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# Video: KS2-M18 • Paper 2 • Question 17
There are 28 pupils in a class. The teacher has 8 liters of orange juice. She pours 225 milliliters of orange juice for every pupil. How much orange juice is left over?
04:57
### Video Transcript
There are 28 pupils in a class. The teacher has eight liters of orange juice. She pours 225 milliliters of orange juice for every pupil. How much orange juice is left over?
The question asks us to calculate how much orange juice is left over. To do this, first, we need to work out how much orange juice the pupils have drunk. We know there are 28 pupils. And the teacher pours 225 milliliters of juice for each pupil. We need to multiply 225 by 28 to find out how much juice the pupils have drunk.
We’re using long multiplication to find this answer. And we’re going to start by multiplying 225 by eight. Five times eight is 40. Two times eight is 16, plus the four we exchanged gives us 20. Two times eight is 16, plus the two we exchanged gives us 18. 225 multiplied by eight is 1800.
Next, we need to multiply 225 by 20. Multiplying by 20 is the same as multiplying by 10 and then by two, which is what we’re going to do. By writing a zero in the ones place, we’re moving the digits one place to the left into the tens column. Now, let’s multiply by two. Five times two is 10. Two times two is four, plus the one we exchanged gives us five. And our final digit, two times two is four. 225 multiplied by 20 is 4500.
Now we just need to add our two totals together to tell us what 225 times 28 is. First, we need to add in the ones column. Zero and zero gives us a total of zero. And we have the same in the tens column giving us a total of zero. Eight plus five is 13. Finally, we need to add in the thousands column. One and four is five, plus the one we exchanged gives us six. 225 multiplied by 28 is 6300. This means that the pupils drank 6300 milliliters of orange juice.
Now, we can work out how much juice is left over. To do this, we need to find the difference between eight liters and 6300 milliliters. Or, we need to subtract 6300 milliliters from eight liters. But first, we need to convert the liters into milliliters. One liter is equal to 1000 milliliters. So eight liters is equal to 8000 milliliters.
To find the difference, we can count on from 6300 to 8000. If we add 700 to 6300, that will take us to 7000. Then, we can add 1000 to take us to 8000. Altogether, we added 1700. So the amount of orange juice leftover is 1700 milliliters, which we could also write as 1.7 liters.
To find the answer, the first step was to calculate how much juice the pupils had drunk by multiplying 225 by 28. Then, we could subtract this amount from the eight liters of juice the teacher had to begin with. But first, we converted liters to milliliters. We subtracted 6300 from 8000 to give us the answer 1700 milliliters of orange juice leftover.
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# You asked: Which property of real numbers is shown below 6?
Contents
## What are the 7 properties of real numbers?
Suppose a, b, and c represent real numbers.
• 1) Closure Property of Addition.
• 2) Commutative Property of Addition.
• 3) Associative Property of Addition.
• 6) Closure Property of Multiplication.
• 7) Commutative Property of Multiplication.
## What is the property of real numbers?
Real numbers are closed under addition, subtraction, and multiplication. That means if a and b are real numbers, then a + b is a unique real number, and a ⋅ b is a unique real number.
## Is 0 a real number?
Real numbers are, in fact, pretty much any number that you can think of. … Real numbers can be positive or negative, and include the number zero. They are called real numbers because they are not imaginary, which is a different system of numbers.
## What does a Commutative Property look like?
The word “commutative” comes from “commute” or “move around”, so the Commutative Property is the one that refers to moving stuff around. For addition, the rule is “a + b = b + a”; in numbers, this means 2 + 3 = 3 + 2. For multiplication, the rule is “ab = ba”; in numbers, this means 2×3 = 3×2.
## What is associative property addition?
The associative property of addition says that changing the grouping of the addends does not change the sum.
THIS IS IMPORTANT: You asked: Can a UK citizen buy a house in Ireland?
## Is any number an integer?
All whole numbers are integers (and all natural numbers are integers), but not all integers are whole numbers or natural numbers. For example, -5 is an integer but not a whole number or a natural number.
## What is the 4 properties of math?
There are four basic properties of numbers: commutative, associative, distributive, and identity. You should be familiar with each of these. It is especially important to understand these properties once you reach advanced math such as algebra and calculus.
## What property is A +(- A )= 0?
The identity property of addition states that the sum of a number and zero is the number. If a is a real number, then a+0=a. The inverse property of addition states that the sum of any real number and its additive inverse (opposite) is zero. If a is a real number, then a+(-a)=0.
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# Area of Irregular Shapes – Definition, Facts, Examples, FAQs
Home » Math Vocabulary » Area of Irregular Shapes – Definition, Facts, Examples, FAQs
## Area of Irregular Shapes
Irregular shapes are polygons with five or more sides of varying lengths. These shapes or figures can be decomposed further into triangles, squares, and quadrilaterals to evaluate the area.
Some examples of irregular shapes are as follows:
Calculating the area of irregular shapes:
The approaches to estimating the area of irregular shape are:
Evaluating area using unit squares
Apply this technique for the shapes with curves apart from perfect circle or semicircles and irregular quadrilaterals. In this method, divide the shape into unit squares. The total number of unit squares falling within the shape determines the total area.
Count the square as “1” if the shaded region covers more than half while calculating the area for a more accurate estimate.
In the following figure, calculate the area by counting the unit squares, which is 6. If we denote each unit square in centimeter, the area will be 6 cm2.
• Dividing the irregular shape in two or more regular shapes
Use this method for irregular shapes, which are a combination of triangles and polygons. Use predefined formulas to calculate the area of such shapes and add them together to obtain the total area.
For example, an irregular shape we divide multiple edges into a triangle and three polygons.
The total area of the figure is given as:
⇒ Area = Area (ABIM) + Area (BCGH) + Area (CDEF) + Area (JKL)
⇒ Area = (AB × BI) + (BC × CG) + (CD × DE) + (12× LJ × KO)
⇒ Area = ( 10 × 5) + (3 × 3) + (2 × 2) + (12× 4 × 4)
⇒ Area = 50 + 9 + 4 + 8
⇒ Area = 71 cm2
• Dividing the irregular shape with curves in two or more regular shapes
In this method, decompose an irregular shape into multiple squares, triangles, or other quadrilaterals. Depending on the shape and curves, a part of the figure can be a circle, semicircle or quadrant as well.
The following figure is an irregular shape with 8 sides, including one curve. Determine the unknown quantities by the given dimensions for the sides. Decompose the figure into two rectangles and a semicircle
The area of the shape ABCDEF is:
Area (ABCDEF) = Area (ABCG) + Area (GDEF) + Area (aob)
Area = (AB × AG) + (GD × DE) + (12 × π × ob2)
Area = (3 × 4) + (10 × 4) + (12 × 3.14 × 12)
Area = 12 + 40 + 1.57
Area = 53.57 cm2
Application
The estimation of area for irregular figures is an essential method for drawing maps, building architecture, and marking agricultural fields. We apply the concept in the cutting of fabrics as per the given design. In higher grades, the technique lays a basis for advanced topics such as calculating volume, drawing conic sections and figures with elliptical shapes.
## Practice Problems On Area of Irregular Shapes
1
### A leaf was traced on a graph paper. It has 10 squares fully covered, 12 squares are covered more than half and 14 squares are covered less than half. What will be the area of the leaf?
29 square units
16 square units
22 square units
23 square units
CorrectIncorrect
Correct answer is: 22 square units
The fully covered squares are counted as it is. More than half-covered squares are counted as 1 square each. Less than half-covered squares are counted as 0 each.
So we have $10 + (1 × 12) + (0 × 14) = 10 + 12 = 22$ square units.
2
### What is the area of a field that is shaped like 2 rectangles with the following measurements: Rectangle 1: l = 5, w = 6 Rectangle 2: l = 8, w = 5
48 square cm
24 square cm
70 square cm
10 square cm
CorrectIncorrect
Correct answer is: 70 square cm
Area of Rectangle 1 = 5 × 6 = 30 sq.cm.
Area of Rectangle 2 = 8 × 5 = 40 sq.cm
Area of Field = Area of Rectangle 1 + Area of Rectangle 2
= 30 + 40 = 70 square cm.
3
### To find the area of an irregular shape, we first break the irregular shape into common shapes. Then we find the area of each shape and ___ them.
Multiply
Add
Subtract
Divide
CorrectIncorrect
Correct answer is: Add
To find the area of an irregular shape, we first break the shape into common shapes. Then we find the area of each shape and add them. For example, if an irregular polygon is made up of a square and a triangle, then: Area of irregular polygon = Area of Square + Area of Triangle.
4
### What is the area of an irregular polygon made of 2 squares with the following measurements? Square 1: side = 5 cm Square 2: side = 3 cm
25 square cm
34 square cm
9 square cm
16 square cm
CorrectIncorrect
Correct answer is: 34 square cm
Area of Square 1 = 5 × 5 = 25 sq. cm.
Area of Square 2 = 3 × 3 = 9 sq. cm.
Area of Irregular polygon = Area of Square 1 + Area of Square 2 = 25 + 9 = 34 square cm.
## Frequently Asked Questions On Area of Irregular Shapes
An irregular shape can be decomposed into known polygons. The area of the irregular shape then is the sum of the area of each of these polygons. If the irregular shape has curved edges and decomposing the entire shape into known polygons is not possible, then estimating the area would be a better approach.
No, there is no general formula to calculate the area of an irregular shape because the sides can be of varying lengths and curves.
We need to find the area of irregular shapes for drawing maps, building architecture, and marking agricultural fields. We also need it for cutting fabrics according to a given design.
If an irregular shape can be drawn or traced on a grid paper, then counting the number of squares covering the entire shape will be the easiest way to find its area.
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The New Illinois Learning Standards for Eighth Grade Statistics and Probability
# The New Illinois Learning Standards for Eighth Grade Statistics and Probability
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## The New Illinois Learning Standards for Eighth Grade Statistics and Probability
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##### Presentation Transcript
1. The New Illinois Learning Standards for Eighth GradeStatistics and Probability Julia Brenson
2. The Four Components of a Statistical Investigation* 1) Formulate a question 2) Design and implement a plan to collect data 3) Analyze the data by measures and graphs 4) Interpret the results in the context of the original question *Guidelines for Assessment and Instruction in Statistics Education (GAISE) Report American Statistical Association http://www.amstat.org/education/gaise/GAISEPreK-12_Full.pdf
3. The New Illinois Learning Standards Eighth Grade
4. Statistics Standards for 8th Grade * See Evidence Statement 8.D.2 (content from Grade 7 including 7.SP.B) and 8.D.3 (Micro-models) from the PARCC Evidence Table – Grade 8 PBA
5. Statistics Standards for 8thGrade Investigate Patterns of Association in Bivariate Data What is bivariate data? bi - means two variate – means variable Bivariate data is data about two variables. If the two variables are numeric, we examine the relationship between the two variables using a scatterplot. If the two variables are categorical, we organize the data in a two-way frequency table and look for an association.
6. Statistics Standards for 8th GradeInvestigate Patterns of Association in Bivariate Data Which is the best interpretation of the scatterplot? A. As heights go up, weight increases. B. As heights go up, weight tends to increase. C. If you get taller, you will get heavier. D. Taller football players tend to be heavier football players. Below is a graph of the heights and weights of currently rostered Chicago Bears (April 2014). http://chicagosports.sportsdirectinc.com/football/nfl-teams.aspx?page=/data/nfl/teams/rosters/roster16.html
7. Statistics Standards for 8thGrade Investigate Patterns of Association in Bivariate Data Example: Shoe Rating and Price From the Engage NY website – Grade 8, module 6 (www.engageny.org) Is there a relationship between price and the quality of athletic shoes? The data in the table below are from the Consumer Reports website. 𝒙= price (in dollars) and 𝒚= Consumer Reports quality rating The quality rating is on a scale of 𝟎 to 𝟏𝟎𝟎, with 𝟏𝟎𝟎 being the highest quality.
8. Statistics Standards for 8thGrade Investigate Patterns of Association in Bivariate Data • Create a scatterplot, choosing appropriate scaling for each axes. • Do you see any pattern in the scatter plot indicating that there is a relationship between price and quality rating for athletic shoes? • Some people think that if shoes have a high price, they must be of high quality. How would you respond?
9. Statistics Standards for 8thGrade Investigate Patterns of Association in Bivariate Data Example: Deep Water From the Engage NY website – Grade 8, module 6 (http://www.engageny.org/sites/default/files/resource/attachments/math-g8-m6-teacher-materials.pdf) Does the current in the water go faster or slower when the water is shallow? The data on the depth and speed of the Columbia River at various locations in Washington State listed on the next slide can help you think about the answer.
10. Statistics Standards for 8thGrade Investigate Patterns of Association in Bivariate Data Example: Deep Water
11. Statistics Standards for 8thGrade Investigate Patterns of Association in Bivariate Data Example: Deep Water The Deep Water Task provides an opportunity for students to work independently through the steps of creating a scatterplot, fitting a line to the data, and interpreting the slope, y-intercept, an ordered pair, and the relationship between depth and velocity in the context of the problem.
12. Statistics Standards for 8thGrade Investigate Patterns of Association in Bivariate Data Example: Florida Alligators The American Alligator can live up to 50 years in the wild. Males have an average length of 11 feet, and females have an average length of 8 feet. The data shown in the scatterplot on the next slide was collected from 25 American alligator captured in Florida. Progressions for the Common Core State Standards in Mathematics (draft) 6-8 Statistics and Probability. Retrieved from http://commoncoretools.files.wordpress.com/2011/12/ccss_progression_sp_68_2011_12_26_bis.pdf
13. Statistics Standards for 8thGrade Investigate Patterns of Association in Bivariate Data Activity: Florida Alligators Describe the relationship between the length in inches and the weight in pounds for this sample of Florida Alligator. Be sure to comment on any interesting features of the scatterplot.
14. Statistics Standards for 8thGrade Investigate Patterns of Association in Bivariate Data The PARCC Evidence Tables for the Eighth Grade PBA indicates that eighth graders may be asked to do a task that requires applying knowledge or skills from 7.SP.B. Bear Hugs, an activity that is shared by Subhash Bagui and Mary Richardson on the Statistics Education Web (STEW), is an activity about arm span and age. Students are first asked to create box plots comparing the arm span of male and female students (7.SP.B). They then proceed to create and interpret a scatterplot of age versus arm span. http://www.amstat.org/education/stew/pdfs/BearHugs.pdf
15. Statistics Standards for 8thGrade Investigate Patterns of Association in Bivariate Data A look at sample activities for 8.SP.1-3 • Oil Changes and Engine Repair Adapted from an NCTM Illuminations’ activity • Bike Weights and Jump Heights NCTM Illuminations • Animal Brains Illustrative Mathematics • US Airports, Assessment Variation Illustrative Mathematics • Patterns in Scatter Plots – Lesson 7 Classwork EngageNY • Determining the Equation of a Line Fit to Data – Lesson 9 Problem Set EngageNY
16. Statistics Standards for 8thGrade Investigate Patterns of Association in Bivariate Data Questions In what order would you use these activities in a unit? What questions did you particularly like? What questions are different than what we might typically ask students? Do you have suggestions for improvement of or additions to these activities?
17. Statistics Standards for 8th Grade Investigate Patterns of Association in Bivariate Data 8.SP.4 Summarize categorical data in two categories Big Ideas: • Review the difference between numeric data and categorical data. • Explain that frequency refers to the count of the data. Relative frequency is a proportion. • Analyze relative frequencies and assess possible associations and trends in the data.
18. Statistics Standards for 8th Grade Investigate Patterns of Association in Bivariate Data Association of two categorical variables • There is an association between two categorical variables if the row (or column) conditional relative frequencies are different from row to row (or column to column) in the table. • The greater the difference between the conditional relative frequencies, the stronger the association.
19. Statistics Standards for 8th Grade Investigate Patterns of Association in Bivariate Data Activity: Music and Sports Adapted from Illustrative Mathematics (http://www.illustrativemathematics.org/illustrations/1098) Is there an association between whether a student plays a sport and whether he or she plays a musical instrument? To investigate this question, each student in your class should answer the following two questions: 1. Do you play a sport? (yes or no) 2. Do you play a musical instrument? (yes or no)
20. Statistics Standards for 8th Grade Investigate Patterns of Association in Bivariate Data Music and Sports Summarize the class data in a two way frequency table. Questions: 1. Of those students who play a sport, what proportion play a musical instrument? 2. Of those students who do not play a sport, what proportion play a musical instrument? 3. Based on the class data, do you think there is an association between playing a sport and playing an instrument?
21. Statistics Standards for 8th Grade Activities • 8-SP Oil Changes and Engine Repair • Bike Weights and Jump Heights NCTM Illuminations • Florida Alligators • Shoe Rating and Price (www.engageny.org ) • Deep Water (www.engageny.org) • 8.SP Animal Brains (http://www.illustrativemathematics.org/illustrations/1520) • 8.SP US Airports, Assessment Variation (http://www.illustrativemathematics.org/illustrations/1370)
22. Statistics Standards for 8th Grade More Activities • Scatter It! (Predict Billy’s Height) Statistics Education Web (STEW) (http://www.amstat.org/education/stew/pdfs/BearHugs.pdf) • Bear Hugs Statistics Education Web (STEW) (http://www.amstat.org/education/stew/pdfs/BearHugs.pdf) • 8-SP-4 Music and Sports Activity adapted from: (http://www.illustrativemathematics.org/illustrations/1370) • Census at School ( http://www.amstat.org/censusatschool/)
23. Statistics Standards for 8th Grade Census at School (http://www.amstat.org/censusatschool/) Statistics Education Web (http://www.amstat.org/education/stew/)
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## THEOREMS AND PROBLEMS ON PARALLELOGRAMS – EXERCISE 4.3.3- Class IX
1. The area of parallelogram is 153.6 cm2. The base measures 19.2 cm. what is the measurement for the height of the parallelogram?
Solution:
Area of the parallelogram = bh = 153 cm2
Base = b = 19.2cm
b x h = 19.2 × h = 153.6
h = 153.6/19.2 = 1536/192 = 8cm
2. In parallelogram ABCD, AD = 25cm and AB = 50 CM. if the altitude from a vertex D on to AB measure 22 cm, what is the altitude from a vertex B on to AD?
Solution:
Area of parallelogram = bh
Area ABCD = 50 × 22cm2
If the altitude from B to AD is h,
And ABCD = 25 × h cm2
25 x h = 50 × 22
\h = 15 ×2225 = 44cmArea of parallelogram = bh
Area ABCD = 50 × 22cm2
If the altitude from B to AD is h,
And ABCD = 25 × h cm2
25 x h = 50 × 22
Therefore, h = (15 ×22)/25 = 44cm
1. In a parallelogram ABCD, AB = 4x and AD = 2x + 1. If the perimeter is 38cm and area 60cm2 find the length of the altitude from D on to AB.
Solution:
Perimeter = 2 (l + b) = 38
l + b = 38/2 = 19
4x + 2x + 1 = 19
6x = 19 – 1 = 18
x = 18/6 = 3
AB = 4x = 4 × 3 = 12cm
Let the altitude DE = h
Area = bh = 12 × h = 60
⸫h = 6012 = 5cm
1. Let ABCD be a parallelogram and consider its diagonal AC. Draw perpendiculars BK and DL to AC.
Prove that BK = DL.
Solution:
Given: ABCD is a parallelogram. BK and DL are perpendiculars drawn
To diagonals AC.
Proof: Statement Reason In quadrilateral. ABCD DL ^ AC, BK ^ AC given ΔADC @ ΔABC Diagonal divides a parallelogram into two congruent triangles ⸫ Area Δ ABC = Area Δ ADC i. e. 1/2 AC × BK = 1/2 AC × DL BK = DL
1. Let ABCD be a parallelogram. Prove that 2 area (ABCD) ≤AC × BD.
Solution:
Given: ABCD is a parallelogram
To prove: 2 areas (ABCD) ≤AC × BD.
Construction: Draw DF and BG perpendicular to AC
Proof:
Area ABCD = area ΔABC + area ΔADC
1/2 AC × DF + 1/2 AC × BG
1/2 AC (DF + BG)
2 area (ABCD) = 2 × 1/2 AC (DF + BG)
= AC (DF + BG)
In right ΔDFE, DF < hypotenuse DE
Similarly in ΔBGE, BG < BE
Thus 2 area ABCD = AC (DF + BG)
≤ AC (DE + BE)
i.e. 2 areas ABCD ≤ AC × BD
2 areas ABCD will be equal to AC × BD if AC ^ BD AC will be perpendicular to BD if ABCD is a rhombus or a square.
⸫ 2 area ABCD = AC × BD in a rhombus or in a square.
6. Prove that the area of triangles standing on the same base or equal bases and between same parallels are equal in area.
Solution:
Given: ABCD and ΔABF stand on the same base and are between the same
Parallel and m
To prove: area ΔABD = area ΔABF
Construction: drew parallelograms ABCD and ABFE
Proof: Statement Reason 1.. Area of parallelogram ABCD and ABFE are equal They stand on the same base and are between the same parallels. 2. Area ΔABD = 1/2 area ΔABCD Area of triangles is equal to half the area of a parallelogram stand base and between the same parallels 3. Similarly area ΔABF = 12 area ABFE 4. Area ΔABD = area ΔABF from (1), (2) and (3)
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# Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 338: 66
The area of the region is $9/2$.
#### Work Step by Step
$y=x^2-2x$ and $y=x$ 1) Draft the graph The graph is enclosed below. From the graph, we choose integration with respect to $x$ to employ here for calculation of the region area. 2) Find the limits of integration: We can find the limits of integration by finding points of intersection between the curve and the line. $$x^2-2x=x$$ $$x^2-3x=0$$ $$x(x-3)=0$$ $$x=0\hspace{1cm}\text{or}\hspace{1cm}x=3$$ So, the upper limit is $3$ and the lower one is $0$. 3) Find the area: Looking at the draft, we see that the region from $x=0$ to $x=3$ is bounded above by $y=x$ and below by $y=x^2-2x$. So, according to definition, the area of the region is $$A=\int^3_{0}[x-(x^2-2x)]dx=\int^3_{0}(-x^2+3x)dx$$ $$A=-\frac{x^3}{3}+\frac{3x^2}{2}\Big]^3_{0}$$ $$A=-\frac{27}{3}+\frac{27}{2}-0$$ $$A=\frac{9}{2}$$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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# Is f(x)=(-5x^3-x^2-3x-11)/(x-3) increasing or decreasing at x=2?
Mar 6, 2017
$\text{increasing at } x = 2$
#### Explanation:
To determine if f(x) is increasing /decreasing at x = a, evaluate f'(a)
• " If f'(a) > 0, then f(x) is increasing at x = a"
• " If f'(a) < 0, then f(x) is decreasing at x = a"
differentiate f(x) using the $\textcolor{b l u e}{\text{quotient rule}}$
$\text{Given "f(x)=(g(x))/(h(x))" then}$
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
$\text{here } g \left(x\right) = - 5 {x}^{3} - {x}^{2} - 3 x - 11$
$\Rightarrow g ' \left(x\right) = - 15 {x}^{2} - 2 x - 3$
$\text{and } h \left(x\right) = x - 3 \Rightarrow h ' \left(x\right) = 1$
$\Rightarrow f ' \left(x\right) = \frac{\left(x - 3\right) \left(- 15 {x}^{2} - 2 x - 3\right) - \left(- 5 {x}^{3} - {x}^{2} - 3 x - 11\right) \left(1\right)}{x - 3} ^ 2$
$\Rightarrow f ' \left(2\right) = \frac{\left(- 1\right) \left(- 67\right) - \left(- 61\right)}{1} = 128$
Since f'(2) > 0, then f(x) is increasing at x = 2
graph{(-5x^3-x^2-3x-11)/(x-3) [-10, 10, -5, 5]}
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Select Page
# Chapter 11 – Work and Energy
## Work and Energy
Q 1.
A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force as shown in figure. Let us take it that the force acts on the object through the displacement. What is the work done in this case?
SOLUTION:
Force, F = 7 N
Displacement, s = 8 m
Then, Work done = 7 N × 8 m
= 56 N m = 56 JQ 2.
When do we say that work is done?SOLUTION:
Work is done by a force on an object if a force acts on the object and the object is displaced from its original position.Q 3.
Write an expression for the work done when a force is acting on an object in the direction of its displacement.SOLUTION:
If F is the constant force acting in the direction of displacement s, then work done by the force, i.e., W = F × s = Fs.Q 4.
Define 1 J of work.SOLUTION:
The amount of work done when a force of 1 N moves a body through a distance of 1 m in the direction of the force is called 1 joule.Q 5.
A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?SOLUTION:
Here, F = 140 N, s = 15 m.
Work done in ploughing the field,
W = F × s = (140 N) × (15 cm) = 2100 JQ 6.
What is the kinetic energy of an object?SOLUTION:
Kinetic energy of an object is the energy possessed by it due to its motion. In fact, kinetic energy of an object moving with a certain velocity is equal to the work done to make it acquire that velocity.Q 7.
Write an expression for the kinetic energy of an object.SOLUTION:
Kinetic energy of an object,
where m = mass of the object and
v = uniform velocity of the objectQ 8.
The kinetic energy of an object of mass, m moving with a velocity of 5 m s–1 is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity increased three times?SOLUTION:
K.E. of an object = 25 J
K.E. of the object with velocity double
Q 9.
What is power?SOLUTION:
The rate of doing work is called power.
Thus,
Power may also be defined as the amount of work done in one unit of time.Q 10.
Define 1 watt of power.SOLUTION:
Power of an object is said to be 1 W if it does 1 joule of work in 1 s, i.e.,
In other words, we say that power is 1 W when the rate of consumption of energy is 1 J s–1.Q 11.
A lamp consumes 1000 J of electrical energy in 10 s. What is its power?SOLUTION:
Energy consumed = 1000 J
So, Time = 10 s
Q 12.
Define average power.SOLUTION:
The concept of average power is useful when the power of an agent varies with time and it does work at different rates during different intervals of time.Q 13.
Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.
• Suma is swimming in a pond.
• A donkey is carrying a load on its back.
• A wind-mill is lifting water from a well.
• A green plant is carrying out photosynthesis.
• An engine is pulling a train.
• Food grains are getting dried in the sun.
• A sailboat is moving due to wind energy.SOLUTION:
(A) Suma is doing work. She is applying force to move horizontally.
(B) Donkey is not doing any work. Here, the displacement and the force are at 90°.
(C) Work is done by the windmill. The water is lifted against force of gravity.
(D) No work is done by a green plant during photosynthesis.
(e) The engine applies a pulling force on the train, and the train moves in the direction of this force. Therefore, engine is doing work.
(f) During drying of food grains in the sun no work is done.
(g) Work is done by the air. The sailboat moves in the direction of the force exerted by wind.Q 14.
An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?SOLUTION:
Since initial and the final positions of the path of the object thrown at a certain angle to the ground lie on the some horizontal plane, the displacement of the object is in the horizontal direction. The force of gravity on the object acts vertically downwards, so no work is said to be done.
Q 15.
A battery lights a bulb. Describe the energy changes involved in the process.SOLUTION:
A battery converts chemical energy into electrical energy. This electrical energy is converted into light energy as the bulb is lighted up, i.e., the sequence of energy changes is as follows :
Chemical energy ? Electrical energy ? Light energy.Q 16.
Certain force acting on a 20 kg mass changes its velocity from 5 m s–1 to 2 m s–1. Calculate the work done by the force.SOLUTION:
Mass of the object, m = 20 kg
Total work done by the force = Change in the kinetic energy of the object
The force will do work equivalent to 210 J. Here the direction of force is opposite to the direction of motion.Q 17.
A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.
SOLUTION:
Here, the mass of 10 kg is moved horizontally from point A to B. The force of gravity acts vertically downwards. Thus, the mass moves in a direction at right angle to the force of gravity. So,
Work done on the object by the force of gravity = FScos? = FScos90° = 0
Therefore, no work is done by the force of gravity on the object.Q 18.
The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?SOLUTION:
As the potential energy of the freely falling object decreases, its kinetic energy increases on account of an increase in its velocity. The sum total of the potential energy and the kinetic energy of the object during its free fall remains the same, i.e., the total mechanical energy (potential energy + kinetic energy) remains constant. Thus, the law of conservation of energy is not violated.Q 19.
What are the various energy transformations that occur when you are riding a bicycle?SOLUTION:
The muscular energy of the cyclist is converted into kinetic energy (rotational) of the pedals of the bicycle which is transferred to its wheels. The kinetic energy of the rotation of the wheels is converted into the kinetic energy of the bicycle and the cyclist.Q 20.
Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?SOLUTION:
No transfer of energy takes place when we push a huge rock unsuccessfully.
The energy is spent for the physical activity of muscles.Q 21.
A certain household has consumed 250 units of energy during a month. How much energy is this in joules?SOLUTION:
Energy consumed = 250 units = 250 kW h
= 250 × 1000 W h = 250 × (1000 W) (3600 s)
= 250 × 1000 × 3600 W s
= 900 × 106 = 9 × 108 J(? 1 W = 1 J s–1)Q 22.
An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.SOLUTION:
Potential energy of the object = m × g × h
= 40 kg × 10 m s–2 × 5 m
= 40 × 10 × 5 kg m s–2 m = 2000 J
Let v be the velocity of the object at half the height i.e., after travelling 2.5 m.
Then v2 – u2 = 2as
v2 – 02 = 2 × 10 m s–2 × 2.5 m = 50 m2 s–2
So, v2 = 50 m2 s–2
Then,
Q 23.
What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.SOLUTION:
When a satellite moves around the earth, its displacement in a short interval is along the tangent to the circular path of the satellite. The gravitational force (F) acting on the satellite due to the earth is along the radius as shown in figure.
Since a tangent is always perpendicular to the radius, the displacement and the force are perpendicular to each other. There is no displacement of the satellite in the direction of the force, i.e., s = 0. Thus, work done by the force of gravity on the satellite is zero as W = F × s = 0.Q 24.
Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.SOLUTION:
Yes. In the absence of any force on the object, i.eF = 0, ma = 0 (as F = ma). Since
m ? 0, a = 0. In such a case, the object is either at rest or in a state of uniform motion in a straight line. In the latter case, there is a displacement of the object without any force acting on it.Q 25.
A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.SOLUTION:
The person holding a bundle of hay on his head gets tired as he experiences muscular fatigue. This fatigue is due to the conversion of chemical energy into thermal energy by the muscular effort. The person has done no work as his effort causes no displacement of the bundle over his head, i.e., s = 0.Q 26.
An electric heater is rated 1500 W. How much energy does it use in 10 hours?SOLUTION:
Here, power of the electric heater, P = 1500 W
Time for which it is used, t = 10 h
Energy used by the electric heater, i.e.,
W = P × t = 1500 W × 10 h = 15000 W h
= 15 kW h = 15 units
Q 27.
Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?SOLUTION:
A small metallic ball (called bob) suspended by a light string (thread) from a frictionless, rigid support is called a simple pendulum. A simple pendulum is shown in figure alongside.
The bob of a simple pendulum swings from one extreme position to the other through its mean position.
The bob is at its highest position at the extreme positions and at lowest at its mean position.
The energy changes which occur during the motion of the bob are shown in figure. So, when a pendulum swings to and fro, its energy changes constantly in the following sequence.
Q 28.
An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?SOLUTION:
Initial kinetic energy of the object,
Q 29.
Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km h–1?SOLUTION:
Here, mass of the car, m = 1500 kg
Initial velocity of the car, u = 60 km h–1
The sign of minus indicates that the work is being done against the moving car.Q 30.
In each of the following a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.
(A)
(B)
(C) SOLUTION:
(A)Here, the force acts at right angle to the displacement. So,
work done by the force = 0
(B) Here, the direction of displacement is the same as that of the force. So,
work done by the force = + ve
(C) Here, the body moves in a direction opposite to the direction of the force. So,
work done by the force = – veQ 31.
A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force as shown in figure. Let us take it that the force acts on the object through the displacement. What is the work done in this case?
SOLUTION:
The acceleration of an object can be zero even if several forces are acting on it provided the resultant force (F) is zero. As F = 0, ma = 0 and accordingly a = 0 (as m ? 0).Q 32.
Find the energy in kW h consumed in 10 hours by four devices of power 500 W each.SOLUTION:
Power of each device = 500 W
Number of devices = 4
Duration of use = 10 h
Total energy consumed = 4 × 500 W × 10 h
= 20000 Wh
= 20 kW hQ 33.
A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?SOLUTION:
When a freely falling body eventually stops on reaching the ground, its kinetic energy appears in the form of :
(i) heat (the body and the ground become warmer due to collision).
(ii) sound (produced due to collision of the body with the ground).
(iii) potential energy of configuration of the body and the ground (the body may be deformed and the ground may be depressed at the place of collision).
This process in which the kinetic energy of a freely falling body is lost in an unproductive chain of energy changes is called dissipation of energy.
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# Modeling Standard, Expanded, & Written Form
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## Objective
SWBAT read and write multi-digit whole numbers using standard form, expanded form, and written form.
#### Big Idea
Being able to understand and explain numbers will help students make sense of multi-digit computation and problem solving.
## Opening
20 minutes
Today's Number Talk
For a detailed description of the Number Talk procedure, please refer to the Number Talk Explanation. For this Number Talk, I am encouraging students to represent their thinking using an array model.
For the first task, 4 x 8, students used Decomposing to Solve 4 x 8. Here, a student shows how to go too far and then come back: 4x8=(5x8)-(1x8)
When we moved on to 14 x 8, students used a range of strategies. This student showed how 14 x 8 = 4[2x2+2x2+5x2+5x2] while another student impressively used Halving & Doubling to solve 14x8. You can tell when students are becoming more comfortable with decomposing and using the array model to multiply when they start Decomposing Over and Over to Find 14x8. This learning experience truly supports Math Practice 8: Look for and express regularity in repeated reasoning.
During the next task, students solved 514x8 by decomposing one or both multiplicands: 514x8=8(300+200+14).
Finally, students applied their understanding of decomposing and arrays to solve an even larger problem! Here, you'll see a student explaining how to use Decomposing to Find 2514x8.
## Guided Practice
50 minutes
Lesson Set Up:
For this lesson, I wanted to provide students with the opportunity to practice representing numbers using the Place Value & Money ModelsStandard FormExpanded Form, and Written Form
Prior to the lesson, I set up stations on the board. For the Place Value Station, I set up large magnetic color-coordinated base 10 blocks pictured below.
For the Money Model Station, I took a Bag of Money (created for a prior lesson using the following sheets of Money) and placed pieces of magnetic tape on the backs of each bill. For the Written Form Station, I projected the following picture and traced it on laminated bulletin board paper:
Finally, to provide a number for student to represent at all the stations, I used this set of Colossal Playing Cards. The kids absolutely loved them!
To begin, I invited students to join me on the carpet. I then introduced the Goal: I can read and write multi-digit whole numbers using standard form, expanded form, and written form.
Forms of Numbers:
I reviewed: Remember, there are three ways to write numbers. Pointing to the Standard Form Poster, I explained: The first way is called standard form. Standard form is a way to write numbers using the digits 0-9. Then, there is expanded form (Expanded Form Poster), a way to write numbers using the value of each digit. Finally, there is written form (Word Form Poster), a way to write numbers using words.
Introducing Stations:
Today we are going to work together to represent numbers using the stations on the board. First, I'll provide a number using these playing cards! For example, I might show you the Number 9 on the Playing Cards. Then, we will represent the number 9 at each of the stations on the board!
Modeling a One-Digit Number:
To further explain each station, I modeled how to represent the number 9 at each of the following stations by asking guiding questions, such as: How would I represent the number 9 using the place value model? Students shared their ideas and I modeled their thinking on the board. Then, I moved on to the next station.
Modeling a Two-Digit Number:
When finished, I created a new number using the playing cards: 31. I then asked for a couple volunteers to model the stations using the number 31. Although 31 is a simpler number to represent, these two students very proudly showed their thinking: Representing 31 Part 1 and Representing 31 Part 2.
While these students shared their thinking, the other students watched closely. I encouraged the class to make sure they agreed with every step these students took.
Modeling a Three-Digit Number:
Next, I created a new number using the playing cards: 961. This time, I asked for six volunteers to model 961 at each of the six stations. As students finished, I asked each volunteer to explain their thinking to the rest of the class: Representing 961 Part 1 and Representing 961 Part 2
Again, while these students modeled each station, the rest the class watched carefully, looking for mistakes they could "respectfully disagree" with. For today's lesson, I only had my 15 fourth graders. Consequently, there were many opportunities for students to participate. To make sure all students were engaged 100% of the time, I could have asked the rest of the class to take on observation roles, specifically assigned to a station. However, the time it took for students to complete each station was short enough that the rest of the students were interested in what was happening!
Modeling a Four-Digit Number:
As students finished, I created a new number using the playing cards: 7,322. In this video, students worked away, Representing 7,322 at each of their stations. To encourage higher-level thinning, I continually asked probing questions.
Modeling a Five-Digit Number:
In order to pull zero into our numbers, I explained to students: Let's say that the face cards (king, queen, jack) all represent zeros. Then, I asked a student to create a 5-digit number Using the Playing Cards. Six new students came to the board: Representing 99,220
Modeling a Six-Digit Number:
Finally, we got to a six digit number! Students were very excited to model this one: Representing 406,000
## Group Practice
30 minutes
Set Up:
At this point, I asked students to return to their desks to continue the same stations in a small-group setting. I passed out Station Posters in page protectors to each group of students. I have the desks in my classroom already arranged in groups so that group-projects are easy to implement.
Each of the six station posters matched one of the stations on the board. Other materials students need to complete this activity included a Bag of MoneyBase-10 Blocks (pictured below), and dry erase markers.
Representing a Three-Digit Number:
To get students started, I created a three-digit number using the playing cards: 803. Each group of 3-4 students went to work representing 803 on each of the station posters at their desks. I didn't give directions on how to split up the work. Some students gravitated toward the manipulatives for a hands-on learning experience. Other students enjoyed writing out the check! No matter what station students worked on within their groups, I encouraged students to continually check on one another to make sure they agreed with the work of other students. I also encouraged student to rotate between the stations to make sure they were able to experience each task. Here, students are Representing 803
Representing a Four-Digit Number:
As students finished, I checked their work before they were able to move on to 2,060. Some groups finished faster than others. To accommodate all levels, I often had up to three numbers represented using the playing cards on the board at one time. This group did a beautiful job Representing 2,060. I loved hearing them construct viable arguments and critique the reasoning of others (Math Practice 3).
Representing a Five-Digit Number:
Next, students moved on to a more complex number, 31,707. I often encouraged students to compare stations in order to check their work: Representing 31,707. Checking work with each other also helped support struggling students.
Representing a Six-Digit Number:
Many students had enough time to represent the last and most challenging number, 607,403. To show the value of the larger numbers, students often found more efficient strategies: Representing 607,403.
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# Whats 30 of 120?
Have you ever found yourself staring at a math problem and wondering, “What in the world does this even mean?” Well, my dear reader, you are not alone. As a matter of fact, I just had to ask myself that very question when posed with the query “What’s 30 of 120?”
So let’s dive right into this head-scratcher and see if we can come out on the other side with some answers (and maybe a few laughs).
## Breaking it down
First things first: let’s break down the language used in this perplexing question.
“What’s” is shorthand for “what is,” meaning we’re being asked for some kind of value or answer. Simple enough.
“30” and “120” are numbers – though that much should be obvious!
The word “of,” however, adds an extra layer of intrigue. It implies some sort of relationship between those two numbers beyond ‘1 + 1 = 2’.
Now that we have dissected each piece on its own, let’s put them together.
## The simple solution
If you aren’t looking for anything too complicated here (wink), then there actually is quite a simple solution available to us using basic arithmetic:
``````(30 /100) x 120 = ?
``````
That’s basically calculating what percentage from `120` represents `30`. This comes directly from applying proportions method in mathematics where fractions `x/100` giving respective percentages can be multiplied by another number representing total quantity.
Simple! Or so it seems…
Because sometimes knowing HOW to do something isn’t enough – especially if your brain hasn’t seen basic maths since high-school…(or primary).
Trying to pick apart algebraic equations while waiting in line at Starbucks can definitely leave one feeling like they might need extra-strength Tylenol.
So here’s numerous ways you can solve What’s 30 of 120? because variety is the spice of life:
## Method One: The Quickie
As we established earlier,
``````(30 /100) x 120 = ? (are you following this?)
``````
Let’s simplify from there just a little bit more:
``````3/10 x 120 = ?
12 × 3/10 = ?
36
``````
The answer to “What’s 30% of `120`?” is `36`. See, wasn’t that easy?
Far easier than string theory anyway.
### Side note:
String theory: an idea in theoretical physics where all matter and energy in our universe is composed of tiny one-dimensional strings vibrating at different frequencies.
Makes your head hurt already! Stick with us for straightforward mathematics rather than going down the rabbit hole filled with incomprehensible jargon.
## Method Two: Calculate by Percentage
We know that “of” means multiplication; therefore,
```(x /100) x y We take a number represented by ‘x’ stored as percentage or decimal and multiply it by some other number ‘y.’```
But what if we were asked for any percentage between two numbers? Not only are percentages used everywhere (51% milk, SPF50 sunblock), but they’re also useful when calculating tips on dinner checks – who needs math class now?!
In our case, we need to calculate for a percentage using the `total amount value given` and some fraction representing the required percentile. Here goes:
First, turn ‘percent’ into decimals:
`30 % ÷ 100 = .3` which becomes our `%`—remember how proportions work?!
Then,
“`.3 × \$120 =\$ ?? (first time using currency!)
\$36
``````
Instead of sweating over bills or arguing with a washing machine over pocket change, use basic percentages and make life easy-peasy.
## Method Three: Fraction Frenzy
Perhaps math never was your strongest point back in high school - we feel you; it's not for everyone.
But maybe Jack’s Pizza only costs \$5 (I mean who can resist that?) but tipping 30% is too rich of a taste to swallow.
Therefore,
``````
(30 /100) x 5 = ?
.3 x 5= ?
\$1.50
``````
Making tip calculation like this so much easier—and remember: always tip on pizza!
An even simpler way to state the answer might be “Thirty out of one hundred twenty" which we already know is equal to `0.25` or more simply put- ‘THREE OUT OF TEN’.
Mathematics doesn't have to give us headaches all day long if we break problems down into friendlier fractions.
# Some Real World Examples
Not sure yet how any of these calculations will come in handy outside the classroom setting?
Let us show you real-world scenarios where 'What's 30 of 120?' isn't just something to study but actually holds importance:
## Sales and Taxes — oh my!
During our discount-shopping-spree time at Checkers Hyper:
"I needed an extra toothbrush, so I got a nice Colgate from there for R38! Got some discount as well."
Calculating tax within stores can be especially useful when you're trying to figure out actual pricing between products after discounts and special offers have been applied.
So let's crack at finding sales-tax percentage rate for that travel-size mouth-cleaner!
First convert Rand `discount value %` into decimals:
```.45 ÷ [total order purchase price] X 100 ==Taxes%
R9/38 X100 = ≈23 %
``````
Hence either you’ll pay at till R47, or your remaining sum will be paid later when buying some groceries.
## More Real World Examples Continued:
If decisions to go ‘green’ are on the cards and being eco-friendly is important in day-to-day life then figuring out estimated carbon footprint from daily activities could translate into coming up with biomass values for example.
Or more commonly used today while watching shows such as “MasterChef” where a recipe calls for measuring ingredients by cups,
### Method Four: Metric Conversions
Metric System – an international decimalised system of measurement covering masses, lengths and other physical phenomena created during the French Revolution – suddenly becomes relevant once again.
For conversions between units:
``````Volume = Mass / Density
``````
So here we simplify…
One cup equals 0.236588 liters of fluid divided by mass density (which changes per kind of ingredient). As much as metrics have never really stuck around comprehension-wise, it’s the way forward to deal with things like converting temp readings from Fahrenheit to Celsius(Add formula no3!).
While maths can seem tedious, there’s something truly special about getting those complex problems just right; it’s a feeling unlike any other!
However at times sticking simple ratios work far better than delving so deep to show off mastery over trigonometry. Our suggestion? Keep calculating everything that comes your way. After all practice makes perfect!
And if you really do need help understanding more complicated equations including algebraic expressions…well that probably means going back to basics honestly (wink).
Happy Problem Solving Everyone!
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# CAHSEE practice: Problems 4-9
## Video transcript
We're on problem 4. Traditions Clothing Store is having a sale. Shirts that were regularly priced at $20 are on sale for$17. What is the percentage of the decrease in the price of the shirts? So what is just the decrease in the price of the shirts? Well, they've gone from $20 to$17. So the decrease is $3. We have a$3 decrease in the price of the shirts. So what is the percentage of decrease? Well, we're starting at $20, and we're decreasing by$3. So the percentage of the decrease-- let me write it in a darker color. We're starting at $20 and we're decreasing it by$3, or decreasing it by 3/20. This is the percent decrease, or the fraction decrease right there. And this is going to be equal to what? This is the same thing. We can divide it out. You might be able to do that in your head. But just in case you can't, let me just write it out like this. If I divide 20 into 3, 20 does not go into 3. Or you could say 20 goes into 3 zero times. 0 times 20 is zero. And then I put a decimal point right there. 20 goes into 30-- maybe I should have done this-- let me do it over here. 20 goes into 3 zero times. 0 times 20 is 0. And then let me put a decimal point like that. And then I'll do 3 minus that 0. So I'll just say you have 3.0, just like that. Maybe I shouldn't even write the decimal. Maybe I'll write a 30 like that. We have the decimal up here. 20 goes into 30 one time. 1 times 20 is 20. 30 minus 20 is 10. And then bring down another 0, this 0 right there. And 20 goes into 100 five times. 5 times 20 is 100. 100 minus 100 is 0. So 3/20 can be rewritten as 0.15. And another way you could think about this is 3/20, is the same thing-- if you multiply the numerator and the denominator by 5-- is the same thing as 15/100. Right? If I multiply the top and the bottom by 5, which we can always do, this is the same thing as 15/100. So that's the way you could have done that in your head, saying, oh, that's the same thing as 0.15. And 0.15 expressed as a percentage is 15%. So the correct answer is B. All right. Problem number 5. Which number equals 2 to the minus fourth power? So this is a little bit of review of exponents. 2 to the minus 4 is the same thing as 1 over 2 to the fourth power. That's all a negative exponent does. It means 1 over essentially the base to the positive exponent. And this is just going to be 1 over 2 times 2 times 2 times 2. And that's what? 2 times 2 is 4, times 2 is 8, times 2 is 16. So it's equal to 1/16. So that is choice C. Problem 6. What is 3/4 minus 1/6? So whenever you add or subtract fractions, you have to find a common denominator. And a good common denominator is the least common multiple of these two guys, or the smallest number that both of these denominators go into. So the smallest number that both of these denominators go into is 12, right? 4 goes into 12 three times, and 6 goes into 12 two times. So let's rewrite these fractions with 12 as a denominator. So something over 12 minus something over 12. So how do we rewrite 3/4 as something over 12? Well, 4 goes into 12 three times. Or we could say 3 times 4 is 12. So 3 times 3 will be the other numerator. You get 9/12. 3/4 and 9/12 are the exact same fraction. This is in kind of its simplest form, when you reduced the numerator and denominator as much as you can. But this is a completely equivalent fraction. To go from there to there, you just multiply the numerator and the denominator by 3. 3 times 3 is 9, 3 times 4 is 12. And the way you can think about it, 4 goes into 12 three times. So just multiply 3 times that numerator. Let's do the same thing with 1/6. 6 goes into 12 two times. So to go from there to there, we have to multiply the 6 by 2. So to go from there to whatever this numerator is, we have to multiply that numerator by 2. So you stick a 2 right there. And now that we have a common denominator, this becomes a simple problem to work out. It is equal to-- we have 9 out of 12 pieces, minus 2 out of 12 pieces, or whatever we're talking about. Or slices of pie. So that's going to be equal to 7/12. And that is choice C. Next problem. Problem 7. Do it in blue again. A salesperson at a clothing store earns a 2% commission on all sales. How much commission does the salesperson earn on a $300 sale? Well, they earn 2% on that. Or you could say 0.02 of the sale. So you could just multiply 300 times 0.02. This is kind of the way to just do it. And I'll maybe give you a little intuition on how you could do this in your head if you're constrained for time. But the easiest way to think about it is just multiply 2 times 300. Or 2 times 0 is 0. 2 times 0 is 0, and 2 times 3 is 6. You could have done that in your head. 2 times 300 is 600. And then you worry about the decimals. So we have two spaces behind the decimal right here. Right? We have one, two. So we're going to have to have two spaces behind the decimal in our answer. So the answer right here is$6. And that is choice A. Now, a way you could have done this in your head, you could have said, look, he makes a 2% commission. So for every $100, he makes$2. Right? 2% is the same thing as 2 out of 100. So every $100, he makes$2. He sold $300, so that would also be 3 times$2 per 100, or \$6. Whatever is easier for you to understand, that's what you should do. Problem number 8. I'll do this in green. Some students attend school 180 of the 365 days in a year. About what part of the year do they attend school? And so the keyword here is about, which tells me that they don't want an exact answer. They don't want me to sit there-- I mean, if I wanted to, they attend 180 out of 365 days. If I wanted the exact percentage, I would have to take 365 and divide it into 180, with some decimals, and I'll get some-- I could work it out, just like I did the decimal division in the past, but it would take some time, and they just want to know about. So what's 180/365 roughly? Well, what's 180 times 2? You could say 180/360, and I picked 360 because that's 180 times 2, that's equal to 50%. This is not that different than this. We just have a small change of the denominator, small relative to how large that denominator is. So the answer here is 50%. If they had a couple of other choices here like 49% or 51%, then you would have to work this out in a little bit more detail. But it's pretty clear that 180/365 is pretty close to 50%. It's nowhere near 18% or 75%, so you can feel pretty good about this answer. If 49% were one of the choices, then I would have to do a little bit more arithmetic right here with the division. Problem number 9. What is the value of 2 to the sixth times 2 to the fourth, divided by 2 to the fifth? Now, you could solve each of these powers. 2 to the sixth is what? It's 64. You could work out each of these exponents, and then multiply them and then divide, but it would take you forever. So what they really want you to do here is use your exponent rules. So when you multiply two exponents and they have the same base-- let me write it this way. 2 to the sixth times 2 to the fourth, all of that over 2 to the fifth. What's that equal to? So let's just simplify the numerator first. So I'll keep the denominator the same. When you multiply two exponents with the same base, you essentially can just add the exponents. So the same base is 2. So this is going to be equal to 2 to the 6 plus 4, or 2 to the 10th power. Now, when you divide exponents that have the same base, you subtract the exponents. So this is going to be equal to 2 to the 10 minus 5 power, which is equal to 2 to the fifth. And that's our answer. And we look at our choices, we don't see 2 to the fifth. So we're actually going to have to multiply it out. And so 2 to the fifth is 2 times 2 times 2 times 2 times 2. 2 times itself five times. That's what 2 to the fifth means. So what is this? 2 times 2 is 4. 4 times 2 is 8. 8 times 2 is 16. 16 times 2 is 32. So that is our answer. D.
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# 3rd Grade - Use Cubes To Find Volume And Area
Grade Level: 3rd Skill: Shapes and Geometry Topic: Use Cubes to Find Volume and Area Goal: Estimate or determine the area and volume of solid figures by covering them with squares or by counting the number of cubes that would fill them. Skill Description: Measurement and Geometry: volume and area of solid figures The ability to determine a number representation of the space inside a shape. The object of this lesson is to show your child a way to calculate the area or volume of a polygon or solid shape without just learning a formula that would perform that calculation. When children use hands-on methods to perform calculations before they are taught a formula (recipe) for completing the task they are more apt to remember how to do it, even when they have not used the information for some time.
### Sample Problems
(1) Define area. (Count of how many square units fit inside a figure as a way of determining area.) (2) Can you use what you learned when you placed the square papers to figure out how we could write a rule for determining area? If we had to cut little squares for everything it would take a long time! Answer: The formula for the area of a rectangle/square is its width times its height. When we used the little squares each had a width of 1 and a height of 1 so we just counted them up. Because each was worth “1” we counted how many there were in each direction and multiplied them together. We could just as easily use the measurement for each direction and multiply those numbers together. . (3) Advanced: How do I find the surface area of a cube? To find the surface area of a cube, find the surface area of one side (the length of a side squared) and then multiply by 6. Children can begin to explore this concept by creating a cube shape out of sugar cubes. (4) What is the formula for the volume of a cube? (a x a x a) (5) What is the difference between the area of a shape and the volume of a solid object? (area refers to the area of the surface (2D) of an object, whereas volume measures the amount of 3D space an object takes up)
### Learning Tips
(1) Have children take an index card and cover it with cut out cm squares. Children can also cover the card with dimes or pennies and a discussion can take place regarding which covers the card better (squares because there are no gaps). Children then count how many squares cover the card = area. Cover other shapes with the squares and children can see that different figures can have the same area. Measure perimeter of each figure, too. Children can see that figures with the same area can also have different perimeters. (2) Use geoboards, square shaped crackers or candy bars to calculate area. Place the items as you would tiles, filling up the space that you have marked. Count how many items it takes to completely fill the space. Children can have fun with the challenge of creating a label for their answer, since the result will not be measured in feet, inches, centimeters, or any other conventional measure. Perhaps the label could be square candies? Square crackers? Children can then measure the space occupied conventionally, calculate the area using paper squares of a known size and then convert their answer to square feet inches or centimeters, depending upon the measure of the paper squares. (3) Mark off (with masking tape) a square section of your kitchen or living room floor. Ask the child to sit on the floor with a package of rulers. Have the child take the rulers and mark off square feet and then calculate the area of the section of floor in square feet. (4) Children can explore volume/area by using base ten one cube blocks or cubes of sugar. Have the child make a cube (many layers) out of ones blocks. It is easier to see the middle of the cube if ones blocks are used. (5) Ask the child to trace a hand on a piece of graph paper and then count the number of squares inside the outline. The child will need to put together ½ and ¼ squares to make a whole and get an accurate count. If the paper is cm graph paper, the number the child gets would be the area in cubic cm. The child can try this activity with their foot or other household objects, too.
### Online Resources
(1) (2) (3) (4) (5)
### Extra Help Problems
(1) The attached worksheet has 24 problems: 12 involve finding area by covering with 1 cm paper squares and 12 involve finding area by pretending that there are additional layers of cubes on top of the 2-dimensional areas calculated on the other page.
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Prove this inequality $\frac{1}{1+a}+\frac{2}{1+a+b}<\sqrt{\frac{1}{a}+\frac{1}{b}}$
Let $a,b>0$ show that
$$\dfrac{1}{1+a}+\dfrac{2}{1+a+b}<\sqrt{\dfrac{1}{a}+\dfrac{1}{b}}$$
It suffices to show that
$$\dfrac{(3a+b+3)^2}{((1+a)(1+a+b))^2}<\dfrac{a+b}{ab}$$
or
$$(a+b)[(1+a)(1+a+b)]^2>ab(3a+b+3)^2$$
this idea can’t solve it to me,are we aware of an elementary way of proving that? Thanks in advance.
Solutions Collecting From Web of "Prove this inequality $\frac{1}{1+a}+\frac{2}{1+a+b}<\sqrt{\frac{1}{a}+\frac{1}{b}}$"
Use Cauchy-Schwarz inequality we have
$$\left(\dfrac{1}{1+a}+\dfrac{2}{1+a+b}\right)^2\le\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\left(\dfrac{a}{(1+a)^2}+\dfrac{4b}{(1+a+b)^2}\right)$$
so suffices to show that
$$\dfrac{a}{(1+a)^2}+\dfrac{4b}{(1+a+b)^2}<1$$
since
$$\dfrac{4b}{(1+a+b)^2}<\dfrac{4b}{4(1+a)b}=\dfrac{1}{1+a}$$
it suffices to show that
$$\dfrac{a}{(1+a)^2}+\dfrac{1}{1+a}<1$$
$$\Longleftrightarrow a+1+a<(1+a)^2\Longleftrightarrow a^2>0$$ it is clear
if op go ahead , he will get from his last step :LHS-RHS$=(4a^4+4a^2+b^3-6a^2b)+(a^2b^2-3ab+b+a)+a^2b^3+ab^3+3a^3b^2+a^2b^2+ab^2+2b^2+3a^4b+a^3b+a^5+6a^3 >0 \iff$
$(4a^4+4a^2+b^3-6a^2b) \ge 0 \cap (a^2b^2-3ab+b+a)\ge 0$
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# Two probabilities of the same thing
Consider this minesweeper situation. The 3 I've circled tells us that there is a mine in one of A, B and C. So, this means the probability that the mine is in A is $\frac{1}{3}$. And the 4 I've circled (and the 2 below it) tells that there is a mine in A or D. This makes the probability of finding a mine in A = $\frac{1}{2}$.
So, how to combine these two probabilities to get the actual probability of finding a mine in A?
I thought of this: There are two possibilities: Either there is a single mine in A or there is one mine in D and one in B or C. These two possibilities are equally likely, I guess. So, the probability that there's a mine in A is $\frac{1}{2}$. If that was correct, then is there any other approach to this?
EDIT: Also, the circled 3 tells that that the probability that a mine is in B or C=$\frac{2}{3}$. But if we consider the 3 to the right of the circled 3, it tells us that there is a single mine in one of B,C,E and F, which makes the probability of finding a mine in B or C=$\frac{1}{2}$. Again, there are two probabilities of finding a mine in B or C.
• Is there only one mine? Can there be 4 mines, one each in ABCD? – Gaurang Tandon Oct 20 '17 at 6:18
• @GaurangTandon In this situation, there can be either one mine or two mines. The numbers tell us how many mines there are in the 8 blocks surrounding the number. For example, the number 4 tells us that there are four mines in the 8 blocks surrounding it. I've determined 3 of those 4 mines (that's where those 3 red flags are). This means the remaining mine is in A or D. – Dove Oct 20 '17 at 6:42
• The total number of possible cases in this situation is so limited you could count them by writing them by down. But, a general solution for a possibly broader situation is more suitable. – Gaurang Tandon Oct 20 '17 at 6:46
• @RobertIsrael The probability of the presence of a mine in ABCD would surely be affected after numbers in blocks surrounding them are uncovered. But, they don't affect the probability as long as they remain covered, at least? We've to base our answer on what we know, not on what we could know. Like, the probability of rain today is 0.75. But, if I come to know later that there is a hurricane approaching, I'd say rain=0.99. But, for now, I'm commenting based on what I know, not what I could know *more*. Willing to be clarified :) – Gaurang Tandon Oct 20 '17 at 7:09
• After doing some googling, the probability model of minesweeper seems to assume they are uniformly random - each mine configuration is equally likely, given that it is consistent with the current information. In your example it is lucky that the group of $6$ grids contains exactly $2$ mines, namely $\{D, B\}, \{D, C\}, \{A, E\}, \{A, F\}$ and thus the probabilities in this group will be independent of the global no of mines and the no of remaining grids. In general you need to count all the possible configurations, but now this case is simple, with $P(D)=P(A)=1/2$ and $P(B)=P(C)=P(E)=P(F)=1/4$ – BGM Oct 20 '17 at 8:58
The inconsistencies in your considerations stem from the fact that you're applying the principle of indifference in cases where it doesn't apply. The events that you're considering as equiprobable actually have different structures, and there is no reason to regard them as equiprobable.
As BGM has pointed out in a comment, you would usually need to take into account non-local information (how many mines are left in all, or with which density the mines were distributed), but in this case you can prove that there are exactly two mines in squares $A$ through $F$, so the local information is enough. Every arrangement of these two mines that is consistent with the displayed numbers is equiprobable, since the game is presumably equally likely to generate any given arrangement of the mines. (This does not imply that any individual properties of the configuration (such as "there is a mine on $A$" or "there is a mine on $B$") are equiprobable -- they may be more or less likely because they may be consistent with more or fewer arrangements; only the arrangements are equiprobable.)
Another interesting question is what is the best strategy to proceed in order to uncover all six labelled squares without dying (assuming for the sake of argument that these are the only relevant squares and the unlabeled squares don't contain mines and can't be clicked). It is not necessarily optimal to uncover the square with the lowest probability of containing a mine, since you also want to gain as much information as possible in order to minimize risks in the future.
If you click on $A$, you die with probability $\frac12$; otherwise you can uncover $B$, $C$ and $D$, but then you still don't know whether the second mine is in $E$ or $F$, so you have another $50\%$ chance of dying, for a total survival probability of $\frac14$.
If you click on $D$, you also die with probability $\frac12$, but in this case, when you survive you have complete information, so the overall survival probability is $\frac12$.
If you click on $B$, you have a $\frac14$ probability to die, a $\frac14$ probability of getting the complete information that the mines are at $C$ and $D$, and a $\frac12$ probability of being left with an unknown mine in $E$ or $F$, with another $50\%$ chance to die, for a total survival probability of $\frac14\cdot0+\frac14\cdot1+\frac12\cdot\frac12=\frac12$.
If you click on $C$, you're certain to get a $1$ if you survive, so that doesn't gain you any information and thus can't be an optimal move.
If you click on $E$ or $F$, you have a $\frac14$ probability to die, but if you survive, you can get complete information without further risks: either the number you get tells you that the remaining two mines must be in $D$ and $B$, or it tells you that $B$ is empty, so you can click it and the number will allow you to distinguish the two remaining possibilities.
Thus, the optimal strategy is to click on $E$ or $F$, and in this case you have a survival probability of $\frac34$ – a stark contrast to the survival probability of only $\frac14$ for the worst strategy, first clicking on $A$.
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Algebra-Lineares Gleichungssystem-Additionsverfahren (2)
Beispiel Nr: 10
$\text{Gegeben:} \\ a1 \cdot x +b1 \cdot y =c1\\ a2 \cdot x +b2 \cdot y =c2 \\ \\ \text{Gesucht:} \\\text{x und y} \\ \\ \textbf{Gegeben:} \\ \\ -1 x +1 y =3\\ \frac{1}{2} x -4 y = 5 \\ \\ \\ \\ \textbf{Rechnung:} \\\begin{array}{l|l} \begin{array}{l} \\I \qquad -1 x +1 y =3\\ II \qquad \frac{1}{2} x -4 y = 5 \\ I \qquad -1 x +1 y =3 \qquad / \cdot\left(-\frac{1}{2}\right)\\ II \qquad \frac{1}{2} x -4 y = 5 \qquad / \cdot\left(-1\right)\\ I \qquad \frac{1}{2} x -\frac{1}{2} y =-1\frac{1}{2}\\ II \qquad -\frac{1}{2} x +4 y = -5 \\ \text{I + II}\\ I \qquad \frac{1}{2} x -\frac{1}{2} x-\frac{1}{2} y +4 y =-1\frac{1}{2} -5\\ 3\frac{1}{2} y = -6\frac{1}{2} \qquad /:3\frac{1}{2} \\ y = \frac{-6\frac{1}{2}}{3\frac{1}{2}} \\ y=-1\frac{6}{7} \\ \text{y in I}\\ I \qquad -1 x +1\cdot \left(-1\frac{6}{7}\right) =3 \\ -1 x -1\frac{6}{7} =3 \qquad / +1\frac{6}{7} \\ -1 x =3 +1\frac{6}{7} \\ -1 x =4\frac{6}{7} \qquad / :\left(-1\right) \\ x = \frac{4\frac{6}{7}}{-1} \\ x=-4\frac{6}{7} \\ L=\{-4\frac{6}{7}/-1\frac{6}{7}\} \end{array} & \begin{array}{l} \\I \qquad -1 x +1 y =3\\ II \qquad \frac{1}{2} x -4 y = 5 \\ I \qquad -1 x +1 y =3 \qquad / \cdot\left(-4\right)\\ II \qquad \frac{1}{2} x -4 y = 5 \qquad / \cdot\left(-1\right)\\ I \qquad 4 x -4 y =-12\\ II \qquad -\frac{1}{2} x +4 y = -5 \\ \text{I + II}\\ I \qquad 4 x -\frac{1}{2} x-4 y +4 y =-12 -5\\ 3\frac{1}{2} x = -17 \qquad /:3\frac{1}{2} \\ x = \frac{-17}{3\frac{1}{2}} \\ x=-4\frac{6}{7} \\ \text{x in I}\\ I \qquad -1 \cdot \left(-4\frac{6}{7}\right) +1y =3 \\ 1 y +4\frac{6}{7} =3 \qquad / -4\frac{6}{7} \\ 1 y =3 -4\frac{6}{7} \\ 1 y =-1\frac{6}{7} \qquad / :1 \\ y = \frac{-1\frac{6}{7}}{1} \\ y=-1\frac{6}{7} \\ L=\{-4\frac{6}{7}/-1\frac{6}{7}\} \end{array} \end{array}$
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# How To Find The Slope Intercept Form With Two Points
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
How To Find The Slope Intercept Form With Two Points – Among the many forms used to illustrate a linear equation one of the most frequently seen is the slope intercept form. You can use the formula of the slope-intercept identify a line equation when you have the slope of the straight line and the yintercept, which is the point’s y-coordinate at which the y-axis meets the line. Read more about this particular line equation form below.
## What Is The Slope Intercept Form?
There are three primary forms of linear equations, namely the standard one, the slope-intercept one, and the point-slope. Though they provide the same results , when used, you can extract the information line produced more quickly using the slope intercept form. Like the name implies, this form utilizes an inclined line, in which its “steepness” of the line is a reflection of its worth.
This formula can be used to discover the slope of a straight line. It is also known as the y-intercept, also known as x-intercept where you can apply different formulas available. The equation for this line in this specific formula is y = mx + b. The straight line’s slope is represented with “m”, while its y-intercept is indicated with “b”. Every point on the straight line can be represented using an (x, y). Note that in the y = mx + b equation formula, the “x” and the “y” must remain as variables.
## An Example of Applied Slope Intercept Form in Problems
For the everyday world, the slope intercept form is frequently used to depict how an object or problem evolves over it’s course. The value that is provided by the vertical axis is a representation of how the equation addresses the degree of change over the value given by the horizontal axis (typically time).
A simple example of the application of this formula is to determine how many people live in a specific area in the course of time. In the event that the population in the area grows each year by a specific fixed amount, the amount of the horizontal line will increase one point at a moment each year and the point worth of the vertical scale is increased to represent the growing population according to the fixed amount.
You may also notice the starting point of a particular problem. The beginning value is located at the y-value of the y-intercept. The Y-intercept is the place at which x equals zero. Based on the example of a previous problem, the starting value would be at the time the population reading begins or when the time tracking starts along with the associated changes.
This is the place at which the population begins to be monitored to the researchers. Let’s suppose that the researcher began to perform the calculation or measure in the year 1995. This year will serve as”the “base” year, and the x = 0 point would occur in the year 1995. Thus, you could say that the population in 1995 corresponds to the y-intercept.
Linear equations that employ straight-line formulas are almost always solved this way. The starting value is expressed by the y-intercept and the rate of change is represented in the form of the slope. The primary complication of this form is usually in the horizontal variable interpretation, particularly if the variable is accorded to one particular year (or any other kind number of units). The most important thing to do is to ensure that you understand the variables’ definitions clearly.
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# What is an example of simplifying fractions?
## What is an example of simplifying fractions?
A fraction is considered simplified if there are no common factors in the numerator and denominator. For example, 23 is simplified because there are no common factors of 2 and 3 . 1015 is not simplified because 5 is a common factor of 10 and 15 .
## Why do we have to simplify fractions?
A fraction is said to be in its simplest form if 1 is the only common factor of its numerator and denominator. We simplify fractions because it is always to work or calculate when the fractions are in the simplest form.
How do you simplify multiple fractions?
The first step when multiplying fractions is to multiply the two numerators. The second step is to multiply the two denominators. Finally, simplify the new fractions. The fractions can also be simplified before multiplying by factoring out common factors in the numerator and denominator.
### What grade do students learn to simplify fractions?
Kids start to learn about fractions in first and second grade. By the end of grade school, many kids understand and can solve basic problems with fractions. Others need more time. Fractions are a difficult math concept, and lots of kids struggle with them.
### What is the rule for simplifying fractions?
The only rule to simplify fractions is that by whatever number you divide the numerator, you must divide the denominator using the same number. There are two ways to simplify fractions: Keep dividing the numerator and denominator by 2, 3, 5, 7, etc. until you cannot go any further.
How to simplify a mixed fraction in factored form?
In order to simplify a mixed fraction, you need to simplify the fractional part only. For that write the numerator and the denominator in factored form and cancel out the common factors. The resultant will be the new numerator and the new denominator of the mixed fraction. For example: Simplify the mixed fraction 3 4 10 3 4 10.
#### How do you simplify a fraction with exponents?
You can simplify the fraction containing exponents in the numerator and denominator. Use the expanded form of exponents in the numerator and denominator to make it easy for you to simplify the fraction with exponents. To make a number easy to read, we sometimes make use of exponents. Suppose we have the fraction 3 5 /3 2
#### How to write a fraction in simplest form?
Easy Ways to Simplify Fractions 1 Step 1: Write the factors of numerator and denominator. 2 Step 2: Determine the highest common factor of numerator and denominator. 3 Step 3: Divide the numerator and denominator by the highest common factor. The fraction so obtained is in the simplest… More
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# NCERT/CBSE MATHS Class 10 Ex- 1.3 Q No.3 Solutions : Prove that the following are irrationals: (i) 1/√2 (ii) 7√5 (iii) 6 + √2
Hy Friends Welcome On NCERT MATHS SOLUTIONS !! Today we are going to solve the Question : ”Prove that the following are irrationals: (i) 1/√2 (ii) 7√5 (iii) 6 + √2 of Exercise 1.3 Question No 2 Solutions (Real Numbers) of Class 10th, which will prove to be very helpful for you.
# CBSE/NCERT MATHS Class 10 Ex- 1.3 Q No.3 Solutions
If you are a student of CBSE, today we are going to give you the CBSE / NCERT Chapter : Real Numbers Exercise 1.3 Question No – 3 Solutions . Hope you like this post about Class 10th Maths Solution.
Here we would like to tell you that there are other websites on which the students demanded the Solution of NCERT MATHS Chapters, we had promised them that we will soon send you the questions of Class 9, And we completed our promise to them.We asked the good knowledgeable teachers of mathematics and asked them to solve the issues of CBSE’s mathematical questions and also said that the answers to our questions are such that the students have no difficulty in solving all those questions.
### Real Numbers CHAPTER : 1
Symbol of real numbers (ℝ) In mathematics, the real number is the value presented to any amount corresponding to the simple line. Actual numbers include all rational numbers such as -5 and fractional numbers such as 4/3 and all irrational numbers such as √2 (1.41421356 …, square root of 2, an unregulated algebraic number). By incorporating the ample numbers in the actual numbers, they can be presented from the eternal points that can be attributed on a line in the form of a real number line.
### (i) 1/√2
Let 1/√2 is rational .
Therefore, we can find two co-prime integers a, ba (b ≠ 0) such that
Or 1/√2 = a/b
√2 = a/b
b/a is rational as a and b are integers.
Therefore, √2 is rational which contradicts to the fact that √2 is irrational.
Hence, our assumption in false and 1/√2 is irrational.
### (ii) 7√5
Let 7√5 is rational.
Therefore, we can find two co-prime integers a, b (b ≠ 0 ) such that
7√5 = a/b
= √5 = a/7b
a/7b is rational as a and b are integers.
Therefore, √5 should be rational. This contradicts the fact that √5 is irrational.
Therefore, our assumption that 7√5 is rational is false. Hence, 7√5 is irrational.
### (iii) 6 + √2
Let ) 6 + √2 be rational.
Therefore, we can find two co-prime integers a, ( b ≠ 0 ) such that
6 + √2 = a/b
√2 = a/b – 6
Since a and b are integers, a/b-6 is also rational and hence, √2 should be rational. This contradicts the fact that√2 is irrational. Therefore, our assumption is false and hence, 6 + √2 is irrational.
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# Leave and Let Dice
Imagine a game with six players, numbered #1 to #6, and one six-sided die. Someone rolls the die and the player who matches the number wins the game. That is, if the die rolls 1, player #1 wins; if the die rolls 2, player #2 wins; and so on. With a fair die, this is a fair game, because each player has exactly a 1/6 chance of winning. You could call it a simultaneous game, because all players are playing at once. It has one rule:
• If the die rolls n, then player #n wins.
Now try a different game with six players and one die. Player #1 rolls the die. If he gets 1, he wins the game. If not, then he leaves the game and player #2 rolls the die. If he gets 2, he wins the game. If not, then he leaves the game and player #3 rolls the die. And so on. You could call this a sequential game, because the players are playing in sequence. It has two rules:
• If player #n rolls n on the die, then he wins.
• If player #n doesn’t roll n, then player n+1 rolls the die.
Is it a fair game? No, definitely not. Player #1 has the best chance of winning. 1/6 or 16.6% of the time he rolls 1 and wins the game. 5/6 of the time, he rolls 2, 3, 4, 5 or 6 and passes the die to player #2. Now player #2 has a 1/6 chance of rolling a 2 and winning. But he has the opportunity to roll the die only 5/6 of the time, so his chance of winning the game is 1/6 * 5/6 = 5/36 = 13.8%. However, if player #2 rolls a 1, 3, 4, 5 or 6, then he loses and player #3 rolls the die. But player #3 has that opportunity only 5/6 * 5/6 = 25/36 of the time. So his chance of winning is 1/6 * 25/36 = 11.57%. And so on.
To put it another way, if the six players play 46656 = 6^6 games under the sequential rules, then on average:
• Player #1 wins 7776 games
• Player #2 wins 6480 games
• Player #3 wins 5400 games
• Player #4 wins 4500 games
• Player #5 wins 3750 games
• Player #6 wins 3125 games
• 15625 games end without a winner.
In other words, player #1 is 20% more likely to win than player #2, 44% more likely than player #3, 72.8% more likely than player #4, 107% more likely than player #5, and 148.8% more likely than player #6. Furthermore, player #2 is 20% more likely to win than player #3, 44% more likely than player #4, 72.8% more likely than player #5, and so on.
But there is a simple way to make the sequential game perfectly fair, so long as it’s played with a fair die. At least, I’ve thought of a simple way, but there might be more than one.
To make the sequential game fair, you add an extra rule:
1. If player #n rolls n on the die, he wins the game.
2. If player #n rolls a number greater than n, he loses and the die passes to player n+1.
3. If player #n rolls a number less than n, then he rolls again.
Let’s run through a possible game to see that it’s fair. Player #1 rolls first. He has a 1/6 chance of rolling a 1 and winning the game. However, 5/6 of the time he loses and passes the die to player #2. If player #2 rolls a 1, he rolls again. In other words, player #2 is effectively playing with a five-sided die, because all rolls of 1 are ignored. Therefore, he has a 1/5 chance of winning the game at that stage.
But hold on: a 1/5 chance of winning is better than a 1/6 chance, which is what player #1 had. So how is the game fair? Well, note the qualifying phrase at the end of the previous paragraph: at that stage. The game doesn’t always reach that stage, because if player #1 rolls a 1, the game is over. Player #2 rolls only if player doesn’t roll 1, which is 5/6 of the time. Therefore player #2’s chance of winning is really 1/5 * 5/6 = 5/30 = 1/6.
However, 4/5 of the time player #2 rolls a 3, 4, 5 or 6 and the die passes to player #3. If player #3 rolls a 1 or 2, he rolls again. In other words, player #3 is effectively playing with a four-sided die, because all rolls of 1 and 2 are ignored. Therefore, he has a 1/4 chance of winning the game at that stage.
A 1/4 chance of winning is better than a 1/5 chance and a 1/6 chance, but the same reasoning applies as before. Player #3 rolls the die only 5/6 * 4/5 = 20/30 = 2/3 of the time, so his chance of winning is really 1/4 * 2/3 = 2/12 = 1/6.
However, 3/4 of the time player #2 rolls a 4, 5 or 6 and the die passes to player #4. If player #4 rolls a 1, 2 or 3, he rolls again. In other words, player #4 is effectively playing with a three-sided die, because all rolls of 1, 2 and 3 are ignored. Therefore, he has a 1/3 chance of winning the game at that stage. 1/3 > 1/4 > 1/5 > 1/6, but the same reasoning applies as before. Player #4 rolls the die only 5/6 * 4/5 * 3/4 = 60/120 = 1/2 of the time, so his chance of winning is really 1/3 * 1/2 = 1/6.
And so on. If the die reaches player #5 and he gets a 1, 2, 3 or 4, then he rolls again. He is effectively rolling with a two-sided die, so his chance of winning is 1/2 * 5/6 * 4/5 * 3/4 * 2/3 = 120/720 = 1/6. If player #5 rolls a 6, he loses and the die passes to player #6. But there’s no need for player #6 to roll the die, because he’s bound to win. He rolls again if he gets a 1, 2, 3, 4 or 5, so eventually he must get a 6 and win the game. If player #5 loses, then player #6 automatically wins.
It’s obvious that this form of the game will get slower as more players drop out, because later players will be rolling again more often. To speed the game up, you can refine the rules like this:
1. If Player #1 rolls a 1, he wins the game. Otherwise…
2. If player #2 rolls a 2, he wins the game. If he rolls a 1, he rolls again. Otherwise…
3. Player #3 rolls twice and adds his scores. If the total is 3, 4 or 5, he wins the game. Otherwise…
4. Player #4 rolls once. If he gets 1 or 2, he wins the game. Otherwise…
5. Player #5 rolls once. If he gets 1, 2 or 3, he wins the game. Otherwise…
6. Player #6 wins the game.
Only player #2 might have to roll more than twice. Player #3 has to roll twice because he needs a way to get a 1/4 chance of winning. If you roll two dice, there are:
• Two ways of getting a total of 3: roll #1 is 1 and roll #2 is 2, or vice versa.
• Three ways of getting a total of 4 = 1+3, 3+1, 2+2.
• Four ways of getting 5 = 1+4, 4+1, 2+3, 3+2.
This means player #3 has 2 + 3 + 4 = 9 ways of winning. But there are thirty-six ways of rolling one die twice. Therefore player #3 has a 9/36 = 1/4 chance of winning. Here are the thirty-six ways of rolling one die twice, with asterisks marking the winning totals for player #3:
01. (1,1)
02. (1,2)*
03. (2,1)*
04. (1,3)*
05. (3,1)*
06. (1,4)*
07. (4,1)*
08. (1,5)
09. (5,1)
10. (1,6)
11. (6,1)
12. (2,2)*
13. (2,3)*
14. (3,2)*
15. (2,4)
16. (4,2)
17. (2,5)
18. (5,2)
19. (2,6)
20. (6,2)
21. (3,3)
22. (3,4)
23. (4,3)
24. (3,5)
25. (5,3)
26. (3,6)
27. (6,3)
28. (4,4)
29. (4,5)
30. (5,4)
31. (4,6)
32. (6,4)
33. (5,5)
34. (5,6)
35. (6,5)
36. (6,6)
# Dice in the Witch House
“Who could associate mathematics with horror?”
John Buchan answered that question in “Space” (1911), long before H.P. Lovecraft wrote masterpieces like “The Call of Cthulhu” (1926) and “Dreams in the Witchhouse” (1933). But Lovecraft’s use of mathematics is central to his genius. So is his recognition of both the importance and the strangeness of mathematics. Weird fiction and maths go together very well.
But weird fiction is about the intrusion or eruption of the Other into the everyday. Maths can teach you that the everyday is already Other. In short, reality is weird — the World is a Witch House. Let’s start with a situation that isn’t obviously weird. Suppose you had three six-sided dice, A, B and C, each with different set of numbers, like this:
Die A = (1, 2, 3, 6, 6, 6)
Die B = (1, 2, 3, 4, 6, 6)
Die C = (1, 2, 3, 4, 5, 6)
If the dice are fair, i.e. each face has an equal chance of appearing, then it’s clear that, on average, die A will beat both die B and die C, while die B will beat die C. The reasoning is simple: if die A beats die B and die B beats die C, then surely die A will beat die C. It’s a transitive relationship: If Jack is taller than Jim and Jim is taller than John, then Jack is taller than John.
Now try another set of dice with different arrangements of digits:
Die A = (1, 2, 2, 5, 6, 6)
Die B = (1, 1, 4, 5, 5, 5)
Die C = (3, 3, 3, 3, 4, 6)
If you roll the dice, on average die A beats die B and die B beats die C. Clearly, then, die A will also beat die C. Or will it? In fact, it doesn’t: the dice are what is called non-transitive. Die A beats die B and die B beats die C, but die C beats die A.
But how does that work? To see a simpler example of non-transitivity, try a simpler set of random-number generators. Suppose you have a triangle with a short rod passing through its centre at right angles to the plane of the triangle. Now imagine numbering the edges of the triangles (1, 2, 3) and throwing it repeatedly so that it spins in the air before landing on a flat surface. It should be obvious that it will come to rest with one edge facing downward and that each edge has a 1/3 chance of landing like that.
In other words, you could use such a spiked triangle as a random-number generator — you could call it a “trie”, plural “trice”. Examine the set of three trice below. You’ll find that they have the same paradoxical property as the second set of six-sided dice above. Trie A beats trie B, trie B beats trie C, but trie C beats trie A:
Trie A = (1, 5, 8)
Trie B = (3, 4, 7)
Trie C = (2, 3, 9)
When you throw two of the trice, there are nine possible outcomes, because each of three edges on one trie can be matched with three possible edges on the other. The results look like this:
Trie A beats Trie B 5/9ths of the time.
Trie B beats Trie C 5/9ths of the time.
Trie C beats Trie A 5/9ths of the time.
To see how this works, here are the results throw-by-throw:
Trie A = (1, 5, 8)
Trie B = (3, 4, 7)
When Trie A rolls 1…
…and Trie B rolls 3, Trie B wins (Trie A has won 0 out of 1)
…and Trie B rolls 4, Trie B wins (0 out of 2)
…and Trie B rolls 7, Trie B wins (0 out of 3)
When Trie A rolls 5…
…and Trie B rolls 3, Trie A wins (1/4)
…and Trie B rolls 4, Trie A wins (2/5)
…and Trie B rolls 7, Trie B wins (2/6)
When Trie A rolls 8…
…and Trie B rolls 3, Trie A wins (3/7)
…and Trie B rolls 4, Trie A wins (4/8)
…and Trie B rolls 7, Trie A wins (5/9)
Trie B = (3, 4, 7)
Trie C = (2, 3, 9)
When Trie B rolls 3…
…and Trie C rolls 2, Trie B wins (Trie B has won 1 out of 1)
…and Trie C rolls 3, it’s a draw (1 out of 2)
…and Trie C rolls 9, Trie C wins (1 out of 3)
When Trie B rolls 4…
…and Trie C rolls 2, Trie B wins (2/4)
…and Trie C rolls 3, Trie B wins (3/5)
…and Trie C rolls 9, Trie C wins (3/6)
When Trie B rolls 7…
…and Trie C rolls 2, Trie B wins (4/7)
…and Trie C rolls 3, Trie B wins (5/8)
…and Trie C rolls 9, Trie C wins (5/9)
Trie C = (2, 3, 9)
Trie A = (1, 5, 8)
When Trie C rolls 2…
…and Trie A rolls 1, Trie C wins (Trie C has won 1 out of 1)
…and Trie A rolls 5, Trie A wins (1 out of 2)
…and Trie A rolls 8, Trie A wins (1 out of 3)
When Trie C rolls 3…
…and Trie A rolls 1, Trie C wins (2/4)
…and Trie A rolls 5, Trie A wins (2/5)
…and Trie A rolls 8, Trie A wins (2/6)
When Trie C rolls 9…
…and Trie A rolls 1, Trie C wins (3/7)
…and Trie A rolls 5, Trie C wins (4/8)
…and Trie A rolls 8, Trie C wins (5/9)
The same reasoning can be applied to the six-sided non-transitive dice, but there are 36 possible outcomes when two of the dice are thrown against each other, so I won’t list them.
Die A = (1, 2, 2, 5, 6, 6)
Die B = (1, 1, 4, 5, 5, 5)
Die C = (3, 3, 3, 3, 4, 6)
Elsewhere other-posted:
At the Mountains of Mathness
Simpson’s Paradox — a simple situation with a very weird outcome
# Live and Let Dice
How many ways are there to die? The answer is actually five, if by “die” you mean “roll a die” and by “rolled die” you mean “Platonic polyhedron”. The Platonic polyhedra are the solid shapes in which each polygonal face and each vertex (meeting-point of the edges) are the same. There are surprisingly few. Search as long and as far as you like: you’ll find only five of them in this or any other universe. The standard cubic die is the most familiar: each of its six faces is square and each of its eight vertices is the meeting-point of three edges. The other four Platonic polyhedra are the tetrahedron, with four triangular faces and four vertices; the octahedron, with eight triangular faces and six vertices; the dodecahedron, with twelve pentagonal faces and twenty vertices; and the icosahedron, with twenty triangular faces and twelve vertices. Note the symmetries of face- and vertex-number: the dodecahedron can be created inside the icosahedron, and vice versa. Similarly, the cube, or hexahedron, can be created inside the octahedron, and vice versa. The tetrahedron is self-spawning and pairs itself. Plato wrote about these shapes in his Timaeus (c. 360 B.C.) and based a mathemystical cosmology on them, which is why they are called the Platonic polyhedra.
Tetrahedron
Hexahedron
Octahedron
Dodecahedron
Icosahedron
They make good dice because they have no preferred way to fall: each face has the same relationship with the other faces and the centre of gravity, so no face is likelier to land uppermost. Or downmost, in the case of the tetrahedron, which is why it is the basis of the caltrop. This is a spiked weapon, used for many centuries, that always lands with a sharp point pointing upwards, ready to wound the feet of men and horses or damage tyres and tracks. The other four Platonic polyhedra don’t have a particular role in warfare, as far as I know, but all five might have a role in jurisprudence and might raise an interesting question about probability. Suppose, in some strange Tycholatric, or fortune-worshipping, nation, that one face of each Platonic die represents death. A criminal convicted of a serious offence has to choose one of the five dice. The die is then rolled f times, or as many times as it has faces. If the death-face is rolled, the criminal is executed; if not, he is imprisoned for life.
The question is: Which die should he choose to minimize, or maximize, his chance of getting the death-face? Or doesn’t it matter? After all, for each die, the odds of rolling the death-face are 1/f and the die is rolled f times. Each face of the tetrahedron has a 1/4 chance of being chosen, but the tetrahedron is rolled only four times. For the icosahedron, it’s a much smaller 1/20 chance, but the die is rolled twenty times. Well, it does matter which die is chosen. To see which offers the best odds, you have to raise the odds of not getting the death-face to the power of f, like this:
3/4 x 3/4 x 3/4 x 3/4 = 3/4 ^4 = 27/256 = 0·316…
5/6 ^6 = 15,625 / 46,656 = 0·335…
7/8 ^8 = 5,764,801 / 16,777,216 = 0·344…
11/12 ^12 = 3,138,428,376,721 / 8,916,100,448,256 = 0·352…
19/20 ^20 = 37,589,973,457,545,958,193,355,601 / 104,857,600,000,000,000,000,000,000 = 0·358…
Those represent the odds of avoiding the death-face. Criminals who want to avoid execution should choose the icosahedron. For the odds of rolling the death-face, simply subtract the avoidance-odds from 1, like this:
1 – 3/4 ^4 = 0·684…
1 – 5/6 ^6 = 0·665…
1 – 7/8 ^8 = 0·656…
1 – 11/12 ^12 = 0·648…
1 – 19/20 ^20 = 0·642…
So criminals who prefer execution to life-imprisonment should choose the tetrahedron. If the Tycholatric nation offers freedom to every criminal who rolls the same face of the die f times, then the tetrahedron is also clearly best. The odds of rolling a single specified face f times are 1/f ^f:
1/4 x 1/4 x 1/4 x 1/4 = 1/4^4 = 1 / 256
1/6^6 = 1 / 46,656
1/8^8 = 1 / 16,777,216
1/12^12 = 1 / 8,916,100,448,256
1/20^20 = 1 / 104,857,600,000,000,000,000,000,000
But there are f faces on each polyhedron, so the odds of rolling any face f times are 1/f ^(f-1). On average, of every sixty-four (256/4) criminals who choose to roll the tetrahedron, one will roll the same face four times and be reprieved. If a hundred criminals face the death-penalty each year and all choose to roll the tetrahedron, one criminal will be reprieved roughly every eight months. But if all criminals choose to roll the icosahedron and they have been rolling since the Big Bang, just under fourteen billion years ago, it is very, very, very unlikely that any have yet been reprieved.
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# Math Assignment Help With Decimal Numbers Ordering
## 2.2 Ordering
The method of placing numbers in increasing or decreasing format is called ordering.
2.2.1 Ordering of number
Numbers are arranged in either Ascending order or Descending order. In case of Ascending order numbers are placed from lowest to highest. The number being the lowest is placed first and the higher number is placed.
For eg: if we asked to place 12, 5, 10, 4, and 3 in ascending order then,
3,4,5,10,12
Similarly, in Descending order numbers are placed from highest to lowest. The number having highest value is placed first and the lower valued number is placed
For eg: taking the similar example, if we are asked to place 12, 5, 10, 4, and 3 in descending order then,
12, 10, 5, 4, 3
2.2.2 Ordering of Decimals
Ordering of decimal is not same as of integers, it can be confusing sometimes. If you take two numbers
0.24 And 0.234
You may say 0.234 is greater than 0.24 as it contains more number of digits. But this is wrong.
0.24 is greater than 0.234. Now we will explain how.
Let us follow a stepwise method.
Take few decimal numbers, say,
0.235, 0.456, 0.237, 0.346, 1.765
Step1: First compare the digit at Units place of all the given decimal numbers. One of them has 1 and rest have 0. So 1.765 is the highest. Put it aside.
Step2: now compare the tenth place of the remaining decimal numbers that is,
0.235, 0.456, 0.237, 0.346
We see 2, 3, and 4 at tenth place, since 4 is greater of them therefore 0.456 is the next highest, the decimals left are
0.235, 0.237, 0.346
Comparing the tenth place again we find 0.346 is the next highest. Here we see two decimals having the same digit at tenth place that is, 2 so we will have to compare the hundredth place as well.
Step3: compare the hundredth place of the remaining two decimals that is,
0.235,0.237
Here digit at hundredth place is also same so compare the thousandth place, 7>5
Therefore 0.237 is the next highest and 0.237 is the lowest.
Hence the correct order is
1.765, 0.456, 0.346, 0.237, 0.235
Note: In decimal number system decimals must be from highest to lowest.
### Email Based Assignment Help in Decimal Numbers Ordering
To Schedule a Decimal Numbers Ordering tutoring session
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# coulomb’s law with math
The interaction between charged objects is a non-contact force that acts over some distance of separation. Charge, charge and distance. Every electrical interaction involves a force that highlights the importance of these three variables. Whether it is a plastic golf tube attracting paper bits, two like-charged balloons repelling or a charged Styrofoam plate interacting with electrons in a piece of aluminum, there is always two charges and a distance between them as the three critical variables that influence the strength of the interaction. In this section of Lesson 3, we will explore the importance of these three variables.
### Force as a Vector Quantity
The electrical force, like all forces, is typically expressed using the unit Newton. Being a force, the strength of the electrical interaction is a vector quantity that has both magnitude and direction. The direction of the electrical force is dependent upon whether the charged objects are charged with like charge or opposite charge and upon their spatial orientation. By knowing the type of charge on the two objects, the direction of the force on either one of them can be determined with a little reasoning. In the diagram below, objects A and B have like charge causing them to repel each other. Thus, the force on object A is directed leftward (away from B) and the force on object B is directed rightward (away from A). On the other hand, objects C and D have opposite charge causing them to attract each other. Thus, the force on object C is directed rightward (toward object D) and the force on object D is directed leftward (toward object C). When it comes to the electrical force vector, perhaps the best way to determine the direction of it is to apply the fundamental rules of charge interaction (opposites attract and likes repel) using a little reasoning.
Electrical force also has a magnitude or strength. Like most types of forces, there are a variety of factors that influence the magnitude of the electrical force. Two like-charged balloons will repel each other and the strength of their repulsive force can be altered by changing three variables. First, the quantity of charge on one of the balloons will affect the strength of the repulsive force. The more charged a balloon is, the greater the repulsive force. Second, the quantity of charge on the second balloon will affect the strength of the repulsive force. Gently rub two balloons with animal fur and they repel a little. Rub the two balloons vigorously to impart more charge to both of them, and they repel a lot. Finally, the distance between the two balloons will have a significant and noticeable effect upon the repulsive force. The electrical force is strongest when the balloons are closest together. Decreasing the separation distance increases the force. The magnitude of the force and the distance between the two balloons is said to be inversely related.
### Coulomb’s Law Equation
The quantitative expression for the effect of these three variables on electric force is known as Coulomb’s law. Coulomb’s law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects. In equation form, Coulomb’s law can be stated as
where Q1 represents the quantity of charge on object 1 (in Coulombs), Q2 represents the quantity of charge on object 2 (in Coulombs), and d represents the distance of separation between the two objects (in meters). The symbol k is a proportionality constant known as the Coulomb’s law constant. The value of this constant is dependent upon the medium that the charged objects are immersed in. In the case of air, the value is approximately 9.0 x 109 N • m2 / C2. If the charged objects are present in water, the value of k can be reduced by as much as a factor of 80. It is worthwhile to point out that the units on k are such that when substituted into the equation the units on charge (Coulombs) and the units on distance (meters) will be canceled, leaving a Newton as the unit of force.
The Coulomb’s law equation provides an accurate description of the force between two objects whenever the objects act as point charges. A charged conducting sphere interacts with other charged objects as though all of its charge were located at its center. While the charge is uniformly spread across the surface of the sphere, the center of charge can be considered to be the center of the sphere. The sphere acts as a point charge with its excess charge located at its center. Since Coulomb’s law applies to point charges, the distance d in the equation is the distance between the centers of charge for both objects (not the distance between their nearest surfaces).
The symbols Q1 and Q2 in the Coulomb’s law equation represent the quantities of charge on the two interacting objects. Since an object can be charged positively or negatively, these quantities are often expressed as “+” or “-” values. The sign on the charge is simply representative of whether the object has an excess of electrons (a negatively charged object) or a shortage of electrons (a positively charged object). It might be tempting to utilize the “+” and “-” signs in the calculations of force. While the practice is not recommended, there is certainly no harm in doing so. When using the “+” and “-” signs in the calculation of force, the result will be that a “-” value for force is a sign of an attractive force and a “+” value for force signifies a repulsive force. Mathematically, the force value would be found to be positive when Q1 and Q2 are of like charge – either both “+” or both “-“. And the force value would be found to be negative when Q1 and Q2 are of opposite charge – one is “+” and the other is “-“. This is consistent with the concept that oppositely charged objects have an attractive interaction and like charged objects have a repulsive interaction. In the end, if you’re thinking conceptually (and not merely mathematically), you would be very able to determine the nature of the force – attractive or repulsive – without the use of “+” and “-” signs in the equation.
### Calculations Using Coulomb’s Law
In physics courses, Coulomb’s law is often used as a type of algebraic recipe to solve physics word problems. Three such examples are shown here.
Example ASuppose that two point charges, each with a charge of +1.00 Coulomb are separated by a distance of 1.00 meter. Determine the magnitude of the electrical force of repulsion between them.
This is not the most difficult mathematical problem that could be selected. It certainly was not chosen for its mathematical rigor. The problem-solving strategy utilized here may seem unnecessary given the simplicity of the given values. Nonetheless, the strategy will be used to illustrate its usefulness to any Coulomb’s law problem.
The first step of the strategy is the identification and listing of known information in variable form. Here we know the charges of the two objects (Q1 and Q2) and the separation distance between them (d). The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the force. So Felect is the unknown quantity. The results of the first two steps are shown in the table below.
Given: Q1 = 1.00 CQ2 = 1.00 C d = 1.00 m Find: Felect = ???
The next and final step of the strategy involves substituting known values into the Coulomb’s law equation and using proper algebraic steps to solve for the unknown information. This step is shown below.
Felect = k • Q1Q2 / d2 Felect = (9.0 x 109 N•m2/C2) • (1.00 C) • (1.00 C) / (1.00 m)2
Felect = 9.0 x 109 N
The force of repulsion of two +1.00 Coulomb charges held 1.00 meter apart is 9 billion Newton. This is an incredibly large force that compares in magnitude to the weight of more than 2000 jetliners.
This problem was chosen primarily for its conceptual message. Objects simply do not acquire charges on the order of 1.00 Coulomb. In fact, more likely Q values are on the order of 10-9 or possibly 10-6 Coulombs. For this reason, a Greek prefix is often used in front of the Coulomb as a unit of charge. Charge is often expressed in units of microCoulomb (µC) and nanoCoulomb (nC). If a problem states the charge in these units, it is advisable to first convert to Coulombs prior to substitution into the Coulomb’s law equation. The following unit equivalencies will assist in such conversions.
1 Coulomb = 106 microCoulomb 1 Coulomb = 109 nanoCoulomb
The problem-solving strategy used in Example A included three steps:
1. Identify and list known information in variable form.
2. List the unknown (or desired) information in variable form.
3. Substitute known values into the Coulomb’s law equation and using proper algebraic steps to solve for the unknown information. (In some cases and for some students, it might be easier to first do the algebra using the variables and then perform the substitution as the last step.)
This same problem-solving strategy is demonstrated in Example B below.
Example BTwo balloons are charged with an identical quantity and type of charge: -6.25 nC. They are held apart at a separation distance of 61.7 cm. Determine the magnitude of the electrical force of repulsion between them.
The problem states the value of Q1 and Q2. Since these values are expressed in units of nanoCoulombs (nC), the conversion to Coulombs must be made. The problem also states the separation distance (d). Since distance is given in units of centimeters (cm), the conversion to meters must also be made. These conversions are required since the units of charge and distance in the Coulomb’s constant are Coulombs and meters. The unknown quantity is the electrical force (F). The results of the first two steps are shown in the table below.
Given: Q1 = -6.25 nC = -6.25 x 10-9 CQ2 = -6.25 nC = -6.25 x 10-9 C d = 61.7 cm = 0.617 m Find: Felect = ???
The final step of the strategy involves substituting known values into the Coulomb’s law equation and using proper algebraic steps to solve for the unknown information. This substitution and algebra is shown below.
Felect = k • Q1Q2 / d2 Felect = (9.0 x 109 N•m2/C2) • (6.25 x 10-9 C) • (6.25 x 10-9 C) / (0.617 m)2
Felect = 9.23 x 10-7 N
Note that the “-” sign was dropped from the Q1 and Q2 values prior to substitution into the Coulomb’s law equation. As mentioned above, the use of “+” and “-” signs in the equation would result in a positive force value if Q1 and Q2 are like charged and a negative force value if Q1 and Q2 are oppositely charged. The resulting “+” and “-” signs on F signifies whether the force is attractive (a “-” F value) or repulsive (a “+” F value).
Example CTwo balloons with charges of +3.37 µC and -8.21 µC attract each other with a force of 0.0626 Newton. Determine the separation distance between the two balloons.
The problem states the value of Q1 and Q2. Since these values are in units of microCoulombs (µC), the conversion to Coulombs will be made. The problem also states the electrical force (F). The unknown quantity is the separation distance (d). The results of the first two steps are shown in the table below.
Given: Q1 = +3.37 µC = +3.37 x 10-6 CQ2 = -8.21 µC = -8.21 x 10-6 C Felect = -0.0626 N (use a – force value since it is attractive) Find: d = ???
As mentioned above, the use of the “+” and “-” signs is optional. However, if they are used, then they have to be used consistently for the Q values and the F values. Their use in the equation is illustrated in this problem.
The final step of the strategy involves substituting known values into the Coulomb’s law equation and using proper algebraic steps to solve for the unknown information. In this case, the algebra is done first and the substitution is performed last. This algebra and substitution is shown below.
Felect = k • Q1Q2 / d2 d2 • Felect = k • Q1Q2
d2 = k • Q1Q2 / Felect
d = SQRT(k • Q1Q2) / Felect
d = SQRT [(9.0 x 109 N•m2/C2) • (-8.21 x 10-6 C) • (+3.37 x 10-6 C) / (-0.0626 N)]
d = Sqrt [ +3.98 m2 ]
d = +1.99 m
### Comparing Electrical and Gravitational Forces
Electrical force and gravitational force are the two non-contact forces discussed in The Physics Classroom tutorial. Coulomb’s law equation for electrical force bears a strong resemblance to Newton’s equation for universal gravitation.
The two equations have a very similar form. Both equations show an inverse square relationship between force and separation distance. And both equations show that the force is proportional to the product of the quantity that causes the force – charge in the case of electrical force and mass in the case of gravitational force. Yet there are some striking differences between these two forces. First, a comparison of the proportionality constants – k versus G – reveals that the Coulomb’s law constant (k) is significantly greater than Newton’s universal gravitation constant (G). Subsequently a unit of charge will attract a unit of charge with significantly more force than a unit of mass will attract a unit of mass. Second, gravitational forces are only attractive; electrical forces can be either attractive or repulsive.
The inverse square relationship between force and distance that is woven into the equation is common to both non-contact forces. This relationship highlights the importance of separation distance when it comes to the electrical force between charged objects. It is the focus of the next section of Lesson 3.
### We Would Like to Suggest …
Sometimes it isn’t enough to just read about it. You have to interact with it! And that’s exactly what you do when you use one of The Physics Classroom’s Interactives. We would like to suggest that you combine the reading of this page with the use of our Coulomb’s Law Interactive. You can find it in the Physics Interactives section of our website. The Coulomb’s Law Interactive allows a learner to explore the effect of charge and separation distance upon the amount of electric force between two charged objects.
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# Name Domain and Range of Functions
Contributor: Ashley Nail. Lesson ID: 13666
Do you know how to find the domain and range of functions? Look at the graph or table, and you will quickly see the relationship between sets of numbers. Find out how!
categories
## Algebra, Algebra I
subject
Math
learning style
Kinesthetic, Visual
personality style
Lion, Otter, Beaver, Golden Retriever
High School (9-12)
Lesson Type
Quick Query
## Lesson Plan - Get It!
Audio:
Jenna is a mechanic, and she needs to order tires for her shop. She wants to make a set of numbers that she can look at any time to determine how many tires to order each week.
• How can we use our knowledge of functions, domains, and ranges to help Jenna?
A function is the relationship between the set of numbers in a domain to the set of numbers in a range.
We also know that each member of the domain relates to exactly one member of the range. In other words, a member of the domain cannot be connected to two values in the range.
This is why the vertical line test proves functions.
If you were to input values into a function and gather a set of output values, you would have a domain and range. You could put those values in a table, then make ordered pairs, and lastly graph the sets of numbers.
As long as you could run a vertical line left to right across the graph and as long as that line never hits more than one point at a time, you have a function!
If you need more review, visit our lesson found under the Additional Resources in the right-hand sidebar.
Now, it is time to learn how to name domain and range sets by looking at these tables, graphs, and equations.
Let's take the example from above:
After inputting values to the function, we have a set of output numbers. These two sets make up the domain and range of this function.
Each member of the domain relates to exactly one member of the range. Every time you input 4, you will always get 9 and no other value.
Now, we can organize this data in a table and then make ordered pairs:
If the ordered pairs and the table were the only data we had, we could name the domain and range.
The domain is the set of inputs. We notate these sets using brackets { } to contain the numbers:
Domain: {1, 4, 6, 7, 10}
The range is the set of outputs. We notate those the same way:
Range: {3, 9, 13, 15, 21}
Notice that the domain and range sets are always listed from least to greatest.
However, with this function, we have more information. We have the equation f(x) = 2x +1.
That means these are not complete domain and range sets. Let's look at the graph.
If we graph the ordered pairs, our coordinate plane looks like this:
But we know we can input even more values into this function.
For example, we could input -2 and get the output value of -3. This means -2 is also a member of the domain, and -3 is a member of the range.
Go ahead and input more values into this function.
• What do you quickly notice?
We could input values all day long. We could input positive and negative integers. We could input fractions and decimals and continue to get output values.
This function makes a linear graph:
Now, we need to update our domain and range.
Looking at this graph, we notice the x-value, or input, can be all real numbers. Therefore, the domain is "x such that x is a member of all real numbers."
Looking at the graph, we also notice the y-value, or output, can be all real numbers:
Let's look at another graph:
If you look at the x-values, the graph starts at 1 and continues on forever, including all numbers greater than or equal to 1.
This means the domain for this graph is {x | x ≥ 1}
If you look at the y-values, the graph starts at -4 and continues on forever, including all numbers greater than or equal to -4.
This means the range for this graph is {y | y ≥ -4}.
• Ready to practice naming the domain and range sets for functions?
Click NEXT to visit the Got It? section to practice naming functions using tables and graphs.
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# Converting Metric Measurement in Word Problems
(27 ratings )
Do you need extra help for EL students? Try the Metric Conversions ConversationsPre-lesson.
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Do you need extra help for EL students? Try the Metric Conversions ConversationsPre-lesson.
Students will be able to solve multi-step word problems that require converting metric units (involving whole numbers and decimals).
The adjustment to the whole group lesson is a modification to differentiate for children who are English learners.
(5 minutes)
• Pass out metric rulers, one per student.
• Direct students to examine their rulers and note the size of the different units.
• Prompt students to share with a partner the names for any units they know (like centimeters and millimeters) and examples of items they could measure with those units.
• Pose the following question to your class: "How might you measure something using the metric system that is longer than a ruler, like their desk, the height of the door, or the distance to the grocery store?"
(10 minutes)
• Direct students to think about our money system and how the units are related. Draw a chart on the board, from right to left with the headings: "100 dollars," "10 dollars," "1 dollar," "one dime," and "one penny" across the top. Under each heading, in the first row, draw and name the coin or bill used for each. Then start a new row and write the value of a dollar (1 penny = .01 dollar; 1 dime = .10 dollar; etc.).
• Note how each value is related to the ones to the right and left of it. Point out that it is 1/10 of the value to the left and 10 times the value of the one to its right. Define this kind of a system as a Base tenSystem.
• Explain that the metric system is also a base ten system, just like our money system. Rather than the unit being a dollar, the metric system uses meters, grams, or liters.
(20 minutes)
• Distribute a copy of the Preparing to Convert Metric Measurement Using Deconstruction worksheet to each student. Review the chart at the top of the page. This sheet will help students think about the different metric units and how they are related.
• Have students work in pairs or small groups to collaborate on Part A. Encourage your class to discuss and explore, and use rulers to assist them.
• Now, go over the example in Part B and model how the measurement is converted from centimeters to meters using deconstruction (they look like number bonds).
• Instruct students to use the model as a guide and try number 3 on their own.
• Review and discuss.
(20 minutes)
• Give students about 15 minutes to work through example number 4.
• Review and discuss the solution. Ask students if they have discovered any other ways to convert metric units. Do they have any shortcuts?
• Show students the method of moving the decimal using the chart on the page. Locate the original units and the units the value is to be converted into on the grid. Note how many units you move from one to the other, and in which direction. This tells you how far to move the decimal and in what direction.
• Distribute the Metric Length Measurement: Word Problems worksheet for students to independently practise converting measurements.
Support:
• View the short instructional videos in the list of supplemental media at the end of this lesson.
• Do some of the Metric Length Measurement problems as a class or small group before independent practise, modeling the movement of the decimal.
Enrichment:
• Have students research to determine what units in the metric system are smaller and larger than the units discussed in class (micro, nano, pico, mega, giga, tera, etc.).
• Have students convert other metric measurements like mass and volume in the Metric Mass and Volume Measurement in Word Problems practise sheet.
(5 minutes)
• Write a metric conversion problem on the board. Have students solve it and write their solution on a scratch paper at their seat. Then you can quickly circulate to check answers. Students who get it wrong should be encouraged to try again. Students who get it right should create a problem of their own and trade with a friend.
(10 minutes)
• Discuss the "Questions for Discussion" at the end of the Preparing to Convert Metric Measurement Using Deconstruction worksheet.
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# 2020 CIME II Problems/Problem 3
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
In a jar there are blue jelly beans and green jelly beans. Then, $15\%$ of the blue jelly beans are removed and $40\%$ of the green jelly beans are removed. If afterwards the total number of jelly beans is $80\%$ of the original number of jelly beans, then determine the percent of the remaining jelly beans that are blue.
## Solution 1
Suppose there are $x$ jelly beans total at the beginning. Suppose further that there are $b$ blue jelly beans and $x-b$ green jelly beans. Then, after the removal, there will be $0.85b$ blue jelly beans and $0.6x-0.6b$ green jelly beans. Because the total number of jelly beans at the end is $80\%$ of the starting number, we can create an equation: $$0.6x+0.25b=0.8x$$ $$0.2x=0.25b$$ $$0.8x=b$$ This tells us there were originally $0.8x$ blue jelly beans and $0.2x$ green jelly beans at the beginning, so now there must be $0.68x$ blue and $0.12x$ green. The percent of the remaining jelly beans that are blue is $$\frac{0.68x}{0.68x+0.12x}=\frac{68}{80}=\frac{85}{100},$$ so the answer is $\boxed{085}$.
|
# How do you differentiate sin^2(x/6)?
Aug 19, 2015
${y}^{'} = \frac{1}{3} \cdot \sin \left(\frac{x}{6}\right) \cdot \cos \left(\frac{x}{6}\right)$
#### Explanation:
You can differentiate this function
$y = \sin \left(\frac{x}{6}\right)$
by using the chain rule twice, once for ${u}_{1}^{2}$, with ${u}_{1} = \sin \left(\frac{x}{6}\right)$, and once for $\sin {u}_{2}$, with ${u}_{2} = \frac{x}{6}$.
Your target derivative will be
$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{d}{{\mathrm{du}}_{1}} \left({u}_{1}^{2}\right) \cdot \frac{d}{\mathrm{dx}} \left({u}_{1}\right)$, with
$\frac{d}{\mathrm{dx}} {u}_{1} = \frac{d}{{\mathrm{du}}_{2}} \cdot \sin {u}_{2} \cdot \frac{d}{\mathrm{dx}} \left({u}_{2}\right)$
This will give you
$\frac{d}{\mathrm{dx}} \left({u}_{1}\right) = \cos {u}_{2} \cdot \frac{d}{\mathrm{dx}} \left(\frac{x}{6}\right)$
$\frac{d}{\mathrm{dx}} \left(\sin \left(\frac{x}{6}\right)\right) = \cos \left(\frac{x}{6}\right) \cdot \frac{1}{6}$
Plug this back into your target derivative to get
${y}^{'} = 2 {u}_{1} \cdot \frac{1}{6} \cdot \cos \left(\frac{x}{6}\right)$
${y}^{'} = \textcolor{g r e e n}{\frac{1}{3} \cdot \sin \left(\frac{x}{6}\right) \cdot \cos \left(\frac{x}{6}\right)}$
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# A box contains cards numbered 3,5,7,9,……35,37. A card is drawn at random from the box. Find the probability that the number on the drawn card is a prime number.
Last updated date: 17th Apr 2024
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Hint: Start with determining how many numbers are there in 3,5,7,9,……35,37 for total number of favorable outcomes. And also how many prime numbers it exhibits.
According to the question, the numbers on the cards represent 3,5,7,9,……35,37. It consists of all the odd numbers up to 37 except 1.
So, total number of cards $= \dfrac{{37 - 3}}{2} + 1 = 17 + 1 = 18$
Therefore, the total number of possible outcomes $= 18$
Now in these, the cards representing prime numbers will be of number 3,5,7,11,13,17,19,23,29,31,37. Thus, there are 18 of these cards.
Therefore the number of favorable outcomes $= 11$.
Let $E$is the event representing the drawing of a prime numbered card. Then the probability is:
$\Rightarrow P\left( E \right) = \dfrac{{{\text{No}}{\text{. of favorable outcomes}}}}{{{\text{Total number of possible outcomes}}}}, \\ \Rightarrow P\left( E \right) = \dfrac{{11}}{{18}}. \\$
Thus, the required probability is $\dfrac{{11}}{{18}}$.
Note: Probability represents the chance of an event to occur. For example in above, the probability is $\dfrac{{11}}{{18}}$. This means that out of 18 trials of drawing the card, there is a chance that 11 of them comes out with a prime number.
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sss Similarity
"SSS" stands for "side-side-side." But what exactly does this mean in the context of triangles? As we''ll soon find out, this represents a special rule we can use to help us determine if two triangles are congruent. But how exactly does this special rule work? What can it teach us about math?
Side-side-side similarity explained
Side-side-side similarity states that:
• When two triangles have corresponding sides of equal length, then we know these triangles are congruent.
Visualizing side-side-side similarity
We can visualize side-side-side congruence more easily by examining the following diagram:
As we can see, these triangles have sides that are shaded in corresponding colors. We know that because all of these sides are shaded with similar colors, they have the same lengths. This means that these two triangles are also congruent.
For triangle similarity, we have a different rule called Side-Side-Side Similarity:
When two triangles have corresponding sides that are proportional then we know these triangles are similar.
In the context of triangles, these ratios are slightly more complex.
For example, we might get two triangles:
One might have side lengths 17, 11.6, and 18.4
The other might have side lengths 8.5, 5.8, and 9.2
These two triangles are similar because their lengths are proportional. Why? Because we can prove that they are proportional with the following formula:
AB/PQ = BC/QR = AC/PR
In this formula, one triangle is ABC while the other is PQR.
Let''s plug in those values:
17/8.5 = 11.6/5.8 = 18.4/9.2
Are they proportional?
Yes, because each fraction can be simplified as 2, and they are all equivalent (or proportional) to one another.
Finding missing values
In some cases, it is possible to find out whether triangles are similar even if you are missing the value of one side. For example, let''s say we have two right triangles with two given sides each. We can use the Pythagorean theorem
(a^2 + b^2 = c^2)
to find the third side and then determine whether the triangles are similar. We also know by the side-angle-side rule that when two sides of a triangle and their included angles are fixed, two proportional sides must also mean that the third side is automatically proportional as well.
Topics related to the sss Similarity
Triangles
Triangle Proportionality Theorem
sss Postulate
Flashcards covering the sss Similarity
Common Core: High School - Geometry Flashcards
Practice tests covering the sss Similarity
Common Core: High School - Geometry Diagnostic Tests
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You are on page 1of 21
# 1
Chapter 2: Circles and Lines
Lesson 2.1: The Rectangular Coordinate System
Each of the pairs of numbers () ( ) and () is an example of an ordered
pair; that is, a pair of numbers written within parentheses in which the order of the numbers is
important. The two numbers are the components of the ordered pair. an ordered pair is
graphed using two real number lines that intersect at right angles at the zero points, as shown
in Figure 2.1. The common zero point is called the origin. The horizontal line, the ,
represents the second. The and the make up a rectangular coordinate
system, or the . The axes form four quadrants, numbered I, II, III, and IV as
shown in Figure 2.1. (A point on an axis is not considered to be in any of the four quadrants.)
Figure 2.1
We locate, or plot, the point on the coordinate system that corresponds to the ordered
pair () by going one unit from zero to the right alone the axis, and then two units up
parallel to the axis. The phrase the point corresponding to the ordered pair () often is
abbreviated the point (). The numbers in an ordered pair are called the coordinates of the
corresponding point. In particular, the first coordinate is referred to as the ,
and the second coordinate is referred to as the . If the point () is labeled
as point , we often write (). The set of all ordered pairs of real numbers is denoted by
; that is
*( ) +.
The parentheses used to represent an ordered pair are also used to represent an open
interval (introduced in an earlier chapter). In general, there should be no confusion between
these symbols because the context of the discussion tells us whether we are discussing ordered
pair or open intervals.
2
The French philosopher Ren Descartes (1595 1650) is the person usually credited
with the invention of the rectangular coordinate system. In honor of Descartes, the rectangular
coordinate system is often referred to as the Cartesian Coordinate System. As a philosopher,
Descartes is responsible for the statement I think, therefore I am. Until Descartes invented his
coordinate system in 1637, algebra and geometry were treated as separate subjects. The
rectangular (or Cartesian) coordinate system allows us to connect algebra and geometry by
associating geometric shapes with algebraic equations. In this sense, the Cartesian coordinate
system is being superimposed on the Euclidean plane so that the definitions, axioms and the
theorems of Euclidean geometry apply.
Figure 2.2
Suppose that we wish to find the distance || between two points (
) and
(
). The Pythagorean Theorem allows us to do this. See figure 2.2. Let be the point with
coordinates (
## ). It is easy to see that || |
| and || |
|. By
Pythagorean Theorem, we have
||
||
||
.
We have just proven the following theorem.
THEOREM 2.1.1 DISTANCE FORMULA
The distance between the points (
) and (
) is given by
|| (
EXAMPLE 2.1.2 Determine whether the points () () () form an isosceles
triangle.
EXAMPLE 2.1.3 Prove that the points () ( ) () are vertices of a right
triangle.
EXAMPLE 2.1.4 Show that the points ( ) () () are collinear.
3
The midpoint of a line segment is the point on the segment that is equidistant from
both endpoints. Given the coordinates of two endpoints of a line segment, it is not difficult to
find the coordinates of the midpoint of the segment.
THEOREM 2.1.5 MIDPOINT FORMULA
The midpoint of the line segment with endpoints (
) and (
) is ( ), where
and
Proof: See Figure 2.3. Let be the midpoint, the midpoint with coordinates (
), the
point with coordinates (
## ), and the point with coordinates (
). Since line is
parallel to line and is the midpoint of the line segment , is the midpoint of
the line segment . Thus, we have
## . Similarly, can be computed.
Figure 2.3
One can also prove Theorem 2.1.5 by noting that, by ASA Congruence Theorem,
.
EXAMPLE 2.1.6 One endpoint of a line segment is () and its midpoint is (). Find the
other endpoint.
EXAMPLE 2.1.7 Find the coordinates of the points that divide the line segment joining the
points ( ) and () into four equal parts.
EXAMPLE 2.1.8 The line segment
## is extended an equal length in both directions so that
the length of the resulting segment is three times that of the original length. Find the
coordinates of the endpoints
and
## of the extended line segment if
( ) and
().
EXAMPLE 2.1.9 If the point () divides the line segment from () to ( ) such that
||
||
## , find the coordinates of point .
(
)
( )
(
)
4
EXAMPLE 2.1.10 Prove that the diagonals of a parallelogram bisect each other.
5
EXERCISES 2.1
In exercises 1 to 2, determine in which quadrant or on what coordinates axis each point lies.
Describes the distance of each point from the and the .
1.
()
()
()
()
()
()
()
()
2.
()
()
()
()
()
()
()
()
In exercises 3 to 14, find the distance between and and find the midpoint of the line
segment from to .
3. () and ()
4. ( ) and ()
5. () and ()
6. ( ) and( )
7. (
) and (
)
8. (
) and (
)
9. ( ) and ( )
10. ( ) and ( )
11. ( ) and ( )
12. ( ) and ( )
13. ( ) and ()
14. () and ( )
15. Find the lengths of the medians of the triangle having vertices () () and
( ).
16. Find the length of the medians of the triangle having the vertices () ( ) and
( ).
17. Determine whether the following points are vertices of a right triangle: () () and
()
18. Determine whether the following points are vertices of a right triangle: ( ) ()
and ( )
19. Determine whether ( ) is on the line segment joining () and ( ).
20. Determine whether ( ) is on the line segment joining ( ) and( )
21. Determine whether the points ( ) () and ( ) are collinear.
22. Determine whether the points () ( ) and () are collinear.
23. If the point ( ) is equidistant from ( ) and (), find .
24. Find the point on the equidistant from ( ) and ().
25. Find the point on the equidistant from ( ) and ().
26. If the point ( ) is equidistant from () and (), find .
27. One endpoint of a line segment is () and its midpoint is (). Find the other point.
6
28. One endpoint of a line segment is () and its midpoint is (
## ). Find the other
endpoint.
29. Show that the triangle with vertices ( ) () and () is isosceles. Find its
area.
30. Show that the triangle with points () () ( ) and ( ) are vertices
of a square. Find its area.
31. Without using the distance formula, find the points and that divide into
congruent segments if the points and are () and ().
32. Find the coordinates of a point which is on the line segment joining the points
( )
and
## ( ) and is three times as far from
as it is from
.
33. Find the point(s) on the line through
() and
() such that
.
34. Let be a point on the line passing through () and () but not on the segment
. Find the coordinates of if it is twice as far from as from .
35. Given ( ) and (), find the point
## of the way from to .
36. Given () and ( ), find the point
## of the way from to .
37. If the point (
point
## of the way from to , find the coordinates of the point
38. If the point () divides the line segment from ( ) to ( ) such that
|
|
|
,
find the coordinates of point .
7
Lesson 2.2 Circles
An application of the Distance Formula (Theorem 2.1.1) leads to one of the most
familiar shapes in geometry. A circle is the set of all points in a plane that lie a fixed distance
from a fixed point. The fixed point is called the center, and the fixed distance is called the
Figure 2.5
Let ( ) be a point on a circle with radius and center at ( ). See figure 2.5. Then,
by the Distance Formula, we have ( )
( )
.
Squaring both sides gives the following theorem.
THEOREM 2.2.1
The equation of a circle of radius with center at ( ) is
( )
( )
In particular, a circle of radius with center at the origin has equation
The first equation in Theorem 2.2.1 is the center radius form or the standard form of
the equation of a circle.
A diameter of a circle is a line segment joining two points of the circle and passing
through the center.
EXAMPLE 2.2.2 Fine an equation of a circle whose diameter has endpoints at
( ) and
().
Solution: The center of the circle is the midpoint of the given diameter. If ( ) is the
center of the circle, then
and
.
8
Thus, the center is at (). The radius of the circle is equal to |
| (or |
|
or
| ).
|
| ( )
( )
Using Theorem 2.2.1, an equation of the circle is
( )
( )
()
Or, simply, ( )
( )
The equation in Example 2.2.2 can also be written in another form by expanding the
squared expressions:
( )
( )
) (
The last equation is the so called general form of the equation of a circle. Every
equation of a circle can be written in the general form
Where are constants, . Note that we can assume that
EXAMPLE 2.2.3 Determine the center and the radius of the circle whose equation is
EXAMPLE 2.2.4 Find all values of such that the circle whose equation is
is tangent to the .
Every circle has an equation in general form. However, the graph of an equation of the
form
Where are constants, is not necessarily a circle. We can explain this statement by
completing the squares, and writing the above equation in the form
( )
( )
Here, the value of will either be positive, zero, or negative.
(i) The graph is a circle if and only if
(ii) The graph is a single point ( ) if and only if
(iii) The graph is an empty set if and only if
EXAMPLE 2.2.5 Determine whether the graph of the following equation is a circle, a point, or an
empty set:
EXAMPLE 2.2.6 Find the values of so that the graph of the equation
Is a circle, a point, or an empty set.
9
EXERCISES 2.2
In exercises 1 6, find the center and the radius of the circle. Then sketch a graph of the circle.
1.
2.
3.
( )
4. (
5. ( )
( )
6. ( )
( )
In exercises 7 12, find the equation of the circle with center at and radius . Write the
equation in both center radius and general forms.
7. ()
8. ()
9. ( )
10. ()
11. ( )
12. ()
In exercises 13 18, find the center and the radius of the circle.
13.
14.
15.
16.
17.
18.
In exercises 19 23, determine whether the graph is a circle, a point, or the empty set. It is a
circle; find its center and radius.
19.
20.
21.
22.
23.
In exercises 24 32, find the equation of the circle(s) satisfying the given conditions.
24. center at ( ) and passes through the point ().
25. center at () and passes through the point ( ).
26. having a diameter whose endpoints are () ().
27. having a diameter whose endpoints are ( ) ().
28. center is at ( ) the midpoint of the segment joining the center and one of the points
on the circle is ()
29. passing through ( ) with (
## of the way from the center
.
10
30. through the points () () ( )
31. through the points () () ()
32. radius is and passing through the points () ()
33. The circle
## passes through the point (). Find the value of , the
radius, and the center of the circle.
34. For what values of and will
## be the equation of circle having
center at ()? Find the radius of this circle.
35. Determine the value(s) of so that the graph of
(a) is a circle.
(b) is a point.
(c) is an empty set.
11
Lesson 2.3 Lines
Two points determine a line. This familiar axiom from high school geometry tells us
that all we need in order to draw a line are two distinct points.
A line also can be determined by a point on the line and some measure of the
steepness of the line. This method can determine a line because, from the given point, the
measure of steepness can determine another point on the line. One way to get a measure of
the steepness of a line is to compare the vertical change in the line (the rise) to the horizontal
change (the run) while moving along the line from one fixed point to another.
Figure 2.7
Suppose that
) and
## ) are two different points on a line . Let be the
point with coordinates (
to
, the value
changes from
to
## , the value of changes from
to
by the amount
. Let
We show that the value of is not affected by the choice of points on . Let
) and
## the point with coordinates (
), and
compute
Since
, we have
|
|
|
|
Thus, the value of is the same number no matter what two points on are selected. This
constant is called the slope of the line.
12
DEFINITION 2.3.1
Consider a line that is not parallel to the . Let
) and
) be two
distinct points on the line. The slope of the line is given by
The slope of a line parallel to the (that is, a vertical line) is not defined.
Figure 2.8
From Definition 2.3.1, a line with a positive slope goes up (rises) from left to right, while
a line with negative slope goes down (falls) from left to right. A horizontal line has zero slope.
Figure 2.8 shows lines of positive, zero, negative, and undefined slope.
EXAMPLE 2.3.2
The slope of a line segment is
## and one endpoint is (). If the other endpoint is on
the what are its coordinates?
EXAMPLE 2.3.3
Find the value(s) of so that the points ( ) ( ) ( ) are collinear.
By an equation of a line we mean an equation that is satisfied by those, and only those, points
on the line.
To find an equation of the non vertical line that passes through a certain point
## ) and has a certain slope , consider a point ( ) on the line,
. The slope
determined by and
## must be the same as the prescribed slope ; that is
Clearing the denominator provides an equation in and , which is called the point slope
form of an equation of the line. The line through (
## ) of slope is described by the
equation
)
13
If two points on a line are known, it is possible to find an equation of the line. First, find
the slope using the slope formula, and then use the slope and one of the given points in the
point slope form.
EXAMPLE 2.3.4
Find an equation of the line passing through the points () and ( )
EXAMPLE 2.3.5
Find an equation of the line passing through the midpoint of the line segment joining
the points () and (), and whose slope is half the slope of the line through and .
.)
If, in the point slope form, we choose the particular point ( ) (that is, the point
where the line intersects the ) for the point (
), we have
( ) or .
The number , the coordinate of the point where the line intersects the , is the
of the line. Consequently, the equation is called the
slope intercept form of an equation of the line. This form is especially useful because
(1) it enables us to find the slope of a line from its equation, and
(2) it expresses the - coordinate of a point on the line explicitly in terms of its
coordinate.
Every equation of a non vertical line has a unique (one and only one) slope intercept form.
EXAMPLE 2.3.6
Find the slope of the line having the equation .
We similarly define the of a (non horizontal) line as the - coordinate
of the point where the line intersects the .
EXAMPLE 2.3.7
The - intercept of a line is three times its intercept. With the coordinate axes, the
line forms a triangle of area two in the first quadrant. Find an equation of the line in slope
intercept form.
EXAMPLE 2.3.8
Determine the value(s) of in the equation so that this line will form a
right triangle with the coordinate axes whose area is square units.
The horizontal line that intersects the - axis at the point ( ) ahs slope .
Therefore, from the slope intercept form, an equation of this line is . However, because
the slope of a vertical line is not defined, we cannot apply the point slope form to obtain its
equation. The vertical line that intersects the - axis at the point ( ) contains those and only
14
those points having the same coordinate. Thus, ( ) is any point on this vertical line if
and only if .
If a line is not vertical, then it has an equation of the form , which can be
written as . If a line is vertical, then it has an equation of the form , which
can also be written as . Each of the above equations is a special case of an equation
of the form , where are constants, and are not both
zero. It implies that every line has an equation of the form . The following
theorem tells us that the converse of this fact is also true.
THEOREM 2.3.9
The equation where are constants, and are not
both zero, is an equation of a line.
Proof: We consider two cases: and .
CASE 1: Suppose , then . The equation becomes , which
gives
,
an equation of a vertical line with - intercept
.
CASE 2: Suppose , then the equation can be written into
which is an equation of a line in slope intercept form.
Therefore, in either cases, the equation gives an equation of a line.
The equation given in Theorem 2.3.9 is the general form of an equation of a line.
We now discuss two applications of the concept of slope in geometry: parallel lines and
perpendicular lines.
In Euclidean geometry, we say that two lines are parallel if they never intersect. Two
distinct vertical lines are always parallel to each other. The following theorem tells us when two
distinct non vertical lines are parallel.
THEOREM 2.3.10
Let
## be two distinct non vertical lines with slopes
, respectively.
Then
is parallel to
if and only if
.
Proof: Let the respective equations of
and
be
and
. Let
) and
## ). Refer to Figure 2.9.
15
Since
and
are non vertical, they intersect with vertical line . Let the line
intersect
at
) and
at
).
The two lines
) (
)
or, simply
Thus,
## are parallel if and only if
.
Figure 2.9
In Euclidean geometry, two lines are perpendicular if they meet at a right angle. A
vertical line and horizontal line are perpendicular to each other. The following theorem tells us
when two distinct non vertical lines are perpendicular.
THEOREM 2.3.11
Let
## be two distinct non vertical lines with slopes
, respectively.
Then
is perpendicular to
if and only if
Figure 2.10
16
Proof: The two lines
## must intersect; otherwise, they are parallel and
.
See Figure 2.10.
We can choose the coordinate axes so that
## intersect at the origin . Then
has an equation of the form
, and
has
.
Since
are non vertical, they intersect with the vertical line . Let the
intersect
at
) and
at
).
The lines
## is right angled at That
is, by the Pythagorean Theorem,
## are perpendicular if and only if
|
.
Applying the Distance Formula (Theorem 2.1.1), we get
|
( )
( )
And
|
( )
Therefore,
## are perpendicular if and only if
) (
)
Or, simply,
.
Because
is equivalent to
and
,
Theorem 2.3.11 states that two non vertical lines are perpendicular if and only if the slope of
one of them is the negative reciprocal of the slope of the other.
EXAMPLE 2.3.12
Given the line having the equation , determine the general form of
an equation of the line through () and
(a) parallel to , and
(b) perpendicular to .
17
EXAMPLE 2.3.13
Find the point in the third quadrant so that the points ( ) () () and
are vertices of a parallelogram.
A tangent line to a circle is a line that intersects the circle at exactly one point, called the
point of tangency. The radius drawn to the point of tangency is perpendicular to the tangent
line. If a line is tangent to a circle, we sometimes say that the circle is tangent to the line.
EXAMPLE 2.3.14
Find the equations (in general form) of the tangent lines to the circle
## at the points of the circle on the
EXAMPLE 2.3.15
Find the equation (in general form) of the circle tangent to the line at
() and whose center in on the .
Before we proceed to the succeeding examples, we need to recall some concepts and
results about circles from high school geometry.
1. A chord of a circle is a line segment joining any two points of the circle. Thus, a
diameter is an example of a chord.
2. A perpendicular bisector of a chord is the line perpendicular to the chord and passing
through the midpoint of the chord. The perpendicular bisector of a chord passes
through the center of the circle.
3. The line through the midpoint of a chord and the center of the circle is perpendicular
to the chord.
4. The line perpendicular to a chord and passing through the center bisects the chord.
5. Three non collinear points determine exactly one circle.
EXAMPLE 2.3.16
Find an equation of the circle containing the point () and tangent to the line
at ().
EXAMPLE 2.3.17
Write an equation of the circle (in general form) containing the points
()
() and
().
18
EXERCISES 2.3
In exercises 1 to 14, find the slope of the line passing through the given points.
1. () ( )
2. () ( )
3. () ()
4. () ()
5. () ( )
6. () ()
7. () ()
8. ( ) ( )
9. ( ) ( )
10.
) (
)
11. (
) (
)
12. (
) (
)
13. () ()
14. () ()
In exercises 15 to 53, find the equation(s) of the lines satisfying the given conditions.
15. slope is and the through the point ().
16. slope is
17. slope is
## and the through the point (
).
18. slope is and through the point (
).
19. slope is and through the point ().
20. slope is and through the point (
).
21. slope is undefined and through the point ( ).
22. slope is undefined through the point (
).
23. passing through the points ( ) and ( ).
24. passing through the points () and ().
25. passing through the points (
) and (
).
26. passing through the points ( ) and ( ).
27. passing through the point () and parallel to the .
28. passing through the point () and parallel to the .
29. passing through the point () and perpendicular to the .
30. passing through the point () and perpendicular to the
31. passing through the point () and parallel to the line
19
32. passing through the point ( ) and parallel to the line .
33. passing through the point () and perpendicular to the line .
34. passing through the point () and perpendicular to the line
35. passing through the point (
## ) and perpendicular to the line containing ( ) and
( )
36. passing through the point (
## ) and perpendicular to the line containing ( ) and
( ).
37. passing through the point ( ) and parallel to the line containing () and ()
38. passing through the point ( ) and parallel to the line containing () and ()
39. perpendicular to and containing the midpoint of the segment joining ( ) and ().
40. perpendicular to and containing the midpoint of the segment joining () and ( )
41. perpendicular to the segment joining ( ) and (), and containing the point
of
the way from .
42. all points on the line are equidistant from () and ().
43. if is any point on the line, then the distance from to ( ) is equal to the distance
from to ().
44. having twice the and half the of the line .
45. having the same as and twice the slope of .
46. through the points of intersection of and
47. through the points of intersection of and
.
48. containing segment where the of the point is and the distance
from to () is
49. containing segment where the of the point is and the distance
from to () is .
50. tangent to the circle
## at the point ().
51. tangent to the circle
## at the point ().
52. tangent to the circle .
52. tangent to the circle
## at the point ().
53. tangent to the circle
at the point ( ).
20
In exercises 54 to 68, find the equation(s) of the circle(s) satisfying the given conditions. Write
54. center at () and tangent to the line .
55. center at () and tangent to the line
.
56. center at () and tangent to the line .
57. center at () and tangent to the line .
58. passes through the points () and (), and whose center is on the line
59. passes through the points ( ) and () and whose center is on the line .
60. tangent to the and axes, and the center is units away from the origin and lies on
the 4
th
61. tangent to the and axes, and the center is units away from the origin and likes
on the 1
st
62. tangent to the line at () and passing through ().
63. tangent to the line at () and passing through ().
64. containing the point () and tangent to the line at the point ().
65. tangent to the line at the point (), and the center is on the line
.
66. tangent to the line at the point (), and the center is on the line
.
67. tangent to the line at the point () and also tangent to the line
at the point ( ).
68. tangent to the line at the point () and also tangent to the line
at the point ().
21
Review for Chapters 1 and 2
1. Solve the following
(a)
(b) | | | |
(c) || | |
2. Find an equation of the circle that passes through the points () and (), and has its
center on the line .
3. Find an equation of the line that contains a diameter of the circle
,
and which is parallel to the line .
4. Prove analytically (i. e. using coordinate geometry) that the diagonals of a rhombus and
perpendicular.
5. Let
and
(
)(
)(
)(
|
# Nets of 3-dimensional shapes
Your dad told you that he is going to have the house repainted next week. He asked you to help him. The first task was to find out how much paint you guys would need in order to paint the whole thing. Are you up for the challenge? Thanks to this chapter on surface area, computing for the approximate amount of paint that you need wouldn’t be that hard to figure out. But how do we know how the surface area of your house?
First off, we need to transform the 3-dimensional house into a 2-dimensional form by looking at its front, top, and sides. In the first part of this chapter, we will learn how to do just that. Note that the key to learning this quickly is either by effectively outlining the edges and vertices of the object or by using different colored pens to denote which sides are found in the top, front and side.
After figuring out the front, top and sides, you will then need to draw the net of the house to see the different shapes that comprise it, such as faces of a cube are in fact 6 squares, and a cylinder is made by a 2 circular bases and a rectangle. We will be able to learn how to do this at ease in the second part of this chapter. Apart from that we will also learn how to transform a net into a 3D shape. If you aren’t too good at drawing 3D objects, then you can learn how to draw 3D objects online.
After finding out all the faces, edges and vertices of your house, we move on to computing the combined area of these shapes. In the third section, we will be looking at how to find the surface are of prisms. We will focus on regular rectangular and triangular prisms.
For the rectangular prisms, we would need to use its width, length and height and plug these values in a formula. For the triangular prisms, we are going to use the same elements, adding the value for the side. If ever the value for the side, height and width are not given, we can use the Pythagorean Theorem to solve for it.
In the last part of this chapter, we will tackle the question of how to find the surface area of a cylinder by using the diameter or radius of the circle and the height of the cylinder.
### Nets of 3-dimensional shapes
It is possible to represent 3-dimensional shapes into 2-dimensional patterns called nets. If you fold the nets back up, they will be reverted to their original 3-dimensional form. It is way easier to find the surface area and perimeter of a 3D object if we know its net!
|
1-D Kinematics - Lesson 6 - Describing Motion with Equations
# Sample Problems and Solutions
Earlier in Lesson 6, four kinematic equations were introduced and discussed. A useful problem-solving strategy was presented for use with these equations and two examples were given that illustrated the use of the strategy. Then, the application of the kinematic equations and the problem-solving strategy to free-fall motion was discussed and illustrated. In this part of Lesson 6, several sample problems will be presented. These problems allow any student of physics to test their understanding of the use of the four kinematic equations to solve problems involving the one-dimensional motion of objects. You are encouraged to read each problem and practice the use of the strategy in the solution of the problem. Then click the button to check the answer or use the link to view the solution.
### Check Your Understanding
1. An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.
See solution below.
2. A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.
See solution below.
3. Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60 seconds, what will be his final velocity and how far will he fall?
See solution below.
4. A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.
See solution below.
5. A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon.
See solution below.
6. Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled is accelerated to a speed of 444 m/s in 1.83 seconds, then what is the acceleration and what is the distance that the sled travels?
See solution below.
7. A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike.
See solution below.
8. An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s2. The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway?
See solution below.
9. A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the car (assume uniform acceleration).
See solution below.
10. A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the kangaroo.
See solution below.
11. If Michael Jordan has a vertical leap of 1.29 m, then what is his takeoff speed and his hang time (total time to move upwards to the peak and then return to the ground)?
See solution below.
12. A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration).
See solution below.
13. A baseball is popped straight up into the air and has a hang-time of 6.25 s. Determine the height to which the ball rises before it reaches its peak. (Hint: the time to rise to the peak is one-half the total hang-time.)
See solution below.
14. The observation deck of tall skyscraper 370 m above the street. Determine the time required for a penny to free fall from the deck to the street below.
See solution below.
15. A bullet is moving at a speed of 367 m/s when it embeds into a lump of moist clay. The bullet penetrates for a distance of 0.0621 m. Determine the acceleration of the bullet while moving into the clay. (Assume a uniform acceleration.)
See solution below.
16. A stone is dropped into a deep well and is heard to hit the water 3.41 s after being dropped. Determine the depth of the well.
See solution below.
17. It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar skidded to a stop with a constant acceleration of -3.90 m/s2, determine the speed of the Jaguar before it began to skid.
See solution below.
18. A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed.
See solution below.
19. A dragster accelerates to a speed of 112 m/s over a distance of 398 m. Determine the acceleration (assume uniform) of the dragster.
See solution below.
20. With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object be thrown to reach a height of 91.5 m (equivalent to one football field)? Assume negligible air resistance.
See solution below.
### Solutions to Above Problems
1. Given:
a = +3.2 m/s2 t = 32.8 s vi = 0 m/s
Find:
d = ??
d = vi*t + 0.5*a*t2
d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s2)*(32.8 s)2
d = 1720 m
Return to Problem 1
2. Given:
d = 110 m t = 5.21 s vi = 0 m/s
Find:
a = ??
d = vi*t + 0.5*a*t2
110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s)2
110 m = (13.57 s2)*a
a = (110 m)/(13.57 s2)
a = 8.10 m/ s2
Return to Problem 2
3. Given:
a = -9.8 m t = 2.6 s vi = 0 m/s
Find:
d = ??
vf = ??
d = vi*t + 0.5*a*t2
d = (0 m/s)*(2.60 s)+ 0.5*(-9.8 m/s2)*(2.60 s)2
d = -33.1 m (- indicates direction)
vf = vi + a*t
vf = 0 + (-9.8 m/s2)*(2.60 s)
vf = -25.5 m/s (- indicates direction)
Return to Problem 3
4. Given:
vi = 18.5 m/s vf = 46.1 m/s t = 2.47 s
Find:
d = ??
a = ??
a = (Delta v)/t
a = (46.1 m/s - 18.5 m/s)/(2.47 s)
a = 11.2 m/s2
d = vi*t + 0.5*a*t2
d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s2)*(2.47 s)2
d = 45.7 m + 34.1 m
d = 79.8 m
(Note: the d can also be calculated using the equation vf2 = vi2 + 2*a*d)
Return to Problem 4
5. Given:
vi = 0 m/s d = -1.40 m a = -1.67 m/s2
Find:
t = ??
d = vi*t + 0.5*a*t2
-1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s2)*(t)2
-1.40 m = 0+ (-0.835 m/s2)*(t)2
(-1.40 m)/(-0.835 m/s2) = t2
1.68 s2 = t2
t = 1.29 s
Return to Problem 5
6. Given:
vi = 0 m/s vf = 444 m/s t = 1.83 s
Find:
a = ??
d = ??
a = (Delta v)/t
a = (444 m/s - 0 m/s)/(1.83 s)
a = 243 m/s2
d = vi*t + 0.5*a*t2
d = (0 m/s)*(1.83 s)+ 0.5*(243 m/s2)*(1.83 s)2
d = 0 m + 406 m
d = 406 m
(Note: the d can also be calculated using the equation vf2 = vi2 + 2*a*d)
Return to Problem 6
7. Given:
vi = 0 m/s vf = 7.10 m/s d = 35.4 m
Find:
a = ??
vf2 = vi2 + 2*a*d
(7.10 m/s)2 = (0 m/s)2 + 2*(a)*(35.4 m)
50.4 m2/s2 = (0 m/s)2 + (70.8 m)*a
(50.4 m2/s2)/(70.8 m) = a
a = 0.712 m/s2
Return to Problem 7
8. Given:
vi = 0 m/s vf = 65 m/s a = 3 m/s2
Find:
d = ??
vf2 = vi2 + 2*a*d
(65 m/s)2 = (0 m/s)2 + 2*(3 m/s2)*d
4225 m2/s2 = (0 m/s)2 + (6 m/s2)*d
(4225 m2/s2)/(6 m/s2) = d
d = 704 m
Return to Problem 8
9. Given:
vi = 22.4 m/s vf = 0 m/s t = 2.55 s
Find:
d = ??
d = (vi + vf)/2 *t
d = (22.4 m/s + 0 m/s)/2 *2.55 s
d = (11.2 m/s)*2.55 s
d = 28.6 m
Return to Problem 9
10. Given:
a = -9.8 m/s2 vf = 0 m/s d = 2.62 m
Find:
vi = ??
vf2 = vi2 + 2*a*d
(0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(2.62 m)
0 m2/s2 = vi2 - 51.35 m2/s2
51.35 m2/s2 = vi2
vi = 7.17 m/s
Return to Problem 10
11. Given:
a = -9.8 m/s2 vf = 0 m/s d = 1.29 m
Find:
vi = ??
t = ??
vf2 = vi2 + 2*a*d
(0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(1.29 m)
0 m2/s2 = vi2 - 25.28 m2/s2
25.28 m2/s2 = vi2
vi = 5.03 m/s
To find hang time, find the time to the peak and then double it.
vf = vi + a*t
0 m/s = 5.03 m/s + (-9.8 m/s2)*tup
-5.03 m/s = (-9.8 m/s2)*tup
(-5.03 m/s)/(-9.8 m/s2) = tup
tup = 0.513 s
hang time = 1.03 s
Return to Problem 11
12. Given:
vi = 0 m/s vf = 521 m/s d = 0.840 m
Find:
a = ??
vf2 = vi2 + 2*a*d
(521 m/s)2 = (0 m/s)2 + 2*(a)*(0.840 m)
271441 m2/s2 = (0 m/s)2 + (1.68 m)*a
(271441 m2/s2)/(1.68 m) = a
a = 1.62*105 m /s2
Return to Problem 12
13. Given:
a = -9.8 m/s2 vf = 0 m/s t = 3.13 s
Find:
d = ??
1. (NOTE: the time required to move to the peak of the trajectory is one-half the total hang time - 3.125 s.)
First use: vf = vi + a*t
0 m/s = vi + (-9.8 m/s2)*(3.13 s)
0 m/s = vi - 30.7 m/s
vi = 30.7 m/s (30.674 m/s)
Now use: vf2 = vi2 + 2*a*d
(0 m/s)2 = (30.7 m/s)2 + 2*(-9.8 m/s2)*(d)
0 m2/s2 = (940 m2/s2) + (-19.6 m/s2)*d
-940 m2/s2 = (-19.6 m/s2)*d
(-940 m2/s2)/(-19.6 m/s2) = d
d = 48.0 m
Return to Problem 13
14. Given:
vi = 0 m/s d = -370 m a = -9.8 m/s2
Find:
t = ??
d = vi*t + 0.5*a*t2
-370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s2)*(t)2
-370 m = 0+ (-4.9 m/s2)*(t)2
(-370 m)/(-4.9 m/s2) = t2
75.5 s2 = t2
t = 8.69 s
Return to Problem 14
15. Given:
vi = 367 m/s vf = 0 m/s d = 0.0621 m
Find:
a = ??
vf2 = vi2 + 2*a*d
(0 m/s)2 = (367 m/s)2 + 2*(a)*(0.0621 m)
0 m2/s2 = (134689 m2/s2) + (0.1242 m)*a
-134689 m2/s2 = (0.1242 m)*a
(-134689 m2/s2)/(0.1242 m) = a
a = -1.08*106 m /s2
(The - sign indicates that the bullet slowed down.)
Return to Problem 15
16. Given:
a = -9.8 m/s2 t = 3.41 s vi = 0 m/s
Find:
d = ??
d = vi*t + 0.5*a*t2
d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s2)*(3.41 s)2
d = 0 m+ 0.5*(-9.8 m/s2)*(11.63 s2)
d = -57.0 m
(NOTE: the - sign indicates direction)
Return to Problem 16
17. Given:
a = -3.90 m/s2 vf = 0 m/s d = 290 m
Find:
vi = ??
vf2 = vi2 + 2*a*d
(0 m/s)2 = vi2 + 2*(-3.90 m/s2)*(290 m)
0 m2/s2 = vi2 - 2262 m2/s2
2262 m2/s2 = vi2
vi = 47.6 m /s
Return to Problem 17
18. Given:
vi = 0 m/s vf = 88.3 m/s d = 1365 m
Find:
a = ??
t = ??
vf2 = vi2 + 2*a*d
(88.3 m/s)2 = (0 m/s)2 + 2*(a)*(1365 m)
7797 m2/s2 = (0 m2/s2) + (2730 m)*a
7797 m2/s2 = (2730 m)*a
(7797 m2/s2)/(2730 m) = a
a = 2.86 m/s2
vf = vi + a*t
88.3 m/s = 0 m/s + (2.86 m/s2)*t
(88.3 m/s)/(2.86 m/s2) = t
t = 30. 8 s
Return to Problem 18
19. Given:
vi = 0 m/s vf = 112 m/s d = 398 m
Find:
a = ??
vf2 = vi2 + 2*a*d
(112 m/s)2 = (0 m/s)2 + 2*(a)*(398 m)
12544 m2/s2 = 0 m2/s2 + (796 m)*a
12544 m2/s2 = (796 m)*a
(12544 m2/s2)/(796 m) = a
a = 15.8 m/s2
Return to Problem 19
20. Given:
a = -9.8 m/s2 vf = 0 m/s d = 91.5 m
Find:
vi = ??
t = ??
First, find speed in units of m/s:
vf2 = vi2 + 2*a*d
(0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(91.5 m)
0 m2/s2 = vi2 - 1793 m2/s2
1793 m2/s2 = vi2
vi = 42.3 m/s
Now convert from m/s to mi/hr:
vi = 42.3 m/s * (2.23 mi/hr)/(1 m/s)
vi = 94.4 mi/hr
Return to Problem 20
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# Square Root Questions
Reviewed by:
Last updated date: 10th Aug 2024
Total views: 390.6k
Views today: 10.90k
## What is Square Root?
A square root is a value that, when multiplied by itself, gives the original number.
### How to Find the Square Root?
We can use two methods to find square roots - Prime factorization and the Long division method.
### What are Squares and Square Roots?
Just the opposite method of squaring a number is the square root. If a number n, such as n2, is squared, then the square root of n2 is equal to the original number n.
### How to Square a Number?
We need to multiply the number by itself to find the square of a number.
For example, 3 multiplied by 3 is equal to 9.
Similarly,
Square of 4 is: 4 multiplied by 4 = 4 × 4 = 16.
### Square Root Problems and Answers
Q1: Can you find out the number of numbers lying between the squares of these following pairs of numbers?
(i) 25 and 26
(ii) 99 and 100
Solution - As we know, between n2 and (n + 1)2, the number of non–perfect square numbers are 2n.
(i) Between 252 and 262 there are 2 × 25 = 50 natural numbers.
(ii) Between 992 and 1002 there are 2 × 99 = 198 natural numbers.
Q2: Write a Pythagorean triplet whose one of the required member is:
(i) 6
(ii) 14
(iii) 16
(iv) 18
Solution -
We know, for any natural number m, 2m, m2 – 1, m2 + 1 is a Pythagorean triplet.
(i) 2m = 6
⇒ m = 6/2 = 3
m2 – 1 = 32 – 1 = 9 – 1 = 8
m2 + 1 = 32 + 1 = 9 + 1 = 10
Therefore, (6, 8, 10) is a Pythagorean triplet.
(ii) 2m = 14
⇒ m = 14/2 = 7
m2 – 1 = 72 – 1 = 49 – 1 = 48
m2 + 1 = 72 + 1 = 49 + 1 = 50
(14, 48, 50) is not a Pythagorean triplet.
(iii) 2m = 16
⇒ m = 16/2 = 8
m2 – 1 = 82 – 1 = 64 – 1 = 63
m2 + 1 = 82 + 1 = 64 + 1 = 65
Therefore, (16, 63, 65) is a Pythagorean triplet.
(iv) 2m = 18
⇒ m = 18/2 = 9
m2 – 1 = 92 – 1 = 81 – 1 = 80
m2 + 1 = 92 + 1 = 81 + 1 = 82
Therefore, (18, 80, 82) is a Pythagorean triplet.
Q3: (n + 1)2 - n2 = ?
Solution -
(n + 1)2 - n2
= (n2 + 2n + 1) – n2
= 2n + 1
Q4: State that the number 121 is the sum of 11 odd natural numbers.
Solution - As 121 = 112
We recognize that n2 is the sum of the first n odd natural numbers.
It shows that 121 = the sum of the first 11 odd natural numbers.
= 1 + 3 + 5+ 7 + 9 + 11 +13 + 15 + 17 + 19 + 21
Q5: Use the identity and find the square of 189.
(a – b)2 = a2 – 2ab + b2
Solution - 189 = (200 – 11)2
= 40000 – 2 × 200 × 11 + 112
= 40000 – 4400 + 121
= 35721
Q6: Find out the square root of 625 using the mathematical identity as stated: (a + b)2 = a2 + b2 + 2ab?
Solution - (625)2
= (600 + 25)2
= 6002 + 2 × 600 × 25 + 252
= 360000 + 30000 + 625
= 390625
### Properties of Square Roots
A square root function is defined in mathematics as a one-to-one function that takes as an input a positive number and returns the square root of the given input number.
f(x) = √x
The following are some of the essential properties of the square root:
• If a number is a perfect square number, a perfect square root exists.
• It may have a square root if a number ends with an even number of zeros (0's).
• It is possible to multiply the two square root values. For instance, √3 can be multiplied by √2, then √6 should be the result.
• If two same square roots are multiplied, then a radical number should be the product. This implies that a non-square root number is a product. When √7 is multiplied by √7, for example, the result obtained is 7.
• It does not define the square root of any negative numbers. And no negative can be the perfect square.
• If a number ends with 2, 3, 7, or 8 (in the digit of the unit), the perfect square root will not exist.
• If in the unit digit, a number ends with 1, 4, 5, 6, or 9, the number would have a square root.
Using the prime factorization process, the square root of a perfect square number is simple to measure. For example -
## Square Root By Prime Factorisation
Number Prime Factorisation Square Root 16 2 × 2 × 2 × 2 √16 = 2 × 2 = 4 144 2x2x2x2x3x3 √144 = 2 × 2 × 3 = 12 169 13 × 13 √169 = 13 256 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 √256 = (2 × 2 × 2 × 2) = 16 576 576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 √576 = 2 × 2 × 2 × 3 = 24
With the help of an example, let us understand this concept:
Example 1: Solve √10 to 2 decimal places.
Solution -
Step 1: Choose any two perfect square roots between which you feel your number may fall.
We know that 22 = 4; 32 = 9, 42 = 16 and 52 = 25
Now, choose 3 and 4 (as √10 lies between these 2 numbers)
Step 2: Divide the given number into one of the square roots chosen.
Divide 10 by 3.
=> 10/3 = 3.33 (round off answer at 2 places)
Step 3: Find the root average and the product of the step above, i.e.
(3 + 3.33)/2 = 3.1667
Verify: 3.1667 × 3.1667 = 10.0279 (Not required)
Repeat step 2 and step 3
Now 10/3.1667 = 3.1579
Average of 3.1667 and 3.1579.
(3.1667+3.1579)/2 = 3.1623
Verify: 3.1623 × 3.1623 = 10.0001 (more accurate)
Stop the process.
Example 2: Find the square roots of whole numbers from 1 to 100 that are perfect squares.
Solution - The perfect squares are - 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
Square Root Result √1 1 √4 2 √9 3 √16 4 √25 5 √36 6 √49 7 √64 8 √81 9 √100 10
Example 3: What is:
1. The square root of 2.
2. The square root of 3.
3. The square root of 4.
4. The square root of 5
Solution - Use a square root list, we have
1. Value of root 2 i.e. √2 = 1.4142.
2. Value of root 3 i.e. √3 = 1.7321.
3. Value of root 4 i.e. √4 = 2.
4. Value of root 5 i.e. √5 = 2.2361.
## FAQs on Square Root Questions
Question 1. Define Square Numbers and Square Roots.
Answer: Square numbers can be defined as the numbers that are produced when a number is multiplied by itself. For example, if n is a number and it is multiplied by itself, then the square of n can be given as n2. Another example is the square of 10, which is 102 = 10 x 10 = 100.
Further, the square root of a number can be explained as the value which when multiplied by itself gives the original number. This value can be represented by the symbol ‘√.’ For example, the square root of 25 is √25 = 5.
Question 2. What do you Understand by Perfect Squares? Provide Examples with Your Answer.
Answer: Perfect squares can be defined as numbers in which the square root of the number provides a whole number. For example, 9 is a perfect square number because its root is a whole number, which means that √9 = 3.
Question 3. What is an Imperfect Square? Provide Examples with Your Answer.
Answer: An imperfect square can be defined as a number in which the square root of the number is a fraction. The value is generated by taking the square root of the imperfect square can also be non-terminating. For example, 3 is an imperfect square because the root of this number is 1.73205080757. This number is a fraction.
Question 4. What is the Main Difference Between a Square and Square Roots?
Answer: The square root of a number provides the root of a particular number that was squared. This is the main difference between a square number and square roots.
Question 5. How to Solve the Square Root Equation.
Answer: We need to follow the below steps to solve the square root equation: Isolate the square to one of the sides (L.H.S or R.H.S). Now solve the remaining equation. Square both sides of the given equation. With examples, let us understand the steps.
Question 6. Is the Square Root of a Negative Number a Whole Number?
Answer: No, negative numbers shouldn't have a square root, as per the square root concept. If we multiply two negative numbers, a positive number will always be the product. Negative number square roots expressed as multiples of I (imaginary numbers).
|
1. ## Arithmetic progression
The first and third term of arithmetic progression are x and y respectively.Find the sixteenth term in term of x and y.
2. Originally Posted by mastermin346
The first and third term of arithmetic progression are x and y respectively.Find the sixteenth term in term of x and y.
The values of the arithmetic progression are calculated by
$a_n=a_1+(n-1)\cdot d$
with
$d = a_{n+1} - a_n$
In your case $d = \frac{y-x}2$.
Plug in the values you know:
$n = 16$
$a_1 = x$
$a_3 = y$
and calculate $a_{16}$.
3. Originally Posted by mastermin346
The first and third term of arithmetic progression are x and y respectively.Find the sixteenth term in term of x and y.
$x + (x+d) + (x+2d) + (x+3d) + ... (x + 15d)$
where $d$ is the constant difference between each successive term.
In our case, we are told that the third term is y. Thus, $x+2d = y$ .
So, $d = \frac{y-x}{2}$
$a_1 = x$
$a_2 = x + \frac{y-x}{2} = \frac{2x}{2} + \frac{y-x}{2} =
\frac{2x+y-x}{2} = \frac{x+y}{2}
$
$a_3 = \frac{x+y}{2} + \frac{y-x}{2}= \frac{2y}{2} = y$
...
...
...
$a_{16} = x + 15d = x + 15\big{(}\frac{y-x}{2}\big{)}$
4. Originally Posted by mastermin346
The first and third term of arithmetic progression are x and y respectively.Find the sixteenth term in term of x and y.
$y = x + 2d \Rightarrow d = \frac{y - x}{2}$. And the first term is x.
You now have everything necessary to do anything you want, including finding the sixteenth term.
|
CBSE Class 9 Maths: Extra Questions - Chapter 1 - Number System (with Answers)
CBSE Class 9 Maths extra Questions for Chapter 1 - Number System (with Answers). These questions are NCERT based important for the preparation of the upcoming CBSE 9th Maths Exam 2020-21. Most of the questions given here are very simple and can be solved in less than 5 minutes. If someone is facing problems in solving these questions then more clarity of basic concepts is required. These extra questions from Class 9 Maths Chapter 1 - Number System are also expected to be asked in CBSE Class 9 Maths test.
CBSE Class 9 Maths Extra Questions from Chapter 1 - Number System (with Answers):
1: Simplify: [{(82 + 92)2}0]2 = ____
2: 2(2/3) x 2(1/3)
3: [2 + (3)1/2] [2 - (3)1/2] = ______
4: Let a > 0 be a real number and p and q be rational numbers. Then which of the following statement(s) is/are correct?
(i) ap . aq = ap + q
(ii) ap bp = (ab)p
(iii) (ap)q = apq
(iv) All of these
5: Find the value of 1/(4)3 ÷ 1/(4)3 + 1/(1)3
6: Divide 10 √15 by 5 √3 .
7. Which of the following given statement(s) is/are correct?
(i) The sum or difference of a rational number and an irrational number is irrational.
(ii) The product or quotient of a non-zero rational number with an irrational number is
irrational.
(iii) If we add, subtract, multiply or divide two irrationals, the result may be rational or
irrational
(iv) All of these statements are correct
(iv) All of these statements are correct.
8. [1/(2 + √5)] can also be written as
(a) (2 - √5)
(b) (√5 - 2)
(c) [1/(2 ÷ √5)]
(d) [1 + (2 + √5)]
[1/(2 + √5)] can be rationalised and can be written as (√5 - 2).
9. π (Pi) is a rational number or irrational number?
|
Math Calculators, Lessons and Formulas
It is time to solve your math problem
mathportal.org
This step-by-step calculator solves quadratic equations using three different methods: the quadratic formula method, completing the square, and the factoring method. Calculator shows all the work and provides detailed explanation on how to solve an equation.
## Solve $\color{blue}{2x^2+3x-2 = 0}$ using the Quadratic Formula.
solution
$$\color{blue}{ x_1 = -2 }~~ \text{and}~~ \color{blue}{ x_2 = \frac{ 1 }{ 2 } }$$
explanation
Step 1: Read the values of $a$, $b$, and $c$ from the quadratic equation: $a$ is the number in front of $x^2$, $b$ is the number in front of $x$, $c$ is the number at the end. In our case:
$$a = 2, \,\, b = 3, \,\, c = -2$$
Step 2: Plug in the values for $a$, $b$, and $c$ into the quadratic formula.
\begin{aligned} x_1,x_2 &= \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\[1 em] x_1,x_2 &= \frac{ -3 \pm \sqrt{ 3 ^2 - 4 \cdot 2 \cdot (-2)} }{ 2 \cdot 2 } \end{aligned}
Step 3: Simplify expression under the square root.
$$x_1,x_2 = \frac{ -3 \pm \sqrt{ 25 } }{ 4 }$$
Step 4: Solve for $x$
\begin{aligned} & \color{blue}{ x_1 = \frac{ -3~-~\sqrt{ 25 } }{ 4 } = -2 } \\\\ & \color{blue}{ x_2 = \frac{ -3~+~\sqrt{ 25 } }{ 4 } = \frac{ 1 }{ 2 } } \end{aligned}
## Report an Error !
Form values: 2 , 1 , 3 , 2 , 2 , qf , g , Solve quadratic equation 2x^2+3x-2 = 0 , , , Solve $2x^2+3x-2 = 0$ using the Quadratic Formula.
Comment (optional)
Enter an equation and choose a solving method, or let the calculator choose one.
help ↓↓ examples ↓↓ tutorial ↓↓
$2x^2 + x - 3 = 0$
$\frac{1}{2}x^2+\frac{3}{4}x - 11 = 0$
$x^2+\sqrt{2}x - 4 = 0$
Factoring Completing the square Quadratic formula
Allow the calculator to select the most efficient method.
Find approximate solution
working...
EXAMPLES
example 1:ex 1:
Solve for $x^2 + 3x - 4 = 0$ by factoring.
example 2:ex 2:
Solve $4x^2 - x - 3 = 0$ by completing the square.
example 3:ex 3:
Solve $-2x^2 - 0.5x + 0.75 = 0$ using the quadratic formula.
example 4:ex 4:
Solve $\frac{2}{3} x^2 - \frac{1}{3} x - 5 = 0$.
Find more worked-out examples in the database of solved problems..
TUTORIAL
## How to use this calculator
The most commonly used methods for solving quadratic equations are:
1. Factoring method
2. Solving quadratic equations by completing the square
In the following sections, we'll go over these methods.
### Method 1A : Factoring method
If a quadratic trinomial can be factored, this is the best solving method.
We often use this method when the leading coefficient is equal to 1 or -1. If this is not the case, then it is better to use some other method.
Example 01: Solve $x^2 \color{red}{-8}x \color{blue}{+ 15} = 0$ by factoring.
Here we see that the leading coefficient is 1, so the factoring method is our first choice.
To factor this equation, we must find two numbers ( $a$ and $b$ ) with a sum is $a + b = \color{red}{8}$ and a product of $a \cdot b = \color{blue}{15}$.
After some trials and errors, we see that $a = 3$ and $b = 5$.
Now we use formula $x^2 - 8x + 15 = (x - a)(x - b)$ to get factored form:
$$x^2 - 8x + 15 = (x - 3)(x - 5)$$
Divide the factored form into two linear equations to get solutions.
\begin{aligned} x^2 - 8x + 15 &= 0 \\ (x - 3)(x - 5) &= 0 \\ x -3 &= 0 ~~ \text{or} ~~ x - 5= 0 \\ x &= 3 ~~ \text{or} ~~ x = 5 \end{aligned}
### Method 1B : Factoring - special cases
Example 02: Solve $x^2 -8x = 0$ by factoring.
In this case, (when the coefficient c = 0 ) we can factor out $\color{blue}{x}$ out of $x^2 - 8x$.
\begin{aligned} x^2 - 8x &= 0 \\ \color{blue}{x} \cdot ( x - 8 ) &= 0 \\ x &= 0 ~~ \text{or} ~~ x - 8 = 0 \\ x &= 0 ~~ \text{or} ~~ x = 8 \end{aligned}
Example 03: Solve $x^2 - 16 = 0$ by factoring.
In this case, ( when the middle term is equal 0) we can use the difference of squares formula.
\begin{aligned} x^2 - 16 &= 0 \\ x^2 - 4^2 &= 0 \text{ use } a^2 - b^2 = (a-b)(a+b) \\ (x - 4)(x+4) &= 0 \\ x - 4 &= 0 ~~ \text{or} ~~ x + 4 = 0 \\ x &= 4 ~~ \text{or} ~~ x = -4 \end{aligned}
### Method 3 : Solve using quadratic formula
This method solves all types of quadratic equations. It works best when solutions contain some radicals or complex numbers.
Example 05: Solve equation $2x^2 + 3x - 2 = 0$ by using quadratic formula.
Step 1: Read the values of $a$, $b$, and $c$ from the quadratic equation. ( $a$ is the number in front of $x^2$ , $b$ is the number in front of $x$ and $c$ is the number at the end)
$$a = 2 ~~ b = 3 ~~ \text{and} ~~ c = -2$$
Step 2:Plug the values for a, b, and c into the quadratic formula and simplify.
\begin{aligned} x_1, x_2 &= \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ x_1, x_2 &= \frac{-3 \pm \sqrt{3^2-4 \cdot 2 \cdot (-2) }}{2 \cdot 2} \\ x_1, x_2 &= \frac{-3 \pm \sqrt{9+ 16 }}{4} \\ x_1, x_2 &= \frac{-3 \pm \sqrt{25}}{4} \\ x_1, x_2 &= \frac{-3 \pm 5}{4} \end{aligned}
Step 3: Solve for $x_1$ and $x_2$
\begin{aligned} x_1 = & \frac{-3 \color{blue}{+} 5}{4} = \frac{2}{4} = \frac{1}{2} \\ x_2 = & \frac{-3 \color{blue}{-} 5}{4} = \frac{-8}{4} = -2 \end{aligned}
### Method 2 : Completing the square
This method can be used to solve all types of quadratic equations, although it can be complicated for some types of equations. The method involves seven steps.
Example 04: Solve equation $2x^2 + 8x - 10= 0$ by completing the square.
Step 1: Divide the equation by the number in front of the square term.
\begin{aligned} 2x^2 + 8x - 10 & = 0 ~~ / ~ \color{orangered}{:2} \\ \frac{2x^2}{2} + \frac{8x}{2} - \frac{10}{2} & = \frac{0}{2} \\ x^2 + 4x - 5 & = 0 \end{aligned}
Step 2: move $-5$ to the right:
$$x^2 + 4x = 5$$
Step 3: Take half of the x-term coefficient $\color{blue}{\dfrac{4}{2}}$, square it $\color{blue}{\left(\dfrac{4}{2} \right)^2}$ and add this value to both sides.
$$x^2 + 4x + \color{blue}{\left(\frac{4}{2} \right)^2} = 5 + \color{blue}{\left(\frac{4}{2} \right)^2}$$
Step 4: Simplify left and right side.
$$x^2 + 4x + 2^2 = 9$$
Step 5: Write the perfect square on the left.
$$\left( x + 2 \right)^2 = 9$$
Step 6: Take the square root of both sides.
\begin{aligned} x + 2 &= \pm \sqrt{9} \\\\ x + 2 &= \pm 3 \end{aligned}
Step 7: Solve for $x_1$ and $x_2$ .
\begin{aligned} x_1 & = +3 - 2 = 1 \\ x_2 & = -3 - 2 = - 5 \end{aligned}
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# Lesson 5May I Have More, Please?Solidify Understanding
## Jump Start
Determine which inequality matches each phrase.
No more than
A minimum of
More than
No less than
At least
A maximum of
Cannot exceed
Fewer than
## Learning Focus
Write and solve inequalities to model real situations.
Write solutions to inequalities using set builder and interval notation.
How can inequalities be used to find solutions to real problem situations?
Technology guidance for today’s lesson:
• Basic Calculations — Square Root:
• Basic Calculations — Exponents:
• Basic Calculations — Add, Subtract, Multiply, and Divide:
• Basic Calculations — Add and Subtract:
## Open Up the Math: Launch, Explore, Discuss
Elvira, the cafeteria manager, needs to be careful with her spending, and she manages the cafeteria budget so that they can serve the best food at the lowest cost. To do this, Elvira keeps good records and analyzes all her budgets.
### 1.
Elvira’s cafeteria has those cute little cartons of milk that are typical of school lunch. The milk supplier charges per carton of milk, in addition to a delivery charge of . What is the maximum number of milk cartons that Elvira can buy if she has budgeted for milk?
#### a.
Write and solve an inequality that models this situation.
#### b.
Describe in words the quantities that would work in this situation.
### 2.
Students love to put ranch dressing on everything, so Elvira needs to keep plenty in stock. The students eat about gallons of ranch each day! Elvira started the school year with gallons of ranch dressing. She needs to have at least gallons left when she reorders to have enough in stock until the new order comes. For how many days will her ranch dressing supply last before she needs to reorder?
#### a.
Write and solve an inequality that models this situation.
#### b.
Describe in words the quantities that would work in this situation.
### 3.
The prices on many of the cafeteria foods change during the year. Elvira finds that she has ordered veggie burgers four times and paid , , , and on the orders. To stay within her budget, Elvira needs to be sure that the average order of veggie burgers is not more than . How much can she spend on the fifth order to keep the average order within her budget?
#### a.
Write and solve an inequality that models this situation.
#### b.
Describe in words the quantities that would work in this situation.
### 4.
Elvira can purchase ready-made pizzas for each. If she makes them in the cafeteria, she can spend on ingredients and per pizza on labor. For how many pizzas is it cheaper for the cafeteria to make the pizzas themselves rather than buy them ready-made?
#### a.
Write and solve an inequality that models this situation.
#### b.
Describe in words the quantities that would work in this situation.
### 5.
Elvira is comparing prices between two different suppliers of fresh lettuce. Val’s Veggies charges for delivery plus per bag of lettuce. Sally’s Salads charges for delivery plus per bag of lettuce. How many bags of lettuce must be purchased for Val’s Veggies to be the cheaper option?
#### a.
Write and solve an inequality that models this situation.
#### b.
Describe in words the quantities that would work in this situation.
### 6.
Each student who buys a school lunch pays . The cafeteria typically brings in between and each day. How many students does the cafeteria usually serve in a day?
#### a.
Model this situation using an inequality.
#### b.
Describe in words the quantities that would work in this situation.
#### c.
River A and River B combine together to form River C. The flow rate of the River C is less than or equal to the sum of the flow rates of the Rivers A and B, but it is greater than or equal to either of the individual flow rates of Rivers A and B. One of the three rivers flows at a rate of . Another river flows at a rate of .
#### a.
Find the minimum flow rate, , of the third river.
#### b.
Find the maximum flow rate, , of the third river.
#### c.
Write an inequality that describes the possible flow rates, , of the third river.
## Takeaways
Using properties of inequalities to solve inequalities:
## Adding Notation, Vocabulary, and Conventions
Compound inequality:
Examples:
## Lesson Summary
In this lesson, we wrote inequalities to model contexts that had a range of solutions. We used the properties of inequalities to solve the inequalities and found that solving inequalities is very similar to solving equations, but we must be careful when multiplying or dividing by a negative number to reverse the inequality sign. We used interval and set notation to write solutions and learned that many of the solutions that we write are compound inequalities.
## Retrieval
### 1.
Graph each equation and find the point where they intersect.
Point of intersection:
|
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# Equation Problems: A Step-by-Step Guide to Solving Them
Whether you’re a student struggling with your math homework or a professional who needs to solve equations for your job, understanding how to solve equation problems is an essential skill. In this guide, we’ll walk you through the steps involved in solving different types of equations, from basic linear equations to more complex systems of equations. We’ll also provide tips for success and common pitfalls to avoid.
## What is an equation?
An equation is a statement that two expressions are equal. It is written in the form of A = B, where A and B are any expressions. For example, the equations 2x + 3 = 7 and x^2 + 2x + 1 = 0 are both valid equations.
### Types of equations
There are many different types of equations, but some of the most common include:
• Linear equations: Linear equations are equations of the first degree, meaning that the highest exponent of any variable is 1. For example, the equation 2x + 3 = 7 is a linear equation.
• Quadratic equations: Quadratic equations are equations of the second degree, meaning that the highest exponent of any variable is 2. For example, the equation x^2 + 2x + 1 = 0 is a quadratic equation.
• Radical equations: Radical equations are equations that contain radicals, which are symbols that represent the square root of a number. For example, the equation √x + 2 = 5 is a radical equation.
• Exponential equations: Exponential equations are equations that contain exponential terms, which are terms of the form x^n, where x is any number and n is a positive integer. For example, the equation 2^x = 8 is an exponential equation.
• Systems of equations: Systems of equations are two or more equations that are solved simultaneously. For example, the system of equations x + y = 5 and 2x – y = 1 is a system of two linear equations.
### How to solve equations
The general steps for solving an equation are as follows:
1. Identify the type of equation. This will help you determine the appropriate method for solving the equation.
2. Isolate the variable. This means moving all of the constant terms to one side of the equation and all of the terms with the variable on the other side.
3. Solve for the variable. This involves using the appropriate mathematical operations to simplify the equation and solve for the variable.
4. Check your answer. Once you have solved for the variable, plug your answer back into the original equation to make sure that it is a solution.
### Solving linear equations
To solve a linear equation, follow these steps:
1. Combine like terms. This means adding or subtracting any terms that have the same variable.
2. Isolate the variable. Move the constant term to the other side of the equation.
3. Solve for the variable. Divide both sides of the equation by the coefficient of the variable.
There are three main methods for solving quadratic equations: factoring, using the quadratic formula, and completing the square.
Factoring: If the quadratic equation can be factored, then it can be solved by setting each factor equal to zero and solving for the variable.
Quadratic formula: The quadratic formula is a general formula that can be used to solve any quadratic equation. It is given by the following equation:
``````x = (-b ± √(b^2 - 4ac)) / 2a
``````
where a, b, and c are the coefficients of the quadratic equation.
Completing the square: Completing the square is a method for solving quadratic equations that involves rewriting the quadratic equation in a way that makes it easier to factor.
1. Isolate the radical. Move all of the other terms to the other side of the equation.
2. Square both sides of the equation. This will eliminate the radical.
3. Solve for the variable. Use the appropriate mathematical operations to simplify the equation and solve for the variable.
### Solving exponential equations
1. Take the log of both sides of the equation. This will convert the exponential equation into a linear equation.
2. Solve for the variable. Use the appropriate mathematical operations to simplify the equation and solve for the variable.
### Solving systems of equations
There are two main methods for solving systems of equations: elimination and substitution.
Elimination: The elimination method involves adding or subtracting the equations in the system in order to eliminate one of the variables. Once one of the variables has been eliminated, the remaining equation can be solved for the other variable.
Substitution: The substitution method involves solving one of the equations in the system for one of the variables and substituting that value into the other equation. Once the value has been substituted, the remaining equation can be solved for the other variable.
### Word problems
Many real-world problems can be modeled using equations. To solve a word problem using an equation, first identify the variables in the problem and write an equation that represents the problem. Once you have written an equation, you can solve it using the steps outlined above.
## Conclusion
Solving equation problems is an essential skill that can be used in many different areas of life. By understanding the concepts and steps involved in solving different types of equations, you can develop the skills you need to solve any equation problem you encounter.
### Tips for success
Here are a few tips for success when solving equation problems:
• Be patient. Solving equation problems can take time and practice. Don’t get discouraged if you don’t get the answer right away.
• Check your answers. Once you have solved an equation problem, plug your answer back into the original equation to make sure that it is a solution.
### Common pitfalls to avoid
Here are a few common pitfalls to avoid when solving equation problems:
• Not isolating the variable. This is one of the most common mistakes students make when solving equation problems. Be sure to isolate the variable before you try to solve for it.
• Dividing by zero. Dividing by zero is not defined. Be sure to check that you are not dividing by zero before you divide.
• Making careless mistakes. When solving equation problems, it is important to be careful and to avoid making careless mistakes. Be sure to check your work carefully before you submit it.
### Real-world applications of equation solving
Equation solving is used in many different areas of life, including:
• Math and science: Equation solving is essential for many math and science courses, such as algebra, geometry, calculus, and physics.
• Engineering: Equation solving is used to design and build bridges, buildings, and other structures.
• Business: Equation solving is used to make financial decisions, such as budgeting and forecasting.
• Everyday life: Equation solving is used in many everyday situations, such as cooking, shopping, and traveling.
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You will be submitting a SINGLE PDF DOCUMENT that contains your entire project. Your project must be typed, though you may choose to hand-write such items as statistical notation (such as μ1or α ) You
### Use the cylindrical shell method to find the volume of the solid generated by revolving the region bounded by = 5 ^3, = 0, and = 1 about the y-axis.
Use the cylindrical shell method to find the volume of the solid generated by revolving the region bounded by = 5 ^3, = 0, and = 1 about the y-axis.
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### Free Online Math Help Chat: Get Unstuck and Ace Your Algebra 1 Problems
Algebra 1 is a foundational math class that covers a wide range of topics, from linear equations to polynomials to quadratic equations. If you’re struggling with Algebra 1, don’t despair! There are many resources available to help you, including free online math help chat. Free online math help chat is
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# What is the Area of Rectangle Formula
Geometry is the specialized branch of mathematics that uses diagrams and figures to do what other mathematics streams do with numbers and notations. In the world of geometry, different shapes are studied. In this article we will study about rectangles and area of rectangle formula.
Among the many different kinds of shapes possible, one way to categorize them is shapes made of straight lines and shapes made of curved lines. An example of a shape with a curved line would be a circle and parabola, and examples of figures with straight lines would be triangles or rectangles. It is also possible to categorize figures into closed shapes or open shapes. For example, the circle is a closed figure, while a parabola has an open shape.
Among the shapes which are closed and made of straight lines, the triangle is the first in the family, which has the least number of lines. One needs to have at least three lines to have a closed shape with straight lines. After the triangle, we have shapes made of four, five, six lines and so on. The shape can be regular or irregular in each category, which means if the geometric shape is made of lines of equal length, then the shape is usually regular.
Else if the different lines are of different lengths, then the geometric shape is irregular. Among the regular category comes a very often found and studied shape of a square formed by four equal lines, and each adjacent line is perpendicular to each other. The closest cousin of the square is a rectangle with four lines and adjacent lines perpendicular to each other, just like a square. However, unlike a square, the adjacent sides are not of equal length in a rectangle while the opposite sides are still equal, just like in a square. To know more about the properties of a rectangle visit website.
Rectangle as a shape is commonly used in day-to-day life and, more often than not, making the building blocks. Bricks are often rectangular in shape, and perhaps it is no surprise that the shape of buildings and rooms is rectangular.
It is not uncommon to see a rectangle-shaped swimming pool or any sports arena like a tennis court is rectangular in shape. In present times, almost all mobile phones are rectangular in shape, so we can conclude that their specific shape has specific properties, making it useful to be used across so many applications.
One of the rectangle’s useful properties as a shape is that it has got a very simple shape, and the way to find the area of a rectangle or perimeter of a rectangle is quite easy. As we discussed, that rectangle has four sides in which adjacent sides are perpendicular to each other and are unequal, and the opposite sides are equal.
Thus to calculate the rectangle’s perimeter, one does not need to know the length of four sides, but only the length of two sides is sufficient to know the rectangle’s perimeter. So, if we have a rectangle with vertices A, B, C and D. Perimeter of the rectangle would be AB + BC + CD + DA, but since AB = CD and BC = DA hence, the perimeter of this rectangle can be simply put as 2 ( AB + BC) or 2 ( CD + DA).
The rectangle area, similar to the perimeter of a rectangle, is easy to calculate. Again, let us take the rectangle with vertices A, B, C, and D. Area of a rectangle would be AB multiplied by BC or CD multiplied by DA. So again, for the rectangle area the same way as the perimeter of a rectangle, one needs only to know the length of adjacent sides. By multiplying them together, one would know the area of a rectangle.
The ability to calculate a rectangle area easily makes it easier in day-to-day life to do many household activities. For example, when painting a rectangular wall, one would need to calculate the wall area to know how much amount of paint would be needed. For more practice on how to calculate the area of a rectangle students can refer to Cuemath website and download relevant rectangle worksheets for their practice.
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# Given a set {-15,1,200,x,1} , for what x would the mean of the set be -15?
Dec 11, 2015
$\textcolor{w h i t e}{\times} - 262$
#### Explanation:
$\textcolor{w h i t e}{\times} \overline{a} = \frac{1}{n} {\sum}_{i = 1}^{n} {a}_{i}$ (the mean of a set of n data points)
$\textcolor{w h i t e}{\times} - 15 = \frac{1}{5} {\sum}_{i = 1}^{5} {a}_{i}$
$\implies \frac{- 15 + 1 + 200 + x + 1}{5} = - 15$
Multiply both sides by $5$
$\textcolor{w h i t e}{\times} \textcolor{red}{5 \cdot} \frac{- 15 + 1 + 200 + x + 1}{5} = \textcolor{red}{5 \cdot} \left(- 15\right)$
$\implies x + 187 = - 75$
Add $- 187$ to both sides
$\textcolor{w h i t e}{\times} x + 187 \textcolor{red}{- 187} = - 75 \textcolor{red}{- 187}$
$\implies x = - 262$
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# EXT: Multiply a single digit number by a three digit number (area model)
Lesson
When we multiply a one-digit number by a three-digit number, there's a great way of using the area of rectangles to help us.
Remember, to find the area of a rectangle, we use the formula:
$\text{Area of a rectangle }=\text{length }\times\text{width }$Area of a rectangle =length ×width
Now take a look at the video to see how you can break your multiplication problem up into smaller steps, using rectangles.
Remember!
• We can break up large rectangles into smaller rectangles and find the areas to solve large multiplication questions.
• One way to break up three-digit numbers is to use the hundreds, tens and unit values.
#### Worked Examples
##### Question 1
We want to find $344\times3$344×3 using the area model.
1. Find the area of the first rectangle.
2. Find the area of the second rectangle.
3. Find the area of the third rectangle.
4. What is the total area of all three rectangles?
5. So what is $344\times3$344×3?
##### question 2
We want to use the area model to find $225\times4$225×4.
1. Fill in the areas of each rectangle.
$200$200 $20$20 $5$5 $4$4 $\editable{}$ $\editable{}$ $\editable{}$
2. What is the total area of all three rectangles?
3. So what is $225\times4$225×4?
##### question 3
We want to use the area model to find $907\times2$907×2.
1. Fill in the areas of each rectangle.
$900$900 $7$7 $2$2 $\editable{}$ $\editable{}$
2. What is the total area of both rectangles?
3. So what is $907\times2$907×2?
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## What is 12×12 in sq. ft?
144 sq ft.
Therefore, 12 X 12 Ft = 144 sq ft. The area of the room will be 144 sq ft.
Is square feet addition or multiplication?
How to find the square footage of a rectangle. Measure the width and length of the area in feet. Multiply your length and width together to get your area.
### How do u figure out square feet?
To find square feet, multiply the length measurement in feet by the width measurement in feet. This yields a product called the area, which is expressed in square feet (or square inches if you are calculating a much smaller space, such as a dollhouse).
How do you calculate the square feet of a house?
Here’s how to calculate your square footage:
1. Sketch a floor plan of the home’s interior.
2. Break down the house into measurable rectangles.
3. Measure the length and width of each rectangle.
4. Calculate the area of each section.
5. Add up the total area.
## How do you calculate an irregular lot size?
How to use irregular area calculator?
1. Step 1: Measure all sides of the area in one unit (Feet, Meter, Inches or any other).
2. Step 2: Enter length of horizontal sides into Length 1 and Length 2. And Width of the vertical sides into Width 1 and Width 2.
3. Step 3: Press calculate button.
4. Our Formula: Area = b × h.
How do you get square feet?
To calculate feet squared (or sq. ft. for short), determine the length and width of the area you are working with, measured in feet. Multiply the length by the width and you’ll have the square feet.
### How do you calculate irregular square feet?
How to Calculate the Square Feet of Odd Shapes
1. Measure all the dimensions or sides of the area.
2. Divide the drawing into shapes. Squares and rectangles are easiest, followed by triangles and then circles.
3. Figure the area of each shape.
4. Add the areas of all the individual shapes to find the total square footage.
How do I figure square feet of a room?
## How do you calculate an uneven area?
To find the Area of Irregular Shapes, first, we need to divide the Irregular Shape into Regular Shapes that you can recognize such as triangles, rectangles, circles, Squares and so forth. Then, find the Area of these individual Shapes and add them to get an Area of Irregular Shapes.
How to calculate square footage?
When you buy a new home, thinking about how to measure its square footage probably isn’t top of mind. And while it may seem like an inconsequential metric, it’s one of the most important factors that determine a property’s value. “If you mismeasure a
### How do you complete a square in a math problem?
Divide all terms by a (the coefficient of x2 ).
• Move the number term ( c/a) to the right side of the equation.
• Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation.
• What is website that solves math problems?
The algebra section allows you to expand,factor or simplify virtually any expression you choose.
• The equations section lets you solve an equation or system of equations.
• The inequalities section lets you solve an inequality or a system of inequalities for a single variable.
• ## How do you do square root math problems?
To solve square root problems, understand that you are finding the number that, when multiplied by itself, equals the number in the square root. For quick recall, memorize the first 10-12 perfect squares, so that you recognize the square root of numbers like 9, 25, 49, or 121.
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# Point of tangency of a circle formula.asp
Oct 25, 2018 · A tangent line is a line that passes by a circle and just touches it at only one point. The word tangent actually means to touch, we get the English word tangible from this same root. If we draw a radius to the point of tangency, then the radius and the tangent line are perpendicular.
Make a conjecture about the angle between the radius and the tangent to a circle at a point on the circle. Complete the sentence: the product of the \(\ldots \ldots\) of the radius and the gradient of the \(\ldots \ldots\) is equal to \(\ldots \ldots\). A circle with centre \(C(a;b)\) and a radius of \(r\) units is shown in the diagram above. Tangents to a circle from a point outside the circle - use of the tangency condition Example: Find the area of the triangle made by points of contact of tangents, drawn from the point: A(15, 12) to the circle (x-5) 2 + (y-2) 2 = 20, and the center S of the circle.
Tangents to a circle from a point outside the circle - use of the tangency condition Example: Find the area of the triangle made by points of contact of tangents, drawn from the point: A(15, 12) to the circle (x-5) 2 + (y-2) 2 = 20, and the center S of the circle. Tangent to a Circle. A tangent to a circle is a straight line, in the plane of the circle, which touches the circle at only one point. The point is called the point of tangency or the point of contact. Tangent to a Circle Theorem: A tangent to a circle is perpendicular to the radius drawn to the point of tangency. Oct 25, 2018 · A tangent line is a line that passes by a circle and just touches it at only one point. The word tangent actually means to touch, we get the English word tangible from this same root. If we draw a radius to the point of tangency, then the radius and the tangent line are perpendicular.
Chapter Test A ~ i .,. .~,.1 . For use after Chapter 10 . The diameter of a circle is given. Find the radius. 1. d = 8 ft 2. d = 9 em 3. d = 2.1 m The radius of 08 is given. Find the diameter of 0B. 4. r = 21 em 5. r = 33 ft 6. r = 2.9 m Using the diagram below, match the notation with the term that best describes it. 7. Chord 8. Point of ... So the center of the circle is at (2, 0). Step 2: find the slope of the tangent line. this is the negative reciprocal of the radius from the circle's center to the point of tangency, because the tangent and the radius are perpendicular: m = - (-1 - 2) / (4 - 0) = 3 / 4. A line is tangent to a circle if it touches it at one and only one point. If a line is tangent to a circle, then it is perpendicular to the radius drawn to the point of tangency. Check out the bicycle wheels in the below figure. In this figure, the wheels are, of course, circles, the spokes are radii, and the ground is a tangent line.
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Quiz 2. The Vector Form For the General Solution / Transpose Matrices. Math 2568 Spring 2017.
Problem 273
(a) The given matrix is the augmented matrix for a system of linear equations.
Give the vector form for the general solution.
$\left[\begin{array}{rrrrr|r} 1 & 0 & -1 & 0 &-2 & 0 \\ 0 & 1 & 2 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 \\ \end{array} \right].$
(b) Let
$A=\begin{bmatrix} 1 & 2 & 3 \\ 4 &5 &6 \end{bmatrix}, B=\begin{bmatrix} 1 & 0 & 1 \\ 0 &1 &0 \end{bmatrix}, C=\begin{bmatrix} 1 & 2\\ 0& 6 \end{bmatrix}, \mathbf{v}=\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}.$ Then compute and simplify the following expression.
$\mathbf{v}^{\trans}\left( A^{\trans}-(A-B)^{\trans}\right)C.$
Solution.
(a) Give the vector form for the general solution
The system of linear equation represented by the augmented matrix is
\begin{align*}
x_1-x_3-2x_5&=0\\
x_2+2x_3-x_5&=0\\
x_4+x_5=0.
\end{align*}
Solving the preceding system, we obtain
\begin{align*}
x_1&=x_3+2x_5\\
x_2&=-2x_3+x_5\\
x_4&=-x_5.
\end{align*}
Here $x_2, x_5$ are free variables and the rest are dependent variables.
Hence the general solution $\mathbf{x}$ can be expressed as
\begin{align*}
\mathbf{x}&=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5
\end{bmatrix}=\begin{bmatrix}
x_3+2x_5 \\
-2x_3+x_5 \\
x_3 \\
-x_5 \\
x_5
\end{bmatrix}\10pt] &=\begin{bmatrix} x_3 \\ -2x_3 \\ x_3 \\ 0 \\ 0 \end{bmatrix}+\begin{bmatrix} 2x_5 \\ x_5 \\ 0 \\ -x_5 \\ x_5 \end{bmatrix}=x_3\begin{bmatrix} 1 \\ -2 \\ 1 \\ 0 \\ 0 \end{bmatrix}+x_5\begin{bmatrix} 2 \\ 1 \\ 0 \\ -1\\ 1 \end{bmatrix}. \end{align*} Therefore, the vector form for the general solution is \[\mathbf{x}=x_3\begin{bmatrix} 1 \\ -2 \\ 1 \\ 0 \\ 0 \end{bmatrix}+x_5\begin{bmatrix} 2 \\ 1 \\ 0 \\ -1\\ 1 \end{bmatrix}.
(b) Matrix product and transpose
First of all, note that by the property of the transpose we have
$(A-B)^{\trans}=A^{\trans}-B^{\trans}.$ Hence we can simply the middle part:
\begin{align*}
A^{\trans}-(A-B)^{\trans}&= A^{\trans}-(A^{\trans}-B^{\trans})\\
&=A^{\trans}-A^{\trans}+B^{\trans}=B^{\trans}.
\end{align*}
Thus the expression becomes
$\mathbf{v}^{\trans}\left( A^{\trans}-(A-B)^{\trans}\right)C=\mathbf{v}^{\trans}B^{\trans}C.$ The transposes of $\mathbf{v}$ and $B$ are
$\mathbf{v}^{\trans}=\begin{bmatrix} 0 & 1 & 0 \\ \end{bmatrix} \text{ and } B^{\trans}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 1 &0 \end{bmatrix}.$ Thus we have
\begin{align*}
&\mathbf{v}^{\trans}\left( A^{\trans}-(A-B)^{\trans}\right)C=\mathbf{v}^{\trans}B^{\trans}C\\
&=\begin{bmatrix}
0 & 1 & 0 \\
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
0 & 1 \\
1 &0
\end{bmatrix}
\begin{bmatrix}
1 & 2\\
0& 6
\end{bmatrix}\\
&=\begin{bmatrix}
0 & 1
\end{bmatrix}\begin{bmatrix}
1 & 2\\
0& 6
\end{bmatrix}
=\begin{bmatrix}
0 & 6
\end{bmatrix}.
\end{align*}
In conclusion, we have obtained
$\mathbf{v}^{\trans}\left( A^{\trans}-(A-B)^{\trans}\right)C=\begin{bmatrix} 0 & 6 \end{bmatrix}.$
Comment.
These are Quiz 2 problems for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017.
List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017
There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions.
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Find All Matrices $B$ that Commutes With a Given Matrix $A$: $AB=BA$
Let $A=\begin{bmatrix} 1 & 3\\ 2& 4 \end{bmatrix}.$ Then (a) Find all matrices $B=\begin{bmatrix} x & y\\ z& w \end{bmatrix}$...
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# Logic and Set Notation
Set theory is a branch of mathematical logic. Therefore, it is natural that logical language and symbols are used to describe sets. In this section, we will look at the basic logical symbols and ways of defining sets.
## Propositional Logic
A proposition is a declarative statement which is either true or false. If a proposition is true, then we say it has a truth value of true. Respectively, if a proposition is false, its truth value is false. So, for example, the following statements have a truth value of true:
• The Earth revolves around the Sun;
• $$10 + 3 = 13;$$
• If $$x$$ is an even integer, then $${x^2}$$ is also even.
Examples of false propositions:
• An electron is heavier than a proton;
• $$1 + 2 \gt 3;$$
• $$6$$ is a prime number.
Not all sentences are propositions:
• $$x \gt 5$$ (This may be true or false depending on $$x$$)
• Is it raining? (This is a question, not a declarative sentence)
• Mondrian paintings are too abstract. (What is abstract and too abstract?)
To represent propositions, we denote them by letters. The most common letters are $$p,$$ $$q,$$ $$r,$$ $$s,$$ $$t.$$
Using logical operators or connectives, we can build compound propositions.
## Logical Operators and Truth Tables
Let $$p$$ and $$q$$ be two propositions. Each of these statements can take two values - true ($$T$$) and false ($$F$$). So there are $$4$$ pairs of input values: $$TT,$$ $$TF,$$ $$FT,$$ and $$FF.$$
Suppose that a new proposition $$r$$ is composed of $$p$$ and $$q.$$ The truth values of the proposition $$r$$ can take different values ($$T$$ or $$F$$) for each pair of input values.
There are total $$2^4 = 16$$ possible output combinations (truth functions) for $$2$$ binary input variables. Each of these combinations is represented by a certain logical operator.
Further we'll look at the most important operators.
### Negation
Negation is a unary logical operator. If $$p$$ is a proposition, then the negation of $$p$$ is called not $$p$$ and is denoted by $$\lnot p.$$
To represent the meaning of a logical expression, it is convenient to use a truth table. Each row of the table contains one possible configuration of the input variables and truth values of the output proposition(s).
In case of the negation operator, the truth table is very simple:
As you can see, the logical negation operator reverses the truth value of the input proposition.
#### Example 1:
$$p:$$ A trapezoid is a quadrilateral (true) $$\lnot p:$$ A trapezoid is not a quadrilateral (false)
#### Example 2:
$$p:$$ $$2$$ is a prime number (true) $$\lnot p:$$ $$2$$ is not a prime number (false)
Сonsider now some binary operators.
### Conjunction
If $$p$$ and $$q$$ are two propositions, then their conjunction means $$p$$ and $$q$$ and is denoted by $$p \land q.$$
The conjunction $$p \land q$$ is true only when both $$p$$ and $$q$$ are true. Otherwise, it is false. Thus, the truth table for conjunction looks as follows:
#### Example 1:
$$p:$$ United Kingdom is a member of European Union (false) $$q:$$ Ireland is a member of European Union (true) $$p \land q:$$ United Kingdom and Ireland are members of European Union (false)
#### Example 2:
$$p:$$ $$3$$ is prime (true) $$q:$$ $$3$$ is odd (true) $$p \land q:$$ $$3$$ is prime and odd (true)
### Disjunction
If $$p$$ and $$q$$ are two propositions, then their disjunction means $$p$$ or $$q$$ and is denoted by $$p \lor q.$$
The disjunction $$p \lor q$$ is true when either $$p$$ is true, $$q$$ is true, or both are true. It is false, if both $$p$$ and $$q$$ are false.
Truth table for disjunction:
#### Example 1:
$$p:$$ A proton has a negative charge (false) $$q:$$ A neutron has a negative charge (false) $$p \lor q:$$ A proton or a neutron has a negative charge (false)
#### Example 2:
$$p:$$ $$\frac{2}{3}$$ is a rational number (true) $$q:$$ $$\frac{\sqrt{2}}{5}$$ is a rational number (false) $$p \lor q:$$ Either $$\frac{2}{3}$$ or $$\frac{\sqrt{2}}{5}$$ or both the numbers are rational (true)
### Material Conditional
The material conditional of $$p$$ and $$q$$ means the statement if $$p$$ then $$q$$ and is denoted by $$p \to q.$$ This logical operator is also called just a conditional statement.
If $$p$$ is true, then the conditional $$p \to q$$ takes the truth value of $$q.$$ If $$p$$ is false, then the conditional $$p \to q$$ is assumed to be true by default.
Here is the truth table for conditional statement:
The conditional $$p \to q$$ can be expressed by different sentences, some of them are listed below:
• $$p$$ implies $$q$$
• $$p$$ is a sufficient condition for $$q$$
• $$q$$ is a necessary condition for $$p$$
• $$q$$ follows from $$p$$
• $$p$$ only if $$q$$
#### Example:
$$p:$$ $$x$$ is divisible by $$2$$ and $$3$$ $$q:$$ $$x$$ is divisible by $$6$$ $$p \to q:$$ If $$x$$ is divisible by $$2$$ and $$3,$$ then $$x$$ is divisible by $$6.$$ $$x = 12:$$ $$p$$ is true, $$q$$ is true, and $$p \to q$$ is true. $$x = 13:$$ $$p$$ is false, $$q$$ is false, and $$p \to q$$ is true.
### Special Conditional Propositions
• The converse of $$p \to q$$ is the proposition $$q \to p$$
• The contrapositive of $$p \to q$$ is the proposition $$\neg q \to \neg p$$
• The inverse of $$p \to q$$ is the proposition $$\neg p \to \neg q$$
If the statement $$p \to q$$ is true, then the contrapositive is also true. If the converse is true, then the inverse is also true.
The special conditional operators are defined by the following truth table:
#### Example:
Statement $$p$$ A quadrilateral is a rectangle (false) Statement $$q$$ The sum of interior angles of a quadrilateral is $$360$$ degrees (true). Conditional statement $$p \to q$$ If a quadrilateral is a rectangle, then the sum of its interior angles is $$360$$ degrees (true). Converse of $$p \to q:$$ $$q \to p$$ If the sum of interior angles of a quadrilateral is $$360$$ degrees, then it is a rectangle (false). Inverse of $$p \to q:$$ $$\neg p \to \neg q$$ If a quadrilateral is a not a rectangle, then the sum of its interior angles is not $$360$$ degrees (false). Contrapositive of $$p \to q:$$ $$\neg q \to \neg p$$ If the sum of interior angles of a quadrilateral is not $$360$$ degrees, then it is not a rectangle (true).
### Material Biconditional
The material biconditional or logical biconditional of $$p$$ and $$q$$ means the statement $$p$$ if and only if $$q$$ and is denoted by $$p \leftrightarrow q.$$
The biconditional statement has the same truth value as the compound logical operator $$\left( {p \to q} \right) \land \left( {q \to p} \right),$$ which means $$p$$ implies $$q$$ and $$q$$ implies $$p.$$
The truth table for biconditional statement has the form
#### Example 1:
$$p:$$ Three vectors are coplanar (true) $$q:$$ The scalar triple product of three vectors is zero (true) $$p \leftrightarrow q:$$ Three vectors are coplanar if and only if their scalar triple product is zero (true)
#### Example 2:
$$p:$$ The tennis match will be played outdoors (false) $$q:$$ It is raining (true) $$p \leftrightarrow q:$$ The tennis match will be played outdoors if and only if it is raining (false)
### Logical Equivalence
Propositions $$p$$ and $$q$$ are said to be logically equivalent if they have the same truth tables. The logical equivalence of $$p$$ and $$q$$ is denoted as $$p \equiv q,$$ or sometimes by $$\Leftrightarrow$$ depending on the notation being used.
## Predicates and Quantifiers
So far, we considered propositions which are defined as statements that can be either true or false.
To extend the simple propositional logic, we introduce the concept of predicate.
A predicate is a logical statement that contains one or more variables or parameters. The predicates are denoted by a capital letter and the variables listed as arguments, like $$P\left( x \right)$$ or $$Q\left( {x,y} \right).$$ The truth value of a predicate depends on the values of its variables.
#### Example 1:
Predicate $$P\left( x \right)$$ $$P\left( x \right):$$ $$x$$ is a planet $$x$$ = Venus True proposition $$P\left( \text{Venus} \right):$$ Venus is a planet $$x$$ = Antares False proposition $$P\left( \text{Antares} \right):$$ Antares is a planet
#### Example 2:
Predicate $$Q\left( {x,y} \right)$$ $$Q\left( {x,y} \right): {x^2} + {y^2} \le 4$$ $$x = 1, y = -1$$ True proposition $$Q\left( {1,-1} \right): {1^2} + {\left( { - 1} \right)^2} \le 4$$ $$x = 1, y = 2$$ False proposition $$Q\left( {1,2} \right): {1^2} + {2^2} \le 4$$
Similarly to propositions, predicate take two values - true or false. Therefore, all operations of logic algebra are applicable to them. Using the operators $$\neg, \land, \lor, \rightarrow,$$ and $$\leftrightarrow,$$ we can form more complex predicates.
There is also an additional operation defined on predicates and called quantification. Quantification allows us to specify the extent of validity of a predicate, that is, the range of values of variables for which the predicate should hold.
There are two types of quantifiers in predicate logic − universal quantifier and existential quantifier.
### Universal Quantifier
The universal quantifier is used to express sentences with words like all or every. It is denoted by the symbol $$\forall.$$ The notation $$\forall x P\left({x}\right)$$ means "for every value of $$x$$ in a particular domain, the predicate $$P\left({x}\right)$$ is true". The domain is called the universe of discourse or domain of discourse.
### Existential Quantifier
The existential quantifier is used to express sentences with words like some or there is a. It is denoted by the symbol $$\exists.$$ The notation $$\exists x P\left({x}\right)$$ means "there is some value of $$x$$ such that $$P\left({x}\right)$$ is true".
Predicates, logical operators, and quantifiers are a powerful set of tools for describing mathematical objects and for modeling the real world.
## Set Builder Notation
The set builder notation is used to specify a set of objects by means of a predicate. The common notation includes $$3$$ parts: a variable $$x,$$ a colon or vertical bar separator, and a logical predicate $$P\left({x}\right):$$
$S = \left\{ {x|P\left( x \right)} \right\},$
where $$S$$ denotes the set of objects.
The single variable $$x$$ can be replaced with a term that may include one or more variables, combined with functions acting on them. The predicate $${P\left( x \right)}$$ may be represented by a complex logical formula.
### Some Other Definitions and Notations
$$\{\;\}$$ is a set $$x \in S$$ $$x$$ is an element or member of $$S$$ $$x \notin S$$ $$x$$ is not an element of $$S$$ $$S \subseteq T$$ $$S$$ is a subset of $$T$$ $$S = T$$ is equivalent to $$\left({S \subseteq T}\right) \land \left({T \subseteq S}\right)$$ $$S \subset T$$ $$S$$ is a proper subset of $$T.$$ This means $$\left({S \subseteq T}\right) \land \left({T \ne S}\right)$$ $$\varnothing$$ the empty set
See solved problems on Page 2.
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Therefore, if you are aiming to tackle even the 160-170 questions on the Q section, you need to have some crafty mental math tricks up your sleeve. One of the most powerful involves the clever use of a famous algebra formula: the difference of two squares formula. See this page — "Quant: Difference of Two Squares" for uses of that formula in general problem solving. Here we will focus on factoring.
## Factoring Big Numbers
In general, factoring a big number can be time-consuming without a calculator. The QUANT might ask a question in which you need to know the factors, or the prime factorization, of a large number. See this page — "Math: Factors" — for more on prime factorizations.
First of all, notice how easy it is to square the multiples of 10. If you can square the numbers from 1 to 10, you can square the multiples of 10 from 10 to 100.
10^2 = 100
20^2 = 400
30^2 = 900
40^2 = 1600
50^2 = 2500
60^2 = 3600
70^2 = 4900
80^2 = 6400
90^2 = 8100
100^2 = 10000
Those should all be recognizable as nice round perfect squares. Now, suppose you are in a situation in which you have to factor, say, 1591, or 3551, or 8099. Notice, all of those are a perfect square less than one of these multiples of ten squared.
1591 = 1600 - 9 = (40)^2 - (3)^2 = (40 + 3)(40 - 3) = 43 \times 37
3551 = 3600 - 49 = (60)^2 - (7)^2 = (60 + 7)(60 - 7) = 67 \times 53
8099 = 8100 - 1 = (90)^2 - (1)^2 = (90 + 1)(90 - 1) = 91 \times 89 = 7 \times 13 \times 89
In general, the QUANT is not going to put you in a situation in which you have to find the prime factorization of a general four digit number. If this situation does arise, you can bet there’s an enormously simplifying trick available, and factoring via the difference of two squares is an awfully likely candidate for that trick.
## Factoring Decimals
Just as the difference of two square can simplify factoring big numbers, it can also simplify factoring decimals. To demonstrate this, I am going to show the solution to a flamboyantly recondite question.
Q1. \dfrac{0.99999999}{1.0001} - \dfrac{0.99999991}{1.0003}=
(A) 10^{-8}
(B) 3(10^{-8})
(C) 3(10^{-4})
(D) 2(10^{-4})
(E) 10^{-4}
Notice that 1.0001 = 1 + \Big(10^{-4}\Big), and 1.0003 = 1 + 3\Big(10^{-4}\Big).
Now, notice that both numerators simplify via difference of two squares formula.
0.99999999 = 1-10^{-8} = 1^2-\Big(10^{-4}\Big)^2 =\Big(1+10^{-4}\Big)\Big(1-10^{-4}\Big)
0.99999991 = 1-9\Big(10^{-8}\Big) = 1^2-\Big[3\Big(10^{-4}\Big)\Big]^2 = \Big[1+3\Big(10^{-4}\Big)\Big] \times \Big[1-3\Big(10^{-4}\Big)\Big]
Thus, the two fractions become
\dfrac{0.99999999}{1.0001} = \Big(1+10^{-4}\Big)\dfrac{\Big(1-10^{-4}\Big)}{\Big(1+10^{-4}\Big)} = 1 - 10^{-4}
\dfrac{0.99999991}{1.0003} =\Big[1+3\Big(10^{-4}\Big)\Big]\dfrac{\Big[1-3\Big(10^{-4}\Big)\Big]}{\Big[1+3\Big(10^{-4}\Big)\Big]} = 1 - 3\Big(10^{-4}\Big)
The difference:
\Big(1 - 10^{-4}\Big) - \Big[1 - 3\Big(10^{-4}\Big)\Big] = 3\Big(10^{-4}\Big)-10^{-4} = 2\Big(10^{-4}\Big)
Q2. \dfrac{49^2 - 35^2}{14} =
(A) 74
(B) 76
(C) 78
(D) 79
(E) 84
Crunching these numbers is cumbersome and, fortunately, unnecessary. Our numerator is a difference of squares, meaning we can break it up by factoring.
This gives us the much simpler \dfrac{(49 + 35)(49 - 35)}{14}.
This simplifies to \dfrac{(84)(14)}{14}.
We can cancel the 14s, which leaves us with 84
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# Algebraic Expressions Objectives: 1)To evaluate algebraic expressions 2)To simplify algebraic expressions.
## Presentation on theme: "Algebraic Expressions Objectives: 1)To evaluate algebraic expressions 2)To simplify algebraic expressions."— Presentation transcript:
Algebraic Expressions Objectives: 1)To evaluate algebraic expressions 2)To simplify algebraic expressions
Definitions Variable: a symbol that represents one of more numbers Algebraic Expression: AKA variable expression. An expression containing one or more variables Evaluate: substituting numbers for variables in an expression and simplifying Term: a number, variable or the product of a number and variable Coefficient: The numerical factor in a term
Properties for Simplifying Algebraic Expressions Definition of Subtraction: a – b = a + (-b) Definition of Division a ÷ b = a/b = a 1/b, b ≠ 0 Distributive Property a(b + c) = ab + ac ; a(b – c) = ab – ac Zero Product Property 0 a = 0
Properties for Simplifying Algebraic Expressions (continued) Multiplicative Inverse: -1 a = -a Opposite of a Sum: -(a + b) = -a + -b Opposite of a Difference -(a – b) = b – a
Properties for Simplifying Algebraic Expressions (continued) Opposite of a Product -(ab) = -a b = a -b Opposite of an Opposite -(-a) = a
Example #1: Evaluating an Algebraic Expression Evaluate 3a – b 2 + ab for a = 3 and b = -1 Replace the variables with numbers but put them in parenthesis 3(3) – (-1 2 ) + (3)(-1) 9 – (1) + (-3) 9 – 1 – 3 8 – 3 5 We’ve simplified all we can, so this is the final answer
Example #2: Combining Like Terms a.3k – k 2k b.5z 2 – 10z – 8z 2 + z 5z 2 – 8z 2 – 10z + z -3z 2 – 9z c.- (m + n) + 2(m – 3n) - m – n + 2m – 6n - m + 2m – n – 6n m – 7n
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# Selina Concise Solutions for Chapter 5 Playing with Numbers Class 8 ICSE Mathematics
### Exercise 5A
1. Write quotient when the sum of 73 and 37 is divided by
(i) 11
(ii) 10
Solution
Sum of 73 and 37 b is to be divided by
(i) 11
(ii) 10
Let ab = 73
and ba = 37
∴ a = 7
and b = 3
(i) The quotient of ab + bc i.e (73 + 37) when divided by 11 is a + b = 7 + 3 = 10
(ii) The quotient of ab + ba i.e. (73 + 37) when divided by 10 (i.e. a + b ) is 11
2. Write the quotient when the sum of 94 and 49 is divide by
(i) 11
(ii) 13
Solution
Sum of 84 and 49 is to be divide by
(i) 11
(ii) 13
Let ab = 94
and ba = 49
∴ a = 9 and b = 4
(i) The quotient of 94 + 49 (i.e 9 + 4 = 13
(ii) The quotient of 94 + 49 (i.e. ab + ba)
When divided by 13 i.e. (a + b) is 11
3. Find the quotient when 73 – 37 is divide by
(i) 9
(ii) 4
Solution
Difference of 73 – 37 is to be divided by
(i) 9
(ii) 4
Let ab = 73 and ba = 37
∴ a = 7 and b = 3
(i) The quotient of 73 – 37 (i.e. ab – bc) when divided by 7 is a - b i.e. 7 – 3 = 4
(ii)
The quotient of 73 – 37 (i.e. ab – ba) when divided by 4 i.e. (a – b) is 9
4. Find the quotient when 94 – 49 is divided by
(i) 9
(ii) 5
Solution
Difference of 94 and 49 is divided by
(i) 9
(ii) 5
Let ab = 94 and ba = 49
∴ a = 9 and b = 4
(i) The quotient of 94 – 49 i.e. (ab – ba) when divided by 9 is (a – b) i.e. 9 – 4 = 5
(ii) The quotient of 94 – 49 i.e. (ab – ba) when divide by 5 i.e.(a – b) is 9
5. Show that 527 + 752 + 275 is exactly divisible by 14.
Solution
Property:
abc = 100a + 106 + c ...(i)
bca = 1006 + 10c + a ...(ii)
and cab = 100c + 10a + b …(iii)
Adding, (i), (ii) and (iii), we get
abc + bca + cab = 111a + 111b + 111c = 111(a + b + c) = 3 × 37 (a + b + c)
Now, let us try this method on
527 + 752 + 275 to check is it exactly divisible by 14
Here, a = 5, 6 = 2, c = 7
527 + 752 + 275 = 3 × 37(5 + 2 + 7) = 3 × 37 × 14
Hence, it shows that 527 + 752 + 275 is exactly divisible by 14.
6. If a = 6, show that abc = bac.
Solution
Given: a = 6
To show: abc = bac
Proof:
abc = 100a + 106 + c ...(i)
(By using property 3)
bac = 1006 + 10a + c …(ii)
(By using property 3)
Since, a = 6
Substitute the value of a = 6 in equation (i) and (ii), we get
abc = 1006 + 106 + c …(iii)
bac = 1006 + 106 + c …(iv)
Subtracting (iv) from (iii) abc – bac = 0
abc = bac
Hence, proved.
7. if a > c; show that abc – cba = 99(a – c).
Solution
Given, a > c
To show : abc – cba = 99(a - c)
Proof:
abc = 100a + 10b + c …(i)
(By using property 3)
cba = 100c + 10b + a ...(i)
(By using property 3)
Subtracting, equation (ii) from (i), we get
abc – cba = 100a + c – 100c – a
⇒ abc – cba = 99a - 99c
⇒ abc – cba = 99(a – c)
Hence, proved.
8. If c > a; show that cba – abc = 99(c – a).
Solution
Given, c > a
To show : cba – abc = 99(c – a)
Proof:
cba = 100c + 106 + a …(i)
(by using property 3)
abc = 100a + 106 + c ...(ii)
(by using property 3)
Subtracting (ii) from (i)
cba – abc = 100c + 106 + a – 100a – 106c
⇒ cba – abc = 99c – 99a
⇒ cba – abc = 99(c – a)
Hence proved.
9. If a = c, show that cba- abc = 0
Solution
Given: a = c
To show: cba - abc = 0
Proof:
cba = 100c + 106 + a …(i)
(By using property 3)
abc = 100a + 106 + c …(ii)
(By using property 3)
Since, a = c,
Substitute the value of a = c in equation (i) and (ii), we get
cba = 100c + 10b + c ...(iii)
abc = 100c + 10b + c …(iv)
subtracting (iv) from (iii), we get
cba – abc – 100c + 106 + c – 100c – 106 – c
⇒ cba – abc = 0
⇒ cba = abc
Hence, proved.
10. Show that 954 – 459 is exactly divisible by 99.
Solution
To show: 954 – 459 is exactly divisible by 399, where a = 9, b = 5, c = 4,
abc = 100a + 10b + c
⇒ 954 = 100 × 9 + 10 × 5 + 4
⇒ 954 = 900 + 50 + 4 ...(i)
And 459 = 100 × 4 + 10 × 5 + 9
⇒ 459 = 400 + 50 + 9 …(ii)
Subtracting (ii) from (i), we get
954 – 59 = 900 + 50 + 4 – 400 – 50 – 9
⇒ 954 – 459 = 500 – 5
⇒ 954 – 459 = 495
⇒ 954 – 458 = 99 × 5
Hence, 954 – 459 is exactly divisible by 99
Hence proved.
### Exercise 5 B
1.
Solution
A = 7 as 7 + 5 = 12.
We want 2 at units place and 1 is carry over. Now 3 + 2 + 1 = 6
B = 6
Hence, A = 7 and B = 6
2.
Solution
A = 5 as 8 + 5 = 13.
We want 3 at units place and 1 is carry over. Now 9 + 4 + 1 = 14.
B = 4 and C = 1
Hence A = 5 and B = 4 and C = 1
3.
Solution
B = 9 as 9 + 1 = 10.
We want 0 at units place and 1 is carry over. Now B – 1 – 1 = A.
∴ A = 9 – 2 = 7
Hence A = 7 and B = 9
4.
Solution
B = 7 as 7 + 1 = 8. We want 8 at unit place.
Now,
7 + A = 11
∴ A = 11 – 7 = 4
Hence A = 4 and B = 7
5.
Solution
A + B = 9
And 2 + A = 10
∴ A = 10 – 2= 8
And 8 + B = 9
∴ B = 9 – 8 = 1
Hence, A = 8 and B = 1
6.
Solution
As we need A at unit place and 9 at ten’s place,
A = 6 as 6 × 6 = 36
7.
Solution
As we need B at unit place and B at ten’s place,
∴ B = 4 as 6 × 4 = 24
Now, we want to find A, 6 × A + 2 = 4 (at unit’s place)
∴ A = 7
8.
Solution
As we need B at unit place and A at ten’s place,
∴ B = 0 as 3 × 0 = 0
Now, we want to find A, 3 × A (at unit’s place)
∴ A = 5, as 3 × 5 = 15
∴ C = 1
9.
Solution
As we need B at unit place and A at ten's place,
B = 0 as 5 × 0 = 0
Now, we want to find A, 5 × A = A (at unit's place)
A = 5, as 5 × 5 = 25
C = 2
10.
Solution
5 + A = 13
And A + 4 = 13
∴ A = 13 – 5 = 8
Hence A = 8
11.
Solution
C + 5 = 11
∴ C = 11 – 5 = 6
And 8 + B + 1 = 15
∴ B = 15 – 9 = 6
And A + 5 + 1 = 13
∴ A = 13 – 6 = 7
And 6 + D + 1 = 9
∴ D = 9 – 7 = 2
Hence A = 7, B = 6, C = 6 and D = 2
### Exercise 5 C
1. Find which of the following numbers are divisible by 2:
(i) 192
(ii) 1660
(iii) 1101
(iv) 2079
Solution
A number having its unit digit 2, 4, 6, 8 or 0 is divisible by 2.
So, Number 192, 1660 are divisible by 2.
2. Find which of the following numbers are divisible by 3:
(i) 261
(ii) 111
(iii) 6657
(iv) 2574
Solution
A number is divisible by 3 if the sum of its digits is divisible by 3,
So, 261, 111 are divisible by 3.
3. Find which of the following numbers are divisible by 4:
(i) 360
(ii) 3180
(iii) 5348
(iv) 7756
Solution
A number is divisible by 4, if the number formed by the last two digits is divisible by 4.
So, number 360, 5348, 7756 are divisible by 4.
4. Find which of the following numbers are divisible by 5 :
(i) 3250
(ii) 5557
(iii) 39255
(iv) 8204
Solution
A number having its unit digit 5 or 0, is divisible by 5.
So, 3250, 39255 are all divisible by 5.
5. Find which of the following numbers are divisible by 4:
(i) 360
(ii) 3180
(iii) 5348
(iv) 7756
Solution
A number having its unit digit 0, is divisible by 10.
So, 5100, 3400 are all divisible by 10.
6. Which of the following numbers are divisible by 11:
(i) 2563
(ii) 8307
(iii) 95635
Solution
A number is divisible by 11 if the difference of the sum of digits at the odd places and sum of the digits at even places is zero or divisible by 11.
So, 2563 is divisible by 11.
### Exercise 5 D
1. For what value of digit x, is :
1 × 5 divisible by 3?
Solution
1 × 5 is divisible by 3
⇒ 1 + x + 5 is a multiple of 3
⇒ 6 + x = 0, 3, 6, 9
Since, x is a digit
x = 0, 3, 6 or 9
2. 31 × 5 is divisible by 3?
Solution
31 × 5 is divisible by 3
⇒ 3 + 1 + x + 5 is a multiple of 3
⇒ 9 + x = 0, 3, 6, 9,
⇒ x = -9 , -6, -3, 0, 3, 6, 9,
Since, x is a digit
x = 0, 3, 6 or 9
3. 28 × 6 is a multiple of 3?
Solution
28 × 6 is a multiple of 3
2 + 8 + x + 6 is a multiple of 3
⇒ 16 + x = 0, 3, 6, 9, 12, 15,18
⇒ x = -18, -5, -2, 0, 2, 5, 8
Since, x is a digit = 2, 5, 8
4. 24x divisible by 6?
Solution
24x is divisible by 6
⇒ 2 + 4 + x is a multiple of 6
⇒ 6 + x = 0, 6, 12
⇒ x = -6, 0, 6
Since, x is a digit
x = 0, 6
5. 3 × 26 a multiple of 6?
Solution
3 × 26 is a multiple of 6
3 + x + 2 + 6 is a multiple of 3
⇒ 11 + x = 0, 3, 6, 9, 12, 15, 18, 21
⇒ x = -11, -8, -5, -2, 1, 4, 7, 10, ……
Since, x is a digit
x = 1, 4 or 7
6. 42 × 8 divisible by 4?
Solution
42 × 8 is divisible by 4
⇒ 4 + 2 + x + 8 is a multiple of 2
⇒ 14 + x = 0, 2, 4, 6, 8,
⇒ x = -8, -6, -4, -2, 2, 4, 6, 8,
Since, x is a digit 2, 4, 6, 8
Question 7: 9142 × a multiple of 4?
Solution 7:
9142 × a multiple of 4
⇒ 9 + 1 + 4 + 2 + x is a multiple of 4
⇒ 16 + x = 0, 4, 8, ……..
x = -8, -4, 0, 4, 8
since, x is a digit 4, 8
8. 7 × 34 divisible by 9?
Solution
7 × 34 is multiple of 9
⇒ 7 + x + 3 + 4 is a multiple of 9
⇒ 14 + x = 0, 9, 18, 27,
⇒ x = -1, 4, 13,
Since, x is a digit x = 4
9. 5 × 555 a multiple of 9?
Solution
Sum of the digits of 5 × 555
= 5 + x + 5 + 5 + 5 = 20 + x
It is multiple by 9
The sum should be divisible by 9
Value of x will be 7
10. 3 × 2 divisible by 11?
Solution
Sum of the digit in even place = x
And sum of the digits in odd place = 3 + 2 = 5
Difference of the sum of the digits in even places and in odd places = x – 5
3 × 2 is a multiple of 11
⇒ x – 5 = 0, 11, 22,
⇒ x = 5, 16, 27
Since, x is a digit x = 5
11. 5 × 2 is a multiple of 11?
Solution
Sum of digit in even place = x
And sum of the digits in odd place = 5 + 2 = 7
Difference of the sum of the digits in even places
And in odd places = x – 7
5 × 2 is a multiple of 11
⇒ x – 7 = 0, 11, 22
⇒ x – 7, 18, 29,
Since, x is a digit x = 7
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# Exercise 1.3Z: Calculating with Complex Numbers II
Considered numbers
in the complex plane
The following three complex quantities are shown in the complex plane to the right:
$$z_1 = 4 + 3\cdot {\rm j},$$
$$z_2 = -2 ,$$
$$z_3 = 6\cdot{\rm j} .$$
Within the framework of this task, the following quantities are to be calculated:
$$z_4 = z_1 \cdot z_1^{\star},$$
$$z_5 = z_1 + 2 \cdot z_2 - {z_3}/{2},$$
$$z_6 = z_1 \cdot z_2,$$
$$z_7 = {z_3}/{z_1}.$$
Hints:
• This exercise belongs to the chapter Calculating with Complex Numbers.
• The topic of this task is also covered in the (German language) learning video
Rechnen mit komplexen Zahlen ⇒ "Arithmetic operations involving complex numbers".
• Enter the phase values in the range of $-\hspace{-0.05cm}180^{\circ} < \phi ≤ +180^{\circ}$.
### Questions
1
Enter the magnitude and phase of $z_1$ .
$|z_1|\ = \$ $\phi_1\ = \$ $\hspace{0.2cm}\text{deg}$
2
What is $z_4 = z_1 \cdot z_1^{\star} = x_4 + \text{j} \cdot y_4$?
$x_4\ = \$ $y_4\ = \$
3
Calculate $z_5 = z_1 + 2 \cdot z_2 - {z_3}/{2} = x_5 + {\rm j} \cdot y_5$ .
$x_5\ = \$ $y_5\ = \$
4
Specify the magnitude and phase of $z_6 = z_1 \cdot z_2$ $($range $\pm 180^{\circ})$.
$|z_6|\ = \$ $\phi_6\ = \$ $\hspace{0.2cm}\text{deg}$
5
What is the phase value of the purely imaginary number $z_3$?
$\phi_3 \ = \$ $\hspace{0.2cm}\text{deg}$
6
Calculate the magnitude and phase of $z_7 = z_3/z_1$ $($range $\pm 180^{\circ})$.
$|z_7| \ = \$ $\phi_7 \ = \$ $\hspace{0.2cm}\text{deg}$
### Solution
#### Solution
(1) The magnitude can be calculated according to the Pythagorean theorem:
$$|z_1| = \sqrt{x_1^2 + y_1^2}= \sqrt{4^2 + 3^2}\hspace{0.15cm}\underline{ = 5}.$$
$$\phi_1 = \arctan \frac{y_1}{x_1}= \arctan \frac{3}{4}\hspace{0.15cm}\underline{ = 36.9^{\circ}}.$$
(2) Multiplying $z_1$ by its conjugate complex $z_1^{\star}$ yields the purely real quantity $z_4$, as the following equations show:
$$z_4 = (x_1 + {\rm j} \cdot y_1)(x_1 - {\rm j} \cdot y_1)= {x_1^2 + y_1^2}= |z_1|^2 = 25,$$
$$z_4 = |z_1| \cdot {\rm e}^{{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}\phi_1} \cdot |z_1| \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \phi_1}= |z_1|^2 = 25\hspace{0.3cm} \Rightarrow\hspace{0.3cm} x_4 \hspace{0.1cm}\underline{= 25}, \hspace{0.25cm}y_4 \hspace{0.15cm}\underline{= 0}.$$
(3) By dividing into real and imaginary part one can write:
$$x_5 = x_1 + 2 \cdot x_2 - {x_3}/{2} = 4 + 2 \cdot(-2) -0 \hspace{0.15cm}\underline{= 0},$$
$$y_5 = y_1 + 2 \cdot y_2 - {y_3}/{2} = 3 + 2 \cdot 0 - \frac{6}{2} \hspace{0.1cm}\underline{=0}.$$
(4) If one writes $z_2$ as magnitude and phase ⇒ $|z_2| = 2, \ \phi_2 = 180^{\circ}$, one obtains for the product:
$$|z_6| = |z_1| \cdot |z_2|= 5 \cdot 2 \hspace{0.15cm}\underline{= 10},$$
$$\phi_6 = \phi_1 + \phi_2 = 36.9^{\circ} + 180^{\circ} = 216.9^{\circ}\hspace{0.15cm}\underline{= -143.1^{\circ}}.$$
(5) The phase is $\phi_3 = 90^{\circ}$ (see graph above). This can be formally proven:
$$\phi_3 = \arctan \left( \frac{6}{0}\right) = \arctan (\infty) \hspace{0.2cm}\Rightarrow \hspace{0.2cm} \phi_3 \hspace{0.15cm}\underline{= 90^{ \circ}}.$$
(6) First, the more inconvenient solution:
$$z_7 = \frac{z_3}{z_1}= \frac{6{\rm j}}{4 + 3{\rm j}} = \frac{6{\rm j}\cdot(4 - 3{\rm j})}{(4 + 3{\rm j})\cdot (4 - 3{\rm j})} = \frac{18 +24{\rm j}}{25} = 1.2 \cdot{\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 53.1^{ \circ}}.$$
• An easier way of solving the problem is:
$$|z_7| = \frac{|z_3|}{|z_1|} = \frac{6}{5}\hspace{0.15cm}\underline{=1.2}, \hspace{0.3cm}\phi_7 = \phi_3 - \phi_1 = 90^{\circ} - 36.9^{\circ} \hspace{0.15cm}\underline{=53.1^{\circ}}.$$
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# Find common units to compare two Fractions
Related Topics:
Lesson Plans and Worksheets for Grade 4
Lesson Plans and Worksheets for all Grades
Examples, solutions, and videos to help Grade 4 students learn how to find common units or number of units to compare two fractions.
Common Core Standards: 4.NF.2
### NYS Common Core Grade 4 Module 5, Lesson 15
Grade 4, Module 5, Lesson 15 Worksheets
Lesson 15 Concept Development
Problem 1: Reason to compare fractions between 1 and 2.
Problem 2: Reason about the size of fractions as compared to 1 1/2.
Problem 3: Reason using benchmarks to compare two fractions.
Lesson 15 Problem Set
1. Draw an area model for each pair of fractions, and use it to compare the two fractions by writing a >, <, or = symbol on the line. The first two have been partly done for you. Each rectangle represents one whole.
a. 1/2 _____ 2/3
b. 4/5 _____ 3/4
c. 2/5 _____ 4/7
d. 3/7 _____ 2/6
e. 5/8 _____ 6/9
f. 2/3 _____ 3/4
2. Rename the fractions, as needed, using multiplication in order to compare the two fractions in each pair by writing a >, <, or =.
a. 3/5 _____ 5/6
b. 2/6 _____ 3/8
c. 7/5 _____ 10/8
d. 4/3 _____ 6/5
3. Use any method to compare the fractions. Record your answer using >, <, or =.
a. 3/4 _____ 7/8
b. 6/8 _____ 3/5
c. 6/4 _____ 8/6
d. 8/5 _____ 9/6
4. Explain two ways you have learned to compare fractions. Provide evidence using words, pictures, and numbers.
Lesson 15 Homework
1. Draw an area model for each pair of fractions, and use it to compare the two fractions by writing a >, <, or &eq; symbol on the line. The first two have been partly done for you. Each rectangle represents one whole.
a. 1/2 _____ 3/5
b. 2/3 _____ 3/4
c. 4/6 _____ 5/8
d. 2/7 _____ 3/5
e. 4/6 _____ 6/9
f. 4/5 _____ 5/6
2. Rename the fractions as needed using multiplication in order to compare the two fractions in each pair by writing a >, <, or &eq;.
a. 2/3 _____ 2/4
b. 4/7 _____ 1/2
c. 5/4 _____ 9/8
d. 8/12 _____ 5/9
3. Use any method to compare the fractions. Record your answer using >, <, or =.
a. 8/9 _____ 2/3
b. 4/7 _____ 4/5
c. 3/2 _____ 9/6
d. 11/7 _____ 5/3
4. Explain which method you prefer to compare fractions. Provide an example using words, pictures, and numbers.
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# Understanding Dependent and Independent Variables
Sep 14, 2022
## Dependent variable vs Independent variable:
Dependent variable changes in response to another variable called an independent variable. An independent variable causes a change in a dependent variable. It is independent because its value is not affected by other variables.
For example, a student grade depends on how much he studies. His grade is a dependent variable and the amount of time spent on studying is an independent variable.
### Dependent and independent variables
Example 1:
The distance a car travels, d, is dependent on the speed, s, at which it travels. Which variable, d, distance or s, speed is the dependent variable? Explain.
Solution:
Speed is the independent variable, and distance is the dependent variable.
In this scenario, the distance, d, traveled depends on the speed of the car, s, so d is the dependent variable.
The speed of the car, s, affects the distance traveled, d, so s is the independent variable.
Example 2:
A baker used a certain number of cups of batter, b, to make p pancakes. Which variable, p, pancakes or b, batter is the dependent variable? Explain.
Solution:
In this scenario, the number of pancakes, p, made depends on the number of cups of batter, b, so p is the dependent variable.
The number of cups of batter used, b, affects the number of pancakes made, p, so b is the independent variable.
### Multiple independent or dependent variables
Example 3:
Annette is preparing for her term-1 examinations. Identify the independent and dependent variables involved in this process.
Solution:
Step 1:
Identify the variables involved in this scenario.
Step 2:
Determine whether the variables are independent or dependent.
Example 4:
The Sunryde bike shop rents sports bikes for a day or for a week. What are the independent and dependent variables involved in the cost of rental?
Solution:
Step 1:
Identify the variables involved in the cost of rental.
Step 2:
Determine whether the variables are independent or dependent.
## Exercise:
1. A(n) variable changes in response to another variable.
2. A(n) variable causes the change in a dependent variable.
3. Mike record the number of miles, m, he bikes to help track the number of calories, c, he burns in an hour. Identify the independent and dependent variables involved in this process.
4. Identify the dependent variable and independent variable for the amount of money, m, earned if t raffle tickets are sold.
5. Identify the dependent variable and independent variable for the number of hours, h, worked, and the amount of money, m, earned.
6. Identify the dependent variable and independent variable for the number of shelves, s, in a bookcase and the number of books, b, the bookcase can hold.
7. Identify the dependent variable and independent variable for the number of pages, p, you read in your book in h hours.
8. Name at least two independent variables that could result in a change in a monthly electric bill.
9. Name at least two independent variables that could result in a change in the price of a basket of apples.
10. The cost of a salad at a restaurant depends on many factors. List at least two independent variables that could affect the cost of a salad.
### What have we learned:
• Understanding and identifying dependent and independent variables.
• Identifying multiple independent or dependent variables for the given scenario.
### Concept Map
#### 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 […]
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# Eighteen% off 356.84 Dollars
How to calculate eighteen % off \$356.84. How to figure out percentages off a price. Using this calculator you will find that the amount after the discount is \$292.6088.
### Inputs
Original Price of the Item: \$
Discount Percent (% off): %
### Results
Amount Saved (Discount): \$
Sale / Discounted Price: \$
Using this calculator you can find the discount value and the discounted price of an item. It is helpfull to answer questions like:
• 1) What is eighteen percent (%) off \$356.84?
• 2) How much will you pay for an item where the original price before discount is \$ 356.84 when discounted eighteen percent (%)? What is the final or sale price?
• 3) \$64.23 is what percent off \$356.84?
• See how to solve these questions just after the Percent-off Calculator (or Discount) below.
## how to work out discounts - Step by Step
To calculate discount it is ease by using the following equations:
• Amount Saved = Orig. Price x Discount % / 100 (a)
•
• Sale Price = Orig. Price - Amount Saved (b)
Now, let's solve the questions stated above:
### 1) What is 18 percent off \$356.84? Find the amount of discount.
Suppose you have a Kohls coupon of \$356.84 and you want to know how much you will save for an item if the discount is 18.
Solution:
Replacing the given values in formula (a) we have:
Amount Saved = Original Price x Discount in Percent / 100. So,
Amount Saved = 356.84 x 18 / 100
Amount Saved = 6423.12 / 100
In other words, a 18% discount for a item with original price of \$356.84 is equal to \$64.2312 (Amount Saved).
Note that to find the amount saved, just multiply it by the percentage and divide by 100.
### 2) How much to pay for an item of \$356.84 when discounted 18 percent (%)? What is item's sale price?
Suppose you have a L.L. Bean coupon of \$356.84 and you want to know the final or sale price if the discount is 18 percent.
Using the formula (b) and replacing the given values:
Sale Price = Original Price - Amount Saved. So,
Sale Price = 356.84 - 64.2312
This means the cost of the item to you is \$292.61.
You will pay \$292.61 for a item with original price of \$356.84 when discounted 18%.
In this example, if you buy an item at \$356.84 with 18% discount, you will pay 356.84 - 64.2312 = \$292.61.
### 3) 64.23 is what percent off \$356.84?
Using the formula (b) and replacing given values:
Amount Saved = Original Price x Discount in Percent /100. So,
64.23 = 356.84 x Discount in Percent / 100
64.23 / 356.84 = Discount in Percent /100
100 x 64.23 / 356.84 = Discount in Percent
6423.12 / 356.84 = Discount in Percent, or
Discount in Percent = 18 (answer).
To find more examples, just choose one at the bottom of this page.
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Probability How likely is an event to occur?
Presentation on theme: "Probability How likely is an event to occur?"— Presentation transcript:
Probability How likely is an event to occur?
What are the chances of that happening??!!
Lesson objectives: To understand the terms ‘equal chance’ and ‘outcome’ when thinking about probability. To complete a maths investigation involving tossing a coin and rolling a dice! - to test the ‘equal chance’ theory!
Mathematical Vocabulary
We can describe the probability or chance of an event happening by saying: It is IMPOSSIBLE. It is UNLIKELY. It is LIKELY. It is CERTAIN.
Some events have an equal chance of happening or not happening
Some events have an equal chance of happening or not happening. Can you think of any?
We discussed these: If you toss a coin you have an equal chance of getting a head or a tail. Heads or tails are the ‘outcomes’. If a baby is born it has an equal chance of being a boy or a girl. Boy or girl are the ‘outcomes’. If you roll a dice you have an equal chance of getting the number 1, 2, 3, 4, 5 or 6! These scores are the ‘outcomes’.
Today’s task: We are going to carry out a maths investigation.
We are going to investigate whether we have an equal number of heads and tails when we toss a coin 30 times. We will also investigate whether we get an equal number of 1s, 2s, 3s, 4, 5s and 6s when we roll a dice 30 times. We will record our findings.
What do you think we will find out?
Perhaps we will have 15 heads and 15 tails when we toss our coin 30 times?? Perhaps we will have each number on the dice 5 times. (five scores of 1, five scores of 2 etc.) It’s up to you to find out!!!!
Did we achieve out learning objectives today?
We understand what is meant by the terms ‘equal chance’ and ‘outcomes’. We have carried out an investigation into ‘equal chance’. We have recorded our findings.
Things to think about! If we rolled 2 coins what possible outcomes could we get? If we rolled more than 1 dice what possible outcomes could we get? What would our chances of getting 2 heads or a 6 be like then?!!!
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"Non-trivial" solutions to equal products of consecutive integers
$$\bullet\ \textbf{Question}$$
One can find equivalent products of consecutive integers such as $$8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14=63\cdot64\cdot65\cdot66.$$ Other solutions of this have been posted already (Equal products of consecutive integers). But the implied meaning of 'trivial' solutions in other posts seems unsatisfactory -- I think I can convince you. There appear to be exactly $$4$$ non-trivial solutions after adjusting our vocabulary. One of which is the previous equation. I'm curious; are their any other non-trivial solutions?
$$\bullet\ \textbf{Triviality}$$
Firstly, let's use the notation $$a^{(b)}=\prod_{k=0}^{b-1}(a+k)$$ to denote rising factorial. The previous equation then becomes $$8^{(7)}=63^{(4)}$$.
All solutions come in pairs -- one overlapping and one disjoint. For example, if we take the solution $$2^{(5)}=2\cdot3\cdot4\cdot5\cdot6=8\cdot9\cdot10=8^{(3)},$$ which we will call disjoint because no integer appears on both sides, we can create an overlapping solution multiplying both sides by $$7$$: $$2^{(6)}=2\cdot3\cdot4\cdot5\cdot6\cdot\textbf{7}=\textbf{7}\cdot8\cdot9\cdot10=7^{(4)}.$$ This duality means that boring solutions like $$2^{(2)}=2\cdot3=6^{(1)}$$ can be disguised as interesting by transforming them from disjoint to overlapping: $$2^{(2)}=6^{(1)}\quad\rightarrow\quad 2^{(4)}=4^{(3)}.$$ For example, two answers on a previous post gave $$3^{(9)}=5^{(8)}$$ as an "interesting" solution. But this is simply the overlapping dual of $$3\cdot 4 = 12$$.
In general, if $$a^{(b)}=c^{(d)}$$ is a solution, then so is its dual: $$a^{(c-a)}=(a+b)^{(c+d-a-b)}.$$
So here we define a trivial solution either 1) having overlap or 2) having only one term on one side (previous inquiries seemed to only have considered #2).
$$\bullet\ \textbf{Non-trivial Solutions}$$
There are $$4$$ known non-trivial solutions under this definition: $$2^{(5)}=8^{(3)},\quad 5^{(3)}=14^{(2)},\quad 19^{(4)}=55^{(3)},\quad\text{and}\quad 8^{(7)}=63^{(4)}.$$ I ran a computer search on equivalent $$a^{(b)}$$ for $$a<100,000$$ and $$b<50$$ and found nothing new (https://repl.it/@onnomc/EqualProductsOfConsecutiveIntegers).
$$\bullet\ \textbf{Proof Attempts}$$
One attempt is to note $$a^{(b)}=\frac{(a+b-1)!}{(a-1)!}$$ and that therefore solutions to $$a^{(b)}=c^{(d)}$$ are also solutions to $$A!B!=C!D!$$. But apparently even $$A!B!=C!$$ has not yet been solved (On the factorial equations $A! B! =C!$ and $A!B!C! = D!$).
I thought also to use Stirling's approximation since $$(a+1)^{(b)}=\frac{(a+b)!}{a!}\approx \Big(\frac{a+b}{a}\Big)^{a+1/2}\Big(\frac{a+b}{e}\Big)^b$$ and ... well, hope for the best. I made no progress here.
I tried also bounding the number of distinct primes dividing $$a^{(b)}$$. We can bound $$c$$ from below since $$a^ba^{b/d}-d+1.$$ And of course, the non-overlapping criteria gives $$c\ge a+b$$. Thus if the prime bound rises fast enough, we can show that we are trying to fit too many prime divisors into $$c(c+1)...(c+d-1)$$. The furthest I got down this trail was finding $$r$$ consecutive integers with a product divisible by at most $$r$$ primes. For example, the largest solution for $$r=5$$ is $$c=24$$ since $$24\cdot25\cdot26\cdot27\cdot28=2^6\cdot3^4\cdot5^2\cdot7\cdot13.$$ The largest solution for $$r=42$$ seems to be $$c=2175$$. But I cannot prove these are the largest such integers -- only that they appear so empirically (https://repl.it/@onnomc/ConsecutiveIntegersOfFewPrimes). Interestingly, the case $$r=2$$ yields $$c=+\infty$$ if there are infinitely many Mersenne primes.
• One of the first issues that strikes me is that the larger string of consecutive numbers can contain no primes. This also seems relevant to the $A!B!=C!D!$ formulation, where the largest prime on the LHS must be present in the RHS as well. So the search is (primarily?) for strings of consecutive numbers that can all be formed from small prime factors, and as the absolute magnitude of the numbers being examined increases, I would expect such strings to become increasingly rare, or very short. Sep 6, 2019 at 19:46
• – Dale
Mar 27, 2020 at 2:36
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Education
Bodmas Rule: Order of Operations with Appropriate Examples
Each alphabet of the word ‘BODMAS’ is the initial of a mathematical operation.
‘B’ stands for Bracket
‘O’ stands for Order
‘D’ stands for Division
‘M’ stands for Multiplication
‘S’ stands for Subtraction
In Basic mathematics, when many operations are present in a numerical equation, we need to have an order which would help us select which operation needs to be performed first.
Imagine a numerical equation that involves addition, division and subtraction. We perform the operations in the same order as in the word ‘BODMAS ‘.
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Let’s have a look at some BODMAS rule questions
Q1. Simplify: 25 130 ÷ 10— 18
A1. We can see 4 mathematical operations. division occurs first in the ‘BODMAS’, solving 130 ÷ 10, we
get 13.
Hence, we have 25 + 13 — 18
The next operation is Addition; since there is no multiplication sign,
Hence, we have 38—18 =20
Q2. Simplify: 165 ÷ 15 × 5 + 3— 7
A2. Since division comes first, we solve 165 ÷ 15= 11
Hence, we have 11 × 5 + 3 — 7 in the next step.
After division comes multiplication, therefore 55 + 3 — 7 is the next step.
Now we have 58—7=51 (subtraction comes last in ‘BODMAS’.
Q3. Simplify: 71 + 23{15 ÷ 5 — 1}
A3. Since the above question involves mostly all the mathematical operations involved,
so by the order of alphabets in the word “B O D M A S ‘,we take up to solve the bracket first.
Inside brackets are operations of division and subtraction. In such equations, we see The signs inside the brackets as a part of brackets and need simplification first.
Solving 15÷5 – 1 first, we get 3 — 1 =2
Getting back to equation, we have 71 23 {2} = 71 _ 23 x 2 =71 + 46= 117
Q4. Simplify:77÷ 11/2 × 9 7 — 3
A4. Since the above question involves all the four basic mathematical operations, so in
accordance with the order of alphabets in the word “B O D M AS ‘, we take up division
First, 77 ÷ 11/2= 7/2
Now ,we have 7/2 × 9 + 7— 3
The next step is to solve multiplication
Hence we have 63/2 + 7 —3
= (63 + 14)/2 —3
=77/2 —3
=71/2
Q5. Simplify:55 of [69 35 ÷5—4}
Since the above question involves mostly all the mathematical operations involved, so
in accordance with the order of alphabets in the word “B O D M A S ‘, we take up to
solve the bracket first.
Applying ‘BODMAS ‘, ’D’ for division comes first
Hence 69 + 7 — 4 =76—4=72
The next step is to get back to the equation
We now have,55 of {72}= 55×72=3960
Conclusion
Therefore, we hope you now have a clear idea of how to use the BODMAS rule. Practice the solved questions among others to implement the concept successfully. In this way, you can easily use this rule to solve numerous problems with ease.
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# What happens when a plus and minus sign are next to each other?
## What happens when a plus and minus sign are next to each other?
The plus–minus sign, ±, is a mathematical symbol with multiple meanings. In mathematics, it generally indicates a choice of exactly two possible values, one of which is obtained through addition and the other through subtraction. In chemistry, the sign is used to indicate a racemic mixture.
What’s the rule for plus and minus?
The Rules:
Rule Example
+(+) Two like signs become a positive sign 3+(+2) = 3 + 2 = 5
−(−) 6−(−3) = 6 + 3 = 9
+(−) Two unlike signs become a negative sign 7+(−2) = 7 − 2 = 5
−(+) 8−(+2) = 8 − 2 = 6
### When you add a negative and a positive what is the answer?
When you have two negative signs, one turns over, and they add together to make a positive. If you have a positive and a negative, there is one dash left over, and the answer is negative.
Is minus Plus Minus Plus?
Numbers can either be positive or negative. Adding and multiplying combinations of positive and negative numbers can cause confusion and so care must be taken. Addition and Subtraction. Two ‘pluses’ make a plus, two ‘minuses’ make a plus.
#### How do you read plus minus sign?
(mathematics) the symbol ±, meaning “plus or minus”, used to indicate the precision of an approximation (as in “The result is 10 ± 0.3”, meaning the result is anywhere in the inclusive range from 9.7 to 10.3), or as a convenient shorthand for a quantity with two possible values of opposing sign and identical magnitude …
What are the rules of addition and subtraction?
Rules of integers for addition and subtraction : 1) If the two numbers have different sign like positive and negative then subtract the two numbers and give the sign of the bigger number. 2) If the two numbers have same sign i.e. either positive or negative signs then add the two numbers and give the common sign.
## What are the rules when adding and subtracting positive and negative numbers?
Two signs
• When adding positive numbers, count to the right.
• When adding negative numbers, count to the left.
• When subtracting positive numbers, count to the left.
• When subtracting negative numbers, count to the right.
How many minus signs make a plus in this equation?
According to sign rules, two minus signs make a plus when they are next to each other. The equation balances, so the answers are correct. The final answers are \\ (a = 7\\) and \\ (b = -2\\).
### What does the plus-minus symbol mean in a quadratic equation?
Quadratic Equation: The use of the plus-minus symbol shows that two equations must be solved to determine which values of x satisfy the quadratic equation. These equations are: (1) Using the “plus” sign: (2) Using the “minus” sign: Margin of Error:
What are the rules for adding and subtracting positive and negative numbers?
Rules for adding and subtracting positive and negative numbers. When two signs are written next to each other, the rules for adding and subtracting numbers are: two signs that are different become a negative sign. two signs that are the same become a positive sign.
#### Are the rules addition and subtraction same as rules for multiplication?
Are the rules addition and subtraction are same as rules for multiplication of integers? Rules for multiplication of integers are different from addition and subtraction rules. Two like signs when multiplied, they result in positive sign. Two unlike signs when multiplied, results in negative sign.
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# 2017 AIME II Problems/Problem 13
## Problem
For each integer $n\geq3$, let $f(n)$ be the number of $3$-element subsets of the vertices of the regular $n$-gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of $n$ such that $f(n+1)=f(n)+78$.
## Solution 1
Considering $n \pmod{6}$, we have the following formulas:
$n\equiv 0$: $\frac{n(n-4)}{2} + \frac{n}{3}$
$n\equiv 2, 4$: $\frac{n(n-2)}{2}$
$n\equiv 3$: $\frac{n(n-3)}{2} + \frac{n}{3}$
$n\equiv 1, 5$: $\frac{n(n-1)}{2}$
To derive these formulas, we note the following: Any isosceles triangle formed by the vertices of our regular $n$-sided polygon $P$ has its sides from the set of edges and diagonals of $P$. Notably, as two sides of an isosceles triangle must be equal, it is important to use the property that same-lengthed edges and diagonals come in groups of $n$, unless $n$ is even when one set of diagonals (those which bisect the polygon) comes in a group of $\frac{n}{2}$. Three properties hold true of $f(n)$:
When $n$ is odd there are $\frac{n(n-1)}{2}$ satisfactory subsets (This can be chosen with $n$ choices for the not-base vertex, and $\frac{n-1}{2}$ for the pair of equal sides as we have $n-1$ edges to choose from, and we must divide by 2 for over-count).*
• Another explanation: For any diagonal or side of the polygon chosen as the base of the isosceles triangle, there is exactly 1 isosceles triangle that can be formed. So, the total number of satisfactory subsets is $\dbinom{n}{2}=\dfrac{n(n-1)}{2}.$
When $n$ is even there are $\frac{n(n-2)}{2}$ satisfactory subsets (This can be chosen with $n$ choices for the not-base vertex, and $\frac{n-2}{2}$ for the pair of equal sides as we have $n-1$ edges to choose from, one of them which is not satisfactory (the bisecting edge), and we must divide by 2 for over-count).
When $n$ is a multiple of three we additionally over-count equilateral triangles, of which there are $\frac{n}{3}$. As we count them three times, we are two times over, so we subtract $\frac{2n}{3}$.
Considering the six possibilities $n \equiv 0,1,2,3,4,5 \pmod{6}$ and solving, we find that the only valid solutions are $n = 36, 52, 157$, from which the answer is $36 + 52 + 157 = \boxed{245}$.
## Solution 2 (elaborates on the possible cases)
In the case that $n\equiv 0\pmod 3$, there are $\frac{n}{3}$ equilateral triangles. We will now count the number of non-equilateral isosceles triangles in this case.
Select a vertex $P$ of a regular $n$-gon. We will count the number of isosceles triangles with their vertex at $P$. (In other words, we are counting the number of isosceles triangles $\triangle APB$ with $A, B, P$ among the vertices of the $n$-gon, and $AP=BP$.)
If the side $AP$ spans $k$ sides of the $n$-gon (where $k<\frac{n}{2}$), the side $BP$ must span $k$ sides of the $n$-gon, and, thus, the side $AB$ must span $n-2k$ sides of the $n$-gon. As $\triangle ABP$ has three distinct vertices, the side $AB$ must span at least one side, so $n-2k \ge 1$. Combining this inequality with the fact that $1\le k<\frac{n}{2}$ and $k\not = \frac{n}{3}$ (as $\triangle ABP$ cannot be equilateral), we find that there are $\lceil\frac{n}{2}\rceil-2$ possible $k$.
As each of the $n$ vertices can be the vertex of a given triangle $\triangle ABP$, there are $\left(\lceil \frac{n}{2} \rceil -2 \right)\cdot n$ non-equilateral isosceles triangles.
Adding in the $\frac{n}{3}$ equilateral triangles, we find that for $n\equiv 0\pmod 3$: $f(n) = \frac{n}{3}+\left(\lceil \frac{n}{2} \rceil -2\right)\cdot n$.
On the other hand, if $n\equiv 1, 2\pmod 3$, there are no equilateral triangles, and we may follow the logic of the paragraph above to find that $f(n)=\left(\lceil \frac{n}{2} \rceil -1\right)\cdot n$.
We may now rewrite the given equation, based on the remainder $n$ leaves when divided by 3.
Case 1: $n\equiv 0\pmod 3$ The equation $f(n+1)=f(n)+78$ for this case is $\left(\lceil \frac{n+1}{2} \rceil -1\right)\cdot (n+1)=\frac{n}{3}+\left(\lceil \frac{n}{2} \rceil -2\right)\cdot n+78$.
In this case, $n$ is of the form $6k$ or $6k+3$, for some integer $k$.
Subcase 1: $n=6k$ Plugging into the equation above yields $k=6\rightarrow n=36$.
Subcase 2: $n=6k+3$ Plugging into the equation above yields $7k=75$, which has no integer solutions.
Case 2: $n\equiv 1\pmod 3$ The equation $f(n+1)=f(n)+78$ for this case is $\left(\lceil \frac{n+1}{2} \rceil -1\right)\cdot (n+1)=\left(\lceil \frac{n}{2} \rceil -1\right)\cdot n+78$.
In this case, $n$ is of the form $6k+1$ or $6k+4$, for some integer $k$.
Subcase 1: $n=6k+1$ In this case, the equation above yields $k=8\rightarrow n=52$.
Subcase 2: $n=6k+4$ In this case, the equation above yields $k=26\rightarrow n=157$.
Case 3: $n\equiv 2\pmod 3$ The equation $f(n+1)=f(n)+78$ for this case is $\frac{n+1}{3}+\left(\lceil \frac{n+1}{2} \rceil -2\right)\cdot (n+1)=\left(\lceil \frac{n}{2} \rceil -1\right)\cdot n+78$.
In this case, $n$ is of the form $6k+2$ or $6k+5$, for some integer $k$.
Subcase 1: $n=6k+2$ The equation above reduces to $5k=77$, which has no integer solutions.
Subcase 2: $n=6k+5$ The equation above reduces to $k=-80$, which does not yield a positive integer solution for $n$.
In summary, the possible $n$ are $36, 52, 157$, which add to $\boxed{245}$.
## Solution 3
We first notice that when a polygon has $s$ sides where $s\not\equiv 0\pmod{3}$, there cannot exist any three vertices that form an equilateral triangle. Also, the parity of $s$ and $s+1$ also matters, since they influence how many isosceles triangles including equilateral triangles exist in the polygon. We can model an equation $2x+y=s$, where the lines that are congruent connect the vertices that are $x$ vertices apart and the other line has vertices that are $y$ vertices apart. If $s$ is even, there are $\frac{s-2}{2}$ solutions for $(x,y)$ which would create a unique type of isosceles triangle. We subtract two since $y$ cannot be zero. If $s$ is odd, there are $\frac{s-1}{2}$ solutions for $(x,y)$. Next, we do casework on the congruence of $s\pmod{3}$ and the parody of $s$ using the information above:
Case 1:
$s\not\equiv 0\pmod{3}$, $s+1\not\equiv 0\pmod{3}$
$s$ is even, $s+1$ is odd
There are $\frac{s-2}{2}$ unique types of isosceles triangles in the polygon with $s$ sides. Each isosceles triangle has a unique point which connects the two congruent sides. Therefore, for each type of isosceles triangle, there exists $s$ of those triangles since the unique point can be any of the $s$ vertices. There are $\frac{s}{2}$ types of isosceles triangles in the polygon with $s+1$ sides and $s+1$ unique points for each type of triangle. Therefore, $\frac{s}{2}\cdot{(s+1)}-\frac{s-2}{2}\cdot{(s)}=78$. Solving, we get $s=52$.
Case 2:
$s\not\equiv 0\pmod{3}$, $s+1\not\equiv 0\pmod{3}$
$s$ is odd, $s+1$ is even
Using similar logic, there are $\frac{s-1}{2}\cdot{(s)}$ possible isosceles triangles in the $s$ sided polygon. There are $\frac{s-1}{2}\cdot{(s+1)}$ possible isosceles triangles in the $s+1$ sided polygon. The difference should be $78$, so $\frac{s-1}{2}\cdot{(s+1)}-\frac{s-1}{2}\cdot{(s)}=78$. Solving gives $s=157$.
For both of these cases, we don't have to worry about equilateral triangles since in both cases $s,s+1\nmid3$.
Case 3:
$s\not\equiv 0\pmod{3}$, $s+1\equiv 0\pmod{3}$
$s$ is even, $s+1$ is odd
There are $\frac{s-2}{2}\cdot{(s)}$ possible isosceles triangles in the $s$ sided polygon. The $s+1$ case is a bit more complicated, as we have to consider equilateral triangles as well. In this case, there is one solution where $x=y$, which would produce an equilateral triangle. Therefore, we subtract that case to calculate only isosceles triangles with 2 congruent sides. Only isosceles triangles with exactly 2 congruent sides have a unique point, while there exist only $\frac{s+1}{3}$ distinct equilateral triangles in a polygon with $s+1$ sides, the rest are equivalent and symmetrical. Therefore, there are $(\frac{s}{2}-1)\cdot{(s+1)}+\frac{s+1}{3}$ isosceles triangles with at least 2 congruent sides in a polygon with $s+1$ sides. Putting it together, $\frac{s}{2}-1\cdot{(s+1)}+(\frac{s+1}{3})-\frac{s-2}{2}\cdot{(s)}=78$. Solving yields $5s=772$ which is impossible since $s$ has to be an integer. Therefore, this case is not valid.
Case 4:
$s\not\equiv 0\pmod{3}$, $s+1\equiv 0\pmod{3}$
$s$ is odd, $s+1$ is even
There are $\frac{s-1}{2}\cdot{(s)}$ isosceles triangles in the $s$ sided polygon. Using the idea above, there are $(\frac{s-1}{2}-1)\cdot{(s+1)}$ isosceles triangles with 2 congruent sides in an $s+1$ sided polygon. There are $\frac{s+1}{3}$ equilateral triangles. Therefore, $(\frac{s-1}{2}-1)\cdot{(s+1)}+(\frac{s+1}{3})-\frac{s-1}{2}\cdot{(s)}=78$. Looking at the left hand side, it is clear that $s$ has to be negative, which is not valid. Therefore, this case is not valid.
Case 5:
$s\equiv 0\pmod{3}$, $s+1\not\equiv 0\pmod{3}$
$s$ is even, $s+1$ is odd
There are a total of $(\frac{s-2}{2}-1)\cdot{(s)}+(\frac{s}{3})$ isosceles triangles in a polygon with $s$ sides. There are a total of $\frac{s}{2}\cdot{(s+1)}$ isosceles triangles in a polygon with $s+1$ sides. Therefore, $\frac{s}{2}\cdot{(s+1)}-(\frac{s-2}{2}-1)\cdot{(s)}-(\frac{s}{3})=78$. Simplifying, we get $s=36$.
Case 6:
$s\equiv 0\pmod{3}$, $s+1\not\equiv 0\pmod{3}$
$s$ is odd, $s+1$ is even
There are a total of $(\frac{s-1}{2}-1)\cdot{(s)}+(\frac{s}{3})$ isosceles triangles in the polygon with $s$ sides. There are $\frac{s-1}{2}\cdot{(s+1)}$ isosceles triangles in the polygon with $s+1$ sides. Therefore,$\frac{s-1}{2}\cdot{(s+1)}-(\frac{s-1}{2}-1)\cdot{(s)}-(\frac{s}{3})=78$. Simplifying, we get $7s=471$, which means $s$ is not an integer. Thus, this case is invalid.
Adding all the valid cases, we obtain $52+157+36=\boxed{245}$
## Video Solution
~MathProblemSolvingSkills.com
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# How do you divide (p^5+5p^3-11p^2-25p+29)div(p+6) using synthetic division?
Jun 25, 2017
The remainder is $= - 9073$ and the quotient is $= {p}^{4} - 6 {p}^{3} + 41 {p}^{2} - 257 p + 1517$
#### Explanation:
Let's perform the synthetic division
$\textcolor{w h i t e}{a a a a}$$- 6$$\textcolor{w h i t e}{a a a a}$$|$$\textcolor{w h i t e}{a a a a}$$1$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a a}$$5$$\textcolor{w h i t e}{a a a a}$$- 11$$\textcolor{w h i t e}{a a a a}$$- 25$$\textcolor{w h i t e}{a a a a}$$29$
$\textcolor{w h i t e}{a a a a a a a a a a a a}$_________
$\textcolor{w h i t e}{a a a a}$$\textcolor{w h i t e}{a a a a a a a}$$|$$\textcolor{w h i t e}{a a a a}$$\textcolor{w h i t e}{a a a}$$- 6$$\textcolor{w h i t e}{a a a a}$$36$$\textcolor{w h i t e}{a a a}$$- 246$$\textcolor{w h i t e}{a a a a}$$1542$$\textcolor{w h i t e}{a a}$$- 9102$
$\textcolor{w h i t e}{a a a a a a a a a a a a}$________
$\textcolor{w h i t e}{a a a a}$$\textcolor{w h i t e}{a a a a a a a}$$|$$\textcolor{w h i t e}{a a a a}$$1$$\textcolor{w h i t e}{a a}$$- 6$$\textcolor{w h i t e}{a a a a}$$41$$\textcolor{w h i t e}{a a a}$$- 257$$\textcolor{w h i t e}{a a a a}$$1517$$\textcolor{w h i t e}{a a}$$\textcolor{red}{- 9073}$
The remainder is $= - 9073$ and the quotient is $= {p}^{4} - 6 {p}^{3} + 41 {p}^{2} - 257 p + 1517$
$\frac{{p}^{5} + 5 {p}^{3} - 11 {p}^{2} - 25 p + 29}{p + 6} = {p}^{4} - 6 {p}^{3} + 41 {p}^{2} - 257 p + 1517 - \frac{9073}{p + 6}$
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# Facts About Parallelograms
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A parallelogram is a two-dimensional quadrilateral -- a shape that has four sides that intersect at four points, also known as vertices. The two opposite sides of a parallelogram are always parallel and congruent -- or equal in length. Rectangles, squares and rhombuses are all examples of parallelograms.
## Opposite Sides
Both pairs of opposite sides of a parallelogram are always parallel, and both pairs of opposite sides of a parallelogram are always congruent. You can find the distance around a parallelogram, also known as the perimeter, by measuring and adding the length of the four sides together. Because opposite sides of a parallelogram are parallel, they will never intersect.
## Diagonal Lines
The diagonals of a parallelogram -- lines that extend from one corner to the opposite corner -- bisect one another. In other words, each diagonal cuts its opposite diagonal into two equal parts. No matter how you reshape a parallelogram, such as making the sides shorter or longer or increasing and decreasing the height, the diagonals will always bisect one another.
## Area of a Parallelogram
Calculate the area of a parallelogram by multiplying the base by the altitude, also known as the height. You can use any side of a parallelogram as the base. The altitude is the perpendicular distance from the base to the opposite side. In some cases, you may need to extend the opposite side of the parallelogram to be able to find and measure the perpendicular distance.
## Interior Angles
Opposite interior angles of a parallelogram are always equal. For example, if one interior angle measures 36 degrees, the opposite interior angle will also measure 36 degrees. Consecutive interior angles in a parallelogram -- angles that are side by side -- are supplementary. That means that when you add two interior consecutive angles together, the total always equals 180 degrees. When you add all four interior angles together, the total always equals 360 degrees.
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# 66666 in words
66666 in words is written as Sixty Six Thousand Six Hundred and Sixty Six. In 66666, the first 6 has a place value of ten thousand, second 6 is in the place value of thousand, third 6 is in the place value of hundred, fourth 6 is in the place value of ten and the last 6 is in the place value of one. The article on Place Value gives more information. The number 66666 is used in expressions that relate to money, distance, social media views, and many more. For example, “In the number Sixty Six Thousand Six Hundred and Sixty Six, the only digit is 6.” Another example “For a festive sale, the smart TV price was reduced to Sixty Six Thousand Six Hundred and Sixty Six rupees.”
66666 in words Sixty Six Thousand Six Hundred and Sixty Six Sixty Six Thousand Six Hundred and Sixty Six in Numbers 66666
## How to Write 66666 in Words?
We can convert 66666 to words using a place value chart. The number 66666 has 5 digits, so let’s make a chart that shows the place value up to 5 digits.
Ten thousand Thousands Hundreds Tens Ones 6 6 6 6 6
Thus, we can write the expanded form as:
6 × Ten thousand + 6 × Thousand + 6 × Hundred + 6 × Ten + 6 × One
= 6 × 10000 + 6 × 1000 + 6 × 100 + 6 × 10 + 6 × 1
= 66666.
= Sixty Six Thousand Six Hundred and Sixty Six.
66666 is the natural number that is succeeded by 66665 and preceded by 66667.
66666 in words – Sixty Six Thousand Six Hundred and Sixty Six.
Is 66666 an odd number? – No.
Is 66666 an even number? – Yes.
Is 66666 a perfect square number? – No.
Is 66666 a perfect cube number? – No.
Is 66666 a prime number? – No.
Is 66666 a composite number? – Yes.
## Solved Example
1. Write the number 66666 in expanded form
Solution: 6 x 10000 + 6 x 1000 + 6 x 100 + 6 x 10 + 6 x 1
We can write 66666 = 60000 + 6000 + 600 + 60 + 6
= 6 x 10000 + 6 x 1000 + 6 x 100 + 6 x 10 + 6 x 1.
## Frequently Asked Questions on 66666 in words
Q1
### How to write the number 66666 in words?
66666 in words is written as Sixty Six Thousand Six Hundred and Sixty Six.
Q2
### Is 66666 a prime number?
No. 66666 is not a prime number.
Q3
### Is 66666 divisible by 10?
No. 66666 is not divisible by 10.
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## Algebra
### Rectangular Coordinate System
Sometimes referred to as the rectangular coordinate system, the Cartesian coordinate system consists of two perpendicular real number lines intersecting at zero. Positions on this grid system are identified using ordered pairs, (xy). The center of the system (0, 0) is called the origin. The x-coordinate indicates horizontal distance from the origin and the y-coordinate indicates vertical distance from the origin.
The horizontal number line, usually called the x-axis, is typically used for the independent variable. The vertical real number line, usually called the y-axis, is used for the dependent variable.
Ordered pairs with 0 as a coordinate do not lie in a quadrant; these points lie on an axis.
Example: Find the distance and midpoint between the two points: (3, 4) and (−1, 2).
Example: Find the distance and midpoint between the two points: (0, 0) and (−3, 4).
Example: Find the distance and midpoint between the two points: (−1,−1) and (1,1).
Example: Find the distance and midpoint between the two points: (−2,−5) and (−4,−3).
Circle Word Problem: If the diameter of a circle is defined by two points (−3, 4) and (7, 4), find the center and radius of the circle. (Hint: diameter = 2*radius)
Area of a Circle: Find the area of a circle given the center (−3, 3) and a point (3, 3) on the circle.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Distance Between Two Points
## Positive length between points.
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Practice Distance Between Two Points
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Distance Between Two Points
What if you were given the coordinates of two points that form either a vertical or horizontal line? How would you determine how far apart those two points are? After completing this Concept, you'll be able to determine the distance between two such points.
### Guidance
Distance is the measure of length between two points. To measure is to determine how far apart two geometric objects are. The most common way to measure distance is with a ruler. Inch-rulers are usually divided up by eighth-inch (or 0.125 in) segments. Centimeter rulers are divided up by tenth-centimeter (or 0.1 cm) segments. Note that the distance between two points is the absolute value of the difference between the numbers shown on the ruler. This implies that you do not need to start measuring at “0”, as long as you subtract the first number from the second.
The segment addition postulate states that if $A$ , $B$ , and $C$ are collinear and $B$ is between $A$ and $C$ , then $AB + BC = AC$ .
You can find the distances between points in the $x$$y$ plane if the lines are horizontal or vertical. If the line is vertical, find the change in the $y-$ coordinates. If the line is horizontal, find the change in the $x-$ coordinates.
#### Example A
What is the distance marked on the ruler below? The ruler is in centimeters.
Subtract one endpoint from the other. The line segment spans from 3 cm to 8 cm. $|8 - 3| = |5| = 5$
The line segment is 5 cm long. Notice that you also could have done $|3 - 8| = |-5| = 5$ .
#### Example B
Make a sketch of $\overline{OP}$ , where $Q$ is between $O$ and $P$ .
Draw $\overline{OP}$ first, then place $Q$ on the segment.
#### Example C
What is the distance between the two points shown below?
Because this line is vertical, look at the change in the $y-$ coordinates.
$|9 - 3| = |6| = 6$
The distance between the two points is 6 units.
### Guided Practice
1. Draw $\overline{CD}$ , such that $CD = 3.825 \ in$ .
2. In the picture from Example B, if $OP = 17$ and $QP = 6$ , what is $OQ$ ?
3. What is the distance between the two points shown below?
Answers:
1. To draw a line segment, start at “0” and draw a segment to 3.825 in.
Put points at each end and label.
2. Use the Segment Addition Postulate.
$OQ + QP & = OP\\OQ + 6 & = 17\\OQ & = 17-6\\OQ & = 11$
3. Because this line is horizontal, look at the change in the $x-$ coordinates.
$|(-4) - 3| = |-7| = 7$
The distance between the two points is 7 units.
### Practice
For 1-4, use the ruler in each picture to determine the length of the line segment.
1. Make a sketch of $\overline{BT}$ , with $A$ between $B$ and $T$ .
2. If $O$ is in the middle of $\overline{LT}$ , where exactly is it located? If $LT = 16 \ cm$ , what is $LO$ and $OT$ ?
3. For three collinear points, $A$ between $T$ and $Q$ :
1. Draw a sketch.
2. Write the Segment Addition Postulate for your sketch.
3. If $AT = 10 \ in$ and $AQ = 5 \ in$ , what is $TQ$ ?
1. For three collinear points, $M$ between $H$ and $A$ :
1. Draw a sketch.
2. Write the Segment Addition Postulate for your sketch.
3. If $HM = 18 \ cm$ and $HA = 29 \ cm$ , what is $AM$ ?
1. For three collinear points, $I$ between $M$ and $T$ :
1. Draw a sketch.
2. Write the Segment Addition Postulate for your sketch.
3. If $IT = 6 \ cm$ and $MT = 25 \ cm$ , what is $AM$ ?
1. Make a sketch that matches the description: $B$ is between $A$ and $D$ . $C$ is between $B$ and $D$ . $AB = 7 \ cm, \ AC = 15 \ cm$ , and $AD = 32 \ cm$ . Find $BC, BD$ , and $CD$ .
2. Make a sketch that matches the description: $E$ is between $F$ and $G$ . $H$ is between $F$ and $E$ . $FH = 4 \ in, \ EG = 9 \ in$ , and $FH = HE$ . Find $FE, HG$ , and $FG$ .
For 12 and 13, Suppose $J$ is between $H$ and $K$ . Use the Segment Addition Postulate to solve for $x$ . Then find the length of each segment.
1. $HJ = 4x + 9, \ JK = 3x + 3, \ KH = 33$
2. $HJ = 5x - 3, \ JK = 8x - 9, \ KH = 131$
For 14-17, determine the vertical or horizontal distance between the two points.
1. Make a sketch of: $S$ is between $T$ and $V$ . $R$ is between $S$ and $T$ . $TR = 6, RV = 23$ , and $TR = SV$ .
2. Find $SV, TS, RS$ and $TV$ from #18.
3. For $\overline{HK}$ , suppose that $J$ is between $H$ and $K$ . If $HJ = 2x + 4, \ JK = 3x + 3$ , and $KH = 22$ , find $x$ .
### Vocabulary Language: English Spanish
distance
distance
Distance is the measure of length between two points.
Absolute Value
Absolute Value
The absolute value of a number is the distance the number is from zero. Absolute values are never negative.
measure
measure
To measure distance is to determine how far apart two geometric objects are by using a number line or ruler.
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# 1 Times Table
Learn here the 1 times table. One times table 1 x 1 = 1, 1 x 2 = 2, 1 x 3 = 3, 1 x 4 = 4, 1 x 5 = 5, 1 x 6 = 6, 1 x 7 = 7, 1 x 8 = 8, 1 x 9 = 9, 1 x 10 = 10.
Understanding basic multiplication tables is crucial for building a strong foundation in mathematics. Among the multiplication tables, the 1 times table might seem trivial at first glance, as multiplying any number by 1 results in the same number. However, mastering the 1 times table is an essential step towards developing number sense, mental math skills, and critical thinking abilities. In this blog, we will explore the significance of the 1 times table and how it lays the groundwork for more complex mathematical concepts.
## 1 Times Table: The Basics:
The 1 times table is the simplest and most straightforward multiplication table. It involves multiplying any number by 1. The pattern is rather intuitive: no matter what number you choose, multiplying it by 1 will always yield the same number as the product. For example, 1 x 2 = 2, 1 x 5 = 5, and so on. This may seem too obvious, but its significance goes beyond the surface.
## Building Number Sense:
The 1 times table serves as a starting point for building number sense, which is the ability to understand and manipulate numbers. By recognizing the patterns of the 1 times table, children begin to comprehend the concept of multiplication. They understand that multiplying by 1 does not change the value of the number and that it acts as an identity element, much like addition's role of the identity element 0. This understanding paves the way for grasping more complex multiplication concepts.
## Mental Math and Efficiency:
Mastering the 1 times table enables children to develop mental math skills. By internalizing the patterns and relationships within the 1 times table, they can quickly and accurately solve multiplication problems involving 1. Mental math is a valuable skill that helps individuals perform calculations efficiently, boosting their confidence and overall mathematical fluency.
## Multiplication Properties:
The 1 times table is closely related to essential multiplication properties. For instance, the commutative property states that changing the order of multiplication does not affect the result. Since multiplying any number by 1 produces the same number, the commutative property is evident within the 1 times table. Additionally, the distributive property, which is vital for more advanced math concepts, can be explored and understood through the 1 times table.
## Real-World Applications:
The 1 times table may seem elementary, but its applications extend into real-world scenarios. Understanding the concept of multiplying by 1 is crucial when dealing with measurements or scaling objects. For example, when converting centimeters to millimeters or dollars to cents, multiplying by 1 is necessary to maintain the same value while changing units. Proficiency in the 1 times table ensures accuracy and precision in everyday calculations.
## Conclusion:
While the 1 times table may appear simple, its significance in mathematics cannot be underestimated. It serves as a foundational step towards understanding multiplication, developing mental math skills, and comprehending key mathematical properties. Mastering the 1 times table builds a solid mathematical framework that students can leverage when tackling more complex concepts and real-world applications. So, let's not overlook the importance of the humble 1 times table and acknowledge its role in nurturing mathematical success.
## One Multiplication Table
Read, Repeat and Learn one times table and Check yourself by giving a test below
Also check 0 times table, 1 times table, 2 times table, 3 times table, 4 times table, 5 times table, 6 times table, 7 times table, 8 times table, 9 times table
## Table of 1
1 Times table Test
## How much is 1 multiplied by other numbers?
PrintableMultiplicationTable.net allows you to create Multiplication Worksheets, in PDF formats for Grade 1, Grade 2, Grade 3, Grade 4 and Grade 5 freely and easily, no registration required.
With Printable Tools, you can create custom worksheets, tables, lined pages, grid pages and download in pdf format.
You can also set the change the heading and subheading before taking print. Each worksheet is generated with a dynamic numbers.
### Printable Tools
Multiplication Worksheets
Subtraction Worksheets
Times Tables
Lined Pages
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# Difference Between Covariance and Correlations
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Covariance and correlation are two very common mathematical ideas both of which are highly applicable in business stats. Both determine the relationship between two random variables and measure the dependence between them. Although they are somewhat similar, they differ from each other between the two mathematical concepts.
Correlation is when one of the items changes may lead to another object changes. Correlation is regarded as the finest means for the quantitative connection between two format factors to be measured and expressed.
Covariance, on the other side, is when two elements differ. To learn the difference between covariance and correlation, just go through this article which describes the literal differences between covariance and correlation.
### Covariance
#### • Definition
Covariance is a statistical word, described as the systematic connection between a two of a kind of random variables where an alteration is reciprocated by an equal change in another variable.
#### • Positive or Negative Covariance
The covariance can hold any consideration in between -∞ to +∞, in which negative valuation is a negative connection measure, while a positive valuation is a positive relation. In addition, the linear relation between factors is also determined using the covariance formula.
#### • Zero or Null Values
Thus, it does not indicate a connection when the value is zero. Furthermore, if all of the variable’s findings are equal, the covariance is null. When we modify the unit of assessment for either or both of the two factors, the power of the relation between two factors does not alter, but it does alter the covariance value.
#### • Covariance Calculator
The calculator of covariance is used by mutating and calculating the average sample, covariance between two different values, for the measurement of both factors (i.e., X and Y).
##### a) Find the mean for X and Y
Mean = Sum of numbers/N Step
b) Subtract all X and Y values with their corresponding middle
c) Multiply the values you get after multiplication with each other
e) Apply covariance formula Cov(X, Y) = ∑(xi−x¯)(yi−y¯) /N
Where xi= Entered X values
X ̄ = Mean X value
X= X values
Yi= Y values
Y ̄ = Y average value
N = entered values number
The covariance calculator can assist you to determine the factor covariance which is a metric of how many two random factors (x, y) vary with each other.
### Correlation
#### • Definition
The correlation is defined as a statistical metric that determines the extent to which two or more random variables pass together. If it was found that the motion in one variable is mutually reciprocated in one variable by an equal motion in one another variable, then the factors should be associated in one manner or another.
#### • Types of Correlation
There are two kinds of correlation, i.e.
a) Negative Correlation
b) Positive Correlation
When these two factors travel in the same path, they are said to be directly or positively correlated. On the other hand, the correlation is either inverse or negative when the two factors travel in contrary directions.
The correlation value resides between-1 and+ 1, where values near + 1 are a high positive correlation and values near to-1 are a powerful negative correlation.
#### • Measure of a Correlation
There are four correlation measures:
i) The scatter diagram
ii) Product-moment correlation coefficient
iii) Ranking correlation coefficient
iv) Coefficient of concurrent deviations
## The following points are noteworthy in regard to the difference between covariance and correlation:
### Major Differences between Covariance and Correlation
1. In the tandem a metric used to specify how far two random variables alter are known as covariance. A correlation is a metric used to show how greatly is two random factorsassociated witheach other.
2. Covariance is nothing more than a correlation metric. Correlation, instead, relates to the scaled covariance.
3. The correlation value is from-1 to+ 1. The value of covariance, on the other hand, resides between -∞ and +∞.
4. Scale changes influence covariance. This affects the calculated covariance of these 2 numbers when, for instance, two variables are multiplied by similar and different constants. However, multiplication by constants does not alter the past outcome by implementing the same correlation system. The reason is that a shift in scale does not influence the correlation.
5. Correlation is dimensional less; i.e., it’s a connection between factors as a unit-free metric. In contrast to covariance, the multiplication of units of the two factors obtains the value.
As discussed above, in contrast to covariance, correlation is a unit-free metric of two variables interdependency. This pro feature facilitates the comparison of calculated correlation scores across 2 factors regardless of their units and directions.
6. Covariance for only 2 factors can be calculated. Instead, for various sets of figures, the correlation can be calculated. This is another factor that makes the correlation in contrast to the covariance more attractive for analysts.
### Common Features
Both measures only a linear connection of two factors, i.e. covariance is also null if the coefficient of correlation is zero.Moreover, the change of location will not affect the two measures.
### Conclusion
Covariance indicates the degree into which two random variables are different from each other. On the other side, correlation measures the power of that connection. The correlation score is anchored at + 1 on the top and-1 on the bottom. It’s, therefore, a certain range.
The spectrum of covariance is nevertheless unlimited. The range is theoretically, -∞ to +∞. It can hold any negative or positive value. You can be confident that the correlation of.5 is larger than.3, and the first number set, with a correlation of.5, is more dependent on each other than the second set.
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Correlation is a unique covariance situation which can be acquired by standardizing the information. In choosing a connection superior for two factors, correlation is chosen over covariance because it is not affected by the shift of place and scale and can also be employed to compare two pairs of factors.
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# Dividing Polynomials.
## Presentation on theme: "Dividing Polynomials."— Presentation transcript:
Dividing Polynomials
3x3 4x2 x 2 1 1 1 1 Dividing by a Monomial
If the divisor only has one term, split the polynomial up into a fraction for each term. divisor Now reduce each fraction. 3x3 4x2 x 2 1 1 1 1
Subtract (which changes the sign of each term in the polynomial)
Long Division If the divisor has more than one term, perform long division. You do the same steps with polynomial division as with integers. Let's do two problems, one with integers you know how to do and one with polynomials and copy the steps. Subtract (which changes the sign of each term in the polynomial) Now multiply by the divisor and put the answer below. x + 11 2 1 Bring down the next number or term Multiply and put below Now divide 3 into or x into 11x First divide 3 into or x into x2 x x2 + 8x - 5 Remainder added here over divisor 64 x2 – 3x subtract 5 8 11x - 5 32 11x - 33 This is the remainder 26 28 So we found the answer to the problem x2 + 8x – 5 x – 3 or the problem written another way:
Let's Try Another One If any powers of terms are missing you should write them in with zeros in front to keep all of your columns straight. Subtract (which changes the sign of each term in the polynomial) y - 2 Write out with long division including 0y for missing term Multiply and put below Bring down the next term Multiply and put below y y2 + 0y + 8 Divide y into y2 Divide y into -2y Remainder added here over divisor y2 + 2y subtract -2y + 8 - 2y - 4 This is the remainder 12
- 3 1 6 8 -2 - 3 - 9 3 1 x2 + x 3 - 1 1 1 Synthetic Division
There is a shortcut for long division as long as the divisor is x – k where k is some number. (Can't have any powers on x). Set divisor = 0 and solve. Put answer here. x + 3 = 0 so x = - 3 1 - 3 Multiply these and put answer above line in next column Multiply these and put answer above line in next column Multiply these and put answer above line in next column Bring first number down below line - 3 Add these up - 9 3 Add these up Add these up 1 x2 + x 3 - 1 1 This is the remainder Put variables back in (one x was divided out in process so first number is one less power than original problem). So the answer is: List all coefficients (numbers in front of x's) and the constant along the top. If a term is missing, put in a 0.
Let's try another Synthetic Division
0 x3 0 x Set divisor = 0 and solve. Put answer here. x - 4 = 0 so x = 4 1 4 Multiply these and put answer above line in next column Multiply these and put answer above line in next column Multiply these and put answer above line in next column Multiply these and put answer above line in next column Bring first number down below line 4 Add these up 16 48 192 Add these up Add these up Add these up 1 x3 + x x + 4 12 48 198 This is the remainder Now put variables back in (remember one x was divided out in process so first number is one less power than original problem so x3). So the answer is: List all coefficients (numbers in front of x's) and the constant along the top. Don't forget the 0's for missing terms.
Let's try a problem where we factor the polynomial completely given one of its factors.
You want to divide the factor into the polynomial so set divisor = 0 and solve for first number. - 2 Multiply these and put answer above line in next column Multiply these and put answer above line in next column Multiply these and put answer above line in next column Bring first number down below line - 8 Add these up 50 Add these up Add these up No remainder so x + 2 IS a factor because it divided in evenly 4 x2 + x - 25 Put variables back in (one x was divided out in process so first number is one less power than original problem). So the answer is the divisor times the quotient: List all coefficients (numbers in front of x's) and the constant along the top. If a term is missing, put in a 0. You could check this by multiplying them out and getting original polynomial
Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. Shawna has kindly given permission for this resource to be downloaded from and for it to be modified to suit the Western Australian Mathematics Curriculum. Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar
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# List Of Multiplication Flash Cards
Studying multiplication after counting, addition, and subtraction is perfect. Young children learn arithmetic using a natural progression. This advancement of understanding arithmetic is usually the adhering to: counting, addition, subtraction, multiplication, and ultimately division. This assertion brings about the issue why discover arithmetic within this pattern? Furthermore, why find out multiplication following counting, addition, and subtraction before department?
## The following information respond to these questions:
1. Kids learn counting very first by associating aesthetic things making use of their fingers. A tangible illustration: How many apples exist in the basket? Far more abstract example is the way outdated are you presently?
2. From counting amounts, another reasonable phase is addition combined with subtraction. Addition and subtraction tables can be extremely helpful instructing assists for children since they are aesthetic instruments creating the cross over from counting less difficult.
3. Which should be learned next, multiplication or department? Multiplication is shorthand for addition. At this stage, children have a organization knowledge of addition. As a result, multiplication will be the following reasonable type of arithmetic to learn.
## Review basic principles of multiplication. Also, evaluate the basics utilizing a multiplication table.
We will evaluation a multiplication case in point. By using a Multiplication Table, increase a number of times about three and have an answer twelve: 4 by 3 = 12. The intersection of row about three and column four of a Multiplication Table is 12; 12 is the answer. For the kids beginning to find out multiplication, this is straightforward. They are able to use addition to eliminate the trouble as a result affirming that multiplication is shorthand for addition. Case in point: 4 by 3 = 4 4 4 = 12. It is an exceptional introduction to the Multiplication Table. The additional gain, the Multiplication Table is aesthetic and demonstrates straight back to discovering addition.
## Where by will we get started understanding multiplication utilizing the Multiplication Table?
1. Very first, get acquainted with the table.
2. Get started with multiplying by 1. Start off at row number one. Relocate to column number 1. The intersection of row one particular and column the first is the perfect solution: 1.
3. Perform repeatedly these actions for multiplying by a single. Grow row a single by posts a single by way of 12. The replies are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 respectively.
4. Repeat these actions for multiplying by two. Increase row two by columns one particular by way of five. The answers are 2, 4, 6, 8, and 10 respectively.
5. Allow us to bounce ahead. Recurring these steps for multiplying by 5. Flourish row five by posts one by means of 12. The solutions are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 correspondingly.
6. Now we will increase the level of problems. Perform repeatedly these steps for multiplying by about three. Flourish row about three by posts 1 via twelve. The answers are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 respectively.
7. Should you be confident with multiplication up to now, try out a check. Resolve the following multiplication difficulties in your head and after that assess your answers towards the Multiplication Table: grow 6 and 2, multiply 9 and about three, increase 1 and 11, flourish 4 and several, and increase 7 as well as two. The situation replies are 12, 27, 11, 16, and 14 respectively.
In the event you got several out from several issues correct, make your personal multiplication tests. Estimate the responses in your head, and check them using the Multiplication Table.
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## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
Seventh grade math covers wide range of topics mostly the basic math which are very important for the higher grade. Number system which include the topics of prime number, prime factorizations multiplicative rules and different divisibility rules. Math also includes finding the least common factor LCM and greatest common multiple GCF. Decimal numbers, integers, fractions and different mathematical operations that can be applied to solve these numbers are covered. Ratio and proportion is one of the other important topics in seventh grade math. The basics of higher level math topics are initiated such as number sequences, linear equalities and inequalities, co-ordinate geometry and statistics.
Example 1: Find the prime factors of the number 28?
Solution: Given is a number 28.
To find its prime factors we divide the given number with the smallest prime number first.
28 ÷ 2 = 14; so, 28 = 14 * 2
Now 14 can be further divided by the prime number 2.
So here we have 14 ÷ 2 = 7. 28 = 2 * 2 * 7.
The number 7 itself is a prime number.
Hence the number 28 = 2 * 2 * 7.
Example 2: Find the x in the proportion x: 4 = 3: 1.
Solution: Here the given proportion is x : 4 = 3 : 1.
It can be expressed as a fraction in the form.
This gives, x / 4 = 3 / 1
Now multiply both sides of the equation by 4.
x = 12/1 = 12.
Hence the value of x for the given proportion is 12.
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Everything to learn better...
Tutor: Toby
# Numbers to 1000
## In a nutshell
Numbers can be made up of one or more digits and can also be written using words. Using place value headings, such as hundreds, tens and ones can help you to understand the value of each digit in a number.
## Place value
When a number is written in digits, the order of the digits determines the value of the number. For example, in the number $512$, the digit $5$ represents $5$ hundreds. However, in the number $145$, the digit $5$ represents $5$ ones, as it is in a different place value column.
Digits should be read aloud from left to right. For numbers below $1000$, the place value columns from left to right are hundreds, tens and ones. You can use a place value chart to show the digits separately.
##### Example 1
Write the number $512$ in a place value chart.
There are three digits, so separate the number into hundreds, tens and ones.
#### Ones
$512$
$5$
$1$
$2$
## Reading numbers up to 1000
Once you can identify the place value for each digit in a number, you can then write the number in words.
##### Example 2
Write the number $781$ in words.
There are $7$ hundreds, $8$ tens and $1$ ones, which can be written as:
Seven hundred and eighty-one.
## Ordering numbers up to 1000
To compare two numbers, place them both in a place value chart. Compare the digit on the left hand side to see which number is bigger. If this digit is the same, move right to the next place value column and see which number is bigger. Continue this process until you reach your answer.
##### Example 3
Which number is bigger: $345$ or $378$?
Firstly, put the numbers into a place value table.
#### Ones
$345$
$3$
$4$
$5$
$378$
$3$
$7$
$8$
Compare the first digit of each number. In this example, the hundreds digit is the same. Therefore, you need to move to the next place value column, the tens. Comparing the digits in the tens column, $7$ is greater than $4$
Therefore $\underline{378 \gt 345}$.
## FAQs - Frequently Asked Questions
### How do you read or write a number in words?
Beta
I'm Vulpy, your AI study buddy! Let's study together.
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# How do you derive y = x^3+(3x)/e^(x^2) using the quotient rule?
Oct 31, 2015
$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 \left({x}^{2} + \setminus \frac{1 - 2 {x}^{2}}{{e}^{{x}^{2}}}\right)$
#### Explanation:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({x}^{3} + \frac{3 x}{{e}^{{x}^{2}}}\right)$
$= \frac{d}{\mathrm{dx}} \left({x}^{3}\right) + \frac{d}{\mathrm{dx}} \left(\frac{3 x}{{e}^{{x}^{2}}}\right)$
$= 3 {x}^{2} + \frac{d}{\mathrm{dx}} \left(\frac{3 x}{{e}^{{x}^{2}}}\right)$
$= 3 {x}^{2} + \frac{{e}^{{x}^{2}} \frac{d}{\mathrm{dx}} \left(3 x\right) - 3 x \frac{d}{\mathrm{dx}} \left({e}^{{x}^{2}}\right)}{{\left({e}^{{x}^{2}}\right)}^{2}}$
$= 3 {x}^{2} + \frac{{e}^{{x}^{2}} \left(3\right) - 3 x \frac{d}{\mathrm{dx}} \left({e}^{{x}^{2}}\right)}{{\left({e}^{{x}^{2}}\right)}^{2}}$
$= 3 {x}^{2} + \frac{{e}^{{x}^{2}} \left(3\right) - 3 x \left(2 x {e}^{{x}^{2}}\right)}{{\left({e}^{{x}^{2}}\right)}^{2}}$
$= 3 {x}^{2} + \frac{{e}^{{x}^{2}} \left(3\right) - 3 x \left(2 x {e}^{{x}^{2}}\right)}{{\left({e}^{{x}^{2}}\right)}^{2}} \frac{{e}^{- {x}^{2}}}{{e}^{- {x}^{2}}}$
$= 3 {x}^{2} + \frac{3 - 3 x \left(2 x\right)}{{e}^{{x}^{2}}}$
$= 3 {x}^{2} + \frac{3 - 6 {x}^{2}}{{e}^{{x}^{2}}}$
$= 3 \left({x}^{2} + \setminus \frac{1 - 2 {x}^{2}}{{e}^{{x}^{2}}}\right)$
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# Playground
The land for the construction of the school playground has the shape of a rectangle with a shorter side of 370 m. Its other side is 260 m longer. How many meters of the fence does the school need to buy on the fencing playground?
Result
x = 2000 m
#### Solution:
$a = 370 \ m \ \\ b = a + 260 = 370 + 260 = 630 \ m \ \\ \ \\ x = 2 \cdot \ (a+b) = 2 \cdot \ (370+630) = 2000 = 2000 \ \text{ m }$
Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!
Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):
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## Next similar math problems:
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Calculate the area of a circle that has the same circumference as the circumference of the rectangle inscribed with a circle with a radius of r 9 cm so that its sides are in ratio 2 to 7.
2. Rectangle
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5. Trapezoid MO
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6. Annular area
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11. Area of a rectangle
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ma006 ยป
We're going to use our knowledge of radicals and factoring to solve quadratic equations. We'll use three different strategies in this unit, and we'll learn them one by one. The three methods are the square root method, completing the square and the quadratic formula. First, we'll learn each of the methods, and then you'll have a chance to pick which method you think is best to use in different cases. For this lesson let's start by focusing on the square root method.
## Square Root Method
If a quadratic equation is in this form, we can simply divide both sides by a, and then take the square root of both sides to find x. If we divide both sides by a, we'll get x squared is equal to c divided by a. And now we take the square root of both sides. When we take the square root of both sides, we need to add a plus or minus symbol in front of the side that doesn't have a variable. And here's why we include this plus or minus sign. Think about if I had x squared equals 1. I know there are really two values that make this true. Positive 1 squared would equal 1, and negative 1 squared would also equal 1. This means when we solve and take the square root of both sides. We need to add the plus or minus symbol to account for both the positive and the negative answer. The square root of x squared is x and the square root of 1 is 1. So really x could be positive 1 like we thought or x could equal negative 1. So you see if you can use the square root method. Try solving this equation by following these steps.
## Square Root Method
First, we divide both sides of our equation by 5, to get x squared equals positive 6. Next, we take the square root of both sides and we indicate the plus or minus sign here. This gives us our answer of positive root 6 and negative root 6. You could have also written the answer like this, root 6, negative root 6. Or switch the order of these 2. So solving this equation isn't that much harder, you just have new symbol of plus or minus.
## Positive and Negative Roots
Like all of our other equations, we can check to make sure that this solution is correct. If we substitute in positive root 6 into our equation, we'll notice that it checks. The square root of 6 squared equals 6, and 5 times 6 equals 30. But remember, we can have the positive answer or the negative answer. So let's check x equal to negative root 6. We plug in negative root 6 in for x, and square it. Squaring a negative number always results in a positive. So we know negative root 6 squared equals positive 6. Then we have 5 times 6, which is 30, so our answer checks. This is always a great way to check your answers. ANd this reinforces why we have the plus or minus symbol. It's because we have two answers.
## Math Quill Plus of Minus Sign
Since we know that taking an even root of an equation creates a positive and a negative result, we need to be able to enter this new sign, this plus or minus. Luckily Math Quill has a way for us to do that. To enter this plus or minus sign, you want to type in a backslash, and then the letters p and m. Of course, p and m here stands for plus or minus. But like most things in math quill, we also need to enter a space in order for this symbol to appear. Try typing these first two expressions using the keystrokes on the screen. And then see if you can figure out these last two. Keep learning, and good luck.
## Math Quill Plus or Minus Sign
For the second one, you need to type in the nthroot and a space. This will create this radical symbol. Your cursor bar will appear here, and you'll need to hit the Right Arrow key and then type in 5 to get what's underneath the radical. These would be the keystrokes for our third expression. Notice that after we type in the square root symbol, we don't need to hit the Right Arrow key. That's because the index for this root is assumed to be a 2, It's rare that you'll ever see it, but we know it's a square root. That's why it's a 2. Math Quill knows it's 2, and so you're just placed back inside the radical. And then you can type in 21. Now for this last expression, you might not have any idea what i is. That's okay. We're going to cover this later. To start this expression, I would type in a forward slash. This would create an entire fraction with a spot for a numerator and the denominator. We type in these keystrokes in order to get our entire numerator. And notice that our cursor bar is here. Once we type in the 6, our cursor bar will appear here, and then we hit the Right Arrow key twice in order to move it down to the denominator. Then, we just type a 4 to wrap up our expression. Now, there are other ways that you can have done this to recreate the fraction bar. You could've created the fraction bar after the 6. But that means, this forward slash would go in between these 2 errors. After you type the 6, you're still inside the radical. So, if you type the forward slash here, you'll create a fraction within the radical. You don't want that to happen, so you hit the Right Arrow key once. Which moves the cursor bar outside of the radical, and then you can type a forward slash to create your fraction, hit the Right Arrow key again to move to the denominator and then enter 4 to finish up. Either way you do this last one is correct, I just want you to get comfortable using math quill.
## Square Root Method Check 1
If that first one gave you some trouble, try this one out. What do you think the solutions are for x? Again, you can either enter your answer with a plus or minus sign, or you can write your two answers separated by a comma.
## Square Root Method Check 1
Here, the answer is plus or minus 5 times root 2. Great thinking if you found these answers. We start by dividing both sides of our equation by 4. This gives us x squared on the left, and 50 on the right. Whenever we use the square root method, we want to isolate x squared. Or there might just be some sort of quantity squared. This allows us to take the square root in this step. Doing that we get x is equal to positive or negative square root of 50. Then we just use our knowledge of simplifying roots to get x is equal to plus or minus 5 times the square root of 2.
## Another Form for the Square Root Method
Our square root method also works for equations in this form. Notice that we just have a quantity over here, squared equal to some number. This means we can take the square root of the left side and the right side. When we take the square root of this side, it undoes the square power, so we're just left with this quantity, ax plus b. So, I have ax plus b on the left side of our equation and plus or minus root c on the right side of our equation. Next we subtract b from both sides, so that way we can start to isolate the X. Notice that this negative b, is not like terms with radical c. All I can do, is list this as negative b plus or minus the square root of c. Finally, to get x by itself, we divide both sides by the coefficient of a. So we know x could equal negative b plus root c divided by a. Or x could equal negative b minus root c, all divided by a. There's really two solutions here.
## Another Form for the Square Root Method
So, using this as your guide, what do you think the solutions would be for this equation? You want to write your answer here using the plus or minus symbol, or you can enter them separated by a comma. Now not every single step on this left side will match this side, but I think you can figure the steps out. Good luck.
## Another Form for the Square Root Method
Our solutions for x are 5 plus or minus 2 times root 3. Great thinking if you got this. Now, I know we haven't done that much practice yet. So that's totally okay if you struggled. I'm sure you'll get the next one. We begin to solve this problem by taking the square root of both sides. Since we have a quantity squared, with a variable inside it, on the left. Taking this square root undoes the square power. Leaving us with x minus 5 on the left hand side of our equation. On the right hand side, we'll just have plus or minus the square root of 12. Next, we'll add 5 to both sides to isolate our x variable. So, we'll have x is equal to 5 plus or minus the square root of 12. And again, here this list them as a sum and difference. We need to simplify the root of 12, so we break it into 4 times 3, this gives us our final answer of 5 plus or minus 2 root 3.
## Square Root Method Check 2
All right. Let's see if you have the right stuff. Try solving this equation and remember to simplify your answer in the end if possible.
## Square Root Method Check 2
The solutions are negative 2 plus 3 root 2, and negative 2 minus 3 root 2. Great thinking if you found those two answers. We start by taking the square root of both sides of our equation. The square root of the quantity x plus 2 squared is just x plus 2. And when we take the square root of 18, we pick up the negative root. So we have plus or minus the square root of 18. Next we isolate x, or get it alone by subtracting 2 from both sides. We don't have light terms here, so we have x equal to negative 2 plus or minus the root of square root of 9 to get our final answer. And do keep in mind there's really two answers here: we have negative 2 plus 3 root 2 and negative 2 minus 3 root
## Square Root Method Check 3
How about this one? What do you think the solutions would be for x? Now there's actually one more step at the end here. Maybe you can get it. As a hint, we no longer have coefficient of 1, in front of x. Now it's a 3.
## Square Root Method Check 3
Here the solutions are negative 5 plus or minus 2 root 6 all divided by 3. Great algebra skills if you got this one correct. You may have also entered this as your answer. This is correct as well. You would of divided 3 into the first term and 3 in the second term. Since there's no common factors we just list these two terms as fractions. With a plus or minus symbol in between. To start this problem, we take the square root of both sides of our equation. And we indicate the plus or minus sign for this root. So we'll have 3x plus 5 on the left. And positive or negative root 24 on the right. Next, we want to isolate this 3x by subtracting 5 from both sides. We'll get 3x equals negative the coefficient 3. This leaves us with 1x equal to negative 5 plus or minus root 24, all divided by 3. We want to look to see if we can simplify any radicals, if possible. The square root of 24 is simplifies to 2 times root 6. So we can replace this here in our final answer.
## Square Root Method Isolate the Square
Sometimes we'll need to perform some extra steps to isolate the square before taking the square root, like in this equation. We have something next to our square. We want to get this square quantity alone, so that way we can take the square root of both sides of our equation. So what do you think we'll need to do here? When you think you have the solutions for x ,enter them here using your plus or minus sign. And don't forget to simplify your root if you can. Good luck.
## Square Root Method Isolate the Square
The solutions for x are 3 plus or minus 4 root 3, all divided by 2 great algebra work for getting that one correct. First we want to isolate the square by adding 48 to both sides of our equation. So, we'll keep this squared quantity on the left and we'll have positive 48 on the right. Next we take the square root of both sides and indicate the plus or minus root here. This square root undoes our square so we're left with the quantity 2x minus three. On the right we still just have plus or minus root 48. We add three to both sides to isolate our 2x, and then we divide both sides by 2, this coefficient. When we divide both sides by 2 we're left with 1x on the left and 3 plus or minus root get this answer. Keep in mind you could divide 2 into each of these terms. This would give us x equal to 3 halves plus or minus 2 root 3. If you list your answers like this, or if you list your answers like this, they're both correct.
## Square Root Method Isolate the Square Check
If the last problem was challenging, then see if you can get this one. Isolate this square first and then continue to solve as usual. When you think you've got it, enter your answer here.
## Square Root Method Isolate the Square Check
The solutions here are 3 plus or minus root 10, all divided by 2. Great work, using that square root method, if you got this one correct. We start solving this equation by adding 10 to both sides. This leaves us with the square isolated on the left and positive 10 on the right. Now that we have something squared alone, we can take the square root of both sides. This means we will have 2x minus 3 on the left. And plus or minus root 10 on the right. We add 3 to both sides to get 2x by itself and then divide both sides by positive 2. This leaves us with 1 x in our answer. And again if you really wanted to, you could split this up into two separate fractions, 3 halves plus or minus root 10 divided by 2. Either form is the same.
## Square Root Method Practice 1
Try solving this first practice problem. What are the possible values for X here? You can either enter the value separated by a comma, or you can use our fancy plus or minus symbol.
## Square Root Method Practice 1
Here our solution is plus or minus 4. We start by taking the square root of both sides of our equation. When we take the square root, we need to indicate plus or minus, since we could have the positive value or the negative value. We take the square root of 16 to get positive 4 or negative 4. These should also make sense because if we square 4, we get 16. And if we square negative 4, we get 16. Great thinking for finding those answers.
## Square Root Method Practice 2
What about this equation, what do you think the values of m could be? For this one you should have two seperate answers. So write one here and the other here, separated by a comma.
## Square Root Method Practice 2
The square quantity is isolated on one side of our equation. So, we can take the square root of this side and this side. The square root undoes the square, leaving us with m plus 2 on the left, and positive or negative root 36 on the right. We know the square root of 36 is just 6. So we have this equation next. We subtract 2 from both sides, to get m is equal to negative 2, plus or minus have 2 equations. M could equal negative 2 plus 6, or m could equal negative 2 minus 6. There's really 2 values of m here. And since these are like terms, we can simplify. So m could equal positive 4 or m could equal negative 8. These were the two solutions.
## Square Root Method Practice 3
For this third practice problem, try finding the solution for r. You can enter your answer here. For this one, you don't want to enter it, separated by commas. You want to use the plus or minus symbol. Good luck solving this one, I know you can do it.
## Square Root Method Practice 3
These are the correct answers. Great algebra solving if you got them. We start by taking the square root of both sides of our equation. So, 3r minus 2 will equal positive or negative root of 27. I'm going to go ahead and simplify the square root of 27 to be 3 root 3. Next we add 2 to both sides of our equation to get 3r is equal to 2 plus or minus 3 root 3. We divide both sides by the coefficient of 3 to get our final result. And if you prefer, you can write your answer as two fractions. 2 3rds plus or minus 3 root 3 divided by 3.
## Square Root Method Practice 4
Alright, how about for this last one? What do you think would be the solutions for x, here? Be careful when you solve, and make sure you check your answers.
## Square Root Method Practice 4
Here our solution is plus or minus three i an imaginary solution, pretty cool. We take the square root of both sides to get x is equal to plus or minus the square root of negative nine. Now we've simplified this radical and remember that negative square roots, need to be simplified using i. We can break this radical down into the square root of 9 times the square root of negative 1. Then we replace the square root of negative 1 with i. We can also check our result. We know the answers are not 3 and negative 3. Since 3 squared would equal positive 9 not negative 9. Also at negative 3 squared would be positive out, lets make sure the plus or minus 3i works in our equation. I'm going to check our positive 3i here in this equation and our negative 3i here in this equation. When we square the quantity of 3i, we get 3 squared which is 9 and i squared. When we square negative 3i negative 3 times negative 3 is positive 9, and i squared is i squared. And remember back from before, i squared equals negative 1. So we can substitute that in here. This effectively changes the sign of these terms. This gives us a check of negative 9 equal to negative 9 in both cases. So we could be confident knowing that our answers are correct.
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# Writing and Graphing Equations of Lines
## Presentation on theme: "Writing and Graphing Equations of Lines"— Presentation transcript:
Writing and Graphing Equations of Lines
3.4 Writing and Graphing Equations of Lines 1 Use the slope-intercept form of the equation of a line. Graph a line by using its slope and a point on the line. Write an equation of a line by using its slope and any point on the line. Write an equation of a line by using two points on the line. Write an equation of a line that fits a data set. 2 3 4 5
Use the slope-intercept form of the equation of a line.
Objective 1 Use the slope-intercept form of the equation of a line. Slide 3.4-3
Use the slope-intercept form of the equation of line.
In Section 3.3, we found the slope of a line by solving for y. In that form, the slope is the coefficient of x. For, example, the slope of the line with equation y = 2x + 3 is 2. So, what does 3 represent? Suppose a line has a slope m and y-intercept (0,b). We can find an equation of this line by choosing another point (x,y) on the line as shown. Then we use the slope formula. Change in y-values Change in x-values Subtract in the denominator. Multiply by x. Add b to both sides. Rewrite. Slide 3.4-4
Use the slope-intercept form of the equation of line.
The result is the slope-intercept form of the equation of a line, because both the slope and the y-intercept of the line can be read directly from the equation. For the line with the equation y = 2x + 3, the number 3 gives the y-intercept (0,3). Slope-Intercept Form The slope-intercept form of the equation of a line with slope m and y-intercept (0,b) is Where m is the slope and b is the y-intercept (0,b). Slide 3.4-5
Identifying Slopes and y-Intercepts
EXAMPLE 1 Identifying Slopes and y-Intercepts Identify the slope and y-intercept of the line with each equation. Solution: Slope: y-intercept: (0,− 6) Slope: − 1 y-intercept: (0,0) Slide 3.4-6
Writing an Equation of a Line
EXAMPLE 2 Writing an Equation of a Line Write an equation of the line with slope −1 and y-intercept (0,5). Solution: Slide 3.4-7
Graph a line by using its slope and a point on the line.
Objective 2 Graph a line by using its slope and a point on the line. Slide 3.4-8
Graph a line by using its slope and a point on the line.
Graphing a Line by Using the Slope and y-Intercept Step 1: Write the equation in slope-intercept form, if necessary, by solving for y. Step 2: Identify the y-intercept. Graph the point (0,b). Step 3: Identify slope m of the line. Use the geometric interpretation of slope (“rise over run”) to find another point on the graph by counting from the y-intercept. Step 4: Join the two points with a line to obtain the graph. Slide 3.4-9
Graphing Lines by Using Slopes and y-Intercepts
EXAMPLE 3 Graphing Lines by Using Slopes and y-Intercepts Graph 3x – 4y = 8 by using the slope and y-intercept. Solution: Slope intercept form Slide
Graphing a Line by Using the Slope and a Point
EXAMPLE 4 Graphing a Line by Using the Slope and a Point Graph the line through (2,−3) with slope Solution: Make sure when you begin counting for a second point you begin at the given point, not at the origin. Slide
Objective 3 Write an equation of a line by using its slope and any point on the line. Slide
Write an equation of a line by using its slope and any point on the line.
We can use the slope-intercept form to write the equation of a line if we know the slope and any point on the line. Slide
Using the Slope-Intercept Form to Write an Equation
EXAMPLE 5 Using the Slope-Intercept Form to Write an Equation Write an equation, in slope-intercept form, of the line having slope −2 and passing through the point (−1,4). Solution: The slope-intercept form is Slide
Write an equation of a line by using its slope and any point on the line.
There is another form that can be used to write the equation of a line. To develop this form, let m represent the slope of a line and let (x1,y1) represent a given point on the line. Let (x, y) represent any other point on the line. Definition of slope Multiply each side by x − x1. Rewrite. Point-Slope Form The point-slope form of the equation of a line with slope m passing through point (x1,y1) is Slope Given point Slide
Using the Point-Slope Form to Write Equations
EXAMPLE 6 Using the Point-Slope Form to Write Equations Write an equation of the line through (5,2), with the slope Give the final answer in slope-intercept form. Solution: Slide
Write an equation of a line by using two points on the line.
Objective 4 Write an equation of a line by using two points on the line. Slide
Write an equation of a line by using two points on the line.
Many of the linear equations in Section 3.1−3.3 were given in the form called standard form, where A, B, and C are real numbers and A and B are not both 0. Slide
Writing the Equation of a Line by Using Two Points
EXAMPLE 7 Writing the Equation of a Line by Using Two Points Find an equation of the line through the points (2,5) and (−1,6). Give the final answer in slope-intercept form and standard form. Solution: Standard form Slope-intercept form The same result would also be found by substituting the slope and either given point in slope-intercept form and then solving for b. Slide
Summary of the forms of linear equations.
Slide
Write an equation of a line that fits a data set.
Objective 5 Write an equation of a line that fits a data set. Slide
Writing an Equation of a Line That Describes Data
EXAMPLE 8 Writing an Equation of a Line That Describes Data Use the points (3, 4645) and (7, 6185) to write an equation in slope-intercept form that approximates the data of the table. How well does this equation approximate the cost in 2005? Solution: The equation gives y = 5415 when x = 5, which is a very good approximation. Slide
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Question
Drag the expressions to the correct functions. Not all expressions will be used.
Consider the functions fand g.
= 4x² + 1
g(x) =
Perform the function compositions:
x² – 3
1. The function composition exists an operation ” ∘ ” that brings two functions f and g, and has a function h = g ∘ f such that h(x) = g(f(x)).
Let the functions be f(x) = 4x² + 1 and g(x) = x² – 3
The correct answer is (f o g)(x) = 4x⁴ – 96x + 37 and
(g o f)(x) = 16x⁴ + 8x² – 2.
### What is composition function?
The function composition exists an operation ” ∘ ” that brings two functions f and g, and has a function h = g ∘ f such that h(x) = g(f(x)). In this operation, the function g exists used for the outcome of applying the function f to x.
Given:
f(x) = 4x² + 1 and g(x) = x² – 3
a) (f o g)(x) = f[g(x)]
f[g(x)] = 4(x² – 3)² + 1
substitute the value of g(x) in the above equation, and we get
= 4(x⁴ – 24x + 9) + 1
simplifying the above equation
= 4x⁴ – 96x + 36 + 1
= 4x⁴ – 96x + 37
(f o g)(x) = 4x⁴ – 96x + 37
b) (g o f)(x) = g[f(x)]
substitute the value of g(x) in the above equation, and we get
g[f(x)] = (4x² + 1)²- 3
= 16x⁴ + 8x² + 1 – 3
simplifying the above equation
= 16x⁴ + 8x² – 2
(g o f)(x) = 16x⁴ + 8x² – 2.
Therefore, the correct answer is (f o g)(x) = 4x⁴ – 96x + 37 and
(g o f)(x) = 16x⁴ + 8x² – 2.
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# Often asked: What Is Relation In Math?
## What is a relation and function in math?
A relation is a set of inputs and outputs, and a function is a relation with one output for each input.
## What is relation and types of relation?
There are 9 types of relations in maths namely: empty relation, full relation, reflexive relation, irreflexive relation, symmetric relation, anti-symmetric relation, transitive relation, equivalence relation, and asymmetric relation.
## What is relation relation?
Both the terms relation and relationship help describe some form of connection or link between groups of people, two individuals, non-living things pr elements. Relation can be used to identify the similarities between two different things, groups of people or individuals. Relation is also used as a reference point.
## What are the 3 types of relation?
The types of relations are nothing but their properties. There are different types of relations namely reflexive, symmetric, transitive and anti symmetric which are defined and explained as follows through real life examples.
## What is relation and example?
A relation between two sets is a collection of ordered pairs containing one object from each set. If the object x is from the first set and the object y is from the second set, then the objects are said to be related if the ordered pair (x,y) is in the relation. A function is a type of relation.
## What is difference between relation and function?
Relation – In maths, the relation is defined as the collection of ordered pairs, which contains an object from one set to the other set. Functions – The relation that defines the set of inputs to the set of outputs is called the functions. In function, each input in the set X has exactly one output in the set Y.
## What are the types of relation?
Types of Relations
• Empty Relation. An empty relation (or void relation) is one in which there is no relation between any elements of a set.
• Universal Relation.
• Identity Relation.
• Inverse Relation.
• Reflexive Relation.
• Symmetric Relation.
• Transitive Relation.
## What is full relation?
The full relation (or universal relation ) between sets X and Y is the set X×Y. The full relation on set E is the set E×E. The full relation is true for all pairs. The identity relation on set E is the set {(x,x) | x∈E}. The identity relation is true for all pairs whose first and second element are identical.
## What is a universal relation?
Universal relation is a relation on set A when A X A ⊆ A X A. In other words, universal – relation is the relation if each element of set A is related to every element of A. For example: Relation on the set A = {1,2,3,4,5,6} by. R = {(a,b) ∈ R: |a -b|≥ 0}
## What is null relation?
The null relation is a relation R in S to T such that R is the empty set: R⊆S×T:R=∅ That is, no element of S relates to any element in T: R:S×T:∀(s,t)∈S×T:¬sRt.
## Who invented relations?
Thomas Aquinas in the Western church noted that in God ” relations are real”, and, echoing Aristotle, claimed that there were indeed three type of relation which give a natural order to the world.
You might be interested: What Is Sets In Math Grade 7?
## What is number relation?
Number Relationships is one of the key mathematical principles or “Big Ideas” in Number Sense and Numeration. It is important to emphasize number relationships with your students to help them learn how numbers are interconnected and how numbers can be used in meaningful ways.
## What does Codomain mean?
The codomain of a function is the set of its possible outputs. In the function machine metaphor, the codomain is the set of objects that might possible come out of the machine. For example, when we use the function notation f:R→R, we mean that f is a function from the real numbers to the real numbers.
## What are four ways to represent a relation?
Relations can be displayed in multiple ways:
• Table: the x-values and y-values are listed in separate columns; each row represents an ordered pair.
• Mapping: shows the domain and range as separate clusters of values.
• Graph: each ordered pair is plotted as a point and can be used to show the relationships between values.
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# What Is Determinant?
Given a system of equations
\begin{align} ax + by & = e \\ cx + dy & = f \end{align}
its solution - if exists - is given by the following expressions
$x = \frac{de - bf}{ad - bc}, y = \frac{ab - ce}{ad - bc}.$
(To obtain $x$, multiply the first equation by $d$, the second by $b$ and take the difference. For $y$, multiply the first equation by $c$, the second by $a$ and take the difference.)
The common denominator $ad - bc$ of the two fractions is known as the determinant of the system of equations, or, more accurately, as the determinant of the matrix $M = \left (\begin{array}{cc} a & b \\ c & d \end{array}\right )$.
The equation has a unique solution only if its determinant is not $0$. The determinant of matrix $M = \left (\begin{array}{cc} a & b \\ c & d \end{array}\right )$ is denoted $|M|$ or, more explicitly, $|M| = \left |\begin{array}{cc} a & b \\ c & d \end{array}\right |$.
With a view to a generalization, observe that in the expression $ad-bc$ there are two products - one with sign plus, the other with sign minus - each of which contains exactly one term from each row and one term from each column! For example, in the term $bc$, $b$ comes from the first row, while $c$ from the second. On the other hand, $c$ comes from the first column and $b$ from the second.
For a $3\times 3$ matrix $M= \left ( \begin{array}{cc} a_{11}&a_{12}&a_{13} \\ a_{21}&a_{22}&a_{23} \\ a_{31}&a_{32}&a_{33} \end{array} \right )$, the determinant $|M| = \left | \begin{array}{cc} a_{11}&a_{12}&a_{13} \\ a_{21}&a_{22}&a_{23} \\ a_{31}&a_{32}&a_{33} \end{array} \right |$ is defined as
$(a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{21}a_{32}a_{13}) - (a_{13}a_{22}a_{31}+a_{12}a_{21}a_{33}+a_{23}a_{32}a_{11}).$
Again, as was the case with a $2\times 2$ matrix, there are equal amounts of products with sign plus and products with the sign minus, each of which contains exactly one number from every row and one number from every column. The diagrams below may help remember which products come with "+" and which with "-":
A $2\times 2$ determinant has a very simple geometric meaning. Indeed, $ad - bc$ is the (signed) area of a parallelogram with vertices at $(0, 0)$, $(a, b)$, $(a+c, b+d)$, $(c, d)$.
The area of the enclosing rectangle is $(a+c)\cdot (b+d)$. To obtain the area of the parallelogram, we need to subtract the area of four triangles and two rectangles. After subtraction, we get exactly $ad - bc$. The determinant $\left |\begin{array}{cc} a & b \\ c & d \end{array}\right |$ changes sign when the rows are swapped. This is why the area was designated "signed". It change signs when the vertices are listed in a different order. (There are other pww of this result: one by dissection and one based on the shearing transform.)
The $3\times 3$ determinant has the meaning of the volume of a parallelopiped defined by three vectors (the rows of the determinant.)
From the definition we may deduce several properties of the determinants:
1. If two rows or two columns are equal, the determinant vanishes:
2. Exchanging two rows or two columns changes the sign of the determinant.
3. Adding one row to another, or one column to another column, does not change the determinant.
4. Multiplying a row or a column by a number changes the value of the determinant by the same factor.
For a $2\times 2$ determinant, we have
1. $\left |\begin{array}{cc} a & b \\ a & b \end{array}\right | = 0.$
2. $\left |\begin{array}{cc} a & b \\ c & d \end{array}\right | = - \left |\begin{array}{cc} c & d \\ a & b \end{array}\right |.$
3. $\left |\begin{array}{cc} a & b \\ c & d \end{array}\right | = \left |\begin{array}{cc} a & b \\ a+c & b+d \end{array}\right |.$
4. $\left |\begin{array}{cc} \lambda a & \lambda b \\ c & d \end{array}\right | = \lambda \left |\begin{array}{cc} a & b \\ c & d \end{array}\right |.$
As a consequence, if $a\ne 0$,
1. $\left |\begin{array}{cc} a & b \\ c & d \end{array}\right | = \left |\begin{array}{cc} 1 & b/a \\ 0 & ad-bc \end{array}\right |,$
which is a combination of "row operation". All five properties are shared by $3\times 3$ and higher dimension (defined shortly) determinants. For $a_{11}\ne 0,$ the row operation 5 shows how to vanish all the other terms in the first column. By this means, calculations of the value of a $N\times N$ determinant are reduced to computing a $(N-1)\times (N-1)$ determinant, and the procedure begs to be repeated, leaving eventually a single number - the sought value of the determinant. (Naturally, there are also "column operations" with similar properties and effects.)
Determinants are defined for all $N\times N$ square matrices $M = (a_{ij})$:
$\left | M \right | = \sum_{\sigma \in S_n} sign(\sigma) \prod_{i=1}^N a_{i,\sigma_i},\$
where $S_n$ is the group of all permutations of the set $\{1, 2, ..., N\}$, $\sigma_i$ is the image of $i$ under the permutation $\sigma$, and $sign(\sigma)=\pm 1$ depending on whether $\sigma$ is even or odd.
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# Solve Second Order Differential Equations - Two distinct real solutions
A tutorial on how to solve second order differential equations with auxiliary equation having 2 distinct real solutions. Examples with detailed solutions are included.
## Examples with Solutions
### Example 1:
Solve the second order differential equation given by
$y'' + 2 y' - 3 y = 0$
Solution to Example 1
The auxiliary equation is given by $k^2 + 2 k - 3 = 0$ Factor the above quadratic equation
$(k + 3)(k - 1) = 0$ Solve for $$k$$ $k_1 = -3 \; \text{and} \; k_2 = 1$ The general solution to the given differential equation is given by
$y = A e^{k_1 x} + B e^{k_2 x} = A e^{-3 x} + B e^x$
where $$A$$ and $$B$$ are constants.
### Example 2:
Solve the second order differential equation given by
$y'' + 3 y' -10 y = 0$
with the initial conditions $$y(0) = 1$$ and $$y'(0) = 0$$
Solution to Example 2
The auxiliary equation is given by $k^2 + 3 k - 10 = 0$ Solve the above quadratic equation to obtain
$$k_1 = 2$$ and $$k_2 = -5$$
The general solution to the given differential equation is given by
$y = A e^{2 x} + B e^{- 5 x}$
where $$A$$ and $$B$$ are constants that may be evaluated using the initial conditions. $$y(0) = 1$$ gives $y(0) = A e^0 + B e^0 = A + B = 1$
$$y'(0) = 0$$ gives
$y'(0) = 2 A e^0 - 5 B e^0 = 2 A - 5 B = 0$ Solve the system of equations $$A + B = 1$$ and $$2 A - 5 B = 0$$ to obtain
$$A = \frac{5}{7}$$ and $$B = \frac{2}{7}$$
The solution to the given equation may be written as
$y = \left(\frac{5}{7}\right) e^{2 x} + \left(\frac{2}{7}\right) e^{- 5 x}$
## Exercises:
Solve the following differential equations.
1. $$y'' + 5 y' - 6 y = 0$$
2. $$y'' + y' - 2 y = 0$$ with the initial conditions $$y(0) = 2$$ and $$y'(0) = 0$$
1. $$y = A e^x + B e^{-6 x}$$, $$A$$ and $$B$$ constant
2. $$y = \left(\frac{4}{3}\right) e^x + \left(\frac{2}{3}\right) e^{-2 x}$$
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# 5.2 Solve systems of equations by substitution (Page 5/5)
Page 5 / 5
Kenneth currently sells suits for company A at a salary of $22,000 plus a$10 commission for each suit sold. Company B offers him a position with a salary of $28,000 plus a$4 commission for each suit sold. How many suits would Kenneth need to sell for the options to be equal?
Kenneth would need to sell 1,000 suits.
Access these online resources for additional instruction and practice with solving systems of equations by substitution.
## Key concepts
• Solve a system of equations by substitution
1. Solve one of the equations for either variable.
2. Substitute the expression from Step 1 into the other equation.
3. Solve the resulting equation.
4. Substitute the solution in Step 3 into one of the original equations to find the other variable.
5. Write the solution as an ordered pair.
6. Check that the ordered pair is a solution to both original equations.
## Practice makes perfect
Solve a System of Equations by Substitution
In the following exercises, solve the systems of equations by substitution.
$\left\{\begin{array}{c}2x+y=-4\hfill \\ 3x-2y=-6\hfill \end{array}$
$\left(-2,0\right)$
$\left\{\begin{array}{c}2x+y=-2\hfill \\ 3x-y=7\hfill \end{array}$
$\left\{\begin{array}{c}x-2y=-5\hfill \\ 2x-3y=-4\hfill \end{array}$
$\left(7,6\right)$
$\left\{\begin{array}{c}x-3y=-9\hfill \\ 2x+5y=4\hfill \end{array}$
$\left\{\begin{array}{c}5x-2y=-6\hfill \\ y=3x+3\hfill \end{array}$
$\left(0,3\right)$
$\left\{\begin{array}{c}-2x+2y=6\hfill \\ y=-3x+1\hfill \end{array}$
$\left\{\begin{array}{c}2x+3y=3\hfill \\ y=\text{−}x+3\hfill \end{array}$
$\left(6,-3\right)$
$\left\{\begin{array}{c}2x+5y=-14\hfill \\ y=-2x+2\hfill \end{array}$
$\left\{\begin{array}{c}2x+5y=1\hfill \\ y=\frac{1}{3}x-2\hfill \end{array}$
$\left(3,-1\right)$
$\left\{\begin{array}{c}3x+4y=1\hfill \\ y=-\frac{2}{5}x+2\hfill \end{array}$
$\left\{\begin{array}{c}3x-2y=6\hfill \\ y=\frac{2}{3}x+2\hfill \end{array}$
$\left(6,6\right)$
$\left\{\begin{array}{c}-3x-5y=3\hfill \\ y=\frac{1}{2}x-5\hfill \end{array}$
$\left\{\begin{array}{c}2x+y=10\hfill \\ -x+y=-5\hfill \end{array}$
$\left(5,0\right)$
$\left\{\begin{array}{c}-2x+y=10\hfill \\ -x+2y=16\hfill \end{array}$
$\left\{\begin{array}{c}3x+y=1\hfill \\ -4x+y=15\hfill \end{array}$
$\left(-2,7\right)$
$\left\{\begin{array}{c}x+y=0\hfill \\ 2x+3y=-4\hfill \end{array}$
$\left\{\begin{array}{c}x+3y=1\hfill \\ 3x+5y=-5\hfill \end{array}$
$\left(-5,2\right)$
$\left\{\begin{array}{c}x+2y=-1\hfill \\ 2x+3y=1\hfill \end{array}$
$\left\{\begin{array}{c}2x+y=5\hfill \\ x-2y=-15\hfill \end{array}$
$\left(-1,7\right)$
$\left\{\begin{array}{c}4x+y=10\hfill \\ x-2y=-20\hfill \end{array}$
$\left\{\begin{array}{c}y=-2x-1\hfill \\ y=-\frac{1}{3}x+4\hfill \end{array}$
$\left(-3,5\right)$
$\left\{\begin{array}{c}y=x-6\hfill \\ y=-\frac{3}{2}x+4\hfill \end{array}$
$\left\{\begin{array}{c}y=2x-8\hfill \\ y=\frac{3}{5}x+6\hfill \end{array}$
(10, 12)
$\left\{\begin{array}{c}y=\text{−}x-1\hfill \\ y=x+7\hfill \end{array}$
$\left\{\begin{array}{c}4x+2y=8\hfill \\ 8x-y=1\hfill \end{array}$
$\left(\frac{1}{2},3\right)$
$\left\{\begin{array}{c}-x-12y=-1\hfill \\ 2x-8y=-6\hfill \end{array}$
$\left\{\begin{array}{c}15x+2y=6\hfill \\ -5x+2y=-4\hfill \end{array}$
$\left(\frac{1}{2},-\frac{3}{4}\right)$
$\left\{\begin{array}{c}2x-15y=7\hfill \\ 12x+2y=-4\hfill \end{array}$
$\left\{\begin{array}{c}y=3x\hfill \\ 6x-2y=0\hfill \end{array}$
Infinitely many solutions
$\left\{\begin{array}{c}x=2y\hfill \\ 4x-8y=0\hfill \end{array}$
$\left\{\begin{array}{c}2x+16y=8\hfill \\ -x-8y=-4\hfill \end{array}$
Infinitely many solutions
$\left\{\begin{array}{c}15x+4y=6\hfill \\ -30x-8y=-12\hfill \end{array}$
$\left\{\begin{array}{c}y=-4x\hfill \\ 4x+y=1\hfill \end{array}$
No solution
$\left\{\begin{array}{c}y=-\frac{1}{4}x\hfill \\ x+4y=8\hfill \end{array}$
$\left\{\begin{array}{c}y=\frac{7}{8}x+4\hfill \\ -7x+8y=6\hfill \end{array}$
No solution
$\left\{\begin{array}{c}y=-\frac{2}{3}x+5\hfill \\ 2x+3y=11\hfill \end{array}$
Solve Applications of Systems of Equations by Substitution
In the following exercises, translate to a system of equations and solve.
The sum of two numbers is 15. One number is 3 less than the other. Find the numbers.
The numbers are 13 and 17.
The sum of two numbers is 30. One number is 4 less than the other. Find the numbers.
The sum of two numbers is −26. One number is 12 less than the other. Find the numbers.
The numbers are −7 and −19.
The perimeter of a rectangle is 50. The length is 5 more than the width. Find the length and width.
The perimeter of a rectangle is 60. The length is 10 more than the width. Find the length and width.
The length is 20 and the width is 10.
The perimeter of a rectangle is 58. The length is 5 more than three times the width. Find the length and width.
The perimeter of a rectangle is 84. The length is 10 more than three times the width. Find the length and width.
The length is 34 and the width is 8.
The measure of one of the small angles of a right triangle is 14 more than 3 times the measure of the other small angle. Find the measure of both angles.
The measure of one of the small angles of a right triangle is 26 more than 3 times the measure of the other small angle. Find the measure of both angles.
The measures are 16° and 74°.
The measure of one of the small angles of a right triangle is 15 less than twice the measure of the other small angle. Find the measure of both angles.
The measure of one of the small angles of a right triangle is 45 less than twice the measure of the other small angle. Find the measure of both angles.
The measures are 45° and 45°.
Maxim has been offered positions by two car dealers. The first company pays a salary of $10,000 plus a commission of$1,000 for each car sold. The second pays a salary of $20,000 plus a commission of$500 for each car sold. How many cars would need to be sold to make the total pay the same?
Jackie has been offered positions by two cable companies. The first company pays a salary of $14,000 plus a commission of$100 for each cable package sold. The second pays a salary of $20,000 plus a commission of$25 for each cable package sold. How many cable packages would need to be sold to make the total pay the same?
80 cable packages would need to be sold.
Amara currently sells televisions for company A at a salary of $17,000 plus a$100 commission for each television she sells. Company B offers her a position with a salary of $29,000 plus a$20 commission for each television she sells. How televisions would Amara need to sell for the options to be equal?
Mitchell currently sells stoves for company A at a salary of $12,000 plus a$150 commission for each stove he sells. Company B offers him a position with a salary of $24,000 plus a$50 commission for each stove he sells. How many stoves would Mitchell need to sell for the options to be equal?
Mitchell would need to sell 120 stoves.
## Everyday math
When Gloria spent 15 minutes on the elliptical trainer and then did circuit training for 30 minutes, her fitness app says she burned 435 calories. When she spent 30 minutes on the elliptical trainer and 40 minutes circuit training she burned 690 calories. Solve the system $\left\{\begin{array}{c}15e+30c=435\hfill \\ 30e+40c=690\hfill \end{array}$ for $e$ , the number of calories she burns for each minute on the elliptical trainer, and $c$ , the number of calories she burns for each minute of circuit training.
Stephanie left Riverside, California, driving her motorhome north on Interstate 15 towards Salt Lake City at a speed of 56 miles per hour. Half an hour later, Tina left Riverside in her car on the same route as Stephanie, driving 70 miles per hour. Solve the system $\left\{\begin{array}{c}56s=70t\hfill \\ s=t+\frac{1}{2}\hfill \end{array}$ .
1. for $t$ to find out how long it will take Tina to catch up to Stephanie.
2. what is the value of $s$ , the number of hours Stephanie will have driven before Tina catches up to her?
$t=2$ hours $s=2\frac{1}{2}$ hours
## Writing exercises
Solve the system of equations
$\left\{\begin{array}{c}x+y=10\hfill \\ x-y=6\hfill \end{array}$
by graphing.
by substitution.
Which method do you prefer? Why?
Solve the system of equations
$\left\{\begin{array}{c}3x+y=12\hfill \\ x=y-8\hfill \end{array}$ by substitution and explain all your steps in words.
## Self check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
After reviewing this checklist, what will you do to become confident for all objectives?
integer greater than 2 and less than 12
2 < x < 12
Felix
I'm guessing you are doing inequalities...
Felix
Actually, translating words into algebraic expressions / equations...
Felix
He charges $125 per job. His monthly expenses are$1,600. How many jobs must he work in order to make a profit of at least $2,400? Alicia Reply at least 20 Ayla what are the steps? Alicia 6.4 jobs Grahame 32 Grahame 1600+2400= total amount with expenses. 4000/125= number of jobs needed to make that min profit of 2400. answer is 32 Orlando He must work 32 jobs to make a profit POP what is algebra Azhar Reply repeated addition and subtraction of the order of operations. i love algebra I'm obsessed. Shemiah hi Krekar One-fourth of the candies in a bag of M&M’s are red. If there are 23 red candies, how many candies are in the bag? Leanna Reply they are 92 candies in the bag POP rectangular field solutions Navin Reply What is this? Donna t muqtaar the proudact of 3x^3-5×^2+3 and 2x^2+5x-4 in z7[x]/ is anas Reply ? Choli a rock is thrown directly upward with an initial velocity of 96feet per second from a cliff 190 feet above a beach. The hight of tha rock above the beach after t second is given by the equation h=_16t^2+96t+190 Usman Stella bought a dinette set on sale for$725. The original price was $1,299. To the nearest tenth of a percent, what was the rate of discount? Manhwa Reply 44.19% Scott 40.22% Terence 44.2% Orlando I don't know Donna if you want the discounted price subtract$725 from $1299. then divide the answer by$1299. you get 0.4419... but as percent you get 44.19... but to the nearest tenth... round .19 to .2 and you get 44.2%
Orlando
you could also just divide $725/$1299 and then subtract it from 1. then you get the same answer.
Orlando
p mulripied-5 and add 30 to it
Tausif
Tausif
how
muqtaar
Can you explain further
p mulripied-5 and add to 30
Tausif
-5p+30?
Corey
p=-5+30
Jacob
How do you find divisible numbers without a calculator?
TAKE OFF THE LAST DIGIT AND MULTIPLY IT 9. SUBTRACT IT THE DIGITS YOU HAVE LEFT. IF THE ANSWER DIVIDES BY 13(OR IS ZERO), THEN YOUR ORIGINAL NUMBER WILL ALSO DIVIDE BY 13!IS DIVISIBLE BY 13
BAINAMA
When she graduates college, Linda will owe $43,000 in student loans. The interest rate on the federal loans is 4.5% and the rate on the private bank loans is 2%. The total interest she owes for one year was$1,585. What is the amount of each loan?
Sean took the bus from Seattle to Boise, a distance of 506 miles. If the trip took 7 2/3 hours, what was the speed of the bus?
66miles/hour
snigdha
How did you work it out?
Esther
s=mi/hr 2/3~0.67 s=506mi/7.67hr = ~66 mi/hr
Orlando
No; 65m/hr
albert
hello, I have algebra phobia. Subtracting negative numbers always seem to get me confused.
what do you need help in?
Felix
Heather
look at the numbers if they have different signs, it's like subtracting....but you keep the sign of the largest number...
Felix
for example.... -19 + 7.... different signs...subtract.... 12 keep the sign of the "largest" number 19 is bigger than 7.... 19 has the negative sign... Therefore, -12 is your answer...
Felix
—12
Thanks Felix.l also get confused with signs.
Esther
Thank you for this
Shatey
ty
Graham
think about it like you lost $19 (-19), then found$7(+7). Totally you lost just $12 (-12) Annushka I used to struggle a lot with negative numbers and math in general what I typically do is look at it in terms of money I have -$5 in my account I then take out 5 more dollars how much do I have in my account well-\$10 ... I also for a long time would draw it out on a number line to visualize it
Meg
practicing with smaller numbers to understand then working with larger numbers helps too and the song/rhyme same sign add and keep opposite signs subtract keep the sign of the bigger # then you'll be exact
Meg
hi
albert
Bruce drives his car for his job. The equation R=0.575m+42 models the relation between the amount in dollars, R, that he is reimbursed and the number of miles, m, he drives in one day. Find the amount Bruce is reimbursed on a day when he drives 220 miles
168.50=R
Heather
john is 5years older than wanjiru.the sum of their years is27years.what is the age of each
46
mustee
j 17 w 11
Joseph
john is 16. wanjiru is 11.
Felix
27-5=22 22÷2=11 11+5=16
Joyce
|
# Difference between revisions of "2007 AMC 12A Problems/Problem 16"
## Problems
How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?
$\mathrm{(A)}\ 96\qquad \mathrm{(B)}\ 104\qquad \mathrm{(C)}\ 112\qquad \mathrm{(D)}\ 120\qquad \mathrm{(E)}\ 256$
## Solution 1
We can find the number of increasing arithmetic sequences of length 3 possible from 0 to 9, and then find all the possible permutations of these sequences.
Common difference Sequences possible Number of sequences 1 $012, \ldots, 789$ 8 2 $024, \ldots, 579$ 6 3 $036, \ldots, 369$ 4 4 $048, \ldots, 159$ 2
This gives us a total of $2 + 4 + 6 + 8 = 20$ sequences. There are $3! = 6$ to permute these, for a total of $120$.
However, we note that the conditions of the problem require three-digit numbers, and hence our numbers cannot start with zero. There are $2! \cdot 4 = 8$ numbers which start with zero, so our answer is $120 - 8 = 112 \Longrightarrow \mathrm{(C)}$.
## Solution 2
Observe that, if the smallest and largest digit have the same parity, this uniquely determines the middle digit. If the smallest digit is not zero, then any choice of the smallest and largest digit gives $3! = 6$ possible 3-digit numbers; otherwise, $4$ possible 3-digit numbers. Hence we can do simple casework on whether 0 is in the number or not.
Case 1: 0 is not in the number. Then there are $\binom{5}{2} + \binom{4}{2} = 16$ ways to choose two nonzero digits of the same parity, and each choice generates $3! = 6$ 3-digit numbers, giving $16 \times 6 = 96$ numbers.
Case 2: 0 is in the number. Then there are $4$ ways to choose the largest digit (2, 4, 6, or 8), and each choice generates $4$ 3-digit numbers, giving $4 \times 4 = 16$ numbers.
Thus the total is $96 + 16 = 112 \Longrightarrow \mathrm{(C)}$. (by scrabbler94)
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# & dding ubtracting ractions.
## Presentation on theme: "& dding ubtracting ractions."— Presentation transcript:
& dding ubtracting ractions
Adding & Subtracting Fractions Teacher Notes:
Use this presentation to teach the steps for adding and subtracting fractions with unlike denominators. It is assumed that students have a thorough understanding of the process for making equivalent fractions. CLICK HERE to view a demonstration with opportunities for students to practice making equivalent fractions.
We need a common denominator to add these fractions. +
+ We need a common denominator to add these fractions. Count by 2's 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 Count by 7's 7, 14, 21, 28, 35…
Count by 2's 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 Count by 7's 7, 14, 21, 28, 35… The first number IN COMMON that appears on both lists becomes the common denominator
7 6 13 = = + 7 + 6 = 13 x 7 X 7 x 2 x 2 Make equivalent fractions.
Add the numerators = + 7 + 6 = 13 x 2 13
+ We need a common denominator to add these fractions. Count by 7's 7, 14, 21, 28, 35, 42, 49, 56, 63 Count by 5's 5, 10, 15, 20, 25, 30, 35, 40, 45
15 7 22 = + = 15 + 7 = 22 x 5 X 5 x 7 x 7 Make equivalent fractions.
Add the numerators = x 7 = 22 22
We need a common denominator to add these fractions. + Count by 7's 7, 14, 21, 28, 35, 42, 49, 56, 63 Count by 8's 8, 16, 24, 32, 40, 48, 56, 64, 72
32 21 53 32 + 21 = 53 = + = x 8 x 8 x 7 x 7 Make equivalent fractions.
Add the numerators x 7 + = 53 = x 7 53
+ We need a common denominator to add these fractions. Count by 3's 3, 6, 9, 12, 15, 18, 21, 24, 27 Count by 5's 5, 10, 15, 20, 25, 30, 35, 40, 45
x 3 9 Add the numerators = X 3 = 19 x 5 10 + = x 5 19
+ We need a common denominator to add these fractions. Count by 3's 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33 Count by 11's 11, 22, 33, 44, 55, 66, 77…
15 22 37 = + = 15 + 22 = 37 x 3 X 3 x 11 x 11 Make equivalent
fractions. = X 3 x 11 Add the numerators 22 + = = 37 x 11 37
+ We need a common denominator to add these fractions. Count by 3's 3, 6, 9, 12, 15, 18, 21, 24, 27 Count by 6's 6, 12, 18, 24,…
4 1 5 4 + 1 = 5 = + = x 2 X 2 x 1 x 1 Make equivalent fractions.
Add the numerators = x 1 4 + 1 = 5 5
+ We need a common denominator to add these fractions. Count by 11's 11, 22, 33, 44, 55, 66, 77… 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72
18 11 29 = + = 18 + 11 = 29 x 6 X 6 x 11 x 11 Make equivalent
fractions. = X 6 x 11 11 + Add the numerators. = = 29 x 11 29
We need a common denominator to add these fractions. +
+ We need a common denominator to add these fractions. Count by 5's 5, 10, 15, 20, 25, 30, 35, 40, 45 Count by 9's 9, 18, 27, 36, 45, 54, 63, 72, 81
Count by 5's 5, 10, 15, 20, 25, 30, 35, 40, 45 Count by 9's 9, 18, 27, 36, 45, 54, 63, 72, 81 The first number IN COMMON that appears on both lists becomes the common denominator
x 9 36 Make equivalent fractions. Make equivalent fractions. = x9 x 5 35 Add the numerators. Add the numerators. + = = 71 = 71 x 5 71
+ We need a common denominator to add these fractions. Count by 12's 12, 24, 36, 48, 60, 72, 84, 96 Count by 8's 8, 16, 24, 32, 40, 48, 56, 64, 72, 79, 80, 88, 96
10 9 19 10 + 9 = 19 = + = x 2 x 2 x 3 x 3 Make equivalent fractions.
Add the numerators. + = = 19 x 3 19
- We need a common denominator to subtract a fraction from another. Count by 5's 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55 Count by 11's 11, 22, 33, 44, 55, 66, 77…
15 11 4 15 - 11 = 4 = = - x 5 x 5 x 11 x 11 Make equivalent fractions.
Subtract. = = 4 - x 11 4
- We need a common denominator to subtract one fraction from another. Count by 2's 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 … Count by 10's 10, 20, 30…
5 3 2 5 – 3 = 2 = - = x 5 x 5 x 1 x 1 Make equivalent fractions.
Subtract. - = 5 – 3 = 2 x 1 2
|
# Sum of Angles in a Polygon
The sum of angles in a polygon depends on the number of vertices it has. As we know, polygons are closed figures, which are made up line-segments in a two-dimensional plane. There are different types of polygons based on the number of sides. They are:
• Triangle (Three-sided polygon)
• Square (Four-sided polygon)
• Pentagon (Five-sided polygon)
• Hexagon (Six-sided polygon)
• Septagon (Seven-sided polygon)
• Octagon (Eight-sided polygon)
• Nonagon (Nine-sided polygon)
• Decagon (Ten-sided polygon)
And so on.
## Angle Sum of Polygons
As we know, by angle sum property of triangle, the sum of interior angles of a triangle is equal to 180 degrees. When we start with a polygon with four or more than four sides, we need to draw all the possible diagonals from one vertex. The polygon then is broken into several non-overlapping triangles.
### Interior Angles Sum of Polygons
The angle sum of this polygon for interior angles can be determined on multiplying the number of triangles by 180°. After examining, we can see that the number of triangles is two less than the number of sides, always.
Hence, we can say now, if a convex polygon has n sides, then the sum of its interior angle is given by the following formula:
S = ( n − 2) × 180°
This is the angle sum of interior angles of a polygon.
### Exterior Angles Sum of Polygons
An exterior angle of a polygon is made by extending only one of its sides, in the outward direction. The angle next to an interior angle, formed by extending the side of the polygon, is the exterior angle.
Hence, we can say, if a polygon is convex, then the sum of the degree measures of the exterior angles, one at each vertex, is 360°.
Therefore, the sum of exterior angles = 360°
Proof: For any closed structure, formed by sides and vertex, the sum of the exterior angles is always equal to the sum of linear pairs and sum of interior angles. Therefore,
S = 180n – 180(n-2)
S = 180n – 180n + 360
S = 360°
Also, the measure of each exterior angle of an equiangular polygon = 360°/n
### How to find the sum of angles of a polygon?
Question 1: Find the sum of interior angles of a regular pentagon.
Solution: A pentagon has five sides.
Therefore, by the angle sum formula we know;
S = ( n − 2) × 180°
Here, n = 5
Hence,
Sum of angles of pentagon = ( 5 − 2) × 180°
S = 3 × 180°
S = 540°
Question 2: Find the measure of each interior angle of a regular decagon.
Solution: A decagon has ten sides.
Therefore, by the angle sum formula we know;
S = ( n − 2) × 180°
Here, n = 10
Hence,
Sum of angles of pentagon = ( 10 − 2) × 180°
S = 8 × 180°
S = 1440°
For a regular decagon, all the interior angles are equal.
Hence, the measure of each interior angle of regular decagon = sum of interior angles/number of sides
Interior angle = 1440/10 = 144°
Quiz on Sum of angles in a polygon
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Intermediate Algebra 2e
# 9.6Graph Quadratic Functions Using Properties
Intermediate Algebra 2e9.6 Graph Quadratic Functions Using Properties
## Learning Objectives
By the end of this section, you will be able to:
• Recognize the graph of a quadratic function
• Find the axis of symmetry and vertex of a parabola
• Find the intercepts of a parabola
• Graph quadratic functions using properties
• Solve maximum and minimum applications
## Be Prepared 9.16
Before you get started, take this readiness quiz.
Graph the function $f(x)=x2f(x)=x2$ by plotting points.
If you missed this problem, review Example 3.54.
## Be Prepared 9.17
Solve: $2x2+3x−2=0.2x2+3x−2=0.$
If you missed this problem, review Example 6.45.
## Be Prepared 9.18
Evaluate $−b2a−b2a$ when a = 3 and b = −6.
If you missed this problem, review Example 1.21.
## Recognize the Graph of a Quadratic Function
Previously we very briefly looked at the function $f(x)=x2f(x)=x2$, which we called the square function. It was one of the first non-linear functions we looked at. Now we will graph functions of the form $f(x)=ax2+bx+cf(x)=ax2+bx+c$ if $a≠0.a≠0.$ We call this kind of function a quadratic function.
## Quadratic Function
A quadratic function, where a, b, and c are real numbers and $a≠0,a≠0,$ is a function of the form
$f(x)=ax2+bx+cf(x)=ax2+bx+c$
We graphed the quadratic function $f(x)=x2f(x)=x2$ by plotting points.
Every quadratic function has a graph that looks like this. We call this figure a parabola.
Let’s practice graphing a parabola by plotting a few points.
## Example 9.42
Graph $f(x)=x2−1.f(x)=x2−1.$
## Try It 9.83
Graph $f(x)=−x2.f(x)=−x2.$.
## Try It 9.84
Graph $f(x)=x2+1.f(x)=x2+1.$
All graphs of quadratic functions of the form f (x) = ax2 + bx + c are parabolas that open upward or downward. See Figure 9.2.
Figure 9.2
Notice that the only difference in the two functions is the negative sign before the quadratic term (x2 in the equation of the graph in Figure 9.2). When the quadratic term, is positive, the parabola opens upward, and when the quadratic term is negative, the parabola opens downward.
## Parabola Orientation
For the graph of the quadratic function f (x) = ax2 + bx + c, if
## Example 9.43
Determine whether each parabola opens upward or downward:
$f(x)=−3x2+2x−4f(x)=−3x2+2x−4$ $f(x)=6x2+7x−9.f(x)=6x2+7x−9.$
## Try It 9.85
Determine whether the graph of each function is a parabola that opens upward or downward:
$f(x)=2x2+5x−2f(x)=2x2+5x−2$ $f(x)=−3x2−4x+7.f(x)=−3x2−4x+7.$
## Try It 9.86
Determine whether the graph of each function is a parabola that opens upward or downward:
$f(x)=−2x2−2x−3f(x)=−2x2−2x−3$ $f(x)=5x2−2x−1.f(x)=5x2−2x−1.$
## Find the Axis of Symmetry and Vertex of a Parabola
Look again at Figure 9.2. Do you see that we could fold each parabola in half and then one side would lie on top of the other? The ‘fold line’ is a line of symmetry. We call it the axis of symmetry of the parabola.
We show the same two graphs again with the axis of symmetry. See Figure 9.3.
Figure 9.3
The equation of the axis of symmetry can be derived by using the Quadratic Formula. We will omit the derivation here and proceed directly to using the result. The equation of the axis of symmetry of the graph of f (x) = ax2 + bx + c is $x=−b2a.x=−b2a.$
So to find the equation of symmetry of each of the parabolas we graphed above, we will substitute into the formula $x=−b2a.x=−b2a.$
Notice that these are the equations of the dashed blue lines on the graphs.
The point on the parabola that is the lowest (parabola opens up), or the highest (parabola opens down), lies on the axis of symmetry. This point is called the vertex of the parabola.
We can easily find the coordinates of the vertex, because we know it is on the axis of symmetry. This means its
x-coordinate is $−b2a.−b2a.$ To find the y-coordinate of the vertex we substitute the value of the x-coordinate into the quadratic function.
## Axis of Symmetry and Vertex of a Parabola
The graph of the function f (x) = ax2 + bx + c is a parabola where:
• the axis of symmetry is the vertical line $x=−b2a.x=−b2a.$
• the vertex is a point on the axis of symmetry, so its x-coordinate is $−b2a.−b2a.$
• the y-coordinate of the vertex is found by substituting $x=−b2ax=−b2a$ into the quadratic equation.
## Example 9.44
For the graph of $f(x)=3x2−6x+2f(x)=3x2−6x+2$ find:
the axis of symmetry the vertex.
## Try It 9.87
For the graph of $f(x)=2x2−8x+1f(x)=2x2−8x+1$ find:
the axis of symmetry the vertex.
## Try It 9.88
For the graph of $f(x)=2x2−4x−3f(x)=2x2−4x−3$ find:
the axis of symmetry the vertex.
## Find the Intercepts of a Parabola
When we graphed linear equations, we often used the x- and y-intercepts to help us graph the lines. Finding the coordinates of the intercepts will help us to graph parabolas, too.
Remember, at the y-intercept the value of x is zero. So to find the y-intercept, we substitute x = 0 into the function.
Let’s find the y-intercepts of the two parabolas shown in Figure 9.4.
Figure 9.4
An x-intercept results when the value of f (x) is zero. To find an x-intercept, we let f (x) = 0. In other words, we will need to solve the equation 0 = ax2 + bx + c for x.
$f(x)=ax2+bx+c0=ax2+bx+cf(x)=ax2+bx+c0=ax2+bx+c$
Solving quadratic equations like this is exactly what we have done earlier in this chapter!
We can now find the x-intercepts of the two parabolas we looked at. First we will find the x-intercepts of the parabola whose function is f (x) = x2 + 4x + 3.
Let $f(x)=0f(x)=0$. Factor. Use the Zero Product Property. Solve. The x-intercepts are $(−1,0)(−1,0)$ and $(−3,0)(−3,0)$.
Now we will find the x-intercepts of the parabola whose function is f (x) = −x2 + 4x + 3.
Let $f(x)=0f(x)=0$. This quadratic does not factor, sowe use the Quadratic Formula. $a=−1,b=4,c=3a=−1,b=4,c=3$ Simplify. The x-intercepts are $(2+7,0)(2+7,0)$ and$(2−7,0)(2−7,0)$.
We will use the decimal approximations of the x-intercepts, so that we can locate these points on the graph,
$(2+7,0)≈(4.6,0)(2−7,0)≈(−0.6,0)(2+7,0)≈(4.6,0)(2−7,0)≈(−0.6,0)$
Do these results agree with our graphs? See Figure 9.5.
Figure 9.5
## Find the Intercepts of a Parabola
To find the intercepts of a parabola whose function is $f(x)=ax2+bx+c:f(x)=ax2+bx+c:$
$y-interceptx-interceptsLetx=0and solve forf(x).Letf(x)=0and solve forx.y-interceptx-interceptsLetx=0and solve forf(x).Letf(x)=0and solve forx.$
## Example 9.45
Find the intercepts of the parabola whose function is $f(x)=x2−2x−8.f(x)=x2−2x−8.$
## Try It 9.89
Find the intercepts of the parabola whose function is $f(x)=x2+2x−8.f(x)=x2+2x−8.$
## Try It 9.90
Find the intercepts of the parabola whose function is $f(x)=x2−4x−12.f(x)=x2−4x−12.$
In this chapter, we have been solving quadratic equations of the form ax2 + bx + c = 0. We solved for x and the results were the solutions to the equation.
We are now looking at quadratic functions of the form f (x) = ax2 + bx + c. The graphs of these functions are parabolas. The x-intercepts of the parabolas occur where f (x) = 0.
For example:
$Quadratic equationQuadratic function x2−2x−15=0(x−5)(x+3)=0x−5=0x+3=0x=5x=−3Letf(x)=0.f(x)=x2−2x−150=x2−2x−150=(x−5)(x+3)x−5=0x+3=0x=5x=−3(5,0)and(−3,0)x-interceptsQuadratic equationQuadratic function x2−2x−15=0(x−5)(x+3)=0x−5=0x+3=0x=5x=−3Letf(x)=0.f(x)=x2−2x−150=x2−2x−150=(x−5)(x+3)x−5=0x+3=0x=5x=−3(5,0)and(−3,0)x-intercepts$
The solutions of the quadratic function are the x values of the x-intercepts.
Earlier, we saw that quadratic equations have 2, 1, or 0 solutions. The graphs below show examples of parabolas for these three cases. Since the solutions of the functions give the x-intercepts of the graphs, the number of x-intercepts is the same as the number of solutions.
Previously, we used the discriminant to determine the number of solutions of a quadratic function of the form $ax2+bx+c=0.ax2+bx+c=0.$ Now we can use the discriminant to tell us how many x-intercepts there are on the graph.
Before you to find the values of the x-intercepts, you may want to evaluate the discriminant so you know how many solutions to expect.
## Example 9.46
Find the intercepts of the parabola for the function $f(x)=5x2+x+4.f(x)=5x2+x+4.$
## Try It 9.91
Find the intercepts of the parabola whose function is $f(x)=3x2+4x+4.f(x)=3x2+4x+4.$
## Try It 9.92
Find the intercepts of the parabola whose function is $f(x)=x2−4x−5.f(x)=x2−4x−5.$
## Graph Quadratic Functions Using Properties
Now we have all the pieces we need in order to graph a quadratic function. We just need to put them together. In the next example we will see how to do this.
## Example 9.47
### How to Graph a Quadratic Function Using Properties
Graph f (x) = x2 −6x + 8 by using its properties.
## Try It 9.93
Graph f (x) = x2 + 2x − 8 by using its properties.
## Try It 9.94
Graph f (x) = x2 − 8x + 12 by using its properties.
We list the steps to take in order to graph a quadratic function here.
## How To
### To graph a quadratic function using properties.
1. Step 1. Determine whether the parabola opens upward or downward.
2. Step 2. Find the equation of the axis of symmetry.
3. Step 3. Find the vertex.
4. Step 4. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
5. Step 5. Find the x-intercepts. Find additional points if needed.
6. Step 6. Graph the parabola.
We were able to find the x-intercepts in the last example by factoring. We find the x-intercepts in the next example by factoring, too.
## Example 9.48
Graph f (x) = –x2 + 6x − 9 by using its properties.
## Try It 9.95
Graph f (x) = −3x2 + 12x − 12 by using its properties.
## Try It 9.96
Graph f (x) = 4x2 + 24x + 36 by using its properties.
For the graph of f (x) = −x2 + 6x − 9, the vertex and the x-intercept were the same point. Remember how the discriminant determines the number of solutions of a quadratic equation? The discriminant of the equation 0 = −x2 + 6x − 9 is 0, so there is only one solution. That means there is only one x-intercept, and it is the vertex of the parabola.
How many x-intercepts would you expect to see on the graph of f (x) = x2 + 4x + 5?
## Example 9.49
Graph f (x) = x2 + 4x + 5 by using its properties.
## Try It 9.97
Graph f (x) = x2 − 2x + 3 by using its properties.
## Try It 9.98
Graph f (x) = −3x2 − 6x − 4 by using its properties.
Finding the y-intercept by finding f (0) is easy, isn’t it? Sometimes we need to use the Quadratic Formula to find the x-intercepts.
## Example 9.50
Graph f (x) = 2x2 − 4x − 3 by using its properties.
## Try It 9.99
Graph f (x) = 5x2 + 10x + 3 by using its properties.
## Try It 9.100
Graph f (x) = −3x2 − 6x + 5 by using its properties.
## Solve Maximum and Minimum Applications
Knowing that the vertex of a parabola is the lowest or highest point of the parabola gives us an easy way to determine the minimum or maximum value of a quadratic function. The y-coordinate of the vertex is the minimum value of a parabola that opens upward. It is the maximum value of a parabola that opens downward. See Figure 9.6.
Figure 9.6
## Minimum or Maximum Values of a Quadratic Function
The y-coordinate of the vertex of the graph of a quadratic function is the
• minimum value of the quadratic equation if the parabola opens upward.
• maximum value of the quadratic equation if the parabola opens downward.
## Example 9.51
Find the minimum or maximum value of the quadratic function $f(x)=x2+2x−8.f(x)=x2+2x−8.$
## Try It 9.101
Find the maximum or minimum value of the quadratic function $f(x)=x2−8x+12.f(x)=x2−8x+12.$
## Try It 9.102
Find the maximum or minimum value of the quadratic function $f(x)=−4x2+16x−11.f(x)=−4x2+16x−11.$
We have used the formula
$h(t)=−16t2+v0t+h0h(t)=−16t2+v0t+h0$
to calculate the height in feet, h , of an object shot upwards into the air with initial velocity, v0, after t seconds .
This formula is a quadratic function, so its graph is a parabola. By solving for the coordinates of the vertex (t, h), we can find how long it will take the object to reach its maximum height. Then we can calculate the maximum height.
## Example 9.52
The quadratic equation h(t) = −16t2 + 176t + 4 models the height of a volleyball hit straight upwards with velocity 176 feet per second from a height of 4 feet.
How many seconds will it take the volleyball to reach its maximum height? Find the maximum height of the volleyball.
## Try It 9.103
Solve, rounding answers to the nearest tenth.
The quadratic function h(t) = −16t2 + 128t + 32 is used to find the height of a stone thrown upward from a height of 32 feet at a rate of 128 ft/sec. How long will it take for the stone to reach its maximum height? What is the maximum height?
## Try It 9.104
A path of a toy rocket thrown upward from the ground at a rate of 208 ft/sec is modeled by the quadratic function of h(t) = −16t2 + 208t. When will the rocket reach its maximum height? What will be the maximum height?
## Media
Access these online resources for additional instruction and practice with graphing quadratic functions using properties.
## Section 9.6 Exercises
### Practice Makes Perfect
Recognize the Graph of a Quadratic Function
In the following exercises, graph the functions by plotting points.
229.
$f ( x ) = x 2 + 3 f ( x ) = x 2 + 3$
230.
$f ( x ) = x 2 − 3 f ( x ) = x 2 − 3$
231.
$y = − x 2 + 1 y = − x 2 + 1$
232.
$f ( x ) = − x 2 − 1 f ( x ) = − x 2 − 1$
For each of the following exercises, determine if the parabola opens up or down.
233.
$f(x)=−2x2−6x−7f(x)=−2x2−6x−7$ $f(x)=6x2+2x+3f(x)=6x2+2x+3$
234.
$f(x)=4x2+x−4f(x)=4x2+x−4$ $f(x)=−9x2−24x−16f(x)=−9x2−24x−16$
235.
$f(x)=−3x2+5x−1f(x)=−3x2+5x−1$ $f(x)=2x2−4x+5f(x)=2x2−4x+5$
236.
$f(x)=x2+3x−4f(x)=x2+3x−4$ $f(x)=−4x2−12x−9f(x)=−4x2−12x−9$
Find the Axis of Symmetry and Vertex of a Parabola
In the following functions, find the equation of the axis of symmetry and the vertex of its graph.
237.
$f ( x ) = x 2 + 8 x − 1 f ( x ) = x 2 + 8 x − 1$
238.
$f ( x ) = x 2 + 10 x + 25 f ( x ) = x 2 + 10 x + 25$
239.
$f ( x ) = − x 2 + 2 x + 5 f ( x ) = − x 2 + 2 x + 5$
240.
$f ( x ) = −2 x 2 − 8 x − 3 f ( x ) = −2 x 2 − 8 x − 3$
Find the Intercepts of a Parabola
In the following exercises, find the intercepts of the parabola whose function is given.
241.
$f ( x ) = x 2 + 7 x + 6 f ( x ) = x 2 + 7 x + 6$
242.
$f ( x ) = x 2 + 10 x − 11 f ( x ) = x 2 + 10 x − 11$
243.
$f ( x ) = x 2 + 8 x + 12 f ( x ) = x 2 + 8 x + 12$
244.
$f ( x ) = x 2 + 5 x + 6 f ( x ) = x 2 + 5 x + 6$
245.
$f ( x ) = − x 2 + 8 x − 19 f ( x ) = − x 2 + 8 x − 19$
246.
$f ( x ) = −3 x 2 + x − 1 f ( x ) = −3 x 2 + x − 1$
247.
$f ( x ) = x 2 + 6 x + 13 f ( x ) = x 2 + 6 x + 13$
248.
$f ( x ) = x 2 + 8 x + 12 f ( x ) = x 2 + 8 x + 12$
249.
$f ( x ) = 4 x 2 − 20 x + 25 f ( x ) = 4 x 2 − 20 x + 25$
250.
$f ( x ) = − x 2 − 14 x − 49 f ( x ) = − x 2 − 14 x − 49$
251.
$f ( x ) = − x 2 − 6 x − 9 f ( x ) = − x 2 − 6 x − 9$
252.
$f ( x ) = 4 x 2 + 4 x + 1 f ( x ) = 4 x 2 + 4 x + 1$
Graph Quadratic Functions Using Properties
In the following exercises, graph the function by using its properties.
253.
$f ( x ) = x 2 + 6 x + 5 f ( x ) = x 2 + 6 x + 5$
254.
$f ( x ) = x 2 + 4 x − 12 f ( x ) = x 2 + 4 x − 12$
255.
$f ( x ) = x 2 + 4 x + 3 f ( x ) = x 2 + 4 x + 3$
256.
$f ( x ) = x 2 − 6 x + 8 f ( x ) = x 2 − 6 x + 8$
257.
$f ( x ) = 9 x 2 + 12 x + 4 f ( x ) = 9 x 2 + 12 x + 4$
258.
$f ( x ) = − x 2 + 8 x − 16 f ( x ) = − x 2 + 8 x − 16$
259.
$f ( x ) = − x 2 + 2 x − 7 f ( x ) = − x 2 + 2 x − 7$
260.
$f ( x ) = 5 x 2 + 2 f ( x ) = 5 x 2 + 2$
261.
$f ( x ) = 2 x 2 − 4 x + 1 f ( x ) = 2 x 2 − 4 x + 1$
262.
$f ( x ) = 3 x 2 − 6 x − 1 f ( x ) = 3 x 2 − 6 x − 1$
263.
$f ( x ) = 2 x 2 − 4 x + 2 f ( x ) = 2 x 2 − 4 x + 2$
264.
$f ( x ) = −4 x 2 − 6 x − 2 f ( x ) = −4 x 2 − 6 x − 2$
265.
$f ( x ) = − x 2 − 4 x + 2 f ( x ) = − x 2 − 4 x + 2$
266.
$f ( x ) = x 2 + 6 x + 8 f ( x ) = x 2 + 6 x + 8$
267.
$f ( x ) = 5 x 2 − 10 x + 8 f ( x ) = 5 x 2 − 10 x + 8$
268.
$f ( x ) = −16 x 2 + 24 x − 9 f ( x ) = −16 x 2 + 24 x − 9$
269.
$f ( x ) = 3 x 2 + 18 x + 20 f ( x ) = 3 x 2 + 18 x + 20$
270.
$f ( x ) = −2 x 2 + 8 x − 10 f ( x ) = −2 x 2 + 8 x − 10$
Solve Maximum and Minimum Applications
In the following exercises, find the maximum or minimum value of each function.
271.
$f ( x ) = 2 x 2 + x − 1 f ( x ) = 2 x 2 + x − 1$
272.
$y = −4 x 2 + 12 x − 5 y = −4 x 2 + 12 x − 5$
273.
$y = x 2 − 6 x + 15 y = x 2 − 6 x + 15$
274.
$y = − x 2 + 4 x − 5 y = − x 2 + 4 x − 5$
275.
$y = −9 x 2 + 16 y = −9 x 2 + 16$
276.
$y = 4 x 2 − 49 y = 4 x 2 − 49$
In the following exercises, solve. Round answers to the nearest tenth.
277.
An arrow is shot vertically upward from a platform 45 feet high at a rate of 168 ft/sec. Use the quadratic function h(t) = −16t2 + 168t + 45 find how long it will take the arrow to reach its maximum height, and then find the maximum height.
278.
A stone is thrown vertically upward from a platform that is 20 feet height at a rate of 160 ft/sec. Use the quadratic function h(t) = −16t2 + 160t + 20 to find how long it will take the stone to reach its maximum height, and then find the maximum height.
279.
A ball is thrown vertically upward from the ground with an initial velocity of 109 ft/sec. Use the quadratic function h(t) = −16t2 + 109t + 0 to find how long it will take for the ball to reach its maximum height, and then find the maximum height.
280.
A ball is thrown vertically upward from the ground with an initial velocity of 122 ft/sec. Use the quadratic function h(t) = −16t2 + 122t + 0 to find how long it will take for the ball to reach its maximum height, and then find the maximum height.
281.
A computer store owner estimates that by charging x dollars each for a certain computer, he can sell 40 − x computers each week. The quadratic function R(x) = −x2 +40x is used to find the revenue, R, received when the selling price of a computer is x, Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue.
282.
A retailer who sells backpacks estimates that by selling them for x dollars each, he will be able to sell 100 − x backpacks a month. The quadratic function R(x) = −x2 +100x is used to find the R, received when the selling price of a backpack is x. Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue.
283.
A retailer who sells fashion boots estimates that by selling them for x dollars each, he will be able to sell 70 − x boots a week. Use the quadratic function R(x) = −x2 +70x to find the revenue received when the average selling price of a pair of fashion boots is x. Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue per day.
284.
A cell phone company estimates that by charging x dollars each for a certain cell phone, they can sell 8 − x cell phones per day. Use the quadratic function R(x) = −x2 +8x to find the revenue received per day when the selling price of a cell phone is x. Find the selling price that will give them the maximum revenue per day, and then find the amount of the maximum revenue.
285.
A rancher is going to fence three sides of a corral next to a river. He needs to maximize the corral area using 240 feet of fencing. The quadratic equation $A(x)=x120−x2A(x)=x120−x2$ gives the area of the corral, A, for the length, x, of the corral along the river. Find the length of the corral along the river that will give the maximum area, and then find the maximum area of the corral.
286.
A veterinarian is enclosing a rectangular outdoor running area against his building for the dogs he cares for. He needs to maximize the area using 100 feet of fencing. The quadratic function $A(x)=x50−x2A(x)=x50−x2$ gives the area, A, of the dog run for the length, x, of the building that will border the dog run. Find the length of the building that should border the dog run to give the maximum area, and then find the maximum area of the dog run.
287.
A land owner is planning to build a fenced in rectangular patio behind his garage, using his garage as one of the “walls.” He wants to maximize the area using 80 feet of fencing. The quadratic function A(x) = x(80 − 2x) gives the area of the patio, where x is the width of one side. Find the maximum area of the patio.
288.
A family of three young children just moved into a house with a yard that is not fenced in. The previous owner gave them 300 feet of fencing to use to enclose part of their backyard. Use the quadratic function $A(x)=x150−x2A(x)=x150−x2$ determine the maximum area of the fenced in yard.
### Writing Exercise
289.
How do the graphs of the functions $f(x)=x2f(x)=x2$ and $f(x)=x2−1f(x)=x2−1$ differ? We graphed them at the start of this section. What is the difference between their graphs? How are their graphs the same?
290.
Explain the process of finding the vertex of a parabola.
291.
Explain how to find the intercepts of a parabola.
292.
How can you use the discriminant when you are graphing a quadratic function?
### Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?
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# Algebra - Expansion of Brackets
### Description
My mind-map on the basics of expanding brackets in algebra.
Mind Map by jerick_hartono, updated more than 1 year ago
Created by jerick_hartono over 8 years ago
655
25
## Resource summary
Algebra - Expansion of Brackets
1. We don't usually write x (times) in an equation when multiplying, we instead just put the number in front of the bracket. (Example: 3y(4-2y), not 3y x (4-2y)
1. Distributive Law
1. a(b+c)
1. We expand it by multiplying the number outside the bracket with both numbers inside the bracket.
2. (a+b)(c+d)
1. we do this by multiplying by using SLIP.
3. Perfect Squares
1. (a+b)^2
1. We get our answer by expanding the brackets, and then simplifying it later,
1. In general:
2. Difference of 2 Squares
1. (a+b)(a-b)
1. Method: We will first have to expand the brackets, which will be a^2 +ab -ab -b^2. We will then need to simplify the terms and it will become a^2 - b^2.
2. When we simplify, the two middle terms cancel out.
1. Examp,e:
3. Simplifying
1. We simplify equations by finding the common factors, and then we simplify.
1. This is so that we can see the equations in a much easier way.
1. Example:
2. Expanding
1. Expanding expression with brackets
1. When there is a negative term outside the bracket, the sign of the multiplied terms change.
1. Example:
2. To expand, we have to multiply every term inside a bracket by the term outside the bracket.
1. Example:
2. Expanding brackets and simplifying
1. Sometimes, we need to multiply out brackets then simplify.
1. Multiplying out both brackets or terms and collecting like terms.
2. Examples:
2. We have to multiply the bracket in order to expand it.
3. We expand expressions by using SLIP
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Chapter 5: Algebra
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Knowing how to calculate the circumference of a circle and, in turn, the length of an arc — a portion of the circumference — is important in pre-calculus because you can use that information to analyze the motion of an object moving in a circle.
An arc can come from a central angle, which is one whose vertex is located at the center of the circle. You can measure an arc in two different ways:
• As an angle. The measure of an arc as an angle is the same as the central angle that intercepts it.
• As a length. The length of an arc is directly proportional to the circumference of the circle and is dependent on both the central angle and the radius of the circle.
If you think back to geometry, you may remember that the formula for the circumference of a circle is
with r representing the radius. Also recall that a circle has 360 degrees. So if you need to find the length of an arc, you need to figure out what part of the whole circumference (or what fraction) you're looking at.
You use the following formula to calculate the arc length: The symbol theta (θ) represents the measure of the angle in degrees, and s represents arc length, as shown in the figure:
The variables involved in computing arc length.
If the given angle theta is in radians,
Time for an example. To find the length of an arc with an angle measurement of 40 degrees if the circle has a radius of 10, use the following steps:
1. Assign variable names to the values in the problem.
The angle measurement here is 40 degrees, which is theta. The radius is 10, which is r.
2. Plug the known values into the formula.
This step gives you
3. Simplify to solve the formula.
You first get
which multiplies to
The figure shows what this arc looks like.
The arc length for an angle measurement of 40 degrees.
Now try a different problem. Find the measure of the central angle of a circle in radians with an arc length of
and a radius of 16. This time, you must solve for theta (the formula is s = rθ when dealing with radians):
1. Plug in what you know to the radian formula.
2. Divide both sides by 16.
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# Question Video: Ordering Given Fractions with like Denominators in Descending Order Mathematics • 3rd Grade
Arrange 10/19, 15/19, 7/19 in descending order.
03:04
### Video Transcript
Arrange ten nineteenths, fifteen nineteenths, and seven nineteenths in descending order.
When we descend something, we go downwards. And so, when we’re asked to arrange some numbers in descending order, or in this case fractions, we need to put them in order so that their values go downwards. In other words, we put them in order from largest to smallest.
Now, in this problem, we’re given three fractions to put in order: ten nineteenths, fifteen nineteenths, and seven nineteenths. Now, one thing we can see about these fractions is that they all have the same denominator, the same bottom number. Now, we know that the denominator in a fraction shows us the number of equal parts that a whole amount has been split into. So, if we use this bar to represent one whole and then split the whole into 19 equal parts, each part will be worth one nineteenth. And because each fraction shows a number of nineteenths, we can compare them really quickly.
If we were to shade ten nineteenths of this whole amount, we need to shade in 10 out of 19 sections. So this shaded area then represents ten nineteenths. If we were to show fifteen nineteenths, we’d need to color more parts. In fact, we’d need to color another five more parts to show 15 out of 19 parts shaded. But you know, we didn’t really need to draw the diagram to work out that fifteen nineteenths was greater than ten nineteenths. We just needed to look at the fractions.
Because all the fractions show a number of nineteenths, we just need to look at the numerator, which is the top number, and compare those. The fraction with the largest numerator is fifteen nineteenths. Out of our three possible fractions, this is the greatest number of nineteenths that we could have. The next largest numerator — remember, we’re putting these in order from largest to smallest — is 10, ten nineteenths. And so, the second fraction in our order, descending order, is ten nineteenths.
And then, finally, we have a numerator of seven in the fraction seven nineteenths. Seven nineteenths is the smallest of our three fractions. And so because each of our fractions has the same denominator, to put them in order, all we had to do was to compare the numerators. To put the fractions in descending order, we had to start with the largest numerator and go through them until we ended with the smallest numerator.
In descending order, the fractions are fifteen nineteenths, ten nineteenths, and seven nineteenths.
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Pre-Calc Exam Notes 131
# Pre-Calc Exam Notes 131 - in QI and its re±ection − 3 θ...
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Solving Trigonometric Equations Section 6.1 131 Example6.5 Solve the equation sin θ = tan θ . Solution: Trying the same method as in the previous example, we get sin θ = tan θ sin θ = sin θ cos θ sin θ cos θ = sin θ sin θ cos θ sin θ = 0 sin θ (cos θ 1) = 0 sin θ = 0 or cos θ = 1 θ = 0 , π or θ = 0 θ = 0 , π , plus multiples of 2 π . So since the above angles are multiples of π , and every multiple of 2 π is a multiple of π , we can combine the two answers into one for the general solution: θ = π k for k = 0, ± 1, ± 2, ... Example6.6 Solve the equation cos 3 θ = 1 2 . Solution: The idea here is to solve for 3 θ first, using the most general solution, and then divide that solution by 3. So since cos 1 1 2 = π 3 , there are two possible solutions for 3 θ : 3 θ = π 3 in QI and its
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Unformatted text preview: in QI and its re±ection − 3 θ =− π 3 around the x-axis in QIV. Adding multiples of 2 π to these gives us: 3 θ = ± π 3 + 2 π k for k = 0, ± 1, ± 2, . .. So dividing everything by 3 we get the general solution for θ : θ = ± π 9 + 2 π 3 k for k = 0, ± 1, ± 2, . .. Example 6.7 Solve the equation sin 2 θ = sin θ . Solution: Here we use the double-angle formula for sine: sin 2 θ = sin θ 2 sin θ cos θ = sin θ sin θ (2 cos θ − 1) = ⇒ sin θ = or cos θ = 1 2 ⇒ θ = 0 , π or θ = ± π 3 ⇒ θ = π k and ± π 3 + 2 π k for k = 0, ± 1, ± 2, . .....
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# Using exactly four 3's and any math symbol, try to write an equivilent expression for each whole...
## Question:
Using exactly four 3's and any math symbol, try to write an equivilent expression for each whole number from 1 through 10.
Example: {eq}1 = \frac {\frac {3*3}{3}}{3} {/eq}.
## Forming an Expression for the Whole Number:
We need to find the arithmetic expression by using the arithmetic operators such as addition, subtraction, multiplication and division. And the number that we are using to find the expression should contain four {eq}3's. {/eq}
The expression for {eq}1 {/eq} can be written as,
\begin{align} 1= \frac{\left(3+3\right)-3}{3} \end{align}
The expression for {eq}2: {/eq}
\begin{align} 2= \frac{\left(3\cdot 3\right)-3}{3} \end{align}
The expression for {eq}3: {/eq}
\begin{align} 3= 3\cdot 3-\left(3+3\right) \end{align}
There is impossible to get the expression for {eq}4 {/eq} using the four {eq}3's. {/eq}
Since the expression for {eq}5: {/eq} can be,
\begin{align} 5=3+3-\frac{3}{3} \end{align}
And the expression for getting {eq}6: {/eq}
\begin{align} 6=3+3-\left(3-3\right) \end{align}
Expression for getting {eq}7: {/eq}
\begin{align} 7=3+3+\frac{3}{3} \end{align}
And the expression for {eq}8: {/eq}
\begin{align} 8=3\cdot 3-\frac{3}{3} \end{align}
The expression for getting {eq}9: {/eq}
\begin{align} 9=3\cdot 3-\left(3-3\right) \end{align}
And finally the expression for {eq}10: {/eq}
\begin{align} 10=3\cdot 3+\left(\frac{3}{3}\right). \end{align}
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# Visualizing The Area of a Triangle Formula
Build an understanding of the area relationship between triangles and rectangles through this set of Area of a Triangle Number Talk Prompts.
## In This Set of Visual Number Talk Prompts…
Students will emerge the formula for the area of a triangle by taking half of the area of a rectangle.
## Intentionality…
This visual math talk and purposeful practice serve to develop a deeper understanding of the following big ideas.
• The area of a rectangle can be determined by multiplying the length of its base by the length of its height;
• The area of a rectangle is used to determine the area formulas for other polygons;
• Using “base” and “height” rather than “length” and “width” can support student understanding when determining the area formulas for triangles and all quadrilaterals;
• A parallelogram is any quadrilateral with two sets of parallel sides;
• A rectangle is a parallelogram;
• The relationships between rectangles, parallelograms, and triangles can be used to determine area formulas;
• Any triangle can be doubled to create a parallelogram;
• Congruent shapes have the same size and shape, but not necessarily the same orientation.
## Visual Number Talk Prompt
Begin playing the visual number talk prompt video and then be ready to ask students:
Calculate the area of each triangle in square metres.
Craft a convincing argument without the use of a calculator.
As you may immediately notice, we have given students two congruent triangles and have conveniently arranged them such that together, they create a rectangle. This is intentional to ensure that all students explicitly see that every rectangle can be partitioned diagonally into two congruent triangles. From this visual math talk prompt, we hope to reiterate that partitioning any rectangle diagonally into two congruent triangles will result in each triangle having half of the area of the rectangle.
Therefore, it may feel intuitive to some students to first find the area of the rectangle and simply half that quantity to determine the area of each of the resulting triangles.
While some students may perform the operation of taking half of the area of the rectangle (or dividing the area by 2), it is likely that many will not naturally write their thinking down symbolically as you see on the right of the screen. The goal here is not necessarily to ensure that all students “copy down” the formula, but rather so they can make a connection between the intuitive actions they had completed to find the area of the triangles to the symbolic representation of that same thinking.
It is important to explicitly highlight that the two triangles that were created by partitioning the rectangle diagonally result in congruent triangles as this may not be obvious to all students immediately.
It could also be helpful to show students that despite taking half of the area of the rectangle (or dividing the area by 2) revealing the area of each congruent triangle, the same operation can be done to determine the area of other congruent shapes that result when partitioning the original rectangle in half. For example, horizontally partitioning the rectangle into halves allows us to complete the same mathematical operation to reveal the same area.
This idea helps to highlight the fact that representing mathematical thinking of area symbolically can strip away information about how the units of area are arranged (i.e.: as a rectangle, a triangle, or some other shape).
## Want to Explore These Concepts & Skills Further?
Two (2) additional number talk prompts are available in Day 2 of the Covering Ground problem based math unit that you can dive into now.
Why not start from the beginning of this contextual 5-day unit of real world lessons from the Make Math Moments Problem Based Units page.
Did you use this in your classroom or at home? How’d it go? Post in the comments!
Math IS Visual. Let’s teach it that way.
#### 1 comment
• Megan says:
There is no sound on the video…
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6.4 Rhombuses, Rectangles, and Squares
Presentation on theme: "6.4 Rhombuses, Rectangles, and Squares"— Presentation transcript:
6.4 Rhombuses, Rectangles, and Squares
Review Find the value of the variables. p + 50° + (2p – 14)° = 180°
52° (2p-14)° 50° 68° p + 50° + (2p – 14)° = 180° p + 2p + 50° - 14° = 180° 3p ° = 180° 3p = 144 ° p = 48 ° 52° + 68° + h = 180° 120° + h = 180 ° h = 60°
Special Parallelograms
Rhombus A rhombus is a parallelogram with four congruent sides.
Special Parallelograms
Rectangle A rectangle is a parallelogram with four right angles.
Special Parallelogram
Square A square is a parallelogram with four congruent sides and four right angles.
Corollaries Rhombus corollary Rectangle corollary Square corollary
A quadrilateral is a rhombus if and only if it has four congruent sides. Rectangle corollary A quadrilateral is a rectangle if and only if it has four right angles. Square corollary A quadrilateral is a square if and only if it is a rhombus and a rectangle.
Example PQRS is a rhombus. What is the value of b? 2b + 3 = 5b – 6
Review In rectangle ABCD, if AB = 7f – 3 and CD = 4f + 9, then f = ___
1 2 3 4 5 7f – 3 = 4f + 9 3f – 3 = 9 3f = 12 f = 4
Example PQRS is a rhombus. What is the value of b? 3b + 12 = 5b – 6
Theorems for rhombus A parallelogram is a rhombus if and only if its diagonals are perpendicular. A parallelogram is a rhombus if and only if each diagonal bisects a pair of opposite angles. L
Theorem of rectangle A parallelogram is a rectangle if and only if its diagonals are congruent. A B D C
Match the properties of a quadrilateral
The diagonals are congruent Both pairs of opposite sides are congruent Both pairs of opposite sides are parallel All angles are congruent All sides are congruent Diagonals bisect the angles Parallelogram Rectangle Rhombus Square B,D A,B,C,D A,B,C,D B,D C,D C
Decide if the statement is sometimes, always, or never true.
A rhombus is equilateral. 2. The diagonals of a rectangle are _|_. 3. The opposite angles of a rhombus are supplementary. 4. A square is a rectangle. 5. The diagonals of a rectangle bisect each other. 6. The consecutive angles of a square are supplementary. Always Sometimes Sometimes Always Always Always Quadrilateral ABCD is Rhombus. 7. If m <BAE = 32o, find m<ECD. 8. If m<EDC = 43o, find m<CBA. 9. If m<EAB = 57o, find m<ADC. 10. If m<BEC = (3x -15)o, solve for x. 11. If m<ADE = ((5x – 8)o and m<CBE = (3x +24)o, solve for x 12. If m<BAD = (4x + 14)o and m<ABC = (2x + 10)o, solve for x. A B E D C 32o 86o 66o 35o 16 26
Coordinate Proofs Using the Properties of Rhombuses, Rectangles and Squares
Using the distance formula and slope, how can we determine the specific shape of a parallelogram? Rhombus – Rectangle – Square - 1. Show all sides are equal distance 2. Show all diagonals are perpendicular. 1. Show diagonals are equal distance 2. Show opposite sides are perpendicular Show one of the above four ways. Based on the following Coordinate values, determine if each parallelogram is a rhombus, a rectangle, or square. P (-2, 3) P(-4, 0) Q(-2, -4) Q(3, 7) R(2, -4) R(6, 4) S(2, 3) S(-1, -3) RECTANGLE RECTANGLE
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Construction | kullabs.com
## Note on Construction
• Note
• Things to remember
• Videos
• Exercise
The following steps will be helpful before drawing the actual figure.
1. Draw a rough sketch of a figure.
2. Mark the given measurement in it.
3. Analyze the figure and plan the steps.
#### 1. Construction of a parallelogram and a triangle having equal area.
Construction of a parallelogram whose area is equal to the area of given triangle when
(a) One angle (b) one side of the parallelogram are given:
(a) Construct a triangle ABC in which AB = 5cm, BC = 6cm and AC = 7cm and construct a parallelogram whose area is equal to the area of given triangle having one angle 60°.
Steps :
1. DrawΔABC with AB = 5cm, BC = 6cm and AC = 7cm.
2. Draw XY parallel to BC through the point A.
3. Take P as mid-point of BC.
4. Draw an angle of60° at P.
5. Cut PC = QR and join the point R and C.
6. Parallelogram PQRC andΔABC are equal in area.
∴ PQRC is the required parallelogram.
(b) Construct a triangle ABC in which AB = 4cm, BC = 5cm and ∠B = 60° and then construct a parallelogram having a side 5.2 cm and equal area to the triangle.
Steps :
1. DrawΔABC with AB = 4cm, BC = 5cm and∠B = 60°.
2. Draw XY parallel to BC through the point A.
3. Take P as mid-point of BC.
4. From P, cut PQ = 5.2cm on XY.
5. CutPC = PQ and join the point R and C.
6. Parallelogram PQRC andΔABC have equal area.
∴ PQRC is the required parallelogram.
#### 2. Construction of rectangle equals in the area to given triangle.
Construct a triangle ABC in which AB = 6.3 cm, BC = 4.5cm and AC = 3.2ccm then construct a rectangle equal area to the triangle.
Steps:
1. DrawΔABC with AB = 6.3 cm BC = 4.5 cm and AC = 3.2 cm.
2. Through A, draw XY//BC.
3. Draw the perpendicular bisector PQ of BC.
4. Draw BP = RQ and join the points R and B.
5. Rectangle BPQR is the required rectangle equal toΔABC.
∴ BPQR is the required rectangle.
#### 3. Construction of two triangles of equal area on the same base and between the same parallels.
Construct a triangle ABC in which AB = 6.3 cm, BC = 7.8 cm and AC = 7.2 cm and construct another triangle PBC equal area toΔABC.
Steps:
1. Draw ΔABC withAB = 6.3 cm, BC = 7.8 cm and AC = 7.2 cm .
2. Through A, draw XY//BC.
3. Take any point P in XY and join P to B and C.
4. ABC and PBC are the triangles of equal area.
∴ PBC is the required triangle.
#### 4. Construction of two parallelograms of equal area on the same base and between the same parallels.
Construct a parallelogram ABCD in which AB = 5.5 cm, BC = 4.8 cm and∠ABC = 75° and construct another parallelogram equal area to the parallelogram ABCD.
Steps:
1. Draw a parallelogram ABCD having AB = 5.5 cm, BC = 4.8 cm and∠ABC =75°.
2. Take two points R and Q in XY such that BC = RQ.
3. Join R to B and Q to C.
4. BCQR is a parallelogram equal in area to parallelogram ABCD.
∴ BCQR is the required parallelogram.
#### 5. Construction of a triangle equal in area to the given quadrilateral.
Construct a quadrilateral ABCDin which AB = 2.8 cm BC = 3.6 cm, AC = 3 cm, CD = 1.7 cm and AD = 2.3 cm and construct a triangle equal area to the quadrilateral ABCD.
Steps:
1. Draw aquadrilateral ABCDin which BC = 3.6 cm, AB = 2.8 cm, AC = 3 cm,AD = 2.3 cm and CD = 1.7 cm.
2. From D, draw DE parallel to AC.
3. Produce BC to E.
4. Join A to E.
5. ABE is a triangle equal area to the quadrilateral ABCD.
∴ ABE is a required triangle.
#### 6. Construction of a quadrilateral equal in area to the given triangle
Construct a triangle ABC in which a = 7.8cm b =7.2 cm and c = 6.3 cm and construct a quadrilateral having an equal area to the triangle ABC.
Steps:
1. DrawΔABC in which BC = a = 7.8 cm, BA = c = 6.3 cm and AC = b = 7.2 cm.
2. Take any point D on BC.
3. Draw DA//CP.
4. Take any point E on CP.
5. ABDE is a quadrilateral equal area toΔABC.
∴ ABDE is the required quadrilateral.
Helpful steps to before drawing actual figure:
1. Draw a rough sketch of the figure.
2. Mark the given measurement in it.
3. Analyze the figure and plan the step.
.
### Very Short Questions
Construct:
Quadrilateral PQRS and $$\triangle$$PSZ
Results:
Area of quadrilteral PQRS = Areaof $$\triangle$$PSZ where;
PQ = 4.2 cm, QR = 6.5 cm, RS = 8 cm, SP = 5.3 cm, SQ = 7.4 cm.
To construct:
Quadrilateral PQRS and $$\\triangle$$PST
Result:
Area of quadrilateral PQRS = Area of $$\triangle$$PST
where, PQ = QR = 5.1 cm, PS = RS = 6.2 cm and QS= 5.6 cm.
To construct:
parallelogram ABCD and $$\triangle$$AEF
Result:
Area of pallelogram ABCD = Area of $$\triangle$$AEF
where, AB = 6 cm, BC = 4 cm, $$\angle$$BAD = 60° and one side of $$\triangle$$AEF = 7.5 cm.
To construct:
quadrilateral ABCD and $$\triangle$$ABE
Result:
Area of quadrilateral ABCD = Area of $$\triangle$$ABE
where,AB = BC = 5.6 cm, CD = AD = 4.9 cm and $$\angle$$BAD = 60°.
Construct:
quadrilateral ABCD and $$\triangle$$ADE
Results:
Area of quadrilateral ABCD = Area of $$\triangle$$ADE
where, Ab = 4.2 cm, BC = 4.8 cm, CD = 5.4 cm, DA = 5.8 cm.
To construct:
Result:
area of quadrilateral ABCD = Area of rectangle ASTU
where, AC = 6.6 cm, BD = 8 cm, AB = 5 cm.
To construct:
quadrilateral PQRS = $$\triangle$$PST
Results:
Area of quadrilateral PQRS = Area of (\triangle\)PST
where,PQ = QR = 5.9 cm, RS = Ps = 6.1 cm and $$\angle$$QPS = 75°.
To construct:
quadrilateral ABCD = $$\triangle$$ADE
Results:
Area of quadrilateral ABCD = Area of $$\triangle$$ADE
where,AB = 8 cm, BC = 3.5 cm, CD = 7 cm, DA = 6 cm and $$\angle$$BAD = 60°.
To construct:
$$\angle$$ABC and parallelogram BEFD
Results:
Area of $$\triangle$$ABC = Area of parallelogram BEFD
where a = 7.8 cm, b = 7.2 cm, c = 6.3 cm and an angle of parallelogram $$\angle$$DBC = 75°.
To construct:
$$\triangle$$PQR and rectangle ARCB
Results:
Area of $$\triangle$$PQR = Area of rectangle ARCB
in whichPQ = 6 cm, QR = 7 cm and RP = 4 cm.
To construct:
$$\triangle$$ABC and $$\triangle$$DBC
Results:
Area of $$\triangle$$ABC = Area of $$\triangle$$DBC
where, AB = 4.5 cm, BC = 6.5 cm, $$\angle$$C = 60°, DB = 8 cm.
Construct:
$$\triangle$$ABC and parallelogram PCQR
Resuts:
Area of $$\triangle$$ABC = Area of parllelogram PCQR
where, a = 7.8 cm, b = 7.2 cm, c = 6.3 cm and $$\angle$$RPC = 75°.
To construct:
$$\triangle$$PQR and parallelogram TQSU
Results:
Area of $$\triangle$$PQR = Area pf parallelogram TSQU
in whichPQ = 7.5 cm, QR = 6.8 cm and RP = 6 cm and TQ = 6.4 cm.
Construct:
$$\triangle$$ABC and parallelogram PBQR
Results:
Area of $$\triangle$$ABC = Area of parallelogram PBQR
where,a = 5 cm, b = 4.8 cm and c = 6.8 cm and $$\angle$$ RPB = 45°.
Result:
Area of parallelogram PQRS = Area of $$\triangle$$PSA
where, QS = 8 cm, PR = 6 cm and PQ = 5 cm.
Results:
Area of $$\triangle$$ABC = Area of rectngle BDEF
where AB = 4 cm, BC = 6.8 cm and CA = 6.5 cm.
Results:
Area of quadrilateral ABCD = Area of $$\triangle$$ ADE
where, AB = 5.2 cm, BC = 5 cm, CD = 4.2 cm, AD = 4 cm and $$\angle$$ABC = 60°.
Results:
Area of quadrilateral PQRS = Area of $$\triangle$$QRT
Results:
Area of rectangle ABCD = Area of $$\triangle$$ABE
0%
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##### Ishwor
Construct a quadrilateral PQRS in which QR= 5 cm, RS = 5cm, PS=5.7, PQ = 6.2 cm and PR = 5.6, then constuct a triangle RQT equal in area to the quadrilateral PQRS.
frgt
##### sanchita
constuction of rectangle frm triangle
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# Class 8 RD Sharma Solutions – Chapter 4 Cubes and Cube Roots – Exercise 4.2
Last Updated : 24 Nov, 2020
### Question 1. Find the cubes of:(i) -11(ii) -12(iii) -21
Solution:
i) Cube of -11 = (-11)3
= -11 × -11 × -11 = -1331
ii) Cube of -12 = (-12)3
= -12 × -12 ×-12 = -1728
iii) Cube of -21 = (-21)3
= -21 × -21 ×-21 = -9261
### Question 2. Which of the following numbers are cubes of negative integers.(i) -64(ii) -1056(iii) -2197(iv) -2744(v) -42875
Solution:
In order to find out there the given negative number is a perfect cube or not, we need to check if its corresponding positive number is a perfect cube.
i) -64
Let’s first check whether 64 is a perfect cube or not.
Prime factorization of 64
64 = 2 × 2 × 2 × 2 × 2 × 2
Also, 64 = (2 × 2 × 2) × (2 × 2 × 2)
Since, 64 can be completely grouped in triplets of the equal factors,
So, 64 is a perfect cube of 4.
Hence, -64 is a perfect cube of negative number i.e -4.
ii) -1056
Let’s first check whether 1056 is a perfect cube or not.
Prime factorization of 1056
1056 = 2 × 2 × 2 × 2 × 2 × 3 × 11
Also, 1056 = (2 × 2 × 2) × 2 × 2 × 3 × 11
Since, 1056 can’t be completely grouped in triplets of the equal factors,
So, 1056 is not a perfect cube.
Hence, -1056 is a not perfect cube of a negative number.
iii) -2197
Let’s first check whether 2197 is a perfect cube or not.
Prime factorization of 2197
2197 = 13 × 13 × 13
Also, 2197 = (13 × 13 × 13)
Since, 2197 can be completely grouped in triplets of the equal factors,
So, 2197 is a perfect cube of 13.
Hence, -2197 is a perfect cube of negative number i.e -13.
iv) -2744
Let’s first check whether 2744 is a perfect cube or not.
Prime factorization of 2744
2744 = 2 × 2 × 2 × 7 × 7 × 7
Also, 2744 = (2 × 2 × 2) × (7 × 7 × 7)
Since, 2744 can be completely grouped in triplets of the equal factors,
So, 2744 is a perfect cube of 14.
Hence, -2744 is a perfect cube of negative number i.e -14.
v) -42875
Let’s first check whether 42875 is a perfect cube or not.
Prime factorization of 42875
42875 = 5 × 5 × 5 × 7 × 7 × 7
Also, 42875 = (5 × 5 × 5) × (7 × 7 × 7)
Since, 42875 can be completely grouped in triplets of the equal factors,
So, 42875 is a perfect cube of 35.
Hence, -42875 is a perfect cube of negative number i.e -35.
### Question 3. Show that the following integers are cubes of negative integers. Also, find the integer whose cube is the given integer :(i) -5832 (ii) -2744000
Solution:
i) -5832
Let’s first check whether 5832 is a perfect cube or not.
Prime factorization of 5832
5832 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
Also, 5832 = (2 × 2 × 2) × (3 × 3 × 3) × (3 × 3 × 3)
Since, 5832 can be completely grouped in triplets of the equal factors,
So, 5832 is a perfect cube of 18.
Hence, -5832 is a perfect cube of negative number i.e -18.
ii) 2744000
Let’s first check whether 2744000 is a perfect cube or not.
Prime factorization of 2744000
2744000 = 2 × 2 × 2 × 7 × 7 × 7 × 2 × 2 × 2 × 5 × 5 × 5
Also, 2744000 = (2 × 2 × 2) × (7 × 7 × 7) × (2 × 2 × 2) × (5 × 5 × 5)
Since, 2744000 can be completely grouped in triplets of the equal factors,
So, 2744000 is a perfect cube of 140.
Hence, -2744000 is a perfect cube of negative number i.e -140.
### Question 4. Find the cube of :(i) 7/9(ii) -8/11(iii) 12/7(iv) -13/8(v) 12/5(vi) 13/4(vii) 0.3(vii) 1/5(ix) 0.08(x) 2.1
Solution:
i) Cube of 7/9 will be 7/9 × 7/9 × 7/9
= 343/729
Hence, the cube of 7/9 is 343/729
ii) Cube of -8/11 will be -8/11 × -8/11 × -8/11
= -512/1331
Hence, the cube of -8/11 is -512/1331
iii) Cube of 12/7 will be 12/7 × 12/7 × 12/7
= 1728/343
Hence, the cube of 12/7 is 1728/343
iv) Cube of -13/8 will be -13/8 × -13/8 × -13/8
= -2197/512
Hence, the cube of -13/8 is -2197/512
v) Cube of 12/5 will be 12/5 × 12/5 × 12/5
= 1728/125
Hence, the cube of 12/5 is 1728/125
vi) Cube of 13/4 will be 13/4 × 13/4 × 13/4
= 2197/64
Hence, the cube of 13/4 is 2197/64
vii) 0.3 = 3/10
So, Cube of 3/10 will be 3/10 × 3/10 × 3/10
= 27/1000 = 0.027
Hence, the cube of 0.3 is 0.027
viii) 1.5 = 15/10
So, Cube of 15/10 will be 15/10 × 15/10 × 15/10
= 3375/1000 = 3.375
Hence, the cube of 1.5 is 3.375
ix) 0.08 = 8/100
So, Cube of 8/100 will be 8/100 × 8/100 × 8/100
= 512/1000000 = 0.000512
Hence, the cube of 0.08 is 0.000512
x) 2.1 = 21/10
So, Cube of 21/10 will be 21/10 × 21/10 × 21/10
= 9261/1000 = 9.261
Hence, the cube of 2.1 is 9.261
### Question 5. Which of the following numbers are cubes of rational numbers :(i) 27/64(ii) 125/128(iii) 0.001331(iv) 0.04
Solution:
i) 27/64
Factorization of 27/64 will be
27/64 = (3 × 3 × 3)/(4 × 4 × 4) = (3/4)3
It means the cube of 3/4 is 27/64
Hence, we can say that 27/64 is a cube of rational number i.e 3/4
ii) 125/128
Factorization of 125/128 will be
125/128 = (5 × 5 × 5)/(2 × 2 × 2) × (2 × 2 × 2) × 2 = 53/23 × 23 × 2
It means 125/128 is not perfect cube
Hence, we can say that 125/158 is not a cube of rational number
iii) 0.001331
0.001331 = 1331/1000000
Factorization of 1331/1000000 will be
1331/1000000 = (11 × 11 × 11)/(10 × 10 × 10) × (10 × 10 × 10) =113/103 × 103
It means the cube of 11/100 is 1331/1000000
Hence, we can say that 0.001331 is a cube of rational number i.e 0.11
iv) 0.04
0.04 = 4/100
Factorization of 4/100 = 2 × 2/ 10 × 10
It means 4/100 is not a perfect cube
Hence, we can say that 0.04 is not a cube of a rational number
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# Mathematics | Graph Isomorphisms and Connectivity
Isomorphism : Consider the following two graphs – Are the graphs and the same? If your answer is no, then you need to rethink it. The graphical arrangement of the vertices and edges makes them look different, but they are the same graph. Also notice that the graph is a cycle, specifically . To know about cycle graphs read Graph Theory Basics. Formally, “The simple graphs and are isomorphic if there is a bijective function from to with the property that and are adjacent in if and only if and are adjacent in .” Example : Show that the graphs and mentioned above are isomorphic. Solution : Let be a bijective function from to . Let the correspondence between the graphs be- [Tex]v2^{\prime} = f(v5) [/Tex][Tex]v4^{\prime} = f(v4) [/Tex]The above correspondence preserves adjacency as- is adjacent to and in , and is adjacent to and in Similarly, it can be shown that the adjacency is preserved for all vertices. Hence, and are isomorphic. Proving that the above graphs are isomorphic was easy since the graphs were small, but it is often difficult to determine whether two simple graphs are isomorphic. This is because there are possible bijective functions between the vertex sets of two simple graphs with vertices. Testing the correspondence for each of the functions is impractical for large values of n. Although sometimes it is not that hard to tell if two graphs are not isomorphic. In order, to prove that the given graphs are not isomorphic, we could find out some property that is characteristic of one graph and not the other. If they were isomorphic then the property would be preserved, but since it is not, the graphs are not isomorphic. Such a property that is preserved by isomorphism is called graph-invariant. Some graph-invariants include- the number of vertices, the number of edges, degrees of the vertices, and length of cycle, etc.
1. Equal number of vertices.
2. Equal number of edges.
3. Same degree sequence
4. Same number of circuit of particular length
In most graphs checking first three conditions is enough. Important Note : The complementary of a graph has the same vertices and has edges between any two vertices if and only if there was no edge between them in the original graph. Consequently, a graph is said to be self-complementary if the graph and its complement are isomorphic.
Connectivity :
Most problems that can be solved by graphs, deal with finding optimal paths, distances, or other similar information. Almost all of these problems involve finding paths between graph nodes. Path – A path of length from to is a sequence of edges such that is associated with , and so on, with associated with , where and . Note : A path is called a circuit if it begins and ends at the same vertex. It is also called a cycle. Connectivity of a graph is an important aspect since it measures the resilience of the graph. “An undirected graph is said to be connected if there is a path between every pair of distinct vertices of the graph.” Connected Component – A connected component of a graph is a connected subgraph of that is not a proper subgraph of another connected subgraph of . For example, in the following diagram, graph is connected and graph is disconnected. Since is connected there is only one connected component. But in the case of there are three connected components. In case the graph is directed, the notions of connectedness have to be changed a bit. This is because of the directions that the edges have. Formally, “A directed graph is said to be strongly connected if there is a path from to and to where and are vertices in the graph. The graph is weakly connected if the underlying undirected graph is connected.” Strongly Connected Component – Analogous to connected components in undirected graphs, a strongly connected component is a subgraph of a directed graph that is not contained within another strongly connected component.
Articulation points –
The removal of a vertex and all the edges incident with it may result in a subgraph that has more connected components than in the original graphs. Such vertices are called articulation points or cut vertices. Analogous to cut vertices are cut edge the removal of which results in a subgraph with more connected components. A cut-edge is also called a bridge. Cut set – In a connected graph , a cut-set is a set of edges which when removed from leaves disconnected, provided there is no proper subset of these edges that disconnects .
Paths and Isomorphisms –
Sometimes even though two graphs are not isomorphic, their graph invariants- number of vertices, number of edges, and degrees of vertices all match. In this case paths and circuits can help differentiate between the graphs.
• Example – Are the two graphs shown below isomorphic?
• Solution – Both the graphs have 6 vertices, 9 edges and the degree sequence is the same. However the second graph has a circuit of length 3 and the minimum length of any circuit in the first graph is 4. Hence the given graphs are not isomorphic.
GATE CS Corner Questions
Practicing the following questions will help you test your knowledge. All questions have been asked in GATE in previous years or GATE Mock Tests. It is highly recommended that you practice them. 1. GATE CS 2013, Question 24 2. GATE CS 2012, Question 26 3. GATE CS 2012, Question 38 4. GATE CS 2014 Set-1, Question 13 5. GATE CS 2014 Set-2, Question 61 6. GATE CS 2015 Set-2, Question 38 7. GATE CS 2015 Set-2, Question 60
References-
Graph Isomorphism – Wikipedia Graph Connectivity – Wikipedia Discrete Mathematics and its Applications, by Kenneth H Rosen If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Level Up Your GATE Prep!
Embark on a transformative journey towards GATE success by choosing Data Science & AI as your second paper choice with our specialized course. If you find yourself lost in the vast landscape of the GATE syllabus, our program is the compass you need.
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# Manhattan GMAT Challenge Problem of the Week – 8 October 2012
by on October 8th, 2012
Here is a new Challenge Problem! If you want to win prizes, try entering our Challenge Problem Showdown. The more people that enter our challenge, the better the prizes!
## Question
If x, y, and z lie between 0 and 1 on the number line, with no two variables equal, then the product of all three variables divided by the sum of all the distinct products of exactly two of the three variables is between
A. 0 and 1/3
B. 1/3 and 2/3
C. 2/3 and 1
D. 1 and 5/3
E. 5/3 and 7/3
## Answer
First, decode the language describing the expression you care about. “The product of all three variables” is xyz. Meanwhile, “the sum of all the distinct products of exactly two of the three variables” is xy + yz + xz. So the expression is xyz/(xy + yz + xz).
At this point, there are two paths forward: doing algebra and picking smart numbers. The latter may turn out to be faster in this case, because the expression is tough to manipulate on its own (we’ll show how in a minute). Say that x = ¼, y = ½, and z = ¾ (all three are supposed to be different). Then the product of all three is 3/32, while xy + yz + xz = (¼)(½) + (½)(¾) + (¼)(¾) = 1/8 + 3/8 + 3/16 = 2/16 + 6/16 + 3/16 = 11/16. So the value you ultimately want equals 3/32 divided by 11/16, or 3/32 times 16/11, which equals 3/22. 3/22 is definitely between 0 and 1/3. Now you know that the question must be well-formed—in other words, there can’t be two right answers. For any legal values of x, y, and z chosen according to the criteria (between 0 and 1, and no two values alike), you must get the same answer to the question. So you only need to check one set of values. The answer is (A).
If you’re interested, here’s one way to manipulate the expression to prove that it must lie between 0 and 1/3. Say that xyz/(xy + yz + xz) = S, to give the expression a variable name. You can’t split denominators that are sums, but you can split numerators. So take the reciprocal of both sides:
(xy + yz + xz)/xyz = 1/S
Now you can split the numerator, making three different fractions. Canceling common terms in each, you get this:
1/z + 1/x + 1/y = 1/S
Since each variable x, y, and z is less than 1 (but still positive), the reciprocal of each of those variables is greater than 1. So the sum of those three reciprocals is greater than 1+1+1=3. Since 1/S is greater than 3, S must be less than 1/3. Meanwhile, S is evidently positive (all the variables are positive and nothing’s being subtracted), so the proper boundaries are 0 < S < 1/3. The two algebra tricks to learn here are (1) naming a whole expression as a new variable, so that you can transform both sides of the equation, and (2) taking the reciprocal of both sides of an equation.
The correct answer is A.
Special Announcement: If you want to win prizes for answering our Challenge Problems, try entering our Challenge Problem Showdown. Each week, we draw a winner from all the correct answers. The winner receives a number of our our Strategy Guides. The more people enter, the better the prize. Provided the winner gives consent, we will post his or her name on our Facebook page.
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# Solving multi step equations
There's a tool out there that can help make Solving multi step equations easier and faster Math can be difficult for some students, but with the right tools, it can be conquered.
## Solve multi step equations
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# SELINA Solutions for Class 9 Maths Chapter 4 - Expansion
Access TopperLearning’s free Selina Solutions for ICSE Class 9 Mathematics Chapter 4 Expansion to learn about finding the square of a given algebraic expression using expansion. Also, go through the solutions to understand expansion of cubes. Highly-experienced Maths experts have created the answers for textbook exercises in a step-wise format to help it easy for students to understand and learn.
With thorough revision using Selina textbook solutions, you will learn to evaluate identities using the right methods. If you need explanation on the basics of expansion, check our ICSE Class 9 Maths concept videos and other chapter resources.
Page / Exercise
## Chapter 4 - Expansion Exercise Ex. 4(A)
Question 1
Find the square of:
(i) 2a + b
(ii) 3a + 7b
(iii) 3a - 4b
(iv)
Solution 1
Question 2
Use identities to evaluate:
(i) (101)2
(ii) (502)2
(iii) (97)2
(iv) (998)2
Solution 2
Question 3
Evalute:
(i)
(ii)
Solution 3
(i)
(ii)
Question 4
Evaluate:
(i)
(ii) (4a +3b)2 - (4a - 3b)2 + 48ab.
Solution 4
(i)Consider the given expression:
(ii)Consider the given expression:
Question 5
If a + b = 7 and ab = 10; find a - b.
Solution 5
Question 6
If a -b = 7 and ab = 18; find a + b.
Solution 6
Question 7
If x + y = and xy = ; find:
(i) x - y
(ii) x2- y2
Solution 7
(i)
(ii)
Question 8
If a - b = 0.9 and ab = 0.36; find:
(i) a + b
(ii) a2 - b2.
Solution 8
(i)
(ii)
Question 9
If a - b = 4 and a + b = 6; find
(i) a2 + b2
(ii) ab
Solution 9
(i)
(ii)
Question 10
If a + = 6 and a ≠ 0 find :
(i)
(ii)
Solution 10
(i)
(ii)
Question 11
If a - = 8 and a ≠0, find :
(i)
(ii)
Solution 11
(i)
(ii)
Question 12
If a2 - 3a + 1 = 0, and a≠ 0; find:
(i)
(ii)
Solution 12
(i)
(ii)
Question 13
If a2 - 5a - 1 = 0 and a ≠ 0; find:
(i)
(ii)
(iii)
Solution 13
(i)
(ii)
(iii)
Question 14
If 3x + 4y = 16 and xy = 4; find the value of 9x2 + 16y2.
Solution 14
Question 15
The number x is 2 more than the number y. If the sum of the squares of x and y is 34, then find the product of x and y.
Solution 15
Given x is 2 more than y, so x = y + 2
Sum of squares of x and y is 34, so x2 + y2 = 34.
Replace x = y + 2 in the above equation and solve for y.
We get (y + 2)2 + y2 = 34
2y2 + 4y - 30 = 0
y2 + 2y - 15 = 0
(y + 5)(y - 3) = 0
So y = -5 or 3
For y = -5, x =-3
For y = 3, x = 5
Product of x and y is 15 in both cases.
Question 16
The difference between two positive numbers is 5 and the sum of their squares is 73. Find the product of these numbers.
Solution 16
Let the two positive numbers be a and b.
Given difference between them is 5 and sum of squares is 73.
So a - b = 5, a2 + b2 = 73
Squaring on both sides gives
(a - b)2 = 52
a2 + b2 - 2ab = 25
But a2 + b2 = 73
So 2ab = 73 - 25 = 48
ab = 24
So, the product of numbers is 24.
## Chapter 4 - Expansion Exercise Ex. 4(B)
Question 1
Find the cube of :
(i) 3a- 2b
(ii) 5a + 3b
(iii)
(iv)
Solution 1
(i)
(ii)
(iii)
(iv)
Question 2
If a2 + = 47 and ≠ 0 find:
(i)
(ii)
Solution 2
(i)
(ii)
Question 3
If a2 + = 18; a ≠ 0 find:
(i)
(ii)
Solution 3
(i)
(ii)
Question 4
If a + = p and a ≠ 0 ; then show that:
Solution 4
Question 5
If a + 2b = 5; then show that:
a3 + 8b3 + 30ab = 125.
Solution 5
Question 6
If and a ≠ 0 ; then show: a3 +
Solution 6
Question 7
If a + 2b + c = 0; then show that:
a3 + 8b3 + c3 = 6abc.
Solution 7
Question 8
Use property to evaluate:
(i) 133 + (-8)3 + (-5)3
(ii)73 + 33 + (-10)3
(iii) 93 - 53 - 43
(iv) 383 + (-26)3 + (-12)3
Solution 8
Property is if a + b + c = 0 then a3 + b3 + c3 = 3abc
(i) a = 13, b = -8 and c = -5
133 + (-8)3 + (-5)3 = 3(13)(-8)(-5) = 1560
(ii) a = 7, b = 3, c = -10
73 + 33 + (-10)3 = 3(7)(3)(-10) = -630
(iii)a = 9, b = -5, c = -4
93 - 53 - 43 = 93 + (-5)3 + (-4)3 = 3(9)(-5)(-4) = 540
(iv) a = 38, b = -26, c = -12
383 + (-26)3 + (-12)3 = 3(38)(-26)(-12) = 35568
Question 9
Solution 9
(i)
(ii)
Question 10
If and a - = 4; find:
(i)
(ii)
(iii)
Solution 10
(i)
(ii)
(iii)
Question 11
If and x + = 2; then show that:
Solution 11
Thus from equations (1), (2) and (3), we have
Question 12
If 2x - 3y = 10 and xy = 16; find the value of 8x3 - 27y3.
Solution 12
Given that 2x - 3y = 10, xy = 16
Question 13
Expand :
(i) (3x + 5y + 2z) (3x - 5y + 2z)
(ii) (3x - 5y - 2z) (3x - 5y + 2z)
Solution 13
(i)
(3x + 5y + 2z) (3x - 5y + 2z)
= {(3x + 2z) + (5y)} {(3x + 2z) - (5y)}
= (3x + 2z)2 - (5y)2
{since (a + b) (a - b) = a2 - b2}
= 9x2 + 4z2 + 2 × 3x × 2z - 25y2
= 9x2 + 4z2 + 12xz - 25y2
= 9x2 + 4z- 25y2 + 12xz
(ii)
(3x - 5y - 2z) (3x - 5y + 2z)
= {(3x - 5y) - (2z)} {(3x - 5y) + (2z)}
= (3x - 5y)2 - (2z)2{since(a + b) (a - b) = a2 - b2}
= 9x2 + 25y2 - 2 × 3x × 5y - 4z2
= 9x2 + 25y2- 30xy - 4z2
= 9x2 +25y2 - 4z2 - 30xy
Question 14
The sum of two numbers is 9 and their product is 20. Find the sum of their
(i) Squares (ii) Cubes
Solution 14
Given sum of two numbers is 9 and their product is 20.
Let the numbers be a and b.
a + b = 9
ab = 20
Squaring on both sides gives
(a+b)2 = 92
a2 + b2 + 2ab = 81
a2 + b2 + 40 = 81
So sum of squares is 81 - 40 = 41
Cubing on both sides gives
(a + b)3 = 93
a3 + b3 + 3ab(a + b) = 729
a3 + b3 + 60(9) = 729
a3 + b3 = 729 - 540 = 189
So the sum of cubes is 189.
Question 15
Two positive numbers x and y are such that x > y. If the difference of these numbers is 5 and their product is 24, find:
(i) Sum of these numbers
(ii) Difference of their cubes
(iii) Sum of their cubes.
Solution 15
Given x - y = 5 and xy = 24 (x>y)
(x + y)2 = (x - y)2 + 4xy = 25 + 96 = 121
So, x + y = 11; sum of these numbers is 11.
Cubing on both sides gives
(x - y)3 = 53
x3 - y3 - 3xy(x - y) = 125
x3 - y3 - 72(5) = 125
x3 - y3= 125 + 360 = 485
So, difference of their cubes is 485.
Cubing both sides, we get
(x + y)3 = 113
x3 + y3 + 3xy(x + y) = 1331
x3 + y3 = 1331 - 72(11) = 1331 - 792 = 539
So, sum of their cubes is 539.
Question 16
If 4x2 + y2 = a and xy = b, find the value of 2x + y.
Solution 16
xy = b ….(i)
4x2 + y2 = a ….(ii)
Now, (2x + y)2 = (2x)2 + 4xy + y2
= 4x2 + y2 + 4xy
= a + 4b ….[From (i) and (ii)]
## Chapter 4 - Expansion Exercise Ex. 4(C)
Question 1
Expand:
(i) (x + 8) (x + 10)
(ii) (x + 8) (x - 10)
(iii) (x - 8) (x + 10)
(iv) (x - 8) (x - 10)
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
If a + b + c = 12 and a2 + b2 + c2 = 50; find ab + bc + ca.
Solution 4
Question 5
If a2 + b2 + c2 = 35 and ab + bc + ca = 23; find a + b + c.
Solution 5
Question 6
If a + b + c = p and ab + bc + ca = q; find a2 + b2 + c2.
Solution 6
Question 7
If a2 + b2 + c2 = 50 and ab + bc + ca = 47, find a + b + c.
Solution 7
Question 8
If x+ y - z = 4 and x2 + y2 + z2 = 30, then find the value of xy - yz - zx.
Solution 8
## Chapter 4 - Expansion Exercise Ex. 4(D)
Question 1
If x + 2y + 3z = 0 and x3 + 4y3 + 9z3 = 18xyz; evaluate:
Solution 1
Given that x3 + 4y3 + 9z3 = 18xyz and x + 2y + 3z = 0
Therefore, x + 2y = - 3z, 2y + 3z = -x and 3z + x = -2y
Now
Question 2
If a + = m and ≠ 0 ; find in terms of 'm'; the value of :
(i)
(ii)
Solution 2
(i)
(ii)
Question 3
In the expansion of (2x2 - 8) (x - 4)2; find the value of
(i) coefficient of x3
(ii) coefficient of x2
(iii) constant term.
Solution 3
Question 4
If x > 0 and find: .
Solution 4
Given that
Question 5
If 2(x2 + 1) = 5x, find :
(i) (ii)
Solution 5
(i)
2(x2 + 1} = 5x
Dividing by x, we have
(ii)
Question 6
If a2 + b2 = 34 and ab = 12; find:
(i) 3(a + b)2 + 5(a - b)2
(ii) 7(a - b)2 - 2(a + b)2
Solution 6
a2 + b2 = 34, ab= 12
(a + b)2 = a2 + b2 + 2ab
= 34 + 2 x 12 = 34 + 24 = 58
(a - b)2 = a2 + b2 - 2ab
= 34 - 2 x 12 = 34- 24 = 10
(i) 3(a + b)2 + 5(a - b)2
= 3 x 58 + 5 x 10 = 174 + 50
= 224
(ii) 7(a - b)2 - 2(a + b)2
= 7 x 10 - 2 x 58 = 70 - 116 = -46
Question 7
If 3x - and x ≠ 0 find : .
Solution 7
Given 3x -
We need to find
Question 8
If x2 + = 7 and x ≠ 0; find the value of:
.
Solution 8
Given that
We need to find the value of
Consider the given equation:
Question 9
If x = and x ≠ 5 find .
Solution 9
By cross multiplication,
=> x (x - 5) = 1 => x2 - 5x = 1 => x2 - 1 = 5x
Dividing both sides by x,
Question 10
If x = and x ≠ 5 find .
Solution 10
By cross multiplication,
=> x (5 - x) = 1 => x2 - 5x =-1 => x2 + 1 = 5x
Dividing both sides by x,
Question 11
If 3a + 5b + 4c = 0, show that:
27a3 + 125b3 + 64c3 = 180 abc
Solution 11
Given that 3a + 5b + 4c = 0
3a + 5b = -4c
Cubing both sides,
(3a + 5b)3 = (-4c)3
=>(3a)3 + (5b)3 + 3 x 3a x 5b (3a + 5b) = -64c3
=>27a3 + 125b3 + 45ab x (-4c) = -64c3
=>27a3 + 125b3 - 180abc = -64c3
=>27a3 + 125b3 + 64c3 = 180abc
Hence proved.
Question 12
The sum of two numbers is 7 and the sum of their cubes is 133, find the sum of their square.
Solution 12
Let a, b be the two numbers
.'. a + b = 7 and a3 + b3 = 133
(a + b)3 = a3 + b3 + 3ab (a + b)
=> (7)3 = 133 + 3ab (7)
=> 343 = 133 + 21ab => 21ab = 343 - 133 = 210
=> 21ab = 210 => ab= 2I
Now a2 + b2 = (a + b)2 - 2ab
=72 - 2 x 10 = 49 - 20 = 29
Question 13
In each of the following, find the value of 'a':
(i) 4x2 + ax + 9 = (2x + 3)2
(ii) 4x2 + ax + 9 = (2x - 3)2
(iii) 9x2 + (7a - 5)x + 25 = (3x + 5)2
Solution 13
(i) 4x2 + ax + 9 = (2x + 3)2
Comparing coefficients of x terms, we get
ax = 12x
so, a = 12
(ii) 4x2 + ax + 9 = (2x - 3)2
Comparing coefficients of x terms, we get
ax = -12x
so, a = -12
(iii) 9x2 + (7a - 5)x + 25 = (3x + 5)2
Comparing coefficients of x terms, we get
(7a - 5)x = 30x
7a - 5 = 30
7a = 35
a = 5
Question 14
If
(i) (ii)
Solution 14
Given
Question 15
The difference between two positive numbers is 4 and the difference between their cubes is 316.
Find:
(i) Their product
(ii) The sum of their squares
Solution 15
Given difference between two positive numbers is 4 and difference between their cubes is 316.
Let the positive numbers be a and b
a - b = 4
a3 - b3 = 316
Cubing both sides,
(a - b)3 = 64
a3 - b3 - 3ab(a - b) = 64
Given a3 - b3 = 316
So 316 - 64 = 3ab(4)
252 = 12ab
So ab = 21; product of numbers is 21
Squaring both sides, we get
(a - b)2 = 16
a2 + b2 - 2ab = 16
a2 + b2 = 16 + 42 = 58
Sum of their squares is 58.
## Chapter 4 - Expansion Exercise Ex. 4(E)
Question 1
Simplify:
(i) (x + 6)(x + 4)(x - 2)
(ii) (x - 6)(x - 4)(x + 2)
(iii) (x - 6)(x - 4)(x - 2)
(iv) (x + 6)(x - 4)(x - 2)
Solution 1
Using identity:
(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc
(i) (x + 6)(x + 4)(x - 2)
= x3 + (6 + 4 - 2)x2 + [6 × 4 + 4 × (-2) + (-2) × 6]x + 6 × 4 × (-2)
= x3 + 8x2 + (24 - 8 - 12)x - 48
= x3 + 8x2 + 4x - 48
(ii) (x - 6)(x - 4)(x + 2)
= x3 + (-6 - 4 + 2)x2 + [-6 × (-4) + (-4) × 2 + 2 × (-6)]x + (-6) × (-4) × 2
= x3 - 8x2 + (24 - 8 - 12)x + 48
= x3 - 8x2 + 4x + 48
(iii) (x - 6)(x - 4)(x - 2)
= x3 + (-6 - 4 - 2)x2 + [-6 × (-4) + (-4) × (-2) + (-2) × (-6)]x + (-6) × (-4) × (-2)
= x3 - 12x2 + (24 + 8 + 12)x - 48
= x3 - 12x2 + 44x - 48
(iv) (x + 6)(x - 4)(x - 2)
= x3 + (6 - 4 - 2)x2 + [6 × (-4) + (-4) × (-2) + (-2) × 6]x + 6 × (-4) × (-2)
= x3 - 0x2 + (-24 + 8 - 12)x + 48
= x3 - 28x + 48
Question 2
Solution 2
Question 3
Using suitable identity, evaluate
(i) (104)3
(ii) (97)3
Solution 3
Using identity: (a ± b)3 = a3 ± b3 ± 3ab(a ± b)
(i) (104)3 = (100 + 4)3
= (100)3 + (4)3 + 3 × 100 × 4(100 + 4)
= 1000000 + 64 + 1200 × 104
= 1000000 + 64 + 124800
= 1124864
(ii) (97)3 = (100 - 3)3
= (100)3 - (3)3 - 3 × 100 × 3(100 - 3)
= 1000000 - 27 - 900 × 97
= 1000000 - 27 - 87300
= 912673
Question 4
Solution 4
Question 5
Solution 5
Question 6
If a - 2b + 3c = 0; state the value of a3 - 8b3 + 27c3.
Solution 6
a3 - 8b3 + 27c3 = a3 + (-2b)3 + (3c)3
Since a - 2b + 3c = 0, we have
a3 - 8b3 + 27c3 = a3 + (-2b)3 + (3c)3
= 3(a)( -2b)(3c)
= -18abc
Question 7
If x + 5y = 10; find the value of x3 + 125y3 + 150xy - 1000.
Solution 7
x + 5y = 10
(x + 5y)3 = 103
x3 + (5y)3 + 3(x)(5y)(x + 5y) = 1000
x3 + (5y)3 + 3(x)(5y)(10) = 1000
= x3 + (5y)3 + 150xy = 1000
= x3 + (5y)3 + 150xy - 1000 = 0
Question 8
Solution 8
Question 9
If a + b = 11 and a2 + b2 = 65; find a3 + b3.
Solution 9
Question 10
Prove that:
x2+ y2 + z2 - xy - yz - zx is always positive.
Solution 10
x2 + y2 + z2 - xy - yz - zx
= 2(x2 + y2 + z2 - xy - yz - zx)
= 2x2 + 2y2 + 2z2 - 2xy - 2yz - 2zx
= x2 + x2 + y2 + y2 + z2 + z2 - 2xy - 2yz - 2zx
= (x2 + y2 - 2xy) + (z2 + x2 - 2zx) + (y2 + z2 - 2yz)
= (x - y)2 + (z - x)2 + (y - z)2
Since square of any number is positive, the given equation is always positive.
Question 11
Find:
(i) (a + b)(a + b)
(ii) (a + b)(a + b)(a + b)
(iii) (a - b)(a - b)(a - b) by using the result of part (ii)
Solution 11
(i) (a + b)(a + b) = (a + b)2
= a × a + a × b + b × a + b × b
= a2 + ab + ab + b2
= a2 + b2 + 2ab
(ii) (a + b)(a + b)(a + b)
= (a × a + a × b + b × a + b × b)(a + b)
= (a2 + ab + ab + b2)(a + b)
= (a2 + b2 + 2ab)(a + b)
= a2 × a + a2 × b + b2 × a + b2 × b + 2ab × a + 2ab × b
= a3 + a2 b + ab2 + b3 + 2a2b + 2ab2
= a3 + b3 + 3a2b + 3ab2
(iii) (a - b)(a - b)(a - b)
In result (ii), replacing b by -b, we get
(a - b)(a - b)(a - b)
= a3 + (-b)3 + 3a2(-b) + 3a(-b)2
= a3 - b3 - 3a2b + 3ab2
### STUDY RESOURCES
REGISTERED OFFICE : First Floor, Empire Complex, 414 Senapati Bapat Marg, Lower Parel, Mumbai - 400013, Maharashtra India.
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In an arithmetic sequence, the first term is -2, the fourth term is 16, and the nth term is 11998, how do you find n and the common difference?
May 12, 2016
common difference, $d = 6$
$n = 2001$
Explanation:
${a}_{n} = a + \left(n - 1\right) d$
Here, $a$ is the first term, $d$ is the common difference and ${a}_{n}$ is the nth term of the sequence.
We are given:
• $a = - 2$
• ${a}_{4} = 16$
• ${a}_{n} = 11998$
$\textcolor{red}{\text{To find "n" and } d .}$
Let's start with finding $d$:
${a}_{n} = a + \left(n - 1\right) d$
$\textcolor{b r o w n}{\text{Put in "a=-2 " and } n = 4}$
${a}_{4} = - 2 + \left(4 - 1\right) d$
$16 = - 2 + \left(4 - 1\right) d$ , $\textcolor{b l u e}{\text{ since } {a}_{4} = 16}$
$16 = - 2 + 3 d$
Add $2$ to both sides:
$16 \textcolor{b l u e}{+ 2} = - 2 \textcolor{b l u e}{+ 2} + 3 d$
$18 = 3 d$
$\textcolor{red}{6 = d}$
Next, calculate $n$:
${a}_{n} = a + \left(n - 1\right) d$
$\textcolor{b r o w n}{\text{Put in "a_n=11998 " , " a=-2 " and } d = 6}$
$11998 = - 2 + \left(n - 1\right) 6$
Add $2$ to both sides:
$11998 \textcolor{b l u e}{+ 2} = - 2 \textcolor{b l u e}{+ 2} + \left(n - 1\right) 6$
$12000 = \left(n - 1\right) 6$
Divide both sides by $6$:
$\frac{12000}{\textcolor{b l u e}{6}} = \frac{\left(n - 1\right) 6}{\textcolor{b l u e}{6}}$
$2000 = n - 1$
Add $1$ to both sides:
$2000 \textcolor{b l u e}{+ 1} = n - 1 \textcolor{b l u e}{+ 1}$
$\textcolor{red}{2001 = n}$
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Finding the nth Term Given the Common Ratio and the First Term
## Sequences where ratio of any two consecutive terms is constant.
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Progress
Practice Finding the nth Term Given the Common Ratio and the First Term
Progress
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Geometric Sequences and Finding the nth Term Given the Common Ratio and the First Term
The following sequence shows the distance (in centimeters) a pendulum travels with each successive swing. Write a general rule for the geometric sequence.
80, 72, 64.8, 58.32, ...
### Guidance
A geometric sequence is a sequence in which the ratio between any two consecutive terms, , is constant. This constant value is called the common ratio. Another way to think of this is that each term is multiplied by the same value, the common ratio, to get the next term.
#### Example A
Consider the sequence
Is this sequence geometric? If so, what is the common difference?
Solution: If we look at each pair of successive terms and evaluate the ratios, we get which indicates that the sequence is geometric and that the common ratio is 3.
### More Guidance
Now let’s see if we can develop a general rule ( term) for this sequence. Since we know that each term is multiplied by 3 to get the next term, let’s rewrite each term as a product and see if there is a pattern.
This illustrates that the general rule is , where is the common ratio. This even works for the first term since .
#### Example B
Write a general rule for the geometric sequence
Solution: From the general rule above we can see that we need to know two things: the first term and the common ratio to write the general rule. The first term is 64 and we can find the common ratio by dividing a pair of successive terms, . The term rule is thus .
#### Example C
Find the term rule for the sequence and hence find the term.
Solution: The first term here is 81 and the common ratio, , is . The term rule is . Now we can find the term . Use the graphing calculator for the last step and MATH > Frac your answer to get the fraction. We could also use the calculator and the general rule to generate terms . Reminder: the function can be found in the LIST ( STAT) Menu under OPS. Be careful to make sure that the entire exponent is enclosed in parenthesis.
Intro Problem Revisit We need to know two things, the first term and the common ratio, to write the general rule. The first term is 80 and we can find the common ratio by dividing a pair of successive terms, . The term rule is thus .
### Guided Practice
1. Identify which of the following are geometric sequences. If the sequence is geometric, find the common ratio.
a.
b.
c.
2. Find the general rule and the term for the sequence
3. Find the term rule and list terms 5 thru 11 using your calculator for the sequence
### Explore More
Identify which of the following sequences are arithmetic, geometric or neither.
Given the first term and common ratio, write the term rule and use the calculator to generate the first five terms in each sequence.
1. and
2. and
3. and
4. and
Find the term rule for each of the following geometric sequences.
Use a geometric sequence to solve the following word problems.
1. Rebecca inherited some land worth $50,000 that has increased in value by an average of 5% per year for the last 5 years. If this rate of appreciation continues, about how much will the land be worth in another 10 years? 2. A farmer buys a new tractor for$75,000. If the tractor depreciates in value by about 6% per year, how much will it be worth after 15 years?
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What Is B In Slope Intercept Form
The Definition, Formula, and Problem Example of the Slope-Intercept Form
What Is B In Slope Intercept Form – One of the many forms employed to represent a linear equation among the ones most commonly encountered is the slope intercept form. You may use the formula for the slope-intercept in order to solve a line equation as long as that you have the straight line’s slope as well as the y-intercept. It is the point’s y-coordinate where the y-axis intersects the line. Find out more information about this particular linear equation form below.
What Is The Slope Intercept Form?
There are three basic forms of linear equations: standard one, the slope-intercept one, and the point-slope. Although they may not yield similar results when used, you can extract the information line that is produced quicker with this slope-intercept form. As the name implies, this form employs a sloped line in which it is the “steepness” of the line indicates its value.
The formula can be used to discover the slope of a straight line, the y-intercept, also known as x-intercept where you can apply different formulas that are available. The equation for this line in this specific formula is y = mx + b. The slope of the straight line is indicated with “m”, while its y-intercept is indicated via “b”. Each point of the straight line is represented as an (x, y). Note that in the y = mx + b equation formula the “x” and the “y” are treated as variables.
An Example of Applied Slope Intercept Form in Problems
For the everyday world In the real world, the “slope intercept” form is commonly used to depict how an object or problem changes in an elapsed time. The value provided by the vertical axis demonstrates how the equation addresses the extent of changes over the value provided by the horizontal axis (typically the time).
An easy example of the application of this formula is to find out how many people live in a particular area in the course of time. Using the assumption that the population in the area grows each year by a specific fixed amount, the point value of the horizontal axis will increase by a single point each year and the point amount of vertically oriented axis will rise to reflect the increasing population by the set amount.
You may also notice the beginning point of a problem. The beginning value is at the y-value of the y-intercept. The Y-intercept is the place at which x equals zero. By using the example of the above problem the beginning point could be when the population reading starts or when the time tracking starts along with the associated changes.
So, the y-intercept is the place that the population begins to be recorded by the researcher. Let’s say that the researcher began to do the calculation or measure in 1995. In this case, 1995 will be”the “base” year, and the x 0 points will occur in 1995. Therefore, you can say that the 1995 population represents the “y”-intercept.
Linear equations that use straight-line formulas can be solved in this manner. The starting point is represented by the yintercept and the change rate is expressed through the slope. The main issue with the slope intercept form is usually in the horizontal variable interpretation especially if the variable is accorded to a specific year (or any other kind number of units). The first step to solve them is to ensure that you know the definitions of variables clearly.
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# 1B Family Update
## Dear Families
We hope you and your family have a wonderful and warm winter break! We will see you in 2017 on January 3rd! Below is an update on what has been going on the last few weeks in 1B...
In reading, we have continued to work on the retelling of a story. This time, we worked to identify the solution of the story. Students realized that sometimes a character doesn't have to resolve the problem. After winter break we will begin our nonfiction reading unit! So far, we've determined what makes a book nonfiction and next we will start exploring the structure of nonfiction texts.
In writing, we started our nonfiction writing unit! So far, we began planning our All About books by selecting a topic that we feel we are an expert on. A way to support us at home is to have us share our topic with you, discuss the 4-5 details or facts we know about the topic, and why we want others to know about it. This will help us to be ready for when we start writing the stories after break.
In math, we wrapped up our unit on geometry. One thing that students struggled with was remembering that a triangle has 3 sides and 3 corners and that a trapezoid has 4 sides and 4 corners. If you have students draw you a trapezoid at home, remind them that we have trapezoid-shaped tables in our classroom. We began our unit on place value, exploring how many ones, tens or even hundreds there are in numbers. For example, in the number 35, there are 3 tens and 5 ones. You can practice this at home by giving your child a number and having them tell you how many tens and ones are in it. Then, have them count to that number by tens and ones. For example, to get to 35, you would count the 3 tens first then switch to ones by saying "10, 20, 30, 31, 32, 33, 34, 35." We have learned that making groups of tens is an efficient or quicker way to represent and count numbers. Having a strong understanding of place value will help us add and subtract two-digit numbers with regrouping when we return from break!
Sincerely,
Amy Czaja and Kendra Thomas
## Materials needed for classroom: Maps
Our next Inquiry unit after Homes will be Maps and Geography, which we will start in January. We are looking for donations of maps for our classroom. These can be subway or museum maps that you pick up or old maps from any location. Thank you!
## Thinker student of the Week: Oola
Oola has demonstrated our learner profile for December: thinker. Thinkers think for themselves to make decisions and be creative when solving problems. Here are examples from his classmates of how Oola showed that he is a thinker:
• How you been a thinker is by solving problems right away like today when we were using the base-10 blocks, our table was very loud and she tried to calm us down a little bit.
• Oola has been a thinker by thinking of people's bodies and feelings like if someone fell, she would say are you okay.
• How Oola has been a thinker is because she is thinking for herself and not just going right up to a teacher. She's deciding if a problem is a teacher-sized problem or a student-sized problem.
• You have been a thinker when I was at recess you thought about how I felt and I felt lonely because I was sitting down on a bench. She came up to me and asked do you want to play with us?
• Oola has been a thinker. Me and Oola sometimes play a little bit, and she is a thinker by stopping talking with me.
• If someone had fell down but no one saw, but she realized the person fell, she would go over and ask if they were okay.
## ANYTIME | ANYWHERE What are you reading?
Two traveling libraries have been created to support our Anytime Anywhere Reading Initiative.
Please Note: Your student may bring home a book with a red or yellow dot taped to the binding. When he or she returns the book to the cart, another book may be checked out. Special books have been purchased for these libraries but have not yet arrived. (We’re making due with overflow from classroom libraries until the curated selection arrives!)
Students track their reading using library cards. Each time they read a book, they write down the title and stamp their library card at one of our vintage time machines. They more they read, the further back in time we can travel!
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# SOLVING POLYNOMIAL EQUATIONS
• PRACTICE (online exercises and printable worksheets)
DEFINITION polynomial equations
A polynomial equation is an equation of the form $\,P(x) = Q(x)\,$, where $\,P\,$ and $\,Q\,$ are polynomials.
If $\,P\,$ has degree $\,m\,$ and $\,Q\,$ has degree $\,n\,$ for positive integers $\,m\,$ and $\,n\,$,
then the degree of the polynomial equation is $\,\max(m,n)\,$.
In other words, the degree of the polynomial equation is the highest power of $\,x\,$ that appears, when looking at both sides of the equation.
There are different situations where you may need to solve a polynomial equation:
1. in a classroom, to demonstrate that you have learned appropriate solution techniques
2. outside of a classroom (in a real-life application!) where you only need to get the solutions—you don't need to know how they were found
In the second situation, jump up to http://www.wolframalpha.com and type in something like this:
solve x^5 - 2 = 2x^3 + 2x^2 + 3x
Try it! You'll be presented with all the complex solutions (which of course include the real solutions), together with appropriate graphs.
Easy, quick, reliable, and free.
If you find yourself in the classroom situation, then this lesson is for you—and it will also be a great review, pulling together ideas from prior sections.
## Solutions of Polynomial Equations: General Concepts
All polynomials in this lesson are assumed to have real number coefficients.
Therefore, instead of saying things like
‘Let $\,P(x) = ax + b\,$ where $\,a\,$ and $\,b\,$ and real numbers and $\,a\ne 0\,$’,
we can more simply say
‘Let $\,P(x) = ax + b\,$, where $\,a\ne 0\,$’.
• Polynomial equations of degree $1$ are called linear equations.
The standard form is $\,ax + b = 0\,$, for $\,a\ne 0\,$.
They are solved using the addition and multiplication properties of equality.
The unique solution is $\,x = -\frac{b}{a}\,$.
• Polynomial equations of degree $2$ are called quadratic equations.
The standard form is $\,ax^2 + bx + c = 0\,$, for $\,a\ne 0\,$.
They are solved using the quadratic formula.
The solutions are: $\displaystyle x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$
The discriminant is: $\,b^2 - 4ac\,$
If the discriminant is positive, there are two different real number solutions—call them $\,r_1\,$ and $\,r_2\,$.
In this case: $ax^2 + bx + c = a(x - r_1)(x - r_2)\,$
If the discriminant is zero, there is exactly one real number solution—call it $\,r\,$.
In this case: $ax^2 + bx + c = a(x - r)^2\,$
If the discriminant is negative, there are no real number solutions.
Instead, there is a complex conjugate pair of solutions, call them $\,z_1\,$ and $\,z_2\,$.
In this case, $ax^2 + bx + c\,$ is irreducible—it can't be factored using real numbers.
However, it can be factored using the complex conjugate pair: $ax^2 + bx + c = a(x - z_1)(x - z_2)\,$.
• Polynomial equations of degree $3$ are called cubic equations.
Polynomial equations of degree $4$ are called quartic equations.
There do exist methods for finding exact solutions of cubic and quartic equations.
However, these methods are hard to use, and they usually give big, complicated radical expressions for the solutions.
Cubic and quartic equations that you run into in math classes usually tend to be a bit contrived
and can be solved using the method discussed in this lesson.
• For polynomial equations of degree $5$ and above, there are no general solution methods.
That is, there are no methods that will always give all the exact solutions.
Again, however, the ones that you run into in math classes can usually be solved by using the method discussed in this lesson.
• Every polynomial equation of degree $n$ (where $n = 1,2,3,...$) with real number coefficients
has exactly $n$ solutions in the set of complex numbers, counting multiplicities.
Any non-real solutions must occur in complex conjugate pairs.
For example, the fifth degree polynomial equation ‘$\,(x-2)^5 = 0\,$’ has exactly one real number solution, $\,x = 2\,$, which has multiplicity $\,5\,$.
Here's a second example: the polynomial equation ‘$\,8(x-1)^2x^3(x^2 + 1) = 0\,$’ has degree $\,7\,$.
It has two real solutions: $1$ and $0$.
Setting the irreducible quadratic $\,x^2 + 1\,$ equal to zero gives two non-real solutions—the complex conjugate pair $\,i\,$ and $\,-i\,$.
The real solution $\,1\,$ has multiplicity $\,\color{red}{2}\,$, from the $\,(x-1)^{\color{red}{2}}\,$.
The real solution $\,0\,$ has multiplicity $\,\color{red}{3}\,$, from the $\,x^{\color{red}{3}}\,$.
The solution $\,i\,$ has multiplicity $\,\color{red}{1}\,$.
The solution $\,-i\,$ has multiplicity $\,\color{red}{1}\,$.
Note that $\,\color{red}{2} + \color{red}{3} + \color{red}{1} + \color{red}{1} = 7\,$.
In factored form over the set of complex numbers, the equation is written as ‘$\,8(x-1)^2(x^3)(x-i)(x+i) = 0\,$’.
## An Approach for finding Exact Solutions of (some) Polynomial Equations of Degree $3^+$
GENERAL PROCEDURE EXAMPLE:
Solve the equation:
$$x^5 - 2 = 2x^3 + 2x^2 + 3x$$
STEP 1:
Write the equation in the form $P(x) = 0$.
That is, get the number zero on the right-hand side;
the polynomial on the left is asigned the named $\,P(x)\,$.
The zeros of $\,P\,$ are the solutions of the equation.
Subtract $\,2x^3 + 2x^2 + 3x\,$ from both sides to get this equivalent equation: $$\underbrace{x^5 - 2x^3 - 2x^2 - 3x - 2}_{P(x)} = 0$$
STEP 2:
Use the Bounding the Real Roots of a Polynomial theorem
to get a bound on the real zeros of $P$.
Call this bound $\,K\,$.
Use a graphing device (such as a graphing calculator
or the WolframAlpha widget supplied below) to graph $\,P\,$
using the initial window $\,-10\le y\le 10\,$ and $\,-K\le x\le K\,$.
You're looking for a ‘nice’ zero—preferably an integer.
Use the behavior of the graph to guess the multiplicity of each zero.
Find all ‘nice’ zeros, re-graphing on tighter intervals as needed.
\begin{align} K &= 1 + \frac{\text{max of absolute values of coefficients}}{\text{absolute value of leading coefficient}}\cr &= 1 + \frac{3}{1} = 4 \end{align} From the graph below
(obtained using the WolframAlpha widget at left),
it's clear that there are only two real zeros: $-1$ and $2$
The number $\,2\,$ is clearly a simple zero.
The number $\,-1\,$ probably has multiplicity two.
STEP 3:
CHECK your (potential) zeros from Step 2.
You don't want to think that (say) $\,2\,$ is a zero, when it's actually $\,1.9\,$!
Use synthetic division and the remainder theorem to do the checks.
Don't forget the $\,0\,$ from the missing $\,x^4\,$ term in $\,P(x)\,$!
$2$ $1$ $0$ $-2$ $-2$ $-3$ $-2$ $2$ $4$ $4$ $4$ $2$ $-1$ $1$ $2$ $2$ $2$ $1$ $0$ $\leftarrow$ so $2$ is a zero $-1$ $-1$ $-1$ $-1$ $-1$ $1$ $1$ $1$ $1$ $0$ $\leftarrow$ so $-1$ is a zero $-1$ $0$ $-1$ $\color{green}{1}$ $\color{green}{0}$ $\color{green}{1}$ $0$ $\leftarrow$ so $-1$ is a zero again
STEP 4:
Use the bottom row of the synthetic division to get the factor that remains
after all factors corresponding to real zeros have been divided out.
The $\ \ \color{green}{1}\ \ \ \color{green}{0}\ \ \ \color{green}{1}\ \$ from the last row of the synthetic division represents: $1x^2 + 0x + 1 = x^2 + 1$
Thus, we have: $$\frac{P(x)}{(x-2)(x+1)(x+1)} = x^2 + 1$$ or, equivalently, $$x^5 - 2x^3 - 2x^2 - 3x - 2 = (x-2)(x+1)^2(x^2 + 1)$$
STEP 5:
Set the last factor (if any) equal to zero, and solve if possible.
$$\begin{gather} x^2 + 1 = 0\cr x^2 = -1\cr x^2 = \pm\sqrt{-1}\cr x = \pm i \end{gather}$$
STEP 6:
Use the results from Step 5 to factor $\,P(x)\,$ as completely as possible.
$$P(x) = (x-2)(x+1)^2(x - i)(x + i)\,$$
STEP 7:
State the solutions to the original equation.
The solutions of the original equation
(where repetition indicates multiplicity) are:
$-1\,$, $-1\,$, $2\,$, $i\,$, and $-i$
Master the ideas from this section
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# Using Integration, Find the Area of the Triangle Whose Vertices Are (2, 3), (3, 5) and (4, 4). - Mathematics
Sum
Using integration, find the area of the triangle whose vertices are (2, 3), (3, 5) and (4, 4).
#### Solution 1
The Vertices of ΔABC are A (2, 3), B (3, 5), and C (4, 4)
Equation of line segment AB is
("y" - 5) = (5-3)/(3-2) ("x" -3)
="y" -5 = 2 ("x" -3)
="y" = 2"x" -1
Equation of line segment BC is
("y" - 5) = (5-3)/(3-4) ("x" -3)
="y" -5 = -1 ("x" -3)
="y" = -"x" + 8
Equation of line segment AC is
("y" - 4) = (4-3)/(4-2) ("x" -4)
="y" -4 = (1)/(2) ("x" -4)
= "y" = ("x")/(2) + 2
∴ Area of ΔABC = int_2^3 [(2"x" -1) - ("x"/2+2)] d"x" + int_3^4 [(-"x" + 8) - ("x"/2 + 2)] . d"x"
= int_2^3 ((3"x")/2 -3) . d"x" + int_3^4 ((-3"x")/2 + 6) . d"x"
= [(3"x"^2)/(4) - 3"x"]_2^3 + [ (-3"x"^2)/4 + 6"x"]_3^4
= (27/4 - 9) - (3 -6) + (-12 + 24) - (-27/4 + 18)
= (3)/(2) "sq. units".
#### Solution 2
The Vertices of ΔABC are A (2, 3), B (3, 5), and C (4, 4)
Equation of line segment AB is
("y" - 5) = (5-3)/(3-2) ("x" -3)
="y" -5 = 2 ("x" -3)
="y" = 2"x" -1
Equation of line segment BC is
("y" - 5) = (5-3)/(3-4) ("x" -3)
="y" -5 = -1 ("x" -3)
="y" = -"x" + 8
Equation of line segment AC is
("y" - 4) = (4-3)/(4-2) ("x" -4)
="y" -4 = (1)/(2) ("x" -4)
= "y" = ("x")/(2) + 2
Area of ΔABC = int_2^3 [(2"x" -1) - ("x"/2+2)] d"x" + int_3^4 [(-"x" + 8) - ("x"/2 + 2)] . d"x"
= int_2^3 ((3"x")/2 -3) . d"x" + int_3^4 ((-3"x")/2 + 6) . d"x"
= [(3"x"^2)/(4) - 3"x"]_2^3 + [ (-3"x"^2)/4 + 6"x"]_3^4
= (27/4 - 9) - (3 -6) + (-12 + 24) - (-27/4 + 18)
= (3)/(2) "sq. units".
Concept: Area of a Triangle
Is there an error in this question or solution?
|
# Taking the Square Root of Both Sides Adapted from Walch EducationAdapted from Walch Education.
## Presentation on theme: "Taking the Square Root of Both Sides Adapted from Walch EducationAdapted from Walch Education."— Presentation transcript:
Taking the Square Root of Both Sides Adapted from Walch EducationAdapted from Walch Education
Complex Numbers The imaginary unit i represents the non-real value. i is the number whose square is –1. We define i so that and i 2 = –1. A complex number is a number with a real component and an imaginary component. Complex numbers can be written in the form a + bi, where a and b are real numbers, and i is the imaginary unit. 5.2.1: Taking the Square Root of Both Sides2
Real Numbers Real numbers are the set of all rational and irrational numbers. Real numbers do not contain an imaginary component. Real numbers are rational numbers when they can be written as, where both m and n are integers and n ≠ 0. Rational numbers can also be written as a decimal that ends or repeats. 5.2.1: Taking the Square Root of Both Sides3
Irrational Numbers Real numbers are irrational when they cannot be written as, where m and n are integers and n ≠ 0. Irrational numbers cannot be written as a decimal that ends or repeats. The real number is an irrational number because it cannot be written as the ratio of two integers. examples of irrational numbers include and π. 5.2.1: Taking the Square Root of Both Sides4
Quadratic Equations A quadratic equation is an equation that can be written in the form ax 2 + bx + c = 0, where a ≠ 0. Quadratic equations can have no real solutions, one real solution, or two real solutions. When a quadratic has no real solutions, it has two complex solutions. Quadratic equations that contain only a squared term and a constant can be solved by taking the square root of both sides. These equations can be written in the form x 2 = c, where c is a constant. 5.2.1: Taking the Square Root of Both Sides5
Quadratic Equations c tells us the number and type of solutions for the equation. 5.2.1: Taking the Square Root of Both Sides6 c Number and type of solutions NegativeTwo complex solutions 0 One real, rational solution Positive and a perfect square Two real, rational solutions Positive and not a perfect square Two real, irrational solutions
Practice # 1 Solve (x – 1) 2 + 15 = –1 for x. 5.2.1: Taking the Square Root of Both Sides7
Solution Isolate the squared binomial. Use a square root to isolate the binomial. 5.2.1: Taking the Square Root of Both Sides8 (x – 1) 2 + 15 = –1Original equation (x – 1) 2 = –16 Subtract 15 from both sides.
Solution, continued Simplify the square root. There is a negative number under the radical, so the answer will be a complex number. 5.2.1: Taking the Square Root of Both Sides9 Equation Write –16 as a product of a perfect square and –1. Product Property of Square Roots Simplify.
Solution, continued Isolate x. The equation (x – 1) 2 + 15 = –1 has two solutions, 1 ± 4i. 5.2.1: Taking the Square Root of Both Sides10 Equation Add 1 to both sides.
Try This One… Solve 4(x + 3) 2 – 10 = –6 for x. 5.2.1: Taking the Square Root of Both Sides11
Thanks for Watching!!!! Ms. Dambreville
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# CAT Quantitative Aptitude Questions | CAT Time Speed and Distance & Races
###### CAT Questions | Speed Time | Relative Speed
The question is from the topic Relative Speed. This is a classic CAT level question testing your understanding in the concept of relative speed from Time Speed and Distance. Time Speed and Distance is a favorite in CAT Exam and appears more often than expected in the CAT Quantitative Aptitude section in the CAT Exam
Question 27 : Two trains left from two stations P and Q towards station Q and station P respectively. 3 hours after they met, they were 675 Km apart. First train arrived at its destination 16 hours after their meeting and the second train arrived at its destination 25 hours after their meeting. How long did it take for the first train to make the whole trip?
1. 18h
2. 36h
3. 25h
4. 48h
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##### Method of solving this CAT Question from Time Speed Distance: Let's start with creating a visual of how the trains cross each other.
Total distance travelled by both the trains before meeting = D. This distance will be covered in the proportion of their speeds. Clarify with the diagram.
3 hours after meeting distance travelled by A = SA * 3 and by B = SB * 3
Therefore, 3 (SA + SB) = 675 => SA + SB = 225.
Now the remaining distance to be covered by first train is DSB/(SA + SB)
Therefore, time taken = $$frac{DS_{B}}{$S_{A} + S_{B}$S_{A}}$ = 16 ---------$1)
Similarly, $$frac{DS_{A}}{$S_{A} + S_{B}$S_{A}}$ = 25 ---------$2)
Dividing equation 1 by 2 -
$$frac{S_A^2}{S_B^2}\\$ = $\frac{25}{16}\\$ => $\frac{S_A}{S_B}\\$ = $\frac{5}{4}\\$. Therefore, SA + $\frac{4}{5}\\$ * SA = 225 => SA = 125 Km/hr and SB = 100 Km/hr From equation 2, $\frac{D}{S_A}\\$ = 16 * $\frac{225}{100}\\$ = 36h, which is the time taken for the first train to complete the journey. The question is "How long did it take for the first train to make the whole trip?" ##### Hence,it took 36h for the first train to make the whole trip. Choice B is the correct answer. ###### 🎯 Huge Discount: 3,500 off on Online CAT Course. Till June 1. Register now! ###### Best Indore IPM & Rohtak IPM CoachingSignup and sample 9 full classes for free. Register now! ###### Already have an Account? ###### CAT Coaching in ChennaiCAT 2020Enroll at 40,000/- Online Classroom Batches Starting Now! ###### Best CAT Coaching in ChennaiRegister Online, get Rs 7000/- off Attend a Demo Class ## CAT Preparation Online | CAT Arithmetic Videos On YouTube #### Other useful sources for Arithmetic Question | Time Speed Distance and Races Sample Questions ##### Where is 2IIM located? 2IIM Online CAT Coaching A Fermat Education Initiative, 58/16, Indira Gandhi Street, Kaveri Rangan Nagar, Saligramam, Chennai 600 093 ##### How to reach 2IIM? Phone:$91) 44 4505 8484
Mobile: (91) 99626 48484
WhatsApp: WhatsApp Now
Email: prep@2iim.com
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