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# Maharashtra Board Class 10 Solutions General Science – Part 1
### Maharashtra Board Class 10 Solutions General Science – Part 1
Maharashtra Board Class 10 Important Question Chapter wise Download at here. Maharashtra Board Class 10 for General Science – Part 1. Maha Board Class 10th Textbook General Science – Part 1.
Maha Board Class 10th General Science – Part 1 Textbook All Questions. Maharashtra State Bureau of Textbook Production and Curriculum Research General Science – Part 1 Book Class 10.
Book Type Text Book Class 10th Medium English Subjects General Science – Part 1
### CHAPTER – 1 – Gravitation
(1) Study the entries in the following table and rewrite them putting the connected items in a single row.
I II III Mass m/s2 Zero at the centre Weight kg Measure of inertia Accelera- tion due to gravity Nm2 /kg2 Same in the entire universe Gravita – tional con- stant N Depends on height
(a) What is the difference between mass and weight of an object. Will the mass and weight of an object on the earth be same as their values on Mars? Why?
(b) What are (i) free fall, (ii) acceleration due to gravity (iii) escape velocity (iv) centripetal force ?
(c) Write the three laws given by Kepler. How did they help Newton to arrive at the inverse square law of gravity?
(d) A stone thrown vertically upwards with initial velocity u reaches a height ‘h’ before coming down. Show that the time taken to go up is same as the time taken to come down.
(e) If the value of g suddenly becomes twice its value, it will become two times more difficult to pull a heavy object along the floor. Why?
(3) Explain why the value of g is zero at the centre of the earth.
(4) Solve the following examples.
(a) An object takes 5 s to reach the grond from a height of 5 m on a planet. What is the value of g on the planet?
(b) The radius of planet A is half the radius of planet B. If the mass of A is MA, what must be the mass of B so that the value of g on B is half that of its value on A?
(c) The mass and weight of an object on earth are 5 kg and 49 N respectively. What will be their values on the moon? Assume that the acceleration due to gravity on the moon is 1/6th of that on the earth.
(d) An object thrown vertically upwards reaches a height of 500 m. What was its initial velocity? How long will the object take to come back to the earth? Assume g = 10 m/s2
(e) A ball falls off a table and reaches the ground in 1 s. Assuming g = 10 m/s2 , calculate its speed on reaching the ground and the height of the table.
(f) The masses of the earth and moon are 6 x 1024 kg and 7.4×1022 kg, respectively. The distance between them is 3.84 x 105 km. Calculate the gravitational force of attraction between the two?
Use G = 6.7 x 10-11 N m2 kg-2
(g) The mass of the earth is 6 x 1024 kg. The distance between the earth and the Sun is 1.5x 1011 m. If the gravitational force between the two is 3.5 x 1022 N, what is the mass of the Sun?
Use G = 6.7 x 10-11 N m2 kg-2
### CHAPTER – 2 – Periodic Classification of Elements
(1) Rearrange the columns 2 and 3 so as to match with the column 1.
Column 1 Column 2 Column 3 i. Triad ii. Octave iii.Atomic number iv. Period v. Nucleus vi. Electron a. Lightest and negatively charged particle in all the atoms b. Concentrated mass and positive charge c. Average of the first and the third atomic mass d. Properties of the eighth element similar to the first e. Positive charge on the nucleus f. Sequential change in molecular formulae 1. Mendeleev 2. Thomson 3. Newlands 4. Rutherford 5. Dobereiner 6. Moseley
(2) Choose the correct option and rewrite the statement.
(a) The number of electrons in the outermost shell of alkali metals is……
(i) 1 (ii) 2 (iii) 3 (iv) 7
(b) Alkaline earth metals have valency 2. This means that their position in the modern periodic table is in …..
(i) Group 2 (ii) Group16 (iii) Period 2 (iv) d-block
(c) Molecular formula of the chloride of an element X is XCl. This compound is a solid having high melting point. Which of the following elements be present in the same group as X.
(i) Na (ii) Mg (iii) Al (iv) Si
(d) In which block of the modern periodic table are the nonmetals found?
(i) s-block (ii) p-block (iii) d-block (iv) f-block
(3) An element has its electron configuration as 2,8,2. Now answer the following questions.
(a) What is the atomic number of this element?
(b) What is the group of this element?
(c) To which period does this element belong?
(d) With which of the following elements would this element resemble? (Atomic numbers are given in the brackets)
N (7), Be (4) , Ar (18), Cl (17)
(4) Write the name and symbol of the element from the description.
(a) The atom having the smallest size.
(b) The atom having the smallest atomic mass.
(c) The most electronegative atom.
(d) The noble gas with the smallest atomic radius.
(e) The most reactive nonmetal.
(5) Write short notes.
(a) Mendeleev’s periodic law.
(b) Structure of the modern periodic table.
(c) Position of isotopes in the Mendeleev’s and the modern periodic table.
(6) Write scientific reasons.
(a) Atomic radius goes on decreasing while going from left to right in a period.
(b) Metallic character goes on decreasing while going from left to right in a period.
(c) Atomic radius goes on increasing down a group.
(d) Elements belonging to the same group have the same valency.
(e) The third period contains only eight elements even through the electron capacity of the third shell is 18 .
(7) Write the names from the description.
(a) The period with electrons in the shellsK, L and M.
(b) The group with valency zero.
(c) The family of nonmetals having valency one.
(d) The family of metals having valency one.
(e) The family of metals having valency two.
(f) The metalloids in the second and third periods.
(g) Nonmetals in the third period.
(h) Two elements having valency 4.
### CHAPTER – 3 – Chemical Reactions and Equations
(1) Choose the correct option from the bracket and explain the statement giving reason.
(Oxidation, displacement, electrolysis, reduction, zinc, copper, double displacement, decomposition)
(a) To prevent rusting, a layer of …….. metal is applied on iron sheets.
(b) The conversion of ferrous sulphate to ferric sulphate is …….. reaction.
(c) When electric current is passed through acidulated water …….. of water takes place.
(d) Addition of an aqueous solution of ZnSO4 to an aqueous solution of BaCl2 is an example of ……. reaction.
(2) Write answers to the following.
(a) What is the reaction called when oxidation and reduction take place simultaneously? Explain with one example.
(b) How can the rate of the chemical reaction, namely, decomposition of hydrogen peroxide be increased?
(c) Explain the term reactant and product giving examples.
(d) Explain the types of reaction with reference to oxygen and hydrogen. Illustratre with examples.
(e) Explain the similarity and difference in two events, namely adding NaOH to water and adding CaO to water.
(3) Explain the following terms with examples.
(a) Endothermic reaction
(b) Combination reaction
(c) Balanced equation d. Displacement reaction
(4) Give scientific reasons.
(a) When the gas formed on heating limstone is passed through freshly prepared lime water, the lime water turns milky.
(b) It takes time for pieces of Shahabad tile to disappear in HCl, but its powder disappears rapidly.
(c)While preparing dilute sulphuric acid from concentrated sulphuric acid in the laboratory, the concentrated sulphuric acid is added slowly to water with constant stirring.
(d) It is reccommended to use air tight container for storing oil for long time.
### CHAPTER – 4 – Effects of electric current
(1) Tell the odd one out. Give proper explanation.
(a) Fuse wire, bad conductor, rubber gloves, generator.
( b) Voltmeter, Ammeter, galvanometer, thermometer.
(c) Loud speaker, microphone, electric motor, magnet.
(2) Explain the construction and working of the follwoing. Draw a neat diagram and label it.
(a) Electric motor
(b) Electric Generator(AC)
(3) Electromagnetic induction means-
(a) Charging of an electric conductor.
(b) Production of magnetic field due to a current flowing through a coil.
(c) Generation of a current in a coil due to relative motion between the coil and the magnet.
(d) Motion of the coil around the axle in an electric motor.
(4) Explain the difference : AC generator and DC generator.
(5) Which device is used to produce electricity? Describe with a neat diagram.
(a) Electric motor
(b) Galvanometer
(c) Electric Generator (DC) d. Voltmeter
(6) How does the short circuit form? What is its effect?
(7) Give Scientific reasons.
(a) Tungsten metal is used to make a solenoid type coil in an electric bulb.
(b) In the electic equipment producing heat e.g. iron, electric heater, boiler, toaster etc, an alloy such as Nichrome is used, not pure metals.
(c) For electric power transmission, copper or aluminium wire is used.
(d) In practice the unit kWh is used for the measurement of electrical energy, rather than joule.
(8) Which of the statement given below correctly describes the magnetic field near a long, straight current carrying conductor?
(a) The magnetic lines of force are in a plane, perpendicular to the conductor in the form of straight lines.
(b) The magnetic lines of force are parallel to the conductor on all the sides of conductor.
(c) The magnetic lines of force are perpendicular to the conductor going radially outword.
(d) The magnetic lines of force are in concentric circles with the wire as the center, in a plane perpendicular to the conductor.
(9) What is a solenoid? Compare the magnetic field produced by a solenoid with the magnetic field of a bar magnet. Draw neat figures and name various components.
(10) Solve the following example.
(a) Heat energy is being produced in a resistance in a circuit at the rate of 100 W. The current of 3 A is flowing in the circuit. What must be the value of the resistance?
(b) Two tungsten bulbs of wattage 100 W and 60 W power work on 220 V potential difference. If they are connected in parallel, how much current will flow in the main conductor?
(c) Who will spend more electrical energy? 500 W TV Set in 30 mins, or 600 W heater in 20 mins?
(d) An electric iron of 1100 W is operated for 2 hrs daily. What will be the electrical consumption expenses for that in the month of April? (The electric company charges Rs 5 per unit of energy).
### CHAPTER – 5 – Heat
(1) Fill in the blanks and rewrite the sentence.
(a) The amount of water vapor in air is determined in terms of its …………
(b) If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their ………………
(c) During transformation of liquid phase to solid phase, the latent heat is ………….
(3) What is meant by specific heat capacity? How will you prove experimentally that different substances have different specific heat capacities?
(4) While deciding the unit for heat, which temperatures interval is chosen? Why?
(5) Explain the following:
(a) What is the role of anomalous behaviour of water in preserving aquatic life in regions of cold climate?
(b) How can you relate the formation of water droplets on the outer surface of a bottle taken out of refrigerator with formation of dew?
(c) In cold regions in winter, the rocks crack due to anomolous expansion of water.
(a) What is meant by latent heat? How will the state of matter transform if latent heat is given off?
(b) Which principle is used to measure the specific heat capacity of a substance?
(c) Explain the role of latent heat in the change of state of a substances? d. On what basis and how will you determine whether air is saturated with vapor or not?
If heat is exchanged between a hot and cold object, the temperature of the cold object goes on increasing due to gain of energy and the temperature of the hot object goes on decreasing due to loss of energy. The change in temperature continues till the temperatures of both the objects attain the same value. In this process, the cold object gains heat energy and the hot object loses heat energy. If the system of both the objects is isolated from the environment by keeping it inside a heat resistant box (meaning that the energy exchange takes place between the two objects only), then no energy can flow from inside the box or come into the box.
(i) Heat is transferred from where to where?
(ii) Which principle do we learn about from this process?
(iii) How will you state the principle briefly? iv. Which property of the substance is measured using this principle?
(8) Solve the following problems:
(a) Equal heat is given to two objects A and B of mass 1 g. Temperature of A increases by 3 o C and B by 5 o C. Which object has more specific heat? And by what factor?
(b) Liquid ammonia is used in ice factory for making ice from water. If water at 20 o C is to be converted into 2 kg ice at 0 o C, how many grams of ammonia are to be evaporated? (Given: The latent heat of vaporization of ammonia= 341 cal/g)
(c) A thermally insulated pot has 150 g ice at temperature 0 oC. How much steam of 100 oC has to be mixed to it, so that water of temperature 50 o C will be obtained? (Given : latent heat of melting of ice = 80 cal/g, latent heat of vaporization of water = 540 cal/g, specific heat of water = 1 cal/g 0C)
(d) A calorimeter has mass 100 g and specific heat 0.1 kcal/ kg oC. It contains 250 gm of liquid at 30 oC having specific heat of 0.4 kcal/kg o C. If we drop a piece of ice of mass 10 g at 0 oC, What will be the temperature of the mixture?
### CHAPTER – 6 – Refraction of light
(1) Fill in the blanks and Explain the completed sentences.
(a ) Refractive index depends on the …………. of light.
(b) The change in ……………. of light rays while going from one medium to another is called refraction.
(2) Prove the following statements.
(a) If the angle of incidence and angle of emergence of a light ray falling on a glass slab are i and e respectively, prove that, i = e.
(b) A raibow is the combined effect of the refraction, dispersion, and total internal reflection of light.
(3) Solve the following examples.
(a) If the speed of light in a medium is 1.5 x 108 m/s, what is the absolute refractive index of the medium?
(b) If the absolute refractive indices of glass and water are 3/2 and 4/3 respectively, what is the refractive index of glass with respect to water?
### CHAPTER – 7 – Lenses
(1) Match the columns in the following table and explain them.
Column 1 Column 2 Column 3 Farsightdness Nearby object can be seen clearly Bifocal lens Presbyopia Far away object can be seen clearly Concave lens Nearsightness Problem of old age Convex lens
(2) Draw a figure explaining various terms related to a lens.
(3) At which position will you keep an object in front of a convex lens so as to get a real image of the same size as the object ? Draw a figure.
(4) Give scientific resons:
(a) Simple microscope is used for watch repairs.
(b) One can sense colours only in bright light.
(c) We can not clearly see an object kept at a distance less than 25 cm from the eye.
(5) Explain the working of an astronomical telescope using refraction of light.
(6) Distinguish between:
(a) Farsightedness and Nearsightedness
(b) Concave lens and Convex Lens
(7) What is the function of iris and the muscles connected to the lens in human eye?
(8) Solve the following examples.
(i) Doctor has prescribed a lens having power +1.5 D. What will be the focal length of the lens? What is the type of the lens and what must be the defect of vision?
(ii) 5 cm high object is placed at a distance of 25 cm from a converging lens of focal length of 10 cm. Determine the position, size and type of the image.
(iii) Three lenses having power 2, 2.5 and 1.7 D are kept touching in a row. What is the total power of the lens combination?
(iv) An object kept 60 cm from a lens gives a virtual image 20 cm in front of the lens. What is the focal length of the lens? Is it a converging lens or diverging lens?
### CHAPTER – 8 – Metallurgy
(1) Write names.
(a) Alloy of sodium with mercury.
(b) Molecular formula of the common ore of aluminium.
(c) The oxide that forms salt and water by reacting with both acid and base.
(d) The device used for grinding an ore.
(e) The nonmetal having electrical conductivity.
( f) The reagent that dissolves noble metals.
(2) Explain the terms.
(a) Metallurgy
(b) Ores
(c) Minerals
(d) Gangue.
(3) Write scientific reasons.
(a) Lemon or tamarind is used for cleaning copper vessels turned greenish.
(b) Generally the ionic compounds have high melting points.
(c) Sodium is always kept in kerosene.
(d) Pine oil is used in froth flotation.
(e) Anodes need to be replaced from time to time during the electrolysis of alumina.
(4) When a copper coin is dipped in silver nitrate solution, a glitter appears on the coin after some time. Why does this happen? Write the chemical equation.
(5) The electronic configuration of metal ‘A’ is 2,8,1 and that of metal ‘B’ is 2,8,2. Which of the two metals is more reactive? Write their reaction with dilute hydrochloric acid.
(6) Draw a neat labelled diagram.
(a) Magnetic separation method.
(b) Froth floatation method.
(c) Electrolytic reduction of alumina.
(d)Hydraulic separation method.
(7) Write chemical equation for the following events. a. Aluminium came
(a) Aluminium came in contact with air.
(b) Iron filings are dropped in aqueous solution of copper sulphate.
(c) A reaction was brought about between ferric oxide and aluminium.
(d) Electrolysis of alumina is done. e. Zinc oxide is dissolved in dilute hydrochloric acid
(8) Complete the following statement using every given options. During the extraction of aluminium…………..
(a)Ingredients and gangue in bauxite.
(b)Use of leaching during the concentration of ore.
(c) Chemical reaction of transformation of bauxite into alumina by Hall’s process.
(d)Heating the aluminium ore with concentrated caustic soda.
(9) Divide the metals Cu, Zn, Ca, Mg, Fe, Na, Li into three groups, namely reactive metals, moderately reactive metals and less reactive metals.
### CHAPTER – 9 – Carbon Compounds
(1) Explain the following terms with example.
(a) Structural isomerism
(b) Covalent bond
(c) Hetero atom in a carbon compound
(d) Functional group
(e) Alkane
(f) Unsaturated hydrocarbon
(g) Homopolumer
(h) Monomer
(i) Reduction
(j) Oxydant
(a) What causes the existance of very large number of carbon compound ?
(b) Saturated hydrocarbons are classified into three types. Write these names giving one example each.
(c) Give any four functional groups containing oxygen as the heteroatom in it. Write name and structural formula of one example each.
(d) Give names of three functional groups containing three different hetero atoms. Write name and structural formula of one example each.
(e) Give names of three natural polymers. Write the place of their occurance and names of monomers from which they are formed.
(f) What is meant by vinegar and gashol? What are their uses ?
(g) What is a catalyst ? Write any one reaction which is brought about by use of catalyst ?
### CHAPTER – 10 – Space Missions
(1) Fill in the blanks and explain the statements with reasoning:
(a) If the height of the orbit of a satellite from the earth surface is increased, the tangential velocity of the satellite will …
(b) The initial velocity (during launching) of the Managalyaan, must be greater than …………..of the earth.
(2) State with reasons whether the following sentences are true or false
(a) If a spacecraft has to be sent away from the influence of earth’s gravitational field, its velocity must be less than the escape velocity.
(b) The escape velocity on the moon is less than that on the earth.
(c) A satellite needs a specific velocity to revolve in a specific orbit.
(d) If the height of the orbit of a satellite increases, its velocity must also increase.
(a) What is meant by an artificial satellite? How are the satellites classified based on their functions?
(b) What is meant by the orbit of a satellite? On what basis and how are the orbits of artificial satellites classified?
(c) Why are geostationary satellites not useful for studies of polar regions?
(d) What is meant by satellite launch vehicles? Explain a satellite launch vehicle developed by ISRO with the help of a schematic diagram.
(e) Why it is beneficial to use satellite launch vehicles made of more than one stage?
(4) Solve the following problems.
(a) If mass of a planet is eight times the mass of the earth and its radius is twice the radius of the earth, what will be the escape velocity for that planet?
(b) How much time a satellite in an orbit at height 35780 km above earth’s surface would take, if the mass of the earth would have been four times its original mass?
Updated: February 12, 2019 — 12:57 pm
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Logic
# Operator Search: Level 2 Challenges
$\large 1 \, \square \, 2 \, \square \, 3 \, \square \, 4 \, \square \, 5 \, \square \, 6 \, \square \, 7 \, \square \, 8 = 9$
There are $$2^7 =128$$ ways in which we can fill the squares with $$+, -$$.
How many ways would make the equation true?
Note: You are not allowed to use parenthesis.
$\large 3 \; \square \; 3 \; \square \; 3 \; \square \; 3$
Fill in the boxes above with any of the four mathematical operators ( $$+, -, \times , \div$$ ). Which of the following cannot be a resultant number?
Note: Order of operations (BODMAS) applied.
$\large 1 \; \square \; 2 \; \square \; 3\; \square \; 4\; \square \; 5 \; \square \; 6 \; \square \; 7 \; \square \; 8 \; \square \; 9 = 33$
Seven of the eight "$$\square$$"s above are filled with "$$+$$", and the other one with "$$-$$".
Before which integer should the "$$-$$" sign be placed to make the equation true?
$\large 18 \ \square \ 12 \ \square \ 4 \ \square \ 5 = 59$
Replace each $$\square$$ with one of $$+,-,\times,\div$$ to make the equation true.
Submit your answer as a 3 digit number, where:
• $$1$$ represents $$+$$
• $$2$$ represents $$-$$
• $$3$$ represents $$\times$$
• $$4$$ represents $$\div$$
Example : If your answer is $$+, -, \times$$ then enter the answer as $$123$$.
$\Huge 1\ \ \ \ 4 \ \ \ \ 9 \ \ = \ \ 16$
Is it possible to make this equation true by inserting the appropriate operations? Any operations and functions can be used.
×
Problem Loading...
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# Find the perimeter of a rectangle with length = 24 cm and diagonal = 25 cm
1 view
asked 6 days ago
Find the perimeter of a rectangle with length = 24 cm and diagonal = 25 cm.
answered 6 days ago by (-2,499 points)
Length of a rectangle (l) = 24 cm Diagonal = 25 cm
Let breadth of the rectangle = b m
Applying Pythagoras Theorem in triangle ABC,
We get, (AC)2 = (AB)2 + (BC)2
(25)= (24)2 + (b)2
625 = 576 + (b)2
625 – 576 = b2
49 = A2
√7x7 = b
∴ b = 7 cm
Now, perimeter of the rectangle
= 2(l + b)
= 2(24 + 7)
= 2(31)
= 62 cm
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# Multiple Towers
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## Objective
SWBAT find common multiples for single-digit numbers.
#### Big Idea
Students will build Multiple Towers using Unifix cubes to compare and analyze common multiples of single-digit numbers.
## 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, students decomposed 48 in a variety of ways. For example, some students decomposed 48 into four twelves: 48:6 = 4 x 12:6. Other students decomposed the 48 into two twenty-fours: 48:6 = 24 x 2:6
For the next task, some students made a connection between task 1 and task 2: Connecting Tasks. Here, a student showed 12:6 = 6 x 2:6. Sometimes I wonder if there is much for my students to gain from such simple problems as 12/6. Then, students like this one, 12:6 = 3x4:6., prove me wrong! I loved how she came up with 4 x 1/2 = 2! We definitely had to share this with the rest of the class!
For 60/6, some students saw the connection between tasks: 60:6 = 48 + 12:6. Others solved this task in other ways: 60:6 = 30 x 2:6
During this task, we didn't take the time to model as most students automatically knew that 600/6 = 100.
For the final task, students decomposed the 648 into multiples of 6: 648:6 = 300 + 300 + 48 :6 and 648:6 = 600 +48:6
Throughout every number talk, I continually model student thinking on the board to inspire other students. This also requires students to use math words to explain their thinking instead of relying on a model to represent the math. As students solved each task, I wrote the answers on the board to encourage students to use prior tasks to solve the more complex tasks: Listed Tasks.
## Teacher Demonstration
50 minutes
Rationale
For this lesson, students will be building Multiple Towers to compare the multiples of single-digit numbers. Eventually, students will line up all of their towers on the board to observe patterns (Students Placed Multiple Towers on the Board). Even though the standards don't specifically state the importance of finding common multiples, it is important for students to analyze multiples on a deeper level than just identification. Also, this lesson will engage students in Math Practice 7: Look for and make use of structure. In addition, if students understand that two and four have common multiples, this builds the foundation for understanding the distributive property: 4 x 5 = 2 (2 x 5).
Goal
To begin today's lesson, I introduced the goal: I can find common multiples for single-digit numbers. I asked: Does anyone remember what common means? Reflecting upon a previous lesson on common factors, students responded, "Shared!" I explained: Today, we will be finding common multiples (under 100) between two numbers, but first, I'd like for you to investigate common multiples with unifix cubes!
Unifix Cubes Activity
Today, we are going to make Multiple Towers! I'd like for each group of two-three students to create a multiple tower for a single-digit number. For example, if you were making a Multiple Tower for 2, you might take two orange cubes and place them together with two green. Then you might add on two blue. Your goal today is to get as close to 50 cubes without going over. Anytime I get out the unifix cubes, students light up! I had one student come running up to me and say, "I love this! Can we do this more often?"
Building Towers
I assigned each group a number between 2 and 10. Even though the focus of our lesson today was on single-digit numbers, I wanted to include 10 as it is a benchmark number that helps students make sense other numbers. Here, students are building a Multiple Tower for 2
Looking for Patterns
Next, Students Placed Multiple Towers on the Board and we all gathered on the carpet to begin Looking for Patterns. I truly wanted student to engage in Math Practice 7: Look for and Make use of Structure. During this time, I also wanted to encourage students to construct viable arguments (Math Practice 3).
Explaining Observations
What happened next was just wonderful! One by one students went to the board and explained their observations. At first, students explained patterns with colors, "It goes red, orange, blue, red orange, blue." The color of the cubes actually wasn't what I was looking for students to point out. However, this is a typical "beginning" observation. Often students begin by pointing out the most noticeable (and often not math-related) patterns first. Then, they build up to observing on a deeper, mathematical level.
Then, students began to see mathematical patterns: Most vs. Least Multiples! Each time a student shared, another student became inspired! Even students that struggle with math were going up to the board!
Another student then pointed out how It's Like a Line Graph. Then she explains a conjecture, "The smaller the number, the more multiples there are under a given number."
Here, the students begin Discussing Divisibility. I loved how students pointed out that 54 is the closet multiple of nine to 50, but 45 is the closest multiple of nine under 50.
Next a student pointed out that the Factors of 10 are Factors of 50. This was such great observation as students were beginning to look at common factors and multiples!
This student was then inspired to begin Comparing Multiples of 5 & 10. She points out that there are 12 groups of 4 in 48 and 6 groups of 8 in 48. Then, she explains that the same pattern works with the multiples of 5 and 10.
Finally, to connect this activity with today's goal, I decided to Rearrange the Multiple Towers. This student explained the connections between the multiple towers perfectly. I knew that this was a great time for students to begin finding common multiples using their Venn Diagrams.
## Guided Practice
35 minutes
Venn Diagram
At this point, I passed out the Number Line Model (inside sheet protectors) to each student. I also passed out a Venn Diagram (found at eduplace.com) to each pair of students.
Then, I reviewed: Remember, a Venn Diagram is used to compare two topics. We have already used this Venn Diagram to compare the factors of two numbers. Today, we are going to compare the multiples of two numbers. Can anyone tell me where the common multiples will go in the diagram? Students responded, "In the middle, where the circles overlap!"
Counting on the Number Line
I created a Venn Diagram on the Board. At first I placed 2 on one side and asked students to count by twos up to 100 on their number lines: Counting by 2s. Next, I placed 6 on the other side of the Venn Diagram on the board. At this point, students counted by 6s up to 100: Counting by 6s. Using this number line model, I wanted students to see how multiples of 2 overlap multiples of 6.
Venn Diagram
At this point, students completed the Venn Diagram for 2 & 6.. As a class, we completed the Venn Diagram on the board. At first students grappled with the idea that there aren't any multiples of 6 that aren't also multiples of 2. After conversing as a class, this idea made more sense to students.
Comparing the Multiples of 3 and 7
We then completed the same process in order to compare the multiples of 3 and 7. Here, Counting by 3s & 7s., students observed how the multiples of 3 and 7 overlap on the number line. Then, with their partners, students constructed a Venn Diagram for 3s & 7s.
Originally, I had planned for students to also compare the multiples of 4 and 8 as well as 5 and 9. However, the rich conversations we had as a class were more important that completing more tasks!
## Closing
5 minutes
To bring closure to this lesson, I celebrated students who were on task and meeting expectations. Students then cleaned up and returned their math materials.
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Q. Three years ago, Rashmi was thrice as old as Nazma. Ten years later, Rashmi will be twice as old as Nazma. How old are Rashmi and Nazma now?
Ans: Let’s consider Rashmi’s present age be X and Nazma’s present age be Y
Hence, Rashmi’s age 3 years ago = X – 3
Nazma’s age 3 years ago = Y – 3
By given first condition, (X – 3) = 3 ( Y – 3)
∴ X – 3 = 3Y – 9
∴ X – 3 Y = – 9 + 3
∴ X – 3 Y = – 6 ………….. (i)
Rashmi’s age 10 years later = X + 10
and Nazma’s age 10 years later = Y + 10
By given second condition, (X + 10) = 2 ( Y + 10)
∴ X + 10 = 2( Y + 10)
∴ X + 10 = 2 Y + 20
∴ X – 2 Y = 20 – 10
∴ X – 2 Y = 10 ………….. (i)
Let’s subtract equation (i) from equation (ii), we get:
(X – 2 Y) – (X – 3 Y) = (10) – (- 6)
∴ X – 2 Y – X + 3 Y = 10 + 6
∴ Y = 16
By substituting value of Y in equation (ii), we get:
X – 2 Y = 10
∴ X – 2 (16) = 10
∴ X = 10 + 32 = 42
Therefore, present ages of Rashmi and Nazma are 42 years and 16 years.
Check:
If Rashmi’s present age is 42 years, her age 3 years ago was 42 – 3 = 39 years
If Nazma’s present age is 16 years, and 5 yher age 3 years ago was 16 – 3 = 13 years
Hence, Rashmi was thrice as old as Nazma.
Rashmi’s age 10 years later : 42 + 10 = 52 years. Nazma’s age 10 years later: 16 + 10 = 26 years.
Hence, Rashmi will be twice as old as Nazma.
Since it meets our given condition, hence our answer is correct.
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×
Get Full Access to Numerical Analysis - 10 Edition - Chapter 3.2 - Problem 3
Get Full Access to Numerical Analysis - 10 Edition - Chapter 3.2 - Problem 3
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# Use Neville's method to approximate \/3 with the following functions and values. a. /(x)
ISBN: 9781305253667 457
## Solution for problem 3 Chapter 3.2
Numerical Analysis | 10th Edition
• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
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Numerical Analysis | 10th Edition
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5
Problem 3
Use Neville's method to approximate \/3 with the following functions and values. a. /(x) 3 X and the values xq = 2, xj = I, X2 0, X3 1, and X4 2. b. /(x) = ^/x and the values xq = 0, xi = I, X2 = 2, X3 = 4, and X4 = 5. c. Compare the accuracy ofthe approximation in parts (a) and (b).
Step-by-Step Solution:
Step 1 of 3
March 2125, 2016 Section 3.2 Suppose f’(x) >0 for all x on an open interval I. Suppose x < x i1 I. 2 en f(x) is continuous on [x1 x2] and it is differentiable on (x 1 x 2 2 x ¿ So by Mean Value Theorem there is c ¿ f( )1f ¿ ' ϵ ( 1x S2)hthen f (c)=¿ Which implies Thus f(x)< 0 is increasing on I. By similar arguments, we can show that f’(x)< 0 on an open interval f is decreasing on I. And if f’(x)=0 on an open interval I, then f(x) is constant on I. Theorem If f’(x)=
Step 2 of 3
Step 3 of 3
##### ISBN: 9781305253667
Since the solution to 3 from 3.2 chapter was answered, more than 327 students have viewed the full step-by-step answer. Numerical Analysis was written by and is associated to the ISBN: 9781305253667. The answer to “Use Neville's method to approximate \/3 with the following functions and values. a. /(x) 3 X and the values xq = 2, xj = I, X2 0, X3 1, and X4 2. b. /(x) = ^/x and the values xq = 0, xi = I, X2 = 2, X3 = 4, and X4 = 5. c. Compare the accuracy ofthe approximation in parts (a) and (b).” is broken down into a number of easy to follow steps, and 66 words. This textbook survival guide was created for the textbook: Numerical Analysis, edition: 10. This full solution covers the following key subjects: . This expansive textbook survival guide covers 76 chapters, and 1204 solutions. The full step-by-step solution to problem: 3 from chapter: 3.2 was answered by , our top Math solution expert on 03/16/18, 03:24PM.
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# College Math Teaching
## December 22, 2015
### Multi leaf polar graphs and total area…
Filed under: calculus, elementary mathematics, integrals — Tags: , — collegemathteaching @ 4:07 am
I saw polar coordinate calculus for the first time in 1977. I’ve taught calculus as a TA and as a professor since 1987. And yet, I’ve never thought of this simple little fact.
Consider $r(\theta) = sin(n \theta), 0 \theta \ 2 \pi$. Now it is well know that the area formula (area enclosed by a polar graph, assuming no “doubling”, self intersections, etc.) is $A = \frac{1}{2} \int^b_a (r(\theta))^2 d \theta$
Now the leaved roses have the following types of graphs: $n$ leaves if $n$ is odd, and $2n$ leaves if $n$ is even (in the odd case, the graph doubles itself).
So here is the question: how much total area is covered by the graph (all the leaves put together, do NOT count “overlapping”)?
Well, for $n$ an integer, the answer is: $\frac{\pi}{4}$ if $n$ is odd, and $\frac{\pi}{2}$ if $n$ is even! That’s it! Want to know why?
Do the integral: if $n$ is odd, our total area is $\frac{n}{2}\int^{\frac{\pi}{n}}_0 (sin(n \theta)^2 d\theta = \frac{n}{2}\int^{\frac{\pi}{n}}_0 \frac{1}{2} + cos(2n\theta) d\theta =\frac{\pi}{4}$. If $n$ is even, we have the same integral but the outside coefficient is $\frac{2n}{2} = n$ which is the only difference. Aside from parity, the number of leaves does not matter as to the total area!
Now the fun starts when one considers a fractional multiple of $\theta$ and I might ponder that some.
## December 16, 2015
### And I deducted points for a “Merry Christmas” math joke
Filed under: pedagogy, recreational mathematics — Tags: , — collegemathteaching @ 11:12 pm
From a student’s final exam in my “Life Contingencies” class (and no, I have no actuarial mathematics training…more on that later)
The student completely ignored the domain considerations for the log function and therefore lost points.
OK, not really. But it makes a better meme to say that.
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# Capacitors discharging
I am a beginner at electronics so excuse my rather simplistic question.
I would like to know regarding capacitors, if a capacitor is connected in a series circuit (as shown below but with a dc supply), will it discharge?
With a DC power supply, the capacitor will charge up to the voltage of the supply.
If you replace the DC supply with a resistor, the capacitor will discharge through the resistor until there is zero volts across the capacitor.
With an AC supply, the capacitor will continually charge and discharge as the AC voltage varies.
• There will never be zero volts, steady state, across the capacitor. ;) May 8, 2016 at 18:43
Since it's connected to a source of AC, it'll charge/discharge (depending on your point of reference) when the signal is one polarity and reverse when the signal changes polarity.
Charge yes. Discharge no.
If you are talking about a DC supply like a battery, the voltage across the capacitor will become the same as the supply voltage over time.
The question is how fast will this happen? This depends on the resistance in the circuit as well as the capacitance of the capacitor. You may say there is no resistance in the circuit, but there is always resistance in a real circuit.
The voltage is given by V = Vs(1 - e^(-t/RC))
V is the voltage as a function of time
Vs is the voltage of the source
e is Euler's number = 2.71828 ...
t is time in seconds
R is the resistance in ohms
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+0
# What is (x+3) (x-1)
0
362
3
What is (x+3) (x-1)
Aug 19, 2015
### Best Answer
#3
+102763
+5
Yes this is algebra.
When ever you use letters inplace of numbers it is algebra :)
Aug 19, 2015
### 3+0 Answers
#1
+102763
+5
(x+3) (x-1)
Take the first term from the first bracket that is x and times it by the second bracket. x(x-1)
now take the second term from the first bracket that is +3 and times it by the second bracket. +3(x-1)
so you have
\$\$\\ x(x-1) +3(x-1)\\
=x^2-1x+3x-3\\
=x^2+2x-3\$\$
.
Aug 19, 2015
#2
+5
Melody is this algebra?
Aug 19, 2015
#3
+102763
+5
Best Answer
Yes this is algebra.
When ever you use letters inplace of numbers it is algebra :)
Melody Aug 19, 2015
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Lesson: Triangles
Exercise 7.1
Question: 1
Which of the following is not a criterion for
congruence of triangles?
(a) SAS
(b) ASA
(c) SSA
(d) SSS
Solution:
c
The rule of congruence of triangles are SSS
(Side-Side-Side), SAS (Side-Angle-Side), ASA
(Angle-Side-Angle), AAS and RHS (Right
angle-Hypotenuse-Side). So, SSA is not a criterion for
congruence of triangles.
Question: 2
If
,AB QR BC PR
and
CA PQ
, then:
(a)
ABC PQR
(b)
CBA PRQ
(c)
BAC RPQ
(d)
PQR BCA
Solution:
b
Since,
,AB QR BC RP
and
CA PQ
A corresponds to Q, B corresponds to R &C
corresponds to P.
i.e.,
ABC QRP
which can also be written as
.CBA PRQ
Question: 3
In
and
50 .B
Then
C
is
equal to:
(a) 40°
(b) 50°
(c) 80°
(d) 130°
Solution:
b
Given: In
,ABC AB AC
and
50B
To find:
C
In
, .ABC AB AC
B C
(
Angles opposite to equal sides are equal).
Since
50B
(Given)
50C
Question: 4
In
,ABC BC AB
and
80 .B
Then
A
is
equal to:
(a) 80°
(b) 40°
(c) 50°
(d) 100°
Solution:
c
Given: In
,ABC BC AB
and
80 .B
To find:
A
In
,ABC BC AB
. iC A
( Angles opposite to equal sides
are equal)
Since, the sum of all the angles of a triangle is 180°
180
80 180 80
A B C
A C B
80 180A A
(Using eqn. (i))
2 180 80 100A
100
2
A
50A
Question: 5
In
,PQR R P
and
4 cmQR
and
5 cm.PR
Then the length of PQ is:
(a) 4 cm
(b) 5 cm
(c) 2 cm
(d) 2.5 cm
Solution:
a
Given:
, , 4 cmPOR R P QR
and
5 cm.PR
In
, .PQR R P
PQ QR
(
Sides opposite to equal angles of a
triangle are equal)
Given
4 cm
4 cm
PQ QR
PQ
Hence, the length of PQ is 4 cm.
Question: 6
D is a point on the side BC of
ABC
bisects
.BAC
Then,
(a)
BD CD
(b)
BA BD
(c)
BD BA
(d)
CD CA
Solution:
b
Given: In
bisects
BAC
and D is a point
on side BC.
In
is an exterior angle.
[
Exterior angle > interior opposites angle]
Using (i) & (ii), we have
BA BD
(In a triangle, side opposite to the greater
angle is greater)
Similarly,
In
is an exterior angle.
[
Exterior angle > interior
opposites angle]
Using (i) & (iii), we have
AC CD
(In a triangle, side opposite to the greater
angle is greater).
Hence, using both the results we conclude that, option
(a), (c) & (d) cannot be true. Hence, the correct answer
is option (b).
Question: 7
It is given that
ABC FDE
and
5 cm,AB
40B
and
80 .A
Which one of the following
is true?
a 5 cm, 60°
b 5 cm, 60°
c 5 cm, 60°
d 5 cm, 0
°
4
DF F
DF E
DE E
DE D
Solution:
b
Given:
ABC FDE
and
5 cmAB ,
40B
and
80A
In ABC,
80A
and
40B
Since, the sum of all the angles of a triangle is 180°
180
80 40 180
180 120
60 i
A B C
C
C
C
Since,
ABC FDE
AB FD
[By CPCT]
5 cm FD
[Given
5 cmAB
]
We can write
5 cmFD
And also,
C E
[By CPCT]
60 E
[Using (i)]
Question: 8
Two sides of a triangle are of lengths 5 cm and 1.5 cm.
The length of the third side of the triangle cannot be:
(a) 3.6 cm
(b) 4.1 cm
(c) 3.8 cm
(d) 3.4 cm
Solution:
d
Let us consider, ABC with sides
5 cmBC
and
1.5 cm.CA
In a triangle, the difference of two sides < third side
and sum of two sides > third side.
BC CA AB
and
BC CA AB
5 1.5 AB
and
5 1.5 AB
3.5 AB
and
6.5 AB
Here, all the options except (d) satisfy the inequality.
Question: 9
In
,PQR
if
R Q
, then:
a
b
c
d
QR PR
PQ PR
PQ PR
QR PR
Solution:
b
In
, .PQR R Q
PQ PR
[Side opposite to greater angle is longer]
Question: 10
In triangles ABC and PQR,
,AB AC C P
and
.B Q
The two triangles are:
(a) Isosceles but not congruent
(b) Isosceles and congruent
(c) Congruent but not isosceles
(d) Neither congruent nor isosceles
Solution:
a
Given, In
, .ABC AB AC
. i C B
[Angles opposite to equal sides
are equal]
So, ABC is an isosceles triangle.
Now, it is also given that
C P
and
.B Q
P Q
[Using eq. (i)]
Now, In
,PQR
P Q
[Proved above]
QR PR
[Sides opposite to equal angles are equal]
So,
PQR
is an isosceles triangle.
Therefore, both triangles are isosceles.
But, corresponding sides of the triangles are not equal
and AAA is not a criterion for the congruence of
triangles. Hence, the triangles are not congruent.
Question: 11
In triangles ABC and
,DEF AB FD
and
.A D
The two triangles will be congruent by SAS axiom if:
a
b
c
d
BC EF
AC DE
AC EF
BC DE
Solution:
b
Given, triangles
ABC
and
,DEF AB FD
and
.A D
Two triangles are congruent by SAS axiom if two
corresponding sides and the included angle are equal.
i.e.,
ABC DEF
Therefore, if
AC DE
, the triangles are congruent by
SAS axiom.
Exercise 7.2
Question: 1
In triangles ABC and
,PQR A Q
and
.B R
Which side of PQR should be equal to side AB of
ABC so that the two triangles are congruent?
Solution:
Given,
∆ABC & ∆PQR,
A Q
and
.B R
AB and QR are the included sides.
So, if
AB QR
, then
ABC QRP
(By ASA axiom).
Question: 2
In triangles ABC and PQR,
A Q
and
.B R
Which side of PQR should be equal to side BC of
ABC so that the two triangles are congruent?
Solution:
Given, in triangles ABC and PQR,
A Q
and
.B R
RP is corresponding side to BC.
So, if
BC RP
, then
ABC QRP
(By AAS axiom).
Question: 3
“If two sides and an angle of one triangle are equal to
two sides and an angle of another triangle, then the
two triangles must be congruent. Is the statement true?
Why?
Solution:
No; because with respect to congruency rule, two
sides and the included angle of one triangle are equal
to two sides and the included angle of another triangle,
which is called SAS rule.
Question: 4
“If two angles and a side of one triangle are equal to
two angles and a side of another triangle, then the two
triangles must be congruent.” Is the statement true?
Why?
Solution:
No, because side of one triangle must be equal to the
corresponding side of the other triangle.
Question: 5
Is it possible to construct a triangle with sides
measuring 4 cm, 3 cm and 7 cm? Give reason for your
Solution:
No; because in a triangle the sum of any two sides is
greater than the third side, but
4 3 7.
Question: 6
It is given that
.ABC RPQ
Is it true to say that
BC QR
? Why?
Solution:
No; because as per the congruency rule, sides &
angles of one triangle should be equal to the
corresponding sides and angle of the other triangle.
Here,
ABC RPQ
, , AB RP BC PQ AC RQ
[By CPCT]
Hence, BC is not necessarily equal to QR.
Question: 7
If
PQR EDF
, then is it true to say that
PR EF
?
Solution:
Yes; because as per the congruency rule, sides &
angles of one triangle should be equal to the
corresponding sides and angle of the other triangle. If
,PQR EDF
, , PQ ED QR DF RP FE
[By CPCT]
Question: 8
In
, 70PQR P
and
30 .R
Which side of this
Solution:
In
,PQR
70 , 30
180
70 30 180
180 70 30
180 100
80
P R
P Q R
Q
Q
Therefore Q is largest angle.
This means PR is the longest. (Side opposite to the
largest angle is longest)
Question: 9
AD is a median of the triangle ABC. Is it true that
Solution:
In triangle ABD,
(Sum of any two sides of a triangle is greater than the
third side)
In triangle ACD,
(Sum of any two sides of a triangle is greater than the
third side)
2
2
2
AB BC AC AD BD CD BC
Hence, it is proved.
Question: 10
M is a point on side BC of a triangle ABC such that
AM is the bisector of
.BAC
Is it true to say that
perimeter of the triangle is greater than 2AM? Give
Solution:
In triangle ABC, AM is the bisector of angle BAC.
In triangle ABM. we have
.. 1AB BM AM
[Sum of any two sides of a
triangle is always greater than its third side]
Similarly, in triangle AMC, we have
AC CM AM
[Sum of any two sides of a triangle
is always greater than its third side] (2)
On adding (1) and (2) we get,
( ) 2
2
AB BM AC CM AM AM
AB BM CM AC AM
AB BC AC AM
Hence, it is proved
Question: 11
Is it possible to construct a triangle with sides
measuring 9 cm, 7 cm and 17 cm? Give reason for
Solution:
No. Because, in any triangle, the sum of any two sides
should be greater than the third side.
But in this case,
9 7 16 17.
Question: 12
Is it possible to construct a triangle with sides
measuring8 cm, 7 cm and4 cm? Give reason for your
Solution:
Yes, because in each case the sum of two sides is
greater than the third side.
. ., 7 4 8,
8 4 7,
7 8 4.
i e
Exercise 7.3
Question: 1
ABC is an isosceles triangle with
AB AC
and
BD
and
CE
are its two medians. Show that
BD CE
.
Solution:
Given:
,AB AC BD
and CE are two medians.
In
ABD
and
,ACE
A A
Common
AB AC
Given
(
D
is midpoint of
,AC
and
E
is midpoint of
.AB
)
ABD ACE
(By SAS property)
BD CE
(By CPCT)
Question: 2
In Fig.7.4, D and E are points on side BC of a ABC
such that
BD CE
and
Show that
.ABD ACE
Solution:
Given: In ABC,
BD EBC
and
(Given)
2 1
(Angles opposite to equal sides of a triangle
are equal.)
1 3 180 ................(i)
2 4 180 ...............(ii)
[Linear pair axiom]
From (i) and (ii)
1 3 2 4
1 2
3 4
In ABD and ACE
(Given)
3 4
(As proved above)
BD CE
(Given)
Δ ΔABD ACE
(By
SAS
property)
Question: 3
CDE is an equilateral triangle formed on a side CD of
a square ABCD (Fig. 7.5). Show that
Solution:
In
,DEC DE CE
(All sides of equilateral are
equal)
...(i)4 3
[Angles opposite to equal sides of a
triangle are equal.]
And,
2 1 90
(ii)
From (i) and (ii),
4 2 1 3
(Sides of square)
by SAS property.
Question 4
In Fig. 7.6,
,BA AC DE DF
such that
BA DE
and
.BF EC
Show that
.ABC DEF
Solution:
In
ABC
and
DEF
90A D
(Given)
BF CE
(Sides opposite to equal angles are equal)
BF CF CE CF
BC EF
AB DE
(Given)
ABC DEF
(By RHS property)
Question: 5
Q is a point on the side SR of a PSR such
that
PQ PR
. Prove that
.PS PQ
Solution:
In
,PRQ
PR PQ
1 R
(Angles opposite to equal sides of a
triangle are equal.)
But,
1 S
(
1
is an exterior angle for
PQS
)
( 1 )R S R
PS PR
(
In any triangle, side opposite to larger
angle is longer)
)( PR PQPS PQ
Hence, the result.
Question: 6
S is any point on side QR of PQR. Show that
2 .PQ QR RP PS
Solution:
In PQS,
... iPQ QS PS
(
Sum of two sides of a is
greater than the third side)
In PSR,
i( i)RS PR PS
(
Sum of two sides of a
is greater than the third side)
Adding eq. (i) and eq. (ii), we get
2
2 ( ).
PQ QS SR PR PS
PQ QR PR PS QS SR QR
Question: 7
D is any point on side AC of a ABC with
.AB AC
Show that
.CD BD
Solution:
In ABC, Join BD
AB AC
(Given)
C ABC
(Angles opposite to equal sides of a
triangle are equal.)
But,
) (CBD ABCAB A CBDC BD
C CBD
(
C ABC
, as proved above)
Then, in DBC,
BD CD
(As the side opposite to greater angle is longer.)
Question: 8
In Fig. 7.7, l ||m and M is the midpoint of a line
segment AB. Show that M is also the midpoint of any
line segment CD, having its end points on l and m,
respectively.
Solution:
Given: l
m, M is midpoint of line segment AB, i.e.,
AM MB
We need to prove that
.MC MD
Let us consider, AMC and BMD.
Since,
l m
and AB is the transversal,
1 2
(Alternate interior angles)
AM BM
(Given)
3 4
(Vertical opposite angles)
AMC BMD Δ
by ASA property
MC MD
(By CPCT)
Question 9
Bisectors of the angles B and C of an isosceles triangle
with
AB AC
intersect each other at O. BO is
produced to a point M. Prove that
.MOC ABC
Solution:
In
OB
and
OC
are the angle bisectors of
B
and
C
respectively.
In
,ABC
AB AC
(Given)
ACB ABC
(Angles opposite to equal sides of
a triangle are equal.)
1 1
2 2
ACB ABC
2 1... i
bisects
bisects
BO ABC
CO ACB
1 2MOC
(Exterior angle property of
triangle.)
2 1MOC
(Using (i))
( 2 1)MOC ABC ABC
Hence, it is proved.
Question 10
Bisectors of the angles B and C of an isosceles triangle
ABC, with
,AB AC
intersect each other at O. Show
that the external angle adjacent to
ABC
is equal to
BOC
Solution:
In
,ABC AB AC
(Given)
ACB ABC
(Angles opposite to equal sides of
a triangle are equal.)
1 1
2 2
ACB ABC
2 1
(
BO
bisects
,ABC CO
bisects
ACB
)
In
,BOC
1 2 180BOC
(Angle sum property of
triangle.)
1 1 180BOC
(
1 2,
as proved
above)
2 1 180BOC
...180 (i)ABC BOC
(
BO
bisects
ABC
)
But,
....180 (ii)ABC DBA
(
DBC is a line segment)
On comparing eqn. (i) and (ii), we get,
DBA BOC
Question 11
In Fig. 7.8, AD is the bisector of
BAC
. Prove
that
.AB BD
Solution:
In ABC, AD is angle bisector of
.BAC
BAC
1 2
But,
[
is an exterior angle to
AB BD
(Side opposite to the larger angle is longer)
Hence, it is proved.
Exercise 7.4
Question: 1
Find all the angles of an equilateral triangle.
Solution:
Given,
,ABC AB AC
is an equilateral triangle.
ACB ABC
(Angle opposite to equal sides of a
triangle is equal.)
1 1
2 2
ACB ABC
2 1
(
BO
bisects
,ABC CO
bisects
ACB
)
In
,BOC
1 2 180BOC
(Angle sum property of a
triangle)
1 1 180BOC
(
1 2,
as proved
above)
2 1 180BOC
...180 (i)ABC BOC
(
BO
bisects
ABC
)
But,
....180 (ii)ABC DBA
(
DBC
is a line segment)
On comparing equations (i) and (ii), we get,
DBA BOC
Question: 2
The image of an object placed at a point A before a
plane mirror LM is seen at the point B by an observer
at D as shown in Fig. 7.12. Prove that the image is as
far behind the mirror as the object is in front of the
mirror.
[Hint: CN is perpendicular to the mirror.
Also, angle of incidence
angle of reflection].
Solution:
To prove:
AE BE
Proof:
&CN LM AB LM
AB NC
... iA i
(Alternate interior angles)
... iiB r
(Corresponding angles)
... iiii r
(Incident angle
reflected angle)
Using (i), (ii) and (iii), we get,
A B
In
CEB
and
CEA
B A
(As proved above)
( )
0
1 2 90 AB LM
CE CE
(Common)
CEB CEA
by AAS rule
BE AE
(By CPCT)
Question: 3
ABC is an isosceles triangle with
AB AC
and D is
a point on BC such that
(Fig. 7.13). To
prove that
a student proceeded as
follows:
In
ABD
and
,ACD
AB AC
(Given)
B C
(Because
AB AC
) and
Therefore,
ABD ACD
(AAS)
So,
(CPCT)
What is the error in the above proof?
(Hint: Recall how
B C
is proved when
AB AC
).
Solution:
AB AC
cannot conclude that
ABD ACD
.
Here is the correct proof.
In
and
( )
0
AB AC
(Given)
(Common)
(By RHS)
(By CPCT)
Question: 4
P is a point on the bisector of
ABC
.
If a line through P, parallel to BA, meet BC at Q,
prove that BPQ is an isosceles triangle.
Solution:
2 3 ... i
(
BX
bisects
ABC
)
BA QP
(Given)
3 1 ... ii
(Alternate interior angles)
From (i) and (ii)
1 2
In
,BPQ
BQ PQ
(Converse of isosceles
property)
BPQ
is an isosceles triangle.
Question: 5
ABCD is a quadrilateral in which
AB BC
and
Show that BD bisects both the angles ABC
Solution:
In
DAB
and
.DCB
AB CB
(Given)
(Given)
DB DB
(Common)
DAB
DCB
by SSS property
1 2
3 4
DB
bisects
D
and
B
Question: 6
ABC is a right triangle with
.AB AC
Bisector of
A
meets
BC
at D. Prove that
Solution:
In right angle
,ABC
AB AC
Also,
BC
is the hypotenuse (
Hypotenuse longest
side)
90BAC
In
and
AC AB
(Given)
1 2
(
is an angle bisector of
A
)
(Common)
by SAS property
CD BD
(By CPCT)
and
....(i)ACD ABD
(By CPCT)
In ABC,
180A B C
90 180B B
(
90A
and using (i))
2 180 90 90B
45 45B C
(Using (i))
Also,
1 45
and
2 45
(
bisects
A
and
90A
)
1 C
and
2 B
In
Similarly, In
(Sides opposite to
equal angles are equal)
Hence,
(Using
above equations)
Therefore,
Question: 7
O is a point in the interior of a square ABCD such that
OAB is an equilateral triangle. Show that
OCD
is
an isosceles triangle.
Solution:
1 2 90 ..... i
(
ABCD
is a square)
3 4 60 ...... ii
(
OAB
is an equilateral
triangle)
Using (i) and (ii)
1 3 2 4 90 60
5 6 30
In
DAO
and
CBO
(Given)
5 6
(As proved above)
AO BO
(Given)
DAO CB
by SAS property
OD OC
OCD
is an isosceles triangle.
Question: 8
ABC and DBC are two triangles on the same base BC
such that A and D lie on the opposite sides of BC,
AB=AC and DB=DC. Show that AD is the
perpendicular bisector of BC.
Solution:
In triangle ABD and ACD
AB AC
(Given)
DB DC
(Given)
(Common)
Therefore,
ABD
is congruent to
.ACD
(By SSS
congruency rule)
So,
By CPCT1 2
Now, in triangle AOB and AOC
AB = AC (Given)
Proved above1 2
AO AO
(Common side)
Therefore,
AOB
is congruent to
AOC
So,
By CPCT3 4
and
BO OC
(By CPCT)
Since
3 4
Therefore,
3 4 90
Hence, proved that AD is the perpendicular bisector of
BC.
Question: 9
ABC
is an isosceles triangle in which
and
BE
are two Altitudes on sides
BC
and
AC
respectively. Prove that
.AE BD
Solution:
In triangles AEB and BDA,
1 2 90
In
,ABC
AC BC
[Given]
4 3
[By isosceles triangle property]
AB BA
(Common)
AEB BDA
(By AAS rule)
AE BD
(By CPCT)
Question: 10
Prove that sum of any two sides of a triangle is greater
than twice the median.
Solution:
Let us consider a ABC such that AD is a median on
BC i.e.
.BD BC
We need to prove that
Let’s produce AD to E such that
and join EC.
Now, in
and
EDC
(By construction)
1 2
(Vertically opposite angle)
DB DC
(Given)
By SAS property
So,
AB EC
(By CPCT)
Now, In
,AEC
AC CE AE
(
AB CE
as proved above
and
by construction)
Hence, it is proved.
Question: 11
, 2 .ABCD AB BC CD DA BD AC
Solution:
In
,AOB
iAO OB AB
(Sum of two sides is greater
than the third side in a triangle)
In
,BOC
ii OB OC BC
(Sum of two sides is greater than
the third side in a triangle)
In
,COD
iiiOC OD CD
(Sum of two sides is greater than
the third side in a triangle)
In
,DOA
(Sum of two sides is greater
than the third side in a triangle)
Adding (i), (ii), (iii) and (iv)
2
( )
2
AO OB OB OC OC OD OD OA
AB BC CD DA
AO OC BO OD AB BC CD DA
AC BD AB BC CD DA
Hence, it is proved.
Question: 12
Show that in a quadrilateral ABCD,
AB+BC+CD+DA>AC+BD.
Solution:
AB BC CD DA AC BD
In
,ABC
iAB BC AC
(Sum of two sides is greater
than the third side in a triangle)
In
,BCD
iiBC CD BD
(Sum of two sides is greater
than the third side in a triangle)
In
,CDA
iiiCD DA AC
(Sum of two sides is greater
than the third side in a triangle)
In
,DAB
(Sum of two sides is greater
than the third side in a triangle)
Adding (i), (ii), (iii) and (iv)
( )
( )
( ) ( )2 2
AB BC BC CD CD DA AD AB
AC BD AC BD
AB BC CD DA AC BD
AB BC CD DA AC BD
Hence, it is proved.
Question: 13
In a triangle ABC, D is the mid-point of side AC such
that
1
.
2
BD AC
Show that
ABC
is a right angle.
Solution:
To prove:
90ABC
1
... i
2
BD AC
(Given)
and
1
..... ii
2
(
D
is midpoint of
AC
)
From (i) and (ii)
In
DAB
1 A
(Angle opposite equal sides are
equal) ... …(iii)
In
Δ ,DBC
BD CD
2C
(Angle opposite equal sides are
equal) ... ….(iv)
In
,ABC
180A ABC C
(Angle sum property of
triangle)
1 2 180ABC
1, 2A C
1 2 180ABC
180ABC ABC
( 1 2 )ABC
2 180
90
ABC
ABC
Hence, it is proved.
Question: 14
In a right triangle, prove that the line-segment joining
the mid-point of the hypotenuse to the opposite vertex
is half the hypotenuse.
Solution:
Let us consider a right
ABC
with
90ABC
and let E be the midpoint of hypotenuse AC.
We need to prove
1
2
BE AC
.
Producing
BE
to
D
s.t.
ED BE
Join A to D and C to D
Since,
=ED EB
(By construction)
and
EA EC
(Given)
ABCD
is a parallelogram
90ABC
Also, in a parallelogram, if one angle is right, then all
angle are right.
Parallelogram
ABCD
rectangle
AC BD
(Diagonals of a rectangle are equal)
1 1
2 2
AC BD
1 1
2 2
AC BE BE ED BD
Hence, it is proved.
Question: 15
Two lines l and m intersect at the point O and P is a
point on a line n passing through the point O such that
P is equidistant from l and m. Prove that n is the
bisector of the angle formed by l and m.
Solution:
We draw
PQ l
and
.PR m
Now, in
OQP
and
,ORP
1 2 90
(By construction)
OP OP
(Common)
PQ PR
(Given P is equidistant from l and m)
OQP ORP
By RHS property
By CPCT
3 4
n
is the bisector of
QOR
Question: 16
Line segment joining the mid-points M and N of
parallel sides AB and DC, respectively of a trapezium
ABCD is perpendicular to both the sides AB and DC.
Prove that
.
Solution:
Joining AN and BN,
in
AMN
and
,BMN
AM BM
(Given)
1 2 90
NM NM
(Common)
AMN BMN
by SAS property
3 4
and,
...(i)AN BN
(By CPCT)
Now,
90DNM CNM
5 3 6 4
....... (5 6 (ii) 3 4)
Now, In
and
,BCN
DN CN
(Given)
5 6
(Using (ii))
AN BN
(Using (i))
by SAS property
(By CPCT)
Question: 17
ABCD is a quadrilateral such that diagonal AC bisects
the angles A and C. Prove that
and
.CB CD
Solution:
Given: In quadrilateral ABCD, diagonal AC bisects
A
and
.C
i.e.,
1 2
and
3 4
Now, in
ABC
and
1 2
(Given)
AC AC
3 4
(Given)
by ASA Property
(By CPCT)
CB CD
(By CPCT)
Question: 18
If ABC is a right angled triangle such that
AB AC
and bisector of angle C intersects the side AB at D.
Prove that
Solution:
Given: In figure ABC is a right angle,
AB AC
, CD
is bisector of
C
Construction: Draw
DE BC
In right angle
,ABC
AB AC
BC is the hypotenuse
90A
In
DAC
and
DEC
3 90A
1 2
(Given)
DC DC
(Common)
DAC DEC
by AAS rule.
... iDA DE
(CPCT)
... iiCA CE
(CPCT)
In
BAC
AB AC
C B
(Angle opposite equal sides are equal)
Now,
180A B C
90 2 180
2 90
45
B B C
B
B
In
BED
5 180 4B
180 45 90
180 135
45
5B
... iiiDE BE
(Sides opposite equal angles are equal)
From (i) and (iii)
... ivDA DE BE
Now,
BC CE BE
CA DA
(Using (ii) and (iii)
Hence, it is proved.
Question: 19
AB and CD are the smallest and largest sides of a
B
and
D
decide
which is greater.
Solution:
AB is the shortest and CD is the longest side of ABCD.
We need to prove
B D
or
D B
.
Joining BD.
Now, in
,ABD
1 3... i
(Angle opposite to the longer side is greater)
In
BCD
CD BC
2 4...... ii
2 4...... ii
(Angle opposite to the longer side is greater)
Adding equation (i) and (ii), we get
1 2 3 4
B D
Question: 20
Prove that in a triangle, other than in an equilateral
triangle, angle opposite the longest side is greater than
2
3
of a right angle.
Solution:
Let us consider
ABC
with BC as the longest side.
We need to prove
2
90
3
BAC
In
,ABC
BC AB
(As BC is the longest side)
..... iA C
(In a
Δ
angle opposite the longer side is greater)
Similarly,
BC AC
(As BC is the longest side)
... iiBA
Adding (i) and (ii) we get,
2 A B C
A
on both the sides,
0
3
3 180
180
3
2
90
3
A A B C
A
A
2
3
A
right angle.
Question: 21
Prove that AC is the perpendicular bisector of BD.
Solution:
In
ABCD
,
and
CB CD
We Join AC and BD.
Let AC and BD intersect at point O.
In
ABC
and
(Given)
BC CD
AC AC
by SSS property
1 2
(CPCT)
In
AOB
and
AOD
(Given)
1 2
(As proved above)
AO AO
AOB AOD
by SAS property
BO DO
(CPCT)
3 4
But,
3 4 180
(Linear pair axiom)
2 3 180
3 90
4 90
Hence,
AC
is perpendicular bisector of
3 90
.B
BO DO
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Peeter Joot [email protected]
Thermodynamic identities Impressed with the clarity of Baez’s entropic force discussion on differential forms [1], let’s use that methodology to find all the possible identities that we can get from the thermodynamic identity (for now assuming N is fixed, ignoring the chemical potential.) This isn’t actually that much work to do, since a bit of editor regular expression magic can do most of the work. Our starting point is the thermodynamic identity ¯ + dW ¯ = TdS − PdV, dU = dQ
(1.1)
0 = dU − TdS + PdV.
(1.2)
or It’s quite likely that many of the identities that can be obtained will be useful, but this should at least provide a handy reference of possible conversions. Differentials in P, V
This first case illustrates the method.
0= dU −TdS + PdV ∂U ∂U ∂S ∂S = dP + dV − T dP + dV + PdV ∂P V ∂V P ∂P ∂V P V ∂U ∂S ∂U ∂S = dP −T + dV −T +P . ∂P V ∂P V ∂V P ∂V P Taking wedge products with dV and dP respectively, we form two two forms ∂U ∂S −T 0 = dP ∧ dV ∂P V ∂P V ∂U ∂S 0 = dV ∧ dP −T +P . ∂V P ∂V P Since these must both be zero we find
∂U ∂P
=T V
1
∂S ∂P
(1.3)
(1.4a) (1.4b)
(1.5a) V
P=−
∂U ∂V
−T P
∂S ∂V
.
(1.5b)
P
Differentials in P, T 0 = dU −TdS + PdV (1.6) ∂U ∂S ∂S ∂V ∂V ∂U dP + dT − T dP + dT + dP + dT, = ∂P T ∂T P ∂P T ∂T P ∂P T ∂T P or
∂U ∂S ∂V 0= −T + ∂P T ∂P T ∂P T ∂U ∂S ∂V 0= −T + . ∂T P ∂T P ∂T P
(1.7a) (1.7b)
Differentials in P, S 0 = dU −TdS + PdV ∂U ∂V ∂V ∂U dP + dS − TdS + P dP + dS , = ∂P S ∂S P ∂P S ∂S P or
∂V ∂U = −P ∂P S ∂P S ∂U ∂V T= +P . ∂S P ∂S P
(1.8)
(1.9a) (1.9b)
Differentials in P, U 0 = dU − TdS + PdV ∂S ∂S ∂V ∂V = dU − T dP + dU + P dP + dU , ∂P U ∂U P ∂P U ∂U P or
∂S ∂U
∂V ∂U P ∂V ∂S =P . T ∂P U ∂P U
0 = 1−T
2
(1.10)
+P
(1.11a) P
(1.11b)
Differentials in V, T 0 = dU −TdS + PdV ∂U ∂U ∂S ∂S = dV + dT − T dV + dT + PdV, ∂V T ∂T V ∂V T ∂T V or
∂S 0= −T +P ∂V T T ∂U ∂S =T . ∂T V ∂T V ∂U ∂V
(1.12)
(1.13a) (1.13b)
Differentials in V, S 0 = dU −TdS + PdV ∂U ∂U dV + dS − TdS + PdV, = ∂V S ∂S V or
∂U P=− ∂V S ∂U T= . ∂S V
(1.14)
(1.15a) (1.15b)
Differentials in V, U 0 = dU − TdS + PdV ∂S ∂V ∂V ∂S dV + dU + P dV + dU = dU − T ∂V U ∂U V ∂V U ∂U V or
∂S ∂V 0 = 1−T +P ∂U V ∂U V ∂S ∂V T =P . ∂V U ∂V U
(1.16)
(1.17a) (1.17b)
Differentials in S, T 0 = dU − TdS + PdV ∂U ∂U ∂V ∂V = dS + dT − TdS + P dS + dT , ∂S T ∂T S ∂S T ∂T S or
∂U ∂V 0= −T+P ∂S T ∂S T ∂U ∂V 0= +P . ∂T S ∂T S
(1.18)
3
(1.19a) (1.19b)
Differentials in S, U 0 = dU − TdS + PdV = dU − TdS + P or
∂V ∂S
dS + U
∂V ∂U
dU
(1.20)
S
∂V 1 =− P ∂U S ∂V T=P . ∂S U
(1.21a) (1.21b)
Differentials in T, U 0 = dU − TdS + PdV ∂S ∂V ∂V ∂S dT + dU + P dT + dU , = dU − T ∂T U ∂U T ∂T U ∂U T or
(1.22)
∂S ∂V 0 = 1−T +P ∂U T ∂U T ∂V ∂S =P . T ∂T U ∂T U
4
(1.23a) (1.23b)
Bibliography [1] John Baez. Entropic forces, 2012. URL http://johncarlosbaez.wordpress.com/2012/02/01/ entropic-forces/. [Online; accessed 07-March-2013]. 1
5
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# Playfair Cipher with Examples
• Difficulty Level : Easy
• Last Updated : 28 Dec, 2022
The Playfair cipher was the first practical digraph substitution cipher. The scheme was invented in 1854 by Charles Wheatstone but was named after Lord Playfair who promoted the use of the cipher. In playfair cipher unlike traditional cipher we encrypt a pair of alphabets(digraphs) instead of a single alphabet.
It was used for tactical purposes by British forces in the Second Boer War and in World War I and for the same purpose by the Australians during World War II. This was because Playfair is reasonably fast to use and requires no special equipment.
Encryption Technique
For the encryption process let us consider the following example:
The Playfair Cipher Encryption Algorithm:
The Algorithm consists of 2 steps:
1. Generate the key Square(5×5):
• The key square is a 5Ă—5 grid of alphabets that acts as the key for encrypting the plaintext. Each of the 25 alphabets must be unique and one letter of the alphabet (usually J) is omitted from the table (as the table can hold only 25 alphabets). If the plaintext contains J, then it is replaced by I.
• The initial alphabets in the key square are the unique alphabets of the key in the order in which they appear followed by the remaining letters of the alphabet in order.
2. Algorithm to encrypt the plain text: The plaintext is split into pairs of two letters (digraphs). If there is an odd number of letters, a Z is added to the last letter.
For example:
```PlainText: "instruments"
After Split: 'in' 'st' 'ru' 'me' 'nt' 'sz'```
1. Pair cannot be made with same letter. Break the letter in single and add a bogus letter to the previous letter.
Plain Text: “hello”
After Split: ‘he’ ‘lx’ ‘lo’
Here ‘x’ is the bogus letter.
2. If the letter is standing alone in the process of pairing, then add an extra bogus letter with the alone letter
Plain Text: “helloe”
AfterSplit: ‘he’ ‘lx’ ‘lo’ ‘ez’
Here ‘z’ is the bogus letter.
Rules for Encryption:
• If both the letters are in the same column: Take the letter below each one (going back to the top if at the bottom).
For example:
```Diagraph: "me"
Encrypted Text: cl
Encryption:
m -> c
e -> l```
•
•
• If both the letters are in the same row: Take the letter to the right of each one (going back to the leftmost if at the rightmost position).
For example:
```Diagraph: "st"
Encrypted Text: tl
Encryption:
s -> t
t -> l```
•
•
• If neither of the above rules is true: Form a rectangle with the two letters and take the letters on the horizontal opposite corner of the rectangle.
For example:
```Diagraph: "nt"
Encrypted Text: rq
Encryption:
n -> r
t -> q```
•
•
For example:
```Plain Text: "instrumentsz"
Encrypted Text: gatlmzclrqtx
Encryption:
i -> g
n -> a
s -> t
t -> l
r -> m
u -> z
m -> c
e -> l
n -> r
t -> q
s -> t
z -> x```
Below is an implementation of Playfair Cipher in C:
## C++
`// C++ program to implement Playfair Cipher` `#include ` `using` `namespace` `std;` `#define SIZE 30` `// Function to convert the string to lowercase` `void` `toLowerCase(``char` `plain[], ``int` `ps)` `{` ` ``int` `i;` ` ``for` `(i = 0; i < ps; i++) {` ` ``if` `(plain[i] > 64 && plain[i] < 91)` ` ``plain[i] += 32;` ` ``}` `}` `// Function to remove all spaces in a string` `int` `removeSpaces(``char``* plain, ``int` `ps)` `{` ` ``int` `i, count = 0;` ` ``for` `(i = 0; i < ps; i++)` ` ``if` `(plain[i] != ``' '``)` ` ``plain[count++] = plain[i];` ` ``plain[count] = ``'\0'``;` ` ``return` `count;` `}` `// Function to generate the 5x5 key square` `void` `generateKeyTable(``char` `key[], ``int` `ks, ``char` `keyT[5][5])` `{` ` ``int` `i, j, k, flag = 0;` ` ``// a 26 character hashmap` ` ``// to store count of the alphabet` ` ``int` `dicty[26] = { 0 };` ` ``for` `(i = 0; i < ks; i++) {` ` ``if` `(key[i] != ``'j'``)` ` ``dicty[key[i] - 97] = 2;` ` ``}` ` ``dicty[``'j'` `- 97] = 1;` ` ``i = 0;` ` ``j = 0;` ` ``for` `(k = 0; k < ks; k++) {` ` ``if` `(dicty[key[k] - 97] == 2) {` ` ``dicty[key[k] - 97] -= 1;` ` ``keyT[i][j] = key[k];` ` ``j++;` ` ``if` `(j == 5) {` ` ``i++;` ` ``j = 0;` ` ``}` ` ``}` ` ``}` ` ``for` `(k = 0; k < 26; k++) {` ` ``if` `(dicty[k] == 0) {` ` ``keyT[i][j] = (``char``)(k + 97);` ` ``j++;` ` ``if` `(j == 5) {` ` ``i++;` ` ``j = 0;` ` ``}` ` ``}` ` ``}` `}` `// Function to search for the characters of a digraph` `// in the key square and return their position` `void` `search(``char` `keyT[5][5], ``char` `a, ``char` `b, ``int` `arr[])` `{` ` ``int` `i, j;` ` ``if` `(a == ``'j'``)` ` ``a = ``'i'``;` ` ``else` `if` `(b == ``'j'``)` ` ``b = ``'i'``;` ` ``for` `(i = 0; i < 5; i++) {` ` ``for` `(j = 0; j < 5; j++) {` ` ``if` `(keyT[i][j] == a) {` ` ``arr[0] = i;` ` ``arr[1] = j;` ` ``}` ` ``else` `if` `(keyT[i][j] == b) {` ` ``arr[2] = i;` ` ``arr[3] = j;` ` ``}` ` ``}` ` ``}` `}` `// Function to find the modulus with 5` `int` `mod5(``int` `a) { ``return` `(a % 5); }` `// Function to make the plain text length to be even` `int` `prepare(``char` `str[], ``int` `ptrs)` `{` ` ``if` `(ptrs % 2 != 0) {` ` ``str[ptrs++] = ``'z'``;` ` ``str[ptrs] = ``'\0'``;` ` ``}` ` ``return` `ptrs;` `}` `// Function for performing the encryption` `void` `encrypt(``char` `str[], ``char` `keyT[5][5], ``int` `ps)` `{` ` ``int` `i, a[4];` ` ``for` `(i = 0; i < ps; i += 2) {` ` ``search(keyT, str[i], str[i + 1], a);` ` ``if` `(a[0] == a[2]) {` ` ``str[i] = keyT[a[0]][mod5(a[1] + 1)];` ` ``str[i + 1] = keyT[a[0]][mod5(a[3] + 1)];` ` ``}` ` ``else` `if` `(a[1] == a[3]) {` ` ``str[i] = keyT[mod5(a[0] + 1)][a[1]];` ` ``str[i + 1] = keyT[mod5(a[2] + 1)][a[1]];` ` ``}` ` ``else` `{` ` ``str[i] = keyT[a[0]][a[3]];` ` ``str[i + 1] = keyT[a[2]][a[1]];` ` ``}` ` ``}` `}` `// Function to encrypt using Playfair Cipher` `void` `encryptByPlayfairCipher(``char` `str[], ``char` `key[])` `{` ` ``char` `ps, ks, keyT[5][5];` ` ``// Key` ` ``ks = ``strlen``(key);` ` ``ks = removeSpaces(key, ks);` ` ``toLowerCase(key, ks);` ` ``// Plaintext` ` ``ps = ``strlen``(str);` ` ``toLowerCase(str, ps);` ` ``ps = removeSpaces(str, ps);` ` ``ps = prepare(str, ps);` ` ``generateKeyTable(key, ks, keyT);` ` ``encrypt(str, keyT, ps);` `}` `// Driver code` `int` `main()` `{` ` ``char` `str[SIZE], key[SIZE];` ` ``// Key to be encrypted` ` ``strcpy``(key, ``"Monarchy"``);` ` ``cout << ``"Key text: "` `<< key << ``"\n"``;` ` ``// Plaintext to be encrypted` ` ``strcpy``(str, ``"instruments"``);` ` ``cout << ``"Plain text: "` `<< str << ``"\n"``;` ` ``// encrypt using Playfair Cipher` ` ``encryptByPlayfairCipher(str, key);` ` ``cout << ``"Cipher text: "` `<< str << ``"\n"``;` ` ``return` `0;` `}` `// This code is contributed by aditya942003patil`
## Java
`// Java program to implement Playfair Cipher ` `import` `java.util.*;` `public` `class` `Solution` `{` ` ``static` `int` `SIZE = ``30``;` ` ``// Function to convert the string to lowercase` ` ``static` `void` `toLowerCase(``char` `plain[], ``int` `ps)` ` ``{` ` ``int` `i;` ` ``for` `(i = ``0``; i < ps; i++) {` ` ``if` `(plain[i] > ``64` `&& plain[i] < ``91``)` ` ``plain[i] += ``32``;` ` ``}` ` ``}` ` ``// Function to remove all spaces in a string` ` ``static` `int` `removeSpaces(``char``[] plain, ``int` `ps)` ` ``{` ` ``int` `i, count = ``0``;` ` ``for` `(i = ``0``; i < ps; i++)` ` ``if` `(plain[i] != ``'\u0000'``)` ` ``plain[count++] = plain[i];` ` ``return` `count;` ` ``}` ` ``// Function to generate the 5x5 key square` ` ``static` `void` `generateKeyTable(``char` `key[], ``int` `ks, ``char` `keyT[][])` ` ``{` ` ``int` `i, j, k, flag = ``0``;` ` ``// a 26 character hashmap` ` ``// to store count of the alphabet` ` ``int` `dicty[] = ``new` `int``[``26``];` ` ``for` `(i = ``0``; i < ks; i++) {` ` ``if` `(key[i] != ``'j'``)` ` ``dicty[key[i] - ``97``] = ``2``;` ` ``}` ` ``dicty[``'j'` `- ``97``] = ``1``;` ` ``i = ``0``;` ` ``j = ``0``;` ` ``for` `(k = ``0``; k < ks; k++) {` ` ``if` `(dicty[key[k] - ``97``] == ``2``) {` ` ``dicty[key[k] - ``97``] -= ``1``;` ` ``keyT[i][j] = key[k];` ` ``j++;` ` ``if` `(j == ``5``) {` ` ``i++;` ` ``j = ``0``;` ` ``}` ` ``}` ` ``}` ` ``for` `(k = ``0``; k < ``26``; k++) {` ` ``if` `(dicty[k] == ``0``) {` ` ``keyT[i][j] = (``char``)(k + ``97``);` ` ``j++;` ` ``if` `(j == ``5``) {` ` ``i++;` ` ``j = ``0``;` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// Function to search for the characters of a digraph` ` ``// in the key square and return their position` ` ``static` `void` `search(``char` `keyT[][], ``char` `a, ``char` `b, ``int` `arr[])` ` ``{` ` ``int` `i, j;` ` ``if` `(a == ``'j'``)` ` ``a = ``'i'``;` ` ``else` `if` `(b == ``'j'``)` ` ``b = ``'i'``;` ` ``for` `(i = ``0``; i < ``5``; i++) {` ` ``for` `(j = ``0``; j < ``5``; j++) {` ` ``if` `(keyT[i][j] == a) {` ` ``arr[``0``] = i;` ` ``arr[``1``] = j;` ` ``}` ` ``else` `if` `(keyT[i][j] == b) {` ` ``arr[``2``] = i;` ` ``arr[``3``] = j;` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// Function to find the modulus with 5` ` ``static` `int` `mod5(``int` `a) { ``return` `(a % ``5``); }` ` ``// Function to make the plain text length to be even` ` ``static` `int` `prepare(``char` `str[], ``int` `ptrs)` ` ``{` ` ``if` `(ptrs % ``2` `!= ``0``) {` ` ``str[ptrs++] = ``'z'``;` ` ``str[ptrs] = ``'\0'``;` ` ``}` ` ``return` `ptrs;` ` ``}` ` ``// Function for performing the encryption` ` ``static` `void` `encrypt(``char` `str[], ``char` `keyT[][], ``int` `ps)` ` ``{` ` ``int` `i;` ` ``int``[] a =``new` `int``[``4``];` ` ``for` `(i = ``0``; i < ps; i += ``2``) {` ` ``search(keyT, str[i], str[i + ``1``], a);` ` ``if` `(a[``0``] == a[``2``]) {` ` ``str[i] = keyT[a[``0``]][mod5(a[``1``] + ``1``)];` ` ``str[i + ``1``] = keyT[a[``0``]][mod5(a[``3``] + ``1``)];` ` ``}` ` ``else` `if` `(a[``1``] == a[``3``]) {` ` ``str[i] = keyT[mod5(a[``0``] + ``1``)][a[``1``]];` ` ``str[i + ``1``] = keyT[mod5(a[``2``] + ``1``)][a[``1``]];` ` ``}` ` ``else` `{` ` ``str[i] = keyT[a[``0``]][a[``3``]];` ` ``str[i + ``1``] = keyT[a[``2``]][a[``1``]];` ` ``}` ` ``}` ` ``}` ` ``// Function to encrypt using Playfair Cipher` ` ``static` `void` `encryptByPlayfairCipher(``char` `str[], ``char` `key[])` ` ``{` ` ``int` `ps; ` ` ``int` `ks;` ` ``char``[][] keyT = ``new` `char``[``5``][``5``];` ` ``// Key` ` ``ks = key.length;` ` ``ks = removeSpaces(key, ks);` ` ``toLowerCase(key, ks);` ` ``// Plaintext` ` ``ps = str.length;` ` ``toLowerCase(str, ps);` ` ``ps = removeSpaces(str, ps);` ` ``ps = prepare(str, ps);` ` ``generateKeyTable(key, ks, keyT);` ` ``encrypt(str, keyT, ps);` ` ``}` ` ``static` `void` `strcpy(``char``[] arr, String s) {` ` ``for``(``int` `i = ``0``;i < s.length();i++){` ` ``arr[i] = s.charAt(i);` ` ``}` ` ``}` ` ``// Driver code` ` ``public` `static` `void` `main(String[] args) {` ` ``char` `str[] = ``new` `char``[SIZE];` ` ``char` `key[] = ``new` `char``[SIZE];` ` ``// Key to be encrypted` ` ``strcpy(key, ``"Monarchy"``);` ` ``System.out.println(``"Key text: "` `+ String.valueOf(key));` ` ``// Plaintext to be encrypted` ` ``strcpy(str, ``"instruments"``);` ` ``System.out.println(``"Plain text: "` `+ String.valueOf(str));` ` ``// encrypt using Playfair Cipher` ` ``encryptByPlayfairCipher(str, key);` ` ``System.out.println(``"Cipher text: "` `+ String.valueOf(str));` ` ``}` `}` `// This code is contributed by karandeep1234`
## C
`// C program to implement Playfair Cipher` `#include ` `#include ` `#include ` `#define SIZE 30` `// Function to convert the string to lowercase` `void` `toLowerCase(``char` `plain[], ``int` `ps)` `{` ` ``int` `i;` ` ``for` `(i = 0; i < ps; i++) {` ` ``if` `(plain[i] > 64 && plain[i] < 91)` ` ``plain[i] += 32;` ` ``}` `}` `// Function to remove all spaces in a string` `int` `removeSpaces(``char``* plain, ``int` `ps)` `{` ` ``int` `i, count = 0;` ` ``for` `(i = 0; i < ps; i++)` ` ``if` `(plain[i] != ``' '``)` ` ``plain[count++] = plain[i];` ` ``plain[count] = ``'\0'``;` ` ``return` `count;` `}` `// Function to generate the 5x5 key square` `void` `generateKeyTable(``char` `key[], ``int` `ks, ``char` `keyT[5][5])` `{` ` ``int` `i, j, k, flag = 0, *dicty;` ` ``// a 26 character hashmap` ` ``// to store count of the alphabet` ` ``dicty = (``int``*)``calloc``(26, ``sizeof``(``int``));` ` ``for` `(i = 0; i < ks; i++) {` ` ``if` `(key[i] != ``'j'``)` ` ``dicty[key[i] - 97] = 2;` ` ``}` ` ``dicty[``'j'` `- 97] = 1;` ` ``i = 0;` ` ``j = 0;` ` ``for` `(k = 0; k < ks; k++) {` ` ``if` `(dicty[key[k] - 97] == 2) {` ` ``dicty[key[k] - 97] -= 1;` ` ``keyT[i][j] = key[k];` ` ``j++;` ` ``if` `(j == 5) {` ` ``i++;` ` ``j = 0;` ` ``}` ` ``}` ` ``}` ` ``for` `(k = 0; k < 26; k++) {` ` ``if` `(dicty[k] == 0) {` ` ``keyT[i][j] = (``char``)(k + 97);` ` ``j++;` ` ``if` `(j == 5) {` ` ``i++;` ` ``j = 0;` ` ``}` ` ``}` ` ``}` `}` `// Function to search for the characters of a digraph` `// in the key square and return their position` `void` `search(``char` `keyT[5][5], ``char` `a, ``char` `b, ``int` `arr[])` `{` ` ``int` `i, j;` ` ``if` `(a == ``'j'``)` ` ``a = ``'i'``;` ` ``else` `if` `(b == ``'j'``)` ` ``b = ``'i'``;` ` ``for` `(i = 0; i < 5; i++) {` ` ``for` `(j = 0; j < 5; j++) {` ` ``if` `(keyT[i][j] == a) {` ` ``arr[0] = i;` ` ``arr[1] = j;` ` ``}` ` ``else` `if` `(keyT[i][j] == b) {` ` ``arr[2] = i;` ` ``arr[3] = j;` ` ``}` ` ``}` ` ``}` `}` `// Function to find the modulus with 5` `int` `mod5(``int` `a) { ``return` `(a % 5); }` `// Function to make the plain text length to be even` `int` `prepare(``char` `str[], ``int` `ptrs)` `{` ` ``if` `(ptrs % 2 != 0) {` ` ``str[ptrs++] = ``'z'``;` ` ``str[ptrs] = ``'\0'``;` ` ``}` ` ``return` `ptrs;` `}` `// Function for performing the encryption` `void` `encrypt(``char` `str[], ``char` `keyT[5][5], ``int` `ps)` `{` ` ``int` `i, a[4];` ` ``for` `(i = 0; i < ps; i += 2) {` ` ``search(keyT, str[i], str[i + 1], a);` ` ``if` `(a[0] == a[2]) {` ` ``str[i] = keyT[a[0]][mod5(a[1] + 1)];` ` ``str[i + 1] = keyT[a[0]][mod5(a[3] + 1)];` ` ``}` ` ``else` `if` `(a[1] == a[3]) {` ` ``str[i] = keyT[mod5(a[0] + 1)][a[1]];` ` ``str[i + 1] = keyT[mod5(a[2] + 1)][a[1]];` ` ``}` ` ``else` `{` ` ``str[i] = keyT[a[0]][a[3]];` ` ``str[i + 1] = keyT[a[2]][a[1]];` ` ``}` ` ``}` `}` `// Function to encrypt using Playfair Cipher` `void` `encryptByPlayfairCipher(``char` `str[], ``char` `key[])` `{` ` ``char` `ps, ks, keyT[5][5];` ` ``// Key` ` ``ks = ``strlen``(key);` ` ``ks = removeSpaces(key, ks);` ` ``toLowerCase(key, ks);` ` ``// Plaintext` ` ``ps = ``strlen``(str);` ` ``toLowerCase(str, ps);` ` ``ps = removeSpaces(str, ps);` ` ``ps = prepare(str, ps);` ` ``generateKeyTable(key, ks, keyT);` ` ``encrypt(str, keyT, ps);` `}` `// Driver code` `int` `main()` `{` ` ``char` `str[SIZE], key[SIZE];` ` ``// Key to be encrypted` ` ``strcpy``(key, ``"Monarchy"``);` ` ``printf``(``"Key text: %s\n"``, key);` ` ``// Plaintext to be encrypted` ` ``strcpy``(str, ``"instruments"``);` ` ``printf``(``"Plain text: %s\n"``, str);` ` ``// encrypt using Playfair Cipher` ` ``encryptByPlayfairCipher(str, key);` ` ``printf``(``"Cipher text: %s\n"``, str);` ` ``return` `0;` `}` `// This code is contributed by AbhayBhat`
## Python3
`# Python program to implement Playfair Cipher` `# Function to convert the string to lowercase` `def` `toLowerCase(text):` ` ``return` `text.lower()` `# Function to remove all spaces in a string` `def` `removeSpaces(text):` ` ``newText ``=` `""` ` ``for` `i ``in` `text:` ` ``if` `i ``=``=` `" "``:` ` ``continue` ` ``else``:` ` ``newText ``=` `newText ``+` `i` ` ``return` `newText` `# Function to group 2 elements of a string` `# as a list element` `def` `Diagraph(text):` ` ``Diagraph ``=` `[]` ` ``group ``=` `0` ` ``for` `i ``in` `range``(``2``, ``len``(text), ``2``):` ` ``Diagraph.append(text[group:i])` ` ``group ``=` `i` ` ``Diagraph.append(text[group:])` ` ``return` `Diagraph` `# Function to fill a letter in a string element` `# If 2 letters in the same string matches` `def` `FillerLetter(text):` ` ``k ``=` `len``(text)` ` ``if` `k ``%` `2` `=``=` `0``:` ` ``for` `i ``in` `range``(``0``, k, ``2``):` ` ``if` `text[i] ``=``=` `text[i``+``1``]:` ` ``new_word ``=` `text[``0``:i``+``1``] ``+` `str``(``'x'``) ``+` `text[i``+``1``:]` ` ``new_word ``=` `FillerLetter(new_word)` ` ``break` ` ``else``:` ` ``new_word ``=` `text` ` ``else``:` ` ``for` `i ``in` `range``(``0``, k``-``1``, ``2``):` ` ``if` `text[i] ``=``=` `text[i``+``1``]:` ` ``new_word ``=` `text[``0``:i``+``1``] ``+` `str``(``'x'``) ``+` `text[i``+``1``:]` ` ``new_word ``=` `FillerLetter(new_word)` ` ``break` ` ``else``:` ` ``new_word ``=` `text` ` ``return` `new_word` `list1 ``=` `[``'a'``, ``'b'``, ``'c'``, ``'d'``, ``'e'``, ``'f'``, ``'g'``, ``'h'``, ``'i'``, ``'k'``, ``'l'``, ``'m'``,` ` ``'n'``, ``'o'``, ``'p'``, ``'q'``, ``'r'``, ``'s'``, ``'t'``, ``'u'``, ``'v'``, ``'w'``, ``'x'``, ``'y'``, ``'z'``]` `# Function to generate the 5x5 key square matrix` `def` `generateKeyTable(word, list1):` ` ``key_letters ``=` `[]` ` ``for` `i ``in` `word:` ` ``if` `i ``not` `in` `key_letters:` ` ``key_letters.append(i)` ` ``compElements ``=` `[]` ` ``for` `i ``in` `key_letters:` ` ``if` `i ``not` `in` `compElements:` ` ``compElements.append(i)` ` ``for` `i ``in` `list1:` ` ``if` `i ``not` `in` `compElements:` ` ``compElements.append(i)` ` ``matrix ``=` `[]` ` ``while` `compElements !``=` `[]:` ` ``matrix.append(compElements[:``5``])` ` ``compElements ``=` `compElements[``5``:]` ` ``return` `matrix` `def` `search(mat, element):` ` ``for` `i ``in` `range``(``5``):` ` ``for` `j ``in` `range``(``5``):` ` ``if``(mat[i][j] ``=``=` `element):` ` ``return` `i, j` `def` `encrypt_RowRule(matr, e1r, e1c, e2r, e2c):` ` ``char1 ``=` `''` ` ``if` `e1c ``=``=` `4``:` ` ``char1 ``=` `matr[e1r][``0``]` ` ``else``:` ` ``char1 ``=` `matr[e1r][e1c``+``1``]` ` ``char2 ``=` `''` ` ``if` `e2c ``=``=` `4``:` ` ``char2 ``=` `matr[e2r][``0``]` ` ``else``:` ` ``char2 ``=` `matr[e2r][e2c``+``1``]` ` ``return` `char1, char2` `def` `encrypt_ColumnRule(matr, e1r, e1c, e2r, e2c):` ` ``char1 ``=` `''` ` ``if` `e1r ``=``=` `4``:` ` ``char1 ``=` `matr[``0``][e1c]` ` ``else``:` ` ``char1 ``=` `matr[e1r``+``1``][e1c]` ` ``char2 ``=` `''` ` ``if` `e2r ``=``=` `4``:` ` ``char2 ``=` `matr[``0``][e2c]` ` ``else``:` ` ``char2 ``=` `matr[e2r``+``1``][e2c]` ` ``return` `char1, char2` `def` `encrypt_RectangleRule(matr, e1r, e1c, e2r, e2c):` ` ``char1 ``=` `''` ` ``char1 ``=` `matr[e1r][e2c]` ` ``char2 ``=` `''` ` ``char2 ``=` `matr[e2r][e1c]` ` ``return` `char1, char2` `def` `encryptByPlayfairCipher(Matrix, plainList):` ` ``CipherText ``=` `[]` ` ``for` `i ``in` `range``(``0``, ``len``(plainList)):` ` ``c1 ``=` `0` ` ``c2 ``=` `0` ` ``ele1_x, ele1_y ``=` `search(Matrix, plainList[i][``0``])` ` ``ele2_x, ele2_y ``=` `search(Matrix, plainList[i][``1``])` ` ``if` `ele1_x ``=``=` `ele2_x:` ` ``c1, c2 ``=` `encrypt_RowRule(Matrix, ele1_x, ele1_y, ele2_x, ele2_y)` ` ``# Get 2 letter cipherText` ` ``elif` `ele1_y ``=``=` `ele2_y:` ` ``c1, c2 ``=` `encrypt_ColumnRule(Matrix, ele1_x, ele1_y, ele2_x, ele2_y)` ` ``else``:` ` ``c1, c2 ``=` `encrypt_RectangleRule(` ` ``Matrix, ele1_x, ele1_y, ele2_x, ele2_y)` ` ``cipher ``=` `c1 ``+` `c2` ` ``CipherText.append(cipher)` ` ``return` `CipherText` `text_Plain ``=` `'instruments'` `text_Plain ``=` `removeSpaces(toLowerCase(text_Plain))` `PlainTextList ``=` `Diagraph(FillerLetter(text_Plain))` `if` `len``(PlainTextList[``-``1``]) !``=` `2``:` ` ``PlainTextList[``-``1``] ``=` `PlainTextList[``-``1``]``+``'z'` `key ``=` `"Monarchy"` `print``(``"Key text:"``, key)` `key ``=` `toLowerCase(key)` `Matrix ``=` `generateKeyTable(key, list1)` `print``(``"Plain Text:"``, text_Plain)` `CipherList ``=` `encryptByPlayfairCipher(Matrix, PlainTextList)` `CipherText ``=` `""` `for` `i ``in` `CipherList:` ` ``CipherText ``+``=` `i` `print``(``"CipherText:"``, CipherText)` `# This code is Contributed by Boda_Venkata_Nikith`
## Javascript
`// JavaScript program to implement Playfair Cipher` ` ``// Function to generate the 5x5 key square` ` ``function` `generateKeyTable(key, ks, keyT) {` ` ``let i, j, k, flag = 0;` ` ``// a 26 character hashmap` ` ``// to store count of the alphabet` ` ``let dicty = ``new` `Array(26).fill(0);` ` ``for` `(i = 0; i < ks; i++) {` ` ``let r = key[i].charCodeAt(0) - 97;` ` ``if` `(key[i] != ``'j'``) {` ` ``dicty[r] = 2;` ` ``}` ` ``}` ` ``dicty[``'j'``.charCodeAt(0) - 97] = 1;` ` ``i = 0;` ` ``j = 0;` ` ``for` `(k = 0; k < ks; k++) {` ` ``let r = key[k].charCodeAt(0) - 97;` ` ``if` `(dicty[r] == 2) {` ` ``dicty[r] -= 1;` ` ``keyT[i][j] = key[k];` ` ``j++;` ` ``if` `(j == 5) {` ` ``i++;` ` ``j = 0;` ` ``}` ` ``}` ` ``}` ` ``for` `(k = 0; k < 26; k++) {` ` ``if` `(dicty[k] == 0) {` ` ``keyT[i][j] = String.fromCharCode(k + 97);` ` ``j++;` ` ``if` `(j == 5) {` ` ``i++;` ` ``j = 0;` ` ``}` ` ``}` ` ``}` ` ``return` `keyT;` ` ``}` ` ``// Function to search for the characters of a digraph` ` ``// in the key square and return their position` ` ``function` `search(keyT, a, b, arr) {` ` ``let i, j;` ` ``if` `(a == ``'j'``)` ` ``a = ``'i'``;` ` ``else` `if` `(b == ``'j'``)` ` ``b = ``'i'``;` ` ``for` `(i = 0; i < 5; i++) {` ` ``for` `(j = 0; j < 5; j++) {` ` ``if` `(keyT[i][j] == a) {` ` ``arr[0] = i;` ` ``arr[1] = j;` ` ``}` ` ``else` `if` `(keyT[i][j] == b) {` ` ``arr[2] = i;` ` ``arr[3] = j;` ` ``}` ` ``}` ` ``}` ` ``return` `arr;` ` ``}` ` ``// Function to find the modulus with 5` ` ``function` `mod5(a) {` ` ``return` `(a % 5);` ` ``}` ` ``// Function to make the plain text length to be even` ` ``function` `prepare(str, ptrs) {` ` ``if` `(ptrs % 2 != 0) {` ` ``str += ``'z'``;` ` ``}` ` ``return` `[str, ptrs];` ` ``}` ` ``// Function for performing the encryption` ` ``function` `encrypt(str, keyT, ps) {` ` ``let i;` ` ``let a = ``new` `Array(4).fill(0);` ` ``let newstr = ``new` `Array(ps);` ` ``for` `(i = 0; i < ps; i += 2) {` ` ``let brr = search(keyT, str[i], str[i + 1], a);` ` ``let k1 = brr[0];` ` ``let k2 = brr[1];` ` ``let k3 = brr[2];` ` ``let k4 = brr[3];` ` ``if` `(k1 == k3) {` ` ``newstr[i] = keyT[k1][(k2 + 1) % 5];` ` ``newstr[i + 1] = keyT[k1][(k4 + 1) % 5];` ` ``}` ` ``else` `if` `(k2 == k4) {` ` ``newstr[i] = keyT[(k1 + 1) % 5][k2];` ` ``newstr[i + 1] = keyT[(k3 + 1) % 5][k2];` ` ``}` ` ``else` `{` ` ``newstr[i] = keyT[k1][k4];` ` ``newstr[i + 1] = keyT[k3][k2];` ` ``}` ` ``}` ` ``let res = ``""``;` ` ``for` `(let i = 0; i < newstr.length; i++) { res += newstr[i]; }` ` ``return` `res;` ` ``}` ` ``// Function to encrypt using Playfair Cipher` ` ``function` `encryptByPlayfairCipher(str, key) {` ` ``let ps, ks;` ` ``let keyT = ``new` `Array(5);` ` ``for` `(let i = 0; i < 5; i++) {` ` ``keyT[i] = ``new` `Array(5);` ` ``}` ` ``str = str.trim();` ` ``key = key.trim();` ` ``str = str.toLowerCase();` ` ``key = key.toLowerCase();` ` ``ps = str.length;` ` ``ks = key.length;` ` ``[str, ps] = prepare(str, ps);` ` ``let kt = generateKeyTable(key, ks, keyT);` ` ``return` `encrypt(str, kt, ps);` ` ``}` ` ``// Driver code` ` ``let key = ``" Monarchy"``;` ` ``let str = ``" instruments"``;` ` ``// Key to be encrypted` ` ``console.log(``"Key text: "` `+ key + ``"
"``);` ` ``console.log(``"Plain text: "` `+ str + ``"
"``);` ` ``// encrypt using Playfair Cipher` ` ``console.log(``"Cipher text: "` `+ encryptByPlayfairCipher(str, key));` ` ` ` ``// This code is contributed by poojaagarwal2`
Output
```Key text: Monarchy
Plain text: instruments
Cipher text: gatlmzclrqtx```
Decryption Technique
Decrypting the Playfair cipher is as simple as doing the same process in reverse. The receiver has the same key and can create the same key table, and then decrypt any messages made using that key.
The Playfair Cipher Decryption Algorithm:
The Algorithm consists of 2 steps:
1. Generate the key Square(5×5) at the receiver’s end:
• The key square is a 5Ă—5 grid of alphabets that acts as the key for encrypting the plaintext. Each of the 25 alphabets must be unique and one letter of the alphabet (usually J) is omitted from the table (as the table can hold only 25 alphabets). If the plaintext contains J, then it is replaced by I.
• The initial alphabets in the key square are the unique alphabets of the key in the order in which they appear followed by the remaining letters of the alphabet in order.
2. Algorithm to decrypt the ciphertext: The ciphertext is split into pairs of two letters (digraphs).
Note: The ciphertext always have even number of characters.
1. For example:
```CipherText: "gatlmzclrqtx"
After Split: 'ga' 'tl' 'mz' 'cl' 'rq' 'tx'```
1. Rules for Decryption:
• If both the letters are in the same column: Take the letter above each one (going back to the bottom if at the top).
For example:
```Diagraph: "cl"
Decrypted Text: me
Decryption:
c -> m
l -> e```
•
•
• If both the letters are in the same row: Take the letter to the left of each one (going back to the rightmost if at the leftmost position).
For example:
```Diagraph: "tl"
Decrypted Text: st
Decryption:
t -> s
l -> t```
•
•
• If neither of the above rules is true: Form a rectangle with the two letters and take the letters on the horizontal opposite corner of the rectangle.
For example:
```Diagraph: "rq"
Decrypted Text: nt
Decryption:
r -> n
q -> t```
•
•
For example:
```Plain Text: "gatlmzclrqtx"
Decrypted Text: instrumentsz
Decryption:
(red)-> (green)
ga -> in
tl -> st
mz -> ru
cl -> me
rq -> nt
tx -> sz```
Below is an implementation of Playfair Cipher Decryption in C:
## C
`#include ` `#include ` `#include ` `#define SIZE 30` `// Convert all the characters` `// of a string to lowercase` `void` `toLowerCase(``char` `plain[], ``int` `ps)` `{` ` ``int` `i;` ` ``for` `(i = 0; i < ps; i++) {` ` ``if` `(plain[i] > 64 && plain[i] < 91)` ` ``plain[i] += 32;` ` ``}` `}` `// Remove all spaces in a string` `// can be extended to remove punctuation` `int` `removeSpaces(``char``* plain, ``int` `ps)` `{` ` ``int` `i, count = 0;` ` ``for` `(i = 0; i < ps; i++)` ` ``if` `(plain[i] != ``' '``)` ` ``plain[count++] = plain[i];` ` ``plain[count] = ``'\0'``;` ` ``return` `count;` `}` `// generates the 5x5 key square` `void` `generateKeyTable(``char` `key[], ``int` `ks, ``char` `keyT[5][5])` `{` ` ``int` `i, j, k, flag = 0, *dicty;` ` ``// a 26 character hashmap` ` ``// to store count of the alphabet` ` ``dicty = (``int``*)``calloc``(26, ``sizeof``(``int``));` ` ``for` `(i = 0; i < ks; i++) {` ` ``if` `(key[i] != ``'j'``)` ` ``dicty[key[i] - 97] = 2;` ` ``}` ` ``dicty[``'j'` `- 97] = 1;` ` ``i = 0;` ` ``j = 0;` ` ``for` `(k = 0; k < ks; k++) {` ` ``if` `(dicty[key[k] - 97] == 2) {` ` ``dicty[key[k] - 97] -= 1;` ` ``keyT[i][j] = key[k];` ` ``j++;` ` ``if` `(j == 5) {` ` ``i++;` ` ``j = 0;` ` ``}` ` ``}` ` ``}` ` ``for` `(k = 0; k < 26; k++) {` ` ``if` `(dicty[k] == 0) {` ` ``keyT[i][j] = (``char``)(k + 97);` ` ``j++;` ` ``if` `(j == 5) {` ` ``i++;` ` ``j = 0;` ` ``}` ` ``}` ` ``}` `}` `// Search for the characters of a digraph` `// in the key square and return their position` `void` `search(``char` `keyT[5][5], ``char` `a, ``char` `b, ``int` `arr[])` `{` ` ``int` `i, j;` ` ``if` `(a == ``'j'``)` ` ``a = ``'i'``;` ` ``else` `if` `(b == ``'j'``)` ` ``b = ``'i'``;` ` ``for` `(i = 0; i < 5; i++) {` ` ``for` `(j = 0; j < 5; j++) {` ` ``if` `(keyT[i][j] == a) {` ` ``arr[0] = i;` ` ``arr[1] = j;` ` ``}` ` ``else` `if` `(keyT[i][j] == b) {` ` ``arr[2] = i;` ` ``arr[3] = j;` ` ``}` ` ``}` ` ``}` `}` `// Function to find the modulus with 5` `int` `mod5(``int` `a)` `{` ` ``if` `(a < 0)` ` ``a += 5;` ` ``return` `(a % 5);` `}` `// Function to decrypt` `void` `decrypt(``char` `str[], ``char` `keyT[5][5], ``int` `ps)` `{` ` ``int` `i, a[4];` ` ``for` `(i = 0; i < ps; i += 2) {` ` ``search(keyT, str[i], str[i + 1], a);` ` ``if` `(a[0] == a[2]) {` ` ``str[i] = keyT[a[0]][mod5(a[1] - 1)];` ` ``str[i + 1] = keyT[a[0]][mod5(a[3] - 1)];` ` ``}` ` ``else` `if` `(a[1] == a[3]) {` ` ``str[i] = keyT[mod5(a[0] - 1)][a[1]];` ` ``str[i + 1] = keyT[mod5(a[2] - 1)][a[1]];` ` ``}` ` ``else` `{` ` ``str[i] = keyT[a[0]][a[3]];` ` ``str[i + 1] = keyT[a[2]][a[1]];` ` ``}` ` ``}` `}` `// Function to call decrypt` `void` `decryptByPlayfairCipher(``char` `str[], ``char` `key[])` `{` ` ``char` `ps, ks, keyT[5][5];` ` ``// Key` ` ``ks = ``strlen``(key);` ` ``ks = removeSpaces(key, ks);` ` ``toLowerCase(key, ks);` ` ``// ciphertext` ` ``ps = ``strlen``(str);` ` ``toLowerCase(str, ps);` ` ``ps = removeSpaces(str, ps);` ` ``generateKeyTable(key, ks, keyT);` ` ``decrypt(str, keyT, ps);` `}` `// Driver code` `int` `main()` `{` ` ``char` `str[SIZE], key[SIZE];` ` ``// Key to be encrypted` ` ``strcpy``(key, ``"Monarchy"``);` ` ``printf``(``"Key text: %s\n"``, key);` ` ``// Ciphertext to be decrypted` ` ``strcpy``(str, ``"gatlmzclrqtx"``);` ` ``printf``(``"Plain text: %s\n"``, str);` ` ``// encrypt using Playfair Cipher` ` ``decryptByPlayfairCipher(str, key);` ` ``printf``(``"Deciphered text: %s\n"``, str);` ` ``return` `0;` `}` `// This code is contributed by AbhayBhat`
Output
```Key text: Monarchy
Plain text: gatlmzclrqtx
Deciphered text: instrumentsz```
1. It is significantly harder to break since the frequency analysis technique used to break simple substitution ciphers is difficult but still can be used on (25*25) = 625 digraphs rather than 25 monographs which is difficult.
2. Frequency analysis thus requires more cipher text to crack the encryption.
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SEARCH HOME
Math Central Quandaries & Queries
Question from brit, a teacher: (5+4)-7*4/28+8
Brit,
In order to solve a question with multiple operations (add/subtract, multiply/divide) there is an order to follow often referred to as 'BEDMAS'
BEDMAS is an acronym that stands for;
B-brackets
E-exponents
DM-multiply or divide (left to right)
AS-add subtract (left to right)
this is to help students remember what order to do the work in.
Here is a similar question,
(3 + 6) - 8 × 3 / 24 + 5
Following BEDMAS, we need to start with B-brackets,
(3 + 6) - 8 × 3 / 24 + 5
= 9 - 8 × 3 / 24 + 5
Then E-exponents, none, followed by DM-divide multiply (left to right),
= 9 - 8 × 3 / 24 + 5 (multiply)
= 9 - 24/24 + 5 (divide)
= 9 - 1 +5
And finish with AS-add subtract,
= 9 - 1 +5
= 8 +5
= 13
Hope this clears things up.
Melanie
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
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# Exploration Guide_ Adding Vectors Gizmo | ExploreLearning
```Exploration Guide: Adding Vectors Gizmo | ExploreLearning
http://www.explorelearning.com/index.cfm?method=cResource...
Describing Vectors
In this activity, you will explore vectors and ways of describing them by magnitude and direction.
tm
1. In the Gizmo , notice that there are two vectors on the graph,
and . You will begin this activity by
exploring .
a. On the graph, how is the initial point of
represented? How is the terminal point represented?
b. Drag the initial point of to the origin. Drag the terminal point of to form the vector 0, 4 . How are
the coordinates of a vector written differently than the coordinates of a point?
c. The magnitude of a vector is the distance from the initial point to the terminal point. What is the
magnitude of ? The direction of a vector is the direction in which the arrow points, from the initial point to
the terminal point. Using the compass directions north, south, east, or west, what is the direction of ?
d. If you drag the terminal point of to form the vector 3, 4 , how has the direction changed? What is the
new magnitude? Click on Click to measure lengths and use the interactive ruler to check your answer. (For
help using the ruler, click on Gizmo help, below the Gizmo.)
e. Turn off the ruler. Drag around by dragging the initial point. As you drag the vector, do the coordinates
of the vector change, or do they remain 0, 4 ? What do you think the coordinates of a vector represent?
Does the magnitude of the vector change? Does the direction change? Explain.
2. Drag the initial point of
to the origin and drag the terminal point of
to form the vector −5, 0 .
a. Because magnitude is the size of the vector, it is always a positive quantity. What is the magnitude of ?
What is the direction of ?
b. Drag the terminal point of
to form the vector 0, −4 . Now what is the direction of ? What is the
magnitude?
c. What is the direction of the vector −4, 4 ? What is the magnitude? To help calculate the magnitude,
sketch a right triangle.
In this activity, you will explore three ways of finding vector sums. The vector sum of
results from
and
is the vector that
+ .
1. Suppose you are in a motorized boat traveling directly north in still waters. Set = 0, 4 to represent the
velocity and direction of the boat. Suppose the boat encounters a strong current in the water, flowing from west
to east. Set = 3, 0 to represent the velocity and direction of the current.
a. Make a conjecture about the effect of this current on your boat. Where do you think your boat will go?
Make a sketch to show your guess.
b. The resulting speed and direction of the boat is represented by the sum, or resultant, of
and . Click
on Show resultant. The resultant is represented by on the graph. What are the coordinates of ? Increase
the wind's force by dragging to 6, 0 . What is the effect on ?
1 of 2
3/5/10 10:30 AM
Exploration Guide: Adding Vectors Gizmo | ExploreLearning
http://www.explorelearning.com/index.cfm?method=cResource...
c. Make a conjecture about how the coordinates of
are obtained from the coordinates of
and . Click
on Show sum computation to check your hypothesis.
d. Vary the coordinates of
and
coordinates of the vectors. Notice that changing the location of
2. Turn off Show resultant. Place the initial point of
of on the terminal point of and set = −4, 1 .
a. Calculate
+
or
does not change what the resultant is.
at the origin and set
= −2, 3 . Place the initial point
and . How does that answer compare to the coordinates
of the terminal point of ? Click on Show resultant and Show sum computation to check your answers.
b. Turn on Show parallelogram. How do
and
relate to the parallelogram? How does the resultant vector
relate to the parallelogram?
c. Drag
and
to make new vectors. Be sure to keep the initial point of
on the origin and the initial
point of
on the terminal point of . Is the vector sum always the same as the coordinates of the terminal
point of ? Is the resultant vector always the diagonal of the parallelogram in which
and
are two of the
sides? Make some sketches representing these findings.
3. Use the Gizmo and what you have learned about vector addition to find the following vector sums.
a. Find 1, 3 + −3, −4 .
b. What is the resultant of
= 0, 4 and
c. Find the sum of
= −3, 0 and
d. What is
+
if
=
and
= −5, 0 ?
= −4, 0 .
= 3, 4 and
= −3, −4 ? This is called a state of equilibrium.
Conditions of Use (7) and our Privacy Policy (8) before using this site. Your use of the site indicates your agreement to be bound by the
Terms & Conditions of Use.
2 of 2
3/5/10 10:30 AM
```
Complex analysis
23 Cards
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Try out our new practice tests completely free!
# Calculus Early Transcendentals Study Set 1
Mathematics
Bookmark
## Quiz 2 : The Derivative
Find f (x) where .
Free
Essay
If , (Do not reduce any trig ratio identities.)
Free
Multiple Choice
B
Find an equation for the tangent line to the graph of y = x sin x at x = .
Free
Multiple Choice
E
Find f (x) where f(x) = x2(sin 8x)6.
Essay
Answer true or false. If , .
True False
y = cot (sin x). Find dy/dx.
Multiple Choice
If , then
Multiple Choice
Find the value of the constant A in y = A sin 3t so that d2y/dt2 + 7y = sin 3t.
Multiple Choice
Answer true or false. If y = cos(x6), d2y/dx2 = -cos(x6).
True False
Use a graphing utility to obtain the graph of . Determine the slope of the tangent line to the graph at x = 0.
Multiple Choice
f(x) = (x4 - 3)38. f (x) =
Multiple Choice
Answer true or false. Given f (x) = 5x and , then if F(x) = f(g(x)).
True False
y = sin3( - 9). Find dy/d.
Multiple Choice
Find dy/dx if y = cos(sin x).
Multiple Choice
Answer true or false. If y = sin 6x - cos x3, d2y/dx2 = 18 cos x.
True False
y = x4 sin(5x). Find dy/dx.
Multiple Choice
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# Mills' constant
In number theory, Mills' constant is defined as the smallest positive real number A such that the floor function of the double exponential function
$\displaystyle{ \lfloor A^{3^{n}} \rfloor }$
is a prime number for all natural numbers n. This constant is named after William Harold Mills who proved in 1947 the existence of A based on results of Guido Hoheisel and Albert Ingham on the prime gaps.[1] Its value is unproven, but if the Riemann hypothesis is true, it is approximately 1.3063778838630806904686144926... (sequence A051021 in the OEIS).
## Mills primes
The primes generated by Mills' constant are known as Mills primes; if the Riemann hypothesis is true, the sequence begins
$\displaystyle{ 2, 11, 1361, 2521008887, 16022236204009818131831320183, }$
$\displaystyle{ 4113101149215104800030529537915953170486139623539759933135949994882770404074832568499, \ldots }$ (sequence A051254 in the OEIS).
If ai denotes the i th prime in this sequence, then ai can be calculated as the smallest prime number larger than $\displaystyle{ a_{i-1}^3 }$. In order to ensure that rounding $\displaystyle{ A^{3^n} }$, for n = 1, 2, 3, …, produces this sequence of primes, it must be the case that $\displaystyle{ a_i \lt (a_{i-1}+1)^3 }$. The Hoheisel–Ingham results guarantee that there exists a prime between any two sufficiently large cube numbers, which is sufficient to prove this inequality if we start from a sufficiently large first prime $\displaystyle{ a_1 }$. The Riemann hypothesis implies that there exists a prime between any two consecutive cubes, allowing the sufficiently large condition to be removed, and allowing the sequence of Mills primes to begin at a1 = 2.
For all a > $\displaystyle{ e^{e^{34}} }$, there is at least one prime between $\displaystyle{ a^3 }$ and $\displaystyle{ (a+1)^3 }$.[2] This upper bound is much too large to be practical, as it is infeasible to check every number below that figure. However, the value of Mills' constant can be verified by calculating the first prime in the sequence that is greater than that figure.
As of April 2017, the 11th number in the sequence is the largest one that has been proved prime. It is
$\displaystyle{ \displaystyle (((((((((2^3+3)^3+30)^3+6)^3+80)^3+12)^3+450)^3+894)^3+3636)^3+70756)^3+97220 }$
and has 20562 digits.[3]
(As of 2015), the largest known Mills probable prime (under the Riemann hypothesis) is
$\displaystyle{ \displaystyle ((((((((((((2^3+3)^3+30)^3+6)^3+80)^3+12)^3+450)^3+894)^3+3636)^3+70756)^3+97220)^3+66768)^3+300840)^3+1623568 }$
(sequence A108739 in the OEIS), which is 555,154 digits long.
## Numerical calculation
By calculating the sequence of Mills primes, one can approximate Mills' constant as
$\displaystyle{ A\approx a(n)^{1/3^n}. }$
Caldwell and Cheng used this method to compute 6850 base 10 digits of Mills' constant under the assumption that the Riemann hypothesis is true.[4] There is no closed-form formula known for Mills' constant, and it is not even known whether this number is rational.[5]
## Fractional representations
Below are fractions which approximate Mills' constant, listed in order of increasing accuracy (with continued-fraction convergents in bold) (sequence A123561 in the OEIS):
1/1, 3/2, 4/3, 9/7, 13/10, 17/13, 47/36, 64/49, 81/62, 145/111, 226/173, 307/235, 840/643, 1147/878, 3134/2399, 4281/3277, 5428/4155, 6575/5033, 12003/9188, 221482/169539, 233485/178727, 245488/187915, 257491/197103, 269494/206291, 281497/215479, 293500/224667, 305503/233855, 317506/243043, 329509/252231, 341512/261419, 353515/270607, 365518/279795, 377521/288983, 389524/298171, 401527/307359, 413530/316547, 425533/325735, 4692866/3592273, 5118399/3918008, 5543932/4243743, 5969465/4569478, 6394998/4895213, 6820531/5220948, 7246064/5546683,7671597/5872418, 8097130/6198153, 8522663/6523888, 8948196/6849623, 9373729/7175358, 27695654/21200339, 37069383/28375697, 46443112/35551055, 148703065/113828523, 195146177/149379578, 241589289/184930633, 436735466/334310211, 1115060221/853551055, 1551795687/1187861266, 1988531153/1522171477, 3540326840/2710032743, 33414737247/25578155953, ...
## Generalisations
There is nothing special about the middle exponent value of 3. It is possible to produce similar prime-generating functions for different middle exponent values. In fact, for any real number above 2.106..., it is possible to find a different constant A that will work with this middle exponent to always produce primes. Moreover, if Legendre's conjecture is true, the middle exponent can be replaced[6] with value 2 (sequence A059784 in the OEIS).
Matomäki showed unconditionally (without assuming Legendre's conjecture) the existence of a (possibly large) constant A such that $\displaystyle{ \lfloor A^{2^{n}} \rfloor }$ is prime for all n.[7]
Additionally, Tóth proved that the floor function in the formula could be replaced with the ceiling function, so that there exists a constant $\displaystyle{ B }$ such that
$\displaystyle{ \lceil B^{r^{n}} \rceil }$
is also prime-representing for $\displaystyle{ r\gt 2.106\ldots }$.[8] In the case $\displaystyle{ r=3 }$, the value of the constant $\displaystyle{ B }$ begins with 1.24055470525201424067... The first few primes generated are:
$\displaystyle{ 2, 7, 337, 38272739, 56062005704198360319209, 176199995814327287356671209104585864397055039072110696028654438846269, \ldots }$
Without assuming the Riemann hypothesis, Elsholtz proved that $\displaystyle{ \lfloor A^{10^{10n}} \rfloor }$ is prime for all positive integers n, where $\displaystyle{ A \approx 1.00536773279814724017 }$, and that $\displaystyle{ \lfloor B^{3^{13n}} \rfloor }$ is prime for all positive integers n, where $\displaystyle{ B \approx 3.8249998073439146171615551375 }$.[9]
## References
1. Dudek, Adrian W. (2016). "An explicit result for primes between cubes". Functiones et Approximatio Commentarii Mathematici 55 (2): 177–197. doi:10.7169/facm/2016.55.2.3.
2. Caldwell, Chris (7 July 2006). "The Prime Database".
3. Caldwell, Chris K.; Cheng, Yuanyou (2005). "Determining Mills' Constant and a Note on Honaker's Problem". Journal of Integer Sequences 8: p. 5.4.1.
4. Finch, Steven R. (2003). "Mills' Constant". Mathematical Constants. Cambridge University Press. pp. 130–133. ISBN 0-521-81805-2.
5. Warren Jr., Henry S. (2013). Hacker's Delight (2nd ed.). Addison-Wesley Professional. ISBN 9780321842688.
6. Matomäki, K. (2010). "Prime-representing functions". Acta Mathematica Hungarica 128 (4): 307–314. doi:10.1007/s10474-010-9191-x.
7. Tóth, László (2017). "A Variation on Mills-Like Prime-Representing Functions". Journal of Integer Sequences 20: p. 17.9.8.
8. Elsholtz, Christian (2020). "Unconditional Prime-Representing Functions, Following Mills". American Mathematical Monthly 127 (7): 639–642. doi:10.1080/00029890.2020.1751560.
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### STAT 200 Final Exam Complete Solutioncorrect answer keys
Donate & Make a Difference
Question Details: #6165
STAT 200 Final Exam Complete Solutioncorrect answer keys
STAT 200 final exam solution correct answer keys
1. True or False. Justify for full credit.
(a) If all the observations in a data set are identical, then the variance for this data set is zero.
(b) If P(A) = 0.4 and P(B) = 0.5, then P(A AND B) = 0.2.
(c) The mean is always equal to the median for a normal distribution.
(d) A 95% confidence interval is wider than a 90% confidence interval of the same parameter.
(e) In a two-tailed hypothesis testing at significance level α of 0.05, the test statistic is calculated as 2. If P(X >2) = 0.03, then we have sufficient evidence to reject the null hypothesis.
Refer to the following frequency distribution for Questions 2, 3, 4, and 5. Show all work. Just the answer, without supporting work, will receive no credit.
A random sample of 100 students was chosen from UMUC STAT 200 classes. The frequency distribution below shows the distribution for study time each week (in hours).
Study Time (in hours)
Frequency
Relative Frequency
0.0 – 4.9
5
5.0 - 9.9
13
10.0 - 14.9
0.22
15.0 -19.9
42
20.0 – 24.9
Total
100
2.
Complete the frequency table with frequency and relative frequency.
3.
What percentage of the study times was at least 15 hours?
4.
5.
Does this distribution have positive skew or negative skew? Why?
Refer to the following information for Questions 6, 7, and 8. Show all work. Just the answer, without supporting work, will receive no credit.
A fair 6-faced die is rolled two times. Let A be the event that the outcome of the first roll is a multiple of 3, and B be the event that the outcome of second roll is greater than 4.
6. How many outcomes are there in the sample space?
7. What is the probability that the outcome of the second roll is greater than 4, given that the first roll is a multiple of 3?
8. Are A and B independent? Why or why not?
Refer to the following situation for Questions 9, 10, and 11.
The five-number summary below shows the grade distribution of two STAT 200 quizzes.
Minimum
Q1
Median
Q3
Maximum
Quiz 1
12
40
60
95
100
Quiz 2
20
35
50
90
100
For each question, give your answer as one of the following: (a) Quiz 1; (b) Quiz 2; (c) Both quizzes have the same value requested; (d) It is impossible to tell using only the given information. Then explain your answer in each case.
9. Which quiz has less interquartile range in grade distribution?
10. Which quiz has the greater percentage of students with grades 90 and over?
11. Which quiz has a greater percentage of students with grades less than 60?
Refer to the following information for Questions 12 and 13. Show all work. Just the answer, without supporting work, will receive no credit.
There are 1000 juniors in a college. Among the 1000 juniors, 300 students are taking STAT200, and 150 students are taking PSYC300. There are 50 students taking both courses.
12. What is the probability that a randomly selected junior is taking at least one of these two courses?
13. What is the probability that a randomly selected junior is taking STAT200, given that he/she is taking PSYC300?
14. There are 8 books in the “Statistics is Fun” series. Mimi would like to choose 2 books from the series for her summer reading. How many different ways can the two books be selected?
15. Let random variable x represent the number of girls in a family of three children.
(a) Construct a table describing the probability distribution.
(b) Determine the mean and standard deviation of x. (Round the answer to two decimal places)
16. Rabbits like to eat the cucumbers in Mimi’s garden. There are 10 cucumbers in her garden which will be ready to harvest in about 10 days. Based on her experience, the probability of a cucumber being eaten by the rabbits before harvest is 0.60. Show all work. Just the answer, without supporting work, will receive no credit.
(a) Let X be the number of cucumbers that Mimi harvests (that is, the number of cucumbers not eaten by rabbits). As we know, the distribution of X is a binomial probability distribution. What is the number of trials (n), probability of successes (p) and probability of failures (q), respectively?
(b) Find the probability that Mimi harvests at least 2 of the 10 cucumbers. (round the answer to 3 decimal places)
(c) How many cucumbers can she expect to harvest?
Refer to the following information for Questions 17, 18, and 19. Show all work. Just the answer, without supporting work, will receive no credit.
The heights of pecan trees are normally distributed with a mean of 10 feet and a standard deviation of 2 feet.
17. What is the probability that a randomly selected pecan tree is between 9 and 12 feet tall?
18. Find the 80th percentile of the pecan tree height distribution.
19. If a random sample of 225 pecan trees is selected, what is the standard deviation of the sample mean?
20. A random sample of 100 SAT scores has a sample mean of 1500. Assume that SAT scores have a population standard deviation of 300. Construct a 95% confidence interval estimate of the mean SAT scores. Show all work. Just the answer, without supporting work, will receive no credit.
21. Consider the hypothesis test given by
H 0 : p H1 : p
0.5
0.5
In a random sample of 100 subjects, the sample proportion is found to be pˆ
0.47 .
(a) Determine the test statistic. Show all work; writing the correct test statistic, without supporting work, will receive no credit.
(b) Determine the p-value for this test. Show all work; writing the correct P-value, without supporting work, will receive no credit.
(c) Is there sufficient evidence to justify the rejection of ? Explain.
22. Mimi was curious if regular excise really helps weight loss, hence she decided to perform a hypothesis test. A random sample of 5 UMUC students was chosen. The students took a 30-minute exercise every day for 6 months. The weight was recorded for each individual before and after the exercise regimen. Does the data below suggest that the regular exercise helps weight loss?
Weight (pounds)
Subject
Before
After
1
159
130
2
170
160
3
185
180
4
165
165
5
200
190
Assume we want to use a 0.01 significance level to test the claim.
(a) Identify the null hypothesis and the alternative hypothesis.
(b) Determine the test statistic. Show all work; writing the correct test statistic, without supporting work, will receive no credit.
(c) Determine the p-value. Show all work; writing the correct critical value, without supporting work, will receive no credit.
(d) Is there sufficient evidence to support the claim that regular exercise helps weight loss? Justify your conclusion.
23. A STAT 200 instructor is interested in whether there is any variation in the final exam grades between her two classes Data collected from the two classes are as follows:
Her null hypothesis and alternative hypothesis are:
• Attachments:
• Solution Details: #6173
STAT 200 Final Exam Complete Solutioncorrect answer keys
Question 1: True, if all the observations in a data set are identical then the mean would be the same value so the deviation from mean would be 0 for all values and thus the variance would be 0. False, if A and B are independent events then only P(A and B) = P(A)*P(B) = 0.5*0.4 = 0.2. True, the normal distribution is symmetric about the mean thus the mean, median and mode are always same. True, the 95% confidence interval uses a larger Z multiplier thus is wider than the 90% confidence interval. False, as this is a two tailed test so P-value = P(|X|> 2) = 2*P(X>2) = 0.06. Which is larger than the significance level and thus the conclusion. Question 2: The complete table is given below, Study Time (in hours) Frequency Relative Frequency 0.0-4.9 5 5/100 = 0.05 5.0-9.9 13 13/100 = 0.13 10.0-14.9 0.22*100 = 22 0.22 15.0-19.9 42 42/100 = 0.42 20.0-24.9 100-(5+13+22+42) = 18 18/100 = 0.18 Total 100 100/100 = 1 Question 3: Required percentage = P(X ≥ 15) = Relative frequency of class 15.0-19.9 + Relative frequency of class20.0-24.9 = 0.42+0.18 = 0.60 = 60% Question 4: There are 100 observations in total out of which the 1st 50 lies on or below the class interval 15.0-19.9. Thus the class interval in which median lie is 15.0-19.9. Question 5: Considering 15.0-19.9 as the median class we can see the class right of this class has higher frequency and smaller in number (on right we have only one class and on left we have 3 classes) as compare to the classes in left. Thus the distribution is left skewed. It can be seen more clearly from the histogram which is given below. Question 6: Note that each roll of a dice has 6 outcomes as 1,2,3,4,5 and 6. Now rolling two times means the same number would appear on the 2nd roll as well. As the rolls are independent hence for each of the outcomes in 1st roll there are 6 outcomes in 2nd roll. As we are rolling a 6 faced dice twice thus the number of elements in the sample space is 6*6 = 36. The sample space would look like, {(1,1); (1,2); (1,3);…(6,5),(6,6)}. Question 7: If the 1st roll is a multiple of 3 then...
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$$\require{cancel}$$
# 5.7: Potential
[ "article:topic", "authorname:tatumj" ]
If work is required to move a mass from point $$\text{A}$$ to point $$\text{B}$$, there is said to be a gravitational potential difference between $$\text{A}$$ and $$\text{B}$$, with $$\text{B}$$ being at the higher potential. The work required to move unit mass from $$\text{A}$$ to $$\text{B}$$ is called the potential difference between $$\text{A}$$ and $$\text{B}$$. In SI units it is expressed in $$\text{J kg}^{−1}$$ .
We have defined only the potential difference between two points. If we wish to define the potential at a point, it is necessary arbitrarily to define the potential at a particular point to be zero. We might, for example define the potential at floor level to be zero, in which case the potential at a height $$h$$ above the floor is $$gh$$; equally we may elect to define the potential at the level of the laboratory bench top to be zero, in which case the potential at a height $$z$$ above the bench top is $$gz$$. Because the value of the potential at a point depends on where we define the zero of potential, one often sees that the potential at some point is equal to some mathematical expression plus an arbitrary constant. The value of the constant will be determined once we have decided where we wish to define zero potential.
In celestial mechanics it is usual to assign zero potential to all points at an infinite distance from any bodies of interest.
Suppose we decide to define the potential at point $$\text{A}$$ to be zero, and that the potential at $$\text{B}$$ is then $$ψ$$ $$\text{J kg}^{−1}$$ . If we move a point mass $$m$$ from $$\text{A}$$ to $$\text{B}$$, we shall have to do an amount of work equal to $$mψ \ \text{J}$$. The potential energy of the mass $$m$$ when it is at $$\text{B}$$ is then $$mψ$$. In these notes, I shall usually use the symbol $$ψ$$ for the potential at a point, and the symbol $$V$$ for the potential energy of a mass at a point.
In moving a point mass from $$\text{A}$$ to $$\text{B}$$, it does not matter what route is taken. All that matters is the potential difference between $$\text{A}$$ and $$\text{B}$$. Forces that have the property that the work required to move from one point to another is route-independent are called conservative forces; gravitational forces are conservative. The potential at a point is a scalar quantity; it has no particular direction associated with it.
If it requires work to move a body from point $$\text{A}$$ to point $$\text{B}$$ (i.e. if there is a potential difference between $$\text{A}$$ and $$\text{B}$$, and $$\text{B}$$ is at a higher potential than $$\text{A}$$), this implies that there must be a gravitational field directed from $$\text{B}$$ to $$\text{A}$$.
$$\text{FIGURE V.22}$$
Figure $$\text{V.22}$$ shows two points, $$\text{A}$$ and $$\text{B}$$, a distance $$δx$$ apart, in a region of space where the gravitational field is $$g$$ directed in the negative $$x$$ direction. We’ll suppose that the potential difference between $$\text{A}$$ and $$\text{B}$$ is $$δψ$$. By definition, the work required to move unit mass from $$\text{A}$$ to $$\text{B}$$ is $$δψ$$. Also by definition, the force on unit mass is $$g$$, so that the work done on unit mass is $$gδx$$. Thus we have
$g = - \frac{dψ}{dx}. \label{5.7.1} \tag{5.7.1}$
The minus sign indicates that, while the potential increases from left to right, the gravitational field is directed to the left. In words, the gravitational field is minus the potential gradient.
This was a one-dimensional example. In a later section, when we discuss the vector operator $$\nabla$$, we shall write equation $$\ref{5.7.1}$$ in its three-dimensional form
$\textbf{g} = - \textbf{grad}ψ = - \nabla ψ. \label{5.7.2} \tag{5.7.2}$
While $$ψ$$ itself is a scalar quantity, having no directional properties, its gradient is, of course, a vector.
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# Math Snap
## 13) $A C=2 x-5, A B=12$, and $B C=x-6$ Find $B C$.
#### STEP 1
Assumptions1. The length of segment $AC$ is $x-5$ . The length of segment $AB$ is $12$ 3. The length of segment $BC$ is $x-6$ 4. $AC$ is the sum of $AB$ and $BC$ because they are on the same line
#### STEP 2
First, we need to set up an equation based on the assumption that $AC$ is the sum of $AB$ and $BC$.$AC = AB + BC$
#### STEP 3
Now, plug in the given values for $AC$, $AB$, and $BC$ to create the equation. $(2x-5) =12 + (x-6)$
#### STEP 4
implify the right side of the equation. $(2x-) = x +6$
#### STEP 5
Subtract $x$ from both sides of the equation to isolate $x$ on one side. $x -5 =$
#### STEP 6
Add $5$ to both sides of the equation to solve for $x$. $x =6 +5$
#### STEP 7
Calculate the value of $x$. $x =11$
#### STEP 8
Now that we have the value of $x$, we can find the length of $BC$ by substituting $x$ into the given expression for $BC$. $BC = x -6$
#### STEP 9
Plug in the value of $x$ to calculate the length of $BC$. $BC =11 -6$
##### SOLUTION
Calculate the length of $BC$. $BC =5$The length of $BC$ is5 units.
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# Worksheet on Factors affecting Interest
We will practice the questions given in the worksheet on factors affecting interest. We know the formula for calculating simple interest (S.I.) is equal to $$\frac{P × R × T}{100}$$.
1. Find the simple interest and amount of:
(i) $500 for 3 years at 4% per annum (ii)$ 2550 for 8 months at 50 cents percent per month
(iii) $2750 for 4.4 years at 12.6 percent per month (iv)$ 7200 for 12$$\frac{1}{2}$$ years at 13$$\frac{1}{3}$$%
(v) $5000 for 1$$\frac{1}{4}$$ years at 7$$\frac{1}{2}$$% 2. Calculate the simple interest on$ 600 for 2$$\frac{1}{2}$$ years at 4% per annum.
3. Find the time when principal is $800 and simple interest is$ 96 at the rate of 6 % per annum.
4. Ron took $2400 from Rex and returned him$ 3000 after 5 years. Find the annum rate of interest in percent.
5. What will be the amount on $900 at 5 % per annum for 6 years? 6. Jack borrowed 4 6000 from a village money lender for 3$$\frac{1}{2}$$ years. Money lender charged interest at the rate of 15 % per annum. At the end of 3$$\frac{1}{3}$$ years. Jack paid$ 5500 cash and one cow to the money lender. Find the price of the cow.
7. Fill in the blanks:
(i) The sum lent is equal __________.
(ii) The money paid for using other’s money is called __________.
(iii) The interest paid on $100 is called __________. (iv) S.I. = $$\frac{P × R × ......}{100}$$ (v) The sum of the principal and interest is called __________. 8. Derek borrowed$ 1800 from his friend Robert. Robert charged interest at the rate of 5 % per annum. Derek returned the money and interest after 8 months. How much money in all did Derek return to Robert?
9. Sally took a loan of $850 from Rita for 3 years at 10 % per annum. After 3 years Sally paid back the loan by giving$ 900 in cash and a new sari. What was the price of the new sari?
Answers for the worksheet on factors affecting interest are given below.
1. (i) $60;$ 560
(ii) $102;$ 2652
(iii) $1524.60;$ 4274.60
(iv) $2400;$ 9600
(v) $468.75;$ 5468.75
2. $60 3. 2 years 4. 5% 5.$ 1170
6. $3650 7. (i) Principal (ii) interest (iii) rate percent (iv) R (v) amount 8.$ 1860
9. \$ 205
Word Problems on Simple Interest.
In Simple Interest when the Time is given in Months and Days.
To find Principal when Time Interest and Rate are given.
To find Rate when Principal Interest and Time are given .
To find Time when Principal Interest and Rate are given.
Worksheet on Simple Interest.
Worksheet on Factors affecting Interest
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# A MAN IS FOUND DEAD ON A SUNDAY MORNING HIS WIFE CALL THE POLICE IMMEDIATELY THE POLICE QUESTION THE WIFE AND STAFF THE WIFE SAID HE WAS ASLEEP THA COOK SAID HE WAS COOKING BREAKFAST THE GARDENER SAID SHE WAS PICKING VEGETABLES THA BUTLER SAID HE WAS C RI RIDDLES WITH ANSWERS TO SOLVE - PUZZLES & BRAIN TEASERS
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## The Train Crash Riddle
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The train fell out of the track and plowed through the soil, opening two hidden graves.
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A microwave
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45 lbs.
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Hogs and Kisses!
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## The Card Trick Riddle
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The answer is very simple. All she had to do is take the fifteen cards from the top and reverse them. This would make another pile out of that and there will be two piles - one of 15 cards and one of 37 cards. Also both of them will have the same number of inverted cards.
Just think about it and if the mathematical explanation will help you understand better, here it is.
Assume that there were p inverted cards initially in the top 15 cards. Then the remaining 37 cards will hold 15-p inverted cards.
Now when she reverses the 15 cards on the top, the number of inverted cards will become 15-p and thus the number of inverted cards in both of the piles will become same.
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## The Policeman And The Boy
Hint:
Actutally, the name of that boy is "Shut Up," and he is playing hide-and-seek with two of his friends who are named "Manners" and "Trouble". "Trouble" is the one who counted and he is looking for the other two boys. "Manners" is hiding up in the tree.
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Because their belt buckles are on their hats!
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Hazel!
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A monster with big hands!
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# Class Note for MATH 1330 with Professor Flagg at UH
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Date Created: 02/06/15
PreCalSection33 Math 1330 Section 33 Laws of Logarithms The Laws of Logarithms Let agt0 andail Let ABC be real numbers with A gt 0 andB gt 0 l loga AB log A log B 2 log log A log B 3 loga AC C loga A 4 loga ac C 5 loga A loga B if and only ifA B The Change of Base Formula Let a b be positive real numbers that are not equal to 1 Let x gt 0 Then 10ga 96 10 x 10g 0 These are the basic rules we need to work with logarithmic equations and simplify logarithmic expressions All 5 of the laws and the change of base formula are derived directly from the de nition of a logarithm given in Section 21 How Let s look at them one at a time First let s call loga A w and loga B y and then write these expressions in their exponential form aw A and ay B 1 In exponential form AB aw ay aw The logarithmic version of this equation is loga AB w y Then substitute for w and y and you get logABlogAlogB w A 2 In exponential form E a y aw The logarithm1c form of this equation is a loga w y Substituting back in for w and y shows log log A log B 3 Remember that log A wtranslates to aw A Raising both sides ofthis exponential equation to the C power gives aw C AC Then remembering that raising to powers corresponds to multiplying exponents this says acm AC The logarithmic form of this equation is loga AC C w C loga A PreCalSection33 2 4 This law is a combination of the third law and the fact that log a l 5 This law is a statement of the fact that the logarithmic functions and exponential functions are onetoone 6 The change of base formula start with logb x k and put into exponential form as bk x Then using Law 5 and base a we can take the logarithm ofboth sides in base a loga bk loga x Then use law 3 to obtain kloga b loga Solving for k and then substituting for k gives the formula Example 1 Rewrite each of the following expressions in a form that has no logarithms of a product quotient or power A log3 x2 x 1 B logoE C x 7 Example 2 Evaluate each of the following expressions using the laws of logarithms A log315 log3 5 B logGE PreCalSection33 C log69log64log6g Example 3 Rewrite each of the following as a single logarithm A lnx 3 lnbc3 l B log x4 2 810g67 x 2 logs x Example 4 Using the Change of Base Formula A Simplify logz 3 10g35 logs 8 B Write logs 5 as a common logarithmic expression C Write log215 as a natural logarithmic expression PreCalSection33 Example 5 Use the laws of logarithms to simplify the following equation lny 3 lnx 2 4lnx ln7 Example 6 Simplify using the laws of logarithms A 10mg 31114 C 30053 54033 7
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Question
# sin (T-x)15, lim
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Solution
## Let the function be , f( x )= sin( π−x ) π⋅( π−x ) We have to find the value of function at limit x→π . So we need to check the function by substituting the value at particular point ( π ), so that it should not be of the form 0 0 . If the condition is true, then we need to simplify the term to remove 0 0 form. f( x )= sin( π−π ) π⋅( π−π ) = 1 π ⋅ sin0 0 = 1 π ⋅ 0 0 = 0 0 Here, we see that the condition is not true and it is in 0 0 form. The given expression f( x )= lim x→π sin( π−x ) π⋅( π−x ) can be solved by assumption. Let the value of ( π−x ) be y . (1) According the given limits of x , x→π , then y→0 Also from equation 1, x=( π−y ) On substituting the value of new limit in terms of y , the new expression is: lim y→0 siny ( π⋅y ) (2) According to the trigonometric theorem, lim x→0 sinx x =1 (3) With the help of equations 2 and 3, we can calculate the value of limits; lim y→0 siny ( π⋅y ) = lim y→0 1 π ⋅ siny ( y ) = 1 π lim y→0 siny ( y ) On further simplification and using equation 3, we get lim y→0 siny ( π⋅y ) = 1 π ⋅1 = 1 π Thus, the value of the given expression lim x→π sin( π−x ) π⋅( π−x ) = 1 π
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Questions & Answers - Ask Your Doubts
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CBSE
Mathematics
Point of intersection of the lines
Determine graphically the vertices of the triangle, the equations of whose sides are given below:
y=x, y=0 and 3x+3y=10
${\text{A}}{\text{. }}\left( {0,0} \right),\left( {\dfrac{{10}}{3},0} \right),\left( {\dfrac{5}{3},\dfrac{5}{3}} \right) \\ {\text{B}}{\text{. }}\left( {0,0} \right),\left( {\dfrac{{17}}{3},0} \right),\left( {\dfrac{5}{3},\dfrac{5}{3}} \right) \\ {\text{C}}{\text{. }}\left( {0,0} \right),\left( {\dfrac{{14}}{3},0} \right),\left( {\dfrac{5}{3},\dfrac{5}{3}} \right) \\ {\text{D}}{\text{. }}\left( {0,0} \right),\left( {\dfrac{{10}}{3},0} \right),\left( {\dfrac{5}{3},\dfrac{8}{3}} \right) \\$
CBSE
Mathematics
Point of intersection of the lines
Locus of the point of the intersection of the lines $xcos\theta = y$ and $cot\theta = a$ is
A) ${x^2} + {y^2} = 2{a^2}$
B) ${x^2} + {y^2} - ax = 0$
C) ${y^2} = 4ax$
D) ${x^2} = {a^2} + {y^2}$
CBSE
Mathematics
Point of intersection of the lines
How do you find the intersection between $y = x - 8$ and ${x^2} + {y^2} = 34$?
CBSE
Mathematics
Point of intersection of the lines
Draw the graph of equation $2y+x=7$ and $2x+y=8$ on the same coordination system. Write the pt of intersection.
CBSE
Mathematics
Point of intersection of the lines
Find the distance of the point $P\left( -1,-5,-10 \right)$ from the point of intersection of the line $\overrightarrow{r}=\left( 2\widehat{i}-\widehat{j}+2\widehat{k} \right)+\lambda \left( 3\widehat{i}+3\widehat{j}+2\widehat{k} \right)$ and the plane $\overrightarrow{r}.\left( \widehat{i}-\widehat{j}+\widehat{k} \right)=5$?
CBSE
Mathematics
Point of intersection of the lines
Which of the following is a real-life example of intersecting lines?
A. orange
B. football
C. hands of the clock
D. none of these
CBSE
Mathematics
Point of intersection of the lines
If lines $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-1}{4}$ and $\dfrac{x-3}{1}=\dfrac{y-k}{2}=\dfrac{z}{1}$ intersect, then find the value of k and hence find the equation of the plane containing these lines.
CBSE
Mathematics
Point of intersection of the lines
Find the points of intersection of the line y = x - 3 and circle (x-32)2 + (y+2)2 = 20.
CBSE
Mathematics
Point of intersection of the lines
If the lines $x + ky + 3 = 0$ and $2x - 5y + 7 = 0$ intersect the coordinate axes in the concyclic points, then k=?
A.$\dfrac{{ - 2}}{5}$
B.$\dfrac{{ - 3}}{5}$
C.$\dfrac{3}{2}$
D.$\dfrac{{ - 5}}{3}$
CBSE
Mathematics
Point of intersection of the lines
A line through origin meets the line $2x = 3y + 13$ at the right angle at point$Q$. Find the absolute difference of coordinates of$Q$.
CBSE
Mathematics
Point of intersection of the lines
The point of intersection of the lines $\dfrac{{x - 6}}{{ - 6}} = \dfrac{{y + 4}}{4} = \dfrac{{z - 4}}{{ - 8}}$ and $\dfrac{{x + 1}}{2} = \dfrac{{y + 2}}{4} = \dfrac{{z + 3}}{{ - 2}}$ is
A. $\left( {0,0, - 4} \right)$
B. $\left( {1,0,0} \right)$
C. $\left( {0,2,0} \right)$
D. $\left( {1,2,0} \right)$
CBSE
Mathematics
Point of intersection of the lines
Draw the graph of the equation $x+3y=15$. Find the co-ordinates of the point where the graph intersects the $x$- axis.
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### Day 05 - Inc/Dec Intervals, Function Operations - 01.13.15
• Quiz 1 today!
• Unit 1 Test will be on Tuesday, Jan. 20th
Bell Ringer
• N/A
Review
• Intro to Functions
• What is a function?
• Function Notation
• Evaluating Functions
• Examples/Counterexamples
• Graph
• Table
• Set
• Domain/Range (using functions students covered in Algebra: linear, constant, square root, etc.)
• Graph
• Table
• Set
• Using Words
• Compound Inequalities
• Interval Notation
• Intercepts
• End Behavior
• Extrema
• Relative Min/Max (Cubic Functions)
• Absolute Min/Max (Quadratic and Absolute Value Functions)
• Using the graphing calculator to calculate mins and maxs
Lesson
• Symmetry
• Axis of Symmetry (Quadratic and Absolute Value Functions)
• Intervals Of Increasing And Decreasing (Quadratic and Absolute Value Functions)
• Evaluating Functions
• Function Operations
• Composition
• Translations
Checkpoints
Exit Ticket
• Posted on board at the end of the block
Lesson Objective(s)
• How is end behavior of a function determined?
• How are extrema found using a graphing calculator?
#### In-Class Help Requests
Standard(s)
• CC.9-12.F.IF.1 Understand the concept of a function and use function notation. Understand that a function from one set (called the domain) to another set (called the range) assigns to each element of the domain exactly one element of the range. If f is a function and x is an element of its domain, then f(x) denotes the output of f corresponding to the input x. The graph of f is the graph of the equation y = f(x).
• CC.9-12.F.IF.2 Understand the concept of a function and use function notation. Use function notation, evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context.
• CC.9-12.F.IF.6 Interpret functions that arise in applications in terms of the context. Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph.*
• CC.9-12.F.IF.7 Analyze functions using different representations. Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases.*
• CC.9-12.F.IF.7a Graph linear and quadratic functions and show intercepts, maxima, and minima.*
• CC.9-12.F.IF.7b Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions.*
• CC.9-12.F.IF.8 Analyze functions using different representations. Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function.
• CC.9-12.F.IF.9 Analyze functions using different representations. Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). For example, given a graph of one quadratic function and an algebraic expression for another, say which has the larger maximum.
• CC.9-12.F.BF.1c (+) Compose functions.
• CC.9-12.F.BF.3 Build new functions from existing functions. Identify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology.
• CC.9-12.F.LE.2 Construct and compare linear, quadratic, and exponential models and solve problems. Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two input-output pairs (include reading these from a table).
Mathematical Practice(s)
• #1 - Make sense of problems and persevere in solving them
• #2 - Reason abstractly and quantitatively
• #4 - Model with mathematics
• #5 - Use appropriate tools strategically
• #7 - Look for and make use of structure
Past Checkpoints
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May 9, 2021
# How to use IMSUM Function in Excel
The IMSUM function returns the sum of two or more complex numbers in x + yi or x + yj text format. When adding complex numbers, the real and imaginary coefficients are added separately i.e., the equation to find the sum of two complex numbers a+bi and c + di is : (a + bi) + (c + di) = (a + c) + (b + d)i
Syntax:= IMSUM(inumber1, [inumber2], …)
The IMSUM function syntax has the following arguments:
• Inumber1, [inumber2], … Inumber1 is required, subsequent numbers are not. 1 to 255 complex numbers to add
Example: Let’s look at some Excel IMSUM function examples and explore how to use the IMSUM function as a worksheet function in Microsoft Excel:
Syntax: =IMSUM(A2,B2)
Result:
Based on the Excel spreadsheet above, the following IMSUM examples would return:
Syntax: =IMSUM(A3,B3)
Result: 4+3i
Syntax: =IMSUM(A4,B4)
Result: 15+3i
Syntax: =IMSUM(A5,B5)
Result: 2i
Syntax: =IMSUM(A6,B6)
Result: 15+2i
Syntax: =IMSUM(A7,B7)
Result: 10+240i
Syntax: =IMSUM(A8,B8)
Result: 18+7i
Syntax: =IMSUM(A9,B9,C9)
Result: 17+2i
Syntax: =IMSUM(A10,B10,C10)
Result: 14-10i
Syntax: =IMSUM(A11,B11,C11)
Result: 10-8i
Syntax: =IMSUM(A12,B12,C12)
Result: 113-35j
Syntax: =IMSUM(A13,B13,C13)
Result: 248+269i
Note:
• The formula returns the #NUM! error if the complex number doesn’t have lower case i or j (iota).
• The formula returns the #VALUE! Error if the complex number doesn’t have correct complex number format.
READ: How to use IMSIN Function in Excel
#### Excel
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http://ma.mathforcollege.com/youtube/02vectors/vectors_04sle_whatisunitvector.html
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INTRODUCTION TO MATRIX ALGEBRA Transforming Matrix Algebra for the STEM Undergraduate HOME | TOPICS | VIDEOS | TEXTBOOK | PPT | QUIZZES | PROBLEM SET | CONTACT | MATH FOR COLLEGE | NUMERICAL METHODS
VECTORS (CHAPTER 2)
What is a unit vector?
By Autar Kaw
TOPIC DESCRIPTION
Learn what the definition of a unit vector is.
This video teaches you what a unit vector is.
ALL VIDEOS FOR THIS TOPIC
What is a vector? [YOUTUBE 2:27 [TRANSCRIPT]
What is a vector? Example [YOUTUBE 1:15] [TRANSCRIPT]
When are two vectors equal? [YOUTUBE 2:02][TRANSCRIPT]
When are two vectors equal? Example [YOUTUBE 2:26][TRANSCRIPT]
What is a null (or zero) vector? [YOUTUBE 1:12] [TRANSCRIPT]
What is a unit vector? [YOUTUBE 1:30][TRANSCRIPT]
What is a unit vector? Example [YOUTUBE 1:49] [TRANSCRIPT]
How do you multiply a vector by a scalar? [YOUTUBE 1:18] [TRANSCRIPT]
How do you multiply a vector by a scalar? Example [YOUTUBE 1:16] [TRANSCRIPT]
What do you mean by linear combination of vectors? [YOUTUBE 1:44] [TRANSCRIPT]
What do you mean by linear combination of vectors? Example [YOUTUBE 2:34][TRANSCRIPT]
What do you mean by vectors being linearly independent? [YOUTUBE 2:28] [TRANSCRIPT]
Are these vectors linearly independent? Example 1 [YOUTUBE 3:17] [TRANSCRIPT]
Are these vectors linearly independent? Example 2 [YOUTUBE 9:13][TRANSCRIPT]
What do you mean by rank of a set of vectors? [YOUTUBE 2:09] [TRANSCRIPT]
Rank of a set of vectors: Example 1 [YOUTUBE 3:53] [TRANSCRIPT]
Rank of a set of vectors: Example 2 [YOUTUBE 2:27] [TRANSCRIPT]
Prove that if a set of vectors contains a null vector, the set of vectors is linearly dependent [YOUTUBE 2:29] [TRANSCRIPT]
Prove that if a set of vectors is linearly independent, then a subset of it is also linearly independent [YOUTUBE 5:42][TRANSCRIPT]
Prove that if a set of vectors is linearly dependent, then at least one vector can be written as a linear combination of others [YOUTUBE 4:06] [TRANSCRIPT]
How can vectors be used to write simultaneous linear equations? [YOUTUBE 5:08] [TRANSCRIPT]
How can vectors be used to write simultaneous linear equations? Example [YOUTUBE 3:13][TRANSCRIPT]
What is the definition of the dot product of two vectors? [YOUTUBE 2:00] [TRANSCRIPT]
Dot product of two vectors Example [YOUTUBE 1:41] [TRANSCRIPT]
COMPLETE RESOURCES
Get in one place the following: a textbook chapter, a PowerPoint presentation, individual YouTube lecture videos, multiple-choice questions, and problem sets on Vectors.
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https://byjus.com/question-answer/indian-cricket-team-won-4-more-matches-than-it-lost-with-new-zealand-if-it-5/
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Question
# Indian cricket team won 4 more matches than it lost with New Zealand If it won $$\displaystyle\frac{3}{5}$$ of its matches how many matches did India play
A
8
B
12
C
16
D
20
Solution
## The correct option is D 20Let the number of matches lost = xNumber of matches won = x + 4Total matches played = x + x +4 = 2x + 4We have,x + 4 = $$\displaystyle\frac{3}{5}\left ( 2x + 4 \right )$$5x + 20 = 6x + 12x = 8$$\displaystyle\therefore$$ total matches played = 2x + 4$$\displaystyle= 2\left ( 8 \right )+ 4= 20$$Mathematics
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# How long is the incline in the following case:A rope is attached to a 50.0-kg crate to pull it up a frictionless incline with a 30 degree angle at a constant speed to a height of 3 meters.
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
The problem states that the incline forms an angle of 30 degree and rises to a height of 3m. We need to find the height of the incline. I do not know what you mean by w parallel and w perpendicular. Actually we do not need of that information.
The length of the incline can be determined just from the information that the incline forms an angle of 30 degrees and the height reached at its end is 3m.
We have a perpendicular triangle here, with the length of the hypotenuse = length of the incline = L and the length of the side opposite the 30 degree angle as 3 m
sin 30 = 0.5 = 3/L
=> L = 3 / 0.5
=> L = 6m
Therefore the incline has a length of 6 m.
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http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Polynomials-and-rational-expressions.faq?hide_answers=1&beginning=1980
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# Questions on Algebra: Polynomials, rational expressions and equations answered by real tutors!
Algebra -> Algebra -> Polynomials-and-rational-expressions -> Questions on Algebra: Polynomials, rational expressions and equations answered by real tutors! Log On
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Algebra: Polynomials, rational expressions and equations Solvers Lessons Answers archive Quiz In Depth
Question 40607: it takes 4 hours for 9 cooks to prepare a school lunch. How long would it take 6 cooks to prepare the lunch? it would take 6 cooks _____hours. is it 2.66? Click here to see answer by AnlytcPhil(1277)
Question 40607: it takes 4 hours for 9 cooks to prepare a school lunch. How long would it take 6 cooks to prepare the lunch? it would take 6 cooks _____hours. is it 2.66? Click here to see answer by stanbon(57377)
Question 40609: if y varies inversly as x and y=24 when x=8 find y when x=4 y=_____ Thank you can sum one teach me this??????please??????????? Click here to see answer by Nate(3500)
Question 40608: the time(t) required to empty a tank varies inverse as the rate(r) of pumping. A pump can empty a tank at the rate of 1200 L/min how will it take the pump to empty the tank at 3000L? it will take____minutes to empty the tank. Click here to see answer by Nate(3500)
Question 40606: if y varies inversly as x and y=5 when x=3 find x when y=15 x=_____ is it 9? Click here to see answer by Nate(3500)
Question 40598: if y varies inversly as x and y=2 when x=1 find x when y=4 x=_____ Thank you Click here to see answer by Nate(3500)
Question 40596: if y varies inversly as x and y=5 when x=2 find x when y=4 x=_____ Thank you Click here to see answer by Nate(3500)
Question 40368: Simplify and write with positive exponents only (x^2)^-3(x^-2) divided by (x^2)-4 Click here to see answer by junior403(76)
Question 40684: Complete the equation of variation where y varies inversly as x and y=8 when x=10. y= ___/x isnt it 80?? Thank you. Click here to see answer by Nate(3500)
Question 40682: if y varies inversly as the square of x and y=4 when x=5 find y when x=2 y=_____ doesnt y=50? Click here to see answer by Nate(3500)
Question 40685: The Nutty Shoppe has 10 kg of mixed pecans and cashews, which sell for \$ 8 per kilogram. Cashews alone sell for \$8 per kilogram, and pecans sell for \$9 per kilogram. How many kilograms of pecans are there in the mix? How many kilograms of cashews are there in the mix? Thanks Click here to see answer by checkley71(8403)
Question 40674: factor each ploynomial using the distributive property. 5x^2 - 15x 34x^4y^3 - 17x^2y^5 2x^2y^2 - 8x^6y^6 please show work if you can, I am trying to learn this stuff! Thanks! Click here to see answer by Nate(3500)
Question 40674: factor each ploynomial using the distributive property. 5x^2 - 15x 34x^4y^3 - 17x^2y^5 2x^2y^2 - 8x^6y^6 please show work if you can, I am trying to learn this stuff! Thanks! Click here to see answer by checkley71(8403)
Question 40663: The speed of sound is 1088 feet per second. Convert the speed of sound to miles per hour. Round your answer to the nearest whole number. Click here to see answer by Nate(3500)
Question 40662: A jogger covers 3.5 miles in 28 minutes. Find the average speed of the jogger in miles per hour. Click here to see answer by Nate(3500)
Question 40661: A car makes a 120 mile trip 10 miles per hours faster than a truck. The truck takes 2 hours longer to make the trip. What are the speeds of the car and the truck? Click here to see answer by checkley71(8403)
Question 40706: Solve. 2y^2 – y – 3 = 0 Thanks! Click here to see answer by stanbon(57377)
Question 40723: A tank can be filled in 18 hours by Pipe A or 24 hours by Pipe B. How long would it take both pipes to fill the tank? _____hours -------------------------------------------------------------------------------- It takes Sarah 8 hours to fill a Widget. Jerry can do the same job in 12 hours. How long would it take them to assemble a Widget if they worked together? _______hours. Thanks Click here to see answer by checkley71(8403)
Question 40725: Christy spent \$500 over her budget on gifts during the holidays. In addition to her regular job, she took a part time job at the Pharmacy that paid \$10 per hour. Her schedule would allow her to work 12 extra hour a week at the pharmacy. How many whole weeks will it take Christy to have enough money to pay off the debt? Thank you Click here to see answer by checkley71(8403)
Question 40722: The area of a rectangle of length x is given by 3x^2 + 5x. Find the width of the rectangle. Click here to see answer by tutorcecilia(2152)
Question 40750: factory the following polynomial completely 14x ^2 - 20x + 6 Click here to see answer by Nate(3500)
Question 40757: INTERSECTIONS AND UNIONS let A={1,2,3,4,5} B={6} C={7,8} Find Intersection A & C i dont understand because there's nothing the same between A and C. Click here to see answer by rajagopalan(158)
Question 40726: Kareem purchased a leather jacket for \$800 with his new credit card. He budgeted \$50 a month to pay toward the debt. When he read the fine print on the credit card statement, he saw that for each month he owed a balance, a finance charge of \$17.50 would be added. How many months will it take Kareem to pay off the leather jacket? Thanks Click here to see answer by rajagopalan(158)
Question 40413: 8. If f(x)= (ax-b)/c and f(f(x))=x for all x, then which of the following must be true? I. |a| = |c| II. a = 0 III. b>0 Answer. I and II only The book has the solution but I can't seem to understand. Perhaps u can explain it in another way? thank u Click here to see answer by venugopalramana(3286)
Question 40761: factor this trinomial 16x^2 + 40xy + 25y^2 Click here to see answer by gsmani_iyer(180)
Question 40764: Rewrite the middle term as the sum of two terms and then factor completely 12w^2 + 19w + 4 Click here to see answer by gsmani_iyer(180)
Question 40765: Rewrite the middle term as the sum of two terms and then factor completely 4z^3 - 18z^2 - 10z Click here to see answer by stanbon(57377)
Question 40727: can someone TEACH me how to find a boundry of a line??????(step by step) for example--- Y>/ x+3 y>/ -x+1 Click here to see answer by stanbon(57377)
Question 40763: Factor this trinomial completly 2a^3 - 52a^2b + 96ab^2 Click here to see answer by mbarugel(146)
Question 40793: A number added to the product of 6 and the reciprocal of the number equals -5. Click here to see answer by venugopalramana(3286)
Question 40795: A semi-truck travels 300 miles through the flatland in the same amount of time that it travels 180 miles through mountains. The rate of the truck is 20 miles per hour slower in the mountains than in the flatland. Find both the flatland rate and the mountain rate. Click here to see answer by josmiceli(9695)
Question 40797: A boat can travel 9 miles upstream in the same amount of time it takes to tarvel 11 miles downstream. If the current of the river is 3miles per hour, complete the chart below and use it to find the speed of the boat in still water. distance = rate * time UPSTREAM 9 r-3 DOWNSTREAM 11 r+3 Click here to see answer by stanbon(57377)
Question 40797: A boat can travel 9 miles upstream in the same amount of time it takes to tarvel 11 miles downstream. If the current of the river is 3miles per hour, complete the chart below and use it to find the speed of the boat in still water. distance = rate * time UPSTREAM 9 r-3 DOWNSTREAM 11 r+3 Click here to see answer by josmiceli(9695)
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### 35th IMO 1994 shortlist
Problem N1
Find the largest possible subset of {1, 2, ... , 15} such that the product of any three distinct elements of the subset is not a square.
Solution
{1, 4, 5, 6, 7, 10, 11, 12, 13, 14} shows that we can have a subset of 10 elements with the required property.
[This is a matter of tiresome verification. Suppose there is a set T of 3 elements with product a square. If 5 is in T, then 10 must be in T (because it is the only other element divisible by 5). Hence the third element must be 2n2, but there is no such element. So 5 is not in T. Hence 10 is not in T. Similarly, if 7 is in T, then 14 must be in T, but then there is no third element of the form 2n2, so neither 7 nor 14 can be in T. 11 cannot be in T because it is the only element divisible by 11. Similarly 13. If 6 or 12 is in T, then the other must be (because they are the only elements divisible by 3). But then 6·12 = 3223, so the third element must be divisible by an odd power of 2 and there are none such (except for 10 and 14, which have already been eliminated). So there are at most two candidates for T (1 and 4) which is not enough.]
If we include 10, 3 and 12 then we must exclude 1 (1·3·12 = 62), 4 (4·3·12 = 122), 9 (9·3·12 = 182), one of 2, 6 (2·6·3 = 62) and one of 5, 15 (3·5·15 = 152), so there are at most 10 elements. If we include 10, and exclude one or more of 3, 12, then we must also exclude at least one of 2, 5 (2·5·10 = 102) and at least one of 1, 4, 9 (1·4·9 = 62) and at least one of 7, 8, 14 (7·8·14 = 282).
Finally, suppose we exclude 10. But we must also exclude one of 1, 4, 9 (product 62), one of 2, 6, 12 (product 122), one of 3, 5, 15 (product 152) and one of 7, 8, 14 (product 282). So there are at most 10 elements.
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https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-12-exponential-functions-and-logarithmic-functions-12-6-solving-exponential-equations-and-logarithmic-equations-12-6-exercise-set-page-826/91
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Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$x\displaystyle \in\{\frac{1}{100,000},\quad 100,000\}$
In order for the equation to be defined, $\left\{\begin{array}{l} x\gt 0\\ x^{\log x}\gt 0 \end{array}\right.\qquad(*)$ apply $\quad\log_{a}M^{p}=p\cdot\log_{a}M$ $\log x\cdot\log x=25\qquad$ ... let $t=\log x$ $t^{2}=25$ $t=\pm 5$ $\left[\begin{array}{lll} \log x=-5 & ...or... & \log x=5\\ x=10^{-5} & & x=10^{5} \end{array}\right]$ Both satisfy (*), so they are both valid solutions $(10^{-5})^{\log 10^{-5}}=(10^{-5})^{-5}=10^{25}\gt 0$ $(10^{5})^{\log 10^{5}}=(10^{5})^{5}=10^{25}\gt 0$
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https://www.chegg.com/homework-help/introduction-to-the-design-and-analysis-of-algorithms-3rd-edition-chapter-5.3-solutions-9780132316811
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# Introduction to the Design and Analysis of Algorithms (3rd Edition) Edit edition 90% (20 ratings) for this chapter’s solutionsSolutions for Chapter 5.3
We have solutions for your book!
Chapter: Problem:
Design a divide-and-conquer algorithm for computing the number of levels in a binary tree. (In particular, the algorithm must return 0 and 1 for the empty and single-node trees, respectively.) What is the time efficiency class of your algorithm?
Step-by-step solution:
Chapter: Problem:
• Step 1 of 3
The divide-and-conquer algorithm divides a problem into sub problems and then solve them individually. Then the results of the sub problems are combined to find the solution of the original problem.
The number of levels in a binary tree can be computed by the dividing the tree into left and right sub trees and computing the height of these sub trees recursively. The maximum of both the values will be the height of the binary tree or the number of levels in the tree.
• Step 2 of 3
Assuming the number of levels for an empty binary tree and a binary tree with a single is 0 and 1 respectively, the algorithm to find the number of levels is as follows:
Algorithm:
Function max(int , int b):
• If the variable a is greater than b, return a.
• Otherwise, return b.
Function find_levels(node* root):
• If the root is null, return 0.
• Otherwise, return the following:
max(find_levels (root->left), find_levels(root->right))+1
• Step 3 of 3
The time complexity of the above algorithm is as follows:
• The function find_levels() is invoked for every node in the binary tree.
• The function max() takes constant time to return the maximum of the 2 values.
• Therefore, the total running time is , where n is the number of nodes in the binary tree.
Corresponding Textbook
Introduction to the Design and Analysis of Algorithms | 3rd Edition
9780132316811ISBN-13: 0132316811ISBN: Anany LevitinAuthors:
Alternate ISBN: 9780133001365
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https://www.gradesaver.com/textbooks/math/calculus/calculus-with-applications-10th-edition/chapter-5-graphs-and-the-derivative-5-3-higher-derivatives-concavity-and-the-second-derivative-test-5-3-exercises-page-283/27
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## Calculus with Applications (10th Edition)
Since the graph of the function lies above its tangent line at each point of $(2,\infty)$. So, concave upward on $$(2,\infty).$$ Since the graph of the function lies below its tangent line at each point of $(-\infty, 2)$ . So, concave downward on $$(-\infty, 2).$$ Since a point where a graph changes concavity is $(2,3).$ So, Inflection point at $$(2,3).$$
Since the graph of the function lies above its tangent line at each point of $(2,\infty)$. So, concave upward on $$(2,\infty).$$ Since the graph of the function lies below its tangent line at each point of $(-\infty, 2)$ . So, concave downward on $$(-\infty, 2).$$ Since a point where a graph changes concavity is $(2,3).$ So, Inflection point at $$(2,3).$$
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You are Here: Home >< Maths
# AQA Maths Core 1+2 May 2013 Watch
1. (Original post by The H)
Sweet
I just left it like that because the mark scheme agrees with it
Cool :P, what angles did you get for tan=-1?
2. (Original post by Alifff)
is that in c1? which question is it?
No it was core 2
3. (Original post by nholliday95)
You might be right I don't think I am, but I was on for ages and couldn't figure out any other answer... can you remember how you did it?
it was a confusing method so i think i got it wrong but anyhow i got (85/2)^0.5 so i made it (170/4)^0.5 and took the quarter out to a half haha
4. (Original post by The H)
No it was core 2
pheww haha i got scared for a moment
5. How many marks was the u3 question worth???
6. (Original post by nimbusquaffle)
on the transformation question, wasn't the x value the same and it was only the y that changed postion. i put a translation (0,-7)
Think about it this way. From (x+1)^2 +y^2 = 49 to (x-5)^2 + (y+7)^2 = 49, the co-ordinates of the centre of the circle move from (-1, 0) to (5, -7). This implies a shift to the right of 6 units in x and 7 down in y. This is represented by the vector [6, -7]. Hope this helps.
7. and there is a question to ask u find that angle in radians, u just make your calculator into ike radian mode and do sin rule dont u
8. (Original post by ash9002)
How many marks was the u3 question worth???
I think it was 2
9. (Original post by wanjianyi)
and there is a question to ask u find that angle in radians, u just make your calculator into ike radian mode and do sin rule dont u
Yep, and because it was an obtuse angle, you had to take the original answer away from 180 degrees. I think i got 1.7 or something like that
10. (Original post by Lauraloo1)
I think it was 2
Will just drop one mark for putting in 3/2 instead of 2/3 then? Got 116 instead of 56 as a result of that :/
11. (Original post by \$oulja)
for my translation, i don't remember what i put... either (-6,7) or (6,-7). i think i'd lose 1 mark there if i got that wrong. thinking of it now, i think i actually did put (6,-7) can only hope for the best now.
By merely stating translation you will get a mark, unless if they specified it was a translation in the question, which I don't think thy did. Hope you get the grade you want man.
12. (Original post by Equality4all)
Think about it this way. From (x+1)^2 +y^2 = 49 to (x-5)^2 + (y+7)^2 = 49, the co-ordinates of the centre of the circle move from (-1, 0) to (5, -7). This implies a shift to the right of 6 units in x and 7 down in y. This is represented by the vector [6, -7]. Hope this helps.
i think you are right as i might have read the question wrong so did the translation backwards and got (-6, 7)
13. (Original post by ash9002)
Will just drop one mark for putting in 3/2 instead of 2/3 then? Got 116 instead of 56 as a result of that :/
aww that sucks hopefully they'll give you a mark for method
14. (Original post by ash9002)
Will just drop one mark for putting in 3/2 instead of 2/3 then? Got 116 instead of 56 as a result of that :/
How did you show that p was 3/2
15. U know for the 9th question of tanx=-1 my teacher said do the inverse of the positive value only so i did and got 45degrees so i got 180 - 45 and 360 + 45, X = 135, 405 Anyone else remember this and then for the very last one i got summit like 14degrees and cant remember the other 2 values U had to factorise i got summit like (3cos(theta)-2) (2cos(theta)+3) Anyone else confirm or better ?
16. In core 2, what was the answer for the transformation question with (2,-0.9) I forgot the numbers :/
17. Hi does anyone remember the 'K' question i remember getting 2^3 at the end but i can't remember the exact Q.. anyone else get 3?
19. (Original post by Alifff)
i think you are right as i might have read the question wrong so did the translation backwards and got (-6, 7)
Unless I misread the question aha :P I had 50 minutes or so to check my work and the wording of the questions, so I do hope I haven't messed that one up. How do you think you did overall?
20. (Original post by Sweatynerd)
Hey guys this is about the c1 exam, can anyone explain the answer to the complete the square one where there was a coefficient in front of the x? Like the question at the end where it said out it in the format 1/2root m (which is an integer)??? Or at least say what they got? Cheers
so complete the square for X^2+3x gives you (x+3)^2-9
so final is: 2(X+3)^2-4=0 as -9+5=-4.
should of gotten 1/2root5.
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https://math.stackexchange.com/questions/2873650/can-a-10-times-10-square-be-entirely-covered-by-25-t-shape-bricks
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Can a $10\times 10$ square be entirely covered by 25 $T$-shape bricks?
Let $ABCD$ be a square in which length of a side is $10$ meters. Suppose that we have $T$-shape brick which consists of $4$ smaller squares in which a side of each smaller square has length of $1$ meter. Can $ABCD$ be entirely covered by 25 $T$-shape bricks?
I have tried but can not figure out where to start. Please give me some hints, not the full solution!
• Use colouring proofs. – Anik Bhowmick Aug 6 '18 at 7:12
• If the grid is coloured like a checkerboard, $25$ T-shapes can never cover the same number of black squares as red. – Robert Israel Aug 6 '18 at 7:20
• Closely related (but not a dupe!) – Jyrki Lahtonen Aug 6 '18 at 7:36
Think to the given square as a $10\times 10$ chessboard with alternate black and white squares and assume that such covering with $T$ pieces exists. Then each of the $25$ $T$ pieces will have 1) $3$ black squares and $1$ white square or 2) $3$ white squares and $1$ black square. Let $b$ the number of $T$ pieces of the first category and $w$ the number of $T$ pieces of the second category. Then the integers $b$ and $w$ should satisfy the following equations: $$3b+w=\frac{10\cdot 10}{2},\quad b+3w=\frac{10\cdot 10}{2}.$$ What may we conclude?
Bonus question: What happens when the given square is $n\times n$?
• Do you know the answer for the $m$-by-$n$ rectangular case? It suffices that both $m$ and $n$ are divisible by $4$, but I cannot prove that this is necessary. I know that one of $m$ and $n$ is divisible by $4$ (or better, just as you showed, $8\mid mn$). – Batominovski Aug 6 '18 at 8:14
• @Batominovski Yes, both $m$ and $n$ are divisible by 4 is also necessary. See jstor.org/stable/pdf/2313337.pdf?seq=1#page_scan_tab_contents – Robert Z Aug 6 '18 at 8:30
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# AP Stats Unit 6 Practice FRQ #2
After completing a sale, a car company likes to send a follow-up survey where customers can indicate their level of satisfaction with their experience. One of the questions in the survey asks “would you recommend our company to a friend looking to purchase a vehicle?” The company wonders if people would answer the question differently based on whether they bought a new or used vehicle. From a list of all 2018 vehicle sales, the company randomly selects 105 customers who bought a new vehicle 120 customers who bought a used vehicle. 88 of the customers who bought new vehicles answered “yes,” while 85 of the customers who bought used vehicles answered “yes.”
At the significance level of 0.05, do the data provide convincing statistical evidence that the proportion of customers who would answer “yes” to the survey question is different for new vs used vehicle sales?
P(yes|new) = 88/105 = 0.838
P(yes|used) = 85/120 = 0.708
margin of error = +/- (1.960)sqrt[(0.838(0.162)/105)+(0.708(0.292)/120)]
margin of error = +/- 0.108
confidence interval = 0.05 +/- 0.108 = (-0.058, 0.113)
No, the data do not provide convincing statistical evidence that the proportion of customers who would answer “yes” to the survey question is different for new vs used vehicle sales. Since 0 is captured in the 95% confidence interval of -0.058 and 0.113, the data shows that the true difference in proportions could be 0.
Hello again!
I’ll give feedback on your work below, but I want to start with noticing that you used a 2-sample confidence interval to answer the question. That is a totally valid strategy for a situation like this, but only because the alternative hypothesis was “different”; had the scenario asked “higher” or “lower” the confidence interval would not work in the same way. Typically, when given a significance level, and asked if there is “convincing statistical evidence” of something, we should be running a hypothesis test. That said, you will still be scored for your work with the confidence interval. The scoring for a “convincing statistical evidence…” scenario includes:
1. Stating null/alternative hypotheses
2. Defining the parameters in the null/alternative hypotheses
3. Choosing an appropriate test/interval by name
4. Checking the conditions to run the chosen test/interval
5. Writing the results from the chosen test/interval
6. Correctly interpreting the results from the chosen test/interval in terms of whether we do or don’t have evidence for the alternative hypothesis.
Given that list (some parts are scored together to create a question with 3-4 scoring components), you can likely see that your work doesn’t have enough there to be earning much of the available credit. You calculate the appropriate margin of error, and therefore obtain a confidence interval, but never name the interval, check conditions (random samples, approximately normal sampling distribution [at least 10 successes/at least 10 failures], 10% condition), or write hypotheses. Additionally, you used “0.05” in the interval, instead of using (0.838 - 0.708 = 0.13) as your difference of proportions to add/subtract the margin of error. That would have led you to a different confidence interval where 0 was not included. Given that your interval did include 0 though, your conclusion that we do not have convincing evidence would get scored as correct, because you interpreted the answer you got correctly. Unfortunately, you would not get credit for the other components of the question.
Whew, that was a lot of writing! I hope I was able to be clear in my explanations though - it’s clear that you know many of the concepts, but need some work in how to address the types of questions the AP exam gives. Having said all of the above, I should note that with the formatting changes of this year’s test, you will likely not need to do the level of calculations that you did. It’s much more likely that you will be given an interval (or p-value, or something else) and be asked to just do the interpretation. Or perhaps choose an appropriate procedure and check conditions, but not actually finish building the interval/obtaining the p-value.
Thanks for reading, and I hope this helps!
~Jerry
p_1 = the proportion of customers who bought a new vehicle and answered yes to the survey question
p_2 = the proportion of customers who bought a new vehicle and answered yes to the survey question
2-sample z test for p_1 - p_2
H_0: p_1 - p_2 = 0, H_a: p_1 - p_2 not equal 0
Conditions:
Random - Stated that the company “randomly selects” customers for the survey
10% Condition for Independence - satisfied since it is safe to assume that there are at least 105(10) = 1050 customers who bought a new vehicle at the car company, and at least 120(10) = 1200 customers who bought a used vehicle at the car company.
Large Counts Condition - satisfied since
n_1p-hat_1 = 1050.838 = 87.99 >=10
n_1*(1-p-hat_1) = 1050.162 = 17.01 >=10
n_2
p-hat_2 = 1200.708 = 84.96 >= 10
n_2
(1-p-hat_2) = 120*0.292 = 35.04 >=10
With Large Counts Condition satisfied, the sampling distribution of p-hat_1 - p-hat_2 is approximately normal.
p-hat_1 = 88/105 = 0.838, n_1=105
p-hat_2 = 85/120 = 0.708, n_2=120
z* = 2.303
P-val = P(z>=2.303 or z<=-2.303) = 0.021247
Since 0.021247 < alpha of 0.05, we reject the null hypothesis, because there is convincing statistical evidence that the proportion of customers who would answer “yes” to the survey question is different for new vs used vehicle sales.
Boys, it’s too bad that we probably won’t get a “straight-up” hypothesis test like this on the exam, because you are READY for them.Strong execution from top to bottom, presented clearly. One thing that you’re going to facepalm about: you defined the parameters p1 and p2 as the exact same thing. “p2” should say “used”
test type- two sample proportional difference hypothesis z-test
formula
1. introduction-
H0= p1-p2=0
Ha=p1-p2 does not = 0
p1=the proportion of all customers from a list of 2018 vehicle sales who bought a new vehicle and would recommend our company to a friend
p2=the proportion of all customers from a list of 2018 vehicle sales who bought a used vehicle and would recommend our company to a friend
1. conditions-
Since we are dealing with a two-sample proportional difference hypothesis test, we will have to pool/combine our proportions.
pc=88+85/105+120=0.75
qc=1-0.75=0.25
*we have to check for the independence of our pooled data --> .75(225)=168.75 >=10 & .25(225)=56.25 >= 10
conditions for new cars:
• simple random sample: stated in the problem- “the company randomly selects 105 customers…”
• independence: 10(105)=1050 Assume that the population of new cars purchased in 2018 is greater than 1050.
• normal: .84(105)=(88 >= 10), .162(105)=(17 >= 10)
All of the conditions for new cars are met.
conditions for old cars:
• simple random sample: stated in the problem- “the company randomly selects…120 customers…”
• independence: 10(120)=1200 Assume that the population of new cars purchased in 2018 is greater than 1200.
• normal: .708(120)=(85 >= 10), .292(120)=(35 >= 10)
All of the conditions for old cars are met.
1. solve-
z=.84-.708/(sqrt.(.25x.75)/105 + (.25x.75)/120) = 2.28 --> *I looked at table z to find the p-value of -2.28, p-value=.0129(2)=.0258
2. conclusion-
(.0258<.05 our significance level) --> Reject the H0 in favor of the Ha. We have significant evidence that the proportion of all customers from a list of 2018 vehicle sales who bought a new vehicle and said they would recommend the company is different than the proportion of all customers from a list of 2018 vehicle sales who bought a used vehicle and said that they would recommend our company to a friend.
QUESTION: In the fiveable livestream about hypothesis tests for proportions, there was an similar example about a pharmaceutical company testing a new headache remedy and testing 2 treatments (old vs. new) given to 2 different sample of people.
For the null hypothesis, we did not do a difference of P_new and P_old (p1-p2) and instead we did P_new= P_old and for the alternative: P_new> P_Old.
For this question could the hypothesis have been Ho : p1=P2, Ha = p1 does not equal p2?
p1= Proportion of customers who would answer “yes” to the survey question
p2= Proportion of customers who would answer “no” to the survey question.
Thank you so much for your help.
p1= Proportion of customers who bought a new car and answers “yes” to the survey question.
p2= Proportion of customers who bought a used car and answered “yes” to the survey question.
Ho= p1-p2=0 Ha= p1-p2 does not equal 0
We are interested in conducting a 2 sample z test for a difference in population proportions.
Conditions:
Random- A random sample of 105 customers who bought a new vehicle and 120 customers who bought a used vehicle is taken
Normal- Sample of new cars: np = 105 * 0.838= 88 is greater than or equal to 10.
n(1-p)= 105(0.162)= 17 is greater than or equal to 10.
Sample of used cars: np= 120* 0.708= 85 is greater than or equal to 10.
n(1-p)= 120(0.292)= 35 is greater than or equal to 10.
Calculator: 2-Prop Z Test {x1=88, n1=105, x2=85, n2=120, p1 does not equal p2} = p: 0.0212
Since the p-value of 0.0212 is less than our alpha level of 0.05, we have convincing statistical evidence to reject the null hypothesis. The proportion of customers who would answer “yes” to the survey question is different for new vs. used vehicle sales.
Hypotheses
H_o: p_1 = p_2
H_a: p_1 ≠ p_2
Where p_1 is the true proportion of customers who bought a new vehicle and answered “yes” to the survey question.
Where p_2 is the true proportion of customers who bought a used vehicle and answered “yes” to the survey question.
Assumptions
Independence:
-We have 2 independent random samples of customers from 2018 vehicle sales.
-Population of new vehicle customers is at least 1050 and the population of used vehicle customers is at least 1200.
Normality:
n_1 * p-hat_1 = 105 * 0.8381 = 88 ≥ 10
n_1 * (1-p-hat_1) = 105* (0.1619) = 16.9995 ≥ 10
n_2 * p-hat_2 = 120 * 0.7083 = 84.996 ≥ 10
n_2 * (1-p-hat_2) = 120 * 0.2917 = 35.004 ≥ 10
Since all 4 are greater than 10, the sampling distribution is approximately normal.
Calculations
p_hat_combined = 105(0.8381) + 120(0.7083) / 105+102 = 0.7689
z = (0.8381 - 0.7083) - 0 / sqrt((0.7689 * (1-0.7689) / 105) + (0.7869 * (1-0.7869) / 120) = 2.3036
p-value = 2*normalcdf(2.3036, 1E99, 0, 1) = 0.0212
alpha = 0.05
p-value<alpha
Conclusion
Since the p-value<alpha, we reject the H_o. There is sufficient evidence to suggest that the proportion of customers who bought a new vehicle and answered “yes” to the survey question is different from the proportion of customers who bought a used vehicle and answered “yes” to the survey question.
This is about as thorough a response as I’ve seen! Very well done - you’ve nailed all of the components. Be ready for just one or two of those components to be explicitly tested this year.
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Nice job! You’ve defined parameters, checked conditions, named the test, obtained appropriate test statistic and p-value, and made an appropriate conclusion. To address your question - we could do p1 = p2 and p1 =/= p2, but not for the parameters you defined. It would need to be p1 = proportion of customers purchasing new cars who would say yes; p2 = proportion of customers purchasing used cars who would say yes
Well done from top to bottom - you’ve got parameters, conditions, appropriate calculations, and appropriate conclusions. You’re ready! Note that it’s likely that you’ll be asked for these components in isolation this year, as opposed to “all at once” like this.
Thank you so much for your help. I had a quick question about the conditions/assumptions of a 2 sample proportion procedure. Is it necessary that we show independence for pooled data? For example, do we have to show that np_hat_c >= 10 and np_hat_c >= 10? If so, why do we need to show this?
It sounds like you’re asking about the “normal” condition (at least 10 successes/failures) and not the “independence” condition (sample is no more than 10% of entire population if selection is done without replacement). If that’s the case, you do not need to check the pooled data - you can do it as you did and check the individual samples.
Identify)
• pn: proportion of customers who bought a new vehicle who would answer “yes” to the survey question.
• pu: proportion of customers who bought a used vehicle who would answer “yes” to the survey question.
• pn-pu: Difference in the proportion of customers who bought a new and used vehicle who would answer “yes” to the survey question.
• Ho: pn=pu
• Ha: pn≠pu
• 2 proportion Z Test
Conditions)
• Random: We were told the that both samples were randomly selected.
• Normal: The sampling distribution is approximately normal because both…
nn×pn>=10 (105×88/105>10) & nn×(1-pn)>=10 (105×17/105>10)
nu×pu>=10 (120×85/120>10) & nu×(1-pu)>=10 (120×35/120>10)
• 10% Condition: There are more than 10×105 new car buyers and 10×120 used car buyers.
• The two samples are Independent
Calculations)
zTest_2Prop(xn=88,nn=105,xu=85,nu=120) p1≠p2=2.30356
x=2.30356
p-value=0.021247
My Teacher told me that for this years exam the actual formulas don’t need to be written and that the calculator function is enough. Have you heard otherwise?
Interpret)
Because the p-value(0.021247) is below a reasonable alpha(0.05), we reject the null hypothesis(Ho). There is sufficient evidence to conclude that the proportion of customers who would answer “yes” to the survey question is different for new vs used vehicle sales.
Nice work! To answer your question: you actually never need to show the formulas on hypothesis test or confidence interval FRQs… assuming you’ve correctly named the interval (which you have here). This year, because the test is being designed to accommodate people without a calculator, it is assumed that you will not need to run a full hypothesis test like this one. It is likely that it will be “broken up” into parts: perhaps you’ll need to write hypotheses and check conditions in one part; perhaps you’ll be given a p-value in another and asked to make a conclusion.
Your work is all correct, by the way.
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2 sample z test for a difference in proportions
Parameters:
p_new = the true proportion of buyers of a new vehicle from the car company who would answer yes to the the question in the survey
p_old = the true proportion of buyers of a used vehicle from the car company who would answer yes to the the question in the survey
Hypotheses:
Ho: p_new = p_old
Ha: p_new is not equal to p_old
Conditions:
1. Random: Random samples of 105 and 120 customers who bought new and used cars, respectively, from the company stated CHECK
2. Success/ Failure: Because the number of success and failures are greater than 10 - CHECK:
n_newp-hat_new =88 >10 and n_newq-hat_new =17>10 and n_used*p-hat_used = 85 >10 and n_used * q-hat_used = 35>10
3. 10% condition (10% * pop > n OR 10*n <population size): It is reasonable to assume that there are more than 105 *10 = 1,050 and 120 * 10 = 1,200 customers who bought cars from the company in 2018 that were new and used, respectively. CHECK
4. Independence: It is reasonable to assume that those customers who buy new and used cars from the company in 2018 are independent of each other. CHECK
Mechanics:
p-hat_new = 0.838095 p-hat_used = 0.708333 p-hat_pooled = 0.768889 n_1 =105 n_2 = 120
z = 2.30356
P-Value( z>2.30356) = 0.021247
Conclusion:
We reject the null hypothesis, Ho, in favor of the alternative because the p value of 0.021247 is less than alpha = 0.05. We have sufficient evidence that the true proportion of customers who would answer yes to the survey question stated that bought new and used cars from the car company in 2018 are different.
Quick question: How do I know which test to use? There are so many options, like 2-sample z test, t test etc. I’m very confused.
Nice work! You’ve defined parameters, chosen the test, checked conditions, made appropriate conclusions… the whole nine yards. Be prepared to do some of these steps in isolation on Friday.
Hi Denise -
That’s a good question! There are 5 procedures you should be familiar with:
1. One-sample z-test for a population proportion
2. Two-sample z-test for a difference in population proportions
3. One-sample t-test for a population mean
4. Two-sample t-test for a difference in population means
5. Matched-pairs t-test for a difference in population means
Options #1 and #2 are only if you have a categorical variable, and thus are using proportions. Options #3-5 are for quantitative variables, and thus means. For a larger breakdown (and some practice questions!), I’d check out this previous stream where I review in-depth the different options and when to use them:
https://app.fiveable.me/ap-stats/unit-7/review-inference-procedures/watch/GaUXKaMPLXkI8jTrCtbM
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*ap® and advanced placement® are registered trademarks of the college board, which was not involved in the production of, and does not endorse, this product.
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NY Regents Exam Teasers IQ Tests Chemistry Biology GK C++ Recipes Search < a href="/cgi/members/home.cgi" class="toplink">Members Sign off
Print Preview
## Arithmetic Homework Help : Decimal Division
Worksheet / Test Paper Quiz Review
Quiz : Type the quotient obtained on dividing the following decimals.
1. 182.5 ÷ 5 =
1825 ÷ 5 = 365
There are 1 decimal places in 182.5, so the quotient must have 1 decimal places.
Thus, 182.5 ÷ 5 = 36.5
2. 184.8 ÷ 2 =
1848 ÷ 2 = 924
There is 1 decimal place in 184.8, so the quotient must have 1 decimal place.
Thus, 184.8 ÷ 2 = 92.4
3. 21.33 ÷ 9 =
2133 ÷ 9 = 237
There are 2 decimal places in 21.33, so the quotient must have 2 decimal places.
Thus, 21.33 ÷ 9 = 2.37
4. 72.27 ÷ 9 =
7227 ÷ 9 = 803
There are 2 decimal places in 72.27, so the quotient must have 2 decimal places.
Thus, 72.27 ÷ 9 = 8.03
5. 86.3 ÷ 10 =
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0
# What fraction of 1 meter is 30 centimeters?
Wiki User
2016-03-25 09:56:01
30 centimeters is 3/10, or 0.3, meters.
1 meter is 100 cm
so 30 cm = 0.3 m
Wiki User
2016-03-25 09:56:01
Study guides
20 cards
## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
➡️
See all cards
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Wiki User
2009-12-13 16:52:47
There are 100 centimetres in one metre. Therefore, 30 centimetres is equal to 3/10, or three tenths, of a metre.
Anonymous
Lvl 1
2020-05-22 10:31:32
0.3m
Anonymous
Lvl 1
2020-07-07 08:03:02
ok
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# Cracking LCM & HCF Remainder Problems: 8 Simple Formulas Explained YouTube Lecture Handouts
Get top class preparation for UGC right from your home: Get detailed illustrated notes covering entire syllabus: point-by-point for high retention.
Watch Video Lecture on YouTube: Cracking LCM & HCF Remainder Problems: 8 Simple Formulas Explained
Cracking LCM & HCF Remainder Problems: 8 Simple Formulas Explained
## Recap
Method:
• Prime Factorization
• Divisibility tests
Problem Keywords
• LCM: Minimum number, least amount, smallest duration etc.
• HCF: Maximum number, most amount, longest duration etc.
4 Types
## Greatest Number Which Divides X, Y and Z Leaves Same Remainder R (Given)?HCF Type 2 (Same Remainder- Given)
Let’s find a number which divides all 14, 20 and 32 leaving remainder 2
## Greatest Number Which Divides X, Y and Z Leaves Same Remainder R (Not Given)?HCF Type 3 (Same Remainder- Not Given)
Number which divides all 14, 20 and 32 leaving same remainder
## Greatest Number Which Divides X, Y and Z, Leaving Remainders a, B and C (Respectively) HCF Type 4 (Different Remainder- Given)
Number which divides 12, 18 and 30 leaving remainder 2, 3 and 0.
## HCF Problems – 4 Types Summary Understand and Remember
• Greatest number which divides x, y and z = HCF (x, y, z)
• Greatest number which divides x, y and z and leaves remainder r = HCF(x - r, y - r, z - r)
• Greatest number which divides x, y and z and leaves same remainder = HCF (|𝑥−𝑦|, |𝑦−𝑧|, |𝑧−𝑥|)
• Greatest number which divides x, y and z and leaves remainder a, b, c = HCF(x - a, y - b, z - c)
4 Types
## Smallest/Largest Number of N Digits Divisible by X, Y, Z? LCM Type 2 (Multiples of LCM)
Smallest/Largest number of 3 digits divisible by 6, 9, 12
## Smallest Number when Divided by X, Y and Z Leaves Same Remainder R (Given)?LCM Type 3 (Same Remainder)
Number divisible by 6, 9, and 12 leaves remainder 2
## Smallest Number when Divided by X, Y and Z Leaves Remainder a, B, C?LCM Type 4 (Different Remainder)
x - a = y - b = z - c = common difference d
Smallest number divided by 2, 3, 4, 5, 6 leaves remainder 1, 2, 3, 4, 5
## LCM Problems – 4 Types SummaryUnderstand and Remember
• Smallest number divisible by x, y and z = LCM(x, y, z)
• Smallest number of n digits divisible by x, y and z = Multiple of LCM(x, y, z)
• Smallest number when divided by x, y and z leaves same remainder r = LCM(x, y, z) + r
• Smallest number when divided by x, y and z leaves remainder a, b, c
• x - a = y - b = z - c = common difference d
• LCM (a, b, c) - d
## Variations of LCM (Understand)
• Smallest/Largest number of n digits when divided by x, y and z leaves same remainder r = Multiple of LCM(x, y, z) + r
• Smallest/Largest number of n digits when divided by x, y and z leaves remainder a, b, c = Multiple of LCM(x, y, z) – d (x-a = y-b = z-c = common difference d)
## Example - 1
Find the greatest number of 5-digits which on being divided by 9, 12, 24 and 45 leaves 3, 6, 18 and 39 as remainders respectively.
## Example - 2
Find the smallest number which, on being added 23 to it, is exactly divisible by 32, 36, 48 and 96.
• When dividing a number by 12, 15 or 48 there will always be a remainder of 10. If the number is the least possible, how many divisors does the number have?
• Number of divisors of (p, q primes) is Application of Combination
## Generalization – Chinese Remainder Theorem Next Class!!
Find the smallest number which when divided by 7, 9, and 11 produces 1, 2, and 3 as reminders
• 7 – 1 = 6
• 9 – 2 = 7
• 11 – 3 = 8
But
• 7 – 2 × 1 = 5
• 9 – 2 × 2 = 5
• 11 – 2 × 3 = 5
Developed by:
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# Homework Help: Average force problem
1. Nov 8, 2004
### ballahboy
A 3.0kg steel ball strikes a massive wall at 10m/s at an angle of 60degrees with the plane of thge wall. It bounces off with the same speed and angle. If the ball is in contact with the wall for .20s, wut is the average force exerted on the ball by the wall?
its is wut i got..
p=mv p=mv
3(10) =3(-10)
=30 =-30
deltaP=mvfinal-mvinitial
=30-(-30)
=60
averageF=deltaP/delta time
=60/.2
=300N
correct or incorrect?
can someone help me?
2. Nov 8, 2004
### ek
Yep, looks good.
To be totally correct though, F*t is really impulse, not momentum, though they are the same magnitudes, so it really doesn't matter. But that's the kind of multiple choice question that can screw you over, so it's worth knowing.
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# AC ANALYSIS
## Time domain analysis
We are surrounded by sinusoidal
signals:
1. Household power supply (50 Hz)
3. Cellular Phones (UHF:400 to
1000 MHz)
4. Television
Less loss in transmission
Carrier waves are modulated by voice
or video signals of lower frequency
Phasors:
We will use the complex
exponential to represent currents
and voltages.
This is known as the phasor
representation.
Reason: The voltage across an
inductor (and the current through a
capacitor), involves derivatives.
If we use sinusoidal functions to
describe we have mixture of sine and
cosine terms: not easy to handle
For a resistor, the current and
voltage (across it) are in phase.
Let the current through the inductor
be given by:
t j
Ie t i
e
= ) (
The voltage across the inductor is
given by:
) ( ) ( t Li j Ie Lj
dt
di
L t v
t j
e e
e
= = =
) 2 / (
) (
t e
e
+
=
t j
LIe t v
Therefore the voltage across the
inductor leads the current by t/2.
For a capacitor, if the voltage across
the capacitor is given by:
t j
Ve t v
e
= ) (
) ( ) ( t Cv j Ve Cj
dt
dv
C t i
t j
e e
e
= = =
) 2 / (
) (
t e
e
+
=
t j
CVe t i
Therefore the current in a capacitor
These relations can be represented
by phasor diagrams.
Imaginary axis
Real axis
I
jeLI
et
Phasor diagram
for inductor
Real axis
Imaginary axis
I
V=I/jeC
et
Phasor diagram
for capacitor
For the resistor, inductor and
capacitor, we get a phasor equation
of the form:
V = Z I
Z is the impedance of the element.
For a resistor: Z
R
= R
For a capacitor: Z
C
= 1/jeC
For an inductor: Z
L
= jeL
Note that the impedance of a
capacitor decreases with increase in
frequency and that of an inductor
increases with increase in frequency.
The reciprocal of impedance is
notation is that we can avoid writing
down differential equations and
solving them.
The source supplies AC given by
V
s
= 6sin2t = 6cos(2t-90
0
)
1/jeC = -j
1 O
~
V
s
=6/-90
0
F
KVL for this circuit can be written
as:
V
s
= I jI
6/-90
0
= (1-j)I
= 2(-45
0
)I
0
0
0
45 2 3
45 2
90 6
Z =
Z
Z
= I
Voltage across capacitor = -jI
0
135 2 3 Z
Real axis
Imaginary axis
V
o
= 32/-135
0
I=32/-45
0
135
0
V
s
=6/-90
0
Once we have obtained the solution
by phasor method, we can write
down the time dependence of the
voltages and currents
V
0
= 32cos(2t-135
0
)
i(t) = 32cos(2t-45
0
)
Application of Thevenins theorem
to AC circuits:
5cos(4t-30
0
)V
5 mF
50 H
Z
L
~
200 O
V
oc
V
Applying KCL at the node labeled
V
oc
:
( )
200 50
30 5
j
V V
j
V
oc oc
=
Z
3V
oc
+V = 20 /-30
0
Applying KCL at node V:
200 200
V
j
V V
oc
=
oc
V
j
V
+
=
1
1
Substituting this in the other
equation, we have:
0
30 20
1
3 Z =
+
+
j
V
V
oc
oc
0
30 20
1
3 4
Z =
|
|
.
|
\
|
+
+
j
j
V
oc
0
30 20
3 4
1
Z
|
|
.
|
\
|
+
+
=
j
j
V
oc
Multiplying the numerator and
denominator by the complex
conjugate of the denominator:
0
30 20
25
7
Z
|
.
|
\
|
+
=
j
V
oc
Convert the complex number in
brackets to polar form.
0 0 0
87 . 21 2 4 30 20 13 . 8
5
2
Z = Z Z =
oc
V
Therefore:
V
oc
= 42cos(4t-21.87
0
)
To find the Thevenin equivalent
impedance, set the independent
source to zero.
Net impedance of the resistor
and the inductor is 200+200j. This is
in parallel with the capacitor.
j j
j j
Z
o
50 200 200
) 50 )( 200 200 (
+
+
=
j
j j
Z
o
3 4
) )( 200 200 (
+
+
=
j
j
j
Z
o
56 8
3 4
) 200 200 (
=
+
+
=
Z
o
= 402/-81.87
0
O
Ex. 4.6
Application of Mesh analysis and
Cramers rule to AC circuits.
9cos(5t)V
1/15 F
~
6 O
V
V
1
- +
3 O
I
1
I
2
For the mesh on the left:
0 ) ( 3 6 9
2 1 1
= + I I j I
0 3 ) ( 9
2 2 1
= + I I I j
For the mesh on the right:
-3v
1
+3I
2
=0
0 3 ) 3 1 (
1 2
= + I I j
Solve using Cramers rule
0
1
49 . 2 3 . 1
53
) 23 ( 3
Z =
+
=
j
I
0
2
6 . 15 24 . 1
53
) 2 7 ( 9
Z =
=
j
I
Power:
If a source is supplying a
sinusoidal voltage, then the average
voltage supplied by the source over
one cycle is zero.
However, the average power supplied
by the source is non-zero.
For an arbitrary element (could be
resistor/capacitor/inductor), the
power absorbed by the element is:
| | ) cos( ) cos( ) ( ) ( ) ( e e + = = t I t V t i t v t p
( ) | | e cos 2 cos
2
+ + = t
VI
}
=
T
ave
dt t p
T
P
0
) (
1
( ) | |
}
+ + =
T
ave
dt t
VI
T
P
0
cos 2 cos
2
1
e
The first integral can be shown to be
equal to zero and the second integral
is non-zero, giving
P
ave
= (VI/2 )cos|
Therefore if the element is:
A resistor: P
ave
= VI/2
an inductor: P
ave
= 0
A capacitor: P
ave
= 0
Recall that, the impedance of an
inductor or a capacitor is complex.
The real part of the impedance is
called resistance and the imaginary
part is called reactance.
Maximum Power transfer:
Given a circuit with a Thevenin
equivalent output impedance
Z
0
= R
o
+jX
o
impedance Z
L
= R
L
+jX
L
maximum transfer of power?
The condition is:
Z
L
= R
L
+jX
L
= R
o
jX
0
= Z
o
*
If Z
L
is restricted to be real (only
is:
2 2
o o L
X R R + =
Effective value of a sinusoid:
We define I
e ,
the effective value
of an ac current, as the dc current
that would yield the same power
absorption (on the average) as the ac
current.
Therefore:
2 2
2
1
RI RI
e
=
I I
e
2
1
=
V V V
e
707 . 0
2
1
= =
For household power supplies, the
usual value (220 Volts) which is
quoted is the effective value and not
the peak value of the sinusoidal
voltage.
Therefore the peak value is
220/0.707 = 311.17 V.
Why is the effective voltage in
India 220 V and 115 V in the US?
The effective value is same as the
RMS value.
The RMS or root mean squared
value of the power dissipated in a
resistor is:
}
=
T
e
dt t Ri
T
RI
0
2 2
) (
1
( )
}
=
T
e
dt t RI
T
RI
0
2 2 2
cos
1
e
2
I
I
e
=
The quantity I
e
V
e
is known as the
apparent power, but we know that
the average power is:
P
ave
= (VI/2 )cos| = V
e
I
e
cos|
We use the unit Volt-Ampere (VA) for
the apparent power to distinguish it from the
actual (average power).
u cos
power apparent
power average
factor Power = =
The angle u is known as the power
factor angle.
If power factor is negative, it
indicates that the load is capacitive
and if pf is positive it indicates an
For an inductive load, the pf is
called lagging (current lags the
voltage)
for a capacitive load, the pf is
Power factor correction:
It is desirable to have a power
factor of around 1 (no lead or lag).
If we have capacitive or
motors etc), then we can apply
power factor correction to
compensate for the lead or lag in the
capacitors or inductors as the case
may be.
1. Regulator 2. Network connection points
3. Fuses 4. Contactors 5. Capacitors
Transformer 400/230 Volts
What is power factor correction?
Ex 4.11
A large consumer of electricity
(like our campus!) requires 10 kW of
power by using 230 V rms at a pf
angle of 60
0
lagging (pf = 0.5).
The current drawn by the load
will be :
A
V
P
I
e
e
87
5 . 0 230
10000
cos
=
= =
u
But the transmission line from the
power station to the campus will
have some resistance (say 0.1 O).
Therefore the power station has to
generate
V
e
= 0.1 I
L
+ V
L
= 8.7/-60
0
+ 230 /0
0
= 234 /-1.84
0
The loss in transmission is
I
e
2
R = 757 W.
If the power factor can be corrected, the
current and hence the transmission loss
will reduce.
What is a Single Phase three-wire
circuit?
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# How to find the third coordinate of a triangle with specific conditions
I am writing a physics paper relating the motion of objects using Loedel diagrams and standard trigonometry.
I am attempting to prove that an object's velocity can be found by using a specific type of triangle, and it has led me to this problem for which I need assistance:
Using a standard Cartesian coordinate system, and for a given triangle $abc$, with angles $A,B,C$, where $a$ is opposite to $A$, $b$ opposite $B$, and $c$ opposite $C$:
If side $c$ of the triangle is bounded at coordinates $(0,0)$ and $(1,0)$, and angles $A$ and $B$ are acute angles, then:
Find the equation for the $x$- and $y$-coordinates of the third point of the triangle such that for any given angle $A$, $a = h/b$, where $h$ is the height of the triangle, and therefore:
$$\sin A = a$$
By the sine law, we have $$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}.$$ Now, since $a = \sin A$, it follows that $c = \sin C$. It is given that $c = 1$, so $C = \frac \pi 2$, i.e., $abc$ s a right-angle triangle.
• The answer is right, and both $A$ and $B$ are acute angles, while $C=90°$. Commented Dec 31, 2017 at 22:45
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## 6045
6,045 (six thousand forty-five) is an odd four-digits composite number following 6044 and preceding 6046. In scientific notation, it is written as 6.045 × 103. The sum of its digits is 15. It has a total of 4 prime factors and 16 positive divisors. There are 2,880 positive integers (up to 6045) that are relatively prime to 6045.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 4
• Sum of Digits 15
• Digital Root 6
## Name
Short name 6 thousand 45 six thousand forty-five
## Notation
Scientific notation 6.045 × 103 6.045 × 103
## Prime Factorization of 6045
Prime Factorization 3 × 5 × 13 × 31
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 6045 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 6,045 is 3 × 5 × 13 × 31. Since it has a total of 4 prime factors, 6,045 is a composite number.
## Divisors of 6045
1, 3, 5, 13, 15, 31, 39, 65, 93, 155, 195, 403, 465, 1209, 2015, 6045
16 divisors
Even divisors 0 16 8 8
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 10752 Sum of all the positive divisors of n s(n) 4707 Sum of the proper positive divisors of n A(n) 672 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 77.7496 Returns the nth root of the product of n divisors H(n) 8.99554 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 6,045 can be divided by 16 positive divisors (out of which 0 are even, and 16 are odd). The sum of these divisors (counting 6,045) is 10,752, the average is 672.
## Other Arithmetic Functions (n = 6045)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 2880 Total number of positive integers not greater than n that are coprime to n λ(n) 60 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 793 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 2,880 positive integers (less than 6,045) that are coprime with 6,045. And there are approximately 793 prime numbers less than or equal to 6,045.
## Divisibility of 6045
m n mod m 2 3 4 5 6 7 8 9 1 0 1 0 3 4 5 6
The number 6,045 is divisible by 3 and 5.
## Classification of 6045
• Arithmetic
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (6045)
Base System Value
2 Binary 1011110011101
3 Ternary 22021220
4 Quaternary 1132131
5 Quinary 143140
6 Senary 43553
8 Octal 13635
10 Decimal 6045
12 Duodecimal 35b9
20 Vigesimal f25
36 Base36 4nx
## Basic calculations (n = 6045)
### Multiplication
n×i
n×2 12090 18135 24180 30225
### Division
ni
n⁄2 3022.5 2015 1511.25 1209
### Exponentiation
ni
n2 36542025 220896541125 1335319591100625 8072006928203278125
### Nth Root
i√n
2√n 77.7496 18.2165 8.81757 5.70531
## 6045 as geometric shapes
### Circle
Diameter 12090 37981.9 1.148e+08
### Sphere
Volume 9.25289e+11 4.59201e+08 37981.9
### Square
Length = n
Perimeter 24180 3.6542e+07 8548.92
### Cube
Length = n
Surface area 2.19252e+08 2.20897e+11 10470.2
### Equilateral Triangle
Length = n
Perimeter 18135 1.58232e+07 5235.12
### Triangular Pyramid
Length = n
Surface area 6.32926e+07 2.60329e+10 4935.72
## Cryptographic Hash Functions
md5 f449d27f42a9b2a25b247ac15989090f 8f24e66720547719862a6a19a2f34c5b925c5d40 476b7e335e8400cd9e420e0c0e3c1c2d6581ff96468c99cba367619cc937bb29 c67caa083bf27a0971ff54ccee660e2498795d5893e5b73176b086b162d59d79be101196edf43ad239561376cc161516292edd8d3e2f2d3f0cd158e65d0e0185 2ecef6930d7e8dc336d4a4f25601b69abc553c28
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# What is a cyclic quadrilateral?
A cyclic quadrilateral is a four-sided polygon whose vertices all lie on a common circle. In other words, it is a closed figure that can be made by tracing a path around a circle. The term "cyclic" means "having the property of being recurrent or periodic." A good way to remember this definition is that the prefix "cycle" comes from the Greek word for circle, which is kyklos. So, a cyclic quadrilateral is basically a four-sided figure whose corners all lie on some circle or other.
Cyclic quadrilaterals have some interesting and important properties that make them useful in mathematical problems. For instance, because all the vertices of a cyclic quadrilateral lie on the same circle, we can infer that the opposite sides of the quadrilateral are parallel to each other. This fact can be very helpful when solving certain types of geometry problems.
The properties of cyclic quadrilaterals can be summarized as follows:
1. All four vertices lie on a common circle.
2. Opposite sides of a cyclic quadrilateral are parallel to each other.
3. The diagonals of a cyclic quadrilateral intersect at two points, which are equidistant from the center of the circle.
4. The sum of the angles of a cyclic quadrilateral is 360 degrees.
5. The altitude (or height) from any vertex to the opposite side intersects that side at its midpoint.
6. A cyclic quadrilateral has two pairs of congruent sides if and only if it is an inscribed rectangle (a rectangle whose vertices all lie on the circumference of a circle).
7. A cyclic quadrilateral has two pairs of opposite angles congruent if and only if it is an inscribed square (a square whose vertices all lie on the circumference of a circle).
In conclusion, we have seen that a cyclic quadrilateral is simply a four-sided figure whose vertices all lie on some circle or other. We have also seen that cyclic quadrilaterals have some interesting and important properties, which make them useful in mathematical problems. Thanks for reading!
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# Equations and Formulae: Age 14-16
This is part of our Secondary Curriculum collection of favourite rich tasks arranged by topic.
### Pick's Theorem
##### Age 14 to 16Challenge Level
Polygons drawn on square dotty paper have dots on their perimeter (p) and often internal (i) ones as well. Find a relationship between p, i and the area of the polygons.
### Warmsnug Double Glazing
##### Age 14 to 16Challenge Level
How have "Warmsnug" arrived at the prices shown on their windows? Which window has been given an incorrect price?
### Which Is Cheaper?
##### Age 14 to 16Challenge Level
When I park my car in Mathstown, there are two car parks to choose from. Can you help me to decide which one to use?
### Arithmagons
##### Age 11 to 16Challenge Level
Can you find the values at the vertices when you know the values on the edges?
### Multiplication Arithmagons
##### Age 14 to 16Challenge Level
Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons?
### How Old Am I?
##### Age 14 to 16Challenge Level
In 15 years' time my age will be the square of my age 15 years ago. Can you work out my age, and when I had other special birthdays?
### Matchless
##### Age 14 to 16Challenge Level
There is a particular value of x, and a value of y to go with it, which make all five expressions equal in value, can you find that x, y pair ?
### Fair Shares?
##### Age 14 to 16Challenge Level
A mother wants to share a sum of money by giving each of her children in turn a lump sum plus a fraction of the remainder. How can she do this in order to share the money out equally?
### CD Heaven
##### Age 14 to 16Challenge Level
All CD Heaven stores were given the same number of a popular CD to sell for £24. In their two week sale each store reduces the price of the CD by 25% ... How many CDs did the store sell at each price?
### Which Is Bigger?
##### Age 14 to 16Challenge Level
Which is bigger, n+10 or 2n+3? Can you find a good method of answering similar questions?
### Terminology
##### Age 14 to 16Challenge Level
Given an equilateral triangle inside an isosceles triangle, can you find a relationship between the angles?
### Training Schedule
##### Age 14 to 16Challenge Level
The heptathlon is an athletics competition consisting of 7 events. Can you make sense of the scoring system in order to advise a heptathlete on the best way to reach her target?
### System Speak
##### Age 16 to 18Challenge Level
Five equations... five unknowns... can you solve the system?
##### Age 16 to 18Challenge Level
What do you get when you raise a quadratic to the power of a quadratic?
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https://heritage.bafta.org/answers/3339123-two-cruise-ships-leave-port-at-the-same-time-ship-a-sails-north-at-a-speed-of-20-mph-while
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# Two cruise ships leave port at the same time. Ship A sails north at a speed of 20 mph while Ship B sails east at a speed of 40 mph. (a) Find an expression in terms of the time t (in hours) giving the distance D between two ships. D(t) = (b) Using the expression obtained in part (a), find the distance between the two ships 4 hr after leaving the port. (Round your answer to two decimal places.) mi
a) D(t) =
b) 178.885 miles
Step-by-step explanation:
Ship A travels north at the rate of 20 mph.
Ship B travels east at the rate of 40 mph.
After t hours, Ship A is at a distance of 20t miles from the origin.
Similarly, Ship B is at a distance of 40t miles from the origin.
(a) Distance D(t) =
=
=
=
(b) Distance between the two ships when t = 4,
=
= miles
= 80 * 2.236
= 178.885 miles
## Related Questions
What is the square root of negative 100?
The square root of negative 100 can be dissected step by step. The square root of 100 is 10 however there is no such thing as square root of something negative, thus the imaginary notion is applied. square root of -1 is equal to i .Hence the answer to this problem is 10 i.
0 + 10i
Step-by-step explanation:
William is reading a book for class. The book is 94 pages long. For the first assignment, William read 25 pages. For the second assignment, he read 37 pages. For the last assignment, he has to finish the book. How many pages does he have left to read?
The book has 94 pages. The following is a break down of the word problem:
94 - 25 = 69
69 - 37 = 32
He reads 32 pages for the last assignment.
Based on the question, we can found that :
25 + 37 + x = 94
62 + x = 94
x = 94 - 62
x = 32 Pages
hope this helps
Fred peeled 9 carrots. Nancy peeled 6 carrot. how many fewer carrots did Nancy peeled than fred
9-6=3 she peeled 3 less than fred Or fred peeled 3 more than Nancy
A lawn sprinkler spins in a circle. The sprinkler covers a radius of 4 feet. Which choice is the closest to the area of lawn that the sprinkler can cover?
The correct answer is 50 feet
Step-by-step explanation:
Area= r^2*π = 4^2*π= 16π.
The final answer is 16π ft^2~
Eduardo counted ten seconds between swing lightning and hearing thunder, and he knew that the lighting was about 2 miles away.if he counted four seconds between the next flash of lighting and thunder, about how far away was the lightning?
The Lightning was about 1/2 mile away from Eduardo.
Consider this function. f(x) = |x – 4| + 6
If the domain is restricted to the portion of the graph with a positive slope, how are the domain and range of the function and its inverse related?
1.Since the domain of the original function is limited to x> 6, the range of the inverse function is y ≤ 6.
2.Since the domain of the original function is limited to x> 4, the range of the inverse function is y ≤ 1.
3.Since the range of the original function is limited to y> 6, the domain of the inverse function is x ≥ 6.
4.Since the range of the original function is limited to y> 4, the domain of the inverse function is x ≥ 1.
3. Since the range of the original function is limited to y> 6, the domain of the inverse function is x ≥ 6.
Step-by-step explanation:
The domain of a function is the range of its inverse, and vice versa. The only answer choice that expresses this relationship is choice 3.
__
The slope of the function is undefined at x=4, so restricting the function domain to the portion with positive slope means the domain restriction of the function is x > 4. That also means the range restriction of the function is y > 6. The domain restriction of the inverse function is the same: x > 6, not x ≥ 6. The answer choice has an error.
3. Since the range of the original function is limited to y> 6, the domain of the inverse function is x ≥ 6.
Step-by-step explanation:
Given absolute function,.
f(x) = |x-4|+6,
Since, an absolute function is defined for all real numbers,
So, the domain of f(x) is the set of all real numbers,
∴ Options (1) and (2) can not be true.
Now, for any real number,
|x-4| ≥ 0,
That is, f(x) ≥ 6,
Hence, the range ( possible value of output of a function ) of f(x) would be all real numbers greater than equal to 6,
I.e. Range of f(x) is, f(x) ≥ 6,
⇒ if y = f(x) ⇒ the range of the original function is limited to y > 6
∴ Option (4) can not be true,
Also, range of = Domain of
Hence, domain of is x ≥ 6.
A spotlight is mounted on the eaves of a house 3030 feet above the ground. a flower bed runs between the house and the sidewalk, so the closest the ladder can be placed to the house is 1616 feet. how long a ladder is needed so that an electrician can reach the place where the light is mounted?
For this case we can model the problem as a rectangle triangle.
We have two sides.
We want to find the hypotenuse of the triangle.
We have then:
h = root ((30) ^ 2 + (16) ^ 2)
h = 34 feet
is needed a ladder of 34 feet long so that an electrician can reach the place where the light is mounted
tina is ordering a pair of shoes that cost 123.99 if the sales tax is 8% and the shipping is 19.99 what is the total cost
\$153.82 is the total cost
256.99 have a nice day sir/maam
A worker has four different job offers, each with a contract for seven years. Assuming the job descriptions are identical, which offer will allow the worker to earn the least over seven years? a starting salary of \$125,000 with a 2% increase each year a starting salary of \$110,000 with a 3% increase each year a starting salary of \$98,000 with a 8% increase each year a starting salary of \$105,000 with a 4% increase each year
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# Making Special Tapes to Measure Tree Diameters!
1,795
15
1
## Introduction: Making Special Tapes to Measure Tree Diameters!
Measurement of the girth of a tree is one of the standard forestry measurements. This measurement is taken at 1.35 meters (4.5 feet) from the ground and known as the Diameter Breast Height (or DBH for short).
This measurement of a tree's diameter can be measured either using calipers, or by using a tool called DBH tape. What is special about DBH tape is it measures the circumference of the tree, and converts this measure to the diameter by dividing by Pi (i.e. the long number 3.14159....). This division is done on the tape, so that every 3.14 cm along the tape, is labeled as one cm. This makes it easy because no calculations are required when using the DBH tape to measure tree diameter.
For this instructable I wanted to show how to make this specialized DBH tape so you could make your own!
You'll need:
• a permanent marker
• a long piece of wood, or similar
• a square tool, or a ruler
• 5 pieces of specialized lumber strapping (flexible plastic that will wrap around trees for making your DBH)
• 2-4 clamps
• a paint pen
• Tuck tape
• String
### Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
## Step 1: Set Up
Lay out a reference board with measurements in pi centimeters (i.e. make a mark every 3.14 cm). Use your square or ruler to mark each interval across the board with your marker.
Then, clamp five lengths of plastic lumber strapping along the reference board.
## Step 2: Marking Your DBH Tape
• Using the reference board as a guide, mark across the strapping at each interval using the paint pen.
• Make sure you leave about 5 cm at each of the ends of the DBH tape.
• Next, divide the first unit (from one mark to the other) into 10 equal parts and make those lines. This will allow you to measure to 1/10 of a Pi cm.
• Number each interval on the DBH tape in order, starting at 1.
• Choose which end will be the starting end. Then, wrap the starting end with TuckTape to secure a string to hold the DBH tape in place when you are making your measurements.
There you have it - a DBH tape of your very own. Now you can go out and wrap them around trees and know exactly what their diameter is! Then you can return in a year and measure the same trees and see if they've grown! Enjoy.
1 88
24
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# 1986 AJHSME Problems/Problem 24
## Problem
The $600$ students at King Middle School are divided into three groups of equal size for lunch. Each group has lunch at a different time. A computer randomly assigns each student to one of three lunch groups. The probability that three friends, Al, Bob, and Carol, will be assigned to the same lunch group is approximately
$\text{(A)}\ \frac{1}{27} \qquad \text{(B)}\ \frac{1}{9} \qquad \text{(C)}\ \frac{1}{8} \qquad \text{(D)}\ \frac{1}{6} \qquad \text{(E)}\ \frac{1}{3}$
## Solution
Let us first assign Al to a group. We want to estimate the probability that Bob and Carol are assigned to the same group as Al. As the groups are large and of equal size, we can estimate that Bob and Carol each have a $\approx \frac{1}{3}$ probability of being assigned to the same group as Al, and that these events are mostly independent of each other. The probability that all three are in the same lunch group is approximately $\left(\frac{1}{3}\right)^2 = \frac{1}{9}$, or $\boxed{\text{(B)}}$.
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https://www.expertsmind.com/library/determine-magnitude-of-acceleration-of-pilot-at-lowest-point-57961.aspx
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### Determine magnitude of acceleration of pilot at lowest point
Assignment Help Physics
##### Reference no: EM137961
Q1. A plate carries a charge of -3.9 µC, while a rod carries a charge of +2.2 µC. How many electrons should be transferred from the plate to the rod, so that both objects have the same charge?
Q2. A 200 g trick baseball is thrown at 78 km/h. It explodes in flight into two pieces, with a 74 g piece continuing straight ahead at 90 km/h.
Q3. At lowest point of vertical dive (radius=0.59 km), an airplane has speed of 300 km/h which is not changing. Determine magnitude of acceleration of pilot at lowest point.
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# Definition: Row Echelon Form of a Matrix A matrix in Row Echelon Form (REF) has the following properties: 1.
All rows consisting entirely of zeros occur at the bottom of the matrix. 2. *Below each leading coefficient (first non-zero entry in a given row, also referred to as a pivot) is a column of zeros. 3. Each pivot lies to the right of the pivot in the row above, producing a ”stair case” pattern. * In some books, they require that the leading coefficient of each non-zero row is 1. Example: Write the following matrix in row echelon form: 9 1 −1 −7 −7 −6 0 9
0 −3 −6 4 −1 −2 −1 3 −2 −3 0 3 1 4 5 −9 ⇓ R1 R4 −9 3 3 4
1 0 0 0
4 2 0 −3
5 4 0 −6 ⇓
−9 −6 0 4
1 4 5 −1 −2 −1 −2 −3 0 0 −3 −6 ⇓ R1 + R2 → R2 1 0 −2 0 4 2 −3 −3 5 4 0 −6
−7 1 −1 9
R3
R4 4 2 −3 0 5 4 −6 0 −9 −6 4 0 −7 −6 9 0
1 0 0 0
−9 −6 3 4
−7 −6 −1 9
⇓ (3/2)R2 + R3 → R3 1 4 5 −9 −7 0 2 4 −6 −6 0 0 0 −5 0 0 0 0 0 0
⇓ 2R1 + R3 → R3 1 4 0 2 0 5 0 −3 5 4 10 −6 −9 −6 −15 4 −7 −6 −15 9
In certain books, the above matrix is in Row Echelon Form. We can carry out one more step to obtain 1’s in the pivot position. ⇓ (1/2)R2 → R2 (−1/5)R3 → R3 1 0 0 0 4 1 0 0 5 −9 −7 2 −3 −3 0 1 0 0 0 0
⇓ (−5/2)R2 + R3 → R3
The above matrix is in Row Echelon Form. The boxed numbers are the pivots.
1
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Pentagon Ratio (Posted on 2013-03-30)
A piece of paper has the precise shape of a regular pentagon. The paper is folded such that a vertex of the pentagon coincides with the midpoint of that side of the pentagon which is farthest from the said vertex.
Determine the ratio of the length of the crease to the length of a side of the pentagon.
See The Solution Submitted by K Sengupta No Rating
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Solution. Comment 2 of 2 |
Call the pentagon ABCDE and let its sides be length 1 unit.
Construct a line parallel to AE through C. Extend sides AB and DE.
Call the intersection of AB and the parallel line point F.
Call the intersection of DE and the parallel line point G.
Call the intersection of AB and DE point H.
Construct the line thought C perpendicular to AE.
Call the intersection of this line and AE point I.
I is the midpoint of AE by symmetry.
If we crease the pentagon as instructed we create the perpendicular bisector of CI.
Call the points of intersection of this line with AB, CI and DE points X, Y and Z respectively.
We seek length XZ.
Since triangles BCF and DCG are isosceles, FC=1 and CG=1 so FG=2
Triangles FHG and AHE are similar with ratio of similitude 2, so I is the midpoint of CH.
Since Y is the midpoint of CI, YH= 3/2 * IH.
Triangles AHE and XHZ are similar with ratio of similitude 3/2, so XZ = 3/2 * AE
The ratio is 3/2.
Posted by Jer on 2013-04-01 13:06:07
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# Thread: Finding the Volume of a Cylinder
1. ## Finding the Volume of a Cylinder
Hello Everyone!
I was wondering if I calculated this problem right.
Diameter of Cylinder
1.580cm
Height of Cylinder
2.140cm
Heres the formula I used: V=π x radius squared x Height
Heres how I started the problem
V= 3.14 x .79(squared) x 2.140 = 4.19370236cm3
V = 4,194cm3 ? Not really sure if thats right
2. $\displaystyle V = \pi r^2 h$
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# Diberi f(x) = 4x+2 Dan g(x)=x2-6x+8 , cari gf(x)
2
dari Izzattywengweng
## Jawapan
• shmh
• Tangan Bantuan
2016-03-07T14:08:33+08:00
Gf(x)=g(4x+2)
=2(4x+2)-6(4x+2)+8
=8x+4-24x-12+8
=-24x+8x+4-12+8
=-16x
2016-03-07T14:28:22+08:00
Gf(x)
= g(4x+2)
= 2(4x+2) -6(4x+2) +8
= 8x+4 -24x-12 +8
= -16x
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## Conversion formula
The conversion factor from cubic feet to tablespoons is 1915.0129870073, which means that 1 cubic foot is equal to 1915.0129870073 tablespoons:
1 ft3 = 1915.0129870073 tbsp
To convert 44.2 cubic feet into tablespoons we have to multiply 44.2 by the conversion factor in order to get the volume amount from cubic feet to tablespoons. We can also form a simple proportion to calculate the result:
1 ft3 → 1915.0129870073 tbsp
44.2 ft3 → V(tbsp)
Solve the above proportion to obtain the volume V in tablespoons:
V(tbsp) = 44.2 ft3 × 1915.0129870073 tbsp
V(tbsp) = 84643.574025723 tbsp
The final result is:
44.2 ft3 → 84643.574025723 tbsp
We conclude that 44.2 cubic feet is equivalent to 84643.574025723 tablespoons:
44.2 cubic feet = 84643.574025723 tablespoons
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 tablespoon is equal to 1.1814245930779E-5 × 44.2 cubic feet.
Another way is saying that 44.2 cubic feet is equal to 1 ÷ 1.1814245930779E-5 tablespoons.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that forty-four point two cubic feet is approximately eighty-four thousand six hundred forty-three point five seven four tablespoons:
44.2 ft3 ≅ 84643.574 tbsp
An alternative is also that one tablespoon is approximately zero times forty-four point two cubic feet.
## Conversion table
### cubic feet to tablespoons chart
For quick reference purposes, below is the conversion table you can use to convert from cubic feet to tablespoons
cubic feet (ft3) tablespoons (tbsp)
45.2 cubic feet 86558.587 tablespoons
46.2 cubic feet 88473.6 tablespoons
47.2 cubic feet 90388.613 tablespoons
48.2 cubic feet 92303.626 tablespoons
49.2 cubic feet 94218.639 tablespoons
50.2 cubic feet 96133.652 tablespoons
51.2 cubic feet 98048.665 tablespoons
52.2 cubic feet 99963.678 tablespoons
53.2 cubic feet 101878.691 tablespoons
54.2 cubic feet 103793.704 tablespoons
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# college algebra
posted by on .
a rectangular lot whose perimeter is 440 feet is fenced along three sides. an expensive fencing along the lot's length cost \$16 per foot, and an inexpensive fencing along the two side widths costs only \$11 per foot. The total cost of the fencing along the three sides comes to \$4060. What are the lot's dimensions?
• college algebra - ,
if width=x and length=y,
2x+2y = 440
2*11x + 16y = 4060
22x + 16(220-x) = 4060
6x = 540
x = 90
y = 130
### Answer This Question
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# evaluating $\int\limits _{0}^{1}\frac{1}{\sqrt{x+\varepsilon}}dx$
I came across this : I'm trying to evaluate it up to $o(\epsilon)$ $$F\left(\varepsilon\right)=\int\limits _{0}^{1}\frac{1}{\sqrt{x+\varepsilon}} \, \mathrm{d}x$$
I've trying considering to look at it as the following $$F\left(\varepsilon\right)=\int\limits _{0}^{\theta}\frac{1}{\sqrt{x+\varepsilon}} \, \mathrm{d}x+\int\limits _{\theta}^{1}\frac{1}{\sqrt{x+\varepsilon}}\, \mathrm{d}x$$
and evaluate each integral separately and hoping that the intermediate region will cancel out (this method is widely used in perturbation theory to evaluate integrals) but I can't seem to find the right way to evaluate the first integral...
Any help?
Thanks
• My idea: it is not that hard to see that $(x+\varepsilon)^{-1/2}=x^{-1/2} -\frac{\varepsilon}{2 x^{3/2}} + o(\varepsilon x^{-3/2})$. This is a useful approximation for large $x$. So choose a splitting point $\theta$. Then approximate the integral on $[\theta,1]$ as I described above, and approximate the integral on $[0,\theta]$ using the simple fact that it is between $0$ and $2\sqrt{\theta}$. Now the part I'm not so sure about (which is why I'm not writing an answer) is how to nicely choose $\theta$ to make this a useful approximation.
– Ian
Commented Jul 3, 2015 at 18:49
• (Cont.) You need $\theta$ so large that $\varepsilon \theta^{-3/2}$ is small, but you also need it so small that $2\sqrt{\theta}$ is a good estimate for the integral on $[0,\theta]$. So there is a balance to be struck. And I'm not sure how that balance is going to get the $-2\sqrt{\varepsilon}$ term from the Taylor approximation of the explicit solution.
– Ian
Commented Jul 3, 2015 at 18:51
• @Ian That a good way of thinking, but as you said, the tricky part is for $[0,\theta]$ for that region im stuck with how to expand the expression to get a solution that is allso $\varepsilon^{2}\theta^{-\frac{3}{2}}$ Commented Jul 3, 2015 at 19:02
• Another thought: as you can see from the explicit solution, $\int_\varepsilon^1 x^{-1/2} dx$ is a $O(\varepsilon)$ approximation. There are two pieces to its error: $\int_0^\varepsilon (x+\varepsilon)^{-1/2}$ and $\int_\varepsilon^1 (x+\varepsilon)^{-1/2}-x^{-1/2} dx$. Somehow there is some cancellation between these two errors, because the first error is on the order of $\varepsilon^{1/2}$, which means the second one must also be of the same order and with a coefficient of the opposite sign.
– Ian
Commented Jul 4, 2015 at 14:07
• Thats exactly what i expect, My problem is just figuring out how to evaluate the first region integral. it cannot be expanded around x=0 ... $\int\limits _{0}^{\theta}\frac{1}{\sqrt{x+\epsilon}}dx = \frac{1}{\sqrt{\varepsilon}}\int\limits _{0}^{\theta}1-\frac{1}{2}\frac{x}{\varepsilon}+\frac{3}{8}\left(\frac{x}{ \epsilon}\right)^{2}+O\left(\left(\frac{x}{\varepsilon}\right)^{3}\right)dx$ Commented Jul 4, 2015 at 18:01
We have that $$\int_0^1 \frac{1}{\sqrt{x+\varepsilon}} \, \mathrm{d}x = 2\left(\sqrt{\varepsilon + 1} - \sqrt{\varepsilon}\right).$$
Since we can write $1/\sqrt{x+\varepsilon} = (x+\varepsilon)^{-1/2}$ and use the reverse chain rule.
• @Simba You can do a series representation of the right side and get the final result that you want. In particular you have $\sqrt{\varepsilon+1}=1+\frac{\varepsilon}{2} + o(\varepsilon)$, so that the integral is $2-2\sqrt{\varepsilon}+\varepsilon+o(\varepsilon)$.
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# Variance and standard deviation
## The complete guide to understand variance and standard deviation
Descriptive statistics are used to describe the basic information about the data. They help us understand some features of the data by giving short summaries about the sample. It’s like the first impression of what the data shows us. In a nutshell, descriptive statistics are metrics and quantitative analysis to briefly describe our sample. And they’re broken down into measures of central tendency and measures of variability. In this article, we’ll discuss the latter and we’ll clear the fog on famous questions about variance and standard deviation.
# Variance
## Definition
Suppose that we have a random variable, by definition, it can take different values. The distribution of this random variable is what determines its range and its variation. Now we want to have a metric to measure how much this variable varies. This is what we call variance: it’s a metric to describe the spread between data set from its mean value. It can also be seen as the measure of the width of a distribution but this is more related to normal distribution as for other distributions.
Mathematically, in a population, we can calculate it as: the mean of the square distance between each point and the mean…
--
--
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# DIVIDE: Google Sheets formulas explained
OK, folks. Who here loves spreadsheets as much as I do? Anyone? Anyone at all?
Alright, I get it. Spreadsheets can be a bit dry and boring. But you know what can make them exciting? Formulas!
Yes, formulas. Those magical little equations that can do all sorts of cool things in Google Sheets (or any other spreadsheet program, for that matter).
Now, I know what some of you might be thinking. "But formulas are so complicated! I could never understand them!"
Well, fear not, my friends. Today, I'm going to give you a little crash course on some of the most useful formulas in Google Sheets. And don't worry, I'll keep it easy to understand (and maybe even a little bit fun).
## The basics
Before we dive into some specific formulas, let's go over some basics.
First of all, when you want to use a formula in Google Sheets, you need to start a cell with the equals sign (=). This tells the program that you're about to enter an equation.
Once you've done that, you can start typing in your formula. There are a ton of different formulas you can use, but I'm going to focus on three key ones today: SUM, AVERAGE, and COUNT.
## SUM
Let's start with SUM. This handy little formula allows you to add up a range of cells. For example, let's say you have a spreadsheet with a bunch of numbers in it, and you want to find the total.
All you have to do is start a cell with an equals sign (=), type "SUM(", and then select the range of cells you want to add up. For example, if your numbers are in cells A2 through A10, you would enter "SUM(A2:A10)".
Google Sheets will automatically add up all of the numbers in that range and display the result in the cell.
## AVERAGE
Next up is AVERAGE. This formula does exactly what you'd expect: it finds the average of a range of cells.
So, let's say you have a list of test scores and you want to find the average. Just start a cell with an equals sign, type "AVERAGE(", and select the range of cells you want to average.
For example, if your test scores are in cells B2 through B10, you would enter "AVERAGE(B2:B10)". Google Sheets will give you the average of those scores in the cell.
## COUNT
Finally, we have COUNT. This formula allows you to count how many cells in a range contain numbers (or text, if you prefer).
Let's say you have a spreadsheet with a bunch of students' grades, and you want to know how many of them got an A. Just start a cell with an equals sign, type "COUNT(", and then select the range of cells you want to count.
For example, if your grades are in cells C2 through C20, you would enter "COUNT(C2:C20)". Google Sheets will tell you how many of those cells contain a value.
## Putting it all together
Now that you know the basics of these three formulas, you can start using them together to do some really cool things.
For example, let's say you have a spreadsheet that tracks your expenses, and you want to know how much you spent on groceries last month. You could use the SUM formula to add up all of the grocery expenses for the month, and then use the AVERAGE formula to find out how much you spent per week.
Or, let's say you have a list of names, and you want to know how many of them start with the letter "A". You could use the COUNT formula to count how many cells start with "A".
## Conclusion
So, there you have it: a quick introduction to some of the most useful formulas in Google Sheets. Whether you're tracking your finances, keeping a list of grades, or just playing around with some numbers, formulas can help you get the most out of your spreadsheets.
And don't worry if you don't get it right away. Formulas can be a bit tricky at first, but with a little practice, you'll be a formula master in no time.
So go forth, my friends, and compute!
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### Ball Bearings
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##### Stage: 3, 4 and 5
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# If two equal and opposing deforming forces are applied parallel to the cross section of the cylinder, then there is a relative displacement between the faces in front of the cylinder. The shear force produced on the unit area due to the applied tangential forces is called the......
A
Tenisile stress
B
Compression stress
C
Shear stress
D
Hydraulic stress
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Verified by Experts
## If two equal and opposite deforming forces are applied parallel to the cross section of the cylinder, then there is a relative displacement between the face in front of the cylinder. The she force produced on the unit are due to the applied tangent forces is called the shear stress The formula to calculate average shear stress is force per unit are Where: τ = the shear stress, : F= the force applied. A = the cross - sectional area material with area parallel the applied force vector .
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Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## If two equal and opposite defirming forces are applied parallel to the cross sectional area of the cylinder as shown in the figure, there is a relative displacement between the oppsite faces of the cylinder. The ratio of Ax to L is known as
Alongitudinal strain
Bvolumetric strain
Cshearing strain
DPoisson's ratio
• Question 2 - Select One
## If two equal and opposite defirming forces are applied parallel to the cross sectional area of the cylinder as shown in the figure, there is a relative displacement between the oppsite faces of the cylinder. The ratio of Ax to L is known as
Aloagitudinal strain
BÂ volumetric strain
CShearing strain
DPoisson's ratio
• Question 3 - Select One
## The internal restoring force acting per unit area of cross-section of the deformed body is called
Astrain
Bstress
Celastic limit
Dvolumetric strain
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+0
# Before and After - Comparing
0
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3
+290
Silas had some money at first. His mother gave him another \$6 and he spent \$20 on a book. As a result, the ratio of the money he had at first to the money he had in the end was 10 : 3.
(a) Find the difference between the amount of money Silas had at first and the amount of money Silas had in the end.
(b) How much money did Silas have at first?
Apr 21, 2021
#1
+66
0
Silas at first = S
S+6-20=S/(10/3)
S=20
Silas had 20 at first and he had 6 in the end.
Apr 21, 2021
#2
0
He did not have 6 in the end
Guest Apr 21, 2021
#3
+33805
+1
m +6 -20 given <==== this is the money at the end
m - 14
m / (m-14) = 10 / 3
3m = 10 m - 140
b) m = 20 dollars originally 20 - 14 = 6 at the end a) difference = 14 dollars
Apr 21, 2021
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# There is a 36-card deck. Three cards are taken without
Author Message
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There is a 36-card deck. Three cards are taken without [#permalink] 06 Jul 2003, 11:47
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Question Stats:
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There is a 36-card deck. Three cards are taken without repetition at random. What is the probability of having any three aces?
Manager
Joined: 18 Jun 2003
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Kudos [?]: 16 [0], given: 0
Not sure how many aces are in a 36-card deck, but assuming there are 4, is it:
(4c3)x(4/36)x(3/35)x(2/34) ?
SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6
Kudos [?]: 76 [0], given: 0
1. (4/36)*(3/35)*(2/34) = 1/1785
OR
2. total 36C3, favorable 4C3
4C3/36C3 = 1/1785
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## How weight is scaled.
Every action has equal and opposite action.
There is gravitational force and it pulls us towards it center and A FORCE FROM EARTH is being applied on our body.This is supporting force or Normal force.
Weight is the force of gravity applied on our body by earth and If Earth pulls by X Newton of energy then a force from Earth is also coming of X Newton.
when we stand on Bathroom Scale than our weight is shown by scale .
HOW???
Gravitational force+ Normal force = 0
Then how weight is scaled.
PhysOrg.com physics news on PhysOrg.com >> Iron-platinum alloys could be new-generation hard drives>> Lab sets a new record for creating heralded photons>> Breakthrough calls time on bootleg booze
Quote by DhruvKumar Every action has equal and opposite action. There is gravitational force and it pulls us towards it center and A FORCE FROM EARTH is being applied on our body.This is supporting force or Normal force. Weight is the force of gravity applied on our body by earth and If Earth pulls by X Newton of energy then a force from Earth is also coming of X Newton. when we stand on Bathroom Scale than our weight is shown by scale . HOW??? Gravitational force+ Normal force = 0 Then how weight is scaled.
I don't know how the bathroom scale works, but you can use a spring and see the length of compression and know your weight since the force applied by the spring is proportional to the compression length
Mr.louis plz read my question and understand it ,then give any answer. According to Newton's Third Law, Gravitational force+ Normal force = 0 Then how weight is scaled. As we know Weight is Gravitational force applied on a body by earth.
## How weight is scaled.
Quote by DhruvKumar Mr.louis plz read my question and understand it ,then give any answer. According to Newton's Third Law, Gravitational force+ Normal force = 0 Then how weight is scaled. As we know Weight is Gravitational force applied on a body by earth.
Taking downward as positive
Gravitational force - Normal force = 0
As explained by Mr. Lois.
The spring in the scale is compress and the force is -kx.
It is scaled according to x.
Mentor
Blog Entries: 1
Quote by DhruvKumar According to Newton's Third Law, Gravitational force+ Normal force = 0
The downward gravitational force (your weight) and the upward normal force are not Newton's third law 'action/reaction' pairs. They happen to add to zero because you are in equilibrium--not accelerating. This is an application of Newton's 2nd law, not the third.
Then how weight is scaled. As we know Weight is Gravitational force applied on a body by earth.
What the scale measures is the normal force it exerts. Which, in this case, happens to equal your weight.
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# If 10 years are subtracted from the present age of Mr. Roy and the remainder divided by 14, then you would get the present age of his grandson Sachin. If Sachin is 9 years younger to Saloni whose age is 14 years, then what is the present age of Mr. Roy? (a) 60 yrs. (b) 70 yrs. (c) 74 yrs. (d) 80 yrs.
## Question ID - 54462 :- If 10 years are subtracted from the present age of Mr. Roy and the remainder divided by 14, then you would get the present age of his grandson Sachin. If Sachin is 9 years younger to Saloni whose age is 14 years, then what is the present age of Mr. Roy? (a) 60 yrs. (b) 70 yrs. (c) 74 yrs. (d) 80 yrs.
3537
Saloni’s age = 14 yrs. Sachin’s age = yrs.
Let the present age of Mr. Roy be yrs.
yrs.
Next Question :
X’s age 3 years ago was three times the present age of Y. At present, Z’s age is twice the age of Y. Also, Z is 12 years younger than X. What of the present age of Z? (a) 15 yrs. (b) 24 yrs. (c) 12 yrs. (d) 18 yrs.
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# 13720 (number)
13,720 (thirteen thousand seven hundred twenty) is an even five-digits composite number following 13719 and preceding 13721. In scientific notation, it is written as 1.372 × 104. The sum of its digits is 13. It has a total of 7 prime factors and 32 positive divisors. There are 4,704 positive integers (up to 13720) that are relatively prime to 13720.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 13
• Digital Root 4
## Name
Short name 13 thousand 720 thirteen thousand seven hundred twenty
## Notation
Scientific notation 1.372 × 104 13.72 × 103
## Prime Factorization of 13720
Prime Factorization 23 × 5 × 73
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 7 Total number of prime factors rad(n) 70 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 13,720 is 23 × 5 × 73. Since it has a total of 7 prime factors, 13,720 is a composite number.
## Divisors of 13720
1, 2, 4, 5, 7, 8, 10, 14, 20, 28, 35, 40, 49, 56, 70, 98, 140, 196, 245, 280, 343, 392, 490, 686, 980, 1372, 1715, 1960, 2744, 3430, 6860, 13720
32 divisors
Even divisors 24 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 32 Total number of the positive divisors of n σ(n) 36000 Sum of all the positive divisors of n s(n) 22280 Sum of the proper positive divisors of n A(n) 1125 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 117.132 Returns the nth root of the product of n divisors H(n) 12.1956 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 13,720 can be divided by 32 positive divisors (out of which 24 are even, and 8 are odd). The sum of these divisors (counting 13,720) is 36,000, the average is 1,125.
## Other Arithmetic Functions (n = 13720)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 4704 Total number of positive integers not greater than n that are coprime to n λ(n) 588 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1624 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 4,704 positive integers (less than 13,720) that are coprime with 13,720. And there are approximately 1,624 prime numbers less than or equal to 13,720.
## Divisibility of 13720
m n mod m 2 3 4 5 6 7 8 9 0 1 0 0 4 0 0 4
The number 13,720 is divisible by 2, 4, 5, 7 and 8.
• Arithmetic
• Abundant
• Polite
• Practical
• Frugal
## Base conversion (13720)
Base System Value
2 Binary 11010110011000
3 Ternary 200211011
4 Quaternary 3112120
5 Quinary 414340
6 Senary 143304
8 Octal 32630
10 Decimal 13720
12 Duodecimal 7b34
20 Vigesimal 1e60
36 Base36 al4
## Basic calculations (n = 13720)
### Multiplication
n×y
n×2 27440 41160 54880 68600
### Division
n÷y
n÷2 6860 4573.33 3430 2744
### Exponentiation
ny
n2 188238400 2582630848000 35433695234560000 486150298618163200000
### Nth Root
y√n
2√n 117.132 23.9397 10.8228 6.72157
## 13720 as geometric shapes
### Circle
Diameter 27440 86205.3 5.91368e+08
### Sphere
Volume 1.08181e+13 2.36547e+09 86205.3
### Square
Length = n
Perimeter 54880 1.88238e+08 19403
### Cube
Length = n
Surface area 1.12943e+09 2.58263e+12 23763.7
### Equilateral Triangle
Length = n
Perimeter 41160 8.15096e+07 11881.9
### Triangular Pyramid
Length = n
Surface area 3.26038e+08 3.04366e+11 11202.3
## Cryptographic Hash Functions
md5 030586fa78fc1abdba85da161d943b08 5b8afbd3a3c4f3b87675b6ac20c4cd179498319b 318e52eb40cfb4324c76d4a8d6ee73e377bc1f8346519a7901d31d502c042045 56a98052d3349995cb08d4417ef082d05b6cd57230e47c69b1e5e758e7a0a24a5865600d473049cf0f98a3e0fc76449d670c4b36ca8a13e6f68882617eff0feb 2b3eac80e04f735f80b5f61e3bdce390e7020c3b
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# Algebra 2
posted by Jackie
Game Theory: Consider a game in which each of two people simultaneously chooses an integer: 1 or 2. Find the expected value for player A and the expected value for player B. Is each game fair.
Question: If the numbers are the same, then player A wins 1 point from player B. If the numbers are different, then player B wins 1 point from player A.
1. MathMate
Find probability of numbers matching (2/4) and not matching (2/4).
Construct table for A:
Outcome value,x P(outcome) x.P(x)
Win..... +1 ...... 0.5 ....... 0.5
Lose.....-1 ...... -0.5 .......-0.5
Sum(x.P(x))=0, so game is fair.
2. MathMate
oops
The probability on the second line should read 0.5 (not negative).
3. Jackie
what if player A wins 2 points from player B, and if the numbers are different, than player B wins 1 point from player A?
4. MathMate
Games are random processes.
All we can say is that if he wins, he gets one point, and if he loses, he loses one point. We also know (or believe) that the probability of winning or losing is each 50%.
The actual outcomes depends naturally on chance. But in the long run, the number of points won or loses will be relatively close to zero, which is our calculated expectation.
## Similar Questions
1. ### math-i have another question
Player A and B invented a new game. The probability for Player A to win a round is 1/3 and the probability that Player B will win a round is 2/3. To make the game fair, Player A will score 3 points when he/she winds a round and Player …
2. ### Statistics
The local lottery sets up a game wherein a player has a chance of collecting \$750. The player must choose a 3-digit number and if it matches, the player receives \$750 It costs \$1 to play the game. What is the expected profit (or loss) …
3. ### Stats
The local lottery sets up a game wherein a player has a chance of collecting \$750. The player must choose a 3-digit number and if it matches, the player receives \$750 It costs \$1 to play the game. What is the expected profit (or loss) …
4. ### PLEASE SOMEBODY Answer this Statistics question!
The local lottery sets up a game wherein a player has a chance of collecting \$750. The player must choose a 3-digit number and if it matches, the player receives \$750 It costs \$1 to play the game. What is the expected profit (or loss) …
5. ### math
Students at Euler Middle School are talking about ways to raise money for a school party. One student suggests a game called Heads or Tails. In this game, a player pays 50 cents and chooses heads or tails. The player then tosses a …
6. ### math
A game consists of tossing a coin and rolling an ordinary die. If the player tosses a head and rolls an even number, the player wins \$3. Otherwise, the player loses \$1. What is the expected value of this game?
7. ### Stats
A game is played where a biased coin is flipped. A head is twice as likely as a tail. It costs \$5 to plat the game and if a head occurs you win &7 but if a tail occurs you pay \$3 which of the following is correct?
8. ### Math
A game is played using one dice. If the dice is rolled and shows 1, the player wins \$1; if it shows 2, the player wins \$2; if it shows a 3, the player wins \$3. If the die shows 4,5,or 6 the player wins nothing. a.) If there is a charge …
9. ### Statistics
In a game of dice, the probability of rolling a 12 is 1/36. The probability of rolling a 9, 10, or 11 is 9/36. The probability of rolling any other number is 26/36. If the player rolls a 12, the player wins \$5. If the player rolls …
10. ### statistics
In a game of dice, the probability of rolling a 12 is 1/36. The probability of rolling a 9, 10, or 11 is 9/36. The probability of rolling any other number is 26/36. If the player rolls a 12, the player wins \$5. If the player rolls …
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# 18.1: Integration- Computing the Area Under a Curve
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
This chapter essentially deals with the problem of computing the area under a curve. First, we will employ a basic approach and form trapezoids under a curve. From these trapezoids, we can calculate the total area under a given curve. This method can be tedious and is prone to errors, so in the second half of the chapter, we will utilize a built-in MATLAB function to carry out numerical integration. (This was reformatted by Carey Smith. Also, the text was modified to show integrating a function that is defined in a separate file in addition to showing the use of an anonymous function.)
## A Basic Approach
There are various methods to calculating the area under a curve, for example, Rectangle Method, Trapezoidal Rule and Simpson's Rule. The following procedure is a simplified method.
Consider the curve below:
Figure $$\PageIndex{1}$$. Numerical integration
Each segment under the curve can be calculated as follows:
$$\frac{1}{2}\left(y_{0}+y_{1}\right) \Delta x+\frac{1}{2}\left(y_{1}+y_{2}\right) \Delta x+\frac{1}{2}\left(y_{2}+y_{3}\right) \Delta x$$
Therefore, if we take the sum of the area of each trapezoid, given the limits, we calculate the total area under a curve. Consider the following example.
Example 1
Given the following data, plot an x-y graph and determine the area under a curve between x=3 and x=30
Index x [m] y [N]
1 3 27.00
2 10 14.50
3 15 9.40
4 20 6.70
5 25 5.30
6 30 4.50
Data Set
First, let us enter the data set.
x=[3,10,15,20,25,30];
y=[27,14.5,9.4,6.7,5.3,4.5];
When you type in [x',y'], you will see the following tabulated result. Here we transpose row vectors with ' and displaying them as columns:
3.0000 27.0000 10.0000 14.5000 15.0000 9.4000 20.0000 6.7000 25.0000 5.3000 30.0000 4.5000
To plot the data type the following:
figure
plot(x,y)
title('Distance-Force Graph')
xlabel('Distance[m]')
ylabel('Force[N]')
grid
The following figure is generated:
Figure $$\PageIndex{2}$$. Distance-Force Graph
To compute dx for consecutive x values, we will use the index for each x value, see the given data in the question.:
dx=[x(2)-x(1)
x(3)-x(2)
x(4)-x(3)
x(5)-x(4)
x(6)-x(5)];
dy is computed by the following command:
dy=[0.5*(y(2)+y(1))
0.5*(y(3)+y(2))
0.5*(y(4)+y(3))
0.5*(y(5)+y(4))
0.5*(y(6)+y(5))];
[dx, dy] % Disply these as a table with 2 columns
7.0000 20.7500 5.0000 11.9500 5.0000 8.0500 5.0000 6.0000 5.0000 4.9000
Our results so far are shown below
x [m] y [N] dx [m] dy [N]
3 27.00
10 14.50 7.00 20.75
15 9.40 5.00 11.95
20 6.70 5.00 8.05
25 5.30 5.00 6.00
30 4.50 5.00 4.90
x, y and corresponding differential elements
If we multiply dx by dy, we find da for each element under the curve. The differential area da=dx*dy, can be computed using the 'term by term multiplication' technique in MATLAB as follows:
da=dx.*dy da = 145.2500
59.7500
40.2500
30.0000
24.5000
Each value above represents an element under the curve or the area of trapezoid.
By taking the sum of array elements, we find the total area under the curve.
sum_da = sum(da) % 299.75
The following illustrates all the steps and results of our MATLAB computation.
x [m] y [N] dx [m] dy [N] dA [Nm] 3 27.00 10 14.50 7.00 20.75 145.25 15 9.40 5.00 11.95 59.75 20 6.70 5.00 8.05 40.25 25 5.30 5.00 6.00 30.00 30 4.50 5.00 4.90 24.50 299.75
Computation of the approximate area under a curve
## The Trapezoidal Rule
Sometimes it is rather convenient to use a numerical approach to solve a definite integral. The trapezoid rule allows us to approximate a definite integral using trapezoids.
### The trapz() Function
Z = trapz(Y) computes an approximation of the integral of Y using the trapezoidal method.
Now, let us see a typical problem.
Example 2
Given Area $$=\int_{2}^{5} x^{2} d x$$, an analytical solution would produce 39.
Use trapz command and solve it
Initialize variable x as a row vector, from 2 with increments of 0.1 to 5:
x=2:.1:5;
Declare variable y as y=x^2;.
Note the following error prompt: ??? Error using ==> mpower Inputs must be a scalar and a square matrix.
This is because x is a vector quantity and MATLAB is expecting a scalar input for y.
Because of that, we need to compute y as a vector and to do that we will use the dot operator as follows: y=x.^2;.
This tells MATLAB to create vector y by taking each x value and raising its power to 2.
Now we can issue the following command to calculate the first area, the output will be as follows:
area1=trapz(x,y) area1 = 39.0050
Notice that this numerical value is slightly off. So let us increase the number of increments and calculate the area again:
x=2:.01:5; y=x.^2; area2=trapz(x,y) area2 = 39.0001
Yet another increase in the number of increments:
x=2:.001:5; y=x.^2; area3=trapz(x,y) area3 = 39.0000
Example 3
Determine the value of the following integral:
$$\int_{0}^{\pi} \sin (x) \mathrm{d} x$$
Initialize variable x as a row vector, from 0 with increments of pi/100 to pi:
x=0:pi/100:pi;
Declare variable y as
y=sin(x);
Issue the following command to calculate the first area, the output will be as follows:
area1=trapz(x,y) area1 = 1.9998
Let us increase the increments, as above:
x=0:pi/1000:pi; y=sin(x); area2=trapz(x,y) area2 = 2.0000
Example 4
A gas expands according to the law, PV1.4=c.
Initially, the pressure is 100 kPa when the volume is 1 m3.
Write a script to compute the work done by the gas in expanding to three times its original volume1.
Recall that PV diagrams can be used to estimate the net work performed by a thermodynamic cycle, see Wikipedia or we can use definite integral to compute the work done (WD) as follows:
$$\mathrm{WD}=\int p \mathrm{d} v$$
If we rearrange the expression pressure as a function of volume, we get:
$$P=\frac{c}{V^{1.4}}$$
By considering the initial state, we can determine the value of c:
\begin{aligned} c &=100 \times 1^{1.4} \\ &=100 \end{aligned}
From the equation and the equation above, we can write:
$$P=\frac{100}{V^{1.4}}$$
By inserting P in WD, we get:
$$\mathrm{WD}=\int_{1}^{3} \frac{100}{v^{1.4}} \mathrm{d} v$$
For MATLAB solution, we will consider P as a function of V and WD. Now, let us apply the three-step approach we have used earlier:
1. Initialize variable volume as a row vector, from 1 with increments of 0.001 to 3:
v=1:0.001:3;
2. Declare variable pressure as
p=100./v.^1.4;
3. Use the trapz function to calculate the work done, the output will be as follows:
WorkDone=trapz(v,p) WorkDone = 88.9015
These steps can be combined in an m-file as follows:
clc disp('A gas expands according to the law, pv^1.4=C') disp('Initial pressure is 100 kPa when the volume is 1 m3') disp('Compute the work done by the gas in expanding') disp('To three times its original volume') disp(' ') % Display blank line v=1:.001:3; % Creating a row vector for volume, v p=100./(v.^1.4); % Computing pressure for volume WorkDone=trapz(v,p) % Integrating p*dv over 1 to 3 cubic meters
Example 5
A body moves from rest under the action of a direct force given by $$F=\frac{15}{x+3}$$
where x is the distance in meters from the starting point. Write a script to compute the total work done in moving a distance 10 m.2
Recall that the general definition of mechanical work is given by the following integral, see Wikipedia:
$$\mathrm{WD}=\int F \mathrm{d} x$$
Therefore we can write:
$$\mathrm{WD}=\int_{0}^{10} \frac{15}{x+3} \mathrm{d} x$$
Applying the steps we followed in the previous examples, we write:
clc disp('A body moves from rest under the action of a direct force given') disp('by F=15/(x+3) where x is the distance in meters') disp('From the starting point.') disp('Compute the total work done in moving a distance 10 m.') disp(' ') % Display blank line x=0:.001:10; % Creating a row vector for distance, x F=15./(x+3); % Computing Force for x WorkDone=trapz(x,F) % Integrating F*dx over 0 to 10 meters.
The output of the above code is:
A body moves from rest under the action of a direct force given
by F=15/(x+3) where x is the distance in meters from the starting point.
Compute the total work done in moving a distance 10 m.
WorkDone = 21.9951
## The integral() Function
As we have seen earlier, trapz gives a good approximation for definite integrals. The integral function streamlines numerical integration even further.
(Note by Carey Smith: the original version introduces anonymous functions. It is not necessary to use anonymous functions. It is simpler to create a separate function file like was done for the fzero() function in Chapter 10. Another alternative for MATLAB, but not for Octave, is to create a subfunction at the bottom of the script. )
Syntax for integral()
To evaluate an integral from a minimum to a maximum value, we specify a function and its minimum and maximum
Z = integral(@fun, xmin, xmax).
@fun is a function handle to the function that you want to integrate.
xmin is the lower bound of the integral and xmax is th eupper bound of the integral.
Example 6
Given y=x^2, evaluate the integral from x=2 to x=5 as we have done it with trapz command.
Define a function called "myfunction.m"
function y = myfunction(x)
y = x.^2;
end
Test integrating this function from x = 2 to 5 with this code:
Z = integral(@myfunction, 2, 5)
You should get a result of Z = 39.
Notice that, unlike in the examples, we did not need to define a vector and change the increments to get an accurate result.
The integral() function determines appropriate point spacings. Typically, the spacing will not be uniform, but are optimized for speed an accuracy.
.
### Anonymous Functions (Optional)
An anonymous function is a function that can be defined in the command window (i.e. it does not need to be stored in a program file). Anonymous functions can accept inputs and return outputs, just as standard functions do such as sqrt(X) or log(X).
To define an anonymous function, first we create a handle with @(x) and type in the function:
myfunction=@(x) x^2+1.
If you want to evaluate myfunction at 1, just type in a=myfunction(1) at the command window and you get the result of 2.
## Summary of Key Points
1. In its simplest form, numerical integration involves calculating the areas of segments that make up the area under a curve,
2. MATLAB has built-in functions to perform numerical integration,
3. Z = trapz(Y) computes an approximation of the integral of Y using the trapezoidal method.
4. Anonymous functions are inline statements that we can define with @(x),
5. Z = integral(fun,xmin,xmax) numerically integrates function fun from xmin to xmax.
## Footnotes
• 1 O. N. Mathematics: 2 by J. Dobinson, Penguin Library of Technology. © 1969, (p. 184)
• 2 O. N. Mathematics: 2 by J. Dobinson, Penguin Library of Technology. © 1969, (p. 183)
This page titled 18.1: Integration- Computing the Area Under a Curve is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Serhat Beyenir via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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# Physics 101: Lecture 1 Notes
Mechanics
Nov 14, 2013 (4 years and 8 months ago)
101 views
Physics 101: Lecture 5, Pg
1
Lecture 5:
Introduction to Physics
PHY101
Chapter 2:
Distance and Displacement, Speed and Velocity
(2.1,2.2)
Acceleration (2.3)
Equations of Kinematics for Constant
Acceleration (2.4)
Physics 101: Lecture 5, Pg
2
Displacement and Distance
Displacement is the vector that points from a body’s
initial position,
x
0
, to its final position,
x
. The length of
the displacement vector is equal to the shortest
distance between the two positions.
x
=
x
x
0
Note:
The length of
x
is (in general) not the same as distance
traveled !
Physics 101: Lecture 5, Pg
3
Average Speed and Velocity
Average speed is a measure of how fast an object
moves on average:
average speed = distance/elapsed time
Average speed does not take into account the direction of
motion from the initial and final position.
Physics 101: Lecture 5, Pg
4
Average Speed and Velocity
Average velocity describes how the displacement of an
object changes over time:
average velocity = displacement/elapsed time
v
av
= (
x
-
x
0
) / (t
-
t
0
) =
x
/
t
Average velocity also takes into account the direction of
motion.
Note:
The magnitude of
v
av
is (in general) not the same as the
average speed !
Physics 101: Lecture 5, Pg
5
Instantaneous Velocity and Speed
Average velocity and speed do not convey any
information about how fast the object moves at a
specific point in time.
The velocity at an instant can be obtained from
the average velocity by considering smaller and smaller
time intervals, i.e.
Instantaneous velocity:
v
= lim
t
-
> 0
x
/
t
Instantaneous speed
is the magnitude of
v
.
Physics 101: Lecture 5, Pg
6
Concept Question
If the average velocity of a car during a trip along a
straight road is positive, is it possible for the
instantaneous velocity at some time during the trip to
be negative?
1
-
Yes
2
-
No
correct
If the driver has to put the car in reverse and back up some time
during the trip, then the car has a negative velocity. However,
since the car travels a distance from home in a certain amount of
time, the average velocity will be positive.
Physics 101: Lecture 5, Pg
7
Acceleration
Average acceleration describes how the velocity
of an object moving from the initial position to
the final position changes on average over time:
a
av
= (
v
-
v
0
) / (t
-
t
0
) =
v
/
t
The acceleration at an instant can be obtained from
the average acceleration by considering smaller and
smaller time intervals, i.e.
Instantaneous acceleration:
a
= lim
t
-
> 0
v
/
t
Physics 101: Lecture 5, Pg
8
Concept Question
If the velocity of some object is not zero, can its
acceleration ever be zero ?
1
-
Yes
2
-
No
correct
If the object is moving at a constant velocity,
then the acceleration is zero.
Physics 101: Lecture 5, Pg
9
Concept Question
Is it possible for an object to have a positive velocity at
the same time as it has a negative acceleration?
1
-
Yes
2
No
correct
An object, like a car, can be moving forward
giving it a positive velocity,
but then brake, causing deccelaration which is
negative.
Physics 101: Lecture 5, Pg
10
Kinematics in One Dimension
Constant Acceleration
Simplifications:
In one dimension all vectors in the previous equations
can be replaced by their scalar component along one axis.
For motion with constant acceleration, average and
instantaneous acceleration are equal.
For motion with constant acceleration, the rate with which
velocity changes is constant, i.e. does not change over time.
The average velocity is then simply given as
v
av
= (v
0
+v)/2
Physics 101: Lecture 5, Pg
11
Kinematics in One Dimension
Constant Acceleration
Consider an object which moves from the initial position x
0
, at time t
0
with velocity v
0
, with constant acceleration along a straight line.
How does displacement and velocity of this object change with time ?
a = (v
-
v
0
) / (t
-
t
0
) =>
v(t) = v
0
+ a (t
-
t
0
)
(1)
v
av
= (x
-
x
0
) / (t
-
t
0
) = (v+v
0
)/2 =>
x(t) = x
0
+ (t
-
t
0
) (v+v
0
)/2
(2)
Use Eq. (1) to replace v in Eq.(2):
x(t) = x
0
+ (t
-
t
0
) v
0
+ a/2 (t
-
t
0
)
2
(3)
Use Eq. (1) to replace (t
-
t
0
) in Eq.(2):
v
2
= v
0
2
+ 2 a (x
-
x
0
)
(4)
Physics 101: Lecture 5, Pg
12
Summary of Concepts
kinematics: A description of motion
position:
displacement:
x
=
change of position
velocity:
rate of change of position
average :
x
/
t
instantaneous:
slope of x vs. t
acceleration:
rate of change of velocity
average:
v
/
t
instantaneous:
slope of v vs. t
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Pythagoras Theorem - Statement Questions
Chapter 6 Class 7 Triangle and its Properties
Concept wise
Introducing your new favourite teacher - Teachoo Black, at only ₹83 per month
### Transcript
Ex 6.5, 5 A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree. Given that tree breaks down Let the part after breaking down be AB Distance of its top from base of tree be BC Since Tree is vertical ∠ B = 90° and AB = 5 m BC = 12 m In Δ ABC By Pythagoras Theorem, 〖(𝐴𝐶)〗^2 = 〖(𝐴𝐵)〗^2 + 〖(𝐵𝐶)〗^2 〖(𝐴𝐶)〗^2 = 〖(12)〗^2 + 〖(5)〗^2 〖(𝐴𝐶)〗^2 = 144 + 25 〖(𝐴𝐶)〗^2 = 169 〖(𝐴𝐶)〗^2 = 〖(13)〗^2 Cancelling square AC = 13 m Original height of tree = AB + AC = 5 + 13 = 18 m ∴ Original height of tree is 18 m
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NCERT Solutions: Electromagnetic Induction
# NCERT Solutions Class 12 Physics Chapter 6 - Electromagnetic Induction
NCERT QUESTION
(Electromagnetic Induction)
Q6.1: Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f).
(a)
(b)
(c)
(d)
(e)
(f)
Ans: The direction of the induced current in a closed loop is given by Lenz’s law. The given pairs of figures show the direction of the induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively.
Using Lenz’s rule, the direction of the induced current in the given situations can be predicted as follows:
(a) The direction of the induced current is along qrpq.
(b) The direction of the induced current is along prqp.
(c) The direction of the induced current is along yzxy.
(d) The direction of the induced current is along zyxz.
(e) The direction of the induced current is along xryx.
(f) No current is induced since the field lines are lying in the plane of the closed loop.
Q6.2: Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19:
(a) A wire of irregular shape turning into a circular shape;
(b) A circular loop being deformed into a narrow straight wire.
Ans: According to Lenz’s law, the direction of the induced emf is such that it tends to produce a current that opposes the change in the magnetic flux that produced it.
(a) When the shape of the wire changes, the flux piercing through the unit surface area increases. As a result, the induced current produces an opposing flux. Hence, the induced current flows along adcb.
(b) When the shape of a circular loop is deformed into a narrow straight wire, the flux piercing the surface decreases. Hence, the induced current flows along
Q6.3 : A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?
Ans: Number of turns on the solenoid = 15 turns/cm = 1500 turns/m
Number of turns per unit length, n = 1500 turns
The solenoid has a small loop of area, A = 2.0 cm2 = 2 × 10−4 m2
Current carried by the solenoid changes from 2 A to 4 A.
Change in current in the solenoid, di = 4 − 2 = 2 A
Change in time, dt = 0.1 s
Induced emf in the solenoid is given by Faraday’s law as:
Where,
= Induced flux through the small loop
= BA ... (ii)
B = Magnetic field
=
μ0 = Permeability of free space
= 4π×10−7 H/m
Hence, equation (i) reduces to:
Hence, the induced voltage in the loop is
Q6.4: A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s−1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?
Ans: Length of the rectangular wire, l = 8 cm = 0.08 m
Width of the rectangular wire, b = 2 cm = 0.02 m
Hence, area of the rectangular loop,
A = lb
= 0.08 × 0.02
= 16 × 10−4 m2
Magnetic field strength, B = 0.3 T
Velocity of the loop, v = 1 cm/s = 0.01 m/s
(a) Emf developed in the loop is given as:
e = Blv
= 0.3 × 0.08 × 0.01 = 2.4 × 10−4 V
Time taken to travel along the width
Hence, the induced voltage is 2.4 × 10−4 V which lasts for 2 s.
(b) Emf developed, e = Bbv
= 0.3 × 0.02 × 0.01 = 0.6 × 10−4 V
Time taken to travel along the length
Hence, the induced voltage is 0.6 × 10−4 V which lasts for 8 s.
Q6.5: A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s−1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.
Ans: Length of the rod, l = 1 m
Magnetic field strength, B = 0.5 T
One end of the rod has zero linear velocity, while the other end has a linear velocity of lω.
Average linear velocity of the rod,
Emf developed between the centre and the ring,
Hence, the emf developed between the centre and the ring is 100 V.
Q6.6: A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s−1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10−4 Wb m−2.
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?
Ans: Length of the wire, l = 10 m
Falling speed of the wire, v = 5.0 m/s
Magnetic field strength, B = 0.3 × 10−4 Wb m−2
(a) Emf induced in the wire,
e = Blv
(b) Using Fleming’s right hand rule, it can be inferred that the direction of the induced emf is from West to East.
(c) The eastern end of the wire is at a higher potential.
Q6.7: Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.
Ans: Initial current, I1 = 5.0 A
Final current, I2 = 0.0 A
Change in current,
Time taken for the change, t = 0.1 s
Average emf, e = 200 V
For self-inductance (L) of the coil, we have the relation for average emf as:
e = L
Hence, the self induction of the coil is 4 H.
Q6.8: A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
Ans: Mutual inductance of a pair of coils, µ = 1.5 H
Initial current, I1 = 0 A
Final current I2 = 20 A
Change in current,
Time taken for the change, t = 0.5 s
Induced emf,
Where is the change in the flux linkage with the coil.
Emf is related with mutual inductance as:
Equating equations (1) and (2), we get
Hence, the change in the flux linkage is 30 Wb.
## Old NCERT Solutions:
Q1: A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 × 10−4 T and the dip angle is 30°.
Ans: Speed of the jet plane, v = 1800 km/h = 500 m/s
Wing spanof jet plane, l = 25 m
Earth’s magnetic field strength, B = 5.0 × 10−4 T
Angle of dip,
Vertical component of Earth’s magnetic field,
BV = B sin
= 5 × 10−4 sin 30°
= 2.5 × 10−4 T
Voltage difference between the ends of the wing can be calculated as:
e = (BV) × l × v
= 2.5 × 10−4 × 25 × 500
= 3.125 V
Hence, the voltage difference developed between the ends of the wings is
3.125 V.
Q2: Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s−1. If the cut is joined and the loop has a resistance of 1.6 Ω how much power is dissipated by the loop as heat? What is the source of this power?
Ans: Sides of the rectangular loop are 8 cm and 2 cm.
Hence, area of the rectangular wire loop,
A = length × width
= 8 × 2 = 16 cm2
= 16 × 10−4 m2
Initial value of the magnetic field,
Rate of decrease of the magnetic field,
Emf developed in the loop is given as:
Where,
= Change in flux through the loop area
= AB
Resistance of the loop, R = 1.6 Ω
The current induced in the loop is given as:
Power dissipated in the loop in the form of heat is given as:
The source of this heat loss is an external agent, which is responsible for changing the magnetic field with time.
Q3: A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s−1 in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10−3 T cm−1 along the negative x-direction (that is it increases by 10− 3 T cm−1 as one moves in the negative x-direction), and it is decreasing in time at the rate of 10−3 T s−1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.
Ans: Side of the square loop, s = 12 cm = 0.12 m
Area of the square loop, A = 0.12 × 0.12 = 0.0144 m2
Velocity of the loop, v = 8 cm/s = 0.08 m/s
Gradient of the magnetic field along negative x-direction,
And, rate of decrease of the magnetic field,
Resistance of the loop,
Rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as:
Rate of change of the flux due to explicit time variation in field B is given as:
Since the rate of change of the flux is the induced emf, the total induced emf in the loop can be calculated as:
∴ Induced current,
Hence, the direction of the induced current is such that there is an increase in the flux through the loop along positive z-direction.
Q4: It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm2 with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 Ω. Estimate the field strength of magnet.
Ans: Area of the small flat search coil, A = 2 cm2 = 2 × 10−4 m2
Number of turns on the coil, N = 25
Total charge flowing in the coil, Q = 7.5 mC = 7.5 × 10−3 C
Total resistance of the coil and galvanometer, R = 0.50 Ω
Induced current in the coil,
Induced emf is given as:
Where,
= Charge in flux
Combining equations (1) and (2), we get
Initial flux through the coil, = BA
Where,
B = Magnetic field strength
Final flux through the coil,
Integrating equation (3) on both sides, we have
But total charge,
Hence, the field strength of the magnet is 0.75 T.
Q5: Figure 6.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mΩ. Assume the field to be uniform.
(a) Suppose K is open and the rod is moved with a speed of 12 cm s−1 in the direction shown. Give the polarity and magnitude of the induced emf.
(b) Is there an excess charge built up at the ends of the rods when
K is open? What if K is closed?
(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.
(d) What is the retarding force on the rod when K is closed?
(e) How much power is required (by an external agent) to keep the rod moving at the same speed (=12 cm s−1) when K is closed? How much power is required when K is open?
(f) How much power is dissipated as heat in the closed circuit?
What is the source of this power?
(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?
Ans: Length of the rod, l = 15 cm = 0.15 m
Magnetic field strength, B = 0.50 T
Resistance of the closed loop, R = 9 mΩ = 9 × 10−3 Ω
(a) Induced emf = 9 mV; polarity of the induced emf is such that end P shows positive while end Q shows negative ends.
Speed of the rod, v = 12 cm/s = 0.12 m/s
Induced emf is given as:
e = Bvl
= 0.5 × 0.12 × 0.15
= 9 × 10−3 v
= 9 mV
The polarity of the induced emf is such that end P shows positive while end Q shows negative ends.
(b) Yes; when key K is closed, excess charge is maintained by the continuous flow of current.
When key K is open, there is excess charge built up at both ends of the rods.
When key K is closed, excess charge is maintained by the continuous flow of current.
(c) Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rod.
There is no net force on the electrons in rod PQ when key K is open and the rod is moving uniformly. This is because magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rods.
(d) Retarding force exerted on the rod, F = IBl
Where,
I = Current flowing through the rod
(e) 9 mW; no power is expended when key K is open.
Speed of the rod, v = 12 cm/s = 0.12 m/s
Hence, power is given as:
When key K is open, no power is expended.
(f) 9 mW; power is provided by an external agent.
Power dissipated as heat = I2 R
= (1)2 × 9 × 10−3
= 9 mW
The source of this power is an external agent.
(g) Zero
In this case, no emf is induced in the coil because the motion of the rod does not cut across the field lines.
Q6: An air-cored solenoid with length 30 cm, area of cross-section 25 cm2 and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10−3 s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.
Ans: Length of the solenoid, l = 30 cm = 0.3 m
Area of cross-section, A = 25 cm2 = 25 × 10−4 m2
Number of turns on the solenoid, N = 500
Current in the solenoid, I = 2.5 A
Current flows for time, t = 10−3 s
Average back emf,
Where,
= Change in flux
= NAB … (2)
Where,
B = Magnetic field strength
Where,
= Permeability of free space = 4π × 10−7 T m A−1
Using equations (2) and (3) in equation (1), we get
Hence, the average back emf induced in the solenoid is 6.5 V.
Q7: (a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig. 6.21.
(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s.
Calculate the induced emf in the loop at the instant when x = 0.2 m.
Take a = 0.1 m and assume that the loop has a large resistance.
Ans: (a) Take a small element dy in the loop at a distance y from the long straight wire (as shown in the given figure).
Magnetic flux associated with element
Where,
dA = Area of element dy = a dy
B = Magnetic field at distance y
I = Current in the wire
= Permeability of free space = 4π × 10−7 T m A−1
y tends from x to .
For mutual inductance M,the flux is given as:
(b) Emf induced in the loop, e = B’av
Given,
I = 50 A
x = 0.2 m
a = 0.1 m
v = 10 m/s
Q8: A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig. 6.22). A uniform magnetic field extends over a circular region within the rim. It is given by,
B = − B0 k (ra; a < R)
= 0 (otherwise)
What is the angular velocity of the wheel after the field is suddenly switched off?
Ans: Line charge per unit length
Where,
r = Distance of the point within the wheel
Mass of the wheel = M
Radius of the wheel = R
Magnetic field,
At distance r,the magnetic force is balanced by the centripetal force i.e.,
∴ Angular velocity,
Q9: A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s−1 in a uniform horizontal magnetic field of magnitude 3.0×10−2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?
Ans: Max induced emf = 0.603 V
Average induced emf = 0 V
Max current in the coil = 0.0603 A
Average power loss = 0.018 W
(Power comes from the external rotor)
Radius of the circular coil, r = 8 cm = 0.08 m
Area of the coil, A = πr2 = π × (0.08)2 m2
Number of turns on the coil, N = 20
Angular speed, ω = 50 rad/s
Magnetic field strength, B = 3 × 10−2 T
Resistance of the loop, R = 10 Ω
Maximum induced emf is given as:
e = AB
= 20 × 50 × π × (0.08)2 × 3 × 10−2
= 0.603 V
The maximum emf induced in the coil is 0.603 V.
Over a full cycle, the average emf induced in the coil is zero.
Maximum current is given as:
Average power loss due to joule heating:
The current induced in the coil produces a torque opposing the rotation of the coil. The rotor is an external agent. It must supply a torque to counter this torque in order to keep the coil rotating uniformly. Hence, dissipated power comes from the external rotor.
The document NCERT Solutions Class 12 Physics Chapter 6 - Electromagnetic Induction is a part of the NEET Course Physics Class 12.
All you need of NEET at this link: NEET
## Physics Class 12
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## FAQs on NCERT Solutions Class 12 Physics Chapter 6 - Electromagnetic Induction
1. What is electromagnetic induction?
Ans. Electromagnetic induction is the process by which a changing magnetic field induces an electromotive force (EMF) or voltage in a conductor, according to Faraday's law of electromagnetic induction.
2. How is electromagnetic induction related to generators?
Ans. Electromagnetic induction is the principle on which generators operate. When a coil of wire rotates in a magnetic field, it experiences a change in magnetic flux, causing an induced current to flow through the wire.
3. What is Lenz's Law in electromagnetic induction?
Ans. Lenz's Law states that the direction of the induced current in a conductor is such that it opposes the change in magnetic flux that produced it. This law helps determine the direction of the induced current in a circuit.
4. How is electromagnetic induction used in transformers?
Ans. Transformers use electromagnetic induction to transfer electrical energy from one circuit to another through mutual induction. When an alternating current passes through the primary coil, it produces a changing magnetic field that induces a voltage in the secondary coil.
5. What are some real-life applications of electromagnetic induction?
Ans. Electromagnetic induction is used in various applications, such as in generators to produce electricity, transformers to step up or step down voltage, induction cooktops for cooking, and metal detectors for security purposes.
## Physics Class 12
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1. Oct 12, 2010
### CalcYouLater
1. The problem statement, all variables and given/known data
I am given that an operator is defined by the following:
$$\hat\pi\psi(x,y)=\psi(y,x)$$
in a 2 variable Hilbert space of functions $$\psi(x,y)$$
I have to show that the operator is linear and Hermitian
2. Relevant equations
An operator is self adjoint if:
$$\left\langle\psi\right|\hat{A}^{\dagger}\left|\right\phi\rangle=\left\langle\phi\right|\hat{A}\left|\right\psi\rangle^{*}$$
3. The attempt at a solution
With the information given, how can I apply the above condition? Should I put x one side and y on the other? Or should I use psi(x,y) on each side. If so, does the fact that one of the psi's is a bra and the other is a ket change anything?
To show that it is linear, I need to show that it is distributive. Is it as simple as saying:
$$\hat{\pi}(A\psi(x,y)+B\psi(x,y))=\hat{\pi}(A\psi(x,y))+\hat{\pi}(B\psi(x,y))$$
If I remember correctly, my professor said that since the space is linear then the operators are distributive. That doesn't make sense to me. If an operator can be anything, then couldn't it be something that isn't linear as well?
2. Oct 12, 2010
### Dick
What's your definition of the inner product? It's an integral, right? Showing it's linear isn't at all hard. And showing it's hermitian I think is just a changes of dummy variables, isn't it? An operator can be anything, but you are asked to show this one is linear. I'm not sure what your last question is about.
3. Oct 12, 2010
### CalcYouLater
Hi. Thanks for responding (and for the help in a different thread).
Inner product:
$$\left\langle\psi\right|\hat{A}\left|\right\psi\rangle=\int_{-\infty}^{\infty}\psi^{*}\hat{\pi}\psi{dx}$$
Hmm, dummy variables. Would that mean that the above integral changes from integrating over dx to over dy?
$$\int_{-\infty}^{\infty}\psi^{*}\hat{\pi}\psi{dy}$$
4. Oct 12, 2010
### Dick
They are functions of two variables, so it's probably a double integral dx*dy, isn't it? Applying pi just flips x and y, doesn't it? Does that affect the integral?
5. Oct 12, 2010
### CalcYouLater
Yes, a double integral does make sense. I think that flipping x & y does not change the integral because it doesn't matter in which order we integrate because the limits are the same for both x and y.
But, does it change the complex conjugation? I say no because I do not see any evidence based off of the information given that complex numbers are affected by this permutation.
6. Oct 12, 2010
### Dick
Right. The limits for x and y are the same. Write out the integral expressions for <psi,pi(phi)> and <pi(psi),phi> and try to convince yourself interchanging the labels 'x' and 'y' makes no difference.
7. Oct 12, 2010
### CalcYouLater
If I say:
$$\left\langle\psi\right|\hat{\pi}\left|\right\phi\rangle=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\psi^{*}\hat{\pi}\phi{dx}{dy}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\psi^{*}{dx}{dy}\hat{\pi}\phi=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\psi^{*}\hat{\pi}\phi{dx}{dy}$$
and likewise:
$$\left\langle\hat{\pi}\psi\right|\left\right\phi\rangle=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\hat{\pi}\psi^{*}\phi{dx}{dy}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\hat{\pi}\psi^{*}{dx}{dy}\phi=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\phi\hat{\pi}\psi^{*}{dx}{dy}$$
Isn't that assuming the operator is linear? I must be missing something here. I was thinking that showing that moving around the integrand does not change the integration proves they're hermitian, but now I think that is based off the idea that (pi*psi*phi)=(psi*pi*phi).
8. Oct 12, 2010
### Dick
integral conjugate(psi(y,x))*phi(x,y)*dx*dy=integral conjugate(psi(x,y))*phi(y,x)*dx*dy. True or false? They only differ by an interchange of x and y, don't they? Do you see what I'm saying?
9. Oct 12, 2010
### CalcYouLater
Ahh. You are good. Does that mean I get a two-fer? Is the fact that the integral is unchanged enough (I think sufficient is the word) to show the operator is linear as well. Or is there a condition for linearity that I am not considering?
10. Oct 13, 2010
### Dick
No, linearity is separate. You want to show stuff like pi(A(x,y)+B(x,y))=pi(A(x,y))+pi(B(x,y)). It's really pretty direct if you think about what pi is doing. It's just switching x and y around.
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# Rolling variance algorithm
I'm trying to find an efficient, numerically stable algorithm to calculate a rolling variance (for instance, a variance over a 20-period rolling window). I'm aware of the Welford algorithm that efficiently computes the running variance for a stream of numbers (it requires only one pass), but am not sure if this can be adapted for a rolling window. I would also like the solution to avoid the accuracy problems discussed at the top of this article by John D. Cook. A solution in any language is fine.
• +1 for mentioning Welford algorithm; I knew it was in Knuth but never knew the original source Feb 28 '11 at 21:14
• Hello, what did you end up doing? Did you adapt Chan's algorithm? Btw, shouldn't kahan sum be able to overcome numerical instabilities when using the "naive" approach (keeping track the sums of the values and their squares)? Sep 29 '12 at 22:38
I've run across this problem as well. There are some great posts out there in computing the running cumulative variance such as John Cooke's Accurately computing running variance post and the post from Digital explorations, Python code for computing sample and population variances, covariance and correlation coefficient. Just could not find any that were adapted to a rolling window.
The Running Standard Deviations post by Subluminal Messages was critical in getting the rolling window formula to work. Jim takes the power sum of the squared differences of the values versus Welford’s approach of using the sum of the squared differences of the mean. Formula as follows:
PSA today = PSA(yesterday) + (((x today * x today) - x yesterday)) / n
• x = value in your time series
• n = number of values you've analyzed so far.
But, to convert the Power Sum Average formula to a windowed variety you need tweak the formula to the following:
PSA today = PSA yesterday + (((x today * x today) - (x yesterday * x Yesterday) / n
• x = value in your time series
• n = number of values you've analyzed so far.
You'll also need the Rolling Simple Moving Average formula:
SMA today = SMA yesterday + ((x today - x today - n) / n
• x = value in your time series
• n = period used for your rolling window.
From there you can compute the Rolling Population Variance:
Population Var today = (PSA today * n - n * SMA today * SMA today) / n
Or the Rolling Sample Variance:
Sample Var today = (PSA today * n - n * SMA today * SMA today) / (n - 1)
I've covered this topic along with sample Python code in a blog post a few years back, Running Variance.
Hope this helps.
Please note: I provided links to all the blog posts and math formulas in Latex (images) for this answer. But, due to my low reputation (< 10); I'm limited to only 2 hyperlinks and absolutely no images. Sorry about this. Hope this doesn't take away from the content.
• In this formula: `Population Var today = (PSA today * n - n * SMA today * SMA today) / n` - why not to remove `n`? `Population Var today = (PSA today - SMA today * SMA today)`. Sep 6 '13 at 17:05
• Due to squaring samples in the formula, this algorithm exhibits the very numerical inaccuracy that OP was trying to avoid. Nov 27 '13 at 10:21
• Yup, not a numerically stable approach. The closest thing to a correct answer is by @DanS below. May 22 '16 at 2:46
• Thanks for the explanation, here's a C# implementation gist.github.com/mattdot/d459b1cb15480fefd953841a1ac70be8 May 11 '18 at 23:50
I have been dealing with the same issue.
Mean is simple to compute iteratively, but you need to keep the complete history of values in a circular buffer.
``````next_index = (index + 1) % window_size; // oldest x value is at next_index, wrapping if necessary.
new_mean = mean + (x_new - xs[next_index])/window_size;
``````
I have adapted Welford's algorithm and it works for all the values that I have tested with.
``````varSum = var_sum + (x_new - mean) * (x_new - new_mean) - (xs[next_index] - mean) * (xs[next_index] - new_mean);
xs[next_index] = x_new;
index = next_index;
``````
To get the current variance just divide varSum by the window size: `variance = varSum / window_size;`
• It might be slightly more stable to do `varSum += (x_new + x_old - mean - new_mean) * (x_new - x_old)`, where `x_old = xs[next_index]`, as you remove a potentially large `mean * new_mean` summand from the two items you subtract to update `varSum`. Other than that, this is the most correct of the answers here, and it's a pity it hasn't gotten more love. May 22 '16 at 1:45
• To clarify Jaime's answer, he did some algebra taking DanS's `varSum` equation and distributing the multiplication. some terms cancel but you also have to perform the trick of adding in `x_new * x_old - x_new * x_old` to arrive at his result Jul 16 '18 at 22:50
• Very late comment: Why are you diving by `window_size` and not `window_size-1`. In other words: Why aren't you using Bessel's correction. I notice that John D. Cook does include Bessel's correction in his running variance code. Aug 9 '18 at 10:09
• couldn't you remove `varSum` entirely by just `variance += (x_new + x_old - mean - new_mean) * (x_new - x_old) / window_size`? Sep 17 '20 at 17:21
If you prefer code over words (heavily based on DanS' post): http://calcandstuff.blogspot.se/2014/02/rolling-variance-calculation.html
``````public IEnumerable RollingSampleVariance(IEnumerable data, int sampleSize)
{
double mean = 0;
double accVar = 0;
int n = 0;
var queue = new Queue(sampleSize);
foreach(var observation in data)
{
queue.Enqueue(observation);
if (n < sampleSize)
{
// Calculating first variance
n++;
double delta = observation - mean;
mean += delta / n;
accVar += delta * (observation - mean);
}
else
{
double then = queue.Dequeue();
double prevMean = mean;
mean += (observation - then) / sampleSize;
accVar += (observation - prevMean) * (observation - mean) - (then - prevMean) * (then - mean);
}
if (n == sampleSize)
yield return accVar / (sampleSize - 1);
}
}
``````
Actually Welfords algorithm can AFAICT easily be adapted to compute weighted Variance. And by setting weights to -1, you should be able to effectively cancel out elements. I havn't checked the math whether it allows negative weights though, but at a first look it should!
I did perform a small experiment using ELKI:
``````void testSlidingWindowVariance() {
MeanVariance mv = new MeanVariance(); // ELKI implementation of weighted Welford!
MeanVariance mc = new MeanVariance(); // Control.
Random r = new Random();
double[] data = new double[1000];
for (int i = 0; i < data.length; i++) {
data[i] = r.nextDouble();
}
// Pre-roll:
for (int i = 0; i < 10; i++) {
mv.put(data[i]);
}
// Compare to window approach
for (int i = 10; i < data.length; i++) {
mv.put(data[i-10], -1.); // Remove
mv.put(data[i]);
mc.reset(); // Reset statistics
for (int j = i - 9; j <= i; j++) {
mc.put(data[j]);
}
assertEquals("Variance does not agree.", mv.getSampleVariance(),
mc.getSampleVariance(), 1e-14);
}
}
``````
I get around ~14 digits of precision compared to the exact two-pass algorithm; this is about as much as can be expected from doubles. Note that Welford does come at some computational cost because of the extra divisions - it takes about twice as long as the exact two-pass algorithm. If your window size is small, it may be much more sensible to actually recompute the mean and then in a second pass the variance every time.
I have added this experiment as unit test to ELKI, you can see the full source here: http://elki.dbs.ifi.lmu.de/browser/elki/trunk/test/de/lmu/ifi/dbs/elki/math/TestSlidingVariance.java it also compares to the exact two-pass variance.
However, on skewed data sets, the behaviour might be different. This data set obviously is uniform distributed; but I've also tried a sorted array and it worked.
Update: we published a paper with details on differentweighting schemes for (co-)variance:
Schubert, Erich, and Michael Gertz. "Numerically stable parallel computation of (co-) variance." Proceedings of the 30th International Conference on Scientific and Statistical Database Management. ACM, 2018. (Won the SSDBM best-paper award.)
This also discusses how weighting can be used to parallelize the computation, e.g., with AVX, GPUs, or on clusters.
• Ported the ELKI MeanVarance.java class to JS, added a value buffer, and used weights of -1 to remove values. I found result precision varies depending on how many values you run through the accumulator. I was seeing ~12 digits of precision after running 1-10M values through it. (I.e. "good enough") Thanks for the tip on using -1 weights! Jul 17 '19 at 21:22
• If you need higher precision than that, you will likely need to use Kahan summation or the Shewchuk algorithm. These use additional floats to store the lost digits, and hence can offer much higher precision. But the implementation gets much more messy and slower. For further details, see the reference I added to the post. Aug 5 '19 at 10:10
Here's a divide and conquer approach that has `O(log k)`-time updates, where `k` is the number of samples. It should be relatively stable for the same reasons that pairwise summation and FFTs are stable, but it's a bit complicated and the constant isn't great.
Suppose we have a sequence `A` of length `m` with mean `E(A)` and variance `V(A)`, and a sequence `B` of length `n` with mean `E(B)` and variance `V(B)`. Let `C` be the concatenation of `A` and `B`. We have
``````p = m / (m + n)
q = n / (m + n)
E(C) = p * E(A) + q * E(B)
V(C) = p * (V(A) + (E(A) + E(C)) * (E(A) - E(C))) + q * (V(B) + (E(B) + E(C)) * (E(B) - E(C)))
``````
Now, stuff the elements in a red-black tree, where each node is decorated with mean and variance of the subtree rooted at that node. Insert on the right; delete on the left. (Since we're only accessing the ends, a splay tree might be `O(1)` amortized, but I'm guessing amortized is a problem for your application.) If `k` is known at compile-time, you could probably unroll the inner loop FFTW-style.
• (Note: it's fine to compute q = 1 - p unless k is stupendously large.) Feb 28 '11 at 22:04
• Okay, this is basically Chan et al.'s parallel algorithm as described on Wikipedia. That's what I get for not scrolling down... Feb 28 '11 at 22:36
• Can you explain in a bit more detail how you would apply this algorithm to variance over a moving window? I'm slightly familiar with the Chan et al approach, but thought of it as a one-pass method for computing a single variance over an entire sample, with the added advantage that the problem can be broken into parts that are run in parallel. Feb 28 '11 at 22:45
• Chan et al gave a way to compute statistics for a concatenation of parts given the statistics of the parts. The high-level idea is to maintain a collection of parts (actually just their statistics) such that any window is the concatenation of O(log k) parts. One way is with a balanced binary tree, but as Rex points out, this is overkill and we can just maintain statistics for aligned parts whose sizes are powers of two (e.g., [0, 1), [1, 2), [0, 2), [2, 3), [3, 4), [2, 4), [0, 4), etc.) Feb 28 '11 at 23:10
I know this question is old, but in case someone else is interested here follows the python code. It is inspired by johndcook blog post, @Joachim's, @DanS's code and @Jaime comments. The code below still gives small imprecisions for small data windows sizes. Enjoy.
``````from __future__ import division
import collections
import math
class RunningStats:
def __init__(self, WIN_SIZE=20):
self.n = 0
self.mean = 0
self.run_var = 0
self.WIN_SIZE = WIN_SIZE
self.windows = collections.deque(maxlen=WIN_SIZE)
def clear(self):
self.n = 0
self.windows.clear()
def push(self, x):
self.windows.append(x)
if self.n <= self.WIN_SIZE:
# Calculating first variance
self.n += 1
delta = x - self.mean
self.mean += delta / self.n
self.run_var += delta * (x - self.mean)
else:
x_removed = self.windows.popleft()
old_m = self.mean
self.mean += (x - x_removed) / self.WIN_SIZE
self.run_var += (x + x_removed - old_m - self.mean) * (x - x_removed)
def get_mean(self):
return self.mean if self.n else 0.0
def get_var(self):
return self.run_var / (self.WIN_SIZE - 1) if self.n > 1 else 0.0
def get_std(self):
return math.sqrt(self.get_var())
def get_all(self):
return list(self.windows)
def __str__(self):
return "Current window values: {}".format(list(self.windows))
``````
• thanks for the idea synthesis in python. I don't like how the size of windows become `WIN_SIZE - 1`in the case where the else block is entered. So if `WIN_SIZE` was 10 when `push` is called and we append, it's still 10 because of the deque constructor option used, then in the `else` block `popleft` reduces the size further to 9. So maybe `maxlen=WIN_SIZE + 1`? Or not use the `maxlen` option. Also, can drop the `n` variable and use `len(self.windows)`. Jul 16 '18 at 23:10
• in the `get_var` method the denominator should be `self.n` or `len(self.windows)` Jul 16 '18 at 23:34
I look forward to be proven wrong on this but I don't think this can be done "quickly." That said, a large part of the calculation is keeping track of the EV over the window which can be done easily.
I'll leave with the question: are you sure you need a windowed function? Unless you are working with very large windows it is probably better to just use a well known predefined algorithm.
I guess keeping track of your 20 samples, Sum(X^2 from 1..20), and Sum(X from 1..20) and then successively recomputing the two sums at each iteration isn't efficient enough? It's possible to recompute the new variance without adding up, squaring, etc., all of the samples each time.
As in:
``````Sum(X^2 from 2..21) = Sum(X^2 from 1..20) - X_1^2 + X_21^2
Sum(X from 2..21) = Sum(X from 1..20) - X_1 + X_21
``````
• I believe this solution is susceptible to the stability problems mentioned in the link in my original post (johndcook.com/standard_deviation.html). In particular, when input values and large and their difference is small than the result could actually be negative. I will have no control over the input, so I would prefer to avoid this approach. Feb 28 '11 at 21:11
• Oh, I see. Is there anything you can say about the input? Intended use? Is it a problem that you can just throw more bits at (64-bit float, arbitary-precision arithmetic, etc.)? Rounding errors go away if you trump the input in significant figures, no?
– John
Feb 28 '11 at 21:16
• agreed -- this has stability issues. Imagine 1000 samples near 1,000,000.0, followed by 20 samples near zero. Feb 28 '11 at 21:20
• @Jason S: The rolling variance is what it is. There might be a lot going on in the transition from 1 million to ~zero, but that's the nature of the beast. That, and the first 980 of the 1000 ~1 million values are out of the picture when the change occurs anyway. My comment suggested that if you had enough significant figures in your calculations, none of this would matter.
– John
Feb 28 '11 at 21:51
• Input could really be anything. Value magnitude could certainly be in the trillions, and while the original data will only have accuracy to a few decimal points, users will be able to transform their data (for instance dividing by any scalar) before calculating the variance. Feb 28 '11 at 21:56
Here's another `O(log k)` solution: find squares the original sequence, then sum pairs, then quadruples, etc.. (You'll need a bit of a buffer to be able to find all of these efficiently.) Then add up those values that you need to to get your answer. For example:
``````||||||||||||||||||||||||| // Squares
| | | | | | | | | | | | | // Sum of squares for pairs
| | | | | | | // Pairs of pairs
| | | | // (etc.)
| |
^------------------^ // Want these 20, which you can get with
| | // one...
| | | | // two, three...
| | // four...
|| // five stored values.
``````
Now you use your standard E(x^2)-E(x)^2 formula and you're done. (Not if you need good stability for small sets of numbers; this was assuming that it was only accumulation of rolling error that was causing issues.)
That said, summing 20 squared numbers is very fast these days on most architectures. If you were doing more--say, a couple hundred--a more efficient method would clearly be better. But I'm not sure that brute force isn't the way to go here.
• "use your standard E(x^2)-E(x)^2 formula" No, don't; it's not even remotely stable. Adapt one of the better algorithms. Feb 28 '11 at 22:34
• @userOVER9000 - Why are you worried about stability over 20 items? Cumulative errors that accumulate over millions of entries are a problem (especially when making a rolling window), but that's not the issue here. Mar 1 '11 at 1:42
• I'm worried about it because it's an issue. Go read the Wikipedia article, and if you're still not convinced, try computing the variance of 20 iid samples of N(1, 1e-10). Mar 1 '11 at 2:12
• I haven't seen this actually be an issue for any realistic data set with sensible units and origin, but fair enough, if that's what the OP wants... Mar 1 '11 at 2:27
For only 20 values, it's trivial to adapt the method exposed here (I didn't say fast, though).
You can simply pick up an array of 20 of these `RunningStat` classes.
The first 20 elements of the stream are somewhat special, however once this is done, it's much more simple:
• when a new element arrives, clear the current `RunningStat` instance, add the element to all 20 instances, and increment the "counter" (modulo 20) which identifies the new "full" `RunningStat` instance
• at any given moment, you can consult the current "full" instance to get your running variant.
You will obviously note that this approach isn't really scalable...
You can also note that there is some redudancy in the numbers we keep (if you go with the `RunningStat` full class). An obvious improvement would be to keep the 20 lasts `Mk` and `Sk` directly.
I cannot think of a better formula using this particular algorithm, I am afraid that its recursive formulation somewhat ties our hands.
This is just a minor addition to the excellent answer provided by DanS. The following equations are for removing the oldest sample from the window and updating the mean and variance. This is useful, for example, if you want to take smaller windows near the right edge of your input data stream (i.e. just remove the oldest window sample without adding a new sample).
``````window_size -= 1; % decrease window size by 1 sample
new_mean = prev_mean + (prev_mean - x_old) / window_size
varSum = varSum - (prev_mean - x_old) * (new_mean - x_old)
``````
Here, x_old is the oldest sample in the window you wish to remove.
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# Math
posted by .
Please can someone show me the steps to do this problem?? Also, please don't tell me to just google the solution because I've already tried and have tried to do this problem by myself for hours now...Thank you so much.
You are measuring trees in a forest. Standing on the ground exactly half -waybetweentwo trees, you measure the angle the top of each tree makes with the horizon: one angle is 67◦, and the other is 82◦. If one tree is 80 feet taller than the other, how far apart (horizontal distance along the ground) are the two trees?
• Math -
make a diagram, showing the two trees at A and B, labeled h for the height of the shorter, and h+80 for the height of the taller.
let P be the midpoint of AB , where AP = BP
Then the angle at P for the taller tree is 82° , and the angle at P for the smaller tree is 67°
then h/AP = tan67° and (h+80)/BP = tan82°
AP = h/tan67 and BP = (h+80)/tan82
but AP and BP are equal, so
h/tan67 = (h+80)/tan82
htan82 = htan67 + 80tan67
htan82 - htan67 = 80tan67
h(tan82-tan67) = 80tan67
h = 80tan67/(tan82-tan67) = .....
You do the button - pushing
once you have h, sub it back into the little equation for AP
(I got AP= 16.8)
• Math -
What does tan67 equal?
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https://www.thenational.academy/teachers/programmes/maths-primary-ks2/units/bridging-100-counting-on-and-back-in-10s-adding-subtracting-multiples-of-10/lessons/bridge-100-by-adding-or-subtracting-a-single-digit-number
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New
New
Year 3
# Bridge 100 by adding or subtracting a single-digit number
I can bridge 100 by adding or subtracting a single-digit number.
New
New
Year 3
# Bridge 100 by adding or subtracting a single-digit number
I can bridge 100 by adding or subtracting a single-digit number.
Share activities with pupils
Share function coming soon...
## Lesson details
### Key learning points
1. Knowledge of number facts to 20 can support calculation across the 100 boundary.
2. If 8 plus 7 is equal to 15 then 98 plus 7 is equal to 90 plus 15 which equals 105
3. Knowledge of place value can help when counting forwards across the 100 boundary.
### Common misconception
Pupils may not partition numbers in efficient ways to add and subtract them.
Encourage pupils to add or subtract to make 100 (e.g. when calculating 103-7, subtract 3 to equal 100, then subtract 4 to equal 96).
### Keywords
• Bridging - A strategy which uses addition or subtraction to cross a number boundary.
• Partition - The act of splitting an object or value down into smaller parts.
Pupils will need to have a good grasp of number pairs that total 10 to access the learning in this lesson. Base 10 blocks may be useful for this lesson, although pupils will need to use this flexibly alongside the images provided.
Teacher tip
### Licence
This content is © Oak National Academy Limited (2024), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).
## Starter quiz
### 6 Questions
Q1.
What is the missing addend? 6 + = 10
Q2.
What is the most efficient way to partition 6 to complete the calculation 7 + 6?
4 and 2
5 and 1
Q3.
What is 1 more than 99?
Q4.
105 − = 100
Q5.
What number is shown by the arrow?
105
115
135
Q6.
Match the base ten representations to the numbers.
111
110
101
## Exit quiz
### 6 Questions
Q1.
What is the missing addend? 96 + = 100
Q2.
What's the most efficient way to partition 7 to complete 99 + 7 = 106?
2 and 5
3 and 4
Q3.
What calculation is represented?
99 + 4 = 103
98 + 3 = 101
Correct answer: 98 + 4 = 102
97 + 5 = 102
Q4.
What addend is missing from the number line representation?
Q5.
What is the missing number? 108 − = 99
Q6.
Match the equations to the answers.
96
Correct Answer:102 − 7 = ,96
96
Correct Answer:99 + 6 = ,105
105
Correct Answer:98 + 8 = ,106
106
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What is the volume of 1.00 mole of H2SO4 at STP
HOME > > What is the volume of 1.00 mole of H2SO4 at STP
# What is the volume of 1.00 mole of H2SO4 at STP
[From: ] [author: ] [Date: 12-12-20] [Hit: ]
and youll have your answer.V = (1 mol)(0.082057 L*atm/mol*K)(273.15K)/1 atm = 22.22.4 L-according the ideal gas law which then would be 22.......
H2SO4 is a liquid at STP, so PV = nRT won't help you here.
1.00 mole of H2SO4 = 98.1 g
V = Density/mass
V = Density/98.1 g
Look up the density of H2SO4 and substitute it into the equation above, and you'll have your answer.
-
Use PV=nRT
Solve for V:
V=nRT/P // now just plug all the STP values in:
V = (1 mol)(0.082057 L*atm/mol*K)(273.15K)/1 atm = 22.41 L
22.4 L
-
according the ideal gas law which then would be 22.4L
but there is not enough information because sulfuric acid is an acid which is usually a aqueous
1
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Equations Using DeMoivre's Theorem
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Equations Using DeMoivre's Theorem
You are given an equation in math class:
$x^4 = 16$
and asked to solve for "x". "Excellent!" you say. "This should be easy. The answer is 2."
"Not quite so fast," says your instructor.
"I want you to find the complex roots as well!"
Can you do this? Read on, and by the end of this Concept, you'll be able to solve equations to find complex roots.
Guidance
We've already seen equations that we would like to solve. However, up until now, these equations have involved solutions that were real numbers. However, there is no reason that solutions need to be limited to the real number line. In fact, some equations cannot be solved completely without the use of complex numbers. Here we'll explore a little more about complex numbers as solutions to equations.
The roots of a complex number are cyclic in nature. This means that when the roots are plotted on the complex plane, the $n^{th}$ roots are equally spaced on the circumference of a circle.
Since you began Algebra, solving equations has been an extensive topic. Now we will extend the rules to include complex numbers. The easiest way to explore the process is to actually solve an equation. The solution can be obtained by using De Moivre’s Theorem.
Example A
Consider the equation $x^5 - 32 = 0$ . The solution is the same as the solution of $x^5 = 32$ . In other words, we must determine the fifth roots of 32.
Solution:
$x^5-32 &= 0 \ \text{and} \ x^5=32.\\r &= \sqrt{x^2+y^2}\\r &= \sqrt{(32)^2+(0)^2}\\r &= 32\\\theta &= \tan^{-1} \left(\frac{0}{32}\right)=0$
Write an expression for determining the fifth roots of $32 = 32 + 0i$
$32^{\frac{1}{5}} &= [32( \cos (0+2\pi k)+i \sin (0+2\pi k)]^{\frac{1}{5}}\\&= 2\left(\cos \frac{2\pi k}{5}+i \sin \frac{2\pi k}{5}\right)k=0, 1, 2, 3, 4\\x_1 &= 2\left(\cos \frac{0}{5}+i \sin \frac{0}{5}\right) \rightarrow 2(\cos 0+i \sin 0)=2 && for \ k=0\\x_2 &= 2\left(\cos \frac{2\pi}{5}+i \sin \frac{2\pi}{5}\right) \approx 0.62 + 1.9i && for \ k=1\\x_3 &= 2\left(\cos \frac{4\pi}{5}+i \sin \frac{4\pi}{5}\right) \approx -1.62 + 1.18i && for \ k=2\\x_4 &= 2\left(\cos \frac{6\pi}{5}+i \sin \frac{6\pi}{5}\right) \approx -1.62-1.18i && for \ k=3\\x_5 &= 2\left(\cos \frac{8\pi}{5}+i \sin \frac{8\pi}{5}\right) \approx 0.62-1.9i && for \ k=4$
Example B
Solve the equation $x^3- 27 = 0$ . This is the same as the equation $x^3 = 27$ .
Solution:
$x^3 = 27\\r &= \sqrt{x^2+y^2}\\r &= \sqrt{(27)^2+(0)^2}\\r &= 27\\\theta &= \tan^{-1} \left(\frac{0}{27}\right)=0$
Write an expression for determining the cube roots of $27 = 27 + 0i$
$27^{\frac{1}{3}} &= [27( \cos (0+2\pi k)+i \sin (0+2\pi k)]^{\frac{1}{3}}\\&= 3\left(\cos \frac{2\pi k}{3}+i \sin \frac{2\pi k}{3}\right)k=0, 1, 2\\x_1 &= 3\left(\cos \frac{0}{3}+i \sin \frac{0}{3}\right) \rightarrow 3(\cos 0 +i \sin 0)=3 && for \ k=0\\x_2 &= 3\left(\cos \frac{2\pi}{3}+i \sin \frac{2\pi}{3}\right) \approx -1.5 + 2.6i && for \ k=1\\x_3 &= 3\left(\cos \frac{4\pi}{3}+i \sin \frac{4\pi}{3}\right) \approx -1.5 - 2.6i && for \ k=2\\$
Example C
Solve the equation $x^4 = 1$
Solution:
$x^4 = 1\\r &= \sqrt{x^2+y^2}\\r &= \sqrt{(1)^2+(0)^2}\\r &= 1\\\theta &= \tan^{-1} \left(\frac{0}{1}\right)=0$
Write an expression for determining the cube roots of $1 = 1 + 0i$
$1^{\frac{1}{4}} &= [1( \cos (0+2\pi k)+i \sin (0+2\pi k)]^{\frac{1}{4}}\\&= 1\left(\cos \frac{2\pi k}{4}+i \sin \frac{2\pi k}{4}\right)k=0, 1, 2, 3\\x_1 &= 1\left(\cos \frac{0}{4}+i \sin \frac{0}{4}\right) \rightarrow 3(\cos 0 +i \sin 0)= 1 && for \ k=0\\x_2 &= 1\left(\cos \frac{2\pi}{4}+i \sin \frac{2\pi}{4}\right) = 0 + i = i && for \ k=1\\x_3 &= 1\left(\cos \frac{4\pi}{4}+i \sin \frac{4\pi}{4}\right) = -1 - 0i = -1 && for \ k=2\\x_4 &= 1\left(\cos \frac{6\pi}{4}+i \sin \frac{6\pi}{4}\right) = 0 - i = -i && for \ k=3\\$
Guided Practice
1. Rewrite the following in rectangular form: $[2(\cos 315^\circ + i \sin 315^\circ)]^3$
2. Solve the equation $x^4 + 1 = 0$ . What shape do the roots make?
3. Solve the equation $x^3-64=0$ . What shape do the roots make?
Solutions:
1. $r &= 2 \ \text{and} \ \theta=315^\circ \ \text{or} \ \frac{7\pi}{4}.\\z^n &= [r(\cos \theta+i \sin \theta)]^n=r^n(\cos n\theta+i \sin n\theta)\\z^3 &= 2^3 \left[(\cos 3 \left(\frac{7\pi}{4}\right)+i \sin 3 \left(\frac{7\pi}{4}\right)\right]\\z^3 &= 8 \left(\cos \frac{21\pi}{4}+i \sin \frac{21\pi}{4}\right)\\z^3 &= 8 \left(-\frac{\sqrt{2}}{2}-i \frac{\sqrt{2}}{2}\right)\\z^3 &= -4\sqrt{2}-4i\sqrt{2}$
$\frac{21\pi}{4}$ is in the third quadrant so both are negative.
2.
$& x^4+1=0 && r=\sqrt{x^2+y^2}\\& x^4=-1 && r=\sqrt{(-1)^2+(0)^2}\\& x^4=-1+0i && r=1\\& \theta = \tan^{-1} \left(\frac{0}{-1}\right)+\pi=\pi$
Write an expression for determining the fourth roots of $x^4 = -1 + 0i$
$(-1+0i)^{\frac{1}{4}} &= [1(\cos (\pi+2\pi k)+i \sin (\pi+2\pi k))]^{\frac{1}{4}}\\(-1+0i)^{\frac{1}{4}} &= 1^{\frac{1}{4}} \left(\cos \frac{\pi+2\pi k}{4}+i \sin \frac{\pi+2\pi k}{4}\right)\\x_1 &= 1 \left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} && \text{for} \ k=0\\x_2 &= 1 \left(\cos \frac{3\pi}{4}+i \sin \frac{3\pi}{4}\right)=-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} && \text{for} \ k=1\\x_3 &= 1 \left(\cos \frac{5\pi}{4}+i \sin \frac{5\pi}{4}\right)=-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2} && \text{for} \ k=2\\x_4 &= 1 \left(\cos \frac{7\pi}{4}+i \sin \frac{7\pi}{4}\right)=\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2} && \text{for} \ k=3$
If a line segment is drawn from each root on the polar plane to its adjacent roots, the four roots will form the corners of a square.
3.
$x^3-64=0 \rightarrow x^3 = 64+0i$
$64 + 0i = 64(\cos (0 + 2 \pi k) + i \sin (0 + 2 \pi k))$
$x = (x^3)^{\frac{1}{3}} &= (64 + 0i)^{\frac{1}{3}}\\ & = \sqrt[3]{64} \left(\cos \left(\frac{0+2\pi k}{3}\right)+i \sin \left(\frac{0+2\pi k}{3}\right)\right)$
$z_1 &=4 \left(\cos \left(\frac{0+2\pi 0}{3}\right)+i \sin \left(\frac{0+2\pi 0}{3}\right)\right)\\ &=4 \cos 0 + 4i \sin 0 \\ &= 4 \text{ for } \ k=0$
$z_2 &=4 \left(\cos \left(\frac{0+2\pi}{3}\right)+i \sin \left(\frac{0+2\pi}{3}\right)\right)\\ &=4 \cos \frac{2\pi}{3}+4i \sin \frac{2\pi}{3}=-2 + 2i \sqrt{3} \text{ for } \ k=1$
$z_3 &=4 \left(\cos \left(\frac{0+4\pi}{3}\right)+i \sin \left(\frac{0+4\pi}{3}\right)\right)\\ &=4 \cos \frac{4\pi}{3}+4i \sin \frac{4\pi}{3}\\ &=-2-2i \sqrt{3} \text{ for } \ k=2$
If a line segment is drawn from each root on the polar plane to its adjacent roots, the three roots will form the vertices of an equilateral triangle.
Concept Problem Solution
Since you want to find the fourth root of 16, there will be four solutions in all.
$x^4=16.\\r &= \sqrt{x^2+y^2}\\r &= \sqrt{(16)^2+(0)^2}\\r &= 16\\\theta &= \tan^{-1} \left(\frac{0}{16}\right)=0$
Write an expression for determining the fourth roots of $16 = 16 + 0i$
$16^{\frac{1}{4}} &= [16( \cos (0+2\pi k)+i \sin (0+2\pi k)]^{\frac{1}{4}}\\&= 2\left(\cos \frac{2\pi k}{4}+i \sin \frac{2\pi k}{4}\right)k=0, 1, 2, 3, 4\\x_1 &= 2\left(\cos \frac{0}{4}+i \sin \frac{0}{4}\right) \rightarrow 2(\cos 0+i \sin 0)=2 && for \ k=0\\x_2 &= 2\left(\cos \frac{2\pi}{4}+i \sin \frac{2\pi}{4}\right) = 2i && for \ k=1\\x_3 &= 2\left(\cos \frac{4\pi}{4}+i \sin \frac{4\pi}{4}\right) = -2 && for \ k=2\\x_4 &= 2\left(\cos \frac{6\pi}{4}+i \sin \frac{6\pi}{4}\right) = -2i && for \ k=3\\$
Therefore, the four roots of 16 are $2, -2, 2i, -2i$ . Notice how you could find the two real roots if you seen complex numbers. The addition of the complex roots completes our search for the roots of equations.
Explore More
Solve each equation.
1. $x^3=1$
2. $x^5=1$
3. $x^8=1$
4. $x^5=-32$
5. $x^4+5=86$
6. $x^5=-1$
7. $x^4=-1$
8. $x^3=8$
9. $x^6=-64$
10. $x^3=-64$
11. $x^5=243$
12. $x^3=343$
13. $x^7=-128$
14. $x^{12}=1$
15. $x^6=1$
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5. Q5b
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Similar Question 1
<p>Use partial factoring to determine the vertex of each function. State if the vertex is a minimum or a maximum.</p><p><code class='latex inline'>f(x) = 0.3x^2 - 3x + 6</code></p>
Similar Question 2
<p>Graph <code class='latex inline'>y = -x^2 + 6x</code> to determine</p><p>a) the equation of the axis of symmetry</p><p>b) the coordinates of the vertex</p><p>c) the y—intercept</p><p>d) the zeros</p>
Similar Question 3
<p>For the quadratic relation,</p> <ul> <li>i) use partial factoring to determine two points that are the same distance from the axis of symmetry</li> <li>ii) determine the coordinates of the vertex</li> <li>iii) express the relation in vertex form</li> <li>iv) sketch the graph</li> </ul> <p><code class='latex inline'>\displaystyle{y=-\frac{1}{2}x^2+2x-3}</code></p>
Similar Questions
Learning Path
L1 Quick Intro to Factoring Trinomial with Leading a
L2 Introduction to Factoring ax^2+bx+c
L3 Factoring ax^2+bx+c, ex1
Now You Try
<p>Without graphing, determine the number of x-intercepts that the relation has.</p><p><code class='latex inline'>\displaystyle y = 2x(x - 5) + 7 </code></p>
<p>Use partial factoring to determine the vertex form of the quadratic relation <code class='latex inline'>y=2x^2-10x+11</code></p>
<p>Determine the maximum or minimum value. Use at least two different methods.</p><p><code class='latex inline'> \displaystyle y = -3x^2 - 12x + 15 </code></p>
<p>Graph <code class='latex inline'>y = -x^2 + 6x</code> to determine</p><p>a) the equation of the axis of symmetry</p><p>b) the coordinates of the vertex</p><p>c) the y—intercept</p><p>d) the zeros</p>
<p>Use graphing technology to graph the parabola for each relation below. Then determine</p> <ul> <li>i) the x—intercepts</li> <li>ii) the equation of the axis of symmetry </li> <li>iii) the coordinates of the vertex</li> </ul> <p><code class='latex inline'> \displaystyle y = 6x^2 + 15x </code></p>
<p>The points <code class='latex inline'>(-3, 8)</code> and <code class='latex inline'>(9, 8)</code> lie on opposite sides of a parabola. Determine the equation of the axis of symmetry.</p>
<p>Each pair of points is located on opposite sides of the same parabola. Determine the equation of the axis of symmetry for each parabola.</p><p> <code class='latex inline'>(-5.25,-2.5),(3.75,-2.5)</code></p>
<p>Use partial factoring to determine the vertex of each function. State if the vertex is a minimum or a maximum.</p><p><code class='latex inline'>f(x) = 0.3x^2 - 3x + 6</code></p>
<p>Use partial factoring to determine the vertex of each function. State if the vertex is a minimum or a maximum.</p><p><code class='latex inline'>\displaystyle f(x)=4 x^{2}-8 x+1 </code></p>
<p>Each pair of points is located on opposite sides of the same parabola. Determine the equation of the axis of symmetry for each parabola.</p><p> <code class='latex inline'>(3,2),(9,2)</code></p>
<p>Write each quadratic relation in vertex form using an appropriate strategy.</p><p><code class='latex inline'>\displaystyle y = x(3x + 12) + 2 </code></p>
<p>For the quadratic relation,</p> <ul> <li>i) use partial factoring to determine two points that are the same distance from the axis of symmetry</li> <li>ii) determine the coordinates of the vertex</li> <li>iii) express the relation in vertex form</li> <li>iv) sketch the graph</li> </ul> <p> <code class='latex inline'>y=2x^2-10x+11</code></p>
<p>Each pair of points is located on opposite sides of the same parabola. Determine the equation of the axis of symmetry for each parabola.</p><p><code class='latex inline'>(-18,3),(7,3)</code></p>
<ol> <li>Use partial factoring to determine the vertex of each function. State if the vertex is a minimum or a maximum.</li> </ol> <p>c) <code class='latex inline'>\displaystyle f(x)=-\frac{1}{2} x^{2}-4 x-3 </code></p>
<p>Use partial factoring to determine the vertex of each function. State if the vertex is a minimum or a maximum.</p><p><code class='latex inline'>f(x) = -0.2x^2 - 2.8x -5.4</code></p>
<ul> <li>i) determine the coordinates of two points on the graph that are the same distance from the axis of symmetry</li> <li>ii) determine the equation of the axis of symmetry </li> <li>iii) determine the coordinates of the vertex</li> <li>iv) write the relation in vertex form</li> </ul> <p> <code class='latex inline'>y=x(x-6)-8</code></p>
<p>Use graphing technology to graph the parabola for each relation below. Then determine</p> <ul> <li>i) the x—intercepts</li> <li>ii) the equation of the axis of symmetry </li> <li>iii) the coordinates of the vertex</li> </ul> <p><code class='latex inline'> \displaystyle y = -x^2 + 18x </code></p>
<ol> <li>Use partial factoring to determine the vertex of each function. State if the vertex is a minimum or a maximum.</li> </ol> <p>d) <code class='latex inline'>\displaystyle f(x)=\frac{4}{7} x^{2}-\frac{8}{7} x+\frac{25}{7} </code></p>
<p>Determine the maximum or minimum value. Use at least two different methods.</p><p><code class='latex inline'> \displaystyle y = 2x^2 + 12x </code></p>
<p>Which equation is the vertex form of the quadratic relation <code class='latex inline'>\displaystyle y=2 x(x-6)-5 ? </code></p><p>A. <code class='latex inline'>\displaystyle y=2(x-3)^{2}-23 </code></p><p>B. <code class='latex inline'>\displaystyle y=2(x-6)^{2}-5 </code></p><p>C. <code class='latex inline'>\displaystyle y=2(x-3)^{2}-5 </code></p><p>D. <code class='latex inline'>\displaystyle y=2(x-3)^{2}+23 </code></p>
<p>Use partial factoring to determine the vertex of each function. State if the vertex is a minimum or a maximum.</p><p><code class='latex inline'>f(x) = \frac{1}{2}x^2 - 3x + 8</code></p>
<p>For the quadratic relation,</p> <ul> <li>i) use partial factoring to determine two points that are the same distance from the axis of symmetry</li> <li>ii) determine the coordinates of the vertex</li> <li>iii) express the relation in vertex form</li> <li>iv) sketch the graph</li> </ul> <p><code class='latex inline'>y=x^2-6x+5</code></p>
<p>Use partial factoring to determine the vertex of each function. State if the vertex is a minimum or a maximum.</p><p><code class='latex inline'>f(x) = 3x^2 -6x + 11</code></p>
<ol> <li>Use partial factoring to determine the vertex of each function. State if the vertex is a minimum or a maximum.</li> </ol> <p>f) <code class='latex inline'>\displaystyle f(x)=-0.4 x^{2}+4 x+1 </code></p>
<p>Determine the maximum or minimum value. Use at least two different methods.</p><p><code class='latex inline'> \displaystyle y = 3x(x - 2) + 5 </code></p>
<p>For the quadratic relation,</p> <ul> <li>i) use partial factoring to determine two points that are the same distance from the axis of symmetry</li> <li>ii) determine the coordinates of the vertex</li> <li>iii) express the relation in vertex form</li> <li>iv) sketch the graph</li> </ul> <p><code class='latex inline'>y=-x^2-6x-13</code></p>
<ul> <li>i) determine the coordinates of two points on the graph that are the same distance from the axis of symmetry</li> <li>ii) determine the equation of the axis of symmetry </li> <li>iii) determine the coordinates of the vertex</li> <li>iv) write the relation in vertex form</li> </ul> <p><code class='latex inline'>y=x(3x+12)+2</code></p>
<p>For the quadratic relation,</p> <ul> <li>i) use partial factoring to determine two points that are the same distance from the axis of symmetry</li> <li>ii) determine the coordinates of the vertex</li> <li>iii) express the relation in vertex form</li> <li>iv) sketch the graph</li> </ul> <p><code class='latex inline'>\displaystyle{y=-\frac{1}{2}x^2+2x-3}</code></p>
<ol> <li>Two points on a parabola are <code class='latex inline'>\displaystyle (4,-1) </code> and <code class='latex inline'>\displaystyle (-10,-1) </code>. What is the equation of the axis of symmetry?</li> </ol>
<p>For the quadratic relation,</p> <ul> <li>i) use partial factoring to determine two points that are the same distance from the axis of symmetry</li> <li>ii) determine the coordinates of the vertex</li> <li>iii) express the relation in vertex form</li> <li>iv) sketch the graph</li> </ul> <p><code class='latex inline'>y=-2x^2+12x-11</code></p>
<p>Find </p> <ul> <li>i. the equation of the axis of symmetry</li> <li>ii. the coordinates of the vertex</li> <li>iii. the y-intercept</li> <li>iv. the zeros</li> </ul> <p>for </p><p><code class='latex inline'> \displaystyle y = x^2 -8x </code></p>
<ol> <li>Use partial factoring to determine the vertex of each function. State if the vertex is a minimum or a maximum.</li> </ol> <p>b) <code class='latex inline'>\displaystyle f(x)=-3 x^{2}-18 x-25 </code></p>
<p>Determine the maximum or minimum value. Use at least two different methods.</p><p><code class='latex inline'> \displaystyle f(x) = x^2 - 8x + 12 </code></p>
<p>Use partial factoring to determine the vertex of each function. State if the vertex is a minimum or a maximum.</p><p><code class='latex inline'>f(x) = -\frac{5}{3}x^2 + 5x - 10</code></p>
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QUIZ
แบบทดสอบหลังเรียน
2 hours ago by
10 questions
Q.
$\left(-8\right)\div\ 2$ = ???
answer choices
4
-4
16
-16
Q.
$15\div\left(-3\right)$ = ???
answer choices
-5
5
6
-9
Q.
$\frac{-20}{5}$ = ???
answer choices
100
-100
-4
4
Q.
$16\div\left(-2\right)$ = ???
answer choices
4
-4
8
-8
Q.
$\left(-8\right)\ \div\ \left(-2\right)\ =\ ???$
answer choices
4
-4
2
-2
Q.
$\frac{-14}{-2}\ =\ ???$
answer choices
7
-7
6
-6
Q.
$\frac{-18}{-3}\ =\ ???$
answer choices
6
-6
3
-3
Q.
$\left(-30\right)\ \div\ \left(-5\right)\ =\ ???$
answer choices
6
-6
3
-3
Q.
$\left(-144\right)\ \div\ \left(-8\right)\ =\ ???$
answer choices
18
-18
13
-13
Q.
$54\ +\ \left(-36\right)\ \div\ 9\ =\ ???$
answer choices
2
-2
58
-58
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## Wednesday, December 10, 2008
### 300
300 = 22 x 3 x 52.
300 is a triangular number, the sum of all positive integers up to 24: 1 + 2 + 3 + . . . + 22 + 23 + 24 = 300.
300 is the sum of twin primes: 149 + 151 = 300.
300 is a number n such that n2 (90000) contains exactly two different digits.
In bowling,
300 is the largest possible score.
Source:
Wolfram MathWorld
#### 1 comment:
Anonymous said...
300 is sum twin prime wow !
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# Different ways of calculating pi
Show 40 post(s) from this thread on one page
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• Nov 28th 2006, 01:10 PM
ThePerfectHacker
Quote:
Originally Posted by TriKri
That random number thing is interesting, cause I don't understand how $\displaystyle \pi$ can be in this kind of discrete probability, if I may say so. I guess that two randomly chosen positive numbers a and b both in the interval $\displaystyle [1, \infty)$ has this chance of not sharing any prime factor: $\displaystyle \prod_{p\text{ prime}} 1 - \frac{1}{p^2}$ because the chance is $\displaystyle \frac{1}{p}$ for a to be divisible by p and hence have prime factor p, and the same is for b. For both a and b to have prime factor p the chance is $\displaystyle \frac{1}{p^2}$. For a and b to not share prime factor p the chance is $\displaystyle 1 - \frac{1}{p^2}$. And for a and b to not share any of the primes as a prime factor the chance is $\displaystyle \prod_{p\text{ prime}} 1 - \frac{1}{p^2}$. Though I have no idea how to connect that formula with $\displaystyle \pi$.
This is the infamous "Euler-Product Formula" that started the entire Riemann hypothesis thing being connected to prime numbers.
The infinite product in this case is the same as the infinite sum of all integral square, namely the zeta function evaluated at 2.
Thus, (and I think you meant this)
$\displaystyle \prod _p(1-p^{-2})^{-1}=\zeta (2)=\frac{\pi^2}{6}$.
Hence, pi appears in this expression.
• Nov 28th 2006, 01:17 PM
CaptainBlack
Quote:
Originally Posted by ThePerfectHacker
I have a question on this. When I first saw this I found it really interesting that $\displaystyle \pi$ appears in the probability (I believe that you have to think of the full area of the needle, which forms a circle, and hence $\displaystyle \pi$). But I was told it does not converge very fast. If randomly done on a computer it shall take a few thousand experiments to get a few decimal points. Your approach for calculating $\displaystyle \pi$ is more elegant but I win this competition because mine is more efficient.
I'm not looking for efficiency I looking for Mad Scientists/Mathemeticians
methods, so I prefer Bob Matthews' and Buffon's methods, so we both win
RonL
• Nov 29th 2006, 05:27 PM
Quick
Quote:
Originally Posted by ThePerfectHacker
Here is an infinite series developed by Ramanjuan.
It has a collasol rate of convergence,
$\displaystyle \frac{1}{\pi} = \frac{\sqrt{8}}{9801}\sum_{n=0}^{\infty} \frac{(4n)!}{(n!)^4} \cdot \frac{1103+26390n}{396^{4n}}$
It certainly is not the nicest looking formula.
But the the most effective.
If you have a computer language, program this and see the results.
Here's my own formula, not nearly as effective (in fact it's reasonably slow) and pretty ugly :o but how many pi formulas have you made :p
And I don't think I'm writing the "sum" part correctly :eek:
$\displaystyle \pi=2\cdot\sum_{a\to n}^{n\to \infty}\frac{\sqrt{2\left(n^2-a^2+a-\sqrt{a^2\left(a-1\right)^2+n^2\left(n^2+2a(1-a)-1\right)}\right)}}{n}$
• Nov 29th 2006, 05:42 PM
Quick
Quote:
Originally Posted by Quick
Here's my own formula, not nearly as effective (in fact it's reasonably slow) and pretty ugly :o but how many pi formulas have you made :p
And I don't think I'm writing the "sum" part correctly :eek:
$\displaystyle \pi=2\cdot\sum_{a\to n}^{n\to \infty}\frac{\sqrt{2\left(n^2-a^2+a-\sqrt{a^2\left(a-1\right)^2+n^2\left(n^2+2a(1-a)-1\right)}\right)}}{n}$
just to clarify.
$\displaystyle n\to \infty$ means that the closer n gets to infinity, the more accurate the number.
$\displaystyle a\to n$ means you add up every integer less than or equal to n.
After 100 terms, the series spits out the number: 3.14129
• Nov 29th 2006, 06:03 PM
ThePerfectHacker
Quote:
Originally Posted by Quick
but how many pi formulas have you made :p
I made one. The most obvious one based on area below a semi-circle.
• Nov 29th 2006, 06:04 PM
topsquark
Quote:
Originally Posted by Quick
Here's my own formula, not nearly as effective (in fact it's reasonably slow) and pretty ugly :o but how many pi formulas have you made :p
And I don't think I'm writing the "sum" part correctly :eek:
$\displaystyle \pi=2\cdot\sum_{a\to n}^{n\to \infty}\frac{\sqrt{2\left(n^2-a^2+a-\sqrt{a^2\left(a-1\right)^2+n^2\left(n^2+2a(1-a)-1\right)}\right)}}{n}$
I believe you are looking for:
$\displaystyle \pi=\lim_{n \to \infty}2\cdot\sum_{a = 0}^n \frac{\sqrt{2\left(n^2-a^2+a-\sqrt{a^2\left(a-1\right)^2+n^2\left(n^2+2a(1-a)-1\right)}\right)}}{n}$
And where the heck does this monstrosity come from?
-Dan
• Nov 29th 2006, 06:16 PM
Quick
Quote:
Originally Posted by topsquark
I believe you are looking for:
$\displaystyle \pi=\lim_{n \to \infty}2\cdot\sum_{a = 0}^n \frac{\sqrt{2\left(n^2-a^2+a-\sqrt{a^2\left(a-1\right)^2+n^2\left(n^2+2a(1-a)-1\right)}\right)}}{n}$
And where the heck does this monstrosity come from?
-Dan
Well this "monstrosity" was a failed attempt at a function :cool:
But it's derived from measuring a quarter of the perimeter of an oddly shaped polygon inscribed in a unit circle ;)
I must say that if you take away the equal sign noone (not even me) would guess that this thing is even remotely related to pi :D
• Nov 29th 2006, 06:19 PM
ThePerfectHacker
Here is another method I came up with when I was Quick's age.
Using the Polygonal method (just like Quick used).
$\displaystyle \lim_{n\to \infty} n\sin \left( \frac{180^o}{n} \right)=\pi$
• Nov 29th 2006, 06:21 PM
Quick
Quote:
Originally Posted by ThePerfectHacker
Here is another method I came up with when I was Quick's age.
Using the Polygonal method (just like Quick used).
$\displaystyle \lim_{n\to \infty} n\sin \left( \frac{180^o}{n} \right)=\pi$
I did that to (remember my quick's quick question?)
Except I thought back and realized that sin might use pi to get the answer :eek:
That's the whole reason for this gigantic new one.
• Nov 30th 2006, 03:51 PM
TriKri
Quote:
Originally Posted by Qiuck
$\displaystyle \lim_{n \to \infty}2\cdot\sum_{a = 0}^n \frac{\sqrt{2\left(n^2-a^2+a-\sqrt{a^2\left(a-1\right)^2+n^2\left(n^2+2a(1-a)-1\right)}\right)}}{n}=\pi$
What does it do? I'm having a hard time following it. :confused:
I can see that it is equivalent to
$\displaystyle \lim_{n \to \infty}\frac{\sqrt{8}}{n}\cdot\sum_{a = 0}^n \sqrt{n^2-a^2+a-\sqrt{a^2\left(a-1\right)^2+n^2(n^2+2a(1-a)-1)}}=\pi$
but I don't know how that's supposed to help :o
• Nov 30th 2006, 06:22 PM
Quick
Quote:
Originally Posted by TriKri
What does it do? I'm having a hard time following it. :confused:
I can see that it is equivalent to
$\displaystyle \lim_{n \to \infty}\frac{\sqrt{8}}{n}\cdot\sum_{a = 0}^n \sqrt{n^2-a^2+a-\sqrt{a^2\left(a-1\right)^2+n^2(n^2+2a(1-a)-1)}}=\pi$
but I don't know how that's supposed to help :o
I've rewritten it (so that it looks better) to:
$\displaystyle \lim_{n\to\infty}\frac{\sqrt8}{n}\cdot\sum_{a=0}^n \sqrt{n^2-a^2+a-\sqrt{a^2\left(a-1\right)^2+2an^2(1-a)+n^2(n^2-1)}}=\pi$
although it still looks pretty ugly :o
If you are asking how it's derived, I can easily show you :)
• Nov 30th 2006, 06:37 PM
Quick
Quote:
Originally Posted by Quick
I've rewritten it (so that it looks better) to:
$\displaystyle \lim_{n\to\infty}\frac{\sqrt8}{n}\cdot\sum_{a=0}^n \sqrt{n^2-a^2+a-\sqrt{a^2\left(a-1\right)^2+2an^2(1-a)+n^2(n^2-1)}}=\pi$
although it still looks pretty ugly :o
If you are asking how it's derived, I can easily show you :)
I'm having a weird problem, hopefully with MSexcel :(
after 100 terms, excel gives: 3.141299
After 200 however, excel gives: 3.194243 :eek:
The way I set up the equation, it should never get above pi...
Can anyone verify excels answer (with mathematica or something)
If it's my equation, then I'll post the method for you guys to see what's wrong (I've gone over the work several times)
• Dec 1st 2006, 04:49 AM
TriKri
> If you are asking how it's derived, I can easily show you
Yes, that's what I meant. :)
• Dec 1st 2006, 02:23 PM
Quick
Quote:
Originally Posted by TriKri
> If you are asking how it's derived, I can easily show you
Yes, that's what I meant. :)
I hope you have a good imagination...
Consider a unit circle surrounding the origin, it's equation is $\displaystyle y^2=1-x^2$
Now my method only uses the first quadrant, so forget all the other ones :eek:
First, put $\displaystyle n$ equally spaced points along the x-axis from 0 to 1.
Now draw vertical lines from the points.
Connect the intersections of the lines and the circle to each other (you'll get a quarter of an oddly shaped polygon)
So try finding the total of those measurements.
To do that you must first solve for the change in y (which I call $\displaystyle \Delta y$ for each increment.
Consider the vertical line at point $\displaystyle a$, it intersects the circle at the y value: $\displaystyle y^2=1-x^2\Longrightarrow y=\sqrt{1-\left(\frac{a}{n}\right)^2}=\sqrt{\frac{n^2}{n^2}-\frac{a^2}{n^2}}=\sqrt{\frac{n^2-a^2}{n^2}}$
So then: $\displaystyle \Delta y=\sqrt{\frac{n^2-a^2}{n^2}}-\sqrt{\frac{n^2-(a-1)^2}{n^2}}$
Then: $\displaystyle \Delta y=\sqrt{\frac{n^2-a^2}{n^2}}-\sqrt{\frac{n^2-(a^2-2a+1)}{n^2}}$
Thus: $\displaystyle \Delta y^2=\frac{n^2-a^2}{n^2}+\frac{n^2-(a^2-2a+1)}{n^2}-2\sqrt{\frac{n^2-a^2}{n^2}}\sqrt{\frac{n^2-(a^2-2a+1)}{n^2}}$
Working through you'll eventually get: $\displaystyle \Delta y^2=\frac{2n^2-2a^2+2a-1-2\sqrt{a^2\left(a-1\right)^2+n^2\left(n^2+2a(1-a)-1\right)}}{n^2}$
Now the value of the hypotenuse of the triangle with $\displaystyle \Delta y$ as the height is: $\displaystyle \sqrt{\frac{2n^2-2a^2+2a-1-2\sqrt{a^2\left(a-1\right)^2+n^2\left(n^2+2a(1-a)-1\right)}}{n^2}+\frac{1}{n^2}}$
And then solve...
Can you do it from here? It wasn't a very good explanation...
• Dec 1st 2006, 03:40 PM
TriKri
I dont get this ... as far as I can see, you calculate $\displaystyle y'$ from the equation $\displaystyle y = \sqrt{1-x^2}$, to put an integral on $\displaystyle y'$, to get ... pi? How is that... :confused:
Isn't it simpler just integrating $\displaystyle y$ directly:
$\displaystyle \sum_{a=0}^n \frac{\sqrt{1-\left(\displaystyle\frac{a}{n}\right)^2}}{n}\ =\ \frac{1}{n}\sum_{a=0}^n \sqrt{\frac{n^2-a^2}{n^2}}\ = \ \frac{1}{n^2}\sum_{a=0}^n \sqrt{n^2-a^2}$
And one thing, I spoted that you set $\displaystyle \Delta y = \sqrt{\Delta y^2}$, that's equal to $\displaystyle |\Delta y|$ for real numbers $\displaystyle \Delta y$.
Note that $\displaystyle \sqrt{\frac{n^2-a^2}{n^2}}-\sqrt{\frac{n^2-(a-1)^2}{n^2}}\ =\$$\displaystyle -\sqrt{\frac{2n^2-2a^2+2a-1-2\sqrt{a^2\left(a-1\right)^2+n^2\left(n^2+2a(1-a)-1\right)}}{n^2}+\frac{1}{n^2}}$
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Prealgebra (7th Edition)
2(z-2)=5z+17 Apply the distributive property 2z-4=5z+17 Add 4 to both sides 2z-4+4=5z+17+4 Simplify 2z=5z+21 Subtract 5z from both sides 2z-5z=5z+21-5z Simplify -3z=21 Divide by- 3 on both sides $\frac{-3z}{-3}$=$\frac{21}{-3}$ z= -7 The solution is -7 Check 2(z-2)=5z+17 Replace z with -7 2(-7-2)=5(-7)+17 Simplify inside parentheses 2(-9)=5(-7)+17 Multiply -18=-35+17 -18=-18
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Take our daily marvellous Primary School Maths Thinker to sharpen your mind and enhance your skills. Find curriculum-aligned maths questions to help your child explore essential concepts. Daily questions will be posted Monday to Thursday for Grade 4 – 7 respectively. Look out for the solutions to these questions every Friday (all grades). The CTSC Education Team will be standing by should you have any questions – please reach out by posting any questions on our Facebook or Instagram page in the comments section. #CTSCMaths
Time for the Grade 7 learners to put their problem-solving caps on.
QUESTIONS:
1. In the report card that Bafo received, he scored 93 in Natural Sciences, 88 in Mathematics, and a score in English that is double his score in Life Orientation. The average score of all 4 courses is 79. What were his scores in English and Life Orientation?
2. There are bicycles and cars in the parking lot. There is a total of 300 wheels including 100 small wheels for bicycles. How many cars and how many bicycles are there?
3. The difference between the two numbers is 17 and their sum is 69. Find the larger of these two numbers.
NUMBER VALUE AND NOTATION – The following questions are aimed at Grade 6 learners.
QUESTIONS:
1. What is 867 rounded to the nearest hundred?
2. Estimate the sum of 467 and 237 by rounding each number to the nearest hundred
3. Estimate the sum of 2.7 and 3.4 by rounding each number to the nearest whole number.
4. What is the product of 34 and 69, after rounding each number to the nearest ten?
5. The iCafe at the Cape Town Science Centre ordered 40 boxes of chocolate bars. In each box, there are 40 individual chocolate bars. How many individual chocolate bars did they order?
6. How many tiles would fill this block if it could fit 600 tiles in its width and 10 tiles in its length? (See image)
7. Which of the following is not a prime number: 2, 47, 31, 55, 61
8. Which number in 5 763 is in the tens position?
9. Comparing the following four numbers, which number represents the lowest value?
– 6 681
– 6 618
– 6 162
– 6 168
10. In which position is number 6 in the following number: 9 610
We are trying out LONG DIVISION to challenge the Grade 5 maths skills.
Question 1, 2 & 3 – See image
4. What is the quotient if the divisor is 30 and the dividend is 9005?
5. Cavendish Movie Theatre can seat 2,060 people in total. There are ten auditorium rooms in the theatre. Each room can hold the same number of people except for the tenth room, which can hold 10 more people than the other rooms. How many people can each room seat
5. 125 learners have signed up to go on a school field trip. The learners will be transported by parents and there are a total of twenty vehicles to transport them. Each vehicle will transport the same number of learners. How many learners will go in each vehicle and will they need another vehicle? Explain
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# Search by Topic
#### Resources tagged with Mathematical reasoning & proof similar to A Story about Absolutely Nothing:
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Broad Topics > Using, Applying and Reasoning about Mathematics > Mathematical reasoning & proof
### Happy Numbers
##### Stage: 3 Challenge Level:
Take any whole number between 1 and 999, add the squares of the digits to get a new number. Make some conjectures about what happens in general.
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Replace each letter with a digit to make this addition correct.
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Look at three 'next door neighbours' amongst the counting numbers. Add them together. What do you notice?
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Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
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##### Stage: 2 Challenge Level:
Are these statements always true, sometimes true or never true?
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Pick the number of times a week that you eat chocolate. This number must be more than one but less than ten. Multiply this number by 2. Add 5 (for Sunday). Multiply by 50... Can you explain why it. . . .
##### Stage: 3 Challenge Level:
Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.
### Geometry and Gravity 2
##### Stage: 3, 4 and 5
This is the second of two articles and discusses problems relating to the curvature of space, shortest distances on surfaces, triangulations of surfaces and representation by graphs.
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##### Stage: 2, 3, 4 and 5
This article invites you to get familiar with a strategic game called "sprouts". The game is simple enough for younger children to understand, and has also provided experienced mathematicians with. . . .
### Largest Product
##### Stage: 3 Challenge Level:
Which set of numbers that add to 10 have the largest product?
### Cows and Sheep
##### Stage: 2 Challenge Level:
Use your logical reasoning to work out how many cows and how many sheep there are in each field.
### Calendar Capers
##### Stage: 3 Challenge Level:
Choose any three by three square of dates on a calendar page...
##### Stage: 1 and 2 Challenge Level:
Who said that adding couldn't be fun?
### Making Pathways
##### Stage: 2 Challenge Level:
Can you find different ways of creating paths using these paving slabs?
### Cycle It
##### Stage: 3 Challenge Level:
Carry out cyclic permutations of nine digit numbers containing the digits from 1 to 9 (until you get back to the first number). Prove that whatever number you choose, they will add to the same total.
### Yih or Luk Tsut K'i or Three Men's Morris
##### Stage: 3, 4 and 5 Challenge Level:
Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots and a little about prime knots, crossing numbers and. . . .
### Impossible Sandwiches
##### Stage: 3, 4 and 5
In this 7-sandwich: 7 1 3 1 6 4 3 5 7 2 4 6 2 5 there are 7 numbers between the 7s, 6 between the 6s etc. The article shows which values of n can make n-sandwiches and which cannot.
### The Triangle Game
##### Stage: 3 and 4 Challenge Level:
Can you discover whether this is a fair game?
### Pythagorean Triples II
##### Stage: 3 and 4
This is the second article on right-angled triangles whose edge lengths are whole numbers.
### Pythagorean Triples I
##### Stage: 3 and 4
The first of two articles on Pythagorean Triples which asks how many right angled triangles can you find with the lengths of each side exactly a whole number measurement. Try it!
##### Stage: 3 Challenge Level:
Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some. . . .
##### Stage: 2 Challenge Level:
Three dice are placed in a row. Find a way to turn each one so that the three numbers on top of the dice total the same as the three numbers on the front of the dice. Can you find all the ways to do. . . .
### Picture Story
##### Stage: 3 and 4 Challenge Level:
Can you see how this picture illustrates the formula for the sum of the first six cube numbers?
### Cross-country Race
##### Stage: 3 Challenge Level:
Eight children enter the autumn cross-country race at school. How many possible ways could they come in at first, second and third places?
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##### Stage: 3 Challenge Level:
Here are three 'tricks' to amaze your friends. But the really clever trick is explaining to them why these 'tricks' are maths not magic. Like all good magicians, you should practice by trying. . . .
### Take One Example
##### Stage: 1 and 2
This article introduces the idea of generic proof for younger children and illustrates how one example can offer a proof of a general result through unpacking its underlying structure.
### Square Subtraction
##### Stage: 2 Challenge Level:
Look at what happens when you take a number, square it and subtract your answer. What kind of number do you get? Can you prove it?
### Same Length
##### Stage: 3 and 4 Challenge Level:
Construct two equilateral triangles on a straight line. There are two lengths that look the same - can you prove it?
### Advent Calendar 2011 - Secondary
##### Stage: 3, 4 and 5 Challenge Level:
Advent Calendar 2011 - a mathematical activity for each day during the run-up to Christmas.
### What Numbers Can We Make Now?
##### Stage: 3 Challenge Level:
Imagine we have four bags containing numbers from a sequence. What numbers can we make now?
### Always, Sometimes or Never?
##### Stage: 1 and 2 Challenge Level:
Are these statements relating to odd and even numbers always true, sometimes true or never true?
### Problem Solving, Using and Applying and Functional Mathematics
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### More Number Sandwiches
##### Stage: 3 and 4 Challenge Level:
When is it impossible to make number sandwiches?
### Always, Sometimes or Never? Shape
##### Stage: 2 Challenge Level:
Are these statements always true, sometimes true or never true?
### What Numbers Can We Make?
##### Stage: 3 Challenge Level:
Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?
### Take Three Numbers
##### Stage: 2 Challenge Level:
What happens when you add three numbers together? Will your answer be odd or even? How do you know?
### Tis Unique
##### Stage: 3 Challenge Level:
This addition sum uses all ten digits 0, 1, 2...9 exactly once. Find the sum and show that the one you give is the only possibility.
##### Stage: 3 and 4 Challenge Level:
Draw some quadrilaterals on a 9-point circle and work out the angles. Is there a theorem?
### Even So
##### Stage: 3 Challenge Level:
Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why?
### Breaking the Equation ' Empirical Argument = Proof '
##### Stage: 2, 3, 4 and 5
This article stems from research on the teaching of proof and offers guidance on how to move learners from focussing on experimental arguments to mathematical arguments and deductive reasoning.
### Elevenses
##### Stage: 3 Challenge Level:
How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results?
### Coins on a Plate
##### Stage: 3 Challenge Level:
Points A, B and C are the centres of three circles, each one of which touches the other two. Prove that the perimeter of the triangle ABC is equal to the diameter of the largest circle.
### Sticky Numbers
##### Stage: 3 Challenge Level:
Can you arrange the numbers 1 to 17 in a row so that each adjacent pair adds up to a square number?
### Unit Fractions
##### Stage: 3 Challenge Level:
Consider the equation 1/a + 1/b + 1/c = 1 where a, b and c are natural numbers and 0 < a < b < c. Prove that there is only one set of values which satisfy this equation.
##### Stage: 2 and 3
A paradox is a statement that seems to be both untrue and true at the same time. This article looks at a few examples and challenges you to investigate them for yourself.
### Shuffle Shriek
##### Stage: 3 Challenge Level:
Can you find all the 4-ball shuffles?
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
### A Chordingly
##### Stage: 3 Challenge Level:
Find the area of the annulus in terms of the length of the chord which is tangent to the inner circle.
### Logic
##### Stage: 2 and 3
What does logic mean to us and is that different to mathematical logic? We will explore these questions in this article.
### Con Tricks
##### Stage: 3
Here are some examples of 'cons', and see if you can figure out where the trick is.
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Physics Solution Manual for 1100 and 2101
When f is large enough the wheel will rise up off the
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Unformatted text preview: t the maximum speed v at which a vehicle can negotiate a curve of radius r is related to the SSF according to v = rg ( SSF ) . No value is given for the radius of the turn. However, by applying this result separately to the sport utility vehicle (SUV) and to the sports car (SC), we will be able to eliminate r algebraically and determine the maximum speed at which the sports car can negotiate the curve without rolling over. SOLUTION Applying v = rg ( SSF ) to each vehicle, we obtain vSUV = rg ( SSF )SUV and vSC = rg ( SSF )SC Dividing these two expressions gives vSC vSUV = rg ( SSF )SC rg ( SSF )SUV or vSC = vSUV (SSF )SC = (18 m/s ) (SSF )SUV 1.4 = 24 m/s 0.80 18. REASONING Since the wheelbarrow is in equilibrium, the net torque acting on it must be zero: Στ = 0 (Equation 9.2). The magnitude of a torque is the magnitude of the force times the lever arm of the force (see Equation 9.1). The lever arm is the perpendicular distance between the line of action of the force and the axis. A torque that tends to produce a counterclockwise rotation about the axis is a positive torque. SOLUTION The lever arms for the forces can be obtained from the distances shown in the text drawing for each design. Equation 9.1 can be used to obtain the magnitude of each torque. We will then write an expression for the zero net torque for each design. These expressions can be solved for the magnitude F of the man’s force in each case: Chapter 9 Problems Left design 447 bg b gb g b gb g 525 + 60.0 b b N g0.400 mgb N g0.600 mg 189 N b F= = Στ = − 525 N 0.400 m − 60.0 N 0.600 m + F 1.300 m = 0 1.300 m Right design bg b gb g 60.0 b b N g0.600 mg 27.7 N F= = Στ = − 60.0 N 0.600 m + F 1.300 m = 0 1.300 m 19. SSM REASONING AND SOLUTION The net torque about an axis through the contact point between the tray and the thumb is 2 2 Στ = F(0.0400 m) − (0.250 kg)(9.80 m/s )(0.320 m) − (1.00 kg)(9.80 m/s )(0.180 m) 2 − (0.200 kg)(9.80 m/s )(0.140 m) = 0 F = 70.6 N, up Similarly, the net torque about an axis through the point of contact between the tray and the finger is 2 2 Στ = T (0.0400 m) − (0.250 kg)(9.80 m/s )(0.280 m) − (1.00 kg)(9.80 m/s )(0.140 m) 2 − (0.200 kg)(9.80 m/s )(0.100 m) = 0 T = 56.4 N, down 20. REASONING The jet is in equilibrium, so the sum of the external forces is zero, and the sum of the external torques is zero. We can use these two conditions to evaluate the forces exerted on the wheels. SOLUTION a. Let Ff be the magnitude of the normal force that the ground exerts on the front wheel. Since the net torque acting on the plane is zero, we have (using an axis through the points of contact between the rear wheels and the ground) Στ = −W lw + Ff lf = 0 where W is the weight of the plane, and lw and lf are the lever arms for the forces W and Ff, respectively. Thus, 448 ROTATIONAL DYNAMICS Στ = −(1.00 × 106 N)(15.0 m − 12.6 m) + Ff (15.0 m) = 0 Solving for Ff gives Ff = 1.60 × 10 5 N . b. Setting the sum of the vertical forces equal to zero yields ΣFy = Ff + 2Fr − W = 0 where the factor of 2 arises because there are two rear wheels. Substituting the data gives ΣFy = 1.60 × 105 N + 2Fr − 1.00 × 106 N = 0 Fr = 4.20 ×105 N 21. REASONING Since the beam is in equilibrium, the sum of the horizontal and vertical forces must be zero: ΣFx = 0 and ΣFy = 0 (Equations 9.4a and b). In addition, the net torque about any axis of rotation must also be zero: Στ = 0 (Equation 9.2). SOLUTION The drawing shows the beam, as well as its weight W, the force P that the pin exerts on the right end of the beam, and the horizontal and vertical forces, H and V, applied to the left end of the beam by the hinge. Assuming that upward and to the right are the positive directions, we obtain the following expressions by setting the sum of the vertical and the sum of the horizontal forces equal to zero: V H P Beam θ θ W Brace Pin (rotational axis) Horizontal forces P cosθ − H = 0 (1) Vertical forces P sin θ + V − W = 0 (2) Using a rotational axis perpendicular to the plane of the paper and passing through the pin, and remembering that counterclockwise torques are positive, we also set the sum of the torques equal to zero. In doing so, we use L to denote the length of the beam and note that the lever arms for W and V are 1 L and L, respectively. The forces P and H create no 2 torques relative to this axis, because their lines of action pass directly through it. Chapter 9 Problems Torques W − c L hVL = 0 1 2 449 ( 3) Since L can be eliminated algebraically, Equation (3) may be solved immediately for V: 1 V = 2W = 1 2 340 = b Ng 170 N Substituting this result into Equation (2) gives 1 P sin θ + 2 W − W = 0 P= W 340 N = = 270 N 2 sin θ 2 sin 39 ° Substituting this result into Equation (1) yields F W Icosθ − H = 0 Gsinθ J HK 2 H= W 340 N = = 210 N 2 tan θ 2 tan 39 ° 22. REASONING AND SOLUTION The net torque about an axis through the elbow joint is Στ = (111 N)(0.300 m) − (22.0 N)(0.150 m) − (0.0250 m)M = 0 or M =...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
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# D1 help
Announcements
#1
Gin is 45% alcohol by volume and tonic is non-alcoholic. Its required to mix gin and tonics which are at least 10% alcohol by volume and contain atleast 200ml of liquid but at most 30ml of alcohol
a) what are the greatest and least amounts of gin a drink can contain
b) what are the greatest and least amounts of tonic a drink can contain
I understand you would first make the constraints x?30, x+y?10, x+y?200 ,
x?0, y?0,
Secondly do you need to draw a graph?
Any help would be appreciated
0
8 years ago
#2
(Original post by Nina8)
Gin is 45% alcohol by volume and tonic is non-alcoholic. Its required to mix gin and tonics which are at least 10% alcohol by volume and contain atleast 200ml of liquid but at most 30ml of alcohol
a) what are the greatest and least amounts of gin a drink can contain
b) what are the greatest and least amounts of tonic a drink can contain
I understand you would first make the constraints x?30, x+y?10, x+y?200 ,
x?0, y?0,
Secondly do you need to draw a graph?
Any help would be appreciated
A graph would be useful, but if you can do it without, fair enough.
I presume x is the amount of gin and y is the amount of tonic.
What constraints did you end up with? Several of those, as they, stands don't look right.
Since the gin is 45% alcohol by volume, then 100ml of gin gives 45 ml of alcohol.
0
#3
(Original post by ghostwalker)
A graph would be useful, but if you can do it without, fair enough.
I presume x is the amount of gin and y is the amount of tonic.
What constraints did you end up with? Several of those, as they, stands don't look right.
Since the gin is 45% alcohol by volume, then 100ml of gin gives 45 ml of alcohol.
I had x>= 0 , y>=0 , 0.45x+y/x+y >= 0.10 (because it says atleast 10 percent alcohol), x+y >= 200 , and Im so confused how to put in atmost 30ml of alcohol.
Im just going in circles right now
0
8 years ago
#4
(Original post by Nina8)
0.45x+y/x+y >= 0.10 (because it says atleast 10 percent alcohol)
That's not correct - the numerator. You need amount of alcohol over total amount >=0.1
There's no alcohol in tonic.
how to put in atmost 30ml of alcohol.
Amount of alcohol is 0.45x, hence 0.45x <= 30
0
#5
[QUOTE=ghostwalker;46479332]That's not correct - the numerator. You need amount of alcohol over total amount >=0.1
There's no alcohol in tonic.
Amount of alcohol is 0.45x, hence 0.45x <= 30[/QUOTE
I meant (0)y which would make it a 0!! Thankyou so much you're a lifesaver ) How would I work to get the greatest and least amount from there?
(sorry for being such a bother)
0
8 years ago
#6
(Original post by Nina8)
...
Do a graph and plot the feasible region.
0
#7
(Original post by ghostwalker)
Do a graph and plot the feasible region.
And find the intersections right? THANKYOU SO MUCH, you're amazing
0
8 years ago
#8
so what should this have been: 0.45x+y/x+y >= 0.10?
0
8 years ago
#9
(Original post by Welbeck)
so what should this have been: 0.45x+y/x+y >= 0.10?
There's no alcohol in the tonic, so:
0.45x/(x+y) >= 0.10
0
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# why is the definition of the determinant so weird? [duplicate]
I learned linear algebra from books of Friedberg, Gilbert & Strang, Anton etc. by myself. I dare say, that I learned all that stuff eagerly.
Studying by myself, I could not intuitively understand the definition of the determinant (its even – odd manner). I could only memorize the definition, and then use it (or try to use it) to solve some related readers' homework problems or other exercises.
As you know, the purpose of a determinant is literally to determine whether a given system of equations has a unique solution or not.
In other words, the "determinant" will determine whether the row vectors (and equivalently, column vectors) of a given square matrix are independent or not.
If those are mutually independent, then they can geometrically represent an $n$-dimensional quantity (for example, area in 2 dimensions or volume in 3D). If not, some of them are dependent, so they cannot form the $n$-dimensional quantity, and correspondingly the determinant is zero.
A multiple of any row can be added to another, this kind of row elementary operation does not change the determinant value. The picture below illustrates an intuitive understanding of that, too.
Writing the sides of the parallelogram as rows or columns of a square matrix, this transformation transforms it to another with the same value of the determinant.
It can be transformed to Gauss–Jordan form, in this case, each of the row / column vectors are orthogonal because their inner products are all zero. (I tried this with the Gram–Schmidt process; however, intuitively, the result is surely the same.)
Those vectors are orthogonal so it is very clear that just multiplication of the diagonal terms should give directly the aforementioned $n$-dimensional quantity, so that's the determinant in such case.
I understand the determinant in this manner, and it makes sense intuitively.
However the textbook definition mentioned above (defined in the "even–odd" manner) looks very weird to me.
What is the motivation of that definition? And can it be generally derived from my intuition about the $n$-dimensional quantities? I succeeded in doing so for the $2\times2$ and $3\times3$ cases, but I cannot see any generalized relation.
It seems to me that the definition of the determinant comes down magically, without enough logic.
I was wondering if you could help me.
• I guess your queries are answered here: math.stackexchange.com/questions/668/…, math.stackexchange.com/questions/250534/…. – StubbornAtom Nov 14 '16 at 17:56
• That image is illegible. Also, could you quote the exact definition? – StubbornAtom Nov 14 '16 at 17:58
• The image shows the 2x2 determinant as the area of the parallelogram formed by the two vectors. – Momo Nov 14 '16 at 18:00
• yeah the image shows the the operation (Add a row to another one multiplied by a number) conserve the "determinant" value – jotkey Nov 14 '16 at 18:01
• You could also watch these videos about the subject. He discusses the determinant about halfway in the series. – Arthur Nov 14 '16 at 18:13
Look at the properties that signed volume has. Think of it as a function $d : \mathbb{R}^n \times \cdots \times \mathbb{R}^n \to \mathbb{R}$.
(i) It is multilinear. (ii) If you swap two parameters, you switch the sign. (iii) $d(e_1,....,e_n) = 1$.
From these properties alone you can derive the textbook formula.
So, if you think of the above as defining the determinant, the definition is far from weird.
Here is a quick sketch of how we obtain the formula:
As Ian noted in the comments (iii) says that the determinant of the identity is one.
From (ii), we can show that if two of the parameters to $d$ are the same then the result is zero.
Take a square matrix $A$. Then the $j$th column is $\sum_{\sigma=1}^n A_{\sigma,j} e_\sigma$.
Using (i) we have $\det A = \sum_{\sigma_1 =1}^n \cdots \sum_{\sigma_n =1}^n A_{\sigma_1,1} \cdots A_{\sigma_n,n} d(e_{\sigma_1},...,e_{\sigma_n})$.
Now note that $d(e_{\sigma_1},...,e_{\sigma_n}) = 0$ whenever any index is repeated. Hence we can replace the sum $\sum_{\sigma_1 =1}^n \cdots \sum_{\sigma_n =1}^n$ by $\sum_{\sigma \in S}$, where $S$ is the set of permutations $\sigma: \{1,...,n\} \to \{1,...,n\}$.
Hence we have $\det A = \sum_{\sigma \in S} A_{\sigma_1,1} \cdots A_{\sigma_n,n} d(e_{\sigma_1},...,e_{\sigma_n})$.
Using (ii) & (iii), we can show that $d(e_{\sigma_1},...,e_{\sigma_n}) = \operatorname{sgn} \sigma$, and we end up with $\det A = \sum_{\sigma \in S} \operatorname{sgn} \sigma A_{\sigma_1,1} \cdots A_{\sigma_n,n}$.
• Sir, I was wondering if you could show me more specific manner? – jotkey Nov 14 '16 at 18:29
• @jotkey: I added a quick summary, you need to finish the details :-). – copper.hat Nov 14 '16 at 18:48
• @copper.hot god bless you america. – jotkey Nov 14 '16 at 18:57
• @jotkey: Glad to help! – copper.hat Nov 14 '16 at 18:59
• @Bungo: Thanks for catching that! – copper.hat Nov 14 '16 at 19:41
Take a system of equations where the coefficients are variables, e.g.
$$a x + b y = e$$ $$c x + d y = f$$
Solve it:
$$x = \frac{d e - b f}{a d - b c}, \qquad y = \frac{a f - c e}{a d - b c}$$
Notice that each expression has the same denominator, namely $a d - b c$. This can be proven to hold for an arbitrary $n \times n$ system, and the thing in the denominator is the determinant of the system. [1]
This can be made into a formal definition; see for instance the expository paper,
• Garibaldi, Skip. “The Characteristic Polynomial and Determinant Are Not Ad Hoc Constructions.” The American Mathematical Monthly, vol. 111, no. 9, 2004, pp. 761–778. http://www.jstor.org/stable/4145188.
[1] To be entirely fair, we could also have chosen the negative of what we normally call the "determinant." If you like to think of the determinant as a signed volume, this choice is equivalent to whether we choose a left-handed or right-handed orientation on space. Of course such a choice is completely arbitrary, but makes little difference as long as we all agree on the same one.
• I think this motivation is more natural than the others, such like signed area/volume. – Eric Nov 15 '16 at 12:22
Another way to interpret the determinant arises naturally from the alternating product construction on vector spaces. Briefly, given a vector space $V$, then $\Lambda^n V$ is a vector space defined to be generated by formal expressions of the form $x_1 \wedge \ldots \wedge x_n$ for $x_1, \ldots, x_n \in V$, subject to the relations:
1. The wedge product is linear in each of the terms, i.e. $$x_1 \wedge \ldots \wedge (\lambda_1 x_i + \lambda_2 x_i') \wedge \ldots \wedge x_n = \lambda_1 (x_1 \wedge \ldots x_i \wedge \ldots x_n) + \lambda_2 (x_1 \wedge \ldots x_i' \wedge \ldots x_n).$$
2. The wedge product is zero if any two adjacent terms are equal: $$x_1 \wedge \ldots \wedge y \wedge y \wedge \ldots x_n = 0.$$ (Note that since also $\ldots \wedge (y+z) \wedge (y+z) \wedge \ldots = 0$, this implies that $$\ldots \wedge y \wedge z \wedge \ldots = -(\ldots \wedge z \wedge y \ldots).$$ This is the reason for the name "alternating product" or "antisymmetric product".)
Now, it turns out that if $V$ is an $n$-dimensional vector space, than $\Lambda^k V$ is an $\binom{n}{k}$-dimensional vector space; and in particular $\Lambda^n V$ is a 1-dimensional vector space. Also, for any linear transformation $T : V \rightarrow W$, it is easy to define a corresponding linear transformation $\Lambda^k T : \Lambda^k V \rightarrow \Lambda^k W$ such that $(\Lambda^k T)(x_1 \wedge \ldots \wedge x_k) = Tx_1 \wedge \ldots Tx_k$.
Now, the interpretation of the determinant is as follows: given a linear operator $T : V \to V$ on an $n$-dimensional vector space, then $\det T$ is simply defined to be the unique scalar such that $\Lambda^n T$ is equal to multiplication by $\det T$. And for a matrix $A \in M_{n \times n}(F)$, $\det A$ is the determinant of the corresponding linear operator on $F^n$.
This definition has some distinct advantages: for example, it's clear from it why the determinant is multiplicative: $\det(T \circ U) = \det(T) \det(U)$. It also gives a relatively natural proof that the determinant of a singular linear operator is 0: just choose a basis including a vector in the null space. On the other hand, actually proving the dimension of $\Lambda^k V$ turns out to be most straightforward using the determinant as a tool - which would make it a circular definition. However, even in that "bootstrapping" phase, keeping this other definition in mind can definitely help in motivating a formulation of the actual initial definition of determinant.
• Great answer. Suggested edit: "Now, it turns out that if $V$ is an $n$-dimensional vector space, then $Λ^kV$ is an $\tbinom{n}{k}$ -dimensional vector space" – justadzr Mar 5 '20 at 5:47
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# Chapter 5. Numerical Integration
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1 Chpter 5. Numericl Integrtion These re just summries of the lecture notes, nd few detils re included. Most of wht we include here is to be found in more detil in Anton. 5. Remrk. There re two topics with similr nmes: Reverse of differentition Indefinite integrl f(x) dx = most generl ntiderivtive for f(x) Definite integrl This is relted to summtion (it is limit of sums of certin kind). The integrl sign ws originlly invented s modified S (for sum). There is no reson to expect connection between these two different things, but there is. See course S, the book by Anton or the ppendix for more detils. Wht we need is the ide of Riemnn sum. 5.2 The Definite Integrl. b b number nd does not involve x. Nottion is convenient. Grphicl interprettion: grph y = f(x), the x-xis nd the verticl lines x =, x = b. f(x) dx (red integrl from to b of the function f). Answer is f(x) dx is the re of the region in the plne bounded by the
2 Mthemtics S3 (Timoney) (picture good for cse f(x) 0) Problems: need wy to compute the re of such shpe. More fundmentlly, need definition of wht you men by the re. 5.3 Nottion. We need nottion to write down formule for these pictures. Deling with b f(x) dx n subdivisions of intervl [, b] Number division point = x 0 < x < x 2 < < x n = b Heights of rectngles y j = f(x j), x j x j x j ny point. Are of jth rectngle = width height = (x j x j )y j Totl re = sum of these = y (x x 0 ) + y 2 (x 2 x ) + + y n (x n x n ) = f(x )(x x 0 ) + f(x 2)(x 2 x ) + + f(x n)(x n x n ) (clled Riemnn sums for the integrl)
3 Numericl Integrtion Definition. b f(x) dx = limit of these Riemnn sums s n nd mx width Theorem (Importnt Theorem). This limit mkes sense if f is continuous on the finite closed intervl [, b] (including end points). Proof of this is too hrd for us. 5.6 Nottion. We cn use Sigm nottion for sums to mke the formule look shorter. For numbers u, u 2,..., u n 5.7 Exmples. u i mens u + u u n i= 5 j 2 = = = 55 j= 25 k= 2k 2 + k = (2 + ) + ( ) + + (2(25) ) 5.8 Remrk. Usul choice for Riemnn sums: ll n subintervls eqully wide. Common width is then h = b n Then x j = + jh for 0 j n Riemnn sum becomes f(x j)(x j x j ) = j= f(x j)h = h f(x j) j= j= Limit of these is the integrl. Limit hrd to find directly s rule, but computer cn find the sum for lrge n. 5.9 Trpezoidl Rule. The trpezoidl rule is technique for finding definite integrls b f(x) dx numericlly. It is one step more clever thn using Riemnn sums. In Riemnn sums, wht we essentilly do is pproximte the grph y = f(x) by step grph nd integrte the step grph. In the Trpezoidl rule, we pproximte y = f(x) by continuous grph mde up of bits of lines.
4 Mthemtics S3 (Timoney) Use n equl divisions. h = b n. x i = + ih. Let y i = f(x i ) (for 0 i n). The trpezoidl rule formul cn be written b ( f(x) dx = h 2 y 0 + y + y 2 + y n + ) 2 y n. Proof (of this formul) uses re of ith trpezoid = h 2 (y i + y i ) 5.0 Exmple. Find 3 e x2 dx pproximtely using the Trpezoidl rule with n = 0. Solution: i x i y i Weight Weight y i / /2 Sum Sum times h is Remrk. A nturl question to sk t this point is: how ccurte is the Trpezoidl rule?
5 Numericl Integrtion 5 Theoreticlly we know tht s n, the trpezoidl rule pproximtion b f(x) dx, but tht does not help us to know how close we re to the limit if we use n = 00 or n = 000. The following theorem gives worst cse scenrio. 5.2 Theorem. Let T n denote the result of using the trpezoidl rule formul with n steps to pproximte b f(x) dx. Then b f(x) dx T n b 2 h2 M 2 where M 2 is the lrgest vlue of f (2) (x) = f (x) for x b. Note: Cn rewrite this in terms of n using h = (b )/n. We won t prove this, or sy nything bout how to show it is true. 5.3 Exmple. (i) In 3 ex2 dx with n = 0 how fr off cn the trpezoidl rule be? Solution: For this we need to know M 2 nd tht involves f (x) with f(x) = e x2. f (x) = 2xe x2 f (x) = 2e x2 + 2x(2x)e x2 = (2 + 4x 2 )e x2 It is firly cler then tht (in this cse) f (x) > 0 lwys (so tht f (x) = f (x)) nd the lrgest vlue for 0 x 3 is M 2 = f (3) = 38e 9 = The theorem tells us then tht the pproximtion T 0 differs from the ctul integrl by t most (or t worst) 3 e x2 dx T 0 b 2 h2 M 2 = 3 2 h2 (30797). The vlue of h is h = b n (0.04)30797 = = 3 0 = 0.2 nd so the error could be s lrge s (ii) How lrge should we chose n so tht the trpezoidl rule pproximtion T n to the sme integrl is certinly within 0.5 of the right vlue? Solution: It will certinly be enough to choose n so tht b 2 h2 M 2 = 3 ( ) 2 3 M 2 < n Rerrnging this, it sys we re sfe if n > So n = 64 will certinly do. 6 (4)M 2/0.5 = = 640 3
6 Mthemtics S3 (Timoney) 5.4 Simpsons Rule. Simpsons Rule is the next most sophisticted method fter the trpezoidl rule. With Riemnn sums we used pproximtion by step grphs (bits of constnt grphs one fter the other), with the trpezoidl rule we used bits of stright lines, nd now we use bits of qudrtic grphs y = x 2 + bx + c. The first problem is tht, while 2 points determine line, we need 3 points to pin down qudrtic grph. Then we lso need formul ( for the re under qudrtic grph (or the integrl of it) nlogous to the formul h 2 y 0 + ) 2 y we used for the re of trpezoid. 5.5 Lemm. If we hve 3 points in the plne with different x-coordintes, then there is exctly one qudrtic grph pssing through them. Proof. (just n ide of how it works) If the points re (x 0, y 0 ), (x, y ) nd (x 2, y 2 ) then one wy to find the qudrtic is to write it down vi the Lgrnge interpoltion formul q(x) = (x x )(x x 2 ) (x 0 x )(x 0 x 2 ) y 0 + (x x 2)(x x 0 ) (x x 2 )(x x 0 ) y + (x x 0)(x x ) (x 2 x 0 )(x 2 x ) y 2 Another pproch is to strt with generl q(x) = x 2 +bx+c nd use the 3 equtions q(x 0 ) = y 0, q(x ) = y, q(x 2 ) = y 2 to find, b, c. 5.6 Lemm. For 3 points eqully spced horizontlly (x 0, y 0 ) = (x h, y 0 ), (x, y ) nd (x 2, y 2 ) = (x + h, y 2 ) nd y = q(x) the qudrtic grph through the 3 points x +h ( q(x) dx = h 3 y y + ) 3 y 2 x h Proof. We ll skip it. You cn find it in Anton. It is just bit messy, not relly complicted. It turns out to mke life lot esier if you ssume x = 0 nd this you cn ssume by shifting the y-xis. 5.7 Simpsons Rule. Ide: For b f(x) dx, choose n even nd preferbly lrge. Divide the intervl [, b] into n equl sections ech of width h = b n, division points x i = + ih (0 i n), corresponding vlues y i = f(x i ). Pir off intervls nd pproximte y = f(x) on ech pir by qudrtic grph. b f(x) dx = sum of integrls of these qudrtics. The formul we get out of this is b ( f(x) dx = h 3 y y y y y n y n + ) 3 y n 5.8 Exmple. Find Solution: 3 e x2 dx pproximtely using Simpsons rule with n = 0.
7 Numericl Integrtion 7 i x i y i Weight Weight y i / / / / / / / / / / /3 Sum Sum times h is (nd this is the pproximte vlue for the integrl given by Simpson s rule). 5.9 Remrk. Now we sk bout the ccurcy of this method. Is it ny better fter the somewht greter compliction? 5.20 Theorem. Let S n denote the result of using Simpsons rule formul with n steps to pproximte b f(x) dx. Then b f(x) dx S n b 80 h4 M 4 where M 4 is the lrgest vlue of f (4) (x) for x b. Note: Cn rewrite this in terms of n using h = (b )/n. 5.2 Exmple. (i) In 3 ex2 dx with n = 0 how fr off cn Simpsons rule be? Solution: (detils not ll here) You cn find tht for f(x) = e x2, the fourth derivtive is f (4) (x) = (6x x 2 + 2)e x2 nd M 4 = 740e 9 = The theorem tells us then tht the pproximtion S 0 differs from the ctul integrl by t most (or t worst) 3 e x2 dx S 0 b 80 h4 M 4 = 3 80 h4 ( ) The vlue of h is h = b = 3 n 0 90 (0.2)4 ( ) = = 0.2 nd so the error could be s lrge s (ii) How lrge should we chose n so tht Simpsons rule pproximtion S n to the bove integrl is certinly within 0.5 of the right vlue? Solution: It will certinly be enough to choose n so tht b 80 h4 M 4 = 3 ( ) 4 3 M 4 < n
8 Mthemtics S3 (Timoney) Rerrnging this, sys we re sfe if n > ( ) /4 90 (6)M 4/0.5 = So n = 48 will certinly do. (Note how much smller this is thn 64 (needed for the trpezoidl rule in the sme sitution.) 5.22 Remrk. The methods we hve discussed were bout finding definite integrls numericlly. We will look lter t methods for finding ntiderivtives in firly systemtic wy. By the fundmentl theorem, if we cn find ntiderivtives we cn find definite integrls b f(x) dx nlyticlly (tht mens we cn come up with n exct formul for the vlue s opposed to numericl pproximte vlue). Our exmple 3 ex2 dx is one tht we will not ever be ble to do nlyticlly Remrk. In the trpezoidl rule, one rrely uses Theorem 5.2 to gurntee the desired ccurcy of the estimte T n. A more prgmtic pproch is to work with T 2, T 2 2 = T 4, etc until we get to vlue of T 2 k which hs stbilised nd where 2 k is resonbly big. There is wy to void mking the sme clcultions over nd over gin nd still to follow this strtegy. The point is tht if we did this for the exmple 3 f(x) dx (sy with the f(x) = ex2 bove), we would be clculting T = 3 2 ( 2 f() + ) 2 f(3) T 2 = 3 2 ( 2 2 f() + f(2) + ) 2 f(3) T 4 = 3 2 ( 4 2 f() + f(.5) + f(2) + f(2.5) + ) 2 f(3) Since we hve to reuse f() nd f(3) ech time, we might be tempted to remember the nswers. Then we need f(2) every time fter T 2 nd so we might like to keep record of the nswer to tht, nd so on. This involves lot of recording (uses up computer memory if we do it on computer) nd there is wy to void so much storge. It turns out tht T 2 = 2 T f(2) T 4 = 2 T (f(.5) + f(2.5)) 4 Thus we cn clculte T 4 using only T 2 nd vlues of f(x) tht were not needed for T 2. This bit is not in the book by Anton, I think.
9 Numericl Integrtion 9 In generl, T 2n = 2 T n + b 2n f i= ( + i ( )) b nd we cn keep finding T, T 2, T 4 = T 2 2, T 8 = T 2 3,... without ny mssive storge requirement (nd without reclculting ny vlues of f(x) we hd to clculte previously). This gives n efficient prgmtic strtegy: Compute T, T 2, T 4 = T 2 2, T 8,... until we get to T n with some resonbly lrge n nd the vlue of T n is roughly equl to the vlue of T n/2. The chnces re tht both re then close to the true vlue of the integrl b f(x) dx. This is n experimentl pproch nd not quite method tht is gurnteed to be 00% right Remrk. As for the trpezoidl rule, one rrely uses Theorem 5.20 to gurntee the desired ccurcy of the estimte S n. A more prgmtic pproch is to work with S 2, S 2 2 = S 4, etc until we get to vlue of S 2 k which hs stbilised nd where 2 k is resonbly big. It turns out tht we cn mke use of the erlier efficient method of working out T, T 2, T 2 2 = T 4, etc together with the formul 2 2n This formul is esy to check. For exmple S 2n = 4 3 T 2n 3 T n T = b ( 2 f() + ) 2 f(b) T 2 = b ( ( 2 2 f() + f + b ) + 2 ) 2 f(b) S 2 = b ( 2 3 f() + 4 ( 3 f + b ) + 3 ) 2 f(b) 4 3 T 2 3 T = b ( 2 2 = S 2 3 f() f ( 3 f() + 3 f(b) 5.25 Exmple. Try out strtegy for our exmple 3 ex2 dx. 2 This bit is not in the book by Anton, I think. ( + b 2 )) ) f(b)
10 Mthemtics S3 (Timoney) Appendix 5A.26 The Indefinite Integrl. We will explin the ide of n indefinite integrl by first giving n exmple. Every time you differentite something you know the ntiderivtive for the result. For exmple d dx (2x4 + 2x 3 7x 2 + x 7) = 8x 3 + 6x 2 4x + nd so if we hppen to wnt to know something tht hs derivtive 8x 3 + 6x 2 4x + we know tht 2x 4 + 2x 3 7x 2 + x 7 is such function. We cll it n n ntiderivtive of 8x 3 +6x 2 4x+ something which when differentited gives us 8x 3 + 6x 2 4x +. Now wht bout other possible ntiderivtives for the sme thing? We cn spot quite esily tht the constnt 7 is not very importnt. Any constnt there insted of 7 would still hve derivtive 0 nd so we relise tht nything of the form 2x 4 + 2x 3 7x 2 + x + C is lso n ntiderivtive for 8x 3 + 6x 2 4x +. In fct this is ll becuse if two functions f(x) nd g(x) hve the sme derivtive, in this cse if f (x) = g (x) = 8x 3 + 6x 2 4x + then d (f(x) g(x)) = 0 dx for ll x. The only functions tht re defined on n intervl nd hve derivtive 0 re constnt functions. So f(x) g(x) = C = constnt nd so f(x) = g(x) + C. This mens tht if we find one ntiderivtive (in our exmple 2x 4 + 2x 3 7x 2 + x 7) for the given function (in our exmple 8x 3 + 6x 2 4x + ) then the most generl one is of the form 2x 4 + 2x 3 7x 2 + x 7 + C. Since C 7 is nother constnt we cn write 8x 3 + 6x 2 4x + dx = 2x 4 + 2x 3 7x 2 + x + C. We cn go bout things bit more systemticlly, by strting with the simplest rules for differentition nd turning them into rules for finding ntiderivtives. (i) We know d dx xn = nx n nd so nx n dx = x n + C. For exmple, 4x 3 dx = x 4 + C. We cn use little ingenuity to find tht n ntiderivtive for x n is n xn (s long s n 0) nd we cn tidy this up to get the rule x n dx = n + xn+ + C (n ) It is perhps interesting to see tht we cn t immeditely write down x dx = /x dx. In fct the ntiderivtive for /x involves the nturl logrithm function ln nd is therefore much more complicted thing tht /x. We will leve this for lter.
11 Numericl Integrtion A detil we should mention, is tht when n < 0, x n is not well-behved t x = 0. So the domin is not n intervl but two intervls x < 0 nd x > 0. So, techniclly, the +C is not dequte in these cses to describe the most generl ntiderivtive. We could switch from one vlue of C to nother s we pss x = 0 nd still hve vlid ntiderivtive (when n < 0). However, people rrely go into this nd probbly you lmost never should encounter this in prctice. Becuse things blow up t x = 0 there should not relly be prcticl problem where x > 0 nd x < 0 re both vlid (when deling with x n nd n < 0). (ii) We know the derivtive of sum is the sum of the derivtives d du (u + v) = dx we cn see tht n ntiderivtive of sum is the sum of the ntiderivtives. Similrly for constnt multiples. + dv dx dx nd so Using these rules, we cn integrte ll polynomils. For exmple ( ) ( ) ( ) ( ) 5x 3 x 2 + 3x + 2 dx = 5 4 x4 3 x x2 + 2 x + C = 5 4 x4 3 x x2 + 2x + C (iii) From the rules for differentiting trigonometric functions d dx sin x = cos x, d dx cos x = sin x, d dx tn x = sec2 x we cn write down rules for integrting some cos x dx = sin x + C, sin x dx = cos x + C, sec 2 x dx = tn x + C With little guesswork we cn figure out some relted integrls like cos 3x dx We might think of sin 3x s possible ntiderivtive but d sin 3x = 3 cos 3x dx is 3 times wht we wnt. Since 3 is constnt, we cn divide cross by it nd we get cos 3x dx = sin 3x + C 3 However, there is no esy wy to do sec x dx (we will see wht the nswer to this is bit lter). We cn write sec x dx = cos x dx
12 Mthemtics S3 (Timoney) nd we know how to integrte cos x but tht does not help. The is no good quotient rule for ntiderivtives. Unlike differentition, where we cn lern smll number of rules tht re enough to differentite lmost ny function we cn esily write down, there re esylooking functions where ntiderivtives re quite hrd to find. We hve seen /x dx, more recently sec x dx nd the exmple cos(x 2 ) dx is one tht is essentilly impossible. It is not tht there is no nswer. There is n ntiderivtive but it is known tht there is no wy to write finite formul for the ntiderivtive of cos(x 2 ) using the fmilir functions (powers, roots, frctions, trig functions, ln, e x ). 5A.27 Exmple. There re very few exmples which cn be worked out directly from the limit of Riemnn sums definition. This is one exmple 4 2 x + 2 dx Solution: Tke n lrge, n eqully wide subintervls of [, b] = [2, 4] with widths h = (b )/n = (4 2)/n = 2/n. This gives subdivision points x i = + ih = 2 + 2i for i = 0,, 2,..., n. To n mke life esy for ourselves we tke x i = x i = 2 + (2i)/n. The Riemnn sum is then ( ) 2i 2 f(x i )(x i x i ) = n + 2 n i= i= ( ) = 2 2i n n + 2 i= i= ( ) ( ) = 4 i n 2 n i= i= = 4 ( ) n(n + ) + 2 n 2 2 n (2n) ( = 2 + ) + 4 n 6 s n Here we used the fct tht i = n i= turns out to be n(n+) 2. This cn be discovered by writing the sum down bckwrds nd dding the two formule verticlly: s = n s = n + n + n s = (n + ) + (n + ) + (n + ) + + (n + ) = n(n + )
13 Numericl Integrtion 3 In most exmples, we will not be ble to write the Riemnn sum s short formul in the wy we did here nd so we would not find the limit. 5A.28 Remrk. There is n mzing connection between definite integrls nd differentition, which we will now stte. It comes bout by considering not just one definite integrl b f(x) dx but whole infinite number of them. Not just 2 f(x) dx but x f(t) dt for 0 x A.29 Theorem (Fundmentl Theorem of Integrl Clculus). Assume tht y = f(x) is continuous for x b. Consider A(x) = x f(t) dt for x b. (A(x) is new function, built from f nd definite integrtion.) Then A(x) is n ntiderivtive for f (tht is A (x) = f(x) for x b). In summry: ( d x ) f(t) dt = f(x) ( x b, if f continuous) dx This is one prt of the Fundmentl theorem, or one wy to stte it. 5A.30 Corollry. There is n ntiderivtive for every continuous function f. 5A.3 Exmple. Find d ( x ( ) ) t 3 cos dt dx t 8 + Solution: By the theorem ( d x ( ) ) t 3 x 3 cos dt = cos dx t 8 + x 8 + 5A.32 Theorem (Other prt of fundmentl theorem). Assume tht y = f(x) is continuous for x b nd suppose g(x) is n ntiderivtive for f(x) (tht is g (x) = f(x) for x b). Then b f(x) dx = g(b) g()= [g(x)] b x=
14 Mthemtics S3 (Timoney) 5A.33 Exmple. Find 4 2 2x + dx Solution: Antiderivtive g(x) = x 2 + x (since g (x) = 2x + ) nd so 4 2 2x + dx = [g(x)] 4 x=2 = [x 2 + x] 4 x=2 = ( ) ( ) = 4 Proof. (of A (x) = f(x) prt of Fund. Thm.) Use first principles x+h x A (x) = lim h 0 A(x + h) A(x) h f(t) dt = f(x)h for h smll. Thus limit is f(x). = lim h 0 Proof. (of b f(x) dx = g(b) g() prt) We know A(x) = x f(t) dt is n ntiderivtive. Hence x+h x f(t) dt h d (A(x) g(x)) = f(x) f(x) = 0. dx Thus A(x) g(x) = c = const. Use x = to find constnt: A() g() = c. But A() = f(t) dt = 0. Thus g() = c nd A(x) g(x) = c = g() for ll x [, b]. Use this for x = b to get A(b) g(b) = g(), A(b) = g(b) g(), tht is b f(t) dt = g(b) g(). 5A.34 Correction. Correction to (or refinement of) grphicl interprettion of definite integrls. b f(x) dx = re under grph is good when f(x) 0 lwys. When f(x) < 0, tht prt of the re is counted with minus sign. Consider negtive terms in Riemnn sum f(x i )(x i x i ) if f(x i ) < 0. i=
15 Numericl Integrtion 5 b f(x) dx = (sum of res where f(x) > 0) (sum where f(x) < 0) 5A.35 Nottion. There is one more thing we hve skimmed over. When > b, there is convention tht llows us to write b f(x) dx (for exmple 2 x dx) even though the limits re upside down. By convention, the mening for this is b f(x) dx = b f(x) dx if > b 3 x2 + This convention comes up in the proof of the fundmentl theorem (the prt where A(x + h) A(x) = x+h f(t) dt, where we hve to be ble to del with h < 0 s well s h > 0). x In fct the fundmentl theorem in the form ( d x ) f(t) dt = f(x) dx is lso vlid for x < s long s f is continuous on n intervl tht includes both nd x. (The importnt thing is to hve ll points in between s well s nd x.) The convention lso fits with the other form of the fundmentl theorem. If g (x) = f(x), then remins true when > b. b 5A.36 Exmples. (i) Find d dx Solution: ( d 0 dx x f(x) dx = [g(x)] b x= = g(b) g() ( 0 x ) t t 6 + dt ) t t 6 + dt = d ( x dx 0 using the bove convention nd the Fundmentl theorem. ) t t 6 + dt = x x 6 +
16 Mthemtics S3 (Timoney) ( ) (ii) Find d x 3 +x cos(t 2 + 4) dt dx Solution: Tke y = x 3 +x cos(t 2 + 4) dt = Chin rule dy dx = dy du nd by the Fundmentl theorem u cos(t 2 + 4) dt where u = x 3 + x. By the du dx = dy du (3x2 + ) dy du = cos(u2 + 4) = cos((x 3 + x) 2 + 4) Putting these together, we get ( ) d x 3 +x cos(t 2 + 4) dt = (3x 2 + ) cos((x 3 + x) 2 + 4) dx ( ) Notice tht the nswer would hve been the sme for d x 3 +x cos(t 2 + 4) dt, s the dx 25 vlue of the lower limit of the integrl does not come into the nswer (s long s it is constnt nd s long s the integrnd is continuous wherever we go). Richrd M. Timoney Mrch 7, 2007
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### Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as
Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph
### Definition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim
Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)
### NUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.
NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with
### Numerical Integration
Chpter 5 Numericl Integrtion Numericl integrtion is the study of how the numericl vlue of n integrl cn be found. Methods of function pproximtion discussed in Chpter??, i.e., function pproximtion vi the
### How can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function?
Mth 125 Summry Here re some thoughts I ws hving while considering wht to put on the first midterm. The core of your studying should be the ssigned homework problems: mke sure you relly understnd those
### SYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus
SYDE 112, LECTURES & 4: The Fundmentl Theorem of Clculus So fr we hve introduced two new concepts in this course: ntidifferentition nd Riemnn sums. It turns out tht these quntities re relted, but it is
### a < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1
Mth 33 Volume Stewrt 5.2 Geometry of integrls. In this section, we will lern how to compute volumes using integrls defined by slice nlysis. First, we recll from Clculus I how to compute res. Given the
### Z b. f(x)dx. Yet in the above two cases we know what f(x) is. Sometimes, engineers want to calculate an area by computing I, but...
Chpter 7 Numericl Methods 7. Introduction In mny cses the integrl f(x)dx cn be found by finding function F (x) such tht F 0 (x) =f(x), nd using f(x)dx = F (b) F () which is known s the nlyticl (exct) solution.
### Riemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties
### 5: The Definite Integral
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### Exam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1
Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH-115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixed-point itertion to converge when solving the eqution
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Contents 7.1 Integrtion by Prts................................... 2 7.2 Trigonometric Integrls.................................. 8 7.2.1 Evluting sin m x cos n (x)......................... 8 7.2.2 Evluting
### p(t) dt + i 1 re it ireit dt =
Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)
### Riemann Sums and Riemann Integrals
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# 4.5: Hand Calculator Use
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## Scientific Notation with a Hand Calculator
To enter numbers in scientific notation into a hand calculator, there is usually a button marked “E” or “EE” used to enter the correct power of ten. For example, to enter the mass of a proton in grams (1.67 x 10-24grams) into a hand calculator, I would enter the following keystrokes:
The [+/-] keystroke changes the sign of the power (24) into a -24. Some calculators allow the use of the subtraction key [-] to do this, but I prefer the “change sign” [+/-] key because its more consistent with the use of that key in other contexts.
If I wanted to enter a negative number in scientific notation into a hand calculator, I would have to be careful how I used the [+/-] key, lest I change the sign of the power and not the significant digit value. Pay attention to this example:
Number to be entered: -3.221 x 10-15:
The first [+/-] keystroke changes the entry from 3.221 to -3.221; the second [+/-] keystroke changes the power from 15 to -15.
## Metric and Scientific Notation Display Modes
Displaying metric and scientific notation on a hand calculator is a different matter. It involves changing the display option from the normal “fixed” decimal point mode to the “scientific” or “engineering” mode. Your calculator manual will tell you how to set each display mode.
These display modes tell the calculator how to represent any number on the numerical readout. The actual value of the number is not affected in any way by the choice of display modes—only how the number appears to the calculator user. Likewise, the procedure for entering numbers into the calculator does not change with different display modes either. Powers of ten are usually represented by a pair of digits in the upper-right hand corner of the display, and are visible only in the “scientific” and “engineering” modes.
The difference between “scientific” and “engineering” display modes is the difference between scientific and metric notation. In “scientific” mode, the power-of-ten display is set so that the main number on the display is always a value between 1 and 10 (or -1 and -10 for negative numbers). In “engineering” mode, the powers-of-ten are set to display in multiples of 3, to represent the major metric prefixes. All the user has to do is memorize a few prefix/power combinations, and his or her calculator will be “speaking” metric!
## Review
• Use the [EE] key to enter powers of ten.
• Use “scientific” or “engineering” to display powers of ten, in scientific or metric notation, respectively.
This page titled 4.5: Hand Calculator Use is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Tony R. Kuphaldt (All About Circuits) via source content that was edited to the style and standards of the LibreTexts platform.
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# Question #86e4f
Apr 6, 2017
Volume of each box is ${4}^{3}$cubic inches $\times 12$ boxes or
$64 \times 12 = 768$ cubic inches.
#### Explanation:
The volume of each box is the product of the lengths of its three sides, or:
$\left(4 \times 4 \times 4\right)$cubic inches $= 64$ cubic inches.
Twelve of these boxes completely fill the carton, so the volume of the carton is the total volume of twelve boxes:
64 cubic inches$\times 12 = 768$ cubic inches.
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https://www.stats4stem.org/r-matrices-and-matrix-computations
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R: Matrices and Matrix Computations
I. Constructing Matrices
Constructing a matrix can be done with the matrix() function as shown below. The default is to fill in the matrix by column. However, one can change this by setting byrow = TRUE which will fill in the matrix by row. To define the number of rows and columns, use the nrow and ncol arguments. You can define both nrow and ncol, but as shown below, if only one is given, R will infer what the other one is by default. Lets begin by making two matrices as shown below:
$$a = \begin{bmatrix} 1 & 4 \\[0.3em] 2 & 4 \\[0.3em] 3 &6 \end{bmatrix}$$
> # Construct a 3 by 2 marix - fill in by column
> a = matrix(c(1,2,3,4,5,6),
+ nrow = 3)
> a
[,1] [,2]
[1,] 1 4
[2,] 2 5
[3,] 3 6
>
> # Construct a 3 by 2 marix - fill in by row
> b = matrix(c(1,2,3,4,5,6),
+ nrow = 3,
+ byrow = TRUE)
> b
[,1] [,2]
[1,] 1 2
[2,] 3 4
[3,] 5 6
>
> # Construct a 2 by 3 matrix - fill in by row
> c = matrix(c(1,2,3,4,5,6),
+ nrow = 2,
+ byrow = TRUE)
> c
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 4 5 6
Adding and subtracting matrices can be done as shown below.
$$m = \begin{bmatrix} 0 & 1 & 2 \\[0.3em] 9 & 8 & 7 \\[0.3em] \end{bmatrix}$$ $$n = \begin{bmatrix} 6 & 5 & 4 \\[0.3em] 3 & 4 & 5 \\[0.3em] \end{bmatrix}$$
> # create two matrices with the same dimensions
> m = matrix(c(0 ,1 ,2, 9, 8, 7),nrow = 2, byrow = 2)
> m
[,1] [,2] [,3]
[1,] 0 1 2
[2,] 9 8 7
> n = matrix(c(6, 5, 4, 3, 4, 5),nrow = 2, byrow = 2)
> n
[,1] [,2] [,3]
[1,] 6 5 4
[2,] 3 4 5
>
> m + n
[,1] [,2] [,3]
[1,] 6 6 6
[2,] 12 12 12
>
> # subtract two matrices
> m - n
[,1] [,2] [,3]
[1,] -6 -4 -2
[2,] 6 4 2
III. Multiplying matrices Matrices
Multiplying matrices can be done using the following code as shown below: %*%
> c = matrix(c(1, 0, -2, 0, 3, -1), nrow = 2, byrow = 2)
> c
[,1] [,2] [,3]
[1,] 1 0 -2
[2,] 0 3 -1
> d = matrix(c(0, 3, -2, -1, 0, 4), nrow = 3, byrow = 2)
> d
[,1] [,2]
[1,] 0 3
[2,] -2 -1
[3,] 0 4
>
> c%*%d
[,1] [,2]
[1,] 0 -5
[2,] -6 -7
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# 1 QM Winter 2009 Winter 2009 Exercises for Test 1 Exercises for Test 1 Chapter s 1- 4 Del Balso.
## Presentation on theme: "1 QM Winter 2009 Winter 2009 Exercises for Test 1 Exercises for Test 1 Chapter s 1- 4 Del Balso."— Presentation transcript:
1 QM Winter 2009 Winter 2009 Exercises for Test 1 Exercises for Test 1 Chapter s 1- 4 Del Balso
2 Probability Sample Any sample that is chosen by chance. Includes Simple Random Sample (SRS) and Stratified Samples. 1. Simple Random Sample Gives each individual in the population an equal chance of being chosen.
3 Use table of random digits (handout) or computer program to choose individuals or samples. Random: every individual or sample has an equal chance of being picked. This ensures the sample is representative of the population. Therefore, we can generalize the results back to the population.
4 Choose a Simple Random Sample (SRS) Step 1. Assign a number to every individual in the population (or to every sample). Assign numbers beginning with 0 and keep going until everyone has a number. Go down columns of lists. Step 2. Use random digits to select the numbers and choose those people.
5 The first two lines of the Random Digit Table p 550: Line 10119223950340575628713 96409125314254482853 10273676471509940001927 27754426488242536290
6 Ex., Choose a Simple Random Sample of 4 from the following past players of the Canadiens. (number down beginning with 0). Begin at line 101. Ribeiro BriseboisMarkov SouraySundstromHainsey RyderRivetHossa ZednikDackellGratton KoivuBouillonLangdon Perreault KilgerTheodore BulisQuintalDwyer Dagenais WardHiggins JuneauBeginPlekanec GaronKomisarek
7 The sample of 4 is: Komisarek (19); Hossa (22); Perreault (05); Dackell (13).
8 Exercises 1. Choose another simple random sample of 4 from the Canadiens hockey players starting at line 103.
9 2.You want to survey people about the recent election in the greater Montreal area. Let’s say there are 25 electoral districts in the greater Montreal area. Choose an SRS of 5 districts using line 110 of the Random Digit table. Explain each step.
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# LeetCode 803. 打砖块
## 题目描述
网格开始为：
[[1,0,0,0]，
[1,1,1,0]]
[[1,0,0,0]
[0,1,1,0]]
[[1,0,0,0],
[0,0,0,0]]
网格开始为：
[[1,0,0,0],
[1,1,0,0]]
[[1,0,0,0],
[1,0,0,0]]
[[1,0,0,0],
[1,0,0,0]]
[[1,0,0,0],
[0,0,0,0]]
## 题解
def getC(t):
c[t] = c[t] if t == f[t] else getC(f[t])
return c[t]
def getC(t):
return c[getF(t)]
## 代码
class Solution:
def hitBricks(self, grid: List[List[int]], hits: List[List[int]]) -> List[int]:
if len(grid) == 0:
return [0] * len(hits)
# import time
# start = time.time()
n = len(grid)
m = len(grid[0])
ngrid = [[grid[x][y] for y in range(m)] for x in range(n)]
for h in hits:
x, y = h
ngrid[x][y] = 0
# print(time.time() - start)
# for x in range(n):
# for y in range(m):
# print(ngrid[x][y], end=" ")
# print()
f = [i for i in range(m * n + 1)]
c = [1 for i in range(m * n + 1)]
def getIdx(x, y):
return x * m + y + 1
def getF(t):
f[t] = f[t] if t == f[t] else getF(f[t])
return f[t]
def getC(t):
# c[t] = c[t] if t == f[t] else getC(f[t])
return c[getF(t)]
def connect(t1, t2):
root1 = getF(t1)
root2 = getF(t2)
if root1 != root2:
c[root2] = c[root1] + c[root2]
c[root1] = c[root2]
f[root1] = root2
for x in range(n):
for y in range(m):
if ngrid[x][y] == 1:
t = getIdx(x, y)
for xx, yy in [(x+1,y), (x,y+1)]:
if 0 <= xx and xx < n and 0 <= yy and yy < m and ngrid[xx][yy] == 1:
tt = getIdx(xx, yy)
connect(t, tt)
# print("connect", t,x,y, tt,xx,yy)
if x == 0:
connect(0, t)
# print(time.time() - start)
# for x in range(n):
# for y in range(m):
# print(getF(getIdx(x, y)), end=" ")
# print()
# for x in range(n):
# for y in range(m):
# print(getC(getIdx(x, y)), end=" ")
# print()
res = [0] * len(hits)
for i, hit in enumerate(reversed(hits)):
x, y = hit
if grid[x][y] == 1:
ngrid[x][y] = 1
beforeNum = getC(0)
# print("insert",x,y)
t = getIdx(x, y)
for xx, yy in [(x-1,y),(x+1,y),(x,y-1),(x,y+1)]:
if 0 <= xx and xx < n and 0 <= yy and yy < m and ngrid[xx][yy] == 1:
tt = getIdx(xx, yy)
connect(t, tt)
# print("connect2", t,x,y, tt,xx,yy)
if x == 0:
connect(0, t)
nowNum = getC(0)
# print("before", beforeNum, "now", nowNum)
res[i] = max(nowNum - beforeNum - 1, 0)
# print(time.time() - start)
res.reverse()
return res
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# Kinetic friction of a wood block
So the problem states:
A 1.4 kg wood block is launched up a wooden ramp that is inclined at a 22 deg angle. The block's initial speed is 14 m/s. Use Uk = 0.20 for the coefficient of kinetic friction for wood on wood.
U = mu...
(a) What vertical height does the block reach above its starting point?
(b) What speed does it have when it slides back down to its starting point?
Okay so this is how I started it...
I drew a force diagram of the object on an inclined ramp at 22 deg. Set the x axis parallel with the object.. so basically at the same angle. 3 forces.. normal, weight, and kinetic friction.
E = summation
M = mass
A = accel
N = normal
Mg = mass * gravity
fk = kinetic friction
E(Fy) = MAy = 0
N - Mg*cos(22) = 0
N = 12.73
E(Fx) = MAx
-fk - Mg*sin(22) = MAx
-Uk*N - Mg*sin(22) = MAx
-0.20*12.73 - 5.14 = MAx
Ax = -7.686 m/s^2
Vf = Vi + a*t
t = 1.95s
Plug that into the x kinematic equation...
xf = xi + vi*t + (1/2)*a*t^2
= 0 + 15*sin(22)*1.95 + (1/2)*(-7.68)(1.95^2)
= 12.507 m
The answer is wrong.. if someone could disect my work and figure out what I'm doing wrong that'd be awesome, thanks!
hage567
Homework Helper
So the problem states:
A 1.4 kg wood block is launched up a wooden ramp that is inclined at a 22 deg angle. The block's initial speed is 14 m/s. Use Uk = 0.20 for the coefficient of kinetic friction for wood on wood.
U = mu...
(a) What vertical height does the block reach above its starting point?
(b) What speed does it have when it slides back down to its starting point?
Okay so this is how I started it...
I drew a force diagram of the object on an inclined ramp at 22 deg. Set the x axis parallel with the object.. so basically at the same angle. 3 forces.. normal, weight, and kinetic friction.
E = summation
M = mass
A = accel
N = normal
Mg = mass * gravity
fk = kinetic friction
E(Fy) = MAy = 0
N - Mg*cos(22) = 0
N = 12.73
E(Fx) = MAx
-fk - Mg*sin(22) = MAx
-Uk*N - Mg*sin(22) = MAx
-0.20*12.73 - 5.14 = MAx
Ax = -7.686 m/s^2
Vf = Vi + a*t
t = 1.95s
Plug that into the x kinematic equation...
xf = xi + vi*t + (1/2)*a*t^2
= 0 + 15*sin(22)*1.95 + (1/2)*(-7.68)(1.95^2)
= 12.507 m
The answer is wrong.. if someone could disect my work and figure out what I'm doing wrong that'd be awesome, thanks!
You didn't actually divide by the M in the final step to get your acceleration.
Can you elaborate on how you found t? Using your value for a, I don't get the same answer that you did. Maybe you made a typo?
Last edited:
Doc Al
Mentor
In addition to what hage567 points out:
Plug that into the x kinematic equation...
xf = xi + vi*t + (1/2)*a*t^2
= 0 + 15*sin(22)*1.95 + (1/2)*(-7.68)(1.95^2)
= 12.507 m
Where do the 15 and the sin(22) come from? The initial speed is 14 m/s parallel to the incline. Once you find the distance up the incline using this formula, you'll have to convert it to height.
hage567
Homework Helper
Plug that into the x kinematic equation...
xf = xi + vi*t + (1/2)*a*t^2
= 0 + 15*sin(22)*1.95 + (1/2)*(-7.68)(1.95^2)
= 12.507 m
Don't forget that the kinematic equations will give you the distance ALONG the incline, not the height that the block was displaced in the vertical direction.
I'm not understanding the 15*sin(22) part. For one thing, the initial velocity is 14 m/s.
EDIT: I'm too slow.
You didn't actually divide by the M in the final step to get your acceleration.
Can you elaborate on how you found t? Using your value for a, I don't get the same answer that you did. Maybe you made a typo?
Ahh good catch. Okay then I get -5.49 m/s^2 for the accel and 2.73 for t. I plugged that into the y formula and get -5.12 which is obviously not right
Okay... so then would just be y = 14*2.73 + (1/2)(-5.49)(2.73^2)?
Actually yeah that's wrong too
I got 17.76 m for the "distance" not sure how to convert to height?
haha nevermind i got it... 17.76*sin(22) = 6.65m for the height thanks guys!
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https://www.spss-tutorials.com/binomial-test/
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SPSS TUTORIALS FULL COURSE BASICS ANOVA REGRESSION FACTOR
Binomial Test – Simple Tutorial
For running a binomial test in SPSS, see SPSS Binomial Test.
A binomial test examines if some
population proportion is likely to be x.
For example, is 50% -a proportion of 0.50- of the entire Dutch population familiar with my brand? We asked a simple random sample of N = 10 people if they are. Only 2 of those -a proportion of 0.2- or 20% know my brand. Does this sample proportion of 0.2 mean that the population proportion is not 0.5? Or is 2 out of 10 a pretty normal outcome if 50% of my population really does know my brand?
The binomial test is the simplest statistical test there is. Understanding how it works is pretty easy and will help you understand other statistical significance tests more easily too. So how does it work?
Binomial Test - Basic Idea
If the population proportion really is 0.5, we can find a sample proportion of 0.2. However, if the population proportion is only 0.1 (only 10% of all Dutch adults know the brand), then we may also find a sample proportion of 0.2. Or 0.9. Or basically any number between 0 and 1. The figure below illustrates the basic problem -I mean challenge- here.
Will the real population proportion please stand up now??
So how can we conclude anything at all about our population based on just a sample? Well, we first make an initial guess about the population proportion which we call the null hypothesis: a population proportion of 0.5 knows my brand.
Given this hypothesis, many sample proportions are possible. However, some outcomes are extremely unlikely or almost impossible. If we do find an outcome that's almost impossible given some hypothesis, then the hypothesis was probably wrong: we conclude that the population proportion wasn't x after all.
So that's how we draw population conclusions based on sample outcomes. Basically all statistical tests follow this line of reasoning. The basic question for now is: what's the probability of finding 2 successes in a sample of 10 if the population proportion is 0.5?
Binomial Test Assumptions
First off, we need to assume independent observations. This basically means that the answer given by any respondent must be independent of the answer given by any other respondent. This assumption (required by almost all statistical tests) has been met by our data.
Binomial Distribution - Formula
If 50% of some population knows my brand and I ask 10 people, then my sample could hold anything between 0 and 10 successes. Each of these 11 possible outcomes and their associated probabilities are an example of a binomial distribution, which is defined as $$P(B = k) = \binom{n}{k} p^k (1 - p)^{n - k}$$ where
• $$n$$ is the number of trials (sample size);
• $$k$$ is the number of successes;
• $$p$$ is the probability of success for a single trial or the (hypothesized) population proportion.
Note that $$\binom{n}{k}$$ is a shorthand for $$\frac{n!}{k!(n - k)!}$$ where $$!$$ indicates a factorial.
For practical purposes, we get our probabilities straight from Google Sheets (it uses the aforementioned formula under the hood but it doesn't bother us with it).
Binomial Distribution - Chart
Right, so we got the probabilities for our 11 possible outcomes (0 through 10 successes) and visualized them below.
If a population proportion is 0.5 and we sample 10 observations, the most likely outcome is 5 successes: P(B = 5) ≈ 0.24. Either 4 or 6 successes are also likely outcomes (P ≈ 0.2 for each).
The probability of finding 2 or fewer successes -like we did- is 0.055. This is our one-sided p-value.
Now, very low or very high numbers of successes are both unlikely outcomes and should both cast doubt on our null hypothesis. We therefore take into account the p-value for the opposite outcome -8 or more successes- which is another 0.055. Like so, we find a 2-sided p-value of 0.11. If we would draw 1,000 samples instead of just 1, then some 11% of those should result in 2(-) or 8(+) successes when the population proportion is 0.5. Our sample outcome should occur in a reasonable percentage of samples. And since 11% is not very unlikely, our sample does not refute our hypothesis that 50% of our population knows our brand.
We ran our example in this simple Google Sheet. It's accessible to anybody so feel free to take a look at it.
Binomial Test - SPSS
Perhaps the easiest way to run a binomial test is in SPSS - for a nice tutorial, try SPSS Binomial Test. The figure below shows the output for our current example. It obviously returns the same p-value of 0.109 as our Google Sheet.
Note that SPSS refers to p as “Exact Sig. (2-tailed)”. Is there a non exact p-value too then? Well, sort of. Let's see how that works.
Binomial Test or Z Test?
Let's take another look at the binomial probability distribution we saw earlier. It kinda resembles a normal distribution. Not convinced? Take a look at the binomial distribution below.
For a sample of N = 100, our binomial distribution is virtually identical to a normal distribution. This is caused by the central limit theorem. A consequence is that -for a larger sample size- a z-test for one proportion (using a standard normal distribution) will yield almost identical p-values as our binomial test (using a binomial distribution).
But why would we prefer a z-test over a binomial test?
• We can always use a 2-sided z-test. However, a binomial test is always 1-sided unless P0 = 0.5.
• A z-test allows us to compute a confidence interval for our sample proportion.
• We can easily estimate statistical power for a z-test but not for a binomial test.
• A z-test is computationally less heavy, especially for larger sample sizes.I suspect that most software actually reports a z-test as if it were a binomial test for larger sample sizes.
So when can we use a z-test instead of a binomial test? A rule of thumb is that P0*n and (1 - P0)*n must both be > 5, where P0 denotes the hypothesized population proportion and n the sample size.
So that's about it regarding the binomial test. I hope you found this tutorial helpful. Thanks for reading!
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• By lydia Okutoyi on July 15th, 2020
This is great
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# FINDING GROSS PAY ASSIGNMENT
Description
LESSON 2-2 FINDING GROSS PAY
INSTRUCTIONS: Gross Pay = Pay per hour times Hours worked
Example: Paid at \$15/hr for 6 hours is \$15 x 6 = \$90 Gross Pay
Exercises
1. Chester Krump worked these hours last week: Monday, 7 hours; Tuesday, 8 hours; Wednesday, 6 hours; Thursday, 7 hours; FridaY,,5 hours. Chester is paid \$12 an hour.
a)How many hours did Chester work last week?
b) What was his gross pay for the week?
2. Rosemary Karzim earns \$9 per hour at her job. The hours she worked during each of the last four weeks are shown at the right. For each week find the gross pay Rosemary earned. Then find her total gross pay for four weeks. Write your answers in the chart.
Week Hours Worked Gross Pay
1 40 \$
2 31 \$
3 37 \$
4 39 \$
Total Gross Pay \$
3. Sandra Milgor has two jobs. Her full-time job pays \$9 per hour. A part-time job pays \$5 an hour. Last week Sandra worked 32 hours at her full-time job and 14 hours at her part-time job. Find the
gross pay she earned at her full-time job.\$
gross pay she earned at her part-time job.\$
total amount she earned from both jobs\$
4. Harrison Cronk earns a weekly salary of \$356.
a)How much would Harrison earn in 4 weeks of work?\$
b)How much would he earn in a 52-week year?\$
LESSON 2-4 FINDING AVERAGE PAY
INSTRUCTIONS: Find simple averages by adding the amounts, then dividing by the number of items added.
Example: Average of 4, 8, and 9 is 4 8 9 = 21, 21/3 = 7 (the average)
Exercises
5. Quinn LeGrow worked at three jobs in three weeks. In the first week, he earned \$310. In the next week, he was paid \$334. In the third week, he was paid \$307.
a) What average pay per week did Quinn earn for these three weeks?
6. The monthly earnings of Cynthia Gragg for May through August were \$1,684, \$1,752, \$1,592, \$1,664.
a)What average amount per month did she earn for these four months?\$
7. Nestor Gorbea worked five days last week and earned these amounts: \$85.63, \$79.66, \$75.65, \$89.28, \$90.58.
a)What average amount per day did Nestor earn during these five days, rounded to the nearest dollar?\$
8. Pauline Morse has worked as a loan officer in a bank for the past three years. During her first year, she earned \$23,400. Her earnings increased to \$26,200 the second year, and to \$29,600 the third year.
a)What were her average annual earnings for the three years she worked for
the bank?\$
Finding Weighted Averages. Example: What is the average rate of pay for workers if 4 workers get \$5.00 per hour, 3 get \$7/hr and 2 get \$10/hour (9 workers).
4 x \$5 = \$20, 3 x \$7 = \$21, 2 x \$11 = \$22; \$20 \$21 \$22 = \$63 / 9 = \$7 per hour average rate of pay
9. Paul Bailey earns extra money by refinishing wood floors. His charge is based on the size and condition of the floor. Last month, he refinished 6 floors for \$80 each, 2 floors for \$68 each, and 1 floor for \$122.
a)What average amount per floor did he earn for the month?\$
10. After working at her new job, Rhonda Bloch found that she had earned the following monthly salaries: \$1,600 monthly for the first 3 months; \$1,720 monthly for the following 2 months; \$1,840 for the sixth month.
a)What average amount per month did Rhonda earn during these six months?\$
11. Four employees of the Micro-Line Company earn \$14 an hour. Another 6 employees earn \$12 an hour, while 4 others are paid \$10 an hour.
a)What is the average pay per hour earned by the employees?\$
Finding an unknown item in a series (see if you can figure these out yourself).
12. By working for 4 weeks at your part-time job you earned these amounts: \$60, \$72, \$53, \$47.
a)What pay will you have to earn in the fifth week to average \$56 a week for five weeks of work?\$
13. The average weekly pay of five employees of the Morris Electrical Supply Company is \$473. The weekly pay of four of the employees is \$385, \$530, \$526, \$495.
a)What is the weekly pay of the other employee? \$
14. On Monday, five employees of the Wexel Cartage Company earned an average gross pay of \$108 for the day. Four of the employees earned these amounts on that day: \$116, \$102, \$119, \$86.
a)How much did the fifth employee earn on Monday?\$
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# Multiplication Chart 100
Discovering multiplication after counting, addition, and subtraction is good. Kids find out arithmetic via a natural progression. This progression of understanding arithmetic is truly the following: counting, addition, subtraction, multiplication, and ultimately division. This assertion results in the question why find out arithmetic in this pattern? Moreover, why learn multiplication soon after counting, addition, and subtraction before department?
## The subsequent information respond to these questions:
1. Youngsters find out counting very first by associating aesthetic things because of their fingers. A tangible example: The amount of apples are there within the basket? Much more abstract illustration is the way outdated have you been?
2. From counting phone numbers, another reasonable phase is addition then subtraction. Addition and subtraction tables are often very useful educating helps for the kids since they are visual tools producing the move from counting much easier.
3. Which ought to be learned following, multiplication or section? Multiplication is shorthand for addition. At this stage, kids have got a organization knowledge of addition. Consequently, multiplication is the next plausible method of arithmetic to find out.
## Review essentials of multiplication. Also, review the fundamentals utilizing a multiplication table.
Let us review a multiplication case in point. By using a Multiplication Table, increase four times 3 and have a solution 12: 4 x 3 = 12. The intersection of row a few and column several of any Multiplication Table is 12; a dozen is the response. For children beginning to learn multiplication, this is certainly simple. They could use addition to fix the trouble thus affirming that multiplication is shorthand for addition. Instance: 4 by 3 = 4 4 4 = 12. It is an excellent summary of the Multiplication Table. The added reward, the Multiplication Table is graphic and mirrors straight back to discovering addition.
## Exactly where can we commence understanding multiplication utilizing the Multiplication Table?
1. Initial, get acquainted with the table.
2. Start with multiplying by a single. Start off at row number 1. Proceed to column number one. The intersection of row a single and column the initial one is the solution: one.
3. Recurring these methods for multiplying by a single. Increase row one particular by columns one by means of 12. The replies are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 respectively.
4. Repeat these steps for multiplying by two. Increase row two by posts a single through 5. The answers are 2, 4, 6, 8, and 10 correspondingly.
5. We will jump ahead of time. Recurring these actions for multiplying by 5. Increase row five by columns one by means of twelve. The replies are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 correspondingly.
6. Now let us improve the quantity of trouble. Replicate these techniques for multiplying by 3. Increase row three by posts 1 by way of a dozen. The replies are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 respectively.
7. Should you be confident with multiplication thus far, consider using a check. Solve the subsequent multiplication problems in your thoughts after which compare your responses towards the Multiplication Table: increase 6 as well as two, flourish nine and about three, grow one and 11, increase a number of and four, and grow six as well as 2. The situation responses are 12, 27, 11, 16, and 14 correspondingly.
If you received a number of out from five issues correct, build your own multiplication checks. Calculate the solutions in your thoughts, and look them while using Multiplication Table.
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• where a is the acceleration, v 0 is the starting velocity, v 1 is the final velocity, and t is the time (acceleration duration or t 1 - t 0). The resulting unit will depend on the units for both time and distance, so if your input was in miles and hours, the acceleration will be in miles/h 2.
• where v is velocity (m/s). Thus, the mass multiplied by the rate of change of the velocity is equal to the net force acting on the body. If the net force is positive, the object will accel- erate. If it is negative, the object will decelerate. If the net force is zero, the object's veloc- ity will remain at a constant level. Example. Consider the expression for final velocity V where U is initial velocity, A is constant acceleration and T is time. Given also that displacement S is given by , show that . Rearrange to make T the subject: Substituting this expression for T into the given expression for S gives: Multiplying both sides by 2A and expanding the brackets ...
• meaning velocity is becoming less positive or more negative. Less positive means slowing down while going up. More negative means speeding up while going down. This acceleration vector is the same on the way u p,a the o and o the w y down! Kinematics Formula Summary (derivations to follow) • v f = v 0 + at • v avg = (v 0 + v f)/2 • Dx = v ...
• Give one example of each of the following situations. (i) Uniformly accelerated motion. (ii) Motion with uniform retardation. (iii) Accelerated motion with uniform magnitude of velocity. (iv)Motion in a direction with acceleration in perpendicular direction. (v) Motion in which v-t graph is a horizontal line parallel to x-axis.
• Instantaneous acceleration: In a velocity-time curve, the instantaneous acceleration is given by the slope of the tangent on the v-t curve at any instant. Positive, negative and zero acceleration. Consider the velocity-time graph shown above.
• In physics, jerk or jolt is the rate at which an object's acceleration changes with respect to time. It is a vector quantity (having both magnitude and direction). Jerk is most commonly denoted by the symbol and expressed in m/s 3 or standard gravities per second (g/s).
• Aug 16, 2010 · One more example: one that is perhaps not as silly as a rocket-powered sled. Imagine a space ship orbiting the Earth. The spaceship has some positive kinetic energy, but it also has some negative potential energy due to its gravitational interaction with the Earth.
• Of these forces, both the normal force and the friction force are coming from the ground, so the sum of those vectors is the correct answer. 19. D Because the lift force is perpendicular to the velocity of the object, lift can have no effect on the magnitude of velocity. The magnitude of velocity is also known as speed, so (D) is correct. 20.
• Acceleration is rate of change of velocity with respect to time. a= dv/dt Or a=(v2 - v1)/(t2 - t1) In general dt or (t2 - t1) is positive. Whenever the change in velocity( dv or (v2 - v1)) is negative for some time interval(t2 - t1), means the acc...
• Velocity and Acceleration. The concepts of velocity and acceleration are linked together, but they are linked incorrectly in many people's minds. Many people think that if an object has a large velocity, it must have a large acceleration - if it has a small velocity, it must have a small acceleration - if its velocity is zero, its acceleration must be zero, too.
• Velocity is a vector quantity—when giving the velocity we must specify the magnitude (the speed) and the direction of travel. For example you might drive 100km/hr (the speed) in a northerly ...
• For example, we could say that the object on the left has a velocity of negative five meters per second, while the object on the right has a velocity of positive five meters per second. And this signifies that the two objects are moving in opposite directions.
• Section 4 Graphing Motion: Distance, Velocity, and Acceleration Physics Words vector: a quantity that has both magnitude and direction. negative acceleration: a decrease in velocity with respect to time. The object can slow down (20 m/s to 10 m/s) or speed up (-20 m/s to -30 m/s). positive acceleration: an increase in velocity with respect to time.
• Give an example where both the velocity and acceleration are negative. 10 11. As a freely falling object speeds up, what is happening to v (m/ s) its acceleration—does it increase, decrease, or stay the same? (a) Ignore air resistance. (b) Consider air resistance. 20 10 12. You travel from point A to point B in a car moving at a constant ...Feb 10, 2018 · Both velocity and acceleration are vector quantities, which have both magnitude and direction. Both the expressions can be positive, negative and zero. Conclusion. The motion of an object can be explained as distance traveled, which can be uniform or non-uniform, depending on the velocity of the object.
• Nov 29, 2020 · Velocity Of Money: The velocity of money is the rate at which money is exchanged from one transaction to another and how much a unit of currency is used in a given period of time. Velocity of ... It means both are vector quantities. If the velocity changes, then from the formula; p = mv; momentum also changes. However, force changes only when the acceleration changes. Even if there is a change in the velocity of an object, but acceleration remains constant, then the force will also remain constant.
• In the case the resultant force F is constant in both magnitude and direction, and parallel to the velocity of the particle, the particle is moving with constant acceleration a along a straight line. The relation between the net force and the acceleration is given by the equation F = ma ( Newton's second law ), and the particle displacement s ...
• Figure 1, acceleration is in the negative direction in the chosen coordinate system , so we say the train is undergoing negative acceleration. If an object in motion has a velocity in the positive direction with respect to a chosen origin and it acquires a constant negative acceleration, the object eventually comes to a rest and reverses direction.
• We have already discussed examples of position functions in the previous section. We now turn our attention to velocity and acceleration functions in order to understand the role that these quantities play in describing the motion of objects. We will find that position, velocity, and acceleration are all tightly interconnected notions.
• The area under the graph of the positive-valued acceleration function for the interval . t. 1 ≤ t ≤ t. 2 . can be found by integrating . a (t) . 4.6.3 Change of Velocity as the Definite Integral of Acceleration . Let . a (t) be the acceleration function over the interval . t ≤ t ≤ t. f . Recall that the velocity . v (t) is an integral ...
• Projectile Motion. Projectile motion occurs when objects are fired at some initial velocity or dropped and move under the influence of gravity. One of the most important things to remember about projectile motion is that the effect of gravity is independent on the horizontal motion of the object. 11. Can an object have a northward velocity and a southward acceleration? Explain. 12. Can the velocity of an object be negative if its acceleration is positive? What about vice versa? Explain. 13. Give an example where both velocity and acceleration are negative. 14. A rock is thrown vertically upward with a speed of v from the edge of a cliff.
• the acceleration and v is the velocity. Thus, the acceleration is positive if dv is positive; the acceleration is negative if dv is negative. (a) An example of one-dimensional motion where the velocity is westward and acceleration is eastward is a car traveling westward and slowing down. (b)An example of one-dimensional motion where the ...
• If a car is moving towards negative direction and car apply brakes due to which it will slow down then its acceleration must be positive 4) Negative acceleration and negative velocity When a ball is falling downwards under the influence of gravity then velocity and acceleration both are positive
• In physics, the sign of an object’s acceleration depends on its direction. If you slow down to a complete stop in a car, for example, and your original velocity was positive and your final velocity was 0, so your acceleration is negative because a positive velocity came down to 0.
• Q. A penny is thrown from a 100.0 m building with an initial velocity of 2.0 m/s down. What is the penny's velocity after 3.00 s? Use an order-of-magnitude estimation to identify the correct choice.
• Note that the angular acceleration as the girl spins the wheel is small and positive; it takes 5 s to produce an appreciable angular velocity. When she hits the brake, the angular acceleration is large and negative. The angular velocity quickly goes to zero. In both cases, the relationships are analogous to what happens with linear motion.
• ©P Chapter 2 Acceleration causes a change in velocity. 31 In both cases, the slope of the graph represents the acceleration of the airplane. Note that Figure 2.4(a) shows zero acceleration. Acceleration is a vector quantity, represented by the variable a , and is defined as the change in velocity per unit time. So, a _ v t.
• Acceleration definition, the act of accelerating; increase of speed or velocity. See more.
• 3) Positive acceleration, but negative velocity . If a car is moving towards negative direction and car apply brakes due to which it will slow down then its acceleration must be positive. 4) Negative acceleration and negative velocity . When a ball is falling downwards under the influence of gravity then velocity and acceleration both are ...However, in physics jargon, acceleration (like velocity) has a more subtle meaning: the acceleration of an object is its rate of change of velocity. From now on, this is what we mean when we say acceleration. At first this might seem to you a nitpicking change of definition -- but it isn't. Remember velocity is a vector.
• Mar 15, 2018 · Thus the initial velocity is negative. The velocity of the object is also negative on the way up but positive on the way down. Note: The convention we use is that upward velocities are negative and downward velocities are positive. Also, displacements above the starting point are negative and those below the starting point are positive.
• Feb 10, 2018 · Both velocity and acceleration are vector quantities, which have both magnitude and direction. Both the expressions can be positive, negative and zero. Conclusion. The motion of an object can be explained as distance traveled, which can be uniform or non-uniform, depending on the velocity of the object.
• In the language of derivatives, the acceleration must have the opposite sign from the velocity. If the acceleration is the same sign as the velocity, then the effect of acceleration is an increase in the speed (the magnitude of velocity). We close this reading by considering the effect of acceleration on the graph of position.
• Dec 03, 2015 · Main Difference – Acceleration vs. Deceleration. Acceleration and deceleration are two of the most basic concepts encountered in mechanics. The main difference between acceleration and deceleration is that acceleration refers to a rate of change of velocity while deceleration refers to a value of acceleration which is negative.
• Velocity is the rate at which an object moves. It has both a magnitude (a value) and a direction. When a velocity is changing as a result of a constant acceleration, the average velocity can be found by adding the initial and final velocities, and dividing by 2. The unit for velocity is meters per second (m/s).
• Velocity and Acceleration Here we will apply particular solutions to find velocity and position functions from an object's acceleration. Example 4: Finding a Position Function Find the position function of a moving particle with the given acceleration, initial position, and initial velocity: Deceleration is the opposite of acceleration. It is the rate at which an object slows down. Deceleration is the final velocity minus the initial velocity, with a negative sign in the result because the velocity is dropping. The formula for acceleration can be used, recognizing that the final result must have a negative sign.
• Mar 13, 2013 · c)make plots of the position,velocity and acceleration as a function of time in an increment of 0.1s for 0<=t<=8 my question is for part c do i just need to use the command plot(x,v,a)? 0 Comments
• Velocity is the rate at which an object moves. It has both a magnitude (a value) and a direction. When a velocity is changing as a result of a constant acceleration, the average velocity can be found by adding the initial and final velocities, and dividing by 2. The unit for velocity is meters per second (m/s).
• Sep 02, 2019 · Problem#4 The position of a particle moving along the x axis is given in centimeters by x = 9.75 + 1.50t 3, where t is in seconds.Calculate (a) the average velocity during the time interval t = 2.00 s to t = 3.00 s; (b) the instantaneous velocity at t = 2.00 s; (c) the instantaneous velocity at t = 3.00 s; (d) the instantaneous velocity at t = 2.50 s; and (e) the instantaneous velocity when ...
Determine the average shear stress in the pin at aBuilt in oven and microwaveReplika something went wrong
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# Question #01fe8
Dec 6, 2016
The equation cannot be verified. It is not true for all values of x.
#### Explanation:
An initial approach is to see if it is a trigonometric identity.
Starting with the left side and converting to sines and cosines,
$L S = S \in X C o s X C o t X$
$L S = S \in X C o s X \left(C o s \frac{X}{S} \in X\right)$
$L S = \left(\cancel{S} \in X\right) \cos X \left(C o s \frac{X}{\cancel{S} \in X}\right)$
$L S = C o {s}^{2} X$
The left side cannot equal the right side for all values of x since
$R S = C s c X$
$R S = \frac{1}{S} \in X$
To prove it is not an identity, it is sufficient to find one value for x which does not satisfy both sides. For example, working in degrees, if x = 30 degrees,
$L S = S \in \left(30\right) C o s \left(30\right) C o t \left(30\right)$
$L S = \left(0.5\right) \left(0.866\right) \left(1.732\right)$
$L S = 0.750 \left(\approx\right)$
$R S = C s c \left(30\right)$
$R S = 2$
Since the left side does not equal the right side for the value x = 30 degrees, the equation is not an identity and cannot be verified.
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The Problem A baseball is hit upward with an initial velocity of = 80 feet per second (or about 55 miles per hour) and leaves the bat with an initial height of = 3 feet. a). Write a formula s(t) that models the height of the baseball after t seconds. b). How high was the baseball after 2 seconds? c). Find the maximum height of the baseball. Support your answer graphically.
a. If the initial velocity of an object is feet per second and its initial height is feet, then the height, in feet, of the object after t seconds can be modeled by the equation .
Now is given to be 80 and is given to be 3. Substituting these values into function s gives .
| 154 | 651 |
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| 3.578125 | 4 |
CC-MAIN-2017-39
|
latest
|
en
| 0.941293 |
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