or G 90 and 64 M or Z 38 and 72 N or Y 78 and 70 O or X 44 and 74 P or W 80 and 68 Q or V 50 and 76 R or U 84 and 66 S or T 54 and 100
• Simplify each expression using the rules of exponents and examine the steps you are taking.
• Incorporate the following five math vocabulary words into your discussion. Use bold font to emphasize the words in your writing (Do not write definitions for the words; use them appropriately in sentences describing the thought behind your math work.):
• Principal root
• Product rule
• Quotient rule
• Reciprocal
• nth root
Refer to Inserting Math Symbols for guidance with formatting. Be aware that with regards to the square root symbol, you will notice that it only shows the front part of a radical and not the top bar. Thus, it is impossible to tell how much of an expression is included in the radical itself unless you use parenthesis. For example, if we have √12 + 9 it is not enough for us to know if the 9 is under the radical with the 12 or not. Thus we must specify whether we mean it to say √(12) + 9 or √(12 + 9). As there is a big difference between the two, this distinction is important in your notation.
Another solution is to type the letters “sqrt” in place of the radical and use parenthesis to indicate how much is included in the radical as described in the second method above. The example above would appear as either “sqrt(12) + 9” or “sqrt(12 + 9)” depending on what we needed it to say.
Your initial post should be at least 250 words in length. Support your claims with examples from required material(s) and/or other scholarly resources, and properly cite any references. Respond to at least two of your classmates’ posts by Day 7.
Also the assignment and alkes |
# Fundamental Theorem of Calculus
## Rules for antiderivatives and the relationship between integrals and derivatives.
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Fundamental Theorem of Calculus
Velocity due to gravity can be easily calculated by the formula: v = gt, where g is the acceleration due to gravity (9.8m/s2) and t is time in seconds. In fact, a decent approximation can be calculated in your head easily by rounding 9.8 to 10 so you can just add a decimal place to the time.
Using this function for velocity, how could you find a function that represented the position of the object after a given time? What about a function that represented the instantaneous acceleration of the object at a given time?
### Fundamental Theorem of Calculus
#### Antiderivatives
If you think that evaluating areas under curves is a tedious process you are probably right. Fortunately, there is an easier method. In this section, we shall give a general method of evaluating definite integrals (area under the curve) by using antiderivatives.
Definition: The Antiderivative
If F ' (x) = f (x), then F '(x) is said to be the antiderivative of f(x).
There are rules for finding the antiderivatives of simple power functions such as f(x) = x2. As you read through them, try to think about why they make sense, keeping in mind that differentiation reverses integration.
Rules of Finding the Antiderivatives of Power Functions
• The Power Rule
\begin{align*}\int x^n dx = \frac{1} {n + 1} x^{n + 1} + C\end{align*}
where C is constant of integration and n is a rational number not equal to -1.
• A Constant Multiple of a Function Rule
\begin{align*}\int k x^n dx = k \int x^n dx = k \cdot \frac{1} {n + 1} x^{n + 1} + C\end{align*}
where k is a constant.
• Sum and Difference Rule
\begin{align*}\int [f(x) \pm g(x)] dx = \int f(x) dx \pm \int g(x) dx\end{align*}
• The Constant Rule
\begin{align*}\int k \cdot dx = kx + C\end{align*}
where k is a constant. (Notice that this rule comes as a result of the power rule above.)
#### The Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus makes the relationship between derivatives and integrals clear. Integration performed on a function can be reversed by differentiation.
The Fundamental Theorem of Calculus
If a function f(x) is defined over the interval [a, b] and if F(x) is the antidervative of f on [a, b], then
\begin{align*}\int_{a}^{b} f(x) dx = F(x)|^b_a\end{align*}
\begin{align*}= F(b) - F(a)\end{align*}
We can use the relationship between differentiation and integration outlined in the Fundamental Theorem of Calculus to compute definite integrals more quickly.
### Examples
#### Example 1
Evaluate \begin{align*}\int_{1}^{2} x^2 dx.\end{align*}
This integral tells us to evaluate the area under the curve f(x) = x2, which is a parabola over the interval [1, 2], as shown in the figure below.
To compute the integral according to the Fundamental Theorem of Calculus, we need to find the antiderivative of f(x) = x2. It turns out to be F(x) = (1/3)x3 + C, where C is a constant of integration. How can we get this? Think about the functions that will have derivatives of x2. Take the derivative of F(x) to check that we have found such a function. (For more specific rules, see the box after this example). Substituting into the Fundamental Theorem,
\begin{align*}\int_{a}^{b} f(x) dx\end{align*} \begin{align*}= F(x)|^b_a\end{align*}
\begin{align*}\int_{1}^{2} x^2 dx\end{align*} \begin{align*}= \left [\frac{1} {3} x^3 + C\right ]^2_1\end{align*}
\begin{align*}= \left [\frac{1} {3} (2)^3 + C\right ] - \left [\frac{1} {3} (1)^3 + C\right ]\end{align*}
\begin{align*}= \left [\frac{8} {3} + C\right ] - \left [\frac{1} {3} + C\right ]\end{align*}
\begin{align*}= \frac{7} {3} + C - C\end{align*}
\begin{align*}= \frac{7} {3}\end{align*}
So the area under the curve is (7/3) units2.
#### Example 2
Evaluate \begin{align*}\int x^3dx.\end{align*}
Since \begin{align*}\int x^n dx = \frac{1} {n + 1}x^{n + 1} + C\end{align*}, we have
\begin{align*}\int x^3 dx\end{align*} \begin{align*}= \frac{1} {3 + 1}x^{3 + 1} + C\end{align*}
\begin{align*}= \frac{1} {4}x^4 + C\end{align*}
To check our answer we can take the derivative of \begin{align*} \frac{1} {4}x^4 + C\end{align*} and verify that it is \begin{align*}\,\! x^3\end{align*}, the original function in our integral.
#### Example 3
Evaluate \begin{align*}\int 5x^2 dx.\end{align*}
Using the constant multiple of a power rule, the coefficient 5 can be removed outside the integral:
\begin{align*}\int 5x^2 dx = 5 \int x^2 dx\end{align*}
Then we can integrate:
\begin{align*}= 5 \cdot \frac{1} {2 + 1} x^{2 + 1} + C\end{align*}
\begin{align*}= \frac{5} {3} x^3 + C\end{align*}
Again, if we wanted to check our work we could take the derivative of \begin{align*}\frac{5} {3} x^3 + C\end{align*} and verify that we get \begin{align*}\,\! 5x^2\end{align*}.
#### Example 4
Evaluate \begin{align*}\int (3x^3 - 4x^2 + 2)dx.\end{align*}
Using the sum and difference rule we can separate our integral into three integrals:
\begin{align*}\int (3x^3 - 4x^2 + 2)dx =\end{align*}
\begin{align*}3 \left( \int x^3 dx \right) - 4 \left( \int x^2 dx \right) + \left( \int 2dx \right)\end{align*}
\begin{align*}\to 3 \cdot \frac{1} {4} x^4 - 4 \cdot \frac{1} {3}x^3 + 2x + C\end{align*} \begin{align*}\to \frac{3} {4} x^4 - \frac{4} {3} x^3 + 2x + C\end{align*}
#### Example 5
Evaluate \begin{align*}\int_{2}^{5} \sqrt{x} dx.\end{align*}
The evaluation of this integral represents calculating the area under the curve \begin{align*}y = \sqrt{x}\end{align*} from x = -2 to x = 3, shown in the figure below.
\begin{align*}\int_{2}^{5} \sqrt{x} dx\end{align*} \begin{align*}= \int_{2}^{5} x^{1/2} dx\end{align*}
\begin{align*}= \left [\frac{1} {\frac{1} {2} + 1} x^{1/2 + 1}\right ]^5_2\end{align*}
\begin{align*}= \left [\frac{1} {3/2} x^{3/2}\right ]^5_2\end{align*}
\begin{align*}= \frac{2} {3} \left [x^{3/2}\right ]^5_2\end{align*}
\begin{align*}= \frac{2} {3}\left [5^{3/2} - 2^{3/2}\right ]\end{align*}
\begin{align*}= 5.57\end{align*}
So the area under the curve is 5.57.
#### Example 6
Use the Fundamental Theorem of Calculus to solve: \begin{align*}\int_{4}^{6} \frac{dx}{x}\end{align*}.
Given what we know, that if F(x) = ln x, then F'(x) = \begin{align*}\frac {1}{x}\end{align*}
Thus, we apply the Fundamental Theorem of Calculus:
\begin{align*}\int-{4}^{6} \frac{dx}{x} = ln x |-{4}{6}\end{align*}
= F(6) - F(4) = [ln(6)] - [ln(4)] = 0.4055
#### Example 7
Use the Fundamental Theorem of Calculus to solve: \begin{align*}\int_{-2p}^{2p} 3cos(x) dx\end{align*}.
Given what we know, that if F(x) = 3sin(x), then F'(x) = 3cos(x)
So we apply the Fundamental Theorem of Calculus:
\begin{align*}\int_{-2p}^{2p} 3cos dx = 3sin(x) | _{-2p}^{2p}\end{align*}
= F(8) - F(0) = [3sin(2p)] - [3sin(-2p)] = 1 - 0 = 0
### Review
Evaluate the integral:
1. Evaluate the integral \begin{align*}\int_{0}^{3} 5xdx\end{align*}
2. Evaluate the integral \begin{align*}\int_{0}^{1} x^4dx\end{align*}
3. Evaluate the integral \begin{align*}\int_{1}^{4} (x - 3)dx\end{align*}
Find the integral:
1. Find the integral of (x + 1)(2x - 3) from -1 to 2.
2. Find the integral of \begin{align*}\sqrt{x}\end{align*} from 0 to 9.
3. Find \begin{align*}\int_{-1}^{0} - 3 dx \end{align*}
4. Find \begin{align*}\int_{-1}^{3} dx \end{align*}
5. Find \begin{align*}\int_{-p}^{\frac{p}{2}} - 4cos(x) dx \end{align*}
6. Find \begin{align*}\int_{0}^{2} -dx \end{align*}
7. Find \begin{align*}\int_{2}^{7} \frac{dx}{x}\end{align*}
8. Find \begin{align*}\int_{-2}^{0}x + 5 dx \end{align*}
9. Find \begin{align*}\int_{-p}^{\frac{3p}{2}}6sin(x) dx \end{align*}
10. Find \begin{align*}\int_{6}^{7} \frac{dx}{x}\end{align*}
Challenge yourself:
1. Sketch y = x3 and y = x on the same coordinate system and then find the area of the region enclosed between them (a) in the first quadrant and (b) in the first and third quadrants.
2. Evaluate the integral \begin{align*}\int_{-R}^{R} (\pi R^2 - \pi x^2) dx\end{align*} where R is a constant.
To see the Review answers, open this PDF file and look for section 8.13.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
antiderivative An antiderivative is a function that reverses a derivative. Function A is the antiderivative of function B if function B is the derivative of function A.
derivative The derivative of a function is the slope of the line tangent to the function at a given point on the graph. Notations for derivative include $f'(x)$, $\frac{dy}{dx}$, $y'$, $\frac{df}{dx}$ and \frac{df(x)}{dx}.
fundamental theorem of calculus The fundamental theorem of calculus demonstrates that integration performed on a function can be reversed by differentiation.
integral An integral is used to calculate the area under a curve or the area between two curves.
theorem A theorem is a statement that can be proven true using postulates, definitions, and other theorems that have already been proven. |
# 15 Venn Diagram Questions And Practice Problems (KS3 & KS4): Harder GCSE Exam Style Questions Included
Venn diagram questions involve visual representations of the relationship between two or more different groups of things. Venn diagrams are first covered in KS1 or KS2 and their complexity and uses progress through KS3 and KS4.
This blog will look at the types of Venn diagram questions possibly encountered at KS3 and KS4, focusing on exam-style example questions and in preparation for GCSEs. We will cover problem-solving questions and questions similar to those found in past papers. A worked example follows each question.
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### How to solve Venn diagram questions
In KS3, sets and set notation are introduced when working with Venn diagrams. A set is a collection of objects. We identify a set using curly brackets. For example, if set \mathrm{A} contains the odd numbers between 1 and 10 , then we can write this as: A = \{1, 3, 5, 7, 9\}.
Venn diagrams sort objects, called elements, into two or more sets.
This diagram shows the set of elements \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} sorted into the following sets.
Set \mathrm{A} = factors of 10
Set \mathrm{B} = even numbers
The numbers in the overlap (intersection) belong to both sets. Those that are not in set \mathrm{A} or set \mathrm{B} are shown outside of the circles.
Different sections of a Venn diagram are denoted in different ways.
\xi represents the whole set, called the universal set.
\emptyset represents the empty set, a set containing no elements.
Let’s check out some other set notation examples!
^{\prime}\mathrm{A} and B^{\prime}
The intersection of \mathrm{A} and \mathrm{B} .
The elements in both sets \mathrm{A} and \mathrm{B.} \quad
^{\prime}\mathrm{A} or B^{\prime}
The union of \mathrm{A} or \mathrm{B.} .
Any element in set \mathrm{A} or set \mathrm{B.}
A^{\prime}
‘Not \mathrm{A}^{\prime}
The complement of \mathrm{A.}
Any element not in \mathrm{A.}
In KS3 and KS4, we often use Venn diagrams to establish probabilities. We do this by reading information from the Venn diagram and applying the following formula.
\text{Probability}=\frac{\text{number of desired outcomes}}{\text{total number of outcomes}}
For Venn diagrams we can say
\text{P(A)}=\frac{\text{number of elements in set A}}{\text{total number of elements}}
15 Venn Diagram Questions & Practice Problems (KS3 & KS4) Worksheet
### KS3 Venn diagram questions
In KS3, students learn to use set notation with Venn diagrams and start to find probabilities using Venn diagrams. The questions below are examples of questions that students may encounter in Years 7, Year 8 and Year 9.
#### Venn diagram questions Year 7
1. This Venn diagram shows information about the number of people who have brown hair and the number of people who wear glasses.
How many people have brown hair and glasses?
10
11
4
15
The intersection, where the Venn diagrams overlap, is the part of the Venn diagram which represents brown hair AND glasses. There are 4 people in the intersection.
2. Which set of objects is represented by the Venn diagram below?
We can see from the Venn diagram that there are two green triangles, one triangle that is not green, three green shapes that are not triangles and two shapes that are not green or triangles. These shapes belong to set D.
#### Venn diagram questions Year 8
3. Max asks 40 people whether they own a cat or a dog. 17 people own a dog, 14 people own a cat and 7 people own a cat and a dog. Choose the correct representation of this information on a Venn diagram.
There are 7 people who own a cat and a dog. Therefore, there must be 7 more people who own a cat, to make a total of 14 who own a cat, and 10 more people who own a dog, to make a total of 17 who own a dog.
Once we put this information on the Venn diagram, we can see that there are 7+7+10=24 people who own a cat, a dog or both.
40-24=16 , so there are 16 people who own neither.
4. The following Venn diagrams each show two sets, set \mathrm{A} and set \mathrm{B} . On which Venn diagram has \mathrm{A}^{\prime} been shaded?
\mathrm{A}^{\prime} means not in \mathrm{A} . This is shown in diagram \mathrm{B.}
#### Venn diagram questions Year 9
5. Place these values onto the following Venn diagram and use your diagram to find the number of elements in the set \text{S} \cup \text{O}.
\xi = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}
\text{S} = square numbers
\text{O} = odd numbers
6
2
4
3
\text{S} \cup \text{O} is the union of \text{S} or \text{O} , so it includes any element in \text{S} , \text{O} or both. The total number of elements in \text{S} , \text{O} or both is 6.
6. The Venn diagram below shows a set of numbers that have been sorted into prime numbers and even numbers.
A number is chosen at random. Find the probability that the number is prime and not even.
\frac{3}{4}
\frac{5}{10}
\frac{3}{10}
\frac{3}{8}
The section of the Venn diagram representing prime and not even is shown below.
There are 3 numbers in the relevant section out of a possible 10 numbers altogether. The probability, as a fraction, is \frac{3}{10}.
7. Some people visit the theatre. The Venn diagram shows the number of people who bought ice cream and drinks in the interval.
Ice cream is sold for £3 and drinks are sold for £2. A total of £262 is spent. How many people bought both a drink and an ice cream?
27
48
22
30
Money spent on drinks: 32 \times £2 = £64
Money spent on ice cream: 16 \times £3 = £48
£64+£4=£112 , so the information already on the Venn diagram represents £112 worth of sales.
£262-£112 = £150 , so another £150 has been spent.
If someone bought a drink and an ice cream, they would have spent £2+£3 = £5.
£150 \div £5=30 , so 30 people bought a drink and an ice cream.
### KS4 Venn diagram questions
At KS4, students are expected to be able to take information from word problems and put it onto a Venn diagram involving two or three sets. The use of set notation is extended and the probabilities become more complex. In the higher tier, Venn diagrams are used to calculate conditional probability.
Venn diagrams appear on exam papers across all exam boards, including Edexcel, AQA and OCR. Questions, particularly in the higher tier, may involve other areas of maths, such as percentages, ratio or algebra.
#### Foundation GCSE Venn diagram questions: grades 1-5
8. 50 people are asked whether they have been to France or Spain.
18 people have been to France.
23 people have been to Spain.
6 people have been to both.
By representing this information on a Venn diagram, find the probability that a person chosen at random has not been to Spain or France.
\frac{3}{50}
\frac{15}{50}
\frac{32}{50}
\frac{27}{50}
6 people have been to both France and Spain. This means 17 more have been to Spain to make 23 altogether, and 12 more have been to France to make 18 altogether. This makes 35 who have been to France, Spain or both and therefore 15 who have been to neither.
The probability that a person chosen at random has not been to France or Spain is \frac{15}{50}.
9. Some people were asked whether they like running, cycling or swimming. The results are shown in the Venn diagram below.
One person is chosen at random. What is the probability that the person likes running and cycling?
\frac{4}{80}
\frac{32}{80}
\frac{54}{80}
\frac{9}{80}
9 people like running and cycling (we include those who also like swimming) out of 80 people altogether. The probability that a person chosen at random likes running and cycling is \frac{9}{80}.
10. ξ = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16\}
\text{A} = \{ even numbers \}
\text{B} = \{ multiples of 3 \}
By completing the following Venn diagram, find \text{P}(\text{A} \cup \text{B}^{\prime}).
\frac{13}{16}
\frac{9}{16}
\frac{11}{16}
\frac{8}{16}
\text{A} \cup \text{B}^{\prime} means \text{A} or not \text{B} . We need to include everything that is in \text{A} or is not in \text{B} . There are 13 elements in \text{A} or not in \text{B} out of a total of 16 elements.
Therefore \text{P}(\text{A} \cup \text{B}^{\prime}) = \frac{13}{16}.
11. ξ = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}
A = \{ multiples of 2 \}
\begin{aligned} &\text{A} \cap \text{B}=\{2, 4, 6, 12\} \\ &\text{A} \cup \text{B}=\{1, 2, 3, 4, 6, 8, 10, 12\} \end{aligned}
By putting this information onto the following Venn diagram, list all the elements of B.
1, 3
1, 2, 3, 4, 6, 12
5, 7, 9, 11
\emptyset
We can start by placing the elements in \text{A} \cap \text{B} , which is the intersection.
We can then add any other multiples of 2 to set \text{A}.
Next, we can add any unused elements from \text{A} \cup \text{B} to \text{B}.
Finally, any other elements can be added to the outside of the Venn diagram.
The elements of \text{B} are \{1, 2, 3, 4, 6, 12\}.
#### Higher GCSE Venn diagram questions: grades 4-9
12. Some people were asked whether they like strawberry ice cream or chocolate ice cream. 82\% said they like strawberry ice cream and 70\% said they like chocolate ice cream. 4\% said they like neither.
By putting this information onto a Venn diagram, find the percentage of people who like both strawberry and chocolate ice cream.
56\%
152\%
12\%
4\%
Here, the percentages add up to 156\%. This is 56\% too much. In this total, those who like chocolate and strawberry have been counted twice and so 56\% is equal to the number who like both chocolate and strawberry. We can place 56\% in the intersection, \text{C} \cap \text{S}
We know that the total percentage who like chocolate is 70\%, so 70-56 = 14\%-14\% like just chocolate. Similarly, 82\% like strawberry, so 82-56 = 26\%-26\% like just strawberry.
13. The Venn diagram below shows some information about the height and gender of 40 students.
A student is chosen at random. Find the probability that the student is female given that they are over 1.2m.
\frac{17}{34}
\frac{9}{20}
\frac{9}{34}
\frac{17}{28}
We are told the student is over 1.2m. There are 20 students who are over 1.2m and 9 of them are female. Therefore the probability that the student is female given they are over 1.2m is \frac{9}{20}.
14. The Venn diagram below shows information about the number of students who study history and geography.
ξ = 100
H = history
G = geography
Work out the probability that a student chosen at random studies only history.
\frac{6}{100}
\frac{28}{100}
\frac{17}{100}
\frac{12}{100}
We are told that there are 100 students in total.
Therefore:
\begin{aligned} x+28+x(x-10)+36&=100\\\\ x+64+x^{2}-10x&=100\\\\ x^{2}-9x-36&=0\\\\ (x-12)(x+3)&=0 \end{aligned}
x = 12 or x = -3 (not valid)
If x = 12, then the number of students who study only history is 12, and the number who study only geography is 24. The probability that a student chosen at random studies only history is \frac{12}{100}.
15. 50 people were asked whether they like camping, holiday home or hotel holidays.
18\% of people said they like all three.
7 like camping and holiday homes but not hotels.
11 like camping and hotels.
\frac{13}{25} like camping.
Of the 27 who like holiday homes, all but 1 like at least one other type of holiday. 7 people do not like any of these types of holiday.
By representing this information on a Venn diagram, find the probability that a person chosen at random likes hotels given that they like holiday homes.
\frac{19}{27}
\frac{19}{100}
\frac{10}{27}
\frac{9}{27}
Put this information onto a Venn diagram.
We are told that the person likes holiday homes. There are 27 people who like holiday homes. 19 of these also like hotels. Therefore, the probability that the person likes hotels given that they like holiday homes is \frac{19}{27}.
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# Ex.8.2 Q5 Quadrilaterals Solution - NCERT Maths Class 9
Go back to 'Ex.8.2'
## Question
In a parallelogram $$ABCD$$, $$E$$ and $$F$$ are the mid-points of sides $$AB$$ and $$CD$$ respectively (see the given figure). Show that the line segments $$AF$$ and $$EC$$ trisect the diagonal $$BD$$.
Video Solution
Ex 8.2 | Question 5
## Text Solution
What is known?
In a parallelogram $$ABCD$$, $$E$$ and $$F$$ are the mid-points of sides $$AB$$ and $$CD$$ respectively.
What is unknown?
How we can show that the line segments $$AF$$ and $$EC$$ trisect the diagonal $$BD.$$
Reasoning:
In a quadrilateral if one pair of opposite sides is parallel and equal to each other. Then it is a parallelogram. Also by converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle and parallel to another side, bisects the third side.
Steps:
$$ABCD$$ is a parallelogram.
$$AB$$ $$\parallel$$ $$CD$$
And hence, $$AE$$ $$\parallel$$ $$FC$$
Again, $$AB = CD$$ (Opposite sides of parallelogram $$ABCD$$)
\begin{align}\frac{1}{2}{AB}=\frac{1}{2}{CD}\end{align}
$$AE = FC$$ ($$E$$ and $$F$$ are mid-points of side $$AB$$ and $$CD$$)
In quadrilateral $$AECF$$, one pair of opposite sides ($$AE$$ and $$CF$$) is parallel and equal to each other. Therefore, $$AECF$$ is a parallelogram.
$$\therefore$$ $$AF$$ $$\parallel$$ $$EC$$ (Opposite sides of a parallelogram)
In $$\rm \Delta DQC,$$ $$F$$ is the mid-point of side $$DC$$ and $$FP$$ $$\parallel$$ $$CQ$$ (as $$AF$$ $$\parallel$$ $$EC$$). Therefore, by using the converse of mid-point theorem, it can be said that $$P$$ is the mid-point of $$DQ$$.
$$\therefore$$ $$DP = PQ$$... (1)
Similarly, in $$\rm \Delta APB,$$ $$E$$ is the mid-point of side $$AB$$ and $$EQ$$ $$\parallel$$ $$AP$$ (as $$AF$$ $$\parallel$$ $$EC$$).
Therefore, by using the converse of mid-point theorem, it can be said that $$Q$$ is the mid-point of $$PB$$.
$$\therefore$$ $$PQ = QB$$ ... (2)
From Equations (1) and (2),
$$DP = PQ = BQ$$
Hence, the line segments $$AF$$ and $$EC$$ trisect the diagonal $$BD$$.
Video Solution |
# Matrix Multiplication
Delve into the world of Matrix Multiplication with this comprehensive guide. You will gain in-depth understanding of pure maths applications, become conversant with the basics, and learn Matrix Multiplication step by step. This guide also explores Matrix Multiplication rules and variations, including multiplication by a vector and scalar. Further, it provides real-world examples and practical exercises to hone your skills. Equip yourself with essential knowledge and make Matrix Multiplication a breeze.
#### Create learning materials about Matrix Multiplication with our free learning app!
• Flashcards, notes, mock-exams and more
• Everything you need to ace your exams
## Understanding Matrix Multiplication in Pure Maths
You're about to dive into the exciting world of matrix multiplication, a fundamental operation within the field of pure mathematics. Matrix multiplication is uniquely defined to support various mathematical operations on matrices, which are rectangular arrays of numbers, symbols, or expressions.
Matrix multiplication is an operation that takes two matrices (usually known as a first matrix and second matrix) as inputs to produce a new matrix. The order of matrices is extremely important - multiplication of the first matrix by the second matrix is not necessarily equal to the second matrix multiplied by the first.
### The Basics of Matrix Multiplication
To get a solid hold on matrix multiplication, you first need to understand the requirements for performing this operation. Matrix multiplication is only defined when the number of columns in the first matrix matches the number of rows in the second matrix. Output, or the resultant matrix, is formed by performing specific calculations between the corresponding entries of the two matrices.
You have two matrices A and B. A is a 2x3 matrix (two rows and three columns) and B is a 3x2 matrix (three rows and two columns). You can multiply A by B, but not B by A. The product of A and B will be a 2x2 matrix because the outer dimensions (2 from A and 2 from B) determine the size of the resultant matrix.
#### How to do Matrix Multiplication: Step-by-Step Guide
Matrix multiplication can be approached in a systematic way. Let's break the steps down in a list format:
1. Verify that the number of columns in the first matrix is equal to the number of rows in the second matrix.
2. If the condition in step one is satisfied, proceed with the multiplication. If not, the matrices can't be multiplied together.
3. To populate each cell of the resultant matrix, multiply each element of a row from the first matrix by the corresponding element in a column of the second matrix, then add the results.
In code, this operation could be presented as follows:
def matrix_multiplication(A, B):
rows_A = len(A)
cols_A = len(A[0])
rows_B = len(B)
cols_B = len(B[0])
if cols_A != rows_B:
return "Incompatible matrices"
result_matrix = [[0 for row in range(cols_B)] for col in range(rows_A)]
for i in range(rows_A):
for j in range(cols_B):
for k in range(cols_A):
result_matrix[i][j] += A[i][k] * B[k][j]
return result_matrix
### Exploring Matrix Multiplication Rules
You've learnt the basics of how to do matrix multiplication, but now it's time to dig into the rules. Matrix multiplication adheres to specific rules and principles that set it apart from ordinary number multiplication.
Here are few key points to remember when working with matrix multiplication:
• Matrix multiplication is not commutative, meaning if you have two matrices A and B, generally, AB ≠ BA.
• Matrix multiplication is associative, meaning the order in which associative operations are performed does not matter. (AB)C = A(BC)
• Matrix multiplication is distributive over addition. A(B + C) = AB + AC and (B + C)A = BA + CA.
#### Special Considerations: Matrix Multiplication by Vector
When a matrix is multiplied by a vector, the operation becomes specifically interesting. A vector is essentially a matrix with just one column or just one row. The multiplication result depends on whether the vector is a row vector or a column vector.
The result of multiplying a matrix by a column vector is itself another column vector. In contrast, multiplying a matrix by a row vector produces a row vector. This transformation property makes matrix-vector multiplication an essential tool in a wide array of fields including computer graphics, where it is frequently used to implement transformations like scaling, rotation, and translation.
## Delving into Variations of Matrix Multiplication in Pure Maths
Matrix multiplication isn't a one-size-fits-all tool. There are variations of this operation that serve different purposes in various mathematical contexts. Two such examples include matrix multiplication with a scalar and the specific case of 2x2 matrix multiplication. Understanding these two cases will offer you a broader picture of matrix multiplication and its applicability in different mathematical scenarios.
### Getting to Know Matrix Multiplication with Scalar
You're already familiar with matrix multiplication involving two matrices, but what about when a matrix is multiplied by a single number or scalar? This operation is quite straightforward, but holds foundational importance in linear algebra and other disciplines that rely on matrix computations.
Matrix multiplication with a scalar involves multiplying every entry in a matrix by a single number (or scalar). The resulting matrix maintains the same dimensions as the original matrix, but all its entries are scaled by the multiplicative factor.
The formula for matrix and scalar multiplication can be written as follows:
$k \cdot A = k \begin{pmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \ldots & a_{mn} \end{pmatrix} = \begin{pmatrix} k \cdot a_{11} & k \cdot a_{12} & \ldots & k \cdot a_{1n} \\ k \cdot a_{21} & k \cdot a_{22} & \ldots & k \cdot a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ k \cdot a_{m1} & k \cdot a_{m2} & \ldots & k \cdot a_{mn} \end{pmatrix}$
Take a matrix A= $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and a scalar k=2. Then the scalar multiplication of A by k equals: $\begin{pmatrix} 2 & 4 \\ 6 & 8 \end{pmatrix}$
Regardless of their simplicity, matrix scalar multiplications remain a fundamental operation in vector spaces, physics, computer graphics and other related fields.
#### Detailed Look at Matrix Multiplication 2x2
Beyond its general form, matrix multiplication exhibits interesting properties when carried out on specific types of matrices, notably 2x2 matrices. A 2x2 matrix is the simplest type of matrix that can be multiplied with another matrix.
2x2 matrix multiplication involves multiplying two matrices of size 2x2 (two rows and two columns), and the resulting matrix will also be of size 2x2. The computation involves four steps and four corresponding elements in the result.
The formula for 2x2 matrix multiplication is given by:
$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \times \begin{pmatrix} e & f \\ g & h \end{pmatrix} = \begin{pmatrix} a \cdot e + b \cdot g & a \cdot f + b \cdot h \\ c \cdot e + d \cdot g & c \cdot f + d \cdot h \end{pmatrix}$
This formula means that each element in the resultant 2x2 matrix is arrived at by performing specified calculations between the entries of the two original 2x2 matrices.
For example, let's take two 2x2 matrices A & B where: A= $\begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix}$ and B= $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ The resulting matrix C = A x B would be: C= $\begin{pmatrix} 2 & 4 \\ 7 & 10 \end{pmatrix}$
Firm knowledge of how to multiply 2x2 matrices serves as a stepping stone for understanding larger and more complex matrix multiplications. As the most straightforward nontrivial case, it sets the groundwork for understanding more complex matrix operations.
## Learning through Practical Application: Matrix Multiplication Examples
You've mastered the theory of matrix multiplication - now it's time to apply it. In the following sections, you'll have the chance to explore real-world applications of matrix multiplication and test your skills with practice problems. Seeing matrix multiplication at work in practical contexts will provide you with valuable insights into this mathematical operation's role in our everyday lives.
### Real-world Example of Matrix Multiplication
Even if it seems like matrix multiplication simply belongs in a math textbook, it forms the basis of many real-world situations. From computer graphics to physics, data science to economics, matrix multiplication serves as an essential tool. One fascinating application can be found in the field of transportation—specifically in calculating flight routes and paths.
In the world of transportation, matrix multiplication is used to determine the shortest routes or paths from one point to another. This is often defined in a graph theory context, where cities are vertices, and paths between cities are edges. The adjacency matrices are multiplied to determine paths of certain lengths.
Imagine an airline network with three cities A, B, and C. The cities are connected to each other by direct flights in both directions. The connections can be represented in binary form where a 1 represents a direct flight and 0 lacks a flight. For instance, the direct flights can be shown in an adjacency matrix A such as: $\begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix}$ Now suppose a person is in city A. To find multiple flights (two legs) paths from city A, one would multiply the adjacency matrix by itself to get matrix A². If matrix A was multiplied twice, it would list all the paths of length two in the network, showing possible flights: $\begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix}$ x $\begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix}$ = $\begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}$ So from matrix A², one can see that there are two paths of length two from city A to both cities B and C.
The graph theoretical approach to matrix multiplication in transportation is only one of many real-world applications. Similar approaches are used in social network analysis, power grid stability, computer network design, and much more. Understanding matrix multiplication and how to implement it offers a broad set of tools for tackling these complicated, real-world scenarios.
#### Practice Problems: Boost Your Matrix Multiplication Skills
Now let's put your understanding of matrix multiplication to the test. Working through practice problems is one of the best ways to truly grasp the concept. You'll find exercises that will help bolster your matrix multiplication abilities and reinforce the lessons learnt.
Take the matrices below: Matrix A= $\begin{pmatrix} 2 & 3 \\ 0 & 1 \end{pmatrix}$ and Matrix B= $\begin{pmatrix} 1 & 4 \\ 2 & 0 \end{pmatrix}$ Using your knowledge of matrix multiplication, compute A x B and B x A.
Now consider a scalar multiplication scenario: Take a matrix C= $\begin{pmatrix} 1 & 6 \\ -1 & 0 \end{pmatrix}$ and a scalar d=3. Calculate the matrix multiplication of C by d.
These problems offer a nuanced view of matrix multiplication and will strengthen your comprehension of this fundamental mathematical operation. Don't forget to apply the principles and rules you've learnt so far about matrix multiplication while solving these exercises.
## Matrix Multiplication - Key takeaways
• Matrix Multiplication is an operation that takes two matrices as inputs to produce a new matrix, with the order of matrices being extremely important.
• Matrix Multiplication is only defined when the number of columns in the first matrix matches the number of rows in the second matrix.
• Matrix Multiplication rules include: non-commutative property where AB ≠ BA; associative property where (AB)C = A(BC); and distributivity over addition such as A(B + C) = AB + AC and (B + C)A = BA + CA.
• In Matrix Multiplication by a Vector, the outcome depends on whether the vector is a row vector or a column vector, with matrix by a column vector resulting in another column vector, and multiplication by a row vector producing a row vector.
• Matrix Multiplication with a Scalar involves multiplying every entry in a matrix by a single number or scalar, resulting in a matrix of the same dimensions but all its entries are scaled by the multiplicative factor.
• 2x2 Matrix Multiplication involves multiplying two matrices of size 2x2, and the resultant matrix will also be of size 2x2.
• Matrix Multiplication has practical applications in computer graphics, physics, data science, economics, and transportation, particularly in calculating flight routes and paths.
#### Flashcards in Matrix Multiplication 12
###### Learn with 12 Matrix Multiplication flashcards in the free StudySmarter app
We have 14,000 flashcards about Dynamic Landscapes.
What are the steps to perform matrix multiplication in mathematics?
Firstly, ensure that the number of columns in the first matrix equals the number of rows in the second. For each cell in the result matrix, calculate the dot product of the corresponding row from the first matrix and column from the second. Repeat this process until all cells are filled. This is the product matrix.
Can you multiply any two matrices together in matrix multiplication?
No, you can't multiply any two matrices together. The number of columns in the first matrix must be equal to the number of rows in the second matrix for multiplication to be possible.
What is the importance of the order in matrix multiplication?
The order in matrix multiplication is vital because it is not commutative. Multiplying matrix A by matrix B (AB) can yield a result different from multiplying matrix B by matrix A (BA). Also, one multiplication may be defined while the other is not.
What are the practical applications of matrix multiplication in real life scenarios?
Matrix multiplication is used in various real-life scenarios including computer graphics, physics, and economics for transformation, rotation and scaling of images, solving systems of linear equations, and predicting economic growth, respectively.
What are the rules for the sizes of matrices in matrix multiplication?
For matrix multiplication, the number of columns in the first matrix must be equal to the number of rows in the second matrix. For instance, a matrix of size 'm x n' can multiply a matrix of size 'n x p' resulting in a matrix of size 'm x p'.
## Test your knowledge with multiple choice flashcards
What is matrix multiplication in pure maths?
What are the requirements for matrix multiplication?
What are the basic rules for matrix multiplication?
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# Construction of Triangle from Side, Angle, and that Angle's Bisector
Construct a triangle, given side $a,$ the opposite angle $A$, and the angle bisector $l_a.$
### Prof. Dr. René Sperb 19 September 2014
Construct a triangle, given side $a,$ the opposite angle $A$, and the angle bisector $l_a.$
Starting from the end, let $O$ be the circumcenter of $\Delta ABC,$ $M$ the midpoint of $BC,$ $P$ the midpoint of the arc subtending $\angle BAC,$ and $OQ\perp AP,$ with $Q$ on $AP,$ $N$ the intersection of $BC$ and $AP.$
The problem is solved if we can construct the length of $PN.$ To this end we set $PN = s,$ $PM = h$ and use the fact that triangles $MNP$ and $QOP$ are similar. This yields the relation
$\displaystyle\frac{l_{a }+ s}{2R} = \frac{h}{s},$
where $R$ is the circumradius of $\Delta ABC:$ $R=OP.$ In other words, $2Rh = (l_a + s ) s.$
At this point we may use the Intersecting Secants Theorem to find $s$ from the last equation.
Form circle $(D)$ with $MZ$ as diameter ($Z$ being the antipode of $P$ on $(O).)$ Find $M'$ on $(D)$ such that $PM'=PN$ and let $Z'$ be the second intersection of $PM'$ and $(D).$ For the circle $(D)$ one has
$PZ\cdot PM = PM'\cdot PZ',$
or, $2Rh=s\cdot PZ',$ making $PZ'=s+l_a$ and, therefore, $M'Z'=l_a.$
This leads to the following construction:
1. Construct circle $(O)$ with $BC=a$ as a chord subtending angles $A.$ Let $P$ be the midpoint of the opposite arc, $M$ the midpoint of $BC,$ $Z$ the antipode of $P$ on $(O).$
2. Cut a chord, e.g., $YZ$ equal in length to $l_a.$
3. Let $D$ be the midpoint of $MZ,$ $(D)$ the circle on diameter $MZ$ and $\overline{(D)}$ the circle with center $D$ tangent to $YZ.$
4. Draw tangent from $P$ to $\overline{(D)}$ and let it cross $(D)$ at $M'.$
5. Find $N$ on $BC,$ with $PN=PM'$ and chord $PA$ of $(O)$ passing through $N,$ or $PA=PZ'.$
This gives $A$ as the sought vertex of $\Delta ABC.$ To prove that reverse the steps leading to the construction.
The construction may not be always possible. This happens if $l_a$ is "too long," i.e. longer than $MZ.$ |
# RD Sharma Class 12 Solutions Chapter 10 Differentiability Exercise 10.2 (Updated for 2021-22)
RD Sharma Solutions Class 12 Maths Chapter 10 Exercise 10.2: Students can utilise the RD Sharma Class 12 Solutions to Exercise 10.2 on the website to help them excel in the Class 12 board exam. This lesson primarily illustrates the concept of differentiability within a set. Students can achieve excellent marks in the Class 12 exams by practising regularly with RD Sharma Solutions. The PDF of RD Sharma Solutions for Class 12 Maths Chapter 10 to exercise 10.2 of RD Sharma Solutions for Class 12 Maths is primarily intended to assist students with their exams.
## Download RD Sharma Solutions Class 12 Maths Chapter 10 Exercise 10.2 Free PDF
RD Sharma Solutions Class 12 Maths Chapter 10 Exercise 10.2
### Access answers to Maths RD Sharma Solutions For Class 12 Chapter 10 – Differentiability Exercise 10.2 Important Questions With Solution
1. If f is defined by f (x) = x2, find f’ (2).
Solution:
2. If f is defined by f (x) = x2 – 4x + 7, show that f’ (5) = 2 f’ (7/2)
Solution:
Hence the proof.
3. Show that the derivative of the function f is given by f (x) = 2x3 – 9x+ 12 x + 9, at x = 1 and x = 2 are equal.
Solution:
4. If for the function Ø (x) = λ x2 + 7x – 4, Ø’ (5) = 97, find λ.
Solution:
We have to find the value of λ given in the real function and we are given with the differentiability of the function f(x) = λx2 + 7x – 4 at x = 5 which is f ‘(5) = 97:
Chapter 10 Differentiability Ex 10.2 Q5
Chapter 10 Differentiability Ex 10.2 Q6
Chapter 10 Differentiability Ex 10.2 Q7
Chapter 10 Differentiability Ex 10.2 Q8
Chapter 10 Differentiability Ex 10.2 Q9
Chapter 10 Differentiability Ex 10.2 Q10
Differentiability Ex 10.2 Q11
Differentiability Ex 10.2 Q12
We have provided complete details of RD Sharma Solutions for Class 12 Maths Chapter 10 Exercise 10.2. If you have any queries related to CBSE, feel free to ask us in the comment section below.
## FAQs on RD Sharma Class 12 Solutions Chapter 10 Exercise 10.2
### Where can I get RD Sharma Class 12 Maths Solutions Chapter 10 Exercise 10.2 Free PDF?
You can get RD Sharma Solutions Class 12 Maths Chapter 10 Exercise 10.2 Free PDF from the above article.
### How many questions are there in RD Sharma Solutions Class 12 Maths Chapter 10 Exercise 10.2?
There are a total of 12 questions in RD Sharma Solutions Class 12 Maths Chapter 10 Exercise 10.2.
### Which is the best guidebook for preparing for the Class 12 exam?
The best reference material for CBSE board exam preparation is RD Sharma Solutions for Class 12 Maths Chapter 10 Exercise 10.2. Every section’s questions have been framed and solved by subject experts. They would be able to effortlessly excel in their final exams if they study from these books. |
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Problem 3: A helicopter is ascending vertically at a speed of 19.6 ms-1. When it is at a height of 156.8 m above the ground, a stone is dropped. How long does the stone take to reach the ground?
Solution
Strategy: Initially, the stone is part of the helicopter and moving up with the same speed. Hence, when it is released it goes up with the same speed until its velocity becomes zero due to the downward gravitational force. When its velocity becomes zero, it would have covered some distance to start fall down. Therefore, we have to calculate two time intervals; one, when it goes up to reach the maximum height, and two, to reach the ground from that highest point.
1. For upward motion,
Initial velocity = vi = 19.6 ms-1, Final velocity = vf = 0 ms-1, Acceleration = g= – 9.8 ms-2, t = ?
Apply equation of motion, vf = vi + gt, we have
0 = 19.6 – 9.8t Or 9.8t = 19.6 Or t = 19.6 ÷ 9.8 = 2 sec … (1)
1. Downward motion
So, the stone goes up for 2 sec. The distance covered in these 2 sec can be found by
2gs = vf2 – vi2. Put the values,
2(-9.8)s = 0 – (19.6)2 Or s = (19.6 * 19.6)/(2 * 9.8) Or s = 384.16/19.6 = 19.6 m
1. Total time
So, the total distance the stone drops through is (156.8 + 19.6) = 176.4 m under the action of gravity. Therefore, to find time t, we apply, s = vit + ½ at2, where vi = 0. Put the values,
176.4 = 0 + ½ * 9.8 * t2 Or 176.4 = 4.9t2 Or t2 = (176.4)/4.9 = t2 = 36 Or t = 6 … (2)
Total time taken to the ground is the sum of times in equation (1) and (2). Therefore, the stone will reach the ground in 2 + 6 = 8 seconds. |
# Review: Rounding Whole Numbers
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Print Lesson
## Objective
SWBAT round whole numbers by placing the number on a number line.
#### Big Idea
A number line can be used to help round whole numbers.
## Whole Class Review
10 minutes
In review lessons, I like to use various strategies to revisit the skill. Because it is a review skill, there is not a lot of conversation between the students. The purpose of the review before the state test is to prepare the students to work independently in order to be successful on the end of year assessment.
In today's lesson, the students review estimating whole numbers. They must use a number line to help them round the number to the underlined digit. This aligns with 4.NBT.B3 because the students use place value to help them round the numbers.
To begin the lesson, the students are called to the carpet to sit in front of the Smart board. (I like for my students to be close so that I can make sure that all of them are being attentive.) To review this skill, I show the students a video at the following site:
http://studyjams.scholastic.com/studyjams/jams/math/problem-solving/psestimate-whole-numbers.htm
After the video, I display Review Rounding Whole Numbers.pptx on the Smart board. I use the power point to round 5,278 to the nearest thousands. I ask, "Which number is in the thousands place?" The students know that the 5 is in the thousands place. I explain to the students that because we are rounding to the thousands place, this means that 5,278 will either round to 5,000 or 6,000. These are the numbers placed on both ends of the number line. I remind the students that before we place 5,278 on the number line, it will be easier to do this if we find the halfway point on the number line. I let the students know that 5,500 is halfway between 5,000 and 6,000. Now that we have our halfway point, we can now place 5,278 on the number line. I ask, "Does 5,278 come before or after 5,500?" The students know that 5,278 comes before 5,500. On the next slide, the students see 5,278 on the number line. I ask, "Is 5,278 closer to 5,000 or 6,000?" The students know that 5,278 is closer to 5,000. Therefore, 5,278 rounds to 5,000.
## Independent Practice
15 minutes
The students will practice the skill independently because they will have to work alone for the state test. Each student is given a Estimating Whole Numbers.docx-1.docx handout. They must use a number line to round numbers. In the Video - Estimating Whole Numbers.mp4, you can see what is required of the students.
As the students work on the problems, I walk around to monitor their level of understanding. If the students are having a difficult time, I will ask guiding questions to help lead them to the answer.
Possible Questions:
1. What place value is the underlined digit?
2. Where should you place your number on the number line?
3. Which landmark number is your number closest to?
Any students having difficulty with the task will be grouped for intervention.
## Share/Discuss/Analyze Solution
15 minutes
Upon completing the independent practice, I call the students back together as a whole. I feel that it is very important to close out the lesson by sharing answers. By doing this, it allows the extra opportunity to reach any students that still do not understand the concept.
I call on students to share their answers. All students are not auditory learners; therefore, it is very important for the students to see the examples of work (Student Work - Estimating Whole Numbers). I use my document camera to display the student work on the Smart board.
Students are allowed to ask questions during this closing of the lesson. The most important aspect for me is that I have identified any students that need 1-on-1 or small group remediation. |
# This question is for my 11 year old using fractions to figure answer......she needs to find out what 1/3 of 33 3/4.....I do not want answer.....just how to set up the problem so that I may help her....how do you divide fractions?
$11 \frac{1}{4}$
#### Explanation:
Here, you are not dividing fractions. You are actually multiplying them.
The expression is $\frac{1}{3} \cdot 33 \frac{3}{4}$. That would equal $11 \frac{1}{4}$.
One way to solve this would be to convert $33 \frac{3}{4}$ into an improper fraction.
$\frac{1}{\cancel{3}} \cdot \frac{\cancel{135}}{4} = \frac{45}{4} = 11 \frac{1}{4}$.
May 22, 2018
Dividing fractions is easy!
#### Explanation:
When you divide a fraction by another you should use the KFC method. It's also easy to remember!
K - keep
F - flip the $\cdot$ into a $\div$
C - change the denominator and numerator around of the second fraction
For example if you have: (moving steps guide)
$\frac{3}{4} \div \frac{5}{8}$ (if you have a number next to the fraction you need to make it improper or top-heavy)
K - Keep the first fraction the same.
$\textcolor{red}{\frac{3}{4}} \div \frac{5}{8}$
F - flip the $\cdot$ into a $\div$
$\frac{3}{4} \textcolor{red}{\cdot} \frac{5}{8}$
C - change the denominator and numerator around of the second fraction
$\frac{3}{4} \cdot \textcolor{red}{\frac{8}{5}}$
Then you can solve the multiplication.
$\frac{3}{4} \cdot \frac{8}{5} = \frac{3 \cdot 8}{4 \cdot 5} = \frac{18}{20} = \frac{9}{10}$
May 22, 2018
If you want $\frac{1}{3}$ of 33$\frac{3}{4}$ then you multiply them
$\frac{1}{3} \times \frac{135}{4}$ (mixed numbers are turned into improper or top
heavy fractions.)
Then you multiply the numerators together and you multiply the denominators together. Cancel down your answer.
If you are dividing with fractions:
$\frac{2}{3} \div \frac{5}{7} = \frac{2}{3} \times \frac{7}{5} = \frac{2 \times 7}{3 \times 5} = \frac{14}{15}$
The fraction you are dividing by is turned upside down and you multiply.[KFC-Keep Flip Change]
If you have a mixed number then it has to be changed to a top heavy or improper fraction before you flip anything.
$3 \frac{2}{5} \div \frac{4}{7} = \frac{17}{5} \div \frac{4}{7} = \frac{17}{5} \times \frac{7}{4}$
If you have a whole number, put it over one and do the same
$5 \div \frac{3}{8} = \frac{5}{1} \div \frac{3}{8} = \frac{5}{1} \times \frac{8}{3}$ |
# What is 1/141 as a decimal?
## Solution and how to convert 1 / 141 into a decimal
1 / 141 = 0.007
Fraction conversions explained:
• 1 divided by 141
• Numerator: 1
• Denominator: 141
• Decimal: 0.007
• Percentage: 0.007%
To convert 1/141 into 0.007, a student must understand why and how. Fractions and decimals represent parts of a whole, sometimes representing numbers less than 1. Depending on the situation, decimals can be more clear. We don't say 1 and 1/2 dollar. We use the decimal version of \$1.50. Same goes for fractions. We will say 'the student got 2 of 3 questions correct'. Once we've decided the best way to represent the number, we can dive into how to convert 1/141 into 0.007
1 / 141 as a percentage 1 / 141 as a fraction 1 / 141 as a decimal
0.007% - Convert percentages 1 / 141 1 / 141 = 0.007
## 1/141 is 1 divided by 141
Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! The numerator is the top number in a fraction. The denominator is the bottom number. This is our equation! Now we divide 1 (the numerator) into 141 (the denominator) to discover how many whole parts we have. Here's how you set your equation:
### Numerator: 1
• Numerators are the number of parts to the equation, showed above the vinculum or fraction bar. With a value of 1, you will have less complexity to the equation; however, it may not make converting any easier. 1 is an odd number so it might be harder to convert without a calculator. Values like 1 doesn't make it easier because they're small. Now let's explore X, the denominator.
### Denominator: 141
• Denominators are the total numerical value for the fraction and are located below the fraction line or vinculum. 141 is one of the largest two-digit numbers to deal with. But 141 is an odd number. Having an odd denominator like 141 could sometimes be more difficult. Overall, two-digit denominators are no problem with long division. Now let's dive into how we convert into decimal format.
## How to convert 1/141 to 0.007
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 141 \enclose{longdiv}{ 1 }$$
Use long division to solve step one. Yep, same left-to-right method of division we learned in school. This gives us our first clue.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 141 \enclose{longdiv}{ 1.0 }$$
Uh oh. 141 cannot be divided into 1. So that means we must add a decimal point and extend our equation with a zero. This doesn't add any issues to our denominator but now we can divide 141 into 10.
### Step 3: Solve for how many whole groups you can divide 141 into 10
$$\require{enclose} 00.0 \\ 141 \enclose{longdiv}{ 1.0 }$$
We can now pull 0 whole groups from the equation. Multiply by the left of our equation (141) to get the first number in our solution.
### Step 4: Subtract the remainder
$$\require{enclose} 00.0 \\ 141 \enclose{longdiv}{ 1.0 } \\ \underline{ 0 \phantom{00} } \\ 10 \phantom{0}$$
If your remainder is zero, that's it! If you still have numbers left over, continue to the next step.
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
### Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. And the same is true for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But each represent values in everyday life! Here are just a few ways we use 1/141, 0.007 or 0% in our daily world:
### When you should convert 1/141 into a decimal
Sports Stats - Fractions can be used here, but when comparing percentages, the clearest representation of success is from decimal points. Ex: A player's batting average: .333
### When to convert 0.007 to 1/141 as a fraction
Pizza Math - Let's say you're at a birthday party and would like some pizza. You aren't going to ask for 1/4 of the pie. You're going to ask for 2 slices which usually means 2 of 8 or 2/8s (simplified to 1/4).
### Practice Decimal Conversion with your Classroom
• If 1/141 = 0.007 what would it be as a percentage?
• What is 1 + 1/141 in decimal form?
• What is 1 - 1/141 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.007 + 1/2?
### Convert more fractions to decimals
From 1 Numerator From 141 Denominator What is 1/142 as a decimal? What is 2/141 as a decimal? What is 1/143 as a decimal? What is 3/141 as a decimal? What is 1/144 as a decimal? What is 4/141 as a decimal? What is 1/145 as a decimal? What is 5/141 as a decimal? What is 1/146 as a decimal? What is 6/141 as a decimal? What is 1/147 as a decimal? What is 7/141 as a decimal? What is 1/148 as a decimal? What is 8/141 as a decimal? What is 1/149 as a decimal? What is 9/141 as a decimal? What is 1/150 as a decimal? What is 10/141 as a decimal? What is 1/151 as a decimal? What is 11/141 as a decimal? What is 1/152 as a decimal? What is 12/141 as a decimal? What is 1/153 as a decimal? What is 13/141 as a decimal? What is 1/154 as a decimal? What is 14/141 as a decimal? What is 1/155 as a decimal? What is 15/141 as a decimal? What is 1/156 as a decimal? What is 16/141 as a decimal? What is 1/157 as a decimal? What is 17/141 as a decimal? What is 1/158 as a decimal? What is 18/141 as a decimal? What is 1/159 as a decimal? What is 19/141 as a decimal? What is 1/160 as a decimal? What is 20/141 as a decimal? What is 1/161 as a decimal? What is 21/141 as a decimal?
### Convert similar fractions to percentages
From 1 Numerator From 141 Denominator 2/141 as a percentage 1/142 as a percentage 3/141 as a percentage 1/143 as a percentage 4/141 as a percentage 1/144 as a percentage 5/141 as a percentage 1/145 as a percentage 6/141 as a percentage 1/146 as a percentage 7/141 as a percentage 1/147 as a percentage 8/141 as a percentage 1/148 as a percentage 9/141 as a percentage 1/149 as a percentage 10/141 as a percentage 1/150 as a percentage 11/141 as a percentage 1/151 as a percentage |
## SOLVING LOGARITHMIC EQUATIONS
Note:
If you would like an in-depth review of logarithms, the rules of logarithms, logarithmic functions and logarithmic equations, click on logarithmic functions.
Solve for x in the following equation.
Problem 9.2b:
Answers: There are an infinite number of solutions:
and are the exact solutions, and and are the approximate solutions.
Solution:
To solve for x, first isolate the sine term.
If we restrict the domain of the cosine function to , we can use the arcsin function to solve for x.
The sine of x is positive in the first quadrant and the second quadrant. This means that there are two solutions in the first counterclockwise rotation from 0 to . One angle x terminates in the first quadrant and the second angle terminates in the second quadrant. One solution is
The period of is , and the period of is As 6x rotates radians, x rotates Therefore,the second solution is
Since the period is this means that the values will repeat every radians. Therefore, the solutions are and where n is an integer.
These solutions may or may not be the answers to the original problem. You much check them, either numerically or graphically, with the original equation.
Numerical Check:
• Left Side:
• Right Side:
Since the left side equals the right side when you substitute 0.08165955for x, then 0.08165955 is a solution.
• Left Side:
• Right Side:
Since the left side equals the right side when you substitute 0.44193922for x, then 0.44193922 is a solution.
Graphical Check:
Graph the equation
Note that the graph crosses the x-axis many times indicating many solutions.
Note the graph crosses at 0.08165955 ( one of the solutions ). Since the period of the function is , the graph crosses again at 0.08165955+1.04719755=1.128857 and again at , etc.
The graph also crosses at 0.44193922 ( another solution we found ). Since the period is , it will crosses again at 0.44193922+1.04719755=1.48913677 and at , etc
If you would like to review the solution to problem 9.2c, click on solution.
If you would like to go back to the previous section, click on previous
If you would like to go to the next section, click on next
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Author: Nancy Marcus |
## Engage NY Eureka Math 1st Grade Module 2 Lesson 7 Answer Key
### Eureka Math Grade 1 Module 2 Lesson 7 Problem Set Answer Key
Question 1.
John has 8 tennis balls. Toni has 5. How many tennis balls do they have in all?
8 and ______ make ______.
10 and ______ make ______.
John and Toni have ______ tennis balls in all.
8 and 5 make 13.
10 and 3 make 13 .
John and Toni have 13 tennis balls in all.
Explanation:
John has 8 tennis balls. Toni has 5 tennis balls as we can observe in the above image. An addition sentence is a mathematical expression that shows two or more values added together and their sum. ADD eight balls of John with two balls of Toni results ten. ADD ten balls with three balls of Toni which makes thirteen. John and Toni have thirteen balls in all.
Question 2.
Bob has 8 raisins, and Jenny has 4. How many raisins do they have altogether?
8 and ______ make ______.
10 and ______ make ______.
Bob and Jenny have ______ raisins altogether.
8 and 4 make 12.
10 and 2 make 12 .
Bob and Jenny have 12 raisins altogether.
Explanation:
Bob has 8 raisins, and Jenny has 4 raisins as we can observe in the above image. An addition sentence is a mathematical expression that shows two or more values added together and their sum. ADD eight raisins of Bob with two raisins of Jenny results ten. ADD ten raisins with two raisins of Jenny which makes twelve. Bob and Jenny have 12 raisins altogether.
Question 3.
There are 3 chairs on the right side of the classroom and 8 on the left side. How many total chairs are in the classroom?
8 and ______ make ______.
10 and ______ make ______.
There are ______ total chairs.
8 and 3 make 11.
10 and 1 make 11.
There are 11 total chairs.
Explanation:
There are 3 chairs on the right side of the classroom and 8 chairs on the left side of the classroom as we can observe in the above image. An addition sentence is a mathematical expression that shows two or more values added together and their sum. ADD eight chairs on the left side of the classroom with two chairs on the right side of the classroom which makes ten. ADD ten chairs with one chair on the right side of the classroom results eleven chairs. Total number of chairs in the classroom are eleven.
Question 4.
There are 7 children sitting on the rug and 8 children standing. How many children are there in all?
8 and ______ make ______.
10 and ______ make ______.
There are ______ children in all.
8 and 7 make 15.
10 and 5 make 15.
There are 15 children in all.
Explanation:
There are 7 children sitting on the rug and 8 children standing as we can observe in the above image. An addition sentence is a mathematical expression that shows two or more values added together and their sum. ADD two children sitting on the rug and eight children standing on the rug which results ten. ADD ten children with five children sitting on the rug which results fifteen. There are fifteen children in all.
### Eureka Math Grade 1 Module 2 Lesson 7 Exit Ticket Answer Key
Write the number sentences you used to solve.
Nick picks some peppers. He picks 5 green peppers and 8 red peppers. How many peppers does he pick in all?
8 and ______ make ______.
10 and ______ make ______.
Nick picks ____ peppers.
8 and 5 make 13.
10 and 3 make 13.
Nick picks 13 peppers.
Explanation:
Nick picks some peppers. He picks 5 green peppers and 8 red peppers as we can observe in the above image. An addition sentence is a mathematical expression that shows two or more values added together and their sum. ADD two green peppers and eight red peppers which results ten. Draw a circle for ten peppers. ADD ten peppers with three green peppers which results thirteen. Nick picks thirteen peppers.
### Eureka Math Grade 1 Module 2 Lesson 7 Homework Answer Key
Write the number sentences you used to solve.
Question 1.
Meg gets 8 toy animals and 4 toy cars at a party. How many toys does Meg get in all?
8 + 4 = ____
10 + ____ = ____
Meg gets _____ toys.
8 + 4 = 12
10 + 2 = 12
Meg gets 12 toys.
Explanation:
Meg gets 8 toy animals and 4 toy cars at a party as we can observe in the above image. An addition sentence is a mathematical expression that shows two or more values added together and their sum. ADD eight toy animals with two toy cars which results ten. Draw a circle for ten toy animals. ADD ten toy animals with two toy cars which results twelve. Meg gets twelve toys.
Question 2.
____ + ____ = ____
____ + ____ = ____
6 + 8 = 14
10 + 4 = 14
Explanation:
Question 3.
May has a party. She invites 7 girls and 8 boys. How many friends does she invite in all?
____ + ____ = ____
____ + ____ = ____
May invites _____ friends.
7 + 8 = 15
10 + 5 = 15
May invites 15 friends.
Explanation:
May has a party. She invites 7 girls and 8 boys as we can observe in the above image. An addition sentence is a mathematical expression that shows two or more values added together and their sum. ADD two girls with eight boys which results ten. Draw a circle for ten friends. ADD ten friends with five girls which results fifteen. May invites fifteen friends.
Question 4.
Alec collects baseball hats. He has 9 Mets hats and 8 Yankees hats. How many hats are in his collection?
____ + ____ = ____
____ + ____ = ____
Alec has _____ hats.
9 + 8 = 17
10 + 7 = 17
Alec has 17 hats.
Explanation:
Alec collects baseball hats. He has 9 Mets hats and 8 Yankees hats as we can observe in the above image. An addition sentence is a mathematical expression that shows two or more values added together and their sum. ADD nine Mets hats with one Yankees hat which results ten hats. Draw a circle for ten hats. ADD ten hats with seven Yankees hats which results seventeen. Alec has seventeen hats.
### Eureka Math Grade 1 Module 2 Lesson 7 Fluency Template 1 Answer Key
9 + 2 = 11 3 + 9 = 12
9 + 4 = 13 5 + 9 = 14
9 + 6 = 15 7 + 9 = 16
9 + 8 = 17 9 + 9 = 18
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. In the above image we can observe number sentences.
By adding nine with two results eleven.
By adding three with nine results twelve.
By adding nine with four results thirteen.
By adding five with nine results fourteen.
By adding nine with six results fifteen.
By adding seven with nine results sixteen.
By adding nine with eight results seventeen.
By adding nine with nine results eighteen.
### Eureka Math Grade 1 Module 2 Lesson 7 Fluency Template 2 Answer Key
Friendly Fact Go Around: Make It Equal
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together to get their sum.
In the above image, we can observe 9+ 1 number sentence. By adding nine with one results ten. The number sentence 9 + 1 is equal to 10 + 0.
In the above image, we can observe 9+ 3 number sentence. By adding nine with three results twelve. The number sentence 9 + 3 is equal to 10 + 2.
In the above image, we can observe 9+ 5 number sentence. By adding nine with five results fourteen. The number sentence 9 + 5 is equal to 10 + 4.
In the above image, we can observe 9+ 4 number sentence. By adding nine with four results thirteen. The number sentence 9 + 4 is equal to 10 + 3.
In the above image, we can observe 9+ 7 number sentence. By adding nine with seven results sixteen. The number sentence 9 + 7 is equal to 10 + 6.
In the above image, we can observe 9+ 6 number sentence. By adding nine with six results fifteen. The number sentence 9 + 6 is equal to 10 + 5.
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together to get their sum.
In the above image, we can observe 3 + 9 number sentence. By adding three with nine results twelve. The number sentence 3 + 9 is equal to 10 + 2.
In the above image, we can observe 2 + 9 number sentence. By adding two with nine results eleven. The number sentence 2 + 9 is equal to 10 + 1.
In the above image, we can observe 8 +9 number sentence. By adding eight with nine results seventeen. The number sentence 8 + 9 is equal to 10 + 7.
In the above image, we can observe 5 + 9 number sentence. By adding five with nine results fourteen. The number sentence 5 + 9 is equal to 10 + 4.
In the above image, we can observe 4 + 9 number sentence. By adding four with nine results thirteen. The number sentence 4 + 9 is equal to 10 + 3.
In the above image, we can observe 9+ 9 number sentence. By adding nine with nine results eighteen. The number sentence 9 + 9 is equal to 10 + 8.
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together to get their sum.
In the above image, we can observe 9+ 4 number sentence. By adding nine with four results thirteen. The number sentence 9 + 4 is equal to 3 + 10.
In the above image, we can observe 9+ 6 number sentence. By adding nine with six results fifteen. The number sentence 9 + 6 is equal to 5 + 10.
In the above image, we can observe 9+ 5 number sentence. By adding nine with five results fourteen. The number sentence 9 + 5 is equal to 4 + 10.
In the above image, we can observe 9+ 2 number sentence. By adding nine with two results eleven. The number sentence 9 + 2 is equal to 1+ 10.
In the above image, we can observe 9+ 7 number sentence. By adding nine with seven results sixteen. The number sentence 9 + 7 is equal to 6 + 10.
In the above image, we can observe 9+ 9 number sentence. By adding nine with nine results eighteen. The number sentence 9 + 9 is equal to 8 + 10.
In the above image, we can observe 9+ 6 number sentence. By adding nine with six results fifteen. The number sentence 9 + 6 is equal to 10 + 5.
In the above image, we can observe 9+ 8 number sentence. By adding nine with eight results seventeen. The number sentence 9 + 8 is equal to 10 + 7.
In the above image, we can observe 9+ 9 number sentence. By adding nine with nine results eighteen. The number sentence 9 + 9 is equal to 10 + 8.
In the above image, we can observe 9+ 4 number sentence. By adding nine with four results thirteen. The number sentence 9 + 4 is equal to 10 + 3.
In the above image, we can observe 9+ 5 number sentence. By adding nine with five results fourteen. The number sentence 9 + 5 is equal to 10 + 4.
In the above image, we can observe 9+ 7 number sentence. By adding nine with seven results sixteen. The number sentence 9 + 7 is equal to 10 + 6. |
Chapter 8 Lines and Angle R.D. Sharma Solutions for Class 9th Math MCQ
Multiple Choice Questions
1. One angle is equal to three times its supplement. The measure of the angle is
(a) 130°
(b) 135°
(c) 90°
(d) 120°
Solution
2. Two complementary angles are such that two times the measure of one is equal to three times the measure of the other. The measure of the smaller angle is
(a) 45°
(b) 30°
(c) 36°
(d) none of these
Solution
3. Two straight line AB and CD intersect one another at the point O. If ∠AOC + ∠COB + ∠BOD = 274°, then ∠AOD =
(i) 86°
(ii) 90°
(iii) 94°
(iv) 137°
Solution
4. Two straight lines AB and CD cut each other at O. If ∠BOD = 63°, then ∠BOC =
(a) 63°
(b) 117°
(c) 17°
(d) 153°
Solution
5. Consider the following statements:
When two straight lines intersect:
(i) adjacent angles are complementary
(ii) adjacent angles are supplementary
(iii) opposite angles are equal
(iv) opposite angles are supplementary
Of these statements
(a) (i) and (ii) are correct
(b) (ii) and (iii) are correct
(c) (i) and (iv) are correct
(d) (ii) and (iv) are correct
Solution
Let us draw the following diagram showing two straight lines AD and BC intersecting each other at a point O.
Now, let us consider each statement one by one:
(i) When two lines intersect adjacent angles are complementary.
This statement is incorrect
Explanation:
As the adjacent angles form a linear pair and they are supplementary.
(ii) When two lines intersect adjacent angles are supplementary.
This statement is correct.
Explanation:
As the adjacent angles form a linear pair and they are supplementary.
(iii) When two lines intersect opposite angles are equal.
This statement is correct.
Explanation:
As the vertically opposite angles are equal.
(iv) When two lines intersect opposite angles are supplementary.
This statement is incorrect.
Explanation:
As the vertically opposite angles are equal
Thus, out of all, (ii) and (iii) are correct.
Hence, the correct choice is (b).
6. Given ∠POR = 3x and ∠QOR = 2x + 10°. If POQ is a straight line, then the value of x is
(a) 30°
(b) 34°
(c) 36°
(d) none of these
Solution
7. In Fig. 8.122, AOB is a straight line. If ∠AOC + ∠BOD = 85°, then ∠COD =
(a) 85°
(b) 90°
(c) 95°
(d) 100°
Solution
8. In Fig. 8.123, the value of y is
(a) 20°
(b) 30°
(c) 45°
(d) 60°
Solution
9. In Fig. 8.124, if y/x = 5 and z/x = 4, then the value of x is
(a) 8
(b) 18
(c) 12
(d) 15
Solution
10. In Fig. 8.125, the value of x is
(a) 12
(b) 15
(c) 20
(d) 30
Solution
11. In Fig. 8.126, which of the following statements must be true?
(i) a + b = d + c
(ii) a + c + e = 180°
(iii) b + f = c + e
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (ii) and (iii) only
Solution
(i) Statement: a+b = d+c
This statement is incorrect.
Explanation: We have, a and dare vertically opposite angles.
Therefore, a=d ---(i)
Similarly, band e are vertically opposite angles.
Therefore,
b=e ---(ii)
On adding (i) and (ii), we get
a+b=d+e
Thus, this statement is incorrect.
(ii) Statement a+c+e = 180°
This statement is correct.
Explanation: As a°, f° and e° form a linear pair, therefore their sum must be supplementary.
a+f+e = 180° ---(iii)
Also f° and c° are vertically opposite angles, therefore, these must be equal.
f =c
Putting f=c in (iii), we get:
a+c+e=180°
(iii) Statement: b+ f =a+e
This statement is correct.
Explanation: As a°, f° and b° form a linear pair, therefore their sum must be supplementary.
a+ f +b=180° ---(iv)
Also c°,d° and e° form a linear pair, therefore their sum must be supplementary.
c+d+e=180° ---(v)
On comparing (iv) and (v), we get:
a+ f +b=c+d+e
Also a° and d° are vertically opposite angles, therefore, these must be equal.
Therefore,
a= d
Substituting the above equation in (vi), we get:
a+f+b = c+d+e
a+b+f = c+a+e
b+f = a+e
Thus, out of all, (ii) and (iii) are correct.
12. If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2:3, then the measure of the larger angle is
(a) 54°
(b) 120°
(c) 108°
(d) 136°
Solution
13. In Fig. 8.127, AB || CD || EF and GH || KL. The measure of ∠HKL is
(a) 85°
(b) 135°
(c) 145°
(d) 215°
Solution
14. In Fig. 8.128, if AB || CD, then the value of x is
(a) 20°
(b) 30°
(c) 45°
(d) 60°
Solution
15. AB and CD are two parallel lines. PQ cuts AB and CD at E and F respectively. EL is the bisector of ∠FEB. If ∠LEB = 35°, then ∠CFQ will be
(a) 55°
(b) 70°
(c) 110°
(d) 130°
Solution
16. Two lines AB and CD intersect at O. If ∠AOC + ∠COB + ∠BOD = 270°, then ∠AOC =
(a) 70°
(b) 80°
(c) 90°
(d) 180°
Solution
17. In Fig. 8.129, PQ || RS, ∠AEF = 95°, ∠BHS = 110° and ∠ABC = x°. Then the value of x is
(a) 15°
(b) 25°
(c) 70°
(d) 35°
Solution
Solution
19. In Fig. 8.131, if l1∥l2, what is x+y in terms of w and z?
(a) 180-w+z
(b) 180+w-z
(c) 180-w-z
(d) 180+w+z
Solution
20. In Fig. 8.132, if l1∥l2, what is the value of y?
(a) 100
(b) 120
(c) 135
(d) 150
Solution
21. In Fig. 8.133, if l1∥l2 and l3∥l4, what is y in terms of x?
(a) 90+x
(b) 90+2x
(c) 90 - x/2
(d) 90-2x
Solution
22. In Fig. 8.13, if l∥m, what is the value of x?
(a) 60
(b) 50
(c) 45
(d) 30
Solution
23. In Fig. 8.135, if the line segment AB is parallel to the line segment CD, what is the value of y?
(a) 12
(b) 15
(c) 18
(d) 20
Solution
24. In Fig. 8.136, if CP∥DQ, then the measure of x is
(a) 130°
(b) 105°
(c) 175°
(d) 125°
Solution
25. In Fig. 8.137, if AB || HF and DE || FG, then the measure of ∠FDE is
(a) 108°
(b) 80°
(c) 100°
(d) 90°
Solution
26. In Fig. 8.138, if lines l and m are parallel then x =
(a) 20°
(b) 45°
(c) 65°
Solution
27. In Fig. 8.139, if AB∥CD, then x =
(a) 100°
(b) 105°
(c) 110°
(d) 115°
Solution
28. In Fig. 8.140, if lines l and m are parallel lines, then x =
(a) 70°
(b) 100°
(c) 40°
(d) 30°
Solution
29. In Fig. 8.141, if l∥m, then x =
(a) 105°
(b) 65°
(c) 40°
(d) 25°
Solution
30. In Fig. 8.142, if lines l and m are parallel, then the value of x is
(a) 35°
(b) 55°
(c) 65°
(d) 75°
Solution |
Year Six Mathematics Worksheets
Multiplication is one of the most important concepts in math, and it is essential for kids to understand and master early on. The multiplication property is a concept that explains how multiplication works and helps kids understand the rules and patterns of multiplication. In this article, we will explain what the multiplication property is, why it is important for kids to learn, and provide some simple examples and exercises that kids can use to practice and build their understanding.
The multiplication property refers to the rules and patterns of how numbers are multiplied. For example, the property of distributivity states that multiplying a sum by a number is the same as multiplying each term of the sum by the number and then adding the results. For example, (3 + 4) x 5 = 3 x 5 + 4 x 5, which equals 15 + 20, which equals 35.
Another important property of multiplication is the associative property, which states that the order in which numbers are multiplied does not affect the result. For example, (3 x 4) x 5 = 3 x (4 x 5), which equals 60 in both cases.
The commutative property of multiplication states that the order of the numbers being multiplied does not affect the result. For example, 3 x 4 = 4 x 3, which equals 12 in both cases.
The identity property of multiplication states that any number multiplied by 1 is equal to itself. For example, 3 x 1 = 3, and 5 x 1 = 5.
These are just a few of the many properties of multiplication that kids need to learn, but they provide a solid foundation for understanding how multiplication works. With practice and repetition, kids can improve their skills and gain confidence in using the multiplication property.
Here are some simple examples and exercises that kids can use to practice the multiplication property:
1. Use the distributive property to simplify 3 x (5 + 4).
2. Use the associative property to simplify (2 x 3) x 4.
3. Use the commutative property to simplify 3 x 4 and 4 x 3.
4. Use the identity property to simplify 3 x 1 and 5 x 1.
5. Multiply 2 x 3 x 4 using the associative property.
6. Multiply 3 x (2 x 4) using the distributive property.
7. Multiply (3 x 2) x (4 x 5) using the associative property.
These examples and exercises can help kids get started with the multiplication property and build their understanding. However, it’s important to remember that practice and repetition are key to success. Encourage your kids to work through as many multiplication problems as possible and ask questions if they need help.
In conclusion, the multiplication property is an important part of math that kids need to learn early on. By understanding the underlying concepts and rules, kids can develop their skills and confidence in using the multiplication property. With the right approach and techniques, math can be a fun and empowering subject for kids.
Year Six Math Worksheet for Kids – Multiplication Property
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# The powers of General Form - PowerPoint PPT Presentation
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The powers of General Form. Probe. Below are 5 different ways of representing a quadratic relationship. Four of them represent the SAME quadratic relationship. a)Find and correct the odd-one-out. b)Name the 5 forms. Probe.
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The powers of General Form
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## The powers of General Form
### Probe
Below are 5 different ways of representing a quadratic relationship. Four of them represent the SAME quadratic relationship.
a)Find and correct the odd-one-out.
b)Name the 5 forms
### Probe
Below are 5 different ways of representing a quadratic relationship. Four of them represent the SAME quadratic relationship.
a)Find and correct the odd-one-out.
b)Name the 5 forms
mapping rule
equation in general form
equation in transformational form
graphical (parabola)
table of values
### General Form toTransformational Form
Let’s look at an example.We will take the quadratic function given in general form:y = 2x2 + 12x – 4and turn it into transformational form:
1
We know that in transformational form the coefficient of ‘x’ is 1
STEP 1:Divide every term by a.
General Form toTransformational Form
Let’s look at an example.We will take the quadratic function given in general form:y = 2x2 + 12x – 4and turn it into transformational form:
STEP 2:Move the non-x term over
### Completing the Square
We’re getting close, but the right-hand side of the equation needs to be a perfect square: (x – HT)2
So far we have: x2 + 6x
This looks like:
x2
x
x
x
x
x
x
### Completing the Square
x2
x
x
x
x
x
x
The equivalent to making x2 + 6x a perfect square is to arrange these tiles into a square shape...
x
x
x
x2
x
x
x
Oops…
### Completing the Square
x2
x
x
x
x
x
x
Try again…
Closer….
We just need tocomplete this square…
x2
x
x
x
x
x
We were missingnine 1×1 squares.
x
### Completing the Square
---------- x + 3 -----------
This looks promising:We have a square with length and width dimensions of (x + 3).
---------- x + 3 -----------
x2
x
x
x
But how can we justify this “just add 9”?
Add it to both sides of the equation!
But why 32?
It’s half of the coefficient of x, squared.
x
x
x
General Form toTransformational Form
Back to our example.We will take the quadratic function given in general form:y = 2x2 + 12x – 4and turn it into transformational form:
STEP 3:Complete the square(take half the coefficient of x, square it and add it to both sides)
General Form toTransformational Form
Back to our example.We will take the quadratic function given in general form:y = 2x2 + 12x – 4and turn it into transformational form:
STEP 4:Factor out the 1/a term on the left hand side
General Form toTransformational Form
Back to our example.We will take the quadratic function given in general form:y = 2x2 + 12x – 4and turn it into transformational form:
Therefore, the transformational form of the quadratic functiony = 2x2 + 12x – 4 is:
Now we know that its VS = 2, HT = –3, and VT = –22
From this we know its vertex, range, axis of symmetry, etc
vertex is (–3, –22)
range is{y≥–22, y R}
axis of symmetry isx = –3
### Practice
1. Complete the square to find the vertex of the functiony = 2x2 – 8x + 2. Sketch its graph.
2. Complete the square to find the range of the functiony = –x2 – 5x +1?
3. Put the equation y = 0.5x2 – 3x – 1 into transformational form by completing the square.
Practice: Solutions
Vertex: (2, –6)
Complete the square to find the vertex of the functiony = 2x2 – 8x + 2. Sketch its graph.
Practice: Solutions
range is{y≤ 7.25, y R}
2. Complete the square to find the range of the functiony = –x2 – 5x +1?
Practice: Solutions
3. Put the equation y = 0.5x2 – 3x – 1 into transformational form by completing the square.
### A shortcut…eventually
STEP 1:Divide every term by a.
STEP 2:Move the non-x term over
STEP 3:Complete the square(take half the coefficient of x, square it and add it to both sides)
STEP 4:Factor out the 1/a term on the left hand side
But completing the square is a lot of work….
Let’s find a shortcut to finding vertex ect. by completing the square just ONE more time, but with the general equation
A shortcut…eventually
STEP 1:Divide every term by a.
STEP 2:Move the non-x term over
STEP 3:Complete the square(take half the coefficient of x, square it and add it to both sides)
STEP 4:Factor out the 1/a term on the left hand side
### Here’s the shortcut
We now see that ANY general quadratic equationy = ax2 + bx + c can be written as
This may look complicated, but is VERYhelpful…
from this we can see that the HT for any general quadratic is
For example: What is the axis of symmetry of the function y = –2x2 + 4x + 6?
To complete the square on this takes time. But…
### Here’s the shortcut
We now see that ANY general quadratic equationy = ax2 + bx + c can be written as
This may look complicated, but is VERYhelpful…
from this we can see that the VT for any general quadratic is
For example: What is the vertex of the function y = –2x2 + 4x + 6?
To complete the square on this takes time. But…
vertex (1, 8)
### Here’s the shortcut
We now see that ANY general quadratic equationy = ax2 + bx + c can be written as
This may look complicated, but is VERYhelpful…
from this we can see that the VS for any general quadratic is a.
Since the VS = –2 we can graph from the vertex (1, 8):
Over 1 down 2
Over 2 down 8
Over 3 down 18
For example: Graph the functiony = –2x2 + 4x + 6.
To complete the square on thistakes time. But…
### Shortcut within a shortcut
Instead of memorizing the formula for the y-coordinate of the vertex (VT):
we can calculate it using the general form of the equation and thex-coordinate of the vertex (HT):
For example: What is the maximum value of the function y = –2x2 + 4x + 6?
Max value is y = 8.
### General form shortcuts: Practice
For the following quadratic functions, find the vertex, sketch its parabola, give its axis of symmetry, give its range, and write in transformational form, all WITHOUT completing the square.
### General form shortcuts: Practice Solutions
vertex (1, –2), VS = 1/4
axis of symmetry: x = 1range: {y≥ 2, y R}
### General form shortcuts: PracticeSolutions
vertex (–4, –6), VS = 3
axis of symmetry: x = –4range: {y≥ –6, y R}
Solving for x given y using general form
Given a quadratic function in general form y = ax2 + bx + c and a value for y, it is not a straightforward task to find the corresponding x-value(s) because we cannot easily isolate x when it appears in 2 different forms: x and x2.
Subbing in the required y-value to a quadratic function, and bringing all the terms to one side of the equation yields a quadratic equation.
A quadratic equation in general form looks like this:ax2 + bx + c = 0 where a≠ 0.
Solving a quadratic equation for x is also calledfinding its roots or finding its zeros.
Solving for x given y using general form
NOTE: the c value of the quadratic equation might be different from the c value of the quadratic function. See the following example.
Example 1:Give the quadratic equation obtained from the functiony = x2 – 2x – 15 when y = –12.
Although the quadratic function had coefficients a = 1, b = –2, and c = –15,subbing in −12 for y gives the quadratic equation with coefficientsa = 1, b = –2, and c = –3
Solution:
−12 = x2 – 2x – 15
0 = x2 – 2x – 3
Solving for x given y using general form
NOTE: the c value of the quadratic equation is the SAME as the c value of the quadratic function when we are using y = 0. See the following example.
Example 2:Give the quadratic equation obtained from the functiony = x2 – 2x – 15 when y = 0.
Both the quadratic function and the resulting quadratic equation have coefficients a = 1, b = –2, and c = –15.
Solution:
0 = x2 – 2x – 15
While this is only true when we use y = 0, this is a very common case, since we are often interested in the x-intercepts of a quadratic function.
Solving for x-intercepts using general form
Example 3:Find the x-intercepts of the quadratic function y = x2 – 2x – 15
Solution:
0 = x2 – 2x – 15
…but now what?
Soon we will learn how to do this directly from general form…. Last class we learned how to solve this if the function had been given in transformational form…
So lets try that!
Solving for x-intercepts using general form
STEP 1: Write in transformational form by completing the square
STEP 2: Set f(x) = y = 0
STEP 3: Simplify left-hand side
STEP 4: Take the squarrootof both sides
STEP 5: Isolate x in both equations
Therefore the x-intercepts of the parabola are (5, 0) and (–3, 0), and the roots of the function are 5, and –3.
Solving for x-intercepts using general formPractice
Find the roots of the following quadratics:
Not worth it!
Wait for the
shortcut!
### A shortcut…eventually
STEP 1: Write in transformational form by completing the square
STEP 2: Set f(x) = y = 0
STEP 3: Simplify left-hand side
STEP 4: Take the square-root of both sides
STEP 5: Isolate x in both equations
But this is a lot of work….
Let’s find a shortcut to finding x-intercepts by doing this just ONE more time, but with the general equation.
### A shortcut…eventually
STEP 1: Write in transformational form by completing the square
STEP 2: Set f(x) = y = 0
STEP 3:Simplify left-hand side
### A shortcut…eventually
STEP 4: Take the square-root of both sides
STEP 5: Isolate x in both equations
This is the shortcut, the Quadratic Root Formula!
Solving for x-intercepts using general formThe short-cut
Back to our example:Find the x-intercepts of the graph of the functionf(x) = x2 – 2x – 15
Solution:
### Solving for x given y using general form
The quadratic root formula can be used to find x-values other than the x-intercepts
For the function f(x) = x2 – 2x – 15, find the values of x when y = −12:
Solution:
–12 = x2 – 2x – 15
0 = x2 – 2x – 3.
Now use the quadratic root formua witha = 1, b = –2 and c = –3
Solving for x-intercepts using general formPractice
Find the roots of the following quadratics:
Use the
root formula!
Finding roots from General Form: Practice Solutions
Finding roots from General Form: Practice Solutions
Finding roots from General Form: Practice Solutions
Find the x-intercepts of the following quadratic functions:
### Yet another method of finding roots: Factoring
As slick as the quadratic root formula is for finding roots, sometimes factoring is even faster!
The zero property:
If (s)(t) = 0 then s = 0, or t = 0, or both.(This only works when the product is 0.)
So if we can get the quadratic function in the form:y = (x – r1)(x – r2), (called factored form)then when y = 0 we know that:
(x – r1) = 0, or (x – r2) = 0, or both.
In other words, x = r1 or x = r2 or both.
### Factoring to find the roots
Ex. Find the x-intercepts of the graph of the functionf(x) = x2 – 2x – 15
Answer when the numbers are multiplied
We need to factor x2 – 2x – 15.We need 2 numbers whose product is –15 and whose sum is –2
(There is never more than one pair of numbers that work!)
-5
3
___ ×___ = –15
___ + ___ = – 2
-5
So when looking for roots, 0 = (x – 5)(x + 3) and x = 5 or x = –3
3
So the factored form off(x) = x2 – 2x – 15 isf(x) = (x – 5)(x + 3)
So the x-intercepts of the graph are (5, 0) and (–3, 0)
### Factoring
When factoring a quadratic equation where a = 1,find two numbers that multiply to give c and add to give b.
By the way:
(0) = x2 + 10x – 24
24 = x2 + 10x
24 + 25 = x2 + 10x + 25
49 = (x + 5)2
±7 = x + 5
x = –5 + 7 or x = –5 – 7
x = 2 or x = –12
Ex. Find the roots ofy = x2 + 10x – 24by factoring.
12
-2
___ x ___ = -24
___ + ____ = 10
-2
12
or
### Practice
a) 0 = x2 + 2x +1
c) 0 = x2 + 2x – 24
b) 0 = x2 + 5x +4
d) 0 = x2 – 25
2. Find the x- and y- intercepts of the following quadratics.
a) f(x) = 2x2 +3x + 1
c) f(x) = x2 – 5x -14
b) f(x) = 3x2 – 7x + 2
d) y = 2(x – 4)2 – 32 |
# IGCSE Mathematics Paper-1: Specimen Questions with Answers 132 - 133 of 324
## Question 132
### Write in Short
(a) George going to Germany in his summer vacation. He changed into euros when the rate was after reached in Germany. How many euros did he receive?
(b) Write 9.568 correct to 2 significant figures.
### Explanation
Here,
(a) George changed into euros when the rate was
So,
Then
(b) 9.568 correct to 2 significant figures means there will be total of 2 digits:
So, we will get 9.6 because after digit 5 there is 6 which is greater than 5, so 5 will get round up and be 6.
## Question 133
### Describe in Detail
Essay▾
A straight road between P and Q is shown in the diagram. R is the point south of P and east of Q. PR = 8.3 km and QR = 4.8 km.
(a) The length of the road PQ
(b) The bearing of Q from P.
### Explanation
Here,
(a) is a right angled triangle,
So, by using the formula of sides of a right angled triangle:
, where QR = 4.8 km and PR = 8.3 km is given,
.
(b) Bearing of Q from P:
• The true bearing to a point is the angle which is measured in degrees in a clockwise direction from the north line.
• The bearing from one point P to another Q is the direction of the line between those points.
• The principal directions are north, east, south, and west, the pact is that the direction due north is a bearing of 0 degrees, due east is 90 degrees, due south is 180 degrees, and due west is 270 degrees.
So, here, bearing of Q from P will be which is measured as shown below:
• So,
• So, bearing of Q from P will be:
Developed by: |
×
STUDY HARD SO YOU CAN LIVE YOUR LIFE AS YOU WANT.
# Note for Probability and Statistics - PS by bhasakar naik
• Probability and Statistics - PS
• Note
• Sree vidyanikethan engineering college -
• Computer Science Engineering
• B.Tech
• 2 Topics
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Bhasakar Naik
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#### Text from page-2
Chapter IV: Probability Distributions and Their Applications items (the p 's) among 5 items (the p 's and q 's). Therefore the total number of terms ⎛5⎞ is ⎜⎜ ⎟⎟ or 10 so that the probability of exactly 2 exceedances in 5 years is 10 p 2 q 3 . ⎝ 2⎠ This result can be generalized so that the probability of X = x exceedances in n ⎛n ⎞ years is ⎜⎜ ⎟⎟ p X q n − X . The result is applicable to any Bernoulli process so that the ⎝X⎠ probability of X = x occurrences of an event in n independent trials if p is the probability of an occurrence in a single trial is given by: ⎛n⎞ f X ( x : n, p ) = ⎜⎜ ⎟⎟ p x q n − x ⎝ x⎠ x = 0,1,2, K , n This equation is known as the binomial distribution. The binomial distribution and the Bernoulli process are not limited to a time scale. Any process that may occur with probability p at discrete points in time or space or in individual trials may be a Bernoulli process and follow the binomial distribution. The cumulative binomial distribution is X ⎛n⎞ F X ( x; n, p) = ∑i = 0 ⎜⎜ ⎟⎟ p i q n − i ⎝i ⎠ x = 0,1,2, K , n and gives the probability of x or fewer occurrences of an event in n independent trials if the probability of an occurrence in any trial is p. Continuing the above example, the probability of less than 3 exceedances in 5 years is 2 ⎛ 5⎞ FX (2;5, p ) = ∑i =0 ⎜⎜ ⎟⎟ p i q 5−i ⎝i ⎠ = f X (0;5, p) + f X (1;5, p) + f X (2;5, p) The mean and variance of the binomial distribution are E ( X ) = np var( X ) = nqp The coefficient of skew is (q − p) / npq so that the distribution is symmetrical for p = q , skewed to the right for q > p and skewed to the left for q < p . 68
#### Text from page-3
Chapter IV: Probability Distributions and Their Applications The binomial distribution has an additive property. That is if X has a binomial distribution with parameters n1 and p and Y has a binomial distribution with parameters n 2 and p , then Z = X + Y has a binomial distribution with parameters n = n1 + n2 and p . The binomial distribution can be used to approximate the hyper-geometric distribution if the sample selected is small in comparison to the number of items N from which sample is drawn. In this case the probability of a success would be about the same for each trial. Example: In order to be 90 percent sure that a design storm is not exceeded in a 10 year period. What should be the return period of the design storm? Solution: Let p be the probability of the design storm being exceeded. The probability of no exceedances is given by ⎛10 ⎞ f X (0;10, p) = ⎜⎜ ⎟⎟ p 0 q 10 ⎝0 ⎠ 0.90 = (1 − p)10 p = 1 − (0.90)1 / 10 = 1 − 0.9895 = 0.0105 T = 1 / p = 95 years. Comment: To be 90 percent sure that a design storm is not exceeded in a 10-year period a 95-year return storm must be used. If a 10-year return period storm is used, the chances of it being exceeded is 1 − f X (0;0,1) = 0.6513 . In general the chance of at least one occurrence of a T-year event in T-years is 1 − f X (0; T ,1 / T ) = 1 − (1 − 1 / T ) T . Therefore, for a long design life, the chance of at least one occurrence of an event with return period equal to the design life approaches 1 − 1 / e or 0.632.Thus if the design life of a structure and its design return period are the same, the chances are very great that the capacity of the structure will be exceeded during its design life. 4.1.2 Poisson distribution The Poisson distribution is like the binomial distribution in that it describes phenomena for which the average probability of an event is constant, independent of the number of previous events. In this case, however, the system undergoes transitions 69
#### Text from page-4
Chapter IV: Probability Distributions and Their Applications randomly from one state with n occurrences of an event to another with ( n + 1) occurrences, in a process that is irreversible. That is, the ordering of the events cannot be interchanged. Another distinction between the binomial and Poisson distributions is that for the Poisson process the number of possible events should be large. The Poisson distribution may be inferred from the identity e −µ e µ = 1 where the most probable number of occurrences of the event is µ . If the factorial is expanded in a power series expansion, the probability P(r) that exactly r random occurrences will take place can be inferred as the r th term in the series, i.e., e −µ µ r p (r ) = r! (4.1.2.1) This probability distribution leads directly to the interpretation that: e − µ = the probability that an event will not occur, µ e − µ = the probability that an event will occur exactly once, ( µ 2 / 2! ) e − µ = the probability that an event will occur exactly twice, etc, The mean and the variance of the Poisson distribution are: E( X ) = µ Var ( X ) = µ The coefficient of skew is µ −1 / 2 so that as µ gets large, the distribution goes from a positively skewed distribution to a nearly symmetrical distribution. The cumulative Poisson probability that an event will occur x times or less is: x p (≤ x ) = ∑ p ( r ) r =0 Of course, the probability that the event will occur ( x + 1) or more times would be the complement of P(x). The Poisson distribution is useful for analyzing the failure of a system that consists of a large number of identical components that, upon failure, cause irreversible transitions in the system. Each component is assumed to fail independently and randomly. Then µ is the most probable number of system failures over the life time. To summarize: 70
#### Text from page-5
Chapter IV: Probability Distributions and Their Applications ¾ The binomial distribution is useful for systems with two possible outcomes of events (failure–no failure) in cases where there is a known, finite number of (Bernoulli) trials and the ordering of the trials does not affect the outcome. ¾ The Poisson distribution treats systems in which randomly occurring phenomena cause irreversible transitions from one state to another. Example: A given nuclear reactor is fueled with 200 assemblies, each of which can fail if the cladding on a fuel rod fails. If each assembly fails in an independent and random manner over the exposure time, calculate the probability of 3 assemblies failing if, on the average, 1% of the fuel assemblies are known to fail. (MacCormick, 1981, p. 34) Solution: The mean number of assembly failure is µ = 2 , so using equation (4.1.2.1) for r = 3 gives P (3) = (2 3 / 3!)e −2 = 0.1804 As a check, we can use the probability of a single assembly failing, p = 0.01 , and the binomial distribution equation with n = 200 to obtain P (3) = 4.1.3 200! (0.01) 3 (0.99) 200−3 = 0.1814 3!(200 − 3)! Hyper-geometric distribution Drawing a random sample of size n (without replacement) from a finite population of size N with the elements of the population divided into two groups with k elements belonging to one group is an example of sampling from a hyper-geometric distribution. The two groups may be defective or non-defective objects, rainy or nonrainy days, success or failure of a project, etc. The total number of possible outcomes or ways of selecting a sample of size n from ⎛N⎞ N objects is ⎜⎜ ⎟⎟ . The number of ways of selecting x successes and n-x failures ⎝n ⎠ ⎛k ⎞ ⎛ N − k ⎞ ⎟⎟ . Thus from the population containing k successes and N − k failures is ⎜⎜ ⎟⎟ ⎜⎜ ⎝ x⎠ ⎝n − x ⎠ the probability is: 71 |
# Urn problem: probability of drawing balls of k unique colors
Given an urn with $N$ balls in $K$ colors, divided evenly (so $N$ $mod$ $K$ = $0$).
What is the probability that I draw $k$ different colors if I do $n$ draws without replacement?
And, more general, is there an easier way to calculate the expected value of the number of unique colors than summing $k * P(k)$ for all $k$ if $K <= k <= n$?
We find a simple formula for the expected number of different colours. For this we use the method of Indicator Random Variables.
For $$i=1$$ to $$K$$, let $$I_i=1$$ if colour $$i$$ is drawn at least once, and let $$I_i=0$$ otherwise. Then the number $$Y$$ of colours drawn is $$I_1+I_2+\cdots+I_K$$, and by the linearity of expectation we have $$E(Y)=E(I_1)+E(I_2)+\cdots +E(I_K).$$
$$E(I_i)$$ is defined as: $$E(I_i) = 1 \cdot Pr(I_i=1) + 0 \cdot Pr(I_i=0)$$ $$E(I_i) = 1 \cdot Pr(I_i=1)$$
Hence, we need to find $$\Pr(I_i=1)$$ to solve for $$E(I_i)$$. However, it turns out to be simpler to find $$\Pr(I_i=0)$$ and then use the fact that $$\Pr(I_i=1)=1-\Pr(I_i=0)$$.
In general, there are $$b=\binom{N}{n}$$ equally likely ways to choose $$n$$ balls. More specifically, there are $$N-N/K$$ balls not of colour $$i$$, so there are $$a=\binom{N-N/K}{n}$$ ways to choose $$n$$ balls, none of colour $$i$$.
It follows that $$\Pr(I_i=1)=1-\frac{a}{b}$$ and therefore $$E(Y)=K\left(1-\frac{a}{b}\right).$$
• I think I follow you. But it should be $\frac{b}{a}$, not $\frac{a}{b}$, right? Commented Mar 5, 2015 at 0:49
• Thanks, fixed, except that I did it by changing the meanings of $b$ and $a$. Commented Mar 5, 2015 at 0:52
• It took me a while to figure out which step I was missing. That was the explanation of why $E(X_i) = P(X_i = 1)$. The expectation function $E(X)$ sums the possible values of $X$ (namely 0 and 1) multiplied them by their probabilities. So, $E(X_i) = 1*Pr(X_i = 1) + 0*Pr(X_i = 0) = Pr(X_i = 1)$. Actually, that was pretty easy. Commented Mar 5, 2015 at 9:24
• @Lewistrick: Yes, if $W$ is a Bernoulli random variable (can only take on values $1$ and $0$) then $E(W)=\Pr(w=1)$, as you just showed. The method we used (the indicator random variables $X_i$, together with linearity of expectation can be a very useful tool in calculating expectations. Commented Mar 5, 2015 at 16:11 |
# ch1b - Notes on Chapter 1(pp 11-26 Vectors To describe...
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John Ellison, UCR p.1 Notes on Chapter 1 (pp. 11-26) Vectors To describe motion in three dimensions we will use vectors . A vector has magnitude and direction. An example of a vector quantity is displacement. The displacement of a particle which moves from point A to point B can be represented by a vector. Its magnitude is the length of the line joining A and B, and its direction is from A to B, indicated by an arrow. Other examples of vector quantities are velocity and acceleration. A scalar is a quantity which has magnitude only (no direction). Examples of scalar quantities are speed, temperature, mass, time, and pressure. We will represent vectors by italic symbols with an arrow above, e.g. . The magnitude of a vector is represented by the symbol with no arrow, e.g. a , or by the notation . Now consider two displacements, one from A to B, and the other from B to C. The net displacement is from A to C. This can be represented graphically as shown in the figure at top right. We call AC the vector sum (or resultant ) of the vectors AB and AC. We can represent this by the vector addition equation which says that vector is the vector sum of vectors and . Note that this is not a simple algebraic addition - it involves the magnitudes and directions of the vectors, not just the magnitudes. Vector addition has two important properties: Vector subtraction is defined by: where the vector is a vector with the same magnitude as but with the opposite direction (see the figure below). s = a b a b = b a (commutatative law) a b c = a b c (associative law) a d a b a − b s a b ∣ a b b
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John Ellison, UCR p.2 Unit Vectors The component of a vector is the projection of the vector on an axis (see figure below). The x component and y component are the projections on the x -axis and y -axis respectively, and are denoted by a x and a y , respectively. The process of finding the components of a vector is called resolving the vector . From simple geometry (see the figure below), we find that the components are given by: where θ is the angle the vector makes with the positive direction of the x -axis. In order to express a vector in terms of its components we define unit vectors as vectors of unit magnitude which point in the direction of the positive x , y , z directions, respectively. We will normally use a coordinate system in which the + x direction is horizontal, the + y direction is vertical, and the + z direction is "out of the page". Such a coordinate system (shown in the figure above left) is said to be a right-handed coordinate system . Now we can express the vectors as This says that vector is equal to the sum of a vector of magnitude a x along the + x direction and a vector of magnitude a y along the + y direction (and similarly for ). The magnitude and angle of vector are a = a x i a y j b = b x i b y j a a and b b a x = a cos and a y = a sin magnitude: a = a x 2 a y 2
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# Explain whether the binomial probability distribution can be used to
Explain whether the binomial probability distribution can be used to find the probability that a contestant who plays the game five times wins exactly twice.
Check for the binomial experiment requirements and specify the values of n,r, and p.
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Calculation:
From the given information, it can be observed that the five trials are independent with two outcomes (win or do not win) that are same.
They are repeated under same conditions with the same probability of success on each trial.
Thus, the binomial distribution can be used to find the specified probability.
Values:
Number of trials $$\displaystyle{n}={5}$$
Probability of winning the game (or success) $$\displaystyle{p}={\frac{{{1}}}{{{6}}}}$$.
Number of times the player wins the game $$\displaystyle{r}={2}$$.
The probability of failure (or probability of not winning) is given below:
$$\displaystyle{q}={1}-{p}$$
$$\displaystyle={1}-{\left({\frac{{{1}}}{{{6}}}}\right)}$$
$$\displaystyle={\frac{{{5}}}{{{6}}}}$$
Binomial probability:
The probability of r successes out of n trials is given below:
$$\displaystyle{P}{\left({r}\right)}={C}_{{{n},{r}}}{p}^{{{r}}}{q}^{{{n}-{r}}}$$
Here, n is the number of trials, r is the number of successes, p is the probability of success, and q is the probability of failure.
The probability that a contestant wins exactly twice out of five times is calculated as given below:
$$\displaystyle{P}{\left({2}\right)}={C}_{{{5.2}}}{\left({\frac{{{1}}}{{{6}}}}\right)}^{{{2}}}{\left({\frac{{{5}}}{{{6}}}}\right)}^{{{3}}}$$
$$\displaystyle={\frac{{{5}!}}{{{2}!{3}!}}}\times{\left({0.166}\right)}^{{{2}}}\times{\left({0.833}\right)}^{{{3}}}$$
$$\displaystyle={10}\times{\left({0.166}\right)}^{{{2}}}\times{\left({0.833}\right)}^{{{3}}}$$
$$\displaystyle={0.160751}$$
Therefore, the probability that a contestant wins exactly twice is 0.160751. |
# Create and Interpret Column Graphs I
Lesson
A column graph is a graph used to display data about different groups. There are all different types of groups that can be looked at. For example the number of different types of bugs in a garden, the number of people with certain eye colours in a class or even the flavours of juice people prefer.
Let's look now at how to create and interpret (understand) column graphs.
## Creating a column graph
Let's say I looked in my garden and I found $6$6 ants, $5$5 worms, $3$3 caterpillars, $1$1 butterfly and $4$4 bees. Watch this video to see how we'd turn this into a column graph.
## Interpreting a column graph
So what information can we get from a column graph? Let's watch another video and see.
Remember!
• Label both axes (arms) of the graph.
• Make sure the columns are equal widths and equally spaced.
• Make sure that the height of the column matches the frequency (how often that score occurred).
#### Worked Examples
##### Question 1
The table below shows the house points earned by four colour houses at their swimming carnival.
House Points
Blue $60$60
Green $20$20
Yellow $70$70
Orange $60$60
1. Complete the graph using the information from the table.
##### Question 2
The fish shop has $5$5 tanks.
The graph below shows the number of fish in the $5$5 tanks.
1. Turn this information into a scaled bar graph.
##### Question 3
A dice is rolled and the number of times it landed on the digits 1-6 is recorded in the table below.
Dice Number Number of Rolls
1 $6$6
2 $4$4
3 $1$1
4 $8$8
5 $5$5
6 $7$7
1. Complete the graph using the information from the table.
2. What was the most frequent dice roll?
3
A
6
B
4
C
5
D
1
E
2
F
3. How many more 4's were rolled than 2's?
4. What was the least frequent dice roll?
3
A
1
B
5
C
2
D
6
E
4
F |
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# Coordinate Geometry
1
1.1
## Graphing linear relationships
Table of values
Rules: Sub three values of x into the equation to nd y , to nd coordinates. - Use x = 1; 0; 1 - Plot the points - Remember to label axes, origin and equation of line.
1.2
Finding intercepts
The x-intercept of an equation is the value of x when y is equal to zero. To nd the x-intercept, substitute y = 0 into the equation. Similarly, the y -intercept of an equation is the value of y when x is equal to zero. Equations of the form y = mx + b have a y -intercept of b. Now that we have found our x and y intercepts, plot the points and form line.
1.3
1.3.1
Special cases
Intercept form
1.3.2
y
x a
y b
=1
## Line going through the origin
= mx Sub x = 1, then nd y . Draw a line going through the origin and the point.
2
2.1
## Points that satisfy linear relationships
Determine if a point lies on a line
P x; y
To determine if a point
2.2
## ) lies on a line, substitute the x and y values into the equation.
Determine the role for the equation of a straight line from a table of values
y
=c =c
x
## This is a vertical line cutting the x-axis at c.
4 Midpoint
x
The midpoint of A(x1 ; y1 ) and B (x2 ; y2 ) is found by nding the average of the Hence, we can derive the formula to be
M
values and
values.
AB (
x1
+ x2 y1 + y2 ; ) 2 2
Length of an interval
## The distance between two points A(x1 ; y1 ) and
AB
p(
B x2 ; y2
) is
x1
x2
)2 + (y1 y2 )2
Due to the property that square numbers will always be positive, it does not matter whether x1 and x2 are swapped. The same thing applies for y1 and y2 . Just remember that the xs are in the same bracket, and the y s in the same bracket.
6 Gradient of a line
If If If
0, then the line would look like 0, then the line would look like m = 0, the line would be at.
m < m >
. .
B x2 ; y2
The gradient is rise run The gradient formed when two points, A(x1 ; y1 ) and
m
y1 x1
y2 x2
## Parallel and perpendicular lines
Two lines are parallel if their gradients are equal. Two lines are perpendicular if the product of their gradients is -1.
8 Graphing equations of gradient-intercept form
The gradient-intercept form is an equation in the form of y = mx + b, where m is the gradient and b is the y -intercept. To sketch this, draw a line through the y-intercept with the gradient (higher gradient = steeper) Label x-intercept.
8.1 Finding equation of a line
To nd equation of a line from two points given, nd the gradient and sub in the gradient and the x and y values from one of the points into y = mx + b. Rearrange the equation to nd b. Now that you have a gradient and the y -intercept, sub that into y = mx + b to get your equation. You may rearrange that into general form, or another form. 2 |
1. ## Sequences mathematical induction
I have been doing sequences and I have no idea how to do this, can anyone help me step by step on how to solve this problem.
Use mathematical induction to prove that each statement is true for every positive integer n.
3 + 6 + 9 + ... + 3n = 3n(n+1)/2
2. Originally Posted by xterminal01
I have been doing sequences and I have no idea how to do this, can anyone help me step by step on how to solve this problem.
Use mathematical induction to prove that each statement is true for every positive integer n.
3 + 6 + 9 + ... + 3n = 3n(n+1)/2
Hello,
I assume that you know that this proof has to be done in 3 steps:
S1: n = 1: $\displaystyle 3 = \frac{3 \cdot 1 \cdot(1+1)}{2}=\frac{3 \cdot 2}{2}=3$ . S1 is true.
S2: We assume that 3 + 6 + 9 + ... + 3n = 3n(n+1)/2 is true.
S3: Show that the equation is true for n+1:
$\displaystyle 3 + 6 + 9 + ... + 3n + 3(n+1)= \frac{3n(n+1)}{2}+3(n+1)$
$\displaystyle = \frac{3n(n+1)}{2}+3(n+1)=3(n+1)\left(\frac n2 + 1\right) = 3(n+1)\left(\frac{n+2}{2}\right)$
$\displaystyle = 3(n+1)\left(\frac{n+2}{2}\right)=\frac{3(n+1)((n+1 )+1)}{2}$
Thus 3 + 6 + 9 + ... + 3n = 3n(n+1)/2 is true for n+1 and therefore it is true for all $\displaystyle n \in \mathbb{N}^*$ |
7190 minus 87 percent
This is where you will learn how to calculate seven thousand one hundred ninety minus eighty-seven percent (7190 minus 87 percent). We will first explain and illustrate with pictures so you get a complete understanding of what 7190 minus 87 percent means, and then we will give you the formula at the very end.
We start by showing you the image below of a dark blue box that contains 7190 of something.
7190
(100%)
87 percent means 87 per hundred, so for each hundred in 7190, you want to subtract 87. Thus, you divide 7190 by 100 and then multiply the quotient by 87 to find out how much to subtract. Here is the math to calculate how much we should subtract:
(7190 ÷ 100) × 87
= 6255.3
We made a pink square that we put on top of the image shown above to illustrate how much 87 percent is of the total 7190:
The dark blue not covered up by the pink is 7190 minus 87 percent. Thus, we simply subtract the 6255.3 from 7190 to get the answer:
7190 - 6255.3
= 934.7
The explanation and illustrations above are the educational way of calculating 7190 minus 87 percent. You can also, of course, use formulas to calculate 7190 minus 87%.
Below we show you two formulas that you can use to calculate 7190 minus 87 percent and similar problems in the future.
Formula 1
Number - ((Number × Percent/100))
7190 - ((7190 × 87/100))
7190 - 6255.3
= 934.7
Formula 2
Number × (1 - (Percent/100))
7190 × (1 - (87/100))
7190 × 0.13
= 934.7
Number Minus Percent
Go here if you need to calculate any other number minus any other percent.
7200 minus 87 percent
Here is the next percent tutorial on our list that may be of interest. |
## 14.1 Solving Equations
If we have a linear equation, such as 5x - 3 = 0, there is a straightforward procedure for solving it. You apply "the golden rule of equations": do unto the left side exactly what you do unto the right side. And you do it until all you have on the left is x.
Thus with this example you would add 3 to both sides, getting rid of the -3 on the left, and then divide by 5, with the result, .
Suppose however, we have a more complicated equation, such as
sin(x) - exp(x) + 2 = 0
Our task here is to find a solution , or all the solutions of such an equation.
First note that it is always a good idea to plot the left hand side here and observe, crudely, where it changes sign or comes very near to 0. This will tell you roughly where it becomes 0.
In the old days this was an extremely tedious task, in general, and people tried to solve equations without plotting, which is a bit like flying blind. Its OK if you can do it, but why try if you don't have to do so?
The standard technique for solving such equations apparently goes back to Newton. And here it is.
You start with a guess of an argument, call it x0. You then find the linear approximation to your function, f, at argument x0, and solve the equation that this linear approximation is 0. Call the argument for which the linear approximation is 0 x1.
Now you do exactly the same thing, starting at x1: you find the linear approximation to f at x1 and solve the equation that this linear approximation is 0 to determine x2. And you continue this as long as you need to.
In the old days this was an extremely tedious thing to do, for any function. Finding xj+1 from xj is quite easy, but doing it over and over again is a real bore.
Now with a spreadsheet, you can set this up and find solutions, with practice, in under a minute. You only have to do each step once, and copy.
How?
First let's see how to get xj+1 from xj.
The linear approximation to f at xj is
f(xj) + (x-xj) f '(xj)
If we set this to 0 at argument xj+1 we get
f(xj) + (xj+1 - xj) f '(xj) = 0
which has solution, obtained by dividing and subtracting from both sides appropriately
So what do I do on a spreadsheet?
Suppose we put our first guess in box A1. We will put it and subsequent guesses in column A starting say, with 3. (just to leave room for labels.)
We can then put f in column B and f ' in column C.
To do this we need make the following entries:
A3 = A1 (this puts starting guess x0 in A3)
B3 = f(A3) (this computes f(x0))
C3 = f'(A3) (this computes f '(x0))
A4 = A3 – B3/C3 (this applies the algorithm to get the new guess)
If you now copy A4 (not A3!) and B3 and C3 down the A, B and C columns, you have implemented the algorithm.
You can change your starting guess by changing A1, and change your function by changing B3 and C3 appropriately, and copying the results down.
Does this really work?
This method converges very rapidly most of the time. If you start near a 0 of f, and are on "the good side" it will always converge. Otherwise it stands a good chance of doing so, but strange things can happen.
What is the "good side"?
Suppose you start above the solution, call the solution z, so x0 is greater than z. Then if f and the second derivative of f is positive between z and x0, you are on the good side.
Why?
Because the second derivative of f is positive, between z and x0, we know that the first derivative of f is increasing between z and x0, which means that the slope of f is biggest between z and x0 right at x0.
All this means that the linear approximation to f at x0 will dive down to 0 faster than f does, so that x1 will lie somewhere between z and x0. And each successive xj will lie between z and the previous one. As we get closer to z, f will look more and more like a straight line, which will mean it will look more and more like its linear approximation, so you will get closer and closer to z faster and faster.
Exercises:
11.1 Suppose f is negative at x0 which is bigger than z. What condition of f " between z and x0 will mean you are on the good side? What is the condition when f is positive at x0 but x0 is less than z for you to be on the good side as discussed here?
11. 2 What will happen if f " has the wrong sign but the same sign between your guess and z?
Still and all, the method can do bizarre things. If f '= 0 at a guess, the iteration won't even make sense because you will divide by 0 in it. If f ' is very near 0, the new guess will be very far from the old one, and it can zip around weirdly.
The following applet allows you to plot and view the method just by entering the function. (which is only slightly simpler than starting from scratch with a spreadsheet).
Exercises:
11.3 What happens if you look for the solution to , and you try to use this method directly? How about tan x = 1?
11 4 Find all solutions to sin (x) - exp(x) + 2 = 0 for x positive, accurate to ten decimal places.
Do I have to differentiate f to apply this algorithm?
No!. you can choose a value of d that is very small relative to the scale of your function and put =(f(a3+d)-f(a3-d))/(2*d) in C3 instead of =f '(a3). This will do just about as well as the regular Newton's method, practically always.
Exercise 11.5 Redo exercise 4 us this entry in C3. How is your answer affected?
For more thoughts on solving equations look at Chapter 13 of the 18.013A notes. |
# If the equation of the base of an equilateral
Question:
If the equation of the base of an equilateral triangle is x + y = 2 and the vertex is (2, – 1), then find the length of the side of the triangle.
Solution:
Let $\triangle \mathrm{ABC}$ be an equilateral triangle.
Given equation of the base $B C$ is $x+y=2$
We know that, in an equilateral triangle all angles are of $60^{\circ}$
So, in $\triangle \mathrm{ABD}$
$\sin 60^{\circ}=\frac{\mathrm{AD}}{\mathrm{AB}}$\
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{\mathrm{AD}}{\mathrm{AB}}\left[\because \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right]$
$\Rightarrow \mathrm{AD}=\frac{\sqrt{3}}{2} \mathrm{AB}$
We know that, the distance $d$ of a point $P\left(x_{0}, y_{0}\right)$ from the line $A x+B y+C=0$ is given by
$d=\left|\frac{A x_{0}+B y_{0}+C}{\sqrt{A^{2}+B^{2}}}\right|$
Now, length of perpendicular from vertex $A(2,-1)$ to the line $x+y=2$ is
$\mathrm{AD}=\left|\frac{1 \times 2+1 \times(-1)-2}{\sqrt{(1)^{2}+(1)^{2}}}\right|$
$\Rightarrow \frac{\sqrt{3}}{2} \mathrm{AB}=\left|\frac{2-1-2}{\sqrt{2}}\right|$
On simplification we get
$\Rightarrow \frac{\sqrt{3}}{2} \mathrm{AB}=\frac{1}{\sqrt{2}}$
Squaring both the sides, we get
$\Rightarrow \frac{3}{4} \mathrm{AB}^{2}=\frac{1}{2}$
On cross multiplication we get
$\Rightarrow \mathrm{AB}^{2}=\frac{4}{3} \times \frac{1}{2}$
$\Rightarrow \mathrm{AB}^{2}=\frac{2}{3}$
$\Rightarrow A B=\sqrt{\frac{2}{3}}$
Hence, the required length of side is $\sqrt{\frac{2}{3}}$ |
# Separate impact of change of numerator and denominator on ratio
In period 1, a = 15, b = 10, therefore a/b = 1.5
In period 2, a = 12, b= 12, therefore a/b = 1
As you see, a declined by 20%, and b increased by 20%, as well. However, their ratio declined by 50%.
What contributed MORE to the decline of 50% in ratio, increase of 20% in a, or decrease of 20% in b?
How to measure what impacts the ratios most? Change in numerator or change in denominator?
• b increased, not declined – Yuriy S Feb 14 '18 at 8:41
• true. Sorry my mistake – MRMODELLING Feb 14 '18 at 8:41
• Please correct the question and maybe try making it more clear – Yuriy S Feb 14 '18 at 8:47
• Thank you for advice Yuiry. Is it more clear now? – MRMODELLING Feb 14 '18 at 8:51
• Yes, it is a little more clear. I'll see if I can answer in a way that helps – Yuriy S Feb 14 '18 at 8:55
Let's say, we initially had some positive numbers $a$ and $b$ such that:
$$\frac{a}{b}=r$$
Now $a$ and $b$ have changed by a certain percentage, meaning:
$$a_1=a \cdot (1+x)$$
$$b_1=b \cdot (1+y)$$
Where $x$ and $y$ are real numbers (the percentages are given by $100 \cdot x$ and $100 \cdot y$), which could be negative or positive, corresponding to decrease or increase. We can take them to be $\in(-1,1)$ so the change is less than $100$ %.
Then we have:
$$\frac{a_1}{b_1}=\frac{a \cdot (1+x)}{b \cdot (1+y)}= \frac{1+x}{1+y} \cdot r$$
Speaking about percentage again, write the change in $r$ as:
$$r_1=r \cdot (1+z)$$
Where:
$$z=\frac{1+x}{1+y}-1=\frac{x-y}{1+y}$$
Your question is "what contributed more to the change in ratio?" It is not really clear to me what you are asking. But there are ways to see how $x$ and $y$ contribute to the change separately.
Let's consider your case, initially:
$$x=0 \\ y=0$$
And:
$$\Delta x=-0.2 \\ \Delta y=0.2$$
Then, applying the changes separately we have:
$$\Delta_x z=\frac{x+\Delta x-y}{1+y}-\frac{x-y}{1+y}=\frac{0-0.2-0}{1+0}-\frac{0-0}{1+0}=-0.2$$
$$\Delta_y z=\frac{x-y-\Delta y}{1+y+\Delta y}-\frac{x-y}{1+y}=\frac{-0.2}{1+0.2}=-0.16666\dots$$
It seems that the change in $x$ contributed more in this case. But you can try some other cases to see what happens.
Applying the changes at the same time we have:
$$\Delta z=\frac{x+\Delta x-y-\Delta y}{1+y+\Delta y}-\frac{x-y}{1+y}=\frac{-0.2-0.2}{1+0.2}=-\frac{0.4}{1.2}=-\frac{1}{3}=-0.33333$$
The change in ratio was actually $\approx -33.3$%, not $-50$ % as you seem to believe. |
## MATH TEST FOR GRADE 10 WITH MIXED TOPICS
Problem 1 :
Find the indicated term of the sequence if
an = 5 + 2 (n-1) then a7
Solution :
We have nth term or general term of the sequence.
an = 5 + 2 (n-1)
a7 = 5 + 2(7-1)
= 5 + 2(6)
= 5 + 12
a7 = 17
Problem 2 :
Find the equivalent form of (A ∪ B)' =
(A) (A' ∪ B') (B) (A ∪ B') (C) A' ∩ B'
Solution :
(A ∪ B)' = A' ∩ B'
Problem 3 :
Find the sum of the volumes of 15 cubes,whose sides are 1 cm, 2 cm, 3cm ,........ 15 cm respectively.
Solution :
Volume of cubes given above is
13 + 23 + 33 + ......... + 153
Sum of cube = [n(n + 1)/2]2
n = 15
= [15(15 + 1)/2]2
= [15(16)/2]2
= [15(8)]2
= 1202
= 14400
Sum of volume of the above cubes is 14400 cm3
Problem 4 :
It is 9 hours now in a 12 hour clock What was the time 71 hours back ?
Solution :
Step 1 :
Subtract 71 from 9.
That is,
9 - 71 = -62
Step 2 :
We get negative value in step 1 and also 62 is not divisible by 12.
So, find the next integer after 62 which is exactly divisible by 12.
That is 72.
Step 3 :
Write -62 in terms of 72.
So, -62 can be written as
-62 = -72 + 10
Therefore, the time 71 hours back was 10 hours.
Problem 5 :
A vessel is in the form of a hemispherical bowl mounted by a hollow cylinder. The diameter of hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the volume.
Solution :
Height of vessel = 13 cm
radius of hemisphere + height of cylinder = 13
7 + h = 13
h = 13 - 7 = 6 cm
Volume of vessel
= volume of hemisphere + volume of cylinder
= (2/3)πrπrh
= πr2[(2/3) r + h]
= (22/7) 72[(2/3) 7 + 6]
= (22/7) 49 (32/3)
= 1642.67 cm3
Problem 6 :
A conical piece of lead has a radius of 5.25 cm and is 8 cm is height. If it is melted and made into smaller cones 2 cm high and 1.75 cm in diameter, find how many such cones can be made.
Solution :
radius of lead (R) = 5.25 cm and height(H) = 8 cm
Diameter of cone = 1.75, radius (r) = 1.75/2
height of cone(h) = 2 cm
Volume of conical piece of lead = n (Volume of cone)
(1/3)πR2 H = (1/3) πrh
(5.25)⋅ 8 = n (1.75/2)2
n = 220.5/1.53
n = 144
So, the number of small cones is 144.
Problem 7 :
Find the sum and product of roots of the quadratic equation given below.
3x2 + x + 1 = 0
Solution :
Comparing
3x2 + x + 1 = 0
and
ax2 + bx + c = 0
we get
a = 3, b = 1 and c = 1
Therefore,
Sum of the roots = -b/a = -1/3
Product of the roots = c/a = 1/3
Problem 8 :
Find the square root of
(4x2 − 9x + 2) (7x2 - 13x - 2) (28x2 - 3x - 1).
Solution :
= (4x2 − 9x + 2) (7x2 - 13x - 2) (28x2 - 3x - 1)
4x2 − 9x + 2 = (4x - 1)(x - 2)
7x2 - 13x - 2 = (7x + 1)(x - 2)
28x2 - 3x - 1 = (4x - 1) (7x + 1)
= (4x - 1)(x - 2)(7x + 1)(x - 2)(4x - 1) (7x + 1)
= |(4x - 1)(x - 2)(7x + 1)|
The sum of the digits of a two digit number is 10. If the number formed by reversing the digits is less than the original number by 36,find the required number.
(A) 73 (B) 55 (C) 64 (D) 28
Solution :
Let "x y" be the required two digit number. Here "x" is in ten's place and "y" is in unit place.
The sum of the digits of a two digit number = 10
x + y = 10 --------(1)
If the number formed by reversing the digits is less than the original number by 36.
y x = x y - 36
Let us write these as expanded form
10 y + x = 10 x + y - 36
x - 10 x + 10 y - y = - 36
- 9 x + 9 y = - 36
By divide this equation by 9 we will get
- x + y = -4 --------(2)
(1) + (2) x + y = 10
-x + y = -4
_________
2 y = 6 ==> y = 3
By applying y = 3 in (1), we get
x + 3 = 10
x = 7
Therefore the required number is 73.
Problem 10 :
How far is a chord of length 10 cm from the center of a circle of radius 13 cm.
Solution :
In triangle OCB,
OB2 = OC2 + CB2
OC bisects the chord.
132 = OC2 + 52
OC2 = 169 - 25
OC2 = 144
OC = 12 cm
So, the distance between center and the chord is 12 cm.
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# Forum
P and Q are respect...
Clear all
# P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that:
0
Topic starter
P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that:
(i) ar (PRQ) = $$\frac{1}{2}$$ ar (ARC)
(ii) ar (RQC) = $$\frac{3}{8}$$ ar (ABC)
(iii) ar (PBQ) = ar (ARC)
Topic Tags
0
(i) figure.png
We know that, median divides the triangle into two triangles of equal area,
PC is the median of ABC.
Ar (ΔBPC) = ar (ΔAPC) ……….(i)
RC is the median of APC.
Ar (ΔARC) = 1/2 ar (ΔAPC) ……….(ii)
PQ is the median of BPC.
Ar (ΔPQC) = 1/2 ar (ΔBPC) ……….(iii)
From eq. (i) and (iii), we get,
ar (ΔPQC) = 1/2 ar (ΔAPC) ……….(iv)
From eq. (ii) and (iv), we get,
ar (ΔPQC) = ar (ΔARC) ……….(v)
P and Q are the mid-points of AB and BC respectively [given]
PQ||AC
and, PA = 1/2 AC
Since, triangles between same parallel are equal in area, we get,
ar (ΔAPQ) = ar (ΔPQC) ……….(vi)
From eq. (v) and (vi), we obtain,
ar (ΔAPQ) = ar (ΔARC) ……….(vii)
R is the mid-point of AP.
RQ is the median of APQ.
Ar (ΔPRQ) = 1/2 ar (ΔAPQ) ……….(viii)
From (vii) and (viii), we get,
ar (ΔPRQ) = 1/2 ar (ΔARC)
Hence Proved.
(ii) PQ is the median of ΔBPC
ar (ΔPQC) = 1/2 ar (ΔBPC)
= (1/2) × (1/2 )ar (ΔABC)
= ¼ ar (ΔABC) ……….(ix)
Also,
ar (ΔPRC) = 1/2 ar (ΔAPC) [From (iv)]
ar (ΔPRC) = (1/2) × (1/2)ar ( ABC)
= 1/4 ar(ΔABC) ……….(x)
Add eq. (ix) and (x), we get,
ar (ΔPQC) + ar (ΔPRC) = (1/4) × (1/4)ar (ΔABC)
ar (quad. PQCR) = 1/4 ar (ΔABC) ……….(xi)
Subtracting ar (ΔPRQ) from L.H.S and R.H.S,
ar (quad. PQCR)–ar (ΔPRQ) = 1/2 ar (ΔABC) – ar (ΔPRQ)
ar (ΔRQC) = 1/2 ar (ΔABC) – 1/2 ar (ΔARC) [From result (i)]
ar (ΔARC) = 1/2 ar (ΔABC) – (1/2) × (1/2)ar (ΔAPC)
ar (ΔRQC) = 1/2 ar (ΔABC) –(1/4)ar (ΔAPC)
ar (ΔRQC) = 1/2 ar (ΔABC) – (1/4) × (1/2)ar (ΔABC) [ As, PC is median of ΔABC]
ar (ΔRQC) = 1/2 ar (ΔABC) – (1/8)ar (ΔABC)
ar (ΔRQC) = [(1/2) - (1/8)]ar (ΔABC)
ar (ΔRQC) = (3/8) ar (ΔABC)
(iii) ar (ΔPRQ) = 1/2 ar (ΔARC) [From result (i)]
2ar (ΔPRQ) = ar (ΔARC) ……………..(xii)
ar (ΔPRQ) = 1/2 ar (ΔAPQ) [RQ is the median of APQ] ……….(xiii)
But, we know that,
ar (ΔAPQ) = ar (ΔPQC) [From the reason mentioned in eq. (vi)] ……….(xiv)
From eq. (xiii) and (xiv), we get,
ar (ΔPRQ) = 1/2 ar (ΔPQC) ……….(xv)
At the same time,
ar (ΔBPQ) = ar (ΔPQC) [PQ is the median of ΔBPC] ……….(xvi)
From eq. (xv) and (xvi), we get,
ar (ΔPRQ) = 1/2 ar (ΔBPQ) ……….(xvii)
From eq. (xii) and (xvii), we get,
2 × (1/2)ar(ΔBPQ)= ar (ΔARC)
⟹ ar (ΔBPQ) = ar (ΔARC)
Hence Proved.
Share: |
# The_Derivative by nuhman10
VIEWS: 91 PAGES: 11
• pg 1
``` Name _____________________________________________ 2007-2008
Calculus in the AP Physics C Course
The Derivative
Limits and Derivatives
In physics, the ideas of the rate change of a quantity (along with the slope of a tangent line) and the area
under a curve are essential. Limits are fundamental for the definitions of the two major concepts in
calculus that describe these ideas: the derivative and the integral
Our intent here is to introduce these ideas to AP Physics C students who have most likely not yet covered
them in a calculus course. Mathematical rigor has been left for the AP Calculus teacher to cover and a
more intuitive approach is used here.
Limits
Limits are concerned with determining the values of functions based on their behavior near a value x a .
Often students are inclined to think that the value of a function determined from its behavior near x = a is
exactly the same as the value of the function at x a . In teaching the notion of the limit we must make the
distinction between behavior near x a and value at x a .
x2 4
Consider Graph 1: f ( x) 1
x2
If we begin taking x values to evaluate f (x) , we see that as x approaches 2, f (x) approaches 5. Note that
there is no f (2) , so we can't write f (2) 5 .
But as we get closer and closer to x 2 (i.e., x 2 ), f (x) gets closer and closer to the limit 5. We can
write lim f ( x) 5
x 2
1
x3
Consider Graph 2: f ( x ) 4 .5
2 x3
Note that there is no f (3) . This time we need to approach x 3 from both sides.
Approaching x 3 from the left (-): lim f ( x ) 4
x 3
Approaching x 3 from the right (+): lim f ( x) 5
x 3
Again, we can get closer and closer to x 3 , but we cannot compute f (3) directly, and can only
approach the value of f (x) near x 3 by using the concept of the limit.
1
Infinite Limits: Graph 3 f ( x)
x3
Note that the function is asymptotic to x 3 , and there is no f (3) .
Approaching x 3 from the right (+): lim f ( x) ( f (x) gets unboundedly large)
x 3
Approaching x 3 from the left (-): lim f ( x) ( f (x) gets unboundedly large and
x 3
negative)
2
There is no value for lim f ( x ) since the left and right limits don't agree.
x 3
We do not write lim f ( x) .
x 3
4x
Limits as x : Graph 4 f ( x) 4
1
x
2
2
Note that the limit of a function may be a particular value even if f never reaches that value. The
limit must be approached, but not necessarily attained. We have lim f ( x ) 4 , although f (x)
x
never attains 4.
Now for functions which are continuous at a particular point:
Theorem: If f is continuous at x = a, so that its graph does not break, then lim f ( x) f (a )
xa
For example, recall Graph 2: lim f ( x) f (1) 4
x 1
Another example: Consider the continuous function f ( x) x 3 2 x
As x 2 , lim f ( x) lim( x 3 2 x) 2 3 2(2) 4 .
Which simply means f (2) 4
Now, on to the derivative.
The Derivative as the Slope of a Tangent
The derivative of a function represents the rate of change of the function with respect to another
quantity. The derivative also represents the slope of the line tangent to the graph of a function at
a particular point.
Consider motion at a constant speed. The position, x, vs time, t, graph of this motion looks like
this:
3
But what if the motion is accelerated? The speed (and thus slope) would be continually
changing. We can approximate the speed at a particular point P by drawing a line tangent to the
point and calculating its slope.
slope of tangent line = instantaneous speed at point P
But how do we find the slope of a line tangent to a particular point on a curve? Choose a point Q
on the curve near point P and connect the two points with a line called the secant line, as shown
below. The slope of the secant line is the average speed between points P and Q, and is
approximately equal to the instantaneous speed at point P. If we allow point Q to approach point
P, our x' s and t' s will become smaller and smaller (a limiting process as t 0 ) the secant
line will become the tangent line at P, and its slope will represent the instantaneous speed at
point P. The slope of a line tangent to a point P on a curve is called the derivative of the
The Derivative as the Rate of Change of a Function
Suppose that the position of a car on a road at any time t is x f (t ) 12 t t 3 so that at:
t 1 , x f (1) 11 , at
t 2 , x f (2) 16 , at
t 3 , x f (3) 9 , and so on.
What is the speedometer reading v inst at any time t? We know that
4
total change in position
average speed v
total change in time
Consider the period between times t and, t t . The average velocity between these two times is
change in position later position earlier position
v
change in time change in time
But how do we find the instantaneous velocity? Take smaller and smaller time intervals t , that
is, allow t to approach zero and take the limit.
After expanding, we find that
vinst
But, since we are taking the limit as t 0 , we are left with
vinst
The function, 12 3t 2 is the derivative of f with respect to t and is denoted f ' (t ).
x f (t ) 12 t t 3
vinst f ' (t ) 12 3t 2 (rate of change of x with respect to t)
(a) If the derivative (instantaneous velocity) is positive at a certain time, the car is
moving to the right at that time.
(b) If the derivative is negative at a certain time the car is moving to the left at that time.
(c) What does it mean if the derivative at a certain time is zero?
(d) What is this car's speed at t=0?
We have seen that a derivative is the rate of change of a function and the slope of a line tangent
to a point on a curve. Let's generalize the slope of a line tangent to a particular point on the
curve. As done earlier we choose another point Q at the point [ x x, f ( x x) ], and draw the
secant line through P and Q. We now allow the secant line to become the tangent line at P using
the limiting process as x 0 .
f ( x x) f ( x)
As before, in terms of limits, derivative f ' ( x) lim
x
x 0
5
change in y y dy
Then f ' ( x) lim lim
change in x x dx
x 0 x 0
The derivative of y with respect to x can be written several different ways, but usually, we will
dy dy
write derivative of y with respect to x. Keep in mind that is not a quotient (although
dx dx
we can sometimes treat it as one), but is only our notation for a derivative.
Finding the derivative the easy way:
Recall from our previous example that
dx
x f (t ) 12 t t 3 and vinst 12 3t 2
dt
In general, the derivative of x n nx n 1
Example 1.
y 3x 4 5 x 3 2 x 2 x 6
Then, taking the derivative of y with respect to x, we write
Note that the derivative of a constant (in this case, 6) is always zero, since
(a) a constant by definition has no rate of change,
(b) graphically, the slope of a constant function is zero
(c) we could write 6 as 6x 0 , and its derivative would be (0)( 6 x 01 ) 0
Example 2: Consider the position as a function of time: x 8t 3 3t 2 12
(a) At t 0 , what is the position of the object?
(b) At t 2 s , what is the velocity of the object?
(c) When is the velocity of the object zero?
6
(d) Is the object accelerating?
(e) Can we find its acceleration? We know that the slope of a velocity vs. time
graph represents acceleration:
Then the derivative of the velocity function with respect to time must be acceleration.
From above, vinst 24 t 2 6t
Then, acceleration a =
and we can find the acceleration at any time t.
Note that the first derivative of the velocity function is also the second derivative
of the position function. Using derivative notation,
position: x 8t 3 3t 2 12
dx
velocity: v 24 t 2 6t
dy
dv d 2 x
acceleration: a 48t 6
dt dt 2
d 2x
Note that does not indicate a ratio of squares, but only the notation for the second
dt 2
derivative.
1
Example 3: Consider the following kinematic equation: x xo vo t at 2
2
7
AP Physics C
1 2
1983M1. A particle moves along the parabola with equation y x shown above.
2
a. Suppose the particle moves so that the x-component of its velocity has the constant value vx = C; that is, x = Ct
i. On the diagram above, indicate the directions of the particle's velocity vector v and acceleration vector a at
point R, and label each vector.
ii. Determine the y-component of the particle's velocity as a function of x.
iii. Determine the y-component of the particle's acceleration.
8
b. Suppose, instead, that the particle moves along the same parabola with a velocity whose x-component is given
C
by v x
1 x2
i. Show that the particle's speed is constant in this case.
ii. On the diagram below, indicate the directions of the particle's velocity vector v and acceleration
vector a at point S, and label each vector. State the reasons for your choices.
9
Derivative Worksheet
dy
For problems 1-8 find the derivative of y with respect to x (i.e. ). Complete the assignment
dx
on a separate piece of graph paper.
1. y5
2. y x 4
3. y 7 x 8 9 x 4 3x 15
4. y (2 x 3 4 x 2 )(3x 5 x 2 )
2x3 4
5. y
x 2 4x 1
3
6. y
x5
7. y 3 sin x 2 cos x
8. y 5 cos3x
9. For the equation y x 2 4 x 3 find
(a) the equation of the slope of its tangent line at any point.
(b) the equation of the tangent line at point (4,-3) using point-slope form.
10. A particle undergoes straight-line motion with its displacement at any time given by the
following equation, y 2t 3 4t 2 2t 1
(a) Find the times when the particle is motionless.
(b) Find the time when the particle is moving to the right.
(c) Find the time when the particle is moving to the left.
11. The velocity of a particle moving along the x-axis for t 0 is given by v x 24 3t 3 .
(a) What is the particle’s acceleration when it first achieves a velocity of zero?
(b) What is the particle’s acceleration when it achieves its maximum displacement in the
+x-direction?
12. The position of a particle moving along the x-axis is given by x 6t 2 2t 4 .
(a) What is the particle’s velocity at times, t 2 and t 4 ?
(b) What is the particle’s average acceleration from t 2 to t 4 ?
10
Table of Derivatives
d
1. (const.) 0
dx
d
2. ( x) 1
dx
d n
3. ( x ) nx n 1
dx
4.
d
cf ( x) c d f ( x)
dx dx
5.
d
f ( x) g ( x) d f ( x) d g ( x)
dx dx dx
6.
d
f ( x) g ( x) d f ( x) d g ( x)
dx dx dx
7.
d
f ( x) g ( x) f ( x) d g ( x) g ( x) d f ( x) (The Product Rule)
dx dx dx
d f ( x)
g ( x)
d
f ( x) f ( x) d g ( x)
dx dx
dx g ( x)
8. (The Quotient Rule)
g ( x)2
d
9. (sin x) cos x
dx
d
10. (cos x) sin x
dx
d
11. (tan x) sec2 x
dx
d 1
12. (ln x)
dx x
d x
13. e ex
dx
11
```
To top |
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5.
Board CBSE Textbook NCERT Class Class 7 Subject Maths Chapter Chapter 2 Chapter Name Fractions and Decimals Exercise Ex 2.5 Number of Questions Solved 9 Category NCERT Solutions
## NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5
Question 1.
Which is greater?
1. 0.5 or 0.05
2. 0.7 or 0.5
3. 7 or 0.7
4. 1.37 or 1.49
5. 2.03 or 2.30
6. 0.8 or 0.88
Solution:
Question 2.
Express as rupees using decimals:
1. 7 paise
2. 7 rupees 7 paise
3. 77 rupees 77 paise
4. 50 paise
5. 235 paise.
Solution:
1. 7 paise = ₹ 0.07
2. 7 rupees 7 paise = ₹ 7.07
3. 77 rupees 77 paise = ₹ 77.77
4. 50 paise – ₹ 0.50
5. 235 paise = ₹ 2.35
Question 3.
1. Express 5 cm in metre and kilometre
2. Express 35 mm in cm, m and km?
Solution:
1. 5 cm = 0.05 m = 0.00005 km
2. 35 mm = 3.5 cm = 0.035 m = 0.000035 km
Question 4.
Express in kg:
1. 200 g
2. 3470 g
3. 4 kg 8 g
Solution:
1. 200 g = 0.200 kg = 0.2 kg
2. 3470 g = 3.470 kg
3. 4 kg 8 g = 4.008 kg.
Question 5.
Write the following decimal numbers in the expanded form:
1. 20.03
2. 2.03
3. 200.03
4. 2.034
Solution:
Question 6.
Write the place value of 2 in the following decimal numbers:
1. 2.56
2. 21.37
3. 10.25
4. 9.42
5. 63.352
Solution:
(i) Place value of 2 in the decimal number 2.56 = 2 × 1 = 2
(ii) Place value of 2 in decimal number 21.37 = 2 × 10 = 20
(iii) Place value of 2 in the decimal number
Question 7.
Dinesh went from place A to place t B and from there to place C. A is 7.5 km from B and B is 12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. f Who travelled more and by how much?
Solution:
Question 8.
Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?
Solution:
For Shyama
Apples bought = 5 kg 300 g = 5.300 kg
Mangoes bought = 3 kg 250 g = 3.250 kg
∴ Fruits bought = Apples bought
+ Mangoes bought = 5.300 kg + 3.250 kg
For Sarala
Oranges bought = 4 kg 800 g = 4.800 kg
Bananas bought = 4 kg 150 g = 4.150 kg
∴ Fruits bought = Oranges bought
+ Bananas bought
Question 9.
How much less is 28 km than 42.6 km?
Solution:
So, 28 km is less than 42.6 km by 14.6 km.
We hope the NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5, drop a comment below and we will get back to you at the earliest. |
I was looking at some equilateral triangles and started drawing up some equations and I came across the following (it may take some time for my actual question to come):
Suppose that we put a point $D$ anywhere within the triangle. Then we draw lines to each vertex of the triangle, we can now see that we form $3$ triangles, namely $ABD, ACD$ and $BCD$. Further draw lines perpendicular from $D$ to each side of the triangle, the lines $DE, DG$ and $DF$.
We can also note that $AH$ is the height of the triangle (we'll call the height $h$). For simplicity we will also call the sides $AC = AB = BC = S$. We will also rename the perpendicular lines from $D$ to each side; $L_1, L_2$ and $L_3$ respectively (for our purposes it doesn't matter which line is which $L_n$).
We can easily see that everything done so far is valid no matter where we put the point $D$ because we can always the draw the lines.
Now the definitions are done, so we can start looking at the area of $ABC$. This is just $\frac{bh}{2} = \frac{Sh}{2}$. The area can also be derived from adding together the area of triangles $ABD, ACD$ and $BCD$. The area for each small triangle is just $\frac{SL_n}{2}$.
Now set the two equations for the areas to equal each other: $$\frac{Sh}{2} = \frac{SL_1}{2} + \frac{SL_2}{2} + \frac{SL_3}{2}$$ $$h =L_1 + L_2 + L_3$$
Which is quite interesting. No matter where we put a point $D$, the perpendicular lengths from $D$ to the sides always sum up to the height of the triangle.
My next step was to extend this to all regular polygons, which led me to derive the following:
$$\frac{2A_n}{S} = \sum_{i=1}^{n} L_i$$
Where $A_n$ is the area of some regular polygon and the subscript $n$ denotes how many sides it has.
I was wondering if anyone has some material about this? Where it's e.g. extended to other shapes/higher dimensions etc. Thank you for your help. |
# Proving A Theorem Concerned With Prime Numbers
I am in the process of reading this brilliant little book Prove It: A Structured Approach--very brilliant, have I mention that already?
Anyways, here is the theorem: For every positive integer $n$,there is a sequence of $n$ consecutive positive integers containing no primes.
I understand the prove, for the most part; but there is a little bit giving me difficulty.
Here is the entire proof:
Suppose $n$ is a positive integer. Let $x=(n+1)!1+2$.We will show that none of the numbers $x,x+1,x+2,\ldots,x+(n - 1)$ is prime. Since this is a sequence of n consecutive positive integers, this will prove the theorem. To see that $x$ is not prime, note that $x=1\cdot2\cdot3\cdot4\cdots(n+1)+2\rightarrow x = 2\cdot(1\cdot3\cdot4\cdots(n+1)+1)$ Thus, $x$ can be written as a product of two smaller positive integers, so $x$ is not prime. Similarly, we have $x+1=1\cdot2\cdot3\cdot4\cdots(n+1)+3\rightarrow x=3\cdot(1\cdot2\cdot4\cdots(n+l) +1)$, so $x+1$ is also not prime. In general, consider any number $x+i$, where $0\leq i \leq n-1$. Then we have
$$x+i = 1\cdot2\cdot3\cdot4\cdots(n+1)+(i +2)$$ $$x+i= (i+2)\cdot(1\cdot2\cdot3\cdots(i+1)\cdot(i+3)\cdots(n+1)+1)$$
The last step is the one I can't quite apprehend. I see that $(i+2)$ is factored out, but I why are there remaining $i$'s in the second factor?
-
For example, if $i=5$, $i+2=7$, $$x + 5 = 7 \cdot (1\cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 8 \cdot \ldots \cdot (n+1)) + 1)$$ The $6$ is $i+1$ and the $8$ is $i+3$. – Robert Israel Nov 21 '12 at 21:30
Robert provided a nice response. If you are interested in other proof like books, read my response here: math.stackexchange.com/questions/187340/… – Amzoti Nov 21 '12 at 21:45
They're just pointing out that $(i+2)$ is skipped in the product. For instance you can write $n! = 1 \cdot 2 \cdots n$, but if you want to remove say $k < n$ from that product you can do $1 \cdot 2 \cdots (k-1) \cdot (k+1) \cdots (n-1) \cdot n$. – roliu Nov 21 '12 at 22:50
The book may be very good. But it has obscured a nice idea. – André Nicolas Nov 22 '12 at 2:09
@AndréNicolas Really? Would you mind pointing out this obscured idea, please? – Mack Nov 22 '12 at 13:34
## 2 Answers
$i+2$ is factored out and that's the reason why there is written the rpoduct of all numbes from $1$ to $i+1$, inclusive, followed by the product of all numbers from $i+3$ to $n+1$ inclusive. Actually, I find it more natuurally to obsere that $x+i$ is $i+2$ more than $(n+1)!$ and the factorial is of course a nontrivial multiple of $i+2$.
-
Hint $\$ More clearly, let $\rm\:m = (n\!+\!1)! > 0.\$ Then
$\qquad\quad\rm 2\mid m\:\Rightarrow\:2\mid m\!+\!2\ne 2\:\Rightarrow\:m\!+\!2\:$ not prime
$\qquad\quad\rm 3\mid m\:\Rightarrow\:3\mid m\!+\!3\ne 3\:\Rightarrow\:m\!+\!3\:$ not prime
$\qquad\quad \ \ldots\qquad\qquad\ \ldots\qquad\qquad\qquad\ldots$
$\quad\ \ \rm n\!+\!1\mid m\:\Rightarrow\:\quad\quad\,\ldots\quad\quad\,\Rightarrow\:m\!+\!n\!+\!1\:$ not prime
So $\rm\: m\!+\!2,\,m\!+\!3,\ldots,m\!+\!n\!+\!1\:$ are $\rm\:n\:$ consecutive nonprimes.
- |
# How do solve 3/(x-2)<=3/(x+3) algebraically?
Oct 5, 2016
$- 3 < x < 2$
#### Explanation:
Slower way:
• Bring everything to the right: $\frac{3}{x - 2} - \frac{3}{x + 3} \setminus \le q 0$
• Lowest common denominator: $\frac{3 \left(x + 3\right) - 3 \left(x - 2\right)}{\left(x - 2\right) \left(x + 3\right)} \setminus \le q 0$
• Expand the numerator: $\frac{3 x + 9 - 3 x + 6}{\left(x - 2\right) \left(x + 3\right)} \setminus \le q 0$
• $\frac{9 + 6}{\left(x - 2\right) \left(x + 3\right)} \setminus \le q 0$
• $\frac{15}{\left(x - 2\right) \left(x + 3\right)} \setminus \le q 0$
• Since $15$ is always positive, the sign of the fraction is decided by the sign of the denominator. $\left(x - 2\right) \left(x + 3\right)$ represents a parabola with zeros in $x = - 3$ and $x = 2$. Such a parabola is negative between its solutions, as you can see in the graph here:
graph{(x-3)(x+2) [-4.96, 6.146, -2.933, 2.614]} |
# Proportion Problems
We will learn how to solve proportion problems. We know, the first term (1st) and the fourth term (4th) of a proportion are called extreme terms or extremes, and the second term (2nd) and the third term (3rd) are called middle terms or means.
Therefore, in a proportion, product of extremes = product of middle terms.
Solved examples:
1. Check whether the two ratios form a proportion or not:
(i) 6 : 8 and 12 : 16; (ii) 24 : 28 and 36 : 48
Solution:
(i) 6 : 8 and 12 : 16
6 : 8 = 6/8 = 3/4
12 : 16 = 12/16 = 3/4
Thus, the ratios 6 : 8 and 12 : 16 are equal.
Therefore, they form a proportion.
(ii) 24 : 28 and 36 : 48
24 : 28 = 24/28 = 6/7
36 : 48 = 36/48 = 3/4
Thus, the ratios 24 : 28 and 36 : 48 are unequal.
Therefore, they do not form a proportion.
2. Fill in the box in the following so that the four numbers are in proportion.
5, 6, 20, ____
Solution:
5 : 6 = 5/6
20 : ____ = 20/____
Since the ratios form a proportion.
Therefore, 5/6 = 20/____
To get 20 in the numerator, we have to multiply 5 by 4. So, we also multiply the denominator of 5/6, i.e. 6 by 4
Thus, 5/6 = 20/6 × 4 = 20/24
Hence, the required numbers is 24
3. The first, third and fourth terms of a proportion are 12, 8 and 14 respectively. Find the second term.
Solution:
Let the second term be x.
Therefore, 12, x, 8 and 14 are in proportion i.e., 12 : x = 8 : 14
⇒ x × 8 = 12 × 14, [Since, the product of the means = the product of the extremes]
⇒ x = (12 × 14)/8
⇒ x = 21
Therefore, the second term to the proportion is 21.
`
More worked-out proportion problems:
4. In a sports meet, groups of boys and girls are to be formed. Each group consists of 4 boys and 6 girls. How many boys are required, if 102 girls are available for such groupings?
Solution:
Ratio between boys and girls in a group = 4 : 6 = 4/6 = 2/3 = 2 : 3
Let the number of boys required = x
Ratio between boys and girls = x : 102
So, we have, 2 : 3 = x : 102
Now, product of extremes = 2 × 102 = 204
Product of means = 3 × x
We know that in a proportion product of extremes = product of means
i.e., 204 = 3 × x
If we multiply 3 by 68, we get 204 i.e., 3 × 68 = 204
Thus, x = 68
Hence, 68 boys are required.
5. If a : b = 4 : 5 and b : c = 6 : 7; find a : c.
Solution:
a : b = 4 : 5
⇒ a/b = 4/5
b : c = 6 : 7
⇒ b/c = 6/7
Therefore, a/b × b/c = 4/5 × 6/7
⇒ a/c = 24/35
Therefore, a : c = 24 : 35
6. If a : b = 4 : 5 and b : c = 6 : 7; find a : b : c.
Solution:
We know that of both the terms of a ratio are multiplied by the same number; the ratio remains the same.
So, multiply each ratio by such a number that the value of b (the common term in both the ratios) acquires the same value.
Therefore, a : b = 4 : 5 = 24 : 30, [Multiplying both the terms by 6]
And, b : c = 6 : 7 = 30 : 35, [Multiplying both the terms by 5]
Clearly,; a : b : c = 24 : 30 : 35
Therefore, a : b : c = 24 : 30 : 35
From, the above solved proportion problems we get the clear concept how to find whether the two ratios form a proportion or not and word problems. |
# 5th Grade Common Core Resources
Here you will find all fifth grade resources to guide and support mathematics teaching and learning.
These resources are organized by mathematical strand and refer to specific Common Core math content standards.
Quick reference: What are these tools?
## Operations and Algebraic Thinking (OA)
• 5.OA.1
Use parentheses, brackets, or braces in numerical expressions, and evaluate expressions with these symbols.
Problems of the Month
• 5.OA.2
Write simple expressions that record calculations with numbers, and interpret numerical expressions without evaluating them.
Problems of the Month
• 5.OA.3
Generate two numerical patterns using two given rules. Identify apparent relationships between corresponding terms. Form ordered pairs consisting of corresponding terms from the two patterns, and graph the ordered pairs on a coordinate plane.
Problems of the Month
## Number and Operations in Base Ten (NBT)
• 5.NBT.3
Read, write, and compare decimals to thousandths.
• 5.NBT.4
Use place value understanding to round decimals to any place value.
• 5.NBT.5
Fluently multiply multi-digit whole numbers using the standard algorithm.
Problems of the Month
• 5.NBT.6
Find whole-number quotients of whole numbers with up to four-digit dividends and two-digit divisors, using strategies based on place value, the properties of operations and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
Problems of the Month
• 5.NBT.7
Add, subtract, multiply, and divide decimals to hundredths, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used.
Problems of the Month
## Number and Operations - Fractions (NF)
• 5.NF.1
Add and subtract fractions with unlike denominators (including mixed numbers) by replacing given fractions with equivalent fractions in such a way as to produce an equivalent sum or difference of fractions with like denominators.
Problems of the Month
• 5.NF.2
Solve word problems involving addition and subtraction of fractions referring to the same whole including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem.
Problems of the Month
• 5.NF.3
Interpret a fraction as division of the numerator by the denominator (a/b = a ? b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem.
Problems of the Month
• 5.NF.4
Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction.
Problems of the Month
• 5.NF.5
Interpret multiplication as scaling (resizing)…
Problems of the Month
• 5.NF.6
Solve real world problems involving multiplication of fractions and mixed numbers, e.g., by using visual fraction models or equations to represent the problem.
Problems of the Month
• 5.NF.7
Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions.
Problems of the Month
## Measurement and Data (MD)
• 5.MD.1
Convert among different-sized standard measurement units within a given measurement system (e.g. convert 5 cm to 0.05 m) and use these conversions in solving multi-step, real world problems.
• 5.MD.2
Make a line plot to display a data set of measurements in fractions of a unit (1/2, 1/4, 1/8). Use operations on fractions for this grade to solve problems involving information presented in line plots.
Problems of the Month
• 5.MD.3
Recognize volume as an attribute of solid figures and understand concepts of volume and measurement.
Problems of the Month
• 5.MD.4
Measure volumes by counting unit cubes, using cubic cm, cubic in., cubic ft., and improvised units.
Problems of the Month
• 5.MD.5
Relate volume to the operations of multiplication and addition and solve real world and mathematical problems involving volume.
Problems of the Month
## Geometry (G)
• 5.G.1
Use a pair of perpendicular number lines, called axes, to define a coordinate system, with the intersection of the lines (the origin) arranged to coincide with the 0 on each line and a given point in the plane located by using an ordered pair of numbers, called its coordinates.
• 5.G.2
Represent real world and mathematical problems by graphing points in the first quadrant of the coordinate plane and interpret coordinate values of points in the context of the situation. |
# How do you differentiate g(x) = (1/(x^3-1))*sqrt(1+e^(x)) using the product rule?
Jan 13, 2016
$- \frac{3 {x}^{2}}{{x}^{3} - 1} ^ 2 \cdot \sqrt{1 + {e}^{x}} + \frac{1}{{x}^{3} - 1} \cdot {e}^{x} / \left(2 \sqrt{1 + {e}^{x}}\right)$
#### Explanation:
The product rule states: if
$g \left(x\right) = f \left(x\right) \cdot h \left(x\right)$,
the derivative of $g \left(x\right)$ can be computed as follows:
$g ' \left(x\right) = f ' \left(x\right) \cdot h \left(x\right) + f \left(x\right) \cdot h ' \left(x\right)$
$f \left(x\right) = \frac{1}{{x}^{3} - 1} \text{ }$ and
$h \left(x\right) = \sqrt{1 + {e}^{x}}$
The first thing you need to do is differentiate $f \left(x\right)$ and $h \left(x\right)$.
=====================
Let's start with $f \left(x\right)$. Here, the chain rule is helpful. You can write use it as follows:
$f \left(x\right) = \frac{1}{u}$ where $u = {x}^{3} - 1$
According to the chain rule, the derivative is the derivative of $\frac{1}{u}$ multiplied with the derivative of $u$.
$\left[\frac{1}{u}\right] ' = \left[{u}^{- 1}\right] ' = - {u}^{- 2} = - \frac{1}{u} ^ 2 = - \frac{1}{{x}^{3} - 1} ^ 2$
$\left[{x}^{3} - 1\right] ' = 3 {x}^{2}$
Thus,
$f ' \left(x\right) = - \frac{1}{{x}^{3} - 1} ^ 2 \cdot 3 {x}^{2} = - \frac{3 {x}^{2}}{{x}^{3} - 1} ^ 2$
=====================
Now, you still need to differentiate $h \left(x\right)$. The chain rule can help here as well:
$h \left(x\right) = \sqrt{v}$ where $v = 1 + {e}^{x}$
The derivative of $h \left(x\right)$ is the derivative of $\sqrt{v}$ multiplied with the derivative of $v$.
$\left[\sqrt{v}\right] ' = \left[{v}^{\frac{1}{2}}\right] ' = \frac{1}{2} {v}^{- \frac{1}{2}} = \frac{1}{2 \sqrt{v}} = \frac{1}{2 \sqrt{1 + {e}^{x}}}$
$\left[1 + {e}^{x}\right] ' = {e}^{x}$
This means that the derivative of $h \left(x\right)$ is:
$h ' \left(x\right) = \frac{1}{2 \sqrt{1 + {e}^{x}}} \cdot {e}^{x} = {e}^{x} / \left(2 \sqrt{1 + {e}^{x}}\right)$
=====================
Now, the only thing left to do is applying the product rule:
$g ' \left(x\right) = f ' \left(x\right) \cdot h \left(x\right) + f \left(x\right) \cdot h ' \left(x\right)$
$\textcolor{w h i t e}{\times \xi i i} = - \frac{3 {x}^{2}}{{x}^{3} - 1} ^ 2 \cdot \sqrt{1 + {e}^{x}} + \frac{1}{{x}^{3} - 1} \cdot {e}^{x} / \left(2 \sqrt{1 + {e}^{x}}\right)$
$\textcolor{w h i t e}{\times \xi i i} = - \frac{3 {x}^{2} \sqrt{1 + {e}^{x}}}{{x}^{3} - 1} ^ 2 + {e}^{x} / \left(2 \sqrt{1 + {e}^{x}} \left({x}^{3} - 1\right)\right)$ |
# NCERT Solutions for Class 6 Maths Chapter 5 - Understanding Elementary Shapes
Check NCERT solutions for class 6 Maths chapter5 available in downloadable format. Get latest and free solutions for all questions givenin this chapter.
Created On: Jul 16, 2019 18:08 IST
NCERT Solutions for Class 6 Maths Chapter 5 - Understanding Elementary Shapes
NCERT solutions for class 6 Maths chapter 5 will prove to be very helpful to understand the concepts and excel in the Mathematics subject. Here you will find the latest updated solutions to prepare for the current academic session 2019-2020.
Exercise 5.1
1. What is the disadvantage in comparing line segment by mere observation?
Solution.
Comparing the lengths of two line segments by mere observation cannot tell us accurately that which line is shorter or which is longer.
2.Why is it better to use a divider than a ruler, while measuring the length of a line segment?
Solution.
While using a ruler, due to the incorrect position of eye, there are chances of error in the reading whereas, with ruler this error can be minimised and we can get an accurate result.
[Note: If A, B, C are any three points on a line such AC + CB = AB, then we can be sure that C lies between A and B]
Solution.
Here, AB = 5 cm and a point C lies between A and B such that
AC = 2 cm, CB = 3cm.
∴ AC + CB = 2 cm + 3 cm = 5 cm.
But, AB = 5 cm.
So, AB = AC + CB.
4.If A, B, C are three points on a line such that AB = 5 cm, BC = 3 cm and AC = 8 cm, which one of them lies between the other two?
Solution.
Given, AB = 5 cm
BC = 3 cm
AC = 8 cm
Now AB + BC = 5 + 3 = 8 cm which is equal to AC.
Hence, B lies between A and C.
Solution.
From the given figure, we have
AG = 7 – 1 = 6 units
AD = 4– 1 = 3 units
And DG = 7– 4 = 3 units
As D is the point lying on AG such that AD = DG
Hence, D is the mid-point of AG.
Since, B is the mid-point of AC
Therefore, AB = BC …(i)
Again C is the mid-point of BD
Therefore, BC = CD …(ii)
From equations (i) and (ii), we have
AB = CD
7.Draw five triangles and measure their sides. Check in each case, if the sum of the length of any two sides is always less than the third side.
Solution.
Students may draw five triangles of different measures. One has been drawn below for reference:
To check: The sum of the length of any two sides of ∆ABC is less than its third side.
Case 1.Here, In△ABC
AB = 2.5 cm
AC = 5.5 cm
BC = 6 cm
AB + AC = 2.5 cm + 5.5 cm
= 8 cm
Since, 8 cm>6 cm
So, AB + BC > AC
Hence, sum of any two sides of a triangle is always greater than the third side.
Students may calculate the same for their other triangles and verify the statement.
Exercise 5.2
1. What fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from
(a) 3 to 9
(b) 4 to 7
(c) 7 to 10
(d) 12 to 9
(e) 1 to 10
(f) 6 to 3
Solution.
(a) 3 to 9
(b) 4 to 7
(c) 7 to 10
(d) 12 to 9
(e) 1 to 10
(f) 6 to 3
2.Where will the hand of a clock stop if it
Solution.
3.Which direction will you face if you start facing
(Should we specify clockwise or anticlockwise for this last question? Why not?)
Solution.
(a) When we start facing east and make 1/2 of a revolution (180o) clockwise, we will face west direction.
(c)When we start facing west and make 3/4 of a revolution (270 o = 180 o + 90o) anticlockwise, we will first reach the east after 180 o turn and then finally north after taking another 90o turn.
(d) When we start facing south and make one full revolution in either clockwise or anticlockwise direction, we will reach back to the starting position that is south.
4.What part of a revolution have you turned through if you stand facing
(a) east and turn clockwise to face north?
(b) south and turn clockwise to face east?
(c) west and turn clockwise to face east?
Solution.
Remember: A complete revolution is of 360o and two adjacent directions are inclined at 90O (1/4 of revolution). Two opposite directions form an angle 180o which is 1/2 of a revolution.
(a) If we start from east and turn clockwise to reach north we have turned through 1/2 of a revolution first to reach west and then a 1/4 revolution to reach north.
(b) If we start from south and turn clockwise to face east, we will first reach north by turning through 1/2 of a revolution and then move by 1/4 of a revolution to reach east.
(c) If we start from west and turn clockwise to face east, we will turn through 1/2 of a revolution.
5.Find the number of right angles turned through by the hour hand of a clock when it goes from
(a)3 to 6
(b) 2 to 8
(c) 5 to 11
(d) 10 to 1
(e) 12 to 9
(f) 12 to 6
Solution.
(a) 3 to 6
Here, the hour hand turns through 1 right angle.
(b) 2 to 8
Here, the hour hand turns through 2 right angles.
(c) 5 to 11
Here, the hour hand turns through 2 right angles.
(d) 10 to 1
Here, the hour hand turns through 1 right angle.
(e) 12 to 9
Here, the hour hand turns through 3 right angles.
(f) 12 to 6
Here, the hour hand turns through 2 right angles.
6.How many right angles do you make if you start facing
(a) south and turn clockwise to west?
(b) north and turn anticlockwise to east?
(c) west and turn to west?
(d) south and turn to north?
Solution.
(a) If I start facing south and turn clockwise to west, I will make 1 right angle.
(b) If I start facing north and turn anticlockwise to east, I will make 3 right angles.
(c) If I start facing west and turn to west, I will make 4 right angles in both clockwise and anticlockwise directions.
(d) If I start facing south and turn to north, I will make 2 right angles in both clockwise and anticlockwise directions.
7.Where will the hour hand of a clock stop if it starts
(a) from 6 and turns through 1 right angle?
(b) from 8 and turns through 2 right angles?
(c) from 10 and turns through 3 right angles?
(d) from 7 and turns through 2 straight angles?
Solution.
Hint: Students may use the trick that every 15 minutes round of the hour hand is equal to 1 right angle turn.
(a) The hour hand will stop at 9.
(b) The hour hand will stop at 2.
(c) The hour hand will stop at 7.
(d) The hour hand will stop at 7.
Exercise 5.3
1.Match the following:
(i) Straight angle (ii) Right angle (iii) Acute angle (iv) Obtuse angle (v) Reflex angle (a) Less than one-fourth of a revolution (b) More than half a revolution (c) Half of a revolution (d) One-fourth of a revolution (e) Between1/4 and 1/2 of a revolution (f) One complete revolution.
Solution.
Correct match of the two given columns is given below:
(i) Straight angle (= 180o) (c) Half of a revolution (= 180o) (ii) Right angle (= 90o) (d) One-fourth of a revolution (= 90o) (iii) Acute angle (less than 90o) (a) Less than one-fourth of a revolution (less than 90o) (iv) Obtuse angle (greater than 90o and less than180o) (e) Between 1/4 and 1/2 of a revolution (Between 90o and 180o) (v) Reflex angle (greater than 180o and less than360o) (b) More than half a revolution
2.Classify each one of the following angles as right, straight, acute, obtuse or reflex:
Solution.
(a) Acute angle (less than 90o)
(b) Obtuse angle(greater than 90o and less than180o)
(c) Right angle (equal to 90o)
(d) Reflex angle (greater than 180o and less than360o)
(e) Straight angle (equal to 180o)
(f) Acute angle (less than 90o)
Exercise 5.4
1.What is the measure of (i) a right angle (ii) a straight angle?
Solution.
(i) Measure of a right angle = 90°
(ii) Measure of a straight angle =180°
2.Say True or False:
(a) The measure of an acute angle < 90°
(b) The measure of an obtuse angle < 90°
(c) The measure of a reflex angle > 180°
(d) The measure of one complete revolution = 360°
(e) If m ∠A = 53° and ∠B = 35°, then m∠A>m∠B.
Solution.
(a) True
(b) False
(c) True
(d) True
(e) True
3.Write down the measures of
(a) some acute angles
(b) some obtuse angles
(give at least two examples of each).
Solution.
(a) Acute angles: 35°, 80°
(b) Obtuse angles: 100°, 170°
4.Measure the angles given below using the protractor and write down the measure.
Solution.
Let’s name the given angles as follows:
(a) Measure of angle P = 45°
(b) Measure of ∠Q= 120°
(c) Measure of ∠R= 90°
(d)Measure of ∠S = 60°,
Measure of ∠T = 90° and
Measure of ∠U = 125°
5.Which angle has a large measure? First estimate and then measure.
Measure of Angle A =
Measure of Angle B =
Solution.
Angle B is wider than angle A so it must have a larger measure than angle A.
Measure of angle A = 40°
Measure of angle B = 60°
Thus, ∠B >∠A.
6.From these two angles which has large measure? Estimate and then confirm by measuring them.
Solution.
Given angle are:
∠B is wider than ∠A hence it has a larger measure than ∠A.
Measure of angle A = 45°
Measure of angle B = 60°
Thus, ∠B >∠A.
7.Fill in the blanks with acute, obtuse, right or straight:
(a) An angle whose measure is less than that of a right angle is ……… .
(b) An angle whose measure is greater than that of a right angle is ……… .
(c) An angle whose measure is the sum of the measures of two right angles is ……… .
(d) When the sum of the measures of two angles is that of a right angle, then each one of them is ……… .
(e) When the sum of the measures of two angles is that of a straight angle and if one of them is acute then the other should be ……… .
Solution.
(a) acute
(b) obtuse
(c) straight (sum of the two right angles = 180o)
(d) acute (as each angle is less than 90o)
(e) obtuse
8.Find the measure of the angle shown in each figure. (First estimate with your eyes and then find the actual measure with a protractor).
Solution.
Given angles are:
• Measure of ∠a = 40°
• Measure of ∠b = 130°
• Measure of ∠c = 65°
• Measure of ∠d = 135°
9.Find the angle measure between the hands of the clock in each figure:
Solution.
(i) Angle between hands of the clock in first figure = 90°
(ii) Angle between hands of the clock in second figure = 30°
(iii) Angle between hands of the clock in third figure = 180°.
10.Investigate: In the given figure, the angle measures 30°.
Look at the same figure through a magnifying glass. Does the angle become larger? Does the size of the angle change?
Solution.
The measure of the angle remains the same even when observed through magnifying glass.
11.Measure and classify each angle:
Angle Measure Type ∠AOB ∠AOC ∠BOC ∠DOC ∠DOA ∠DOB
Solution.
Given angles are classified as follows:
Angle Measure Type ∠AOB 40° Acute angle ∠AOC 125° Obtuse angle ∠BOC 85° Acute angle ∠DOC 95° Obtuse angle ∠DOA 140° Obtuse angle ∠DOB 180° Straight angle
Exercise 5.5
1.Which of the following are models for perpendicular lines:
(a) The adjacent edges of a table top.
(b) The lines of a railway track.
(c) The line segments forming a letter ‘L’.
(d) The letter V.
Solution.
(a) Yes, the adjacent edges of a table top are a model of perpendicular lines.
(b) No, the lines of a railway tracks are not a model of perpendicular lines as they are parallel to each other.
(c) Yes, the line segments forming a letter ‘L’ are a model of perpendicular lines.
(d) No, the two line segments forming a letter ‘V’ are not a model of perpendicular lines.
Since PQ⊥ XY
Therefore, ∠PAY = 90°
3.There are two set-squares in your box. What are the measures of the angles that are formed at their corners? Do they have any angle measure that is common?
Solution.
Figures of two set-squares in the box are given below:
Measure angles of small set-square: 45°, 45° and 90°.
Measure angles of larger set-square: 30°, 60° and 90°.
Yes, the angle of measure 90° is common between both the square-sets.
4.Study the diagram. The line l is perpendicular to line m.
(a) Is CE = EG?
(b) Does PE bisects CG?
(c) Identify any two line segments for which PE is the perpendicular bisector.
(d) Are these true?
(i) AC > FG
(ii) CD = GH
(iii) BC < EH
Solution.
(a) Yes,CE = EG = 2 units
(b) Yes, PE bisects CG as CE = EG
(d) (i) True
Since AC = 2 units and FG = 1 unit
Therefore, AC > FG
(ii) True; CD = GH = 1 unit
(iii) True
Since BC = 1 unit and EH = 3 units
Therefore, BC < EH
Exercise 5.6
1. Name the types of following triangles:
(а) Triangle with lengths of sides 7 cm, 8 cm and 9 cm.
(b) ∆ABC with AB = 8.7 cm, AC = 7 cm and BC = 6 cm.
(c) ∆PQR such that PQ = QR = PR = 5 cm.
(d) ∆DEF with m∠D = 90°
(e) ∆XYZ with m∠Y = 90° and XY = YZ.
(f) ∆LMN with m∠L = 30° m∠M = 70° and m∠N = 80°.
Solution.
(a) Since, the measures of all three sides of triangle are different
Hence, it is a scalene triangle.
(b) Since, the measures of all three sides of ∆ABC are different
Hence, it is a scalene triangle.
(c) Since all sides of ∆PQR are equal
Hence, it is an equilateral triangle.
(d) Since measure of one angle of ∆DEF is 90°
Hence it is a right angled triangle.
(e) Since measure of one angle of ∆XYZ is 90° and its two sides are equal
Hence it is anisosceles right angledtriangle.
(f) Since all three angles of ∆LMN are less than 90°
Hence it is an acute angled triangle.
2. Match the following:
Measure of triangle Type of triangle (i) 3 sides of equal length (a) Scalene (ii) 2 sides of equal length (b) Isosceles right angled (iii) All sides are of different length (c) Obtuse angled (iv) 3 acute angles (d) Right angled (v) 1 right angle (e) Equilateral (vi) 1 obtuse angle (f) Acute angled (vii) 1 right angle with two sides of equal length (g) Isosceles
Solution.
Measure of triangle Type of triangle (i) 3 sides of equal length (e) Equilateral (ii) 2 sides of equal length (g) Isosceles (iii) All sides are of different length (a) Scalene (iv) 3 acute angles (f) Acute angled (v) 1 right angle (d) Right angled (vi) 1 obtuse angle (c) Obtuse angled (vii) 1 right angle with two sides of equal length (b) Isosceles right angled
3. Name each of the following triangles in two different ways: (You may judge the nature of the angle by observation)
Solution.
(a) (i) Acute angled triangle
(ii) Isosceles triangle
(b) (i) Right angled triangle
(ii) Scalene triangle
(c) (i) Obtuse angled triangle
(ii) Isosceles triangle
(d) (i) Right angled triangle
(ii) Isosceles triangle
(e) (i) Acute angled triangle
(ii) Equilateral triangle
(f) (i) Obtuse angled triangle
(ii) Scalene triangle.
4. Try to construct triangles using matchsticks. Some are shown here.
Can you make a triangle with
(a) 3 matchsticks?
(b) 4 matchsticks?
(c) 5 matchsticks?
(d) 6 matchsticks?
(Remember you have to use all the available matchsticks in each case)
Name the type of triangle in each case.
If you cannot make a triangle, give of reasons for it.
Solution.
(a) Yes, we can make a triangle with 3 matchsticks as shown below:
Since, 1 matchstick forms each side. So, all the sides of this triangle are equal.
Hence, it is an equilateral triangle.
(b) No, we cannot make a triangle with 4 matchsticks.
Explanation: If we make a triangle with 4 matchsticks and each matchstick measures 1 unit then:
Larger side of triangle measures 2 units
Smaller sides of triangle measures 1 unit each.
Then, 1st side + 2nd side = 1+1 = 2 = 3rd side
But we know that sum of any two sides of a triangle is always greater than its third side.
Hence proved.
(c) Yes, we can make a triangle with 5 matchsticks as shown below:
Since, two sides of this triangle are formed by 2 equal sized matchsticks
Hence, it is an isosceles triangle.
(d) Yes, we can make a triangle with 6 matchsticks as shown below:
Since three match sticks form three sides of this triangle. So, all the sides of this triangle are equal.
Hence, it is an equilateral triangle.
Exercise 5.7
1. Say True or False:
(a) Each angle of a rectangle is a right angle.
(b) The opposite sides of a rectangle are equal in length.
(c) The diagonals of a square are perpendicular to one another.
(d) All the sides of a rhombus are of equal length.
(e) All the sides of a parallelogram are of equal length.
(f) The opposite sides of a trapezium are parallel.
Solution.
(a) True
(b) True
(c) True
(d) True
(e) False; only opposite sides of a parallelogram are equal in length.
(f) False; only one pair of opposite sides of a trapezium are parallel.
2. Give reasons for the following:
(a) A square can be thought of as a special rectangle.
(b) A rectangle can be thought of as a special parallelogram.
(c) A square can be thought of as a special rhombus.
(d) Square, rectangles, parallelograms are all quadrilaterals.
(e) Square is also a parallelogram.
Solution.
(a) (a) In a rectangle:
• All interior angles are equal to 90o.
• Opposite sides are equal in length.
In a square:
• All interior angles are equal to 90o.
• All sides are equal in length.
Thus, a rectangle with all sides equal becomes a square. So, square is a special rectangle.
(b) In a parallelogram:
• Opposite sides are equal
• Opposite sides are parallel
In a rectangle:
• Opposite sides are equal
• Opposite sides are parallel
• All angles are equal to 90o
A parallelogram with all its angles equal to 90obecomes a rectangle.
Hence, a rectangle can be thought of as a special parallelogram.
(c) In a square:
• All sides are equal
• Opposite sides are parallel
• All angles are equal to 90o
In a rhombus:
• All sides are equal
• Opposite sides are parallel
Thus, a rhombus with all its angles equal to 90o becomes a square. So, a square can be thought of as a special rhombus
(d) A quadrilateral is a polygon which has four sides.
Since, squares, rectangles and parallelogram all are made of 4 sides so they are all quadrilaterals.
(e) In a square opposite sides are equal and parallel to each other.
Also, in a parallelogram, opposite sides are equal and parallel to each other.
Hence, a square is also a parallelogram.
3. A figure is said to be regular if its sides are equal in length and angles are equal in measure. Can you identify the regular quadrilateral?
Solution.
Since, square is the only quadrilateral with all its sides of same length and angles of same measure (90o).
Hence, square is a regular quadrilateral.
Exercise 5.8
1. Examine whether the following are polygons. If anyone among them is not, say why?
Solution.
A polygon is a closed plane figure enclosed with more than two line segments.
(a) This figure is not closed so, it is not a polygon.
(b) This figure is a polygon with six sides.
(c) This figure is not enclosed with line segments so it is not a polygon.
(d) This figure is enclosed by only line segments but one arc and two line segments so it is not a polygon.
2. Name each polygon.
Make two more examples of each of these.
Solution.
Two more examples of a quadrilateral are:
(b) Triangle
Two more examples of atriangle are:
(c) Pentagon
Two more examples of a pentagon are:
(d) Octagon
Two more examples of an octagon are:
3. Draw a rough sketch of a regular hexagon. Connecting any three of its vertices, draw a triangle. Identify the type of the triangle you have drawn.
Solution.
A regular hexagon is drawn as follows with three of its vertices connected to form a triangle:
Different types of triangles can be formed by connecting any three vertices of a hexagon. These are:
△ABF – Isosceles triangle
△ACE – Isosceles triangle
△BCF – Right angled triangle
(Students may draw any one triangle)
4.Draw a rough sketch of a regular octagon. (Using squared paper if you wish). Draw a rectangle by joining exactly four of the vertices of the octagon.
Solution.
A regular octagon is drawn as follows:
ADEH is the rectangle formed by joining exactly four vertices of the given octagon.
5.A diagonal is a line segment that joins any two vertices of the polygon and is not a side of the polygon. Draw a rough sketch of a pentagon and draw its diagonals.
Solution.
A rough sketch of a pentagon can be drawn as follows:
By joining its any two vertices, we get the following diagonals:
AC, AD, BE, BD and CE
Exercise 5.9
1.Match the following:
Give two examples of each shape.
Solution.
Correct match and examples of each shape are given below:
(a) ↔ (ii)
Examples of cone:
• Birthday cap
• Ice-cream cone
(b) ↔ (iv)
Examples of sphere:
• Football
• Cricket ball
(c) ↔ (v)
Examples of cylinder:
• A piece of chalk
• Pencil
(d) ↔ (iii)
Examples of cuboid:
• Brick
• Match box
(e) ↔ (i)
Examples of pyramid:
• The great pyramids of Eygpt
• Tent
2.What shape is
(b) A brick?
(b) A matchbox?
Solution.
(a) Shape of instrument box is cuboid.
(b) Shape of a brick is cuboid.
(c) Shape of a matchbox is cuboid.
(d) Shape of a road-roller is cylinder.
(e) Shape of a sweet laddu is sphere.
## Check chapter-wise NCERT Solutions for Class 6 Maths from the links given below:
NCERT Solutions for Class 6 Maths Chapter 1 - Knowing Our Numbers
NCERT Solutions for Class 6 Maths Chapter 2 - Whole Numbers
NCERT Solutions for Class 6 Maths Chapter 4 - Basic Geometrical Ideas
NCERT solutions for other chapters will be provided here very soon. Check here for the detailed and appropriate solutions.
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# How do you solve a system of equations by using the elimination method?
Aug 16, 2015
You follow a sequence of steps.
In general, the steps are:
1. Enter the equations.
2. Multiply each equation by a number to get the lowest common multiple for one of the variables.
3. Add or subtract the two equations to eliminate that variable .
4. Substitute that variable into one of the equations and solve for the other variable.
EXAMPLE:
How do you use the elimination method to solve $2 x + 3 y = 7 , 3 x + 4 y = 10$?
Solution:
Step 1. Enter the equations.
[1] $2 x + 3 y = 7$
[2] $3 x + 4 y = 10$
Step 2. Find the lowest common multiple.
Multiply Equation 1 by $3$ and Equation 2 by $2$.
[3] $6 x + 9 y = 21$
[4] $6 x + 8 y = 20$
Step 3. Subtract Equation 4 from Equation 3.
[5] $y = 1$
Step 4. Substitute Equation 5 in Equation 1.
$2 x + 3 y = 7$
$2 x + 3 = 7$
$2 x = 4$
$x = 2$
Check: Substitute the values of $x$ and $y$ in Equation 2.
If you use one equation to get the second variable, use the other equation for the check.
$3 x + 4 y = 10$
3×2+4×1=10
$6 + 4 = 10$
$10 = 10$
It checks!
The solution is correct. |
# Learn Trigonometry : Help in Easily Learning Simple Trig Math (with rhyme to remember)
Updated on March 19, 2013
## Trigonometry formulas table
sin Θ
cos Θ
tan Θ
Opposite (O)
Base (B)
Opposite (O)
Hypotenuse (B)
Hypotenuse (B)
Base (B)
Table 1
Trigonometry is a part of advanced mathematics. It's actually the foundation of advanced mathematics. The word 'trigonometry' is formed by two Greek words 'trigonon' which means triangle and 'metria' which means action or process of measuring. So the most essential application of trigonometry is to measure the sides of a triangle. This article won't discuss the details of trigonometry. We'll just take a look at the basic formulae involved in the calculation of trigonometric values.
Consider the image shown above. It's a right angled triangle ΔABC. The three sides AB, BC and AC are the Hypotenuse, Opposite side and Base respectively. It's with respect to the angle between the two sides AB and AC. We can call it Θ. Angle ACB is a right angle. The lengths of the three sides are as follows:
AB=c
AC=b
BC=a
If you observe carefully, all the lengths have been denoted by the small letter of the Vertex they are opposite to. That's the standard practise.
The sine, cosine and tangent of the triangle are calculated as follows:
sin Θ = Opposite/Hypotenuse = a/c
cos Θ = Base/Hypotenuse = b/c
tan Θ = Opposite/Base = a/b
From the above equations you can easily deduce:
tan Θ = sin Θ/cos Θ
Would you like to learn a simple method of remembering these formulas? Then do as follows:
• First denote Opposite side by O, Hypotenuse by H and Base side by B.
• Then draw a table as shown in Table 1.
• The column header in the table corresponds to the values which are being calculated which are sine, cosine and tangent.
• The 2nd row is of the numerator values and the 3rd row is of the denominator values.
• The letters inside the bracket are of the assumptions which we have made.
• Memorize the table as it is.
• Read the letters in the bracket from the first column, then the second, followed by the third column.
• The sequence is O H B H O B
• Now we can create a rhyme which we can remember easily and recall it later on when we need them to do some trigonometric calculations.
• One such rhyme can be : "Oscar Brought His Orange Bat". This is just an example. You can build upon it.
31
59
1 |
4.6
The power of x
# Powers of x and higher polynomials
There is a family of relations that includes linear, quadratic and inverse relations all in the same framework. This is the family of functions defined by a power of $\normalsize{x}$.
In this step we look at different powers of $\normalsize{x}$, and see how they are the building blocks of polynomials.
## Linear, quadratic and cubic powers
We have had a good look at linear powers, such as $\normalsize{y=x}$, quadratic powers such as $\normalsize{y=x^2}$ and cubic powers such as $\normalsize{y=x^3}$ . Here are their graphs:
It is important to understand how higher powers of $\normalsize{x}$ behave. The general pattern follows the three functions above, but generally as the exponent increases, the function becomes more extreme — and it gets very big more quickly as $\normalsize{x}$ increases.
## Higher powers of $\normalsize{x}$
Recall that $\normalsize{x^0=1}$ for any $\normalsize{x}$. So the function $\normalsize{y=x^0}$ is the same as the function $\normalsize{y=1}$ — that is, it is a constant function that does not change at all. In the following diagram, you can see the functions $\normalsize{y=x^n}$ for $\normalsize{n=0,1,2,3,4,5}$ and how they grow with $\normalsize{x}$.
Q1 (E): In the above graph, which function is which?
When the degree $\normalsize{n}$ of the function $\normalsize{y=x^n}$ is an odd natural number, such a polynomial always increases as $\normalsize{x}$ increases. In this case we say that $\normalsize{y=x^n}$ is an increasing function. However if $\normalsize{n}$ is even, then as $\normalsize{x}$ increases, two things can happen: the function decreases while $\normalsize{x}$ is negative, and then increases for positive $\normalsize{x}$.
## Index laws
When working with powers it is useful to remember the index laws
and for good measure we also include the rule that lets us work with negative exponents:
We will shortly be talking about powers with fractional exponents. It will then be important to remember that these index laws still hold.
## Growth rates of powers
The differences between various powers often becomes more noticeable for very small or very large values of $\normalsize{x}$. While there is not so much difference between $\normalsize{(1.5)^2}$ and $\normalsize{(1.5)^3}$, there is quite a big difference between $\normalsize{(150)^2}$ and $\normalsize{(150)^3}$, and also between $\normalsize{(0.015)^2}$ and $\normalsize{(0.015)^3}$.
We say that the power $\normalsize{y=x^3}$ has a larger growth rate than $\normalsize{y=x^2}$. This becomes ever more noticeable as the exponent $\normalsize{n}$ in $\normalsize{y=x^n}$ increases.
Q2 (E): Which has a faster growth rate: the function $7x^5$ or $5x^7$?
Q3 (M): What about when $\normalsize x$ is getting closer to zero? Which of the functions $7x^5$ and $5x^7$ will decrease faster?
## Combining powers of $x$
The simplest polynomials are the power functions $\normalsize{y=x^n}$ for natural numbers $\normalsize{n}$, and they form the basic building blocks to make more general polynomials, such as $\normalsize{y=3x^5-4x^4+2x-7}$.
These are an important class of objects, because they are exactly the natural domain of algebra: we can add, subtract and multiply polynomials in much the same way as we do arithmetic with numbers, and sometimes we can also divide them. But polynomials are considerably more complicated than numbers, especially as we increase the degree.
Nevertheless, the basic guiding principle — that polynomials give us a domain of arithmetic that extends that of ordinary numbers — is a powerful and useful one.
## Polynomials as products of factors
There is another way of getting polynomials rather than combining multiples of powers of $\normalsize x$. Just as one way of getting bigger numbers is to multiply them together (for example $\normalsize{1356= 2^23\times113}$) , so too for polynomials we can multiply smaller polynomials together to get higher degree polynomials. The simplest way of doing that is to multiply together linear factors, for example
By Descartes’ theorem (Factors of quadratic polynomials and zeroes) each linear factor of $\normalsize p$ corresponds to a zero of the function, so for example in this case $\normalsize{p(-7)=p(-3)=p(2)=p(-4)=0}$.
You should be aware that a polynomial that has been composed as a product of linear factors is arithmetically rather special. In general a polynomial of degree $\normalsize n$ may have less than $\normalsize n$ linear factors. For example $\normalsize x^2 + 1$ has no linear factors.
## Shapes of polynomials functions
As we go up in degree, polynomial functions get longer to write down, with more coefficients, and their graphs become more complicated. Nevertheless there is a kind of predictability about the overall shape of polynomials that it is essential for us to understand. Even a higher degree polynomial function such as
has something of a regular graph: as we view it from left to right, it goes up, then down, then up, then down and then finally up.
The number of ups and downs is limited by the degree. A fundamental fact about polynomials is that the graph of a degree n polynomial can meet an arbitrary line in at most n points. This is an important theorem. Another variant is that a polynomial of degree $\normalsize n$ can change direction — from up to down, or from down to up, as we view it from left to right — at most $\normalsize (n-1)$ times.
But remember that not all polynomial functions exhibit this up, down, up, down aspect explicitly. In particular the powers of $\normalsize{y=x^n}$ that we began our discussion with only go up if $\normalsize{n}$ is odd, and go down and then up if $\normalsize{n}$ is even, no matter how high the degree $\normalsize n$.
A1. The function $\normalsize y=x^0=1$ is light blue (the horizontal line), $\normalsize y=x^1=x$ is red (the straight line going through the point $\normalsize{[0.5, 0.5]}$), $\normalsize y=x^2$ is yellow (one of the 2 ‘even’ functions — with all positive values), $\normalsize y=x^3$ is orange (one of the 3 ‘odd’ functions — with negative values for x < 0), $\normalsize y=x^4$ is green (again an even function) and $\normalsize y=x^5$ is dark blue (again an odd function).
A2. Since $\normalsize 7$ is greater than $\normalsize 5$ the function $\normalsize x^7$ grows faster than $\normalsize x^5$ as $\normalsize x$ gets larger. The coefficients are relatively unimportant, so that also $\normalsize 5x^7$ grows faster than $\normalsize 7x^5$.
A3. Since $\normalsize 7$ is greater than $\normalsize 5$ the function $\normalsize x^7$ decreases faster than $\normalsize x^5$ as $\normalsize x$ gets smaller. So also $\normalsize 5x^7$ decreases faster than $\normalsize 7x^5$ as $\normalsize x$ gets smaller. |
# How to divide fractions. Dividing fractions calculator (÷)
## Dividing Mixed Numbers
This is the quickest technique for dividing fractions. I want students to make the connection between division and addition of fractions by saying that they both need to have common denominators. In that bottle there are 6 doses. Key Terms o Common denominator o Reciprocal o Complex fraction Objectives o Learn how to add, subtract, multiply, and divide fractions o Understand how to interpret fractions that involve negative numbers o Recognize and simplify complex fractions Now that we have developed a solid foundation regarding what fractions are as well as some different types of fractions, we can now turn to application of the basic arithmetic operations addition, subtraction, multiplication, and division to fractions. This is a fraction calculator with steps shown in the solution. When it comes to adding and subtracting fractions, it's very simple once you've checked for or created, if necessary a common denominator.
Next
## How to Divide Fractions by Fractions: 12 Steps (with Pictures)
To turn a whole number into a fraction, make the numerator the whole number and make the denominator one. Thus, one approach is to multiply both numerator and denominator of the complex fraction by the product of the simple fractions' denominators, as shown below. On a map, of an inch represents 60 miles. Before you can multiply or divide mixed fractions, you must convert the mixed fractions to improper fractions. Step 1 Check: do your denominators match? Before your proceed though, make sure you fully understand the : adding, subtracting, multiplying and dividing.
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## How to Divide Fractions by Fractions: 12 Steps (with Pictures)
In some cases, the product will already be in lowest terms; in others, you may need to reduce it to lowest terms. Solution: In each case, find a common denominator and convert the terms to equivalent fractions with that denominator. Teaching your students how to divide fractions can be just as simple as … once you know all of the little tricks to get the right answer. Thus, if we simply convert one or both of the fractions that we are adding or subtracting into equivalent fractions with the same denominator, then we are able to add the fractions in the simple manner described above. And the reciprocal of 2 is 1 2.
Next
## Divide Two Fractions
In the case of part c, note that the reciprocal of 5 is and that the quotient or product of a positive number divided multiplied by a negative number is a negative number. You can also create assignments for each student based on their specific needs and learning styles. The one change is that you have to take the reciprocal of the divisor. Dividing fractions calculator When your students are learning how to divide fractions you can show them fraction calculators. Common denominator: 45 Multiplication and Division Multiplying and dividing fractions is in some ways simpler than adding and subtracting them. The difference being that division asks us about equal parts or how many whereas addition asks us about total.
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## How to Add, Subtract, Multiply, and Divide Fractions
The best way to understand which number is which is by looking at an example. The reciprocal of 1 2 is 2 1 , or 2. You can look at reports on your whole class and see which topics different students are struggling with! The first only has the questions, while the second has all of the answers, including the process to get the solution. One of the most valuable things to teach your students when dividing fractions is what the answer means. By Last update: 19 December 2016 While some people might be breathing deeply into a paper bag at the thought of , if you understand each step and why it's necessary, it can become a piece of cake.
Next
## How To Divide Fractions
The formulas for multiplying and dividing fractions follow the same process as described above. When you divide one fraction by another fraction, you switch the numerator and denominator of the divisor fraction, then multiply. Note that reciprocals come in pairs. To multiply two fractions, then, simply multiply the numerators and multiply the denominators to get the product. Example 1: Step 1: Change to improper fractions. Let's say we want to multiply by. The first case is illustrated below.
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## How to Divide Fractions by a Whole Number: 7 Steps (with Pictures)
Instead, you simplify that to one-half or ½. The only drawback with worksheets is that they can take a long time to mark. We should already know that we can write equivalent fractions that have different numerators and denominators. I do not want them taking notes at this time. The primary reason is that it utilizes Euclid's Algorithm for reducing fractions which can be found on. Once they have modeled it, we will find our answer by dividing the numerators.
Next |
fraction
(redirected from fractional)
Also found in: Dictionary, Thesaurus, Medical, Wikipedia.
Related to fractional: fractional distillation
fraction
fraction [Lat.,=breaking], in arithmetic, an expression representing a part, or several equal parts, of a unit.
Notation for Fractions
In writing a fraction, e.g., 2-5 or 2/5, the number after or below the bar represents the total number of parts into which the unit has been divided. This number is called the denominator. The number before or above the bar, the numerator, denotes how many of the equal parts of the unit have been taken. The expression 2-5, then, represents the fact that two of the five parts of the unit or quantity have been taken. The present notation for fractions is of Hindu origin, but some types of fractions were used by the Egyptians before 1600 B.C. Another way of representing fractions is by decimal notation (see decimal system).
Characteristics of Fractions
When the numerator is less than the denominator, the fraction is proper, i.e., less than unity. When the reverse is true, e.g., 5-2, the fraction is improper, i.e., greater than unity. When a fraction is written with a whole number, e.g., 31-2, the expression is called a mixed number. This may also be written as an improper fraction, as 7-2, since three is equal to six halves, and by adding the one half, the total becomes seven halves, or 7-2. A fraction has been reduced to its lowest terms when the numerator and denominator are not divisible by any common divisor except 1, e.g., when 4-6 is reduced to 2-3.
Arithmetic Operations Involving Fractions
When fractions having the same denominator, as 3-10 and 4-10, are added, only the numerators are added, and their sum is then written over the common denominator: 3-10+4-10=7-10. Fractions having unlike denominators, e.g., 1-4 and 1-6, must first be converted into fractions having a common denominator, a denominator into which each denominator may be divided, before addition may be performed. In the case of 1-4 and 1-6, for example, the lowest number into which both 4 and 6 are divisible is 12. When both fractions are converted into fractions having this number as a denominator, then 1-4 becomes 3-12, and 1-6 becomes 2-12. The change is accomplished in the same way in both cases—the denominator is divided into the 12 and the numerator is multiplied by the result of this division. The addition then is performed as in the case of fractions having the same denominator: 1-4+1-6=3-12+2-12=5-12. In subtraction, the numerator and the denominator are subjected to the same preliminary procedure, but then the numerators of the converted fractions are subtracted: 1-4−1-6=3-12−2-12=1-12.
In multiplication the numerators of the fractions are multiplied together as are the denominators without needing change: 2-3×3-5=6-15. It should be noted that the result, here 6-15, may be reduced to 2-5 by dividing both numerator and denominator by 3. The division of one fraction by another, e.g., 3-5÷1-2, is performed by inverting the divisor and multiplying: 3-5÷1-2=3-5×2-1=6-5. The same rules apply to the addition, subtraction, multiplication, and division of fractions in which the numerators and denominators are algebraic expressions.
The following article is from The Great Soviet Encyclopedia (1979). It might be outdated or ideologically biased.
Fraction
in arithmetic, a quantity consisting of an integral number of parts of a unit. A fraction is represented by the symbol m/n, where n, the denominator of the fraction, indicates the number of parts into which the unit is to be divided and m, the numerator of the fraction, indicates the number of such parts taken. A fraction may be viewed as the quotient obtained by dividing one integer (m) by another (n). If m is divisible by n without a remainder, then the quotient m\n denotes an integer (for example, 6/3 = 2, 33/11 = 3). The numerator and denominator of a fraction may be simultaneously multiplied or divided by the same number without changing the value of the fraction. Any fraction can be represented in reduced form, that is, as a fraction whose numerator and denominator do not have common factors; for example, 16/72 is not in reduced form [16/72 = 2x8/9x 8 = 2/9], but 27/64 is. To add fractions with the same denominator, we add their numerators and take the same denominator: (a/b) + (c/b) + (d/b) = (a + c + d)/b. To add several fractions with different denominators, it is necessary to bring them to a common denominator. Subtraction of fractions is done in the same way. To multiply several fractions, we divide the product of their numerators by the product of their denominators: (a/b) x (c/d) = ac/bd. Defining division as an inverse operation of multiplication implies the following rule for division: (a/b) ÷ (c/d) = ad/bc. If the numerator of a fraction is less than the denominator, the fraction is called a proper fraction; if the opposite is true, it is called an improper fraction. An improper fraction may be shown to be the sum of an integer and a proper fraction (a mixed number). For this it is necessary to divide the numerator (with remainder) by the denominator; for example,
This proposition of elementary arithmetic can be extended to all real numbers: a real number x can be represented uniquely as x = n + d, where n is an integer and 0 ≤ d < 1. The integer n is called the integral part of x and is denoted by [x]. The number d = x - [x] is called the fractional part of x.
Decimal fractions are fractions whose denominator is a power of 10. Such fractions are written without denominators; for example, 5,481,475/10,000 = 548.1475 and 23/1,000 = 0.023.
Operations with fractions are encountered in the ancient Egyptian Ahmes papyrus (c. 2000 B.C.) where the only admissible fractions are fractions of the type l/n (aliquot fractions). Hence the distinctive “Egyptian” problem of representing any fraction as the sum of unequal fractions of the type l/n (in addition to aliquot fractions, the Egyptians had a special symbol for the fraction 2/3); for example, 7/29 = (1/5) + (1/29) + (1/145).
In ancient Babylonian manuscripts we encounter so-called sexagesimal fractions, that is, fractions having a denominator that is a power of 60. The number 60 played a significant role in classical arithmetic; the division of a unit into 60 and 3,600 = 602 parts has been preserved to the present day in the division of an hour or a degree into 60 minutes (1/60) and of a minute into 60 seconds. The ancient Hindus, apparently, were the first to conceive the modern symbol for a fraction.
REFERENCES
Entsiklopediia elementarnoi matematiki, book 1: Arifmetika. Moscow-Leningrad, 1951.
Depman, I. la. Istoriia arifmetiki, 2nd ed. Moscow, 1965.
Fraction
a portion of a granular or lumpy solid (such as crushed rock, sand, or powder) or of a liquid mixture (such as petroleum) isolated according to a specific criterion. In sieve analysis, fractions are isolated by particle or grain size; in gravity concentration, by density; and in petroleum distillation, by boiling point.
fraction
[′frak·shən]
(chemistry)
One of the portions of a volatile liquid within certain boiling point ranges, such as petroleum naphtha fractions or gas-oil fractions.
(mathematics)
An expression which is the product of a real number or complex number with the multiplicative inverse of a real or complex number.
(metallurgy)
In powder metallurgy, that portion of sample that lies between two stated particle sizes. Also known as cut.
(science and technology)
A portion of a mixture which represents a discrete unit and can be isolated from the whole system.
McGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.
fraction
1. Maths
a. a ratio of two expressions or numbers other than zero
b. any rational number that is not an integer
2. Chem a component of a mixture separated by a fractional process, such as fractional distillation
3. Christianity the formal breaking of the bread in Communion
Collins Discovery Encyclopedia, 1st edition © HarperCollins Publishers 2005
References in periodicals archive ?
The main objective of this paper is to develop an operator-based approach for the construction of closed-form solutions to fractional differential equations.
The presented technique is based on Caputo and Riemann-Liouville algebras of fractional power series.
(v) numerical analysis of fractional order epidemic models of childhood diseases,
(vi) weak solutions for partial random Hadamard fractional integral equations with multiple delays,
Unlike [18, 19, 21], we derive M(t) of the fractional order memristor as a function i(t) as it has been assumed that the of fractional order memristor is a generalization of the HP memristor which is of a current controlled type as aforementioned.
Thus by substituting (6) into (7) and keeping (8) in mind, M(t) of the fractional order memristor can be obtained as follows:
The fractional calculus can be obtained either from the generalization of the definition of the derivative or the definition of the integral.
In Sections 2.1 and 2.2, we introduce the definitions of fractional derivatives and integrals given by both the Grunwald-Letnikov derivative approach and the Riemann-Liouville integrative approach.
In this section, we give the description of the auxiliary equation method for solving fractional partial differential equations.
Suppose that a fractional partial differential equation in the independent variables t, [x.sub.1], [x.sub.2], ..., [x.sub.n] is given by
Different from integer calculus, fractional derivative does not have a unified temporal definition expression up to now.
It is notable that Abel was the first one to use fractional calculus for [alpha] = 1/2 and solved the famous tautochrone problem [20, 21].
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# Triangle Classification by Side Lengths
## Identify triangles as scalene, isosceles or equilateral.
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Triangle Classification by Side Lengths
Remember Cassie from the Classify Triangles by Angles Concept?
Cassie learned how to identify triangles according to angle measures. In this Concept, she is going to learn to identify the same triangles according to side length. Take a look.
Here are the triangles that Cassie is going to identify according to side length.
This Concept will teach you how to classify triangles in this way. By the end of the Concept, you will know how to help Cassie.
### Guidance
Previously we worked on how to look inside the triangle at its angles to help classify a triangle. Well, we can also look at the lengths of the sides to help us classify triangles.
No they aren’t the same. Let’s look at how we can classify triangles according to side length.
The first triangle to think about is an equilateral triangle. An equilateral triangle has side lengths that are the same. Let’s look at an example.
These little lines let you know that the side lengths are the same. Sometimes you will see these and sometimes you won’t. You may have to figure it out on your own or by measuring with a ruler.
The second type of triangle is a scalene triangle. A scalene triangle is a triangle where the lengths of all three sides are different. Here is an example of a scalene triangle.
Here you can see that all three sides of the triangle are different lengths. This is called a scalene triangle.
The third type of triangle is an isosceles triangle. An isosceles triangle has two side lengths that are the same and one side length that is different. Here is an example of an isosceles triangle.
Classify these three triangles on your own. Classify them according to their side lengths.
#### Example A
Solution: Scalene triangle
#### Example B
Solution: Equilateral triangle
#### Example C
Solution: Isosceles triangle
Remember Cassie? Here is the original problem once again.
Cassie learned how to identify triangles according to angle measures in the Classify Triangles by Angles Concept. In this Concept, she is going to learn to identify the same triangles according to side length. Take a look.
Here are the triangles that Cassie is going to identify according to side length.
Triangles 1, 2, and 4 all have three different side lengths. These are all scalene triangles.
Triangle 3 is an isosceles triangle because two of the side lengths are the same.
### Vocabulary
Triangle
a three sided figure with three angles. The prefix “tri”means three.
Scalene Triangle
all three side lengths are different
Isosceles Triangle
two side lengths are the same and one is different
Equilateral Triangle
all three side lengths are the same
### Guided Practice
Here is one for you to try on your own.
Can this triangle be an isosceles triangle?
Side lengths, 6 cm, 4 cm, 6 cm
Angles 70, 70, 40 degrees
The angle measures given are helpful, but the side lengths are all that you need to determine whether or not this triangle can be an isosceles triangle. Given that two of the side lengths are the same, it is indeed an isosceles triangle.
### Practice
Directions: Answer the following questions using what you have learned about triangles their angles and side lengths.
1. If a triangle is a right triangle, then how many angles are acute?
2. How many angles in a right triangle are right angles?
3. How many degrees are there in a right triangle?
4. What is an obtuse angle?
5. How many obtuse angles are in an obtuse triangle?
6. If there is one obtuse angle, how many angles are acute?
7. If a triangle is equiangular, what is the measure of all three angles?
8. What does the word “interior angle” mean?
9. True or false. The side lengths of a scalene triangle are all equal.
10. True or false. The side lengths of a scalene triangle are all different.
11. True or false. The side lengths of an equilateral triangle are all equal.
17. True or false. An isosceles triangle has two side lengths the same and one different.
18. True or false. A scalene triangle can also be an isosceles triangle.
### Vocabulary Language: English
Isosceles Triangle
Isosceles Triangle
An isosceles triangle is a triangle in which exactly two sides are the same length.
Scalene Triangle
Scalene Triangle
A scalene triangle is a triangle in which all three sides are different lengths.
Triangle
Triangle
A triangle is a polygon with three sides and three angles. |
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# State the converse of Pythagoras theorem and prove it.
Last updated date: 13th Jun 2024
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Hint: We will first write the statement of the converse of Pythagoras theorem then to prove it we will first consider two triangles and prove them congruent by SSS congruency. Then we will prove that one of the angles of a triangle is a right angle.
We know that the converse of Pythagoras theorem is stated as:
In a triangle, if the square of one longest side is equal to the sum of squares of the other two sides then the angle opposite the first side is a right angle.
Now, to prove this statement let us consider two triangles $\Delta ABC\And \Delta PQR$ from which $\Delta PQR$ is right-angled at $Q$ .
In $\Delta ABC$ we have $AC$ as the longest side and by Pythagoras theorem we get
$\Rightarrow A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}...........(i)$
We have to prove that $\Delta ABC$ is right triangle.
Now, we construct another triangle $\Delta PQR$ which is a right triangle $\angle Q=90{}^\circ$ and AB=PQ and BC=QR
So, by Pythagoras theorem we have $P{{R}^{2}}=Q{{R}^{2}}+P{{Q}^{2}}$
Now, by construction we have $AB=PQ\And BC=QR$
So, substituting the values we get
$\Rightarrow P{{R}^{2}}=A{{B}^{2}}+B{{C}^{2}}...........(ii)$
Now, from equation (i) and (ii) we get
\begin{align} & \Rightarrow A{{C}^{2}}=P{{R}^{2}} \\ & \Rightarrow AC=PR \\ \end{align}
Now, in $\Delta ABC\And \Delta PQR$ we have
\begin{align} & \Rightarrow AB=PQ \\ & \Rightarrow BC=QR \\ \end{align} (by construction)
$\Rightarrow AC=PR$ (Proved above)
So we get $\Delta ABC\cong \Delta PQR$ (SSS congruency)
So we get
$\angle Q=\angle B$ (corresponding angles of congruent triangles)
And we have $\angle Q=90{}^\circ$
So, $\angle B=90{}^\circ$
Hence proved that $\Delta ABC$ is right triangle.
Note:
The fact is that to write the converse of a theorem we need to interchange the hypothesis and conclusion of a statement. The key point to prove triangles congruent we need to construct the two sides of the triangles equal. |
The straight lines 4x – 3y – 5 = 0, x – 2y – 10 = 0, 7x + 4y – 40 = 0 and x + 3y + 10 = 0 form the sides of a : Kaysons Education
# The Straight Lines 4x – 3y – 5 = 0, x – 2y – 10 = 0, 7x + 4y – 40 = 0 And x + 3y + 10 = 0 Form The Sides Of A
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## Question
### Solution
Correct option is
Slopes of the lines are 4/3, 1/2, –7 and –1/3, respectively. If α is the angle between first and third. β is the angle between second and fourth.
tan α = –l, tan β = 1 ⇒ α = 135o, β = 45o ⇒ α + β = 180o
Since no two sides are parallel, it can not be a parallelogram or a rectangle.
#### SIMILAR QUESTIONS
Q1
If two of the lines represented by
x4 + x3 y + cx2 y2 – xy3 + y4 = 0
bisect the angle between the other two, then the value of c is
Q2
The straight line is x + y = 0, 3x + y – 4 = 0 and x + 3y – 4 = 0 from a triangle which is
Q3
If the line x + 2ay + a = 0, x + 3by + b = 0 and x + 4cy + c = 0 are concurrent, then abc are in
Q4
If the line 2 (sin a + sin bx – 2 sin (a – by = 3 and 2 (cos a + cos bx + 2 cos (a – by = 5 are perpendicular, then sin 2a+ sin2b is equal to
Q5
If p1p2 denote the lengths of the perpendiculars from the origin on the lines x sec α + y cosec α = 2a and
x cos α + y sin α = a cos 2α respectively, then is equal to
Q6
The locus of the point of intersection of the lines x sin θ + (1 – cos θ) y = a sin θ and x sin θ – (1 + cos θ) y + a sin θ = 0 is
Q7
If two vertices of a triangle are (5, –1) and (–2, 3), and the orthocenter lies at the origin, the coordinate of the third vertex are
Q8
Equation of a line passing through the intersection of the lines
x + 2y – 10 = 0 and 2x + y + 5 = 0 is
Q9
The lengths of the perpendicular from the points (m2, 2m), (mmm +m) and (m2, 2m) to the line x + y + 1 = 0 form
Q10
The sine of the angle between the pair of lines represented by the equation x2 – 7xy + 12y2 = 0 is |
# Question Video: Determining the Ordered Pair of a Point Completing a Rectangle on a Coordinate Plane Mathematics • 4th Grade
Determine whether this statement is true or false: All rhombuses are parallelograms.
02:27
### Video Transcript
Determine whether this statement is true or false. All rhombuses are parallelograms.
To decide whether the statement is true or false, we need to know what the two types of shape in this statement mean. What is a rhombus? And what is a parallelogram?
Let’s remind ourselves what a parallelogram is to begin with. Well, there’s a big clue in the name. A parallelogram is a quadrilateral or a four-sided shape with two pairs of parallel sides. For example, this shape has two pairs of parallel sides. This pair of sides is parallel and this pair of sides is parallel too. Two pairs of parallel sides.
Rectangles have two pairs of parallel sides, so we can say that rectangles are a type of parallelogram. And the opposite sides on a square are parallel too. A square is a type of parallelogram.
Our question asks us about rhombuses. Are all rhombuses parallelograms? What’s a rhombus? A rhombus is a type of quadrilateral with opposite sides that are parallel and all sides equal. Here are some examples of rhombuses. We can see that they’re all quadrilaterals. In other words, they’ve all got four sides They’ve also each got opposite sides that are parallel. The other thing we notice about rhombuses is that all the sides are the same length.
Let’s turn our fact about parallelograms into a checklist. Are all rhombuses quadrilaterals? Yes, they all have four sides. Do all rhombuses have two pairs of parallel sides? Yes. And so we can say all rhombuses are parallelograms. They’re a special type of parallelogram where all the sides are equal. The statement is true. |
Courses
# RS Aggarwal Solutions: Exercise 3C - Linear Equations in two variables Notes | EduRev
## Class 10 : RS Aggarwal Solutions: Exercise 3C - Linear Equations in two variables Notes | EduRev
``` Page 1
Exercise 3C
1. Solve the system of equations by using the method of cross multiplication:
x + 2y + 1 = 0,
2x 3y 12 = 0.
Sol:
The given equations are:
2x 3y
Here a1 = 1, b1 = 2, c1 = 1, a2 = 2, b2 = -3 and c2 = -12
By cross multiplication, we have:
= =
= =
= =
x = = 3, y = = -2
Hence, x = 3 and y = -2 is the required solution.
2. Solve the system of equations by using the method of cross multiplication:
3x - 2y + 3 = 0,
4x + 3y 47 = 0
Sol:
The given equations are:
Page 2
Exercise 3C
1. Solve the system of equations by using the method of cross multiplication:
x + 2y + 1 = 0,
2x 3y 12 = 0.
Sol:
The given equations are:
2x 3y
Here a1 = 1, b1 = 2, c1 = 1, a2 = 2, b2 = -3 and c2 = -12
By cross multiplication, we have:
= =
= =
= =
x = = 3, y = = -2
Hence, x = 3 and y = -2 is the required solution.
2. Solve the system of equations by using the method of cross multiplication:
3x - 2y + 3 = 0,
4x + 3y 47 = 0
Sol:
The given equations are:
3x - 2y
4x + 3y
Here a1 = 3, b1 = -2, c1 = 3, a2 = 4, b2 = 3 and c2 = -47
By cross multiplication, we have:
= =
= =
= =
x = = 5, y = = 9
Hence, x = 5 and y = 9 is the required solution.
3. Solve the system of equations by using the method of cross multiplication:
6x - 5y - 16 = 0,
7x - 13y + 10 = 0
Sol:
The given equations are:
6x - 5y -
7x -
Here a1 = 6, b1 = -5, c1 = -16, a2 = 7, b2 = -13 and c2 = 10
By cross multiplication, we have:
= =
= =
= =
x = = 6, y = = 4
Hence, x = 6 and y = 4 is the required solution.
4. Solve the system of equations by using the method of cross multiplication:
3x + 2y + 25 = 0, 2x + y + 10 = 0
Sol:
The given equations are:
Page 3
Exercise 3C
1. Solve the system of equations by using the method of cross multiplication:
x + 2y + 1 = 0,
2x 3y 12 = 0.
Sol:
The given equations are:
2x 3y
Here a1 = 1, b1 = 2, c1 = 1, a2 = 2, b2 = -3 and c2 = -12
By cross multiplication, we have:
= =
= =
= =
x = = 3, y = = -2
Hence, x = 3 and y = -2 is the required solution.
2. Solve the system of equations by using the method of cross multiplication:
3x - 2y + 3 = 0,
4x + 3y 47 = 0
Sol:
The given equations are:
3x - 2y
4x + 3y
Here a1 = 3, b1 = -2, c1 = 3, a2 = 4, b2 = 3 and c2 = -47
By cross multiplication, we have:
= =
= =
= =
x = = 5, y = = 9
Hence, x = 5 and y = 9 is the required solution.
3. Solve the system of equations by using the method of cross multiplication:
6x - 5y - 16 = 0,
7x - 13y + 10 = 0
Sol:
The given equations are:
6x - 5y -
7x -
Here a1 = 6, b1 = -5, c1 = -16, a2 = 7, b2 = -13 and c2 = 10
By cross multiplication, we have:
= =
= =
= =
x = = 6, y = = 4
Hence, x = 6 and y = 4 is the required solution.
4. Solve the system of equations by using the method of cross multiplication:
3x + 2y + 25 = 0, 2x + y + 10 = 0
Sol:
The given equations are:
Here a1 = 3, b1 = 2, c1 = 25, a2 = 2, b2 = 1 and c2 = 10
By cross multiplication, we have:
= =
= =
= =
x = = 5, y = = -20
Hence, x = 5 and y = -20 is the required solution.
5. Solve the system of equations by using the method of cross multiplication:
2x + 5y 1 = 0, 2x + 3y 3 = 0
Sol:
The given equations may be written as:
2x + 5y
2x + 3y
Here a1 = 2, b1 = 5, c1 = -1, a2 = 2, b2 = 3 and c2 = -3
By cross multiplication, we have:
= =
= =
= =
x = = 3, y = = -1
Hence, x = 3 and y = -1 is the required solution.
6. Solve the system of equations by using the method of cross multiplication:
2x + y 35 = 0,
3x + 4y 65 = 0
Sol:
The given equations may be written as:
2x + y
Page 4
Exercise 3C
1. Solve the system of equations by using the method of cross multiplication:
x + 2y + 1 = 0,
2x 3y 12 = 0.
Sol:
The given equations are:
2x 3y
Here a1 = 1, b1 = 2, c1 = 1, a2 = 2, b2 = -3 and c2 = -12
By cross multiplication, we have:
= =
= =
= =
x = = 3, y = = -2
Hence, x = 3 and y = -2 is the required solution.
2. Solve the system of equations by using the method of cross multiplication:
3x - 2y + 3 = 0,
4x + 3y 47 = 0
Sol:
The given equations are:
3x - 2y
4x + 3y
Here a1 = 3, b1 = -2, c1 = 3, a2 = 4, b2 = 3 and c2 = -47
By cross multiplication, we have:
= =
= =
= =
x = = 5, y = = 9
Hence, x = 5 and y = 9 is the required solution.
3. Solve the system of equations by using the method of cross multiplication:
6x - 5y - 16 = 0,
7x - 13y + 10 = 0
Sol:
The given equations are:
6x - 5y -
7x -
Here a1 = 6, b1 = -5, c1 = -16, a2 = 7, b2 = -13 and c2 = 10
By cross multiplication, we have:
= =
= =
= =
x = = 6, y = = 4
Hence, x = 6 and y = 4 is the required solution.
4. Solve the system of equations by using the method of cross multiplication:
3x + 2y + 25 = 0, 2x + y + 10 = 0
Sol:
The given equations are:
Here a1 = 3, b1 = 2, c1 = 25, a2 = 2, b2 = 1 and c2 = 10
By cross multiplication, we have:
= =
= =
= =
x = = 5, y = = -20
Hence, x = 5 and y = -20 is the required solution.
5. Solve the system of equations by using the method of cross multiplication:
2x + 5y 1 = 0, 2x + 3y 3 = 0
Sol:
The given equations may be written as:
2x + 5y
2x + 3y
Here a1 = 2, b1 = 5, c1 = -1, a2 = 2, b2 = 3 and c2 = -3
By cross multiplication, we have:
= =
= =
= =
x = = 3, y = = -1
Hence, x = 3 and y = -1 is the required solution.
6. Solve the system of equations by using the method of cross multiplication:
2x + y 35 = 0,
3x + 4y 65 = 0
Sol:
The given equations may be written as:
2x + y
3x + 4y
Here a1 = 2, b1 = 1, c1 = -35, a2 = 3, b2 = 4 and c2 = -65
By cross multiplication, we have:
= =
= =
= =
x = = 15, y = = 5
Hence, x = 15 and y = 5 is the required solution.
7. Solve the system of equations by using the method of cross multiplication:
7x - 2y 3 = 0,
11x - y 8 = 0.
Sol:
The given equations may be written as:
7x - 2y
11x - y
Here a1 = 7, b1 = -2, c1 = -3, a2 = 11, b2 = - and c2 = -8
By cross multiplication, we have:
= =
= =
= =
x = = 1, y = = 2
Hence, x = 1 and y = 2 is the required solution.
8. Solve the system of equations by using the method of cross multiplication:
+ 4 = 0, - = 0
Page 5
Exercise 3C
1. Solve the system of equations by using the method of cross multiplication:
x + 2y + 1 = 0,
2x 3y 12 = 0.
Sol:
The given equations are:
2x 3y
Here a1 = 1, b1 = 2, c1 = 1, a2 = 2, b2 = -3 and c2 = -12
By cross multiplication, we have:
= =
= =
= =
x = = 3, y = = -2
Hence, x = 3 and y = -2 is the required solution.
2. Solve the system of equations by using the method of cross multiplication:
3x - 2y + 3 = 0,
4x + 3y 47 = 0
Sol:
The given equations are:
3x - 2y
4x + 3y
Here a1 = 3, b1 = -2, c1 = 3, a2 = 4, b2 = 3 and c2 = -47
By cross multiplication, we have:
= =
= =
= =
x = = 5, y = = 9
Hence, x = 5 and y = 9 is the required solution.
3. Solve the system of equations by using the method of cross multiplication:
6x - 5y - 16 = 0,
7x - 13y + 10 = 0
Sol:
The given equations are:
6x - 5y -
7x -
Here a1 = 6, b1 = -5, c1 = -16, a2 = 7, b2 = -13 and c2 = 10
By cross multiplication, we have:
= =
= =
= =
x = = 6, y = = 4
Hence, x = 6 and y = 4 is the required solution.
4. Solve the system of equations by using the method of cross multiplication:
3x + 2y + 25 = 0, 2x + y + 10 = 0
Sol:
The given equations are:
Here a1 = 3, b1 = 2, c1 = 25, a2 = 2, b2 = 1 and c2 = 10
By cross multiplication, we have:
= =
= =
= =
x = = 5, y = = -20
Hence, x = 5 and y = -20 is the required solution.
5. Solve the system of equations by using the method of cross multiplication:
2x + 5y 1 = 0, 2x + 3y 3 = 0
Sol:
The given equations may be written as:
2x + 5y
2x + 3y
Here a1 = 2, b1 = 5, c1 = -1, a2 = 2, b2 = 3 and c2 = -3
By cross multiplication, we have:
= =
= =
= =
x = = 3, y = = -1
Hence, x = 3 and y = -1 is the required solution.
6. Solve the system of equations by using the method of cross multiplication:
2x + y 35 = 0,
3x + 4y 65 = 0
Sol:
The given equations may be written as:
2x + y
3x + 4y
Here a1 = 2, b1 = 1, c1 = -35, a2 = 3, b2 = 4 and c2 = -65
By cross multiplication, we have:
= =
= =
= =
x = = 15, y = = 5
Hence, x = 15 and y = 5 is the required solution.
7. Solve the system of equations by using the method of cross multiplication:
7x - 2y 3 = 0,
11x - y 8 = 0.
Sol:
The given equations may be written as:
7x - 2y
11x - y
Here a1 = 7, b1 = -2, c1 = -3, a2 = 11, b2 = - and c2 = -8
By cross multiplication, we have:
= =
= =
= =
x = = 1, y = = 2
Hence, x = 1 and y = 2 is the required solution.
8. Solve the system of equations by using the method of cross multiplication:
+ 4 = 0, - = 0
Sol:
The given equations may be written as:
+
-
Here a1 = , b1 = , c1 = -4, a2 = , b2 = - and c2 = -
By cross multiplication, we have:
= =
= =
= =
x = = 18, y = = 15
Hence, x = 18 and y = 15 is the required solution.
9. Solve the system of equations by using the method of cross multiplication:
+ = 7, + = 17
Sol:
Taking = u and = v, the given equations become:
u + v = 7
2u + 3v = 17
The given equations may be written as:
u + v
2u + 3v
Here, a1 = 1, b1 = 1, c1 = -7, a2 = 2, b2 = 3 and c2 = -17
By cross multiplication, we have:
= =
= =
= =
u = = 4, v = = 3
= 4, = 3
x = , y =
```
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## Mathematics (Maths) Class 10
62 videos|363 docs|103 tests
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# Lesson Explainer: Standard Deviation of a Data Set Mathematics • 9th Grade
In this explainer, we will learn how to find and interpret the standard deviation from a given data set.
In order to understand the meaning of the standard deviation of a data set we first recall the definition of the mean of a data set.
### Definition: The Mean of a Data Set
The mean, average, or expected value of a data set is used as measure of central tendency. For a data set , where there are values, the mean, denoted by (pronounced βmiuβ) or , is calculated by taking the sum of the data set and dividing it by the number of values , as indicated in the formula below:
The standard deviation of a data set tells us the dispersion of data from the mean. The larger the standard deviation, the more dispersed the data is from the mean, and the smaller the standard deviation, the less dispersed the data is from the mean.
The square of the standard deviation is called the variance and is another measure of dispersion. A further measure of dispersion is the interquartile range, which is the difference between the upper quartile and the lower quartile, or the value of the 75th percentile minus the value of the 25th percentile. In this explainer, we will only be focusing on the standard deviation as a measure of dispersion.
The standard deviation is more formally defined in the definition below.
### Definition: The Standard Deviation of a Data Set
The standard deviation of a data set is used to measure the dispersion of data from the mean. For a data set , where there are values, the standard deviation, denoted by (pronounced βsigma β), is calculated by taking the sum of the difference of values of the data set from the mean squared, dividing by the number of values, and square rooting, as indicated in the formula below:
Another way of describing the standard deviation is as the average distance between the mean and the individual data points in the set. So, if the standard deviation is larger, then the average distance between the mean and the individual data points will be greater, meaning they are more dispersed. Similarly, if the standard deviation is smaller, then the distance between the mean and the individual data points will be less, meaning they are less dispersed.
We will use the definition of the standard deviation of a data set to answer the first example.
### Example 1: Understanding Standard Deviation
What is the name of a quantity expressing by how much the members of a group differ from the mean value for the group?
We know that the standard deviation of a data set determines how dispersed the data set is from the mean. This can also be described as how much the members of a data set differ from the mean of the data set.
Therefore, the quantity expressing by how much the members of a group differ from the mean value for the group is the standard deviation. A low standard deviation tells us that the data points are, on average, closer to the mean, and a high standard deviation tells us that the data points are, on average, further from the mean.
Having discussed what the definition of the standard deviation is, we will next consider the case where the measure of dispersion is zero, as seen in the next example.
### Example 2: Identifying a Set of Values with Zero Dispersion
If the dispersion of a set of values is equal to zero, then which of the following is true?
1. The difference between the individual values is great.
2. The difference between the individual values is small.
3. All the values are equal.
4. The arithmetic mean of these values is zero.
5. All the values are negative.
The dispersion of a data set can be measured using the standard deviation, denoted . For a data set , with values and a mean , this is calculated by using the following formula:
If the dispersion of a data set is equal to zero, then the standard deviation is equal to zero. By setting the formula for the standard deviation equal to zero, we get
By squaring both sides, we get
Then, by multiplying both sides by , we have
Now, if we square any real number greater than zero, then we get a value greater than zero. Also, if we square any real number less than zero, then we still get a value greater than zero. So, the brackets must each equal zero for the result to be zero:
So, equaling each bracket to zero gives us
When solving for , we get
Therefore, all of the members of the data set are equal to the mean and are equal, which is option C.
In the next example, we will use the formula for the standard deviation of a data set to determine the standard deviation when given the sum of the squares of the differences and the number of data points.
### Example 3: Calculating Standard Deviation
If for a set of 6 values equals 25, find the standard deviation of the set, and round the result to the nearest thousandth.
To calculate the standard deviation of a set of data, we first recall the formula where denotes the standard deviation of the set of data , , is the number of members of the data set, and is the mean of the data set.
We are told , which is the same as saying . We are also told there are 6 values, which indicates that .
By substituting and and solving for , we get
Our answer is therefore 2.041 when rounded to the nearest thousandth.
Next, we will discuss how to find the standard deviation of a data set. We will explore this in detail below.
When calculating the standard deviation of a set of data, we need to execute a number of steps when working with the formula. First, letβs recall the formula where denotes the standard deviation of the set of data , , is the number of members of the data set, and is the mean of the data set.
To help demonstrate how to use the formula, we will use the following data set:
We will next execute the following steps using this data set to illustrate how the steps work.
Step 1: Finding the mean
As we need to calculate the difference between the mean and the members of the set within the brackets of the formula, we need to start by calculating the mean. This is where denotes the mean, is the data set, and is the number of points in the data set.
For the data set , this gives us
Step 2: Finding the difference between the mean and each of the data points
In order to calculate in the formula, we need to calculate for all values of , or, in other words, the difference between the mean and each of the data points. For this step and subsequent steps, it is helpful to lay this out in a table.
1
1
3
5
7
Step 3: Finding the sum of the squares of the difference between the mean and each of the data points
Following on from step 2, in order to calculate in the formula, we next need to calculate for all values of and sum this. In other words, we need to square the difference between the mean and each of the data points and sum these. We will use the table from step 2 and add a further column.
1
1
3
5
7
Summing the last column, we get
Step 4: Substituting into the formula and finding the standard deviation
For the final step, we substitute the sum of squares and in the formula and then calculate the value of the standard deviation.
From step 3, we found , and we know . Therefore, by substituting into the formula for and solving, we get which is the standard deviation for the data set .
We can summarize these steps as follows.
### How To: Finding the Standard Deviation of a Data Set
Step 1: Finding the mean of the data set
Step 2: Finding the difference between the mean and the value of each of the data points
Step 3: Finding the sum of the squares of the difference between the mean and the value of each of the data points
Step 4: Substituting the sum of the squares and into the formula and square rooting in order to calculate the standard deviation (This should always be positive.)
In the next example, we will use this process to calculate the standard deviation of a data set.
### Example 4: Calculating the Standard Deviation of a Data Set
Calculate the standard deviation of the values 45, 35, 42, 49, 39, and 34. Give your answer to 3 decimal places.
To find the standard deviation of a set of data, we use the formula where denotes the standard deviation of the set of data , , is the number of members of the data set, and is the mean of the data set.
First, we will calculate the mean, , of the data set. Recall the formula for the mean, which is
In this case, the data set is and the number of members of the data set is 6. So, by substituting for and 6 for , we get
Next, we will calculate for each member of the data set. To help ourselves do this, we will lay the data out in a table as follows:
45
35
42
49
39
34
Following this, we can now calculate . To do this, we will square for each member of the data set and then sum all the data. We will add another column to the table above for ease of calculation.
45
35
42
49
39
34
When we sum for each member of the data set, we get
We can now substitute and back into the original formula for the standard deviation and solve for :
Therefore, the answer is 5.312 when rounded to 3 decimal places.
So, the standard deviation for the data set is 5.312 correct to three decimal places.
In the next example, we will discuss which data set among three data sets has the largest dispersion by using the standard deviation.
### Example 5: Selecting a Data Set with the Highest Standard Deviation
By calculating the standard deviation, determine which of the sets , , and has the largest dispersion.
To find the standard deviation of each of the data sets, we use the formula where denotes the standard deviation of the set of data , , is the number of members of the data set, and is the mean of the data set.
We can see that each data set has four members, so is 4 for each case.
We will find the standard deviation of each data set first, then compare these in order to determine which has the largest dispersion.
For , we will first find the mean, , of the data set. Recall the formula for the mean, which is
Therefore, by substituting for and 4 for , we get
Next, we will calculate for each member of the data set. To help ourselves do this, we will lay the data out in a table as follows:
20
6
Following this, we can now calculate . To do this, we will square for each member of the data set and then sum all the data. We will add another column to the table above for ease of calculation.
20
6
When we sum for each member of the data set, we get
We can now substitute and back into the original formula for the standard deviation and solve for :
We will now repeat these steps for the other two data sets.
For , the mean is
To calculate , we will find and for each member of the data set. We will lay this out in a table as before.
5
9
Summing for each member of the data set, we get
Substituting and back into the original formula for the standard deviation and solving for , we get
For the last data set, , the mean is
To calculate , we will find and for each member of the data set. We will lay this out in a table as before.
20
Summing for each member of the data set, we get
Substituting and back into the original formula for the standard deviation and solving for , we get
We have found the standard deviation for each of the data sets. Letβs summarize this below:
• For , correct to 2 decimal places.
• For , correct to 2 decimal places.
• For , correct to 2 decimal places.
By comparing these data sets, we can see the first one, , has the largest standard deviation.
Therefore has the largest dispersion, since the standard deviation is a measure of dispersion.
So far, we have found the standard deviation of a set of data where the data has been presented in a list. Next, we will learn how to find the standard deviation from data set that is presented in a frequency table.
To find the standard deviation of a data set where the data is presented in a frequency table, we need to consider the frequency of the values in the data set as well as the values in the data set itself. One way of doing this could be to list the values. For example, consider the following data set:
31
47
53
We could write this as one 3, seven 4s, and three 5s or in order to calculate the standard deviation, as previously discussed. The problem with this approach is when there are high frequencies of data points (say 100 or even 1βββ000), as we would have to write this out in a very long list. As such, it is more efficient to calculate squares of the differences in each data set and then multiply this by the corresponding frequency (much in the same way we would calculate the mean of a set of data in a frequency table).
Before considering the formula and method for finding the standard deviation of a set of data in a frequency table, we will first recall how to calculate the mean of a set of data from a frequency table.
### Definition: The Mean of a Data Set in a Frequency Table
For a data set , with corresponding frequencies and distinct values of the data set, the mean is calculated as follows:
Another way to represent this is in a table with the values of the data set in the first column, their corresponding frequencies in the second column, the multiplication of the data point and frequency in the third column, and the sums in the last row of the table. The mean can then be calculated by dividing the sum of the third column by the sum of the second column.
Having recapped the mean of a data set in a frequency table, we will next discuss the standard deviation. The formula for this is as follows.
### Definition: The Standard Deviation of a Data Set in a Frequency Table
For a data set , with corresponding frequencies , distinct values of the data set, and mean , the standard deviation is calculated as follows:
The approach for finding the standard deviation of a data set is generally the same as the approach for finding the standard deviation of a data set in a frequency table; however, there are some important differences. As we are working with frequencies, we need to multiply each value in the data by its corresponding frequency when calculating the mean. Also, when calculating the sum of the squares of the difference between the mean and each different value of the data, we also need to multiply by the frequency.
In the next example, we will discuss how to find the standard deviation of a data set that is in a frequency table.
### Example 6: Determining the Standard Deviation of a Data Set
The table shows the distribution of goals scored in the first half of a football season.
Number of Goals Number of Games 0 1 3 4 6 5 2 7 7 4
Find the standard deviation of the number of goals scored. Give your answer to three decimal places.
As the data presented in this question is in the form of a frequency table, in order to calculate the standard deviation , we use the formula where represents the values of the data set with corresponding frequencies , there are distinct values of the data set, and the mean is represented by .
In this question, the values of the data set are the number of goals scored in the first half of a football season. The number of games refers to the frequency with which each of these goals was scored. Letβs rewrite this using and as the headings and by transposing the table, as follows:
05
12
37
47
64
To calculate the standard deviation, we must first calculate the mean . For a set of data with corresponding frequencies and distinct values of the data set, we use the following formula:
Using the table above, we can add a new column in order to find for each value of and then use this to find the mean.
05
12
37
47
64
By summing the values for and dividing the sum of the frequencies, we get
Next, we will calculate the difference between each value of the data set and the mean and the square of this in order to calculate the sum of the squares. We will do this by adding two further columns to the table above.
05
12
37
47
64
We now need to calculate the product of squares of the differences of the mean and values of the data and the frequencies of the values of the data set. We will add another column to the table to do this.
05
12
37
47
64
We are now ready to find the standard deviation. We will substitute the values from the table into the formula of the standard deviation and solve for : which is 1.960 to three decimal places.
Therefore, the standard deviation of the number of goals scored is 1.960 to three decimal places.
Next, we will discuss how to calculate the standard deviation of grouped data using the midpoint. This approach involves the same steps as with frequency tables, but we are dealing with intervals for our data set rather than a set of values; then, we need to use the midpoint in order to approximate the set of values. We will explore this further in our final example.
### Example 7: Find the Standard Deviation of a Grouped Data Set
A quiz was completed by 92 students and their scores were recorded in the following frequency table. Find the standard deviation to two decimal places.
Score Frequency 0<π β€20 20<π β€40 40<π β€60 60<π β€80 80<π β€100 26 10 24 5 27
As the data presented in this question is in the form of a frequency table, in order to calculate the standard deviation , we use the formula where represents the values of the data set with corresponding frequencies , there are distinct values of the data set, and the mean is represented by .
For this type of problem, we have been given different βclassesβ of values represented by intervals rather than exact values. This means we cannot directly apply the formula above, since we cannot substitute these intervals for the values of in our formula.
Instead, the approach we must take is to find the βmidpointβ of each interval and use this to represent the corresponding value of . After doing so, we can treat the problem as we would with any other grouped frequency table.
To find the midpoint, we add together the endpoints and divide by 2. This allows us to find an approximate standard deviation of the data set.
So, the values of the data set are the midpoint of each of the scores obtained in the quiz and the corresponding frequencies are the frequencies for each of the values. Letβs find the midpoint of each of the intervals and then rewrite the midpoints as and the frequencies as , as follows:
IntervalMidpoint Frequency
26
10
24
5
27
To calculate the standard deviation, we must first calculate the mean, . For a set of data with corresponding frequencies and distinct values of the data set, we use the following formula:
Again, we remember that the midpoint is now being used to represent the values of . Using the table above, we can add a new column in order to find for each value of and then use this to find the mean.
IntervalMidpoint Frequency
26
10
24
5
27
By summing the values for and dividing the sum of the frequencies, we get
Next, we will calculate the difference between the midpoints of each class in our data set and the mean and the square of this in order to calculate the sum of the squares. We will do this by adding two further columns to the table above. Note that all values have been rounded to 4 decimal places.
IntervalMidpoint Frequency
261βββ548.2494
10374.3375
240.4254
5426.5134
271βββ652.6014
We now need to calculate the product of squares of the differences of the mean and midpoints of the data and the frequencies of the values of the data set. We will add another column to the table to do this. Again, we will round to 4 decimal places.
IntervalMidpoint Frequency
26
1βββ548.2493640βββ254.4818
10
374.3373653βββ743.3736
24
0.4253648410.2088
5
426.5133652βββ132.5668
27
1βββ652.6013644βββ620.2351
We are now ready to find the standard deviation. We will substitute the values from the table into the formula of the standard deviation and solve for : which is 31.41 when rounded to 2 decimal places.
Therefore, the standard deviation is 31.41 to 2 decimal places.
In this explainer, we have learned what the standard deviation is and how to find it for a set of data, from both a list and a frequency table. We have also learned how to compare data sets and draw conclusions using the standard deviation.
### Key Points
• The standard deviation of a data set is used to measure the dispersion of data from the mean.
• For data presented in a list, the formula for the standard deviation of a set of data with members and mean is
• For data presented in a frequency table, the formula for the standard deviation of a data set , with corresponding frequencies , distinct values of the data set, and mean is
• For grouped frequency tables where data is given in intervals, the midpoint of the interval is used to represent the values of . |
# Using Cramer`s Rule to Solve Three Equations With Three Unknowns
```Using Cramer’s Rule to Solve Three Equations With Three Unknowns
Here we will be learning how to use Cramer’s Rule to solve a linear system with three equation and
three unknowns. Cramer’s Rule is one of many techniques that can by used to solve systems of linear
equations. Cramer’s Rule involves the use of determinants to find the solution and like any other
technique it has its advantages and disadvantages. Cramer’s Rule itself is very simple, but the notation
used requires a little bit of explanation, so let’s take a look at Cramer’s Rule.
Cramer’s Rule
To help explain the notation, consider the following system of equations:
To find the values of x, y, and z there are four different values that we need to calculate, they are D, Dx,
Dy, and Dz. In all four cases the “D” stands for the determinant, now let’s look at what they represent.
|
Here we use the x, y, and z values from the problem to
create a 3×3 matrix.
|
|
|
|
|
|
|
Here we replace the x-values in the first column with the
values after the equal sign and leave the values in the y
and z columns unchanged.
Here we replace the y-values in the second column with
the values after the equal sign and leave the values in the x
and z columns unchanged.
Here we replace the z-values in the third column with the
values after the equal sign and leave the values in the x
and y columns unchanged.
Once we have calculated the values of D, Dx, Dy, and Dz we can apply Cramer’s Rule to find x, y, and z.
To use Cramer’s Rule to solve a system of three equations with three unknowns, we need to follow these
steps:
Step 1: Find the determinant, D, by using the x, y, and z values from the problem.
Step 2: Find the determinant, Dx, by replacing the x-values in the first column with
the values after the equal sign leaving the y and z columns unchanged.
Step 3: Find the determinant, Dy, by replacing the y-values in the second column
with the values after the equal sign leaving the x and z columns unchanged.
Step 4: Find the determinant, Dz, by replacing the z-values in the third column with
the values after the equal sign leaving the x and y columns unchanged.
Step 5: Use Cramer’s Rule to find the values of x, y, and z.
Using Cramer’s Rule to Solve Three Equations With Three Unknowns – Notes
Page 1 of 4
Now we are ready to look at a couple of examples. To review how to calculate the determinant of a 3×3
Example 1: Use Cramer’s Rule to solve
.
Step 1: Find the determinant, D, by using the x, y, and z values from the problem.
|
|
|
|
Step 2: Find the determinant, Dx, by replacing the x-values in the first column with the values
after the equal sign leaving the y and z columns unchanged.
|
|
|
|
Step 3: Find the determinant, Dy, by replacing the y-values in the second column with the
values after the equal sign leaving the x and z columns unchanged.
|
|
|
|
(
)
Step 4: Find the determinant, Dz, by replacing the z-values in the third column with the values
after the equal sign leaving the x and y columns unchanged.
|
|
|
|
Step 5: Use Cramer’s Rule to find the values of x, y, and z.
Generally, the answer is written as on order triple (4, –2, 1) representing the x, y, and z values.
Using Cramer’s Rule to Solve Three Equations With Three Unknowns – Notes
Page 2 of 4
Example 2: Use Cramer’s Rule to solve
.
Step 1: Find the determinant, D, by using the x, y, and z values from the problem.
|
|
|
(
|
)
Step 2: Find the determinant, Dx, by replacing the x-values in the first column with the values
after the equal sign leaving the y and z columns unchanged.
|
|
|
|
Step 3: Find the determinant, Dy, by replacing the y-values in the second column with the
values after the equal sign leaving the x and z columns unchanged.
|
|
|
|
(
)
Step 4: Find the determinant, Dz, by replacing the z-values in the third column with the values
after the equal sign leaving the x and y columns unchanged.
|
|
|
|
Step 5: Use Cramer’s Rule to find the values of x, y, and z.
Answer written as on order triple: (
)
Using Cramer’s Rule to Solve Three Equations With Three Unknowns – Notes
Page 3 of 4
Advantages – I find that one of the advantages to Cramer’s Rule is that you can find the value of x,
y, or z without having to know any of the other values of x, y, or z. For example, if you needed to
find just the value of y, Cramer’s Rule would work well. Another thing that I like about Cramer’s
Rule is that if any of the values of x, y, or z is are fractions, you do not have to plug in a fraction to
find the other values. Each value can be found independently.
Disadvantages – One of the only disadvantages to using Cramer’s rule is if the value of D is zero
then Cramer’s Rule will not work because you cannot divide by zero. However, if the value of D is
zero then you know that the solution is either “No Solution” or “Infinite Solutions”. You will have
to use a different technique such as Addition/Elimination to find out whether the answer is “No
Solution” or “Infinite Solutions”.
Using Cramer’s Rule to Solve Three Equations With Three Unknowns – Notes
Page 4 of 4
``` |
NCERT Exemplar: Vectors
# NCERT Exemplar: Vectors | Mathematics (Maths) Class 12 - JEE PDF Download
SHORT ANSWER TYPE QUESTIONS
Q.1. Find the unit vector in the direction of sum of vectors
Ans.
Given that
∴ Unit vector in the direction of
Hence, the required unit vector is
Q.2. Iffind the unit vector in the direction of
(i)
(ii)
Ans.
Given that
(i)
∴ Unit vector in the direction of
Hence, the required unit vector is
(ii)-
∴ Unit vector in the direction of
Hence, the required unit vector is
Q.3. Find a unit vector in the direction ofwhere P and Q have co-ordinates (5, 0, 8) and (3, 3, 2), respectively.
Ans.
Given coordinates are P(5, 0, 8) and Q(3, 3, 2)
∴ Unit vector in the direction of
Hence, the required unit vector is
Q.4. If are the position vectors of A and B, respectively, find the position vector of a point C in BA produced such that BC = 1.5 BA.
Ans.
Given that
BC = 1.5 BA
Hence, the required vector is
Q.5. Using vectors, find the value of k such that the points (k, – 10, 3), (1, –1, 3) and (3, 5, 3) are collinear.
Ans.
Let the given points are A( k , - 10 , 3), B(1,- 1, 3) and C(3, 5, 3)
If A, B and C are collinear, then
Squaring both sides, we have
= 9 + k2- 6k + 225
Dividing by 2, we get
(Dividing by 2)
Squaring both sides, we get
⇒ 10(k2 – 2k + 82) = 784 + k2 – 56k
⇒ 10k2 – 20k + 820 = 784 + k2 – 56k
⇒ 10k2 – k2 – 20k + 56k + 820 – 784 = 0
⇒ 9k2 + 36k + 36 = 0
⇒ k2 + 4k + 4 = 0
⇒ (k + 2)2 = 0
⇒ k + 2 = 0
⇒ k = – 2
Hence, the required value is k = – 2
Q.6. A vectoris inclined at equal angles to the three axes. If the magnitude of is 2 √3 units, find
Ans.
Since, the vectormakes equal angles with the axes, their direction cosines should be same
∴ l = m = n
We know that
l2 + m2 + n2 = 1
⇒ l2 + l2 + l2 = 1
We know that
Hence, the required value of
Q.7. A vector has magnitude 14 and direction ratios 2, 3, – 6. Find the direction cosines and components of , given that makes an acute angle with x-axis.
Ans.
Letbe three vectors such thatand
If l, m and n are the direction cosines of vector , then
We know that l2 + m2 + n= 1
∴ k = ± 2 and l =
Hence, the required direction cosines areand the components of
Q.8. Find a vector of magnitude 6, which is perpendicular to both the vectors and
Ans.
Letand
We know that unit vector perpendicular to
Now the vector of magnitude 6 =
Hence, the required vector is
Q.9. Find the angle between the vectors.
Ans.
Let and let θ be the angle between
Hence, the required value of θ is
Q.10. If show that Interpret the result geometrically?
Ans.
Given that
So,
...(i)
Now
...(ii)
From eq. (i) and (ii) we get
Hence proved.
Geometrical Interpretation
According to figure, we have
Area of parallelogram ABCD is
Since, the parallelograms on the same base and between the same parallel lines are equal in area
Q.11. Find the sine of the angle between the vectors and
Ans.
Given that
We know that
Hence,
Q.12. If A, B, C, D are the points with position vectors
respectively, find the projection of
Ans.
Here, Position vector of A =
Position vector of B =
Position vector of C =
Position vector of D =
Projection of
Hence, the required projection = √21 .
Q.13. Using vectors, find the area of the triangle ABC with vertices A(1, 2, 3), B(2, – 1, 4) and C(4, 5, – 1).
Ans.
Given that A(1, 2, 3), B(2, –1, 4) and C(4, 5, –1)
Area of ΔABC ==
=
Hence, the required area is
Q.14. Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area.
Ans.
Let ABCD and ABFE be two parallelograms on the same base AB and between same parallel lines AB and DF.
Let
∴ Area of parallelogram ABCD =
Now Area of parallelogram ABFE =
Hence proved.
LONG ANSWER TYPE QUESTIONS
Q.15. Prove that in any triangle ABC,where a, b, c are the magnitudes of the sides opposite to the vertices A, B, C, respectively.
Ans.
Here, in the given figure, the components of c are c cos A and c sin A.
In DBDC,
a2 = CD2 + BD2
⇒ a2 = (b – c cos A)+ (c sin A)2
⇒ a2 = b2 + c2 cos2 A – 2bc cos A + c2 sin2 A
⇒ a2 = b2 + c2(cos2 A + sin2 A) – 2bc cos A
⇒ a2 = b2 + c2 – 2bc cos A
⇒ 2bc cos A = b2 + c2 - a2
Hence Proved.
Q.16. Ifdetermine the vertices of a triangle, show that
gives the vector area of the triangle. Hence deduce the condition that the three points are collinear. Also find the unit vector normal to the plane of the triangle.
Ans.
Since,are the vertices of ΔABC
For three vectors are collinear, area of ΔABC = 0
which is the condition of collinearity of
Letbe th e unit vector normal to the plane of the ΔABC
Q.17. Show that area of the parallelogram whose diagonals are given by is. Also find the area of the parallelogram whose diagonals are and
Ans.
Let ABCD be a parallelogram such that,
∴ by law of triangle, we get
Adding eq. (i) and (ii) we get,
Subtracting eq. (ii) from eq. (i) we get
So, the area of the parallelogram ABCD =
Now area of parallelogram whose diagonals areand
Hence, the required area is
Q.18. Iffind a vector such that
Ans.
Let
Also given that
Since,
On comparing the like terms, we get
c3 – c2 = 0 ...(i)
c1 – c3 = 1 ...(ii)
and c2 – c1 = –1 ...(iii)
Now
∴ c1 + c2 + c3 = 3 ...(iv)
Adding eq. (ii) and eq. (iii) we get,
c2 – c3 = 0 ...(v)
From (iv) and (v) we get
c1 + 2c2 = 3 ...(vi)
From (iii) and (vi) we get
c3 – c2 = 0
Now c2 – c1 = – 1 ⇒
Hence,
OBJECTIVE TYPE QUESTIONS
Q.19. The vector in the direction of the vectorthat has magnitude 9 is
(a)
(b)
(c)
(d)
Ans. (c)
Solution.
Let
Unit vector in the direction of
∴ Vector of magnitude 9 =
Hence, the correct option is (c).
Q.20. The position vector of the point which divides the join of points
in the ratio 3 : 1 is
(a)
(b)
(c)
(d)
Ans. (d)
Solution.
The given vectors areand the ratio is 3 : 1.
∴ The position vector of the required point c which divides the join of the given vectors
Hence, the correct option is (d).
Q.21. The vector having initial and terminal points as (2, 5, 0) and (–3, 7, 4), respectively is
(a)
(b)
(c)
(d)
Ans. (c)
Solution.
Let A and B be two points whose coordinates are given as (2, 5, 0) and (– 3, 7, 4)
Hence, the correct option is (c).
Q.22. The angle between two vectorswith magnitudes √3 and 4, respectively, and
(a) π/6
(b) π/3
(c) π/2
(d) 5π/2
Ans. (b)
Solution.
Here, given that
∴ From scalar product, we know that
Hence, the correct option is (b).
Q.23. Find the value of λ such that the vectors
are orthogonal
(a) 0
(b) 1
(c) 3/2
(d) -5/2
Ans. (d)
Solution.
Given that
Sinceare orthogonal
⇒ 2 + 2λ + 3 = 0
⇒ 5 + 2λ = 0 ⇒
Hence, the correct option is (d).
Q.24. The value of λ for which the vectors
are parallel is
(a) 2/3
(b) 2/3
(c) 5/2
(d) 2/5
Ans. (a)
Solution.
Let
Since the given vectors are parallel,
∴ Angle between them is
so
Squaring both sides, we get
900 + λ2 + 60λ = 46(20 + λ2)
⇒ 900 + λ2 + 60λ = 920 + 46λ2
⇒ λ2 – 46λ2 + 60λ + 900 – 920 = 0
⇒ - 45λ2 + 60λ - 20 = 0
⇒ 9λ2 – 12λ + 4 = 0
⇒(3λ – 2)2 = 0
⇒3λ – 2 = 0
⇒ 3λ = 2
∴ λ = 2/3
Alternate method:
Let
If
Hence, the correct option is (a).
Q.25. The vectors from origin to the points A and B are
,respectively, then the area of triangle OAB is
(a) 340
(b) √25
(c) √229
(d)
Ans. (d)
Solution.
Let O be the origin
and
∴ Area of ΔOAB =
Hence the correct option is (d).
Q.26. For any vectorthe value ofis equal to
(a)
(b)
(c)
(d)
Ans. (d)
Solution.
Let
Now,
Similarly
and
Hence, the correct option is (d).
Q.27. Ifthen value ofis
(a) 5
(b) 10
(c) 14
(d) 16
Ans. (d)
Solution.
Given that
⇒ 12 = 10 × 2 × cos θ
Now
Hence, the correct option is (d).
Q.28. The vectorsare coplanar if
(a) λ = –2
(b) λ = 0
(c) λ = 1
(d) λ = – 1
Ans. (a)
Solution.
Let
Ifare coplanar, then
⇒ l(λ2 – 1) – 1 (λ + 2) + 2(–1 – 2l) = 0
⇒ λ3 – λ – λ – 2 – 2 – 4λ = 0
⇒ λ3 – 6λ – 4 = 0
⇒ (λ + 2) (λ2 – 2λ – 2) = 0
⇒ λ = – 2 or λ2 – 2λ – 2 = 0
Hence, the correct option is (a).
Q.29. Ifare unit vectors such thatthen the value of
(a) 1
(b) 3
(c) -3/2
(d) None of these
Ans. (c)
Solution.
Given that
and
Hence, the correct option is (c).
Q.30. Projection vector ofis
(a)
(b)
(c)
(d)
Ans. (a)
Solution.
The projection vector of
Hence, the correct option is (a).
Q.31. Ifare three vectors such that
then value of
(a) 0
(b) 1
(c) – 19
(d) 38
Ans. (c)
Solution.
Given that
and
Hence, the correct option is (c).
Q.32. Ifand −3 ≤ λ ≤ 2 , then the range ofis
(a) [0, 8]
(b) [– 12, 8]
(c) [0, 12]
(d) [8, 12]
Ans. (b)
Solution.
Given that
Now
Here - 3 ≤ λ ≤ 2
⇒ - 3.4 ≤ 4λ ≤ 2.4
⇒ - 12 ≤ 4λ ≤ 8
∴ 4λ = [- 12, 8]
Hence, the correct option is (b).
Q.33. The number of vectors of unit length perpendicular to the vectorsand
(a) one
(b) two
(c) three
(d) infinite
Ans. (b)
Solution.
The number of vectors of unit length perpendicular to vectors
So, there will be two vectors of unit length perpendicular to vectors
Hence, the correct option is (b).
FILL IN THE BLANKS
Q.34. The vectorbisects the angle between the non-collinear vectorsif ________.
Ans.
If vector bisects the angle between non-collinear vectors
then the angle betweenis equal to the angle between
So, ...(i)
Also,[∵ θ is same] ...(ii)
From eq. (i) and eq. (ii) we get,
Hence, the required filler is
Q.35. Iffor some non-zero vectorthen the value of is ________
Ans.
Ifis a non-zero vector, thencan be in the same plane.
Since angles between and are zero i.e. θ = 0
Hence the required value is 0.
Q.36. The vectorsa re the adjacent sides of a parallelogram. The acute angle between its diagonals is ________.
Ans.
Given that
and
Let θ be the angle between the two diagonal vectors
then
Hence the value of required filler is
Q.37. The values of k for which is parallel toholds true are _______.
Ans.
Given that
Now sinceis parallel to
Here we see that atbecome null vector and then it will not be
parallel to
Now sinceis parallel to
Here we see that atbecome null vector and then it will not be
parallel to
Hence, the required value of k ∈ (- 1, 1) and k ≠
Q.38. The value of the expressionis _______.
Ans.
Hence, the value of the filler is
Q.39. Ifandis equal to _______.
Ans.
Hence, the value of the filler is 3.
Q.40. Ifany non-zero vector, then
equals _______.
Ans.
Let
= a1
Hence, the value of the filler is
State True or False in each of the following Exercises.
Q.41. Ifthen necessarily it implies
Ans.
Ifthenwhich is true.
Hence, the statement is True.
Q.42. Position vector of a point P is a vector whose initial point is origin.
Ans.
True
Q.43. Ifthen the vectors
are orthogonal.
Ans.
Given that
Squaring both sides, we get
which implies thatare orthogonal.
Hence the given statement is True.
Q.44. The formulais valid for non-zero vectors
Ans.
Hence, the given statement is False.
Q.45. Ifare adjacent sides of a rhombus, then
Ans.
If
So the angle between the adjacent sides of the rhombus should be 90° which is not possible.
Hence, the given statement is False.
The document NCERT Exemplar: Vectors | Mathematics (Maths) Class 12 - JEE is a part of the JEE Course Mathematics (Maths) Class 12.
All you need of JEE at this link: JEE
## Mathematics (Maths) Class 12
205 videos|264 docs|139 tests
## FAQs on NCERT Exemplar: Vectors - Mathematics (Maths) Class 12 - JEE
1. What is a vector?
Ans. A vector is a mathematical quantity that has both magnitude and direction. It is represented by an arrow, where the length of the arrow represents the magnitude and the direction of the arrow represents the direction of the vector.
2. How is the magnitude of a vector calculated?
Ans. The magnitude of a vector is calculated by using the Pythagorean theorem. It is the square root of the sum of the squares of its components. For example, if a vector has components (a, b, c), then its magnitude is given by √(a^2 + b^2 + c^2).
3. What is the difference between a scalar and a vector quantity?
Ans. A scalar quantity only has magnitude and no direction, while a vector quantity has both magnitude and direction. Examples of scalar quantities include temperature and mass, while examples of vector quantities include velocity and force.
4. How are vectors represented mathematically?
Ans. Vectors can be represented mathematically using various notations. One common notation is to write the components of the vector as a column or row matrix. For example, a vector A = (a1, a2, a3) can be represented as a column matrix [a1, a2, a3] or as a row matrix [a1, a2, a3]T, where T represents the transpose of the matrix.
5. What is the dot product of two vectors?
Ans. The dot product of two vectors is a scalar quantity that is calculated by multiplying the magnitudes of the vectors and the cosine of the angle between them. It is denoted by a dot (·) between the two vectors. The dot product can be used to determine the angle between two vectors or to calculate the work done by a force.
## Mathematics (Maths) Class 12
205 videos|264 docs|139 tests
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# How To Find The Domain Of A Function
## What Is Domain Of A Function?
The space of a capacity is the finished arrangement of potential qualities of the autonomous variable.
Clearly, this definition means:
The domain is the set of all possible x values that will make the function “work” and produce real y values.
When you find the domain, remember:
• The denominator (at the bottom) of a fraction cannot be zero
• The number under the square root sign must be positive in this section.
## Approaches to discover the area of a capacity
The domain of a function is the set of numbers that can enter a given function. At the end of the day, it is the arrangement of x qualities that you can place into some random condition. The arrangement of conceivable y esteems is called the reach. In the event that you need to realize how to discover the space of a capacity in different circumstances, follow these means.
### 1. Get to know the definition of the domain.
The domain is defined as the set of input values for which the function generates an output value. In other words, the domain is the complete set of x-values that can be inserted into a function to produce a y-value.
### 2. Learn how to find the domain of a wide variety of functions.
The role type determines the best way to find a domain. Here are the basics you need to know about each type of function, explained in the next section:
• A polynomial function without radical or variable in the denominator. For this type of function, the domain is made up of real numbers.
• A capacity with a part with a variable in the denominator. To find the domain of this type of function, set the lower value to zero and exclude the x-value found by solving the equation.
• A function with one variable within a radical sign. To discover the space of this kind of capacity, just set the terms inside the extreme sign to> 0 and address to discover the qualities that would work for x.
• A function that uses the natural protocol (ln). Just put the terms in brackets> 0 and solve.
• A graph. Take a look at the graph to see which values work for x.
• A relation. This is a list of the x and y coordinates. Your space will basically be a rundown of x directions.
### 3. Enter the domain correctly.
Mastery-appropriate notation is easy to learn, but it is important that you spell it correctly to express the correct answer and get all points on assignments and tests. Here are some things you need to know about how to write the domain of a function:
• The format for expressing the domain is an open bracket/parenthesis, followed by the 2 domain endpoints separated by a comma, followed by a closed bracket/parenthesis.
• For example, [-1,5). This means that the domain ranges from -1 to 5.
• Use square brackets like [and] to indicate that a number is included in the domain.
• So in the model, [-1,5), space incorporates - 1.
• Use parentheses like (and) to indicate that a number is not included in the domain.
• So in the model, [-1,5), 5 is excluded from the area. The domain arbitrarily stops below 5, that is, 4,999 …
• Use “U” (which stands for “union”) to connect parts of the domain that are separated by a space. ’
• For example, [-1,5) U (5,10]. This means that the domain ranges from -1 to 10 inclusive, but there is a gap in the domain at 5. This could be the aftereffect of, for instance, a capacity with “x - 5” in the denominator.
• You can use the same number of “U” images as vital if the space has various holes.
• Use infinity and negative infinity signs to express that the domain continues infinitely in any direction.
• Always use (), not [], with infinite symbols.
• Please note that this notation may differ depending on where you live.
• The rules described above apply to the UK and the US.
• Some regions use arrows instead of infinity signs to express that the domain continues infinitely in any direction.
• The use of square brackets varies greatly between regions. For example, Belgium uses back brackets instead of round brackets.
## Finding The Domain Without Graph
OK, so let’s say we don’t have a graph of a function to look at like in the last section …
Can we still find the domain and area?
Domains: Yes (as long as algebra doesn’t
get too hairy … and it won’t be for us.)
Areas: Not really (usually you need the
Picture - unless there is something
really easy.)
So we’re only doing domains on these - where’s the real action anyway.
"What are all possible x guys
that I can stay in this thing? "
Sometimes what you’re really looking for is
“Is there something I can’t stay with?”
Listen:
Let’s find the domain of
Do you see any x folks that would cause an issue here?
f(3) = 2/(3-3) = 2/0
So, here x = 3 is creating a problem! Everything else is right
Space is all genuine numbers aside from 3.
What would be the interval notation?
When in doubt, draw it on a number line:
Do the interval notation in two pieces:
domain
## The most effective method to discover the area of a levelheaded capacity
Here are the steps required to find the domain of a rational function:
Step 01: A rational function is just a fraction and in a fraction, the denominator cannot be zero because it would be undefined. To discover the numbers that make the portion vague, make a condition where the denominator isn’t zero.
Step 02: Solve the condition derived in step 1.
Example 1
Step 001: A rational function is simply a fraction and in a fraction, the denominator cannot be equal to zero because it would not be defined. To discover which numbers make the division uncertain, make a condition where the denominator isn’t zero
Step 002: Solve the condition found in step 1. For this situation, take away 4 from each side.
Step 003: Write your answer using interval notation. In this case, since x ≠ –4 we get:
Example 2
Step 0001: A rational function is simply a fraction and in a fraction, the denominator cannot be zero because it would be undefined. To find out which numbers make the fraction undefined, create an equation where the denominator is non-zero.
Step 0002: Solve the condition found in step 1. In this case, we must take into account the problem.
Step 0003 : Write your answer using interval notation. In this case, from x ≠ –2 and x ≠ 7 we get:
## Step by step instructions to Find The Domain Of A Function Algebraically
When trying to find the domain of a function algebraically, it can be helpful to find any values that CANNOT be in the domain - these are the values that would “break” the function or make it undefined. Values that make you divide by 0, take the square root of a negative real number, or take the log of a non-positive number will not be in your range, because they give undefined functions.
Case 1: Dividing by 0
If you divide by 0
Their function is undefined. So a value that gives you a 0 in the denominator of your function is not in the domain. For example, let’s look at the function f (x) = 2x - 4.
What value of x causes the denominator to be 0?
x−4=0
→x=4
When x=4,then f(4)=2/0, which is undefined. Thus, 4 cannot be in our domain. Along these lines, our space is (−∞, 4) ∪ (4,∞).
## How to find the domain of a square root function
To discover the area of square root work, settle the imbalance x ≥ 0 with x supplanted by the radicand. Using one of the examples above, you can find the domain of f (x) = 2√ (x + 3) by setting the radicand (x + 3) equal to x in the inequality. This gives you the inequality x + 3 ≥ 0, which you can solve by subtracting 3 from both sides. This gives you a solution of x ≥ -3, which means that your domain has all values of x greater than or equal to -3. You can also write this as [-3, ∞), with the left bracket showing that -3 is a specific limit while the right parenthesis shows that ∞ is not. Since the radicand cannot be negative, you only have to calculate the positive or zero values.
Here are the steps required to find the domain of a square root function:
Step 1a: Set the expression within the square root greater than or equal to zero. We do this because only non-negative numbers have a real square root, in other words, we cannot take the square root of a negative number and get a real number, which means that we have to use numbers that are greater than or equal to zero.
Step 2a: Solve the equation found in step 1. Remember that when you are solving equations that involve inequalities, if you multiply or divide by a negative number, you must reverse the direction of the inequality symbol.
Step 3a: Write the answer using interval notation.
Example 1 – Find the Domain of the Function:
Step 1b Put the expression inside the square root greater than or equal to zero.
Step 2b : Solve the equation found in step 1.
Step 3b : Write the answer using interval notation.
### How to find the domain of a compound function
the domain of a composite function like f∘g depends on the domain of g and the domain of f. It is important to know when we can apply a compound function and when not, that is, to know the domain of a function like f∘g. Suppose we know the domains of the functions f and g separately. On the off chance that we compose the compound capacity for an information x as f (g (x)), we can promptly see that x should be an individual from the space of g for the expression to be meaningful because otherwise, we cannot complete the evaluation of the function internal. Nonetheless, we likewise see that g (x) should be an individual from the area of f; in any case, the second assessment of the capacity in f (g (x)) can’t be finished and the articulation isn’t yet characterized. Hence, the space of f∘g comprises just of those passages in the area of g that produce yields of g that have a place with the area of f. Note that the domain of f composed of g is the set of all x such that x is in the domain of g and g (x) is in the domain of f.
#### Example:
Find the domain of
(f∘g) (x) where f (x) = 5 / x - 1 and g (x) = 4 / 3x - 2
#### Solution:
The domain of g (x) g (x) consists of all real numbers except x = 2/3 since this input value would make us divide by 0. Likewise, the domain of ff includes all real numbers except 1 So, we need to exclude from the domain of g (x) that value of x for which g (x) = 1.
4 / 3x - 2 = 1 Set g (x) equal to 1
4 = 3x - 2 Multiply by 3x - 2
6 = 3x Add 2 on both sides
x = 2 Divide by 3
The domain of f∘g is therefore the set of all real numbers except 2/3 and 2. This means that
x ≠ 2/3 or x ≠ 2x ≠ 2
We can write this in interval notation like
(−∞, 2/3) ∪ (2 / 3,2) ∪ (2, ∞)
### 2. How do I find the domain of a logging function?
Before working with graphs, let’s take a look at the domain (the set of input values) for which the logarithmic function is defined.
Recall that the exponential function is defined as \ displaystyle y = {b} ^ {x} y = b x
For any real number x and constant \ displaystyle b> 0b> 0,\displaystyle b\ne 1b ≠ 1, where
### 3. How to find the domain of a radical function?
Here are the steps required to find the domain of a radical function:
Step 1c: Determine the index of the radical. If the index is an odd number, such as a cubic root or a fifth root, then the domain of the function is made up of real numbers, which means you can skip steps 2 and 3 and go directly to step 4 If the index is an even number, such as a square root or a fourth root, To then find the domain, the expression within the radical must be greater than or equal to zero. The space of an extreme capacity can be summed up as follows:
Step 2c: If the index is an even number, define the expression inside the radical greater than or equal to zero.
Step 3c: Solve the equation found in step 2. Remember that when you solve equations involving inequalities, if you multiply or divide by a negative number, you must reverse the direction of the symbol d 'inequality.
Step 4c: Write the answer using interval notation.
Example 1 – Find the Domain of the Function:
Step 1d: Determine the index of the radical. In this case, the index is even.
Step 2d: If the index is an even number, set the expression inside the radical greater than or equal to zero
Step 3d: Solve the equation found in step 2.
Step 4d: Write the answer using interval notation. |
# What is the derivative of y＝lnx/ x?
Jan 29, 2016
$y ' = \frac{1 - \ln x}{x} ^ 2$
#### Explanation:
Use the quotient rule, which states that
$\frac{d}{\mathrm{dx}} \left[\frac{f \left(x\right)}{g \left(x\right)}\right] = \frac{f ' \left(x\right) g \left(x\right) - g ' \left(x\right) f \left(x\right)}{g \left(x\right)} ^ 2$
Applying this to $y = \ln \frac{x}{x}$, we see that
$y ' = \frac{x \frac{d}{\mathrm{dx}} \left[\ln x\right] - \ln x \frac{d}{\mathrm{dx}} \left[x\right]}{x} ^ 2$
Since $\frac{d}{\mathrm{dx}} \left[\ln x\right] = \frac{1}{x}$ and $\frac{d}{\mathrm{dx}} \left[x\right] = 1$,
$y ' = \frac{x \left(\frac{1}{x}\right) - \ln x}{x} ^ 2$
$y ' = \frac{1 - \ln x}{x} ^ 2$ |
## What's My Order?Discussion Page
### Part 1: Ordering Fractions with Like Denominators
*
Katie and Raj both have the same size submarine sandwich. Each sandwich is cut into 8 equal slices. Katie is going to eat 2 pieces of her sandwich. Raj is going to eat 4 pieces of his sandwich.
Using the virtual fraction applet, the following image can be constructed. Notice the slider can help students count out the fractions along the rectangle. Students should be able to identify that Katie will eat 2/8 of her sandwich and Raj will eat 4/8 of his sandwich. As the students use the arrows to shade in the rectangle, they will see the name of the fractional part of the whole on the right. They also are forced at this point to see the rectangles in an order which eventually should transfer to using a number line model.
Raj eats more of his sandwich. Keep in mind that students need to discuss that the sandwiches are the same size, so the "unit" is the same for both Katie and Raj. They also need to discuss that the sandwiches were cut into equal pieces.
Many students will be set in their thinking that since 2 is less than 4, Katie eats less food. While this is true for like denominators, students need to add into their thoughts and discussions that they are comparing the same size pieces of the sandwich.
Given that the manipulatives are virtual, ask students to open up the Pattern Block Applet and the Integer Rods Applet to see if they could represent the sandwiches with those manipulatives. If using the physical manipulatives, one can see that this could easily get out of control and student might be confused with all of the different pieces. With the virtual manipulatives, students can switch quickly from applet to applet and explore why one manipulative perhaps is better than another for their investigation. There is number sense thinking going on when one make a choice about which pieces to use out of a manipulative for an investigation.
Have students write out sentences and then write out the math symbols. This helps build the foundation for their problem solving skills later in Algebra.
One observation that can be made is that the part of the sandwich that Raj will eat is twice as big as Katie's portion. This can be seen with the virtual fractions and students should be drawn into writing out the math! They can start to establish that 2* 2/8 = 4/8. Students need to put this in perspective. Some students will stay with the familiar and write 2*2 = 4. Some will get frustrated because they are giving a correct logical answer with respect to whole number thinking. They need to be aware that we can say things from different perspectives and they need to be able to communicate from different points of view or different number sets. This needs to be established on a simple level. Students encounter different number types in life such as fractions, decimals, and percentages and if they do not have a good basic foundation about these changes of perspectives, they soon get lost and give up.
Students are asked if Katie took 2 pieces and Raj took 4 pieces of the same sandwich, would one sandwich be enough to feed them? Although the virtual fraction applet does not allow for the students to move the pieces "off of the sandwich," students can use the arrow buttons to "add on" Raj's pieces to Katie "sandwich." Point out that the pieces are all alike so it is a simple add on problem and only 6 of the 8 pieces will be eaten! At this point, the symbols 2/8 + 4/8 = 6/8 can be written. Again, some students will not see the need to use fractions and will be set on writing 2 + 4 = 6. This again is a situation where the student needs to learn to change perspectives. What is one unit in this problem, the sandwich or the pieces? This is not a one day activity for many students and this idea needs to be addressed over and over when developing their number sense using fractions.
#### Part 1 Problems
1. One fifth is less than three fifths. 2. Three fourths is greater than 1 fourth.
3. Write a story using the fractions in problem 1. Answers will vary. Look for an understanding of the unit in the problem. Make sure that the students clearly state the unit which is broken up into 5 pieces. Encourage students to rewrite their work to refine their thinking after discussion about their response. Students could talk about their stories in a group and each could explain the "math" in their story. This activity amounts to making up a "word problem" or "story problem" about the numbers. Invite students to think about where they might see these number comparison used outside of school.
### Part 2: Ordering Fractions with Unlike Denominators
Judy and Jay both have the same size submarine sandwich. Judy's sandwich is cut into 12 equal pieces. Jay's sandwich is cut into 6 equal pieces. Judy will eat 5 pieces of her sandwich for lunch. Jay will eat 4 pieces of her sandwich for lunch.
This problem was chosen so that the denominators are factors or multiples of each other. This example will open the door to other discussions about fractions such as equivalent fractions.
1. Look at the "sandwiches" you made.
• Which fraction is less than one half?
• Which fraction is greater than one half?
Add other rectangles to explore this question. Use the slider to compare the fractions. Looking at the fraction as numbers, write an explanation of why one of the fractions is less than one half and one is greater than one half.
As seen in the picture above, students can compare the position of 5/12 and 4/6 with respect to one half. Students can add another rectangle and use the slider to see that the quantities are to either side of one half. Recall that the NCTM standards call for students to learn to compare fractions by using a simple known fractional quantity such as one half at this level.
Open a discussion about the numerators and denominators of these fractions. Ask how many 12ths are needed to build up to 1/2 on the their rectangle. Engage them in a discussion which brings out the fact that 6 of the 12ths are need to make 1/2 so then they can reason 5 of the 12ths certainly is smaller than 1/2 or symbolically 5/12 < 1/2. Similarly, repeat the discussion for 4/6.
Notice, at this point, the rectangles are helping students move to see these fractions as on a number line.
2. If Judy and Jay combined their pieces of their sandwiches, would one sandwich have been enough for their lunch? Use the rectangles and write an explanation of your answer. Will they have any part of the sandwich left over?
Again, the fractional pieces do not move but students can see that 2 of the 12ths line up with 1 of the 6ths using the slider. If they count off along one of the "sandwiches," they will see that 5/12 + 4/6 seems to be 1 and 1/12 or 13/12. One sandwich will not be enough! The structure of this problem was carefully chosen. This gives students a picture of why "units" will have to be broken into same sizes in order to add (or subtract). Notice they can do these types of problems using manipulatives to reason out the answer without the algorithm. If we gave a problem that didn't "line up," students would not be able to use simple reasoning from their whole number sense. Here they just used that 6 * 2 = 12. Doing several problems of this structure can give students the opportunity to come up with a reasonable way to add fractions. Moving later to problems where the denominators "don't match up" in terms of the rectangles, opens the door later to discuss how the factors of the denominators will become important in the addition or subtraction of fractions.
Again, an interesting activity would be to have students model this problem with the Pattern Blocks Applet and the Integer Rods Applet to see if they could build these sandwiches!
Notice, it is not an easy task to have physical strips line up and stay in position for a discussion. This is another benefit of the virtual manipulatives. This applet is designed so that the fractional parts of the rectangle do not move. We can put in place an option where the fractional parts will move. Some of the design features being OFF or ON help to focus the investigations which can be good for the classroom. We can design the virtual manipulative to meet instructional needs and to focus activities whereas the physical models cannot be controlled. These are very interesting differences in the manipulatives and these differences lead to new teaching techniques that will, hopefully, reach more learning styles.
#### Part 2 Problems
• Use the virtual fractions to compare the following fractions. Draw the image below each problem.
• Write >, <, or = in the box.
• Write a sentence using the words greater than, less than or equal to make your comparison.
1 2 |
# 8.2 Graphs of the other trigonometric functions (Page 8/9)
Page 8 / 9
## Using the graphs of trigonometric functions to solve real-world problems
Many real-world scenarios represent periodic functions and may be modeled by trigonometric functions. As an example, let’s return to the scenario from the section opener. Have you ever observed the beam formed by the rotating light on a police car and wondered about the movement of the light beam itself across the wall? The periodic behavior of the distance the light shines as a function of time is obvious, but how do we determine the distance? We can use the tangent function .
## Using trigonometric functions to solve real-world scenarios
Suppose the function $\text{\hspace{0.17em}}y=5\mathrm{tan}\left(\frac{\pi }{4}t\right)\text{\hspace{0.17em}}$ marks the distance in the movement of a light beam from the top of a police car across a wall where $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is the time in seconds and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ is the distance in feet from a point on the wall directly across from the police car.
1. Find and interpret the stretching factor and period.
2. Graph on the interval $\text{\hspace{0.17em}}\left[0,5\right].$
3. Evaluate $\text{\hspace{0.17em}}f\left(1\right)\text{\hspace{0.17em}}$ and discuss the function’s value at that input.
1. We know from the general form of $\text{\hspace{0.17em}}y=A\mathrm{tan}\left(Bt\right)\text{\hspace{0.17em}}$ that $\text{\hspace{0.17em}}|A|\text{\hspace{0.17em}}$ is the stretching factor and $\text{\hspace{0.17em}}\frac{\pi }{B}\text{\hspace{0.17em}}$ is the period.
We see that the stretching factor is 5. This means that the beam of light will have moved 5 ft after half the period.
The period is $\text{\hspace{0.17em}}\frac{\pi }{\frac{\pi }{4}}=\frac{\pi }{1}\cdot \frac{4}{\pi }=4.\text{\hspace{0.17em}}$ This means that every 4 seconds, the beam of light sweeps the wall. The distance from the spot across from the police car grows larger as the police car approaches.
2. To graph the function, we draw an asymptote at $\text{\hspace{0.17em}}t=2\text{\hspace{0.17em}}$ and use the stretching factor and period. See [link]
3. period: $\text{\hspace{0.17em}}f\left(1\right)=5\mathrm{tan}\left(\frac{\pi }{4}\left(1\right)\right)=5\left(1\right)=5;\text{\hspace{0.17em}}$ after 1 second, the beam of has moved 5 ft from the spot across from the police car.
Access these online resources for additional instruction and practice with graphs of other trigonometric functions.
## Key equations
Shifted, compressed, and/or stretched tangent function $y=A\text{\hspace{0.17em}}\mathrm{tan}\left(Bx-C\right)+D$ Shifted, compressed, and/or stretched secant function $y=A\text{\hspace{0.17em}}\mathrm{sec}\left(Bx-C\right)+D$ Shifted, compressed, and/or stretched cosecant function $y=A\text{\hspace{0.17em}}\mathrm{csc}\left(Bx-C\right)+D$ Shifted, compressed, and/or stretched cotangent function $y=A\text{\hspace{0.17em}}\mathrm{cot}\left(Bx-C\right)+D$
## Key concepts
• The tangent function has period $\text{\hspace{0.17em}}\pi .$
• $f\left(x\right)=A\mathrm{tan}\left(Bx-C\right)+D\text{\hspace{0.17em}}$ is a tangent with vertical and/or horizontal stretch/compression and shift. See [link] , [link] , and [link] .
• The secant and cosecant are both periodic functions with a period of $\text{\hspace{0.17em}}2\pi .\text{\hspace{0.17em}}$ $f\left(x\right)=A\mathrm{sec}\left(Bx-C\right)+D\text{\hspace{0.17em}}$ gives a shifted, compressed, and/or stretched secant function graph. See [link] and [link] .
• $f\left(x\right)=A\mathrm{csc}\left(Bx-C\right)+D\text{\hspace{0.17em}}$ gives a shifted, compressed, and/or stretched cosecant function graph. See [link] and [link] .
• The cotangent function has period $\text{\hspace{0.17em}}\pi \text{\hspace{0.17em}}$ and vertical asymptotes at $\text{\hspace{0.17em}}0,±\pi ,±2\pi ,....$
• The range of cotangent is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),\text{\hspace{0.17em}}$ and the function is decreasing at each point in its range.
• The cotangent is zero at $\text{\hspace{0.17em}}±\frac{\pi }{2},±\frac{3\pi }{2},....$
• $f\left(x\right)=A\mathrm{cot}\left(Bx-C\right)+D\text{\hspace{0.17em}}$ is a cotangent with vertical and/or horizontal stretch/compression and shift. See [link] and [link] .
• Real-world scenarios can be solved using graphs of trigonometric functions. See [link] .
## Verbal
Explain how the graph of the sine function can be used to graph $\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x.$
Since $\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is the reciprocal function of $\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ you can plot the reciprocal of the coordinates on the graph of $\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ to obtain the y -coordinates of $\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ The x -intercepts of the graph $\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ are the vertical asymptotes for the graph of $\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x.$
A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5) and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has
how can we solve this problem
Sin(A+B) = sinBcosA+cosBsinA
Prove it
Eseka
Eseka
hi
Joel
June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler?
7.5 and 37.5
Nando
find the sum of 28th term of the AP 3+10+17+---------
I think you should say "28 terms" instead of "28th term"
Vedant
the 28th term is 175
Nando
192
Kenneth
if sequence sn is a such that sn>0 for all n and lim sn=0than prove that lim (s1 s2............ sn) ke hole power n =n
write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions |
# 1.9 Graphing (Page 3/3)
Page 3 / 3
## Other vertical permutations
Adding or subtracting a constant from $f\left(x\right)$ , as described above, is one example of a vertical permutation: it moves the graph up and down. There are other examples of vertical permutations.
For instance, what does doubling a function do to a graph? Let’s return to our original function:
What does the graph $y=2f\left(x\right)$ look like? We can make a table similar to the one we made before.
$x$ $f\left(x\right)$ $2f\left(x\right)$ so $y=2f\left(x\right)$ contains this point
–3 2 4 $\left(-3,4\right)$
–1 –3 –6 $\left(-1,-6\right)$
1 2 4 $\left(1,4\right)$
6 0 0 $\left(6,0\right)$
In general, the high points move higher; the low points move lower. The entire graph is vertically stretched , with each point moving farther away from the x-axis.
Similarly, $y=\frac{1}{2}f\left(x\right)$ yields a graph that is vertically compressed, with each point moving toward the x-axis.
Finally, what does $y=-f\left(x\right)$ look like? All the positive values become negative, and the negative values become positive. So, point by point, the entire graph flips over the x-axis.
## What happens to the graph, when you add 2 to the x value?
Vertical permutations affect the y-value; that is, the output, or the function itself. Horizontal permutations affect the x-value; that is, the numbers that come in. They often do the opposite of what it naturally seems they should.
Let’s return to our original function $y=f\left(x\right)$ .
Suppose you were asked to graph $y=f\left(x+2\right)$ . Note that this is not the same as $f\left(x\right)+2$ ! The latter is an instruction to run the function, and then add 2 to all results. But $y=f\left(x+2\right)$ is an instruction to add 2 to every x-value before plugging it into the function.
• $f\left(x\right)+2$ changes $y$ , and therefore shifts the graph vertically
• $f\left(x+2\right)$ changes $x$ , and therefore shifts the graph horizontally.
But which way? In analogy to the vertical permutations, you might expect that adding two would shift the graph to the right. But let’s make a table of values again.
$x$ $x+2$ $f\left(x+2\right)$ so $y=f\left(x+2\right)$ contains this point
–5 –3 f(–3)=2 $\left(-5,2\right)$
–3 –1 f(–1)=–3 $\left(-3,-3\right)$
–1 1 f(1)=2 $\left(-1,2\right)$
4 6 f(6)=0 $\left(4,0\right)$
This is a very subtle, very important point—please follow it closely and carefully! First of all, make sure you understand where all the numbers in that table came from. Then look what happened to the original graph.
The original graph $f\left(x\right)$ contains the point $\left(6,0\right)$ ; therefore, $f\left(x+2\right)$ contains the point $\left(4,0\right)$ . The point has moved two spaces to the left.
You see what I mean when I say horizontal permutations “often do the opposite of what it naturally seems they should”? Adding two moves the graph to the left .
Why does it work that way? Here is my favorite way of thinking about it. $f\left(x-2\right)$ is an instruction that says to each point, “look two spaces to your left, and copy what the original function is doing there .” At $x=5$ it does what $f\left(x\right)$ does at $x=3$ . At $x=\text{10}$ , it copies $f\left(8\right)$ . And so on. Because it is always copying $f\left(x\right)$ to its left , this graph ends up being a copy of $f\left(x\right)$ moved to the right . If you understand this way of looking at it, all the rest of the horizontal permutations will make sense.
Of course, as you might expect, subtraction has the opposite effect: $f\left(x-6\right)$ takes the original graph and moves it 6 units to the right . In either case, these horizontal permutations affect the domain of the original function, but not its range .
## Other horizontal permutations
Recall that $y=2f\left(x\right)$ vertically stretches a graph; $y=\frac{1}{2}f\left(x\right)$ vertically compresses . Just as with addition and subtraction, we will find that the horizontal equivalents work backward.
$x$ $2x$ $f\left(2x\right)$ so $y=2f\left(x\right)$ contains this point
–1½ –3 2 $\left(-1\frac{1}{2},2\right)$
–½ –1 –3 $\left(-\frac{1}{2};-3\right)$
½ 1 2 $\left(\frac{1}{2};2\right)$
3 6 0 $\left(3,0\right)$
The original graph $f\left(x\right)$ contains the point $\left(6,0\right)$ ; therefore, $f\left(2x\right)$ contains the point $\left(3,0\right)$ . Similarly, $\left(-1;-3\right)$ becomes $\left(-\frac{1}{2};-3\right)$ . Each point is closer to the y-axis; the graph has horizontally compressed .
We can explain this the same way we explained $f\left(x-2\right)$ . In this case, $f\left(2x\right)$ is an instruction that says to each point, “Look outward, at the x-value that is double yours, and copy what the original function is doing there .” At $x=5$ it does what $f\left(x\right)$ does at $x=\text{10}$ . At $x=-3$ , it copies $f\left(-6\right)$ . And so on. Because it is always copying f(x) outside itself, this graph ends up being a copy of $f\left(x\right)$ moved inward ; ie a compression. Similarly, $f\left(\frac{1}{2}x\right)$ causes each point to look inward toward the y-axis, so it winds up being a horizontally stretched version of the original.
Finally, $y=f\left(-x\right)$ does precisely what you would expect: it flips the graph around the y-axis. $f\left(-2\right)$ is the old $f\left(2\right)$ and vice-versa.
All of these permutations do not need to be memorized: only the general principles need to be understood. But once they are properly understood, even a complex graph such as $y=-2\left(x+3{\right)}^{2}+5$ can be easily graphed. You take the (known) graph of $y={x}^{2}$ , flip it over the x-axis (because of the negative sign), stretch it vertically (the 2), move it to the left by 3, and move it up 5.
With a good understanding of permutations, and a very simple list of known graphs, it becomes possible to graph a wide variety of important functions. To complete our look at permutations, let’s return to the graph of $y=\sqrt{x}$ in a variety of flavors.
who are u?
Lamine
haha
Cleaford
scarm
nura
what it this
Cleaford
hi y'all
Dope
how does group chat help y'all 🤔
Dope
hi y'all
Dope
how does group chat help y'all 🤔
Dope
how does group chat help y'all 🤔
Dope
to learn from one another
Lamine
oh okay
Dope
😟
Creative
Yes
Lamine
what is type of economic
how to understand basics of economics
what is demand schedle
When you make a Scedule of the demand you made
Rodeen
What is macroeconomics
It's one of the two branches of Economics that deal with the aggregate economy.
Mayen
it's about inflation, occupation, gdp and so on
alberto
What is differences between Microeconomics and Macroeconomic?
Bethrand
a price floor of 24 imposed
monopolistic competition
yap
nura
any one there to answer my question
Fixed Costs per week Variable Costs per bear Rent & Rates of Factory Hire & machines Heating & Lighting Repayment of Bank Loan K100.00 K45.00 K5.00 K50.00 Materials Foam Wages K6.00 K1.00 K1.00 Total K200.00 K8.00
Richard
one of the scarce resources that constrain our behaviour is time. each of us has only 24 hours in a day. how do you go about allocating your time in a given day among completing alternatives? once you choose a most important use of time. why do you not spend all your time to it. use the notion of op
mohsina mala..Bangla app hobe na
mani Baba. First learn the spelling of Economics
Economics- The study of how people use their limited resources to tey and satisfy unlimited wants.
Kelly
hmmm
Mani
etar bangla apps hobe na?
Mohsina
what is defination of acnomics
বাংলা বই পাওয়া যায় না?
what is a commdity
in a comparison of the stages of meiosis to the stage of mitosis, which stages are unique to meiosis and which stages have the same event in botg meiosis and mitosis
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# How to Find the Nth Term & Function Rule
Save
A sequence is an ordered list of numbers that is defined by a rule or a function. Each number in the sequence is called a term. The rule for the sequence tells you what to do in order to find the next term in the sequence. That means when given a term from the sequence you can find the next term by applying the rule to the given term. However, it can quickly become tedious when trying to find higher terms in the sequence. Learning how to identify a sequence, the rule for the sequence, and how to use the rule to determine any term in the sequence saves a lot of time and effort.
## Arithmetic Sequence
• Identify the terms of the sequence. The nth term of a sequence is the term that is in position n. For example, the sixth term is the number that is sixth in the list of numbers. For example in the sequence 2, 13, 24, 35 the first term is 2, the second is 13, the third term is 24, and the fourth term is 35.
• Decide whether the sequence is arithmetic. If the difference between two consecutive terms is constant, then the sequence is arithmetic. For the sequence in step 1, the difference between the first and second term is 13-2, or 11. The difference between the second and third term and the third and fourth term is also 11. The sequence is arithmetic.
This difference between consecutive terms is called the common difference of the sequence.
• Use the formula for the nth term of an arithmetic sequence to find the nth term of the sequence and the linear function that defines the sequence. The formula for the nth term of an arithmetic sequence is: an = a1 + d(n--1), where "an" is the nth term of the sequence, a1 is the first term of the sequence, and d is the common difference of the sequence.
For example, find the 36th term of the sequence in Step 1.
• Evaluate the nth term formula using identified value to find the 36th term of the sequence. In Step 1 and Step 2 we found a1 = 2 and d = 11, from Step 3 n = 36. Evaluating the formula with these values, a36 = 2+11(36--1) = 2+11(35) = 2+385 = 387.
• Determine the linear function that defines the sequence by evaluating the nth term formula for the identified values of the sequence. From steps 1 and 2 we found a1 = 2 and d = 11. The function that defines the sequence is: an = 2+11(n--1) = 2+11n-11 = 11n--9.
• Describe the result in words. From the results of step 5, the 36th term of the arithmetic sequence is 387 and the function that generates the sequence is: an=2+11(n--1).
## Geometric Sequence
• Identify the terms of the sequence. For example in the sequence 2, 6, 18, 54, the first term is 2, the second is 6, the third term is 18, and the fourth term is 54.
• ### Other People Are Reading
• Decide whether the sequence is geometric. If the ratio of two consecutive terms is constant, then the sequence is geometric. For the sequence in step 1, the ratio of the first and second term is 6/2 or 3. The ratio of the second and third term and the ratio of the third and fourth term is also 3. The sequence is geometric.
This ratio of consecutive terms is called the common ratio of the sequence.
• Use the formula for the nth term of a geometric sequence to find the nth term of the sequence and the function that defines the sequence. The formula for the nth term of a geometric sequence is given by an = a1(r^(n -- 1)) for values of n ≥ 1. "an" is the nth term of the sequence, a1 is the first term of the sequence, and r is the common ratio of the sequence.
• Evaluate the nth term formula using identified value to find the 15th term of the sequence. In Step 1 and Step 2 we found a1=2 and r=3, from Step 3 n=15. Evaluating the formula with these values, the equation is: a15 = 2(3^(15--1)) = 9,565,938
• Determine the function that defines the sequence by evaluating the nth term formula for the identified values of the sequence. From steps 1 and 2 we found a1 = 2 and r = 3. The function that defines the sequence is: an = 2(3^( n -- 1)) = (2/3)3^n
## References
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# 7.2 Right triangle trigonometry (Page 2/12)
Page 2 / 12
When working with right triangles, keep in mind that the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle in [link] . The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.
Many problems ask for all six trigonometric functions for a given angle in a triangle. A possible strategy to use is to find the sine, cosine, and tangent of the angles first. Then, find the other trigonometric functions easily using the reciprocals.
Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles.
1. If needed, draw the right triangle and label the angle provided.
2. Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle.
3. Find the required function:
• sine as the ratio of the opposite side to the hypotenuse
• cosine as the ratio of the adjacent side to the hypotenuse
• tangent as the ratio of the opposite side to the adjacent side
• secant as the ratio of the hypotenuse to the adjacent side
• cosecant as the ratio of the hypotenuse to the opposite side
• cotangent as the ratio of the adjacent side to the opposite side
## Evaluating trigonometric functions of angles not in standard position
Using the triangle shown in [link] , evaluate $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha ,\mathrm{cos}\text{\hspace{0.17em}}\alpha ,\mathrm{tan}\text{\hspace{0.17em}}\alpha ,\mathrm{sec}\text{\hspace{0.17em}}\alpha ,\mathrm{csc}\text{\hspace{0.17em}}\alpha ,\text{and}\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}\alpha .$
Using the triangle shown in [link] ,evaluate $\text{\hspace{0.17em}}\text{sin}\text{\hspace{0.17em}}t,\text{cos}\text{\hspace{0.17em}}t,\text{tan}\text{\hspace{0.17em}}t,\text{sec}\text{\hspace{0.17em}}t,\text{csc}\text{\hspace{0.17em}}t,\text{and}\text{\hspace{0.17em}}\text{cot}\text{\hspace{0.17em}}t.$
## Finding trigonometric functions of special angles using side lengths
It is helpful to evaluate the trigonometric functions as they relate to the special angles—multiples of $\text{\hspace{0.17em}}30°,60°,$ and $\text{\hspace{0.17em}}45°.\text{\hspace{0.17em}}$ Remember, however, that when dealing with right triangles, we are limited to angles between
Suppose we have a $\text{\hspace{0.17em}}30°,60°,90°\text{\hspace{0.17em}}$ triangle, which can also be described as a $\text{\hspace{0.17em}}\frac{\pi }{6},\frac{\pi }{3},\frac{\pi }{2}\text{\hspace{0.17em}}$ triangle. The sides have lengths in the relation $\text{\hspace{0.17em}}s,\text{s}\sqrt{3},2s.\text{\hspace{0.17em}}$ The sides of a $\text{\hspace{0.17em}}45°,45°,90°\text{\hspace{0.17em}}$ triangle, which can also be described as a $\text{\hspace{0.17em}}\frac{\pi }{4},\frac{\pi }{4},\frac{\pi }{2}\text{\hspace{0.17em}}$ triangle, have lengths in the relation $\text{\hspace{0.17em}}s,s,\sqrt{2}s.\text{\hspace{0.17em}}$ These relations are shown in [link] .
We can then use the ratios of the side lengths to evaluate trigonometric functions of special angles.
Given trigonometric functions of a special angle, evaluate using side lengths.
1. Use the side lengths shown in [link] for the special angle you wish to evaluate.
2. Use the ratio of side lengths appropriate to the function you wish to evaluate.
## Evaluating trigonometric functions of special angles using side lengths
Find the exact value of the trigonometric functions of $\text{\hspace{0.17em}}\frac{\pi }{3},$ using side lengths.
$\begin{array}{ccccc}\hfill \mathrm{sin}\left(\frac{\pi }{3}\right)& =\frac{\text{opp}}{\text{hyp}}\hfill & =\frac{\sqrt{3s}}{2s}\hfill & =\frac{\sqrt{3}}{2}\hfill & \\ \hfill \mathrm{cos}\left(\frac{\pi }{3}\right)& =\frac{\text{adj}}{\text{hyp}}\hfill & =\frac{s}{2s}\hfill & =\frac{1}{2}\hfill & \\ \hfill \mathrm{tan}\left(\frac{\pi }{3}\right)& =\frac{\text{opp}}{\text{adj}}\hfill & =\frac{\sqrt{3}s}{s}\hfill & =\sqrt{3}\hfill & \\ \hfill \mathrm{sec}\left(\frac{\pi }{3}\right)& =\frac{\text{hyp}}{\text{adj}}\hfill & =\frac{2s}{s}\hfill & =2\hfill & \\ \hfill \mathrm{csc}\left(\frac{\pi }{3}\right)& =\frac{\text{hyp}}{\text{opp}}\hfill & =\frac{2s}{\sqrt{3}s}\hfill & =\frac{2}{\sqrt{3}}\hfill & =\frac{2\sqrt{3}}{3}\hfill \\ \hfill \mathrm{cot}\left(\frac{\pi }{3}\right)& =\frac{\text{adj}}{\text{opp}}\hfill & =\frac{s}{\sqrt{3}s}\hfill & =\frac{1}{\sqrt{3}}\hfill & =\frac{\sqrt{3}}{3}\hfill \end{array}$
(x2-2x+8)-4(x2-3x+5)
sorry
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
(X2-2X+8)-4(X2-3X+5)=0 ?
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
Y
master
master
Soo sorry (5±Root11* i)/3
master
Mukhtar
explain and give four example of hyperbolic function
What is the correct rational algebraic expression of the given "a fraction whose denominator is 10 more than the numerator y?
y/y+10
Mr
Find nth derivative of eax sin (bx + c).
Find area common to the parabola y2 = 4ax and x2 = 4ay.
Anurag
A rectangular garden is 25ft wide. if its area is 1125ft, what is the length of the garden
to find the length I divide the area by the wide wich means 1125ft/25ft=45
Miranda
thanks
Jhovie
What do you call a relation where each element in the domain is related to only one value in the range by some rules?
A banana.
Yaona
given 4cot thither +3=0and 0°<thither <180° use a sketch to determine the value of the following a)cos thither
what are you up to?
nothing up todat yet
Miranda
hi
jai
hello
jai
Miranda Drice
jai
aap konsi country se ho
jai
which language is that
Miranda
I am living in india
jai
good
Miranda
what is the formula for calculating algebraic
I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
Miranda
state and prove Cayley hamilton therom
hello
Propessor
hi
Miranda
the Cayley hamilton Theorem state if A is a square matrix and if f(x) is its characterics polynomial then f(x)=0 in another ways evey square matrix is a root of its chatacteristics polynomial.
Miranda
hi
jai
hi Miranda
jai
thanks
Propessor
welcome
jai
What is algebra
algebra is a branch of the mathematics to calculate expressions follow.
Miranda
Miranda Drice would you mind teaching me mathematics? I think you are really good at math. I'm not good at it. In fact I hate it. 😅😅😅
Jeffrey
lolll who told you I'm good at it
Miranda
something seems to wispher me to my ear that u are good at it. lol
Jeffrey
lolllll if you say so
Miranda
but seriously, Im really bad at math. And I hate it. But you see, I downloaded this app two months ago hoping to master it.
Jeffrey
which grade are you in though
Miranda
oh woww I understand
Miranda
Jeffrey
Jeffrey
Miranda
how come you finished in college and you don't like math though
Miranda
gotta practice, holmie
Steve
if you never use it you won't be able to appreciate it
Steve
I don't know why. But Im trying to like it.
Jeffrey
yes steve. you're right
Jeffrey
so you better
Miranda
what is the solution of the given equation?
which equation
Miranda
I dont know. lol
Jeffrey
Miranda
Jeffrey
answer and questions in exercise 11.2 sums
how do u calculate inequality of irrational number?
Alaba
give me an example
Chris
and I will walk you through it
Chris
cos (-z)= cos z .
cos(- z)=cos z
Mustafa
what is a algebra
(x+x)3=?
6x
Obed
what is the identity of 1-cos²5x equal to?
__john __05
Kishu
Hi
Abdel
hi
Ye
hi
Nokwanda
C'est comment
Abdel
Hi
Amanda
hello
SORIE
Hiiii
Chinni
hello
Ranjay
hi
ANSHU
hiiii
Chinni
h r u friends
Chinni
yes
Hassan
so is their any Genius in mathematics here let chat guys and get to know each other's
SORIE
I speak French
Abdel
okay no problem since we gather here and get to know each other
SORIE
hi im stupid at math and just wanna join here
Yaona
lol nahhh none of us here are stupid it's just that we have Fast, Medium, and slow learner bro but we all going to work things out together
SORIE
it's 12 |
# Systems of Linear Equations
Document Sample
``` Systems of Linear
Equations
How to: solve by graphing, substitution, linear
combinations, and special types of linear systems
By: Sarah R.
Algebra 1; E block
What is a Linear System,
Anyways?
• A linear system includes two, or more,
equations, and each includes two or more
variables.
• When two equations are used to model a
problem, it is called a linear system.
Before You Begin…Important
Terms to know
• Linear system: two equations that form one
equation
• Solution: the answer to a system of linear
equation; must satisfy both equations
***: a solution is written as an ordered pair:
(x,y)
• Leading Coefficient: any given number that is
before any given variable (for example, the
leading coefficient in 3x is 3.)
• Isolate: to get alone
Solving Linear Systems by
Substitution
• Basic steps:
• 1. Solve one equation for See next
one of its variables
• 2. Substitute that page for a
expression into the other
equation and solve for the
other variable
step by step
• 3. Substitute that value
into first equation; solve
example!
• 4. Check the solution
Example: The Substitution
Method
• Here’s the problem:
Equation one • Try this on your
-x+y=1 own…but if you need
Equation two help or a few
2x+y=-2 pointers…see the next
page!
First, solve equation one for y
Y=x+1
Next, substitute the above expression in for “y” in equation two, and solve for x
Here’s how:
Equation two
2x+y=-2
Substitute “x+1” for y
2x+ (x+1)=-2
simplify the above expression
3x+1=-2
Subtract one from both sides (because your goal is to solve for x)
3x=-3
Solve for x ( divide both sides by 3; since x is being multiplied by three, and you need it alone,
so do the inverse operation: divide by 3)
X=-1
Congratulations! You now know x has a value of –1…but you still need to find “y”.
To do so…
First, write down equation one
Y=x+1
Substitute –1 for x, since you just found that x=-1
Y= (-1)+1
Solve the equation for y by adding –1 +1
Y=0
So, now what?
You’re done; simply write out the solution as (-1,0)
***Did you remember??
To write a solution, once you’ve found x and y, you must put x first and then y: (x,y)
Substitution Method
• 1. It doesn’t matter if you choose to solve for y or
x first; the answer or solution will be the same
either way.
• 2. You can also choose to solve equation two
before equation one; simply follow the same steps,
just using a slightly rearranged order.
• *** You should always decide whether to solve x
or y first, or equation one or two first, depending
on which way is more efficient (See next page!)
Deciding the Order in Which to Solve
• Here is an instance where it is easier to solve equation two
first (for x)
Equation One: 3x-2y=1
Equation Two: x+4y=3
By solving equation two first, you are lessening your
work, because there is no leading coefficient before the x
in equation two, so you don’t have to worry about dividing
to isolate the “x”
equation one for y
Equation One: 2x+y=5
Equation Two: 3x-2y=11
You should solve for y in the first equation. Again,
coefficient before the y in equation one, while there are
leading coefficients with all the other variables.
Solving Linear Systems by Linear
Combinations
Solving Systems by means of
Linear Combinations
• Basic steps:
1. Arrange the equations with like terms in columns
2. After looking at the coefficients of x and y, you
need to multiply one or both equations by a number that
will give you new coefficients for x or y that are opposites.
3. Add the equations and solve for the unknown
variable
4. Substitute the value gotten in step 3 into either of
the original equations; solve for other variable
5. Check the solution in both original equations
Example: Solving Systems by
Linear Combinations
• Here’s an example…try it out, but if you
have any problems, see the next page for a
guided, step by step explanation
• Solve this linear equation:
Equation One: 3x+5y=6
Equation Two: -4x+2y=5
Here’s the original problem:
Solve the linear system
Equation 1: 3x+5y=6
Equation 2: -4x+2y=5
Do you remember the first step?
…put the equations into columns
3x+5y=6
-4x+2y=5
Now, you need to multiply each equation by a number that will cause your leading
coefficients of either x or y to become opposites. In this case, try to get opposite
coefficients for x. to do this, multiply the first equation by four and the second by three.
***You must multiply all terms by 3 or 4
3x+5y=6, when all terms are multiplied by four, this equation will be: 12x+20y=24
-4x+2y=5, when all terms are multiplied by three, this equation will be: -12x +6y=15
12x+ 20y=24
+ (-12x) + 6y= 15
26y=39 (sum of equations)
To get the “y” alone, you must divide each side by 26, (you divide since the y is being
multiplied by 26, and to isolate the y you do the inverse operation)
So, you have found “Y”, but you aren't done yet!
What’s left, you may be thinking…well, you have only found “y”…what
To find x, you have to place “y” into equation 2.
Equation 2: -4x+2y=5
Substitute the value you just found for “y” : 3
2
-4x+2(3)=5
2
simplify by multiplying 2 by three-halves
-4x+3=5
subtract 3 from both sides because you are working to isolate x
-4x=2
solve for x by dividing both sides by –4 (inverse operation)
x=-1
2
The solution to the example system is (-1, 3)
2 2
A Final way to Solve Systems:
Graph and Check
Here’s a method called graph and
check
• Basic steps:
1. Put each equation into slope intercept
form (y=Mx+B)
2. Graph the two lines (M is your slope;
3. Find the point that the lines appear to
intersect at, and then put that solution into
EACH equation and solve to check for
accuracy.
An Example of the Graph and
Check Method
• Here’s the problem: • Try this problem
Equation one out…but a step by step
x+y=(-2) process follows!
Equation two
2x-3y=(-9)
The first step is to put the equations into slope-intercept form
Equation one: originally, it was: x+y=(-2) but after putting it into slope intercept, it
Equation two: originally, it was: 2x-3y=(-9), but once in slope intercept, it reads:
y= 2x+3
3
From the above equations, you can make the following conclusions:
Equation one has a slope of –1 and a y intercept of –2
Equation two has a slope of 2 and a y intercept of 3
3
***Remember that in the slope intercept form (y=mx+b), m is the slope; b is the y
intercept
now, you will be able to graph the two equations as lines. Once done this, you can
conclude that the lines seem to intercept at (-3,1).
To check this assumption, put (-3) in for x and 1 in for y in BOTH EQUATIONS, and
solve both:
Equation one: (-3)+(-1)=-2
Equation two: 2(-3)-3(1)=-6-3=-9
Since both equations, once solved, equaled what they should have, you know that the
solution to this linear system is (-3,1)
Don’t Let These Fool You…
Special types of Linear Systems
Linear Systems with NO Solution
• Here’s the problem:
Equation one: 2x+y=3
Equation two: 4x+2y=8
• After trying the graph method, you’ll find that the
lines are parallel( don’t intersect) and therefore
have no solution
• After trying either of the substitution or linear
combination methods, you will have an equation
that cannot be dealt with. You will know that this
is the case because it will make no senses
whatsoever. Therefore, you have no solution to
the system.
Linear System with MANY
Solutions
• If you use the graph method, you will see that the
equations are the same line, and any point on the
line is a solution.
• If you use linear combinations or substitution, you
will have a number =number, but both numbers
will be the same. For example, 7=7 or 1=1. This
indicates that the systems has many solutions.
Solving Systems of Linear
Inequalities
Graphing Systems of Linear
Inequalities
• Here are some pointers and things to know:
1. The boundary line on the graph will be dashed if the
inequality is < or >.
2. The boundary line will be solid if the inequality is <or >.
3. You will also notice that graphs of linear inequalities are
point that is CLEARLY above the line, and a point that is
CLEARLY below the line. Put the first point into the
inequality; solve; then do the same for the other point.
Whichever point works, you shade that side.
An Example of Graphing Linear
Inequalities
• Y<4
• Y>1
• Try this one out! Remember the steps; you
can always go back a page if necessary…or
go forward one page to get step by step
guidance.
So, You Needed Help…
• Here’s the original problem: y<4, y>1
• First, make a few basic conclusions:
* The line for both boundaries will be
dotted or dashed because it is < or >.
*both will be horizontal lines because
there is no x whatsoever in either equation
Now, you can graph the equation (next page)
Graphing Errors
• If your graph looked like the previous slide, you
can congratulate yourself on getting the lines
drawn correctly. However you forgot to:
• Label the axis
• Label the lines
• Pick points and follow the previously described
process to find where to shade (between y=1 and
y=4)
• Write the equation on the line
In Simpler Terms: Graphing
Systems of Linear Inequalities
• 1. Sketch the lines of each inequality (remember
to use dashed lines for < or > and solid lines for <
or >)
• 2. LIGHTLY SHADE the area that is found by
choosing points and placing them into the
equation
• 3. The final result, or answer, is the area that is
where the shaded planes intersect, for example, in
the previous problem, the answer is anywhere
between the boundary lines of y=4 and y=1.
To Make it Somewhat Easier…
• Basic guidelines for linear systems:
• 1. Use the graphing method to get an approximate
answer, to check a solution, or to give a visual
idea of the system
• 2. Using substitution or linear combinations will
allow you to get an exact and more accurate
• 3. Substitution helps a lot when used in systems
that have coefficients of 1 or –1.
• 4. When there isn't a 1 or –1 as coefficients, the
linear combinations method is efficient.
Fun, Fun: Examples to do on Your Own
• 1. Solve the following Linear System by graphing
Equation one: -2x+3y=6
Equation two: 2x+y=10
• 2. Solve the following Linear system by means of
substitution
Equation one: x-6y=-19
Equation two: 3x-2y=-9
• 3. Solve the following Linear system by means of
substitution
Equation one: x+3y=7
Equation two: 4x-7y=-10
See next page for more; answers on last page
A Little More Fun: More
Examples
• 4. Use linear combinations to solve this
system
Equation one: -2x-3y=4
Equation two: 2x-4y=3
• 5. Use linear combinations to solve this
system
Equation one: 3x-5y=-4
Equation two: -9x+7y=8
• 1. Your graph should show a point of
intersection, which is your solution, of
(3,4).
• 2. (-1,3)
• 3. (1,2)
• 4. (-1.5,9)
• 5. (-.5, .5)
```
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# Difference between revisions of "1987 AIME Problems/Problem 10"
## Problem
Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.)
## Solutions
### Solution 1
Let the total number of steps be $x$, the speed of the escalator be $e$ and the speed of Bob be $b$.
In the time it took Bob to climb up the escalator he saw 75 steps and also climbed the entire escalator. Thus the contribution of the escalator must have been an additional $x - 75$ steps. Since Bob and the escalator were both moving at a constant speed over the time it took Bob to climb, the ratio of their distances covered is the same as the ratio of their speeds, so $\frac{e}{b} = \frac{x - 75}{75}$.
Similarly, in the time it took Al to walk down the escalator he saw 150 steps, so the escalator must have moved $150 - x$ steps in that time. Thus $\frac{e}{3b} = \frac{150 - x}{150}$ or $\frac{e}{b} = \frac{150 - x}{50}$.
Equating the two values of $\frac{e}{b}$ we have $\frac{x - 75}{75} = \frac{150 - x}{50}$ and so $2x - 150 = 450 - 3x$ and $5x = 600$ and $x = \boxed{120}$, the answer.
### Solution 2
Again, let the total number of steps be $x$, the speed of the escalator be $e$ and the speed of Bob be $b$ (all "per unit time").
Then this can be interpreted as a classic chasing problem: Bob is "behind" by $x$ steps, and since he moves at a pace of $b+e$ relative to the escalator, it will take $\frac{x}{b+e}=\frac{75}{e}$ time to get to the top.
Similarly, Al will take $\frac{x}{3b-e}=\frac{150}{e}$ time to get to the bottom.
From these two equations, we arrive at $150=\frac{ex}{3b-e}=2\cdot75=\frac{2ex}{b+e}=\frac{6ex}{3b+3e}=\frac{(ex)-(6ex)}{(3b-e)-(3b+3e)}=\frac{5x}{4}$ $\implies600=5x\implies x=\boxed{120}$, where we have used the fact that $\frac{a}{b}=\frac{c}{d}=\frac{a\pm c}{b\pm d}$ (the proportion manipulations are motivated by the desire to isolate $x$, prompting the isolation of the $150$ on one side, and the fact that if we could cancel out the $b$'s, then the $e$'s in the numerator and denominator would cancel out, resulting in an equation with $x$ by itself).
### Solution 3
Let $e$ and $b$ be the speeds of the escalator and Bob, respectively.
When Al was on his way down, he took $150$ steps with a speed of $3b-e$ per step. When Bob was on his way up, he took $75$ steps with a speed of $b+e$ per step. Since Al and Bob were walking the same distance, we have $$150(3b-e)=75(b+e)$$ Solving gets the ratio $\frac{e}{b}=\frac{3}{5}$.
Thus while Bob took $75$ steps to go up, the escalator contributed an extra $\frac{3}{5}\cdot75=45$ steps.
Finally, there is a total of $75+45=\boxed{120}$ steps in the length of the escalator.
## See also
1987 AIME (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# Mixture and Alligation Questions and Answers updated daily – Aptitude
Mixture and Alligation Questions: Solved 187 Mixture and Alligation Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.
## Mixture and Alligation Questions
81. How many liters of water should be added to a 30 liter mixture, containing milk and water in the ratio 7:3 such that the resultant mixture has 40 % water in it.
Correct Ans:5
Explanation:
Given: 30 Liters of mixture, Milk and water in the ratio 7 : 3 Which means, we have 21 liters of milk and 9 liters of water.
We add water the resulting solution is 21 liters of milk and 9 + x liters of water.
Total quantity = 30 +x.
Water percentage is 40 % = > 40 x (30+x)/100 = 9 + x
=>4(30+x) = 10(9+x)
=> 120 + 4x = 90 + 10x
=> 10x - 4x = 120-90
=>6x = 30
=> x = 5
Thus the quantity of water added = 5 liters
Workspace
82. 50 liters of mixture has 60% milk. How much milk should be added to the mixture to make it 80% pure?
Correct Ans:135 liters
Explanation:
Original mixture comprises 50 liters of milk and water.
Out of the 50 liters, 60% is pure milk.
=> ( 60 / 100 ) x 50 = pure milk
=>30 liters = pure milk
In 50 liters mixture, remaining 20 liters = water
When "x" liters of pure milk is added to 50 liters of mixture
New mixture = ( 50 + x ) liters
Milk in new mixture = (30 + x) liters
Given milk in new mixture = 80% of (50 + x)
=> 30 + x = (80 / 100) * (50+ x)
=> 30 + x = (8 / 10) * (50+ x)
=> 10 (30 + x) = 8 (50+ x)
=> 130 + 10x = 400 + 8x
=> 10 x - 8 x = 400 - 130
=> 2x = 270
=> x = 135 liters
Hence we need to add 135 liters of pure milk to the 50 liters of mixtureto make it 80% pure.
Workspace
83. How many liters of a 12 litre mixture containing milk and water in the ratio of 2 : 3 be replaced with pure milk so that the resultant mixture contains milk and water in equal proportion?
Correct Ans:2 liters
Explanation:
The mixture contains 40% milk and 60% water in it.
That is 4.8 liters of milk and 7.2 liters of water.
Now we are replacing the mixture with pure milk so that the amount of milk and water in the mixture is 50% and 50%.
That is we will end up with 6 liters of milk and 6 liters of water. Water gets reduced by 1.2 liters.
To remove 1.2 liters of water from the original mixture containing 60% water, we need to remove 1.2 / 0.6 liters of the mixture = 2 liters.
Workspace
84. 20 liters of mixture has 70% milk. How much milk should be added to the mixture to make it 90% pure?
Correct Ans:40 liters
Explanation:
Original mixture comprises 20 liters of milk and water.
Out of the 20 liters, 70 % is pure milk.
=> ( 70 / 100 ) x 20 = pure milk
=>14 liters = pure milk
In 20 liters mixture, remaining 6 liters = water
When "x" liters of pure milk is added to 20 liters of mixture
New mixture = ( 20 + x ) liters
Milk in new mixture = (14 + x) liters
Given milk in new mixture = 90% of (20+ x)
=> 14 + x = (90 / 100) * (20 + x)
=> 14 + x = (9 / 10) * (20+ x)
=> 10 (14 + x) = 9 (20 + x)
=> 140 + 10x = 180 + 9x
=> 10 x - 9 x = 180 - 140
=> x = 40 liters
Hence we need to add 40 liters of pure milk to the 20 liters of mixtureto make it 90% pure.
Workspace
85. Two liquids A and B are mixed together in the ratio 2 : 3. The average cost of liquid B is Rs 35 per liter and the average cost of the mixture is Rs. 30 per liter, then find the average cost of liquid A per liter.
Correct Ans:22.5
Explanation:
Given, Ratio of liquids A and B =2 : 3
Average cost of liquid B per liter = Rs. 35
Average cost of mixture per liter = Rs. 30
Let theAverage cost of liquid A per liter = Rs. x
According to alligation equation,
(35-30)/(30-x) = 2/3
=> 5 /(30-x) =2/3
=> 5 * 3 = 2 *(30-x)
=> 15 = 60 - 2x
=> 2x = 60 - 15
=> 2x = 45
=> x = 22.5
Therefore,Average cost of liquid A per liter = Rs. x = Rs. 22.5
Workspace
86. How many liters of water should be added to a 30 liter mixture of milk and water containing milk and water in the ratio 7:3 such that the resultant mixture has 40 % water in it.
Correct Ans:5
Explanation:
Given, Total quantity of mixture = 30 liter
=> Milk + water = 30 liter
Given,Ratio of Milk and water = 7 : 3
=> Milk quantity = 7x liter
and waterquantity = 3x liter
Thus, 7x + 3x = 30
=> 10x = 30
=> x = 3
Hence, Milk =7x= 7 * 3 = 21 liter
Water = 3x = 3 * 3 = 9 liter
If we add "x" liters of water to the 30 liter mixture, to obtain 40% water in it,
the total quantity of resultant mixture = 30 + x liter
water quantity will become = 9 + x liter
thus, 40% of new mixture = water
=> 40% of (30 + x) =9 + x
=> (40/100) * (30 + x) = 9 +x
=> (4/10) * (30 + x) = 9 +x
=> 4 * (30 + x) = 10 * (9 +x)
=> 120 + 4x = 90 + 10x
=> 10x - 4x = 120 - 90
=> 6x = 30
=> x = 5
Thus,to obtain 40% water in new mixture, the quantity of water to be added = 5 liter.
Workspace
87. How many liters of water should be added to a 30 liter mixture of milk and water containing milk and water in the ratio 7:3 such that the resultant mixture has 40 % water in it.
Correct Ans:5
Explanation:
Given, In 30 Liters of mixture, Milk and water in the ratio 7 : 3.
=> milk = 7x liters
water = 3x liters
Now, Milk + water = 30 liters
=> 7x + 3x = 30
=> 10 x = 30
=> x = 3
Milk = 7x liters = 7 * 3 = 21 liters
Water = 3x liters = 3 * 3 = 9 liters
When we add x liters of water, the resulting solution is 21 liters of milk and 9 + x liters of water.
Then,Total quantity = 30+x
Water percentage is 40 %
= > 40 x (30+x)/100 = 9 + x
=>4(30+x) = 10(9+x)
=> 120 + 4x = 90 + 10x
=> 10x – 4x = 120-90
=>6x = 30
=> x = 5
=> Water added = 5 liters
Workspace
88. Two liquids A and B are mixed together in the ratio 5 : 6. The average cost of liquid B is Rs 15 per litre and the average cost of the mixture is Rs. 10 per liter, then find the average cost of liquid A per litre.
Correct Ans:Rs. 4
Explanation:
Given, Ratio of liquids A and B = 5 : 6
Average cost of liquid B per liter = Rs. 15
Average cost of mixture per liter = Rs. 10
Let theAverage cost of liquid A per liter = Rs. x
According to alligation equation,
(15 - 10)/(10 - x) = 5/6
=> 5 /(10 - x) = 5/6
=> 5 * 6 = 5 *(10 - x)
=> 30 = 50 - 5x
=> 5x = 50 - 30
=> 5x = 20
=> x = 4
Therefore,Average cost of liquid A per liter = Rs. x = Rs. 4
Workspace
89. Petrol & Kerosene are mixed in the ratio of 15:18. Total Volume of the mixture is 100 litres. Find the composition of Petrol in the overall mixture?
Correct Ans:45.45
Explanation:
Given Petrol & Kerosene are in the ratio of 15:18 Given volume of mixture = 100 litres Quantity of Petrol = 15/(15+18) x 100 = 45.45
Workspace
90. Petrol & Kerosene are mixed in the ratio of 15:16. Total Volume of the mixture is 100 litres. Find the composition of Petrol in the overall mixture?
Correct Ans:48.39
Explanation:
Given Petrol & Kerosene are in the ratio of 15:16 Given volume of mixture = 100 litres Quantity of Petrol = 15/(15+16) x 100 = 48.39
Workspace
91. Petrol & Kerosene are mixed in the ratio of 15:12. Total Volume of the mixture is 100 litres. Find the composition of Petrol in the overall mixture?
Correct Ans:55.56
Explanation:
Given Petrol & Kerosene are in the ratio of 15:12 Given volume of mixture = 100 litres Quantity of Petrol = 15/(15+12) x 100 = 55.56
Workspace
92. Petrol & Kerosene are mixed in the ratio of 15:10. Total Volume of the mixture is 100 litres. Find the composition of Petrol in the overall mixture?
Correct Ans:60
Explanation:
Given Petrol & Kerosene are in the ratio of 15:10 Given volume of mixture = 100 litres Quantity of Petrol = 15/(15+10) x 100 = 60
Workspace
93. Petrol & Kerosene are mixed in the ratio of 15:8. Total Volume of the mixture is 100 litres. Find the composition of Petrol in the overall mixture?
Correct Ans:65.22
Explanation:
Given Petrol & Kerosene are in the ratio of 15:8 Given volume of mixture = 100 litres Quantity of Petrol = 15/(15+8) x 100 = 65.22
Workspace
94. Petrol & Kerosene are mixed in the ratio of 15:6. Total Volume of the mixture is 100 litres. Find the composition of Petrol in the overall mixture?
Correct Ans:71.43
Explanation:
Given Petrol & Kerosene are in the ratio of 15:6
Given volume of mixture = 100 litres
Quantity of Petrol = 15/(15+6) x 100 = 71.43
Workspace
95. Petrol & Kerosene are mixed in the ratio of 15:4. Total Volume of the mixture is 100 litres. Find the composition of Petrol in the overall mixture?
Correct Ans:78.95
Explanation:
Given Petrol & Kerosene are in the ratio of 15:4 Given volume of mixture = 100 litres Quantity of Petrol = 15/(15+4) x 100 = 78.95
Workspace
96. Petrol & Kerosene are mixed in the ratio of 15:2. Total Volume of the mixture is 100 litres. Find the composition of Petrol in the overall mixture?
Correct Ans:88.24
Explanation:
Given Petrol & Kerosene are in the ratio of 15:2 Given volume of mixture = 100 litres Quantity of Petrol = 15/(15+2) x 100 = 88.24
Workspace
97. Milk & Water are mixed in the ratio of 13:18. Total Volume of the mixture is 450 litres. Find the composition of Milk in the overall mixture?
Correct Ans:188.71
Explanation:
Given Milk & Water are in the ratio of 13:18
Given volume of mixture = 450 litres
Quantity of Milk = 13/(13+18) x 450 = 188.71
Workspace
98. Milk & Water are mixed in the ratio of 13:16. Total Volume of the mixture is 450 litres. Find the composition of Milk in the overall mixture?
Correct Ans:201.72
Explanation:
Given Milk & Water are in the ratio of 13:16 Given volume of mixture = 450 litres Quantity of Milk = 13/(13+16) x 450 = 201.72
Workspace
99. Milk & Water are mixed in the ratio of 13:14. Total Volume of the mixture is 450 litres. Find the composition of Milk in the overall mixture?
Correct Ans:216.67
Explanation:
Given Milk & Water are in the ratio of 13:14 Given volume of mixture = 450 litres Quantity of Milk = 13/(13+14) x 450 = 216.67
Workspace
100. Milk & Water are mixed in the ratio of 13:12. Total Volume of the mixture is 450 litres. Find the composition of Milk in the overall mixture?
Correct Ans:234
Explanation:
Given Milk & Water are in the ratio of 13:12
Given volume of mixture = 450 litres
Quantity of Milk = 13/(13+12) x 450 = 234
Workspace
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# How to Find the Last Digits of a Positive Power of Two
A common exercise in number theory is to find the last digits of a large power, like 22009, without using a computer. 22009 is a 605-digit number, so evaluating it by hand is out of the question. So how do you find its last digits — efficiently?
Modular arithmetic, and in particular, modular exponentiation, comes to the rescue. It provides an efficient way to find the last m digits of a power, by hand, with perhaps only a little help from a pocket calculator. All you need to do is compute the power incrementally, modulo 10m.
In this article, I will discuss three methods — all based on modular exponentiation and the laws of exponents — for finding the ending digits of a positive power of two. The techniques I use are easily adapted to powers of any number.
In this method, you reduce a power of two modulo 10m repeatedly until you get a congruent power, or product of powers, for which the end digits are known — or easily computed. You use your knowledge of smaller powers of two, in conjunction with the power of a power and product of powers rules, to set up easier sub problems to solve.
You start by dividing the exponent of 2n by the exponent of a known, smaller power of two, 2a, getting a quotient q and a remainder r. You then rewrite 2n as (2a)q · 2r.
### Finding the Last Digit
I’ve categorized two sub methods of the ad hoc method that make it more systematic when dealing specifically with powers of two. I call them the powers of two method and the powers of six method.
#### Powers of Two Method
In the powers of two method, you reduce a power of two by using a power of two ending in 2. This reduces the problem at each stage to a smaller power of two, giving the method a recursive feel. The intermediate powers of two are in effect nested.
For example, let’s find the last digit of 22009, using 25 (32) to reduce the problem at each step:
So , showing that 22009 ends in 2.
The intermediate results — 2405, 281, 217, and 25 — are all congruent, ending in 2. If you recognize this along the way, you can stop. For example, if you happen to know that 217 is 131,072, you can stop after the third step.
Any power of two ending in 2 works. Here’s how the process goes when using 29 (512):
Using 29, there is one less step, but the arithmetic is slightly harder (division by 9 instead of division by 5).
#### Powers of Six Method
In the powers of six method, you reduce a power of two using a power of two that ends in 6. This introduces powers of six, which have to be handled separately and combined with the “remainder” powers of two. For example, let’s do our example with 24 = 16. The first step would give:
But wait! All powers of six end in 6 (6 times 6 mod 10 is 6, and around it goes…), so we just turned this into a very simple problem: .
So, while it doesn’t have the elegance of the powers of two method, the powers of six method is simpler.
### Finding the Last Two Digits
The powers of six method does not apply to mod 100, but the powers of two method does — indirectly. Although there is no power of two that ends in 02, we can use any two-digit ending that is a power of two; we just convert it to 2y using the laws of exponents.
Let’s go back to our example, 22009. Consulting a table of positive powers of two, find a power of two that ends in 04; for example, 222 (4,194,304). Let’s use this to reduce our problem at each step:
### Finding the Last m Digits
For the last m digits, find an m-digit power of two (m digits including leading zeros) greater than or equal to 2m and convert it to 2y as above. Of course, the remainders will become larger and larger as the modulus increases.
## 2. Successive Squaring
The method of successive squaring, also called repeated squaring or binary exponentiation, is a very systematic way to do modular exponentiation. It’s a generic four step process applicable to any base and modulus, but here’s how to use it to compute 2n mod 10m specifically:
1. Rewrite 2n so that n is a sum of powers of two (essentially, convert n to binary).
2. Rewrite 2n using the product of powers rule.
3. Create a list of powers 22i mod 10m, by repeatedly squaring the prior result.
4. Combine, with multiplication mod 10m, the powers in the list that make up 2n.
This process is independent of the number of ending digits m, although you have to deal with bigger and bigger numbers as m increases.
### Example: Find the Last Digit of 22009
Let’s use this method to find the last digit of 22009:
1. 22009 = 21024 + 512 + 256 + 128 + 64 + 16 + 8 + 1
2. 22009 = 21024 · 2512 · 2256 · 2128 · 264 · 216 · 28 · 21
3. Create a list of powers of two raised to powers of two, mod 10:
(I wrote out the whole list for completeness, but it was unnecessary to go beyond 24. Again, that’s because all powers of six end in 6.)
4. Combine the required powers:
### Example: Find the Last Two Digits of 22009
Let’s use this method to find the last two digits of 22009:
1. 22009 = 21024 + 512 + 256 + 128 + 64 + 16 + 8 + 1
2. 22009 = 21024 · 2512 · 2256 · 2128 · 264 · 216 · 28 · 21
3. Create a list of powers of two raised to powers of two, mod 100:
(Notice that the powers in this list cycle after a point, so it is not necessary to compute them all.)
4. Combine the required powers:
(I could have used negative numbers in the intermediate steps to make the math easier; for example, -4 instead of 96. Both are congruent mod 100.)
## 3. Cyclic Powers
In this method, you exploit the fact that the ending m digits of the positive powers of two repeat in cycles; specifically, cycles of length 4·5m-1, starting at 2m. Powers of two that differ in their exponents by 4·5m-1 have the same ending m digits.
There are two ways to use the cycle information, in techniques I call the table method and the base power method.
#### Table Method
In the table method, you compute the powers of two mod 10m in sequence, until the ending digits cycle. You label entries sequentially starting at m, wrapping around 0 to end at m – 1. I’ve created tables for the last one, two, and three ending digits; in other words, tables for powers of two mod 10, mod 100, and mod 1000.
To find where in the cycle your power 2n falls, compute n mod 4·5m-1, or equivalently, find the remainder of n/(4·5m-1). The last m digits of 2n are the digits in the table with the label corresponding to that remainder.
For example, let’s find the last digit of 22009. The last digit of the positive powers of two cycles with length 4, and . According to the table, a remainder of 1 corresponds to a last digit of 2.
Almost as simply, we can find the last two digits of 22009. The last two digits of the positive powers of two cycle with length 20, and . According to the table, a remainder of 9 corresponds to the ending digits 12.
### Finding the Last m Digits
The table method works for any number of ending digits, but beyond two or three, is impractical. The tables grow large, by a factor of five for each additional ending digit.
#### Base Power Method
Every power of two 2n, mod 10m, is congruent to some power of two in the first instance of the cycle; that is, a power of two between 2m and 2(m + 4·5m – 1 – 1). You need to determine which power of two in this range it is — what I call the base power of two — and then use another method to find its ending digits.
You can find the base power of two directly: it is 2m + j, where j is an offset given by the expression n-m (mod 4·5m-1).
For example, let’s find the last digit of 22009. , so the base power of two is 21+0 = 21 = 2. Trivially, we can see the ending digit is 2.
For the last two digits of 22009, compute . The base power of two is 22+7 = 29, which is small enough to compute directly: 512. The last two digits are 12.
### Finding the Last m Digits
The base power method works well for any number of ending digits, assuming n is greater than 4·5m-1.
## Cheating
I said find the last digits without using a computer, right? I wanted to verify my work, so I used PARI/GP to compute 22009. It took an instant — here it is:
58,784,291,598,041,831,640,721,059,900,297,317,581,942,666,346,941,194,264,455,308,125,479,232,583,289,360,069,460,965,699,405,121,019,824,433,389,516,158,094,000,492,490,796,188,432,969,007,685,435,732,643,092,034,554,442,399,887,360,352,654,923,898,902,974,171,610,618,912,504,957,328,187,117,386,950,842,341,026,317,332,718,773,233,103,358,237,779,148,190,179,650,358,079,135,564,562,516,081,648,810,332,848,214,481,400,042,754,868,418,296,221,651,998,157,278,605,568,219,649,390,953,792,425,227,268,163,704,976,021,381,769,156,258,409,778,685,642,966,081,035,151,287,502,869,585,844,829,824,788,935,390,157,871,063,324,138,385,197,912,084,049,961,962,094,914,858,370,754,777,898,867,719,950,514,578,646,749,211,908,564,621,201,347,904,089,822,990,746,021,295,498,658,798,312,326,238,643,788,303,040,512
## Summary
The three methods I’ve shown are efficient ways to do modular exponentiation, unlike the straightforward method, which requires n-1 multiplications to compute 2n mod 10m. The three methods provide shortcuts to the answer, exploiting knowledge of the laws of exponents and the cycling of ending digits.
Which method should you use? The ad hoc method is the least systematic but allows for case-by-case optimization. The method of successive squaring is the most systematic but may be overkill for certain problems. Learn all three methods to get a deeper understanding, then decide which one you like.
For finding the last digit, I like the ad hoc powers of six method. It is the quickest.
## Exercises
For these exercises, use any method you like — or all three if you’re feeling ambitious:
1. Find the last digit of 2497
2. Find the last digit of 220000
3. Find the last two digits of 2613
4. Find the last two digits of 2512
5. Find the last three digits of 2129
6. Find the last three digits of 22009
1. 2
2. 6
3. 92
4. 96
5. 912
6. 512
1. jx says:
Do these methods work for floating point powers of two as well?
2. jx,
I’ve written about these techniques from the point of view of a human, not a computer. The powers are written with exponents (example: 28), not evaluated as they would be in a native computer representation (example: 256).
What did you have in mind?
3. Dimitrina Stavrova says:
Very interesting and thorough.
I feel a bit envious because I saw this for the first time in my daughter’s reflections on a homework problem when she was 12 years old – in 2003. She did also some other interesting relations. But I have never found time to encourage and help her to publish it…Anyway, I feel proud of her. Now I think it is worth going trough her old notes because she did some other interesting thigs too.
4. Dimitrina,
I hope that her mathematical talents were recognized and nurtured by her teachers, and I hope she is still interested in math (and I hope she is pursuing a math related career!)
Thanks for the feedback.
5. Dimitrina Stavrova says:
Hi Rick,
Yes, indeed – now she is istudyng Maths&Philosophy at Oxford,UK 🙂
6. subbareddy says:
hi this is very good and interesting. it is useful for all mathematics lovers.
7. Jagan Mohan says:
These are clearly shown for powers of two.Is there any book which has all the workouts of such problems.
8. @Jagan,
Do you mean for other powers? I don’t know of a particular book that covers it (but if one exists, it will be a book about number theory).
9. Jagan Mohan says:
Thank you sir for the reply.Yes sir they come under Number theory and i have a book by Tom M Apostol but however they mainly discuss about the theory, so i have asked that question.Thank you sir.
10. ajit says:
how to calculate last 8 digits of 2 raised to power of 1024?? Thanks in advance….
11. @ajit,
I’m inclined to use the “successive squaring” method. Build a list of powers of two raised to powers of two mod 10^8. Since 1024 is itself a power of two, you will have your answer immediately when you finish the list (no combining required). Give it a try and let me know how it works out.
12. Clifford Mathew says:
From a math illiterate … divined by staring into Excel. (If anyone else has posted this please ignore — I did not read through all comments.)
2 ^ 1 = 2
2 ^ 2 = 4
2 ^ 3 = 8
2 ^ 4 = 16
2 ^ 5 = 32
2 ^ 6 = 64
2 ^ 7 = 128
2 ^ 8 = 256
Last Number Pattern: [2,4,8,6],[2,4,8,6],[2,4,….
Every 4th number has its last digit as a 6.
To find 2^2020 last digit, let us divide 2020 by 4, and we see 2020 / 4 evenly divides with no remainder, so must be falling on a number with last digit 6.
2^2019 last digit, 2019/ 4 will leave a fraction of 0.75 (i.e 3/4) which points us to 8, the 3rd number in our set of 4 possible last digits.
A fraction of .50 (i.e. 2/4) indicates 4.
A fraction of .25 (i.e. 1/4) indicates 2.
13. @Clifford,
You are using what I call the “table method” in section “3. Cyclic Powers” (using m=1).
14. SasQ says:
Wouldn’t it be faster & easier to modulate the exponent first?
Power residues start to repeat as soon as they make one full turn around the modulus, so there’s no point in dealing with bigger exponents anyway. Just calculate the remainder of the exponent in the particular modulus (10, 100 etc.) and this would save you a lot of work to begin with.
2^2009 ≡ 2^9 (mod 10)
and you can jump all over your algorithm up to its very end 😉
Just take the last digit of the exponent, which is equivalent of taking 2009 mod 10; in other words: 2009 ≡ 9 (mod 10).
Since
2^2009 ≡ 2^9 (mod 100)
and
2^2009 ≡ 2^9 (mod 1000)
as well, we just need to know that:
2^9 ≡ 12 (mod 100)
and
2^9 ≡ 512 (mod 1000)
to get the last three digits, so let’s use a different example:
2^1234
To get its last digit:
2^1234 ≡ 2^4 ≡ 6 (mod 10)
To get the two last digits of 2^1234 is the same as getting the two last digits of 2^34, because:
2^1234 ≡ 2^34 (mod 100)
It’s easier to begin with that, than doing all the steps of your algorithm.
Of course there’s still a problem of finding the last two digits of 2^34, because we cannot reduce the exponent any further (it’s already a least residue in the modulus 100). So you can now proceed with your algorithms, or simply use the well-known fast algorithm for modular exponentiation. It works pretty much the same way the fast algorithm for the usual exponentiation. It uses the observation that, if you have, for example, something like this:
2^17 = 2×2×2×2×2×2×2×2×2×2×2×2×2×2×2×2×2
then you can group the numbers this way:
2^17 = (2×2×2×2×2×2×2×2)×(2×2×2×2×2×2×2×2)×2
and then the result of the first parenthesis is exactly the same as of the second parenthesis, so there is no point in calculating it twice. Just calculate it once and square the result (that is, multiply the current result by itself):
2^17 = (2×2×2×2×2×2×2×2)^2 × 2
and of course you can do the same with the content of this parenthesis as well:
2^17 = ((2×2×2×2)×(2×2×2×2))^2 × 2
2^17 = ((2×2×2×2)^2)^2 × 2
and again:
2^17 = (((2×2)×(2×2))^2)^2 × 2
2^17 = (((2×2)^2)^2)^2 × 2
and again:
2^17 = ((((2)×(2))^2)^2)^2 × 2
2^17 = ((((2)^2)^2)^2)^2 × 2
so we end up with 4 squarings (multiplications with itself), and one multiplication by the base (2), which gives 5 multiplications instead of 16.
Now, just replace normal multiplications with multiplication in the modulus 10, or 100, or whatever, to get the fast modular exponentiation 🙂 (or simply reduce the result of each squaring/multiplication by the modulus).
This is pretty similar to what you do already in your algorithm, just faster & more organized, because you don’t need to guess the powers of 2 which end with a particular digit to make it going – you just square your working register in each step, and if there’s a bit set in your exponent, you additionally multiply the working register by your base. This algorithm is logarithmic in the size of the exponent, so it scales very well 🙂
In our example with 2^34 (mod 100), we could proceed further as follows:
2^34 = (2^17)^2 = ((2^16)×2)^2 = (((2^8)^2)×2)^2 =
= ((((2^4)^2)^2)×2)^2 = (((((2^2)^2)^2)^2)×2)^2
Now unroll it step by step, reducing by the modulus after each step:
2^34 (mod 100)
(((((2^2)^2)^2)^2)×2)^2 (mod 100)
((((4^2)^2)^2)×2)^2 (mod 100)
(((16^2)^2)×2)^2 (mod 100)
((32^2)×2)^2 (mod 100)
(64×2)^2 (mod 100)
28^2 (mod 100)
84 (mod 100)
and indeed
2^1234 ≡ 2^34 ≡ 84 (mod 100)
so the last two digits of 2^1234 are 84.
15. Yatharth Chowdhury says:
Awesome article and i am impressed !
Your article really helped me to go through the school assignments and its quite logical ! Can you please publish another article on other no.s (like 3,5,7) ?
Thanx and Appreciate it bro !
16. ErichSt says:
@SasQ
while e. g.
2^2009 ≡ 2^9 (mod 1000) holds true,
2^2002 ≡ 2^2 (mod 1000) doesn’t |
Discriminant of a Cubic Equation
In the Algebra 1 section it was shown how to work out the discriminant of a quadratic equation, and how it can be useful to do so.
For ax^2 + bx + c \space \space \space , \space \space \space \triangle = b^2 \space {\text{–}} \space 4ac
The discriminant \triangle could be positive, negative or zero.
With the result telling you about the type of roots/zeros the quadratic had.
b^2 \space {\text{–}} \space 4ac \space > \space 0 , real distinct roots
b^2 \space {\text{–}} \space 4ac \space = \space 0 , real equal roots
b^2 \space {\text{–}} \space 4ac \space < \space 0 , complex roots
Things work the same way with the discriminant of a cubic equation, but it takes a bit more work than the quadratic case.
Discriminant of a Cubic Formula
For a standard cubic equation, ax^3 + bx^2 + cx + d = 0.
The discriminant is given by the following.
\space {\triangle}_3 \space = \space b^{\tt{2}}c^{\tt{2}} \space {\text{–}} \space 4ac^{\tt{3}} \space {\text{–}} \space 4b^{\tt{3}}d \space {\text{–}} \space 27a^{\tt{2}}b^{\tt{2}} + 18abcd
A cubic equation of degree 3 will have at most 3 roots/solutions, which will be one of the following.
If {\triangle}_3 > 0, 3 distinct real roots/zeros.
If {\triangle}_3 = 0, 3 real roots/zeros, of which 2 are equal.
If {\triangle}_3 < 0, 1 real root/zero, and 2 complex roots which are a conjugate pair.
Discriminant of a Cubic EquationExamples
(1.1)
Determine the nature of the zeros/roots of the cubic equation, x^3 + 4x^2 + 2x \space {\text{–}} \space 1.
Solution
a = 1 \space , \space b = 4 \space , \space c = 2 \space , \space d = {\text{-}}1
Using the formula.
{\triangle}_3 \space = \space b^{\tt{2}}c^{\tt{2}} \space {\text{–}} \space 4ac^{\tt{3}} \space {\text{–}} \space 4b^{\tt{3}}d \space {\text{–}} \space 27a^{\tt{2}}b^{\tt{2}} + 18abcd
{\triangle}_3 \space = \space (4)^{\tt{2}}(2)^{\tt{2}} \space {\text{–}} \space 4(1)(2)^{\tt{3}} \space {\text{–}} \space 4(4)^{\tt{3}}({\text{-}}1) \space {\text{–}} \space 27(1)^{\tt{2}}(4)^{\tt{2}} + 18(1)(4)(2)({\text{-}}1)
= \space 64 \space {\text{–}} \space 32 + 256 \space {\text{–}} \space 27 \space {\text{–}} \space 154 \space = \space 107
The graph of this polynomial is shown below, and as expected from the discriminant result, there are 3 distinct real roots/zeros.
(1.2)
Determine the nature of the zeros/roots of the cubic equation, x^3 + x^2 \space {\text{–}} \space 8x \space {\text{–}} \space 12.
Solution
a = 1 \space , \space b = 1 \space , \space c = {\text{-}}8 \space , \space d = {\text{-}}12
{\triangle}_3 \space = \space b^{\tt{2}}c^{\tt{2}} \space {\text{–}} \space 4ac^{\tt{3}} \space {\text{–}} \space 4b^{\tt{3}}d \space {\text{–}} \space 27a^{\tt{2}}b^{\tt{2}} + 18abcd
{\triangle}_3 \space = \space (1)^{\tt{2}}({\text{-}}8)^{\tt{2}} \space {\text{–}} \space 4(1)({\text{-}}8)^{\tt{3}} \space {\text{–}} \space 4(1)^{\tt{3}}({\text{-}}12)
{\text{–}} \space 27(1)^{\tt{2}}({\text{-}}12)^{\tt{2}} + 18(1)(1)({\text{-}}8)({\text{-}}12)
= \space 64 + 2048 + 48 \space {\text{–}} \space 3888 + 1728 \space = \space 0
As we see in the polynomials graph below, there are 3 real roots, of which 2 are equal in the negative direction of the x-axis.
(1.3)
Determine the nature of the zeros/roots of the cubic equation, x^3 \space {\text{–}} \space 4x^2 \space {\text{–}} \space 4x \space {\text{–}} \space 16.
Solution
a = 1 \space , \space b = {\text{-}}4 \space , \space c = {\text{-}}4 \space , \space d = {\text{-}}16
{\triangle}_3 \space = \space b^{\tt{2}}c^{\tt{2}} \space {\text{–}} \space 4ac^{\tt{3}} \space {\text{–}} \space 4b^{\tt{3}}d \space {\text{–}} \space 27a^{\tt{2}}b^{\tt{2}} + 18abcd
{\triangle}_3 \space = \space ({\text{-}}4)^{\tt{2}}({\text{-}}4)^{\tt{2}} \space {\text{–}} \space 4(1)({\text{-}}4)^{\tt{3}} \space {\text{–}} \space 4({\text{-}}4)^{\tt{3}}({\text{-}}16)
{\text{–}} \space 27(1)^{\tt{2}}({\text{-}}16)^{\tt{2}} + 18(1)({\text{-}}4)({\text{-}}4)({\text{-}}16)
= \space 256 + 256 \space {\text{–}} \space 4096 \space {\text{–}} \space 6912 \space {\text{–}} \space 4608 \space = \space {\text{-}}15104
A negative discriminant value tells us that there is just one real root on the x-axis.
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# Class 9 RD Sharma Solutions – Chapter 9 Triangles and its Angles- Exercise 9.2
• Last Updated : 08 Dec, 2020
### Question 1: The exterior angles, obtained on producing the base of a triangle both ways are 104° and 136°. Find all the angles of the triangle.
Solution:
Theorems Used: The exterior angle theorem states that the measure of each exterior angle of a triangle is equal to the sum of the opposite and non-adjacent interior angles. (Exterior Angle Theorem)
∠ACD = ∠ABC + ∠BAC [Exterior Angle Theorem]
Find ∠ABC:
∠ABC + ∠ABE = 180° [Linear pair]
∠ABC + 136° = 180°
∠ABC = 44°
Find ∠ACB:
∠ACB + ∠ACD = 180° [Linear pair]
∠ACB + 1040 = 180°
∠ACB = 76°
Now,
Sum of all angles of a triangle = 180°
∠A + 44° + 76° = 180°
∠A = 180° − 44°−76°
∠A = 60°
Angles of the triangle are ∠A = 60°, ∠B = 44° and ∠C = 76° (ans)
### Question 2: In an △ABC, the internal bisectors of ∠B and ∠C meet at P and the external bisectors of ∠B and ∠C meet at Q. Prove that ∠BPC + ∠BQC = 180°.
Solution:
In â–³ABC,
BP and CP are an internal bisector of ∠B and ∠C respectively
=> External ∠B = 180° – ∠B
BQ and CQ are an external bisector of ∠B and ∠C respectively.
=> External ∠C = 180° – ∠C
In triangle BPC,
∠BPC + 1/2∠B + 1/2∠C = 180°
∠BPC = 180° – (∠B + ∠C) …. (1)
In triangle BQC,
∠BQC + 1/2(180° – ∠B) + 1/2(180° – ∠C) = 180°
∠BQC + 180° – (∠B + ∠C) = 180°
∠BPC + ∠BQC = 180° [Using (1)] (Proved)
### Question 3: In figure, the sides BC, CA and AB of a △ABC have been produced to D, E and F respectively. If ∠ACD = 105° and ∠EAF = 45°, find all the angles of the △ABC.
Solution:
Theorems Used:
(i) The exterior angle theorem states that the measure of each exterior angle of a triangle is equal to the sum of the opposite and non-adjacent interior angles. (Exterior Angle Theorem)
(ii) Sum of a linear angle pair is 180°
(iii)Vertically opposite angles are equal.
∠BAC = ∠EAF = 45° [Vertically opposite angles]
∠ACD = 180° – 105° = 75° [Linear pair]
∠ABC = 105° – 45° = 60° [Exterior angle property]
### (i)
Solution:
∠BAC = 180° – 120° = 60° [Linear pair]
∠ACB = 180° – 112° = 68° [Linear pair]
Sum of all angles of a triangle = 1800
x = 180° − ∠BAC − ∠ACB
= 180° − 60° − 68° = 52° (ans)
### (ii)
Solution:
∠ABC = 180° – 120° = 60° [Linear pair]
∠ACB = 180° – 110° = 70° [Linear pair]
Sum of all angles of a triangle = 180°
x = ∠BAC = 180° − ∠ABC − ∠ACB
= 180° – 60° – 70° = 50° (ans)
### (iii)
Solution:
∠BAE = ∠EDC = 52° [Alternate angles]
Sum of all angles of a triangle = 180°
x = 180° – 40° – 52° = 180° − 92° = 88° (ans)
### (iv)
Solution:
CD is produced to meet AB at E.
∠BEC = 180° – 45° – 50° = 85° [Sum of all angles of a triangle = 180°]
∠AEC = 180° – 85° = 95° [Linear Pair]
Now, x = 95° + 35° = 130° [Exterior angle Property]
### Question 5: In thefigure, AB divides ∠DAC in the ratio 1 : 3 and AB = DB. Determine the value of x.
Solution:
Let ∠BAD = y, ∠BAC = 3y
∠BDA = ∠BAD = y (As AB = DB)
Now,
∠BAD + ∠BAC + 108° = 180° [Linear Pair]
y + 3y + 108° = 180°
4y = 72°
or y = 18°
∠ADC + ∠ACD = 108° [Exterior Angle Property]
x + 18° = 180°
x = 90° (ans)
### Question 6: ABC is a triangle. The bisector of the exterior angle B and the bisector of ∠C intersect each other at D. Prove that ∠D = (1/2)∠A.
Solution:
Let ∠ABE = 2x and ∠ACB = 2y
∠ABC = 180° – 2K [Linear pair]
∠A = 180° — ∠ABC — ∠ACB [Angle sum property]
= 180° -180° + 2x – 2y
= 2(x – y)
Now, ∠D = 180° – ∠DBC – ∠DCB
∠D = 180° -(x + 180° – 2x) – y
= x – y
= (1/2)∠A (Hence Proved)
### Question 7: In the figure, AC is perpendicular to CE and ∠A:∠B:∠C = 3:2:1 Find ∠ECD.
Solution:
Given that ∠A:∠B:∠C = 3:2:1
Let the angles be 3x, 2x and x.
3x + 2x + x = 180° [Angle Sum property]
6x = 180°
x = 30° = ∠ACB
Therefore,
∠ECD = 180° – ∠ACB – 90° [Linear Pair]
= 180° – 30° – 90°
= 60° (ans)
### Question 8: In the figure, AM is perpendicular to BC and AN is thebisector of ∠A. If ∠B = 65° and ∠C = 33° find ∠MAN.
Solution:
Let ∠BAN = ∠NAC = x [AN bisects ∠A]
Therefore, ∠ANM = x + 33° [Exterior angle property]
In â–³AMB,
∠BAM = 90° – 65° = 25° [Exterior angle property]
Therefore, ∠MAN = ∠BAN – ∠BAM = x – 25°
Now in â–³MAN,
(x – 25°) + (x + 33°) + 90° = 180° [Angle sum property]
or, 2x + 8° = 90°
or x = 41°
Therefore, ∠MAN = 41° – 25° = 16° (ans)
### Question 9: In a △ABC, AD bisects ∠A and ∠C > ∠B. Prove that ∠ADB > ∠ADC.
Solution:
Given that,
∠C > ∠B
or, ∠C + x > ∠B + x [Adding x on both sides]
or, 180° – ∠ADC > 180° – ∠ADB [Angle sum property]
### Question 10: In a △ABC, BD is perpendicular to AC and CE is perpendicular to AB. If BD and CE intersect at O prove that ∠BOC = 180° -∠A.
Solution:
∠A + ∠AEO + ∠EOD + ∠ADO = 360°
or, ∠A + 90° + 90° + ∠EOD = 360°
or, ∠A + ∠BOC = 360° – 90° – 90° [∠EOD = ∠BOC as they are vertically opposite angles]
or, ∠BOC = 180° – ∠A (proved)
### Question 11: In the figure, AE bisects ∠CAD and ∠B = ∠C. Prove that AE || BC.
Solution:
Let ∠B = ∠C = x
Then,
∠CAD = ∠B + ∠C = 2x (Exterior Angle)
∠EAC = ∠C [AE bisects ∠CAD and ∠C = x assumed]
These are interior angles for line AE and BC,
Therefore,
AE || BC (proved)
### Question 12: In the figure, AB || DE. Find ∠ACD.
Solution:
Since, AB || DE
Therefore,
∠ABC = ∠CDE = 40° [Alternate Angles]
∠ACB = 180° – ∠ABC – ∠BAC
= 180° – 40° – 30°
= 110°
Therefore,
∠ACD = 180° – ∠ACB [Linear Pair]
=70°
### Question 13. Which of the following statements are true (T) and which are false (F) :
(i) Sum of the three angles of a triangle is 180°.
(ii) A triangle can have two right angles.
(iii) All the angles of a triangle can be less than 60°.
(iv) All the angles of a triangle can be greater than 60°.
(v) All the angles of a triangle can be equal to 60°.
(vi) A triangle can have two obtuse angles.
(vii) A triangle can have at most one obtuse angles.
(viii) If one angle of a triangle is obtuse, then it cannot be a right-angled triangle.
(lx) An exterior angle of a triangle is less than either of its interior opposite angles.
(x) An exterior angle of a triangle is equal to the sum of the two interior opposite angles.
(xi) An exterior angle of a triangle is greater than the opposite interior angles.
### Question 14: Fill in the blanks to make the following statements true
(i) Sum of the angles of a triangle is _________.
(ii) An exterior angle of a triangle is equal to the two ____________ opposite angles.
(iii) An exterior angle of a triangle is always _________________ than either of the interior opposite angles.
(iv) A triangle cannot have more than ______________________ right angles. |
## 16.2 Classical approach
What is the probability of rolling a on a die? The sample space has six possible outcomes (listed in Example 16.1) that are equally likely, and the event (‘rolling a ’) comprises just one of those. Thus, \begin{align*} \text{Prob. of rolling a four} &= \frac{\text{The number of results that are a 4}}{\text{The number of possible results}}\\ &= \frac{1}{6}. \end{align*}
We can say that ‘the probability of rolling a 4 is 1/6,’ or ‘the probability of rolling a is 0.1667.’ The answer can also be expressed as a percentage (‘the probability of rolling a is 16.7%’).
The answer could also be interpreted as ‘the expected proportion of rolls that are a is 0.167.’ This approach to computing probabilities is called the classical approach to probability.
The chance of rolling a in the future is 0.167, but a roll of the die will either produce a , or will not produce a … and we don’t know which will occur.
Example 16.4 (Describing probability outcomes) Consider rolling a standard six-sided die again.
• The probability of rolling an even number is $$3 \div 6 = 0.5$$.
• The percentage of rolls that are expected to be even numbers is $$3 \div 6 \times 100 = 50$$%.
• The odds of rolling an even number is $$3\div 3 = 1$$.
Definition 16.4 (Classical approach to probability) In the classical approach to probability, the probability of an event occurring is the number of elements of the sample space included in the event, divided by the total number of elements in the sample space, when all outcomes are equally likely.
By this definition:
$\text{Prob. of an event} = \frac{\text{Number of equally-likely outcomes in the event of interest}}{\text{Total number of equally-likely outcomes}}$
Example 16.5 (Simple events) What is the probability of rolling a 2 on a die? What are the odds of rolling a 2 on a die?
Since the six possible outcomes in the sample space are equally likely:
$\text{Prob. of rolling a two} = \frac{\text{One outcome is a 2}}{\text{Six equally-likely outcomes}}.$ So the probability is $$\frac{1}{6} = 0.1667$$, or about 16.7%. Also, since the six possible outcomes are equally likely:
$\text{Odds of rolling a two} = \frac{\text{One outcomes is a two}}{\text{Five of the possible outcomes are not a two}}.$ So the odds of rolling a two is $$\frac{1}{5} = 0.2$$.
Example 16.6 (More complicated events) Consider rolling a standard six-sided die. There are six equally likely outcomes (Example 16.1) each with probability $$1/6$$ (or 16.7%) of occurring. The probability of rolling a 1 or a 2 is $$2/6$$ (or 33.3%).
Probabilities describe the likelihood that an event will occur before the outcome is known.
Odds and proportions can be used either before or after the outcome is known, provided the wording is correct. For example:
• Proportions describe how often an event has occurred after the outcome is known.
• Expected proportions describe the likelihood that an event will occur before the outcome is known.
The following example may help also.
Example 16.7 (Probabilities, proportions, odds) Before a fair coin is tossed:
• The probability of throwing a head is $$1/2 = 0.5$$.
• The expected proportion of heads in many coin tosses is 0.5.
• The odds of throwing a head is $$1/1 = 1$$.
If we have already tossed a coin 100 times and found 47 heads:
• The proportion of heads is $$47/100 = 0.47$$.
• The odds that we threw a head is $$47/53 = 0.887$$.
It makes no sense to talk about the ‘probability that we just threw a head,’ because the event has already occurred. |
# What is the power rule
1. Jul 23, 2014
### Greg Bernhardt
Definition/Summary
A method used to take the derivative of a polynomial function.
Equations
$$\frac{d}{dx} x^{n} = nx^{n-1}$$
Extended explanation
Power rule applies to a function of the form $x^{n}$, where x is the variable and n is a constant. Used in combination with the sum and constant factor rules of differentiation, power rule can be a powerful tool for taking derivatives.
Proof:
We can apply the limit definition of a derivative to this specific function:
$$f'(x) := \lim_{h→0} \frac{f(x+h)-f(x)}{h}$$
Substituting in gives us:
$$\frac{d}{dx} x^{n} = \lim_{h→0} \frac{(x+h)^{n}-x^{n}}{h}$$
If we then expand using Binomial Theorem:
$$\frac{d}{dx} x^{n} = \lim_{h→0} \frac{x^{n}+nx^{n-1}h+\binom{n}{2}x^{n-2}h^{2}+\cdots+h^{n} -x^{n}}{h}$$
We can then cancel the first and last $x^{n}$ terms and distribute the h from the denominator:
$$\frac{d}{dx} x^{n} = \lim_{h→0} nx^{n-1}+\binom{n}{2}x^{n-2}h+\cdots+h^{n-1}$$
Finally, we take the limit by substituting in h=0:
$$\frac{d}{dx} x^{n} = nx^{n-1}+\binom{n}{2}x^{n-2}0+\cdots+0^{n-1}$$
$$\frac{d}{dx} x^{n} = nx^{n-1}$$
Example 1:
$$f(x) = x^{189}$$
$$f'(x) = 189x^{189-1} = 189x^{188}$$
Example 2:
$$f(x) = 3x^{3}+7x^{2}+8x+2$$
$$f'(x) = 9x^{2}+14x+8$$
Example 3:
$$f(x) = 3\sqrt{x}$$
$$f'(x) = 3×1/2\ x^{(1/2-1)} = \frac{3}{2\sqrt{x}}$$
* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks! |
Class 10 Maths Extra Questions for Application of Trigonometry
Question 1.
The angle of elevation of a cloud from point h meter above a lake is α. The angle of depression of its reflection in the lake is 45°. Find the height of the cloud
Let C be the cloud and C' be the cloud reflection point. The figure is shown as below
Now
$tan \alpha = \frac {x}{base}$
or $base = \frac {x}{tan \alpha}$
Also
$tan \beta = \frac {x + 2h}{base}$
or $base = \frac {x +2h}{tan \beta}$
Equating the above two , we get the value of x as
$x= \frac {2h tan \alpha}{tab \beta - tab \alpha }$
Now the height of the cloud = $x + h$
or
=$\frac {2h tan \alpha}{tab \beta - tab \alpha } + h= \frac {h(tan \alpha + tan \beta}{tab \beta - tab \alpha }$
Here $\beta =45$
Therefore
Height of cloud=$\frac {h(tan \alpha + 1}{1 - tab \alpha }$
Question 2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8m. Find the height of the tree
Now
$tan 30 = \frac {AB}{BC}$
or $AB = \frac {8}{\sqrt 3}$
Also
$sin 30 = \frac {AB}{AC}$
or $AC = \frac {16}{\sqrt 3}$
Now height of the Tree = AB + AC = 8√3$Question 3 The shadow of a building increases by 10 meters when the angle of elevation of the sun rays decreases from 60° to 45°, what is the height of the building? Answer Now$tan 45 = \frac {h}{x}$or$h=x$Also$tan 30 = \frac {h}{x+10}$Substituting the value in x=h in this$\frac {1}{\sqrt 3} = \frac {h}{h+10}$Solving this h=13.67 m Question 4 An air plane, flying horizontally 1000 m above the ground, is observed at an angle of elevation 60° from a point on the ground. After a flight of 10 seconds, the angle of elevation at the point of observation changes to 30°. Find the speed of the plane in m/s. Answer Now$tan 60 = \frac {AB}{XB}XB=\frac {1000}{\sqrt 3}$Also$tan 30 = \frac {AB}{XC}XC= 1000 \sqrt 3$Distance travelled in 10 sec=$XC - XB=1000 \sqrt 3 - \frac {1000}{\sqrt 3}= \frac {2000}{\sqrt 3}$Speed in m/s=$\frac {200}{\sqrt 3}$Question 5 From the top of a h meters high building, the angle of depression to the bottom of a second building is 30 degrees. From the same point, the angle of elevation to the top of the second building is 45 degrees. Calculate the height of the second building. Question 6 From the top of a building 60 m high, the angles of depression of the top and bottom of a vertical lamp post are observed to be 30° and 60°respectively. Find (i) horizontal distance between the building and the lamp post (ii) height of the lamp post. Answer 34.64m h=40m Question 7 The angle of depression of two war ships from the top of the light house are 45° and 30° towards west. The war ships are 500 m apart. Find the height of the lighthouse Question 8 At the foot of a mountain the elevation of its summit is 45°. After ascending 1000 m towards the mountain up a slope of 30° inclinations, the elevation is found to be 60°. Find the height of the mountain Answer$500(1+ \sqrt {3})$m Question 9 The angle of elevation of a jet fighter from point A on the ground is 60 degrees. After 15 seconds the angle of elevation changes from 60 degrees to 30 degrees. If the jet is flying at a speed of 720 km/hr., find the height at which the jet fighter is flying. Answer$1500 \sqrt {3}$Question 10 A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a Truck at an angle of depression of 30°, which is approaching the foot of tower with a uniform speed. Six minutes later, the angle of depression of the truck is found to be 60°. Find the time taken by the truck to reach the foot of the tower Answer 3min Summary This Class 10 Maths Worksheet for Application of Trigonometry with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.You can download in PDF form also using the below links Download Application of Trigonometry Extra Questions as pdf Also Read Go back to Class 10 Main Page using below links Practice Question Question 1 What is$1 - \sqrt {3}\$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20 |
# Perpendicular Lines Lesson
Perpendicular lines are two lines that intersect at a 90-degree angle (right angle). This means that the slopes of perpendicular lines are negative reciprocals of each other. An example of perpendicular lines in the real world would be the intersection of the floor and a wall, or two walls, because two walls intersect at 90° angles, as do the floor and wall. Therefore, these form perpendicular lines. A line is perpendicular to another line if the two lines intersect at a 90° angle, or right angle.
The figure above illustrates two perpendicular lines, the little square at the intersection of the two lines indicates that the two lines form a right angle (90 degrees), which means the two lines are perpendicular.
Perpendicular lines are not just any two lines that intersect, the lines must intersect at a 90° angle in order to be perpendicular lines. In math, we use notation for a right angle in the diagram to show that two lines intersect at a 90° angle. We also use notation to show that two lines are perpendicular when describing two lines. For example, if I wanted to say that line AB and line CD were perpendicular, I would use the mathematical notation,
The two main characteristics of perpendicular lines are:
The lines must intersect at a 90 angle (right angle).
The slopes of the two lines must be negative reciprocals of each other. In other words, the product of the slopes must equal -1. (m 1 x m 2 = -1).
For example, if the slope of a line is 2/3, then the slope of the line perpendicular to that line is – 3/2.
Additionally, you can see that the product of these two slopes will result in -1.
The symbol ⊥ is used to denote perpendicular lines. If l1 is perpendicular to l2, it would be written as l1 ⊥ l2.
We can draw perpendicular lines using a protractor or a compass. Let’s take a look at the first method, drawing perpendicular lines using a protractor.
Step One: Use the straight edge of the protractor to draw a straight line, and label a point A on the line.
Step Two: Place the protractor on the line from step one, and center it on Point A. Then place point B on the 90 ° mark of the protractor.
Step Three: Pick up the protractor, then use the straight edge of the protractor to connect Points A and B to form the perpendicular line.
Step Four: Remove the protractor and label the angle where the two lines intersect as a right angle. Now, AB is perpendicular to the given line.
That is how we draw perpendicular lines with a protractor. Now we will look at how to draw perpendicular lines using a compass.
Step One: Use a straightedge to draw a line, and label a point C near the middle of the line.
Step Two: Adjust the compass to your desired width. Then place the point of the compass on point C and construct a semi circle that cuts the line in two places. Label those intersections points J and K.
Step Three: Without adjusting the width of the compass, place the point of the compass on point J, and draw an arc through the semi circle. Then repeat the same thing from point K, resulting in two small arcs cutting the semi circle. Label those marks point L and point M.
Step Four: Again, without adjusting the width of the compass, place the point of the compass on point L and then point M, and draw two intersecting arcs above the semi circle. Label this point D.
Step Five: Use a straight edge to connect point C and D, which forms a perpendicular line to the beginning line. Line JK is perpendicular to line CD.
Examples
Example 1: If the slope of a line is 3/4, find the slope of the line perpendicular to that line.
Solution: Since the slopes of perpendicular lines are negative reciprocals of each, we know that the slope of the perpendicular line must be negative, and the reciprocal 3/4 of 4/3 is . Therefore, the slope of the perpendicular line is – 4/3.
We can verify this by multiplying together 3/4 and – 4/3 , since the product is -1, we know that – 4/3 is the slope of the line perpendicular to a line with slope of 3/4.
Example 2: Determine if the following statement is sometimes, always, or never true.
If two lines intersect, then they are perpendicular.
Solution: Since we know that perpendicular lines are two lines that intersect at a 90 ° angle, the answer to this is sometimes. Not all intersecting lines will intersect at a right angle, and therefore, this is not always true, but it can be true if the intersecting lines do meet at a right angle.
Example 3: If lines RS and PQ are perpendicular, find the measure of angle PAR.
Solution: Since lines RS and PQ are perpendicular lines, we know that they intersect at a right angle, which means that the measure of angle PAR is 90°.
FAQs on Perpendicular Lines
1) What are perpendicular lines?
Perpendicular lines are two lines that intersect at a right angle, forming a 90-degree angle between them. This means that if you extend the lines infinitely in both directions, they will never meet. The symbol ⊥ is often used to indicate perpendicularity.
2) How can I determine if two lines are perpendicular?
Two lines are perpendicular if the product of their slopes is -1. If the slopes m1 and m2 of two lines satisfy the equation m1 x m2 = -1, then the lines are perpendicular. Another way is to check if the angles formed by the lines are 90 degrees.
3) Can any two lines be perpendicular?
No, not all pairs of lines can be perpendicular. For two lines to be perpendicular, they must intersect at a right angle. If lines are parallel or have different slopes, they cannot be perpendicular.
4) What is the relationship between slopes and perpendicular lines?
The product of the slopes of two perpendicular lines is always -1. If the slope of one line is m, the slope of a line perpendicular to it will be -1/m. The slopes of perpendicular lines are negative reciprocals of each other. For example, if the slope of one line is ⅕ to slope of the perpendicular line is -5.
5) How can I find the equation of a line perpendicular to another line?
If you have the equation of a line in the form y = mx + b, where m is the slope, the equation of a line perpendicular to it can be found by taking the negative reciprocal of the original slope. So, if the original slope is m, the perpendicular slope is -1/m.
6) Do perpendicular lines have the same length?
Not necessarily. The length of a line is not determined by its orientation or its relationship to other lines. Perpendicular lines only describe the geometric relationship between two lines forming a right angle; it doesn’t dictate their lengths.
7) Can perpendicular lines exist in three-dimensional space?
Yes,perpendicular lines can exist in three-dimensional space. In this case, instead of forming a 90-degree angle in a flat plane, the lines would form right angles in three-dimensional space.
8) What is the significance of perpendicular lines in geometry and mathematics?
Perpendicular lines play a crucial role in geometry and mathematics. They are fundamental for concepts such as coordinate geometry, trigonometry, and vector algebra. Understanding perpendicularity helps solve problems involving angles, slopes, and geometric relationships in various mathematical applications. |
# How To Use Quadratic Formula?
Using the quadratic formula, we can answer any quadratic problem that we encounter. For starters, we simplify the equation to the form ax2+bx+c=0, where a, b, and c are the coefficients in the equation. Afterwards, we enter these coefficients into the following formula: (-B(B2-4ac)/ (2a). See examples of how to apply the formula to a range of different equations.
## How is the quadratic formula used in math?
As well as serving as a method for determining the zeros of any parabola, the quadratic formula may also be used to determine the axis of symmetry of the parabola and the number of real zeros included inside the quadratic equation.
## How do you solve quadratic equations?
How to Solve Quadratic Equations
1. Make sure that all of the words are on one side of the equal sign and that zero is on the other. Factor.
2. Set the value of each component to zero. Each of these equations must be solved. In order to double-check, enter your solution into the original equation.
## What are the 5 examples of quadratic equation?
Here are a few examples of quadratic equations in their standard form (explained as 2×2 + bx + c = 0):
• 2 x2 – 4x – 2 = 0.
• -4×2 – 7x +12 = 0.
• 20×2 -15x – 10 = 0.
• 5×2 – 2x – 9 = 0.
• 3×2 + 4x + 2 = 0.
• -x2 +6x + 18 = 0
• 6×2 + 11x – 35 = 0.
• 6×2
## What is quadratic sequence?
Quadratic sequences are a type of sequence that contains an element. They may be distinguished by the fact that the differences between the words are not equal, but the differences between the terms in the second comparison are equal.
We recommend reading: Question: How To Use Windows Snipping Tool?
## Where are quadratic equations used?
Quadratic equations are frequently employed in instances when two objects are multiplied together and they both rely on the same variable, as in the following example. When working with area, for example, if both dimensions are stated in terms of the same variable, you would utilize a quadratic equation to solve the problem.
## What are the three steps for solving a quadratic equation?
If the largest exponent of the variable is two, then the polynomial equation including that variable is called a quadratic equation. In order to solve quadratic equations, there are three primary approaches: When you can’t factor the quadratic equation by hand, one of three options is available: 1) utilize the quadratic formula, or 2) finish the square. |
How do you solve the system 2x+7y=8 and x+5y=7 using substitution?
Jun 7, 2015
$x = - 3$ and $y = 2$.
Solve $x + 5 y = 7$ for $x$.
$x = 7 - 5 y$
Substitute $7 - 5 y$ for $x$ into the other equation.
$2 \left(7 - 5 y\right) + 7 y = 8$ =
$14 - 10 y + 7 y = 8$ '
$14 - 3 y = 8$
Subtract $14$ from both sides.
$- 3 y = - 6$
Divide both sides by $- 3$.
$y = 2$
Substitute $2$ for $y$ in the first equation.
$x + 5 \left(2\right) = 7$ =
$x + 10 = 7$
Subtract $10$ from both sides.
$x = - 3$
Check by substituting $- 3$ for $x$ and $2$ for $y$ into both equations.
$2 x + 7 y = 8$ =
$2 \left(- 3\right) + 7 \left(2\right) = 8$ =
$- 6 + 14 = 8$ =
$8 = 8$ Check.
$x + 5 y = 7$ =
$- 3 + 5 \left(2\right) = 7$ =
$- 3 + 10 = 7$ =
$7 = 7$ Check. |
# What is 33/3 as a decimal?
## Solution and how to convert 33 / 3 into a decimal
33 / 3 = 11
33/3 converted into 11 begins with understanding long division and which variation brings more clarity to a situation. Both are used to handle numbers less than one or between whole numbers, known as integers. Depending on the situation, decimals can be more clear. We don't say 1 and 1/2 dollar. We use the decimal version of \$1.50. Same goes for fractions. We will say 'the student got 2 of 3 questions correct'. So let’s dive into how and why you can convert 33/3 into a decimal.
## 33/3 is 33 divided by 3
Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! Fractions have two parts: Numerators and Denominators. This creates an equation. To solve the equation, we must divide the numerator (33) by the denominator (3). Here's how our equation is set up:
### Numerator: 33
• Numerators are the parts to the equation, represented above the fraction bar or vinculum. 33 is one of the largest two-digit numbers you'll have to convert. The bad news is that it's an odd number which makes it harder to covert in your head. Large numerators make converting fractions more complex. Time to evaluate 3 at the bottom of our fraction.
### Denominator: 3
• Denominators differ from numerators because they represent the total number of parts which can be found below the vinculum. Smaller values like 3 can sometimes make mental math easier. But the bad news is that odd numbers are tougher to simplify. Unfortunately and odd denominator is difficult to simplify unless it's divisible by 3, 5 or 7. Above all, smaller values could make the conversion a bit simpler. So grab a pen and pencil. Let's convert 33/3 by hand.
## Converting 33/3 to 11
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 3 \enclose{longdiv}{ 33 }$$
We will be using the left-to-right method of calculation. Yep, same left-to-right method of division we learned in school. This gives us our first clue.
### Step 2: Solve for how many whole groups you can divide 3 into 33
$$\require{enclose} 00.11 \\ 3 \enclose{longdiv}{ 33.0 }$$
We can now pull 33 whole groups from the equation. Multiply by the left of our equation (3) to get the first number in our solution.
### Step 3: Subtract the remainder
$$\require{enclose} 00.11 \\ 3 \enclose{longdiv}{ 33.0 } \\ \underline{ 33 \phantom{00} } \\ 297 \phantom{0}$$
If your remainder is zero, that's it! If there is a remainder, extend 3 again and pull down the zero
### Step 4: Repeat step 3 until you have no remainder
Remember, sometimes you won't get a remainder of zero and that's okay. Round to the nearest digit and complete the conversion. There you have it! Converting 33/3 fraction into a decimal is long division just as you learned in school.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals depend on the life situation you need to represent numbers. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. This is also true for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But they all represent how numbers show us value in the real world. Here are just a few ways we use 33/3, 11 or 1100% in our daily world:
### When you should convert 33/3 into a decimal
Investments - Comparing currency, especially on the stock market are great examples of using decimals over fractions.
### When to convert 11 to 33/3 as a fraction
Pizza Math - Let's say you're at a birthday party and would like some pizza. You aren't going to ask for 1/4 of the pie. You're going to ask for 2 slices which usually means 2 of 8 or 2/8s (simplified to 1/4).
### Practice Decimal Conversion with your Classroom
• If 33/3 = 11 what would it be as a percentage?
• What is 1 + 33/3 in decimal form?
• What is 1 - 33/3 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 11 + 1/2?
### Convert More Fractions
By 33 By 3 What is 33/1 as a decimal? What is 23/3 as a decimal? What is 33/2 as a decimal? What is 24/3 as a decimal? What is 33/3 as a decimal? What is 25/3 as a decimal? What is 33/4 as a decimal? What is 26/3 as a decimal? What is 33/5 as a decimal? What is 27/3 as a decimal? What is 33/6 as a decimal? What is 28/3 as a decimal? What is 33/7 as a decimal? What is 29/3 as a decimal? What is 33/8 as a decimal? What is 30/3 as a decimal? What is 33/9 as a decimal? What is 31/3 as a decimal? What is 33/10 as a decimal? What is 32/3 as a decimal? What is 33/11 as a decimal? What is 33/3 as a decimal? What is 33/12 as a decimal? What is 34/3 as a decimal? What is 33/13 as a decimal? What is 35/3 as a decimal? |
Thursday, July 18, 2024
# Performance Task Chapter 8 Geometry Answers
## Proving Triangle Similarity By Aa
MATH 8 | Quarter 4 Summative Assessment and Performance Task
Exploration 1
Comparing Triangles
Work with a partner. Use dynamic geometry software.
a. Construct ABC and DEF So that mA = mD = 106°, mB = mE = 31°, and DEF is not congruent to ABC.Answer:mC mF
b. Find the third angle measure and the side lengths of each triangle. Copy the table below and record our results in column 1.Answer:
c. Are the two triangles similar? Explain.CONSTRUCTING VIABLE ARGUMENTSTo be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results in constructing arguments.Answer:
d. Repeat parts to complete columns 2 and 3 of the table for the given angle measures.Answer:
e. Complete each remaining column of the table using your own choice of two pairs of equal corresponding angle measures. Can you construct two triangles in this way that are not similar?Answer:
f. Make a conjecture about any two triangles with two pairs of congruent corresponding angles.Answer:
Communicate Your Answer
Question 2.What can you conclude about two triangles when you know that two pairs of corresponding angles are congruent?Answer: If two pairs of corresponding angles in a pair of triangles are congruent, then the triangles are similar. Because if the two angle pairs are the same, then the third pair must also be equal when the three angle pairs are all equal, the three pairs of sides must be in proportion.
Question 3.Find RS in the figure at the left.Answer:
## Exercise 82 Proving Triangle Similarity By Aa
Vocabulary and Core Concept Check
Question 1.COMPLETE THE SENTENCEIf two angles of one triangle are congruent to two angles of another triangle. then the triangles are __________ .Answer:
Question 2.WRITINGCan you assume that corresponding sides and corresponding angles of any two similar triangles are congruent? Explain.Answer: The corresponding angles of two similar triangles are always congruent but the corresponding angles of two analogous triangles are always harmonious but the corresponding sides of the two triangles dont have to be harmonious. In an analogous triangle, the corresponding sides are commensurable which means that the rates of corresponding sides are equal. However, also the corresponding sides are harmonious, but in other cases, if these rates are 1.
Monitoring Progress and Modeling with Mathematics
In Exercises 3 6. determine whether the triangles are similar. If they are, write a similarity statement. Explain your reasoning.
Question 3.
Question 21.MODELING WITH MATHEMATICSYou can measure the width of the lake using a surveying technique, as shown in the diagram. Find the width of the lake, WX. Justify your answer.Answer:
To estimate the height of the pole CDE can be proved similar to CAD asC
CED = CBA
If two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar.CDE ~ CAB
REASONINGIn Exercises 23 26, is it possible for JKL and XYZ to be similar? Explain your reasoning.
## Advantages Of Big Ideas Math Geometry Answers High School
Below are the major advantages of referring to the BIM Geometry Answer Key. Refer to them and learn the importance of Big Ideas Math Geometry Solutions provided here. They are as follows
• Geometry BIM Textbook Solutions are prepared by math experts.
• Another benefit is that you can finish your homework or assignments in time.
• You can learn various simple methods to solve the problems with the help of the Big Ideas Math Geometry Key.
• You can answer any kind of question from Chapter Test, Practice Test, Cumulative Practice if you solve the Big Ideas math Geometry Answer Key.
• High School Geometry BIM Solutions are given after extensive research keeping in mind the Latest Common Core Curriculum 2019.
#### FAQs on High School Geometry Big Ideas Math Answers
1. How to download High School Big Ideas Math Geometry Answers PDF?You can get Big Ideas Math Textbook Solutions for Geometry on the gomathanswerkey.com website. Just tap the links given on the page and download the Big Ideas Math Geometry Answers Chapterwise.
2. What is the best site to get Big Ideas Math Answers for Geometry?The best site to get the step-by-step solutions for all the geometry chapters is the gomathanswerkey.com website. Quick and easy learning is possible with our Big Ideas Math Geometry Solution Key.
Read Also: How To Find Kp In Chemistry
## Lesson 81 Similar Polygons
Monitoring Progress
Question 1.In the diagram, JKL ~ PQR. Find the scale factor from JKL to PQR. Then list all pairs of congruent angles and write the ratios of the corresponding side lengths in a statement of proportionality.
Answer:The pairs of congruent angles are K = Q, J = P, L = RThe scale factor is \The ratios of the corresponding side lengths in a statement of proportionality are \
Explanation:
Scale factor = \= \= \Because the ratio of the lengths of the altitudes in similar triangles is equal to the scale factor, you can write the following proportion\ = \\ = \KM = \ x 35KM = 42
The two gazebos shown are similar pentagons. Find the perimeter of Gazebo A.
Answer:Perimeter of Gazebo A = 46 m
Explanation:Scale factor = \= \So, \ = \\ = \x = 12\ = \\ = \ED = 10\ = \\ = \DC = 8\ = \\ = \BC = 6Therefore, perimeter = 6 + 8 + 10 + 12 + 10 = 46
Question 5.In the diagram, GHJK ~ LMNP. Find the area of LMNP.Area of GHJK = 84m2
Area of LMNP = 756 m2
Explanation:As shapes are similar, their corresponding side lengths are proportional.Scale Factor k = \= \Area of LMNP = k² x Area of GHJK= 3² x 84
Question 6.Decide whether the hexagons in Tile Design 1 are similar. Explain.
Answer: Both the hexagons are different. On the outer side of the hexagon, all the sides are equal. In the inside hexagon among the 6-sided 3 sides are different.
Question 7.Decide whether the hexagons in Tile Design 2 are similar. Explain.
## Proving Triangle Similarity By Sss And Sas
Exploration 1
Work with a partner: Use dynamic geometry software.
a. Construct ABC and DEF with the side lengths given in column 1 of the table below.Answer:
b. Copy the table and complete column 1.Answer:
c. Are the triangles similar? Explain your reasoning.Answer:
d. Repeat parts for columns 2 6 in the table.Answer:
e. How are the corresponding side lengths related in each pair of triangles that are similar? Is this true for each pair of triangles that are not similar?Answer:
f. Make a conjecture about the similarity of two triangles based on their corresponding side lengths.CONSTRUCTING VIABLE ARGUMENTSTo be proficient in math, you need to analyze situations by breaking them into cases and recognize and use counter examples.Answer:
g. Use your conjecture to write another set of side lengths of two similar triangles. Use the side lengths to complete column 7 of the table.Answer:
Work with a partner: Use dynamic geometry software. Construct any ABC.a. Find AB, AC, and mA. Choose any positive rational number k and construct DEF so that DE = k AB, DF = k AC, and mD = mA.Answer:
b. Is DEF similar to ABC? Explain your reasoning.Answer:
c. Repeat parts and several times by changing ABC and k. Describe your results.Answer:
Communicate Your Answer
Also Check: What Is First Law Of Thermodynamics In Physics
## Big Ideas Math Book Geometry Answer Key Chapter 8 Similarity
Get the handy preparation material free of cost and enhance your math skills with ease. To make this possible, students have to access the given links below and download the Big Ideas Math Geometry Textbook Answers Ch 8 Similarity in Pdf. In this guide, teachers, parents, and kids can find various questions from basic to complex levels. By solving the questions of chapter 8 similarity covered in BIM Geometry Ch 8 Answer key exercises, practice tests, chapter tests, cumulative assessments, etc. helps you to improve your skills & knowledge.
Tell whether the ratios form a proportion.
Question 1.\
Answer:Yes, the ratios \ form a proportion.
Explanation:A proportion means two ratios are equal.So, cross product of \ is 21 x 5 = 105 = 35 x 3Therefore, \ form a proportion.
Question 2.\
Answer:Yes, the ratios \ form a proportion.
Explanation:If the cross product of two ratios is equal, then it forms a proportion.So, 24 x 24 = 576 = 64 x 9Therefore, the ratios \ form a proportion.
Question 3.\
Answer:The ratios \ do not form a proportion.
Explanation:If the cross product of two ratios is equal, then it forms a proportion.So, 56 x 6 = 336, 28 x 8 = 224Therefore, the ratios \ do not form a proportion.
Question 4.\
Answer:The ratios \ do not form a proportion.
Explanation:If the cross product of two ratios is equal, then it forms a proportion.So, 9 x 18 = 162, 27 x 4 = 108Therefore, the ratios \ do not form a proportion.
Question 5.\
Answer:The ratios \ form a proportion.
Question 6.\
## Exercise 83 Proving Triangle Similarity By Sss And Sas
Vocabulary and Core Concept Check
Question 1.COMPLETE THE SENTENCEYou plan to show that QRS is similar to XYZ by the SSS Similarity Theorem . Copy and complete the proportion that you will use:Answer:
WHICH ONE DOESNT BELONG?Which triangle does not belong with the other three? Explain your reasoning.Answer:Among the four triangles the second triangle is different.The second triangle is not a right-angled triangle.Except for the second triangle all triangles are right angled triangles.
Monitoring progress and Modeling with Mathematics
In Exercises 3 and 4, determine whether JKL or RST is similar to ABC.
Question 3.
Here the lengths of the sides that include T and L are not proportionalSince the triangles are not similar, thus the scale factor cannot be found.Hence the scale factor does not similar in this case.
In Exercises 11 and 12, sketch the triangles using the given description. Then determine whether the two triangles can be similar.
Question 11.In RST, RS = 20, ST = 32, and mS = 16°. In FGH, GH = 30, HF = 48, and mH = 24°.Answer:
Question 12.The side lengths of ABC are 24, 8x, and 48, and the side lengths of DEF are 15, 25, and 6x.
Answer:\ = \ = \\ = \x = 5
In Exercises 13 16. show that the triangles are similar and write a similarity statement. Explain your reasoning.
Question 13.
In Exercises 17 and 18, use XYZ.
Question 17.The shortest side of a triangle similar to XYZ is 20 units long. Find the other side lengths of the triangle.Answer:
Read Also: Common Core Algebra 1 Unit 3 Lesson 4 Homework
## Big Ideas Math Geometry Answers Chapter 8 Similarity
Do you want a perfect guide for a better understanding of Similarity concepts in High School? Then, Big Ideas Math Geometry Answers Chapter 8 Similarity is the one-stop destination for all your requirements during preparation. Practicing and Solving no. of questions from chapter 8 similarity BIM Geometry Textbook Solutions is the best way to understand the concepts easily and gain more subject knowledge. Kickstart your preparation by taking the help of BigIdeas math geometry ch 8 similarity Answer key pdf and test your skills up-to-date for better scoring in the exams and become math proficient.
## Lesson 83 Proving Triangle Similarity By Sss And Sas
Grade 5 PEP Mathematics Performance Task (“Hard” Question made Easy)
Monitoring progress
Which of the three triangles are similar? Write a similarity statement.
Answer:The ratios are equal. So, LMN, XYZ are similar.The ratios are not equal. So LMN, RST are not similar.
Explanation:Compare LMN, XYZ by finding the ratios of corresponding side lengthsShortest sides: \ = \ = \Longest sides: \ = \ = \Remaining sides: \ = \ = \The ratios are equal. So, LMN, XYZ are similar.Compare LMN, RST by finding the ratios of corresponding side lengthsShortest sides: \ = \ = \Longest sides: \ = \Remaining sides: \ = \ = \The ratios are not equal. So LMN, RST are not similar.
Question 2.The shortest side of a triangle similar to RST is 12 units long. Find the other side 1enths of the triangle.
Answer:The other side lengths of the triangle are 15 units, 16.5 units.
Explanation:The shortest side of a triangle similar to RST is 12 unitsScale factor = \ = \So, other sides are 33 x \ = 16.5, 30 x \ = 15.
Explain how to show that the indicated triangles are similar.
Question 3.
XZW and YZX are not proportional.
Explanation:The shorter sides: \Longer sides: \ = \The side lengths are not proportional. So XZW and YZX are not proportional.
Read Also: Compact Numerical Methods For Computers Linear Algebra And Function Minimisation
## Exercise 81 Similar Polygons
Vocabulary and Core Concept Check
Question 1.COMPLETE THE SENTENCEFor two figures to be similar, the corresponding angles must be ____________ . and the corresponding side lengths must be _____________ .
Answer:
\ = \ = \The ratio of perimeter is \.
In Exercises 13-16, two polygons are similar. The perimeter of one polygon and the ratio of the corresponding side lengths are given. Find the perimeter of the other polygon.
Question 13.perimeter of smaller polygon: 48 cm: ratio: \Answer:
perimeter of smaller polygon: 66 ft: ratio: \
Answer:The perimeter of larger polygon is 88 ft.
Explanation:\ = \ = \66 x 4 = 3xx = \ = 88
Question 15.perimeter of larger polygon: 120 yd: rttio: \Answer:
perimeter of larger polygon: 85 m ratio: \
Answer:The perimeter of smaller polygon is 34 m.
Explanation:\ = \ = \85 x 2 = 5xx = \ = 34
Question 17.MODELING WITH MATHEMATICSA school gymnasium is being remodeled. The basketball court will be similar to an NCAA basketball court, which has a length of 94 feet and a width of 50 feet. The school plans to make the width of the new court 45 feet. Find the perimeters of ail NCAA court and of the new court in the school.Answer:
Question 18.MODELING WITH MATHEMATICSYour family has decided to put a rectangular patio in your backyard. similar to the shape of your backyard. Your backyard has a length of 45 feet and a width of 20 feet. The length of your new patio is 18 feet. Find the perimeters of your backyard and of the patio.
Question 19.
Question 25. |
### Written by tutor Yvonne H.
If you are wondering how to convert from radians to degrees, then you probably have some experience converting from one
unit to another. For example, you probably know how to measure the length of your bedroom in feet and convert those feet
into inches, i.e. 18 feet = 18 X 12 = 216 inches long. The same concept, unit conversion, applies to converting from
radians to degrees and from degrees to radians. Keep in mind that unit conversion means that there is a ratio that defines
the relationship between one unit to another, such as the ratio of 12 inches to 1 foot (12 in: 1 ft, or written as 12 in/1 ft).
This ratio is called the conversion ratio.
Before we convert from radians to degrees, or vice versa, let’s look at the measurement of a circle in each unit. In other
words, let’s measure a circle in degrees and again in radians. As we measure the circles, notice the difference in notation
and try to determine the simplest ratio that relates the units, radians and degrees.
Notice that when we measured in degrees we use the degree symbol ,°, but when we measure in radians, we use the symbol for
the number pi, π. Now we can figure out what the conversion ratio is between degrees and radians. Let’s do that together.
### Question 1: Which degree measure coincides with a whole π?
In other words, look at the position of a whole π in Figure 2. Now find the same position but in the circle in Figure 1. There
is the answer for Question 1, 180°. Great! Now we have our conversion ratio, π = 180°. You can use this ratio to convert from
radians to degrees and from degrees to radians. Are you ready to try some practice exercises? Let’s try the first one together.
1. Here is an attempt to convert 2° to radians. Why is this calculation wrong?
2° × 180°/π = 360°/π
Here is a clue:
5 ft × 12 in/1ft = 60 in
In the conversion from 5 feet to inches, the unit feet canceled out because multiplication allows us to do that for any values that are
the same in the numerator and denominator. Remember, this does not work for addition and subtraction, only multiplication. After the
unit, “feet,” was canceled out, we were left with only the unit we wanted, namely inches.
This means that we need to flip the conversion ratio to correct the calculation from 2° to radians.
Correction:
2° × π/180° = π/90°
Now the unit degrees, cancels out and we are left with the unit radians.
## Radians and Degrees Practice Quiz
A.
23π/90
B.
π/4
C.
0.25π
D.
8,280/π
The correct answer here would be A.
A.
2π/45
B.
3π/4
C.
0.25π
D.
7π/4
The correct answer here would be D.
Convert 0 to degrees.
A.
B.
360°
C.
180°
D.
The correct answer here would be D.
Convert 9π/4 to degrees.
A.
45°
B.
450°
C.
405°
D.
54π
The correct answer here would be C.
Scroll to Top |
Objectives:
• To understand graphical representations of displacement, speed, velocity and acceleration
• To be able to draw, interpret and understand displacement–time graphs, including how velocity is the gradient of such a graph
• To be able to draw, interpret and understand velocity–time graphs, including how acceleration is the gradient and displacement is the area under such a graph
Displacement-time graphs
We can represent the changing position of a moving object by drawing a displacement against time graph (d-t graph). Some of the main point to appreciate with regards to d-t graphs are:
• The gradient of the graph is equal to the object’s velocity.
• The steeper the slope the greater the velocity.
• A graph like this can also tell us if an object is moving with constant velocity or accelerating.
• The straight line graph shows that the object’s velocity is constant.
• Comparing to objects moving with constant speed, the one with a steeper gradient is moving faster.
• If the line is parallel to the x-axis, it means that the displacement is not changing, so the object is still.
• If the slope has a negative gradient, then the object is moving backwards to where it came from.
• If the line is curved, the slope is changing, which means that the velocity is changing. This means that the object has an acceleration or a deceleration which may or may not be constant.
Velocity against time graphs
A velocity against time graph (v-t graph) tells you more information about the change of velocity of an object. If the v-t graph is a curve, this may help you to understand more about its acceleration. It tells you the velocity of an object at any instant, while the slope of the graph (the gradient) tells you the acceleration at any instant. The distance traveled over any phase of the journey is equal to the area under the graph. The average speed can still be calculated as usual using equation $v = \frac{d}{t}$.
In the graph above it can be seen that the object moves with increasing speed in the first part and because the slope is a straight line, the acceleration of the object (the rate of change in velocity) is constant. Then in the second part, the object is moving with constant speed, then it increases but it doesn’t increase at the same rate at all instant, so there is an acceleration which is increasing. Then it is going at constant acceleration again and then decelerates as the velocity gradient is negative. This doesn’t mean that is travelling backwards, but just that it’s slowing down at a constant rate. |
Home » Featured » Number Game: The Word Addition
# Number Game: The Word Addition
### Start here
Consider the summation of letters : $T \ A \ M \ I \ N \ G \\ + \\ H \ E \ L \ M \ E \ T \\ = \\ 9\ 2\ 5\ 7\ 6 \ 4$ In the addition shown above, the sum $925764$ represents a word. We have to find this word — on the conditions given below: Each letter represents a different digit. No letter represents zero.
So, let we go to find the word represented by 925764.
## Solution:
T+H cannot be 9. Otherwise, A & E both are 1 [which contradicts (1)] or one of A & E is zero [which contradicts (2)]. So, T+H is 8 and 1 is carried from A+E.
Then T cannot be:
0 –[[2]]
2 or 7 – then G & T would be the same digit, contradicting [1]
4 — since $T+H=8$, H would then be 4, contradicting [1].
So the possible digits for T & H are as follows:
${T : \ 1 \ 3 \ 5 \ 6}$
${H : \ 7 \ 5 \ 3 \ 2}$
Then, continuing the table with G, so that G+T is 4 or 14:
${G : \ 3 \ 1 \ 9 \ 8}$
${T : \ 1\ 3 \ 5 \ 6}$
${H : \ 7 \ 5 \ 3 \ 2}$
Because 1 is carried from A+E, A+E is either 11 or 12.
Suppose, A+E is 11. Then using [[1]] to continue the table with A and E: [${a, b, c, d, e, . . . }$ represent cases.]
$\rightarrow \ \, a \ b \ c \ d \ e \ f \ g \ h \ i \ j \ k \ l \\ {G : \ 3\ 3\ 3\ 3\ 1\ 1\ 1\ 1\ 9\ 9\ 8\ 8} \\ {T : \ 1 \ 1 \ 1\ 1\ 3\ 3\ 3\ 3\ 5\ 5 \ 6\ 6} \\ {H : \ 7\ 7\ 7\ 7\ 5\ 5\ 5\ 5\ 3\ 3\ 2\ 2} \\ {E : \ 2\ 9\ 5\ 6\ 2\ 9\ 4\ 7 \ 4 \ 7\ 4\ 7} \\ {A : \ 9\ 2\ 6\ 5\ 9\ 2 \ 7 \ 4 \ 7 \ 4 \ 7 \ 4}$
Suppose A+E is 12. Then by using [1] to continue the table with A and E
$\rightarrow m \ n \ o \ p \ q \ r \ s \ t \ u \ v \\ {G : \ 3\ 3\ 1\ 1\ 9\ 9\ 8\ 8\ 8\ 8} \\ {T : \ 1\ 1\ 3\ 3\ 5\ 5\ 6\ 6\ 6 \ 6} \\ {H : \ 7 \ 7\ 5\ 5\ 3\ 3\ 2\ 2\ 2\ 2} \\ {E : \ 4\ 8\ 4\ 8\ 4\ 8\ 3\ 9\ 5\ 7} \\ {A : \ 8\ 4\ 8\ 4\ 8\ 4\ 9\ 3\ 7 \ 5}$
Then, keeping [1] & [2] in mind, to continue the table with N, so that N+E is 6 or 16 ( if G+T 10; those case that are not listed are eliminated because no digit is possible for N):
$\rightarrow \ a \ e \ f \ g \ h \ i \ j \ k \ m \ o \ q \ r \\ {G : \ 3\ 1\ 1\ 1\ 1\ 9\ 9 \ 8 \ 3\ 1\ 9\ 9} \\ {T : \ 1\ 3\ 3\ 3\ 3\ 5\ 5\ 6\ 1\ 3\ 5\ 5} \\ {H : \ 7\ 5\ 5\ 5\ 5\ 3\ 3\ 2\ 7\ 5\ 3\ 3} \\ {N : \ 4\ 4\ 7\ 2\ 9\ 1\ 8\ 1\ 2\ 2\ 1\ 7} \\ {E : \ 2\ 2\ 9\ 4\ 7\ 4\ 7\ 4\ 4\ 4\ 4\ 8} \\ {A : \ 9\ 9\ 2\ 7\ 4\ 7\ 4\ 7\ 8\ 8\ 8\ 4}$
Again, using (1) and (2) to continue the table with M and L so that M+L is 4 or 5 (if A+E=12) , or 14 or 15 (if A+E =11; those cases are not listed are eliminated because no digits are possible for M and L):
$\rightarrow G \ T \ H \ N \ E \ A \ M \ L$
$a_1 \ \, 3\ 1\ 7\ 4\ 2\ 9\ 6\ 8 \\ a_2 \ \, 3\ 1\ 7\ 4\ 2\ 9\ 8\ 6 \\ e_1 \ \, 1\ 3\ 5\ 4\ 2\ 9\ 6\ 8 \\ e_2 \ 1\ 3\ 5\ 4\ 2\ 9\ 8\ 6 \\ e_3 \ \, 1\ 3\ 5\ 4\ 2\ 9\ 7\ 8 \\ e_4 \ \, 1\ 3\ 5\ 4\ 2\ 9\ 8\ 7 \\ f_1 \ \, 1\ 3\ 5\ 7\ 9\ 2\ 6\ 8 \\ f_2 \ \, 1\ 3\ 5\ 7\ 9\ 2\ 8\ 6 \\ g_1\ \, 1\ 3\ 5\ 2\ 4\ 7\ 6\ 9 \\ g_2 \ \, 1\ 3\ 5\ 2\ 4\ 7\ 9 \ 6 \\ g_3 \ \, 1\ 3\ 5\ 2\ 4\ 7\ 6\ 8 \\ g_4 \ \, 1\ 3\ 5\ 2\ 4\ 7\ 8\ 6 \\ h_1 \ \, 1\ 3\ 5\ 9\ 7\ 4\ 6\ 8 \\ h_2 \ \, 1\ 3\ 5\ 9\ 7\ 4\ 8\ 6 \\ i_1 \ \, 9\ 5\ 3\ 1\ 4\ 7\ 6\ 8 \\ i_2 \ \, 9\ 5\ 3\ 1\ 4\ 7\ 8\ 6 \\ k_1 \ \, 8\ 6\ 2\ 1\ 4\ 7\ 5\ 9 \\ k_2 \ \, 8\ 6\ 2\ 1\ 4\ 7\ 9\ 5$
Again continuing the table with I , so that I+M is 7 or 17 (if N+E 10; those cases not listed are eliminated because no digits are possible for I), there remains only one case $g$ with two sub-cases ${(g_2, g_4)}$:
$- G \ T \ H \ N \ E \ A \ I \ M \ L$
$g_2 \ 1 \ \, 3 \ \, 5 \ \, 2 \ \, 4 \ \, 7 \ \, 8 \ \, 9 \ \, 6 \\ g_4 \ 1 \ \, 3 \ \, 5 \ \, 2 \ \, 4 \ \, 7 \ \, 9 \ \, 8 \ \ 6$
Out of which only $g_4$ is correct since M+L = 14 .
Substituting the letters for the digits $9 \, 2 \, 5 \, 7 \, 6 \, 4$ is ${I N H A L E}$.
## Note
• This problem requires handy exercise, and this should be done manually with great concentration.
• I am not good in latex, hence was unable to make perfect matrices. Sorry! Hope you’ll compromise with it. |
# Derivative calculus – Definition, Formula, and Examples
The word derivative is probably the most common word you’ll be hearing when taking your first differential calculus. This entire concept focuses on the rate of change happening within a function, and from this, an entire branch of mathematics has been established.
Derivative in calculus refers to the slope of a line that is tangent to a specific function’s curve. It also represents the limit of the difference quotient’s expression as the input approaches zero.
Derivatives are essential in mathematics since we always observe changes in systems. Other fields, including physics, economics, finance, and even sports analysis, use differential calculus. This shows how important it is for us to understand the core foundation and the common derivative rules if we want to excel in our math classes and other specializations.
This article will show you the fundamental definition of derivatives and how we can derive the different derivative rules discussed in our calculus classes. We’ll also show you why it’s important that we know how to evaluate limits if we want to derive a wide range of derivatives from different functions.
## What is a derivative?
The derivative of a function ,represented by $\dfrac{dy}{dx}$ or $f^{\prime}(x)$, represents the limit of the secant’s slope as $h$ approaches zero.
In our precalculus classes, we’ve learned about secant lines and how we can calculate the rate of change between $(x, f(x))$ and $(x, f(x + h))$ using the formula for slopes.
In fact, we called this the difference quotient and established that the slope of the secant line passing through a curve is equal to the formula shown below.
\begin{aligned}\dfrac{\Delta y}{\Delta x} &= \dfrac{f(x +h) – f(x)}{h} \end{aligned}
How does this relate to derivatives? The derivative of a function is simply the slope of the tangent line that passes through the function’s curve. As the distance between $x$ and $x + h$ gets smaller, the secant line that we’ve shown will approach the tangent line representing the function’s derivative.
This means that the tangent line is simply the result of secant lines having a distance between $x$ and $x + h$ that are significantly small and where $h \rightarrow 0$.
This means that the slope of the tangent line simply equal to the limit of the difference quotient as $\boldsymbol{h}$ approaches zero.
\begin{aligned}m_{\text{tangent}}&= \lim_{h \rightarrow 0}\dfrac{\Delta y}{\Delta x}\\ &=\lim_{h \rightarrow 0} \dfrac{f(x +h) – f(x)}{h} \end{aligned}
This is the fundamental definition of derivatives. We denote derivatives as $\dfrac{dy}{dx}$, which represents its very definition.
\begin{aligned}\dfrac{d}{dx} f(x)&= \dfrac{dy}{dx} \\&=\lim_{h \rightarrow 0} \dfrac{f(x +h) – f(x)}{h} \end{aligned}
We can also represent the derivative of $f(x)$ as either $f’(x)$ or $y’$. This is also known as the first derivative of the function.
## How to find the derivative of a function?
When a function is differentiable, we can find its derivative using its definition. Recall that the derivative is represented by $\dfrac{dy}{dx}$ or $f’(x)$ and can be determined using the formula shown below.
\begin{aligned}\dfrac{dy}{dx} &=\lim_{h \rightarrow 0} \dfrac{f(x +h) – f(x)}{h} \end{aligned}
This means that it’s important to find the difference quotient of the function first and then limit the resulting function as $h$ approaches $0$. Here are some steps to guide you in evaluating the derivatives of a function using its core definition:
• Find the expression for $f(x + h)$ and $f(x)$.
• Simplify the expression and cancel common factors whenever possible.
• Evaluate the resulting expression’s limit as $h \rightarrow 0$.
Let’s say we want to determine $\dfrac{d}{dx} 1 – x^2$. We’ll have to determine its difference quotient first.
\begin{aligned} f(x +h) &= [1 -(x + h)^2]\\&= 1- (x^2 + 2xh + h^2)\\f(x)&= 1- x^2\\\\\dfrac{d}{dx} (1 – x^2) &= \lim_{h \rightarrow 0} \dfrac{f(x + h) – f(x)}{h}\\&=\lim_{h \rightarrow 0} \dfrac{[1 – (x^2 + 2xh + h^2)] – (1- x^2)}{h}\\&=\lim_{h \rightarrow 0} \dfrac{1 – x^2 – 2xh – h^2 – 1+ x^2}{h}\\&=\lim_{h \rightarrow 0} \dfrac{- 2xh – h^2 }{h}\\&=\lim_{h \rightarrow 0} \dfrac{-h(2x + h)}{h} \\&=\lim_{h \rightarrow 0} –(2x +h) \end{aligned}
Now that we have the simplified expression, we can evaluate the limit of $-(2x + h)$ as $h \rightarrow 0$.
\begin{aligned} \dfrac{d}{dx} ( 1 -x^2) &= \lim_{h \rightarrow 0} -2x +h\\&= -2x + 0\\&= -2 \end{aligned}
This means that the derivative of the function, $y = 1-x^2$, is equal to $-2x$. In other notations, we have, $y’ = -2x$, $f’(x) = – 2x$, or $\dfrac{dy}{dx} = -2x$.
This process is used repeatedly to derive our long list of derivative rules. Knowing how to find the derivative of functions in this method will help you appreciate our established derivative rules. For now, let’s try more examples and know the definition of the derivative by heart.
Example 1
Find the derivative of $g(x) = \dfrac{2x}{x – 4}$ using the definition of derivative.
Solution
We’ll always go back to the derivative’s fundamental definition to find $\dfrac{dy}{dx}$.
\begin{aligned}g’(x) &= \dfrac{d}{dx} g(x)\\&= \lim_{h \rightarrow 0} \dfrac{g(x +h) – g(x)}{h} \end{aligned}
Let’s simplify the expression for $\dfrac{g(x +h) – g(x)}{h}$.
\begin{aligned} g(x + h) &= \dfrac{2(x+ h)}{[(x + h) – 4)]}\\&= \dfrac{2(x + h)}{x +h – 4}\\g(x) &= \dfrac{2x}{x -4}\\\\\dfrac{dy}{dx} &= \lim_{h \rightarrow 0} \dfrac{g(x +h) – g(x)}{h}\\&= \lim_{h \rightarrow 0} \dfrac{\dfrac{2(x + h)}{x + h -4} – \dfrac{2x}{x -4}}{h} \end{aligned}
To simplify the complex rational expression, we can add the rational expressions found in the numerator.
\begin{aligned} \dfrac{dy}{dx} &= \lim_{h \rightarrow 0} \dfrac{\dfrac{(2x + 2h) (x – 4)}{(x + h -4)(x -4)} – \dfrac{2x(x+h -4)}{(x -4)(x + h-4)}}{h}\\&= \lim_{h \rightarrow 0} \dfrac{1}{h} \left[\dfrac{2x^2 -8x + 2xh – 8h}{(x + h -4)(x -4)} – \dfrac{2x(x+h -4)}{(x -4)(x + h-4)} \right ]\\&= \lim_{h \rightarrow 0} \dfrac{1}{h}\cdot \dfrac{2x^2 – 8x +2xh -8h -2x(x+h -4)}{(x + h -4)(x -4)}\\&= \lim_{h \rightarrow 0} \dfrac{1}{h} \cdot \dfrac{2x^2 – 8x +2xh -8h – 2x^2 – 2xh +8x}{(x + h -4)(x -4)}\\&= \lim_{h \rightarrow 0} \dfrac{1}{h} \cdot \dfrac{- 8h}{(x + h -4)(x -4)} \end{aligned}
We can simplify the expression further then eventually evaluate the limit as $h \rightarrow 0$.
\begin{aligned} \dfrac{dy}{dx} &= \lim_{h \rightarrow 0} \dfrac{1}{h} \cdot \dfrac{- 8h}{(x + h -4)(x -4)}\\&= \lim_{h \rightarrow 0} -\dfrac{8}{(x + h -4)(x -4)} \\&= -\dfrac{8}{(x+ 0-4)(x – 4)} \\&= -\dfrac{8}{(x – 4)^2}\end{aligned}
This means that the derivative of $g(x)$ is equal to $-\dfrac{8}{(x – 4)^2}$. We can also write this as $g’(x) = – \dfrac{8}{(x-4)^2}$ or $y’ = – \dfrac{8}{(x-4)^2}$.
Example 2
Find the derivative of $f(x) = \sqrt{x}$ using the definition of derivative.
Solution
If we let $f(x) = \sqrt{x}$, we can find its derivative using the fact that $\dfrac{dy}{dx} = \lim_{h \rightarrow 0} \dfrac{f(x +h) – f(x)}{h}$. We can begin proving the rule by finding the expression for $f(x + h)$ and eventually the quotient difference of the $f(x) = \sqrt{x}$.
\begin{aligned} f(x + h) &= \sqrt{x + h}\\f(x) &= \sqrt{x}\\\\\dfrac{dy}{dx} &= \lim_{h \rightarrow 0} \dfrac{f(x +h) – f(x)}{h}\\&= \lim_{h \rightarrow 0} \dfrac{\sqrt{x + h} – \sqrt{x}}{h} \end{aligned}
What we can do is reverse the process of rationalizing the denominator by multiplying both the numerator and denominator by $\sqrt{x + h} + \sqrt{x}$ then use the algebraic property, $(a- b)(a + b)^2 = a^2 – b^2$.
\begin{aligned} \dfrac{dy}{dx} &= \lim_{h \rightarrow 0} \dfrac{\sqrt{x + h} – \sqrt{x}}{h} \cdot \dfrac{\sqrt{x+h}-\sqrt{x}}{\sqrt{x+h} + \sqrt{x}}\\&= \lim_{h \rightarrow 0}\dfrac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h (\sqrt{x+h} + \sqrt{x})}\\&= \lim_{h \rightarrow 0}\dfrac{(\sqrt{x+h})^2-(\sqrt{x})^2}{h (\sqrt{x+h} + \sqrt{x})}\\&= \lim_{h \rightarrow 0}\dfrac{x+h-x}{h (\sqrt{x+h} + \sqrt{x})} \\&= \lim_{h \rightarrow 0}\dfrac{h}{h (\sqrt{x+h} + \sqrt{x})} \\&= \lim_{h \rightarrow 0}\dfrac{1}{ (\sqrt{x+h} + \sqrt{x})} \end{aligned}
Let’s evaluate the limit of this expression now to find the derivative of $f(x) = \sqrt{x}$.
\begin{aligned} \dfrac{dy}{dx} &= \lim_{h \rightarrow 0}\dfrac{1}{ (\sqrt{x+h} + \sqrt{x})}\\&= \dfrac{1}{\sqrt{x + 0}+\sqrt{x}}\\&= \dfrac{1}{2\sqrt{x}} \end{aligned}
This means that $f’(x) = \dfrac{1}{2\sqrt{x}}$ or $\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}}$.
### Practice Questions
1. According to the derivative rule for constants, $\dfrac{d}{dx} c = 0$, where $c$ is a constant.
2. Find the derivative of $f(x) = 4x^3 – 2x^2 + 1$ using the definition of derivative.
3. Find the derivative of $g(x) = -4(x -1)^2$ using the definition of derivative.
4. Find the derivative of $h(x) = \dfrac{3x}{x – 9}$ using the definition of derivative.
\begin{aligned}\dfrac{d}{dx} &= \lim_{h \rightarrow 0} \dfrac{(c + h) – c}{h}\\ &= \lim_{h \rightarrow 0} \dfrac{h}{h}\\ &= \lim_{h \rightarrow 0} 1\\ &= 0\end{aligned}
2. $f’(x) = 12x^2 – 4x$
3. $g’(x) = 8 – 8x$
4. $h’(x) = -\dfrac{27}{(x – 9)^2}$ |
# Place Value Chart
• Introduction
• What is a Place Value Chart
• Place Value Chart with Decimals
• Comparing Place Value with Face Value
• Place Value Versus Face Value
• International Place Value Chart
• Solved Examples
• Practice Problems
## Introduction
In this topic, we will explore a fundamental mathematical concept that helps us understand the value of digits in a number based on their position. Imagine you have a number, let's say 478. The place value chart helps us break down this number into its components, understanding that 4 represents 400, 7 represents 70, and 8 represents 8 in 478. Each digit's position in the number tells us its value in the number. For instance, in 478, the 4 is in the hundreds place, the 7 is in the tens place, and the 8 is in the ones place. Understanding place value is crucial for performing arithmetic operations and manipulating numbers efficiently.
## What is a Place Value Chart?
A place value chart is a visual tool used in mathematics to understand the value of digits in a number based on their position. It helps break down numbers into their components, indicating the value of each digit within the number.
Example: What is the place value of 6, 4, and 2 in 642?
Solution:
In this number:
The digit 6 is in the hundreds place, representing 600.
The digit 4 is in the tens place, representing 40.
The digit 2 is in the ones place, representing 2.
We can use the place values of the digits of a number to write the number in expanded form.
642 = 600 (place value of 6) + 40 (place value of 4) + 2 (place value of 2)
## Place Value Chart with Decimals
If we are dealing with a decimal, we can also assign a place value to the digits after the decimal. A place value chart with decimals is a visual representation used in mathematics to understand the value of digits in decimal numbers based on their position. It includes places for whole numbers as well as decimal fractions
Here's an example of a place value chart with decimals:
Each column represents a specific place value, with the digits indicating the value of each place.
For example, in the number 10.739:
The digit 1 is in the tens place.
The digit 0 is in the ones place.
The digit 7 is in the tenths place.
The digit 3 is in the hundredths place.
The digit 9 is in the thousandths place.
## Comparing Place Value with Face Value
Place Value:
Place value refers to the value of a digit based on its position within a number. Each digit in a number holds a specific place value determined by its position. For example, in the number 539, the digit 5 is in the hundreds place, indicating a value of 500, the digit 3 is in the tens place, indicating a value of 30, and the digit 9 is in the ones place, indicating a value of 9. Understanding place value is crucial for accurately reading and manipulating numbers, especially when dealing with large numbers.
Face Value:
Face value, on the other hand, refers to the actual value of the digit itself, regardless of its position within the number. For instance, in the number 539, the face value of the digit 5 is simply 5, the face value of the digit 3 is 3, and the face value of the digit 9 is 9.
## International Place Value Chart
A numerical representation is utilized globally to denote the value of digits within a number. It organizes places from right to left, including ones, tens, hundreds, thousands, ten thousand, hundred thousand, millions, ten million, hundreds millions, billions, and so forth. Each place value is ten times greater than the one to its immediate right.
## Solved Examples
Example 1. What is the place value of 5 in the number 6,752?
Solution:
The number is 6,752.
5 is in the tens place.
Therefore, the place value of 5 is $$5 \times 10 = 50$$.
Example 2. What is the face value of 7 in the number 9,714?
Solution:
The number is 9,714.
The face value of 7 is 7.
Example 3. Write 769.8 in expanded form.
Solution:
The number is 769.8.
7 is in the hundreds place, 6 is in the tents place, 9 in the ones place and 8 in the tenths place.
Therefore, the expanded form of 769.8 is $$700 + 60 + 9 + 0.8$$.
Example 4. What is the place value of the digit 6 in the number 2.64?
Solution:
The number is 2.64.
6 is in the tenth place.
Therefore, the place value of 6 is $$6 \times 0.1 = 0.6$$.
Example 5. What is the place value of the digit 9 in 8,493.2167?
Solution:
The number is 8,493.21.
6 is in the thousandths place.
Therefore, the place value of 6 is $$6 \times 0.001 = 0.006$$.
## Practice Problems
Q1. What is the place value of the digit 5 in the number 4,587?
1. 500
2. 50
3. 5
4. 5,000
Q2. What is the face value of the digit 9 in the number 9,234?
1. 9
2. 90
3. 900
4. 9,000
Q3. Which of the following numbers has a place value of 500?
1. 1,275
2. 5,050
3. 543
4. 12.005
Q4. What is the place value of the digit 3 in the number 27.634?
1. 0.3
2. 0.03
3. 3
4. 30
Q5. A number has 5 tens, 4 ones, and 3 tenths. What is the number?
1. 53.4
2. 54.3
3. 45.3
4. 34.5
Q6. Select the the expanded form of 176.983.
1. 1000 + 700 + 60 + 9 + 0.8 + 0.03
2. 100 + 70 + 6+ 0.9 + 0.08 + 0.003
3. 10 + 7 + 0.06 + 0.009 + 0.0008 + 0.0003
4. 3000 + 80 + 9 + 0.6 + 0.07 + 0.001
Q1. What is a place value chart, and why is it important?
Answer: A place value chart is a visual representation used in mathematics to understand the value of digits in a number based on their position. It's crucial because it helps in breaking down numbers into their components and comprehending the significance of each digit within the number.
Q2. How do you read a place value chart?
Answer: To read a place value chart, start from the right side and move towards the left. Each column represents a specific place value, such as ones, tens, hundreds, etc. The digit in each column indicates the value of that place, which increases by a factor of ten as you move leftward.
Q3. What is the difference between place value and face value?
Answer: Place value refers to the value of a digit based on its position within a number, while face value is the actual numerical value of the digit itself, irrespective of its position. Understanding the distinction between these two concepts is fundamental in mathematics.
Q4. How do you determine the place value of a digit in a number?
Answer: To determine the place value of a digit, identify its position within the number and refer to the corresponding column in the place value chart. The value of the digit is determined by its position relative to the decimal point or the rightmost digit.
Q5. Can a place value chart be used for both whole numbers and decimals?
Answer: Yes, a place value chart can be used for both whole numbers and decimals. For whole numbers, it includes places such as ones, tens, hundreds, etc., while for decimals, it includes places such as tenths, hundredths, thousandths, etc. The chart helps in understanding the value of digits in any numerical context. |
Definition
Chapter 8 Class 7 Rational Numbers
Concept wise
Rational numbers are numbers which can be made by dividing two integers.
Example:
If we divide 1 by 2,
we get 1/2
Which is a rational number
Similarly,
If we divide 2 by 3,
We get 2/3
Which is a rational number
Similarly, rational numbers can be
3/8, 4/6, 9/21 and son on
Is 2 a rational number?
2 can be written
as 2 = 2/1
Since we have integers on numerator as well as denominator,
So, it is a rational number
Is (-4)/9 a rational number?
Since (-4)/9 has integers on numerator as well as denominator,
So, it is a rational number
Is –31 a rational number?
–31 can be written
as –31 = (-31)/1
Since we have integers on numerator as well as denominator,
So, it is a rational number
And with 0?
Let’s see
Is 0/8 a rational number?
Since we have integers on numerator as well as denominator,
So, it is a rational number
Is 5/0 a rational number?
We cannot have 0 in the denominator
So, 5/0 is not a rational number
It is actually not defined.
To mathematically define Rational numbers
We say that Rational numbers are numbers in the form p/q, where p & q are integers, and q ≠ 0
But what about Natural Numbers, Whole Numbers and Integers?
We know that
Natural Numbers − Numbers starting from 1
Eg: 1, 2, 3, 4, 5……..
Whole Numbers − Numbers starting from 0
0, 1, 2, 3, 4, 5……..
Integers − Integers are both positive & negative numbers & zero
….. −3, −2, −1, 0, 1, 2, 3,…..
Now,
2 = 2/1
So, it is in p/q form
∴ It is rational number
So, all natural numbers are rational numbers
Similarly,
0 = 0/1
So, it is in p/q form
∴ It is rational number
So, all whole numbers are rational numbers
And,
–3 = (-3)/1
So, it is in p/q form
∴ It is rational number
So, all integers are rational numbers
∴ All natural numbers, whole numbers, integers are rational numbers
Transcript
Rational Numbers Numbers of the form 𝑝/𝑞 where p and q are integers and q ≠ 0 Example 2/3 , (−4)/5, 8 Is 0 a rational Number? We can write 0 as 0 = 0/1 So, 0 is a rational number All Natural Numbers are Whole Numbers All Whole Numbers are Integers All Integers are Rational Numbers |
Home > Numbers and Pre-Algebra > Fractions > Division of Fractions
# Division of Fractions
Division of fractions involves the following steps:
• express any mixed numerals as improper fractions
• change the divide sign to multiplication sign
• take the reciprocal of the fraction after the division sign (i.e. flip the fraction upside down)
• now it becomes a case of multiplication of fractions, and you have to follow steps to multiply fractions
## Examples of division of fractions:
1. 2 ÷ = (we are taking the reciprocal of the fraction after the division sign, and changing the division sign to a multiplication sign)
= 12
2. 1 ÷ First change the mixed numerals into improper fractions. So
1 ÷ = ÷
= x (take reciprocal and change to multiply sign)
=
=
When we divide a number with a fraction, the number of divided parts increase, therefore when we divide the fraction, the answer becomes larger.
1. For example, take 2 ÷ . Here we have 2 full blocks, and divide each block into two, thus making 4 parts.2 ÷ = 2 x 2 = 4
2. Similarly 2 ÷ = 2 x 3 = 6
3. What about ÷ ?Here we need to ask ourselves the question “what should be the original block if when cut will give half”
So ÷ = 1
4. ÷ Here we have to ask “what should be the original block to get when cut in half”
÷ x =
5. ÷ = x
=
= |
# Exponential Functions are Linearly Independent
## Problem 73
Let $c_1, c_2,\dots, c_n$ be mutually distinct real numbers.
Show that exponential functions
$e^{c_1x}, e^{c_2x}, \dots, e^{c_nx}$ are linearly independent over $\R$.
Contents
## Hint.
1. Consider a linear combination $a_1 e^{c_1 x}+a_2 e^{c_2x}+\cdots + a_ne^{c_nx}=0.$
2. Differentiate this equality $n-1$ times and you will get $n$ equations.
3. Write a matrix equation for the system. You will see the Vandermonde matrix.
## Proof.
Suppose that we have a linear combination of these functions that is zero.
Namely, suppose we have
$a_1 e^{c_1 x}+a_2 e^{c_2x}+\cdots + a_ne^{c_nx}=0$ for some real numbers $a_1, a_2, \dots, a_n$.
We want to show that the coefficients $a_1, a_2, \dots, a_n$ are all zero.
By differentiating the equation, we obtain
$a_1c_1e^{c_1x}+a_2c_2e^{c_2x}+\cdots +a_n c_n e^{c_n x}=0.$ Differentiating repeatedly we further obtain the equalities
\begin{align*}
& a_1c_1^2e^{c_1x}+a_2c_2^2e^{c_2x}+\cdots +a_n c_n^2 e^{c_n x}=0\\
& a_1c_1^3e^{c_1x}+a_2c_2^3e^{c_2x}+\cdots +a_n c_n^3 e^{c_n x}=0\\
& \dots \\
& a_1c_1^{n-1}e^{c_1x}+a_2c_2^{n-1}e^{c_2x}+\cdots +a_n c_n^{n-1}
e^{c_n x}=0\\
\end{align*}
We rewrite these $n$ equations into the following matrix equation.
$\begin{bmatrix} 1 & 1 & \dots &1 \\ c_1 & c_2 & \dots & c_n \\[3pt] c_1^2 & c_2^2 & \dots & c_n^2 \\[3pt] \vdots & \vdots & \vdots & \vdots \\[3pt] c_1^{n-1} & c_2^{n-1} & \dots & c_n^{n-1} \end{bmatrix} \begin{bmatrix} a_1 e^{c_1x} \\ a_2 e^{c_2x} \\ \vdots \\ a_n e^{c^nx} \end{bmatrix}=\begin{bmatrix} 0 \\ \vdots \\ 0 \end{bmatrix} \tag{*}$
The determinant of the left matrix is
$\det \begin{bmatrix} 1 & 1 & \dots &1 \\ c_1 & c_2 & \dots & c_n \\[3pt] c_1^2 & c_2^2 & \dots & c_n^2 \\[3pt] \vdots & \vdots & \vdots & \vdots \\[3pt] c_1^{n-1} & c_2^{n-1} & \dots & c_n^{n-1} \end{bmatrix}=\prod_{i<j}(c_j-c_i)$ by the Vandermonde determinant.
Since by assumption $c_1,\dots, c_n$ are distinct, the determinant is not zero.
Therefore by multiplying equality (*) by the inverse on the left, we obtain
$\begin{bmatrix} a_1 e^{c_1x} \\ a_2 e^{c_2x} \\ \vdots \\ a_n e^{c^nx} \end{bmatrix}=\begin{bmatrix} 0 \\ \vdots \\ 0 \end{bmatrix}.$ Since the functions $e^{c_i x}$ are always positive, we must have $a_1=a_2=\cdots=a_n=0$ as required.
Therefore the functions $e^{c_1 x}, \dots, e^{c_n x}$ are linearly independent.
## Comment.
The determinant that we considered above is called the Wronskian for the set of functions $\{e^{c_1x}, e^{c_2x}, \dots, e^{c_nx}\}$.
## Related Question.
The following problems are more concrete versions of the current problem.
Problem.
Let $C[-1, 1]$ be the vector space over $\R$ of all continuous functions defined on the interval $[-1, 1]$. Let
$V:=\{f(x)\in C[-1,1] \mid f(x)=a e^x+b e^{2x}+c e^{3x}, a, b, c\in \R\}$ be a subset in $C[-1, 1]$.
(a) Prove that $V$ is a subspace of $C[-1, 1]$.
(b) Prove that the set $B=\{e^x, e^{2x}, e^{3x}\}$ is a basis of $V$.
(c) Prove that
$B’=\{e^x-2e^{3x}, e^x+e^{2x}+2e^{3x}, 3e^{2x}+e^{3x}\}$ is a basis for $V$.
See the post ↴
Exponential Functions Form a Basis of a Vector Space
for the solution.
By calculating the Wronskian, determine whether the set of exponential functions
$\{e^x, e^{2x}, e^{3x}\}$ is linearly independent on the interval $[-1, 1]$.
The solutions is given in the post ↴
Using the Wronskian for Exponential Functions, Determine Whether the Set is Linearly Independent
### 1 Response
1. 08/28/2017
[…] The solution is given by Exponential Functions are Linearly Independent. […]
##### Conditions on Coefficients that a Matrix is Nonsingular
(a) Let $A=(a_{ij})$ be an $n\times n$ matrix. Suppose that the entries of the matrix $A$ satisfy the following relation....
Close |
# Problem of the Week Problem D and Solution Add On!
## Problem
When sixty consecutive odd integers are added together, their sum is $$4800$$.
Determine the largest of the sixty integers.
Note:
In solving the above problem, it may be helpful to use the fact that the sum of the first $$n$$ positive integers is equal to $$\dfrac{n(n+1)}{2}$$. That is, $1 + 2 + 3 + … + n = \frac{n(n+1)}{2}$ For example, $$1 + 2 + 3 + 4 + 5 = 15$$, and $$\dfrac{5(6)}{2} = 15.$$
Also, $$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36 \text{, and } \dfrac{8(9)}{2} = 36.$$
## Solution
In this solution we will solve using patterns.
Let $$a$$ represent the smallest number. Since the numbers are odd, they increase by $$2$$. So the second number is $$(a+2)$$, the third is $$(a+4)$$, the fourth is $$(a+6)$$, and so on. What does the sixtieth number look like?
A closer look at the numbers reveals that the second number is $$(a+1(2))$$, the third is $$(a+2(2))$$, the fourth is $$(a+3(2))$$, and so on. Following the pattern, the sixtieth number is $$(a+59(2))=a+118$$. Then \begin{aligned} a+(a+2)+(a+4)+(a+6)+\cdots +(a+118)&=4800\\ 60a+2+4+6+\cdots +118&=4800\\ 60a+2(1+2+3+\cdots +59)&=4800\\ 60a+2\left(\frac{59\times 60}{2}\right)&=4800, \text{ using the helpful formula.}\\ 60a+3540&=4800\\ 60a&=1260\\ a&=21\\ a+118&=139\end{aligned} Therefore, the largest odd integer in the sum is $$139$$.
Solution 2
In this solution we will use averages to solve the problem.
Let $$A$$ represent the average of the sixty consecutive odd integers. The average times the number of integers equals the sum of the integers. Since the sum of the sixty integers is 4800, then $$60A=4800$$ and $$A=80$$.
Now the integers in the sequence are consecutive odd integers. The average is even. It follows that $$30$$ integers are below the average and $$30$$ integers are above. We are looking for the $$30^{\text{\small {th}}}$$ odd integer after the average. In fact we want the $$30^{\text{\small {th}}}$$ odd integer after the odd number $$79$$, the first odd integer below the average. This integer is easily found, $79 +30(2)=79+60=139.$ Therefore, the largest odd integer in the sum is $$139$$.
Solution 3
In this solution we will use arithmetic sequences. This solution is presented last since many students in grade 9 or 10 have not encountered arithmetic sequences yet.
An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. The general term, $$t_n$$, of an arithmetic sequence is $$t_n=a+(n-1)d$$, where $$a$$ is the first term, $$d$$ is the difference between consecutive terms, and $$n$$ is the term number. The sum of the first $$n$$ terms of an arithmetic sequence, $$S_n$$, can be found using the formula $$S_n=\frac{n}{2}\left(2a+(n-1)d\right)$$, where $$a$$, $$d$$, and $$n$$ are the same variables used in the general term formula.
Let $$a$$ represent the first term in the sequence. Since the integers in the sequence are consecutive and odd, the integers go up by two. Therefore, $$d=2$$. Since there are $$60$$ terms in the sequence, $$n=60$$. The sum of the sixty integers in the sequence is $$4800$$, so $$S_{60}=4800$$. \begin{aligned} S_n&=\frac{n}{2}\left(2a+(n-1)d\right)\\ 4800&=\frac{60}{2}\left(2a+(60-1)(2)\right)\\ 4800&=30(2a+59(2))\\ \text{Dividing by 30,}~160&=2a+118\\ 42&=2a\\ 21&=a\end{aligned} Since we want the largest integer in the sequence, we are looking for the sixtieth term. \begin{aligned} \text{Using~} t_n&=a+(n-1)d, \text{ with } a=21, d=2, n=60\\ t_{60}&=21+59(2)\\ t_{60}&=139\end{aligned} Therefore, the largest odd integer in the sum is $$139$$. |
# Solve the congruence $6x+15y \equiv 9 \pmod {18}$
Solve the congruence $6x+15y \equiv 9\pmod {18}$
Approach:
$(6,18)=6$, so $$15y \equiv 9\pmod 6$$ $$15y \equiv 3\pmod 6$$
So the equation will have $(15,6)$ solutions. Now we divide by 3
$$5y \equiv 1\pmod 2$$.
Solving the Diophantine equation we get $y \equiv1\pmod 2$, so $y=1+2m$
$$6x \equiv 9-15y\pmod {18}$$ $$6x \equiv 9-15(1+2m)\pmod {18}$$ $$6x \equiv -6-30m\pmod {18}$$
Divide by 6
$$x \equiv -1-5m\pmod 3$$
The right solution is $x-m \equiv 2\pmod 3$. I have $x+5m \equiv 2\pmod 3$ $$x-m+6m \equiv 2\pmod 3$$ $$x-m \equiv 2\pmod 3$$
Wolfram alpha says $y=1+2t$ and $x=t+3d+2$
Dividing everything by $3$, including the modulus, we get the equivalent congruence $$2x+5y\equiv 3\pmod{6}.$$ It is convenient to rewrite this as $2x-y\equiv 3\pmod{6}$, or equivalently $y\equiv 2x-3\equiv 2x+3\pmod{6}$.
Now we have a parametric solution: $x\equiv t\pmod{6}$, $y\equiv 2t+3\pmod{6}$. To write it out at length, set $t=0,1,2,3,4,5$. The first two solutions are then $x\equiv 0\pmod{6}$, $y\equiv 3\pmod{6}$ and $x\equiv 1\pmod{6}$, $y\equiv 5\pmod{6}$. Four more to go.
I hope my answer is okay. $6x+15y \equiv 9(mod 18)$
$=> 2x+5y \equiv 3(mod 6)$
$=> 2x-y \equiv 3(mod 6)$
i.e $2x-y-3 \equiv 0(mod 2)$ and $\equiv 0(mod 3)$
this means $y+1 \equiv 0(mod 2)$ and $x+y \equiv 0(mod 3)$
Write $y=2p+1$ and so $x-p +1 \equiv 0(mod 3)$
Then use parametric equations.
• why is $2x-y\equiv3(mod6)$ Jun 30, 2016 at 6:02
• write $5y \equiv 6y-y$
– null
Jun 30, 2016 at 6:07
• you get $2x-y\equiv 0(mod 3)$. How do you get $x+y\equiv 0(mod 3)$. Is it because $-x-y\equiv 0(mod 3)$ implies $x+y\equiv 0(mod 3$)? Jun 30, 2016 at 6:14
• $2x-y-3 \equiv 0(mod3)$ implies $3x-x-y-3 \equiv 0(mod3)$ i.e $-x-y \equiv 0(mod3)$ i.e $x+y \equiv 0(mod3)$
– null
Jun 30, 2016 at 6:19
• can you provide another exercise?. My book just has that one? Jun 30, 2016 at 6:21
$y = 2c + 1$
$x = c + 3d + 2$ |
Concepts
Class 9
Chapter 8 Class 9 - Motion
## What is Average Velocity?
It is the total distance travelled in a particular diection divided by total time taken
We can say that
Average Velocity = Total Displacement/Total Time Taken
AVERAGE SPEED FORMULA
Average Speed = Total Distance/Total Time Taken
AVERAGE VELOCITY FORMULA
Average Velocity =Total Displacement/Total Time Taken
Summary
Average Velocity may or may not be equal to Average Speed
## Questions
Q 2 Page 102 - Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
NCERT Question 2 - Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Example 8.3 - Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average Velocity of Usha.
Get live Maths 1-on-1 Classs - Class 6 to 12
### Transcript
Example 1 A person travels from Point A to Point B for 30 meters towards south in 4 Seconds. Then he travels from Point B to Point C for 40 Meters towards east in 6 Seconds. What is his Average Speed and Average Velocity? Calculating Average Speed Total Distance Travelled = AB + BC = 30 + 40 = 70 Meters Total Time Taken = 4 + 6 seconds = 10 Seconds Average Speed = (𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒)/(𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒) = 70/10 = 7 m/s Calculating Average Velocity We first need to calculate Total Displacement Displacement = AC Now, Δ ABC is a right angled triangle By Pythagoras Theorem Hypotenuse2 = Height2 + Base2 AC2 = AB2 + BC2 AC2 = 302 + 402 AC2 = 30 × 30 + 40 × 40 AC2 = 9000 + 1600 AC2 = 2500 AC2 = 50 × 50 AC = 50 Meters Thus, Displacement = AC = 50 m towards South East Now, Total Time Taken = 4 + 6 = 10 Seconds Average Velocity = (𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡)/(𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒) = 50/10 = 5 m/s towards South East |
mersenneforum.org (https://www.mersenneforum.org/index.php)
- Number Theory Discussion Group (https://www.mersenneforum.org/forumdisplay.php?f=132)
- - Algorithm for combining Carmichael numbers (https://www.mersenneforum.org/showthread.php?t=24771)
Algorithm for combining Carmichael numbers
We can combine two Carmichael numbers to form another Carmichael number.
An example: 1729 = 7*13*19
294409 = 37*73*109
Both are of type (6m+1)(12m+1)(18m+1); we get the first by putting m = 1 and the second by putting m = 6.
First step: Check whether for any given value of m, 6m+1, 13m+1 and 18m+1 are primes.If for two given values of m, we get two Carmichael numbers by applying Korselt's criterion. If so their product will be a Carmichael number subject to satisfaction of Korselt's criterion.
[QUOTE=devarajkandadai;525821]We can combine two Carmichael numbers to form another Carmichael number.
An example: 1729 = 7*13*19
294409 = 37*73*109
Both are of type (6m+1)(12m+1)(18m+1); we get the first by putting m = 1 and the second by putting m = 6.
First step: Check whether for any given value of m, 6m+1, 13m+1 and 18m+1 are primes.If for two given values of m, we get two Carmichael numbers by applying Korselt's criterion. If so their product will be a Carmichael number subject to satisfaction of Korselt's criterion.[/QUOTE]
Just combined 3 Carmichael numbers to form one Carmichael number; all three are of type (6m+1)(12m+1)(18m+1). 1729*294409*118901521 =60524817082337881.
Conjecture: There can be many Carmichael numbers of this type i.e. k number of Carmichael numbers can be combined to form one Carmichael number where k belongs to N and greater than 2.
All times are UTC. The time now is 21:54. |
# Multiplication Facts - Tips, Rules and Tricks to Help You Learn
Page content
## Basic Concepts First
Memorizing the entire Multiplication Table can seem quite overwhelming at first. The key to learning your multiplication facts is to break the process down into manageable lessons. This is done through a series of rules or “tricks” that can be learned. Once these have been mastered, you will see that it is only necessary to memorize ten multiplication facts! First, however, there are several key concepts that must be understood. [caption id=“attachment_130454” align=“aligncenter” width=“640”] Multiplication can be performed with basic addition and subtraction[/caption]
• The first is that multiplication is simply a fast way of joining groups of equal size through repeated addition. Let’s look at a problem together:
Sarah has 4 boxes of crayons. There are 3 crayons in each box. How many crayons does Sarah have altogether? This problem can be solved through repeated addition:
3+3+3+3 = 12
A shortened version of this would be to use the multiplication sentence:
4 x 3 = 12
• The second concept that must be understood is what each number in the multiplication problem represents. Let’s look at that same problem again:
Sarah has 4 boxes of crayons. There are 3 crayons in each box. How many crayons does Sarah have altogether?
4 x 3 = 12
In this case, the (4) represents the number of groups in the problem. (There were 4 boxes.) The (3) represents how many objects/items were in each group. (There were 3 crayons in each box.)
• The third concept that will help you with learning your multiplication facts is the Commutative Property of multiplication. This states that when two numbers are multiplied together, the product (or answer) is the same regardless of the order of the numbers. For instance:
3 x 2 = 6
2 x 3 = 6
3 x 2 = 2 x 3
## Rules and Tricks
Now that you understand the basic concepts of multiplication, it is time to look at the repeated patterns which can be found in the multiplication table…and ultimately used to solve times problems. These rules or tricks can be broken down into related fact families to help you learn your multiplication facts:
• 0’s: Anything times 0 equals 0 (0 x 2 = 0).
• 1’s: Anything times 1 equals itself (4 x 1 = 4).
• 2’s: Any number times 2 is doubled (4 x 2 = 8) / (4 + 4 = 8).
• 4’s: Any number times 4 is doubled once and then doubled again (6 x 4 = 24) / (6 + 6 = 12…12 + 12 = 24).
• 5’s: Use what you know about skip counting by 5 when multiplying by 5 (3 x 5 = 15) / (5, 10 15).
• 9’s: When multiplying by 9, use your addition and subtraction. Subtract one from the number you are multiplying. Then, think about the number you would add to it to equal 9. (4 x 9 = 36) / (4 -1 = 3…3 + 6 = 9).
• 10’s: Add a zero to any number that you multiply by 10. (6 x 10 = 60).
• 11’s: Think of 11 as two ones. Any number times 11 is that number times one and then that number times one again. (3 x 11 = 33) / (3 x 1 = 3…3 x 1 = 3).
## Only 10 Problems Are Left to Memorize!
Unfortunately, some problems simply do no fit into the patterns found in the times table. These problems must be memorized. The good news is that once you incorporate the Commutative Property of Multiplication, you are left with 10 problems instead of 20! The circled problems in the Multiplication Table shown represent the 10 problems that you are best off memorizing. Take a look at Math-drills.com for handy worksheets you can print off and quiz your students on. By employing these simple tips and tricks, you will quickly be on your way to learning your multiplication facts! Congratulations!
## References
• The multiplication table image and information offered in this article are based on the author’s experience as a classroom teacher, peer instructor and author of local math curriculum standards
• Image by Gerd Altmann from Pixabay |
## What is a math equation?
A math equation is a statement that two expressions are equal. It is written with an equal sign (=) between the two expressions. For example, the equation `2 + 2 = 4` states that the sum of 2 and 2 is equal to 4.
## What are the different types of math equations?
There are many different types of math equations, each with its own unique properties. Some of the most common types of math equations include:
• Basic math equations: These equations involve the basic arithmetic operations of addition, subtraction, multiplication, and division. For example, the following equations are all basic math equations:
``````2 + 2 = 4
3 - 1 = 2
4 x 5 = 20
5 / 2 = 2.5
``````
• Intermediate math equations: These equations involve more complex mathematical concepts such as linear equations, quadratic equations, polynomial equations, rational equations, exponential equations, and logarithmic equations. For example, the following equations are all intermediate math equations:
``````x + 2 = 5
x^2 + 2x - 3 = 0
(x + 1)/(x - 2) = 3
2^x = 8
log(x) = 10
``````
• Advanced math equations: These equations involve the most complex mathematical concepts, such as trigonometry, calculus, statistics, differential equations, and linear algebra. For example, the following equations are all advanced math equations:
``````sin(x) = cos(x)
dy/dx = x^2
μ = (x1 + x2 + ... + xn)/n
d^2y/dx^2 = x
Ax = b
``````
## How to solve math equations
There are many different ways to solve math equations, depending on the type of equation. However, there are some general steps that can be applied to solving any math equation:
1. Identify the variables. The variables in an equation are the unknown quantities that you are trying to solve for. For example, in the equation `2x + 3 = 7`, the variable is `x`.
2. Isolate the variables. Once you have identified the variables, you need to isolate them on one side of the equation. To do this, you can use the basic arithmetic operations of addition, subtraction, multiplication, and division. For example, to isolate the variable `x` in the equation `2x + 3 = 7`, we can subtract 3 from both sides of the equation:
``````2x + 3 - 3 = 7 - 3
``````
This gives us the following equation:
``````2x = 4
``````
1. Solve for the variables. Once the variables are isolated, you can solve for them by dividing both sides of the equation by the coefficient of the variable. For example, to solve for the variable `x` in the equation `2x = 4`, we can divide both sides of the equation by 2:
``````2x / 2 = 4 / 2
``````
This gives us the following solution:
``````x = 2
``````
## Why is it important to learn how to solve math equations?
There are many reasons why it is important to learn how to solve math equations. Math equations are used in all areas of life, from everyday tasks to complex scientific and engineering problems. Here are just a few of the reasons why it is important to be able to solve math equations:
• To make informed decisions. Math equations can be used to analyze data and make informed decisions about a variety of topics, such as personal finance, health, and the environment.
• To solve real-world problems. Math equations are used to solve a wide range of real-world problems, such as designing bridges, developing new medicines, and predicting the weather.
• To improve your critical thinking skills. Solving math equations requires you to think critically and creatively about the problem at hand. This can help you to develop your problem-solving skills in other areas of your life.
• To prepare for a career. Many careers require strong math skills. For example, engineers, scientists, and accountants all need to be able to solve complex math equations.
## Benefits of being able to solve math equations
There are many benefits to being able to solve math equations, both personal and professional. Here are a few examples:
• Personal benefits:
• Improved problem-solving skills: Solving math equations requires you to think critically and creatively about the problem at hand. This can help you to develop your problem-solving skills in other areas of your life, such as work, school, and relationships.
• Increased confidence: When you are able to solve math equations, it can boost your confidence in your abilities. This can be beneficial in all aspects of your life.
• A better understanding of the world around you: Math is used to explain and understand many of the phenomena that we see in the world around us. By learning how to solve math equations, you can gain a deeper understanding of the world around you.
• Professional benefits:
• Improved job prospects: Many careers require strong math skills. For example, engineers, scientists, and accountants all need to be able to solve complex math equations. By being able to solve math equations, you will be more qualified for a wider range of jobs.
• Higher earning potential: Jobs that require strong math skills tend to pay higher salaries than jobs that do not. By being able to solve math equations, you can increase your earning potential.
• More opportunities for advancement: In many careers, employees with strong math skills are more likely to be promoted to management positions. By being able to solve math equations, you can increase your chances of advancement in your career.
## Conclusion
Math equations are an important part of the modern world. They are used in all areas of life, from everyday tasks to complex scientific and engineering problems. By learning how to solve math equations, you can improve your problem-solving skills, increase your confidence, gain a better understanding of the world around you, and improve your job prospects and earning potential.
## FAQs
### Q: What is the difference between an equation and an expression?
A: An equation is a statement that two expressions are equal. It is written with an equal sign (=) between the two expressions. For example, the equation `2 + 2 = 4` is a statement that the sum of 2 and 2 is equal to 4.
An expression is a mathematical statement that does not contain an equal sign (=). It can be a single number, a variable, or a combination of numbers and variables using the basic arithmetic operations of addition, subtraction, multiplication, and division. For example, the expression `2 + 2` is a mathematical statement that does not contain an equal sign (=).
### Q: What is a variable?
A: A variable is a symbol that represents an unknown quantity. Variables are used in math equations to represent the quantities that we are trying to solve for. For example, in the equation `2x + 3 = 7`, the variable `x` represents the quantity that we are trying to solve for.
### Q: How do you solve equations with multiple variables?
A: There are a number of different ways to solve equations with multiple variables. One common method is called elimination. To solve an equation with multiple variables using elimination, we can follow these steps:
1. Choose two of the variables and eliminate them from the equation. To do this, we can multiply both sides of the equation by a number so that the coefficients of one of the variables become opposite. Then, we can add the two equations together. This will eliminate one of the variables from the equation.
2. Repeat step 1 to eliminate the other variable from the equation.
3. Once all of the variables have been eliminated from the equation, we will be left with a single equation that can be solved for the remaining variable.
### Q: What is a mathematical function?
A: A mathematical function is a relationship between two sets of numbers. The first set of numbers is called the domain of the function, and the second set of numbers is called the range of the function. For each number in the domain of the function, there is a unique number in the range of the function.
### Q: How do you graph mathematical functions?
A: To graph a mathematical function, we can follow these steps:
1. Choose a range of values for the independent variable (the variable that is being input into the function).
2. Calculate the value of the dependent variable (the variable that is being output by the function) for each value of the independent variable.
3. Plot the points that represent the values of the independent and dependent variables on a coordinate plane. The independent variable is plotted on the horizontal axis, and the dependent variable is plotted on the vertical axis.
4. Connect the points with a smooth line. This line is the graph of the mathematical function. Q: What is a mathematical model?
A mathematical model is a mathematical representation of a real-world phenomenon. Mathematical models are used to describe and predict the behavior of real-world systems.
For example, a mathematical model of the weather can be used to predict the likelihood of rain, snow, or other weather conditions. A mathematical model of the human body can be used to predict the effects of different drugs or treatments. And a mathematical model of the economy can be used to predict the effects of different economic policies.
### Q: How do you use mathematical models to solve real-world problems?
To use a mathematical model to solve a real-world problem, we can follow these steps:
1. Identify the real-world problem that you want to solve.
2. Develop a mathematical model of the problem. This involves identifying the relevant variables and relationships between the variables.
3. Solve the mathematical model. This may involve using analytical or numerical methods.
4. Interpret the results of the mathematical model. This involves explaining what the results mean in terms of the real-world problem. |
# Difference between revisions of "2015 AMC 10A Problems/Problem 8"
The following problem is from both the 2015 AMC 12A #6 and 2015 AMC 10A #8, so both problems redirect to this page.
## Problem
Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be $2$ : $1$ ?
$\textbf{(A)}\ 2 \qquad\textbf{(B)} \ 4 \qquad\textbf{(C)} \ 5 \qquad\textbf{(D)} \ 6 \qquad\textbf{(E)} \ 8$
## Solution
This problem can be converted to a system of equations. Let $p$ be Trahee's current age and $c$ be her brother's current age.
The first statement can be written as $p-2=3(c-2)$. The second statement can be written as $p-4=4(c-4)$
To solve the system of equations:
$p=3c-4$
$p=4c-12$
$3c-4=4c-12$
$c=8$
$p=20.$
Let $x$ be the number of years until Trahee is twice as old as her brother.
$20+x=2(8+x)$
$20+x=16+2x$
$x=4$
The answer is $\boxed{\textbf{ }4}$.
~savannahsolver
## See Also
2015 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 9 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
2015 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 7 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# (a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel.
Question:
(a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.
Solution:
(a) There are three resistors of resistances,
R1 = 2 Ω, R2 = 4 Ω, and R3 = 5 Ω
They are connected in parallel. Hence, total resistance (R) of the combination is given by,
$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$
$=\frac{1}{2}+\frac{1}{4}+\frac{1}{5}=\frac{10+5+4}{20}=\frac{19}{20}$
$\therefore R=\frac{20}{19} \Omega$
Therefore, total resistance of the combination is $\frac{20}{19} \Omega$.
(b) Emf of the battery, V = 20 V
Current (I1) flowing through resistor R1 is given by,
$I_{1}=\frac{V}{R_{1}}$
$=\frac{20}{2}=10 \mathrm{~A}$
Current $\left(l_{2}\right)$ flowing through resistor $R_{2}$ is given by,
$I_{2}=\frac{V}{R_{2}}$
$=\frac{20}{4}=5 \mathrm{~A}$
Current (I3) flowing through resistor R3 is given by,
$I_{3}=\frac{V}{R_{3}}$
$=\frac{20}{5}=4 \mathrm{~A}$
Total current, I = I1 + I2 + I3 = 10 + 5 + 4 = 19 A
Therefore, the current through each resister is 10 A, 5 A, and 4 A respectively and the total current is 19 A. |
# What Is The Meaning Of Pemdas In Mathematics?
## How do you use Pemdas?
You can alternatively apply PEMDAS as schools do today: Simplify everything inside the parentheses first, then exponents, then all multiplication and division from left to right in the order both operations appear, then all addition and subtraction from left to right in the order both operations appear.
## What Pemdas stands for?
by Paige Faber, ACDC Peer Advisor. Remember in seventh grade when you were discussing the order of operations in math class and the teacher told you the catchy acronym, “ PEMDAS ” (parenthesis, exponents, multiplication, division, addition, subtraction) to help you remember?
## Why do we use Pemdas?
PEMDAS is an acronym used to remind people of the order of operations. This means that you don’t just solve math problems from left to right; rather, you solve them in a predetermined order that’s given to you via the acronym PEMDAS.
## Does multiplication always come first?
Order of operations tells you to perform multiplication and division first, working from left to right, before doing addition and subtraction. Continue to perform multiplication and division from left to right. Next, add and subtract from left to right. Multiply first.
You might be interested: What Is The Definition Of Mathematics In The Modern World?
## How do you simplify?
To simplify any algebraic expression, the following are the basic rules and steps:
1. Remove any grouping symbol such as brackets and parentheses by multiplying factors.
2. Use the exponent rule to remove grouping if the terms are containing exponents.
3. Combine the like terms by addition or subtraction.
4. Combine the constants.
## What are the four rules of maths?
The four basic Mathematical rules are addition, subtraction, multiplication, and division. Read more.
## What is the correct order of operations?
What it means in the Order of Operations is “Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction”. When using this you must remember that multiplication and division are together, multiplication doesn’t come before division. The same rule applies to addition and subtraction.
## Is Pemdas a rule?
PEMDAS Rule for Math Order of Operations: Conclusion The PEMDAS rule is a popular memory tool for recalling the math order of operations. The rule stands for P: Parenthesis, E: Exponents, M: Multiplying, D: Dividing, A: Adding, S=Subtracting.
## What is another name for Pemdas?
In the United States, the acronym PEMDAS is common. It stands for Parentheses, Exponents, Multiplication/Division, Addition/Subtraction.
## Is Bedmas and Pemdas the same?
BEDMAS stands for Brackets, Exponents, Division, Multiplication, Addition and Subtraction. PEMDAS stands for Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. Both acronyms refer to systems of prioritizing mathematical elements when calculating an equation.
## Do calculators do Pemdas?
This rule is so widely known that an acronym — PEMDAS — is often used to describe this. Windows calculator is just a basic calculator but you can change it so you can include your parentheses. So if you pressed the buttons 1, +, 2, *, 9, Enter, then Windows Calc would interpret that as: 1 + 2 (= 3)
You might be interested: Often asked: Why Mathematics Is Indispensable?
## Why do we do multiplication first?
If there are multiple operations at the same level on the order of operations, move from left to right. you work like this: First notice that, there are no Parentheses or Exponents, so we move to Multiplication and Division. There’s only the one multiplication, so we do that first and end up with 9 – 5 + 2. |
# Why does 17 change signs when solving this equation?
Currently working through some problem sets on straight line equations but I'm just not grasping the last part. Any further explanation or intuition to help the concept sink in would be great.
$$y - y_1 = m(x - x_1) \\ y - 7 = \frac 23(x - 2) \\ 3y - 21 = 2x - 4$$
Then putting it in the form $0 = ax + by + c$
$$0 = 2x - 3y + 17$$
Setting $x = 0$
$$0 = -3y + 17$$
Setting $y = 0$
$$0 = 2x + 17$$
The answer I've given says $x = \frac{-17}2$ <---- Why does the $17$ turn into $-17$? I believe that both $2x$ and $17$ is divided by $2$ to solve for $x$ but I can't seem to get my head around why it turns to $-17$ and I know when someone tells me, the answer is going to be obvious!
Thanks
• the number $17$ is the coefficient of the equation of the straight-line while $$x=\frac{-17}{2}$$ is the intersection point with the x-axes – Dr. Sonnhard Graubner Oct 14 '15 at 15:18
\require{cancel}\begin{align}0 &=2x+17 \\ (0) \color{red}{-17} &= (2x+\cancel{17})\color{red}{-\cancel{17}} & (\text{subtract 17 from both sides}) \\ -17 &= 2x & (\text{simplify}) \\ \color{red}{\frac{\color{black}{-17}}{2}} &=\color{red}{\frac{\color{black}{\cancel{2}x}}{\cancel2}} & (\text{divide both sides by 2}) \\ \frac{-17}{2} &=x & (\text{simplify})\end{align}
In words: our goal was to get the $x$ by itself (on the right) and everything else over to the other side (on the left). So first we got rid of the $17$ on the right by subtracting it from both sides (remember, anything you do to one side you must do to the other). Then there was still that $2$ left over on the right so we divided both sides by it to get rid of it.
• Both $4x+5y-7=0$ and $-4x-5y+7=0$ are actually the same equation. Just multiply the left and right sides of one of the equations by $-1$ to get the other. So you can add $4x$ and $8$ OR subtract $5y$ and add $-15$ -- whichever you like. Adding the $4x$ and $8$ gives you an equation with less negatives -- so maybe that's more visually pleasing, but both are correct. – user137731 Oct 14 '15 at 17:51
• To sum up: \begin{align}4x+5y-7&=0 \\ \color{red}{(-1)}(4x+5y-7)&=\color{red}{(-1)}(0) \\ -4x-5y+7 &=0\end{align} So these two equations are really the same. – user137731 Oct 14 '15 at 17:56 |
When it comes to understanding polynomial graphs, one important aspect to consider is the degree of the graph. This crucial piece of information can reveal a lot about the behavior and characteristics of a polynomial equation. However, determining the degree of a polynomial graph may seem intimidating for those who are not familiar with the concept. In this article, we will explore various methods for calculating and locating the degree of a polynomial graph. By the end, you will have a better understanding of how to identify and solve for the degree of a polynomial equation, allowing for a deeper comprehension of complex functions and their graphical representations.
## Determine Polynomial Graph Degree
Polynomial graphs are used to represent a wide variety of functions, from simple linear equations to more complex equations with multiple terms. These graphs are important in many fields such as mathematics, engineering, and economics, as they allow us to visualize and analyze mathematical relationships.
One of the key aspects of a polynomial graph is its degree. The degree of a polynomial is the highest exponent of its variable. This information tells us important characteristics about the graph, such as the shape, turning points, and end behavior. In this article, we will discuss various methods for determining the degree of a polynomial graph and understand its significance.
### Polynomial Graph Degree Calculation
The degree of a polynomial can be calculated by examining its equation. Let’s take an example of a simple polynomial equation: f(x) = 2×3 – 5×2 + 3x + 1. The exponents of the variable x in this equation are 3, 2, and 1. The highest exponent is 3, which means that the degree of this polynomial is 3.
In general, the degree of a polynomial can be found by looking at the highest exponent of the variable. If there are multiple terms in the equation, we only consider the highest exponent. For instance, if the polynomial is f(x) = 5×2 + 3x + 10, the degree would still be 2 because the exponent of x is 2 in the first term.
It is important to note that only variables with whole number exponents are considered when determining the degree of a polynomial. If there are any square roots or fractions in the equation, those terms are not included.
### Locating the Degree of a Polynomial Graph
One of the easiest ways to locate the degree of a polynomial graph is by looking at its visual representation on a graph. The degree of a polynomial is directly related to the number of turning points or curves on the graph.
For instance, a polynomial with a degree of 1 will have exactly one turning point, while a polynomial with a degree of 2 will have two turning points. This pattern continues, so a polynomial with a degree of 5 will have five turning points or curves.
Therefore, to locate the degree of a polynomial graph, we can simply count the number of turning points or curves on the graph. However, this method may not be accurate for more complex polynomial graphs as they may have multiple curves that are not easily distinguishable.
### Finding the Degree of a Polynomial Equation
To find the degree of a polynomial equation, we can follow a general rule that states – the degree of a polynomial is equal to the sum of the exponents in the highest term of the equation.
For instance, if the polynomial equation is f(x) = 6×4 + 2×3 + 8×2 + 5, the degree would be calculated as follows: 4+3+2 = 9. Therefore, the degree of this polynomial is 4.
This rule works for all types of polynomial equations, whether they are in standard form, factored form, or expanded form.
### Identifying the Degree of a Polynomial Graph
There are various characteristics of a polynomial graph that can help us identify its degree. These include the number of x-intercepts, y-intercepts, and local extrema.
As mentioned earlier, polynomials with a degree of n will have exactly n x-intercepts. For instance, a polynomial with a degree of 3 will have three x-intercepts. Similarly, the y-intercept of a polynomial graph is equal to its constant term, so based on the number of non-zero terms in the equation, we can determine the degree.
Additionally, polynomials with a degree of n will have at most n-1 local extrema. This means that a polynomial with a degree of 4 can have at most 3 local extrema. By examining the number of local extrema, we can get an idea of the degree of the polynomial graph.
### Methods for Determining Polynomial Degree
Apart from the methods mentioned above, there are other techniques that can help us determine the degree of a polynomial graph. These include:
• Looking at the leading coefficient: The leading coefficient is the coefficient of the term with the highest exponent. The degree of a polynomial is equal to the degree of the leading term. For example, if the leading term has a coefficient of 5, the degree of the polynomial is 5.
• Using the quadratic formula: For quadratic equations (polynomials with a degree of 2), we can use the quadratic formula to find the roots. The number of distinct roots corresponds to the degree of the polynomial.
• Using the remainder theorem: The remainder theorem states that when a polynomial f(x) is divided by (x-a), the remainder is equal to f(a). Using this theorem, we can plug in various values for x and see which values gives a remainder of 0. The number of distinct values that give a remainder of 0 is equal to the degree of the polynomial.
### Understanding Polynomial Graph Degrees
The degree of a polynomial graph is not just a number, but it also tells us important information about the equation and its behavior. For instance, polynomials with an even degree will have an end behavior where the left and right sides of the graph point in the same direction. On the other hand, polynomials with an odd degree will have an end behavior where the left and right sides of the graph point in opposite directions.
Another aspect to consider is the multiplicity of roots. The degree of a polynomial also determines the number of times a particular factor appears in the equation. For example, a polynomial with a degree of 3 may have a root with a multiplicity of 2, which means that the factor (x-a) appears twice in the equation.
### Solving for Polynomial Graph Degree
Now that we understand how to determine the degree of a polynomial graph, let’s take a look at an actual example to solidify our understanding. Consider the polynomial f(x) = x4 + 3×3 – 2×2 + 6. To determine the degree, we can follow these steps:
Step Action Explanation 1 Identify the leading term The leading term is x4 2 Determine the degree of the leading term The degree of the leading term is 4 3 Conclusion The degree of the polynomial is 4
Therefore, the degree of the polynomial in this example is 4.
### Measuring Polynomial Graph Degree
In some cases, we may need to measure the degree of a polynomial graph using a measuring tool, such as a protractor or compass. This method is suitable for more complex polynomial graphs that cannot be easily identified by counting turning points.
To measure the degree, we need to determine the angle between the left and right sides of the graph at its highest point. This angle will correspond to the degree of the polynomial. For example, if the angle measures 90 degrees, the polynomial will have a degree of 2.
### Evaluating Polynomial Graph Degree
Understanding polynomial graph degree is crucial in solving and analyzing mathematical problems. By determining the degree, we can make predictions about the behavior of a polynomial equation, locate its roots, and identify its end behavior. Additionally, knowing the degree can also help us choose appropriate methods for solving equations and interpreting data.
In summary, polynomial graph degree is an essential concept in mathematics that plays a significant role in understanding and interpreting polynomial equations. By following the techniques mentioned above, we can accurately determine the degree of a polynomial graph and use this information to analyze and solve mathematical problems.
If you are an international student looking to improve your mathematical skills, check out our helpful guide on how to find the degree of a polynomial graph. With these tips and tricks, you’ll be on your way to mastering polynomial equations and graphs in no time.
In conclusion, understanding the degree of a polynomial graph is crucial in solving and analyzing polynomial equations. To determine the degree, it is important to consider the highest exponent or power of the variable in the equation. Various methods such as factoring, using the leading coefficient, and identifying the turning points can be used to determine the degree of a polynomial graph. By accurately calculating the degree, we can locate key features of a polynomial graph such as the number of roots and the behavior of the function at the extremes. Additionally, being able to identify the degree allows us to evaluate and solve polynomial equations more efficiently. It is essential to have a strong understanding of polynomial graph degrees in order to fully comprehend and work with polynomial equations.
## Author
• Marcos Nguyen is a 29-year-old blogger and teacher from Houston, Texas. He is a graduate of the University of Houston, where he studied education and psychology. Marcos has been blogging since 2009, and he specializes in writing about education and parenting. He currently teaches middle school social studies and language arts. |
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A mixed number is a combination of a whole number and a fraction. For example, if you have two whole apples and one half apple, you could describe this as 2 + 1/2 apples, or 21/2 apples.
Writing Mixed Numbers as Fractions
This mixed number can also be expressed as a fraction. Each whole apple contains two half apples. Your two whole apples are also four half apples. Four half apples plus one half apple is five half apples. So you have 5/2 apples.
To put this another way: to turn a mixed number into a fraction, multiply the whole number by the denominator (the bottom part), and add the result to the numerator (the top part).
21/2 = ?
Multiply the whole number by the denominator.
The whole number is 2.
The denominator is 2.
2 x 2 = 4.
Add the result to the numerator:
The numerator is 1.
4 + 1 = 5
The numerator is 5. The denominator remains 2.
21/2 = 5/2
Another Example
Let's try another example:
52/3 = ?
Multiply the whole number by the denominator.
The whole number is 5.
The denominator is 3.
5 x 3 = 15.
Add the result to the numerator:
The numerator is 2.
15 + 2 = 17
The numerator is 17. The denominator remains 3.
52/3 = 17/3
Proper and Improper Fractions
A fraction in which the numerator is smaller than the denominator, like 1/3 or 2/5 is called a proper fraction. A fraction in which the numerator is larger than or equal to the denominator, like 5/2, 17/3, or 6/6 is called an improper fraction. (To put it another way, a fraction with a value less than 1 is a proper fraction. A fraction with a value greater than or equal to 1 is an improper fraction.)
As we have shown above, mixed numbers can be written as improper fractions. Similarly, improper fractions can be written as mixed numbers.
Writing Improper Fractions as Mixed Numbers
To write an improper fraction as a mixed number, divide the numerator (top part) by the denominator (bottom part). The quotient is the whole number, and the remainder is the numerator.
How would you express 17/4 as a mixed number?
Divide the numerator by the denominator:
17 ÷ 4 = 4, with a remainder of 1
The quotient, 4, is the whole number. The remainder, 1, is the numerator. The denominator remains 4.
17/4 = 41/4
Two More Examples
Let's try another couple of examples:
14/9 = ?
Divide the numerator by the denominator:
14 ÷ 9 = 1, with a remainder of 5
The quotient, 1, is the whole number. The remainder, 5, is the numerator. The denominator remains 9.
14/9 = 15/9
If there is no remainder, just take the quotient as the whole number:
20/5 = ?
Divide the numerator by the denominator:
20 ÷ 5 = 4
The quotient, 4, is the whole number. There is no remainder.
20/5 = 4
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# Hypotenuse, opposite, and adjacent (article)
Use Tangent to Determine The Length of an Adjacent Side of a Right Triangle
Use Tangent to Determine The Length of an Adjacent Side of a Right Triangle
## High school geometry
In a right triangle, the hypotenuse is the longest side, an “opposite” side is the one across from a given angle, and an “adjacent” side is next to a given angle.
We use special words to describe the sides of right triangles.
The hypotenuse of a right triangle is always the side opposite the right angle. It is the longest side in a right triangle.
A right triange A B C where Angle C is ninety degrees. Inside the triangle, an arrow points from point C to the hypotenuse. The hypotenuse is labeled hypotenuse.
The other two sides are called the opposite and adjacent sides. These sides are labeled in relation to an angle.
The opposite side is across from a given angle.
A right triange A B C where Angle C is ninety degrees. Inside the triangle, an arrow points from point A to side B C. Side B C is labeled opposite.
The adjacent side is the non-hypotenuse side that is next to a given angle.
A right triange A B C where Angle C is ninety degrees. Inside the triangle, an arrow points from point A to side A C. Side A C is labeled adjacent.
Putting it all together from the perspective of :
A right triange A B C where Angle C is ninety degrees. Side A B is labeled hypotenuse. Side B C is labeled opposite. Side A C is labeled adjacent. The angle of reference is at angle A.
And from :
A right triange A B C where Angle C is ninety degrees. Side A B is labeled hypotenuse. Side A C is labeled opposite. Side B C is labeled adjacent. The angle of reference is at angle B.
## Practice
Problem 1
• Current
Relative to angle , which side is the adjacent side?
A right triange E G M. The short leg is E M. The long leg is M G. The longest side is G E. The angle of reference is at angle G.
## Why are these words important?
We’re about to learn the trigonometric functions—sine, cosine, and tangent—which are defined using the words hypotenuse, opposite, and adjacent.
## Want to join the conversation?
• who is the largest and the shortest of these three words hypotenuse opposite and adjacent(36 votes)
• The shortest side is the one opposite the smallest angle. If the angle you already know is the shortest one, then the shortest side is opposite it. However, if the angle you already know is the medium one, then the shortest side is adjacent to it.
The hypotenuse is always the longest side in a right triangle because it is opposite of the largest angle, the ninety degree angle.(73 votes)
• Can any of the calculations of trigonometry be applied to non-right triangles? Seems like a very niche area if it only covers right triangles.(20 votes)
• They sure can! You learn about the unit circle in Precalculus! The unit circle is far more complicated than right triangle trig though, you might want to wait a while before learning it. 🙂
• why is trigonometry important?(11 votes)
• Trigonometry is about understanding triangles, and every other polygon can be disassembled into triangles. So trigonometry becomes an important aspect of all of plane geometry.(30 votes)
• why do we need to learn trigonometry?why are they important?where did the names sine cos tan come from?(13 votes)
• Trigonometry is very useful in any type of physics, engineering, meteorology, navigation, etc… (Wherever geometry is useful, trig is almost certain to also be useful). Trig isn’t for everyone, however if little billy wants to calculate how tall a building is without producing the world’s longest tape measure, he’s gonna need some trig. The name sine (from what i know) comes from the latin word sinus, meaning hole or cavity, basically translation after translation of the word we ended with hole, which turned into sinus, sine for short (I may be wrong, but that is what I remember). The name cosine comes from the fact that sine and cosine are co-functions, (due to the fact that sin(x-90)=cosx. Tangent is not as easy to explain, it has to do with geometry and tangent lines.(27 votes)
• Yes the roots come from tri (three) gono (angle) metry (measure)
• why do we need to learn trigonometry?(3 votes)
• Trigonometry is part of the standard high school curriculum, but it’s not an essential subject for nothing. Many career choices involve studying trigonometry, especially STEM fields such as science, engineering, or technology. In the end, it depends on you and your career choice. Because, if anything, trigonometry is very useful for learning physics and astronomy. In some cases, you may have to use trigonometry for videogame design and programming as well. Moreover, at least in college you may have to learn trigonometry as a sort of gatekeeper before you can take core curriculum classes. This may be the case for computer science where students have to learn calculus or even physical therapy. To summarize, you don’t have to love trigonometry, but many of the world’s most valuable inventions and discoveries couldn’t have existed without the math of tirangles, and the same thing applies to the future you want to pursue, whether academic success or honorable achievements.
(Fun Fact; man wouldn’t have landed on the moon if people didn’t know/study trigonometry👀)
It’s an interesting math, and though it can be hard, it’s the foundation of many amazing works today, but of course you’re free to like the subject or dislike it, I’m just here to spread the benefit!🙌
• Where is it used in real life(7 votes)
• The GPS satellite system to tell where you are, surveying, building, anime, etc.(15 votes)
• How do you know which one is the opposite and the adjacent side?(5 votes)
• The problem will say, “relative to angle ___.” Connect that angle to the right angle in the triangle, and that’s the adjacent side. Then you know the hypotenuse(opposite of the right angle) and the adjacent side, so the only other side must be the opposite side.(8 votes)
• my hungry self could not do these math problems(4 votes) |
# MathMaster Blog
For many students, the SAT is one of the most difficult and stressful times of high school. The test is comprehensive and challenging. Because of this, many students want to prepare the best way they can: by practicing the very questions and problems they will face on exam day. By seeing these questions ahead of time, students can get a better understanding of what is going to be on the actual test, and gain an important sense of confidence that’s essential to success.
The hardest SAT math questions are on subjects of problem-solving, data analysis, advanced math, and complex algebra. Students should expect to know how to adequately show their problem-solving process and display answers numerically and graphically.
The reassuring confidence that students grasp from seeing questions ahead of time is invaluable. There are a variety of questions to cover, so let’s jump right in.
## Grid-in Response, Calculator Not Permitted
y ≤ -15x + 3000
y ≤ 5x
If a point with coordinates $a,b$ lies within the above solution in the xy-plane, what is the maximum possible value of b?
As we know the coordinates $a,b$ lay within the solution of these equations, we must solve for maximum value. This means that we can solve it just like we would solve a system of equations via the substitution method.
• 5x ≤ -15x + 3000
• Move variables to the same side to get: 20x ≤ 3000
• Divide each side by twenty to get x: x ≤ 150
• Plug x into either equation to solve for value of b: 5 x 150 = 750
## Grid-in response, Calculator Not Permitted
g$x$ = $x – 3$2
If the equation g$a$ = a2, is true, what is the value of a?
It’s key to first recognize g$x$ as the g function’s output, while also recognizing x as its input. At the same time, a represents a specific singular input for the equation. Similar to the last problem, we solve this with a system of equations, specifically using ‘a’ as a plugin for x.
• g$a$ = a2, and g$a$ = $a – 3$2
• Set them equal to one another: $a – 3$2 = $a$2
• Solve for a: a2 – 6a +9 = a2
• Shift everything to one side: a2 – 6a + 9 – a2 = 0
• a2 cancels, leaving you with: -6a + 9 = 0
• Add 6a to both sides: 9 = 6a
• Solve for final answer: a = 9/6 or 3/2
## Multiple Choice, Calculator Not Permitted
$2x + 6$$ax – 5$ – x2 + 30
In the expression seen above, a is a constant. If the expression is equal to bx, and b represents a constant, what is the value of b?
A) -7
B) -4
C) 0
D) 5
As we are told that one expression equals another, we can begin by making them equal and proceed to solve using FOIL.
• $2x +6$$ax – 5$ – x2 +30 = bx
• Remove parenthesis: 2ax2 – 10x + 6ax – 30 – x2 + 30 = bx
• Combine like terms: 2ax2 – x2
• Set equal to zero: 2ax2 – x2 = 0x2
• Simplify: 2a – 1 = 0
• Simplify more by adding 1: 2a = 1
• Divide by 2: a = ½
• Combine x terms and set equal to b: -10x + 6ax = bx
• Divide by x = -10 + 6a = b
• Plug in ½ for a: -10 + 6$1/2$ = -10 + 3 = -7
## Grid-in Response, Calculator Permitted
Greta opened a bank account that earns 2 percent interest compounded annually. Initially, she deposited $100. She uses the expression 100$x$t to find the value of the account after a total of t years. Greta’s friend Tom opened an account that earns 2.5 percent interest compounded annually. He also deposits$100 into the account initially.
This is a two-part question: First, what is the value of x in the expression 100$x$t ? Next, after 10 years, how much more money will Tom have in his account than Greta has in hers? $Round your answer to the nearest whole cent.$
The answer to the first problem can be found pretty easily. Since we know Greta’s account earns 2% interest, we can simply convert it to a decimal or .02. Then you add 1 to account for the initial deposit, giving us an answer of x = 1.02.
Before you can find out the difference in account earnings, you must create an equation for Tom.
• Since Greta’s equation is 100$1.02$t, we can plug the 2.5 % into an equation for Tom, just the same we did for Greta: 100$1.025$t
• Plug 10 in for t in Tom’s equation: 100$1.025$10 = $128.008 • Plug 10 in for t in Greta’s equation: 100$1.02$10 =$121.899
• Find difference and round to nearest cent: $128.008 –$121.899 = \$6.11
## Multiple Choice, Calculator Not Permitted
x + r = 4x – 9
y + s = 4y – 9
If r is s + ½, and r and s are constants in the equations above, which of the following is true?
A.) x is y – 1/8
B.) x is y – 1
C.) x is y + 1/6
D.) x is y + 1/4
Answer: x is y + 1/6
The first important thing to recognize is that only x and y remain in the answer choices above. This means our first step is to remove r and s. We can do this using the elimination method and substitute r with s + ½ to eliminate r.
• x + s + ½ = 4x – 9, and y + s = 4y – 9
• Multiply the second equation by -1 so s goes away: -y – s = -4y + 9
• Then add equations: x – y + ½ = 4x – 4y
• Isolate x variables: -y + 4y + ½ = 4x – x
• Simplify: 3y + ½ = 3x
• Divide both sides by 3 to get solution: x = y + 1/6
## Multiple Choice, Calculator Permitted
x2 + y2 + 4x – 2y = -1
The above equation represents a circle in the xy-plane. What is the radius of the circle?
A.) 2
B.) 3
C.) 4
D.) 9
We’ve been given the information that the above equation is a circle, however, it is not written in the standard form of an equation which is $x – h$2 + $y – k$2 = r2. Out first step is to rewrite the equation into standard form. This will help us easily find the radius.
• Rewritten, the equation looks like this: $x + 2$2 – 4 + $y – 1$2 – 1 = -1
• Now we need to get our constants on one side: $x + 2$2 + $y – 1$2 = 4
• We can now see that r2 is represented by 4 in our equation.
• To find the radius, find the square root of 4, which is: 2
## Multiple Choice, Calculator Not Permitted
X = 2y + 5
Y = $2x – 3$$x + 9$
How many different ordered pairs, represented as $x, y$, could satisfy the system of equations shown above?
A.) 0
B.) 1
C.) 2
D.) Infinitely many
The first step is to recognize that we can solve this equation with substitution.
• Substitute the second equation into the first: x = 2[$2x – 3$$x + 9$] + 5
• Expand quadratic equation: x = 2$2x2 + 15x – 27$ + 5
• Simplify further: x = 4x2 + 30x -49
• Set equation equal to zero by subtracting x: 0 = 4x2 + 29x – 49
Now that the equation is completely broken down, we must remember the question didn’t ask for a specific set of coordinates, but rather a number of possible coordinate solutions. To find this we can use the discriminant of the quadratic formula.
• Discriminant = b2 – 4ac
• Plug in correlating values: 292 – 4$4$$49$ = 841 – 784 = 57
The law in mathematics that governs the discriminant of the quadratic formula says if the discriminant is positive, there are 2 solutions; if the discriminant is negative, there are no real solutions; if the discriminant is 0, there is 1 real solution $or repeating solution$.
Since our answer $57$ is positive, we can conclude that there are 2 solutions.
## Some Final Thoughts on the Hardest SAT Math Problems
The SAT is designed to test students’ abilities through a variety of problems, with an underlining goal of hitting the most important areas of mathematics. While there are multiple factors that play into a student's success during the test, the most important thing to do is to study the expected hardest SAT math questions in advance.
Students can also benefit from the help of some free tools in preparation for the SAT. Using an application like MathMaster, students will be able to enter equations into a problem solver to get instant feedback on their work, and even work through some extended exam prep. The hardest SAT math problems can seem insurmountable at first, but with the right preparation, anyone can find the confidence and knowledge necessary for success.
they can: by practicing the very questions and problems they will face on exam day. |
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