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# The Progression of Fractions
## Exploring Fraction Constructs and Proportional Reasoning
Fractions are a beast of a concept that causes struggles for many adults and students alike. While we all come to school with some intuition to help us with thinking fractionally and proportionally, the complexity quickly begins to increase as we move from concrete, to visual, to symbolic and from identifying, to comparing, to manipulating. Fractions are formally introduced in the Ontario Math Curriculum when students begin dividing whole objects into pieces and identify these pieces using fractional names (e.g.: halves; fourths or quarters) and continues to promote the development fractional fluency concretely through each grade. Interestingly enough, it is often said that students struggle much more with numbers represented in fraction notation than those represented in decimal notation, yet the word “fraction” appears in the curriculum document 99 times beginning in grade 1, while the word “decimal” appears only 69 times beginning three years later in the 4th grade. While it might be true that fractions tend to intimidate, I wonder if our dependence on the calculator has tricked us into believing we are more fluent with quantities represented in decimal form than is reality.
## Paying Attention to Fractions
As I did with the Progression of Proportional Reasoning, I’d like to reference the Paying Attention To Mathematics series released by the Ontario Ministry of Education Literacy and Numeracy Secretariat (LNS) called Paying Attention to Fractions. These guides are a great start to help you wrap your head around big ideas in mathematics and thus, this post will attempt to expand on the ideas shared in this particular document.
## Why Fractions?
As if the struggles our students experience when working with fractions aren’t enough justification, I like this quote shared in the guide:
“No area of elementary school mathematics is as mathematically rich, cognitively complicated, and difficult to teach as fractions, ratios, and proportionality. These ideas all express mathematical relationships: fractions and ratios are ‘relational’ numbers. They are the first place in which students encounter numerals like ‘ 3/4 ’ that represent relationships between two discrete or continuous quantities, rather than a single discrete (‘three apples’) or continuous quantity (‘4 inches of rope’).”
(Litwiller & Bright, 2002, p. 3)
## What Is A Fraction?
A fraction is a number.
While fractional notation is typically used to represent quantities that are not whole, it is possible for all quantities to be represented as a fraction.
While these descriptions are simple on the surface, they do not appropriately communicate the complex constructs that lie within this big idea.
## Connecting Proportional Reasoning and Fractions
If you recall from my last post, proportional reasoning is takes place when one compares two numbers in relative terms rather than absolute terms. Thus, it can be said that a fraction provides a means for representing that relationship between two numbers.
It is reasonable to believe that young learners begin working towards proportional reasoning even before they enter school when they make comparisons between two quantities such as:
• this bag is heavier than that one;
• I’m “this much” shorter than you are; and,
• there are 5 more red candies than green.
While these early comparisons are likely absolute (i.e.: additive) in nature, we are one step closer to comparing these quantities relatively (i.e.: multiplicatively).
Let’s take some time to explore four common fraction constructs.
## Counting Cartons: Part-Whole Relationships
Have a look at the image below:
I might ask students to determine how many eggs there are. This question is quite clear that I would like to know the quantity of eggs, however they get to decide how they would like to represent that quantity. While many may simply choose to represent the number of eggs as 18, others may get creative and unitize (i.e.: 3 groups of 6; 9 groups of 2; etc.) or even reference the number of eggs relative to egg cartons.
After students share out with a partner, I might ask students:
how many cartons of eggs are there?
After students have some time to think independently, I would have them share out with their table groups. Some interesting responses are sure to come up such as:
• 18/2
• 18/1.5
• 3/2
• 1.5
• 2
You may notice that by asking students to state how many cartons of eggs there are, there is room for interpretation. Some students might simply state that there are two egg cartons without considering the number of eggs present, some might give a quantity relative to two egg cartons, while others might state the number of eggs relative to a single carton.
After students have been given an opportunity to defend their thinking, I might ask students to state how many cartons are full of eggs?
This fraction construct would be an example of a part-whole relationship. Students might consider how many whole and partial cartons there are in the form of a mixed fraction (1 and 1/2), improper fraction (3/2) or in decimal form (1.5). Some common representations for these different fractional forms might be using a set model, area model or number line. The following animated gif shows each of these representations, sometimes more than one simultaneously:
You might have noticed the double-number line included in the animated gif. This idea can be very useful when making the jump from thinking fractionally to proportional reasoning.
## Doritos Roulette: Part-Part Relationships
In the Doritos Roulette 3 Act Math Task, students are asked questions that stem from the relationship of “hot” chips to “not hot” chips based on the image on the front of the bag:
I might ask students to consider the image above and ask them this question:
What fraction of hot chips to not hot chips are there in a bag of Doritos Roulette?
Here are some common responses I see when I ask students (and teachers) this question:
• 1/6
• 1/7
• 1:6
• 1:7
Despite the explicit request for a part-part relationship between “hot” chips (1) and “not hot” chips (6), we often see the relationship between “hot” chips (1) and the whole (7).
Something I might ask the group to discuss with their neighbours is whether we are “allowed” to represent a part-part relationship as a fraction. Great (and sometimes heated) discussions can arise from this question, which makes this exciting to facilitate.
The discourse this question can promote is a great way to address the fact that part-part relationships are commonly represented as a ratio, but that both part-part relationships and ratios can also be represented as a fraction. In this case, the numerator represents the number of “hot” chips; the denominator represents the number of “not hot” chips; and the sum of each part, the numerator and denominator, is the whole:
After exploring part-whole and part-part fraction constructs, you may begin to notice how important clarity around what a given fraction represents prior to attempting any comparison or manipulation of that fraction.
While the use of a set model and area model remain fairly similar regardless of whether you are dealing with a part-whole or part-part relationship, the use of a number line changes significantly.
Recall how the number line used in the part-whole counting cartons example made identifying the part and the whole fairly intuitive:
When we attempt to use a number line with a part-part relationship, it becomes more difficult to determine the size of the second part without modifying how we use this representation:
Note that in the representation above, two parts are clearly shown and the sum of those two parts represents the whole.
Although the number line representation might not be as intuitive in the part-part case, it does help you visualize both parts proportionally:
Extending from a single number line to a double-number line can help shine a light on the meaning behind the part-part relationship. In this case, there are one-sixth as many “hot” chips as there are “not hot” chips. Alternatively, there are 6 times as many “not hot” chips as there are “hot” chips.
## Baking Brownies: Fraction as Quotient
When considering fractions as quotient, we are referring to the times when we are dividing two numbers. The Paying Attention to Fractions document uses an example of kids fair sharing some brownies. Let’s keep the context the same, but I’ll add some visuals to enhance the experience.
Suppose there are 6 brownies that are to be shared amongst 4 friends fairly. How much should each person receive?
Like many situations in life, there are multiple ways to share these brownies fairly. Let’s represent a few of these methods.
### Partitioning Brownies into Fourths
Since there are 4 people to share the 6 brownies, we can divide the 6 brownies into fourths and then pass them out, one-by-one:
After splitting the brownies into quarters, you’ll notice that each of the four people should receive 6 quarters. Hence why we can represent this fraction as quotient as the improper fraction, 6/4.
### Sharing Full Brownies Equally, Then Partitioning Remainder
Some people might find it useful to share whole brownies equally and then when there are not enough whole brownies to share fairly, consider partitioning up the remaining brownies appropriately:
In the case with 6 brownies, we can share one brownie each and then partition the remaining brownies into four halves to share. This method would be equivalent to the mixed fraction of 1 and 1/2 brownies per person.
### Partition Brownies Into Four Equal Portions
This method involves considering all 6 brownies and where cuts must be made to create four equal groupings:
This method would be another representation of the mixed fraction, 1 and 1/2 brownies for each person.
For each of the methods above, we could also consider the use of a number line to support the partitioning of the brownies to distribute a fair amount to each person. Here are a few examples:
## How Many Rolos: Fraction as Operator
Another common construct is the fraction as operator.
Let’s have a look at another problem. We mind as well keep the context as food, because that is just yummy.
There are 7 pieces in every full roll of Rolo chocolate.
Two partially eaten rolls are found in a drawer; one with 5 pieces and the other with 4 pieces.
How many full rolls of Rolo are there?
If we use concrete manipulatives or visuals, like we will in this instance, it is much easier to see how to add and subtract fractions. In this case, we’ll show a few different representations including how using a number line can be helpful:
So in this particular example, you can see that 5 pieces plus 4 pieces will yield 9 pieces, which is more than a full 7 piece roll; 9/7 or 1 and 2/7.
## Fraction As Operator: Multiplication
If you’re a 3 act math fan, you might be familiar with my Gimme a Break 3 act math task involving the use of fractions as operators. I should note that while there are some useful pieces in that task, it has really evolved over time and I just haven’t had the time to update the post yet. Here’s some of the pieces I’ve added along the way.
The task begins with an opportunity for students to notice/wonder after watching a video of me opening a KitKat bar.
Later in the task, the following video is shown asking students to consider what operation could be taking place:
Since the video is open ended, I’m always hoping that all students will be able to share something mathematical about the situation. For example, we might see some responses like these:
However, the new learning I’m hoping to spend some time on is the multiplication of a fraction by a whole number. In this case, a student might think of this situation as 4 groups of one quarter of a whole bar:
Alternatively, we could think of this situation as one quarter groups of four whole KitKat bars:
In either case, the use of arrays and/or area models can be very useful when trying to help build an understanding of multiplication with fractions. In the following example, we extend this idea to the multiplication of two fractions:
While I find that multiplication of fractions seems to be the one operator I hear the least noise about, my gut tells me that it is more about being able to remember the algorithm and not due to any conceptual understanding. In my experience, many students can manage to remember to “multiply the tops, multiply the bottoms” without necessarily having any real conceptual understanding to fall back on in times of doubt.
## Fraction As Operator: Division
On the other hand, division of fractions seems to be a sore spots for many students (and adults) because the algorithm is a bit more complicated and less intuitive. Pair that with a very low level of conceptual understanding as to why the algorithm actually works and you’ll end up with confusion and frustration for many. The gap between the pure memorization of steps and the conceptual understanding of what dividing fractions really means is often so large that many educators simply follow suit and focus purely on the algorithm. While I believe it is so important to build the conceptual understanding behind a complex idea like dividing fractions, I know that the task can be quite difficult and many teachers aren’t so sure how to approach it. When we rush to an algorithm before constructing conceptual understanding, students (and adults) are often very difficult to engage in the heavy thinking required to connect the dots. This leaves many teachers believing that successful use of the algorithm is enough (or all they have time for) and they just move on to the next topic.
Let’s look at the Gimme a Break context in terms of division of fractions. With manipulatives on the table such as pattern blocks or relational rods, I might ask students:
What would 1 ÷ 1/4 “look like” in terms of the KitKat task?
Can you show a neighbour what that looks like to you?
Hopefully students come up with some representation to model what a whole KitKat bar would look like an a quarter of a KitKat bar:
Then, I would ask them to determine how many quarter pieces there are in a whole KitKat bar. With this in mind, students should be able to “see” and manipulate their representation on the table to come up with an answer of 4.
Then, I might consider asking students to represent a couple more similar situations using whole numbers divided by a certain number of quarter pieces to build some confidence.
For example, 2 divided by 1/4:
3 divided by 3/4:
When ready, students might be ready to start stepping outside of whole numbers to situations like 3/4 divided by 2/4:
Then, once students are really comfortable dividing fractions with common denominators, you might start moving to situations where the fractions have uncommon denominators.
Note that context is always very important for students to build a conceptual understanding. If we are trying to explain abstraction through abstraction (i.e.: explaining a rule algebraically based on another algebraic rule), then it is unlikely to stick. However, if we, the educators, spend the time necessary to understand mathematics conceptually, it will be a huge pay off for the understanding of our students. Only once students are able to visualize what mathematics “looks like” does it make sense for us to move towards more efficient methods like algorithms or more abstract problems where context is stripped away.
Throughout this post we have tackled the idea of fraction constructs by using a similar approach to how I introduce most big ideas in math class: using tasks that are contextual, visual and concrete in order to build a conceptual understanding.
This post is by no means the “fractions rulebook”, but rather a journal outlining my own journey to understanding fractions conceptually. As a secondary mathematics teacher, much of what our curriculum is built on is the assumed understanding of topics like fractions. However, the more I reflect on my own mathematical understanding, the more I realize that I never truly had a conceptual understanding of most concepts that I was teaching.
Let’s avoid the rush to the algorithm and slow down to let kids truly experience and understand mathematics. I’d appreciate any pieces I should add to this post in the comments section.
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## About Kyle Pearce
I’m Kyle Pearce and I am a former high school math teacher. I’m now the K-12 Mathematics Consultant with the Greater Essex County District School Board, where I uncover creative ways to spark curiosity and fuel sense making in mathematics. Read more.
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# Problem of the Week Problem E and Solution Fine Line
## Problem
Suppose that $$f(x) = ax + b$$ and $$g(x) = f^{-1}(x)$$ for all values of $$x$$. That is, $$g$$ is the inverse of the function $$f$$.
If $$f(x) - g(x) = 2022$$ for all values of $$x$$, determine all possible values for $$a$$ and $$b$$.
## Solution
Since $$f(x) = ax+b$$, we can determine an expression for $$g(x) = f^{-1}(x)$$ by letting $$y = f(x)$$ to obtain $$y = ax + b$$. We then interchange $$x$$ and $$y$$ to obtain $$x = ay + b$$, which we solve for $$y$$ to obtain $$ay = x - b$$ or $$y = \dfrac{x}{a} - \dfrac{b}{a}$$.
Therefore, $$f^{-1}(x) = \dfrac{x}{a} - \dfrac{b}{a}$$. Note that $$a \neq 0$$. (This makes sense since the function $$f(x) = b$$ has a graph which is a horizontal line, and so cannot be invertible.)
Therefore, the equation $$f(x) - g(x) = 2022$$ becomes \begin{aligned} (ax + b) - \left( \dfrac{x}{a} - \dfrac{b}{a} \right) &= 2022\\ \left( a - \dfrac{1}{a} \right) x + \left(b + \dfrac{b}{a} \right)& = 2022 \end{aligned}
This is true for all $$x$$.
From here, we will present two approaches for determining the possible values for $$a$$ and $$b$$.
• Approach 1: Comparing coefficients
Since the equation $\left( a - \dfrac{1}{a} \right) x + \left(b + \dfrac{b}{a} \right) = 2022 = 0x + 2022$ is true for all $$x$$, then the coefficients of the linear expression on the left side must match the coefficients of the linear expression on the right side.
Therefore, $$a - \dfrac{1}{a} = 0$$ and $$b + \dfrac{b}{a} = 2022$$.
From the first of these equations, we obtain $$a = \dfrac{1}{a}$$ or $$a^2 = 1$$, which gives $$a = 1$$ or $$a = -1$$.
If $$a = 1$$, the equation $$b + \dfrac{b}{a} = 2022$$ becomes $$b + b = 2022$$, which gives $$b = 1011$$.
If $$a = -1$$ the equation $$b + \dfrac{b}{a} = 2022$$ becomes $$b - b = 2022$$ which is not possible.
Therefore, we must have $$a=1$$ and $$b = 1011$$, and so $$f(x) = x + 1011$$.
• Approach 2: Trying specific values for $$x$$
Since the equation $\left( a - \dfrac{1}{a} \right) x + \left(b + \dfrac{b}{a} \right) = 2022$ is true for all values of $$x$$, then it must be true for any specific values of $$x$$ that we choose.
Choosing $$x = b$$, we obtain \begin{align*} \left( a - \dfrac{1}{a} \right) b + \left(b + \dfrac{b}{a} \right) &= 2022\\ ab + b &= 2022 \tag{1} \end{align*}
Choosing $$x = 0$$, we obtain \begin{aligned} 0 + b + \dfrac{b}{a} &=2022\\ b + \dfrac{b}{a} &= 2022\\ \dfrac{ab+b}{a} &= 2022 \end{aligned} Then, substituting $$ab+b=2022$$ from equation $$(1)$$, we obtain \begin{aligned} \dfrac{ab + b}{a} &= 2022\\ \dfrac{2022}{a} &= 2022\\ a &=1\end{aligned}
Since $$a = 1$$, then $$ab + b = 2022$$ gives $$2b = 2022$$ or $$b = 1011$$.
Thus, $$f(x) = x + 1011$$.
In summary, the only possible values for $$a$$ and $$b$$ for which the given equation is true for all $$x$$ are $$a=1$$ and $$b=1011$$. |
## A rancher has 800 feet of fencing to put around a rectangular field and then subdivide the field into 2 identical smaller rectangular plots
Question
A rancher has 800 feet of fencing to put around a rectangular field and then subdivide the field into 2 identical smaller rectangular plots by placing a fence parallel to one of the field’s shorter sides. Find the dimensions that maximize the enclosed area. Write your answers as fractions reduced to lowest terms.
in progress 0
1 day 2021-09-15T18:59:27+00:00 1 Answer 0
The dimensions of enclosed area are 200 and 400/3 feet
Step-by-step explanation:
* Lets explain how to solve the problem
– There are 800 feet of fencing
– We will but it around a rectangular field
– We will divided the field into 2 identical smaller rectangular plots
by placing a fence parallel to one of the field’s shorter sides
– Assume that the long side of the rectangular field is a and the
shorter side is b
∵ The length of the fence is the perimeter of the field
∵ We will fence 2 longer sides and 3 shorter sides
∴ 2a + 3b = 800
– Lets find b in terms of a
∵ 2a + 3b = 800 ⇒ subtract 2a from both sides
∴ 3b = 800 – 2a ⇒ divide both sides by 3
⇒ (1)
– Lets find the area of the field
∵ The area of the rectangle = length × width
∴ A = a × b
– To find the dimensions of maximum area differentiate the area with
respect to a and equate it by 0
⇒ Add 4/3 a to both sides
⇒ multiply both sides by 3
∴ 800 = 4a ⇒ divide both sides by 4
∴ 200 = a
– Substitute the value of a in equation (1)
* The dimensions of enclosed area are 200 and 400/3 feet |
# If the line segment joining the points A(a,b) and B(c,d) subtends an angle θ at the origin, then cosθ is equal to
A
ab+cd(a2+b2)(c2+d2)
B
ac+bd(a2+b2)(c2+d2)
C
acbd(a2+b2)(c2+d2)
D
Video Solution
Text Solution
Generated By DoubtnutGPT
## To solve the problem, we need to find the value of cosθ where θ is the angle subtended at the origin by the line segment joining the points A(a,b) and B(c,d).1. Identify Points and Distances: - Let O be the origin (0,0), A(a,b), and B(c,d). - We need to find the lengths of the segments OA, OB, and AB.2. Calculate Distances: - The distance OA from the origin to point A is given by: OA=√a2+b2 - The distance OB from the origin to point B is given by: OB=√c2+d2 - The distance AB between points A and B is given by: AB=√(c−a)2+(d−b)23. Use the Cosine Rule: - According to the cosine rule in triangle OAB: cosθ=OA2+OB2−AB22⋅OA⋅OB4. Substitute the Values: - Substitute the distances into the cosine formula: cosθ=(OA2)+(OB2)−(AB2)2⋅OA⋅OB - This becomes: cosθ=(a2+b2)+(c2+d2)−((c−a)2+(d−b)2)2⋅√a2+b2⋅√c2+d25. Expand and Simplify: - Expand (c−a)2+(d−b)2: (c−a)2+(d−b)2=(c2−2ac+a2)+(d2−2bd+b2)=c2+d2+a2+b2−2ac−2bd - Substitute back into the equation: cosθ=(a2+b2)+(c2+d2)−(c2+d2+a2+b2−2ac−2bd)2⋅√a2+b2⋅√c2+d2 - Simplifying gives: cosθ=2ac+2bd2⋅√a2+b2⋅√c2+d26. Final Result: - Thus, we arrive at: cosθ=ac+bd√a2+b2⋅√c2+d2
|
Updated on:7/8/2024
### Knowledge Check
• Question 1 - Select One
## If the point C divides the line segment joining the points A(4,−2,5)andB(−2,3,7) externally in the ratio 8:5, then points C is
A(12,343,313)
B(12,313,343)
C(12,143,313)
D(12,313,143)
• Question 2 - Select One
## If the point C divides the line segment joining the points A(2,−1,−4)andB(3,−2,5) externally in the ratio 3:2, then points C is
A(5,4,23)
B(5,4,23)
C(5,4,23)
D(5,4,23)
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# Help solve algebra equations
This Help solve algebra equations provides step-by-step instructions for solving all math problems. We will also look at some example problems and how to approach them.
## The Best Help solve algebra equations
Keep reading to learn more about Help solve algebra equations and how to use it. These are the building blocks of all other math problems. Once you've mastered these skills, try more advanced problems like addition and multiplication of fractions, decimals and percentages. One of the best ways to increase your chances of success is to break a geometric sequence into smaller pieces. This will make it easier for you to understand what each part represents and how they relate to each other. When you solve a geometric sequence, the order in which you do each step doesn't matter as much as the number of steps you take (and the order in which you take them). So don't get bogged down by trying to figure out the exact order in which you should solve each problem. Just take it one step at a time and remember that every step counts!
Quadratic formula, or the quadratic equation y = ax^2 + bx + c, is one of the most important equations in algebra. It’s used to solve for two unknown values in a system of equations. In other words, it helps you find out where one number comes from another number. It’s also a very useful tool in math and science. The quadratic formula is especially important when solving problems that have a variable with a significant amount of value. One type of problem that often has a variable with a high value is the area under a curve. If you want to find the area underneath a graph that shows how many times something happened during a certain time period, then you can use the quadratic formula to get an accurate answer. Another example is finding the volume of a cube. If you want to find out how deep a box is, the quadratic formula can help you do that as well.
Solving is a process of finding the answer to a problem by logical reasoning. Solving problems involves many steps, but the first step is to identify the problem. Once you have identified the problem, you must figure out why it is happening. Once you know why a problem is happening, you can start to solve it. Solving problems may involve doing research, brainstorming ideas, or finding solutions through trial and error. Solving problems is an active process that requires both time and effort. However, with enough time and effort, you can solve almost any problem! Solving problems is part of everyday life; whether it's figuring out how to fix a broken appliance or solving math problems at school. Everyone deals with problems every day - from paying bills to making friends - so everyone can learn how to solve them eventually! It's never too late to start learning how to solve problems!
A composition of functions solver can be a useful tool for solving mathematical problems. In mathematics, function composition is the operation of combining two functions to produce a third function. For example, if f(x) = 2x + 1 and g(x) = 3x - 5, then the composition of these two functions, denoted by g o f, is the function defined by (g o f)(x) = g(f(x)) = 3(2x + 1) - 5 = 6x + 8. The composition of functions is a fundamental operation in mathematics and has many applications in science and engineering. A composition of functions solver can be used to quickly find the composition of any two given functions. This can be a valuable tool for students studying mathematics or for anyone who needs to solve mathematical problems on a regular basis. Thanks to the composition of functions solver, finding the composition of any two given functions is now quick and easy. |
# What is 1/359 as a decimal?
## Solution and how to convert 1 / 359 into a decimal
1 / 359 = 0.003
Fraction conversions explained:
• 1 divided by 359
• Numerator: 1
• Denominator: 359
• Decimal: 0.003
• Percentage: 0.003%
To convert 1/359 into 0.003, a student must understand why and how. Fractions and decimals represent parts of a whole, sometimes representing numbers less than 1. In certain scenarios, fractions would make more sense. Ex: baking, meal prep, time discussion, etc. While decimals bring clarity to others including test grades, sale prices, and contract numbers. Now, let's solve for how we convert 1/359 into a decimal.
1 / 359 as a percentage 1 / 359 as a fraction 1 / 359 as a decimal
0.003% - Convert percentages 1 / 359 1 / 359 = 0.003
## 1/359 is 1 divided by 359
Converting fractions to decimals is as simple as long division. 1 is being divided by 359. For some, this could be mental math. For others, we should set the equation. The numerator is the top number in a fraction. The denominator is the bottom number. This is our equation! We use this as our equation: numerator(1) / denominator (359) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. This is how we look at our fraction as an equation:
### Numerator: 1
• Numerators are the number of parts to the equation, showed above the vinculum or fraction bar. Comparatively, 1 is a small number meaning you will have less parts to your equation. 1 is an odd number so it might be harder to convert without a calculator. Ultimately, having a small value may not make your fraction easier to convert. Let's take a look below the vinculum at 359.
### Denominator: 359
• Denominators are the total numerical value for the fraction and are located below the fraction line or vinculum. 359 is a large number which means you should probably use a calculator. But 359 is an odd number. Having an odd denominator like 359 could sometimes be more difficult. Have no fear, large two-digit denominators are all bark no bite. Let's start converting!
## How to convert 1/359 to 0.003
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 359 \enclose{longdiv}{ 1 }$$
Use long division to solve step one. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 359 \enclose{longdiv}{ 1.0 }$$
Uh oh. 359 cannot be divided into 1. So we will have to extend our division problem. Add a decimal point to 1, your numerator, and add an additional zero. Now 359 will be able to divide into 10.
### Step 3: Solve for how many whole groups you can divide 359 into 10
$$\require{enclose} 00.0 \\ 359 \enclose{longdiv}{ 1.0 }$$
Since we've extended our equation we can now divide our numbers, 359 into 10 (remember, we inserted a decimal point into our equation so we we're not accidentally increasing our solution) Multiply this number by 359, the denominator to get the first part of your answer!
### Step 4: Subtract the remainder
$$\require{enclose} 00.0 \\ 359 \enclose{longdiv}{ 1.0 } \\ \underline{ 0 \phantom{00} } \\ 10 \phantom{0}$$
If you don't have a remainder, congrats! You've solved the problem and converted 1/359 into 0.003 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, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. 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 1/359 and 0.003 bring clarity and value to numbers in every day life. Here are just a few ways we use 1/359, 0.003 or 0% in our daily world:
### When you should convert 1/359 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 0.003 to 1/359 as a fraction
Time - spoken time is used in many forms. But we don't say It's '2.5 o'clock'. We'd say it's 'half passed two'.
### Practice Decimal Conversion with your Classroom
• If 1/359 = 0.003 what would it be as a percentage?
• What is 1 + 1/359 in decimal form?
• What is 1 - 1/359 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.003 + 1/2?
### Convert more fractions to decimals
From 1 Numerator From 359 Denominator What is 1/360 as a decimal? What is 2/359 as a decimal? What is 1/361 as a decimal? What is 3/359 as a decimal? What is 1/362 as a decimal? What is 4/359 as a decimal? What is 1/363 as a decimal? What is 5/359 as a decimal? What is 1/364 as a decimal? What is 6/359 as a decimal? What is 1/365 as a decimal? What is 7/359 as a decimal? What is 1/366 as a decimal? What is 8/359 as a decimal? What is 1/367 as a decimal? What is 9/359 as a decimal? What is 1/368 as a decimal? What is 10/359 as a decimal? What is 1/369 as a decimal? What is 11/359 as a decimal? What is 1/370 as a decimal? What is 12/359 as a decimal? What is 1/371 as a decimal? What is 13/359 as a decimal? What is 1/372 as a decimal? What is 14/359 as a decimal? What is 1/373 as a decimal? What is 15/359 as a decimal? What is 1/374 as a decimal? What is 16/359 as a decimal? What is 1/375 as a decimal? What is 17/359 as a decimal? What is 1/376 as a decimal? What is 18/359 as a decimal? What is 1/377 as a decimal? What is 19/359 as a decimal? What is 1/378 as a decimal? What is 20/359 as a decimal? What is 1/379 as a decimal? What is 21/359 as a decimal?
### Convert similar fractions to percentages
From 1 Numerator From 359 Denominator 2/359 as a percentage 1/360 as a percentage 3/359 as a percentage 1/361 as a percentage 4/359 as a percentage 1/362 as a percentage 5/359 as a percentage 1/363 as a percentage 6/359 as a percentage 1/364 as a percentage 7/359 as a percentage 1/365 as a percentage 8/359 as a percentage 1/366 as a percentage 9/359 as a percentage 1/367 as a percentage 10/359 as a percentage 1/368 as a percentage 11/359 as a percentage 1/369 as a percentage |
These worksheets will give your students practice with equations for the locus of different points.
#### A locus (in geometric terms) is a series of points that is determined by specific conditions. When we work with loci in math, it makes me feel like I am placing a pin in corkboard. We place that pin based on the instructions that are provided. In most cases, we will just moving up and down the y-axis or left and right across the x-axis. In these worksheets you will be the one placing the pin in the corkboard.
What is a locus at a fixed distance? A locus is the arrangement of all focuses which fulfill a specific condition. The locus at a fixed separation, d, from point P is a hover with the given point P as its inside and d as its span. The locus at a fixed separation, d, from a line m, is a couple of equal lines a good way off of d from line m and situated on either side of m. The locus is equidistant from two focuses. An and B is the opposite bisector of the line fragment joining the two focuses. The locus equidistant from two parallel lines, m1 and m2, is a line corresponding to both m1 and m2 and somewhere between them. These worksheets explain how to find the locus of two points at a fixed distance and write its equation.
# Print Locus at a Fixed Distance Worksheets
## Finding the Locus Lesson
This worksheet explains how to find the locus of two points at a fixed distance. A sample problem is solved, and two practice problems are provided.
## Worksheet
Students will write the equation of the locus described. Ten problems are provided.
## Practice
Students will describe the locus indicated. Example problem: A wooden block is 200 feet long and 40 feet wide. It is planned to cut it 28 feet from the center of the block. Describe where it will be cut. Ten problems are provided.
## Review and Practice
Students review how to find these measures. Here is a sample problem: A book-cover is 12 feet long and 6 feet wide. It is planned to color it 5 foot from the center of the book-cover. Describe where it will be painted. Six practice problems are provided.
## Quiz
Students will demonstrate their proficiency this skill. Example: A railway track is 125 feet long and 50 feet wide. It is planned to cover it with mud 30 feet from the center of the track. Describe where it will be covered with mud. Ten problems are provided.
## Check
Students will find the locus described and write its equation. Example: Describe the locus of points 6 units from the line y = -14. Three problems are provided, and space is included for students to copy the correct answer when given.
## Equation of the Locus Lesson
This worksheet explains how to find and write the equation of the locus. A sample problem is solved, and two practice problems are provided.
## Worksheet
Example: A tree A is 50 feet from another tree B. Shadowing range of A is 30 feet and that of B is 25 feet. Draw a diagram showing the areas where each tree shadows. Is any area shadowed by both the trees?
## Practice Worksheet
Given two points, students will find and write the equation of the locus. Ten problems are provided.
## Review and Practice
A workshop is located at the coordinates (5, 12) on a coordinate grid. The delivery service extends for 5 km. Write the equation of the locus which represents the outer edge of the delivery service area.
## Quiz
Students will demonstrate their proficiency in finding and writing the equation of the locus. Ten problems are provided.
## Skills Check
Students will find the locus and write its equation. Three problems are provided, and space is included for students to copy the correct answer when given. |
Exploration 2: Oh Where, Oh Where Has My Parabola Gone?
by
Elizabeth Gieseking
In this exploration we will be focusing on the equations of parabolas. A parabola is defined as the set of points which is equidistant from a fixed point called the focus and a fixed line called the directrix. In the figure below, we have a parabola with a focus at (0, p) and a directrix y= - p.
We know all points (x, y) on the parabola are the same distance from the directrix and the focus. Thus we can use the Pythagorean distance formula to derive the equation of a parabola.
If we instead have a vertical directrix, the equation of a parabola with the vertex at the origin is
When the vertex of the parabola is not at the origin, the equations can be translated to the new vertex (h, k).
The two most commonly used forms for the equation of a parabola are the standard form: and the vertex form: . In this exploration we will look at what happens to the graph when we vary each of these coefficients and then we will examine the relationship between the two forms of the equations.
The coefficient a: We will start by looking at the simplest equation of a parabola and we will vary the coefficient a.
We see from this set of graphs that if a is positive, the parabola opens to the top and if a is negative, the parabola opens to the bottom. We also see that as a increases, the parabola rises more quickly, making it appear narrower. This is considered a similiarity transformation. The size of the parabola changes, but not its shape. Since a parabola extends infinitely from the vertex, when we change the value of a, it is if we are zooming in or out from the vertex.
Next we will examine the role of the coefficient b. This set of graphs will use the equation in which b is varied from -4 to 4.
When b is zero, the graph is centered around the y-axis, with the vertex of the parabola at (0, 1). When b is positive, the graph shifts downward and to the left. When b is negative, the graph shifts downward and to the right. All of the graphs intercept the y-axis at the point (0, 1).
Now we will examine the role of the constant c. We will graph the equation and vary the constant c from -5 to 5.
From these graphs we see that as c increases, the vertex of the parabola moves upwards along the line x = -1. When c < 1, the parabola crosses the x-axis in two distinct points, resulting in two real zeros for these equations. When c = 1, the parabola touches the x-axis at one point, (-1, 0). Looking at the equation, we see that it is a perfect square. . Finally, when c > 1, we see that the parabola does not cross the x-axis at all, meaning there are no real zeros.
Both b and c change the location of the parabola but not its size. These transformations are both translations of the parabola.
We will also examine h and k from the vertex form of the equation. The vertex of the parabola is at (h, k). When we vary h, we see that the parabola moves left and right. When h is negative, the vertex of the parabola is left of the y-axis. When h is positive, the vertex is right of the y-axis.
Varying k is very similar to varying c in the standard form of the equation because it results in a vertical shift.
The question now arises, how is the standard form of the equation of a parabola related to the vertex form? We will now look at how we can convert the standard form to the vertex form by completing the square.
Thus in our vertex form, , we see that . Although it takes some effort to convert an equation in standard form to vertex form, it provides many benefits.
The shapes of all parabolas are the same - the only things that change are the vertex and the scaling factor, a. Thus, a parabola in vertex form is much easier to graph. Once you have the vertex, you can simply step up the sides to find the next points.
These equations also help explain the effect of changing b in the standard form of the quadratic equation. When we changed a, only the size of the parabola was affected. When we changed c, only the vertical position of the vertex changed. However, when we changed b both the horizontal and vertical positions of the vertex were affected. We see in our equations above that both h and k are affected by a change in b, resulting in both horizontal and vertical shifts of the graph.
The vertex form can also make it easier to find the zeros of the quadratic.
From this equation, we see that x will have no real roots if k > 0, one real root if k = 0, and two roots if k < 0. |
# What is a prime number? Definition, Types, Sample Problems
• Last Updated : 17 Aug, 2021
The method used to represent and work with numbers is known as the number system. A number system is a system of writing to represent numbers. It is the mathematical notation used to represent numbers of a given set by using digits or other symbols. It allows operating arithmetic operations such as division, multiplication, addition, subtraction. Some important number systems are Decimal Number System, Binary Number System, Octal Number System, Hexadecimal Number System.
### Decimal Number System
The decimal number system consists of ten digits i.e. from 0 to 9. The base of the decimal number system is 10. These digits can be used to represent or express any numeric value. For example, the decimal number 153 consists of the digit 3 in one place, the digit 5 in the tens place, and the digit 1 in hundreds place which can be represented as,
(1×102 ) + (5 × 101) + (3 × 100)
= (1 × 100) + (5 × 10) + (3 × 1)
{where, 100 = 1}
= 100 + 50 + 3
= 153
There are different types in decimal number systems based on the different characteristics, for instance, there are whole numbers, natural numbers, prime numbers, composite numbers, etc. Let’s learn about Prime numbers in detail,
### What is a Prime number?
In the number system, Prime Numbers are those numbers that have only two factors that is 1 and the number itself. In other words, a prime number is that number that is exactly divisible by 1 and the number itself.
• A Prime number should contain exactly two factors.
• A prime number should be divisible 1 and the number itself.
Let’s assume p is a prime number then p has only 2 factors that are 1 and p itself. Any number which does not follow this is termed a composite number. For example factors of 8 are 1, 2, 4, and 8, which are four factors in total. But factors of 5 are 1 and 5 itself, totally two factors. Hence, 5 is a prime number but 8 is not a prime no, instead, it is a composite number.
First Ten Natural Prime Numbers are – 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
1. Factors of 1 are =1 ( Not Prime Number because it has only one factor)
2. Factors of 2 are = 1 and 2 ( Prime Number because it has only two factors )
3. Factors of 3 are =1 and 3 ( Prime Number because it has only two factors )
4. Factors of 4 are =1, 2, and 4 ( Not Prime Number because it has three factors )
5. Factors of 5 are =1 and 5 ( Prime Number because it has only two factors )
6. Factors of 6 are =1, 2, 3, and 6 ( Not Prime Number because it has four factors )
7. Factors of 7 are =1 and 7 ( Prime Number because it has only two factors )
8. Factors of 8 are =1, 2, 4, and 8 ( Not Prime Number because it has four factors )
9. Factors of 9 are =1, 3, and 9 ( Not Prime Number because it has three factors )
10. Factors of 10 are =1, 2, 5, and 10 ( Not Prime Number because it has four factors )
NOTE: 1 is a non-prime number because according to the definition, a prime number should contains only two factors but 1 has only one factor. Therefore 1 is not a prime number.
List of Prime Number between 1 to 100
It is known, the prime numbers are the numbers that have only two factors which are 1 and the number itself. The above are the prime numbers that are present between 1 and 100.
Even Prime Numbers
Even prime numbers are the numbers that are evenly divisible by 2. Therefore, 2 is the only prime number in the number system that is even so 2 is known as an even prime number and the remaining prime numbers are the odd numbers, therefore they are called odd prime numbers
• Even Prime Numbers = 2
• Odd Prime Numbers = 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 etc.
Twin Prime Numbers
Twin prime numbers are the numbers that have only one composite number between them are known as twin prime numbers or twin primes. The other words of twin prime numbers are the pair of prime numbers whose difference between two prime numbers is 2 only. Few examples are,
• (5, 7) = 7 – 5 = 2
• (11, 13) = 13 – 11 = 2
• (17, 19) = 19 – 17 = 2
• (41, 43) = 43 – 41 = 2
### Sample Problems
Question 1: Is 51 a Prime Number ?
No, because the only factors 51 are 1, 3, 17, and 51. According to the definition, a prime number should contain only two factors. Factors of 51 = 1, 3, 17, 51 ( 4 Factors ). So, 51 is not a prime number.
Question 2: Is 1 a Prime Number?
1 is not a prime number because according to the definition, a prime number should exactly two factors But, number 1 has one and only one factor which is 1 itself. Thus, 1 is not considered a Prime number. Factors of 1 = 1 ( 1 Factor). So, 1 is not a prime number
Question 3: Is 11 is a Prime Number?
Yes, 11 is a prime number because it is only divisible by 2 numbers that is 1 and the number itself (11). Its has two factors 1 and 11 only. Factors of 11 = 1, 11 ( 2 Factors ). So, 11 is a prime number.
Question 4: Find all the prime numbers form the following numbers 1, 22, 3, 51, 75, 88, 65, 63, 19, 7, 39, 47, 60, 100, 12, 10, 5 ?
All the prime number from the given numbers are – 3, 19, 7, 39, 47, 5
Explanation,
• Factor of 1 are = 1
• Factor of 22 are = 1, 2, 11, 22
• Factor of 3 are = 1, 3 ( only two factors )
• Factor of 51 are = 1, 3, 17, 51
• Factor of 75 are = 1, 3, 5, 15, 25, 75.
• Factor of 88 are = 1, 2, 4, 8, 11, 22, 44, 88
• Factor of 65 are = 1, 5, 65
• Factor of 63 are = 1, 3, 7, 9 , 21, 63
• Factor of 19 are = 1, 19 ( only two factors )
• Factor of 7 are = 1, 7 ( only two factors )
• Factor of 39 are = 1, 39 ( only two factors )
• Factor of 47 are = 1, 47 ( only two factors )
• Factor of 60 are = 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
• Factor of 100 are = 1, 2, 4, 5,10, 20, 25, 50, 100
• Factor of 12 are = 1, 2, 3, 4, 6, 12
• Factor of 10 are = 1, 2, 5, 10
• Factor of 5 are = 1, 5 ( only two factors )
Question 5: Which is the smallest Prime number in the number system? |
## INTRODUCTION
Trigonometric functions are based on relationships present between the sides and angles of triangles. Many functions are derived from a right triangle—a triangle containing a right (90°) angle. Consider the following right triangle with sides A, B, and C, and angles α, β, and γ.
Side C, which is the longest side and the side opposite the right angle, is known as the hypotenuse of the triangle.
A commonly used trigonometric relationship for right triangles is the Pythagorean theorem. The Pythagorean theorem is an expression of the relationship between the hypotenuse and the other two sides of a right triangle:
The sum of the squares of the lengths of the two sides of a right triangle is equal to the square of the length of the hypotenuse.
Using the sides of the labeled triangle yields the following:
A2 + B2 = C2
Suppose that sides A and B are 3 and 4 units long, respectively. The Pythagorean theorem can be used to solve for the length of side C:
Three trigonometric relationships are based on the ratios of the lengths of the sides of a right triangle. The sine (abbreviated sin) of an angle is defined as the ratio of the length of the side of the triangle opposite the angle to the length of the hypotenuse. Using the labeled triangle yields the following:
With A = 3, B = 4, and C = 5:
The cosine (abbreviated cos) of an angle is defined as the ratio of the length of the side of the triangle adjacent to the angle to the length of the hypotenuse. Using the labeled triangle yields the following:
With A = 3, B = 4, and C = 5:
The third function, the tangent (abbreviated tan) of an angle, is defined as the ratio of the length of the side of the triangle opposite the angle to that of the side adjacent to the angle. Using the labeled triangle yields the following:
With A = 3, B = 4, and C = 5:
Two useful trigonometric relationships are applicable to all triangles. The first is known as the law of sines:
The ratio between the length of any side of a triangle and the angle opposite that side is equal to the ratio between the length of any other side of the triangle and the angle opposite that side.
With respect to the labeled triangle, this may be stated as ...
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# Equivalence Relationship
The axiom of equivalence states that all numbers are equal to themselves. The same is true for any pair of equalities, as well as for any set of equalities. This is the key to the existence of equivalence relations. They also capture three of the four properties of a mathematical relationship. These properties make equivalence more than just a mathematical property – they also make the relationships more than just a simple metric measure.
For example, let’s consider the relationship between a set of integers and a set of real numbers. An element in a set cannot be contained in another. Therefore, a set of integers X and a subset S are equivalent. A similar example is given below. Using this relationship, we can derive two sets of elements. One set is a prime number, while the other is a real number.
The relationship between two integers is an equivalence relation. For example, if x and y have the same remainder, y and z are related. For example, a pair of integers x and y is equivalent to a set of prime numbers. A similar relation applies to an equivalence class. The inverse of an equivalence class is a multiplication rule.
As an example, suppose that a set X is equal to a group X. This is an equivalence relationship. In this case, a set X is an equivalence relation of x and y. A set X is the same as a set Y. In this case, x = y. And vice-versa. Once you have a set of equivalence relations, you can write a formula to find the equivalence between the two sets.
A bijection is another example of an equivalence relation. This type of equivalence shows that a set is equal to an object. For instance, two sets can only be compared if their attributes are the same. It also means that they are the same. As such, an equivalence relation between two sets is an equivalence between those two sets. It is possible to compare a set and an equivalence between two different types of objects.
An equivalence relation is defined as a relationship that has the same value as another. It is the most general type of equivalence. When a pair is similar, they have the same function. For example, X is an equivalence between two types of objects. This is the same. A related quantity is a product. If a product is not the same, then it is not the same.
If an object is an equivalence between two types of objects, an equivalence relation is a set that has a common value. A set is a function of two different types. The equivalence between two types of elements is a function of their similarity. If two sets are equivalent, their properties are the same. It is a type of equivalence.
An equivalence relation is a mathematical relation between two variables. Its definition is that a set can be similar to another object, if two sets are equivalent. For example, a set can be equivalent to itself. Similarly, a pair of objects may have a common feature. These objects are equivalences. They are both related to each other.
A set of equal values is a corresponding subset. In this case, X is an equivalence relation. An equivalence relation is a relationship between two objects that are equivalent. For instance, if two sets X is equal, Y is the same as it is. If a set Y is the same as a given pair, then sim Y is an equivalence relation between the two classes.
The best equivalence relation is the equality equivalence relation. This relation holds for all pairs of elements. It is an equivalence relation over a set. A pair of functions is deemed equivalent when the set of their fixpoints has the same cardinality. In a permutation, a set of elements with the same cardinality is equal. This relationship holds for the same types of values.
## How to Prove an Equivalence Relationship
Equivalence relations are properties of sets and functions. Each set has a certain function. Its equivalence relation is called ‘congruence modulo n (U)’. An equivalence relation is one in which the value of one object equals that of another. This property is very important in mathematical models. Using it correctly will help you find the equivalence between two objects.
The definition of an equivalence relation is a property of two sets. It states that the two elements belong to the same component of a set. It also says that a set is symmetrical if a partition is symmetrical. Therefore, an equivalence relation is symmetric and reflexive. It can be used in mathematical models. Several types of equivalence relations are known, and this knowledge is useful in evaluating them.
A common example of an equivalence relation is a pair. Suppose that you have two pairs of numbers. You can write the values of each pair separately. Then, divide the result by two to find the equivalence. A pair of numbers may have the same or opposite sign. For example, a pair of numbers can have the same value as a single number. If the pair is equal, then it is equal to two.
An equivalence relation can be expressed in two ways: as a pair of integers or as a product of two numbers. The equivalence relation of two integers, for instance, is a transitive relationship. The result is the same. However, there are many cases where an equivalence relation is not transitive. The equivalence of two numbers is asymmetric.
An equivalence relation between two numbers is a relation that binds two sets. If two numbers have the same value, then they are equivalent. For example, if a pair of integers is equivalent to each other, then the pair is asymmetric. Its equivalence relationship is asymmetric. It is not asymmetric. This means that the set is asymmetric.
An equivalence relation between two numbers is defined by a set’s properties. The equivalence between two objects is a function that relates the two sets. The equivalence between two functions is a type of dependency. A connection between two variables has a symmetric relationship. The same is true of a subset’s properties. An equivalence between objects is asymmetric.
If a set and a function are identical, then they are equivalent. For instance, if a set is homogeneous, then its equivalence relation with another is non-homogeneous. This is asymmetric. If the two objects are similar, then they are not equivalent. They are mutually exclusive. If the two sets are similar, then they are orthogonal.
When a set and a function are symmetric, they are equivalences. This is a transitive relation between two sets. Asymmetric and transversal relationships are transitive, meaning they are related to each other. Asymmetric and transitive relations are equivalent when two sets are not. These two types of equivalences are mutually exclusive. If a pair is not equivalent, it is asymmetric.
An equivalence relation between two sets can be either orthogonal or similar. An equivalence relation between a set and a function is a symmetric relationship. This means that two sets are equivalent in terms of their elements. If a pair of sets is symmetrically equivalent, then it is an equivalence between those two sets. A symmetric relation is an equivalence relation between sets.
An equivalence relation between two sets is transitive if it is disjoint. Its members are symmetric if the set is symmetrical. When a set is disjoint, it is disjoint. Its equivalence is asymmetric on the set. An equivalence relation between two classes can be considered asymmetric, which means that they are mutually unrelated.
The equivalence relation between two sets is transitive, reflexive, and symmetric. It is the equivalence of two elements. A class is an equivalence of two objects. A class is asymmetric when it is a symmetric set. Similarly, a symmetry of a set is asymmetric. Its equivalence between two sets is asymmetric. |
# Lesson 5
Represent Products as Areas
## Warm-up: How Many Do You See: One More (10 minutes)
### Narrative
The purpose of this How Many Do You See is for students to subitize or use grouping strategies to describe the images they see. The arrangement of the groups of dots encourages students to see 5 groups of dots in the first image and then 6 groups of dots in the next image. When students use equal groups and a known quantity to find an unknown quantity, they are looking for and making use of structure. (MP7).
### Launch
• Groups of 2
• “How many do you see? How do you see them?”
• Flash the image.
• 30 seconds: quiet think time
### Activity
• Display the image.
• “Discuss your thinking with your partner.”
• 1 minute: partner discussion
• Record responses.
• Repeat for each image.
### Student Facing
How many do you see? How do you see them?
### Student Response
Teachers with a valid work email address can click here to register or sign in for free access to Student Response.
### Activity Synthesis
• “How did the first image help you find the number of dots in the second image?” (I know that 5 groups of 3 is 15, and one more group of 3 would be 18.)
• “How did the first and second images help you find the number of dots in the third image?” (I figured out 5 groups of 4 pretty quickly, then added another group of 4.)
## Activity 1: Match Expressions and Areas (15 minutes)
### Narrative
The purpose of this activity is for students to directly connect multiplication expressions to equal groups they see within rectangular areas. Students may decompose the rectangular areas in various ways to see equal groups, but they should relate the rows and columns to the factors of a multiplication expression. This will be helpful in future activities when students multiply side lengths to find the area.
### Required Materials
Materials to Copy
• Match Expressions and Areas
### Launch
• Groups of 3–4
• Sketch a 5-by-3 gridded rectangle, as shown.
• “What is one way you could describe this rectangle?” (It has 3 rows of 5 squares. There are 5 groups of 3. Its area is 15 square units. There are 15 squares.)
• Share and record responses. Save responses for discussion after the next activity.
• Display rectangles from the blackline master around the room.
### Activity
• “Match each expression to one of the rectangles posted around the room. Be ready to explain your reasoning.”
• 5–7 minutes: group work time
### Student Facing
Your teacher has posted images of rectangles around the room. Match each expression with a rectangle that can represent it. Be prepared to explain your reasoning.
1. $$9 \times 5$$
2. $$8 \times 2$$
3. $$7 \times 10$$
4. $$3 \times 3$$
5. $$2 \times 6$$
6. $$8 \times 4$$
7. $$5 \times 7$$
### Student Response
Teachers with a valid work email address can click here to register or sign in for free access to Student Response.
### Advancing Student Thinking
If students don’t mention the groups in the rows and columns of squares, consider asking:
• “How did you decide which rectangle matched each expression?”
• “Where do we see equal groups in the rectangles?”
### Activity Synthesis
• “How do you see each factor in the rectangle?” (I can see one factor in the number of squares in a row. I can see the other factor as the number of rows. I see one factor as the number of squares in a column. The other factor is the number of columns. It’s like I see the factors in an array, only it’s squares, not dots.)
• “How do you see the product in the rectangles?” (If we count the squares in each rectangle, it gives us the same number as the product of the factors. The product is the same as the total number of squares in each rectangle.)
• “Why does multiplication give the same number as counting one by one?”
## Activity 2: Create from Expressions (20 minutes)
### Narrative
The purpose of this activity is for students to represent multiplication expressions as rectangular areas. Students use a grid to draw the rectangular area that represents a multiplication expression. In the synthesis, students explain how they interpret the multiplication expression, specifically how they see the equal groups in the rows and columns of the rectangular area. Give students access to square tiles if needed. When students draw and relate area diagrams to multiplication expressions they are reasoning abstractly and quantitatively (MP2).
Engagement: Provide Access by Recruiting Interest. Leverage choice around perceived challenge. Invite students to select at least 3 of the 5 problems to complete.
Supports accessibility for: Organization, Attention, Social-emotional skills
### Required Materials
Materials to Gather
### Launch
• Groups of 2
• Give students access to inch tiles.
• “Now you’re going to draw rectangles that match some multiplication expressions.”
• 1 minute: quiet think time
### Activity
• 7–10 minutes: partner work time
• Monitor for a rectangle that two students oriented differently.
### Student Facing
1. The numbers in each expression represent the number of rows (or columns) in a rectangle and how many squares are in each row (or column).
On the grid, draw each rectangle, label it with the numbers, and find its area.
1. $$3 \times 4$$
2. $$4 \times 6$$
3. $$6 \times 3$$
4. $$7 \times 4$$
5. $$3 \times 2$$
2. Explain why multiplying the numbers in each expression gives us the area of the rectangle.
### Student Response
Teachers with a valid work email address can click here to register or sign in for free access to Student Response.
### Activity Synthesis
• Have 2–3 students share a rectangle for each expression.
• For each student sample ask:
• “How does the area of this rectangle match the expression?” (I see 4 equal groups of 6 because each row has 6 squares. I see 4 equal groups because each column has 6 squares.)
• Consider asking:
• “Did anyone draw a different rectangle for this expression?”
• Display a rectangle that two students oriented differently.
• “How can both of these rectangles match the same expression?” (They have the same number of squares. They have the same side lengths, just switched. I see the groups in the rows in the first rectangle and in the columns in the second rectangle.)
• Display the 3-by-5 rectangle and descriptions from the launch in the first activity.
• “Which way of describing a rectangle was the most helpful to you as you drew rectangles in this activity?” (It was helpful to describe how many rows were in the rectangle and how many squares were in each row. It was helpful to think about one factor as the number of columns and the other factor as the number of squares in each column.)
## Lesson Synthesis
### Lesson Synthesis
Display or sketch a 2-by-7 gridded rectangle with the side lengths labeled 2 and 7, as shown:
“How could you figure out the total number of squares in this rectangle?” (Count by one. Count by 2. Count by 7. Multiply $$2 \times 7$$. Multiply $$7 \times 2$$.)
“How are rectangular areas similar to other ways we’ve shown multiplication?” (We can see rows and columns like arrays. We can see equal groups in the rows. We can see equal groups in the columns.)
“How are rectangular areas different from other ways we’ve shown multiplication?” (We’re counting spaces instead of objects.)
## Cool-down: Create a Rectangular Area (5 minutes)
### Cool-Down
Teachers with a valid work email address can click here to register or sign in for free access to Cool-Downs. |
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# December 4, 2012 - PowerPoint PPT Presentation
December 4, 2012. AIM : How do we find the derivative of products? Can we find derivatives of derivatives?. Do Now : If f(x) = 2cosx – 3sinx + 4, find f’(x) Find g’(x) if g(x) = . HW2.3a Pg. 126 - 128 #1 – 5 odd, 13, 63, 93, 94, 97. How can we find the derivative of
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Presentation Transcript
December 4, 2012
AIM: How do we find the derivative of products? Can we find derivatives of derivatives?
Do Now:
If f(x) = 2cosx – 3sinx + 4, find f’(x)
Find g’(x) if g(x) =
HW2.3a Pg. 126 - 128 #1 – 5 odd, 13, 63, 93, 94, 97
h(x) = 3x2(5x + 1)?
How can we find the derivative of
h(x) = 3x2(5x + 1)?
We can use the Product Rule:
If f and g are differentiable functions, then fg is a differentiable and
(fg)’(x) = f(x)g’(x) + g(x)f’(x)
“The derivative of a product is equal to the first function times the derivative of the second plus the second function times the derivative of the first”
Use the product rule to find the derivative of
h(x) = 3x2(5x + 1)
Step 1: Define f(x) and g(x)
Step 2: Find f’(x) and g’(x)
Step 3: Plug into formula: f(x)g’(x) + g(x)f’(x) and simplify
• Find
• Find if
• Find
• Find f’(x) if f(x) = xsinx
• Find g’(x) if g(x) = excosx
• Find h’(x) if h(x) = sinxcosx
Find the derivative of:
Now find the second derivative:
And the third derivative:
• Find if f(x) = 4x3 – 2x + x-1
• Find f (3)(x) if f(x) = xex |
# What is 1/2 + 2/7?
Here's how you add
1 2
+
2 7
## Step 1
We can't add two fractions with different denominators (the bottom number). So you need to get a common denominator - both bottom numbers need to match. To do this, you'll multiply the denominators times each other... but the numerators have to change, too. They get multiplied by the other term's denominator.
So we multiply 1 by 7, and get 7, then we multiply 2 by 7 and get 14.
Do the same for the second term. We multiply 2 by 2, and get 4, then multiply 2 by 7 and get 14.
So now our fractions look like this:
7 14
+
4 14
## Step 2
Since our denominators match, we can add the numerators.
7 + 4 = 11
Now we have an answer.
11 14
## Step 3
Last of all, we need to simplify the fraction, if possible. Can it be reduced to a simpler fraction?
To find out, we try dividing it by 2...
Nope! So now we try the next greatest prime number, 3...
Nope! So now we try the next greatest prime number, 5...
Nope! So now we try the next greatest prime number, 7...
Nope! So now we try the next greatest prime number, 11...
Nope! So now we try the next greatest prime number, 13...
No good. 13 is larger than 11. So we're done reducing.
There you have it! Here's the final answer to 1/2 + 2/7
1 2
+
2 7
=
11 14
© 2014 Randy Tayler |
# Solving and Graphing Absolute Value Inequalities- Practice Problems
/ / درس 3
### توضیح مختصر
Solving and graphing absolute value inequalities brings a lot of different skills together in one place. The practice problems in this video will give you a good chance to see more examples of absolute value inequalities but will also test your general algebraic knowledge.
• زمان مطالعه 8 دقیقه
• سطح خیلی سخت
### دانلود اپلیکیشن «زوم»
این درس را میتوانید به بهترین شکل و با امکانات عالی در اپلیکیشن «زوم» بخوانید
## Inequality Review
There is one big difference between absolute value equations and absolute value inequalities that we should quickly review before jumping into some practice problems, and that is: when we split an inequality into two different ones in order to undo an absolute value, we need to remember to flip the sign on the one we match with the negative. Other than that, all the rules are the same and we should be good to go.
I also want to mention that because this is a practice problem video and we’ll be doing about four practice problems, I encourage you to pause the video when you see the problem. Try it on your own and see how far you get. If you get stuck, or if you finish the whole thing and want to check your answer, watch the video to see if I did it the same way you did it and if you got the right answer. That way you’ll be able to focus in on what you did wrong, or you can skip through it quickly if you already know you got it right.
## Graph of an OR compound inequality
We’ll practice that one difference in this first question here: Solve and graph x + 4 > 5.
Because there is nothing going on the outside of the absolute value, we can begin by splitting the inequality up to undo the absolute value. That leaves us with two inequalities: one, x + 4 > 5, and another, x + 4 < -5. So not only did we set it to -5, we also flipped it around to a ‘less than’ sign. We still need to solve each inequality for x , which means undoing the + 4 on both of them. You can undo addition with subtraction, and doing that for both inequalities gives us our solved inequality, x > 1 or x < -9. This is now a compound inequality, because there are two inequalities in one problem.
Graphing this compound inequality is as easy as putting both graphs on the same number line one at a time. We can begin by putting an open circle at 1 and drawing an arrow to the right. It’s an open circle because it’s only ‘greater than,’ not ‘equal to’, and the arrow goes to the right because we want all numbers that are bigger than 1. Putting the other one on there means an open circle at -9 and an arrow going to the left. We see that our graph is complete, and it looks like it’s an OR compound inequality because the graphs are going in opposite directions. That means that either being bigger than 1 or being less than -9 is enough to satisfy this inequality. It does not have to be both; it can be one or the other. We actually could have known that it was an OR inequality from the beginning just by realizing first that all one-variable absolute value inequalities give us compound inequalities and that problems where the absolute value is greater than something lead us to OR examples.
## AND Compound Inequality
Our second problem might at first seem like it will also be an OR compound inequality, but there is a difference that will make it end up not being exactly as it might initially seem. The problem, solve and graph -2 x -1 > -9, also has an absolute value and a > symbol, but this time there are mathematical operations on the outside of the absolute value. This means before we split the inequality into two, we need to undo the -1 and the times -2 on the outside.
The outermost step is the -1, so we undo that with addition. We next undo multiplication of -2 with division of -2, and now we must remember the rule of inequalities that tells us to flip the symbol whenever multiplying or dividing by a negative number. This leaves us with the resulting inequality as x < 4, and now we realize that this will end up being an AND compound inequality because the absolute value is now less than 4 instead of greater than like in the beginning. Splitting the inequality and flipping one of the symbols leaves us with our solved inequality: x < 4 and x > -4.
To graph this, I can again do one at a time and put a closed circle (because it’s ‘or equal to’) at 4 and draw an arrow to the left, then put another closed circle at -4 and draw an arrow to the right. Because the arrows are pointing toward each other, we can just connect the two dots and we end up with our AND compound inequality graph looking like so.
## Negative Numbers and Inequalities
If at any point you are solving a problem like one of these two and you end up with a statement where an absolute value of anything is either greater than or less than a negative number, you know something is up. For example, take 7 x - 1 > -5. It doesn’t matter what is happening on the inside of the absolute value, because we know that it will eventually be turned into a positive number. Therefore, this absolute value will always be bigger than -5; the inequality will always be true no matter what we substitute in for x , and therefore there are an infinite number of solutions to this problem. But on the other hand, if instead we had 7 x - 1 < -5, the same logic tells us that it is impossible for us to make an absolute value smaller than -5, and therefore there are no solutions to this one.
## System of Inequalities
We’ll end this practice problem video with a system of two-variable absolute value inequalities. Graph y > (1/2) x - 5 and y < 2.
This is a system of inequalities because there is more than one inequality, and at least one of them has two variables. We can solve this problem simply by putting one inequality on the graph at a time and then figuring out where they overlap. First, let’s start with graphing y > (1/2) x - 5. To do this, we’ll have to remember how to use translations and also how to graph an absolute value. All absolute value graphs look more or less like ‘V’s, but this one is going to have a few differences. First off, the -5 on the end will pull the vertex of the graph down five places, so instead of it being at the origin (0,0), it will be pulled down to the point (0,-5). Secondly, the 1/2 in front of the absolute value will make the slope of the ‘V’ 1/2 instead of 1. That means that, starting from our vertex, we’ll go up one and then over two in each direction to determine how steep the ‘V’ is, and we can sketch it in like this. We can leave the line of the ‘V’ solid because it is ‘or equal to’ from the inequality, but we still need to determine which part of the graph to shade. Substituting in (0,0) give us the inequality 0 > (1/2)(0) - 5, which can simplify down to 0 > -5, which is true. That means we can shade the area of the graph with (0,0) in it, and we fill in everything on the inside of the ‘V’ to get that two-variable inequality.
## Graphing a system of inequalities
But this was a system of inequalities, so we’ve still got another piece to add to this graph, y < 2. Just like we began the first half of this problem by graphing the line where y was equal to the absolute value to get our ‘V’ and then determining which side to shade, we’ll start graphing y < 2 by graphing where y =2 and then deciding which side of that line to shade. Graphing y = lines on a coordinate plane leaves us with horizontal lines, so y =2 looks like this. We need to make it a dotted line to indicate that it is strictly ‘less than,’ not ‘equal to,’ and then shade below it to indicate that y =0 works when I substitute it into the inequality (0 < 2). Now we’ve got a graph with a ton of shading everywhere. Because the solution to a system of inequalities is only where the shaded regions overlap, when we lay the y < 2 graph on top of our absolute value ‘V’, we find that our solution is only the triangular region in the middle of the graph, inside the ‘V’ but below the horizontal line.
## Lesson Summary
To review, when splitting an absolute value inequality into two new ones to undo an absolute value, you must flip the inequality symbol on the one with the negative. Operations outside an absolute value must be undone before undoing the absolute value itself. Finally, systems of inequalities can be done with absolute values just like other lines, one graph at a time, where the solution is only the area where the shading overlaps.
## Lesson Objectives
When you complete this lesson you’ll be able to split and unsplit absolute value inequalities and give systems of inequalities for absolute values.
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# IB Math Analysis & Approaches Questionbank-Topic: SL 2.11 Transformations of graphs SL Paper 1
## Question
Let $$f(x) = 3{x^2} – 6x + p$$. The equation $$f(x) = 0$$ has two equal roots.
Write down the value of the discriminant.
[2]
a(i).
Hence, show that $$p = 3$$.
[1]
a(ii).
The graph of $$f$$has its vertex on the $$x$$-axis.
Find the coordinates of the vertex of the graph of $$f$$.
[4]
b.
The graph of $$f$$ has its vertex on the $$x$$-axis.
Write down the solution of $$f(x) = 0$$.
[1]
c.
The graph of $$f$$Â has its vertex on the $$x$$-axis.
The function can be written in the form $$f(x) = a{(x – h)^2} + k$$. Write down the value of $$a$$.
[1]
d(i).
The graph of $$f$$ has its vertex on the $$x$$-axis.
The function can be written in the form $$f(x) = a{(x – h)^2} + k$$. Write down the value of $$h$$.
[1]
d(ii).
The graph of $$f$$ has its vertex on the $$x$$-axis.
The function can be written in the form $$f(x) = a{(x – h)^2} + k$$. Write down the value of $$k$$.
[1]
d(iii).
The graph of $$f$$ has its vertex on the $$x$$-axis.
The graph of a function $$g$$ is obtained from the graph of $$f$$ by a reflection of $$f$$ in the $$x$$-axis, followed by a translation by the vector $$\left( \begin{array}{c}0\\6\end{array} \right)$$. Find $$g$$, giving your answer in the form $$g(x) = A{x^2} + Bx + C$$.
[4]
e.
## Markscheme
correct value $$0$$, or $$36 – 12p$$ Â Â A2 Â Â N2
[2 marks]
a(i).
correct equation which clearly leads to $$p = 3$$ Â Â A1
eg   $$36 – 12p = 0,{\text{ }}36 = 12p$$
$$p = 3$$ Â Â AG Â Â N0
[1 mark]
a(ii).
METHOD 1
valid approach   (M1)
eg   $$x = – \frac{b}{{2a}}$$
correct working   A1
eg   $$– \frac{{( – 6)}}{{2(3)}},{\text{ }}x = \frac{6}{6}$$
correct answers   A1A1   N2
eg   $$x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)$$
METHOD 2
valid approach   (M1)
eg   $$f(x) = 0$$, factorisation, completing the square
correct working   A1
eg   $${x^2} – 2x + 1 = 0,{\text{ }}(3x – 3)(x – 1),{\text{ }}f(x) = 3{(x – 1)^2}$$
correct answers   A1A1   N2
eg   $$x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)$$
METHOD 3
valid approach using derivative   (M1)
eg   $$f'(x) = 0,{\text{ }}6x – 6$$
correct equation   A1
eg   $$6x – 6 = 0$$
correct answers   A1A1   N2
eg   $$x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)$$
[4 marks]
b.
$$x = 1$$ Â Â A1 Â Â N1
[1 mark]
c.
$$a = 3$$ Â Â A1 Â Â N1
[1 mark]
d(i).
$$h = 1$$ Â Â A1 Â Â N1
[1 mark]
d(ii).
$$k = 0$$ Â Â A1 Â Â N1
[1 mark]
d(iii).
attempt to apply vertical reflection   (M1)
eg    $$– f(x),{\text{ }} – 3{(x – 1)^2}$$, sketch
attempt to apply vertical shift 6 units up   (M1)
eg    $$– f(x) + 6$$, vertex $$(1, 6)$$
transformations performed correctly (in correct order) Â Â (A1)
eg    $$– 3{(x – 1)^2} + 6,{\text{ }} – 3{x^2} + 6x – 3 + 6$$
$$g(x) =Â – 3{x^2} + 6x + 3$$ Â Â A1 Â Â N3
[4 marks]
e.
## Question
The following diagram shows part of the graph of a quadratic function $$f$$.
The vertex is at $$(1,{\text{ }} – 9)$$, and the graph crosses the yaxis at the point $$(0,{\text{ }}c)$$.
The function can be written in the form $$f(x) = {(x – h)^2} + k$$.
Write down the value of $$h$$ and of $$k$$.
[2]
a.
Find the value of $$c$$.
[2]
b.
Let $$g(x) =Â – {(x – 3)^2} + 1$$. The graph of $$g$$ is obtained by a reflection of the graph of $$f$$ in the $$x$$-axis, followed by a translation of $$\left( {\begin{array}{*{20}{c}} p \\ q \end{array}} \right)$$.
Find the value of $$p$$ and of $$q$$.
[5]
c.
Find the x-coordinates of the points of intersection of the graphs of $$f$$ and $$g$$.
[7]
d.
## Markscheme
$$h = 1,{\text{ }}k =Â – 9\;\;\;\left( {{\text{accept }}{{(x – 1)}^2} – 9} \right)$$ Â Â A1A1 Â Â N2
[2 marks]
a.
METHOD 1
attempt to substitute $$x = 0$$ into their quadratic function   (M1)
eg$$\;\;\;f(0),{\text{ }}{(0 – 1)^2} – 9$$
$$c =Â – 8$$ Â Â A1 Â Â N2
METHOD 2
attempt to expand their quadratic function   (M1)
eg$$\;\;\;{x^2} – 2x + 1 – 9,{\text{ }}{x^2} – 2x – 8$$
$$c =Â – 8$$ Â Â A1 Â Â N2
[2 marks]
b.
evidence of correct reflection   A1
eg$$\;\;\; – \left( {{{(x – 1)}^2} – 9} \right)$$, vertex at $$(1,{\text{ }}9)$$, y-intercept at $$(0,{\text{ }}8)$$
valid attempt to find horizontal shift   (M1)
eg$$\;\;\;1 + p = 3,{\text{ }}1 \to 3$$
$$p = 2$$ Â Â A1 Â Â N2
valid attempt to find vertical shift   (M1)
eg$$\;\;\;9 + q = 1,{\text{ }}9 \to 1,{\text{ }} – 9 + q = 1$$
$$q =Â – 8$$ Â Â A1 Â Â N2
Notes: Â Â An error in finding the reflection may still allow the correct values of $$p$$ and $$q$$ to be found, as the error may not affect subsequent working. In this case, award A0 for the reflection, M1A1 for $$p = 2$$, and M1A1 for $$q =Â – 8$$.
If no working shown, award N0 for $$q = 10$$.
[5 marks]
c.
valid approach (check FT from (a)) Â Â M1
eg$$\;\;\;f(x) = g(x),{\text{ }}{(x – 1)^2} – 9 =Â – {(x – 3)^2} + 1$$
correct expansion of both binomials   (A1)
eg$$\;\;\;{x^2} – 2x + 1,{\text{ }}{x^2} – 6x + 9$$
correct working   (A1)
eg$$\;\;\;{x^2} – 2x – 8 =Â – {x^2} + 6x – 8$$
correct equation   (A1)
eg$$\;\;\;2{x^2} – 8x = 0,{\text{ }}2{x^2} = 8x$$
correct working   (A1)
eg$$\;\;\;2x(x – 4) = 0$$
$$x = 0,{\text{ }}x = 4$$ Â Â A1A1 Â Â N3
[7 marks]
Total [16 marks]
d.
## Question
The following diagram shows the graph of a function $$f$$, for −4 ≤ x ≤ 2.
On the same axes, sketch the graph of $$f\left( { – x} \right)$$.
[2]
a.
Another function, $$g$$, can be written in the form $$g\left( x \right) = a \times f\left( {x + b} \right)$$. The following diagram shows the graph of $$g$$.
Write down the value of a and of b.
[4]
b.
## Markscheme
A2 N2
[2 marks]
a.
recognizing horizontal shift/translation of 1 unit   (M1)
eg  b = 1, moved 1 right
recognizing vertical stretch/dilation with scale factor 2Â Â Â (M1)
eg  a = 2,  y ×(−2)
a = −2, b = −1   A1A1 N2N2
[4 marks]
b.
## Question
The following diagram shows the graph of a function $$f$$.
Find $${f^{ – 1}}( – 1)$$.
[2]
a.
Find $$(f \circ f)( – 1)$$.
[3]
b.
On the same diagram, sketch the graph of $$y = f( – x)$$.
[2]
c.
## Markscheme
valid approach   (M1)
eg$$\;\;\;$$horizontal line on graph at $$– 1,{\text{ }}f(a) =Â – 1,{\text{ }}( – 1,5)$$
$${f^{ – 1}}( – 1) = 5$$ Â Â A1 Â Â N2
[2 marks]
a.
attempt to find $$f( – 1)$$ Â Â (M1)
eg$$\;\;\;$$line on graph
$$f( – 1) = 2$$ Â Â (A1)
$$(f \circ f)( – 1) = 1$$ Â Â A1 Â Â N3
[3 marks]
b.
   A1A1   N2
Note: Â Â The shape must be an approximately correct shape (concave down and increasing). Only if the shape is approximately correct, award the following for points in circles:
A1 for the $$y$$-intercept,
A1 for any two of these points $$( – 5,{\text{ }} – 1),{\text{ }}( – 2,{\text{ }}1),{\text{ }}(1,{\text{ }}2)$$.
[2 marks]
Total [7 marks]
c.
## Question
Let $$f(x) = 3{(x + 1)^2} – 12$$ .
Show that $$f(x) = 3{x^2} + 6x – 9$$ .
[2]
a.
For the graph of f
(i)Â Â Â Â write down the coordinates of the vertex;
(ii)Â Â Â write down the equation of the axis of symmetry;
(iii) Â write down the y-intercept;
(iv)Â Â find both x-intercepts.
[8]
b(i), (ii), (iii) and (iv).
Hence sketch the graph of f .
[2]
c.
Let $$g(x) = {x^2}$$ . The graph of f may be obtained from the graph of g by the two transformations:
a stretch of scale factor t in the y-direction
followed by a translation of $$\left( {\begin{array}{*{20}{c}} p\\ q \end{array}} \right)$$ .
Find $$\left( {\begin{array}{*{20}{c}} p\\ q \end{array}} \right)$$Â and the value of t.
[3]
d.
## Markscheme
$$f(x) = 3({x^2} + 2x + 1) – 12$$Â Â Â Â A1
$$= 3{x^2} + 6x + 3 – 12$$Â Â Â Â Â A1
$$= 3{x^2} + 6x – 9$$Â Â Â Â Â AGÂ Â Â Â N0
[2 marks]
a.
(i) vertex is $$( – 1{\text{, }} – 12)$$ Â Â Â A1A1Â Â Â Â N2
(ii) $$x = – 1$$ (must be an equation)Â Â Â Â A1 Â Â N1
(iii) $$(0{\text{, }} – 9)$$ Â Â Â A1Â Â Â Â N1
(iv) evidence of solving $$f(x) = 0$$Â Â Â Â (M1)
e.g. factorizing, formula,
correct working    A1
e.g. $$3(x + 3)(x – 1) = 0$$ , $$x = \frac{{ – 6 \pm \sqrt {36 + 108} }}{6}$$
$$( – 3{\text{, }}0)$$, $$(1{\text{, }}0)$$ Â Â Â A1A1 Â Â N1N1
[8 marks]
b(i), (ii), (iii) and (iv).
    A1A1    N2
Note: Award A1 for a parabola opening upward, A1 for vertex and intercepts in approximately correct positions.
[2 marks]
c.
$$\left( {\begin{array}{*{20}{c}} p\\ q \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { – 1}\\ { – 12} \end{array}} \right)$$
, $$t = 3$$ (accept $$p = – 1$$ , $$q = – 12$$ , $$t = 3$$ )Â Â Â Â A1A1A1Â Â Â Â N3
[3 marks]
d.
## Question
Part of the graph of a function f is shown in the diagram below.
On the same diagram sketch the graph of $$y = – f(x)$$ .
[2]
a.
Let $$g(x) = f(x + 3)$$ .
(i)Â Â Â Â Find $$g( – 3)$$ .
(ii)Â Â Â Describe fully the transformation that maps the graph of f to the graph of g.
[4]
b(i) and (ii).
## Markscheme
    M1A1    N2
Note: Award M1 for evidence of reflection in x-axis, A1 for correct vertex and all intercepts approximately correct.
a.
(i) $$g( – 3) = f(0)$$Â Â Â Â Â (A1)
$$f(0) = – 1.5$$Â Â Â Â A1Â Â Â Â N2
(ii) translation (accept shift, slide, etc.) of $$\left( {\begin{array}{*{20}{c}} { – 3}\\ 0 \end{array}} \right)$$Â Â Â Â
A1A1Â Â Â Â N2
[4 marks]
b(i) and (ii).
## Question
Let $$f(t) = a\cos b(t – c) + d$$ , $$t \ge 0$$ . Part of the graph of $$y = f(t)$$ is given below.
When $$t = 3$$ , there is a maximum value of 29, at M.
When $$t = 9$$ , there is a minimum value of 15.
Â
(i)Â Â Â Â Find the value of a.
(ii) Â Â Show that $$b = \frac{\pi }{6}$$ .
(iii)Â Â Find the value of d.
(iv) Â Write down a value for c.
[7]
a(i), (ii), (iii) and (iv).
The transformation P is given by a horizontal stretch of a scale factor of $$\frac{1}{2}$$ , followed by a translation of $$\left( {\begin{array}{*{20}{c}} 3\\ { – 10} \end{array}} \right)$$ .
Let $${M’}$$ be the image of M under P. Find the coordinates of $${M’}$$ .
[2]
b.
The graph of g is the image of the graph of f under P.
Find $$g(t)$$ in the form $$g(t) = 7\cos B(t – c) + D$$ .
[4]
c.
The graph of g is the image of the graph of f under P.
Give a full geometric description of the transformation that maps the graph of g to the graph of f .
[3]
d.
## Markscheme
(i) attempt to substitute    (M1)
e.g. $$a = \frac{{29 – 15}}{2}$$
$$a = 7$$ (accept $$a = – 7$$Â )Â Â Â Â A1Â Â Â Â N2
(ii) $${\text{period}} = 12$$Â Â Â Â (A1)
$$b = \frac{{2\pi }}{{12}}$$Â Â Â A1
$$b = \frac{\pi }{6}$$Â Â Â AGÂ Â Â N0
(iii) attempt to substitute    (M1)
e.g. $$d = \frac{{29 + 15}}{2}$$
$$d = 22$$Â Â Â Â A1Â Â Â Â N2
(iv) $$c = 3$$Â (accept $$c = 9$$Â from $$a = – 7$$Â )Â Â Â Â A1 Â Â N1
Note: Other correct values for c can be found, $$c = 3 \pm 12k$$ , $$k \in \mathbb{Z}$$ .
[7 marks]
a(i), (ii), (iii) and (iv).
stretch takes 3 to 1.5Â Â Â Â (A1)
translation maps $$(1.5{\text{, }}29)$$ to $$(4.5{\text{, }}19)$$ (so $${M’}$$ is $$(4.5{\text{, }}19)$$)Â Â Â Â A1Â Â Â Â N2
[2 marks]
b.
$$g(t) = 7\cos \frac{\pi }{3}\left( {t – 4.5} \right) + 12$$Â Â Â A1A2A1 Â Â N4
Note: Award A1 for $$\frac{\pi }{3}$$Â , A2 for 4.5, A1 for 12.
Other correct values for c can be found, $$c = 4.5 \pm 6k$$ , $$k \in \mathbb{Z}$$ .
[4 marks]
c.
translation $$\left( {\begin{array}{*{20}{c}} { – 3}\\ {10} \end{array}} \right)$$Â Â Â Â (A1)
horizontal stretch of a scale factor of 2Â Â Â Â (A1)
completely correct description, in correct order    A1   N3
e.g. translation $$\left( {\begin{array}{*{20}{c}} { – 3}\\ {10} \end{array}} \right)$$Â then horizontal stretch of a scale factor of 2
[3 marks]
d.
## Question
Let $$f(x) = {x^2}$$ and $$g(x) = 2{(x – 1)^2}$$ .
The graph of g can be obtained from the graph of f using two transformations.
Give a full geometric description of each of the two transformations.
[2]
a.
The graph of g is translated by the vector $$\left( {\begin{array}{*{20}{c}} 3\\ { – 2} \end{array}} \right)$$ to give the graph of h.
The point $$( – 1{\text{, }}1)$$ on the graph of f is translated to the point P on the graph of h.
Find the coordinates of P.
[4]
b.
## Markscheme
in any order
translated 1 unit to the right    A1   N1
stretched vertically by factor 2Â Â Â Â A1 Â Â N1
[2 marks]
a.
METHOD 1
finding coordinates of image on g    (A1)(A1)
e.g. $$– 1 + 1 = 0$$ , $$1 \times 2 = 2$$ , $$( – 1{\text{, }}1) \to ( – 1 + 1{\text{, }}2 \times 1)$$ , $$(0{\text{, }}2)$$
P is (3, 0)Â Â Â Â A1A1Â Â Â Â N4
METHOD 2
$$h(x) = 2{(x – 4)^2} – 2$$Â Â Â Â (A1)(A1)
P is $$(3{\text{, }}0)$$ Â Â Â A1A1Â Â Â Â N4
b.
## Question
Let $$f(x) = \frac{{ax}}{{{x^2} + 1}}$$ , $$– 8 \le x \le 8$$ , $$a \in \mathbb{R}$$ .The graph of f is shown below.
The region between $$x = 3$$ and $$x = 7$$ is shaded.
Show that $$f( – x) = – f(x)$$ .
[2]
a.
Given that $$f”(x) = \frac{{2ax({x^2} – 3)}}{{{{({x^2} + 1)}^3}}}$$ , find the coordinates of all points of inflexion.
[7]
b.
It is given that $$\int {f(x){\rm{d}}x = \frac{a}{2}} \ln ({x^2} + 1) + C$$ .
(i)Â Â Â Â Find the area of the shaded region, giving your answer in the form $$p\ln q$$ .
(ii)Â Â Â Find the value of $$\int_4^8 {2f(x – 1){\rm{d}}x}$$Â .
[7]
c.
## Markscheme
METHOD 1
evidence of substituting $$– x$$ for $$x$$Â Â Â Â Â (M1)
$$f( – x) = \frac{{a( – x)}}{{{{( – x)}^2} + 1}}$$Â Â Â Â A1
$$f( – x) = \frac{{ – ax}}{{{x^2} + 1}}$$ $$( = – f(x))$$Â Â Â Â AG Â Â N0
METHOD 2
$$y = – f(x)$$ is reflection of $$y = f(x)$$ in x axis
and $$y = f( – x)$$ is reflection of $$y = f(x)$$ in y axis    (M1)
sketch showing these are the same    A1
$$f( – x) = \frac{{ – ax}}{{{x^2} + 1}}$$ $$( = – f(x))$$Â Â Â Â AGÂ Â Â Â N0
[2 marks]
a.
evidence of appropriate approach    (M1)
e.g. $$f”(x) = 0$$
to set the numerator equal to 0Â Â Â Â (A1)
e.g. $$2ax({x^2} – 3) = 0$$ ; $$({x^2} – 3) = 0$$
(0, 0) , $$\left( {\sqrt 3 ,\frac{{a\sqrt 3 }}{4}} \right)$$ , $$\left( { – \sqrt 3 , – \frac{{a\sqrt 3 }}{4}} \right)$$ (accept $$x = 0$$ , $$y = 0$$ etc)Â Â Â Â Â A1A1A1A1A1Â Â Â Â N5
[7 marks]
b.
(i) correct expression    A2
e.g. $$\left[ {\frac{a}{2}\ln ({x^2} + 1)} \right]_3^7$$ , $$\frac{a}{2}\ln 50 – \frac{a}{2}\ln 10$$ , $$\frac{a}{2}(\ln 50 – \ln 10)$$
area = $$\frac{a}{2}\ln 5$$    A1A1    N2
(ii) METHOD 1
recognizing the shift that does not change the area    (M1)
e.g. $$\int_4^8 {f(x – 1){\rm{d}}x}Â = \int_3^7 {f(x){\rm{d}}x}$$ , $$\frac{a}{2}\ln 5$$
recognizing that the factor of 2 doubles the area    (M1)
e.g. $$\int_4^8 {2f(x – 1){\rm{d}}x = } 2\int_4^8 {f(x – 1){\rm{d}}x}$$ $$\left( { = 2\int_3^7 {f(x){\rm{d}}x} } \right)$$
$$\int_4^8 {2f(x – 1){\rm{d}}x = a\ln 5}$$ (i.e. $$2 \times$$ their answer to (c)(i))Â Â Â Â A1Â Â Â Â N3
METHOD 2
changing variable
let $$w = x – 1$$ , so $$\frac{{{\rm{d}}w}}{{{\rm{d}}x}} = 1$$
$$2\int {f(w){\rm{d}}w = } \frac{{2a}}{2}\ln ({w^2} + 1) + c$$Â Â Â Â Â (M1)
substituting correct limits
e.g. $$\left[ {a\ln \left[ {{{(x – 1)}^2} + 1} \right]} \right]_4^8$$ , $$\left[ {a\ln ({w^2} + 1)} \right]_3^7$$ , $$a\ln 50 – a\ln 10$$    (M1)
$$\int_4^8 {2f(x – 1){\rm{d}}x = a\ln 5}$$Â Â Â Â A1 Â Â N3
[7 marks]
c.
## Question
The diagram below shows the graph of a function $$f(x)$$ , for $$– 2 \le x \le 4$$ .
Let $$h(x) = f( – x)$$ . Sketch the graph of $$h$$ on the grid below.
[3]
a.
Let $$g(x) = \frac{1}{2}f(x – 1)$$ . The point $${\text{A}}(3{\text{, }}2)$$ on the graph of $$f$$ is transformed to the point P on the graph of $$g$$ . Find the coordinates of P.
[3]
b.
## Markscheme
    A2    N2
[2 marks]
a.
evidence of appropriate approach    (M1)
e.g. reference to any horizontal shift and/or stretch factor, $$x = 3 + 1$$ , $$y = \frac{1}{2} \times 2$$
P is $$(4{\text{, }}1)$$ (accept $$x = 4$$ , $$y = 1$$)Â Â Â Â A1A1Â Â Â Â N3
[3 marks]
b.
## Question
Let $$f(x) = \frac{1}{2}{x^3} – {x^2} – 3x$$Â . Part of the graph of f is shown below.
There is a maximum point at A and a minimum point at B(3, − 9) .
Find the coordinates of A.
[8]
a.
Write down the coordinates of
(i) Â Â the image of B after reflection in the y-axis;
(ii)Â Â Â the image of B after translation by the vector $$\left( {\begin{array}{*{20}{c}} { – 2}\\ 5 \end{array}} \right)$$ ;
(iii)Â Â the image of B after reflection in the x-axis followed by a horizontal stretch with scale factor $$\frac{1}{2}$$ .
[6]
b(i), (ii) and (iii).
## Markscheme
$$f(x) = {x^2} – 2x – 3$$Â Â Â Â A1A1A1
evidence of solving $$f'(x) = 0$$Â Â Â Â (M1)
e.g. $${x^2} – 2x – 3 = 0$$
evidence of correct working    A1
e.g. $$(x + 1)(x – 3)$$ , Â $$\frac{{2 \pm \sqrt {16} }}{2}$$
$$x =Â – 1$$ (ignore $$x = 3$$Â )Â Â Â Â (A1)
evidence of substituting their negative x-value into $$f(x)$$Â Â Â Â Â (M1)
e.g. $$\frac{1}{3}{( – 1)^3} – {( – 1)^2} – 3( – 1)$$ , $$– \frac{1}{3} – 1 + 3$$
$$y = \frac{5}{3}$$Â Â Â Â A1
coordinates are $$\left( { – 1,\frac{5}{3}} \right)$$Â Â Â Â Â N3
[8 marks]
a.
(i) $$( – 3{\text{, }} – 9)$$Â Â Â Â Â A1Â Â Â Â N1
(ii) $$(1{\text{, }} – 4)$$Â Â Â Â Â A1A1Â Â Â N2
(iii) reflection gives $$(3{\text{, }}9)$$Â Â Â Â Â (A1)
stretch gives $$\left( {\frac{3}{2}{\text{, }}9} \right)$$Â Â Â Â Â A1A1Â Â Â Â N3
[6 marks]
b(i), (ii) and (iii).
## Question
Let $$f(x) = {x^2} + 4$$ and $$g(x) = x – 1$$ .
Find $$(f \circ g)(x)$$ .
[2]
a.
The vector $$\left( {\begin{array}{*{20}{c}} 3\\ { – 1} \end{array}} \right)$$ translates the graph of $$(f \circ g)$$ to the graph of h .
Find the coordinates of the vertex of the graph of h .
[3]
b.
The vector $$\left( {\begin{array}{*{20}{c}} 3\\ { – 1} \end{array}} \right)$$ translates the graph of $$(f \circ g)$$ to the graph of h .
Show that $$h(x) = {x^2} – 8x + 19$$ .
[2]
c.
The vector $$\left( {\begin{array}{*{20}{c}} 3\\ { – 1} \end{array}} \right)$$ translates the graph of $$(f \circ g)$$ to the graph of h .
The line $$y = 2x – 6$$ is a tangent to the graph of h at the point P. Find the x-coordinate of P.
[5]
d.
## Markscheme
attempt to form composition (in any order)Â Â Â Â (M1)
$$(f \circ g)(x) = {(x – 1)^2} + 4$$Â Â Â $$({x^2} – 2x + 5)$$Â Â Â Â Â A1Â Â Â Â N2
[2 marks]
a.
METHOD 1
vertex of $$f \circ g$$ at (1, 4)Â Â Â Â (A1)
evidence of appropriate approach    (M1)
e.g. adding $$\left( {\begin{array}{*{20}{c}} 3\\ { – 1} \end{array}} \right)$$ to the coordinates of the vertex of $$f \circ g$$
vertex of h at (4, 3)Â Â Â Â A1 Â Â N3
METHOD 2
attempt to find $$h(x)$$Â Â Â Â (M1)
e.g. $${((x – 3) – 1)^2} + 4 – 1$$ , $$h(x) = (f \circ g)(x – 3) – 1$$
$$h(x) = {(x – 4)^2} + 3$$Â Â Â Â Â (A1)
vertex of h at (4, 3)Â Â Â Â A1 Â Â N3
[3 marks]
b.
evidence of appropriate approach    (M1)
e.g. $${(x – 4)^2} + 3$$ ,$${(x – 3)^2} – 2(x – 3) + 5 – 1$$
simplifying    A1
e.g. $$h(x) = {x^2} – 8x + 16 + 3$$ , $${x^2} – 6x + 9 – 2x + 6 + 4$$
$$h(x) = {x^2} – 8x + 19$$Â Â Â Â Â AG Â Â N0
[2 marks]
c.
METHOD 1
equating functions to find intersection point    (M1)
e.g. $${x^2} – 8x + 19 = 2x – 6$$ , $$y = h(x)$$
$${x^2} – 10x + 25 + 0$$Â Â Â Â A1
evidence of appropriate approach to solve    (M1)
appropriate working    A1
e.g. $${(x – 5)^2} = 0$$
$$x = 5$$Â $$(p = 5)$$Â Â Â Â Â A1Â Â Â Â N3
METHOD 2
attempt to find $$h'(x)$$Â Â Â Â Â (M1)
$$h(x) = 2x – 8$$Â Â Â Â A1
recognizing that the gradient of the tangent is the derivative    (M1)
e.g. gradient at $$p = 2$$
$$2x – 8 = 2$$Â $$(2x = 10)$$Â Â Â Â Â A1
$$x = 5$$Â Â Â Â Â A1 Â Â N3
[5 marks]
d.
## Question
Let $$f(x) = 3\ln x$$ and $$g(x) = \ln 5{x^3}$$ .
Express $$g(x)$$ in the form $$f(x) + \ln a$$ , where $$a \in {{\mathbb{Z}}^ + }$$ .
[4]
a.
The graph of g is a transformation of the graph of f . Give a full geometric description of this transformation.
[3]
b.
## Markscheme
attempt to apply rules of logarithms    (M1)
e.g. $$\ln {a^b} = b\ln a$$ , $$\ln ab = \ln a + \ln b$$
correct application of $$\ln {a^b} = b\ln a$$ (seen anywhere)Â Â Â Â A1
e.g. $$3\ln x = \ln {x^3}$$
correct application of $$\ln ab = \ln a + \ln b$$ (seen anywhere)Â Â Â Â A1
e.g. $$\ln 5{x^3} = \ln 5 + \ln {x^3}$$
so $$\ln 5{x^3} = \ln 5 + 3\ln x$$
$$g(x) = f(x) + \ln 5$$ (accept $$g(x) = 3\ln x + \ln 5$$ ) Â Â A1Â Â Â Â N1
[4 marks]
a.
transformation with correct name, direction, and value    A3
e.g. translation by $$\left( {\begin{array}{*{20}{c}} 0\\ {\ln 5} \end{array}} \right)$$ , shift up by $$\ln 5$$ , vertical translation of $$\ln 5$$
[3 marks]
b.
## Examiners report
This question was very poorly done by the majority of candidates. While candidates seemed to have a vague idea of how to apply the rules of logarithms in part (a), very few did so successfully. The most common error in part (a) was to begin incorrectly with $$\ln 5{x^3} = 3\ln 5x$$Â . This error was often followed by other errors.
a.
In part (b), very few candidates were able to describe the transformation as a vertical translation (or shift). Many candidates attempted to describe numerous incorrect transformations, and some left part (b) entirely blank.
b.
## Question
The diagram below shows the graph of a function $$f(x)$$ , for $$– 2 \le x \le 3$$ .
Â
Sketch the graph of $$f( – x)$$ on the grid below.
[2]
a.
The graph of f is transformed to obtain the graph of g . The graph of g is shown below.
The function g can be written in the form $$g(x) = af(x + b)$$ . Write down the value of a and of b .
[4]
b.
## Markscheme
    A2   N2
[2 marks]
a.
$$a = – 2,b = – 1$$Â Â Â Â A2A2Â Â Â Â N4
Note: Award A1 for $$a = 2$$Â , A1 for $$b = 1$$Â .
[4 marks]
b.
## Question
Let $$f(x) = 3{x^2} – 6x + p$$. The equation $$f(x) = 0$$ has two equal roots.
Write down the value of the discriminant.
[2]
a(i).
Hence, show that $$p = 3$$.
[1]
a(ii).
The graph of $$f$$has its vertex on the $$x$$-axis.
Find the coordinates of the vertex of the graph of $$f$$.
[4]
b.
The graph of $$f$$ has its vertex on the $$x$$-axis.
Write down the solution of $$f(x) = 0$$.
[1]
c.
The graph of $$f$$Â has its vertex on the $$x$$-axis.
The function can be written in the form $$f(x) = a{(x – h)^2} + k$$. Write down the value of $$a$$.
[1]
d(i).
The graph of $$f$$ has its vertex on the $$x$$-axis.
The function can be written in the form $$f(x) = a{(x – h)^2} + k$$. Write down the value of $$h$$.
[1]
d(ii).
The graph of $$f$$ has its vertex on the $$x$$-axis.
The function can be written in the form $$f(x) = a{(x – h)^2} + k$$. Write down the value of $$k$$.
[1]
d(iii).
The graph of $$f$$ has its vertex on the $$x$$-axis.
The graph of a function $$g$$ is obtained from the graph of $$f$$ by a reflection of $$f$$ in the $$x$$-axis, followed by a translation by the vector $$\left( \begin{array}{c}0\\6\end{array} \right)$$. Find $$g$$, giving your answer in the form $$g(x) = A{x^2} + Bx + C$$.
[4]
e.
## Markscheme
correct value $$0$$, or $$36 – 12p$$ Â Â A2 Â Â N2
[2 marks]
a(i).
correct equation which clearly leads to $$p = 3$$ Â Â A1
eg   $$36 – 12p = 0,{\text{ }}36 = 12p$$
$$p = 3$$ Â Â AG Â Â N0
[1 mark]
a(ii).
METHOD 1
valid approach   (M1)
eg   $$x = – \frac{b}{{2a}}$$
correct working   A1
eg   $$– \frac{{( – 6)}}{{2(3)}},{\text{ }}x = \frac{6}{6}$$
correct answers   A1A1   N2
eg   $$x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)$$
METHOD 2
valid approach   (M1)
eg   $$f(x) = 0$$, factorisation, completing the square
correct working   A1
eg   $${x^2} – 2x + 1 = 0,{\text{ }}(3x – 3)(x – 1),{\text{ }}f(x) = 3{(x – 1)^2}$$
correct answers   A1A1   N2
eg   $$x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)$$
METHOD 3
valid approach using derivative   (M1)
eg   $$f'(x) = 0,{\text{ }}6x – 6$$
correct equation   A1
eg   $$6x – 6 = 0$$
correct answers   A1A1   N2
eg   $$x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)$$
[4 marks]
b.
$$x = 1$$ Â Â A1 Â Â N1
[1 mark]
c.
$$a = 3$$ Â Â A1 Â Â N1
[1 mark]
d(i).
$$h = 1$$ Â Â A1 Â Â N1
[1 mark]
d(ii).
$$k = 0$$ Â Â A1 Â Â N1
[1 mark]
d(iii).
attempt to apply vertical reflection   (M1)
eg    $$– f(x),{\text{ }} – 3{(x – 1)^2}$$, sketch
attempt to apply vertical shift 6 units up   (M1)
eg    $$– f(x) + 6$$, vertex $$(1, 6)$$
transformations performed correctly (in correct order) Â Â (A1)
eg    $$– 3{(x – 1)^2} + 6,{\text{ }} – 3{x^2} + 6x – 3 + 6$$
$$g(x) =Â – 3{x^2} + 6x + 3$$ Â Â A1 Â Â N3
[4 marks]
e.
## Question
The following diagram shows the graph of a function $$f$$.
Find $${f^{ – 1}}( – 1)$$.
[2]
a.
Find $$(f \circ f)( – 1)$$.
[3]
b.
On the same diagram, sketch the graph of $$y = f( – x)$$.
[2]
c.
## Markscheme
valid approach   (M1)
eg$$\;\;\;$$horizontal line on graph at $$– 1,{\text{ }}f(a) =Â – 1,{\text{ }}( – 1,5)$$
$${f^{ – 1}}( – 1) = 5$$ Â Â A1 Â Â N2
[2 marks]
a.
attempt to find $$f( – 1)$$ Â Â (M1)
eg$$\;\;\;$$line on graph
$$f( – 1) = 2$$ Â Â (A1)
$$(f \circ f)( – 1) = 1$$ Â Â A1 Â Â N3
[3 marks]
b.
   A1A1   N2
Note: Â Â The shape must be an approximately correct shape (concave down and increasing). Only if the shape is approximately correct, award the following for points in circles:
A1 for the $$y$$-intercept,
A1 for any two of these points $$( – 5,{\text{ }} – 1),{\text{ }}( – 2,{\text{ }}1),{\text{ }}(1,{\text{ }}2)$$.
[2 marks]
Total [7 marks]
c.
## Question
The following diagram shows part of the graph of a quadratic function $$f$$.
The vertex is at $$(1,{\text{ }} – 9)$$, and the graph crosses the yaxis at the point $$(0,{\text{ }}c)$$.
The function can be written in the form $$f(x) = {(x – h)^2} + k$$.
Write down the value of $$h$$ and of $$k$$.
[2]
a.
Find the value of $$c$$.
[2]
b.
Let $$g(x) =Â – {(x – 3)^2} + 1$$. The graph of $$g$$ is obtained by a reflection of the graph of $$f$$ in the $$x$$-axis, followed by a translation of $$\left( {\begin{array}{*{20}{c}} p \\ q \end{array}} \right)$$.
Find the value of $$p$$ and of $$q$$.
[5]
c.
Find the x-coordinates of the points of intersection of the graphs of $$f$$ and $$g$$.
[7]
d.
## Markscheme
$$h = 1,{\text{ }}k =Â – 9\;\;\;\left( {{\text{accept }}{{(x – 1)}^2} – 9} \right)$$ Â Â A1A1 Â Â N2
[2 marks]
a.
METHOD 1
attempt to substitute $$x = 0$$ into their quadratic function   (M1)
eg$$\;\;\;f(0),{\text{ }}{(0 – 1)^2} – 9$$
$$c =Â – 8$$ Â Â A1 Â Â N2
METHOD 2
attempt to expand their quadratic function   (M1)
eg$$\;\;\;{x^2} – 2x + 1 – 9,{\text{ }}{x^2} – 2x – 8$$
$$c =Â – 8$$ Â Â A1 Â Â N2
[2 marks]
b.
evidence of correct reflection   A1
eg$$\;\;\; – \left( {{{(x – 1)}^2} – 9} \right)$$, vertex at $$(1,{\text{ }}9)$$, y-intercept at $$(0,{\text{ }}8)$$
valid attempt to find horizontal shift   (M1)
eg$$\;\;\;1 + p = 3,{\text{ }}1 \to 3$$
$$p = 2$$ Â Â A1 Â Â N2
valid attempt to find vertical shift   (M1)
eg$$\;\;\;9 + q = 1,{\text{ }}9 \to 1,{\text{ }} – 9 + q = 1$$
$$q =Â – 8$$ Â Â A1 Â Â N2
Notes: Â Â An error in finding the reflection may still allow the correct values of $$p$$ and $$q$$ to be found, as the error may not affect subsequent working. In this case, award A0 for the reflection, M1A1 for $$p = 2$$, and M1A1 for $$q =Â – 8$$.
If no working shown, award N0 for $$q = 10$$.
[5 marks]
c.
valid approach (check FT from (a)) Â Â M1
eg$$\;\;\;f(x) = g(x),{\text{ }}{(x – 1)^2} – 9 =Â – {(x – 3)^2} + 1$$
correct expansion of both binomials   (A1)
eg$$\;\;\;{x^2} – 2x + 1,{\text{ }}{x^2} – 6x + 9$$
correct working   (A1)
eg$$\;\;\;{x^2} – 2x – 8 =Â – {x^2} + 6x – 8$$
correct equation   (A1)
eg$$\;\;\;2{x^2} – 8x = 0,{\text{ }}2{x^2} = 8x$$
correct working   (A1)
eg$$\;\;\;2x(x – 4) = 0$$
$$x = 0,{\text{ }}x = 4$$ Â Â A1A1 Â Â N3
[7 marks]
Total [16 marks]
d.
## Question
Let $$f'(x) = \frac{{6 – 2x}}{{6x – {x^2}}}$$, for $$0 < x < 6$$.
The graph of $$f$$ has a maximum point at P.
The $$y$$-coordinate of P is $$\ln 27$$.
Find the $$x$$-coordinate of P.
[3]
a.
Find $$f(x)$$, expressing your answer as a single logarithm.
[8]
b.
The graph of $$f$$ is transformed by a vertical stretch with scale factor $$\frac{1}{{\ln 3}}$$. The image of P under this transformation has coordinates $$(a,{\text{ }}b)$$.
Find the value of $$a$$ and of $$b$$, where $$a,{\text{ }}b \in \mathbb{N}$$.
[[N/A]]
c.
## Markscheme
recognizing $$f'(x) = 0$$Â Â Â (M1)
correct working   (A1)
eg$$\,\,\,\,\,$$$$6 – 2x = 0$$
$$x = 3$$ Â Â A1 Â Â N2
[3 marks]
a.
evidence of integration   (M1)
eg$$\,\,\,\,\,$$$$\int {f’,{\text{ }}\int {\frac{{6 – 2x}}{{6x – {x^2}}}{\text{d}}x} }$$
using substitution   (A1)
eg$$\,\,\,\,\,$$$$\int {\frac{1}{u}{\text{d}}u}$$ where $$u = 6x – {x^2}$$
correct integral   A1
eg$$\,\,\,\,\,$$$$\ln (u) + c,{\text{ }}\ln (6x – {x^2})$$
substituting $$(3,{\text{ }}\ln 27)$$ into their integrated expression (must have $$c$$) Â Â (M1)
eg$$\,\,\,\,\,$$$$\ln (6 \times 3 – {3^2}) + c = \ln 27,{\text{ }}\ln (18 – 9) + \ln k = \ln 27$$
correct working   (A1)
eg$$\,\,\,\,\,$$$$c = \ln 27 – \ln 9$$
EITHER
$$c = \ln 3$$ Â Â (A1)
attempt to substitute their value of $$c$$ into $$f(x)$$Â Â Â (M1)
eg$$\,\,\,\,\,$$$$f(x) = \ln (6x – {x^2}) + \ln 3$$Â Â Â A1 Â Â N4
OR
attempt to substitute their value of $$c$$ into $$f(x)$$Â Â Â (M1)
eg$$\,\,\,\,\,$$$$f(x) = \ln (6x – {x^2}) + \ln 27 – \ln 9$$
correct use of a log law   (A1)
eg$$\,\,\,\,\,$$$$f(x) = \ln (6x – {x^2}) + \ln \left( {\frac{{27}}{9}} \right),{\text{ }}f(x) = \ln \left( {27(6x – {x^2})} \right) – \ln 9$$
$$f(x) = \ln \left( {3(6x – {x^2})} \right)$$ Â Â A1 Â Â N4
[8 marks]
b.
$$a = 3$$ Â Â A1 Â Â N1
correct working   A1
eg$$\,\,\,\,\,$$$$\frac{{\ln 27}}{{\ln 3}}$$
correct use of log law   (A1)
eg$$\,\,\,\,\,$$$$\frac{{3\ln 3}}{{\ln 3}},{\text{ }}{\log _3}27$$
$$b = 3$$ Â Â A1 Â Â N2
[4 marks]
c.
## Question
The following diagram shows the graph of a function $$f$$, for −4 ≤ x ≤ 2.
On the same axes, sketch the graph of $$f\left( { – x} \right)$$.
[2]
a.
Another function, $$g$$, can be written in the form $$g\left( x \right) = a \times f\left( {x + b} \right)$$. The following diagram shows the graph of $$g$$.
Write down the value of a and of b.
[4]
b.
## Markscheme
A2 N2
[2 marks]
a.
recognizing horizontal shift/translation of 1 unit   (M1)
eg  b = 1, moved 1 right
recognizing vertical stretch/dilation with scale factor 2Â Â Â (M1)
eg  a = 2,  y ×(−2)
a = −2, b = −1   A1A1 N2N2
[4 marks]
b.
## Question
Let $$f(x) = 3{(x + 1)^2} – 12$$ .
Show that $$f(x) = 3{x^2} + 6x – 9$$ .
[2]
a.
For the graph of f
(i)Â Â Â Â write down the coordinates of the vertex;
(ii)Â Â Â write down the y-intercept;
(iii)Â Â find both x-intercepts.
[7]
b(i), (ii) and (iii).
Hence sketch the graph of f .
[3]
c.
Let $$g(x) = {x^2}$$ . The graph of f may be obtained from the graph of g by the following two transformations
a stretch of scale factor t in the y-direction,
followed by a translation of $$\left( \begin{array}{l} p\\ q \end{array} \right)$$ .
Write down $$\left( \begin{array}{l} p\\ q \end{array} \right)$$ and the value of t .
[3]
d.
## Markscheme
$$f(x) = 3({x^2} + 2x + 1) – 12$$Â Â Â Â Â A1
$$= 3{x^2} + 6x + 3 – 12$$Â Â Â Â Â A1
$$= 3{x^2} + 6x – 9$$Â Â Â Â Â AGÂ Â Â Â N0
[2 marks]
a.
(i) vertex is $$( – 1, – 12)$$Â Â Â Â Â A1A1Â Â Â Â N2
(ii) $$y = – 9$$ , or $$(0, – 9)$$Â Â Â Â A1 Â Â N1
(iii) evidence of solving $$f(x) = 0$$Â Â Â Â Â M1
e.g. factorizing, formula
correct working    A1
e.g. $$3(x + 3)(x – 1) = 0$$ , $$x = \frac{{ – 6 \pm \sqrt {36 + 108} }}{6}$$
$$x = – 3$$ , $$x = 1$$ , or $$( – 3{\text{, }}0){\text{, }}(1{\text{, }}0)$$Â Â Â Â Â A1A1 Â Â N2
[7 marks]
b(i), (ii) and (iii).
    A1A1A1   N3
Note: Award A1 for a parabola opening upward, A1 for vertex in approximately correct position, A1 for intercepts in approximately correct positions. Scale and labelling not required.
[3 marks]
c.
$$\left( \begin{array}{l} p\\ q \end{array} \right) = \left( \begin{array}{l} – 1\\ – 12 \end{array} \right)$$ , $$t = 3$$Â Â Â Â A1A1A1Â Â Â Â N3
[3 marks]
d. |
## Linear Regression (3 of 4)
### Learning Objectives
• For a linear relationship, use the least squares regression line to model the pattern in the data and to make predictions.
Let’s quickly revisit the list of our data analysis tools for working with linear relationships:
• Use a scatterplot and r to describe direction and strength of the linear relationship.
• Find the equation of the least-squares regression line to summarize the relationship.
• Use the equation and the graph of the least-squares line to make predictions.
• Avoid extrapolation when making predictions.
Now we focus on the equation of a line in more detail. Our goal is to understand what the numbers in the equation tell us about the relationship between the explanatory variable and the response variable.
Here are some of the equations of lines that we have used in our discussion of linear relationships:
Predicted distance = 576 − 3 * Age
Predicted height = 39 + 2.7 * forearm length
Predicted monthly car insurance premium = 97 − 1.45 * years of driving experience
Notice that the form of the equations is the same. In general, each equation has the form
Predicted y = a + b * x
When we find the least-squares regression line, a and b are determined by the data. The values of a and b do not change, so we refer to them as constants.
In the equation of the line, the constant a is the prediction when x = 0. It is called initial value. In a graph of the line, a is the y-intercept.
In the equation of the line, the constant b is the rate of change, called the slope. In a graph of the least-squares line, b describes how the predictions change when x increases by one unit. More specifically, b describes the average change in the response variable when the explanatory variable increases by one unit.
We can write the equation of the line to reflect the meaning of a and b:
Predicted y = a + b * x
Predicted y-value = (initial value) + (rate of change)*x
Predicted y-value = (y-intercept) + (slope)*x
The constants a and b are shown in the graph of the line below.
### The algebra of a line
The general form for the equation of a line is Y = a + bX. The constants “a” and “b” can be either positive or negative. The constant “a” is the y-intercept where the line crosses the y-axis. The constant “b” is the slope. It describes the steepness of the line. In algebra we describe the slope as “rise over run”. The slope is the amount that Y increases (or decreases) for each 1-unit increase in X.
## 1
Consider the line $Y=1+\frac{1}{3}X$. The intercept is 1. The slope is 1/3, and the graph of this line is, therefore:
## 2
Consider the line $Y=1-\frac{1}{3}X$. The intercept is 1. The slope is -1/3, and the graph of this line is, therefore:
The simulation below allows you to see how changing the values of the slope and y-intercept changes the line. The slider on the left controls the y-intercept, a. The slider on the right controls the slope, b.
Use the simulation to draw the following lines:
Y = 3 + 0.67X
Y = 5 – X (which can also be written Y = 5 – 1.0X)
Y = 2X (which can also be written Y = 0 + 2X)
Y = 5 – 2X
Use the following graphs in the next activity to investigate the equation of lines.
### Interpreting the Slope and Intercept
The constants in the equation of a line give us important information about the relationship between the predictions and x. In the next examples, we focus on how to interpret the meaning of the constants in the context of data.
## Highway Sign Visibility Data
Recall that from a data set of 30 drivers, we see a strong negative linear relationship between the age of a driver (x) and the maximum distance (in feet) at which a driver can read a highway sign. The least-squares regression line is
Predicted y-value = (starting value) + (rate of change)*x
Predicted distance = 576 − 3 * Age
Predicted distance = 576 + (−3 * Age)
The value of b is −3. This means that a 1-year increase in age corresponds to a predicted 3-foot decrease in maximum distance at which a driver can read a sign. Another way to say this is that there is an average decrease of 3 feet in predicted sign visibility distance when we compare drivers of age x to drivers of age x + 1.
The 576 is the predicted value when x = 0. Obviously, it does not make sense to predict a maximum sign visibility distance for a driver who is 0 years old. This is an example of extrapolating outside the range of the data. But the starting value is an important part of the least-squares equation for predicting distances based on age.
The equation tells us that to predict the maximum visibility distance for a driver, start with a distance of 576 feet and subtract 3 feet for every year of the driver’s age.
## Body Measurements
In the body measurement data collected from 21 female community college students, we found a strong positive correlation between forearm length and height. The least-squares regression line is
Predicted height = 39 + 2.7 * forearm length
The value of b is 2.7. This means that a 1-inch increase in forearm length corresponds to a predicted 2.7-inch increase in height. Another way to say this is that there is an average increase of 2.7-inches in predicted height when we compare women with forearm length of x to women with forearm length of x + 1.
The 39 is the predicted value when x = 0. Obviously, it does not make sense to predict the height of a woman with a 0-inch forearm length. This is another example of extrapolating outside the range of the data. But 39 inches is the starting value in the least-squares equation for predicting height based on forearm length.
The equation tells us that to predict the height of a woman, start with 39 inches and add 2.7 inches for every inch of forearm length.
In the graph below, we see the slope b represented by a triangle. An 8-inch increase in foreman length corresponds to a 21.6-inch increase in predicted height. b = 21.6 / 8 = 2.7. An arrow points to the starting value a = 39. This is the point with x = 0. |
# Frank Solutions for Chapter 1 Irrational Numbers Class 9 Mathematics ICSE
### Exercise 1.1
1. State which of these fractions have a terminating decimal.
(a) (3/ 5)
(b) (5 / 7)
(c) (25 / 49)
(d) (37 / 40)
(e) (57 / 64)
(f) (59 / 75)
(g) (89 / 125)
(h) (125 / 213)
(i) (147 / 160)
(a) (3/5)
5 = 1 × 5
⇒ 5 = 20 × 51
i.e, 5 can be expressed as 2m × 5n
Therefore,
(3/5) has terminating decimal representation
(b) (5/7)
7 = 1 × 7
i.e, 7 cannot be expressed a 2m × 5n
Therefore,
(5/7) does not have terminating decimal representation
(c) (25/49)
49 = 7 × 7
i.e, 49 cannot be expressed as 2m × 5n
Therefore,
(25/49) does not have terminating decimal representation
(d) (37/40)
40 = 2 × 2 × 2 × 5
⇒ 40 = 23 × 51
i.e, 40 can be expressed as 2m × 5n
Therefore,
(37/40) has terminating decimal representation
(e) (57 / 64)
64 = 2 × 2 × 2 × 2 × 2 × 2
⇒ 64 = 26 × 50
i.e, 64 can be expressed as 2m × 5n
Therefore,
(57/64) has terminating decimal representation
(f) (59/75)
75 = 5 × 5 × 3
⇒ 75 = 52 × 31
i.e, 75 cannot be expressed as 2m × 5n
Therefore,
(59/75) does not have terminating decimal representation
(g) (89/125)
125 = 5 × 5 × 5
⇒ 125 = 20 × 53
i.e, 125 can be expressed as 2m × 5n
Therefore,
(89/125) has terminating decimal representation
(h) (125/213)
213 = 3 × 71
i.e, 213 cannot be expressed as 2m × 5n
Therefore,
(125/213) does not have terminating decimal representation
(i) (147 / 160)
160 = 2 × 2 × 2 × 2 × 2 × 5
⇒ 160 = 2× 51
i.e, 160 can be expressed as 2m × 5n
Therefore,
(147/160) has terminating decimal representation
2. Express each of the following decimals as a rational number
(a) 0.93
(b) 4.56
(c) 0.614
(d) 21.025
(a) 0.93 = 93 / 100
Hence,
The rational number of decimal 0.93 is (93 / 100)
(b) 4.56 = (456 / 100)
= (456 ÷ 4) / (100 ÷ 4)
We get,
= (114 / 25)
Hence,
The rational number of decimal 4.56 is (114 / 25)
(c) 0.614 = (614/1000)
= (614 ÷ 2)/(1000 ÷ 2)
We get,
= (307/500)
Hence,
The rational number of decimal 0.614 is (307/500)
(d) 21.025 = (21025/1000)
= (21025 ÷ 25)/(1000 ÷ 25)
We get,
= (841/40)
Hence,
The rational number of decimal 21.025 is (841/40)
3. Convert the following fractions into decimals:
(i) (3/5)
(ii) (8/11)
(iii) (-2/7)
(iv) (12/21)
(v) (13/25)
(vi) (2/3)
(i) (3/5)
(3/5) = 0.6
Hence,
The decimal form of (3/5) is 0.6
(ii) (8/11)
(8/11) = 0.72727272…..
(iii) (-2/7)
(-2/7) = -0.285714285714….
(iv) (12/21)
(12/21) = 0.571428571428……
(v) (13/25)
(13/25) = 0.52
Hence,
The decimal form of (13/25) is 0.52
(vi) (2/3)
(2/3) = 0.6666…..
⇒ (2/3) = 0.6
Hence,
The decimal form of (2/3) is 0.6
4. Express each of the following decimals as a rational number.
(a) 0.7
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(a) 0.7
Let x = 0.7
Then,
x = 0.7777 ...(1)
Here, the number of digits recurring is only 1,
So, we multiply both sides of the equation (1) by 10
We get,
10x = 10 × 0.7777 …(2)
⇒ 10x = 7.777…..
On subtracting (1) from (2),
We get,
9x = 7
⇒ x = (7/9)
⇒ 0.7 = (7/9)
Therefore,
0.7 = (7/9)
(b)
(c)
Here,
The number of digits recurring is 2,
So we multiply both sides of the equation (1) by 100
We get,
100x = 100 × 0.898989……
= 89.8989 …(2)
On subtracting (1) from (2),
We get,
99x = 89
(d)
(e)
(f)
(g)
Here, Only numbers 724 is being repeated.
So first, we need to remove 6 which proceeds 724
We multiply by 10 so that only the recurring digits remain after decimal
Thus,
10x = 10 × 4.6724724…….
⇒ 10x = 46. 724724 …(1)
The number of digits recurring in equation (1) is 3
Hence, we multiply both sides of the equation (1) by 1000
10000x = 1000 × 46.724724 = 46724.724 ...(2)
On subtracting (1) from (2), we get,
9990x = 46678
⇒ x = 46678 / 9990
We get,
x = 23339 / 4995
(h)
Here, only numbers 17 is being repeated, so first, we need to remove 0 which proceeds 17
We multiply by 10 so that only the recurring digits remain after the decimal,
Hence,
10x = 0.1717…(1)
The number of digits recurring in equation (1) is 2,
So we multiply both sides of the equation (1) by 100,
Hence,
1000x = 100 × 0.1717……..
= 17.1717….(2)
On subtracting (1) from (2),
We get,
990x = 17
(i)
Here, only number 7 is being repeated, so first, we need to remove 02 which proceeds 7
We multiply by 100 so that only the recurring digits remain after decimal
Hence,
100x = 1702.7777…(1)
The number of digits recurring in equation (1) is 1,
So we multiply both sides of the equation (1) by 10
Hence,
1000x = 10 × 1702.7777……
= 17027.777…(2)
On subtracting (1) from (2),
We get,
900x = 15325
x = (15325 / 900)
5. Insert a rational number between:
(a) 2/5 and 3/4
(b) 3/4 and 5/7
(c) 4/3 and 7/5
(d) 5/9 and 6/7
(a) (2/5) and (3/4)
= [{(2/5) + (3/4)}/2]
On further calculation, we get,
= [{(8+15)/20}/2]
= {(23/20)/2}
We get,
= (23/40)
Therefore,
A rational number lying between (2/5) and (3/4) is (23/40)
(b) (3/4) and (5/7)
= [{(3/4) + (5/7)}/2]
On further calculation, we get,
= [{(21+20)/28}/2]
= {(41/28)/2}
We get,
= (41/56)
Therefore,
A rational number lying between (3/4) and (5/7) is (41/56)
(c) (4/3) and (7/5)
= [{(4/3) + (7/5)}/2]
On further calculation, we get,
= [{(20 + 21)/15}/2]
= {(41/15)/2}
We get,
= (41/30)
Therefore,
A rational number lying between (4/3) and (7/5) is (41/30)
(d) (5/9) and (6/7)
= [{(5/9) + (6/7)}/2]
On further calculation, we get,
= [{(35 + 54)/63}/2]
= {(89/63)/2}
We get,
= (89/126)
Therefore,
A rational number lying between (5/9) and (6/7) is (89/126)
6. State, whether the following numbers are rational or irrational:
(a) (3 + √3)2
(b) (5 – √5)2
(c) (2 + √2) (2 – √2)
(d) {(√5) / (3√2)}2
(a) (3 + √3)2
= (3)2 + (√3)2 + 2×3×√3
On calculating further, we get,
= 9 + 3 + 6√3
= 12 + 6√3
which is a rational number
Therefore,
(3 + √3)2 is a rational number
(b) (5 – √5)2
= (5)2 + (√5)2 – 2×5×√5
On further calculation, we get,
= 25 + 5 – 10√5
= 30 – 10√5
which is a irrational number
Therefore,
(5 – √5)2 is an irrational number
(c) (2 + √2) (2 – √2)
= (2)2 – (√2)2
= 4 – 2
= 2
which is a rational number
Therefore,
(2 + √2) (2 – √2) is a rational number
(d) {(√5)/(3√2)}2
= {(5)/(9×2)}
We get,
= 5/18
which is a rational number
Therefore,
{√5/(3√2)}2 is a rational number
7. Check whether the square of the following is rational or irrational:
(a) 3√2
(b) 3 + √2
(c) (3√2) / 2
(d) √2 + √3
(a) 3√2
(3√2)2
= 9 × 2
= 18
which is a rational number
Hence,
The square of (3√ 2) is a rational number
(b) 3 + √2
(3 + √2)2
= (3)2 + (√2)2 + 2×3×√2
On further calculation, we get,
= 9 + 2 + 6√2
= 11 + 6√2
which is irrational number
Hence,
The square of (3 + √2) is an irrational number
(c) (3√2)/2
{(3√2)/2}2
= (9×2)/4
= (9/2)
which is a rational number
Hence,
The square of {(3√2)/2} is a rational number
(d) √2 + √3
(√2+√3)2
= (√2)2 + (√3)2 + 2×√2×√3
= 2 + 3 + 2√6
We get,
= 5 + 2√6
which is irrational number
Hence,
The square of (√2 + √3) is an irrational number
8. Show that √5 is an irrational number. (Use division method)
Here,
Clearly, √5 = 2.23606…… is an irrational number
Therefore,
√5 is an irrational number
9. Without using division method show that √7 is an irrational number
Let √7 be a rational number
Hence,
√7 = (a / b)
On squaring both sides, we get,
7 = (a2 / b2)
a2 = 7b2
Since, a2 is divisible by 7, a is also divisible by 7…(1)
Let a = 7c
On squaring both sides, we get,
a2 = 49c2
Substituting a2 = 7b2
We get,
7b2 = 49c2
b2 = 7c2
Since, b2 is divisible by 7, b is also divisible by 7…(2)
From (1) and (2) we can observe that both a and b are divisible by 7
i.e, a and b have a common factor 7
This contradicts our assumption that (a/b) is rational number
i.e, a and b do not have any common factor other than unity (1)
Hence,
(a/b) is not rational number
√7 is not rational number
Therefore,
√7 is an irrational number
10. Write a pair of irrational numbers
(a) (√3 + 5) and (√5 – 3) whose sum is irrational
(b) (√3 + 5) and (4 – √3) whose sum is rational
(c) (√3 + 2) and (√2 – 3) whose difference is irrational
(d) (√5 – 3) and (√5 + 3) whose difference is rational
(e) (5 + √2) and (√5 – 2) whose product is irrational
(f) (√3 + √2) and (√3 – √2) whose product is rational
(a) Given
(√3 + 5) and (√5 – 3) are irrational numbers whose sum is irrational
Thus,
We have,
(√3 + 5) + (√5 – 3)
= √3 + 5 + √5 – 3
We get,
= √3 + √5 + 2
which is irrational number
(b) Given,
(√3 + 5) and (4 – √3) are irrational numbers whose sum is rational
Thus,
We have,
(√3 + 5) + (4 – √3)
= √3 + 5 + 4 – √3
We get,
= 9
which is a rational number
(c) Given
(√3 + 2) and (√2 – 3) are irrational numbers whose difference is irrational
Thus,
We have,
(√3 + 2) – (√2 – 3)
= √3 + 2 – √2 + 3
We get,
= √3 – √2 + 5
which is irrational
(d) Given
(√5 – 3) and (√5 + 3) are irrational numbers whose difference is rational
Thus,
We have,
(√5 – 3) – (√5 + 3)
= √5 – 3 – √5 – 3
We get,
= -6
which is a rational number
(e) Given
(5 + √2) and (√5 – 2) are irrational numbers whose product is irrational
Thus,
We have,
(5 + √2) (√5 – 2)
= 5 (√5 – 2) + √2 (√5 – 2)
We get,
= 5√5 – 10 + √10 – 2√2
-which is irrational numbers
(f) Given
(√3 + √2) and (√3 – √2) are irrational numbers whose product is rational
Thus,
We have,
(√3 + √2) (√3 – √2)
= (√3)2 – (√2)2
= 3 – 2
We get,
= 1
which is a rational number
11. Simplify by rationalizing the denominator in each of the following:
(a) (3√2) / √5
(b) {1 / (5 + √2)}
(c) {1 / (√3 + √2)}
(d) {2 / (3 + √7)}
(e) {5 / (√7 – √2)}
(f) {42 / (2√3 + 3√2)}
(g) {(√3 + 1) / (√3 – 1)}
(h) (√5 – √7) / √3
(i) (3 – √3) / (2 + √2)
(a) (3√2)/√5
= {(3√2)/√5} × (√5)/(√5)
On simplification, we get,
= {(3√2) × √5}/(√5)2
We get,
= (3√10)/5
(b) {1/(5 + √2)}
= {1/(5 + √2)} × {(5 – √2)/(5 – √2)}
On simplification, we get,
= (5 – √2) / (5)2 – (√2)2
= (5 – √2) / (25 – 2)
We get,
= (5 – √2)/23
(c) {1/(√3 + √2)}
= {1/(√3 + √2)} × (√3 – √2)/(√3 – √2)
On simplification, we get,
= (√3 – √ 2)/(√3)2 – (√2)2
= (√3 – √2)/(3 – 2)
= (√3 – √2)/1
We get,
= (√3 – √2)
(d) {2/(3 + √7)}
= {2/(3 + √7)} × (3 – √7)/(3 – √7)
On further calculation, we get,
= {2 (3 – √7)}/(3)2 – (√7)2
= {2(3 – √7)}/(9 – 7)
= {2(3 – √7)}/2
We get,
= (3 – √7)
(e) {5/(√7 – √2)}
= {5/(√7 – √2)} × (√7 + √2)/(√7 + √2)
On further calculation, we get,
= {5(√7 + √2)}/(√7)2 – (√2)2
= {5(√7 + √2)}/(7 – 2)
= {5(√7 + √2)}/5
We get,
= (√7 + √2)
(f) {42/(2√3 + 3√2)}
= {42/(2√3 + 3√2)} × {(2√3 – 3√2)/(2√3 – 3√2)}
= {42 (2√3 – 3√2)}/(2√3)2 – (3√2)2
On further calculation, we get,
= (84√3 – 126√2)/(4 × 3) – (9 × 2)
= (84√3 – 126√2)/(12 – 18)
= (84√3 – 126√2)/-6
= -14√3 + 21√2
= 21√2 – 14√3
We get,
= 7(3√2 – 2√3)
(g) {(√3 + 1)/(√3 – 1)}
= (√3 + 1)/(√3 – 1) × (√3 + 1)/(√3 + 1)
On further calculation, we get,
= (√3 + 1)2/(√3)2 – (1)2
= {(√3)2 + 2×√3×1 + (1)2}/(3 – 1)
= (3 + 2√3 + 1)/2
= (4 + 2√3)/2
We get,
= 2 + √3
(h) (√5 – √7)/√3
= {(√5 – √7)/√3} × (√3)/(√3)
= (√5×√ 3 – √7×√ 3)/(√3)2
We get,
= (√15 – √21)/3
(i) (3 – √3)/(2 + √2)
= (3 – √3)/(2 + √2) × (2 – √2)/(2 – √2)
= {3(2 – √2) – √3 (2 – √2)}/(2)2 – (√2)2
On further calculation, we get,
= (6 – 3√2 – 2√3 + √6)/(4 – 2)
= (6 – 3√2 – 2√3 + √6)/2
12. Simplify by rationalizing the denominator in each of the following:
(i) (5 + √6)/(5 – √6)
(ii) (4 + √8)/(4 – √8)
(iii) (√15 + 3)/(√15 – 3)
(iv) (√7 – √5)/(√7 + √5)
(v) (3√5 + √7)/(3√5 – √7)
(vi) (2√3 – √6)/(2√3 + √6)
(vii) (5√3 – √15)/(5√3 + √15)
(viii) (2√6 – √5)/(3√5 – 2√6)
(ix) (7√3 – 5√2)/(√48 + √18)
(x) (√12 + √18)/(√75 – √50)
(i) (5 + √6)/(5 – √6)
= (5 + √6)/(5 – √6) × (5 + √6)/(5 + √6)
On further calculation, we get,
= (5 + √6)2/(5)2 – (√6)2
= {(5)2 + 2×5×√6 + (√6)2}/(25 – 6)
= (25 + 10√6 + 6)/19
We get,
= (31 + 10√6)/19
(ii) (4 + √8)/(4 – √8)
= (4 + √8)/(4 – √8) × (4 + √8)/(4 + √8)
On further calculation, we get,
= (4 + √8)2/(4)2 – (√8)2
= {(4)2 + 2×4×√8 + (√8)2}/(16 – 8)
= (16 + 8√8 + 8)/(16 – 8)
= (24 + 8√8)/8
We get,
= 3 + √8
(iii) (√15 + 3)/(√15 – 3)
= (√15 + 3)/(√15 – 3) × (√15 + 3)/(√15 + 3)
= (√15 + 3)2/(√15)2 – (3)2
= {(√15)2 + 2×√15×3 + (3)2}/(15 – 9)
On further calculation, we get,
= (15 + 6√15 + 9)/6
= (24 + 6√15)/6
We get,
= 4 + √15
(iv) (√7 – √5)/(√7 + √5)
= (√7 – √5)/(√7 + √5) × (√7 – √5)/(√7 – √5)
= (√7 – √5)2/(√7)2 – (√5)2
On further calculation, we get,
= (7 + 5 – 2√35)/(7 – 5)
= (12 – 2√35)/2
We get,
= 6 – √35
(v) (3√5 + √7)/(3√5 – √7)
= (3√5 + √7)/3√5 – √7) × (3√5 + √7)/(3√5 + √7)
= (3√5 + √7)2/(3√5)2 – (√7)2
On further calculation, we get,
= {(3√5)2 + (√7)2 + 2 × 3√5 × √7}/(45 – 7)
= (45 + 7 + 6√35)/38
= (52 + 6√35)/38
We get,
= (26 + 3√35)/19
(vi) (2√3 – √6)/(2√3 + √6)
= (2√3 – √6)/(2√3 + √6) × (2√3 – √6)/(2√3 – √6)
= (2√3 – √6)2/(2√3)2 – (√6)2
On further calculation, we get,
= {(2√3)2 + (√6)2 – 2×2√3 ×√6}/(4×3 – 6)
= (12 + 6 – 4√18)/(12 – 6)
= (18 – 4√18)/6
= (9 – 2√18)/3
= (9 – 6√2)/3
We get,
= 3 – 2√2
(vii) (5√3 – √15)/(5√3 + √15)
= (5√3 – √15)/5√3 + √15) × (5√3 – √15)/(5√3 – √15)
= (5√3 – √15)2/(5√3)2 – (√15)2
On further calculation, we get,
= (75 + 15 – 10√45)/(75 – 15)
= (90 – 10√45)/60
= (9 – 1√45)/6
= (9 – 3√5)/6
We get,
= (3 – √5)/2
(viii) (2√6 – √5)/(3√5 – 2√6)
= (2√6 – √5)/(3√5 – 2√6) × (3√5 + 2√6)/(3√5 + 2√6)
On simplification, we get,
= (6√30 + 24 – 15 – 2√30)/(3√5)2 – (2√6)2
= (6√30 + 9 – 2√30)/(45 – 24)
We get,
= (4√30 + 9)/21
(ix) (7√3 – 5√2)/(√48 + √18)
= (7√3 – 5√2)/(√48 + √18) × (√48 – √18)/(√48 – √18)
On simplification, we get,
= (7√144 – 7√54 – 5√96 + 5√36)/(√48)2 – (√18)2
= (84 – 21√6 – 20√6 + 30)/(48 – 18)
We get,
= (114 – 41√6)/30
(x) (√12 + √18)/(√75 – √50)
= (√12 + √18)/(√75 – √50) × (√75 + √50)/(√75 + √50)
On further calculation, we get,
= {(2√3 + 3√2) (5√3 + 5√2)}/ (√75)2 – (√50)2
= (30 + 10√6 + 15√6 + 30)/(75 – 50)
= (60 + 25√6)/25
We get,
= (12 + 5√6)/5
13. Simplify each of the following:
(i) 3 / (5 – √3) + 2 / (5 + √3)
(ii) (4 + √5) / (4 – √5) + (4 – √5) / (4 + √5)
(iii) (√5 – 2)/(√5 + 2) – (√5 + 2)/(√5 – 2)
(iv) (√7 – √3)/(√7 + √3) – (√7 + √3)/(√7 – √3)
(v) (√5 + √3)/ √5 – √3) + (√5 – √3) / √5 + √3)
(i) 3/(5 – √3) + 2/(5 + √3)
= {3(5 + √3) + 2(5 – √3)}/(5 – √3) (5 + √3)
On simplification, we get,
= (15 + 3√3 + 10 – 2√3)/(5)2 – (√3)2
= (25 + √3)/(25 – 3)
We get,
= (25 + √3)/22
(ii) (4 + √5)/(4 – √5) + (4 – √5)/(4 + √5)
= {(4 + √5)2 + (4 – √5)2}/(4 – √5) (4 + √5)
On simplification, we get,
= (16 + 5 + 8√5 + 16 + 5 – 8√5)/(4)2 – (√5)2
= (21 + 21)/(16 – 5)
We get,
= (42/11)
(iii) (√5 – 2)/(√5 + 2) – (√5 + 2)/(√5 – 2)
= (√5 – 2)2 – (√5 + 2)2}/(√5 + 2) (√5 – 2)
On simplification, we get,
= (5 + 4 – 4√5 – 5 – 4 – 4√5)/(√5)2 – (2)2
= –8√5 /(5 – 4)
We get,
= –8√5
(iv) (√7 – √3)/(√7 + √3) – (√7 + √3)/(√7 – √3)
= (√7 – √3)2 – (√7 + √3)2/(√7 + √3) (√7 – √3)
On simplification, we get,
= (7 + 3 – 2√21 – 7 – 3 – 2√21)/(√7)2 – (√3)2
= – 4√21/(7 – 3)
= (- 4√21)/4
We get,
= –√21
(v) (√5 + √3)/ √5 – √3) + (√5 – √3)/√5 + √3)
= (√5 + √3)2 + (√5 – √3)2/(√5 – √3) (√5 + √3)
On simplification, we get,
= (5 + 3 + 2√15 + 5 + 3 – 2√15)/(5 – 3)
= 16/2
We get,
= 8
14Simplify the following:
(i) √6/(√2 + √3) + 3√2/(√6 + √3) – 4√3/(√6 + √2)
(ii) 3√2/(√6 – √3) – 4√3/(√6 – √2) + 2√3/(√6 + 2)
(iii) 6/(2√3 – √6) + √6/(√3 + √2) – 4√3/(√6 – √2)
(iv) 7√3/(√10 + √3) – 2√5/(√6 + √5) – 3√2/(√15 + 3√2)
(v) 4√3/(2 – √2) – 30/(4√3 – 3√2) – 3√2/(3 + 2√3)
(i) √6/(√2 + √3) + 3√2/(√6 + √3) – 4√3/(√6 + √2)
Rationalizing the denominator of each term, we have
= {√6 (√2 – √3)/(√2 + √3) (√2 – √3)} + {(3√2 (√6 – √3)/(√6 + √3) (√6 – √3)}- {(4√3 (√6 – √2)/(√6 + √2) (√6 – √2)}
On further calculation, we get,
= {(√12 – √18)/(2 – 3)} + {(3√12 – 3√6)/(6 – 3)} – {(4√18 – 4√6)/(6 – 2)}
= {(√12 – √18)/-1} + {(3√12 – 3√6)/3} – {(4√18 – 4√6)/4}
= √18 – √12 + √12 – √6 – √18 + √6
We get,
= 0
(ii) 3√2/(√6 – √3) – 4√3/(√6 – √2) + 2√3/(√6 + 2)
Rationalizing the denominator of each term, we have,
= {3√2(√6 + √3)/(√6 – √3) (√6 + √3)} – {(4√3 (√6 + √2)/(√6 – √2) (√6 + √2)} + {(2√3 (√6 – 2)/(√6 + 2) (√6 – 2)}
On further calculation, we get,
= {(3√12 + 3√6)/(6 – 3)} – {(4√18 + 4√6)/(6 – 2)} + {(2√18 – 4√3)/(6 – 4)}
= {(3√12 + 3√6)/3} – {(4√18 + 4√6)/4} + {(2√18 – 4√3)/2}
= √12 + √6 – √18 – √6 + √18 – 2√3
= √12 – 2√3
= 2√3 – 2√3
We get,
= 0
(iii) 6 / (2√3 – √6) + √6 / (√3 + √2) – 4√3 / (√6 – √2)
Rationalizing the denominator of each term, we have
= {6 (2√3 + √6)/(2√3 – √6) (2√3 + √6)} + {(√6 (√3 – √2)/(√3 + √2) (√3 – √2)} – {(4√3 (√6 + √2)/(√6 – √2) (√6 + √2)
On simplification, we get,
= (12√3 + 6√6) / (12 – 6) + (√18 – √12)/(3 – 2) – (4√18 + 4√6)/(6 – 2)
= {(12√3 + 6√6)/6} + {(√18 – √12)/1} – {(4√18 + 4√6)/4}
= 2√3 + √6 + √18 – √12 – √18 – √6
= 2√3 – √12
= 2 √3 – 2√3
We get,
= 0
(iv) 7√3/(√10 + √3) – 2√5/(√6 + √5) – 3√2/(√15 + 3√2)
Rationalizing the denominator of each term, we have,
= {7√3 (√10 – √3)/(√10 + √3) (√10 – √3)} – {2√5 (√6 – √5)/(√6 + √5) (√6 – √5)} – {3√2 (√15 – 3√2)/(√15 + 3√2) (√15 – 3√2)
= {(7√30 – 21)/(10 – 3)} – {(2√30 – 10)/(6 – 5)} – {(3√30 – 18)/(15 – 18)}
= (7√30 – 21)/7 – (2√30 – 10)/1 – (3√30 – 18)/-3
= (7√30 – 21)/7 – (2√30 – 10)/1 + (3√30 – 18)/3
= √30 – 3 – 2√30 + 10 + √30 – 6
We get,
= 1
(v) 4√3/(2 – √2) – 30/(4√3 – 3√2) – 3√2/(3 + 2√3)
Rationalizing the denominator of each term, we have,
= {(4√3 (2 + √2)/(2 – √2) (2 + √2)} – {30 (4√3 + 3√2)/(4√3 – 3√2) (4√3 + 3√2)} – {(3√2 (3 – 2√3)/(3 + 2√3) (3 – 2√3)
= {(8√3 + 4√6)/(4 – 2)} – {(120√3 + 90√2)/(48 – 18)} – {(9√2 – 6√6)/(9 – 12)}
= {(8√3 + 4√6)/2} – {120√3 + 90√2)/30} – {(9√2 – 6√6)/-3}
= {(8√3 + 4√6)/2} – {(120√3 + 90√2)/30} + {(9√2 – 6√6)/3}
= 4√3 + 2√6 – 4√3 – 3√2 + 3√2 – 2√6
We get,
= 0
15. If (√2.5 – √0.75)/(√2.5 + √0.75) = p + q√30, find the values of p and q.
Given
(√2.5 – √0.75)/(√2.5 + √0.75)
= {(√2.5 – √0.75)/(√2.5 + √0.75)} × {(√2.5 – √0.75)/(√2.5 – √0.75)}
= (√2.5 – √0.75)2/(√2.5)2 – (√0.75)2
On simplification, we get,
= (2.5 – 2×√2.5×√0.75 + 0.75)/(2.5 – 0.75)
= (3.25 – 2×√0.25×10×√0.25×3)/1.75
= (3.25 – 2 × 0.25√30)/1.75
= (3.25 – 0.5√30)/1.75
= (3.25)/(1.75) – (0.5)/(1.75) √30
= (325/175) – (50/175) √30
= (13/7) – (2/7)√30
= (13/7) + (- 2/7) √30
= p + q√30
Therefore,
The value of p = (13/7) and q = (-2/7)
### Exercise 1.2
1. State, whether the following numbers are rational or irrational.
(a) (3 + √3)2
(b) (5 - √5)2
(c) (2 + √2)(2 - √2)
(d) [√5/(3√2)]2
(a)
(b)
(c)
(d)
2. Check whether the square of the following is rational or irrational.
(a) 3√2
(b) 3 + √2
(c) (3√2)/2
(d) √2 + √3
(a)
(b)
(c)
(d)
3. Show that √5 is an irrational numbers. [Use division method]
4. Without using division method show that √7 is an irrational numbers.
5. (a) Write a pair of irrational numbers whose sum is irrational.
(b) Write a pair of irrational whose sum is rational.
(c) Write a pair of irrational numbers whose difference is irrational.
(d) Write a pair of irrational numbers whose difference is rational.
(e) Write a pair of irrational number whose product is irrational.
(f) Write a pair of irrational numbers whose product is rational.
(a) (√3 + 5) and (√5 – 3) are irrational numbers whose sum is irrational.
Thus, we have
(√3 + 5) + (√5 - 3)
= √3 + 5 + √5 – 3
= √3 + √5 + 2, which is irrational.
(b)
(c)
(d)
(e)
(f)
6. Compare the following:
(a) ∜12 and ∛15
(b) ∛48 and √36
(a)
(b)
7. (a) Write the following in ascending order:
2√5, √3 and 5√2
(b) Write the following in ascending order.
2∛3, 4∛3 and 3∛3
(c) Write the following in ascending order:
5√7, 7√5 and 6√2
(d) Write the following in ascending order:
7∛5, 6∛4, 5∛6
(a)
(b)
(c)
(d)
8. (a) Write the following in descending order:
√2, ∛5 and ∜10
(b) Write the following in descending order:
5√3, √15 and 3√5
(c) Write the following in descending order:
√6, ∛8 and ∜3
(a)
(b)
(c)
9. Insert two irrational numbers between 3 and 4.
10. Insert five irrational numbers between 2√3 and 3√5.
11. Write two numbers between √3 and √7.
12. Write four rational numbers between √2 and √3.
13. State which of the following are surds. Give reasons.
(a) 150
(b) ∛4
(c) ∛50.∛20
(d) ∛-27
(e) √(2+√3)
(f)
(a)
(b)
(c)
(d)
(e)
(f)
14. Represent the number √7 on the number line
.
### Exercise 1.3
1. Simplify by rationalizing the denominator in each of the following:
(a) (3√2)/√5
(b) 1/(5+√2)
(c) 1/(√3+√2)
(d) 2/(3+√7)
(e) 5/(√7-√2)
(f) 42/(2√3 + 3√2)
(g) (√3+1)/(√3–1)
(h) (√5-√7)/√3
(i) (3-√3)/(2+√2)
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
2. Simplify by rationalizing the denominator in each of the following:
(i) (5 + √6)/(5 - √6)
(ii) (4 + √8)/(4 - √8)
(iii) (√15+3)/(√15-3)
(iv) (√7 - √5)/(√7 + √5)
(v) (3√5 + √7)/(3√5 - √7)
(vi) (2√3 - √6)/(2√3 + √6)
(vii) (5√3 - √15)/5√3 + √15)
(viii) (2√6 - √5)/(3√5 - 2√6)
(ix) (7√3 - 5√2)/(√48 + √18)
(x) (√12+√18)/(√75-√50)
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
3. Simplify each of the following:
(i) 3/(5 - √3) + 2/(5 + √3)
(ii) (4 + √5)/(4 - √5) + (4 - √5)/(4 + √5)
(iii) (√5 – 2)/(√5 + 2) – (√5 + 2)/(√5 – 2)
(iv) [(√7 - √3)/(√7 + √3)] – [(√7 + √3)/(√7 - √3)]
(v) (√5 + √3)/(√5 - √3) + (√5 - √3)/(√5 + √3)
(i)
(ii)
(iii)
(iv)
(v)
4. Simplify the following:
(i) √6/(√2+√3) + 3√2/(√6+√3) - 4√3/(√6+√2)
(ii) 3√2/(√6-√3) - 4√3/(√6-√2) + 2√3/(√6 + 2)
(iii) 6/(2√3-√6) + √6/(√3+√2) - 4√3/(√6-√2)
(iv) 7√3/(√10+√3) - 2√5/(√6+√5) - 3√2/(√15 + 3√2)
(v) 4√3/(2-√2) – 30/(4√3 - 3√2) - 3√2/(3 + 2√3)
(i)
(ii)
(iii)
(iv)
(v)
5. If (√2.5 - √0.75)/(√2.5 + √0.75) = p+ q√30, find the values of p and q.
6. In each of the following, find the values of a and b.
(a) (√3 – 1)/(√3 + 1) = (a + b√3)
(b) (3 + √7)/(3 - √7) = a + b√7
(c) (5 + 2√3)/(7 + 4√3) = a + b√3
(d) 1/(√5 - √3) = (8√5 - b√3)
(e) (√3 – 2)/(√3 + 2) = 8√3 + b
(f) (√11 - √7)/(√11 + √7) = a - b√77
(g) (7√3 - 5√2)/(4√3 + 3√2) = a - b√6
(h) (√2 + √3)/(3√2 - 2√3) = (a - b√6)
(i) (7 + √5)/(7 - √5) – (7 - √5)/(7 + √5) = a + b√5
(j) (√3 – 1)/(√3 + 1) + (√3 + 1)(√3 – 1) = a + b√3
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
7. If x = (7 + 4√3), find the values of
(i) √x + 1/√x
(ii) x2 + 1/x2
(iii) x3 + 1/x3
(iv) If x = (7 + 4√3), find the values of: (x + 1/x)2
(i)
(ii)
(iii)
(iv)
8. If x = (4 - √15), find the values of
(i) 1/x,
(ii) x + 1/x
(iii) x2 + 1/x2
(iv) x3 + 1/x3
(v) If x = (4 - √15), find the values of: (x + 1/x)2
(i)
(ii)
(iii)
(iv)
(v)
9. If x = (2 + √5)/(2 - √5) and y = (2 - √5)/(2 + √5), show that (x2 – y2) = 144
5.
10. If x = (√3 + 1)/(√3 – 1) and y = (√3 – 1)/(√3 + 1), find the values of
(i) x2 + y2,
(ii) x3 + y3, and
(iii) x2 – y2 + xy
(i)
(ii)
(iii)
11. If x = 1/(3 - 2√2) and y = 1/(3 + 2√2), find the values of
(i) x2 + y2,
(ii) x3 + y3 |
# 4900 in Words
In English, the number 4900 is written as Four Thousand Nine Hundred. If you bought a jacket worth Rs. 4900, you can say, “I have bought a jacket worth Rupees Four Thousand Nine Hundred”. The number 4900 is a cardinal number. In this article, students are provided to learn how to read and write numbers in English in a comprehensive manner.
4900 in words Four Thousand Nine Hundred Four Thousand Nine Hundred in numerical form 4900
## 4900 in English Words
Generally, numbers in words are expressed using the English alphabet. Thus, we can read 4900 in English as Four Thousand Nine Hundred.
## How to Write 4900 in Words?
Let us have a look at the place value chart mentioned below for the conversion of the number 4900 to words.
Thousands Hundreds Tens Ones 4 9 0 0
Thus, the expanded form can be written as:
4 x Thousand + 9 x Hundred + 0 × Ten + 0 × One
= 4 x 1000 + 9 x 100 + 0 x 10 + 0 x 1
= 4000 + 900 + 0 + 0
= 4000 + 900
= 4900
= Four Thousand Nine Hundred
Thus, 4900 in words is written as Four Thousand Nine Hundred
Interesting way of writing 4900 in words
4 = Four
49 = Forty-Nine
490 = Four Hundred and Ninety
4900 = Four Thousand Nine Hundred
Hence, the word form of the number 4900 is Four Thousand Nine Hundred
4900 is a natural number that precedes 4901 and succeeds 4899
• 4900 in words – Four Thousand Nine Hundred
• Is 4900 an odd number? – No
• Is 4900 an even number? – Yes
• Is 4900 a perfect square number? – Yes
• Is 4900 a perfect cube number? – No
• Is 4900 a prime number? – No
• Is 4900 a composite number? – Yes
## Frequently Asked Questions on 4900 in Words
Q1
### How do you write 4900 in words?
4900 in words is written as Four Thousand Nine Hundred.
Q2
### Write the value of 3600 + 1300 in words.
3600 + 1300 = 4900 Thus, the value of 3600 + 1300 in words is written as Four Thousand Nine Hundred.
Q3
### Is 4900 a composite number?
Yes, 4900 is a composite number. |
# Verifying Trigonometric Identities
Trigonometric identities can be of varied forms. We have already seen different types of standard identities such as double angle formulas, sum of angles, difference of angles, reciprocal identities, etc. These are the standard identities that form the basis to the other identities which are derived from them. These identities which can be broken down into fundamental identities are often referred to as secondary identities.
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In fact, a vast number of identities can be formed from fundamental identities. To verify such identities, we use the same type of procedure that we use to solve different equations i.e. we solve for the common identity.
## Trigonometric Identities Verification Rules
1. If there is an identity consisting of different Trigonometric functions such as an identity consisting of tangent and sine, try to separate them.
2. Try to get tangent on one side and sine on the other hand of the equation, this makes it easier to apply the basic identities.
3. Now the most important thing is to know which fundamental identity one has to apply.
4. Convert the equation into the same function i.e. for instance, either convert all the tangents to sine or all the sines to tangents.
5. After doing all the above steps, if the right-hand side (RHS) of the equation is equal to the left-hand side (LHS) then the identity is true: $LHS=RHS$
## How to prove a Trigonometric Identity Algebraically?
We can prove the trigonometric identities algebraically by solving LHS and RHS separately. Here first we solve the complex side of the equation and check if both the LHS and RHS are equal or not. Let's look at a few examples.
Verify that${\mathrm{cot}}^{2}\mathrm{\theta }-{\mathrm{cos}}^{2}\mathrm{\theta }={\mathrm{cot}}^{2}\theta {\mathrm{cos}}^{2}\mathrm{\theta }$ is a consistent identity.
Solution:
Step 1: Adding ${\mathrm{cos}}^{2}\mathrm{\theta }$ on both sides of the equation, we get,
${\mathrm{cot}}^{2}\mathrm{\theta }={\mathrm{cos}}^{2}\mathrm{\theta }+{\mathrm{cos}}^{2}\mathrm{\theta }{\mathrm{cot}}^{2}\mathrm{\theta }\phantom{\rule{0ex}{0ex}}={\mathrm{cos}}^{2}\mathrm{\theta }\left(1+{\mathrm{cot}}^{2}\mathrm{\theta }\right)\phantom{\rule{0ex}{0ex}}$
Step 2: It can be seen that $1+{\mathrm{cot}}^{2}\mathrm{\theta }$ is a fundamental identity, which is
$\mathrm{cos}e{c}^{2}\mathrm{\theta }=1+{\mathrm{cot}}^{2}\mathrm{\theta }$
Step 3: Substituting for this identity, we get
${\mathrm{cot}}^{2}\mathrm{\theta }={\mathrm{cos}}^{2}\mathrm{\theta }{\mathrm{cosec}}^{2}\mathrm{\theta }$
Step 4: Now, we will use a reciprocal identity for cosecant,$\mathrm{cosec\theta }=\frac{1}{\mathrm{sin\theta }}$ and squaring on both sides, we get
${\mathrm{cosec}}^{2}\mathrm{\theta }=\frac{1}{{\mathrm{sin}}^{2}\mathrm{\theta }}$
Step 5: Substituting the above identity, we get
${\mathrm{cot}}^{2}\mathrm{\theta }=\frac{{\mathrm{cos}}^{2}\mathrm{\theta }}{{\mathrm{sin}}^{2}\mathrm{\theta }}$
Step 6: We already know that $\mathrm{cot\theta }=\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}$, and after squaring it, the above step is obtained.
Hence, $LHS=RHS$. Thus the identity is verified.
Verify the identity $\frac{1+\mathrm{cot\varphi }}{\mathrm{cosec\varphi }}=\mathrm{sin\varphi }+\mathrm{cos\varphi }$ where $\mathrm{\varphi }$ is defined over the appropriate domain.
Solution:
The LHS of the equation seems a bit complicated compared to the RHS of the equation. It is always easier to simplify the more complicated side of the equation. So let us simplify the LHS to get the RHS.
Step 1: Writing the denominator separately, we get
LHS =$\frac{1}{\mathrm{cosec\varphi }}+\frac{\mathrm{cot\varphi }}{\mathrm{cosec\varphi }}$
Step 2: Using the reciprocal identity,
$\mathrm{sin\varphi }=\frac{1}{\mathrm{cosec\varphi }}$
we get
LHS =$\mathrm{sin\varphi }+\frac{\mathrm{cot\varphi }}{\mathrm{cosec\varphi }}$
Step 3: Using the same identity again, and $\mathrm{cot}\varphi =\frac{\mathrm{cos\varphi }}{\mathrm{sin\varphi }}$, we get
LHS =$\mathrm{sin\varphi }+\mathrm{cos\varphi }$
Hence we can see that $LHS=RHS$.
Thus the identity is proved.
### When is a Trigonometric identity invalid?
A trigonometric identity is true if and only if it satisfies all the values for which the function is defined. To demonstrate this, consider the identity,
${\mathrm{sin}}^{2}\mathrm{x}=\mathrm{cosx}$
This identity does not hold for all values of x in the domain of sine and cosine. It is only true for certain values of x. To prove this identity is false, one can algebraically show that it is invalid or give a counterexample. Here one counterexample is
$LHS={\mathrm{sin}}^{2}45°=\frac{1}{2}$
$RHS=\mathrm{cos}45°=\frac{1}{\sqrt{2}}$
Where it can be seen that $LHS\ne RHS$.
So far we have seen how a Trigonometric identity can be proved algebraically using the fundamental identities. It is important to note that certain identities may seem true but actually be false. This has to do with one function being undefined for a certain value in the domain while the other function is defined. In that case, when we divide by something, the denominator may tend to 0 and that identity would be false. This is a rare scenario but one should be aware about it.
Solving an identity algebraically is not the only way to prove it, an alternative way is to prove it is graphically, lets take a look at it.
## How to prove a Trigonometric Identity Graphically?
For proving an identity graphically, we verify LHS and RHS separately. Firstly, the graph of LHS is plotted, then the graph of RHS is plotted. After examining the graphs, if the graphs are identical i.e. exactly the same for every value in its domain, then we say that $LHS=RHS$. The main drawback of using this method is that only simple identities can be proved. Identities with multiple and higher powers can be very difficult to graph. This is the reason why the algebraic method is preferred over this one. Let's prove an identity to get a grasp of this method.
Prove that $\mathrm{sinx}=\frac{1}{\mathrm{cosecx}}$.
Solution:
Step 1: Plot the graph of LHS, here it is sinx, and its graph looks like:
The graph of y=sinx, StudySmarter Originals
Step 2: Plot the graph of$y=\frac{1}{\mathrm{cos}ecx}$versus x, which looks like:
The graph of reciprocal of cosecx, StudySmarter Originals
It can be seen that the above graphs are exactly the same and hence we deduce that $\mathrm{sin}x=\frac{1}{\mathrm{cos}ecx}$.
## Examples on Verifying Trigonometric Identities
Prove that ${\mathrm{cos}}^{4}\mathrm{\theta }-{\mathrm{sin}}^{4}\mathrm{\theta }=\mathrm{cos}2\mathrm{\theta }$ is a valid trigonometric identity.
Solution:
Step 1: In this particular example, the LHS seems more complicated. Hence we will try to simplify LHS in order to arrive at RHS.
Step 2: The term${\mathrm{cos}}^{4}\mathrm{\theta }-{\mathrm{sin}}^{4}\mathrm{\theta }$ can also be written as ${\left({\mathrm{cos}}^{2}\mathrm{\theta }\right)}^{2}-{\left({\mathrm{sin}}^{2}\mathrm{\theta }\right)}^{2}$ and now it can be written as a product of two terms using the algebraic identity ${\mathrm{a}}^{2}-{\mathrm{b}}^{2}=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)$. Hence, we have:
${\left({\mathrm{cos}}^{2}\mathrm{\theta }\right)}^{2}-{\left({\mathrm{sin}}^{2}\mathrm{\theta }\right)}^{2}=\left({\mathrm{cos}}^{2}\mathrm{\theta }+{\mathrm{sin}}^{2}\mathrm{\theta }\right)\left({\mathrm{cos}}^{2}\mathrm{\theta }-{\mathrm{sin}}^{2}\mathrm{\theta }\right)$
Step 3: Now, we can apply the Pythagorean identity, ${\mathrm{cos}}^{2}\mathrm{\theta }+{\mathrm{sin}}^{2}\mathrm{\theta }=1$,
${\mathrm{cos}}^{4}\mathrm{\theta }-{\mathrm{sin}}^{4}\mathrm{\theta }={\mathrm{cos}}^{2}\mathrm{\theta }-{\mathrm{sin}}^{2}\mathrm{\theta }$
Step 4: Applying the Half angle formula $\mathrm{cos}2\mathrm{\theta }={\mathrm{cos}}^{2}\mathrm{\theta }-{\mathrm{sin}}^{2}\mathrm{\theta }$, we get
${\mathrm{cos}}^{4}\mathrm{\theta }-{\mathrm{sin}}^{4}\mathrm{\theta }=\mathrm{cos}2\mathrm{\theta }$
which is what we are asked to prove.
Thus, $LHS=RHS$.
Is $\left(1+\mathrm{sinx}\right)\left(1-\mathrm{sinx}\right)=\mathrm{cosx}$ a valid trigonometric identity?
Solution:
Step 1: The LHS seems trickier than the RHS so let’s simplify LHS and check whether we arrive at RHS or not.
Step 2: It can be seen that LHS can be simplified using the algebraic identity.
$\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)={\mathrm{a}}^{2}-{\mathrm{b}}^{2}$
Applying this identity respectively, we get
$\left(1+\mathrm{sinx}\right)\left(1-\mathrm{sinx}\right)=1-{\mathrm{sin}}^{2}\mathrm{x}$
Step 3: Now applying the Pythagorean identity for sine and cosine, we get
$1-{\mathrm{sin}}^{2}\mathrm{x}={\mathrm{cos}}^{2}\mathrm{x}$
which cannot be simplified further.
It can be observed that $LHS\ne RHS$ for every value of x. But for an identity to be valid, it should satisfy each value for which the function is defined.
Thus, the given trigonometric identity is false.
## Verifying Trigonometric Identities - Key takeaways
• A Trigonometric identity is true if and only if it satisfies all the values for which the domain is defined.
• If an identity is only true for certain values then those values are called the solutions to that equation.
• There are two ways to prove a Trigonometric identity: Algebraically and Graphically.
• To prove an identity algebraically, simplify one of the sides of the identity by reducing it to simpler terms with the help of fundamental identities.
• For the graphical method, plot the graphs of both sides of the equation and if both of them are exactly the same, then the identity is true.
• It is always more convenient to prove an identity using the Algebraic method as it is more effective and easy.
#### Flashcards in Verifying Trigonometric Identities 9
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How many ways are there to verify trigonometric identities?
There are two ways to verify a trigonometric identity: Algebraically and Graphically.
How to graphically verify trigonometric identities?
Plot both the sides of the identity as different graphs. If both the graphs are exactly the same for all values in its domain then the identity is true.
What are the rules when verifying trigonometric identities?
Ensure that the identity is true for each and every value in its domain. Start with the side of the identity which is complicated and try to reduce it to simpler terms.
How can trigonometric identities be verified mathematically?
By using different fundamental identities and carefully simplifying the terms, an identity can be proved mathematically.
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Polyhedron – Definition, Examples, Practice Problems, FAQs
Home » Math Vocabulary » Polyhedron – Definition, Examples, Practice Problems, FAQs
What is a Polyhedron?
We know that a polygon is a flat, plane, two-dimensional closed shape bounded by line segments. Common examples of polygons are square, triangle, pentagon, etc.
Now, can you imagine a three dimensional figure with faces in the shape of a polygon? Such a three-dimensional figure is known as a Polyhedron. Examples you must be familiar with are diamonds, rubik’s cube, pyramids, etc.
Polyhedron Definition
A three-dimensional shape with flat polygonal faces, straight edges, and sharp corners or vertices is called a polyhedron.
Common examples are cubes, prisms, pyramids. However, cones, and spheres are not polyhedrons since they do not have polygonal faces.
The plural of a polyhedron is called polyhedra or polyhedrons.
Faces, Edges, and Vertices
The parts of a polyhedron are classified as faces, edges, and vertices.
• Polyhedron Faces: The flat surfaces of a polyhedron are termed as its faces, which are basically polygons.
• Polyhedron Edges: Edges are the line segments where two faces meet.
• Polyhedron Vertices: The point of intersection of two edges is known as a vertex.
The figure given below will give us a better view:
Prisms, Pyramids, and Platonic Solids
Polyhedrons can be broadly classified as prisms, pyramids, and Platonic solids. Let’s understand each type.
1. Prism
A prism has its both ends (base and top) as identical polygons and its side faces are flat (rectangles or parallelograms). Prisms are named after its base, which can be a triangle, square, rectangle, or any n-sided polygon.
2. Pyramid
A pyramid has its base as any polygon and its side faces are triangles with a common vertex (known as apex). If the base of a pyramid is an n-sided polygon, then it has (n+1) faces, (n+1) vertices, and 2n edges.
3. Platonic Solids
In Platonic solids, all the faces are congruent regular polygons and the same number of faces meet at each vertex.
Examples in Real Life
We can observe several polyhedra in our daily existence such as a Rubik’s cube, dice, buckyball, pyramids, and so on. A diamond is one of the real-life examples of polyhedron.
Types of Polyhedron
Polyhedra are mainly divided into two types based on the polygonal faces and the base — regular polyhedrons and irregular polyhedrons. Platonic solids are regular polygons. Prisms and pyramids are irregular polyhedrons.
Regular Polyhedron
Regular polyhedrons are made up of regular polygons. They are also known as “Platonic solids.” They have all their faces, edges, and angles congruent. The following is the list of the five regular polyhedrons:
Irregular Polyhedron
An irregular polyhedron has polygonal faces that are not congruent to each other. It is made up of polygons having different shapes. So, all its components are not the same.
Polyhedrons can also be divided into convex and concave categories, just like polygons.
Convex Polyhedron
A convex polyhedron is similar to a convex polygon. If a line segment that joins any two points on the surface lies inside the polyhedron, then it is known as a convex polyhedron. All platonic solids are convex.
Concave Polyhedron
A concave polyhedron is similar to a concave polygon. If a line segment that joins any two points on the surface lies outside the polyhedron, it is known as a concave polyhedron.
Euler’s Formula
There is a relationship between the number of faces, edges, and vertices in a polyhedron, which can be presented by a math formula known as “Euler’s Formula.”
$\text{F} + \text{V}$ $–$ $\text{E} = 2$
where,
$\text{F} =$ number of faces
$\text{V} =$ number of vertices
$\text{E} =$ number of edges
If we know any two values among F, V, or E, we can find the third missing value using Euler’s formula. So, the questions like ‘How many edges does a polyhedron have?’ or ‘How many faces does a polyhedron have?’ can be answered easily if the other two values are known. We can also check if a polyhedron with the given number of parts exists or not.
For example, a cube has 8 vertices, 6 faces, and 12 edges.
$\text{F} = 6, \text{V} = 8, \text{E} = 12$
Applying Euler’s formula, we get $\text{F} + \text{V}$ $–$ $\text{E} = 2$
Substituting the values in the formula: $6 + 8$ $–$ $12 = 2 \Rightarrow 2 = 2$.
Hence, the cube is a polyhedron.
Solved Examples on Polyhedron
1. How many types of regular polygons are there?
Solution: There are 5 types of regular polygons: tetrahedron, cube, octahedron, dodecahedron, and icosahedron.
2. Check if the polyhedron with 10 vertices, 8 edges, and 4 faces exists or not.
Solution: We will apply Euler’s formula:
$\text{V} = 10, \text{E} = 8$, and $\text{F} = 4$
$\text{F} + \text{V}$ $–$ $\text{E} = 4 + 10$ $–$ $8 = 6 ≠ 2$.
The polyhedron with the given dimensions does not exist.
3. How many faces does a polyhedron with 12 vertices and 18 edges have?
Solution: $\text{V} = 12$, $\text{E} = 18$, and $\text{F} =$ ?
By applying Euler’s formula, we get
$\text{F} + \text{V}$ $–$ $\text{E} = 2$
$\Rightarrow\text{F} + 12$ $–$ $18 = 2$
$\Rightarrow\text{F} = 8$
Number of faces = 8
Practice Problems on Polyhedron
1
Which of the following is not a polyhedron?
Cube
Cone
Cuboid
Triangular Prism
CorrectIncorrect
Cone is not a polyhedron as the faces of the cone are curved.
2
Which of the following is a concave polyhedron?
A
B
C
D
CorrectIncorrect
If we draw a line from one face to another, the line will lie outside the polyhedron.
3
Which of the following combinations will not form a polyhedron?
$\text{F} = 4, \text{V} = 10, \text{E} = 12$
$\text{F} = 8, \text{V} = 12, \text{E} = 18$
$\text{F} = 7, \text{V} = 13, \text{E} = 15$
$\text{F} = 9. \text{V} = 15, \text{E} = 22$
CorrectIncorrect
Correct answer is: $\text{F} = 7, \text{V} = 13, \text{E} = 15$
$\text{F} + \text{V}$ $–$ $\text{E} = 7 + 13 – 15 = 5 ≠ 2$
4
Which of the following is not a regular polyhedron?
Cube
Tetrahedron
Both A and B
Hexagonal pyramid
CorrectIncorrect |
# CLASS 10 MATH NCERT SOLUTIONS FOR CHAPTER – 2 POLYNOMIALS EX – 2.3
## Polynomials
Exercise 2.3 Page: 36
1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
(i) p(x) = x3-3x2+5x–3 , g(x) = x2–2
Solution:
Given,
Dividend = p(x) = x3-3x2+5x–3
Divisor = g(x) = x2– 2
Therefore, upon division we get,
Quotient = x–3
Remainder = 7x–9
(ii) p(x) = x4-3x2+4x+5 , g(x) = x2+1-x
Solution:
Given,
Dividend = p(x) = x– 3x+ 4x +5
Divisor = g(x) = x2 +1-x
Therefore, upon division we get,
Quotient = x+ x–3
Remainder = 8
(iii) p(x) =x4–5x+6, g(x) = 2–x2
Solution:
Given,
Dividend = p(x) =x4 – 5x + 6 = x+0x2–5x+6
Divisor = g(x) = 2–x2 = –x2+2
Therefore, upon division we get,
Quotient = -x2-2
Remainder = -5x + 10
2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
(i) t2-3, 2t+3t3-2t2-9t-12
Solutions:
Given,
First polynomial = t2-3
Second polynomial = 2t+3t3-2t-9t-12
As we can see, the remainder is left as 0. Therefore, we say that, t2-3 is a factor of 2t2+3t+4.
(ii)x2+3x+1 , 3x4+5x3-7x2+2x+2
Solutions:
Given,
First polynomial = x2+3x+1
Second polynomial = 3x4+5x3-7x2+2x+2
As we can see, the remainder is left as 0. Therefore, we say that, x2 + 3x + 1 is a factor of 3x4+5x3-7x2+2x+2.
(iii) x3-3x+1, x5-4x3+x2+3x+1
Solutions:
Given,
First polynomial = x3-3x+1
Second polynomial = x5-4x3+x2+3x+1
As we can see, the remainder is not equal to 0. Therefore, we say that, x3-3x+1 is not a factor of x5-4x3+x2+3x+1 .
3. Obtain all other zeroes of 3x4+6x3-2x2-10x-5, if two of its zeroes are √(5/3) and – √(5/3).
Solutions:
Since this is a polynomial equation of degree 4, hence there will be total 4 roots.
√(5/3) and – √(5/3) are zeroes of polynomial f(x).
∴ (x –√(5/3)) (x+√(5/3) = x2-(5/3) = 0
(3x2−5)=0, is a factor of given polynomial f(x).
Now, when we will divide f(x) by (3x2−5) the quotient obtained will also be a factor of f(x) and the remainder will be 0.
Therefore, 3x+6x−2x−10x–5 = (3x–5)(x2+2x+1)
Now, on further factorizing (x2+2x+1) we get,
x2+2x+1 = x2+x+x+1 = 0
x(x+1)+1(x+1) = 0
(x+1)(x+1) = 0
So, its zeroes are given by: x= −1 and x = −1.
Therefore, all four zeroes of given polynomial equation are:
√(5/3),- √(5/3) , −1 and −1.
4. On dividing x3-3x2+x+2 by a polynomial g(x), the quotient and remainder were x–2 and –2x+4, respectively. Find g(x).
Solution:
Given,
Dividend, p(x) = x3-3x2+x+2
Quotient = x-2
Remainder = –2x+4
We have to find the value of Divisor, g(x) =?
As we know,
Dividend = Divisor × Quotient + Remainder
∴ x3-3x2+x+2 = g(x)×(x-2) + (-2x+4)
x3-3x2+x+2-(-2x+4) = g(x)×(x-2)
Therefore, g(x) × (x-2) = x3-3x2+x+2
Now, for finding g(x) we will divide x3-3x2+x+2 with (x-2)
Therefore, g(x) = (x2–x+1)
5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Solutions:
According to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x)≠0. Then we can find the value of quotient q(x) and remainder r(x), with the help of below given formula;
Dividend = Divisor × Quotient + Remainder
∴ p(x) = g(x)×q(x)+r(x)
Where r(x) = 0 or degree of r(x)< degree of g(x).
Now let us proof the three given cases as per division algorithm by taking examples for each.
(i) deg p(x) = deg q(x)
Degree of dividend is equal to degree of quotient, only when the divisor is a constant term.
Let us take an example, 3x2+3x+3 is a polynomial to be divided by 3.
So, (3x2+3x+3)/3 = x2+x+1 = q(x)
Thus, you can see, the degree of quotient is equal to the degree of dividend.
Hence, division algorithm is satisfied here.
(ii) deg q(x) = deg r(x)
Let us take an example , p(x)=x2+x is a polynomial to be divided by g(x)=x.
So, (x2+x)/x = x+1 = q(x)
Also, remainder, r(x) = 0
Thus, you can see, the degree of quotient is equal to the degree of remainder.
Hence, division algorithm is satisfied here.
(iii) deg r(x) = 0
The degree of remainder is 0 only when the remainder left after division algorithm is constant.
Let us take an example, p(x) = x2+1 is a polynomial to be divided by g(x)=x.
So,( x2+1)/x= x=q(x)
And r(x)=1
Clearly, the degree of remainder here is 0.
Hence, division algorithm is satisfied here. |
# Recurrence relation explained
In mathematics, a recurrence relation is an equation that recursively defines a sequence or multidimensional array of values, once one or more initial terms are given; each further term of the sequence or array is defined as a function of the preceding terms.
The term difference equation sometimes (and for the purposes of this article) refers to a specific type of recurrence relation. However, "difference equation" is frequently used to refer to any recurrence relation.
## Definition
A recurrence relation is an equation that expresses each element of a sequence as a function of the preceding ones. More precisely, in the case where only the immediately preceding element is involved, a recurrence relation has the form
un=\varphi(n,un-1)forn>0,
where
\varphi:N x X\toX
is a function, where is a set to which the elements of a sequence must belong. For any
u0\inX
, this defines a unique sequence with
u0
as its first element, called the initial value.[1]
It is easy to modify the definition for getting sequences starting from the term of index 1 or higher.
This defines recurrence relation of first order. A recurrence relation of order has the form
un=\varphi(n,un-1,un-2,\ldots,un-k)forn\gek,
where
\varphi:N x Xk\toX
is a function that involves consecutive elements of the sequence.In this case, initial values are needed for defining a sequence.
## Examples
### Factorial
The factorial is defined by the recurrence relation
n!=n(n-1)!forn>0,
and the initial condition
0!=1.
### Logistic map
An example of a recurrence relation is the logistic map:
xn+1=rxn(1-xn),
with a given constant r; given the initial term x0 each subsequent term is determined by this relation.
Solving a recurrence relation means obtaining a closed-form solution: a non-recursive function of n.
### Fibonacci numbers
The recurrence of order two satisfied by the Fibonacci numbers is the archetype of a homogeneous linear recurrence relation with constant coefficients (see below). The Fibonacci sequence is defined using the recurrence
Fn=Fn-1+Fn-2
with initial conditions
F0=0
F1=1.
Explicitly, the recurrence yields the equations
F2=F1+F0
F3=F2+F1
F4=F3+F2
etc.
We obtain the sequence of Fibonacci numbers, which begins
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
The recurrence can be solved by methods described below yielding Binet's formula, which involves powers of the two roots of the characteristic polynomial t2 = t + 1; the generating function of the sequence is the rational function
t 1-t-t2
.
### Binomial coefficients
A simple example of a multidimensional recurrence relation is given by the binomial coefficients
\tbinom{n}{k}
, which count the number of ways of selecting k elements out of a set of n elements.They can be computed by the recurrence relation
\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k},
with the base cases
\tbinom{n}{0}=\tbinom{n}{n}=1
. Using this formula to compute the values of all binomial coefficients generates an infinite array called Pascal's triangle. The same values can also be computed directly by a different formula that is not a recurrence, but that requires multiplication and not just addition to compute:
\binom{n}{k}= n! k!(n-k)!
.
## Relationship to difference equations narrowly defined
\left\{an\right\}
infty n=1
of real numbers: the first difference
\Delta(an)
is defined as
\Delta(an)=an+1-an.
The second difference
2(a \Delta n)
is defined as
2(a \Delta n)
=\Delta(an+1)-\Delta(an),
which can be simplified to
2(a \Delta n)
=an+2-2an+1+an.
More generally: the k-th difference of the sequence an written as
k(a \Delta n)
is defined recursively as
k(a \Delta n)
=\Deltak-1(an+1)-\Deltak-1(an)=\sum
k t=0
\binom{k}{t}(-1)tan+k-t.
(The sequence and its differences are related by a binomial transform.) The more restrictive definition of difference equation is an equation composed of an and its kth differences. (A widely used broader definition treats "difference equation" as synonymous with "recurrence relation". See for example rational difference equation and matrix difference equation.)
Actually, it is easily seen that,
an+k={k\choose0}an+{k\choose1}\Delta(an)++{k\choosek}
k(a \Delta n).
Thus, a difference equation can be defined as an equation that involves an, an−1, an−2 etc. (or equivalentlyan, an+1, an+2 etc.)
Since difference equations are a very common form of recurrence, some authors use the two terms interchangeably. For example, the difference equation
2(a 3\Delta n)
+2\Delta(an)+7an=0
is equivalent to the recurrence relation
3an+2=4an+1-8an.
Thus one can solve many recurrence relations by rephrasing them as difference equations, and then solving the difference equation, analogously to how one solves ordinary differential equations. However, the Ackermann numbers are an example of a recurrence relation that do not map to a difference equation, much less points on the solution to a differential equation.
See time scale calculus for a unification of the theory of difference equations with that of differential equations.
Summation equations relate to difference equations as integral equations relate to differential equations.
### From sequences to grids
Single-variable or one-dimensional recurrence relations are about sequences (i.e. functions defined on one-dimensional grids). Multi-variable or n-dimensional recurrence relations are about n-dimensional grids. Functions defined on n-grids can also be studied with partial difference equations.[2]
## Solving
### Solving homogeneous linear recurrence relations with constant coefficients
See main article: Constant-recursive sequence and Linear difference equation.
#### Roots of the characteristic polynomial
An order-d homogeneous linear recurrence with constant coefficients is an equation of the form
an=c1an-1+c2an-2+ … +cdan-d,
where the d coefficients ci (for all i) are constants, and
cd\ne0
.
A constant-recursive sequence is a sequence satisfying a recurrence of this form. There are d degrees of freedom for solutions to this recurrence, i.e., the initial values
can be taken to be any values but then the recurrence determines the sequence uniquely.
The same coefficients yield the characteristic polynomial (also "auxiliary polynomial")
p(t)=td-
d-1 c 1t
-
d-2 c 2t
- … -cd
whose d roots play a crucial role in finding and understanding the sequences satisfying the recurrence. If the roots r1, r2, ... are all distinct, then each solution to the recurrence takes the form
an=k1
n r 1
+k2
n r 2
++kd
n, r d
where the coefficients ki are determined in order to fit the initial conditions of the recurrence. When the same roots occur multiple times, the terms in this formula corresponding to the second and later occurrences of the same root are multiplied by increasing powers of n. For instance, if the characteristic polynomial can be factored as (xr)3, with the same root r occurring three times, then the solution would take the form
an=k1rn+k2nrn+k3n2rn.
[3]
As well as the Fibonacci numbers, other constant-recursive sequences include the Lucas numbers and Lucas sequences, the Jacobsthal numbers, the Pell numbers and more generally the solutions to Pell's equation.
For order 1, the recurrence
an=ran-1
has the solution an = rn with a0 = 1 and the most general solution is an = krn with a0 = k. The characteristic polynomial equated to zero (the characteristic equation) is simply t − r = 0.
Solutions to such recurrence relations of higher order are found by systematic means, often using the fact that an = rn is a solution for the recurrence exactly when t = r is a root of the characteristic polynomial. This can be approached directly or using generating functions (formal power series) or matrices.
Consider, for example, a recurrence relation of the form
an=Aan-1+Ban-2.
When does it have a solution of the same general form as an = rn? Substituting this guess (ansatz) in the recurrence relation, we find that
rn=Arn-1+Brn-2
must be true for all n > 1.
Dividing through by rn−2, we get that all these equations reduce to the same thing:
r2=Ar+B,
r2-Ar-B=0,
which is the characteristic equation of the recurrence relation. Solve for r to obtain the two roots λ1, λ2: these roots are known as the characteristic roots or eigenvalues of the characteristic equation. Different solutions are obtained depending on the nature of the roots: If these roots are distinct, we have the general solution
an=
n Cλ 2
while if they are identical (when), we have
an=Cλn+Dnλn
This is the most general solution; the two constants C and D can be chosen based on two given initial conditions a0 and a1 to produce a specific solution.
In the case of complex eigenvalues (which also gives rise to complex values for the solution parameters C and D), the use of complex numbers can be eliminated by rewriting the solution in trigonometric form. In this case we can write the eigenvalues as
λ1,λ2=\alpha\pm\betai.
Then it can be shown that
an=
n Cλ 2
can be rewritten as[4]
an=2Mn\left(E\cos(\thetan)+F\sin(\thetan)\right)=2GMn\cos(\thetan-\delta),
where
\begin{array}{lcl} M=\sqrt{\alpha2+\beta2}&\cos(\theta)=\tfrac{\alpha}{M}&\sin(\theta)=\tfrac{\beta}{M}\\ C,D=E\mpFi&&\\ G=\sqrt{E2+F2}&\cos(\delta)=\tfrac{E}{G}&\sin(\delta)=\tfrac{F}{G} \end{array}
Here E and F (or equivalently, G and δ) are real constants which depend on the initial conditions. Using
λ1+λ2=2\alpha=A,
λ1
2+\beta λ 2=\alpha
2=-B,
one may simplify the solution given above as
an=
n 2
(-B)
\left(E\cos(\thetan)+F\sin(\thetan)\right),
where a1 and a2 are the initial conditions and
\begin{align} E&=
-Aa1+a2 B
\\ F&=-i
A2a1-Aa2+2a1B B\sqrt{A2+4B
} \\\theta &=\arccos \left (\frac \right)\end
In this way there is no need to solve for λ1 and λ2.
In all cases—real distinct eigenvalues, real duplicated eigenvalues, and complex conjugate eigenvalues—the equation is stable (that is, the variable a converges to a fixed value [specifically, zero]) if and only if both eigenvalues are smaller than one in absolute value. In this second-order case, this condition on the eigenvalues can be shown[5] to be equivalent to |A| < 1 − B < 2, which is equivalent to |B| < 1 and |A| < 1 − B.
The equation in the above example was homogeneous, in that there was no constant term. If one starts with the non-homogeneous recurrence
bn=Abn-1+Bbn-2+K
with constant term K, this can be converted into homogeneous form as follows: The steady state is found by setting bnbn−1bn−2b* to obtain
b*=
K 1-A-B
.
Then the non-homogeneous recurrence can be rewritten in homogeneous form as
[bn-
*]=A[b b n-1
*]+B[b -b n-2
-b*],
which can be solved as above.
The stability condition stated above in terms of eigenvalues for the second-order case remains valid for the general nth-order case: the equation is stable if and only if all eigenvalues of the characteristic equation are less than one in absolute value.
Given a homogeneous linear recurrence relation with constant coefficients of order d, let p(t) be the characteristic polynomial (also "auxiliary polynomial")
td-
d-1 c 1t
-
d-2 c 2t
--cd=0
such that each ci corresponds to each ci in the original recurrence relation (see the general form above). Suppose λ is a root of p(t) having multiplicity r. This is to say that (t−λ)r divides p(t). The following two properties hold:
1. Each of the r sequences
λn,nλn,nn,...,nr-1λn
satisfies the recurrence relation.
1. Any sequence satisfying the recurrence relation can be written uniquely as a linear combination of solutions constructed in part 1 as λ varies over all distinct roots of p(t).
As a result of this theorem a homogeneous linear recurrence relation with constant coefficients can be solved in the following manner:
1. Find the characteristic polynomial p(t).
2. Find the roots of p(t) counting multiplicity.
3. Write an as a linear combination of all the roots (counting multiplicity as shown in the theorem above) with unknown coefficients bi.
an=\left(b
n 1
+b2nλ
n 1
+
n+ … +b b r
nr-1
n λ 1
\right)+ … +\left(bd-q+1
n λ *
++bdnq-1
n λ *
\right)
This is the general solution to the original recurrence relation. (q is the multiplicity of λ*)
1. Equate each
from part 3 (plugging in n = 0, ..., d into the general solution of the recurrence relation) with the known values
from the original recurrence relation. However, the values an from the original recurrence relation used do not usually have to be contiguous: excluding exceptional cases, just d of them are needed (i.e., for an original homogeneous linear recurrence relation of order 3 one could use the values a0, a1, a4). This process will produce a linear system of d equations with d unknowns. Solving these equations for the unknown coefficients
b1,b2,...,bd
of the general solution and plugging these values back into the general solution will produce the particular solution to the original recurrence relation that fits the original recurrence relation's initial conditions (as well as all subsequent values
a0,a1,a2,...
of the original recurrence relation).
The method for solving linear differential equations is similar to the method above—the "intelligent guess" (ansatz) for linear differential equations with constant coefficients is eλx where λ is a complex number that is determined by substituting the guess into the differential equation.
This is not a coincidence. Considering the Taylor series of the solution to a linear differential equation:
infin \sum n=0
f(n)(a) n!
(x-a)n
it can be seen that the coefficients of the series are given by the nth derivative of f(x) evaluated at the point a. The differential equation provides a linear difference equation relating these coefficients.
This equivalence can be used to quickly solve for the recurrence relationship for the coefficients in the power series solution of a linear differential equation.
The rule of thumb (for equations in which the polynomial multiplying the first term is non-zero at zero) is that:
y[k]\tof[n+k]
and more generally
xm*y[k]\ton(n-1)...(n-m+1)f[n+k-m]
Example: The recurrence relationship for the Taylor series coefficients of the equation:
(x2+3x-4)y[3]-(3x+1)y[2]+2y=0
is given by
n(n-1)f[n+1]+3nf[n+2]-4f[n+3]-3nf[n+1]-f[n+2]+2f[n]=0
or
-4f[n+3]+2nf[n+2]+n(n-4)f[n+1]+2f[n]=0.
This example shows how problems generally solved using the power series solution method taught in normal differential equation classes can be solved in a much easier way.
Example: The differential equation
ay''+by'+cy=0
has solution
y=eax.
The conversion of the differential equation to a difference equation of the Taylor coefficients is
af[n+2]+bf[n+1]+cf[n]=0.
It is easy to see that the nth derivative of eax evaluated at 0 is an.
#### Solving via linear algebra
A linearly recursive sequence y of order n
yn+k-cn-1yn-1+k-cn-2yn-2+k+-c0yk=0
is identical to
yn=cn-1yn-1+cn-2yn-2++c0y0.
Expanded with n−1 identities of kind
yn-k=yn-k
, this n-th order equation is translated into a matrix difference equation system of n first-order linear equations,
yn=\begin{bmatrix}yn\yn-1\\vdots\\vdots\y1\end{bmatrix}= \begin{bmatrix} cn-1&cn-2&&&c0\\ 1&0&&&0\\ 0&\ddots&\ddots&&\vdots\\ \vdots&\ddots&\ddots&\ddots&\vdots\\ 0&&0&1&0\end{bmatrix} \begin{bmatrix}yn-1\\yn-2\\vdots\\vdots\y0\end{bmatrix}=Cyn-1=Cny0.
Observe that the vector
yn
can be computed by n applications of the companion matrix, C, to the initial state vector,
y0
. Thereby, n-th entry of the sought sequence y, is the top component of
yn,yn=yn[n]
.
Eigendecomposition,
yn=Cny0=a
ne 1
+a
ne 2
++anλ
ne n
into eigenvalues,
λ1,λ2,\ldots,λn
, and eigenvectors,
e1,e2,\ldots,en
, is used to compute
yn
. Thanks to the crucial fact that system C time-shifts every eigenvector, e, by simply scaling its components λ times,
Cei=λiei=C\begin{bmatrix}ei,n\\ei,n-1\\vdots\ei,1\end{bmatrix}=\begin{bmatrix}λiei,n\ λiei,n-1\\vdots\ λiei,1\end{bmatrix}
that is, time-shifted version of eigenvector, e, has components λ times larger, the eigenvector components are powers of λ,
ei=
n-1 \begin{bmatrix}λ i
&&
2 λ i
&λi&1\end{bmatrix}T,
and, thus, recurrent homogeneous linear equation solution is a combination of exponential functions,
yn=
n \sum 1
{ciλ
ne i}
. The components
ci
can be determined out of initial conditions:
y0=\begin{bmatrix}y0\y-1\\vdots\y-n+1\end{bmatrix}=
n \sum i=1
{ciλ
0e i}
=\begin{bmatrix}e1&e2&&en\end{bmatrix}\begin{bmatrix}c1\c2\\vdots\cn\end{bmatrix}=E\begin{bmatrix}c1\c2\\vdots\cn\end{bmatrix}
Solving for coefficients,
\begin{bmatrix}c1\c2\\vdots\cn\end{bmatrix}=E-1y0=
n-1 \begin{bmatrix}λ 1
&
n-1 λ 2
&&
n-1 λ n
\\vdots&\vdots&\ddots&\vdots\ λ1&λ2&&λn\ 1&1&&1\end{bmatrix}-1\begin{bmatrix}y0\y-1\\vdots\y-n+1\end{bmatrix}.
This also works with arbitrary boundary conditions
\underbrace{ya,yb,\ldots}n
, not necessary the initial ones,
\begin{bmatrix}ya\yb\\vdots\end{bmatrix}=\begin{bmatrix}ya[n]\yb[n]\\vdots\end{bmatrix}=
n \begin{bmatrix}\sum i=1
{ciλ
ae i[n]}
n \ \sum i=1
{ciλ
be i[n]}
\\vdots\end{bmatrix}
n =\begin{bmatrix}\sum i=1
{ciλ
n-1 i
} \\ \sum_^n \\ \vdots\end
=\begin{bmatrix}\sum{ciλ
a+n-1 i
} \\ \sum \\ \vdots\end = \begin\lambda_1^ & \lambda_2^ & \cdots & \lambda_n^ \\ \lambda_1^ & \lambda_2^ & \cdots & \lambda_n^ \\ \vdots & \vdots & \ddots & \vdots \end\,\beginc_1 \\ c_2 \\ \vdots \\ c_n\end.
This description is really no different from general method above, however it is more succinct. It also works nicely for situations like
\begin{cases} an=an-1-bn-1\\ bn=2an-1+bn-1. \end{cases}
where there are several linked recurrences.[6]
#### Solving with z-transforms
Certain difference equations - in particular, linear constant coefficient difference equations - can be solved using z-transforms. The z-transforms are a class of integral transforms that lead to more convenient algebraic manipulations and more straightforward solutions. There are cases in which obtaining a direct solution would be all but impossible, yet solving the problem via a thoughtfully chosen integral transform is straightforward.
### Solving non-homogeneous linear recurrence relations with constant coefficients
If the recurrence is non-homogeneous, a particular solution can be found by the method of undetermined coefficients and the solution is the sum of the solution of the homogeneous and the particular solutions. Another method to solve a non-homogeneous recurrence is the method of symbolic differentiation. For example, consider the following recurrence:
an+1=an+1
This is a non-homogeneous recurrence. If we substitute nn+1, we obtain the recurrence
an+2=an+1+1
Subtracting the original recurrence from this equation yields
an+2-an+1=an+1-an
or equivalently
an+2=2an+1-an
This is a homogeneous recurrence, which can be solved by the methods explained above. In general, if a linear recurrence has the form
an+k=λk-1an+k-1+λk-2an+k-2++λ1an+1+λ0an+p(n)
where
λ0,λ1,...,λk-1
are constant coefficients and p(n) is the inhomogeneity, then if p(n) is a polynomial with degree r, then this non-homogeneous recurrence can be reduced to a homogeneous recurrence by applying the method of symbolic differencing r times.
If
P(x)=
infty \sum n=0
pnxn
is the generating function of the inhomogeneity, the generating function
A(x)=
infty \sum n=0
a(n)xn
of the non-homogeneous recurrence
an=
s \sum i=1
cian-i+pn,n\genr,
with constant coefficients is derived from
\left
i (1-\sum ix
\right
nr-1 )A(x)=P(x)+\sum n=0
[an-p
s i=1
nr-i-1 c n=0
n. a nx
If P(x) is a rational generating function, A(x) is also one. The case discussed above, where pn = K is a constant, emerges as one example of this formula, with P(x) = K/(1−x). Another example, the recurrence
an=10an-1+n
with linear inhomogeneity, arises in the definition of the schizophrenic numbers. The solution of homogeneous recurrences is incorporated as p = P = 0.
### Solving first-order non-homogeneous recurrence relations with variable coefficients
Moreover, for the general first-order non-homogeneous linear recurrence relation with variable coefficients:
an+1=fnan+gn, fn0,
there is also a nice method to solve it:[7]
an+1-fnan=gn
an+1
n \prod fk k=0
-
fnan
n \prod fk k=0
=
gn
n \prod fk k=0
an+1
n \prod fk k=0
-
an
n-1 \prod fk k=0
=
gn
n \prod fk k=0
Let
An=
an
n-1 \prod fk k=0
,
Then
An+1-An=
gn
n \prod fk k=0
n-1 \sum m=0
(Am+1-Am)=An-A0=
n-1 \sum m=0
gm
m \prod fk k=0
an
n-1 \prod fk k=0
=A0+
n-1 \sum m=0
gm
m \prod fk k=0
an=
n-1 \left(\prod k=0
fk\right)\left(A0+
n-1 \sum m=0
gm
m \prod fk k=0
\right)
If we apply the formula to
an+1=(1+hfnh)an+hgnh
and take the limit h→0, we get the formula for first order linear differential equations with variable coefficients; the sum becomes an integral, and the product becomes the exponential function of an integral.
### Solving general homogeneous linear recurrence relations
Many homogeneous linear recurrence relations may be solved by means of the generalized hypergeometric series. Special cases of these lead to recurrence relations for the orthogonal polynomials, and many special functions. For example, the solution to
Jn+1=
2n z
Jn-Jn-1
is given by
Jn=Jn(z),
the Bessel function, while
(b-n)Mn-1+(2n-b-z)Mn-nMn+1=0
is solved by
Mn=M(n,b;z)
the confluent hypergeometric series. Sequences which are the solutions of linear difference equations with polynomial coefficients are called P-recursive. For these specific recurrence equations algorithms are known which find polynomial, rational or hypergeometric solutions.
### Solving first-order rational difference equations
See main article: Rational difference equation.
A first order rational difference equation has the form
wt+1=\tfrac{awt+b}{cwt+d}
. Such an equation can be solved by writing
wt
as a nonlinear transformation of another variable
xt
which itself evolves linearly. Then standard methods can be used to solve the linear difference equation in
xt
.
## Stability
### Stability of linear higher-order recurrences
The linear recurrence of order d,
an=c1an-1+c2an-2+ … +cdan-d,
has the characteristic equation
λd-c1λd-1-c2λd-2--cdλ0=0.
The recurrence is stable, meaning that the iterates converge asymptotically to a fixed value, if and only if the eigenvalues (i.e., the roots of the characteristic equation), whether real or complex, are all less than unity in absolute value.
### Stability of linear first-order matrix recurrences
See main article: Matrix difference equation.
In the first-order matrix difference equation
[xt-x*]=A[xt-1-x*]
with state vector x and transition matrix A, x converges asymptotically to the steady state vector x* if and only if all eigenvalues of the transition matrix A (whether real or complex) have an absolute value which is less than 1.
### Stability of nonlinear first-order recurrences
Consider the nonlinear first-order recurrence
xn=f(xn-1).
This recurrence is locally stable, meaning that it converges to a fixed point x* from points sufficiently close to x*, if the slope of f in the neighborhood of x* is smaller than unity in absolute value: that is,
|f'(x*)|<1.
A nonlinear recurrence could have multiple fixed points, in which case some fixed points may be locally stable and others locally unstable; for continuous f two adjacent fixed points cannot both be locally stable.
A nonlinear recurrence relation could also have a cycle of period k for k > 1. Such a cycle is stable, meaning that it attracts a set of initial conditions of positive measure, if the composite function
g(x):=f\circf\circ\circf(x)
with f appearing k times is locally stable according to the same criterion:
|g'(x*)|<1,
where x* is any point on the cycle.
In a chaotic recurrence relation, the variable x stays in a bounded region but never converges to a fixed point or an attracting cycle; any fixed points or cycles of the equation are unstable. See also logistic map, dyadic transformation, and tent map.
## Relationship to differential equations
When solving an ordinary differential equation numerically, one typically encounters a recurrence relation. For example, when solving the initial value problem
y'(t)=f(t,y(t)), y(t0)=y0,
with Euler's method and a step size h, one calculates the values
y0=y(t0), y1=y(t0+h), y2=y(t0+2h), ...
by the recurrence
yn+1=yn+hf(tn,yn),tn=t0+nh
Systems of linear first order differential equations can be discretized exactly analytically using the methods shown in the discretization article.
## Applications
### Biology
Some of the best-known difference equations have their origins in the attempt to model population dynamics. For example, the Fibonacci numbers were once used as a model for the growth of a rabbit population.
The logistic map is used either directly to model population growth, or as a starting point for more detailed models of population dynamics. In this context, coupled difference equations are often used to model the interaction of two or more populations. For example, the Nicholson–Bailey model for a host-parasite interaction is given by
Nt+1=λNt
-aPt e
Pt+1=
-aPt N t(1-e
),
with Nt representing the hosts, and Pt the parasites, at time t.
Integrodifference equations are a form of recurrence relation important to spatial ecology. These and other difference equations are particularly suited to modeling univoltine populations.
### Computer science
Recurrence relations are also of fundamental importance in analysis of algorithms.[8] [9] If an algorithm is designed so that it will break a problem into smaller subproblems (divide and conquer), its running time is described by a recurrence relation.
A simple example is the time an algorithm takes to find an element in an ordered vector with
n
elements, in the worst case.
A naive algorithm will search from left to right, one element at a time. The worst possible scenario is when the required element is the last, so the number of comparisons is
n
.
A better algorithm is called binary search. However, it requires a sorted vector. It will first check if the element is at the middle of the vector. If not, then it will check if the middle element is greater or lesser than the sought element. At this point, half of the vector can be discarded, and the algorithm can be run again on the other half. The number of comparisons will be given by
c1=1
cn=1+cn/2
the time complexity of which will be
O(log2(n))
.
### Digital signal processing
In digital signal processing, recurrence relations can model feedback in a system, where outputs at one time become inputs for future time. They thus arise in infinite impulse response (IIR) digital filters.
For example, the equation for a "feedforward" IIR comb filter of delay T is:
yt=(1-\alpha)xt+\alphayt,
where
xt
is the input at time t,
yt
is the output at time t, and α controls how much of the delayed signal is fed back into the output. From this we can see that
yt=(1-\alpha)xt+\alpha((1-\alpha)xt-T+\alphayt)
yt=(1-\alpha)xt+(\alpha-\alpha2)xt-T+\alpha2yt
etc.
### Economics
See also: time series analysis and simultaneous equations model. Recurrence relations, especially linear recurrence relations, are used extensively in both theoretical and empirical economics.[10] [11] In particular, in macroeconomics one might develop a model of various broad sectors of the economy (the financial sector, the goods sector, the labor market, etc.) in which some agents' actions depend on lagged variables. The model would then be solved for current values of key variables (interest rate, real GDP, etc.) in terms of past and current values of other variables.
## References
### Bibliography
• Book: Paul M. . Batchelder. An introduction to linear difference equations. 1967. Dover Publications.
• Book: Kenneth S. . Miller. Linear difference equations. 1968 . W. A. Benjamin.
• News: Jay P. . Fillmore . Morris L. . Marx. Linear recursive sequences . SIAM Rev. . 10. 3. 324–353 . 2027658. 1968.
• Book: Alfred . Brousseau. Linear Recursion and Fibonacci Sequences. 1971. Fibonacci Association.
• Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. Introduction to Algorithms, Second Edition. MIT Press and McGraw-Hill, 1990. . Chapter 4: Recurrences, pp. 62–90.
• Book: Ronald L. . Graham . Donald E. . Knuth. Oren . Patashnik. Concrete Mathematics: A Foundation for Computer Science. 2. Addison-Wesley. 1994 . 0-201-55802-5.
• Book: 2010 . Applied Econometric Times Series . Walter . Enders . 3 . dead . https://web.archive.org/web/20141110221941/http://eu.wiley.com/WileyCDA/WileyTitle/productCd-EHEP000338.html . 2014-11-10.
• Book: Paul . Cull . Mary . Flahive. Mary Flahive. Robbie . Robson. Difference Equations: From Rabbits to Chaos. Difference Equations: From Rabbits to Chaos. Springer . 2005 . 0-387-23234-6 . chapter 7.
• Book: Ian . Jacques. Mathematics for Economics and Business. limited . Fifth. Prentice Hall. 2006. 0-273-70195-9. 551–568. Chapter 9.1: Difference Equations.
• Web site: Tang . Minh . Tan . Van To. Using generating functions to solve linear inhomogeneous recurrence equations. 399–404 . 2006. Proc. Int. Conf. Simulation, Modelling and Optimization, SMO'06.
• Web site: Andrei D. . Polyanin. Difference and Functional Equations: Exact Solutions. at EqWorld - The World of Mathematical Equations.
• Web site: Andrei D. . Polyanin. Difference and Functional Equations: Methods. at EqWorld - The World of Mathematical Equations.
• Xiang-Sheng . Wang. Roderick . Wong. Asymptotics of orthogonal polynomials via recurrence relations. Anal. Appl. . 2012 . 10 . 2. 215–235 . 10.1142/S0219530512500108 . 1101.4371. 28828175.
• Web site: OEIS Index Rec. OEIS index to a few thousand examples of linear recurrences, sorted by order (number of terms) and signature (vector of values of the constant coefficients)
## Notes and References
1. Jacobson, Nathan, Basic Algebra 2 (2nd ed.), § 0.4. pg 16. |
## Solving Inequations
Inequations can be solved just like equations, except you need to learn one new important rule:
· If you multiply both sides of an inequation by ‘–1’, you need to swap the direction of the inequality symbol.
Swapping the direction of the inequality symbol means you change a “larger than” symbol to a “smaller than symbol” and vice versa.
> becomes < < becomes > ≤ becomes ≥ ≥ becomes ≤
When you solve inequations, the answer you get isn’t an exact number. Instead your answer is usually something like, “x is smaller than 5,” or, “y is larger than or equal to.”
Solve the following inequation for x: Solution So in our final answer we basically want x all by itself on one side of the inequation. To get to that stage we’ll need a few steps. At the moment, the x is on the left hand side of the equation. There’s also a 5 on that side – let’s get rid of the 5 by subtracting it from both sides: Now it’s a bit better, but we’ve still got problems. There’s a negative sign in front of the term with the x in it. Let’s deal with the negative sign. You can change the sign of a term by multiplying it by ‘–1’. Let’s do that to both sides of the equation: Uh-oh. What have I forgotten to do? Well, we’ve just multiplied both sides of the equation by ‘–1’ – we need to swap the inequality symbol around. So we need to change it from a “larger than or equal to” symbol to a “smaller than or equal to” symbol: OK, that’s better. Now it’s almost in the right format, we just would like to have the x completely by itself. At the moment it’s got a 3 in front of it. We can get rid of the 3 by dividing both sides by 3:
It may seem obvious to you, but a lot of students forget to think about what they’ve learnt in the weeks or months before an exam. Chances are that most of the questions in an exam are going to require you to use some of the stuff you’ve recently learnt to solve them.
Say that it’s been six weeks since your last exam, and for those last six weeks all you’ve done is study algebra, algebra and more algebra. You’re now sitting in a new exam and starting to read the first question. It’s pretty likely that to solve a lot of the questions in this exam, you’ll need to use some of that algebra. So remember that when you’re reading the questions.
Sometimes you’ll come across a question which you think you can solve without using stuff you’ve recently learnt. This is fine – often there are many different ways to solve a problem. However, your teacher probably wants you to demonstrate that you can use the material you’ve learnt recently. So, if you can, try and solve the problems using stuff you’ve just learnt. If you can’t do this, then any other way of solving the problem will do just fine.
### Using set notation to solve inequations
We’ve already learnt how to solve inequations, such as the following one:
It’s the same process as we use for normal equations, except we have to remember that when we multiply or divide both sides by ‘–1’, we need to swap the direction of the inequality symbol.
Now, there is a more formal mathematical way of writing down an inequality question, using sets. For instance, the last question would be written something like this:
You can read this as “Find the set of x, given that two x plus three is smaller than or equal to seven.” So we could solve this as we normally do:
### Expressing solutions to inequations on a number line
There is a graphical way of representing the answer to an inequation on the number line. By drawing an arrow along the number line, you can indicate the range of values that the variable can have. For instance, we would draw an arrow starting at 2, pointing in the negative direction:
Notice how I’ve used a small, solid black circle at the start of the arrow, above the ‘2’ on the number line. This tells the reader that the solution includes the number ‘2’. If the solution had been , then you’d draw the same arrow, starting at ‘2’, but you’d use a hollow circle to indicate that the set of x does not actually include ‘2’ itself:
The arrow also tells the reader that the set of x includes all the values in the direction the arrow points – so in this case all the negative values. We can also have a solution where the set of x is a fixed range, like this:
The number line representation for this solution won’t have any arrows in it, but will look like this:
Notice how the line is a finite length – it doesn’t have an arrow on either end pointing for ever in either direction. Instead, we have a solid circle above the ‘–2’ on the number line, telling us that the set of ‘x’ does include ‘–2’. The line runs towards the positive numbers, ending at above ‘3’ on the number line. The circle above ‘3’ is a hollow circle, telling us that the set of ‘x’ does not include ‘3’ itself, but the values leading up to it. |
# Solving Problems Involving Division of Fractions.pptx
28 de May de 2023
1 de 14
### Solving Problems Involving Division of Fractions.pptx
• 1. MATHEMATICS 5 Solving Problems Involving Division of Fractions WEEK 9
• 2. -Solve routine or non-routine problems involving division without or with any of the other operations of fractions and whole numbers using appropriate problem solving strategies and tools. (MELC-Math 5)
• 3. Review
• 4. Dividing Simple Fractions 𝟒 𝟓 ÷ 𝟐 𝟑 = 𝟒 𝟓 CCF Method Copy the dividend Change to multiplication x Flip (Get the reciprocal) 𝟑 𝟐 = 𝟏𝟐 𝟏𝟎 Copy-Change-Flip = 𝟏 𝟐 𝟏𝟎 = 𝟏 𝟏 𝟓
• 5. What are the four main steps in solving math problems? 1. Understand 2. Plan 3. Solve 4. Check 1. Study 2. Think 3. Act 4. Review
• 6. Division To check if division is the operation involved in solving a problem, you need to know the following key terms: quotient, divided by, cut, average, split into.
• 7. Romeo is celebrating his birthday! To prepare his lasagna, Romeo has 𝟑 𝟒 gallon of tomato sauce that needs to be divided into 5 bowls of the same size. How many gallons of tomato sauce should Romeo put into each bowl? UNDERSTAND 1. What is asked in the problem? The number of gallons of tomato sauce should Romeo put into each bowl. 2. What are given? 𝟑 𝟒 gallon of tomato sauce 5 bowls
• 8. PLAN 1. What is the operation will be used? Division 2. What is the mathematical sentence? 𝟑 𝟒 ÷ 5 = N Romeo is celebrating his birthday! To prepare his lasagna, Romeo has 𝟑 𝟒 gallon of tomato sauce that needs to be divided into 5 bowls of the same size. How many gallons of tomato sauce should Romeo put into each bowl?
• 9. SOLVE 𝟑 𝟒 ÷ 5= N Answer: 𝟑 𝟐𝟎 gallon of tomato sauce Romeo is celebrating his birthday! To prepare his lasagna, Romeo has 𝟑 𝟒 gallon of tomato sauce that needs to be divided into 5 bowls of the same size. How many gallons of tomato sauce should Romeo put into each bowl? 𝟑 𝟒 ÷ 𝟓 𝟏 = 𝟑 𝟒 x 𝟏 𝟓 = 𝟑 𝟐𝟎
• 10. CHECK Answer: 𝟑 𝟐𝟎 gallon of tomato sauce Romeo is celebrating his birthday! To prepare his lasagna, Romeo has 𝟑 𝟒 gallon of tomato sauce that needs to be divided into 5 bowls of the same size. How many gallons of tomato sauce should Romeo put into each bowl? 3 20 𝑥 5 = 3 20 𝑥 5 1 = 15 20 = 3 4
• 11. Five boxes of pizza were ordered. 3 𝟓 𝟖 of the pizza ordered were given to the visitors and the remaining slices of pizza were equally shared by 5 siblings. How much did each sibling get? UNDERSTAND 1. What is asked in the problem? The amount of pizza each sibling got. 2. What are given? Five boxes of pizza, 3 𝟓 𝟖 of the pizza, 5 siblings
• 12. PLAN 1. What are the operations will be used? Subtraction and Division 2. What is the mathematical sentence? (5 - 3 𝟓 𝟖 ) ÷ 5 = N Five boxes of pizza were ordered. 3 𝟓 𝟖 of the pizza ordered were given to the visitors and the remaining slices of pizza were equally shared by 5 siblings. How much did each sibling get?
• 13. SOLVE (5 - 3 𝟓 𝟖 ) ÷ 5 = N Answer: 𝟏𝟏 𝟒𝟎 of pizza each sibling got. 5 - 𝟑 𝟓 𝟖 =𝟒 𝟖 𝟖 -𝟑 𝟓 𝟖 = 𝟏𝟏 𝟖 x 𝟏 𝟓 = 𝟏𝟏 𝟒𝟎 Five boxes of pizza were ordered. 3 𝟓 𝟖 of the pizza ordered were given to the visitors and the remaining slices of pizza were equally shared by 5 siblings. How much did each sibling get? =1 𝟑 𝟖 ÷ 5
• 14. CHECK Answer: 𝟏𝟏 𝟒𝟎 of pizza each sibling got 11 40 𝑥 5= 11 40 𝑥 5 1 = 11 8 =1 3 8 Five boxes of pizza were ordered. 3 𝟓 𝟖 of the pizza ordered were given to the visitors and the remaining slices of pizza were equally shared by 5 siblings. How much did each sibling get? 1 8 +3 5 8 =4 8 8 =5 |
# Trigonometry
• Difficulty Level : Medium
• Last Updated : 30 May, 2022
Hipparchus, a Greek mathematician, introduced the idea of trigonometry, which is one of the most important branches of mathematics. Trigonometry is basically applicable to the Right-angled triangle. In this article, we’ll look at the relationship between the sides and angles of a right-angled triangle. The fundamentals of trigonometry describe six functions: sin, cos, tan, sec, cosec, and cot.
Trigonometry uses trigonometric ratios (which is the ratio of two sides of the triangle)to determine the angles and incomplete sides of a triangle. Angles are measured in either radians or degrees. Angles in trigonometry that are widely used are 0°, 30°, 45°, 60°, 90°, 15 °, and 18°.
### Basic Trigonometry
Basic Trigonometry covers, tangent, sine, and cosine these are the names given to the three ratios. It can then be expanded to include other ratios and trigonometry in the Cartesian Plane. These trigonometry studies will cover the following topics: Trigonometry in the Cartesian Plane, Graphs of Trigonometric Functions, and Trigonometric Identities are all topics covered in this course. Following are the articles that cover all the topics the same:
### Inverse Trigonometry
Inverse trigonometric functions are essentially the inverse functions of the fundamental trigonometric functions sine, cosine, tangent, cotangent, secant, and cosecant. These functions are also known as arcs functions, anti-trigonometric functions, and cyclometer functions. These trigonometry inverse functions are used to calculate the angle for each of the trigonometry ratios. In physics, mechanics, geometry, and navigation, inverse trigonometry functions are widely used.
### Applications of Trigonometry
Trigonometry is used to calculate the height of a structure or a mountain. Using trigonometric functions, the distance of a structure from the perspective and the elevation angle can easily calculate the height of a building. It does not have particular applications in solving functional problems, but it is used in a variety of fields. For example, trigonometry is used in the development of computer music: as you might know, sound moves in the form of waves, and this wave pattern is used in the development of computer music by passing it through a sine or cosine function. Below are a few examples of how trigonometry and its functions can be used:
My Personal Notes arrow_drop_up
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# Algebra
What is the length of the original rectangle?
The perimeter of a rectangle is equal to 40. If the length is halved and the width is divided by 3, the new perimeter is decreased by 24.
Select one:
a. 12
b. 8
c. 4
d. 16
1. 👍 0
2. 👎 0
3. 👁 148
1. P = Original perimeter = 40
P = 2 w + 2 l = 2 ( w + l )
40 = 2 ( w + l )
Divide both sides by 2
20 = w + l
w + l = 20
Subtract w to both sides
w + l - w = 20 - w
l = 20 - w
If the length is halved and the width is divided by 3 mean:
New lenth l1 = l / 2 = ( 20 - w ) / 2 = 10 - w / 2
New width w1 = w / 3
The new perimeter is decreased by 24 mean:
P1 = New perimeter = 40 - 24 = 16
P1 = 2 w1 + 2 l1 = 2 ( w1 + l1 )
16 = 2 ( w1 + l1 )
Divide both sides by 2
8 = w1 + l1
w1 + l1 = 8
w / 3 + 10 - w / 2 = 8
Subtract 10 to both sides
w / 3 + 10 - w / 2 - 10 = 8 - 10
w / 3 - w / 2 = - 2
2 w / 6 - 3 w / 6 = - 2
- w / 6 = - 2
Multiply both sides by - 6
( - 6 ) ∙ ( - w / 6 ) = ( - 2 ) ∙ ( - 6 )
w = 12
l = 20 - w = 20 - 12 = 8
Proof:
Original perimeter:
P = 2 w + 2 l = 2 ( w + l ) = 2 ∙ ( 12 + 8 ) = 2 ∙ 20 = 40
New lenth:
l1 = l / 2 = 8 / 2 = 4
New width:
w1 = w / 3 = 12 / 3 = 4
New rectangle will be the square. ( the square is a special case of the rectangle )
New perimeter:
P1 = 2 w1 + 2 l1 = 2 ( 4 + 4 ) = 2 ∙ 8 = 16
The length of the original rectangle:
l = 8
Answer b
1. 👍 0
2. 👎 0
posted by Bosnian
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## SBT Solution Lesson 3: Isosceles Triangle (C8 SBT Math 7 Horizons) – Math Book
SBT Solution Lesson 3: Isosceles triangle (C8 SBT Math 7 Horizon)
============
### Solve problem 1 page 49 SBT Math 7 Creative horizon episode 2 – CTST
Given triangle MNP is isosceles at M. Name the sides, base, vertex angle, and base angle of that isosceles triangle.
Detailed instructions for solving Lesson 1
Solution method
Use the definition of an isosceles triangle to name the sides, base, top angle, and base angle.
Detailed explanation
Triangle MNP bound at M has: side sides MN and MP, base side NP, top angle $$\widehat M$$, base angle $$\widehat N$$ and $$\widehat P$$
–>
— *****
### Solve problem 2 page 49 SBT Math 7 Creative horizon episode 2 – CTST
a) Is a triangle with two angles equal to $${60^o}$$ a triangle or not? Find the remaining angle of this triangle.
b) Is a triangle with two angles equal to $${45^o}$$ an isosceles triangle or not? Find the remaining angles of this triangle.
Detailed instructions for solving Lesson 2
Solution method
Use the definition of an isosceles triangle.
– Use the theorem that the sum of the three angles in a triangle is equal to $${180^o}$$ to calculate the remaining angles of the triangle.
Detailed explanation
a) The remaining angle is equal to: $${180^o} – {2.60^o} = {60^o}$$. This triangle is both an equilateral triangle and an isosceles triangle at its three vertices.
b) The remaining angle is equal to: $${180^o} – {2.45^o} = {90^o}$$. This triangle is both an isosceles triangle and a right triangle and is called an isosceles right triangle for short.
–>
— *****
### Solution 3 page 49 SBT Math 7 Creative horizon episode 2 – CTST
In Figure 6, calculate angle B and angle C knowing $$\widehat {{A^{}}} = {138^o}$$
Detailed instructions for solving Lesson 3
Solution method
Use the isosceles triangle property to find the measure of the angle you want to find
Detailed explanation
We have: $$\widehat B = \widehat C = \frac{{{{180}^o} – {{80}^o}}}{2} = {50^o}$$(because of triangle ABC weigh at A)
–>
— *****
### Solution 4 page 49 SBT Math 7 Creative horizon episode 2 – CTST
Given figure 7, know AB = AC and BE is the bisector of $$\widehat {ABC}$$, CF is the bisector of angle $$\widehat {ACB}$$. Prove that:
a) $$\Delta ABE = \Delta ACF$$
b) Isosceles triangle OEF
Detailed instructions for solving Lesson 4
Solution method
– Use the property of bisectors to show that two angles are congruent from proving two triangles are congruent.
– Proving OE = OF should deduce triangle OEF isosceles
Detailed explanation
a) we have AB = AC, so triangle ABC is isosceles at A, so $$\widehat B = \widehat C$$
Otherwise: $$\widehat {FCA} = \frac{{\widehat C}}{2}$$ (because CF is the bisector of angle $$\widehat {ACB}$$)
$$\widehat {EBA} = \frac{{\widehat B}}{2}$$ (because BE is the bisector of $$\widehat {ABC}$$)
So $$\widehat {FCA} = \widehat {EBA}$$
Consider triangle ACF and triangle ABE with:
$$\widehat {{A^{}}}$$general
AC = AB
$$\widehat {FCA} = \widehat {EBA}$$
Reason: $$\Delta ACF = \Delta ABE(g – c – g)$$
b) We have: $$\Delta ACF = \Delta ABE(g – c – g)$$ infer: BE = CF (1)
We have an isosceles triangle OBC at O, so OB = OC (2)
From (1) and (2) deduce BE – OB = CF – OC, so OE = OF
So triangle OEF is isosceles at O.
–>
— *****
### Solution 5 page 49 SBT Math 7 Creative horizon episode 2 – CTST
Given triangle MEF isosceles at M, there are $$\widehat M = {80^o}$$
a) Calculate $$\widehat E{,^{}}\widehat F$$
b) Let N, P be the midpoints of ME, MF, respectively. Prove that triangle MNP is isosceles.
c) Prove that NP // EF
Detailed instructions for solving Lesson 5
Solution method
Use the isosceles triangle property to find the measures of the angles
– Prove that MN = MP deduce the triangle MNP is isosceles at M
– Prove two angles $$\widehat {MNP} = \widehat {{\rm{NEF}}}$$ infer NP // EF
Detailed explanation
a) Since the triangle MEF is isosceles at M, $$\widehat E = \widehat F = \frac{{{{180}^o} – {{80}^o}}{2} = {50^o}$$
b) We have an isosceles triangle MEF at M so ME = MF.
Derive: $$MN = \frac{{ME}}{2} = \frac{{MF}}{2} = MP$$
So triangle MNP is isosceles at M.
c) In isosceles triangle MNP we have: $$\widehat N = \widehat P = \frac{{{{180}^o} – {{80}^o}}}{2} = {50^o}$$
so $$\widehat {MNP} = \widehat {{\rm{NEF}}} = {50^o}$$
Derive NP // EF (since two isotopic angles are equal)
–>
— *****
### Solution 6 page 50 SBT Math 7 Creative horizon episode 2 – CTST
Let ABC be an isosceles right triangle at A. The bisector of angle B intersects AC at N, the bisector of angle C intersects AB at M. Let O be the intersection of BN and CM.
a) Calculate the measure of angles OBC, OCB.
b) Prove that triangle OBC is isosceles.
c) Calculate the measure of angle BOC.
Detailed instructions for solving Lesson 6
Solution method
Use the bisector of an angle to calculate the angle measure.
– Prove that $$\widehat {OBC} = \widehat {OCB}$$ deduces triangle OBC isosceles at O.
– Use the theorem that the sum of the three angles in a triangle is equal to $${180^o}$$ to calculate the remaining angles of the triangle.
Detailed explanation
a) Since triangle ABC is right-angled at A, $$\widehat B = \widehat C = {45^o}$$
We have: $$\widehat {OBC} = \widehat {OCB} = \frac{{{{45}^o}}}{2} = 22,{5^o}$$
b) Triangle OBC has $$\widehat {OBC} = \widehat {OCB}$$ so triangle OBC is isosceles at O.
c) we have: $$\widehat {BOC} = {180^o} – \left( {\widehat {OBC} + \widehat {OCB}} \right) = {180^o} – {45^o} = {135^o}$$
–>
— ***** |
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# Unit 8: Circle Geometry
Introduction: Definitions
Tangent
Diameter
## the distance across a circle, measured through its
center; or the line segment that joins two points on
the circle and passes through the center.
Point of Tangency
## the distance or line segment from the center of a
circle to any point on the circle.
Central Angle
Chord(s)
Arc
B
Inscribed Angle
## An angle in a circle with its vertex and endpoints
of its arms on the circle.
Minor Arc
The shorter of two arcs between two points on a circle.
For example: AB
For example,
PQR
## Try to find the missing angles in the following diagrams
A)
B)
A tangent to a circle is perpendicular to the radius
at the point of tangency.
APO = BPO = 900
Example Problems
A)
## Point O is the center of a circle and AB is tangent to the circle. In
AOB = 550. Determine the measure of OBA.
OAB,
x = 48o
Since
A = 900 and
0 = 550
Then 90 + 55 = 145
The three angles in a triangle
145 = 350.
x = 76o
x = 180
Application Example
Since AC is a tangent
BDA = BDC = 900
Find x
x + 90 + 57 = 180
x + 147 = 180
x + 147 - 147 = 180 - 147
x = 33o
Find y
y + 90 + 35 = 180
y + 125 = 180
y + 125 - 125 = 180 - 125
y = 55o
1.
## 3. An airplane is cruising at an altitude of 9000m. A cross section of the earth is a
circle with a radius approximately 6400km. A passenger wonders how far she is from a
point H on the horizon she sees outside the window. Calculate the distance to the
nearest kilometer.
## Since BM is a tangent we know
that OBM = 900.
a 2 + b 2 = c2
d2 + 64002 = 64092
d2 + 40960000 = 41075281
d2 = 41075281 - 40960000
d2 = 115281
d= 115281
d = 339.5 km
a 2 + b 2 = c2
82 + b2 = 102
64 + b2 = 100
b2 = 100 64
b2 = 36
b= 36
b = 6 cm
## Try this one!
Since BM is a tangent we know
that OBM = 900.
2.
a 2 + b 2 = c2
122 + b2 = 162
144 + b2 = 256
b2 = 256 144
b2 = 112
b= 112
b = 10.6 cm
## 8.2 Properties of Chords in a Circle
In any circle with center O and chord AB:
x If OC bisects AB, then OC AB
x If OC AB, then AC = CB
x The perpendicular bisector of AB goes through the center O.
Remember:
Perpendicular means there is a 90o angle.
Bisector means it is divided into 2 equal parts
## If AC = 10cm, then BC =10cm
Example # 1
Example # 3 A chord MN is 24cm. The radius of a circle is 20cm. Find the length of x.
## Since the chord is 24cm, half it is 12cm. Use the
Pythagorean Theorem to find the missing side of
the triangle.
## Solution: Use the Pythagorean Theorem to solve
for BC
a 2 + b 2 = c2
62 + BC2 = 102
36 + BC2 = 100
BC2 = 100 36
b2 = 64
BC= 64
BC = 8 cm
a 2 + b 2 = c2
122 + b2 = 202
144 + b2 = 400
b2 = 400 144
b2 = 256
b= 256
b = 16 cm
AC= BC = 8cm
So the length of AB is 2 x 8cm = 16cm
Radius is 20 cm ALL the way around the circle!
Example # 2
## The length of x must be 20 cm 16 cm = 4 cm
The diameter of a circle is 18cm. A chord JK is 5cm from the center. Find the length of
the chord.
a 2 + b 2 = c2
52 + b 2 = 9 2
25 + b2 = 81
b2 = 81 25
b2 = 56
b= 56
b = 7.5 cm
If b = 7.5 cm, then the chord JK is 2 x 7.5cm =
15cm
Example # 4:
Finding Angle Measurements
x , y and z.
## Section 8.3 Properties of Angles in a Circle
Solution
Since OC bisects chord AB, OC is perpendicular to AB. Therefore, x = 90o
## The 3 angles in a triangle must add up to 180o.
The measure of a central angle is twice the measure of an inscribed angle subtended
by the same arc.
y + 30 + 90 = 180
y + 120 = 180
y + 120 - 120 = 180 - 120
y = 60o
Examples
Since radii are equal (OA = OB) and OAB is isosceles, z = 30o.
Remember that in an isosceles triangle the 2 base angles are equal.
#1
#1
#2
#3
Try These
A).
B).
x = 350
y = 360
## Inscribed Angles Property
Inscribed angles subtended by the same arc are equal.
Examples #1
## Angles in a Semicircle Property
ACB and ADB are inscribed angles
subtended by the same arc AB. So,
ACB = ADB. x = 220.
Central angle
angle
## Inscribed angles subtended by a
semicircle (half the circle) are right
angles. This means these angles
use the diameter.
Example #1 Find the missing angle measures.
AB.
AOB = 2 ACB
y = 2 22 y = 440
MIN is an inscribed angle subtended by a semicircle.
So, x = 900.
Since three angles in a triangle add to 1800,
#2
y + 90 + 40 = 180
y + 130 = 180
y + 130 130 = 180 130
y = 500
ACB and ADB are inscribed angles
subtended by the same arc AB. So,
ACB = ADB. x = 140.
Central angle
angle
AB.
AOB = 2 ACB
y = 2 14 y = 280
#3
## MIN is an inscribed angle subtended by a semicircle.
So, x = 900.
Since three angles in a triangle add to 1800,
## Since both inscribed angles are subtended
from the same arc as the central angle
ACB = |
###### Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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# Powers of i - Problem 1
Carl Horowitz
###### Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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Simplifying powers of i. So when we’re simplifying powers of i what we need to do is figure out what multiples of 4 are close to this, because we know that i to the 4th and any multiple of 4 there after are going to be the same thing as 1. There’s two ways of doing that. When we’re dealing with a simple number that we know we can just figure out what the closest multiple of 4 is. So in the case of 120 I know that 4 times 30, sorry, in the case of 121, I went a little bit a head of myself, I know that 4 times 30 is 120. Therefore I know that i to the 120 is just equal to 1 because it’s a multiple of 4. I have more i after that so this is just i to the 121 is just going to be 1 times i or i.
The other way to do this is we'll pretend that I didn’t know that 4 times 30 was 120 so therefore we’re close to 121. What you can do is just look at 121 divided by 4. Divide this out long division, 4 goes into 12 three times and we’re left with zero 1 and so doesn’t go into it at all, subtract zero leaves this with a remainder of 1. That remainder is the power of i we’re still concerned with. So this tells me, this remainder of 1 tells me i to the 121 is the same thing as i to the first which I know to be just i.
If I had a remainder of 3 I would know that we had, remainder of 3 would imply that we’re dealing with i to the third which I can remember is –i.
So two different ways if you have a number that is close to a number that you know is a multiple of 4, you can just take the difference, if you don’t know it, just divide it up and the remainder is the power of i you’re concerned with. |
Posted by: Alexandre Borovik | June 17, 2018
## DIalogue about the augmented matrix of a system of linear equations
This short dialogue started with someone posting on LinkedIn group Math, Math Education, Math Culture a link to the following blogpost:
Matrices are a common tool used in algebra. They provide a way to deal with equations that have commonly held variables. In this post, we learn some of the basics of developing matrices.
From Equation to Matrix
Using a matrix involves making sure that the same variables and constants are all in the same column in the matrix. This will allow you to do any elimination or substitution you may want to do in the future. Below is an example
Above we have a system of equations to the left and an augmented matrix to the right. If you look at the first column in the matrix it has the same values as the x variables in the system of equations (2 & 3). This is repeated for the y variable (-1 & 3) and the constant (-3 & 6).
The number of variables that can be included in a matrix is unlimited. Generally, when learning algebra, you will commonly see 2 & 3 variable matrices. The example above is a 2 variable matrix below is a three-variable matrix.
If you look closely you can see there is nothing here new except the z variable with its own column in the matrix.
Row Operations
When a system of equations is in an augmented matrix we can perform calculations on the rows to achieve an answer. You can switch the order of rows as in the following.
You can multiply a row by a constant of your choice. Below we multiple all values in row 2 by 2. Notice the notation in the middle as it indicates the action performed.
You can also add rows together. In the example below row 1 and row 2, are summed to create a new row 1.
You can even multiply a row by a constant and then sum it with another row to make a new row. Below we multiply row 2 by 2 and then sum it with row 1 to make a new row 1.
The purpose of row operations is to provide a way to solve a system of equations in a matrix. In addition, writing out the matrices provides a way to track the work that was done. It is easy to get confused even the actual math is simple
Conclusion
System of equations can be difficult to solve. However, the use of matrices can reduce the computational load needed to solve them. You do need to be careful with how you modify the rows and columns and this is where the use of row operations can be beneficial
What follows is a short exchange of comments:
AB: What is missed in this discussion of augmented [matrix] is the key issue: why do row operations lead to a solution of the original system of simultaneous linear equation? This crucial fact is stated, but not explained in any way.
Response: Well i wasn’t trying to be that deep. My goal is to support students who are learning algebra. Perhaps you can explain here why row operations work.
AB: It is a really simple but fundamental fact: if you have a system of simultaneous linear equations, then the following operations:
* swapping two equations
* adding a multiple of one equation to another equation
* multiplying an equation by a non-zero number
do not change (the set of) solutions; it is useful to start discussion of system of linear equations by proving this statement and emphasising that these operations are reversible; the proof crucially depends on this observation. Row operations on the augmented matrix are operations on the system of equations. If we do some row operations on the augmented matrix and then write a system of equations corresponding to the resulting matrix, the new system of equation in equivalent to the original one, that is, the two systems have exactly the same solutions. |
# Prove that the sum of the squares of the first n integers is 1/24 (2n) (2n+1) (2n+2)?
Relevance
now sum = 1/6(n*(n+1)* (2n+1) = 1/24(2n* (2n+2) * (2n+ 1) = 1/24 (2n) (2n+1) (2n+2)
Proved
extending it we get ((2n+2)P3)/4
• Raj K
Lv 7
∑n² = 1² +2²+3²+4²+................+n²
Using the identity
n³ −(n−1)³= n³ −(n ³−3n²+3n−1)
→n³ −(n−1)³= 3n²−3n+1
→(n−1)³ −(n−2)³= 3(n−1)²−3(n−1)+1
....................................
.................................
.......................................
→(1)³ −(0)³= 3(1)²−3(1)+1
n³ = 3∑n²−3∑n+∑1
→ 3∑n²= n³+3∑n−∑1 Since
=n³+3n(n+1)/2−n Since ∑n=n(n+1)/2 and ∑1=n
=n{n²+3(n+1)/2−1} Simplifying
=n(2n²+3n+3−1)/2 Simplifying
=n(2n²+3n+2)/2
=n(2n+1)(n+1)/2
or ∑n²=n(2n+1)(n+1)/6
or ∑n²=4×n(2n+1)(n+1)/(6×4) Multiply and divide by 4
=2n(2n+1)(2n+2)/(24)
=(2n)(2n+1)(2n+2)/(24)
=(1/24)(2n)(2n+1)(2n+2)/(24)
As before, we set up as follows:
sumSquares1.gif
Saying the sum to n is one term less than the sum to (n+1)
We expand the sums:
sumSquares2.gif
As expected, the cubic terms cancel, and we rearrange the formula to have the sum of the squares on the left:
sumSquares3.gif
Expanding the cube and summing the sums:
sumSquares4.gif |
# Trachtenberg system
The Trachtenberg system is a system of rapid mental calculation. The system consists of a number of readily memorized operations that allow one to perform arithmetic computations very quickly. It was developed by the Russian Jewish engineer Jakow Trachtenberg in order to keep his mind occupied while being held in a Nazi concentration camp. The rest of this article presents some methods devised by Trachtenberg. These are for illustration only. To actually learn the method requires practice. Students begin learning the Trachtenberg system using multiplication algorithms. These initial algorithms are discussed first followed by a more general method for multiplication. Even if you know well how to do arithmetic, the Trachtenberg method can be faster. It is also a method you may use to check work done by more traditional methods.
## Beginning to Multiply Using the Trachtenberg Method
When performing any of these multiplication algorithms the following "steps" should be applied.
Write a zero before the number to be multiplied.
The answer must be found one digit at a time starting at the least significant digit ie the digit to the far right. Move left one digit at the time to work. The last calculation is on the leading zero of the multiplicand. ( In 0128 x 6, the 0128 is the multiplicand and the 6 is the multiplier. In 0382 x 28, the 0382 is the multiplicand and the 28 is the multiplier. The answer may be referred to also as the "product".)
When working with the multiplicand, one digit at a time has your primary focus. The digit immediately to that digit's right is that digit's neighbor. In the Trachtenberg method, never use the digit immediately to the left as the neighbor. The rightmost digit does not have a neighbor as there is no digit to its right. Some algorithms ask for a value of a digit's neighbor. When the focus is on the last digit to the right, the value used for the neighbor is always zero.
The 'halve' operation has a particular meaning to the Trachtenberg system. If a digit is even take half this value. If a digit is odd, mentally subtract one before taking half the value. Alternatively half the odd digit may be taken discarding any decimal, fraction or remainder. For speed reasons people following the Trachtenberg system are encouraged to make this halving process instantaneous. Instead of thinking "half of seven is three and a half, so three" it's suggested that one thinks "seven, three". This speeds up calculation considerably.
The Trachtenberg system also uses something called the tens complement. In this same way the tables for subtracting digits from 10 or 9 are to be memorized. This requires subtracting digits from nine except the last digit to the right of a number which is subtracted from 10. The suggestion is to be able to look at the digits 0-9 and immediately know the result if this digit were subtracted from 9 or subtracted from 10.
Whenever the rule calls for adding half of the neighbor, always add 5 if the current digit is odd.
There will be more discussion on the reasons and details of these methods below.
### Multiplying by 10
It is easier to simply add a final zero to multiply by 10, however it is important to understand also how to multiply by 10 using a Trachtenberg type of technique. This technique will be used when Trachtenberg multiplies by 11, 12, 8, and 9.
We will use the multiplicand ab as an example. In the Trachtenberg method place a leading zero on the multiplicand. We will write the answer just below the multiplicand. Think of 0ab and the answer below it in columns. You will do a procedure to a digit in the multiplicand and this will give the digit in the answer directly below this digit in the multiplicand. The first column considered is that on the far right. In this case this column has a "b" in the multiplicand. The process proceeds from right to left.
0ab x 10 =
ab0
Rule: For each digit in the multiplicand take zero times that digit and add the neighboring digit immediately to the right in the multiplicand.
(Carry as you normally would in addition problems to the next column to the left if the value exceeds 9 ex if the answer is 10 write down the zero and carry the one in the next column to the left.)
In the first column on the right, start with zero x b. There is no neighbor to the right of b so consider the neighbor value to be zero. (0xb)+0=0 . Write down zero underneath the b in the answer.
In the second column from the right, use zero times a. The neighbor to the right is b. ( 0xa )+b= b so write b in this position ie in the second column in the answer.
In the third column from the right, which is the leading zero in the multiplicand, 0x the digit itself =0x0 = 0. The neighbor to the right is a. o+a=a . Write a as the answer in this position.
Notice that if done correctly the answer should be as expected the multiplicand moved one place to the left and with a zero in the ones position. This is likely how you were taught to multiply by 10 in early grades. 327x10 = 3270. 4268x10=42680
Notice we have made a big deal in that it is zero times the digit first. Why? Later the multiplier for each digit may not be zero so we are making sure to be clear that for this case it is in fact zero.
Now that we know how to multiply by 10 by the Trachtenberg method, what is the pay off? Taking one times the digit in the multiplicand and also following the multiplication by 10 rule i. e. adding the number to the right would result in multiplying by (1+10). This would be 11 times the multiplicand. Doubling each original digit then following the rule for multiplying by 10 would be equivalent to multiplying by (2+10). This would be 12 times the multiplicand. This is discussed in the next 2 sections.
It is would be possible to triple the original digit and then follow the 10 rule to multiply by 13, and quadruple the digit and then follow the 10 rule to multiply by 14. It brings up the possibility of 22 times the multiplicand as 2 times the digit itself plus 2 times the neighbor to the right. For these beginning multiplication algorithms Trachtenberg limits his memorized math facts down to multiplying by 2. After the beginner algorithms, Trachtenberg will go more in depth regarding extending his method to a general method for multiplication. Look for that in the later section on general multiplication.
Just realize with this algorithm that adding one times the neighboring digit to the right means adding 10 times your original multiplicand to your answer.
Example:
0 3 4 2 5 x 10 =
3 4 2 5 0
(= (0x0)+3) (=(0x3)+4) (= (0x4)+2) (= (0x2)+5) (= (0x5)+0)
### Multiplying by 11
Rule: Add the digit to its neighbor. (By "neighbor" we mean the digit on the right.)
Example:
03425 x 11 =
37675
Each of these digits comes from one times the digit in the multiplicand plus the neighbor ie the digit to the right. Read these from right to left.
(=0+3) (=3+4) (=4+2) (=2+5) (=5+0)
Why does this work? Because 11 = 1+10 . Using the digit itself and the moving the digit to the right over one column to the left (effectively multiplying that digit times ten) accomplishes (1+10 times the multiplier)
Thus,
03425+34250 = 37675
Note: after learning the basic method for the algorithms from right to left, some students practice also from left to right. This entails backing up and correcting for carrying so is mentally more difficult but it is possible if the problems involve higher numbers requiring carrying in the addition.
### Multiplying by 12
Rule: double each digit and add the neighbor. (The "neighbor" is the digit on the right.)
If the answer is greater than a single digit, simply carry over the extra digit (which will be a 1 or 2) to the next operation. The remaining digit is one digit of the final result.
Example:
0316 x12 =
3792
Determine neighbors in the multiplicand 0316:
• digit 6 has no right neighbor
• digit 1 has neighbor 6
• digit 3 has neighbor 1
• digit 0 (the prefixed zero) has neighbor 3
(6x2)+0= 12 Write the (2 and carry the 1)
(1x2)+6+1=9 Where did that final 1 come from? This is the 1 you carried from 12 in the column immediately to the right.
(3x2)+1= 7
(0x2)+3= 3
### Multiplying by 5
• Rule: to multiply by 5: Take half of the neighbor, then, if the current digit is odd, add 5.
An odd number = (1+an even number). 5 x an odd number = 5 x (1+ an even number ) = (5x1) + (5 x the even number) = 10 x 1/2 x an even number) .
An even number = (0 + an even number). 5 times an even number = 5 x ( 0 + an even number) = (5x0) + (5 x the even number) = 10 x 1/2 x the even number.
Notice that (10 x 1/2) x an even number may equivalently be calculated as 10 x (1/2 x an even number)
Handle the 1 portion of odd numbers when multiplying by 5 by adding 5 to the answer in that column. Handle the even portion of the number by waiting until your focus has moved one column to the left then add half the neighbor to the right. This last bit may also be described as adding half the neighbor, discarding any decimal, fraction or remainder. This fraction is not being thrown away. This additional portion was handled one column to the right from your current focus when you added 5 if the number was odd. You do not wish to count this portion twice.
Examples:
042×5=
210
Half of 2's neighbor, the trailing zero, is 0.
Half of 4's neighbor is 1.
Half of the leading zero's neighbor is 2.
043×5 =
215
Half of 3's neighbor is 0. Remember also to add 5 because 3 is odd
Half of 4's neighbor is 1. Remember to discard any fraction when you divide by 2 for this method.
Half of the leading zero's neighbor is 2.
093×5=
465
Half of 3's neighbor is 0, plus 5 because 3 is odd, is 5.
Half of 9's neighbor is 1, plus 5 because 9 is odd, is 6. Remember to discard any remainder when dividing by 2.
Half of the leading zero's neighbor is 4. Remember to discard any remainder or fraction when dividing by 2.
### Multiplying by 6
6= 1 + 5
One times the digit, and add in the multiplication by 5 rule.
• Rule: to multiply by 6: Take the digit, add 5 if the digit is odd, then add half of the neighbor to each digit, .
Example:
0357 × 6 =
2142
Working right to left,
7 + 5 (since 7 is odd)+0 since 7 has no neighbor to the right = 12. Write 2, carry the 1.
5 + 5 (since the starting digit 5 is odd) +half of 7 (3) + 1 (carried) = 14. Write 4, carry the 1.
3 + 5 (since 3 is odd) + half of 5 (2) + 1 (carried) = 11. Write 1, carry 1.
0 + 0 (since 0 is even) + half of 3 (1) + 1 (carried) = 2. Write 2.
Note: remember when taking half of a number in the Trachtenberg method to discard any remainder or fraction.
### Multiplying by 7
7 = 2 + 5
Double each digit and also add in the multiplication by 5 rule.
Rule: to multiply by 7:
1. Double each digit.
2. Add half of its neighbor.
3. If the digit is odd, add 5.
Example:
0523 × 7 =
3661
Double the 3, add half the neighbor (zero as there is no neighbor, and add 5 as 3 is odd. 11 so write 1 and carry 1.
(2x3) +(1/2 x 0) + 5 = 11
Double the 2, add half of 3 discarding any fraction. Two is even. Add in the one carried from the 11 total in the prior column.
(2x2) + (1/2 x 3) +(0) +(1) = 6
Double the 5, add half of 2, add 5 since the 5 is odd. 16 so record 6 and carry 1.
(2x5)+ (1/2 x 2)+(5) = 16
Double the 0, add half of 5 discarding any fraction. Zero is even. Add the carried 1 from the 16 total in the prior column.
(2x0) + (1/2 x 5)+ (0) + (1)=3
### Multiplying by 9
Multiplying by 9 is equivalent to multiplying by (-1 + 10).
This method also relies on the fact that adding a number to a problem then subtracting that same number results in no net change to the problem.
Lets multiply qrstx9 where q,r,s,and t are each digits in a 4 digit number. A leading zero is placed in front of qrst giving us 0qrst. Always remember to place a leading zero in front of the multiplicand (the number you are going to multiply).
0qrstx9 = (-1 x 0qrst)+ (10 x 0qrst)
How will you work with -1 x 0qrst? Try adding a multiple of 10 that is just larger than qrst ie a one followed by as many zeros as there are digits in your multiplicand excluding its leading zero. In this case this means we will use 10000 as qrst is a 4 digit number. If you add in 10000 this will change the problem unless you subtract this back off later.
10000
- 1 x 0qrst
+ 10 x 0qrst
-10000
_______
9 x 0qrst
In the initial step, you can see why each digit in qrst is subtracted from 9 except the right most digit which is subtracted from 10. This process completes 10000 - (1xqrst). This step is called finding the "tens complement". This procedure is not unique to the Trachtenberg method. See "Method of Complements" in Wikipedia.
Adding the digit to the right in the multiplicand 0qrst is equivalent to multiplying by ten and adding this to your answer.
Finally when working under the leading zero of the factor 0qrst, DO NOT try to create some sort of tens complement for this zero. Remember this is just a place holder and not part of your original multiplicand. Add the neighbor ie q of qrst minus one. You must remember to use q as a neighbor, otherwise you havent completely multiplied qrst by 10. Subtracting 1 in this final step is the equivalent of subtracting 10000. You added 10000 to perform the problem. It must be removed or the problem will be incorrect.
0qrst x 9 = q-1___(9-q)+r___(9-r)+s____(9-s)+t____(10-t)
Editors note: Read the underlines as separating the different digits.
Rule:
1. Subtract the right-most digit from 10.
1. Subtract the remaining digits from 9.
2. Add the neighbor.
3. For the leading zero, subtract 1 from the neighbor.
For rules 9, 8, 4, and 3 only the first digit ( the digit to the far right ) is subtracted from 10. After that each digit is subtracted from nine instead.
Example: 02,130 × 9 = 19,170
Working from right to left:
• (10 − 0) + 0 = 10. Write 0, carry 1.
• (9 − 3) + 0 + 1 (carried) = 7. Write 7.
• (9 − 1) + 3 = 11. Write 1, carry 1.
• (9 − 2) + 1 + 1 (carried) = 9. Write 9.
• 2 − 1 = 1. Write 1.
### Multiplying by 8
Rule:
1. Subtract right-most digit from 10.
1. Subtract the remaining digits from 9.
2. Double the result.
3. Add the neighbor.
4. For the leading zero, subtract 2 from the neighbor.
To understand what is going on here, understand that 8 = -2+10.
Again let's use qrst as an example 4 digit number.
- 1 x 0qrst
- 1 x 0qrst
+ 10 x 0qrst
__________
8 x 0qrst
How do you work with - 0qrst? You work with the tens complement. This means adding in a multiple of 10 with as many zeros as digits in your original multiplicand i. e. 4 for 0qrst. Since your working with -0qrst twice you must add this multiple of 10 twice. This is only to make the working easier. You must remove these later if your answer is to remain correct.
This is equivalent to:
+ 10000
- 1 x 0qrst
+ 10000
- 1 x 0qrst
+ 10 x 0qrst
-20000
__________
8 x 0qrst
This simplifies to:
2 x ( 10000 - 0qrst )
+ 10 x 0qrst
- 20000
__________
8 x 0qrst
So you find the 10s complement of a digit and double it. Add the neighbor to the right i. e. use the multiply by 10 procedure and add that. When working under the leading zero of the multiplicand, get the neighboring digit to the right in the multiplicand and subtract 2. This step subtracts the 20000 you added to work the problem. Use this neighbor to the right minus 2 as your left most digit in your answer.
0qrst x 8 = q-2 ____ {2 x (9-q)}+r ____ {2 x (9-r)}+s ____ {2 x (9-s)}+t ____ {2x(10-t)}
Editors note: Read the underlines as separating the different digits.
Example: 456 × 8 = 3648
Working from right to left:
• (10 − 6) × 2 + 0 = 8. Write 8.
• (9 − 5) × 2 + 6 = 14, Write 4, carry 1.
• (9 − 4) × 2 + 5 + 1 (carried) = 16. Write 6, carry 1.
• 4 − 2 + 1 (carried) = 3. Write 3.
### Multiplying by 4
4 = -1 + 5
-1 times a number means in the Trachtenberg method a tens complement is used.
5 times the number should suggest to you the multiply by 5s rule.
-1 X 0qrst
+5 x 0qrst
___________
4 x 0qrst
Your working this as
10000
- 0qrst
+5 x 0qrst
- 10000
_______________
4 x 0qrst
The 10000 - 0qrst is the tens complement. Subtract each digit from nine unless it is the digit to the far right. For the digit to the far right subtract it from 10. The leading zero in 0qrst is not used to find the tens complement. In other words, this is the same tens complement procedure you have been using.
Add to your answer i. e. to your tens complement, the multiplication by 5 procedure. Remember you add 5 if the number is odd and zero if even. Add half the number to the right discarding any remainder or fraction.
While working under the leading zero of your factor decrease your answer for half the neighbor by 1. You added 10000 to make the work easier. You must subtract that back out at the end or you will have changed the problem and end up with a wrong answer.
0qrst x 4 = (q-1) __(9-q) +5 if q is odd + (r/2) __ (9-r) +5 if r is odd + (s/2) __ (9-s) +5 if s is odd + (t/2) __ (10-t)+ 5 if t is odd
Remember in set of equations for the digits in this example when a number divided by 2 is written, discard any remainder or fraction.
Rule:
1. Subtract the right-most digit from 10.
1. Subtract the remaining digits from 9.
2. Add half of the neighbor, plus 5 if the digit is odd.
3. For the leading 0, subtract 1 from half of the neighbor.
Example: 346 * 4 = 1384
Working from right to left:
• (10 − 6) + Half of 0 (0) = 4. Write 4.
• (9 − 4) + Half of 6 (3) = 8. Write 8.
• (9 − 3) + Half of 4 (2) + 5 (since 3 is odd) = 13. Write 3, carry 1.
• Half of 3 (1) − 1 + 1 (carried) = 1. Write 1.
### Multiplying by 3
The general idea for multiplying by 3 is that 3 = -2 + 5.
- 2 x 0qrst
+ 5 x 0qrst
__________________
3 x 0qrst
Work this as follows:
2 x ( 10000
- 0qrst)
+ 5 x 0qrst
- 20000
__________________
3 x 0qrst
Find the tens complement and double it. Then follow the 5s rule. Remember that doubling the tens complement means you added 10000 twice to make the math easier. You must subtract that when calculating from the leading zero in 0qrst. This means you must subtract 2 from that leading number in your answer before you record it. This is usually expressed as taking 2 from half the neighbor to the right when calculating that last digit.
0qrst x 3 = (q-2) __2x(9-q)+5 if q is odd +(r/2) __ 2x(9-r) +5 if r is odd+(s/2) __ 2x(9-s)+5 if s is odd + (t/2) __ 2x(10-t)+ 5 if t is odd
Remember in set of equations for the digits in this example when a number divided by 2 is written, discard any remainder or fraction.
Rule:
1. Subtract the rightmost digit from 10.
1. Subtract the remaining digits from 9.
2. Double the result.
3. Add half of the neighbor, plus 5 if the digit is odd.
4. For the leading zero, subtract 2 from half of the neighbor.
Example: 492 × 3 = 1476
Working from right to left:
• (10 − 2) × 2 + Half of 0 (0) = 16. Write 6, carry 1.
• (9 − 9) × 2 + Half of 2 (1) + 5 (since 9 is odd) + 1 (carried) = 7. Write 7.
• (9 − 4) × 2 + Half of 9 (4) = 14. Write 4, carry 1.
• Half of 4 (2) − 2 + 1 (carried) = 1. Write 1.
### Multiplying by 2
• Rule: to multiply by 2, double each digit.
Further Note: If you really understand why you are doing the steps, you can create new procedures for yourself such as for multiplying by 15 by using the 10 rule plus the 5 rule ie add 5 if the number is odd and add one and a half times the neighbor discarding any fraction or remainder.
## General multiplication
The method for general multiplication is a method to achieve multiplications $a\times b$ with low space complexity, i.e. as few temporary results as possible to be kept in memory. This is achieved by noting that the final digit is completely determined by multiplying the last digit of the multiplicands. This is held as a temporary result. To find the next to last digit, we need everything that influences this digit: The temporary result, the last digit of $a$ times the next-to-last digit of $b$, as well as the next-to-last digit of $a$ times the last digit of $b$. This calculation is performed, and we have a temporary result that is correct in the final two digits.
In general, for each position $n$ in the final result, we sum for all $i$:
$a \text{ (digit at } i\text{ )} \times b \text{ (digit at } (n-i)\text{)}.$
People can learn this algorithm and thus multiply four digit numbers in their head – writing down only the final result. They would write it out starting with the rightmost digit and finishing with the leftmost.
Trachtenberg defined this algorithm with a kind of pairwise multiplication where two digits are multiplied by one digit, essentially only keeping the middle digit of the result. By performing the above algorithm with this pairwise multiplication, even fewer temporary results need to be held.
Example: $123456 \times 789$
To find the first digit of the answer:
The units digit of $9 \times 6 = 4$.
To find the second digit of the answer, start at the second digit of the multiplicand:
The units digit of $9 \times 5$ plus the tens digit of $9 \times 6$ plus
The units digit of $8 \times 6$.
$5 + 5 + 8 = 18$.
The second digit of the answer is $8$ and carry $1$ to the third digit.
To find the fourth digit of the answer, start at the fourth digit of the multiplicand:
The units digit of $9 \times 3$ plus the tens digit of $9 \times 4$ plus
The units digit of $8 \times 4$ plus the tens digit of $8 \times 5$ plus
The units digit of $7 \times 5$ plus the tens digit of $7 \times 6$.
$7 + 3 + 2 + 4 + 5 + 4 = 25 + 1$ carried from the third digit.
The fourth digit of the answer is $6$ and carry $2$ to the next digit.
2 Finger method
Trachtenberg called this the 2 Finger Method. The calculations for finding the fourth digit from the example above are illustrated at right. The arrow from the nine will always point to the digit of the multiplicand directly above the digit of the answer you wish to find, with the other arrows each pointing one digit to the right. Each arrow head points to a UT Pair, or Product Pair. The vertical arrow points to the product where we will get the Units digit, and the sloping arrow points to the product where we will get the Tens digits of the Product Pair. If an arrow points to a space with no digit there is no calculation for that arrow. As you solve for each digit you will move each of the arrows over the multiplicand one digit to the left until all of the arrows point to prefixed zeros.
A method of adding columns of numbers is presented in the book by Trachtenberg. His addition method uses intermediate totals. These are used in an L-shaped algorithm to check for accuracy without repeating the initial procedure. This method allows the precise column in which an error occurs to be identified. For the procedure to be effective, the different operations used in each stages must be kept distinct, otherwise there is a risk of interference.
## Division
Setting up for Division
Division in the Trachtenberg System is done much the same as in multiplication but with subtraction instead of addition. Splitting the dividend into smaller Partial Dividends, then dividing this Partial Dividend by only the left-most digit of the divisor will provide the answer one digit at a time. As you solve each digit of the answer you then subtract Product Pairs (UT pairs) and also NT pairs (Number-Tens) from the Partial Dividend to find the next Partial Dividend. The Product Pairs are found between the digits of the answer so far and the divisor. If a subtraction results in a negative number you have to back up one digit and reduce that digit of the answer by one. With enough practice this method can be done in your head.
## Publications
• Rushan Ziatdinov, Sajid Musa. Rapid mental computation system as a tool for algorithmic thinking of elementary school students development. European Researcher 25(7): 1105-1110, 2012 [1].
• The Trachtenberg Speed System of Basic Mathematics by Jakow Trachtenberg, A. Cutler (Translator), R. McShane (Translator), was published by Doubleday and Company, Inc. Garden City, New York in 1960.[1]
## References
1. ^ Trachtenberg, Jakow (1960). The Trachtenberg Speed System of Basic Mathematics. Translated by A. Cutler, R. McShane. Doubleday and Company, Inc. p. 270. |
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# 2.7: Exact Differential Equations
Consider the equation
$f(x,y) = C$
$f_x(x,y)\hat{\textbf{i}} + f_y(x,y)\hat{\textbf{j}} = 0$
We can write this equation in differential form as
$f_x(x,y)\, dx+ f_y(x,y)\, dy = 0$
Now divide by $$dx$$ (we are not pretending to be rigorous here) to get
$f_x(x,y)+ f_y(x,y) \dfrac{dy}{dx} = 0$
Which is a first order differential equation. The goal of this section is to go backward. That is if a differential equation if of the form above, we seek the original function $$f(x,y)$$ (called a potential function). A differential equation with a potential function is called exact. If you have had vector calculus, this is the same as finding the potential functions and using the fundamental theorem of line integrals.
Example 1
Solve
$4xy + 1 + (2x^2 + \cos y)y' = 0$
Solution
We seek a function $$f(x,y)$$ with
$f_x(x,y) = 4xy + 1 \;\;\; \text{and} f_y(x,y) = 2x^2 + \cos y$
Integrate the first equation with respect to $$x$$ to get
$f(x,y) = 2x^2y + x + C(y)$
Notice since $$y$$ is treated as a constant, we write $$C(y)$$. Now take the partial derivative with respect to y to get
$f_y(x,y) = 2x^2 + C'(y)$
We have two formulae for $$f_y(x,y)$$ so we can set them equal to eachother.
$2x^2 + \cos y = 2x^2 + C'(y)$
That is
$C'(y) = \cos\, y$
or
$C(y) = \sin \, y$
Hence
$f(x,y) = 2x^2y + x + \sin \, y$
The solution to the differential equation is
$2x^2y + x + \sin \, y = C$
Does this method always work? The answer is no. We can tell if the method works by remembering that for a function with continuous partial derivatives, the mixed partials are order independent. That is
$f_{xy} = f_{yx}$
If we have the differential equation
$M(x,y) + N(x,y)y' = 0$
then we say it is an exact differential equation if
$M_y(x,y) = N_x(x,y)$
Theorem (Solutions to Exact Differential Equations)
Let $$M$$, $$N$$, $$M_y$$, and $$N_x$$ be continuous with
$M_y = N_x$
Then there is a function $$f(x,y)$$ with
$$f_x = M$$ and $$f_y = N$$
such that
$f(x,y) = C$
is a solution to the differential equation
$M(x,y) + N(x,y)y' = 0$
Example 2
Solve the differential equation
$y + (2xy - e^{-2y})y' = 0$
#### Solution
We have
$M(x,y) = y \text{and} N(x,y) = 2xy - e^{-2y}$
Now calculate
$$M_y = 1$$ and $$N_x = 2y$$
Since they are not equal, finding a potential function $$f$$ is hopeless. However there is a glimmer of hope if we remember how we solved first order linear differential equations. We multiplied both sides by an integrating factor $$m$$. We do that here to get
$mM + mN_y' = 0$
For this to be exact we must have
$(mM)_y = (mN)_x$
Using the product rule gives
$m_yM + mM_y = m_xN + mN_x$
We now have a new differential equation that is unfortunately more difficult to solve than the original differential equation. We simplify the equation by assuming that either m is a function of only $$x$$ or only $$y$$. If it is a function of only $$x$$, then $$m_y = 0$$ and
$mM_y = m_xN + mN_x$
Solving for $$m_x$$, we get
$m_x = \dfrac{M_y-N_x}{N}$
If this is a function of $$y$$ only, then we will be able to find an integrating factor that involves $$y$$ only. If it is a function of only $$y$$, then $$m_x = 0$$ and
$m_yM + mM_y = mN_x$
Solving for $$m_y$$, we get
$m_y = \dfrac{N_x-M_y}{M} m$
If this is a function of $$y$$ only, then we will be able to find an integrating factor that involves $$y$$ only.
For our example
$m_y = \dfrac{N_x - M_y }{M} m = \dfrac{2y-1}{y} m = (2-\frac{1}{y})m$
Separating gives
$\dfrac{dm}{m} = (2-\frac{1}{y}) \,dy$
Integrating gives
$ln \, m = 2y - ln\, y$
$m = e^{2y - ln\, y} = y ^{-1}e^{2y}$
Multiplying both sides of the original differential equation by $$m$$ gives
$y(y ^{-1}e^{2y}) + (y ^{-1}e^{2y})(2xy - e^{-2y})y' = 0$
$e^{2y} + (2xe^{2y} - \frac{1}{y})y' = 0$
Now we see that
$M_y = 2e^{2y} = N_x$
Which tells us that the differential equation is exact. We therefore have
$f_x (x,y) = e^{2y}$
Integrating with respect to $$x$$ gives
$f(x,y) = xe^{2y} + C(y)$
Now taking the partial derivative with respect to y gives
$f_y(x,y) = 2xe^{2y} + C'(y) = 2xe^{2y} - \frac{1}{y}$
So that
$C'(y) = \frac{1}{y}$
Integrating gives
$C(y) = ln\, y$
The final solution is
$xe^{2y} + ln\, y = 0$ |
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## High school geometry
### Course: High school geometry>Unit 2
Lesson 1: Rigid transformations overview
# Finding measures using rigid transformations
Finding measures after a rigid transformation, like a reflection, is pretty simple! Since the shape and size stay the same, the lengths of corresponding sides and angle measures remain unchanged. Area and perimeter depend on the side lengths, so they stay the same too. So, if we know the measures of the original figure, we can use those same measures for the transformed figure.
## Want to join the conversation?
• Mind if I ask what the Pythagorean Theorem is? I don't recall Khan Academy explaining it,
so I'm a bit confused. Thanks for the help!
• First, let's define a few terms. In a right triangle, the two sides forming the right angle are called legs, and the side opposite the right angle is called the hypotenuse.
The Pythagorean Theorem states that for any right triangle, the sum of the squares of the two legs equals the square of the hypotenuse. This is quite an important theorem in geometry!
The theorem is usually expressed as the formula a^2 + b^2 = c^2, where a and b are the legs of the right triangle, and c is the hypotenuse of the right triangle.
Note well: this does not mean that a + b = c.
Example: if the legs of a right triangle are 3 and 4, then we can find the hypotenuse c by solving the equation 3^2 + 4^2 = c^2, which gives 9 + 16 = c^2, which gives 25 = c^2, which gives c = +-5. Since the hypotenuse c should not be negative, we discard c = -5. So the hypotenuse is c = 5.
• why do i not understand this
• Start from the fundamentals(very basics) and then go on higher level.
• If there are four angles and I have to find one of the angles, and I add those three angles equal over 180, what do I do?
• If a polynomial has 4 angles, it is a quadrilateral. As such, the angles inside of a quadrilateral add up to be 180(n - 2) where n is number of sides (or angles), so 180 (4-2) = 360. Thus, if you know 3 angles, add them together and subtract that total from 360.
• a little tip for those who don't know, if a right triangle has 3(x) and 4(x), the length of the hypotenuse will always be 5(x) (x can be any real number). I figured this out long ago before I started using khan academy.
• You're correct, of course. This is because you can prove the similarity of any two right triangles if they have two sides that are the same ratio apart and an included angle that is the same. Here, the right angle will be the same for every right triangle, and it is between the "3" side and the "4" side, so you can establish the similarity. From there, you can scale the sides and the hypotenuse however you want.
Nice thinking!
• What is the Pythagorean Theorem?
• The Pythagorean Theorem is the formula:
a^2 + b^2 = c^2
Where a and b are the legs of a right triangle and c is the hypotenuse of that triangle.
• why does he call the letters prime
• In the video:
ΔABC is reflected across line ℓ to form ΔA'B'C'
ΔABC is read "triangle A B C"
-and-
ΔA'B'C' is read "triangle A-prime B-prime C-prime"
See how the original triangle is called ΔABC, and the transformed triangle is called ΔA'B'C'?
We keep the point names (A B & C) the same, so that we can easily see how each point from the original triangle (ΔABC) corresponds to the points of the transformed triangle (ΔA'B'C').
And when we are speaking, we differentiate between the two triangles by using the suffix "prime" after each of the transformed triangle's points: for example, "A-prime B-prime C-prime". That way it is perfectly clear which triangle we are talking about.
Also, when you look at the graph, you will know that the original triangle is the one without the prime markings (ΔABC), and the transformed triangle is the one with the prime markings (ΔA'B'C').
Hope this helps!
• I have no clue how to doo this.
• its simple enough when you get the hang of it! let me walk you through it!
when you use rigid transformations, such as reflections, lengths and angles do not change. using this, we know that the reflection of triangle ABC will have the same lengths and angles as triangle A'B'C'. then, you can use given measures to figure out the questions! i suggest watching the video again to see how Sal goes through it!
hope this helped :)
• At , why did Sal refer to A'B'C' to 'a prime b prime c prime'? Is that how you pronounce that?
• Yes. The " ' " means "prime" or " 's image" in the context of "A's image"
• I didn't get how he found the measure with only one number available? And how do i do this on a triangle that most tests have that has no rights angles?
All the angles of a triangle will always add up to 180°. The measure of angle A is a right angle (90°), and the measure of angle C is 53°. Therefore, angle B is (180-90-53), which is 37°, and as angle B and angle B' are the same, the measure of angle B' is 37°. |
Identifying roots and critical pointsneed to editlesson 15. Stretch of y x 2 narrower than the graph of fx x, refl ected over the xaxis, translated up 5000 units. You can use the skills in this chapter to determine the maximum height of a ball thrown into the air. Many quadratic equations can be solved by factoring when the equation has a leading coefficient of 1 or if the equation is a difference of squares.
The table shows the linear and quadratic parent functions. Quadratic functions vocabulary quadratic function is a polynomial function with the highest degree of 2 for the variable x. Decreasing compresses the graph vertically and widens it. Transformation of quadratic functions day 2, video narrative, warm up. Characteristics of quadratic functions fill in the blanks and the y column of the chart. In lesson 51 you learned to identify linear functions. Learning from students voices a dissertation presented by jennifer suzanne stokes parent to the faculty of the graduate college of the university of vermont in partial fulfillment of the requirements for the degree of doctor of education specializing in educational leadership and policy studies. The xcoordinate of the vertex can be found using the formula b2a, and the ycoordinate of the vertex can be found by substituting the xcoordinate of the vertex into the function for x. Write quadratic functions in standard form and use the results. Identify the values of a, b, and c in the quadratic function y 3 x2. Transformations include reflections, translations both vertical and horizontal, expansions, contractions, and rotations.
If we know what the parent graph looks like, we can use transformations to graph any graph in that family. Teacher defines the parent function of a quadratic as begin mathsize. Quadratic functions also provide models for the shape of suspension bridge cables, television dish antennas, and the graphs of revenue and profit functions in business. The basics the graph of a quadratic function is a parabola. Such a function is characterized graphically as a parabola. Some quadratic equations will have complex solutions. Quadratic functions this unit investigates quadratic functions. You can also graph quadratic functions by applying transformations to the graph of the parent function fx x2. A parabola is a special, symmetrical curve which is one of the conic sections. The teacher may use the sheet as is or photocopied onto colored paper. In example 1, note that the coefficient a determines how. Learn vocabulary, terms, and more with flashcards, games, and other study tools. |
Factors
# Factors of 67 | Prime Factorization of 67 | Factor Tree of 67
Written by Prerit Jain
Updated on: 09 Jun 2023
## Factors of 67
Calculate Factors of
The Factors are
https://wiingy.com/learn/math/factors-of-67/
## What are the factors of 67
A factor of a number is a number that can divide the given number evenly. To find the factors of 67, you can divide 67 by each of the numbers that can be evenly divided into it without a remainder and a decimal point in the quotient.
67 is a prime number, which means that it has only two factors: 1 and itself. This means that the only numbers that can be evenly divided into 67 are 1 and 67.
Thus, the factors of 67 are 1 and 67.
## How to Find Factors of 67
Here are four methods you can use to find the factors of a number:
1. Factors of 67 using the Multiplication Method
2. Factors of 67 using the Division Method
3. Prime Factorization of 67
4. Factor tree of 67
## Factors of 67 Using the Multiplication Method
1. To find the factors of 67 using the multiplication method, you can start by listing the numbers from 1 to 67. Then, for each number in the list, you can multiply it by all of the other numbers to see if the product is equal to 67.
2. For example, you can start by multiplying 1 by 67. If the product is 67, then 1 and 67 are both factors of 67. You can then move on to the next number in the list and repeat the process.
3. Using this method, you will find that the only two numbers that multiply to equal 67 are 1 and 67. Therefore, the factors of 67 using the multiplication method are 1 and 67.
## Factors of 67 Using the Division Method
To find the factors of 67 using the division method, you can start by dividing 67 by each of the numbers that can be evenly divided into it. You can continue this process until you reach a number that 67 cannot be divided by evenly.
For example, if you start by dividing 67 by 1 and the result is an integer (a whole number), then 1 is a factor of 67. You can then move on to the next number in the list and repeat the process.
Using this method, you will find that the only two numbers that 67 can be evenly divided by are 1 and 67. Therefore, the factors of 67 using the division method are 1 and 67.
## Prime Factorization of 67
Calculate Prime Factors of
The Prime Factors of 67 =
67
https://wiingy.com/learn/math/factors-of-67/
The prime factorization of 67 is the expression of 67 as the product of its prime factors. Because 67 is a prime number, its prime factorization is simply 67 itself. This means that the prime factorization of 67 is written as 67.
To find the prime factorization of a number, you can express the number as the product of its prime factors. For example, the prime factorization of 15 is 3 x 5, because 15 can be expressed as the product of the prime numbers 3 and 5 (3 x 5 = 15).
The prime factorization of a number is written as the product of its prime factors. For example, the prime factorization of 67 is written as 67 x 1.
## Factor tree of 67
https://wiingy.com/learn/math/factors-of-67/
To create a factor tree for 67, you can start by finding two factors of 67 that multiply to equal 67. Since 67 is a prime number, it only has two factors: 1 and itself.
We can represent these factors as branches on a tree, like this:
67
/
1 67
This shows that 67 can be expressed as the product of 1 and 67. Since 1 and 67 are both prime numbers, we cannot find any other factors for them, so we can stop there. Our final tree would look like this:
67
/
1 67
This is the complete factor tree for 67.
## Factor Pairs of 67
Calculate Pair Factors of
1 x 67=67
So Pair Factors of 67 are
(1,67)
https://wiingy.com/learn/math/factors-of-67/
A factor pair of a number is a set of two factors that multiply together to produce that number. For 67, the only factor pair is (1, 67), since these are the only two factors of 67.
The factors of a number are the numbers that can be divided evenly into that number. For 67, the only factors are 1 and 67, since these are the only numbers that can be evenly divided into 67.
In general, the factor pairs of a number can be found by taking all of the factors of that number and pairing them up in all possible combinations. For 67, the only factor pair is (1, 67).
## Factors of 67 – Quick Recap
• Factors of 67: 1, 67.
• Negative Factors of 67: -1, -67.
• Prime Factors of 67: 1 and 67
• Prime Factorization of 67: 1 and 67
## Factors of 67 – Fun Facts
1. 67 is a prime number, which means it has only two factors: 1 and itself. This means that 67 is only divisible by 1 and 67.
2. The prime factorization of 67 is 67, which means that 67 is a prime number and cannot be written as the product of any other numbers.
3. The sum of the factors of 67 is 68, which is 1 + 67.
4. There are no even factors of 67 since 67 is an odd number and can only be divided by odd numbers.
5. 67 is not a perfect square, so it does not have any square factors.
Also Check: Multiples, Square Root, and LCM
## Solved Examples of Factor of 67
Q.1: What pairs of numbers multiplied together would equal sixty-seven?
Solution: Two numbers which when multiplied together would equal sixty-seven is 1×67; 1×67=67 only.
Q.2: If you divide sixty-seven by three, what will the remainder be?
Solution:
The remainder when dividing sixty-seven by three is one.
Q.3: How many even numbers remain between one and sixty-seven when all odd numbers are removed?
Solution:
Thirty-two even numbers remain between one and sixty-seven when all odd numbers are removed; these would include 2, 4, 6, 8 10, 12 14, 16 18, 20 22, 24 26, 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 & 66.
Q.4: Find the prime factorization of sixty-seven.
Solution:
The prime factorization of sixty-seven is 67 = 67 x 1; both being prime numbers.
Q.5: Chris needs to multiply three unequal numbers together in order to generate a total of eighty-four, which combination can he use?
Solution:
Three number combinations that can be used to multiply together to total eighty-four are 2 x 3 x 14 = 84, 1 x 4 x 21 = 84, and 6 x 7 x 2 = 84.
Q.6: Find the greatest common factor for twenty-nine and thirty-one.
Solution:
The greatest common factor for twenty-nine and thirty-one is one as neither can be divided evenly without a remainder over another number apart from themselves or one (29/31= 0.93548387).
Q.10:What pair of prime numbers can only be divided evenly with themselves and one in order to produce a total that equals fifty-three?
Solution:
Two prime numbers that can only be divided evenly with themselves and one in order to produce a total that equals fifty-three are 53 & 1; 53 x1=53and neither can be divided evenly with another number apart from themselves or one in order to equal fifty-three.
## Frequently Asked Questions on Factors of 67
### What is the Greatest Common Factor (GCF) of 67?
The greatest common factor (GCF) of 67 is 1, it’s the largest number which both can be divided without a remainder.
### How many factors does sixty-seven have?
Sixty-seven has two different factors; these include 1 and 67.
### Is 30 a multiple or a factor of 67?
30 is not a multiple or factor of sixty-seven as it cannot be divided evenly with no remainder (30/67 = 0.4477611940).
### How many odd numbers remain between 1-67 when all even numbers are removed?
Thirty-two odd numbers remain between one and sixty-seven when all even numbers are removed; these would include 1, 3, 5, 7, 9 11, 13 15, 17 19, 21 23, 25 27 29, 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 & 67.
### How many pairs of factors are needed in order to multiply together in order to generate fifty-three?
One pair of factors need multiplying together in order to generate fifty-three; these would include (53,1)
Written by by
Prerit Jain
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Examples
Chapter 14 Class 11 Probability
Serial order wise
Transcript
Example 10 Find the probability that when a hand of 7 cards is drawn from a well shuffled deck of 52 cards, it contains (i) all Kings 7 cards are to be chosen from 52 cards Total number of combinations (hands) possible = 52C7 = 52!/7!(52 −7)! = 52!/7!45! Let A be the event that all kings are selected There are only 4 kings in a pack of 52 cards Hence if 7 cards are chosen 4 kings to be chosen out of 4 and 3 others to be chosen out of remaining 48 Hence total number of combinations n(A) = 4C4 × 48C3 = 4!/4!0! × 48!/3!(48 −3)! = 1× 48/(3! 45!) = 48!/(3! 45!) Hence P (A) = (𝑛(𝐴))/(𝑛(𝑆)) = 48!/(3! 45!) ÷ 52!/(7! 45!) = (48! × 7!)/(3! × 52!) = 𝟏/𝟕𝟕𝟑𝟓 Example 10 Find the probability that when a hand of 7 cards is drawn from a well shuffled deck of 52 cards, it contains (ii) 3 Kings Let B be that event that 3 king are selected There are only 4 king in a pack of 52 cards Hence if 7 cards are chosen, 3 king to be chosen out of 4 and 4 other to be chosen out of remaining 48 Hence, number of combination n(B) = 4C3 × 48C4 = 4!/3!(4 −3)! × 48!/4!(48 −4)! = 4 × 48!/4!44! Hence, P(B) = (𝑛(𝐵))/(𝑛(𝑆)) = (4 × 48!/4!44!)/(52!/7!45!) = (4 × 48!)/4!44! × 7!45!/52! = (4 × 48! × 7! × 45!)/(4! × 44! × 52!) = (4 × 48! × 7! × 45!)/(4! × 44! × 52 × 51 × 50 × 49 × 48!) = (4 × 7! × 45)/(4! × 52 × 51 × 50 × 49 ) = (4 × 7 × 6 × 5 × 4! × 45)/( 52 × 51 × 50 × 49 × 4!) = 𝟗/𝟏𝟓𝟒𝟕 Example 10 Find the probability that when a hand of 7 cards is drawn from a well shuffled deck of 52 cards, it contains (iii) at least 3 Kings. Atleast 3 kings are selected means either 3 kings are selected or 4 kings are selected So, P(at least 3 king) = P(3 King) + ( 4 King) We know that, P(3 Kings) = 9/1547 P(4 Kings) = 1/7735 (calculated in (ii) part) (calculated in (i) part) P(at least 3 king) = 9/1547 + 1/7735 = 𝟒𝟔/𝟕𝟕𝟑𝟓 |
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### Number
An alternative method for finding a common denominator is to determine the least common multiple (LCM) for the denominators, then add or subtract the numerators as one would an integer. Using the least common multiple can be more efficient and is more likely to result in a fraction in simplified form. In the example above, the denominators were 4, 6, and 2. The least common multiple is the first shared multiple of these three numbers.The key thing to carrying out the addition of fractions correctly is to always keep in mind the most important part of the fraction is the number under the line, known as the denominator. If we have a situation where the denominators in the fractions involved in the addition process are the same, then we merely add the numbers that are above the separation line or as a mathematician would put it: "Adding the numerators only". We can have a look at an example of adding two fractions like 3⁄7 and 4⁄7. The expression would look like this: 3⁄7 + 4⁄7 = 7⁄7. In the case when the nominator is equal to the denominator, like in the foregoing example, it can also be equated to 1.
However, this was one of the easiest examples of adding fractions. The process may become slightly more difficult if we face a situation when the denominators of the fractions involved in the calculation are different. Nonetheless, there is a rule that allows us to carry out this type of calculations effectively. Remember the first thing: when adding the fractions, the denominators must always be the same, or, to put it in mathematicians language - the fractions should have a common denominator. In order to do that, we need to look at the denominator that we have. Here is an example: 2⁄3 + 3⁄5. So, we do not have a common denominator yet. Therefore, we use the multiplication table to find the number that is the product of the multiplication of 5 by 3. This is 15. So, the common denominator for this fraction will be 15. However, this is not the end. If we divide 15 by 3 we get 5. So, now we need to multiply the first fraction's numerator by 5 which gives us 10 (2 x 5). Also, we multiply the second fraction's denominator by 3 because 15⁄5 = 3. We get 9 (3 x 3 = 9). Now we can input all these numbers into the expression: 10⁄15 + 9⁄15 = 19⁄15 However, this was one of the easiest examples of subtracting fractions. The process may become slightly more difficult if we face a situation when the denominators of the fractions involved in the calculation are different. Nonetheless, there is a rule that allows us to carry out this type of calculations effectively. Remember the first thing: when subtracting the fractions, the denominators must always be the same, or, to put it in mathematicians language - the fractions should have a common denominator. In order to do that, we need to look at the denominator that we have. Here is an example: 3⁄3 - 2⁄5. So, we do not have a common denominator yet. Therefore, we use the multiplication table to find the number that is the product of the multiplication of 5 by 3. This is 15. So, the common denominator for this fraction will be 15. However, this is not the end. If we divide 15 by 3 we get 5. So, now we need to multiply the first fraction's numerator by 5 which gives us 10 (2 x 5). Also, we multiply the second fraction's denominator by 3 because 15⁄5 = 3. We get 9 (3 x 3 = 9). Now we can input all these numbers into the expression: 9⁄15 - 10⁄15 = -1⁄15 (Source: goodcalculators.com)
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Polls and Margin of Error Section 4.5 Public opinion polls are used to gather information about a population’s views, expectations, likes and dislikes. Sampling and Inferential Statistics. The larger the sample, the more likely it will reflect the population. However, costs and feasibility are major factors in determining size of a sample. If 100 people polled and 79 responded yes to the question, “Is there too much violence on TV?” how confident is a pollster with his or her results? Do the sample data reflect the general population? Does that mean that 79% of the entire population think there is too much violence on TV? By gathering data from different samples, another 100 might answer 72 yes or 68 yes to violence on TV. The set of all sample proportions and probabilities can be represented by a bell-shaped curve. A sample estimate is not 100% accurate, although it comes close to the true population, there is an error associated with the sample. Using the bell curve, we can predict the probable error of a sample estimate. ______________________|_______________________ x/n True Population Sample Proportion proportion In order to predict error, there is new terminology used In a standard normal distribution, we use a z-distribution body table. Alpha = a = area from P(0 < z < z a ) z a is the value of z that yields area a . P(0 < z < z a ) = a Using a differentiates sample proportion from other normal distributions. Example 1: Find a) z0.3810 means find P(0 < z < z0.3810 ) = 1.18 on the body table b) z0.47 The body table does not have 0.47 exactly, but does have 0.4699, so we use this value of z = 1.88 c) z0.4954 .4954 comes between .4953 and .4955 so half-way between z = 2.60 and z = 2.61 will yield z = 2.605 Sample proportions x/n vary from sample to sample; some will have small errors and some large errors. Knowing that errors are inherent with sample estimates, the margin of error (MOE) is a prediction of errors with sample estimates. MOE = z a /2 The reason we use a/2 is that the body table uses .5 data. 2 MOE depends on two factors, sample size, n, and level of confidence a . Example 2: a) Find MOE when 500 people are polled and level of confidence is 90%. N = 500, a = .90 , a /2 = .45 z .45 = 1.645 MOE = 1.645 = .0372 ~ 3.7% Margin of Error 2 b) Find MOE when 1000 people are polled and level of confidence is 90%. N = 1000, a = .90, a /2 = .45 z .45 = 1.645 MOE = 1.645 = .0260 ~ 2.6% Margin of Error 2 Notice the larger n is (sample size), the smaller the MOE. Example 3: In a survey of 500 undergraduate students, 410 responded that they will graduate in 4 years. a) Find the sample proportion that responded to graduating in 4 years. 410/500 = 82% b) At 95% confidence level, find MOE. a = .95 a /2 = .475 z.475 = 1.96 n = 500 MOE = 1.96 = .0438 ~ 4.4% 2 So we are 95% confident that 82% 4.4% believe they will graduate in 4 years. From 77.6% to 86.4% believe they will graduate in 4 years. Back to Statistics Main Page Back to the Survey of Math Ideas Home Page e-mail Questions and Suggestions |
# Solve distance formula
It’s important to keep them in mind when trying to figure out how to Solve distance formula. We can solve math problems for you.
## Solving distance formula
The best way to Solve distance formula is to eliminate as many options as possible. In order to solve for slope, you need to use the formula: One of the most common problems with slope is that people lose track of the units. The formula is easy to remember once you realize that it is just like a proportion: % change divided by 100. So if your house value increased by \$100, then your slope would be 50%. If your house value decreased by \$100, then your slope would be -50%. In the case of your house value increasing or decreasing by \$100, you'd have a slope of 0%. 0% slope means no change in value. Of course, in real life there are many other factors that might contribute to value changes, so this simple formula only gives you a rough estimate of how much your house has changed relative to the rest of the area.
The problem solver is a value that is “solved” by the user. It can be a numeric value, or a set of values that are represented by text. If a numeric value is used as the problem solver, it should always be positive. If a set of values is used as the problem solver, then the values must be separated by commas. For example: ProblemSolver>3.2/ProblemSolver>. The problem solver is an important part of any mobile app because it allows users to interact with your app in a way that best suits their current level of skill and knowledge. By setting up the problem solver in this way, you can help users solve problems they might be having while they’re using your app, which will hopefully increase engagement and retention rates.
Solve for x right triangles by using a Pythagorean formula. This calculator is useful for determining the length of a side of a right triangle, known as the hypotenuse. The Pythagorean relationship between sides x and y is: The ratio or proportion between sides x and y is given by: Substituting this into the above equation gives: or in other words: This can be simplified further as shown below: Therefore, solving for "x" right triangles involves applying this formula to any right triangle with lengths equal to 1, 2, and 3. If the hypotenuse (AB) is known then "x" can be determined from the equation. For example, if AB = 9 then "x" = 9. On the other hand, if AB = 16 then "x" = 16. For example, if AB = 12 then "x" = 12.
A good math help calculator is one that you can use to solve math problems and find out the answer right away. These calculators are often very simple, with just a few buttons and a display screen. They can be used to solve basic math problems, like addition and subtraction, as well as more advanced calculations like multiplication and division. TIP: If you're new to using a calculator, start with simple operations like adding two numbers together or subtracting one number from another. This will help you learn how to use the different buttons on the calculator accurately. Once you know how to use it, move on to more complex calculations like multiplication and division. This will allow you to work on your math skills in small steps and give you an idea of what it's like to solve real-world mathematical problems.
Good but needs the ability to go through and understand word problems. May be difficult to add but would significantly increase performance of the app. Overall well designed and function excellently and has assisted me for years. 5 Stars.
Flore Turner
It is a very useful app and helped me understand some of the questions I would get caught on. Also helps with remembering ones if I forget. Please can you add concise math 8 standard text book. I have voted many times buttonlike is my text book so please add it
Natalia Nelson
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# What is a multiplication fact table?
## What is a multiplication fact table?
A multiplication chart is a table that shows the products of two numbers. Usually, one set of numbers is written on the left column and another set is written as the top row. The products are listed as a rectangular array of numbers. Multiplication is repeated addition. There are 3 groups of 4 butterflies each.
What are the 12 multiplication facts?
12 Times Table
• 12 Times Table Chart:
• 12 Times Table Chart Up To 10: 12 x 1 = 12. 12 x 2 = 24. 12 x 3 = 36. 12 x 4 = 48. 12 x 5 = 60. 12 x 6 = 72. 12 x 7 = 84. 12 x 8 = 96. 12 x 9 = 108.
• 12 Times Table Chart Up To 20. 12 x 11 = 132. 12 x 12 = 144. 12 x 13 = 156. 12 x 14 = 168. 12 x 15 = 180. 12 x 16 = 192. 12 x 17 = 204. 12 x 18 = 216.
### What is a multiplication fact in math?
What Is A Multiplication Fact? A multiplication fact is the product of two specific numbers. And the order in which the numbers are presented does not change the product. For example, 2×3=6 and 3×2=6. Nowadays, multiplication facts are often taught as fact families with their opposite operation, division.
How do you teach 12 multiplication facts?
There are a few ways to look at these multiplications:
1. Adding 12 each time is a common method: 12+12= 24, 24 + 12 = 36, 36+12 = 48.
2. Also notice the pattern in the ones columns: 12 24 36 48 60 72 84 96 108 120 the 0, 2, 4, 6, 8, 0 pattern repeats through all the 12 x tables.
## How do you write a 12 table?
What is a 12 times table? The table of 12 is given by: 12 times 1 is 1, 12 times 2 is 24, 12 times 3 is 36, 12 times 4 is 48, 12 times 5 is 60, 12 times 6 is 72, 12 times 7 is 84, 12 times 8 is 96, 12 times 9 is 108 and 12 times 10 is 120.
What is a fact family for multiplication?
Fact family: It is a set of four related multiplication and division facts that use the same three numbers. For example: The fact family for 3, 8 and 24 is a set of four multiplication and division facts. |
Online Calculator Resource
# LCD Calculator - Least Common Denominator
Least Common Denominator (LCD)
LCD = 8
Equivalent Fractions with the LCD
3/4
=
6/8
3/8
=
3/8
Solution:
Rewriting input as fractions if necessary:
3/4, 3/8
For the denominators (4, 8) the least common multiple (LCM) is 8.
LCM(4, 8)
Therefore, the least common denominator (LCD) is 8.
Rewriting the original inputs as equivalent fractions with the LCD:
6/8, 3/8
## Calculator Use
Use this Least Common Denominator Calculator to find the lowest common denominator (LCD) of fractions, integers and mixed numbers. Finding the LCD is important because fractions need to have the same denominator when you are doing addition or subtraction math with fractions.
## What is the Least Common Denominator?
The least common denominator (LCD) is the smallest number that can be a common denominator for a set of fractions. Also known as the lowest common denominator, it is the lowest number you can use in the denominator to create a set of equivalent fractions that all have the same denominator.
## How to Find the LCD of Fractions, Integers and Mixed Numbers:
To find the least common denominator first convert all integers and mixed numbers (mixed fractions) into fractions. Then find the lowest common multiple (LCM) of the denominators. This number is same as the least common denominator (LCD).You can then write each term as an equivalent fraction with the same LCD denominator.
### Steps to find the LCD of fractions, integers and mixed numbers
1. Convert integers and mixed numbers to improper fractions
2. Find the LCD of all the fractions
3. Rewrite fractions as equivalent fractions using the LCD
### Example Using the Lowest Common Denominator Calculator
Find the LCD of: 1 1/2, 3/8, 5/6, 3
• Convert integers and mixed numbers to improper fractions.
3/8 and 5/6 are already fractions so we can use those as they are written.
1 1/2 is the same as (1/1) + (1/2). Using the formula for adding fractions, ((n1*d2)+(n2*d1)) / (d1*d2), we get ((1*2)+(1*1)) / (1*2) = 3/2.
3 can be rewritten as a fraction as 3/1
• Equivalent fractions are: 3/2, 3/8, 5/6, 3/1
• Now find the least common denominator (LCD) (or the least common multiple (LCM) of the denominators)
• LCD = 24
• Rewriting the fractions as equivalent fractions using the LCD
• 36/24, 9/24, 20/24, 72/24
## Related Calculators
We also have calculators for least common multiple, math with fractions, simplifying fractions, math with mixed numbers, and comparing fractions.
Cite this content, page or calculator as:
Furey, Edward "LCD Calculator - Least Common Denominator"; from https://www.calculatorsoup.com - Online Calculator Resource. |
# SSS Congruence Theorem and Its Proof
Many high textbooks consider the congruence theorems (SSS Congruence Theorem, SAS Congruence Theorem, ASA Congruence Theorem) as postulates. This is because their proofs are complicated for high school students. However, let us note that strictly speaking, in Euclidean Geomtery (the Geometry that we learn in high school), there are only five postulates and no others. All of other postulates mentioned in textbooks aside from these five are really theorems without proofs.
In this post, we are going to prove the SSS Congruence Theorem. Recall that the theorem states that if three corresponding sides of a triangle are congruent, then the two triangles are congruent.
Before proving the SSS Congruence theorem, we need to understand several concepts that are pre-requisite to its proof. These concepts are isometries particulary reflection and translation, properties of kites, and the transitive property of congruence. If you are familiar with these concepts, you can skip them and go directly to the proof.
##### Isometry
In the figure below, $\triangle MNP$ is slid to the right forming $\triangle M^\prime N^\prime P^\prime$. Clearly, when you side a figure, the size and shape are preserved, so clearly, the two triangles are congruent.
Sliding or translation is a form of isometry, a type of mapping that preserves distance. In the isometry above, the preimage $\triangle MNP$ is mapped onto the image $\triangle M^\prime N^\prime P^\prime$. Notice that there is a 1-1 mapping between the objects in the preimage and the objects in the image. Each object in the preimage has exactly one image. Also, each object in the image has exactly one preimage.
##### Properties of Kites
A kite is a polygon with two distinct pairs of congruent sides. In the figure below, $ABCD$ is a kite with $\overline{AB} \cong \overline{AD}$ and $\overline{BC} \cong \overline{CD}$.
The diagonal $\overline{AC}$ is a line of symmetry of the kite. This means that $\triangle ABC$ mirrors $\triangle ADC$. Mirroring an image or reflection preserves distance. This means that $\triangle ABC$ and $\triangle ADC$ congruent. Thus, we say that a kite is reflection-symmetric.
##### The Transitive Properties of Congruence
Let us recall the transitive property of equality of real numbers. It says that for any real numbers $a$, $b$, and $c$, if $a = b$ and $b = c$, then $a = c$. This is also true in congruence. For any figure $P$, $Q$ and $R$,
If $P \cong Q$ and $Q \cong R$, then $P \cong R$.
Now that we finished the prerequisite, we now prove the theorem.
##### The SSS Congruence Theorem
If in two triangles, three sides of one are congruent to three sides of the other, then the two triangles are congruent.
Proof
Given triangles $ABC$ and $DEF$ with $AB \cong DE$, $AC \cong DF$ and $AC \cong DF$.
In proving the theorem, we will use the transitive property of congruence. We show that if a third triangle exists, and $\triangle ABC$ is congruent to it, then $\triangle DEF$ is also congruent to it. Let the third triangle be $\triangle A^\prime B^\prime C^\prime$, an image of $\triangle ABC$ under an isometry.
To begin, since $\overline{AB} \cong \overline{DE}$, there is an isometry that maps $\overline{AB}$ to $\overline{DE}$. So, there is a triangle $A^\prime B^\prime C^\prime$ which is an image of $\triangle ABC$ that has a common side with $\triangle DEF$.
Now, $\overline{AC} \cong \overline{A^\prime C^\prime}$ and $\overline{BC} \cong \overline{B^\prime C^\prime}$. By the transitive property of congruence, $\overline {A^\prime C^\prime} \cong \overline{DF}$ and $\overline{B^\prime C^\prime} \cong \overline{EF}$. Therefore, $\triangle A^\prime B^\prime C^\prime$ and $\triangle DEF$ form a kite. Since this kite is reflection-symmetric over line $DE$, $\triangle DEF$ is a reflection of $\triangle A^\prime B^\prime C^\prime$ which means that $\triangle A^\prime B^\prime C^\prime \cong \triangle DEF$.
So, if the three pairs of sides of $\triangle ABC$ can be mapped onto $\triangle DEF$ by an isometry, by the definition of congruence, $\triangle ABC \cong \triangle DEF$. This prove the SSS Congruence Theorem.
Reference: An old edition of Geometry (University of Chicago School Mathematics Project) |
# The set of values of a for which the inequality, x2+ax+a2+6a<0 is satisfied for all xbelongs(1,2) lies in the interval:
A
(1,2)
B
[1,2]
C
[7,4]
D
none of these
Video Solution
Text Solution
Generated By DoubtnutGPT
## To solve the inequality x2+ax+a2+6a<0 for all x in the interval (1,2), we will follow these steps:Step 1: Define the functionLet f(x)=x2+ax+a2+6a.Step 2: Evaluate the function at the endpoints of the intervalWe need to ensure that f(x)<0 for all x in (1,2). Therefore, we will evaluate f(1) and f(2).Evaluate f(1):f(1)=12+a(1)+a2+6a=1+a+a2+6a=a2+7a+1We require:a2+7a+1<0(1)Evaluate f(2):f(2)=22+a(2)+a2+6a=4+2a+a2+6a=a2+8a+4We require:a2+8a+4<0(2)Step 3: Solve the inequalitiesNow we will solve both inequalities (1) and (2).Solve inequality (1):The quadratic inequality a2+7a+1<0 can be solved by finding its roots using the quadratic formula:a=−b±√b2−4ac2a=−7±√72−4⋅1⋅12⋅1=−7±√49−42=−7±√452=−7±3√52The roots are:a1=−7−3√52,a2=−7+3√52The quadratic opens upwards (since the coefficient of a2 is positive), so a2+7a+1<0 is satisfied between the roots:−7−3√52<a<−7+3√52Solve inequality (2):Similarly, for a2+8a+4<0:a=−8±√82−4⋅1⋅42⋅1=−8±√64−162=−8±√482=−8±4√32=−4±2√3The roots are:b1=−4−2√3,b2=−4+2√3The quadratic opens upwards, so a2+8a+4<0 is satisfied between the roots:−4−2√3<a<−4+2√3Step 4: Find the intersection of the intervalsWe need to find the intersection of the two intervals:1. (−7−3√52,−7+3√52)2. (−4−2√3,−4+2√3)Step 5: ConclusionThe set of values of a for which the inequality x2+ax+a2+6a<0 is satisfied for all x∈(1,2) lies in the intersection of these two intervals.
|
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# If the product of two consecutive even integers is 76 less than 10 times their sum, how do you find the two integers?
Apr 25, 2016
There are two possible solutions - $4$ and $6$ or $14$ and $16$
#### Explanation:
First put the statement into mathematical terms. Let $x$ be the first integer, then the second one is $x + 2$.
Then the product $x \left(x + 2\right) = 10 \left(x + \left(x + 2\right)\right) - 76$
Expanding,
${x}^{2} + 2 x = 20 x + 20 - 76$
${x}^{2} - 18 x + 56 = 0$
$\left(x - 14\right) \left(x - 4\right) = 0$
Then $x = 14$ or $x = 4$ and the second integer is $16$ or $6$ |
# Circumcircle of a Triangle
The circumcircle of a triangle is the circle that passes through all three vertices of the triangle. The construction first establishes the circumcenter and then draws the circle. circumcenter of a triangle is the point where the perpendicular bisectors of the sides intersect. This page shows how to construct (draw) the circumcircle of a triangle with compass and straightedge or ruler. This construction assumes you are already familiar with Constructing the Perpendicular Bisector of a Line Segment.
## Printable step-by-step instructions
The above animation is available as a printable step-by-step instruction sheet, which can be used for making handouts or when a computer is not available.
## Proof
The image below is the final drawing above with the red labels added.
Note: This proof is almost identical to the proof in Constructing the circumcenter of a triangle.
Argument Reason
1 JK is the perpendicular bisector of AB. By construction. For proof see Constructing the perpendicular bisector of a line segment
2 Circles exist whose center lies on the line JK and of which AB is a chord. (* see note below) The perpendicular bisector of a chord always passes through the circle's center.
3 LM is the perpendicular bisector of BC. By construction. For proof see Constructing the perpendicular bisector of a line segment
4 Circles exist whose center lies on the line LM and of which BC is a chord. (* see note below) The perpendicular bisector of a chord always passes through the circle's center.
5 The point O is the circumcenter of the triangle ABC, the center of the only circle that passes through A,B,C. O is the only point that lies on both JK and LM, and so satisfies both 2 and 4 above.
5 The circle O is the circumcircle of the triangle ABC. The circle passes through all three vertices A, B, C
* Note
Depending where the center point lies on the bisector, there is an infinite number of circles that can satisfy this. Two of them are shown on the right. Steps 2 and 4 work together to reduce the possible number to just one.
## Try it yourself
Click here for a printable worksheet containing two triangle circumcircle problems. When you get to the page, use the browser print command to print as many as you wish. The printed output is not copyright. |
# Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.4
Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 3 Differentiation Ex 3.4 Questions and Answers.
## Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.4
1. Find $$\frac{d y}{d x}$$ if:
Question 1.
√x + √y = √a
Solution:
√x + √y = √a
Differentiating both sides w.r.t. x, we get
Question 2.
x3 + y3 + 4x3y = 0
Solution:
x3 + y3 + 4x3y = 0
Differentiating both sides w.r.t. x, we get
Question 3.
x3 + x2y + xy2 + y3 = 81
Solution:
x3 + x2y + xy2 + y3 = 81
Differentiating both sides w.r.t. x, we get
2. Find $$\frac{d y}{d x}$$ if:
Question 1.
y.ex + x.ey = 1
Solution:
y.ex + x.ey = 1
Differentiating both sides w.r.t. x, we get
Question 2.
xy = e(x-y)
Solution:
xy = e(x-y)
∴ log xy = log e(x-y)
∴ y log x = (x – y) log e
∴ y log x = x – y …..[∵ log e = 1]
∴ y + y log x = x
∴ y(1 + log x) = x
∴ y = $$\frac{x}{1+\log x}$$
Question 3.
xy = log(xy)
Solution:
xy = log (xy)
∴ xy = log x + log y
Differentiating both sides w.r.t. x, we get
3. Solve the following:
Question 1.
If x5 . y7 = (x + y)12, then show that $$\frac{d y}{d x}=\frac{y}{x}$$
Solution:
x5 . y7 = (x + y)12
∴ log(x5 . y7) = log(x + y)12
∴ log x5 + log y7 = log(x + y)12
∴ 5 log x + 7 log y = 12 log (x + y)
Differentiating both sides w.r.t. x, we get
Question 2.
If log(x + y) = log(xy) + a, then show that $$\frac{d y}{d x}=\frac{-y^{2}}{x^{2}}$$
Solution:
log (x + y) = log (xy) + a
∴ log(x + y) = log x + log y + a
Differentiating both sides w.r.t. x, we get
Question 3.
If ex + ey = e(x+y), then show that $$\frac{d y}{d x}=-e^{y-x}$$.
Solution:
ex + ey = e(x+y) ……….(1)
Differentiating both sides w.r.t. x, we get |
# Chapter 10 - Exponents and Radicals - 10.1 Radical Expressions and Functions - 10.1 Exercise Set - Page 634: 91
$g(19)=2 ,\\\\ g(-13)=\text{does not exist} ,\\\\ g(1)=\text{does not exist} ,\\\\ g(84)=3$
#### Work Step by Step
$\bf{\text{Solution Outline:}}$ Substitute the given function value in $g(t)=\sqrt[4]{t-3} .$ $\bf{\text{Solution Details:}}$ If $t=19 ,$ then \begin{array}{l}\require{cancel} g(t)=\sqrt[4]{t-3} \\\\ g(19)=\sqrt[4]{19-3} \\\\ g(19)=\sqrt[4]{16} \\\\ g(19)=2 .\end{array} If $t=-13 ,$ then \begin{array}{l}\require{cancel} g(t)=\sqrt[4]{t-3} \\\\ g(-13)=\sqrt[4]{-13-3} \\\\ g(-13)=\sqrt[4]{-16} \text{ (not a real number)} .\end{array} If $t=1 ,$ then \begin{array}{l}\require{cancel} g(t)=\sqrt[4]{t-3} \\\\ g(1)=\sqrt[4]{1-3} \\\\ g(1)=\sqrt[4]{-2} \text{ (not a real number)} .\end{array} If $t=84 ,$ then \begin{array}{l}\require{cancel} g(t)=\sqrt[4]{t-3} \\\\ g(84)=\sqrt[4]{84-3} \\\\ g(84)=\sqrt[4]{81} \\\\ g(84)=3 .\end{array} Hence, \begin{array}{l}\require{cancel} g(19)=2 ,\\\\ g(-13)=\text{does not exist} ,\\\\ g(1)=\text{does not exist} ,\\\\ g(84)=3 .\end{array}
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. |
ML Aggarwal Solutions for Chapter 6 Factorization Class 10 Maths ICSE
Here, we are providing the solutions for Chapter 6 Factorization from ML Aggarwal Textbook for Class 10 ICSE Mathematics. Solutions of the sixth chapter has been provided in detail. This will help the students in understanding the chapter more clearly. Class 10 Chapter 6 Factorization ML Aggarwal Solutions for ICSE is one of the most important chapter for the board exams which is based on the simplifying algebraic expressions, factorizing quadratic polynomial and factorizing cubic polynomials.
1. Find the remainder (without divisions) on dividing f(x) by x – 2, where
(i) f(x) = 5x2 – 1x + 4
(ii) f(x) = 2x3 – 7x2 + 3
Let x – 2 = 0, then x = 2
(i) Substituting value of x in f(x)
f(x) = 5x2 – 7x + 4
⇒ f(2) = 5(2)2 – 7(2) + 4
⇒ f(2) = 20 – 14 + 4 = 10
Hence Remainder = 10
(ii) f(x) = 2x3 – 7x2 + 3
∴ f(2) = 2(2)3 – 7(2)2 + 3
= 16 – 28 + 3
Hence remainder = - 9
2. using remainder theorem, find the remainder on dividing f(x) by (x + 3) where
(i) f(x) = 2x2 – 5x + 1
(ii) f(x) = 3x3 + 7x2 – 5x + 1
Let x + 3 = 0
⇒ x = - 3
Substituting the value of x in f(x),
(i) f(x) = 2x2 – 5x + 1
∴ f(-3) = 2(-3)2 – 5(-3) + 1
= 18 + 15 + 1
= 34
Hence Remainder = 34
(ii) f(x) = 3x3 + 7x2 – 5x + 1
= 3(-3)3 + 7(-3)2 – 5(-3) + 1
= - 81 + 63 + 15 + 1
= - 2
Hence Remainder = - 2
3. Find the remainder (without division) on dividing f(x) by (2x + 1) where
(i) f(x) = 4x2 + 5x + 3
(ii) f(x) = 3x2 – 7x2 + 4x + 11
Let 2x + 1 = 0, then x = - (1/2)
Substituting the value of x in f(x):
(i) f(x) = 4x2 + 5x + 3
= 4(- 1/2)2 + 5 × (-1/2) + 3
= 4 × 1/4 – 5/2 +3
= 1 – 5/2 + 3
= 4 – 5/2
= 3/2
∴ Remainder = 3/2
(ii) f(x) = 3x3 – 7x2 + 4x + 11
= - 3(- 1/2)3 – 7(- 1/2)2 + 4(- 1/2) + 11
= 3(- 1/8) – 7(1/4) + 4(- 1/2) + 11
= - 3/8 – 7/4 – 2 + 11
= (- 3 – 14 – 16 + 88)/8
= 55/8
= 6.7/8
4. (i) Find the remainder (without division) when 2x3 – 3x2 + 7x – 8 is divided by x – 1
(ii) Find the remainder (without division) on dividing 3x2 + 5x – 9 by (3x + 2)
(i) Let x – 1 = 0, then x = 1
Substituting value of x in f(x)
f(x) = 2x3 – 3x2 + 7x – 8
= 2(1)3 – 3(1)2 + 7(1) – 8
= 2 × 1 – 3 × 1 + 7 × 1 – 8
= 2 – 3 + 7 – 8
= - 2
∴ Remainder = 2
(ii) Let 3x + 2 = 0, then 3x = - 2
⇒ x = -2/3
Substituting the value of x in f(x)
f(x) = 3x2 + 5x – 9
= 3(-2/3)2 + 5{-(2/3)} – 9
= 3 × 4/9 – 5 × 2/3 – 9
= 4/3 – 10/3 – 9
= - 6/3 – 9
= - 2 – 9
= - 11
∴ Remainder = - 11
5. Using remainder theorem, find the value of k if on dividing 2x2 + 3x2 – kx + 5 by x – 2, leaves a remainder 7.
f(x) = 2x2 + 3x2 – kx + 5
g(x) = x – 2, if x – 2 = 0, then x = 2
Dividing f(x) by g(x) the remainder will be
f(2) = 2(2)3 + 3(2)2 – k × 2 + 5
= 16 + 12 – 2k + 5
= 33 – 2k
Remainder = 7
∴ 33 – 2k = 0
⇒ 33 – 7 = 2k
⇒ 2k = 26
⇒ k = 26/2 = 13
∴ k = 13
6. Using remainder theorem, find the value of a if the division of x3 + 5x2 – ax + 6 by (x – 1) leaves the remainder 2a.
Let x – 1 = 0, then x = 1
Substituting the value of x in f(x)
f(x) = x3 + 5x2 – ax + 6
= (1)3 + 5(1)2 – a(1) + 6
= 1 + 5 – a + 6
= 12 – a
∵ Remainder = 2a
∴ 12 – a = 2a
⇒ 12 = a + 2a
⇒ 3a = 12
∴ a = 4
7. (i) What number must be subtracted from 2x2 – 5x so that the resulting polynomial leaves the remainder 2, when divided by 2x + 1 ?
(ii) What number must be added to 2x3 – 7x2 + 2x so that the resulting polynomial leaves the remainder – 2 when divided by 2x – 3?
(i) Let a be subtracted from 2x2 – 5x,
Dividing 2x2 – 5x by 2x + 1,
Here remainder is (3 – a) but we are given that remainder is 2.
∴ 3 – a = 2
⇒ - a = 2 – 3 = - 1
⇒ a = 1
Hence, 1 is to be subtracted.
(ii) Let a be added to 2x3 – 7x2 + 2x dividing by 2x – 3, then
But remainder is – 2, then
a – 6 = - 2
⇒ a = - 2 + 6
⇒ a = 4
Hence, 4 is to be added.
8. (i) When divided by x – 3 the polynomials x2 – px2 + x + 6 and 2x3 – x2 – (p + 3)x – 6 leave the same remainder. Find the value of ‘p’
(ii) Find ‘a’ if the two polynomials ax3 + 3x2 – 9 and 2x3 + 4x + a, leaves the same remainder when divided by x + 3.
(i) By dividing x3 – px2 + x + 6
And 2x3 – x2 – (p + 3)x – 6
By x – 3, the remainder is same Let x – 3 = 0, then x = 3
Now by Remainder Theorem,
Let p(x) = x3 – px2 + x + 6
p(3) = (3)3 – p(3)2 + 3 + 6
= 27 – 9p + 9
= 36 – 9p
and q(x) = 2x3 – x2 – (p + 3)x – 6
q(3) = 2(3)2 – (3)2 – (p + 3) × 3 – 6
= 2 × 27 – 9 – 3p – 9 – 6
= 54 – 24 – 3p – 9 – 6
= 54 – 24 – 3p
= 30 – 3p
∵ The remainder in each case is same
∴ 36 – 9p = 30 – 3p
36 – 30 = 9p – 3p
⇒ 6 = 6p
⇒ p = 6/6 = 1
∴ p = 1
(ii) Find ‘a’ if the two polynomials ax3 + 3x2 – 9 and 2x3 + 4x + a, leaves the same remainder when divided by x + 3.
The given polynomials are ax3 + 3x2 – 9 and 2x3 + 4x + a
Let p(x) = ax3 + 3x2 – 9
and q(x)= 2x3 + 4x + a
Given that p(x) and q(x) leave the same remainder when divided by (x + 3),
Thus by Remainder Theorem, we have
p(-3) = q(-3)
⇒ a(-3)3 + 3(-3)2 – 9 = 2(-3)3 + 4(-3) + a
⇒- 27a + 27 – 9 = - 54 – 12 + a
⇒ - 27a – a = - 66 – 18
⇒ - 28a = - 84
⇒ a = 84/28
∴ a = 3
9. By factor theorem show that (x + 3) and (2x – 1) are factors of 2x2 + 5x – 3.
Let x + 3 = 0 then x = - 3
Substituting the value of x in f(x)
f(x) = 2x2 + 5x – 3 = 2(- 3)2 + 5(- 3) – 3
f(- 3) = 18 – 15 – 3 = 0
∵ Remainder = 0, then x + 3 is a factor
Again let 2x – 1 = 0, then x = 1/2
Substituting the value of x in f(x),
f(x) = 2x2 + 5x – 3
f(1/2) = 2(1/2)2 + 5(1/2) – 3
= 2 × 1/4 + 5/2 – 3
= 1/2 + 5/2 – 3 = 0
∵ Remainder = 0,
∴ 2x – 1 is also a factor
Hence, proved
10. Without actual division, prove that x4 + 2x3 – 2x2 + 2x – 3 is exactly divisible by x2 + 2x – 3.
x2 + 2x – 3
= x2 + 3x – x – 3
= x(x + 3) – 1(x + 3)
= (x + 3)(x – 1)
Let p(x) = x4 + 2x3 – 2x2 + 2x – 3
We see that
p(-3) = (- 3)4 + 2(- 3)3 – 2(-3)2 + 2(-3) – 3
= 81 – 54 – 18 – 6 – 3
= 0
Hence by converse of factor theorem, (x + 3) is a factor of p(x).
Also we see that
p(1) = (1)4 + 2(1)3 - 2(1)2 + 2(1) – 3 = 0
Hence by converse of factor theorem, (x – 1) is a factor of p(x).
From above, we see that
(x + 3)(x – 1), i.e., x2 + 2x – 3 is a factor of p(x)
⇒ p(x) is exactly divisible by (x2 + 2x – 3)
11. Show that (x – 2) is a factor of 3x2 – x – 10. Hence factorise 3x2 – x – 10.
Let x – 2 = 0, then x = 2
Substituting the value of x in f(x),
f(x) = 3x2 – x – 10
= 3(2)2 – 2 – 10
= 12 – 2 – 10 = 0
∵ Remainder is zero
∴ x – 2 is a factor of f(x).
Dividing 3x2 – x – 10 by x – 2, we get
∴ 3x2 – x – 10 = (x – 2)(3x + 5)
12. (i) Show that (x – 1) is a factor of x3 – 5x2 – x + 5. Hence, factorise x3 – 5x2 – x + 5.
(ii) Show that (x – 3) is a factor of x3 – 7x2 + 15x – 9. Hence, factorise x3 – 7x2 + 15x – 9
(i) Let x – 1 = 0, then x = 1
Substituting the value of x in f(x),
f(x) = x3 – 5x2 – x + 5
= (1)3 – 5(1)2 – 1 + 5
= 1 – 5 – 1 + 5
= 0
∵ Remainder = 0
∴ x – 1 is a factor of x3 – 5x2 – x + 5
Now dividing f(x) by x – 1, we get
∴ x3 – 4x2 – x + 5
= (x – 1)(x2 – 4x – 5) = (x – 1)[x2 – 5x + x – 5]
= (x – 1)[x(x – 5) + 1(x – 5)]
= (x – 1)(x + 1)(x – 5)
(ii) Let x – 3 = 0, then x = 3,
Substituting the value of x in f(x), f(x) = x3 – 7x2 + 15x – 9
= (3)3 – 7(3)2 + 15(3) – 9
= 27 – 63 + 45 – 9
= 72 – 72
= 0
∵ Remainder = 0,
∴ x – 3 is a factor of x3 – 7x2 + 15x – 9
Now dividing it by x – 3, we get
∴ x3 – 7x2 + 15x – 9
= (x – 3)(x2 – 4x + 3) = (x – 3)[x2 – x – 3x + 3]
= (x – 3)[x(x – 1) – 3(x – 1)]
= (x – 3)(x – 1)(x – 3) = (x – 3)2(x – 1)
13. Show that (2x + 1) is a factor of 4x3 + 12x2 + 11x + 3. Hence factorise 4x3 + 12x3 + 11x + 3.
Let 2x + 1 = 0,
Then x = - 1/2
Substituting the value of x in f(x),
f(x) = 4x3 + 12x2 + 11x + 3
f(-1/2) = 4(-1/2)3 + 12(-1/2)2 + 11(-1/2) + 3
= 4(- 1/8)
= - 1/2 + 3 – 11/2 + 3
= (6) – (6) = 0
∵ Remainder = 0,
∴ 2x + 1 is a factor of
4x3 + 12x2 + 11x + 3
Now dividing f(x) by 2x + 1, we get
∴ 4x3 + 12x2 + 11x + 3
= (2x + 1)(2x2 + 5x + 3)
= (2x + 1)[2x2 + 2x + 3x + 3]
(2x + 1)[2x(x + 1) + 3(x + 1)]
= (2x + 1)[(x + 1)(2x + 3)]
= (2x + 1)(x + 1)(2x + 3)
14. Show that 2x + 7 is a factor of 2x3 + 5x2 – 11x – 14. Hence factorise the given expression completely, using the factor theorem.
Let 2x + 7 = 0, then 2x = - 7
x = (-7)/2
Substituting the value of x in f(x),
f(x) = 2x3 + 5x2 – 11x – 14
f(- 7/2) = 2(- 7/2)3 + 5(-7/2)2 – 11(- 7/2) – 14
= (-343)/4 + 245/4 + 77/2 – 14
= (- 343 = 245 + 154 – 56)/4
= (- 399 + 399)/4
= 0
Hence, (2x + 7) is a factor of f(x).
Now, 2x3 + 5x2 – 11x – 14 = (2x + 7)(x2 – x – 2)
= (2x + 7)[x2 – 2x + x – 2]
= (2x + 7)[x(x – 2) + 1(x – 2)]
= (2x + 7)(x + 1)(x – 2)
15. Use factor theorem to factorise the following polynomials completely.
(i) x3 + 2x2 – 5x – 6
(ii) x3 – 13x – 12
(i) Let f(x) = x3 + 2x2 – 5x – 6
Factors of (∵ 6 = ± 1; ± 2, ± 3, ± 6)
Let x = - 1, then
f(-1) = (-1)3 + 2(-1)2 – 5(-1) – 6
= - 1 + 2(1) + 5 – 6
= - 1 + 2 + 5 – 6
= 7 – 7
= 0
∵ f(-1) = 0
∴ x + 1 is a factor of f(x)
Now, dividing f(x) by x + 1, we get
f(x) = (x + 1) (x2 + x – 6)
= (x + 1)(x2 + 3x – 2x – 6)
= (x + 1){x(x + 3) – 2(x + 3)}
= (x + 1)(x + 3)(x – 2)
(ii) f(x) = x3 – 13x – 12
Let x = 4, then
f(x) = (4)3 – 13(4) – 12
= 64 – 52 – 12
= 64 – 64 = 0
∵ f(x) = 0
∴ x – 4 is a factor of f(x).
Now dividing f(x) by (x – 4), we get,
f(x) = (x – 4)(x2 + 4x + 3)
= (x – 4)(x2 + 3x + x + 3)
= (x – 4)[x(x + 3) + 1(x + 3)]
Question 16: Use the remainder Theorem to factorise the following expression
(i) 2x3 + x2 – 13x + 6.
(ii) 3x2 + 2x2 – 19x + 6
(iii) 2x3 + 3x2 – 9x – 10
(i) Let f(x) = 2x3 + x2 – 13x + 6
Factors of 6 are ±1, ±2, ±3, ±6
Let x = 2, then
f(2) = 2(2)3 + (2)2 – 13 × 2 + 6
= 16 + 4 – 26 + 6
= 26 – 26
= 0
∵ f(2) = 0
∴ x – 2 is the factor of f(x) (By Remainder Theorem)
Dividing f(x) by x – 2, we get
∴ f(x) = (x – 2)(2x2 + 5x – 3)
= (x – 2){2x2 + 6x – x – 3}
= (x – 2){2x(x + 3) – 1(x + 3)}
= (x – 2)(x + 3)(2x – 1)
(ii) P(x) = 3x3 + 2x2 – 19x + 6
P(1) = 3 + 2 – 19 + 6
= - 8 ≠ 0
P(-1)= - 3 + 2 + 19 + 6
= - 24 ≠ 0
P(2) = 24 + 8 – 38 + 6 = 0
Hence, (x – 2) is a factor of p(x)
∴ P(x) = 3x3 + 2x2 – 19x + 6
= 3x3 – 6x2 + 8x2 – 16x – 3x + 6
= 3x2(x – 2) + 8x(x – 2) – 3(x – 2)
= (x – 2)(3x2 + 8x – 3)
= (x – 2)(3x2 + 9x – x – 3)
= (x – 2){3x(x + 3) – 1(x + 3)
= (x – 2)(x + 3)(3x – 1)
(iii) f(x) = 2x3 + 3x2 – 9x - 10
f(-1) = 2(-1)3 + 3(-1)2 – 9(-1) – 10
∴ f(-1) = - 2 + 3 + 9 – 10 = 0
∴ (x + 1) is a factor.
∴ 2x2 + x – 10 = 2x2 + 5x – 4x – 10
= x(2x + 5) – 2(2x + 5) – (2x + 5)(x – 2)
∴ factors are (x + 1) (x – 2)(2x + 5)
17. Using the Remainder and factor Theorem, factorise the following polynomial:
x3 + 10x2 – 37x + 26.
f(x) = x3 + 10x2 – 37x + 26
f(1) = (1)3 + 10(1)2 – 37(1) + 26
= 1 + 10 – 37 + 26
= 0
x = 1
x – 1 is factor of f(x)
∴ f(x) = (x – 1)(x2 + 11x – 26)
= (x – 1)(x2 + 13x – 2x – 26)
= (x – 1)[x(x + 13) – 2(x + 13)]
= (x – 1)[(x – 2)(x + 13)]
18. If (2x + 1) is a factor of 6x3 + 5x2 + ax – 2. Find the value of a.
Let 2x + 1 = 0, then x = - (1/2)
Substituting the value of x in f(x),
f(x) = 6x3 + 5x2 + ax – 2
f(-1/2) = 6(- 1/2)3 + 5(- 1/2)2 + a(-1/2) – 2
= 6(-1/8) + 5(1/4) + a(- 1/2) – 2
= - 3/4 + 5/4 – a/2 – 2
= (-3 + 5 – 2a – 8)/4
= (- 6 – 2a)/4
∵ 2x + 1 is a factor of f(x)
∴ Remainder = 0
∴ (- 6 – 2a)/4 = 0
⇒ - 6 – 2a = 0
⇒ 2a = - 6
⇒ a = - 3
∴ a = - 3
19. If (3x – 2) is a factor of 3x3 – kx2 + 21x – 10, find the value of k.
Let 3x – 2 = 0, then 3x = 2
⇒ x = 2/3
Substituting the value of x in f(x),
f(x) = 3x3 – kx2 + 21x – 10
f(2/3) = 3(2/3)3 – k(2/3)2 + 21(2/3) – 10
= 3 × 8/27 – k × 4/9 + 21 × 2/3 – 10
= 8/9 – 4k/9 + 14 – 10
= (8 – 4k)/9 + 4
∵ Remainder is 0
∴ (8 – 4k)/9 + 4 = 0
⇒ 8 – 4k + 36 = 0
⇒ - 4k + 44 = 0
⇒ 4k = 44
∴ k = 11
20. If ( x – 2) is a factor of 2x3 –x2 + px – 2, then
(i) find the value of p.
(ii) with this value of p, factorize the above expression completely
(i) Let x – 2 = 0, then x = 2
Now f(x) = 2x3 – x2 + px – 2
∴ f(2) = 2(3)3 – (2)2 + p × 2 – 2
= 2 × 8 – 4 + 2p – 2
= 16 – 4 + 2p – 2 = 10 + 2p
(ii) ∴ f(2) = 0, then 10 + 2p = 0
⇒ 2p = -10
⇒ p = -5
Now, the polynomial will be
2x3 – x2 – 5x – 2
= (x – 2)(2x2 + 3x + 1)
= (x – 2)[2x2 + 2x + x + 1]
= (x – 2)[2x(x + 1) + 1(x + 1)]
= (x – 2)(x + 1)(2x + 1)
21. Find the value of ‘K’ for which x = 3 is a solution of the quadratic equation, (k + 2)x2 – Kx + 6 = 0. Also, find the other root of the equation.
(K + 2)x2 – Kx + 6 = 0 …(1)
Substitute x = 3 in equation (1)
(- 4 + 2)x2 – (- 4)x + 6 = 0
⇒ - 2x2 + 4x + 6 = 0
⇒ x2 – 2x – 3 = 0 (Dividing by 2)
⇒ x3 – 3x + x – 3 = 0
⇒ x2 – 3x + x – 3 = 0
⇒ x(x – 3) + 1(x – 3) = 0
⇒ (x + 1)(x – 3) = 0
So, the roots are x = - 1and x = 3
Thus, the other root of the equation is x = - 1
22. What number should be subtracted from 2x3 – 5x2 + 5x so that the resulting polynomial has 2x – 3 as a factor?
Let the number to be subtracted be k and the resulting polynomial be f(x), then
f(x) = 2x3 – 5x2 + 5x – k
Since, 2x – 3 is a factor of f(x),
Now, converting 2x – 3 to factor theorem
f(3/2) = 0
⇒ 2x3 – 5x2 + 5x – k = 0
⇒ 2(3/2)3 – 5(3/2)2 + 5(3/2) – k = 0
⇒ 2 × 27/8 – 5 × 9/4 + 5 × 3/2 – k = 0
⇒ 27/4 – 45/4 + 15/2 – k = 0
⇒ 27 – 45 + 30 – 4k = 0
⇒ - 4k + 12 = 0
⇒ k = - 12/-4
⇒ k = 3
23. (i) Find the value of the constant a and b, if (x – 2) and (x + 3) are both factors of the expression x3 + ax2 + bx – 12.
(ii) If (x + 2) and (x + 3) are factors of x3 + ax + b, Find the values of a and b.
Let x – 2 = 0, then x = 0
Substituting value of x in f(x)
f(x) = x3 + ax2 + bx – 12
f(2) = (2)3 + a(2)2 + b(2) – 12
= 8 + 4a + 2b – 12
= 4a + 2b – 4
∵ x – 2 is a factor
∴ 4a + 2b – 4 = 0
⇒ 4a + 2b = 4
⇒ 2a + b = 2
Again let x + 3 = 0, then x = - 3
Substituting the value of x in f(x)
f(x) = x3 + ax2 + bx – 12
= (-3)3 + a(-3)2 + b(-3) – 12
⇒ - 27 + 9x – 3b – 12
= - 39 + 9a – 3b
∵ x + 3 is a factor of f(x)
∴ - 39 + 9a – 3b = 0
⇒ 9a – 3b = 39
⇒ 3a – b = 13 ….(ii)
5a = 15
⇒ a = 3
Substituting the value of a in (i)
2(3) + b = 2
⇒ 6 + b = 2
⇒ b = 2 – 6
∴ b = - 4
Hence a = 3, b = - 4
(ii) Since (x + 2) = 0
f(x) = x3 + ax + b
f(-2) = (-2)3 + a(-2) + b
Since x + 2 is a factor, by factor theorem
- 8 – 2a + b = 0
∴ - 2a + b = 8 …..(i)
Since x + 3 = 0
∴ x = - 3
f(x) = x3 + ax + b
∴ f(-3) = (-3)3 + a(-3) + b
∴ f(-3) = - 27 – 3a + b
By factor theorem, - 27 – 3a + b = 0
∴ - 3a + b = 27 ….(ii)
Subtracting eq. (ii) from eq. (i)
- 2a + b = 8
- 3a + b = 27
a = - 19
Now, substituting the value of a in eq. (i)
- 2(- 19) + b = 8
⇒ 38 + b = 8
b = - 30
24. If (x + 2) and (x – 3) are factors of x3 + ax + b, find the values of a and b, With these values of a and b, factorize the given expression.
Let x + 2 = 0, then x = - 2
Substituting the value of x in f(x),
f(x) = x3 + ax + b
f(-2) = (-2)3 + a(-2) + b
= - 8 – 2a + b
∵ x + 2 is a factor
∴ Remainder is zero.
∴ - 8 – 2a + b = 0
⇒ - 2a + b = 8
∴ 2a – b = - 8 ...(i)
Again let x – 3 = 0, then x = 3, Substituting the value of x in f(x),
f(x) = x3 + ax +b
f(3) = (3)3 + a(3) + b
= 27 + 3a + b
∵ x – 3 is a factor
∴ Remainder = 0
⇒ 27 + 3a + b = 0
⇒ 3a + b = - 27 ...(ii)
5a = - 35
⇒ a = - 35/5
= - 7
Substituting value of a in (i)
2(-7) – b = - 8
∴ - 14 – b = - 8
- b = - 8 + 14
⇒ - b = 6
∴ b = - 6
Hence a= - 7, b = - 6
(x + 2) and (x – 3) are the factors of x2 + ax + b
x3 + ax + b ⇒ x3 + 7x - 6
Now dividing x3 – 7x – 6
Now dividing x3 – 7x – 6 by (x + 2)
(x – 3) or x2 – x – 6, we get
∴ Factors are (x + 2), (x – 3) and (x + 1)
25. (x – 2) is a factor of the expression x3 + ax2 + bx + 6. When this expression is divided by (x – 3), it leaves the remainder 3. Find the values of a and b.
As, x – 2 is a factor of
f(x) = x3 + ax2 + bx + 6
∴ f(2) = 0
∴ (2)3 + a(2)2 + b(2) + 6 = 0
⇒ 8 + 4a + 2b + 6 = 0
⇒ 4a + 2b = - 14
⇒ 2a + b = - 7 ...(i)
As on dividing f(x) by x – 3
Remainder = 3
∴ f(3) = 3
∴ (3)3 + a(3)2 + b(3) + 6 = 3
⇒ 27 + 9a + 3b + 6 = 3
⇒ 9a + 3b = - 30
⇒ 3a + b = - 10 ...(ii)
Solving simultaneously equation (i) and (ii),
∴ 2a + b = - 7
3a + b = - 10
Subtracting,
- a = 3
a = - 3
Substituting value of a in equation (i)
2(-3) + b = - 7
∴ 6 + b = - 7
∴ b = - 1
∴ a = - 3, b = - 1
26. If (x – 2) is a factor of the expression 2x3 + ax2 + bx – 14 and when the expression is divide by (x – 3), it leaves a remainder 52, find the values of a and b.
f(x) = 2x3 + ax2 +bx – 14
∴ (x – 2) is factor of f(x)
f(2) = 0
2(2)3 + a(2)2 + b(2) – 14 = 0
16 + 4a + 2b – 14 = 0
⇒ 4a + 2b = - 2
2a + b = - 1 ….(i)
Also, (x – 3) it leaves remainder = 52
∴ f(3) = 52
2(3)3 + a(3)2 + b(3) – 14 = 52
⇒ 54 + 9a + 3b – 14 = 52
⇒ 9a + 3b = 52 – 40
9a + 3b = 12
3a + b = 4 ...(ii)
From (i) and (ii)
2a + b = - 1
3a + b = 4
Subtracting – a = - 5
∴ a = 5 put in (i)
∴ 2(5) + b = - 1
⇒ b = - 1 – 10
⇒ b = - 11
∴ a = 5, b = - 11
⇒ - 27a/8 + 27/4 – 3b/2 – 3 = 0
⇒ - 27a + 54 – 12b – 24 = 0 (Multiplying by 8)
⇒ - 27a – 12b + 30 = 0
⇒ - 27 a – 12b = - 30
⇒ 9a + 4b = 10 [Dividing by (-3)]
9a + 4b = 10 …..(i)
Again let x + 2 = 0 then x = - 2
Substituting the value of x in f(x)
f(x) = ax3 + 3x2 + bx – 3
f(-2) = a(-2)3 + 3(-2)2 + b(-2) – 3
= - 8a + 12 – 2b – 3
= - 8a – 2b + 9
∵ Remainder = - 3
∴ - 8a – 2b + 9 = - 3
⇒ - 8a – 2b = - 3 – 9
⇒ - 8a -2b = - 12 (Dividing by 2)
⇒ 4a + b = 6 …(ii)
Multiplying (ii) by 4
16a + 4b = 24
9a + 4b = 10
Subtracting, 16a + 4b = 24
7a = 14
7a = 14
⇒ a = 14/7 = 2
Subtracting the value of a in (i)
9(2) + 4b = 10
⇒ 18 + 4b = 10
⇒ 4b = 10 – 18
⇒ 4b = - 8
∴ b = - 8/4 = - 2
Hence a = 2, b = - 2
∴ f(x) = ax3 + 3x2 + bx – 3
= 2x3 + 3x2 – 2x – 3
∵ 2x + 3 is a factor
∴ Dividing f(x) by x + 2
∴ 2x3 + 3x2 – 2x – 3
= (2x + 3)(x2 – 1) = (2x + 3)[(x)2 – (1)2]
= (2x + 3)(x + 1)(x – 1)
27. If ax3 + 3x2 + bx – 3 has a factor (2x + 3) and leaves remainder – 3 when divided by (x + 2), find the values of a and 6. With these values of a and 6, factorise the given expression.
Let 2x + 3 = 0 then 2x = - 3
⇒ x = (-3)/2
Substituting the value of x in f(x),
f(x) = ax3 + 3x2 + 6x – 3
f(-3/2) = a(-3/3)3 + 3(-3/2)2 + b(-3/2) – 3
= a(- 27/8) + 3(9/4) + b(-3/2) - 3
= - 27a/8 + 27/4 – 3b/2 – 3
∵ 2x + 3 is a factor of f(x)
∴ Remainder = 0
28. Given f(x) = ax2 + bx + 2 and g(x) = bx2 + ax + 1. If x – 2 is a factor of f(x) but leaves the remainder – 15 when it divides g(x), find the values of a and b. With these values of a and b, factorise the expression. f(x) + g(x) + 4x2 + 7x.
f(x) = ax2 + bx + 2
g(x) = bx2 + ax + 1
x – 2 is a factor of f(x)
Let x – 2 = 0
⇒ x = 2
∴ f(2) = a(2)2 + b × 2 + 2
= 4a + 2b + 2
∴ 4a + 2b + 2 = 0 (∵ x – 2 is its factor)
⇒ 2a + b + 1 = 0 ….(i) (Dividing by 2)
Dividing g(x) by x – 2, remainder = - 15
Let x – 2 = 0
⇒ x = 2
∴ g(2) = b(2)2 + a × 2 + 1
= 4b + 2a + 1
∵ Remainder is – 15
∴ 4b + 2a + 1 = - 15
⇒ 4b + 2a + 1 + 15 = 0
⇒ 4b + 2a + 16 = 0
⇒ 2b + a + 8 = 0 (Dividing by 2)
⇒ a + 2b + 8 = 0 …(ii)
Multiplying (i) by 2 and (ii) by 1
4a + 2b + 2 = 0
a + 2b + 8 = 0
3a – 6 = 0
⇒ 3a = 6
⇒ a = 6/3
∴ a = 2
Substituting the value of a in (i)
2 × 2 + b + 1 = 0
⇒ 4 + b + 1 = 0
⇒ b + 5 = 0
⇒ b = - 5
Hence a = 2, b = - 5
Now f(x) + g(x) = 4x2 + 7x
= 2x2 – 5x + 2 + (-5x2 + 2x + 1) + 4x2 + 7x
= 2x2 – 5x + 2 – 5x2 + 2x + 1 + 4x2 + 7x
= 6x2 – 5x2 – 5x + 2x + 7x + 2 + 1
= x2+ 2x + 3
= x2 + x + 3x + 3
= x(x + 1) + 3(x + 1)
= (x + 1)(x + 3)
Multiple Choice Questions
Choose the correct answer from the given four options (1to 5):
1. When x3 – 3x2 + 5x – 7 is divided by x – 2, then the remainder is. .
(a) 0
(b) 1
(c) 2
(d) – 1
(d) – 1
f(x) = x3 – 3x2 + 5x – 7
g(x) = x – 2, if x – 2 = 0, then x = 2
Remainder will be
∴ f(2) = (2)2 – 3(2)2 + 5 × 2 – 7
= 8 – 12 + 10 – 7
= 18 – 19
= - 1
∴ Remainder = - 1
2. When 2x3 – x2 – 3x + 5 is divided by 2x + 1, then the remainder is
(a) 6
(b) – 6
(c) – 3
(d) 0
(a) 6
f(x) = 2x3 – x2 – 3x + 5
g(x) = 2x + 1
Let 2x + 1 = 0, then x = -1/2
Then remainder will be
f(-1/2) = 2(-1/2)3 – (-1/2)2 – 3(-1/2) + 5
= 2 × -1/8 – 1/4 + 3/2 + 5
= -1/4 – 1/4 + 3/2 + 5
= (- 1 – 1 + 6 + 20)/4
= 24/4
= 6
∴ Remainder = 6
3. If on dividing 4x2 – 3kx + 5 by x + 2, the remainder is – 3 then the value of k is
(a) 4
(b) – 4
(c) 3
(d) – 3
(b) – 4
f(x) = 4x2 – 3kx + 5
g(x) = x + 2
Remainder = - 3
Let x + 2 = 0, then x = - 2
Now remainder will be
f(-2) = 4(-2)2 – 3k(-2) + 5
= 16 + 6k + 5
= 21 + 6k
∴ 21 + 6k = - 3
⇒ k = - 24/6
= - 4
∴ k = - 4
4. If on dividing 2x3 + 6x2 – (2k – 7)x + 5 by x + 3, the remainder is k – 1 then the value of k is
(a) 2
(b) – 2
(c) – 3
(d) 3
(d) 3
f(x) = 2x3 + 6x2 – (2k – 7)x + 5
g(x) = x + 3
Remainder = k – 1
If x + 3 = 0, then x = - 3
∴ Remainder will be
f(-3) = 2(-2)3 + 6(-3)2 – (2k – 7) (-3) + 5
= -54 + 54 + 3(2k – 7) + 5
= -54 + 54 + 6k – 21 + 5
= 6k – 16
∴ 6k – 16 = k – 1
6k – k = - 1 + 16
⇒ 5k = 15
⇒ k = 15/5 = 3
∴ k = 3
5. If x + 1 is a factor of 3x3 + kx2 + 7x + 4, then the value of k is
(a) – 1
(b) 0
(c) 6
(d) 10
(c) 6
f(x) = 3x3 + kx2 + 7x + 4
g(x) = x + 1
Remainder = 0
Let x + 1 = 0, then x = - 1
f(-1) = 3(-1)3 + k(-1)2 + 7(-1) + 4
= - 3 + k – 7 + 4
= k – 6
∴ Remainder = 0
∴ k – 6 = 0
⇒ k = 6
Chapter Test
1. Find the remainder when 2x3 – 3x2 + 4x + 7 is divided by
(i) x – 2
(ii) x + 3
(iii) 2x + 1
f(x) = 2x3 – 3x2 + 4x + 7
(i) Let x – 2 = 0, then x = 2
Substituting value of x in f(x)
f(2) = 2(2)3 – 3(2)2 + 4(2) + 7
= 2 × 8 – 3 × 4 + 4 × 2 + 7
= 16 – 12 + 8 + 7
= 19
Remainder = 19
(ii) Let x + 3 = 0, then x = - 3
Substituting the value of x in f(x)
f(-3) = 2(-3)3 – 3(-3)2 + 4(-3) + 7
= 2 ×(-27) – 3(9) + 4(-3) + 7
= - 54 – 27 – 12 + 7
= - 93 + 7
= - 86
∴ Remainder = - 86
(ii) Let 2x + 1 = 0, then 2x = - 1
⇒ x = - (1/2)
Now substituting the value of x in f(x)
f(- 1/2) = 2(- 1/2)3 – 3(- 1/2)2 + 4(- 1/2) + 7
= 2 ( - 1/8) – 3(1/4) + 4(- 1/2) + 7
= - 1/4 – 3/4 – 2 + 7
= - 1 – 2 + 7
= 4
∴ Remainder = 4
2. When 2x3 – 9x2 + 10x – p is divided by (x + 1), the remainder is – 24. Find the value of p.
Let x + 1 = 0 then x = - 1
Substituting the value of x in f(x)
f(x) = 2x3 – 9x2 + 10x – p
f(-1) = 2(-1)3 – 9(-1)2 + 10(-1) – p
= - 2 – 9 – 10 – p
= - 21 – p
∵ - 21 – p = - 24
⇒ - p = - 24 + 21 = - 3
∴ p = 3
3. If (2x – 3) is a factor of 6x2 + x + a, find the value of a. With this value of a, factorise the given expression.
Let 2x – 3 = 0 then 2x = 3
⇒ x = 3/2
Substituting the value of x in f(x)
f(x) = 6x2 + x + a
f(3/2) = 6(3/2)2 + 3/2 + a
= 6 × 9/4 + 3/2 + a
= 27/2 + 3/2 + a
= 30/2 + a
= 15 + a
∴ 2x – 3 is the factor
∴ Remainder = 0
∴ 15 + a = 0
⇒ a = - 15
Now f(x) will be 6x2 + x – 15 s
Dividing 6x2 + x – 15 by 2x – 3, we get
∴ 6x2 + x – 15 = (2x – 3)(3x + 5)
4. When 3x2 – 5x + p is divided by (x – 2), the remainder is 3. Find the value of p. Also the polynomial 3x2 – 5x + p – 3.
f(x) = 3x2 – 5x + p
Let (x – 2) = 0, then x = 2
f(2) = 3(2)2 – 5(2) + p
= 3 × 4 – 10 + p,
= 12 – 10 + p
= 2 + p
∵ Remainder = 3
∴ 2 + p = 3
⇒ p = 3 – 2 = 1
Hence p = 1
Now f(x) = 3x2 – 5x + p – 3
= 3x2 – 5x + 1 – 3
= 3x2 – 5x – 2
Dividing by (x – 2), we get
3x2 – 5x – 2 = (x – 2)(3x + 1)
5. Prove that (5x + 4) is a factor of 5x3 + 4x2 – 5x – 4. Hence factorize the given polynomial completely.
f(x) = 5x3 + 4x2 – 5x – 4
Let 5x + 4 = 0, then 5x = - 4
⇒ x = - 4/2
∴ f( - 4/5) = 5(- 4/5)3 + 4(- 4/5)2 – 5( - 4/5) – 4
= 5 × ( - 64/125) + 4 × 16/25 + 4 – 4
= - 64/25 + 64/25 + 4 – 4
= 0
∵ f(-4/5) = 0
∴ (5x + 4) is a factor of f(x)
Now, dividing f(x) by 5x + 4, we get
5x3 + 4x2 – 5x – 4
= (5x + 4)(x2 – 1)
= (5x + 4){(x)2 – (1)2}
= (5x + 4)(x + 1)(x – 1)
6. Use factor theorem to factorise the following polynomials completely:
(i) 4x3 + 4x2 – 9x – 9
(ii) x – 19x – 30
(i) f(x) = 4x3 + 4x2 – 9x – 9
Let x = - 1, then
f(-1) = 4(1)3 + 4(-1)2 – 9(1) – 9
= - 4 + 4 + 9 – 9
= 13 – 13
= 0
∴ (x + 1) is a factor of f(x)
Now dividing f(x) by x + 1, we get
f(x) = 4x3 + 4x2 – 9x – 9
= (x + 1)(4x2 – 9)
= (x + 1){(2x)2 – (3)2}
= (x + 1)(2x + 3)(2x – 3)
(ii) f(x) = x3 – 19x – 30
Let x = - 2, then
f(-2) = (-2)2 – 19(-2) – 30
= - 8 + 38 – 30
= - 38 – 38
= 0
∴ (x + 2) is a factor of f(x).
Now dividing f(x) by (x + 2), we get
f(x) = x3 – 19x – 30
= (x + 2) (x2 – 2x – 15)
= (x + 2){(x2 – 5x + 3x – 15)}
= (x + 2){x(x – 5) + 3(x – 5)}
= (x + 2)(x – 5)(x + 3)
7. If x3 – 2x2 + px + q has a factor (x + 2) and leaves a remainder 9, when divided by (x + 1), find the values of p and q. With these values of p and q, find the values of p and q. With these values of p and q, factorize the given polynomial completely.
f(x) = x3 – 2x2 + px + q
(x + 2) is a factor
f(-2) = (-2)3 – 2(-2)2 + p(-2) + q
= - 8 – 2 × 4 – 2p + q
= - 8 – 8 – 2p + q
= - 16 – 2p + q
∵ (x + 2) is a factor of f(x)
∴ f(-2) = 0
⇒ - 16 – 2p + q = 0
⇒ 2p – q = - 16 ...(i)
Again, let x + 1 = 0, then x = - 1
∴ f(-1) = (- 1)3 – 2(-1)2 + p(-1) + q
= - 1 – 2 × 1 – p + q
= - 1 – 2 – p + q
= - 3 – p + q
∵ Remainder = 9, then
- 3 – p + q = 9
⇒ - p + q = 9 + 3 = 12
- p + q = 12 ...(ii)
p = - 4
Substituting the value of p in (ii)
- (- 4) + q = 12
4 + q = 12
⇒ q = 12 – 4 = 8
∴ p = - 4, q = 8
∴ f(x) = x3 – 2x2 – 4x + 8
Dividing f(x) by (x + 2), we get
f(x) = (x + 2)(x2 – 4x + 4)
= (x + 2){(x)2 – 2 × x(-2) + (2)2}
= (x + 2)(x – 2)2
8. If (x + 3) and (x – 4) are factors of x3 + ax2 – bx + 24, find the values of a and b: with these values of a and b, factorise the given expression.
f(x) = x3 + ax2 – bx + 24
Let x + 3 = 0, then x = - 3
Substituting the value of x in f(x)
f(-3) = (-3)3 + a(- 3)2 – b(-3) + 24,
= - 27 + 9a + 3b + 24
= 9a + 3b – 3
∵ Remainder = 0,
∴ 9a + 3b – 3 = 0
⇒ 3a + b – 1 = 0 (Dividing by 3)
⇒ 3a + b = 1 ...(i)
Again Let x – 4 = 0, then x = 4
Substituting the value of x in f(x)
f(x) = (4)3 + a(4)2 – b(4) + 24
= 64 + 16a- 4b + 24
= 16a – 4b + 88
∵ x – 4 is a factor
∴ Remainder = 0
16a – 4b + 88 = 0
⇒ 16a – 4b = - 88 (Dividing by 4)
⇒ 4a – b = - 22
7a = - 21,
⇒ a = - 3
Substituting the value of a in (i)
3(-3) + b = 1
⇒ - 9 + b = 1
⇒ b = 1 + 9 = 10
Now f(x) will be
f(x) = x3 – 3x2 – 10x + 24
∵ x + 3 and x – 4 are factors of f(x)
∴ Dividing f(x) by (x + 3)(x – 4)
or x2 – x – 12
x3 – 3x2 – 10x + 24
= (x2 – x – 12)(x – 2)
= (x + 3)(x – 4)(x – 2)
9. If 2x3 + ax2 – 11x + b leaves remainder 0 and 42 when divided by (x – 2) and (x – 3) respectively, find the values of a and b. With these values of a and b, factorize the given expression.
f(x) = 2x3 + ax2 – 11x + b
Let x – 2 = 0, then x = 2,
Substituting the value of x in f(x)
f(2) = 2(2)3 + a(2)2 – 11(2) + b
= 2 × 8 + 4a – 22 + b
= 16 + 4a – 22 + b
= 4a + b – 6
∵ Remainder = 0,
∴ 4a + b – 6 = 0
⇒ 4a + b = 6 …(i)
Again let x – 3 = 0, then x = 3
Substituting the value of x is f(x)
f(3) = 2(3)3 + a(3)2 – 11 × 3 + b
= 2 × 27 + 9a – 33 + b
= 54 + 9a – 33 + b
⇒ 9a + b + 21
∵ Remainder = 42
∴ 9a + b + 21 = 42
⇒ 9a + b = 42 – 21
⇒ 9a + b = 21 ….(ii)
Subtracting (i) from (ii)
5a = 15
⇒ a = 15/5 = 3
Substituting the value of a is (i) 4(3) + b = 6
⇒ 12 + b = 6
⇒ b = 6 – 12
⇒ b = - 6
∴ f(x) will be 2x3 + 3x2 – 11x – 6
∵ x – 2 is a factor (as remainder = 0)
∴ Dividing f(x) by x – 2, we get
∴ 2x3 + 3x2 – 11x – 6
= (x – 2)(2x2 + 7x + 3)
= (x – 2)[2x2 + 6x + x + 3]
= (x – 2)[2x(x + 3) + 1(x + 3)]
= (x – 2)(x + 3)(2x + 1)
10. If (2x + 1) is a factor of both the expressions 2x2 – 5x + p and 2x2 + 5x + q, find the value of p and q. Hence find the other factors of both the polynomials.
Let 2x + 1 = 0, then 2x = - 1
x = - (1/2)
Substituting the value of x in
f(x) = 2x2 – 5x + p
f(-1 /2) = 2(-1/2)2 – 5(-1/2) + p
= 2 × 1/4 + 5/2 + p
= 1/2 + 5/2 + p
= 3 + p
∵ 2x + 1 is the factor of p(x)
∴ Remainder = 0
⇒ 3 + p = 0
⇒ p = - 3
Again substituting the value of x in q(x)
q(x) = 2x2 + 5x + q
q(- 1/2) = 2(- 1/2)2 + 5(- 1/2) + q
= 2 × 1/4 – 5/2 + q
= 1/2 – 5/2 + q
= - 4/2 + q
= q – 2
∵ 2x + 1 is the factor of q(x)
∴ Remainder = 0
⇒ q – 2 = 0
⇒ q = 2
Hence p = - 3, q = 2
Now (i) ∵ 2x + 1 is the factor of p(x)
= 2x2 – 5x – 3
∴ Dividing p(x) by 2x + 1,
∴ 2x2 – 5x – 3 = (2x + 1)(x – 3)
(ii) ∵ 2x + 1 is the factor of q(x) = 2x2 + 5x + 2
∴ Dividing q(x) by 2x + 1,
∴ 2x2 + 5x + 2 = (2x + 1)(x + 2)
11. If a polynomial f(x) = x4 – 2x3 – 3x2 – ax – b leaves remainder 5 and 19 when divided by (x – 1) and (x + 1) respectively, Find the values of a and b. Hence determined the remainder when f(x) is divided by (x – 2).
f(x) = x4 – 2x3 + 3x2 – ax + b
f(1) = 5 and f(-1) = 19
∴ (1)4 – 2(1)3 + 3(1)2 – a (1) + b = 5
And (-1)4 – 2(-1)3 + 3(-1)2 – a(-1) + b = 19
⇒ 1 – 2 + 3 – a + b = 5
and 1 + 2 + 3 + a + b = 19
⇒ - a + b = 5 – 2 and a + b = 19 – 6
⇒ a+ b = 3 ….(i)
and a + b = 13 ….(ii)
On subtracting (i) and (2), we het
a + b – (- a + b) = 13 – 3
a + b + a – b = 10
2a = 10
a = 5
putting a = 5 in equation 1, we get
= 5 + b = 3, b = 8
a = 5, b = 8
12. When a polynomial f(x) is divided by (x – 1), the remainder I 5 and when it is, divided by (x – 2), the remainder is 7. Find – the remainder when it is divided by (x – 1)(x – 2).
When f(x) id divided by (x – 1),
Remainder = 5
Let x – 1 = 0
⇒ x = 1
∴ f(1) = 5
When divided by (x – 2), Remainder = 7
Let x – 2 = 0
⇒ x = 2
∴ f(2) = 7
Let f(x) = (x – 1)(x – 2)q(x) + ax + b
Where q(x) is the quotient and ax + b is remainder
Putting x = 1, we get:
f(1) = (1 – 1) (1 – 2) q(1) + a × 1 + b
= 0 + a + b
= a + b
And x = 2, then
f(2) = (2 – 1)(2 – 2) q(2) + a ×1 + b
= 0 + 2a + b
= 2a + b
∴ a + b = 5 …(i)
2a + b = 7 ….(ii)
Subtracting, we get
- a = - 2
⇒ a = 2
Substituting the value of a in (i)
2 + b = 5
⇒ b = 5 – 2 = 3
∴ a = 2, b = 3
∴ Remainder = ax + b
= 2x + 3
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# algebra
the third term of an arithmetic sequence is 14 and the ninth term is -1. Find the first four terms of the sequence
1. 👍 0
2. 👎 0
3. 👁 211
1. ninth -1, third 14
it goes now by 15 in six steps; step 15/6 or 2.5
third through tenth terms
14, 11.5, 9, 6.5, 4, 1.5, -1, -3.5,...
check that
1. 👍 0
2. 👎 0
2. third term is 14 and ninth term is -1
14 - 2.5 = 11.5 > 4th term
11.5 - 2.5 = 9 > 5th term
9 - 2.5 = 6.5 > 6th term
6.5 - 2.5 = 4 > 7th term
4 - 2.5 = 1.5 > 8th term
1.5 - 2.5 = -1 > 9th term
The terms are decreasing by 2.5 > (-2.5)
To go from one term to the next, subtract 2.5.
The common difference is -2.5
d = -2.5
Arithmetic Sequence Formula:
Tn = Tn + d(n-1)
a = 1st term
n = nth term
d = common difference
We don't know the first term yet!
Tn = T1 + d(n-1)
Substitute 3 for n, and -2.5 for d
T3 = T1 + -2.5(3-1)
Now substitute 14 for T3
14 = T1 + -2.5(2)
14 = T1 -5
14 + 5 = T1 -5 +5
19 = T1
T1 = 19
T2 = 16.5
T3 = 14
T4 = 11.5
Now to find a term:
Tn = T1 + d(n-1)
T2 = 19 + -2.5(2-1) T3 = 19 + -2.5(3-1)
T2 = 19 + -2.5(1) T3 = 19 + -2.5(2)
T2 = 19 + -2.5 T3 = 19 + -5
T2 = 16.5 T3 = 14
__________________________
You can also use this formula:
Tn = dn + 21.5
T1 = -2.5(1) + 21.5
T1 = 19
T4 = -2.5(4) + 21.5
T4 = 11.5
T9 = -2.5(9) + 21.5
T9 = -1
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2. 👎 0
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2.1: Back in Time?
Difficulty Level: At Grade Created by: CK-12
This activity is intended to supplement Algebra I, Chapter 1, Lesson 5.
ID: 11572
Time Required: 15 minutes
Activity Overview
Students will explore the definition of a function graphically, with a set of ordered pairs, and by using an input-output model with the graphing calculator. This model dynamically allows students to discover the function by experimenting with input values that produce the desired output. Function notation is also reinforced.
Topic: Functions & Relations
• Function definition & notation
• Using ordered pairs, function machines and graphs.
Teacher Preparation and Notes
Associated Materials
Problem 1 – Graphical
The students are asked to consider a graph of position vs. time. They are asked if this is a function. Since there is not one unique output of $d$ for each value of $t$, it is not a function.
Students then discuss with their neighbor why the answer is ‘no’ based on the definition, and under what circumstance could the graph occur. It could only occur if Marty went back in time.
Students then redraw the graph so that it is a function. An example of a solution is to the right.
Use this opportunity to discuss the concept of the vertical line test.
Problem 2 – Set of ordered pairs
To review the notation of a set of ordered pairs this section begins with a multiple choice question with multiple solutions. Students should realize that as long as there is only one unique output for each input then it describes a function, otherwise it can only be referred to as a relation.
Another question that can be asked is “What are some examples of relations that are not functions?” Answers include inequalities, like $y > 3x, x$ as a function of $y$ when $y = x^2$, and $x$ as a function of $y$ when $y = |x + 2|$.
Students will calculate the output values for given input values using the formula $d=\frac{1}{2} \cdot a \cdot t^2$ for when $a = 12 \ ft/s^2$. The formula can be simplified to $d=6 \cdot t^2$.
To compute $d$, students can substitute the $t-$values directly into the formula as shown in the first screenshot.
To compute $t$, students need to either solve for $t \left( t=\pm \sqrt{\frac{d}{6}} \right)$ and then substitute the $d-$values into the formula as shown in the second screenshot.
Or they can use the solve command.
The solve command has parameters
solve(expression, variable, guess, {low bound, high bound}).
Note that the low bound and high bound are option, but if used must be placed in braces.
The expression needs to be equal to zero, but low and high bounds do not have to entered.
Press $2^{nd}$ [CATALOG][LN] and scroll down to select solve(.
4. Functions are $A, B$ and $D$ because for every input value there is only one output value.
5. $(0, 0), (1, 6), (2, 24), (6, 216)$
6. $(0, 0), \left( \frac{2}{3}, \frac{1}{3}\right), \left(\frac{2}{3}, - \frac{1}{3}\right), (6, 1), (6, -1)$
7. The first set is a function because it has only one output for every input.
8. A. $d$ is a function of $t$
Problems 3 – Function notation
In this problem, students will explore using function notation on the calculator entering input values to find the output value.
Students will enter the function next to $Y1$ on the $Y=$ screen. Then to find $f(4)$, they will enter $Y1(4)$ on the Home screen. For Question 10, students will need to update the function next to $Y1$.
9. $f(4) = 4^2 - 2(4) + 3 = 11$
10. $f(2) = 3(2)^2 + 5(2) + 3 = 25$
Problem 4 – Function Machine
For this problem, students will need the program MACHINE. This program has 3 options, each option is a different function. When students enter an input value, it will return the output value.
Students need to determine which input will give the goal output and then determine the correct function that determines the given output.
11. $17$
12. $f(x) = 0.5x$
13. $-4$
14. $f(x) = x + 10$
15. $20$
16. $f(x) = 4x + 3$
Feb 22, 2012
Oct 31, 2014 |
# Complex Conjugates Theorem
How to use the complex conjugates theorem to find the missing zeros: theorem, example, and its solution.
## Theorem
If (a + bi) is the zero of f(x),
then its complex conjugate, (a - bi),
is also the zero of f(x).
Complex conjugates
So both [x - (a + bi)] and [x - (a - bi)]
are the factors of the f(x).
Factor theorem
## Example
The given zeros are 2 and 3 + i.
So, by the complex conjugate theorem,
the complex conjugate of (3 + i), (3 - i),
is also the zero of the function.
So x = 2, 3 + i, 3 - i.
The highest degree term is x3.
So this function has at most 3 zeros.
So these are the zeros.
x = 2, 3 + i, 3 - i.
The highest degree term is x3.
So, by the factor theorem,
f(x) = (x - 2)[x - (3 + i)][x - (3 - i)].
Factor theorem
To expand [x - (3 + i)][x - (3 - i)],
find the sum and the product of the zeros.
Sum and product of the roots of a quadratic equation
Then (3 + i) + (3 - i) = 6.
Multiply the zeros.
Then (3 + i)(3 - i) = 32 + 12.
Complex conjugates
32 = 9
12 = 1
9 + 1 = 10
(3 + i) + (3 - i) = 6
(3 + i)(3 - i) = 10
So [x - (3 + i)][x - (3 - i)]
= x2 - 6x + 10.
Sum and product of the roots of a quadratic equation
So f(x) = (x - 2)(x2 - 6x + 10)
Solve (x - 2)(x2 - 6x + 10).
x(x2 - 6x + 10)
= x3 - 6x2 + 10x
-2(x2 - 6x + 10)
= -2x2 + 12x - 20
Multiplying polynomials
x3 = x3
-6x2 - 2x2 = -8x2
+10x + 12x = +22x
-20 = -20
So f(x) = x3 - 8x2 + 22x - 20. |
## What does closed under division mean?
2. To complement the previous answer, the set of integers is closed under addition because if you take two integers and add them, you will always get another integer. The set of integers is not closed under division, because if you take two integers and divide them, you will not always get an integer.
## Is the set of integers closed under division?
b) The set of integers is not closed under the operation of division because when you divide one integer by another, you don’t always get another integer as the answer. For example, 4 and 9 are both integers, but 4 ÷ 9 = 4/9.
## Is Q closed under division?
No. The integers are not closed under division for reasons other than the fact that 1/0 is undefined.
## Which of the following sets is not closed under division?
Answer: Integers, Irrational numbers, and Whole numbers none of these sets are closed under division. Let us understand the concept of closure property. Thus, Integers are not closed under division. Thus, Irrational numbers are not closed under division.
## How do you know if a set is closed?
One way to determine if you have a closed set is to actually find the open set. The closed set then includes all the numbers that are not included in the open set. For example, for the open set x < 3, the closed set is x >= 3. This closed set includes the limit or boundary of 3.
## Why is division not closed?
First you should know that Any number Divided by 0 is not a number… … ( in rational number denominator should be non zero…) So Division is not closed for rational numbers… (Note : If you gake denominator other than zero , then Division operation will be closed….but here we have to check for all rational number…
## Are rationals closed under division?
Rational number is not closed under division.
## What are rationals closed under?
Closure property
We can say that rational numbers are closed under addition, subtraction and multiplication.
## Is W closed under subtraction and division?
Thus the set of whole numbers, W is closed under addition and multiplication. … This property does not hold true in the case of subtraction and division operations on whole numbers. As, 0 and 2 are whole numbers, but 0 – 2 = -2, which is not a whole number. Similarly, 2/0 is not defined.
## Why are whole numbers not closed under subtraction?
Whole numbers are not closed under subtraction operation because when we consider any two numbers, then one number is subtracted from the other number. it is not necessary that the difference so obtained is a whole number.
## What sets are closed under subtraction?
The operation we used was subtraction. If the operation on any two numbers in the set produces a number which is in the set, we have closure. We found that the set of whole numbers is not closed under subtraction, but the set of integers is closed under subtraction.
## What are irrational numbers closed under?
Irrational Numbers: The irrational numbers are the set of number which can NOT be written as a ratio (fraction). Decimals which never end nor repeat are irrational numbers. Irrational numbers are “not closed” under addition, subtraction, multiplication or division.
## Which of the following sets is not closed under subtraction?
Answer: The set that is not closed under subtraction is b) Z. A set closed means that the operation can be performed with all of the integers, and the resulting answer will always be an integer.
## What is a closed set in math?
The point-set topological definition of a closed set is a set which contains all of its limit points. Therefore, a closed set is one for which, whatever point is picked outside of , can always be isolated in some open set which doesn’t touch .
## What is a closure of a set?
The closure of a set is the smallest closed set containing . Closed sets are closed under arbitrary intersection, so it is also the intersection of all closed sets containing . Typically, it is just. with all of its accumulation points. The term “closure” is also used to refer to a “closed” version of a given set.
## Are polynomials closed under division?
Polynomials and Closure:
Polynomials form a system similar to the system of integers, in that polynomials are closed under the operations of addition, subtraction, and multiplication. CLOSURE: Polynomials will be closed under an operation if the operation produces another polynomial.
## What is closed set with example?
Examples of closed sets
of real numbers is closed. … The Cantor set is an unusual closed set in the sense that it consists entirely of boundary points and is nowhere dense. Singleton points (and thus finite sets) are closed in Hausdorff spaces.
## Is Za closed set?
Note that Z is a discrete subset of R. Thus every converging sequence of integers is eventually constant, so the limit must be an integer. This shows that Z contains all of its limit points and is thus closed. |
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# Show that only 286 four letter words can be formed out of the letter of the word ‘infinite’.
Last updated date: 13th Jul 2024
Total views: 449.7k
Views today: 5.49k
Verified
449.7k+ views
Hint: We are going to use formulae of permutations and combinations with repetition of letters.
In the word ‘infinite’, we have a total 8 letters out of which are 3 i’s, 2 n’s and 3 different letters. 0
Number of different arrangements of ‘n’ letters where ${n_1}$ repeated letters, ${n_2}$ repeated letters ... ${n_k}$ repeated letters, is equal to $$\dfrac{{n!}}{{{n_1}!{n_2}!...{n_k}!}}$$.
From the word ‘infinite’, we are taking 4 letters to form all possible four lettered words.
(i). When we take 4 letters in which all are different, then
Number of words formed = $^5{C_4} \cdot 4! = 120$
[We have 5 unique letters in the word ‘infinite’, so we took $^5{C_4}$.]
(ii). When we take 2 alike and 2 different letters, then
Number of words formed = $$^2{C_1}{ \cdot ^4}{C_2} \cdot \dfrac{{4!}}{{2!}} = 144$$
[We have ‘n’ is repeated two times, ‘i’ is repeated three times, selecting two letters from these repeated letters is $$^2{C_1}$$]
(iii).When we take 2 alike and 2 alike letters, then
Number of words formed = $^2{C_2} \cdot \dfrac{{4!}}{{2!2!}} = 6$
[Taking two similar and two other similar letters from the given word is $^2{C_2}$.]
(iv).When we take 3 alike and 1 different letter, then
Number of words formed = $^1{C_1}{ \cdot ^4}{C_1} \cdot \dfrac{{4!}}{{3!}} = 16$
Total number of four digit letters = 120 + 144 + 6 +16 = 286.
Hence proved.
Note: In the given word ‘infinite’, we can observe this 8 lettered word, it actually consists of only 5 different digits i.e., (i, n, f, t, e). Two letters are repeated. So we used the formula of arrangement of n letters with repetition of letters. |
Question Video: Using an Area Model to Deduce the Distributive Property of Multiplication | Nagwa Question Video: Using an Area Model to Deduce the Distributive Property of Multiplication | Nagwa
Question Video: Using an Area Model to Deduce the Distributive Property of Multiplication Mathematics
The diagram shows a rectangle of sides π and π + π. Its area is thus π Γ (π Γ π), also written as π(π + π). Work out the area of the two rectangles that make up the bigger rectangle to find an equivalent expression to π(π + π).
01:49
Video Transcript
The diagram shows a rectangle of sides π and π plus π. Its area is thus π times parenthesis π times π parenthesis, also written as π parenthesis π plus π parenthesis. Work out the area of the two rectangles that make up the bigger rectangle to find an equivalent expression to π times parenthesis π plus π parenthesis.
So here we have side π and side π plus π of this larger triangle. The area of a rectangle is equal to length times width. So for the entire rectangle, its area would be π times π plus π, which can also be written as π parenthesis π plus π parenthesis.
Now weβre told to work out the area of the two rectangles that make up the bigger rectangle. So we have this one and this one. So letβs look at the striped one first. Its area will be π times π, which could also just be written as ππ.
Now the polka dot rectangle, this length must also be π. So its length times width formula, weβll actually be also be using the same π. So its area would be π times π, or ππ.
So to have the area of the entire rectangle, we will just need to add these areas together. So we would have ππ plus ππ. And this should make sense because if we would distribute the π to the π plus π, we would have π times π, which is ππ, plus π times π. And thatβs what we have: ππ plus ππ. So this will be our final answer.
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### Course: AP®︎/College Calculus AB>Unit 5
Lesson 6: Determining concavity of intervals and finding points of inflection: graphical
# Analyzing concavity (graphical)
Sal walks through an exercise where you are asked to recognize the concavity of a function in certain regions. Created by Sal Khan.
## Want to join the conversation?
• Is there an algebraic method for determining concavity?
• Yes. However, it will often be less work (and less prone to error) to use the Calculus method. Anyway here is how to find concavity without calculus.
Step 1: Given f(x), find f(a), f(b), f(c), for x= a, b and c, where a < c < b
Where a and b are the points of interest. C is just any convenient point in between them.
Step 2: Find the equation of the line that connects the points found for a and b.
Step 3: Find that y-coordinate of the line from Step 2 at point C.
Step 4: If the value found in Step 3 is greater than f(c) of the original function, then you have a positive concavity. If less, you have a negative concavity. If equal, you have an average concavity of zero.
Example: f(x) = x⁵ - 5x³ + 8x² - 2 at x = 3 and 5
f(3) = 178 ; f(5) = 2698
For a point in between, I pick 4 (could be any point between 3 and 5, though). f(4) = 830
Step 2: The line connecting (3, 178), (5, 2698) is y = 1260x-3602
Step 3: At x=4, the line's y-coordinate is y = 1438
Step 4: 1438 > 830. Therefore, this function as a positive concavity between f(3) and f(5).
It should be obvious, though, that this method can make mistakes (particularly, if there is an inflection point between A and B).
• if the 2nd derivative means concavity, then doesn't the 3rd derivative mean the frequency? (i.e. rate of change of concavity)
• Typically in mathematics and natural sciences, we don't deal too much with the 3rd derivative. I think the easiest way to understand the 3rd derivative is through physics, in which it is the rate of change of acceleration. If you have ever driven a car and felt that your body was being sucked into the seat, then that is the feeling of acceleration, but if you can feel that you are progressively getting sucked into the seat harder or slower, then that is the jerk (3rd derivative). It's much easier to visualize in that sense, but it mathematically, the rate of change of the concavity is equivalent. Frequency typically deals with waves, so we save it for there.
• If there are two critical points where the slope is 0 -let's call them f(a) and f(b)- Can we always make the assumption that f''((a+b)/2) = 0? In other words, can we assume that the second derivative of a function at the average x value of two critical points always be 0? I'm just wondering because all graphs I've seen Sal use appear to have this property.
• Interesting question – I was sure this wasn't true, but it was harder than I expected to come up with a counter example.
Try this:
`f(x) = sin(eˣ)f'(x) = eˣ•cos(eˣ)f"(x) = eˣ•[cos(eˣ) - eˣ•sin(eˣ)]`
`f'(x) = 0: eˣ•cos(eˣ) = 0 cos(eˣ) = 0 — since eˣ is never equal to zero eˣ = π/2 + n•pi — where n can be any integer x = ln(π/2 + c•pi) — where c is any non-negative integer`
So, if we take the average of the first two of these and plug them into the equation above for f"(x):
`a = ln(π/2) b = ln(3π/2) (a + b) / 2 ~= 1.00088885 f"(1.00088885) ~= -5.50761219357`
I suspect that the graphs you are talking about are third order (cubic) polynomials, which thus have a constant third derivative. I believe that this constant rate of change in the second derivative leads to your observation.
Based on this, if you play around with fourth order polynomials I think you will find that your rule doesn't hold true for them either ...
EDIT:
Checked for `g(x) = -x⁴ + x²` – I get h"[(a+b)/2] = 0.5 ...
• What is the the general purpose of the inflection point and the second derivative (Calculus 1)?
What are the limitations of the second derivative?
One that comes to mind, though I don't fully understand, is I tried to use f'' to to find a local max or min, but it didn't work...
Thanks
• The second derivative is one of the higher order derivatives. One example is the second derivative of the position function, s(t), which is equivalent to taking the derivative of the velocity function, v(t). The result of doing so will yield the acceleration function, a(t). This especially useful in problems such as those concerning the gravitational forces of the earth and the moon, and finding the ratio.
To find the local maximum and minimum, I believe you must take the first derivative of the function f(x), and then set this derivative equal to zero. For example,
if f(x) = x^2-4x
then f'(x) = 2x-4
When you set this equal to zero,
2x-4 = 0
x=2
Substitution of this x value into f(x) will produce the corresponding y value, which will be -4.
Thus one can conclude that there is a minimum of -4, at x=2
• " f''(x)>0 means the slope of f(x) is increasing "
is the meaning of that if the slope is positive it become more positive and if the slope is negative it become more negative??
• First half is correct, but if the slope is negative and f"(x) > 0, then the slope still increases – this means it becomes less negative, i.e. becomes closer to zero.
• I have a general question, so i learned that in an inflow/ outflow problem, when it asks for the maximum amount, you graph the rate and the constant inflow and find the intersection. Can someone explain the reason behind this?
• So if the 2nd derivative is negative that doesnt necessarilly mean that the 1st derivative is also negative, but that it becomes less positive, or more negative?
(1 vote)
• Yep! That's correct. The second derivative being negative only implies that the first derivative is decreasing (becoming less positive or more negative) |
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Best place for relax your brain
# What are the four 4 basic operations of math?
## What are the four 4 basic operations of math?
Basic Operations. The basic arithmetic operations for real numbers are addition, subtraction, multiplication, and division.
## What are the 5 operations in math?
There are five fundamental operations in mathematics: addition, subtraction, multiplication, division, and modular forms.
How do you find basic operations?
The basic operation is the comparison executed on each repetition of the loop. In this algorithm, the number of comparisons is the same for all arrays of size n. Depending on the value of the Key K, the number of comparisons performed can range from 1 to n. basic operation as O(n), rather than Θ(n).
### What are the four operations of functions?
Functions behave exactly as one would expect with regard to the four basic operations of algebra (addition, subtraction, multiplication, and division).
### What is operational fraction?
A fraction compares two numbers by division. To conduct basic operations, keep in mind that any number except 0 divided by itself is 1, and 1 times any number is itself. To add fractions that have the same, or a common, denominator, simply add the numerators, and use the common denominator. …
What are the 4 order of operations?
The order of operations is a rule that tells the correct sequence of steps for evaluating a math expression. We can remember the order using PEMDAS: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right). Created by Sal Khan.
## What is its basic operation?
Basic Operation: The operation contributing the most to the total running time of an algorithm. – It is typically the most time consuming operation in the algorithm’s innermost loop. • Examples: Key comparison operation; arithmetic operation (division being. the most time-consuming, followed by multiplication)
## What are all the operations in math?
What are the 5 operations of functions?
We can add, subtract, multiply and divide functions! The result is a new function.
### What is function operation?
A function operation are rules that are followed in order to solve functions. Learn the definition of a function operation including the rules for addition, subtraction, multiplication, and division.
### How do you solve Order of operations with fractions?
Order of Operations with Fractions. The order of operations is the order in which you solve the problem. If numbers are in parentheses, you do them first. Second, you solve for any numbers with exponents. Then, you go from left to right and do the multiplication and division. Finally, you go back and do the addition and subtraction – again, from left to right.
How do you work with fractions?
Convert compound fractions by multiplying the whole number by the denominator of the fraction and then adding the numerator. Write a new fraction with the total as the numerator and the same number as the denominator. For example, 2 1/3 becomes 7/3: 2 times 3, plus 1.
## What is the Order of operations with integers?
Students learn to use the following order of operations when simplifying problems that involve integers: Parentheses, Exponents, Multiplication/Division, and Addition/Subtraction (PEMDAS).
## How do you solve a fraction problem?
To solve fraction problems using basic math, understand the operations necessary to solve different fraction problems, such as simplifying fractions, understanding equivalent fractions, adding fractions, subtracting fractions and multiplying fractions. |
# How do you write an equation in slope-intercept form for a line with points (-3, 1) and (-2, -5)?
Mar 18, 2017
See the entire solution process below:
#### Explanation:
First, we need to determine the slope of the line passing through these two points. The slope can be found by using the formula: $m = \frac{\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}}{\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}}$
Where $m$ is the slope and ($\textcolor{b l u e}{{x}_{1} , {y}_{1}}$) and ($\textcolor{red}{{x}_{2} , {y}_{2}}$) are the two points on the line.
Substituting the values from the points in the problem gives:
$m = \frac{\textcolor{red}{- 5} - \textcolor{b l u e}{1}}{\textcolor{red}{- 2} - \textcolor{b l u e}{- 3}} = \frac{\textcolor{red}{- 5} - \textcolor{b l u e}{1}}{\textcolor{red}{- 2} + \textcolor{b l u e}{3}} = - \frac{6}{1} = - 6$
Next, we can write an equation in point-slope form. The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$
Where $\textcolor{b l u e}{m}$ is the slope and $\textcolor{red}{\left(\left({x}_{1} , {y}_{1}\right)\right)}$ is a point the line passes through. Substituting the slope we calculated and the first point gives:
$\left(y - \textcolor{red}{1}\right) = \textcolor{b l u e}{- 6} \left(x - \textcolor{red}{- 3}\right)$
$\left(y - \textcolor{red}{1}\right) = \textcolor{b l u e}{- 6} \left(x + \textcolor{red}{3}\right)$
Now, solve for $y$ to put the equation in slope-intercept form. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$
Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.
$y - \textcolor{red}{1} = \left(\textcolor{b l u e}{- 6} \times x\right) + \left(\textcolor{b l u e}{- 6} \times \textcolor{red}{3}\right)$
$y - \textcolor{red}{1} = - 6 x - 18$
$y - \textcolor{red}{1} + 1 = - 6 x - 18 + 1$
$y - 0 = - 6 x - 17$
$y = \textcolor{red}{- 6} x - \textcolor{b l u e}{17}$
Mar 18, 2017
$y = - 6 \cdot x - 17$
#### Explanation:
y = mx + b Calculate the slope, m, from the given point values, solve for b by using one of the point values, and check your solution using the other point values.
A line can be thought of as the ratio of the change between horizontal (x) and vertical (y) positions. Thus, for any two points defined by Cartesian (planar) coordinates such as those given in this problem, you simply set up the two changes (differences) and then make the ratio to obtain the slope, m.
Vertical difference “y” = y_2 – y_1 = -5 – 1 = -6
Horizontal difference “x” = x_2 – x_1 = -2 – -3 = 1
Ratio = “rise over run”, or vertical over horizontal$= \left(\frac{- 6}{1}\right) = - 6$ for the slope, m.
A line has the general form of y = mx + b, or vertical position is the product of the slope and horizontal position, x, plus the point where the line crosses (intercepts) the x-axis (the line where z is always zero.) So, once you have calculated the slope you can put any of the two points known into the equation, leaving us with only the intercept 'b' unknown.
1 = (-6)*(-3) + b ; 1 = 18 + b ; 1 – 18 = b ; -17 = b
Thus the final equation is $y = - 6 \cdot x - 17$
We then check this by substituting the other known point into the equation:
-5 = (-6)*(-2) - 17 ; -5 = 12 - 17 ; -5 = -5 CORRECT! |
## 30 is what percentage of 43?
30 is 69.767% of 43
# Percentage calculators
Use these percentage calculators to work out percentages quickly. If you want to know how to work them out yourself, follow the examples below to learn how.
## Calculate what percent one value is of another
out of is how many percent?
## Calculate the percentage of a value
What is percent of ?
# The calculations explained
## How do I calculate what percent 10 is of 400?
To calculate 10 of 400 as a percentage, first calculate what one percent of 400 is. To do so, divide 400 by 100.
400 ÷ 100 = 4
Now divide 10 by 4 to get the answer.
10 ÷ 4 = 2%
## How do I calculate what 10% of 600 is?
First calculate what one percent of 600 is by dividing it by 100.
600 ÷ 100 = 6
Now multiply 10 by 6 to get the answer.
10 × 6 = 60
## Exercises
### Question
If Arthur takes 331 phones from a collection of 400 phones, what percent of the phones does Arthur have?
Arthur has 331 phones. There are 400 phones altogether. 1 percent of 400 is 400 ÷ 100 = 4. So, divide the amount Arthur has by one percent. 331 ÷ 4 = 82.75% . Arthur has 82.75% of the phones.
### Question
87 percent of 303 is what?
To work out 87% of 303, first divide 303 by 100 to work out one percent: 303 ÷ 100 = 3.03. Now times that by 100 to get the answer. 87 × 3.03 = 263.61
### Question
48 percent of 464 is what?
To calculate 48 percent of 464, first divide 464 by 100 to get 1 percent: 464 ÷ 100 = 4.64. Now times that by 100 to get the answer. 48 × 4.64 = 222.72
### Question
There is a trunk with 75 cakes from which Kevin helps themself to 24. What percentage of cakes does Kevin have?
Kevin has 24 cakes. There are 75 cakes in total. 1% of 75 is 75 ÷ 100 = 0.75. Next, divide the amount Kevin has by one percent. 24 ÷ 0.75 = 32% . Kevin has 32% of the cakes.
### Question
Find 76 percent of 455.
To work out 76 percent of 455, first divide 455 by 100 to determine 1%: 455 ÷ 100 = 4.55. Now times that by 100 to get the answer. 76 × 4.55 = 345.8
### Question
Out of 200 sweets, if Paul takes 37, what percent of the sweets does Paul have?
Paul has 37 sweets. There are 200 sweets altogether. One % of 200 is 200 ÷ 100 = 2. Now, divide the amount Paul has by one percent. 37 ÷ 2 = 18.5% . Paul has 18.5% of the sweets.
### Question
What is 40% of 477?
To get the 40 percent of 477, first divide 477 by 100 to determine one percent: 477 ÷ 100 = 4.77. Now times that by 100 to get the answer. 40 × 4.77 = 190.8
### Question
There is a total of 40 toys. Jo holds 2. What percent of all the toys does Jo have?
Jo has 2 toys. There are 40 toys in total. One % of 40 is 40 ÷ 100 = 0.4. Next, divide the amount Jo has by one percent. 2 ÷ 0.4 = 5% . Jo has 5% of the toys.
### Question
A collection contains 125 sweets. Abigail is given 44. How many percent of the 125 sweets does Abigail have?
Abigail has 44 sweets. There are 125 sweets altogether. 1 percent of 125 is 125 ÷ 100 = 1.25. Now, divide the amount Abigail has by one percent. 44 ÷ 1.25 = 35.2% . Abigail has 35.2% of the sweets.
### Question
From a bag of 250 pears, Laura takes 210 pears. What percent of the total pears does Laura have?
Laura has 210 pears. There are 250 pears combined. One % of 250 is 250 ÷ 100 = 2.5. Next, divide the amount Laura has by one percent. 210 ÷ 2.5 = 84% . Laura has 84% of the pears.
### Question
There is a bag with 88 cakes from which Irene helps themself to 48. What percentage of cakes does Irene have?
Irene has 48 cakes. There are 88 cakes as a whole. 1% of 88 is 88 ÷ 100 = 0.88. Now, divide the amount Irene has by one percent. 48 ÷ 0.88 = 54.55% . Irene has 54.55% of the cakes.
### Question
Find 88% of 447.
To calculate 88 percent of 447, first divide 447 by 100 to determine one%: 447 ÷ 100 = 4.47. Now times that by 100 to get the answer. 88 × 4.47 = 393.36
### Question
If Viv takes 303 fidget-spinners from a bucket of 600 fidget-spinners, what percent of the fidget-spinners does Viv have?
Viv has 303 fidget-spinners. There are 600 fidget-spinners in total. 1 percent of 600 is 600 ÷ 100 = 6. Now, divide the amount Viv has by one percent. 303 ÷ 6 = 50.5% . Viv has 50.5% of the fidget-spinners.
### Question
Find 54 percent of 302.
To work out 54 percent of 302, first divide 302 by 100 to get 1 percent: 302 ÷ 100 = 3.02. Now times that by 100 to get the answer. 54 × 3.02 = 163.08
### Question
59% of 145 is what?
To calculate 59% of 145, first divide 145 by 100 to work out one percent: 145 ÷ 100 = 1.45. Now multiply that by 100 to get the answer. 59 × 1.45 = 85.55
### Question
Find 73% of 138.
To get the 73 percent of 138, first divide 138 by 100 to work out one%: 138 ÷ 100 = 1.38. Now times that by 100 to get the answer. 73 × 1.38 = 100.74
### Question
There is a total of 50 sweets. Kylie holds 40. What percent of all the sweets does Kylie have? |
### Round and Round the Circle
What happens if you join every second point on this circle? How about every third point? Try with different steps and see if you can predict what will happen.
### Pumpkin Pie Problem
Peter wanted to make two pies for a party. His mother had a recipe for him to use. However, she always made 80 pies at a time. Did Peter have enough ingredients to make two pumpkin pies?
### The Moons of Vuvv
The planet of Vuvv has seven moons. Can you work out how long it is between each super-eclipse?
# Four Go
##### Stage: 2 Challenge Level:
This was a good solution that was sent in this time.
Abigail and Amber from Harrison Primary School wrote this very comprehensive reply;
We first started to think about what numbers could and couldn't be made with the digits and operations given. We took out all of the prime numbers (momentarily) as we know that they have no factors (2, 3, 5, 7, 11, 13, 17). However as some of those numbers could be made by multiplying two other numbers together we re-added some of them in (2, 3, 5,). Then we found 4 numbers in row that could be made and went for them for the highest chance of winning.
In our opinion we think it doesn't matter if you go first or second ( as they both have there advantages). We thought about it logically and found all easy possibilities of a sequence of 4 in a row. To do this we took out prime numbers that had no factors and couldn't be made by multiplying 2 numbers together then we looked at 4 numbers in a row that could be made.
Now we looked at would it matter if you went 1st or second? As we started to dig deeper we realize that they both had there pros and cons. When you went first you had the choice of picking what row you wanted to go for, however then the other person could see what angle you were going for and block you immediately. Then its was the same for going 2nd, you could
easily block your opponent yet they got the first chose of there 4 in the row sequence.
Here are some oter solutions to consider from when the game has been "live" before.
Arti wrote to us to say:
There are two things that are not defined which need a definition:
- If player A marked a number in the number line, can player B mark it later?
- If player A used two numbers from the square, can player B use one or both those numbers?
These are good questions, Arti. What did you decide in the game/s you played? Did that work well? You could try out both versions and decide which works better.
Rowena from Christ Church Primary told us:
I played this game with my Mum and neither of us won. We played it again and my Mum let me win!
We decided to list all the possible whole number answers. They were 2, 3, 4, 5, 6, 8, 9, 10, 12, 15 and 20. Once we knew these, it was easy to choose numbers to block the opponent and not let them get four in a row.
You can only win if your opponent makes a mistake or lets you win!
Thank you, Rowena - that was a good idea to make a list of the whole number answers.
I wonder whether you could change the game to make it a better game? |
# What is 62 80 as a grade?
Now we can see that our fraction is 77.5/100, which means that 62/80 as a percentage is 77.5%. And there you have it! Two different ways to convert 62/80 to a percentage.
## Can you simplify 64 81?
As you can see, 64/81 cannot be simplified any further, so the result is the same as we started with.
## What’s 65 80 as a percentage?
Solution and how to convert 65 / 80 into a percentage 0.81 times 100 = 81.25. That’s all there is to it!
## What’s 59 80 as a percentage?
Solution and how to convert 59 / 80 into a percentage 0.74 times 100 = 73.75. That’s all there is to it!
## Can a simplified fraction be a mixed number?
In order to simplify a fraction there must be: A number that will divide evenly into both the numerator and denominator so it can be reduced, or. The numerator must be greater than the denominator, (an improper fraction), so it can be converted to a mixed number.
## What percent is 64 out of 81?
Now we can see that our fraction is 79.012345679012/100, which means that 64/81 as a percentage is 79.0123%.
## What are all real square roots of 81?
Explanation: 81=9⋅9 then the square root of √81=9 . Because the double multiplication for the same sign is always positive, the square root is also valid with the other sign 81=(−9)⋅(−9) then √81=−9 and we can say that √81=±9 . What if we do not know the value?
## How do you calculate 80 percent marks?
20/100 x 100 = called as 20 percent = 20% 50/100 x 100 = called as 50 percent = 50% 80/100 x 100 = called as 80 percent = 80% 90/100 x 100 = called as 90 percent = 90%
## What is a 95 grade?
Thus, an A is a 95, halfway between 90 and 95. An A- is a 91.25, halfway between 90 and 92.5. Etc. Here is the conversion more precisely: Letter -> Number Conversion.
## What is 6 61 as a percentage?
What is this? Now we can see that our fraction is 9.8360655737705/100, which means that 6/61 as a percentage is 9.8361%.
## Can you reduce 7 40?
Find Out More About The Free Tools on Visual Fractions As you can see, 7/40 cannot be simplified any further, so the result is the same as we started with.
## What is 22.5 as a fraction?
2. What is 22.5% in the fraction form? 22.5% in the fraction form is 22.5/100. If you want you can simplify it further as 9/40.
## Can you simplify 25 81?
Find Out More About The Free Tools on Visual Fractions As you can see, 25/81 cannot be simplified any further, so the result is the same as we started with. Not very exciting, I know, but hopefully you have at least learned why it cannot be simplified any further!
## What is the percentage of 60 out of 80?
60/80 x 100 = 6/8 = 3/4 =75% of 100! that’s how you work out the percentage. you have 20 left out of 80 when you take 60.
## What percent is 75 out of 85?
Now we can see that our fraction is 88.235294117647/100, which means that 75/85 as a percentage is 88.2353%.
## Is a 59 passing?
This is an above-average score, between 80% and 89% C – this is a grade that rests right in the middle. D – this is still a passing grade, and it’s between 59% and 69% F – this is a failing grade.
## Do Grade 11 marks matter?
Yes, Grade 11 marks matter, but only for the time between Grade 11 and 12. Low Grade 11 marks will prevent you from applying early. |
# P(a, b) is a point in the first quadrant. Circles are drawn through P touching the coordinate axes, such that the length of common chord of these circles is maximum, find the ratio a : b?
## while solving the question, i realized that there will always be two such circles that can be drawn through P and satisfying the above criterion, with radius: r1 = a+b+(2ab)^0.5 and r2 = a+b-(2ab)^0.5 the length of the common chord then turned out to be something like L = root2*modulus(a-b) if the question were to minimize this length, a : b would have easily been 1. but for maximising, woudn't the answer be infinity?
Jun 27, 2018
$a : b = 3 \pm 2 \sqrt{2}$
#### Explanation:
In order for a circle to be touching both coordinate axes in the first quadrant, it must have its center located on the line $y = x$, as seen below:
The common chord is going to be $P Q$, where $Q$ is the symmetric point (a.k.a. the reflection) of $P$ with respect to the line $y = x$, hence $Q$ has coordinates $\left(b , a\right)$.
In order for the common chord to be the maximised, it has to be the diameter of the circle where $P$ and $Q$ are on its "high ends":
In our particular example, it almost overlaid one of the original circles.
Anyway, the center of this new circle is the midpoint of $P Q$, which we will denote $L$.
$L \left(\frac{a + b}{2} , \frac{a + b}{2}\right)$
The equation of the circle is:
${\left(x - \frac{a + b}{2}\right)}^{2} + {\left(y - \frac{a + b}{2}\right)}^{2} = {\left(\frac{P Q}{2}\right)}^{2}$
The lenght of $P Q$ is, as calculated by you, $\sqrt{2} | a - b |$.
After some arithmetic, we reach the equation
(x−a)(x−b)+(y−a)(y−b)=0
The circle with diameter $P Q$ must touch the axes as well, hence the points $\left(\left(a + b\right) \text{/} 2 , 0\right)$ and $\left(0 , \left(a + b\right) \text{/} 2\right)$ must be on it as well.
If we substitute the coordinates of one of these points (doesn't matter which; it gives us the same answer) into the circle equation we get the condition
${\left(a + b\right)}^{2} = 8 a b$
${a}^{2} + 2 a b + {b}^{2} = 8 a b$
${a}^{2} - 6 a b + {b}^{2} = 0$
Divide both sides by ${b}^{2}$:
${a}^{2} / {b}^{2} - 6 \frac{a}{b} + 1 = 0$
${\left(\frac{a}{b}\right)}^{2} - 6 \left(\frac{a}{b}\right) + 1 = 0$
This is a quadratic equation with the indeterminate the ratio of $a$ and $b$, exactly what we want!
The solutions of this equation are
$a : b = 3 \pm 2 \sqrt{2}$
These are the ratios we desire. |
Mixed Numbers and Fraction
How well do you understand Mixed Numbers and Fractions? To understand the relationship between the two we will look at how they are used to represent the same value, and how to convert from one form to another. Like all concepts in math, it is important to visualize the situation in different way. To do this we will use both fraction tiles and number lines to help us understand the problems.
Mixed numbers and fractions
Let’s look at the number 4$\frac{2}{3}$. We call this a mixed number, but why?
Simply put, a mixed number is a mixture of a whole number and a fraction. The 4 represents a whole number and the $\frac{2}{3}$ is a fraction that lets us know that we are dealing with a number that is little more than the original 4. In fact, we could write this as 4 + $\frac{2}{3}$
Therefore, we are looking at a number between 4 and 5 on a number line.
But where exactly would it lie on a number line?
Looking at the denominator of our fraction, we see that the whole was divided into thirds. So, we need to divide the spaces in between each number into three equal parts. Now we need to mark where two thirds would be located.
Is there any reason we couldn’t have just written the number as a fraction? Not really.
Counting each space on the number line, we can see that 1 is the same as $\frac{3}{3}$, 2 is the same as $\frac{6}{3}$, 3 is the same as $\frac{9}{3}$, and 4 is the same as $\frac{12}{3}$. Now we need to add the additional $\frac{2}{3}$, giving us $\frac{14}{3}$.
So, the mixed number 4$\frac{2}{3}$ represents the same amount as $\frac{14}{3}$.
Some people call $\frac{14}{3}$ an improper fraction, but there is nothing improper about this fraction. Depending on the situation, sometimes it is more useful to think of this as a mixed number and sometimes it is more useful to think of it as fraction.
So, now that we know what a mixed number is:
How do we convert mixed numbers to fractions?
Let’s start with 2$\frac{3}{5}$.
We can think of this as 2 + $\frac{3}{5}$, or 1 + 1 + $\frac{3}{5}$. One is the same as $\frac{5}{5}$ so this would be $\frac{5}{5}$ + $\frac{5}{5}$ + $\frac{3}{5}$. For a total of $\frac{13}{5}$. We could change any mixed number to a fraction by repeated addition, but that may become a problem if the whole number was much larger.
Instead of repeated addition, we could think of the problem as $\frac{5}{5}$ two times.
\begin{align*}
2 + \tfrac{3}{5} &= \\
\left(1 * 2\right)+ \tfrac{3}{5} &= \\
\left(\tfrac{5}{5}*2\right)+ \tfrac{3}{5} &= \\ \left(\tfrac{5}{5} * \tfrac{2}{1}\right) + \tfrac{3}{5} &= \\ \tfrac{10}{5} + \tfrac{3}{5} &= \tfrac{13}{5}
\end{align*}
Now let’s look at the reverse situation:
What if we had a fraction and we wanted to turn that into a mixed number?
$\frac{23}{7}$ = ?
We could using receptive adding until we got to $\frac{23}{7}$
\begin{align*}
\tfrac{7}{7} + \tfrac{7}{7} + \tfrac{7}{7} + \tfrac{2}{7} & = \tfrac{23}{7} \\
1 + 1 + 1 + \tfrac{2}{7} &= 3\tfrac{2}{7}
\end{align*}
OR we can remember that a fraction also represents division.
$\frac{23}{7} = 23 \div 7 = 3\tfrac{2}{7}$ |
# 1.6 Rational expressions
Page 1 / 6
In this section students will:
• Simplify rational expressions.
• Multiply rational expressions.
• Divide rational expressions.
• Add and subtract rational expressions.
• Simplify complex rational expressions.
A pastry shop has fixed costs of $\text{\hspace{0.17em}}\text{}280\text{\hspace{0.17em}}$ per week and variable costs of $\text{\hspace{0.17em}}\text{}9\text{\hspace{0.17em}}$ per box of pastries. The shop’s costs per week in terms of $\text{\hspace{0.17em}}x,$ the number of boxes made, is $\text{\hspace{0.17em}}280+9x.\text{\hspace{0.17em}}$ We can divide the costs per week by the number of boxes made to determine the cost per box of pastries.
$\frac{280+9x}{x}$
Notice that the result is a polynomial expression divided by a second polynomial expression. In this section, we will explore quotients of polynomial expressions.
## Simplifying rational expressions
The quotient of two polynomial expressions is called a rational expression . We can apply the properties of fractions to rational expressions, such as simplifying the expressions by canceling common factors from the numerator and the denominator. To do this, we first need to factor both the numerator and denominator. Let’s start with the rational expression shown.
$\frac{{x}^{2}+8x+16}{{x}^{2}+11x+28}$
We can factor the numerator and denominator to rewrite the expression.
$\frac{{\left(x+4\right)}^{2}}{\left(x+4\right)\left(x+7\right)}$
Then we can simplify that expression by canceling the common factor $\text{\hspace{0.17em}}\left(x+4\right).$
$\frac{x+4}{x+7}$
Given a rational expression, simplify it.
1. Factor the numerator and denominator.
2. Cancel any common factors.
## Simplifying rational expressions
Simplify $\text{\hspace{0.17em}}\frac{{x}^{2}-9}{{x}^{2}+4x+3}.$
Can the $\text{\hspace{0.17em}}{x}^{2}\text{\hspace{0.17em}}$ term be cancelled in [link] ?
No. A factor is an expression that is multiplied by another expression. The $\text{\hspace{0.17em}}{x}^{2}\text{\hspace{0.17em}}$ term is not a factor of the numerator or the denominator.
Simplify $\text{\hspace{0.17em}}\frac{x-6}{{x}^{2}-36}.$
$\frac{1}{x+6}$
## Multiplying rational expressions
Multiplication of rational expressions works the same way as multiplication of any other fractions. We multiply the numerators to find the numerator of the product, and then multiply the denominators to find the denominator of the product. Before multiplying, it is helpful to factor the numerators and denominators just as we did when simplifying rational expressions. We are often able to simplify the product of rational expressions.
Given two rational expressions, multiply them.
1. Factor the numerator and denominator.
2. Multiply the numerators.
3. Multiply the denominators.
4. Simplify.
## Multiplying rational expressions
Multiply the rational expressions and show the product in simplest form:
$\frac{\left(x+5\right)\left(x-1\right)}{3\left(x+6\right)}\cdot \frac{\left(2x-1\right)}{\left(x+5\right)}$
Multiply the rational expressions and show the product in simplest form:
$\frac{{x}^{2}+11x+30}{{x}^{2}+5x+6}\cdot \frac{{x}^{2}+7x+12}{{x}^{2}+8x+16}$
$\frac{\left(x+5\right)\left(x+6\right)}{\left(x+2\right)\left(x+4\right)}$
## Dividing rational expressions
Division of rational expressions works the same way as division of other fractions. To divide a rational expression by another rational expression, multiply the first expression by the reciprocal of the second. Using this approach, we would rewrite $\text{\hspace{0.17em}}\frac{1}{x}÷\frac{{x}^{2}}{3}\text{\hspace{0.17em}}$ as the product $\text{\hspace{0.17em}}\frac{1}{x}\cdot \frac{3}{{x}^{2}}.\text{\hspace{0.17em}}$ Once the division expression has been rewritten as a multiplication expression, we can multiply as we did before.
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 |
A Program of The Actuarial Foundation. Aligned with Common Core State and NCTM Standards.
What is an actuary? An actuary is an expert in statistics who works with businesses, governments, and organizations to help them plan for the future. Actuarial science is the discipline that applies math and statistical methods to assess risk.
4-6
DURATION
45
UNIT PLAN
Bars, Lines, & Pies!
Looking Through Line Graphs
In this lesson and activity, students understand how to use and read line graphs.
OBJECTIVE
Students will understand:
• that line graphs show how two pieces of information are related and how data changes over time;
• the dependent variable of a line graph typically appears on the Y-axis, and the independent variable appears on the X-axis;
• that line graphs are used to analyze the nature of changes in quantities.
MATERIALS
Reproducible Activity 3, rulers, colored pencils, calculators
REPRODUCIBLES
DIRECTIONS
1. Draw and label an X-axis and a Y-axis on the board. Tell students that they will be learning about line graphs in this activity. Explain that a line graph uses points and lines to examine changes over time. Line graphs are often used when examining relationships between two types of information.
2. Tell students that, like the bar graph, the line graph has an X- and a Y-axis. The dependent variable is plotted on the Y-axis and usually measures quantity (percentage, dollars, liters, etc.). The independent variable is plotted on the X-axis and usually measures time. Use the following data to complete your line graph on the board:
• Y-axis: \$0.50, \$0.75, \$1.00, \$1.25
• X-axis: year: 2005, 2006, 2007, 2008
3. Write the title “Cost of Milk at School” above the graph. Ask students where the first point in the graph should go if the cost of milk was 50 cents in 2005. Mark this point on the graph. Have students point out where other points should be marked. When done, connect the points with a line.
4. Distribute Reproducible Activity 3. Read the first paragraph as a class. When you are finished, point out the data group that students will be using to make their graph.
5. Direct students to the “Line It Up” question on the reproducible. Instruct them to use the data from the table to create a line graph in the space provided. Tell students they will need to create a frequency scale on the Y-axis to illustrate the frequency of each group of data. Remind students to include a title and labels on their graph. Once students have finished their graphs, instruct them to move on to the questions on the reproducible.
6. Once complete, review the answers to the reproducible as a class and invite students to share their line graphs with the class.
7. Ask the class when a line graph would be chosen to illustrate a data set. Have a volunteer give an example of when percentages might be illustrated using a line graph rather than a pie chart.
LESSON EXTENSION
Real-World Math:
1. Ask students to think of graphs that they have seen in the real world. For what purposes were they used? Have students hunt for examples in books, in magazines, on the Internet, in newspapers, and in business documents.
2. Review with students the definition of actuary on the poster. How can statistics help someone plan for the future? How might an actuary use graphs and math in the following real-world situations?
--Help a school principal plan a recycling program. How could math and graphs show what the school has used in the past, and how much could be saved in the future by recycling? (Use past data to figure out future data [extrapolate], and compare results in a graph.)
--Help the manager of a city plan for a second landfill. How much space would be needed for the new landfill? (Use past data from the first landfill, as well as data that reflects current use and extrapolate for future data. Display the findings in a graph.)
--Help the manager of a company figure out how much money could be saved by recycling over a period of 10 years. (A line graph would reflect the increase of money saved over a period of time.)
EMAIL THIS |
# solve the right triangle abc with c 90 degrees
A link to the app was sent to your phone. Solve the right triangle ABC, where C = 90°. Then, degrees Using the sine law, where a, b, and c are the sides of the triangle, So, and Simplifying, and solving for a and b, the measure of each side is a= 9.58 units b=35.74 units See if you an finish that one and then use a similar method to find b. Solve the right triangle ABC, with Upper C equals 90 degrees.
Get a free answer to a quick problem. Type integers or decimals.) Answer to: This problem refers to the right triangle ABC with C = 90 degrees.
Thus, in this type of triangle…
\tan \theta & = \frac{\text{Perpendicular}}{\text{Base}} \\[0.2 cm] This problem refers to the right triangle {eq}ABC $$,$$\displaystyle \tan \theta = \frac{\text{Perpendicular}}{\text{Base}}
30°-60°-90° triangle: The 30°-60°-90° refers to the angle measurements in degrees of this type of special right triangle. o B= You can use the basic trig. (Round to the nearest minute as needed.) Substitute in angle A and the hypotenuse c and you will be able to solve for side a. Whatever yards are. $$,$$\displaystyle \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} To find a, notice that a is opposite the angle A and 8.4 is the hypotenuse.
{/eq}. \tan B & = \frac{b}{a} \\[0.2 cm] a = 11 m, c= 29 m b A= m (Round to the nearest whole number as needed.). $$.$$. & = \color {bluee} { 67.52 ^\circ}. & = \color {bluee} { 22.48 ^\circ}. Type an integer … A = 57 Degrees 16', c = 8.4 ft. n the right triangle ABC, with C = 90 .
The {eq}ABC {/eq}. Earn Transferable Credit & Get your Degree, Get access to this video and our entire Q&A library, Who is Euclid? (Round to the nearest minute as needed.)
A = angle A B = angle B C = angle C a = side a b = side b c = side c P = perimeter s = semi-perimeter K = area r = radius of inscribed circle R = radius of circumscribed circle $$,$$\begin{align} In the right triangle ABC, c^{2} & = a^{2} + b^{2} \\[0.2 cm]
Sciences, Culinary Arts and Personal \\[0.2 cm] C=90°. © copyright 2003-2020 Study.com. \end{align} angleB therefore =90-27.9=62.1 degrees #tan A = a/b = a/29.7 #--> # a = 29.7.tan (31.5) = 29.7(0.61) = 18.20#, #cos A = 0.85 = b/c = 31.5/c# --> #c = 31.5/(0.85) = 37.06#, 10815 views Draw the right triangle and put the numbers you have in your diagram. $$\displaystyle (\text{Hypotenuse})^{2} = (\text{Base})^{2} + (\text{Perpendicular })^{2} Triangle ABC is a right triangle with sides of lengths a, b, and c and a right angle at C. Find the indicated trigonometric function value of A. & = \color {blue} { 94.15} ~\rm ft. \\[0.2 cm] How do you find the missing angles when the sides of a right triangle are given? For Free, Deriving Trig Identities with Euler's Formula. How do you find leg length, BC, to the nearest tenth if in a right triangle ABC, the hypotenuse... How do you solve right triangles using a graphing calculator? A & = \tan ^{-1} \left(\frac{36}{87}\right) \\[0.2 cm] Services, Thales & Pythagoras: Early Contributions to Geometry, Working Scholars® Bringing Tuition-Free College to the Community. Start here or give us a call: (312) 646-6365, © 2005 - 2020 Wyzant, Inc. - All Rights Reserved, a Question C=90°. & = 36^{2} + 87^{2} 7\\[0.2 cm] In this type of right triangle, the sides corresponding to the angles 30°-60°-90° follow a ratio of 1:√ 3:2. - Definition, Shapes & Angles, Balanced Chemical Equation: Definition & Examples, Research Variables: Dependent, Independent, Control, Extraneous & Moderator, Eukaryotic and Prokaryotic Cells: Similarities and Differences, High School Algebra II: Homework Help Resource, High School Algebra I: Homeschool Curriculum, NY Regents Exam - Geometry: Test Prep & Practice, AP Calculus AB & BC: Homeschool Curriculum, TExES Mathematics 7-12 (235): Practice & Study Guide, GED Math: Quantitative, Arithmetic & Algebraic Problem Solving, High School Trigonometry: Help and Review, High School Algebra I: Homework Help Resource, Introduction to Statistics: Help and Review, Biological and Biomedical Solve right triangle knowing A = 31.5, b = 29.7 B = 58.5 deg Find angle B, sides c and a. What are inverse trigonometric functions and when do you use it? Inv tanA=39.9/75.4=27.9 degrees. B = 32 degrees 44' What is is the length of side a = and of side b = & = 1296 + 7569 \\[0.2 cm] Choose an expert and meet online. \\[0.2 cm] B = 58.5 deg, B = 90 - A = 90 - 31.5 = 58.5 deg Our experts can answer your tough homework and study questions. Rationalize denominators containing radicals. {/eq} with {eq}C = 90 ^{\circ} - Biography, Contribution & Theorems, Thales the Philosopher: Theory & Contributions to Philosophy, Undefined Terms of Geometry: Concepts & Significance, Similar Triangles: Definition, Formula & Properties, Blaise Pascal: Contributions, Inventions & Facts, The Theory of Forms by Plato: Definition & Examples, What is a Fraction? Special Right Triangles. All rights reserved. What will be the length of the shadow of the tower, correct to the nearest meter, on a day that... What is the angle θ of the ramp if monster truck drives off a ramp in order to jump on to a row... How do I find the sides of a right triangle using only the trig functions sin cos and tan? No packages or subscriptions, pay only for the time you need.$$, \begin{align} \end{align} Since angle C is 90 degrees and angle A is 15 degrees. {/eq}. B & = \tan ^{-1} \left(\frac{87}{36}\right) \\[0.2 cm] Type an integer or a decimal. See all questions in Solving Right Triangles. around the world., $$\displaystyle \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} If side {eq}a = 36 All other trademarks and copyrights are the property of their respective owners. \tan \theta & = \frac{\text{Perpendicular}}{\text{Base}} \\[0.2 cm] - Definition and Types, 30-60-90 Triangle: Theorem, Properties & Formula, The Differences Between Inductive and Deductive Reasoning, What is a Polygon? {/eq}, and side {eq}c Give angles in degrees and minutes. The sum of the interior angles of a triangle is 180 degrees. = m (Simplity your answer. {/eq}, angle {eq}B answered 06/15/20, Highly Experienced, Patient High School, College, SAT/ACT Math Tutor. Solve the right triangle ABC, with Upper C equals 90 degrees. c & = \sqrt{8865} \\[0.2 cm] B = 90 - A = 90 - 31.5 = 58.5 deg In the right triangle ABC, tan A = a/b = a/29.7 --> a = 29.7.tan (31.5) = 29.7(0.61) = 18.20 cos A = 0.85 = b/c = 31.5/c --> c = 31.5/(0.85) = 37.06 What is is the length of side a = and of side b =, Tom S. Solve right triangle knowing A = 31.5, b = 29.7 Solve the right triangle ABC, with C = 90° B = 30° 14. c = 0.6295 m ws A-' (Simplify your answers. According to this theorem, in a right triangle, the area of the square of the hypotenuse side is always equal to the sum of the areas of the individual squares having the base and the perpendicular as the side. Find an angle theta between 0 degree and 90 degree having the specified function value. How do you find all the missing angles, if you know one of the acute angles of a right triangle? Round your answer to the nearest hundredth of a degree. Solve for all the missing parts using the given information. Now applying the Pythagoras theorem in the given right triangle:$$\begin{align}
Round to four decimal places as needed) bm (Simplify your answer. (Round your answer to the nearest whole number.).
{/eq} ft and side {eq}b = 87 functions to find the two missing sides. How do you know what trigonometric function to use to solve right triangles? Most questions answered within 4 hours. {/eq} is a right triangle with {eq}\angle C = 90^\circ \tan A & = \frac{a}{b} \\[0.2 cm] \end{align} c=(75.4^2+39.9^2)^1/2 yards. (\text{Hypotenuse})^{2} &= (\text{Base})^{2} + (\text{Perpendicular })^{2} \\[0.2 cm] What do you mean by dinding?
{/eq} ft, find angle {eq}A Solve for all the missing parts using the given information. In Euclidean geometry, the Pythagoras theorem is applicable in the right triangles. |
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# 10. Vector Calculus
## Rate of Change of Variable Vectors
Rate of Change of a Position Vector in Elliptical Motion
We saw in Variable Vectors how vectors can vary with time. In this section, we learn how to find the rate of change of such varying vectors.
To find the time rate of change of a vector, we simply differentiate each component.
### Example 1
Let's consider the 2-dimensional force vector example from before:
F = (3t2 + 5) i + 4t j
The time rate of change of this vector is given by the derivative with respect to t of each component.
(dbb{text(F)})/dt=6t\ bb{text(i)}+4\ bb{text(j)}
At time t = 5, the rate of change of the vector F is the vector 30 i + 4 j. The units will be N/s.
### Example 2
The 3-D acceleration vector we met earlier in Example 2, Variable Vectors was given by
A = (5t) i + (2t + 3) j + (t2 + 10) k
The rate of change of this vector is given by differentiating each term, as follows:
(dbb{text(A)})/dt=5\ bb{text(i)}+2\ bb{text(j)} +2t\ bb{text(k)}
The units will be m/s3.
At the specific time t = 4 s, the rate of change of the vector will be
5 i + 2 j + 8 k m/s3
## Rate of Change of a Position Vector in Elliptical Motion
Next we consider the case when the terminal point of a vector is moving in an ellipse.
### Example 3
The following vector (units in m) follows such a pattern at time t (units in s):
P = (3 cos t) i + (sin t) j
This expression is based on the expression for a circle , P = (cos t) i + (sin t) j. The 3 in the i term stretches the circle into an ellipse. Of course, the time t is measured in radians, not degrees.
Following is the graph of the motion of the terminal point of the ellipse.
The terminal point starts at t = 0 s at the position (3, 0) and proceeds in an anti-clockwise direction. Its position at various times is indicated on the graph.
This idea of a curve being generated as a point moves in time is the same concept as Parametric Equations that we came across before.
The resulting vector has initial point at the origin as above. The vector for the cases t = 0 s (magnitude 3 m, direction horizontal, to the right), t = 1 s and t = 2 s are shown below:
To find the time rate of change of the position vector in elliptical motion, we differentiate the terms as we did earlier.
Since
P = 3 cos t i + sin t j
then
(dbb{text(P)})/dt=-3\ sint\ bb{text(i)}+cost\ bb{text(j)}
Note that the magnitude of the vector changes as time goes on because the terminal point is moving around the ellipse.
To find the rate of change at particular times, we substitute in values of t.
At t = 0 s,
rate of change = 0 i + j = j
Considering our diagram above, this answer makes sense. When the terminal point starts to move, it is in the vertical direction only (no horizontal component is present).
At t = 1 s, we expect a negative horizontal component and a positive vertical component (the terminal point is moving up and to the left at t = 1 s) . Substituting t = 1 (in radians, of course) gives us:
rate of change = (-3 sin 1) i + (cos 1) j = -2.52 i + 0.54 j
The x-value is negative and the y-value is positive, as expected.
At t = 6 s, we expect a positive horizontal component (the terminal point is moving to the right) and a positive vertical component (the terminal point is moving up).
rate of change = (-3 sin 6) i + (cos 6) j = 0.84 i + 0.96 j
Both components are positive, as expected. |
How do you find the vertical, horizontal or slant asymptotes for y = (8 x^2 + x - 2)/(x^2 + x - 72)?
Oct 23, 2016
The vertical asymptotes are $x = 8$ and $x = - 9$
and the horizontal asymptote is $y = 8$
Explanation:
We start by factorising the denominator
${x}^{2} + x - 72 = \left(x - 8\right) \left(x + 9\right)$
As we cannot divide by so $x \ne 8$ and $x \ne - 9$
So the vertical asymptotes are $x = 8$ and $x = - 9$
As the degree of the polynomial of the numerator and denominator are the same, there is no slant asymptote.
To find the horizontal asymptote, we find the limit as $x \to \pm \infty$
We take the highest order of the polynomials
$\lim y = \frac{8}{1}$
$x \to - \infty$
And
$\lim y = \frac{8}{1}$
$x \to \infty$
So the horizontal asymptote is $y = 8$ |
Mathematics NCERT Grade 8, Chapter 15: Introduction to Graphs- The chapter includes a sectional explanation of the concepts related to graphs
The chapter focuses on topics like A Bar graphA Pie graphA histogram, A line graph
• A bar graph is used to show a comparison among categories. It may contain parallel vertical or horizontal bars or rectangles.
• A pie graph is used to compare parts of a whole. It is also referred to as circle graphs.
• A histogram is a bar graph that shows data in intervals. It consists of adjacent bars over the intervals.
• A line graph displays data that change continuously over periods of times
Apart from this, the emphasis is laid upon Linear graphs and Some Applications
• A line graph which is a whole unbroken line is called a linear graph.
For a linear graph, different aspects are explained in this chapter.
• Location of a point
• Coordinates: Vertical Axis is called the Y-axis. Horizontal Axis is called the x-axis.
• For fixing a point on the graph sheet we need, x-coordinate and y-coordinate.
The concept of Dependent and Independent variables is also stated in this chapter- Introduction to Graphs
• The relation between the dependent variable and the independent variable is explained with the help of a graph.
Applications of graphs are explained through various examples on quantity and cost, principal and simple interest, time and distance.
Understanding of each topic of the chapter, Introduction to Graphs is made easier by adding examples, activities, and diagrams.
In addition to this, the topic-wise assessment of the understanding is done with the help of questions provided at the end of each topic.
A summary of all the important points of the chapter, Introduction to Graphs is provided at the end followed by a set of exercises that tests the overall understanding of concepts mentioned in the chapter.
#### Question 1:
The following graph shows the temperature of a patient in a hospital, recorded every hour.
(a) What was the patient’s temperature at 1 p.m.?
(b) When was the patient’s temperature 38.5°C?
(c) The patient’s temperature was the same two times during the period given. What were these two times?
(d) What was the temperature at 1.30 p.m? How did you arrive at your answer?
(e) During which periods did the patient’s temperature show an upward trend?
(a) At 1 p.m., the patient’s temperature was 36.5°C.
(b) The patient’s temperature was 38.5°C at 12 noon.
(c) The patient’s temperature was same at 1 p.m. and 2 p.m.
(d) The graph between the times 1 p.m. and 2 p.m. is parallel to the x-axis. The temperature at 1 p.m. and 2 p.m. is 36.5°C. So, the temperature at 1:30 p.m. is 36.5°C.
(e) During the following periods, the patient’s temperature showed an upward trend.
9 a.m. to 10 a.m., 10 a.m. to 11 a.m., 2 p.m. to 3 p.m.
##### Video Solution for introduction to graphs (Page: 236 , Q.No.: 1)
NCERT Solution for Class 8 maths - introduction to graphs 236 , Question 1
#### Question 2:
The following line graph shows the yearly sales figure for a manufacturing company.
(a) What were the sales in (i) 2002 (ii) 2006?
(b) What were the sales in (i) 2003 (ii) 2005?
(c) Compute the difference between the sales in 2002 and 2006.
(d) In which year was there the greatest difference between the sales as compared to its previous year?
(a)
(i) In 2002, the sales were Rs 4 crores.
(ii) In 2006, the sales were Rs 8 crores.
(b)
(i) In 2003, the sales were Rs 7 crores.
(ii) In 2005, the sales were Rs 10 crores.
(c)
(i) In 2002, the sales were Rs 4 crores and in 2006, the sales were Rs 8 crores.
Difference between the sales in 2002 and 2006
= Rs (8 − 4) crores = Rs 4 crores
(d) Difference between the sales of the year 2006 and 2005
= Rs (10 − 8) crores = Rs 2 crores
Difference between the sales of the year 2005 and 2004
= Rs (10 − 6) crores = Rs 4 crores
Difference between the sales of the year 2004 and 2003
= Rs (7 − 6) crore = Rs 1 crore
Difference between the sales of the year 2003 and 2002
= Rs (7 − 4) crores = Rs 3 crores
Hence, the difference was the maximum in the year 2005 as compared to its previous year 2004.
#### Question 3:
For an experiment in Botany, two different plants, plant A and plant B were grown under similar laboratory conditions. Their heights were measured at the end of each week for 3 weeks. The results are shown by the following graph.
(a) How high was Plant A after (i) 2 weeks (ii) 3weeks?
(b) How high was Plant B after (i) 2 weeks (ii) 3weeks?
(c) How much did Plant A grow during the 3rd week?
(d) How much did Plant B grow from the end of the 2nd week to the end of the 3rd week?
(e) During which week did Plant A grow most?
(f) During which week did Plant B grow least?
(g) Were the two plants of the same height during any week shown here? Specify.
(a)
(i) After 2 weeks, the height of plant A was 7 cm.
(ii) After 3 weeks, the height of plant A was 9 cm.
(b)
(i) After 2 weeks, the height of plant B was 7 cm.
(ii) After 3 weeks, the height of plant B was 10 cm.
(c) Growth of plant A during 3rd week = 9 cm − 7 cm = 2 cm
(d) Growth of plant B from the end of the 2nd week to the end of the 3rd week
= 10 cm − 7 cm = 3 cm
(e) Growth of plant A during 1st week = 2 cm − 0 cm = 2 cm
Growth of plant A during 2nd week = 7 cm − 2 cm = 5 cm
Growth of plant A during 3rd week = 9 cm − 7 cm = 2 cm
Therefore, plant A grew the most, i.e. 5 cm, during the 2nd week.
(f) Growth of plant B during 1st week = 1 cm − 0 cm = 1 cm
Growth of plant B during 2nd week = 7 cm − 1 cm = 6 cm
Growth of plant B during 3rd week = 10 cm − 7 cm = 3 cm
Therefore, plant B grew the least, i.e. 1 cm, during the 1st week.
(g) At the end of the 2nd week, the heights of both plants were same.
#### Question 4:
The following graph shows the temperature forecast and the actual temperature for each day of a week.
(a) On which days was the forecast temperature the same as the actual temperature?
(b) What was the maximum forecast temperature during the week?
(c) What was the minimum actual temperature during the week?
(d) On which day did the actual temperature differ the most from the forecast temperature?
(a) The forecast temperature was same as the actual temperature on Tuesday, Friday, and Sunday.
(b) The maximum forecast temperature during the week was 35°C.
(c) The minimum actual temperature during the week was 15°C.
(d) The actual temperature differs the most from the forecast temperature on Thursday.
#### Question 5:
Use the tables below to draw linear graphs.
(a) The number of days a hill side city received snow in different years.
Year 2003 2004 2005 2006 Days 8 10 5 12
(b) Population (in thousands) of men and women in a village in different years.
Year 2003 2004 2005 2006 2007 Number of men 12 12.5 13 13.2 13.5 Number of women 11.3 11.9 13 13.6 12.8
(a) By taking the years on x-axis and the number of days on y-axis and taking scale as 1 unit = 2 days on y-axis and 2 unit = 1 year on x-axis, the linear graph of the given information can be drawn as follows.
(b) By taking the years on x-axis and population on y-axis and scale as 1 unit = 0.5 thousand on y-axis and 2 unit = 1 year on x-axis, the linear graph of the given information can be drawn as follows.
#### Question 6:
A courier-person cycles from a town to a neighboring suburban area to deliver a parcel to a merchant. His distance from the town at different times is shown by the following graph.
(a) What is the scale taken for the time axis?
(b) How much time did the person take for the travel?
(c) How far is the place of the merchant from the town?
(d) Did the person stop on his way? Explain.
(e) During which period did he ride fastest?
(a) Scale taken for the time axis is 4 units = 1 hour
(b) The person travelled during the time 8 a.m. − 11:30 a.m.
Therefore, the person took hours to travel.
(c) The merchant is 22 km far from the town.
(d) Yes, the person stopped on his way from 10 a.m. to 10: 30 a.m. This is indicated by the horizontal part of the graph.
(e) From the graph, it can be observed that during 8 a.m. to 9 a.m., the person travelled the maximum distance. Thus, the person’s ride was the fastest between 8 a.m. and 9 a.m.
#### Question 7:
Can there be a time temperature graph as follows? Justify you’re answer:
(i) (ii) (iii) (iv)
(i) This can be a time−temperature graph, as the temperature can increase with the increase in time.
(ii) This can be a time−temperature graph, as the temperature can decrease with the decrease in time.
(iii) This cannot be a time−temperature graph since different temperatures at the same time are not possible.
(iv) This can be a time−temperature graph, as same temperature at different times is possible.
##### Video Solution for introduction to graphs (Page: 239 , Q.No.: 7)
NCERT Solution for Class 8 maths - introduction to graphs 239 , Question 7
#### Question 1:
Plot the following points on a graph sheet. Verify if they lie on a line
(a) A(4, 0), B(4, 2), C(4, 6), D(4, 2.5)
(b) P(1, 1), Q(2, 2), R(3, 3), S(4, 4)
(c) K(2, 3), L(5, 3), M(5, 5), N(2, 5)
(a) We can plot the given points and join the consecutive points on a graph paper as follows.
From the graph, it can be observed that the points A, B, C, and D lie on the same line.
(b) We can plot the given points and join the consecutive points on a graph paper as follows.
Hence, points P, Q, R, and S lie on the same line.
(c) We can plot the given points and join the consecutive points on a graph paper as follows.
Hence, points K, L, M, and N are not lying on the same line.
#### Question 2:
Draw the line passing through (2, 3) and (3, 2). Find the coordinates of the points at which this line meets the x-axis and y-axis.
From the graph, it can be observed that the line joining the points (2, 3) and (3, 2) meets the x-axis at the point (5, 0) and the y-axis at the point (0, 5).
##### Video Solution for introduction to graphs (Page: 243 , Q.No.: 2)
NCERT Solution for Class 8 maths - introduction to graphs 243 , Question 2
#### Question 3:
Write the coordinates of the vertices of each of these adjoining figures.
The coordinates of the vertices in the given figure are as follows.
O (0, 0), A (2, 0), B (2, 3), C (0, 3)
P (4, 3), Q (6, 1), R (6, 5), S (4, 7)
K (10, 5), L (7, 7), M (10, 8)
##### Video Solution for introduction to graphs (Page: 243 , Q.No.: 3)
NCERT Solution for Class 8 maths - introduction to graphs 243 , Question 3
#### Question 4:
State whether True or False. Correct those are false.
(i) A point whose x coordinate is zero and y-coordinate is non-zero will lie on the y-axis.
(ii) A point whose y coordinate is zero and x-coordinate is 5 will lie on y-axis.
(iii) The coordinates of the origin are (0, 0).
(i) True
(ii) False
The point whose y-coordinate is zero and x-coordinate is 5 will lie on x-axis.
(iii) True
##### Video Solution for introduction to graphs (Page: 243 , Q.No.: 4)
NCERT Solution for Class 8 maths - introduction to graphs 243 , Question 4
#### Question 1:
Draw the graphs for the following tables of values, with suitable scales on the axes.
(a) Cost of apples
Number of apples 1 2 3 4 5 Cost (in Rs) 5 10 15 20 25
(b) Distance travelled by a car
Time (in hours) 6 a.m. 7 a.m. 8 a.m. 9 a.m. Distance (in km) 40 80 120 160
(i) How much distance did the car cover during the period 7.30 a.m. to 8 a.m.?
(ii) What was the time when the car had covered a distance of 100 km since its start?
(c) Interest on deposits for a year:
Deposit (in Rs) 1000 2000 3000 4000 5000 Simple interest (in Rs) 80 160 240 320 400
(i) Does the graph pass through the origin?
(ii) Use the graph to find the interest on Rs 2500 for a year:
(iii) To get an interest of Rs 280 per year, how much money should be deposited?
(a) Taking a suitable scale (for x-axis, 1 unit = 1 apple and for y-axis, 1 unit = Rs 5), we can mark the number of apples on x-axis and the cost of apples on y-axis. A graph of the given data is as follows.
(b) Taking a suitable scale (for x-axis, 2 units = 1 hour and for y-axis, 2 units = 40 km), we can represent the time on x-axis and the distance covered by the car on y-axis. A graph of the given data is as follows.
(i) During the period 7:30 a.m. to 8 a.m., the car covered a distance of 20 km.
(ii) The car covered a distance of 100 km at 7:30 a.m. since its start.
(c) Taking a suitable scale,
For x-axis, 1 unit = Rs 1000 and for y-axis, 1 unit = Rs 80
We can represent the deposit on x-axis and the interest earned on that deposit on y-axis. A graph of the given data is obtained as follows.
From the graph, the following points can be observed.
(i) Yes. The graph passes through the origin.
(ii) The interest earned in a year on a deposit of Rs 2500 is Rs 200.
(iii) To get an interest of Rs 280 per year, Rs 3500 should be deposited.
##### Video Solution for introduction to graphs (Page: 247 , Q.No.: 1)
NCERT Solution for Class 8 maths - introduction to graphs 247 , Question 1
#### Question 2:
Draw a graph for the following.
(i)
Side of square (in cm) 2 3 3.5 5 6 Perimeter (in cm) 8 12 14 20 24
Is it a linear graph?
(ii)
Side of square (in cm) 2 3 4 5 6 Area (in cm2) 4 9 16 25 36
Is it a linear graph?
(i) Choosing a suitable scale,
For x-axis, 1 unit = 1 cm and for y-axis, 1 unit = 4 cm
We can represent the side of a square on x-axis and the perimeter of that square on y-axis. A graph of the given data is drawn as follows.
It is a linear graph.
(ii)Choosing a suitable scale,
For x-axis, 1 unit = 1 cm and for y-axis, 1 unit = 4 cm2
We can represent the side of a square on the x-axis and the area of that square on y-axis. A graph of the given data is as follows.
It is not a linear graph.
View NCERT Solutions for all chapters of Class 8 |
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# In the Figure; Pa is a Tangent to the Circle, Pbc is Secant and Ad Bisects Angle Bac. Show that Triangle Pad is an Isosceles Triangle. Also, Show That: ∠Cad =1/2(∠Pba-∠Pab) - ICSE Class 10 - Mathematics
ConceptTangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments
#### Question
In the figure; PA is a tangent to the circle, PBC is secant and AD bisects angle BAC. Show that triangle PAD is an isosceles triangle. Also, show that:
∠CAD =1/2(∠PBA-∠PAB)
#### Solution
i) PA is the tangent and AB is a chord
∴ ∠PAB = ∠C …….. (i) ( angles in the alternate segment)
AD is the bisector of ∠BAC
∴ ∠1 = ∠2 ……….(ii)
Ext. ∠ADP = ∠C + ∠1
⇒ Ext ∠ADP = ∠PAB + ∠2 = ∠PAD
Therefore, Δ PAD is an isosceles triangle.
ii) In ΔABC,
Ext. ∠PBA = ∠C + ∠BAC
∠BAC = ∠PBA - ∠C
⇒ ∠1 + ∠2 = ∠PBA - ∠PAB
(fom (i) part)
2∠1 = ∠PBA - ∠PAB
∠1=1/2(∠PBA - ∠PAB)
⇒∠CAD = 1/2 (∠PBA - ∠ PAB)
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [1]
Solution In the Figure; Pa is a Tangent to the Circle, Pbc is Secant and Ad Bisects Angle Bac. Show that Triangle Pad is an Isosceles Triangle. Also, Show That: ∠Cad =1/2(∠Pba-∠Pab)` Concept: Tangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments.
S |
# To prove cos3theta +2cos5theta +cos7theta/costheta +2cos3theta +cos5theta=cos2theta-sin2theta×tan3theta ?
Jul 22, 2018
Please refer to a Proof in Explanation.
#### Explanation:
$\text{The Expression} = \frac{\cos 3 \theta + 2 \cos 5 \theta + \cos 7 \theta}{\cos \theta + 2 \cos 3 \theta + \cos 5 \theta}$.
$\text{The Nr.} = \left(\cos 3 \theta + 2 \cos 5 \theta + \cos 7 \theta\right)$
$= \left(\cos 7 \theta + \cos 3 \theta\right) + 2 \cos 5 \theta$,
$= 2 \cos \left(\frac{7 \theta + 3 \theta}{2}\right) \cos \left(\frac{7 \theta - 3 \theta}{2}\right) + 2 \cos 5 \theta$,
$= 2 \cos 5 \theta \cos 2 \theta + 2 \cos 5 \theta$,
$\therefore \text{ The Nr.} = 2 \cos 5 \theta \left(\cos 2 \theta + 1\right)$.
$\text{On similar lines the Dr.} = 2 \cos 3 \theta \left(\cos 2 \theta + 1\right)$.
$\therefore \text{The Exp.} = \frac{\cos 5 \theta}{\cos 3 \theta}$,
$= \frac{\cos \left(3 \theta + 2 \theta\right)}{\cos 3 \theta}$,
$= \frac{\cos 3 \theta \cos 2 \theta - \sin 3 \theta \sin 2 \theta}{\cos 3 \theta}$,
$= \frac{\cos 3 \theta \cos 2 \theta}{\cos 3 \theta} - \frac{\sin 3 \theta \sin 2 \theta}{\cos 3 \theta}$,
$= \cos 2 \theta - \left(\sin 2 \theta\right) \left\{\frac{\sin 3 \theta}{\cos 3 \theta}\right\}$,
$= \cos 2 \theta - \sin 2 \theta \cdot \tan 3 \theta$, as desired! |
Algebra 1 : Absolute Value Inequalities
Example Questions
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Example Question #1 : Absolute Value Inequalities
Solve for :
The inequality has no solution.
The inequality has no solution.
Explanation:
The absolute value of a number must always be nonnegative, so can never be less than . This means the inequality has no solution.
Example Question #1 : Absolute Value Inequalities
Solve the inequality .
Explanation:
First, we can simplify this inequality by subtracting 7 from both sides. This gives us
Next, however, we need to make two separate inequalities due to the presence of an absolute value expression. What this inequality actually means is that
and
(Be careful with the inequality signs here! The second sign must be switched to allow for the effect of absolute value on negative numbers. In other words, the inequality must be greater than because, after the absolute value is applied, it will be less than 7.) When we solve the two inequalities, we get two solutions:
and
For the original statement to be true, both of these inequalities must be fulfilled. We're left with a final answer of
Example Question #1 : Absolute Value Inequalities
Solve the inequality:
(no solution)
(no solution)
Explanation:
The inequality compares an absolute value function with a negative integer. Since the absolute value of any real number is greater than or equal to 0, it can never be less than a negative number. Therefore, can never happen. There is no solution.
Example Question #1 : Solving Inequalities
Solve this inequality.
Explanation:
Split the inequality into two possible cases as follows, based on the absolute values.
First case:
Second case:
Let's find the inequality of the first case.
Multiply both sides by x + 6.
Subtract x from both sides, then subtract 3 from both sides.
Divide both sides by 3.
Let's find the inequality of the second case.
Multiply both sides by x + 6.
Simplify.
Add x to both sides, then subtract 3 from both sides.
Divide both sides by 5.
So the range of x-values is and .
Example Question #1 : Solving Absolute Value Equations
Solve for :
Explanation:
Solve for positive values by ignoring the absolute value. Solve for negative values by switching the inequality and adding a negative sign to 7.
Example Question #1 : Solving Absolute Value Equations
Give the solution set for the following equation:
Explanation:
First, subtract 5 from both sides to get the absolute value expression alone.
Split this into two linear equations:
or
The solution set is
Example Question #81 : Linear Inequalities
Solve for in the inequality below.
No solutions
All real numbers
Explanation:
The absolute value gives two problems to solve. Remember to switch the "less than" to "greater than" when comparing the negative term.
or
Solve each inequality separately by adding to all sides.
or
This can be simplified to the format .
Explanation:
Example Question #3 : Absolute Value Inequalities
Solve the inequality.
Explanation:
Remove the absolute value by setting the term equal to either or . Remember to flip the inequality for the negative term!
Solve each scenario independently by subtracting from both sides.
Solve for : |
# LCM and GCD Exercises Set 2
Here are some Civil Service exam exercises on GCD and LCM.
1.) What is the GCD of 8, 20, and 28?
2.) What is the GCD of 21, 35, and 56?
3.) What is 18/54 in lowest terms?
4.) What is 38/95 in lowest terms?
5.) What is the LCM of 6 and 8?
6.) What is the LCM of 5, 6, and 12?
7.) What is the product of the LCM and the GCD of 4, 8, and 20?
8.) There are 18 red marbles and 27 blue marbles to be distributed among children. What is the maximum number of children that can receive the marbles if each kid receives the same number of marbles for each color and no marble is to be left over?
9.) In a school sportsfest, there are 60 Grade 4 pupils, 48 Grade 5 pupils and 36 Grade 6 pupils. What is the largest number of teams that can be formed if the pupils in each Grade level are equally distributed and no pupil is left without a team?
10.) In a disco, the red lights blink every 3 seconds and the blue lights blink every 5 seconds. If the two colored lights blink at the same time if you turn them on, they will blink at the same time every ___ seconds.
Solution:
Divisors of 8 – 1, 2, 4, 8
Divisors of 20 – 1, 2, 4, 5, 10, 20
Divisors of 28 – 1, 2, 4, 7, 14, 28
Solution:
Divisors of 21 – 1, 3, 7, 21
Divisors of 35 -1, 3, 5, 7, 35
Divisors of 56 -1, 2, 7, 8, 14, 28, 56
Note: Reducing fractions to lowest terms is one of the applications of GCD.
Solution:
Divisors of 18 – 1, 2, 6, 9, 18
Divisors of 54 – 1, 2, 3, 9, 18, 27, 54
Numerator = 18 divided by 18 (GCD) = 1
Denominator = 54 divided by 18 (GCD) = 3
Solution:
Divisors of 38 – 1, 2, 19, 38
Divisors of 95 – 1, 2, 3, 19, 95
Solution:
Multiples of 6 – 6, 12, 18, 24
Multiples of 8 – 8, 16, 24
Solution:
Multiples of 5 – 5, 10, 15, 20, 25, 30 … 55, 60
Multiples of 6 – 6, 12, 18, 24, 30, 36, 42, 48, 54, 60
Multiples of 12 – 12, 24, 36, 48, 60
Solution
GCD of 4, 6, and 20 is 4.
LCM of 4, 6, and 20 is 40.
4 x 40 = 160.
Solution:
Divisors of 18 – 1, 2, 3, 6, 9, 18
Divisors of 27 – 1, 2, 3, 9, 27 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 4.1: Trigonometric Identities
Difficulty Level: At Grade Created by: CK-12
This activity is intended to supplement Trigonometry, Chapter 3, Lesson 1.
ID: 9848
Time required: 45 minutes
## Activity Overview
Students will verify, prove, and explore trigonometric identities symbolically, numerically, and graphically.
Topic: Trigonometric Identities
• Use the Pythagorean Theorem to prove the trigonometric identities sin2θ+cos2θ=1\begin{align*}\sin^2 \theta + \cos^2 \theta = 1\end{align*} and sec2θ=1+tan2θ\begin{align*}\sec^2 \theta = 1 + \tan^2 \theta\end{align*}.
• Verify trigonometric identities by graphing.
Teacher Preparation and Notes
• Prior to beginning the activity, students should download the VERITRIG program to their handhelds. This investigation has students exploring, proving, and verifying trigonometric identities.
• Students should already be familiar with right triangle trigonometric properties, similar triangles, and the Pythagorean Theorem since much of the activity is based on these fundamental theorems.
• Students should begin by clearing out any functions from the Y=\begin{align*}Y=\end{align*} screen and turning off all Stat Plots.
Associated Materials
## Problem 1 – Prove cos2θ+sin2θ=1\begin{align*}\cos^2 \theta + \sin^2 \theta = 1\end{align*}
In this problem, students will prove two of the basic trigonometric identities using the Pythagorean Theorem.
Press PRGM to access the Program menu.
Then choose the VERITRIG program and press ENTER. Select 1: PROVE ID 1.
Students will see a circle with a radius of 1 with a right triangle in the first quadrant.
Students will label the right triangle using the Text tool. Press 2nd\begin{align*}2^{nd}\end{align*} [PRGM] and arrow down to select 0:Text.
Now students can position the cursor near the middle of the hypotenuse and enter in the radius. Then press 2nd\begin{align*}2^{nd}\end{align*} [APLHA] to lock the alpha key on and enter in x\begin{align*}x\end{align*}, y\begin{align*}y\end{align*}, and θ\begin{align*}\theta\end{align*} in their respective places in a right triangle.
If students make a mistake, they can press 2nd\begin{align*}2^{nd}\end{align*} [PRGM] and select ClrDraw. Then they need to run the program again.
On the worksheet, students will prove: cos2θ+sin2θ=1\begin{align*}\cos^2 \theta + \sin^2 \theta = 1\end{align*}.
1. x2+y2=1\begin{align*}x^2 + y^2 = 1\end{align*}
2. cosθ=x1\begin{align*}\cos\theta = \frac{x}{1}\end{align*} and sinθ=y1\begin{align*}\sin\theta = \frac{y}{1}\end{align*}
3. Substituting x=cosθ\begin{align*}x = \cos\theta\end{align*} and y=sinθ\begin{align*}y = \sin\theta\end{align*} into x2+y2=1\begin{align*}x^2 + y^2 = 1\end{align*} gives cos2θ+sin2θ=1\begin{align*}\cos^2\theta + \sin^2\theta = 1\end{align*}.
When finished, students can press 2nd\begin{align*}2^{nd}\end{align*} [QUIT] to exit the drawing.
## Problem 2 – Prove sec2θ=1+tan2θ\begin{align*}\sec^2\theta = 1 + \tan^2\theta\end{align*}
Students need to run the VERITRIG program and select 2: PROVE ID 2.
Students will see a new version of the unit circle with the original right triangle and then a similar triangle drawn to incorporate the original triangle.
Again, they need to use the Text tool and label the original triangle.
Students are told on the worksheet that the length of the base of the large triangle is 1, the height is Y\begin{align*}Y\end{align*} and the hypotenuse is X\begin{align*}X\end{align*}. Students need to use the Text tool to label the large similar triangle accordingly.
Using the ratio given on the worksheet, students will prove sec2θ=1+tan2θ\begin{align*}\sec^2\theta = 1 + \tan^2\theta\end{align*}. You may need to explain to students that (small \begin{align*}\vartriangle\end{align*}) and (large \begin{align*}\vartriangle\end{align*}) reference where that part of the ratio came from and is not actually part of the ratio.
1cosθ1X=X1=(cosθ)(X)=1cosθ=secθYsinθ=1cosθ(Y)(cosθ)=(sinθ)(1)Y=sinθcosθ=tanθ\begin{align*}\frac{1}{\cos\theta} &= \frac{X}{1} && \frac{Y}{\sin\theta} = \frac{1}{\cos\theta}\\ 1 &= (\cos\theta)(X) && (Y)(\cos\theta) = (\sin\theta)(1)\\ X &= \frac{1}{\cos\theta} = \sec\theta && Y = \frac{\sin\theta}{\cos\theta} = \tan\theta\end{align*}
Substituting Y=tanθ\begin{align*}Y = \tan\theta\end{align*} and X=secθ\begin{align*}X = \sec\theta\end{align*} into the Pythagorean Theorem based on the larger triangle: 12+Y2=X21+(tanθ)2=(secθ)21+tan2θ=sec2θ\begin{align*}1^2 + Y^2 = X^2 \rightarrow1+ (\tan\theta)^2 = (\sec\theta)^2 \qquad \qquad 1 + \tan^2\theta = \sec^2\theta \end{align*}
## Problem 3 – Numerical verification
Now that students have proved the two identities, they will numerically verify the two identities.
Students need to run the VERITRIG program and select 3: NUM EXPLORE.
Students will see the same circle, but with three triangles in the first quadrant. When they press TRACE, they will see the x\begin{align*}x-\end{align*}values for cosine and the y\begin{align*}y-\end{align*}values for sine for each angle measurement.
Students should observe that the values for cosine and sine repeat but change signs depending on what quadrant the angle is in.
These values are stored in the lists of the calculator with the angle measurements in L1\begin{align*}L_1\end{align*}, the cosine values in L2\begin{align*}L_2\end{align*}, and the sine values in L3\begin{align*}L_3\end{align*}.
To numerically verify cos2θ+sin2θ=1\begin{align*}\cos^2\theta + \sin^2\theta = 1\end{align*}; students are to enter L22+L32\begin{align*}L2^2 + L3^2\end{align*} at the top of L4\begin{align*}L4\end{align*}.
Students will discover that they need to enter “1cosθ\begin{align*}\frac{1}{\cos\theta}\end{align*}” to get secθ\begin{align*}\sec\theta\end{align*}. The value of secθ\begin{align*}\sec\theta\end{align*} is undefined at 0 and 3π2\begin{align*}\frac{3 \pi}{2}\end{align*} , and the calculator will not evaluate these values within a list. The same thing will happen to tanθ\begin{align*}\tan\theta\end{align*} , which is equal to sinθcosθ\begin{align*}\frac{\sin\theta}{\cos\theta}\end{align*}.
However, students should arrive at the fact that they can graph the identities to verify the identity as the screen shots illustrate. They should set the calculator to the trig window by pressing ZOOM and select 7:ZTrig. (Note: Students will need to use the ClrDraw command to erase the unit circle and turn off Plot1.)
## Problem 4 – Verifying trig identities using graphing.
The previous problem leads into utilizing graphing techniques to verify various trig identities.
Students are to determine whether the identity
sin2x=1cos2x\begin{align*}\sin^2x = 1 - \cos^2x\end{align*} is true by entering sin2x\begin{align*}\sin^2x\end{align*} into Y1\begin{align*}Y1\end{align*} and 1cos2x\begin{align*}1 - \cos^2x\end{align*} into Y2\begin{align*}Y2\end{align*}. To set the bubble animation for the second graph, students need to press ENTER to the left of the equals sign.
Students need to select ZTrig for a trig window.
If the bubble runs over the same graph as the first, the equation is a trig identity.
Explain to students that they can check any equation using this method.
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Back EQUATIONS OF SCIENCE Lesson 2 - Page 3 Next
Fibonacci
Fibonacci Sequence
The Fibonacci Sequence is made up of the Fibonacci numbers. This number series was named for Leonardo of Pisa, son of Bonacci (figlio (son in Italian) of Bonnacci) or Fibonacci (1180-1228). He first wrote about the sequence in Liber Abaci by giving the solution to how many rabbits can breed in a year. The French mathematician Edouard Lucas (1842-1891) named the Fibonacci sequence in honor of the inventor.
Fibonacci started to think about how many pairs of rabbits could be breed in 12 months. In order to figure it out, Fibonacci had to establish some rules. In the first month there is just one newly born pair of baby rabbits. So he averaged that in one month a pair would have two baby rabbits and it would take 1 month for the rabbits to become an adult and have babies of their own. None of the rabbits could die during the year. He discovered a sequence that is based on the idea that each number is the sum of the two preceding numbers.
So you start with two pairs (1 and 1), then you add the preceding two numbers so you get the total number of rabbits each month. The number of rabbits is derived from 1 (starting), 1=(0 +1) (this gives us 1 pair); 2=(1+1), 3=(1+2), 5=(2+3), 8=(3+5), 13=(5+8), 21(8 +13), 34(13 + 21); 55 (21 + 34), 89 (34 +55), 144 (55+89). So in one year, there would be 144 rabbits.
The Fibonacci Sequence is 1,1,2,3,5,8,13,21,34,55,89,144. If you graph these numbers as squares, you get an interesting configuration called the Golden Rectangle. If you draw a line from corner to corner you get what is called the Golden Spiral, which we see in nature. The Nautilus shell is a good example of this figure.
Click on drawings for example Squaring the Fibonacci Numbers and you build a rectangle arrangement Draw a line and you get a golden spiral
Back Next |
The story begins with a coding problem:
Given a positive integer $$n$$, find the least number of perfect square numbers (for example, $$1, 4, 9, 16$$, …) which sum to $$n$$.
## A DP Algorithm
The above problem can be solved with a simple DP algorithm:
int solve(int n)
{
vector<int> v(n + 1, INT_MAX);
v[0] = 0;
for (int i = 1;i <= n;i++)
for (int j = 1;j * j <= i;j++)
v[i] = min(v[i], v[i - j * j] + 1);
return v[n];
}
Not much surprise. But what if you run it through $$[1, 100]$$, or even more? The result is always between 1 and 4.
## Theorems
In fact there is a Lagrange’s four-square theorem which states:
Every natural number can be represented as the sum of four integer squares.
In fact there is also a Legendre’s three-square theorem which states:
A natural number can be represented as the sum of three squares of integers if and only if $$n$$ is not of the form $$n = 4^a(8b + 7)$$ for integers $$a$$ and $$b$$.
## A Faster Algorithm
With these two theorems we have a faster solution to the above problem:
int solve(int n)
{
while (n % 4 == 0)
n /= 4;
if (n % 8 == 7)
return 4;
if (int(sqrt(n)) * int(sqrt(n)) == n)
return 1;
for(int i = 1;i * i < n;i++) {
int j = n - i * i;
if (int(sqrt(j)) * int(sqrt(j)) == j)
return 2;
}
return 3;
}
We now give an analysis of this algorithm. If not otherwise noted, all integers we consider in the analysis below are positive integers.
### Analysis
1. Lagrange’s four-square theorem shows that the answer to any number $$n$$ is no more than 4. Legendre’s three-square theorem shows that the answer is greater than 3 when $$n$$ is not of the form $$n = 4^a(8b + 7)$$. So the answer must be exactly 4 when $$n$$ is of the form $$n = 4^a(8b + 7)$$.
2. In all other cases the answers can only be 1, 2 or 3. We first determine whether the answer is 1 by checking whether $$n$$ is a perfect square. However, the code is actually not checking $$n$$ but $$m = 8b + 7$$. We need to prove $$n$$ is a perfect square if and only if $$m$$ is a perfect square.
• If part. If $$m = x^2$$ is a perfect square, then $$n = 4^am = (2^ax)^2$$ is also a perfect square.
• Only-If part. If $$n = 4^am = y^2$$ is a perfect square, then $$(2^a / y)^2 = m$$. So $$2^a / y = \sqrt{m}$$. Since $$m$$ is a positive integer, if $$m$$ is not a perfect square then $$\sqrt{m}$$ is irrational. But $$2^a / y$$ is rational. This is a contradiction.
3. If $$n$$ is not a perfect square then we further check whether it is a sum of two perfect squares. We can still check $$m$$ instead of $$n$$ since we can prove $$n$$ is a sum of two perfect squares if and only if $$m$$ is a sum of two perfect squares.
• If part. If $$m = x^2 + y^2$$ is a sum of two perfect squares, then $$n = 4^am = (2^ax)^2 + (2^ay)^2$$ is also a sum of two perfect squares.
• Only-If part. We first prove the following lemma:
If an even number $$4w$$ is a sum of two perfect squares, then $$w$$ is also a sum of two perfect squares.
Proof. Let $$4w = u^2 + v^2$$. Then $$w = (u / 2)^2 + (v / 2)^2$$. Since $$4w \equiv 0 \bmod 4$$, it must be that $$u$$ and $$v$$ are both even:
• Since $$4w$$ is even, it must be that $$u$$ and $$v$$ are both odd or both even.
• If $$u = 2s - 1$$ and $$v = 2t - 1$$ are both odd, then $$(u^2 + v^2) \equiv (4s^2 - 4s + 1 + 4t^2 - 4t + 1) \equiv 2 \bmod 4$$. This is a contradiction.
Therefore $$u$$ and $$v$$ are both even and $$u / 2$$ and $$v / 2$$ are both integers. So $$w$$ is a sum of two perfect squares.
Repeated use of the lemma on $$n = 4^am$$ will finish the proof.
4. If $$n$$ is neither a perfect square nor a sum of two perfect squares, then the answer is 3.
## Acknowledgment
Legendre Lagrange |
During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.
# Difference between revisions of "2012 AMC 12B Problems/Problem 21"
## Problem 21
Square $AXYZ$ is inscribed in equiangular hexagon $ABCDEF$ with $X$ on $\overline{BC}$, $Y$ on $\overline{DE}$, and $Z$ on $\overline{EF}$. Suppose that $AB=40$, and $EF=41(\sqrt{3}-1)$. What is the side-length of the square?
$\textbf{(A)}\ 29\sqrt{3} \qquad\textbf{(B)}\ \frac{21}{2}\sqrt{2}+\frac{41}{2}\sqrt{3} \qquad\textbf{(C)}\ \ 20\sqrt{3}+16 \qquad\textbf{(D)}\ 20\sqrt{2}+13 \sqrt{3} \qquad\textbf{(E)}\ 21\sqrt{6}$
## Solution
Extend $AF$ and $YE$ so that they meet at $G$. Then $\angle FEG=\angle GFE=60^{\circ}$, so $\angle FGE=60^{\circ}$ and therefore $AB$ is parallel to $YE$. Also, since $AX$ is parallel and equal to $YZ$, we get $\angle BAX = \angle ZYE$, hence $\triangle ABX$ is congruent to $\triangle YEZ$. We now get $YE=AB=40$.
Let $a_1=EY=40$, $a_2=AF$, and $a_3=EF$.
Drop a perpendicular line from $A$ to the line of $EF$ that meets line $EF$ at $K$, and a perpendicular line from $Y$ to the line of $EF$ that meets $EF$ at $L$, then $\triangle AKZ$ is congruent to $\triangle ZLY$ since $\angle YZL$ is complementary to $\angle KZA$. Then we have the following equations:
$$\frac{\sqrt{3}}{2}a_2 = AK=ZL = ZE+\frac{1}{2} a_1$$ $$\frac{\sqrt{3}}{2}a_1 = YL =ZK = ZF+\frac{1}{2} a_2$$
The sum of these two yields that
$$\frac{\sqrt{3}}{2}(a_1+a_2) = \frac{1}{2} (a_1+a_2) + ZE+ZF = \frac{1}{2} (a_1+a_2) + EF$$ $$\frac{\sqrt{3}-1}{2}(a_1+a_2) = 41(\sqrt{3}-1)$$ $$a_1+a_2=82$$ $$a_2=82-40=42.$$
So, we can now use the law of cosines in $\triangle AGY$:
$$2AZ^2 = AY^2 = AG^2 + YG^2 - 2AG\cdot YG \cdot \cos 60^{\circ}$$ $$= (a_2+a_3)^2 + (a_1+a_3)^2 - (a_2+a_3)(a_1+a_3)$$ $$= (41\sqrt{3}+1)^2 + (41\sqrt{3}-1)^2 - (41\sqrt{3}+1)(41\sqrt{3}-1)$$ $$= 6 \cdot 41^2 + 2 - 3 \cdot 41^2 + 1 = 3 (41^2 + 1) = 3\cdot 1682$$ $$AZ^2 = 3 \cdot 841 = 3 \cdot 29^2$$
Therefore $AZ = 29\sqrt{3} ... \framebox{A}$
2012 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 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 |
# Derivatives of Logarithmic Functions
Sub Topics
A logarithm with the natural number 'e' as the base is called the natural logarithm and is written as "ln".
The general rules of a logarithm apply for natural logarithms as well:
Ln(xy) = ln(x) + ln(y) Product rule for logarithms
Ln$(\frac{x}{y})$=ln(x)-ln(y) Quotient rule for logarithms
Ln(xn) = n.ln(x) Power rule for logarithms
When considered as functions natural logarithmic function ln(x) is the inverse of the exponential function $e^{x}$. The graph given below demonstrates this. We can see the two graphs, y = ln(x) and y = $e^{x}$ are mirror images of each other over the line y = x. Thus the composition of ln and the exponential function's output 'x'.
$Ln(e^{x})=x$ $e^{in}x=x$
The derivative of the ln function is generally understood as the derivative of the logarithmic function.
Derivative of natural log (ln) function
for x> 0, $\frac{d}{dx}$ log(x)= $\frac{1}{x}$
There are many approaches to Prove$\frac{d}{dx}$ log(x)= $\frac{1}{x}$. One would be to use Taylor's expansion for ln function:
ln(1+x) =$x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-...........$
Proof:
Using the limit definition of a derivative,
$\frac{d}{dx}$ log(x) =$\lim_{h\rightarrow 0}$ $\frac{Iog(x+h)-Iog(x)}{h}$
=$\lim_{h\rightarrow 0}$ $\frac{Iog(\frac{x+h}{x})}{h}$ Using the quotient rule for logarithms,
=$\lim_{h\rightarrow 0}$ Iog (1+$\frac{h}{x}$)
= udv + vdu
= $\lim_{h \rightarrow o}$ $\frac{\left ( \frac{h}{x}-\frac{(\frac{h}{x})^{2}}{2}+\frac{(\frac{h}{x})^{3}}{3} -\right )}{h}$ Using Taylor's Series for natural logarithm,
= $\lim_{h \rightarrow o}$ $\left ( \frac{1}{x}-\frac{h}{2x^{2}}+\frac{h^{2}}{3x^{3}}-......... \right )$ Dividing each term by h
= $\frac{1}{x}$ as the rest of the terms containing h approach zero.
x can assume only positive values as ln(x) is defined only for x > 0. Graphically, only the portion in the first quadrant of the reciprocal function represents the derivative of a natural logarithm function.
In differentiation problems, the derivative of the ln function is used along with general rules of differentiation like the product rule, the quotient rule and the chain rule. The following examples show differentiation of functions involving a logarithmic function.
## Derivative of Log Function
1. Find the derivative of x.ln(x)
The function can be differentiated using the product rule for derivatives.
u(x) = x and v(x) = log(x) .Hence u’(x) = 1 and v’(x) =$\frac{1}{x}$
$\frac{d}{dx}$
(x.log(x)) = udv + vdu =x($\frac{1}{x}$) + log(x)(1) = 1 + log(x)
2. Find the derivative of log(sin x)
This function is differentiated using the chain rule,
f(x) = ln(sin x)
f(x) = $\frac{1}{\sin x}.\frac{d}{dx}$ (sin x)
= $\frac{1}{\sin x}$ .cos x = cot x
3. Find the derivative of $\frac{In(x)}{x+1}$
The quotient rule is to be used:
f(x) =$\frac{\log(x)}{x+1}$ where u(x) = ln(x) and v(x) = x+1. Hence u’(x) =$\frac{1}{x}$ and v’(x) = 1
f(x) = $\frac{vdu-udv}{v^{2}}$ = $\frac{(x+1).\frac{1}{x}-In(x).1}{(x+1)^{2}}$ = $\frac{x+1-xIn(x)}{x(x+1)^{2}}$
Finding the derivatives of logarithmic functions is relatively difficult when compared with computing the derivatives of exponential functions. The steps and simplifications are to be done with care, since for complicated functions the differentiation would be involved.
4. Differentiate:$\sqrt[Iog]{\frac{1+\sin x}{1-\sin x}}$
This function looks scary at first sight. We can identify four functions here, namely logarithmic, radical, quotient and sine functions. If the chain rule is to be used in succession, the formidable derivatives of logarithmic and root functions will render it highly difficult to manage the simplification. We apply one algebraic technique coupled with trigonometric identities to simplify the function first.
y = $\sqrt[Iog]{\frac{1+\sin x}{1-\sin x}}$
y = $\sqrt[Iog]{\frac{(1+\sin (x))(1+\sin (x))}{(1-\sin x)(1+\sin (x))}}$ Multiply and divide the quotient by 1+sinx
y = $\frac{Iog\sqrt(1+\sin x)^2}{(1-sin^2x)}$
y = Iog $\left ( \frac{1+\sin x}{\cos x} \right )$ 1- sin2x = cos2x
y=Iog $(\sec x+\tan x)$ This is helpful in avoiding the use of the quotient rule
differentiating using the derivative of ln function and the chain rule
y' =$\frac{1}{\sec x+\tan x}$ $(\sec x\tan x+\sec ^{2}x)$ $\frac{\sec x(\sec x+\tan (x))}{\sec x+\tan x}$ = $\sec x$
The derivative of the complicated function is unexpectedly a simple one.
## Logarithmic Differentiation
Even though differentiation using the derivative of a logarithmic function can be complicated, logarithms provide an easy method to differentiate complicated products and quotient functions. This method makes use of the properties of logarithms to write products and quotients as sums and products.
Example:
Find the derivative of $\frac{\sqrt{x}(x+4)^{\frac{3}{2}}}{(4x-3)^{\frac{4}{3}}}$
Solution:
Given: y = $\frac{\sqrt{x}(x+4)^{\frac{3}{2}}}{(4x-3)^{\frac{4}{3}}}$
y = $\frac{x^{\frac{1}{2}}(x+4)^{\frac{3}{2}}}{(4x-3)^{\frac{4}{3}}}$ The radical sign is replaced with a rational index.
Taking a natural logarithm on either side of the equation,
ln y =$\frac{1}{2}$ Iog x + $\frac{3}{2}$ Iog(x+4) - $\frac{4}{3}$ Iog(4x-3) Using the product and quotient rules for logarithms
Differentiating using the derivative of ln
$\frac{1}{y}\frac{dy}{dx}$ = $\frac{1}{2}\frac{1}{x}$ + $\frac{3}{2}\frac{1}{(x+4)}$-$\frac{4}{3}\frac{4}{(4x-3)}$ = $\frac{1}{2x}$+$\frac{3}{2(x+4)}$-$\frac{16}{3(4x-3)}$
$\frac{dy}{dx}$ =y $\frac{1}{2x}$ + $\frac{3}{2(x+4)}$-$\frac{16}{3(4x-3)}$
$\frac{dy}{dx}$ = $\frac{x^{\frac{1}{2}}(x+4)^{\frac{3}{2}}}{(4x-3)^{\frac{4}{3}}\left [ \frac{1}{2x}+\frac{3}{2(x+4)}-\frac{16}{3(4x-3)} \right ]}$
Derivative of an infinite Series
If y=$(\sin [(x)]^{\sin [(x)]^{\sin [(x)]^{\sin [(x)]^{\sin [(x)........]}}}}$ prove that
$\frac{dy}{dx}$
= $\frac{y^{2}\cot x}{(1-y\log \sin (x)}$
Can we take a chain of logarithms to solve this problem? That is not feasible here. The infinite series not affected if we remove one chain here. So this function can be rewritten as
y=$(\sin (x))^{y}$. Now the method of logarithmic differentiation can be applied.
$log y$ = y = $Iog(\sin (x))$ $\frac{1}{y}\frac{dy}{dx}$ = y. $\frac{\cos x}{\sin x}$
+$Iog(\sin (x))$ $\frac{dy}{dx}$ Used the product rule to differentiate
$\frac{dy}{dx}$
= $y^{2}\cot x+ y Iog(\sin (x))$ $\frac{dy}{dx}$ Multiplied the equation by y
$\frac{dy}{dx}$ (1-y Iog (sin x)) = $y^{2}\cot x$
$\frac{dy}{dx}$ = $\frac{y^{2}\cot x}{(1-yIn(\sin (x))}$
Thus the derivative of the infinite series is found in a few steps.
## Derivative of Log Base a of x
The derivative of a logarithmic function with any base can be found from the derivative of the ln function using the change of base rule.
The change of base rule for a logarithmic when the base is changed to ‘b’ from ‘a’ is explained below:
New base = a and old base = b = $\log_{b}x$ = $\frac{\log_{a}x}{\log_{a}b}$
If f(x) = $\log_{a} x$ then the base could be changed to ‘e’ as the log base e is the natural logarithm function ln.
f(x)=$\frac{\log_{e}x}{\log_{e}a}$
$\log_{e}a$ is a constant, which can be evaluated using a calculator.
Differentiating with respect to x,
f(x) = $\frac{1}{\log_{e}ax}\frac{1}{x}$ = $\frac{1}{x\log_{e}a}$ = $\frac{1}{x Iog a}$
Derivative of log base a $\frac{d}{dx}$ $\log_{a}x$ = $\frac{1}{x Iog a}$
## Derivative of Log Base 10
The derivative of a log base 10 can be found by substituting a=10 in the above formula.
$\frac{d}{dx}$
$\log_{10}x$ = $\frac{1}{x Iog10}$ and $\log 10 \approx 2.303$
$\frac{d}{dx}$ $\log_{10}x \approx$ $\frac{1}{2.303x}$
The derivative of a log base 10 $\frac{d}{dx}$ $\log_{10}x$ = $\frac{1}{x \log 10}$
Example using a log with another base
Find the derivative of 4x + $\log_{4}x-Inx$
Differentiating using the relevant formulas:
f(x) = $In(4)4^{x}$ + $\frac{1}{xIn4}\frac{1}{-x}$ simplifying by taking the common denominator,
f(x) = $\frac{x(In (4)^{2})4^{x} - In 4+1}{X In 4}$ |
Hong Kong
Stage 1 - Stage 3
# Units of Capacity II
Lesson
As we have been learning, everything in maths that relates to the ‘real world’ has units. If an amount represents a real life quantity, it has units attached to it. The units that we use depend on what we are measuring.
There are some units we need to be familiar with, and know how to convert between.
LENGTH/DISTANCE mm, cm, m, km mm2, cm2, m2, km2 mm3, cm3, m3, km3 mL, L, kL, ML mg, g, kg, metric tonne sec, mins, hrs, days, weeks, months, years
## Units of capacity
Capacity is the amount of liquid that a container can hold. Capacity is usually measured using one of the following units:
• millilitres (mL
• litres (L
• kilolitres (kL
• megalitres (ML)
You would be used to most of these through previous experiences in the size of your milk container, measuring ingredients when cooking, measuring medicines or even in science experiments you may have done at school.
$1$1 = $1000$1000 mL
$1$1 kL = $1000$1000 L = $1000000$1000000 mL ($1000\times1000$1000×1000)
$1$1 ML = $1000$1000 kL = $1000000$1000000 L = $1000000000$1000000000 mL Thats about 200 million full teaspoons!
To move from larger capacity units to smaller capacity units we multiply at each step.
To move from smaller capacity units to larger capacity units we divide at each step.
Do you see some patterns when looking at changing units of capacity? For example, notice that there is a factor of 1000 between each step.
## The prefixes
You see here that we also use the prefixes of milli and kilo again. Remember that the prefix milli means $\frac{1}{1000}$11000 th of something. So we can see here that a millilitre is $\frac{1}{1000}$11000th of a litre, which means that there are $1000$1000 mL in $1$1 litre. Also the prefix kilomeans $1000$1000 lots of something, so a kilolitre is $1000$1000 litres.
#### Example
##### Question 1
Question: Change $6.732$6.732kL into mL.
Think: Think about the steps needed to move from kL to mL, (kL -> L -> mL) and identify the multiplication amounts for each step. I suggest moving through one step at a time.
Do:
First convert to litres: $6.732\times1000=6732$6.732×1000=6732 L
Then convert to millilitres: $6732\times1000=6732000$6732×1000=6732000 mL
It really doesn't matter if you think about it like I did, or if you do it differently. What is important is to keep track of your steps. See how my units changed at each calculation.
Here is another:
#### Example
##### Question 2
Question: Convert $468296$468296 mL into litres
Think: Think about the steps needed to move from ml to litres, (mL -> L ) and identify the division amounts for each step.
Do:
$468296\div1000$468296÷1000 $=$= $468.296$468.296 L
#### More Worked Examples
##### QUESTION 3
Convert $3\frac{4}{5}$345 litres to millilitres.
1. $3\frac{4}{5}$345 litres = $\editable{}$ millilitres
##### QUESTION 4
Convert each of the following volumes to the indicated unit:
1. $2000$2000 mL = $\editable{}$ L
2. $5$5 kL = $\editable{}$ L
3. $240$240 mL = $\editable{}$ L
4. $0.4$0.4 L= $\editable{}$ mL
##### QUESTION 5
How many $150$150 millilitre jugs of soda water will be needed to fill a $2.4$2.4 litre container? |
# 009A Sample Midterm 1, Problem 5
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
The displacement from equilibrium of an object in harmonic motion on the end of a spring is:
${\displaystyle y={\frac {1}{3}}\cos(12t)-{\frac {1}{4}}\sin(12t)}$
where ${\displaystyle y}$ is measured in feet and ${\displaystyle t}$ is the time in seconds.
Determine the position and velocity of the object when ${\displaystyle t={\frac {\pi }{8}}.}$
Foundations:
What is the relationship between position ${\displaystyle s(t)}$ and velocity ${\displaystyle v(t)}$ of an object?
${\displaystyle v(t)=s'(t)}$
Solution:
Step 1:
To find the position of the object at ${\displaystyle t={\frac {\pi }{8}},}$
we need to plug ${\displaystyle t={\frac {\pi }{8}}}$ into the equation ${\displaystyle y.}$
Thus, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {y{\bigg (}{\frac {\pi }{8}}{\bigg )}}&=&\displaystyle {{\frac {1}{3}}\cos {\bigg (}{\frac {12\pi }{8}}{\bigg )}-{\frac {1}{4}}\sin {\bigg (}{\frac {12\pi }{8}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {1}{3}}\cos {\bigg (}{\frac {3\pi }{2}}{\bigg )}-{\frac {1}{4}}\sin {\bigg (}{\frac {3\pi }{2}}{\bigg )}}\\&&\\&=&\displaystyle {0-{\frac {1}{4}}(-1)}\\&&\\&=&\displaystyle {{\frac {1}{4}}{\text{ foot}}.}\end{array}}}$
Step 2:
Now, to find the velocity function, we need to take the derivative of the position function.
Thus, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {v(t)}&=&\displaystyle {y'}\\&&\\&=&\displaystyle {{\frac {-1}{3}}\sin(12t)(12)-{\frac {1}{4}}\cos(12t)(12)}\\&&\\&=&\displaystyle {-4\sin(12t)-3\cos(12t).}\end{array}}}$
Therefore, the velocity of the object at time ${\displaystyle t={\frac {\pi }{8}}}$ is
${\displaystyle {\begin{array}{rcl}\displaystyle {v{\bigg (}{\frac {\pi }{8}}{\bigg )}}&=&\displaystyle {-4\sin {\bigg (}{\frac {3\pi }{2}}{\bigg )}-3\cos {\bigg (}{\frac {3\pi }{2}}{\bigg )}}\\&&\\&=&\displaystyle {-4(-1)+0}\\&&\\&=&\displaystyle {4{\text{ feet/second}}.}\end{array}}}$
position is ${\displaystyle {\frac {1}{4}}{\text{ foot}}.}$
velocity is ${\displaystyle 4{\text{ feet/second}}.}$ |
Mediant (mathematics)
In mathematics, the mediant of two fractions, generally made up of four positive integers
${\displaystyle {\frac {a}{c}}\quad }$ and ${\displaystyle \quad {\frac {b}{d}}\quad }$ is defined as ${\displaystyle \quad {\frac {a+b}{c+d}}.}$
That is to say, the numerator and denominator of the mediant are the sums of the numerators and denominators of the given fractions, respectively. It is sometimes called the freshman sum, as it is a common mistake in the early stages of learning about addition of fractions.
Technically, this is a binary operation on valid fractions (nonzero denominator), considered as ordered pairs of appropriate integers, a priori disregarding the perspective on rational numbers as equivalence classes of fractions. For example, the mediant of the fractions 1/1 and 1/2 is 2/3. However, if the fraction 1/1 is replaced by the fraction 2/2, which is an equivalent fraction denoting the same rational number 1, the mediant of the fractions 2/2 and 1/2 is 3/4. For a stronger connection to rational numbers the fractions may be required to be reduced to lowest terms, thereby selecting unique representatives from the respective equivalence classes.
The Stern–Brocot tree provides an enumeration of all positive rational numbers via mediants in lowest terms, obtained purely by iterative computation of the mediant according to a simple algorithm.
Properties
• The mediant inequality: An important property (also explaining its name) of the mediant is that it lies strictly between the two fractions of which it is the mediant: If ${\displaystyle a/c and ${\displaystyle c\cdot d>0}$, then ${\displaystyle {\frac {a}{c}}<{\frac {a+b}{c+d}}<{\frac {b}{d}}.}$ This property follows from the two relations ${\displaystyle {\frac {a+b}{c+d}}-{\frac {a}{c}}={{bc-ad} \over {c(c+d)}}={d \over {c+d}}\left({\frac {b}{d}}-{\frac {a}{c}}\right)}$ and ${\displaystyle {\frac {b}{d}}-{\frac {a+b}{c+d}}={{bc-ad} \over {d(c+d)}}={c \over {c+d}}\left({\frac {b}{d}}-{\frac {a}{c}}\right).}$
• Componendo and Dividendo Theorems: If ${\displaystyle a/c=b/d}$ and ${\displaystyle c\neq 0,\ d\neq 0}$, then[1] ${\displaystyle {\frac {a}{c}}={\frac {b}{d}}={\frac {a+b}{c+d}}}$
${\displaystyle {\frac {a+c}{c}}={\frac {b+d}{d}}}$
${\displaystyle {\frac {a-c}{c}}={\frac {b-d}{d}}}$
• Assume that the pair of fractions a/c and b/d satisfies the determinant relation ${\displaystyle bc-ad=1}$. Then the mediant has the property that it is the simplest fraction in the interval (a/c, b/d), in the sense of being the fraction with the smallest denominator. More precisely, if the fraction ${\displaystyle a'/c'}$ with positive denominator c' lies (strictly) between a/c and b/d, then its numerator and denominator can be written as ${\displaystyle a'=\lambda _{1}a+\lambda _{2}b}$ and ${\displaystyle c'=\lambda _{1}c+\lambda _{2}d}$ with two positive real (in fact rational) numbers ${\displaystyle \lambda _{1},\,\lambda _{2}}$. To see why the ${\displaystyle \lambda _{i}}$ must be positive note that ${\displaystyle {\frac {\lambda _{1}a+\lambda _{2}b}{\lambda _{1}c+\lambda _{2}d}}-{\frac {a}{c}}=\lambda _{2}{{bc-ad} \over {c(\lambda _{1}c+\lambda _{2}d)}}}$ and ${\displaystyle {\frac {b}{d}}-{\frac {\lambda _{1}a+\lambda _{2}b}{\lambda _{1}c+\lambda _{2}d}}=\lambda _{1}{{bc-ad} \over {d(\lambda _{1}c+\lambda _{2}d)}}}$ must be positive. The determinant relation ${\displaystyle bc-ad=1\,}$ then implies that both ${\displaystyle \lambda _{1},\,\lambda _{2}}$ must be integers, solving the system of linear equations ${\displaystyle a'=\lambda _{1}a+\lambda _{2}b}$ ${\displaystyle c'=\lambda _{1}c+\lambda _{2}d}$ for ${\displaystyle \lambda _{1},\lambda _{2}}$. Therefore, ${\displaystyle c'\geq c+d.}$
• The converse is also true: assume that the pair of reduced fractions a/c < b/d has the property that the reduced fraction with smallest denominator lying in the interval (a/cb/d) is equal to the mediant of the two fractions. Then the determinant relation bcad = 1 holds. This fact may be deduced e.g. with the help of Pick's theorem which expresses the area of a plane triangle whose vertices have integer coordinates in terms of the number vinterior of lattice points (strictly) inside the triangle and the number vboundary of lattice points on the boundary of the triangle. Consider the triangle ${\displaystyle \Delta (v_{1},v_{2},v_{3})}$ with the three vertices v1 = (0, 0), v2 = (ac), v3 = (bd). Its area is equal to ${\displaystyle {\text{area}}(\Delta )={{bc-ad} \over 2}\,.}$ A point ${\displaystyle p=(p_{1},p_{2})}$ inside the triangle can be parametrized as ${\displaystyle p_{1}=\lambda _{1}a+\lambda _{2}b,\;p_{2}=\lambda _{1}c+\lambda _{2}d,}$ where ${\displaystyle \lambda _{1}\geq 0,\,\lambda _{2}\geq 0,\,\lambda _{1}+\lambda _{2}\leq 1.}$ The Pick formula ${\displaystyle {\text{area}}(\Delta )=v_{\mathrm {interior} }+{v_{\mathrm {boundary} } \over 2}-1}$ now implies that there must be a lattice point q = (q1, q2) lying inside the triangle different from the three vertices if bcad > 1 (then the area of the triangle is ${\displaystyle \geq 1}$). The corresponding fraction q1/q2 lies (strictly) between the given (by assumption reduced) fractions and has denominator ${\displaystyle q_{2}=\lambda _{1}c+\lambda _{2}d\leq \max(c,d) as ${\displaystyle \lambda _{1}+\lambda _{2}\leq 1.}$
• Relatedly, if p/q and r/s are reduced fractions on the unit interval such that |ps − rq| = 1 (so that they are adjacent elements of a row of the Farey sequence) then ${\displaystyle ?\left({\frac {p+r}{q+s}}\right)={\frac {1}{2}}\left(?\left({\frac {p}{q}}\right)+{}?\left({\frac {r}{s}}\right)\right)}$ where ? is Minkowski's question mark function.
In fact, mediants commonly occur in the study of continued fractions and in particular, Farey fractions. The nth Farey sequence Fn is defined as the (ordered with respect to magnitude) sequence of reduced fractions a/b (with coprime a, b) such that b ≤ n. If two fractions a/c < b/d are adjacent (neighbouring) fractions in a segment of Fn then the determinant relation ${\displaystyle bc-ad=1}$ mentioned above is generally valid and therefore the mediant is the simplest fraction in the interval (a/cb/d), in the sense of being the fraction with the smallest denominator. Thus the mediant will then (first) appear in the (c + d)th Farey sequence and is the "next" fraction which is inserted in any Farey sequence between a/c and b/d. This gives the rule how the Farey sequences Fn are successively built up with increasing n.
Graphical determination of mediants
A positive rational number is one in the form ${\displaystyle a/b}$ where ${\displaystyle a,b}$ are positive natural numbers; i.e. ${\displaystyle a,b\in \mathbb {N} ^{+}}$. The set of positive rational numbers ${\displaystyle \mathbb {Q} ^{+}}$ is, therefore, the Cartesian product of ${\displaystyle \mathbb {N} ^{+}}$ by itself; i.e. ${\displaystyle \mathbb {Q} ^{+}=(\mathbb {N} ^{+})^{2}}$. A point with coordinates ${\displaystyle (b,a)}$ represents the rational number ${\displaystyle a/b}$, and the slope of a segment connecting the origin of coordinates to this point is ${\displaystyle a/b}$. Since ${\displaystyle a,b}$ are not required to be coprime, point ${\displaystyle (b,a)}$ represents one and only one rational number, but a rational number is represented by more than one point; e.g. ${\displaystyle (4,2),(60,30),(48,24)}$ are all representations of the rational number ${\displaystyle 1/2}$. This is a slight modification of the formal definition of rational numbers, restricting them to positive values, and flipping the order of the terms in the ordered pair ${\displaystyle (b,a)}$ so that the slope of the segment becomes equal to the rational number.
Two points ${\displaystyle (b,a)\neq (d,c)}$ where ${\displaystyle a,b,c,d\in \mathbb {N} ^{+}}$ are two representations of (possibly equivalent) rational numbers ${\displaystyle a/b}$ and ${\displaystyle c/d}$. The line segments connecting the origin of coordinates to ${\displaystyle (b,a)}$ and ${\displaystyle (d,c)}$ form two adjacent sides in a parallelogram. The vertex of the parallelogram opposite to the origin of coordinates is the point ${\displaystyle (b+d,a+c)}$, which is the mediant of ${\displaystyle a/b}$ and ${\displaystyle c/d}$.
The area of the parallelogram is ${\displaystyle bc-ad}$, which is also the magnitude of the cross product of vectors ${\displaystyle \langle b,a\rangle }$ and ${\displaystyle \langle d,c\rangle }$. It follows from the formal definition of rational number equivalence that the area is zero if ${\displaystyle a/b}$ and ${\displaystyle c/d}$ are equivalent. In this case, one segment coincides with the other, since their slopes are equal. The area of the parallelogram formed by two consecutive rational numbers in the Stern–Brocot tree is always 1.[2]
Generalization
The notion of mediant can be generalized to n fractions, and a generalized mediant inequality holds,[3] a fact that seems to have been first noticed by Cauchy. More precisely, the weighted mediant ${\displaystyle m_{w}}$ of n fractions ${\displaystyle a_{1}/b_{1},\ldots ,a_{n}/b_{n}}$ is defined by ${\displaystyle {\frac {\sum _{i}w_{i}a_{i}}{\sum _{i}w_{i}b_{i}}}}$ (with ${\displaystyle w_{i}>0}$). It can be shown that ${\displaystyle m_{w}}$ lies somewhere between the smallest and the largest fraction among the ${\displaystyle a_{i}/b_{i}}$. |
5.3: Difference of Two Means
In this section we consider a difference in two population means, $$\mu_1 - \mu_2$$, under the condition that the data are not paired. The methods are similar in theory but different in the details. Just as with a single sample, we identify conditions to ensure a point estimate of the difference $$\bar {x}_1 - \bar {x}_2$$ is nearly normal. Next we introduce a formula for the standard error, which allows us to apply our general tools from Section 4.5.
We apply these methods to two examples: participants in the 2012 Cherry Blossom Run and newborn infants. This section is motivated by questions like "Is there convincing evidence that newborns from mothers who smoke have a different average birth weight than newborns from mothers who don't smoke?"
Point Estimates and Standard Errors for Differences of Means
We would like to estimate the average difference in run times for men and women using the run10Samp data set, which was a simple random sample of 45 men and 55 women from all runners in the 2012 Cherry Blossom Run. Table $$\PageIndex{2}$$ presents relevant summary statistics, and box plots of each sample are shown in Figure 5.6.
Table $$\PageIndex{2}$$: Summary statistics for the run time of 100 participants in the 2009 Cherry Blossom Run.
men women
$$\bar {x}$$ 87.65 102.13
$$s$$ 12.5 15.2
$$n$$ 45 55
The two samples are independent of one-another, so the data are not paired. Instead a point estimate of the difference in average 10 mile times for men and women, $$\mu_w - \mu_m$$, can be found using the two sample means:
$\bar {x}_w - \bar {x}_m = 102.13 - 87.65 = 14.48$
Figure $$\PageIndex{1}$$: A histogram of time for the sample Cherry Blossom Race data.
Because we are examining two simple random samples from less than 10% of the population, each sample contains at least 30 observations, and neither distribution is strongly skewed, we can safely conclude the sampling distribution of each sample mean is nearly normal. Finally, because each sample is independent of the other (e.g. the data are not paired), we can conclude that the difference in sample means can be modeled using a normal distribution. (Probability theory guarantees that the difference of two independent normal random variables is also normal. Because each sample mean is nearly normal and observations in the samples are independent, we are assured the difference is also nearly normal.)
Conditions for normality of $$\bar {x}_1 - \bar {x}_2$$
If the sample means, $$\bar {x}_1$$ and $$\bar {x}_2$$, each meet the criteria for having nearly normal sampling distributions and the observations in the two samples are independent, then the difference in sample means, $$\bar {x}_1 - \bar {x}_2$$, will have a sampling distribution that is nearly normal.
We can quantify the variability in the point estimate, $$\bar {x}_w - \bar {x}_m$$, using the following formula for its standard error:
$SE_{\bar {x}_w - \bar {x}_m} = \sqrt {\dfrac {\sigma^2_w}{n_w} + \dfrac {\sigma^2_m}{n_m}}$
We usually estimate this standard error using standard deviation estimates based on the samples:
\begin{align} SE_{\bar {x}_w-\bar {x}_m} &\approx \sqrt {\dfrac {s^2_w}{n_w} + \dfrac {s^2_m}{n_m}} \\[6pt] &= \sqrt {\dfrac {15.2^2}{55} + \dfrac {12.5^2}{45}} \\&= 2.77 \end{align}
Because each sample has at least 30 observations ($$n_w = 55$$ and $$n_m = 45$$), this substitution using the sample standard deviation tends to be very good.
Distribution of a difference of sample means
The sample difference of two means, $$\bar {x}_1 - \bar {x}_2$$, is nearly normal with mean $$\mu_1 - \mu_2$$ and estimated standard error
$SE_{\bar {x}_1-\bar {x}_2} = \sqrt {\dfrac {s^2_1}{n_1} + \dfrac {s^2_2}{n_2}} \label{5.4}$
when each sample mean is nearly normal and all observations are independent.
Confidence Interval for the Difference
When the data indicate that the point estimate $$\bar {x}_1 - \bar {x}_2$$ comes from a nearly normal distribution, we can construct a confidence interval for the difference in two means from the framework built in Chapter 4. Here a point estimate, $$\bar {x}_w - \bar {x}_m = 14.48$$, is associated with a normal model with standard error SE = 2.77. Using this information, the general confidence interval formula may be applied in an attempt to capture the true difference in means, in this case using a 95% confidence level:
$\text {point estimate} \pm z^*SE \rightarrow 14.48 \pm 1.96 \times 2.77 \times (9.05, 19.91)$
Based on the samples, we are 95% confident that men ran, on average, between 9.05 and 19.91 minutes faster than women in the 2012 Cherry Blossom Run.
Exercise $$\PageIndex{1}$$
What does 95% confidence mean?
Solution
If we were to collected many such samples and create 95% confidence intervals for each, then about 95% of these intervals would contain the population difference, $$\mu_w - \mu_m$$.
Exercise $$\PageIndex{2}$$
We may be interested in a different confidence level. Construct the 99% confidence interval for the population difference in average run times based on the sample data.
Solution
The only thing that changes is z*: we use z* = 2:58 for a 99% confidence level. (If the selection of $$z^*$$ is confusing, see Section 4.2.4 for an explanation.) The 99% confidence interval:
$14.48 \pm 2.58 \times 2.77 \rightarrow (7.33, 21.63).$
We are 99% confident that the true difference in the average run times between men and women is between 7.33 and 21.63 minutes.
Hypothesis tests Based on a Difference in Means
A data set called baby smoke represents a random sample of 150 cases of mothers and their newborns in North Carolina over a year. Four cases from this data set are represented in Table $$\PageIndex{2}$$. We are particularly interested in two variables: weight and smoke. The weight variable represents the weights of the newborns and the smoke variable describes which mothers smoked during pregnancy. We would like to know if there is convincing evidence that newborns from mothers who smoke have a different average birth weight than newborns from mothers who don't smoke? We will use the North Carolina sample to try to answer this question. The smoking group includes 50 cases and the nonsmoking group contains 100 cases, represented in Figure $$\PageIndex{2}$$.
Table $$\PageIndex{2}$$: Four cases from the baby smoke data set. The value "NA", shown for the first two entries of the first variable, indicates that piece of data is missing.
fAge mAge weeks weight sexBaby smoke
1 NA 13 37 5.00 female nonsmoker
2 NA 14 36 5.88 female nonsmoker
3 19 15 41 8.13 male smoker
$$\vdots$$ $$\vdots$$ $$\vdots$$ $$\vdots$$ $$\vdots$$ $$\vdots$$ $$\vdots$$
150 45 50 36 9.25 female nonsmoker
Figure $$\PageIndex{2}$$: The top panel represents birth weights for infants whose mothers smoked. The bottom panel represents the birth weights for infants whose mothers who did not smoke. Both distributions exhibit strong skew.
Example $$\PageIndex{1}$$
Set up appropriate hypotheses to evaluate whether there is a relationship between a mother smoking and average birth weight.
Solution
The null hypothesis represents the case of no difference between the groups.
• H0: There is no difference in average birth weight for newborns from mothers who did and did not smoke. In statistical notation: $$\mu_n - \mu_s = 0$$, where $$\mu_n$$ represents non-smoking mothers and $$\mu_s$$ represents mothers who smoked.
• HA: There is some difference in average newborn weights from mothers who did and did not smoke ($$\mu_n - \mu_s \ne 0$$).
Summary statistics are shown for each sample in Table $$\PageIndex{3}$$. Because the data come from a simple random sample and consist of less than 10% of all such cases, the observations are independent. Additionally, each group's sample size is at least 30 and the skew in each sample distribution is strong (Figure $$\PageIndex{2}$$). However, this skew is reasonable for these sample sizes of 50 and 100. Therefore, each sample mean is associated with a nearly normal distribution.
Table $$\PageIndex{3}$$: Summary statistics for the baby smoke data set.
smoker nonsmoker
mean 6.78 7.18
st. dev. 1.43 1.60
samp. size 50 100
Exercise $$\PageIndex{3}$$
1. What is the point estimate of the population difference, $$\mu_n - \mu_s$$?
2. Can we use a normal distribution to model this difference?
3. Compute the standard error of the point estimate from part (a)
Solution
(a) The difference in sample means is an appropriate point estimate: $$\bar {x}_n - \bar {x}_s = 0.40$$.
(b) Because the samples are independent and each sample mean is nearly normal, their difference is also nearly normal.
(c) The standard error of the estimate can be estimated using Equation \ref{5.4}:
$SE = \sqrt {\dfrac {\sigma^2_n}{n_n} + \dfrac {\sigma^2_s}{n_s}} \approx \sqrt {\dfrac {s^2_n}{n_n} + \dfrac {s^2_s}{n_s}} = \sqrt {\dfrac {1.60^2}{100} + \dfrac {1.43^2}{50}} = 0.26$
The standard error estimate should be sufficiently accurate since the conditions were reasonably satisfied.
Example $$\PageIndex{2}$$
If the null hypothesis from Exercise 5.8 was true, what would be the expected value of the point estimate? And the standard deviation associated with this estimate? Draw a picture to represent the p-value.
Solution
If the null hypothesis was true, then we expect to see a difference near 0. The standard error corresponds to the standard deviation of the point estimate: 0.26. To depict the p-value, we draw the distribution of the point estimate as though H0 was true and shade areas representing at least as much evidence against H0 as what was observed. Both tails are shaded because it is a two-sided test.
Example $$\PageIndex{3}$$
Compute the p-value of the hypothesis test using the figure in Example 5.9, and evaluate the hypotheses using a signi cance level of $$\alpha = 0.05.$$
Solution
Since the point estimate is nearly normal, we can nd the upper tail using the Z score and normal probability table:
$Z = \dfrac {0.40 - 0}{0.26} = 1.54 \rightarrow \text {upper tail} = 1 - 0.938 = 0.062$
Because this is a two-sided test and we want the area of both tails, we double this single tail to get the p-value: 0.124. This p-value is larger than the signi cance value, 0.05, so we fail to reject the null hypothesis. There is insufficient evidence to say there is a difference in average birth weight of newborns from North Carolina mothers who did smoke during pregnancy and newborns from North Carolina mothers who did not smoke during pregnancy.
Exercise $$\PageIndex{4}$$
Does the conclusion to Example 5.10 mean that smoking and average birth weight are unrelated?
Solution
Absolutely not. It is possible that there is some difference but we did not detect it. If this is the case, we made a Type 2 Error.
Exercise $$\PageIndex{5}$$
If we made a Type 2 Error and there is a difference, what could we have done differently in data collection to be more likely to detect such a difference?
Solution
We could have collected more data. If the sample sizes are larger, we tend to have a better shot at finding a difference if one exists.
Summary for inference of the difference of two means
When considering the difference of two means, there are two common cases: the two samples are paired or they are independent. (There are instances where the data are neither paired nor independent.) The paired case was treated in Section 5.1, where the one-sample methods were applied to the differences from the paired observations. We examined the second and more complex scenario in this section.
When applying the normal model to the point estimate $$\bar {x}_1 - \bar {x}_2$$ (corresponding to unpaired data), it is important to verify conditions before applying the inference framework using the normal model. First, each sample mean must meet the conditions for normality; these conditions are described in Chapter 4 on page 168. Secondly, the samples must be collected independently (e.g. not paired data). When these conditions are satisfied, the general inference tools of Chapter 4 may be applied.
For example, a confidence interval may take the following form:
$\text {point estimate} \pm z^*SE$
When we compute the confidence interval for $$\mu_1 - \mu_2$$, the point estimate is the difference in sample means, the value $$z^*$$ corresponds to the confidence level, and the standard error is computed from Equation \ref{5.4}. While the point estimate and standard error formulas change a little, the framework for a confidence interval stays the same. This is also true in hypothesis tests for differences of means.
In a hypothesis test, we apply the standard framework and use the specific formulas for the point estimate and standard error of a difference in two means. The test statistic represented by the Z score may be computed as
$Z = \dfrac {\text {point estimate - null value}}{SE}$
When assessing the difference in two means, the point estimate takes the form $$\bar {x}_1- \bar {x}_2$$, and the standard error again takes the form of Equation \ref{5.4}. Finally, the null value is the difference in sample means under the null hypothesis. Just as in Chapter 4, the test statistic Z is used to identify the p-value.
Examining the Standard Error Formula
The formula for the standard error of the difference in two means is similar to the formula for other standard errors. Recall that the standard error of a single mean, $$\bar {x}_1$$, can be approximated by
$SE_{\bar {x}_1} = \dfrac {s_1}{\sqrt {n_1}}$
where $$s_1$$ and $$n_1$$ represent the sample standard deviation and sample size.
The standard error of the difference of two sample means can be constructed from the standard errors of the separate sample means:
$SE_{\bar {x}_1- \bar {x}_2} = \sqrt {SE^2_{\bar {x}_1} + SE^2_{\bar {x}_2}} = \sqrt {\dfrac {s^2_1}{n_1} + \dfrac {s^2_2}{n_2}} \label {5.13}$
This special relationship follows from probability theory.
Exercise $$\PageIndex{6}$$
Prerequisite: Section 2.4. We can rewrite Equation \ref{5.13} in a different way:
$SE^2_{\bar {x}_1 - \bar {x}_2} = SE^2_{\bar {x}_1} + SE^2_{bar {x}_2}$
Explain where this formula comes from using the ideas of probability theory.10 |
# Co-Calculus Homework 2 Solutions
Find all real zeros of the following polynomials. Write each polynomial in factored
form.
6. f(x) = x3 − 6x2 + 11x − 6
Solution :
First of all, since f is a polynomial of degree 3, we know that f has at most 3
real zeros . Second, the coefficients in front of the powers of x in f(x) change sign 3
times
(1 → −6 → 11 → −6), so f(x) has either 3 positive real zeros or 3-2=1 positive
real zero. Note that
so we see that the coefficients of f (−x) have NO sign changes, so f(x) has no negative
real real zeros.
To find the potential rational zeros of f, we use the Rational zeros theorem, which
says that, if x is a zero of f, then it must be of the form , where p is a factor of the
coefficient
in front of x0 = 1, and q is a factor of the coefficient in front of xn, where
n is the highest power of x in the polynomial . In this case, possibilities for p include
1,-1,2,-2,3,-3, 6, or -6, and possibilities for q include 1 and -1. So possible RATIONAL
zeros of f include 1,-1,2,-2,3,-3, 6, or -6. Let’s test x = 1. Plugging x = 1 into f(x)
gives f(1) = 1 − 6 + 11 − 6 = 0. Great news! This means 1 is a zero of f, so x − 1 is
a factor of f. So f(x) = (x − 1)q(x), where q(x) is a polynomial. To solve for q (x),
we use long division:
This means f(x) = (x − 1)(x2 − 5x + 6). To find other zeros of f, we can just
find the other factors of f, and to do this, we simply factor (x2 − 5x + 6). We get
(x2 − 5x + 6) = (x − 2)(x − 3) (you can use the quadratic formula to do this) . So,
finally,
f(x) = (x − 1)(x − 2)(x − 3),
and the zeros of f are given by 1, 2, and 3. Again this follows because (x−1), (x−2),
and (x − 3) are factors of f!
9. f(x) = x4 − 10x3 + 35x2 − 50x + 24
Solution:
First of all, since f is a polynomial of degree 4, it has at most 4 real zeros. Since
the coefficients of f(x) change sign 4 times, f has either 4, 4-2=2, or 4-4=0 positive
real zeros; that is, f has either 4,2, or 0 positive real zeros (recall it is the number
of sign changes or the number of sign changes minus an even integer). To determine
the possible number of negative real zeros, we write
Since coefficients of f(−x) never change sign, we conclude that f has NO negative
real zeros.
Using the Rational zeros theorem, we list the potential rational zeros of f. Using
the same strategy as in problem 5 above, we get that potential rational zeros are:
1, -1, 2, -1, 3, -3, 4, -4, 6, -6, 8, -8, 12, -12, 24, -24. We can eliminate all negative
numbers from the list since f has no negative real zeros, so possible rational
zeros of f include 1, 2, 3, 4, 6, 8, 12, and 24. We test x = 1, and we find that
f(1) = 0. So (x − 1) is a factor of f, which means f(x) = (x − 1)q(x), where q(x)
is a polynomial. To determine q(x), we use long division (which I leave to you),
and we get f(x) = (x − 1)(x3 − 9x2 + 26x − 24). Unfortunately, we cannot factor
q(x) = x3 − 9x2 + 26x − 24 in our heads (at least I can’t), so we need to factor q(x)
by determining its zeros. Note that the rational zeros for q must come from the list
of potential rational zeros for our original polynomial f, because if x−c is a factor of
q(x), then it is also a factor of f(x) (meaning if c is a zero of q, then it is also a zero
of f). So we go back to our list and test x = 1 AGAIN to see if x = 1 is a zero of q(x)
(its possible that x−1 is a factor that shows up twice in f(x)). In this case q(1) ≠ 0.
Luckily, q(2) = 0, so x−2 is a factor of q(x). Now, q(x) = (x−2)r(x), where r(x) is a
polynomial. To determine r(x), we use long division, and we get r(x) = x2 −7x+12,
and q(x) = (x−2)(x2−7x+12), so f(x) = (x−1)q(x) = (x−1)(x−2)(x2−7x+12).
To finish finding zeros of f and factoring f, we just need to factor x2 − 7x + 12 =
(x − 4)(x − 3). So
(1) f(x) = (x − 1)(x − 2)(x − 4)(x − 3)
and the zeros of f are 1, 2, 3, and 4.
11. f(x) = 9x6 + 44x4 + 31x2 − 4
Solution:
First of all, f is a polynomial of degree 6, so it has at most 6 real zeros. Using
Descartes rule of signs, we see only one sign change in coefficients of f(x), so f has
exactly one positive real zero. Note that
which, again, has only one sign change, so f has exactly one NEGATIVE real zero.
This is very useful information, because now we know that f has exactly two real
zeros: one positive and one negative. Once we find them, we will know that we can
stop working. Also, since f(x) = f(−x), we can conclude that if x = c is a zero of
f(x), then x = −c is a zero of f(x), so we really only need to find one zero of f.
Using the rational zeros theorem, we see that potential rational zeros include 1, -1, 2,
. Certainly, 1, -1, 2, -2, 4, and -4
will not work, because when you plug them in, you will get a large number minus 4.
After some testing, we find that works, and by the logic above , works as well.
We have found all of the real zeros. We know by the factor theorem that and
are factors of f, that is,
where q(x) cannot be factored (since f has no other real zeros). To find q(x), we use
long division to divide into f(x). We get q(x) = 9x4+45x2+36.
So, finally,
We cannot factor f any further. The real zeros of f are and .
Prev Next |
Solving Equations with Variables On Both Sides Worksheet 8th Grade
When Solving Equations with Variables on Both Sides Worksheet 8th Grade, one of the more interesting and challenging problems I have found recently is an algebra equation where you can solve it only by solving both equations. That means that the algebraic equation must be multiplied by the derivative to solve the entire equation. It is not impossible, but it can be tricky if you are not a great student. Since algebra is such a challenging subject, it can be very satisfying to find a simple way to solve a difficult problem.
Here is a question for you, how can you solve a problem like this one? You can start by looking at the traditional method of solving equations with variables on both sides. You will probably know the solution because you can solve any other problem in the same form.
You need to be familiar with the fundamental theorem of calculus and know that the derivatives of a function concerning a given function are simply the sum of the derivatives of the two functions. This means that the derivative of a function concerning a given function will be equal to the sum of the derivatives of the two functions.
Now for the important step, you will have to convert one side of the equation to a number that is less than the other side. The easiest way to do this is to multiply the number you would like to find by the right-hand side of the equation. So if you have written the equation as
then you can multiply it by – one to get the answer. It is important to remember that if the two sides of the equation are equal, then they are both positive.
If one side is negative then you may want to check whether you have the initial value of the function at the beginning of the substitution. If not, then you must choose another side. Then you have to multiply by the derivative of the function to find the answer. You will also have to be aware that the final answer to this equation is negative, so make sure that you are working out the derivative right before you finally find the answer.
If you have a calculator then you could use the og.com substitution to solve this equation, but unless you have a calculator with a special function for this type of substitution then you might have to use a word processor to work out the equation. For many students, solving problems like this one is not an easy task, but if you practice with the problems and remember what you are doing, then you will get better and more accurate answers. |
# 2023 AMC 12B Problems/Problem 15
The following problem is from both the 2023 AMC 10B #18 and 2023 AMC 12B #15, so both problems redirect to this page.
## Problem
Suppose $a$, $b$, and $c$ are positive integers such that$$\frac{a}{14}+\frac{b}{15}=\frac{c}{210}.$$Which of the following statements are necessarily true?
I. If $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both, then $\gcd(c,210)=1$.
II. If $\gcd(c,210)=1$, then $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both.
III. $\gcd(c,210)=1$ if and only if $\gcd(a,14)=\gcd(b,15)=1$.
$\textbf{(A)}~\text{I, II, and III}\qquad\textbf{(B)}~\text{I only}\qquad\textbf{(C)}~\text{I and II only}\qquad\textbf{(D)}~\text{III only}\qquad\textbf{(E)}~\text{II and III only}$
## Solution 1 (Guess and check + Contrapositive)
We examine each of the conditions.
The first condition is false. A simple counterexample is $a=3$ and $b=5$. The corresponding value of $c$ is $17\cdot15=255$. Clearly, $\gcd(3,14)=1$ and $\gcd(5,15)=5$, so condition $I$ would imply that $\gcd(c,210)=1.$ However, $\gcd(255,210)$ is clearly not $1$ (they share a common factor of $5$). Obviously, condition $I$ is false, so we can rule out choices $A,B,$ and $C$.
We are now deciding between the two answer choices $D$ and $E$. What differs between them is the validity of condition $II$, so it suffices to simply check $II$.
We look at statement $II$'s contrapositive to prove it. The contrapositive states that if $\gcd(a,14)\neq1$ and $\gcd(b,15)\neq1$, then $\gcd(c,210)\neq1.$ In other words, if $a$ shares some common factor that is not $1$ with $14$ and $b$ shares some common factor that is not $1$ with $15$, then $c$ also shares a common factor with $210$. Let's say that $a=a'\cdot n$, where $a'$ is a factor of $14$ not equal to $1$. (So $a'$ is the common factor.)
We can rewrite the given equation as $15a+14b=c\implies15(a'n)+14b=c.$ We can express $14$ as $a'\cdot n'$, for some positive integer $n'$ (this $n'$ can be $1$). We can factor $a'$ out to get $a'(15n+14n')=c.$
We know that all values in this equation are integers, so $c$ must be divisible by $a'$. Since $a'$ is a factor of $14$, $a'$ must also be a factor of $210$, a multiple of $14$. Therefore, we know that $c$ shares a common factor with $210$ (which is $a'$), so $\gcd(c,210)\neq1$. This is what $II$ states, so therefore $II$ is true.
Thus, our answer is $\boxed{\textbf{(E) }\text{II and III only}}.$ ~Technodoggo
## Solution 2
The equation given in the problem can be written as $$15 a + 14 b = c. \hspace{1cm} (1)$$
$\textbf{First, we prove that Statement I is not correct.}$
A counter example is $a = 1$ and $b = 3$. Thus, ${\rm gcd} (c, 210) = 3 \neq 1$.
$\textbf{Second, we prove that Statement III is correct.}$
First, we prove the if part.
Suppose ${\rm gcd}(a , 14) = 1$ and ${\rm gcd}(b, 15) = 1$. However, ${\rm gcd} (c, 210) \neq 1$.
Thus, $c$ must be divisible by at least one factor of 210. W.L.O.G, we assume $c$ is divisible by 2.
Modulo 2 on Equation (1), we get that $2 | a$. This is a contradiction with the condition that ${\rm gcd}(a , 14) = 1$. Therefore, the if part in Statement III is correct.
Second, we prove the only if part.
Suppose ${\rm gcd} (c, 210) \neq 1$. Because $210 = 14 \cdot 15$, there must be one factor of 14 or 15 that divides $c$. W.L.O.G, we assume there is a factor $q > 1$ of 14 that divides $c$. Because ${\rm gcd} (14, 15) = 1$, we have ${\rm gcd} (q, 15) = 1$. Modulo $q$ on Equation (1), we have $q | a$.
Because $q | 14$, we have ${\rm gcd}(a , 14) \geq q > 1$.
Analogously, we can prove that ${\rm gcd}(b , 15) > 1$.
$\textbf{Third, we prove that Statement II is correct.}$
This is simply a special case of the only if part of Statement III. So we omit the proof.
All analyses above imply $\boxed{\textbf{(E) II and III only}}.$
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
It can easily be shown that statement I is false (a counterexample would be $a=1, b=5, c=85$), meaning the only viable answer choices are D and E. Since both of these answer choices include statement III, this means III is true. Since III is true, this actually implies that statement II is true, as III is just a stronger version of statement II (or it's contrapositive, to be precise). Therefore the answer is $\boxed{\textbf{(E) II and III only}}.$
~SpencerD. ~edited by A_MatheMagician
## Video Solution
~MathProblemSolvingSkills.com
~megahertz13
## Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) |
# Samacheer Kalvi 11th Maths Solutions Chapter 12 Introduction to Probability Theory Ex 12.2
## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 12 Introduction to Probability Theory Ex 12.2
Question 1.
If A and B are mutually exclusive events P(A) = $$\frac{3}{8}$$ and P (B) = $$\frac{1}{8}$$, then find
(i) P$$(\overline{\mathrm{A}})$$
(ii) P(A ∪ B)
(iii) P($$\overline{\mathrm{A}}$$ ∩ B
(iv) P$$(\overline{\mathrm{A}} \cup \overline{\mathrm{B}})$$
Solution:
Question 2.
If A and B are two events associated with a random experiment for which P(A) = 0.35, P(A or B) = 0.85, and P(A and B) = 0.15.
Find (i) P(only B)
(ii) P$$(\overline{\mathrm{B}})$$
(iii) P(only A)
Solution:
Given P(A) = 0.35
P(A ∪ B) = 0.85
P(A ∩ B) = 0.15
We know P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
(i.e.,) 0.85 = 0.35 + P(B) – 0.15
⇒ 0.85 – 0.2 = P(B)
(i.e.,) P(B) = 0.65
(i) P(only B ) = P(B) – P(A ∩ B)
= 0.65 – 0.15 = 0.50
(ii) P$$(\overline{\mathrm{B}})$$ = 1 – P(B) = 1 – 0.65 = 0.35
(iii) P(A only) = P(A) – P(A ∩ B) = 0.35 – 0.15 = 0.20
Question 3.
A die is thrown twice. Let Abe the event, ‘First die shows 5’ and B be the event, ‘second die shows 5’. Find P(A ∪ B).
Solution:
When a die is throw twice
n(s) = 62 = 36
Let A be the event that first die shows 5 and B be the event that second die shows 5 Now A = {(5, 1), (5, 2) (5, 3), (5, 4), (5, 5) (5, 6}
n(A) = 6 ⇒ P(A) = $$\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{6}{36}$$
and B = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)}
n(B) = 6 ⇒ P(B) = $$\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{6}{36}$$
Also A ∩ B = {(5, 5)} ⇒ n (A ∩ B) = 1
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= $$\frac{6}{36}+\frac{6}{36}-\frac{1}{36}=\frac{11}{36}$$
Question 4.
The probability of an event A occurring is 0.5 and B occurring is 0.3. If A and B are mutually exclusive events, then find the probability of
(i) P(A ∪ B)
(ii) P(A ∩ $$\overline{B}$$)
(iii)P($$\overline{A}$$ ∩ B)
Solution:
Given P(A) = 0.5, P(B) = 0.3
Also A and B are mutually exclusive. ∴ A ∩ B = Φ
n(A ∩ B) = 0
(i) P (A ∪ B)
P(A ∪B) = P(A) + P(B) – P(A ∩ B)
P(A ∪ B) = P(A) + P(B) – 0 = 0.5 + 0.3 = 0.8
(ii) P(A ∩ B̅)
P(A ∩ B̅) = P(A) – P(A ∩ B) = P(A) – 0
since A and B are mutually exclusive = 0.5
(iii) P(A̅ ∩ B)
P(A̅ ∩ B) = P(B) – P(A ∩ B) = P(B) – 0
since A and B are mutually exclusive = 0.3
Question 5.
A town has 2 fire engines operating independently. The probability that a fire engine is available when needed is 0.96.
(i) What is the probability that a fire engine is available when needed?
(ii) What is the probability that neither is available when needed?
Solution:
(i) Probability that a fire engine is available when needed = P (availability of the first fire engine A or availability of the second fire engine B)
= P(A ∪ B)
= P(A) + P(B) – P (A ∩ B)
= P (A) + P ( B) – P (A) . P (B)
since A and B are independent events
= 0.96 + 0.96 – 0.96 × 0.96
= 2 × 0.96 – (0.96)2
= 0.96 (2 – 0.96)
= 0.96 × 1.04 = 0.9984
(ii) Probability of neither of the fire engine available when needed = Probability of non-availability of the first fire engine and Probability of non-availability of the second fire engine.
= P(A̅ ∩ B̅)
= P(A̅) . P(B̅)
since A and B are independent.
= 0.04 × 0.04
= 0.0016
Question 6.
The probability that a new railway bridge will get an award for its design is 0.48, the probability that it will get an award for the efficient use of materials is 0.36, and that it will get both awards is 0.2. What is the probability, that (i) it will get atleast one of the two awards 00 it will get only one of the awards?
Solution:
Given P(A) = 0.48, P(B) = 0.36 and P(A ∩ B) = 0.2
(i) it will get at least one of the two awards
Given the probability that a new railway bridge will get an award for its design is 0.48.
P (A) = 0.48
The probability that it will get an award for the efficient use of materials is 0.36
P (B) = 0.36
The probability that it will get both awards is 0.2
P(A ∩ B) = 0.2
P(atleast one of the two awards) = P(A or B)
= P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= 0.48 + 0.36 – 0.2
= 0.84 – 0.20
= 0.64
(ii) P (only one of the awards) = P (only A or only B)
= P [(A ∩ B̅) ∪ (A̅ ∩ B)]
= P (A ∩ B̅) + P (A̅ ∩ B)
A ∩ B and A̅ ∩ B are mutually exclusive
= [P(A) – P(A ∩ B)] + [P(B) – P(A ∩ B)]
= [0.48 – 0.2] + [0.36 – 0.2]
= 0.28 + 0.16
= 0.44
### Samacheer Kalvi 11th Maths Solutions Chapter 12 Introduction to Probability Theory Ex 12.2 Additional Problems
Question 1.
A and B are two events associated with random experiment for which P(A) = 0.36, P(A or B) = 0.90 and P(A and B) = 0.25. Find
(i) P(B)
(ii) P$$(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})$$
Solution:
(i) Given P(A) = 0.36, P(A ∪ B) = 0.09, P(A ∩ B) = 0.25
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
(i.e.,) 0.90 = 0.36 + P(B) – 0.25
0.90 = 0.11 + P(B)
∴ P(B) = 0.90 – 0.11 = 0.79
(ii) P$$(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})$$ = P{(A’ ∪ B)’} (Demorgan Law)
P(A ∪ B)’ = 1 – P(A ∪ B) = 1 – 0.90 = 0.1.
Question 2.
Given P(A) = 0.5, P(B) = 0.6 and P(A ∩ B) = 0.24. Find
(i) P(A ∪ B)
(ii) P($$\overline{\mathrm{A}}$$ ∩ B)
(iii) P(A ∩ $$\overline{\mathrm{B}}$$)
(iv) P($$\overline{\mathrm{A}} \cup \overline{\mathrm{B}})$$
(v) P$$(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})$$
Solution:
(i) P(A) = 0.5, P(B) = 0.6, P(A ∩ B) = 0.24
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
(i.e.,) P(A ∪ B) = 0.5 + 0.6 – 0.24
= 1.1 – 0.24
= 0.86
∴ P(A ∪ B) = 0.86
(ii) P($$\overline{\mathrm{A}}$$ ∩ B) = P(B) – P(A ∩ B)
= 0.6 – 0.24
= 0.36
(iii) P(A ∩ $$\overline{\mathrm{B}}$$) = P(A) – P(A ∩ B)
= 0.5 – 0.24
= 0.26
(iv) P($$\overline{\mathrm{A}} \cup \overline{\mathrm{B}})$$ = P {(A ∩ B)’} = 1 – P(A ∩ B)
= 1 – 0.24
= 0.76
(v) P$$(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})$$ = P{A ∪ B)’} = 1 – P(A ∪ B)
= 1 – 0.86
= 0.14.
Question 3.
The probability of an event A occurring is 0.5 and B occurring is 0.3. If A and B are mutually exclusive events, then find the probability of neither A nor B occurring.
Solution:
Given A and B are mutually exclusive and P(A) = 0.5, P(B) = 0.3
∴ P(A ∪ B) = P(A) + P(B) = 0.5 + 0.3 = 0.8
So, P(A’ ∩ B’) = P{(A ∪ B)’} = 1 – P(A ∪ B)
= 1 – 0.8
= 0.2
Question 4.
The probability that a new ship will get an award for its design is 0.25, the probability that it will get an award for the efficient use of materials is 0.35 and that it will get both awards is 0.15. What is the probability, that (/) it will get atleast one of the two awards (ii) it will get only one of the awards?
Solution:
Probability of getting the award for its design = P(A) = 0.25
Probability of getting the award for the efficient use of materials = P(B) = 0.35
Probability of getting both awards = P(A ∩ B) = 0.15
Now P(A) = 0.25; P(B) = 0.35 and P(A ∩ B) = 0.15
∴ (i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.25 + 0.35 – 0.15
= 0.60 – 0.15
= 0.45
(ii) P(A’ ∩ B’ or B ∩ A’) = P(A ∩ B’) + P(A’ ∩ B)
P(A ∩ B’) = P(A) – P(A ∩ B)
= 0.25 – 0.15
= 0.10
P(A’ ∩ B) = P(B) – P(A ∩ B)
= 0.35 – 0.15
= 0.20
∴ P(A ∩ B’) + P(A’ ∩ B) = 0.10 + 0.20 = 0.30. |
Engage NY Eureka Math 4th Grade Module 3 Lesson 17 Answer Key
Eureka Math Grade 4 Module 3 Lesson 17 Problem Set Answer Key
Show the division using disks. Relate your model to long division. Check your quotient and remainder by using multiplication and addition.
Question 1.
5 ÷ 2
quotient = ____2______
remainder = __1________
Explanation:
Shown the division 5 ÷ 2 using disks. Related my work on the place value chart to long division. Checking my quotient and remainder by using multiplication and addition as shown above.
Question 2.
50 ÷ 2
quotient = _____25_____
remainder = ____0______
50 ÷ 2 =
Explanation:
Shown the division 5 ÷ 2 using disks. Related my work on the place value chart to long division. Checking my quotient and remainder by using multiplication and addition as shown above.
Question 3.
7 ÷ 3
quotient = _____2_____
remainder = ____1______
7 ÷ 3 =
Explanation:
Shown the division 7 ÷ 3 using disks. Related my work on the place value chart to long division. Checking my quotient and remainder by using multiplication and addition as shown above.
Question 4.
75 ÷ 3
quotient = ____25____
remainder = ___0____
75 ÷ 3 =
Explanation:
Shown the division 75 ÷ 3 using disks. Related my work on the place value chart to long division. Checking my quotient and remainder by using multiplication and addition as shown above.
Question 5.
9 ÷ 4
quotient = ___2_______
remainder = ____1______
9 ÷ 4 =
Explanation:
Shown the division 9 ÷ 4 using disks. Related my work on the place value chart to long division. Checking my quotient and remainder by using multiplication and addition as shown above.
Question 6.
92 ÷ 4
quotient = ____23____
remainder = ___0_____
92 ÷ 4 =
Explanation:
Shown the division 9 ÷ 4 using disks. Related my work on the place value chart to long division. Checking my quotient and remainder by using multiplication and addition as shown above.
Eureka Math Grade 4 Module 3 Lesson 17 Exit Ticket Answer Key
Show the division using disks. Relate your model to long division. Check your quotient by using multiplication and addition.
Question 1.
5 ÷ 4
quotient = ____1______
remainder = ____1______
5 ÷ 4 =
Explanation:
Shown the division 5 ÷ 4 using disks. Related my work on the place value chart to long division. Checking my quotient and remainder by using multiplication and addition as shown above.
Question 2.
56 ÷ 4
quotient = ____14______
remainder = ____0______
Explanation:
Shown the division 56 ÷ 4 using disks. Related my work on the place value chart to long division. Checking my quotient and remainder by using multiplication and addition as shown above.
Show the division using disks. Relate your model to long division. Check your quotient by using multiplication and addition.
Question 1.
7 ÷ 2
quotient = ____3_____
remainder = ___1_______
7 ÷ 2 =
Explanation:
Shown the division 7 ÷ 2 using disks. Related my work on the place value chart to long division. Checking my quotient and remainder by using multiplication and addition as shown above.
Question 2.
73 ÷ 2
quotient = ____24______
remainder = ___1_______
73 ÷ 2 =
Explanation:
Shown the division 73 ÷ 2 using disks. Related my work on the place value chart to long division. Checking my quotient and remainder by using multiplication and addition as shown above.
Question 3.
6 ÷ 4
quotient = ____1______
remainder = ____2______
Explanation:
Shown the division 6 ÷ 4 using disks. Related my work on the place value chart to long division. Checking my quotient and remainder by using multiplication and addition as shown above.
Question 4.
62 ÷ 4
quotient = _____15_____
remainder = ____2______
Explanation:
Shown the division 62 ÷ 4 using disks. Related my work on the place value chart to long division. Checking my quotient and remainder by using multiplication and addition as shown above.
Question 5.
8 ÷ 3
quotient = ____2______
remainder = ____2______
Explanation:
Shown the division 8 ÷ 3 using disks. Related my work on the place value chart to long division. Checking my quotient and remainder by using multiplication and addition as shown above.
Question 6.
84 ÷ 3
quotient = ____28______
remainder = ____3______ |
# Solve the system -2x-y=-9 5x-2y=18
hala718 | Certified Educator
-2x - y = -9.............(1)
5x - 2y = 18..............(2)
We will use the substitution method to solve the system.
First, we will rewrite equation (1).
==> -2x - y = -9
==> y = -2x + 9 .................(3)
Now we will substitute (3) into (2).
==> 5x - 2y = 18
==> 5x - 2 ( -2x + 9) = 18
==> 5x + 4x - 18 = 18
==> 9x = 36
Now we will divide by 9.
==> x= 4
Now to find y, we will substitute into equation (3);
==> y= -2x + 9
==> y= -2(4) + 9
= -8 + 9 = 1
==> y= 1
Then, the solution is the pair ( 4,1)
justaguide | Certified Educator
We have to solve
-2x-y=-9 ...(1)
5x-2y=18 ...(2)
(2) - 2*(1)
=> 5x - 2y + 4x + 2y = 18 + 18
=> 9x = 36
=> x = 4
Substitute x = 4 in (1)
-2x-y=-9
=> y = 9 - 2*4
=> y = 9 - 8
=> y = 1
Therefore x is 4 and y is 1.
giorgiana1976 | Student
We'll solve the system using substitution. We'll change the first equation.
-2x-y=-9
-y = 2x - 9
We'll multiply by -1:
y = -2x + 9 (1)
5x-2y=18 (2)
We'll substitute (1) in (2):
5x - 2(-2x + 9) = 18
We'll remove the brackets:
5x + 4x - 18 = 18
We'll combine like terms:
9x = 18 + 18
x = 2*18/9
x = 4
We'll substitute x = 4 in (1):
y = -2*4 + 9
y = 9 - 8
y = 1
The solution of the system is: {4 ; 1}. |
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# Mind Games, Trigonometric Functions and Fun
Jul 15, 2013 by admin
Hi Guys,
How are you?
I hope you are great.
I have a little challenge for you today.
Please tell me which one of the circles in the picture below is rotating.
Are you sure? Look at one of the circles closely.
Today we are going to learn about Trigonometric Functions.
Whenever we want to find an unknown side or angle in a right angle triangle we can use one of the trigonometric equations below.
EXPLANATION OF TRIGONOMETRIC
The Sin represents the word Sine and θ represents the angle. Sin is a function on your calculator. Take out your calculator and press Sin 90. When you do this you will get 1 for your answer. In the trigonometric equation the value of Sinθ is equal to opposite / hypotenuse which is the opposite side of the angle divided by the hypotenuse.
Well sometimes you are given questions like the one below, where you are asked to find the OPPOSITE side and you are only given the angle and the hypotenuse. You cannot use Pythagoras theorem because you don’t know the value for TWO sides and remember that you must know the value of two sides for you use Pythagoras theorem.
EXAMPLE 1
What is the value of the Opposite Side(x)?
Step 1: Understand the Question
The first thing that you must do is understand what the question is asking. Ask yourself, ‘What is the question really asking me to find?’. In our case, the question is asking us to find the value for x, which is the Opposite side of the angle.
Step 2: Write down what you know and what you don’t know
Ok now, let’s write down what we know and what we don’t know.
Step 3: Decide which Trigonometric Equation to use
Ok we are looking for a trigonometric function that have opposite, hypotenuse and Angle(θ).The trigonometric equation that has all these variables is Sinθ = opposite / hypotenuse.
Step 4: Put the values in the equation and solve
Well guys that’s it for today. Tomorrow we will do Cosθ and Tanθ.
Have a great day guys and Keep Being Cool
You can click on the “Books” button below for some great math books.
-Peta |
# Problems based on Kirchhoff’s laws – Problems and solutions
Kirchhoff postulated two basic laws called Kirchhoff’s laws. The two laws are Kirchhoff’s Current Law and Kirchhoff’s Voltage Law. The problems below are based on Kirchhoff’s laws.
## Problems based on Kirchhoff’s laws
Q1: From the figure below find the value Resistance “R1
Solution: First name all nodes, assign currents to all branches and ground the node as shown in figure below \\\\ we have assumed \\Current I in branch “ab” \\ I1 in branch “be” \\ I2 in branch “bc” and \\Ground node “e” as shown in above figure.
Now Apply Kirchhoff’s Voltage Law to loop “fabef” starting from node “f” and moving in direction “fabef”\\ 100 – 60I – I1R1– 20 = 0\\ 100 – 20= 60I + I1R1 \\ 80= 60I + I1R1 \;\;\;\;\;\;\;\;(1)\\
Now Apply Kirchoff’s current Law (KCL) at node “b”\\I = I1+ I2 \\ I2 = 0 because of open circuit \\ thus \\ I = I1 \;\;\;\;\;\;\;\;(2)
The Voltage across Resistance R1 \\70 – 20 = 50 \\thus\\\;\\50 = I1 R1\;\;\;\;\;\;\;\;(3)\\ Putting the value of I1R1 from equation (3) into equation (1) \\thus \\\;\\ 80 = 60I +50 \\\;\\ 80 – 50 =60I\\\;\\30 = 60I\\\;\\I=\frac{30}{60}=\frac{1}{2}\;\;\;\;\;\;\;\;\;(4)\\\;\\
From equation (2)\\\;\\I_{1}=I=\frac{1}{2}\\\;\\ Thus equation (3) becomes\\\;\\50=\frac{R_{1}}{2}\\\;\\R_{1}=100
Q2: In the circuit shown below find the value of V1 \\\;\;\;\;\;\;\;\;
Assign Names to each node. Assuming mesh current “I1“, “I2“, “I3“, and “I4” to loop “I”, “II”, “III”, and “IV” respectively, as shown in figure below\\\;\\\;\;\;\;\;\\Applying Kirchhoff’s Voltage law to loop “I” and move in the direction of arrow starting from node “a” .
-7-2(I_{1}-I_{2})+V_{1}=0\\\;\\ V_{1}=2I_{1}-2I_{2}+7\;\;\;\;\;\;\;\;\;(1)
Now Apply Kirchhoff’s Voltage law to Loop “II” and moving in the direction of arrow, starting from node “c” \\\;\\5-2(I_{2}-I_{1})-6=0\\\;\\ -1=2I_{2}-2I_{1} \\\;\\1=2I_{1}-2I_{2}\;\;\;\;\;\;\;(2)\\\;\\ Putting the value of \;2I_{1}-2I_{2}\; in equation (1) from equation (2) Thus \\\;\\V_{1}=1+7=8V
Q3: In the circuit shown in figure find the value of voltage Vab \\\;\;\;\;\;\;\;\;
Assign names of each node and loops as shown in figure. \\\;\\\;\;\;\;\;\\Let us assume Vab is the voltage across 0.2I1 current source in branch “ab” , “I2” is current through 3Ω resistance, “I3” and “I4” are the currents through branch “ec” and “ac” respectively.
Applying Kirchhoff’s current Law (KCL) at node “d” \\\;\\8=i_{1}+2\\\;\\i_{1}=6
Apply Kirchhoff’s current law (KCL) at node “b” \\\;\\i_{1}=0.2i_{1}+0.3i_{1}+i_{2}\\\;\\ i_{2}=0.5i_{1} \\\;\\ Putting the value of “I1” in above equation \\\;\\i_{2}=3
Applying KCL at node “a” \\\;\\2+i_{4}+0.2i_{1}=0\\\;\\ putting the value “I1” in above equation \\\;\\i_{4}=-3.2\\\;\\
Now Applying Kirchhoff’s Voltage Law to loop “II” i.e. “abca” and move in dircetion “a” to “b” to “c” to “a” starting from node “a” \\\;\\V_{ab}-3i_{2}-2i_{4}=0\\\;\\ Putting the values of “I2” and “I4\\\;\\V_{ab}-3\left ( 3 \right )-2\left ( -3.2 \right )=0\\\;\\V_{ab}=2.6 V
Q4: Consider the circuit shown in figure below find the relationship between voltage V1 and V2 \\\;\;\;\;\;\;\;\;
Assign names of each node as shown in figure below.\\\;\\\;\;\;\;\;\\\;\\ Applying KVL to branch “ab” \\\;\\V_{1}-6i_{a}-8-V_{2}=0\\\;\\V_{1}=6i_{a}+8+V_{2} |
# Math Snap
## Let $x$ be a random variable that represents the percentage of successful free throws a professional basketball player makes in a season. Let y be a random variable that represents the percentage of successful field goals a professional basketball player makes in a season. The table below shows the percentage of successful free throws and field goals a professional basketball player makes in a season: \begin{tabular}{|c|c|c|c|c|c|c|} \hline $\mathrm{x}$ & 58 & 52 & 61 & 59 & 69 & 90 \\ \hline $\mathrm{y}$ & 72 & 75 & 80 & 90 & 79 & 97 \\ \hline \end{tabular} Suppose you want to predict the percentage of successful field goals a professional basketball player using the percentage of successful free throws, what is the percentage of successful field goals if the percentage of successful free throws is 81 ? Round your answers to the nearest hundredths.
#### STEP 1
Assumptions 1. The variable $x$ represents the percentage of successful free throws. 2. The variable $y$ represents the percentage of successful field goals. 3. We have the following data points for $x$ and $y$: $\begin{array}{|c|c|c|c|c|c|c|} \hline x & 58 & 52 & 61 & 59 & 69 & 90 \\ \hline y & 72 & 75 & 80 & 90 & 79 & 97 \\ \hline \end{array}$ 4. We need to find the percentage of successful field goals ($y$) when the percentage of successful free throws ($x$) is 81. 5. We will use linear regression to predict $y$ from $x$.
#### STEP 2
First, calculate the means of $x$ and $y$. $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$ $\bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i$ where $n$ is the number of data points.
#### STEP 3
Calculate the mean of $x$: $\bar{x} = \frac{58 + 52 + 61 + 59 + 69 + 90}{6}$
#### STEP 4
Perform the arithmetic to find $\bar{x}$: $\bar{x} = \frac{58 + 52 + 61 + 59 + 69 + 90}{6} = \frac{389}{6} \approx 64.83$
#### STEP 5
Calculate the mean of $y$: $\bar{y} = \frac{72 + 75 + 80 + 90 + 79 + 97}{6}$
#### STEP 6
Perform the arithmetic to find $\bar{y}$: $\bar{y} = \frac{72 + 75 + 80 + 90 + 79 + 97}{6} = \frac{493}{6} \approx 82.17$
#### STEP 7
Calculate the covariance of $x$ and $y$: $\text{Cov}(x, y) = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})(y_i - \bar{y})$
#### STEP 8
Calculate the variance of $x$: $\text{Var}(x) = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2$
#### STEP 9
Calculate the covariance of $x$ and $y$: $\text{Cov}(x, y) = \frac{1}{6} \left[ (58 - 64.83)(72 - 82.17) + (52 - 64.83)(75 - 82.17) + (61 - 64.83)(80 - 82.17) + (59 - 64.83)(90 - 82.17) + (69 - 64.83)(79 - 82.17) + (90 - 64.83)(97 - 82.17) \right]$
#### STEP 10
Perform the arithmetic to find $\text{Cov}(x, y)$: $\text{Cov}(x, y) = \frac{1}{6} \left[ (-6.83)(-10.17) + (-12.83)(-7.17) + (-3.83)(-2.17) + (-5.83)(7.83) + (4.17)(-3.17) + (25.17)(14.83) \right]$ $\text{Cov}(x, y) = \frac{1}{6} \left[ 69.42 + 91.91 + 8.31 + (-45.71) + (-13.22) + 373.27 \right]$ $\text{Cov}(x, y) = \frac{1}{6} \left[ 484.98 \right] \approx 80.83$
#### STEP 11
Calculate the variance of $x$: $\text{Var}(x) = \frac{1}{6} \left[ (58 - 64.83)^2 + (52 - 64.83)^2 + (61 - 64.83)^2 + (59 - 64.83)^2 + (69 - 64.83)^2 + (90 - 64.83)^2 \right]$
#### STEP 12
Perform the arithmetic to find $\text{Var}(x)$: $\text{Var}(x) = \frac{1}{6} \left[ (-6.83)^2 + (-12.83)^2 + (-3.83)^2 + (-5.83)^2 + (4.17)^2 + (25.17)^2 \right]$ $\text{Var}(x) = \frac{1}{6} \left[ 46.63 + 164.64 + 14.66 + 33.99 + 17.39 + 633.58 \right]$ $\text{Var}(x) = \frac{1}{6} \left[ 910.89 \right] \approx 151.82$
#### STEP 13
Calculate the slope $b$ of the regression line: $b = \frac{\text{Cov}(x, y)}{\text{Var}(x)}$
#### STEP 14
Plug in the values for $\text{Cov}(x, y)$ and $\text{Var}(x)$: $b = \frac{80.83}{151.82} \approx 0.53$
#### STEP 15
Calculate the intercept $a$ of the regression line: $a = \bar{y} - b \bar{x}$
#### STEP 16
Plug in the values for $\bar{y}$, $b$, and $\bar{x}$: $a = 82.17 - 0.53 \times 64.83$
#### STEP 17
Perform the arithmetic to find $a$: $a = 82.17 - 34.36 \approx 47.81$
#### STEP 18
Write the equation of the regression line: $y = a + bx$ $y = 47.81 + 0.53x$
#### STEP 19
Use the regression equation to predict $y$ when $x = 81$: $y = 47.81 + 0.53 \times 81$
##### SOLUTION
Perform the arithmetic to find $y$: $y = 47.81 + 42.93 = 90.74$ The percentage of successful field goals if the percentage of successful free throws is 81 is approximately 90.74%. |
Дисциплины:
# POINT 3. IRRATIONAL FUNCTIONS
POINT 1. RATIONAL FUNCTIONS (RATIONAL FRACTIONS)
Def. 1. Rational function (rationas fraction) is called a function which can be represented as a ratio of two polynomials
. ( 1 )
Def. 2. Rational function (rationas fraction) (1) is called proper one if and improper otherwise ( ).
Theorem 1 (extraction of integer part of an improper rational function).Every improper rational function can be represented as a sum of some polynomial (so-called integer part) and a proper rational function.
■Let . Dividing the numerator by the denominator we get
where polynomials are a quotient and a remainder respectively and
is a proper rational function.■
Ex. 1. Extract an integer part of an improper rational fraction
a) The first (theoretical) way. After division of by we get
b) The second way. Subtracting and adding 1 in the numerator we’ll have
There are partial [simplest, elementary] fractions of 1- 4 types.
1. ;
2. ;
3. ;
4. .
We’ve integrated the fractions 1, 3 in the Point 2 of preceding lecture (ex. 14, 16, formulas (7), (10) and (11)). To integrate the fraction 2 we can put . Integration of the fraction 4 with the help of the substitution
leads to a linear combination of a simple integral
and the next one
.
As to evaluation of this latter see textbooks. For small values of k ( ) one can use the next substitution: .
Ex. 2.
.
Thus we can say that we able to integrate partial fractions 1 – 4.
Theorem 2(a partial decomposition of a proper rational function).Every proper rational function can be represented as a linear combination of partial fractions.
For example
Here are some unknown numbers (undetermined coefficients), which one can find by so-called method of undetermined coefficients.
Corollary.Every rational function can be integrated by virtue of the linear property of indefinite integral.
Rule of integration of a rational function.To integrate a rational function it’s necessary:
1. To extract an integer part of the function if it is improper one.
2. To factorize the denominator of obtained proper function into a product of polynomials of degree not higher than two.
3. To make a partial decomposition of the proper function.
4. To integrate all the terms of the obtained algebraic sum.
Ex. 3.
.
Ex. 4. Evaluate the indefinite integral .
1 step (factorizing the denominator of the proper rational function).
.
2 step (a partial decomposition of the proper function).
Let’s assign two particular values to x in (*), namely .
3 step (integration of the given function by integration of its partial decomposition).
Ex. 5. Calculate the next indefinite integral
1 step (a partial decomposition of the proper rational function with factorized denominator).
,
. (**)
Assigning tree arbitrary values to x in (**), for example , we get a system of linear equations in ,
2 step (integration of all the terms of obtained partial decomposition of the integrand) |
Ex 9.3
Chapter 9 Class 11 Straight Lines
Serial order wise
### Transcript
Ex 9.3, 4 Find the points on the x-axis, whose distances from the line 𝑥/3 + 𝑦/4 = 1 are 4 units. We need to find point on the x-axis Let any point on x-axis be P(x, 0) Given that perpendicular distance from point P(x, 0) from given line 𝑥/3 + 𝑦/4 = 1 is 4 Simplifying equation of line 𝑥/3 + 𝑦/4 = 1 (4𝑥 + 3𝑦 )/12 = 1 4x + 3y = 12 4x + 3y – 12 = 0 We know that Perpendicular distance from point (x, y) to the line Ax + By + C = 0 is d = |𝐴𝑥1 + 𝐵𝑦1 + 𝑐|/√(𝐴^2 + 𝐵^2 ) Given perpendicular distance of point P(x, 0) from line 4x + 3y – 12 = 0 is 4 Here x1 = x, y1 = 0 & A = 4 , B = 3 , C = − 12 & d = 4 Putting values 4 = |4(𝑥) + 3(0) − 12|/√(〖(4)〗^2 + 〖(3)〗^2 ) 4 = |4𝑥 − 12|/√(16 + 9) 4 = |4𝑥 − 12|/√25 4 = |4𝑥 − 12|/5 4 × 5 = |4𝑥−12| 20 = |4𝑥−12| |4𝑥−12| = 20 4x – 12 = ± 20 Thus, 4x − 12 = 20 or 4x − 12 = − 20 4x – 12 = 20 4x = 20 + 12 4x = 32 x = 32/4 x = 8 4x − 12 = −20 4x = −20 + 12 4x = −8 x = (−8)/4 x = −2 4x − 12 = −20 4x = −20 + 12 4x = −8 x = (−8)/4 x = −2 Thus, x = 8 or x = −3 Hence the required points on x-axis are (8, 0) & (−2, 0) |
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# Solve for x: $\dfrac{2}{{3{x^2}}} - \dfrac{1}{{3x}} - 1 = 0$
Last updated date: 17th Sep 2024
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Hint: In this question, we are given an algebraic expression in which the unknown variables are in the denominator so we will first simplify the expression by taking the least common factor of the denominators of the three fractions ( LCM is the smallest number divisible by both the numbers}. After finding the common denominator for the three fractions, the numerators are multiplied with the quotients obtained on dividing the LCM by their denominator and then we have to perform the given arithmetic operation like addition, subtraction, multiplication and division in the numerator.
Complete step-by-step solution:
We have to solve $\dfrac{2}{{3{x^2}}} - \dfrac{1}{{3x}} - 1 = 0$
It can be rewritten as –
$\Rightarrow \dfrac{2}{{3{x^2}}} - \dfrac{1}{{3x}} - \dfrac{1}{1} = 0 \\ \Rightarrow \dfrac{{2 - x - 3{x^2}}}{{3{x^2}}} = 0 \\$
Taking the negative side common and then taking -1 and $3{x^2}$ to the other side, we get –
$3{x^2} + x - 2 = 0$
The obtained equation is a quadratic equation and can be solved by factorization as follows –
$\Rightarrow 3{x^2} + 3x - 2x - 2 = 0 \\ \Rightarrow 3x(x + 1) - 2(x + 1) = 0 \\ \Rightarrow (3x - 2)(x + 1) = 0 \\ \Rightarrow x = \dfrac{2}{3},\,x = - 1 \\$
Hence, when $\dfrac{2}{{3{x^2}}} - \dfrac{1}{{3x}} - 1 = 0$ , we get $x = \dfrac{2}{3}$or $x = - 1$ .
Note: The highest exponent is 2 so the given equation is a polynomial equation of degree 2, hence it is a quadratic equation and has exactly two solutions. We have to solve this equation, that is, we have to find its solutions. $a{x^2} + bx + c = 0$ is the standard form of a quadratic equation. To find the factors of the given equation, we compare the given equation and the standard equation and get the values of a, b and c. Then we will try to write b as a sum of two numbers such that their product is equal to the product of a and c, that is, ${b_1} \times {b_2} = a \times c$ , this method is known as factorization. We find the value of ${b_1}$ and ${b_2}$ by hit and trial. |
# Sec squared formula
## Formula
$\sec^2{\theta} \,=\, 1+\tan^2{\theta}$
The square of secant function equals to the addition of one and square of tan function is called the secant squared formula. It is also called as the square of secant function identity.
### Introduction
The secant functions are sometimes involved in trigonometric expressions and equations in square form. The expressions and equations can be simplified by only transforming the secant squared functions into its equivalent form. Hence, it is must for learning the square of secant function identity for studying the advanced trigonometry further.
#### Usage
The secant squared trigonometric identity is sometimes used as a formula in two cases mostly.
1. The square of secant function is expanded as the summation of one and the tangent squared function.
2. The sum of one and the tan squared function is simplified as the square of secant function.
#### Popular forms
The secant squared function law is also expressed popularly in two forms in trigonometric mathematics.
1. $\sec^2{x} \,=\, 1+\tan^2{x}$
2. $\sec^2{A} \,=\, 1+\tan^2{A}$
Therefore, you can write the square of secant function formula in terms of any angle in this way in mathematics.
#### Proof
Let theta be a symbol, which represents an angle of a right triangle. The secant and tan functions are written as $\sec{\theta}$ and $\tan{\theta}$ respectively in mathematics. Mathematically, the relationship between secant and tan functions can be written in the following mathematical form according to the Pythagorean identity of secant and tan functions.
$\sec^2{\theta}-\tan^2{\theta} \,=\, 1$
$\,\,\, \therefore \,\,\,\,\,\,$ $\sec^2{\theta} \,=\, 1+\tan^2{\theta}$
Therefore, it has derived successfully that the square of secant function is equal to the addition of one and square of tan function.
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# AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.1
AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.1 Textbook Questions and Answers.
## AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.1
Question 1.
Using divisibility rules, fmd which of the following numbers are divisible by 2,5,10 ( say
yes or no ) in the given table. What do you observe?
Solution:
Question 2.
Using divisibility tests, determine which of following numbers are divisible by 2
(a) 2144 (b) 1258 (c) 4336 (d) 633 (e) 1352
Solution:
If a number is divisible by 2 then the units digit of the number be 0, 2, 4, 6, 8.
∴ a) 2144, b) 1258, c) 4336 e) 1352 are divisible by ‘2’.
Question 3.
Using divisibility tests, determine which of the following numbers are divisible by 5
(a) 438750 (b) 179015 (c) 125 (d) 639210 (e) 17852
Solution:
If a number is divisible by 5 its units digit be either ‘0’ or 5.
∴ a) 438750, b) 179015, c) 125 d) 639210 are divisible by 5.
Question 4.
Using divisibility tests, determine which of the following numbers are divisible by 10:
(a) 54450 (b) 10800 (c) 7138965 (d) 7016930 (e) 10101010
Solution:
If a number is divisible by 10 then its units digit must be 0’.
a) 54450, b) 10800, d) 7016930, e) 1010100 are divisible by 10.
Question 5.
Write the number of factors of the following’?
(a) 18 (b) 24 (e) 45 (d) 90 (e) 105
Solution:
Number Factors , No.of factors a) 18 1,2,3,6,9,18 6 b) 24 1, 2, 3, 4, 6, 8, 12, 24 8 c) 45 1,3,5,9,15,45 6 d) 90 1, 2, 3, 5, 6, 9,10, 15, 18, 30, 45, 90 12 e) 105 1,3, 5, 7,15,21,35,105 8
Question 6.
Write any 5 numbers which are divisible by 2,5 and 10.
Solution:
10, 20, 30, 40. are divisible by 2, 5 and 10
[∵ The L.C.M. of 2, 5, 10 is 10]
Question 7.
A number 34A is exactly divisible by 2 and leaves a remainder 1, when divided by 5, find A.
Solution:
If 34A Is divisible by 2 then the remainder should be equal to 0.
∴ A should be equal to 0, 2, 4, 6, 8.
∴ 340, 342, 344, 346, 348 are divisible by 2 and gives the remainder ‘0’.
Among these 346 is divisible by 5 and gives the remainder 1.
∴ 346 → $$\frac{6}{5}$$ (R = 1)
∴ The value of A = 6. |
# What is a quotient in algebraic expressions?
## What is a quotient in algebraic expressions?
When you divide one number by another, the result is called the quotient. This is true for rational expressions too! When you divide one rational expression by another, the result is called the quotient.
What is a quotient of a number and 3?
the quotient of three and a number n translates to 3 divided by the number n, given the quotient indicates division or a fraction.
Which algebraic expression represents the quotient of 5 plus d and 12 minus W?
“5 plus d ” will be expressed as 5+d , “12 minus w ” will be expressed as 12−w , Therefore, “the quotient of 5 plus d and 12 minus w ”, will be expressed as: 5+d12−w .
### How do you add and subtract algebraic?
To add two or more monomials that are like terms, add the coefficients; keep the variables and exponents on the variables the same. To subtract two or more monomials that are like terms, subtract the coefficients; keep the variables and exponents on the variables the same. on the variables the same.
What’s the quotient of 14 and 7?
2
When 14 is divided by 7 the quotient is 2. The number 14 is the dividend and the number 7 is the divisor and we have 14 ÷ 7 = 2.
What is the quotient of 39 and 3?
39 ÷ 3 = 13.
#### Which algebraic expression represents a number decreased 7?
Algebraic Expressions With Example Problems and Interactive Exercises
Phrase Expression
fourteen decreased by a number p 14 – p
seven less than a number t t – 7
the product of 9 and a number n 9 · n or 9n
thirty-two divided by a number y 32 ÷ y or
What is quotient in Algebra?
Quotient indicates division. Divide the first number n by the second number 5. More challenging exercises about writing an algebraic expression. Let n be the number. Five more than twice the difference of a number and thirteen. Key words : more than, twice, and difference. Twice = two times.
What is algebraic expression in math?
Algebraic Expressions. Just like any language, math has a way to communicate ideas. An algebraic expression is a compact way of describing mathematical objects using a combination of numbers, variables (letters), and arithmetic operations namely addition, subtraction, multiplication, and division.
## What is the quotient of N and 8?
The quotient is the result of division between two numbers. The quotient of a number (I used n) and 8 would be n 8
How to solve simplifying algebraic expression?
Solution: 1 Remove parentheses by multiplying factors. = (x * x) + (1 * x) + (2 * x) + (2 * 1) 2 Combine like terms by adding coefficients. (x * x) = x 2 (1 * x) = 1x (2* x) = 2x 3 Combine the constants. (2 * 1) = 2 4 Therefore, Simplifying Algebraic Expression is solved as |
# Two Numbers Differ by 5. If their Product is 336, then the Sum of the Two Numbers is?
### Computer MCQs Series for PPSC, FPSC – Most Repeated MCQs | Set 6
What are you looking for? Let’s dig in quickly
## Explanation
• Two numbers differ by 5.
• Their product is 336.
• The sum of the two numbers will be figure out in this way.
Let suppose the numbers are y and z.
The difference between two numbers is 5
y – z = 5 ________ (i)
The product of these numbers is 336.
yz = 336 ________ (ii)
By solving equation (i) and (ii) simultaneously, we can easily find out the value of y (y = 21) and z (z = 16) in order to find out the sum (y + z = 21 + 16 = 37).
Sum = ?
## Solution
Let suppose
Integers are y and z.
According to the given conditions
y – z = 5 ________ (i)
yz = 336
y = 336/z ________ (ii)
Putting y = 336/z in equation (i)
336/z – z = 5
336 – z2 = 5z
z2 + 5z – 336 = 0
z2 + 21z – 16z – 336 = 0
z(z + 21) – 16(z + 21) = 0
(z – 16)(z + 21) = 0
z – 16 = 0 & z + 21 = 0
z = 16 (possible) & z = -21 (not possible)
Put z = 16 in equation (i)
y – 16 = 5
y = 5 + 16
y = 21
Required Sum = 16 + 21 = 37answer
## Conclusion
Two numbers differ by 5. If their product is 336, then the numbers are 21 and 16 and there sum will be 37. |
By accessing our 180 Days of Math for Sixth Grade Answers Key Day 5 regularly, students can get better problem-solving skills.
Directions Solve each problem.
Question 1.
_________
20+12 = 32.
Explanation:
The addition of 20 and 12 is 20+12 = 32.
Question 2.
9 × 9 = _____
9 × 9 = 81.
Explanation:
The multiplication of 9 by 9 is 9 × 9 = 81.
Question 3.
35÷5 = 7.
Explanation:
By dividing 35 by 5 we will get 35÷5 = 7.
Question 4.
What is the numeral for fifteen?
_______________________
XV.
Explanation:
The numeral of fifteen is XV.
Question 5.
$$\frac{1}{8}$$ of 32 is ____
4.
Explanation:
Given that $$\frac{1}{8}$$ of 32 which is
= $$\frac{1}{8}$$ × 32
= 4.
Question 6.
Find the missing number.
3,296; 3,286; ____; 3,266
3,296; 3,286; 3,276; 3,266.
Explanation:
Here, the pattern follows countback off 10. So the missing digit 3,286-10 = 3,276. So the pattern is 3,296; 3,286; 3,276; 3,266.
Question 7.
140 ÷ = 10
The missing digit is 14.
Explanation:
Let the missing digit be X, So 140÷X = 10
X = $$\frac{140}{10}$$
= 14.
Question 8.
Solve for x when x + 9 = 49. x = ______
X = 40.
Explanation:
Given the expression is x + 9 = 49
X = 49 – 9
= 40.
Question 9.
Calculate the area of a square with 7 cm sides.
_________________
14 cm sq.
Explanation:
The area of the square is 7×7
= 14 cm sq.
Question 10.
Use a protractor to measure the angle. Then identify it as either acute, reflex, obtuse, or straight.
Size 120 degree Name Obtuse
Question 11.
What is the mean of this set of data?
45, 31, 26,46
The mean of the data set is 37.
Explanation:
The mean for this data set is $$\frac{45+31+26+46}{4}$$
= $$\frac{148}{4}$$
= 37.
Question 12.
A snail is climbing a brick wall that is 15 m high. It climbs up 3 m every day but slides down 1 m every night. How many days will it take to reach the top of the wall?
_______________________ |
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