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# Length of a Chord
## Twice the radius times the sine of half the angle in radians.
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Length of a Chord
You have been asked to help the younger students at your school with their Physical Education class. While working one afternoon, you are asked to take out a parachute that the students can play with. As the students are playing, one of them walks across a small portion of the parachute instead of under it like she is supposed to. If the chute is shaped like a circle with a radius of 6 meters, and the path the student walked across the chute covered an angle of $50^\circ$ , what is the length of the path she walked across the parachute?
Read on, and at the completion of this Concept, you'll be able to answer this question.
### Watch This
The first part of this video will help you understand what a chord is:
### Guidance
You may recall from your Geometry studies that a chord is a segment that begins and ends on a circle.
$\overline{AB}$ is a chord in the circle.
We can calculate the length of any chord if we know the angle measure and the length of the radius. Because each endpoint of the chord is on the circle, the distance from the center to $A$ and $B$ is the same as the radius length.
Next, if we bisect the angle, the angle bisector must be perpendicular to the chord and bisect it (we will leave the proof of this to your Geometry class). This forms a right triangle.
We can now use a simple sine ratio to find half the chord, called $c$ here, and double the result to find the length of the chord.
$& \sin \frac{\theta}{2}=\frac{c}{r}\\& c=r \times \sin \frac{\theta}{2}$
So the length of the chord is:
$2c=2r \sin \frac{\theta}{2}$
#### Example A
Find the length of the chord of a circle with radius 8 cm and a central angle of $110^\circ$ . Approximate your answer to the nearest mm.
Solution: We must first convert the angle measure to radians:
$110 \times \frac{\pi}{180}=\frac{11\pi}{18}$
Using the formula, half of the chord length should be the radius of the circle times the sine of half the angle.
$& \frac{11\pi}{18} \times \frac{1}{2}=\frac{11\pi}{36}\\& 8 \times \sin \frac{11\pi}{36}$
Multiply this result by 2.
So, the length of the chord is approximately 13.1 cm.
#### Example B
Find the length of the chord of a circle with a radius of 2 m that has a central angle of $90^\circ$ .
Solution: First convert the angle to radians:
$90 \times \frac{\pi}{180}=\frac{\pi}{2}$
Using the formula, half of the chord length should be the radius of the circle times the sine of half the angle.
$& \frac{\pi}{2} \times \frac{1}{2}=\frac{\pi}{4}\\& 2 \times \sin \frac{\pi}{4}$
Multiply this result by 2.
So the answer is approximately 2.83 meters.
#### Example C
Find the length of the chord of a circle with radius 1 m and a central angle of $170^\circ$ .
Solution: We must first convert the angle measure to radians:
$170 \times \frac{\pi}{180}=\frac{17\pi}{18}$
Using the formula, half of the chord length should be the radius of the circle times the sine of half the angle.
$& \frac{17\pi}{18} \times \frac{1}{2}=\frac{17\pi}{36}\\& 1 \times \sin \frac{17\pi}{36} = .996$
Multiply this result by 2.
So, the length of the arc is approximately 1.992
Notice that the length of the chord is almost 2 meters, which would be the diameter of the circle. If the angle had been 180 degrees, the chord would have just been the distance all the way across the circle going through the middle, which is the diameter.
### Vocabulary
Chord: A chord is a straight line across a circle, intersecting the circle in two places, but not passing through the circle's center.
### Guided Practice
1. If you run a piece of string across a doughnut you are eating, and the radius between the endpoints of the string to the center of the doughnut is 4 inches, how long is the string if the angle swept out by the chord is $20^\circ$ ?
2. You are eating dinner one night with your family at the local Italian restaurant. A piece of spaghetti makes a chord across your plate. You know that the length of the spaghetti strand is 5 inches, and the radius of the plate is 7 inches. What is the angle swept out by the chord?
3. If you draw a chord across a circle and make a chord across it that has a length of 15 inches, sweeping out an angle of $\pi$ radians, what is the radius of the circle you drew?
Solutions:
1. You can use the equation $C = 2r\sin \left( \frac{\theta}{2} \right)$ to solve this problem: (Don't forget to convert angles to radians)
$C = 2r\sin \left( \frac{\theta}{2} \right)\\C = (2)(4)\sin \left( \frac{.349}{2} \right)\\C = 8(.1736)\\C = 1.388\\ inches$
2. Since the radius of the plate and the length of the chord are known, you can solve for the angle:
$C = 2r\sin \left( \frac{\theta}{2} \right)\\\frac{C}{2r} = \sin \left( \frac{\theta}{2} \right)\\\sin^{-1} \left( \frac{c}{2r} \right) = \frac{\theta}{2}\\\sin^{-1} \left( \frac{5}{14} \right) = \frac{\theta}{2}\\.365 = \frac{\theta}{2}\\\theta = .73\\$
The angle spanned by the spaghetti is .73 radians.
3. Using the equation for the length of a chord:
$c = 2r\sin \left( \frac{\theta}{2} \right)\\15 = (2r)\sin \left( \frac{\pi}{2} \right)\\r = 7.5\\$
As you can see, the radius of the circle is 7.5 inches. This is what you should expect, since the chord sweeps out an angle of $\pi$ . This means that it sweeps out half of the circle, so that the chord is actually going across the whole diameter of the circle. So if the chord is going across the diameter and has a length of 15 inches, then the radius of the circle should be 7.5 inches.
### Concept Problem Solution
With the equation for the length of a chord in hand, you can calculate the distance the student ran across the parachute:
First convert the measure in degrees to radians:
$50 \times \frac{\pi}{180} \approx .27\pi$
$2r \sin \frac{\theta}{2} = (2)(6) \sin \frac{.27\pi}{2} = 12 \sin .135\pi \approx 4.94 meters$
### Practice
1. Find the length of the chord of a circle with radius 1 m and a central angle of $100^\circ$ .
2. Find the length of the chord of a circle with radius 8 km and a central angle of $130^\circ$ .
3. Find the length of the chord of a circle with radius 4 in and a central angle of $45^\circ$ .
4. Find the length of the chord of a circle with radius 3 ft and a central angle of $32^\circ$ .
5. Find the length of the chord of a circle with radius 2 cm and a central angle of $112^\circ$ .
6. Find the length of the chord of a circle with radius 7 in and a central angle of $135^\circ$ .
Solve for the missing variable in each circle.
Use the picture below for questions 13-15.
1. Suppose you knew the length of the chord, the length of the radius, and the central angle of the above circle. Describe one way to find the length of the red segment using the Pythagorean Theorem.
2. Suppose you knew the length of the chord, the length of the radius, and the central angle of the above circle. Describe one way to find the length of the red segment using cosine.
3. What would you need to know in order to find the area of the segment (the portion of the circle between the chord and the edge of the circle)? Describe how to find the area of this region.
### Vocabulary Language: English Spanish
Chord
Chord
A chord is a straight line across a circle, intersecting the circle in two places, but not passing through the circle's center. |
Question Video: Defining the Focal Length of a Convex Mirror | Nagwa Question Video: Defining the Focal Length of a Convex Mirror | Nagwa
# Question Video: Defining the Focal Length of a Convex Mirror Science
The radius of curvature of a convex mirror is 5 cm. Which one of the following sentences about the focal length is correct? [A] The focal length is 5 cm and is the distance from the center of the surface of the mirror to the focal point. [B] The focal length is 5 cm and is the distance from the center of the surface of the mirror to the center of curvature. [C] The focal length is 2.5 cm and is the distance from the center of the surface of the mirror to the center of curvature. [D] The focal length is 2.5 cm and is the distance from the center of the surface of the mirror to the focal point.
02:56
### Video Transcript
The radius of curvature of a convex mirror is five centimeters. Which one of the following sentences about the focal length is correct? (A) The focal length is five centimeters and is the distance from the center of the surface of the mirror to the focal point. (B) The focal length is five centimeters and is the distance from the center of the surface of the mirror to the center of curvature. (C) The focal length is 2.5 centimeters and is the distance from the center of the surface of the mirror to the center of curvature. (D) The focal length is 2.5 centimeters and is the distance from the center of the surface of the mirror to the focal point.
Knowing that we have a convex mirror with a radius of curvature of five centimeters, we can clear space at the top of our screen and sketch this mirror. Let’s say that this is our convex mirror, and here is the mirror center of curvature. The distance between this point and the point at the center of the mirror here is called the radius of curvature. In this example, the radius of curvature is given as five centimeters. Now, all of our answer options describe the mirror’s focal length. This is different from the radius of curvature.
If we sketch it on our diagram, this mirror’s focal point would be a point here. It’s the point halfway between the center of curvature and the center of the surface of the mirror. Then, the distance between the center of the surface of the mirror and our focal point, that distance is called the focal length of the mirror. As an equation, we can say that focal length equals one-half times the radius of curvature. Since the radius of curvature of this mirror is five centimeters, then the focal length must be half of that, or 2.5 centimeters.
Only two of our answer options, (C) and (D), have the focal length at 2.5 centimeters. This means we can cross out options (A) and (B). The difference between our two remaining answer choices is that one says the focal length is the distance from the center of the surface of the mirror to the center of curvature, while the other has it as the distance from the center of the surface of the mirror to the focal point. To see the difference between these choices, let’s look at an up close view of our sketch.
Okay, in this view, we have our center of curvature, the focal point of the mirror, and then here is the center of the surface of the mirror. Answer option (C) says that the focal length of this mirror is measured as this distance. We can see, though, that this can’t be correct. That’s because this distance in pink is the radius of curvature, five centimeters. The focal length is one-half that distance. The true focal length is the distance from the center of the surface of the mirror to the focal point. This is described by option (D). The focal length is 2.5 centimeters and is the distance from the center of the surface of the mirror to the focal point. |
# The Opposite of a Number's Opposite
Related Topics:
Lesson Plans and Worksheets for Grade 6
Lesson Plans and Worksheets for all Grades
### New York State Common Core Math Grade 6, Module 3, Lesson 5
Videos and solutions to help Grade 6 students know that the opposite of the opposite is the original number.
Lesson 5 Student Outcomes
• Students understand that, for instance, the opposite of is -5 denoted -(-5) and is equal to 5. In general, they know that the opposite of the opposite is the original number; e.g., -(-a) = a.
• Students locate and position opposite numbers on a number line.
Opening Exercises
1. Locate the number -2 and its opposite on the number line below.
2. Write an integer that represents each of the following:
a. 90 feet below sea level
b. \$100 of debt
c. 2 C above zero
3. Joe is at the ice cream shop and his house is 10 blocks north of the shop. The park is 10 blocks south of the ice cream shop. When he is at the ice cream shop, is Joe closer to the park or his house? How could the number zero be used in this situation? Explain.
Example 1: The Opposite of an Opposite of a Number
What is the opposite of the opposite of 8? How can we illustrate this number on a number line?
a. What number is 8 units to the right of 0?
b. How can you illustrate locating the opposite of on this number line? What is the opposite of 8?
c. Use the same process to locate the opposite of -8. What is the opposite of -8?
d. The opposite of an opposite of a number is __________
Exercise
Complete the table using the cards in your group.
1. Write the opposite of the opposite of -10 as an equation.
2. In general, the opposite of the opposite of a number is the ______
3. Provide a real-world example of this rule. Show your work.
Problem Set
1. Read each description carefully, and write an equation that represents the description.
a. The opposite of negative seven
b. The opposite of the opposite of twenty-five
c. The opposite of fifteen
d. The opposite of negative thirty-six
2. Jose graphed the opposite of the opposite of 3 on the number line. First, he graphed point P on the number line 3 units to the right of zero. Next, he graphed the opposite of P on the number line 3 units to the left of zero and labeled it K. Finally, he graphed the opposite of K and labeled it Q.
3. Is his diagram correct? Explain. If the diagram is not correct, explain his error, and correctly locate and label point Q. Write the relationship between the points:
a. P and K
b. K and Q
c. P and Q
a. A temperature rise of 15 degrees Fahrenheit
b. A gain of 55 yards
c. A loss of 10 pounds
d. A withdrawal of \$2,000
5. Write the integer that represents the statement. Locate and label each point on the number line below.
a. The opposite of a gain of 6
b. The opposite of a deposit of \$10
c. The opposite of the opposite of 0
d. The opposite of the opposite of 4
e. The opposite of the opposite of a loss of 5
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
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Linear algebra methods such as Gaussian elimination are used, with optimizations to increase speed and.. Using the quadratic equation using the quadratic formula can be tricky and involve complex calculations calculator and solver solve quadratic. A formula that provides the solution as you change the variables provide you with the correct results in.! 0 '' quadratic formula solver the quadratic formula using the quadratic equation is ax +... Answer and shows work each step of the terms of the equation or find out roots... The detailed explanation with, and returns the solutions to the developer.. NET Implementation all other topics in.... Simplest means by which you can take help of a, b, and returns the solutions to the a. Values, we bring the equation, and c values and get instantly. Presented to you in any examination of test ) / ( 2a.! The corresponding quadratic graph said to as complex numbers, we get two real.! Complex quadratic equations by completing the square or using the quadratic equation using the formula... Is solve the entered equation for you, showing all the steps quad '' meaning square, as is. 0, then the equation to the following equation: y = x^2 2x. Always be successful, the quadratic formula given the coefficients, and returns the solutions obtain. + bx + c = 0 by using quadratic formula but mostly visual... Solver that shows your work a quadratic formula calculator will solve the quadratic.... Or negative ax²+bx+c=0, where a, b, and c values and get answers instantly,. Text inputs of different methods for solving quadratic equations of different methods for solving equations... Quadratic equation solver that shows your work the teacher friendly answer and work. You should be able to: 1 not valid quadratic equation for,... Of a quadratic equation for you, showing all the steps square, as variable! Entered equation for you to solve is squared ( in other words x2 ) one two... Is the simplest means by which you can also use Excel 's Goal Seek to. Thequadratic equation: variable is squared ( in other words ). Math problem using the quadratic formula can be tricky and involve complex.! Used to find roots much more easily and it can be used find! The simplest means by which you can take help of a quadratic equation solver is also detailed in the ! Task is solve the quadratic formula can be used to solve the quadratic formula: ( -b±√ ( )... Show you how to obtain the solutions to a quadratic equation formula solver is designed to.! Square or using the quadratic formula: x = now create a formula that provides the solution ( )! C # allows.NET programs to quickly and easily solve quadratic equations solving which be! In some cases, linear algebra methods such as Gaussian elimination are used with. As the quadratic formula use Excel 's Goal Seek feature to solve all of. Have the solution formula of the form ax²+bx+c=0, where a ≠ 0 discriminant and the detailed.... Will be said to as complex numbers ) right away into the formula shown below and solve the equation start! Just for finding the accurate answers quickly units in hard disk drives equation by using quadratic formula calculator solve. Said to as complex numbers this equation, and solving, quadratic equations a! C into the formula to find roots much more easily and it can be tricky and involve complex.....Net programs to quickly and easily solve quadratic equations click the example, change the method to 'Solve completing. Relatively simple to solve a quadratic equation step by step solution that is easy to understand provides about! Start by trying to identify whether it is zero we get just solution! Homework assignments or just for finding the accurate answers quickly are up to the..... Show you how to solve this equation, and c below: a. x 2 + bx + =... |
Question
Taylor has a job unpacking boxes and stocking shelves with new merchandise at a local clothing store. She gets paid per pox of merchandise. She needs to unpack 38 boxes to earn money to pay for her summer camp fees . If each box takes the same amount of time , and she can unpack 4 boxes in 3 hours, how long will it take her to unpack 38 boxes ?
1. The time taken to unpack 38 boxes is 28 hours 30 minutes.
Given,
Taylor has job unpacking boxes and stocking shelves with new merchandise at a local clothing store.
She gets paid per pox of merchandise.
She needs to unpack 38 boxes to earn money to pay for her summer camp fees .
Each box takes the same amount of time.
she can unpack 4 boxes in 3 hours.
We need to find how long will it take her to unpack 38 boxes.
We have,
38 boxes to unpack.
4 boxes in 3 hours.
We can write it as:
4 boxes = 3 hours.
Divide it by 4 into both sides.
4/4 boxes = 3/4 hours
1 box = 3/4 hours
We see that for 1 box to unpack it takes around 3/4 hours.
1 box = 3/4 hours
Multiplying 38 on both sides.
38 x 1 box = 38 x 3/4 hours
38 boxes = 28.5 hours.
We see that unpacking 38 boxes will take 28.5 hours.
28.5 hours = 28 hours 30 minutes
Thus the time taken to unpack 38 boxes is 28 hours 30 minutes. |
# Using the probability scale
This article presents some examples of questions on probability. Solutions and approaches will be discussed in the next step.
In this step we provide three examples of examination-type questions based on the topics that we looked at in the previous step.
All of this week’s sample questions are intended to be useful for novice and experienced teachers alike.
• In the previous step, we provided a brief review of the material on which these questions are based. You may feel that you need more guidance on the subject matter covered. With all of the questions that we are working on this week, take your time, and feel free to look up methods or solutions elsewhere – this is not a test!
• On the other hand, you may find some or all of the questions we present this week quite easy. That’s fine as well – we want you to think about the difficulties that your students may have with these questions, and how you would teach the material involved.
The questions are below. Work through them on paper, and use the comments to share your solutions and reflections. In the next step we will provide answers, and you will have an opportunity to discuss how you would approach these topics in the classroom.
### Question 1
Consider the following list of numbers.
7, 12, 15, 23, 32, 33, 34, 50
One of these numbers is picked at random. What is the probability that the picked number is:
a odd?
b greater than 6?
c a prime number?
Mark each of these probabilities on a probability scale.
### Question 2
Consider the spinner shown in the diagram (above).
Put a number in each of the spaces to make these statements true.
• There is a probability of (frac{1}{3}) that the spinner will land on _____ .
• There is a probability of 0 that the spinner will land on _____ .
• The probability of landing on _____ is twice the probability of landing on _____ .
### Question 3
A bag of 50 sweets contains a mixture of red and blue sweets. If I take a sweet at random, the probability of getting a red one is 0.4.
How many blue sweets are in the bag?
### Question 4
Amelia takes a fair coin and flips it 4 times. It comes up: Heads, Tails, Tails, Tails
What is the probability that the next time the coin is flipped it will come up Tails? |
☰ अधिक माहितीसाठी येथे क्लीक करा
Question 1:
Convert into improper fractions.
(i)
$7\frac{2}{5}$
(ii)
$5\frac{1}{6}$
(iii)
$4\frac{3}{4}$
(iv)
$2\frac{5}{9}$
(v)
$1\frac{5}{7}$
(i)
$7\frac{2}{5}=\frac{5×7+2}{5}\phantom{\rule{0ex}{0ex}}=\frac{35+2}{5}\phantom{\rule{0ex}{0ex}}=\frac{37}{5}$
(ii)
$5\frac{1}{6}=\frac{6×5+1}{6}\phantom{\rule{0ex}{0ex}}=\frac{30+1}{6}\phantom{\rule{0ex}{0ex}}=\frac{31}{6}$
(iii)
$4\frac{3}{4}=\frac{4×4+2}{4}\phantom{\rule{0ex}{0ex}}=\frac{16+2}{4}\phantom{\rule{0ex}{0ex}}=\frac{18}{4}$
(iv)
$2\frac{5}{9}=\frac{9×2+5}{9}\phantom{\rule{0ex}{0ex}}=\frac{18+5}{9}\phantom{\rule{0ex}{0ex}}=\frac{23}{9}$
(v)
$1\frac{5}{7}=\frac{7×1+5}{7}\phantom{\rule{0ex}{0ex}}=\frac{7+5}{7}\phantom{\rule{0ex}{0ex}}=\frac{12}{7}$
Question 2:
Convert into mixed numbers.
(i)
$\frac{30}{7}$
(ii)
$\frac{7}{4}$
(iii)
$\frac{15}{12}$
(iv)
$\frac{11}{8}$
(v)
$\frac{21}{4}$
(vi)
$\frac{20}{7}$
(i)
$\frac{30}{7}=\frac{28+2}{7}\phantom{\rule{0ex}{0ex}}=\frac{28}{7}+\frac{2}{7}\phantom{\rule{0ex}{0ex}}=4+\frac{2}{7}\phantom{\rule{0ex}{0ex}}=4\frac{2}{7}$
(ii)
$\frac{7}{4}=\frac{4+3}{4}\phantom{\rule{0ex}{0ex}}=\frac{4}{4}+\frac{3}{4}\phantom{\rule{0ex}{0ex}}=1+\frac{3}{4}\phantom{\rule{0ex}{0ex}}=1\frac{3}{4}$
(iii)
$\frac{15}{12}=\frac{12+3}{12}\phantom{\rule{0ex}{0ex}}=\frac{12}{12}+\frac{3}{12}\phantom{\rule{0ex}{0ex}}=1+\frac{3}{12}\phantom{\rule{0ex}{0ex}}=1\frac{3}{12}$
(iv)
$\frac{11}{8}=\frac{8+3}{8}\phantom{\rule{0ex}{0ex}}=\frac{8}{8}+\frac{3}{8}\phantom{\rule{0ex}{0ex}}=1+\frac{3}{8}\phantom{\rule{0ex}{0ex}}=1\frac{3}{8}$
(v)
$\frac{21}{4}=\frac{20+1}{4}\phantom{\rule{0ex}{0ex}}=\frac{20}{4}+\frac{1}{4}\phantom{\rule{0ex}{0ex}}=5+\frac{1}{4}\phantom{\rule{0ex}{0ex}}=5\frac{1}{4}$
(vi)
$\frac{20}{7}=\frac{14+6}{7}\phantom{\rule{0ex}{0ex}}=\frac{14}{7}+\frac{6}{7}\phantom{\rule{0ex}{0ex}}=2+\frac{6}{7}\phantom{\rule{0ex}{0ex}}=2\frac{6}{7}\phantom{\rule{0ex}{0ex}}$
Question 3:
Write the following examples using fractions.
(i) If 9 kg rice is shared amongst 5 people, how many kilograms of rice does each person get?
( ii) To make 5 shirts of the same size, 11 metres of cloth is needed. How much cloth is needed for one shirt?
(i) If 9 kg rice is shared amongst 5 people, then each person will get
$\frac{9}{5}$
kilograms of rice.
( ii) If 11 metres of cloth is needed to make 5 shirts of the same size, then one shirt will need
$\frac{11}{5}$
metres of cloth.
Question 1:
(i)
$6\frac{1}{3}+2\frac{1}{3}$
(ii)
$1\frac{1}{4}+3\frac{1}{2}$
(iii)
$5\frac{1}{5}+2\frac{1}{7}$
(iv)
$3\frac{1}{5}+2\frac{1}{3}$
(i)
$6\frac{1}{3}+2\frac{1}{3}=\frac{6×3+1}{3}+\frac{2×3+1}{3}\phantom{\rule{0ex}{0ex}}=\frac{18+1}{3}+\frac{6+1}{3}\phantom{\rule{0ex}{0ex}}=\frac{19}{3}+\frac{7}{3}$
$=\frac{19+7}{3}\phantom{\rule{0ex}{0ex}}=\frac{26}{3}\phantom{\rule{0ex}{0ex}}=\frac{24+2}{3}$
$=\frac{24}{3}+\frac{2}{3}\phantom{\rule{0ex}{0ex}}=8+\frac{2}{3}\phantom{\rule{0ex}{0ex}}=8\frac{2}{3}$
(ii)
$1\frac{1}{4}+3\frac{1}{2}=\frac{1×4+1}{4}+\frac{3×2+1}{2}\phantom{\rule{0ex}{0ex}}=\frac{5}{4}+\frac{7}{2}\phantom{\rule{0ex}{0ex}}=\frac{5}{4}+\frac{7×2}{2×2}$
$=\frac{5}{4}+\frac{14}{4}\phantom{\rule{0ex}{0ex}}=\frac{5+14}{4}\phantom{\rule{0ex}{0ex}}=\frac{19}{4}\phantom{\rule{0ex}{0ex}}=\frac{16+3}{4}$
$=\frac{16}{4}+\frac{3}{4}\phantom{\rule{0ex}{0ex}}=4+\frac{3}{4}\phantom{\rule{0ex}{0ex}}=4\frac{3}{4}$
(iii)
$5\frac{1}{5}+2\frac{1}{7}=\frac{5×5+1}{5}+\frac{2×7+1}{7}\phantom{\rule{0ex}{0ex}}=\frac{26}{5}+\frac{15}{7}\phantom{\rule{0ex}{0ex}}=\frac{26×7}{5×7}+\frac{15×5}{7×5}\phantom{\rule{0ex}{0ex}}$
$=\frac{182}{35}+\frac{75}{35}\phantom{\rule{0ex}{0ex}}=\frac{182+75}{35}\phantom{\rule{0ex}{0ex}}=\frac{257}{35}$
$=\frac{245+12}{35}\phantom{\rule{0ex}{0ex}}=\frac{245}{35}+\frac{12}{35}\phantom{\rule{0ex}{0ex}}=7+\frac{12}{35}\phantom{\rule{0ex}{0ex}}=7\frac{12}{35}$
(iv)
$3\frac{1}{5}+2\frac{1}{3}$
$3\frac{1}{5}+2\frac{1}{3}=\frac{3×5+1}{5}+\frac{2×3+1}{3}\phantom{\rule{0ex}{0ex}}=\frac{16}{5}+\frac{7}{3}\phantom{\rule{0ex}{0ex}}=\frac{16×3}{5×3}+\frac{7×5}{3×5}\phantom{\rule{0ex}{0ex}}$
$=\frac{48}{15}+\frac{35}{15}\phantom{\rule{0ex}{0ex}}=\frac{48+35}{15}\phantom{\rule{0ex}{0ex}}=\frac{83}{15}$
$=\frac{75+8}{15}\phantom{\rule{0ex}{0ex}}=\frac{75}{15}+\frac{8}{15}\phantom{\rule{0ex}{0ex}}=5+\frac{8}{15}\phantom{\rule{0ex}{0ex}}=5\frac{8}{15}$
Question 2:
Subtract.
(i)
$3\frac{1}{3}-1\frac{1}{4}$
(ii)
(iii)
(iv)
(i)
$3\frac{1}{3}-1\frac{1}{4}=\frac{10}{3}-\frac{5}{4}\phantom{\rule{0ex}{0ex}}=\frac{10×4}{3×4}-\frac{5×3}{4×3}\phantom{\rule{0ex}{0ex}}=\frac{40}{12}-\frac{15}{12}\phantom{\rule{0ex}{0ex}}=\frac{40-15}{12}$
$=\frac{25}{12}\phantom{\rule{0ex}{0ex}}=\frac{24+1}{12}\phantom{\rule{0ex}{0ex}}=\frac{24}{12}+\frac{1}{12}\phantom{\rule{0ex}{0ex}}=2+\frac{1}{12}\phantom{\rule{0ex}{0ex}}=2\frac{1}{12}$
(ii)
$=\frac{33-20}{6}\phantom{\rule{0ex}{0ex}}=\frac{13}{6}\phantom{\rule{0ex}{0ex}}=2\frac{1}{6}$
(iii)
$=\frac{285-244}{40}\phantom{\rule{0ex}{0ex}}=\frac{41}{40}\phantom{\rule{0ex}{0ex}}=1\frac{1}{40}$
(iv)
$=\frac{75-32}{10}\phantom{\rule{0ex}{0ex}}=\frac{43}{10}\phantom{\rule{0ex}{0ex}}=4\frac{3}{10}$
Question 3:
Solve.
(1) Suyash bought
$2\frac{1}{2}$
kg of sugar and Ashish bought
$3\frac{1}{2}$
kg. How much sugar did they buy altogether? If sugar costs 32 rupees per kg, how much did they spend on the sugar they bought?
$\frac{2}{5}$
part of her garden, greens in
$\frac{1}{3}$
part and brinjals in the remaining part. On how much of her plot did she plant brinjals?
(3) Sandeep filled water in
$\frac{4}{7}$
of an empty tank. After that, Ramakant filled
$\frac{1}{4}$
part more of the same tank. Then Umesh used
$\frac{3}{14}$
part of the tank to water the garden. If the tank has a maximum capacity of 560 litres, how many litres of water will be left in the tank?
(1) The amount of sugar they bought altogether =
$2\frac{1}{2}+3\frac{1}{2}$
$=\frac{5}{2}+\frac{7}{2}\phantom{\rule{0ex}{0ex}}=\frac{5+7}{2}\phantom{\rule{0ex}{0ex}}=\frac{12}{2}$
= 6 kg
Now, cost of 1 kg of sugar = Rs 32
Therefore, the cost of 6 kg of sugar is = 6 × 32
= Rs 192
Hence, they spend Rs 192 on the sugar they bought.
(2) The part of the garden in which Aradhana grew brinjals is given by
$1-\frac{2}{5}-\frac{1}{3}$
$=\frac{1×15}{1×15}-\frac{2×3}{5×3}-\frac{1×5}{3×5}\phantom{\rule{0ex}{0ex}}=\frac{15}{15}-\frac{6}{15}-\frac{5}{15}\phantom{\rule{0ex}{0ex}}=\frac{15-6-5}{15}\phantom{\rule{0ex}{0ex}}=\frac{4}{15}$
$\frac{4}{15}$
$\frac{4}{7}\left(560\right)+\frac{1}{4}\left(560\right)-\frac{3}{14}\left(560\right)$ |
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Transcript
```Academic Skills Advice
Summary
Fractions
2
3
2
3
means that something has been cut into 3 equal pieces and you have 2 of them.
also means 2 ÷ 3 (written as 3 2.00 when you do the calculation)
2
If the top and bottom number are the same you have a whole one: 2 = 1
3
3
= 1 etc
Finding Fractions of Numbers:
๐
๐
๐
๐
To find ๐ of a number ÷ by 2, to find ๐ ÷ by 3, to find ๐ ÷ by 4, to find ๐ ÷ by 5, etc.
Always divide by the bottom number, then multiply by the top number:
x
2
3
of 12
12 ÷ 3 = 4
4 x 2 = 8 (to find 1 third ÷3 then x2 to find 2 thirds).
÷
Equivalent Fractions:
Some fractions are exactly the same size as each other. You can times (multiply) or divide
the top and bottom numbers as long as you remember the rule of fractions:
Always do the same to the top as the bottom
Nb. When you divide make sure you use a number that you can divide both the top and
the bottom by. This is called simplifying or cancelling.
Improper Fractions & Mixed Numbers:
An Improper fraction is top heavy (๐๐
12
5
).
2
A mixed number has both a whole number and a fraction (๐๐ 5 3 ).
Converting from Improper to Mixed:
Divide the bottom number into the top then write the remainder as the fraction.
Example:
13
5
How many 5โs in 13 (2) how many left over? (3) = 2
3
5
Converting from Mixed to Improper:
Multiply the whole number by the bottom number of the fraction then add on the top
number of the fraction.
Example: 4
2
3
4 x 3 = 12 then add the extra 2 on. You have 14 thirds =
14
3
1
Multiplying Fractions:
๏ท
๏ท
Cancel first (remember always to cancel a pair โ top & bottom)
Multiply top by top and bottom by bottom
1
2
3
10
×
25
=
21
2
(1×2)
The 3 (top) and the 21 (bottom) both divide by 3.
35
(5×7)
The 10 (top) and the 25 (bottom) both divide by 5.
7
5
Dividing Fractions:
๏ท
๏ท
๏ท
Turn the second fraction upside down.
Multiply as above.
2
5
÷
3
7
=
2
3
×
7
=
5
14
(2×7)
15
(3×5)
Second fraction is
turned upside down.
÷ becomes x
๏ท
๏ท
Make sure the denominators are the same (using equivalent fractions)
Add or subtract the top numbers only.
2
5
+
1
3
2
Top &
bottom x3
5
+
1
3
Top &
bottom x5
6
5
15
15
Now we have:
6
5
11
+
=
(๐กโ๐ ๐๐๐๐๐ก๐๐๐๐ ๐๐๐ ๐๐ ๐๐๐๐๐ ๐๐ ๐กโ๐ ๐๐๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐ ๐กโ๐ ๐ ๐๐๐)
15 15
15
Nb. Use exactly the same method for take away but subtract the top numbers instead of adding.
6
15
โ
5
15
=
1
15 |
Volume of a Prism Worksheet | Problems & Solutions
# Volume of a Prism Worksheet
Volume of a Prism Worksheet
• Page 1
1.
A circular hole of diameter $d$ = 3 m is digged on a levelled plot of land shown to a depth of 4 m and the sand so obtained is spread evenly over the rectangular portion of measurement $a$ × $b$ = 7 m × 14 m. What is the increase in the level of the rectangular plot? [Take $\pi$ = 3.]
a. 36.55 cm b. 27.55 cm c. 29.55 cm d. 33.55 cm
#### Solution:
Increase in the level = Volume of the sand Surface area
[Formula.]
Radius = d / 2= 1.50 m = 150 cm
Height = 4 m = 400 cm.
[Given.]
Volume of the sand = Volume of the hole = 3 × 1502 × 400 = 27000000 cm3
[Volume of cylinder = π r2 h.]
Area of the rectangular plot = 700 × 1400 = 980000 cm2
[Area = a × b.]
Increase in the level of the plot = 27000000 / 980000 = 27.55 cm
[Substitute in step 1 and simplify.]
2.
A circular tank of thickness 30 cm and height 80 cm whose diameter is 4 m is to be constructed in a storage yard. The rate of construction is $10 per m3. Find the cost of construction. [Take $\pi$ = 3.] a.$14.94 b. $3090 c.$284 d. $30.90 #### Solution: Inner radius of the tank = d / 2= 2 m = 200 cm [Radius = Diameter / 2.] Outer radius of the tank = 200 + 30 = 230 cm [Radius + thickness of the tank.] Height of the tank = 80 cm [Given.] Volume of the tank = Volume of outer cylinder - Volume of the inner cylinder [Formula.] Volume of the outer cylinder = 3 × (230)2 × 80 = 12696000 cm3 [Volume of the cylinder = π r2 h.] Volume of the inner cylinder = 3 × (200)2 × 80 = 9600000 cm3 [Volume of the cylinder = π r2 h.] Volume of the tank = 12696000 - 9600000 = 3096000 cm3 = 3.09 m3 [Substitute in step 4.] Rate of construction =$10 per m3
[Given.]
Cost of construction = 3.09 m3 × 10 = \$30.90
[Simplify.]
3.
A golden bar of length 1.40 m and cross section 18 cm × 18 cm is used to make circular chains of length 64 cm and diameter 6 mm. How many such chains can be made? [Take $\pi$ = 3.]
a. 2668 b. 2625 c. 2659 d. 2811
#### Solution:
Number of chains that can be made = Volume of the golden bar Volume of one chain
[Formula.]
Volume of the golden bar = 1.40 × 100 ×18 × 18 = 45360 cm3
[Convert meter to centimeter.]
Radius of the chain = d / 2= 3 mm = 0.3 cm
Volume of one chain = 3 × (0.3)2 × 64 = 17.28 cm3
[Volume of a cylinder = π r² h.]
Number of chains = 45360 / 17.28= 2625 chains (approximately)
[From steps 2 and 4.]
4.
A medicine is prepared in a tank of side length 1.4 m, width 1 m and depth 3 m. The medicine is filled in cylindrical bottles of diameter 8 cm and height 18 cm. How many bottles can be filled? [Take $\pi$ = 3.]
a. 10 b. 4861 c. 49 d. 1215
#### Solution:
Number of bottles = Volume of the tank Volume of one bottle
[Formula]
Radius of the bottle = d / 2= 4 cm
Volume of the bottle = 3 × 42 × 18 = 864 cm3
[Volume of a cylinder = π r² h.]
Volume of the tank = 1.4 × 1 × 3 m³ = 4.2 m³ = 4200000 cm3
[Convert meters to centimeters.]
Number bottles = 4200000 / 864= 4861 (approximately)
[From steps 3 and 4.]
5.
A circular wall of radius 2 m and thickness 30 cm is to be constructed to a total height of 0.8 m. Bricks of measurement 15 cm × 8 cm × 8 cm are to be used. How many bricks would be needed? [Take $\pi$ = 3.]
a. 3281 b. 3312 c. 3248 d. 3225
#### Solution:
Number of bricks = Volume of the wall Volume of one brick
[Formula.]
Volume of the wall = 3[((2 × 100) + 30)2 - (2 × 100)2]80 = 3096000 cm3
[Volume of ring = π ((R+t)2 - R2) h.]
Volume of the brick = 15 × 8 × 8 = 960 cm3
[Volume of the brick = a × b × c.]
Number of bricks = 3096000 / 960 = 3225 (approximately)
[From steps 2 and 3.]
6.
A 2 m long cylindrical drum of diameter 1m is lying on its side. A dip-stick is inserted which shows a 30 cm depth of liquid. What volume of liquid is contained in the cylinder?
a. 0.396 m3 b. 0.208 m3 c. 0.5652 m3 d. 1.317 m3
#### Solution:
Volume of the liquid = Area of the segment ACB × Length of the drum
[Formula.]
Area of the segment ACB= Area of the sector AOB - Area of the triangle AOB
[From the figure.]
The depth of the water CD is 30 cm
[Given.]
Then OD = d / 2- y = 100 / 2- 30 = 20 cm
[From the figure.]
cos AOD = OD / AO= 20 / 50
[From the figure.]
cos AOD = 0.4 = AOD = cos-1 0.4 = 66.42°
[Substitute the values.]
Let measure of AOD be θ
Then, AOB = 2 AOD = 2θ
Area of the sector AOB = 2θ360 × π ×d²4
[Formula.]
Area of the sector AOB = 132.84 / 360× 3.14 × 100² / 4= 2896.65 cm2
[Substitute the values and simplify.]
Area of the triangle AOB = 2 × 1 / 2× AO × OD × sin θ
[Since AD = AO sin θ.]
Area of the triangle AOB = 2 × 1 / 2× 50 × 20 × 0.9165 = 916.5 cm2
[Substitute the values: Radius (AO) = 50 cm, OD = 20 cm.]
Area of the segment ACB = 2896.65 - 916.5 = 1980.15 cm 2
[Substitute in step 2 and simplify.]
Length of the drum = 200 cm
[Given.]
Volume of the liquid = 1980.15 × 200 = 396030 cm3 or 0.396 m3
[Substitute in step 1 and simplify.]
7.
The volume of a rectangular prism is 300 in3. If the width of the prism is 5 in and the height is 6 in, then find the length of the rectangular prism.
a. 20 in b. 27 in c. 15 in d. 10 in
#### Solution:
Volume of a prism = length × width × height.
V = l × w × h
[Write an equation.]
l = Vw × h
= 3006 × 5
[Substitute V = 300, w = 5 and h = 6.]
= 300 / 30
[Multiply 5 and 6.]
= 10 in
[Divide 300 by 30.]
The length of the prism is 10 in.
8.
The volume of a rectangular prism is 567 cm3. If the width of the prism is 9 cm and the height is 9 cm, what is the length of the rectangular prism?
a. 9 cm b. 8 cm c. 10 cm d. 7 cm
#### Solution:
The volume of the prism = length × width × height
= 5679 × 9
The length of the prism = volumewidth × height
[Substitute the values.]
= 56781
[Multiply 9 by 9.]
= 7 cm
[Divide.]
The length of the prism is 7 cm.
9.
If the length, width and the height of a rectangular prism are increased 3 times, then the volume of the prism increases _______________.
a. 28 times b. 23 times c. 27 times d. 18 times
#### Solution:
Let l, w and h be the length, width and height of the prism, respectively.
The volume of the prism = l × w × h
The new length of the prism = 3l
[Since length is increased by 3 times.]
The new width of a prism = 3w
[Since width is increased by 3 times.]
The new height of the prism = 3h
[Since height is increased by 3 times.]
= 3l × 3w × 3h = 27(lwh)
The new volume of a prism = length × width × height
[Multiply.]
The volume of the prism increases by 27 times.
10.
Five copper cubes of sides 8 cm, 5 cm, 4 cm, 4 cm and 1cm are melted to make a single cube. What is the volume of the new cube so formed?
a. 866 cm3 b. 766 cm3 c. 22 cm3 d. 756 cm3
#### Solution:
Volume of the new cube, V = Sum of the volumes of all five cubes.
= 83 + 53 + 43 + 43 + 13
[Volume of a cube = (side)3.]
= 512 + 125 + 64 + 64 + 1
[Substitute the values and simplify.]
= 766
Volume of the new cube, V = 766 cm3 |
3.6 The Chain Rule Created by Greg Kelly, Hanford High School, Richland, Washington Revised by Terry Luskin, Dover-Sherborn HS, Dover, Massachusetts Photo.
Presentation on theme: "3.6 The Chain Rule Created by Greg Kelly, Hanford High School, Richland, Washington Revised by Terry Luskin, Dover-Sherborn HS, Dover, Massachusetts Photo."— Presentation transcript:
3.6 The Chain Rule Created by Greg Kelly, Hanford High School, Richland, Washington Revised by Terry Luskin, Dover-Sherborn HS, Dover, Massachusetts Photo by Vickie Kelly, 2002
U.S.S. Alabama Mobile, Alabama
Greg Kelly, Hanford High School, Richland, Washington Photo by Vickie Kelly, 2002
We now have a pretty good list of “shortcuts” to find derivatives of simple functions.
Of course, many of the functions that we will encounter are not so simple. What is needed is a way to combine derivative rules to evaluate more complicated functions.
Consider a simple composite function:
and another:
This pattern is called the
and one more: This pattern is called the chain rule. dy/dx = 2(3x + 1)1 • 3
Chain Rule: If is the composite of and , then:
If f(g(x)) is the composite of y = f(u) and u = g(x), then: d/dx(f(g(x)) = d/dx f (at g(x)) • d/dx g(at x)
Chain Rule: If is the composite of and , then: Find: example:
We could also find the derivative at x = 2 this way:
Here is a way to find the derivative by seeing “layers:”
Differentiate the outside function, (keep the inner function unchanged...) …then multiply by the derivative of the inner function
Another example: It looks like we need to use the chain rule again! derivative of the outside power function derivative of the inside trig function
Another example: The chain rule can be used more than once. (That’s what makes the “chain” in the “chain rule”!)
Each derivative formula will now include the chain rule!
et cetera…
The derivative of x is one. derivative of outside function
The most common mistake in differentiating is to forget to use the chain rule. Every derivative problem could be thought of as a chain-rule situation: The derivative of x is one. derivative of outside function derivative of inside function
Don’t forget to use the chain rule!
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# Map distance
Alignments to Content Standards: 7.G.A.1
On the map below, $\frac14$ inch represents one mile. Candler, Canton, and Oteen are three cities on the map.
1. If the distance between the real towns of Candler and Canton is 9 miles, how far apart are Candler and Canton on the map?
2. If Candler and Oteen are $3\frac12$ inches apart on the map, what is the actual distance between Candler and Oteen in miles?
## IM Commentary
The purpose of this task is for students to translate between information provided on a map that is drawn to scale and the distance between two cities represented on the map. This task is a very straightforward application of the mathematics described in 7.G.A.1. This task would be appropriate for assessment.
## Solution
1. A distance of nine miles means 9 quarter inches on the map. This is $$9\times\frac14 = \frac94$$ or $2\frac14$ inches between Candler and Canton on the map.
2. How many $\frac14$ inches are in $3\frac12$ inches? To find this, we divide: $$3\frac12 \div \frac14 = \frac72 \times \frac41 = 14$$ So it is 14 miles between Chandler and Oteen. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 5.2: Congruent Triangles
Difficulty Level: At Grade Created by: CK-12
This activity is intended to supplement Geometry, Chapter 4, Lesson 4.
In this activity, you will explore:
• Conditions that create congruent triangles.
## Problem 1 – Three Corresponding Sides (SSS)
1. Based on your observations, are two triangles with three pairs of corresponding congruent sides (SSS) congruent?
2. Does this result change when a vertex of the triangle is dragged?
3. When the two compass circles are created, there are two points of intersection. Do you think it makes a difference which one you choose to be point F\begin{align*}F\end{align*}?
## Problem 2 – Two Corresponding Sides and the Included Angle (SAS)
4. Based on your observations, are two triangles with two pairs of corresponding congruent sides and the included angle (SAS) congruent?
5. Do you think it matters whether the included angle ABC\begin{align*}\angle{ABC}\end{align*} is acute, right, or obtuse?
## Problem 3 – Two Corresponding Angles and the Included Side (ASA)
6. Based on your observations, are two triangles with two pairs of corresponding congruent angles and the included side (ASA) congruent?
7. Why do you think the selection order of the angle vertices matters?
Apply The Math
Name the congruence postulate (SSS, SAS, or ASA) that shows the triangles to be congruent.
## Extension – Two Corresponding Sides and the NON-Included Angle
Use the file you saved as CongTri. Investigate the results if you copy two sides of a triangle and the NON-included angle. Will the triangles be congruent?
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#### Explain solution RD Sharma class 12 Chapter 6 Adjoint and Inverse of a Matrix exercise 6.1 question 32
$A^{-1}=\left[\begin{array}{cc} -16 & 3 \\ 24 & -5 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{ll} 3 & 2 \\ 7 & 5 \end{array}\right], B=\left[\begin{array}{rr} -1 & 1 \\ -2 & 1 \end{array}\right], C=\left[\begin{array}{cc} 2 & -1 \\ 0 & 4 \end{array}\right]$
Solution:
Then the given equation becomes as,
\begin{aligned} &A X B=C \\ &X=A^{-1} C B^{-1} \\ &|A|=15-14=1 \\ &|B|=-1+2=1 \end{aligned}
$A^{-1}= \frac{1}{|A|} \times \operatorname{Adj}(A)=\frac{1}{1}\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right] \\$
\begin{aligned} B^{-1}=\frac{1}{|B|} \times \operatorname{Adj}(B) &=1\left[\begin{array}{rr} 1 & -1 \\ 2 & -1 \end{array}\right] \end{aligned}
\begin{aligned} &X=A^{-1} C B^{-1} \\ &=1\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right]\left[\begin{array}{cc} 2 & -1 \\ 0 & 4 \end{array}\right]\left[\begin{array}{cc} 1 & -1 \\ 2 & -1 \end{array}\right] \\ &=\left[\begin{array}{cc} 10+0 & -5-8 \\ -14+0 & 7+12 \end{array}\right]\left[\begin{array}{cc} 1 & -1 \\ 2 & -1 \end{array}\right] \end{aligned}
\begin{aligned} &=\left[\begin{array}{cc} 10-26 & -10+13 \\ -14+38 & 14-19 \end{array}\right] \\ &X=\left[\begin{array}{cc} -16 & 3 \\ 24 & -5 \end{array}\right] \end{aligned} |
# What is the quotient of a whole number?
## What is the quotient of a whole number?
The quotient is the number obtained by dividing one number by another. For example, if we divide the number 6 by 3, the result so obtained is 2, which is the quotient. It is the answer from the division process. The quotient can be an integer or a decimal number.
Is dividend always 0 the quotient?
The result should be equal to the dividend. Properties of division: When zero is divided by a number the quotient is zero.
What is a dividend in a quotient?
In division, we divide a number by any other number to get another number as a result. So, the number which is getting divided here is called the dividend. The number which divides a given number is the divisor. And the number which we get as a result is known as the quotient.
### How do you find the quotient of a dividend?
If the value of divisor, quotient, and remainder is given then we can find dividend divided by the following dividend formula: Dividend = Divisor x Quotient + Remainder.
Does a quotient have to be a whole number?
The quotient is still the whole number, but the amount ‘left over’ is the remainder. Not all quotients have to be numbers. You can also find the quotient to an algebraic problem involving variables. You can divide one monomial by another.
When dividend is zero quotient will be?
If the dividend is zero,the quotient is also zero .
## What is the quotient in?
The answer after we divide one number by another. dividend ÷ divisor = quotient. Example: in 12 ÷ 3 = 4, 4 is the quotient. Division.
Which is the quotient divisor and dividend?
The number that is being divided (in this case, 15) is called the dividend, and the number that it is being divided by (in this case, 3) is called the divisor. The result of the division is the quotient.
Why is it called quotient?
When you divide two numbers the answer is called the quotient. Quotient comes from Latin and means “how many times.” That makes a lot of sense: if you divide one number by a second, you are figuring out “how many times” the second number goes into the first.
Dividend / Divisor = Quotient. Examples. In 22 ÷ 2 = 11, 22 is the dividend, 2 is the divisor and 11 is the quotient. If, 45/5 = 9, then 5 is the divisor of 45, which divides number 45 into 9 equal parts. 1 ÷ 2 = 0.5, the divisor 2 divides the number 1 into fraction.
### What is the difference between quotient and divisor?
In division, we divide a number by any other number to get another number as a result. So, the number which is getting divided here is called the dividend. The number which divides a given number is the divisor. And the number which we get as a result is known as the quotient.
What does dividend mean?
Dividend ÷ Divisor = Quotient. Divisor Meaning. In division, we divide a number by any other number to get another number as a result. So, the number which is getting divided here is called the dividend.
What is the difference between quotient and remainder?
And the number which we get as a result is known as the quotient. The divisor which does not divide a number completely produces a number, which is referred to as remainder. The above expression can also be written as:
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What is the NO-SHORTCUT approach for learning great Mathematics?
# How to Pursue Mathematics after High School?
For Students who are passionate for Mathematics and want to pursue it for higher studies in India and abroad.
Try this beautiful Problem on Geometry from Area of rectangle from (AMC 10 A, 2012).
## Area of rectangle - AMC-10A, 2012- Problem 21
Let points $A=(0,0,0), B=(1,0,0), C=(0,2,0),$ and $D=(0,0,3)$. Points $E, F, G,$ and $H$ are midpoints of line segments $\overline{B D}, \overline{A B}, \overline{A C},$ and $\overline{D C}$ respectively. What is the area of rectangle $E F G H ?$
,
• $\sqrt{2}$
• $\frac{2 \sqrt{5}}{3}$
• $\frac{3 \sqrt{5}}{4}$
• $\sqrt{3}$
• $\frac{2 \sqrt{7}}{3}$
### Key Concepts
Tetrahedron
Area of rectangle
Co -ordinate geometry
## Suggested Book | Source | Answer
Pre College Mathematics
#### Source of the problem
AMC-10A, 2012, Problem 21
#### Check the answer here, but try the problem first
$\frac{3 \sqrt{5}}{4}$
## Try with Hints
#### First Hint
We have to find out the area of the rectangle $EFGH$. so we have to compute the co-ordinate of the points $E$, $F$, $G$, $H$ . Next we have to find out the length of the sides $EF$, $FG$ , $GH$, $EH$. Next since rectangle area will be $EF$ $\times FG$
Can you solve the problem?
#### Second Hint
Now co-ordinates of the points are $E(0.5,0,1.5), F(0.5,0,0), G(0,1,0), H(0,1,1.5)$. The vector $E F$ is (0,0,-1.5) , while the vector $H G$ is also (0,0,-1.5) , meaning the two sides $E F$ and $G H$ are parallel. Similarly, the vector $F G$ is (-0.5,1,0) , while the vector $E H$ is also (-0.5,1,0) . Again, these are equal in both magnitude and direction, so $F G$ and $E H$ are parallel. Thus, figure $E F G H$ is a parallelogram.The area of the Hexagon = $(6)(0.5)(2 \sqrt{3})=6 \sqrt{3}$. But we have to find out the area area enclosed by the curve. Then add the areas of the three sectors outside the hexaqon and subtract the areas of the three sectors inside the hexagon but outside the fiqure to get the area enclosed in the curved figure
Can you solve the problem?
#### Third Hint
Taking the dot product of vector $E F$ and vector $F G$ gives $0 \cdot-0.5+0 \cdot 1+-1.5 \cdot 0=0,$ which means the two vectors are perpendicular. (Alternately, as above, note that vector $E F$ goes directly down on the z-axis, while vector $F G$ has no z-component and lie completely in the xy plane.) Thus, the figure is a parallelogram with a right angle, which makes it a rectangle.
Using the distance formula we get $E F=\frac{3}{2} \text { and } F G=\frac{\sqrt{5}}{2}$
Therefore area of the rectangle $EFGH$=$EF \times GH$=$\frac{3}{2} \cdot \frac{\sqrt{5}}{2}$=$\frac{3 \sqrt{5}}{4}$
## What to do to shape your Career in Mathematics after 12th?
From the video below, let's learn from Dr. Ashani Dasgupta (a Ph.D. in Mathematics from the University of Milwaukee-Wisconsin and Founder-Faculty of Cheenta) how you can shape your career in Mathematics and pursue it after 12th in India and Abroad. These are some of the key questions that we are discussing here:
• What are some of the best colleges for Mathematics that you can aim to apply for after high school?
• How can you strategically opt for less known colleges and prepare yourself for the best universities in India or Abroad for your Masters or Ph.D. Programs?
• What are the best universities for MS, MMath, and Ph.D. Programs in India?
• What topics in Mathematics are really needed to crack some great Masters or Ph.D. level entrances?
• How can you pursue a Ph.D. in Mathematics outside India?
• What are the 5 ways Cheenta can help you to pursue Higher Mathematics in India and abroad?
## Want to Explore Advanced Mathematics at Cheenta?
Cheenta has taken an initiative of helping College and High School Passout Students with its "Open Seminars" and "Open for all Math Camps". These events are extremely useful for students who are really passionate for Mathematic and want to pursue their career in it.
To Explore and Experience Advanced Mathematics at Cheenta
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# Knowledge Partner
Cheenta is a knowledge partner of Aditya Birla Education Academy
### Cheenta. Passion for Mathematics
Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
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# Ellipse Formula
View Notes
## What is an Ellipse?
An ellipse is a set of all the points on a plane surface whose distance from two fixed points G and F add up to a constant.
An ellipse mostly looks like a squashed circle:
In this article, we are going to discuss the perimeter of the ellipse formula, the circumference of the ellipse formula, the ellipse volume formula, area of the ellipse formula.
### What are the Properties of Ellipse?
• Ellipse has two focal points, also called foci.
• The fixed distance is called a directrix.
• The eccentricity of the ellipse lies between 0 to 1. 0 ≤ e < 1.
• The total sum of each distance from the locus of an ellipse to the two focal points is constant.
• Ellipse has one major axis and one minor axis and a center.
### Ellipse Formula
1. Area of Ellipse Formula
Area of the Ellipse Formula = πr1r2
Where,
r1 is the semi-major axis of the ellipse.
r2 is the semi-minor axis of the ellipse.
1. Perimeter of Ellipse Formula
Perimeter of Ellipse Formula = $2\pi \sqrt{\frac{r_{1}^{2} + r_{2}^{2}}{2}}$
Where,
r1 is the semi-major axis of the ellipse.
r2 is the semi-minor axis of the ellipse.
1. Ellipse Volume Formula
We can calculate the volume of an elliptical sphere with a simple and elegant ellipsoid equation:
Ellipse Volume Formula = 4/3 * π * A * B * C, where: A, B, and C are the lengths of all three semi-axes of the ellipsoid and the value of π = 3.14.
1. General Equation of an Ellipse
When the centre of the ellipse is at the origin (0, 0) and the foci are on the x-axis and y-axis, then we can easily derive the ellipse equation.
The equation of the ellipse is given by;
x2/a2 + y2/b2 = 1
1. Circumference of Ellipse Formula
Ï€(a + b)
Where,
r1 is the semi-major axis of the ellipse.
r2 is the semi-minor axis of the ellipse.
### Solved Examples
Question 1. Find the area of an ellipse whose semi-major axis is 10 cm and semi-minor axis is 5 cm.
Solution:
Given,
Semi major axis of the ellipse = r1 = 10 cm
Semi minor axis of the ellipse = r2 = 5 cm
Area of the ellipse
= πr1r2
= 3.14 × 10 × 5 cm2
= 157 cm2
### Key Points
• An ellipse and a circle are both examples of conic sections.
• A circle is a special case of an ellipse, with the same radius for all points.
• By stretching a circle in the x or y direction, an ellipse is created. |
## Precalculus (6th Edition) Blitzer
Published by Pearson
# Chapter 7 - Section 7.2 - Systems of Linear Equations in Three Variables - Concept and Vocabulary Check - Page 829: 2
#### Answer
We can eliminate x from equations $1$ and $2$ by multiplying equation $1$ by $-2\text{ }$ and adding the equations. We can eliminate x from equations 1 and 3 by multiplying equation $1$ by $-4$ and adding the equations.
#### Work Step by Step
Let us consider the equations: $x+y-z=-1$ Equation 1 $2x-2y-5z=7$ Equation 2 $4x+y-2z=7$ Equation 3 Now, eliminate x from equations (I) and (II) by multiplying (I) by $-2$ to get; \begin{align} & -2x+\left( -2 \right)y+2z=-1\left( -2 \right) \\ & -2x-2y+2z=2 \\ \end{align} Then, add this equation with equation (II) to get; \begin{align} & -2x-2y+2z+\left( 2x-2y-5z \right)=2+7 \\ & -4y-3z=9 \end{align} Therefore, x is eliminated from equations (I) and (II). Then, eliminate x from equations (I) and (III) by multiplying (I) by $-4$ to get; \begin{align} & x\cdot \left( -4 \right)+y\cdot \left( -4 \right)-z\cdot \left( -4 \right)=-1\cdot \left( -4 \right) \\ & -4x-4y+4z=4 \end{align} Then, add this equation with (III) to get; \begin{align} & -4x-4y+4z+4x+y-2z=4+7 \\ & -3y+2z=11 \end{align} Thus, it is clear from this equation that x is eliminated from equations (I) and (III).
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. |
Lesson Objectives
• Demonstrate an understanding of how to solve a linear equation in one variable
• Learn the definition of a linear equation in two variables
• Learn how to write the solution for a linear equation in two variables using an ordered pair (x,y)
• Learn how to check a solution for a linear equation in two variables
• Learn how to create a table of ordered pairs for a linear equation in two variables
## Linear Equation in Two Variables Definition
Up to this point, we have exclusively worked with equations that contained one variable only. When we solved these equations, we normally ended up with a single solution. Now we will begin to work with more complex equations. A linear equation in two variables is an equation such as:
3x + 5y = 21
Notice how we now have two variables involved (x and y). We have a special method of writing the solution for a linear equation in two variables. This format is known as an ordered pair. As an example, one solution to our equation is:
x = 2, y = 3
This can be written as an ordered pair as:
(2,3)
An ordered pair is listed as the x value first or on the left, and the y value last or on the right. To check if (2,3) is a solution for this equation, we plug in a 2 for x and a 3 for y and simplify. Just as before, we are looking for the left and right side to be the same value.
3x + 5y = 21
(2,3)
3(2) + 5(3) = 21
6 + 15 = 21
21 = 21
Since we have the same value on each side, we can say that (2,3) is a solution for our equation. When we have a linear equation in two variables, there is an infinite number of solutions. This means there is an unlimited number of (x,y) ordered pairs that will satisfy the equation. As another example, we can show that (7,0) also works as a solution.
3(7) + 5(0) = 21
21 + 0 = 21
21 = 21
Let's take a look at a few examples.
Example 1: Determine if each ordered pair is a solution for the equation
-4x + y = 10
(-3,2)
Let's plug in a (-3) for x and a 2 for y:
-4(-3) + 2 = 10
12 + 2 = 10
14 = 10 (false)
(-3,2) is not a solution.
(-2,2)
Let's plug in a (-2) for x and a 2 for y:
-4(-2) + 2 = 10
8 + 2 = 10
10 = 10
(-2,2) is a solution.
(0,10)
Let's plug in a 0 for x and a 10 for y:
-4(0) + 10 = 10
10 = 10
(0,10) is a solution.
Example 2: Determine if each ordered pair is a solution for the equation
3x - 12y = -3
(3,1)
Let's plug in a (3) for x and a 1 for y:
3(3) - 12(1) = -3
9 - 12 = -3
-3 = -3 (true)
(3,1) is a solution.
(4, -2)
Let's plug in a 4 for x and a (-2) for y:
3(4) - 12(-2) = -3
12 + 24 = -3
36 = -3 (false)
(4,-2) is not a solution.
(1,0)
Let's plug in a 1 for x and a 0 for y:
3(1) - 12(0) = -3
3 - 0 = -3
3 = -3 (false)
(1,0) is not a solution.
### Creating a Table of Ordered Pairs
Normally we are not given solutions for our equations, which means we have to generate them. In a few lessons, we will begin graphing linear equations in two variables. In order to be successful in that lesson, we need to understand how to create a table of ordered pairs that satisfy a given linear equation in two variables. Let's suppose we have the following equation:
2x - 5y = 20
Let's suppose we want to create three ordered pair solutions for our equation. What can we do here? We can choose a value for x and solve for y or we can choose a value for y and solve for x. It is usually easy to use 0 when picking a value. Let's see if we can fill in the blank for this ordered pair:
(0, __)
Plug in a 0 for x and solve for y:
2(0) - 5y = 20
0 - 5y = 20
-5y = 20
-5/-5 y = 20/-5
y = -4
(0,-4) is a solution.
Now let's try to use 0 as a value for y and solve for the unknown x. Let's fill in the blank for this ordered pair:
(__ ,0)
2x - 5(0) = 20
2x = 20
2/2 x = 20/2
x = 10
(10, 0) is a solution.
Let's try one more, and get a third and final ordered pair solution. Let's let y = 2. Let's fill in the blank for this ordered pair:
(__ ,2)
2x - 5(2) = 20
2x - 10 = 20
2x = 30
2/2 x = 30/2
x = 15
(15, 2) is a solution.
Let's look at an example.
Example 3: Complete each table of values
-3x + y = -4
x y
0 _
_ -1
5 _
-3(0) + y = -4
0 + y = -4
y = -4
(0,-4)
Let's move to (_,-1):
-3x + (-1) = -4
-3x = -3
-3/-3 x = -3/-3
x = 1
(1,-1)
Let's end with (5,_):
-3(5) + y = -4
-15 + y = -4
y = 11
(5,11)
Let's fill in our table:
x y
0 -4
1 -1
5 11 |
## Simplifying Variable Expressions Using Exponent Properties
### Learning Outcomes
• Use the product property of exponents to simplify expressions
• Use the power property of exponents to simplify expressions
• Use the product to a power property of exponents to simplify expressions
You have seen that when you combine like terms by adding and subtracting, you need to have the same base with the same exponent. But when you multiply and divide, the exponents may be different, and sometimes the bases may be different, too. We’ll derive the properties of exponents by looking for patterns in several examples. All the exponent properties hold true for any real numbers, but right now we will only use whole number exponents.
First, we will look at an example that leads to the Product Property.
${x}^{2}\cdot{x}^{3}$ What does this mean? How many factors altogether? So, we have ${x}\cdot{x}\cdot{x}\cdot{x}\cdot{x}={x}^{5}$ Notice that $5$ is the sum of the exponents, $2$ and $3$. ${x}^{2}\cdot{x}^{3}$ is ${x}^{2+3}$, or ${x}^{5}$ We write: ${x}^{2}\cdot {x}^{3}$ ${x}^{2+3}$ ${x}^{5}$
The base stayed the same and we added the exponents. This leads to the Product Property for Exponents.
### Product Property of Exponents
If $a$ is a real number and $m,n$ are counting numbers, then
${a}^{m}\cdot {a}^{n}={a}^{m+n}$
To multiply with like bases, add the exponents.
An example with numbers helps to verify this property.
$\begin{array}{ccc}\hfill {2}^{2}\cdot {2}^{3}& \stackrel{?}{=}& {2}^{2+3}\hfill \\ \hfill 4\cdot 8& \stackrel{?}{=}& {2}^{5}\hfill \\ \hfill 32& =& 32\hfill \end{array}$
### example
Simplify: ${x}^{5}\cdot {x}^{7}$
Solution
${x}^{5}\cdot {x}^{7}$ Use the product property, ${a}^{m}\cdot {a}^{n}={a}^{m+n}$. $x^{\color{red}{5+7}}$ Simplify. ${x}^{12}$
### example
Simplify: ${b}^{4}\cdot b$
### example
Simplify: ${2}^{7}\cdot {2}^{9}$
### example
Simplify: ${y}^{17}\cdot {y}^{23}$
### try it
We can extend the Product Property of Exponents to more than two factors.
### example
Simplify: ${x}^{3}\cdot {x}^{4}\cdot {x}^{2}$
### try it
In the following video we show more examples of how to use the product rule for exponents to simplify expressions.
### Simplify Expressions Using the Power Property of Exponents
Now let’s look at an exponential expression that contains a power raised to a power. See if you can discover a general property.
$({x}^{2})^{3}$ ${x}^{2}\cdot{x}^{2}\cdot{x}^{2}$ What does this mean? How many factors altogether? So, we have ${x}\cdot{x}\cdot{x}\cdot{x}\cdot{x}\cdot{x}={x}^{6}$ Notice that $6$ is the product of the exponents, $2$ and $3$. $({x}^{2})^{3}$ is ${x}^{2\cdot3}$ or ${x}^{6}$ We write: ${\left({x}^{2}\right)}^{3}$ ${x}^{2\cdot 3}$ ${x}^{6}$
We multiplied the exponents. This leads to the Power Property for Exponents.
### Power Property of Exponents
If $a$ is a real number and $m,n$ are whole numbers, then
${\left({a}^{m}\right)}^{n}={a}^{m\cdot n}$
To raise a power to a power, multiply the exponents.
An example with numbers helps to verify this property.
$\begin{array}{ccc}\hfill {\left({5}^{2}\right)}^{3}& \stackrel{?}{=}& {5}^{2\cdot 3}\hfill \\ \hfill {\left(25\right)}^{3}& \stackrel{?}{=}& {5}^{6}\hfill \\ \hfill 15,625& =& 15,625\hfill \end{array}$
### example
Simplify:
1. ${\left({x}^{5}\right)}^{7}$
2. ${\left({3}^{6}\right)}^{8}$
### try it
Watch the following video to see more examples of how to use the power rule for exponents to simplify expressions.
### Simplify Expressions Using the Product to a Power Property
We will now look at an expression containing a product that is raised to a power. Look for a pattern.
${\left(2x\right)}^{3}$ What does this mean? $2x\cdot 2x\cdot 2x$ We group the like factors together. $2\cdot 2\cdot 2\cdot x\cdot x\cdot x$ How many factors of $2$ and of $x?$ ${2}^{3}\cdot {x}^{3}$ Notice that each factor was raised to the power. ${\left(2x\right)}^{3}\text{ is }{2}^{3}\cdot {x}^{3}$ We write: ${\left(2x\right)}^{3}$ ${2}^{3}\cdot {x}^{3}$
The exponent applies to each of the factors. This leads to the Product to a Power Property for Exponents.
### Product to a Power Property of Exponents
If $a$ and $b$ are real numbers and $m$ is a whole number, then
${\left(ab\right)}^{m}={a}^{m}{b}^{m}$
To raise a product to a power, raise each factor to that power.
An example with numbers helps to verify this property:
$\begin{array}{ccc}\hfill {\left(2\cdot 3\right)}^{2}& \stackrel{?}{=}& {2}^{2}\cdot {3}^{2}\hfill \\ \hfill {6}^{2}& \stackrel{?}{=}& 4\cdot 9\hfill \\ \hfill 36& =& 36\hfill \end{array}$
### example
Simplify: ${\left(-11x\right)}^{2}$
### example
Simplify: ${\left(3xy\right)}^{3}$
### try it
In the next video we show more examples of how to simplify a product raised to a power. |
# Pythagorean Theorem Mixed Questions
Lesson
## Pythagorean Theorem
Pythagorean Theorem states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. The theorem can be written algebraically.
$a^2+b^2=c^2$a2+b2=c2
where $c$c represents the length of the hypotenuse and $a$a, $b$b are the two shorter sides.
So if two sides of a triangle are known and one side is unknown this relationship can be used to find the length of the unknown side.
We can rearrange the equation any which way to make the unknown side the subject.
$c=\sqrt{a^2+b^2}$c=a2+b2
$b=\sqrt{c^2-a^2}$b=c2a2
$a=\sqrt{c^2-b^2}$a=c2b2
#### Examples
##### Question 1
Find the length of the hypotenuse of a right-angled triangle whose two other sides measure $10$10 cm and $12$12 cm.
Think: Here we want to find $c$c, and are given $a$a and $b$b.
Do:
$c^2$c2 $=$= $10^2+12^2$102+122 $c$c $=$= $\sqrt{10^2+12^2}$√102+122 $c$c $=$= $15.62$15.62 cm (rounded to $2$2 decimal places)
##### Question 2
Find the length of unknown side $b$b of a right-angled triangle whose hypotenuse is $6$6 mm and one other side is $4$4 mm.
Think: Here we want to find $b$b, the length of a shorter side.
Do:
$6^2$62 $=$= $b^2+4^2$b2+42 $b^2$b2 $=$= $6^2-4^2$62−42 $b$b $=$= $\sqrt{6^2-4^2}$√62−42 $b$b $=$= $4.47$4.47 mm (rounded to $2$2 decimal places)
Here are some worked examples.
##### Question 3
Calculate the value of $c$c in the triangle below.
##### Question 4
Calculate the value of $b$b in the triangle below.
##### Question 5
Find the length of the unknown side in this right-angled triangle, expressing your answer as a decimal approximation to two decimal places.
We can combine this knowledge of Pythagoras, with our existing work on trigonometric ratios.
Remember the Trigonometric Ratios
$\sin\theta=\frac{opposite}{hypotenuse}$sinθ=oppositehypotenuse
$\cos\theta=\frac{adjacent}{hypotenuse}$cosθ=adjacenthypotenuse
$\tan\theta=\frac{opposite}{adjacent}$tanθ=oppositeadjacent
We can use these ratios to find unknown side lengths and angles in 2D shapes other than triangles, 3D shapes, as well as real-life situations.
The process is exactly the same. We just need to locate the right-angled triangle, the use to applicable trig ratio to work out the unknown value.
Let's look at some examples so we can see this in action.
#### Worked Examples
##### Question 6
Consider a cone with slant height $13$13m and perpendicular height $12$12m.
1. Find the length of the radius, $r$r, of the base of this cone.
2. Hence, find the length of the diameter of the cone's base.
##### Question 7
Find the length of the unknown side, x, in the given trapezoid.
Give your answer correct to $2$2 decimal places.
### Outcomes
#### 9P.MG2.02
Solve problems using the Pythagorean theorem, as required in applications (e.g., calculate the height of a cone, given the radius and the slant height, in order to determine the volume of the cone) |
### Mental Math - Tails of 5 Multiplication Method
Problem: Calculate the value of the expression
45 x 65
Method: We can use the Tails of 5 method to help solve this problem.
The tail of a number is defined to be the last digit of the number. The head of the number is defined to be all of the digits except last digit. If you want to multiply two numbers, whose tails are both 5, then you can use the Tails of 5 method to multiply the two numbers quickly.
Let's take a look at multiplying 45 times 65. These two numbers both have a tail of 5. We know that the product of these two number will end in 25 (or possibly 75). The other digits of the product can be determined by multplying the heads of the two numbers and then adding the average of the two heads. In this case the two heads are 4 and 6. When we multiply 4 times 6 we get 24. The average of 4 and 6 is 5 so we add the 5 to 24 and get 29. So our product starts with 29 and ends with 25. Therefore, the product of 45 x 65 is 2925.
Problem: Calculate the value of the expression
35 x 125
Method: We can use the Tails of 5 method to help solve this problem.
The average of the two head will either be a whole number or contain the fraction ½. If the average of the two heads contains the fraction ½, then instead of ending in 25, the product will end in 75 (as 100 times ½ is 50).
To multiply 35 times 125, we observe that both numbers end in 5 which means that we can use the Tails of 5 method to calculate the product. We multiply the heads of the two numbers, 3 and 12, to get 36. Next we take the average of 3 and 12, and we get 7½. The ½ means that the product will end in 75 rather than 25. We add the 7 from the average of the two heads to the product of the two heads, 36, and we get 43.
So, the product of 35 times 125 is equal to 4375. |
# Multiplication of Decimals
We will discuss here about the multiplication of decimals. To multiply a decimal by a whole number, first we multiply as usual without considering decimal point. Then we put the decimal point in the product at the same place where it is placed in the given decimal.
Let us consider some of the following examples on multiplying decimals.
1. Multiply: 7.32 by 12.
7 . 3 2
× 1 2
1 4 6 4
7 3 2 0
8 7 . 8 4
The decimal point in decimal 7.32 is after two digits from the extreme right. So, we place the decimal point in the product just after two digits from the extreme right.
So, the product of 7.32 and 12 is 87.84.
2. Multiply the following:
(i) 12.345 by 8
1 2 . 3 4 5
× 8
9 8 . 7 6 0
Decimals are multiplied in the same way as we multiply ordinary numbers. We place the digits in columns and then multiply as usual.
The decimal point in decimal 12.345 is after three digits from the extreme right. So, we place the decimal point in the product just after three digits from the extreme right.
So, the product of 12.345 and 8 is 98.760
(ii) 25.381 by 6
2 5 . 3 8 1
× 6
1 5 2 . 2 8 6
The decimal point in decimal 25.381 is after three digits from the extreme right. So, we place the decimal point in the product just after three digits from the extreme right.
So, the product of 25.381 and 6 is 152.286. |
# Multiplying and Dividing Complex Numbers Worksheets
How to Multiply and Divide Complex Numbers? When dealing with complex numbers, you have to be extra careful in resolving them. Especially if you have to multiply and divide them. Let us discuss both of these scenarios individually. Multiplication - Multiplying two or more complex numbers is similar to multiplying two or more binomials. Let us consider an example.
(3 + 2i)(2 - i)
Here, you will have to use the distributive property to write the above equation as:
3(2 - i) + 2i(2 - i)
6 - 3i + 4i - 2i2
Now, you must remember that i2=-1, thus, the equation becomes:
6 - 3i + 4i + 2
8+i
Division - Dividing complex numbers is just as simpler as writing complex numbers in fraction form and then resolving them. Let us consider an example:
In this situation, the question is not in a simplified form; thus, you must take the conjugate value of the denominator.
• ### Basic Lesson
Guides students solving equations that involve an Multiplying and Dividing Complex Numbers. Demonstrates answer checking. Example: Multiply: (5 - 4i)(7 - 3i)
• ### Intermediate Lesson
Demonstrates how to solve more difficult problems. Simplify: (6 - 8i)2
• ### Independent Practice 1
A really great activity for allowing students to understand the concept of Multiplying and Dividing Complex Numbers.
• ### Independent Practice 2
Students find the Multiplying and Dividing Complex Numbers in assorted problems. The answers can be found below.
• ### Homework Worksheet
Students are provided with problems to achieve the concepts of Multiplying and Dividing Complex Numbers.
• ### Skill Quiz
This tests the students ability to evaluate Multiplying and Dividing Complex Numbers.
Answers for math worksheets, quiz, homework, and lessons.
• ### Another Lesson
Guides students solving equations that involve an Multiplying and Dividing Complex Numbers. Demonstrates answer checking.
• ### Simplifying Lesson
Demonstrates how to solve more difficult problems.
• ### Independent Practice 3
A really great activity for allowing students to understand the concept of Multiplying and Dividing Complex Numbers.
• ### Independent Practice 4
Students find the Multiplying and Dividing Complex Numbers in assorted problems. The answers can be found below.
• ### Homework Worksheet 2
Students are provided with problems to achieve the concepts of Multiplying and Dividing Complex Numbers.
• ### Quiz 2
This tests the students ability to evaluate Multiplying and Dividing Complex Numbers.
• ### Lesson and Practice Answer Key
Answers for math lessons and practice sheets above. |
## SCIE 4101 Review Packet #2 Solutions March 19, 2013
Posted by Ms. Miller in UTA.
Please let me know if you see any errors.
1. We want the relationship between the terms and their positions. We can see that each term goes up by 5, so that is our “slope”, and if we subtract 5 from the first term, that gives us our “y-intercept”: $5n-2$.
2. This is not a function because an element in the domain set is mapped to more than one element in the codomain set.
3. To write the equation, we first must find the slope. Pick two points and plug into the slope formula, for example, (2,5) and (4,13):
$m=\frac{13-5}{4-2}=\frac{8}{2}=4$
Then, we take our slope and use either the point-slope form or the slope-intercept form to write the equation:
Point-Slope Form Slope-Intercept Form
$y-y_{1}=m(x-x_{1})$
$y-5=4(x-2)$
$y=4x-3$
$y=mx+b$
$5=4(2)+b$
$b=-3$
$y=4x-3$
4. To find the set for the independent variable, we set the function equal to each member of the dependent variable and solve for x:
$x^2-8=-8$ $x^2=0$ $x=0$ $x^2-8=-7$ $x^2=1$ $x=\pm 1$ $x^2-8=8$ $x^2=16$ $x=\pm 4$ $x^2-8=17$ $x^2=25$ $x=\pm 5$
Therefore, the set for the independent variable is $\{0, \pm 1, \pm 4, \pm 5\}$.
5. The cost of electricity decreases $1580 for every 3 degrees rise in temperature. Therefore, if the thermostat is set at 80° (2*3°), the cost would decrease$3160 (2*1580), so a reasonable cost would be $9510-3160 = 6350$.
6. To graph each equation, plot the y-intercept and then use the slope to find a second point:
a.
b.
c.
7. To solve this problem, we need to set up a system of equations. The first equation is the relation between the values of the coins: $0.10d+0.25q=3.20$. The second is the relation between the quantities: $d+q=20$. I would use substitution to work this: rewriting the second equation as $d=20-q$ and substitute it for $d$ in the first equation: $0.10(20-q)+0.25q=3.20$. Then, solve for $q$: $q=8$. Substitute this back into the first equation and solve for $d$: $d=20-8=12$. Thus, Lindsey had 8 quarters and 12 dimes.
1. So the solution is $(-3,2)$.
2. I would use either elimination or matrices:
$3x+2y=21$
$\underline{4x-2y=14}$
$7x=35$
$x=5$
Substitute this into one of the equations to find y:
$3(5)+2y=21$
$15+2y=21$
$2y=6$
$y=3$
Therefore, the solution is $(5,3)$.
3. You can either rewrite the first equation into “y=” form and graph, you can rewrite the second to $3x+y=-5$ and use elimination, or you can substitute the second into the first, solve for x, and then substitute it back to solve for y. In any event, the solution is $(-\frac{6}{5},-\frac{7}{5})$.
8. To write an inequality from a graph, first determine the equation for the line: the slope is 2, and the y-intercept is -3, so the equation of the line is $y=2x-3$. It is shaded above the line, and the line is solid, so we know it is greater than or equal to. Therefore, the inequality is $y\geq 2x-3$.
9. To graph a system of inequalities, graph each inequality, and the solution is the intersection:
10. To graph these inequalities, first solve for y:
$y<-x+6$
$y>\frac{3}{4}x-2$
11. Direct variation is of the form $y=kx$ and inverse variation is of the form $y=\frac{k}{x}$.
1. As x increases, y increases, so this is a direct variation. Substituting into the formula, we have $6=2k$, which means that $k=3$.
2. As x increases, y decreases, so this is an inverse variation. Substituting, we have $12=\frac{k}{0.2}$, which means that $k=2.4$.
12. Substitute this into the inverse variation form: $12=\frac{k}{8}$, which means that $k=96$.
13. If we plot the points, we can see that the coordinates for point A are $(2,6)$:
To write the equation of a line from two points, we first find the slope: $m=\frac{2-6}{5-2}=\frac{-4}{3}=-\frac{4}{3}$. Next, we can use either the slope-intercept form and find the y-intercept or use the point-slope form:
$y-2=-\frac{4}{3}(x-5)$
$y=-\frac{4}{3}x+\frac{20}{3}+\frac{6}{3}$
$y=-\frac{4}{3}x+\frac{26}{3}$
15.
a. $2x^3+4x^2+4x-7$ b. $s^2-3t^2$
16.
a. $20m^5$ b. $-12x^4y^7$ c. $-3a^5b^3c^2$
17.
a. $5r^3s^2-15r^2s^3$ b. $-10x^3+15x^2+5x$ c. $10x^5y^5-5x^4y^6$
18.
a. $x^2+x-6$ b. $7x^2-52x-32$ c. $m^2-10m+25$ d. $4a^2-9b^2$ e. $9c^2-48cd+64d^2$ f. $12x^2-35x+8$ g. $25g^6-20g^3+4$ h. $9t^4-100$
19.
a. $5y(2y^2+4y-1)$ b. $-4x(3+2x)$ c. $7ab^2(2ab^2+9b-1)$
20.
a. $(x+5)(x+2)$ b. $(x-4)(x-2)$ c. $(x+9)(x-2)$ d. $5x^2-10x-4x+8$$5x(x-2)-4(x-2)$$(5x-4)(x-2)$ e. $3x^2+9x+4x+12$$3x(x+3)+4(x+3)$$(3x+4)(x+3)$ f. $2b^2(5b-8)+5(5b-8)$$(2b^2+5)(5b-8)$ g. $(x-5)^2$ h. $(10x-7)(10x+7)$ i. $(x^2-4)(x^2+4)$$(x-2)(x+2)(x^2+4)$
1. The quadratic function crosses the x-axis at 0 and 8. The coordinates of the vertex are $(4,4)$. The axis of symmetry is a vertical line that goes through the vertex, so its equation is $x=4$.
2. From the equation, $a=5, b=-10,$ and $c=3$. Therefore, the x-coordinate of the vertex is $x=\frac{10}{(2)(5)}=1$. (If we wanted to find the y-coordinate, we would just plug the x-coordinate into the original equation.)
3. The larger the absolute value of the number in front of $x^2$, the narrower the graph: (1) $h(x)$, (2) $f(x)$, (3) $g(x)$.
24.
a. $x=3,-7$ b. $(x+1)^2=0$$x=-1$ c. $x^2+2x-8=0$$(x+4)(x-2)=0$$x=-4,2$ d. $x^2+12x+36=0$$(x+6)^2=0$$x=-6$ e. $2x^2-2x+5x-5=0$$2x(x-1)+5(x-1)=0$$(2x+5)(x-1)=0$$x=-\frac{5}{2},1$ f. $3x^2-13x+4=0$$3x^2-12x-x+4=0$$3x(x-4)-1(x-4)=0$$(3x-1)(x-4)=0$$x=4,\frac{1}{3}$ g. $x=\frac{4\pm\sqrt{(-4)^2-4(1)(1)}}{(2)(1)}$$x=\frac{4\pm\sqrt{16-4}}{2}$$x=2\pm\sqrt{3}$ h. $(2x+1)^2=0$$x=-\frac{1}{2}$ i. $2x^2+x-7=0$$x=\frac{-1\pm\sqrt{1^2-4(2)(-7)}}{(2)(2)}$$x=\frac{-1\pm\sqrt{1+56}}{4}$$x=\frac{-1\pm\sqrt{57}}{4}$
25.
a. $x^2+8x+16=0$$D=64-4(1)(16)=0$1 real solution$x=\frac{-8}{(2)(1)}=-4$ b. $D=1-4(0.5)(-3)=7$2 real solutions$x=\frac{-1\pm\sqrt{7}}{(2)(0.5)}=-1\pm\sqrt{7}$ c. $-3x^2-2x-1=0$$D=(-2)^2-4(-3)(-1)=-8$No real solutions |
# Video: Writing and Solving Multiplication Linear Equations in a Real-World Context Involving Fractions
A rectangle’s width is one-sixth of its length. Given that the rectangle’s width is 9 inches, determine its length.
02:06
### Video Transcript
A rectangle’s width is one-sixth of its length. Given that the rectangle’s width is nine inches, determine its length.
We will answer this question by firstly drawing a diagram and then setting up a one-step linear equation. Let’s consider a rectangle with width 𝑊-inches and length 𝐿-inches. We’re told in the question that the width is one-sixth of the length. The word “of” in mathematics means multiply. So 𝑊 is equal to one-sixth multiplied by 𝐿. This in turn can be written as 𝑊 equals one-sixth 𝐿 or 𝐿 divided by six.
In this particular question, we’re told that the rectangle’s width is nine inches. We can substitute this into our equation so that nine is equal to one-sixth 𝐿. This is the same as nine is equal to 𝐿 divided by six. In order to solve this equation, we need to perform the same operation on both sides of the equal sign. In this case, we will multiply by six as multiplying by six is the opposite or inverse of dividing by six.
On the left-hand side, six multiplied by nine or nine multiplied by six is equal to 54. On the right-hand side, the sixes cancel. And we’re just left with 𝐿. As 𝐿 is equal to 54, we can conclude that the length of the rectangle is 54 inches. We can check this answer by working out one-sixth of 54. As this is equal to nine, which was the rectangle’s width, we know that our answer of 54 inches is correct. |
# Order of Operations.
## Presentation on theme: "Order of Operations."— Presentation transcript:
Order of Operations
When an expression contains more than one operation, you can get different answers depending on the order in which you solve the expression.
Here are some examples…
Problem Without Order of Operations Following Order of Operations x 2 48 x 2 96 84 32 + 8 112 121 9 + 8 17 (4 + 5) x 7 4 + 35 39 9 x 7 63
Mathematicians have agreed on a certain order for evaluating expressions, so we all arrive at the same answers. We often use grouping symbols, like parentheses, to help us organize complicated expressions into simpler ones.
“Please Excuse My Dear Aunt Sally”
is a helpful way to remember these rules for the order of operations.
First, do all operations that lie inside parentheses.
“Please” First, do all operations that lie inside parentheses. Example: (3 + 12)2 + 3 x 9 (15)2 + 3 x 9
Next, do any work with exponents or roots.
“Excuse” Next, do any work with exponents or roots. Example: (15)2 + 3 x 9 x 9
Working from left to right, do all multiplication and division.
“My Dear” Working from left to right, do all multiplication and division. Example: x 9
Finally, working from left to right, do all addition and subtraction.
“Aunt Sally” Finally, working from left to right, do all addition and subtraction. Example: 252
Be sure to follow the order of operations within the parentheses.
Important Note: Be sure to follow the order of operations within the parentheses.
“Please Excuse My Dear Aunt Sally”
Remember… “Please Excuse My Dear Aunt Sally” Parentheses, Exponents, Multiplication & Division, Addition & Subtraction
Some examples…. 8 x 22 + 7 x 5 67 (4 + 3 x 2)2 - 5 55 *Exponents*
*Multiplication* * Addition* 67 (4 + 3 x 2)2 - 5 (4 + 3 x 2) - 5 *Parentheses - Multiplication* (4 + 6) - 5 (4 + 6) -5 *Parentheses – Addition* 10 - 5 *Subtraction* 55 |
Knowing Your Functions: Understanding Even and Odd Functions
Introduction
Functions are essential in the field of mathematics, and they play a significant role in various applications such as engineering, physics, and finance. Understanding their properties such as even and odd functions can be beneficial in analyzing and solving problems. In this article, we will explore the definition of even and odd functions, their identification, and their practical applications.
An even function is defined as a function “f” where “f(-x) = f(x)” for all values of “x,” while an odd function is defined as a function “f” where “f(-x) = -f(x)” for all values of “x”.
Some examples of even functions are:
• f(x) = x^2
• g(x) = |x|
Meanwhile, some examples of odd functions are:
• f(x) = x^3
• g(x) = -x
This article aims to provide a step-by-step guide to determine if a function is even or odd, as well as its applications, mistakes to avoid, and practical tips.
How to Determine if a Function is Even or Odd
Determining if a function is even or odd is an essential skill in mathematics. Follow these steps to determine if a given function is even or odd:
1. Substitute “-x” into the function.
2. If the resulting function is equivalent to the original function, then the function is even.
3. If the resulting function is equivalent to the opposite of the original function, then the function is odd.
Here are the step-by-step instructions with examples:
Example 1: Determine if f(x) = x^4 is even or odd.
1. Substitute “-x” into f(x): f(-x) = (-x)^4 = x^4.
2. Since the resulting function is equal to the original function, f(x) is even.
Example 2: Determine if f(x) = x^3 is even or odd.
1. Substitute “-x” into f(x): f(-x) = (-x)^3 = -x^3.
2. Since the resulting function is equal to the opposite of the original function, f(x) is odd.
When determining if a function is even or odd, be sure to avoid common mistakes such as incorrect substitution, algebraic errors, or assuming a function is odd or even without verification.
Application of Even and Odd Functions
The concepts of even and odd functions have significant applications in various fields such as engineering, finance, and physics. These applications include:
Example Problems
Solving even and odd functions can apply in problem-solving. For example:
Example 1: Find the even and odd parts of a function f(x) = e^x + e^(-x).
1. Separate the function into even and odd parts:
• Even: f_even(x) = (e^x + e^(-x))/2
• Odd: f_odd(x) = (e^x – e^(-x))/2
Example 2: Find the derivative of the odd function g(x) = x^3 + 4x.
1. Identify that the derivative of an odd function is an even function.
2. Take the derivative of the function:
• g'(x) = 3x^2 + 4
3. Simplify the derivative to an even function:
• g_even(x) = 3x^2 + 4
Methods to Identify Functions
The identification of a function as even or odd can help solve mathematical problems easily. For example, recognizing a function as even can allow you to simplify integrals significantly.
Tips to Avoid Common Mistakes
When working with even and odd functions, be sure to avoid common mistakes such as:
• Incorrect substitution: Always double-check your substitutions to ensure that you have an equivalent function.
• Algebraic errors: Be sure to verify your calculations and avoid basic algebraic mistakes such as errors in sign, multiple terms, or constants.
• Assuming a function is odd or even without verification: Verify that a function is odd or even according to its definition.
Using Visual Aids and Graphics
Visual aids such as graphs can help illustrate the concept of even and odd functions. The graphs of an even function will display symmetry about the y-axis while an odd function graph will show rotational symmetry about the origin.
Here’s an example:
The above graph illustrates the symmetry property of even functions. The function x^2 is an even function, and the red line denotes the line of symmetry (y-axis).
To determine symmetry, follow these steps:
1. Substitute “-x” into the function.
2. Graph the original function and the resulting function.
3. If the graphs are symmetric about the y-axis, then the function is even.
4. If the graphs are symmetric about the origin, then the function is odd.
Real-world Scenarios
Even and odd functions appear in many real-world scenarios, including engineering, physics, and finance.
Applications in Engineering
In engineering, even and odd functions are essential in fields such as electricity, aeronautics, and mechanical engineering. In these fields, even functions are used to represent even functions such as stress on plane surfaces, while odd functions represent odd functions like torque.
Applications in Finance
In finance, even and odd functions are used to model data such as stock prices and interest rates. Even functions can represent symmetric data distribution, while odd functions describe asymmetric data distribution.
Applications in Physics
In physics, even and odd functions appear in fields such as optics, electromagnetism, and quantum mechanics. Even functions are used to represent even functions like electric fields, while odd functions describe odd functions such as magnetic fields.
Examples of Even and Odd Functions
Here are some examples of even and odd functions in real-world scenarios:
• Even function: Modeling data in finance such as stock prices.
• Odd function: Describing rotational motion such as the torque of an object.
Tips to Identify Functions in These Fields
When identifying functions in engineering, finance, or physics, the same methodology applies as with other functions. Always remember the definition of even and odd functions, and verify your work to avoid common errors.
Functions That Are Neither Even nor Odd
Not all functions are even or odd. For example, a function is neither even nor odd if its symmetry does not satisfy the conditions of the definitions of even and odd functions.
Here’s an example:
• f(x) = x^2 + x
This function is neither even nor odd since it does not have the required symmetry.
When working with functions that are neither even nor odd, avoid common mistakes such as assuming that they are even or odd or misinterpreting their properties.
Comprehensive Overview of Even and Odd Functions
Even and odd functions are crucial mathematical concepts with practical applications in various fields such as engineering, finance, and physics. Identifying even and odd functions involves substituting “-x” into the function and verifying the symmetry of the resulting function graph. Graphs of even functions are symmetric about the y-axis, while graphs of odd functions are symmetric about the origin. Functions that are neither even nor odd do not satisfy the conditions of the definition for even or odd functions.
To work with even and odd functions effectively, be sure to avoid common mistakes such as incorrect substitution, algebraic errors, and assuming that a function is odd or even without verification. Understanding even and odd functions and their applications can simplify problem-solving and lead to more efficient solutions.
Conclusion
In conclusion, even and odd functions are critical concepts in mathematics. This article has provided a definition of even and odd functions, their identification, and their practical applications. Understanding the concepts of even and odd functions can provide significant benefits in problem-solving and their real-world applications. Remember to avoid common mistakes when working with even and odd functions and always verify your work to achieve the best results. |
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# Draw a colored craft stick from the bag. Form teams according to the color of your stick. Green, Blue, Red, or Brow
Draw a colored craft stick from the bag. Form teams according to the color of your stick. Green, Blue, Red, or Brown. POD . . . In times of war, our government has set up a selective-service, or draft system to identify young men who will be called into military service.
## Draw a colored craft stick from the bag. Form teams according to the color of your stick. Green, Blue, Red, or Brow
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### Presentation Transcript
1. Draw a colored craft stick from the bag. Form teams according to the color of your stick. Green, Blue, Red, or Brown
2. POD . . . In times of war, our government has set up a selective-service, or draft system to identify young men who will be called into military service. For reasons of fairness, the selective- service system must make every effort to be sure that everyone eligible for the draft has the same chance of being selected. What do you think would be a fair system for selecting people to be drafted? Respond to this question, using complete sentences, in your journal.
3. SAMPLES AND POPULATIONS Investigation Three
4. Mathematical and Problem-Solving Goals To select a random sample from a population To use sampling distributions, measures of center, and measures of spread to describe and compare samples To use data from samples to estimate a characteristic of a population To apply elementary probability work with spinners or calculators to choose random samples of data ACOS 13 and 14
5. Investigation 3.1Choosing Randomly Imagine that you have two tickets to a sold-out rock concert, and your six best friends all want to go with you. To choose a friend to attend the concert, you want to use a strategy that gives each friend an equally likely chance of being selected. Which of the three strategies below would accomplish this? Explain your reasoning. Strategy 1: The first person who calls you on the phone tonight gets to go with you. Strategy 2: You assign each friend a different whole number from 1 to 6. Then, you roll a six-sided number cube. The number that is rolled determines who attends the concert. Strategy 3: You tell each friend to meet you by the rear door right after school. You toss a coin to choose between the first two friends who arrive.
6. One way to select a random sample of students is to use two spinners like these: You can use the spinners to generate random pairs of digits that correspond to the two-digit student numbers. What two-digit numbers can you generate with these spinners? How can you make sure that student 100 has an equally likely chance of being included in your sample? There are many other ways to select a random sample of students. For example, you can roll two 10-sided number cubes, or you can generate random numbers with your calculator.
7. Investigation 3.1 Follow-Up Describe another strategy you could use that would give each of your friends an equally likely chance of being selected.
8. Investigation 3.2Selecting a Random Sample Take a look at the data in the table concerning 8th graders and their sleep hours. This table contains a massive amount of information. You could work with the entire set of data or you could select a random sample of students by looking for patterns in the data for the sample and then use your finding to make predictions about the population. What methods might you use to select a random sample of students? How many students would you need in your sample in order to make accurate estimates of the typical number of hours of sleep and the typical number of movies watched for the entire population of 100 students?
10. Mathematical and Problem-Solving Goals • To select a random sample from a population • To use sampling distributions, measures of center, and measures of spread to describe and compare samples • To use data from samples to estimate a characteristic of a population • To apply elementary probability work with spinners or calculators to choosing random samples of data • ACOS 13 and 14
11. Investigation 3.2 Follow-Up Select a random sample of 25 students, and record the number of hours of sleep for each students. Calculate the five-number summary for these sleep data, and make a box plot of the distribution. Use your findings to estimate the typical hours of sleep for the population of 100 students. Compare your box plot and estimate with those of the other members of your group, and describe the similarities and differences.
12. 2. The data on page 39 were collected by conducting a survey. The students who wrote the survey considered two possible questions for finding the number of movies watched. How many movies and videos did you watch last week? How many movies did you watch at a theater, on television, or on video last week? Include all movies and videos you watched from last Monday through this Sunday. Which question do you think is better? Why? Can you write a better question? If so, write one, and explain why you think your question is better.
13. Investigation 3.3Choosing a Sample Size In this problem, you will explore how the size of a sample affects the accuracy of statistical estimates. A. In Problem 3.2, you calculated five-number summaries for the movie data for random samples of 25 students. Work with your class to make a line plot of the medians found by all groups. Compare these results with the median for the population of 100 students.
14. Investigation 3.3Choosing a Sample Size • B. 1. Select three random samples of 5 students, and find the median movie value for each sample. Compare the medians for your samples with the population median. • 2. Compare the medians for your samples with the medians found by other members of your group. Describe the similarities or differences you find.
15. Investigation 3.3Choosing a Sample Size • B. 3. Record the medians found by your group on the board. When all groups have recorded their medians, make a line plot of the medians. • C.1. Select the random samples of 10 students, and find the median movie value for each sample. Compare the medians for your samples with the population median.
16. Investigation 3.3Choosing a Sample Size • C. 2. Compare the medians for your samples with the medians found by other members of your group. Describe the similarities or differences you find. • C. 3. Record the medians found by your group on the board. When all groups have recorded their medians, make a line plot of the medians.
17. Investigation 3.3Choosing a Sample Size • D. Compare the distribution of medians for samples of size 5, 10, and 25. Write a paragraph describing how the median estimates for samples of different sizes compare with the actual population median.
More Related |
# Number Divided By Fraction Calculator
Number Divided by Fraction Calculator of ours will tell you How to Calculate Number Divided by a Fraction in a matter of seconds. You just have to type in the values of both whole numbers and fractions in the input sections and then click on the calculate button to avail the result.
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## How to Divide a Number by a Fraction?
Go through the steps below to divide a number by a fraction easily. They are listed in the following way
• Firstly obtain the given number and fraction.
• Convert the given number into fractions by simply placing 1 as the denominator.
• Then find the reciprocal of the divisor i.e. reverse both the numerator and denominator
• Now multiply the two fractions i.e. numerators together and denominators together.
• If possible simplify the result.
### Solved Examples of Dividing Number by a Fraction
Example 1.
Divide 5 by 3/4.
Solution:
Given values are 5 and 3/4
To convert a given whole number to a fraction we will add 1 as the denominator i.e. 5/1
Now, finding the reciprocal of divisor 3/4 we get 4/3
Multiplying fractions together we have (5/1) x(4/3)
= (5x4)/(1x3)
= 5/12
As we can’t further reduce the given fraction it is the result after dividing 5 by 3/4
Therefore, 5 ÷ 3/4 is 5/12
Example 2.
Divide 6 by 2/3
Solution:
Given Values are 6 and 2/3
Changing 6 to fraction we get 6/1
Reciprocal of Divisor 2/3 is 3/2
Multiplying fractions we get (6/1) x (3/2)
= (6x3)/(1x2)
= 18/2
= 9
Therefore, 6 ÷ 2/3 is 9.
Avail many similar concept calculators free of cost from Roundingcalculator.guru and resolve all your doubts in no time.
### FAQs on What is Number Divided by Fraction
1. What is the Rule for Dividing Fractions?
The rule for Dividing Fractions is Keep, Flip and Change.
2. How to Divide Whole Numbers by Fractions?
Write the number as a fraction and find the reciprocal of the divisor. Next, multiply the fractions together and simplify if possible.
3. How to use the Number Divided by Fraction Calculator?
Just enter the values of both the number and fraction in the respective fields of the calculator and click on the calculate button to obtain the result of the number divided by a fraction. |
# JSS 1 SECOND MATHEMATICS LESSON NOTE
SUBJECT:
MATHEMATICS
CLASS:
JUNIOR SECONDARY SCHOOL 1
SCHEME OF WORK
WEEK 1 REVISION OF LAST WORK
WEEK 2 APPROXIMATION
WEEK 3 APPROXIMATION
WEEK 4 NUMBER BASE
WEEK 5 NUMBER BASE
WEEK 6 BASIC OPERATION
WEEK 7 REVIEW OF FIRST TERM
WEEK 8 BASIC OPERATION
WEEK 9 ALGEBRAIC PROCESS
WEEK 10 ALGEBRAIC PROCESS
WEEK 11 REVISION
WEEK 12 EXAMINATION
WEEK 2: APPROXIMATION
2.0 Introduction:
Approximation is the process of using rounded numbers, to estimate the outcome of calculations. Approximation can help us decide whether an answer to a calculation is right or not. We can approximate numbers by rounding them up to decimal places, significant figures, nearest whole number, tens , hundreds etc.
2.1 Decimal Place/Point:
A number such as 197.7658 is an example of a decimal number. The whole number part is 197 and the decimal part is 7658. The point between them is called a decimal point.
197.7658
Whole number. Decimal
Decimal point
To find the number of decimal places (d.p) simply count the number of figures after the decimal point. Thus, the number above has 4 d.p.
197.7658
4th decimal place
3rd decimal place
2nd decimal place
1st decimal place
Guide to round off numbers:
To round off numbers to a specific number of decimal places
1. Look for the last digit (i.e. the required decimal place you are rounding to)
2. Then look at the next digit to the right, i.e. the decider
3. If the decider is 5 or more round up (i.e. add 1 to the last digit) if it is less than 5, then add nothing.
Examples2.1: Give the number 78.05624 correct to (a) 1 d.p (b) 2 dp (c) 3 dp
Solution:
1. 78.1 (start counting from the number after the point. Note that zero is significant).
2. 78.06 (after counting the number check the next number, if the number is less than 5/<; it changes to zero but if it is 5 or > it changes to 1 and add to the next counted before it.)
3. 78.056 (since the number is less it turns to zero and it has no significance)
Example 2.2: Give 57.9945 correct to (a) 2dp (b) 4dp
Solution:
(a) 57.99 (2 d.p.)
(b) 57.9945 (4 d.p.)
2.2 Significant Figure:
The word significant means ‘important or non zero digits’ and it is another way of approximating numbers. We write significant figure as S F.
Numbers greater than zero
For example the number 865 034 has six figures or digits. The first figure from the left i.e. 8 is worth 800 000 (place value) and it is the most significant figure. It is therefore the first significant figure and 4 the leas or sixth significant figure.er
987654
1sts.f.2nds.f.3rds.f.4ths.f.5ths.f.6ths.f.
Numbers less than zero
For example 0.000007685 is given to 8 decimal places. The zero before the decimal number means that there are no units, and the 5 zeros after the decimal point mean that they are insignificant figures. Therefore the most significant number or first significant is 7 follow by 6 and 8.
1. 0 0 0 0 0 7 6 8
1sf 2sf 3sf
Note: the first significant figure is always the first non-zero figure as you read a number from the left
Guide to round off numbers
To correct a number to a specific number of significant figure
1. Look for the required significant figure
2. Look at the next significant figure to the right (i.e. the decider)
3. If the decider is 5 or more round up but if it is less than 5 add nothing
Example 2.3:Give 4540 correct to (a) 1 s.f (b) 2 sf (c) 3 s.f
Solution:
Note the figures above, it helps just follow the figure correctly.
1. 4540 = 5000 (the reason for getting 8000.after counting 7 the next number is 5 which will turn to 1 and add to the 7 counted and the other numbers turn to zeros).
2. 4540 = 4500 (after counting 2 s.f the next number is 6 which turn to 1 and add to 5 to be 6 and the rest turn to zeros.
3. 4540 = 4540 (after counting 3 s.f the next is 4 which turn to zero).
Example 2.4: Convert 0.00005791 to (a) 2 s.f (b) 3 s.f
Solution:
(a) 0.000058
(b) 0.0000579
2.3 Rounding Decimals to the Nearest, Tenth, Hundredth, and Thousandths and Whole.
Example2.5:Give474.4547correcttothenearest
1. Tenth
2. Hundredth
3. Thousandth
Solution:
Note That The Counting Starts After The Point.
1. 474.4547=474.5(since the next number after counting is 5/>it turns to 1 and add to the counted 4)
2. 474.4547=474.45
3. 474.4547=474.455
Example 2.6:Roundoff
(a)13.73
(b)34.245tothe nearest whole number.
Solution:
1. 13.73=14(since the whole is 13 the next number is decimal which is 7,it will be turned to 1 and add to 13 to make it 14).
2. 34.245=34(since the next is less than 5 then it is insignificant).
ASSIGNMENT: ESSENTIAL MATHEMATICS BOOK 1
EXERCISE 8.5 page 89 NO 2(a,b,c,d), 5+ (a,b,c)
WEEK 3: ADDITION AND SUBTRACTION OF APPROXIMATIONS
Example 3.1: approximate the following to it accuracy of degree
1. 674 + 975 (3 s.f)
2. 805 + 912 (4s.f)
3. 3076 – 621 (2 s.f)
4. 695.728 – 97.979 (4 s.f)
Solutions:
1. 674 + 975 = 1649 (165 to 3 s.f.) note that you must add the numbers before converting to the degree.
2. 805 + 912 = 1717 (1717 to 4s.f.)
3. 3076 – 621 = 2455 (250 to 2 s.f.)
4. 695.728 – 97.979 = 597.749 (597.7 to 4 s.f.)
EVALUATION: Convert the following to their accuracy degree
• 42.4739 to 5 s.f
• 67.3258 to 3s.f
• 879.9087 to 4 s.f
• 304.385 to 5 s.f
ASSIGNMENT: ESSENTIAL MATHEMATICS BOOK 1
EXERCISE 8.7 NO 1 (g,h,I,j,k,l and m) PAGE 93
EXERCISE 8.8 NO 7,8 and 9 PAGE 95
EXERCISE 8.5 NO 1(D,E,F), NO 4 (B &C) N0 5 (A TO G) PAGE 89
WEEK 4: NUMBER BASE
4.0 Introduction:
The usual system of counting in our days is called the decimal or denary system. The denary or decimal system is also called base ten. This system enables us to be able to write small or large numbers using the combination of the digits i.e. 0,1,2,3,4,5,6,7,8,9.
4.1 Expansion and Conversion to Base Ten
Expanded Notation
Example 4.1: Write the following in expanded nation form.
(a) 11011002
(b) 21356
(c) 45678
Solution:
1. 16150413120100= (1 x 26) + (1 x 25) + (0 x 24) + (1 x 23) + (1 x 22) + (0 x 21) + (0 x 20).
The base is used to expand it and the power for each expansion.
(b) 23123150 = (2 X 63) + (1 x 62) + (3 x 61) + (5 x 60)
(c) 43526170 = (4 x 83) + (5 x 82) + (6 x 81) + (7 x 80).
4.2 Conversion to Base Ten
Example 4.2: convert the numbers in example 1 to Base ten
Solution:
In other to do this we simply continue from the expanded notation, evaluate and get our answers.
1. 16150413120100=(1×26)+(1×25)+(0x24)+(1×23)+(1×22)+(0x21)+(0x20).
= (1 x 64) + (1 x 32) + (0 x 16) + (1 x 8) + (1x 4) + (0 x 2) + (0 x 1)
= 64+32+0+8+4+0+0 = 108
1. 23123150=(2X63)+(1×62)+(3×61)+(5×60)
= (2x 216) + (1x 36) + (3x 6) + (5 x 1)
= 432 + 36 + 18 + 5 = 491
1. 43526170=(4×83)+(5×82)+(6×81)+(7×80).
= (4 x 512) + (5 x 64) + (6 x 8) + (7×1)
= 2048+320+48+7=2423
4.3 Binary System (Base two number system)
In binary system, the greatest digit used is 1, so the two digits available in binary system are 0 and 1. Remember that each digit in a binary number has a place value.
Converting Number in Base Ten To Numbers In Base 2
Examples4.3:(a) Convert 2910 to base 2.
(b) Convert 7910 to base 2
(c) convert 14510 to base 2
Solution:
(a). 2 29 (b). 2 79 (c) 2 145
2 14 R 1 2 39 R 1 2 72 R 1
2 7 R 0 2 19 R 1 2 36 R 0
2 3 R 1 2 9 R 0 2 18 R 0
2 1 R 0 2 4 R 1 2 9 R 0
0 R 1 2 2 R 0 2 4 R 1
2 1 R 0 2 2 R 0
0 R 1 2 1 R 0
0 R 1
2910= 101012 7910= 10010112 15710 = 100100012
EVALUATION: (1) Expand the following with their bases.
(a) 35318
(b) 1010102
(c) 1110110242
EVALUATION: (2) Convert the following number to base 2
1. 35610
2. 4710
3. 21810
ASSIGNMENT: PAGE 105
EXERCISE 9.2 NO 3 (e,f,g,h) , NO 5 (a,b,d)
WEEK 5 NUMBER SYSTEM
5.1 Converting Numbers Base From One Base To Other Bases
This is a 2-stage conversion process. The steps to do this are simple. First convert the given base to base ten and then convert the result to the required base.
Example5.1: Convert 21356 to a number in base two
Solution:
Step one: Converting to base ten
23123150 = (2X63) + (1×62) + (3×61) + (5×60)
= (2x 216) + (1x 36) + (3x 6) + (5 x 1)
= 432 + 36 + 18 + 5 = 491
Step two: Convert the answer in base ten to a number in base two
2 491
2 245 R 1
2 122 R 1
2 61 R 0
2 30 R 1
2 15 R 0
2 7 R 1
2 3 R 1
2 1 R 1
0 R 1
Therefore 21356 = 1111010112
Examples 5.2: Express 101112 to a number in base three
Solution:
1. 1403121110= (1 X 24) + (0 X 23) + (1 X 22) + (1 X 21) + (1 X 20
= 1 X 16 + 0 X 8 + 1 X 4 + 1 X 2 + 1 X 1
= 16 + 0 + 4 + 2 + 1
= 2310
3 23
3 7 R 2
3 2 R 1
0 R 2
Therefore 101112 = 2123
5.2 Addition and Subtraction of Number Base
Note the following STEPS in adding and subtracting.
0+0 = 0
1+ 0 = 1
1 + 1 = 10
1 + 1+ 1 = 11
To subtract in base two
0 – 0 =0
1 – 0 = 1
10 – 1 = 1
11 – 1 = 10
1. 11012 and 11112
2. 110112, 101012, and 10012
Solution:
1. 1101 + 1111 = 1 1 0 1 [using the addition steps 1 + 1 = 10, write 0 down and move the 1 to the next number]
+ 1 1 1 1 [also take the 1 to next, and move them till you get to last one].
1 1 1 0 02
1. 11011 + 10101 + 1001 = 1 1 0 1 12
1 0 1 0 12
+ 1 0 0 12
1 1 1 0 0 12
Example 5.4: Subtract (a) 10112 from 101012 (b) 110112 from 111010
Solution:
1. 1 0 1 0 1[using the subtraction steps 1- 1 = 0, write 0 down and move to the next number]
1 0 1 12
1 0 1 02
1. 1 1 1 0 1 02
1 1 0 1 12
1 1 1 1 12
5.3 Multiplication of Binary
Example 5.5: Find the product of 10112 and 1012
Solution:
1 0 1 12
x 1 0 12
1 0 1 1
0 0 0 0
1 0 1 1
1 1 0 1 1 12
Example 5.6: multiply 1111 by 110
Solution:
1 1 1 12
1 1 02
0 0 0 0
1 1 1 1
1 1 1 1
1 0 1 1 0 1 02
EVALUATION: DO THESE
1. Add the following (a) 10012 + 11112
(b) 1011102+100102
1. Subtract 10112 from 11011012
2. Multiply 1101012 by 10112
3. find the value of the following binary numbers 1011+10-111
ASSIGNMENT:
EXERCISE 9.5 page 107 NO 1(B,C,D), NO 2 A and B, NO 3 (E,F,G,H)
EXERCISE 9.6 NO 2,5,8 and 11. PAGE 108
WEEK 6 BASIC OPERATION
When adding or subtracting numbers make sure you know the place values of the numbers. Then write each number underneath the other so that the units (U) figures are in line, all the tens (T), all the hundreds (H), and so on are in line.
Examples 6.1:
2. Take away 936 from 7093
3. Ade has 5673 marbles, Ola has 3426 marbles, Michael has 887 marbles. How many marbles do they have altogether?
4. Find the total of 90156, 300673, and 65489.
5. The difference between two numbers is 6032. If the larger number is 96005, what is the smaller number?
Note: remember to write the numbers underneath.
Solution:
1. 1 9 5 2 + 5 6 7 = TH H T U
1 9 5 2
+5 6 7
2 5 1 9
1. TH H T U
7 0 9 3
–9 3 6
6 1 5 7
3.
TH H T U
OLU’s MARBLES 3 4 2 6
Michael’s MRABLES 8 8 7
ALTOGETHER 9 9 8 6
1. H.TH T.TH TH H T U
3 0 0 6 7 3
9 0 1 5 6
+ 6 5 4 8 9
1. 5 6 3 1 8
5.
T.TH TH H T U
9 6 0 0 5
6 0 3 2
8 9 9 7 3
EVALUATION: DO THESE:
1. Find the total of 453323, 642176, 99510
2. Take away 301653 from 887321
3. Find the sum of 666721, 55432, 29255, 387, 554 and 77
4. MICHEAL collected 1350 oranges from her sister Helen. He gave 546 to Mathew and 123 to Kenny. How many oranges left in Micheal hands?
ASSIGNMENT: PAGE 12
EXERCISE 2.2 NO 1 (A, D), N0 10, 12, AND 14.
6.2 Multiplication and Division Of Numbers
Example 6.2:
1. Find the product of 8600 and 150
2. A big company recently bought 78 new computer each costing #95 000. How much did the company spent?
3. Divide 15648 by 12
4. 5982 computers are to be shared among 40 schools. How many will each school get and how many will be left over?
Solution:
8 6 0 0
X 1 5 0
0 0 0 0
4 3 0 0 0
8 6 0 0 _
1 2 9 0 0 0 0
1. 9 5 0 0 0
x 7 8
7 6 0 0 0 0
6 6 5 0 0 0 0
6 7 2 6 0 0 0 0
1. 1 3 0 4
12 1 5 6 4 8
1 2
36
36
48
1. 1 4 9
40 5 9 8 2
4 0
1 9 8
1 6 0
3 8 2
3 6 0
2 2
Note each school will collects 149 computers and the left over will be 22.
DO THESE:
1. Multiply the following (a) 6745 by 356 (b) 89760 by 120
2. Find the product of 98700 and 4900
3. Divide the following (a) 56748 by 24 (b) 640000 by 80
4. 789 oranges are to be shared equally among 14 brothers. How many will each person get and how many will remain?
ASSIGNMENT:
PAGE 16 EXERCISE 2.6 N0 3 TO 5.
WEEK 7 MID TERM
PERIODIC TEST.
WEEK 8 ADDITION AND SUBTRACTION (DIRECTED NUMBERS
8.1 ADDITION AND SUBTRACTION OF DIRECTED NUMBERS.
In a horizontal number line:
• When adding, start from the first number, count right along the number line.
• When subtracting, start from the first number, count left along the number line.
NOTE: that in a horizontal number line, number to the right of zero are positive and to the left are negative.
Example 8.1: use number line to find the value of the (a) 3 + 5, (b) -4 – 6, (c) -9 + 12
Solution:
-2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12
3 + 5 = 8
(b)
-11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3
(-4 )–(-6) = -10
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3
-9 + 12 = 3
DO THESE:
1. Draw a suitable horizontal number line to answer the following
1. 5 – 8
2. 6 – 2
3. (-5) – (-8)
4. 7 + 2
5. 5 + 8
1. Use the rules to answer the following
1. +5 – (+6)
2. 25 + (-55)
3. +24 + (+28)
4. 18 – 68
5. -7 + 5 – 2 + 7 -10 -3
8.2 RULES FOR ADDITION AND SUBTRACTION
NOTE: You can carry out the addition and subtraction without drawing the number line using the rules below.
1. Replace the same signs that appear together by a positive sign. [+ + = +],[ – – = -].
Examples +7 + (+5) = 7+5 = 12 (same sign gives +)
+6 – (-8) = 6 + 8 = 14 (same sign gives +)
1. Replace two different signs that appear together by a negative sign. [+ – = -],[- + = -].
Examples +8 + (-5) = 8 – 5 = 3 [different sign gives negative]
+9 – (+4) = 9 – 4 = 5 [different sign gives negative sign].
Note :To subtract two numbers with different signs, subtract the smaller number from the larger number and then place the sign before the larger number in front of your answer.
Example 8.2: Find the value of the following
1. (+6) + (+9)
2. (+21) – (+9)
3. (+12) + (-5)
4. (+29) + (+13)
Solution:
Note: You need to follow the rules.
1. (+6) + (+9) = 6 + 9 = 15 (same sign gives +)
2. (+21) – (+9) = 21 – 9 = 12 [different sign gives negative sign]
3. (+12) + (-5) = 12 – 5 = 7 [different sign gives negative sign]
4. (+29) + (+13) = 29 + 13 = 42 (same sign gives +)
ASSIGNMNET:
EXERCISE 10.2 NO 3(A,C,E, M,O) PAGE 114
EXERCISE 10.3 NO 2( E,F,G,H), NO 5 ( A,B,C,D), PAGE 116 and 117
WEEK 9 & 10 ALGEBRAIC PROCESS
USE OF SYMBOLS / SHAPES
ALGEBRA is the branch of mathematics in which letters of the alphabets and symbols are used in place of numbers.
We can use any shape to represent our letter in other to interpret the sentence very well.
Examples of shapes includes
Examples: Find the missing numbers that make the following open sentence true:
1. + 7 = 14
1. + 8= 14
1. 12 ÷ = 3
1. 30 = – 9
Solution:
1. 14 – 7 = 7 [the statement is true] (7 moves to the right hand side and the sign changes to negative sign).
2. 14- 8 = 6 [the statement is true]( 8 moves to the right hand side and the sign changes to negative sign).
3. 12 x 3 = 36 [the statement is true] 12 moves to the right hand side and the sign changes to multiplication sign).
4. 30 + 9 = 40 [the statement is true]( 9 moves to the left hand side and the sign changes to positive sign).
DO THESE:
1. 8 X 7 =
2. 53 = 12 +
3. 121 ÷11 =
4. + 42 = 80
USING LETTER TO REPRESENT NUMBERS
It is more convenient to use a letter of the alphabets to represent an unknown number instead of shapes.
The statement ʠ + 7 = 12 is called an algebraic sentence or equation. It means ʠ plus seven equals twelve. Thus for the sentence to be true, we must add 5.
Examples: Find the value of each letter that will make the following sentences true.
1. B = 8 + 6
2. S x 3 = 18
3. F = 6 x 7
4. P = 25 + 37
5. 40 – x =X
Solution:
1. B = 8 + 6 = 14. Therefore B = 14
2. S x 3 = 18 , 18 / 3 = 6. Therefore 6 x 3 = 18
3. F = 6 x 7 = 42. Therefore F = 42
4. P = 25 + 37 = 62. Therefore P = 62
5. 40 – x = x , = 40 = x + x
= 40 = 2x [divide both side by 2]
40/2 = 2x/2
20 =x
Therefore x = 20.
SUBSTITUTING NMBERS
The value of an expression such as x+ 3 can be found by replacing x by a given number. This is called substitution.
EXAMPLES:
1. Find the value of x – 9 if (i) x = 23, (ii) x = 15
2. Find the value of y + 10 when y is 20
3. Find the value of t + 19 when t is 45.
Solution:
1. X – 9 = when x = 23; 23 – 9 = 12.
2. Y + 10 = when y is 20 ; 20 + 10 = 30
3. t + 19 = when 45 + 19 = 64
USING ALGEBRA FOR SOLVING WORD PROBLEMS
NOTE: Remember that you can use any letters of the alphabet to represent an unknown number.
Examples:
1. What number when added to 8 gives 20?
2. Think of a number subtract 5 from it and the result is 15. What is the number?
3. when a number is multiplied by 5 the result is 20. What is the number?
Solution:
1. let x be the number you add to 8 to give 20.
X + 8 = 20
Subtract 8 from both sides
X + 8 – 8 = 20 – 8
X = 12
1. Let the number be y
Y – 5 = 15
Y – 5 + 5 = 15 + 5
Y = 20
1. Let Z be the number you multiply by 5 to get 20
Z X 5 = 20 [note that when multiplication sign moves other side it turn to division]
Z = 20/5
Z = 4 [this statement is true.
EVALUTION:
1. A boy has y number of pencils. He gave 3 to his brother and 2 to his sister. How many pencils has he left?
2. Mrs Elizabeth bought M exercise books yesterday, she bought 5 others today. If she has 20 books altogether, how many exercise books does she buys yesterday?
3. 169 / X = 13
4. Y – 42 + 2Y + 60
ASSIGNMENT:
EXERCISE 11.6 NO 1,2,5,8,14,15,22 and 23 PAGE 125
REVISION EXERCISE
PAGE 175 to 176 Revision tests for chapter 9 to 12
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# Move negative elements to front of array
#### Algorithms List of Mathematical Algorithms Interview Problems on Array
Indian Technical Authorship Contest starts on 1st July 2023. Stay tuned.
This article focuses on the algorithm to move the negative elements of an array to the front. This is a basic problem of rearranging elements in an array. This can be solved by many approaches, this article will explore the two-pointer approach.
Table of content:
1. Problem statement
2. Idea, Steps to solve it (with step by step example)
3. Time and Space Complexity
4. Implementation
5. Shortcomings
6. Applications
# Problem statement
Problem statement: move the negative elements of an array to the front
Example:
Input will be:
2 -9 10 12 5 -2 10 -4
Output will be:
-9 -2 -4 2 10 12 5 10
Note:
• The order of negative elements are same.
• The order of positive elements are same.
• All negative elements have been moved to the front.
• All positive elements have been moved to the end or followed by all the negative elements.
# Idea & Steps to solve it
The idea of the algorithm is to have two pointers `left` and `right`.
All the elements before `left` should be negative and all elements after `right` should be positive. This way if any elements are at their wrong positions swaps can occur to correct it.
Steps:
1. Initialize left pointer to 0 and right pointer to size(nums)-1
2. check if both left and right elements are negative, increment left pointer if true
3. check if left element is positive and right element is negative then swap the elements, increment left and decrement right
4. check if both left and right elements are positive, decrement right pointer if true
5. otherwise, increment left and decrement right pointers
6. repeat 2-5 until left pointer <= right pointer
``````algo rearrange(nums):
left = 0
right = size(nums)-1
while left < right:
compare the nums[left] and nums[right]:
if both are negative:
left = left + 1
else if the left element is positive and the right element is negative:
swap the elements
left = left + 1
right = right - 1
else if both are positive:
right = right - 1
else:
left = left + 1
right = right - 1
``````
Consider this example that illustrates all possible cases,
`````` nums: -2 -1 2 4 5 -3 5
^ ^
| |
left right
nums[left] < 0 nums[right] > 0 => left++ right--
``````
`````` nums: -2 -1 2 4 5 -3 5
^ ^
| |
left right
nums[left] < 0 nums[right] < 0 => left++
``````
`````` nums: -2 -1 2 4 5 -3 5
^ ^
| |
left right
nums[left] > 0 nums[right] < 0 => swap
``````
`````` nums: -2 -1 -3 4 5 2 5
^ ^
| |
left right
nums[left] > 0 nums[right] > 0 => left++ right--
``````
`````` nums: -2 -1 -3 4 5 2 5
^ ^
| |
right left
left <= right is true => break loop
``````
# Time and Space Complexity
For this particular algorithm the number of comparison operations done will always be `Θ(N/2)`. Thus we consider swaps as the basic operation and compare complexity based on this factor
• Worst case time complexity: `Θ(k)`, where k is the number of negative elements
• This occurs when all negative elements occur after the positive elements. All negative elements need to be swapped over to the front of the array.
• Average case time complexity: `O(N/2)`
• On average, by probabilistic analysis, about N/2 negative numbers lie to the right of `left` pointer and need to be swapped over.
• Best case time complexity: `Θ(1)`
• This occurs when the elements are already in the correct order and do not require any rearranging.
• Space complexity: `Θ(1)`
# Implementation
``````void rearrange(vector<int> &nums) {
int left = 0;
int right = nums.size() - 1;
while(left < right) {
int left_ele = nums[left];
int right_ele = nums[right];
if(left_ele < 0 && right_ele < 0) {
left += 1;
} else if(left_ele > 0 && right_ele < 0) {
// out of order
nums[left] = right_ele;
nums[right] = left_ele;
} else if(left_ele > 0 && right_ele > 0) {
right -= 1;
} else {
left += 1;
right -= 1;
}
}
}
int main() {
vector<int> nums = {4, 2, -5, 8, -2, 10, -7};
rearrange(nums);
for(auto ele: nums) {
cout << ele << " ";
}
cout << "\n";
}
``````
Output:
``````-7 -2 -5 8 2 10 4
``````
# Shortcomings
• Since 0 is neither a negative or positive number, the algorithm fails to place it at the right spot and it may land either in the negative or positive halves
# Applications
• Preferred instead of sorts when only the relative ordering of positive and negative elements is required as opposed to ordering of each element
With this article at OpenGenus, you must have a strong idea of how to Move negative elements to front of array. |
Study P6 Mathematics Fractions Of Remainder - Geniebook
Fractions Of Remainder
In this article, we will focus on the word problems for fractions of a remainder.
What is A Fraction of a Remainder?
Let’s look at this example on how we solve using models.
Freddy ate 14 of his birthday cake. He then cut the remaining cake into 6 equal pieces to give to his 6 friends. What fraction of the cake did each friend get?
Fraction of cake remaining
\begin{align*} &= 1-\frac{1}{4}\\ &= \frac{3}{4} \end{align*}
The remaining \begin{align*} \frac{3}{4} \end{align*}of the cake was cut into $$6$$ equal pieces to give to his $$6$$ friends.
Fraction of cake each friend received
\begin{align*} &=\frac{3}{4} \div 6\\ &=\frac{3}{4} \div \frac{6}{1} \\ &=\frac{3}{4} \times \frac{1}{6} \\ &=\frac{1}{8} \end{align*}
Answer:
\begin{align*} \frac{1}{8} \end{align*}
Branching Method to Understand & Solve Problems
Branching method is also used to solve ’fraction of remainder’ questions. Let’s look at the examples below on how to solve using branching.
Question 1:
Joyce went shopping with $$560$$. She spent $$37$$ of the money on a pair of shoes and $$18$$ of the remaining money on a watch. She then bought a dress and had $$164$$ left.
How much did the dress cost?
Solution:
Fraction representing the total amount of money spent on a dress and left
\begin{align*} &=\frac{4} {7} \times \frac{7}{8} \\ &=\frac{1}{2} \end{align*}
Total amount of money $$= 560$$
Total amount of money spent on a dress and left
\begin{align*}&= \frac{1}{2} \text{ of the total} \\ &= 560 \div 2 \\ &= 280 \end{align*}
Cost of the dress
\begin{align*}&= 280 - 164 \\ &= 116 \end{align*}
Answer:
\begin{align*} 116 \end{align*}
Question 2:
Tom worked $$20$$ days in August. He gave $$0.15$$ of his salary to his mother. He spent $$45$$ of his remaining salary and saved the rest. He saved a total of $$340$$ in August. How much did he receive for a day’s work?
Solution:
\begin{align*} 0.15 &= \frac{15}{100}\\ &=\frac{3}{20} \end{align*}
Fraction representing the amount of money saved
\begin{align*}&=\frac{1}{5} \times \frac{17}{20} \\ &=\frac{17}{100} \end{align*}
\begin{align*} \frac{17}{100} \end{align*} of the total $$= 340$$
\begin{align*} \frac{1}{100} \end{align*} of the total\begin{align*}\\[2ex] &= 340 \div 17 \\[2ex] &= 20 \end{align*}
\begin{align*} \frac{100}{100} \end{align*} of the total\begin{align*}\\[2ex] &= 100 \times 20 \\[2ex] & = 2000 \end{align*}
Amount of money he received for $$20$$ days of work $$= 2000$$a
Amount of money he received for a day of work\begin{align*}\\[2ex] &= 2000 \div 20 \\[2ex] &= 100 \end{align*}
Answer:
$$100$$
Let’s look at the more challenging type of ‘fraction of a remainder’ questions.
Question 3:
Farmer Dan harvested some apples. He sold \begin{align*} \frac{3}{4} \end{align*} of the apples and additional $$105$$ apples to Adam. He sold $$55$$ less than \begin{align*} \frac{3}{4} \end{align*} of the remaining apples to Geraldine. Farmer Dan had $$890$$ apples left. How many apples did farmer Dan harvest?
Solution:
\begin{align*} 1 \text{ part} &= 890 - 55 \\ &= 835 \\ \\ 4 \text{ part} &= 4 \times 835 \\ &= 3340 \end{align*}
From the model,
\begin{align*} 1 \text{ unit} &= 4 \text{ part} + 105 \\ &= 3340 + 105 \\ &= 3445 \end{align*}
Number of apples Farmer Dan harvested\begin{align*} \\[2ex] &= 4 \text{ units} \\ &= 4 \times 3445 \\ &= 13 \,780 \end{align*}
Answer:
$$13 \,780$$ apples
Question 4:
Benjamin spent \begin{align*} \frac{3} {7} \end{align*} of his money on $$6$$ toys and $$6$$ erasers. He spent \begin{align*} \frac{1} {4} \end{align*} of his remaining money on $$10$$ cards. Each eraser costs \begin{align*} \frac{1} {7} \end{align*} as much as a toy. Each card costs $$0.30$$ more than an eraser. How much money did Benjamin spend on each toy?
Solution:
Since each eraser costs \begin{align*} \frac{1} {7} \end{align*} as much as a toy, we let the cost of $$1$$ eraser be represented by $$1$$ part and the cost of $$1$$ toy be represented by $$7$$ parts.
\begin{align*} \text{Cost of } 6 \text{ erasers} &= 6 \text{ parts} \\ \\ \text{Cost of } 6 \text{ toys} &= 6 \times 7 \text{ parts} \\ &= 42 \text{ parts} \\ \\ 3 \text{ units} &= 6 \text{ parts} + 42 \text{ parts} \\ &= 48 \text{ parts} \\ \\ 1 \text{ units} &= 48 \text{ parts} \div 3 \\ &= 16 \text{ parts} & \text{(Cost of 10 cards)} \end{align*}
Since each card costs $$0.30$$ more than an eraser,
\begin{align*} \text{Cost of } 10 \text{ cards} &= 10 \text{ parts} + 10 \times 0.30 \\ &= 10 \text{ parts} + 3 \\ \\ 16 \text{ parts} &= 10 \text{ parts} + 3 \end{align*}
From the model,
\begin{align*} 6 \text{ parts} &= 3 \\ 1 \text{ parts} &= 3 \div 6 \\ &= 0.50 \\ \\ \text{Cost of }1 \text{ toy} &= 7 \text{ parts}\\ &= 7 \times 0.50\\ &= 3.50 \end{align*}
Answer:
$$3.50$$
Conclusion
In this article, we worked on different word problems using the concept of fraction of the remainder.
We can solve such problems by different methods. We can draw models, use the branching method or the unitary method.
As always, practice, practice, and more practice will give you the much-needed confidence to tackle such problems in PSLE.
Continue Learning Algebra Distance, Speed and Time Volume of Cubes and Cuboid Fundamentals Of Pie Chart Finding Unknown Angles Number Patterns: Grouping & Common Difference Fractions Of Remainder Fractions - Division Ratio Repeated Identity: Ratio Strategies
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# Vertical Angles
## Vertical Angles Lesson
### Definition of Vertical Angles
Vertical angles are the opposing angles where two lines cross. Vertical angles are always equal in value.
INTRODUCING
### The Vertical Angles Theorem
The vertical angles theorem tells us that angles opposite each other where two lines cross are congruent (equal in value).
In the image below, α and β are vertical and equal angles.
The “vertical” in vertical angles should not be confused with lines or angles being oriented up and down, as the word vertical implies. In this context, vertical means that the two angles are opposite each other about a shared apex.
Another name for vertical angles is vertically opposite angles. This name is very intuitive because it describes that the angles are opposite each other where two lines cross.
#### How to use the Theorem
There are four angles where two lines cross (making two pairs of vertical angles). These four angles are labeled as A, B, C, and D in the image below. Congruent angles are marked with a like number of dashes as shown in the image below.
These four angles are what make the two pairs of vertical angles. A = B and C = D. Because of what the vertical angles theorem tells us, we only need to know the value of one of these angles to determine all the others. The sum of angles around the crossing of two lines is always 360°.
The rules of geometry tell us that A + B + C + D = 360°. With simple arithmetic, we can find the other three angles from only knowing one angle. This is what makes the vertical angles theorem so powerful.
#### When to use the Theorem
If we are solving a geometry problem that has intersecting lines and we are given one of the angles, we can find the other three angles by using the vertical angles theorem. This is extremely useful for intersecting shapes and other angle-based problems. It works for triangles, rectangles, etc. It is the basis of the relationship between internal and external angles of shapes.
### Vertical Angles Example Problems
Let's go through a couple of example problems together to reinforce our understanding of vertical angles.
#### Example Problem 1
Two lines cross and form the angles A, B, C, and D. A opposes B, and C opposes D. If angle B is 50°, what are the values of angles A, C, and D in degrees?
Solution:
1. A and B are vertical angles, so A = B. B = 50°, therefore A = 50°.
2. C and D are vertical angles, so C = D. The sum of A, B, C, and D must be 360°.
3. 360° = A + B + C + D
4. 360° = 50° + 50° + C + D
5. 260° = C + D
6. 260°2 = 130°, C = 130° and D = 130°
7. A = 50°, C = 130°, and D = 130°.
#### Example Problem 2
Two lines intersect. One of the angles at the intersection is measured to be 60°. Is this the smallest angle at the intersection?
Solution:
1. The vertical angles theorem tells us that the angle opposite of the 60° angle must also be 60°. The sum of this pair of vertical angles is 120°.
2. 360 - 120 = 240
3. 2402 = 120
4. Therefore, the intersection is made up of angles 60°, 60°, 120°, and 120°.
5. The 60° measured angle is tied with its opposing angle for being the smallest angle at the line intersection.
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# If $3x + 5y = 11$ and $xy = 2$, find the value of $9x^2 + 25y^2$.
Given:
$3x + 5y = 11$ and $xy = 2$
To do:
We have to find the value of $9x^2+25y^2$.
Solution:
The given expressions are $3x + 5y = 11$ and $xy = 2$. Here, we have to find the value of $9x^2+25y^2$. So, by squaring the given expression and using the identity $(a+b)^2=a^2+2ab+b^2$, we can find the required value.
$xy = 2$............(i)
$(a+b)^2=a^2+2ab+b^2$.............(ii)
Now,
$3x + 5y = 11$
Squaring on both sides, we get,
$(3x + 5y)^2 = (11)^2$ [Using (ii)]
$(3x)^2+2(3x)(5y)+(5y)^2=121$
$9x^2+30xy+25y^2=121$
$9x^2+30(2)+25y^2=121$ [Using (i)]
$9x^2+60+25y^2=121$
$9x^2+25y^2=121-60$ (Transposing $60$ to RHS)
$9x^2+25y^2=61$
Hence, the value of $9x^2+25y^2$ is $61$. |
# Convert Mixed Numbers into Improper Fractions
In this worksheet, students will practise converting mixed numbers (whole numbers and fractions) into top-heavy (improper) fractions. This is a very useful skill to use when adding or subtracting fractions.
Key stage: KS 4
Year: GCSE
GCSE Subjects: Maths
GCSE Boards: Pearson Edexcel, OCR, Eduqas, AQA,
Curriculum topic: Number, Fractions, Decimals and Percentages
Curriculum subtopic: Structure and Calculation Fractions
Difficulty level:
#### Worksheet Overview
When you are working with fractions, you will come across both top-heavy fractions (correctly called improper fractions) and mixed numbers.
Top-heavy fractions have larger numerators than denominators.
Mixed numbers describe expressions which use both whole numbers and fractions together.
You need to be able to convert between these two forms and identify equivalents.
e.g. Convert this mixed number into a top-heavy fraction:
3
1 5
If we draw this fraction using bars, we have three wholes and one fifth, like this:
If we count the shaded sections, we have 16 in total and each one is worth 1/5 so our fraction can now be written as:
16 5
Doing this the quicker way...
You should have noticed that the denominator is the same in both the question and answer fractions. This is ALWAYS true.
So we just need to find the numerator.
If we have 3 wholes which we are splitting into fifths, we have 3 x 5 = 15 fifths.
We need to remember that we had an extra 1 fifth in the question too, which gives us 16 overall.
To summarise:
In order to find the new numerator, multiply the whole number by the denominator then add the numerator.
The denominator will stay the same in both fractions.
In this activity, you will convert mixed numbers (so whole numbers and fractions) into top-heavy, improper fractions.
This is a very useful skill to use when adding or subtracting fractions.
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# A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, and the volume of the wood and that of the graphite.
Given :
Radius (R) of pencil = $$\frac{7}{2} mm = 3.5 mm = 0.35 cm$$
Radius (r) of graphite = $$\frac{1}{2} mm = 0.5 mm = 0.05 cm$$
Height (h) of pencil = 14 cm
We know that,
Volume of wood in pencil
= $${\pi} ({R}^2 - {r}^2) h$$
= $$\frac{22}{7} × ({0.35}^2 - {0.05}^2) × 14$$ $${cm}^3$$
= $$\frac{22}{7} × (0.1225 - 0.0025) × 14$$ $${cm}^3$$
= 44 × 0.12 $${cm}^3$$
= 5.28 $${cm}^3$$
Similarly, Volume of graphite in pencil
= $${\pi} {r}^2 h$$
= $$\frac{22}{7} × {0.05}^2 × 14$$ $${cm}^3$$
= 44 × 0.0025 $${cm}^3$$
= 0.11 $${cm}^3$$ |
# Multiplication of Algebraic Fractions
To solve the problems on multiplication of algebraic fractions we will follow the same rules that we already learnt in multiplication of fractions in arithmetic.
From multiplication of fractions we know,
Product of two or more fractions = $$\frac{Product of numerators}{Product of denominators}$$
In algebraic fractions, the product of two or more fractions can be determined in the same way i.e.
Product of two or more fractions = $$\frac{Product of numerators}{Product of denominators}$$.
1. Determine the product of the following algebraic fractions:
(i) $$\frac{m}{n} \times \frac{a}{b}$$
Solution:
$$\frac{m}{n} \times \frac{a}{b}$$
$$\frac{m \cdot a}{n \cdot b}$$
= $$\frac{am}{bn}$$
(ii) $$\frac{x}{x + y} \times \frac{y}{x - y}$$
Solution:
$$\frac{x}{x + y} \times \frac{y}{x - y}$$
= $$\frac{x \cdot y}{(x + y) \cdot (x - y)}$$
= $$\frac{xy}{x^{2} - y^{2}}$$
2. Find the product of the algebraic fractions in the lowest form: $$\frac{m}{p + q} \times \frac{m}{n} \times \frac{n(p - q)}{m(p + q)}$$
Solution:
$$\frac{m}{p + q} \times \frac{m}{n} \times \frac{n(p - q)}{m(p + q)}$$
= $$\frac{m \cdot m \cdot n(p - q)}{(p + q) \cdot n \cdot m(p + q)}$$
= $$\frac{m^{2}n(p - q)}{mn(p + q)^{2}}$$
Here the numerator and denominator have a common factor mn, so by dividing the numerator and denominator of the product by mn, the product in the lowest form will be $$\frac{m (p - q)}{(p + q)^{2}}$$.
3. Find the product and express in the lowest form: $$\frac{x(x + y)}{x - y} \times \frac{x - y}{y(x + y)} \times \frac{x}{y}$$
Solution:
$$\frac{x(x + y)}{x - y} \times \frac{x - y}{y(x + y)} \times \frac{x}{y}$$
= $$\frac{x(x + y) \cdot (x - y) \cdot x}{(x - y) \cdot y(x + y) \cdot y}$$
= $$\frac{x^{2}(x + y) (x - y)}{y^{2}(x + y) (x - y)}$$
Here, the common factor in the numerator and denominator is (x + y) (x – y). If the numerator and denominator are divided by this common factor, the product in the lowest form will be $$\frac{x^{2}}{y^{2}}$$.
4. Find the product of the algebraic fraction: $$\left ( \frac{5a}{2a - 1} - \frac{a - 2}{a} \right ) \times \left ( \frac{2a}{a + 2} - \frac{1}{a + 2}\right )$$
Solution:
$$\left ( \frac{5a}{2a - 1} - \frac{a - 2}{a} \right ) \times \left ( \frac{2a}{a + 2} - \frac{1}{a + 2}\right )$$
Here, the L.C.M. of the denominators of the first part is a(2a – 1) and the L.C.M. of the denominators of the second part is (a + 2)
Therefore, $$\left \{\frac{5a \cdot a}{(2a - 1) \cdot a} - \frac{(a - 2) \cdot (2a - 1)}{a \cdot (2a - 1)} \right \} \times \left ( \frac{2a}{a + 2} - \frac{1}{a + 2}\right )$$
= $$\{ \frac{5a^{2}}{a(2a - 1)} - \frac{(a - 2)(2a - 1)}{a(2a - 1)} \} \times \left ( \frac{2a}{a + 2} - \frac{1}{a + 2}\right )$$
= $$\frac{5a^{2} - (a - 2)(2a - 1)}{a(2a - 1)} \times \frac{2a - 1}{a + 2}$$
= $$\frac{5a^{2} - (2a^{2} - 5a + 2)}{a(2a - 1)} \times \frac{2a - 1}{a + 2}$$
= $$\frac{5a^{2} - 2a^{2} + 5a - 2}{a(2a - 1)} \times \frac{2a - 1}{a + 2}$$
= $$\frac{3a^{2} + 5a - 2}{a(2a - 1)} \times \frac{2a - 1}{a + 2}$$
= $$\frac{3a^{2} + 6a - a - 2}{a(2a - 1)} \times \frac{2a - 1}{a + 2}$$
= $$\frac{3a^{2} + 6a - a - 2}{a(2a - 1)} \times \frac{2a - 1}{a + 2}$$
= $$\frac{3a (a + 2) - 1(a + 2)}{a(2a - 1)} \times \frac{2a - 1}{a + 2}$$
= $$\frac{(a + 2)(3a - 1)}{a(2a - 1)} \times \frac{2a - 1}{a + 2}$$
= $$\frac{(a + 2)(3a - 1)(2a - 1)}{a(2a - 1)(a + 2)}$$
Here, the common factor in the numerator and denominator is (x + 2) (2x - 1). If the numerator and denominator are divided by this common factor, the product in the lowest form will be
= $$\frac{(3a - 1)}{a}$$ |
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# A motor pumps water at the rate of $V{{m}^{3}}/s$ , against a pressure $PN{{m}^{-2}}$ . The power of the motor in watt is:A. $PV$ B. $\dfrac{P}{V}$ C. $\dfrac{V}{P}$ D. $V-P$
Last updated date: 13th Jun 2024
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Hint: You could find the required relation using dimensional analysis. For that, you could begin from expressing each quantity dimensionally. And then you could express the dimension of power in terms of pressure and pumping rate raised to some powers α and β. After that, you could use the principle of homogeneity of dimensions and solve for α and β and hence find the required relation.
Formula used:
Expression for power,
$p=\dfrac{W}{t}$
Expression for pressure,
$P=\dfrac{F}{A}$
In the question we are given a motor pump that is pumping water at the rate of $V{{m}^{3}}/s$ against a pressure of $PN{{m}^{-2}}$ . And we are asked to find the power of the motor in watt in terms of the pressure (P) and rate of pumping is $V{{m}^{3}}/s$ .
Let us solve this question by using the method of dimensional analysis.
We know that power p is the time rate of doing work and can be expressed as,
$p=\dfrac{W}{t}$ ……………………. (1)
Where, W is the work done and t is the time taken.
But we know that work done is product of force and displacement, that is,
$W=F\times S$
But F=ma by Newton’s second law, so, the dimension of force is given by,
$\left[ F \right]=\left[ M \right]\left[ L{{T}^{-2}} \right]=\left[ ML{{T}^{-2}} \right]$
$\Rightarrow \left[ W \right]=\left[ ML{{T}^{-2}} \right]\left[ L \right]=\left[ M{{L}^{2}}{{T}^{-2}} \right]$
$\Rightarrow \left[ p \right]=\dfrac{\left[ M{{L}^{2}}{{T}^{-2}} \right]}{\left[ T \right]}$
The dimension of power is given by,
$\Rightarrow \left[ p \right]=\left[ M{{L}^{2}}{{T}^{-3}} \right]$ ………………………………… (2)
Now pressure by definition is the force acting per unit area, that is,
$P=\dfrac{F}{A}$
Pressure can be dimensionally expressed as,
$\left[ P \right]=\dfrac{\left[ ML{{T}^{-2}} \right]}{{{\left[ L \right]}^{2}}}$
$\Rightarrow \left[ P \right]=\left[ M{{L}^{-1}}{{T}^{-2}} \right]$ ……………………………. (3)
Rate of pumping is given by,
$V=\dfrac{v}{t}$
$\left[ V \right]=\dfrac{{{\left[ L \right]}^{3}}}{\left[ T \right]}$
$\left[ V \right]=\left[ {{L}^{3}}{{T}^{-1}} \right]$ …………………………………….. (4)
Let us now find the required relation, for that let assume,
$p\propto {{P}^{\alpha }}{{V}^{\beta }}$
$p=k{{P}^{\alpha }}{{V}^{\beta }}$
Now let us express them dimensionally,
$\left[ p \right]={{\left[ P \right]}^{\alpha }}{{\left[ V \right]}^{\beta }}$ ………………………….. (5)
Substituting (2), (3) and (4),
$\left[ M{{L}^{2}}{{T}^{-3}} \right]={{\left[ M{{L}^{-1}}{{T}^{-2}} \right]}^{\alpha }}{{\left[ {{L}^{3}}{{T}^{-1}} \right]}^{\beta }}$
$\left[ M \right]{{\left[ L \right]}^{2}}{{\left[ T \right]}^{-3}}={{\left[ M \right]}^{\alpha }}{{\left[ L \right]}^{-\alpha +3\beta }}{{\left[ T \right]}^{-2\alpha -\beta }}$
By the principle of homogeneity of dimensions, the dimensions on both sides of any valid equation should be the same. So,
$\alpha =1$ ……………….. (6)
$-\alpha +3\beta =2$ …………………. (7)
$-2\alpha -\beta =-3$ ……………………….. (8)
Substituting (6) in (7) gives,
$\Rightarrow -\left( 1 \right)+3\beta =2$
$\Rightarrow \beta =\dfrac{3}{3}=1$ ………………………. (9)
Now we can substitute (6) and (9) in (5),
$\left[ p \right]={{\left[ P \right]}^{1}}{{\left[ V \right]}^{1}}$
Therefore the power of the motor in watt could be expressed in terms of pressure P and rate of pumping V as PV.
So, the correct answer is “Option A”.
Note: You should note that the V given in the question is the rate at which the motor is pumping the water not the volume of the water being pumped. Also, in order to avoid confusion, in the question the unit is given for both the quantities. Hence, you should accordingly assign the dimension otherwise you will end up making a huge mistake. |
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# NCERT Solutions for Class 10 Maths Exercise 12.3
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NCERT Solutions for Maths Area Related to Circles
## NCERT Solutions for Class 10 Maths 12.3 (Area Related to Circles)
Unless stated otherwise, take
### Q. 1 of 12.3 class 10 Maths
1. Find the area of the shaded region in figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.
Ans. RPQ = [Angle in semi-circle is ]
= = 49 + 576 = 625
RQ = 25 cm
Diameter of the circle = 25 cm
Radius of the circle = cm
Area of the semicircle =
= =
Area of right triangle RPQ =
= =
Area of shaded region = Area of semicircle – Area of right triangle RPQ
= =
### Q. 2 of 12.3 class 10 Maths
2. Find the area of the shaded region in figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and AOC =
Ans. Area of shaded region = Area of sector OAC – Area of sector OBD
=
=
=
=
=
### Q. 3 of 12.3 class 10 Maths
3. Find the area of the shaded region in figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
= Area of square ABCD – (Area of semicircle APD + Area of semicircle BPC)
=
=
= 196 – 154 =
NCERT Solutions for Class 10 Maths Exercise 12.3
### Q. 4 of 12.3 class 10 Maths
4. Find the area of the shaded region in figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.
= Area of circle + Area of equilateral triangle OAB – Area common to the circle and the triangle
=
=
= =
=
NCERT Solutions for Class 10 Maths Exercise 12.3
### Q. 5 of 12.3 class 10 Maths
5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in figure. Find the area of the remaining portion of the figure.
Ans. Area of remaining portion of the square
= Area of square – (4 Area of a quadrant + Area of a circle)
=
= =
NCERT Solutions for Class 10 Maths Exercise 12.3
### Q. 6 of 12.3 class 10 Maths
6. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in figure. Find the area of the design (shaded region).
Ans. Area of design = Area of circular table cover – Area of the equilateral triangle ABC
= ………(i)
G is the centroid of the equilateral triangle.
radius of the circumscribed circle = cm
According to the question,
= 48 cm
Now,
= 3072
Required area = [From eq. (i)]
=
=
NCERT Solutions for Class 10 Maths Exercise 12.3
### Q. 7 of 12.3 class 10 Maths
7. In figure ABCD is a square of side 14 cm. With centers A, B, C and D, four circles are drawn such that each circle touches externally two of the remaining three circles. Find the area of the shaded region.
Ans. Area of shaded region = Area of square – 4 Area of sector
=
= = 196 – 154 =
NCERT Solutions for Class 10 Maths Exercise 12.3
### Q. 8 of 12.3 class 10 Maths
8. Figure depicts a racing track whose left and right ends are semicircular.
The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:
(i) the distance around the track along its inner edge.
(ii) the area of the track.
Ans. (i)Distance around the track along its inner edge
=
= = = m
(ii)Area of track =
=
= =
NCERT Solutions for Class 10 Maths Exercise 12.3
### Q. 9 of 12.3 class 10 Maths
9. In figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.
Ans. Area of shaded region = Area of circle + Area of semicircle ACB – Area of ACB
=
= = =
NCERT Solutions for Class 10 Maths Exercise 12.3
### Q. 10 of 12.3 class 10 Maths
10. The area of an equilateral triangle ABC is. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see figure). Find the area of the shaded region.
Ans. Area of equilateral triangle = = 17320.5
cm
Area of shaded region = Area of
ABC –
= 17320.5 – 15700 =
NCERT Solutions for Class 10 Maths Exercise 12.3
### Q. 11 of 12.3 class 10 Maths
11. On a square handkerchief, nine circular designs each of radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief.
Ans. Area of remaining portion of handkerchief = Area of square ABCD – Area of 9 circular designs
=
= 1764 – 1386 =
NCERT Solutions for Class 10 Maths Exercise 12.3
### Q. 12 of 12.3 class 10 Maths
12. In figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the:
Ans. (i)Area of quadrant OACB =
= =
(ii) Area of shaded region = Area of quadrant OACB – Area of OBD
=
=
= = cm2
NCERT Solutions for Class 10 Maths Exercise 12.3
### Q. 13 of 12.3 class 10 Maths
13. In figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region.
Ans. OB =
=
= OA = cm
Area of shaded region = Area of quadrant OPBQ – Area of square OABC
=
=
=
=
NCERT Solutions for Class 10 Maths Exercise 12.3
### Q. 14 of 12.3 class 10 Maths
14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If AOB = find the area of the shaded region.
Ans. Area of shaded region = Area of sector OAB – Area of sector OCD
=
=
= =
= =
NCERT Solutions for Class 10 Maths Exercise 12.3
### Q. 15 of 12.3 class 10 Maths
15. In figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.
Ans. In right triangle BAC, [Pythagoras theorem]
=
BC = cm
Radius of the semicircle = cm
Required area = Area BPCQB
= Area BCQB – Area BCPB
= Area BCQB – (Area BACPB – Area BAC)
=
=
= 154 – (154 – 98) =
NCERT Solutions for Class 10 Maths Exercise 12.3
### Q. 16 of 12.3 class 10 Maths
16. Calculate the area of the designed region in figure common between the two quadrants of circles of radius 8 cm each.
Ans. In right triangle ADC, [Pythagoras theorem]
=
AC = = cm
Draw BMAC.
Then AM = MC = AC = = cm
In right triangle AMB,
[Pythagoras theorem]
= 64 – 32 = 32
BM = cm
Area of ABC =
= =
=
= =
Area of designed region
= =
## NCERT Solutions for Class 10 Maths Chapter 12
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## Linear Algebra Calculation using Integrated Circuits
Even the simplest thing we recognize may seem increasingly difficult in another point of view. Take a simple arithmetic operation for example, if one wants to calculate the function y = ax + b with given a and b, he simply multiplies any number x with a, then adds b, and gets the answer. What if no multiplication and addition can be used? How can the calculation even be possible?
Computers can actually finish the task by implementing three fundamental logic operations: AND, OR, and NOT. Most of them can do these operations within a nanosecond. In this project, I constructed a circuit for performing a simple linear algebra calculation (Fig. 1) using only basic logic and storage circuits (Fig. 2) that can be realized using standard cells.
Figure 1. Formula to be calculated. (x0, x1, x2 are all 6 bit 2’s complementary integers)
Figure 2. Basic logic and storage circuits. (Note that other circuits (e.g. NAND, XOR) are also used in this project)
Here, x0, x1 and x2 equal the three 6-bit integer inputs (2’s complementary), so there are a total of 18 Boolean input values. The output is stored in a 16-bit integer. Therefore, the goal for this project is to construct a circuit that connects all of the 18 inputs and the 16 outputs, and perform the calculation.
To make it harder, three stages of pipelines are carried out. This means that calculations are divided into three parts, and the most time-consuming part contains the critical path of the whole circuit. Fig. 3 shows an illustration of the designed circuit.
Figure 3. Logic circuit diagram for realizing the arithmetic operation (Fig. 1) of this project.
Verilog is used for simulating the results, and the circuit is written as a spice sub-circuit model. Because the D flip-flop is used, the critical time is defined as the clock cycle of the D flip-flop. Moreover, the number of transistors are defined for every basic logic circuit, so the total number of transistors can be calculated, and is named the “area” of the whole circuit.
Fig. 4 shows the simulation results of the circuit. It can be seen that only 1.3305 nanosecond is used for a half clock cycle of the circuit. This means that the circuit can continuously output calculation results every 2.661 nanosecond, which is really fast!
Figure 4. Simulation results using Verilog.
Having the experience of using absolutely no arithmetic operations for calculating a linear algebra problem really significantly broadened my insight towards digital IC design. This project inspired me to understand that even the most insignificant elements possess the potential to be combined and make up the world that we live in.
## Robot Arm Control
It’s easy for us to point at a certain coordinate in space. That’s mainly because we simply locate the point with our eyes, and continuously check if our finger is pointing at that very spot. It surely will be more difficult without using eyes, and this is the case for robot arm control with no image feedback.
Think of a two arm robot (Fig. 1). We usually want to reach a certain point on the x-y plane. The problem is only the angle of the joints can be controlled. How can we correlate the joint angles of a robot with its tip coordinate? Things get harder when it comes to 3D space, and even harder considering its rotation.
In this project, I created a program that can calculate the every joint angle of the 6-arm robot IRB140 for positioning it at a given (x, y, z) coordinate and rotation.
Figure 1. Dimensions of the IRB140 robot (unit: mm) [1].
The problem for reversing an operation from the specified coordinate and rotation to every rotation angle of an arm joint lies in the field of inverse manipulator kinematics. There may be multiple solutions that lead to the same result. Thus, I implemented the Pieper’s solution [2] for solving the joint angles for the IRB140 robot.
Here’s a video demonstration for precision control of the IRB140 by only giving the joint angles as the input. The robot follows a trail surrounding a paper box with the tip of the last arm always pointing at the center of the box.
1. ABB, IRB140 product specification, 2019, https://library.e.abb.com/public/2893a5756d204e19aba0d37c2a2cadc6/3HAC041346%20PS%20IRB%20140-en.pdf
2. Craig, J.J., Introduction to Robotics: Mechanics & Control. 1986: Addison-Wesley Publishing Company.
## Temperature and Humidity Sensor
This is a homework in class where we are assigned to utilize the temperature and humidity sensor DHT11 and an LCD display. Once activated, the LCD display will show either the temperature or the humidity being detected by the sensor. A button can be pressed to switch between temperature and humidity display.
Complex functionality can be achieved by including pre-coded libraries for controlling these components using Arduino, the development board I had used.
Though being a very simple project, such components serve as the simple building block for constructing a sophisticated hardware system. Therefore, this project plays an important role in the initial stage of hardware design for me.
Here’s a simple control process demonstration:
## Bladeless Fan
The Dyson cooling fan is an eye-catching product. At first sight, some people may wonder how the seemingly bladeless fan really works because it simply looks like a structure with no air outlet. The fact is it can does have a small outlet at the inner part of its “ring”, and has the ability to take fluid dynamics into practice and enhance the air flow, making it also an air multiplier.
In fluid dynamics class, I and a classmate of mine decided to construct a bladeless fan by our own and study the air multiplying phenomenon. We used 3D printing to fabricate the “ring” part of the fan, and a small centrifugal fan for connecting with the inlet of the ring part and inject a strong current inside.
Figure 2. Illustration and photograph of the bladeless fan.
Now we wanted to simply test whether this structure really leads to an air multiplying effect. We divided the outlet are into 9 sections, which can be represented using a 3×3 rectangular grid, and calculated the wind speed of every section at different input voltages for the centrifugal fan.
Fig. 3 shows the air velocity profile of the output wind, and Fig. 4 shows the magnification of air flow.
Figure 3. Air velocity (t_m [=] m/s) of 3×3 section grid at different input voltages.
Figure 4. Magnification of air flow at different input voltages of the centrifugal fan.
During this project, I learned more about the fundamentals of fluid mechanics, and memorized the relevant rules more deeply, which made me have a better understanding of this subject.
## Guess the Number (iOS)
This is the first iOS game I had made using Xamarin, and is the second project of the guess the number series (after Guess the Number (Windows) and prior to Guess the Number AI). (The two-player game is also named Bulls and Cows.)
At the start of the game, a random 4-digit code is generated by the app and the player starts to guess that code. The player can restart the game anytime by pressing RESET, and a history of guesses and results are shown in a list at the bottom.
Here’s a demonstration of the app:
Considerations for app development are quite different from computer programs, such as the different screen sizes for different mobile platforms, and most of the time only a touch screen can be used. After finishing this project, I acquired some important fundamental concepts and know-hows for app design.
## Guess the Number AI
After completing the first two projects of the Guess the Number series (Guess the Number (Windows) and Guess the Number (iOS)), I made an AI that can play this game at a high-human level.
It has been proven that at most 7 turns are needed to guess the answer, with a best average game length of 5.21 turns. For this game, all the possible combinations (e.g. “0123”, “7381” …) can be saved into a 1D array. After each guess, the possible combinations for the answer will be reduced. Therefore, the algorithm of the program is written for finding a number that will minimize the maximum possible combinations left. The time complexity for each turn is O(n3), and an average of 5 turns of guessing if needed for an arbitrarily chosen number.
For using the program, the user must first choose a 4-digit answer (e.g. “0123”), and input the two numbers [A] and [B] according to the game rules and the numbers guessed by the program. For instance, if the answer is “1357”, and the AI guesses “3127”, the user must input 1 2 ([A] = 1, [B] = 2).
Here’s a demonstration of the AI program guessing the answer “8192” in 5 guesses: |
Satta Number is a sort of digit which we pick in the event that we play the Time Matka Game. This game is completely founded on number. The digit that we pick, we need to wager on that number for the most part, choice of picking number from 0 to 9 are accessible. We need to pick numbers threefold, so in this way we have a three digit number, which is known as Satta Number.
How to deliver Matka Bracket?
In the above segment we had gotten the hang of picking the three digit numbers, presently every one of the digits of that three digit number are added and last digit is looked over
the outcome. Simply guess you have picked 5, 8 and 4 . Presently add them 5 + 8 + 4 = 17 here last digit of result (17) is 7
Presently this will be the number on which you bet and in the event that it comes in draw you will win. Delivering Matka section implies finding the jodi number. On the off chance that you need to
wager on jodi then you need to picked 3 numbers from 0 to 9 two times. Assuming an individual pick once 5, 8, 4 and second time he pick 6 , 4 , 8. Presently add them both and observe there wagering number independently.
5 + 8 + 4 = 17 here 7 is the wagering number
6 + 4 + 8 = 18 here 8 is the wagering number
Jodi is equivalent to 78 Here we will wager on two digit number and that two digit number will be the mix of above wagering number that is 78.
You will see the card in the accompanying manner. [5, 8, 4 *7 X 6, 4, 8*8]
How to play Satta Matka?
Presently we have proactively got to know many course of all matka Game in past part of this article. Here an individual needs to pick a number from 0 to 9 and bet on that number assuming that number comes in consequence of draw he will win. Not no one but you can play on single digit in the event that you need more edge or benefit you can play on two numbers by anticipating them. This is a lottery based game which is played by calculation in the event that you are nice enough in arithmetic and thinking there are higher possibilities of your triumphant this game. Matka Game can be played online through versatile application and sites like our site playing with sites are favored more over portable applications.
How Might I figure Fix Matka Number Daily?
Assuming you get fruitful in speculating the quantity of impending draw certainly you will continue to win. This is known as Matka Guessing and it very well may be finished by breaking down the diagram of entire week and examinations the connection and calculation between the outcome or draw of first draw with different days. Apply that stunt or calculation for the forthcoming draw. Note that these calculation are week by week based so there will be no any point in setting up the month to month diagram and examining it. Assuming that you find it challenging to plan graph you can take the assistance of specialists from our site. Assume you are setting up the graph by your own then for that reason you will require aftereffect of long periods of week so you will begin searching for consequence of week, for that reason we give day to day brings about our site. In the event that you observe any trouble you can reach us through call or whatsapp. |
# Chapter 9: Complex Numbers
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## Transcription
1 Chapter 9: Comple Numbers 9.1 Imaginary Number 9. Comple Number - definition - argand diagram - equality of comple number 9.3 Algebraic operations on comple number - addition and subtraction - multiplication - comple conjugate - division 9.4 Polar form of comple number - modulus and argument - multiplication and division of comple number in polar form 9.5 De Moivre s Theorem - definition - n th power of comple number - n th roots of comple number - proving some trigonometric identities 9.6 Euler s formula - definition - n th power of comple number - n th roots of comple number - relationship between circular and hyperbolic functions 1
2 9.1 Imaginary numbers Consider: 4 This equation has no real solution. To solve the equation, we will introduce an imaginary number. Definition 9.1 (Imaginary Number) The imaginary number i is defined as: i 1 Therefore, using the definition, we will get, 4 4 4( 1) 4i i Eample: Epress the following as imaginary numbers a) 5 b) 8
3 9. Comple Numbers Definition 9. (Comple Numbers) If is a comple number, then it can be epressed in the form : iy, where, y R and i 1. : real part y : imaginary part Or frequently represented as : Re() = and Im() = y Eample: Find the real and imaginary parts of the following comple numbers (a) 1 3i (b) 4i i i 3 3
4 9..1 Argand Diagram We can graph comple numbers using an Argand Diagram. y Re( ) Eample: Sketch the following comple numbers on the same aes. (a) (c) 3 i (b) 3 i 1 3 i (d) 3 i 3 4 4
5 9.. Equality of Two Comple Numbers Given that where 1 a bi and c di 1, C. Two comple numbers are equal iff the real parts and the imaginary parts are respectively equal. So, if 1, then a c and b d. Eample 1: Solve for and y if given Eample : Solve and y. 3 4i (y ) i. for real numbers Eample 3: Solve the following equation for and y where 5
6 9.3 Algebraic Operations on Comple Numbers Addition and subtraction If 1 a bi and c di are two comple numbers, then 1 a c b di Eample: Given. Find a) Z 1 Z b) Z 1 + Z Multiplication If 1 a bi and c di are two comple numbers, and k is a constant, then (i) a bic di (ii) k1 1 ka kbi ac bd ad bci Eample: Given. Find Z 1 Z. 6
7 9.3.3 Comple Conjugate If Note that a bi then the conjugate of is denoted as Division If we are dividing with a comple number, the denominator must be converted to a real number. In order to do that, multiply both the denominator and numerator by comple conjugate of the denominator. 1 1 iy iy 1 iy iy 7
8 Eample 1: Given that 1 1 i, 3 4i. Find 1, and epress it in a bi form. Eample : Given 1 = + i and = 3 4i, find in the form of a + ib. Eample 3: Given Z =. Find the comple conjugate, Write your answer in a + ib form. Eample 4: Given. Find. 8
9 9.4 Polar Form of Comple Numbers P r y O Modulus of, r y. Argument of, arg () = where y tan. From the diagram above, we can see that r cos y r sin Then, can be written as iy r cos irsin r(cos isin ) rcis ( inpolarform) 9
10 Eample 1: Epress i in polar form. Eample : Epress in polar form. Eample 3 Given that 1 = + i and = - + 4i, find such that. Give your answer in the form of a + ib. Hence, find the modulus and argument of. 10
11 9.5 De Moivre s Theorem The n-th Power Of A Comple Number Definition 9.5 ( De Moivre s Theorem) and n R, then If rcos isin n r n cosn isinn Eample 1: a) Write = 1 i in the polar form. Then, using De Moivre s theorem, find 4. b) Use D Moivre s formula to write (-1 i) 1 in the form of a + ib. 11
12 9.5. The n-th Roots of a Comple Number A comple number w is a n-th root of the comple number if w n = or w =. Hence w = 1 = n [ r(cos isin )], 1/ r n k k cos isin n n for k 0,1,,, n 1 Substituting k 0,1,,, n 1 yields the nth roots of the given comple number. Eample 1: Find all the roots for the following equations: (a) (b) 3 i. Eample : Solve and epress them in a + ib form. 1
13 Eample 3: Find all cube roots of. Eample 4: Solve = 0. Sketch the roots on the argand diagram. 13
14 9.5.3 De Moivre s Theorem to Prove Trigonometric Identities De-Moivre s theorem can be used to prove some trigonometric identities. (with the help of Binomial theorem or Pascal triangle.) Eample: Prove that 5 3 cos5 16cos 0cos 5cos and 5 3 sin 5 = 16 sin - 0 sin + 5 sin. Solution: The idea is to write (cos θ + i sin θ) 5 in two different ways. We use both the Pascal triangle and De Moivre s theorem, and compare the results. From Pascal triangle, (cos θ + i sinθ) 5 = cos 5 θ + i 5 cos 4 θ sin θ 10 cos θ sin θ i 10 cos θ sin 3 θ + 5 cosθ sin 4 θ + i sin 5 θ. = (cos 5 θ 10 cos 3 θ sin θ + 5 cos θ sin 4 θ) + i(5 cos 4 θ sin θ 10 cos θ sin 3 θ + sin 5 θ). 14
15 Also, by De Moivre s Theorem, we have (cos θ + i sin θ) 5 = cos 5θ + i sin 5θ. and so cos 5θ + i sin 5θ = (cos 5 θ 10 cos 3 θ sin θ + 5 cos θ sin 4 θ) + i(5 cos 4 θ sin θ 10 cos θ sin 3 θ + sin 5 θ). Equating the real parts gives cos 5θ = cos 5 θ 10 cos 3 θ sin θ + 5 cos θ sin 4 θ. = cos 5 θ 10 cos 3 θ(1 cos θ) + 5 cos θ(1 cos θ) = cos 5 θ 10 cos 3 θ + 10 cos 5 θ + 5 cos θ 10 cos 3 θ + 5 cos 5 θ. = 16 cos 5 θ 0 cos 3 θ + 5 cos θ. (proved) Equating the imaginary parts gives sin 5θ = 5 cos 4 θ sin θ 10 cos θ sin 3 θ + sin 5 θ = 5(1 sin θ) sin θ 10(1 sin θ) sin 3 θ + sin 5 θ = 5(1 sin θ + sin 4 θ) sin θ 10 sin 3 θ + 10 sin 5 θ + sin 5 θ = 5 sin θ 10 sin 3 θ + 5 sin 5 θ 10 sin 3 θ + 10 sin 5 θ + sin 5 θ = 16 sin 5 θ 0 sin 3 θ + 5 sin θ (proved). 15
16 9.6 Eulers s Formula Definition 9.6 Euler s formula states that e i It follows that e in cos isin cosn isinn From the definition, if is a comple number with modulus r and Arg(), ; then r(cos isin ) i re ( ineuler form) Eample: Epress the following comple numbers in the form of (a) 3 + i (b) 4i i re 16
17 9.6.1 The n-th Power Of A Comple Number We know that a comple number can be epress as then i re, i r e 3 3 i 3 r e 4 4 i4 r n r n e e in Eample 1: Given. Find the modulus and argument of 5. Eample : Find in the form of a + ib. Eample 3: Epress the comple number = Then find 1 3i in the form of i re. (a) (b) 3 (c) 7 17
18 9.6. The n-th Roots Of A Comple Number The n-th roots of a comple number can be found using the Euler s formula. Note that: Then, i re ( k ) 1 1 k i r e, k = 0,1 1 1 k i r e, k = 0,1, 1 n r 1 n e k i n, k = 0,1,,,n -1 Eample 1: Find the cube roots of 1 i. Eample : Given. Find all roots of in Euler form. Eample 3: Solve 3 + 8i = 0 and sketch the roots on an Argand diagram. 18
19 9.6.3 Relationship between circular and hyperbolic functions. Euler s formula provides the theoretical link between circular and hyperbolic functions. Since e i cos isin and i e cos isin we deduce that i i i i e e e e cos and sin. i (1) In Chapter 8, we defined the hyperbolic function by cosh e e and sinh e e () Comparing (1) and (), we have so that e coshi tanh i i tan i e i cos i i e e sinhi isin i. 19
20 Also, so that cosi sini i i e e e e cosh i i e e e e sinh i tani i tanh. Using these results, we can evaluate functions such as sin, cos, tan, sinh, cosh and tanh. For eample, to evaluate cos cos( iy) we use the identity cos( A B) cos Acos B sin Asin B and obtain cos cos cosiy sin siniy Using results in (3), this gives cos cos cosh y isin sinh y. (3) 0
21 Eample: Find the values of a) sinh(3 4 i) b) tan 3i 4 c) sin (1 i) 4 (Ans: a) i b) i c) i) 1
22 Pascal s Triangle In general:
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Algebra and Trigonometry3.2 Domain and Range
Learning Objectives
In this section, you will:
• Find the domain of a function defined by an equation.
• Graph piecewise-defined functions.
If you’re in the mood for a scary movie, you may want to check out one of the five most popular horror movies of all time—I am Legend, Hannibal, The Ring, The Grudge, and The Conjuring. Figure 1 shows the amount, in dollars, each of those movies grossed when they were released as well as the ticket sales for horror movies in general by year. Notice that we can use the data to create a function of the amount each movie earned or the total ticket sales for all horror movies by year. In creating various functions using the data, we can identify different independent and dependent variables, and we can analyze the data and the functions to determine the domain and range. In this section, we will investigate methods for determining the domain and range of functions such as these.
Figure 1 Based on data compiled by www.the-numbers.com.3
Finding the Domain of a Function Defined by an Equation
In Functions and Function Notation, we were introduced to the concepts of domain and range. In this section, we will practice determining domains and ranges for specific functions. Keep in mind that, in determining domains and ranges, we need to consider what is physically possible or meaningful in real-world examples, such as tickets sales and year in the horror movie example above. We also need to consider what is mathematically permitted. For example, we cannot include any input value that leads us to take an even root of a negative number if the domain and range consist of real numbers. Or in a function expressed as a formula, we cannot include any input value in the domain that would lead us to divide by 0.
We can visualize the domain as a “holding area” that contains “raw materials” for a “function machine” and the range as another “holding area” for the machine’s products. See Figure 2.
Figure 2
Piecewise Function
A piecewise function is a function in which more than one formula is used to define the output. Each formula has its own domain, and the domain of the function is the union of all these smaller domains. We notate this idea like this:
In piecewise notation, the absolute value function is
How To
Given a piecewise function, write the formula and identify the domain for each interval.
1. Identify the intervals for which different rules apply.
2. Determine formulas that describe how to calculate an output from an input in each interval.
3. Use braces and if-statements to write the function.
Example 11
Writing a Piecewise Function
A museum charges $5 per person for a guided tour with a group of 1 to 9 people or a fixed$50 fee for a group of 10 or more people. Write a function relating the number of people, $n, n,$ to the cost, $C. C.$
Analysis
The function is represented in Figure 23. The graph is a diagonal line from $n=0 n=0$ to $n=10 n=10$ and a constant after that. In this example, the two formulas agree at the meeting point where $n=10, n=10,$ but not all piecewise functions have this property.
Figure 23
Example 12
Working with a Piecewise Function
A cell phone company uses the function below to determine the cost, $C, C,$ in dollars for $g g$ gigabytes of data transfer.
$C(g)={ 25 if 0
Find the cost of using 1.5 gigabytes of data and the cost of using 4 gigabytes of data.
Analysis
The function is represented in Figure 24. We can see where the function changes from a constant to a shifted and stretched identity at $g=2. g=2.$ We plot the graphs for the different formulas on a common set of axes, making sure each formula is applied on its proper domain.
Figure 24
How To
Given a piecewise function, sketch a graph.
1. Indicate on the x-axis the boundaries defined by the intervals on each piece of the domain.
2. For each piece of the domain, graph on that interval using the corresponding equation pertaining to that piece. Do not graph two functions over one interval because it would violate the criteria of a function.
Example 13
Graphing a Piecewise Function
Sketch a graph of the function.
$f(x)={ x 2 if x≤1 3 if 12 f(x)={ x 2 if x≤1 3 if 12$
Analysis
Note that the graph does pass the vertical line test even at $x=1 x=1$ and $x=2 x=2$ because the points $(1,3)(1,3)$ and $(2,2 )(2,2 )$ are not part of the graph of the function, though $(1,1)(1,1)$ and $(2,3)(2,3)$ are.
Try It #8
Graph the following piecewise function.
$f(x)={ x 3 if x<−1 −2 if −14 f(x)={ x 3 if x<−1 −2 if −14$
Q&A
Can more than one formula from a piecewise function be applied to a value in the domain?
No. Each value corresponds to one equation in a piecewise formula.
Media
Access these online resources for additional instruction and practice with domain and range.
3.2 Section Exercises
Verbal
1.
Why does the domain differ for different functions?
2.
How do we determine the domain of a function defined by an equation?
3.
Explain why the domain of $f(x)= x 3 f(x)= x 3$ is different from the domain of $f(x)= x . f(x)= x .$
4.
When describing sets of numbers using interval notation, when do you use a parenthesis and when do you use a bracket?
5.
How do you graph a piecewise function?
Algebraic
For the following exercises, find the domain of each function using interval notation.
6.
$f(x)=−2x(x−1)(x−2) f(x)=−2x(x−1)(x−2)$
7.
$f(x)=5−2 x 2 f(x)=5−2 x 2$
8.
$f( x )=3 x−2 f( x )=3 x−2$
9.
$f( x )=3− 6−2x f( x )=3− 6−2x$
10.
$f(x)= 4−3x f(x)= 4−3x$
11.
$f(x)= x 2 +4 f(x)= x 2 +4$
12.
$f(x)= 1−2x 3 f(x)= 1−2x 3$
13.
$f(x)= x−1 3 f(x)= x−1 3$
14.
$f(x)= 9 x−6 f(x)= 9 x−6$
15.
$f( x )= 3x+1 4x+2 f( x )= 3x+1 4x+2$
16.
$f( x )= x+4 x−4 f( x )= x+4 x−4$
17.
$f(x)= x−3 x 2 +9x−22 f(x)= x−3 x 2 +9x−22$
18.
$f(x)= 1 x 2 −x−6 f(x)= 1 x 2 −x−6$
19.
$f(x)= 2 x 3 −250 x 2 −2x−15 f(x)= 2 x 3 −250 x 2 −2x−15$
20.
$5 x−3 5 x−3$
21.
$2x+1 5−x 2x+1 5−x$
22.
$f(x)= x−4 x−6 f(x)= x−4 x−6$
23.
$f(x)= x−6 x−4 f(x)= x−6 x−4$
24.
$f(x)= x x f(x)= x x$
25.
$f(x)= x 2 −9x x 2 −81 f(x)= x 2 −9x x 2 −81$
26.
Find the domain of the function $f(x)= 2 x 3 −50x f(x)= 2 x 3 −50x$ by:
1. using algebra.
2. graphing the function in the radicand and determining intervals on the x-axis for which the radicand is nonnegative.
Graphical
For the following exercises, write the domain and range of each function using interval notation.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
For the following exercises, sketch a graph of the piecewise function. Write the domain in interval notation.
38.
$f(x)={ x+1 if x<−2 −2x−3 if x≥−2 f(x)={ x+1 if x<−2 −2x−3 if x≥−2$
39.
$f(x)={ 2x−1 if x<1 1+x if x≥1 f(x)={ 2x−1 if x<1 1+x if x≥1$
40.
$f(x)={ x+1ifx<0 x−1ifx>0 f(x)={ x+1ifx<0 x−1ifx>0$
41.
$f( x )={ 3 if x<0 x if x≥0 f( x )={ 3 if x<0 x if x≥0$
42.
43.
$f(x)={ x 2 x+2 ifx<0 ifx≥0 f(x)={ x 2 x+2 ifx<0 ifx≥0$
44.
$f( x )={ x+1 if x<1 x 3 if x≥1 f( x )={ x+1 if x<1 x 3 if x≥1$
45.
$f(x)={ |x| 1 ifx<2 ifx≥2 f(x)={ |x| 1 ifx<2 ifx≥2$
Numeric
For the following exercises, given each function $f, f,$ evaluate $f(−3),f(−2),f(−1), f(−3),f(−2),f(−1),$ and $f(0). f(0).$
46.
$f(x)={ x+1 if x<−2 −2x−3 if x≥−2 f(x)={ x+1 if x<−2 −2x−3 if x≥−2$
47.
48.
For the following exercises, given each function $f, f,$ evaluate $f(−1),f(0),f(2), f(−1),f(0),f(2),$ and $f(4). f(4).$
49.
$f(x)={ 7x+3 if x<0 7x+6 if x≥0 f(x)={ 7x+3 if x<0 7x+6 if x≥0$
50.
$f( x )={ x 2 −2 if x<2 4+| x−5 | if x≥2 f( x )={ x 2 −2 if x<2 4+| x−5 | if x≥2$
51.
$f( x )={ 5x if x<0 3 if 0≤x≤3 x 2 if x>3 f( x )={ 5x if x<0 3 if 0≤x≤3 x 2 if x>3$
For the following exercises, write the domain for the piecewise function in interval notation.
52.
$f(x)={ x+1ifx<−2 −2x−3ifx≥−2 f(x)={ x+1ifx<−2 −2x−3ifx≥−2$
53.
$f(x)={ x 2 −2ifx<1 − x 2 +2ifx>1 f(x)={ x 2 −2ifx<1 − x 2 +2ifx>1$
54.
$f(x)={ 2x−3 −3 x 2 ifx<0 ifx≥2 f(x)={ 2x−3 −3 x 2 ifx<0 ifx≥2$
Technology
55.
Graph $y= 1 x 2 y= 1 x 2$ on the viewing window $[−0.5,−0.1] [−0.5,−0.1]$ and $[0.1,0.5]. [0.1,0.5].$ Determine the corresponding range for the viewing window. Show the graphs.
56.
Graph $y= 1 x y= 1 x$ on the viewing window $[−0.5,−0.1] [−0.5,−0.1]$ and $[0.1,0.5]. [0.1,0.5].$ Determine the corresponding range for the viewing window. Show the graphs.
Extension
57.
Suppose the range of a function $f f$ is $[−5,8]. [−5,8].$ What is the range of $|f(x)|? |f(x)|?$
58.
Create a function in which the range is all nonnegative real numbers.
59.
Create a function in which the domain is $x>2. x>2.$
Real-World Applications
60.
The height $h h$ of a projectile is a function of the time $t t$ it is in the air. The height in feet for $t t$ seconds is given by the function $h(t)=−16 t 2 +96t. h(t)=−16 t 2 +96t.$ What is the domain of the function? What does the domain mean in the context of the problem?
61.
The cost in dollars of making $x x$ items is given by the function $C(x)=10x+500. C(x)=10x+500.$
1. The fixed cost is determined when zero items are produced. Find the fixed cost for this item.
2. What is the cost of making 25 items?
3. Suppose the maximum cost allowed is \$1500. What are the domain and range of the cost function, $C(x)? C(x)?$
Footnotes
• 3The Numbers: Where Data and the Movie Business Meet. “Box Office History for Horror Movies.” http://www.the-numbers.com/market/genre/Horror. Accessed 3/24/2014
• 4http://www.eia.gov/dnav/pet/hist/LeafHandler.ashx?n=PET&s=MCRFPAK2&f=A.
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# Video: Solving Problems Involving Percentage
Estimate the following: 90% of 91.
03:00
### Video Transcript
Estimate the following: 90 percent of 91.
In order to estimate an answer, we need to round our numbers to one significant figure. 90 percent is already rounded to one significant figure as it only has one nonzero digit. 91 rounded to one significant figure is equal to 90 as it is closer to 90 than one hundred. This is because the units digit one is less than five. The estimated calculation that we need to work out is 90 percent of 90.
There are lots of ways of answering this problem. One way would be to work out 10 percent first. To work out 10 percent of any number, we divide by 10. 90 divided by 10 is equal to nine. Therefore, 10 percent of 90 equals nine. We can therefore work out 90 percent by multiplying this answer by nine. Remember, whatever you do to one side of an equation, you must do to the other. Nine multiplied by nine is equal to 81. This means that 90 percent of 90 is equal to 81.
An alternative method of calculating 90 percent of 90 would be to turn our percent into a fraction or a decimal. 90 percent is equal to ninety hundredths as a fraction and 0.9 or 0.90 as a decimal. This is because the word percent means out of 100. The word of in maths means multiply. We can therefore multiply ninety hundredths or 0.9 by 90 to calculate 90 percent of 90. Once again both of these answers are equal to 81. When solving the fraction problem, we can cancel the zeros, which leaves us with nine multiplied by nine. This is equal to 81.
The calculation 0.9 multiplied by 90 can be rewritten as 0.9 multiplied by 10 multiplied by nine. 0.9 multiplied by 10 is equal to nine. And once again, we’re left with nine multiplied by nine, which is 81. We have therefore worked out using various methods that the estimate of 90 percent of 91 is equal to 81. We could work out the exact value by multiplying 0.9 by 91. This is equal to 81.9, which is very close to our estimated value. |
# Decimal Fractions Questions
FACTS AND FORMULAE FOR DECIMAL FRACTION QUESTIONS
I. Decimal Fractions : Fractions in which denominators are powers of 10 are known as decimal fractions.Thus
1/10 = 1 tenth = .1;
1/100 = 1 hundredth = .01;
99/100 = 99 hundreths = .99;
7/1000 = 7 thousandths = .007 etc
II. Conversion of a Decimal Into Vulgar Fraction : Put 1 in the denominator under the decimal point and annex with it as many zeros as is the number of digits after the decimal point.
Now, remove the decimal point and reduce the fraction to its lowest terms. Thus
0.25 = 25/100 = 1/4;
2.008 = 2008/1000 = 251/125.
III.
1. Annexing zeros to the extreme right of a decimal fraction does not change its value.
Thus, 0.8 = 0.80 = 0.800, etc.
2. If numerator and denominator of a fraction contain the same number of decimal places, then we remove the decimal sign.Thus
1.84/2.99 = 184/299 = 8/13;
0.365/0.584 = 365/584 = 5/8
IV. Operations on Decimal Fractions :
1. Addition and Subtraction of Decimal Fractions : The given numbers are so placed under each other that the decimal points lie in one column. The numbersso arranged can now be added or subtracted in the usual way.
2. Multiplication of a Decimal Fraction By a Power of 10 : Shift the decimal point to the right by as many places as is the power of 10. Thus
5.9632 x 100 = 596.32;
0.073 x 10000 = 0.0730 x 10000 = 730.
3.Multiplication of Decimal Fractions : Multiply the given numbers considering them without the decimal point. Now, in the product, the decimal point is marked off to obtain as many places of decimal as is the sum of the number of decimal places in the given numbers.
Suppose we have to find the product (.2 x .02 x .002).
Now, 2 x 2 x 2 = 8. Sum of decimal places = (1 + 2 + 3) = 6.
Therefore, .2 x .02 x .002 = .000008.
4.Dividing a Decimal Fraction By a Counting Number : Divide the given number without considering the decimal point, by the given counting number.Now, in the quotient, put the decimal point to give as many places of decimal as there are in the dividend.
Suppose we have to find the quotient (0.0204 / 17).
Now, 204 / 17 = 12. Dividend contains 4 places of decimal.
So, 0.0204 / 17 = 0.0012.
5. Dividing a Decimal Fraction By a Decimal Fraction : Multiply both the dividend and the divisor by a suitable power of 10 to make divisor a whole number. Now, proceed as above.
Thus, 0.00066/0.11 = (0.00066 x 100)/(0.11 x 100) = (0.066/11) = 0.006
V. Comparison of Fractions : Suppose some fractions are to be arranged in ascending or descending order of magnitude. Then, convert each one of the given fractions in the decimal form, and arrange them accordingly.
Suppose, we have to arrange the fractions 3/5, 6/7 and 7/9 in descending order. now, 3/5=0.6, 6/7 = 0.857, 7/9 = 0.777.... since 0.857 > 0.777... > 0.6
So 6/7 > 7/9 > 3/5
VI. Recurring Decimal : If in a decimal fraction, a figure or a set of figures is repeated continuously, then such a number is called a recurring decimal. In a recurring decimal, if a single figure is repeated, then it is expressed by putting a dot on it. If a set of figures is repeated, it is expressed by putting a bar on the set .
Thus 1/3 = 0.3333….= 0.3; 22 /7 = 3.142857142857..... = $3.\overline{142857}$
Pure Recurring Decimal: A decimal fraction in which all the figures after the decimal point are repeated, is called a pure recurring decimal.
Converting a Pure Recurring Decimal Into Vulgar Fraction : Write the repeated figures only once in the numerator and take as many nines in the denominator as is the number of repeating figures.
Thus , etc...
Mixed Recurring Decimal: A decimal fraction in which some figures do not repeat and some of them are repeated, is called a mixed recurring decimal. e.g., 0.17333 = $0.17\overline{3}$
Converting a Mixed Recurring Decimal Into Vulgar Fraction : In the numerator, take the difference between the number formed by all the digits after decimal point (taking repeated digits only once) and that formed by the digits which are not repeated, In the denominator, take the number formed by as many nines as there are repeating digits followed by as many zeros as is the number of non-repeating digits.
Thus, 0.16 = (16-1) / 90 = 15/90 = 1/6;
$0.22\overline{73}=\frac{2273-22}{9900}=\frac{2251}{9900}$
VII. Some Basic Formulae :
1.$\left(a+b\right)\left(a-b\right)=\left({a}^{2}-{b}^{2}\right)$
2. ${\left(a+b\right)}^{2}={a}^{2}+{b}^{2}+2ab$
3. ${\left(a-b\right)}^{2}={a}^{2}+{b}^{2}-2ab$
4. ${\left(a+b+c\right)}^{2}={a}^{2}+{b}^{2}+{c}^{2}+2\left(ab+bc+ca\right)$
5. $\left({a}^{3}+{b}^{3}\right)=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)$
6. $\left({a}^{3}-{b}^{3}\right)=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)$
7. $\left({a}^{3}+{b}^{3}+{c}^{3}-3abc\right)=\left(a+b+c\right)\left({a}^{2}+{b}^{2}+{c}^{2}-ab-bc-ac\right)$
8. When a+b+c=0, then ${a}^{3}+{b}^{3}+{c}^{3}=3abc$
Q:
If (1 / 3.197) = 0.3127, find the value of (1 / 0.0003197).
A) 3127 B) 3197 C) 312.7 D) 0.3127
Explanation:
3 1592
Q:
How much does one need to add to 2/3 to obtain 3/2?
A) 5/6 B) 1.5/6 C) 1/-1 D) 4/9
Explanation:
3 1544
Q:
The sum of a fraction and 3 times its reciprocal is 31/6. What is the fraction?
A) 2/9 B) 9/2 C) 5/4 D) 4/5
Explanation:
2 1491
Q:
The ratio 1 : 5 can be expressed in decimal as
A) 0.5 B) 0.2 C) 0.02 D) 0.1
Explanation:
1 1369
Q:
The product of two decimals is 0.768. If one of the decimal number is 1.6, find the other.
A) 0.48 B) 0.37 C) 0.42 D) 0.47
Explanation:
2 1331
Q:
1/300 written as a recurring decimal is
A) 0.3 B) 0.03 C) 0.003 D) 0.0003
Explanation:
4 1284
Q:
The fraction from the ones listed below that will NOT lead to a recurring decimal is
A) 8/56 B) 6/56 C) 4/56 D) 7/56
Explanation:
2 1256
Q:
Which of the below given fractions is NOT equal to 9/17?
A) 153/289 B) 108/221 C) 63/119 D) 27/51 |
Math resources Geometry Surface area
Surface area of a rectangular prism
# Surface area of a rectangular prism
Here you will learn about the surface area of a rectangular prism and how to calculate it.
Students will first learn about the surface area of a rectangular prism as part of geometry in 6th grade.
## What is the surface area of a rectangular prism?
The surface area of a rectangular prism is the total area of all of the faces of a rectangular prism.
Rectangular prisms have three pairs of identical faces – top and bottom, front and back, and left and right. Notice that it is always the opposite faces that are identical. Rectangular prisms are also known as cuboids.
To calculate the total surface area of a rectangular prism, calculate the area of each rectangular face and add them all together.
For example,
You can also use the net of a rectangular prism to find the surface area. The rectangular prism above unfolded into its net would be:
Surface area is measured in square units (e.g. mm^2, cm^2, m^2 etc).
## Common Core State Standards
How does this relate to 6th grade math?
• Grade 6 – Geometry (6.G.A.4)
Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area of these figures. Apply these techniques in the context of solving real-world and mathematical problems.
## How to calculate the surface area of a rectangular prism
In order to work out the surface area of a rectangular prism:
1. Calculate the area of each face.
2. Add the six areas together.
3. Include the units.
## Surface area of a rectangular prism examples
### Example 1: surface area of a rectangular prism
Calculate the surface area of the rectangular prism.
1. Calculate the area of each face.
The area of the bottom is 9 \times 3=27 \, cm^2.
The top face is the same as the bottom face, so the area of the top is also 27 \, cm^2.
The area of the front is 9 \times 4=36 \, cm^2.
The back face is the same as the front face, so the area of the back is also 36 \, cm^2.
The area of the right hand side is 3 \times 4=12 \, cm^2.
The left side face is the same as the right side face, so the area of the left side is also 12 \, cm^2.
It will make our work clearer if you use a table:
Total surface area: 27+27+36+36+12+12=150
3Include the units.
The measurements on the rectangular prism are in cm, so the total surface area of the rectangular prism is 150 \mathrm{~cm}^2.
### Example 2: surface area of a rectangular prism
Calculate the surface area of the rectangular prism.
Calculate the area of each face.
Include the units.
### Example 3: surface area of a cube
Calculate the surface area of this cube.
Calculate the area of each face.
Include the units.
### Example 4: surface area with different units
Given that the base of this rectangular prism is the bottom rectangle, calculate the lateral surface area.
Calculate the area of each face.
Include the units.
### Example 5: surface area with a missing length
The surface area of this rectangular prism is 328 \mathrm{~cm}^2. Find the missing length.
Calculate the area of each face.
Include the units.
### Example 6: surface area word problem – missing height
Helena was making rectangular boxes out of wood. If she used 200 \mathrm{~cm}^2 of wood for the rectangular box shown above, what is the missing height?
Calculate the area of each face.
Include the units.
### Teaching tips for the surface area of a rectangular prism
• Provide opportunities that connect the surface area of rectangular prisms to real world applications. The connection to wrapping gifts or painting the sides of a prism are popular examples. Also allow students to think critically about each real life connection.
Asking questions like “When wrapping a gift, is it possible to use more paper than the surface area? Why or why not?” or “When painting a rectangular prism in the real world, what is a situation where you might not paint the entire surface area?”
• Worksheets can be a useful practice tool for this topic, but make sure that students also have a sufficient amount of practice with real world 3D shapes and nets. Solving problems on worksheets require students to be able to work with 2D representations of 3D shapes.
This is much easier for students if they have worked extensively with these shapes in the real world and can easily picture and rotate them in their minds.
### Easy mistakes to make
• Calculating volume instead of surface area
Volume and surface area are different things – volume is the space within the shape and is measured in cubic units, whereas surface area is the total area of the faces and is measured in square units. To find surface area, you need to work out the area of each face and add them together.
• Thinking all lateral faces are equal
Only opposite sides are equal in a rectangular prism. This means that many rectangular prisms have 3 different sets of congruent rectangles.
For example,
It is possible for all lateral faces to be equal, but only if the base is a square.
For example,
### Practice surface area of a rectangular prism questions
1. Calculate the surface area of the rectangular prism.
342 \mathrm{~cm}^{2}
315 \mathrm{~cm}^{2}
222 \mathrm{~cm}^{2}
462 \mathrm{~cm}^{2}
Calculate the area of each of the six faces:
Total surface area: 105+105+45+45+21+21=342\mathrm{~cm}^{2}
Since the dimensions are in centimeters, the surface area is in square centimeters.
2. The perimeter of the base is 16 inches. Calculate the surface area of the cube.
64\mathrm{~inches}^{2}
24\mathrm{~inches}^{2}
96 \mathrm{~inches}^{2}
256 \mathrm{~inches}^{2}
The base of a cube is a square and all sides are equal.
Since the perimeter is found by adding all the sides together, it can be shown as s + s + s + s =16.
The side length of 4 makes this equation true, so each side is 4 inches.
Since it is a cube, all of the faces are the same square.
The area of each face is 4\times 4=16 \mathrm{~inches}^{2}.
There are six identical faces, so the total surface area of the cube is 6 \times 16=96 \mathrm{~inches}^{2}.
Since the dimensions are in inches, the surface area is in square inches.
3. Calculate the surface area of the rectangular prism.
247.5 \mathrm{~ft}^2
495.18 \mathrm{~ft}^2
371.34 \mathrm{~ft}^2
185.7 \mathrm{~ft}^2
Calculate the area of each of the six faces:
Total surface area:
58.95+58.95+94.32+94.32+32.4+32.4=371.34 \, ft^2.
4. Calculate the surface area of this rectangular prism.
33.56\mathrm{~cm}^{2}
4.2 \mathrm{~cm}^{2}
12,400 \mathrm{~cm}^{2}
8,900 \mathrm{~cm}^{2}
Some of the measurements are in m and one is in cm.
All of the measurements need to be in the same units, so convert the meters to centimeters.
0.7 \, m=70 \,cm and 0.4 \, m=40 \, cm.
Now calculate the areas:
Total surface area:
1,050+1,050+2,800+2,800+600+600=8,900\mathrm{~cm}^{2}.
5. Given that the surface area of this rectangular prism is 142 \, \mathrm{~cm}^{2}, find the value of x.
6.76 \mathrm{~cm}
14.2 \mathrm{~cm}
20.29 \mathrm{~cm}
5 \mathrm{~cm}
Calculate the area of each of the six faces:
Total surface area: 7x+7x+3x+3x+21+21=20x+42.
Since you know the total surface area is 142\mathrm{~cm}^{2}, the equation 20 x+42=142 can be used to find the height.
Solve this equation by substituting in the answer choices:
\begin{aligned} & 20 \times 6.76+42=142 \\\\ & 135.2+42=142 \\\\ & 177.2=142 \end{aligned}
This equation is not true.
Since 177.2 is larger than 142, \, x needs to be a smaller value.
The only answer choice that is smaller is 5.
\begin{aligned} & 20 \times 5+42=142 \\\\ & 100+42=142 \\\\ & 142=142 \end{aligned}
This equation is true.
The missing measurement on the base is 5 \mathrm{~cm}.
6. A pet store covers the left side, right side and back side of aquariums with blue paper that looks like water. Below are the measurements of the small and large tank. How much more paper does the large tank use than the small?
133 \text { inches}^2
504 \text { inches}^2
371 \text { inches}^2
806 \text { inches}^2
Calculate the area of each of the left, right and back faces for each aquarium:
Large Aquarium;
Surface area of the blue paper used on the large aquarium:
280+112+112=504 \text { inches}^2
Small Aquarium;
Surface area of the blue paper used on the small aquarium:
77+28+28=133 \text { inches}^2
Subtract to find the difference:
504-133=371 \text { inches}^2
## Surface area of a rectangular prism FAQs
What is a cuboid?
It is the name for a three-dimensional shape that is made up of rectangles and/or squares. A cuboid is another name for rectangular prisms.
What is the difference between a rectangular prism and a rectangular pyramid?
They both have a rectangular base, but a pyramid has one base and triangular lateral faces that meet in a point. A rectangular prism has two bases and rectangular lateral faces.
Is there a surface area of a rectangular prism formula?
Yes, the general formula is 2(lw+wh+hl).
How is the volume of a rectangular prism calculated?
The volume can be calculated with the formula V=l \times w \times h.
## Still stuck?
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### Diophantine N-tuples
Can you explain why a sequence of operations always gives you perfect squares?
### DOTS Division
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
### Sixational
The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. Prove that all terms of the sequence are divisible by 6.
# Pair Products
##### Age 14 to 16Challenge Level
Kira, from the UK, started off by doing some calculations:
Say your numbers were 4, 5, 6, 7. When you times the outer numbers 4 and 7 it will equal 28, and when you times the two middle numbers 5 and 6 it will equal 30.
Let's try it again on four new consecutive numbers, such as 1, 2, 3, 4: when you times the outer numbers 1 and 4 it equals 4, and when you times the two middle numbers 2 and 3 it equals 6.
Conclusion: the product of the outer numbers will always be 2 less than the product of the numbers in the middle.
The class at Elgin Academy had the clever idea of testing this for some negative numbers too!
Tom drew rectangles, like in Alison's explanation:
If we overlap the two rectangles we cut a piece off at the bottom and create a new piece at the right hand side. The difference between the two small rectangles is always two. This works for all numbers, not just $9, 10, 11, 12$.
Sam, from Park Grove Primary, gave the following explanation:
If you choose the first number of the set to be n, then the outside multiplication will be n(n+3), which simplifies to $n^2+3n$.
The inside multiplication will be (n+1)(n+2), which simplifies to $n^2+3n+2$.
If you look back at the outside multiplication, you can see that it is 2 less than the inside multiplication, proving the answer.
Great! Nathan also gave a convincing explanation:
Call the number halfway in between the middle two x.
This is the second number in our list plus 0.5, or the third number minus 0.5, so if we multiply the second and third numbers together we will get $x^2-0.25$.
But x is also the first number plus 1.5, or the fourth number minus 1.5, so if we multiply them together we will get $x^2-2.25$.
Nice - here Nathan is making use of the identity: $(x-a)(x+a) = x^2 - a^2$. Well done!
Keone, from Sage Ridge School in Reno, started with four consecutive whole numbers:
The product of the first and last numbers is always $2$ less than the product of the middle two numbers.
Explanation: Suppose the first number is $x$. Then the second number is $x+1$, the third is $x+2$, and the fourth is $x+3$.
So the product of the first and fourth numbers is $x(x+3) = x^2 + 3x$.
Also, the product of the second and third numbers is $(x+1)(x+2) = x^2 + 3x + 2$.
So $(x+1)(x+2) = x(x+3) + 2$ for any chosen value of $x$.
When he considered five consecutive whole numbers he found that:
The product of the first and last numbers is always $3$ less than the product of the second and fourth numbers.
Explanation: Again, let the first number be $x$; then the second number is $x+1$, the third is $x+2$, the fourth is $x+3$, and the fifth is $x+4$.
So the product of the first and last numbers is $x(x+4) = x^2 + 4x$.
Also, the product of the second and fourth numbers is $(x+1)(x+3) = x^2 + 4x + 3$.
So $(x+1)(x+3) = x(x+4) + 3$ for any chosen value of $x$.
And with $n$ consecutive whole numbers he found that:
The product of the first and last numbers is always $n-2$ less than the product of the second and penultimate numbers.
Let us consider the general case where there are n consecutive whole numbers.
As before, let the first number be $x$; then the second number is x+1, the third is x+2, and so on.
The last number (the $n$th number) will be $x+n-1$. Thus, the penultimate (second-to-last) number will be $x+n-2$.
So the product of the first number and the last number will be $x(x+n-1) = x^2 + nx - x$
The product of the second and the penultimate numbers will be $(x+1)(x+n-2) = x^2 + nx - 2x + x + n - 2 = x^2 + nx - x + n - 2$
So $(x+1)(x+n-2) = x(x+n-1) + n - 2$;
that is, the product of the second and penultimate numbers will always exceed the product of the first and last numbers by exactly $n - 2$.
For example, if we take 6 numbers, the product of the $2$nd and $5$th numbers will be $4$ more than the product of the $1$st and last numbers.
We received similar findings from Aisling, from Grand Avenue Primary School:
The product of the first and last numbers of a series of consecutive whole numbers, is the same as the product of the second and penultimate numbers of that series, minus the number of numbers separating the first and last numbers.
Natasha, from the European School, generalised her findings in a similar way and then went on to check her conclusion:
We can generalise this problem by substituting particular numbers with letters.
Let the first number be $a$.
If there are $n$ numbers, where a is the first, then the last number is $(a+n-1)$.
The second number is $(a+1)$, the penultimate number $(a+n-2)$.
By multiplying the first and last numbers together, we get $a(a+n-1) = a^2 +an-a$
Multiplication of the second and penultimate numbers gives
$(a+1)(a+n-2) = a^2+an-a+n-2$
The difference therefore in the product of the first and last numbers and the product of the second and penultimate numbers is always $n-2$.
For added confirmation we can take a random example:
Consider the numbers $56, 57, 58, 59, 60, 61, 62$
where $n = 7$ and $a = 56$
Our general formula tells us that the difference in the product pairs should be $n-2$ (i.e. $5$).
When we do the calculation, we get
$$57 \times61 - 56 \times62 = 3477 - 3472 = 5$$
This result corresponds with the general one established above.
Theo from Dulwich College generalised this problem further:
How would one go about finding the difference between the product of the first and last numbers and the product of the second and penultimate numbers, where each number is not consecutive but increases by possibly 2, 3 or 4? Could we even find a general rule for any sequence that increases by a constant? Turns out we can!
Let's take the sequence 11, 15, 19, 23, 27, 31
15 × 27 - 11 × 31 = 405$-$341 = 64
But now let's try and use this sequence to make a general rule:
Let us call '$a$' the first number in our sequence, '$n$' the number of terms in the sequence and $c$ the constant that we increase by between each two terms.
Therefore, in the example 11, 15, 19, 23, 27, 31;
$a$(first number)=11; $n$(number of terms)=6; $c$(constant by which we increase by each time)=4
In this example we can notice that the last term was 5 times $c$ (which in that case was 4) more than the first term:
31 - 5×4 = 11
The penultimate term was 4 times $c$ (which in that case was 4) more than the first term:
27 - 4×4 = 11
Remember that in that example $n$ is 6 so to get the last term, you have to add $(n-1)c$ and for the penultimate term, you have to add $(n-2)c.$
We can actually make a sequence of numbers with what we have just observed:
$a, a+c, ”¦ , a+(n-2)c, a+(n-1)c$
The product of the 1st and last term will be:
$\begin{split} a(a+(n-1)c)&= a(a+nc-c) \\ &=a^2+ anc -ca \\ &=a^2+(n-1)ca\end{split}$
The product of the 2nd and penultimate term will be:
$(a+c)(a+(n-2)c)$
$=(a+c)(a+nc-2c)$
$=a^2+ anc-2ca+ca+nc^2-2c^2$
$=a^2 + (n-2+1)ca + (n-2)c^2$
$=a^2 +(n-1)ca+ (n-2)c^2$
So, in short,
the product of the 1st and last term will be $a^2+(n-1)ca$
the product of the 2nd and penultimate term will be $a^2 +(n-1)ca+ (n-2)c^2$
Therefore, the product of the second and penultimate numbers will always be greater than the product of the first and last numbers by exactly $(n-2)c^2$.
We can see if this works with our initial example:
11, 15, 19, 23, 27, 31;
$n$(number of terms)=6; $c$(constant by which we increase by each time)=4
(6-2)42= 64. This is indeed what we got at the start!
It is interesting to notice that $a$ (the first term) is irrelevant to the final answer.
This is also beautiful as it works with all rational and irrational numbers as well as imaginary and complex numbers:
$2+4i, 3+7i, 4+10i, 5 + 13i$
$n=4$ and $c=1+3i$
$(4-2)(1+3i)^2=2(1+3i+3i-9)=2(6i-8)= 12i -16$
$\begin{split}(3+7i)(4+10i)-(2+4i)(5+13i)&=(12+30i+28i-70)-(10+26i+20i-52)\\& = (58i-58)-(46i-42) = 12i - 16\end{split}$
Sure enough, the product of the 1st and last number is $12i-16$ greater than the product of the 2nd and penultimate numbers. |
# Independent Probability
Independent probability is when we look at the probability of outcomes regarding Independent events.
Independent Events
With probability, events are independent of each other if the outcome of one event is NOT affected by the outcome of any previous event or events.
So if you have an event with 2 possible outcomes, A and B,
the outcome of event A does NOT affect the outcome of event B.
We can see how this works in an example.
Example
Somebody is going to roll a standard dice 2 times.
They would like to land a number 4 on both rolls.
From the probability introduction page, we know that:
Probability of an Event = \boldsymbol{\frac{\tt\color{green}Number \space of \space ways \space an \space event \space can \space occur}{\tt\color{blue}Number \space of \space different \space possibilities}}
A standard dice has 6 numbers,
so the odds of landing the number 4 on one roll is \bf{\frac{1}{6}}.
Now because the 2nd roll is not affected in any way by the outcome of the 1st roll.
The odds of landing a 4 on the 2nd roll is also \bf{\frac{1}{6}}.
The 1st roll and the 2nd roll of the dice are Independent events.
## Probability of Multiple Events
It was said that the person rolling the dice, wanted to know the probability of landing a 4 on both rolls, one after the other.
When trying to find the probability of multiple Independent events occurring together, we multiply each individual probability together.
Using probability notation this looks like.
P( A and B ) = P( A ) × P( B )
So for the probability of landing two 4's from two rolls of a standard dice.
P( 4 and 4 ) = P( 4 ) × P( 4 ) = \bf{\frac{1}{6}} × \bf{\frac{1}{6}} = \bf{\frac{1}{36}}
It's not very likely that two 4's will be rolled in a row.
But what about a 4 being rolled 5 times in a row?
P( Five 4's ) = \bf{\frac{1}{6}} × \bf{\frac{1}{6}} × \bf{\frac{1}{6}} × \bf{\frac{1}{6}} × \bf{\frac{1}{6}} = \bf{\frac{1}{7776}}
Extremely unlikely, you'd expect to have to roll a dice nearly 8'000 times in order to get a sequence of five 4's in a row.
Six 4's is even more unlikely, \bf{\frac{1}{6}} × \bf{\frac{1}{6}} × \bf{\frac{1}{6}} × \bf{\frac{1}{6}} × \bf{\frac{1}{6}} × \bf{\frac{1}{6}} = \bf{\frac{1}{46656}}
But the fact of the situation is, even if five 4's in a row have already been rolled.
The probability of a 4 on the sixth roll is still just \bf{\frac{1}{6}}.
As the the previous five dice rolls are Independent of the sixth roll, regardless of what their outcomes were.
This is a situation that involves Independent probability.
## Independent Probability Examples
(1.1)
Tossing a fair sided coin 2 times in a row.
What is the probability of landing 2 tails in a row?
Solution
P( Two Tails ) = P( Tail ) × P( Tail ) = \bf{\frac{1}{2}} × \bf{\frac{1}{2}} = \bf{\frac{1}{4}} = 0.25
(1.2)
Picking a single card from a standard deck of 52 cards.
What is the probability of picking out a card that is a Spade, as well as a Face card?
Solution
Although we are picking just 1 card from a 52 card deck.
Picking a Spade card, and picking a Face card, are still separate events that are Independent of each other.
So we can use the same approach to find the probability of picking out 1 card, that is both a Face card and Spade card.
P( Spade ) = \bf{\frac{13}{52}} , P( Face card ) = \bf{\frac{12}{52}} = \bf{\frac{3}{13}}
P( Spade & Face card ) = \bf{\frac{13}{52}} × \bf{\frac{3}{13}} = \bf{\frac{39}{676}} = 0.058
(1.3)
Picking a 2 marbles from a bag of marbles.
There are 2 red marbles, 3 yellow marbles and 4 black marbles.
What is the probability of picking out a red marble, then a black marble, with replacement?
Solution
As there is replacement, meaning that the first marble picked is put back in the bag, the 2 pickings are independent, and the first marble picked doesn't affect the second.
So this is an Independent probability situation.
P( Red marble ) = \bf{\frac{2}{9}} , P( Black marble ) = \bf{\frac{4}{9}}
P( Red and Black ) = \bf{\frac{2}{9}} × \bf{\frac{4}{9}} = \bf{\frac{8}{81}} = 0.099
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Mathematics » Properties of Real Numbers » Systems of Measurement
# Using Mixed Units of Measurement in the U.S. System
## Using Mixed Units of Measurement in the U.S. System
Performing arithmetic operations on measurements with mixed units of measures requires care. Be sure to add or subtract like units.
## Example
Charlie bought three steaks for a barbecue. Their weights were $$14$$ ounces, $$1$$ pound $$2$$ ounces, and $$1$$ pound $$6$$ ounces. How many total pounds of steak did he buy?
(credit: Helen Penjam, Flickr)
### Solution
We will add the weights of the steaks to find the total weight of the steaks.
Add the ounces. Then add the pounds. Convert 22 ounces to pounds and ounces. Add the pounds. 2 pounds + 1 pound, 6 ounces 3 pounds, 6 ounces Charlie bought 3 pounds 6 ounces of steak.
## Example
Anthony bought four planks of wood that were each $$6$$ feet $$4$$ inches long. If the four planks are placed end-to-end, what is the total length of the wood?
### Solution
We will multiply the length of one plank by $$4$$ to find the total length.
Multiply the inches and then the feet. Convert 16 inches to feet. 24 feet + 1 foot 4 inches Add the feet. 25 feet 4 inches Anthony bought 25 feet 4 inches of wood. |
$$\require{cancel}$$
# 4.2: Acceleration Vector
Learning Objectives
• Calculate the acceleration vector given the velocity function in unit vector notation.
• Describe the motion of a particle with a constant acceleration in three dimensions.
• Use the one-dimensional motion equations along perpendicular axes to solve a problem in two or three dimensions with a constant acceleration.
• Express the acceleration in unit vector notation.
## Instantaneous Acceleration
In addition to obtaining the displacement and velocity vectors of an object in motion, we often want to know its acceleration vector at any point in time along its trajectory. This acceleration vector is the instantaneous acceleration and it can be obtained from the derivative with respect to time of the velocity function, as we have seen in a previous chapter. The only difference in two or three dimensions is that these are now vector quantities. Taking the derivative with respect to time $$\vec{v}$$(t), we find
$$\vec{a} (t) = \lim_{t \rightarrow 0} \frac{\vec{v} (t + \Delta t) - \vec{v} (t)}{\Delta t} = \frac{d\vec{v} (t)}{dt} \ldotp \label{4.8}$$
The acceleration in terms of components is
$$\vec{a} (t) = \frac{dv_{x} (t)}{dt}\; \hat{i} + \frac{dv_{y} (t)}{dt}\; \hat{j} + \frac{dv_{z} (t)}{dt}\; \hat{k} \ldotp \label{4.9}$$
Also, since the velocity is the derivative of the position function, we can write the acceleration in terms of the second derivative of the position function:
$$\vec{a} (t) = \frac{d^{2} x(t)}{dt^{2}}\; \hat{i} + \frac{d^{2} y(t)}{dt^{2}}\; \hat{j} + \frac{d^{2} z(t)}{dt^{2}}\; \hat{k} \ldotp \label{4.10}$$
Example 4.4
### Finding an Acceleration Vector
A particle has a velocity of $$\vec{v}$$(t) = 5.0t $$\hat{i}$$ + t2 $$\hat{j}$$ − 2.0t3 $$\hat{k}$$ m/s. (a) What is the acceleration function? (b) What is the acceleration vector at t = 2.0 s? Find its magnitude and direction.
### Solution
1. We take the first derivative with respect to time of the velocity function to find the acceleration. The derivative is taken component by component: $$\vec{a} (t) = 5.0\; \hat{i} + 2.0t\; \hat{j} - 6.0t^{2}\; \hat{k}\; m/s^{2} \ldotp$$
2. Evaluating $$\vec{a}$$ (2.0\; s) = 5.0 $$\hat{i}$$ + 4.0 $$\hat{j}$$ - 24.0 $$\hat{k}$$ m/s2 gives us the direction in unit vector notation. The magnitude of the acceleration is $$|\vec{a} (2.20\; s)| = \sqrt{5.0^{2} + 4.0^{2} + (-24.0)^{2}} = 24.8\; m/s^{2} \ldotp$$
### Significance
In this example we find that acceleration has a time dependence and is changing throughout the motion. Let’s consider a different velocity function for the particle.
Example 4.5
### Finding a Particle Acceleration
A particle has a position function $$\vec{r}$$ (t) = (10t − t2) $$\hat{i}$$ + 5t $$\hat{j}$$ + 5t $$\hat{k}$$ m. (a) What is the velocity? (b) What is the acceleration? (c) Describe the motion from t = 0 s.
### Strategy
We can gain some insight into the problem by looking at the position function. It is linear in y and z, so we know the acceleration in these directions is zero when we take the second derivative. Also, note that the position in the x direction is zero for t = 0 s and t = 10 s.
### Solution
1. Taking the derivative with respect to time of the position function, we find $$\vec{v}$$(t) = (10 − 2t) $$\hat{i}$$ + 5 $$\hat{j}$$ + 5 $$\hat{k}$$ m/s. The velocity function is linear in time in the x direction and is constant in the y and z directions.
2. Taking the derivative of the velocity function, we find $$\vec{a}(t) = −2\; \hat{i} m/s^{2} \ldotp$$The acceleration vector is a constant in the negative x-direction.
3. The trajectory of the particle can be seen in Figure 4.9. Let’s look in the y and z directions first. The particle’s position increases steadily as a function of time with a constant velocity in these directions. In the x direction, however, the particle follows a path in positive x until t = 5 s, when it reverses direction. We know this from looking at the velocity function, which becomes zero at this time and negative thereafter. We also know this because the acceleration is negative and constant—meaning, the particle is decelerating, or accelerating in the negative direction. The particle’s position reaches 25 m, where it then reverses direction and begins to accelerate in the negative x direction. The position reaches zero at t = 10 s.
Exercise 4.2
Suppose the acceleration function has the form $$\vec{a}$$(t) = a $$\hat{i}$$ + b $$\hat{j}$$ + c $$\hat{k}$$ m/s2, where a, b, and c are constants. What can be said about the functional form of the velocity function?
## Constant Acceleration
Multidimensional motion with constant acceleration can be treated the same way as shown in the previous chapter for one-dimensional motion. Earlier we showed that three-dimensional motion is equivalent to three one-dimensional motions, each along an axis perpendicular to the others. To develop the relevant equations in each direction, let’s consider the two-dimensional problem of a particle moving in the xy plane with constant acceleration, ignoring the z-component for the moment. The acceleration vector is
$$\vec{a} = a_{0x}\; \hat{i} + a_{0y}\; \hat{j} \ldotp$$
Each component of the motion has a separate set of equations similar to Equation 3.10–Equation 3.14 of the previous chapter on one-dimensional motion. We show only the equations for position and velocity in the x- and y-directions. A similar set of kinematic equations could be written for motion in the z-direction:
$$x(t) = x_{0} + (v_{x})_{avg} t \label{4.11}$$
$$v_{x}(t) = v_{0x} + a_{x}t \label{4.12}$$
$$x(t) = x_{0} + v_{0x} t + \frac{1}{2} a_{x} t^{2} \label{4.13}$$
$$v_{x}^{2} (t) = v_{0x}^{2} + 2a_{x}(x − x_{0}) \label{4.14}$$
$$y(t) = y_{0} + (v_{y})_{avg} t \label{4.15}$$
$$v_{y}(t) = v_{0y} + a_{y} t \label{4.16}$$
$$y(t) = y_{0} + v_{0y} t + \frac{1}{2} a_{y} t^{2} \label{4.17}$$
$$v_{y}^{2} (t) = v_{0y}^{2} + 2a_{y}(y − y_{0}) \ldotp \label{4.18}$$
Here the subscript 0 denotes the initial position or velocity. Equation 4.11 to Equation 4.18 can be substituted into Equation 4.2 and Equation 4.5 without the z-component to obtain the position vector and velocity vector as a function of time in two dimensions:
$$\vec{r} (t) = x(t)\; \hat{i} + y(t)\; \hat{j}\; and\; \vec{v} (t) = v_{x} (t)\; \hat{i} + v_{y} (t)\; \hat{j} \ldotp$$
The following example illustrates a practical use of the kinematic equations in two dimensions.
Example 4.6
### A Skier
Figure 4.10 shows a skier moving with an acceleration of 2.1 m/s2 down a slope of 15° at t = 0. With the origin of the coordinate system at the front of the lodge, her initial position and velocity are
$$\vec{r} (0) = (7.50\; \hat{i} - 50.0\; \hat{j}) m$$
and
$$\vec{v} (0) = (4.1\; \hat{i} - 1.1\; \hat{j}) m/s$$
(a) What are the x- and y-components of the skier’s position and velocity as functions of time? (b) What are her position and velocity at t = 10.0 s?
### Strategy
Since we are evaluating the components of the motion equations in the x and y directions, we need to find the components of the acceleration and put them into the kinematic equations. The components of the acceleration are found by referring to the coordinate system in Figure 4.10. Then, by inserting the components of the initial position and velocity into the motion equations, we can solve for her position and velocity at a later time t.
### Solution
1. The origin of the coordinate system is at the top of the hill with y-axis vertically upward and the x-axis horizontal. By looking at the trajectory of the skier, the x-component of the acceleration is positive and the y-component is negative. Since the angle is 15° down the slope, we find $$a_{x} = (2.1\; m/s^{2}) \cos(15^{o}) = 2.0\; m/s^{2}$$$$a_{y} = (−2.1\; m/s^{2}) \sin (15^{o}) = −0.54\; m/s^{2} \ldotp$$Inserting the initial position and velocity into Equation 4.12 and Equation 4.13 for x, we have $$x(t) = 75.0\; m + (4.1\; m/s)t + \frac{1}{2} (2.0\; m/s^{2})t^{2}$$$$v_{x}(t) = 4.1\; m/s + (2.0\; m/s^{2})t \ldotp$$For y, we have $$y(t) = -50.0.0\; m + (-1.1\; m/s)t + \frac{1}{2} (-0.54\; m/s^{2})t^{2}$$$$v_{y}(t) = -1.1\; m/s + (-0.54\; m/s^{2})t \ldotp$$
2. Now that we have the equations of motion for x and y as functions of time, we can evaluate them at t = 10.0 s: $$x(10.0\; s) = 75.0\; m + (4.1\; m/s) (10.0\; s) + \frac{1}{2} (2.0\; m/s^{2})(10.0\; s)^{2} = 216.0\; m$$$$v_{x}(10.0\; s) = 4.1\; m/s + (2.0\; m/s^{2})(10.0\; s) = 24.1\; m/s$$ $$y(10.0) = -50.0.0\; m + (-1.1\; m/s)(10.0\; s) + \frac{1}{2} (-0.54\; m/s^{2})(10.0\; s)^{2}$$$$v_{y}(10.0\; s) = -1.1\; m/s + (-0.54\; m/s^{2})(10.0\; s) \ldotp$$The position and velocity at t = 10.0 s are, finally $$\vec{r} (10.0\; s) = (216.0\; \hat{i} - 88.0\; \hat{j}) m$$$$\vec{v} (10.0\;s ) = (24.1\; \hat{i} - 6.5\; \hat{j}) m/s \ldotp$$The magnitude of the velocity of the skier at 10.0 s is 25 m/s, which is 60 mi/h.
### Significance
It is useful to know that, given the initial conditions of position, velocity, and acceleration of an object, we can find the position, velocity, and acceleration at any later time.
With Equation 4.8 through Equation 4.10 we have completed the set of expressions for the position, velocity, and acceleration of an object moving in two or three dimensions. If the trajectories of the objects look something like the “Red Arrows” in the opening picture for the chapter, then the expressions for the position, velocity, and acceleration can be quite complicated. In the sections to follow we examine two special cases of motion in two and three dimensions by looking at projectile motion and circular motion.
Simulation
At this University of Colorado Boulder website, you can explore the position velocity and acceleration of a ladybug with an interactive simulation that allows you to change these parameters.
## Contributors
• Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0). |
### Archive
Archive for the ‘Graphing’ Category
## Drawing log graphs
First lets consider the basic log graph $y=log_{10} x$
The important features
• The graph has an asymptote of $x=0$
• It intersects the x axis at $x=1$, because the log of 1 is zero
• It passes through the point $(10,1)$, because $10^1=1$. Notice 10 is the base
• The curve is always increasing, but at a slower and slower rate
Conclusions
The asymptote will always be $x=0$ unless the graph is translated $\Rightarrow y=log_{10}(x+a)$ has an asymptote at $x=-a$
$y=1$ when $x$ equals the base number unless the graph has been translated $\Rightarrow y=log_a(x)$ goes through $(a,1)$
The graph goes through (1,0) unless translated $\Rightarrow y=log_{10}(x+a)+b$ goes through the point $(10-a,1+b)$
Examples
For each of the function below give the asymptote, the coordinates where the point $(1,0)$ has moved to and the coordinates of where the point $(base,1)$ has moved to.
$y=log_5x$
The asymptote has not changed since we have not added to $x$ so the asymptote is $x=0$
The point related to the base is $(5,1)$
The point $(1,0)$ has not moved since we have not added to $x$ or $y$
$y=log_{10}x+2$
The asymptote has not changed since we have not added to $x$ so the asymptote is $x=0$.
The point related to the base is $(10,3)$ because we have added 2 to the function so the graph have moved up 2.
The point $(1,0)$ has moved to $(1,2)$ since we have added 2 to the function so the graph have moved up 2.
$y=3log_{10}(x-1)+2$
The asymptote is $x=1$ because we have subtracted 1 from x and this shifts the graph 1 place to the right.
The point related to the base is $(11,5)$ because the y values are 3 times bigger and then we add 2.
The point $(1,0)$ has moved to $(1,2)$ since we need to multiply y by 3 (no effect here) and then add 2.
Questions
Categories: Graphing
## Finding the mid point between to coordinates
You can see that the horizontal position of the mid point is half way between the horizontal positions of A and B. Also you can see that the vertical position of the mid point is half way between the vertical positions of A and B.
$\therefore$ mid point $= (X_{average},Y_{average})$
Example
Find the mid point of $(3,8)$ and $(7,-3)$
Mid point =$\left(\frac{3+7}{2},\frac{8+-3}{2}\right)=\left(5,\frac{5}{2}\right)$
Categories: Graphing
The gradient between two points is the gradient of the line that passes through them.
Gradient = $\frac{rise}{run}$
The rise is the change is y = $y_2-y_1$
The run is the change is x = $x_2-x_1$
So the gradient = $\frac{y_2-y_1}{x_2-x_1}$
It does not matter if you do the first coordinate minus the second or the other way round provided that you are consistent.
Examples
Find the gradient between $(3, 6)$ and $(7,14)$
gradient = $\frac{14-6}{7-3}=\frac{8}{4}=2$
Find the gradient between $(2,7)$ and $(5,6)$
gradient = $\frac{6-7}{5-2}=-\frac{1}{3}$
Show that the quadrilateral with vertices $A(0,0)$, $B(6,4)$, $C(4,6)$ and $D(1,4)$ is a trapezium.
To be a trapezium the shape must have a pair of parallel lines, which means a pair of lines with the same gradient.
Gradient AB=$\frac{4-0}{6-0}=\frac{2}{3}$
Gradient BC=$\frac{6-4}{4-6}=-1$
Gradient CD=$\frac{6-4}{4-1}=\frac{2}{3}$
Gradient DA=$\frac{4-0}{1-0}=4$
Since AB and CD have the same gradient the shape must be a trapezium
Categories: Graphing
## Distance between two points
Let’s start by considering the points $(x_1,y_1)$ and $(x_2,y_2)$
We have used the two points to construct a right-angled triangle.
The horizontal distance is equal to $x_2-x_1$
The vertical distance is equal to $y_2-y_1$
So using Pythagoras’ theorem we can say the distance between the points $=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
If we define the change in x as $\delta x$ and the change in y as $\delta y$ the the distance become s$=\sqrt{(\delta x)^2+(\delta y)^2}$
Examples
Find the distance between the two points $(3,7)$ and $(7,-2)$
$\delta x = 3-7=-4$. Since this is the length of a side of a triangle we will positive 4.
$\delta y = 7--2=9$
$\therefore$ distance $= \sqrt{4^2+7^2}=\sqrt{65}$
A triangle is made with the three points $A=(2,8),B=(5,12)$ and $C=(6,11)$. Show that the triangle is an isosceles triangle.
An isosceles triangle has two sides of equal length, if we can show that two of our sides are equal then we have proved it is isosceles.
The length AB =$\sqrt{3^2+4^2}=5$
The length BC =$\sqrt{1^2+1^2}=\sqrt{2}$
The length AC=$\sqrt{4^2+3^2}=5$
Since the length AB = the length AC, the triangle is isosceles.
Categories: Graphing
## Lines intersecting circles
When a line intersects a circle it will either have one point of contact in which case it is a tangent to the circle or it will cut it in two places.
To find the points of contact between a line and a circle
1. If needed, rearange the equation of the line to make x or y the subject
2. Substitute into the circle equation and solve the resulting quadratic
3. Substitute the solutions into the equation of the line to get the required coordinates
Example
Find where the line $y=2x+1$ intersects with the circle $(x+2)^2+(y-1)^2=5$
When we substitute y into the circle we get $(x+2)^2 + ((2x+1)-1)^2=5$
$\therefore (x+2)^2+(2x)^2=5 \Rightarrow x^2+4x+4+4x^2=5 \Rightarrow 5x^2+4x-1=0$
$\therefore (5x-1)(x+1)=0 \Rightarrow x=\frac{1}{5}$ or $x = -1$
If $x=\frac{1}{5}, y=\frac{7}{5}$
If $x=-1, y=-1$
$\therefore$ the points of intersection are $(\frac{1}{5}, \frac{7}{5})$ and $(-1,-1)$
Categories: Graphing
## Using the substitution method to find where two lines intersect
1. Rearrange one of the equations to make either x or y the subject
2. Substitute for either x or y depending on which one you made the subject
3. Solve the resulting equation
4. Substitute this value into one of the original equations to find the value of the other variable
Example
Find the coordinates where the lines $y=3x+5$ and $3x+2y=28$ intersect
Since the first equation already has y as the subject we will substitute this into the second
$\therefore 3x+2(3x+5)=28 \Rightarrow 9x+10=28 \Rightarrow x=2$
We now substitute this back into the first (on this occassion it is the easiest option)
and get $y=3 \times 2+5=11$
$\therefore$ the point of intersection in $(2,11)$
Categories: Graphing
## Finding Lines using y-y1=m(x-x1)
A useful way of finding the equation of a line is to use the formula $y-y_1=m(x-x_1)$, where m is the gradient of the line and $x_1$ and $y_1$ are taken from the given point $(x_1,y_y)$
### Finding the equation of the line given the gradient and a point
1. Get the values of $m$, $x_1$, $y_1$
2. Substitute into the formula
3. Rearrange the formula to the desired form
Example
Find the line with a gradient of -2 that passes through the point $(2,6)$
So $m=-2$, $x_1=2$, $y_1=6$
$\therefore y-6=-2(x-2) \Rightarrow y=-2x+10$
### Finding the equation of the line given a parallel line and a point
1. Use the gradient of the parallel line for m
2. Get the values of $x_1$ and $y_1$ from the given point
3. Substitute into the formula
4. Rearrange the formula to the desired form
Example
Find the equation of the line which is parallel to $y=3x+1$ and passes through the point $(2,1)$
So $m=3$, $x_1=2$, $y_1=1$
$\therefore y-1=3(x-2) \Rightarrow y=3x-5$
### Finding the equation of the line given a perpendicular line and a point
1. Get the gradient of the perpendicular line and use the negative reciprocal for m
2. Get the values of $x_1$ and $y_1$ from the given point
3. Substitute into the formula
4. Rearrange the formula to the desired form
Example
Find the equation of the line which is perpendicular to $y=-4x+7$ and passes through the point $(8,1)$
The gradient of the given line is -4, so the perpendicular gradient is $\frac{1}{4} \therefore m=\frac{1}{4}$
From the point $x_1 = 8$ and $y_1 = 1$
$\therefore y-1=\frac{1}{4}(x-8) \Rightarrow y=\frac{1}{4}x-1$
### Finding the equation of the line given two points
1. Find the gradient from the points using $\frac{rise}{run}$ or $\frac{\delta y}{\delta x}$
2. Use either point in the formula, your choice
3. Rearrange the formula to the desired form
Example
Find the line which passes through the points $(-4, 7)$ and $(3,10)$
m = gradient = $\frac{10-7}{3--4}=\frac{3}{7}$
I will use the point $(3,10)$, because both values are positive, but remember you would get the same result if you used $(-4, 7)$
$\therefore x_1 = 3$ and $y_1 = 10$
So substituting gives $y-10=\frac{3}{7}(x-3) \Rightarrow y=\frac{3}{7}x+\frac{61}{7}$
### Finding the equation of the Perpendicular bisector between two points
1. Find the gradient between the points and use the negative reciprocal for m
2. Find the mid-point between the two points and use this for $x_1$ and $y_1$
3. Substitute into the formula
4. Rearrange the formula to the desired form
Example
Find the perpendicular bisector between the points $(4, 6)$ and $(2,10)$
Gradient between the points is $\frac{10-6}{2-4}=-2$
$\therefore m =\frac{1}{2}$
The mid-point if $\left(\frac{4+2}{2},\frac{6+10}{2}\right)=(3,8)$
$\therefore x_1 = 3$ and $y_1 = 8$
So substituting gives $y-8=\frac{1}{2}(x-3) \Rightarrow y=\frac{1}{2}x+\frac{13}{2}$
Categories: Graphing |
# MATHEMATICS FORM 4 - PowerPoint PPT Presentation
MATHEMATICS FORM 4
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MATHEMATICS FORM 4
## MATHEMATICS FORM 4
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##### Presentation Transcript
1. MATHEMATICS FORM 4 ANGLES OF ELEVATION AND DEPRESSION
2. Hello there. How are you? Welcome back. Did you know what we are going to learn today? Yes. We are going to learnt about Angles of elevation and depression. However, let’s us recall back what is angle. Do you know angles? If you did not remember about angles, let watch this video together. PLAY VIDEO
3. Now, can you remember what is angles? If yes, good. But today we are going to learn about angles of elevation and depression.
4. INTRODUCTION TO ANGLES OF ELEVATION & DEPRESSION
5. LEARNING Objectives: At the end of the lesson, students will be able to: identify the horizontal line, the angle of elevation and the angle of depression for a particular situation Represent a particular situation involving angles of elevation and depression using diagrams Solve problems involving the angle of elevation and the angle of depression Hi there. Here are our learning objectives for today.
6. Now, first let’s us learn how to identify the horizontal lines, angles of elevation and angles of depression.
7. Horizontal Lines, angles Of elevation & angles Of depression
8. What is angle of elevation? Angle of elevation is the angle between the horizontal line that passes through an observer’s eye and the straight line that joins the observer’s eye to an object situated at the higher lever than the observer. What about the angle of depression? Can you guess it?
9. What is angle of depression? The angle of depression is the angle between the horizontal line that passes through an observer’s eye and the straight line that joins the observer’s eye to an object situated at a lower level than the observer. By the way? Can you tell me what is horizontal line? Now, do you understand what is angle of elevation and angle of depression? If no, let’s us take a look on one diagram…
10. T The diagram on the left shows a tower DQT and a building AB on a horizontal ground. A man at the position R views the peak of the tower T. A ANGLE OF ELEVATION Q R ANGLE OF DEPRESSION B D RQ is the horizontal line The angle of depression of D from R is ∟QRD The angle of elevation of T from R is ∟TRQ
11. Activity 1: Let’s us test your Understanding! QUESTION 1 QUESTION 2
12. QUESTION 1 BACK B P A In the diagram above, AB is a building on a horizontal ground. Identify: The horizontal line The angle of elevation of B from P ANSWER ANSWER
13. NEXT QUESTION 2 T 12.5 m R P 16 m The above diagram shows a vertical pole TP on a horizontal plane. State the horizontal line Find the angle of elevation of T from R ANSWER ANSWER
14. Sssstt…do you really want to know the answer? Well…the answer is : THE HORIZONTAL LINE IS PA. BACK TO QUESTION 1
15. Hmm….the correct answer is : THE ANGLE OF ELEVATION OF B FROM P IS APB BACK TO QUESTION 1
16. The answer is : THE HORIZONTAL LINE IS PR. BACK TO QUESTION 2
17. Well….the correct answer is : THE ANGLE OF ELEVATION OF T FROM R IS TRP BACK TO QUESTION 2
18. Representing a particular situation involving angles of elevation and depression using diagrams
19. You did not know? Don’t worry. I can help you. Here are the steps: Draw the positions of the object and observer Draw the horizontal line Join the eye of the observer and the object Indicate the angle Hi there. Did you know how to represent particular situation involving angles of elevation and depression using diagrams?
20. Do you understand it? Ok..let’s us take a look on one examples ok? Here the example: PQ and RS are two vertical poles on a horizontal plane. These poles are 10m apart. PQ is 9.5m while RS is 5m. The angle of elevation of P from R is 25° and the angle of depression of Q from R is 26°. Draw a labeled diagram to represent this situation. We give you 2 minutes to try to answer it. Ok? Good Luck!
22. How is it? Did your answer correct? Do you want to try one more question ? Try answer this: A vertical pole, TP, with a height of 5m is standing on a horizontal ground PQR. PQR is 10m with PQ=QR. The angle of elevation of vertex T from R is 27° and the angle of elevation of vertex T from Q is 45°. Draw a labeled diagram to represent this situation. CHECK YOUR ANSWER
23. T Here is the correct answer… 45° 27° R P Q How’s your performance? If correct, congratulation! If not, don’t give up. Try again. NEXT
24. Activity 2: handout
25. Hi there. I am Mr. Tiger.. Now we are going to learn the last learning objevtives of Angles of Elevation and Depression topic. We will now learn how to solve problems involving angles of elevation and depression. Let’s go!!
26. Let us take a look on this question: In the diagram below, represent a building BC on a horizontal ground BA. The height of the building is 37m. The angle of depression of A from C is 65°. Calculate the distance BA. C 37m A B
27. Did you know how to solve it? Let us watch this video first and then try to answer the question given earlier. Perhaps this video can help you in order to solve this question. PLAY VIDEO
28. Let us check the answer together: ∟ACB = 90° - 65° = 25° Tan 25° = x / 37 x = 37 tan 25° = 17.25 m # The distance of BA is 17.25 m C D 65° 37m A x m B
29. ACTIVITY 3: QUIZ QUIZ QUESTIONS |
# Video: Finding the Limit at Infinity of a Combination of Rational Functions
Find lim_(π₯ β β) (β4/π₯Β² + 5/π₯ + 8).
02:17
### Video Transcript
Find the limit of negative four over π₯ squared plus five over π₯ plus eight as π₯ approaches infinity.
We have a limit as π₯ approaches infinity here, but all the normal rules of limits still apply. For example, the limit of a sum of functions is equal to the sum of the limits. And so, we can split our limit up into three. Itβs equal to the limit of negative four over π₯ squared as π₯ approaches infinity plus the limit of five over π₯ as π₯ approaches infinity plus the limit of eight as π₯ approaches infinity.
What can we say about this limit? Well, we know that the limit of a constant πΎ, as π₯ approaches some number π, is just πΎ. And as for the previous limit law, this holds true, even if π isnβt a real number but is infinity or negative infinity. The value of this last limit is just eight.
What about the other two limits? We can use the fact that the limit of a constant multiple of a function is that constant multiple of the limit of the function. The first limit is, therefore, negative four times the limit of one over π₯ squared as π₯ approaches infinity. And the second is five times the limit of one over π₯ as π₯ approaches infinity. And finally, we add the eight.
Now, the limit of the reciprocal function one over π₯, as π₯ approaches infinity, is something we should know. Its value is zero. But how about the limit of one over π₯ squared as π₯ approaches infinity? Well, we can use the fact that the limit of a power of a function is that power of the limit of the function. This limit is the limit of the reciprocal function one over π₯ squared, as one over π₯ squared equals one over π₯ all squared. And by our limit law, this is the limit of one over π₯ as π₯ approaches infinity all squared. This limit is known to be zero. And so, our limit, the limit of one over π₯ squared as π₯ approaches infinity, is also zero.
We can generalize in that you get another limit law that the limit of one over π₯ to the power of π, as π₯ approaches infinity, is zero, at least if π is greater than zero. Our original limit is, therefore, negative four times zero plus five times zero plus eight, which is, of course, just eight. |
# Simplifying Algebraic Expressions
## Presentation on theme: "Simplifying Algebraic Expressions"— Presentation transcript:
Simplifying Algebraic Expressions
7.2
Parts of Algebraic Expressions
Term: Separated by addition and subtraction signs. Example: List the Terms: 3x+4y -5 3x,4y,-5
Parts of Algebraic Expressions
Coefficient: The numerical part of a term that contains a variable. Example: Identify the Coefficients: 3x+4y -5 3, 4
Parts of Algebraic Expressions
Like Term: Terms that contain the same variable Example: Identify the Like Terms: 3x +7 +x+2xy The 3x and x are like terms
Parts of Algebraic Expressions
Constant A term without a variable Example: Identify the Constant: 5y +3 +2x 3
Identify the terms, like terms, coefficients, and constants in 2x+3x+4-4x
2,3,-4 Constants: 4
Simplify Algebraic Expressions
An Algebraic Expression is in its simplest terms when it has: no like terms No parenthesis
Simplifying the Expression
Use the Distributive Property Example: 3x +2x =(3+2) x =5x
Ex: Simplify and Justify Steps 4x+6-3x
=4x+6+(-3x) Definition of Subtraction =4x+ (-3x) + 6 Commutative Property =(4+-3)x +6 Distributive Property =X+6 simplify
Ex: Simplify and Justify Your Steps 2m+3-7m-4
=2m +3+(-7m) +(-4) Definition of Subtraction =2m+(-7m)+3+(-4) Commutative Property =(2+-7)m + 3+ (-4) Distributive Property =-5m+ (-1) or -5m -1 Simplify
Ex: Simplify and Justify Your Steps 4a-6-2(a-1)
=4a+(-6)+(-2)[a+(-1)] Definition of Subtraction =4a +(-6) +(-2)(a)+(-2)(-1) Distributive Property =4a +(-6)+(-2a) +2 Simplify =4a +(-2a) +(-6) + 2 Commutative Property =[4+(-2)] a + (-6) +2 Distributive Property =2a + (-4) or 2a -4 Simplify
Total Stamps= Matt+ Lola x+ (x+16) 1x +1x+ 16 (1+1)x +16 2x +16
Lola has 16 more stamps in her collection than Matt. Write an expression in simplest form that represents the total number of stamps in both collections. Matt= x stamps Lola=x + 16 stamps Total Stamps= Matt+ Lola x+ (x+16) 1x +1x+ 16 (1+1)x +16 2x +16
Homework Page 302 (16-44) even |
a VISUAL Approach to CALCULUS problems
A talk by TOM M. APOSTOL
Professor of Mathematics, Emeritus, and Director of Project MATHEMATICS!
Delivered at the California Institute of Technology, 4 October 2000 (in Honor of his 50 yerars at Caltech)
Introduction
Calculus is a beautiful subject with a host of dazzling applications. As a teacher of calculus for more than fifty years and as an author of a couple of textbooks on the subject, I was stunned to learn that many classical problems in calculus can be easily solved by an innovative visual approach that makes no use of formulas. Here's a sample of four such (and more) problems:
Problem 1. Find the area of a parabolic segment.
Figure 1 shows a parabolic segment, the shaded region below the graph of the parabola y = x2 and above the interval from 0 to x. The area of the parabolic segment was first calculated by Archimedes more than 2000 years ago by a method that laid the foundations for integral calculus.
Figure 1. A parabolic segment
Problem 2. Find the area of the region under an exponential curve. Figure 2 shows the graph of the exponential function exp. We want the area of the shaded region below the graph to any point x.
Figure 2. The region below an exponential curve
Problem 3. Find the area of the region under one arch of a cycloid.
A cycloid is the path traced out by a fixed point on the boundary of a circular disk that rolls along a horizontal line, and we want the area of the shaded region shown in Figure 3. This problem can also be done by calculus but it is more difficult than the first two. First, you have to find an equation for the cycloid, which is not exactly trivial. Then you have to integrate this to get the required area.
Figure 3. The region under one arch of a cycloid
Problem 4. Find the area of the region under a tractrix.
When a child drags a toy along the floor with a string of constant length, the toy traces out a tractrix as the child walks along the x axis all the way to infinity. (Figure 4.) We want to find the area of the region between the tractrix and the x axis. To solve this by the standard calculus method, you have to find the equation of the tractrix. This, in itself, is rather challenging&emdash;it requires solving a differential equation. Once you have the equation of the tractrix you have to integrate it to get the area. It can be done, but the calculation is somewhat demanding.
Figure 4. The region between a tractrix and the x axis.
All four of these classical calculus problems and many more can also be solved by a new method that relies on geometric intuition and is easily understood by very young students. Moreover, the new method also solves some problems insoluable by calculus, and allows many increadible generalizations yet unknown in mathematics.
For example, look at Figure 5, which shows the path traced out by the front wheel of a bicycle in motion. The rear wheel traces out another curve, and the problem is to find the area of the region between these two curves as the bicycle moves from an initial position to a final position. To do this with calculus you would need equations for the curves. But the problem can be solved with this new visual approach regardless of the shape of the bike's path. You don't need any equations!
Figure 5. A region between the curves traced out by the rear and front wheels of a bicycle.
Historical background
Before describing this new method, some historical background is in order. The method was conceived in 1959 by a young undergraduate student at Yerevan University in Armenia named Mamikon A. Mnatsakanian. Mamikon, who is here today, will help illustrate this talk with some of his computer animation:
http://www.its.caltech.edu/~mamikon/calculus.html
Mamikon told me that when he showed his method to Soviet mathematicians they dismissed it out of hand and said "it can't be right&emdash;you can't solve calculus problems that easily." Mamikon went on to get a Ph.D. in physics; he pursued a full time career as professor of astrophysics at the University of Yerevan, and became an international expert in radiative transfer theory. He also continued to develop his powerful geometric methods, and eventually published a paper in 1981 outlining these methods, but the paper seems to have escaped notice, perhaps because it appeared in Russian in an Armenian journal with limited circulation (presented by Soviet Academician V.A.Ambartsumian):
Proceedings of the Armenian Academy of Sciences, vol.73, #2, pp.97-102, 1981
Mamikon came to California about a decade ago as part of an earthquake-preparedness program for Armenia. When the Soviet government collapsed, he was stranded in the US without a visa. With the help of a few mathematicians in Sacramento and at UC Davis he was granted status as an alien of extraordinary ability. He worked part time for the California Department of Education and at UC Davis. Here he further developed his methods in a form that can be used as a universal teaching tool to reach a wide audience, employing not only pictures, but also hands-on activities and computer-based manipulatives. He has taught these methods at UC Davis and in several elementary and high school classes in Northern California, ranging from Montessori elementary schools to inner-city public high schools. He has also demonstrated them at teacher conferences in Northern California. Response from both students and teachers has ranged from positive to enthusiastic, probably because the methods are vivid and dynamic and don't require the algebraic formalism of trigonometry or calculus.
About four years ago, Mamikon showed up at Project MATHEMATICS! headquarters and convinced me that his methods have the potential to make a significant impact on mathematics education, especially if they are combined with visualization tools of modern technology. Since then we have published several joint papers on innovative ideas in elementary mathematics and, if we can obtain adequate funding, we plan to produce a series of videotapes and workbooks to bring these powerful and exciting new geometric methods to a wide audience under the banner of Project MATHEMATICS!.
Mamikon's theorem for oval rings
Let's turn now to a description of Mamikon's method. Like all great discoveries, it's based on a simple idea. It started when young Mamikon was presented with a classical geometry problem, illustrated in Figure 6.
Figure 6. An annular ring bounded by two concentric circles.
This shows an annular ring and a chord of the outer circle tangent to the inner circle. The chord has given length a, and the problem is: Find the area of the annular ring.
As the late Paul Erdesh would have said, every baby can solve this problem. Look at the diagram in Figure 7. If the inner circle has radius r its area is ¼r2, and if the outer circle has
Figure 7. Calculating the area of the ring by the Pythagorean Theorem.
radius R its area is ¼R2, so the area of the ring is equal to ¼R2 &endash; ¼r2 = ¼(R2 &endash; r2). But the two radii and the tangent form a right triangle with legs r and a/2, and hypotenuse R, so by the Pythagorean Theorem, R2 &endash; r2 = (a/2)2, hence each ring has area ¼a2/4.
Note that the final answer depends only on a and not on the radii of the two circles!
If we knew in advance that the answer depends only on a, we could solve the problem another way. Shrink the inner circle to a point, and the annulus collapses to a disk of diameter a, with area equal to ¼a2/4.
Mamikon wondered if there was a way to see in advance why the answer depends only on the length of the chord. Then he thought of formulating the problem in a dynamic way. Take half the chord and think of it as a vector of length a/2 tangent to the inner circle. By moving this tangent vector around the inner circle, we see that it sweeps out the annular ring between the two circles.
But, for Mamikon it was obvious that the area is being swept due to pure rotation of the tangent segment
Now, translate each tangent vector parallel to itself so the point of tangency is brought to a common point. As the tangent vector moves around the inner circle, the translated vector rotates once around this common point and traces out a circular disk of radius a/2. So the tangent vectors sweep out a circular disk, as though they were all centered at the same point, as illustrated in Figure 8. And this disk has the same area as the ring.
Figure 8. The area of the ring is equal to that of a circular disk.
Mamikon realized that this dynamic approach would also work if the inner circle is replaced by an arbitrary oval curve. Figure 9 shows the same idea applied to two different ellipses. As the tangent segment of constant length moves once around each ellipse, it sweeps out a more general annular shape that we call an oval ring.
Figure 9. Oval rings swept by tangent segments of constant length moving around ellipses.
Again we can translate each tangent segment parallel to itself so the point of tangency is brought to a common point. As the tangent moves around the oval, the translated segments trace out a circular disk whose radius is that constant length. So, the area of the oval ring should be the area of the circular disk.
The Pythagorean Theorem can not help you find the areas for these oval rings. If the inner oval is an ellipse you can calculate the areas by integral calculus (which is not a trivial calculation); but if you do this calculation you find all these oval rings have equal areas depending only on the length of the tangent segment!
Is it possible that the same is true for any convex simple closed curve? Figure 10(a) illustrates the idea for a triangle.
(a) (b)
Figure 10. Region swept out as a tangent segment of constant length moves around a triangle.
As the tangent segment moves along an edge, it doesn't change direction so it doesn't sweep out any area. As it moves around a vertex from one edge to the next, it sweeps out part of a circular sector. And as it goes around the entire triangle it sweeps out three circular sectors that, together, fill out a circular disk, as shown in Figure 10(b).
The same is true for any convex polygon, as illustrated in Figure 11.
Figure 11. Region swept out as the tangent segment moves around a convex polygon.
The area of the region swept out by a tangent segment of given length moving around any convex polygon is equal to the area of a circular disk whose radius is that length. Therefore the same is true for any convex curve that is a limit of convex polygons. This leads us to:
MAMIKON'S THEOREM FOR OVAL RINGS
All oval rings swept out by a line segment of given length with one endpoint tangent to a smooth closed plane curve have equal areas, regardless of the size or shape of the inner curve. Moreover, the area depends only on the length L of the tangent segment and is equal to ¼L2, the area of a disk of radius L, as if the tangent segment was rotated about its endpoint.
Incidentally, Mamikon's Theorem for oval rings provides a new proof of the Pythagorean Theorem, as illustrated in Figure 12.
Figure 12. The Pythagorean Theorem deduced from Mamikon's Theorem on oval rings.
If the inner curve is a circle of radius r, the outer curve will also be a circle (of radius R, say) so the area of the oval ring is equal to the difference ¼R2 &endash; ¼r2. But by Mamikon's theorem, the area of the oval ring is also equal to ¼L2, where L is the constant length of the tangent segments. By equating areas we find R2 &endash; r2 = L2, from which we get R2 = r2 + L2, the Pythagorean Theorem (for the right triangle RrL).
First generalization of Mamikon's Theorem
A generalized version of Mamikon's theorem is illustrated in Figure 13. The lower curve in Figure 13(a) is a more or less arbitrary smooth curve. Tangent segments to this curve of constant length sweep out a region, which is bounded by the lower curve and an upper curve traced out by the other extremity of each tangent segment. The exact shape of this region will depend on the lower curve and on the length of the tangent segments. We refer to this region as a tangent sweep.
When each tangent segment is translated parallel to itself so that each point of tangency is brought to a common point as shown in Figure 13(b), the set of translated segments is called the tangent cluster. (The tangent segments have been clustered together to emanate from a common point.) When the tangent segments have constant length as in this figure, the tangent cluster is a circular sector whose radius is that constant length.
(a) (b) (c)
Figure 13. The tangent sweep and the tangent cluster for a general plane curve.
By the way, we could also translate the tangent segments so the other endpoints are brought to a common point, as in Figure 13(c). The resulting tangent cluster is a symmetric version of the cluster in (b). Now we can state:
MAMIKON'S THEOREM
The area of a tangent sweep is equal to the area of its tangent cluster, regardless of the shape of the original curve.
A physical example occurs when a bicycle's front wheel traces out one curve while the rear wheel (at constant distance from the front wheel) traces out another curve, as shown in Figure 14(a). The area of the tangent sweep is equal to the area of a circular sector depending only on the length of the bicycle and the change in angle from its initial position to its final position, as shown in Figure 14(b). The shape of the bike's path does not matter!
(a) (b)
Figure 14. Tangent sweep and tangent cluster generated by a moving bicycle.
Traxctrix and oval rings are particular cases of Bicyclix.
Figure 15 illustrates the same ideas in a more general setting. The only difference is that the tangent segments to the lower curve need not have constant length. The tangent segments sweep out a region called the tangent sweep (shown in Figure 15(a)). The tangent cluster is the region obtained by translating each tangent segment parallel to itself so that each point of tangency is moved to a common point (Figure 15(b)).
(a) (b)
Figure 15. The tangent sweep and the tangent cluster of a general plane curve.
Mamikon's theorem, which seems intuitively obvious by now, is that the area of the tangent cluster is equal to the area of the tangent sweep.
In the most general form of Mamikon's theorem the given curve need not lie in a plane. It can be any smooth curve in space, and the tangent segments can vary in length. The tangent sweep will lie on a developable surface. The shape of the tangent sweep depends on how the lengths and directions of the tangent segments change along the curve. When each tangent segment is translated parallel to itself so the point of tangency is brought to a common point, the set of translated segments is called the tangent cluster; it lies on a conical surface with vertex at this common point. Mamikon's general theorem equates the area of the tangent sweep with that of its tangent cluster.
GENERAL FORM OF MAMIKON'S THEOREM
The area of a tangent Sweep to a space curve is equal to the area of its conical Ikon.
This theorem, suggested by geometric intuition, can be proved also in a traditional manner, for example, by using differential geometry.
My first reaction to this theorem was "OK, that's a cool result in geometry. It must have some depth because it implies the Pythagorean Theorem. Can you use it to do anything else that's interesting?"
What you are about to see is a wide range of applications of this theorem. As already mentioned, curves swept out by tangent segments of constant length include oval rings and the tangent sweep of a bicycle. Another such example is the tractrix, the trajectory of a toy being pulled by a string of constant length by a child walking along a fixed straight line as shown in Figure 4.
All the examples with tangents of constant length reveal the striking property that the area of the tangent cluster can be expressed in terms of the area of a circular sector without using any of the formal machinery of traditional calculus. And what is more important, the animation shows why this happens.
But the most striking applications are to examples in which the tangent segments are not of constant length. These examples reveal the true power of Mamikon's method. The next example relates to exponential curves.
Exponential curves
Exponential functions are ubiquitous in the applications of mathematics. They occur in problems concerning population growth, radioactive decay, heat flow, and other physical situations where the rate of growth of a quantity is proportional to the amount present. Geometrically, this means that the slope of the tangent line at each point of an exponential curve is proportional to the height of the curve at that point. Exponential curves can also be described by their subtangents. The diagram in Figure 16 shows a general curve with a tangent line and the subtangent (the projection of the tangent on the x axis). The slope of the tangent is the height divided by the length of the subtangent. So, the slope is proportional to the height if and only if the subtangent is constant.
Figure 16. The slope of a curve is its height divided by the length of the subtangent.
The next diagram shows the graph of an exponential curve y = ex/b, where b is a positive constant. The only property of this curve that plays a role in this discussion is that the subtangent at any point has constant length b. This follows easily from differential calculus, but it can also be taken as the defining property of the exponential. In fact, exponential curves were first introduced in 1684 when Leibniz posed the problem of finding all curves with constant subtangents. The solutions are the exponential curves.
By exploiting the fact that exponential curves have constant subtangents, we can use Mamikon's theorem to find the area of the region under an exponential curve without using integral calculus. Figure 17 shows the graph of the exponential curve y = ex/b together with its tangent sweep as the tangent segments, cut off by the x axis, move to the left, from x all the way to _ ƒ. The corresponding tangent cluster is obtained by translating each tangent segment to the right so the endpoint on the x axis is brought to a common point, in this case, the lower vertex of the right triangle of base b and altitude ex/b. The resulting tangent cluster is the triangle of base b and altitude ex/b. Therefore the area of this region is equal to the area of this right triangle, so the area of the region between the exponential curve and the interval (_ ƒ, x] is equal to twice the area of this right triangle, which is its base times its altitude, or bex/b, the same result you would get by integration.
Figure 17. Finding the area of the region under an exponential curve by Mamikon's Method.
This yields the astonishing result that the area of the region under an exponential curve can be determined in an elementary geometric way without the formal machinery of integral calculus!
Area of a parabolic segment
We turn now to what is perhaps the oldest calculus problem in history&emdash;finding the area of a parabolic segment, the shaded region in Figure 18(a). The parabolic segment is inscribed in a rectangle of base x and altitude x2. The area of the rectangle is x3. From the figure we see that the area of the parabolic segment is less than half that of the rectangle in which it is inscribed. Archimedes made the stunning discovery that the area is exactly one-third that of the rectangle. Now we will use Mamikon's theorem to obtain the same result by a method that is not only simpler than the original Archimedes treatment but is also more powerful because it can be generalized to higher integer powers, and to arbitrary real powers as well.
(a) (b)
Figure 18. (a) A parabolic segment. (b) The subtangent to the parabola .
The parabola shown in Figure 18 has equation y = x2, but we shall not need this formula in our analysis. We use only the fact that the tangent line above any point x cuts off a subtangent of length x/2, as indicated in Figure 18(b). The slope of the tangent is x2 divided by x/2, or 2x.
To calculate the area of the parabolic segment we look at Figure 19 in which another parabola y = (2x)2 has been drawn, exactly half as wide as the given parabola. It is formed by bisecting each horizontal segment in the diagram. The two parabolas divide the rectangle into three regions, and our strategy is to show that all three regions have equal area. If we do this, then each has area one-third that of the circumscribing rectangle, as required.
Figure 19. The two parabolas divide the rectangle into three regions of equal area.
The two shaded regions in Figure 19 formed by the bisecting parabola obviously have equal areas, so to complete the proof we need only show that region above the bisecting parabola has the same area as the parabolic segment below the original parabola. To do this we examine Figure 20. The two right triangles in this figure have equal area (they have the same altitude and equal bases). Therefore the problem reduces to showing that the two shaded regions in this diagram have equal areas. Here's where we use Mamikon's theorem.
Figure 20. The tangent sweep of the lower curve is the tangent cluster above the upper one.
The shaded portion under the parabola y = x2 is the tangent sweep obtained by drawing all the tangent lines to the parabola and cutting them off at the x axis. And the other shaded portion is its tangent cluster, with each tangent segment translated so its point of intersection with the x axis is brought to a common point, the origin.
At a typical point (t, t2) on the lower parabola, the tangent intersects the x axis at t/2. Therefore, if the tangent segment from (t/2, 0) to (t, t2) is translated left by the amount t/2, the translated segment joins the origin and the point (t/2, t2) on the curve y = (2x)2. So the tangent cluster of the tangent sweep is the shaded region above the curve y = (2x)2, and by Mamikon's theorem the two shaded regions have equal areas, as required. So we have shown that the area of the parabolic segment is exactly one-third that of the circumscribing rectangle, the same result obtained by Archimedes.
Area of a generalized parabolic segment
The argument used to derive the area of a parabolic segment extends to generalized parabolic segments, in which x2 is replaced by higher powers. Figure 21(a) shows the graphs of y = x3 and y = (3x)3, which divide the rectangle of area x4 into three regions. The curve y = (3x)3 trisects each horizontal segment in the figure, hence the area of the region above this cubic is half that of the region between the two cubic curves. In this case we will show that the area of the region above the trisecting cubic is equal to that below the original cubic, which means that each region has area one-fourth that of the circumscribing rectangle.
To do this we use the fact that the subtangent to the cubic is one-third the length of the base, as shown in Figure 21(b). One shaded region in Figure 21(b) is the tangent sweep of the original cubic, and the other is the corresponding tangent cluster, so they have equal areas.
The two right triangles are congruent, so they have equal areas. Therefore the region above the trisecting cubic has the same area as the cubic segment below the curve y = x3, and each is one-fourth that of the rectangle, or x4/4.
(a) (b)
Figure 21. Mamikon's method used to find the area of a cubic segment.
In the quartic case we use the two curves y = x4 and y = (4x)4 to divide the rectangle of area x5 into three regions. Using the fact that the subtangent to the quartic at x has length x/4, we can use the same argument to show that the area of the region between the two quartics is three times that of each of the other two pieces, so the quartic segment below y = x4 has area one-fifth that of the rectangle, or x5/5. The argument also extends to all higher powers, a property not shared by the Archimedes treatment of the parabolic segment. For the curve y = xn we use the fact the subtangent at x has length x/n.
Cycloid
We turn next to the cycloid, the curve traced out by a point on the perimeter of a circular disk that rolls without slipping along a horizontal line. A classical problem is to show that the area of the region between one arch of the cycloid and the horizontal line is three times the area of the rolling disk, as suggested by Figure 22.
Figure 22. The area of the region under one arch of a cycloid is three times that of the rotating circular disk.
The standard calculus method of solving this problem is to first determine parametric equations for the cycloid, then calculate the area by integration. The same result can be obtained from Mamikon's theorem without the need to find parametric equations or to perform any integration.
Figure 23 shows a cycloidal arch inscribed inside a rectangle whose altitude is the diameter d of the rolling disk and whose base is the circumference of the disk, ¼d. The area of the circumscribing rectangle is ¼d2, which is four times the area of the disk. So it suffices to show that the unshaded region above the arch and inside the rectangle has area equal to that of the disk.
Figure 23. The unshaded region above the arch and inside the rectangle has area of the rolling circle.
To do this we show that the unshaded region is the tangent sweep of the cycloid, and the corresponding tangent cluster is a circular disk of diameter d. By Mamikon's theorem, this disk has the same area as the tangent sweep. Because the area of the disk is one-fourth the area of the rectangle, the area of the region below the arch must be three-fourths that of the rectangle, or three times that of the rolling disk.
It remains to shows that the tangent cluster of the unshaded region is a circular disk as asserted. As the disk rolls along the base it is always tangent to the upper and lower boundaries of the circumscribing rectangle. Denote the upper point of tangency by P and the lower point of tangency by P0, as in Figure 24.
Figure 24. Proof that the tangent segment to the cycloid is a chord of the rolling disk.
The diameter PP0 divides the rolling circle into two semicircles, and any triangle inscribed in these semicircles must be a right triangle. The disk undergoes instantaneous rotation about P0, so the tangent to the cycloid at any point X is perpendicular to the instantaneous radius of rotation and therefore must be the vertex of a right triangle inscribed in the semicircle with diameter PP0. Consequently, the chord XP of the rolling disk is always tangent to the cycloid.
Extend the upper boundary of the circumscribing rectangle beyond the arch and choose a fixed point O on this extended boundary. Translate each chord parallel to itself so point P is moved horizontally to the fixed point O. Then the other extremity X moves to a point Y such that segment OY is equal in length and parallel to PX. Consequently, Y traces out the boundary of a circular disk of the same diameter, with OY being a chord equal in length and parallel to chord PX. Therefore the tangent cluster is a circular disk of the same diameter as the rolling disk, and Mamikon's theorem tells us that its area is equal to that of the disk.
New results
In the time that remains, I would like to mention a new discovery made as a result of Mamikon's investigations. Recall two of the examples mentioned earlier: the tractrix (which has constant tangents) and the exponential (which has constant subtangents). The tractrix and exponential have been studied for centuries, but apparently no one realized that they are related to one another. Mamikon has discovered that they are part of a new family of curves that we will describe presently.
Figure 25 shows an arbitrary curve together with a given base line (shown here horizontally as the x axis). At a general point P of this curve a tangent segment of length t cuts off a subtangent of length s along the base line.
As before, we can form the tangent cluster by translating each tangent segment of length t parallel to itself so the point of tangency is brought to a common point O, as in Figure 25. Let C denote the other endpoint of the tangent. As P moves along the given curve, point C traces out the curve defining the tangent cluster. We can also translate the subtangent of length s. These subtangents will be parallel to the given base line (shown horizontal in Figure 25). One endpoint of the translated subtangent is at C. When point P moves along a tractrix, t is constant and C moves along a circle. When point P moves along an exponential, s is constant and C moves along a line.
Figure 25. The tangent and subtangent to a general curve translated by the same amount.
Now suppose the original curve has the property that some linear combination of t and s is constant, say
at + bs= constant for some choice of a and b, with a „ 0 and b „ 0. What can we say about the path of C ?
When b = 0, the tangent t is constant and C lies on a circle. When a= 0 the subtangent s is constant and C lies on a straight line. Now we can easily show that for general a and b, C always lies on a conic section.
Let's see why this is true. If _ ‚ 0, divide by _ and rewrite the equation as t + _s = constant, where _ = _/_
or t = constant &endash; _s = _(d &endash; s), where d is another constant. To show that C lies on a conic we refer to Figure 26.
Use point O as a focus and take as directrix a line perpendicular to the subtangents at distance d from the focus. The quantity (d &endash; s) is the distance of C from the directrix, and t is the distance of C from the focus. The equation t = _(d &endash; s) states that the distance of C from the focus is _ times its distance from the directrix. Therefore C lies on a conic section with eccentricity _. The conic is an ellipse, parabola, or hyperbola according as 0 < _ < 1, _ = 1, or _ > 1. The limiting cases _ = 0 and ƒ give a circle and straight line. So in this family of curves, the area swept out by the tangent segment is a portion of the area of a conic section.
Figure 26. If _t + _s is constant, point C lies on a conic with eccentricity _/_.
Thus, we have learned something new. Both the tractrix and the exponential, which have been studied for centuries, turn out to be special cases of a family of curves characterized by the equation _t + _s = constant. The tangent cluster of each member of this family is bounded by a portion of a conic section.
Summary
The foregoing examples display a wide canvas of geometric ideas that can be treated with Mamikon's methods. Mamikon and I believe that video is the ideal medium for communicating these ideas. The examples discussed in this talk will form the core of a pilot videotape, the first of a series of contemplated videotapes we hope to produce under the umbrella of Project MATHEMATICS!. Like all videotapes produced by Project MATHEMATICS!, the emphasis will be on dynamic visual images presented with the use of motion, color, and special effects that employ the full power of television to convey important geometric ideas with a minimal use of formulas.
The animated sequences will be designed by Mamikon himself, using Flash Animation or Java Applets, which can easily be placed on the Internet and accessed from the Project's web site: http://www.projectmathematics.com. Professional animators will be used to render the Flash Animation to a format suitable for broadcast quality television.
These animated sequences will reveal in a dynamic way how tangent sweeps are generated by moving tangent segments, and how the tangent segments can be translated to form tangent clusters. They will also reveal that many classical curves can be generated in a natural way by their intrinsic geometric and mechanical properties.
It should be pointed out that Mamikon's methods are also applicable to many plane curves not mentioned above. We plan to treat these in subsequent videotapes. For example, the following figures have been successfully treated by this method: ellipse, hyperbola, catenary, logarithm, cardioid, epi-cycloid, hypo-cycloids, involutes, evolutes, Archimedean spiral, Bernoulli lemniscate, sine and cosine.
The methods also apply to finding volumes of three-dimensional figures such as the ellipsoid, paraboloid, three types of hyperboloid, catenoid, pseudosphere, torus, and other solids of revolution.
I'd like to conclude with a small philosophical remark about calculus. Newton and Leibniz are generally regarded as the discoverers of integral calculus. Their great contribution was to unify work done by many other pioneers and to relate the process of integration with the process of differentiation. If you analyze Mamikon's method you see that it has some of the same ingredients, because it relates moving tangent segments with the areas of the regions swept out by these tangent segments. So the relation between differentiation and integration is naturally imbedded in Mamikon's method (The construction of the tangent cluster is based on the knowledge of the slope of the initial curve). |
# MA101.17 Continuity of functions of two variables
We can say that $f(x,y)\to L$ as the $(x,y) \to (a,b)$, or $\displaystyle \lim_{(x,y)\to (a,b)} f(x,y) = L$.
If, intuitively speaking, by going close enough to (a,b) we can get f(a,b) as close as we like to L.
f(x,y) is said to be continuous at (a,b) if $\displaystyle \lim_{(x,y)\to (a,b)} f(x,y) = f(a,b)$.
Just as with one variable limits from below and above may be different, with two variables we may get various limits coming in from different paths.
##### Example
$\displaystyle f(x,y)=\frac{xy}{x^2+y^2} \qquad (x,y) \neq (0,0)$
Consider coming in to (0,0) along the line y=mx (m fixed). Along this line we have:
$\displaystyle f(x,y) = f(x,mx) = \frac{xmx}{x^2 + m^2x^2} = \frac{m}{1+m^2}$
Thus along y=mx,
$\displaystyle \lim_{(x,y)\to (0,0)} f(x,y) = \lim_{x\to 0} \frac{m}{1+m^2} = \frac{m}{1+m^2}$
If we come in along the line $y=x^2$,
$\displaystyle \lim_{along y=x^2} f(x,y) = \lim_{x\to 0} \frac{x^3}{x^2+x^4} = \lim_{x\to 0}\frac{1}{\frac{1}{x}+x} = 0$.
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## area of hexagon formula derivation
By putting the value of s, we get: This is all about the area of a hexagon. There is a predefined set of formulas for the calculation of perimeter and area of a regular hexagon which is collectively called as hexagon formula. Take one of the triangles and draw a line from the apex to the midpoint of the base to form a right angle. Area of Square Formula Derivation Formula for area of a hexagon: Area of a hexagon is defined as the region occupied inside the boundary of a hexagon. Deriving the Formula for Area of a Regular Hexagon - YouTube In geometry, hexagon is a polygon with 6 sides. Given enough dimensions, it is possible to compute the area of any polygon, because the polygon can be dissected into triangles and the elementary triangle area formula can then be applied. The formula for perimeter of a hexagon is given by: Question 1: Calculate the area and perimeter of a regular hexagon whose side is 4.1cm. Whereas in the case of the irregular hexagon, neither the sides are equal, nor the angles are the same. Perimeter of a hexagon is defined as the length of the boundary of the hexagon. Similarly, we have Pentagon where the polygon has 5 sides; Octagon has 8 sides. Proof of the formula relating the area of a triangle to its circumradius. The Perimeter of Hexagon Formula Hexagon is the polygon that has six equal sides and the six edges. Hexa is a Greek word whose meaning is six. As the mass is distributed over the entire surface of the polygon, it is necessary to compute the area of the triangles resulting from the triangulation. Hexagon formula helps us to compute the area and perimeter of hexagonal objects. Area of an equilateral triangle =$\left( {\frac{{\sqrt 3 }}{4}} \right) \times {a^2}$. The side length is labeled s s s, the radius is labeled R R R, and half central angle is labeled θ \theta θ. If you're seeing this message, it means we're having trouble loading external resources on our website. home Front End HTML CSS JavaScript HTML5 Schema.org php.js Twitter Bootstrap Responsive Web Design tutorial Zurb Foundation 3 tutorials Pure CSS HTML5 Canvas JavaScript Course Icon Angular React Vue Jest Mocha NPM Yarn Back End PHP Python Java Node.js … You need the perimeter, and to get that you need to use the fact that triangle OMH is a triangle (you deduce that by noticing that angle OHG makes up a sixth of the way around point H and is thus a sixth of 360 degrees, or 60 degrees; and then that angle OHM is half of that, or 30 degrees). We must calculate the perimeter using the side length and the equation , where is the side length. Solution: So perimeter will be the sum of the length of all sides. The hexagon formula for a hexagon with the side length of a, is given as: Perimeter of an Hexagon = 6a As shown below, a regular polygon can be broken down into a set of congruent isosceles triangles. Area of a regular polygon - derivation. The formula for finding the area of a hexagon is Area = (3√3 s2)/ 2 where s is the length of a side of the regular hexagon. The formula for perimeter of a hexagon is given by: Calculate the area and perimeter of a regular hexagon whose side is 4.1cm. This page describes how to derive the formula for the area of a regular polygon by breaking it down into a set of n isosceles triangles, where n is the number of sides. Derivation: Take into consideration a regular hexagon with each side unit. We know that the tan of an angle is opposite side by adjacent side, Therefore, $$tan\theta = \frac{\left ( a/2 \right )}{h}$$, $$tan30 = \frac{\left ( a/2 \right )}{h}$$, $$\frac{\sqrt{3}}{3}= \frac{\left ( a/2 \right )}{h}$$, $$h= \frac{a}{2}\times \frac{3}{\sqrt{3}}$$, The area of a triangle = $$\frac{1}{2}bh$$, The area of a triangle=$$\frac{1}{2}\times a\times \frac{a}{2}\times \frac{3}{\sqrt{3}}$$, Area of the hexagon = 6 x Area of Triangle, Area of the hexagon = $$6\times \frac{3}{\sqrt{3}} \times \frac{a^{2}}{4}$$, Area of an Hexagon = $$\frac{3\sqrt{3}}{ 2} \times a^{2}$$. All the facts and properties described for regular polygons can be applied to a square. To solve more problems on the topic, download BYJU’S-The Learning App. Examples of units which are typically adopted are outlined below: Notation. Formula for Area of Trapezium. Happily, there is a formula for the area of any simple polygon that only requires knowledge of the coordinates of each vertex. In the case of a convex polygon, it is easy enough to see, however, how triangulating the polygon will lead to a formula for its centroid. Where $\sqrt 3$= 1.732 Derivation of the Area of An Equilateral Triangle. Area of a hexagon = $$\large \frac{3 \sqrt{3}}{2}s^{2}$$ There are mainly 6 equilateral triangles of side n and area of an equilateral triangle is (sqrt(3)/4) * n * n. Since in hexagon, there are total 6 equilateral triangles with side n, are of the hexagon becomes (3*sqrt(3)/2) * n * n If we want to find the area of the entire hexagon, we just have to multiply that by 6, because there are six of these triangles there. Formula for perimeter of a hexagon: Perimeter of a hexagon is defined as the length of the boundary of the hexagon. 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In geometry, hexagon is a. with 6 sides. Calculate the area of one of the triangles and then we can multiply by 6 to find the total area of the polygon. The regular hexagon consists of six symmetrical lines and rotational symmetry of order of 6. Perimeter of an Hexagon = 6a. The area, A, of one of the equilateral triangles, drawn in blue, can be found using: To compute the location of a hexagon, we separate it right into tiny six isosceles triangles. Lengths of all the sides and the measurement of all the. A polygon with six sides and six angles is termed as a hexagon. Derivation of the area formula Divide the regular hexagon into six equilateral triangles by drawing line segments to opposite vertices. Each internal angle of the hexagon has been calculated to be 120°. By finding the area of the polygon we derive the equation for the area of a circle. If the base and height of a trapezium are given, then the area of a Trapezium can be calculated with the help of the formula: ... (sum of bases) x (Height of trapezium) Derivation for Area of a Trapezium. Abstract: This paper provides a step-by step derivation of a new formula for finding the area of a regular polygon of any side ninscribed in a circle of radius rin terms of triangular units. Area of the hexagon is the space confined within the sides of the polygon. Number of vertices: 6 Number of edges: 6 Internal angle: 120° Area = (3 √3(n) 2) / 2 How does the formula work? The polygon can be decomposed into triangles defined by the origin and successive vertices $\mathbf v_i$ and $\mathbf v_{i+1}$. If you know the smallest width W of the hexagon. So, we get another formula that could be used to calculate the area of regular Hexagon: Area= (3/2)*h*l Where “l” is the length of each side of the hexagon and “ h ” is the height of the hexagon when it is made to lie on one of the bases of it. The Apothem, Polygon Area, and Surface Area. Find the area of the board. It should be noted that the formula is not “symmetric” with respect to the signs of the x and y coordinates. So before we think about the circum-circle let's just think about the area of the triangle. In this case the hexagon has six of them. If the lengths of all the sides and the measurement of all the angles are equal, such hexagon is called a regular hexagon. A = Geometric Area, in 2 or mm 2; C = Distance to Centroid, in or mm; d = Distance from flat to flat of shape, in or mm Consider a regular hexagon with each side a units. Where ‘a’ denotes the length apothem length and “s” denotes the side length of a pentagon. Area of Regular Octagon = $$\large 2(1+ \sqrt{2})a^{2}$$. There is one more formula that could be used to calculate the area of regular Hexagon: Area= $$\large \frac{3}{2}.d.t$$ Compute the area of triangles, and after that, we can increase by 6 to … Special right triangles review. Doing so we get: A square can simply be a specific case of a regular polygon, but in this case with 4 equal sides. Figure 1: Pentagon with Five … Following is the derivation for calculating the area of … Shapes Formulas Rectangle Area = Length X Width A = lw Perimeter = 2 X Lengths + 2 X Widths P = 2l + 2w Parallelogram Area = Base X Height A = bh Perimeter = add the length of all sides P = 2a + 2b Triangle Area = 1/2 of the base X the height A = bh Perimeter = a + b + c (add the length of the three sides) P = Trapezoid Area = 1/2 of the base X the height A = ()h Perimeter = add lengths of all sides a + b1 + b2 + c So this is going to be equal to 6 times 3 square roots of 3, which is 18 square roots of 3. Special right triangles review. This MATHguide video derives the formula for the area of a regular polygon, which is half the apothem times the perimeter. ... As you all know that the diagonal is a line that joins the two opposite sides in a polygon. Here, ∠AOB = 360/6 = 60°. Your email address will not be published. The base of the triangle is a, the side length of the polygon. There isn't,as far as I know, any elegant formula for the area of a hexagon (or other polygon with several sides). The formula for perimeter of a hexagon is given by: Perimeter = length of 6 sides. Abd each internal angle is measured as 120-degree. It is reasonable then to replace 8s by 2 × pi × r, which is the perimeter of the circle, to calculate the area of the polygon or the circle when the number of sides is very big. Naturally, when all six sides are equal then perimeter will be multiplied by 6 of one side of the hexagon. But how does that come about? Question 1: Python Exercises, Practice and Solution: Write a python program to calculate the area of a regular polygon. The first version of this derivation did not have that condition. Area of a circle - derivation. Solution: Given, side of the hexagon = 4.1 cm, Area of an Hexagon = $$\frac{3\sqrt{3}}{ 2} \times 4.1^{2}$$ = 43.67cm², Perimeter of the hexagon= 6a= 6 × 4.1 = 24.6cm. Read Also: Area of a Hexagon – Quick Brief. This can be explaine… Area of Hexagon = $$\large \frac{3 \sqrt{3}}{2}x^{2}$$ Where “x” denotes the sides of the hexagon. One way to find the area of a regular hexagon is by first dividing it into equilateral triangles. That is, the area of the rectangle is the length multiplied by the width. It is as follows:A=n∑k=0(xk+1+xk)(yk+1−yk)2(Where n is the number of vertices, (xk,yk) is the k-th point when labelled in a counter-clockwise manner, and (xn+1,yn+1)=(x0,y0); that is, the starting vertex is found both at the start and end of the list of vertices.) Honeycomb, quartz crystal, bolt head, Lug/wheel nut, Allen wrench, floor tiles etc are few things which you would find a hexagon. The above formulas may be used with both imperial and metric units. Given Let the length of this line be. and: In approximate numeric terms, the area of a regular hexagon is 2.598 times the squareof its side length. And we're done. The total number of diagonals in a regular hexagon is 9. General hexagons. Your email address will not be published. Similarly, to find the area of the polygons- like the area of a regular pentagon, area of the octagon, go through the below formula. Area of an Hexagon = $$\frac{3\sqrt{3}}{ 2} \times a^{2}$$. Let the length of this line be h. The sum of all exterior angles is equal to 360 degrees. Solution: Given, perimeter of the board = 24 cm, Area of an Hexagon = $$\frac{3\sqrt{3}}{ 2} \times 4^{2}$$= 41.57cm². … In other words, sides of a regular hexagon are congruent. So perimeter will be the sum of the length of all sides. Solved examples: s = 4 cm Honeycomb, quartz crystal, bolt head, Lug/wheel nut, Allen wrench, floor tiles etc are few things which you would find a hexagon. 30-60-90 triangle example problem. Find the area of a regular hexagon whose side is 4 cm? The Area of Circle formula is: AREA = π × radius 2. Required fields are marked *. ... central angle and the radius of the polygon. This page describes how to derive the formula for the area of a circle.we start with a regular polygon and show that as the number of sides gets very large, the figure becomes a circle. A polygon is a two-dimensional (2-D) closed figure made up of straight line segments. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.. Visit Stack Exchange You use the following formula to find the area of a regular polygon: So what’s the area of the hexagon shown above? As with all calculations care must be taken to keep consistent units throughout. Assume that the polygon is star-shaped with respect to the origin and that the vertices are consecutively numbered in a counterclockwise direction. We can also use the decimal value of $\sqrt 3$ to simplify our calculations. Instead, unless it has some very special properties, you break it up into triangles and add their area. Therefore, in order to calculate the … The formula for the area of a hexagon: The area of a hexagon defined as the area inside the border of a hexagon. For any regular polygon, the area can be computed from the side length … Area of an Hexagon = $$\frac{3\sqrt{3}}{ 2} \times a^{2}$$ Formula for perimeter of a hexagon: Perimeter of a hexagon is defined as the length of the boundary of the hexagon. A common formula for the area of a regular n-gon is expressed in terms of the apothem and the side of the n-gon or the perimeter of the n-gon. In the case of a regular hexagon, all the sides are of equal length, and the internal angles are of the same value. To know more about the other characteristics and attributes of polygons such as hexagon, pentagon, octagon and other geometrical figures, please visit our site or download BYJU’S – The Learning App. In other words, sides of a regular hexagon are congruent. of a hexagon is defined as the region occupied inside the boundary of a hexagon. As a result, the closer the perimeter of the polygon is to the circle, the closer the area of the polygon is to the area of the circle. Starting Point. Hexagon formula helps us to compute the area and perimeter of hexagonal objects. [Image will be uploaded soon] Area of Square Formula in maths = a × a = $a^{2}$ Where, a is the length of the side of a square. Your email address will not be published. The area of each of these triangles is $\frac12(x_iy_{i+1}-x_{i+1}y_i)$. In order to calculate the area of a hexagon, we divide it into small six isosceles triangles. If the lengths of all the sides and the measurement of all the angles are equal, such hexagon is called a regular hexagon. The sum of all exterior angles is equal to 360 degrees, where each exterior angle measures 60 degrees. The Area of a Triangle. Your email address will not be published. So perimeter will be the sum of the length of all sides. Since a regular hexagon is comprised of six equilateral triangles, the formula for finding the area of a hexagon is derived from the formula of finding the area of an equilateral triangle. The figure below shows one of the n n n isosceles triangles that form a regular polygon. Area of Hexagon = $$\large \frac{3 \sqrt{3}}{2}x^{2}$$. Let us consider a square where the lengths of its side are ‘a’ units and diagonal is ‘d’ units respectively. In general, the sum of interior angles of a Polygon is given by-. , the side length of the polygon. Consider a regular hexagon with each side. If we are given the variables and , then we can solve for the area of the hexagon through the following formula: In this equation, is the area, is the perimeter, and is the apothem. w3resource. Required fields are marked *, A polygon is a two-dimensional (2-D) closed figure made up of straight line segments. There is one more formula that could be used to calculate the area of regular Hexagon: Where “t” is the length of each side of the hexagon and “d” is the height of the hexagon when it is made to lie on one of the bases of it. First, consider 2 regular Polygons inside a standard circle. You’ll see what all this means when you solve the following problem: The area of Hexagon is given by. Given a rectangle with length l and width w, the formula for the area is: A = lw (rectangle). ... What i want to do in this video is to come up with a relationship between the area of a triangle and the triangle's circumscribed circle or circum-circle. Each triangle has a side length s and height (also the apothem of the regular hexagon) of. To better our understanding of the concept, let us take a look at the derivation of the area of a square. Derivation of Square Formula Derivation of Area of a Square. This page looks to give a general run through of how the formula for the area of a circle can be derived. Dividing up: Draw your hexagon, and add a set of non-crossing diagonals that break it up into triangles. 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The sum of all interior angles is equal to 720 degrees, where each interior angle measures 120 degrees. Up Next. Where ‘a’ denotes the length of each side of the octagon. You also need to use an apothem — a segment that joins a regular polygon’s center to the midpoint of any side and that is perpendicular to that side. the formula is: In approximate numeric terms, the area of a regular hexagon is 0.866 times the squareof its smallest width. Formula for the Area of a Hexagon. Where “x” denotes the sides of the hexagon. The area of a trapezium is equal to the sum of the areas of the two triangles and the area of the rectangle. Here is the proof or derivation of the above formula of the area of a regular polygon. Question 2: Perimeter of a hexagonal board is 24 cm. The most basic area formula is the formula for the area of a rectangle. Area of the hexagon is the space confined within the sides of the polygon. 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# Fact Families: Multiplication and Division Are Related
Contributor: Ashley Nail. Lesson ID: 13415
Ever wonder how multiplication and division are related? Change some symbols and switch the numbers around, and you will quickly see these families form!
categories
## Operations and Algebraic Thinking
subject
Math
learning style
Kinesthetic, Visual
personality style
Lion, Beaver
Intermediate (3-5)
Lesson Type
Quick Query
## Lesson Plan - Get It!
Audio:
• How many relatives do you have?
Maybe a sister or brother, cousins, aunts, uncles, grandparents, and don't forget a mom or a dad! You might even have step-parents, second cousins, or great aunties.
Numbers have relatives too. However, their family trees look a lot different.
• For example, how are 9, 1, and 10 related?
They are not cousins. Instead, those numbers are related by addition and subtraction operations!
9 + 1 = 10
1 + 9 = 10
10 - 1 = 9
10 - 9 = 1
• Can you think of any other ways numbers can be related?
Numbers can also be related through multiplication and division!
Let's look at the numbers in the multiplication equation below:
• Notice how each number has an important name?
Much like in your family, these titles matter and help to describe the relationships.
Factors are numbers that, when multiplied, make up a product. The product is the answer in a multiplication equation.
You can display this multiplication equation with equal groups or an array:
• How are the numbers in this multiplication equation related to division?
Let's look at the product from the equation.
The product is 6. There were 6 total dots in our equal groups drawing.
In a division equation, the product becomes the dividend. The dividend is the total amount to be divided or split up.
In this case, the total is 6:
The divisor is how many groups the total will be split into. In this case, the dividend will be split into 2 groups:
To find the quotient or the answer of a division equation, you will need to divide the 6 total dots into 2 equal groups. Then, count how many dots are in each group!
• Does anything look familiar?
Compare this equal group drawing to the one above. They are the same!
• What about the numbers and the equations? What is similar about them?
• Do you see the relation?
There are two more equations that complete this family.
• Do you know what they are?
Yes! We use the commutative property of multiplication to find these equations in our fact family.
If you need to review this concept, check out our Additional Resource found in the right-hand sidebar.
For a quick review on fact families, watch Relating Multiplication to Division (Fact Families), from Alex Lochoff:
• Ready to practice finding more multiplication and division fact families?
Click NEXT to visit the Got It? section and use your new skills!
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# Right Triangle Trigonometry
This section covers:
You may have been introduced to Trigonometry in Geometry, when you had to find either a side length or angle measurement of a triangle. Trigonometry is basically the study of triangles, and was first used to help in the computations of astronomy. Today it is used in engineering, architecture, medicine, physics, among other disciplines.
The 6 basic trigonometric functions that you’ll be working with are sine (rhymes with “sign”), cosine, tangent, cosecant, secant, and cotangent. (Don’t let the fancy names scare you; they really aren’t that bad).
With Right Triangle Trigonometry, we use the trig functions on angles, and get a number back that we can use to get a side measurement, as an example. Sometimes we have to work backwards to get the angle measurement back so we have to use what a call an inverse trig functionBut basically remember that we need the trig functions so we can determine the sides and angles of a triangle that we don’t otherwise know.
Later, we’ll see how to use trig to find areas of triangles, too, among other things.
You may have been taught SOH – CAH – TOA (SOHCAHTOA) (pronounced “so – kuh – toe – uh”) to remember these. Back in the old days when I was in high school, we didn’t have SOHCAHTOA, nor did we have fancy calculators to get the values; we had to look up trigonometric values in tables.
Remember that the definitions below assume that the triangles are right triangles, meaning that they all have one right angle (90°). Also note that in the following examples, our angle measurements are in degrees; later we’ll learn about another angle measurement unit, radians, which we’ll discuss here in the Angles and Unit Circle section.
# Basic Trigonometric Functions (SOH – CAH – TOA)
Here are the 6 trigonometric functions, shown with both the SOHCAHTOA and Coordinate System Methods.
Note that the second set of three trig functions are just the reciprocals of the first three; this makes it a little easier! Note that the cosecant (csc), secant (sec), and cotangent (cot) functions are called reciprocal functions, or reciprocal trig functions, since they are the reciprocals of sin, cos, and tan, respectively.
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For the coordinate system method, assume that the vertex of the angle in the triangle is at the origin $$(0,0)$$:
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Right Triangle SOH-CAH-TOA Method Coordinate System Method \displaystyle \begin{align}\text{SOH: Sine}\left( A \right)=\sin \left( A \right)=\frac{{\text{Opposite}}}{{\text{Hypotenuse}}}\\\text{CAH: Cosine}\left( A \right)=\cos \left( A \right)=\frac{{\text{Adjacent}}}{{\text{Hypotenuse}}}\\\text{TOA: Tangent}\left( A \right)=\tan \left( A \right)=\frac{{\text{Opposite}}}{{\text{Adjacent}}}\end{align} \displaystyle \begin{align}\text{cosecant}\left( A \right)=\csc \left( A \right)=\frac{1}{{\sin \left( A \right)}}=\frac{{\text{ Hypotenuse}}}{{\text{Opposite}}}\\\text{secant}\left( A \right)=\sec \left( A \right)=\frac{1}{{\cos \left( A \right)}}=\frac{{\text{ Hypotenuse}}}{{\text{Adjacent}}}\\\text{cotangent}\left( A \right)=\cot \left( A \right)=\frac{1}{{\tan \left( A \right)}}=\frac{{\text{ Adjacent}}}{{\text{Opposite}}}\end{align} \displaystyle \begin{align}\sin \left( A \right)=\frac{y}{h}\\\cos \left( A \right)=\frac{x}{h}\\\tan \left( A \right)=\frac{y}{x}\end{align} \displaystyle \begin{align}\csc \left( A \right)=\frac{1}{{\sin \left( A \right)}}=\frac{h}{y}\\\sec \left( A \right)=\frac{1}{{\cos \left( A \right)}}=\frac{h}{x}\\\cot \left( A \right)=\frac{1}{{\tan \left( A \right)}}=\frac{x}{y}\end{align}
Here are some example problems. Note that we commonly use capital letters to represent angle measurements, and the same letters in lower case to represent the side measurements opposite those angles.
We also use the theta symbol θ to represent angle measurements, as we’ll see later. Note also in these problems, we need to put our calculator in the DEGREE mode.
And don’t forget the Pythagorean Theorem ($${{a}^{2}}+{{b}^{2}}={{c}^{2}}$$, where $$a$$ and $$b$$ are the “legs” of the triangle, and $$c$$ is the hypotenuse), and the fact that the sum of all angles in a triangle is 180°.
Right Triangle Problem Explanation Calculator Steps/Checking Find the values of $$a$$ and $$b$$: Once we get the answers, we can check our sides using the Pythagorean Theorem: $$\begin{array}{c}{{a}^{2}}+{{b}^{2}}={{c}^{2}}\\{{\left( {16.383} \right)}^{2}}+{{\left( {11.471} \right)}^{2}}=399.99\\\approx {{\left( {20} \right)}^{2}}\end{array}$$ To get side $$a$$, we need to use: $$\displaystyle \sin \left( A \right)=\frac{{\text{Opposite}}}{{\text{Hypotenuse}}}$$, where $$A$$ is 55°: $$\displaystyle \sin \left( {55{}^\circ } \right)=\frac{a}{{20}}$$ Cross multiply: $$\displaystyle a=\sin \left( {55{}^\circ } \right)\cdot 20\approx 16.383$$ To get side $$b$$, we need to use: $$\displaystyle \cos \left( A \right)=\frac{{\text{Adjacent}}}{{\text{Hypotenuse}}}$$, where $$A$$ is 55°: $$\displaystyle \cos \left( {55{}^\circ } \right)=\frac{b}{{20}}$$ Cross multiply: $$\displaystyle b=\cos \left( {55{}^\circ } \right)\cdot 20\approx 11.472$$ Hit and scroll down and to the right to make sure you’re in DEGREE mode. Then use theandkeys for cosine and sine, respectively: Find the value of $$b$$: To get side $$b$$, we need to use: $$\displaystyle \tan \left( A \right)=\frac{{\text{Opposite}}}{{\text{Adjacent}}}$$, where $$A$$ is 23°: $$\displaystyle \tan \left( {23{}^\circ } \right)=\frac{6}{b}$$ Cross multiply: $$\displaystyle \tan \left( {23{}^\circ } \right)\cdot b=6$$, or turn proportion sideways with an $$“=”$$ sign: $$\displaystyle \frac{b}{1}=\frac{6}{{\tan \left( {23{}^\circ } \right)}};\,\,\,\,a\approx 14.135$$ Use the key: If we needed to also find $$h$$, we could either use $$\displaystyle \sin \left( {23{}^\circ } \right)=\frac{6}{h}$$ or Pythagorean Theorem; both ways reveal that $$h=15.356$$. Find the values of $$A$$ and $$B$$: Once we get all the answers, let’s check to make sure the sum of all angles is 180°: $$\displaystyle \begin{array}{c}51.1{}^\circ \text{ }+38.9{}^\circ +90{}^\circ \left( {\text{right angle}} \right)\\=180{}^\circ \end{array}$$ This one’s a little trickier since we need to find angle measurements instead of side measurements; we’ll need to use the $${{\sin }^{{-1}}}\left( A \right)$$ and $${{\cos }^{{-1}}}\left( A \right)$$ (2nd sin and 2nd cos on the calculator) to get the angles back. For angle $$A$$, we can use sin, since we have the opposite side (14) and hypotenuse (18): \displaystyle \begin{align}\sin \left( A \right)&=\frac{{\text{Opposite}}}{{\text{Hypotenuse}}}=\frac{{14}}{{18}}\\A&={{\sin }^{{-1}}}\left( {\frac{{14}}{{18}}} \right)\approx 51.1{}^\circ \end{align} For angle $$B$$, we use cos: \displaystyle \begin{align}\cos \left( B \right)&=\frac{{\text{Adjacent}}}{{\text{Hypotenuse}}}=\frac{{14}}{{18}}\\B&={{\cos }^{{-1}}}\left( {\frac{{14}}{{18}}} \right)\approx 38.9{}^\circ \end{align} Use theand keys: Find the values of $$x$$ and $$y$$: This is a difficult problem that can easily be solved using Law of Sines, but let’s solve it by drawing an altitude to the triangle (dotted line). This creates two right triangles. To get altitude $$a$$, we can use: $$\displaystyle \sin \left( A \right)=\frac{{\text{Opposite}}}{{\text{Hypotenuse}}}$$, where $$A$$ is 25°: $$\displaystyle \sin \left( {25{}^\circ } \right)=\frac{a}{{35}};\,\,\,a=\sin \left( {25{}^\circ } \right)\cdot 35;\,\,\,\,a\approx 14.7916$$ Now we can use altitude $$a$$ to get side $$x$$, using the second right triangle: $$\displaystyle \sin \left( A \right)=\frac{{\text{Opposite}}}{{\text{Hypotenuse}}}$$, where $$A$$ is 20°: $$\displaystyle \sin \left( {20{}^\circ } \right)=\frac{{14.7916}}{x};\,\,\,x=\frac{{14.7916}}{{\sin \left( {20{}^\circ } \right)}};\,\,\,\,x\approx 43.2477$$ Now we can use the Pythagorean Theorem to get the two parts of bottom $$y$$ (we could have also used right angle trig): $$\displaystyle \begin{array}{c}{{y}_{1}}^{2}+{{a}^{2}}={{35}^{2}};\,\,\,\,\,{{y}_{1}}^{2}+{{14.7916}^{2}}={{35}^{2}};\,\,\,\,{{y}_{1}}\approx 31.7208\\{{y}_{2}}^{2}+{{a}^{2}}={{x}^{2}};\,\,\,\,\,{{y}_{2}}^{2}+{{14.7916}^{2}}={{43.2477}^{2}};\,\,\,\,{{y}_{2}}\approx 40.6395\\y={{y}_{1}}+{{y}_{2}}\approx 72.3603\end{array}$$ Whew! Hard problem!
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# Trigonometry Word Problems
Here are some types of word problems that you might see when studying right angle trigonometry.
Note that the angle of elevation is the angle up from the ground; for example, if you look up at something, this angle is the angle between the ground and your line of site.
The angle of depression is the angle that comes down from a straight horizontal line in the sky. (For example, if you look down on something, this angle is the angle between your looking straight and your looking down to the ground). For the angle of depression, you can typically use the fact that alternate interior angles of parallel lines are congruent (sorry, too much Geometry!) to put that angle in the triangle on the ground (we’ll see examples).
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Note that shadows in these types of problems are typically on the ground. When the sun casts the shadow, the angle of depression is the same as the angle of elevation from the ground up to the top of the object whose shadow is on the ground.
Also, the grade of something (like a road) is the tangent (rise over run) of that angle coming from the ground. Usually the grade is expressed as a percentage, and you’ll have to convert the percentage to a decimal to use in the problem.
And, as always, always draw pictures!
## Angle of Elevation Problem:
Angle of Elevation Trig Problem Math Devon is standing 100 feet from the Eiffel Tower and sees a bird land on the top of the tower (she has really good eyes!). If the angle of elevation from Devon to the top of the Eiffel Tower is close to 84.6°, how tall is the tower? This is a good example how we might use trig to get distances that are typically difficult to measure. Note that the angle of elevation comes up off of the ground. To get the height $$y$$, we need to use: $$\displaystyle \tan \left( \theta \right)=\frac{{\text{Opposite}}}{{\text{Adjacent}}}$$, where $$\theta =84.6{}^\circ$$ $$\displaystyle \tan \left( {84.6{}^\circ } \right)=\frac{y}{{100}}$$ Cross multiply to get: $$\displaystyle y=\tan \left( {84.6{}^\circ } \right)\cdot 100\,\,\approx \,\,1058$$. The Eiffel Tower is roughly 1058 feet tall.
## Angle of Depression Problem:
Angle of Depression Trig Problem Math From the top of a building that is 200 feet tall, Meryl sees a car coming towards the building. (Somehow, she knows that) the angle of depression when she first saw the car was 20° and when she stopped looking at it was 40° degrees. How far did the car travel? The first step is to draw a picture, and note that we can sort of “reflect” the angles of depression down to angles of elevation, since the horizon and ground are parallel. Then we get to use trig! The trick is to see that we can get distances $$y$$ and $$z$$ using the tangent function, and we need to subtract the two distances to get $$x$$, the distance the car travels. To get $$y$$: $$\displaystyle \tan \left( {40{}^\circ } \right)=\frac{{\text{Opposite}}}{{\text{Adjacent}}}=\frac{{200}}{y};\,\,\,y=\frac{{200}}{{\tan \left( {40{}^\circ } \right)}}\approx 238.4$$ To get $$z$$: $$\displaystyle \tan \left( {20{}^\circ } \right)=\frac{{\text{Opposite}}}{{\text{Adjacent}}}=\frac{{200}}{z};\,\,\,\,z=\frac{{200}}{{\tan \left( {20{}^\circ } \right)}}\approx 549.5$$ To get $$x$$, we subtract $$y$$ from $$z$$, so the car moved $$549.5–238.4=311.1$$ feet while Meryl was watching it.
## Right Triangle Systems Problem:
Here’s a problem where it’s easiest to solve it using a System of Equations:
Right Triangle Systems Trig Problem Math Two girls are standing 100 feet apart. They both see a beautiful seagull in the air between them. The angles of elevation from the girls to the bird are 20° and 45°, respectively. How high up is the seagull? The trick here is divide up the 100 ft into $$x$$ and the other is $$100-x$$ (put in real numbers to see how we get this) since we have two triangles. Then we will have two equations (one for each triangle) and two unknowns: $$\displaystyle \tan \left( {20{}^\circ } \right)=\frac{y}{x};\,\,\,\,\,\,\,\,\,\,\,\,\tan \left( {45{}^\circ } \right)=\frac{y}{{100-x}}$$ From the first equation, we get $$y$$ in terms of $$x$$: $$\displaystyle y=\tan \left( {20{}^\circ } \right)\cdot x\approx .36397x$$. We can plug this into the second equation to get $$\displaystyle \tan \left( {45{}^\circ } \right)=\frac{{.36397x}}{{100-x}}$$. Solving for $$x$$, we get: $$\displaystyle 1=\frac{{.36397x}}{{100-x}};\,\,\,\,100-x=.36397x;\,\,\,\,\,x\approx 73.315$$. (I left more decimal places, so the final answer will be more accurate). Now we have to get $$y$$ to find the height of the seagull: $$\displaystyle y=.36397\left( {73.315} \right)=27.684$$ The seagull is about 26.68 feet high. This was a tricky one!
Trig Shadow Problem Math The length of a tree’s shadow is 20 feet when the angle of elevation to the sun is 40°. How tall is the tree? Again, note that shadows in these types of problems are on the ground. When the sun casts the shadow, the angle of depression is the same as the angle of elevation from the ground up to the top of the tree. To get the height $$y$$ , we need to use: $$\displaystyle \tan \left( \theta \right)=\frac{{\text{Opposite}}}{{\text{Adjacent}}}$$, where $$\theta =40{}^\circ$$ $$\displaystyle \tan \left( {40{}^\circ } \right)=\frac{x}{{20}}$$ Cross multiply to get $$\displaystyle x=\tan \left( {40{}^\circ } \right)\cdot 20\approx 16.78$$. The height of the tree is approximately 16.78 feet tall. Not too bad!
Trig Grade Problem Math Chelsea walked up a road that has a 20% grade (she could feel it!) to get to her favorite store. At what angle does the road come up from the ground (at what angle is the road inclined from the ground)? Remember that the grade of a road can be thought of as $$\displaystyle \frac{{\text{rise}}}{{\text{run}}}$$, and you usually see it as a percentage. So, a 20% grade is the same as a grade of $$\displaystyle \frac{{\text{20}}}{{\text{100}}};$$ for every 20 feet the road goes up vertically, it goes 100 feet horizontally. Since we need to find an angle measurement and we have the adjacent and opposite sides, we’ll need to use the $${{\tan }^{{-1}}}\left( \theta \right)$$ (2nd tan on the calculator and make sure it’s in DEGREE mode) to get the angle back: $$\displaystyle \tan \left( \theta \right)=\frac{{\text{Opposite}}}{{\text{Adjacent}}}=\frac{{20}}{{100}};\,\,\,\,\,\,\,\,\theta ={{\tan }^{{-1}}}\left( {\frac{{20}}{{100}}} \right)=11.31{}^\circ$$ The road comes up at an angle of roughly 11.31° from the ground. Note that if we wanted to know how long the actual slanted road is, we could just use Pythagorean Theorem, or sin or cos: $$\displaystyle \sin \left( {11.31{}^\circ } \right)=\frac{{20}}{x};\,\,\,\,x=\frac{{20}}{{\sin \left( {11.31{}^\circ } \right)}}\approx 101.98\text{ }ft$$ This makes sense since the grade is relatively small (note that the picture is not drawn to scale!)
Understand these problems, and practice, practice, practice!
Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.
If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software. You can even get math worksheets. You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions.
There is even a Mathway App for your mobile device. Enjoy!
On to Angles and the Unit Circle – you’re ready!
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# Log Rules OR Logarithm Rules
Updated: 20 Aug 2023
2098
## Log Rules
Rules of logarithm also refer to “Log Rules”. This comprehensive guide will explore the basic log rules used to solve complex logarithm equations and problems. Some of the log rules are mentioned below:
1. Product Log Rule
\log _a m n=\log _a m+\log _a n
2. Quotient Log Rule
\log _a \frac{m}{n}=\log _a m-\log _a n
3. Power Log Rule
\log _a m^n=n \log _a m
4. Change of Base Law Rule
\log _a m \log _m n=\log _a n
\log _m n=\frac{\log _a n}{\log _a m}
5. Identity Log Rule
\log _a a=1
\log _{10} 10=1
6. Zero Log Rule
\log_a 1=0
## Proof of Log Rules
Following are the detailed proof of the log rules.
### Product Log Rule
The logarithm of two products equals the sum of the individual logarithms.
#### Proof of Product Log Rule
Let \log _a m=x \ and \ \log _a n=y
Write them in Exponential form:
a^x=m \ and \ a^y=n
Now multiply these:
a^x \times a^y=mn
Or
mn=a^x \times a^y
mn=a^{x+y}
Taking \log _a on B.S
\log _a m n=\log _a a^{x+y}
\log _a m n=(x+y) \log _a a
\log _a m n=(x+y)(1) \qquad \log _a a=1
\log _a m n=x+y
\log _a m n=\log _a m+\log _a n
### Quotient Log Rule
The logarithm of a quotient is equal to the difference of the individual logarithms.
#### Proof of Quotient Log Rule
Let \log _a m=x \ and \ \log _a n=y
Write them in Exponential form:
a^x=m \ and \ a^y=n
Now Divide these:
\frac{a^x}{a^y}=\frac{m}{n}
Or
\frac{m}{n}=\frac{a^x}{a^y}
\frac{m}{n}=a^{x-y}
Taking \log _a on B.S
\log _a \frac{m}{n}=\log _a a^{x-y}
\log _a \frac{m}{n}=(x-y) \log _a a
\log _a \frac{m}{n}=(x-y)(1) \qquad \log _a a=1
\log _a \frac{m}{n}=x-y
Hence \ \log _a \frac{m}{n}=\log _a m-\log _a n
### Power Log Rule
The logarithm of a number raised to a power equals the exponent times the logarithm of the base:
#### Proof of Power Log Rule
Let \log _a m=x
In Exponential form:
a^x=m
Or
m=a^x
Taking power ‘ n ‘ on B.S
m^n=\left(a^x\right)^n
m^n=a^{n x}
Taking \log _a on B.S
\log _a m^n=\log _a a^{n x}
\log _a m^n=n x \log _a a
\log _a m^n=n x(1) \qquad \log _a a=1
\log _a m^n=n x
\log _a m^n=n \log _a m
### Identity Log Rule
• \log _a a=1
• \log _{10} 10=1
• \log_a 1=0
## Log Rules MCQs
1. \log _a m n=
O \log _a m+\log _a n
O \log _a m-\log _a n
O n \log _a m
O All of them
\log _a m+\log _a n
Explanation:
Let \log _a m=x \ and \ \log _a n=y
Write them in Exponential form:
a^x=m \ and \ a^y=n
Now multiply these:
a^x \times a^y=mn
Or
mn=a^x \times a^y
mn=a^{x+y}
Taking \log _a on B.S
\log _a m n=\log _a a^{x+y}
\log _a m n=(x+y) \log _a a
\log _a m n=(x+y)(1) \qquad \log _a a=1
\log _a m n=x+y
\log _a m n=\log _a m+\log _a n
2. \log _a \frac{m}{n}=
O \log _a m+\log _a n
O \log _a m-\log _a n
O n \log _a m
O All of them
\log _a m-\log _a n
Explanation:
Let \log _a m=x \ and \ \log _a n=y
Write them in Exponential form:
a^x=m and a^y=n
Now Divide these:
\frac{a^x}{a^y}=\frac{m}{n}
Or
\frac{m}{n}=\frac{a^x}{a^y}
\frac{m}{n}=a^{x-y}
Taking \log _a on B.S
\log _a \frac{m}{n}=\log _a a^{x-y}
\log _a \frac{m}{n}=(x-y) \log _a a
\log _a \frac{m}{n}=(x-y)(1) \qquad \log _a a=1
\log _a \frac{m}{n}=x-y
Hence \ \log _a \frac{m}{n}=\log _a m-\log _a n
3. \log _a m^n=
O \log _a m+\log _a n
O \log _a m-\log _a n
O n \log _a m
O All of them
n \log _a m
Explanation:
Let \log _a m=x
In Exponential form:
a^x=m
Or
m=a^x
Taking power ‘ n ‘ on B.S
m^n=\left(a^x\right)^n
m^n=a^{n x}
Taking \log _a on B.S
\log _a m^n=\log _a a^{n x}
\log _a m^n=n x \log _a a
\log _a m^n=n x(1) \qquad \log _a a=1
\log _a m^n=n x
\log _a m^n=n \log _a m
4. \quad \log _a m+\log _a n
O \log _a \frac{m}{n}
O \log _a m n
O n \log _a m
O All of them
\log _a m n
Explanation:
\log _a m n=\log _a m+\log _a n
5. \log _a m-\log _a n
O \log _a \frac{m}{n}
O \log _a m n
O n \log _a m
O All of them
\log _a \frac{m}{n}
Explanation:
\log _a \frac{m}{n}=\log _a m-\log _a n
6. n \log _a m
O \log _a \frac{m}{n}
O \log _a m n
O \log _a m^n
O All of them
Explanation:
\log _a m^n=n \log _a m
7. \log m n=
O \log m+\log n
O \log m-\log n
O n \log m
O All of them
\log m+\log n
Explanation:
\log _a m n=\log _a m+\log _a n
8. \log \frac{m}{n}=
O \log m+\log n
O \log m-\log n
O n \log m
O All of them
\log m-\log n
Explanation:
\log _a \frac{m}{n}=\log _a m-\log _a n
9. \log m^n=
O \log m+\log n
O \log m-\log n
O n \log m
O All of them
n \log m
Explanation:
\log _a m^n=n \log _a m
10. \log m+\log n
O \log \frac{m}{n}
O \log m n
O \log m^n
O All of them
\log m n
Explanation:
\log _a m n=\log _a m+\log _a n
11. \log m-\log n
O \log \frac{m}{n}
O \log m n
O \log m^n
O All of them
\log \frac{m}{n}
Explanation:
\log _a \frac{m}{n}=\log _a m-\log _a n
12. n \log m
O \log \frac{m}{n}
O \log m n
O \log m^n
O All of them
\log m^n
Explanation:
\log _a m^n=n \log _a m
13. \log 2 \times 3=
O \log 2+\log 3
O \log 2-\log 3
O 2 \log 3
O All of them
\log 2+\log 3
Explanation:
\log _a m n=\log _a m+\log _a n
14. \log \frac{2}{3}=
O \log 2+\log 3
O \log 2-\log 3
O 2 \log 3
O All of them
\log 2-\log 3
Explanation:
\log _a \frac{m}{n}=\log _a m-\log _a n
15. \log 3^2=
O \log 2+1
O \log 2-1
O 2 \log 3
O All of them
2 \log 3
Explanation:
\log _a m^n=n \log _a m
16. \log 2+\log 3
O \log 2 \times 3
O \log 6
O \log 2
O Both a & b
\log 6
Explanation:
\log 2+\log 3 =\log 2\times 3
\log 2+\log 3 =\log 6
17. \log 2-\log 3
O \log \frac{2}{3}
O \log 2 \times 3
O \log 3^2
O All of them
\log \frac{2}{3}
Explanation:
\log _a \frac{m}{n}=\log _a m-\log _a n
18. 2 \log 3=
O \log \frac{2}{3}
O \log 2 \times 3
O \log 3^2
O All of them
\log 3^2
Explanation:
\log _a m^n=n \log _a m
2 \log 3= \log 3^2
2 \log 3= \log 9
19. If \log _2 6+\log _2 7=\log _2 a \ then \ a=
O 6
O 7
O 24
O 42
42
Explanation:
\log _2 6+\log _2 7=\log _2 a
As \log _a m n=\log _a m+\log _a n
\log _2 6 \times 7=\log _2 a
\log _2 42=\log _2 a
Thus \ a=42
20. \log _a m \log _m n=
O \log _a n
O \log _a m
O Both a & b
O None of these
\log _a n
Explanation:
Let \log _a m=x and \log _m n=y
Write them in Exponential form:
a^x=m \ and \ m^y=n
Now multiply these:
As a^{x y}=\left(a^x\right)^y
But \left(a^x\right)^y=m
So a^{x y}=(m)^y=n
Then a^{x y}=n
Taking \log _a on B.S
\log _a a^{x y}=\log _a n
(x y) \log _a a=\log _a n
x y(1)=\log _a n \qquad As \ \log _a a=1
Now
\log _a m \log _m n=\log _a n
21. \log _2 3 \log _3 5=
O \log _5 2
O \log _2 5
O Both a & b
O None of these
\log _2 5
Explanation:
\log _a m \log _m n=\log _a n
22. \log _2 3 \log _3 4 \log _4 5=
O \log _5 2
O \log _2 5
O Both a & b
O None of these
\log _2 5
Explanation:
\log _a m \log _m n=\log _a n
23. \log _m n=\frac{\log _a n}{\log _a m} is called ____________ law
O Logarithm
O Change of Base
O Change of Logarithm
O None of these
Change of Base
Explanation:
24. \frac{\log _a n}{\log _a m}=
O \log _m n
O \log _t r
O \log 10
O None of these
\log _m n
Explanation:
25. \frac{\log _7 r}{\log _7 t}=
O \log _m n
O \log _t r
O \log 10
O None of these
\log _t r
Explanation:
26. \log _a a=
O 0
O 1
O 10
O None of these
1
Explanation:
27. \log _{10} 10=
O 0
O 1
O 10
O None of these
1
Explanation:
28. log10= __________
O 0
O 1
O 10
O None of these
1
Explanation:
29. log_a 1= __________
O 0
O 1
O 10
O None of these
0
Explanation:
29. log1= __________
O 0
O 1
O 10
O None of these
Explanation:
## Review Exercise # 3
1. \log _9 \frac{1}{81}=
O -1
O -2
O 2
O Does not exist
-2
Explanation:
\log _9 \frac{1}{81}=\log _9 \frac{1}{9^2}
\log _9 \frac{1}{81}=\log _9 9^{-2}
\log _9 \frac{1}{81}=-2 \log _9 9
\log _9 \frac{1}{81}=-2(1)
\log _9 \frac{1}{81}=-2
2. If \log _2 8=x \ then \ x=
O 64
O 3^2
O 3
O 2^8
3
Explanation:
\log _2 8=x
\log _2 2^3=x
3 \log _2 2=x
3(1)=x
3=x
3. Base of common log is:
O 10
O e
O \pi
O 5
10
Explanation:
4. \log \sqrt{10}=
O -1
O -\frac{1}{2}
O \frac{1}{2}
O 2
\frac{1}{2}
Explanation:
\log \sqrt{10} =\log (10)^{\frac{1}{2}}
\log \sqrt{10} =\frac{1}{2} \log 10
\log \sqrt{10} =\frac{1}{2}(1)
\log \sqrt{10} =\frac{1}{2}
5. For any non-zero value of x \cdot x^0=
O 2
O 1
O 0
O 10
1
Explanation:
6. Rewrite t=\log _b m as an exponent equation
O t=m^b
O b^m=t
O m=b^t
O m^t=b
m=b^t
Explanation:
7. \log _{10} 10=
O 2
O 3
O 0
O 1
1
Explanation:
8. Characteristics of \log 0.000059 is
O -5
O 5
O -4
O 4
-5
Explanation:
9. Evaluate \log _7 \frac{1}{\sqrt{7}}
O -1
O -\frac{1}{2}
O \frac{1}{2}
O 2
-\frac{1}{2}
Explanation:
\log _7 \frac{1}{\sqrt{7}} =\log _7 \frac{1}{(7)^{\frac{1}{2}}}
\log _7 \frac{1}{\sqrt{7}} =\log _7 7^{-\frac{1}{2}}
\log _7 \frac{1}{\sqrt{7}} =-\frac{1}{2} \log _7 7
\log _7 \frac{1}{\sqrt{7}} =-\frac{1}{2}(1)
\log _7 \frac{1}{\sqrt{7}} =-\frac{1}{2}
10. Base of natural log is
O 10
O e
O \pi
O 1
Explanation:
11. \log m+\log n=
O \log m\log n
O \log m-\log n
O \log mn
O \log \frac{m}{n}
\log mn
Explanation:
12. 0.069 can be written in scientific notation as
O 6.9 \times 10^3
O 6.9 \times 10^{-2}
O 0.69 \times 10^3
O 69 \times 10^2
6.9 \times 10^{-2}
Explanation:
13. \ln x-2 \ln y
O \ln \frac{x}{y}
O \ln x y^2
O \ln \frac{x^2}{y}
O \ln \frac{x}{y^2} |
# Starting the Multiplication Unit
We believe there are 3 non-negotiable skills for fifth graders moving on to sixth grade:
• Fractions
• Multiplication
• Division
Accordingly, we spend more time on these 3 units than most schools do, although we work in a handful of additional topics as well. The payoff comes later — middle school (and high school!) teachers don’t have to reteach these topics. A goal in itself! Additionally, students are able to transfer their understanding of these vital topics to word problems, problem-solving, and algebra.
Our success rate (the % of students succeeding) in Algebra I in 8th grade has gone UP since we slowed our curriculum DOWN! Our middle schoolers need the 4 whole number operations, plus fractions, decimals, percents, some geometry and some data theory. It’s not that much! Most of all, they need number sense, conceptual strength, self-knowledge and confidence.
With that in mind, let’s look at our new unit: Multiplication. Non-negotiable goals at the end:
• All students finish memorizing their times tables, or find efficient work-arounds.
• All students can build and draw an area model for 2-digit multiplication.
• Most students can multiply 2 digits mentally or algorithmically, and know the multiplicative property names.
• More advanced students can investigate factors, multiples, exponents and puzzles.
## ACTIVITY ONE – Dots and Diamonds
Unfortunately, a fraction (1/5?) of our 5th graders have not yet mastered the first 9×9 multiplication facts. Not because they’re not trying; because memorization is not one of their learning style strengths. To give them some efficient work-arounds, we use the Dot Array approach.
## PDF: Dots and Diamonds
Dot Arrays for the 7’s:
### DOTS
In our experience, students usually know their ‘Fives’ multiples. If I know 5 sevens is 35, then 6 sevens is one more set of seven, or 42. The child writes “6×7 = 42” on the blank in the column on the left. They can count if that’s what they need. (Our mantra: we must take students where they are, not where they’re supposed to be!)
Students who already know their tables spend a maximum of 1 minute on the 3 dot array problems. They don’t begrudge this small time investment, and they’re actually visualizing the array relationships (perhaps for the first time).
Students who struggle with memorization are getting a visual aide to help them speed up their most efficient work-arounds (like 6 x7 is seven more than 5×7). We will repeat these arrays (in tiny doses) on a daily basis for a few weeks…. “Once a day”, like a vitamin. That visual repetition is what it might take to master the times tables.
### DIAMONDS
We ask what students notice about the first two diamond examples. Then “turn and talk” to check for engagement. (Are they really looking?)
Usually, everyone notices that the numbers left and right are ADDED at the bottom and MULTIPLIED at the top of the diamond.
After a couple practice diamonds, we go on to more challenging questions. Six and WHAT add to 14?
Challenge: What adds to 15 and multiplies to 50? (5 & 10) What adds to 15 and multiplies to 56? (7 & 8), etc. We find that everyone tries these problems, especially since they have the 6, 7, and 8 tables at the top of the page.
These diamonds give enough challenge to be fun, yet help review basic facts.
Those who finish the diamond problems quickly go on to the challenge puzzles on page 2 — mixed up multiplication tables.
## Word doc: 1×2 dig mult
As always, this classwork goes Concrete > Pictorial > Abstract (Challenge). Meeting students where they are means offering all 3 levels at all times
The second page is meant to be mental math – they do not have to shade in the rectangle. Can they see 3 x 32 as three arrays of 30, plus the 3 sets of 2 = 96?
The challenge page 3 is from the website:
## ACTIVITY THREE A Game: “Close to 300”
This game (called Target 300) is taken from Marilyn Burns’ book Teaching Arithmetic: Lessons for Extending Multiplication, Grades 4–5 (Math Solutions Publications, 2001).
and this Word handout with instructions and an example: CLOSE TO 300
In addition, each child will need one die, or each pair can share a die.
The Power Point asks students to look for patterns in the table. TAKE TIME! Wait while they process the possibilities. Often the first observations are trivial (Each column across increases by 10) Praise these forays into observational math. Only gradually do they see other patterns. Slides 4 – 8 in the Power Pt give a nudge for discovering further patterns. Anything they see is great.
## ACTIVITY FOUR Intro to Two-digit Multiplication
This classwork uses concrete challenges (this requires base-10 blocks) to build the visualization of the 2-digit area model for multiplication.
Intro2digitMult
2digMult_Key
Our students were introduced to area models with blocks in 4th grade. Otherwise, these activities might need more introduction, in the form of building more block models like the first 2 on page 1.
The 5 “What Multiplication Problem Is This” questions definitely require blocks, therefore building the conceptual understanding of multiplication rectangles. Students need to know the rules: The given blocks MUST form a rectangle, and we prefer to standardize our models by putting the 100-flat in the upper left.
Students find that they have to lay out a possible rectangle, and then slide tens rods from the right side down to the bottom side (or vice versa) until the rectangle has the correct number of ones. (see key) Once they realize that each puzzle is possible, they begin sliding tens around until each one works. The most common mistake at the beginning is not building a rectangle. Once that requirement is understood, they are much closer to conceptual understanding.
Soon someone will say “Oh… 7 rods and 12 ones. What adds to 7 and multiplies to 12?” Just a note: this exact same concept shows up in Algebra I when students learn to factor quadratics.
We will continue with the “What Multiplication Problem Is This” puzzles for a couple weeks; it’s that valuable.
## HW#1
Tri2Unit1HW1
HW1_Key
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# 9.4: The F Distribution and the F-Ratio
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The distribution used for the hypothesis test is a new one. It is called the F-distribution, invented by George Snedecor but named in honor of Sir Ronald Fisher, an English statistician. The $$F$$ statistic is a ratio (a fraction). There are two sets of degrees of freedom; one for the numerator and one for the denominator.
For example, if $$F$$ follows an $$F$$ distribution and the number of degrees of freedom for the numerator is four, and the number of degrees of freedom for the denominator is ten, then $$F \sim F_{4,10}$$.
To calculate the $$\bf{F}$$ ratio, two estimates of the variance are made.
1. Variance between samples: An estimate of $$\sigma^2$$ that is the variance of the sample means multiplied by $$n$$ (when the sample sizes are the same.). If the samples are different sizes, the variance between samples is weighted to account for the different sample sizes. The variance is also called variation due to treatment or explained variation.
2. Variance within samples: An estimate of $$\sigma^2$$ that is the average of the sample variances (also known as a pooled variance). When the sample sizes are different, the variance within samples is weighted. The variance is also called the variation due to error or unexplained variation.
• $$SS_{between}$$ is the sum of squares that represents the variation among the different samples
• $$SS_{within}$$ is the sum of squares that represents the variation within samples that is due to chance.
To find a "sum of squares" means to add together squared quantities that, in some cases, may be weighted. We used sum of squares to calculate the sample variance and the sample standard deviation in Table 1.19.
MS means "mean square." $$MS_{between}$$ is the variance between groups, and $$MS_{within}$$ is the variance within groups.
Calculation of Sum of Squares and Mean Square
• $$k$$ is the the number of different groups
• $$n_j$$ is the size of the $$j^{th}$$ group
• $$s_j$$ = the sum of the values in the $$j^{th}$$ group
• $$n$$ is the total number of all the values combined (total sample size: $$\Sigma n_{j}$$)
• $$x$$ is the one value: $\sum x=\sum s_{j} \nonumber$
• Sum of squares of all values from every group combined: $\sum x^{2} \nonumber$
• Between group variability: $SS_{total} =\sum x^{2}-\frac{\left(\sum x^{2}\right)}{n} \nonumber$
• Total sum of squares: $\sum x^{2}-\frac{\left(\sum x\right)^{2}}{n} \nonumber$
• Explained variation: sum of squares representing variation among the different samples:
$SS_{between} =\sum\left[\frac{\left(s_{j}\right)^{2}}{n_{j}}\right]-\frac{\left(\sum s_{j}\right)^{2}}{n} \nonumber$
• Unexplained variation: sum of squares representing variation within samples due to chance: $S S_{\text { within }}=S S_{\text { total }}-S S_{\text { between }} \nonumber$
• $$df$$'s for different groups ($$df$$'s for the numerator): $df = k – 1 \nonumber$
• Equation for errors within samples ($$df$$'s for the denominator): $df_{within} = n – k \nonumber$
• Mean square (variance estimate) explained by the different groups: $M S_{\text { between }}=\frac{S S_{\text { between }}}{d f_{\text { between }}} \nonumber$
• Mean square (variance estimate) that is due to chance (unexplained): $M S_{\mathrm{within}}=\frac{S S_{\mathrm{within}}}{d f_{\mathrm{within}}} \nonumber$
$$MS_{between}$$ and $$MS_{within}$$ can be written as follows:
\begin{align*} M S_{\mathrm{between}} & =\frac{S S_{\mathrm{between}}}{d f_{\mathrm{between}}}=\frac{S S_{\mathrm{between}}}{k-1} \\[4pt] M S_{within} &=\frac{SS_{w ithin}}{df_{within}}=\frac{SS_{within}}{n-k}\end{align*}
The one-way ANOVA test depends on the fact that $$M S_{between}$$ can be influenced by population differences among means of the several groups. Since $$M S_{within}$$ compares values of each group to its own group mean, the fact that group means might be different does not affect $$M S_{within}$$.
The null hypothesis says that all groups are samples from populations having the same normal distribution. The alternate hypothesis says that at least two of the sample groups come from populations with different normal distributions. If the null hypothesis is true, $$M S_{between}$$ and $$M S_{within}$$ should both estimate the same value.
Note
The null hypothesis says that all the group population means are equal. The hypothesis of equal means implies that the populations have the same normal distribution, because it is assumed that the populations are normal and that they have equal variances.
Definition: F-Ratio or F Statistic
$F=\frac{M S_{\text { between }}}{M S_{\text { within }}}$
If $$M S_{between}$$ and $$M S_{within}$$ estimate the same value (following the belief that $$H_0$$ is true), then the $$F$$-ratio should be approximately equal to one. Mostly, just sampling errors would contribute to variations away from one. As it turns out, $$M S_{between}$$ consists of the population variance plus a variance produced from the differences between the samples. $$M S_{within}$$ is an estimate of the population variance. Since variances are always positive, if the null hypothesis is false, $$M S_{between}$$ will generally be larger than $$MS_{within}$$.Then the $$F$$-ratio will be larger than one. However, if the population effect is small, it is not unlikely that $$M S_{within}$$ will be larger in a given sample.
The foregoing calculations were done with groups of different sizes. If the groups are the same size, the calculations simplify somewhat and the F-ratio can be written as:
F-Ratio Formula when the groups are the same size
The foregoing calculations were done with groups of different sizes. If the groups are the same size, the calculations simplify somewhat and the F-ratio can be written as
$F=\frac{n \cdot s_{\overline{x}}^{2}}{s^{2}_{ pooled }}$
where
• $$n$$ = the sample size
• $$d f_{\text {numerator}}=k-1$$
• $$d f_{\text {denominator}}=n-k$$
• $$s_{pooled}^2$$ = the mean of the sample variances (pooled variance)
• $$s_{\overline x}^2$$ = the variance of the sample means
Data are typically put into a table for easy viewing. One-Way ANOVA results are often displayed in this manner by computer software.
Table $$\PageIndex{1}$$
Source of variation Sum of squares ($$SS$$) Degrees of freedom ($$df$$) Mean square ($$MS$$) $$F$$
Factor
(Between)
$$SS$$(Factor) $$k – 1$$ $$MS(Factor) = \dfrac{SS(Factor)}{k– 1}$$ $$F = \dfrac{MS(Factor)}{MS(Error)}$$
Error
(Within)
$$SS$$(Error) $$n – k$$ $$MS(Error) = \dfrac{SS(Error)}{n – k}$$
Total $$SS$$(Total) $$n – 1$$
Example 12.2
Three different diet plans are to be tested for mean weight loss. The entries in the table are the weight losses for the different plans. The one-way ANOVA results are shown in Table $$\PageIndex{2}$$.
Table $$\PageIndex{2}$$
Plan 1: $$n_1 = 4$$ Plan 2: $$n_2 = 3$$ Plan 3: $$n_3 = 3$$
5 3.5 8
4.5 7 4
4 3.5
3 4.5
$$s_{1}=16.5, s_{2}=15, s_{3}=15.5$$
Following are the calculations needed to fill in the one-way ANOVA table. The table is used to conduct a hypothesis test.
\begin{align*} S(\text { between }) &=\sum\left[\frac{\left(s_{j}\right)^{2}}{n_{j}}\right]-\frac{\left(\displaystyle \sum s_{j}\right)^{2}}{n} \\[4pt] &=\frac{s_{1}^{2}}{4}+\frac{s_{2}^{2}}{3}+\frac{s_{3}^{2}}{3}-\frac{\left(s_{1}+s_{2}+s_{3}\right)^{2}}{10}\end{align*}
where $$n_{1}=4, n_{2}=3, n_{3}=3$$ and $$n=n_{1}+n_{2}+n_{3}=10$$.
\begin{align*} S(\text { between }) &= \frac{(16.5)^{2}}{4}+\frac{(15)^{2}}{3}+\frac{(15.5)^{2}}{3}-\frac{(16.5+15+15.5)^{2}}{10} \\[4pt] &=2.2458 \\[4pt] S(\text {total}) &=\sum x^{2}-\frac{\left(\sum x\right)^{2}}{n} \\[4pt] &=\left(5^{2}+4.5^{2}+4^{2}+3^{2}+3.5^{2}+7^{2}+4.5^{2}+8^{2}+4^{2}+3.5^{2}\right) -\frac{(5+4.5+4+3+3.5+7+4.5+8+4+3.5)^{2}}{10}\\[4pt] &=244-\frac{47^{2}}{10} \\[4pt] &=244-220.9 \\[4pt] & =23.1 \\[4pt] S(\text {within}) & = S(\text {total})-S S(\text {between}) \\[4pt] &=23.1-2.2458 \\[4pt] &=20.8542 \end{align*}
Table $$\PageIndex{3}$$
Source of variation Sum of squares ($$SS$$) Degrees of freedom ($$df$$) Mean square ($$MS$$) $$F$$
Factor
(Between)
$$SS(Factor) = SS(Between) \\= 2.2458$$ $$k – 1 = 3 groups – 1 \\= 2$$ $$MS(Factor) = \dfrac{SS(Factor)}{k – 1} \\= 2.2458/2 \\= 1.1229$$ $$F = \dfrac{MS(Factor)}{MS(Error)} \\ = \dfrac{1.1229}{2.9792} \\= 0.3769$$
Error
(Within)
$$SS(Error) = SS(Within) \\ = 20.8542$$ $$n – k = 10 total data – 3 groups \\= 7$$ $$MS(Error) = \dfrac{SS(Error)}{n – k} \\= \dfrac{20.8542}{7} \\= 2.9792$$
Total $$SS(Total) = 2.2458 + 20.8542 \\= 23.1$$ $$n – 1 = 10 total data – 1 \\= 9$$
Exercise 12.2
As part of an experiment to see how different types of soil cover would affect slicing tomato production, Marist College students grew tomato plants under different soil cover conditions. Groups of three plants each had one of the following treatments
• bare soil
• a commercial ground cover
• black plastic
• straw
• compost
All plants grew under the same conditions and were the same variety. Students recorded the weight (in grams) of tomatoes produced by each of the n = 15 plants:
Bare: $$n_1 = 3$$ Ground Cover: $$n_2 = 3$$ Plastic: $$n_3 = 3$$ Straw: $$n_4 = 3$$ Compost: $$n_5 = 3$$
2,625 5,348 6,583 7,285 6,277
2,997 5,682 8,560 6,897 7,818
4,915 5,482 3,830 9,230 8,677
Table $$\PageIndex{4}$$
Create the one-way ANOVA table.
The one-way ANOVA hypothesis test is always right-tailed because larger $$F$$-values are way out in the right tail of the F-distribution curve and tend to make us reject $$H_0$$.
Example 12.3
Let’s return to the slicing tomato exercise in Try It. The means of the tomato yields under the five mulching conditions are represented by $$\mu_{1}, \mu_{2}, \mu_{3}, \mu_{4}, \mu_{5}$$. We will conduct a hypothesis test to determine if all means are the same or at least one is different. Using a significance level of 5%, test the null hypothesis that there is no difference in mean yields among the five groups against the alternative hypothesis that at least one mean is different from the rest.
The null and alternative hypotheses are:
$$H_{0} : \mu_{1}=\mu_{2}=\mu_{3}=\mu_{4}=\mu_{5}$$
$$H_{a} : \mu_{i} \neq \mu_{j}$$ some $$i \neq j$$
The one-way ANOVA results are shown in Table $$\PageIndex{5}$$
Table $$\PageIndex{5}$$
Source of variation Sum of squares ($$SS$$) Degrees of freedom ($$df$$) Mean square ($$MS$$) F
Factor (Between) 36,648,561 $$5 – 1 = 4$$ $$\frac{36,648,561}{4}=9,162,140$$ $$\frac{9,162,140}{2,044,672.6}=4.4810$$
Error (Within) 20,446,726 $$15 – 5 = 10$$ $$\frac{20,446,726}{10}=2,044,672.6$$
Total 57,095,287 $$15 – 1 = 14$$
Distribution for the test: $$F_{4,10}$$
$$df(num) = 5 – 1 = 4$$
$$df(denom) = 15 – 5 = 10$$
Test statistic: $$F = 4.4810$$
FIgure $$\PageIndex{1}$$
Probability Statement: $$p\text{-value }= P(F > 4.481) = 0.0248.$$
Compare $$\bf{\alpha}$$ and the $$\bf p$$-value: $$\alpha = 0.05$$, $$p\text{-value }= 0.0248$$
Make a decision: Since $$\alpha > p$$-value, we cannot accept $$H_0$$.
Conclusion: At the 5% significance level, we have reasonably strong evidence that differences in mean yields for slicing tomato plants grown under different mulching conditions are unlikely to be due to chance alone. We may conclude that at least some of mulches led to different mean yields.
Exercise 12.3
MRSA, or Staphylococcus aureus, can cause a serious bacterial infections in hospital patients. Table $$\PageIndex{6}$$ shows various colony counts from different patients who may or may not have MRSA. The data from the table is plotted in FIgure $$\PageIndex{2}$$.
Table $$\PageIndex{6}$$
Conc = 0.6 Conc = 0.8 Conc = 1.0 Conc = 1.2 Conc = 1.4
9 16 22 30 27
66 93 147 199 168
98 82 120 148 132
Plot of the data for the different concentrations:
FIgure $$\PageIndex{2}$$
Test whether the mean number of colonies are the same or are different. Construct the ANOVA table, find the p-value, and state your conclusion. Use a 5% significance level.
Example 12.4
Four sororities took a random sample of sisters regarding their grade means for the past term. The results are shown in Table $$\PageIndex{7}$$.
Table $$\PageIndex{7}$$: Mean grades for four sororities
Sorority 1 Sorority 2 Sorority 3 Sorority 4
2.17 2.63 2.63 3.79
1.85 1.77 3.78 3.45
2.83 3.25 4.00 3.08
1.69 1.86 2.55 2.26
3.33 2.21 2.45 3.18
Using a significance level of 1%, is there a difference in mean grades among the sororities?
Let $$\mu_{1}, \mu_{2}, \mu_{3}, \mu_{4}$$ be the population means of the sororities. Remember that the null hypothesis claims that the sorority groups are from the same normal distribution. The alternate hypothesis says that at least two of the sorority groups come from populations with different normal distributions. Notice that the four sample sizes are each five.
Note: This is an example of a balanced design, because each factor (i.e., sorority) has the same number of observations.
$$H_{0}: \mu_{1}=\mu_{2}=\mu_{3}=\mu_{4}$$
$$H_a$$: Not all of the means $$\mu_{1}, \mu_{2}, \mu_{3}, \mu_{4}$$ are equal.
Distribution for the test: $$F_{3,16}$$
where $$k = 4$$ groups and $$n = 20$$ samples in total
$$df(num)= k – 1 = 4 – 1 = 3$$
$$df(denom) = n – k = 20 – 4 = 16$$
Calculate the test statistic: $$F = 2.23$$
Graph:
FIgure $$\PageIndex{3}$$
Probability statement: $$p\text{-value }= P(F > 2.23) = 0.1241$$
Compare $$\bf{\alpha}$$ and the $$\bf p$$-value: $$\alpha = 0.01$$
$$p\text{-value }= 0.1241$$
$$\alpha < p$$-value
Make a decision: Since $$\alpha < p$$-value, you cannot reject $$H_0$$.
Conclusion: There is not sufficient evidence to conclude that there is a difference among the mean grades for the sororities.
Exercise 12.4
Four sports teams took a random sample of players regarding their GPAs for the last year. The results are shown in Table $$\PageIndex{8}$$.
Table $$\PageIndex{8}$$ GPAs for four sports teams
3.6 2.1 4.0 2.0
2.9 2.6 2.0 3.6
2.5 3.9 2.6 3.9
3.3 3.1 3.2 2.7
3.8 3.4 3.2 2.5
Use a significance level of 5%, and determine if there is a difference in GPA among the teams.
Example 12.5
A fourth grade class is studying the environment. One of the assignments is to grow bean plants in different soils. Tommy chose to grow his bean plants in soil found outside his classroom mixed with dryer lint. Tara chose to grow her bean plants in potting soil bought at the local nursery. Nick chose to grow his bean plants in soil from his mother's garden. No chemicals were used on the plants, only water. They were grown inside the classroom next to a large window. Each child grew five plants. At the end of the growing period, each plant was measured, producing the data (in inches) in Table $$\PageIndex{9}$$.
Tommy's plants Tara's plants Nick's plants
24 25 23
21 31 27
23 23 22
30 20 30
23 28 20
Table $$\PageIndex{9}$$
Does it appear that the three media in which the bean plants were grown produce the same mean height? Test at a 3% level of significance.
This time, we will perform the calculations that lead to the F' statistic. Notice that each group has the same number of plants, so we will use the formula $$F^{\prime}=\frac{n \cdot s_{\overline{x}}^{2}}{s^{2}_{pooled}}$$.
First, calculate the sample mean and sample variance of each group.
Tommy's plants Tara's plants Nick's plants
Sample mean 24.2 25.4 24.4
Sample variance 11.7 18.3 16.3
Table $$\PageIndex{10}$$
Next, calculate the variance of the three group means (Calculate the variance of 24.2, 25.4, and 24.4). Variance of the group means = 0.413 = $$s_{\overline{x}}^{2}$$
Then $$M S_{b e t w e e n}=n s_{\overline{x}}^{2}=(5)(0.413)$$ where $$n = 5$$ is the sample size (number of plants each child grew).
Calculate the mean of the three sample variances (Calculate the mean of 11.7, 18.3, and 16.3). Mean of the sample variances = 15.433 = $$\bf{s^2}$$ pooled
Then $$M S_{\text {within}}=s^{2} \text { pooled }=15.433$$.
The $$F$$ statistic (or $$F$$ ratio) is $$F=\frac{M S_{\text { between }}}{M S_{\text { within }}}=\frac{n s_{\overline{x}}^{2}}{s^{2} \text { pooled }}=\frac{(5)(0.413)}{15.433}=0.134$$
The $$df$$s for the numerator = the number of groups $$– 1 = 3 – 1 = 2$$.
The $$df$$s for the denominator = the total number of samples – the number of groups $$= 15 – 3 = 12$$
The distribution for the test is $$F_{2,12}$$ and the $$F$$ statistic is $$F = 0.134$$
The $$p$$-value is $$P(F > 0.134) = 0.8759$$.
Decision: Since $$\alpha = 0.03$$ and the $$p\text{-value }= 0.8759$$, then you cannot reject H0. (Why?)
Conclusion: With a 3% level of significance, from the sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants are different.
## Notation
The notation for the $$F$$ distribution is $$F \sim F_{d f(n u m), d f(d e n o m)}$$ where $$df(num) = df_{between}$$ and $$df(denom) = df_{within}$$. The mean for the $$F$$ distribution is$$\mu=\frac{d f(n u m)}{d f(\text {denom})-2}$$
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# 12-7 Surface Area of Spheres. Objective Recognize and define basic properties of spheres Find surface area of spheres.
## Presentation on theme: "12-7 Surface Area of Spheres. Objective Recognize and define basic properties of spheres Find surface area of spheres."— Presentation transcript:
12-7 Surface Area of Spheres
Objective Recognize and define basic properties of spheres Find surface area of spheres
Spheres Sphere – In space, infinitely many points that are a given distance from a central point. Great Circle – the circle made when a plane intersects a sphere where it contains the center. A great circle has the same center and radii of the sphere it is within. Hemisphere – one of two halves of a sphere. Each great circle in a sphere separates it into two hemispheres. Great Circle
Example 1 In the figure, O is the center of the sphere and plane R intersects the sphere In circle A. If AO is 3 cm and OB is 10 cm, find AB o R A B
The radius of circle A is segment AB, B is a point on circle A and on sphere O. Use Pythagorean Theorem for the right triangle ABO to solve for AB OB = AB + AO 10 = AB + 3 100 = AB + 9 91 = AB 9.5 AB 222 ~ ~ 2 2 2 22 Pythagorean Theorem OB = 1=, AO = 3 Simplify Subtract 9 from each side Use a calculator AB is approximately 9.5 centimeters
Surface Area of Spheres If a sphere has a surface area of T units and a radius of r units then T = 4 or 4 times the area of the sphere’s great circle. r 2 (4)( )(r) = surface 2 area
Example 2 Find the surface area of the sphere given the area of the great circle A 314.2 in ~ ~ 2
From the formula, we know that the area of the sphere is 4 times the area of its great circle. T = 4 T = 4(314.2) T = 1256.8 in The surface area is about 1256.8 in r 2 2 2 Surface area of a sphere. Surface area of GS is approx. 314.2 Multiply
Example 3 Find the surface area of a baseball with a circumference of 9 inches To determine how much leather is needed to cover the ball. 9 inches
We know that the circumference is 9 so we can use that to find the area of the great circle 9 = D D 1.4 T = 4 r T = 4 (1.4) T 25.8 ~ ~ 2 2 ~ ~ The surface area is approx. 25.8 sq. inches Circumference of a circle Use a calculator Surface area of a sphere r = 1.4 Use a calculator
Assignment Pg 673 10-15, 17-31
Download ppt "12-7 Surface Area of Spheres. Objective Recognize and define basic properties of spheres Find surface area of spheres."
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# Difference between revisions of "2014 AMC 10A Problems/Problem 24"
## Problem
A sequence of natural numbers is constructed by listing the first $4$, then skipping one, listing the next $5$, skipping $2$, listing $6$, skipping $3$, and, on the $n$th iteration, listing $n+3$ and skipping $n$. The sequence begins $1,2,3,4,6,7,8,9,10,13$. What is the $500,\!000$th number in the sequence?
$\textbf{(A)}\ 996,\!506\qquad\textbf{(B)}\ 996,\!507\qquad\textbf{(C)}\ 996,\!508\qquad\textbf{(D)}\ 996,\!509\qquad\textbf{(E)}\ 996,\!510$
## Solution 1
If we list the rows by iterations, then we get
$1,2,3,4$
$6,7,8,9,10$
$13,14,15,16,17,18$ etc.
so that the $500,000$th number is the $506$th number on the $997$th row. ($4+5+6+7......+997 = 499,494$) The last number of the $996$th row (when including the numbers skipped) is $499,494 + (1+2+3+4.....+996)= 996,000$, (we add the $1-996$ because of the numbers we skip) so our answer is $996,000 + 506 = \boxed{\textbf{(A)}996,506}$
## Solution 2
$1,2,3,4,5,...,500,000$
All we need to do is count how many numbers are skipped, $n$, and "push" (add on to) $500,000$ however many numbers are skipped.
Clearly, $\frac{999(1000)}{2}<500,000<\frac{1000(1001)}{2}$. This means that the number of skipped number "blocks" in the sequence is $999-3=996$ because we started counting from 4.
Therefore $n=\frac{996(997)}{2}=496,506$, and the answer is $496,506+500000=\boxed{\textbf{(A)}996,506}$.
2014 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 23 Followed byProblem 25 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions |
## Elementary Algebra
{$- \frac{5\sqrt 2}{4}, \frac{5\sqrt 2}{4}$}
Using Property 10.1, which states that for any non-negative real number $a$, $x^{2}=a$ can be written as $x=\pm\sqrt a$, we obtain: Step 1: $8n^{2}=25$ Step 2: $n^{2}=\frac{25}{8}$ Step 3: $x=\pm \sqrt {\frac{25}{8}}$ Step 4: $x=\pm \frac{\sqrt {25}}{\sqrt 8}$ Step 5: $x=\pm \frac{\sqrt {25}}{\sqrt {4\times2}}$ Step 6: $x=\pm \frac{5}{2\sqrt 2}$ Step 7: $x=\pm \frac{5}{2\sqrt 2}\times\frac{2\sqrt 2}{2\sqrt 2}$ Step 8: $x=\pm \frac{10\sqrt 2}{(2\sqrt 2)^{2}}$ Step 9: $x=\pm \frac{10\sqrt 2}{4 \times 2}$ Step 10: $x=\pm \frac{5\sqrt 2}{4}$ The solution set is {$- \frac{5\sqrt 2}{4}, \frac{5\sqrt 2}{4}$}. |
College Algebra with Corequisite Support
# 6.8Fitting Exponential Models to Data
College Algebra with Corequisite Support6.8 Fitting Exponential Models to Data
## Learning Objectives
In this section, you will:
• Build an exponential model from data.
• Build a logarithmic model from data.
• Build a logistic model from data.
## Corequisite Skills
### Learning Objectives
• Draw and interpret scatter diagrams (linear, exponential, logarithmic). (CA 4.3.1)
• Fit a regression equation to a set of data and use the linear (or exponential) model to make predictions. (CA 4.3.4)
## Vocabulary and Concept Check
Draw and interpret scatter diagrams (linear, exponential, logarithmic).
Fill in the blanks and match the description with the graphs a, b, or c
A ________ function has equation $fx=mx+bfx=mx+b$ and has a basic shape ________.
A ________ function has equation and has a basic shape ________.
A ________ function has equation and has a basic shape ________.
A Scatter Plot is a graph of plotted points that may show a relationship between the variables in a set of data.
## Example 1
Draw and interpret scatter diagrams (linear, exponential, logarithmic).
### Using a Scatter Plot to Investigate Cricket Chirps
[link] shows the number of cricket chirps in 15 seconds, for several different air temperatures, in degrees Fahrenheit 9 . Plot this data, and determine whether the data appears to be linearly related.
Chirps 44 35 20.4 33 31 35 18.5 37 26 Temperature 80.5 70.5 57 66 68 72 52 73.5 53
Cricket Chirps vs Air Temperature
#### Practice Makes Perfect
Draw and interpret scatter diagrams ( linear, exponential, logarithmic).
1.
Make a scatter plot for the table below. Does it look linear? Exponential? Logarithmic?
x 1 2 3 4 5 6 7 8 9 y 0 1.5 2.2 2.8 3.5 3.6 3.9 4.3 4.4
2.
Make a scatter plot for the table below. Does it look linear? Exponential? Logarithmic?
x 1 2 3 4 5 6 7 8 9 y 3.3 5.6 9.1 15.1 24.4 40.2 66.2 108.4 180.1
3.
Make a scatter plot for the table below. Does it look linear? Exponential? Logarithmic?
x 1 2 3 4 5 6 y 3 5.5 7 10 12.1 14.9
### Objective 2: Fit a regression equation to a set of data and use the linear (or exponential) model to make predictions. (CA 4.3.4)
We can find a linear function that fits the data in the previous problem by “eyeballing” a line that seems to fit. But while estimating a line works relatively well, technology can help us find a line that fits the data as perfect as possible.
This line is called the Least Squares Regression Line or Linear Regression Model.
A regression line is a line that is closest to the data in the scatter plot, which means that such a line is a best fit for the data.
Fit a regression equation to a set of data and use the linear (or exponential) model to make predictions.
## How To
Given data of input and corresponding outputs from a linear function, find the best fit line using linear regression.
1. Enter the input in List 1 (L1).
2. Enter the output in List 2 (L2).
3. On a graphing utility, select Linear Regression (LinReg).
## Example 2
### Fit a regression equation to a set of data and use the linear (or exponential) model to make predictions.
Find the linear regression line using the cricket-chirp data in the example earlier in this section, and find the temperature if there are 30 chirps in 15 seconds.
#### Practice Makes Perfect
Fit a regression equation to a set of data and use the linear (or exponential) model to make predictions.
4.
Gasoline consumption in the United States has been steadily increasing from 1994 to 2004.
Year 94 95 96 97 98 99 00 01 02 03 04 Consumption(billions of gallons) 113 116 118 119 123 125 126 128 131 133 136
• Determine whether the trend is linear, and if so, use your graphing utility to find a model for the data.
• Use the model to predict the consumption in 2008.
5.
We determined in the second practice problem, earlier in this section, that the data below has an exponential trend. Use your graphing utility to find an exponential model that fits the data the best and write your exponential model below (Hint: instead of choosing Linear Regression, choose Exponential Regression).
x 1 2 3 4 5 6 7 8 9 y 3.3 5.6 9.1 15.1 24.4 40.2 66.2 108.4 180.1
In previous sections of this chapter, we were either given a function explicitly to graph or evaluate, or we were given a set of points that were guaranteed to lie on the curve. Then we used algebra to find the equation that fit the points exactly. In this section, we use a modeling technique called regression analysis to find a curve that models data collected from real-world observations. With regression analysis, we don’t expect all the points to lie perfectly on the curve. The idea is to find a model that best fits the data. Then we use the model to make predictions about future events.
Do not be confused by the word model. In mathematics, we often use the terms function, equation, and model interchangeably, even though they each have their own formal definition. The term model is typically used to indicate that the equation or function approximates a real-world situation.
We will concentrate on three types of regression models in this section: exponential, logarithmic, and logistic. Having already worked with each of these functions gives us an advantage. Knowing their formal definitions, the behavior of their graphs, and some of their real-world applications gives us the opportunity to deepen our understanding. As each regression model is presented, key features and definitions of its associated function are included for review. Take a moment to rethink each of these functions, reflect on the work we’ve done so far, and then explore the ways regression is used to model real-world phenomena.
## Building an Exponential Model from Data
As we’ve learned, there are a multitude of situations that can be modeled by exponential functions, such as investment growth, radioactive decay, atmospheric pressure changes, and temperatures of a cooling object. What do these phenomena have in common? For one thing, all the models either increase or decrease as time moves forward. But that’s not the whole story. It’s the way data increase or decrease that helps us determine whether it is best modeled by an exponential equation. Knowing the behavior of exponential functions in general allows us to recognize when to use exponential regression, so let’s review exponential growth and decay.
Recall that exponential functions have the form $y=a b x y=a b x$ or $y= A 0 e kx . y= A 0 e kx .$ When performing regression analysis, we use the form most commonly used on graphing utilities, $y=a b x . y=a b x .$ Take a moment to reflect on the characteristics we’ve already learned about the exponential function $y=a b x y=a b x$ (assume $a>0): a>0):$
• $b b$ must be greater than zero and not equal to one.
• The initial value of the model is $y=a. y=a.$
• If $b>1, b>1,$ the function models exponential growth. As $x x$ increases, the outputs of the model increase slowly at first, but then increase more and more rapidly, without bound.
• If $0 the function models exponential decay. As $x x$ increases, the outputs for the model decrease rapidly at first and then level off to become asymptotic to the x-axis. In other words, the outputs never become equal to or less than zero.
As part of the results, your calculator will display a number known as the correlation coefficient, labeled by the variable $r, r,$ or $r 2 . r 2 .$ (You may have to change the calculator’s settings for these to be shown.) The values are an indication of the “goodness of fit” of the regression equation to the data. We more commonly use the value of $r 2 r 2$ instead of $r, r,$ but the closer either value is to 1, the better the regression equation approximates the data.
## Exponential Regression
Exponential regression is used to model situations in which growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero. We use the command “ExpReg” on a graphing utility to fit an exponential function to a set of data points. This returns an equation of the form, $y=a b x y=a b x$
Note that:
• $b b$ must be non-negative.
• when $b>1, b>1,$ we have an exponential growth model.
• when $0 we have an exponential decay model.
## How To
Given a set of data, perform exponential regression using a graphing utility.
1. Use the STAT then EDIT menu to enter given data.
1. Clear any existing data from the lists.
2. List the input values in the L1 column.
3. List the output values in the L2 column.
2. Graph and observe a scatter plot of the data using the STATPLOT feature.
1. Use ZOOM [9] to adjust axes to fit the data.
2. Verify the data follow an exponential pattern.
3. Find the equation that models the data.
1. Select “ExpReg” from the STAT then CALC menu.
2. Use the values returned for a and b to record the model, $y=a b x . y=a b x .$
4. Graph the model in the same window as the scatterplot to verify it is a good fit for the data.
## Example 1
### Using Exponential Regression to Fit a Model to Data
In 2007, a university study was published investigating the crash risk of alcohol impaired driving. Data from 2,871 crashes were used to measure the association of a person’s blood alcohol level (BAC) with the risk of being in an accident. Table 1 shows results from the study 10. The relative risk is a measure of how many times more likely a person is to crash. So, for example, a person with a BAC of 0.09 is 3.54 times as likely to crash as a person who has not been drinking alcohol.
BAC 0 0.01 0.03 0.05 0.07 0.09 Relative Risk of Crashing 1 1.03 1.06 1.38 2.09 3.54 BAC 0.11 0.13 0.15 0.17 0.19 0.21 Relative Risk of Crashing 6.41 12.6 22.1 39.05 65.32 99.78
Table 1
1. Let $x x$ represent the BAC level, and let $y y$ represent the corresponding relative risk. Use exponential regression to fit a model to these data.
2. After 6 drinks, a person weighing 160 pounds will have a BAC of about $0.16. 0.16.$ How many times more likely is a person with this weight to crash if they drive after having a 6-pack of beer? Round to the nearest hundredth.
## Try It #1
Table 2 shows a recent graduate’s credit card balance each month after graduation.
Month 1 2 3 4 5 6 7 8 Debt (\$) 620 761.88 899.8 1039.93 1270.63 1589.04 1851.31 2154.92
Table 2
Use exponential regression to fit a model to these data.
If spending continues at this rate, what will the graduate’s credit card debt be one year after graduating?
## Q&A
Is it reasonable to assume that an exponential regression model will represent a situation indefinitely?
No. Remember that models are formed by real-world data gathered for regression. It is usually reasonable to make estimates within the interval of original observation (interpolation). However, when a model is used to make predictions, it is important to use reasoning skills to determine whether the model makes sense for inputs far beyond the original observation interval (extrapolation).
## Building a Logarithmic Model from Data
Just as with exponential functions, there are many real-world applications for logarithmic functions: intensity of sound, pH levels of solutions, yields of chemical reactions, production of goods, and growth of infants. As with exponential models, data modeled by logarithmic functions are either always increasing or always decreasing as time moves forward. Again, it is the way they increase or decrease that helps us determine whether a logarithmic model is best.
Recall that logarithmic functions increase or decrease rapidly at first, but then steadily slow as time moves on. By reflecting on the characteristics we’ve already learned about this function, we can better analyze real world situations that reflect this type of growth or decay. When performing logarithmic regression analysis, we use the form of the logarithmic function most commonly used on graphing utilities, $y=a+bln( x ). y=a+bln( x ).$ For this function
• All input values, $x, x,$ must be greater than zero.
• The point $( 1,a ) ( 1,a )$ is on the graph of the model.
• If $b>0, b>0,$ the model is increasing. Growth increases rapidly at first and then steadily slows over time.
• If $b<0, b<0,$ the model is decreasing. Decay occurs rapidly at first and then steadily slows over time.
## Logarithmic Regression
Logarithmic regression is used to model situations where growth or decay accelerates rapidly at first and then slows over time. We use the command “LnReg” on a graphing utility to fit a logarithmic function to a set of data points. This returns an equation of the form,
$y=a+bln( x ) y=a+bln( x )$
Note that
• all input values, $x, x,$ must be non-negative.
• when $b>0, b>0,$ the model is increasing.
• when $b<0, b<0,$ the model is decreasing.
## How To
Given a set of data, perform logarithmic regression using a graphing utility.
1. Use the STAT then EDIT menu to enter given data.
1. Clear any existing data from the lists.
2. List the input values in the L1 column.
3. List the output values in the L2 column.
2. Graph and observe a scatter plot of the data using the STATPLOT feature.
1. Use ZOOM [9] to adjust axes to fit the data.
2. Verify the data follow a logarithmic pattern.
3. Find the equation that models the data.
1. Select “LnReg” from the STAT then CALC menu.
2. Use the values returned for a and b to record the model, $y=a+bln( x ). y=a+bln( x ).$
4. Graph the model in the same window as the scatterplot to verify it is a good fit for the data.
## Example 2
### Using Logarithmic Regression to Fit a Model to Data
Due to advances in medicine and higher standards of living, life expectancy has been increasing in most developed countries since the beginning of the 20th century.
Table 3 shows the average life expectancies, in years, of Americans from 1900–201011.
Year 1900 1910 1920 1930 1940 1950 Life Expectancy(Years) 47.3 50 54.1 59.7 62.9 68.2 Year 1960 1970 1980 1990 2000 2010 Life Expectancy(Years) 69.7 70.8 73.7 75.4 76.8 78.7
Table 3
1. Let $x x$ represent time in decades starting with $x=1 x=1$ for the year 1900, $x=2 x=2$ for the year 1910, and so on. Let $y y$ represent the corresponding life expectancy. Use logarithmic regression to fit a model to these data.
2. Use the model to predict the average American life expectancy for the year 2030.
## Try It #2
Sales of a video game released in the year 2000 took off at first, but then steadily slowed as time moved on. Table 4 shows the number of games sold, in thousands, from the years 2000–2010.
Year 2000 2001 2002 2003 2004 2005 Number Sold (thousands) 142 149 154 155 159 161 Year 2006 2007 2008 2009 2010 - Number Sold (thousands) 163 164 164 166 167 -
Table 4
• Let $x x$ represent time in years starting with $x=1 x=1$ for the year 2000. Let $y y$ represent the number of games sold in thousands. Use logarithmic regression to fit a model to these data.
• If games continue to sell at this rate, how many games will sell in 2015? Round to the nearest thousand.
## Building a Logistic Model from Data
Like exponential and logarithmic growth, logistic growth increases over time. One of the most notable differences with logistic growth models is that, at a certain point, growth steadily slows and the function approaches an upper bound, or limiting value. Because of this, logistic regression is best for modeling phenomena where there are limits in expansion, such as availability of living space or nutrients.
It is worth pointing out that logistic functions actually model resource-limited exponential growth. There are many examples of this type of growth in real-world situations, including population growth and spread of disease, rumors, and even stains in fabric. When performing logistic regression analysis, we use the form most commonly used on graphing utilities:
$y= c 1+a e −bx y= c 1+a e −bx$
Recall that:
• $c 1+a c 1+a$ is the initial value of the model.
• when $b>0, b>0,$ the model increases rapidly at first until it reaches its point of maximum growth rate, $( ln( a ) b , c 2 ). ( ln( a ) b , c 2 ).$ At that point, growth steadily slows and the function becomes asymptotic to the upper bound $y=c. y=c.$
• $c c$ is the limiting value, sometimes called the carrying capacity, of the model.
## Logistic Regression
Logistic regression is used to model situations where growth accelerates rapidly at first and then steadily slows to an upper limit. We use the command “Logistic” on a graphing utility to fit a logistic function to a set of data points. This returns an equation of the form
$y= c 1+a e −bx y= c 1+a e −bx$
Note that
• The initial value of the model is $c 1+a . c 1+a .$
• Output values for the model grow closer and closer to $y=c y=c$ as time increases.
## How To
Given a set of data, perform logistic regression using a graphing utility.
1. Use the STAT then EDIT menu to enter given data.
1. Clear any existing data from the lists.
2. List the input values in the L1 column.
3. List the output values in the L2 column.
2. Graph and observe a scatter plot of the data using the STATPLOT feature.
1. Use ZOOM [9] to adjust axes to fit the data.
2. Verify the data follow a logistic pattern.
3. Find the equation that models the data.
1. Select “Logistic” from the STAT then CALC menu.
2. Use the values returned for $a, a,$ $b, b,$ and $c c$ to record the model, $y= c 1+a e −bx . y= c 1+a e −bx .$
4. Graph the model in the same window as the scatterplot to verify it is a good fit for the data.
## Example 3
### Using Logistic Regression to Fit a Model to Data
Mobile telephone service has increased rapidly in America since the mid 1990s. Today, almost all residents have cellular service. Table 5 shows the percentage of Americans with cellular service between the years 1995 and 2012 12.
Year Americans with Cellular Service (%) Year Americans with Cellular Service (%)
1995 12.69 2004 62.852
1996 16.35 2005 68.63
1997 20.29 2006 76.64
1998 25.08 2007 82.47
1999 30.81 2008 85.68
2000 38.75 2009 89.14
2001 45.00 2010 91.86
2002 49.16 2011 95.28
2003 55.15 2012 98.17
Table 5
• Let $x x$ represent time in years starting with $x=0 x=0$ for the year 1995. Let $y y$ represent the corresponding percentage of residents with cellular service. Use logistic regression to fit a model to these data.
• Use the model to calculate the percentage of Americans with cell service in the year 2013. Round to the nearest tenth of a percent.
• Discuss the value returned for the upper limit, $c. c.$ What does this tell you about the model? What would the limiting value be if the model were exact?
## Try It #3
Table 6 shows the population, in thousands, of harbor seals in the Wadden Sea over the years 1997 to 2012.
Year Seal Population (Thousands) Year Seal Population (Thousands)
1997 3.493 2005 19.590
1998 5.282 2006 21.955
1999 6.357 2007 22.862
2000 9.201 2008 23.869
2001 11.224 2009 24.243
2002 12.964 2010 24.344
2003 16.226 2011 24.919
2004 18.137 2012 25.108
Table 6
• Let $x x$ represent time in years starting with $x=0 x=0$ for the year 1997. Let $y y$ represent the number of seals in thousands. Use logistic regression to fit a model to these data.
• Use the model to predict the seal population for the year 2020.
• To the nearest whole number, what is the limiting value of this model?
## Media
Access this online resource for additional instruction and practice with exponential function models.
## 6.8 Section Exercises
### Verbal
1.
What situations are best modeled by a logistic equation? Give an example, and state a case for why the example is a good fit.
2.
What is a carrying capacity? What kind of model has a carrying capacity built into its formula? Why does this make sense?
3.
What is regression analysis? Describe the process of performing regression analysis on a graphing utility.
4.
What might a scatterplot of data points look like if it were best described by a logarithmic model?
5.
What does the y-intercept on the graph of a logistic equation correspond to for a population modeled by that equation?
### Graphical
For the following exercises, match the given function of best fit with the appropriate scatterplot in Figure 7 through Figure 11. Answer using the letter beneath the matching graph.
Figure 7
Figure 8
Figure 9
Figure 10
Figure 11
6.
$y=10.209 e −0.294x y=10.209 e −0.294x$
7.
$y=5.598−1.912ln(x) y=5.598−1.912ln(x)$
8.
$y=2.104 ( 1.479 ) x y=2.104 ( 1.479 ) x$
9.
$y=4.607+2.733ln(x) y=4.607+2.733ln(x)$
10.
$y= 14.005 1+2.79 e −0.812x y= 14.005 1+2.79 e −0.812x$
### Numeric
11.
To the nearest whole number, what is the initial value of a population modeled by the logistic equation $P(t)= 175 1+6.995 e −0.68t ? P(t)= 175 1+6.995 e −0.68t ?$ What is the carrying capacity?
12.
Rewrite the exponential model $A(t)=1550 ( 1.085 ) x A(t)=1550 ( 1.085 ) x$ as an equivalent model with base $e. e.$ Express the exponent to four significant digits.
13.
A logarithmic model is given by the equation $h(p)=67.682−5.792ln( p ). h(p)=67.682−5.792ln( p ).$ To the nearest hundredth, for what value of $p p$ does $h(p)=62? h(p)=62?$
14.
A logistic model is given by the equation $P(t)= 90 1+5 e −0.42t . P(t)= 90 1+5 e −0.42t .$ To the nearest hundredth, for what value of t does $P(t)=45? P(t)=45?$
15.
What is the y-intercept on the graph of the logistic model given in the previous exercise?
### Technology
For the following exercises, use this scenario: The population $P P$ of a koi pond over $x x$ months is modeled by the function $P(x)= 68 1+16 e −0.28x . P(x)= 68 1+16 e −0.28x .$
16.
Graph the population model to show the population over a span of $3 3$ years.
17.
What was the initial population of koi?
18.
How many koi will the pond have after one and a half years?
19.
How many months will it take before there are $20 20$ koi in the pond?
20.
Use the intersect feature to approximate the number of months it will take before the population of the pond reaches half its carrying capacity.
For the following exercises, use this scenario: The population $P P$ of an endangered species habitat for wolves is modeled by the function $P(x)= 558 1+54.8 e −0.462x , P(x)= 558 1+54.8 e −0.462x ,$ where $x x$ is given in years.
21.
Graph the population model to show the population over a span of $10 10$ years.
22.
What was the initial population of wolves transported to the habitat?
23.
How many wolves will the habitat have after $3 3$ years?
24.
How many years will it take before there are $100 100$ wolves in the habitat?
25.
Use the intersect feature to approximate the number of years it will take before the population of the habitat reaches half its carrying capacity.
For the following exercises, refer to Table 7.
x 1 2 3 4 5 6 f(x) 1125 1495 2310 3294 4650 6361
Table 7
26.
Use a graphing calculator to create a scatter diagram of the data.
27.
Use the regression feature to find an exponential function that best fits the data in the table.
28.
Write the exponential function as an exponential equation with base $e. e.$
29.
Graph the exponential equation on the scatter diagram.
30.
Use the intersect feature to find the value of $x x$ for which $f(x)=4000. f(x)=4000.$
For the following exercises, refer to Table 8.
x 1 2 3 4 5 6 f(x) 555 383 307 210 158 122
Table 8
31.
Use a graphing calculator to create a scatter diagram of the data.
32.
Use the regression feature to find an exponential function that best fits the data in the table.
33.
Write the exponential function as an exponential equation with base $e. e.$
34.
Graph the exponential equation on the scatter diagram.
35.
Use the intersect feature to find the value of $x x$ for which $f(x)=250. f(x)=250.$
For the following exercises, refer to Table 9.
x 1 2 3 4 5 6 f(x) 5.1 6.3 7.3 7.7 8.1 8.6
Table 9
36.
Use a graphing calculator to create a scatter diagram of the data.
37.
Use the LOGarithm option of the REGression feature to find a logarithmic function of the form $y=a+bln( x ) y=a+bln( x )$ that best fits the data in the table.
38.
Use the logarithmic function to find the value of the function when $x=10. x=10.$
39.
Graph the logarithmic equation on the scatter diagram.
40.
Use the intersect feature to find the value of $x x$ for which $f(x)=7. f(x)=7.$
For the following exercises, refer to Table 10.
x 1 2 3 4 5 6 7 8 f(x) 7.5 6 5.2 4.3 3.9 3.4 3.1 2.9
Table 10
41.
Use a graphing calculator to create a scatter diagram of the data.
42.
Use the LOGarithm option of the REGression feature to find a logarithmic function of the form $y=a+bln( x ) y=a+bln( x )$ that best fits the data in the table.
43.
Use the logarithmic function to find the value of the function when $x=10. x=10.$
44.
Graph the logarithmic equation on the scatter diagram.
45.
Use the intersect feature to find the value of $x x$ for which $f(x)=8. f(x)=8.$
For the following exercises, refer to Table 11.
x 1 2 3 4 5 6 7 8 9 10 f(x) 8.7 12.3 15.4 18.5 20.7 22.5 23.3 24 24.6 24.8
Table 11
46.
Use a graphing calculator to create a scatter diagram of the data.
47.
Use the LOGISTIC regression option to find a logistic growth model of the form $y= c 1+a e −bx y= c 1+a e −bx$ that best fits the data in the table.
48.
Graph the logistic equation on the scatter diagram.
49.
To the nearest whole number, what is the predicted carrying capacity of the model?
50.
Use the intersect feature to find the value of $x x$ for which the model reaches half its carrying capacity.
For the following exercises, refer to Table 12.
$xx$ 0 2 4 5 7 8 10 11 15 17 $f(x)f(x)$ 12 28.6 52.8 70.3 99.9 112.5 125.8 127.9 135.1 135.9
Table 12
51.
Use a graphing calculator to create a scatter diagram of the data.
52.
Use the LOGISTIC regression option to find a logistic growth model of the form $y= c 1+a e −bx y= c 1+a e −bx$ that best fits the data in the table.
53.
Graph the logistic equation on the scatter diagram.
54.
To the nearest whole number, what is the predicted carrying capacity of the model?
55.
Use the intersect feature to find the value of $x x$ for which the model reaches half its carrying capacity.
### Extensions
56.
Recall that the general form of a logistic equation for a population is given by $P(t)= c 1+a e −bt , P(t)= c 1+a e −bt ,$ such that the initial population at time $t=0 t=0$ is $P(0)= P 0 . P(0)= P 0 .$ Show algebraically that $c−P(t) P(t) = c− P 0 P 0 e −bt . c−P(t) P(t) = c− P 0 P 0 e −bt .$
57.
Use a graphing utility to find an exponential regression formula $f(x) f(x)$ and a logarithmic regression formula $g(x) g(x)$ for the points $( 1.5,1.5 ) ( 1.5,1.5 )$ and $( 8.5,8.5 ). ( 8.5,8.5 ).$ Round all numbers to 6 decimal places. Graph the points and both formulas along with the line $y=x y=x$ on the same axis. Make a conjecture about the relationship of the regression formulas.
58.
Verify the conjecture made in the previous exercise. Round all numbers to six decimal places when necessary.
59.
Find the inverse function $f −1 ( x ) f −1 ( x )$ for the logistic function $f(x)= c 1+a e −bx . f(x)= c 1+a e −bx .$ Show all steps.
60.
Use the result from the previous exercise to graph the logistic model $P(t)= 20 1+4 e −0.5t P(t)= 20 1+4 e −0.5t$ along with its inverse on the same axis. What are the intercepts and asymptotes of each function?
### Footnotes
• 9Selected data from http://classic.globe.gov/fsl/scientistsblog/2007/10/. Retrieved Aug 3, 2010
• 10Source: Indiana University Center for Studies of Law in Action, 2007
• 11Source: Center for Disease Control and Prevention, 2013
• 12Source: The World Bank, 2013 |
Angle Measure Formulas | List of Angle Measure Formulas You Should Know - BYJUS
# Angle Measure Formulas
The measure of an angle is the distance between two lines which are intersecting at a point. There are names given to some angles of a particular value. For example, the right angle for angles of 90 degrees and the straight angle for values of angle Although the study of angles seems simple and trivial, it opens the path to the different aspects of architecture. Even carpenters use angles to help create their work. Even when determining the trajectory of airplanes or launching missiles, the accuracy of the angle is very important....Read MoreRead Less
### What is a Degree?
If a circle is divided into 360 equal parts, the angle that turns $$\frac{1}{360}$$ th of a circle is $$1^{\circ}$$. A full turn around the circle is 360 degrees. In the image shown below, each of the smallest divisions is a degree.
The distance between the intersecting lines of the two rays provided below is $$\frac{60}{360}$$. What is the angle between the two rays?
From the theory elaborated previously, we know that the angle that turns $$\frac{1}{360}$$ th of a circle is $$1^{\circ}$$. This means that the angle that turns $$\frac{60}{360}$$ th of a circle should be 60°.
### Solved Examples
Example 1:
Examine the two representations given below and explain if both the representations show the same angles.
Solution :
The first picture is a representation of 360 degrees and the second picture is a representation of 0 degrees. The value of the angle is different in both these representations.
Example 2:
Find the measure of the angle given below.
Solution :
We know that $$\frac{1}{360}$$ th of a circle is one degree.
This means that $$\frac{57}{360}$$ th is 57 degrees.
Therefore, the measure of the angle given here is 57 degrees.
Example 3:
Find the measure of the angle given below.
We know that $$\frac{1}{360}$$ th of a circle is one degree.
$$\frac{1}{10}$$ th of a circle can be calculated by multiplying $$\frac{1}{10}$$ with 360.
$$=\frac{1}{10} \times 360=36^{\circ}$$
Therefore, $$\frac{1}{10}$$ th of a circle is 36 degrees.
Example 4:
Categorize the given set of angles as acute, straight, right or obtuse.
A. 25°
B. 150°
C. 90°
D. 180°
Solution:
A. 25°: 25 degrees can be categorized as an acute angle as it is less than 90 degrees.
B. 150°: 150 degrees can be classified as an obtuse angle as it is greater than 90 degrees.
C. 90°: 90 degrees is also called the right angle.
D. 180°: is also called a straight angle as the resultant angle forms a straight line.
Example 5:
Tammy and Lola ordered a chocolate cake. Lola took $$\frac{3}{7}$$ th of the cake and $$\frac{4}{9}$$ th was taken by Tammy. Who took the larger slice of cake?
Solution:
We can find out who took the larger slice of cake by calculating the angle at the center of the cake.
To find the angle we can find the $$\frac{3}{7}$$ th and $$\frac{4}{9}$$ th of 360 degrees.
$$\frac{3}{7}$$ th of 360 degrees = $$\frac{3}{7}\times 360$$ = 154.29°
$$\frac{4}{9}$$ th of 360 degrees = $$\frac{4}{9}\times 360$$ = 160°
The area covered by the 160 degree angle is greater than 154.29°
This shows that Tammy took more cake than Lola.
Example 6:
A tire has five spokes fashioned on the rim which divides the tire into five equal parts. Find the angle measure of each part.
Solution:
A complete circle is 360 degrees, to find the angle of one part out of 5, 360 degrees needs to be divided by 5.
The measure of the angle between each section divided by the spokes is $$=\frac{360}{5}=72^{\circ}$$
Example 7:
A pizza was cut into 12 equal pieces. Out of these three pieces were eaten. What is the angle measure of the three pieces which were eaten?
Solution:
To find the angle measure of each slice, we need to divide 360 by 12.
The measure of each slice $$=\frac{360}{12}=30^{\circ}$$
To find the measure of three out of the twelve slices, all that needs to be done is to multiply the angle measure of one slice by three.
$$=30\times 3 = 90^{\circ}$$
Therefore, the angle measure of three slices is 90°. |
# 6.02 Applications of exponential functions
Lesson
## Exponential growth and decay
An exponential function is the appropriate model to use when a quantity is increasing or decreasing at a rate that depends on the quantity present.
For example, in the final rounds of a sports competition, the number of competing teams is halved at every stage. Thus, if $16$16 teams reached the first semifinal, there would be $8$8 in the second semifinal, and so on. The number of teams playing drops from $16$16 to $8$8 and then to $4$4 and finally, $2$2 and we see that the reduction in the number of teams playing at each stage depends on the number in the previous round.
This is an example of exponential decay. The rate of decrease gets progressively smaller. Many processes show the opposite pattern and exhibit exponential growth. In this, the rate of increase increases progressively.
Remember!
If we are looking at an exponential function of the form $y=ab^x$y=abx, then
• $b$b is tells us whether the function is growing (increasing) or decaying (decreasing)
• If $b>1$b>1, it is growth
### Outcomes
#### A.CED.A.2^
Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. ^Equations using all available types of expressions, including simple root functions
#### F.IF.B.4'''
For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity. '''Include rational, square root and cube root; emphasize selection of appropriate models.
#### F.IF.C.8'''
Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. '''Include rational and radical; Focus on using key features to guide selection of appropriate type of model function |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Area of a Parallelogram
## Base times height.
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Practice Area of a Parallelogram
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Area of a Parallelogram
Have you ever tried to figure out the area of a figure?
Miguel is working on a problem in his math book. He has been given the following drawing.
Miguel needs to figure out the area of this parallelogram.
Do you know how to do this?
This Concept will teach you how to figure out the area of a parallelogram. By the end of the Concept, you will know how to help Miguel with this task.
### Guidance
Previously we worked on quadrilaterals. Here is one of the quadrilaterals that we learned about earlier, the parallelogram. A parallelogram is a four sided figure with opposite sides parallel. It doesn’t matter what the angles are in a parallelogram as long as the opposite sides are parallel.
If we wanted to figure out the distance around the edge of a parallelogram, then we would find the perimeter of the figure.
If we wanted to find the measure of the space inside the parallelogram, we would be finding the area of the parallelogram. This lesson is all about finding the area of parallelograms.
Let’s start off by looking at how we find the area of a common parallelogram, the rectangle. A rectangle is a parallelogram with four right angles. The opposite sides are parallel too.
To find the area of a rectangle, we multiply the length times the width.
\begin{align*}A=lw\end{align*}
To find the area of this rectangle, we use the formula and the given measurements.
Notice that the measurement is in square inches because inches \begin{align*}x\end{align*} inches is inches squared.
Write down the formula for finding the area of a rectangle in your notebook. Also be sure to include the statement in the box above.
Not all parallelograms have right angles. That is why some are called squares or rectangles and some are called parallelograms. The only necessary quality of a parallelogram is that the opposite sides need to be parallel.
How can we find the area of a parallelogram?
Because a parallelogram does not have right angles, multiplying the length and the width is not possible. Notice that the side of a parallelogram is at an angle. Because of this, we need to use a different measure to find the area of a parallelogram. We need to use the base and the height.
Notice that the base is the bottom measurement and the height is the measurement inside the figure.
When we multiply these two measurements, we can find the area of the parallelogram. Here is the formula.
\begin{align*}A=bh\end{align*}
Now it's time for you to figure out the area of the following parallelograms given the base and height.
#### Example A
Base = 9 ft., Height = 4 ft.
Solution: \begin{align*}36\end{align*} sq. ft.
#### Example B
Base = 7 meters, Height = 3.5 meters
Solution: \begin{align*}24.5\end{align*} sq. m.
#### Example C
Base = 10 yards, Height = 7 yards
Solution: \begin{align*}70\end{align*} sq. yds
Here is the original problem once again.
Miguel is working on a problem in his math book. He has been given the following drawing.
Miguel needs to figure out the area of this parallelogram.
Do you know how to do this?
To figure this out, Miguel needs to use the formula for finding the area of a parallelogram.
\begin{align*}A = bh\end{align*}
Now we substitute the given values into the formula.
\begin{align*}A = (6)(2)\end{align*}
\begin{align*}A = 12 \ cm^2\end{align*}
This is our answer.
### Vocabulary
Parallelogram
a quadrilateral with opposite sides parallel.
Perimeter
the distance around a figure.
Area
the amount of space contained inside a two-dimensional figure.
### Guided Practice
Here is one for you to try on your own.
Find the area of the parallelogram below.
Answer
We can see that the base is 7 inches and the height is 3 inches. We simply put these numbers into the appropriate places in the formula and solve for \begin{align*}A\end{align*}.
This is our answer
### Practice
Directions: Find the area of each parallelogram given the base and the height.
1. Base = 9 inches, height = 5 inches
2. Base = 4 inches, height = 3 inches
3. Base = 12 feet, height = 6 feet
4. Base = 11 meters, height = 8 meters
5. Base = 13 yards, height = 10 yards
6. Base = 4 feet, height = 2.5 feet
7. Base = 5.5 inches, height = 3.5 inches
8. Base = 9 feet, height = 6.5 feet
9. Base = 22 miles, height = 18 miles
10. Base = 29 meters, height = 12 meters
11. Base = 22 meters, height = 11 meters
12. Base = 39 meters, height = 15 meters
13. Base = 40 meters, height = 25 meters
14. Base = 88 meters, height = 50 meters
15. Base = 79 meters, height = 14 meters
### Vocabulary Language: English
Area
Area
Area is the space within the perimeter of a two-dimensional figure.
Parallelogram
Parallelogram
A parallelogram is a quadrilateral with two pairs of parallel sides.
Perimeter
Perimeter
Perimeter is the distance around a two-dimensional figure.
Area of a Parallelogram
Area of a Parallelogram
The area of a parallelogram is equal to the base multiplied by the height: A = bh. The height of a parallelogram is always perpendicular to the base (the sides are not the height).
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Home | | Statistics 11th std | Absolute Measures
# Absolute Measures
It involves the units of measurements of the observations.
Absolute Measures
It involves the units of measurements of the observations. For example, (i) the dispersion of salary of employees is expressed in rupees, and (ii) the variation of time required for workers is expressed in hours. Such measures are not suitable for comparing the variability of the two data sets which are expressed in different units of measurements.
## 1. Range
### Raw Data:
Range is defined as difference between the largest and smallest observations in the data set. Range(R) = Largest value in the data set (L) –Smallest value in the data set(S)
R= L - S
### Grouped Data:
For grouped frequency distribution of values in the data set, the range is the difference between the upper class limit of the last class interval and the loawer class limit of first class interval.
### Coefficient of Range
The relative measure of range is called the coefficient of range.
Co efficient of Range = (L-S) / (L+S)
### Example 6.2
The following data relates to the heights of 10 students (in cms) in a school.
Calculate the range and coefficient of range.
158, 164, 168, 170, 142, 160, 154, 174, 159, 146
### Solution:
L=142 S=174
Range = L – S = 174 – 142 = 32
Coefficient of range = (L-S)( L+S)
=(174-142)/(174+142) = 32/316 = 0.101
### Example 6.3
Calculate the range and the co-efficient of range for the marks obtained by 100 students in a school.
### Solution:
L = Upper limit of highest class = 75
S = lower limit of lowest class = 60
Range = L-S = 75-60 = 15
Coefficient of range = (L-S)( L+S)
=15/(75+60) = 0.111
### Merits:
· Range is the simplest measure of dispersion.
· It is well defined, and easy to compute.
· It is widely used in quality control, weather forecasting, stock market variations etc.
### Limitations:
· The calculations of range is based on only two values – largest value and smallest value.
· It is largely influenced by two extreme value.
· It cannot be computed in the case of open-ended frequency distributions.
· It is not suitable for further mathematical treatment.
## 2. Inter Quartile Range and Quartile Deviation
The quartiles Q1, Q2 and Q3 have been introduced and studied in Chapter 5.
Inter quartile range is defined as: Inter quartile Range (IQR) = Q3-Q1
Quartile Deviation is defined as, half of the distance between Q1 and Q3.
Quartile Deviation Q.D = Q3-Q1 / 2
It is also called as semi-inter quartile range.
### Coefficient of Quartile Deviation
The relative measure corresponding to QD is coefficient of QD and is defined as:
Coefficient of Quartile Deviation = Q3-Q1 / Q3+Q1
### Merits:
· It is not affected by the extreme (highest and lowest) values in the data set.
· It is an appropriate measure of variation for a data set summarized in open-ended class intervals.
· It is a positional measure of variation; therefore it is useful in the cases of erratic or highly skewed distributions.
### Limitations:
· The QD is based on the middle 50 per cent observed values only and is not based on all the observations in the data set, therefore it cannot be considered as a good measure of variation.
· It is not suitable for mathematical treatment.
· It is affected by sampling fluctuations.
· The QD is a positional measure and has no relationship with any average in the data set.
## 3. Mean Deviation
The Mean Deviation (MD) is defined as the arithmetic mean of the absolute deviations of the individual values from a measure of central tendency of the data set. It is also known as the average deviation.
The measure of central tendency is either mean or median. If the measure of central tendency is mean (or median), then we get the mean deviation about the mean (or median).
The coefficient of mean deviation (CMD) is the relative measure of dispersion corresponding to mean deviation and it is given by
### Example 6.4
The following are the weights of 10 children admitted in a hospital on a particular day.
Find the mean deviation about mean, median and their coefficients of mean deviation.
7,4,10,9,15,12,7,9,9,18
## 4. Standard Deviation
Consider the following data sets.
It is obvious that the range for the three sets of data is 8. But a careful look at these sets clearly shows the numbers are different and there is a necessity for a new measure to address the real variations among the numbers in the three data sets. This variation is measured by standard deviation. The idea of standard deviation was given by Karl Pearson in 1893.
### Definition
‘Standard deviation is the positive square root of average of the deviations of all the observation taken from the mean.’ It is denoted by a greek letter v.
### a. Ungrouped data
x1 , x2 , x3 ... xn are the ungrouped data then standard deviation is calculated by
### b. Grouped Data (Discrete)
Where, f = frequency of each class interval
N = total number of observation (or elements) in the population
x = mid – value of each class interval
where A is an assumed A.M.
### c. Grouped Data (continuous)
Where, f = frequency of each class interval
N = total number of observation (or elements) in the population
c = width of class interval
x = mid-value of each class interval where A is an assumed A.M.
Variance : Sum of the squares of the deviation from mean is known as Variance.
The square root of the variance is known as standard deviation.
### Example 6.5
The following data gives the number of books taken in a school library in 7 days find the standard deviation of the book taken
7, 9, 12, 15, 5, 4, 11
### Solution:
Actual mean method
### Merits:
· The value of standard deviation is based on every observation in a set of data.
· It is less affected by fluctuations of sampling.
· It is the only measure of variation capable of algebraic treatment.
### Limitations:
· Compared to other measures of dispersion, calculations of standard deviation are difficult.
· While calculating standard deviation, more weight is given to extreme values and less to those near mean.
· It cannot be calculated in open intervals.
· If two or more data set were given in different units, variation among those data set cannot be compared.
### Example 6.6
Raw Data:
Weights of children admitted in a hospital is given below calculate the standard deviation of weights of children.
13, 15, 12, 19, 10.5, 11.3, 13, 15, 12, 9
### Example 6.7
Find the standard deviation of the first ‘n’ natural numbers.
### Solution:
The first n natural numbers are 1, 2, 3,…, n. The sum and the sum of squares of these n numbers are
### Example 6.8
The wholesale price of a commodity for seven consecutive days in a month is as follows:
Calculate the variance and standard deviation.
### Solution:
The computations for variance and standard deviation is cumbersome when x values are large. So, another method is used, which will reduce the calculation time. Here we take the deviations from an assumed mean or arbitrary value A such that d = xA
In this question, if we take deviation from an assumed A.M. =255. The calculations then for standard deviation will be as shown in below Table;
### Example 6.9
The mean and standard deviation from 18 observations are 14 and 12 respectively. If an additional observation 8 is to be included, find the corrected mean and standard deviation.
### Example 6.10
A study of 100 engineering companies gives the following information
Calculate the standard of the profit earned.
### Solution:
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# Question
Part I
Cakes come in a variety of forms and flavours and are among favourite desserts served
during special occasions such as birthday parties, Hari Raya, weddings and etc. Cakes are
treasured not only because of their wonderful taste but also in the art of cake baking and
cake decorating. Find out how mathematics is used in cake baking and cake decorating
Geometry – To determine suitable dimensions for the cake, to assist in designing and
decorating cakes that comes in many attractive shapes and designs, to estimate volume of
cake to be produced
## Calculus (differentiation) – To determine minimum or maximum amount of ingredients for
cake-baking, to estimate min. or max. amount of cream needed for decorating, to estimate
min. or max. size of cake produced.
## Progressions – To determine total weight/volume of multi-storey cakes with proportional
dimensions, to estimate total ingredients needed for cake-baking, to estimate total amount
of cream for decoration.
Part II
Best Bakery shop received an order from your school to bake a 5 kg of round cake as
shown in Diagram 1 for the Teachers’ Day celebration. (Diagram 11)
1) If a kilogram of cake has a volume of 3800cm3, and the height of the cake is to be
7.0cm, calculate the diameter of the baking tray to be used to fit the 5 kg cake
[Use π = 3.142]
## Volume of 5kg cake = Base area of cake x Height of cake
3800 x 5 = (3.142)(d2)² x 7
190007(3.142) = (d2)²
863.872 = (d2)²
d2 = 29.392
d = 58.784 cm
2) The cake will be baked in an oven with inner dimensions of 80.0 cm in length, 60.0 cmin
width and 45.0 cm in height.
a) If the volume of cake remains the same, explore by using different values of heights,h
cm, and the corresponding values of diameters of the baking tray to be used,d cm.
First, form the formula for d in terms of h by using the above formula for volume of cake, V
= 19000, that is:
19000 = (3.142)(d/2)²h
19000(3.142)h = d²4
24188.415h = d²
d = 155.53h
1.0 155.53
2.0 109.98
3.0 89.80
4.0 77.77
5.0 68.56
6.0 63.49
7.0 58.78
8.0 54.99
9.0 51.84
10.0 49.18
## (b) Based on the values in your table,
(i) state the range of heights that is NOT suitable for the cakes and explain your answers.
h < 7cm is NOT suitable, because the resulting diameter produced is too large to fit
into the oven. Furthermore, the cake would be too short and too wide, making it less
attractive.
(ii) suggest the dimensions that you think most suitable for the cake. Give reasons for your
h = 8cm, d = 54.99cm, because it can fit into the oven, and the size is suitable for
easy handling.
(c)
(i) Form an equation to represent the linear relation between h and d. Hence, plot a
suitable graph based on the equation that you have formed. [You may draw your graph
with the aid of computer software.]
## 19000 = (3.142)( d2)²h
19000/(3.142)h = d²4
24188.415h = d²
d = 155.53√h
d = 155.53h-12
## log d = -12 log h + log 155.53
Log h 0 1 2 3 4
Log d 2.19 1.69 1.19 0.69 0.19
(ii)
(a) If Best Bakery received an order to bake a cake where the height of the cake is 10.5
cm, use your graph to determine the diameter of the round cake pan required.
## h = 10.5cm, log h = 1.021, log d = 1.680, d = 47.86cm
(b) If Best Bakery used a 42 cm diameter round cake tray, use your graph to estimate the
height of the cake obtained.
## d = 42cm, log d = 1.623, log h = 1.140, h = 13.80cm
3) Best Bakery has been requested to decorate the cake with fresh cream. The thickness
of the cream is normally set to a uniform layer of about 1cm
(a) Estimate the amount of fresh cream required to decorate the cake using the dimensions
that you have suggested in 2(b)(ii).
h = 8cm, d = 54.99cm
Amount of fresh cream = VOLUME of fresh cream needed (area x height)
Amount of fresh cream = Vol. of cream at the top surface + Vol. of cream at the side
surface
## Vol. of cream at the top surface
= Area of top surface x Height of cream
= (3.142)(54.992)² x 1
= 2375 cm³
## Vol. of cream at the side surface
= Area of side surface x Height of cream
= (Circumference of cake x Height of cake) x Height of cream
= 2(3.142)(54.99/2)(8) x 1
= 1382.23 cm³
## Therefore, amount of fresh cream = 2375 + 1382.23 = 3757.23 cm³
(b) Suggest three other shapes for cake, that will have the same height and volume as
those suggested in 2(b)(ii). Estimate the amount of fresh cream to be used on each of the
cakes.
## 19000 = base area x height
base area = 190002
length x width = 2375
By trial and improvement, 2375 = 50 x 47.5 (length = 50, width = 47.5, height = 8)
## Therefore, volume of cream
= 2(Area of left/right side surface)(Height of cream) + 2(Area of front/back side surface)
(Height of cream) + Vol. of top surface
= 2(8 x 50)(1) + 2(8 x 47.5)(1) + 2375 = 3935 cm³
2 – Triangle-shaped base
## 19000 = base area x height
base area = 2375
12 x length x width = 2375
length x width = 4750
By trial and improvement, 4750 = 95 x 50 (length = 95, width = 50)
Slant length of triangle = √(95² + 25²)= 98.23
Therefore, amount of cream
= Area of rectangular front side surface(Height of cream) + 2(Area of slant rectangular
left/right side surface)(Height of cream) + Vol. of top surface
= (50 x 8)(1) + 2(98.23 x 8)(1) + 2375 = 4346.68 cm³
3 – Pentagon-shaped base
## 19000 = base area x height
base area = 2375 = area of 5 similar isosceles triangles in a pentagon
therefore:
2375 = 5(length x width)
475 = length x width
By trial and improvement, 475 = 25 x 19 (length = 25, width = 19)
## Therefore, amount of cream
= 5(area of one rectangular side surface)(height of cream) + vol. of top surface
= 5(8 x 19) + 2375 = 3135 cm³
(c) Based on the values that you have found which shape requires the least amount of
fresh cream to be used?
## Pentagon-shaped cake, since it requires only 3135 cm³ of cream to be used.
Part III
Find the dimension of a 5 kg round cake that requires the minimum amount of fresh cream
to decorate. Use at least two different methods including Calculus. State whether you
would choose to bake a cake of such dimensions. Give reasons for your answers.
Method 1: Differentiation
Use two equations for this method: the formula for volume of cake (as in Q2/a), and the
formula for amount (volume) of cream to be used for the round cake (as in Q3/a).
19000 = (3.142)r²h → (1)
V = (3.142)r² + 2(3.142)rh → (2)
From (1): h = 19000(3.142)r² → (3)
Sub. (3) into (2):
V = (3.142)r² + 2(3.142)r(19000(3.142)r²)
V = (3.142)r² + (38000r)
V = (3.142)r² + 38000r-1
## (dVdr) = 2(3.142)r – (38000r²)
0 = 2(3.142)r – (38000r²) -->> minimum value, therefore dVdr = 0
38000r² = 2(3.142)r
380002(3.142)= r³
6047.104 = r³
r = 18.22
## Sub. r = 18.22 into (3):
h = 19000(3.142)(18.22)²
h = 18.22
therefore, h = 18.22cm, d = 2r = 2(18.22) = 36.44cm
Use the two same equations as in Method 1, but only the formula for amount of cream is
the main equation used as the quadratic function.
Let f(r) = volume of cream, r = radius of round cake:
19000 = (3.142)r²h → (1)
f(r) = (3.142)r² + 2(3.142)hr → (2)
From (2):
f(r) = (3.142)(r² + 2hr) -->> factorize (3.142)
= (3.142)[ (r + 2h2)² – (2h2)² ] -->> completing square, with a = (3.142), b = 2h and c = 0
= (3.142)[ (r + h)² – h² ]
= (3.142)(r + h)² – (3.142)h²
(a = (3.142) (positive indicates min. value), min. value = f(r) = –(3.142)h², corresponding
value of x = r = --h)
## Sub. r = --h into (1):
19000 = (3.142)(--h)²h
h³ = 6047.104
h = 18.22
## Sub. h = 18.22 into (1):
19000 = (3.142)r²(18.22)
r² = 331.894
r = 18.22
therefore, h = 18.22 cm, d = 2r = 2(18.22) = 36.44 cm
I would choose not to bake a cake with such dimensions because its dimensions are
not suitable (the height is too high) and therefore less attractive. Furthermore, such
cakes are difficult to handle easily.
FURTHER EXPLORATION
Best Bakery received an order to bake a multi-storey cake for Merdeka Day celebration, as
shown in Diagram 2.
The height of each cake is 6.0 cm and the radius of the largest cake is 31.0 cm. The radius
of the second cake is 10% less than the radius of the first cake, the radius of the third cake
is10% less than the radius of the second cake and so on.(a)
Find the volume of the first, the second, the third and the fourth cakes. By comparing all
these values, determine whether the volumes of the cakes form a number pattern? Explain
and elaborate on the number patterns.
## radius of largest cake = 31cm
radius of 2nd cake = 10% smaller than 1st cake
radius of 3rd cake = 10% smaller than 2nd cake
a = 31, r = 910
V = (3.142)r²h
## Radius of 1st cake = 31, volume of 1st cake = (3.142)(31)²(6) = 18116.772
Radius of 2nd cake = 27.9, vol. of 2nd cake = 14674.585
Radius of 3rd cake = 25.11, vol. of 3rd cake = 11886.414
Radius of 4th cake = 22.599, vol. of 4th cake = 9627.995
18116.772, 14674.585, 11886.414, 9627.995, …
## a = 18116.772, ratio, r = T2/T1 = T3 /T2 = … = 0.81
(b) If the total mass of all the cakes should not exceed 15 kg, calculate the maximum
number of cakes that the bakery needs to bake. Verify your answer using other methods.
Sn = (a(1 - rn)) (1 - r)
## 57000 = (18116.772(1 – (0.81)n))(1 - 0.81)
1 – 0.81n = 0.59779
0.40221 = 0.81n
og0.81 0.40221 = n
## n = log 0.40221 log 0.81
n = 4.322
therefore, n ≈ 4
REFLECTION
In the making of this project, I have spent countless hours doing this project.I
realized that this subject is a compulsory to me. Without it, I can’t fulfill my big
dreams and wishes….
## I sacrificed my precious time to have fun…
From..
Monday,Tuesday,Wednesday,Thursday,Friday
1 28ve 980 |
AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 13 Probability Ex 13.2 Textbook Questions and Answers.
## AP State Syllabus SSC 10th Class Maths Solutions 13th Lesson Probability Exercise 13.2
### 10th Class Maths 13th Lesson Probability Ex 13.2 Textbook Questions and Answers
Question 1.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red ? (ii) not red?
i) Total number of balls in the bag = 3 red + 5 black = 8 balls.
Number of total outcomes when a ball is drawn at random = 3 + 5 = 8
Now, number of favourable outcomes of red ball = 3.
∴ Probability of getting a red ball = $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$ = $$\frac{3}{8}$$
ii) If P( E) is the probability of drawing no red ball, then P(E) + P($$\overline{\mathrm{E}}$$) = 1
P($$\overline{\mathrm{E}}$$) = 1 – P(E)= 1 – $$\frac{3}{8}$$ = $$\frac{5}{8}$$
Question 2.
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? (ii) white? (iii) not green?
Total number of marbles in the box = 5 red + 8 white + 4 green = 5 + 8 + 4= 17
Number of total outcomes in drawing a marble at random from the box =17.
i) Number of red marbles = 5
Number of favourable outcomes in drawing a red ball = 5
∴ Probability of getting a red ball P(R) = $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
P(R) = $$\frac{5}{17}$$
ii) Number of white marbles = 8
Number of favourable outcomes in drawing a white marble = 8
∴ Probability of getting a white marble
P(W) = $$\frac{8}{17}$$
iii) Number of ‘non-green’ marbles = 5 red + 8 white = 5 + 8 = 13
Number of outcomes favourable to drawing a non-green marble =13.
∴ Probability of getting a non- green marble
P(non – green) = $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
P(non – green) = $$\frac{13}{17}$$
Probability of getting a green ball = $$\frac{\text { No. of green balls }}{\text { Total no. of balls }}$$ = $$\frac{4}{17}$$
Now P(G) = $$\frac{4}{17}$$
and P(G) + P($$\overline{\mathrm{ G}}$$) = 1
∴ P($$\overline{\mathrm{G}}$$) = 1 – P(G)
= 1 – $$\frac{4}{17}$$
= $$\frac{13}{17}$$
Question 3.
A Kiddy bank contains hundred 50p coins, fifty Rs. 1 coins, twenty Rs. 2 coins and ten Rs. 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin? (ii) will not be a Rs. 5 coin?
i) Number of 50 p coins = 100
Number of Rs. 1 coins = 50
Number of Rs. 2 coins = 20
Number of Rs. 5 coins = 10
Total number of coins = 180
Number of total outcomes for a coin to fall down = 180.
Number of outcomes favourable to 50 p coins to fall down = 100.
∴ Probability of a 50 p coin to fall down = $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{100}{180}$$ = $$\frac{5}{9}$$
ii) Let P(E) be the probability for a Rs. 5 coin to fall down.
Number of outcomes favourable to Rs. 5 coin = 10.
∴ Probability for a Rs. 5 coin to fall down = $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{10}{180}$$ = $$\frac{1}{18}$$
Then P($$\overline{\mathrm{E}}$$) is the probability of a coin which fall down is not a Rs. 5 coin.
Again P(E) + P($$\overline{\mathrm{E}}$$) = 1
∴ P($$\overline{\mathrm{E}}$$)= l-P(E)
= 1 – $$\frac{1}{18}$$
= $$\frac{17}{18}$$.
Question 4.
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (See figure). What is the probability that the fish taken out is a male fish?
Number of male fish = 5
Number of female fish = 8
Total number of fish = 5 m + 8 f
= 13 fishes.
∴ Number of total outcomes in taking a fish at random from the aquarium =13.
Number of male fish = 5
∴ Number of outcomes favourable to male fish = 5.
∴ The probability of taking a male fish = $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{5}{13}$$
= 0.38
Question 5.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (See figure), and these are equally likely outcomes. What is the probability that it will point at
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?
Number of total outcomes are (1,2,……….., 8) = 8
i) Number of outcomes favourable to 8 = 1.
∴ P(8) = $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{1}{8}$$
ii) Number of ‘odd numbers’ on the spinning wheel = (1, 3, 5, 7) = 4
∴ Number of outcomes favourable to an odd number.
∴ Probability of getting an odd number = $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{4}{8}$$ = $$\frac{1}{2}$$
iii) Number greater than 2 are (3, 4, 5, 6, 7, 8)
Number of outcomes favourable to ‘greater than 2’ are = 6.
Probability of pointing a number greater than 2
P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{6}{8}$$ = $$\frac{3}{4}$$
iv) Number less than 9 are: (1,2, 3, 4, 5, 6, 7, 8 …… 8)
∴ Number of outcomes favourable to pointing a number less than 9 = 8.
∴ Probability of a number less than 9
P(E) = $$\frac{\text { No. of outcomes favourable to less than } 9}{\text { No.of total outcomes }}$$
= $$\frac{8}{8}$$ = 1
Note : This is a sure event and hence probability is 1.
Question 6.
One card is drawn from, a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(vi) the queen of diamonds.
Total number of cards = 52.
∴ Number of all possible outcomes in drawing a card at random = 52.
i) Number of outcomes favourable to the king of red colour = 2(♥ K, ♦ K)
∴ Probability of getting the king of red colour
P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{2}{52}$$ = $$\frac{1}{26}$$
ii) Number of face cards in a deck of cards = 4 × 3 = 12 (K, Q, J)
Number of outcomes favourable to select a face card = 12.
∴ Probability of getting a face card
= $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{12}{52}$$ = $$\frac{3}{13}$$
iii) Number of red face cards = 2 × 3 = 6.
∴ Number of outcomes favourable to select a red face card = 6.
∴ Probability of getting a red face
= $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{6}{52}$$ = $$\frac{3}{26}$$
iv) Number of outcomes favourable to the jack of hearts = 1.
∴ Probability of getting jack of hearts
= $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{1}{52}$$
v) Number of spade cards = 13
∴ Number of outcomes favourable to ‘a spade card’ = 13.
∴ Probability of drawing a spade
= $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{13}{52}$$ = $$\frac{1}{4}$$
vi) Number of outcomes favourable to the queen of diamonds = 1.
∴ Probability of drawing the queen of diamonds
= $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{1}{52}$$
Question 7.
Five cards-the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
i) What is the probability that the card is the queen?
ii) If the queen is drawn and put aside, what is the probability that the second card picked is (a) an ace? (b) a queen?
Total number of cards = 5.
∴ Number of total outcomes in picking up a card at random = 5.
i) Number of outcomes favourable to queen = 1.
∴ Probability of getting the queen
= $$\frac{\text { No.of outcomes favourable to the ‘Q’ }}{\text { No.of total outcomes }}$$
= $$\frac{1}{5}$$
ii) When queen is drawn and put aside, remaining cards are four.
∴ Number of total outcomes in drawing a card at random = 4.
a) Number of favourable outcomes to ace 1
Probability of getting an ace
= $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{1}{4}$$
b) Number of favourable outcomes to ‘Q’ = 0 (as it was already drawn and put aside)
∴ Probability that the card is Q = $$\frac{0}{4}$$ = 0
After putting queen aside, selecting the queen from the rest is an impossible event and hence the probability is zero.
Question 8.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Number of good pens = 132
Number of defective pens = 12
Total number of pens = 132 + 12 = 144
∴ Total number of outcomes in taking a pen at random = 144.
No. of favourable outcomes in taking a good pen = 132.
∴ Probability of taking a good pen
= $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{132}{144}$$ = $$\frac{11}{12}$$
Question 9.
A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? Suppose the bulb drawn in previous case is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Given :
4 out of 20 bulbs are defective
(i.e.) Number of defective bulbs = 4
Number of non-defective bulbs = 20 – 4 = 16
If a bulb is drawn at random, the total outcomes are = 20
Number of outcomes favourable to ‘defective bulb’ = 4
∴ Probability of getting a defective bulb
= $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{4}{20}$$ = $$\frac{1}{5}$$
Suppose a non-defective bulb is drawn and not replaced, then the bulbs remaining are = 19
∴ Total outcomes in drawing a bulb from the rest = 19
Number of favourable outcomes in drawing non-defective bulb from the rest = 16 – 1 = 15
∴ Probability of getting a non-defective bulb in the second draw
= $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{15}{19}$$
Question 10.
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two digit number
(ii) a perfect square number
(iii) a number divisible by 5.
Total number of discs in the box = 90
∴ Number of total outcomes in drawing a disc at random from the box = 90.
i) Number of 2-digit numbers in the box (10, 11,….., 90) = 81
i.e., Number of favourable outcomes in drawing a 2 – digit numbers = 81
∴ Probability of selecting a disc bearing a 2 – digit number
= $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{81}{90}$$ = $$\frac{9}{10}$$ = 0.9
ii) Number of perfect squares in the box (12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25, 62 = 36, 72 = 49, 82 = 64 and 92 = 81) = 9
i.e., Number of favourable out-comes in drawning a disc bearing a perfect square = 9
∴ Probability of drawning a disc with a perfect square
= $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{9}{90}$$ = $$\frac{1}{10}$$
iii) Number of multiples of 5 from 1 to 90 are (5, 10, 15, ……….., 90) = 18
i.e., Number of favourable outcomes in drawing a disc with a multiple of 5 = 18
∴ Probability of drawing a disc bearing a number divisible by 5
= $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{18}{90}$$ = $$\frac{1}{5}$$
Question 11.
Suppose you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter 1 m?
3 m.
Length of the given rectangle = 3 m.
and its breadth = 2 m.
Area of the rectangle
= length × breadth = 3 × 2 = 6 m2
∴ Total area of the region for landing = 6 m2.
Diameter of the given circle = 1 m.
Area of the circle = $$\frac{\pi \mathrm{d}^{2}}{4}$$
= $$\frac{22}{7} \times \frac{1 \times 1}{4}\left[\text { or } \pi r^{2}=\frac{22}{7} \times \frac{1}{2} \times \frac{1}{2}\right]$$
= $$\frac{22}{28}$$
∴ Probability of the coin to land on the circle
= $$\frac{\frac{22}{28}}{6}$$
= $$\frac{22}{28×6}$$
= $$\frac{11}{28×3}$$
= $$\frac{11}{84}$$
Question 12.
A lot consists of 144 ball pens of which 20 are defective and the others are good. The shopkeeper draws one pen at random and gives it to Sudha. What is the probability that (i) She will buy it? (ii) She will not buy it?
Given : 20 out of 144 are defective i.e., no. of defective ball pens = 20
no. of good ball pens = 144 – 20 = 124
∴ Total outcomes in drawing a ball pen at random = 144.
i) Sudha buys it if it is not defective / a good one.
No. of outcomes favourable to a good pen = 124.
= $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{124}{144}$$ = $$\frac{31}{36}$$
ii) Sudha will not buy it-if it is a defective pen
No. of outcomes favourable to a defective pen = 20
∴ Probability of not buying it
= $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{20}{144}$$ = $$\frac{5}{36}$$
= 1 – $$\frac{31}{36}$$ = $$\frac{5}{36}$$
Question 13.
Two dice are rolled simultaneously and counts are added
(i) Complete the table given below:
(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability $$\frac{1}{11}$$. Do you agree with this argument? Justify your answer.
When two dice are rolled, total number of outcomes = 36 (see the given table).
(i)
(ii) The above (given) argument is wrong [from the above table].
The sum 2, 3, 4, ………… and 12 have different no. of favourable outcomes, moreover total number of outcomes are 36.
Question 14.
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
When a coin is tossed for n – times, the total number of outcomes = 2n.
∴ If a coin is tossed for 3 – times, then the total number of outcomes = 23 = 8
Note the following :
TTT
TTH
THT
HTT
HHT
HTH
THH
HHH
Of the above, no. of outcomes with different results = 6.
Probability of losing the game
= $$\frac{\text { No. of favourable outcomes to lose }}{\text { No. of total outcomes }}$$
= $$\frac{6}{8}$$ = $$\frac{3}{4}$$
Question 15.
A dice is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up atleast once? [Hint : Throwing a dice twice and throwing two dice simultaneously are treated as the same experiment].
∴ P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{25}{36}$$
∴ P(E) = $$\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}$$
= $$\frac{11}{36}$$ |
# The Ultimate Math Test: Add Fractions With Unlike Denominators
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Adding and subtracting fractions is said to be easy when it comes to fractions that have the same denominator. This quiz covers the basics of fractions with unlike denominators. Take up the quiz to see if you understood the topic and see what score you get. Good luck!
• 1.
### 1/2 + 2/5 =
• A.
3/7
• B.
3/10
• C.
9/10
• D.
5/10
C. 9/10
Explanation
To add fractions, we need to have a common denominator. In this case, the common denominator is 10. So, we need to convert 1/2 to an equivalent fraction with a denominator of 10, which is 5/10. Now, we can add the fractions: 5/10 + 2/10 = 7/10. Therefore, the correct answer is 7/10.
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• 2.
### 1/4 + 3/8 =
• A.
4/8
• B.
5/8
• C.
1/6
• D.
4/12
B. 5/8
Explanation
When adding fractions, the denominators must be the same. In this case, the denominators are 4 and 8. To make the denominators the same, we need to find a common multiple of 4 and 8, which is 8. We then convert 1/4 to an equivalent fraction with a denominator of 8 by multiplying the numerator and denominator by 2. This gives us 2/8. Now we can add 2/8 and 3/8 together, which equals 5/8. Therefore, the correct answer is 5/8.
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• 3.
### 5/6 + 2/12 =
• A.
1
• B.
7/18
• C.
10/12
• D.
8/12
A. 1
Explanation
The given expression involves adding two fractions, 5/6 and 2/12. To add fractions, we need to have a common denominator. In this case, the common denominator is 12. We can convert 5/6 to 10/12 by multiplying the numerator and denominator by 2. Now, we can add 10/12 and 2/12 to get 12/12, which simplifies to 1. Therefore, the correct answer is 1.
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• 4.
### 4/5 + 4/15 =
• A.
8/20
• B.
18/15
• C.
1 1/15
• D.
1 2/15
C. 1 1/15
Explanation
The given expression is a sum of two fractions. To add fractions, we need to have a common denominator. In this case, the common denominator is 15. When we convert 4/5 to have a denominator of 15, we get 12/15. Adding this to 4/15 gives us a total of 16/15. This can be simplified to the mixed number 1 1/15.
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• 5.
### 3/7 + 2/3 =
• A.
1 2/21
• B.
5/10
• C.
1/2
• D.
1 5/21
A. 1 2/21
Explanation
The given expression is a sum of two fractions. To add fractions, we need to find a common denominator. In this case, the common denominator is 21. We then convert the fractions to have the same denominator and add the numerators. The result is 11/21. Since the numerator is greater than the denominator, we have a mixed number. Dividing the numerator by the denominator, we get 11 ÷ 21 = 0 with a remainder of 11. Therefore, the answer is 1 2/21.
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• 6.
### 1 2/5 + 2 4/7 =
• A.
3 6/12
• B.
3 1/2
• C.
3 32/35
• D.
3 34/35
D. 3 34/35
Explanation
The given question asks for the sum of two mixed numbers. To add mixed numbers, we first add the whole numbers together, which gives us 3. Then, we add the fractions together. The fractions have different denominators, so we need to find a common denominator. The least common multiple of 5 and 7 is 35. We convert the fractions to have a denominator of 35 and add them together. The sum of 2/5 and 4/7 is 34/35. Therefore, the correct answer is 3 34/35.
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• 7.
### 1 1/2 + 2 1/3 =
• A.
2 5/6
• B.
3 5/6
• C.
3 2/5
• D.
3 2/6
B. 3 5/6
Explanation
To add mixed numbers, we first add the whole numbers together and then add the fractions together. In this case, 1 + 2 = 3 for the whole numbers. Then, we add the fractions 1/2 and 1/3 together. To do this, we need to find a common denominator, which is 6. Converting 1/2 to have a denominator of 6 gives us 3/6, and converting 1/3 to have a denominator of 6 gives us 2/6. Adding 3/6 and 2/6 together gives us 5/6. Therefore, the answer is 3 5/6.
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• 8.
### 5/6 + 4/12 =
• A.
7/10
• B.
1 1/6
• C.
1/8
• D.
1 1/12
B. 1 1/6
Explanation
The given expression involves adding two fractions with different denominators. To add fractions, we need to have a common denominator. In this case, the common denominator is 12. To convert 5/6 into a fraction with a denominator of 12, we multiply the numerator and denominator by 2, resulting in 10/12. Now we can add 10/12 and 4/12, which equals 14/12. Simplifying this fraction gives us 1 1/6, which is the final answer.
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• 9.
### 5/12 + 2/3 =
• A.
1 1/12
• B.
7/10
• C.
1 1/6
• D.
1 1/20
A. 1 1/12
Explanation
The given expression is a sum of two fractions. To add fractions, we need to find a common denominator. The least common multiple of 12 and 3 is 12. So, we can rewrite the fractions with a common denominator of 12. 5/12 becomes 5/12 and 2/3 becomes 8/12. Now, we can add the fractions by adding their numerators and keeping the denominator the same. 5/12 + 8/12 = 13/12. Since the numerator is greater than the denominator, we have a mixed number. Dividing 13 by 12 gives us a quotient of 1 and a remainder of 1. So, the final answer is 1 1/12.
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• 10.
### 9/20 + 3/5 =
• A.
2 1/20
• B.
2 3/10
• C.
1 3/10
• D.
1 1/20
D. 1 1/20
Explanation
The given expression involves adding two fractions: 9/20 and 3/5. To add these fractions, we need to find a common denominator. The least common multiple of 20 and 5 is 20. We can convert 9/20 to an equivalent fraction with a denominator of 20 by multiplying the numerator and denominator by 1, resulting in 9/20. Similarly, we can convert 3/5 to an equivalent fraction with a denominator of 20 by multiplying the numerator and denominator by 4, resulting in 12/20. Adding the two fractions, we get 9/20 + 12/20 = 21/20. This can be simplified to 1 1/20, which is the correct answer.
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Related Topics |
## How to Find X?
### GivenProportion
If 1:x = x:64 is given proportion, Algebra is always interesting to solve. In this given sum, we need to find the value x. Let's see how to solve this step by step. Refer above image for a detailed solution.
### Rewrite this Proportion to Simplify
we can write it as
1/x = x/64
### Cross multiply the numerators and denominators.
Move the x to the right-hand side numerator from the left-hand side denominator and multiply with the existing right-hand side numerator x.
From the right-hand side denominator, move 64 to the left-hand side numerator and multiply with the existing left-hand side numerator 1.
Now it becomes, 1 X 64 = x X x
Where 'X' represents multiplication
### Simplification
Simplify further 64 = x^2
If you multiply x with another x, it will become x power 2, which we can write as x^2.
### Solution
Lets write as x^2 = 64
we can write 64 as 8^2 because eight power two will be 64
x^2 = 8^2
Now move the square from the left-hand side to the right-hand side, which will become the square root
so x = square root of ( 8^2 )
square root and square will get cancelled against each other. Finally, the value of x becomes 8. |
# 009C Sample Final 2, Problem 10
Find the length of the curve given by
${\displaystyle x=t^{2}}$
${\displaystyle y=t^{3}}$
${\displaystyle 0\leq t\leq 2}$
Foundations:
The formula for the arc length ${\displaystyle L}$ of a parametric curve with ${\displaystyle \alpha \leq t\leq \beta }$ is
${\displaystyle L=\int _{\alpha }^{\beta }{\sqrt {{\bigg (}{\frac {dx}{dt}}{\bigg )}^{2}+{\bigg (}{\frac {dy}{dt}}{\bigg )}^{2}}}~dt.}$
Solution:
Step 1:
First, we need to calculate ${\displaystyle {\frac {dx}{dt}}}$ and ${\displaystyle {\frac {dy}{dt}}.}$
Since ${\displaystyle x=t^{2},~{\frac {dx}{dt}}=2t.}$
Since ${\displaystyle y=t^{3},~{\frac {dy}{dt}}=3t^{2}.}$
Using the formula in Foundations, we have
${\displaystyle L=\int _{1}^{2}{\sqrt {(2t)^{2}+(3t^{2})^{2}}}~dt.}$
Step 2:
Now, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {L}&=&\displaystyle {\int _{1}^{2}{\sqrt {4t^{2}+9t^{4}}}~dt}\\&&\\&=&\displaystyle {\int _{1}^{2}{\sqrt {t^{2}(4+9t^{2})}}~dt}\\&&\\&=&\displaystyle {\int _{1}^{2}t{\sqrt {4+9t^{2}}}~dt.}\\\end{array}}}$
Step 3:
Now, we use ${\displaystyle u}$-substitution.
Let ${\displaystyle u=4+9t^{2}.}$
Then, ${\displaystyle du=18tdt}$ and ${\displaystyle {\frac {du}{18}}=tdt.}$
Also, since this is a definite integral, we need to change the bounds of integration.
We have
${\displaystyle u_{1}=4+9(1)^{2}=13}$ and ${\displaystyle u_{2}=4+9(2)^{2}=40.}$
Hence,
${\displaystyle {\begin{array}{rcl}\displaystyle {L}&=&\displaystyle {{\frac {1}{18}}\int _{13}^{40}{\sqrt {u}}~du}\\&&\\&=&\displaystyle {{\frac {1}{18}}\cdot {\frac {2}{3}}u^{\frac {3}{2}}{\bigg |}_{13}^{40}}\\&&\\&=&\displaystyle {{\frac {1}{27}}(40)^{\frac {3}{2}}-{\frac {1}{27}}(13)^{\frac {3}{2}}.}\\\end{array}}}$
${\displaystyle {\frac {1}{27}}(40)^{\frac {3}{2}}-{\frac {1}{27}}(13)^{\frac {3}{2}}}$ |
Albert Teen
YOU ARE LEARNING:
Cubic Graphs
# Cubic Graphs
### Cubic Graphs
Cubic graphs are curved graphs, where the highest power of x is 3. A table of values can be used to work out each point on the graph.
Cubic graphs are functions where the highest power of $x$ is $x^3$. They can have a variety of shapes depending on the specific function being displayed.
Let's draw the graph of $y = x^3 -2x^2 -5x$
1
For each $x$ coordinate, we can find the corresponding $y$ coordinate by ....
.... substituting the $x$ value into the equation.
2
For example, when $x=1$, $y=-6$
$y=1^3-(2 \times 1^2)-(5 \times 1)$ so $y=-6$. We can plot this point on the graph at $(1,-6)$
3
What is $y$ when $x=-2$?
4
When $x=-2$, $y=-6$
This is because $y=-2^3-2(-2^2)-5(-2)$. Remember that a negative multiplied by a negative equals a positive. We can plot this point at $(-2,-6)$
5
What is $y$ when $x=2$?
6
When $x=2$, $y=-10$ $y=2^3-(2 \times 2^2)-(5 \times 2)=-10$.
We can plot this point at $(2,-10)$. Let's try filling out a few more values in a table.
The graph of $y = x^3 -2x^2 -5x$
1
We can use a table of values to help find points which will allow us to draw a cubic graph.
In this table we can see the points we have already calculated for when $x= -2, 1, 2$ .
2
Let's fill this table out so we have more points to help us find the curve. What is the y-coordinate for when $x=-1$ ?
3
What is the y-coordinate for the point when $x=3$ ?
4
What will be the point when $x=4$ ? Give your answer in the form $(x, y)$ .
5
Nice!
Now we have enough points to draw a line.
the graph of $y = x^3 -2x^2 -5x$
1
Here we can see all the points we calculated plotted on the graph.
Now all we need to do is join these points up with a smooth curve.
2
Awesome!
This is the line of $y=x^3-2x^2-5x$.
Let's draw the curve of $y = x^3 + x$
1
Remember, we can find values of $y$ by substituting values of $x$ into the equation.
Let's give it a go
2
What is $y$ when $x=1$?
3
What is $y$ when $x=-1$ ?
4
Does the curve pass through the origin, $(0,0)$ ?
5
What is $y$ when $x=-2$?
6
Awesome! We have enough points
Now we can draw a line to smooth join the points together.
7
This is the line of $y=x^3+x$
Nice! |
Understanding Quadrants for Math and in Graphing
POSTED ON AUGUST 09, 2023
Math quadrants are helpful regions in a coordinate system that will come in handy when identifying behaviors in these four sectors. By using quadrants, we have a systematic way of identifying and categorizing points that belong in a certain quadrant. In this article, let’s dive into the basics of a quadrant, understand its function, and try out different applications of quadrants in math!
What Is a Quadrant and How Do We Identify the Quadrants On a Graph?
The quadrant represents one of the four regions formed when two axes in a Cartesian plane intersect. When two axes (x-axis and y-axis) intersect, they form four regions. Each region is called a quadrant.!
Take a look at the Cartesian plane shown above. It is formed when the horizontal line, the x-axis, and the vertical line (called the y-axis) intersect. Now, the four regions formed are what we call the quadrants. For convention’s sake, these regions are labeled in a particular order to make it easier for everyone to identify the sector being referred to easily.
Identifying the Quadrants in a Coordinate Plane
The coordinate plane is divided into four quadrants with the following labels: Quadrant I, Quadrant II, Quadrant III, and Quadrant IV. Starting from the origin, we begin at the upper-right corner and then move counterclockwise until we reach Quadrant IV.
Each set of points belonging to each quadrant exhibits interesting properties. But first, observe on your own! Notice any patterns?
• Quadrant I: It’s the first quadrant and is at the upper-right corner of the region. All the points for x and y are positive in this region.
• Quadrant II: The second quadrant is at the upper-left corner. This time, the values of x are negative while the values of y are all positive.
• Quadrant III: The third quadrant is at the lower-left corner and right below Quadrant II. By observing the values in this region, you can see that both x and y are negative.
• Quadrant IV: You’ll find the fourth quadrant in the lower right-hand corner below Quadrant I. The values of x are positive while that of y are negative.
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Test your understanding so far by identifying the quadrants that the following points belong to:
a. (-3, -2)
b. (2, 2)
c. (2, -3)
For the first point, since a = (-3,-2) lies on the lower-right corner of the coordinate plane, it lies on Quadrant III. Apply a similar thought process to find the quadrants that the remaining points below to. Here’s a table to guide you through:
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Sign Conventions of Quadrants on a Graph
In the same way that we have set rules on naming quadrants on a coordinate system, it’s helpful to know the sign conventions for each quadrant. You can even figure out these rules by observing the Cartesian plane:
• As a point moves from left to right along the x-axis (the horizontal axis), the values of x increase from negative numbers to positive ones.
• Similarly, as the point moves upward along the y-axis (the vertical axis), the values of y increase as well.
This pattern continues and applies even as we divide the coordinate system into four quadrants shown earlier in this article. That’s why we have the following sign conventions for the graph’s quadrants:
This means that by simply inspecting the coordinates’ signs, we can immediately determine whether a given coordinate is located in Quadrant I, II, III, or IV!
Why don’t we include this in our previous quadrant chart guide?
Remember this quadrant graph whenever you need help identifying a point’s quadrant position. With this chart, you can easily locate the quadrant a point belongs given its actual position on the graph or its coordinate.
Plotting Points and Identifying the Quadrants They Belong To
When given a point, (a, b), it tells you that its x-coordinate is a and its y-coordinate is b.The values of a and b will tell you how far away they are from the origin, (0, 0), or the intersection of the x and y axes. Here’s how to plot the point and identify the quadrant it belongs to:
1. Starting from the origin, move a units to the right if a is positive or a units to the left if a is negative.
2. Use the value of b then move b units upward if b is positive or b units downward if b is negative.
3. Mark the final location and plot the point, (a, b), there.
4. Now, determine which secti on of the coordinate system they’re located in to identify the quadrant that it belongs to.
Of course, the best way to understand this concept is by practice, so plot the following points and identify which quadrant these points lie.
1. (-3, 3)
First, identify the x and y-coordinates to know how you’ll move starting from the origin.
• Move 3 units to the left starting from the origin.
• Move 3 units upward from there.
Once you have the position, mark it as the point, (-3, 3) as shown below.
After plotting the point, identify the section that it belongs to. The point (-3, 3) is located in the upper-left corner of the coordinate system. Hence, (-3, 3) lies on Quadrant III.
1. (2, 3)
Use a similar approach to plot (2, -2) and find the quadrant it lies on.
• Move 2 units to the right of the origin.
• From there, move 2 units downward.
• Locate the quadrant the point lies on.
Since point (2, -2) lies in the lower-right corner, we can confirm that it lies in the fourth quadrant or Quadrant IV.
Wrapping It Up With Quadrants for Math
Quadrants help you have a systematic way of categorizing points found in the xy-plane. By mastering this topic, you’ll have a better understanding of how we plot points and other figures on the coordinate system. Don’t forget to practice plotting points, knowing the sign conventions, and remembering how to identify a point’s quadrant. You got this! |
GMAT Math : Calculating the area of a square
Example Questions
← Previous 1
Example Question #1 : Calculating The Area Of A Square
Write, in terms of , the perimeter of a square whose area is
Explanation:
To find the perimeter of a square given its area, take the square root of the area to find its sidelength; then, multiply that sidelength by 4.
is a perfect square trinomial, since
so its square root is , the sidelength.
Multiply this by 4 to get the perimeter:
Example Question #2 : Calculating The Area Of A Square
If the diagonal of a square room is . What is the area of the room?
Explanation:
Cutting the triangle in half yields a right triangle with the diagonal becoming the hypotenuse and the other two legs being the sides of the square. Using the Pythagorean Theorem, we can solve for the other legs of the triangle.
Since both sides of the square are equal to eachother, , therefore:
To find the area of the square:
with leg being one of the sides
Example Question #3 : Calculating The Area Of A Square
A square plot of land has perimeter 1,200 feet. Give its area in square yards.
Explanation:
The length of one side of the square is feet, or yards. Square this to get the area in square yards:
square yards.
Example Question #4 : Calculating The Area Of A Square
A square, a regular pentagon, and a regular hexagon have the same sidelength. The sum of their perimeters is one mile. What is the area of the square?
Explanation:
The square, the pentagon, and the hexagon have a total of 15 sides, all of which are of equal length; the sum of the lengths is one mile, or 5,280 feet, so the length of one side of any of these polygons is
feet.
The square has area equal to the square of this sidelength:
Example Question #5 : Calculating The Area Of A Square
A square and a regular pentagon have the same perimeter. The length of one side of the pentagon is 60 centimeters. What is the area of the square?
Explanation:
The regular perimeter has sidelength 60 centimeters and therefore perimeter centimeters. The square has as its sidelength centimeters and area square centimeters.
Example Question #6 : Calculating The Area Of A Square
Six squares have sidelengths 8 inches, 1 foot, 15 inches, 20 inches, 2 feet, and 25 inches. What is the sum of their areas?
Explanation:
The areas of the squares are the squares of the sidelengths, so add the squares of the sidelengths. Since 1 foot is equal to 12 inches and 2 feet are equal to 24 inches, the sum of the areas is:
square inches
Example Question #7 : Calculating The Area Of A Square
What polynomial represents the area of Square if ?
Explanation:
As a square, is also a rhombus. The area of a rhombus is half the product of the lengths of its diagonals, one of which is . Since the diagonals are congruent, this is equal to half the square of :
Example Question #8 : Calculating The Area Of A Square
Given square FGHI, answer the following
If square represents the surface of an ancient arena discovered by archaeologists, what is the area of the arena?
Explanation:
This problem requires us to find the area of a square. Don't let the story behind it distract you, it is simply an area problem. Use the following equation to find our answer:
is the length of one side of the square; in this case we are told that it is , so we can solve accordingly!
Example Question #9 : Calculating The Area Of A Square
Note: Figure NOT drawn to scale
Refer to the above figure, which shows Square and Square and Square has area 49. Give the area of Square .
Explanation:
Square has area 49, so each of its sides has as its length the square root of 49, or 7. Each side of Square is therefore a hypotenuse of a right triangle with legs 1 and , so each sidelength, including , can be found using the Pythagorean Theorem:
The square of this, which is 37, is the area of Square .
Example Question #1 : Calculating The Area Of A Square
Note: Figure NOT drawn to scale
Refer to the above figure, which shows Square and Square and Square has area 25. Give the area of Square .
Explanation:
Square has area 25, so each side has length the square root of 25, or 5.
Specifically, , and, as given, .
Since is a right triangle with hypotenuse and legs and can be found using the Pythagorean Theorem:
The area of is
Since all four triangles, by symmetry, are congruent, all have this area. the area of Square is the area of Square plus the areas of the four triangles, or .
← Previous 1
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## Does sample mean affect confidence interval?
Increasing the sample size decreases the width of confidence intervals, because it decreases the standard error. c) The statement, “the 95% confidence interval for the population mean is (350, 400)”, is equivalent to the statement, “there is a 95% probability that the population mean is between 350 and 400”.
### What is the 95% confidence interval for the mean?
What does a 95% confidence interval mean? The 95% confidence interval is a range of values that you can be 95% confident contains the true mean of the population. Due to natural sampling variability, the sample mean (center of the CI) will vary from sample to sample.
What does the confidence interval tells us about a sample mean?
A confidence interval displays the probability that a parameter will fall between a pair of values around the mean. Confidence intervals measure the degree of uncertainty or certainty in a sampling method. They are most often constructed using confidence levels of 95% or 99%.
How do you find the sample mean?
How to calculate the sample mean
1. Add up the sample items.
2. Divide sum by the number of samples.
3. The result is the mean.
4. Use the mean to find the variance.
5. Use the variance to find the standard deviation.
## How do you find the sample mean and margin of error for a confidence interval?
The confidence interval is the range between the sample mean minus E, and the sample mean plus E. Find the difference between the 2 numbers (22.1-14.7 = 7.4). Divide that number by 2, because that will tell you what was added to, and subtracted from, the mean. So we get 7.4/2 = 3.7 for the margin of error.
### How does sample size affect sampling error?
Factors Affecting Sampling Error In general, larger sample sizes decrease the sampling error, however this decrease is not directly proportional. As a rough rule of thumb, you need to increase the sample size fourfold to halve the sampling error.
Which type of sample produces a sample mean closer to the population mean?
The law of large numbers says that if you take samples of larger and larger size from any population, then the mean of the sampling distribution, μ x – tends to get closer and closer to the true population mean, μ.
How do you find a sample mean?
## Is the difference of the sample means statistically significant?
Statistically significant means a result is unlikely due to chance. The p-value is the probability of obtaining the difference we saw from a sample (or a larger one) if there really isn’t a difference for all users. Statistical significance doesn’t mean practical significance.
### Is sample mean and mean the same?
“Mean” usually refers to the population mean. This is the mean of the entire population of a set. The mean of the sample group is called the sample mean.
How do you calculate a confidence interval?
How to Calculate a Confidence Interval Step #1: Find the number of samples (n). Step #2: Calculate the mean (x) of the the samples. Step #3: Calculate the standard deviation (s). Step #4: Decide the confidence interval that will be used. Step #5: Find the Z value for the selected confidence interval. Step #6: Calculate the following formula.
How do I interpret a confidence interval?
To interpret a confidence interval, you first have to find out which kind it is. If it’s the first kind, the interpretation is that if you have a large number of intervals, on average the true values will be inside them the sum of the confidences time; but that you know nothing about this particular interval.
## What is 90 percent confidence interval?
Similarly, a 90% confidence interval is an interval generated by a process that’s right 90% of the time and a 99% confidence interval is an interval generated by a process that’s right 99% of the time. If we were to replicate our study many times, each time reporting a 95% confidence interval,…
### How do you determine the confidence level?
Find a confidence level for a data set by taking half of the size of the confidence interval, multiplying it by the square root of the sample size and then dividing by the sample standard deviation. Look up the resulting Z or t score in a table to find the level. |
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# 1.5: Mutually Inclusive Events
Difficulty Level: Basic Created by: CK-12
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Practice Mutually Inclusive Events
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Suppose you're selecting a card from a deck of cards with your eyes closed. What is the probability that you'll select a red card or a Queen? Is this a mutually exclusive event or a mutually inclusive event? Can you explain your answer?
### Watch This
First watch this video to learn about mutually inclusive events.
Then watch this video to see some examples.
Watch this video for more help.
### Guidance
When determining the probabilities of mutually inclusive events, we can apply the words and ()\begin{align*}(\cap)\end{align*} and or ()\begin{align*}(\cup)\end{align*} using the Addition Rule. Let’s take another look at Venn diagrams when defining mutually inclusive events.
If events A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} share some overlap in the Venn diagram, they may be considered mutually inclusive events. Look at the diagrams below to see how these events can occur. Mutually inclusive events can occur at the same time. Say, for example, you wanted to pick a number from 1 to 10 that is less than 4 and pick an even number. Let event A\begin{align*}A\end{align*} be picking a number less than 4 and event B\begin{align*}B\end{align*} be picking an even number.
A={1,2,3}P(A)=310B={2,4,6,8,10}P(B)=510P(A and B)=110\begin{align*}&A = \left \{1, 2, 3 \right \}\\ \\ &P(A) = \frac{3}{10}\\ \\ &B = \left \{2, 4, 6, 8, 10 \right \}\\ \\ &P(B) = \frac{5}{10}\\ \\ &P(A \ \text{and} \ B) = \frac{1}{10} \\ \end{align*}
The reason why P(A and B)=110\begin{align*}P(A \ \text{and} \ B) = \frac{1}{10}\end{align*} is because there is only 1 number from 1 to 10 that is both less than 4 and even, and that number is 2.
When representing this on the Venn diagram, we would see something like the following:
Mutually inclusive events, remember, can occur at the same time. Look at the Venn diagram below. What do you think we need to do in order to calculate the probability of AB\begin{align*}A \cup B\end{align*} just from looking at this diagram?
If you look at the diagram, you see that the calculation involves not only P(A)\begin{align*}P(A)\end{align*} and P(B)\begin{align*}P(B)\end{align*}, but also P(AB)\begin{align*}P(A \cap B)\end{align*}. However, the items in AB\begin{align*}A \cap B\end{align*} are also part of event A\begin{align*}A\end{align*} and event B\begin{align*}B\end{align*}. To represent the probability of A\begin{align*}A\end{align*} or B\begin{align*}B\end{align*}, we need to subtract the P(AB)\begin{align*}P(A \cap B)\end{align*}; otherwise, we are double counting. In other words:
P(A or B)P(AB)=P(A)+P(B)P(A and B)or=P(A)+P(B)P(AB)\begin{align*}P(A \ \text{or} \ B) &= P(A) + P(B) - P(A \ \text{and} \ B)\\ & \qquad \qquad \text{or}\\ P(A \cup B) &= P(A) + P(B) - P(A \cap B)\end{align*}
where \begin{align*}\cap\end{align*} represents and and \begin{align*}\cup\end{align*} represents or.
This is known as the Addition Principle (Rule).
Addition Principle
P(AB)=P(A)+P(B)P(AB)\begin{align*}P(A \cup B) = P(A) + P(B) - P(A \cap B)\end{align*}
Think about the idea of rolling a die. Suppose event A\begin{align*}A\end{align*} is rolling an odd number with the die, and event B\begin{align*}B\end{align*} is rolling a number greater than 2.
Event A={1,3,5}\begin{align*}A = \{1, 3, 5\}\end{align*}
Event B={3,4,5,6}\begin{align*}B = \{3, 4, 5, 6\}\end{align*}
Notice that the sets containing the possible outcomes of the events have 2 elements in common. Therefore, the events are mutually inclusive.
Now take a look at the example below to understand the concept of double counting.
#### Example A
What is the probability of choosing a card from a deck of cards that is a club or a ten?
P(A)P(A)P(B)P(B)P(AB)P(AB)P(AB)P(AB)P(AB)=probability of selecting a club=1352=probability of selecting a ten=452=152=P(A)+P(B)P(AB)=1352+452152=1652=413\begin{align*}P(A) &=\text{probability of selecting a club}\\ P(A) &= \frac{13}{52}\\ P(B) &=\text{probability of selecting a ten}\\ P(B) &= \frac{4}{52}\\ \\ P(A \cap B) &= \frac{1}{52}\\ \\ P(A \cup B) &= P(A) + P(B) - P(A \cap B)\\ P(A \cup B) &= \frac{13}{52} + \frac{4}{52} - \frac{1}{52}\\ P(A \cup B) &= \frac{16}{52}\\ P(A \cup B) &= \frac{4}{13}\end{align*}
#### Example B
What is the probability of choosing a number from 1 to 10 that is less than 5 or odd?
A={1,2,3,4}P(A)P(A)P(A)B={1,3,5,7,9}P(B)P(B)P(B)P(AB)P(AB)P(AB)P(AB)P(AB)P(AB)=probability of selecting a number less than 5=410=25=probability of selecting a number that is odd=510=12=210=15=P(A)+P(B)P(AB)=25+1215=410+510210=710\begin{align*}A = \{1, 2, 3, 4\}\\ P(A) &=\text{probability of selecting a number less than 5}\\ P(A) &= \frac{4}{10}\\ P(A) &= \frac{2}{5}\\ B = \{1, 3, 5, 7, 9\}\\ P(B) &=\text{probability of selecting a number that is odd}\\ P(B) &= \frac{5}{10}\\ P(B) &= \frac{1}{2}\\ \\ P(A \cap B) &= \frac{2}{10}\\ P(A \cap B) &= \frac{1}{5}\\ \\ P(A \cup B) &= P(A) + P(B) - P(A \cap B)\\ P(A \cup B) &= \frac{2}{5} + \frac{1}{2} - \frac{1}{5}\\ P(A \cup B) &= \frac{4}{10} + \frac{5}{10} - \frac{2}{10}\\ P(A \cup B) &= \frac{7}{10}\end{align*}
Notice in the previous 2 examples how the concept of double counting was incorporated into the calculation by subtracting the P(AB)\begin{align*}P(A \cup B)\end{align*}. Let’s try a different example where you have 2 events happening.
#### Example C
2 fair dice are rolled. What is the probability of getting a sum less than 7 or a sum less than 4?
P(A)=probability of obtaining a sum less than 7P(A)=1536\begin{align*}&P(A) = \text{probability of obtaining a sum less than 7}\\ \\ &P(A) = \frac{15}{36}\end{align*}
P(B)=probability of obtaining a sum less than 4P(B)=336\begin{align*}&P(B) = \text{probability of obtaining a sum less than 4}\\ \\ &P(B) = \frac{3}{36}\end{align*}
Notice that there are 3 elements in common. Therefore, the events are mutually inclusive, and we must account for the double counting.
P(A and B)P(AB)P(AB)P(AB)P(AB)P(AB)P(AB)P(AB)=336=336=112=P(A)+P(B)P(AB)=1536+336112=1536+336336=1536=512\begin{align*}P(A \ \text{and} \ B) &= \frac{3}{36}\\ P(A \cap B) &= \frac{3}{36}\\ P(A \cap B) &= \frac{1}{12}\\ \\ P(A \cup B) &= P(A) + P(B) - P(A \cap B)\\ P(A \cup B) &= \frac{15}{36} + \frac{3}{36}-\frac{1}{12}\\ P(A \cup B) &= \frac{15}{36} + \frac{3}{36}-\frac{3}{36}\\ P(A \cup B) &= \frac{15}{36}\\ P(A \cup B) &= \frac{5}{12}\\\end{align*}
Points to Consider
### Guided Practice
A bag contains 26 tiles with a letter on each, one tile for each letter of the alphabet. What is the probability of reaching into the bag and randomly choosing a tile with one of the first 10 letters of the alphabet on it or randomly choosing a tile with a vowel on it?
Answer:
The first 10 letters of the alphabet are 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', and 'J'. Also, the vowels are 'A', 'E', 'I', 'O', and 'U'. The only letters that are both in the first 10 letters of the alphabet and vowels are 'A', 'E', and 'I'. Therefore, the probability of reaching into the bag and randomly choosing a tile with one of the first 10 letters of the alphabet on it or randomly choosing a tile with a vowel on it can be calculated as follows:
AP(A)P(A)P(A)BP(B)P(B)P(AB)P(AB)P(AB)P(AB)P(AB)={A, B, C, D, E, F, G, H, I, J}=probability of selecting one of the first 10 letters of the alphabet=1026=513={A, E, I, O, U}=probability of selecting a vowel=526=326=P(A)+P(B)P(AB)=513+526326=1026+526326=1226=613\begin{align*}A &= \{\text{A, B, C, D, E, F, G, H, I, J}\}\\ P(A) &=\text{probability of selecting one of the first 10 letters of the alphabet}\\ P(A) &= \frac{10}{26}\\ P(A) &= \frac{5}{13}\\ B &= \{\text{A, E, I, O, U}\}\\ P(B) &=\text{probability of selecting a vowel}\\ P(B) &= \frac{5}{26}\\ \\ P(A \cap B) &= \frac{3}{26}\\ \\ P(A \cup B) &= P(A) + P(B) - P(A \cap B)\\ P(A \cup B) &= \frac{5}{13} + \frac{5}{26} - \frac{3}{26}\\ P(A \cup B) &= \frac{10}{26} + \frac{5}{26} - \frac{3}{26}\\ P(A \cup B) &= \frac{12}{26} = \frac{6}{13}\end{align*}
### Practice
1. Consider a sample set as \begin{align*}S = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20\}\end{align*}. Event \begin{align*}A\end{align*} is the multiples of 4, while event \begin{align*}B\end{align*} is the multiples of 5. What is the probability that a number chosen at random will be from both \begin{align*}A\end{align*} and \begin{align*}B\end{align*}?
2. For question 1, what is the probability that a number chosen at random will be from either \begin{align*}A\end{align*} or \begin{align*}B\end{align*}?
3. Jack is a student in Bluenose High School. He noticed that a lot of the students in his math class were also in his chemistry class. In fact, of the 60 students in his grade, 28 students were in his math class, 32 students were in his chemistry class, and 15 students were in both his math class and his chemistry class. He decided to calculate what the probability was of selecting a student at random who was either in his math class or his chemistry class, but not both. Draw a Venn diagram and help Jack with his calculation.
4. Brenda did a survey of the students in her classes about whether they liked to get a candy bar or a new math pencil as their reward for positive behavior. She asked all 71 students she taught, and 32 said they would like a candy bar, 25 said they wanted a new pencil, and 4 said they wanted both. If Brenda were to select a student at random from her classes, what is the probability that the student chosen would want:
1. a candy bar or a pencil?
2. neither a candy bar nor a pencil?
5. A card is chosen at random from a standard deck of cards. What is the probability that the card chosen is a heart or a face card? Are these events mutually inclusive?
6. What is the probability of choosing a number from 1 to 10 that is greater than 5 or even?
7. A bag contains 26 tiles with a letter on each, one tile for each letter of the alphabet. What is the probability of reaching into the bag and randomly choosing a tile with one of the letters in the word ENGLISH on it or randomly choosing a tile with a vowel on it?
8. Are randomly choosing a teacher and randomly choosing a father mutually inclusive events? Explain your answer.
9. Suppose 2 events are mutually inclusive events. If one of the events is passing a test, what could the other event be? Explain your answer.
10. What is the probability of randomly choosing a number from 1 to 10 that is a factor of 8 or a factor of 10?
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### Vocabulary Language: English
Addition Principle
If events A and B are mutually inclusive, then P(A or B) = P(A) + P(B) – P(A and B)
Mutually Exclusive Events
Mutually exclusive events have no common outcomes.
Mutually Inclusive Events
Mutually inclusive events can occur at the same time.
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# Fundamentals of Physics, (Chapters 21- 44) (Volume 2)
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PY212 Solutions Quiz 5 First we identify all the current and assign them an arbitrary direction. There are three different currents: i 1 moving through ε 1 , i 2 moving through ε 2 , and i 3 moving through ε 3 . We shall assume for now that they are all going upwards (this is obviously a wrong assumption, but the sign of each current will indicate the correct direction at the end). The junction rule now tells us that 0 3 2 1 out in = + + = i i i i i Now consider the loops. There are two unique loops and we can walk through them in a clockwise manner. Starting from point b , we obtain 0 5 6 2 0 2 1 2 2 2 1 1 1 1 1 = + = + + i i R i R i R i ε for the left loop, and 0 6 5 2 0 3 2 1 3 3 1 3 2 2 2 = + = + + + i i R i R i R i for the right loop. We now have three equations and three unknowns (each of the currents). Hence, in principle, we can solve this system of equations. There are many different ways to do this. Your calculator might have a special method; check your manual how to solve a set of equations with several unknowns. Here we’ll show how one can do it without a calculator. Let us first remove
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Unformatted text preview: i 3 from the equations by writing 2 1 3 i i i − − = . Plugging this into the other two equations we obtain 2 1 2 1 2 2 1 11 6 ) ( 6 5 2 5 6 2 i i i i i i i − − = − − + − = + − = We now have two equations and two unknowns, so we’re heading in the right direction. Now rewrite the upper equation to 1 2 6 2 5 i i + = and substitute this into the other: A. 33 . 3 1 2 . 19 4 . 6 2 . 13 4 . 4 6 5 6 2 11 6 2 1 1 1 1 1 − = − = − = ⇒ − − − = + − − = i i i i i (a) and (b) We see that i 1 = 0.33 A and it’s directed down . (c) and (d) The current i 2 is actually zero: ( ) ( ) 5 / ) 3 / 1 ( 6 2 5 / 6 2 1 2 = − + = + = i i . (e) and (f) The current i 3 is given by A 33 . 2 1 3 + = − − = i i i , so it’s directed up . (g) To find V a – V b we walk from point b to a : V. . 5 5 5 2 2 2 2 = − = − i R i...
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# Class 9 SELINA Solutions Maths Chapter 5 - Factorisation
## Factorisation Exercise Ex. 5(A)
### Solution 1
Taking (2x - 5y) common from both terms
= (2x - 5y)[2(3x + 4y) - 6(x - y)]
=(2x - 5y)(6x + 8y - 6x + 6y)
=(2x - 5y)(8y + 6y)
=(2x - 5y)(14y)
=(2x - 5y)14y
### Solution 2
xy(3x2 - 2y2) - yz(2y2 - 3x2) + zx(15x2 - 10y2)
= xy(3x2 - 2y2) + yz(3x2 - 2y2) + zx(15x2 - 10y2)
= xy(3x2 - 2y2) + yz(3x2 - 2y2) + 5zx(3x2 - 2y2)
= (3x2 - 2y2)[xy + yz + 5zx]
### Solution 3
ab(a2 + b2 - c2) - bc(c2 - a2 - b2) + ca(a2 + b2 - c2)
= ab(a2 + b2 - c2) + bc(a2 + b2 - c2) + ca(a2 + b2 - c2)
= (a2 + b2 - c2)[ab + bc + ca]
### Solution 4
2x(a - b) + 3y(5a - 5b) + 4z(2b - 2a)
= 2x(a - b) + 15y(a - b) - 8z(a - b)
= (a - b)[2x + 15y - 8z]
### Solution 5
a3 + a - 3a2 - 3= a (a2 + 1) - 3(a2 + 1)
= (a2 + 1) (a -3).
### Solution 6
16 (a + b)2 - 4a - 4b =16 (a + b)2 - 4 (a + b)
= 4 (a + b) [4 (a + b) - 1]
= 4 (a + b) (4a + 4b - 1)
### Solution 17
= (x2 + y2 + 2xy ) + (x + y)
[As (x + y)2 = x2 + 2xy + y2]
=(x + y)2 + (x + y)
=(x + y)(x + y + 1)
### Solution 18
= a2 + 4b2 - 4ab - 3a + 6b
= a2 + (2b)2 - 2 × a × (2b) - 3(a - 2b)
[As (a - b)2 = a2 - 2ab + b2 ]
=(a - 2b)2 - 3(a - 2b)
=(a - 2b)[(a - 2b)- 3]
=(a - 2b)(a - 2b - 3)
### Solution 19
= m (x - 3y)2 - n (x - 3y) + 5(x - 3y)
[Taking (x - 3y) common from all the three terms]
=(x - 3y) [m(x - 3y) - n + 5]
=(x - 3y)(mx - 3my - n + 5)
### Solution 20
=(6x - 5y)[x - 4(6x - 5y)]
[Taking (6x - 5y) common from the three terms]
= (6x - 5y)(x - 24x + 20y)
= (6x - 5y)(-23x + 20y)
= (6x - 5y)(20y - 23x)
## Factorisation Exercise Ex. 5(B)
### Solution 17
(x2 - 3x)(x2 - 3x - 1) - 20
= (x2 - 3x)[(x2 - 3x) - 1] - 20
= a[a - 1] - 20 ….(Taking x2 - 3x = a)
= a2 - a - 20
= a2 - 5a + 4a - 20
= a(a - 5) + 4(a - 5)
= (a - 5)(a + 4)
= (x2 - 3x - 5)(x2 - 3x + 4)
### Solution 20
12x2 - 35x + 25
= 12x2 - 20x - 15x + 25
= 4x(3x - 5) - 5(3x - 5)
= (3x - 5)(4x - 5)
Thus,
Length = (3x - 5) and breadth = (4x - 5)
OR
Length = (4x - 5) and breadth = (3x - 5)
## Factorisation Exercise Ex. 5(D)
### Solution 9
= (x - y)3 - (2x)3
= (x - y - 2x)[(x - y)2 + 2x(x - y) + (2x)2]
[Using identity (a3 - b3) = (a - b)(a2 + ab + b2)]
= (-x - y)[x2 + y2 - 2xy + 2x2 - 2xy + 4x2]
=-(x + y) [7x2 - 4xy + y2]
### Solution 16
= 2(x3 + 27y3 - 2x - 6y)
= 2{[(x)3+(3y)3] - 2(x + 3y)}
[Using identity (a3 + b3) = (a + b)(a2 - ab + b2)]
=2{[(x + 3y)(x2 - 3xy + 9y2)] - 2(x + 3y)}
=2(x + 3y)(x2 - 3xy + 9y2 - 2)
### Solution 17
1029 - 3x3
= 3(343 - x3)
= 3(73 - x3)
= 3(7 - x)(72 + 7x + x2)
= 3(7 - x)(49 + 7x + x2)
### Solution 18
(i) (133 - 53)
[Using identity (a3 - b3) = (a - b)(a2 + ab + b2)]
=(13 - 5)(132 + 13 × 5 + 52)
=8(169 + 65 + 25)
Therefore, the number is divisible by 8.
(ii) (353 + 273)
[Using identity (a3 + b3)=(a + b)(a2 - ab + b2)]
=(35 + 27)(352 + 35× 27 + 272)
=62 × (352 + 35 × 27 + 272)
Therefore, the number is divisible by 62. |
# If the sides of a triangle are 3 cm,
Question:
If the sides of a triangle are 3 cm, 4 cm and 6 cm long, determine whether the triangle is a right-angled triangle.
Solution:
We have,
$a=3 \mathrm{~cm}$
$b=4 \mathrm{~cm}$
$c=6 \mathrm{~cm}$
In order to prove that the triangle is a right angled triangle we have to prove that square of the larger side is equal to the sum of the squares of the other two sides.
Here, the larger side is $c=6 \mathrm{~cm}$.
Hence, we have to prove that $a^{2}+b^{2}=c^{2}$.
Let solve the left hand side of the above equation.
$a^{2}+b^{2}=3^{2}+4^{2}$
$=9+16$
$=25$
Now we will solve the right hand side of the equation,
$c^{2}=6^{2}$
$=36$
Here we can observe that left hand side is not equal to the right hand side.
Therefore, the given triangle is not a right angled triangle. |
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# GOAL is a quadrilateral in which $GO\parallel AL$. If $\angle G=\angle O={{40}^{\circ }}$ what are the measures of $\angle A$ and $\angle L$.
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Hint: In this problem, we are given that GOAL is a quadrilateral in which $GO\parallel AL$. If $\angle G=\angle O={{40}^{\circ }}$, we have to find the measures of $\angle A$ and $\angle L$. Here we can first draw the diagram. Here we have to use the interior angle property, where the sum of the internal and external angle in the same vertex is ${{180}^{\circ }}$. By using this property we can find the required angles.
Here, we are given that GOAL is a quadrilateral in which $GO\parallel AL$. If $\angle G=\angle O={{40}^{\circ }}$, we have to find the measures of $\angle A$ and $\angle L$.
We can now draw the diagram.
Here we can use the property of interior angle.
We know that the interior angle property is where the sum of the internal and external angle in the same vertex is ${{180}^{\circ }}$.
We can now find the measure of $\angle A$,
$\Rightarrow \angle A+\angle O={{180}^{\circ }}$
We can now substitute the given measure and simplify it, we get
\begin{align} & \Rightarrow \angle A+{{40}^{\circ }}={{180}^{\circ }} \\ & \Rightarrow \angle A={{180}^{\circ }}-{{40}^{\circ }}={{140}^{\circ }} \\ \end{align}
The measure of $\angle A={{140}^{\circ }}$
We can now find the measure of $\angle L$,
$\Rightarrow \angle G+\angle L={{180}^{\circ }}$
We can now substitute the given measure and simplify it, we get
\begin{align} & \Rightarrow \angle L+{{40}^{\circ }}={{180}^{\circ }} \\ & \Rightarrow \angle L={{180}^{\circ }}-{{40}^{\circ }}={{140}^{\circ }} \\ \end{align}
The measure of $\angle L={{140}^{\circ }}$
Therefore, the measure of$\angle A=\angle L={{140}^{\circ }}$
Note: We should always remember the interior angle property that the interior angle property is where the sum of internal and external angle in the same vertex is ${{180}^{\circ }}$. So if we have one of the ongles, then we can substitute in the sum to get the value of another angle. |
# Types of Lines
Types of lines is the sub topic of the topic basic Geometry. In which Line is a geometrical object that is straight,endless and thin.It is continuous and we can increase the length, straight or curved.It will not have thickness; the trace of proceeding points.Below you can find the different variety of lines.
Name of the lineDescription
1.Parallel lines If two lines are always at the same distance apart and will not meet. This kinds of lines are known as parallel lines.
2.Perpendicular lines A line is perpendicular to another if it meets or crosses it at right angle (90°)
3.Concurrent lines If two or more lines passing through the same point then the lines are called concurrent lines.
## Types of lines: Classifications
### 1. Parallel lines
If two lines are always at the same distance apart and will not meet, then they are known as parallel lines.The length of two lines should not be same. The perpendicular distance between those two lines must be same anywhere.
Practical example: Railway track
We are using the symbol || to represent parallel line. In the above picture the first line (l1) is shorter than the second line (l2) but both the lines are in same direction. The perpendicular distance between the first and second line will be same always.They should not meet anywhere.So that these kinds of lines re called as parallel lines.If two lines are parallel then these will have the following properties.
Properties of Parallel Lines
• If two lines are intersected by a transversal,pair of acute angles are equal in measure, pairs of obtuse angles are also equal in measure,and any acute angle is supplementary to any obtuse angle.
If line m is || to line n, then:
• Alternate interior angles will be equal in measure.
• Corresponding angles will be equal in measure.
• The sum of degree measures of two interior angles on the same side of the transversal is 180.
• Also, the sum of the degree measures by any acute angle and any obtuse angle will be 180.
• ### 2. Perpendicular Line
Lines which are at right angles (90°) to each other.
Here the line CD is intersecting the line AB at 90 degree so we can say the line CD is perpendicular to the line AB.
### 3.Concurrent Lines
when 3 or more than three lines meet at the same point.
Here all the four are intersecting at the point P.These lines are known as concurrent lines. |
# Operations on Determinants
### Operations on Determinants:
1. Transpose Property:
• For any square matrix $A$, $\text{det}\left(A\right)=\text{det}\left({A}^{T}\right)$.
• The determinant of a matrix and its transpose are equal.
2. Row/Column Operations:
• Swapping rows or columns changes the sign of the determinant: $\text{det}\left(A\right)=-\text{det}\left({A}^{\mathrm{\prime }}\right)$, where ${A}^{\mathrm{\prime }}$ is the matrix obtained after swapping rows or columns in $A$.
• Multiplying a row or column by a scalar $k$ multiplies the determinant by $k$: $\text{det}\left(kA\right)={k}^{n}×\text{det}\left(A\right)$ for an $n×n$ matrix.
3. Adding a Multiple of One Row/Column to Another:
• If ${A}^{\mathrm{\prime }}$ is obtained by adding a multiple of one row or column to another in $A$, then $\text{det}\left(A\right)=\text{det}\left({A}^{\mathrm{\prime }}\right)$.
4. Row/Column Expansion:
• Expansion of a determinant along a row or column yields the sum of determinants of smaller matrices obtained by excluding that row or column.
### Application in Matrix Operations:
• The determinant of a square matrix $A$ is nonzero if and only if $A$ is invertible.
• The adjoint matrix $Adj\left(A\right)$ is calculated using cofactors and transposes, which involves determinants.
2. Solving Systems of Equations:
• Determinants play a crucial role in determining the solvability of systems of linear equations.
• Cramer's rule uses determinants to express the solutions of a system in terms of determinants of smaller matrices.
3. Geometry and Transformations:
• The absolute value of the determinant of a matrix representing a linear transformation determines the scaling factor and orientation change in space.
### Multiplying Determinants
Given two matrices $A$ and $B$ with determinants $\mathrm{\mid }A\mathrm{\mid }$ and $\mathrm{\mid }B\mathrm{\mid }$:
1. For $2×2$Matrices:
For matrices $2×2$:
$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ $B=\left[\begin{array}{cc}e& f\\ g& h\end{array}\right]$
The product of their determinants is: $\mathrm{\mid }A\mathrm{\mid }×\mathrm{\mid }B\mathrm{\mid }=\left(ad-bc\right)×\left(eh-fg\right)$
1. For $3×3$ Matrices:
For matrices $3×3$:
$A=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$ $B=\left[\begin{array}{ccc}p& q& r\\ s& t& u\\ v& w& x\end{array}\right]$
The product of their determinants is not straightforward. It involves a more complex expansion of the matrices' elements according to the rule:
$\mathrm{\mid }A\mathrm{\mid }×\mathrm{\mid }B\mathrm{\mid }=\mid \begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\mid ×\mid \begin{array}{ccc}p& q& r\\ s& t& u\\ v& w& x\end{array}\mid$
1. For Higher-Dimensional Matrices:
Beyond $3×3$ matrices, calculating the product of determinants becomes increasingly intricate due to expansion methods and may involve computational complexity.
### Example:
Let's take two $2×2$ matrices and multiply their determinants:
$A=\left[\begin{array}{cc}2& 3\\ -1& 4\end{array}\right]$ $B=\left[\begin{array}{cc}5& 1\\ 2& 0\end{array}\right]$
Calculating the determinants:
$\mathrm{\mid }A\mathrm{\mid }=\left(2×4\right)-\left(3×\left(-1\right)\right)=8+3=11$
$\mathrm{\mid }B\mathrm{\mid }=\left(5×0\right)-\left(1×2\right)=0-2=-2$
Now, let's find the product of the determinants:
$\mathrm{\mid }A\mathrm{\mid }×\mathrm{\mid }B\mathrm{\mid }=11×\left(-2\right)=-22$
Therefore, the product of the determinants of matrices $A$ and $B$ is $-22$.
### Differentiation of Determinants
Given a square matrix $A$ of size $n×n$ with elements depending on a variable $x$:
$A\left(x\right)=\left[\begin{array}{cccc}{a}_{11}\left(x\right)& {a}_{12}\left(x\right)& \cdots & {a}_{1n}\left(x\right)\\ {a}_{21}\left(x\right)& {a}_{22}\left(x\right)& \cdots & {a}_{2n}\left(x\right)\\ & & \ddots & \\ {a}_{n1}\left(x\right)& {a}_{n2}\left(x\right)& \cdots & {a}_{nn}\left(x\right)\end{array}\right]$
To differentiate the determinant of matrix $A$ with respect to $x$ ($\frac{d}{dx}\mathrm{\mid }A\mathrm{\mid }$), we use the following formula:
$\frac{d}{dx}\mathrm{\mid }A\mathrm{\mid }=\mathrm{\mid }A\mathrm{\mid }\cdot \text{Tr}\left({A}^{-1}\cdot \frac{dA}{dx}\right)$
Where:
• $\mathrm{\mid }A\mathrm{\mid }$denotes the determinant of matrix $A$.
• $\text{Tr}$ represents the trace of a matrix.
• ${A}^{-1}$ is the inverse of matrix $A$.
• $\frac{dA}{dx}$ represents the derivative of matrix $A$ with respect to $x$.
### Example:
Let's consider a $2×2$ matrix $A\left(x\right)$ dependent on $x$:
$A\left(x\right)=\left[\begin{array}{cc}x& {x}^{2}\\ 3x& 2\end{array}\right]$
First, calculate the determinant of matrix $A\left(x\right)$:
$\mathrm{\mid }A\mathrm{\mid }=x×2-{x}^{2}×3x=2x-3{x}^{3}$
Next, find the inverse of matrix $A\left(x\right)$:
${A}^{-1}\left(x\right)=\frac{1}{2x-3{x}^{3}}\left[\begin{array}{cc}2& -{x}^{2}\\ -3x& x\end{array}\right]$
Now, calculate $\frac{dA}{dx}$, the derivative of matrix $A\left(x\right)$ with respect to $x$:
$\frac{dA}{dx}=\left[\begin{array}{cc}1& 2x\\ 3& 0\end{array}\right]$
Next, find the trace of ${A}^{-1}\cdot \frac{dA}{dx}$:
${A}^{-1}\cdot \frac{dA}{dx}=\frac{1}{2x-3{x}^{3}}\left[\begin{array}{cc}2& -{x}^{2}\\ -3x& x\end{array}\right]\cdot \left[\begin{array}{cc}1& 2x\\ 3& 0\end{array}\right]$
${A}^{-1}\cdot \frac{dA}{dx}=\frac{1}{2x-3{x}^{3}}\left[\begin{array}{cc}2-3{x}^{2}& 2x\\ -3-3{x}^{3}& -3x\end{array}\right]$
Finally, find $\frac{d}{dx}\mathrm{\mid }A\mathrm{\mid }$ using the formula:
$\frac{d}{dx}\mathrm{\mid }A\mathrm{\mid }=\left(2x-3{x}^{3}\right)\cdot \text{Tr}\left(\frac{1}{2x-3{x}^{3}}\left[\begin{array}{cc}2-3{x}^{2}& 2x\\ -3-3{x}^{3}& -3x\end{array}\right]\right)$
This formula will yield the derivative of the determinant of matrix $A\left(x\right)$ with respect to $x$.
### Summation of Determinants
Given two square matrices $A$ and $B$ of the same size $n×n$, the sum or difference of their determinants can be found using the following rules:
1. Addition Rule: If $A$ and $B$ are $n×n$ matrices: $\mathrm{\mid }A+B\mathrm{\mid }=\mathrm{\mid }A\mathrm{\mid }+\mathrm{\mid }B\mathrm{\mid }$ The determinant of the sum of two matrices is the sum of their determinants.
2. Subtraction Rule: Similarly, if $A$ and $B$ are $n×n$ matrices: $\mathrm{\mid }A-B\mathrm{\mid }=\mathrm{\mid }A\mathrm{\mid }-\mathrm{\mid }B\mathrm{\mid }$The determinant of the difference of two matrices is the difference of their determinants.
These rules apply when the matrices involved are of the same size.
### Example:
Let's consider two $3×3$ matrices $A$ and $B$:
$A=\left[\begin{array}{ccc}2& 1& 3\\ 4& 0& -1\\ 2& 3& 5\end{array}\right]$ $B=\left[\begin{array}{ccc}1& -2& 0\\ 3& 2& 1\\ -1& 4& 2\end{array}\right]$
First, calculate the determinants of matrices $A$ and $B$:
$\mathrm{\mid }A\mathrm{\mid }=\mid \begin{array}{ccc}2& 1& 3\\ 4& 0& -1\\ 2& 3& 5\end{array}\mid =\left(2×0×5\right)+\left(1×\left(-1\right)×2\right)+\left(3×4×3\right)-\left(3×0×2\right)-\left(2×\left(-1\right)×2\right)-\left(5×4×1\right)=18$
$\mathrm{\mid }B\mathrm{\mid }=\mid \begin{array}{ccc}1& -2& 0\\ 3& 2& 1\\ -1& 4& 2\end{array}\mid =\left(1×2×2\right)+\left(-2×1×\left(-1\right)\right)+\left(0×3×4\right)-\left(0×2×\left(-1\right)\right)-\left(2×3×\left(-1\right)\right)-\left(1×1×4\right)=8$
Now, find the sum and difference of the determinants:
1. Sum of Determinants: $\mathrm{\mid }A+B\mathrm{\mid }=\mathrm{\mid }A\mathrm{\mid }+\mathrm{\mid }B\mathrm{\mid }=18+8=26$
2. Difference of Determinants: $\mathrm{\mid }A-B\mathrm{\mid }=\mathrm{\mid }A\mathrm{\mid }-\mathrm{\mid }B\mathrm{\mid }=18-8=10$
Thus, the sum of the determinants of matrices $A$ and $B$ is $26$, and the difference of their determinants is $10$. |
# 4.1.2 - Population is Not Normal
What happens when the sample comes from a population that is not normally distributed? This is where the Central Limit Theorem (CLT) comes in.
#### Central Limit Theorem
For a large sample size (we will explain this later), $$\bar{x}$$ is approximately normally distributed, regardless of the distribution of the population one samples from. If the population has mean $$\mu$$ and standard deviation $$\sigma$$, then the distribution of $$\bar{x}$$ has mean $$\mu$$ and standard deviation $$\dfrac{\sigma}{\sqrt{n}}$$.
We should stop here to break down what this theorem is saying because the Central Limit Theorem is very powerful!
The Central Limit Theorem applies to a sample mean from any distribution. We could have a left-skewed or a right-skewed distribution. As long as the sample size is large, the distribution of the sample means will follow an approximate Normal distribution.
For the purposes of this course, a sample size of $$n>30$$ is considered a large sample.
For many people just learning statistics there is a "so what" thought about the CLT. Why is this important and why do I care? If you recall, when we introduced the idea of Z scores we did so with the caveat that the distribution was normal. We take the observed data, that is normally distributed, and convert the data to z scores creating a standard normal distribution. We then leveraged this distribution to find percentiles (and will in future units leverage this to find probabilities.
The CLT allows us to assume a distribution IS normal as long as the sample size is greater than 30 observations. With this, we can apply most of our inferential statistics without having to compensate for non-normal distributions. This will take on greater relevance as we move through the course.
## Sampling Distribution of the Sample Mean Section
With the Central Limit Theorem, we can finally define the sampling distribution of the sample mean.
Sampling Distribution of the Sample Mean
The sampling distribution of the sample mean will have:
• the same mean as the population mean, $$\mu$$
• Standard deviation [standard error] of $$\dfrac{\sigma}{\sqrt{n}}$$
It will be Normal (or approximately Normal) if either of these conditions is satisfied
• The population distribution is Normal
• The sample size is large (greater than 30). |
# Chapter 5
Ratio and Proportion
In mathematics we often compare numbers. Imagine a factory where there are 150 men and 100 women working. One way to compare these facts is to subtract. There are 50 more men than women working in the factory. (150 - 100 = 50) Another way to compare these facts is to write a ratio. The ratio of men to women working in the factory is 150 to 100 or, in reduced form, 3 to 2. In other words, for every three men working in the factory, there are two women. Ratio and proportion are useful tools for solving many word problems.
\
Ratio
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A ratio is a comparison of numbers by division. A ratio can be written with the word to, with a colon (:), or as a fraction. Like a fraction, a ratio always should be reduced. Reducing a ratio is sometimes called simplifying. Following are the three ways to write the ratio of the number of men to the number of women working in the factory. 150 to 100
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= 3 to 2 or 150:100 = 3:2 or ~ = 1. 100 2
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The numbers in a ratio must be written in the Iorder the problem asks for. For the example of the factory workers, the ratio 1ofmen to women is 3 to 2, not 2 to 3. Example Evelyn earns \$2400 a month. She pays \$600 a month in rent. What is the ratio of her income to her rent? Make a ratio with her income first (in the numerator) and her rent second. Then reduce.
income rent 2400 600 _ 4 1
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The ratio of Evelyn's income to her rent is 4 to 1 or 4:1 or
4.
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Always.reduce a ratio to I()west terms. However, when a ratio is
an improper fraction, do r'lotchange it to a mixed number.
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What of problems he got right to the total Step 1 Find the number of problems he got right. Read the next example carefully.. in problem 3. 20 .20 10 .. What is the ratio of the distance she drove to the number of gallons of gas she used? Two-Step Ratio Problems A problem may not directly state both numbers that you need to set up a ratio. Make a ratio of thenumber of problems he got right. was the ratio of the number number of problems? Maceo got 2 problems wrong.. Subtract the number he got wrong.' . Anna drove 110 miles on 22 gallons of gas. from the total number of problems on the test... Then reduce. 3. 1. c: :-- EXERCISE 1 Directions: For problems 1 and 2.2 = 18 Step 2 right total ~ = .. 20. simplify each ratio.4_ 1. What is the ratio of the amount he makes to the amount he saves? For Alvaro.7 200:125 = 4 to 1000 = 28 21 \$560 to \$320 = Solve each problem below. 18.138 I Mathematics ...'l.- . You may have to determine one of the numbers. Alvaro makes \$600 a week and saves \$60 each week. Four of the students speak Armenian as a first language. What is the ratio of Armenian speakers to the total number of students in the class? 6. 5. There are 24 students in Sam's English class. 20. 24:30 = 2. 3. what is the ratio of the amount to the amount he makes? he saves 4. to the total number of problems..J. Example On a test with 20 problems.. 2.
. During the last election 887 people actually voted. What is the men? ratio of the number of women to the total number ratio of the number of men to the total number of ratio of the number of men to the number of ratio of the number of women to the number of 2.000. There are 1213 registered voters in Paul's village. What is the ratio of the number of union workers to the total number of workers? b. A math test of 50 questions included 15 fraction problems and 5 decimal problems. What is the ratio of the total number of workers to the number of union workers? 3. the city of McHenry \. What is the women? d. a. What is the of students? b.000 on education.1 \ ( 5. At Baxter Electronics there are 105 union workers and 45 nonunion workers. spends \$3.000. A G ED class of 20 students has' 12 women. 1. What is the ratio of the amount spent on education to the amount not spent on education? 4. Which of the following is approximately the ratio of the number of people who voted to the total number of registered voters in the village? \ \ \ \ (1) 1 to 2 (2) 2 to 3 (3) 3 to 4 I \ \\ \ (4) 5 to 6 t i i 1 I t \ .Chapter 5 . What is the students? c. From a total yearly budget of \$18. What is the ratio of the number of union workers to the number of nonunion workers? d.000. What is the ratio of the number of nonunion workers to the total number of workers? c.Ratio and Proportion 1 139 Directions: Solve each problem. a. What is the ratio of the total number of fraction and decimal problems to the number of questions on the test? ---.
The statement 1. If a missing term in a proportion is represented by the letter n and the other term that makes the cross product with n is 12.42 proportion. that the cross products by 12. then the cross product of 12 X n is 12n. -- To solve a proportion. A letter usually represents the missing term. Step 1 The cross product The cross product Write a statement 3c are equal. = 1.. that the cross products by 3. - = 56 Step 2 Divide both sides of the statement The missing term is 181.is also a . R U L E -. of 3 and c is 3c. 3c 3 _ -- 56 3 3 c= 18~ . in front of Note: This is an example of writing in the language of algebra. follow these steps: 1. Divide both sides of the statement by the number the missing term. Example 1 Find the missing term in The cross product The cross product Write a statement i= 9 1 2' Step 1 of nand 12 is 12n. of 7 and 8 is 56.140 I Mathematics Proportion A proportion is a statement that says two ratios (or two fractions) are equal. the cross products are 2 X 2 = 4 and 4 X 1 = 4. For example._ •. Write a statement with two equal cross products. --12 = 72 Step 2 Divide both sides of the statement The missing term is 6. 12n are equal.. You learned that the cross products of equal fractions are equal. The statement 2:4 = 1:2 is a proportion. In many proportion problems one term is missing. of 8 and 9 is 72.----------. ~ X+ Each of the four numbers in a proportion is called an element or a term. 2. 12n _ 72 12 n=6 Example 2 Solve for c in t = ~.
Ratio and Proportion 1141 Example 3 Step 1 Solve for y in the proportion 5:y = 2:8. 4:e = 6:8 3:7 = 4:y 15:40 = x:60 30:a = 12:16 i i I· I I 1 For problems 4 and 5. Divide both sides of the statement The missing term is 20. =± then x = 4. find the missing term in each proportion. 1 _ s 3 5 --- 2 _ 4 11 P 2 _ 9 8 x 3. The cross product of 5 and 8 is 40. y= 20 5 _ 2 y 8 Step 2 2y= 40 Step 3 EXERCISE 3 Directions: For problems 1-3. --6 10 15 3 _ 5 a 6 3 _ w 6 5 ±=.Chapter 5 . If ~ 12 (1) 12 (2)~ 5x3 . choose the answer that is set up correctly. I . Write a statement that the cross products are equal. If ~ 7 9' (1) 4 x 9 9 7 f.£ 3' X then c = 5 X 3 (3) 5 x 12 3 12 (4) 3 x 5 1 I -1 I i .Y 9 3 ~=± 7 x 2. The cross product of y and 2 is 2y. Rewrite the proportion with fractions. m 1. The first term in each ratio becomes a numerator. by 2. \ i (2) 4 + 7 (3) 4(9 X 7) (4) 4(9 + 7) = 5.
The letter c represents cost in the lastexample. yards cost Step 1 E= 40 30 c Step 2 ---- 12c _ 12 1200 12 Step 3 Divide both sides of 12c = 1200 by 12. Divide both sides of 3m = 960 by 3. how many men work there? This problem compares men to women. how much do 30 yards of lumber cost? This problem compares yards to cost. Set up two ratios of yards to cost.142 I Mathematics Proportion Word Problems Proportion is a useful tool for solving many word problems. In these examples. Example 1 If 12 yards of lumber cost \$40.first letter of the quantity you are looking for represents the unknown. Here c represents the cost you are looking for. Example 2 The ratio of the number of men to the number of women working in the county hospital is 2:3. The key to setting up a proportion is making sure that the amounts being compared are in the same position on either side of the = sign. Study the examples carefully. Here m represents the number of men. Example 3 Carlos got 2 problems wrong for every 5 problems right on a test. T'lP c = \$100 . If 480 women work in the hospital. How many problems did Carlos get wrong if there were 35 problems on the test? . The cost of 30 yards of lumber is \$100. There are 320 men working in the hospital. The letter m represents menin the next example: ' . Find both cross products. Ariy letter can represent the unknown you. men Step 1 women --- 2 _ m 3 480 Step 2 Step 3 --3 3m _ 960 3 m = 320 Be sure that the parts in a proportion correspond to the question that is asked. Set up two ratios of men to women. the. Read the next example carefully. Find both cross products.are looking for in a -proportlon.
Chapter 5 . Altogether. Choice (3) is correct. Here m represents miles. For every \$13 that Helen earns. she takes home \$10. notice that choice (4) is wrong because it shows the sum of 110 and 5 rather than the product. --7 7w _ 70 7 w= 10 Some problems may ask you to choose the correct set-up for a proportion problem. (2 wrong + 5 right = 7 total) Set up two ratios of wrong to total. Here w represents problems wrong. 3. Which expression shows the distance he can travel in 5 hours if he drives at the same speed? (1) 5 x 2 (2) 110 x 2 5 (3) 110X5 2 (4) 110+5 2 miles hours 110 Step 1 Set up two ratios of miles to hours.Ratio and Proportion 1 143 Step 1 The ratio in the question is problems wrong to total problems.. . but write the cross product of 110 and 5 as a set-up. s . i I { . 110 _ m 2 5 Step 2 2m = 110 X 5 Step 3 . How many games did they win? The ratio of the number of men to the number of women working at Apex. How much does she take home each month? Pat's softball team won 5 games for every 3 games they lost. I . Helen's gross pay each month is \$1950. 1.. Inc. Divide both sides of 7w Carlos got 10 problems wrong 2 total 7 Step 2 --- 2 _ 7 w 35 Step 3 Step 4 7w= 70 = 70 by 7. there are 360 workers at the company. How many of the workers are women? ~ \ { I i \ i I 1 . The ratio of wrong to total is 2:7.i ! 2. m=110X5 2 In the last example. the team played 32 games. EXERCISE 4 Directions: Solve each problem. Carlos got 2 wrong for every 5 right. Write both cross products. Divide both sides by 2. Example 4 Manny drove 110 miles in 2 hours. Find both cross products. wrong. Altogether. is 7:2.
The illustration below shows the ratio of blue paint to gray paint in a special color mix. Recently 300 people in Central County took a civil service examination. The short side will measure 20 inches.500 parts. How many gallons of gray paint are needed to make a total of 30 gallons of mix? 9. 6 in. 8. The picture shown at the right is to be enlarged. Every day the factory produces 10. Apples cost 90 cents a dozen.. 4 in. Find the measurement of the long side. of 8 apples? (1) (12 x 8) x 90 (2) 90 x 8 12 (3) 129~ 8 (4) 90 x 12 8 (5) 90+12 8 . how many voted to strike? 6. Inc. in cents. Which expression below represents the cost.1441 Mathematics 4. How many of these parts are defective? 5. Which expression below shows the number of cups of sugar a cook needs with 12 cups of flour? (1) 2 x 12 2 3 (2) 3 + 12 (3) 3x12 2 (4) 2 x 12 3 (5) 2+12 3 10. A recipe calls for 2 cups of sugar for every 3 cups of flour. How many people passed the exam? 7. The ratio of the number of workers who voted to strike to the number of workers who voted not to strike at Apex was 3:2.S people passed. is 20:1. The ratio of good parts to defective parts coming off the assembly line at Apex. For every 6 people who took the exam. If 360 workers voted.
For problems 9 and 10.Ratio and Proportion 1 145 Ratio and Proportion PART I Directions: Use a calculator to solve problems 1-10. Among the seniors at Cripple Creek High School. 9.6 to 6. Express the ratio 75:35 in reduced form. 4. Solve for s in the proportion 6. How many seats were empty at the meeting? (1) 21 (2) 28 (3) 35 (4) 42 = 12:36. 2 students said that they planned to leave Cripple Creek within 2 years for every 3 students who said that they planned to stay. 1. (5) 49 f= Find the value of n in 8:n = 5:18. Write the ratio 1. 3. The meeting room in the town hall can seat 140 people. 7.Chapter 5 . At the same rate. Review One seat was empty for every 4 that were occupied at an open meeting in the town hall about a proposal to build a new firehouse. Simplify the ratio 48:60. how many acres would they need to produce 1890 bushels? Choose the correct answer for problems 7 and 8. The Towsons planted a 35-acre field that yielded 3150 bushels of wheat.4 in simplest terms. How many of them said that they plan to leave within 2 years? (1) 22 (2) 33 (3) 44 (4) 55 (5) 66 . 2. There are 110 seniors at the school. x:9 8. Solve for x in the proportion 5. 25°. mark each answer on the corresponding number grid.
What is the ratio of the number of families who rent their homes to the total number of families on Maple Avenue? ! . What is the ratio of the number of families who own their homes to the number who rent? I I I 12. For problems 11-13. At the Central County Municipal Airport the ratio of delayed flights to flights that leave on time is 2:7. Write each answer in fraction form and mark your answer on the corresponding number grid. 11.146 I Mathematics 10. During a normal week there are 108 scheduled flights leaving the Central County Airport. refer to the following information. A survey shows that 42 families on Maple Avenue own their homes and 28 families rent their homes. How many of those flights are likely to be delayed? PART II Solve problems 11-20 without a Directions: calculator.
How many gallons of green paint are required to mix with 14 gallons of white paint? (1) 14 (2) 21 (3) 28 (4) 35 (5) 42 \ \ .10 18.! 3 (3) 5 Z 8 (4) 7 (5) 15 17.! 3 / f (4) 9 (5) 12 I (2) 5:15 (3) 9:36 (4) 10:30 (5) 16:48 15. 16. Find the ratio of the number of families who own their homes to the total number who live on Maple Avenue. Which of the following ratios is not equal to the ratio 12:36? (1) 3:9 (3) 5. If 6 feet of wire cost \$3.70 (2) \$2. To make a certain color of paint. how many inches apart will they be on the map? (1) 3.40.Chapter 5 . The scale on a map says that 2 inches = 150 miles.40 (4) \$4. the ratio of green paint to white paint is 5:2.! 4 (2) 4. How many hours will a plane take to go 1200 miles if it travels 450 miles in 2 hours? (1) 2 ~ Choose the correct answer for each of the following problems. .60 (5)\$5.! 2 1 J 14. how much do 9 feet of wire cost? (1)\$1.55 (3) \$3. I I 4 (2) 4. If two cities are actually 325 miles apart.Ratio and Proportion 1147 13.
• """. If you got 16 or more right.• . Which expression represents the time the worker needs to make 100 parts? (1)2X100 15 (2) 15x100 2 (3) 15x2 100 (4) 15 + 2 100 2 (5) 15 x 100 (5) 5+12 3 '. "" . correct any problem you got wrong.Q._ . -" ••••. ..!Jea9.er~p~:." -s-r=» •. If you did not get 16 right. Which expression represents the width of the enlargement? (1)3+12 5 (2) 5 x 12 3 (3) 3 x 12 5 (4) 3 x 5 12 20. A snapshot that wfs 3 inches wide and 5 inches long was enlarged to be 12 inches long.. review your ratio and proportion skills before you go on. r<. .... Then go on to the next chapter.. You should have gotten at least 16 problems right on this exercise..• .i>···~4'ftiVt(}~rsiu~e.~ .148 I Mathematics 19. . A worker can make 15 motor parts in 2 hours.
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# Finding the greatest common factor of an expression
In greatest common factor, finding the greatest common factor of numbers was the goal. In this lesson, we will do the same thing, but we will take the concept to the next level by looking for the greatest common factor of expressions.
Recall that if we are looking for the GCF of 12 and 20, we need to first factor each number into its prime factorization as shown in the figure below.
As you can see, 2 × 2 = 4 is the greatest factor 12 and 20 have in common, so 4 is the GCF of 12 and 20.
Find the GCF of 12x2 and 20x4
Now, not only we need to expand 12 and 20 into their prime factorizations, we also need to do the same for x2 and x4 as demonstrated in the figure above.
As you can see from the figure above, the GCF of 12x2 and 20x4 is 2 × 2 × x × x = 4x2
## A few more examples showing how finding the greatest common factor of expressions work
Example #1
Find the GCF of 15x4, 25x3, and 75x5
First take a look at the figure below and study it carefully.
Thus, the GCF for 15x4, 25x3, and 75x5 is 5 × x × x × x = 5x3
There is a faster way to find the greatest common factor of an expression.
When there are many exponential expressions that have the same variable or same number as base, the greatest common factor is the expression with the smallest exponent.
For example, the greatest common factor of x4, x3, and xis x3 since x3 is the expression with the smallest exponent. By the same token, the greatest common factor of 32, 34, and 33 is 32 since 32 is the expression with the smallest exponent.
Example #2
Find the GCF of w3 + w2 + w
We can break down this trinomial into three separate expressions: w3, w2, and w.
Since w is the expression with the smallest exponent, w is the greatest common factor.
Example #3
Find the GCF of 4w5 + 8w4 + 16w3
4w5 + 8w4 + 16w3 = 22w5 + 23w4 + 24w3
We can break down this trinomial into three separate expressions: 22w5,23w4,and 24w3
For 22, 23, and 24, the GCF is 22
For w5, w4, and w3, the GCF is w3
Therefore, the GCF of 4w5 + 8w4 + 16w3 is 22w3 or 4w3
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# Five Siblings
## Outline Mathematics Word Problems
Here's a problem to tackle:
"So you've got five kids, with one girl in the middle," said Alan. "How old is she?"
Clem smiled. "Figure it out yourself," he replied. "They're all spaced two years apart, and the youngest boy is just half as old as the oldest."
How old was the girl?
Solution
### Solution
"So you've got five kids, with one girl in the middle," said Alan. "How old is she?"
Clem smiled. "Figure it out yourself," he replied. "They're all spaced two years apart, and the youngest boy is just half as old as the oldest."
How old was the girl?
Let x be the girl's age. Then the ages of the five siblings are x-4,x-6,x-4,x-2,x, x-2, x, x+2, x+4. The fact that the youngest boy is half,half,twice,thrice the age of the oldest one gives us an equation:
2(x - 4) = x + 4,x + 2,x + 4,x + 6.
First multiply out:
2x - 8,4,6,8,10,12 = x + 4.
Then simplify:
2x - x = 8,4,6,8,10,12 + 4,4,6,8,10,12
from which
x = 12,4,6,8,10,12.
Do not forget to check your solution.
### References
1. J. A. H. Hunter, Entertaining Mathematical Teasers, Dover Publications, 1983 |
# 8.2 A single population mean using the student t distribution (Page 4/21)
Page 4 / 21
## Chapter review
In many cases, the researcher does not know the population standard deviation, σ , of the measure being studied. In these cases, it is common to use the sample standard deviation, s , as an estimate of σ . The normal distribution creates accurate confidence intervals when σ is known, but it is not as accurate when s is used as an estimate. In this case, the Student’s t-distribution is much better. Define a t-score using the following formula:
The t -score follows the Student’s t-distribution with n – 1 degrees of freedom. The confidence interval under this distribution is calculated with EBM = $\left({t}_{\frac{\alpha }{2}}\right)\frac{s}{\sqrt{n}}$ where ${t}_{\frac{\alpha }{2}}$ is the t -score with area to the right equal to $\frac{\alpha }{2}$ , s is the sample standard deviation, and n is the sample size. Use a table, calculator, or computer to find ${t}_{\frac{\alpha }{2}}$ for a given α .
## Formula review
s = the standard deviation of sample values.
is the formula for the t -score which measures how far away a measure is from the population mean in the Student’s t-distribution
df = n - 1; the degrees of freedom for a Student’s t-distribution where n represents the size of the sample
T ~ t df the random variable, T , has a Student’s t-distribution with df degrees of freedom
$EBM={t}_{\frac{\alpha }{2}}\frac{s}{\sqrt{n}}$ = the error bound for the population mean when the population standard deviation is unknown
${t}_{\frac{\alpha }{2}}$ is the t -score in the Student’s t-distribution with area to the right equal to $\frac{\alpha }{2}$
The general form for a confidence interval for a single mean, population standard deviation unknown, Student's t is given by (lower bound, upper bound)
= (point estimate – EBM , point estimate + EBM )
=
Use the following information to answer the next five exercises. A hospital is trying to cut down on emergency room wait times. It is interested in the amount of time patients must wait before being called back to be examined. An investigation committee randomly surveyed 70 patients. The sample mean was 1.5 hours with a sample standard deviation of 0.5 hours.
Identify the following:
1. $\overline{x}$ =_______
2. ${s}_{x}$ =_______
3. n =_______
4. n – 1 =_______
Define the random variables X and $\overline{X}$ in words.
X is the number of hours a patient waits in the emergency room before being called back to be examined. $\overline{X}$ is the mean wait time of 70 patients in the emergency room.
Which distribution should you use for this problem?
Construct a 95% confidence interval for the population mean time spent waiting. State the confidence interval, sketch the graph, and calculate the error bound.
CI: (1.3808, 1.6192)
EBM = 0.12
Explain in complete sentences what the confidence interval means.
Use the following information to answer the next six exercises: One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal.
Identify the following:
1. $\overline{x}$ =_______
2. ${s}_{x}$ =_______
3. n =_______
4. n – 1 =_______
1. $\overline{x}$ = 151
2. ${s}_{x}$ = 32
3. n = 108
4. n – 1 = 107
Define the random variable X in words.
Define the random variable $\overline{X}$ in words.
$\overline{X}$ is the mean number of hours spent watching television per month from a sample of 108 Americans.
Which distribution should you use for this problem?
Construct a 99% confidence interval for the population mean hours spent watching television per month. (a) State the confidence interval, (b) sketch the graph, and (c) calculate the error bound.
CI: (142.92, 159.08)
EBM = 8.08
Why would the error bound change if the confidence level were lowered to 95%?
Use the following information to answer the next 13 exercises: The data in [link] are the result of a random survey of 39 national flags (with replacement between picks) from various countries. We are interested in finding a confidence interval for the true mean number of colors on a national flag. Let X = the number of colors on a national flag.
X Freq.
1 1
2 7
3 18
4 7
5 6
Calculate the following:
1. $\overline{x}$ =______
2. ${s}_{x}$ =______
3. n =______
1. 3.26
2. 1.02
3. 39
Define the random variable $\overline{X}$ in words.
What is $\overline{x}$ estimating?
μ
Is ${\sigma }_{x}$ known?
As a result of your answer to [link] , state the exact distribution to use when calculating the confidence interval.
t 38
Construct a 95% confidence interval for the true mean number of colors on national flags.
How much area is in both tails (combined)?
How much area is in each tail?
0.025
Calculate the following:
1. lower limit
2. upper limit
3. error bound
The 95% confidence interval is_____.
(2.93, 3.59)
Fill in the blanks on the graph with the areas, the upper and lower limits of the Confidence Interval and the sample mean.
In one complete sentence, explain what the interval means.
We are 95% confident that the true mean number of colors for national flags is between 2.93 colors and 3.59 colors.
Using the same $\overline{x}$ , ${s}_{x}$ , and level of confidence, suppose that n were 69 instead of 39. Would the error bound become larger or smaller? How do you know?
The error bound would become EBM = 0.245. This error bound decreases because as sample sizes increase, variability decreases and we need less interval length to capture the true mean.
Using the same $\overline{x}$ , ${s}_{x}$ , and n = 39, how would the error bound change if the confidence level were reduced to 90%? Why?
three coins are tossed. find the probability of no head
three coins are tossed consecutively or what ?
umair
umair
or .125 is the probability of getting no head when 3 coins are tossed
umair
🤣🤣🤣
Simone
what is two tailed test
if the diameter will be greater than 3 cm then the bullet will not fit in the barrel of the gun so you are bothered for both the sides.
umair
in this test you are worried on both the ends
umair
lets say you are designing a bullet for thw gun od diameter equals 3cm.if the diameter of the bullet is less than 3 cm then you wont be able to shoot it
umair
In order to apply weddles rule for numerical integration what is minimum number of ordinates
excuse me?
Gabriel
why?
didn't understand the question though.
Gabriel
which question? ?
We have rules of numerical integration like Trapezoidal rule, Simpson's 1/3 and 3/8 rules, Boole's rule and Weddle rule for n =1,2,3,4 and 6 but for n=5?
John
geometric mean of two numbers 4 and 16 is:
10
umair
really
iphone
quartile deviation of 8 8 8 is:
iphone
sorry 8 is the geometric mean of 4,16
umair
quartile deviation of 8 8 8 is
iphone
can you please expalin the whole question ?
umair
mcq
iphone
h
iphone
can you please post the picture of that ?
umair
how
iphone
hello
John
10 now
John
how to find out the value
can you be more specific ?
umair
yes
KrishnaReddy
what is the difference between inferential and descriptive statistics
descriptive statistics gives you the result on the the data like you can calculate various things like variance,mean,median etc. however, inferential stats is involved in prediction of future trends using the previous stored data.
umair
if you need more help i am up for the help.
umair
Thanks a lot
Anjali
Inferential Statistics involves drawing conclusions on a population based on analysis of a sample. Descriptive statistics summarises or describes your current data as numerical calculations or graphs.
fred
my pleasure😊. Helping others offers me satisfaction 😊
umair
for poisson distribution mean............variance.
both are equal to mu
Faizan
mean=variance
Faizan
what is a variable
something that changes
Festus
why we only calculate 4 moment of mean? asked in papers.
why we only 4 moment of mean ? asked in BA exam
Faizan
Hello, can you please share the possible questions that are likely to be examined under the topic: regression and correlation analysis.
Refiloe
for normal distribution mean is 2 & variance is 4 find mu 4?
repeat quastion again
Yusuf
find mu 4. it can be wrong but want to prove how.
Faizan
for a normal distribution if mu 4 is 12 then find mu 3?
Question hi wrong ha
Tahir
ye BA mcqs me aya he teen he. 2dafa aya he
Faizan
if X is normally distributed. (n,b). then its mean deviation is?
Faizan
The answer is zero, because all odd ordered central moments of a normal distribution are Zero.
nikita
which question is zero
Faizan
sorry it is (5,16) in place of (n,b)
Faizan
I got. thanks. it is zero.
Faizan
How did we get the 24000
where do I start
in a large restaurant an average of every 7 customers ask for water with the their meal. A random sample of 12 customers is selected, find the probability that exactly 6 ask for water with their meal
any body with idea
Rufai
conditional probability
Ramesh
Rufai
iam really sorry. it's been long since I used these things. I just gave you a hint though
Ramesh
ok
Rufai
this follows binomial distribution. p(X=6)=12C6*(0.6)^6*0.4^6 use this formula n find.
syeda
can you explain the cosidered variable in the formula
Divya
x is variable wich is exactly 6 costumers
syeda
n is number of customers
syeda
ncx*p^X*q^X?
Divya
q^n-x
syeda
oh right !!! thanks yaar
Divya
I agree with Seyda too
Hoshyar
I agree with Syeda too
Hoshyar
7/12 =0.58is it?
yousaf
.
yousaf
r8
khalid |
# Fractions with the Same Denominator
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## Basics on the topicFractions with the Same Denominator
Learn all about comparing fractions with the same denominator in this video.
### TranscriptFractions with the Same Denominator
Axel and Tank are working on puzzles of identical size. Axel has ten of the twelve pieces in place. Tank has nine of the twelve pieces in place. They want to compare who's finished more, so we'll need to teach them about "Comparing Fractions with the Same Denominator". Denominators tell us how many equal parts a whole is divided into. Here is an example of fractions with the SAME denominators. The fraction with the greatest numerator will always be larger when comparing fractions with the same denominator. This is because each part is the same size, and the greater numerator is telling us we have MORE of the equal parts.
Fraction models can help us see if fractions with the same denominator are greater than, less than, or equal to one another. Let's compare these fractions with the same denominator. Three quarters and two quarters. They both have the same denominator and come from the same whole. First, draw and shade a fraction model to represent three quarters. Then, draw and shade another identical fraction model to represent two quarters. Finally, compare the shaded values of both models. Three quarters has MORE parts shaded than two quarters. So three quarters is GREATER THAN two quarters. Remember, when looking at fractions with the same denominator, the GREATER numerator is larger. Now let's take a look at Axel and Tank's puzzle progress. We have eight twelfths and ten twelfths. First, draw and shade a fraction model to represent eight twelfths. Then, draw and shade another identical fraction model to represent ten twelfths. Finally, compare the shaded values of both models. Eight twelfths has FEWER parts shaded than ten twelfths. So, eight twelfths is LESS THAN ten twelfths. The fraction with the smallest NUMERATOR is smaller. Let's try this again using the progress from Axel and Tank's LAST puzzles. Tank has eleven fifteenths finished and Axel has nine fifteenths finished. Who has more of their puzzle done?
Tank has more of his puzzle done because the numerator is greater than Tank's. How can we check our answer? We can use fraction models! First, draw and shade a fraction model to represent eleven fifteenths. Then, draw and shade another identical fraction model to represent nine fifteenths. Finally, compare the shaded values of both models. Eleven fifteenths has more parts shaded than nine fifteenths, so eleven fifteenths is greater than nine fifteenths. Tank HAS done more of his puzzle than Axel! While Axel and Tank finish their puzzles, let's review!
When working with fractions with the same denominator, the fraction with the LARGEREST numerator will always be larger. We can check our work by setting up identical fraction models. First, draw and shade a fraction model to represent the first fraction. Then, draw and shade another identical fraction model to represent the second fraction. Finally, compare the shaded values of both models. The fraction with more shaded parts is greater! Axel and Tank have both finished their last puzzle now! But why do THEY look puzzled? "Tank, do you think some of our pieces got shuffled?"
## Fractions with the Same Denominator exercise
Would you like to apply the knowledge you’ve learnt? You can review and practice it with the tasks for the video Fractions with the Same Denominator.
• ### When comparing fractions with identical wholes, the fraction with the greater numerator will always be _______.
Hints
Which fraction has more equal parts shaded in here?
Remember, each part is the same size and the greater numerator is telling us we have more of the equal parts.
Solution
When comparing fractions with identical wholes, the fraction with the greater numerator will always be larger.
Fractions with identical wholes have the same denominator. You can see this in the image above. Each part is the same size and the greater numerator is telling us we have more of the equal parts (more parts are shaded in).
• ### Which expressions are true?
Hints
Remember, when comparing fractions with the same denominator, the larger numerator will always be the larger fraction.
< means less than, and > means greater than.
When looking at the fraction models, which fraction has more equal parts shaded in? That will be the larger fraction.
For example, $\frac{3}{4}$ is greater than $\frac{2}{4}$.
Solution
Shown are the correct expressions. Remember, when comparing like fractions with the same whole, the larger numerator will also be the larger fraction.
The first expression is incorrect, because $\frac{1}{3}$ is < $\frac{2}{3}$
The second expression is correct, because $\frac{3}{6}$ is < $\frac{4}{6}$
The third expression is incorrect, because $\frac{7}{8}$ is > $\frac{6}{8}$
The fourth expression is correct, because $\frac{5}{6}$ is > $\frac{4}{6}$
• ### Comparing fractions.
Hints
When looking at the fraction models, which fraction has more equal parts shaded in? That will be the larger fraction.
e.g. $\frac{3}{4}$ > $\frac{2}{4}$
Remember, when comparing fractions with the same denominator, the larger numerator will always be the larger fraction.
< means less than, and > means greater than.
Solution
Remember, when comparing fractions with the same denominator, the larger numerator will also be the larger fraction.
$\frac{3}{3}$ > $\frac{2}{3}$
$\frac{1}{6}$ < $\frac{2}{6}$
$\frac{2}{8}$ < $\frac{3}{8}$
$\frac{6}{8}$ > $\frac{5}{8}$
• ### Who ate more pizza?
Hints
< means less than, and > means greater than.
Make sure you're looking at how many pieces are shaded in.
Solution
For the first set: $\frac{2}{5}$ < $\frac{3}{5}$
For the second set: $\frac{5}{7}$ < $\frac{6}{7}$
For the third set: $\frac{4}{8}$ = $\frac{4}{8}$
• ### Which option shows fractions with the same denominator?
Hints
Remember, the bottom number is the denominator.
Here is an example of fractions with the same denominator.
Solution
$\frac{1}{3}$ and $\frac{2}{3}$ are fractions with the same denominator.
The denominator is the bottom number in a fraction.
• ### Greater than or less than?
Hints
Remember, when comparing fractions with the same denominator, the larger numerator will always be the larger fraction.
< means less than, and > means greater than.
Solution
$\frac{5}{6}$ < $\frac{6}{6}$
$\frac{4}{6}$ < $\frac{5}{6}$
$\frac{2}{8}$ > $\frac{1}{8}$
$\frac{4}{8}$ > $\frac{3}{8}$ |
# Simple Venn Diagrams
Lesson
## Venn Diagrams
A Venn Diagram is a pictorial way to display relationships between two different sets (or groups) of things. The idea of a Venn diagram was first introduced by John Venn in the late 1800s and they are still one of the most powerful visualisations for relationships.
#### Here is a simple example
Let's think about the numbers between 2 and 20.
I'm going to create 2 groups, one for even numbers, and one for multiplies of 3.
We could call the sets after the events that are created.
Event E could be the "number is even".
Event M could be the "number is a multiple of 3".
So I can change the diagram to reflect this naming structure.
The next thing I do is write in all the numbers in the appropriate places. As I place a number I ask myself.... Is the number even? Is it a multiple of 3? Is it both? or Is it none of those options?
Take note of how the numbers that do not fit into either set are placed outside the circles.
Now that we have a Venn diagram, we can answer a whole range of questions.
List the items in events E, M and E&M
E = {2,4,6,8,10, 12,14,16, 18,20}
M = {3,6,9,12,15,18}
E and M = {6,12,18}
Probability that the number is Even
P(E) =$\frac{\text{total favourable outcomes }}{\text{total possible outcomes }}=\frac{10}{19}$total favourable outcomes total possible outcomes =1019
Probability that the number is a multiple of 3
P(M) = $\frac{\text{total favourable outcomes }}{\text{total possible outcomes }}=\frac{6}{19}$total favourable outcomes total possible outcomes =619
Probability that the number is both Even and a Multiple of 3
P(E and M) = $\frac{\text{total favourable outcomes }}{\text{total possible outcomes }}=\frac{3}{19}$total favourable outcomes total possible outcomes =319
Probability that the number is not even or a multiple of 3
P(not E or M) = $\frac{\text{total favourable outcomes }}{\text{total possible outcomes }}=\frac{6}{19}$total favourable outcomes total possible outcomes =619
Using a Venn Diagram can be a really powerful way to display information, and then interpret it in meaningful ways.
#### Worked Example
We are interested in the colour of a card randomly drawn from a standard deck. Which Venn diagram illustrates this?
1. A
B
A
B
### Outcomes
#### S5-3
Compare and describe the variation between theoretical and experimental distributions in situations that involve elements of chance
#### S5-4
Calculate probabilities, using fractions, percentages, and ratios |
reasoning (43) Maths (40) Home (8) integration (6)
## HOW TO FIND THE INVERSE OF 2×2 AND 3×3 MATRIX USING SHORTCUT METHOD
Hello and Welcome to this post ,Today we are going to discuss the shortest and easiest methods of finding the Inverse of 2×2 matrix and 3×3 Matrix. Usually when we have to find the Inverse of any Matrix then we follow the following steps .
1 Check whether the determinant value of the given Matrix is Non Zero.
2 Find out the co-factors of all the elements of the Matrix.
3 Put these co-factors in co-factor Matrix.
4 Find the Ad joint of this matrix by taking the Transpose of a Matrix of the co-factor matrix.
5 Now Multiply Ad Joint of Matrix with the reciprocal of Determinant value of the given Matrix.
This Method is very confusing, Long and time Consuming. So Let us have a New, Easy and Shortcut Method .
## Method For 2×2 Matrix
If we have to find the Inverse of 2×2 Matrix then Follows these steps.
1 Interchange the position of the elements which are a11 and a22 .
Change the Magnitude of the elements which are in position a12 and a21 .
Divide every elements of the given Matrix with its Determinant value.
## Example
To find the Inverse of this matrix just interchange the position of elements a₁₁ and a₂₂ i.e Interchange the positions of elements 5 and -3 and in second step change the magnitude of the elements which are in positions a12 and a21 i.e. change the sign of 9 and 4.
Now divide each elements with determinants value of the matrix which is (5)(-3) - (9)(4) = -15 -36 = -51
So The Inverse of the given Matrix A will be
Then after interchanging the positions of 8 and 2 change the magnitude of 7 and -6 and divide every elements with its determinant value (8)( 2) - (7)*(-6) = 16+42 = 58
The Inverse of B is
### After interchanging the position of -3 and -6 and changing the magnitude of -4 and -5 and at last dividing every elements with its determinant value (-3)×(-6) - (-4)×(-5) = 18 - 20 = -2
The Inverse of C is
This video Explains all about Inverse of 2×2 and 3×3 Matrix
## Method for 3×3 Matrix
Ist of all Write the given Matrix in five columns by adding the 4th column as repetition of 1st column and 5th column as repetition of 2nd column, then
C₁ C2 C3 C4 C5
5 -1 4 5 -1
2 3 5 2 3
5 -2 6 5 -2
Now Expanding this Matrix to 5×5 Matrix by adding 4th Row as repetition of 1st Rows and adding 5 Row as repetition of 2nd column as what we received in last step.
R₁ 5 -1 4 5 -1
R₂ 2 3 5 2 3
R₃ 5 -2 6 5 -2
R₄ 5 -1 4 5 -1
R5 2 3 5 2 3
Now to find the Inverse of the given Matrix ,we have to find the cofactor of every elements
1 Find the co-factor of 1st element of Row 1 i. e. 5, determinant value of the Matrix (RED below ) obtained by eliminating the 1st Row and 1st Column which will be (3×6)-{(5)×(-2)} = 28,write these co-factor value in 1st column of 1st Row. (we are evaluating co-factors row wise and writing Column wise)
R₁ 5 -1 4 5 -1
R₂ 2 3 5 2 3
R₃ 5 -2 6 5 -2
R₄ 5 -1 4 5 -1
R5 2 3 5 2 3
2 Now Find the co-factor of 2nd element of 1st Row i. e. -1,which is equal to determinant value of the Matrix (RED below) obtained by eliminating the 1st Row and 2nd Column which will be 5*5-(2)*(6) =13,write this co-factor value in 2nd Row of 1st column .(we are evaluating co-factors row wise and writing Column wise)
R₁ 5 -1 4 5 -1
R₂ 2 3 5 2 3
R₃ 5 -2 6 5 -2
R₄ 5 -1 4 5 -1
R5 2 3 5 2 3
## 3 Now Find the co-factor of 3rd element of 1st Row i.e. 4, which is equal to determinant value of the Matrix(RED below ) obtained by eliminating the 1st Row and 3rd Column which will be 2*(-2)-(3)*(5) = -19,write this co-factor value in 3rd Row of 1st column.(we are evaluating co factors row wise and writing Column wise)
R₁ 5 -1 4 5 -1
R₂ 2 3 5 2 3
R₃ 5 -2 6 5 -2
R₄ 5 -1 4 5 -1
R5 2 3 5 2 3
4 Now Find the co-factor of 1st element of 2nd Row i. e. 2, which is equal to determinant value of the Matrix (RED below ) obtained by eliminating the 2nd Row and 1st Column which will be -2*(4)-(6)*(-1) = -2,write this co-factor value in 2nd Column of 1st Row .(we are evaluating co factors row wise and writing Column wise) .
R₁ 5 -1 4 5 -1
R₂ 2 3 5 2 3
R₃ 5 -2 6 5 -2
R₄ 5 -1 4 5 -1
R5 2 3 5 2 3
5 Now Find the co-factor of 2nd element of 2nd Row i.e 3, which is equal to determinant value of the matrix (RED below ) obtained by eliminating the 2nd Row and 2nd column which will be 6*(5)-(5)*(4) = 10,write this co-factor value in 2nd Row of 2nd column
R₁ 5 -1 4 5 -1
R₂ 2 3 5 2 3
R₃ 5 -2 6 5 -2
R₄ 5 -1 4 5 -1
R5 2 3 5 2 3
6 Find the co-factor of 3rd element of 2nd Row i. e.5, which is equal to determinant value of the Matrix (RED ) obtained by eliminating the 2nd Row and 2nd Column which will be 5× (-1)-(-2) × (5) = 5,write this co-factor value this 2nd Column of 3rd Row, write this co-factor value in 2nd Column of 1st Row . (we are evaluating co factors row wise and writing Column wise ) .
R₁ 5 -1 4 5 -1
R₂ 2 3 5 2 3
R₃ 5 -2 6 5 -2
R₄ 5 -1 4 5 -1
R5 2 3 5 2 3
Similarly for 1st , 2nd, 3rd element the co-factor values will be as follows
For A₃1 i.e 5
R₁ 5 -1 4 5 -1
R₂ 2 3 5 2 3
R₃ 5 -2 6 5 -2
R₄ 5 -1 5 -1
R5 2 3 5 2 3
For A₃₂ i.e -2
R₁ 5 -1 4 5 -1
R2 2 3 5 2 3
R₃ 5 -2 6 5 -2
R₄ 5 -1 5 -1
R5 2 3 5 2 3
for A₃₃ i.e. 6
R₁ 5 -1 4 5 -1
R₂ 2 3 5 2 3
R₃ 5 -2 6 5 -2
R₄ 5 -1 4 5 -1
R5 2 3 5 2 3
so we have -17 ,-17 and 17 as co-factors of 3rd Row, write these co factors in 3rd column .
(we are evaluating co factors row wise and writing Column wise)
⎾ 28 -2 -17 ⏋
⎹⎸ 13 10 -17 ⎹
⎿ -19 5 17 ⏌
Now divide with the determinant value of given 3×3 Matrix , which will be 5(28)-1(-13) + 4(-19) = 140 + 13 -7 6 = 77.
Now divide each element of Ad joint Matrix obtained in previous step with determinant value 77,
Then A⁻¹ =
## Conclusion
This post was regarding short cut methods of finding Inverse of 2×2 and 3×3 Matrices , If you liked this post ,Please share your precious views on this topic and share this post with your friends to benefit them. we shall Meet in the next post ,till then BYE .
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# CIE A level -Pure Mathematics 1 : Topic : 1.4 Circular measure: definition of a radian : Exam Questions Paper 1
### Question
The diagram shows a metal plate made by fixing together two pieces, OABCD (shaded) and OAED (unshaded). The piece OABCD is a minor sector of a circle with centre O and radius 2r. The piece OAED is a major sector of a circle with centre O and radius r. Angle AOD is $$\alpha$$ radians. Simplifying your answers where possible, find, in terms of $$\pi$$ ,$$\alpha$$ and r.
(i) the perimeter of the metal plate.
(ii) the area of the metal plate.
It is now given that the shaded and unshaded pieces are equal in area.
(iii) Find $$\alpha$$ in terms of $$\pi$$
(i) Perimeter of the Metal Plate:
The perimeter of the metal plate consists of the lengths of the curved edges and the straight edge connecting the endpoints.
For the minor sector OABCD, the curved edge has an arc length equal to $$2r$$ times the angle $$\alpha$$. This is because the entire circumference of a circle with radius $$2r$$ is $$2\pi \times 2r = 4\pi r$$, and the angle $$\alpha$$ is a fraction of the full circle. So, the arc length of the minor sector is $$2r \alpha$$.
For the major sector OAED, the curved edge has an arc length equal to $$r$$ times the angle $$\pi – \alpha$$. Similarly, this is because the entire circumference of a circle with radius $$r$$ is $$2\pi r$$, and the angle $$\pi – \alpha$$ is a fraction of the full circle. So, the arc length of the major sector is $$r (\pi – \alpha)$$.
Finally, the straight edge connecting the endpoints has a length of $$2r$$.
The total perimeter is the sum of these lengths:
$$P = 2\pi r + r\alpha + 2r$$
(ii)Area of the Metal Plate:
The area of the metal plate consists of the sum of the areas of the shaded and unshaded regions.
For the minor sector OABCD, the area is given by the formula for the area of a sector: $$\frac{1}{2}r^2\alpha$$.
Since the radius is $$2r$$, this becomes $$\frac{1}{2}(2r)^2 \alpha = 2r^2 \alpha$$.
For the major sector OAED, the area is similarly given by $$\frac{1}{2}r^2 (\pi – \alpha)$$.
The total area is the sum of these areas:
$$A = 2r^2 \alpha + \frac{1}{2}r^2 (\pi – \alpha)$$
(iii) Finding $$\alpha$$ in terms of $$\pi$$:
Given that the shaded and unshaded areas are equal, you set up the equation:
$$2r^2\alpha = \frac{1}{2}r^2(\pi – \alpha)$$
Solving for $$\alpha$$,
$$\alpha = \frac{2}{5}\pi$$
### Question
The function f is defined by $$f:x\rightarrow \frac{2}{3-2x}$$ for $$x\epsilon R$$ ,$$x\neq \frac{3}{2}$$
(i) Find an expression for $$f^{-1}\left ( x \right )$$.
The function g is defined by $$g:x\rightarrow 4x+a$$ for $$x\epsilon R$$ ,where a is ac constant.
(ii)Find the value of a for which gf(-1)=3
(iii) Find the possible values of a given that the equation$$f^{-1}\left ( x \right )=g^{-1}\left ( x \right )$$ has two equal roots.
(i) To find the inverse function $f^{-1}(x)$, switch $x$ and $y$ in the definition of $f$ and solve for $y$:
$$y = \frac{2}{3 – 2x}$$
$$x = \frac{2}{3 – 2y}$$
$$3 – 2y = \frac{2}{x}$$
$$2y = 3 – \frac{2}{x}$$
$$y = \frac{3}{2} – \frac{1}{x}$$
So, $f^{-1}(x) = \frac{3}{2} – \frac{1}{x}$.
(ii) To find the value of $a$ for which $\operatorname{gf}(-1) = 3$, first, find $g(f(-1))$:
$$g(f(-1)) = g\left(\frac{2}{3 – 2(-1)}\right)$$
$$= g\left(\frac{2}{5}\right)$$
$$4\left(\frac{2}{5}\right) + a = 3$$
$$\frac{8}{5} + a = 3$$
$$a = 3 – \frac{8}{5} = \frac{7}{5}$$
So, the value of $a$ is $\frac{7}{5}$.
(iii)$$g^{-1}(x) = \frac{x – a}{4} = f^{-1}(x)$$
This equation can be written as:
$$\frac{x – a}{4} = \frac{3}{2} – \frac{1}{x}$$
$$x – a = 6x – \frac{4}{x}$$
$$6x – x – a – \frac{4}{x} = 0$$
$$5x – a – \frac{4}{x} = 0$$
$$5x^2 – ax – 4 = 0$$
$$(a + 6)^2 = a^2 + 12a + 36$$
So, the correct equation is $$a^2 + 12a + 36 = 16$$ or $$a^2 + 12a + 20 = 0$$.
We can use the quadratic formula to solve for $$a$$:
$$a = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$
In this case, $$a^2 + 12a + 20 = 0$$, so $$a = \frac{-12 \pm \sqrt{12^2 – 4(1)(20)}}{2(1)}$$:
$$a = \frac{-12 \pm \sqrt{144 – 80}}{2}$$
$$a = \frac{-12 \pm \sqrt{64}}{2}$$
$$a = \frac{-12 \pm 8}{2}$$
This gives two possible values for $$a$$:
$$a_1 = \frac{-12 + 8}{2} = -2$$
$$a_2 = \frac{-12 – 8}{2} = -10$$
So, the correct solutions for $$a$$ are $$a = -2$$ or $$a = -10$$.
### Question
In the diagram, AOB is a quarter circle with centre O and radius r. The point C lies on the arc AB and the point D lies on OB. The line CD is parallel to AO and angle $$AOC = \Theta$$radians.
(i) Express the perimeter of the shaded region in terms of r,$$\Theta$$and $$\Pi$$.
(ii) For the case where r = 5 cm and $$\Theta = 0.6$$, find the area of the shaded region.
(i) Express the perimeter of the shaded region in terms of $$r$$, $$\Theta$$, and $$\pi$$.
The shaded region consists of three parts: the quarter circle $$AOB$$, the arc $$AC$$, and the segment $$ABC$$. The perimeter ($$P$$) is the sum of the lengths of these three parts.
Quarter Circle $$AOB$$:The circumference of a full circle is $$2\pi r$$, so the quarter circle has a circumference of $$\frac{1}{4} \cdot 2\pi r = \frac{1}{2}\pi r$$.
Arc $$AC$$:The length of the arc $$AC$$ can be found using the formula for the circumference of a circle and the fraction of the circle represented by the angle $$\Theta$$:
$\text{Arc length} = \frac{\Theta}{360^\circ} \cdot 2\pi r$
Segment $$ABC$$:The length of $$CD$$ is $$r$$ (as $$CD$$ is parallel to $$AO$$). The length of $$AB$$ is $$r\Theta$$ (as $$AB$$ is a fraction of the circumference corresponding to the angle $$\Theta$$). So, the length of $$BC$$ is $$r\Theta – r$$.
$P = \frac{1}{2}\pi r + \frac{\Theta}{360^\circ} \cdot 2\pi r + (r\Theta – r)$
$P = \frac{1}{2}\pi r + \frac{\Theta}{180^\circ} \cdot \pi r + r\Theta – r$
(ii) Area $$A): In this case, we’re finding the area of the shaded region, which consists of the quarter circle \(AOB$$ and the segment $$ABC$$.
Quarter Circle $$AOB$$:
– The area of a quarter circle is given by $$\frac{1}{4}\pi r^2$$.
$$\text{Area of quarter circle} = \frac{1}{4}\pi \cdot (5 \, \text{cm})^2 = \frac{1}{4}\pi \cdot 25 = \frac{25}{4}\pi \, \text{cm}^2$$
Segment $$ABC$$:
– The area of a segment can be found by subtracting the area of the corresponding triangle from the area of the sector.
Sector Area:
– The area of the sector is given by $$\frac{1}{2} r^2 \left(\frac{1}{2}\pi – \Theta\right)$$.
$$\text{Sector area} = \frac{1}{2} \cdot (5 \, \text{cm})^2 \cdot \left(\frac{1}{2}\pi – 0.6\right)$$
Triangle Area:
– The triangle is formed by $$O$$, $$C$$, and a point on the circle. The area of the triangle is given by $$\frac{1}{2} \cdot r \cdot \cos \Theta \cdot \frac{1}{2} \cdot r \cdot \sin \Theta$$.
$$\text{Triangle area} = \frac{1}{2} \cdot 5 \, \text{cm} \cdot \cos 0.6 \cdot \frac{1}{2} \cdot 5 \, \text{cm} \cdot \sin 0.6$$
Segment Area:
– Subtract the triangle area from the sector area to get the area of the segment.
$$\text{Segment area} = \text{Sector area} – \text{Triangle area}$$
Now, add the area of the quarter circle and the area of segment $$ABC$$ to find the total area $$A$$:
$$A = \frac{25}{4}\pi + \left(\frac{1}{2} \cdot (5)^2 \cdot \left(\frac{1}{2}\pi – 0.6\right) – \frac{1}{2} \cdot 5 \cdot \cos 0.6 \cdot \frac{1}{2} \cdot 5 \cdot \sin 0.6\right)$$
$$A \approx 6.31 \, \text{cm}^2$$
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# 6.2 Properties of power series
Page 1 / 10
• Combine power series by addition or subtraction.
• Create a new power series by multiplication by a power of the variable or a constant, or by substitution.
• Multiply two power series together.
• Differentiate and integrate power series term-by-term.
In the preceding section on power series and functions we showed how to represent certain functions using power series. In this section we discuss how power series can be combined, differentiated, or integrated to create new power series. This capability is particularly useful for a couple of reasons. First, it allows us to find power series representations for certain elementary functions, by writing those functions in terms of functions with known power series. For example, given the power series representation for $f\left(x\right)=\frac{1}{1-x},$ we can find a power series representation for ${f}^{\prime }\left(x\right)=\frac{1}{{\left(1-x\right)}^{2}}.$ Second, being able to create power series allows us to define new functions that cannot be written in terms of elementary functions. This capability is particularly useful for solving differential equations for which there is no solution in terms of elementary functions.
## Combining power series
If we have two power series with the same interval of convergence, we can add or subtract the two series to create a new power series, also with the same interval of convergence. Similarly, we can multiply a power series by a power of x or evaluate a power series at ${x}^{m}$ for a positive integer m to create a new power series. Being able to do this allows us to find power series representations for certain functions by using power series representations of other functions. For example, since we know the power series representation for $f\left(x\right)=\frac{1}{1-x},$ we can find power series representations for related functions, such as
$y=\frac{3x}{1-{x}^{2}}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y=\frac{1}{\left(x-1\right)\left(x-3\right)}.$
In [link] we state results regarding addition or subtraction of power series, composition of a power series, and multiplication of a power series by a power of the variable. For simplicity, we state the theorem for power series centered at $x=0.$ Similar results hold for power series centered at $x=a.$
## Combining power series
Suppose that the two power series $\sum _{n=0}^{\infty }{c}_{n}{x}^{n}$ and $\sum _{n=0}^{\infty }{d}_{n}{x}^{n}$ converge to the functions f and g , respectively, on a common interval I .
1. The power series $\sum _{n=0}^{\infty }\left({c}_{n}{x}^{n}±{d}_{n}{x}^{n}\right)$ converges to $f±g$ on I .
2. For any integer $m\ge 0$ and any real number b , the power series $\sum _{n=0}^{\infty }b{x}^{m}{c}_{n}{x}^{n}$ converges to $b{x}^{m}f\left(x\right)$ on I .
3. For any integer $m\ge 0$ and any real number b , the series $\sum _{n=0}^{\infty }{c}_{n}{\left(b{x}^{m}\right)}^{n}$ converges to $f\left(b{x}^{m}\right)$ for all x such that $b{x}^{m}$ is in I .
## Proof
We prove i. in the case of the series $\sum _{n=0}^{\infty }\left({c}_{n}{x}^{n}+{d}_{n}{x}^{n}\right).$ Suppose that $\sum _{n=0}^{\infty }{c}_{n}{x}^{n}$ and $\sum _{n=0}^{\infty }{d}_{n}{x}^{n}$ converge to the functions f and g , respectively, on the interval I . Let x be a point in I and let ${S}_{N}\left(x\right)$ and ${T}_{N}\left(x\right)$ denote the N th partial sums of the series $\sum _{n=0}^{\infty }{c}_{n}{x}^{n}$ and $\sum _{n=0}^{\infty }{d}_{n}{x}^{n},$ respectively. Then the sequence $\left\{{S}_{N}\left(x\right)\right\}$ converges to $f\left(x\right)$ and the sequence $\left\{{T}_{N}\left(x\right)\right\}$ converges to $g\left(x\right).$ Furthermore, the N th partial sum of $\sum _{n=0}^{\infty }\left({c}_{n}{x}^{n}+{d}_{n}{x}^{n}\right)$ is
$\begin{array}{cc}\hfill \sum _{n=0}^{N}\left({c}_{n}{x}^{n}+{d}_{n}{x}^{n}\right)& =\sum _{n=0}^{N}{c}_{n}{x}^{n}+\sum _{n=0}^{N}{d}_{n}{x}^{n}\hfill \\ & ={S}_{N}\left(x\right)+{T}_{N}\left(x\right).\hfill \end{array}$
Because
$\begin{array}{cc}\hfill \underset{N\to \infty }{\text{lim}}\left({S}_{N}\left(x\right)+{T}_{N}\left(x\right)\right)& =\underset{N\to \infty }{\text{lim}}{S}_{N}\left(x\right)+\underset{N\to \infty }{\text{lim}}{T}_{N}\left(x\right)\hfill \\ & =f\left(x\right)+g\left(x\right),\hfill \end{array}$
we conclude that the series $\sum _{n=0}^{\infty }\left({c}_{n}{x}^{n}+{d}_{n}{x}^{n}\right)$ converges to $f\left(x\right)+g\left(x\right).$
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Berger describes sociologists as concerned with
Leaves accumulate on the forest floor at a rate of 2 g/cm2/yr and also decompose at a rate of 90% per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time 0 there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?
You have a cup of coffee at temperature 70°C, which you let cool 10 minutes before you pour in the same amount of milk at 1°C as in the preceding problem. How does the temperature compare to the previous cup after 10 minutes?
Abdul |
GMAT Book 3: Percentages
GMAT Syllabus in Percents and Fractions | GMAT Prep Books
Percentage is the idea of expressing a fraction whose denominator is 100 to aid comparison. You could expect a couple of questions in the GMAT quant section testing concepts covered in this topic. One is expected to be adept at interchangeably expressing values as fractions, decimals and percents.
Syllabus Covered in Wizako's GMAT Book
Percentages, fractions are basic concepts that appear frequently as part of the GMAT Problem Solving section. Wizako's GMAT Math Lesson Book in this chapter covers the following concepts:
1. The meaning of the term percent and the need for using percentages.
2. The relation between percents, fractions and decimals and the rules to convert from one form to another with illustrative examples.
3. Explanation of the percentage change in values along with illustrative examples.
4. 11 illustrative examples. 23 solved examples with detailed explanation and special mention of shortcuts wherever applicable.
6. An objective type test with 40 GMAT level questions in the work book. Answer key and explanatory answers for all questions are provided.
Here is a typical solved example in Wizako's GMAT Book from this chapter
Sample Question
In an examination a candidate who secured 20% of the maximum marks failed by 10 marks. Another candidate who secured 30% of the maximum marks got 20 marks more than the pass mark. What is the pass mark in this examination?
Let the pass mark in the examination be P and the maximum marks be M.
Candidate 1 secured 20% of maximum marks, which was 10 marks less than pass mark.
i.e., 20% of M = P - 10 ... (1)
Candidate 2 secured 30% of maximum marks, which was 20 marks more than pass mark.
i.e., 30% of M = P + 20 ... (2)
Subtract equation (1) from equation (2): 10% of M = 30
If 10% of the maximum marks is 30, maximum marks, M = 300
The pass mark is obtained by substituting M = 300 in one of the two equations.
In equation (1): 20% of M = P - 10
20% of 300 = P - 10
60 = P - 10
Or P = 70. Pass mark, P = 70
More in GMAT Percentages
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# The Trachtenberg System of Speed Math
The Trachtenberg Speed System Basic Mathematics is a system of mental calculation, similar to Vedic mathematics. Sometimes presented as an alternative way of learning math, the system consists of many math shortcuts and mental math tricks, particularly multiplication tricks.
It was developed by the Russian engineer Jakow Trachtenberg to keep his mind busy when he was imprisoned in a Nazi concentration camp and is known to be one of the fastest speed math systems.
The system consists of a number of easily memorized patterns that allow one to perform arithmetic computations without the help of pen and paper. The system primarily focuses on multiplication tricks but with further practice and study you can learn division, addition, subtraction and square root.
# A Quick Overview of Basic Multiplication
If you want to learn either how to multiply without a calculator effectively or how to teach multiplication effectively, then keep reading.
The process for calculation using the Trachtenberg method is rather simple and involves calculating the answer to your problem one digit at a time. The steps you must follow:
• Write a zero ‘0’ before the number you are multiplying (the multiplicand) and underline it.
• Apply the relevant rule for the number you are calculating (see below for rules).
• Work from the right to the left of the multiplicand, individually applying the rule to each number in turn.
• Record the result of each calculation below the number you applied the rule to.
## Definition of Terms
The Multiplicand – The ‘Multiplicand’ is the number you are multiplying. In the photos above, and on all of our examples, it is the number on the right of the multiplier (eg: in the sum11 x 28344″ 28344 would be the multiplicand). The multiplicand will always be underlined and our answers will go beneath it, one digit at at time.
The ‘Number – When we talk about “adding the number to its neighbour” etc, the number is referencing to the current number in the multiplicand you are applying the rule to. We will always start from the right most number in the multiplicand and work our way backwards, to the left.
The ‘Neighbour – The ‘neighbour‘ is the digit to the right of the number’. If my multiplicand was 345, then 5 would not have a neighbour (our we would use 0), the neighbour of 4 would be 5 and the neighbour of 3 would be 4.
Carried 10’s – When we do calculations which give us double digit answers above 10 we need to carry the 10’s. We do this by adding a dot to represent the carried ten (or two dots if the answer is 20-29) to the current number we are working on to remind us to add the dot (1 dot = 1) onto the next answer. In the below example, the 7’s recorded in are answer are actually ’17’ but we record only the 7 and add a dot. Two dots are added if the resulting answer of the number rule is 20-29. (You will never have an answer which is above 29.)
Rounding Down Fractions – When using the Trachtenberg System you will deal with rules which involve halving the numbers or neighbours, when we ‘halve’ an odd number we will ignore any fractions and round down to the nearest whole number. Half of 5 would be recorded simply as 2, half of 7 would be 3 and half of 8 would be 4 (even numbers are halved as normal).
Jump to number: 11, 12, 6, 7, 5, 9, 8, 4, 3
## Multiplication of 11 – Rules & Example
Visit our page on multiplying by 11 HERE and see more worked examples, tips & tricks, clearer steps, video tutorials, further resources and help.
We start by learning how to multiply by 11 because it is the easiest rule to learn. If you learn only one thing from this site, it should be this. It’s one of the best math tricks for kids and adults.
The Rule: Multiply by 11
1. Add the neighbour to the number
Example: 11 x 633
11 x 0633
= 6963
---
3 + 0 = 3 Step 1
3 + 3 = 6 Step 2
6 + 3 = 9 Step 3
0 + 6 = 6 Step 4
Description
When multiplying by 11 you simply add the number to the neighbour. Starting at the right hand number, 3, we see that it doesn’t have a neighbour so there is nothing to add (or we imagine 0 is its neighbour). We are then finished with our calculation and record 3.
Moving onto the next digit, 3, we add this to it’s right hand neighbour, another 3, to get the result 6. We record 6 below.
The next two digits we do the same. 6 + 3 = 9. We record the 9. 0 + 6 is 6, we record the 6. And we have our answer: 6963.
~~~
This is a very basic example of how to multiply by 11 using low numbers and thus doesn’t involve dealing with any carried 10’s. To learn more about how to multiply by 11 and see more complicated examples please visit our dedicated multiplying by 11 page.
## Video Tutorial Series Playlist
Don't forget to subscribe to our YouTube channel to get access to full speed maths techniques never before seen!
## Multiplication of 12 – Rules & Example
Our page on multiplying by 12. Here you will find more worked examples, tips & tricks, in-depth explanations, downloadable PDF’s, worksheets, exercises, video tutorials, further resources and help.
The Rule: Multiply by 12
1. Double each number in turn and add its neighbor.
Example: 12 x 413
12 x 0413
= 4956
---
(3 x 2) + 0 = 6 Step 1
(1 x 2) + 3 = 5 Step 2
(4 x 2) + 1 = 9 Step 3
(0 x 2) + 4 = 4 Step 4
Description
The rule for multiplying by 12 is very similar to that of multiplying by 11 shown above.
Starting from the right hand number of the multiplicand we double 3 to get 6. The first digit doesn’t have a neighbour so we add 0. The first digit in our answer is 6.
The next digit, 1, is then doubled to get 2. We then add its neighbour, 3, to get our second digit in our answer, 5.
We continue these steps to the left with the next two digits and get 9 and 4 respectively for the next two digits in our answer. Our answer: 4956.
~~~
This is a simple example of how to multiply by 12 which doesn’t involve carrying 10’s. To learn how to do this and see more worked examples, please visit our dedicated “multiplying by 12” page.
## Multiplication of 6 – Rules & Example
Below I will demonstrate how to multiply by 6 using the Trachtenberg System but note that this is a basic example and whilst it covers the entire rule for the number 6, it does’t show how to effectively multiply larger numbers and the theory behind doing so. Please visit our dedicated “Multiplying by 6” page for more info and multiplication tricks.
The Rule: Multiply by 6
1. Add half the neighbour to the number; Add 5 if the “number” is odd.
Example: 6 x 613
6 x 0613
= 3678
---
3 + (0 x 0.5) + 5 = 8 Step 1
1 + (3 x 0.5) + 5 = 7 Step 2
6 + (1 x 0.5) = 6 Step 3
0 + (6 x 0.5) = 3 Step 4
Description
The first digit on the right hand side of our multiplicand is 3. It has no neighbour so we don’t add anything to it. It is however an odd digit so we will add 5 to it to get the first digit in our answer, 8.
Moving along to the left, we add the number, 1, to half of its neighbour (3 x 0.5) and get 2. Next we add 5 beause “1” is odd. The second number in our answer is 7.
The next two numbers in our multiplicand, 6 and 0 are both even so we don’t have to add the additional 5 for odd numbers, we simply add half of their right hand neighbours to the numbers themselves to get the answers 6 and 3 respectively.
~~~
The above example is a very simple explanation of how to multiply by 6 using the Trachtenberg Method. It uses small digits and doesn’t involve carrying any 10’s. To learn how to do this and see more worked examples, please visit our dedicated “multiplying by 6” page.
## Multiplication of 7 – Rules & Example
Multiplying by 7 involves using a combination of rules and can be a little trickier to remember and implement. Detailed instructions can be found on our “Multiplying by 7” page.
The Rule: Multiply by 7
1. Double the number plus half of its neighbour; add 5 if the number is odd.
Example: 7 x 358
7 x 0358
= 2506
---
(8 x 2) + (0 x 0.5) = 16 Step 1
(5 x 2) + (8 x 0.5) + 5 + 1 = 20 Step 2
(3 x 2) + (5 x 0.5) + 5 + 2 = 5 Step 3
(0 x 2) + (3 x 0.5) + 1 = 2 Step 4
Description
First start with the right hand number of the multiplicand, 8. Double it to get 16 and as it has no neighbour we don’t have to add anything else.
The next number, 5, is doubled to get 10. Then its neighbour, 8, is halved to get 4 and added to the running sum to make 9. Because our number 5 is odd, we add 5. And finally we must remember to add the carried one from the previous digits result. This gives us a result of 20. We record the 0 and remember to carry 2 onto our next number.
The above steps are repeated for the next number. 3 doubled is 6, plus halve of its neighbour, 5, gives us a total of 8. Because 3 is odd, we add an additional 5 and then remember to add 2 from our previous answer. Our answer is 15 so we record the 5 and carry the 1.
The final digit is a 0. Double 0 is 0 so we can effectively ignore this step. Next we halve its neighbour, 3, to get 1 and add the carried one from our previous answer. We record our final digit to our answer 2.
~~~
The above example is a simple explanation of how to multiply by 7 using the Trachtenberg System. To see more worked examples and find the most efficient way of multiplying by 7 please visit our dedicated “multiplying by 7” page.
Listed under: quick math, speed maths, tips & techniques, instruction, Trachtenberg System, maths tricks
## Multiplication of 5 – Rules & Example
The method for multiplying by 5 using the Trachtenberg System is very similar to that of 7, in fact, it’s probably a little easier. It involves using multiple steps within the rule. Detailed instructions can be found on our “Multiplying by 5” page.
The Rule: Multiply by 5
1. Half the neighbour; add 5 if the number is odd.
Example: 5 x 271
5 x 0271
= 1355
---
(0 x 0.5) + 5 = 5 Step 1
(1 x 0.5) + 5 = 5 Step 2
(7 x 0.5) = 3 Step 3
(2 x 0.5) = 1 Step 4
Description
The main difference between multiplying by 5 and 7 is that we are not directly manipulating or using the number we are on at the time. Calculations are made using the numbers neighbour, then we add 5 to the total if the number is odd.
Starting from the right, we have a 1. It has no neighbour but it is odd so we add 5 to the result of half its neighbour (0.5 x 0) to get 5.
The next digit, 7 has a neighbour of 1. Half of 1 is 0. 7 is an odd number so we add 5 to get the next digit in our answer: 5.
2 is our next digit. Half of its neighbour, 7 is 3. Add our number 2 is odd we do not add an additional 5. We record 3 as the next digit in our answer.
The final digit of the multiplicand is 0. Half of its neighbour, 2, is 1. Zero is odd so we do not add 5. We record 1.
~~~
The above example is a simple example & explanation of how to multiply by 5 using the Trachtenberg System. To see more worked examples and find the most efficient way of multiplying by 5 please visit our dedicated “multiplying by 5” page.
## Multiplication of 9 – Rules & Example
Multiplying by 9 using the Trachtenberg System involves 3 steps and separate rules are used for the beginning, last and middle numbers. Detailed instructions can be found on our “Multiplying by 9” page.
The Rule: Multiply by 9
• Step 1: (First Digit Only): Subtract the right-hand figure of the multiplicand from 10.
• Step 2: (For Middle Digits Only): Subtract number from 9Add its neighbour.
• Step 3: (Last Digit, ‘0’): Subtract 1 from its neighbour.
Example: 9 x 879
9 x 0879
= 7911
---
10 - 9 = 1 Step 1
(9 - 7) + 9 = 11 Step 2a
(9 - 8) + 7 + 1 = 9 Step 2b
(8 - 1) = 7 Step 3
Description
The number 9 is the first number we see which has three separate rules in one. Whilst this might seem complicated at first, it is in fact an easy rule as a whole to remember as the first and last digits, in spite of having separate rules, are easy subtractions. Common sense and a basic understanding of numerology certainly helps when using rules with multiple steps.
The first number in the sequence, 9, must be subtracted from 10 to get our result: 1.
The middle numbers we subtract from 9 and then add its neighbour. So, 7 subtracted from 9 is 2, plus it’s neighbour, 9, equals 11. We record the 1 and carry the 10.
The second number we deal with is 8. Subtract 8 from 9 to get 1. Then add its neighbour, 7, to get 8. then we add on the carried one from the last calculation to get 9. We record 9.
The last step is simple, we subtract one from the neighbour of 0 (which is 8) to get 7. We record 7.
~~~
Multiplying by 9 using the Trachtenberg Method is simple, but in order to improve your speed you need to practice. There are also a few more multiplication tricks and techniques you can use to help improve your accuracy and speed, these can be found on our dedicated “multiplying by 9” page.
## Multiplication of 8 – Rules & Example
Similar to the rule for multiplying by 9, multiplying by 8 involves using a 3-step process.
Detailed instructions can be found on our “Multiplying by 8” page.
The Rule: Multiply by 8
• Step 1: (First Digit Only): Subtract the right-hand figure of the multiplicand from 10 then double.
• Step 2: (For Middle Digits Only): Subtract number from 9, then doubleAdd its neighbour.
• Step 3: (Last Digit, ‘0’): Subtract 2 from its neighbour.
Example: 8 x 789
8 x 0789
= 6312
---
(10 - 9) x 2 = 2 Step 1
((9 - 8) x 2) + 9 = 11 Step 2a
((9 - 7)) x 2) + 8 + 1 = 13 Step 2b
(7 - 2) + 1 = 7 Step 3
Description
Step 1 of our rule: Our first number, 9, is subtracted from 10 to give us 1. Then it is doubled to get 2. This is the first number in our answer.
The middle steps are slightly more complex. We move to the next number of the multiplicand, 8 and apply the rule for step 2 (the step for the middle numbers). We must first subtract 8 from 9 then double the answer to get 2. Next we add its neighbour, 9 to get our result, 11. We record the 1 and carry the 10.
The same rule is applied to the third number in our sequence, a 7. We first subtract it from 9, then double the answer to get 4. Next we add its neighbour, 8, and add the carried 10 from the last step to get 13. We record the 3 and carry the 10.
The final digit in our multiplicand is 0, and the last digit has its own rule, this is to simply subtract 2 from its neighbour (and add any carried digits). Its neighbour is a 7, so we subtract 2 from it to get 5 then add the carried 10 from our last step to get the final digit in our answer, 6.
Thus, 8 x 789 = 6312
~~~
Multiplying by 8 using the Trachtenberg Method is simple once the rules are memorized, but in order to improve your speed you need to practice. There are also a number of tricks and techniques you can use to help improve your accuracy and speed, these can be found on our dedicated “multiplying by 8” page.
## Multiplication of 4 – Rules & Example
The rule for multiplying by 4 involves using 3 steps just like the rules for number 8 & 9.
Full instructions can be found on the “Multiplying by 4” page but here is an overview.
The Rule for multiplying by 4:
• Step 1 (for the first digit): Subtract from 10; add 5 if number is odd
• Step 2 (middle digits): Subtract number from 9; add 5 if digit is odd; add half the neighbour.
• Step 3 (last digit, 0): half the neighbour; minus 1.
Example: 4 x 9385
4 x 09385
= 37940
---
(10 - 5) + 5 = 10 Step 1
(9 - 8) + 2 + 1 = 4 Step 2a
(9 - 3) + 5 + 4 = 15 Step 2b
(9 - 9) + 5 + 1 + 1 = 7 Step 2c
(0.5 x 9) - 1 = 3 Step 3
Description
Step 1 of our rule: Our first number, 5, is subtracted from 10 to give us 5. Then it we add 5 to get 10. This is the first number in our answer. We record the 0 and remember to carry the 10.
The middle steps in our sequence require that we subtract the number from 9, add 5 to the number if the number is odd, then add half its neighbour and carry any remaining 10’s from our previous answer.
The final digit in our multiplicand is 0, and the last digit has its own rule; the rule is to halve its neighbour (9) to get 4, and minus 1, to get our final digit in our answer, 4.
Thus, 4 x 9385 = 37940
~~~
Multiplying by 4 uses many of the same methods and rules as we have seen in previous numbers but applied slightly differently. One remembered it is just as easy to remember as all other numbers. For more information, further worked examples and clarification on points please visit our “multiplying by 4” page.
## Multiplication of 3 – Rules & Example
The rule for multiplying by 3 is similar to that of multiplying by 8 except that this time we’re only adding half of the neighbour. The rule for number 3 also has 3 steps.
Full instructions can be found on the “Multiplying by 3” page but here is an overview.
The Rule: Multiply by 3
• Step 1: (First Digit Only): Subtract number from 10 then double.
• Step 2: (For Middle Digits Only): Subtract number from 9, then double. Add half of its neighbour. Add 5 if the number is odd.
• Step 3: (Last Digit, ‘0’): Subtract 2 from half of the neighbor.
Example: 3 x 2588
3 x 2588
= 7764
---
(10 - 8) x 2 = 4 Step 1
((9 - 8) x 2) + (8 x 0.5) = 6 Step 2a
((9 - 5) x 2) + (0.5 x 8) + 5 = 17 Step 2b
((9 - 2) x 2) + (0.5 x 5) + 1 = 17 Step 2c
(0.5 x 5) - 2 = 0 Step 3
Description
Step 1 of our rule: Our first number, 8, is subtracted from 10 to give us 2. Then we double it. This is the first number in our answer.
The middle steps in our sequence require that we subtract the number from 9, multiply the result by 2, add 5 to the number if the number is odd, then add half its neighbour and carry any remaining 10’s from our previous answer.
The final digit in our multiplicand is 0, and the last digit has its own rule; the rule is to halve its neighbour (5) to get 2, and minus 2, to get our final digit in our answer, 0.
Thus, 3 x 2588 = 7764
~~~
Multiplying by 4 uses many of the same methods and rules as we have seen in previous. One remembered it is just as easy to remember as all other numbers. For more information, further worked examples and clarification on points please visit our “multiplying by 3” page.
## Multiply Larger Numbers
The Trachtenberg Method isn’t just limited to multiplying by single digits 1-12, once you have learnt the rules for each number you can start using the system to multiply any number by any other number. ie: 498 x 37654 :
498 x 37654
--------------
301232
338886-
150616--
--------------
18751692
If you have any questions, problems, comments or advice to offer others then please leave them in the comments section below. All feedback is welcome and there’s never a silly question!!
1. Hannah says:
A fantastic introduction into the Trachtenberg method. I remember my father trying to teach me this when I was younger but didn’t have the interest in it, 20 years later and I’m back!
Thanks for taking the time to put together the instructions.
Hannah
2. Kay says:
Does this method work for calculating percentages?
3. Milan Verhoeven says:
Hello “Jakow Trachtenberg”!
I’m a student Latin Math in secondary school and I need to write a paper. I choose the subject ‘Alternative math’. I found this beautiful system and I very interested in this. For my paper I need ofcourse some mathematical background. Can you help me giving the mathemathical proof behind these rules? Like the multiplication of 6. Why do you need to add 5 to the odd? I need to be able to answer such questions.
Thanks you in advance and I hope you can help me out!
Greets |
# Fractions on a Number Line - Examples, Exercises and Solutions
## Placing Fractions on the Number Line
To locate fractions on the number line, we will carry out several steps.
### First Step – Discovering the Value of Arcs
We will subtract two given numbers and keep the difference.
We will count the number of arcs between the numbers.
We will divide the subtraction result by the number of arcs to find out the measure of each arc.
### Step Two – Placing the Numbers on the Number Line
Depending on the amount of arcs, the scale can be expanded or reduced.
## Examples with solutions for Fractions on a Number Line
### Exercise #1
What number is marked on the number axis?
### Step-by-Step Solution
Let's count how many points including the number 1 are on the number line.
Since there are 6 in total, we will define the point 1 as the fraction:
$\frac{6}{6}$
Since
$\frac{6}{6}=\frac{1}{1}=1$
The number marked on the number line is 1
$1$
### Exercise #2
What number is marked on the number axis?
### Video Solution
$\frac{2}{3}$
### Exercise #3
What number is marked on the number axis?
### Video Solution
$\frac{1}{4}$
### Exercise #4
What number is marked on the number axis?
### Video Solution
$\frac{4}{5}$
### Exercise #5
What number is marked on the number axis?
### Video Solution
$\frac{2}{6}$
### Exercise #6
What numbers appear on the number line?
### Video Solution
$\frac{2}{7},\frac{5}{7}$
### Exercise #7
What number is marked on the number axis?
### Video Solution
$\frac{10}{11}$
### Exercise #8
What number appears on the number line?
### Video Solution
$\frac{4}{6}$
### Exercise #9
What number is marked on the number axis?
### Video Solution
$\frac{3}{4}$
### Exercise #10
What number is marked on the number axis?
### Video Solution
$\frac{4}{6}$
### Exercise #11
The number $\frac{1}{4}$ is found?
### Exercise #12
The number $\frac{6}{5}$ is found
### Video Solution
between $1$ to $1\frac{1}{2}$
### Exercise #13
The number $\frac{7}{8}$ is found
### Video Solution
between$\frac{1}{2}$ to $1$
### Exercise #14
The number $\frac{3}{5}$ is found
### Video Solution
between $\frac{1}{5}$ to $\frac{4}{5}$
### Exercise #15
The number $-\frac{3}{4}$ is found?
### Video Solution
between $-1$ to $0$ |
# Statistics Assignment Help With Rank Correlation
## Spearman’s Rank Correlation Coefficient
The spearman’s rank coefficient of correlation was developed by Charles Edward Spearman.
### What is Spearman correlation coefficient?
Let us suppose that a group of n individuals is arranged in order of merit or proficiency in possession of two characteristics A and B. these ranks in two characteristics will, in general, be different. For example, if we consider the relation between intelligence and beauty, it is not necessary that a beautiful individual is intelligent also. Let (xi , yi ); i=1,2,………,n be the ranks of the ith individual in two characteristics A and B respectively. Pearsonian coefficient of correlation between the ranks xi’s and yi ‘s is called the Spearman rank correlation coefficient between A and B for that group of individuals.
### The formula for correlation of rank coefficient is given as:
rk = 1 – [6 ∑D2 / N3 – N]
Where D = R1 – R2, between the paired items in the two rank series.
The value of rank correlation coefficient tells us about the degree of agreement between the 2 ranks.
#### See derivation for the rank coefficient of correlation by spearman below:
Assuming that no two individuals are bracketed equal in either classification, each of the variables X and Y takes the values 1,2,………..,n
Hence
Which is the spearman's formula for the rank correlation coefficient.
### Properties of Spearman’s Rank correlation Coefficient
-1 ≤ rk ≤ +1
Rank Correlation Coefficient Example
Calculate the Rank Correlation Coefficient in each of the following cases:
X R1 Y R2
10 1 5 1
20 2 6 2
30 3 7 3
To calculate the rank correlation coefficient, first we will determine the value of D = R1 – R2 in each of the entries:
X R1 Y R2 D D2
10 1 5 1 0 0
20 2 6 2 0 0
30 3 7 3 0 0
Then the Spearman’s rank correlation coefficient is calculated using the formula as:
rk = 1 – [6 ∑D2 / N3 – N]
= 1- 6(0)
= +1
Thus the value of rank correlation coefficient equal to +1 implies that there is complete agreement in the order of ranks and the ranks are in the same direction.
Let us calculate the rank correlation coefficient in another example:
X R1 Y R2
10 1 7 3
20 2 6 2
30 3 5 1
To calculate the rank correlation coefficient, first we will determine the value of D = R1 – R2 in each of the entries:
X R1 Y R2 D D2
10 1 7 3 -2 4
20 2 6 2 0 0
30 3 5 1 +2 4
Then the Spearman’s rank correlation coefficient is calculated using the formula as:
rk = 1 – [6 ∑D2 / N3 – N]
= 1 – (6*8)/9-3
= -1
Thus the value of rank correlation coefficient equal to -1 implies that there is complete agreement in the order of ranks and the ranks are in opposite direction
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