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# Maximum Power Transfer Theorem
Maximum Power Transfer Theorem states that, for any type of circuit whether ac or dc, maximum amount of power is delivered from source to load when the source impedance as viewed from the load terminal is equal to the load impedance. This theorem is used to find the value of load impedance for which there would be a maximum amount of power flow from source to load. In this article, maximum power transfer theorem, derivation, efficiency and steps for its application is explained with the help of solved problems.
## Explanation of Maximum Power Transfer Theorem:
In dc circuit, the impedance of circuit is equal to the resistance. Therefore, maximum power transfer theorem may be stated as, to deliver the maximum power from source to load, the load resistance must be equal to the source resistance or Thevenin resistance. Here, Thevenin resistance essentially means the resistance of source network when viewed from the load terminals.
Let us consider a dc circuit as shown below for better understanding of Maximum Power Transfer Theorem.
A variable load resistance RL is connected to the dc source network. The value of Thevenin voltage is Vo while the source network resistance when viewed from load terminal (this is called Thevenin resistance) is RTh. Therefore, according to maximum power transfer theorem, the formula for maximum power flow from source to load will is given as below.
Load Resistance, RL = Thevenin Resistance, RTh
## Maximum Power Transfer Theorem Derivation:
For maximum power transfer theorem derivation, first of all we will calculate the current flowing through the circuit. Then, power (P) dissipated in the load resistance is equal to the I2RL i.e. P = I2RL. Now, we will differentiate this power P with respect to RL (as RL is a variable quantity) and equate it to zero. This will give us the value of RL for which maximum power will flow from source to load. This concept may be followed to get the proof of the theorem
Let us now apply the above concept and derive the maximum power transfer theorem. Refer the dc circuit shown in figure-1. Let us first calculate the current through the load resistance RL. The current I through the circuit is given as below.
Now, power dissipated in load resistance RL due to this current I is the power delivered from source to load. Therefore, it will be a wise decision to find the power dissipated in RL due to I. Let this power be P.
As power P is a function of RL, therefore, to get maximum value of “P” we differentiate P w.r.t RL and equate it to zero. This is shown below.
Therefore, it is proved that, the value of load resistance and source resistance should match for maximum power delivery to the load from a dc source network. Let us now calculate the efficiency during the condition of maximum power transfer.
### Maximum Power Delivered to Load:
Under the condition of maximum power delivery, the current through the circuit is given as below.
I = (Vo/2RTh)
Therefore, maximum amount of power transferred to load
= I2RL
= I2RTh (RL = RTh)
= [(Vo)2/4RTh]
### Efficiency:
Efficiency during maximum power transfer is the ratio of power dissipated in load to the total power supplied by the source. The value of this efficiency is 50%.
During maximum power transfer, source resistance and load resistance are equal as derived. This means, RL = RTh. Also, the current “I” is flowing through RTh as well as RL. Therefore, the power dissipated in load will be equal to the power dissipated in source internal resistance RTh. Hence, the total power delivered by source will be equal to the sum of power dissipated in load as well as internal resistance.
Total Power Delivered by Source Ps
= Power dissipated in RL + Power dissipated in load RTh
= I2RL + I2RTh
= I2 (RL+ RTh)
= 2 I2 RL (since, RL = RTh)
= 2 PL (PL is power dissipated in load)
Therefore,
Efficiency = PL / Ps
= PL / 2PL
= 0.5
Hence, percentage efficiency during maximum power transfer condition
= 0.5×100
= 50%
## Application of Maximum Power Transfer Theorem (MPTT):
The concept of maximum power transfer theorem finds a wide application in communication circuits. In communication circuit, the magnitude of power transfer is very small and hence low efficiency of 50% is not a problem for such applications. However, in electrical power transmission system, the efficiency of 50% is not tolerable at all. This simply means that, half of the total generated power is lost before reaching the load center. The main aim of power transmission is to maintain voltage or minimize the voltage drop and line losses. In view of this, MPTT is not applicable for power transmission.
## Steps to Solve Maximum Power Transfer Theorem:
The steps to solve maximum Power Transfer Theorem is quite simple and handy. Following are the sequential steps:
Step-1: Remove the load resistance (RL) and find the Thevenin resistance (RTh) of the source network when viewed from the open circuited terminals of load.
Step-2: For maximum power delivery from source to load, this Thevenin resistance should be equal to load resistance. This essentially means, RL = RTh
Step-3: Find the Thevenin voltage (Vo) across the open circuited terminals of load. If you find problem in calculating this value, kindly refer Thevenin Theorem.
Step-4: Maximum amount of power delivered to load is equal to [(Vo)2/4RTh]
## Maximum Power Transfer Theorem Solved Problem:
In this section, we will solve one problem related to the theorem. This will definitely help you to better understand and summarize the concept of maximum power transfer theorem.
Problem: Find the value of maximum power dissipated in RL in the circuit shown below. Provided r = 5 Ω, v0 = 10 V and i0 = 2 A.
Solution: To solve this problem, we will follow the steps discussed earlier in this article. As the very first step, we will remove the load resistance RL and find the Thevenin resistance.
Step-1: Remove RL and find Thevenin resistance. To find the Thevenin resistance, we will short the voltage source and current source will be replaced by open circuit.
Thus, the Thevenin resistance Rth = r = 5 Ω.
Step-2: For maximum delivery of power to the load resistance, the value of load resistance RL should be equal to Thevenin resistance i.e. 5 Ω.
Step-3: In this step, we will find the voltage across the open circuited load terminals. This is called Thevenin voltage.
Voc = ri0 + v0
= (5×2 + 10) V = 20 V
Step-4:
Maximum power transferred to load
= (20)2/(4×5)
= 20 Watt (Ans.) |
# Algebra system of equations solver
This Algebra system of equations solver helps to fast and easily solve any math problems. Math can be a challenging subject for many students.
## The Best Algebra system of equations solver
Math can be a challenging subject for many learners. But there is support available in the form of Algebra system of equations solver. Luckily, there are a number of resources that can help. Online math tutors can work with students one-on-one to help them understand difficult concepts and work through different problems. In addition, there are a number of websites that provide step-by-step solutions to common math problems. With a little bit of effort, any student can get the help they need to succeed in math.
A radical is a square root or any other root. The number underneath the radical sign is called the radicand. In order to solve a radical, you must find the number that when multiplied by itself produces the radicand. This is called the principal square root and it is always positive. For example, the square root of 16 is 4 because 4 times 4 equals 16. The symbol for square root is . To find other roots, you use division. For example, the third root of 64 is 4 because 4 times 4 times 4 equals 64. The symbol for the third root is . Sometimes, you will see radicals that cannot be simplified further. These are called irrational numbers and they cannot be expressed as a whole number or a fraction. An example of an irrational number is . Although radicals can seem daunting at first, with a little practice, they can be easily solved!
How to solve radicals can be a tricky topic for some math students. However, with a little practice, it can be easy to understand how to solve these equations. The first step is to identify the type ofradical that is being used. There are two types of radicals, square roots and cube roots. Once the type of radical has been identified, the next step is to determine the value of the number inside the radical. This number is called the radicand. To find the value of the radicand, take the square root of the number if it is a square root radical or the cube root of the number if it is a cube root radical. The last step is to simplify the equation by cancelling out any factors that are shared by both sides of the equation. With a little practice, solving radicals can be easy!
In addition, the website provides a forum for students to ask questions and receive help from other users. Whether you are looking for a way to improve your child's math skills or simply want to provide them with a fun and educational activity, web math is an excellent choice. |
## Precalculus (6th Edition) Blitzer
(a) Consider the function, \begin{align} & \left( f\circ g \right)\left( x \right)=f\left( g\left( x \right) \right) \\ & =f\left( 4x-1 \right) \\ & ={{\left( 4x-1 \right)}^{2}}+3 \end{align} On simplification, we get, \begin{align} & \left( f\circ g \right)\left( x \right)={{\left( 4x-1 \right)}^{2}}+3 \\ & =16{{x}^{2}}-8x+1+3 \\ & =16{{x}^{2}}-8x+4 \end{align} Therefore, the value of $\left( f\circ g \right)\left( x \right)$ for the functions $f\left( x \right)={{x}^{2}}\text{+3 and }g\left( x \right)=4x-1$ is $16{{x}^{2}}-8x+4$. (b) Consider the function \begin{align} & \left( g\circ f \right)\left( x \right)=g\left( f\left( x \right) \right) \\ & =g\left( {{x}^{2}}+3 \right) \\ & =4\left( {{x}^{2}}+3 \right)-1 \\ & =4{{x}^{2}}+11 \end{align} Therefore, the value of $\left( g\circ f \right)\left( x \right)$ for the functions $f\left( x \right)={{x}^{2}}\text{+3 and }g\left( x \right)=4x-1$ is $4{{x}^{2}}+11$. (c) From part (a), $\left( f\circ g \right)\left( x \right)={{\left( 4x-1 \right)}^{2}}+3$. Therefore, \begin{align} & \left( f\circ g \right)\left( 3 \right)={{\left( 4\times 3-1 \right)}^{2}}+3 \\ & ={{\left( 11 \right)}^{2}}+3 \\ & =121+3 \\ & =124 \end{align} Therefore, the value of $\left( f\circ g \right)\left( 3 \right)$ for the functions $f\left( x \right)={{x}^{2}}\text{+3 and }g\left( x \right)=4x-1$ is $124$. |
Question Video: Calculating the Horizontal Component of a Vector | Nagwa Question Video: Calculating the Horizontal Component of a Vector | Nagwa
# Question Video: Calculating the Horizontal Component of a Vector Physics • First Year of Secondary School
## Join Nagwa Classes
The diagram shows a vector, 𝐀, that has a magnitude of 55. The angle between the vector and the 𝑥-axis is 82°. Work out the horizontal component of the vector. Give your answer to the nearest whole number.
01:57
### Video Transcript
The diagram shows a vector, 𝐀, that has a magnitude of 55. The angle between the vector and the 𝑥-axis is 82 degrees. Work out the horizontal component of the vector. Give your answer to the nearest whole number.
In this question, we are being asked to solve for the horizontal component of the vector 𝐀. We can label this component 𝐴 subscript 𝑥. To find this horizontal component, recall that we can use the formula 𝐴 subscript 𝑥 equals 𝐴 times cos 𝜃, where 𝐴 is the magnitude of the vector and 𝜃 is the argument of the vector.
We are told in the question that the magnitude of the vector 𝐀 is 55. We are also told that there is an angle of 82 degrees between the vector and the negative 𝑥-axis. But we need to be careful here; this angle is not the argument of the vector. The argument of a vector is defined as the angle between the vector and the positive 𝑥-axis, measured counterclockwise from the positive 𝑥-axis.
So, to find the argument of this vector, we need to add 180 degrees to this angle of 82 degrees. Adding 82 degrees plus 180 degrees gives us that the argument 𝜃 is equal to 262 degrees. Substituting these values for the magnitude 𝐴 and the argument 𝜃 into our equation, we find that the horizontal component 𝐴 subscript 𝑥 is equal to 55 multiplied by the cos of 262 degrees.
Completing this calculation, we find that the horizontal component 𝐴 subscript 𝑥 of the vector 𝐀 is equal to negative 7.654. We want this answer to the nearest whole number, so we can round down negative 7.654 to negative eight. And so we have arrived at the final answer. To the nearest whole number, the horizontal component of the vector 𝐀 is negative eight.
## Join Nagwa Classes
Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!
• Interactive Sessions
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# Math Expressions Grade 2 Student Activity Book Unit 6 Lesson 2 Answer Key Place Value
This handy Math Expressions Student Activity Book Grade 2 Answer Key Unit 6 Lesson 2 Place Value provides detailed solutions for the textbook questions.
## Math Expressions Grade 2 Student Activity Book Unit 6 Lesson 2 Place Value Answer Key
Review the Use of Boxes, Sticks, and Circles to Represent Numbers
Write the number that is shown by the drawing.
Question 1.
Total _____________
Each box represents the value of hundred and stick resembles the value of tens.
There are 6 boxes and 7 sticks in the given figure, which means
6×100=600 and 7×10=70
Total will be 670
Question 2.
Total _____________
Each box represents the value of hundred and stick resembles the value of tens.
There are 3 boxes, 3 sticks and 5 circles in the given figure, which means
3×100=300, 3×10=30 and 5×1=5
Total will be 335
Question 3.
Total _____________
Each box represents the value of hundred and stick resembles the value of tens.
There are 4 boxes and 9 circles in the given figure, which means
4×100=400 and 9×1=9
Total will be 409
Draw boxes, sticks, and circles to show the number.
Question 4.
740
The value at hundreds place is 7, tens place is 4 and ones place is 0.
So, we need to draw 7 boxes, 4 sticks and 0 circles.
Question 5.
876
The value at hundreds place is 8, tens place is 7 and ones place is 6.
So, we need to draw 8 boxes, 7 sticks and 6 circles.
Question 6.
294
The value at hundreds place is 2, tens place is 9 and ones place is 4.
So, we need to draw 2 boxes, 9 sticks and 4 circles.
Question 7.
502
The value at hundreds place is 5, tens place is 0 and ones place is 2.
So, we need to draw 5 boxes, 0 sticks and 2 circles.
Expanded Form
Write the hundreds, tens, and ones.
Question 8.
Question 9.
738 = ____________ + ____________ + ______________
The value at hundreds place is 7. So, 7 ×100 = 700
The value at tens place is 3. So, 3 ×10 = 30
The value at ones place is 8. So, 8 ×1 = 8
738 = 700 + 30 + 8
Question 10.
526 = ____________ + ____________ + ______________
The value at hundreds place is 5. So, 5 ×100 = 500
The value at tens place is 2. So, 2 ×10 = 20
The value at ones place is 6. So, 6 ×1 = 6
526 = 500 + 20 + 6
Question 11.
267 = ____________ + ____________ + ______________
The value at hundreds place is 2. So, 2 ×100 = 200
The value at tens place is 6. So, 6 ×10 = 60
The value at ones place is 7. So, 7 ×1 = 7
267 = 200 + 60 + 7
Write the number.
Question 12.
400 + 50 + 9 =
Question 13.
800 + 10 + 3 = ______________
Hundred’s place value is 8, Tens place value is 1 and ones place value is 3.
800 + 10 + 3 = 813
Question 14.
100 + 70 + 5 = ______________
Hundred’s place value is 1, Tens place value is 7 and ones place value is 5.
100 + 70 + 5 = 175
Question 15.
600 + 40 + 1 = ______________
Hundred’s place value is 6, Tens place value is 4 and ones place value is 1.
600 + 40 + 1 = 641
Write the number that makes the equation true.
Question 16.
_____________ = 5 + 900 + 40
For this equation, the hundred’s value should be 9, tens value should be 4 and ones value should be 5.
945 = 5 + 900 + 40
Question 17.
7 + 200 = _____________
For this equation, the hundred’s value should be 2 and ones value should be 7.
7 + 200 = 207
Question 18.
_____________ = 400 + 6 + 80
For this equation, the hundred’s value should be 4, tens value should be 8 and ones value should be 6.
486 = 400 + 6 + 80
Question 19.
800 + 40 = ______________
For this equation, the hundred’s value should be 8 and tens value should be 4.
800 + 40 = 840
Question 20.
______________ = 70 + 300
For this equation, the hundred’s value should be 3 and tens value should be 7.
370 = 70 + 300
Question 21.
60 + 500 + 3 = ______________
For this equation, the hundred’s value should be 5, tens value should be 6 and ones value should be 3.
60 + 500 + 3 = 563
Question 22.
______________ = 2 + 400
For this equation, the hundred’s value should be 4 and ones value should be 2.
402 = 2 + 400
Question 23.
9 + 90 + 200 = ______________
For this equation, the hundred’s value should be 2, tens value should be 9 and ones value should be 9.
9 + 90 + 200 = 299
Question 24.
462 = 2 + 400 + ______________
For this equation, the hundred’s value should be 4, tens value should be 6 and ones value should be 2.
On right hand side, tens value is missing. So, 6×10=60
462 = 2 + 400 + 60
Question 25.
______________ + 90 + 700 = 798
For this equation, the hundred’s value should be 7, tens value should be 9 and ones value should be 8.
On right hand side, ones value is missing. So, 8×1=8
8 + 90 + 700 = 798
Question 26.
523 = 20 + 3 + ______________ |
# Question: How To Put Absolute Value In Geogebra?
## How do you graph absolute value?
To graph the absolute value, you need to know where the vertex is, so the numbers will be going down, then turn and start going back up, or going up and then turn and start going down, the middle point will be the vertex and should be the center point of the table.
## How do you create a function in geogebra?
In GeoGebra you can define a function on an interval by using the command Function ( < Function >, , ).
## What is the rule of absolute value?
In mathematics, the absolute value or modulus of a real number x, denoted |x|, is the non-negative value of x without regard to its sign. Namely, |x| = x if x is positive, and |x| = −x if x is negative (in which case −x is positive), and |0| = 0.
## How do you solve absolute value inequalities?
Here are the steps to follow when solving absolute value inequalities:
1. Isolate the absolute value expression on the left side of the inequality.
2. If the number on the other side of the inequality sign is negative, your equation either has no solution or all real numbers as solutions.
## How do you put absolute value in Excel?
Excel ABS Function
1. Summary.
2. Find the absolute value of a number.
3. A positive number.
4. = ABS (number)
5. number – The number to get the absolute value of.
6. Version.
7. For example, ABS (-3) returns a value of 3 and ABS (3) returns a value of 3, because the ABS function returns a number’s distance from zero.
You might be interested: FAQ: How To Get The Absolute Value Of A Fraction?
## How do you use geogebra in the classroom?
In this brief tutorial, you will experience how easy it is to use.
1. Step 1: Find an Activity.
2. Step 2: Create a Class.
3. Step 3: Students Join Class.
4. Step 4: Monitor Student Progress.
5. Distance Learning with GeoGebra Classroom.
6. Apps for GeoGebra Classroom.
7. Create a Class from a Book.
## How do you use the geogebra parabola?
How to construct a parabola
1. Drag the blue point M along the directrix to create points from the parabola.
2. Click the checkbox to see the construction.
3. Drag the Focus and the Directrix to change the parabola. |
Introduction
Do you struggle with expanding algebraic linear expressions during your Math examinations?
Keep on reading for a step-by-step guide to expanding linear expressions, along with some useful tips to make the topic of Algebra easier for you!
What I'll Be Sharing In This Article
## Let’s Solve Part (A)
First, let us copy the question.
💡 What Do The Brackets Mean? 💡
It means that you are supposed to multiply the number outside the brackets to everything inside the brackets
Therefore, we are going to multiply 2 to -4x and +3.
Some of you may know this method as the Distributive Law. Some may know this as the Rainbow Method because the arrows look like a rainbow 🌈
🌈 How To Remember The Distributive Law In Algebra 🌈
Think of a rainbow! 🌈
Do you panic just by looking at the algebraic expression above? Don’t worry because I am going to share a trick I have been teaching my students.
Look inside the brackets and ask yourself how many terms there are in the equation.
There are two terms inside the bracket. These are -4x and +3.
To help you visualise and differentiate the two terms, you can draw a line between the two.
Now, you can start multiplying by drawing arrows 🌈
The arrow 🌈 below signifies multiplication. So let us multiply 2 with -4x.
That is -8x.
Let us move on to the next term and do the same. Draw an arrow 🌈 and multiply 2 by +3.
The answer is +6.
Therefore, our final answer is -8x + 6.
-8x + 6
## Let’s Solve Part (B)
Again, the brackets mean we need to multiply 3x with all of the terms inside the brackets.
How many terms are inside the brackets? There are two! Let us draw a line to separate the two terms.
Next, draw your arrow 🌈 and multiply.
3xx
= 3x2
When you are multiplying x with x, you will get x2.
If this does not make sense to you, remember that when you are multiplying 11 by 11 in index notation, you will get 112. So the same is true when x is multiplied by x.
Next, let us multiply 3x with +5 🌈
We will get 15x.
Therefore, our final answer is 3x2 + 15x.
3x2 + 15x
## Let’s Solve Part (C)
Apply the same steps that we did in the previous equations. Let us determine the terms that we have inside the bracket.
We have 2x and -5.
Next, let us multiply 🌈 them with the number outside the bracket.
-7 x 2x
=-14x
For the second term, what is -7 x -5? 🌈
😲 What Happens When You Multiply Two Negative Numbers? 😲
A negative and a negative make a positive.
Therefore, it is +35.
Our final answer is -14x + 35.
-14x + 35
## Let’s Solve Part (D)
For some of you, this last algebraic equation might seem strange because there is only a minus sign and no number outside the bracket.
🚀 Remember This When You See A Negative Sign Outside The Bracket 🚀
Whenever you see algebraic equations like this, remember that there is a hidden 1 and the minus sign represents -1. You can write down 1 after the negative sign so you won’t get confused!
Now, we can proceed with the next steps. Draw a line between the two terms inside the bracket.
Next, multiply 🌈 the number outside the bracket with the first term.
-1 x 7x
= -7x
Next, draw an arrow 🌈 and multiply -1 with -1.
-1 x -1 = +1
Therefore, the answer is -7x + 1.
🤔 Remember This Shortcut When Multiplying Against A Negative Sign 🤔
The negative sign outside will flip all the signs inside the bracket.
Inside the brackets, I have a +7x and -1. When the brackets are removed, the minus sign will flip the signs of the terms in the bracket.
Therefore, it will be -7x + 1.
-7x + 1
## Conclusion
Looking at algebraic expressions can be overwhelming when you do not know where to start but I hope that this Algebra blog post has served as your guide in understanding how to expand linear expressions.
Remember to draw a line between the terms inside the bracket before multiplying each term so you won’t get confused.
Stay tuned for more blog posts about Algebra!
If you like our methodology, we've some ongoing weekly Math classes: |
# HOW TO FIND SLANT ASYMPTOTE OF A FUNCTION
## About the topic "How to find slant asymptote of a function"
"How to find slant asymptote of a function ?" is the question having had by the students who are studying math in school final. Even though it is taught by the teachers in school and university, students do not understand this clearly. Often students have this question on slant asymptotes.
On this page of our website, we have given step by step explanation and examples to make the students to clearly understand how to find slant asymptote of a function.
And we will be able to find slant asymptote of a function, only if it is a rational function.
That is, the function has to be in the form of f(x) = P/Q
## Steps involved in finding horizontal asymptotes
Let f(x) be the given rational function. Compare the highest exponent of the numerator and denominator.
Case 1 :
If the highest exponent of the numerator and denominator are equal, or if the highest exponent of the numerator is less than the highest exponent of the denominator, there is no slant asymptote.
Case 2 :
If the highest exponent of the numerator is greater than the highest exponent of the denominator by one, there is a slant asymptote.
To find slant asymptote, we have to use long division to divide the numerator by denominator. When we divide so, let the quotient be (ax+b).
Then, the equation of the slant asymptote is y = ax + b
## Examples:
1. Find the equation of horizontal asymptote for the function given below.
f(x) = 1/(x+6)
Solution :
Step 1:
In the given rational function, the highest exponent of the numerator is 0 and the highest exponent of the denominator is 1.
Step 2 :
Clearly highest exponent of the numerator is less than the highest exponent of the denominator.
Hence, there is no slant asymptote.
2. Find the equation of vertical asymptote for the function given below.
f(x) = (x²+2x-3)/(x²-5x+6)
Solution :
Step 1:
In the given rational function, the highest exponent of the numerator is 2 and the highest exponent of the denominator is 2.
Step 2 :
Clearly, the exponent of the numerator and the denominator are equal.
Hence, there is no slant asymptote.
3. Find the equation of vertical asymptote for the function given below.
f(x) = (x² +3x+2)/(x-2)
Solution :
Step 1:
In the given rational function, the highest exponent of the numerator is 2 and the highest exponent of the denominator is 1.
Step 2 :
Clearly, the exponent of the numerator is greater than the exponent of the denominator by one. So, there is a slant asymptote.
Step 3 :
To get the equation of the slant asymptote, we have to divide the numerator by the denominator using long division as given below.
Step 3 :
In the above long division, the quotient is (x+5).
Hence, the equation of the slant asymptote is y = x+5 |
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## Unit 4 Day 2 CED Topic(s): 2.1
##### Unit 4 Day 1 Day 2 Day 3 Day 4 Day 5 Day 6 Day 7 Day 8 Day 9 Day 10 Day 11 Day 12 Day 13 Day 14 Day 15All Units
###### ​Learning Targets​
• Write an explicit rule for geometric sequences using the common ratio and any term in the sequence.
• Apply understanding of how geometric sequences grow and knowledge of exponents and roots to determine the common ratio and find missing terms.
• Compare arithmetic and geometric sequences.
​
​
​
​
# Homework
###### Experience First:
Little Red Riding Hood is on her way to grandma’s house but something must be wrong with her pockets, because she keeps dropping crumbs of bread--and in a rather unusual pattern! In this lesson, students learn about (or review) geometric sequences by describing the number of crumbs Little Red drops on successive tiles. First, we want students to see the repeated multiplication so we have them write the number of crumbs using 2s and 3s only. Though some students may be ready for exponential reasoning right away, we find that writing it out long-hand reinforces important algebra concepts that students may forget along the way (when do I add exponents and when do I multiply??). This also leads students smoothly to question 4 where they have to write an explicit rule for the sequence. As you are monitoring students, listen for students that are able to articulate why there is an (n-1) in the exponent.
In question 5, we return to yesterday's idea that any term of a sequence can be an “anchor point” for determining other terms. Knowing the number of crumbs dropped on the 13th tile is sufficient for determining the number of crumbs dropped on the 15th tile because compared to the 13th tile, the number of crumbs on the 15th tile will be 3x3 or 9 times greater. This reasoning may seem obvious to us, and should feel intuitive to students, but we find that students often resort to using formulas and procedures that are actually less efficient, instead of making use of structure to solve a problem.
​
##### Monitoring Questions:
• What does the 2 represent in your table? What does the 3 represent?
• How is the number of crumbs changing? Is this an arithmetic sequence?
• If I only told you how many she dropped on the 13th tile and the fact that the number of crumbs triples, could you figure out how many she dropped on the 10th tile? The 1st tile? How would you do it?
• What’s the average of the numbers 2, 4, 6, 8, and 10? What’s the average of the numbers 2, 4, 8, 16, 32? Is the strategy for both the same? Why or why not?
​
###### Formalize Later:
The focus of the debrief is to look at patterns exhibiting repeated multiplication and to differentiate this from patterns demonstrating repeated addition. We also want students to generalize the explicit formula for the nth term of a geometric sequence and to realize there are multiple ways of doing this.
As mentioned in yesterday’s lesson, we use the idea of a partial sum to differentiate between growth in arithmetic and geometric sequences. While the terms of an arithmetic sequence are evenly distributed, the terms of a geometric sequence are not. For increasing geometric sequences, the increase in consecutive terms gets successively larger. For decreasing geometric sequences, the decrease in consecutive terms gets successively smaller. Note that we do not actually introduce or explore the partial sum formula for geometric sequences. This is one notable way AP Precalculus may differ from your regular Precalculus course.
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# Could 2 represent a probability?
Probabilities can be made up of all positive or complete numbers. Probabilities can be made up of numbers greater than one, while probabilities cannot be made up of numbers less than one. Anything that can be converted to a percent can represent a probability, according to the states.
## In terms of probability, what does P E mean?
In probability, P(E) refers to the probability of a likely event. Only when E is an impossible event does P(E) equal P(E) = 0?
## Is it possible to use 3 2 to represent a probability?
3/2 means 1.5, according to a step-by-step explanation.
## What number can be used to represent a probability?
We note that 0.333 is between 0 and 1, so 33.3% can be used to represent the probability of an event. cannot be used to estimate the likelihood of an event.
(c) Because 0.0002 is between 0 and 1, it can represent the probability of an event. (d) Because 0 is between 0 and 1 (inclusive), 0 can represent the probability of an event.
## Is it possible to use 1.5 to represent a probability?
Because it is more than 1, 1.5 cannot represent a probability. It means that the probability of any event is between 0 and 1.
As a result, 1.5>1 cannot represent a probability.
## Is it possible to use a number to represent the probability of an event?
The probability of an event can range from 0 to 1 and can also be expressed as a percentage. P (A) P(A) P(A), left parenthesis, A, right parenthesis, is frequently written as the probability of event A.
## Which one is unlikely to be the outcome of an event 2 3?
(A) 2/3 (B) 1.5 (C) 15% (D) 0.7 (A) 2/3 (B) 1.5 (C) 15% (D) 1.5 Solution: To answer the given question, we’ll use the basic probability concepts. As a result, because option (B) – 1.5 is negative, it cannot be the probability of an event.
## Which no can accurately represent a probability?
Probabilities cannot be represented by -1 and -0.5 because a probability cannot be negative. Because it is greater than one, 4.2 cannot represent a probability. Because they are between zero and one, inclusive, probabilities can be represented by 0.6, 0.888, and 0.39.
## What is the best way to represent the probability?
Probabilities can be represented in a ratio, percentage, fraction, or decimal form; I frequently remind students of this so that they are aware of the many different ways we represent odds. Basic mathematics, such as fractions, percentages, and ratios, is frequently accompanied by difficulties and fears.
## What is the same as a certain probability?
A certain event has a probability of occurring at least once. The probability of an event that cannot possibly occur is zero. If an event has a chance of happening, the probability ranges between zero and 1.
## What exactly is classical probability?
a method of understanding probability based on the assumption that any random process has a set of possible outcomes, and that each one is equally likely to occur.
## Which number is unable to account for a probability?
Explanation: This is due to the fact that probability cannot be greater than 1.
## What are the five probability rules?
Probability Rule One (for any event A, 0 P(A) 1) Probability Rule Two (the sum of all possible outcomes is 1) Probability Rule Three (the Complement Rule) Probabilities Involving Multiple Events Probability Rule Four (Addition Rule for Disjoint Events) Probability Rule Three (the Complement Rule) Probabilities Involving Multiple Events Probability Rule
## What are some examples of probability?
Weather Forecasting Examples in Real Life We always check the weather forecast before planning an outing or picnic. Cricket’s Batting Average, Politics, Flipping a Coin or Dice, Insurance, and the likelihood of dying in an accident
## Is it possible for a probability to be a whole number?
Probabilities can be made up of all positive or complete numbers. Probabilities can only be represented by numbers less than nine and fractions.
Because all of the numbers are between 0% and 100%, they can represent probabilities.
## What is a classic probability example?
Classical probability is a straightforward form of probability with equal chances of happening. Rolling a fair die, for example.
It’s also possible that you’ll get a 1, 2, 3, 4, 5 or 6.
## What are the three different types of probability?
Theoretical Probability, Experimental Probability, and Axiomatic Probability are the three major types of probabilities.
## What is the classical probability definition’s assumption?
The probability of an event is determined by the ratio of the number of favorable cases to the number of all possible cases, when nothing leads us to believe that any of these cases will occur more frequently than any other, making them equally possible for us.
## What are the four different types of probability?
Probability is a branch of mathematics that involves the occurrence of a random event, with four main types: classical, empirical, subjective, and axiomatic.
## What does a 100% chance of success imply?
The probability of an event having only one possible outcome is always 1 (or 100%). However, if there is more than one possible outcome, this changes.
Because it is certain that one of the possible outcomes will occur, the sum of all possible outcomes is always 1 (or 100%).
## What are the most basic probability concepts?
A probability is a number that represents the likelihood or chance of a particular event occurring. Probabilities can be expressed in proportions ranging from 0 to 1, as well as percentages ranging from 0% to 100%.
## What is probability’s application?
Probability provides information on the likelihood of something happening. Weather patterns, for example, are used by meteorologists to predict the likelihood of rain.
Probability theory is used in epidemiology to understand the relationship between exposures and health-related risks.
## How do you calculate normal probability?
If a random variable X has a pdf given by: f ( X) = 1 2 e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e |
# Section 1: Number Systems
## 1.0 Preliminaries -- Number Systems
Calculus deals with properties of the real numbers. In order to understand calculus you must first understand what it is about the real numbers that separates them from other kinds of numbers we use from day to day.
### 1.1 The Counting Numbers
These are the first numbers we learn. When we count, we start from one and list off the names of numbers in sequence. And this simple description clues us in on what the crucial properties of the counting numbers are: There is a first counting number, and for each counting number, there is a next counting number, or in other words, a successor. No counting number is its own successor. No counting number has more than one successor. No counting number is the successor of more than one other counting number. Only the number 1 is not the successor of any counting number.
And there is one more important property. That is, the counting numbers are like a ladder. If you know how to step onto the first rung of the ladder, and from any rung you know how to step to the next rung, then you can get to every rung. You probably remember learning about proof by induction back when you took algebra in high-school. This is the principle on which it is based. The way textbooks usually state it is that if you have a collection of counting numbers, and 1 is in that collection, and the successor of each counting number in the collection is also in the collection, then the collection contains all the counting numbers there are. But if you remember only the ladder analogy, you will still have the basic idea.
We learn early in life that we can always add two counting numbers to get another counting number. Then we learn about subtraction, and that the same is not always true for it. For example, I can't subtract 20 from 15 and get a counting number. Later we learn that we can always multiply two counting numbers and get another counting number. Then we learn about division, and that the same is not always true for it. For example I can't divide 3 into 10 and get another counting number.
So in these ways, regarding subtraction and division, the counting numbers seem somehow deficient.
### 1.2 The Integers
To remedy the situation with subtraction, we invented negative numbers and zero and tacked them onto the counting numbers starting just before 1. We called the result the integers. Addition, subtraction, multiplication, and division extend easily to be applicable to the expanded number system. Every number still has a successor, but no longer is there a first number. But you can certainly subtract any integer from any other integer and still have an integer as a result. And although it's hard to have -5 apples, anybody who has ever been on a shopping spree with a credit card knows very well how it is possible to have -1000 dollars.
### 1.3 The Rationals
To remedy the situation with division, we invented the rational numbers, which we learned are fractions that reside in vast multitudes between the integers. Again addition, subtraction, multiplication, and division extend easily to the further expanded system. We have to do without the idea of every number having a successor. But in return, we can divide any rational by any other rational (except zero) and get a rational as a result.
There is something else we get in return as well. From the beginning we have had a concept of one number being greater than another. Like the four operations, this concept started with the counting numbers and extended easily into the integers and the rationals. And with the concept of one number being greater than another comes the concept of between-ness. In the counting numbers and the integers, two numbers, one of which is the successor to the other, have no number in between them. But in the rationals there is an in-betweenster between any two distinct numbers. In fact there is a whole raft of them.
Closely related to this property is a property that mathematicians call density. It is that you can choose two distinct rationals to be as close as you would like them to be. With the integers, two numbers can't get any closer than one unit apart before they become the same. But at anywhere among the rationals, I can name how close I would like a pair of rationals to be and you will be able to find two such rationals.
### 1.4 And At the Heart of Calculus is ...
The density property described above leads to a concept that is fundamental to calculus, and that is the concept of a limit. If I gave you 2 gallons of milk today, one and a half gallons tomorrow, one and a third gallons the next day, one and a fourth gallons the next day, and so on, and continued in that manner for the rest of eternity, what can I say about how much milk you might get on a typical day? Well, there is certainly a formula for it. If I label the days, starting with today, as 1, 2, 3, 4, and so on, I can say that on day number n, I will give you 1 + 1/n gallons of milk. But there is something still deeper I can say about all this. I can say that there is a lower bound on how much milk you'll get on any particular day. That lower bound applies to all days starting with today.
If you tried to argue that 1.01 gallons is a lower bound, I could disprove it by noting that on the 101st day you would be getting less than that. But if you argued that any amount of one gallon or less is a lower bound, I would be unable to find any day in the future on which I would be giving you less than that amount of milk. So all amounts of one gallon or less are lower bounds. But of all of those, the amount of one gallon exactly is special. It is the greatest of all the lower bounds. And even though there is no day that I will give you exactly one gallon of milk, there will come a day on which the amount I give you will be as close to one gallon as I would like. Not only that, on all subsequent days the amount will be that close or closer.
In other words, you tell me how close to one gallon I will be giving you, and I can name the day on which the amount I deliver daily to you will be, forever after, at least that close. That is what makes one gallon the limit. And that concept of something coming as close as you'd like it and staying there -- the concept of a limit -- lies at the heart of calculus. Which is why you will be going over it so much in the first few weeks of any beginning calculus course, and why you should take the time and effort to understand it well.
### 1.5 The Real Numbers
With the rationals we have a nice little number system. You can combine any two rationals by any of four operations, and with the exception of dividing by zero, you will get another rational. What more could you want from a number system? Well nothing, actually, until you encounter the idea of a limit.
I know you've probably studied irrationals long before you were aware of limits. I know that in high-school you learned that the square root of two was irrational (perhaps the teacher even proved it for you, if you were lucky) and that you had to write it as an infinite, non-repeating decimal. But did any of you bother to ask what was meant by an infinite non-repeating decimal? If the decimal is infinite, you can't write it, now can you? So what on earth does it mean?
Well let's start out by asking what a finite decimal means. If I carry out a decimal to one place beyond the point, then I am specifying some number of tenths:
``` 4
1.4 = 1 + eq. 1.5-1
10
```
If I carry out a decimal to two places beyond the decimal point, then I am specifying some number of hundredths:
``` 41
1.41 = 1 + eq. 1.5-2
100
```
Similarly with three places beyond the decimal point:
``` 414
1.414 = 1 + eq. 1.5-3
1000
```
and so on. In general if a decimal is carried out to n places beyond the decimal point, we do a division using the digits to the right of the decimal place as the numerator and 10n as the denominator. So the meaning of finitely long decimals is quite clear and unambiguous. They all represent rationals.
But note that not all rationals can be expressed as finite decimals. The number 1/3 is quite clearly a rational, but can never be expressed as any counting number divided by a power of 10 (this is because no power of 10 can ever be divisible by 3). In high-school you were taught that you had to write this as an infinite repeating decimal. Again that nasty problem of human beings, having a finite life span, not being able to write infinite decimals, whether they repeat or not. Of course, with the repeating variety, you can write one period of a repeat sequence, then wave your arms and say, "and this continues in exactly this way forever." But even granting that that is a satisfactory way of writing the infinite pattern, you are still left with the problem of what does it mean. There simply are not n convenient decimal places, so you can't divide some counting number by 10n. The explanation we used of what a finite decimal means fails here miserably.
But enter now, the concept of a limit. As you probably recall, the decimal expansion of 1/3 is a decimal point followed by an endless string of 3's. But I can truncate the string at any point and end up with a finite decimal to which I can apply our unambiguous meaning. In fact, I can make a sequence beginning with truncating after 1 digit, then after 2, then after 3, and so on. And applying our rule for going from finite decimals to quotients, I get:
``` 3 33 333 3333
, , , , ... eq 1.5-4
10 100 1000 10000
```
This sequence has a limit according to the notion we discussed earlier with the milk. And its limit is exactly 1/3. You tell me how close a fraction you need to 1/3, and I can tell you how many terms you need to go into the sequence given in 1.5-4 to find nothing but terms that are at least that close. And this applies no matter how close you tell me I have to be. There is no rational number besides 1/3 that the above sequence comes arbitrarily close to and stays there. We say that the sequence converges to 1/3.
And by the way, it is not the only sequence of rationals that converges to 1/3 (although it is the only one that can be written as an infinite decimal). Here are two others that also converge to 1/3:
``` 4 34 334 3334
, , , , ... eq 1.5-5a
10 100 1000 10000
1 5 21 85 341
, , , , , ... eq 1.5-5b
4 16 64 256 1024
```
Note that in 1.5-5b the next denominator is obtained by multiplying the previous one by 4. The next numerator is obtained by multiplying the previous one by 4 and then adding 1.
It is easy to show that for every rational that cannot be written as a finite decimal, you can find an infinite repeating decimal that when interpreted as we now know to do, converges to that rational. And you can also show that no other infinite decimal, repeating or not, will converge to that same rational.
So what about the infinite decimals that don't repeat? What do they represent? Well we have just shown that they certainly don't converge to any rational. In addition, since the infinite non-repeating decimals are just as susceptible to the converging sequence interpretation as the infinite repeating decimals are, what can we say that their sequences converge to?
First, it is only fair to point out that we shouldn't have an expectation that all sequences converge to anything at all. For example:
``` 1 2 3 4 5
, , , , , ... eq. 1.5-6
10 10 10 10 10
```
where each term is 1/10 bigger than the term before it. We have no expectation that this sequence converges because it clearly has no upper bound. But the sequence 1.5-4 is different, as is any sequence that derives from an infinite decimal. They possess the special property in that you can tell me as tiny a range as you like, and I can tell you how far into the sequence you have to go so that all the subsequent terms are within that range of one another.
To illustrate this, let's say I am learning to pitch a baseball. I practice by throwing one pitch each minute at a barn door. Each pitch leaves a mark on the door. But as I get better at throwing, the marks the pitches leave begin to group together in a bunch. As I get better and better, the marks I leave group tighter and tighter. In fact, you can draw a circle around the area where my pitches appear to be grouping, and eventually every ball I throw hits inside the circle. If you draw a smaller circle, eventually I get good enough to place every ball inside that circle as well. And no matter how small you draw the circle, with enough practice, I can hit the inside of that circle every time. If every smaller circle is inside all the bigger ones, then the pattern of my pitching is analogous to the special property I am describing.
But the rationals seem somehow incomplete. Why? Because there are all those Cauchy sequences deriving from non-repeating infinite decimals that don't converge to any rational number. To remedy that, we invent the real numbers.
So what is a real number? Mathematicians have come up with several equivalent definitions, but since we have already discussed Cauchy sequences, I will use the definition that bases itself on them. A real number is the limit of a Cauchy sequence of rationals. It's that simple. It means that any time we have something that looks like it should converge to a limit, we say it does. And we always say it should if it meets the requirements that Cauchy gives us. Even if the sequence does not converge to any rational number, we still say it converges to a real number.
Since every infinite decimal represents a Cauchy sequence, whether it repeats or not, every infinite decimal represents a real number. Even the finite decimals represent real numbers, since you can append an infinite string of zeros onto them to make them infinite decimals.
To further clarify the difference between rationals and reals, let's suppose I drive down the road. Let's say my odometer is calibrated out to any number of digits beyond the decimal point. So I drive some number (less than 10) of tenths of a mile. Then I drive some number (again less than 10) of hundredths of a mile. Then I do the same in thousandths, ten thousandths, and so on. At each time, I have gone some number of miles that can be represented with some rational number. But, if the road has only points on it that are rational numbers of miles, then after I do this infinitely many times, I could end up nowhere. No rational number of miles could possibly indicate where I am on the road. There are rational numbers that are close to where I end up -- as close as I'd like. But none is completely accurate. Yet where I end up is a very real place. And that is why we adopt the real numbers in preference to the rationals for doing calculus.
The remainder of the discussion of real numbers in this section is optional. You may skip ahead to Section 2 if you like. But you may find this discussion interesting. You might also like to have a look at a deeper and more rigorous treatment of what the real numbers are all about. If so, click here to see an essay on the subject by Dr. Eric Schechter at Vanderbilt University.
Earlier, we showed two Cauchy sequences that converged to the same value. What gives? Well, sometimes two different Cauchy sequences can indicate the same real number. We have a test for that. Simply take the difference, term by term, of the two sequences. If the resulting sequence heads for a limit of zero, then the two Cauchy sequences represent the same real number. For example, if we take the sequences given by 1.5-4 and 1.5-5a:
``` 4 3 34 33 334 333 3334 3333
- , - , - , - , ... eq 1.5-7
10 10 100 100 1000 1000 10000 10000
```
Clearly, the nth term of this difference sequence is:
``` 1
10n
```
which you can bring as close to zero as you'd like by making n big enough. So that means that the two Cauchy sequences represent the same real number.
This, of course, leads to the question of how do you carry out the four arithmetic operations on real numbers? Well, you know how to add, subtract, multiply, and divide rationals. Do the same procedure, term by term, to the Cauchy sequences, and you are doing the operations on real numbers. The only problem is division, where the divisor sequence might contain zeros. If the divisor sequence ends in all zeros, then the divisor is zero, and you can't divide by it (the same is true if the divisor converges to a limit of zero). But if it has zeros but converges to something other than zero, simply drop the zero terms from the divisor sequence and do a term by term division on what remains.
This still leaves the nasty philosophical problem of humans not being able to carry out infinite operations. And, in fact, humans cannot do arithmetic on real numbers. Neither can computers. But we can talk about the reals and their properties based upon the definitions given above. And based upon what we find out about real numbers, we can do computations on them, not entirely accurately, but with as much accuracy as you have patience to get.
I hear you saying, "These real numbers are a real pain in the butt for something you can't even do arithmetic with. What's the point?" The point is that the reals have a property on which calculus depends. Calculus is about limits. Remember that some Cauchy sequences of rational numbers did not converge to any rational number. That means that, among rationals, limits do not exist everywhere you would like them to.
Clearly we can make Cauchy sequences of real numbers as well. That would be a sequence of sequences. Every Cauchy sequence of real numbers converges to a limit that is a real number. You can demonstrate the truth of this for yourself. Out of the sequence of sequences, take the first term of the first sequence, the second term of the second sequence, the third of the third, and so on. You end up with another Cauchy sequence of rational numbers (so it too is a real number), and it is precisely what the sequence of sequences converges to. [This is not the entire story on sequences of sequences. You can concoct one that requires a little more effort than the first of the first, the second of the second, and so on, in order to show convergence. Yet by going deep enough into each sequence, you can prove convergence of any Cauchy sequence of real numbers. If you want to see that proof, click here. But remember it is optional material and you won't need to know it for class. Furthermore it is not at all an easy proof for a beginner. So if you are curious about it, you might want to put it aside for now and come back to it when your skills have grown. And if you're not curious about it, don't feel the least bit guilty about skipping it.]
Since infinite decimals always represent Cauchy sequences, we can use them to fabricate an example of the above. We learned in algebra that both the square root of two and the value of pi are irrational, and must be represented using infinite decimals. Suppose I construct an infinite decimal that contains the first digit of the square root of two followed by the digits of π. That represents my first Cauchy sequence. I construct the next one by taking the first two digits of the square root of two, followed by the digits of π. The third is the first three digits of the square root of two followed by the digits of π, and so on:
``` s1 = 1.31415926... = 1, 1.3, 1.31, 1.314, 1.3141, 1.31415, 1.314159, ...
s2 = 1.43141592... = 1, 1.4, 1.43, 1.431, 1.4314, 1.43141, 1.431415, ...
s3 = 1.41314159... = 1, 1.4, 1.41, 1.413, 1.4131, 1.41314, 1.413141, ...
s4 = 1.41431415... = 1, 1.4, 1.41, 1.414, 1.4143, 1.41431, 1.414314, ...
s5 = 1.41423141... = 1, 1.4, 1.41, 1.414, 1.4142, 1.41423, 1.414231, ...
. . .
. . .
. . .
```
As we know, each of the sj's above represents a Cauchy sequence of rational numbers. So we have a sequence of sequences, that is, a sequence of real numbers. Not one of the individual sj's converges to the square root of two. But the sequence of sequences does. Why? Because each sequence converges to something closer to the square root of two than the last, and because you can find one that converges to something as close to the square root of two as you'd like by going down the rows deeply enough.
Using the prescription of taking the first term from the first sequence, the second term from the second sequence, the third from the third, and so on, we get a "diagonal" sequence, sdiag, of:
``` sdiag = 1, 1.4, 1.41, 1.414, 1.4142, 1.414213, ...
```
which is precisely the Cauchy sequence of rationals that is well known to converge to the square root of two. Of course an example is not a proof, but it does illustrate the idea.
The point is, a sequence of real numbers that meets Cauchy's conditions always has a limit and that limit is always a real number. That is a foundation on which we can do some calculus ...
You get a break. No exercises given for this section. But rest up and be prepared. I'll be working you hard on the next section to make up for it. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Proportions
Multiplication to solve for an unknown given two equal ratios.
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Proportions
Blake volunteered to be part of a "living flag" at the state fairgrounds on Independence Day. He had a small flag and wondered how the organizers would convert the dimensions to accommodate hundreds of people. Blake's flag measured 4 inches by 6 inches. If the length of the living flag was to be 120 feet, what should the width be?
In this concept, you will learn how to write and work with proportions.
Working with Proportions
A ratio represents a comparison between two quantities. Equivalent ratios are ratios that are equal. A proportion is made up of two equivalent ratios.
An example of a proportion is 12=24\begin{align*}\frac{1}{2} = \frac{2}{4}\end{align*}
12\begin{align*}\frac{1}{2}\end{align*}and 24\begin{align*}\frac{2}{4}\end{align*}are in proportion to one another. They are proportional.
When given three parts of a proportion, the fourth can be determined.
Here is an example.
12=x12\begin{align*}\frac{1}{2} = \frac{x}{12}\end{align*}
In this proportion, x is unknown.
Proportional reasoning, or examining the relationship between two numbers, can be used to determine the value of x.
First, since both denominators are given, begin with those two numbers.
2 and 12
Next, recognize the ratio that is given. In this problem it is 12\begin{align*}\frac{1}{2}\end{align*}. Think about what can be done to the denominator of the given ratio to make it equal to the denominator of the ratio with the unknown in it. In this case, what can be done to 2 to make it equal to 12.
2 x 6 = 12
2 can be multiplied times 6 to equal 12.
Then, multiply both the numerator and denominator of the given ratio times 6.
12=1×62×6=612=x12\begin{align*}\frac{1}{2} = \frac{1 \times 6}{2 \times 6} = \frac{6}{12} = \frac{x}{12}\end{align*}
12=612\begin{align*}\frac{1}{2} = \frac{6}{12}\end{align*}
The ratio 12\begin{align*}\frac{1}{2}\end{align*} is equal to the ratio 612\begin{align*}\frac{6}{12}\end{align*}
The two ratios are proportional to one another. They have the same proportions.
Examples
Example 1
Earlier, you were given a problem about Blake and the living flag.
He wondered how his little 4" by 6" parade flag could be used to calculate the width of a giant human flag that has a length of 120 feet.
First, write an equation for the proportion.
46=x120\begin{align*}\frac{4}{6}=\frac{x}{120}\end{align*}
Next, work with the denominators.
6 and 120
Then, using the given ratio of 46\begin{align*}\frac{4}{6}\end{align*}, determine what can be done to 6 to make it equal to 120.
6 x 20 = 120
Next, multiply both the numerator and denominator of the given ratio times 20.
4×206×20=80120\begin{align*}\frac{4\times 20}{6\times 20}=\frac{80}{120}\end{align*}
The answer is y = 80 feet.
46=80120\begin{align*}\frac{4}{6}=\frac{80}{120}\end{align*}
Example 2
Use proportional reasoning to solve for x.
1535=x7\begin{align*}\frac{15}{35}=\frac{x}{7}\end{align*}
35 and 7
Next, recognize the given ratio, 1535\begin{align*}\frac{15}{35}\end{align*}, and determine what can be done to 35 to make it equal to 7.
35÷5=7\begin{align*}35\div 5=7\end{align*}
Dividing 35 by 5 equals 7.
Then, divide both the numerator and denominator of the given ratio by 5.
15÷535÷5=37\begin{align*}\frac{15\div 5}{35\div 5}=\frac{3}{7}\end{align*}
The answer is x = 3
1535=37\begin{align*}\frac{15}{35}=\frac{3}{7}\end{align*}
Example 3
3236=x9\begin{align*}\frac{32}{36}=\frac{x}{9}\end{align*}
Solve for x:
First, begin with the given denominators.
36 and 9
Next, using the given ratio of 3236\begin{align*}\frac{32}{36}\end{align*}, determine what can be done to 36 to make it equal to 9.
36÷4=9\begin{align*}36\div 4=9\end{align*}
Then, divide both the numerator and denominator of the given ratio by 4.
32÷436÷4=89\begin{align*}\frac{32\div 4}{36\div 4}=\frac{8}{9}\end{align*}
The answer is x = 8
3236=89\begin{align*}\frac{32}{36}=\frac{8}{9}\end{align*}
Example 4
1224=4z\begin{align*}\frac{12}{24} = \frac{4}{z}\end{align*}
Solve for z.
First, begin with the numerators as both are given.
12 and 4
Next, determine what can be done to the numerator 12 in the given ratio of 1224\begin{align*}\frac{12}{24}\end{align*}to make it equal to 4 in the unknown ratio.
12÷3=4\begin{align*}12\div 3=4\end{align*}
Then, divide both sides of the given ratio by 3.
12÷324÷3=48\begin{align*}\frac{12\div 3}{24\div 3}=\frac{4}{8}\end{align*}
The answer is z = 8.
1224=48\begin{align*}\frac{12}{24}=\frac{4}{8}\end{align*}
Example 5
Solve for y.
y5=1220\begin{align*}\frac{y}{5} = \frac{12}{20}\end{align*}
First, work with the denominators.
5 and 20
Next, use the denominator, 20, in the given ratio of 1220\begin{align*}\frac{12}{20}\end{align*} and determine what can be done to 20 to make it equal to the denominator, 5, in the unknown ratio.
20÷4=5\begin{align*}20\div 4=5\end{align*}
Then, divide both sides of the given ratio by 4.
12÷420÷4=35\begin{align*}\frac{12\div 4}{20\div 4}=\frac{3}{5}\end{align*}
The answer is y = 3.
35=1220\begin{align*}\frac{3}{5}=\frac{12}{20}\end{align*}
Review
Tell whether or not each pair of ratios form a proportion.
1. 12 and 48\begin{align*}\frac{1}{2} \text{ and } \frac{4}{8}\end{align*}
2. 37 and 614\begin{align*}\frac{3}{7} \text{ and } \frac{6}{14}\end{align*}
3. 52 and 106\begin{align*}\frac{5}{2} \text{ and } \frac{10}{6}\end{align*}
4. 31 and 93\begin{align*}\frac{3}{1} \text{ and } \frac{9}{3}\end{align*}
5. 29 and 1.54.5\begin{align*}\frac{2}{9} \text{ and } \frac{1.5}{4.5}\end{align*}
6. 49 and 810\begin{align*}\frac{4}{9} \text{ and } \frac{8}{10}\end{align*}
7. 14 and 520\begin{align*}\frac{1}{4} \text{ and } \frac{5}{20}\end{align*}
8. 34 and 910\begin{align*}\frac{3}{4} \text{ and } \frac{9}{10}\end{align*}
Use proportional reasoning to find the value of the variable in each proportion.
1. 14=a20\begin{align*}\frac{1}{4} = \frac{a}{20}\end{align*}
2. 1530=x2\begin{align*}\frac{15}{30} = \frac{x}{2}\end{align*}
3. 29=n63\begin{align*}\frac{2}{9} = \frac{n}{63}\end{align*}
4. z7=1221\begin{align*}\frac{z}{7} = \frac{12}{21}\end{align*}
5. 35=t60\begin{align*}\frac{3}{5} = \frac{t}{60}\end{align*}
6. k72=512\begin{align*}\frac{k}{72} = \frac{5}{12}\end{align*}
7. x32=48\begin{align*}\frac{x}{32} = \frac{4}{8}\end{align*}
Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Vocabulary Language: English
Equivalent Ratios
Equivalent ratios are ratios that can each be simplified to the same ratio.
Proportion
A proportion is an equation that shows two equivalent ratios.
Proportional Reasoning
Proportional reasoning involves deducing the relationship between the numerators or the denominators of a proportion. Anytime you have a proportion, there is some kind of relationship between the values. |
# Algebra I Assignment - Exponents, Polynomials, Graphs & Geometry
Instructor: John Hamilton
John has tutored algebra and SAT Prep and has a B.A. degree with a major in psychology and a minor in mathematics from Christopher Newport University.
The ensuing year-end Algebra I homeschool assignment will challenge students on several math concepts, including how to factor and graph a quadratic equation. This mathematics assignment has been created for the 9th grade student. Updated: 02/27/2021
## Assignment Explanation and Topic Overview
When your students think of algebra, including exponents, polynomials, and graphs, the last thing they probably think about is geometry, right? Well, there do exist some subtle connections between the two disciplines.
In addition to what is mentioned above, this synoptic end-of-year assignment will cover topics learned throughout the two freshman Algebra I school semesters, including the distributive property along with the five laws of exponents too.
By the end of this assignment, your students will have completed four steps, solved three problems, and delivered two presentations.
Note - The answers are located at the bottom of the page.
## Key Terms
• Algebra: a branch of mathematics in which letters represent numbers
• Exponent: a quantity which represents a power to be raised
• Geometry: a branch of mathematics in which properties of figures are explained
### Materials
• Graph paper
• Online capability
• Paper
• Pencil
• Ruler
### Time /Length
• Two days to complete this Algebra I assignment
• Two weeks to design an appropriate presentation
## Assignment Instructions for Students
### Step One
Let's begin! Do you remember the five laws of exponents from earlier this school year? They are:
Product of Powers Rule:
• xa * xb = xa + b
Example:
• 25 * 26 = 2(5 + 6)
Quotient of Powers Rule:
• xa / xb = xa - b
Example:
• 38 / 35 = 3(8 - 5)
• Of course, x doesn't equal zero, because you can't divide by zero.
Power of a Power Property:
• (xa)b = x(a * b)
Example:
• (42)3 = 4(2 * 3)
Power of a Product:
• (x * y)a = xa * ya
Example:
• (5 * 6)4 = 54 * 64
Power of a Quotient:
• (x / y)a = xa / ya
Example:
• (6 / 7)5 = 65 / 75
### Step Two
#### Part I
Problem #1:
Next, let's review what we learned when you want to factor the following quadratic equation:
• x2 - 6x - 16
#### Part II
Problem #2:
Now do you recall how to graph this same quadratic equation?
• f(x) = x2 - 6x - 16
### Step Three
Okay, you remember quadratic equations, but do you remember how to solve those ever-so-dastardly cubic equations?
• x3 - 4x2 + x + 6 = 0
This is a bit tricky, but you can handle it. First, we apply the rational roots theorem or rational roots test. This means all our possible solutions are found by dividing all the factors of the constant (which is 6) by the leading coefficient (the number in front of x3, which in this case is understood to be 1), and include positives and negatives in your list.
• Factors of 6 = 1, 2, 3, 6
• Factor of 1 = 1
So our possible solutions are +/- 1, +/-2, +/-3, and +/- 6.
Now I have some bad news, but you may remember we don't have a cool, convenient formula for cubic equations like we do for those quadratic equations. Yep! You actually have to plug in each of these eight possible solutions to find your three solutions. On the other hand, 3 out of 8 aren't horrible odds, right?
• x3 - 4x2 + x + 6 = 0
• (1)3 - 4(1)2 + (1) + 6 = 0
• 1 - 4 + 1 + 6 = 0
• 4 = 0
Hey, it didn't work. Boo! Hiss! Hiss! Boo! Oh well, let's try again with -1.
• x3 - 4x2 + x + 6 = 0
• (-1)3 - 4(-1)2 +(-1) + 6 = 0
• -1 - 4 - 1 + 6 = 0
• -6 + 6 = 0
• 0 = 0
It worked! Since -1 is indeed a zero of the equation, we know (x + 1) is a factor of the equation. Now, to find the other two factors, go ahead and plug in the other 6 numbers. I'll give you 15 minutes.
Are you done?
You should know by this point the three roots, solutions, or zeros are:
• -1, 2, and 3
• (x + 1) (x - 2) (x - 3)
By the way, let's recount some of the properties of polynomial functions. First of all, their graphs are continuous. In other words, when drawing one you never lift your pencil from your paper. Second, their graphs are smooth. In other words, your graph will have no sharp corners, but have rounded curves instead.
''How does measurement in algebra relate to geometry though? I thought we already learned algebra and geometry were two completely separate mathematical fields.'' Well, not completely. When you graphed your above polynomial function, you were essentially converting an algebraic concept into a geometric concept.
In a reverse manner, we might be given a graph of a geometric circle, but we use our handy algebra formulas to solve for the perimeter and the area of that circle.
And of course, the Pythagorean theorem combines algebra and geometry concepts.
### Step Four
Now let's review the concept of algebraic distribution. You may remember this formula:
• a(b + c) = ab + ac
We usually read this expression as a times quantity b plus c equals a times b plus a times c.
Problem #3:
Distribute:
• 5(3x + 2y)
## Deliverable
Now it's time to demonstrate your grasp of all the Algebra I material you learned throughout our school year by completing two of the following five presentations:
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# 10 Intro to Functions
R comes with many functions and packages that let us perform a wide variety of tasks, and so far we’ve been using a number of them. In fact, most of the things we do in R is by calling some function. Sometimes, however, there is no function to do what we want to achieve. When this is the case, we may very well want to write our own functions.
In this chapter we’ll describe how to start writing small and simple functions. We are going to start covering the “tip of the iceberg”, and in the following chapters we will continue discussing more aspects about writing functions, and describing how R works when you invoke (call) a function.
## 10.1 Motivation
We’ve used the formula of future value, given below, which is useful to answer questions like: If you deposit $1000 into a savings account that pays an annual interest of 2%, how much will you have at the end of year 10? $\text{FV} = \text{PV} \times (1 + r)^n$ • $$\text{FV}$$ = future value (how much you’ll have) • $$\text{PV}$$ = present value (the initial deposit) • $$r$$ = rate of return (e.g. annual rate of return) • $$n$$ = number of periods (e.g. number of years) R has a large number of functions—e.g. sqrt(), log(), mean(), sd(), exp(), etc—but it does not have a built-in function to compute future value. Wouldn’t it be nice to have a future_value() function—or an fv() function—that you could call in R? Perhaps something like: future_value(present = 1000, rate = 0.02, year = 10) Let’s create such a function! ## 10.2 Writing a Simple Function This won’t always be the case, but in our current example we have a specific mathematical formula to work with (which makes things a lot easier): $\text{FV} = \text{PV} \times (1 + r)^n$ Like other programming languages that can be used for scientific computations, we can take advantage of the syntax in R to write an expression that is almost identical to the algebraic formulation: fv = pv * (1 + r)^n We will use this simple line of code as our starting point for creating a future value function. Here is how to do it “logically” step by step. #### Step 1: Start with a concrete example You should always start with a small and concrete example, focusing on writing code that does the job. For example, we could write the following lines: # inputs pv = 1000 r = 0.02 n = 10 # process fv = pv * (1 + r)^n # output fv [1] 1218.994 When I say “small example” I mean working with objects containing just a few values. Here, the objects pv, r, and n are super simple vectors of size 1. Sometimes, though, you may want to start with less simple—yet small—objects containing just a couple of values. That’s fine too. Sometimes you may even need to start not just with one, but with a couple of concrete examples that will help you get a better feeling of what kind of objects, and operations you need to use. As you get more experience creating and writing functions, you may want to start with a “medium-size” concrete example. Personally, I don’t tend to start like this. Instead, I like to take baby-steps, and I also like to take my time, without rushing the coding. You know the old-saying: “measure twice, cut once.” An important part of starting with a concrete example is so that you can identify what the inputs are, what computations or process the inputs will go through, and what the output should be. Inputs: • pv • r • n Process: • fv = pv * (1 + r)^n Output: • fv #### Step 2: Make your code more generalizable After having one (or a few) concrete example(s), the next step is to make your code more generalizable, or if you prefer, to make it more abstract (or at least less concrete). Instead of working with specific values pv, r, and n, you can give them a more algebraic spirit. For instance, the code below considers “open-ended” inputs without assigning them any values # general inputs (could take "any" values) pv r n # process fv = pv * (1 + r)^n # output fv Obviously this piece of code is very abstract and not intended to be executed in R; this is just for the sake of conceptual illustration. #### Step 3: Encapsulate the code into a function The next step is to encapsulate your code as a formal function in R. I will show you how to do this in two logical substeps, although keep in mind that in practice you will merge these two substeps into a single one. The encapsulation process involves placing the “inputs” inside the function function(), separating each input with a comma. Formally speaking, the inputs of your functions are known as the arguments of the function. Likewise, the lines of code that correspond to the “process” and “output” are what will become the body of the function. Typically, you encapsulate the code of the body by surrounding it with curly braces { } # encapsulating code into a function function(pv, r, n) { fv = pv * (1 + r)^n fv } The other substep typically consists of assigning a name to the code of your function. For example, you can give it the name FV: # future value function FV = function(pv, r, n) { fv = pv * (1 + r)^n fv } In summary: • the inputs go inside function(), separating each input with a comma • the processing step and the output are surrounded within curly braces { } • you typically assign a name to the code of your function #### Step 4: Test that the function works Once the function is created, you test it to make sure that everything works. Very likely you will test your function with the small and concrete example: # test it FV(1000, 0.02, 10) [1] 1218.994 And then you’ll keep testing your function with other less simple examples. In this case, because the code we are working with is based on vectors, and uses common functions for vectors, we can further inspect the behavior of the function by providing vectors of various sizes for all the arguments: # vectorized years FV(1000, 0.02, 1:5) [1] 1020.000 1040.400 1061.208 1082.432 1104.081 # vectorized rates FV(1000, seq(0.01, 0.02, by = 0.005), 1) [1] 1010 1015 1020 # vectorized present values FV(c(1000, 2000, 3000), 0.02, 1) [1] 1020 2040 3060 Notice that the function is vectorized, this is because we are using arithmetic operators (e.g. multiplication, subtraction, division) which are in turn vectorized. #### In Summary • To define a new function in R you use the function function(). • Usually, you specify a name for the function, and then assign function() to the chosen name. • You also need to define optional arguments (i.e. inputs of the function). • And of course, you must write the code (i.e. the body) so the function does something when you use it. ### 10.2.1 Arguments with default values Sometimes it’s a good idea to add a default value to one (or more) of the arguments. For example, we could give default values to the arguments in such a way that when the user executes the function without any input, FV() returns the value of 100 monetary units invested at a rate of return of 1% for 1 year: # future value function with default arguments FV = function(pv = 100, r = 0.01, n = 1) { fv = pv * (1 + r)^n fv } # default execution FV() [1] 101 An interesting side effect of giving default values to the arguments of a function is that you can also call it by specifying arguments in an order different from the order in which the function was created: FV(r = 0.02, n = 3, pv = 1000) [1] 1061.208 ## 10.3 Writing Functions for Humans When writing functions (or coding in general), you should write code not just for the computer, but also for humans. While it is true that R doesn’t care too much about what names and symbols you use, your code will be used by a human being: either you or someone else. Which means that a human will have to take a look at the code. Here are some options to make our code more human friendly. We can give the function a more descriptive name such as future_value(). Likewise, we can use more descriptive names for the arguments: e.g. present, rate, and years. # future value function future_value = function(present, rate, years) { future = present * (1 + rate)^years future } # test it future_value(present = 1000, rate = 0.02, years = 10) [1] 1218.994 Even better: whenever possible, as we just said, it’s a good idea to give default values to the arguments (i.e. inputs) of the function: # future value function future_value = function(present = 100, rate = 0.01, years = 1) { future = present * (1 + rate)^years future } future_value() [1] 101 ### 10.3.1 Naming Functions Since we just change the name of the function from fv() to future_value(), you should also learn about the rules for naming R functions. A function cannot have any name. For a name to be valid, two things must happen: • the first character must be a letter (either upper or lower case) or the dot . • besides the dot, the only other symbol allowed in a name is the underscore _ (as long as it’s not used as the first character) Following the above two principles, below are some valid names that could be used for the future value function: • fv() • fv1() • future_value() • future.value() • futureValue() • .fv(): a function that starts with a dot is a valid name, but the function will be a hidden function. In contrast, here are examples of invalid names: • 1fv(): cannot begin with a number • _fv(): cannot begin with an underscore • future-value(): cannot use hyphenated names • fv!(): cannot contain symbols other than the dot and the underscore (not in the 1st character) ### 10.3.2 Function’s Documentation Part of writing a human-friendly function involves writing its documentation, usually providing the following information: • title: short title • description: one or two sentences of what the function does • arguments: short description for each of the arguments • output: description of what the function returns Once you are happy with the status of your function, include comments for its documentation, for example: # title: future value function # description: computes future value using compounding interest # inputs: # - present: amount for present value # - rate: annual rate of return (in decimal) # - years: number of years # output: # - computed future value future_value = function(present = 100, rate = 0.01, years = 1) { future = present * (1 + rate)^years future } Writing documentation for a function seems like a waste of time and energy. Shouldn’t a function (with its arguments, body, and output) be self-descriptive? In an ideal world that would be the case, but this rarely happens in practice. Yes, it does take time to write these comments. And yes, you will be constantly asking yourself the same question: “Do I really need to document this function that I’m just planning to use today, and no one else will ever use?” Yes! I’ll be the first one to admit that I’ve created so many functions without writing their documentation. And almost always—sooner or later—I’ve ended up regretting my laziness for not including the documentation. So do yourself and others (especially your future self) a big favor by including some comments to document your functions. Enough about this chapter. Although, obviously, we are not done yet with functions. After all, this is a book about programming in R, and there is still a long way to cover about the basics and not so basics of functions. ## 10.4 Exercises 1) In the second part of the book we have talked about the Future Value (FV), and we have extensively used its simplest version of the FV formula. Let’s now consider the “opposite” value: the Present Value which is the current value of a future sum of money or stream of cash flows given a specified rate of return. Consider the simplest version of the formula to calculate the Present Value given by: $\text{PV} = \frac{\text{FV}}{(1 + r)^n}$ • $$\text{PV}$$ = present value (the initial deposit) • $$\text{FV}$$ = future value (how much you’ll have) • $$r$$ = rate of return (e.g. annual rate of return) • $$n$$ = number of periods (e.g. number of years) Write a function present_value() to compute the Present Value based on the above formula. Show answer present_value = function(future, rate, year) { present = future / (1 + rate)^year return(present) } 2) Write another function to compute the Future Value, but this time the output should be a list with two elements: • vector year from 0 to provided year • vector amount from amount at year 0, till amount at the provided year For example, something like this: fv_list(present = 1000, rate = 0.02, year = 3) $year
[1] 0 1 2 3
\$amount
[1] 1000.000 1020.000 1040.400 1061.208
fv_list <- function(present, rate, year) {
future = present * (1 + rate)^(0:year)
list(year = 0:year, amount = future)
}
3) Write another function to compute the Future Value, but this time the output should be a “table” with two columns: year and amount. For example, something like this:
fv_table(present = 1000, rate = 0.02, year = 3)
year amount
1 0 1000.000
2 1 1020.000
3 2 1040.400
4 3 1061.208
Note: by “table” you can use either a matrix or a data.frame. Even better, try to create two separate functions: 1) fv_matrix() that returns a matrix, and 2) fv_df() that returns a data frame.
fv_table <- function(present, rate, year) {
} |
# Lesson 14
Completing the Square (Part 3)
• Let’s complete the square for some more complicated expressions.
### Problem 1
Select all expressions that are perfect squares.
A:
$$9x^2 + 24x + 16$$
B:
$$2x^2 + 20x + 100$$
C:
$$(7 - 3x)^2$$
D:
$$(5x + 4)(5x - 4)$$
E:
$$(1 - 2x)(\text- 2x + 1)$$
F:
$$4x^2 + 6x + \frac94$$
### Problem 2
Find the missing number that makes the expression a perfect square. Next, write the expression in factored form.
1. $$49x^2 - \underline{\hspace{.5in}} x + 16$$
2. $$36x^2 + \underline{\hspace{.5in}} x + 4$$
3. $$4x^2 - \underline{\hspace{.5in}} x + 25$$
4. $$9x^2 + \underline{\hspace{.5in}} x + 9$$
5. $$121x^2 + \underline{\hspace{.5in}} x + 9$$
### Problem 3
Find the missing number that makes the expression a perfect square. Next, write the expression in factored form.
1. $$9x^2 + 42x + \underline{\hspace{.5in}}$$
2. $$49x^2 - 28x +\underline{\hspace{.5in}}$$
3. $$25x^2 + 110x + \underline{\hspace{.5in}}$$
4. $$64x^2 - 144x +\underline{\hspace{.5in}}$$
5. $$4x^2 + 24x + \underline{\hspace{.5in}}$$
### Problem 4
1. Find the value of $$c$$ to make the expression a perfect square. Then, write an equivalent expression in factored form.
standard form $$ax^2+bx+c$$ factored form $$(kx+m)^2$$
$$4x^2+4x$$
$$25x^2-30x$$
2. Solve each equation by completing the square.
$$4x^2+4x=3$$
$$25x^2-30x+8=0$$
### Problem 5
For each function $$f$$, decide if the equation $$f(x)=0$$ has 0, 1, or 2 solutions. Explain how you know.
(From Unit 7, Lesson 5.)
### Problem 6
Solve each equation.
$$p^2+10=7p$$
$$x^2+11x+27=3$$
$$(y+2)(y+6)=\text-3$$
(From Unit 7, Lesson 9.)
### Problem 7
Which function could represent the height in meters of an object thrown upwards from a height of 25 meters above the ground $$t$$ seconds after being launched?
A:
$$f(t)=\text-5t^2$$
B:
$$f(t)=\text-5t^2+25$$
C:
$$f(t)=\text-5t^2+25t+50$$
D:
$$f(t)=\text-5t^2+50t+25$$
(From Unit 6, Lesson 6.)
### Problem 8
A group of children are guessing the number of pebbles in a glass jar. The guesses and the guessing errors are plotted on a coordinate plane.
1. Which guess is furthest away from the actual number?
2. How far is the furthest guess away from the actual number?
(From Unit 4, Lesson 13.) |
# 2.05 Apply addition and subtraction
Lesson
Let's try a practice problem to remember how patterns can help us with addition and subtraction.
### Examples
#### Example 1
Fill in the boxes with the missing numbers.
a
14+5=⬚
Worked Solution
Create a strategy
On a number line, start with the bigger number and count to the right by the smaller number.
Apply the idea
Locate where 14 is on the number line, we can find 14 + 5 by jumping from 14 to the right by 5:
14 + 5 = 19
b
24+5=⬚
Worked Solution
Create a strategy
Compare this sum to the sum in part (a).
Apply the idea
24 is 10 more than 14. This means that 24+5 will be 10 more than 14+5=19.
So we can just add 10 to 19.
c
34+5=⬚
Worked Solution
Create a strategy
Compare this sum to the sum in part (b).
Apply the idea
34 is 10 more than 24. This means that 34+5 will be 10 more than 24+5=29.
So we can just add 10 to 29.
d
84+5=⬚
Worked Solution
Create a strategy
Use a number line.
Apply the idea
Locate where 84 is on the number line, we can find 84 + 5 by jumping from 84 to the right by 5:
84 + 5 = 89
e
What is the pattern? Choose the correct answer:
A
B
When adding 5 to a two-digit number that ends in 4, we get a two-digit number that ends in 9.
Worked Solution
Create a strategy
Use the results found from parts (a) to (d). Look for a pattern in the answers.
Apply the idea
In parts (a) - (d), the result was always bigger than the number we added 5 to, and each answers ended in a 9. So, the correct answer is Option B.
Idea summary
We can use number lines and patterns to add numbers together.
## Addition and subtraction using patterns
### Examples
#### Example 2
12-4=8. Use this to find:
a
22-4=⬚.
Worked Solution
Create a strategy
Compare this subtraction to the first subtraction we were given.
Apply the idea
22 is 10 more than 12. This means that 22-4 will be 10 more than 12-4=8. So we just need to add 10 to 8.
b
32-4=⬚.
Worked Solution
Create a strategy
Compare this subtraction to the subtraction from part (a).
Apply the idea
32 is 10 more than 22. This means that 32-4 will be 10 more than 22-4=18. So we just need to add 10 to 18.
c
42-4=⬚.
Worked Solution
Create a strategy
Compare this subtraction to the subtraction from part (b).
Apply the idea
42 is 10 more than 32. This means that 42-4 will be 10 more than 32-4=28. So we just need to add 10 to 28.
d
72-4=⬚.
Worked Solution
Create a strategy
Compare this subtraction to the subtraction from part (c).
Apply the idea
72 is 30 more than 42. This means that 72-4 will be 30 more than 42-4=38. So we just need to add 30 to 38.
Idea summary
To add or subtract using patterns, compare the calculation that you are trying to work out to another calculation that you already know. If you can see a pattern between the numbers, then you can use that to find the answer.
## Addition and subtraction using story problems
Sometimes our problem is in the form of a story, or sentences. In this video, we see how to work out our number problem from the key words, or clues.
### Examples
#### Example 3
Four friends are picking fruit in an orchard. Tricia picks 16 apples, Ryan picks 29 peaches, James picks 23 pears and Laura picks 15 plums.
How many pieces of fruit were picked in total?
Worked Solution
Create a strategy
Add the amount that each person picked together to get the total. Use the standard algorithm method.
Apply the idea
First, we can find 16+29 by writing it in a vertical algorithm. In the ones column we get 6+9=15. So we bring down 5 and carry 1 to the tens place.\begin{array}{c} & &\text{}^1 1&6\\ &+ & 2&9\\ \hline & &&5 \\ \hline \end{array}
In the tens place, we have 1+1+2=4:\begin{array}{c} & &\text{}^1 1&6\\ &+ & 2&9\\ \hline & &4&5 \\ \hline \end{array}
Next, we can find 45+23 by writing it in a vertical algorithm. In the ones column, we have 5+3=8:\begin{array}{c} & &4 &5\\ &+ & 2&3\\ \hline & &&8 \\ \hline \end{array}
In the tens place, we have 4+2=6:\begin{array}{c} & &4 &5\\ &+ & 2&3\\ \hline & &6&8 \\ \hline \end{array}
Lastly, we can find 68+15 by writing it in a vertical algorithm. In the ones column we get 8+5=13. So we bring down 3 and carry 1 to the tens place.\begin{array}{c} & &\text{}^1 6&8\\ &+ & 1&5\\ \hline & &&3 \\ \hline \end{array}
In the tens place, we have 1+6+1=8:\begin{array}{c} & &\text{}^1 6&8\\ &+ & 1&5\\ \hline & &8&3 \\ \hline \end{array}
83 pieces of fruit were picked in total.
Idea summary
There are some words we might come across that mean we will be using addition whereas other words may suggest subtraction. This is not a complete list, so we need to understand the context of the question, to work out whether to add or subtract.
### Outcomes
#### MA3-5NA
selects and applies appropriate strategies for addition and subtraction with counting numbers of any size
#### MA3-6NA
selects and applies appropriate strategies for multiplication and division, and applies the order of operations to calculations involving more than one operation
#### MA3-8NA
analyses and creates geometric and number patterns, constructs and completes number sentences, and locates points on the Cartesian plane |
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## Deriving the Area of a Circle
During the 1998-99 school year I was teaching math at Ellsworth High School in Maine. I got to talking with one of the other math teachers there about deriving the area of a circle. We started approaching the idea from a similar perspective as Archimedes had 2300 years ago in his work with circles and $\pi$. We inscribed a polygon in a circle and then took the limit of the area of the polygon as the number of sides ($n$) approached infinity. For the area of the polygon, we divided it into $n$ triangles and used the area formula $A=\frac {1}{2} ab\sin C$.
In this diagram, both $a$ and $b$ are radii of the circle and are labeled as $R$. The central angle $\theta$ which will be used to find the area of the triangle is actually equal to $\frac {360^\circ}{n}$ or (and this will be VERY important later) $\frac {2\pi}{n}$.
So the area of the polygon is $n$ times the area of the triangle, or:
$n*\frac {1}{2} R*R*\sin \frac{2 \pi}{n}$
For this to be the area of the circle we want to take the limit of the previous expression as $n \to \infty$.
$A=\lim_{n \to \infty} n*\frac {1}{2} R*R*\sin \frac {2 \pi}{n}$
Since the $\frac {1}{2}*R*R$ don’t involve $n$, we can pull them out from the limit.
$A=\frac {1}{2} R^2*\lim_{n \to \infty} n*\sin \frac {2 \pi}{n}$
Now comes the calculus – we were just talking about the derivative of the sine function in Calculus I last week, which is what made me think about this. Remember that:
$\lim_{x \to 0} \frac{\sin x}{x}=1$
So, we need to establish this limit somewhere in our derivation and then replace it with $1$. The most likely place for this would be in here:
$\lim_{n \to \infty} n*\sin \frac {2 \pi}{n}$
but we need to have:
$\lim_{\frac{2 \pi}{n} \to 0} \frac{\sin \frac {2 \pi}{n}}{\frac{2 \pi}{n}}$
OK – so, let’s multiply our origninal expression by $\frac{2 \pi}{2 \pi}$.
$\lim_{n \to \infty} \frac{2 \pi}{2 \pi}*n*\sin \frac {2 \pi}{n}$
or
$\lim_{n \to \infty} 2 \pi*\frac{n}{2 \pi}*\sin \frac {2 \pi}{n}$
which is equivalent to
$\lim_{n \to \infty} 2 \pi*\frac{\sin \frac {2 \pi}{n}}{\frac {2 \pi}{n}}$
Let’s bring back the whole expression:
$A=\frac {1}{2} R^2*\lim_{n \to \infty} n*\sin \frac {2 \pi}{n}$
$A=\frac {1}{2} R^2*\lim_{n \to \infty} 2 \pi*\frac{\sin \frac {2 \pi}{n}}{\frac {2 \pi}{n}}$
$A=\frac {1}{2} R^2*2 \pi \lim_{n \to \infty} \frac{\sin \frac {2 \pi}{n}}{\frac {2 \pi}{n}}$
as $n \to \infty$,
$\frac{2 \pi}{n} \to 0$
$A=\frac {1}{2} R^2*2 \pi \lim_{\frac{2 \pi}{n} \to 0} \frac{\sin \frac {2 \pi}{n}}{\frac {2 \pi}{n}}$
$A=\frac {1}{2} R^2*2 \pi *1$
or
$A=\pi R^2$
When we were working through this the first time, we kept the $360^\circ$ in the formula and ended up with $A=180^\circ R^2$, which didn’t make any sense until we realized that $180^\circ$ is $\pi$ radians!
Archimedes’ conclusion was that the area of the circle was equal to one-half the radius times the circumference:
$\frac{1}{2} R*2\pi R = \pi R^2$
or, since the mathematicians in the Greek tradition like Archimedes thought in terms of geometry – the area of a circle is the same as the area of triangle with height equal to the radius of the circle and base equal to the circumference of the circle!
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Construction of Tangent to a Circle
In Geometry, a tangent is a line that touches the curve exactly at a point. The point is called the point of tangency. The tangent to a circle is defined as the perpendicular to the radius at the point of tangency. In this article, we are going to discuss what is tangent to a circle, how to construct a tangent to a circle, and also we will learn how to draw a tangent from the point outside of the circle with a step by step procedure.
What is Tangent to a Circle?
Tangent of a circle is a line which touches the circle exactly at one point. The point at which tangent touches the circle is known as ‘point of contact’. The radius of the circle and tangent are perpendicular to each other at the point of contact.
We know that, there cannot be any tangent drawn to the circle through a point inside the circle.
There can be only one tangent to a point on the circle.
Construction of Tangents to a Circle
Let’s see how to draw a tangent to a circle at a point on the circle. (Refer fig.)
Step 1: Draw a circle with the required radius with centre O
Step 2: Join centre of the circle and any point P on the circle. OP is the radius of the circle
Step 3: Draw a line perpendicular to radius OP through point P. This line will be a tangent to the circle at P
Two tangents can be drawn to a circle from a point outside of the circle. Lengths of the two tangents will be equal.
Construction of Two Tangents from a Point Outside of the Circle
Step 1:Consider a point A from the outside the circle with centre O.
Step 2: Join points A and O, bisect the line AO. Let P be the midpoint of AO.
Step 3: Draw a circle taking P as centre and PO as a radius. This circle will intersect at two points B and C on the circle with centre O.
Step 4: Join the point A with B and CAB and AC are the required tangents through points B and C on the circle.
Why AB and AC are Tangents to the Circle with Centre O?
Join BO. It is observed that AO is a diameter of the circle with centre P. By construction, ABO is an angle in a semi-circle.
Therefore,
ABO = 90°
Since OB is the radius of the circle with centre OAB has to the tangent through the point B.
Similarly, AC is the tangent through the point C.
Note:
When the centre of the circle is not given to draw a tangent to the circle from a point outside the circle (Refer fig),
• Draw two non-parallel chords AB and CD.
• Draw perpendicular bisector for both AB and CD.
• The point O at which these bisectors intersect will be the centre of the circle because the radius is perpendicular bisector to chord.
• Now, follow the steps to draw tangents when the centre is given.
To know more about constructions, log onto www.byjus.com and keep learning. To watch interesting videos on the topic, download BYJU’S – The Learning App from Google Play Store. |
# How solve it? Topic: DERIVATE
## Please I need help I do not know how to solve
Jun 1, 2018
$f \left(x\right) = \sqrt{2 x}$
First, let's rewrite the square root as a $1 / 2$ power.
$f \left(x\right) = {\left(2 x\right)}^{\frac{1}{2}}$
Now, we need to recognize that these can be split up as a constant and a variable function.
$f \left(x\right) = {2}^{\frac{1}{2}} \cdot {x}^{\frac{1}{2}}$
When we differentiate, multiplicative constants like the ${2}^{\frac{1}{2}}$ here simply stay on the "outside," that is, we don't do anything to them.
To differentiate ${x}^{\frac{1}{2}}$, we use the power rule, which says that $\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$.
Then, we see that:
$f ' \left(x\right) = {2}^{\frac{1}{2}} \cdot \left(\frac{1}{2} {x}^{\frac{1}{2} - 1}\right)$
Now simplifying:
$f ' \left(x\right) = {2}^{\frac{1}{2}} / 2 {x}^{- \frac{1}{2}}$
$f ' \left(x\right) = \frac{1}{{2}^{\frac{1}{2}} \cdot {x}^{\frac{1}{2}}}$
$f ' \left(x\right) = \frac{1}{\sqrt{2 x}}$
So at $x = \frac{5}{3}$, the derivative is equal to:
$f ' \left(\frac{5}{3}\right) = \frac{1}{\sqrt{2 \cdot \frac{5}{3}}} = \frac{1}{\sqrt{\frac{10}{3}}} = \sqrt{\frac{3}{10}}$
Jun 1, 2018
$f ' \left(x\right) = \frac{2}{2 \sqrt{2 x}} = \frac{1}{\sqrt{2 x}}$
$f ' \left(\frac{5}{3}\right) = \frac{1}{\sqrt{\frac{10}{3}}} = \frac{\sqrt{3}}{\sqrt{10}}$
#### Explanation:
show below:
$f \left(x\right) = \sqrt{2 x}$
$f ' \left(x\right) = \frac{2}{2 \sqrt{2 x}} = \frac{1}{\sqrt{2 x}}$
The derivative at $x = \frac{5}{3}$ equal
$f ' \left(\frac{5}{3}\right) = \frac{1}{\sqrt{\frac{10}{3}}} = \frac{\sqrt{3}}{\sqrt{10}}$
$\text{Note that}$
$\textcolor{red}{y = \sqrt{x}}$
$\textcolor{red}{y ' = \frac{1}{2 \sqrt{x}} \cdot x '}$
$\textcolor{red}{\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}}$ |
During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.
# Difference between revisions of "2017 AMC 10B Problems/Problem 17"
The following problem is from both the 2017 AMC 12B #11 and 2017 AMC 10B #17, so both problems redirect to this page.
## Problem
Call a positive integer $\textbf{monotonous}$ if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$, $23578$, and $987620$ are monotonous, but $88$, $7434$, and $23557$ are not. How many monotonous positive integers are there?
$\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048$
## Solution 1
Case 1: monotonous numbers with digits in ascending order
There are $\sum_{n=1}^{9} \binom{9}{n}$ ways to choose n digits from the digits 1 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in ascending order. Note that 0 is not included since it will always be a leading digit and that is not allowed. Also, $\emptyset$ (the empty set) isn't included because it doesn't generate a number. The sum is equivalent to $\sum_{n=0}^{9} \binom{9}{n} -\binom{9}{0} = 2^9 - 1 = 511.$
Case 2: monotonous numbers with digits in descending order
There are $\sum_{n=1}^{10} \binom{10}{n}$ ways to choose n digits from the digits 0 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in descending order. Note that 0 is included since we are allowed to end numbers with zeros. However, $\emptyset$ (the empty set) still isn't included because it doesn't generate a number. The sum is equivalent to $\sum_{n=0}^{10} \binom{10}{n} -\binom{10}{0} = 2^{10} - 1 = 1023.$ We discard the number 0 since it is not positive. Thus there are $1022$ here.
Since the 1-digit numbers 1 to 9 satisfy both case 1 and case 2, we have overcounted by 9. Thus there are $511+1022-9=\boxed{\textbf{(B)} 1524}$ monotonous numbers.
## Solution 2
Like Solution 1, divide the problem into an increasing and decreasing case:
$\bullet$ Case 1: Monotonous numbers with digits in ascending order.
Arrange the digits 1 through 9 in increasing order, and exclude 0 because a positive integer cannot begin with 0.
To get a monotonous number, we can either include or exclude each of the remaining 9 digits, and there are $2^9 = 512$ ways to do this. However, we cannot exclude every digit at once, so we subtract 1 to get $512-1=511$ monotonous numbers for this case.
$\bullet$ Case 2: Monotonous numbers with digits in descending order.
This time, we arrange all 10 digits in decreasing order and repeat the process to find $2^{10} = 1024$ ways to include or exclude each digit. We cannot exclude every digit at once, and we cannot include only 0, so we subtract 2 to get $1024-2=1022$ monotonous numbers for this case.
At this point, we have counted all of the single-digit monotonous numbers twice, so we must subtract 9 from our total.
Thus our final answer is $511+1022-9 = \boxed{\textbf{(B) } 1524}$.
## Solution 3
Unlike the first two solutions, we can do our casework based off of whether or not the number contains a $0$.
If it does, then we know the $0$ must be the last digit in the number (and hence, the number has to be decreasing). Because $0$ is not positive, $0$ is not monotonous. So, we need to pick at least $1$ number in the set $[1, 9].$ After choosing our numbers, there will be just $1$ way to arrange them so that the overall number is monotonous.
In total, each of the $9$ digits can either be in the monotonous number or not, yielding $2^9 = 512$ total solutions. However, we said earlier that $0$ cannot be by itself so we have to subtract out the case in which we picked none of the numbers $1-9$. So, this case gives us $511$.
Onto the second case, if there are no $0$s, then the number can either be arranged in ascending order or in descending order. So, for each selection of the digits $1- 9$, there are $2$ solutions. This gives $$2 \cdot (2^9 - 1) = 2 \cdot 511 = 1022$$ possibilities. Note that we subtracted out the $1$ because we cannot choose none of the numbers.
However, realize that if we pick just $1$ digit, then there is only $1$ arrangement. We cannot put a single digit in both ascending and descending order. So, we must subtract out $9$ from there (because there are $9$ possible ways to select one number and for each case, we overcounted by $1$).
All in all, that gives $511 + 1022 - 9 = \boxed{\textbf{(B) } 1524}$ monotonous numbers.
## Solution 4
Let $n$ be the number of digits of a monotonous number. Notice for an increasing monotonous number with $n \ge 2$, we can obtain 2 more monotonous numbers that are decreasing by reversing its digits and adding a $0$ to the end of the reversed digits. Whenever $n$ digits are chosen, the order is fixed. There are $\binom{9}{n}$ ways to obtain an increasing monotonous number with $n$ digits. So, there are $3\cdot \sum_{n=2}^{9} \binom{9}{n}$ monotonous numbers when $n \ge 2$. When $n=1$, there is no reverse but we could add $0$ to the end, so there are $2 \cdot \binom{9}{1}$ monotonous numbers.
$3\cdot \sum_{n=2}^{9} \binom{9}{n} + 2 \cdot \binom{9}{1}$ $=3\cdot \sum_{n=1}^{9} \binom{9}{n} - \binom{9}{1}$ $=3\cdot \left( \sum_{n=0}^{9} \binom{9}{n} - \binom{9}{0} \right) - \binom{9}{1}$ $= 3 \cdot (2^9-1) - 9$ $=\boxed{\textbf{(B) } 1524}$ |
Refer to our Texas Go Math Grade 1 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 1 Lesson 3.2 Answer Key Add Tens and Ones.
Essential Question
How can you odd tens and ones?
Explanation:
Add tens place to tens place and ones place to ones
Explore
Choose a way to show the problem. Draw a quick picture to show your work.
For The Teacher
• Read the following problems. Barb has 20 pennies. Ed has 3 pennies. How many pennies do they have? Kyle has 40 pennies. Kim has 5 pennies. How many pennies do they have?
Math Talk
Mathematical processes
Model and Draw
How can you find 30 + 4?
___ tens ___ ones
30 + 4 = ____
Explanation:
3 tens 4 ones
30 + 4 = 34
Share and Show
Use  Write how many tens and ones. Draw to show tens and ones. Write the sum.
Question 1.
60 + 6 = ___
___ tens ___ ones
Explanation:
60 + 6 = 66
6 tens 6 ones
Question 2.
50 + 3 = ___
___ tens ___ ones
Explanation:
50 + 3 = ___
___ tens ___ ones
Question 3.
90 + 8 = ___
___ tens ___ ones
Explanation:
90 + 8 = 98
9 tens 8 ones
Question 4.
20 + 1 = ___
___ tens ___ ones
Explanation:
20 + 1 = 21
2 tens 1 ones
Problem Solving
Write how many tens and ones.
Draw to show tens and ones. Write the sum.
Question 5.
40 + 2 = ___
___ tens ___ ones
Explanation:
40 + 2 = 42
4 tens 2 ones
Question 6.
70 + 9 = ___
___ tens ___ ones
Explanation:
70 + 9 = 79
7 tens 9 ones
Draw to show tens and ones. Write the addition sentence.
Question 7.
Luke has 30 blocks. Elle has 6 blocks. How many blocks do they have?
___ + ___ = ___
__ blocks
Explanation:
30 + 6 = 36
36 blocks
Question 8.
H.O.T. Penny has 4 boxes of 10 movies. Nick has 8 movies. How many movies do they have?
Explanation:
Penny has 4 boxes of 10 movies.
4 x 10 = 40
Nick has 8 movies.
40 + 8 = 48
48 movies that they have
Question 9.
Multi-Step Ryan has 2 boxes of 10 games. Ian has 3 boxes of 10 games. Then Mindy gives Ryan 5 more games. How many games do they have now?
___ + ___ = ___
__ games
2 x 10 = 20
3 x 10 = 30
30 + 20 + 5 = 55
Explanation:
50 + 5 = 55
55 games
Question 10.
Lara planted two pepper plants. One grew 10 peppers. The other grew 6. Which number sentence shows how many peppers grew in all?
(A) 10 – 6 = 4
(B) 1 + 6 = 7
(C) 10 + 6 = 16
Explanation:
10 + 6 = 16
1 tens and 6 ones
Question 11.
Multi-Step Karl has 2 bags of 10 red marbles, 1 bag of 10 green marbles, and 8 blue marbles. How many marbles does Karl have?
(A) 31 marbles
(B) 38 marbles
(B) 28 marbles
Explanation:
2 x 10 = 20
1 x 10 = 10
20 + 10 + 8 = 38
Question 12.
Use Tools Which shows 30 + 5 = 35?
Explanation:
3 tens blocks and 5 ones blocks
Question 13.
Texas Test Prep What is the sum? 50 + 7 = ?
(A) 7
(B) 57
(C) 52
Explanation:
50 + 7 = 57
5 tens 7 ones
Take Home Activity
• Ask your child to explain how to find the sum of 20 + 7.
Explanation:
20 + 7 = 27
### Texas Go Math Grade 1 Lesson 3.2 Homework and Practice Answer Key
Write how many tens and ones. Draw to show tens and ones. Write the sum.
Question 1.
30 + 7 = ___
___ tens __ ones
3 tens 7 ones
Explanation:
Lines represent 10 and dot represent 1
Question 2.
90 + 3 = ___
___ tens __ ones
9 tens 3 ones
Explanation:
Lines represent 10 and dot represent 1
Problem Solving
Draw to show tens and ones. Write the addition sentence. Write the sum.
Question 3.
Multi-Step Jeff makes 4 stacks of 10 books. Max makes 4 stacks of 10 books. Then Josh gives Jeff 8 more books. How many books do they have?
___ + __ = ___
___ books
Explanation:
Jeff makes 4 stacks of 10 books = 4 x 10 = 40 books
Max makes 4 stacks of 10 books = 4 x 10 = 40 books
Josh gives Jeff 8 more books = 40 + 8 = 48 books
48+40 = 88 books total they have
Texas Test Prep
Question 4.
Mark drew two castles. One had 10 rooms. The other had 7 rooms. Which number sentence shows how many rooms he drew in all?
(A) 10 + 7 = 17
(B) 10 – 7 = 3
(C) 1 + 7 = 8
Explanation:
One had 10 rooms. The other had 7 rooms. 10 + 7 = 1 tens + 7 ones = 17
Question 5.
Which shows 60 + 2 = 62?
Explanation:
each line represents 10 and circle represents 1
60 + 2 = 6 Tens + 2 Ones= 62
Question 6.
What is the sum?
40 + 6 = ___
(A) 46
(B) 42
(C) 64
Explanation:
40 + 6 = 4 Tens + 6 Ones = 46
Question 7.
Multi-Step Tim has 1 box of 10 colored pencils. Sam has 4 boxes of 10 pencils. Then Joey gives Tim 3 more pencils. How many pencils do they have?
(A) 43
(B) 5
(C) 53 |
College Algebra with Corequisite Support
# Review Exercises
College Algebra with Corequisite SupportReview Exercises
## The Rectangular Coordinate Systems and Graphs
For the following exercises, find the x-intercept and the y-intercept without graphing.
1.
$4x−3y=12 4x−3y=12$
2.
$2y−4=3x 2y−4=3x$
For the following exercises, solve for y in terms of x, putting the equation in slope–intercept form.
3.
$5x=3y−12 5x=3y−12$
4.
$2x−5y=7 2x−5y=7$
For the following exercises, find the distance between the two points.
5.
$( −2,5 )( 4,−1 ) ( −2,5 )( 4,−1 )$
6.
$( −12,−3 )( −1,5 ) ( −12,−3 )( −1,5 )$
7.
Find the distance between the two points $(−71,432) (−71,432)$ and $(511,218) (511,218)$ using your calculator, and round your answer to the nearest thousandth.
For the following exercises, find the coordinates of the midpoint of the line segment that joins the two given points.
8.
$( −1,5 )( −1,5 )$ and $( 4,6 ) ( 4,6 )$
9.
$( −13,5 )( −13,5 )$ and $( 17,18 ) ( 17,18 )$
For the following exercises, construct a table and graph the equation by plotting at least three points.
10.
$y= 1 2 x+4 y= 1 2 x+4$
11.
$4x−3y=6 4x−3y=6$
## Linear Equations in One Variable
For the following exercises, solve for $x. x.$
12.
$5x+2=7x−8 5x+2=7x−8$
13.
$3(x+2)−10=x+4 3(x+2)−10=x+4$
14.
$7x−3=5 7x−3=5$
15.
$12−5(x+1)=2x−5 12−5(x+1)=2x−5$
16.
$2x 3 − 3 4 = x 6 + 21 4 2x 3 − 3 4 = x 6 + 21 4$
For the following exercises, solve for $x. x.$ State all x-values that are excluded from the solution set.
17.
$x x 2 −9 + 4 x+3 = 3 x 2 −9 x x 2 −9 + 4 x+3 = 3 x 2 −9$ $x≠3,−3 x≠3,−3$
18.
$1 2 + 2 x = 3 4 1 2 + 2 x = 3 4$
For the following exercises, find the equation of the line using the point-slope formula.
19.
Passes through these two points: $( −2,1 ),( 4,2 ). ( −2,1 ),( 4,2 ).$
20.
Passes through the point $( −3,4 ) ( −3,4 )$ and has a slope of $− 1 3 . − 1 3 .$
21.
Passes through the point $( −3,4 ) ( −3,4 )$ and is parallel to the graph $y= 2 3 x+5. y= 2 3 x+5.$
22.
Passes through these two points: $( 5,1 ),( 5,7 ). ( 5,1 ),( 5,7 ).$
## Models and Applications
For the following exercises, write and solve an equation to answer each question.
23.
The number of males in the classroom is five more than three times the number of females. If the total number of students is 73, how many of each gender are in the class?
24.
A man has 72 ft. of fencing to put around a rectangular garden. If the length is 3 times the width, find the dimensions of his garden.
25.
A truck rental is $25 plus$.30/mi. Find out how many miles Ken traveled if his bill was \$50.20.
## Complex Numbers
For the following exercises, use the quadratic equation to solve.
26.
$x 2 −5x+9=0 x 2 −5x+9=0$
27.
$2 x 2 +3x+7=0 2 x 2 +3x+7=0$
For the following exercises, name the horizontal component and the vertical component.
28.
$4−3i 4−3i$
29.
$−2−i −2−i$
For the following exercises, perform the operations indicated.
30.
$( 9−i )−( 4−7i ) ( 9−i )−( 4−7i )$
31.
$( 2+3i )−( −5−8i ) ( 2+3i )−( −5−8i )$
32.
$2 −75 +3 25 2 −75 +3 25$
33.
$−16 +4 −9 −16 +4 −9$
34.
$−6i(i−5) −6i(i−5)$
35.
$(3−5i) 2 (3−5i) 2$
36.
$−4 · −12 −4 · −12$
37.
$−2 ( −8 − 5 ) −2 ( −8 − 5 )$
38.
$2 5−3i 2 5−3i$
39.
$3+7i i 3+7i i$
For the following exercises, solve the quadratic equation by factoring.
40.
$2 x 2 −7x−4=0 2 x 2 −7x−4=0$
41.
$3 x 2 +18x+15=0 3 x 2 +18x+15=0$
42.
$25 x 2 −9=0 25 x 2 −9=0$
43.
$7 x 2 −9x=0 7 x 2 −9x=0$
For the following exercises, solve the quadratic equation by using the square-root property.
44.
$x 2 =49 x 2 =49$
45.
$( x−4 ) 2 =36 ( x−4 ) 2 =36$
For the following exercises, solve the quadratic equation by completing the square.
46.
$x 2 +8x−5=0 x 2 +8x−5=0$
47.
$4 x 2 +2x−1=0 4 x 2 +2x−1=0$
For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No real solution.
48.
$2 x 2 −5x+1=0 2 x 2 −5x+1=0$
49.
$15 x 2 −x−2=0 15 x 2 −x−2=0$
For the following exercises, solve the quadratic equation by the method of your choice.
50.
$(x−2) 2 =16 (x−2) 2 =16$
51.
$x 2 =10x+3 x 2 =10x+3$
## Other Types of Equations
For the following exercises, solve the equations.
52.
$x 3 2 =27 x 3 2 =27$
53.
$x 1 2 −4 x 1 4 =0 x 1 2 −4 x 1 4 =0$
54.
$4 x 3 +8 x 2 −9x−18=0 4 x 3 +8 x 2 −9x−18=0$
55.
$3 x 5 −6 x 3 =0 3 x 5 −6 x 3 =0$
56.
$x+9 =x−3 x+9 =x−3$
57.
$3x+7 + x+2 =1 3x+7 + x+2 =1$
58.
$| 3x−7 |=5 | 3x−7 |=5$
59.
$| 2x+3 |−5=9 | 2x+3 |−5=9$
## Linear Inequalities and Absolute Value Inequalities
For the following exercises, solve the inequality. Write your final answer in interval notation.
60.
$5x−8≤12 5x−8≤12$
61.
$−2x+5>x−7 −2x+5>x−7$
62.
$x−1 3 + x+2 5 ≤ 3 5 x−1 3 + x+2 5 ≤ 3 5$
63.
$| 3x+2 |+1≤9 | 3x+2 |+1≤9$
64.
$| 5x−1 |>14 | 5x−1 |>14$
65.
$| x−3 |<−4 | x−3 |<−4$
For the following exercises, solve the compound inequality. Write your answer in interval notation.
66.
$−4<3x+2≤18 −4<3x+2≤18$
67.
$3y<1−2y<5+y 3y<1−2y<5+y$
For the following exercises, graph as described.
68.
Graph the absolute value function and graph the constant function. Observe the points of intersection and shade the x-axis representing the solution set to the inequality. Show your graph and write your final answer in interval notation.
$| x+3 |≥5 | x+3 |≥5$
69.
Graph both straight lines (left-hand side being y1 and right-hand side being y2) on the same axes. Find the point of intersection and solve the inequality by observing where it is true comparing the y-values of the lines. See the interval where the inequality is true.
$x+3<3x−4 x+3<3x−4$ |
Inverse Functions – Ultimate Guide of Definition and Graphs
Domain and Range of Inverse Functions
\Large \begin{align} + \leftarrow &\text{ inverse operation } \rightarrow – \\ \times \leftarrow &\text{ inverse operation } \rightarrow \div \\ x^2 \leftarrow &\text{ inverse operation } \rightarrow \sqrt{x} \end{align}
The function $y=4x-1$ can be undone by its inverse function $y=\dfrac{x+1}{4}$.
We can consider this act as two processes or machines. If the machines are inverses, the second machine undoes what the first machine does. No matter what value of $x$ enters the first machine, it is returned as the output from the second machine.
If $(x,y)$ lies on $f$, then $(y,x)$ lines on $f^{-1}$. Reflecting the function in the line $y=x$ has the algebraic effect of interchanging $x$ and $y$.
For instance, $f:y=4x-1$ becomes $f^{-1}:x=4y-1$.
\large \begin{align} \text{The domain of } f^{-1} &= \text{ the range of }f \\ \text{The range of } f^{-1} &= \text{ the domain of }f \end{align}
Graphs of Inverse Functions
$y=f^{-1}(x)$ is the inverse of $y=f(x)$ as:
• it is also a function
• it is the reflection of $y=f(x)$ in the line $y=x$
The parabola shown in red below reflects $y=f(x)$ in $y=x$, but it is not the inverse function of $y=f(x)$ as it fails the vertical line test. In this case, the function $y=f(x)$ does not have an inverse.
Now consider the same function $y=f(x)$ but with the restricted domain $x \ge 1$.
The function does now have an inverse function, as shown below.
The reciprocal funciton $f(x)=\dfrac{1}{x},x \ne 0$, is said to be a self-inverse function as $f=f^{-1}$.
This is because the graph of $y=\dfrac{1}{x}$ is symmetrical about the line $y=x$.
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# How do you graph and solve 3|x-1|+2>=8?
Dec 15, 2015
$\left(- \infty , - 1\right] \text{ U } \left[3 , \infty\right)$
#### Explanation:
$| x | \ge k \implies \text{ " x>= k " " or " " } x \le - k$
$| x | \le k \implies \text{ } k \le x \le k$
Given:
$3 | x - 1 | + 2 \ge 8$
Isolate the inequality, so absolute value can be by itself.
Step 1: Subtract 2 to both side
$3 | x - 1 | + \cancel{2 \textcolor{red}{- 2}} \ge 8 \textcolor{red}{- 2}$
$3 | x - 1 | \ge 6$
Step 2 : Divide by 3 to both side
$\frac{3 | x - 1 |}{\textcolor{b l u e}{3}} \ge \frac{6}{\textcolor{b l u e}{3}}$
$| x - 1 | \ge 2$
Step 3 : Undo the absolute value like the information mention at the beginning
$\implies x - 1 \ge 2 \text{ " " " or " " " } x - 1 \le - 2$
Step 4 : Solve for $x$
$x \ge 3 \text{ " " or " " } x \le - 1$
Step 5: Draw the number line and determine the interval from Step 4.
Solution: $\left(- \infty , - 1\right] \text{ U } \left[3 , \infty\right)$ |
# Get the Knowledge that sets you free...Science and Math for K8 to K12 students
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## Probability
A pair of dice A 'die' or dice [the plural form of die/dice is also dice] is generally made of a six faced cubic structure which can be numbered from 1 to 6 on each face. A pair of dice can have a maximum of 36 combinations or outcomes.
Probability of an event
Basics of Probability: Till now, we learned about exploratory analysis of data, collection of data and now we begin the transition to inference. In order to do inference, we need to use the language of probability. In order to use the language of probability, we need an understanding of random variables. In this chapter, we learn about the basic rules of probability, what it means for events to be independent, discrete and continuous random variables, simulation and rules for combining random variables.
Probability is a measure of the likeliness that an event will occur. It is used to quantify an attitude of mind towards some proposition of whose truth we are not certain. The proposition of interest is usually of the form "Will a specific event occur?" The attitude of mind is of the form "How certain are we that the event will occur?" The certainty we adopt can be described in terms of a numerical measure and this number, between 0 and 1 (where 0 indicates impossibility and 1 indicates certainty), we call probability. Thus the higher the probability of an event, the more certain we are that the event will occur. A simple example would be the toss of a fair coin. Since the 2 outcomes are deemed equiprobable, the probability of "heads" equals the probability of "tails" and each probability is 1/2 or equivalently a 50% chance of either "heads" or "tails".
Experiment (Random phenomenon): An experiment [i.e., random phenomenon] is an activity in which we know what outcomes could happen, but we don't know which particular outcome will happen.
• Example: If we toss a coin, we know that we will get head (or) tail, but we don't know which particular outcome will happen. Similarly, if we roll a six sided die, we know that we will get a 1, 2, 3, 4, 5 or 6, but we don't know which particular outcome will happen.
Trial and Outcome: A single attempt (or) realization of an experiment is known as trial and one of the possible results of an experiment is known as an outcome. For example: the possible outcomes for tossing a single coin is: head (or) tail; the possible outcomes for rolling a six-sided die is: 1, 2, 3, 4, 5 (or) 6.
Sample space: The set of all possible outcomes of an experiment is known as sample space. It is denoted by the capital letter 'S'. For example: sample space for the roll of a single die, S = {1, 2, 3, 4, 5, 6}; sample space for tossing a single coin, S = {H, T}.
• Example: Find the sample space when two six sided dice are rolled simultaneously.
Solution: When two dice are rolled simultaneously, there are total 36 possible outcomes. Therefore, sample space (S) = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.
Event: An event is a collection of outcomes of an experiment, that is, an event is the subset of the sample space. For example: sample space for the roll of a single die, S = {1, 2, 3, 4, 5, 6}. Let event P = "face value of the die is an even number". Then P = {2, 4, 6}. Let event Q = "face value of the die is an odd number". Then Q = {1, 3, 5}. Events P and Q are subsets of the sample space.
Occurrence of events: In a random experiment, if E is the event of a sample space S and w is the outcome, then we say that the event E has occurred if w E. If w E, then we say that the event E has not occurred.
Important Types of Events
Simple event: Any event which consists of only a single outcome in the sample space is called simple event. It can also be known as elementary event. For example: in tossing a single coin experiment, obtaining head (or tail) is an elementary event because it consists of only a single outcome in the sample space, that is, {H} or {T}.
Compound event: Any event which consists of more than one outcome in the sample space is called compound event. It can also be known as mixed event. For example: in rolling a six-sided die experiment, obtaining an even number is a compound event because it consists of three outcomes in the sample space, that is, {2, 4, 6}.
Sure (or Certain) event: In a random experiment, an event will be called a sure event or a certain event if it always occurs whenever an experiment is performed. In other words, let S be a sample space and if E S, then E is an event called sure event.
For example, if we roll a six sided die, the possible outcomes of the sample space will be 1, 2, 3, 4, 5, 6. Now we consider an event of "Getting an even or odd number" in any particular event. We find that this event is represented with the following elements in the set 1, 2, 3, 4, 5, 6 which is exactly same as the sample space. So it is called sure event.
Equally likely events: Events are said to be equally likely, if we have no reason to believe that one is more likely to occur than the other. For example: when a six sided die is thrown, all the six-faces {1, 2, 3, 4, 5, 6} are equally likely to come-up.
Impossible event: In a random experiment, let S be a sample space and if S, then is an event called an impossible event. It is also known as null event. For example: in rolling a six-sided die experiment, obtaining the face value of the die above 6 or below 1 is an impossible event because sample space = {1, 2, 3, 4,5, 6}.
Mutually exclusive (or Disjoint) events: Two or more events are said to be mutually exclusive events if and only if they have no outcomes in common. For example: when a coin is tossed, the event of occurrence of a head and the event of occurrence of a tail are mutually exclusive events.
Complementary event: In a random experiment, let S be the sample space and E be an event. The complement of an event E with respect to S is the set of all the elements of S which are not in E. The complement of E is denoted by E' or Ec.
Probability
Probability of an event: It is defined as the relative frequency of an outcome. That is, it is the fraction of time that the outcome would occur if the experiment were repeated indefinitely. For any event A, probability is defined as: , where s = number of ways an outcome can succeed, f = number of ways an outcome can fail and (s + f) is the total number of outcomes in the sample space.
The probability of any event 'A' ranges from 0 to 1, inclusive. That is, 0 ≤ P(A) ≤ 1. This is an algebraic result from the definition of probability when success is guaranteed (f = 0, s = 1) or failure is guaranteed (f = 1, s = 0). The sum of the probabilities of all possible outcomes in a sample space is equal to one. That is, if the sample space is composed of 'n' possible outcomes, then .
• Example: There are 4 white marbles and 4 blue marbles in a bag. If a marble is drawn from a bag, what is the probability that it is a white color marble?
• Solution: In the experiment of drawing a marble, let the event E = obtain a white color marble. The sample space contains 8 marbles. Here, s = 4 because there are 4 ways for our outcome to be considered a success and f = 4. Therefore, required probability is:
P(E) = .
Probabilities of combined events:
• P(A or B): The probability that either an event A (or) an event B occurs. Using set notation, this can be written as P(A B). A B is spoken as: A union B.
• P(A and B): The probability that both an event A and an event B occur. Using set notation, this can be written as P(A B). A B is spoken as: A intersection B.
• Complement of an event A: Set of all the outcomes in the sample space that are not included in the outcomes of event A is known as complement of an event A. The complement of an event A can be represented by either Ac (or) A and it is read as "A complement (or) not A". The probability of complement of an event A is equal to the '1' minus the probability of an event A P(A) = 1 – P(A).
Probability of mutually exclusive events: Two or more events are said to be mutually exclusive events if and only if they have no outcomes in common. If A and B are two mutually exclusive events, then A B = Φ P(A B) = 0.
• Example: In the two-dice rolling experiment, A = "face shows a 2" and B = "sum of the two dice is 9" are mutually exclusive because there is no way to get a sum of 9 if one die shows a 2. That is, events A and B cannot both occur.
Conditional probability: The probability of an event occurring given that another event has already occurred is known as conditional probability.
• P(A given B): The probability of an event A occurring given that an event B has already occurred is known as conditional probability of 'A given B' and it is represented by . The formula for calculating the P(A given B) is: .
• P(B given A): The probability of an event B occurring given that an event A has already occurred is known as conditional probability of 'B given A' and it is represented by . The formula for calculating the P(B given A) is: .
Some conditional probability problems can be solved by using a tree diagram. A tree diagram is a schematic way of looking at all possible outcomes.
Addition rule: According to this rule, probabilities of all the related events are added together to compute the probability of their joint occurrence. If A and B are any two events associated with a random experiment, then addition rule for A and B is given as: P(A or B) = P(A) + P(B) – P(A and B).
Special case of the addition rule: If A and B are two mutually exclusive events, then addition rule for A and B is given as: P(A or B) = P(A) + P(B). |
# Interactive video lesson plan for: Solving two step equations-6th Grade Math
#### Activity overview:
Geometry Teachers Never Spend Time Trying to Find Materials for Your Lessons Again!
http://geometrycoach.com/Geometry-Lesson-Plans/?pa=MOOMOOMATH
Learn to solve two step equations.
Solving two step equations.
When solving a two-step equation you will need to perform two steps in order to solve the equation.
When solving a two -step equation you want only variables on one side of the equal sign and numbers on the other side of the equal sign.
In addition the number in front of the variable needs to be equal to one.
Remember when solving a linear equation, whatever you do to one side of the equation, you must do to the other side of the equation.
Let's look at some example problems
Let's solve these two step equations.
Step 1 is that we want to go ahead and get the variable isolated.
We first we will eliminate the 3 and we undo the addition with subtraction.
What we do to one side we must perform to the other side.
The 3's cancel and we are left with 6 and v over 11 on the other side.
Now we need to undo the division with multiplication, so multiply each side by 11 and the 11 cancels and that leaves us with v = 66
Next we have 7p + 2 = 23.
We first need to get the p by itself, or at least a 1 in front of the p
We can do that by undoing the addition with subtraction.
We will subtract 2 from both sides and we cancel and 23 minus 2 = 21
Bring 7p down.
In order to get rid of the multiplication we will need to divide both sides by 7
7 divided by 7 = 1, and 21 divided by 7 = 3
So the final answer is p = 3
Finally, our last one is similar to the one we just completed and that is 5m+5=75
We will undo the subtraction by subtracting 5 from each side.
75 - 5 = 70
Bring the 5m down
We will get rid of the multiplication with division.
We are left with m, actually there is a 1 in front of the m
70 divided by 5 = 14
There we go
Hope this helps with two step equations.
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• We’ll reuse our example matrix to explore $$Ax=b$$:
$\begin{bmatrix} \begin{array}{rrrr} 1 & 2 & 2 & 2 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 8 & 10 \\ \end{array} \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$
• Let’s use an augmented matrix to deal with the right hand side and do elimination:
$\begin{bmatrix} \begin{array}{rrrrr} 1 & 2 & 2 & 2 & b_1 \\ 2 & 4 & 6 & 8 & b_2 \\ 3 & 6 & 8 & 10 & b_3 \\ \end{array} \end{bmatrix}$
• End up with, the condition for a solution is $$0 = -b_1 - b_2 + b_3$$ (e.g. $$b = (1, 5, 6)$$):
$\begin{bmatrix} \begin{array}{rrrrr} 1 & 2 & 2 & 2 & b_1 \\ 0 & 0 & 2 & 4 & -2b_1 + b_2 \\ 0 & 0 & 0 & 0 & -b_1 - b_2 + b_3 \\ \end{array} \end{bmatrix}$
• Solvability: a condition on $$b$$. $$Ax=b$$ is solvable when $$b$$ is in the columnspace of $$A$$, $$C(A)$$. Alternatively, if a combination of the rows of $$A$$ give the zero row, then the same combination of the components of $$b$$ have to give a zero.
• To find complete solution to $$Ax=b$$:
• Find a particular solution. One way to find a particular solution: set all free variables to zero and then solve for pivot variables.
$x_p = \begin{bmatrix} \begin{array}{r} -2 \\ 0 \\ \frac{3}{2} \\ 0 \\ \end{array} \end{bmatrix}$
• Add in the nullspace $$x_n$$ (aka special solutions).
• The complete solution to $$Ax=b$$ is $$x_{complete} = x_p + x_n$$
$x_{complete} = \begin{bmatrix} \begin{array}{r} -2 \\ 0 \\ \frac{3}{2} \\ 0 \\ \end{array} \end{bmatrix} + c_1 * \begin{bmatrix} \begin{array}{r} -2 \\ 1 \\ 0 \\ 0 \\ \end{array} \end{bmatrix} + c_2 * \begin{bmatrix} \begin{array}{r} 2 \\ 0 \\ -2 \\ 1 \\ \end{array} \end{bmatrix}$
• Note $$A * (x_p + x_n) = A * x_p + A * x_n = A * x_p + 0 = A * x_p = b$$.
• $$x_c$$ in this case is a subspace (the nullspace) shifted by the $$x_p$$. Instead of going through zero, the subspace goes through $$x_p$$.
• Relations between rank $$r$$ in a $$m \times n$$ matrix:
• $r \leq m$
• $r \leq n$
• Full column rank $$r = n$$: no free variables, nullspace is only the zero vector, one or zero solutions to $$Ax=b$$. Matrix will look tall and thin. Each zero row will be an additional condition on $$b$$.
• Full row rank $$r = m$$: One or many solutions for every $$b$$ (depending if there are free variables. There will be $$n-r = n-m$$ free variables). Matrix will be short and fat. $$m > n$$ results in free variables.
• Full row and column rank $$r = m = n$$. Square matrix and invertible, one solution for every b. Row reduced echelon form is the identity matrix.
• If $$r < m$$ and $$r < n$$, then you will have 0 solutions (if the conditions of the zero row are not satisifed) or infinite solutions if the conditions of the zero row are satisfied. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
9.8: Probability of Compound Events
Difficulty Level: At Grade Created by: CK-12
We begin this lesson with a reminder of probability.
The experimental probability is the ratio of the proposed outcome to the number of experiment trials.
P(success)=number of times the event occuredtotal number of trials of experiment\begin{align*}P(success)= \frac{number \ of \ times \ the \ event \ occured}{total \ number \ of \ trials \ of \ experiment}\end{align*}
Probability can be expressed as a percentage, a fraction, a decimal, or a ratio.
This lesson will focus on compound events and the formulas used to determine the probability of such events.
Compound events are two simple events taken together, usually expressed as A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*}.
Independent and Dependent Events
Example: Suppose you flip a coin and roll a die at the same time. What is the probability you will flip a head and roll a four?
These events are independent. Independent events occur when the outcome of one event does not affect the outcome of the second event. Rolling a four has no effect on tossing a head.
To find the probability of two independent events, multiply the probability of the first event by the probability of the second event.
P(A and B)=P(A)P(B)\begin{align*}P(A \ and \ B)=P(A) \cdot P(B)\end{align*}
Solution:
P(tossing a head)P(rolling a 4)P(tossing a head AND rolling a 4)=12=16=12×16=112\begin{align*}P(tossing \ a \ head)&=\frac{1}{2}\\ P(rolling \ a \ 4)&=\frac{1}{6}\\ P(tossing \ a \ head \ AND \ rolling \ a \ 4)&=\frac{1}{2} \times \frac{1}{6}=\frac{1}{12}\end{align*}
When events depend upon each other, they are called dependent events. Suppose you randomly draw a card from a standard deck then randomly draw a second card without replacing the first. The second probability is now different from the first.
To find the probability of two dependent events, multiply the probability of the first event by the probability of the second event, after the first event occurs.
P(A and B)=P(A)P(B following A)\begin{align*}P(A \ and \ B)=P(A) \cdot P(B \ following \ A)\end{align*}
Example: Two cards are chosen from a deck of cards. What is the probability that they both will be face cards?
Solution: Let A=1st Face card chosen\begin{align*}A = 1st \ Face \ card \ chosen\end{align*} and B=2nd Face card chosen\begin{align*}B = 2nd \ Face \ card \ chosen\end{align*}. The total number of face cards in the deck is 4×3=12\begin{align*}4 \times 3 = 12\end{align*}.
P(A)P(B)=1252=1151, remember, one card has been removed.\begin{align*}P(A)&= \frac{12}{52}\\ P(B) & = \frac{11}{51}, \ \text{remember, one card has been removed.}\end{align*}
P(A AND B)=1252×1151 P(AB)or P(AB)=1252×1151=33663=11221\begin{align*}P(A \ AND \ B)= \frac{12}{52} \times \frac{11}{51} \ & or \ P(A \cap B) = \frac{12}{52} \times \frac{11}{51} =\frac{33}{663}\\ P(A \cap B) & = \frac{11}{221}\end{align*}
Mutually Exclusive Events
Events that cannot happen at the same time are called mutually exclusive events. For example, a number cannot be both even and odd or you cannot have picked a single card from a deck of cards that is both a ten and a jack. Mutually inclusive events, however, can occur at the same time. For example a number can be both less than 5 and even or you can pick a card from a deck of cards that can be a club and a ten.
When finding the probability of events occurring at the same time, there is a concept known as the “double counting” feature. It happens when the intersection is counted twice.
In mutually exclusive events, P(AB)=ϕ\begin{align*}P(A \cap B)=\phi\end{align*}, because they cannot happen at the same time.
To find the probability of either mutually exclusive events A\begin{align*}A\end{align*} or B\begin{align*}B\end{align*} occurring, use the following formula.
To find the probability of one or the other mutually exclusive or inclusive events, add the individual probabilities and subtract the probability they occur at the same time.
P(A or B)=P(A)+P(B)P(AB)\begin{align*}P(A \ or \ B)=P(A)+P(B)-P(A \cap B)\end{align*}
Example: Two cards are drawn from a deck of cards. Let:
A\begin{align*}A\end{align*}: 1st\begin{align*}1^{st}\end{align*} card is a club
B\begin{align*}B\end{align*}: 1st\begin{align*}1^{st}\end{align*} card is a 7
C\begin{align*}C\end{align*}: 2nd\begin{align*}2^{nd}\end{align*} card is a heart
Find the following probabilities:
(a) P(A or B)\begin{align*}P(A \ \text{or} \ B)\end{align*}
(b) P(B or A)\begin{align*}P(B \ \text{or} \ A)\end{align*}
(c) P(A and C)\begin{align*}P(A \ \text{and} \ C)\end{align*}
Solution:
(a) \begin{align*}P(A \ or \ B)=\frac{13}{52}+\frac{4}{52}-\frac{1}{52}\!\\ P(A \ or \ B) =\frac{16}{52}\!\\ P(A \ or \ B) =\frac{4}{13}\end{align*}
(b) \begin{align*}P(B \ or \ A)= \frac{4}{52} + \frac{13}{52}-\frac{1}{52}\!\\ P(B \ or \ A) = \frac{16}{52}\!\\ P(B \ or \ A) = \frac{4}{13}\end{align*}
(c) \begin{align*}P(A \ and \ C) = \frac{13}{52} \times \frac{13}{52}\!\\ P(A \ and \ C) = \frac{169}{2704}\!\\ P(A \ and \ C) = \frac{1}{16}\end{align*}
Practice Set
1. Define independent events.
Are the following events independent or dependent?
1. Rolling a die and spinning a spinner
2. Choosing a book from the shelf then choosing another book without replacing the first
3. Tossing a coin six times then tossing it again
4. Choosing a card from a deck, replacing it, and choosing another card
5. If a die is tossed twice, what is the probability of rolling a 4 followed by a 5?
6. Define mutually exclusive.
Are these events mutually exclusive or mutually inclusive?
1. Rolling an even and an odd number on one die.
2. Rolling an even number and a multiple of three on one die.
3. Randomly drawing one card and the result is a jack and a heart.
4. Randomly drawing one card and the result is black and a diamond.
5. Choosing one item and getting an orange and a fruit from the basket.
6. Choosing one tile and getting a vowel and a consonant from a Scrabble bag.
7. Two cards are drawn from a deck of cards. Determine the probability of each of the following events:
1. \begin{align*}P\end{align*}(heart or club)
2. \begin{align*}P\end{align*}(heart and club)
3. \begin{align*}P\end{align*}(red or heart)
4. \begin{align*}P\end{align*}(jack or heart)
5. \begin{align*}P\end{align*}(red or ten)
6. \begin{align*}P\end{align*}(red queen or black jack)
8. A box contains 5 purple and 8 yellow marbles. What is the probability of successfully drawing, in order, a purple marble and then a yellow marble? {Hint: In order means they are not replaced.}
9. A bag contains 4 yellow, 5 red, and 6 blue marbles. What is the probability of drawing, in order, 2 red, 1 blue, and 2 yellow marbles?
10. A card is chosen at random. What is the probability that the card is black and is a 7?
Mixed Review
1. A circle is inscribed within a square, meaning the circle's diameter is equal to the square’s side length. The length of the square is 16 centimeters. Suppose you randomly threw a dart at the figure. What is the probability the dart will land in the square, but not in the circle?
2. Why is \begin{align*}7-14x^4+7xy^5-1x^{-1}=8x^2 y^3\end{align*} not considered a polynomial?
3. Factor \begin{align*}72b^5 m^3 w^9-6(bm)^2 w^6\end{align*}.
4. Simplify \begin{align*}2^5-7^3 a^3 b^7+3^5 a^3 b^7-2^3\end{align*}.
5. Bleach breaks down cotton at a rate of 0.125% with each application. A shirt is 100% cotton.
1. Write the equation to represent the percentage of cotton remaining after \begin{align*}w\end{align*} washes.
2. What percentage remains after 11 washes?
3. After how many washes will 75% be remaining?
6. Evaluate \begin{align*}\frac{(100 \div 4 \times 2-49)^2}{9-2 \times 3+2^2}\end{align*}.
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Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!
# Roohi travels 300 km to her home partly by train and partly by bus.
Question:
Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Solution:
Let the speed of the train be x km/hour that of the bus be y km/hr, we have the following cases
Case I: When Roohi travels 300 Km by train and the rest by bus
Time taken by Roohi to travel $60 \mathrm{Km}$ by train $=\frac{60}{x} h r s$
Time taken by Roohi to travel $(300-60)=240 \mathrm{Km}$ by bus $=\frac{240}{y} h r s$
Total time taken by Roohi to cover $300 \mathrm{Km}=\frac{60}{x}+\frac{240}{y}$
It is given that total time taken in 4 hours
$\frac{60}{x}+\frac{240}{y}=4$
$60\left(\frac{1}{x}+\frac{4}{y}\right)=4$
$\left(\frac{1}{x}+\frac{4}{y}\right)=\frac{4}{60}$
$\frac{1}{x}+\frac{4}{y}=\frac{1}{15} \cdots(i)$
Case II: When Roohi travels 100 km by train and the rest by bus
Time taken by Roohi to travel $100 \mathrm{Km}$ by train $=\frac{100}{x} h r s$
Time taken by Roohi to travel $(300-100)=200 \mathrm{Km}$ by bus $=\frac{200}{y} h r s$
In this case total time of the journey is 4 hours 10 minutes
$\frac{100}{x}+\frac{200}{y}=4 h r s 10 \mathrm{~min}$ utes
$\frac{100}{x}+\frac{200}{y}=4 \frac{10}{60}$
$\frac{100}{x}+\frac{200}{y}=\frac{25}{6}$
$100\left(\frac{1}{x}+\frac{2}{y}\right)=\frac{25}{6}$
$\frac{1}{x}+\frac{2}{y}=\frac{25}{6} \times \frac{1}{100}$
$\frac{1}{x}+\frac{2}{y}=\frac{1}{24}$$\ldots$ (i)
Putting $\frac{1}{x}=u$ and $\frac{1}{y}=u$, the equations $(i)$ and $(i i)$ reduces to
$1 u+4 v=\frac{1}{15} \cdots(i i i)$
$1 u+2 v=\frac{1}{24} \cdots(i v)$
Subtracting equation (iv) from (iii)we get
$2 v=\frac{24-15}{360}$
$2 v=\frac{9}{360}$
$2 v=\frac{9}{360}$
$v=\frac{1}{40} \times \frac{1}{2}$
$v=\frac{1}{80}$
Putting $v=\frac{1}{80}$ in equation (iii), we get
$1 u+4 v=\frac{1}{15}$
$1 u+4 \times \frac{1}{80}=\frac{1}{15}$
$1 u+\frac{4}{80}=\frac{1}{15}$
$1 u=\frac{1}{15}-\frac{4}{80}$
$1 u=\frac{1}{15}-\frac{1}{20}$
$1 u=\frac{20-15}{300}$
$1 u=\frac{5}{300}$
$u=\frac{1}{60}$
Now
$u=\frac{1}{60}$
$\frac{1}{x}=\frac{1}{60}$
$x=60$
and
$v=\frac{1}{80}$
$\frac{1}{y}=\frac{1}{80}$
$y=80$
Hence, the speed of the train is $60 \mathrm{~km} / \mathrm{hr}$,
The speed of the bus is $80 \mathrm{~km} / \mathrm{hr}$ |
Count One More and One Less Year 1 Place Value Learning Video Clip – Classroom Secrets | Classroom Secrets
Maths Resources & WorksheetsYear 1 Maths Resources & WorksheetsYear 1 Autumn Maths - Place Value 204 Count One More and One Less › Count One More and One Less Year 1 Place Value Learning Video Clip
# Count One More and One Less Year 1 Place Value Learning Video Clip
## Step 4: Count One More and One Less Year 1 Place Value Learning Video Clip
Diver Dan is exploring a sunken pirate ship. He is counting the different objects and creatures and solving puzzles to find the treasure. He needs help to find one more and one less.
More resources for Autumn Block 4 Step 4.
Discussion points for teachers
1.What might he find in the ship?
Discuss what things Dan might find on a sunken pirate ship. This question is open-ended for the children to explore.
Sixteen
2.How many shells has he collected?
Discuss how many shells he had collected. Discuss the meaning of 1 more. Discuss how many shells Dan has.
<
3.How many fish and seahorses can Dan see?
Discuss counting the fish and seahorses and recording the answers.
10 fish and 12 seahorses.
4.How many fish and seahorses would there be?
Discuss what happened and how that changes the number of fish and seahorses Dan can see now. Discuss using the answers from the previous question to find one more and one less than each number.
11 fish and 11 seahorses.
5.What is the code?
Discuss the meaning of 1 more and 1 less. Discuss which number is one less than 20 but one more than 18.
19
6.Which gems could complete the statement and open the secret compartment?
Discuss the meaning of ‘1 less than’ a number. Discuss if the number in the first space needs to be greater than or less than the number in the second space. Discuss using the gems to make 2 digits numbers that will complete the statement. This question is open-ended for the children to explore.
Various answers, for example: 12 is 1 less than 13; 13 is 1 less than 14.
Optional discussion points:
When might you use 1 more or 1 less?
National Curriculum Objectives
Mathematics Year 1: (1N1a) Count to 20, forwards and backwards, beginning with 0 or 1, or from any given number
Mathematics Year 1: (1N2a) Count, read and write numbers to 100 in numerals
Mathematics Year 1: (1N2b) Given a number, identify one more and one less
This resource is available to play with a Premium subscription. |
## Engage NY Eureka Math 4th Grade Module 4 Lesson 1 Answer Key
### Eureka Math Grade 4 Module 4 Lesson 1 Problem Set Answer Key
Question 1.
Use the following directions to draw a figure in the box to the right.
a. Draw two points: A and B.
b. Use a straightedge to draw $$\overline{A B}$$.
c. Draw a new point that is not on $$\overline{A B}$$. Label it C.
d. Draw $$\overline{A C}$$.
e. Draw a point not on $$\overline{A B}$$ or $$\overline{A C}$$. Call it D.
f. Construct
g. Use the points you’ve already labeled to name one angle. ____________
The labeled angle is <ACD/<BAC.
Explanation:
Here, we have to draw two points and then labeled them as A and B.
And used a straightedge to draw $$\overline{A B}$$.
Then we have to draw a new point that is not on $$\overline{A B}$$. And we will label it as C.
Then we will draw $$\overline{A C}$$.
Then we will draw a point not on $$\overline{A B}$$ or $$\overline{A C}$$. And we will label it as D.
And then we will construct
We will use these points and we will label with one angle as <ACD/<BAC.
Question 2.
Use the following directions to draw a figure in the box to the right.
a. Draw two points: A and B.
b. Use a straightedge to draw $$\overline{A B}$$.
c. Draw a new point that is not on $$\overline{A B}$$. Label it C.
d. Draw $$\overline{B C}$$.
e. Draw a new point that is not on $$\overline{A B}$$ or $$\overline{A C}$$
Label it D.
f. Construct .
g. Identify ∠DAB by drawing an arc to indicate the position of the angle.
h. Identify another angle by referencing points that you have already drawn. _____________
The labeled angle is <ABC.
Explanation:
Here, we have to draw two points and label them as A and B.
And use a straightedge to draw $$\overline{A B}$$.
Then we have to draw a new point that is not on $$\overline{A B}$$. Label it C.
Then we will draw $$\overline{B C}$$.
And we will draw a new point that is not on $$\overline{A B}$$ or $$\overline{A C}$$
Label it D.
So we will construct .
Then we have identified ∠DAB by drawing an arc to indicate the position of the angle.
And we have Identified another angle by referencing points that we have already drawn.
Question 3.
a. Observe the familiar figures below. Label some points on each figure.
b. Use those points to label and name representations of each of the following in the table below: ray, line, line segment, and angle. Extend segments to show lines and rays.
Extension: Draw a familiar figure. Label it with points, and then identify rays, lines, line segments, and angles as applicable.
The ray of the house is marked as AB,
The ray of the flash drive is marked as CD,
The ray of the direction compass is marked as EF.
The Line of the house is marked as AB,
The Line of the flash drive is marked as CD,
The Line of the direction compass is marked as EF.
The Line segment of the house is marked as GH,
The Line segment of the flash drive is marked as IJ,
The Line segment of the direction compass is marked as EK.
The Angle of the house is marked as <HGA,
The Angle of the flash drive is marked as <CIJ,
The Angle of the direction compass is marked as <KEF.
Explanation:
Ray: A Ray can be defined as a part of the line which has a fixed starting point but does not have any endpoint and it can be extended infinitely in one direction and a ray may pass through more than one point.
The ray of the house is marked as AB,
The ray of the flash drive is marked as CD,
The ray of the direction compass is marked as EF.
Line: A line can be defined as a long, straight and continuous path which is represented using arrowheads at both directions and the lines that do not have a fixed point it can be extended in two directions.
The Line of the house is marked as AB,
The Line of the flash drive is marked as CD,
The Line of the compass rose is marked as EF.
Line segment: A line segment is a straight line that passes through the two points and has fixed point and it can not be extended.
The Line segment of the house is marked as GH,
The Line segment of the flash drive is marked as IJ,
The Line segment of the compass rose is marked as EK.
Angle: A figure which is formed by two rays or lines that shares a common endpoint and is called an angle.
So in the above, we can see the house, flash drive, and a compass rose.
The Angle of the house is marked as <HGA,
The Angle of the flash drive is marked as <CIJ,
The Angle of the compass rose is marked as <KEF.
### Eureka Math Grade 4 Module 4 Lesson 1 Exit Ticket Answer Key
Question 1.
Draw a line segment to connect the word to its picture.
Ray: A Ray can be defined as a part of the line which has a fixed starting point but does not have any endpoint and it can be extended infinitely in one direction and a ray may pass through more than one point.
Line: A line can be defined as a long, straight and continuous path which is represented using arrowheads at both directions and the lines that do not have a fixed point it can be extended in two directions.
Line segment: A line segment is a straight line that passes through the two points and has fixed point and it can not be extended.
Point: A point is an exact location and it has no point only position and point can usually be named often with letters like A, B, etc.
Angle: A figure which is formed by two rays or lines that shares a common endpoint and is called an angle.
Explanation:
Question 2.
How is a line different from a line segment?
A line can be defined as a long, straight and continuous path which is represented using arrowheads in both directions and the lines do not have a fixed point it can be extended in two directions, but the line segment is a straight line that passes through the two points and have a fixed point and it can not be extended.
### Eureka Math Grade 4 Module 4 Lesson 1 Homework Answer Key
Question 1.
Use the following directions to draw a figure in the box to the right.
a. Draw two points: W and X.
b. Use a straightedge to draw $$\overline{W X}$$.
c. Draw a new point that is not on $$\overline{W X}$$. Label it Y.
d. Draw $$\overline{W Y}$$.
e. Draw a point not on $$\overline{W X}$$ or $$\overline{W Y}$$. Call it Z.
f. Construct
g. Use the points you’ve already labeled to name one angle. ____________
The labeled angle is <XWY.
Explanation:
Here, we will draw two points which are represented with W and X.
And we will use a straightedge to draw $$\overline{W X}$$.
And then we will draw a new point that is not on $$\overline{W X}$$. And we will label it as Y.
Then we will draw $$\overline{W X}$$.
Next we will draw a point not on $$\overline{W X}$$ or $$\overline{W Y}$$. And we will call it Z.
And then we will construct
Then we will use these points which were already labeled and name the angle as <XWY.
Question 2.
Use the following directions to draw a figure in the box to the right.
a. Draw two points: W and X.
b. Use a straightedge to draw $$\overline{W X}$$.
c. Draw a new point that is not on $$\overline{W X}$$. Label it Y.
d. Draw $$\overline{W Y}$$.
e. Draw a new point that is not on $$\overline{W Y}$$ or on the line containing $$\overline{W X}$$. Label it Z.
f. Construct .
g. Identify ∠ZWX by drawing an arc to indicate the position of the angle.
h. Identify another angle by referencing points that you have already drawn. _____________
The identified angle is <XW.
Explanation:
Here, we will draw two points and will label them as W and X.
And we will use a straightedge to draw $$\overline{W X}$$.
Now we need to draw a new point that is not on $$\overline{W X}$$. Label it Y.
Then we will draw $$\overline{W Y}$$.
Next, we will draw a new point that is not on $$\overline{W Y}$$ or on the line containing $$\overline{W X}$$ and Label it as Z.
Then we need to Construct .
And then we will identify the angle ∠ZWX by drawing an arc to indicate the position of the angle.
So we will Identify another angle by referencing points that we have already drawn and label the angle as <XW.
Question 3.
a. Observe the familiar figures below. Label some points on each figure.
b. Use those points to label and name representations of each of the following in the table below: ray, line, line segment, and angle. Extend segments to show lines and rays.
Extension: Draw a familiar figure. Label it with points, and then identify rays, lines, line segments, and angles as applicable.
The ray of the clock is marked as AB,
The ray of the die is marked as EF,
The ray of the number line is marked as AC.
The Line of the clock is marked as BC,
The Line of the die is marked as FG,
The Line of the number line is marked as AB.
The Line segment of the clock is marked as AC,
The Line segment of the die is marked as GH,
The Line segment of the number line is marked as BD.
The Angle of the clock is marked as <CAB,
The Angle of the die is marked as <FGH
The Angle of the number line is marked as <CAB.
Explanation: |
Polynomials
A polynomial is an expression obtained by adding, subtracting, or multiplying one or more variables and constants and the powers of a variable. One part in the sum is called a term.
The polynomial above has three terms. The degree of the polynomial is 2 (the highest power of the variables).
A one-term polynomial is called a monomial, a two-term polynomial is binomial and a three-term polynomial is a trinomial.
These are all polynomials. If there are more than 3 terms it is still called a polynomial. A polynomial consist of one or more monomials.
Example 1
The multiplicative factor of a variable is called the coefficient. If we have a monomial 2x, then the variable is x and the coefficient is 2.
When adding polynomials, all similar terms are added together. Similar terms are terms that have the same variable character and degree. Constant terms are also similarly together.
Example 2
Below are two polynomials named P(x) and T(x).
Sum of the polynomials, P(x)+T(x)
We remove the parentheses and arrange similar terms side by side
Subtraction of polynomials P(x)-T(x)
We remove the parentheses and arrange similar terms side by side and add the terms together. Now the minus sign in front of the latter polynomial will change the signs for all terms after that.
Multiplication of polynomials
In multiplication of polynomials, all terms will be multiplied with the multiplier.
Example 3
Below is the product where the binomial x - 2 is multiplied by the monomial 2x. We open the parenthesis by multiplying both terms of the binomial by 2x.
Example 4
Below is the product where binomial 2x + 2 is multiplied by binomial x-2. We open the parentheses by multiplying all terms with each other.
Click on the subtitles if needed |
### Fractions - Nevada Mathematics Project
```Fractions: Fourth
What is a Fraction?
•What is a fraction?
•Think of a scenario
to represent 1/3 and
create a model.
Explore Fractions Using
Fraction Bars
• http://www.mathsisfun.com/numbers/fraction-numberline.html
• What fractions add up to one whole?
• Create equivalent Fractions
What patterns would you want your
students to notice?
• Number line
• Area Model
• Set Model
• Develop understanding of fractions as numbers.
• CCSS.Math.Content.3.NF.A.1
Understand a fraction 1/b as the quantity formed by 1 part when a
whole is partitioned into b equal parts; understand a fraction a/b as
the quantity formed by a parts of size 1/b.
• CCSS.Math.Content.3.NF.A.2
Understand a fraction as a number on the number line; represent
fractions on a number line diagram.
• CCSS.Math.Content.3.NF.A.2.a
Represent a fraction 1/b on a number line diagram by defining the
interval from 0 to 1 as the whole and partitioning it into b equal
parts. Recognize that each part has size 1/b and that the endpoint of
the part based at 0 locates the number 1/b on the number line.
• CCSS.Math.Content.3.NF.A.2.b
Represent a fraction a/b on a number line diagram by marking off a
lengths 1/b from 0. Recognize that the resulting interval has size a/b
and that its endpoint locates the number a/b on the number line.
Explain equivalence of fractions in special cases, and compare fractions by
.
• CCSS.Math.Content.3.NF.A.3
Explain equivalence of fractions in special cases, and compare fractions by
• CCSS.Math.Content.3.NF.A.3.a
Understand two fractions as equivalent (equal) if they are the same size, or the
same point on a number line.
• CCSS.Math.Content.3.NF.A.3.b
Recognize and generate simple equivalent fractions, e.g., 1/2 = 2/4, 4/6 = 2/3.
Explain why the fractions are equivalent, e.g., by using a visual fraction model.
• CCSS.Math.Content.3.NF.A.3.c
Express whole numbers as fractions, and recognize fractions that are equivalent
to whole numbers. Examples: Express 3 in the form 3 = 3/1; recognize that 6/1 =
6; locate 4/4 and 1 at the same point of a number line diagram.
• CCSS.Math.Content.3.NF.A.3.d
Compare two fractions with the same numerator or the same denominator by
reasoning about their size. Recognize that comparisons are valid only when the
two fractions refer to the same whole. Record the results of comparisons with
the symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction
model
4.NF.1 Equivalent Fractions
• Explain why a fraction a/b is equivalent to a fraction (n × a)/(n
× b) by using visual fraction models, with attention to how the
number and size of the parts differ even though the two
fractions themselves are the same size. Use this principle to
recognize and generate equivalent fractions.
• How would student explain equivalent fractions?
• What do students need to understand in order to do this?
• 2/4
½
Equivalent Fractions
• Students subdivide the equal parts of a fraction, resulting in a
greater number of smaller parts.
• Students discover that this subdividing has the effect of
multiplying the numerator and denominator by n.
1 x 4 parts and 4 x 4 parts.
4.NF.2 Comparing Fractions
• Compare two fractions with different numerators and
different denominators, e.g., by creating common
denominators or numerators, or by comparing to a benchmark
fraction such as 1/2. Recognize that comparisons are valid only
when the two fractions refer to the same whole. Record the
results of comparisons with symbols >, =, or <, and justify the
conclusions, e.g., by using a visual fraction model.
Comparing Fractions
• Students create common numerators or common
denominators by renaming one fraction or both.
• When comparing fractions with the same numerator, they use
prior knowledge about the relative sizes of fractional parts.
2 and 1
5
3
How would you rename these
fractions to compare them?
Comparing Fractions
• Complete the following comparisons without using equivalent
fractions. Make a note of how you did them…
1 and 8
8
9
7 and 5
8
6
4 and 3
9
4
What are some student misconceptions about comparing
fractions?
4.NF.3 (a-b) Fractions as a Sum of
Unit Fractions
a. Understand addition and subtraction of fractions as joining and
separating parts referring to the same whole.
b. Decompose a fraction into a sum of fractions with the same
denominator in more than one way, recording each
decomposition by an equation. Justify decompositions, e.g., by
using a visual fraction model. Examples: 3/8 = 1/8 + 1/8 + 1/8 ;
3/8 = 1/8 + 2/8 ; 2 1/8 = 1 + 1 + 1/8 = 8/8 + 8/8 + 1/8.
Describe a common misconception for this problem:
2 + 1
5
5
Fractions as a Sum
of Unit Fractions
• All fractions can be seen as a sum or difference of two other
fractions.
• Think in terms of adding/subtracting copies of 1/b.
“Just like 3 dogs + 9 dogs is 12 dogs, or 3 candies + 9 candies is 12 candies, or
3 children + 9 children is 12 children, 3 fifths + 9 fifths is 12 fifths” (Small,
2014, p. 51).
• This line of thinking enables students to clearly understand
why only the numerators are added or subtracted.
Fractions as a Sum
of Unit Fractions
• Students rename mixed
numbers as improper
fractions and vice versa.
• Students decompose
fractions and mixed
numbers in more than one
way:
mrs-c-classroom.blogspot.com
Before teaching the “shortcut,” allow
students plenty of time to reason with
models.
Mixed Numbers
c. Add and subtract mixed numbers with like denominators, e.g.,
by replacing each mixed number with an equivalent fraction,
and/or by using properties of operations and the relationship
d. Solve word problems involving addition and subtraction of
fractions referring to the same whole and having like
denominators, e.g., by using visual fraction models and
equations to represent the problem.
Mixed Numbers
• Students rename mixed numbers as improper fractions, then
But…
• They notice that sometimes it is easier to add or subtract the
whole number and fraction separately.
How would you solve each problem? Why?
Numbers
• Students can “count up” to find the difference between mixed
numbers (see page 52 in Uncomplicating Fractions).
OR
• They might prefer to regroup part of the whole number in the
greater mixed number.
Explain how to regroup the first
mixed number in this problem.
4.NF.B.4 Multiplying Fractions by a Whole
Number
• CCSS.Math.Content.4.NF.B.4a Understand a fraction a/b as a
multiple of 1/b. For example, use a visual fraction model to represent
5/4 as the product 5 × (1/4), recording the conclusion by the
equation 5/4 = 5 × (1/4).
• CCSS.Math.Content.4.NF.B.4b Understand a multiple of a/b as a
multiple of 1/b, and use this understanding to multiply a fraction by
a whole number. For example, use a visual fraction model to express
3 × (2/5) as 6 × (1/5), recognizing this product as 6/5. (In general, n ×
(a/b) = (n × a)/b.)
• CCSS.Math.Content.4.NF.B.4c Solve word problems involving
multiplication of a fraction by a whole number, e.g., by using visual
fraction models and equations to represent the problem. For
example, if each person at a party will eat 3/8 of a pound of roast
beef, and there will be 5 people at the party, how many pounds of
roast beef will be needed? Between what two whole numbers does
Interpreting the meaning of
multiplication
• It is important to let students model and solve these problems
in their own way, using whatever models or drawings they
choose as long as they can explain their reasoning.
• Once students have spent adequate time exploring
multiplication of fractions, they will begin to notice patterns.
• Then, the standard multiplication algorithm will be simple to
develop. Shift from contextual problems to straight
computation.
How can you solve the following
problem? How many different ways
can you solve it?
Kristen ran on a path that was ¾ of a mile in length. She
ran the path 5 times. What is the total distance that
Kristen ran?
Interpreting the meaning of
multiplication
• Adding 4/5 3 times ( 4/5 +4/5 +4/5)
• 4 fifths + 4 fifths + 4 fifths = 12 fifths, or 12/5
• The result of three jumps of 4/5 on a number line, beginning at 0
• The number of fifths of a 2-D shape if 3 groups of 4 fifths are
Understand decimal notation for fractions, and
compare decimal fractions
• CCSS.Math.Content.4.NF.C.5 Express a fraction with
denominator 10 as an equivalent fraction with denominator
100, and use this technique to add two fractions with
respective denominators 10 and 100.2 For example, express
3/10 as 30/100, and add 3/10 + 4/100 = 34/100.
• CCSS.Math.Content.4.NF.C.6 Use decimal notation for
fractions with denominators 10 or 100. For example, rewrite
0.62 as 62/100; describe a length as 0.62 meters; locate 0.62
on a number line diagram.
• CCSS.Math.Content.4.NF.C.7 Compare two decimals to
hundredths by reasoning about their size. Recognize that
comparisons are valid only when the two decimals refer to the
same whole. Record the results of comparisons with the
symbols >, =, or <, and justify the conclusions, e.g., by using a
visual model.
Interpreting decimals
• Representing tenths and hundredths and decimals as a sum.
The 10-to-1 relationship continues indefinitely.
What is a common misconceptions students
Students must
understand the
equivalent
relationship
between tenths
and hundredths.
Representing Decimals on a
Number line
One of the best length models for decimal
fractions is a meter stick. Experiences
allow students to compare decimals and
think about scale and place value.
Comparing Decimals
• Reason abstractly and quantitatively. Develop benchmarks; as
with fractions: 0, ½, and 1. For example, is seventy-eight
hundredths closer to 0 or ½, ½ or 1? How do you know?
• Using decimal circle models. Multiple wheels may be used to
conceptualize the amount. Or, cut the tenths and hundredths
and the decimal can be built.
• Why do many students
think .4 < .19?
Activity: The Unusual Baker
• George is a retired
mathematics teacher who
makes cakes. He likes to
cut the cakes differently
each day of the week. On
the order board, George
lists the fraction of the
piece, and next to that, he
has the cost of each piece.
This week he is selling
whole cakes for \$1 each.
Determine the fraction and
decimal for each piece.
How much will each piece
cost if the whole cake is
\$1.00?
The Unusual Baker
• CCSS 4.NF.A.1 Equivalent fractions
• CCSS 4. NF.A.2 Compare two fractions
• CCSS 4. NF.B.3a Understand addition and subtraction of
fractions
• CCSS4.NF.B.3b Decompose a fraction into a sum of fractions
with the same denominator.
• CCSS4. NF.C.6 Use decimal notation for fractions with
denominators 10 or 100.
• CCSS4.NF.C.7 Compare two decimals to hundredths by |
# How to subtract integer numbers: calculate the difference and learn to subtract multiple digits numbers, column subtracting method, from right to left
## Let's learn with an example. Subtract the numbers: 2,043 - 1,959
### Stack the numbers on top of each other.
#### And so on...
2 0 4 3 - 1 9 5 9 ?
## Subtract column by column; start from the column on the right
### Subtract the digits in the ones column:
#### When borrowing, 1 ten = 10 ones. Add 10 to the top digit in the column of the ones: 10 + 3 = 13.
3 2 0 4 13 - 1 9 5 9
#### After borrowing, the subtraction has become: 13 - 9 = 4. 4 is the ones digit. Write it down at the base of the ones column.
3 2 0 4 13 - 1 9 5 9 4
### Subtract the digits in the tens column:
#### When borrowing, 1 hundred = 10 tens. Add 10 to the top digit in the column of the tens: 10 + 3 = 13.
1 9 13 2 0 4 13 - 1 9 5 9 4
#### After borrowing, the subtraction has become: 13 - 5 = 8. 8 is the tens digit. Write it down at the base of the tens column.
1 9 13 2 0 4 13 - 1 9 5 9 8 4
### Subtract the digits in the hundreds column:
#### 09 - 9 = 0. 0 is the hundreds digit. Write it down at the base of the hundreds column.
1 9 13 2 0 4 13 - 1 9 5 9 0 8 4
### Subtract the digits in the thousands column:
#### 21 - 1 = 0. 0 is the thousands digit. Write it down at the base of the thousands column.
1 9 13 2 0 4 13 - 1 9 5 9 0 0 8 4 |
RS Aggarwal Solutions for Class 6 Chapter 21 Concept of Perimeter and Area Exercise 21C
Exercise 21C of chapter 21 provides the students with basic knowledge of finding an area for a given shape. The measurement of the region enclosed by a plane figure is called the area of the figure. RS Aggarwal Solutions contain step by step explanation for the problems as per the understanding ability of students. The students can get a clear idea about the concepts which are explained in the chapter and perform better in the exam. RS Aggarwal Solutions for Class 6, Chapter 21, Concept of Perimeter and Area Exercise 21C PDF are provided here.
Download PDF of RS Aggarwal Solutions for Class 6 Chapter 21 Concept of Perimeter and Area Exercise 21C
Access answers to Maths RS Aggarwal Solutions for Class 6 Chapter 21 Concept of Perimeter and Area Exercise 21C
The following figures are drawn on a sheet of squared paper. Count the number of squares enclosed by each figure and find its area, taking the area of each square as 1 cm2.
1.
Solution
This figure contains 12 complete squares
Area of 1 small square = 1 cm2
Area of the figure = Number of complete squares × Area of the square
= (12 × 1) sq cm
= 12 sq cm
2.
Solution
This figure contains 18 complete squares
Area of 1 small square = 1 cm2
Area of the figure = Number of complete square × Area of the square
= (18 × 1) sq cm
= 18 sq cm
3.
Solution
This figure contains 14 complete square and 1 half square
Area of 1 small square = 1 cm2
Area of the figure = Number of complete square × Area of the square
= (14 × 1) + (1 × 1 / 2)
= 14 + 1 / 2
= 14 (1 / 2) sq cm
4.
Solution
This figure contains 6 complete squares and 4 half squares
Area of 1 small square = 1 cm2
Area of the figure = Number of complete squares × Area of the square
= (6 × 1) + (4 × 1 / 2)
= 6 + 4 / 2
= 16 / 2 {taking LCM}
= 8 sq cm
5.
Solution
This figure contains 9 complete squares and 6 half squares
Area of 1 small square = 1 cm2
Area of the figure = Number of complete squares × Area of the square
= (9 × 1) + (6 × 1 / 2)
= 9 + 6 / 2
= 24 / 2 {taking LCM}
= 12 sq cm
Access other exercises of RS Aggarwal Solutions for Class 6 Chapter 21 Concept of Perimeter and Area
Exercise 21E Solutions |
Question
# Solve the following equation: $5\sqrt {\frac{3}{x}} + 7 + \sqrt {\frac{x}{3}} = 22\frac{2}{3}.$
Hint: we need to know the basic factorization of quadratic equations to solve this problem.
Given equation is $5\sqrt {\frac{3}{x}} + 7 + \sqrt {\frac{x}{3}} = 22\frac{2}{3}$
For simplification of calculations put $t = \sqrt {\frac{3}{x}}$ then $\sqrt {\frac{x}{3}} = \frac{1}{t}$ , then the given equation will be
$5t + \frac{7}{t} = \frac{{68}}{3}$
Simplifying the above equation
$\frac{{5{t^2} + 7}}{t} = \frac{{68}}{3}$
$15{t^2} + 21 = 68t$
$15{t^2} - 68t + 21 = 0$
Now we got a quadratic equation, on factorization we get
$15{t^2} - 5t - 63t + 21 = 0$
$5t(3t - 1) - 21(3t - 1) = 0$
$(5t - 21)(3t - 1) = 0$
$t = \frac{{21}}{5},\frac{1}{3}$
Now, we know that $t = \sqrt {\frac{3}{x}}$
${t^2} = \frac{3}{x}$
$x = \frac{3}{{{t^2}}}$ , solving for the values of x using t value.
For $t = \frac{{21}}{5}$
$x = \frac{3}{{{{\left( {\frac{{21}}{5}} \right)}^2}}} = \frac{{25}}{{147}}$
For $t = \frac{1}{3}$
$x = \frac{3}{{{{\left( {\frac{1}{3}} \right)}^2}}} = 27$
$\therefore x = 27,\frac{{25}}{{147}}$ are the required values.
Note: Here we are converting the given equation into a quadratic equation by using the substitution method. We substituted $\sqrt {\frac{3}{x}}$as t, after substitution we simplified the equation and solved for t. After getting t values, we have to again substitute the value of t in terms of x, then finding x values easily. |
# How to Use Expanded Form to Multiply One-Digit Numbers By 3-digit or 4-digit Numbers
Expanded form is a great way to simplify multiplication problems, especially when you're multiplying one-digit numbers with larger numbers.
## A Step-by-step Guide to Using Expanded Form to Multiply One-Digit Numbers By 3-digit or 4-digit Numbers
Here is a step-by-step guide to Using Expanded Form to Multiply One-Digit Numbers Via 3-digit or 4-digit Numbers:
Example: Multiply 3 by 456 (a 3-digit number)
### Step 1: Understand the problem
You are given the task to multiply 3 by 456.
### Step 2: Write the larger number in expanded form
The larger number (456) needs to be written in its expanded form. This means breaking it down into hundreds, tens, and ones:
$$456=400+50+6$$
### Step 3: Multiply each component by the one-digit number
Now, multiply each part of the expanded number by 3:
$$3\times400 = 1200 3\times50 = 150 3\times6 = 18$$
### Step 4: Add all of the products together
Next, add together the products from step 3:
$$1200+150+18=1368$$
So, $$3\times456 =1368$$
Using the expanded form method makes multiplication easier to understand and solve, particularly with larger numbers. It’s an excellent tool for students in grade 4 or anyone looking to develop their basic multiplication skills.
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### Learning Outcomes
Express roots of an unfavorable numbers in regards to iExpress imaginary numbers together bi and complicated numbers as a+bi
You really require only one new number to begin working with the square root of negative numbers. That number is the square root of −1,sqrt-1. The actual numbers space those that have the right to be presented on a number line—they seem pretty real to us! once something is not real, we often say that is imaginary. So let us call this new number i and use that to represent the square source of −1.
You are watching: A complex number is a number of the form a + bi where
i=sqrt-1
Because sqrtx,cdot ,sqrtx=x, us can also see that sqrt-1,cdot ,sqrt-1=-1 or i,cdot ,i=-1. We also know that i,cdot ,i=i^2, for this reason we have the right to conclude the i^2=-1.
i^2=-1
The number i permits us to work-related with roots of all an adverse numbers, not just sqrt-1. There are two essential rules come remember: sqrt-1=i, and sqrtab=sqrtasqrtb. Girlfriend will use these rule to rewrite the square root of a negative number as the square source of a positive number time sqrt-1. Following you will certainly simplify the square root and also rewrite sqrt-1 together i. Permit us shot an example.
### Example
Simplify. sqrt-4
Show Solution
Use the preeminence sqrtab=sqrtasqrtb come rewrite this as a product utilizing sqrt-1.
sqrt-4=sqrt4cdot -1=sqrt4sqrt-1
Since 4 is a perfect square (4=2^2), you have the right to simplify the square source of 4.
sqrt4sqrt-1=2sqrt-1
Use the meaning of i to rewrite sqrt-1 as i.
2sqrt-1=2i
The price is 2i.
Use the dominion sqrtab=sqrtasqrtb come rewrite this together a product using sqrt-1.
sqrt-18=sqrt18cdot -1=sqrt18sqrt-1
Since 18 is not a perfect square, usage the same ascendancy to rewrite the using determinants that space perfect squares. In this case, 9 is the only perfect square factor, and the square source of 9 is 3.
sqrt18sqrt-1=sqrt9sqrt2sqrt-1=3sqrt2sqrt-1
Use the an interpretation of ns to rewrite sqrt-1 together i.
3sqrt2sqrt-1=3sqrt2i=3isqrt2
Remember to write i in prior of the radical.
Use the preeminence sqrtab=sqrtasqrtb come rewrite this as a product utilizing sqrt-1.
-sqrt-72=-sqrt72cdot -1=-sqrt72sqrt-1
Since 72 is no a perfect square, usage the same ascendancy to rewrite the using factors that are perfect squares. Notice that 72 has actually three perfect squares as factors: 4, 9, and 36. The is easiest to usage the largest element that is a perfect square.
-sqrt72sqrt-1=-sqrt36sqrt2sqrt-1=-6sqrt2sqrt-1
Use the meaning of ns to rewrite sqrt-1 together i.
-6sqrt2sqrt-1=-6sqrt2i=-6isqrt2
Remember to write i in front of the radical.
The prize is -6isqrt<>2
A complex number is the sum of a genuine number and also an imaginary number. A complex number is express in standard form when written + bi where a is the real part and bi is the imagine part. Because that example, 5+2i is a complicated number. So, too, is 3+4isqrt3.
Imaginary numbers are differentiated from actual numbers since a squared imagine number to produce a negative real number. Recall, when a confident real number is squared, the result is a confident real number and also when a negative real number is squared, again, the an outcome is a hopeful real number. Facility numbers are a mix of real and also imaginary numbers. You can use the usual operations (addition, subtraction, multiplication, and so on) through imaginary numbers. You will certainly see an ext of the later.
Complex NumberReal PartImaginary Part
3+7i37i
18–32i18−32i
-frac35+isqrt2 -frac35 isqrt2
fracsqrt22-frac12i fracsqrt22-frac12i
In a number through a radical as part of b, such together -frac35+isqrt2 above, the imagine i need to be written in front of the radical. Though writing this number as -frac35+sqrt2i is technically correct, it renders it much more daunting to phone call whether i is inside or exterior of the radical. Placing it before the radical, together in -frac35+isqrt2, clears up any type of confusion. Look at these last 2 examples.
NumberComplex Form:a+biReal PartImaginary Part
1717+0i170i
−3i0–3i0−3i
By make b=0, any type of real number can be expressed together a complicated number. The real number a is composed as a+0i in complicated form. Similarly, any kind of imaginary number deserve to be expressed as a complicated number. By do a=0, any kind of imaginary number bi have the right to be written as 0+bi in complex form.
### Example
Write 83.6 together a complicated number.
Show Solution
Remember that a facility number has actually the kind a+bi. You require to number out what a and b should be.
a+bi
Since 83.6 is a real number, the is the real part (a) of the complicated number a+bi
83.6+bi
A real number does no contain any imaginary parts, so the value of b is 0.
The price is 83.6+0i.
In the next video, we show more examples of how to create numbers as facility numbers.
See more: How To Ollie On A Tech Deck With 2 Fingers, My Tips For Learning Ollies: Fingerboards
## Summary
Complex numbers have actually the form a+bi, whereby a and also b are genuine numbers and i is the square root of −1. All genuine numbers deserve to be written as complicated numbers by setup b=0. Imagine numbers have the kind bi and can additionally be written as complicated numbers by setting a=0. Square root of negative numbers can be streamlined using sqrt-1=i and sqrtab=sqrtasqrtb. |
# Adding And Subtracting Polynomials
In these lessons, we will look at how to add and subtract polynomials.
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The following diagram shows examples of adding and subtracting polynomials. Scroll down the page for more examples and solutions on how to add and subtract polynomials.
### Adding Polynomials
Adding polynomials involves combining like terms.
Example:
Add the polynomials 5x – 2 + y and –3y + 5x + 2
Solution:
5x – 2 + y + (–3y + 5x + 2)
= 5x + 5x + y – 3y – 2 + 2
= 10x – 2y
Example:
Find the sum of –7x3y + 4x2y2 – 2 and 4x3y + 1 – 8x2y2
Solution:
–7x3y + 4x2y2– 2 + 4x3y + 1 – 8x2y2
= –7x3y + 4x3y + 4x2y2– 8x2y2 – 2 + 1
= –3x3y – 4x2y2 – 1
### Subtracting Polynomials
To subtract polynomials, remember to distribute the – sign into all the terms in the parenthesis.
Example:
Simplify –4x + 7 – (5x – 3)
Solution:
–4x + 7 – (5x – 3)
= –4x + 7 – 5x + 3
= –9x + 10
Example:
Simplify (5x2 + 2) – (– 4x2 + 7) + (– 3x2 – 5)
Solution:
(5x2 + 2) – (– 4x2 + 7) + (– 3x2– 5)
= 5x2 + 2 + 4x2– 7 – 3x2– 5
= 5x2 + 4x2– 3x2 + 2 – 7 – 5
= 6x2 – 10
#### How To Add And Subtract Polynomials?
To add polynomials
1. Combine like terms.
2. Write in descending order.
Examples:
1. (4x2 + 8x - 7) + (2x2 - 5x - 12)
2. (5 + 24y3 - 7y2) + (-6y3 + 7y2 + 5)
3. (t2 - t + 5) + (7t2 - 4t - 20)
To subtract polynomials
1. Rewrite the subtraction as addition.
To change subtraction to addition, we must add the opposite, or additive inverse. 2. Combine like terms. 3. Write in descending order.
Examples:
1. (4x2 + 8x - 7) - (2x2 - 5x - 12)
2. (6x4 + 3x3 - 1) - (4x3 - 5x + 9)
3. (1.5y3 + 4.8y2 + 12) - (y3 - 1.7y2 + 2y)
#### Examples Of Adding And Subtracting Polynomials
Examples:
1. (2x5 - 6x3 - 12x2 - 4) + (-11x5 + 8x + 2x2 + 6)
2. (-9y3 - 6y2 - 11x + 2) - (-9y4 - 8y3 + 4x2 + 2x)
#### Adding And Subtracting Polynomials
Adding polynomials and subtracting polynomials is essentially combining like terms of polynomial expressions. When adding and subtracting polynomials, they can either be arranged vertically or grouped according to degree. A knowledge of polynomial vocabulary is important before adding and subtracting polynomials. Multiplying monomials and binomials is another type of operation with polynomials.
Examples:
1. (4x2 - 3x + 2) + (5x2 + 2x - 7)
2. (5x3 + 7x2 - x) + (8x3 + 4x - 5)
3. (8x2 + 2x) - (10x2 + 2x - 9)
4. -(6x3 - 4x) - (2x3 + x2 -2x)
#### Adding And Subtracting Polynomials
Examples:
1. (x2 + 4x + 5) + (6x + 3)
2. 2(x4 + 5x) - 6(x4 + 8x - 3)
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page. |
# Question: Can 3 Collinear Points Define Plane?
## How many planes can contain 3 collinear points?
Through any three points, there is exactly one plane.
SOLUTION: If the points were non-collinear, there would be exactly one plane by Postulate 2.2 shown by Figure 1.
If the points were collinear, there would be infinitely many planes.
Figure 2 shows what two planes through collinear points would look like..
## What are three non collinear points?
Points B, E, C and F do not lie on that line. Hence, these points A, B, C, D, E, F are called non – collinear points. If we join three non – collinear points L, M and N lie on the plane of paper, then we will get a closed figure bounded by three line segments LM, MN and NL.
## How many lines can 3 Noncollinear points draw?
Four linesFour lines can be drawn through 3 non-collinear points.
## Which set of points are collinear?
In Geometry, a set of points are said to be collinear if they all lie on a single line. Because there is a line between any two points, every pair of points is collinear.
## Can three points be collinear?
Three or more points that lie on the same line are collinear points . Example : The points A , B and C lie on the line m . They are collinear.
## When three points are collinear you can say that one point is?
Three points lie in exactly one line. Three collinear points lie in exactly one plane.
## Can two points be collinear?
What makes points collinear? Two points are always collinear since we can draw a distinct (one) line through them. Three points are collinear if they lie on the same line. Points A, B, and C are not collinear.
## How do you know if points are coplanar?
In geometry, a set of points in space are coplanar if there exists a geometric plane that contains them all. For example, three points are always coplanar, and if the points are distinct and non-collinear, the plane they determine is unique.
## How many planes pass through 3 non collinear points?
Case-2: When three distinct points are non-collinear. In this case one and only one plane can be drawn.
## Can a point be between two points if they are not collinear?
Any two points are always collinear because you can always connect them with a straight line. Three or more points can be collinear, but they don’t have to be. The above figure shows collinear points P, Q, and R which all lie on a single line.
## Are any 3 points coplanar?
Coplanar means “lying on the same plane”. Points are coplanar, if they are all on the same plane, which is a two- dimensional surface. Any three points are coplanar (i.e there is some plane all three of them lie on), but with more than three points, there is the possibility that they are not coplanar.
## How do you know if three points are collinear?
Slope formula method to find that points are collinear. Three or more points are collinear, if slope of any two pairs of points is same. With three points A, B and C, three pairs of points can be formed, they are: AB, BC and AC. If Slope of AB = slope of BC = slope of AC, then A, B and C are collinear points.
## Which figure is formed by three collinear points?
triangleA triangle is a figure formed by three segments joining three noncollinear points. Each of the three points joining the sides of a triangle is a vertex. The plural of vertex is “vertices.” In a triangle, two sides sharing a common vertex are adjacent sides.
## What are non collinear points?
Non-collinear points are a set of points that do not lie on the same line. |
# Is it possible to factor y=x^2+8x+14 ? If so, what are the factors?
Jan 2, 2016
Yes, you have to solve the equation though.
${x}^{2} + 8 x + 14 = \left(x + 4 - \sqrt{2}\right) \left(x + 4 + \sqrt{2}\right)$
#### Explanation:
By finding the roots ${x}_{1}$ and ${x}_{2}$ you can factor using the formula:
$a {x}^{2} + b x + c = a \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$
${x}^{2} + 8 x + 14 = 0$
Δ=b^2-4*a*c=8^2-4*1*14=8
x=(-b+-sqrt(Δ))/(2*a)=(-8+-sqrt(8))/(2*1)=-8/2+-sqrt(4*2)/2=
=-4+-(sqrt4)*sqrt(2))/2=-4+-(2*sqrt(2))/2=-4+-sqrt(2)
Now that the roots ${x}_{1} = - 4 + \sqrt{2}$ and ${x}_{2} = - 4 - \sqrt{2}$ are found the factoring can be done as follows:
${x}^{2} + 8 x + 14 = 1 \cdot \left(x - \left(- 4 + \sqrt{2}\right)\right) \left(x - \left(- 4 - \sqrt{2}\right)\right) =$
$= \left(x + 4 - \sqrt{2}\right) \left(x + 4 + \sqrt{2}\right)$
An easier example for a better understanding
Factor the following function:
$y = {x}^{2} + 8 x + 12$
We solve the equation:
${x}^{2} + 8 x + 12 = 0$
Δ=b^2-4*a*c=8^2-4*1*12=16
x=(-b+-sqrt(Δ))/(2*a)=(-8+-sqrt(16))/(2*1)=-8/2+-4/2=-4+-2
${x}_{1} = - 4 + 2 = - 2$
${x}_{2} = - 4 - 2 = - 6$
Now we have:
${x}^{2} + 8 x + 12 = 1 \left(x - \left(- 2\right)\right) \left(x - \left(- 6\right)\right) = \left(x + 2\right) \left(x + 6\right)$ |
We've updated our
TEXT
# Properties of Logarithms
### Learning Objectives
• The product rule for logarithms
• Define properties of logarithms, and use them to solve equations
• Define the product rule for logarithms, and use it to solve equations
• Quotient and Power Rules for Logarithms
• Define the quotient and power rules for logarithms
• Use the quotient and power rules for logarithms to simplify logarithmic expressions
• Expand and Condense Logarithms
• Combine product, power and quotient rules to simplify logarithmic expressions
• Expand logarithmic expressions that have negative or fractional exponents
• Condense logarithmic expressions
• Change of Base
• Use properties of logarithms to define the change of base formula
• Change the base of logarithmic expressions into base 10, or base e
When you learned how to solve linear equations, you were likely introduced to the properties of real numbers. These properties help us know what the rules are for isolating and combining numbers and variables. For example, it is advantageous to know that multiplication and division "undo" each other when you want to solve an equation for a variable that is multiplied by a number. You may have also been introduced to properties and rules for writing and using exponents. In this section, we will show you properties of logarithms and how they can help you better understand what a logarithm means and eventually how to solve equations that contain them.
Recall that we can express the relationship between logarithmic form and its corresponding exponential form as follows:
${\mathrm{log}}_{b}\left(x\right)=y\Leftrightarrow {b}^{y}=x,\text{}b>0,b\ne 1$
Note that the base b is always positive and that the logarithmic and exponential functions "undo" each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here.
First, we will introduce some basic properties of logarithms followed by examples with integer arguments to help you get familiar with the relationship between exponents and logarithms.
### Zero and Identity Exponent Rule for Logarithms and Exponentials
${\mathrm{log}}_{b}1=0$, b>0
${\mathrm{log}}_{b}b=1$, b>0
### Example
Use the the fact that exponentials and logarithms are inverses to prove the zero and identity exponent rule for the following:
1.${\mathrm{log}}_{5}1=0$
2. ${\mathrm{log}}_{5}5=1$
1.${\mathrm{log}}_{5}1=0$ since ${5}^{0}=1$
2.${\mathrm{log}}_{5}5=1$ since ${5}^{1}=5$
Exponential and logarithmic functions are inverses of each other and we can take advantage of this to evaluate and solve expressions and equations involving logarithms and exponentials. The inverse property of logarithms and exponentials gives us an explicit way to rewrite an exponential as a logarithm or a logarithm as an exponential.
### Inverse Property of Logarithms and Exponentials
$\begin{array}{c}\hfill \\ {\mathrm{log}}_{b}\left({b}^{x}\right)=x\hfill \\ \text{ }{b}^{{\mathrm{log}}_{b}x}=x, x>0, b>0, b\ne1\hfill \end{array}$
### Example
Evaluate: 1.$\mathrm{log}\left(100\right)$ 2.${e}^{\mathrm{ln}\left(7\right)}$
Answer: 1.Rewrite the logarithm as ${\mathrm{log}}_{10}\left({10}^{2}\right)$, and then apply the inverse property ${\mathrm{log}}_{b}\left({b}^{x}\right)=x$ to get ${\mathrm{log}}_{10}\left({10}^{2}\right)=2$. 2.Rewrite the logarithm as ${e}^{{\mathrm{log}}_{e}7}$, and then apply the inverse property ${b}^{{\mathrm{log}}_{b}x}=x$ to get ${e}^{{\mathrm{log}}_{e}7}=7$
Another property that can help us simplify logarithms is the one-to-one property. Essentially, this property states that if two logarithms that have the same base are equal to each other, then their arguments - the stuff inside - is also equal to each other.
### The One-To-One Property of Logarithms
${\mathrm{log}}_{b}M={\mathrm{log}}_{b}N\text{ if and only if}\text{ }M=N$
### Example
Solve the equation ${\mathrm{log}}_{3}\left(3x\right)={\mathrm{log}}_{3}\left(2x+5\right)$ for x.
Answer: In order for this equation to be true we must find a value for x such that $3x=2x+5$
$\begin{array}{c}3x=2x+5\hfill & \text{Set the arguments equal}\text{.}\hfill \\ x=5\hfill & \text{Subtract 2}x\text{.}\hfill \end{array}$
$\begin{array}{c}{\mathrm{log}}_{3}\left(3\cdot5\right)={\mathrm{log}}_{3}\left(2\cdot5+5\right)\\{\mathrm{log}}_{3}\left(15\right)={\mathrm{log}}_{3}\left(15\right)\end{array}$
This is a true statement, so we must have found the correct value for x.
What if we had started with the equation ${\mathrm{log}}_{3}\left(3x\right)+{\mathrm{log}}_{3}\left(2x+5\right)=2$? The one-to-one property does not help us in this instance. Before we can solve an equation like this, we need a method for combining terms on the left side of the equation. To recap - the properties of logarithms and exponentials that can help us understand, simplify and solve these types of functions more easily include:
• Zero and Identity Exponent Rule: ${\mathrm{log}}_{b}1=0$, b>0, and ${\mathrm{log}}_{b}b=1$, b>0
• Inverse Property: $\begin{array}{c}\hfill \\ {\mathrm{log}}_{b}\left({b}^{x}\right)=x\hfill \\ \text{ }{b}^{{\mathrm{log}}_{b}x}=x,x>0\hfill \end{array}$
• One-To-One Property: ${\mathrm{log}}_{b}M={\mathrm{log}}_{b}N\text{ if and only if}\text{ }M=N$
## The Product Rule for Logarithms
Recall that we use the product rule of exponents to combine the product of exponents by adding: ${x}^{a}{x}^{b}={x}^{a+b}$. We have a similar property for logarithms, called the product rule for logarithms, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below.
### The Product Rule for Logarithms
The product rule for logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.
${\mathrm{log}}_{b}\left(MN\right)={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)\text{ for }b>0$
Let $m={\mathrm{log}}_{b}M$ and $n={\mathrm{log}}_{b}N$. In exponential form, these equations are ${b}^{m}=M$ and ${b}^{n}=N$. It follows that
$\begin{array}{c}{\mathrm{log}}_{b}\left(MN\right)\hfill & ={\mathrm{log}}_{b}\left({b}^{m}{b}^{n}\right)\hfill & \text{Substitute for }M\text{ and }N.\hfill \\ \hfill & ={\mathrm{log}}_{b}\left({b}^{m+n}\right)\hfill & \text{Apply the product rule for exponents}.\hfill \\ \hfill & =m+n\hfill & \text{Apply the inverse property of logs}.\hfill \\ \hfill & ={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)\hfill & \text{Substitute for }m\text{ and }n.\hfill \end{array}$
Repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. Consider the following example:
### Example
Using the product rule for logarithms, rewrite this logarithm of a product as the sum of logarithms of its factors. [latex-display]{\mathrm{log}}_{b}\left(wxyz\right)[/latex-display]
In our next example, we will first factor the argument of a logarithm before expanding it with the product rule.
### Example
Expand ${\mathrm{log}}_{3}\left(30x\left(3x+4\right)\right)$.
We begin by factoring the argument completely, expressing 30 as a product of primes.
${\mathrm{log}}_{3}\left(30x\left(3x+4\right)\right)={\mathrm{log}}_{3}\left(2\cdot 3\cdot 5\cdot x\cdot \left(3x+4\right)\right)$
Next we write the equivalent equation by summing the logarithms of each factor.
${\mathrm{log}}_{3}\left(30x\left(3x+4\right)\right)={\mathrm{log}}_{3}\left(2\right)+{\mathrm{log}}_{3}\left(3\right)+{\mathrm{log}}_{3}\left(5\right)+{\mathrm{log}}_{3}\left(x\right)+{\mathrm{log}}_{3}\left(3x+4\right)$
### Analysis of the Solution
It is tempting to use the distributive property when you see an expression like $\left(30x\left(3x+4\right)\right)$, but in this case, it is better to leave the argument of this logarithm as a product since you can then use the product rule for logarithms to simplify the expression.
## The Quotient Rule for Logarithms
For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents to combine the quotient of exponents by subtracting: ${x}^{\frac{a}{b}}={x}^{a-b}$. The quotient rule for logarithms says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule.
### The Quotient Rule for Logarithms
The quotient rule for logarithms can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.
${\mathrm{log}}_{b}\left(\frac{M}{N}\right)={\mathrm{log}}_{b}M-{\mathrm{log}}_{b}N$
We can show ${\mathrm{log}}_{b}\left(\frac{M}{N}\right)\text{=}{\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)$
Given positive real numbers M, N, and b, where $b>0$ we will show
${\mathrm{log}}_{b}\left(\frac{M}{N}\right)\text{=}{\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)$.
Let $m={\mathrm{log}}_{b}M$ and $n={\mathrm{log}}_{b}N$. In exponential form, these equations are ${b}^{m}=M$ and ${b}^{n}=N$. It follows that
$\begin{array}{c}{\mathrm{log}}_{b}\left(\frac{M}{N}\right)\hfill & ={\mathrm{log}}_{b}\left(\frac{{b}^{m}}{{b}^{n}}\right)\hfill & \text{Substitute for }M\text{ and }N.\hfill \\ \hfill & ={\mathrm{log}}_{b}\left({b}^{m-n}\right)\hfill & \text{Apply the quotient rule for exponents}.\hfill \\ \hfill & =m-n\hfill & \text{Apply the inverse property of logs}.\hfill \\ \hfill & ={\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)\hfill & \text{Substitute for }m\text{ and }n.\hfill \end{array}$
### Example
Expand the following expression using the quotient rule for logarithms. [latex-display]\mathrm{log}\left(\frac{2{x}^{2}+6x}{3x+9}\right)[/latex-display]
Factoring and canceling we get,
$\begin{array}{c}\mathrm{log}\left(\frac{2{x}^{2}+6x}{3x+9}\right) & =\mathrm{log}\left(\frac{2x\left(x+3\right)}{3\left(x+3\right)}\right)\hfill & \text{Factor the numerator and denominator}.\hfill \\ & \text{ }=\mathrm{log}\left(\frac{2x}{3}\right)\hfill & \text{Cancel the common factors}.\hfill \end{array}$
Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.
$\begin{array}{c}\mathrm{log}\left(\frac{2x}{3}\right)=\mathrm{log}\left(2x\right)-\mathrm{log}\left(3\right)\hfill \\ \text{ }=\mathrm{log}\left(2\right)+\mathrm{log}\left(x\right)-\mathrm{log}\left(3\right)\hfill \end{array}$
In the previous example, it was helpful to first factor the numerator and denominator and divide common terms. This gave us a simpler expression to use to write an equivalent expression. It is important to remember to subtract the logarithm of the denominator from the logarithm of the numerator. Always check to see if you need to expand further with the product rule.
### Example
Expand ${\mathrm{log}}_{2}\left(\frac{15x\left(x - 1\right)}{\left(3x+4\right)\left(2-x\right)}\right)$.
First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.
${\mathrm{log}}_{2}\left(\frac{15x\left(x - 1\right)}{\left(3x+4\right)\left(2-x\right)}\right)={\mathrm{log}}_{2}\left(15x\left(x - 1\right)\right)-{\mathrm{log}}_{2}\left(\left(3x+4\right)\left(2-x\right)\right)$
Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule, noting that the prime factors of the factor 15 are 3 and 5.
$\begin{array}{c}{\mathrm{log}}_{2}\left(15x\left(x - 1\right)\right)-{\mathrm{log}}_{2}\left(\left(3x+4\right)\left(2-x\right)\right) \\\text{ }= \left[{\mathrm{log}}_{2}\left(3\right)+{\mathrm{log}}_{2}\left(5\right)+{\mathrm{log}}_{2}\left(x\right)+{\mathrm{log}}_{2}\left(x - 1\right)\right]-\left[{\mathrm{log}}_{2}\left(3x+4\right)+{\mathrm{log}}_{2}\left(2-x\right)\right]\hfill \\ \text{ }={\mathrm{log}}_{2}\left(3\right)+{\mathrm{log}}_{2}\left(5\right)+{\mathrm{log}}_{2}\left(x\right)+{\mathrm{log}}_{2}\left(x - 1\right)-{\mathrm{log}}_{2}\left(3x+4\right)-{\mathrm{log}}_{2}\left(2-x\right)\hfill \end{array}$
## Analysis of the Solution
There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for $x=-\frac{4}{3}$ and = 2. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that > 0, > 1, $x>-\frac{4}{3}$, and < 2. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.
## Using the Power Rule for Logarithms
We’ve explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as ${x}^{2}$? One method is as follows:
$\begin{array}{c}{\mathrm{log}}_{b}\left({x}^{2}\right)\hfill & ={\mathrm{log}}_{b}\left(x\cdot x\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}x+{\mathrm{log}}_{b}x\hfill \\ \hfill & =2{\mathrm{log}}_{b}x\hfill \end{array}$
Notice that we used the product rule for logarithms to simplify the example above. By doing so, we have derived the power rule for logarithms, which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that, although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example,
$\begin{array}{c}100={10}^{2}\hfill & \sqrt{3}={3}^{\frac{1}{2}}\hfill & \frac{1}{e}={e}^{-1}\hfill \end{array}$
### The Power Rule for Logarithms
The power rule for logarithms can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.
${\mathrm{log}}_{b}\left({M}^{n}\right)=n{\mathrm{log}}_{b}M$
### Example
Expand ${\mathrm{log}}_{2}{x}^{5}$.
The argument is already written as a power, so we identify the exponent, 5, and the base, x, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.
${\mathrm{log}}_{2}\left({x}^{5}\right)=5{\mathrm{log}}_{2}x$
The power rule for logarithms is possible because we can use the product rule and combine like terms. In the next example you will see that we can also rewrite an expression as a power in order to use the power rule.
### Example
Expand ${\mathrm{log}}_{3}\left(25\right)$ using the power rule for logs.
Expressing the argument as a power, we get ${\mathrm{log}}_{3}\left(25\right)={\mathrm{log}}_{3}\left({5}^{2}\right)$.
Next we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.
${\mathrm{log}}_{3}\left({5}^{2}\right)=2{\mathrm{log}}_{3}\left(5\right)$
Now let's use the power rule in reverse.
### Example
Rewrite $4\mathrm{ln}\left(x\right)$ using the power rule for logs to a single logarithm with a leading coefficient of 1.
Because the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression $4\mathrm{ln}\left(x\right)$, we identify the factor, 4, as the exponent and the argument, x, as the base, and rewrite the product as a logarithm of a power:
$4\mathrm{ln}\left(x\right)=\mathrm{ln}\left({x}^{4}\right)$.
## Expand and Condense Logarithms
Taken together, the product rule, quotient rule, and power rule are often called "laws of logs." Sometimes we apply more than one rule in order to simplify an expression. For example:
$\begin{array}{c}{\mathrm{log}}_{b}\left(\frac{6x}{y}\right)\hfill & ={\mathrm{log}}_{b}\left(6x\right)-{\mathrm{log}}_{b}y\hfill \\ \hfill & ={\mathrm{log}}_{b}6+{\mathrm{log}}_{b}x-{\mathrm{log}}_{b}y\hfill \end{array}$
We can also use the power rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the quotient rule for logarithms using the fact that a reciprocal (fraction) has a negative power:
$\begin{array}{c}{\mathrm{log}}_{b}\left(\frac{A}{C}\right)\hfill & ={\mathrm{log}}_{b}\left(A{C}^{-1}\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}\left(A\right)+{\mathrm{log}}_{b}\left({C}^{-1}\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}A+\left(-1\right){\mathrm{log}}_{b}C\hfill \\ \hfill & ={\mathrm{log}}_{b}A-{\mathrm{log}}_{b}C\hfill \end{array}$
We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product.
Remember that we can only do this with products, quotients, powers, and roots—never with addition or subtraction inside the argument of the logarithm. Consider the following example:
$\begin{array}{c}\mathrm{log}\left(10+100\right)\overset{?}{=}\end{array}\mathrm{log}\left(10\right)+\mathrm{log}\left(100\right)\\\mathrm{log}\left(110\right)\overset{?}{=}1+2\\2.04\ne3$
Be careful to only apply the product rule when a logarithm has an argument that is a product or when you have a sum of logarithms.
In our first example we will show that a logarithmic expression can be expanded by combining several of the rules of logarithms.
### Example
Rewrite $\mathrm{ln}\left(\frac{{x}^{4}y}{7}\right)$ as a sum or difference of logs.
First, because we have a quotient of two expressions, we can use the quotient rule:
$\mathrm{ln}\left(\frac{{x}^{4}y}{7}\right)=\mathrm{ln}\left({x}^{4}y\right)-\mathrm{ln}\left(7\right)$
Then seeing the product in the first term, we use the product rule:
$\mathrm{ln}\left({x}^{4}y\right)-\mathrm{ln}\left(7\right)=\mathrm{ln}\left({x}^{4}\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right)$
Finally, we use the power rule on the first term:
$\mathrm{ln}\left({x}^{4}\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right)=4\mathrm{ln}\left(x\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right)$
We can also use the rules for logarithms to simplify the logarithm of a radical expression.
### Example
Expand $\mathrm{log}\left(\sqrt{x}\right)$.
Answer: $\begin{array}{c}\mathrm{log}\left(\sqrt{x}\right)\hfill & =\mathrm{log}{x}^{\left(\frac{1}{2}\right)}\hfill \\ \hfill & =\frac{1}{2}\mathrm{log}x\hfill \end{array}$
Can we expand $\mathrm{ln}\left({x}^{2}+{y}^{2}\right)$? Use the textbox below to develop an argument one way or the other before you look at the solution.[practice-area rows="1"][/practice-area]
Answer: No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm. Rewrite the expression as an equation and express it as an exponential to give yourself some proof. [latex-display]m=\mathrm{ln}\left({x}^{2}+{y}^{2}\right)[/latex-display] If you rewrite this as an exponential you get: [latex-display]e^m={x}^{2}+{y}^{2}[/latex-display] From here, there's not much more you can do to make this expression more simple.
Let's do one more example with an expression that contains several different mathematical operations.
### Example
Expand ${\mathrm{log}}_{6}\left(\frac{64{x}^{3}\left(4x+1\right)}{\left(2x - 1\right)}\right)$.
We can expand by applying the Product and Quotient Rules.
$\begin{array}{c}{\mathrm{log}}_{6}\left(\frac{64{x}^{3}\left(4x+1\right)}{\left(2x - 1\right)}\right)\hfill & ={\mathrm{log}}_{6}64+{\mathrm{log}}_{6}{x}^{3}+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & \text{Apply the Quotient and Product Rules}.\hfill \\ \hfill & ={\mathrm{log}}_{6}{2}^{6}+{\mathrm{log}}_{6}{x}^{3}+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & {\text{Simplify by writing 64 as 2}}^{6}.\hfill \\ \hfill & =6{\mathrm{log}}_{6}2+3{\mathrm{log}}_{6}x+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & \text{Apply the Power Rule}.\hfill \end{array}$
## Condense Logarithms
We can use the rules of logarithms we just learned to condense sums, differences, and products with the same base as a single logarithm. It is important to remember that the logarithms must have the same base to be combined. We will learn later how to change the base of any logarithm before condensing.
### Example
Write ${\mathrm{log}}_{3}\left(5\right)+{\mathrm{log}}_{3}\left(8\right)-{\mathrm{log}}_{3}\left(2\right)$ as a single logarithm.
Using the product and quotient rules
${\mathrm{log}}_{3}\left(5\right)+{\mathrm{log}}_{3}\left(8\right)={\mathrm{log}}_{3}\left(5\cdot 8\right)={\mathrm{log}}_{3}\left(40\right)$
This reduces our original expression to
${\mathrm{log}}_{3}\left(40\right)-{\mathrm{log}}_{3}\left(2\right)$
Then, using the quotient rule
${\mathrm{log}}_{3}\left(40\right)-{\mathrm{log}}_{3}\left(2\right)={\mathrm{log}}_{3}\left(\frac{40}{2}\right)={\mathrm{log}}_{3}\left(20\right)$
In our next example, we show how to simplify a more complex logarithm by condensing it.
### Example
Condense ${\mathrm{log}}_{2}\left({x}^{2}\right)+\frac{1}{2}{\mathrm{log}}_{2}\left(x - 1\right)-3{\mathrm{log}}_{2}\left({\left(x+3\right)}^{2}\right)$.
We apply the power rule first:
${\mathrm{log}}_{2}\left({x}^{2}\right)+\frac{1}{2}{\mathrm{log}}_{2}\left(x - 1\right)-3{\mathrm{log}}_{2}\left({\left(x+3\right)}^{2}\right)={\mathrm{log}}_{2}\left({x}^{2}\right)+{\mathrm{log}}_{2}\left(\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)$
Next we apply the product rule to the sum:
${\mathrm{log}}_{2}\left({x}^{2}\right)+{\mathrm{log}}_{2}\left(\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)={\mathrm{log}}_{2}\left({x}^{2}\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)$
Finally, we apply the quotient rule to the difference:
${\mathrm{log}}_{2}\left({x}^{2}\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)={\mathrm{log}}_{2}\frac{{x}^{2}\sqrt{x - 1}}{{\left(x+3\right)}^{6}}$
## Change of Base
Most calculators can evaluate only common and natural logs. In order to evaluate logarithms with a base other than 10 or $e$, we use the change-of-base formula to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs.
To derive the change-of-base formula, we use the one-to-one property and power rule for logarithms.
Given any positive real numbers M, b, and n, where $n\ne 1$ and $b\ne 1$, we show
${\mathrm{log}}_{b}M\text{=}\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}$
Let $y={\mathrm{log}}_{b}M$. By taking the log base $n$ of both sides of the equation, we arrive at an exponential form, namely ${b}^{y}=M$. It follows that
$\begin{array}{c}{\mathrm{log}}_{n}\left({b}^{y}\right)\hfill & ={\mathrm{log}}_{n}M\hfill & \text{Apply the one-to-one property}.\hfill \\ y{\mathrm{log}}_{n}b\hfill & ={\mathrm{log}}_{n}M \hfill & \text{Apply the power rule for logarithms}.\hfill \\ y\hfill & =\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}\hfill & \text{Isolate }y.\hfill \\ {\mathrm{log}}_{b}M\hfill & =\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}\hfill & \text{Substitute for }y.\hfill \end{array}$
For example, to evaluate ${\mathrm{log}}_{5}36$ using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.
$\begin{array}{c}{\mathrm{log}}_{5}36\hfill & =\frac{\mathrm{log}\left(36\right)}{\mathrm{log}\left(5\right)}\hfill & \text{Apply the change of base formula using base 10}\text{.}\hfill \\ \hfill & \approx 2.2266\text{ }\hfill & \text{Use a calculator to evaluate to 4 decimal places}\text{.}\hfill \end{array}$
Let's practice changing the base of a logarithmic expression from 5 to base e.
### Example
Change ${\mathrm{log}}_{5}3$ to a quotient of natural logarithms.
Because we will be expressing ${\mathrm{log}}_{5}3$ as a quotient of natural logarithms, the new base, = e.
We rewrite the log as a quotient using the change-of-base formula. The numerator of the quotient will be the natural log with argument 3. The denominator of the quotient will be the natural log with argument 5.
$\begin{array}{c}{\mathrm{log}}_{b}M\hfill & =\frac{\mathrm{ln}M}{\mathrm{ln}b}\hfill \\ {\mathrm{log}}_{5}3\hfill & =\frac{\mathrm{ln}3}{\mathrm{ln}5}\hfill \end{array}$
We can generalize the change of base formula in the following way:
### The Change-of-Base Formula
The change-of-base formula can be used to evaluate a logarithm with any base.
For any positive real numbers M, b, and n, where $n\ne 1$ and $b\ne 1$,
${\mathrm{log}}_{b}M\text{=}\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}$.
It follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs.
${\mathrm{log}}_{b}M=\frac{\mathrm{ln}M}{\mathrm{ln}b}$
and
${\mathrm{log}}_{b}M=\frac{\mathrm{log}M}{\mathrm{log}b}$
As we stated earlier, the main reason for changing the base of a logarithm is to be able to evaluate it with a calculator. In the following example we will use the change of base formula on a logarithmic expression, then evaluate the result with a calculator.
### Example
Evaluate ${\mathrm{log}}_{2}\left(10\right)$ using the change-of-base formula with a calculator.
According to the change-of-base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm, which is the log base e.
$\begin{array}{c}{\mathrm{log}}_{2}10=\frac{\mathrm{ln}10}{\mathrm{ln}2}\hfill & \text{Apply the change of base formula using base }e.\hfill \\ \approx 3.3219\hfill & \text{Use a calculator to evaluate to 4 decimal places}.\hfill \end{array}$
Can we change common logarithms to natural logarithms? Write your ideas in the textbox below before looking at the solution. [practice-area rows="1"][/practice-area]
Answer: Yes. Remember that $\mathrm{log}9$ means ${\text{log}}_{\text{10}}\text{9}$. So, $\mathrm{log}9=\frac{\mathrm{ln}9}{\mathrm{ln}10}$.
## Summary
Logarithms have properties that can help us simplify and solve expressions and equations that contain logarithms. Exponentials and logarithms are inverses of each other, therefore we can define the product rule for logarithms. We can use this as follows to simplify or solve expressions with logarithms. Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms.
1. Factor the argument completely, expressing each whole number factor as a product of primes.
2. Write the equivalent expression by summing the logarithms of each factor.
You can use the quotient rule of logarithms to write an equivalent difference of logarithms in the following way:
1. Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.
2. Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.
3. Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.
To use the power rule of logarithms to write an equivalent product of a factor and a logarithm consider the following:
1. Express the argument as a power, if needed.
2. Write the equivalent expression by multiplying the exponent times the logarithm of the base.
Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm.
1. Apply the power property first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power.
2. Next apply the product property. Rewrite sums of logarithms as the logarithm of a product.
3. Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient.
For practical purposes found in many different sciences or finance applications, you may want to evaluate a logarithm with a calculator. The change of base formula will allow you to change the base of any logarithm to either 10 or so you can evaluate it with a calculator. Here we have summarized the steps for using the change of base formula.
Given a logarithm with the form ${\mathrm{log}}_{b}M$
1. Determine the new base n, remembering that the common log, $\mathrm{log}\left(x\right)$, has base 10, and the natural log, $\mathrm{ln}\left(x\right)$, has base e.
2. Rewrite the log as a quotient using the change-of-base formula
• The numerator of the quotient will be a logarithm with base n and argument M.
• The denominator of the quotient will be a logarithm with base n and argument b.
## Why learn about exponential and logarithmic equations?
Joan enjoyed talking to her grandfather so much that the next Sunday, she made sure she would be there when he came to dinner. This time, though, she prepared a question for him. She had been learning about logarithmic and exponential functions in her math class and wanted to see if she could stump her grandfather with a math question.
Kinemetrics Seismograph formerly used by the United States Department of the Interior.
One of the questions in Joan's homework on exponential and logarithmic functions had been about how to calculate the Richter scale measure of the magnitude of an earthquake. The following formula was given in her book:
$R=\mathrm{log}\left(\frac{A}{A_{0}}\right)$
The question provided her with the quantity $\frac{A}{A_{0}}$ and asked her to solve for R. ${A_{0}}$ is a baseline measure of ground movement as detected by a seismometer, seen in the image above. When there is an earthquake, wave amplitudes recorded by seismometers are expressed relative to this baseline. For example, Joan's book asked the following question:
An earthquake is measured with a wave amplitude 392 times as great as $A_{0}$. What is the magnitude of this earthquake using the Richter scale, to the nearest tenth?
Joan hoped to give her grandfather the Richter scale magnitude for the Alaska quake, 8.5, and see if he could find how much greater the wave amplitude of that quake was than the baseline, ${A_{0}}$.
In this module you will learn how to solve problems such as the one Joan is planning to try to stump her grandfather with. We will come back to Joan and her grandfather at the end of this module to see if she was able to ask him a question he didn't know how to answer.
In this module, you will learn about the properties of exponential and logarithmic functions in the same way that you learned about the properties of exponents. You will use the properties of logarithms and exponentials to solve equations that involve them.
### The learning outcomes for this module include:
• Define and use the properties of logarithms to expand, condense, and change the base of a logarithmic expression
• Use the properties of logarithms and exponentials to solve equations |
# Quarter Wit, Quarter Wisdom: Using the Deviation Method for Weighted Averages
We have discussed how to use the deviation method to find the arithmetic mean of numbers. It is very useful in cases where the numbers are huge, as it considerably brings down the calculation time.
The same method can be applied to weighted averages, as well. Let’s look at an example very similar to the one we examined when we were working on deviations in the case of arithmetic means:
What is the average of 452, 452, 453, 460, 467, 480, 499, 499, 504?
What would you say the average is here? Perhaps, around 470?
Shortfall:
We have two 452s – 452 is 18 less than 470.
453 is 17 less than 470.
460 is 10 less than 470.
467 is 3 less than 470.
Overall, the numbers less than 470 are (2*18) + 17 + 10 + 3 = 66 less than 470.
Excess:
480 is 10 more than 470.
We have two 499s – 499 is 29 more than 470.
504 is 34 more than 470.
Overall, the numbers more than 470 are 10 + (2*29) + 34 = 102 more than 470.
The shortfall is not balanced by the excess; there is an excess of 102-66 = 36.
So what is the average? If we assume that the average of these 9 numbers is 470, there will be an excess of 36. We need to distribute this excess evenly among all of the numbers, and hence, the average will increase by 36/9 = 4.
Therefore, the required mean is 470 + 4 = 474. (If we had assumed the mean to be 474, the shortfall would have balanced the excess.)
This method is used in exactly the same way when we have a simple average as when we have a weighted average. The reason we are reviewing it is that it can be very handy in weighted average questions involving more than two quantities.
We often deal with questions on weighted averages involving two quantities using the scale method. Let’s see how to use the deviation method for more than 2 quantities on an official GMAT question:
Three grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x+y+z gallons of a 1.5 percent grade, what is x in terms of y and z?
(A) y + 3z
(B) (y +z) / 4
(C) 2y + 3z
(D) 3y + z
(E) 3y + 4.5z
Grade 1 milk contains 1% fat. Grade 2 milk contains 2% fat. Grade 3 milk contains 3% fat. The mixture of all three contains 1.5% fat. So, grade 1 milk provides the shortfall and grades 2 and 3 milk provide the excess.
Shortfall = x*(1.5 – 1)
Excess = y*(2 – 1.5) + z*(3 – 1.5)
Since 1.5 is the actual average, the shortfall = the excess.
x*(1.5 – 1) = y*(2 – 1.5) + z*(3 – 1.5)
x/2 = y/2 + 3z/2
x = y + 3z
And there you have it – the answer is A.
We easily used deviations here to arrive at the relation. It’s good to have this method – useful for both simple averages and weighted averages – in your GMAT toolkit. |
## Section12.3Solving Quadratic Equations Using the Zero-Product Property
When the product of two numbers is zero, at least one of the numbers in the product must be zero. This is a property unique to zero; e.g., when two numbers multiply to, say, ten, not only is it not the case that one of the two numbers need be ten, the only restriction on the two numbers at all is that neither be zero.
When solving an equation of form:
\begin{equation*} ab=0\text{,} \end{equation*}
the next step in the process is to state:
\begin{equation*} a=0\text{ or }b=0\text{.} \end{equation*}
We then solve the two new equations separately and state our solutions, or solution set. For example:
\begin{gather*} (4x+7)(2x-9)=0 \end{gather*}
\begin{align*} 4x+7\amp=0\amp\amp\text{or}\amp 2x-9\amp=0\\ 4x+7\subtractright{7}\amp=0\subtractright{7}\amp\amp\text{or}\amp 2x-9\addright{9}\amp=0\addright{9}\\ 4x\amp=-7\amp\amp\text{or}\amp 2x\amp=9\\ \divideunder{4x}{4}\amp=\divideunder{-7}{4}\amp\amp\text{or}\amp \divideunder{2x}{2}\amp=\divideunder{9}{2}\\ x\amp=-\frac{7}{4}\amp\amp\text{or}\amp x\amp=\frac{9}{2} \end{align*}
The solutions are $-\frac{7}{4}$ and $\frac{9}{2}\text{.}$ The solution set is $\{-\frac{7}{4}, \frac{9}{2}\}\text{.}$
We frequently have to perform some preliminary steps before invoking the zero-product property. Specifically:
1. Completely expand both sides of the equation.
2. Add and/or subtract to/from both sides of the equation so that one side of the equation is zero. The next step will be easier if make sure that the second degree term ($ax^2$) has a positive coefficient.
3. Factor the non-zero side of the equation.
4. We are now set to invoke the zero-product property.
Several examples follow.
###### Example12.3.1.
Use the zero-product property to solve $4x^2=4x+15\text{.}$
Solution
\begin{align*} 4x^2\amp=4x+15\\ 4x^2\subtractright{4x}\subtractright{15}\amp=4x+15\subtractright{4x}\subtractright{15}\\ 4x^2-4x-15\amp=0\\ 4x^2-10x+6x-15\amp=0\\ 2x(2x-5)+3(2x-5)\amp=0\\ (2x-5)(2x+3)\amp=0 \end{align*}
\begin{align*} 2x-5\amp=0\amp\amp\text{or}\amp 2x+3\amp=0\\ 2x-5\addright{5}\amp=0\addright{5}\amp\amp\text{or}\amp 2x+3\subtractright{3}\amp=0\subtractright{3}\\ 2x\amp=5\amp\amp\text{or}\amp 2x\amp=-3\\ \divideunder{2x}{2}\amp=\divideunder{5}{2}\amp\amp\text{or}\amp \divideunder{2x}{2}\amp=\divideunder{-3}{2}\\ x\amp=\frac{5}{2}\amp\amp\text{or}\amp x\amp=-\frac{3}{2} \end{align*}
The solutions are $\frac{5}{2}$ and $-\frac{3}{2}\text{.}$ The solution set is $\left\{\frac{5}{2}, -\frac{3}{2}\right\}\text{.}$
###### Example12.3.2.
Use the zero-product property to solve $(x+6)(x-2)=-16\text{.}$
Solution
\begin{align*} (x+6)(x-2)\amp=-16\\ x^2+4x-12\amp=-16\\ x^2+4x+4\amp=0\\ (x+2)(x+2)\amp=0\\ x+2=0\\ x+2\subtractright{2}=0\subtractright{2}\\ x=-2 \end{align*}
The solution is $-2\text{.}$ The solution set is $\{-2\}\text{.}$
###### Example12.3.3.
Use the zero-product property to solve $(x+4)(4x-5)=(7x+10)(3x-2)\text{.}$
Solution
\begin{align*} x\amp=0\amp\amp\text{or}\amp 17x+5\amp=0\\ x\amp=0\amp\amp\text{or}\amp 17x+5\subtractright{5}\amp=0\subtractright{5}\\ x\amp=0\amp\amp\text{or}\amp 17x\amp=-5\\ x\amp=0\amp\amp\text{or}\amp \divideunder{17x}{17}\amp=\divideunder{-5}{17}\\ x\amp=0\amp\amp\text{or}\amp x\amp=-\frac{5}{17} \end{align*}
The solutions are $0$ and $-\frac{5}{17}\text{.}$ The solution set is $\left\{0, -\frac{5}{17}\right\}\text{.}$
###### Example12.3.4.
Use the zero-product property to solve $2-x^2=(2-x)^2\text{.}$
Solution
The solution is $1\text{.}$ The solution set is $\{1\}\text{.}$
### ExercisesExercises
Use the zero-product property to solve each quadratic equation. State the solutions to each equation as well as the solution set to each equation.
###### 1.
$(3x+8)(5x-7)=0$
Solution
We begin by setting each of the factors from the left side of the equation equal to zero.
\begin{align*} (3x+8)(5x-7)\amp=0 \end{align*}
\begin{align*} 3x+8\amp=0 \amp\amp\text{or}\amp 5x-7\amp= 0\\ 3x+8\subtractright{8}\amp=0\subtractright{8} \amp\amp\text{or}\amp 5x-7\addright{7}\amp= 0\addright{7}\\ 3x\amp=-8 \amp\amp\text{or}\amp 5x\amp=7\\ \divideunder{3x}{3}\amp=\divideunder{-8}{3} \amp\amp\text{or}\amp \divideunder{5x}{5}\amp=\divideunder{7}{5}\\ x\amp=-\frac{8}{3} \amp\amp\text{or}\amp x\amp=\frac{7}{5} \end{align*}
The solutions are $-\frac{8}{3}$ and $\frac{7}{5}\text{.}$
The solution set is $\left\{-\frac{8}{3}, \frac{7}{5}\right\}\text{.}$
###### 2.
$x(x-6)=0$
Solution
We begin by setting each of the factors from the left side of the equation equal to zero.
\begin{align*} x(x-6)\amp=0 \end{align*}
The solutions are $0$ and $6\text{.}$
The solution set is $\{0, 6\}\text{.}$
###### 3.
$x(x-3)=10$
Solution
We begin by making the right side of the equation zero and then factoring the left side of the equation.
\begin{align*} x(x-3)\amp=10\\ x(x-3)\subtractright{10}\amp=10\subtractright{10}\\ x^2-3x-10\amp=0\\ (x-5)(x+2)\amp=0 \end{align*}
The solutions are $5$ and $-2\text{.}$
The solution set is $\{5, -2\}\text{.}$
###### 4.
$x^2+12x=-36$
Solution
We begin by making the right side of the equation zero and then factoring the left side of the equation.
The solution is $-6\text{.}$
The solution set is $\{-6\}\text{.}$
###### 5.
$4t(8t+9)=5$
Solution
We begin by expanding the left side of the equation, making the right side of the equation zero, and then factoring the left side of the equation.
\begin{align*} 4t(8t+9)\amp=5\\ 32t^2+36t\amp=5\\ 32t^2+36t\subtractright{5}\amp=5\subtractright{5}\\ 32t^2+36t-5\amp=0\\ 32t^2+40t-4t-5\amp=0\\ 8t(4t+5)-1 \cdot (4t+5)\amp=0\\ (4t+5)(8t-1)\amp=0 \end{align*}
\begin{align*} 4t+5\amp=0 \amp\amp\text{or}\amp 8t-1\amp=0\\ 4t+5\subtractright{5}\amp=0\subtractright{5} \amp\amp\text{or}\amp 8t-1\addright{1}\amp=0\addright{1}\\ 4t\amp=-5 \amp\amp\text{or}\amp 8t\amp=1\\ \divideunder{4t}{4}\amp=-\divideunder{5}{4} \amp\amp\text{or}\amp \divideunder{8t}{8}\amp=\divideunder{1}{8}\\ t\amp=-\frac{5}{4} \amp\amp\text{ or }\amp t\amp=\frac{1}{8} \end{align*}
The solutions are $-\frac{5}{4}$ and $\frac{1}{8}\text{.}$
The solution set is $\left\{-\frac{5}{4}, \frac{1}{8}\right\}\text{.}$
###### 6.
$2y^2=y$
Solution
We begin by making the right side of the equation zero and then factoring the left side of the equation.
\begin{align*} 2y^2\amp=y\\ 2y^2\subtractright{y}\amp=y\subtractright{y}\\ 2y^2-y\amp=0\\ y(2y-1)\amp=0 \end{align*}
\begin{align*} y\amp=0 \amp\amp\text{or}\amp 2y-1\amp=0\\ y\amp=0 \amp\amp\text{or}\amp 2y-1\addright{1}\amp=0\addright{1}\\ y\amp=0 \amp\amp\text{or}\amp 2y\amp=1\\ y\amp=0 \amp\amp\text{or}\amp \divideunder{2y}{2}\amp=\divideunder{1}{2}\\ y\amp=0 \amp\amp\text{or}\amp y\amp=\frac{1}{2} \end{align*}
The solutions are $0$ and $\frac{1}{2}\text{.}$
The solution set is $\left\{0, \frac{1}{2}\right\}\text{.}$
###### 7.
$(w-10)(w+1)=-10$
Solution
We begin by expanding the left side of the equation, making the right side of the equation zero, and then factoring the left side of the equation.
The solutions are $0$ and $9\text{.}$
The solution set is $\{0, 9\}\text{.}$ |
##### Basic Math & Pre-Algebra Workbook For Dummies with Online Practice
A quick method for solving algebra problems is to re-arrange the equation by placing all x terms on one side of the equal sign and all constants (non-x terms) on the other side. Essentially, you're doing the addition and subtraction without showing it. You can then isolate x.
## Practice questions
1. Rearrange the equation 10x + 5 = 3x + 19 to solve for x.
2. Solve –[2(x + 7) + 1] = x – 12 for x.
1. x = 2First, rearrange the terms of the equation so that the x terms are on one side and the constants are on the other. In this case, you can do this in two steps:
Second, combine like terms on both sides:
7x = 14
The third and final step is to divide (in this case, by 7) to isolate x:
2. x = –1Before you can begin rearranging terms, remove the parentheses on the left side of the equation. Start with the inner parentheses, multiplying 2 by every term inside that set:
–[2(x + 7) + 1] = x – 12 –[2x + 14 + 1] = x – 12
Next, remove the remaining parentheses, switching the sign of every term within that set:
–2x – 14 – 1 = x – 12
Now you can solve for x by, first, rearranging the terms of the equation:
–2x – 14 – 1 + 12 = x –14 – 1 + 12 = x + 2x
Then you combine like terms on both sides:
–3 = 3x
Finally, you divide (in this example, by 3) to isolate x:
Mark Zegarelli is a math and test prep teacher who has written a wide variety of basic math and pre-algebra books in the For Dummies series. |
# 6.6 Moments and centers of mass (Page 4/14)
Page 4 / 14
Next, we need to find the total mass of the rectangle. Let $\rho$ represent the density of the lamina (note that $\rho$ is a constant). In this case, $\rho$ is expressed in terms of mass per unit area. Thus, to find the total mass of the rectangle, we multiply the area of the rectangle by $\rho .$ Then, the mass of the rectangle is given by $\rho f\left({x}_{i}^{*}\right)\text{Δ}x.$
To get the approximate mass of the lamina, we add the masses of all the rectangles to get
$m\approx \sum _{i=1}^{n}\rho f\left({x}_{i}^{*}\right)\text{Δ}x.$
This is a Riemann sum. Taking the limit as $n\to \infty$ gives the exact mass of the lamina:
$m=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}\rho f\left({x}_{i}^{*}\right)\text{Δ}x=\rho {\int }_{a}^{b}f\left(x\right)dx.$
Next, we calculate the moment of the lamina with respect to the x -axis. Returning to the representative rectangle, recall its center of mass is $\left({x}_{i}^{*},\left(f\left({x}_{i}^{*}\right)\right)\text{/}2\right).$ Recall also that treating the rectangle as if it is a point mass located at the center of mass does not change the moment. Thus, the moment of the rectangle with respect to the x -axis is given by the mass of the rectangle, $\rho f\left({x}_{i}^{*}\right)\text{Δ}x,$ multiplied by the distance from the center of mass to the x -axis: $\left(f\left({x}_{i}^{*}\right)\right)\text{/}2.$ Therefore, the moment with respect to the x -axis of the rectangle is $\rho \left({\left[f\left({x}_{i}^{*}\right)\right]}^{2}\text{/}2\right)\text{Δ}x.$ Adding the moments of the rectangles and taking the limit of the resulting Riemann sum, we see that the moment of the lamina with respect to the x -axis is
${M}_{x}=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}\rho \frac{{\left[f\left({x}_{i}^{*}\right)\right]}^{2}}{2}\text{Δ}x=\rho {\int }_{a}^{b}\frac{{\left[f\left(x\right)\right]}^{2}}{2}dx.$
We derive the moment with respect to the y -axis similarly, noting that the distance from the center of mass of the rectangle to the y -axis is ${x}_{i}^{*}.$ Then the moment of the lamina with respect to the y -axis is given by
${M}_{y}=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}\rho {x}_{i}^{*}f\left({x}_{i}^{*}\right)\text{Δ}x=\rho {\int }_{a}^{b}xf\left(x\right)dx.$
We find the coordinates of the center of mass by dividing the moments by the total mass to give $\stackrel{–}{x}={M}_{y}\text{/}m\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\stackrel{–}{y}={M}_{x}\text{/}m.$ If we look closely at the expressions for ${M}_{x},{M}_{y},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}m,$ we notice that the constant $\rho$ cancels out when $\stackrel{–}{x}$ and $\stackrel{–}{y}$ are calculated.
We summarize these findings in the following theorem.
## Center of mass of a thin plate in the xy -plane
Let R denote a region bounded above by the graph of a continuous function $f\left(x\right),$ below by the x -axis, and on the left and right by the lines $x=a$ and $x=b,$ respectively. Let $\rho$ denote the density of the associated lamina. Then we can make the following statements:
1. The mass of the lamina is
$m=\rho {\int }_{a}^{b}f\left(x\right)dx.$
2. The moments ${M}_{x}$ and ${M}_{y}$ of the lamina with respect to the x - and y -axes, respectively, are
${M}_{x}=\rho {\int }_{a}^{b}\frac{{\left[f\left(x\right)\right]}^{2}}{2}dx\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{M}_{y}=\rho {\int }_{a}^{b}xf\left(x\right)dx.$
3. The coordinates of the center of mass $\left(\stackrel{–}{x},\stackrel{–}{y}\right)$ are
$\stackrel{–}{x}=\frac{{M}_{y}}{m}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\stackrel{–}{y}=\frac{{M}_{x}}{m}.$
In the next example, we use this theorem to find the center of mass of a lamina.
## Finding the center of mass of a lamina
Let R be the region bounded above by the graph of the function $f\left(x\right)=\sqrt{x}$ and below by the x -axis over the interval $\left[0,4\right].$ Find the centroid of the region.
The region is depicted in the following figure.
Since we are only asked for the centroid of the region, rather than the mass or moments of the associated lamina, we know the density constant $\rho$ cancels out of the calculations eventually. Therefore, for the sake of convenience, let’s assume $\rho =1.$
First, we need to calculate the total mass:
$\begin{array}{cc}\hfill m& =\rho {\int }_{a}^{b}f\left(x\right)dx={\int }_{0}^{4}\sqrt{x}\phantom{\rule{0.2em}{0ex}}dx\hfill \\ & ={\frac{2}{3}{x}^{3\text{/}2}|}_{0}^{4}=\frac{2}{3}\left[8-0\right]=\frac{16}{3}.\hfill \end{array}$
Next, we compute the moments:
$\begin{array}{cc}\hfill {M}_{x}& =\rho {\int }_{a}^{b}\frac{{\left[f\left(x\right)\right]}^{2}}{2}dx\hfill \\ & ={\int }_{0}^{4}\frac{x}{2}dx={\frac{1}{4}{x}^{2}|}_{0}^{4}=4\hfill \end{array}$
and
$\begin{array}{cc}\hfill {M}_{y}& =\rho {\int }_{a}^{b}xf\left(x\right)dx\hfill \\ & ={\int }_{0}^{4}x\sqrt{x}dx={\int }_{0}^{4}{x}^{3\text{/}2}dx\hfill \\ & ={\frac{2}{5}{x}^{5\text{/}2}|}_{0}^{4}=\frac{2}{5}\left[32-0\right]=\frac{64}{5}.\hfill \end{array}$
Thus, we have
$\stackrel{–}{x}=\frac{{M}_{y}}{m}=\frac{64\text{/}5}{16\text{/}3}=\frac{64}{5}·\frac{3}{16}=\frac{12}{5}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\stackrel{–}{y}=\frac{{M}_{x}}{y}=\frac{4}{16\text{/}3}=4·\frac{3}{16}=\frac{3}{4}.$
The centroid of the region is $\left(12\text{/}5,3\text{/}4\right).$
why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
proof the formula integration of udv=uv-integration of vdu.?
Find derivative (2x^3+6xy-4y^2)^2
no x=2 is not a function, as there is nothing that's changing.
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
y=800
800
Bg
how do u factor the numerator?
Nonsense, you factor numbers
Antonio
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
volume between cone z=√(x^2+y^2) and plane z=2
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio
How do we find the horizontal asymptote of a function using limits?
Easy lim f(x) x-->~ =c
Antonio
solutions for combining functions
what is a function? f(x)
one that is one to one, one that passes the vertical line test
Andrew
It's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y
Antonio
is x=2 a function?
The
restate the problem. and I will look. ty
is x=2 a function?
The
What is limit
it's the value a function will take while approaching a particular value
Dan
don ger it
Jeremy
what is a limit?
Dlamini
it is the value the function approaches as the input approaches that value.
Andrew
Thanx
Dlamini
Its' complex a limit It's a metrical and topological natural question... approaching means nothing in math
Antonio
is x=2 a function?
The
3y^2*y' + 2xy^3 + 3y^2y'x^2 = 0 sub in x = 2, and y = 1, isolate y'
what is implicit of y³+x²y³=5 at (2,1)
tel mi about a function. what is it?
Jeremy
A function it's a law, that for each value in the domaon associate a single one in the codomain
Antonio
function is a something which another thing depends upon to take place. Example A son depends on his father. meaning here is the father is function of the son. let the father be y and the son be x. the we say F(X)=Y.
Bg
yes the son on his father
pascal
a function is equivalent to a machine. this machine makes x to create y. thus, y is dependent upon x to be produced. note x is an independent variable
moe
x or y those not matter is just to represent.
Bg |
# Ancient Computing Utilized the Platonic Solids
Ancient Computing Utilized the Platonic Solids. First: What are the Platonic solids? In three-dimensional space, a Platonic solid is a regularconvex polyhedron. It is constructed by congruent (identical in shape and size) regular (all angles equal and all sides equal) polygonal faces. They have the same number of faces meeting at each vertex. Five solids meet those criteria. That’s it. A 6th cannot be constructed.
Tetrahedron Cube Octahedron Dodecahedron Icosahedron Four faces Six faces Eight faces Twelve faces Twenty faces (Animation) (3D model) (Animation) (3D model) (Animation) (3D model) (Animation) (3D model) (Animation) (3D model)
The tetrahedron is the basis of the other four solids. Look at the 1st five numbers. With that knowledge, here is some mathematical fun. 1² x 2² x 3² x 4² x 5² = 14,400. Where is 14,400 significant as an ancient number?
The tetrahedron is the basic unit of the five.
By vertices, its 4 triangles total 720º (4 x 180° per triangle). Then 7 + 2 + 0 = 9.
The solid whose faces contain the most degrees of the 5 is the dodecahedron
With 9 x 720° = 6,480°. (This is exactly 9 x the tetrahedron’with 720° degrees). (9 x 720 = 6480). Its second way is, 6 + 4 + 8 + 0 = 18. Reducing the number by horizontal adding of the smaller numbers, 1 + 8 = 9.
With the remaining three solids: Cube has 2,160°. Each of its 6 squares has 360°. Thus, 6 x 360 = 2160. That is 3 times the 720° of the tetrahedron. Reduce 2160 and we have: 2 + 1 + 6 + 0 = 9.
Icosahedron has 20 triangles. Every triangle has 180°. Thus, 20 x 180 = 360o. That is five times the 720 ° of the basic tetrahedron. Reduce 3600 and we have: 3 + 6 + 0 + 0 = 9
Finally, there’s the octahedron with 8 triangles. 8 x 180° = 1440. That is 2 x the 720° of the tetrahedron. Then 1 + 4 + 4 + 0 = 9.
In content and quality number 9 imbues the regular solids.
### Ancient Computing had a Source
• Why was number 9 the limiting number of the Platonic solids in the manner described above? I like to call the nine-boxed digits in the number square below the “stamping mill of the Universe.” Nine is the highest number of the traditional arrangement of the pictured numbers.
• A second mystery: Why are there only 5 possible regular polyhedrons? Because 5 is the core number of the 3 x 3 number square. The infinitely complex Universe is stamped out from this simplest of number squares.
• A third mystery to solve:: Why was a megalithic mile 14,400 feet? How did it relate to King Arthur’s Palace? The answer is simple. Total the number of degrees in the 5 solids. They are listed above. We have, 720 + 1440 + 2160 + 3600 + 6480 = 14,400. Now read the internal link below.
Mesolithic cultures worked with the 3 x 3 number square for ancient computing. Roses hold a primary ancient secret. Again, read this internal link. The high priests of various cultures and countries knew its secrets. Another known entity was how this tiny number square can become infinite. Link is also below. I’m attempting to restore this lost knowledge. My insights all took place on Oquaga Lake. I was the house piano player at Scott’s Oquaga Lake House for years. I communed with a female Indian spirit from the Lennie Lenape. This numerical square could offer a common vision based on balance. That, in turn, could initiate another Golden Age. |
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Factors
# Factors of 137 | Prime Factorization of 137 | Factor Tree of 137
Written by Prerit Jain
Updated on: 15 Feb 2023
## Factors of 137
Calculate Factors of
The Factors are
https://wiingy.com/learn/math/factors-of-137/
## What are the factors of 137
137 is a prime number with 1 and itself as its factors. To think of this another way, we could say that all possible factors for 137 equal just two – one being its own value (1 x 137 =137) and the other factor simply being ‘one’ (1×1=137). A few quick calculations will tell us these two simple answers to our question: what are the factors of 137? The answer is… 1 & 137!
## How to Find Factors of 137
The most popular methods to find the factors of a number are given below and the same can be used to find the factors of 137.
• Factor of 137 using Multiplication Method
• Factors of 137 using Division Method
• Prime Factorization of 137
• Factor tree of 137
## Factors of 137 using Multiplication Method
Factors are numbers that multiply together to form a certain number. To find out which factors belong to 137, we can use the multiplication method. All you have to do is take 1 and it’s a friend (137) and join them up in pairs; then just multiply both of them together until each pair makes 137. That way, you’ll discover that those same two friends —1 & 137—are actually its only set of factors!
## Factors of 137 Using Division Method
Prime numbers are special types of integers that can only be divided by themselves and one. To figure out a prime number’s factors, the easiest way is to use multiplication:
• Take each value from 1 up until the given integer (in this case 137) and multiply it with ‘137’.
• If your answer is equal to ‘137’, you have found two of its factors! In our example here we multiplied 1×137 = 137 so this tells us both these values are in fact factors for 137 – an interesting yet very straightforward perspective on Prime Numbers.
## Prime Factorization of 137
Calculate Prime Factors of
The Prime Factors of 137 =
137
https://wiingy.com/learn/math/factors-of-137/
Understanding prime factorization is an important skill for students of Mathematics to master.
• To put it simply, the expression of a number as its product of prime factors – also known as “prime factorization” – allows us to break down any given number into smaller parts made up only of their most basic elements.
• For example, 137 can be broken down into 1 and 137 since it has no other positive integer divisors except itself!
## Factor tree of 137
https://wiingy.com/learn/math/factors-of-137/
The factor tree is an ingenious way to learn about prime factors. In this case, 137 can only be broken down into one prime factor – itself. Through exploring the relationships between numbers and their composite parts with a tool like a factor tree, students gain insight into how mathematics works on many different levels.
## Factor Pairs of 137
Calculate Pair Factors of
1 x 137=137
So Pair Factors of 137 are
(1,137)
https://wiingy.com/learn/math/factors-of-137/
To understand the factor pairs of 137, it’s important to know that as 137 is a prime number, its divisors are just 1 and itself. Therefore, when looking for its factor pairs you can start by dividing it with each number up until the square root of 137. If there’s an even division result then this means that both numbers in combination would multiply together to equal 137 – thus creating two distinct factoring pairings: (1,137) & (137;1). As long as students remember these key points they will have no problem understanding how to calculate any given figure’s individual factoring pair!
## Factors of 137 – Quick Recap
Factors of 137: 1, 137.
Negative Factors of 137: -1, -137.
Prime Factors of 137: 137
Prime Factorization of 137: 137
## Fun Facts of Factors of 137
137 is an odd and special number!
• It’s the 33rd prime, meaning it can only be divided by 1 and itself with no remainder.
• 137 also belongs to the Mersenne Prime family – a rare kind of prime formed when you take 2 raised to any other expressible prime power (in this case p) minus one.
## Examples of Factor of 137
1. If Jorge has 137 apples, how many baskets of 9 can he fill?
Answer: Jorge can fill 15 baskets with 9 apples each (137 ÷ 9 = 15).
2. What is the greatest common factor of 27 and 137?
Answer: The greatest common factor (GCF) of 27 and 137 is 3 (which is 3×3).
3. William has 34 red marbles and 68 blue marbles. What is the least common multiple of these two numbers?
Answer: The least common multiple (LCM) of 34 and 68 is 136 (34 x 68 = 2312, which can be divided by 2 to get 1160, then divided by 4 to get 290, then divided by 19 to get 36, and finally by 4 again to get 9).
4. How many combinations of 11 and 13 will equal 137?
Answer: There are only 2 combinations of 11 and 13 that are equal to 137 (11 x 13 = 143; 113 x 1 = 113).
5. How many pairs of 17 do you need for a total sum of 137?
Answer: You need 8 pairs of 17 for a total sum of 137 (17 x 8 = 136).
6. If I divide 137 into groups of 11, how many groups will I have left over?
Answer: You will have 3 groups left over if you divide 137 into groups of 11 (137 ÷ 11 = 12 with remainder 5).
7. What number do you multiply by 18 in order to get a product of 244?
Answer: To get a product of 244 when multiplying by 18, you would need to multiply by 14(244/18=13.6 with remainder2so 14*18=252>244).
8. If Kyle collects coins that are worth 0.25 each, how much money does he have in 56 quarters?
Answer: Kyle has 14 dollars in 56 quarters (56 x 0.25 = 14).
9. Find the prime factors for the number 137 using exponential notation for any prime factors that appear more than once in the factor tree.
Answer: The prime factors for 137 using exponential notation are 3²×17 or 3⁰×3²×17.
10. How many unique factors does 137 have?
Answer: There are 7 unique factors for the number 137 (1, 3, 9, 11, 27, 33, and 137).
## Frequently Asked Questions on Factors of 137
### What is a factor of 137?
A factor of 137 is any number that can divide137 without leaving a remainder. The factors of 137 are 1, 3, 9, 11, 27, 33 and 137.
### Is 18 a factor of 137?
No, 18 is not a factor of 137 because there will be a remainder (137 ÷ 18 = 7 with remainder 13).
### How many factors does 137 have?
There are 7 factors for the number 137 (1, 3, 9 ,11 ,27 ,33 and 137).
### How can I find the prime factors of 137?
To find the prime factors of 137, you need to list all its factors and identify which ones are prime numbers (such as 2, 3 or 5). The prime factors of 137 are 3 x 47 or 3 x 11 x 13 .
### What is the greatest common factor of 136 and 137?
The greatest common factor (GCF) of 136 and 137 is 1.
### Can 135 be divided evenly by 132?
No, 135 cannot be divided evenly by 132 because there will be a remainder equal to 3 (135 ÷ 132 = 1 with remainder 3).
### What is the least common multiple(LCM)of 136 and 137?
The least common multiple(LCM)of 136 and137 is 1352 (136×137=1352 ).
### What two numbers can you multiply together to get the product 135?
You can multiply 45 x3 or 15 x 9 to get the product 135.
### If someone has 56 quarters, how much money do they have in total?
56 quarters equals 14 dollars (\$14; 56*0.25=14 ).
### Is one a factor of every number?
Yes, one is considered an “identity” factor for every number as multiplying it by one will not change its value.
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Prerit Jain
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McGraw Hill My Math Grade 5 Chapter 9 Lesson 12 Answer Key Subtract Mixed Numbers
All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 9 Lesson 12 Subtract Mixed Numbers will give you a clear idea of the concepts.
McGraw-Hill My Math Grade 5 Answer Key Chapter 9 Lesson 12 Subtract Mixed Numbers
Math in My World
Example 1
One King crab weighs 2$$\frac{3}{4}$$ pounds. A second King crab weighs 1$$\frac{1}{4}$$– pounds. How much more does the one King crab weigh? Use models to find the difference.
Find 2$$\frac{3}{4}$$ – 1$$\frac{1}{4}$$
Estimate 3 – 1 = ____
Model 2$$\frac{3}{4}$$ using fraction tiles.
Subtract 1$$\frac{1}{4}$$ by crossing out 1 whole and one $$\frac{1}{4}$$-tile.
There is one whole and two $$\frac{1}{4}$$-tiles left,
which is 1$$\frac{2}{4}$$, or
So, 2$$\frac{3}{4}$$ – 1$$\frac{1}{4}$$ = .
The first King crab weighs pounds more than the second.
Check for Reasonableness ____ ≈
we need to find 2 3/4 – 1 1/4
estimate: 3 – 1 = 2
simplify the equation:
= 2 – 1 – 3/4 – 1/4
= 1 – 2/4
= 1 – 1/2
= 1 1/2.
so, 2$$\frac{3}{4}$$ – 1$$\frac{1}{4}$$ =
The first crab weighs, pounds more than the second.
check for reasonable: 2 ≈
Example 2
Find 6$$\frac{11}{16}$$ – 2$$\frac{5}{8}$$
Estimate 7 – 3 = 4
1. Write an equivalent fraction for 2$$\frac{5}{8}$$ so that the fractions have the same denominator. The LCD is 16.
2. Subtract the wholes. Then subtract the fractions.
So, 6$$\frac{11}{16}$$ – 2$$\frac{5}{8}$$ =
Check for Reasonableness ___4_____ ≈
Talk Math
Describe the steps you would take to find 3$$\frac{5}{8}$$ – 2$$\frac{3}{8}$$.
The above-given equation:
3 5/8 – 2 3/8
In the first step, we separate the whole numbers and the fractions.
The equation is: 3 – 2 – 5/8 – 3/8
In the second step:
simplify the equation and check whether the denominators are the same.
denominators are equal so we can subtract directly.
= 1 – 2/8
= 1 – 1/4
= 1 1/4.
Therefore, 3 5/8 – 2 3/8 = 1 1/4.
Guided Practice
Estimate, then subtract. Write each difference in simplest form.
Question 1.
The above-given mixed fractions:
4 2/3 – 2 1/3
Here the denominators are equal so we can subtract directly.
= 4 – 2 – 2/3 – 1/3
= 2 – 1/3
This can be written as 2 1/3.
Therefore,
Question 2.
The above-given mixed fractions:
5 4/5 – 3 2/5
= 5 – 3 – 4/5 – 2/5
= 2 – 4/5 – 2/5
Here the denominators are equal so we can subtract directly.
= 2 – 2/5
This can also be written as 2 2/5.
Therefore,
Independent Practice
Estimate, then subtract. Write each difference in simplest form.
Question 3.
The above-given mixed fractions:
5 3/4 – 2 1/2
= 5 – 2 – 3/4 – 1/2
= 3 – 3/4 – 1/2
Here denominators are not equal so we have to make the denominators equal.
– Find the common denominator.
– 4 is the least common multiple of denominators 4 and 2. Use it to convert to equivalent fractions with this common denominator.
= 3 – (3 – 2)/4
= 3 – 1/4
This can also be written as 3 1/4
Therefore,
Question 4.
The above-given mixed fractions:
6 5/7 – 3 3/7
Here the denominators are equal so we can subtract directly.
= 6 – 3 – 5/7 – 3/7
= 3 – 2/7
This can also be written as 3 2/7.
Therefore,
Question 5.
The above-given mixed fractions:
7 8/9 – 5 1/3
= 7 – 5 – 8/9 – 1/3
here the denominators are not equal so we have to make the denominators equal.
– Find the common denominator.
– 9 is the least common multiple of denominators 9 and 3. Use it to convert to equivalent fractions with this common denominator.
= 2 – (8 – 3)/9
= 2 – 5/9
This can also be written as 2 5/9.
Question 6.
The above-given mixed fractions:
15 11/12 – 4 1/3
= 15 – 4 – 11/12 – 1/3
here the denominators are not equal so we have to make the denominators equal.
– Find the common denominator.
– 12 is the least common multiple of denominators 12 and 3. Use it to convert to equivalent fractions with this common denominator.
= 11 – (11 – 4)/12
= 11 – 7/12
This can also be written as 11 7/12.
Question 7.
The above-given mixed fractions:
13 9/10 – 4 2/5
= 13 – 4 – 9/10 – 2/5
here the denominators are not equal so we have to make the denominators equal.
– Find the common denominator.
– 10 is the least common multiple of denominators 10 and 5. Use it to convert to equivalent fractions with this common denominator.
= 9 – (9 – 4)/10
= 9 – 5/10
= 9 – 1/2
This can also be written as 9 1/2.
Question 8.
The above-given mixed fractions:
12 5/6 – 7 1/3
= 12 – 7 – 5/6 – 1/3
= 5 – 5/6 – 1/3
here the denominators are not equal so we have to make the denominators equal.
– Find the common denominator.
– 6 is the least common multiple of denominators 6 and 3. Use it to convert to equivalent fractions with this common denominator.
= 5 – (5 – 2)/6
= 5 – 3/6
= 5 – 1/2
= 5 1/2
Therefore,
Question 9.
8$$\frac{3}{8}$$ – 2$$\frac{1}{4}$$ = _____
The above-given mixed fractions:
8 3/8 – 2 1/4
= 8 – 2 – 3/8 – 1/4
= 6 – 3/8 – 1/4
here the denominators are not equal so we have to make the denominators equal.
– Find the common denominator.
– 8 is the least common multiple of denominators 8 and 4. Use it to convert to equivalent fractions with this common denominator.
= 6 – (3 – 2)/8
= 6 – 1/8
This can also be written as 6 1/8
Therefore, 8$$\frac{3}{8}$$ – 2$$\frac{1}{4}$$ = 6 1/8
Question 10.
7$$\frac{7}{8}$$ – 4$$\frac{1}{2}$$ = _____
The above-given mixed fractions:
7 7/8 – 4 1/2
= 7 – 4 – 7/8 – 1/2
= 3 – 7/8 – 1/2
here the denominators are not equal so we have to make the denominators equal.
– Find the common denominator.
– 8 is the least common multiple of denominators 8 and 2. Use it to convert to equivalent fractions with this common denominator.
= 3 – (7 – 4)/8
= 3 – 2/8
= 3 – 1/4
This can also be written as 3 1/4
Therefore, 7$$\frac{7}{8}$$ – 4$$\frac{1}{2}$$ = 3 1/4
Question 11.
12$$\frac{7}{10}$$ – 7$$\frac{2}{5}$$ = _____
The above-given mixed fractions:
12 7/10 – 7 2/5
= 12 – 7 – 7/10 – 2/5
= 5 – 7/10 – 2/5
here the denominators are not equal so we have to make the denominators equal.
– Find the common denominator.
– 10 is the least common multiple of denominators 10 and 2. Use it to convert to equivalent fractions with this common denominator.
= 5 – (7 – 4)/10
= 5 – 3/10
This can also be written as 5 3/10.
Therefore, 12$$\frac{7}{10}$$ – 7$$\frac{2}{5}$$ = 5 3/10.
Mathematical Practice 2 Use Algebra Find each unknown.
Question 12.
11$$\frac{11}{12}$$ – 2$$\frac{1}{12}$$ = x
x = _____
The above-given mixed fractions:
11 11/12 – 2 1/12 = x
we need to find out the x value.
11 – 2 – 11/12 – 1/12
Here denominators are the same. So we can subtract directly.
x = 9 – (11 – 1)/12
x = 9 – 10/12 { in 2nd table 2 x 5 = 10; 2 x 6 = 12}
x = 9 – 5/6
This can also be written as 9 5/6
Therefore, the value of x is 9 5/6.
Question 13.
14$$\frac{9}{14}$$ – 5$$\frac{2}{7}$$ = c
c = _____
The above-given mixed fractions:
14 9/14 – 5 2/7 = c
we need to find the c value.
c = 14 – 5 – 9/14 – 2/7
c = 9 – 9/14 – 2/7
here the denominators are not equal so we have to make the denominators equal.
– Find the common denominator.
– 14 is the least common multiple of denominators 14 and 7. Use it to convert to equivalent fractions with this common denominator.
c = 9 – (14 – 4)/14
c = 9 – 10/14
c = 9 – 5/7
This can also be written as 9 5/7.
Therefore, the value of c is 9 5/7.
Question 14.
18$$\frac{11}{15}$$ – 9$$\frac{2}{5}$$ = n
n = _____
The above-given mixed fractions:
18 11/15 – 9 2/5 = n
we need to find the n value.
n = 18 – 9 – 11/15 – 2/5
Here the denominators are the same so we can subtract directly.
n = 9 – (11 – 2)/5
n = 9 – 9/5
This can also be written as 9 9/5.
Therefore, the value of n is 9 9/5.
Problem Solving
Question 15.
The length of Mr Cho’s garden is 8$$\frac{5}{6}$$ feet. Find the width of Mr Cho’s garden if it is 3$$\frac{1}{6}$$ feet less than the length.
The above-given:
The length of the Cho’s garden in feet = 8 5/6
we need to find the width. Let it be w.
The other fraction is 3 1/6
Based on the given conditions, formulate:
8 5/6 – 3 1/6
= 8 – 3 – 5/6 – 1/6
= 5 – 5/6 – 1/6
Here denominators are the same so we can subtract directly.
w = 5 – (5 – 1)/6
w = 5 – 4/6
w = 5 – 2/3
This can also be written as 5 2/3.
Therefore, the width is 5 2/3 feet.
Question 16.
Timberly spent 3$$\frac{4}{5}$$ hours and Misty spent 2$$\frac{1}{10}$$ hours at gymnastics practice over the weekend. How many more hours did Timberly spend than Misty at gymnastics practice?
The above-given:
The number of hours Timberly spent at gymnastics = 3 4/5
The number of hours Misty spent at gymnastics = 2 1/10
The number of more hours Timberly spent than Misty = g
g = 3 4/5 – 2 1/10
g = 3 – 2 – 4/5 – 1/10
g = 1 – 4/5 – 1/10
here the denominators are not equal so we have to make the denominators equal.
– Find the common denominator.
– 10 is the least common multiple of denominators 5 and 10. Use it to convert to equivalent fractions with this common denominator.
g = 1 – (8 – 1)/10
g = 1 – 7/10
This can also be written as 1 7/10
Therefore, Timberly spent 1 7/10 hours more than Misty.
Question 17.
Mathematical PRACTICE 1 Make Sense of Problems Warner lives 9$$\frac{1}{4}$$– blocks away from the ocean. Shelly lives 12$$\frac{7}{8}$$ blocks away from the ocean. How many more blocks does Shelly live away from the ocean than Warner?
The above-given:
The number of blocks away from Warner lives = 9 1/4
The number of blocks away from the ocean Shelly lives = 12 7/8
The number of blocks Shelly lives away from the ocean than Warner = o
o = 12 7/8 – 9 1/4
o = 12 – 9 – 7/8 – 1/4
o = 3 – 7/8 – 1/4
here the denominators are not equal so we have to make the denominators equal.
– Find the common denominator.
– 8 is the least common multiple of denominators 8 and 4. Use it to convert to equivalent fractions with this common denominator.
o = 3 – (7 – 2)/8
o = 3 – 5/8
This can also be written as 3 5/8
Therefore, she lives 3 5/8 blocks away from the ocean than a warner.
HOT Problems
Question 18.
Mathematical PRACTICE 4 Model Math Write a real-world problem involving the subtraction of two mixed numbers whose difference is less than 2$$\frac{1}{2}$$. Then solve.
The two mixed numbers are x and y
x – y < 2 1/2
One King crab weighs 2$$\frac{3}{4}$$ pounds. A second King crab weighs 1$$\frac{1}{4}$$– pounds. How much more does the one King crab weigh?
Now subtract 2 3/4 and 1 1/4
2 3/4 – 1 1/4
= 2 – 1 – 3/4 – 1/4
= 1 – 2/4
= 1 – 1/2
This can also be written as 1 1/2
1 1/2 < 2 1/2
Therefore, 2 3/4 – 1 1/4 < 2 1/2.
Question 19.
? Building on the Essential Question How can number sense help to know if I have subtracted two mixed numbers correctly?
You could first convert each to an improper fraction. If they don’t have common denominators, then find a common denominator and use it to rewrite each fraction. Then subtract the fractions and compare the given mixed fractions. So that we can easily find out.
McGraw Hill My Math Grade 5 Chapter 9 Lesson 12 My Homework Answer Key
Practice
Estimate, then subtract. Write each difference in simplest form.
Question 1.
The above-given mixed fractions:
6 5/8 – 2 3/8
= 6 – 2 – 5/8 – 3/8
Here denominators are the same so that we can subtract directly.
= 4 – (5 – 3)/8
= 4 – 2/8
= 4 – 1/4
This can also be written as 4 1/4
Therefore, 6 5/8 – 2 3/8 = 4 1/4
Question 2.
The above- is given:
9 3/4 – 1 1/3
= 9 – 1 – 3/4 – 1/3
here the denominators are not equal so we have to make the denominators equal.
– Find the common denominator.
– 12 is the least common multiple of denominators 4 and 3. Use it to convert to equivalent fractions with this common denominator.
= 8 – (9 – 4)/12
= 8 – 5/12
This can also be written as 8 5/12.
Question 3.
The above-given mixed fractions:
4 5/6 – 4 1/3
= 4 – 4 – 5/6 – 1/3
here the denominators are not equal so we have to make the denominators equal.
– Find the common denominator.
– 6 is the least common multiple of denominators 6 and 3. Use it to convert to equivalent fractions with this common denominator.
= 0 – (5 – 2)/6
= 3/6
= 1/2
Problem Solving
Question 4.
Mrs Gabel bought 7$$\frac{5}{6}$$ gallons of punch for the class party. The students drank 4$$\frac{1}{2}$$ gallons of punch. How much punch was left at the end of the party? Write in simplest form.
The above-given:
The number of gallons of punch Gabel bought for the party = 7 5/6
The number of gallons of punch students drank = 4 1/2
The number of gallons of punch left = p
p = 7 5/6 – 4 1/2
p = 7 – 4 – 5/6 – 1/2
p = 3 – 5/6 – 1/2
here the denominators are not equal so we have to make the denominators equal.
– Find the common denominator.
– 6 is the least common multiple of denominators 6 and 2. Use it to convert to equivalent fractions with this common denominator.
p = 3 – (5 – 3)/6
p = 3 – 2/6
p = 3 – 1/3
This can also be written as 3 1/3
Therefore, 3 1/3 gallons of punch was left at the end of the party.
Question 5.
Bella is 10$$\frac{5}{12}$$ years old. Franco is 12$$\frac{7}{12}$$ years old. What is the difference in their ages? Write in simplest form.
The above-given:
The number of years old the Bella is = 10 5/12
The number of years old Franco was = 12 7/12
The difference in their ages = a
a = 12 7/12 – 10 5/12
Here denominators are the same so we have to subtract directly.
a = 12 – 10 – 7/12 – 5/12
a = 2 – (7 – 5)/12
a = 2 – 2/12
a = 2 – 1/6
a = 2 1/6.
Therefore, the difference is 2 1/6
Question 6.
In one week, the fifth grade class recycled 9$$\frac{2}{3}$$ pounds of glass and 12$$\frac{3}{4}$$ pounds of newspaper. How many more pounds of a newspaper than glass did the class recycle?
The above-given:
The number of pounds of glass recycled by the fifth grade = 9 2/3
The number of pounds of newspapers recycled = 12 3/4
The number of more pounds of newspaper recycled than glass = R
R = 12 3/4 – 9 2/3
R = 12 – 9 – 3/4 – 2/3
R = 3 – 3/4 – 2/3
here the denominators are not equal so we have to make the denominators equal.
– Find the common denominator.
– 12 is the least common multiple of denominators 3 and 4. Use it to convert to equivalent fractions with this common denominator.
R = 3 – (9 – 8)/12
R = 3 – 1/12
R = 3 1/12
Therefore, 3 1/4 pounds of the newspaper is recycled than glass.
Question 7.
Mathematical PRACTICE 2 Use Number Sense A snack mix recipe calls for 5$$\frac{3}{4}$$ cups of cereal and 3$$\frac{5}{12}$$ cups less of raisins. How many cups of raisins are needed? Write in simplest form.
The above-given:
The number of cups of cereal mixed for a snack = 5 3/4
The number of cups of raisins less = 3 5/12
The number of cups of raisins needed = c
c = 5 3/4 – 3 5/12
c = 5 – 3 – 3/4 – 5/12
c = 2 – 3/4 – 5/12
here the denominators are not equal so we have to make the denominators equal.
– Find the common denominator.
– 12 is the least common multiple of denominators 4 and 12. Use it to convert to equivalent fractions with this common denominator.
c = 2 – (9 – 5)/12
c = 2 – 4/12
c = 2 – 1/3
c = 2 1/3
Therefore, 2 1/3 cups of raisins are needed.
Test Practice
Question 8.
What is the difference between the two weights?
A. $$\frac{1}{2}$$ ounce
B. $$\frac{7}{8}$$ ounce
C. 1$$\frac{3}{8}$$ ounces
D. 1$$\frac{7}{8}$$ ounces
Explanation:
The above-given weights:
The weight of the phone = 4 1/2
The weight of the remote = 3 1/8
The difference between the objects = d
d = 4 1/2 – 3 1/8
d = 4 – 3 – 1/2 – 1/8
d = 1 – 1/2 – 1/8
here the denominators are not equal so we have to make the denominators equal.
– Find the common denominator.
– 8 is the least common multiple of denominators 2 and 8. Use it to convert to equivalent fractions with this common denominator.
d = 1 – (4 – 1)/8
d = 1 – 3/8
d = 1 3/8
Therefore, the difference between the weights is 1 3/8 ounces.
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# 11.1 Systems of linear equations: two variables (Page 7/20)
Page 7 / 20
Meal tickets at the circus cost $\text{\hspace{0.17em}}\text{}4.00\text{\hspace{0.17em}}$ for children and $\text{\hspace{0.17em}}\text{}12.00\text{\hspace{0.17em}}$ for adults. If $\text{\hspace{0.17em}}1,650\text{\hspace{0.17em}}$ meal tickets were bought for a total of $\text{\hspace{0.17em}}\text{}14,200,$ how many children and how many adults bought meal tickets?
700 children, 950 adults
Access these online resources for additional instruction and practice with systems of linear equations.
## Key concepts
• A system of linear equations consists of two or more equations made up of two or more variables such that all equations in the system are considered simultaneously.
• The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. See [link] .
• Systems of equations are classified as independent with one solution, dependent with an infinite number of solutions, or inconsistent with no solution.
• One method of solving a system of linear equations in two variables is by graphing. In this method, we graph the equations on the same set of axes. See [link] .
• Another method of solving a system of linear equations is by substitution. In this method, we solve for one variable in one equation and substitute the result into the second equation. See [link] .
• A third method of solving a system of linear equations is by addition, in which we can eliminate a variable by adding opposite coefficients of corresponding variables. See [link] .
• It is often necessary to multiply one or both equations by a constant to facilitate elimination of a variable when adding the two equations together. See [link] , [link] , and [link] .
• Either method of solving a system of equations results in a false statement for inconsistent systems because they are made up of parallel lines that never intersect. See [link] .
• The solution to a system of dependent equations will always be true because both equations describe the same line. See [link] .
• Systems of equations can be used to solve real-world problems that involve more than one variable, such as those relating to revenue, cost, and profit. See [link] and [link] .
## Verbal
Can a system of linear equations have exactly two solutions? Explain why or why not.
No, you can either have zero, one, or infinitely many. Examine graphs.
If you are performing a break-even analysis for a business and their cost and revenue equations are dependent, explain what this means for the company’s profit margins.
If you are solving a break-even analysis and get a negative break-even point, explain what this signifies for the company?
This means there is no realistic break-even point. By the time the company produces one unit they are already making profit.
If you are solving a break-even analysis and there is no break-even point, explain what this means for the company. How should they ensure there is a break-even point?
Given a system of equations, explain at least two different methods of solving that system.
You can solve by substitution (isolating $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ ), graphically, or by addition.
## Algebraic
For the following exercises, determine whether the given ordered pair is a solution to the system of equations.
#### Questions & Answers
write down the polynomial function with root 1/3,2,-3 with solution
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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
the polar co-ordinate of the point (-1, -1)
prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x
tanh`(x-iy) =A+iB, find A and B
B=Ai-itan(hx-hiy)
Rukmini
what is the addition of 101011 with 101010
If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
Jack
extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3 |
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Activity Discussion Math How to find LCM and HCF?
• # How to find LCM and HCF?
Posted by on May 8, 2021 at 7:46 am
How to find LCM and HCF?
4 Members · 3 Replies
• 3 Replies
• ### 19 BCS 081
Member
May 8, 2021 at 12:26 pm
0
LCM stands for ‘Least Common Multiple’. It is thus the smallest number which gets exactly divisible by the given two or more numbers. If we had to find the then LCM of two numbers say a and b. Then L(a,b) is the least positive integer which gets divisible by both a and b.
For example:- Find the LCM of 4 and 12.
Solution:-
Multiples of 4= 4, 8, 12,16, 20, 24, 28, 32, 36, ……
Multiples of 12= 12, 24, 36, 48, 60, 72, 84, ……
Common Multiples= 12, 24, 36, …….
Least Common Multiple= L(4, 12)= 12
HCF stand for ‘Highest Common Factor’. Hence, it is the greatest factor that exists between given two or more than two numbers. Or as we may say, it is the highest positive integer H(a, b) dividing two or more than two numbers (here, a and b) without leaving any remainder.
For example:- Find the HCF of 144 and 104.
Solution:-
Factors of 144= 2 x 2 x 2 x 2 x 3 x 3
Factors of 104= 2 x 2 x 2 x 13
Highest Common Factor= H(144, 104)= 2 x 2 x 2 = 8
• ### Aruja
Member
May 9, 2021 at 2:49 pm
0
Full Form of LCM is ‘ Least common multiple’.
That means the smallest number (C) that can be divided by both given integers in the question (a,b). The LCM of two number comes when we take out the smallest common multiple from the tables of the two given numbers in the question.
Example- Find the LCM of 8, 16
so first we will take out the factor of
8= 8, 16, 24, 32, 40, 48,…..
16= 16, 32, 48,……
The common multiple that we get from both the tables is 16.
Ans) LCM of (8,16)= 16
Full Form of HCF is ‘ Highest Common Factor ‘.
That means the highest common factor basically the number which can divide both the number leaving number in denominator or which completely divide without leaving a remainder. It comes when we pick as many common numbers from both the numbers factors.
Example- Find HCF of 6,16
Factors of 6= 2 * 3
Factors of 16= 2 * 2 * 2 * 2
Now we will pick the common number from both of the factor = 2.
So HCF of (6,16)= 2
• ### Jyothi krishna
Member
May 12, 2021 at 11:46 am
0
Highest Common Factor (H.C.F) : It is also called
Greatest common Diviser (G.C.D). When a greatest number
divides perfectly the two or more given numbers then that
number is called the H.C.F. of two or more given numbers.
e.g.
The H.C.F of 10, 20, 30 is 10 as they are perfectly
divided by 10,5 and 2 and 10 is highest or greatest of
them.
Least common Multiple (L.C.M.) : The least number
which is divisible by two or more given numbers, that least
number is called L.C.M. of the numbers.
L.C.M. of 3,5,6 is 30, because all 3 numbers divide
30, 60, 90, …… and so on perfectly and 30 is minimum of
them.
Factor and Multiple : If a number m, divides perfectly
second number n, then m is called the factor of n and n is
called the multiple of m.
Rule 1 : 1st number × 2nd number = L.C. M. × H.C.F.
l There are two methods for calculating the H.C.F
and L.C.M.
(i) Factor Method
(ii) Division Method
l If the ratio of two numbers is a:b, (lowest form i.e.
indivisible to each other) then
Numbers are ak and bk, where k is a constant and
hence,
H.C.F. is K and L.C.M. is abk
Rule 2 :
L.C.M of fractions=L.C.M.of numerators/
H.C.F.of denominators
Rule 3 :
H.C.F. of fractions=H.C.F of numerators/
L.C.M.of denominators
l If there is no common factor between two numbers,
then L.C.M. will be the product of both numbers.
l If there are ‘n’ numbers in a set and H.C.F. of any two
numbers is H and L.C.M. of all ‘n’ numbers is L, then
product of all ‘n’ numbers is [(H)^n-1×L].
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# Primary Mathematics/Fractions
Primary Mathematics
← Negative numbers Fractions Working with fractions →
## Learning to use fractions (Visually)
Fractions or rational numbers are in essence the same as division, however we use them more often to express numbers less than one - for instance a half or a quarter. Fractions have a numerator (on the top) and a denominator (on the bottom). If a fraction is larger than 1 then the numerator will be larger.
### Modern methods to teach fractions
Today's modern methods of teaching math and fractions are drastically different than how they were taught just 10 years ago. The difference between these methods is that the later method explores the visual evidence for certain ways of manipulating fractions and whereas the earlier approach simply used variables from the beginning. Tiles of different colors sorted into groups can be useful in representing fractions visually.
### Origami and Fractions
There is perhaps little emphasis these days which shows the elaborate "visualization" that math requires. Students used to be taught merely by equation, but to my understanding, by teaching them such methods they tend to take the "cooking approach" to problems in that they have an inadequate sense of "visualizing" the concept the problem in their mind. What i want to do is emphasize the creative aspect of fractions while at the same time exploring the richness of why such problems are true.
Why i call this section Origami and math is because they are very much related to each other. The thing about Origami which is rich in math is that essentially folding a piece of paper proves that such a fraction exists!
in order to do this experiment, you need the following materials:
1. Square piece of paper
2. pencil
Steps to seeing some nice fractions
First, we have to say to ourselves, "This piece of paper is 1 piece of paper"
Now, we will explore fractions by seeing how "much" of the remaining paper we see as we fold. Every time we see the paper, we will write down the fraction of the paper on the front of it. By the time we get done, we'll have lots of fractions written on a piece of paper.
Step 1: Obtain square piece of paper. Write "1" on it.
Step 2: fold piece of paper in half. Write "1/2" on it.
Step 3: Fold again in half. write "1/4"
Step 4: Fold again in half. write "1/8"
Step 5: Unfold the piece of paper and write lines in where there are folds in the paper
Notice: if there is a 1 on top, then whatever number is on the bottom, it takes that many "pieces" to make a "whole". For example, 1/8 needs 8 pieces to become 1 whole.
### The square model vs. the circle model
A square divided into fractions A circle divided into fractions
### The money approach: 1, 1/4, 1/2, 1/10, 1/20, 1/100
A very practical way to learn fractions is the use of money, as we use it everyday.
Questions:
1. How many quarters are in a dollar?
2. How many dimes are in a dollar?
3. How many nickels are in a dollar?
4. How many pennies are in a dollar?
As said from the previous section, if there is a one on top and some number on bottom, that means it needs that much pieces to make it a whole (a whole means 1 by the way)
For example, 1/10 is the value of a dime (10 cents). You need 10 dimes (10 x 10 = 1 dollar) to get one full dollar.
### Whole number fractions
First of all, as always, instead of looking at complicated variable jargon, we will instead look at certain ways to "view" certain types of number. Just like art, you don't need to be an accomplished artist to draw, but rather you just need to know how to look at things better (in this case, numbers)
Since fractions have both a top (called a numerator. think "topinator") and a bottom (denominator, which "downominator" which is divided by a bar, we have to "adjust our thinking" so that we can recognize what our friendly fractions might look like.)
For example,
Q: 5 ÷ 5 reads "5 divided by 5". What does that look in Fraction form?
Well, since we are prospective mathematicians (and artists...) we will look at the magic of what the divided sign actually means:
o === o (my divided sign) What it actually means is that dots tell you to "Make me a number!" (because Zeroes often become lonely...they want to have value in life...) and you see that bridge which devides them too says that, "In order to separate such ZEROES in life and making more zeroes, we say always divide your numbers from each other."
A: $\frac{5}{5} = 1$, which also reads "5 divided by 5".
Since we know how to recognize some fractions, we have to "see" with our naked eyes what some fractions actaully mean. The best preparation that you will have in our "artist" training is that whole numbers actually have "1" on the bottom. You may ask yourself, "How can that be? Is that even possible?"
Well Just for fun, we will go through this little exercise to show that there is indeed ones on the bottom when you have whole numbers.
Example: express "5" in a fraction form.
Look at our previous example about the placement of things. We now know what a denominator is and a numerator is. Since the whole number is the number of top, all we have to do is see what number is on bottom, i wonder what it is.
$\frac{5}{?}$ In order to figure out what goes on bottom, we can figure it out many ways. Fractions are in ways actually like ratios. Such as if there were two ice creams for me and none for you, my ratio would be 2:1 and yours would be 0:1...Sad isn't it? Getting back to the point, we must think about why and how it becomes a whole number. $\frac{(X)(X)(X)(X)(X)}{(X)}$ The upper part represents five oranges. the lower represents 1. $=\frac{5}{1}$
So our lesson is, if there is any whole number, "1" will always go on the bottom. So if I say, "What's the number that goes on the bottom if the number is 1,2,3,4,5,6,7, or any other whole number?"
Of course, your answer should be "1", "1", "1", "1", "1", "1", "1" and always "1." I want you to remember that!
## Multiplying fractions
In general, multiplying fractions involves a simple formula:
$\frac{A}{B} * \frac{C}{D} = \frac{A*C}{B*D}$
Where A, B, C, and D represent the numbers within the fractions. If this seems intimidating, there are methods used to tackle the multiplication more easily.
1. Know what the fractions are. Write it out.
2. If you wrote it out correctly, ignore the fractions part and look at it this way.
For example:
$\frac{5}{3} * \frac{5}{3} = \frac{?}{?}$
If you wrote it correctly, this is what it should look like. Also, it would be helpful if you wrote it equal to the mysteriously equal question marks, "?/?." They're there so you can see where to write your answers.
3. Using your finger or card, cover or hide the bottom (called the denominator) so that you can concentrate on multiplying the top. Focus solely on multipling the upper portion:
$\frac{5}{} * \frac{5}{} = \frac{?}{}$
5 x 5 = ? (what is 5 times 5? what is the value (in cents) of 5 nickels? why 25 of course!) So if 25 is equals ?, than simply "erase" the "?" and write "25." Its that simple!
Again, since you know whats on top, do the same on the bottom:
$\frac{}{3} * \frac{}{3} = \frac{}{?}$
Again, what is 3 x 3? Of course, it is 9. So, again, erase the "?" and write down "9"
Once you remove the finger or card, you are presented with the answer: $\frac{5}{3} * \frac{5}{3} = \frac{25}{9}$
### Multiplying Fractions with nice pictures
Another good way to prove that answers are actually correct is to use a diagram or tiles to prove that it is correct. So how we do that is first write out one of our fractions on one side and the other fraction on the other side. To see what i mean, here is an example:
Here is our little lesson. We will multiply 3/4 x 1/2 = ?.
1/4 1/4 1/4 1/4
So just imagine the figure to the right is one whole 4 story "chocobar" (i will remind you that one chocobar that has 4 stories still is one chcobar) You will see that each story represents a fraction. 1/4 means its needs 4 pieces to make a whole.
0000 __________________
000000/ |
OOOO Mr. Anaconda | "CHOBARS ARE YUMMY!"
\ _______ | "I hope I eat a lot
| / \ | of them!"
| / \ |
| / \ |
| / \ /
\-/ \-/
But the thing is, Mr. Anaconda only has enough energy to eat once a week. His mouth is 1/2-full with chocbar and the deepest he can eat is 3/4 a chocbar. How much of the chocbar has he eaten?
The fraction is set up like this:
$\frac{3}{4} * \frac{1}{2} = \frac{?}{?}$
Visually, it will look like this:
1/2 1/2
1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8
1/4
1/4
1/4
1/4
So, simplified from our little illustration is our little chocobar broken up into nice convenient blocks so that Mr. Anaconda can eat it.
1/2 1/2
1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8
1/4
1/4
1/4
1/4
From the image, you can count 3/8ths of the chocobar was eaten.
### Dividing fractions
As mentioned earlier, a fraction is one number divided by another. Because of textual limitations, fractions here will be represented using numbers and slashes. For example, the fraction one half will be represented as 1/2. The fraction two thirds will be represented as 2/3.
Now, on to some specifics. You may already know how to multiply fractions. Remember all you have to do is multiply the two top numbers together and put them above your two bottom numbers multiplied together. If know how to multiply fractions, dividing fractions will also be easy, and only requires one extra step.
In order to divide two fractions, you invert the second fraction and multiply. This means that you switch around the numerator (top) and denominator (bottom) of the fraction before multiplication.
Let's look at examples using our two fractions. First, let's divide the fraction 1/2 by the fraction 1/2. Since we are dividing a number by itself, we should expect to get an answer of 1.
Here is our problem.
$\frac{1}{2}\div\frac{1}{2} = \frac{1}{2} * \frac{2}{1} = \frac{1*2}{2*1} = \frac{2}{2} = 1$
Now let's try another example.
$\frac{1}{2}\div\frac{2}{3} = \frac{1}{2} * \frac{3}{2} = \frac{1*3}{2*2} = \frac{3}{4}$
Following these simple steps will let you solve any fraction division problem.
There is also an alternate way to divide fractions using an equivalent fraction approach. You obtain this by multiplying the numerator and denominator so that they're equivalent, cancelling out the denominator, and creating a fraction from the two remaining numerators.
$\frac{1}{2}\div\frac{2}{3} = \frac{3}{6}\div\frac{4}{6} = \frac{3}{4}$
Here is another example:
$\frac{2}{5}\div\frac{3}{4} = \frac{8}{20}\div\frac{15}{20} = \frac{8}{15}$
It works because the common denominator will always be $\frac{n}{n} = 1$
Fractions that have the same or "Common" denominator are called "Like" fractions.
1/3, 2/3
(one-third, two-thirds)
To add Like fractions together such as these:
$\frac{1}{3} + \frac{2}{3} = ?$
1. Add the numerators (the top numbers):
• 1 + 2 = 3 (one plus two equals three)
2. Use the common denominator (the bottom numbers):
• /3
All together it looks like this:
$\frac{1}{3} + \frac{2}{3} = \frac{3}{3}$
3. Simplify the answer as much as you can by dividing the denominator into the numerator.
• $\frac{3}{3} = 1$
If the numerator is now larger than the denominator it might look like this:
4/5 + 3/5 = 7/5 (four-fifths plus three-fifths equals seven-fifths)
This is simplified by creating a mixed number:
• 5 goes into 7 one time, with 2 left over. Put the remainder (2) over the denominator (5).
• $\frac{7}{5} = 1 \frac{2}/{5}$
To add or subtract fractions with different denominators, you must first convert them to equivalent fractions with common denominators.
### Multiplying to get equivalent fractions
In some cases, you may need to add 1/2 to 1/3:
$\frac{1}{2}+\frac{1}{3}=?$
You will first have to convert the fractions into like fractions. The easiest means of doing so is to multiply the top and bottom of the left fraction by the bottom on the right, and the top and bottom of the right fraction by the bottom on the left:
$\frac{1*3}{2*3}+\frac{1*2}{3*2}=?$
$\frac{3}{6}+\frac{2}{6}=?$
The addition can now take place as expected:
$\frac{3}{6}+\frac{2}{6}=\frac{5}{6}$
Primary Mathematics
← Negative numbers Fractions Working with fractions → |
# Difference between revisions of "2005 AIME I Problems/Problem 4"
## Problem
The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have.
## Solution
### Solution 1
If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$. If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$. Thus, if the number of columns is $n$, the number of students is $n(n + 7)$ which must be 5 more than a perfect square, so $n \leq 14$. In fact, when $n = 14$ we have $n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5$, so this number works and no larger number can. Thus, the answer is 294.
### Solution 2
Define the number of rows/columns of the square formation as $s$, and the number of rows of the rectangular formation $r$ (so there are $r - 7$ columns). Thus, $s^2 + 5 = r(r-7) \Longrightarrow r^2 - 7r - s^2 - 5 = 0$. The quadratic formula yields $r = \frac{7 \pm \sqrt{49 - 4(1)(-s - 5)}}{2} = \frac{7 \pm \sqrt{4s^2 + 69}}{2}$. $\displaystyle \sqrt{4s^2 + 69}$ must be an integer, say $x$. Then $4s^2 + 69 = x^2$ and $(x + 2s)(x - 2s) = 69$. The factors of $69$ are $(1,69), (3,23)$; $x$ is maximized for the first case. Thus, $x = \frac{69 + 1}{2} = 35$, and $r = \frac{7 \pm 35}{2} = 21, -14$. The latter obviously can be discarded, so there are $21$ rows and $21 - 7 = 14$ columns, making the answer $294$. |
# Linear and Absolute Value Functions Linear Functions Are
• Slides: 30
Linear and Absolute Value Functions
Linear Functions •
Are these equations linear?
Finding Slope • To find slope (rate of change): – From 2 points—use point-slope formula – From a graph—use rise/run – From an equation—put in slope intercept form – Slope formula:
Find the slope given the following 2 points
Find the slope from the graph
Do these ordered pairs represent linear functions? •
Can a linear equation be used to solve the following problems? 1. A phone company charges \$0. 85 for 5 minutes and \$1. 10 for 10 minutes. How much will a 20 minute phone call cost? 2. If a projectile is traveling at an initial velocity of 96 ft/sec, how many seconds after launch is the projectile 128 feet above the ground. 3. The hypotenuse of a right triangle is 4 in long. One leg is 1 in longer than the other. Find the lengths of the legs.
Writing Equations of Lines • To write the equation of a line, you must know a point and the slope • Use the point slope formula to write the equation • Put the final equation in slope intercept form
Write the equation of the line • From a point and the slope
Write the equation of the line • From 2 points—must find the slope first
Parallel lines • Parallel lines have the same slope • To write the equation of a line given the parallel line, use the same slope
Parallel lines • To write the equation of a line given the parallel line, use the same slope
Perpendicular lines • Perpendicular lines have opposite reciprocal slopes • To write the equation of a line given the perpendicular line, flip the slope and change the sign
Perpendicular lines • To write the equation of a line given the perpendicular line, flip the slope and change the sign
Perpendicular lines • To write the equation of a line given the perpendicular line, flip the slope and change the sign
Absolute Value Functions • Functions that contain absolute value expressions • Must do two cases to solve—once the abs value is by itself • Absolute value can never equal a negative number because it represents the distance from zero on a number line • Check solutions!!
Solve
Solve
Solve
Graph and Solve
Graph and Solve •
Graph and Solve •
Graph and Solve
Absolute Value Inequalities • Inequalities containing an absolute value expression • Solve just like an absolute value equation • Just remember < “and” > “or” • Special cases when < or > a negative #
Solve and graph. Describe the solution set. •
Solve and graph. Describe the solution set.
Solve and graph. Describe the solution set.
Solve and graph. Describe the solution set.
Solve and graph. Describe solution sets. |
×
# PLYNUM – Editorial
Author: Arkapravo Ghosh
Tester: Arkapravo Ghosh
Editorialist: Arkapravo Ghosh
EASY-MEDIUM
Graph, BFS
# PROBLEM:
Given a number A, you have to reach another number B using the minimum number of operations, given you can only perform the following operations:-
1) Decrement the current number by 1
2) Increment the current number by 3
3) Multiply the current number by 2
# QUICK EXPLANATION:
We can simply consider this as a graph problem. We can consider A as a node in a graph and start a BFS from that node till we reach B. Here each node will have three outgoing edges.
# EXPLANATION:
We can consider this problem as a graph problem. Here each number will represent a node in the graph. Each node will have three outgoing edges. We can first start BFS from node A till we reach node B. The three nodes outgoing nodes from a node numbered N will be N-1, N+3 and N*2. But the graph now has infinite number of nodes. However we can limit the range of the node numbering with a simple observation that we don’t need to go below 0 and above B + (B-A)/3. In the worst case we can reach B from A in (B-A)/3 steps, since we can increment by 3 in each step. Again from B + (B-A)/3, we can reach B in (B-A)/3 steps decrementing by 1 each time, since it is the only way of decrementing a number. So, going upto B + (B-A)/3 from A and then decrementing by 1 for (B-A)/3 times is as good as incrementing A to B in (B-A)/3 steps. So we can limit the range of nodes from 0 to B+(B-A)/3.
If B is less than A, then decrementing by 1 in each step is the only way to reach B from A. So it will take (A-B) steps.
The time complexity of this algorithm is proportional to the number of vertices and edges in the graph. In the worst case the no of vertices and edges is (B + (B-A)/3) * 3 = O(B), thus worst case time for BFS will be O(B) i.e. linear time.
# AUTHOR'S AND EDITORIALIST'S SOLUTIONS:
Author's and editorialist’s solution can be found here.
3436
accept rate: 21%
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1 I didn't use this limit since the space of values was small enough already, but the path from $A$ to a larger $B$ doesn't need to cross approximately $8B/7$, since stepping from below $B$ to above by doubling to an equal number of steps away $s$, gives $2(B-3s) = (B+s) => s = B/7$. I would add 3 to that limit: 8B/7 + 3 to allow for detailed adjustment. Thus the best path from say 4 to 91 shouldn't need to go above 107. On further reflection, if the better option in the above analysis is to come down to $B$ from above, it makes more sense to do the subtractions before the doubling, since only half as many will be required. This simplifies all the above analysis to expecting that the best path from $A$ to a larger $B$ can be achieved without going above $B+2$. On the lower end, there's clearly no point in going below $A/2$. answered 07 Aug '18, 18:47 5★joffan 948●8 accept rate: 13%
0 How is the max no of steps required (B-A)/3 ? Please explain :) I am not able to get it from the above explanation. answered 07 Aug '18, 13:58 318●1●10 accept rate: 1% 2 Suppose you have two numbers 0 and 100. So according to the given problem, you can increment by 3 every time instead of using the multiplication operation(which will make the no of steps less, as you increment fast). If you increment by 3 every time, then you will take (100 - 0)/3 i.e. ~33 steps. So this is the worst case, as you can simply multiply by 2 in the intermediate numbers and reduce the no. of steps. So, the maximum number of steps will be O((B-A)/3). I hope this clears your doubt. If not, feel free to ask again. :) (07 Aug '18, 14:33)
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## Circumference of a Circle: Definition, Formula, Properties, and Examples
The circumference of a circle gives the length of the outer border of a circular shape. It is similar to the perimeter of any other two-dimensional figure. The term perimeter is specifically referred to as circumference in the case of a circle.
The circumference is an important measurement since it is directly related to the size of any circle. It is essential for anyone studying geometry or any field that involves circular objects.
In this blog, we will discuss the definition of the circumference with an explanation. We will explain both the mathematical formula and properties of the circumference. We will include examples for our readers in the example section to clarify this concept.
## What is the Circumference of a Circle?
The circumference of a circle is the distance of its outside boundary. It measures the entire length of the border of the circle. Let's learn this concept through real-life examples. Imagine a girl running around a circular park and completing a single rotation.
The distance covered by her in this circular track in one trip is called the circumference of the park.
The circumference of a circle is a basic concept in geometry. We can apply it in our daily life to compute the perimeter of a circular racetrack, determine the belt length required to encircle a pulley or calculate the distance traveled by a wheel during one complete round.
## Explanation of Circumference of Circle
The circumference of a circle can be defined as the length of the boundary that encloses a 2D circular shape. If we cut a circle and create a straight line, the total length of this line would be the circumference of a given circle.
Circle circumferences are measured in length units such as feet, inches, centimeters, meters, miles, or kilometers. Let’s learn about a few important terms associated with circles to understand how to find the circumference of a circle.
### Center of Circle
A circle is a set of all points that are a constant distance from a fixed point, known as the center of the circle.
### Diameter
The diameter is a straight-line segment that passes through the center of the circle and connects two points on the boundary of the circle.
The radius of a circle is the distance from the center of the circle to any point on its circumference. The radius (r) is equivalent to half of the diameter (i.e. r = d/2).
## Formula for Calculating Circumference of Circle
The circumference measures the overall length of the outer border of the circle. It can be evaluated using the formula:
C = 2 π r
In this formula:
• C expresses the circumference or perimeter of a circle.
• π (pi) is a mathematical constant approximately equal to 22/7 (or 3.14159 in decimal).
• r denotes the radius of the circle which is the distance from the center of the circle to the point on its outer edge.
We can also calculate the Circumference using the diameter with the following formula:
C = π d
Where:
• C is the Circumference.
• π (Pi) is a constant.
• d expresses the diameter of the circle
Note that the choice of the formula depends on the given data.
## Properties of Circumference of Circle
The circumference has the following essential characteristics:
• The circumference of a circle is directly proportional to its radius or diameter. The perimeter will also be doubled if the radius or diameter is doubled.
• The ratio of the circumference of a circle to its diameter is referred to as pi (i.e. π = C/d).
• The circumference of a circle always gives a positive value.
• The unit of the determined circumference of a circle is the same as the unit of radius or diameter.
## Steps to Find the Circumference of Circle
The Circumference of any 2D circular shape can be evaluated in a few simple steps:
1. Find the measurement of the radius or diameter of the circular shape.
2. Substitute the measured radius or diameter into the general formula.
3. Calculate the circumference by multiplying the appropriate value (2πr or πd) with the measured radius or diameter.
4. The result of the calculation is the circumference of the circle.
5. Write down the appropriate unit of measurement with the final answer.
## Examples of Circumference of a Circle
The following examples illustrate how to calculate the circumference of a circle manually.
Example 1:
If the circle has a radius of 7 cm, find its circumference.
Solution:
Given Data:
Radius (r) = 7cm
∴ π ≈ 3.14
The circumference of a circular shape can be calculated using the following formula.
C = 2πr
Substitute these given values in the above formula, we get
C = 2 × 3.14 × 7
Circumference = 43.96 cm
Example 2:
Calculate the length of fabric required to make a 36-inch-radius round dining tablecloth.
Solution:
Here;
r = 36 inches
An estimated value of pi (π) is 3.14.
C = 2πr
Substitute the given values into the formula:
Circumference = 2 (3.14) (36)
C = 226.08 inches
The tablecloth requires around 226.08 inches of fabric.
Example 3:
Calculate the circumference for the circular region with a 10-meter diameter.
Solution:
Give:
Diameter = d = 10
∴ π ≈ 3.14
The formula for the circumference when d is given:
C = πd
Put the value of d and pi in the formula of circumference.
C = 3.14 × 10 m
Circumference = 31.43 m
Thus, the circumference with a radius of 10m is approximately 31.43m.
To solve problems of calculating the circumference of a circle, Use our circumference calculator for precise results quickly. |
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## Section6.1Using Definite Integrals to Find Area and Length
###### Motivating Questions
• How can we use definite integrals to measure the area between two curves?
• How do we decide whether to integrate with respect to $x$ or with respect to $y$ when we try to find the area of a region?
• How can a definite integral be used to measure the length of a curve?
Early on in our work with the definite integral, we learned that if we have a nonnegative velocity function, $v\text{,}$ for an object moving along an axis, the area under the velocity function between $a$ and $b$ tells us the distance the object traveled on that time interval. Moreover, based on the definition of the definite integral, that area is given precisely by $\int_a^b v(t) \, dt\text{.}$ Indeed, for any nonnegative function $f$ on an interval $[a,b]\text{,}$ we know that $\int_a^b f(x) \, dx$ measures the area bounded by the curve and the $x$-axis between $x = a$ and $x = b\text{.}$
Through our upcoming work in the present section and chapter, we will explore how definite integrals can be used to represent a variety of different physically important properties. In Preview Activity 6.1.1, we begin this investigation by seeing how a single definite integral may be used to represent the area between two curves.
###### Preview Activity6.1.1
Consider the functions given by $f(x) = 5-(x-1)^2$ and $g(x) = 4-x\text{.}$
1. Use algebra to find the points where the graphs of $f$ and $g$ intersect.
2. Sketch an accurate graph of $f$ and $g$ on the axes provided, labeling the curves by name and the intersection points with ordered pairs.
3. Find and evaluate exactly an integral expression that represents the area between $y = f(x)$ and the $x$-axis on the interval between the intersection points of $f$ and $g\text{.}$
4. Find and evaluate exactly an integral expression that represents the area between $y = g(x)$ and the $x$-axis on the interval between the intersection points of $f$ and $g\text{.}$
5. What is the exact area between $f$ and $g$ between their intersection points? Why?
### Subsection6.1.1The Area Between Two Curves
Through Preview Activity 6.1.1, we encounter a natural way to think about the area between two curves: the area between the curves is the area beneath the upper curve minus the area below the lower curve. For the functions $f(x) = (x-1)^2 + 1$ and $g(x) = x+2\text{,}$ shown in Figure 6.1.2,
we see that the upper curve is $g(x) = x+2\text{,}$ and that the graphs intersect at $(0,2)$ and $(3,5)\text{.}$ Note that we can find these intersection points by solving the system of equations given by $y = (x-1)^2 + 1$ and $y = x+2$ through substitution: substituting $x+2$ for $y$ in the first equation yields $x+2 = (x-1)^2 + 1\text{,}$ so $x+2 = x^2 - 2x + 1 + 1\text{,}$ and thus
\begin{equation*} x^2 - 3x = x(x-3) = 0\text{,} \end{equation*}
from which it follows that $x = 0$ or $x = 3\text{.}$ Using $y = x+2\text{,}$ we find the corresponding $y$-values of the intersection points.
On the interval $[0,3]\text{,}$ the area beneath $g$ is
\begin{equation*} \int_0^3 (x+2) \, dx = \frac{21}{2}\text{,} \end{equation*}
while the area under $f$ on the same interval is
\begin{equation*} \int_0^3 [(x-1)^2 + 1] \, dx = 6\text{.} \end{equation*}
Thus, the area between the curves is
$$A = \int_0^3 (x+2) \, dx - \int_0^3 [(x-1)^2 + 1] \, dx = \frac{21}{2} - 6 = \frac{9}{2}\text{.}\label{E-DiffOfInt}\tag{6.1.1}$$
A slightly different perspective is also helpful here: if we take the region between two curves and slice it up into thin vertical rectangles (in the same spirit as we originally sliced the region between a single curve and the $x$-axis in Section 4.2), then we see that the height of a typical rectangle is given by the difference between the two functions. For example, for the rectangle shown at left in Figure 6.1.3,
we see that the rectangle's height is $g(x) - f(x)\text{,}$ while its width can be viewed as $\Delta x\text{,}$ and thus the area of the rectangle is
\begin{equation*} A_{\text{rect} } = (g(x) - f(x)) \Delta x\text{.} \end{equation*}
The area between the two curves on $[0,3]$ is thus approximated by the Riemann sum
\begin{equation*} A \approx \sum_{i=1}^{n} (g(x_i) - f(x_i)) \Delta x\text{,} \end{equation*}
and then as we let $n \to \infty\text{,}$ it follows that the area is given by the single definite integral
$$A = \int_0^3 (g(x) - f(x)) \, dx\text{.}\label{E-IntOfDiff}\tag{6.1.2}$$
In many applications of the definite integral, we will find it helpful to think of a “representative slice” and how the definite integral may be used to add these slices to find the exact value of a desired quantity. Here, the integral essentially sums the areas of thin rectangles.
Finally, whether we think of the area between two curves as the difference between the area bounded by the individual curves (as in (6.1.1)) or as the limit of a Riemann sum that adds the areas of thin rectangles between the curves (as in (6.1.2)), these two results are the same, since the difference of two integrals is the integral of the difference:
\begin{equation*} \int_0^3 g(x) \, dx - \int_0^3 f(x) \, dx = \int_0^3 (g(x) - f(x)) \, dx\text{.} \end{equation*}
Moreover, our work so far in this section exemplifies the following general principle.
If two curves $y = g(x)$ and $y = f(x)$ intersect at $(a,g(a))$ and $(b,g(b))\text{,}$ and for all $x$ such that $a \le x \le b\text{,}$ $g(x) \ge f(x)\text{,}$ then the area between the curves is $A = \int_a^b (g(x) - f(x)) \, dx\text{.}$
###### Activity6.1.2
In each of the following problems, our goal is to determine the area of the region described. For each region, (i) determine the intersection points of the curves, (ii) sketch the region whose area is being found, (iii) draw and label a representative slice, and (iv) state the area of the representative slice. Then, state a definite integral whose value is the exact area of the region, and evaluate the integral to find the numeric value of the region's area.
1. The finite region bounded by $y = \sqrt{x}$ and $y = \frac{1}{4}x\text{.}$
2. The finite region bounded by $y = 12-2x^2$ and $y = x^2 - 8\text{.}$
3. The area bounded by the $y$-axis, $f(x) = \cos(x)\text{,}$ and $g(x) = \sin(x)\text{,}$ where we consider the region formed by the first positive value of $x$ for which $f$ and $g$ intersect.
4. The finite regions between the curves $y = x^3-x$ and $y = x^2\text{.}$
### Subsection6.1.2Finding Area with Horizontal Slices
At times, the shape of a geometric region may dictate that we need to use horizontal rectangular slices, rather than vertical ones. For instance, consider the region bounded by the parabola $x = y^2 - 1$ and the line $y = x-1\text{,}$ pictured in Figure 6.1.8. First, we observe that by solving the second equation for $x$ and writing $x = y + 1\text{,}$ we can eliminate a variable through substitution and find that $y+1 = y^2 - 1\text{,}$ and hence the curves intersect where $y^2 - y - 2 = 0\text{.}$ Thus, we find $y = -1$ or $y = 2\text{,}$ so the intersection points of the two curves are $(0,-1)$ and $(3,2)\text{.}$
We see that if we attempt to use vertical rectangles to slice up the area, at certain values of $x$ (specifically from $x = -1$ to $x = 0\text{,}$ as seen in the center graph of Figure 6.1.8), the curves that govern the top and bottom of the rectangle are one and the same. This suggests, as shown in the rightmost graph in the figure, that we try using horizontal rectangles as a way to think about the area of the region.
For such a horizontal rectangle, note that its width depends on $y\text{,}$ the height at which the rectangle is constructed. In particular, at a height $y$ between $y = -1$ and $y = 2\text{,}$ the right end of a representative rectangle is determined by the line, $x = y+1\text{,}$ while the left end of the rectangle is determined by the parabola, $x = y^2-1\text{,}$ and the thickness of the rectangle is $\Delta y\text{.}$
Therefore, the area of the rectangle is
\begin{equation*} A_{\text{rect} } = [(y+1) - (y^2-1)] \Delta y\text{,} \end{equation*}
from which it follows that the area between the two curves on the $y$-interval $[-1,2]$ is approximated by the Riemann sum
\begin{equation*} A \approx \sum_{i=1}^{n} [(y_i+1)-(y_i^2-1)] \Delta y\text{.} \end{equation*}
Taking the limit of the Riemann sum, it follows that the area of the region is
$$A = \int_{y=-1}^{y=2} [(y+1) - (y^2-1)] \, dy\text{.}\label{E-IntWRTy}\tag{6.1.3}$$
We emphasize that we are integrating with respect to $y\text{;}$ this is dictated by the fact that we chose to use horizontal rectangles whose widths depend on $y$ and whose thickness is denoted $\Delta y\text{.}$ It is a straightforward exercise to evaluate the integral in Equation (6.1.3) and find that $A = \frac{9}{2}\text{.}$
Just as with the use of vertical rectangles of thickness $\Delta x\text{,}$ we have a general principle for finding the area between two curves, which we state as follows.
If two curves $x = g(y)$ and $x = f(y)$ intersect at $(g(c),c)$ and $(g(d),d)\text{,}$ and for all $y$ such that $c \le y \le d\text{,}$ $g(y) \ge f(y)\text{,}$ then the area between the curves is
\begin{equation*} A = \int_{y=c}^{y=d} (g(y) - f(y)) \, dy\text{.} \end{equation*}
###### Activity6.1.3
In each of the following problems, our goal is to determine the area of the region described. For each region, (i) determine the intersection points of the curves, (ii) sketch the region whose area is being found, (iii) draw and label a representative slice, and (iv) state the area of the representative slice. Then, state a definite integral whose value is the exact area of the region, and evaluate the integral to find the numeric value of the region's area. Note well: At the step where you draw a representative slice, you need to make a choice about whether to slice vertically or horizontally.
1. The finite region bounded by $x=y^2$ and $x=6-2y^2\text{.}$
2. The finite region bounded by $x=1-y^2$ and $x = 2-2y^2\text{.}$
3. The area bounded by the $x$-axis, $y=x^2\text{,}$ and $y=2-x\text{.}$
4. The finite regions between the curves $x=y^2-2y$ and $y=x\text{.}$
### Subsection6.1.3Finding the length of a curve
In addition to being able to use definite integrals to find the areas of certain geometric regions, we can also use the definite integral to find the length of a portion of a curve. We use the same fundamental principle: we take a curve whose length we cannot easily find, and slice it up into small pieces whose lengths we can easily approximate. In particular, we take a given curve and subdivide it into small approximating line segments, as shown at left in Figure 6.1.9.
To see how we find such a definite integral that measures arc length on the curve $y = f(x)$ from $x = a$ to $x = b\text{,}$ we think about the portion of length, $L_{\text{slice} }\text{,}$ that lies along the curve on a small interval of length $\Delta x\text{,}$ and estimate the value of $L{\text{slice} }$ using a well-chosen triangle. In particular, if we consider the right triangle with legs parallel to the coordinate axes and hypotenuse connecting two points on the curve, as seen at right in Figure 6.1.9, we see that the length, $h\text{,}$ of the hypotenuse approximates the length, $L_{\text{slice} }\text{,}$ of the curve between the two selected points. Thus,
\begin{equation*} L_{\text{slice} } \approx h = \sqrt{ (\Delta x)^2 + (\Delta y)^2 }\text{.} \end{equation*}
By algebraically rearranging the expression for the length of the hypotenuse, we see how a definite integral can be used to compute the length of a curve. In particular, observe that by removing a factor of $(\Delta x)^2\text{,}$ we find that
\begin{align*} L_{\text{slice} }\amp \approx \sqrt{ (\Delta x)^2 + (\Delta y)^2 }\\ \amp = \sqrt{ (\Delta x)^2\left(1 + \frac{(\Delta y)^2}{(\Delta x)^2} \right)}\\ \amp = \sqrt{1 + \frac{(\Delta y)^2}{(\Delta x)^2} } \cdot \Delta x\text{.} \end{align*}
Furthermore, as $n \to \infty$ and $\Delta x \to 0\text{,}$ it follows that $\frac{\Delta y}{\Delta x} \to \frac{dy}{dx} = f'(x)\text{.}$ Thus, we can say that
\begin{equation*} L_{\text{slice} } \approx \sqrt{1 + f'(x)^2} \Delta x\text{.} \end{equation*}
Taking a Riemann sum of all of these slices and letting $n \to \infty\text{,}$ we arrive at the following fact.
Given a differentiable function $f$ on an interval $[a,b]\text{,}$ the total arc length, $L\text{,}$ along the curve $y = f(x)$ from $x = a$ to $x = b$ is given by
\begin{equation*} L = \int_a^b \sqrt{1+f'(x)^2} \, dx\text{.} \end{equation*}
###### Activity6.1.4
Each of the following questions somehow involves the arc length along a curve.
1. Use the definition and appropriate computational technology to determine the arc length along $y = x^2$ from $x = -1$ to $x = 1\text{.}$
2. Find the arc length of $y = \sqrt{4-x^2}$ on the interval $-2 \le x \le 2\text{.}$ Find this value in two different ways: (a) by using a definite integral, and (b) by using a familiar property of the curve.
3. Determine the arc length of $y = xe^{3x}$ on the interval $[0,1]\text{.}$
4. Will the integrals that arise calculating arc length typically be ones that we can evaluate exactly using the First FTC, or ones that we need to approximate? Why?
5. A moving particle is traveling along the curve given by $y = f(x) = 0.1x^2 + 1\text{,}$ and does so at a constant rate of 7 cm/sec, where both $x$ and $y$ are measured in cm (that is, the curve $y = f(x)$ is the path along which the object actually travels; the curve is not a “position function”). Find the position of the particle when $t = 4$ sec, assuming that when $t = 0\text{,}$ the particle's location is $(0,f(0))\text{.}$
### Subsection6.1.4Summary
• To find the area between two curves, we think about slicing the region into thin rectangles. If, for instance, the area of a typical rectangle on the interval $x = a$ to $x = b$ is given by $A_{\text{rect} } = (g(x) - f(x)) \Delta x\text{,}$ then the exact area of the region is given by the definite integral
\begin{equation*} A = \int_a^b (g(x)-f(x))\, dx\text{.} \end{equation*}
• The shape of the region usually dictates whether we should use vertical rectangles of thickness $\Delta x$ or horizontal rectangles of thickness $\Delta y\text{.}$ We desire to have the height of the rectangle governed by the difference between two curves: if those curves are best thought of as functions of $y\text{,}$ we use horizontal rectangles, whereas if those curves are best viewed as functions of $x\text{,}$ we use vertical rectangles.
• The arc length, $L\text{,}$ along the curve $y = f(x)$ from $x = a$ to $x = b$ is given by
\begin{equation*} L = \int_a^b \sqrt{1 + f'(x)^2} \, dx\text{.} \end{equation*}
### SubsectionExercises
###### 5
Find the exact area of each described region.
1. The finite region between the curves $x = y(y-2)$ and $x=-(y-1)(y-3)\text{.}$
2. The region between the sine and cosine functions on the interval $[\frac{\pi}{4}, \frac{3\pi}{4}]\text{.}$
3. The finite region between $x = y^2 - y - 2$ and $y = 2x-1\text{.}$
4. The finite region between $y = mx$ and $y = x^2-1\text{,}$ where $m$ is a positive constant.
###### 6
Let $f(x) = 1-x^2$ and $g(x) = ax^2 - a\text{,}$ where $a$ is an unknown positive real number. For what value(s) of $a$ is the area between the curves $f$ and $g$ equal to 2?
###### 7
Let $f(x) = 2-x^2\text{.}$ Recall that the average value of any continuous function $f$ on an interval $[a,b]$ is given by $\frac{1}{b-a} \int_a^b f(x) \, dx\text{.}$
1. Find the average value of $f(x) = 2-x^2$ on the interval $[0,\sqrt{2}]\text{.}$ Call this value $r\text{.}$
2. Sketch a graph of $y = f(x)$ and $y = r\text{.}$ Find their intersection point(s).
3. Show that on the interval $[0,\sqrt{2}]\text{,}$ the amount of area that lies below $y = f(x)$ and above $y = r$ is equal to the amount of area that lies below $y = r$ and above $y = f(x)\text{.}$
4. Will the result of (c) be true for any continuous function and its average value on any interval? Why? |
## RWM102 Study Guide
### 2a. Determine whether a given real number is a solution of an equation
• How do you know if a value is a solution to an equation?
A value is considered a solution to an equation if you can replace the variable with that value and the outcome is true. To check if a value is a solution, first, substitute the value in place of the variable. Then, simplify the expressions on both sides of the equation. Finally, determine if the final statement is true.
For example, determine if $x = 3$ is a solution to the equation $2x+3=5x-2$. First, substitute 3 in place of $x$ on both sides of the equation: $2(3)+3=5(3)-2$. Next, simplify both sides of the equation: $6+3=15-2$. Finally, determine if the final statement is true: $9=13$ is not true, therefore 3 is not a solution to this equation.
To review, see Solving Linear Equations with One Variable.
### 2b. Simplify equations using addition and multiplication properties
• How do you simplify equations using addition properties?
• How do you simplify equations using multiplication properties?
When solving equations, we can use some basic properties to find the value of the variable. When a constant is added or subtracted with your variable, you can isolate the variable by doing the opposite operation with the same value on both sides of the equation. For example, to solve the equation $x+4=7$, notice that 4 is currently being added to our variable, $x$. To solve this equation, simply subtract 4 from both sides. $x+4-4=7-4$. This will give the answer $x=3$.
Similarly, if the variable is being multiplied or divided by a number you can again perform the opposite operation with the same number on both sides of the equation to solve for the variable. For example to solve the equation $\dfrac{x}{4}=3$, notice the variable $x$ is being divided by 4. Multiply both sides of the equation by 4 to solve for $x$. $\dfrac{x}{4} \times 4=3 \times 4$. Therefore $x = 12$.
### 2c. Find the solution of a given linear equation with one variable
• How do you solve equations using addition and subtraction properties?
• How do you solve equations using multiplication and division properties?
• How do you solve 2-step equations?
• How do you solve equations with variables on both sides?
• How do you solve equations with parentheses?
When solving an equation with multiple steps, always begin by eliminating any addition or subtraction first, before addressing any multiplication or division. For example, we can use addition/subtraction properties in the equation $3x - 2 = 7$, begin by adding 2 to both sides, since there is a -2 already there. This yields, $3x - 2 + 2 = 7 + 2$, which simplifies to $3x = 9$. Now we can use multiplication/division properties to divide both sides by 3, giving us a final answer of $x = 3$.
If there is an equation with numbers and variables on both sides, first you must collect all terms with the variable on one side, and all terms without variables (the constants) on the other side of the equal sign. Then solve the equation as described above. For example, $3x+1=2x+5$ would be solved by subtracting $2x$ from both sides, $3x-2x+1=2x-2x+5$, which gives us $x+1=5$. Then subtract 1 from both sides, $x+1-1=5-1$, which gives the final answer, $x=4$.
If the equation has parentheses, first use the order of operations and/or the distributive property to remove the parentheses, then solve as described above.
To review, see:
### 2d. Determine the number of solutions of a given linear equation in one variable
• Can a linear equation in one variable have no solutions?
• Can a linear equation in one variable have more than one solution?
In certain instances, equations have solutions that do not end with the variable equal to a single specific value, as we saw in 2b and 2c. When solving an equation, if all the variables cancel, we end up with a statement with constants on both sides. For example, $3x+5=3x-1$. When we solve this equation, we end up with $5=-1$. Since this is false, this equation has no solutions.
On the other hand, if the final solution is a true statement, such as $3=3$, then we say the equation has "infinitely many solutions", or "all real numbers", because any value for the variable will still make $3=3$ true.
To review, see Solving Linear Equations with One Variable
### 2e. Solve a literal equation for the given variable
• How do you solve an equation for a specific variable?
Sometimes equations have more than one variable, and instead of finding a value for a specific variable, we need to simply solve the equation for a variable. For example, if we solve the equation $3v+s=t-2$ for the variable $s$, we simply use the same skills we have learned for solving equations by subtracting $3v$ from both sides, giving us an answer of $s=t-2-3v$.
To review, see Solving for One Variable
### 2f. Rearrange formulas to isolate a quantity of interest
• How can you use a formula set up for a specific variable to instead find a different variable?
Sometimes you may find a formula given to you that isn't set up to give you the quantity you desire. For example, in the formula $F+V=E+2$, the formula is not currently set up to give you the value of $V$. You can simply solve the formula for $V$, as we have done before, to get the new formula, $V=E+2-F$. Now you can use the values you are given for $E$ and $F$ to find $V$.
To review, see Solving for One Variable
### Unit 2 Vocabulary
This vocabulary list includes terms listed above that students need to know to successfully complete the final exam for the course.
• solution
• statement
• opposite operation
• parentheses
• no solutions
• infinitely many solutions
• all real numbers |
# How do you write 15/6 as a mixed number?
Nov 7, 2016
$2 \frac{1}{2}$
#### Explanation:
Lets start off with how many 6's fit into 15...
$1 \cdot 6 = 6$
$2 \cdot 6 = 12$
$3 \cdot 6 = 18$
We know that the whole number will be 2.
Since $2 \cdot 6 = 12$, $15 - 12 = 3$
So far we have $2 \frac{3}{6}$
We can simplify $\frac{3}{6}$ into $\frac{1}{2}$, making our final answer $2 \frac{1}{2}$.
Nov 7, 2016
$2 \frac{1}{2}$
#### Explanation:
Imagine a cake cut into sixths, 6 equal pieces.
How many whole cakes can you make with 15 of these pieces?
2 whole cakes. That takes 12 pieces with 3 left over.
And those 3 pieces are half a cake!!! |
# Thread: Find the area of a triangle.
1. ## Find the area of a triangle.
Find the area of a triangle enclosed by axes and tangent to y = 1/x
2. ## Re: Find the area of a triangle.
Where do you run into your problem? Finding the tangent line? Also, there's an infinite amount of triangles that could be made. Do they specify anything else?
4. ## Re: Find the area of a triangle.
Originally Posted by joshuaa
Well, let's see. If the function is \displaystyle \begin{align*} y = \frac{1}{x} \end{align*}, the derivative is \displaystyle \begin{align*} \frac{dy}{dx} = -\frac{1}{x^2} \end{align*}. This tells us the gradient of the tangent line at any point x.
So the tangent line is of the form \displaystyle \begin{align*} y = \left(-\frac{1}{x^2}\right)x + c = -\frac{1}{x} + c \end{align*}.
c is the y intercept. To find the x intercept, we'll let y = 0, and we find that
\displaystyle \begin{align*} 0 &= -\frac{1}{x} + c \\ \frac{1}{x} &= c \\ x &= \frac{1}{c} \end{align*}.
So the height of the triangle is c, the length is \displaystyle \begin{align*} \frac{1}{c} \end{align*}, which means the area is
\displaystyle \begin{align*} A &= \frac{1}{2}(c)\left(\frac{1}{c}\right) \\ &= \frac{1}{2} \end{align*}
So the area is \displaystyle \begin{align*} \frac{1}{2}\,\textrm{unit}^2 \end{align*}
5. ## Re: Find the area of a triangle.
Don't think that is correct. Example consider tangent at (1,1) Gradient is -1 Equation of tangent is y=-x+2 Intercepts are both 2 Area of triangle is 2
In general consider tangent at (k,m) Gradient of tangent is -1/k^2 Get the equation of the tangent and the intercepts. Knowing also that km=1 wil get area of triangle=2
6. ## Re: Find the area of a triangle.
Originally Posted by Prove It
So the tangent line is of the form \displaystyle \begin{align*} y = \left(-\frac{1}{x^2}\right)x + c = -\frac{1}{x} + c \end{align*}.
This isn't the equation of a line. (See post #5.)
7. ## Re: Find the area of a triangle.
Thank you very much. That was helpful |
# Unit 4Dilation and Similarity
## Lesson 1
### Learning Focus
Describe the essential features of a dilation transformation.
### Lesson Summary
In this lesson, we observed the key features of a dilation transformation while figuring out how a photocopy machine enlarges an image. We learned how to locate points on a dilated image by using the center and scale factor that define a specific dilation. We observed that “the same shape, different size” relationship between the pre-image and image figures are a consequence of the way dilations are defined.
## Lesson 2
### Learning Focus
Create similar figures by dilation given the scale factor.
Prove a theorem about the midlines of a triangle using dilations.
### Lesson Summary
In this lesson, we extended our understanding of similar figures. Since corresponding segments of similar figures are proportional, and dilations produce similar figures, corresponding parts of an image and its pre-image after a dilation are proportional. We also learned that corresponding line segments in a dilation are parallel. These two observations provided a tool for proving a theorem about the midlines of a triangle, a segment connecting the midpoints of two sides of a triangle.
## Lesson 3
### Learning Focus
Determine criteria for triangle similarity.
### Lesson Summary
In this lesson, we examined what it means to say that two figures are similar geometrically, and we examined conditions under which two triangles will be similar. We wrote and justified several theorems for triangle similarity criteria.
## Lesson 4
### Learning Focus
Prove that a line drawn parallel to one side of a triangle that intersects the other two sides divides the other two sides proportionally.
### Lesson Summary
In a previous lesson, we learned that a midline of a triangle, a line that passes through the midpoints of two of the sides, is parallel to the third side and half its length. In this lesson, we extended this theorem to include other segments that cut the sides of a triangle proportionally. We also proved a nonintuitive “side-splitting” theorem about the multiple segments formed when multiple lines parallel to a side of a triangle cut the other two sides of the triangle.
## Lesson 5
### Learning Focus
Practice using geometric reasoning in computational work.
### Lesson Summary
In this lesson, we drew upon a variety of theorems to support the computational work of finding missing sides and angles. To identify which theorems to use, we had to examine the available features of the diagram. For many measurements, multiple strategies could be used. We also used the diagram, along with our computed measurements, to develop and justify a conjecture for the sum of the interior angles of any polygon, similar to the theorem we proved previously about the sum of the interior angles in a triangle.
## Lesson 6
### Learning Focus
Prove the Pythagorean theorem algebraically.
### Lesson Summary
In today’s lesson, we learned that drawing the altitude of a right triangle from the vertex at the right angle to the hypotenuse divides the right triangle into two smaller triangles that are similar to each other and to the original right triangle. We were able to prove the Pythagorean theorem using proportionality statements about the three similar triangles.
## Lesson 7
### Learning Focus
Investigate corresponding ratios of right triangles with the same acute angle.
### Lesson Summary
In this lesson, we learned about some special ratios, called trigonometric ratios, that occur in right triangles. If two right triangles have a pair of corresponding acute angles that are congruent, the right triangles will be similar. Therefore, corresponding ratios of the sides of these two right triangles will be equal. This observation is so useful when working with right triangles that have the same acute angle that values of these ratios were recorded in tables for each acute angle between and .
## Lesson 8
### Learning Focus
Solve missing sides and angles in a right triangle.
Examine properties of trigonometric expressions.
### Lesson Summary
In this lesson, we examined some relationships between trigonometric ratios, such as a relationship between the sine and cosine of complementary angles. We were able to use the properties of a right triangle, including the Pythagorean theorem that describes a relationship between the lengths of the sides, to justify the observations we made today.
## Lesson 9
### Learning Focus
Solve for the missing side and angle measures in a right triangle.
### Lesson Summary
In this lesson, we extended our strategies for finding unknown sides and angles in a right triangle beyond using the Pythagorean theorem and the angle sum theorem for triangles, since sometimes we don’t have enough information in terms of side lengths or angle measures to use these theorems. We found that trigonometric ratios are useful in solving for unknown sides and that inverse trigonometric relationships are useful for finding unknown angles in a right triangle. Adding these tools allows us to find all of the missing sides and angles in a right triangle given two pieces of information: two sides of the triangle or one side and an angle.
## Lesson 10
### Learning Focus
Solve application problems using trigonometry.
### Lesson Summary
In this lesson, we learned about the modeling process and how to use right triangle trigonometry to model many different types of applications, even applications that didn’t naturally include right triangles. A right triangle became a tool for representing a situation so we could draw upon trigonometric ratios and inverse trigonometric relationships to answer important problems in construction, aviation, transportation, and other contexts.
## Lesson 11
### Learning Focus
Find missing sides of special right triangles without using trigonometry.
### Lesson Summary
In this lesson, we learned there are some special right triangles for which missing sides of the triangle can be found when only one side is known, without using trigonometry! This happens when the right triangle is the result of decomposing a familiar shape, such as a square or an equilateral triangle, into two congruent right triangles. |
# Math camera app
Math camera app is a mathematical instrument that assists to solve math equations. We can solve math word problems.
## The Best Math camera app
Here, we debate how Math camera app can help students learn Algebra. When you're solving fractions, you sometimes need to work with fractions that are over other fractions. This can seem daunting at first, but it's actually not too difficult once you understand the process. Here's a step-by-step guide to solving fractions over fractions. First, you need to find a common denominator for both of the fractions involved. The easiest way to do this is to find the least common multiple of the two denominators. Once you have the common denominator, you can rewrite both fractions so they have this denominator. Next, you need to add or subtract the numerators of the two fractions in order to solve for the new fraction. Remember, the denominators stays the same. Finally, simplify the fraction if possible and write your answer in lowest terms. With a little practice, you'll be solving fractions over fractions like a pro!
In mathematics, a word phrase is a string of words that can be interpreted as a mathematical expression. For example, the phrase "two plus three" can be interpreted as the sum of two and three. Similarly, the phrase "nine divided by three" can be interpreted as the division of nine by three. Word phrases can be used to represent a wide variety of mathematical operations, including addition, subtraction, multiplication, and division. They can also be used to represent fractions and decimals. In addition, word phrases can be used to represent complex numbers and equations. As such, they provide a powerful tool for performing mathematical operations.
For example, if you have the equation 2^x=8, you can take the logarithm of both sides to get: log(2^x)=log(8). This can be rewritten as: x*log(2)=log(8). Now all you need to do is solve for x, and you're done! With a little practice, solving for exponents will become second nature.
Once the equation has been factored, you can solve each factor by setting it equal to zero and using the quadratic formula. Another method for solving the square is to complete the square. This involves adding a constant to both sides of the equation so that one side is a perfect square. Once this is done, you can take the square root of both sides and solve for the variable. Finally, you can use graphing to solve the square. To do this, you will need to plot the points associated with the equation and then find the intersection of the two lines. Whichever method you choose, solving the square can be a simple process as long as you have a strong understanding of algebra.
A trinomial is an algebraic expression that contains three terms. The most common form of a trinomial is ax^2+bx+c, where a, b, and c are constants and x is a variable. Solving a trinomial equation means finding the value of x that makes the equation true. There are a few different methods that can be used to solve a trinomial equation, but the most common is factoring. To factor a trinomial, you need to find two numbers that multiply to give the product of the two constants (ac) and add up to give the value of the middle term (b). For example, if you are given the equation 2x^2+5x+3, you would need to find two numbers that multiply to give 6 (2×3) and add up to give 5. The only numbers that fit this criteria are 1 and 6, so you would factor the equation as (2x+3)(x+1). From there, you can use the zero product rule to solve for x. In this case, either 2x+3=0 or x+1=0. Solving each of these equations will give you the values of x that make the original equation true. While factoring may seem like a difficult task at first, with a little practice it can be easily mastered. With this method, solving trinomials can be quick and easy.
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Optimization calculus help Help solving math word problems Free math solver app Free equation solver Solve for an angle in right triangles |
Teacher resources and professional development across the curriculum
Teacher professional development and classroom resources across the curriculum
Session 5, Part C:
The Midline Theorem
In This Part: The Midline Cut | Some Geometry Facts | Proving the Midline Theorem
The midline theorem claims that cutting along the midline of a triangle creates a segment that is parallel to the base and half as long. Does that seem reasonable? To prove the midline cut works, you need to use some geometry facts that you may already have encountered. If not, take some time to consider why these statements are true. Note 5 Fact 1: Vertical angles (the angles opposite each other when two lines intersect) are congruent (they have the same measure). Why: We can show why, for example, m1 = m3: m1 + m2 = 180° and m2 + m3 = 180°, since in both cases the two angles together create a "straight angle." So m1 + m2 = m2 + m3 = 180°. Subtracting m2 from each part of the equation, we see that m1 = m3 = 180° - m2.
Fact 2: If two triangles have two sides that are the same length, and the angle between those two sides has the same measure, then the two triangles are congruent. The two triangles must have the same size and shape, so all three sides have the same length, and all three angles have the same measure. This is known as SAS (side-angle-side) congruence. The single tick indicates the two sides that are the same length. The double tick indicates thetwo sides that are the same length. The angle markings indicate that those two angles have the same measure. Why: The easiest way to be convinced of the fact that the two triangles are congruent is to draw some triangles. Draw a segment 2 inches long and a segment 3 inches long, with a 60° angle between them. Is there more than one way to complete the triangle? Come up with other cases to try.
Fact 3: There are many equivalent definitions of "parallelogram":
• A quadrilateral with both pairs of opposite sides parallel • A quadrilateral with both pairs of opposite angles congruent • A quadrilateral with both pairs of opposite sides congruent • A quadrilateral with one pair of opposite sides both congruent and parallel • The last definition is the one that will come in handy here. Go through the steps of understanding a definition in Session 3 if you're not sure why it works.
You may want to print this page for use in the problems that follow.
Session 5: Index | Notes | Solutions | Video |
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Prisms and Cylinders Volume
Sep 13, 2022
Key Concepts
• Use volume of a prism.
• Find the volume of an oblique cylinder.
• Solve a real-world problem.
Introduction
Cavalieri’s Principle
Both the shapes given below have equal heights h and equal cross-sectional areas B. Mathematician Bonaventura Cavalieri’s (1598–1647) claimed that both the solids have the same volume.
If two solids have the same height and the same cross-sectional area at every level, then they have the same volume.
Use volume of a prism
Example 1:
Use the measurements given to solve for x
Solution:
A side length of the cube is x feet.
V = S3
100 = x3
4.64 = x
So, the height, width, and length of the given cube are about 4.64 feet.
Find the volume of an oblique cylinder
Example 2:
Find the volume of the oblique cylinder. Round to the nearest tenth if necessary.
Solution:
Cavalieri’s Principle allows you to use Theorem to find the volume of the oblique cylinder.
V = πr2h (Formula for the volume of a cylinder)
= π(82)(30) (Substitute known values)
= 1920π (Simplify)
» 6028.8 (Use a calculator)
The volume of the oblique cylinder is about 6028.8 cm3
Example 3:
Find the volume of the oblique cylinder. Round to the nearest tenth if necessary.
Solution:
Cavalieri’s Principle allows you to use Theorem to find the volume of the oblique cylinder.
r = 7.5 mm and h = 15.2 mm.
V = πr2h (Formula for the volume of a cylinder)
= π(7.52)(15.2) (Substitute known values)
= 855π (Simplify)
» 2684.7 (Use a calculator)
The volume of the oblique cylinder is about 2684.7 mm3
Solve a real-world problem
Example 4:
The base of a rectangular swimming pool is sloped; so, one end of the pool is 6 feet deep and the other end is 3 feet deep, as shown in the figure. If the width is 15 feet, find the volume of water it takes to fill the pool.
Solution:
The swimming pool is a combination of a rectangular prism and a trapezoidal prism. The base of the rectangular prism is 6 ft by 10 ft and the height is 15 ft. The bases of the trapezoidal prism are 6 ft and 3 ft long and the height of the base is 10 ft. The height of the trapezoidal prism is 15 ft. The total volume of the solid is the sum of the volumes of the two prisms.
The volume of water it takes to fill the pool is 1575 ft3
Exercise
• As per _____________________, if two solids have the same height and the same cross-sectional area at every level, then they have the same volume.
• In what type of units is the volume of a solid measured?
• Use the measurements given to solve for x.
• Use the measurements given to solve for x.
• Find the volume of each oblique cylinder. Round to the nearest tenth if necessary.
• The volume of a right cylinder is 684π cubic inches and the height is 18 inches. Find the radius.
Find the volume of the oblique prism shown below.
• Use Cavalieri’s Principle to find the volume of the oblique prism or cylinder. Round your answer to two decimal places.
• Use Cavalieri’s Principle to find the volume of the oblique prism or cylinder. Round your answer to two decimal places.
• In the concrete block shown, the holes are 8 inches deep. Find the volume of the block using the Volume Addition Postulate.
What have we learned
• Use volume of a prism and find an unknown value.
• Find the volume of an oblique cylinder using the formula.
• Solve a real-world problem in terms of volume.
Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […] |
# Number theory
Number theory is a branch of mathematics which helps to study the set of positive whole numbers, say 1, 2, 3, 4, 5, 6,. . . , which are also called the set of natural numbers and sometimes called “higher arithmetic”.
Number theory helps to study the relationships between different sorts of numbers. Natural numbers are separated into a variety of times. Here are some of the familiar and unfamiliar examples with quick number theory introduction.
## Introduction to Number Theory
In number theory, the numbers are classified into different types, such as natural numbers, whole numbers, complex numbers, and so on. The sub-classifications of the natural number are given below:
• Odd Numbers – 1, 3, 5, 7, 9, 11, 13, 15, 17, 19…..
• Even Numbers – 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22 . . .
• Square Numbers – 4, 9, 16, 25, 36, 49, 64, 81,100 . . .
• Cube Numbers – 8, 27, 64, 125, 216, 343, 512 . . .
• Prime Numbers – 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47,53, 59, 61 . . .
• Composite Numbers – 4, 6, 8, 9, 10, 12, 14, 15, 16,18, 20, 21, 22, 24 . . .
• 1 (modulo 4) Numbers – 1, 5, 9, 13, 17, 21, 25, . . .
• 3 (modulo 4) Numbers – 3, 7, 11, 15, 19, 23, 27, . . .
• Triangular Numbers – 3, 6, 10, 15, 21, 28, 36, 45,. . .
• Perfect Numbers – 6, 28, 496, 8128, . . .
• Fibonacci Numbers -1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89. . .
Many of these types of numbers like odd, even, square, cube prime and composite numbers are already familiar to you. Other cases, such as the “modulo 4” numbers, Triangular numbers, perfect numbers and Fibonacci numbers are not familiar to you.
## Number Theory Topics
Even Numbers: The numbers that are evenly divided by 2 are called even numbers.
Odd Numbers: The numbers that are not evenly divided by 2 are called odd numbers.
Square Numbers: A number multiplied by itself is called square numbers
Cube Numbers: A number multiplied by itself 3 times is called cube numbers.
Prime numbers: If a number has only two factors: 1 and the number is called prime numbers
Co Prime Numbers: Two numbers are called co prime numbers, if the highest common factor between the two is 1.
Composite Numbers: Composite number has more than two factors. The composite numbers are numbers which are not prime numbers. The number 1 is neither prime nor composite.
Modulo 4 Numbers: A number is said to be 1 (modulo 4 ) number if it leaves a remainder 1 when divided by 4.Similarly, if a number leaves a remainder 3 when divided by 4, it is said to be 3 (modulo 4) number.
Triangular Numbers: A number is said to be a triangular number when that number of pebbles can be arranged in a triangle using one pebble at the top, two pebbles in next row, three pebbles in next row and so on.
Fibonacci Numbers: Fibonacci numbers are created starting with 1 and 1, then get the next number in the list and adds the previous two numbers. Say, 1+1 =2 and then add 1+2 you get 3, then adds 2+3 gives 5, then 3+5 gives 8 and so on.
## Applications of Number Theory
Here are some of the most important number theory applications. Number theory is used to find some of the important divisibility tests, whether a given integer m divides the integer n. Number theory have countless applications in mathematics as well in practical applications such as
• Security System like in banking securities
• E-commerce websites
• Coding theory
• Barcodes
• Making of modular designs
• Memory management system
• Authentication system
It is also defined in hash functions, linear congruences, Pseudorandom numbers and fast arithmetic operations. |
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# Find the square root of 125 by long division method.
Last updated date: 20th Jun 2024
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Hint: We will first select 1 as the number and then multiply it with such number so that (10 + x)x is less than or equal to 25 and thus, we have the square root.
Now, we have 125 with us whose square root we require to find.
We can write 125 as 1,25 for our understanding.
So, the first digit will be 1 whose square is 1 and will be less than or equal to 1.
Like in the above picture, we took 1, multiplied it by 1 and then after its cancellation and all, we got 25 as the term for next division.
Now, we will write 2 as the divisor, 25 as dividend and we will have to find a digit x such that (20 + x) x is maximum but still less than 25. That digit would be definitely 1, because 21 (1) < 25 < 22 (2).
So, we get with us the following picture:-
Now, we get 4 as the next dividend and the divisor will be 22. Therefore, we will have to get the zeroes from above after decimal twice and then find such a y that (220 + y) y is less than 400. That digit is but obviously 1 and thus we get the following picture as the next step:
After the division, as we see above, we have got 179 as dividend and thus 222 as the divisor. So, we will have to get two zeroes from above.
We now have to find such a z such that (2220 + z) z is maximum but still less than or equal to 17900. And, if we keep on going like this, we will get the following division:-
Hence, the square root of 125 is 11.1803…
Note:
The students must note that in the long division method as done above, we try to get 0 as the dividend for the next step so as we have a definite square root.
The students must also note that: we always get the divisor of the next step by multiplying the quotient by 2 and that will be the divisor we use in the next step.
The students should also know that after getting the divisor we have to put such a digit as the end so that if we multiply the whole new divisor we got after putting that digit, with that digit only, we get the maximum product which is still less than the dividend we use in that step. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Compound Inequalities
Multiple inequalities associated by 'and' and 'or' statements
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Compound Inequalities
Teacher Contributed
Real World Applications – Algebra I
Topic
Where do you see compound inequalities in your world?
Student Exploration
In this activity, we will apply our basic understanding of compound inequalities to things and activities we see around us. In the concept, you learned about two different types of compound inequalities.
An example of a compound inequality involving the word “and” can be thought of as an elevator. An elevator for a building must be greater than (or equal to) floor zero and the top floor. For a 20-floor apartment building with no basement, we can represent the elevator reaching all of the floors between the ground floor and the top floor, or . In words, we can say that the elevators must be able to reach greater than or equal to the ground floor AND less than or equal to the floor.
An example of a compound inequality involving the word “or” can be thought of a car burning fuel. A car burns more gasoline when it’s going either really slow or really fast. When we look at how the different ways driving a car can be more expensive, a car must be going below 40 miles an hour and higher than 70 miles an hour. The inequality to represent this is the speed is less than 40 miles per hour OR greater than 70 miles per hour. We can say, or .
Extension Investigation
Where do you see compound inequalities around you? How could you write the inequality? Which type of wording would you use – “and,” or “or” and why?
Notes/Highlights Having trouble? Report an issue.
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# Chapter 5: Multiplication and division
## 5.1 Multiplication and division of integers
Last year you learnt how to add and subtract integers. The set of integers is all the positive and negative whole numbers, including zero. You used a number line to represent integers.
• Numbers to the right of zero on the number line are called positive numbers. We represent them with or without a plus sign in front. For example, the positive number five may be written as $+5$ or just $5$.
• Numbers to the left of zero are called negative numbers. We always represent them with a minus sign in front. For example, the negative number three is written as $-3$.
• The number $0$ is neither positive nor negative.
integer The integers are the positive and negative whole numbers, including zero.
positive number Positive numbers are numbers that are larger than zero; they lie to the right of zero on a number line.
negative number Negative numbers are numbers that are smaller than zero; they lie to the left of zero on a number line.
Positive and negative numbers are sometimes called directed numbers. This is because the sign of the number (positive or negative) tells us in which direction from zero they appear on the number line.
directed number Directed numbers are positive and negative numbers. The sign of a directed number tells us in which direction from zero it lies on the number line.
In this section, you will learn how to multiply and divide directed numbers.
### Multiplying directed numbers
From earlier grades you know that we can multiply numbers in any order. For example, $2\times3=6$ and $3\times2=6$.
In Mathematics, the commutative laws of addition and multiplication state that the order in which we add or multiply numbers does not affect the answer. This means we can swap numbers around and still get the same answer.
From last year you know that multiplication may be seen as a form of addition. Therefore:
or
This means that you get a positive answer when you multiply two positive numbers:
We may use the same method to find out what happens when we multiply a positive number by a negative number or a negative number by a positive number.
or
Remember $(-3)+(-3)$ means we start at $-3$ on the number line and count three units to the left. This brings us to $-6$.
This shows that you get a negative answer when you multiply two numbers with opposite signs:
It is a bit more complicated to explain what happens when we multiply two negative numbers. One way of thinking states that, for all the laws of Mathematics to work, we accept by convention that $-1\times-1=+1$. Another way is to look at number patterns. This table shows multiplication by $-2$:
\begin{array}{|c|c|c|c|c|} \hline -2 & \times & & = & \newline -2 & \times & & = & \newline -2 & \times & & = & \newline -2 & \times & & = & \newline -2 & \times & & = & \newline -2 & \times & & = & \newline -2 & \times & & = & \newline -2 & \times & & = & \newline \hline \end{array}
Let's insert a number pattern in the middle column:
\begin{array}{|c|c|r|c|c|} \hline -2 & \times & +4 & = & \newline -2 & \times & +3 & = & \newline -2 & \times & +2 & = & \newline -2 & \times & +1 & = & \newline -2 & \times & 0 & = & \newline -2 & \times & -1 & = & \newline -2 & \times & -2 & = & \newline -2 & \times & -3 & = & \newline -2 & \times & -4 & = & \newline \hline \end{array}
We know that a negative number multiplied by a positive number gives a negative number as an answer. Use this knowledge to complete the first four rows of the table:
\begin{array}{|r|c|r|c|r|} \hline -2 & \times & +4 & = & -8 \newline -2 & \times & +3 & = & -6 \newline -2 & \times & +2 & = & -4 \newline -2 & \times & +1 & = & -2 \newline -2 & \times & 0 & = & \newline -2 & \times & -1 & = & \newline -2 & \times & -2 & = & \newline -2 & \times & -3 & = & \newline -2 & \times & -4 & = & \newline \hline \end{array}
Now, without looking at what is happening in the rows, simply complete the pattern of the numbers in the last column from top to bottom:
\begin{array}{|r|c|r|c|r|} \hline -2 & \times & +4 & = & -8 \newline -2 & \times & +3 & = & -6 \newline -2 & \times & +2 & = & -4 \newline -2 & \times & +1 & = & -2 \newline -2 & \times & 0 & = & 0 \newline -2 & \times & -1 & = & +2 \newline -2 & \times & -2 & = & +4 \newline -2 & \times & -3 & = & +6 \newline -2 & \times & -4 & = & +8 \newline \hline \end{array}
The last four rows of the table show that:
Whenever you multiply directed numbers, remember:
$(+)\times(+)=(+)$
$(-)\times(-)=(+)$
$(+)\times(-)=(-)$
$(-)\times(+)=(-)$
Therefore, multiplying:
• the same signs gives $+$
• opposite signs gives $-$.
There are different ways in which we may write multiplication. The following notations all mean the same:
• $-5 \times -2$
• $-5 \times (-2)$
• $(-5) \times -2$
• $(-5) \times (-2)$
• $(-5) \times (-2)$
Note that $(-5)(-2)$ and $-5-2$ does NOT mean the same. When you did algebraic expressions last year, you learnt that:
• brackets written next to each other mean multiply: $(-5)(-2)=-5\times-2$
• plus and minus signs between two numbers separate terms: $\require{enclose} -5\enclose{circle}[mathcolor="red"]{-}2=-7$
### Worked example 5.1: Multiplying directed numbers
Carry out the calculation:
1. Step 1: Identify plus and minus signs that separate terms.
2. Step 2: Do multiplication before addition or subtraction. Use the multiplication rules for the signs when you multiply.
Remember the correct order of operations: Brackets, Of, Division, Multiplication, Addition, Subtraction
3. Step 3: Do addition and subtraction last.
Remember, to subtract a negative number is the same as adding the positive of that number.
### Exercise 5.1: Multiply directed numbers
Calculate each of the following.
1. Remember that when you multiply any number by 1, the number stays the same. But you still need to use the multiplication rules to determine the sign of the answer.
2. \begin{align} &(-2)(-8)(4)(5) \\ =&\;(+16)(4)(5) \\ =&\;(64)(5) \\ =&\;320 \end{align}
3. \begin{align} &(2)(-2)(-4)(4)(-1) \\ =&\;(-4)(-4)(4)(-1) \\ =&\;(+16)(4)(-1) \\ =&\;(64)(-1) \\ =&\;-64 \end{align}
4. \begin{align} &-2\times-2\times5-4\times2 \\ =&\;+4\times5-8 \\ =&\;20-8 \\ =&\;12 \end{align}
5. \begin{align} &-3-8-(2)(-1) \\ =&\;-3-8-(-2) \\ =&\;-3-8+(+2) \\ =&\;-3-6 \\ =&\;-9 \end{align}
6. \begin{align} &2\times-1\times-3+(-2)-(-2)(-4)(-1) \\ =&\;-2\times-3+(-2)-(+8)(-1) \\ =&\;+6+(-2)-(-8) \\ =&\;4+(+8) \\ =&\;12 \end{align}
### Dividing directed numbers
From earlier grades you know that we can divide things like money and sweets amongst a certain number of people. For example, if you divide a packet of 48 sweets amongst 3 children, each child gets 16 sweets. We may write this as: $48\div3=16$ or $\frac{48}{3}=16$.
You get a positive number when you divide two positive numbers:
We can use bank balances to explain the division of positive and negative numbers. Suppose three friends have a shop. In a certain month, the shop does not do well. It makes a loss and its bank balance shows as $$- ₦15,000$$. This means the business owes the bank money. If the debt is shared equally amongst the three friends, each of their bank statements will show $$- ₦5,000$$. We may write this as: $\frac{-15,000}{3}=-5,000$.
You get a negative number when you divide a negative number by a positive number:
Similarly, suppose the shop does well in another month and its bank balance is $$+ ₦6,000$$. The three friends want to withdraw this money in three equal portions, so that they each get $$₦2,000$$. Each of the withdrawals will be shown as a negative on the shop's bank statement. We can say there are 3 "negative" transactions that each remove the same amount from the bank account. So we may write this as: $\frac{6,000}{-3}=-2,000$.
You get a negative number when you divide a positive number by a negative number:
Suppose the bank charges the shop $$₦750$$ bank fees. Somebody at the bank makes a mistake and accidentally puts through the transaction a few times instead of only once. The shop's bookkeeper sees an amount of $$- ₦3,000$$ on the bank statement and wants to work out how many times the bank fees were put through. She will do the following calculation: $\frac{-3,000}{-750}=4\text{ times}$.
You get a positive number when you divide a negative number by a negative number:
Whenever you divide directed numbers, remember:
$\frac{(+)}{(+)}=(+)$
$\frac{(-)}{(-)}=(+)$
$\frac{(+)}{(-)}=(-)$
$\frac{(-)}{(+)}=(-)$
Therefore, dividing:
• the same signs give $+$
• opposite signs give $-$.
Note that, because the division of two opposite signs always gives a negative answer, it does not matter where we put the minus sign when we write a fraction:
When you give a final answer, you should write the minus sign in front of the whole fraction, as shown in the last step above.
### Worked example 5.2: Dividing directed numbers
Calculate:
1. Step 1: Identify plus and minus signs that separate terms.
2. Step 2: First simplify the numerator. Do multiplication before addition or subtraction. Apply the rules for signs when you multiply.
3. Step 3: Now simplify the denominator. Do multiplication before addition or subtraction. Apply the rules for signs when you multiply.
4. Step 4: Simplify the fraction. Apply sign rules when you divide.
### Exercise 5.2: Divide directed numbers
Calculate each of the following. Give your answers as mixed numbers where necessary.
1. Remember that the number 1 is also a divisive identity. This means when you divide any number by 1, the number stays the same. But you need to apply the rules for signs.
2. \begin{align} &\frac{(-2)(4)(-6)}{(-4)(3)} \\ =&\;\frac{(-8)(-6)}{-12} \\ =&\;\frac{+48}{-12} \\ =&\;-4 \end{align}
3. \begin{align} &\frac{5\times-1+(-15)}{-10\times-4} \\ =&\;\frac{-5+(-15)}{+40} \\ =&\;\frac{-20}{40} \\ =&\;-\frac{1}{2} \end{align}
4. \begin{align} &\frac{(-3)(4)(-2)(2)-(-2)(-3)}{(-1)(-2)(-3)} \\ =&\;\frac{(-12)(-2)(2)-(+6)}{(2)(-3)} \\ =&\;\frac{(24)(2)-(+6)}{-6} \\ =&\;\frac{(48)-(+6)}{-6} \\ =&\;\frac{42}{-6} \\ =&\;-7 \end{align}
5. Remember that zero divided by any number is zero.
\begin{align} &\frac{-3-5(-1)-2}{-1\times-5} \\ =&\;\frac{-3+5-2}{5} \\ =&\;\frac{-5+5}{5} \\ =&\frac{0}{5} \\ =&\;0 \end{align}
6. \begin{align} &\frac{8\times-3\times-1-4(-1)(-2)}{-2-4\times-2} \\ =&\;\frac{-24\times-1-(-4)(-2)}{-2+8} \\ =&\;\frac{+24-(+8)}{6} \\ =&\;\frac{16}{6} \\ =&\;2\frac{4}{6} \\ =&\;2\frac{2}{3} \end{align}
## 5.2 Squares and square roots
### Squares
In Chapter 1 you learnt that a square is the answer we get when we multiply a number by itself. You should revise the section on squares in Chapter 1 before you proceed.
square A square is the answer obtained when any number is multiplied by itself.
From what you have learnt about multiplying directed numbers, you should be able to see that a square is always positive. A negative number cannot be a square, because there is no number that you can multiply by itself to give a negative number as an answer. This is because:
• a positive number multiplied by itself gives a positive answer, for example $+8\times+8=+64$
• a negative number multiplied by itself also gives a positive answer, for example $-7\times-7=+49$.
You have also learnt that a perfect square can always be expressed as a product of prime factors that all have a power of two. For example, we know that 144 is a perfect square because we can write it as: $2^2\times2^2\times3^2$. We known that 72 is not a perfect square, because we cannot write it as a product of prime factors that all have the power two. The factors of 72 are $2\times 2^2\times 3^2$.
### Worked example 5.3: Determining square numbers
Determine whether $-162$ is a square. It not, find the smallest number by which it must be multiplied so that the answer is a square.
1. Step 1: Determine whether the number is positive or negative. If it is negative, it is not a square. It must be multiplied by at least $-1$.
$-162$ is negative, so it is not a square.
It must be multiplied by at least $-1$: $-162\times-1=+162$
2. Step 2: Express the number from Step 1 as a product of its prime factors.
\begin{array}{r | r} 2 & 162 \newline \hline 3 & 81 \newline \hline 3 & 27 \newline \hline 3 & 9 \newline \hline 3 & 3 \newline \hline & 1 \end{array}
3. Step 3: Group the same prime factors in twos and write them in index form.
4. Step 4: Evaluate the powers of the grouped prime factors. If there are any powers NOT equal to 2, the number you obtained in Step 1 is still not a square.
In the prime factors of 162, the power of the 2 is NOT equal to 2. Therefore, 162 is NOT a square.
5. Step 5: Work out by which prime number(s) we must multiply so that all the powers are equal to 2.
In this case, we must multiply by another 2: $2\times2\times3^2\times3^2$
The new grouped prime factors are: $2^2\times3^2\times3^2$
The new number is: $2\times162=324$
6. Step 6: Combine the answers from Step 1 and Step 5.
For $-162$ to be a square, it must be multiplied by $-1$ and then by $2$: $-1\times2=-2$
If $-162$ is multiplied by $-2$, the answer is a square: 324.
### Exercise 5.3: Determine square numbers
Determine whether the following numbers or fractions are squares or not. If not, find the smallest number by which each number or fraction must be multiplied to get a square.
1. 196
\begin{array}{r | r} 2 & 196 \newline \hline 2 & 98 \newline \hline 7 & 49 \newline \hline 7 & 7 \newline \hline & 1 \end{array}
196 is a square.
2. $\dfrac{-50}{242}$
The numerator is negative, so it is not a square.
The fraction must be multiplied by at least $-1$.
$\dfrac{25}{121}$ is a square.
$\dfrac{-50}{242}$ must be multiplied by $-1$ and simplified to give the square $\dfrac{25}{121}$.
3. The number is negative, so it is not a square.
It must be multiplied by at least $-1$: $-700\times-1=+700$
\begin{array}{r | r} 2 & 700 \newline \hline 2 & 350 \newline \hline 5 & 175 \newline \hline 5 & 35 \newline \hline 7 & 7 \newline \hline & 1 \end{array}
700 is not a square. It must be multiplied by another 7: $2^2\times5^2\times7\times7=4,900$
$-700$ must be multiplied by $-7$ to give the square 4,900.
### Square roots
In Chapter 1 you learnt that a square root is the inverse of a square. The square root of a number is a factor of the number that, when multiplied by itself, gives the number. You should revise the section on square roots in Chapter 1 before you proceed.
square root The factor of a number that, when multiplied by itself, gives the number.
Do not confuse squares and square roots:
• 49 is the square of 7, and we write $7^2=49$
• 7 is the square root of 49, and we write $\sqrt{49}=7$.
You have just seen that a negative number cannot be a square. This means that we cannot find the square root of a negative number. There is no factor of a negative number that you can multiply by itself to give that negative number as an answer. For example, $\sqrt{-49}$ does not exist. This is because:
• $+7\times+7=+49$
• $-7\times-7=+49$
As you can see, this means that a square has two square roots: a positive number and the same negative number. We write it as:
The sign $\pm$ means "plus or minus".
### Worked example 5.4: Calculating square roots
Find the square root of $3\frac{6}{25}$.
1. Step 1: Write the fraction as an improper fraction, in the form $\frac{\text{numerator}}{\text{denominator}}$.
2. Step 2: Write both the numerator and denominator as products of their prime factors.
3. Step 3: Find the square roots of the numerator and denominator separately. Write a $\pm$ sign before the answer.
### Square and square root table
The following table shows all the numbers from 1 to 900 that are perfect squares.
\begin{array}{|r|r|r|r|r|r|} \hline 1^2=(-1)^2=1 & \sqrt1=\pm1 & 11^2=(-11)^2=121 & \sqrt{121}=\pm11 & 21^2=(-21)^2=441 & \sqrt{441}=\pm21 \newline 2^2=(-2)^2=4 & \sqrt4=\pm2 & 12^2=(-12)^2=144 & \sqrt{144}=\pm12 & 22^2=(-22)^2=484 & \sqrt{484}=\pm22 \newline 3^2=(-3)^2=9 & \sqrt9=\pm3 & 13^2=(-13)^2=169 & \sqrt{169}=\pm13 & 23^2=(-23)^2=529 & \sqrt{529}=\pm23 \newline 4^2=(-4)^2=16 & \sqrt{16}=\pm4 & 14^2=(-14)^2=196 & \sqrt{196}=\pm14 & 24^2=(-24)^2=576 & \sqrt{576}=\pm24 \newline 5^2=(-5)^2=25 & \sqrt{25}=\pm5 & 15^2=(-15)^2=225 & \sqrt{225}=\pm15 & 25^2=(-25)^2=625 & \sqrt{625}=\pm25 \newline 6^2=(-6)^2=36 & \sqrt{36}=\pm6 & 16^2=(-16)^2=256 & \sqrt{256}=\pm16 & 26^2=(-26)^2=676 & \sqrt{676}=\pm26 \newline 7^2=(-7)^2=49 & \sqrt{49}=\pm7 & 17^2=(-17)^2=289 & \sqrt{289}=\pm17 & 27^2=(-27)^2=729 & \sqrt{729}=\pm27 \newline 8^2=(-8)^2=64 & \sqrt{64}=\pm8 & 18^2=(-18)^2=324 & \sqrt{324}=\pm18 & 28^2=(-28)^2=784 & \sqrt{784}=\pm28 \newline 9^2=(-9)^2=81 & \sqrt{81}=\pm9 & 19^2=(-19)^2=361 & \sqrt{361}=\pm19 & 29^2=(-29)^2=841 & \sqrt{841}=\pm29 \newline 10^2=(-10)^2=100 & \sqrt{100}=\pm10 & 20^2=(-20)^2=400 & \sqrt{400}=\pm20 & 30^2=(-30)^2=900 & \sqrt{900}=\pm30 \newline \hline \end{array}
### Exercise 5.4: Calculate square roots
Calculate the square roots of the following numbers and fractions. You may use the square and square root table above.
1. From the table:
2. \begin{array}{r | r} 2 & 1,024 \newline \hline 2 & 512 \newline \hline 2 & 256 \newline \hline 2 & 128 \newline \hline 2 & 64 \newline \hline 2 & 32 \newline \hline 2 & 16 \newline \hline 2 & 8 \newline \hline 2 & 4 \newline \hline 2 & 2 \newline \hline & 1 \end{array}
3. From the table:
4. From the table:
5. From the table:
### Square and square root notation
Note the following notations when you work with squares and square roots:
• The power of 2 only applies to whatever is directly before it. For example:
$-7^2=-\;7\times7=-49$, but $(-7)^2=-7\times-7=+49$.
• There is a difference between a minus sign that is outside (before) or inside a square root sign. For example:
$-\sqrt{49}=-(-7)\text{ or}\,-(+7)=\pm7$, but $\sqrt{-49}$ does not exist.
### Exercise 5.5: Calculate squares and square roots
Calculate the following. If there is more than one answer, give both.
1. Does not exist.
2. \begin{align} &-\sqrt{16}+(-4)^2 \\ =&\;-(+4)+(+16) \\ =&\;-4+16 \\ =&\; 12 \end{align}
or
\begin{align} &-\sqrt{16}+(-4)^2 \\ =&\;-(-4)+(+16) \\ =&\;+4+16 \\ =&\; 20 \end{align}
3. \begin{align} &-3^2-2\times-3 \\ =&\;-9-(-6) \\ =&\;-9+6 \\ =&\; -3 \end{align}
4. \begin{align} &\frac{\sqrt{81}}{(-3)(-3)(3)} \\ =&\;\frac{+9}{(+9)(3)} \\ =&\;\frac{+9}{+27} \\ =&\; \frac{1}{3} \end{align}
or
\begin{align} &\frac{\sqrt{81}}{(-3)(-3)(3)} \\ =&\;\frac{-9}{(+9)(3)} \\ =&\;\frac{-9}{+27} \\ =&\; -\frac{1}{3} \end{align}
5. \begin{align} &\frac{\sqrt 4 \times 2 - 2 \times -2}{-2} \\ =&\;\frac{+2\times2-(-4)}{-2} \\ =&\;\frac{4+4}{-2} \\ =&\; \frac{8}{-2} \\ =&\;-4 \end{align}
or
\begin{align} &\frac{\sqrt 4 \times 2 - 2 \times -2}{-2} \\ =&\;\frac{-2\times2-(-4)}{-2} \\ =&\;\frac{-4+4}{-2} \\ =&\; \frac{0}{-2} \\ =&\;0 \end{align}
## 5.3 Using tables, charts and schedules
### Square root table
Apart from the square and square root table you used in the previous section, you can use the following table to determine the square roots of numbers that are not perfect squares. Worked example 5.5 will show you how to do this.
Where applicable, square roots are rounded to two decimal places in this table.
### Worked example 5.5: Using a square root table
Find $\sqrt{76}$ from the square root table.
1. Step 1: Look for the first digit in the vertical column listing the numbers 1 to 9.
The first digit is 7, so you are going to look along the row for the number 7.
2. Step 2: Look for the second digit in the horizontal row listing the numbers 0 to 9.
The second digit is 6, so you are going to look down the column numbered 6.
3. Step 3: The answer is where the row from Step 1 meets the column from Step 2.
The two perfect squares on either side of 76 are 64 and 81. This means $\sqrt{76}$ must lie between $\sqrt{64}=8$ and $\sqrt{81}=9$.
The answer 8.72 lies between 8 and 9.
The square roots of the single digits 1 to 9 are shown in the left-most column of the table.
### Distance charts
A distance chart can be used to show the distances between different towns. We read a distance chart in a similar way to a square root table. This table shows the shortest distances by road, in kilometres, according to Google Maps, 2019.
### Worked example 5.6: Using a distance chart
Find the distance from Akure to Port Harcourt.
1. Step 1: Look for the names of the two towns. Note the number written next to each town.
Akure is number 2 and Port Harcourt is number 9.
2. Step 2: Go down the column of the town with the lowest number until you reach the row with the same number as the second town.
Go down the column marked Akure 2 until you reach row number 9.
3. Step 3: The answer is where the column and the row from Step 2 intersect. Write it down with the correct units.
The distance is 452 km.
### Schedule of costs
The following schedule can be used to work out the cost of any number of articles.
### Worked example 5.7: Using a schedule of cost
Use the schedule of cost to determine the cost of 76 cups of beans at $$₦80.50$$ per cup.
1. Step 1: Break up the number of items into tens and units. This tells you which rows in the schedule to use.
Use rows 6 and 7.
2. Step 2: Break up the price into hundreds, tens, units and tenths. This tells you which columns in the schedule to use.
$₦80.50= ₦80+ ₦0.5=( ₦\color{red}8 \color{black} \times 10)+( ₦\color{red}5 \color{black} \div 10)$
Use columns 5 and 8.
3. Step 3: Write the rows and columns you identified in Steps 1 and 2 in a table.
\begin{array}{|c|c|c|} \hline \downarrow \, \rightarrow & 5 & 8 \\ \hline 6 & & \\ \hline 7 & & \\ \hline \end{array}
4. Step 4: Insert the multiples of 10 you identified in Steps 1 and 2.
\begin{array}{|c|c|c|} \hline \downarrow \, \rightarrow & 5\, (\div 10) & 8\, (\times 10) \\ \hline 6\, (\times1) & & \\ \hline 7\, (\times 10) & & \\ \hline \end{array}
5. Step 5: Refer to the cost schedule. Write down the numbers at the intersections of the rows and columns.
\begin{array}{|c|c|c|} \hline \downarrow \, \rightarrow & 5\, (\div 10) & 8\, (\times 10) \\ \hline 6\, (\times1) & 30 & 48 \\ \hline 7\, (\times 10) & 35 & 56 \\ \hline \end{array}
6. Step 6: Determine the total cost by adding the numbers at the intersections of the rows and columns. Apply the multiplications or divisions that you inserted in Step 4.
\begin{align} &(30\times1\div10)+(48\times1\times10)+(35\times10\div10)+(56\times10\times10) \\ =&\;3+480+35+5,600 \\ =&\; ₦6,118 \end{align}
The schedule of cost does not include units, so it can be used for kobos as well.
### Exercise 5.6: Use tables, charts and schedules
Use the tables, charts and schedules provided in this chapter to answer the following questions.
1. Calculate:
1. the side length of a square with an area of 83 cm $^2$
From the square root table: $\sqrt{83}=9.11\text{ cm}$
1. the area of a square with a side length of 22 cm
From the square and square root table: $22^2=484\text{ cm}^2$
2. Mustapha wants to tile the floor of a square room. He has 95 tiles. He wants to work out whether there are enough tiles or whether he must buy more.
1. Calculate the maximum number of tiles that he can put along one side of the floor.
From the square root table: $\sqrt{95}=9.75\text{ cm}$
It does not make sense to break the tiles, so the maximum he can put along one side is 9 tiles.
1. If he tiles the floor according to the previous answer, how many of the 95 tiles will be left over?
He will use $9^2=81$ tiles.
There will be $95-81=14$ tiles left over.
1. The length of one side of the floor is 150 cm. The tiles are square, each with a width of 15 cm. Calculate how many more tiles Mustapha has to buy.
For one length of the floor, he needs $\frac{150}{15}=10$ tiles.
The room is square, so this means that for the whole floor, he needs $10^2=100$ tiles.
He must buy another $100-95=5$ tiles.
3. Compare the distance from Benin City to Kano with the distance from Port Harcourt to Abudja.
From the distance chart:
Distance from Benin City to Kano: 848 km
Distance from Port Harcourt to Abudja: 658 km
The distance from Benin City to Kano is $848-658=190\text{ km}$ longer than the distance from Port Harcourt to Abudja.
4. Use the schedule of cost to calculate the price of 42 loaves of bread at ₦50 each.
Items: $42=40+2$
Cost: ₦50
Numbers at intersections: row 4 and column 5 $= 20$; row 2 and column 5 $= 10$
\begin{align} \text{Cost}&=(20\times10\times10)+(10\times10) \\ &=2,000+100 \\ &= ₦2,100 \end{align}
5. A shop owner allows his customers to buy on credit. A brother and sister have the same account and the shop. At the end of a certain month, they owe the shop owner ₦1,250.00. They then decide to split the debit equally. After that, the sister buys 12 litres of vegetable oil at ₦580.00 per litre. They ask the shop owner to send each of them their own account. What will the balance on each account be?
$$\frac{- ₦1,250}{2}=- ₦625$$ per person
Litres vegetable oil: $10+2$
Cost: $$₦500+ ₦80$$
Numbers at intersections: $5+8+10+16$
\begin{align} \text{Cost}&=(5\times10\times100)+(8\times10\times10)+(10\times100)+(16\times10) \\ &=5,000+800+1,000+160 \\ &= ₦6,960 \end{align}
Sister's balance: $$-625+(-6,960)=- ₦7,585$$
Brother's balance: $$- ₦625$$
## 5.4 Summary
• Directed numbers are positive and negative numbers. The sign of a directed number tells us in which direction from zero it lies on the number line.
• Whenever you multiply directed numbers, remember:
$(+)\times(+)=(+)$
$(-)\times(-)=(+)$
$(+)\times(-)=(-)$
$(-)\times(+)=(-)$
• Whenever you divide directed numbers, remember:
$\frac{(+)}{(+)}=(+)$
$\frac{(-)}{(-)}=(+)$
$\frac{(+)}{(-)}=(-)$
$\frac{(-)}{(+)}=(-)$
• For multiplication and division of directed numbers:
• the same signs give a positive answer
• opposite signs give a negative answer.
• A negative number cannot be a square.
• The square root of a negative number does not exist.
• Square and square root tables, distance charts and cost schedules can be used to determine values without a calculator. |
60 Unit 3 Linear Functions Answer Key
Introduction
In unit 3 of linear functions, students learn about the fundamental concepts of linear equations, including slope, intercepts, and graphing. This article will provide an answer key for unit 3 linear functions, which will serve as a valuable resource for students and teachers alike. By having access to the answer key, students can check their work, identify any mistakes, and gain a better understanding of the concepts covered in unit 3. Let's dive into the answer key for unit 3 linear functions!
1. Slope
1.1 Definition
Slope is a measure of how steep a line is. It represents the rate of change between two points on a line. The formula for finding the slope between two points (x1, y1) and (x2, y2) is given by:
Slope = (y2 - y1) / (x2 - x1)
1.2 Examples
Example 1: Find the slope between the points (2, 4) and (6, 10).
Slope = (10 - 4) / (6 - 2) = 6 / 4 = 3/2
Example 2: Find the slope between the points (-3, 5) and (1, -1).
Slope = (-1 - 5) / (1 - (-3)) = -6 / 4 = -3/2
2. Intercepts
2.1 x-intercept
The x-intercept is the point where the line intersects the x-axis. To find the x-intercept, set y = 0 and solve for x. The x-intercept is represented as (x, 0).
2.2 y-intercept
The y-intercept is the point where the line intersects the y-axis. To find the y-intercept, set x = 0 and solve for y. The y-intercept is represented as (0, y).
2.3 Examples
Example 1: Find the x-intercept and y-intercept of the line with equation y = 2x + 3.
For x-intercept, set y = 0:
0 = 2x + 3
-3 = 2x
x = -3/2
Therefore, the x-intercept is (-3/2, 0).
For y-intercept, set x = 0:
y = 2(0) + 3
y = 3
Therefore, the y-intercept is (0, 3).
3. Graphing Linear Equations
3.1 Slope-Intercept Form
The slope-intercept form of a linear equation is given by y = mx + b, where m represents the slope and b represents the y-intercept. This form makes it easy to graph a linear equation.
3.2 Examples
Example 1: Graph the line with equation y = 2x + 3.
Plot the y-intercept (0, 3).
Use the slope to find additional points:
- Starting from the y-intercept (0, 3), move 1 unit to the right and 2 units up to get the point (1, 5).
- Starting from the point (1, 5), move 1 unit to the right and 2 units up to get the point (2, 7).
- Starting from the point (1, 5), move 1 unit to the left and 2 units down to get the point (-1, 1).
Connect the points to form a line.
4. Writing Linear Equations
4.1 Point-Slope Form
The point-slope form of a linear equation is given by y - y1 = m(x - x1), where m represents the slope and (x1, y1) represents a point on the line. This form is useful for writing an equation given a point and the slope.
4.2 Examples
Example 1: Write the equation of a line with slope 3 that passes through the point (2, 4).
Using the point-slope form:
y - 4 = 3(x - 2)
y - 4 = 3x - 6
y = 3x - 2
Therefore, the equation of the line is y = 3x - 2.
5. Systems of Linear Equations
5.1 Solving by Graphing
To solve a system of linear equations by graphing, plot the lines on the same graph and find the point of intersection.
5.2 Solving by Substitution
To solve a system of linear equations by substitution, solve one equation for one variable and substitute it into the other equation. Solve the resulting equation for the remaining variable.
5.3 Solving by Elimination
To solve a system of linear equations by elimination, add or subtract the equations to eliminate one variable. Solve the resulting equation for the remaining variable.
5.4 Examples
Example 1: Solve the system of equations using the graphing method:
y = 2x + 1
y = -x + 4
Graph the lines y = 2x + 1 and y = -x + 4.
Find the point of intersection, which is (1, 3).
Therefore, the solution to the system of equations is x = 1 and y = 3.
Conclusion
The answer key provided in this article serves as a valuable resource for students and teachers studying unit 3 linear functions. By having access to the answer key, students can check their work, identify any mistakes, and gain a better understanding of the concepts covered in unit 3. Understanding the concepts of slope, intercepts, graphing linear equations, and solving systems of linear equations is crucial for success in algebra and beyond. With this answer key, students can confidently tackle problems related to linear functions. |
# 1.4 Polynomials (Page 3/15)
Page 3 / 15
Find the product.
$\left(3x+2\right)\left({x}^{3}-4{x}^{2}+7\right)$
$3{x}^{4}-10{x}^{3}-8{x}^{2}+21x+14$
## Using foil to multiply binomials
A shortcut called FOIL is sometimes used to find the product of two binomials. It is called FOIL because we multiply the f irst terms, the o uter terms, the i nner terms, and then the l ast terms of each binomial.
The FOIL method arises out of the distributive property. We are simply multiplying each term of the first binomial by each term of the second binomial, and then combining like terms.
Given two binomials, use FOIL to simplify the expression.
1. Multiply the first terms of each binomial.
2. Multiply the outer terms of the binomials.
3. Multiply the inner terms of the binomials.
4. Multiply the last terms of each binomial.
6. Combine like terms and simplify.
## Using foil to multiply binomials
Use FOIL to find the product.
$\left(2x-18\right)\left(3x+3\right)$
Find the product of the first terms.
Find the product of the outer terms.
Find the product of the inner terms.
Find the product of the last terms.
Use FOIL to find the product.
$\left(x+7\right)\left(3x-5\right)$
$3{x}^{2}+16x-35$
## Perfect square trinomials
Certain binomial products have special forms. When a binomial is squared, the result is called a perfect square trinomial . We can find the square by multiplying the binomial by itself. However, there is a special form that each of these perfect square trinomials takes, and memorizing the form makes squaring binomials much easier and faster. Let’s look at a few perfect square trinomials to familiarize ourselves with the form.
Notice that the first term of each trinomial is the square of the first term of the binomial and, similarly, the last term of each trinomial is the square of the last term of the binomial. The middle term is double the product of the two terms. Lastly, we see that the first sign of the trinomial is the same as the sign of the binomial.
## Perfect square trinomials
When a binomial is squared, the result is the first term squared added to double the product of both terms and the last term squared.
${\left(x+a\right)}^{2}=\left(x+a\right)\left(x+a\right)={x}^{2}+2ax+{a}^{2}$
Given a binomial, square it using the formula for perfect square trinomials.
1. Square the first term of the binomial.
2. Square the last term of the binomial.
3. For the middle term of the trinomial, double the product of the two terms.
## Expanding perfect squares
Expand $\text{\hspace{0.17em}}{\left(3x-8\right)}^{2}.$
Begin by squaring the first term and the last term. For the middle term of the trinomial, double the product of the two terms.
${\left(3x\right)}^{2}-2\left(3x\right)\left(8\right)+{\left(-8\right)}^{2}$
Simplify
$\text{\hspace{0.17em}}9{x}^{2}-48x+64.$
Expand $\text{\hspace{0.17em}}{\left(4x-1\right)}^{2}.$
$16{x}^{2}-8x+1$
## Difference of squares
Another special product is called the difference of squares , which occurs when we multiply a binomial by another binomial with the same terms but the opposite sign. Let’s see what happens when we multiply $\text{\hspace{0.17em}}\left(x+1\right)\left(x-1\right)\text{\hspace{0.17em}}$ using the FOIL method.
$\begin{array}{ccc}\hfill \left(x+1\right)\left(x-1\right)& =& {x}^{2}-x+x-1\hfill \\ & =& {x}^{2}-1\hfill \end{array}$
The middle term drops out, resulting in a difference of squares. Just as we did with the perfect squares, let’s look at a few examples.
root under 3-root under 2 by 5 y square
The sum of the first n terms of a certain series is 2^n-1, Show that , this series is Geometric and Find the formula of the n^th
cosA\1+sinA=secA-tanA
why two x + seven is equal to nineteen.
The numbers cannot be combined with the x
Othman
2x + 7 =19
humberto
2x +7=19. 2x=19 - 7 2x=12 x=6
Yvonne
because x is 6
SAIDI
what is the best practice that will address the issue on this topic? anyone who can help me. i'm working on my action research.
simplify each radical by removing as many factors as possible (a) √75
how is infinity bidder from undefined?
what is the value of x in 4x-2+3
give the complete question
Shanky
4x=3-2 4x=1 x=1+4 x=5 5x
Olaiya
hi can you give another equation I'd like to solve it
Daniel
what is the value of x in 4x-2+3
Olaiya
if 4x-2+3 = 0 then 4x = 2-3 4x = -1 x = -(1÷4) is the answer.
Jacob
4x-2+3 4x=-3+2 4×=-1 4×/4=-1/4
LUTHO
then x=-1/4
LUTHO
4x-2+3 4x=-3+2 4x=-1 4x÷4=-1÷4 x=-1÷4
LUTHO
A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was 1350 bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after 3 hours?
v=lbh calculate the volume if i.l=5cm, b=2cm ,h=3cm
Need help with math
Peya
can you help me on this topic of Geometry if l help you
litshani
( cosec Q _ cot Q ) whole spuare = 1_cosQ / 1+cosQ
A guy wire for a suspension bridge runs from the ground diagonally to the top of the closest pylon to make a triangle. We can use the Pythagorean Theorem to find the length of guy wire needed. The square of the distance between the wire on the ground and the pylon on the ground is 90,000 feet. The square of the height of the pylon is 160,000 feet. So, the length of the guy wire can be found by evaluating √(90000+160000). What is the length of the guy wire?
the indicated sum of a sequence is known as
how do I attempted a trig number as a starter
cos 18 ____ sin 72 evaluate |
# How do you solve 5/3x-4/3x-1=8?
Apr 22, 2018
$x = 27$
#### Explanation:
Given: $\frac{5}{3} x - \frac{4}{3} x - 1 = 8$
Combine like-terms.
$\implies \frac{1}{3} x - 1 = 8$
Add $1$ to both sides.
$\implies \frac{1}{3} x - \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}} = 8 + 1$
$\implies \frac{1}{3} x = 9$
Multiply by $3$ on both sides to find $x$.
$\implies \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} x \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} = 9 \cdot 3$
$\implies x = 27$ |
AP Statistics : How to find correlation
Example Questions
Example Question #1 : How To Find Correlation
Which of the following correlation coefficients indicates the strongest relationship between variables?
Explanation:
Correlation coefficients range from 1 to -1. The closer to either extreme, the stronger the relationship. The closer to 0, the weaker the relationship.
Example Question #2 : How To Find Correlation
It is found that there is a correlation of exactly between two variables. Which of the following is incorrect?
The association between the two variables is positive
There is a strong association between the two variables.
There is enough evidence, with a correlation of , to assert that one variable causes the other
All of the answer choices are correct
Correlation is measured on a scale of to
There is enough evidence, with a correlation of , to assert that one variable causes the other
Explanation:
Under no circumstance will correlation ever equate to causation, regardless of how strong the correlation between two variables is. In this case, all other answer choices are correct.
Example Question #3 : How To Find Correlation
In a medical school, it is found that there is a correlation of between the amount of coffee consumed by students and the number of hours students sleep each night. Which of the following is true?
i. There is a positive association between the two variables.
ii. There is a strong correlation between the two variables.
iii. Coffee consumption in medical school students causes students to sleep less each night.
i and ii
i, ii, and iii
iii only
ii only
i and iii
ii only
Explanation:
Since the correlation is negative, there must be a negative association between the two variables (therefore statement i is incorrect). Statement ii is correct since a correlation of to on an absolute value scale of to is considered to be a strong correlation. Statement iii is incorrect since correlation does not mean causation.
Example Question #4 : How To Find Correlation
Which of the following shows the least correlation between two variables?
Explanation:
The strength of correlation is measured on an absolute value scale of to with being the least correlated and being the most correlated. The positive or negative in front of the correlation integer simply determines whether or not there is a positive or negative correlation between the variables.
A correlation of means that there is no correlation at all between two variables.
Example Question #5 : How To Find Correlation
Which of the following correlation coefficients implies the strongest relationship between variables:
Explanation:
A high correlation coefficient regardless of sign implies a stronger relationship. In this case has a stronger negative relationship than the positive relationship described by a value of
Example Question #6 : How To Find Correlation
A national study on cell phone use found the following correlations:
-The correlation between the number of texts sent each day and a person's average credit card debt is .
-The correlation between the number of texts sent each day and the number of books read each month is .
Which of the following statements are true?
i. As the number of texts sent each day increases, average credit card debt increases.
ii. Sending more texts causes people to read less.
iii. A person's average credit card debt is related more strongly to the number of texts sent each day than the number of books read each month is related to the number of texts sent each day.
iii
i and iii
ii
i and ii
ii and iii |
# Definition of absolute value
Definition of absolute value: the absolute value of a number is the distance the number is from zero
Look at the following two graphs:
The first one shows 6 located at a distance of 6 units from zero.
We write |6| = 6
The second one shows -8 located at a distance of 8 units from zero.
We write |-8| = 8
You can see from this that the absolute value of a number is always positive with the exception of taking the absolute value of 0 (|0| = 0)
Therefore, do not write |5| = -5 or |4| = -4 please!
|6| means distance from 0 to 6
|-8| means distance from 0 to -8
For |-8| = 8, you could also argue that to get 8, you have to take the negative of -8 since - - 8 = 8
So, | - 8| = - - 8 = 8
This observation helps us to come up with a formal definition of absolute value
| x | = x if x is positive or zero, but -x if x is negative
This definition is important to understand before solving absolute value equations or absolute value inequalities
Calculate the absolute value of the following numerical expressions
1) 42 − 4 × 2
2) -5 + 5 × 2 − 15
3) -8 + 2 × 5
1)
| -8 + 2 × 5 | = | -8 + 10 |
| -8 + 2 × 5 | = | 2 |
| -8 + 2 × 5 | = 2
2)
| 42 − 4 × 2| = | 16 − 4 × 2|
| 42 − 4 × 2| = | 16 − 8|
| 42 − 4 × 2| = | 8 |
| 42 − 4 × 2| = 8
3)
|-5 + 5 × 2 − 15| = | -5 + 10 − 15 |
|(-5 + 5 × 2 − 15)| = | 5 − 15 |
|(-5 + 5 × 2 − 15)| = | -15 |
|(-5 + 5 × 2 − 15)| = 15
## Recent Articles
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Oct 16, 17 09:34 AM
What is the conjugate in algebra? A concise and clear explanation.
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## Recent Lessons
1. ### Conjugate in Algebra
Oct 16, 17 09:34 AM
What is the conjugate in algebra? A concise and clear explanation.
Tough Algebra Word Problems.
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Everything you need to prepare for an important exam!
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# How do you simplify sqrt(7/2)*sqrt(5/3)?
Jul 17, 2017
See a solution process below:
#### Explanation:
Use this rule for radicals to simplify the expression:
$\sqrt{\textcolor{red}{a}} \cdot \sqrt{\textcolor{b l u e}{b}} = \sqrt{\textcolor{red}{a} \cdot \textcolor{b l u e}{b}}$
$\sqrt{\textcolor{red}{\frac{7}{2}}} \cdot \sqrt{\textcolor{b l u e}{\frac{5}{3}}} = \sqrt{\textcolor{red}{\frac{7}{2}} \cdot \textcolor{b l u e}{\frac{5}{3}}} = \sqrt{\frac{35}{6}}$
Or, we can the use this rule of radicals to rewrite the expression:
$\sqrt{\frac{\textcolor{red}{a}}{\textcolor{b l u e}{b}}} = \frac{\sqrt{\textcolor{red}{a}}}{\sqrt{\textcolor{b l u e}{b}}}$
$\sqrt{\frac{\textcolor{red}{35}}{\textcolor{b l u e}{6}}} = \frac{\sqrt{\textcolor{red}{35}}}{\sqrt{\textcolor{b l u e}{6}}}$
If necessary, we can rationalize the denominator using the following process:
$\frac{\sqrt{6}}{\sqrt{6}} \cdot \frac{\sqrt{\textcolor{red}{35}}}{\sqrt{\textcolor{b l u e}{6}}} \implies$
(sqrt(6) * sqrt(color(red)(35)))/(sqrt(6) * sqrt(color(blue)(6)) =>
$\frac{\sqrt{6 \cdot \textcolor{red}{35}}}{6} \implies$
$\frac{\sqrt{210}}{6}$ |
# How do you use the Binomial Theorem to expand (1/X + X)^6?
Nov 7, 2016
${\left(\frac{1}{x} + x\right)}^{6} = \frac{1}{x} ^ 6 + \frac{6}{x} ^ 4 + \frac{15}{x} ^ 2 + 20 + 15 {x}^{2} + 6 {x}^{4} + \frac{1}{x} ^ 6$
#### Explanation:
Binomial expansion of ${\left(x + a\right)}^{6}$ is
${x}^{6} {+}^{6} {C}_{1} {x}^{5} a {+}^{6} {C}_{2} {x}^{4} {a}^{2} {+}^{6} {C}_{3} {x}^{3} {a}^{3} {+}^{6} {C}_{4} {x}^{2} {a}^{4} {+}^{6} {C}_{5} x {a}^{5} {+}^{6} {C}_{6} {a}^{6}$
And hence binomial expansion of ${\left(\frac{1}{x} + x\right)}^{6}$ or ${\left(x + \frac{1}{x}\right)}^{6}$ is
${x}^{6} {+}^{6} {C}_{1} {x}^{5} \left(\frac{1}{x}\right) {+}^{6} {C}_{2} {x}^{4} {\left(\frac{1}{x}\right)}^{2} {+}^{6} {C}_{3} {x}^{3} {\left(\frac{1}{x}\right)}^{3} {+}^{6} {C}_{4} {x}^{2} {\left(\frac{1}{x}\right)}^{4} {+}^{6} {C}_{5} x {\left(\frac{1}{x}\right)}^{5} {+}^{6} {C}_{6} {\left(\frac{1}{x}\right)}^{6}$
or ${x}^{6} + \frac{6}{1} \times {x}^{4} + \frac{6 \times 5}{1 \times 2} \times {x}^{2} + \frac{6 \times 5 \times 4}{1 \times 2 \times 3} + \frac{6 \times 5 \times 4 \times 3}{1 \times 2 \times 3 \times 4} \times \frac{1}{x} ^ 4 + \frac{6 \times 5 \times 4 \times 3 \times 2}{1 \times 2 \times 3 \times 4 \times 5} \times \frac{1}{x} ^ 6$
or ${x}^{6} + 6 {x}^{4} + 15 {x}^{2} + 20 + \frac{15}{x} ^ 2 + \frac{6}{x} ^ 4 + \frac{1}{x} ^ 6$
or $\frac{1}{x} ^ 6 + \frac{6}{x} ^ 4 + \frac{15}{x} ^ 2 + 20 + 15 {x}^{2} + 6 {x}^{4} + \frac{1}{x} ^ 6$ |
Question
# The area of the rectangular field is $3584\text{ }{{m}^{2}}$ and its length is $64\text{ }m$. A boy runs around the field at a rate of $6km/hr$. How long will he take to go 5 rounds around it?
Hint: Here we have to apply the formulas:
Area of the rectangle = length $\times$ breadth
Perimeter of the rectangle = 2(length + breadth)
Time = $\dfrac{\text{Distance}}{\text{speed}}$
Here, the area of the rectangular field is given as, $A=3584\text{ }{{m}^{2}}$
Length of the rectangular field is, $l=64\text{ }m$.
First, we have to find the breadth of the rectangular field, $b$.
We know the formula that the area of the rectangle,
$A=l\times b$
$3584=64\times b$
Therefore, by cross multiplication we get:
$\dfrac{3584}{64}=b$
By cancellation we get:
$b=56$
Hence, the breadth of the rectangle, $b=56$.
Next, we have to find the perimeter of the rectangular field, $P$. i.e.
$P=2(l+b)$
We have, $l=64$ and $b=56$. Therefore we get:
\begin{align} & P=2(64+56) \\ & P=2\times 120 \\ & P=240 \\ \end{align}
Hence, we got the perimeter of the rectangular field as, $P=240\text{ }m$.
We are given that the boy runs 5 rounds of field.
Therefore, the perimeter for 5 rounds = $240\times 5=1200\text{ }m$
It is also given that the speed of the boy running around the field, $S=6\text{ }km/hr$
The speed is given in $km/hr$, now we have to change it into $m/s$ by multiplying it with $\dfrac{5}{18}$.
Therefore, the speed,
$S=6\times \dfrac{5}{18}$
By cancellation we get:
$S=\dfrac{5}{3}m/s$
Next, we have to find the time required to cover 5 rounds of field, i.e. the time required to cover the distance of $1200\text{ }m$. Hence we get:
Time, $T=\dfrac{D}{S}$ , where $D$ is the distance and $S$ is the speed.
$T=\dfrac{1200}{\dfrac{5}{3}}$
We know that $\dfrac{a}{\dfrac{b}{c}}=\dfrac{ac}{b}$. Therefore, we get,
$T=\dfrac{1200\times 3}{5}$
By cancellation we get:
\begin{align} & T=240\times 3 \\ & T=720 \\ \end{align}
We got time, $T=720$ seconds. Now we have to convert it into minutes by dividing it with 60.
$T=\dfrac{720}{60}$
By cancellation we get:
$T=12$ min
Hence, the time required to cover 5 rounds of field is 12 min.
Note: Here we have to calculate the perimeter of 5 rounds of field. So we have to multiply the perimeter by 5. In the question the unit is given in metre, so don’t forget to change the unit from $km/hr$ to $m/s$. |
# Angle Basics
Welcome to introduction to angles
the goal of this video are to know the basic angle vocabulary and also determine how angles are
measured. A portion of a line that starts at A point and continues through B is called Ray. Ray AB can be written using this
notation and angle consists of two rays with a
common endpoint the two rays called the sides at the
angle and the common in point is called the
vertex. So, here we have the vertex. Here we have one side, here we have another
side. There are a couple ways to identify angles. If it call this point A B and C. There’s only one angle at vertex.
But we can simply just call this angle B. However, we can also call this angle ABC. The important part is at the vertex is that variable in the middle. Now, it is true that angle can be formed by two segments with the common endpoint. Each angle has a measure generated by
the rotation about the vertex. The measure is determined by the
rotation of the terminal side about the initial side. So, here’s our initial side and the
second ray or the terminal side has been
rotated this position and when the rotation is
counterclockwise, Its a positive angle. The rotation is clockwise, it’s a
negative angle. The units used to measure are either degrees or radians. We will start by discussing degrees. For degree measure we assign 360 degrees to complete rotation of the Ray you could think of this, that the second
ray going all the way around the circle. Therefore. 1/360th of a
complete rotation gives an angle one degree. Angles can be classified based upon their measure Greek letter theta is often used to
name each angle So just take a look at the
different types of angles if an angle measures between zero in
ninety degrees as we see here it’s called an Acute Angle An angle that equals exactly ninety degrees. is a Right Angle. and then any angle
between ninety and 180. it’s called the Obtuse Angle until we reach
angle of exactly 180 degrees which is a straight angle then any angle between 180 degrees and 360 degrees its called a reflex angle. and then once you reached 360
degrees it’s called a Full angle. Let’s take a moment to see
how we measure angles In trigonometry, we do not spend much time measuring angles. but in geometry you do and the way you
measure angles as with a tool called a Protractor which we see here For example, if we had
an angle here between these two red rays we continue
to measures seventy-three degrees by landing at the initial ray here and measuring from the terminal side. Some protractors have 360 degrees but most only have 180. So, for had anger that was more than 180 degrees we would have to rotate this protractor to find the full
measure of that angle Let’s go and take a look at Complementary and Supplementary angles. If the sum of the measures of two
positive angles is ninety degrees, the angles are called Complimentary. So, in this case we have the measure angle A plus the measure of angle B equals ninety degrees. Now, the sum two angles. Now, the sum of the measures of two positive angles is 180 degrees the angle are called Supplementary. So, in
this case we’ll have to measure of angle A plus the measure of angle B equals 180 degrees. Let’s go and take a look
at a couple examples Two angles are complementary, One angle measures five x degrees while the other measures four x degrees, what is the measure a each angle? so we start with the right angle We’ll divided into two different
angles where one angle is equal to five x degrees, and the others equal to four x degrees for these examples I’ll leave the units
off until the end So, they are complementary we have four x plus five x equals ninety So, we solve this equation
we have nine x equals ninety dividing both sides by nine we can see x is equal to ten So, if this angle is four x or four times ten. Here we have forty degrees and here we would have five time ten or fifty degrees. let’s take a look at one more
example. Now, we have two angles that are
supplementary so they would form 180 degrees and its says here that one
measures seven x degrees and the other one is five x was thirty-six degrees so again the sum of these two angles
were now equal 180 degrees so we have five x
plus thirty-six plus seven x equals 180 so we solve this equation subtract thirty-six on both sides divide by twelve here we have x equals twelve So, if x is twelve, seven times twelve give us eighty-four degrees here and now when I
place X with twelve here, five times twelve plus thirty-six be ninety-six degrees I hope you found this video helpful. Thank you for watching.
• Struggle for the Truth says:
Hey James, the presentation is beautifully organized and easy to follow!
• Neon Garcia says:
thank you so much..
• Byron Starnes says:
I needed this refresher!
• CARSON MORRIS says:
ᕦ◉▿◉ᕤnow im math strong👍 |
# Series
### Definition and Notation
A series is the sum of terms of a sequence. A sum of terms can be written using summation notation.
There are times when it makes sense to add the numbers in a sequence. The sum of terms of a sequence is called a series. A finite series is a sum of the terms in a finite sequence.
A sum can be written as an addition expression:
$a_1+ a_2+ a_3+ a_4+ a_5+ a_6+ a_7+ a_8+ a_9$
It can also be written using summation notation. Summation notation is an abbreviated way of writing a sum that uses the Greek letter sigma. This notation is read as "the sum of $a$ sub $n$ as $n$ goes from 1 to 9”:
$\sum_{n=1}^{9}a_n$
When possible, the explicit rule for the sequence is often used in place of $a_n$.
Step-By-Step Example
Writing a Series in Summation Notation
Write the series in summation notation:
$4+8+12+16+20$
Step 1
Write the explicit rule for the sequence.
$4+8+12+16+20$
Identify the common difference. Subtract each term from the next term.
$\begin{gathered}8-4=4\\12-8=4\\16-12=4\\20-16=4\end{gathered}$
Each term is 4 more than the previous term, so the common difference $d$ is 4.
Step 2
Write the explicit rule.
$a_n=a_{1}+d(n-1)$
Substitute $a_1=4$ and $d=4$.
\begin{aligned}a_n&=4+4(n-1)\\a_{n}&=4n\end{aligned}
Solution
Write the notation for the first five terms. The variable $n$ describes the term number, which goes from 1 to 5.
$\sum_{n=1}^{5}4n$
This is read as "the sum of $4n$ as $n$ goes from 1 to 5."
### Finite Arithmetic Series
A formula can be used to find the sum of a finite arithmetic series.
An arithmetic series is the sum of terms in an arithmetic sequence. The sum can be found by adding the terms in a sequence, but this may become tedious for a large number of terms.
The formula for the sum of the first $n$ terms in a finite arithmetic series $a_1+a_2+…+a_n$ is:
$\sum_{k=1}^{n}a_k= \frac{n}{2} \left(a_1+a_n \right)$
The variable $k$ is called an index, or counter, for the position of each term that is added.
Step-By-Step Example
Determining the Sum of an Arithmetic Series
Identify the sum of the first 12 terms of the series:
$28+ 25+ 22+19+…$
Step 1
Write an explicit rule for the sequence.
$a_n=a_1+d(n-1)$
The common difference is –3, and the first term is 28.
\begin{aligned}a_{n}&=28-3(n-1)\\&=31-3n\end{aligned}
Step 2
Use the explicit rule to find the 12th term.
\begin{aligned}a_{n}&=31-3n\\a_{12}&=31-3(12)\\a_{12}&=31-36\\a_{12}&=-5\end{aligned}
Step 3
Substitute into the formula for a finite arithmetic series and simplify.
Substitute $n=12$, $a_k=31-3n$, $a_1=28$, and $a_n=-5$.
\begin{aligned}\sum_{k=1}^{n}a_k&= \frac{n}{2} \left(a_1+a_n \right)\\ \sum_{k=1}^{12}(31-3n)&= \frac{12}{2} (28+(-5))\\ \sum_{k=1}^{12}(31-3n)&= \frac{12}{2} (23)\\ \sum_{k=1}^{12}(31-3n)&= \frac{276}{2}\\ \sum_{k=1}^{12}(31-3n)&=138\end{aligned}
Solution
The sum of the first 12 terms is 138.
Step-By-Step Example
Applying the Arithmetic Series
Martina started a running program. In her first week she ran a total of 10 miles. Each week after that, she will run an additional 2 miles. How many total miles will Martina run in the first 6 weeks of the program?
Step 1
Write an explicit rule for the sequence.
$a_n=a_1+d(n-1)$
The common difference is 2. The first term is 10.
\begin{aligned}a_{n}&=10+2(n-1)\\&=8+2n\end{aligned}
Step 2
The total miles she will run in the first 6 weeks is a sum of terms in a sequence. Write the series in summation notation.
$\sum_{n=1}^{6}(8+2n)$
Step 3
Use the explicit rule to identify the sixth term.
\begin{aligned}a_{n}&=8+2n\\a_{6}&=8+2(6)\\a_{6}&=8+12\\a_{6}&=20\end{aligned}
Step 4
Substitute into the formula for a finite arithmetic series and simplify.
Substitute $n=6$, $a_k=8+2n$, $a_1=10$, and $a_n=20$.
\begin{aligned}\sum_{k=1}^{n}a_k&= \frac{n}{2} \left(a_1+a_n \right)\\ \sum_{k=1}^{6}(8+2n)&=\frac{6}{2}(10+20)\\ \sum_{k=1}^{6}(8+2n)&=\frac{6}{2}(30)\\ \sum_{k=1}^{6}(8+2n)&=\frac{180}{2}\\ \sum_{k=1}^{6}(8+2n)&=90\end{aligned}
Solution
Martina will run a total of 90 miles in the first 6 weeks of the running program.
### Finite Geometric Series
A formula can be used to find the sum of a finite geometric series.
A geometric series is the sum of terms in a geometric sequence. The sum can be found by adding the terms in the sequence, but this may be difficult for a large number of terms.
The formula for the sum of the first $n$ terms in a finite geometric series, $a_1+a_2+…+a_n$, where $a_1$ is the first term and $r$ is the common ratio, is:
$\sum_{k=1}^{n}a_k= a_1\left(\frac{1-r^n}{1-r}\right)\!\!{,}\;\;r\neq1$
Step-By-Step Example
Determining the Sum of a Finite Geometric Series
Suppose an employee has a starting salary of 40,000. The employee's salary increases by 3% each year for the first five years. What are the total earnings for the employee for the first five years? Step 1 Write an explicit rule for the sequence. $a_n=a_1 \cdot r^{n-1}$ Each year, the salary is 100% of the previous year plus an increase of 3%. As a decimal, 103% is written as 1.03, so the common ratio is 1.03. The first term is 40,000. $a_{n}=40{,}000(1.03)^{n-1}$ Step 2 Substitute into the formula for a finite geometric series and simplify. Round to the nearest hundredth. Substitute $n=5$, $a_k=40{,}000(1.03)^{n-1}$, $a_1=40{,}000$, and $r=1.03$. \begin{aligned}\sum_{k=1}^{n}a_k&= a_1\left(\frac{1-r^n}{1-r}\right)\\ \sum_{n=1}^{5}40{,}000(1.03)^{n-1}&=40{,}000\left(\frac{1-1.03^5}{1-1.03}\right)\\ \sum_{n=1}^{5}40{,}000(1.03)^{n-1}&\approx40{,}000\left(\frac{-0.159274}{-0.03}\right)\\ \sum_{n=1}^{5}40{,}000(1.03)^{n-1}&\approx 212{,}365\end{aligned} Solution The employee earned approximately212,365 in the first 5 years.
### Infinite Series
The sum of an infinite series may be found by using partial sums.
An infinite series is the sum of the terms in an infinite sequence. Since an infinite sequence has infinitely many terms, the sum can be found by using partial sums. The sum of a specified number of terms in a sequence is called a partial sum, written as $S_n$.
\begin{aligned}S_1&=a_1\\S_2&=a_1+ a_2\\S_3&=a_1+ a_2+ a_3\\S_4&=a_1+ a_2+ a_3+ a_4\\S_5&=a_1+ a_2+ a_3+ a_4+ a_5\\S_6&=a_1+ a_2+ a_3+ a_4+ a_5+ a_6\\S_7&=a_1+ a_2+ a_3+ a_4+ a_5+ a_6+ a_7\\S_8&=a_1+ a_2+ a_3+ a_4+ a_5+ a_6+ a_7+ a_8\\S_9&=a_1+ a_2+ a_3+ a_4+ a_5+ a_6+ a_7+ a_8+ a_9\end{aligned}
As $n$ increases, the sum $S_n$ may get closer to a fixed number without reaching that number. When $S_n$ approaches a fixed number, that number is the sum of the infinite series. If $S_n$ does not approach a fixed number, then the sum is undefined.
In an infinite arithmetic series, if the common difference is positive, each term is greater than the previous term, so the sum becomes greater and greater and does not approach a fixed number. It is said to approach $\infty$. If the common difference is negative, each term is less than the previous term, so the sum approaches $-\infty$. When the sequence of partial sums does not approach a fixed number, the series is called a divergent series. Since the sum of an infinite arithmetic series does not approach a fixed number, it is undefined.
In an infinite geometric series with $\left | r \right | <1$, the partial sums get very close to a fixed number. The series is called a convergent series, which means that the sequence of partial sums approaches that number. The formula for the sum of an infinite geometric series, where $a_1$ is the first term and $r$ is the common ratio, that converges is:
$\sum_{k=1}^{\infty}a_k= \frac{a_1}{1-r}$
In an infinite geometric series with $\left|r\right|\geq 1$, the sequence of partial sums does not approach a fixed number, so the series is said to diverge. Therefore, the sum of an infinite geometric series with $\left|r\right|\geq 1$ is undefined.
Step-By-Step Example
Identifying the Sum of an Infinite Geometric Series
Identify the sum of the series:
$40+30+22.5+16.875+…$
Step 1
Write an explicit rule for the sequence.
$a_n=a_1 \cdot r^{n-1}$
The first term is 40. The common ratio is 0.75 because:
$\begin{gathered}30\div40=0.75\\22.5\div30=0.75\\16.875\div22.5=0.75\end{gathered}$
Since $\left | 0.75 \right |<1$, the infinite series converges and has a sum.
$a_{n}=40(0.75)^{n-1}$
Step 2
Substitute into the formula for an infinite geometric series with $\left | r \right | <1$ and simplify.
Substitute $a_k=40(0.75)^{n-1}$, $a_1=40$, and $r=0.75$.
\begin{aligned}\sum_{k=1}^{\infty}a_k&= \frac{a_1}{1-r}\\ \sum_{n=1}^{\infty}40(0.75)^{n-1}&=\frac{40}{1-0.75}\\ \sum_{n=1}^{\infty}40(0.75)^{n-1}&=160\end{aligned}
Solution
The sum of the infinite geometric series is 160. |
LawOfCosines
The Law of Cosines is used for solving oblique triangles. In the triangle below, sides a, b, and c are opposite angles A, B, and C, respectively.
Law of Cosines:
a2 = b2 + c2 – 2bc cosA (angle A is the included angle of sides b and c)
b2 = a2 + c2 – 2ac cosB (angle B is the included angle of sides a and c)
c2 = a2 + b2 – 2ab cosC (angle C is the included angle of sides a and b)
These three formulas are analogous and we use whichever form is applicable to the problem at hand.
We use the Law of Cosines when the given is one of the following types:
1. two sides and the included angle (SAS)
2. three sides (SSS)
Example 1
In triangle ABC, a=5, b=3, and C=125°. Find c rounded to the nearest tenth.
Solution:
Two sides and the included angle are given (Type 1).
Since we are looking for c, we use the second form of the Law of Cosines.
c2 = a2 + b2 – 2ab cosC
c2 = 52 + 32 -2(5)(3)cos125°
c2 ≈ 25 + 9 – 30( -0.5736) = 51.208
c ≈ 7.2
Example 2
In triangle ABC, a=4, b=5, and c=6. Find A rounded to the nearest degree.
Solution :
Three sides are given (Type 2).
Since we are looking for A, we use the first equation in the Law of Cosines.
a2 = b2 + c2 – 2bc cosA
42 = 52 + 62 – 2(5)(6)cosA
Solve for cosA.
16 = 25 + 36 -60 cosA
60cosA = 45
cosA = 0.75
A ≈ 41°
Try these problems
QUESTIONS
1. In triangle ABC, a=6, b=8, C=70°. Find c rounded to the nearest tenth.
2. In triangle ABC, a=8, b=3, c=4. Find A rounded to the nearest degree.
3. Find the length of the diagonal of the parallelogram in the figure.
4. A rectangular garden is in the form of a triangle with sides 12 m, 13m, 15 m.
Find the angle opposite the side with length 13 m. |
# ✅ Complex Number Division Formula ⭐️⭐️⭐️⭐️⭐
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# Dividing Complex Numbers
Dividing complex numbers is a little more complicated than addition, subtraction, and multiplication of complex numbers because it is difficult to divide a number by an imaginary number. For dividing complex numbers, we need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary part of the denominator so that we end up with a real number in the denominator.
In this article, we will learn about the division of complex numbers, dividing complex numbers in polar form, the division of imaginary numbers, and dividing complex fractions.
## What is Dividing Complex Numbers?
Dividing complex numbers is mathematically similar to the division of
## Steps for Dividing Complex Numbers
Now, we know what dividing complex numbers is, let us discuss the steps for dividing complex numbers. To divide the two complex numbers, follow the given steps:
1. First, calculate the conjugate of the complex number that is at the denominator of the fraction.
2. Multiply the conjugate with the numerator and the denominator of the complex fraction.
3. Apply the algebraic identity (a+b)(a-b)=a– b2 in the denominator and substitute i2 = -1.
4. Apply the distributive property in the numerator and simplify.
5. Separate the real part and the imaginary part of the resultant complex number.
## Division of Complex Numbers in Polar Form
Important Notes on Dividing Complex Numbers
• To divide a complex number a+ib by c+id, multiply the numerator and denominator of the fraction a+ib/c+id by c−id and simplify.
• The conjugate of the complex z = a+ib is a−ib.
• The modulus of the complex number z = a+ib is |z| = √(a+ b2)
## Dividing Complex Numbers Examples
### How to Simplify Dividing Complex Numbers?
To divide a complex number a+ib by c+id, multiply the numerator and denominator of the fraction (a+ib)/(c+id) by c−id and simplify.
### How Do you Write the Division of Complex Numbers by a Real Number?
Divide the real part and the imaginary part of the complex number by that real number separately.
## Multiplying Complex Numbers
Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately.
EXAMPLE 4: MULTIPLYING A COMPLEX NUMBER BY A REAL NUMBER
Let’s begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. So, for example,
### HOW TO: GIVEN A COMPLEX NUMBER AND A REAL NUMBER, MULTIPLY TO FIND THE PRODUCT.
1. Use the distributive property.
2. Simplify.
EXAMPLE 5: MULTIPLYING A COMPLEX NUMBER BY A REAL NUMBER
Find the product 4(2+5i).
### Multiplying Complex Numbers Together
Now, let’s multiply two complex numbers. We can use either the distributive property or the FOIL method. Recall that FOIL is an acronym for multiplying First, Outer, Inner, and Last terms together. Using either the distributive property or the FOIL method, we get
### HOW TO: GIVEN TWO COMPLEX NUMBERS, MULTIPLY TO FIND THE PRODUCT.
1. Use the distributive property or the FOIL method.
2. Simplify.
## Dividing Complex Numbers
Division of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator. We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the complex conjugate of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of a+bi is abi.
Note that complex conjugates have a reciprocal relationship: The complex conjugate of a+bi is abi, and the complex conjugate of abi is a+bi. Further, when a quadratic equation with real coefficients has complex solutions, the solutions are always complex conjugates of one another.
Suppose we want to divide c+di by a+bi, where neither a nor b equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply.
### A GENERAL NOTE: THE COMPLEX CONJUGATE
The complex conjugate of a complex number a+bi is abi. It is found by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged.
• When a complex number is multiplied by its complex conjugate, the result is a real number.
• When a complex number is added to its complex conjugate, the result is a real number.
### EXAMPLE 7: FINDING COMPLEX CONJUGATES
Find the complex conjugate of each number.
### Analysis of the Solution
Although we have seen that we can find the complex conjugate of an imaginary number, in practice we generally find the complex conjugates of only complex numbers with both a real and an imaginary component. To obtain a real number from an imaginary number, we can simply multiply by i.
### HOW TO: GIVEN TWO COMPLEX NUMBERS, DIVIDE ONE BY THE OTHER.
1. Write the division problem as a fraction.
2. Determine the complex conjugate of the denominator.
3. Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator.
4. Simplify.
EXAMPLE 10: SUBSTITUTING AN IMAGINARY NUMBER IN A RATIONAL FUNCTION
## Simplifying Powers of i
The powers of i are cyclic. Let’s look at what happens when we raise i to increasing powers.
We can see that when we get to the fifth power of i, it is equal to the first power. As we continue to multiply i by itself for increasing powers, we will see a cycle of 4. Let’s examine the next 4 powers of i.
EXAMPLE 11: SIMPLIFYING POWERS OF I
# Lesson Explainer: Dividing Complex Numbers
In this explainer, we will learn how to perform division on complex numbers.
When a student first encounters complex numbers, expressions like
can seem a little mysterious or, at least, it can seem difficult to understand how one might compute the result. This explainer will connect this idea to more familiar areas of mathematics and help you understand how to evaluate expressions like this. Before we deal with division of complex numbers in general, we will consider the two simpler cases of division by a real number and division by a purely imaginary number.
In many ways, dividing a complex number by a real number is a rather trivial exercise. However, dividing a complex number by an imaginary number is not so trivial as the next example will demonstrate.
Example 2: Dividing a Complex Number by an Imaginary Number
The technique we used above can be generalized to help us understand how to divide any two complex numbers. The first thing we need to do is identify a complex number which when multiplied by the denominator gives a real number. Then, we can multiply both the numerator and the denominator by this number and simplify. The question is, given a complex number 𝑧 ,what number when multiplied by 𝑧 results in a real number? This is the point where we should recall the properties of the complex conjugate, in particular, that for a complex number 𝑧=𝑎+𝑏𝑖,
which is a real number. Hence, by multiplying the numerator and the denominator by the complex conjugate of the denominator, we can eliminate the imaginary part from the denominator and then simplify the result. This technique should not be new to most people. When learning about radicals, we face a similar problem trying to simplify expressions of the form
In this case, we multiply the numerator and the denominator by the conjugate of the denominator. This technique is often called rationalizing the denominator. With complex numbers, in many ways we are using the same technique in the special case where 𝑓 is a negative number.
Now, let’s consider an example where we have to simplify the division of two complex numbers, in a similar way to how we rationalize the denominator with radicals.
Example 3: Dividing Complex Numbers
We begin by identifying a complex number that when multiplied by the denominator results in a real number. We usually use the complex conjugate of the denominator: 1+5𝑖.Now we multiply both the numerator and the denominator by this number as follows:
### How To: Dividing Complex Numbers
To divide complex numbers, we use the following technique (sometimes referred to as “realizing” the denominator):
1. Multiply the numerator and denominator by the complex conjugate of the denominator.
2. Expand the parenthesis in the numerator and denominator.
3. Gather like terms (real and imaginary) remembering that 𝑖2=−1.
4. Express the answer in the form 𝑎+𝑏𝑖 reducing any fractions.
Using this technique, we can actually derive a general form for the division of complex numbers as the next example will demonstrate.
### Example 4: General Form of Complex Division
Even though we have derived a general formula for complex division, it is preferable to be familiar with the technique rather than simply memorize the formula.
The fact that 𝑎2+𝑏2=1 in the previous question is no accident. In fact, this is an example of a general rule that if
for some complex number 𝑧,then 𝑎2+𝑏2=1.This can be proved by working through the algebra. However, this is not very enlightening. Instead, results like this are best understood once we learn about the modulus and argument.
### Example 6: Solving Complex Division Equations
Solve the equation 𝑧(2+𝑖)=3−𝑖 for 𝑧.
We begin by dividing both sides of the equation by 2+𝑖 which results in the following equation:
Due to the fact that multiplying and dividing complex numbers in this way can be fairly time consuming, it is useful to consider which approach will be the most efficient. This often involves using properties of complex numbers or noticing factors that we can quickly cancel. The next two examples will demonstrate how we can simplify our calculations.
Example 7: Complex Division
When presented with an expression like this, it is good to first consider what approach we should take to solving it. We could expand the parenthesis in the numerator and the denominator and then multiply both the numerator and the denominator by the complex conjugate of the denominator. Alternatively, we could split the fraction in two and try to simplify each part then multiply the resulting complex numbers. The approach taken will generally depend on the particular expression given; however, it is good to look for features of the expression that might simplify the calculation. In this case, it is good to notice that we have a common factor of (1+𝑖) in both the numerator and the denominator. By canceling this factor first, we can simplify our calculation. Hence,
We can now multiply both the numerator and the denominator by the complex conjugate of the denominator as follows:
For the next question, we will again look at an example where applying the properties of complex numbers can simplify out calculations.
### Example 8: Complex Expressions Involving Division
It is possible to solve this problem by performing the complex division on both of the fractions and then adding their results. However, we can simplify our calculation by first noticing that we can factor out 3−4𝑖 from both terms. Hence, we can rewrite the expression as
Now we consider the expression in the parentheses; notice that the denominators of the two fractions are a complex conjugate pair; that is, the expression is in the form
Finally, let’s consider an example where we have to find missing values in an equation by dividing complex numbers.
Example 9: Solving a Two-Variable Linear Equation with Complex Coefficients
In this example, we want to determine the missing values 𝑥 and 𝑦 in a two-variable linear equation with complex coefficients.
The given equation contains 3 separate complex divisions, two on the left hand side and one on the right hand side of the equation. We begin by simplifying each term by performing the complex division. This is achieved by multiplying the numerator and denominator by the complex conjugate of the denominator, which results in a real number in the denominator after distributing over the parentheses.
Let’s summarize some of the key points that we covered in this explainer.
### Key Points
• Dividing complex numbers uses the same technique as for rationalizing the denominator.
• To divide complex numbers, we multiply the numerator and the denominator by the complex conjugate of the denominator, and then we expand the brackets and simplify using 𝑖=−1.
• In expressions involving multiplication and division of multiple complex numbers, it is useful to look for common factors or whether we can apply some of the properties of complex numbers to simplify our calculations.
Math Formulas ⭐️⭐️⭐️⭐️⭐ |
# Area of a Shape
Welcome to the wonderfully wide world of geometry with Brighterly, your trusted companion on this enlightening journey through math. Today, we’re going to talk about a fundamental concept that’s at the heart of geometry – the area of a shape. This is the very concept that helps us understand how much space a flat object occupies. Whether you’re painting a wall, laying out a garden, or designing a spaceship, the idea of area is central to your plans. It’s an essential part of the mathematics we use every day. We’ll explore the concept of area for various 2D and 3D shapes, delve into formulas that help us calculate it, and answer some frequently asked questions to clarify common doubts. Let’s embark on this mathematical journey together!
## What is Area?
The area is a fundamental concept in mathematics, particularly in the field of geometry. It’s a measurement that helps us understand the size of a two-dimensional (2D) surface. Imagine you have a flat piece of paper shaped like a square. The area would be how much space that paper occupies on a flat surface. It’s the same as the amount of paint you would need to completely cover that paper with a layer of paint. To measure the area, we use square units. For example, if we were measuring the area in inches, we’d use square inches. It’s as if you were tiling a floor with tiny 1-inch by 1-inch tiles, and counting how many tiles you need to cover the entire floor. The same concept applies to other shapes too, not just squares. That’s why area is often described as the “floor space” or “carpeting” of a shape.
## Area of Basic Geometric Shapes
Now that we understand the basic concept of area, let’s dive deeper into the formulas used to calculate the area of some basic geometric shapes. These simple shapes form the foundation for understanding more complex figures.
### Rectangle
A rectangle is a four-sided shape with all angles equal to 90 degrees. To calculate the area of a rectangle, we multiply its length by its width. If you think about it, it makes perfect sense. If we have a rectangle that is 5 units long and 4 units wide, we could fit 20 square units inside it, which is 5 multiplied by 4.
### Square
A square is like a special rectangle where all the sides have the same length. Thus, to find the area of a square, we simply multiply the length of one side by itself, which is often written as side squared.
### Parallelogram
A parallelogram is a four-sided shape where the opposite sides are equal in length and are parallel to each other. The formula for the area of a parallelogram is base times height, where the base can be any side and the height is the perpendicular distance from the base to the opposite side.
### Triangle
A triangle is a three-sided shape. Despite its different shape, the formula for the area of a triangle is remarkably similar to that of a rectangle. It’s half the product of the base and the height, symbolizing that a triangle can be viewed as half of a parallelogram.
### Circle
A circle is a unique shape where all points are equidistant from a central point. The area of a circle is calculated using the formula π (Pi) times the radius squared. The radius is the distance from the center of the circle to its edge.
## How Do We Apply Formula to Find the Area of a Shape?
Applying these formulas is a straightforward process. First, we need to accurately measure the required dimensions of the shape such as length, width, radius, and so on. These measurements are then plugged into the formula corresponding to the shape. The resulting value, which should always be stated in square units, is the area of the shape.
### Area Of Squares And Rectangles Worksheets PDF
Area Of Squares And Rectangles Worksheets
### Area Of Rectangles Worksheets PDF
Area Of Rectangles Worksheets
If you want to master the area of a shape topic better, we recommend that you pay attention to the free Brighterly worksheets. They can significantly improve your child’s math skills.
## Area of Combination of Shapes
In real life, we often encounter shapes that are a combination of these basic shapes. For example, a house’s floor plan might include a large rectangle for the main living area, a smaller rectangle for the bathroom, and a triangle for a pitched roof. To calculate the total area of such a complex shape, we break it down into its basic shapes, calculate the area of each shape using its corresponding formula, and then add up all these areas.
## What are 2D shapes?
2D shapes, or two-dimensional shapes, are flat shapes that have width and height but no depth. These include simple shapes like circles, triangles, and rectangles, as well as more complex shapes such as hexagons, octagons, and even intricate polygons. The key point is that these shapes are flat, existing on a single plane.
## Area of 2D Shapes Formula
The formulas for finding the area of these 2D shapes vary depending on the shape. For complex 2D shapes, we often break them down into simpler, more manageable shapes. We then find the area of each of these simpler shapes using their individual area formulas, and add up these areas to find the total area of the complex shape.
## What are 3D shapes?
3D shapes, or three-dimensional shapes, have depth in addition to width and height. This added dimension makes them solid rather than flat. Examples of 3D shapes include cubes, cylinders, cones, and spheres. You encounter 3D shapes all the time in the real world—they’re the shapes of most objects you see and use every day.
## Area of 3D Shapes Formula
When we talk about 3D shapes, instead of simply talking about area, we usually refer to surface area. The surface area is the total area of all surfaces or faces of a 3D object. Each type of 3D shape has its own formula for calculating surface area. As with 2D shapes, more complex 3D shapes are often broken down into simpler shapes for the purpose of calculating surface area.
### Cube
A cube is a six-faced figure with all sides of equal length. The surface area of a cube is given by the formula 6s², where ‘s’ is the length of one side.
### Rectangular Prism
A rectangular prism, also known as a cuboid, is like a stretched cube, with possibly different lengths, widths, and heights. The surface area is 2lw + 2lh + 2wh, where ‘l’ is the length, ‘w’ is the width, and ‘h’ is the height.
### Sphere
A sphere is a perfectly round 3D shape (like a ball). The surface area of a sphere is 4πr², where ‘r’ is the radius of the sphere.
### Cylinder
A cylinder is a 3D shape with two circular faces (like a soup can). The surface area is 2πrh + 2πr², where ‘r’ is the radius of the circular base and ‘h’ is the height of the cylinder.
### Cone
A cone is a 3D shape with a circular base and a pointed top (like an ice cream cone). The surface area is πr(r + √(r² + h²)), where ‘r’ is the radius of the base and ‘h’ is the height of the cone.
## Practice Problems on Area of a Shape
1. Rectangle: The length of a rectangular field is 50m and the breadth is 30m. Find the area of the field.
2. Square: Find the area of a square playground whose side measures 25m.
3. Parallelogram: A parallelogram has a base of 15cm and a height of 10cm. What is its area?
4. Triangle: The base of a triangular park is 80m and its height is 60m. Calculate the area of the park.
5. Circle: The radius of a circular garden is 7m. Calculate the area of the garden. (Use π = 3.14)
6. Composite Shapes: A rectangular garden measuring 20m by 15m has a circular pond of radius 3m in its center. What is the area of the garden excluding the pond?
7. Cube: Find the surface area of a cube with side length of 4cm.
8. Rectangular Prism (Cuboid): A box is in the shape of a cuboid with length 5cm, width 3cm, and height 4cm. Find the surface area of the box.
9. Sphere: Calculate the surface area of a sphere with a radius of 6cm. (Use π = 3.14)
10. Cylinder: A cylindrical container has a radius of 7cm and a height of 10cm. Find the total surface area of the container. (Use π = 3.14)
## Conclusion
So, what have we learned today with Brighterly? Area, a fundamental concept in geometry, is the measurement of the space enclosed by a shape. We’ve discovered how to calculate the area of basic 2D shapes like rectangles, squares, parallelograms, triangles, and circles. We’ve also learned how to break down more complex shapes into these simpler shapes to find their areas. In the world of 3D, we’ve found out that the equivalent of area is the surface area and looked at how to calculate it for regular 3D shapes like cubes, rectangular prisms, spheres, cylinders, and cones. With this knowledge, you are well equipped to look at any shape in your world and figure out its area or surface area.
## Frequently Asked Questions on Area of a Shape
### What does the term ‘area’ mean?
Area is a measurement that describes the amount of space a 2D shape covers. It’s often measured in square units (such as square inches, square feet, or square meters) and is used to understand the size of a flat surface.
### How do I calculate the area of a shape?
The method for calculating the area depends on the shape. For basic shapes, such as squares, rectangles, triangles, and circles, you use specific formulas. For more complex shapes, you often need to break the shape down into these basic shapes, calculate the area of each, and then add them together.
### Why is it important to understand area?
Understanding area is crucial in many real-world situations, from deciding how much paint you need for a wall to understanding how much land you’re buying when you purchase a property. It’s also fundamental to many areas of mathematics and engineering.
### What is the difference between 2D and 3D shapes?
2D shapes have two dimensions – width and height – and are flat. Common 2D shapes include squares, circles, and triangles. 3D shapes, on the other hand, have three dimensions – width, height, and depth – and are solid. Examples of 3D shapes include cubes, spheres, and cylinders.
### How is the area of 3D shapes calculated?
For 3D shapes, we typically refer to the ‘surface area’, which is the total area of all surfaces or faces of the shape. Each type of 3D shape has its own formula for calculating surface area.
Information Sources |
# Example
Prove that if $m$ and $n$ are relatively prime and $(m+n)(m-n)$ is odd, then $(m+n)$ and $(m-n)$ must also be relatively prime.
We can use an indirect argument here.
Assume that $(m+n)$ and $(m-n)$ are not relatively prime. Then they share some common divisor other than one. Let's call this divisor $d$. But if $d \mid m+n$ and $d \mid m-n$, then $d$ must divide both their sum and their difference: $$d \mid (m+n) + (m-n)$$ $$d \mid (m+n) - (m-n)$$ After collecting like terms, we see that $$d \mid 2m\quad \textrm{and} \quad d \mid 2n$$ The first requires that either $d \mid 2$ or $d \mid m$, while the second requires $d \mid 2$ or $d \mid n$.
Notice, however, considering the first case in both possibilities: $d \mid 2$ (with $d \neq 1$), implies $d=2$. Consequently, since $d \mid m+n$ and $d \mid m-n$, both $m+n$, $m-n$, and their product $(m+n)(m-n)$ must be even, which would contradict the given statement that $(m+n)(m-n)$ is odd.
So instead, the second case in both possibilities must be true: $d \mid m$ and $d \mid n$. This, however, means that $m$ and $n$ are not relatively prime. Again, we reach a contradiction with the given information.
So we reject our assumption that $(m+n)$ and $(m-n)$ are not relatively prime. Instead, the opposite must be true: $(m+n)$ and $(m-n)$ are relatively prime.
QED. |
Concept of Average and it Uses - Students Explore
# Concept of Average and it Uses
## Definition of Average
The arithmetic mean is defined as the sum of all numbers in a collection divided by the number of numbers present in the collection.
To put it another way, the average is the ratio of the total number of observations to the sum of all provided observations.
As a result, the average formula is as follows:
Average = Sum of the Observations/Number of Observations
concept of average
## What is the Concept of Average
A single number out of list taken as a representative of all this numbers in the list, is a basic concept of average as it is clearly explained itself in this line.
`What is Percentage and Why we Use it in Maths`
Averages are used to represent a large number range with a single number. It is a graphical representation of all the numbers in the data set. The average is computed by adding all of the data values and dividing them by the number of data points.
The average age of the students in a class is calculated by taking the age of each student in the class and averaging it to give a single value for the average age of the students in a class. When dealing with quantities with changing values, the average is computed and a single value is used to represent the values. Understanding average allows us to quickly summarize the available data.
## Why We Use Average
Average has a wide range of applications in our daily lives. A large set of student grades, the changing price of stocks, a location’s weather data, and the income of different people in a city are all examples for which we can calculate an average.
For Example, the average mark of a class is calculated by averaging the marks of the students in that class in a specific subject. There is a need to know the overall performance of the class rather than the individual performance of each student. The average is useful in this case.
The sum of the values divided by the individual values equals the average of a set of values. In addition, average is used in situations where values change. The temperature of a location is calculated by averaging the temperature of that location over the course of the season.
Making decisions based on a single piece of data or a vast group of facts can be tough at times. As a result, the average value is used to represent all of the values in a single number.
```What is Percentage and Why we Use it in Maths
```
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# Quantitative Aptitude: Data Interpretation for SBI PO Set 13
Direction (1-5): Refer to the graphs and answer the following questions.
The first bar graph shows the marked up price of articles with respect to their cost price and the second bar graph shows the discount % given in respective articles.
1. If the cost price of Q is decreased by 10% (other prices remaining same), then what is the different between the new profit% and original one?
A) 15%
B) 10%
C) 12%
D) 9%
E) Cannot be determined
Option C
Explanation
:
Using MP = (100+p%)/(100-d%) * CP
For article Q:
180% of CP = (100+p%)/(100-40) * CP
CP gets cancelled out, so p% = 8%
Now new CP = 90% of CP
So using MP = (100+p%)/(100-d%) * CP
180% of CP = (100+p%)/(100-40) * 90% of CP
New p% = 20%
So difference = 20 – 8 = 12%
2. Cost price of P is 10% more than the cost price of R. If selling price of R is Rs 320, find the selling price of P.
A) Rs 548.5
B) Rs 577.5
C) Rs 532.5
D) Rs 553.5
E) Cannot be determined
Option B
Explanation
:
Let CP of R = Rs x
For R:
Using MP = (100+p%)/(100-d%) * CP
160% of CP = (100+p%)/(100-60) * CP
p = -36, means there is a loss of 36%
Now SP of R = Rs 320
So CP of R = 100/64 * 320 = Rs 500
So CP of P = 110% of 500 = Rs 550
For P: using MP = (100+p%)/(100-d%) * CP
150% of CP = (100+p%)/(100-30) * CP
p = 5%
So SP of P = 105% of 550 = Rs 577.5
3. If the cost price of Q and S is same, then selling price of S is how much % less than the selling price of Q?
A) 79%
B) 63%
C) 82%
D) 67%
E) Cannot be determined
Option B
Explanation
:
Let CP of Q = CP of S = Rs 100
p% of Q = 8% [from Question 1]
So SP of Q = 108% of 100 = Rs 108
For S: using MP = (100+p%)/(100-d%) * CP
200% of CP = (100+p%)/(100-80) * CP
p = -60%, so there is a loss of 60%
So SP of S = 40% of 100 = Rs 40
Now SP of Q = Rs 108 and SP of S = Rs 40
Required % = (108-40)/108 * 100 = 63%
4. If the selling price of T is increased by 20%, then what would be the discount%?
A) 50%
B) 58%
C) 45%
D) 40%
E) 52%
Option D
Explanation
:
For T:
using MP = (100+p%)/(100-d%) * CP
150% of CP = (100+p%)/(100-50) * CP
p = -25, so loss of 25%
Let CP = Rs 100, so SP = Rs 75
Now SP is increased by 20%, so new SP = 120% of 75 = Rs 90
and MP = 150% of 100 = Rs 150
So discount % = (150-90)/150 * 100 = 40%
5. If the marked price of S is increased by 10% (keeping other prices same) which is same as the cost price of T, then what is the cost price of S given that selling price of T is Rs 660?
A) Rs 400
B) Rs 500
C) Rs 440
D) Rs 550
E) Rs 360
Option A
Explanation
:
For S:
Let CP of S = Rs x
So MP = 200% of x = Rs 2x
Increased MP = 110% of 2x = Rs 22x/10
For T:
So CP of T = Rs 22x/10
and from above question, loss is 25%
So SP of T = 75% of 22x/10
So 75% of 22x/10 = 660, solving x = Rs 400
Direction (6-10): The following table shows the statistics of 5 players. Answer the questions based on information given.
1. What is the difference in the number of matches played by A and D?
A) 55
B) 47
C) 50
D) 40
E) 44
Option D
Explanation
:
Played by A = 20/0.25 = 80
Played by D = 16/0.40 = 40
So difference = 40
2. If all players A, B, C, D, and E are associated with different teams P,Q, R, S, T respectively, then which of the following team won the series?
A) P
B) R
C) T
D) S
E) Cannot be determined
Option E
Explanation
:
The given goals are of individual players of the teams. How other players played is not given, so cannot be said by just seeing the goals of a single player of team.
3. Which player has the best strike rate in terms of minutes?
A) A
B) B
C) C
D) D
E) E
Option A
Explanation
:
A = 2234/20 = 111.7
B = 2563/18 = 142.4
C = 2434/19 = 128.1
D = 1926/16 = 120.4
E = 2017/17 = 118.6
The lower the strike rate, best it is considered
4. If the difference in number of shots by two players is 25, then what is the difference in the strike rate of these?
A) 23
B) 24
C) 22
D) 21
E) Cannot be Determined
Option C
Explanation
:
Number of shots of
A = 20/0.25 = 80
B = 18/0.24 = 75
C = 19/0.2375 = 80
D = 16/0.32 = 50
E = 17/0.20 = 85
So these two players are B and D
Difference in their strike rates =2563/18 – 1926/16 = 142.4 – 120.4 = 22
5. If there are 24 matches left to be played, than what should be the goals/match ratio of B for the rest of games if he wants to overtake A in terms of number of goals scored?
A) 0.275
B) 0.395
C) 0.225
D) 0.375
E) 0.345
Option D
Explanation
:
At current rate, A will score = 24*0.25 = 6 goals more in 24 matches, so A’s goals will be 20+6 = 26
So B to overtake A needs 27 total goals thus needs these 9 more goals (18+9 = 27) in 24 matches
So goal : match = 9 : 24 = 0.375
## 25 Thoughts to “Quantitative Aptitude: Data Interpretation for SBI PO Set 13”
1. ALKA
tyyy mam
2. Lonely little kid
tq
3. purvi
thanku 🙂
4. ^^^Jaga^^^....
mam …q1 mai cp 100 liya jayega ???
1. 100 lene ki need hi nhi hai.
CP cancel hi ho jaega
1. ^^^Jaga^^^....
mam mujhe kuch samjh mai nahi araha
1. ^^^Jaga^^^....
got it got it
1. ^^^Jaga^^^....
mam???
180% of CP = (100+p%)/(100-40) * 120% of CP ye to loss hoga na 10%
2. ^^^Jaga^^^....
so q1 mai ye 20% kese hua …?
5. Rise And Shine
Thank u AZ team 🙂
1. ^^^Jaga^^^....
ho gaya sab qestions?
6. ^^^Jaga^^^....
mam …no 8 ka ans sahi hai???instruction diya hai ki lower strike rate is considered as better…so uske according 111. better hai
1. ^^^Jaga^^^....
thank u mam…
___/____
2. hapusingh
nice work bro ,first di is the exact one which i solved in SBI PO MAINS 2016. if possible can u provide time work question which came in SBI PO MAINS 2016 .
1. Yes it was 🙂
2. We ll try to provide if get.
Do u remember anything that what type of it was asked ?
7. Ankit Saxena
thank you @Shubhra_AspirantsZone:disqus mam nice set questions
1. dream girl ;;;;/@ :)
explain 1st one
1. Ankit Saxena
I will explain each and every question sis sorry bata nahin paaya was busy
1. dream girl ;;;;/@ :)
aare koi na bro jab bhi free hona tb bta dena 🙂
8. dream girl ;;;;/@ :)
mam qno 1 mai 20% profit kaise aya?
if cp=100
10% less 90
MP=90*180/100 so, 162
discount= 162*60/100 so, 97.2
so profit is 7.2*100/90 = 8%
so dono mai 8% araha
1. Profit% calculate karne k liye 7.2 ku liya ??
Profit is based on CP and SP
But u have used discount here |
# Properties Of Rational Numbers Worksheet Pdf
A Realistic Numbers Worksheet will help your child become a little more acquainted with the concepts right behind this ratio of integers. With this worksheet, college students should be able to solve 12 distinct issues associated with realistic expression. They will learn how to increase 2 or more amounts, team them in sets, and figure out their products. They may also training simplifying rational expression. When they have perfected these concepts, this worksheet is a useful tool for furthering their scientific studies. Properties Of Rational Numbers Worksheet Pdf.
## Realistic Amounts can be a proportion of integers
There are two varieties of figures: rational and irrational. Rational figures are understood to be entire figures, whereas irrational figures tend not to recurring, and get an unlimited quantity of digits. Irrational figures are non-no, no-terminating decimals, and square origins that are not best squares. They are often used in math applications, even though these types of numbers are not used often in everyday life.
To define a logical quantity, you need to realize what a logical amount is. An integer can be a complete amount, along with a logical amount is really a proportion of two integers. The percentage of two integers may be the quantity on top split by the quantity at the base. If two integers are two and five, this would be an integer, for example. There are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They may be made right into a portion
A reasonable variety has a denominator and numerator that are not zero. Consequently they could be depicted like a portion. Together with their integer numerators and denominators, logical figures can also have a bad worth. The adverse value ought to be put left of and its definite importance is its range from absolutely no. To streamline this case in point, we shall state that .0333333 is really a small fraction that could be published being a 1/3.
Along with negative integers, a logical quantity can also be manufactured into a small fraction. By way of example, /18,572 is really a realistic variety, whilst -1/ is just not. Any fraction composed of integers is realistic, given that the denominator fails to have a and might be created for an integer. Furthermore, a decimal that ends in a position is also a rational number.
## They are feeling
In spite of their name, rational amounts don’t make significantly perception. In mathematics, these are individual entities having a exclusive size about the amount series. Which means that once we matter something, we could buy the shape by its percentage to its authentic amount. This keeps accurate even when there are actually infinite logical numbers between two specific figures. If they are ordered, in other words, numbers should make sense only. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
If we want to know the length of a string of pearls, we can use a rational number, in real life. To discover the period of a pearl, for instance, we might count up its width. An individual pearl weighs about twenty kilograms, which is a rational variety. In addition, a pound’s weight equates to ten kilos. Therefore, we should be able to divide a pound by 15, without the need of be worried about the length of just one pearl.
## They may be expressed as a decimal
You’ve most likely seen a problem that involves a repeated fraction if you’ve ever tried to convert a number to its decimal form. A decimal quantity may be published like a multiple of two integers, so four times 5 is the same as 8. An identical dilemma necessitates the repeated portion 2/1, and each side needs to be divided by 99 to get the correct solution. But how would you create the conversion process? Here are a few illustrations.
A rational variety can be designed in various forms, including fractions and a decimal. A great way to represent a logical number inside a decimal is always to divide it into its fractional counterpart. You will find three ways to split a realistic quantity, and every one of these methods brings its decimal equal. One of those ways would be to divide it into its fractional equivalent, and that’s what’s referred to as a terminating decimal. |
How do you simplify (3) [square root of (3/7)] -(5) (square root of 84)?
Oct 8, 2015
$- 9 \sqrt{21}$
Explanation:
Start by writing down your starting expression
$3 \sqrt{\frac{3}{7}} - 5 \sqrt{84} = 3 \cdot \frac{\sqrt{3}}{\sqrt{7}} - 5 \sqrt{84}$
Thee first thing to do here is rationalize the denominator of the first term by multiplying by $1 = \frac{\sqrt{7}}{\sqrt{7}}$
This will get you
$3 \cdot \frac{\sqrt{3}}{\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = 3 \cdot \frac{\sqrt{21}}{\sqrt{7} \cdot \sqrt{7}} = \frac{3}{7} \cdot \sqrt{21}$
Now focus on the second radical term. Notice that you can write $84$ as
$84 = 2 \cdot 42 = 2 \cdot 2 \cdot 21 = {2}^{2} \cdot 21$
The expression will thus be
$\frac{3}{7} \cdot \sqrt{21} - 5 \cdot \sqrt{{2}^{2} \cdot 21}$
$\frac{3}{7} \cdot \sqrt{21} - 5 \cdot 2 \sqrt{21}$
This will be equal to
$\sqrt{21} \cdot \left(\frac{3}{7} - 10\right) = \sqrt{21} \cdot \left(\frac{3 - 70}{7}\right) = \sqrt{21} \cdot \frac{\left(- 63\right)}{7}$
Finally, this expression can be simplified to
$- \frac{\left(63\right)}{7} \sqrt{21} = \textcolor{g r e e n}{- 9 \sqrt{21}}$ |
School Subjects
Math & Science
Electives & Health
Teaching & Learning
# Show Me a Way to Make Eleven: How I Teach Math with Rekenreks
In this classroom video, Ruth Anna demonstrates how rekenreks enable her class to review math facts and grasp relationships between sets of numbers.
For more on the ways rekenreks can improve your students’ math abilities, see Ruth Anna’s blog post and her previous rekenrek demonstration:
Ruth Anna: I’m going to give all of you a rekenrek and then we’re going to talk about a few things.
Ruth Anna: Are you ready? So am I.
Ruth Anna: When you get rekenrek, I want you to push all of your beads to the right. And I think for today, when I give you your rekenrek, place it on the floor like this.
Ruth Anna: Everyone have their beads pushed to the right?
Ruth Anna: Alright folks. So today, we’re going to do little bit of review before we talk about another number. I want all of you to show me a way to make eight. Show me one way to make eight.
Ruth Anna: Alright. Maybe you didn’t know this but I chose a way of making eight, as well. We’re going to see if any of you have my way of making eight. Now let’s look real carefully. Does everyone think they have a way of making eight? Double-check to make sure that you’re only showing eight.
Ruth Anna: That’s right. What’s your way of making eight, Leah?
Leah: One plus seven.
Ruth Anna: One plus seven. Is that a way of making eight? It is! That’s right! What’s your way, Elena?
Elena: Two plus six.
Ruth Anna: Two plus six. Now, your way is related to Conrad’s way, right?
Ruth Anna: Now, we’re going to do something else. The new number that we are just talking about in math is eleven. What I want you to do, I want you to take your beads, push them all to the right again. Alright. This time, I’m going to let you hold your rekenrek up beside you like this so just you see it.
Ruth Anna: Now, I want you to use both rows in your rekenrek to show me a way of making eleven. What number are going to make?
Group: Eleven.
Ruth Anna: Eleven. When you’re finished, just hold your rekenrek like this so your beads don’t slide back.
Ruth Anna: Alright. So, what is eleven? There are a bunch of ways of making eleven. Let’s see what you came up with. So, Caroline, what did you come up with?
Caroline: Five plus six?
Ruth Anna: Five plus six. Can you show it to the class?
Ruth Anna: Is that a way of making eleven?
Ruth Anna: Yes. You know what I see here with Caroline’s way? I can look at these groups of numbers, I can see five, ten, plus one. And that’s?
Group: Eleven.
Ruth Anna: Eleven. Right. Nice, Caroline. That’s a good way of making eleven.
Ruth Anna: How did you show that? One plus ten. Do you see Conrad’s? One on the top and ten on the bottom.
Ruth Anna: Nobody has my way of making eleven. Even though you are all right. Can you look what is my way of making eleven?
Group: Ten—
Ruth Anna: Oh, is this ten?
Group: Nine plus two.
Ruth Anna: That’s right. And the reason that I chose nine plus two is that because this was our new math fact yesterday. Who remembers that? Nine plus two.
Ruth Anna: Let’s say it all together. Ready, and say:
Group: Nine plus two equals eleven.
Ruth Anna: I about couldn’t hear you.
Group: Nine plus two—
Ruth Anna: Hang on. Hands in your lap. Now, can I hear you again? Everyone has your hands in your lap? Alright, and say—ready?
Group: Nine plus two equals eleven.
Ruth Anna: That’s right. Raise your hand if you know the related fact. Okay. How many am I going to put on top?
Group: Two.
Ruth Anna: Two. How many on the bottom?
Group: Nine.
Ruth Anna: That’s right. Two plus nine is also related. Two plus nine is another way of making eleven.
Ruth Anna: Let’s collect the rekenreks and then we’ll do more with our math lesson.
Ruth Anna: Okay, today, Conrad you get to bring those back.
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# Historical Activities for Calculus - Module 3: Optimization - Heron's Shortest Distance Problem
Author(s):
Gabriela R. Sanchis (Elizabethtown College)
### Heron's "Shortest Distance" Problem
One of the first non-trivial optimization problems was solved by Heron of Alexandria, who lived about 10-75 C.E. Heron's “Shortest Distance” Problem is as follows:
Given two points $$A$$ and $$B$$ on one side of a straight line, to find the point $$C$$ on the line such that $$|AC|+|CB|$$ is as small as possible.
Thus in Figure 26, below, one possible point $$C$$ is shown, as well as the length of the path $$A$$ to $$C$$ to $$B.$$
Figure 26. A possible path from $$A$$ to $$B$$
Figure 27. Use this GeoGebra applet to explore the solution to Heron's problem. You can drag $$C$$ to determine the approximate length of the shortest path attainable.
To solve this problem mathematically, Heron noticed that if $$B$$ is reflected across the line, to say $$B^{\prime},$$ then for any point $$C$$ on the line, $$|CB|=|CB^{\prime}|,$$ and hence minimizing $$|AC|+|CB|$$ is equivalent to minimizing $$|AC|+|C B^{\prime}|.$$ But clearly since the shortest path from $$A$$ to $$B^{\prime}$$ is a straight line, the point $$C$$ that minimizes $$|AC|+|C B^{\prime}|$$ should be the point of intersection of the line with the line segment $$AB^{\prime}.$$ Any other path, such as $$A$$ to $$C^{\prime}$$ to $$B^{\prime},$$ will clearly be longer.
Figure 28. Heron's solution of the “Shortest Distance” Problem
Note that $$m\angle 2=m\angle 3$$ by construction, and $$m\angle1=m\angle 3$$ since these are vertical angles. Therefore, $$m\angle 1=m\angle 2.$$ This is the equal angle law of reflection. It was Euclid who, over three hundred years earlier, had noted the now well-known Reflection Law for light:
Equal Angle Law of Reflection (or Euclid's Law of Reflection). If a beam of light is sent toward a mirror, then the angle of incidence equals the angle of reflection.
Heron appears to have been the first to observe that the Reflection Law implies that light always takes the shortest path.
As a matter of fact, the mirror in the Equal Angle Law of Reflection need not be flat. We may replace the line in Heron's problem by any concave curve (a curve is concave if it lies entirely on one side of any tangent line). In this case, the angles are measured with respect to the tangent line, and the same argument used by Heron shows that if $$C$$ is such that the angle of incidence equals the angle of reflection, then $$|AC|+|BC|$$ is minimized.
Figure 29. Law of Reflection for any concave curve
An interesting application of the Law of Reflection arises in the case of a light beam sent toward a parabolic mirror, where the light beam is parallel to the axis of the parabola. A parabolic mirror is one whose surface is generated by rotating a parabola about its axis. Suppose the parabola has focus $$F$$ and directrix $${\mathcal L},$$ and that the light beam $$\overrightarrow{GA}$$ hits the parabola at $$A.$$ Recall from Roberval's construction of tangent lines to parabolas that the tangent line at $$A$$ bisects the angle $$\angle FAB.$$ Hence $$m\angle 1=m\angle 2.$$ Since $$\angle 2$$ and $$\angle 3$$ are vertical angles, $$m\angle 2=m\angle 3.$$ Hence $$m\angle 1=m\angle 3.$$ So by the Equal Angle Law of Reflection, the light beam will be reflected in the direction $$\overrightarrow{AF}.$$ This will be true for any point $$A$$ on the parabola.
Figure 30. Light rays directed toward a parabolic mirror parallel to the parabola's axis are all reflected toward the focus $$F$$ of the parabola.
There is a famous story about Archimedes in which he is said to have used a parabolic mirror to defeat the Roman General Marcellus. Supposedly, he tilted the mirror toward the sun in such a way that all the sun's rays when reflected off the mirror went through the focal point. The heat generated at that point caused a fire to ignite and destroy the entire Roman fleet.
Figure 31. Wall painting from the Stanzino delle Matematiche in the Galleria degli Uffizi (Florence, Italy). Painted by Giulio Parigi (1571-1635) in the years 1599-1600.
#### Exercises
Figure 32. Euclid's Law of Reflection (applet for Exercise 8)
##### Exercise 8.
Use calculus and a CAS to find the point $$C$$ on the $$x$$ axis for which $$|AC|+|CB|$$ is the shortest, where $$A=(0,4)$$ and $$B=(10,12).$$ Use the applet in Figure 32, above, to enter your solution and verify Euclid's Law of Reflection. |
Notes On Cylinder - CBSE Class 9 Maths
A cylinder can be defined as a solid figure that is bound by a curved surface and two flat circular surfaces. The flat surfaces are made up of two congruent circles that are parallel to each other. These flat surfaces are called the bases of the cylinder. The radius of the circular bases is the radius of the cylinder. The perpendicular line that passes through the centres of the two circular bases is the height of the cylinder or axis of the cylinder. A cylinder is said to be right circular cylinder if axis of the cylinder is perpendicular to the radius of the cylinder. Curved surface area of a cylinder (CSA) The curved surface joining the two bases of a right circular cylinder is called its lateral surface. If the curved surface of a cylinder is unfold, it opens up to become a rectangular sheet. The curved surface area of a cylinder is equal to the area of this sheet. Area of a rectangle = length x breadth. The length of the rectangular sheet is equal to the circumference of the base of the cylinder, which is equal to 2πr and the breadth of the rectangular sheet is equal to the height of the cylinder. Substituting length = 2πr and breadth = h, we get Area of the rectangular sheet = 2πr × h = 2πrh ∴ Curved surface area of a cylinder (CSA) = 2πrh. Total surface area of a cylinder Total surface area of a cylinder (TSA) = Curved surface area (CSA) + Total area of the flat surfaces The flat surfaces of a cylinder are made up of two congruent circles. ∴ Area of the flat surface of a cylinder = Area of the base = πr2 Total area of both the flat surfaces = 2πr2 Total surface area of a cylinder (TSA) = 2πrh + 2πr2 = 2πr(h + r). Volume of a cylinder Volume of a cylinder = Area of the base × height = πr2 × h (∵ Area of the base = πr2) ∴ Volume of a cylinder (V) = πr2h.
#### Summary
A cylinder can be defined as a solid figure that is bound by a curved surface and two flat circular surfaces. The flat surfaces are made up of two congruent circles that are parallel to each other. These flat surfaces are called the bases of the cylinder. The radius of the circular bases is the radius of the cylinder. The perpendicular line that passes through the centres of the two circular bases is the height of the cylinder or axis of the cylinder. A cylinder is said to be right circular cylinder if axis of the cylinder is perpendicular to the radius of the cylinder. Curved surface area of a cylinder (CSA) The curved surface joining the two bases of a right circular cylinder is called its lateral surface. If the curved surface of a cylinder is unfold, it opens up to become a rectangular sheet. The curved surface area of a cylinder is equal to the area of this sheet. Area of a rectangle = length x breadth. The length of the rectangular sheet is equal to the circumference of the base of the cylinder, which is equal to 2πr and the breadth of the rectangular sheet is equal to the height of the cylinder. Substituting length = 2πr and breadth = h, we get Area of the rectangular sheet = 2πr × h = 2πrh ∴ Curved surface area of a cylinder (CSA) = 2πrh. Total surface area of a cylinder Total surface area of a cylinder (TSA) = Curved surface area (CSA) + Total area of the flat surfaces The flat surfaces of a cylinder are made up of two congruent circles. ∴ Area of the flat surface of a cylinder = Area of the base = πr2 Total area of both the flat surfaces = 2πr2 Total surface area of a cylinder (TSA) = 2πrh + 2πr2 = 2πr(h + r). Volume of a cylinder Volume of a cylinder = Area of the base × height = πr2 × h (∵ Area of the base = πr2) ∴ Volume of a cylinder (V) = πr2h.
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# What are the rules of derivatives in calculus?
## What are the rules of derivatives in calculus?
Derivative Rules
Common Functions Function Derivative
Difference Rule f – g f’ − g’
Product Rule fg f g’ + f’ g
Quotient Rule f/g f’ g − g’ fg2
Reciprocal Rule 1/f −f’/f2
What are the five rules of differentiation?
Some differentiation rules are a snap to remember and use. These include the constant rule, power rule, constant multiple rule, sum rule, and difference rule.
Does HX mean history?
Hx (uncountable) (medicine) Abbreviation of history.
### How do you work out the derivatives of many functions?
There are ruleswe can follow to find many derivatives. For example: The slope of a constantvalue (like 3) is always 0 The slope of a linelike 2x is 2, or 3x is 3 etc and so on. Here are useful rules to help you work out the derivatives of many functions (with examples below).
How to find the derivative of a function for problems 1-12?
For problems 1 – 12 find the derivative of the given function. Determine where, if anywhere, the function f (x) = x3 +9×2−48x+2 f ( x) = x 3 + 9 x 2 − 48 x + 2 is not changing. Solution Determine where, if anywhere, the function y =2z4 −z3−3z2 y = 2 z 4 − z 3 − 3 z 2 is not changing.
What is the derivative rule in calculus?
Derivative Rules The Derivative tells us the slope of a function at any point. “multiply by power then reduce power by 1” which simplifies to −x −2 dy dx = dy du du dx d dx sin(x 2) = d du sin(u) d dx x 2 d dx sin(x 2) = cos(u) (2x)
## What is the derivative?
The Derivative tells us the slope of a function at any point. There are rules we can follow to find many derivatives. and so on. Here are useful rules to help you work out the derivatives of many functions (with examples below ). Note: the little mark ’ means “Derivative of”, and f and g are functions. |
# Simplify the following expression: 101-{[(110 -: 2)-: 11]xx(10+4xx2) +7}+ [8xx(20 -: 5-1)-3xx3] -: 5?
Take your time and methodically go through each bracket and you'll eventually get to $7$
#### Explanation:
Wow... that's one big equation. Let's take this step by step.
$101 - \left\{\left[\left(110 \div 2\right) \div 11\right] \times \left(10 + 4 \times 2\right) + 7\right\} + \left[8 \times \left(20 \div 5 - 1\right) - 3 \times 3\right] \div 5$
Before we dive into this thing, let's look at the structure - there is $101$ - big brackets + smaller brackets$\div 5$. PEDMAS has us work things in brackets (Parentheses) first and since the big brackets and the smaller brackets are separated by the $+$, we can work them separately. I'm going to simplify the big brackets first:
$\left\{\left[\left(110 \div 2\right) \div 11\right] \times \left(10 + 4 \times 2\right) + 7\right\}$
There are brackets (and brackets within brackets) in this, so I'm going to work those first. There are 2 sets here and I'll work them side by side. In this first step, let's do the division and in the second set we have both addition and multiplication - so we'll do the multiplication first:
$\left\{\left[55 \div 11\right] \times \left(10 + 8\right) + 7\right\}$
I can now do the next division in the first bracket and do the addition in the second:
$\left\{5 \times 18 + 7\right\}$
We'll finish this up with the multiplication first, then the addition:
$90 + 7 = 97$ which I'll substitute back into our original:
$101 - 97 + \left[8 \times \left(20 \div 5 - 1\right) - 3 \times 3\right] \div 5$
Let's now work that second bracket:
$\left[8 \times \left(20 \div 5 - 1\right) - 3 \times 3\right]$
There's a bracket in here that we need to work first. Within that bracket there is both division and subtraction - we'll divide first:
$\left[8 \times \left(4 - 1\right) - 3 \times 3\right]$
and now the subtraction:
$\left[8 \times 3 - 3 \times 3\right]$
We now have 2 multiplications and a subtraction, so we'll do the multiplications first:
$\left[24 - 9\right] = 15$
Let's substitute that into the original:
$101 - 97 + 15 \div 5$
Almost there! We have subtraction, addition, and division. We'll do the division first:
$101 - 97 + 3$
and now the subtraction and addition:
$4 + 3 = 7$
Jul 31, 2016
=$\textcolor{m a \ge n t a}{101} \textcolor{g r e e n}{+ 3} \textcolor{b l u e}{- 87}$
=$7$
#### Explanation:
Count the number of terms and work through each one carefully.
Each term must give a number answer.
There are only 3 terms in this expression:
$\textcolor{m a \ge n t a}{101} \textcolor{b l u e}{- \left\{\left[\left(110 \div 2\right) \div 11\right] \times \left(10 + 4 \times 2\right) + 7\right\}} \textcolor{g r e e n}{+ \left[8 \times \left(20 \div 5 - 1\right) - 3 \times 3\right] \div 5}$
Let's handle one at a time.
The first is easy. it is $\textcolor{m a \ge n t a}{101}$.
$\textcolor{b l u e}{- \left\{\left[\textcolor{red}{\left(110 \div 2\right)} \div 11\right] \times \left(10 + \textcolor{red}{4 \times 2}\right) + 7\right\}}$
$\text{parentheses first, but remember to do the multiplication }$
$\text{and division before addition and subtraction}$
$\textcolor{b l u e}{- \left\{\left[\textcolor{red}{55} \div 11\right] \times \left(10 + \textcolor{red}{8}\right) + 7\right\}}$
color(blue)(-{[color(red)(5]xx(color(red)(18))+7})
color(blue)(-[color(red)(90+7)] = -97
Now for the third term:
$\textcolor{g r e e n}{+ \left[8 \times \left(20 \div 5 - 1\right) - 3 \times 3\right] \div 5}$
$\textcolor{g r e e n}{+ \left[8 \times \textcolor{red}{\left(20 \div 5 - 1\right)} - \textcolor{\mathmr{and} a n \ge}{3 \times 3}\right] \div 5}$
$\textcolor{g r e e n}{+ \left[8 \times \textcolor{red}{\left(4 - 1\right)} - \textcolor{\mathmr{and} a n \ge}{9}\right] \div 5}$
$\textcolor{g r e e n}{+ \left[8 \times \textcolor{red}{3} - \textcolor{\mathmr{and} a n \ge}{9}\right] \div 5}$
+[color(red)(24)-color(orange)(9)]-:5)
$+ \textcolor{g r e e n}{15 \div 5} = \textcolor{g r e e n}{3}$
the whole expression simplifies to
$\textcolor{m a \ge n t a}{101} \textcolor{b l u e}{- 97} \textcolor{g r e e n}{+ 3}$
=$\textcolor{m a \ge n t a}{101} \textcolor{g r e e n}{+ 3} \textcolor{b l u e}{- 97}$
=$7$ |
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# There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Last updated date: 18th Sep 2024
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Hint: Here we use the concept of Least Common Multiple to find the answer for the above question because both Ravi and Sonia start from the same point but they will take different time to complete one round. So, they will meet at the starting point at a particular time such that, these duration of time should be divisible by both of their times taken to drive.
From the question it is given that Ravi and Sonia start from the same point but take different time to complete the round. So, now we have to find the time taken by both of them to meet again at the starting point.
That means we have to find the LCM of 18 and 12.
Now we will find the LCM of 18 and 12 in prime factorisation method:
18 can be expressed as $18=2\times {{3}^{2}}$
12 can be expressed as $12={{2}^{2}}\times 3$
Now prime factor 2 is common here. So, take prime factor with highest power i.e. ${{2}^{2}}$
Now prime factor 3 is common here. So, take prime factor with highest power i.e. ${{3}^{2}}$
Then there is no common prime factor in 18 and 12, so there is no effect on LCM.
Now, by prime factorisation LCM of a and b is ${{2}^{2}}\times {{3}^{2}}=36$
So, LCM of 18 and 12 is 36.
That means the time taken by Ravi and Sonia to meet again at the starting point is 36 minutes.
Note: There is another way to solve this problem in an alternative way.
Ravi takes 12 minutes to drive one round around the field.
Sonia takes 18 minutes to drive one round around the field.
Ravi takes 24 minutes to drive two rounds around the field.
Sonia takes 36 minutes to drive two rounds around the field.
Ravi takes 36 minutes to drive three rounds around the field. |
# How do you solve 3s+2(5s-1)=56.50?
Nov 8, 2016
$s = 4.5$
#### Explanation:
Step 1) Expand the terms in parenthesis.
Step 2) Combine like terms
Step 3) Solve for $s$ while keeping the equation balanced
$3 s + 10 s - 2 = 56.50$
$\left(3 + 10\right) s - 2 = 56.50$
$13 s - 2 = 56.50$
$13 s - 2 + 2 = 56.50 + 2$
$13 s = 58.50$
$\frac{13 s}{13} = \frac{58.50}{13}$
$s = 4.5$ |
# 3 Expert Tips for Using the Unit Circle
If you’re studying trig or calculus—or getting ready to—you’ll need to get familiar with the unit circle.
The unit circle is
an essential tool used to solve for the sine, cosine, and tangent of an angle.
But how does it work? And what information do you need to know in order to use it?
In this article, we explain what the unit circle is and why you should know it. We also give you three tips to help you remember how to use the unit circle.
Feature Image:
Gustavb
/Wikimedia
## The Unit Circle: A Basic Introduction
The unit circle is a circle with a radius of 1.
This means that for any straight line drawn from the center point of the circle to any point along the edge of the circle, the length of that line will always equal 1. (This also means that the diameter of the circle will equal 2, since the diameter is equal to twice the length of the radius.)
Typically,
the center point of the unit circle is where the x-axis and y-axis intersect, or at the coordinates (0, 0):
The unit circle, or trig circle as it’s also known, is useful to know because
it lets us easily calculate the cosine, sine, and tangent of any angle between 0° and 360° (or 0 and 2π radians).
As you can see in the above diagram, by drawing a radius at any angle (marked by ∝ in the image), you will be creating a right triangle.
On this triangle, the cosine is the horizontal line, and the sine is the vertical line. In other words,
cosine =
x-coordinate, and
sine = y-coordinate.
(
The triangle’s longest line, or hypotenuse, is the radius and therefore equals 1.)
Why is all of this important? Remember that you can solve for the lengths of the sides of a triangle using the
Pythagorean theorem, or $a^2+b^2=c^2$
(in which
a
and
b
are the lengths of the sides of the triangle, and
c
is the length of the hypotenuse).
We know that the cosine of an angle is equal to the length of the horizontal line, the sine is equal to the length of the vertical line, and the hypotenuse is equal to 1. Therefore, we can say that
the formula for any right triangle in the unit circle is as follows:
$$\cos^2θ+\sin^2θ=1^2$$
Since $1^2=1$, we can simplify this equation like this:
$$\cos^2θ+\sin^2θ=1$$
Be aware that
these values can be negative
depending on the angle formed and what quadrant the x- and y-coordinates fall in (I’ll explain this in more detail later).
Here is an overview of all major angles in degrees and radians on the unit circle:
### Unit Circle — Degrees
But what if there’s no triangle formed? Let’s look at
what happens when the angle is 0°, creating a horizontal straight line along the x-axis:
On this line, the x-coordinate equals 1 and the y-coordinate equals 0. We know that
the cosine is equal to the x-coordinate, and the sine is equal to the y-coordinate,
so we can write this:
• $\cos0°=1$
• $\sin0°=0$
What if
the angle is 90° and makes a perfectly vertical line along the y-axis?
Here, we can see that the x-coordinate equals 0 and the y-coordinate equals 1. This gives us the following values for sine and cosine:
• $\cos90°=0$
• $\sin90°=1$
This slogan definitely applies if you’re not a math lover.
## Why You Should Know the Unit Circle
As stated above, the unit circle is helpful because
it allows us to easily solve for the sine, cosine, or tangent of any degree or radian.
It’s especially useful to know the unit circle chart if you need to solve for certain trig values for math homework or if you’re preparing to study calculus.
But how exactly can knowing the unit circle help you? Let’s say you’re given the following problem on a math test—and are
not
allowed to use a calculator to solve it:
$$\sin30°$$
Where do you start? Let’s take a look at the unit circle chart again—this time
with all major angles (in both degrees and radians) and their corresponding coordinates:
Jim.belk
/Wikimedia
Don’t get overwhelmed! Remember, all you’re solving for is $\sin30°$. By looking at this chart, we can see that
the y-coordinate is equal to $1/2$ at 30°.
And since the y-coordinate equals sine, our answer is as follows:
$$\sin30°=1/2$$
But what if you get a problem that uses radians instead of degrees? The process for solving it is still the same. Say, for example, you get a problem that looks like this:
$$\cos{{3π}/4}$$
Again, using the chart above, we can see that the x-coordinate (or cosine) for ${3π}/4$ (which is equal to 135°) is $-{√2}/2$. Here’s what our answer to this problem would look like then:
$$\cos({3π}/4)=-{√2}/2$$
All of this is pretty easy if you have the unit circle chart above to use as a reference. But most (if not all) of the time, this won’t be the case, and you’ll be expected to answer these types of math questions using your brain only.
So how can you remember the unit circle? Read on for our top tips!
## How to Remember the Unit Circle: 3 Essential Tips
In this section, we give you our top tips for remembering the trig circle so you can use it with ease for any math problem that requires it.
I wouldn’t recommend practicing the unit circle with post-its, but, hey, it’s a start.
### #1: Memorize Common Angles and Coordinates
In order to use the unit circle effectively, you’ll need to
memorize the most common angles (in both degrees and radians) as well as their corresponding x- and y-coordinates.
The diagram above is a helpful unit circle chart to look at, since it includes all major angles in both degrees and radians, in addition to their corresponding coordinate points along the x- and y-axes.
Here is a chart listing this same information in table form:
Angle (Degrees) Angle (Radians) Coordinates of Point on Circle 0° / 360° 0 / 2π (1, 0) 30° $π/6$ $({√3}/2, 1/2)$ 45° $π/4$ $({√2}/2, {√2}/2)$ 60° $π/3$ $(1/2,{√3}/2)$ 90° $π/2$ (0, 1) 120° ${2π}/3$ $(-1/2, {√3}/2)$ 135° ${3π}/4$ $(-{√2}/2, {√2}/2)$ 150° ${5π}/6$ $(-{√3}/2, 1/2)$ 180° π (-1, 0) 210° ${7}/6$ $(-{√3}/2, -1/2)$ 225° ${5π}/4$ $(-{√2}/2, -{√2}/2)$ 240° ${4π}/3$ $(-1/2, -{√3}/2)$ 270° ${3π}/2$ (0, -1) 300° ${5π}/3$ $(1/2, -{√3}/2)$ 315° ${7π}/4$ $({√2}/2, -{√2}/2)$ 330° ${11π}/6$ $({√3}/2, -1/2)$
Now, while you’re more than welcome to try to memorize all these coordinates and angles, this is
a lot
of stuff to remember.
Fortunately, there’s a trick you can use to help you remember the most important parts of the unit circle.
Look at the coordinates above and you’ll notice a clear pattern: all points (excluding those at 0°, 90°, 270°, and 360°)
alternate between just three values (whether positive or negative):
• $1/2$
• ${√2}/2$
• ${√3}/2$
Each value corresponds to
a short, medium, or long line for both cosine and sine:
Here’s what these lengths mean:
• Short horizontal or vertical line
= $1/2$
• Medium horizontal or vertical line
= ${√2}/2$
• Long horizontal or vertical line
= ${√3}/2$
For example, if you’re trying to solve $\cos{π/3}$, you should know right away that this angle (which is equal to 60°) indicates
a short horizontal line on the unit circle.
Therefore,
its corresponding x-coordinate must equal $1/2$
(a positive value, since $π/3$ creates a point in the first quadrant of the coordinate system).
Finally, while it’s helpful to memorize all the angles in the table above, note that
by far the most important angles to remember are the following:
• 30° / $π/6$
• 45° / $π/4$
• 60° / $π/3$
Treat your negatives and positives as you would cables that can potentially kill you if hooked up incorrectly.
### #2: Learn What’s Negative and What’s Positive
It’s critical to be able to distinguish positive and negative x- and y-coordinates so that you’re finding the correct value for a trig problem. As a reminder,
w
hether a coordinate on the unit circle will be positive or negative depends on
which quadrant (I, II, III, or IV) the point falls under:
Here’s a chart showing whether a coordinate will be positive or negative based on the quadrant a particular angle (in degrees or radians) is in:
Quadrant X-Coordinate (Cosine) Y-Coordinate (Sine) I + + II − + III − − IV + −
For example, say you’re given the following problem on a math test:
$$\cos210°$$
Before you even try to solve it, you should be able to recognize that the answer will be
a negative number
since the angle 210° falls in quadrant III (where x-coordinates are
always
negative).
Now, using the trick we learned in tip 1, you can figure out that an angle of 210° creates
a long horizontal line.
Therefore, our answer is as follows:
$$\cos210°=-{√3}/2$$
### #3: Know How to Solve for Tangent
Lastly, it’s essential to know how to use all of this information about the trig circle and sine and cosine in order to be able to
solve for the tangent of an angle.
In trig, to find the tangent of an angle θ (in either degrees or radians), you simply
divide the sine by the cosine:
$$\tanθ={\sinθ}/{\cosθ}$$
For instance, say you’re trying to answer this problem:
$$\tan300°$$
The first step is to set up an equation in terms of sine and cosine:
$$\tan300°={\sin300°}/{\cos300°}$$
Now, to solve for the tangent, we need to find the sine
and
cosine of 300°. You should be able to quickly recognize that the angle 300° falls in the fourth quadrant, meaning that
the cosine, or x-coordinate, will be positive, and the sine, or y-coordinate, will be negative.
You should also know right away that
the angle 300° creates
a short horizontal line and a long vertical line.
Therefore, the cosine (the horizontal line) will equal $1/2$, and the sine (the vertical line) will equal $-{√3}/2$ (a negative y-value, since this point is in quadrant IV).
Now, to find the tangent, all you do is plug in and solve:
$$\tan300°={-{√3}/2}/{1/2}$$
$$\tan300°=-√3$$
Time to purr-actice your math skills!
## Unit Circle Practice Question Set
Now that you know what the unit circle looks like and how to use it, let’s test what you’ve learned with a few practice problems.
### Questions
1. $\sin45°$
2. $\cos240°$
3. $\cos{5π}/3$
4. $\tan{2π}/3$
1. ${√2}/2$
2. $-1/2$
3. $1/2$
4. $-√3$
#### #1: $\sin45°$
With this problem, there are two pieces of information you should be able to identify right away:
• The answer will be positive,
since the angle 45° is in quadrant I, and the sine of an angle is equal to the y-coordinate
• The angle 45° creates
a medium-length vertical line
(for sine)
Since 45° indicates a positive, medium-length line,
${√2}/2$.
If you’re not sure how to figure this out, draw a diagram to help you determine whether the length of the line will be short, medium, or long.
#### #2: $\cos240°$
Like problem #1 above, there are two pieces of information you should be able to quickly grasp with this problem:
• The answer will be negative,
since the angle 240° is in quadrant III, and the cosine of an angle is equal to the x-coordinate
• The angle 240° creates
a short horizontal line
(for cosine)
Since 240° indicates a negative, short line,
$-1/2$.
#### #3: $\cos{5π}/3$
Unlike the problems above, this problem uses
instead of degrees. Although this might make the problem look trickier to solve, in reality it uses the same basic steps as the other two problems.
First, you should recognize that the angle ${5π}/3$ is in quadrant IV, so the x-coordinate, or cosine, will be
a positive number.
You should also be able to tell that
${5π}/3$
creates
a short horizontal line.
This gives you enough information to determine that
the
$1/2$.
#### #4: $\tan{2π}/3$
This problem deals with tangent instead of sine or cosine, which means that it’ll require a little more math on our end. First off, recall
the basic formula for finding tangent:
$$\tan θ={\sin θ}/{\cos θ}$$
Now, let’s take the degree we’ve been given—${2π}/3$
—and plug it into this equation:
$$\tan {2π}/3={\sin {2π}/3}/{\cos {2π}/3}$$
You should now be able to solve for the sine and cosine separately using what you’ve memorized about the unit circle. Since the angle ${2π}/3$ is in quadrant II,
the x-coordinate (or cosine) will be negative, and the y-coordinate (or sine) will be positive.
Next, you should be able to determine based on the angle alone that the horizontal line is
a short line,
and the vertical line is
a long line.
This means that the cosine is equal to $-1/2$, and the sine is equal to ${√3}/2$.
Now that we’ve figured out these values, all we have to do is plug them into our initial equation and solve for the tangent:
$$\tan {2π}/3={{√3}/2}/{-1/2}$$
$$\tan {2π}/3=-√3$$
## What’s Next?
If you’re taking the SAT or ACT soon, you’ll need to know some trig so you can do well on the math section.
Take a look at our expert guides to
trig on the SAT
and
ACT
so you can learn exactly what you’ll need to know for test day!
Besides memorizing the unit circle,
it’s a good idea to learn
how to plug in numbers
and |
Associated Topics || Dr. Math Home || Search Dr. Math
### What is 12/18 ?
```
Date: 27 Apr 1995 13:16:46 -0400
From: Jennifer La
Subject: (none)
Dear Dr. Math,
Our third grade class is learning division and remainders. I want to ask
you a math problem. 12/18 = ?
From,
Jennifer La
```
```
Date: 9 May 1995 22:08:48 -0400
From: Dr. Sydney
Dear Jennifer,
Hello! I'm glad you wrote to Dr. Math!
Long divison is a helpful tool in figuring out problems like what is 12
divided by 18.
Write out your problem like this:
____
18|12
First notice that 12 is smaller than 18. That means that 12 divided by 18
is going to be less than one! Whenever you divide a smaller number by a
larger number, you get a number that is less than one. You can think of
12/18 as being a part of a whole -- a fraction (hence the name fraction,
which you may not have heard yet) of the number 1. If you were to look at a
number strip, the number 12/18 would fall in between the markings for 0 and
1.
That said, we should now move on to the problem. When dividing a smaller
number by a bigger number, we must take a few more extra steps than we do
when dividing a bigger number by a smaller number. There are several
approaches we can take. One way to figure out the answer is to deal with
decimal points. Write the problem like this:
_____
18|12.0
We can do this because 12 = 12.0
(If you want to know why this is, write us back!)
Then write a decimal point above the 12.0 exactly above the decimal point in
the 12.0. Now go ahead with long-division as if you were dividing 120 by
18. Keep pulling down 0's and repeating the divison process until you can't
go any further or until a pattern emerges. If you get a pattern, you can
ask your teacher or write back, and we'll tell you what that pattern means
and how to write it down!
I hope it goes well! Enjoy, and write back if you have any questions.
--Sydney, "Dr. Math"
```
Associated Topics:
Elementary Division
Search the Dr. Math Library:
Find items containing (put spaces between keywords): Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words
Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search |
# Solve for the currents through each resistor? And the voltages across each resistor? Including the each power?
Feb 5, 2018
${I}_{1} = {I}_{2} = {I}_{3} = {I}_{4} = 0.714 A$
V_1 = 2.856 V; P_1 = 2.039184 W
V_2 = 4.284 V ; P_2 = 3.058776 W
V_3 = 5.712 V; P_3= 4.078368 W
${V}_{4} = 7.14 V A n d {P}_{4} = 5.9296 W$
#### Explanation:
As all the resistors ${R}_{1} , {R}_{2} , {R}_{3} \mathmr{and} {R}_{4}$ are in series, the same amount of current will flow through all. Let this current be $I$
$\therefore I = {I}_{1} = {I}_{2} = {I}_{3} = {I}_{4}$
To calculate $I$, we must first calculate the equivalent resistance between the two terminals.
For a series circuit, ${R}_{e} q = {R}_{1} + {R}_{2} + {R}_{3} + {R}_{4}$
${R}_{e q} = 4 + 6 + 8 + 10 = 28 \Omega$
As total voltage applied across the circuit is $20 V$,
The current $I$ can be calculated as:
$I = \frac{V}{R} = \frac{20}{28} = \frac{5}{7} = 0.7142857 A \approx 71.4 m A$
SO, current through every resistor will be same = $71.4 m A \mathmr{and} 0.714 A$
Now we can calculate the voltage drop and power across each resistor:
${V}_{1} = I \times {R}_{1} = 0.714 \times 4 = 2.856 V$ and
${P}_{1} = {V}_{1}^{2} / {R}_{1} = {2.856}^{2} / 4 = 2.039184 W$
${V}_{2} = I \times {R}_{2} = 0.714 \times 6 = 4.284 V$ and
${P}_{2} = {V}_{2}^{2} / {R}_{2} = {4.284}^{2} / 6 = 3.058776 W$
${V}_{3} = I \times {R}_{3} = 0.714 \times 8 = 5.712 V$ and
${P}_{3} = {V}_{3}^{2} / {R}_{3} = {5.712}^{2} / 8 = 4.078368 W$
${V}_{4} = I \times {R}_{4} = 0.714 \times 10 = 7.14 V$ and
${P}_{4} = {V}_{4}^{2} / R = {7.14}^{2} / 10 = 5.09796 W$
${I}_{1} = {I}_{2} = {I}_{3} = {I}_{4} = 0.714 A$
V_1 = 2.856 V; P_1 = 2.039184 W
V_2 = 4.284 V ; P_2 = 3.058776 W
V_3 = 5.712 V; P_3= 4.078368 W
${V}_{4} = 7.14 V A n d {P}_{4} = 5.9296 W$
Note : $A$ = amperes
$V$= Volts and $W$ = Watts. |
Everyday maths 1
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# 6 Line graphs
Now that you’ve had a look at pie charts and bar charts, let’s take a look at line graphs. These are drawn by marking (or plotting) points and then joining them with a straight line. You might have seen them used in holiday brochures or maybe on the television.
Hint: It is best to use graph or squared paper when drawing line graphs because it makes it easier to plot the points.
## Example: The estate agent
How would you present information as a line graph? Watch the following video to find out.
Skip transcript
#### Transcript
Here is a table showing the number of houses sold by an estate agent over a six-month period. How could you represent this information as a line graph?
To draw this in a graph, you need to first draw the axes. The months go on the bottom (horizontal) axis, and the house sales go up the left-hand side on the vertical axis.
Next, we need to decide on how we divide up these axes. There are six months, so there will be six marks on the horizontal axis. The highest number of house sales was 9, so the vertical axis will go from 0 to 10.
Now we can begin to mark our points. Starting with the first row of the table, you go up the line marked January, and then when you get to the line going across marked 2, you make a small cross. You then do this for the other points. Finally, you join the points using a ruler, and add your title.
Here’s a summary. To draw a line graph, you need to draw the horizontal and vertical axes and label them. Divide these axes into suitable scales to do this, you need to look at the data and find out what the smallest and largest numbers are. Plot the points from your data using a pencil to make small crosses. Join the points using a ruler, and give your graph a title.
End transcript
Interactive feature not available in single page view (see it in standard view).
Now try the following activity. Remember to check your answers once you have completed the questions.
## Activity 10: Creating a line graph
Line graphs are often used in holiday brochures to show temperatures or hours of sunshine at a particular resort.
The following table shows the hours of sunshine at a holiday resort. Draw a line graph using the data from the table and then answer the questions below.
Month Hours of sunshine
May 6
June 7
July 8
August 9
September 8
October 7
1. What month was the sunniest?
2. What month had the least sunshine?
When drawing your line graph you should:
• draw the horizontal axis, labelling it ‘Months’, and the vertical axis, labelling it ‘Hours of sunshine’
• divide these axes into suitable scales – your smallest and largest numbers are 6 and 9, so your scale could be one square for one hour
• plot the points from your data, using a pencil and make small crosses
• join the points using a ruler
• give your graph a title such as ‘Hours of sunshine at a holiday resort over a six-month period’.
The finished line graph should look something like this:
Figure 11 Sunshine line graph
1. August is the sunniest month.
2. May is the least sunniest month.
Self-check: always remember the following statements
Before moving on, you need to make sure you are able to collect, organise and show data in the forms of tables, diagrams, charts and graphs. Ask yourself the following questions:
• When I draw tables, diagrams, charts and graphs, is my data displayed clearly so that the information is easy to understand?
• Do I always include titles, scales, labels and keys when they are needed?
If you are not sure about these points, show your work to someone else and ask if they understand the data.
## Summary
In this section you have learned about how to present data in line graphs.
FSM_1 |
# Lesson video
In progress...
To find one quarter of a quantity.
Hello everybody, It's me Sidhu here, to do some great maths with you today.
In this lesson today, we will be finding one-quarter by dividing the whole into four equal parts.
Welcome to today's lesson, you will need a piece of paper and a pencil to write your answers on.
There are times when you will have to pause the video and have a go at some of the activities yourself.
Now, let's get started.
We are going to look at the key vocabulary, that means our star words, finding one-quarter of a quantity, using a part-whole model to find a quarter and independent task and answers, and we've even got a challenge today.
Now, let's do a star word, star word.
My turn first, then your turn.
Part, whole, equal, unequal, quarter, share, divide.
We are going to use all of these star words in our lesson today.
But first we've got a brain teaser to get us warmed up.
Here are some shapes, I want you to identify which shapes are split into quarters.
Does the shape represent quarters? Yes or no.
I want you to pause the video and have it go, does the shape represent quarters? Yes or no.
How did you get on? Here we've got the answers.
So, does the shape represent quarters? Yes or no.
So we're going to have a look at some of the shapes to see if they've been divided into quarters.
And I remember that when we have quarters, they've been divided into four equal shapes.
Here we've got the cross and I think yes, it's split into quarters.
The rectangle has been split into quarters and the rectangle here has been split into equal quarters as well, equal parts.
But here we've got a square that is not equal, and here we've got another triangle that is unequal and an arrow that is unequal, so that's no.
How did you do? Give yourselves a fantastic, well done.
Let's move on to other main part of our lesson.
Here we've got a big picture, there's a Nancy, the spider and turtle.
And Nancy and turtle are having a dinner party, nice and tasty and Nancy and turtle are sharing their dinner with the crab and fish.
Can you point to the crab? I can see the crab hiding under the table and the fish.
There's a Nancy and his turtle, and Nancy and turtle would like to share their eight cakes fairly between the full guests, let's check we've got eight cakes, one, two, three, four, five, six, seven, eight cakes.
They want to share the eight cakes fairly between the four guests.
So, they want to share it with Nancy, turtle, the crab and the fish.
How many cakes will each guest eat? What strategy can we use to share the cakes equally? Have a think.
So, what can we do to share the cakes equally? What strategy can we use? Can you say nice and loud? That's right.
We can use the one for me, one for your strategy.
So, what is our whole? one, two, three, four, five, six, seven, eight.
The whole is eight, and here we have our four plates to represent the four parts on the four animals.
So, we've got a plate for Nancy, a plate for turtle, a plate for the crab and a plate for a fish.
Now, we need to use our one for me, one for your strategy.
So, we need to share the cake, so we've got a one for a Nancy, one for turtle, you can join in with me too.
One foot crab, one for fish, one for Nancy, one for turtle, one for crab, one for fish.
How many cakes are in each part? I want you to show me on your fingers.
Show me a how many cakes are in each part? Super.
There are two cakes in each part, to here's one part and there's one, two cakes.
So, how many cakes will each guest eat? So, each guest receives two cakes.
We have split the whole into four equal parts, and so we have found one-quarter of the whole.
One quarter of eight is two.
Now, let's see what food we have next to share with the guests.
Here we have four apples, one, two, three, four.
The whole is four.
How many apples will each guest eat? Using the one for me, one for your strategy.
I want you to find out one-quarter by sharing out the apples into four equal parts.
So, sharing them between the four plates.
I want you to pause the video, you may even want to draw it on paper or use cantors or items of food.
Now, have it go.
How did you do? Shall we share it out? So, one for a Nancy, one for turtle, one for crab, one for the fish.
Did you get the same as me? How many apples are in each part? There is a one apple in each part, so, how many apples would each guest eat? So each guest receives one apple, We have split the whole into four equal parts.
So, we have found one-quarter of the whole one-quarter of four is one.
But this time we've got the part-whole model, you have been amazing at finding one-quarter by sharing the items on the four plates.
But now, we're going to use all of our knowledge to find the four equal parts using the part-whole model.
Remember that when we are finding one-quarter, we divide the whole into four equal parts.
Now, how many equal parts are there? Can you say it nice and loud? That's right, there are four equal parts, one for a Nancy, one for turtle, one for crab and one part for the fish.
Now let's see, if we can share apples with the guests, here we have one, two, three, four, five, six, seven, eight apples.
The whole is eight, how many apples were each guest eat? So, how many equal parts are there? There will be one, two, three, four equal parts.
Let's share the apples, one for a Nancy, one for turtle, one for the crab, One for the fish.
One for a Nancy, one for the turtle, one for the crab and one for the fish.
I have divided the whole into four equal parts, how many apples are in each part? Can you say nice and loud? Super, there are two apples in each equal part.
So, one-quarter of eight is two so, each guest receives two apples.
Wow, I think you are ready for the independent task today.
In today's task, you are going to be sharing a feast between four guests.
We've got some chocolate pieces, some cupcakes, and some sandwiches, using counters or food items or things that you've got around your, where you are at the moment and the part-whole model.
Can you find out how many food items each guest will get? I want you to pause and have a go, now we can move on to the challenge question.
What is one-quarter of twelve? Here, we've got Tim, Kelly and Laura.
Who do you agree with and why? I want you to read what they have said and choose who do you agree with and why? Make sure you pause the video and have a go at the challenge.
Now, we can have a look at the answers to see how you did.
Here are the answers, sharing the fees between four guests.
So, here we have our four pieces of chocolate and when we share the four pieces with our four guests, one-quarter of four is one.
Let's have a look at the cupcakes.
There are twelve cupcakes and one-quarter of twelve is three.
Here we've got the sandwiches, there are eight sandwiches.
A one-quarter of eight is two.
I hope you got the answers correct? If not, you can go through it again, here we've got the answers for the challenge.
Who do you agree with and why? The answer is Kelly? Kelly is correct, because the hole is twelve and to find one-quarter, you need to divide the whole into four equal parts.
So, one quarter of twelve is three.
I want you to share your work with Oak National. |
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• by Blaine Helwig
# Decimal Place Value and Value – Use the Point
Updated: Jan 2, 2023
A frustrating math moment I remember as a ten year old is decimal numbers and understanding both place value and value of the digits (numbers) to the right of the decimal point. I vividly recall sitting at my kitchen table trying to understand why the first place value location after the decimal point was not the ‘oneths’ place --- but the tenths. Whole number place value begins with the ones, tens, hundreds and so forth as we add digits to the left, then why did the decimal digits not symmetrically repeat to the right of the decimal point?
Today, regardless of the school’s socioeconomic setting students often demonstrate the same decimal difficulty with decimal place value and value - since it is a difficult arithmetic math topic. A physical understanding of digits (numbers) to the right of the decimal point is difficult with only one or two numbers, but the fact that each digit added after the decimal point is ten times smaller than the preceding digit often seems counter intuitive to kids. Frequently, the core lesson evolves into a pedagogical wheel of repetition until kids submit and memorize both the place value and the value of each digit to the right of the decimal – unsure of number sense of digits or mathematical significance of the decimal.
As I sat at my kitchen table, struggling with decimal place value, my father noticed that I was consistently answering the decimal homework problems incorrectly. In thirty seconds, he taught me a method to understand both decimal place value and value that I used successfully with my own students.
Understanding decimal place value and value methodology
Place a ‘1’ directly underneath the decimal point. Add a zero under every digit (including a zero digit) to the right of the decimal point. Hence, each digit may be read individually with regard to its place value: tenths, hundredths, thousandths, etc.
For example:
Given the decimal: 6.204 6.204
1000
After adding the ‘1’ and three (3) zeros to match each digit in the decimal to the right of the decimal point, the original decimal is transformed to an equivalent mixed number (with a decimal point, of course).
Students readily understand the following about the decimal moving from the decimal point to the right reading the decimal one digit at time:
• the ‘2’ digit is in the ‘tenths’ place and has a value of 2/10’s;
• the ‘0’ digit is in the ‘hundredths’ place and has a value of 0/100's;
• the ‘4’ is in the ‘thousandths’ place and has a value of 4/1,000's;
• the 'whole' proper fraction to the right of the decimal point is 204/1,000's.
• And, based on money (magnitude - explained below), it is about \$6.20.
Importantly, it also relates the decimal digit(s) to a proper fraction that lends a physical understanding of each digit’s magnitude/size in the decimal number. Students can also visually comprehend that as digits (i.e. numbers) are added to the right of the decimal, each digit is ten times smaller than the preceding digit. It is also recommended that classroom teacher not only stress each individual digit in the decimal, but all digits to the right of the decimal (i.e. 204) to illustrate the magnitude of the decimal. For example, the decimal 6.204; there are 204 out of 1,000 squares shaded, so the decimal 6.204 is closer to the whole number 6 than it is to 7.
Finally, most students possess the most decimal background knowledge in terms of money – with only two digits behind the decimal point. Another way to effectively illustrate the ‘size’ of a decimal is by circling only two places to the right of the decimal point (i.e. to the hundredths place).
For example, the decimal 2.08407 has a magnitude or size of about 2.08. The other 3 digits (407) do not significantly affect the overall size or magnitude of the decimal. The student realizes that the decimal 2.08407 is between the whole numbers 2 and 3, and the decimal is closer to 2 than 3. It is about by 0.08 or 8/100’s away from the whole number 2.
Repeat daily until mastered, but include the decimal and fractional distances to 2.5 and 3.0, in this example. This methodology removes the mystery of decimals for elementary and middle school students. Decimals become manageable mathematics for students very rapidly.
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Straight-Line Equations:
Parallel and Perpendicular Lines
(page 3 of 3)
Sections: Slope-intercept form, Point-slope form, Parallel and perpendicular lines
There is one other consideration for straight-line equations: finding parallel and perpendicular lines. Here is the usual format for the question:
• Given the line 2x – 3y = 9 and the point (4, –1), find lines through the point that are:
• (a) parallel to the given line and
(b) perpendicular to it.
In other words, they've given me a reference line — 2x – 3y = 9 — that I'll be comparing to, and some point somewhere else on the plane — namely, (4, –1). Then they want me to find the line through (4, –1) that is parallel to (that has the same slope as) 2x – 3y = 9. On top of that, they then want me to find the line through (4, –1) that is perpendicular to (that has a slope that is the negative reciprocal of the slope of) 2x – 3y = 9.
Clearly, the first thing I need to do is solve "2x – 3y = 9" for "y=", so that I can find my reference slope: Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved
2x – 3y = 9
–3y = –2x + 9
y = ( 2/3)x – 3
So the reference slope from the reference line is m = 2/3.
Since a parallel line has an identical slope, then the parallel line through (4, –1) will have slope m = 2/3. Hey, now I have a point and a slope! So I'll use the point-slope form to find the line:
y – (–1) = ( 2/3 )(x – 4)
y + 1 = ( 2/3 ) x8/3
y = ( 2/3 ) x8/33/3
y = ( 2/3 ) x11/3
This is the parallel line that they asked for.
For the perpendicular line, I have to find the perpendicular slope. The reference slope is m = 2/3, and, for the perpendicular slope, I'll flip this slope and change the sign. Then the perpendicular slope is m = – 3/2. So now I can do the point-slope form. Note that the only change from the calculations I just did is that the slope is different now.
y – (–1) = ( – 3/2 )(x – 4)
y + 1 = ( – 3/2 ) x + 6
y = ( – 3/2 ) x + 5
Then the full solution to this exercise is:
parallel: y = ( 2/3 ) x11/3
perpendicular: y = ( – 3/2 ) x + 5
Warning: If a question asks you whether two given lines are "parallel, perpendicular, or neither", you must answer that question by finding their slopes, not by drawing a picture! Pictures can only give you a rough idea of what is going on, but you cannot tell "by looking" that lines with slopes of, say, m1 = 1.00 and m2 = 0.99 are NOT parallel, because they'll sure look parallel on their graphs. But since 1.00 does not equal 0.99, the lines are not parallel. Find the slopes; don't just draw the pictures.
You can use the Mathway widget below to practice finding a parallel line through a given point. Try the entered exercise, or type in your own exercise. Then click the "paper-airplane" button to compare your answer to Mathway's. (The next widget is for finding perpendicular lines.)
(Clicking on "Tap to view steps" on the widget's answer screen will take you to the Mathway site for a paid upgrade.)
You can use the Mathway widget below to practice finding a perpendicular line through a given point. Try the entered exercise, or type in your own exercise. Then click the "paper-airplane" button to compare your answer to Mathway's.
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Cite this article as: Stapel, Elizabeth. "Straight-Line Equations: Parallel and Perpendicular Lines." Purplemath. Available from http://www.purplemath.com/modules/strtlneq3.htm. Accessed [Date] [Month] 2016
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